488 free, high-quality past-paper-style questions on Number and algebra for IB Maths AA SL, each with a full mark scheme. Covers: Standard form; Arithmetic sequences and series; Financial applications; and more. For when you have finished the past papers, or want to practise one topic at a time.Topic 1 at Higher Level →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.3,347 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1AASL1-q00011.1 Standard formPaper 1Easy4 marks
Let \(a = 3.6\times10^{5}\) and \(b = 1.2\times10^{-2}\). Give each answer in the form \(k\times10^{n}\), where \(1\le k\lt 10\) and \(n\in\mathbb{Z}\).
(a)
Find \(ab\).
(2)
(b)
Find \(\dfrac{a}{b}\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(3.6\times1.2=4.32\) and \(10^{5}\times10^{-2}=10^{3}\)
(M1)
Award for either product of the decimal parts or correct addition of indices.
\(4.32\times10^{3}\)
A1
(b)
\(3.6\div1.2=3\) and \(10^{5}\div10^{-2}=10^{7}\)
(M1)
\(3\times10^{7}\)
A1
Accept \(3.0\times10^{7}\). Award A0 for \(30\times10^{6}\) or calculator notation such as 3E7.
Write \(0.000\,407\) in the form \(a\times10^{k}\), where \(1\le a\lt 10\) and \(k\in\mathbb{Z}\).
(1)
(b)
Write \(52.3\times10^{4}\) in the form \(a\times10^{k}\), where \(1\le a\lt 10\) and \(k\in\mathbb{Z}\).
(1)
(c)
Write the following numbers in order of size, starting with the smallest. \(3.1\times10^{-2}\qquad 0.0298\qquad 31.2\times10^{-3}\qquad 2.99\times10^{-2}\)
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(4.07\times10^{-4}\)
A1
(b)
\(5.23\times10^{5}\)
A1
(c)
writing all four in a common form, e.g. \(0.031,\ 0.0298,\ 0.0312,\ 0.0299\)
5AASL1-q00101.2 Arithmetic sequences and seriesPaper 1Easy7 marks
Consider the arithmetic sequence \(23,\ 17,\ 11,\ 5,\ \ldots\)
(a)
Write down the common difference.
(1)
(b)
Find an expression for \(u_{n}\) in the form \(an+b\), where \(a,b\in\mathbb{Z}\).
(2)
(c)
Find the value of \(n\) for which \(u_{n}=-121\).
(2)
(d)
Determine whether \(-100\) is a term of the sequence.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(d=-6\)
A1
(b)
\(u_{n}=23+(n-1)(-6)\)
(M1)
\(u_{n}=29-6n\)
A1
(c)
\(29-6n=-121\)
(M1)
FT from (b).
\(n=25\)
A1
(d)
\(29-6n=-100\Rightarrow n=21.5\)
M1
\(n\) is not an integer, so \(-100\) is not a term
R1
Do not award R1 without a valid method.
Answers: (a) \(-6\) · (b) \(u_{n}=29-6n\) · (c) \(n=25\) · (d) not a term
6AASL1-q00141.4 Financial applicationsPaper 1Easy6 marks
Mira deposits $4000 in an account that pays simple interest at a fixed annual rate. The value of the account at the end of each year forms an arithmetic sequence. After 6 years, the value of the account is $4780.
(a)
Find the interest paid each year.
(2)
(b)
Find the annual rate of interest.
(2)
(c)
Find the number of complete years after which the value of the account first exceeds $6000.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(4000+6d=4780\)
(M1)
$130
A1
(b)
\(\dfrac{130}{4000}\times100\)
(M1)
FT from (a).
\(3.25\%\)
A1
(c)
\(4000+130n\gt 6000\Rightarrow n\gt 15.38\ldots\)
(M1)
16 years
A1
Answers: (a) $130 · (b) 3.25% · (c) 16 years
7AASL1-q00151.4 Financial applicationsPaper 1Easy6 marks
A machine is bought for $48 000. Its value decreases by the same amount at the end of each year. After 3 years the machine is worth $37 200.
(a)
Find the amount by which the value decreases each year.
(2)
(b)
Find the value of the machine after 10 years.
(1)
(c)
Find the number of complete years after which the value of the machine is first less than 10% of the price paid for it.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{48\,000-37\,200}{3}\)
(M1)
$3600
A1
(b)
\(48\,000-10(3600)=12\,000\), so $12 000
A1
FT from (a).
(c)
10% of the price is $4800
(A1)
\(48\,000-3600n\lt 4800\Rightarrow n\gt 12\)
(M1)
After 12 years the value is exactly $4800.
13 years
A1
Award A0 for 12.
Answers: (a) $3600 · (b) $12 000 · (c) 13 years
8AASL1-q00231.3 Geometric sequences and seriesPaper 1Easy7 marks
Consider the geometric sequence \(2187,\ 729,\ 243,\ 81,\ \ldots\)
(a)
Write down the common ratio.
(1)
(b)
Find \(u_{8}\).
(2)
(c)
Given that \(2187=3^{7}\), show that \(u_{n}=3^{8-n}\).
(2)
(d)
Hence find the value of \(n\) for which \(u_{n}=\dfrac{1}{81}\).
17AASL1-q00641.7 Rational exponents and laws of logarithmsPaper 1Easy6 marks
(a)
Find the value of \(2\log_{6}3+\log_{6}4\).
(3)
(b)
Write \(3\ln2-\ln4+\ln5\) in the form \(\ln k\), where \(k\in\mathbb{Z}\).
(2)
(c)
Write down the value of \(\log_{5}1+\log_{5}5\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(2\log_{6}3=\log_{6}9\)
M1
\(\log_{6}9+\log_{6}4=\log_{6}36\)
M1
\(2\)
A1
(b)
\(\ln8-\ln4+\ln5=\ln\dfrac{8\times5}{4}\)
(M1)
Award for at least two correct uses of the laws.
\(\ln10\)
A1
(c)
\(0+1=1\)
A1
Answers: (a) 2 · (b) \(\ln10\) · (c) 1
18AASL1-q00661.7 Rational exponents and laws of logarithmsPaper 1Easy8 marks
The temperature, \(T\) °C, of a cup of tea \(t\) minutes after it is poured is modelled by \(T=20+70\mathrm{e}^{-kt}\), \(t\ge0\), where \(k\) is a positive constant. After 10 minutes the temperature of the tea is 55 °C.
(a)
Write down the temperature of the tea when it is poured.
(1)
(b)
Show that \(k=\dfrac{\ln2}{10}\).
(3)
(c)
Find the time taken for the temperature of the tea to fall to 37.5 °C.
(3)
(d)
Explain why, according to the model, the temperature of the tea never falls to 20 °C.
(1)
Show mark scheme
Working
Mark
Notes
(a)
90 °C
A1
(b)
\(20+70\mathrm{e}^{-10k}=55\)
M1
\(\mathrm{e}^{-10k}=\tfrac{1}{2}\)
A1
\(-10k=\ln\tfrac{1}{2}=-\ln2\)
A1
\(k=\dfrac{\ln2}{10}\)
AG
(c)
\(70\mathrm{e}^{-kt}=17.5\), i.e. \(\mathrm{e}^{-kt}=\tfrac{1}{4}\)
(M1)
\(-\dfrac{\ln2}{10}t=\ln\tfrac{1}{4}=-2\ln2\)
(M1)
Or \(\mathrm{e}^{-10k}=\tfrac{1}{2}\), so \(\tfrac{1}{4}=\left(\mathrm{e}^{-10k}\right)^{2}=\mathrm{e}^{-20k}\).
\(t=20\) minutes
A1
(d)
\(\mathrm{e}^{-kt}\gt 0\) for all \(t\), so \(T\gt 20\)
R1
Accept \(70\mathrm{e}^{-kt}=0\) has no solution.
Answers: (a) 90 °C · (c) 20 minutes
19AASL1-q00741.8 Infinite geometric seriesPaper 1Easy5 marks
Consider the infinite geometric series \(24-12+6-3+\cdots\).
(a)
Write down the common ratio, \(r\).
(1)
(b)
Find the sum to infinity of the series.
(2)
(c)
A different infinite geometric series has sum to infinity 45 and common ratio \(\tfrac{1}{3}\). Find its first term.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(r=-\tfrac{1}{2}\)
A1
(b)
\(\dfrac{24}{1-\left(-\frac{1}{2}\right)}\)
(M1)
\(16\)
A1
(c)
\(\dfrac{u_{1}}{1-\frac{1}{3}}=45\)
(M1)
\(u_{1}=30\)
A1
Answers: (a) \(-\tfrac{1}{2}\) · (b) 16 · (c) 30
20AASL1-q00831.9 The binomial theoremPaper 1Easy5 marks
(a)
Write down the row of Pascal’s triangle that gives the coefficients in the expansion of \((a+b)^{5}\).
(1)
(b)
Use the formula for \(\dbinom{n}{r}\) to show that \(\dbinom{7}{3}=35\).
(2)
(c)
Write down the value of \(\dbinom{7}{4}\).
(1)
(d)
Write down the term in \(a^{2}b^{5}\) in the expansion of \((a+b)^{7}\).
21IBMA2S-044-04-011.9 The binomial theoremPaper 1Easy4 marks
No calculator
Find the coefficient of \(x^3\) in the expansion of \((3-2x)^5\).
(a)
Find the coefficient of \(x^3\) in the expansion of \((3-2x)^5\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the general term \(\binom{5}{r}3^{5-r}(-2x)^r\) with \(r=3\) A1 \(\binom{5}{3}=10\) A1 correct powers \(3^2\) and \((-2)^3\), i.e. \(10\times 9\times(-8)\) A1 coefficient \(=-720\)
✓ 4
Common error: Raise the whole of \(-2x\) to the power, not just \(x\). Writing \(-2x^3\) instead of \((-2x)^3\) loses both the factor 8 and the sign.
22IBMA2S-044-04-021.9 The binomial theoremPaper 1Easy4 marks
No calculator
Find the term independent of \(x\) in the expansion of \(\left(x^2+\frac{3}{x}\right)^6\).
(a)
Find the term independent of \(x\) in the expansion of \(\left(x^2+\frac{3}{x}\right)^6\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 general term \(\binom{6}{r}(x^2)^{6-r}\left(\frac{3}{x}\right)^r=\binom{6}{r}3^r x^{12-3r}\) M1 setting the power of \(x\) to zero: \(12-3r=0\) A1 \(r=4\) A1 term \(=\binom{6}{4}3^4=15\times 81=1215\)
✓ 4
Exam tip: Combine all the powers of \(x\) in the general term into a single power before setting it equal to zero. This makes the value of \(r\) immediate.
23IBMA2S-044-04-041.9 The binomial theoremPaper 1Easy4 marks
No calculator
In the expansion of \((1+2x)^n\), where \(n\) is a positive integer, the coefficient of \(x\) is 24.
(a)
Find the value of \(n\).
(2)
(b)
Hence find the coefficient of \(x^2\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\binom{n}{1}\times 2=24\), i.e. \(2n=24\) A1 \(n=12\)
✓ 2
Part (b)
M1 \(\binom{12}{2}(2)^2\) A1 \(66\times4=264\)
✓ 2
Common error: Squaring only the \(x\) and forgetting \(2^2=4\) leads to 66 instead of 264. Bracket \((2x)^2\) every time.
24IBMA2S-044-07-011.6 Deductive proofPaper 1Easy4 marks
No calculator
A tiler lays a rectangular panel of \((2n-1)\) rows by \((2n+1)\) columns of identical square tiles, where \(n\) is a positive integer. She then adds one extra tile and claims that all the tiles can always be rearranged into a single square.
Panel of 2n − 1 rows by 2n + 1 columns (n = 2 shown) and one extra tile. Not to scale
(a)
Prove that \((2n-1)(2n+1)+1\) is a perfect square for every positive integer \(n\), stating the side length of the square in terms of \(n\).
(3)
(b)
A panel is 19 tiles by 21 tiles. State the side length of the square formed when one extra tile is added.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 expands \((2n-1)(2n+1)\) to \(4n^2-1\)
✓ 1
A1 adds 1 to obtain \(4n^2\)
✓ 1
A1 writes \(4n^2=(2n)^2\), so the tiles form a square of side \(2n\) tiles
✓ 1
Part (b)
A1 \(n=10\), so the side length is 20 tiles
✓ 1
Exam tip: In a 'prove' question, finish with a clear statement that links your algebra to the claim, e.g. 'which is \((2n)^2\), a perfect square'.
25IBMA2S-044-07-021.6 Deductive proofPaper 1Easy5 marks
No calculator
Let \(n\) be a positive integer.
(a)
Prove that \(\frac{1}{n}-\frac{1}{n+1}\equiv\frac{1}{n(n+1)}\).
(2)
(b)
Hence find the exact value of \(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\dots+\frac{1}{24\times25}\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes the left-hand side over the common denominator \(n(n+1)\)
✓ 1
A1 numerator \((n+1)-n=1\), giving \(\frac{1}{n(n+1)}\), which is the right-hand side
✓ 1
Part (b)
M1 rewrites each term using the identity proved in the first part, e.g. \(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\dots+\left(\frac{1}{24}-\frac{1}{25}\right)\)
✓ 1
M1 recognises that all intermediate terms cancel, leaving \(1-\frac{1}{25}\)
✓ 1
A1 \(\frac{24}{25}\)
✓ 1
Exam tip: Start from one side only and work to the other. Do not write the identity down and manipulate both sides at once.
26IBMA2S-044-06-011.1 Standard formPaper 1Easy4 marks
No calculator
A transistor on a modern microchip has a gate length of 5×10−9 m. A human hair has a width of 0.00008 m.
(a)
Write the width of the hair in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ.
(1)
(b)
Find how many transistor gate lengths, placed end to end, would span the width of the hair. Give your answer in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 8×10−5 m
✓ 1
Part (b)
M1 dividing the hair width by the gate length: (8×10−5) ÷ (5×10−9) M1 dealing with the parts separately: 8 ÷ 5 = 1.6 and 10−5−(−9) = 104 A1 1.6×104
✓ 3
Common error: Subtracting a negative index is where marks are lost: 10⁻⁵ ÷ 10⁻⁹ = 10⁻⁵⁺⁹ = 10⁴, not 10⁻¹⁴.
27IBMA2S-044-06-031.7 Rational exponents and laws of logarithmsPaper 1Easy4 marks
No calculator
A hospital prescribes an antibiotic at a dose of 2.4×10−4 grams per kilogram of body mass. A patient has a mass of 7.5×101 kg.
(a)
Find the total dose for this patient, in grams. Give your answer in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ.
(3)
(b)
Write the total dose in milligrams.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 multiplying the dose rate by the mass: (2.4×10−4)×(7.5×101) M1 2.4×7.5 = 18 and 10−4×101 = 10−3, giving 18×10−3 A1 1.8×10−2 g
✓ 3
Part (b)
A1 18 mg
✓ 1
Common error: 18×10⁻³ is not in standard form because 18 is not between 1 and 10. Adjust it to 1.8×10⁻² by raising the index by one.
28IBMA2S-044-01-011.9 The binomial theoremPaper 1Easy5 marks
No calculator
Answer the following without using a calculator.
(a)
Find the exact value of \(32^{-\frac{3}{5}}\).
(2)
(b)
Simplify \(\dfrac{(27x^{6})^{\frac{2}{3}}}{(4x^{-2})^{\frac{1}{2}}}\), where \(x>0\), giving your answer in the form \(ax^{n}\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognises \(32^{\frac{1}{5}}=2\), or writes \(32=2^5\)
✓ 1
A1 \(2^{-3}=\frac{1}{8}\)
✓ 1
Part (b)
M1 applies the power to both the coefficient and the power of \(x\) in at least one bracket
✓ 1
A1 numerator \(9x^{4}\) and denominator \(2x^{-1}\)
✓ 1
A1 \(\frac{9}{2}x^{5}\)
✓ 1
Common error: A negative exponent gives a reciprocal, not a negative number. \(32^{-3/5}\) is \(\frac{1}{8}\), not \(-8\). Apply a fractional power to every factor inside the bracket, including the number.
29IBMA2S-044-01-021.9 The binomial theoremPaper 1Easy6 marks
No calculator
Let \(\log_b 5 = p\) and \(\log_b 2 = q\), where \(b>0\), \(b\neq 1\). Give each answer in terms of \(p\) and \(q\).
(a)
Find an expression for \(\log_b 40\).
(2)
(b)
Find an expression for \(\log_b\left(\dfrac{\sqrt{5}}{16}\right)\).
(2)
(c)
Find an expression for \(\log_b 0.4\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(40=2^{3}\times 5\) and uses the product or power law
✓ 1
A1 \(3q+p\)
✓ 1
Part (b)
M1 uses the quotient law with \(\sqrt{5}=5^{\frac{1}{2}}\) and \(16=2^{4}\)
✓ 1
A1 \(\frac{1}{2}p-4q\)
✓ 1
Part (c)
M1 writes \(0.4=\frac{2}{5}\) (or \(\frac{4}{10}=\frac{2^2}{2\times 5}\)) and uses the quotient law
✓ 1
A1 \(q-p\)
✓ 1
Exam tip: Before using any log law, write the number as a product of the primes you have been given. Decimals such as 0.4 become fractions first.
30IBMA2S-044-00-011.2 Arithmetic sequences and seriesPaper 1Easy5 marks
No calculator
As part of a rehabilitation programme after knee surgery, a patient walks 250 m on day 1. Each day after that she walks 40 m more than on the previous day.
(a)
Find the first day on which she walks more than 1 km.
(3)
(b)
Find the total distance she walks over the first 20 days of the programme.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets up \(250+40(n-1)>1000\)
✓ 1
A1 \(n>19.75\)
✓ 1
A1 day 20
✓ 1
Part (b)
M1 \(S_{20}=\frac{20}{2}(2\times250+19\times40)\)
✓ 1
A1 12 600 m (12.6 km)
✓ 1
Common error: Solving \(250+40n>1000\) (forgetting the \(n-1\)) gives day 19, where she walks only 970 m. Check the day you find by working out that day's distance.
31IBMA2S-044-00-021.8 Infinite geometric seriesPaper 1Easy5 marks
No calculator
In a shock-absorber test, a heavily damped pendulum is released. The tip of the pendulum moves 54 cm along its first swing. Each swing after that is a fixed fraction of the length of the previous swing. If the motion carried on forever, the total distance moved by the tip would be 81 cm.
(a)
Find the common ratio of the lengths of the swings.
Common error: The fourth term is \(u_1r^3\), not \(u_1r^4\). The power of \(r\) is always one less than the term number.
32IBMA2S-044-00-031.2 Arithmetic sequences and seriesPaper 1Easy5 marks
No calculator
A vineyard is planted on a hillside cut into 30 terraces. The numbers of vines on successive terraces form an arithmetic sequence. Terrace 5 has 38 vines and terrace 14 has 65 vines.
(a)
Find the common difference and the number of vines on terrace 1.
(3)
(b)
Find the total number of vines on all 30 terraces.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 forms \(u_1+4d=38\) and \(u_1+13d=65\), or uses \(9d=27\)
✓ 1
A1 \(d=3\)
✓ 1
A1 \(u_1=26\)
✓ 1
Part (b)
M1 \(S_{30}=\frac{30}{2}(2\times26+29\times3)\)
✓ 1
A1 2085
✓ 1
Exam tip: From terrace 5 to terrace 14 there are 14 − 5 = 9 steps of size \(d\), not 10. Count the gaps between the terms, not the terms themselves.
33IBMA2S-044-02-011.3 Geometric sequences and seriesPaper 1Easy5 marks
No calculator
A short video is posted online. The number of new shares it receives in hour \(n\) after posting follows a geometric sequence. There are 6 new shares in hour 2 and 162 new shares in hour 5.
(a)
Find the common ratio and the number of new shares in hour 1.
(3)
(b)
Find the total number of new shares in the first 6 hours.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 forms \(u_1r=6\) and \(u_1r^4=162\) and divides to get \(r^3=27\)
✓ 1
A1 \(r=3\)
✓ 1
A1 \(u_1=2\)
✓ 1
Part (b)
M1 \(S_6=\frac{2(3^6-1)}{3-1}\) OR lists and adds the six terms \(2+6+18+54+162+486\)
✓ 1
A1 728
✓ 1
Exam tip: Dividing one term by another cancels \(u_1\) and leaves a power of \(r\). Here the power is \(5-2=3\), not 5.
34IBMA2S-044-02-031.3 Geometric sequences and seriesPaper 1Easy6 marks
No calculator
Consider the series \(\displaystyle\sum_{k=1}^{n}3\times2^{k}\).
(a)
Write down the first term and the common ratio of the series.
(2)
(b)
Show that \(\displaystyle\sum_{k=1}^{n}3\times2^{k}=6(2^{n}-1)\).
(2)
(c)
Given that \(\displaystyle\sum_{k=1}^{n}3\times2^{k}=1530\), find the value of \(n\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 first term 6
✓ 1
A1 common ratio 2
✓ 1
Part (b)
M1 substitutes into \(S_n=\frac{6(2^n-1)}{2-1}\)
✓ 1
A1 simplifies to \(6(2^n-1)\)
✓ 1
Part (c)
M1 \(2^n-1=255\), so \(2^n=256\)
✓ 1
A1 \(n=8\)
✓ 1
Common error: The first term is found by substituting \(k=1\), which gives \(3\times2=6\). It is not 3.
35IBMA2S-044-05-011.8 Infinite geometric seriesPaper 1Easy6 marks
No calculator
A data centre contains \(4^{12}\) identical storage drives. Each drive holds \(8^{5}\) bytes of data.
(a)
Show that the total storage of the data centre is \(2^{39}\) bytes.
(2)
(b)
A backup file has size \(16^{3}\) bytes. Find the number of such files that the data centre can store, giving your answer in the form \(2^{k}\).
(2)
(c)
The total storage can also be written as \(32^{n}\) bytes. Find the exact value of \(n\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(4^{12}=(2^2)^{12}=2^{24}\) and \(8^5=(2^3)^5=2^{15}\)
✓ 1
A1 \(2^{24}\times 2^{15}=2^{39}\)
✓ 1
Part (b)
M1 \(16^3=2^{12}\) and divides \(2^{39}\div 2^{12}\)
✓ 1
A1 \(2^{27}\)
✓ 1
Part (c)
M1 \(32^n=2^{5n}\) so \(5n=39\)
✓ 1
A1 \(n=\frac{39}{5}\)
✓ 1
Exam tip: Rewrite every base as a power of the same prime before applying the index laws; then compare exponents.
36IBMA2S-044-05-021.5 Exponents and introduction to logarithmsPaper 1Easy6 marks
No calculator
The sound level \(L\) decibels (dB) of a noise of intensity \(I\) is given by \(L=10\log_{10}\left(\frac{I}{I_0}\right)\), where \(I_0\) is a fixed reference intensity.
(a)
A vacuum cleaner produces a noise of intensity \(10^{6}I_0\). Find its sound level.
(2)
(b)
A drone produces a sound level of 85 dB. Express its intensity in the form \(10^{p}I_0\).
(2)
(c)
A stadium crowd's sound level rises by 20 dB when a goal is scored. Find the factor by which the sound intensity increases.
Common error: Multiply the whole exponent by 2: \((3^2)^{x+1}=3^{2x+2}\), not \(3^{2x+1}\).
38IBMA2S-044-03-021.4 Financial applicationsPaper 1Easy6 marks
No calculator
A collector buys a rare sports trading card for $6400. Two years later the card is valued at $10 000. Assume the value of the card increases by a fixed percentage, r%, each year.
(a)
Show that r = 25.
(3)
(b)
Find the value of the card 4 years after it was bought.
(2)
(c)
Assuming the same rate of growth applied before the purchase, find the value of the card one year before it was bought.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets up 6400(1 + r/100)2 = 10 000 A1 (1 + r/100)2 = \(\frac{10000}{6400} = \frac{25}{16}\) A1 1 + r/100 = \(\frac{5}{4}\), rejecting the negative root because the growth factor must be positive (the value increases), so r = 25
Exam tip: Write the growth factor as an exact fraction such as 5/4. Powers of fractions are much easier to work out by hand than powers of decimals.
39IBMA3S-1-16r10-011.1 Standard formPaper 1Easy5 marks
No calculator
A geostationary communications satellite is 3.6 × 107 m above a ground station. Radio signals travel at 3 × 108 m s−1.
(a)
Find the time for a signal to travel from the ground station to the satellite. Give your answer in the form \(a \times 10^k\) where \(1 \le a < 10\) and \(k \in \mathbb{Z}\).
(2)
(b)
Write down the time for a signal to travel to the satellite and back to the ground station.
(1)
(c)
A message of \(4.8 \times 10^{6}\) bits is sent at \(1.2 \times 10^{6}\) bits per second. Find the time taken to send it.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to divide distance by speed
✓ 1
A1 \(1.2 \times 10^{-1}\) s
✓ 1
Part (b)
A1 \(2.4 \times 10^{-1}\) s
✓ 1
Part (c)
M1 \(\frac{4.8 \times 10^6}{1.2 \times 10^6}\)
✓ 1
A1 4 seconds
✓ 1
Common error: When dividing numbers in standard form, divide the coefficients and subtract the powers separately, then check that the coefficient lies in [1, 10).
40IBMA3S-1-16r10-031.8 Infinite geometric seriesPaper 1Easy7 marks
No calculator
In a fermentation tank, 12 litres of carbon dioxide are released in the first hour. In each later hour the volume released is \(\frac{1}{3}\) of the volume in the previous hour, so the volumes form a geometric sequence.
(a)
Find the volume released in the 3rd hour.
(2)
(b)
Find the total volume that would be released if the process continued indefinitely.
(2)
(c)
Find the exact total volume released in the first 4 hours.
Common error: The sum to infinity formula \(\frac{u_1}{1-r}\) only applies when \(|r| < 1\); check this before using it.
41IBMA3S-1-16r11-011.3 Geometric sequences and seriesPaper 1Easy5 marks
No calculator
A library adds new books to its shelves each week. The number of new books added in week \(n\) forms an arithmetic sequence. In week 4, 23 books are added and in week 9, 53 books are added.
(a)
Find the common difference.
(2)
(b)
Find the number of books added in week 1.
(1)
(c)
Find the total number of books added in the first 20 weeks.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(u_9-u_4=5d\)
✓ 1
A1 \(d=6\)
✓ 1
Part (b)
A1 \(u_1=23-3\times6=5\)
✓ 1
Part (c)
M1 \(S_{20}=\frac{20}{2}(2\times5+19\times6)\)
✓ 1
A1 1240 books
✓ 1
Common error: The 9th term is 5 steps from the 4th term, not 4. Count the number of differences, not the number of terms.
42IBMA3S-1-16r11-031.8 Infinite geometric seriesPaper 1Easy4 marks
No calculator
After the power to a wind turbine is cut, the rotor slows. In the first second it turns through \(80^\circ\). In each following second it turns through \(\frac{3}{4}\) of the angle it turned through in the previous second.
(a)
Find the total angle the rotor turns through, in degrees, if it keeps turning indefinitely under this model.
(2)
(b)
Hence show that the rotor never completes one full revolution.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(S_\infty=\frac{u_1}{1-r}\) with \(u_1=80,\ r=\frac34\)
✓ 1
A1 \(\frac{80}{1/4}=320^\circ\)
✓ 1
Part (b)
R1 the sum of any number of terms is less than \(S_\infty=320\) (all terms positive)
✓ 1
R1 \(320<360\), so it never completes a full turn
✓ 1
Common error: The sum to infinity formula needs \(|r|<1\). Check this first, then substitute carefully using the first term.
43IBMA3S-1-16r13-011.1 Standard formPaper 1Easy4 marks
No calculator
A turbine in an aircraft engine rotates \(1.5\times10^{4}\) times per minute. Give each answer in the form \(a\times10^{k}\) where \(1\le a<10\).
(a)
Find the number of rotations in \(4\times10^{2}\) minutes.
(2)
(b)
The engine is overhauled after \(2.4\times10^{3}\) hours of running. Find the number of rotations between overhauls.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to multiply \(1.5\times10^{4}\) by \(4\times10^{2}\)
✓ 1
A1 \(6\times10^{6}\)
✓ 1
Part (b)
M1 attempt to convert hours to minutes, for example \(1.5\times10^{4}\times60\times2.4\times10^{3}\)
✓ 1
A1 \(2.16\times10^{9}\) (= 2160000000)
✓ 1
Common error: When converting hours to minutes, multiply by 60. Rewrite the final product so the number before the power of ten is between 1 and 10.
44IBMA3S-1-16r13-021.2 Arithmetic sequences and seriesPaper 1Easy4 marks
No calculator
The first row of an orchard has 7 trees. Each later row has 4 more trees than the row before, so the numbers of trees form an arithmetic sequence.
Each dot represents one tree. Diagram not to scale
(a)
Find the number of trees in the 15th row.
(2)
(b)
Find the total number of trees in the first 15 rows.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_{15}=7+14\times4\)
✓ 1
A1 63
✓ 1
Part (b)
M1 \(S_{15}=\frac{15}{2}(7+63)\) (ECF from (a))
✓ 1
A1 525
✓ 1
Exam tip: Find the last term first, then use the sum formula with the first and last terms.
45IBMA3S-1-16r13-031.9 The binomial theoremPaper 1Easy4 marks
No calculator
In one tuning system, the interval from a note of frequency \(f_1\) Hz to a note of frequency \(f_2\) Hz is \(n=12\log_2\left(\frac{f_2}{f_1}\right)\) semitones.
(a)
Find the interval from a note of frequency 220 Hz to a note of frequency 1760 Hz.
(2)
(b)
A second note is 24 semitones below a note of frequency 220 Hz, that is, the interval from 220 Hz to this note is \(-24\). Find its frequency.
M1 \(\log_2\left(\frac{f}{220}\right)=-2\) or \(\frac{f}{220}=2^{-2}\)
✓ 1
A1 \(f=55\) Hz
✓ 1
Common error: Remember \(\log_2 8=3\) because \(2^3=8\); do not divide 8 by 2.
46IBMA3S-1-16r13-041.6 Deductive proofPaper 1Easy4 marks
No calculator
Prove that the sum of the squares of two consecutive integers is always odd.
(a)
Write a proof.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 let the integers be \(n\) and \(n+1\) and form \(n^2+(n+1)^2\)
✓ 1
A1 \(=2n^2+2n+1\)
✓ 1
M1 write as \(2(n^2+n)+1\)
✓ 1
R1 \(n^2+n\) is an integer, so the expression has the form \(2k+1\) and is odd
✓ 1
Exam tip: State what each letter represents and finish with a sentence that links back to the claim.
47IBMA3S-1-17r13-011.1 Standard formPaper 1Easy4 marks
No calculator
A beekeeper models the number of bees in a colony using \(N = 5000 \times 3^{t/4}\), where \(t\) is the time in weeks since the colony was established.
(a)
Find the number of bees after 2 weeks. Give your answer in exact form.
(2)
(b)
Find the value of \(t\) for which the colony has 45 000 bees.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute \(t=2\) to get \(N = 5000\times 3^{1/2}\)
Common error: Writing \(3^{1/2}\) as \(\frac{3}{2}\). A power of \(\frac12\) means a square root.
48IBMA3S-1-17r13-031.8 Infinite geometric seriesPaper 1Easy4 marks
No calculator
A hospital pump delivers a medicine in hourly doses. In the first hour it delivers 12 ml. In each later hour it delivers \(\frac{2}{3}\) of the volume it delivered in the previous hour.
(a)
Find the volume delivered in the 5th hour.
(2)
(b)
The pump runs for a very long time. Find the limit that the total volume delivered approaches.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(12\times\left(\frac{2}{3}\right)^4\)
✓ 1
A1 \(\frac{64}{27}\) ml (≈ 2.37 ml)
✓ 1
Part (b)
M1 \(S_\infty = \frac{12}{1-\frac{2}{3}}\)
✓ 1
A1 36 ml
✓ 1
Exam tip: The sum to infinity formula needs \(|r|<1\). Quote this condition when you use it.
49IBMA3S-1-17r13-051.9 The binomial theoremPaper 1Easy5 marks
No calculator
Let \(\log_5 x = p\) and \(\log_5 y = q\), where \(x\) and \(y\) are positive.
(a)
Express \(\log_5 (x^2 y)\) in terms of \(p\) and \(q\).
(2)
(b)
Express \(\log_5\left(25\sqrt{\dfrac{x}{y^3}}\right)\) in terms of \(p\) and \(q\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\log_5 x^2 + \log_5 y\)
✓ 1
A1 \(2p + q\)
✓ 1
Part (b)
M1 \(\log_5 25 + \frac{1}{2}(\log_5 x - 3\log_5 y)\)
✓ 1
A1 \(\log_5 25 = 2\)
✓ 1
A1 \(2 + \frac{p}{2} - \frac{3q}{2}\)
✓ 1
Common error: Writing \(\log_5(x^2y)\) as \(2\log_5 x \times \log_5 y\). Products inside a logarithm become sums outside it.
50IBMA3S-1-16r14-021.3 Geometric sequences and seriesPaper 1Easy5 marks
No calculator
The amount of a drug in a patient's bloodstream, in mg, at the start of each successive hour forms a geometric sequence. At the start of the first hour there are 640 mg and the common ratio is \(\frac{1}{2}\).
(a)
Find the amount of the drug at the start of the fifth hour.
(2)
(b)
At the start of hour n there are 5 mg of the drug. Find the value of n.
Exam tip: Always check \(|r|<1\) before using the sum to infinity formula.
53IBMA3S-1-17r15-011.8 Infinite geometric seriesPaper 1Easy4 marks
No calculator
A train begins braking in a tunnel. In the first minute of braking it travels 1600 m. In each later minute it travels 0.2 times the distance of the previous minute. Treat the distances in successive minutes as an infinite geometric sequence.
(a)
Find the total distance the train travels before it comes to rest, according to this model.
(2)
(b)
Find the distance travelled in the 4th minute.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(S_\infty=\frac{u_1}{1-r}\) with \(u_1=1600\), \(r=0.2\)
✓ 1
A1 \(\frac{1600}{0.8}=2000\) m
✓ 1
Part (b)
M1 \(u_4=1600\times0.2^{3}\)
✓ 1
A1 \(12.8\) m
✓ 1
Exam tip: Check that \(|r|<1\) before using the sum to infinity formula and say so in your working.
54IBMA3S-1-17r15-031.9 The binomial theoremPaper 1Easy5 marks
No calculator
Let \(p=\log_3 5\) and \(q=\log_3 2\). Express each of the following in terms of \(p\) and \(q\).
(a)
\(\log_3 10\)
(2)
(b)
\(\log_3\left(\dfrac{45}{8}\right)\)
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\log_3(5\times2)=\log_3 5+\log_3 2\)
✓ 1
A1 \(p+q\)
✓ 1
Part (b)
M1 \(\log_3 45=\log_3 9+\log_3 5=2+p\)
✓ 1
M1 \(\log_3 8=3\log_3 2=3q\) and quotient law
✓ 1
A1 \(2+p-3q\)
✓ 1
Exam tip: Break each number into factors you know the logarithm of, such as 9, 5 and 2 here, before applying the laws.
55IBMA3S-1-17r15-131.8 Infinite geometric seriesPaper 1Easy6 marks
No calculator
The recurring decimal \(0.272727\ldots\) can be written as the infinite geometric series \(0.27+0.0027+0.000027+\ldots\)
(a)
Write down the common ratio.
(1)
(b)
Hence show that \(0.272727\ldots=\frac{3}{11}\).
(3)
(c)
Hence write \(3.272727\ldots\) as a fraction in the form \(\frac{a}{b}\), where \(a,b\) are integers.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(r=0.01\)
✓ 1
Part (b)
M1 \(S_\infty=\frac{0.27}{1-0.01}\)
✓ 1
A1 \(\frac{27}{99}\)
✓ 1
A1 \(\frac{3}{11}\)
✓ 1
Part (c)
M1 \(3+\frac{3}{11}\)
✓ 1
A1 \(\frac{36}{11}\) (\(=3.272727\ldots\))
✓ 1
Exam tip: For a recurring decimal, the common ratio is a power of 0.1 matching the length of the repeating block.
56IBMA3S-1-17r16-021.6 Deductive proofPaper 1Easy4 marks
No calculator
Two swimmers are placed in neighbouring lanes of a competition pool. The lane numbers are consecutive positive integers. A coach claims that the sum of the squares of two consecutive integers is always an odd number.
(a)
Prove that the coach's claim is true for all consecutive integers.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 let the integers be n and n + 1
✓ 1
M1 sum = n2 + (n + 1)2 = 2n2 + 2n + 1
✓ 1
A1 = 2(n2 + n) + 1
✓ 1
R1 2(n2 + n) is even since n2 + n is an integer, so the sum is even + 1 = odd; hence the claim is true
✓ 1
Exam tip: In a proof, define your variables, show every algebraic step and finish with a sentence that states the conclusion in words.
57IBMA3S-1-17r16-081.1 Standard formPaper 1Easy4 marks
No calculator
A wind farm produces 6.4 × 108 kWh of electricity in a year. The farm supplies 1.6 × 105 households equally.
(a)
Find the energy supplied to each household in one year. Give your answer in the form a × 10k, where 1 ≤ a < 10 and k ∈ ℤ.
(2)
(b)
Each household in a different town uses 2.5 × 103 kWh per year. Find how many such households the wind farm could supply. Give your answer in the form a × 10k.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (6.4 × 108) ÷ (1.6 × 105)
✓ 1
A1 4 × 103 kWh
✓ 1
Part (b)
M1 (6.4 × 108) ÷ (2.5 × 103)
✓ 1
A1 2.56 × 105
✓ 1
Common error: When dividing, subtract the powers of ten: 10⁸ ÷ 10⁵ = 10³, not 10¹³ or 10¹.⁶.
58IBMA3S-1-17r16-101.1 Standard formPaper 1Easy6 marks
M1 correct use of the power \(\frac23\) on a term, e.g. \(27^{\frac23}=9\) or \((x^9)^{\frac23}=x^6\)
✓ 1
A1 \((27x^9)^{\frac23}=9x^6\)
✓ 1
A1 \((9x^{-2})^{\frac12}=3x^{-1}\)
✓ 1
A1 \(a=3\), \(b=7\)
✓ 1
Part (b)
M1 \(3x^7=384\), so \(x^7=128\)
✓ 1
A1 \(x=2\)
✓ 1
Common error: A power applied to a bracket applies to every factor inside, including the number: \((27x^9)^{\frac23}\) is \(9x^6\), not \(27x^6\).
62IBMA3S-1-16r20-011.6 Deductive proofPaper 1Easy4 marks
No calculator
Consider three consecutive integers.
(a)
Prove that the sum of any three consecutive integers is divisible by 3.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 let the three integers be \(n\), \(n+1\) and \(n+2\), where \(n\) is an integer
✓ 1
A1 sum \(= n+(n+1)+(n+2)\)
✓ 1
A1 \(=3n+3\)
✓ 1
R1 \(=3(n+1)\), and \(n+1\) is an integer, so the sum is a multiple of 3 (divisible by 3)
✓ 1
Exam tip: A proof needs a general statement (let \(n\) be any integer), not a check of particular numbers, and a closing sentence that states the conclusion.
63IBMA3S-1-16r20-021.9 The binomial theoremPaper 1Easy5 marks
No calculator
Consider the expansion of \((3+x)^5\).
(a)
Find the coefficient of \(x^3\).
(3)
(b)
Hence, or otherwise, find the coefficient of \(x^3\) in the expansion of \((3+2x)^5\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(\binom{5}{3}=10\)
✓ 1
A1 \(3^2=9\)
✓ 1
A1 coefficient \(=90\)
✓ 1
Part (b)
M1 multiplies by \(2^3\) (replaces \(x\) by \(2x\)): \(8\times 90\)
✓ 1
A1 \(720\)
✓ 1
Common error: When \(x\) is replaced by \(2x\), the factor \(2\) is cubed along with \(x\); writing \(2\times 90\) is a common slip.
64IBMA3S-1-16r20-031.8 Infinite geometric seriesPaper 1Easy5 marks
No calculator
(a)
Evaluate \(81^{\frac{3}{4}}\).
(2)
(b)
Solve \(4^{x+1}=32\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(81^{\frac14}=3\)
✓ 1
A1 \(27\)
✓ 1
Part (b)
M1 writes both sides as powers of 2: \(2^{2(x+1)}=2^5\)
✓ 1
A1 \(2x+2=5\)
✓ 1
A1 \(x=\frac{3}{2}\)
✓ 1
Exam tip: Writing both sides of an exponential equation with the same base lets you equate the powers without using logarithms.
65IBMA3S-1-17r20-011.9 The binomial theoremPaper 1Easy4 marks
No calculator
Consider the expansion of \((2x+3)^5\) in descending powers of \(x\).
(a)
Write down the number of terms in the expansion.
(1)
(b)
Find the coefficient of \(x^3\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 6 terms
✓ 1
Part (b)
M1 attempt to use the binomial theorem, for example \(\binom{5}{3}(2x)^3(3)^2\)
✓ 1
A1 correct term \(10\times 8x^3\times 9\)
✓ 1
A1 720
✓ 1
Common error: When the first bracket contains a coefficient such as 2x, remember to raise the whole 2 to the power, not just the x.
66IBMA3S-1-17r20-031.5 Exponents and introduction to logarithmsPaper 1Easy5 marks
No calculator
Solve the equation \(\log_3 x+\log_3 (x-8)=2\).
(a)
Solve the equation \(\log_3 x+\log_3 (x-8)=2\).
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\log a+\log b=\log ab\), giving \(\log_3 x(x-8)=2\)
✓ 1
M1 convert to exponential form, \(x(x-8)=3^2\)
✓ 1
A1 \(x^2-8x-9=0\)
✓ 1
A1 \(x=9\) and \(x=-1\)
✓ 1
A1 \(x=9\) only, since \(\log_3(-1)\) is not defined
✓ 1
Common error: Solving the quadratic is not the end: substitute each root back and reject any that make a logarithm's argument zero or negative.
67IBMA3S-1-17r20-041.6 Deductive proofPaper 1Easy4 marks
No calculator
Prove that the difference between the squares of two consecutive odd integers is always a multiple of 8.
(a)
Prove that the difference between the squares of two consecutive odd integers is always a multiple of 8.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 represent two consecutive odd integers algebraically, for example \(2n-1\) and \(2n+1\) where \(n\) is an integer
✓ 1
M1 form and expand \((2n+1)^2-(2n-1)^2=(4n^2+4n+1)-(4n^2-4n+1)\)
✓ 1
A1 \(=8n\)
✓ 1
R1 \(8n\) is a multiple of 8 because \(n\) is an integer, so the result holds for every such pair
✓ 1
Exam tip: A proof must work for a general integer: testing a few pairs such as 3 and 5 earns no credit on its own.
68IBMA3S-1-17r22-021.3 Geometric sequences and seriesPaper 1Easy5 marks
No calculator
In a video game, the treasure chest on each level contains coins, and the numbers of coins in the chests on levels 1, 2, 3, … form a geometric sequence. The level 3 chest contains 18 coins and the level 6 chest contains 486 coins.
(a)
Find the common ratio and the number of coins in the level 1 chest.
(3)
(b)
Find the total number of coins in the chests on levels 1 to 8.
Exam tip: Dividing two terms of a geometric sequence gives a power of r; take the correct root of the ratio.
69IBMA3S-1-17r22-041.6 Deductive proofPaper 1Easy5 marks
No calculator
Consider integers.
(a)
Prove that the sum of any three consecutive integers is a multiple of 3.
(3)
(b)
A student claims that the sum of any four consecutive integers is a multiple of 4. Show that the claim is false.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 let the integers be \(n\), \(n+1\), \(n+2\)
✓ 1
A1 sum \(=3n+3\)
✓ 1
R1 \(=3(n+1)\), which is a multiple of 3 since \(n+1\) is an integer
✓ 1
Part (b)
A1 a valid counterexample, for example \(1+2+3+4=10\)
✓ 1
R1 10 is not a multiple of 4, so the claim is false
✓ 1
Exam tip: To disprove a general statement one counterexample is enough; to prove it you must use algebra for all integers.
70IBMA3S-1-16r25-011.8 Infinite geometric seriesPaper 1Easy4 marks
No calculator
A geometric sequence has first term \(u_1 = 6\), third term \(u_3 = 54\) and a positive common ratio \(r\).
(a)
Find the value of \(r\).
(2)
(b)
Find the sum of the first five terms of the sequence.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(6r^2 = 54\), so \(r^2 = 9\)
✓ 1
A1 \(r = 3\)
✓ 1
Part (b)
M1 \(S_5 = \frac{6(3^5 - 1)}{3 - 1}\)
✓ 1
A1 \(S_5 = 726\)
✓ 1
Exam tip: Use the information that r is positive to reject the negative root as soon as you solve for r.
71IBMA3S-1-16r25-031.7 Rational exponents and laws of logarithmsPaper 1Easy6 marks
No calculator
(a)
Solve \(2^{x+1} = 32\).
(2)
(b)
Find the exact value of \(\log_2 40 - \log_2 5\).
(2)
(c)
Solve \(3^{2x} = 27^{x-1}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(32 = 2^5\), so \(x + 1 = 5\)
✓ 1
A1 \(x = 4\)
✓ 1
Part (b)
M1 \(\log_2\left(\frac{40}{5}\right) = \log_2 8\)
✓ 1
A1 3
✓ 1
Part (c)
M1 \(3^{2x} = 3^{3(x-1)}\), so \(2x = 3x - 3\)
✓ 1
A1 \(x = 3\)
✓ 1
Exam tip: Write both sides of an exponential equation with the same base before comparing the powers.
72IBMA3S-1-16r25-071.1 Standard formPaper 1Easy6 marks
No calculator
A space probe travels at a constant speed of \(1.8 \times 10^{4}\) m s−1. The probe is launched towards a target that is \(1.26 \times 10^{12}\) m away.
(a)
Find the distance travelled by the probe in \(5.0 \times 10^{7}\) seconds. Give your answer in the form \(a \times 10^k\) where \(1 \leq a < 10\) and \(k \in \mathbb{Z}\).
(2)
(b)
Find the distance remaining to the target at this time, giving your answer in the form \(a \times 10^k\).
(2)
(c)
Find the further time, in seconds, needed to reach the target at the same speed. Give your answer in the form \(a \times 10^k\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1.8 \times 10^4 \times 5.0 \times 10^7\)
✓ 1
A1 \(9.0 \times 10^{11}\) m
✓ 1
Part (b)
M1 \(1.26 \times 10^{12} - 9.0 \times 10^{11}\)
✓ 1
A1 \(3.6 \times 10^{11}\) m
✓ 1
Part (c)
M1 \(\frac{3.6 \times 10^{11}}{1.8 \times 10^4}\) (ECF from (b))
✓ 1
A1 \(2.0 \times 10^{7}\) s (that is, 20000000 s)
✓ 1
Exam tip: Convert numbers to the same power of ten before adding or subtracting them in standard form.
73IBMA3S-1-18r29-011.5 Exponents and introduction to logarithmsPaper 1Easy4 marks
No calculator
Each turbine on a wind farm produces 2.4 × 106 W. The farm has 125 turbines, all working at full power.
(a)
Find the total power of the farm. Give your answer in the form a × 10k where 1 ≤ a < 10 and k ∈ ℤ.
(2)
(b)
Find the energy produced by the farm in one hour, in joules. Give your answer in the form a × 10k where 1 ≤ a < 10 and k ∈ ℤ.
Common error: When a product such as 10.8 × 10¹¹ is not in standard form, adjust the power of ten: 10.8 × 10¹¹ = 1.08 × 10¹².
74IBMA3S-1-18r29-031.8 Infinite geometric seriesPaper 1Easy6 marks
No calculator
A swimmer practises a drill in which she swims 64 m in the first minute. In each later minute she swims 3/4 of the distance she swam in the previous minute.
(a)
Find the distance she swims in the 4th minute.
(2)
(b)
Find the total distance she would swim if the pattern continued indefinitely.
(2)
(c)
Find the total distance she swims in the first 4 minutes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of u4 = 64(3/4)3
✓ 1
A1 27 m
✓ 1
Part (b)
M1 use of S∞ = 64 ÷ (1 − 3/4)
✓ 1
A1 256 m
✓ 1
Part (c)
M1 use of S4 = 64(1 − (3/4)4) ÷ (1 − 3/4), or 64 + 48 + 36 + 27
✓ 1
A1 175 m
✓ 1
Exam tip: The sum to infinity formula is only valid when |r| < 1.
75IBMA3S-1-18r29-101.9 The binomial theoremPaper 1Easy6 marks
No calculator
Consider the expansion of (1 + 3x)4.
(a)
Write down the expansion in ascending powers of x, simplifying each coefficient.
(4)
(b)
By substituting a suitable value of x, use the first three terms of your expansion to estimate 1.034.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of binomial coefficients 1, 4, 6, 4, 1 with powers of 3x
✓ 1
A1 1 + 12x
✓ 1
A1 + 54x2 + 108x3
✓ 1
A1 + 81x4
✓ 1
Part (b)
M1 substitute x = 0.01 into 1 + 12x + 54x2
✓ 1
A1 1.1254
✓ 1
Exam tip: Check your expansion by setting x = 1: the coefficients must add to 4⁴ = 256.
76IBMA3S-1-19r29-021.5 Exponents and introduction to logarithmsPaper 1Easy4 marks
No calculator
Evaluate \(\dfrac{25^{3/2}\times 8^{-2/3}}{4^{-1/2}}\), giving your answer as a fraction or decimal.
Exam tip: Split a rational exponent as root then power: \(a^{m/n}=(\sqrt[n]{a})^m\). Take the root first to keep numbers small.
77IBMA3S-1-19r29-041.7 Rational exponents and laws of logarithmsPaper 1Easy6 marks
No calculator
Let \(x\) and \(y\) be positive real numbers with \(p=\log_{10}x\) and \(q=\log_{10}y\).
(a)
Express \(\log_{10}\left(100x\sqrt{y}\right)\) in terms of \(p\) and \(q\).
(3)
(b)
Given that \(\log_{10}\left(\dfrac{x^3}{y}\right)=7\) and \(\log_{10}(xy)=5\), find the value of \(p\) and the value of \(q\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\log(ab)=\log a+\log b\)
✓ 1
A1 \(\log_{10}100=2\)
✓ 1
A1 \(2+p+\frac{q}{2}\)
✓ 1
Part (b)
M1 forming \(3p-q=7\) and \(p+q=5\)
✓ 1
A1 \(p=3\)
✓ 1
A1 \(q=2\)
✓ 1
Common error: \(\log(a+b)\) is not \(\log a+\log b\). Only products turn into sums, and \(\sqrt y\) becomes the power \(\frac12\) before the log law is applied.
78IBMA3S-1-18r31-031.8 Infinite geometric seriesPaper 1Easy6 marks
No calculator
In a computer game, a ball is dropped from a height of 10 m onto the ground. After each bounce it rises to \(\frac{3}{5}\) of the height from which it fell.
(a)
Find the height the ball reaches after the third bounce. Give your answer as a fraction.
(2)
(b)
Find the total vertical distance the ball travels before it comes to rest.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(10 \times \left(\frac{3}{5}\right)^{3}\)
✓ 1
A1 \(\frac{54}{25}\)
✓ 1
Part (b)
M1 recognises the downward distances after the first drop and the upward distances as geometric series with \(r=\frac35\)
✓ 1
M1 total \(= 10 + 2\left(\frac{6}{1-\frac35}\right)\) (upward journeys \(6+\frac{18}{5}+\dots\), each counted twice)
✓ 1
A1 sum to infinity \(\frac{6}{1-\frac35}=15\)
✓ 1
A1 total distance 40 m
✓ 1
Common error: After the first drop each rebound height is travelled twice, once up and once down. Counting the upward journeys once gives 25 m instead of 40 m.
79IBMA3S-1-17r32-031.6 Deductive proofPaper 1Easy5 marks
No calculator
Two athletes in a relay squad wear consecutive odd bib numbers. The smaller bib number is \(2n+1\), where \(n\) is a non-negative integer.
(a)
Prove that the sum of the two bib numbers is always divisible by 4.
(3)
(b)
Two athletes have bib numbers 2023 and 2025. Use the result of part (a) to find the sum of their bib numbers divided by 4.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 the next odd number is \(2n+3\), so sum \(=(2n+1)+(2n+3)\)
✓ 1
A1 \(=4n+4\)
✓ 1
R1 \(=4(n+1)\), where \(n+1\) is an integer, so the sum is divisible by 4
✓ 1
Part (b)
M1 \(2n+1=2023\) so \(n=1011\)
✓ 1
A1 quotient \(=n+1=1012\)
✓ 1
Exam tip: In a proof, define your symbols (what \(n\) is) and finish with a sentence that states the conclusion in words.
80IBMA3S-1-18r38-011.5 Exponents and introduction to logarithmsPaper 1Easy4 marks
No calculator
In telecoms, the signal level in decibels is L = 10 log10(P/P0), where P is the signal power and P0 is a fixed reference power.
(a)
Find L when P = 1000P0.
(2)
(b)
The power is multiplied by 105. Find the change in L.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 L = 10 log10 1000
✓ 1
A1 L = 30
✓ 1
Part (b)
M1 new L = 10 log10(105P/P0) = 10(5 + log10(P/P0))
✓ 1
A1 L increases by 50
✓ 1
Exam tip: Use log(ab) = log a + log b to separate a multiplying factor; it turns multiplication of power into addition of level.
81IBMA3S-1-18r38-031.8 Infinite geometric seriesPaper 1Easy5 marks
No calculator
A telecoms signal produces a sequence of echoes. The amplitudes, in millivolts, are 40, 10, 2.5, … and form a geometric sequence.
(a)
Find the common ratio, and the amplitude of the fifth echo as a fraction.
(3)
(b)
Find the sum of the amplitudes of all the echoes, if the sequence continued indefinitely.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 1/4
✓ 1
M1 40 × (1/4)4
✓ 1
A1 5/32 mV
✓ 1
Part (b)
M1 S∞ = 40/(1 − 1/4)
✓ 1
A1 160/3 mV
✓ 1
Exam tip: A sum to infinity exists only when |r| < 1; say so before using the formula.
82IBMA3S-1-18r41-141.1 Standard formPaper 1Easy6 marks
No calculator
The power output \(P\) kW of a wind turbine at wind speed \(v\) m/s is modelled by \(P=kv^{3}\). At a wind speed of 4 m/s the power output is 20 kW.
(a)
Find the value of \(k\).
(2)
(b)
Find the power output when the wind speed is 8 m/s.
(2)
(c)
Find the wind speed at which the power output is 540 kW.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(20=k\times4^{3}\)
✓ 1
A1 \(k=\frac{5}{16}\) (or 0.3125)
✓ 1
Part (b)
M1 \(\frac{5}{16}\times8^{3}\)
✓ 1
A1 160 kW
✓ 1
Part (c)
M1 \(540=\frac{5}{16}v^{3}\), so \(v^{3}=1728\)
✓ 1
A1 \(v=12\) m/s
✓ 1
Common error: To undo a cube, take the cube root, not the square root. Write v³ = 1728 and recognise 1728 = 12³.
83IBMA3S-1-17r42-011.6 Deductive proofPaper 1Easy4 marks
No calculator
Four huts along a ridge trail are numbered with the consecutive integers \(n\), \(n+1\), \(n+2\) and \(n+3\).
(a)
Show that the sum of the four hut numbers is \(4n+6\).
(2)
(b)
Hence prove that the sum of four consecutive integers is never a multiple of 4.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 adds the four expressions \(n+(n+1)+(n+2)+(n+3)\)
✓ 1
A1 collects like terms to obtain \(4n+6\) (AG)
✓ 1
Part (b)
M1 rewrites the sum as \(4(n+1)+2\)
✓ 1
R1 \(4(n+1)\) is a multiple of 4 for any integer \(n\), so the sum always leaves remainder 2 when divided by 4; hence it is never a multiple of 4
✓ 1
Common error: Substituting particular values of n shows the result is true for those cases only; a proof needs a general expression in n.
84IBMA3S-1-17r42-031.7 Rational exponents and laws of logarithmsPaper 1Easy5 marks
No calculator
Telecoms engineers use laws of exponents to simplify signal-gain expressions.
(a)
Simplify \(\left(27x^{-6}\right)^{-\frac{2}{3}}\), giving your answer in the form \(\dfrac{x^{p}}{q}\).
(2)
(b)
Solve \(4^{2x+1}=8^{x-2}\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(27^{-2/3}=\left(3^{3}\right)^{-2/3}=3^{-2}=\dfrac19\) and \(\left(x^{-6}\right)^{-2/3}=x^{4}\)
✓ 1
A1 \(\dfrac{x^{4}}{9}\)
✓ 1
Part (b)
M1 writes both sides as powers of 2: \(2^{2(2x+1)}=2^{3(x-2)}\)
✓ 1
M1 equates exponents: \(4x+2=3x-6\)
✓ 1
A1 \(x=-8\)
✓ 1
Common error: Writing 27^(-2/3) as -18 confuses the negative exponent with a negative number; a negative exponent means a reciprocal.
85IBMA3S-1-17r42-111.6 Deductive proofPaper 1Easy4 marks
No calculator
Consider the expression \((2n+1)^{2}-(2n-1)^{2}\), where \(n\) is an integer.
(a)
Show that \((2n+1)^{2}-(2n-1)^{2}=8n\).
(2)
(b)
Hence find the value of \(1001^{2}-999^{2}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 expands both squares: \(4n^{2}+4n+1-\left(4n^{2}-4n+1\right)\)
✓ 1
A1 \(=8n\) (AG)
✓ 1
Part (b)
M1 \(2n+1=1001\), so \(n=500\)
✓ 1
A1 \(8\times500=4000\)
✓ 1
Exam tip: When a question says 'hence', use the result you have just shown rather than starting a fresh calculation.
86IBMA3S-1-17r43-011.9 The binomial theoremPaper 1Easy5 marks
No calculator
Consider the expansion of \((1+2x)^4\) in ascending powers of \(x\).
(a)
Expand \((1+2x)^4\) in full.
(3)
(b)
Hence find the coefficient of \(x^2\) in the expansion of \((1+2x)^4(3-x)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 binomial coefficients 1, 4, 6, 4, 1 (or Pascal's triangle) combined with powers of \(2x\)
✓ 1
A1 at least three terms correct
✓ 1
A1 \(1+8x+24x^2+32x^3+16x^4\)
✓ 1
Part (b)
M1 attempt to combine \(x^2\) terms: \(3\times 24\) and \((-1)\times 8\)
✓ 1
A1 \(72-8=64\)
✓ 1
Common error: When multiplying two brackets, students often take only the product of the two \(x^2\) terms and forget the cross terms such as \(x\times x\).
87IBMA3S-1-17r43-031.9 The binomial theoremPaper 1Easy5 marks
No calculator
Answer both parts without a calculator.
(a)
Find the exact value of \(27^{-\frac{2}{3}}\times 4^{\frac{1}{2}}\).
(3)
(b)
Find the value of \(\log_5 75-\log_5 3\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(27^{1/3}=3\)
✓ 1
A1 \(27^{-2/3}=\frac{1}{9}\)
✓ 1
A1 \(\frac{1}{9}\times2=\frac{2}{9}\)
✓ 1
Part (b)
M1 \(\log_5\frac{75}{3}=\log_5 25\)
✓ 1
A1 2
✓ 1
Common error: A negative index means a reciprocal, not a negative answer: \(27^{-2/3}\) is positive.
88IBMA3S-1-19r44-011.2 Arithmetic sequences and seriesPaper 1Easy4 marks
No calculator
A beekeeper counts the bees in one hive at the start of three consecutive weeks. The counts are 2560, 3200 and 4000, and they form a geometric sequence.
(a)
Find the common ratio.
(2)
(b)
Find the number of bees at the start of week 5.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to divide consecutive terms, e.g. 3200 ÷ 2560 or 4000 ÷ 3200
✓ 1
A1 r = 1.25 (accept 5/4)
✓ 1
Part (b)
M1 attempt to use u5 = 2560 × 1.254, or continue the sequence two more terms from 4000
✓ 1
A1 6250
✓ 1
Exam tip: Check a ratio against a second pair of terms; if 3200 ÷ 2560 and 4000 ÷ 3200 disagree, the sequence is not geometric.
89IBMA3S-1-19r44-031.6 Deductive proofPaper 1Easy5 marks
No calculator
In a Formula 1 garage, bays on one side are numbered with consecutive odd integers. Let the numbers of two adjacent bays be \(2n+1\) and \(2n+3\), where \(n\) is a positive integer.
(a)
Prove that the sum of the numbers of two adjacent bays is always a multiple of 4.
(3)
(b)
Prove that the product of the numbers of two adjacent bays is always odd.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to add, (2n + 1) + (2n + 3)
✓ 1
A1 4n + 4
✓ 1
R1 = 4(n + 1), and n + 1 is an integer, so the sum is a multiple of 4
✓ 1
Part (b)
M1 expand, (2n + 1)(2n + 3) = 4n2 + 8n + 3
✓ 1
R1 = 2(2n2 + 4n + 1) + 1, which is of the form 2k + 1 with k an integer, so odd
✓ 1
Exam tip: A proof must end with a sentence linking the algebra to the statement, such as 'this is 4 times an integer, so it is a multiple of 4'.
90IBMA3S-1-19r44-131.2 Arithmetic sequences and seriesPaper 1Easy4 marks
No calculator
In a hospital emergency call tree, 5 staff members are called first. Each of them then calls 3 new staff members, each of whom calls 3 new staff members, and so on. The numbers of staff called at each stage are 5, 15, 45, ... and form a geometric sequence.
Only the calls made by one of the first 5 staff are shown. Diagram not to scale
(a)
Find the number of staff called at stage 8.
(2)
(b)
At one stage 3645 staff members are called. Find which stage this is. (You may use the fact that \(3^6 = 729\).)
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 u8 = 5 × 37
✓ 1
A1 10 935
✓ 1
Part (b)
M1 set 5 × 3n−1 = 3645, so 3n−1 = 729
✓ 1
A1 n = 7
✓ 1
Exam tip: When the unknown is the position in a sequence, form the equation $u_n = $ given value first, then use the known power.
91IBMA3S-1-19r46-031.9 The binomial theoremPaper 1Easy5 marks
No calculator
Solutions to this question must be shown without a calculator.
(a)
Solve \(8^{x-1}=4^{x+2}\).
(3)
(b)
Write \(\log_2 40-\log_2 5\) as an integer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 write both sides as powers of 2: \(2^{3(x-1)}=2^{2(x+2)}\)
✓ 1
M1 equate the exponents: \(3x-3=2x+4\)
✓ 1
A1 \(x=7\)
✓ 1
Part (b)
M1 use of the quotient law: \(\log_2\frac{40}{5}\)
✓ 1
A1 3
✓ 1
Exam tip: To solve an exponential equation without a calculator, look for a common base before thinking about logarithms.
92IBMA3S-1-17r47-011.5 Exponents and introduction to logarithmsPaper 1Easy4 marks
No calculator
A wind farm delivers 5.4 × 109 J of energy to the grid in 3.6 × 103 s.
(a)
Calculate the average power output of the wind farm, in W. Give your answer in the form a × 10k where 1 ≤ a < 10 and k is an integer. (Power = energy ÷ time.)
(2)
(b)
Each home supplied uses 2.5 × 103 W on average. Find the number of homes the wind farm can supply.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to divide, e.g. (5.4 ÷ 3.6) × 109−3
✓ 1
A1 1.5 × 106 W
✓ 1
Part (b)
M1 1.5 × 106 ÷ (2.5 × 103) (ECF from (a))
✓ 1
A1 600 homes
✓ 1
Common error: When dividing numbers in standard form, divide the coefficients and subtract the powers of ten separately; subtracting the coefficients is a frequent slip.
93IBMA3S-1-17r47-021.2 Arithmetic sequences and seriesPaper 1Easy6 marks
No calculator
The number of members of a swimming club is modelled by \(M = 5 \times 2^{t/3}\), where \(t\) is the time in years since the club opened.
(a)
State the number of members when the club opened.
(1)
(b)
Find the number of members when \(t = 6\).
(2)
(c)
Find the value of \(t\) when the club has 80 members.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 5
✓ 1
Part (b)
M1 \(5 \times 2^{2}\)
✓ 1
A1 20
✓ 1
Part (c)
M1 \(2^{t/3} = 16\)
✓ 1
M1 write 16 as \(2^4\) (or equivalent use of logarithms)
✓ 1
A1 \(t = 12\)
✓ 1
Exam tip: In a no-calculator question, rewrite both sides of an exponential equation with the same base before comparing the powers.
94IBMA3S-1-17r47-031.1 Standard formPaper 1Easy6 marks
No calculator
Given that \(\ln 3 = p\) and \(\ln 5 = q\), write each of the following in terms of \(p\) and \(q\).
(a)
\(\ln 15\)
(1)
(b)
\(\ln 0.6\)
(2)
(c)
\(\ln \sqrt{45}\)
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(p + q\)
✓ 1
Part (b)
M1 write 0.6 as \(\frac{3}{5}\) and use the quotient law
✓ 1
A1 \(p - q\)
✓ 1
Part (c)
M1 write \(45 = 3^2 \times 5\) and use the product law, e.g. \(\ln 45 = 2p + q\)
✓ 1
M1 use the power law with exponent \(\frac{1}{2}\)
116IBMA2S-044-04-061.9 The binomial theoremPaper 1Medium6 marks
No calculator
The profile of a skate-park ramp is modelled by a polynomial of the form \((1+kx)(2-x)^5\), where \(k\) is a constant.
(a)
Find the first three terms, in ascending powers of \(x\), of the expansion of \((2-x)^5\).
(3)
(b)
The coefficient of \(x^2\) in the expansion of \((1+kx)(2-x)^5\) is \(-160\). Find the value of \(k\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of binomial coefficients with powers of 2 and \(-x\) A1 \(32-80x\) A1 \(+80x^2\)
✓ 3
Part (b)
M1 identifying both contributions: \(1\times80x^2\) and \(kx\times(-80x)\) A1 \(80-80k=-160\) A1 \(k=3\)
✓ 3
Common error: The \(x^2\) term of the product comes from two pairings. Multiplying only \(1\times80x^2\) misses the \(kx\times(-80x)\) contribution.
117IBMA2S-044-04-071.9 The binomial theoremPaper 1Medium6 marks
No calculator
In the expansion of \((2+x)^n\), where \(n\ge3\) is an integer, the coefficient of \(x^2\) is equal to the coefficient of \(x^3\).
(a)
Find the value of \(n\).
(4)
(b)
Find this common coefficient.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writing both coefficients: \(\binom{n}{2}2^{n-2}\) and \(\binom{n}{3}2^{n-3}\) M1 equating and expanding: \(\frac{n(n-1)}{2}2^{n-2}=\frac{n(n-1)(n-2)}{6}2^{n-3}\) A1 simplifying to \(1=\frac{n-2}{6}\) A1 \(n=8\)
✓ 4
Part (b)
M1 \(\binom{8}{2}2^6\) or \(\binom{8}{3}2^5\) A1 \(28\times64=1792\)
✓ 2
Exam tip: Divide both sides by the common factor \(n(n-1)2^{n-3}\) early on. This reduces the equation to a linear one in \(n\).
118IBMA2S-044-04-101.9 The binomial theoremPaper 1Medium11 marks
No calculator
This question investigates powers of \(1+\sqrt2\).
(a)
Use the binomial theorem to express \((1+\sqrt2)^4\) in the form \(a+b\sqrt2\), where \(a,b\in\mathbb{Z}\).
(4)
(b)
Write down \((1-\sqrt2)^4\) in the form \(a+b\sqrt2\).
(1)
(c)
Show that \(0<17-12\sqrt2<1\).
(3)
(d)
Hence find the greatest integer less than \((1+\sqrt2)^4\).
M1 uses part (a) to write the left-hand side as \(\log_x 2+\log_x 3+\log_x 5\)
✓ 1
M1 uses the product law to obtain \(\log_x 30\)
✓ 1
A1 uses part (a) again: \(\log_x 30=\frac{1}{\log_{30} x}\), the right-hand side
✓ 1
Part (c)
M1 obtains \(\log_x 30=2\), so \(x^2=30\)
✓ 1
A1 \(x=\sqrt{30}\), rejecting \(-\sqrt{30}\) because \(x>0\)
✓ 1
Common error: \(\log_2 x+\log_3 x\) cannot be combined with the product law, because the bases differ. Rewrite the terms with a common base first.
120IBMA2S-044-07-061.6 Deductive proofPaper 1Medium6 marks
No calculator
In a round-robin hockey league, every pair of teams plays each other exactly once, so a league of \(n\) teams has \(\binom{n}{2}\) matches.
(a)
Prove algebraically that \(\binom{n}{1}+\binom{n}{2}\equiv\binom{n+1}{2}\) for integers \(n\ge2\).
(3)
(b)
Explain the identity in part (a) in terms of a league of \(n\) teams that is joined by one new team.
(1)
(c)
A league grows from 15 teams to 16 teams. Use part (a) to find the total number of matches in the new league.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes the left-hand side as \(n+\frac{n(n-1)}{2}\)
✓ 1
M1 uses a common denominator: \(\frac{2n+n^2-n}{2}\)
✓ 1
A1 \(=\frac{n(n+1)}{2}=\binom{n+1}{2}\)
✓ 1
Part (b)
R1 the new team plays each of the \(n\) existing teams once, adding \(\binom{n}{1}=n\) matches to the original \(\binom{n}{2}\). This total must equal \(\binom{n+1}{2}\), the number of matches in a league of \(n+1\) teams
Exam tip: Write \(\binom{n}{2}\) as \(\frac{n(n-1)}{2}\) before combining terms. Factorial forms are harder to simplify by hand.
121IBMA2S-044-07-071.6 Deductive proofPaper 1Medium6 marks
No calculator
A ticket machine at a ski lift prints consecutive integer ticket numbers. Four consecutive tickets are taken, and the smallest number is \(n\).
(a)
Show that the sum of the four ticket numbers is \(4n+6\).
(2)
(b)
Deduce that the sum of any four consecutive integers is never divisible by 4.
(1)
(c)
Prove that the product of the two middle numbers minus the product of the smallest and largest numbers is always 2.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes the numbers as \(n\), \(n+1\), \(n+2\), \(n+3\)
✓ 1
A1 sum \(=4n+6\)
✓ 1
Part (b)
R1 \(4n+6=4(n+1)+2\), so the sum always leaves remainder 2 on division by 4
✓ 1
Part (c)
M1 expands \((n+1)(n+2)=n^2+3n+2\)
✓ 1
M1 expands \(n(n+3)=n^2+3n\)
✓ 1
A1 difference \(=2\) for every integer \(n\)
✓ 1
Common error: Using \(n\), \(m\), \(p\), \(q\) for the four numbers loses the fact that they are consecutive. Write every number in terms of a single variable.
122IBMA2S-044-07-081.6 Deductive proofPaper 1Medium8 marks
No calculator
A patient takes a 20 mg tablet every morning. By the following morning, exactly half of the drug in the body remains. Let \(A_n\) be the amount of drug in the body, in mg, immediately after the \(n\)th tablet.
(a)
Prove that \((1-r)(1+r+r^2+\dots+r^{n-1})\equiv1-r^n\).
(2)
(b)
Explain why \(A_n=20(1+0.5+0.5^2+\dots+0.5^{n-1})\) and hence show that \(A_n=40(1-0.5^n)\).
(3)
(c)
Prove that the amount of drug in the body never reaches 40 mg.
(1)
(d)
Find the exact amount of drug in the body immediately after the 6th tablet.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 expands to \((1+r+\dots+r^{n-1})-(r+r^2+\dots+r^n)\)
✓ 1
A1 all terms except \(1\) and \(-r^n\) cancel, giving \(1-r^n\)
✓ 1
Part (b)
R1 the newest tablet contributes 20 mg, and each earlier tablet has been halved once for every day since it was taken
✓ 1
M1 applies part (a) with \(r=0.5\): \(1+0.5+\dots+0.5^{n-1}=\frac{1-0.5^n}{0.5}\)
✓ 1
A1 \(A_n=\frac{20(1-0.5^n)}{0.5}=40(1-0.5^n)\)
✓ 1
Part (c)
R1 \(0.5^n>0\) for all \(n\), so \(1-0.5^n<1\) and therefore \(A_n<40\)
✓ 1
Part (d)
M1 \(A_6=40\left(1-\frac{1}{64}\right)\)
✓ 1
A1 \(\frac{315}{8}\) mg \(=39.375\) mg
✓ 1
Exam tip: For 'never reaches' arguments, point to the term that is always positive and state the inequality that follows.
123IBMA2S-044-07-091.8 Infinite geometric seriesPaper 1Medium6 marks
No calculator
A delivery drone follows a programmed flight path. Its height above level ground, in metres, \(t\) seconds after a manoeuvre begins is \(h(t)=t^2-6t+11\), for \(0\le t\le6\).
(a)
Express \(h(t)\) in the form \((t-a)^2+b\).
(2)
(b)
Hence prove that the drone never touches the ground during the manoeuvre.
(2)
(c)
Write down the time at which the drone is lowest and its minimum height.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempts to complete the square, e.g. \((t-3)^2-9+11\)
✓ 1
A1 \((t-3)^2+2\)
✓ 1
Part (b)
R1 \((t-3)^2\ge0\) for all real \(t\)
✓ 1
A1 so \(h(t)\ge2>0\), hence the height is never zero and the drone is always above the ground
✓ 1
Part (c)
A1 \(t=3\) seconds
✓ 1
A1 minimum height 2 m
✓ 1
Common error: Writing only 'b = 2 is positive, so h(t) > 0' leaves the argument incomplete. The proof also needs the fact that \((t-3)^2\ge0\) for every \(t\), which gives \(h(t)\ge2\).
124IBMA2S-044-06-051.5 Exponents and introduction to logarithmsPaper 1Medium6 marks
No calculator
In this question, 1 terabyte (TB) = 1012 bytes. A smartphone photo uses 4×106 bytes of storage. Give all answers in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ.
(a)
Find the number of photos that can be stored on a 2 TB drive.
(2)
(b)
A data centre stores 8×1017 bytes of data using only 2 TB drives, each filled completely. Find the number of drives needed.
(2)
(c)
Each drive uses 6 W of power. Find the total power used by the drives in the data centre.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (2×1012) ÷ (4×106) = 0.5×106 A1 5×105 photos
✓ 2
Part (b)
M1 (8×1017) ÷ (2×1012) A1 4×105 drives
✓ 2
Part (c)
M1 6 × their number of drives, e.g. 6 × 4×105 = 24×105 A1 2.4×106 W
✓ 2
Common error: 0.5×10⁶ and 24×10⁵ are correct values but are not in standard form. Rewrite them as 5×10⁵ and 2.4×10⁶ to gain the final mark.
125IBMA2S-044-06-071.7 Rational exponents and laws of logarithmsPaper 1Medium6 marks
No calculator
Let x = 6×10p and y = 4×10q, where p, q ∈ ℤ.
(a)
Write xy in the form a×10k, where 1 ≤ a < 10, giving k in terms of p and q.
(2)
(b)
Write \(\frac{x}{y}\) in the form a×10k, where 1 ≤ a < 10, giving k in terms of p and q.
(1)
(c)
Given that xy = 2.4×107 and \(\frac{x}{y}\) = 1.5×102, find the value of p and the value of q.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 6×4 = 24, giving 24×10p+q A1 2.4×10p+q+1
✓ 2
Part (b)
A1 1.5×10p−q
✓ 1
Part (c)
M1 forming both equations: p + q + 1 = 7 and p − q = 2 M1 solving the equations simultaneously, e.g. adding them to get 2p = 8 A1 p = 4, q = 2
✓ 3
Common error: Writing xy = 2.4×10p⁺q forgets that 24 = 2.4×10. If you miss the extra 1 in the index, every value in part (c) is wrong.
126IBMA2S-044-06-101.1 Standard formPaper 1Medium7 marks
No calculator
An adult has a blood volume of 5×106 mm3. Each mm3 of blood contains 5.2×106 red blood cells.
(a)
Find the total number of red blood cells in this adult's blood. Give your answer in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ.
(2)
(b)
Each red blood cell lasts 1.3×102 days before it is replaced. Assuming the number of cells stays constant, find the number of cells replaced each day.
(2)
(c)
The volume of one red blood cell is 9×10−8 mm3. Find the total volume of all the red blood cells, in mm3.
(2)
(d)
Find the percentage of the blood volume taken up by red blood cells.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (5.2×106)×(5×106) = 26×1012 A1 2.6×1013
✓ 2
Part (b)
M1 (2.6×1013) ÷ (1.3×102) A1 2×1011
✓ 2
Part (c)
M1 (2.6×1013)×(9×10−8) = 23.4×105 A1 2.34×106 mm3
✓ 2
Part (d)
A1 (2.34×106 ÷ 5×106)×100 = 46.8 %
✓ 1
Exam tip: Without a calculator, multiply the leading numbers and combine the powers of ten separately. Write the result in standard form as your final step.
127IBMA2S-044-01-031.7 Rational exponents and laws of logarithmsPaper 1Medium6 marks
No calculator
Solve the equation \(\log_4(x+3)=\log_2(x-3)\).
(a)
Solve the equation, giving clear reasons for any value of \(x\) that you reject.
R1 rejects \(x=1\) because \(x-3<0\), so \(\log_2(x-3)\) is undefined; \(x=6\)
✓ 1
Exam tip: Change every logarithm to the same base before combining them. A log of a negative number is undefined, so always check your solutions in the original equation.
128IBMA2S-044-01-061.7 Rational exponents and laws of logarithmsPaper 1Medium7 marks
No calculator
Answer the following without using a calculator.
(a)
Solve \(2^{2x+1}-9\times 2^{x}+4=0\).
(4)
(b)
Solve \(x^{\frac{2}{3}}-5x^{\frac{1}{3}}+4=0\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(2^{2x+1}=2\times(2^{x})^{2}\)
✓ 1
M1 lets \(y=2^{x}\) to obtain \(2y^{2}-9y+4=0\)
✓ 1
A1 \((2y-1)(y-4)=0\), so \(y=\frac{1}{2}\) or \(y=4\)
✓ 1
A1 \(x=-1\) or \(x=2\)
✓ 1
Part (b)
M1 lets \(u=x^{\frac{1}{3}}\) to obtain \(u^{2}-5u+4=0\)
✓ 1
A1 \((u-1)(u-4)=0\), so \(u=1\) or \(u=4\)
✓ 1
A1 \(x=1\) or \(x=64\)
✓ 1
Exam tip: Look for one power that is the square of another, such as \(2^{2x}=(2^{x})^{2}\) or \(x^{2/3}=(x^{1/3})^{2}\). A substitution then gives a quadratic. Remember to convert back to \(x\) at the end.
129IBMA2S-044-01-091.7 Rational exponents and laws of logarithmsPaper 1Medium7 marks
No calculator
Answer the following without using a calculator.
(a)
Find the exact value of \(\log_8 32\).
(2)
(b)
Find the exact value of \(\log_2 3\times\log_3 5\times\log_5 64\).
(3)
(c)
Solve \(\log_x 27=\frac{3}{2}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(8^{x}=32\) as \(2^{3x}=2^{5}\), or uses change of base to base 2
✓ 1
A1 \(\frac{5}{3}\)
✓ 1
Part (b)
M1 writes each factor as a quotient of logs in a common base, e.g. \(\dfrac{\ln 3}{\ln 2}\times\dfrac{\ln 5}{\ln 3}\times\dfrac{\ln 64}{\ln 5}\)
✓ 1
M1 cancels to obtain \(\dfrac{\ln 64}{\ln 2}\), i.e. \(\log_2 64\)
✓ 1
A1 6
✓ 1
Part (c)
M1 converts to exponential form: \(x^{\frac{3}{2}}=27\)
✓ 1
A1 \(x=27^{\frac{2}{3}}=9\)
✓ 1
Exam tip: A product of logs with linked bases collapses by change of base. Write every factor in the same base, and the middle terms cancel like a telescope.
130IBMA2S-044-00-061.8 Infinite geometric seriesPaper 1Medium7 marks
No calculator
A patient takes a 200 mg dose of a drug at the same time each day. By the time of the next dose, a fraction \(r\) of the drug that was present just after the previous dose remains in the body. In the long run, the amount of drug present just after a dose approaches \(200+200r+200r^2+\dots\), and this amount is 320 mg.
(a)
Find the value of \(r\).
(3)
(b)
Find the long-run amount of drug present just before a dose is taken.
(2)
(c)
The value of \(r\) stays the same. Find the largest daily dose for which the long-run amount just after a dose does not exceed 400 mg.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{200}{1-r}=320\)
✓ 1
A1 \(1-r=\frac{5}{8}\)
✓ 1
A1 \(r=\frac{3}{8}\)
✓ 1
Part (b)
M1 \(320\times\frac{3}{8}\) or \(320-200\)
✓ 1
A1 120 mg
✓ 1
Part (c)
M1 \(\frac{D}{1-\frac{3}{8}}\le400\)
✓ 1
A1 \(D=250\) mg
✓ 1
Exam tip: The long-run level is directly proportional to the dose. With \(r\) fixed, \(\frac{400}{320}\times200=250\) gives the same answer and is a useful check.
131IBMA2S-044-00-081.8 Infinite geometric seriesPaper 1Medium7 marks
No calculator
A geometric series has a sum to infinity equal to four times its first term. The third term of the series is 27.
(a)
Find the common ratio.
(2)
(b)
Find the first term.
(2)
(c)
Find the sum of all the terms from the fourth term onwards.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{u_1}{1-r}=4u_1\)
✓ 1
A1 \(r=\frac{3}{4}\)
✓ 1
Part (b)
M1 \(u_1\left(\frac{3}{4}\right)^2=27\)
✓ 1
A1 \(u_1=48\)
✓ 1
Part (c)
M1 \(S_\infty=\frac{48}{1/4}=192\) or \(u_4=27\times\frac{3}{4}=\frac{81}{4}\)
✓ 1
M1 \(192-(48+36+27)\) or \(\frac{u_4}{1-r}=\frac{81/4}{1/4}\) or \(r^3S_\infty=\frac{27}{64}\times192\)
✓ 1
A1 81
✓ 1
Exam tip: The terms from \(u_4\) onwards form a geometric series of their own, with first term \(u_4\) and the same ratio. Working it out two ways is a quick check.
132IBMA2S-044-02-051.6 Deductive proofPaper 1Medium7 marks
No calculator
The first three terms of a geometric sequence are \(x\), \(x+6\) and \(8x+16\), where \(x>0\).
A1 \(x^2+12x+36=8x^2+16x\), which gives \(7x^2+4x-36=0\)
✓ 1
Part (b)
M1 factorises as \((7x+18)(x-2)=0\)
✓ 1
A1 \(x=2\), rejecting \(x=-\frac{18}{7}\) because \(x>0\)
✓ 1
Part (c)
A1 \(r=4\)
✓ 1
Part (d)
M1 \(\frac{2(4^5-1)}{4-1}\)
✓ 1
A1 682
✓ 1
Exam tip: For three consecutive geometric terms \(a,b,c\), the condition \(b^2=ac\) avoids working with fractions.
133IBMA2S-044-02-091.8 Infinite geometric seriesPaper 1Medium6 marks
No calculator
A tennis knockout tournament starts with 128 players. In each round, every remaining player is paired into a match, and only the winners go through to the next round. Round 1 has 64 matches.
(a)
Find an expression, in terms of \(k\), for the number of matches in round \(k\).
(2)
(b)
Find the number of rounds in the tournament.
(2)
(c)
Use a geometric series to find the total number of matches played.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognises a geometric sequence with \(u_1=64\) and \(r=\frac12\)
✓ 1
A1 \(u_k=64\left(\frac12\right)^{k-1}\), or equivalent such as \(2^{7-k}\)
✓ 1
Part (b)
M1 uses the fact that the final round has one match, e.g. \(2^{7-k}=1\), or lists 64, 32, 16, 8, 4, 2, 1, or halves 128 repeatedly down to 1
✓ 1
A1 7 rounds
✓ 1
Part (c)
M1 \(S_7=\frac{64\left(1-\left(\frac12\right)^7\right)}{1-\frac12}\) or equivalent
✓ 1
A1 127 matches
✓ 1
Exam tip: Check the total another way: every match knocks out exactly one player, and 127 of the 128 players must be knocked out.
134IBMA2S-044-05-071.7 Rational exponents and laws of logarithmsPaper 1Medium6 marks
No calculator
Consider the equation \(2^{2x+1}-5(2^{x})+2=0\).
(a)
Show that the substitution \(u=2^{x}\) transforms the equation into \(2u^{2}-5u+2=0\).
(2)
(b)
Hence solve the equation \(2^{2x+1}-5(2^{x})+2=0\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(2^{2x+1}=2\times(2^x)^2\)
✓ 1
A1 obtains \(2u^2-5u+2=0\) (AG)
✓ 1
Part (b)
M1 factorises \((2u-1)(u-2)=0\) or uses the quadratic formula
✓ 1
A1 \(u=\frac{1}{2}\), \(u=2\)
✓ 1
M1 sets \(2^x=\frac{1}{2}\) and \(2^x=2\)
✓ 1
A1 \(x=-1\), \(x=1\)
✓ 1
Common error: \(2^{2x+1}\) is \(2\times 4^x\), not \(4^x+2\) or \(2^{2x}\times 1\); split off the extra power of 2 carefully.
135IBMA2S-044-03-081.4 Financial applicationsPaper 1Medium6 marks
No calculator
An engineering firm buys an industrial 3D printer. Its value decreases by a fixed percentage of its value at the start of each year. At the end of the second year the printer is worth €6400, and at the end of the third year it is worth €5120.
(a)
Find the annual percentage rate of depreciation.
(2)
(b)
Find the original price of the printer.
(2)
(c)
Find the value of the printer at the end of the fifth year.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5120 ÷ 6400 A1 multiplier 0.8, so 20% per year
✓ 2
Part (b)
M1 V0 × 0.82 = 6400 A1 €10 000
✓ 2
Part (c)
M1 10 000 × 0.85 or 5120 × 0.82 A1 €3276.80
✓ 2
Exam tip: The ratio of the values in consecutive years is the multiplier. Starting from the latest known value in part (c) cuts the working down to 5120 × 0.64.
136IBMA3S-1-16r10-051.2 Arithmetic sequences and seriesPaper 1Medium5 marks
No calculator
The altitudes of flags placed at equal intervals along a mountain trail form an arithmetic sequence. The 3rd flag is at 1400 m and the 8th flag is at 1650 m.
(a)
Find the common difference of the sequence.
(2)
(b)
Find the altitude of the 1st flag.
(1)
(c)
Find the number of the flag that is at an altitude of 2000 m.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(5d = 1650 - 1400\)
✓ 1
A1 \(d = 50\) m
✓ 1
Part (b)
A1 \(1400 - 2(50) = 1300\) m
✓ 1
Part (c)
M1 \(1300 + (n-1)(50) = 2000\)
✓ 1
A1 \(n = 15\)
✓ 1
Exam tip: Subtracting two terms gives a multiple of d: \(u_8 - u_3 = 5d\).
137IBMA3S-1-16r10-071.3 Geometric sequences and seriesPaper 1Medium6 marks
No calculator
A telecoms network has layers of nodes. Layer 1 has 3 nodes and layer 4 has 81 nodes. The number of nodes in successive layers forms a geometric sequence.
(a)
Find the common ratio.
(2)
(b)
Find the number of the first layer that has more than 10 000 nodes.
(2)
(c)
Find the total number of nodes in the first 6 layers.
Common error: With first term 3 and ratio 3, the \(n\)th term is \(3 \times 3^{n-1} = 3^n\); do not write \(3^{n-1}\) alone.
138IBMA3S-1-16r11-061.6 Deductive proofPaper 1Medium5 marks
No calculator
The aisles in a library are numbered with odd integers. Two adjacent aisles have the numbers \(n\) and \(n+2\), where \(n\) is an odd integer.
(a)
Prove that the product of the numbers of two adjacent aisles is always one less than a multiple of 4.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 let \(n=2k+1\) where \(k\) is an integer, so the aisles are \(2k+1\) and \(2k+3\)
✓ 1
M1 product \(=(2k+1)(2k+3)\)
✓ 1
A1 \(=4k^2+8k+3\)
✓ 1
A1 \(=4(k^2+2k+1)-1\)
✓ 1
R1 \(k^2+2k+1\) is an integer, so the product is one less than a multiple of 4
✓ 1
Common error: A proof needs a general case. Showing that the result works for 3 and 5 does not prove it for every odd integer.
139IBMA3S-1-16r11-071.2 Arithmetic sequences and seriesPaper 1Medium6 marks
No calculator
The total number of pool lengths a swimmer has completed in her first \(n\) days of training is \(S_n=2n^2+3n\).
(a)
Find \(u_1\) and \(u_2\), the number of lengths swum on days 1 and 2.
(2)
(b)
Show that the number of lengths swum on day \(n\) forms an arithmetic sequence.
(3)
(c)
Find the number of lengths swum on day 40.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_1=S_1\) and \(u_2=S_2-S_1\)
✓ 1
A1 \(u_1=5,\ u_2=9\)
✓ 1
Part (b)
M1 \(u_n=S_n-S_{n-1}\)
✓ 1
A1 \(u_n=2n^2+3n-2(n-1)^2-3(n-1)\)
✓ 1
R1 \(u_n=4n+1\), so \(u_n-u_{n-1}=4\) is constant
✓ 1
Part (c)
A1 \(u_{40}=4(40)+1=161\)
✓ 1
Exam tip: To get a single term from a sum, use \(u_n=S_n-S_{n-1}\) and take care that the first term is \(S_1\).
140IBMA3S-1-16r11-111.2 Arithmetic sequences and seriesPaper 1Medium5 marks
No calculator
The distances (in metres) swum by a swimmer on successive days of a training block form an arithmetic sequence \(u_1,u_2,u_3,\dots\)
(a)
Prove that \(u_3+u_{11}=u_5+u_9\).
(3)
(b)
In one block, \(u_3=320\), \(u_5=360\) and \(u_9=440\). Find \(u_{11}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_3+u_{11}=(a+2d)+(a+10d)\) with first term \(a\) and common difference \(d\)
✓ 1
A1 \(=2a+12d\)
✓ 1
R1 \(u_5+u_9=(a+4d)+(a+8d)=2a+12d\), so the two sums are equal
✓ 1
Part (b)
M1 \(u_{11}=u_5+u_9-u_3\)
✓ 1
A1 480 m
✓ 1
Exam tip: Start a proof from the general form \(u_n=a+(n-1)d\) and simplify both sides to the same expression.
141IBMA3S-1-16r11-141.2 Arithmetic sequences and seriesPaper 1Medium6 marks
No calculator
A library runs a fine amnesty. The amounts (in £) it recovers in successive weeks form a geometric sequence. The amount recovered in week 2 is £12. If the amnesty could continue indefinitely, the total recovered would be £50.
(a)
Show that the common ratio \(r\) satisfies \(25r^2-25r+6=0\).
(3)
(b)
Find the two possible values of \(r\).
(2)
(c)
Find the largest possible amount recovered in week 1.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_1r=12\), so \(u_1=\frac{12}{r}\)
✓ 1
M1 \(\frac{12/r}{1-r}=50\)
✓ 1
A1 \(12=50r(1-r)\) leading to \(25r^2-25r+6=0\)
✓ 1
Part (b)
M1 \((5r-2)(5r-3)=0\)
✓ 1
A1 \(r=0.4\) or \(r=0.6\)
✓ 1
Part (c)
A1 \(u_1=\frac{12}{0.4}=\) £30
✓ 1
Common error: Both roots satisfy \(|r|<1\), so both are valid. Check the condition before discarding any solution.
142IBMA3S-1-16r13-051.9 The binomial theoremPaper 1Medium6 marks
No calculator
(a)
Evaluate \(16^{3/4}\times27^{-2/3}\), giving your answer as a fraction.
(3)
(b)
Solve \(9^{x}=27^{x-2}\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use fractional powers, for example \((\sqrt[4]{16})^3\) or \((\sqrt[3]{27})^{-2}\)
✓ 1
A1 \(16^{3/4}=8\) and \(27^{-2/3}=\frac19\)
✓ 1
A1 \(\frac89\)
✓ 1
Part (b)
M1 write both sides as powers of 3: \(3^{2x}=3^{3(x-2)}\)
✓ 1
A1 \(2x=3x-6\)
✓ 1
A1 \(x=6\)
✓ 1
Common error: A negative power means a reciprocal; the denominator of the fractional power is the root.
143IBMA3S-1-16r13-081.7 Rational exponents and laws of logarithmsPaper 1Medium5 marks
No calculator
Solve \(\log_3(x+2)+\log_3(x-4)=3\).
(a)
Solve the equation.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use the product law: \(\log_3\big((x+2)(x-4)\big)=3\)
✓ 1
M1 convert: \((x+2)(x-4)=3^3=27\)
✓ 1
A1 \(x^2-2x-35=0\)
✓ 1
A1 \(x=7\) or \(x=-5\)
✓ 1
R1 reject \(x=-5\) since \(\log_3(x-4)\) is undefined, so \(x=7\)
✓ 1
Common error: Always check solutions against the domain: the argument of a logarithm must be positive.
144IBMA3S-1-16r13-111.7 Rational exponents and laws of logarithmsPaper 1Medium7 marks
No calculator
Let \(\log_2 x=p\) and \(\log_2 y=q\), where \(x\) and \(y\) are positive frequency ratios.
(a)
Write \(\log_2\left(\frac{x^3}{\sqrt y}\right)\) in terms of \(p\) and \(q\).
(2)
(b)
Write \(\log_4(xy)\) in terms of \(p\) and \(q\).
(2)
(c)
Given that \(\log_2\left(\frac{x^3}{\sqrt y}\right)=4\) and \(\log_4(xy)=3\), find \(p\) and \(q\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the laws of logarithms: \(3\log_2x-\frac12\log_2y\)
✓ 1
A1 \(3p-\frac q2\)
✓ 1
Part (b)
M1 change of base: \(\frac{\log_2(xy)}{\log_2 4}\)
✓ 1
A1 \(\frac{p+q}{2}\)
✓ 1
Part (c)
M1 form two equations: \(3p-\frac q2=4\) and \(p+q=6\)
✓ 1
M1 attempt to eliminate one variable
✓ 1
A1 \(p=2\), \(q=4\)
✓ 1
Exam tip: Reduce every logarithm to base 2 before combining: \(\log_4 t=\frac{\log_2 t}{2}\).
145IBMA3S-1-17r13-061.7 Rational exponents and laws of logarithmsPaper 1Medium5 marks
No calculator
A Formula 1 team models the mass \(M\) grams of material worn from a brake disc after \(n\) laps using \(M = 0.8\,n^{3/2}\).
(a)
Find the mass worn after 16 laps.
(2)
(b)
After a long run the mass worn is 274.4 g. Find the number of laps \(n\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.8\times 16^{3/2} = 0.8 \times 64\)
✓ 1
A1 51.2 g
✓ 1
Part (b)
M1 \(n^{3/2} = \frac{274.4}{0.8} = 343\)
✓ 1
M1 \(n = 343^{2/3}\)
✓ 1
A1 \(n = 49\)
✓ 1
Exam tip: To undo a power of \(\frac32\), raise both sides to the power \(\frac23\).
146IBMA3S-1-17r13-091.6 Deductive proofPaper 1Medium4 marks
No calculator
Prove that the sum of any three consecutive odd integers is an odd multiple of 3.
(a)
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 let the integers be \(2n-1\), \(2n+1\), \(2n+3\) where \(n\) is an integer
✓ 1
A1 sum \(= 6n+3\)
✓ 1
A1 \(= 3(2n+1)\)
✓ 1
R1 \(2n+1\) is an odd integer, so the sum is an odd multiple of 3
✓ 1
Common error: Using \(n, n+1, n+2\) for odd integers. Consecutive odd integers differ by 2, not 1.
147IBMA3S-1-17r13-121.9 The binomial theoremPaper 1Medium6 marks
No calculator
A Formula 1 team models the time \(T\) seconds taken to complete a corner section of radius \(r\) metres by \(T = a r^{k}\), where \(a\) and \(k\) are constants. The model gives \(T = 4\) when \(r = 16\) and \(T = 13.5\) when \(r = 81\).
(a)
Show that \(\left(\dfrac{81}{16}\right)^k = \dfrac{27}{8}\).
(2)
(b)
Hence find \(k\).
(2)
(c)
Find \(a\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 divide the two equations: \(\frac{13.5}{4} = \frac{a\,81^k}{a\,16^k}\)
✓ 1
A1 \(\frac{13.5}{4} = \frac{27}{8}\) (AG)
✓ 1
Part (b)
M1 \(\frac{81}{16} = \left(\frac32\right)^4\) and \(\frac{27}{8} = \left(\frac32\right)^3\), so \(\left(\frac32\right)^{4k} = \left(\frac32\right)^3\)
✓ 1
A1 \(k = \frac34\)
✓ 1
Part (c)
M1 \(4 = a\times 16^{3/4} = 8a\)
✓ 1
A1 \(a = 0.5\)
✓ 1
Exam tip: Write both bases as powers of the same number so the exponents can be equated.
148IBMA3S-1-16r14-041.9 The binomial theoremPaper 1Medium5 marks
No calculator
Find the constant term in the expansion of \(\left(x^2 + \frac{2}{x}\right)^6\).
(a)
Find the constant term in the expansion of \(\left(x^2 + \frac{2}{x}\right)^6\).
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 general term \(\binom{6}{r}(x^2)^{6-r}\left(\frac{2}{x}\right)^r\)
✓ 1
A1 \(\binom{6}{r}2^r x^{12-3r}\)
✓ 1
M1 set the power of \(x\) equal to 0
✓ 1
A1 \(r = 4\)
✓ 1
A1 constant term \(= \binom{6}{4} \times 2^4 = 240\)
✓ 1
Common error: Combine the powers of x in both brackets (including the negative power from the fraction) before setting the total equal to zero.
149IBMA3S-1-16r14-071.6 Deductive proofPaper 1Medium6 marks
No calculator
Consider two consecutive odd integers \(2n+1\) and \(2n+3\), where \(n \in \mathbb{Z}\).
(a)
Prove that the difference between the squares of these two integers is divisible by 8.
(4)
(b)
Hence find the value of \(1001^2 - 999^2\) without using a calculator.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \((2n+3)^2 - (2n+1)^2\)
✓ 1
M1 expand, or use the difference of two squares
✓ 1
A1 \(8n + 8 = 8(n+1)\)
✓ 1
R1 \(n+1\) is an integer, so the difference is a multiple of 8
✓ 1
Part (b)
M1 recognise \(2n+1 = 999\) so \(n = 499\)
✓ 1
A1 \(8(499+1) = 4000\)
✓ 1
Exam tip: After proving a general result, substitute into it for the 'hence' part rather than calculating from scratch.
150IBMA3S-1-16r14-111.9 The binomial theoremPaper 1Medium12 marks
No calculator
Consider the expansion of \((2 + kx)^6\) in ascending powers of \(x\), where \(k \neq 0\).
(a)
Find the first three terms of the expansion, giving the coefficients in terms of \(k\).
(3)
(b)
The coefficient of \(x^2\) is 5 times the coefficient of \(x\). Find the value of \(k\).
(4)
(c)
Using this value of \(k\), find the coefficient of \(x^3\) in the expansion.
(2)
(d)
Hence find the coefficient of \(x^2\) in the expansion of \((1+x)(2+4x)^6\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the binomial expansion, e.g. \(\binom{6}{1}2^5(kx)\)
✓ 1
A1 \(64 + 192kx\)
✓ 1
A1 \(+ 240k^2x^2\)
✓ 1
Part (b)
M1 \(240k^2 = 5 \times 192k\)
✓ 1
M1 divide by \(k\) (as \(k \neq 0\)), or factorise \(240k^2 - 960k = 0\)
✓ 1
A1 \(k = 4\)
✓ 1
R1 reason for rejecting \(k = 0\) stated
✓ 1
Part (c)
M1 \(\binom{6}{3} \times 2^3 \times 4^3\)
✓ 1
A1 10 240
✓ 1
Part (d)
M1 coefficient of \(x^2\) is (coefficient of \(x\)) + (coefficient of \(x^2\)) from \((2+4x)^6\)
✓ 1
A1 \(768 + 3840\)
✓ 1
A1 4608
✓ 1
Common error: In the product, the x² term comes from two sources: 1 × (x² term) and x × (x term); do not forget the cross term.
151IBMA3S-1-16r14-121.3 Geometric sequences and seriesPaper 1Medium6 marks
No calculator
A Formula 1 team posts a lap highlights clip online. The number of viewers, in thousands, in each successive hour forms a geometric sequence. In hour 2 there are 12 thousand viewers and in hour 5 there are 96 thousand viewers.
(a)
Find the common ratio.
(3)
(b)
Find the number of viewers, in thousands, in hour 1.
(1)
(c)
Find the total number of viewers, in thousands, in the first 6 hours.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_5 = u_2 r^3\), so \(96 = 12r^3\)
✓ 1
A1 \(r^3 = 8\)
✓ 1
A1 \(r = 2\)
✓ 1
Part (b)
A1 \(\frac{12}{2} = 6\)
✓ 1
Part (c)
M1 \(S_6 = \frac{6(2^6-1)}{2-1}\)
✓ 1
A1 378 thousand
✓ 1
Exam tip: To find the ratio from two non-adjacent terms, divide them: the result is r raised to the difference of their positions.
152IBMA3S-1-16r15-071.8 Infinite geometric seriesPaper 1Medium6 marks
No calculator
A basketball is dropped from a height of 4 m. After each bounce it rises to \(\frac{3}{4}\) of the height it reached before.
(a)
Find the height the ball reaches after the third bounce.
(2)
(b)
Find the total distance the ball travels before it comes to rest.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(4\left(\frac{3}{4}\right)^{3}\)
✓ 1
A1 \(\frac{27}{16}\) m
✓ 1
Part (b)
M1 recognise that after the first drop the ball travels each height twice (up and down)
✓ 1
A1 geometric series with first term 3 and \(r=\frac{3}{4}\)
✓ 1
M1 \(4+2\times\frac{3}{1-\frac{3}{4}}\)
✓ 1
A1 28 m
✓ 1
Common error: After the first drop, count each rebound height twice: once going up and once coming down.
153IBMA3S-1-16r15-101.2 Arithmetic sequences and seriesPaper 1Medium5 marks
No calculator
A geometric sequence has second term 6 and fifth term 48.
(a)
Find the common ratio.
(2)
(b)
Find the first term.
(1)
(c)
Find the sum of the first 8 terms.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(r^{3}=\frac{48}{6}\)
✓ 1
A1 \(r=2\)
✓ 1
Part (b)
A1 \(u_1=3\)
✓ 1
Part (c)
M1 \(\frac{3(2^{8}-1)}{2-1}\)
✓ 1
A1 765
✓ 1
Exam tip: Divide two terms to cancel the first term and find the common ratio.
154IBMA3S-1-17r15-061.8 Infinite geometric seriesPaper 1Medium6 marks
No calculator
The infinite geometric series \(12+12(2x-1)+12(2x-1)^{2}+\ldots\) is considered.
(a)
Find the set of values of \(x\) for which the series converges.
(3)
(b)
Given that the sum to infinity is 20, find the value of \(x\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(|2x-1|<1\)
✓ 1
A1 \(-1<2x-1<1\)
✓ 1
A1 \(0<x<1\)
✓ 1
Part (b)
M1 \(\frac{12}{1-(2x-1)}=20\)
✓ 1
A1 \(2-2x=0.6\)
✓ 1
A1 \(x=0.7\) (consistent with part (a))
✓ 1
Exam tip: After solving for the unknown, check your answer lies inside the convergence interval from part (a).
155IBMA3S-1-17r15-081.7 Rational exponents and laws of logarithmsPaper 1Medium5 marks
No calculator
Solve the equation \(\log_2(x+1)+\log_2(x-2)=2\), giving only values of \(x\) that are valid.
(a)
Solve the equation.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\log a+\log b=\log ab\): \(\log_2[(x+1)(x-2)]=2\)
✓ 1
M1 convert to \((x+1)(x-2)=2^{2}\)
✓ 1
A1 \(x^{2}-x-6=0\)
✓ 1
A1 \(x=3\) or \(x=-2\)
✓ 1
R1 reject \(x=-2\) because \(\log_2(x-2)\) is undefined for \(x<2\)
✓ 1
Exam tip: Always substitute back, or check the domain, when a logarithmic equation reduces to a quadratic.
156IBMA3S-1-17r16-091.6 Deductive proofPaper 1Medium6 marks
No calculator
Books in a library reading room are numbered with odd integers. Let 2k − 1 and 2k + 1 be two consecutive odd integers, where k ∈ ℤ.
(a)
Prove that the product of any two consecutive odd integers is one less than a multiple of 4.
(4)
(b)
Hence, by choosing a suitable value of k, find the value of 99 × 101.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 product = (2k − 1)(2k + 1)
✓ 1
M1 expansion attempt
✓ 1
A1 4k2 − 1
✓ 1
R1 = 4(k2) − 1 where k2 is an integer, so one less than a multiple of 4
✓ 1
Part (b)
M1 2k − 1 = 99 so k = 50
✓ 1
A1 4 × 502 − 1 = 9999
✓ 1
Exam tip: 'Hence' means you must use the result of the previous part; calculating 99 × 101 directly will not earn the marks.
157IBMA3S-1-17r16-121.7 Rational exponents and laws of logarithmsPaper 1Medium5 marks
No calculator
Solve \(\log_3(x+2) + \log_3(x-6) = 2\).
(a)
Find the value of x, justifying your answer.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\log_3\big((x+2)(x-6)\big)\)
✓ 1
M1 \((x+2)(x-6) = 3^2\)
✓ 1
A1 \(x^2 - 4x - 21 = 0\)
✓ 1
A1 \(x = 7\) or \(x = -3\)
✓ 1
R1 reject \(x = -3\) since \(\log_3(x-6)\) is undefined; \(x = 7\)
✓ 1
Common error: After solving a logarithmic equation always test each root in the original equation; arguments of logarithms must be positive.
158IBMA3S-1-16r19-051.8 Infinite geometric seriesPaper 1Medium6 marks
No calculator
The first term of an infinite geometric series is 18 and its sum to infinity is 54.
(a)
Find the common ratio.
(2)
(b)
Find the third term of the series.
(2)
(c)
Find the sum of all the terms of the series from the third term onwards.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(S_\infty=\frac{u_1}{1-r}\): \(\frac{18}{1-r}=54\)
✓ 1
A1 \(r=\frac23\)
✓ 1
Part (b)
M1 use of \(u_3=u_1r^2\)
✓ 1
A1 \(8\)
✓ 1
Part (c)
M1 attempt to subtract the first two terms from the sum to infinity, or use \(\frac{u_3}{1-r}\)
✓ 1
A1 \(24\)
✓ 1
Exam tip: The terms from the third onwards form their own geometric series with first term \(u_3\) and the same ratio.
159IBMA3S-1-16r19-071.7 Rational exponents and laws of logarithmsPaper 1Medium5 marks
No calculator
Solve the equation \(4^{x}-6\times2^{x}+8=0\).
(a)
Solve the equation \(4^{x}-6\times2^{x}+8=0\).
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writing \(4^x=(2^x)^2\)
✓ 1
A1 \((2^x)^2-6(2^x)+8=0\)
✓ 1
M1 attempt to solve the quadratic in \(2^x\): \((2^x-2)(2^x-4)=0\)
✓ 1
A1 \(2^x=2\) or \(2^x=4\)
✓ 1
A1 \(x=1\) or \(x=2\)
✓ 1
Exam tip: Use a substitution such as \(u=2^x\) to turn the equation into a quadratic, then convert back.
160IBMA3S-1-16r19-081.7 Rational exponents and laws of logarithmsPaper 1Medium6 marks
No calculator
Solve \(\log_2(x+1)+\log_2(x-2)=2\).
(a)
Solve \(\log_2(x+1)+\log_2(x-2)=2\).
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the product law: \(\log_2\big((x+1)(x-2)\big)\)
✓ 1
M1 rewriting as \((x+1)(x-2)=2^2\)
✓ 1
A1 \(x^2-x-6=0\)
✓ 1
M1 attempt to solve: \((x-3)(x+2)=0\)
✓ 1
A1 \(x=3\) or \(x=-2\)
✓ 1
R1 reject \(x=-2\) because \(\log_2(x-2)\) is not defined for \(x<2\), so \(x=3\)
✓ 1
Common error: Always check solutions of log equations in the original equation; arguments of logarithms must be positive.
161IBMA3S-1-16r20-061.7 Rational exponents and laws of logarithmsPaper 1Medium6 marks
No calculator
Let \(\log_a 2=p\) and \(\log_a 3=q\), where \(a>0\) and \(a\neq 1\).
(a)
Express \(\log_a 18\) in terms of \(p\) and \(q\).
(2)
(b)
Solve \(\log_2 x+\log_2 (x-2)=3\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(18=2\times 3^2\), so \(\log_a 18=\log_a 2+\log_a 3^2\)
✓ 1
A1 \(p+2q\)
✓ 1
Part (b)
M1 combines logarithms: \(\log_2 [x(x-2)]=3\)
✓ 1
M1 \(x(x-2)=2^3\) and forms \(x^2-2x-8=0\)
✓ 1
A1 \(x=4\) or \(x=-2\)
✓ 1
R1 rejects \(x=-2\) since \(\log_2(-2)\) is undefined, so \(x=4\)
✓ 1
Common error: After solving a logarithmic equation, check each root in the original equation; a root that makes a logarithm's argument negative must be rejected.
162IBMA3S-1-16r20-071.6 Deductive proofPaper 1Medium5 marks
No calculator
(a)
Prove that the square of any odd integer can be written in the form \(8k+1\), where \(k\) is an integer.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 let the odd integer be \(2n+1\) where \(n\) is an integer
✓ 1
M1 expands \((2n+1)^2=4n^2+4n+1\)
✓ 1
A1 \(=4n(n+1)+1\)
✓ 1
R1 \(n(n+1)\) is a product of consecutive integers so is even, \(n(n+1)=2k\)
✓ 1
R1 hence \((2n+1)^2=8k+1\)
✓ 1
Exam tip: If one factor of a product must be even, say why (two consecutive integers always include an even number) rather than just asserting it.
163IBMA3S-1-16r20-091.9 The binomial theoremPaper 1Medium5 marks
No calculator
The coefficient of \(x^2\) in the expansion of \((2+kx)^5\) is 720, where \(k>0\).
(a)
Find the value of \(k\).
(3)
(b)
Find the coefficient of \(x^4\) in the expansion.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\binom{5}{2}\,2^3k^2\) as the \(x^2\) term
✓ 1
M1 \(80k^2=720\), so \(k^2=9\)
✓ 1
A1 \(k=3\)
✓ 1
Part (b)
M1 \(\binom{5}{4}\times 2\times k^4\) with \(k=3\) (ECF from (a))
✓ 1
A1 \(810\)
✓ 1
Common error: In \((2+kx)^5\) the powers of both \(2\) and \(k\) count: the \(x^2\) term contains \(2^3\) and \(k^2\).
164IBMA3S-1-17r20-061.9 The binomial theoremPaper 1Medium5 marks
No calculator
The expansion of \((2+ax)^6\), where \(a>0\), has a coefficient of \(x^2\) equal to 960.
(a)
Find the value of \(a\).
(3)
(b)
Hence find the coefficient of \(x^3\) in the expansion.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find the \(x^2\) term: \(\binom{6}{2}2^4(ax)^2\)
✓ 1
A1 \(240a^2=960\)
✓ 1
A1 \(a=2\)
✓ 1
Part (b)
M1 \(\binom{6}{3}2^3a^3\) with their \(a\)
✓ 1
A1 1280 (ECF from (a))
✓ 1
Common error: The coefficient of \(x^2\) includes \(a^2\) and the power \(2^4\) from the constant term; forgetting either gives the wrong equation.
165IBMA3S-1-17r20-071.8 Infinite geometric seriesPaper 1Medium7 marks
No calculator
An infinite geometric series has first term 12 and common ratio \(\dfrac{x-2}{3}\).
(a)
Find the set of values of \(x\) for which the series converges.
(3)
(b)
Show that the sum to infinity is \(\dfrac{36}{5-x}\).
(2)
(c)
The sum to infinity is 9. Find the value of \(x\) and verify that the series converges for this value.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(|r|<1\)
✓ 1
A1 \(|x-2|<3\)
✓ 1
A1 \(-1<x<5\)
✓ 1
Part (b)
M1 \(S_\infty=\dfrac{12}{1-\frac{x-2}{3}}\)
✓ 1
A1 \(\dfrac{12}{\frac{5-x}{3}}=\dfrac{36}{5-x}\)
✓ 1
Part (c)
M1 \(\dfrac{36}{5-x}=9\) so \(5-x=4\)
✓ 1
A1 \(x=1\), which lies in \(-1<x<5\) so the series converges
✓ 1
Exam tip: After solving for a parameter, always check the answer lies inside the convergence interval found earlier.
166IBMA3S-1-17r20-081.7 Rational exponents and laws of logarithmsPaper 1Medium6 marks
No calculator
Consider the equation \(2^{2x}-5\times2^{x}+4=0\).
(a)
Using the substitution \(u=2^x\), show that the equation can be written as \(u^2-5u+4=0\).
(2)
(b)
Hence solve the equation for \(x\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2^{2x}=(2^x)^2\)
✓ 1
A1 \(u^2-5u+4=0\) (AG)
✓ 1
Part (b)
M1 attempt to solve, for example \((u-1)(u-4)=0\)
✓ 1
A1 \(u=1\) or \(u=4\)
✓ 1
M1 \(2^x=1\) and \(2^x=4\)
✓ 1
A1 \(x=0\) and \(x=2\)
✓ 1
Common error: Do not stop at the values of \(u\): the question asks for \(x\), so convert back with \(2^x=u\).
167IBMA3S-1-17r20-111.9 The binomial theoremPaper 1Medium6 marks
No calculator
Consider the expansion of \(\left(x+\dfrac{2}{x}\right)^6\).
(a)
Show that the general term can be written as \(\binom{6}{k}2^k x^{6-2k}\).
(2)
(b)
Find the constant term.
(2)
(c)
Find the coefficient of \(x^2\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 general term \(\binom{6}{k}x^{6-k}\left(\dfrac{2}{x}\right)^k\)
Exam tip: To find a specific power, set the exponent in the general term equal to the target and solve for \(k\) before evaluating anything.
168IBMA3S-1-17r22-071.8 Infinite geometric seriesPaper 1Medium7 marks
No calculator
A water-softening cartridge absorbs 12 g of minerals in week 1. In each later week it absorbs a fixed fraction \(r\) of the mass absorbed in the previous week. The maximum total mass the cartridge can ever absorb is 30 g.
A1 \(x = 4\), rejecting \(x = -2\) because \(\log_2(-2)\) is undefined
✓ 1
Common error: After solving a logarithmic equation, check each solution in the original equation because the argument of a logarithm must be positive.
173IBMA3S-1-16r25-141.9 The binomial theoremPaper 1Medium6 marks
No calculator
(a)
Show that \(8^x \times 4^{x+1} = 2^{5x+2}\).
(2)
(b)
Hence solve \(8^x \times 4^{x+1} = 128\).
(2)
(c)
Prove that \(8^x \times 4^{x+1}\) is a multiple of 4 for every positive integer \(x\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(8^x = 2^{3x}\) and \(4^{x+1} = 2^{2x+2}\)
✓ 1
A1 \(2^{3x} \times 2^{2x+2} = 2^{5x+2}\)
✓ 1
Part (b)
M1 \(128 = 2^7\), so \(5x + 2 = 7\)
✓ 1
A1 \(x = 1\)
✓ 1
Part (c)
M1 \(2^{5x+2} = 4 \times 2^{5x}\)
✓ 1
R1 \(2^{5x}\) is an integer when \(x\) is a positive integer, so the expression is 4 times an integer
✓ 1
Exam tip: Convert every power to the same base first; products of powers then reduce to adding indices.
174IBMA3S-1-18r29-041.8 Infinite geometric seriesPaper 1Medium6 marks
No calculator
The extra yearly energy produced by a wind farm after a series of upgrades forms a geometric sequence. The extra energy in the second year is 12 GWh, and the total extra energy over all future years is 64 GWh. Find the two possible values of the common ratio r, and the first term corresponding to each.
(a)
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 u1r = 12 and u1/(1 − r) = 64
✓ 1
M1 eliminate u1: 64r(1 − r) = 12
✓ 1
A1 16r2 − 16r + 3 = 0
✓ 1
M1 attempt to solve the quadratic, e.g. (4r − 3)(4r − 1) = 0
✓ 1
A1 r = 3/4 or r = 1/4
✓ 1
A1 u1 = 16 GWh for r = 3/4 and u1 = 48 GWh for r = 1/4
✓ 1
Common error: Check each value of r satisfies |r| < 1 before using the sum to infinity.
175IBMA3S-1-18r29-051.8 Infinite geometric seriesPaper 1Medium5 marks
No calculator
The recurring decimal 0.454545… can be written as an infinite geometric series.
(a)
Write down the first term and the common ratio of this series.
(2)
(b)
Hence show that 0.454545… = 5/11.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 first term 0.45
✓ 1
A1 common ratio 0.01
✓ 1
Part (b)
M1 use of S∞ = u1/(1 − r)
✓ 1
A1 0.45 ÷ 0.99
✓ 1
A1 45/99 = 5/11 (AG)
✓ 1
Exam tip: For a 'show that' question, write each step clearly and finish with the result stated in the question.
176IBMA3S-1-18r29-061.8 Infinite geometric seriesPaper 1Medium6 marks
No calculator
A geometric series has first term 5 and common ratio (2x − 1)/6.
(a)
Find the set of values of x for which the series converges.
(4)
(b)
When x = 2, find the sum to infinity of the series.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 convergence requires |r| < 1, so |2x − 1| < 6
✓ 1
M1 −6 < 2x − 1 < 6
✓ 1
A1 x < 7/2
✓ 1
A1 x > −5/2
✓ 1
Part (b)
M1 r = 3/6 = 1/2 and S∞ = 5 ÷ (1 − 1/2)
✓ 1
A1 10
✓ 1
Common error: |r| < 1 gives two inequalities; do not stop after solving only one of them.
177IBMA3S-1-18r29-111.9 The binomial theoremPaper 1Medium5 marks
No calculator
Find the term independent of x in the expansion of (x2 + 2/x)6.
(a)
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 general term C(6, r)(x2)6−r(2/x)r
✓ 1
A1 power of x is 12 − 3r
✓ 1
M1 set 12 − 3r = 0
✓ 1
A1 r = 4
✓ 1
A1 term = 15 × 16 = 240
✓ 1
Common error: The powers of x in each term come from both brackets: (x²)⁶⁻r and (1/x)r.
178IBMA3S-1-19r29-061.8 Infinite geometric seriesPaper 1Medium7 marks
No calculator
The step lengths of a walking robot form an infinite geometric sequence with common ratio \(r>0\). The second step is 6 cm long and the sum to infinity of the sequence is 27 cm.
(a)
Show that \(9r^2-9r+2=0\).
(3)
(b)
Find the two possible values of \(r\).
(2)
(c)
The first step is longer than 12 cm. Find the length of the first step.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_1r=6\) and \(\frac{u_1}{1-r}=27\)
✓ 1
M1 eliminating \(u_1\): \(\frac{6}{r(1-r)}=27\) or \(27r(1-r)=6\)
✓ 1
A1 rearranging to \(9r^2-9r+2=0\)
✓ 1
Part (b)
M1 attempt to factorise or use the quadratic formula, \((3r-1)(3r-2)=0\)
A1 \(0.1818\ldots=\frac{2}{11}\) or \(0.0181818\ldots=\frac{1}{55}\)
✓ 1
A1 \(\frac{3}{10}+\frac{1}{55}=\frac{7}{22}\)
✓ 1
Exam tip: Split a decimal with a non-repeating start into the non-repeating part plus a scaled recurring part, then apply the sum to infinity.
183IBMA3S-1-18r31-041.6 Deductive proofPaper 1Medium6 marks
No calculator
In a game, a player earns bonus points in rounds. In round 1 the player earns 6 points. The points in the rounds form an infinite geometric sequence \(6,\ 4x,\ \frac{8x^2}{3},\ \dots\), where \(x>0\).
(a)
Find the range of values of \(x\) for which the sum of all the points the player earns is finite.
(3)
(b)
The sum of all the points earned is 18. Find the value of \(x\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 common ratio \(r=\frac{4x}{6}=\frac{2x}{3}\)
✓ 1
M1 uses \(|r|<1\), giving \(\frac{2x}{3}<1\)
✓ 1
A1 \(0<x<\frac32\)
✓ 1
Part (b)
M1 \(\frac{6}{1-r}=18\)
✓ 1
M1 \(r=\frac23\) so \(\frac{2x}{3}=\frac23\)
✓ 1
A1 \(x=1\)
✓ 1
Exam tip: Always check that your value of \(x\) lies inside the range for which an infinite geometric series converges. Here \(x=1\) is inside \(0<x<\frac32\).
184IBMA3S-1-18r31-051.6 Deductive proofPaper 1Medium5 marks
No calculator
Let \(n\) be any integer. Two consecutive odd integers can be written as \(2n+1\) and \(2n+3\).
(a)
Expand and simplify \((2n+1)^2 + (2n+3)^2\).
(2)
(b)
Hence prove that the sum of the squares of any two consecutive odd integers is always 2 more than a multiple of 8.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 expands both brackets, \(4n^2+4n+1\) and \(4n^2+12n+9\)
✓ 1
A1 \(8n^2+16n+10\)
✓ 1
Part (b)
M1 rewrites as \(8(n^2+2n+1)+2\) (or \(8(n^2+2n)+10\))
✓ 1
R1 \(n^2+2n+1\) is an integer for any integer \(n\), so the expression has the form \(8k+2\) with \(k\) an integer
✓ 1
A1 states the conclusion: the sum is always 2 more than a multiple of 8
✓ 1
Exam tip: A proof must work for every integer. Use a letter such as \(n\) and never test particular numbers, which only shows the result is true for those cases.
185IBMA3S-1-18r31-131.2 Arithmetic sequences and seriesPaper 1Medium10 marks
No calculator
A geometric series has first term \(a\) and common ratio \(r\), where \(r\neq1\). The sum of the first \(n\) terms is \(S_n\).
(a)
By writing out \(S_n\) and \(rS_n\), show that \(S_n=\frac{a(r^n-1)}{r-1}\).
(4)
(b)
In a game, the points for the first six levels are \(3,\ 12,\ 48,\ \dots\) Find the total number of points for the six levels.
(2)
(c)
Consider the geometric series \(1+4+16+\dots+4^{n-1}\). By using part (a), prove that \(4^n-1\) is divisible by 3 for every positive integer \(n\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(S_n=a+ar+ar^2+\dots+ar^{n-1}\)
✓ 1
M1 \(rS_n=ar+ar^2+\dots+ar^{n}\)
✓ 1
M1 subtracts, so \(S_n-rS_n=a-ar^{n}\) (the middle terms cancel)
✓ 1
A1 \(S_n(1-r)=a(1-r^n)\), so \(S_n=\frac{a(r^n-1)}{r-1}\)
✓ 1
Part (b)
M1 \(r=4\) and \(S_6=\frac{3(4^6-1)}{4-1}\)
✓ 1
A1 4095 points
✓ 1
Part (c)
M1 uses \(a=1\), \(r=4\) to write \(S_n=\frac{4^n-1}{3}\)
✓ 1
R1 every term of the series is an integer, so \(S_n\) is an integer
✓ 1
M1 rearranges to \(4^n-1=3S_n\)
✓ 1
R1 \(4^n-1\) is 3 times an integer, so it is divisible by 3 for every positive integer \(n\)
✓ 1
Common error: In the proof, do not check a few values of \(n\). The argument works because \(S_n\) is an integer for every \(n\), and \(4^n-1=3S_n\).
186IBMA3S-1-17r32-081.6 Deductive proofPaper 1Medium6 marks
No calculator
Two runners in a school race are given consecutive whole-number bib numbers, \(n\) and \(n+1\). A stewarding code is formed by adding the squares of the two bib numbers.
(a)
Prove that the stewarding code is always an odd number.
(4)
(b)
A steward claims that the code is always divisible by 3. Show that this claim is false.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum \(=n^{2}+(n+1)^{2}\)
✓ 1
A1 \(=2n^{2}+2n+1\)
✓ 1
M1 \(=2(n^{2}+n)+1\)
✓ 1
R1 \(n^{2}+n\) is an integer, so \(2(n^{2}+n)\) is even and the code is odd
✓ 1
Part (b)
M1 choose a counterexample, e.g. \(n=1\): \(1^{2}+2^{2}=5\)
✓ 1
R1 5 is not divisible by 3, so the claim is false
✓ 1
Exam tip: One counterexample is enough to disprove a general claim, but a list of examples that work never proves one.
187IBMA3S-1-17r32-091.1 Standard formPaper 1Medium7 marks
No calculator
For bodies orbiting the Sun, the orbital period \(T\) in years is related to the average distance \(a\) from the Sun in astronomical units (AU) by \(T=a^{3/2}\).
(a)
Find the period of an asteroid with \(a=9\) AU.
(2)
(b)
A comet has a period of 64 years. Find its average distance from the Sun.
(3)
(c)
Show that if the average distance of an orbit is multiplied by 4, the period is multiplied by 8.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(9^{3/2}=(\sqrt9)^{3}\)
✓ 1
A1 27 years
✓ 1
Part (b)
M1 \(a^{3/2}=64\) so \(a=64^{2/3}\)
✓ 1
M1 \((\sqrt[3]{64})^{2}=4^{2}\)
✓ 1
A1 16 AU
✓ 1
Part (c)
M1 \((4a)^{3/2}=4^{3/2}a^{3/2}\)
✓ 1
A1 \(4^{3/2}=8\), so new period \(=8T\) (AG)
✓ 1
Exam tip: For \(x^{m/n}\), take the root first and then the power, as in \(64^{2/3}=(\sqrt[3]{64})^{2}\), to keep the numbers small.
188IBMA3S-1-17r32-101.7 Rational exponents and laws of logarithmsPaper 1Medium9 marks
No calculator
Let \(p=\log_{10}2\) and \(q=\log_{10}3\).
(a)
Express \(\log_{10}72\) in terms of \(p\) and \(q\).
(2)
(b)
Express \(\log_{10}0.75\) in terms of \(p\) and \(q\).
(2)
(c)
Solve \(\log_{2}x+\log_{2}(x-2)=3\).
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(72=2^{3}\times3^{2}\) and use of \(\log(ab)=\log a+\log b\)
✓ 1
A1 \(3p+2q\)
✓ 1
Part (b)
M1 \(0.75=\frac{3}{4}=3\times2^{-2}\)
✓ 1
A1 \(q-2p\)
✓ 1
Part (c)
M1 \(\log_{2}\big(x(x-2)\big)=3\)
✓ 1
M1 \(x(x-2)=2^{3}=8\)
✓ 1
M1 \(x^{2}-2x-8=0\) and attempt to factorise or solve
✓ 1
R1 reject \(x=-2\) since \(\log_{2}(-2)\) is undefined
✓ 1
A1 \(x=4\)
✓ 1
Common error: After solving a logarithmic equation, check every solution in the original equation: a root that makes a logarithm's argument negative must be rejected.
189IBMA3S-1-18r38-051.6 Deductive proofPaper 1Medium7 marks
No calculator
The breaking strength S kN of a rope of diameter d mm is modelled by S = kd3/2. A cord of diameter 4 mm has breaking strength 6.4 kN.
(a)
Find the value of k.
(2)
(b)
Find the breaking strength of a rope of diameter 9 mm.
(2)
(c)
A haul line has breaking strength 51.2 kN. Find its diameter.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 6.4 = k × 43/2 = 8k
✓ 1
A1 k = 0.8
✓ 1
Part (b)
M1 0.8 × 93/2 = 0.8 × 27
✓ 1
A1 21.6 kN
✓ 1
Part (c)
M1 d3/2 = 51.2/0.8 = 64
✓ 1
M1 d = 642/3
✓ 1
A1 16 mm
✓ 1
Common error: Rational exponents: am/ⁿ = (ⁿ√a)m; take the root first to keep the numbers small.
190IBMA3S-1-18r38-061.7 Rational exponents and laws of logarithmsPaper 1Medium6 marks
Hence solve the original equation, justifying your choice of solution.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 log2((x + 5)(x − 1)) = 4
✓ 1
M1 (x + 5)(x − 1) = 24 = 16
✓ 1
A1 x2 + 4x − 5 = 16, so x2 + 4x − 21 = 0 (AG)
✓ 1
Part (b)
M1 (x + 7)(x − 3) = 0
✓ 1
A1 x = 3 or x = −7
✓ 1
R1 reject x = −7 since log2(x − 1) is undefined; x = 3
✓ 1
Common error: Always check solutions of log equations against the domain: each argument must be positive.
191IBMA3S-1-18r38-091.2 Arithmetic sequences and seriesPaper 1Medium7 marks
No calculator
On a long ridge, a mountaineer's daily height gains, in km, form a geometric sequence with common ratio r, where 0 < r < 1. The gain on day 2 is 1.2 km. If the sequence continued indefinitely, the total gain would be 6.4 km.
(a)
Show that 16r2 − 16r + 3 = 0.
(3)
(b)
Find the two possible values of r.
(2)
(c)
The gain on day 1 was more than 3 km. Find the gain on day 1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 u1 = 1.2/r
✓ 1
M1 S∞ = u1/(1 − r) = 6.4
✓ 1
A1 1.2 = 6.4r(1 − r), which rearranges to 16r2 − 16r + 3 = 0 (AG)
✓ 1
Part (b)
M1 (4r − 1)(4r − 3) = 0
✓ 1
A1 r = 1/4 or r = 3/4
✓ 1
Part (c)
R1 u1 = 1.2/r gives 4.8 km for r = 1/4 and 1.6 km for r = 3/4; only r = 1/4 gives more than 3
✓ 1
A1 4.8 km
✓ 1
Common error: Two values of r can both satisfy |r| < 1; use the extra condition in the question to choose between them.
192IBMA3S-1-18r41-061.2 Arithmetic sequences and seriesPaper 1Medium6 marks
No calculator
A shop arranges tins in a display of rows. The numbers of tins in successive rows form an arithmetic sequence. Row 3 has 17 tins and row 8 has 42 tins.
(a)
Find the number of tins in row 1 and the common difference.
(4)
(b)
The display has 20 rows. Find the total number of tins in the display.
Exam tip: After solving a logarithmic equation, check that each solution keeps every logarithm's argument positive.
196IBMA3S-1-17r43-051.9 The binomial theoremPaper 1Medium5 marks
No calculator
In the expansion of \(\left(x^{2}+\frac{2}{x}\right)^{6}\), the general term is \(T_{r+1}\), where \(r=0,1,\dots,6\) and \(r\) is the number of factors of \(\frac{2}{x}\) chosen.
A1 \(\binom{6}{r}2^r x^{12-2r-r}=\binom{6}{r}2^r x^{12-3r}\), so shown
✓ 1
Part (b)
M1 set \(12-3r=0\)
✓ 1
A1 \(r=4\)
✓ 1
A1 \(\binom{6}{4}\times2^4=15\times16=240\)
✓ 1
Exam tip: For a constant term, set the power of \(x\) in the general term to zero and solve for \(r\) before substituting.
197IBMA3S-1-17r43-061.9 The binomial theoremPaper 1Medium5 marks
No calculator
The coefficient of \(x^3\) in the expansion of \((1+kx)^7\), where \(k\in\mathbb{R}\), is 280.
(a)
Find the value of \(k\).
(3)
(b)
Hence find the coefficient of \(x^2\) in the expansion.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\binom{7}{3}=35\) used with \(k^3\)
✓ 1
A1 \(35k^3=280\) so \(k^3=8\)
✓ 1
A1 \(k=2\)
✓ 1
Part (b)
M1 \(\binom{7}{2}k^2=21k^2\) (ECF from (a))
✓ 1
A1 84
✓ 1
Common error: The coefficient of \(x^r\) in \((1+kx)^n\) is \(\binom{n}{r}k^r\); forgetting to raise \(k\) to the power \(r\) gives the wrong equation.
198IBMA3S-1-17r43-101.7 Rational exponents and laws of logarithmsPaper 1Medium6 marks
No calculator
Let \(\log_2 x=p\) and \(\log_2 y=q\), where \(x,y>0\). Write each of the following in terms of \(p\) and \(q\).
(a)
\(\log_2\left(x^3y\right)\)
(2)
(b)
\(\log_2\left(\dfrac{\sqrt{x}}{y^2}\right)\)
(2)
(c)
\(\log_8 x\)
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\log_2 x^3+\log_2 y\)
✓ 1
A1 \(3p+q\)
✓ 1
Part (b)
M1 \(\frac12\log_2 x-2\log_2 y\)
✓ 1
A1 \(\frac{p}{2}-2q\)
✓ 1
Part (c)
M1 change of base: \(\frac{\log_2 x}{\log_2 8}\)
✓ 1
A1 \(\frac{p}{3}\)
✓ 1
Exam tip: Split products and quotients into sums and differences first, then bring powers down as multipliers.
199IBMA3S-1-19r44-041.9 The binomial theoremPaper 1Medium5 marks
No calculator
A hospital data analyst models the effect of a drug using the expression \((3+2x)^5\), where \(x\) is a scaled measure of time.
(a)
Find the coefficient of \(x^3\) in the expansion of \((3+2x)^5\).
(3)
(b)
The coefficient of \(x^2\) in the expansion is \(c\). Find the value of \(\dfrac{720}{c}\), giving your answer as a fraction in its simplest form.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the binomial theorem with a term in x3, e.g. C(5,3)(3)2(2x)3
✓ 1
A1 10 × 9 × 8
✓ 1
A1 720
✓ 1
Part (b)
A1 c = 10 × 27 × 4 = 1080
✓ 1
A1 720/1080 = 2/3
✓ 1
Common error: In \((a+bx)^n\) the constant \(b\) is raised to the same power as \(x\); writing \(2x^3\) instead of \((2x)^3 = 8x^3\) loses the accuracy mark.
200IBMA3S-1-19r44-071.9 The binomial theoremPaper 1Medium5 marks
No calculator
A beekeeper models the temperature rise in a hive with the function \(P(x) = (1+ax)^6\), where \(a>0\). The coefficient of \(x^2\) in the expansion of \(P(x)\) is 240.
(a)
Find the value of \(a\).
(3)
(b)
Find the coefficient of \(x^3\) in the expansion of \(P(x)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the binomial theorem, C(6,2)(ax)2 = 15a2x2
✓ 1
M1 solve 15a2 = 240, so a2 = 16
✓ 1
A1 a = 4 (reject −4 since a > 0)
✓ 1
Part (b)
M1 C(6,3)(4)3 = 20 × 64
✓ 1
A1 1280
✓ 1
Exam tip: When a question gives a condition such as \(a>0\), use it explicitly to reject the other root.
201IBMA3S-1-19r44-081.6 Deductive proofPaper 1Medium6 marks
No calculator
Beds in a hospital ward are numbered with odd integers. Let two consecutive odd integers be \(2n+1\) and \(2n+3\), where \(n\) is a positive integer.
(a)
Prove that the difference between the squares of two consecutive odd integers is always a multiple of 8.
(4)
(b)
Hence find the value of \(1001^2 - 999^2\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 write (2n + 3)2 − (2n + 1)2
✓ 1
M1 expand both squares, (4n2 + 12n + 9) − (4n2 + 4n + 1), or factorise as a difference of squares
✓ 1
A1 8n + 8
✓ 1
R1 = 8(n + 1), and n + 1 is an integer, so a multiple of 8
✓ 1
Part (b)
M1 identify n = 499 (so 999 = 2n + 1 and 1001 = 2n + 3)
✓ 1
A1 8 × 500 = 4000
✓ 1
Exam tip: Use the structure of the statement: writing consecutive odd numbers as \(2n+1\) and \(2n+3\) is what makes the multiple of 8 visible.
202IBMA3S-1-19r46-081.9 The binomial theoremPaper 1Medium5 marks
No calculator
Consider the expansion of \(\left(2x-\frac{1}{x^2}\right)^6\).
(a)
Show that the term in the expansion containing \(\left(-\frac{1}{x^2}\right)^r\) has \(x\) to the power \(6-3r\).
(2)
(b)
Hence find the term that is independent of \(x\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 general term \(\binom{6}{r}(2x)^{6-r}\left(-\frac{1}{x^2}\right)^r\)
✓ 1
A1 power of \(x\): \((6-r)-2r=6-3r\)
✓ 1
Part (b)
M1 set \(6-3r=0\) to get \(r=2\)
✓ 1
A1 \(\binom{6}{2}=15\) and \(2^4=16\) with \(\left(-1\right)^2=1\)
✓ 1
A1 240
✓ 1
Common error: A negative power in the second bracket still counts towards the total power of \(x\): \(\frac{1}{x^2}\) contributes \(-2\) for each factor.
203IBMA3S-1-17r47-071.9 The binomial theoremPaper 1Medium9 marks
No calculator
This question is about exponents.
(a)
Evaluate \(27^{-\frac{2}{3}}\).
(2)
(b)
Solve \(4^{x+1} = 8^{x-2}\).
(3)
(c)
Given that \(x > 0\), solve \(2^{2x} - 5 \times 2^{x} + 4 = 0\).
Must see an intermediate simplified step; \(6n-5\) is given.
\(u_{n+1}-u_{n}=6\) is constant (independent of \(n\)), so the sequence is arithmetic
R1
(c)
\(6n-5\gt 500\Rightarrow n\gt 84.1\ldots\)
(M1)
\(n=85\)
A1
Answers: (a) \(u_{1}=1,\ u_{2}=7\) · (c) \(n=85\)
207AASL1-q00201.2 Arithmetic sequences and seriesPaper 1Hard14 marks
An arithmetic sequence has first term 4 and common difference \(d\), where \(d\neq0\). The first, third and ninth terms of this arithmetic sequence are, in that order, the first three terms of a geometric sequence.
(a)
Show that \(d=4\).
(4)
(b)
Find the common ratio of the geometric sequence.
(1)
(c)
Show that the sum of the first \(n\) terms of the arithmetic sequence is \(2n^{2}+2n\).
(2)
(d)
Find an expression for the sum of the first \(n\) terms of the geometric sequence, in the form \(a\left(3^{n}-1\right)\), where \(a\in\mathbb{Z}\).
(2)
(e)
Find which term of the arithmetic sequence is equal to the fifth term of the geometric sequence.
(3)
(f)
Prove that every term of the geometric sequence is also a term of the arithmetic sequence.
(2)
Show mark scheme
Working
Mark
Notes
(a)
third term \(4+2d\), ninth term \(4+8d\)
A1
\(\dfrac{4+2d}{4}=\dfrac{4+8d}{4+2d}\), or \((4+2d)^{2}=4(4+8d)\)
M1
\(16+16d+4d^{2}=16+32d\), so \(4d^{2}-16d=0\)
A1
\(4d(d-4)=0\) and \(d\neq0\)
R1
Must reject \(d=0\).
\(d=4\)
AG
(b)
\(r=3\)
A1
Terms 4, 12, 36.
(c)
\(\dfrac{n}{2}\big(2(4)+4(n-1)\big)\)
M1
\(\dfrac{n}{2}(4n+4)\)
A1
\(=2n^{2}+2n\)
AG
(d)
\(\dfrac{4\left(3^{n}-1\right)}{3-1}\)
(M1)
\(2\left(3^{n}-1\right)\)
A1
(e)
fifth term of the geometric sequence \(=4\times3^{4}=324\)
(A1)
\(4+4(n-1)=324\), i.e. \(4n=324\)
(M1)
the 81st term
A1
(f)
\(n\)th arithmetic term is \(4n\); \(k\)th geometric term is \(4\times3^{k-1}\)
M1
\(4\times3^{k-1}=4n\) with \(n=3^{k-1}\), which is a positive integer for every \(k\in\mathbb{Z}^{+}\); so each geometric term is the \(\left(3^{k-1}\right)\)th arithmetic term
R1
Answers: (b) \(r=3\) · (d) \(2\left(3^{n}-1\right)\) · (e) the 81st term
208AASL1-q00211.2 Arithmetic sequences and seriesPaper 1Hard13 marks
Two arithmetic sequences are defined as follows.
Sequence A: \(5,\ 12,\ 19,\ 26,\ \ldots\)
Sequence B: \(3,\ 8,\ 13,\ 18,\ \ldots\)
(a)(i)
Find an expression for the \(n\)th term of sequence A.
(1)
(a)(ii)
Find an expression for the \(m\)th term of sequence B.
(1)
(b)
Find the sum of the first 50 terms of sequence A.
(2)
(c)
Show that 33 is a term of both sequences.
(2)
(d)
The numbers that are terms of both sequences form a third sequence, C, with first term 33. Explain why C is arithmetic with common difference 35.
(2)
(e)
Find the number of terms of sequence C that are less than 1000.
(3)
(f)
Find the sum of the terms of sequence C that are less than 1000.
(2)
Show mark scheme
Working
Mark
Notes
(a)(i)
\(7n-2\)
A1
(a)(ii)
\(5m-2\)
A1
(b)
\(\dfrac{50}{2}\big(2(5)+49(7)\big)\)
(M1)
\(8825\)
A1
(c)
\(7n-2=33\Rightarrow n=5\), an integer
A1
\(5m-2=33\Rightarrow m=7\), an integer
A1
(d)
\(7n-2=5m-2\Rightarrow7n=5m\), so \(n\) must be a multiple of 5 (as 5 and 7 have no common factor)
R1
Or: a common term must increase by a multiple of both 7 and 5 to reach the next common term.
common terms are \(7(5k)-2=35k-2\), so consecutive common terms differ by 35
R1
(e)
\(33+35(k-1)\lt 1000\), i.e. \(35k-2\lt 1000\)
(M1)
\(k\lt 28.6\ldots\), or 28th term 978 and 29th term 1013
Award the final A1 only if \(x=-4\) is rejected (since \(\log_{2}x\) requires \(x\gt 0\)) or not given.
Answers: (b) \(x=6\)
212AASL1-q00681.7 Rational exponents and laws of logarithmsPaper 1Hard7 marks
Let \(x\gt 0\), \(x\neq1\).
(a)
Show that \(\log_{x}2=\dfrac{1}{\log_{2}x}\).
(2)
(b)
Hence solve \(\log_{2}x+3\log_{x}2=4\).
(5)
Show mark scheme
Working
Mark
Notes
(a)
\(\log_{x}2=\dfrac{\log_{2}2}{\log_{2}x}\)
M1
Change of base to base 2 (or to any common base).
\(\log_{2}2=1\)
A1
\(\log_{x}2=\dfrac{1}{\log_{2}x}\)
AG
(b)
\(\log_{2}x+\dfrac{3}{\log_{2}x}=4\)
M1
letting \(y=\log_{2}x\): \(y^{2}-4y+3=0\)
M1
\(y=1\) or \(y=3\)
A1
\(x=2^{1}\) or \(x=2^{3}\)
(M1)
\(x=2\) or \(x=8\)
A1
Both values required.
Answers: (b) \(x=2\) or \(x=8\)
213AASL1-q00701.7 Rational exponents and laws of logarithmsPaper 1Hard10 marks
Consider the simultaneous equations
\(2^{x}\times4^{y}=64\) and \(\log_{2}x+\log_{2}y=2\), where \(x,y\gt 0\).
(a)
Show that \(x+2y=6\).
(2)
(b)
Show that \(xy=4\).
(2)
(c)
Hence find all pairs \((x,y)\) that satisfy both equations.
(4)
(d)
The second equation is replaced by \(\log_{2}x+\log_{2}y=3\). Explain why the new pair of equations has no solution.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(4^{y}=2^{2y}\)
M1
\(2^{x+2y}=2^{6}\)
A1
\(x+2y=6\)
AG
(b)
\(\log_{2}(xy)=2\)
M1
\(xy=2^{2}\)
A1
\(xy=4\)
AG
(c)
substituting, e.g. \((6-2y)y=4\)
M1
\(2y^{2}-6y+4=0\), i.e. \(y^{2}-3y+2=0\)
A1
Or \(x^{2}-6x+8=0\).
\(y=1\) or \(y=2\)
A1
\((4,\,1)\) and \((2,\,2)\)
A1
Both pairs required.
(d)
\(xy=8\) leads to \((6-2y)y=8\), i.e. \(y^{2}-3y+4=0\)
M1
discriminant \(=9-16=-7\lt 0\), so there are no real solutions
R1
Answers: (c) \((4,\,1)\) and \((2,\,2)\)
214AASL1-q00721.7 Rational exponents and laws of logarithmsPaper 1Hard16 marks
A geometric sequence has first term \(u_{1}=2\) and common ratio \(8\). A second sequence is defined by \(v_{n}=\log_{2}u_{n}\), for \(n\in\mathbb{Z}^{+}\).
(a)
Write down the values of \(u_{2}\) and \(u_{3}\).
(1)
(b)
Show that \(v_{n}=3n-2\).
(3)
(c)
Hence explain why the sequence \(v_{1},v_{2},v_{3},\ldots\) is arithmetic, and state its common difference.
(2)
(d)
Find \(\displaystyle\sum_{n=1}^{20}v_{n}\).
(2)
(e)
Hence find the value of \(\log_{2}\left(u_{1}u_{2}u_{3}\cdots u_{20}\right)\).
(2)
(f)
Prove that, for any geometric sequence with first term \(a\gt 0\) and common ratio \(r\gt 0\), the sequence \(\ln u_{1},\ln u_{2},\ln u_{3},\ldots\) is arithmetic.
(2)
(g)
A geometric sequence \(w_{1},w_{2},w_{3},\ldots\) has positive terms, with \(\log_{10}w_{1}=5\) and \(\log_{10}w_{6}=-5\). Find the sum to infinity of this sequence.
216AASL1-q00761.8 Infinite geometric seriesPaper 1Hard10 marks
An infinite geometric series has sum to infinity 9 and second term 2.
(a)
Show that the common ratio, \(r\), satisfies \(9r^{2}-9r+2=0\).
(3)
(b)
Find the two possible values of \(r\).
(2)
(c)
Find the first term corresponding to each value of \(r\).
(2)
(d)
Explain why both values of \(r\) give a series whose sum to infinity exists.
(1)
(e)
Find the positive difference between the sums of the first three terms of the two series.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{u_{1}}{1-r}=9\), so \(u_{1}=9(1-r)\)
M1
\(u_{1}r=2\), so \(9r(1-r)=2\)
M1
\(9r-9r^{2}=2\)
A1
\(9r^{2}-9r+2=0\)
AG
(b)
\((3r-1)(3r-2)=0\)
(M1)
\(r=\tfrac{1}{3}\) or \(r=\tfrac{2}{3}\)
A1
(c)
\(r=\tfrac{1}{3}\): \(u_{1}=6\)
A1
\(r=\tfrac{2}{3}\): \(u_{1}=3\)
A1
(d)
\(\left|\tfrac{1}{3}\right|\lt 1\) and \(\left|\tfrac{2}{3}\right|\lt 1\)
R1
(e)
\(6+2+\tfrac{2}{3}=\tfrac{26}{3}\) and \(3+2+\tfrac{4}{3}=\tfrac{19}{3}\)
(M1)
Award for either sum correct.
\(\tfrac{7}{3}\)
A1
Answers: (b) \(\tfrac{1}{3}\) or \(\tfrac{2}{3}\) · (c) 6 and 3 · (e) \(\tfrac{7}{3}\)
217AASL1-q00791.8 Infinite geometric seriesPaper 1Hard15 marks
Square \(S_{1}\) has sides of length 8 cm. Square \(S_{2}\) is formed by joining the midpoints of the sides of \(S_{1}\). Square \(S_{3}\) is formed by joining the midpoints of the sides of \(S_{2}\), and the process continues indefinitely, as shown in the diagram.
(a)
Show that the length of a side of \(S_{2}\) is \(4\sqrt{2}\) cm.
(2)
(b)
Show that the areas of \(S_{1},S_{2},S_{3},\ldots\) form a geometric sequence with common ratio \(\tfrac{1}{2}\).
(2)
(c)
Find the area of \(S_{6}\).
(2)
(d)
Find the sum to infinity of the areas of the squares.
(2)
(e)
The perimeters of \(S_{1},S_{2},S_{3},\ldots\) also form a geometric sequence. Find the sum to infinity of the perimeters, giving your answer in the form \(a+b\sqrt{2}\), where \(a,b\in\mathbb{Z}\).
(4)
(f)
Find the least value of \(n\) for which the perimeter of \(S_{n}\) is less than 1 cm.
(3)
Show mark scheme
Working
Mark
Notes
(a)
each half-side of \(S_{1}\) is 4 cm; using Pythagoras: \(\sqrt{4^{2}+4^{2}}\)
M1
\(\sqrt{32}=\sqrt{16\times2}\)
A1
\(=4\sqrt{2}\) cm
AG
(b)
each side is \(\dfrac{1}{\sqrt{2}}\) times the previous side (by the same argument as part (a))
M1
Or \(\dfrac{4\sqrt{2}}{8}=\dfrac{1}{\sqrt{2}}\) applied generally.
so each area is \(\left(\dfrac{1}{\sqrt{2}}\right)^{2}=\dfrac{1}{2}\) times the previous area: \(64,\,32,\,16,\ldots\)
A1
geometric with common ratio \(\tfrac{1}{2}\)
AG
(c)
\(64\left(\tfrac{1}{2}\right)^{5}\)
(M1)
\(2\) cm2
A1
(d)
\(\dfrac{64}{1-\frac{1}{2}}\)
(M1)
\(128\) cm2
A1
(e)
first term 32, common ratio \(\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
\(12-3r=5\) gives \(r=\tfrac{7}{3}\), which is not an integer
R1
Answers: (b) 240
219AASL1-q00821.9 The binomial theoremPaper 1Hard6 marks
In the expansion of \((1+kx)^{n}\), where \(n\in\mathbb{Z}^{+}\) and \(k\gt 0\), the coefficient of \(x\) is 12 and the coefficient of \(x^{2}\) is 60.
\(\left(1-\sqrt{2}\right)^{5}=-\left(\sqrt{2}-1\right)^{5}\), and \(0\lt\left(\sqrt{2}-1\right)^{5}\lt 1\), so \(\left(1+\sqrt{2}\right)^{5}=82+\left(\sqrt{2}-1\right)^{5}\)
R1
\(N=82\)
A1
Answers: (b) \(41-29\sqrt{2}\) · (c) \(N=82\)
221IBMA2S-044-04-121.9 The binomial theoremPaper 1Hard11 marks
No calculator
Consider the expansion of \(\left(2x-\frac{1}{x^2}\right)^n\), where \(n\) is a positive integer.
(a)
Show that the expansion can contain a term independent of \(x\) only if \(n\) is a multiple of 3.
(3)
(b)
For \(n=6\), find the term independent of \(x\).
(3)
(c)
For \(n=6\), find the coefficient of \(x^{-3}\).
(2)
(d)
Hence find the term independent of \(x\) in the expansion of \((1+x^3)\left(2x-\frac{1}{x^2}\right)^6\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 general term \(\binom{n}{r}(2x)^{n-r}(-1)^r x^{-2r}\) M1 power of \(x\) is \(n-3r\) R1 a constant term needs \(n-3r=0\), so \(n=3r\) with \(r\) an integer, hence \(n\) is a multiple of 3
M1 recognising two contributions: \(1\times\)(constant term) and \(x^3\times\)(term in \(x^{-3}\)) A1 \(240+(-160)\) A1 80
✓ 3
Common error: In a 'show that' part, finding \(r\) for one particular value of \(n\) is not enough. The argument must hold for every \(n\).
222IBMA2S-044-07-111.6 Deductive proofPaper 1Hard13 marks
No calculator
A software developer writes a quick check that rejects large numbers that cannot be perfect squares. The check is based on the following results. In this question, \(n\), \(k\), \(m\), \(p\) and \(q\) are integers.
(a)
Prove that \(n^2+n\) is even for every integer \(n\).
(3)
(b)
Hence prove that the square of any odd integer can be written in the form \(8m+1\), where \(m\) is an integer.
(4)
(c)
Prove that the difference between the squares of any two odd integers is a multiple of 8.
(3)
(d)
Hence, or otherwise, show that \(1\,000\,003\) is not the square of an integer.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 factorises to \(n(n+1)\)
✓ 1
R1 \(n\) and \(n+1\) are consecutive integers, so one of them is even
✓ 1
A1 hence the product \(n(n+1)\) is even
✓ 1
Part (b)
M1 writes the odd integer as \(2k+1\)
✓ 1
A1 \((2k+1)^2=4k^2+4k+1\)
✓ 1
M1 writes this as \(4k(k+1)+1\) and uses the result that \(k^2+k\) is even to write \(k(k+1)=2m\)
✓ 1
A1 \((2k+1)^2=8m+1\)
✓ 1
Part (c)
M1 uses the \(8m+1\) result to write the squares as \(8p+1\) and \(8q+1\)
✓ 1
A1 difference \(=8p+1-(8q+1)=8(p-q)\)
✓ 1
R1 \(p-q\) is an integer, so the difference is a multiple of 8
✓ 1
Part (d)
R1 \(1\,000\,003\) is odd, and the square of an even integer is even, so any integer square root would have to be odd
✓ 1
M1 EITHER compares with the odd square \(1^2\): \(1\,000\,003-1=1\,000\,002=8\times125\,000+2\), OR writes \(1\,000\,003=8\times125\,000+3\)
✓ 1
R1 EITHER \(1\,000\,002\) is not a multiple of 8, so by the difference-of-squares result \(1\,000\,003\) is not an odd square, OR the remainder on division by 8 is 3, not 1, so by the \(8m+1\) result it is not an odd square; hence it is not a perfect square
✓ 1
Exam tip: When a question says 'hence', quote the earlier result explicitly, for example 'since \(k^2+k\) is even, \(k(k+1)=2m\)'.
223IBMA2S-044-07-121.6 Deductive proofPaper 1Hard14 marks
No calculator
An engineer designing components for a space probe uses an inequality to find optimal dimensions and speeds. In this question, \(a\) and \(b\) are positive real numbers.
(a)
Prove that \((a+b)^2-4ab\equiv(a-b)^2\).
(2)
(b)
Hence prove that \(\frac{a+b}{2}\ge\sqrt{ab}\), and state when equality holds.
(4)
(c)
A rectangular solar panel on the probe has a perimeter of 12 m. Use part (b) to prove that its area cannot exceed 9 m2, and state the dimensions that give this maximum.
(3)
(d)
The fuel used by a manoeuvre, in tonnes, is modelled by \(F(x)=x+\frac{4}{x}\), where \(x>0\) is the speed in km s−1. Prove that \(F(x)\ge4\) for all \(x>0\).
(3)
(e)
Find the speed at which the fuel used is a minimum.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 expands \((a+b)^2-4ab=a^2+2ab+b^2-4ab\)
✓ 1
A1 \(=a^2-2ab+b^2=(a-b)^2\)
✓ 1
Part (b)
R1 \((a-b)^2\ge0\) for all real \(a\), \(b\)
✓ 1
A1 so by part (a), \((a+b)^2\ge4ab\)
✓ 1
R1 \(a+b>0\) and \(ab>0\), so taking positive square roots gives \(a+b\ge2\sqrt{ab}\), i.e. \(\frac{a+b}{2}\ge\sqrt{ab}\)
✓ 1
A1 equality holds if and only if \(a=b\)
✓ 1
Part (c)
M1 with sides \(a\) and \(b\), \(a+b=6\), and applies part (b): \(\sqrt{ab}\le3\)
✓ 1
A1 area \(ab\le9\) m2
✓ 1
A1 the maximum occurs when \(a=b=3\), i.e. a 3 m by 3 m square
✓ 1
Part (d)
M1 applies part (b) with \(a=x\) and \(b=\frac{4}{x}\), both positive
✓ 1
A1 \(\sqrt{x\times\frac{4}{x}}=\sqrt{4}=2\)
✓ 1
A1 \(\frac{x+\frac{4}{x}}{2}\ge2\), so \(F(x)\ge4\)
✓ 1
Part (e)
M1 uses the equality condition \(x=\frac{4}{x}\), so \(x^2=4\)
✓ 1
A1 \(x=2\) km s−1, since \(x>0\)
✓ 1
Common error: Taking square roots of both sides of an inequality is only valid when both sides are non-negative. State that \(a+b>0\) before doing so.
224IBMA2S-044-01-121.7 Rational exponents and laws of logarithmsPaper 1Hard15 marks
No calculator
This question uses laws of logarithms and exponents. Give all answers exactly.
(a)
Show that \(\log_9 x=\frac{1}{2}\log_3 x\) for \(x>0\).
Solve the simultaneous equations \(3^{x}\times9^{y}=3^{7}\) and \(\log_2 x+\log_2 y=\log_2 6\).
(5)
(d)
Solve \(x^{\log_3 x}=81\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses change of base: \(\log_9 x=\dfrac{\log_3 x}{\log_3 9}\)
✓ 1
A1 uses \(\log_3 9=2\) to obtain \(\frac{1}{2}\log_3 x\) (AG)
✓ 1
Part (b)
M1 writes every term in base 3, using \(\log_{27}x=\frac{1}{3}\log_3 x\)
✓ 1
A1 \(\frac{11}{6}\log_3 x=\frac{11}{2}\)
✓ 1
A1 \(\log_3 x=3\)
✓ 1
A1 \(x=27\)
✓ 1
Part (c)
M1 writes \(9^{y}=3^{2y}\) and equates powers
✓ 1
A1 \(x+2y=7\)
✓ 1
A1 uses the product law to obtain \(xy=6\)
✓ 1
M1 substitutes to form \(2y^{2}-7y+6=0\)
✓ 1
A1 \(x=4,\ y=\frac{3}{2}\) and \(x=3,\ y=2\) (both valid because \(x,y>0\))
✓ 1
Part (d)
M1 takes \(\log_3\) of both sides and uses the power law: \(\log_3 x\times\log_3 x=\log_3 81\)
✓ 1
A1 \((\log_3 x)^{2}=4\)
✓ 1
A1 \(x=9\)
✓ 1
A1 \(x=\frac{1}{9}\)
✓ 1
Common error: When you take a square root of \((\log_3 x)^2=4\), keep both signs. The negative root gives the valid solution \(x=\frac{1}{9}\), because \(x\) is still positive.
225IBMA2S-044-00-111.2 Arithmetic sequences and seriesPaper 1Hard10 marks
No calculator
An arithmetic sequence has first term \(a\) and common difference \(d\), where \(d\neq0\). The 1st, 4th and 6th terms of this arithmetic sequence are, in that order, the first three terms of a geometric sequence.
(a)
Show that \(a=-9d\).
(3)
(b)
Find the common ratio of the geometric sequence.
(2)
(c)
The sum to infinity of the geometric sequence is 54. Find the values of \(d\) and \(a\).
(3)
(d)
Find the value of \(n\) for which the sum of the first \(n\) terms of the arithmetic sequence is zero.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(\frac{a+3d}{a}=\frac{a+5d}{a+3d}\) or \((a+3d)^2=a(a+5d)\)
✓ 1
A1 \(a^2+6ad+9d^2=a^2+5ad\), so \(ad+9d^2=0\)
✓ 1
R1 \(d(a+9d)=0\) and \(d\neq0\), so \(a=-9d\)
✓ 1
Part (b)
M1 terms are \(-9d,\,-6d,\,-4d\), so \(r=\frac{-6d}{-9d}\)
✓ 1
A1 \(r=\frac{2}{3}\)
✓ 1
Part (c)
M1 \(\frac{-9d}{1-\frac{2}{3}}=54\)
✓ 1
A1 \(-27d=54\), so \(d=-2\)
✓ 1
A1 \(a=18\)
✓ 1
Part (d)
M1 \(\frac{n}{2}(36-2(n-1))=n(19-n)=0\)
✓ 1
A1 \(n=19\) (rejecting \(n=0\))
✓ 1
Common error: Do not divide by \(d\) without stating that \(d\neq0\). Factorising \(d(a+9d)=0\) and then using the given condition earns the reasoning mark.
226IBMA2S-044-02-111.2 Arithmetic sequences and seriesPaper 1Hard12 marks
No calculator
A geometric sequence has first term \(a\) and common ratio \(r\), where \(a\ne0\) and \(r\ne1\). The sum of the first six terms is 28 times the sum of the first three terms. Also, \(u_4-u_2=48\).
(a)
Show that \(\frac{S_6}{S_3}=1+r^3\).
(3)
(b)
Find the value of \(r\).
(1)
(c)
Find the value of \(a\).
(2)
(d)
Find the smallest value of \(n\) for which \(S_n>1000\).
(3)
(e)
Find the value of \(\displaystyle\sum_{k=1}^{4}u_{2k}\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(S_6=\frac{a(r^6-1)}{r-1}\) and \(S_3=\frac{a(r^3-1)}{r-1}\)
M1 recognises that \(u_2,u_4,\dots\) is geometric with first term 6 and ratio 9
✓ 1
M1 \(\frac{6(9^4-1)}{9-1}\)
✓ 1
A1 4920
✓ 1
Exam tip: Taking every second term of a geometric sequence gives another geometric sequence with ratio \(r^2\).
227IBMA2S-044-05-111.7 Rational exponents and laws of logarithmsPaper 1Hard10 marks
No calculator
Let \(x=\log_2 3\).
(a)
Write down the value of \(2^{x}\).
(1)
(b)
Express \(\log_2 24\) in terms of \(x\).
(2)
(c)
The equation \(2^{t}=3^{t-2}\) has exactly one solution. Show that \(t=\frac{2x}{x-1}\).
(3)
(d)
Show that \(1<x<2\).
(2)
(e)
Hence show that \(t>4\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(2^x=3\)
✓ 1
Part (b)
M1 \(\log_2 24=\log_2 8+\log_2 3\), with \(\log_2 8=3\)
✓ 1
A1 \(3+x\)
✓ 1
Part (c)
M1 takes \(\log_2\) of both sides to get \(t=(t-2)\log_2 3=(t-2)x\). Also accept taking ln or lg, giving \(t\ln 2=(t-2)\ln 3\), then using \(\frac{\ln 3}{\ln 2}=x\)
✓ 1
M1 rearranges to \(tx-t=2x\), i.e. \(t(x-1)=2x\)
✓ 1
A1 \(t=\frac{2x}{x-1}\) (AG)
✓ 1
Part (d)
R1 \(2^1=2<3<4=2^2\), i.e. \(2<2^x<2^2\)
✓ 1
R1 \(2^x\) is increasing (or \(\log_2\) is increasing), so \(1<x<2\)
✓ 1
Part (e)
M1 writes \(t=2+\frac{2}{x-1}\). Also accept setting up \(\frac{2x}{x-1}>4 \Leftrightarrow 2x>4(x-1)\), using \(x-1>0\)
✓ 1
R1 \(0<x-1<1\) so \(\frac{2}{x-1}>2\), hence \(t>4\). Also accept \(2x>4x-4 \Leftrightarrow x<2\), which is true, hence \(t>4\)
✓ 1
Exam tip: Rewriting \(\frac{2x}{x-1}\) as \(2+\frac{2}{x-1}\) leaves \(x\) in only one place, so the inequality is easy to justify. If you multiply out an inequality instead, state that \(x-1>0\).
228IBMA3S-1-16r10-111.8 Infinite geometric seriesPaper 1Hard10 marks
No calculator
A geometric series has second term 6 and sum to infinity 27. Let the first term be \(a\) and the common ratio be \(r\). This series models the successive echoes of a signal in a telecoms cable.
(a)
Show that \(9r^2 - 9r + 2 = 0\).
(3)
(b)
Hence find the two possible values of \(r\).
(3)
(c)
Find the two possible values of the first term.
(2)
(d)
Given that the first term is greater than 12, find the sum of the first three terms.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(ar = 6\) and \(\frac{a}{1-r} = 27\)
✓ 1
M1 eliminate \(a\): \(\frac{6}{r(1-r)} = 27\)
✓ 1
A1 \(27r - 27r^2 = 6\)
✓ 1
Part (b)
M1 attempt to factorise or use the quadratic formula
✓ 1
A1 \((3r - 1)(3r - 2) = 0\)
✓ 1
A1 \(r = \frac{1}{3}\) or \(r = \frac{2}{3}\)
✓ 1
Part (c)
A1 \(a = 18\) (when \(r = \frac{1}{3}\))
✓ 1
A1 \(a = 9\) (when \(r = \frac{2}{3}\))
✓ 1
Part (d)
M1 \(18 + 6 + 2\) or \(S_3\) formula with \(a = 18\), \(r = \frac{1}{3}\)
✓ 1
A1 26
✓ 1
Common error: Both values of \(r\) satisfy \(|r| < 1\), so neither can be rejected without further information; the extra condition is needed to choose.
229IBMA3S-1-16r10-131.2 Arithmetic sequences and seriesPaper 1Hard10 marks
No calculator
The heights of the masts in a mountain valley telecoms link form an arithmetic sequence with first term \(u_1\) and non-zero common difference \(d\). The heights of masts 1, 2 and 6 are, in that order, consecutive terms of a geometric sequence.
(a)
Show that \(d = 3u_1\).
(5)
(b)
Mast 1 is 20 m high. Find \(d\) and the common ratio of the geometric sequence.
(2)
(c)
Find the total of the heights of the first 8 masts.
By letting \(u = 2^{x}\), show that the equation can be written as \(u^2 - 5u + 4 = 0\).
(2)
(b)
Hence find the values of x that satisfy the equation.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(4^x = (2^2)^x = (2^x)^2\)
✓ 1
A1 \(u^2 - 5u + 4 = 0\)
✓ 1
Part (b)
M1 factorise: \((u-1)(u-4)=0\)
✓ 1
A1 \(u = 1\) or \(u = 4\)
✓ 1
M1 \(2^x = 1\) or \(2^x = 4\)
✓ 1
A1 \(x = 0\) or \(x = 2\)
✓ 1
Exam tip: If a question gives a substitution, write the new equation in terms of that variable, solve it, and then convert back to the original variable.
237IBMA3S-1-16r19-111.8 Infinite geometric seriesPaper 1Hard7 marks
No calculator
An infinite geometric series has second term 6 and sum to infinity 27. The common ratio \(r\) satisfies \(|r|<1\).
(a)
Show that \(9r^2-9r+2=0\).
(3)
(b)
Find the two possible values of \(r\).
(2)
(c)
Find the greater of the two possible values of the first term.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_1=\frac{6}{r}\)
✓ 1
M1 substituting into \(S_\infty=\frac{u_1}{1-r}\): \(\frac{6}{r(1-r)}=27\)
✓ 1
A1 \(27r(1-r)=6\), leading to \(9r^2-9r+2=0\) (AG)
✓ 1
Part (b)
M1 attempt to solve: \((3r-1)(3r-2)=0\)
✓ 1
A1 \(r=\frac13\) or \(r=\frac23\)
✓ 1
Part (c)
M1 use of \(u_1=\frac{6}{r}\)
✓ 1
A1 \(18\)
✓ 1
Exam tip: When the second term is given, express the first term as \(u_2\div r\) before using the sum to infinity.
238IBMA3S-1-17r20-141.7 Rational exponents and laws of logarithmsPaper 1Hard9 marks
No calculator
Let \(x\) and \(y\) be positive real numbers. Let \(p=\log_5 x\) and \(q=\log_5 y\).
(a)
Express \(\log_5\left(\dfrac{25x^2}{\sqrt{y}}\right)\) in terms of \(p\) and \(q\).
(3)
(b)
Given that \(\log_5(xy)=4\) and \(\log_5\left(\dfrac{x}{y^2}\right)=1\), find the values of \(x\) and \(y\).
(5)
(c)
Hence find the exact value of \(\log_5\left(\dfrac{25x^2}{\sqrt{y}}\right)\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the laws of logarithms for product, quotient and powers
✓ 1
A1 \(\log_525=2\) and \(\log_5x^2=2p\)
✓ 1
A1 \(2+2p-\dfrac{q}{2}\)
✓ 1
Part (b)
M1 \(p+q=4\)
✓ 1
M1 \(p-2q=1\)
✓ 1
A1 \(q=1\)
✓ 1
A1 \(p=3\)
✓ 1
A1 \(x=125\), \(y=5\)
✓ 1
Part (c)
A1 \(2+6-\dfrac12=7.5\) (ECF from (b))
✓ 1
Common error: \(\log_5(x+y)\) is not \(\log_5x+\log_5y\): only a product inside the logarithm splits into a sum.
239IBMA3S-1-17r22-121.8 Infinite geometric seriesPaper 1Hard10 marks
No calculator
An infinite geometric series has first term \(u_1\) and common ratio \(r\). The sum to infinity of the series is 54 and the sum of the first two terms is 30.
(a)
Show that \(r^2=\frac49\).
(4)
(b)
Find the two possible values of \(u_1\).
(2)
(c)
For the series with a negative common ratio, find the sum of the terms \(u_2+u_4+u_6+\dots\)
M1 the even-numbered terms form a geometric series with common ratio \(r^2=\frac49\)
✓ 1
M1 \(S_\infty=\frac{-60}{1-\frac49}\)
✓ 1
A1 \(-108\)
✓ 1
Exam tip: Terms in alternate positions of a geometric series form another geometric series whose common ratio is r².
240IBMA3S-1-17r22-141.2 Arithmetic sequences and seriesPaper 1Hard11 marks
No calculator
An arithmetic sequence has first term \(a\) and common difference \(d\), where \(d\ne0\). The 2nd, 5th and 11th terms of this sequence are, in that order, consecutive terms of a geometric sequence.
(a)
Show that \(a=2d\).
(5)
(b)
The sum of the first 12 terms of the arithmetic sequence is 360. Find \(a\) and \(d\).
(3)
(c)
Find the sum of the first 6 terms of the geometric sequence whose first three terms are the 2nd, 5th and 11th terms of the arithmetic sequence.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_2=a+d\), \(u_5=a+4d\), \(u_{11}=a+10d\)
✓ 1
M1 \((a+4d)^2=(a+d)(a+10d)\)
✓ 1
A1 \(a^2+8ad+16d^2=a^2+11ad+10d^2\)
✓ 1
A1 \(6d^2=3ad\)
✓ 1
R1 divide by \(3d\) since \(d\ne0\), giving \(a=2d\)
✓ 1
Part (b)
M1 \(S_{12}=6(2a+11d)\)
✓ 1
A1 \(90d=360\)
✓ 1
A1 \(d=4\) and \(a=8\)
✓ 1
Part (c)
M1 terms are 12, 24, 48 with \(r=2\)
✓ 1
M1 \(S_6=\frac{12(2^6-1)}{2-1}\)
✓ 1
A1 756
✓ 1
Exam tip: For three terms to be consecutive in a geometric sequence, the middle term squared equals the product of the outer two.
241IBMA3S-1-18r29-121.9 The binomial theoremPaper 1Hard7 marks
No calculator
In the expansion of (2 + kx)6, where k > 0, the coefficient of x2 is equal to the coefficient of x3.
(a)
Find the value of k.
(3)
(b)
Using this value of k, find the coefficient of x4 in the expansion of (1 + x)(2 + kx)6.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 coefficient of x2 is C(6,2) × 24 × k2 = 240k2 and of x3 is C(6,3) × 23 × k3 = 160k3
✓ 1
M1 set 240k2 = 160k3
✓ 1
A1 k = 3/2
✓ 1
Part (b)
M1 coefficient of x4 = (coefficient of x4) + (coefficient of x3) in (2 + kx)6
✓ 1
A1 coefficient of x4 is 15 × 4 × (3/2)4 = 1215/4
✓ 1
A1 coefficient of x3 is 160 × (3/2)3 = 540
✓ 1
A1 3375/4 (= 843.75)
✓ 1
Exam tip: When a second bracket such as (1 + x) multiplies the expansion, collect every pair of terms whose powers add to the one you need.
242IBMA3S-1-19r29-131.7 Rational exponents and laws of logarithmsPaper 1Hard7 marks
No calculator
Let \(N>0\) with \(N\ne1\).
(a)
Prove that \(\dfrac{1}{\log_2N}+\dfrac{1}{\log_3N}+\dfrac{1}{\log_5N}=\dfrac{1}{\log_{30}N}\).
(5)
(b)
Hence find the exact value of \(\dfrac{1}{\log_2900}+\dfrac{1}{\log_3900}+\dfrac{1}{\log_5900}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\frac{1}{\log_aN}=\log_Na\) (change of base)
✓ 1
A1 left side \(=\log_N2+\log_N3+\log_N5\)
✓ 1
M1 use of \(\log a+\log b=\log(ab)\)
✓ 1
A1 \(=\log_N30\)
✓ 1
R1 \(\log_N30=\frac{1}{\log_{30}N}\), so the right side is reached
✓ 1
Part (b)
M1 \(=\log_{900}30\) or \(\frac{1}{\log_{30}900}\)
✓ 1
A1 \(900=30^2\), so the value is \(\frac12\)
✓ 1
Exam tip: The identity \(\log_ab=\frac{1}{\log_ba}\) turns reciprocals of logs into logs with a common base, where sums become products.
243IBMA3S-1-18r31-061.6 Deductive proofPaper 1Hard6 marks
No calculator
Prove that the product of any two consecutive even integers is divisible by 8.
(a)
Write your proof in full, justifying each step.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 represents two consecutive even integers as \(2k\) and \(2k+2\) where \(k\) is an integer
✓ 1
A1 product \(=2k(2k+2)=4k(k+1)\)
✓ 1
R1 \(k\) and \(k+1\) are consecutive integers, so exactly one of them is even
✓ 1
M1 \(k(k+1)=2m\) for some integer \(m\)
✓ 1
A1 product \(=4\times 2m = 8m\)
✓ 1
R1 \(m\) is an integer, so the product is a multiple of 8 for every integer \(k\)
✓ 1
Common error: Showing that \(4k(k+1)\) is a multiple of 4 is not enough. The key step is to explain why \(k(k+1)\) is always even.
244IBMA3S-1-17r32-131.6 Deductive proofPaper 1Hard11 marks
No calculator
Three consecutive rail carriages carry the whole-number labels \(n-1\), \(n\) and \(n+1\), where \(n\ge2\). A maintenance code is the sum of the squares of the three labels.
(a)
Show that the code equals \(3n^{2}+2\).
(2)
(b)
Hence prove that the code is never divisible by 3.
(3)
(c)
The code for one set of three carriages is 245. Find the labels of the three carriages.
(2)
(d)
Four consecutive carriages are labelled \(m\), \(m+1\), \(m+2\) and \(m+3\), where \(m\) is a positive integer. Prove that the sum of the squares of these four labels is never divisible by 4.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \((n-1)^{2}+n^{2}+(n+1)^{2}\) with expansion attempted
✓ 1
A1 \(n^{2}-2n+1+n^{2}+n^{2}+2n+1=3n^{2}+2\) (AG)
✓ 1
Part (b)
M1 \(3n^{2}+2=3(n^{2})+2\)
✓ 1
R1 \(3n^{2}\) is a multiple of 3, so the remainder on division by 3 is 2
✓ 1
R1 this holds for every integer \(n\), so the code is never divisible by 3
✓ 1
Part (c)
M1 \(3n^{2}+2=245\) so \(n^{2}=81\)
✓ 1
A1 \(n=9\) (since \(n\ge2\)), labels 8, 9 and 10
✓ 1
Part (d)
M1 sum \(=m^{2}+(m+1)^{2}+(m+2)^{2}+(m+3)^{2}\)
✓ 1
A1 \(=4m^{2}+12m+14\)
✓ 1
M1 \(=4(m^{2}+3m+3)+2\)
✓ 1
R1 \(4(m^{2}+3m+3)\) is a multiple of 4, so the remainder is 2, so the sum is never divisible by 4
✓ 1
Exam tip: To prove something is never divisible by \(k\), write the expression as a multiple of \(k\) plus a fixed non-zero remainder.
245IBMA3S-1-18r38-121.7 Rational exponents and laws of logarithmsPaper 1Hard5 marks
No calculator
Consider the equation 32x + 1 − 10 × 3x + 3 = 0.
(a)
Let y = 3x. Show that the equation can be written as 3y2 − 10y + 3 = 0.
(2)
(b)
Hence find the positive solution of the original equation.
R1 \(n=-40\) rejected since \(n\) must be positive
✓ 1
Exam tip: A sum equation for an arithmetic series becomes a quadratic in n. Reject any root that is negative or not a whole number.
247IBMA3S-1-17r42-101.9 The binomial theoremPaper 1Hard7 marks
No calculator
Consider the expansion of \(\left(2x^{2}-\dfrac1x\right)^{6}\).
(a)
Show that the general term can be written \(\binom6k\,2^{6-k}(-1)^{k}x^{12-3k}\).
(2)
(b)
Find the constant term.
(2)
(c)
Find the coefficient of \(x^{3}\) in the expansion of \(\left(1-x^{3}\right)\left(2x^{2}-\dfrac1x\right)^{6}\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 general term \(\binom6k\left(2x^{2}\right)^{6-k}\left(-\dfrac1x\right)^{k}\)
✓ 1
A1 powers of \(x\): \(x^{2(6-k)}\times x^{-k}=x^{12-3k}\) (AG)
✓ 1
Part (b)
M1 \(12-3k=0\), so \(k=4\)
✓ 1
A1 \(\binom64\times2^{2}\times(-1)^{4}=60\)
✓ 1
Part (c)
M1 identifies two contributions: \(1\times\)(the \(x^{3}\) term) and \(-x^{3}\times\)(the constant term)
✓ 1
A1 the \(x^{3}\) term has \(k=3\): \(\binom63\times2^{3}\times(-1)^{3}=-160\)
✓ 1
A1 coefficient \(=-160-60=-220\) (ECF from (b))
✓ 1
Exam tip: When a bracket multiplies an expansion, list every pair of terms whose powers combine to the one you need.
248IBMA3S-1-17r43-111.9 The binomial theoremPaper 1Hard11 marks
No calculator
The expansion of \((3+bx)^n\), where \(n\in\mathbb{Z}^+\), begins \(243-810x+\ldots\)
(a)
Find the value of \(n\).
(2)
(b)
Find the value of \(b\).
(3)
(c)
Find the sum of all the coefficients in the expansion of \((3+bx)^n\).
(2)
(d)
Find the coefficient of \(x^2\) in the expansion of \((3+bx)^n\).
(2)
(e)
Hence find the coefficient of \(x^2\) in the expansion of \((1+x)(3+bx)^n\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(3^n=243\)
✓ 1
A1 \(n=5\)
✓ 1
Part (b)
M1 coefficient of \(x\) is \(n\times3^{n-1}\times b\)
✓ 1
A1 \(5\times81\times b=-810\) i.e. \(405b=-810\)
✓ 1
A1 \(b=-2\)
✓ 1
Part (c)
M1 substitute \(x=1\): \((3-2)^5\)
✓ 1
A1 1
✓ 1
Part (d)
M1 \(\binom{5}{2}\times3^3\times(-2)^2\)
✓ 1
A1 1080
✓ 1
Part (e)
M1 \(1080+1\times(-810)\) (ECF)
✓ 1
A1 270
✓ 1
Common error: Use the given first two terms to find \(n\) and \(b\) separately; the constant term fixes \(n\), and the \(x\) term then fixes \(b\).
249IBMA3S-1-17r43-131.7 Rational exponents and laws of logarithmsPaper 1Hard9 marks
No calculator
Answer both parts without a calculator.
(a)
Solve \(\log_2(x-1)+\log_2(x+2)=2\).
(5)
(b)
Solve \(9^{x}-4\times3^{x}-45=0\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\log_2\big((x-1)(x+2)\big)=2\)
✓ 1
M1 \((x-1)(x+2)=2^2=4\)
✓ 1
A1 \(x^2+x-6=0\)
✓ 1
A1 \(x=2\) or \(x=-3\)
✓ 1
R1 reject \(x=-3\) because \(\log_2(x-1)\) is undefined, so \(x=2\)
✓ 1
Part (b)
M1 write \(9^x=(3^x)^2\) and let \(u=3^x\)
✓ 1
A1 \((u-9)(u+5)=0\)
✓ 1
R1 \(3^x>0\), so \(u=-5\) is rejected
✓ 1
A1 \(x=2\)
✓ 1
Common error: After solving a logarithmic or exponential equation by factorising, always check each root against the domain: a logarithm needs a positive argument and \(3^x\) is always positive.
250IBMA3S-1-19r44-121.9 The binomial theoremPaper 1Hard7 marks
No calculator
An engineer expands two expressions, \(\left(x^2+\dfrac{2}{x}\right)^9\) and \((1+x)(2-3x)^5\), in a model of aerodynamic drag.
(a)
Find the term independent of \(x\) in the expansion of \(\left(x^2+\dfrac{2}{x}\right)^9\).
(4)
(b)
Find the coefficient of \(x^3\) in the expansion of \((1+x)(2-3x)^5\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 general term C(9,r)(x2)9−r(2/x)r = C(9,r)2rx18−3r
✓ 1
M1 set 18 − 3r = 0
✓ 1
A1 r = 6
✓ 1
A1 84 × 64 = 5376
✓ 1
Part (b)
M1 identify that the x3 term comes from 1 × (x3 term) and x × (x2 term) of (2 − 3x)5
✓ 1
A1 x3 coefficient −1080 and x2 coefficient 720
✓ 1
A1 −1080 + 720 = −360
✓ 1
Exam tip: For a product with a bracket such as \((1+x)\), list every pairing of terms that gives the required power before adding.
251IBMA3S-1-19r46-141.9 The binomial theoremPaper 1Hard7 marks
No calculator
The expansion of \((1+ax)^n\), where \(n\) is a positive integer and \(a>0\), begins \(1+12x+60x^2+\ldots\)
(a)
Write down two equations connecting \(n\) and \(a\).
(2)
(b)
Hence find the value of \(n\) and the value of \(a\).
(3)
(c)
Find the coefficient of \(x^3\) in the expansion.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(na=12\)
✓ 1
A1 \(\frac{n(n-1)}{2}a^2=60\)
✓ 1
Part (b)
M1 eliminate \(a\), e.g. \(n(n-1)a^2=120\) and \(n^2a^2=144\), giving \(\frac{n-1}{n}=\frac{5}{6}\)
✓ 1
A1 \(n=6\)
✓ 1
A1 \(a=2\)
✓ 1
Part (c)
M1 \(\binom{6}{3}\times2^3\)
✓ 1
A1 160
✓ 1
Exam tip: Match the coefficient of each power of \(x\) to the binomial term, then divide one equation by the square of the other to eliminate the unknown you do not need.
252IBMA3S-1-17r47-121.7 Rational exponents and laws of logarithmsPaper 1Hard10 marks
No calculator
The positive real numbers \(x\) and \(y\) satisfy the simultaneous equations \(\log_2 x + \log_2 y = 7\) \(\log_2 x - \log_4 y = 1\).
(a)
Show that \(\log_4 y = \frac{1}{2}\log_2 y\).
(2)
(b)
Find the value of \(x\) and the value of \(y\).
(5)
(c)
Hence find the exact value of \(\log_8(xy)\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 let \(\log_4 y = k\), so \(y = 4^k = 2^{2k}\) (or use the change of base formula)
✓ 1
A1 \(\log_2 y = 2k\), so \(\log_4 y = \frac{1}{2}\log_2 y\) (AG)
✓ 1
Part (b)
M1 let \(p = \log_2 x\), \(q = \log_2 y\) and form \(p + q = 7\), \(p - \frac{q}{2} = 1\)
✓ 1
M1 eliminate one variable, e.g. subtract to get \(\frac{3q}{2} = 6\)
✓ 1
A1 \(q = 4\), \(p = 3\)
✓ 1
A1 \(x = 2^3 = 8\)
✓ 1
A1 \(y = 2^4 = 16\)
✓ 1
Part (c)
M1 \(xy = 128 = 2^7\) (ECF from (b))
✓ 1
M1 \(\log_8 2^7 = \frac{7}{\log_2 8}\) or change of base to base 2
✓ 1
A1 \(\frac{7}{3}\)
✓ 1
Exam tip: Write every logarithm to a common base first; the equations then become ordinary linear equations in \(\log_2 x\) and \(\log_2 y\).
253IBMA3S-1-17r47-131.8 Infinite geometric seriesPaper 1Hard8 marks
No calculator
The first three terms of a geometric sequence are \(27\), \(3k\) and \(5k - 18\), where \(k\) is a constant.
(a)
Find the two possible values of \(k\).
(4)
(b)
Show that the sum to infinity of the sequence exists for only one of these values of \(k\).
(2)
(c)
Hence find the sum to infinity of the sequence for the valid value of \(k\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\frac{u_2}{u_1} = \frac{u_3}{u_2}\), i.e. \(\frac{3k}{27} = \frac{5k - 18}{3k}\)
R1 the sum to infinity requires \(|r| < 1\), so only \(k = 6\) is valid
✓ 1
Part (c)
M1 \(S_\infty = \frac{27}{1 - \frac{2}{3}}\)
✓ 1
A1 81
✓ 1
Exam tip: If a question asks for a sum to infinity after you have found more than one possible ratio, check \(|r| < 1\) for each before using the formula.
254AASL1-q00061.1 Standard formPaper 2Easy5 marks
One light year is \(9.46\times10^{15}\) m. The distance from the Sun to a particular star is \(4.0\times10^{16}\) m.
(a)
Find this distance in light years.
(2)
(b)
A space probe travels at a constant speed of \(1.7\times10^{4}\) m s−1. Taking one year to be \(3.15\times10^{7}\) s, find the time, in years, that the probe would take to travel this distance. Give your answer in the form \(a\times10^{k}\), where \(1\le a\lt 10\) and \(k\in\mathbb{Z}\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{4.0\times10^{16}}{9.46\times10^{15}}\)
(M1)
\(4.23\) light years
A1
\(4.22832\ldots\)
(b)
time \(=\dfrac{4.0\times10^{16}}{1.7\times10^{4}}\;(=2.35\times10^{12}\text{ s})\)
(M1)
dividing their time in seconds by \(3.15\times10^{7}\)
(M1)
\(7.47\times10^{4}\) years
A1
\(74\,696.5\ldots\) Accept \(7.46\times10^{4}\) from premature rounding. Award A0 for 74 700 or 7.47E4 (not in the required form).
Answers: (a) 4.23 light years · (b) \(7.47\times10^{4}\) years
255AASL1-q00171.2 Arithmetic sequences and seriesPaper 2Easy5 marks
A finite arithmetic sequence is \(12.5,\ 15.2,\ 17.9,\ \ldots,\ 282.5\).
(a)
Write down the common difference.
(1)
(b)
Find the number of terms in the sequence.
(2)
(c)
Find the sum of all the terms of the sequence.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(d=2.7\)
A1
(b)
\(12.5+(n-1)(2.7)=282.5\)
(M1)
\(n=101\)
A1
(c)
\(\dfrac{101}{2}(12.5+282.5)\) or \(\dfrac{101}{2}\big(2(12.5)+100(2.7)\big)\)
(M1)
FT from (b).
\(14\,897.5\)
A1
Accept 14 900.
Answers: (a) 2.7 · (b) 101 · (c) 14 897.5
256AASL1-q00281.3 Geometric sequences and seriesPaper 2Easy5 marks
After a clean-up programme, the concentration of a pollutant in a lake decreases by 30% each year. At the start of the programme the concentration is 2.4 mg per litre.
(a)
Find the concentration after 5 years.
(2)
(b)
Find the least number of complete years needed for the concentration to fall below 0.05 mg per litre.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(2.4\times0.7^{5}\)
(M1)
Award M0 for \(0.3^{5}\).
\(0.403\) mg per litre
A1
\(0.403368\)
(b)
\(2.4\times0.7^{n}\lt 0.05\)
(M1)
\(n\gt 10.85\ldots\), or after 10 years \(0.0678\); after 11 years \(0.0475\)
(A1)
11 years
A1
Answers: (a) 0.403 mg per litre · (b) 11 years
257AASL1-q00361.4 Financial applicationsPaper 2Easy5 marks
A car is bought for £28 500. Its value depreciates by 14% each year.
(a)
Find the value of the car after 5 years.
(2)
(b)
Find the number of complete years after which the value of the car first falls below £10 000.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(28\,500\times0.86^{5}\)
(M1)
£13 407.17
A1
Accept £13 400.
(b)
\(28\,500\times0.86^{n}\lt 10\,000\)
(M1)
\(n\gt 6.94\), or values £11 530.17 after 6 years and £9915.94 after 7 years
(A1)
7 years
A1
Answers: (a) £13 407.17 · (b) 7 years
258AASL1-q00381.4 Financial applicationsPaper 2Easy5 marks
Kenji invests ¥800 000 in an account that pays a nominal annual interest rate of 3.6%, compounded half-yearly.
(a)
Find the value of the investment after 7 years. Give your answer to the nearest yen.
(2)
(b)
Find the interest earned in these 7 years.
(1)
(c)
Find the least number of years, given as a multiple of 0.5, after which the value of the investment is at least ¥1 000 000.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(800\,000\left(1+\dfrac{3.6}{200}\right)^{14}\), or TVM with \(N=14,\ I\%=3.6,\ PV=-800\,000,\ P/Y=2,\ C/Y=2\)
259AASL1-q00501.5 Exponents and introduction to logarithmsPaper 2Easy4 marks
The sound intensity level, \(L\) decibels (dB), of a sound of intensity \(I\) W m−2 is given by \(L=10\log_{10}\left(\dfrac{I}{I_{0}}\right)\), where \(I_{0}=10^{-12}\) W m−2.
(a)
Find \(L\) for a sound of intensity \(3.2\times10^{-5}\) W m−2.
(2)
(b)
Find the intensity of a sound with a level of 85 dB.
\(\log_{10}\left(\dfrac{I}{10^{-12}}\right)=8.5\), so \(I=10^{-12}\times10^{8.5}\)
(M1)
\(3.16\times10^{-4}\) W m−2
A1
Accept 0.000316.
Answers: (a) 75.1 dB · (b) \(3.16\times10^{-4}\) W m−2
260IBMA2S-044-04-031.9 The binomial theoremPaper 2Easy6 marks
Calculator
A smartphone battery keeps 99.8% of its capacity after each full charge cycle. After 40 cycles, the fraction of the original capacity remaining is \((0.998)^{40}\).
(a)
Write down the first three terms, in ascending powers of \(x\), of the expansion of \((1-x)^{40}\).
(3)
(b)
Use your answer to the previous part to estimate the percentage of the original capacity remaining after 40 cycles.
(2)
(c)
Explain why using only three terms gives a good estimate in this case.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 first term 1 A1 \(-40x\) A1 \(+780x^2\)
✓ 3
Part (b)
M1 substituting \(x=0.002\): \(1-40(0.002)+780(0.002)^2=1-0.08+0.00312\) A1 \(0.92312\), so 92.3%
✓ 2
Part (c)
R1 \(x=0.002\) is very small, so the terms in \(x^3\) and higher are negligible. The next term is \(-9880(0.002)^3\approx-8\times10^{-5}\), which does not change the answer to 3 s.f.
✓ 1
Exam tip: When the question says 'use your answer', you must show the substitution into the expansion. Typing \(0.998^{40}\) into a calculator gives \(0.92304\), which is useful as a check, but on its own it does not earn the method mark.
261IBMA2S-044-07-041.1 Standard formPaper 2Easy4 marks
Calculator
A laboratory culture starts with \(P_0\) cells, where \(P_0\) is a positive integer. The number of cells triples every hour, so after \(n\) hours there are \(P_n=P_0\times3^n\) cells, where \(n\) is a non-negative integer.
(a)
Prove that \(P_{n+2}-P_n\) is always a multiple of 8.
(3)
(b)
Given that \(P_0=120\), find the increase in the number of cells between \(n=4\) and \(n=6\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(P_0\times3^{n+2}-P_0\times3^n\) and factorises out \(P_0\times3^n\)
✓ 1
A1 obtains \(P_0\times3^n(9-1)=8P_0\times3^n\)
✓ 1
R1 states that \(P_0\times3^n\) is an integer (since \(P_0\) is a positive integer and \(n\ge0\)), so \(P_{n+2}-P_n\) is a multiple of 8
✓ 1
Part (b)
A1 77760 cells, from \(8\times120\times3^4\) or from \(120\times3^6-120\times3^4=87480-9720\)
✓ 1
Exam tip: To prove divisibility, show the expression equals the divisor multiplied by something, and state that this something is an integer.
262IBMA2S-044-06-021.1 Standard formPaper 2Easy5 marks
Calculator
Radio signals travel at 2.998×108 m s−1. At one close approach, the distance from Earth to Mars is 5.46×1010 m. A rover on Mars receives a command sent from Earth and immediately sends a reply.
(a)
Find the time, in seconds, for the command to travel from Earth to Mars. Give your answer in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ, with a correct to 3 significant figures.
(3)
(b)
Find the total time, in minutes, between sending the command and receiving the reply.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 time = distance ÷ speed: (5.46×1010) ÷ (2.998×108) A1 182.12… s A1 1.82×102 s
✓ 3
Part (b)
M1 doubling the one-way time and dividing by 60 A1 6.07 minutes
✓ 2
Exam tip: Keep the unrounded one-way time (182.12… s) in your calculator for part (b). Doubling a rounded value can change the third significant figure.
263IBMA2S-044-06-041.1 Standard formPaper 2Easy5 marks
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A marathon race is 42 195 m long. In one city marathon, 38 000 runners finished the race. The mean distance from the Earth to the Moon is 3.84×108 m.
(a)
Write 42 195 in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ.
(1)
(b)
Find the total distance run by all the finishers. Give your answer in metres in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ, with a correct to 3 significant figures.
(2)
(c)
Find the ratio of the total distance in part (b) to the mean Earth–Moon distance. Give your answer correct to 3 significant figures.
M1 dividing their total from (b) by 3.84×108 A1 4.18 (from the unrounded total 1.60341×109); accept 4.17 (from 1.60×109)
✓ 2
Exam tip: Part (a) asks for an exact conversion, so give all the significant figures (4.2195). Keep the full unrounded total in your calculator for part (c) and round only at the end.
264IBMA2S-044-01-041.3 Geometric sequences and seriesPaper 2Easy7 marks
Calculator
After an injection, the concentration of a drug in a patient's blood is modelled by \(C(t)=12\times 2^{-\frac{t}{5.5}}\), where \(C\) is in mg L−1 and \(t\) is the time in hours after the injection.
(a)
Find the concentration 8 hours after the injection.
(2)
(b)
Find the time taken for the concentration to fall to 1.5 mg L−1.
(3)
(c)
A second patient receives a larger dose, so that the initial concentration is 24 mg L−1. The same half-life applies. Find how much longer it takes this patient's concentration to fall to 1.5 mg L−1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \(t=8\): \(12\times 2^{-8/5.5}\)
✓ 1
A1 4.38 mg L−1
✓ 1
Part (b)
M1 sets up \(12\times 2^{-t/5.5}=1.5\)
✓ 1
M1 obtains \(2^{-t/5.5}=\frac{1}{8}\) and uses logs or powers of 2 to give \(\frac{t}{5.5}=3\)
✓ 1
A1 \(t=16.5\) hours
✓ 1
Part (c)
M1 solves \(24\times 2^{-t/5.5}=1.5\), giving \(2^{-t/5.5}=\frac{1}{16}\) and \(t=22\), or argues that doubling the dose adds one half-life
✓ 1
A1 5.5 hours longer
✓ 1
Exam tip: If the ratio is a power of the base, as with \(\frac{1}{8}=2^{-3}\), spot it. You then get an exact answer with no rounding.
265IBMA2S-044-01-071.7 Rational exponents and laws of logarithmsPaper 2Easy7 marks
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For planets orbiting the Sun, the orbital period \(T\) (in Earth years) and the mean orbital radius \(a\) (in astronomical units, AU) are related by \(T=a^{\frac{3}{2}}\).
(a)
Mars has a mean orbital radius of 1.524 AU. Find its orbital period.
(2)
(b)
An asteroid has an orbital period of 4.6 years. Find its mean orbital radius.
(3)
(c)
Show that \(\ln T=\frac{3}{2}\ln a\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes: \(1.524^{1.5}\)
✓ 1
A1 1.88 years
✓ 1
Part (b)
M1 sets up \(a^{\frac{3}{2}}=4.6\)
✓ 1
M1 raises both sides to the power \(\frac{2}{3}\)
✓ 1
A1 \(a=2.77\) AU
✓ 1
Part (c)
M1 takes natural logs of both sides: \(\ln T=\ln\left(a^{\frac{3}{2}}\right)\)
✓ 1
A1 uses the power law to obtain \(\ln T=\frac{3}{2}\ln a\) (AG)
✓ 1
Common error: To undo a power of \(\frac{3}{2}\), raise both sides to the power \(\frac{2}{3}\). Dividing by \(\frac{3}{2}\) does not undo a power.
266IBMA2S-044-00-051.2 Arithmetic sequences and seriesPaper 2Easy6 marks
Calculator
A language-learning app rewards a daily streak with coins. On day 1 a user earns 5 coins, and on each day after that she earns 3 more coins than on the previous day.
(a)
Show that the total number of coins earned in the first \(n\) days is \(\frac{n}{2}(3n+7)\).
(2)
(b)
Find the least number of days needed for her total to reach at least 2000 coins.
(3)
(c)
Find the number of coins she earns on that final day.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes into \(S_n=\frac{n}{2}(2u_1+(n-1)d)\): \(\frac{n}{2}(10+3(n-1))\)
✓ 1
A1 simplifies to \(\frac{n}{2}(3n+7)\)
✓ 1
Part (b)
M1 \(\frac{n}{2}(3n+7)\ge2000\), or \(3n^2+7n-4000\ge0\)
✓ 1
A1 \(n\ge35.4\) (35.367…)
✓ 1
A1 36 days
✓ 1
Part (c)
A1 \(5+35\times3=110\) coins
✓ 1
Exam tip: When \(n\) must be a whole number, round up so that the condition is met. A quick check helps: \(S_{35}=1960\) and \(S_{36}=2070\).
267IBMA2S-044-02-021.2 Arithmetic sequences and seriesPaper 2Easy7 marks
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A cyclist rides 40 km in week 1 of a training plan. Each week after that, the distance ridden is 8% more than in the week before.
(a)
Find the distance ridden in week 10.
(2)
(b)
Find the first week in which the distance ridden is more than 100 km.
(3)
(c)
Find the total distance ridden in the first 12 weeks.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(40\times1.08^{9}\)
✓ 1
A1 80.0 km (79.96…)
✓ 1
Part (b)
M1 sets up \(40\times1.08^{n-1}>100\)
✓ 1
M1 solves using logs or GDC, giving \(n-1>11.9\)
✓ 1
A1 week 13
✓ 1
Part (c)
M1 \(S_{12}=\frac{40(1.08^{12}-1)}{1.08-1}\)
✓ 1
A1 759 km
✓ 1
Common error: Week \(n\) is \(40\times1.08^{n-1}\), not \(40\times1.08^{n}\). Using power \(n\) shifts every answer by one week.
268IBMA2S-044-02-041.3 Geometric sequences and seriesPaper 2Easy5 marks
Calculator
The concentration of a drug in a patient's blood is measured every hour, at \(t=0,1,2,\ldots\) hours. Each reading is 15% lower than the reading one hour earlier. The reading at \(t=6\) is 4.2 mg L−1.
(a)
Write down the common ratio of the sequence of readings.
(1)
(b)
Find the concentration at \(t=0\).
(2)
(c)
Find the concentration at \(t=10\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(r=0.85\)
✓ 1
Part (b)
M1 \(C_0\times0.85^{6}=4.2\)
✓ 1
A1 11.1 mg L−1 (11.136…)
✓ 1
Part (c)
M1 \(4.2\times0.85^{4}\) or \(11.136\times0.85^{10}\)
✓ 1
A1 2.19 mg L−1
✓ 1
Common error: A 15% decrease means multiplying by 0.85, not by 0.15.
269IBMA2S-044-05-031.1 Standard formPaper 2Easy6 marks
Calculator
In a laboratory culture, the number of bacteria \(t\) hours after a sample is prepared is modelled by \(N=500e^{0.35t}\).
(a)
Write down the number of bacteria when the sample is prepared.
(1)
(b)
Find the number of bacteria after 6 hours.
(2)
(c)
The sample must be discarded once it contains 20 000 bacteria. Find the time at which this happens.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 500
✓ 1
Part (b)
M1 substitutes \(t=6\): \(500e^{2.1}\)
✓ 1
A1 4083 (accept 4080 or 4083.08)
✓ 1
Part (c)
M1 sets \(500e^{0.35t}=20000\)
✓ 1
M1 \(e^{0.35t}=40\), so \(0.35t=\ln 40\)
✓ 1
A1 \(t=10.54\ldots\), so \(t=10.5\) hours
✓ 1
Exam tip: Divide by the coefficient before taking logarithms. The logarithm of a product is not the product of the logarithms.
270IBMA2S-044-03-011.4 Financial applicationsPaper 2Easy6 marks
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A hospital buys a new MRI scanner for $1 850 000. Its value depreciates by 12% of its value at the start of each year.
(a)
Find the value of the scanner at the end of 6 years.
(3)
(b)
Find the number of complete years after which the value of the scanner first falls below $500 000.
(2)
(c)
Find the value of the scanner at the end of the year found in part (b).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognises the annual multiplier 0.88 M1 substitutes into 1 850 000 × 0.886 A1 $859 000 (859 147.5…)
✓ 3
Part (b)
M1 attempts to solve 1 850 000 × 0.88n < 500 000, e.g. by logarithms or a table, giving n > 10.2 A1 11 years
✓ 2
Part (c)
A1 1 850 000 × 0.8811 = $453 000 (453 399.5…)
✓ 1
Common error: The rate is 12%, so the multiplier is 0.88, not 0.12 or 1.12. In part (b), n > 10.2 means the value is still above $500 000 after 10 years, so the answer is 11.
271IBMA2S-044-03-031.4 Financial applicationsPaper 2Easy5 marks
Calculator
A student is saving to buy an e-bike. She deposits €2500 into a savings account that pays a nominal annual interest rate of 3.6%, compounded monthly. She makes no further deposits or withdrawals.
(a)
The e-bike costs €2950. Find the minimum number of complete months needed for the amount in the account to exceed €2950.
(3)
(b)
Find the amount in the account after 3 years.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses a monthly rate of 0.3%, i.e. a multiplier of 1.003 M1 sets up 2500 × 1.003n > 2950, or uses a TVM solver with correct inputs, giving n > 55.3 A1 56 months
✓ 3
Part (b)
M1 2500 × 1.00336 A1 €2784.67
✓ 2
Common error: With monthly compounding, divide the annual rate by 12 and count the time in months. Using 1.036³ gives the wrong answer.
272IBMA2S-044-03-041.4 Financial applicationsPaper 2Easy6 marks
Calculator
A photography drone was bought for $1450. Four years later it was worth $620. Its value decreased by r% of its value at the start of each year.
(a)
Find the value of r.
(4)
(b)
Assuming the same rate continues, find the value of the drone 2 years after it was worth $620.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets up 1450(1 − r/100)4 = 620 A1 (1 − r/100)4 = 0.42758… M1 takes the fourth root: 1 − r/100 = 0.80864… A1 r = 19.1
✓ 4
Part (b)
M1 620 × 0.80864…2, or 620 × (1 − r/100)2 using their r from (a) A1 $405 . Accept $406 as a follow-through answer from r = 19.1.
✓ 2
Exam tip: Keep the unrounded multiplier 0.80864… in your calculator for part (b). Using 0.809, or 1 − 0.191, gives $406 rather than $405.
273IBMA3S-1-16r10-021.2 Arithmetic sequences and seriesPaper 2Easy6 marks
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A mountaineer climbs 300 m in the first hour of an ascent. In each later hour she climbs 20 m less than in the previous hour, so the distances climbed each hour form an arithmetic sequence.
(a)
Find the distance she climbs in the 5th hour.
(2)
(b)
Find the total distance she climbs in the first 8 hours.
(2)
(c)
Find the number of the first hour in which she climbs less than 100 m.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_5 = 300 + 4(-20)\)
✓ 1
A1 220 m
✓ 1
Part (b)
M1 \(S_8 = \frac{8}{2}(2(300) + 7(-20))\)
✓ 1
A1 1840 m
✓ 1
Part (c)
M1 \(300 + (n-1)(-20) < 100\)
✓ 1
A1 \(n = 12\)
✓ 1
Exam tip: For 'first term below a value' questions, form the inequality and check the terms either side of the boundary, since equality (100 m in hour 11) does not satisfy 'less than'.
274IBMA3S-1-16r11-041.4 Financial applicationsPaper 2Easy5 marks
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A library receives a donation of €15 000, which it invests in an account paying a nominal annual interest rate of 3.6%, compounded monthly. Give all money answers to the nearest cent.
(a)
Find the value of the investment after 4 years.
(3)
(b)
Find the interest earned over the 4 years.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(15000\left(1+\frac{0.036}{12}\right)^{n}\)
✓ 1
M1 \(n=48\)
✓ 1
A1 €17 319.53
✓ 1
Part (b)
M1 subtract the original amount: \(17\,319.53-15\,000\)
✓ 1
A1 €2319.53
✓ 1
Exam tip: Monthly compounding means the rate is divided by 12 and the number of periods is multiplied by 12.
275IBMA3S-1-17r13-021.2 Arithmetic sequences and seriesPaper 2Easy5 marks
Calculator
In a Formula 1 race the lap times of a car form an arithmetic sequence because the car gets lighter as fuel is burnt. The first lap takes 95.40 s and each lap is 0.08 s faster than the one before.
M1 takes logarithms: \(t=\frac{\ln 1.75}{\ln 1.04}\) (or uses a solver)
✓ 1
A1 \(t=14.3\)
✓ 1
Exam tip: To solve an equation with the unknown in the exponent, isolate the power first, then take logarithms of both sides.
282IBMA3S-1-17r20-021.8 Infinite geometric seriesPaper 2Easy6 marks
Calculator
A cyclist stops pedalling at the top of a hill and coasts to a halt. In the first minute she travels 100 m. In each later minute she travels 80% of the distance she travelled in the previous minute.
(a)
Find the distance she travels in the 4th minute.
(2)
(b)
Find the total distance she travels in the first 8 minutes, giving your answer to 3 significant figures.
(2)
(c)
Find the total distance she travels before she stops.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(u_n=u_1r^{n-1}\) with \(r=0.8\), for example \(100\times0.8^3\)
✓ 1
A1 51.2 m
✓ 1
Part (b)
M1 use of \(S_n=\dfrac{u_1(1-r^n)}{1-r}\), for example \(\dfrac{100(1-0.8^8)}{0.2}\)
✓ 1
A1 416 m
✓ 1
Part (c)
M1 use of \(S_\infty=\dfrac{u_1}{1-r}\) with \(|r|<1\), for example \(\dfrac{100}{1-0.8}\)
✓ 1
A1 500 m
✓ 1
Exam tip: Check \(|r|<1\) before using the sum to infinity formula, and say so in your working.
283IBMA3S-1-17r22-011.3 Geometric sequences and seriesPaper 2Easy6 marks
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A patient is prescribed a course of a drug. The dose on day 1 is 40 mg and the dose increases by 12 mg each day, so the daily doses form an arithmetic sequence.
(a)
Find the dose on day 9.
(2)
(b)
The course continues until the next dose would exceed 300 mg. Find the number of days on which a dose is taken.
(2)
(c)
Find the total mass of drug taken over the whole course.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(u_n=u_1+(n-1)d\) with \(u_1=40\), \(d=12\)
✓ 1
A1 136 mg
✓ 1
Part (b)
M1 \(40+(n-1)\times12\le300\) (or equation) leading to \(n\le22.67\)
✓ 1
A1 22 days
✓ 1
Part (c)
M1 \(S_{22}=\frac{22}{2}(2\times40+21\times12)\) (ECF from (b) for \(n\))
✓ 1
A1 3652 mg
✓ 1
Common error: Remember that n must be a whole number of days: round the solution of the inequality down, not to the nearest integer.
284IBMA3S-1-17r22-031.3 Geometric sequences and seriesPaper 2Easy4 marks
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A village well delivers 600 litres of water on day 1. On each later day it delivers 0.8 times the volume delivered on the previous day.
(a)
Find the volume delivered on day 4.
(2)
(b)
Find the total volume the well could deliver if it continued indefinitely.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(600\times0.8^3\)
✓ 1
A1 307.2 litres
✓ 1
Part (b)
M1 \(S_\infty=\frac{600}{1-0.8}\) (valid as \(|r|<1\))
✓ 1
A1 3000 litres
✓ 1
Common error: Use S∞ = u₁/(1 − r) only when |r| < 1; here r = 0.8, so the series converges.
285IBMA3S-1-16r25-021.1 Standard formPaper 2Easy5 marks
Calculator
A space probe is \(2.4 \times 10^{12}\) m from Earth. Radio signals travel at \(3.0 \times 10^{8}\) m s−1.
(a)
Find the time, in seconds, for a signal from the probe to reach Earth. Give your answer in the form \(a \times 10^k\) where \(1 \leq a < 10\) and \(k \in \mathbb{Z}\).
(2)
(b)
Write your answer to part (a) in hours, correct to 3 significant figures.
(1)
(c)
A second probe is \(7.5 \times 10^{11}\) m from Earth. Find how many times longer the signal from the first probe takes to arrive than the signal from the second probe.
M1 ratio of distances \(\frac{2.4 \times 10^{12}}{7.5 \times 10^{11}}\)
✓ 1
A1 3.2
✓ 1
Common error: Check that the number in front of the power of ten lies from 1 up to (but not including) 10 before giving a standard form answer.
286IBMA3S-1-18r29-021.4 Financial applicationsPaper 2Easy5 marks
Calculator
A swimming club invests €8000 in an account that pays a nominal annual interest rate of 3.2%, compounded quarterly. No money is added or withdrawn.
(a)
Find the value of the investment after 6 years, to the nearest euro.
(3)
(b)
Find the interest earned over the 6 years. Use the unrounded value of the investment from (a), and give your answer to the nearest euro.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of compound interest formula 8000(1 + 3.2/(100×4))4×6
✓ 1
A1 correct substitution of 1.008 and 24
✓ 1
A1 €9686
✓ 1
Part (b)
M1 interest = final value − 8000
✓ 1
A1 €1686
✓ 1
Exam tip: With quarterly compounding the rate per period is r/4 and the number of periods is 4 × years.
287IBMA3S-1-19r29-011.8 Infinite geometric seriesPaper 2Easy7 marks
Calculator
A robotic arm sorts parcels. On its first sweep it moves 48 cm, and each later sweep is 75% as long as the one before, so the sweep lengths form a geometric sequence.
(a)
Find the total distance the arm would move if it made an unlimited number of sweeps.
(2)
(b)
Find the least number of sweeps for which the total distance moved exceeds 190 cm.
(3)
(c)
Find the distance, in cm to 3 significant figures, that remains to be moved after 16 sweeps to reach the total in part (a). Use unrounded values.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(S_\infty=\frac{u_1}{1-r}\) with \(u_1=48\), \(r=0.75\)
✓ 1
A1 \(S_\infty=192\) cm
✓ 1
Part (b)
M1 \(192(1-0.75^n)>190\)
✓ 1
M1 \(0.75^n<\frac{2}{192}\) and use of logarithms, \(n>\frac{\ln(2/192)}{\ln 0.75}=15.9\)
✓ 1
A1 \(n=16\)
✓ 1
Part (c)
M1 \(192-S_{16}\) or \(192\times0.75^{16}\)
✓ 1
A1 1.92 cm
✓ 1
Common error: When solving \(0.75^n<k\) the inequality reverses on dividing by \(\ln 0.75\), because \(\ln 0.75\) is negative. Check by substituting n and n − 1.
288IBMA3S-1-19r29-031.3 Geometric sequences and seriesPaper 2Easy6 marks
Calculator
The stock of cod in a fishery is modelled by \(N=8500\times0.92^{t}\), where \(N\) is the mass in tonnes and \(t\) is the time in years since monitoring began.
(a)
Find the stock after 5 years, to the nearest tonne.
(2)
(b)
State what the value 0.92 tells you about the yearly change in stock.
(1)
(c)
Find the time for the stock to fall to 4000 tonnes. Give your answer in years to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(8500\times0.92^5\)
✓ 1
A1 5602 tonnes
✓ 1
Part (b)
A1 the stock falls by 8% each year
✓ 1
Part (c)
M1 \(8500\times0.92^t=4000\), so \(0.92^t=\frac{4000}{8500}\)
✓ 1
M1 \(t=\frac{\ln(4000/8500)}{\ln0.92}\)
✓ 1
A1 \(t=9.04\) years
✓ 1
Exam tip: Isolate the exponential term first, then take logarithms of both sides. Write the line \(t=\ln(\ldots)/\ln(\ldots)\) before using the calculator.
289IBMA3S-1-18r31-011.3 Geometric sequences and seriesPaper 2Easy5 marks
Calculator
A patient is given a 400 mg dose of a drug. Each hour, 15% of the drug in the bloodstream is eliminated, so the amount remaining after \(n\) hours is \(400 \times 0.85^{n}\) mg.
(a)
Find the first whole number of hours after which the amount of drug is less than 50 mg.
(3)
(b)
Find the amount of drug remaining after this time. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 forms the inequality or equation \(400 \times 0.85^{n} < 50\) (or \(0.85^n<0.125\))
✓ 1
M1 attempts to solve, e.g. by logarithms or a table of values, giving \(n>12.8\)
✓ 1
A1 \(n = 13\) hours
✓ 1
Part (b)
M1 \(400 \times 0.85^{13}\) (using the value of \(n\) from (a))
✓ 1
A1 48.4 mg
✓ 1
Common error: When solving an inequality with \(0.85^n\), the answer must be a whole number of hours: 12.8 rounds up to 13, because after 12 hours the amount is still above 50 mg.
290IBMA3S-1-17r32-011.1 Standard formPaper 2Easy4 marks
Calculator
The star Proxima Centauri is a distance of \(4.02\times10^{13}\) km from Earth. Light travels at \(3.00\times10^{5}\) km s−1.
(a)
Calculate the time, in seconds, taken by light to travel from Proxima Centauri to Earth. Give your answer in the form \(a\times10^{k}\), where \(1\le a<10\) and \(k\) is an integer.
(2)
(b)
One year is \(3.15\times10^{7}\) s. Convert your answer to part (a) into years, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 time = distance ÷ speed \(=\frac{4.02\times10^{13}}{3.00\times10^{5}}\)
✓ 1
A1 \(1.34\times10^{8}\) s
✓ 1
Part (b)
M1 \(\frac{1.34\times10^{8}}{3.15\times10^{7}}\)
✓ 1
A1 4.25 years
✓ 1
Exam tip: Enter powers of ten with the calculator's exponent key (EXP), not by typing ×10 then a power, to avoid bracket errors.
291IBMA3S-1-17r32-041.5 Exponents and introduction to logarithmsPaper 2Easy4 marks
Calculator
A train of mass \(4.8\times10^{5}\) kg travels at 83.3 m s−1. Its kinetic energy is given by \(E=\frac{1}{2}mv^{2}\).
(a)
Calculate the kinetic energy of the train in joules. Give your answer in the form \(a\times10^{k}\) where \(1\le a<10\) and \(k\in\mathbb{Z}\), with \(a\) correct to 3 significant figures.
(2)
(b)
Given that 1 kilowatt-hour \(=3.6\times10^{6}\) J, use the unrounded kinetic energy to find the kinetic energy of the train in kilowatt-hours, correct to 3 significant figures.
Common error: The speed must be squared before multiplying by the mass: using \(\frac12(mv)^{2}\) or \(\frac12 mv\times2\) gives an energy that is wrong by a large factor.
292IBMA3S-1-18r38-021.4 Financial applicationsPaper 2Easy4 marks
Calculator
A mountaineering club invests €8000 for an expedition fund in an account paying a nominal annual interest rate of 2.8%, compounded monthly.
(a)
Find the value of the investment after 5 years, to the nearest euro.
(3)
(b)
Hence find the interest earned over the 5 years.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of 8000(1 + 0.028/12)n
✓ 1
A1 n = 60
✓ 1
A1 €9201
✓ 1
Part (b)
A1 €1201 (€9201 − €8000)
✓ 1
Common error: Divide the nominal rate by 12 for monthly compounding, and use 12 × 5 = 60 compounding periods.
293IBMA3S-1-18r38-041.4 Financial applicationsPaper 2Easy5 marks
Calculator
In a food-science experiment, the number of bacteria in a sample is modelled by N = 250 × 1.35t, where t is the time in hours since the start.
(a)
Find the number of bacteria after 4 hours, to 3 significant figures.
(2)
(b)
Find the time, in hours, for the number of bacteria to reach 5000. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 250 × 1.354
✓ 1
A1 830
✓ 1
Part (b)
M1 250 × 1.35t = 5000, so 1.35t = 20
✓ 1
M1 t = ln 20 / ln 1.35 (or log1.35 20)
✓ 1
A1 9.98 hours
✓ 1
Exam tip: Isolate the exponential term before taking logarithms: divide by 250 first.
294IBMA3S-1-18r41-041.4 Financial applicationsPaper 2Easy7 marks
Calculator
A retailer exchanges £4500 for euros at an exchange rate of £1 = €1.16. The bank charges a commission of 1.5% of the euros received.
(a)
Find the number of euros the retailer receives after commission has been taken off.
(3)
(b)
The retailer later changes all of the euros from part (a) back into pounds at a rate of €1 = £0.86, with no commission. Find the amount in pounds, to the nearest penny.
(2)
(c)
Find the percentage of the original £4500 that has been lost. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(4500\times1.16=5220\)
✓ 1
M1 reduce by 1.5%: \(5220\times0.985\) (or \(5220-0.015\times5220\))
✓ 1
A1 €5141.70
✓ 1
Part (b)
M1 \(5141.70\times0.86\)
✓ 1
A1 £4421.86
✓ 1
Part (c)
M1 \(\frac{4500-4421.86}{4500}\times100\)
✓ 1
A1 1.74%
✓ 1
Common error: Commission is taken as a percentage of the amount received, so multiply by (1 − 0.015) = 0.985 rather than subtracting 1.5 from the amount.
295IBMA3S-1-17r42-021.1 Standard formPaper 2Easy5 marks
Calculator
A food scientist starts a yeast culture with 800 cells at time \(t=0\). The number of cells is modelled by \(N = 800\times 2^{t/1.5}\), where \(t\) is the time in hours.
(a)
Find the number of cells after 6 hours.
(2)
(b)
Find the time taken for the culture to reach 50 000 cells. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(N=800\times 2^{6/1.5}=800\times 2^{4}\)
✓ 1
A1 \(N=12\,800\)
✓ 1
Part (b)
M1 \(50\,000=800\times 2^{t/1.5}\), so \(2^{t/1.5}=62.5\)
Exam tip: When the unknown is in an exponent, isolate the power first, then take logarithms of both sides.
296IBMA3S-1-17r43-141.1 Standard formPaper 2Easy4 marks
Calculator
A Formula 1 circuit has lap length \(5.4\times10^{3}\) m. A race is 56 laps long and the winner's total race time is \(5.2\times10^{3}\) s.
(a)
Calculate the race distance in metres. Give your answer in the form \(a\times10^{k}\) where \(1\le a<10\) and \(a\) is given to 3 significant figures.
(2)
(b)
Calculate the winner's average speed in m s−1, to 3 significant figures. Use the exact race distance of 302 400 m.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(5.4\times10^{3}\times56\)
✓ 1
A1 \(3.02\times10^{5}\) m
✓ 1
Part (b)
M1 \(\frac{302400}{5200}\)
✓ 1
A1 58.2 m s−1
✓ 1
Exam tip: Use the exact value of the distance in the second part rather than the rounded standard-form value.
297IBMA3S-1-19r44-021.4 Financial applicationsPaper 2Easy5 marks
Calculator
A hospital charity invests €32 000 in an account that pays a nominal annual interest rate of 2.8%, compounded quarterly. No money is added or withdrawn.
(a)
Find the value of the investment after 7 years. Give your answer to the nearest euro.
(3)
(b)
Find the total interest earned over the 7 years. Give your answer to the nearest euro.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the compound interest formula with k = 4 and n = 7
M1 subtract the initial investment from the value found, 38 902.29 − 32 000
✓ 1
A1 €6902
✓ 1
Common error: Dividing the rate by 4 but forgetting to multiply the number of years by 4 gives the value for a much shorter period.
298IBMA3S-1-19r46-011.4 Financial applicationsPaper 2Easy4 marks
Calculator
An archaeological trust places €15 000 in a restoration fund that pays 3.6% per annum nominal interest, compounded quarterly. No money is added or withdrawn.
(a)
Find the value of the fund after 8 years, giving your answer to the nearest euro.
(2)
(b)
Find the interest earned over the 8 years, to the nearest euro.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of compound interest formula with 32 compounding periods, e.g. \(15000\left(1+\frac{3.6}{400}\right)^{32}\)
✓ 1
A1 €19 981 (19980.60… to the nearest euro)
✓ 1
Part (b)
M1 subtract the original €15 000 from the value in (a) (accept the unrounded value)
✓ 1
A1 €4981 (ECF from (a))
✓ 1
Common error: Dividing the annual rate by 4 but then using 8 compounding periods instead of 32 (4 × 8).
299IBMA3S-1-19r46-041.3 Geometric sequences and seriesPaper 2Easy5 marks
Calculator
A racing bike is bought new for £2400. Its value falls by 15% each year.
(a)
Find the value of the bike after 3 years, to the nearest pound.
(2)
(b)
Find how long, in years to 3 significant figures, it takes for the value of the bike to fall to £500.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2400\times 0.85^3\)
✓ 1
A1 £1474 (1473.90)
✓ 1
Part (b)
M1 \(2400\times 0.85^n=500\) (or equivalent equation in the time)
✓ 1
M1 use of logarithms or a GDC solver to isolate \(n\), e.g. \(n=\frac{\ln(500/2400)}{\ln 0.85}\)
✓ 1
A1 9.65 years
✓ 1
Common error: A fall of 15% means multiplying by 0.85 each year, not by 0.15.
300IBMA3S-1-17r47-041.1 Standard formPaper 2Easy4 marks
Calculator
Each blade of a wind turbine has length 45 m, so the blades sweep out a circle of radius 45 m.
(a)
Find the area swept by the blades. Give your answer in the form a × 10k where 1 ≤ a < 10, correct to three significant figures.
(2)
(b)
A wind farm has 16 identical turbines. Using the unrounded answer to part (a), find the total area swept by all 16 turbines. Give your answer in the form a × 10k where 1 ≤ a < 10, correct to three significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\pi \times 45^2\)
✓ 1
A1 6.36 × 103 m2
✓ 1
Part (b)
M1 \(16 \times \pi \times 45^2\)
✓ 1
A1 1.02 × 105 m2
✓ 1
Exam tip: Keep the full calculator value between parts and round only the final answer, then write it in standard form.
301AASL1-q00071.1 Standard formPaper 2Medium6 marks
A red blood cell is modelled as a sphere of radius \(2.8\times10^{-6}\) m.
(a)
Find the volume of one red blood cell, according to the model. Give your answer in the form \(a\times10^{k}\), where \(1\le a\lt 10\) and \(k\in\mathbb{Z}\).
(2)
(b)
A blood sample has a volume of \(5.0\times10^{-6}\) m3. Red blood cells make up 45% of the volume of the sample. Use the model to find the number of red blood cells in the sample. Give your answer in the form \(a\times10^{k}\), where \(1\le a\lt 10\) and \(k\in\mathbb{Z}\).
(2)
(c)
An adult has \(2.5\times10^{13}\) red blood cells. The body produces new red blood cells at a rate of \(2.4\times10^{6}\) cells per second. Find the number of days the body takes to produce \(2.5\times10^{13}\) cells.
302AASL1-q00161.2 Arithmetic sequences and seriesPaper 2Medium6 marks
The table shows the number of members of a cycling club at the start of each year from 2019 to 2023.
Year
2019
2020
2021
2022
2023
Members
412
437
465
488
514
(a)
Show that the numbers of members do not form an exact arithmetic sequence.
(1)
(b)
Using the values for 2019 and 2023, find an estimate for the common difference.
(2)
(c)
The club models the number of members with an arithmetic sequence, with \(u_{1}=412\) in 2019 and the common difference found in part (b). Use the model to predict the number of members in 2030.
(2)
(d)
Comment on the reliability of the prediction in part (c).
(1)
Show mark scheme
Working
Mark
Notes
(a)
differences are 25, 28, 23, 26, which are not all equal
A1
Must see at least two different differences.
(b)
\(\dfrac{514-412}{4}\)
(M1)
\(25.5\)
A1
(c)
\(u_{12}=412+11(25.5)\)
(M1)
FT from (b). Award M0 for \(u_{11}\) or \(10d\).
\(692.5\), so about 692 or 693 members
A1
Accept 692, 692.5 or 693.
(d)
not reliable: 2030 is far outside the range of the data (extrapolation), and the differences are not constant
R1
Accept any valid comment that refers to extrapolation or to the varying differences.
Answers: (b) 25.5 · (c) about 693 members
303AASL1-q00181.2 Arithmetic sequences and seriesPaper 2Medium8 marks
A cyclist follows a training plan. She rides 18 km on day 1, and on each following day she rides 2 km more than on the previous day.
(a)
Find the distance she rides on day 25.
(2)
(b)
Find the least number of days needed for the total distance ridden to be at least 1500 km.
(3)
(c)
The plan is changed so that, once her daily distance reaches 50 km, she rides exactly 50 km each day after that. Find the total distance she rides in the first 30 days under the changed plan.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(18+24(2)\)
(M1)
\(66\) km
A1
(b)
\(\dfrac{n}{2}\big(36+2(n-1)\big)\ge1500\), i.e. \(n^{2}+17n-1500\ge0\)
(M1)
\(n\ge31.2\), or \(S_{31}=1488\) and \(S_{32}=1568\)
(A1)
32 days
A1
(c)
\(18+2(n-1)=50\Rightarrow\) day 17
(M1)
\(\dfrac{17}{2}(18+50)+13\times50\)
(M1)
\(578+650\)
\(1228\) km
A1
Answers: (a) 66 km · (b) 32 days · (c) 1228 km
304AASL1-q00191.2 Arithmetic sequences and seriesPaper 2Medium6 marks
Two friends start saving in the same week.
Leila saves $2.50 in week 1 and increases the amount she saves by $0.40 each week.
Tomás saves $5.00 in week 1 and increases the amount he saves by $0.25 each week.
(a)
Find the first week in which Leila saves more than Tomás in that week.
(3)
(b)
Find the first week at the end of which Leila’s total savings are greater than Tomás’s total savings.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(2.5+0.4(n-1)\gt 5+0.25(n-1)\)
(M1)
\(n-1\gt 16.7\ldots\), or week 17: $8.90 and $9.00; week 18: $9.30 and $9.25
\(n-1\gt 33.3\ldots\), or week 34: $309.40 and $310.25; week 35: $325.50 and $323.75
(A1)
week 35
A1
Answers: (a) week 18 · (b) week 35
305AASL1-q00271.3 Geometric sequences and seriesPaper 2Medium7 marks
In the first month after its launch, an app had 2400 new downloads. In each following month, the number of new downloads was 15% greater than in the previous month.
(a)
Find the number of new downloads in the sixth month.
(2)
(b)(i)
The total number of downloads in the first 12 months.
Write down an expression, using sigma notation, for this total.
(1)
(b)(ii)
Find this total.
(2)
(c)
Find the month in which the total number of downloads first exceeds 200 000.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(2400\times1.15^{5}\)
(M1)
\(4827\)
A1
\(4827.26\ldots\) Accept 4830.
(b)(i)
\(\displaystyle\sum_{n=1}^{12}2400(1.15)^{n-1}\)
A1
Accept equivalent forms, e.g. \(\sum_{n=0}^{11}2400(1.15)^{n}\).
(b)(ii)
\(\dfrac{2400(1.15^{12}-1)}{1.15-1}\)
(M1)
\(69\,604\)
A1
\(69\,604.0\ldots\) Accept 69 600.
(c)
\(\dfrac{2400(1.15^{n}-1)}{0.15}\gt 200\,000\), or table: \(S_{18}=182\,007\), \(S_{19}=211\,708\)
306AASL1-q00291.3 Geometric sequences and seriesPaper 2Medium8 marks
The population of a town was 24 500 at the start of 2015 and 28 100 at the start of 2020. The population at the start of each year is modelled by a geometric sequence.
(a)
Find the common ratio of the sequence, giving your answer correct to four significant figures.
(3)
(b)
Use the model to estimate the population at the start of 2027.
(2)
(c)
Find the first year in which the population at the start of the year exceeds 40 000, according to the model.
\(34\,045.7\ldots\) with unrounded \(r\); \(34\,126\) with \(r=1.028\). Accept answers from 34 000 to 34 130.
(c)
\(24\,500r^{n}\gt 40\,000\)
(M1)
\(n\gt 17.9\) (or 17.8 with \(r=1.028\)); or start of 2032: 39 048, start of 2033: 40 134
(A1)
2033
A1
Answers: (a) 1.028 · (b) about 34 000 · (c) 2033
307AASL1-q00301.3 Geometric sequences and seriesPaper 2Medium9 marks
The table shows the number of active users of a new online game in each of its first five weeks.
Week
1
2
3
4
5
Active users
1200
1440
1730
2070
2490
(a)
Show that the numbers of users are approximately, but not exactly, a geometric sequence.
(2)
(b)
Use the values for weeks 1 and 5 to estimate the common ratio.
(2)
(c)
The numbers of users are modelled by a geometric sequence with first term 1200 and common ratio 1.2. Use this model to find the number of users in week 10.
(2)
(d)
Use the model in part (c) to find the first week in which the number of users exceeds 20 000.
(2)
(e)
Suggest one reason why the model may not be valid for predicting the number of users after several months.
(1)
Show mark scheme
Working
Mark
Notes
(a)
finding the ratios of consecutive terms: \(1.2,\ 1.201,\ 1.197,\ 1.203\)
M1
At least three ratios.
the ratios are all close to (but not exactly equal to) 1.2
e.g. the number of potential players is limited, so growth cannot continue at 20% per week indefinitely; interest in the game may fall
R1
Accept any reasonable comment about extrapolation or limits on growth.
Answers: (b) 1.20 · (c) 6192 · (d) week 17
308AASL1-q00311.3 Geometric sequences and seriesPaper 2Medium9 marks
In its first year, a mine produces 52 000 tonnes of ore. In each following year, the output is 6% less than in the previous year. The mine closes at the end of the last year in which its output is at least 20 000 tonnes.
(a)
Find the output of the mine in its 8th year.
(2)
(b)
Find the total output of the mine in its first 8 years.
(2)
(c)
Find the number of years for which the mine operates.
\(n\le16.4\); or year 16: \(20\,555\); year 17: \(19\,322\)
(A1)
16 years
A1
(d)
\(\dfrac{52\,000\left(1-0.94^{16}\right)}{0.06}\)
(M1)
FT from (c).
\(545\,000\) tonnes
A1
\(544\,635.6\ldots\)
Answers: (a) 33 700 t · (b) 338 000 t · (c) 16 years · (d) 545 000 t
309AASL1-q00351.4 Financial applicationsPaper 2Medium6 marks
Ana invests €5000 in an account that pays simple interest at a rate of 4.5% per year. Ben invests €5000 in an account that pays a nominal annual interest rate of 3.8%, compounded monthly.
(a)
Find the value of Ana’s investment after 10 years.
(2)
(b)
Find the value of Ben’s investment after 10 years. Give your answer to the nearest cent.
(2)
(c)
Find the number of complete years after which the value of Ben’s investment first exceeds the value of Ana’s investment.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(5000+10(0.045\times5000)\) (arithmetic sequence with \(d=225\))
(M1)
€7250
A1
(b)
\(5000\left(1+\dfrac{3.8}{1200}\right)^{120}\), or TVM with \(N=120,\ I\%=3.8,\ PV=-5000,\ P/Y=12,\ C/Y=12\)
(M1)
€7307.03
A1
(c)
comparing \(5000+225n\) with \(5000\left(1+\frac{3.8}{1200}\right)^{12n}\), e.g. table or graph
(M1)
e.g. after 8 years: 6800 and 6773.09; after 9 years: 7025 and 7035.00
9 years
A1
Answers: (a) €7250 · (b) €7307.03 · (c) 9 years
310AASL1-q00371.4 Financial applicationsPaper 2Medium7 marks
Sofia invests $12 000 in an account that pays a nominal annual interest rate of 5.2%, compounded quarterly.
(a)
Find the value of the investment after 8 years. Give your answer to the nearest dollar.
(2)
(b)
Find the minimum number of complete years needed for the investment to at least double in value.
(2)
(c)
The annual rate of inflation is 2.5% for the 8 years. Find the real value of the investment after 8 years, in terms of today’s money. Give your answer to the nearest dollar.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(12\,000\left(1+\dfrac{5.2}{400}\right)^{4\times8}\), or TVM with \(N=32,\ I\%=5.2,\ PV=-12\,000,\ P/Y=4,\ C/Y=4\)
(M1)
$18 142
A1
$18 141.93…
(b)
\(12\,000\left(1+\dfrac{5.2}{400}\right)^{4n}\ge 24\,000\), or TVM giving \(N=53.7\) quarters (13.4 years)
(M1)
14 years
A1
(c)
attempt to adjust for inflation, e.g. dividing by \(1.025^{n}\) or using \(5.2-2.5\)
(M1)
\(\dfrac{18\,141.93}{1.025^{8}}\)
(A1)
$14 890
A1
$14 889.93… Also accept the real-interest-rate method, \(12\,000\left(1+\frac{5.2-2.5}{400}\right)^{32}=14\,882\). Accept answers between $14 882 and $14 890.
Answers: (a) $18 142 · (b) 14 years · (c) $14 890
311AASL1-q00391.4 Financial applicationsPaper 2Medium6 marks
An investment of £15 000 grows to £20 000 in exactly 6 years, with interest compounded monthly at a fixed nominal annual rate of \(r\%\).
(a)
Find the value of \(r\).
(2)
(b)
Find the annual interest rate, compounded annually, that would give the same growth over the 6 years.
(2)
(c)
The investment remains in the account at the same rate \(r\%\), compounded monthly. Find its value after a total of 10 years.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(15\,000\left(1+\dfrac{r}{1200}\right)^{72}=20\,000\), or TVM with \(N=72,\ PV=-15\,000,\ FV=20\,000,\ P/Y=12,\ C/Y=12\)
Account B: interest rate 4.2% per year, compounded annually.
(a)
Find the amount Lena must invest now in Account A. Give your answer to the nearest cent.
(2)
(b)
Find the amount Lena must invest now in Account B. Give your answer to the nearest cent.
(2)
(c)
Explain why the amount in part (a) is smaller than the amount in part (b).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(PV\left(1+\dfrac{4.2}{400}\right)^{40}=50\,000\), or TVM with \(N=40,\ I\%=4.2,\ FV=50\,000,\ P/Y=4,\ C/Y=4\)
(M1)
$32 924.36
A1
(b)
\(PV(1.042)^{10}=50\,000\)
(M1)
$33 135.45
A1
(c)
with quarterly compounding, interest is added more often and itself earns interest sooner, so the same nominal rate gives greater growth
R1
Accept equivalent: the effective annual rate of A (4.27%) is greater than 4.2%.
Answers: (a) $32 924.36 · (b) $33 135.45
313AASL1-q00411.4 Financial applicationsPaper 2Medium8 marks
A piece of factory equipment was bought for $64 000. It depreciates by a fixed percentage each year. After 4 years it is valued at $31 000.
(a)
Find the annual rate of depreciation, giving your answer as a percentage correct to three significant figures.
(3)
(b)
Find the value of the equipment after 8 years.
(2)
(c)
The equipment is replaced at the end of the first year in which its value falls below 10% of its purchase price. Find the number of years for which the equipment is used.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(64\,000\,x^{4}=31\,000\), where \(x\) is the annual multiplier
(M1)
\(x=0.83424\ldots\)
(A1)
\(16.6\%\)
A1
\(16.575\ldots\%\)
(b)
\(64\,000(0.83424\ldots)^{8}\), or \(31\,000\times\dfrac{31\,000}{64\,000}\)
(M1)
$15 016
A1
\(15\,015.625\). Accept $14 980 (from 16.6%) to $15 016.
(c)
\(64\,000(0.83424\ldots)^{n}\lt 6400\)
(M1)
\(n\gt 12.7\); or after 12 years $7273, after 13 years $6068
(A1)
13 years
A1
Answers: (a) 16.6% · (b) $15 016 · (c) 13 years
314AASL1-q00511.5 Exponents and introduction to logarithmsPaper 2Medium8 marks
Let \(N=7^{200}\). The number \(N\) is too large to be evaluated directly on a calculator.
(a)
Find the value of \(\log_{10}N\), giving your answer correct to three decimal places.
(2)
(b)
Hence write \(N\) in the form \(a\times10^{k}\), where \(1\le a\lt 10\) and \(k\in\mathbb{Z}\), giving the value of \(a\) correct to three significant figures.
(3)
(c)
Write down the number of digits in \(N\).
(1)
(d)
Find the value of \(p\) for which \(N=\mathrm{e}^{p}\), giving your answer correct to one decimal place.
315AASL1-q00521.5 Exponents and introduction to logarithmsPaper 2Medium13 marks
The magnitude, \(M\), of an earthquake on the Richter scale is given by \(M=\log_{10}A\), where \(A\) is the maximum amplitude, in micrometres, recorded by a seismometer at a standard distance from the centre of the earthquake.
(a)
Find the magnitude of an earthquake for which \(A=2.5\times10^{4}\).
(2)
(b)
An earthquake has magnitude 6.3. Find the value of \(A\). Give your answer in the form \(a\times10^{k}\), where \(1\le a\lt 10\) and \(k\in\mathbb{Z}\).
(2)
(c)
Two earthquakes have amplitudes \(A_{1}\) and \(A_{2}\) and magnitudes \(M_{1}\) and \(M_{2}\). Show that if \(M_{2}=M_{1}+1\) then \(A_{2}=10A_{1}\).
(2)
(d)
Earthquake X has magnitude 7.1. The amplitude of earthquake Y is 40 times the amplitude of earthquake X. Find the magnitude of earthquake Y.
(3)
(e)
The energy, \(E\) joules, released by an earthquake of magnitude \(M\) satisfies \(\log_{10}E=4.8+1.5M\). Find the energy released by an earthquake of magnitude 6.3.
(2)
(f)
Find how many times more energy is released by earthquake Y than by earthquake X.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\log_{10}\left(2.5\times10^{4}\right)\)
(M1)
\(M=4.40\)
A1
\(4.39794\ldots\)
(b)
\(A=10^{6.3}\)
(M1)
\(2.00\times10^{6}\)
A1
\(1\,995\,262.3\ldots\) Accept \(1.995\times10^{6}\). Award A0 for 2E6 or for an answer not in the required form.
316AASL1-q00711.7 Rational exponents and laws of logarithmsPaper 2Medium7 marks
The mass, \(m\) grams, of a radioactive isotope remaining after \(t\) years is given by \(m=120\left(\tfrac{1}{2}\right)^{\frac{t}{28.8}}\), \(t\ge0\).
(a)
Find the mass remaining after 50 years.
(2)
(b)
Find the time taken for the mass to fall to 20 grams.
(3)
(c)
The model can also be written as \(m=120\mathrm{e}^{-kt}\). Find the value of \(k\).
321IBMA2S-044-04-081.9 The binomial theoremPaper 2Medium7 marks
Calculator
In the expansion of \(\left(x+\frac{k}{x^2}\right)^9\), where \(k>0\), the constant term is 500.
(a)
Find the value of \(k\).
(5)
(b)
Find the coefficient of \(x^3\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 general term \(\binom{9}{r}x^{9-r}k^r x^{-2r}\) M1 setting the power to zero: \(9-3r=0\) A1 \(r=3\) A1 \(84k^3=500\) A1 \(k=1.81\) (1.8123…)
✓ 5
Part (b)
M1 \(9-3r=3\) gives \(r=2\), so the coefficient is \(\binom{9}{2}k^2=36k^2\) A1 118 (118.24…); accept 118 from \(k=1.81\)
✓ 2
Exam tip: Keep the unrounded value of \(k\) in your calculator memory for part (b). Rounding too early can change the answer: using \(k=1.8\) gives 116.6 instead of 118.
322IBMA2S-044-04-091.8 Infinite geometric seriesPaper 2Medium10 marks
Calculator
A precision-engineering firm machines steel cubes with a nominal edge length of 5 cm. Because of manufacturing variation, the actual edge length of a cube is \((5+h)\) cm, where \(h\) is small. The volume of the cube is \(V\) cm3.
(a)
Use the binomial theorem to expand \((5+h)^3\).
(2)
(b)
A customer requires \(V\le130\). Using the linear approximation \(V\approx125+75h\), find the largest allowable value of \(h\).
(2)
(c)
Using the exact expression for \(V\), find the largest allowable value of \(h\).
(2)
(d)
Explain why, for \(h>0\), the linear approximation overestimates the largest allowable value of \(h\).
(2)
(e)
Find the percentage error in the linear estimate of the largest allowable \(h\), relative to the exact value. Use unrounded values from parts 2 and 3.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of coefficients 1, 3, 3, 1 with decreasing powers of 5 A1 \(V=125+75h+15h^2+h^3\)
✓ 2
Part (b)
M1 \(125+75h=130\) A1 \(h=\frac{1}{15}=0.0667\)
✓ 2
Part (c)
M1 \((5+h)^3=130\), so \(h=\sqrt[3]{130}-5\) A1 \(h=0.0658\) (0.065797…)
✓ 2
Part (d)
R1 for \(h>0\), the neglected terms \(15h^2+h^3\) are positive, so the true \(V\) is greater than \(125+75h\) R1 hence the true \(V\) reaches 130 at a smaller value of \(h\)
Exam tip: In a percentage error, always divide by the exact value, not the approximate one. Keep full calculator values for both before rounding, because rounding early can noticeably change a small difference.
323IBMA2S-044-07-101.2 Arithmetic sequences and seriesPaper 2Medium7 marks
A cyclist trains so that the distance ridden each day increases by the same amount every day. The daily distances form an arithmetic sequence with first term \(a\) km and common difference \(d\) km. Let \(S_k\) be the total distance ridden in the first \(k\) days.
(a)
Prove that \(S_{3n}=3(S_{2n}-S_n)\) for any positive integer \(n\).
(4)
(b)
The cyclist rides a total of 150 km in the first 10 days and 400 km in the first 20 days. Find the total distance ridden in the first 30 days.
(2)
(c)
Hence find the distance ridden from day 21 to day 30 inclusive.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(S_{2n}=n(2a+(2n-1)d)\) and \(S_n=\frac{n}{2}(2a+(n-1)d)\)
✓ 1
M1 subtracts using the common factor \(\frac{n}{2}\): \(\frac{n}{2}[4a+(4n-2)d-2a-(n-1)d]\)
✓ 1
A1 \(S_{2n}-S_n=\frac{n}{2}(2a+(3n-1)d)\)
✓ 1
A1 \(3(S_{2n}-S_n)=\frac{3n}{2}(2a+(3n-1)d)=S_{3n}\), so the result holds
✓ 1
Part (b)
M1 uses \(S_{30}=3(S_{20}-S_{10})=3(400-150)\), OR solves \(2a+9d=30\) and \(2a+19d=40\) to get \(a=10.5\), \(d=1\) and then finds \(S_{30}=15(21+29)\)
✓ 1
A1 750 km
✓ 1
Part (c)
A1 \(S_{30}-S_{20}=750-400=350\) km
✓ 1
Exam tip: Factor out \(\frac{n}{2}\) early. Two series formulas with the same bracketed structure are then much easier to compare.
324IBMA2S-044-06-061.7 Rational exponents and laws of logarithmsPaper 2Medium6 marks
Calculator
A laboratory culture starts with 3.5×103 bacteria. The number of bacteria doubles every 20 minutes. The number of bacteria, N, after t minutes is modelled by \(N = 3.5\times10^{3}\times 2^{t/20}\).
(a)
Find the number of bacteria after 6 hours. Give your answer in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ, with a correct to 3 significant figures.
(3)
(b)
Find the time, in minutes, at which the number of bacteria first exceeds 1.0×1012. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 using t = 360, so t/20 = 18 M1 3.5×103×218 A1 9.18×108
✓ 3
Part (b)
M1 setting up 3.5×103×2t/20 = 1.0×1012 (or >) M1 using logarithms or GDC to obtain t/20 = 28.09… A1 t = 562 minutes
✓ 3
Common error: The doubling period is given in minutes, so time must be in minutes too. Using t = 6 hours directly in 2t/²⁰ gives a far too small answer. Convert 6 hours to 360 minutes first.
325IBMA2S-044-06-081.3 Geometric sequences and seriesPaper 2Medium6 marks
Calculator
The mass of the Sun is 1.989×1030 kg and the mass of the Earth is 5.972×1024 kg. Give all answers in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ, with a correct to 3 significant figures.
(a)
Find how many times greater the mass of the Sun is than the mass of the Earth.
(2)
(b)
The Sun loses mass at a rate of 4.26×109 kg every second. Taking one year as 365.25 days, find the mass the Sun loses in one year.
(2)
(c)
Find the mass lost in one year as a percentage of the Sun's mass. Give your answer as a percentage in the form stated above.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (1.989×1030) ÷ (5.972×1024) A1 3.33×105
✓ 2
Part (b)
M1 4.26×109 × 365.25 × 24 × 3600 A1 1.34×1017 kg
✓ 2
Part (c)
M1 (their mass lost in one year from the previous part ÷ 1.989×1030) × 100 A1 6.76×10−12 % (accept 6.74×10−12 % from using the rounded value 1.34×1017; follow through from their previous answer)
✓ 2
Exam tip: Use 365.25 × 24 × 3600 = 31 557 600 seconds as one step on your calculator. Keep the unrounded value from the previous part in your calculator memory, because rounding in stages can change the third significant figure.
326IBMA2S-044-06-091.1 Standard formPaper 2Medium10 marks
Calculator
When a steel bridge of length L metres is heated by ΔT °C, its length increases by ΔL metres, where ΔL = αLΔT. For steel, α = 1.2×10−5 per °C. A steel road bridge has length 1280 m.
(a)
Write 1280 in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ.
(1)
(b)
Between winter and summer the temperature of the bridge rises by 35 °C. Find ΔL, giving your answer in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ, with a correct to 3 significant figures.
(2)
(c)
The bridge has expansion joints. Each joint can absorb at most 4.5×10−2 m of expansion. Find the minimum number of joints needed for the 35 °C temperature rise.
(3)
(d)
Find the temperature rise that would make the bridge 1.00 m longer.
(2)
(e)
A copper cable is fixed along one section of the bridge. For copper, α = 1.7×10−5 per °C. The cable lengthens by 0.25 m when its temperature rises by 35 °C. Find the length of the cable, giving your answer in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ, with a correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 1.28×103
✓ 1
Part (b)
M1 substituting: 1.2×10−5 × 1.28×103 × 35 A1 5.38×10−1 m
✓ 2
Part (c)
M1 dividing their ΔL by 4.5×10−2 A1 11.9… (accept 11.95 or 11.96) R1 rounding up because 11 joints cannot absorb all the expansion, so 12 joints
✓ 3
Part (d)
M1 1.00 = 1.2×10−5 × 1280 × ΔT A1 ΔT = 65.1 °C
✓ 2
Part (e)
M1 L = 0.25 ÷ (1.7×10−5 × 35) A1 4.20×102 m
✓ 2
Common error: In part (c), rounding 11.9 to the nearest whole number does not earn the reasoning mark. Say why you round up: 11 joints would absorb only 0.495 m.
327IBMA2S-044-01-051.5 Exponents and introduction to logarithmsPaper 2Medium7 marks
Calculator
The sound intensity level \(L\), in decibels (dB), of a sound with intensity \(I\) W m−2 is given by \(L=10\log_{10}\left(\dfrac{I}{I_0}\right)\), where \(I_0=1\times10^{-12}\) W m−2.
(a)
At a football stadium, the crowd noise has intensity 0.25 W m−2. Find the sound intensity level of the crowd.
(2)
(b)
Show that doubling the intensity of any sound increases its sound intensity level by \(10\log_{10}2\) dB.
(2)
(c)
A single vuvuzela produces a sound intensity level of 95 dB. Assuming intensities add, find the sound intensity level produced by 40 identical vuvuzelas sounding together.
M1 writes \(10\log_{10}\left(\frac{2I}{I_0}\right)\) and uses the product law to split off \(10\log_{10}2\)
✓ 1
A1 obtains \(L+10\log_{10}2\), so the increase is \(10\log_{10}2\) dB (AG)
✓ 1
Part (c)
M1 recognises that the total intensity is \(40I\)
✓ 1
M1 uses the product law: \(L=95+10\log_{10}40\)
✓ 1
A1 111 dB
✓ 1
Common error: Decibel levels do not add. Forty sources at 95 dB do not give 3800 dB. Multiply the intensity, then use the product law of logarithms.
328IBMA2S-044-01-101.7 Rational exponents and laws of logarithmsPaper 2Medium7 marks
Calculator
A password security app estimates the strength of a password by its entropy \(E\) bits, where \(E=L\log_2 N\). Here \(L\) is the number of characters in the password and \(N\) is the number of possible symbols for each character.
(a)
Find the entropy of an 8-character password that uses only the 26 lowercase letters.
(2)
(b)
Show that \(E=\dfrac{L\ln N}{\ln 2}\).
(2)
(c)
A company requires 10-character passwords to have an entropy of at least 64 bits. Find the least number of possible symbols \(N\) that each character must be chosen from.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes: \(8\log_2 26\)
✓ 1
A1 37.6 bits
✓ 1
Part (b)
M1 uses change of base: \(\log_2 N=\dfrac{\ln N}{\ln 2}\)
✓ 1
A1 substitutes to obtain \(E=\dfrac{L\ln N}{\ln 2}\) (AG)
✓ 1
Part (c)
M1 sets up \(10\log_2 N=64\) or \(10\log_2 N\ge 64\) (accept \(\dfrac{10\ln N}{\ln 2}=64\))
✓ 1
M1 obtains \(N=2^{6.4}\approx 84.4\)
✓ 1
A1 \(N=85\), because \(N\) must be a whole number and 84 symbols gives less than 64 bits
✓ 1
Common error: From \(\log_2 N=6.4\), the next step is \(N=2^{6.4}\), not \(N=6.4^{2}\). The base of the log becomes the base of the power. A number of symbols must be a whole number, so round up to 85. Rounding down to 84 would not reach 64 bits.
329IBMA2S-044-01-111.7 Rational exponents and laws of logarithmsPaper 2Medium12 marks
Calculator
In a laboratory, two strains of bacteria are grown. The number of bacteria of strain A after \(t\) hours is modelled by \(A=500\times1.35^{t}\). The number of bacteria of strain B is modelled by \(B=2000\times1.12^{t}\).
(a)
Find the time taken for the population of strain A to double.
(3)
(b)
The model for strain A can be written as \(A=500e^{kt}\). Find the exact value of \(k\), and give its value correct to 3 significant figures.
(2)
(c)
Show that the two populations are equal when \(t=\dfrac{\ln 4}{\ln 1.35-\ln 1.12}\).
(3)
(d)
Find the value of \(t\) at which the populations are equal.
(2)
(e)
Find the number of bacteria in each strain at this time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets up \(1.35^{t}=2\)
✓ 1
M1 takes logs: \(t=\dfrac{\ln 2}{\ln 1.35}\)
✓ 1
A1 2.31 hours
✓ 1
Part (b)
M1 writes \(e^{k}=1.35\)
✓ 1
A1 \(k=\ln 1.35\approx0.300\)
✓ 1
Part (c)
M1 equates the models and rearranges: \(\left(\frac{1.35}{1.12}\right)^{t}=4\)
✓ 1
M1 takes logs and uses the quotient law: \(t(\ln 1.35-\ln 1.12)=\ln 4\)
✓ 1
A1 obtains the given expression for \(t\) (AG)
✓ 1
Part (d)
M1 evaluates the expression given in the previous part
✓ 1
A1 \(t=7.42\) hours
✓ 1
Part (e)
M1 substitutes their \(t\) into either model, e.g. \(500\times1.35^{7.422}\)
✓ 1
A1 4640 bacteria
✓ 1
Exam tip: Collect the terms containing \(t\) on one side and the constants on the other before you take logs. Use your unrounded value of \(t\) in the final part to avoid rounding errors.
330IBMA2S-044-00-041.3 Geometric sequences and seriesPaper 2Medium7 marks
Calculator
An oil well produces 52 000 barrels in its first year. In each year after that, production is 8% less than in the previous year.
(a)
Find the production in the sixth year.
(2)
(b)
Find the least number of complete years after which the total production first exceeds 500 000 barrels.
(3)
(c)
Show that, under this model, the total production can never exceed 650 000 barrels.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(52000\times0.92^5\)
✓ 1
A1 34 300 barrels (34 272.2…)
✓ 1
Part (b)
M1 \(\frac{52000(1-0.92^n)}{1-0.92}>500000\) (equation or inequality)
✓ 1
A1 \(n>17.6\) (17.586…)
✓ 1
A1 18 years
✓ 1
Part (c)
M1 \(S_\infty=\frac{52000}{1-0.92}=650000\) OR \(S_n=\frac{52000(1-0.92^n)}{0.08}=650000(1-0.92^n)\)
✓ 1
R1 since \(0<0.92^n\) for all \(n\), \(S_n=650000(1-0.92^n)<650000\) (or: \(|r|<1\) and all terms positive, so the partial sums increase towards 650 000 but never reach it)
✓ 1
Common error: An 8% decrease means a ratio of 0.92, not 0.08. The 0.08 appears only as \(1-r\) in the denominator.
331IBMA2S-044-00-071.2 Arithmetic sequences and seriesPaper 2Medium7 marks
Calculator
An 800 m runner does a session of 12 intervals. The first interval is 250 m and each interval is a fixed distance longer than the one before. The total distance of the 12 intervals is 5.64 km.
(a)
Find the increase in length from one interval to the next.
(3)
(b)
In a later session she uses the same first interval and the same increase, but adds intervals until her total distance first exceeds 10 km. Find the number of intervals in this session and the total distance she runs.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{12}{2}(2\times250+11d)=5640\)
✓ 1
A1 \(500+11d=940\)
✓ 1
A1 \(d=40\) m
✓ 1
Part (b)
M1 \(\frac{n}{2}(500+40(n-1))>10000\)
✓ 1
A1 \(n>17.3\), so 18 intervals
✓ 1
M1 \(S_{18}=\frac{18}{2}(500+17\times40)\)
✓ 1
A1 10 620 m (10.6 km)
✓ 1
Common error: Convert 5.64 km to 5640 m before substituting, because the other distances are in metres. Mixing units gives a meaningless value of \(d\).
332IBMA2S-044-00-091.3 Geometric sequences and seriesPaper 2Medium13 marks
Calculator
A rowing coach compares two weekly training programmes. Programme A: 40 km in week 1, then each week 3.5 km more than the previous week. Programme B: 60 km in week 1, then each week 90% of the previous week's distance.
(a)
Find the distance rowed in week 10 of Programme A.
(2)
(b)
Find the first week in which the weekly distance for Programme B is less than the weekly distance for Programme A. Justify your answer.
(2)
(c)
Find the total distance rowed in the first 10 weeks of Programme A.
(2)
(d)
Find the least number of weeks after which the total distance for Programme A first exceeds 600 km.
(3)
(e)
(i) Find the total distance rowed in the first 10 weeks of Programme B. (ii) Show that the total distance for Programme B can never exceed 600 km.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(40+9\times3.5\)
✓ 1
A1 71.5 km
✓ 1
Part (b)
M1 compares \(40+3.5(n-1)\) with \(60(0.9)^{n-1}\), showing B > A in week 3 (A = 47, B = 48.6) and B < A in week 4 (A = 50.5, B = 43.74)
✓ 1
A1 week 4
✓ 1
Part (c)
M1 \(\frac{10}{2}(40+71.5)\) or \(\frac{10}{2}(80+9\times3.5)\)
✓ 1
A1 557.5 km
✓ 1
Part (d)
M1 \(\frac{n}{2}(80+3.5(n-1))>600\), or a table of sums showing \(S_{10}=557.5\) and \(S_{11}=632.5\)
✓ 1
A1 \(n>10.6\) (10.57…), or the correct values of \(S_{10}\) and \(S_{11}\)
✓ 1
A1 11 weeks
✓ 1
Part (e)
(i) M1 \(\frac{60(1-0.9^{10})}{1-0.9}\)
✓ 1
A1 391 km (390.79…)
✓ 1
(ii) M1 \(S_n=\frac{60(1-0.9^n)}{1-0.9}=600(1-0.9^n)\), or \(S_\infty=\frac{60}{1-0.9}=600\)
✓ 1
R1 since \(0.9^n>0\) for all \(n\), \(S_n<600\); or, since all terms are positive and \(|r|<1\), the partial sums increase towards 600 km but never reach it
✓ 1
Exam tip: The second part compares the distances in single weeks, while the later parts compare running totals. Read each part carefully to see whether it asks about one week's distance or the total so far.
333IBMA2S-044-00-121.1 Standard formPaper 2Medium13 marks
Calculator
Engineers test two models of the emergency braking of an autonomous car. In both models the car travels 25 m in the first second after braking begins. Model G: in each second, the car travels 80% of the distance it travelled in the previous second. Model A: in each second, the car travels 3 m less than in the previous second. The model applies only while the distance travelled in a second is positive.
(a)
For Model G, find the distance travelled in the 5th second.
(2)
(b)
For Model G, find the theoretical total stopping distance.
(2)
(c)
For Model A, find the number of seconds for which the car moves and the total stopping distance.
(3)
(d)
A pedestrian is 120 m ahead of the car when braking begins. Determine which model predicts that the car stops before reaching the pedestrian. For the other model, find during which second the car reaches the pedestrian's position.
(4)
(e)
For Model G, find the distance travelled in the first 10 seconds.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(25\times0.8^4\)
✓ 1
A1 10.24 m
✓ 1
Part (b)
M1 \(\frac{25}{1-0.8}\)
✓ 1
A1 125 m
✓ 1
Part (c)
M1 \(25-3(n-1)>0\), or a list showing the last positive term is \(u_9=1\)
✓ 1
A1 9 seconds
✓ 1
A1 \(\frac{9}{2}(25+1)=117\) m
✓ 1
Part (d)
R1 Model A: \(117<120\), so the car stops in time; Model G: \(125>120\), so it does not
✓ 1
M1 \(\frac{25(1-0.8^n)}{0.2}\ge120\), i.e. \(0.8^n\le0.04\), OR evaluating cumulative sums \(S_n\) for Model G
✓ 1
A1 \(n\ge14.4\) (14.42…), OR \(S_{14}\approx119.5<120\) and \(S_{15}\approx120.6>120\)
✓ 1
A1 during the 15th second
✓ 1
Part (e)
M1 \(\frac{25(1-0.8^{10})}{1-0.8}\)
✓ 1
A1 112 m (111.58…)
✓ 1
Exam tip: In context, interpret \(n\ge14.4\) carefully. The car has not reached the pedestrian after 14 full seconds but has done so by the end of the 15th, so the answer is 'during the 15th second'.
334IBMA2S-044-02-061.5 Exponents and introduction to logarithmsPaper 2Medium7 marks
Calculator
A new smartphone battery has a maximum capacity of 4500 mAh. After each full charge cycle, the maximum capacity is multiplied by a constant \(r\). After 500 cycles, the maximum capacity is 3600 mAh.
(a)
Find the value of \(r\), giving your answer to 6 decimal places.
(2)
(b)
Find the number of complete cycles after which the maximum capacity first falls below 3000 mAh.
(3)
(c)
Find the maximum capacity after 1000 cycles.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(4500r^{500}=3600\), so \(r^{500}=0.8\)
✓ 1
A1 \(r=0.999554\)
✓ 1
Part (b)
M1 \(4500r^{n}<3000\)
✓ 1
M1 solves using logs or GDC, giving \(n>908.5\)
✓ 1
A1 909 cycles
✓ 1
Part (c)
M1 \(4500r^{1000}=4500\times0.8^{2}\)
✓ 1
A1 2880 mAh
✓ 1
Exam tip: Store \(r\) in full on your GDC. Rounding it early can shift a cycle count like 909 by one.
335IBMA2S-044-02-071.8 Infinite geometric seriesPaper 2Medium11 marks
Calculator
A pile driver hammers a steel post into the ground. The first blow drives the post 50 cm into the ground. Each blow after that drives the post 90% as far as the blow before.
(a)
Find the distance the post is driven by the 5th blow.
(2)
(b)
Find the total depth of the post after 10 blows.
(2)
(c)
The post must reach a depth of at least 4 m. Find the minimum number of blows needed.
(4)
(d)
A heavier hammer also drives the post 90% as far on each blow as on the blow before. With this hammer, the total depth after 10 blows is exactly 400 cm. Find the distance the first blow drives the post.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(50\times0.9^{4}\)
✓ 1
A1 32.8 cm
✓ 1
Part (b)
M1 \(\frac{50(1-0.9^{10})}{1-0.9}\)
✓ 1
A1 326 cm
✓ 1
Part (c)
M1 \(\frac{50(1-0.9^{n})}{0.1}\ge400\)
✓ 1
A1 \(0.9^{n}\le0.2\)
✓ 1
M1 solves using logs or GDC, giving \(n\ge15.3\)
✓ 1
A1 16 blows
✓ 1
Part (d)
M1 \(\frac{a(1-0.9^{10})}{0.1}=400\)
✓ 1
M1 rearranges to \(a=\frac{40}{1-0.9^{10}}\)
✓ 1
A1 61.4 cm
✓ 1
Common error: In part (c), the answer must be rounded up. Fifteen blows give only about 397 cm.
336IBMA2S-044-02-081.8 Infinite geometric seriesPaper 2Medium7 marks
Calculator
A patient takes a 200 mg tablet at the same time every day. Just before each tablet is taken, a fixed fraction \(p\) of the drug present just after the previous tablet remains in the body. Just after the second tablet, the amount of drug in the body is 260 mg.
(a)
Find the value of \(p\).
(2)
(b)
Show that the amount of drug in the body just after the \(n\)th tablet is \(\frac{2000}{7}(1-0.3^{n})\) mg.
(3)
(c)
Find the amount of drug in the body just after the 7th tablet.
(2)
Show mark scheme
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Notes
Part (a)
M1 \(200p+200=260\)
✓ 1
A1 \(p=0.3\)
✓ 1
Part (b)
M1 writes the amount as \(200+200(0.3)+200(0.3)^2+\dots+200(0.3)^{n-1}\)
✓ 1
M1 uses the geometric sum with \(a=200\), \(r=0.3\) and \(n\) terms
Exam tip: The most recent tablet is the full 200 mg. Older tablets carry increasing powers of 0.3, so the first term is 200.
337IBMA2S-044-02-121.3 Geometric sequences and seriesPaper 2Medium7 marks
Calculator
A solar farm generates 2.4 GWh of electricity in its first year. Because the panels degrade, the output each year after that is \(p\)% lower than in the year before. The total output over the first 25 years is 55.2 GWh.
(a)
Let \(r=1-\frac{p}{100}\). Write down an equation in \(r\) for the total output over the first 25 years.
(2)
(b)
Find the value of \(p\).
(3)
(c)
A newer type of panel loses 0.4% of its output each year. A farm using these panels also generates 2.4 GWh in its first year. Find its total output over the first 25 years.
(2)
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Mark
Notes
Part (a)
M1 uses the geometric sum with \(a=2.4\) and \(n=25\)
✓ 1
A1 \(\frac{2.4(1-r^{25})}{1-r}=55.2\)
✓ 1
Part (b)
M1 solves using a GDC (graph or solver)
✓ 1
A1 \(r=0.992968\ldots\)
✓ 1
A1 \(p=0.703\)
✓ 1
Part (c)
M1 \(\frac{2.4(1-0.996^{25})}{1-0.996}\)
✓ 1
A1 57.2 GWh
✓ 1
Common error: The GDC also gives the solution \(r=1\), but it is not valid because \(1-r\) appears in the denominator. Reject it and use the other root.
338IBMA2S-044-05-051.7 Rational exponents and laws of logarithmsPaper 2Medium7 marks
Calculator
The concentration of a drug in a patient's blood, \(t\) hours after an injection, is modelled by \(C=C_0e^{-kt}\) mg L−1, where \(C_0\) and \(k\) are positive constants. The concentration halves every 6 hours.
(a)
Show that \(k=\frac{\ln 2}{6}\).
(2)
(b)
For one dose, \(C_0=80\). Find the concentration 10 hours after the injection.
(2)
(c)
A further dose is given when the concentration falls to 15 mg L−1. Find how long after the first injection this happens.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 sets \(e^{-6k}=\frac{1}{2}\)
✓ 1
A1 \(-6k=\ln\frac{1}{2}=-\ln 2\), so \(k=\frac{\ln 2}{6}\) (AG)
Exam tip: For a 'show that' result, start from the given condition and end exactly at the printed result; do not substitute the answer.
339IBMA2S-044-05-061.4 Financial applicationsPaper 2Medium7 marks
Calculator
The maximum capacity of a smartphone battery, as a percentage of its original capacity, after \(n\) full charge cycles is modelled by \(B=100(0.9996)^{n}\).
(a)
Find the capacity after 500 charge cycles.
(2)
(b)
The battery is replaced free of charge once its capacity falls below 80%. Find the least number of complete charge cycles after which this happens.
(3)
(c)
A rival battery loses \(r\%\) of its current capacity each cycle and still has exactly 80% capacity after 800 cycles. Find \(r\).
Common error: Dividing by \(\ln 0.9996\) (a negative number) reverses an inequality; check the answer by substituting back.
340IBMA2S-044-05-081.1 Standard formPaper 2Medium7 marks
Calculator
During the first part of a race, a sprinter's speed \(t\) seconds after the start is modelled by \(v=10.5\left(1-e^{-0.8t}\right)\) m s−1.
(a)
Write down the speed that the sprinter approaches according to this model, justifying your answer.
(2)
(b)
Find the sprinter's speed 2 seconds after the start.
(2)
(c)
Find the time taken for the sprinter to reach 95% of the speed that the sprinter approaches.
(3)
Show mark scheme
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Mark
Notes
Part (a)
R1 as \(t\) increases, \(e^{-0.8t}\to 0\)
✓ 1
A1 10.5 m s−1
✓ 1
Part (b)
M1 substitutes \(t=2\): \(10.5(1-e^{-1.6})\)
✓ 1
A1 8.38 m s−1
✓ 1
Part (c)
M1 sets \(10.5(1-e^{-0.8t})=0.95\times10.5\), i.e. \(1-e^{-0.8t}=0.95\)
✓ 1
M1 \(e^{-0.8t}=0.05\), so \(t=\frac{\ln 20}{0.8}\) (or a correct GDC solution of \(v=9.975\))
✓ 1
A1 \(t=3.74\) s
✓ 1
Exam tip: Working with the proportion 0.95 rather than the speed 9.975 m s⁻¹ avoids rounding errors before taking logarithms.
341IBMA2S-044-05-091.4 Financial applicationsPaper 2Medium7 marks
Calculator
A space probe is powered by a radioisotope generator. Its power output \(t\) years after launch is modelled by \(P=P_0b^{t}\) watts. At launch the output is 470 W and after 6 years it is 400 W.
(a)
Find the value of \(b\).
(2)
(b)
Find the power output 20 years after launch.
(2)
(c)
A scientific instrument needs at least 250 W to operate. Find the time after launch at which the instrument stops operating.
(3)
Show mark scheme
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Notes
Part (a)
M1 sets \(470b^6=400\)
✓ 1
A1 \(b=0.973\)
✓ 1
Part (b)
M1 substitutes \(470b^{20}\) using unrounded \(b\)
✓ 1
A1 275 W
✓ 1
Part (c)
M1 sets \(470b^t=250\)
✓ 1
M1 \(t=\frac{\ln(250/470)}{\ln b}\)
✓ 1
A1 \(t=23.5\) years
✓ 1
Exam tip: Store the unrounded value of \(b\) in your calculator; using 0.973 changes later answers enough to lose accuracy.
342IBMA2S-044-05-101.7 Rational exponents and laws of logarithmsPaper 2Medium10 marks
Calculator
Two music-streaming platforms launch at the same time. The number of subscribers, in millions, \(t\) years after launch is modelled by Platform A: \(S_A=2.4\times 1.15^{t}\) Platform B: \(S_B=5.1e^{0.06t}\).
(a)
Find the number of subscribers to Platform A after 3 years.
(2)
(b)
Find the time taken for Platform B to reach 8 million subscribers.
(3)
(c)
Find the time at which the two platforms have the same number of subscribers.
(3)
(d)
Find the number of subscribers each platform has at this time.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 substitutes \(t=3\): \(2.4\times1.15^3\)
✓ 1
A1 3.65 million
✓ 1
Part (b)
M1 sets \(5.1e^{0.06t}=8\)
✓ 1
M1 valid method: \(t=\frac{1}{0.06}\ln\left(\frac{8}{5.1}\right)\), or uses a GDC solver or graph intersection with \(y=8\)
✓ 1
A1 \(t=7.50\) years
✓ 1
Part (c)
M1 sets \(2.4\times1.15^t=5.1e^{0.06t}\)
✓ 1
M1 valid method: takes logs to obtain \(t=\frac{\ln(5.1/2.4)}{\ln1.15-0.06}\), or uses a GDC solver or the intersection of the two graphs
✓ 1
A1 \(t=9.45\) years
✓ 1
Part (d)
M1 substitutes the unrounded \(t\) into either model
✓ 1
A1 8.99 million
✓ 1
Common error: Do not combine \(2.4\times1.15^t\) into \((2.4\times1.15)^t\). Only the 1.15 is raised to the power \(t\).
343IBMA2S-044-03-051.4 Financial applicationsPaper 2Medium7 marks
Calculator
A physiotherapist wants to invest £15 000 for 5 years to buy clinic equipment. She compares two accounts. Account A pays 4.2% per year, compounded annually. Account B pays a nominal annual rate of 4.1%, compounded monthly.
(a)
Find the value of the investment after 5 years in Account A.
(2)
(b)
Find the value of the investment after 5 years in Account B.
(2)
(c)
Find the nominal annual rate, compounded monthly, that Account B would need to offer for both accounts to give the same value after 5 years.
M1 sets up (1 + i/12)12 = 1.042, or 15 000(1 + i/12)60 = 18 425.95 M1 attempts to solve, e.g. 1 + i/12 = 1.0421/12 A1 i = 4.12%
✓ 3
Exam tip: More frequent compounding does not always win. Compare the final values: a slightly lower nominal rate compounded monthly can still give less than a higher rate compounded annually.
344IBMA2S-044-03-061.4 Financial applicationsPaper 2Medium7 marks
Calculator
An astronomy club invests $3200 in an account paying a nominal annual interest rate of r%, compounded quarterly. After 6 years the investment is worth $4100. No further deposits or withdrawals are made.
(a)
Find the value of r.
(3)
(b)
The club wants to buy a telescope costing $6000. Find the minimum number of complete quarters after the initial investment until the account first exceeds $6000.
(2)
(c)
Find the amount the club would have needed to invest initially, at the same rate, to have exactly $6000 after 6 years.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets up 3200(1 + r/400)24 = 4100 M1 attempts to solve, e.g. 1 + r/400 = 1.281251/24 = 1.01038… A1 r = 4.15
✓ 3
Part (b)
M1 sets up 3200(1.01038…)n > 6000, giving n > 60.9 A1 61 quarters
✓ 2
Part (c)
M1 P × 1.28125 = 6000, or P(1.01038…)24 = 6000 A1 $4682.93
✓ 2
Common error: With quarterly compounding, 6 years is 24 periods and the rate per period is r/4 percent. Using 6 as the number of periods makes r about four times too large.
345IBMA2S-044-03-071.4 Financial applicationsPaper 2Medium10 marks
Calculator
A professional footballer deposits $5000 into a retirement account at the start of every year. The account pays 3% interest per year, compounded annually, and interest is added at the end of each year. Let Vn be the value of the account at the end of year n.
(a)
Show that V1 = 5150.
(1)
(b)
Show that \(V_n = \frac{515000}{3}(1.03^n - 1)\).
(3)
(c)
Find the minimum number of years for the value of the account to exceed $100 000.
(3)
(d)
Find the total interest earned by the end of year 10.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 5000 × 1.03 = 5150
✓ 1
Part (b)
M1 recognises Vn = 5000(1.03) + 5000(1.03)2 + … + 5000(1.03)n M1 uses the geometric series sum with u1 = 5150 and ratio 1.03 A1 \(\frac{5150(1.03^n - 1)}{0.03} = \frac{515000}{3}(1.03^n - 1)\)
✓ 3
Part (c)
M1 sets up \(\frac{515000}{3}(1.03^n - 1) > 100000\) A1 n > 15.5 A1 16 years
✓ 3
Part (d)
M1 substitutes n = 10 into the formula for Vn A1 V10 = $59 039 (59 038.98…) A1 interest = 59 039 − 50 000 = $9039
✓ 3
Common error: Each deposit is made at the start of the year, so even the last deposit earns one year of interest. The first term of the series is 5150, not 5000.
346IBMA2S-044-03-091.4 Financial applicationsPaper 2Medium7 marks
Calculator
An electric car costs $42 000 when new. Its value falls by 15% each year, so at the end of each year it is worth 85% of its value at the end of the previous year. A petrol car costs $30 000 when new. Its value falls by 9% each year in the same way. Both cars are bought new on the same day.
(a)
Find the value of the electric car at the end of 3 years.
(2)
(b)
Find the number of complete years after which the electric car is first worth less than the petrol car.
(3)
(c)
Find the difference between the values of the two cars at the end of the year found in part (b).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 42 000 × 0.853 A1 $25 800 (25 793.25)
✓ 2
Part (b)
M1 sets up 42 000 × 0.85n < 30 000 × 0.91n, or compares the values year by year A1 n > 4.93, or after 4 years the values are 21 924 > 20 572 and after 5 years 18 636 < 18 721 A1 5 years
✓ 3
Part (c)
M1 finds both values at n = 5: 18 635.62 and 18 720.96 A1 $85.3
✓ 2
Exam tip: For part (b), rearrange to (0.85/0.91)ⁿ < 30/42 and take logarithms, or graph both value functions and find where they intersect. Then check n = 4 and n = 5 to confirm the first whole year that works.
347IBMA2S-044-03-101.4 Financial applicationsPaper 2Medium5 marks
Calculator
A medical research team needs USD 250 000 in 8 years' time to buy a new centrifuge. They plan to make a single deposit now into an account paying a nominal annual interest rate of 5.4%, compounded monthly.
(a)
Find the amount they must deposit now. Give your answer to the nearest dollar.
(3)
(b)
Find the effective annual interest rate, that is, the equivalent rate if interest were compounded once a year. Give your answer as a percentage correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses a monthly rate of 0.45% (0.054/12) and 96 compounding periods M1 sets up P(1.0045)96 = 250 000, or uses an equivalent TVM solver entry A1 USD 162 460
Common error: Part (a) asks for the present value, so divide 250 000 by the growth factor. Multiplying 250 000 by (1.0045)⁹⁶ finds a future value instead.
348IBMA3S-1-16r10-061.2 Arithmetic sequences and seriesPaper 2Medium6 marks
Calculator
A food scientist adds yeast to a starter culture. On day 1 she adds 2.0 g, and on each later day she adds 0.5 g more than on the previous day. The daily masses form an arithmetic sequence.
(a)
Find the mass of yeast added on day 12.
(2)
(b)
The total mass added reaches more than 100 g on day \(N\) for the first time. Find the total mass of yeast added by the end of day \(N\).
(4)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(2.0 + 11(0.5)\)
✓ 1
A1 7.5 g
✓ 1
Part (b)
M1 \(S_n = \frac{n}{2}(4 + 0.5(n-1))\), set greater than 100
✓ 1
M1 attempt to solve, e.g. \(S_{16} = 92\), \(S_{17} = 102\)
✓ 1
A1 \(N = 17\)
✓ 1
A1 102 g
✓ 1
Exam tip: Check the sum for the integer values either side of the solution to the inequality, as the solution of the equation is rarely a whole number.
349IBMA3S-1-16r10-091.2 Arithmetic sequences and seriesPaper 2Medium6 marks
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On the first day of an attempt on a ridge, a team climbs 600 m. On each later day they climb 85% of the distance climbed on the previous day, so the daily distances form a geometric sequence.
(a)
Find the greatest total distance the team could climb if they continued indefinitely.
(2)
(b)
The summit is 3800 m along the ridge from the start. Find the first day on which the total distance climbed exceeds 3800 m, and the total distance climbed by the end of that day, to the nearest metre.
(4)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(S_\infty = \frac{600}{1 - 0.85}\)
✓ 1
A1 4000 m
✓ 1
Part (b)
M1 \(4000(1 - 0.85^n) > 3800\)
✓ 1
M1 attempt to solve, e.g. \(0.85^n < 0.05\) or checking \(S_{18} = 3785\)
✓ 1
A1 day 19
✓ 1
A1 3818 m
✓ 1
Exam tip: Because the sum to infinity is 4000 m, the summit at 3800 m is reachable, which is why the inequality has a solution.
350IBMA3S-1-16r10-101.3 Geometric sequences and seriesPaper 2Medium11 marks
Calculator
A telecoms customer is comparing two tariffs. With Plan A the cost in month 1 is £15 and the cost increases by £2 each month. With Plan B the cost in month 1 is £10 and the cost increases by 15% each month. Let \(A_n\) and \(B_n\) be the costs, in £, in month \(n\).
(a)
Find \(A_{12}\).
(2)
(b)
Find the total cost of Plan A over the first 12 months.
(2)
(c)
Find the first month in which the monthly cost of Plan B is greater than the monthly cost of Plan A.
(3)
(d)
Find the total cost of Plan B over the first 12 months, to 2 decimal places.
(2)
(e)
Find how much less Plan B costs than Plan A over the first 12 months, to 2 decimal places.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(15 + 11(2)\)
✓ 1
A1 £37
✓ 1
Part (b)
M1 \(S_{12} = \frac{12}{2}(2(15) + 11(2))\)
✓ 1
A1 £312
✓ 1
Part (c)
M1 \(10(1.15)^{n-1} > 15 + 2(n-1)\)
✓ 1
M1 attempt to compare values, e.g. \(B_9 = 30.59 < A_9 = 31\)
✓ 1
A1 month 10 (\(B_{10} = 35.18\), \(A_{10} = 33\))
✓ 1
Part (d)
M1 \(S_{12} = \frac{10(1.15^{12} - 1)}{0.15}\)
✓ 1
A1 £290.02
✓ 1
Part (e)
M1 \(312 - 290.02\)
✓ 1
A1 £21.98
✓ 1
Common error: In geometric sequences the \(n\)th term uses \(r^{n-1}\), but the sum formula uses \(r^n\) for \(n\) terms.
351IBMA3S-1-16r10-121.2 Arithmetic sequences and seriesPaper 2Medium11 marks
Calculator
The total data, in GB, used by a customer in the first \(n\) days of a month is given by \(S_n = 2n^2 + 3n\). The daily usage forms an arithmetic sequence \(u_1, u_2, u_3, \ldots\).
(a)
Find \(u_1\).
(2)
(b)
Find \(u_2\) and the common difference, and write down an expression for \(u_n\).
(3)
(c)
Find the first day on which the daily usage is greater than 60 GB.
(2)
(d)
Find the first day on which the total usage exceeds 1000 GB, and the total usage by the end of that day.
(4)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(u_1 = S_1\)
✓ 1
A1 5 GB
✓ 1
Part (b)
M1 \(u_2 = S_2 - S_1 = 14 - 5\)
✓ 1
A1 \(u_2 = 9\), \(d = 4\)
✓ 1
A1 \(u_n = 4n + 1\)
✓ 1
Part (c)
M1 \(4n + 1 > 60\)
✓ 1
A1 \(n = 15\)
✓ 1
Part (d)
M1 \(2n^2 + 3n > 1000\)
✓ 1
M1 attempt to solve or compare, e.g. \(S_{21} = 945\)
✓ 1
A1 day 22
✓ 1
A1 1034 GB
✓ 1
Exam tip: The \(n\)th term of any sequence is \(u_n = S_n - S_{n-1}\), and \(u_1 = S_1\) separately.
352IBMA3S-1-16r11-091.4 Financial applicationsPaper 2Medium6 marks
Calculator
A library pays €250 into a savings account at the end of every month for 5 years (60 payments). The account pays a nominal annual interest rate of 4.8%, compounded monthly. Give money answers to the nearest cent.
(a)
Write down the monthly interest rate as a decimal and the number of payments.
(2)
(b)
Find the value of the account immediately after the last payment.
(3)
(c)
Find the total interest earned.
(1)
Show mark scheme
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Mark
Notes
Part (a)
A1 monthly rate \(=0.004\)
✓ 1
A1 60 payments
✓ 1
Part (b)
M1 payments grow for 59, 58, ..., 0 months: \(250(1.004^{59}+1.004^{58}+\dots+1)\)
✓ 1
M1 sum of a geometric series \(250\times\frac{1.004^{60}-1}{0.004}\)
✓ 1
A1 €16 915.04
✓ 1
Part (c)
A1 \(16\,915.04-60\times250=\) €1915.04
✓ 1
Exam tip: The first payment is made at the end of month 1, so it earns interest for 59 months, not 60.
353IBMA3S-1-16r11-101.2 Arithmetic sequences and seriesPaper 2Medium7 marks
Calculator
A wind farm is built in rows. The first row has 6 turbines, and each row has 4 more turbines than the row before it.
Each dot represents one turbine. Diagram not to scale
(a)
Find the number of turbines in row 7.
(2)
(b)
Show that the total number of turbines in the first \(n\) rows is \(2n^2+4n\).
(2)
(c)
Find the least number of rows needed for the farm to have at least 500 turbines.
(2)
(d)
Find the number of turbines in the 15th row.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_7=6+6\times4\)
✓ 1
A1 30
✓ 1
Part (b)
M1 \(S_n=\frac n2(2\times6+(n-1)\times4)\)
✓ 1
A1 \(=\frac n2(4n+8)=2n^2+4n\)
✓ 1
Part (c)
M1 \(2n^2+4n\ge500\)
✓ 1
A1 \(n=15\) (since \(S_{14}=448\) and \(S_{15}=510\))
✓ 1
Part (d)
A1 \(6+14\times4=62\)
✓ 1
Common error: Solve \(S_n\ge 500\) and then test whole numbers. Check both \(S_{14}\) and \(S_{15}\) rather than rounding blindly.
354IBMA3S-1-16r13-061.2 Arithmetic sequences and seriesPaper 2Medium7 marks
Calculator
The lengths of the bars of a xylophone form an arithmetic sequence. The longest bar is 42 cm and the 8th bar is 28 cm.
(a)
Find the common difference.
(2)
(b)
Find the total length of the first 14 bars.
(3)
(c)
Each bar must be at least 10 cm long. Find the greatest number of bars the xylophone can have.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(42+7d=28\)
✓ 1
A1 \(d=-2\)
✓ 1
Part (b)
M1 \(u_{14}=42+13(-2)=16\)
✓ 1
M1 \(S_{14}=\frac{14}{2}(42+16)\)
✓ 1
A1 406 cm
✓ 1
Part (c)
M1 \(42-2(n-1)\ge10\)
✓ 1
A1 \(n\le17\), so 17 bars
✓ 1
Exam tip: For 'greatest number' questions, write the inequality and check the boundary value, here a bar of exactly 10 cm is allowed.
355IBMA3S-1-16r13-071.1 Standard formPaper 2Medium7 marks
Calculator
A rice farm has \(1.8\times10^{5}\) plants. On average each plant produces \(9.5\times10^{2}\) grains and each grain has mass \(2.4\times10^{-2}\) g.
(a)
Find the total number of grains. Give your answer in the form \(a\times10^{k}\), where \(1\le a<10\).
(2)
(b)
Find the total mass of the grains in tonnes, correct to 3 significant figures. (1 tonne = \(10^{6}\) g.)
(3)
(c)
The rice sells for €310 per tonne. Using the unrounded mass from part (b), find the total income, to the nearest euro.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1.8\times10^{5}\times9.5\times10^{2}\)
✓ 1
A1 \(1.71\times10^{8}\)
✓ 1
Part (b)
M1 \(1.71\times10^{8}\times2.4\times10^{-2}\)
✓ 1
A1 \(4.104\times10^{6}\) g
✓ 1
A1 4.10 tonnes
✓ 1
Part (c)
M1 \(4.104\times310\)
✓ 1
A1 €1272
✓ 1
Common error: Do not use a rounded intermediate value when the question says to use the unrounded one; round only the final answer.
356IBMA3S-1-16r13-091.2 Arithmetic sequences and seriesPaper 2Medium6 marks
Calculator
In a vineyard, the numbers of vines in successive rows form an arithmetic sequence with common difference \(d\). Let \(u_k\) be the number of vines in row \(k\).
(a)
Show that \(u_{k-2}+u_{k-1}+u_k+u_{k+1}+u_{k+2}=5u_k\).
(5)
(b)
Five consecutive rows contain 135 vines in total. Write down the number of vines in the middle row of these five.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 write the terms in terms of \(u_k\) and \(d\)
✓ 1
A1 \((u_k-2d)+(u_k-d)+u_k+(u_k+d)+(u_k+2d)\)
✓ 1
M1 add the terms
✓ 1
A1 the \(d\) terms cancel, giving \(5u_k\)
✓ 1
R1 conclude that the sum equals \(5u_k\) for any \(k\) and any \(d\)
✓ 1
Part (b)
A1 \(135\div5=27\) vines
✓ 1
Exam tip: In a proof, use the middle term and the common difference so that terms cancel in pairs.
357IBMA3S-1-16r13-101.2 Arithmetic sequences and seriesPaper 2Medium11 marks
Calculator
After take-off, an aircraft gains 900 m of altitude in the first minute. In each following minute it gains 40 m less than in the previous minute, so the altitude gains form an arithmetic sequence.
(a)
Find the altitude gained in the 5th minute.
(2)
(b)
Find the total altitude gained in the first 10 minutes.
(2)
(c)
Find the least whole number of minutes after which the total altitude gained is at least 10000 m.
(4)
(d)
The model is valid only while the altitude gained in each minute is positive. Find the greatest total altitude gain predicted by the model.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(900+4(-40)\)
✓ 1
A1 740 m
✓ 1
Part (b)
M1 \(S_{10}=\frac{10}{2}\big(2(900)+9(-40)\big)\) or \(\frac{10}{2}(900+540)\)
✓ 1
A1 7200 m
✓ 1
Part (c)
M1 use of \(S_n=\frac n2\big(1800-40(n-1)\big)\), which simplifies to \(920n-20n^2\)
✓ 1
A1 \(S_{17}=9860\)
✓ 1
A1 \(S_{18}=10080\)
✓ 1
R1 \(S_{17}<10000\le S_{18}\), so 18 minutes
✓ 1
Part (d)
M1 solve \(900-40(n-1)>0\)
✓ 1
A1 \(u_{23}=20>0\) and \(u_{24}=-20<0\), so the last valid minute is the 23rd
✓ 1
A1 \(S_{23}=10580\) m
✓ 1
Exam tip: For 'at least' questions, evaluate the sum at the two neighbouring whole numbers and state which one crosses the target.
358IBMA3S-1-17r13-041.2 Arithmetic sequences and seriesPaper 2Medium6 marks
Calculator
During a vaccination drive, the number of patients seen each day by a hospital outpatient clinic forms an arithmetic sequence. On day 5 the clinic sees 34 patients and on day 12 it sees 76 patients.
(a)
Find the common difference and the number of patients seen on day 1.
(3)
(b)
The clinic's cumulative total of patients first exceeds 1000 on day \(N\). Find the number of patients seen on day \(N\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 form two equations, e.g. \(u_1+4d=34\) and \(u_1+11d=76\)
✓ 1
A1 \(d = 6\)
✓ 1
A1 \(u_1 = 10\)
✓ 1
Part (b)
M1 \(S_n = \frac{n}{2}(20+6(n-1)) = n(3n+7)\) and attempt to compare with 1000
✓ 1
A1 \(S_{17}=986\) and \(S_{18}=1098\), so \(N=18\)
✓ 1
A1 \(u_{18} = 10 + 17\times 6 = 112\)
✓ 1
Common error: Stopping at \(n\) when the question asks for a term or a sum at that \(n\). Reread what is asked for.
359IBMA3S-1-17r13-071.4 Financial applicationsPaper 2Medium7 marks
Calculator
A beekeeper records the honey harvested from a hive each year. The harvests form a geometric sequence. The harvest in year 1 is 36 kg, and the total of all the harvests, taken over all future years, is predicted to be 144 kg.
(a)
Find the common ratio of the sequence.
(2)
(b)
Find the harvest in year 6, correct to 3 significant figures.
(2)
(c)
Find the harvest in the first year in which it falls below 4.5 kg. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{36}{1-r} = 144\)
✓ 1
A1 \(r = 0.75\)
✓ 1
Part (b)
M1 \(36\times 0.75^5\)
✓ 1
A1 8.54 kg
✓ 1
Part (c)
M1 \(36\times 0.75^{n-1} < 4.5\)
✓ 1
M1 attempt to solve, e.g. \(0.75^{n-1}<0.125\) using logarithms or a table
✓ 1
A1 year 9 (\(u_8 = 4.81 > 4.5\) and \(u_9 < 4.5\)) with harvest \(36\times 0.75^8\) = 3.60 kg
✓ 1
Common error: Using \(n\) instead of \(n-1\) as the power: the first term is \(ar^0\).
360IBMA3S-1-17r13-081.2 Arithmetic sequences and seriesPaper 2Medium7 marks
Calculator
A Formula 1 team uploads sensor records after every lap. After \(n\) laps the total number of records uploaded is \(S_n = 3n^2 + 5n\).
(a)
Find the number of records uploaded in lap 1 and in lap 2.
(2)
(b)
Show that the number of records uploaded in lap \(n\) forms an arithmetic sequence and state its common difference.
R1 \(u_n - u_{n-1} = 6\), constant, so arithmetic with \(d=6\)
✓ 1
Part (c)
M1 \(6n + 2 = 98\)
✓ 1
A1 \(n = 16\)
✓ 1
Exam tip: The \(n\)th term of a sequence is the difference of consecutive partial sums, \(S_n - S_{n-1}\), for \(n\geq 2\).
361IBMA3S-1-17r13-111.3 Geometric sequences and seriesPaper 2Medium6 marks
Calculator
The concentration of an antibiotic in a patient's blood, in mg/L, is modelled by \(C = 80\times 2^{-t/6}\), where \(t\) is the time in hours after the dose is given.
(a)
Find the concentration after 9 hours. Give your answer to 3 significant figures.
(2)
(b)
Show that \(C = 15\) when \(t = 24 - 6\log_2 3\).
(3)
(c)
Find this value of \(t\) to 3 significant figures.
Exam tip: To solve for an exponent, isolate the power first and then take logarithms to the matching base.
362IBMA3S-1-16r14-031.3 Geometric sequences and seriesPaper 2Medium7 marks
Calculator
In autumn, the mass of nectar, in kg, that a bee colony collects on successive days forms a geometric sequence. On day 1 the colony collects 2.4 kg and each day it collects 90% of the amount collected on the previous day.
(a)
Find the mass of nectar collected on day 10. Give your answer to three significant figures.
(2)
(b)
Find the total mass of nectar collected during the first 10 days. Give your answer to three significant figures.
(3)
(c)
Find the total mass of nectar the colony would collect if the pattern continued indefinitely.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2.4 \times 0.9^{9}\)
✓ 1
A1 0.930 kg
✓ 1
Part (b)
M1 use of \(S_n = \frac{u_1(1-r^n)}{1-r}\)
✓ 1
A1 \(\frac{2.4(1-0.9^{10})}{1-0.9}\)
✓ 1
A1 15.6 kg
✓ 1
Part (c)
M1 \(S_\infty = \frac{u_1}{1-r}\), valid as \(|r|<1\)
✓ 1
A1 \(\frac{2.4}{0.1} = 24\) kg
✓ 1
Exam tip: Check that |r| < 1 before using the sum to infinity, and quote the condition in your working.
363IBMA3S-1-16r14-051.9 The binomial theoremPaper 2Medium7 marks
Calculator
The coefficient of \(x^2\) in the expansion of \((1+2x)^n\), where \(n \in \mathbb{N}\), \(n \geq 2\), is 112.
(a)
Find the value of \(n\).
(3)
(b)
Find the coefficient of \(x^3\) in the expansion.
(2)
(c)
By substituting a suitable value of \(x\) into the first three terms of the expansion, estimate \(1.02^8\). Give your answer to four decimal places.
Exam tip: To estimate (1 + a)ⁿ choose x so that (1 + 2x) equals the required base, then truncate after the stated number of terms.
364IBMA3S-1-16r14-091.9 The binomial theoremPaper 2Medium6 marks
Calculator
During a Formula 1 race the lap times of a car increase as its tyres wear. The lap times form a geometric sequence. Lap 1 takes 82.0 s and lap 11 takes 84.0 s.
(a)
Find the percentage increase in lap time from one lap to the next. Give your answer to three significant figures.
(3)
(b)
Find the total time, in seconds, for the first 30 laps. Use the unrounded value of \(r\) and give your answer to the nearest second.
M1 attempt to solve using logarithms or technology
✓ 1
A1 7.49 years
✓ 1
Common error: With monthly compounding, divide the annual rate by 12 and multiply the number of years by 12.
366IBMA3S-1-16r15-061.4 Financial applicationsPaper 2Medium6 marks
Calculator
A new train is bought for €2.4 million. Its value falls by 12% each year.
(a)
Find the value of the train after 5 years. Give your answer in € million to 3 significant figures.
(3)
(b)
Find the number of years, to 3 significant figures, for the value of the train to fall to €1 million.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 multiplier 0.88
✓ 1
A1 \(2.4\times0.88^{5}\)
✓ 1
A1 €1.27 million
✓ 1
Part (b)
M1 \(2.4\times0.88^{n}=1\)
✓ 1
M1 attempt to solve using logarithms or technology
✓ 1
A1 6.85 years
✓ 1
Exam tip: A fall of 12% means multiplying by 0.88 each year.
367IBMA3S-1-16r15-081.1 Standard formPaper 2Medium6 marks
Calculator
The mass of the Sun is \(1.99\times10^{30}\) kg and the mass of the Earth is \(5.97\times10^{24}\) kg. The Earth can be modelled as a sphere of radius \(6.37\times10^{6}\) m.
(a)
Find how many times greater the mass of the Sun is than the mass of the Earth. Give your answer in the form \(a\times10^{k}\) where \(1\le a<10\), to 3 significant figures.
(2)
(b)
Find the average density of the Earth, in kg m−3. Give your answer in the form \(a\times10^{k}\) to 3 significant figures.
Exam tip: Keep the unrounded volume in your calculator rather than retyping a rounded value.
368IBMA3S-1-16r15-091.3 Geometric sequences and seriesPaper 2Medium7 marks
Calculator
A new rail line carries 2400 passengers in month 1. The number of passengers each month forms a geometric sequence with common ratio 1.05.
(a)
Find the smallest month number \(n\) in which the number of passengers exceeds 5000.
(4)
(b)
Find the total number of passengers carried in the first 12 months. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2400\times1.05^{n-1}>5000\)
✓ 1
M1 attempt to solve using logarithms or technology
✓ 1
A1 \(n>16.04\dots\)
✓ 1
A1 \(n=17\)
✓ 1
Part (b)
M1 use of \(S_{12}=\frac{u_1(r^{12}-1)}{r-1}\)
✓ 1
A1 \(\frac{2400(1.05^{12}-1)}{0.05}\)
✓ 1
A1 38200
✓ 1
Common error: The \(n\)th term uses \(r^{n-1}\), not \(r^{n}\).
369IBMA3S-1-16r15-111.4 Financial applicationsPaper 2Medium7 marks
Calculator
An athlete pays £1200 into a savings account at the start of each year for 10 years. The account pays 3% compound interest per year, added at the end of each year.
Timeline not to scale
(a)
Find the value of the first deposit at the end of the 10 years. Give your answer to the nearest penny.
(2)
(b)
Show that the total value of all deposits at the end of the 10 years is the sum of a geometric series, and state its first term and common ratio.
(2)
(c)
Find the total value of the account at the end of the 10 years. Give your answer to the nearest penny.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1200\times1.03^{10}\)
✓ 1
A1 £1612.70
✓ 1
Part (b)
A1 deposits are worth \(1200\times1.03^{10}\), \(1200\times1.03^{9}\), … , \(1200\times1.03\)
✓ 1
A1 first term \(1200\times1.03\) and common ratio 1.03 (accept first term 1236 and ratio 1.03)
✓ 1
Part (c)
M1 \(S_{10}=\frac{1236(1.03^{10}-1)}{1.03-1}\)
✓ 1
A1 correct substitution
✓ 1
A1 £14169.35
✓ 1
Common error: Each deposit earns interest for a different number of years; list the powers before summing.
370IBMA3S-1-17r15-021.3 Geometric sequences and seriesPaper 2Medium7 marks
Calculator
A star survey catalogues 240 stars in its first year. Each year after that, the number of stars catalogued in that year is 15% greater than in the previous year. The numbers catalogued each year form a geometric sequence.
(a)
Find the number of stars catalogued in year 5, giving your answer to the nearest whole number.
(2)
(b)
Find the first year in which more than 1000 stars are catalogued.
(3)
(c)
Find the total number of stars catalogued in the first 10 years, to the nearest whole number.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(240\times1.15^{4}\)
✓ 1
A1 \(419.76\ldots\) so \(420\)
✓ 1
Part (b)
M1 \(240\times1.15^{n-1}>1000\) or equation
✓ 1
A1 \(n-1>10.2\ldots\) (or value in year 11 is \(970.9\) and in year 12 is \(1116.6\))
✓ 1
A1 year 12
✓ 1
Part (c)
M1 \(\frac{240(1.15^{10}-1)}{0.15}\)
✓ 1
A1 \(4872.9\ldots\) so \(4873\)
✓ 1
Common error: Writing \(1.15^{n}\) instead of \(1.15^{n-1}\) for the \(n\)th term shifts the answer by one year.
371IBMA3S-1-17r15-051.2 Arithmetic sequences and seriesPaper 2Medium8 marks
Calculator
A cyclist follows a training plan. In week 1 she cycles 20 km. In each later week she cycles 8% further than in the previous week. Give all answers to 3 significant figures.
(a)
Find the distance she cycles in week 10.
(2)
(b)
Find the total distance she cycles in the first 12 weeks.
(2)
(c)
In week \(k\) the distance she cycles first exceeds 60 km. Find the distance she cycles in week \(k\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(20\times1.08^{9}\)
✓ 1
A1 \(40.0\) km
✓ 1
Part (b)
M1 \(\frac{20(1.08^{12}-1)}{0.08}\)
✓ 1
A1 \(380\) km
✓ 1
Part (c)
M1 \(20\times1.08^{k-1}>60\) or equation
✓ 1
A1 \(k-1>14.27\ldots\)
✓ 1
A1 \(k=16\)
✓ 1
A1 \(20\times1.08^{15}=63.4\) km
✓ 1
Exam tip: When a question asks for a term 'first exceeding' a value, find the integer \(k\) first, then substitute it back to get the required quantity.
372IBMA3S-1-17r15-071.7 Rational exponents and laws of logarithmsPaper 2Medium7 marks
Calculator
The electrical power, \(P\) watts, from a space probe's power source \(t\) years after launch is modelled by \(P=470\times2^{-t/87.7}\).
(a)
Find the power after 20 years, to 3 significant figures.
(2)
(b)
Find the time at which the power first falls to 300 W. Give your answer to 3 significant figures.
(3)
(c)
The model can be written as \(P=470e^{-kt}\). Find the value of \(k\), giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(470\times2^{-20/87.7}\)
✓ 1
A1 \(401\) W
✓ 1
Part (b)
M1 \(470\times2^{-t/87.7}=300\)
✓ 1
M1 use of logarithms or GDC, e.g. \(t=87.7\log_2\frac{470}{300}\)
✓ 1
A1 \(56.8\) years
✓ 1
Part (c)
M1 \(2^{-t/87.7}=e^{-t\ln2/87.7}\) so \(k=\frac{\ln 2}{87.7}\)
✓ 1
A1 \(0.00790\)
✓ 1
Common error: Rewriting a base-2 exponential as base \(e\) means multiplying the exponent by \(\ln 2\), not by 2.
373IBMA3S-1-17r15-101.8 Infinite geometric seriesPaper 2Medium8 marks
Calculator
The pointing corrections, in arc-seconds, applied to a telescope form a geometric sequence. The 3rd correction is 32 and the 6th correction is \(-4\).
(a)
Find the common ratio and the first term.
(4)
(b)
Find the sum to infinity of the sequence, giving your answer as a fraction.
(2)
(c)
Find the sum of the first 10 corrections.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(r^{3}=\frac{-4}{32}\)
✓ 1
A1 \(r=-\frac12\)
✓ 1
M1 \(u_1=\frac{32}{r^{2}}\)
✓ 1
A1 \(u_1=128\)
✓ 1
Part (b)
M1 \(\frac{128}{1-(-\frac12)}\)
✓ 1
A1 \(\frac{256}{3}\)
✓ 1
Part (c)
M1 \(\frac{128(1-(-0.5)^{10})}{1-(-0.5)}\)
✓ 1
A1 \(85.25\)
✓ 1
Exam tip: A negative common ratio gives terms that alternate in sign, so \(r^{3}\) must have the same sign as the ratio of the two terms.
374IBMA3S-1-17r16-011.1 Standard formPaper 2Medium4 marks
Calculator
A wind farm has 48 turbines. On a particular day each turbine delivers an average power of 3.5 × 106 W.
(a)
Find the total average power of the wind farm. Give your answer in the form a × 10k, where 1 ≤ a < 10 and k ∈ ℤ.
(2)
(b)
Calculate the energy delivered by the wind farm in this day (86 400 s). Give your answer in joules, correct to 3 significant figures, in the form a × 10k.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 48 × 3.5 × 106
✓ 1
A1 1.68 × 108 W
✓ 1
Part (b)
M1 1.68 × 108 × 86 400
✓ 1
A1 1.45 × 1013 J
✓ 1
Common error: When a calculator shows 1.45152E13, the standard form is 1.45 × 10¹³; do not copy the calculator display into your answer.
375IBMA3S-1-17r16-031.4 Financial applicationsPaper 2Medium4 marks
Calculator
A town library places €15 000 in an account that pays a nominal annual interest rate of 3.2%, compounded quarterly. No money is added or withdrawn.
(a)
Find the value of the account after 6 years. Give your answer to the nearest euro.
(3)
(b)
Calculate the interest earned over the 6 years. Give your answer to the nearest euro.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 15 000(1 + 0.032/4)4×6
✓ 1
A1 correct substitution of n = 24 and 0.008
✓ 1
A1 €18 161
✓ 1
Part (b)
A1 €3161 (ECF from (a); 18 161 − 15 000)
✓ 1
Common error: With quarterly compounding the rate per period is 3.2 ÷ 4 and the number of periods is 4 × 6, not 6.
376IBMA3S-1-17r16-041.4 Financial applicationsPaper 2Medium5 marks
Calculator
A library fund of €8000 is invested in an account that pays r% interest per year, compounded annually. The library wants the fund to grow to €10 000 after exactly 5 years.
(a)
Find the value of r, correct to 3 significant figures.
(3)
(b)
A different account offers 4.2% per year, compounded annually. Find the value of the €8000 after 5 years in this account, to the nearest euro, and state whether it reaches €10 000.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 8000(1 + r/100)5 = 10 000
✓ 1
M1 1 + r/100 = 1.251/5 (or use of solver)
✓ 1
A1 r = 4.56
✓ 1
Part (b)
M1 8000 × 1.0425
✓ 1
A1 €9827, which is less than €10 000, so it does not reach the target
✓ 1
Exam tip: To undo a power of 5 raise both sides to the power 1/5; do not divide by 5.
377IBMA3S-1-17r16-051.4 Financial applicationsPaper 2Medium6 marks
Calculator
A wind-farm operator pays €250 000 into a decommissioning fund that earns 2.5% interest per year, compounded annually.
(a)
Find the value of the fund after 5 years, to the nearest euro.
(2)
(b)
Find the smallest whole number of years after which the fund first exceeds €300 000.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 250 000 × 1.0255
✓ 1
A1 €282 852
✓ 1
Part (b)
M1 250 000 × 1.025n > 300 000
✓ 1
M1 attempt to solve using logarithms or a graph/table
✓ 1
A1 n > 7.38
✓ 1
A1 n = 8 years
✓ 1
Common error: An answer such as 7.38 years must be rounded up: after 7 years the fund is still below the target.
378IBMA3S-1-17r16-061.4 Financial applicationsPaper 2Medium6 marks
Calculator
A wind turbine is bought for €1.8 million. Its value falls by 8% each year.
(a)
Find the value of the turbine after 6 years. Give your answer correct to 3 significant figures.
(2)
(b)
Find the number of complete years after which the value of the turbine first falls below €900 000.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 1.8 × 106 × 0.926
✓ 1
A1 €1 090 000 (1.09 × 106)
✓ 1
Part (b)
M1 1.8 × 106 × 0.92n < 900 000
✓ 1
M1 attempt to solve by logarithms or graph
✓ 1
A1 n > 8.31
✓ 1
A1 9 years
✓ 1
Exam tip: A fall of 8% per year means multiplying by 0.92 each year, not by 0.08.
379IBMA3S-1-16r19-031.3 Geometric sequences and seriesPaper 2Medium8 marks
Calculator
A swimmer trains every day. On day 1 she swims 400 m. On each following day she swims 5% further than on the previous day, so the daily distances form a geometric sequence.
(a)
Find the distance she swims on day 10. Give your answer to the nearest metre.
(2)
(b)
Find the first day on which she swims more than 1000 m.
(3)
(c)
Find the total distance she swims in the first 10 days. Give your answer to the nearest metre.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(u_n=400\times1.05^{n-1}\), with \(n=10\)
✓ 1
A1 \(621\) m
✓ 1
Part (b)
M1 \(400\times1.05^{n-1}>1000\)
✓ 1
M1 attempt to solve, e.g. using logarithms: \(n-1>18.78\)
✓ 1
A1 day \(20\)
✓ 1
Part (c)
M1 use of \(S_n=\frac{u_1(r^n-1)}{r-1}\) with \(n=10\), \(r=1.05\)
✓ 1
A1 \(\frac{400(1.05^{10}-1)}{0.05}\) seen or implied
✓ 1
A1 \(5031\) m
✓ 1
Common error: The 10th term uses the power \(n-1=9\), not \(10\).
380IBMA3S-1-16r19-041.4 Financial applicationsPaper 2Medium5 marks
Calculator
A library invests €8000 in an account that pays a nominal annual interest rate of 4.2%, compounded quarterly. No money is added or withdrawn.
(a)
Find the smallest number of complete years after which the investment is worth more than €12 000.
(3)
(b)
Find the value of the investment after this number of years. Give your answer to the nearest cent (2 decimal places).
M1 attempt to solve, e.g. by logarithms or a graph: \(4n>38.8\)
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A1 \(10\) years
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Part (b)
M1 \(8000\times1.0105^{40}\) (ECF from (a))
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A1 €12 149.06
✓ 1
Exam tip: Quarterly compounding means \(4n\) periods at a quarter of the annual rate, here \(\frac{4.2}{4}=1.05\)% per quarter.
381IBMA3S-1-16r19-061.3 Geometric sequences and seriesPaper 2Medium8 marks
Calculator
A wind farm produces 24.0 GWh of electricity in its first year. Because of wear on the turbines, its output falls by 1.5% each year, so the yearly outputs form a geometric sequence.
(a)
Find the output in year 8. Give your answer to 3 significant figures.
(2)
(b)
Find the first year in which the output is below 20.0 GWh.
(3)
(c)
Find the total output over the first 20 years. Give your answer to 3 significant figures.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(24.0\times0.985^{7}\)
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A1 \(21.6\) GWh
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Part (b)
M1 \(24\times0.985^{n-1}<20\)
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M1 attempt to solve, e.g. by logarithms: \(n-1>12.06\)
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A1 year \(14\)
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Part (c)
M1 use of the sum formula with \(r=0.985\) and \(n=20\)
✓ 1
A1 \(\frac{24(1-0.985^{20})}{0.015}\) seen or implied
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A1 \(417\) GWh
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Common error: A fall of 1.5% each year means a common ratio of \(0.985\), not \(0.015\).
382IBMA3S-1-16r19-091.7 Rational exponents and laws of logarithmsPaper 2Medium7 marks
Calculator
The concentration of chlorine in a swimming pool is 3.0 mg/L at 08:00. The concentration falls by 12% each hour, so the concentration \(C\) mg/L, \(t\) hours after 08:00, is modelled by \(C=3.0\times0.88^{t}\).
(a)
Find the concentration at 13:00. Give your answer to 3 significant figures.
(2)
(b)
The pool must be treated when the concentration first falls below 1.2 mg/L. Find how many hours after 08:00 this happens. Give your answer to 3 significant figures.
(3)
(c)
The model can be written as \(C=3.0e^{-kt}\). Find the value of \(k\) to 3 significant figures.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 substituting \(t=5\)
✓ 1
A1 \(1.58\) mg/L
✓ 1
Part (b)
M1 \(3.0\times0.88^{t}=1.2\)
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M1 attempt to solve using logarithms or a graph: \(t=\frac{\ln 0.4}{\ln 0.88}\)
✓ 1
A1 \(7.17\) hours
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Part (c)
M1 equating \(0.88=e^{-k}\) or \(0.88^t=e^{-kt}\)
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A1 \(k=-\ln 0.88=0.128\)
✓ 1
Exam tip: A 12% fall each hour is a multiplier of 0.88; to find a time, take logarithms of both sides.
383IBMA3S-1-16r20-051.1 Standard formPaper 2Medium7 marks
Calculator
During a disease outbreak, the number of people infected, \(N\), is modelled by \(N=25\times 2^{\frac{t}{3}}\), where \(t\) is the number of days since the first report.
(a)
Find the value of \(N\) when \(t=12\).
(2)
(b)
Find the value of \(N\) when \(t=20\). Give your answer to the nearest whole number.
(2)
(c)
Find the value of \(t\) when \(N=10\,000\). Give your answer to 3 significant figures.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(25\times 2^{4}\)
✓ 1
A1 \(N=400\)
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Part (b)
M1 \(25\times 2^{\frac{20}{3}}\)
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A1 \(N=2540\)
✓ 1
Part (c)
M1 \(2^{\frac{t}{3}}=400\)
✓ 1
M1 takes logarithms: \(\frac{t}{3}=\frac{\ln 400}{\ln 2}\) (or uses a solver)
✓ 1
A1 \(t=25.9\)
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Common error: Do not divide \(10\,000\) by \(25\) and then divide by \(2\) to find \(t\); the unknown is in the exponent, so logarithms are needed.
384IBMA3S-1-16r20-081.9 The binomial theoremPaper 2Medium6 marks
Calculator
The power output of a wind turbine is proportional to the cube of the wind speed. The wind speed increases by a fraction \(x\) of its original value, so the power is multiplied by \((1+x)^3\).
(a)
Expand \((1+x)^3\).
(2)
(b)
For small \(x\), the first two terms give a good approximation. Use them to estimate the percentage increase in power when the wind speed increases by \(2\%\).
(2)
(c)
Find the exact percentage increase in power when the wind speed increases by \(2\%\). Give your answer to 3 significant figures.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 uses the binomial theorem or multiplies out
✓ 1
A1 \(1+3x+3x^2+x^3\)
✓ 1
Part (b)
M1 \(x=0.02\) in \(3x\) (increase \(\approx 3x\))
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A1 \(6\%\)
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Part (c)
M1 \(1.02^3-1\) (or \(0.06+3(0.02)^2+(0.02)^3\))
✓ 1
A1 \(6.12\%\)
✓ 1
Exam tip: Check an approximation against the exact value; the difference here comes from the terms in \(x^2\) and \(x^3\) that were ignored.
385IBMA3S-1-16r20-101.4 Financial applicationsPaper 2Medium6 marks
Calculator
The annual revenue, in thousands of euros, of two retailers \(t\) years after 2020 is modelled by Store A: \(R_A=5000\times 1.03^{t}\) and Store B: \(R_B=3200\times 1.07^{t}\).
(a)
Find the revenue of Store A when \(t=4\). Give your answer to the nearest thousand euros.
(2)
(b)
Find the value of \(t\) for which the two retailers have equal revenue. Give your answer to 3 significant figures.
(4)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(5000\times 1.03^4\)
✓ 1
A1 €5628 thousand
✓ 1
Part (b)
M1 \(5000\times 1.03^t=3200\times 1.07^t\)
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M1 rearranges to \(\left(\frac{1.07}{1.03}\right)^t=1.5625\)
Exam tip: To compare two exponential models, divide one by the other so that the unknown appears in a single exponent.
386IBMA3S-1-16r20-141.7 Rational exponents and laws of logarithmsPaper 2Medium6 marks
Calculator
Consider the equation \(9^{x}-4\times 3^{x}-21=0\).
(a)
Show that the substitution \(y=3^{x}\) transforms the equation into \(y^2-4y-21=0\).
(2)
(b)
Hence solve the equation for \(x\). Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(9^x=(3^x)^2=y^2\)
✓ 1
A1 \(y^2-4y-21=0\)
✓ 1
Part (b)
M1 factorises \((y-7)(y+3)=0\)
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A1 \(y=7\) or \(y=-3\)
✓ 1
R1 rejects \(y=-3\) because \(3^x>0\) for all real \(x\)
✓ 1
A1 \(x=\log_3 7=1.77\)
✓ 1
Common error: Always state why a root of the quadratic in \(y\) is rejected: an exponential such as \(3^x\) can never be negative.
387IBMA3S-1-17r20-051.3 Geometric sequences and seriesPaper 2Medium6 marks
Calculator
A survey of a young star cluster counts \(N\) detected exoplanets after \(t\) years. The model \(N=12\times1.3^{t}\) is proposed, where \(t\ge0\).
(a)
Use the model to find \(N\) when \(t=5\), giving your answer to 3 significant figures.
(2)
(b)
Find the value of \(t\) for which \(N=500\). Give your answer to 3 significant figures.
(3)
(c)
Give one reason why the model may be unsuitable for large values of \(t\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute \(t=5\): \(12\times1.3^5\)
✓ 1
A1 44.6
✓ 1
Part (b)
M1 \(12\times1.3^t=500\), so \(1.3^t=\dfrac{500}{12}\)
✓ 1
M1 take logarithms: \(t=\dfrac{\ln(500/12)}{\ln 1.3}\)
✓ 1
A1 \(t=14.2\)
✓ 1
Part (c)
R1 for example: the number of planets in the cluster is finite, so growth cannot continue indefinitely
✓ 1
Exam tip: Isolate the exponential term before taking logarithms; take care not to take the log of the 12 on its own.
388IBMA3S-1-17r20-091.4 Financial applicationsPaper 2Medium6 marks
Calculator
A rail operator holds a maintenance reserve fund of €25000. The value of the fund is modelled by \(V=25000\times1.032^{t}\) euros, where \(t\) is the time in years and need not be a whole number.
(a)
Find the value of the fund after 10 years, to the nearest euro.
(2)
(b)
Find how many years it takes for the fund to reach €40000. Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(25000\times1.032^{10}\)
✓ 1
A1 €34256
✓ 1
Part (b)
M1 \(25000\times1.032^t=40000\)
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M1 \(1.032^t=1.6\)
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M1 take logarithms: \(t=\dfrac{\ln1.6}{\ln1.032}\)
✓ 1
A1 \(t=14.9\) years
✓ 1
Exam tip: Write the equation, isolate the power, then take logs; a calculator solver is acceptable but show the equation first.
389IBMA3S-1-17r22-051.2 Arithmetic sequences and seriesPaper 2Medium7 marks
Calculator
A game gives a player a login bonus of points each day. The bonuses on day 1, day 2, day 3, … form an arithmetic sequence. The bonus on day 5 is 34 points and the total of the first 12 bonuses is 516 points.
(a)
Find the bonus on day 1 and the daily increase.
(4)
(b)
Find the value of the first bonus that exceeds 200 points.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_1+4d=34\)
✓ 1
M1 \(\frac{12}{2}(2u_1+11d)=516\), so \(2u_1+11d=86\)
✓ 1
A1 \(d=6\) (from \(3d=18\))
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A1 \(u_1=10\)
✓ 1
Part (b)
M1 \(10+6(n-1)>200\) (or equation)
✓ 1
A1 \(n=33\) (since \(n>32.67\))
✓ 1
A1 202 points
✓ 1
Common error: Subtracting two linear equations eliminates u₁ neatly: double the first equation so both contain 2u₁.
390IBMA3S-1-17r22-061.2 Arithmetic sequences and seriesPaper 2Medium6 marks
Calculator
A patient takes a 100 mg tablet every 8 hours. Just before each tablet is taken, 35% of the drug that was in the body immediately after the previous tablet is still present. Let \(A_n\) be the amount of drug in the body, in mg, immediately after the \(n\)th tablet is taken. It can be shown that \(A_n=100+100(0.35)+100(0.35)^2+\dots+100(0.35)^{n-1}\).
(a)
Show that \(A_3=147.25\).
(2)
(b)
Use the formula for the sum of a geometric series to find \(A_6\). Give your answer to 4 significant figures.
(2)
(c)
Find the value that \(A_n\) approaches as \(n\) becomes very large. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(A_2=100+0.35\times100=135\)
✓ 1
A1 \(A_3=100+0.35\times135=147.25\)
✓ 1
Part (b)
M1 \(A_6=\frac{100(1-0.35^6)}{1-0.35}\)
✓ 1
A1 153.6 mg
✓ 1
Part (c)
M1 \(S_\infty=\frac{100}{1-0.35}\)
✓ 1
A1 154 mg (153.846…)
✓ 1
Exam tip: 'Very large n' signals the sum to infinity, which is valid because the common ratio 0.35 lies between −1 and 1.
391IBMA3S-1-17r22-101.3 Geometric sequences and seriesPaper 2Medium7 marks
Calculator
A crew is laying a water pipeline of total length 504 m. On day 1 they lay 90 m of pipe, and on each following day they lay 6 m less than on the previous day. The pipeline is completed on day \(N\) exactly, with \(N\) a whole number.
(a)
Find the value of \(N\).
(5)
(b)
Find the length of pipe laid on the last day.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(S_N=\frac N2(2\times90+(N-1)(-6))=504\)
✓ 1
A1 \(N^2-31N+168=0\) (or equivalent)
✓ 1
M1 solve the quadratic
✓ 1
A1 \(N=7\) or \(N=24\)
✓ 1
R1 reject \(N=24\) since the daily length would become negative (day 17 onwards)
✓ 1
Part (b)
M1 \(90-6\times(7-1)\) (ECF)
✓ 1
A1 54 m
✓ 1
Common error: A quadratic from a sum formula often gives two solutions; check each against the context (here, lengths cannot be negative).
392IBMA3S-1-17r22-111.3 Geometric sequences and seriesPaper 2Medium8 marks
Calculator
A community saves for a new water tank. It deposits €400 in month 1, and each later monthly deposit is 5% larger than the deposit in the previous month.
(a)
Find the deposit in month 12. Give your answer to the nearest cent.
(2)
(b)
Find the smallest number of months after which the total amount deposited exceeds €10 000.
(4)
(c)
Find the total amount deposited after that number of months. Give your answer to the nearest euro.
M1 \(\frac{400(1.05^{17}-1)}{0.05}\) using \(n=17\)
✓ 1
A1 €10336
✓ 1
Exam tip: Solving an inequality for n: take logarithms of both sides, then round up to the next whole number of months.
393IBMA3S-1-16r25-041.7 Rational exponents and laws of logarithmsPaper 2Medium5 marks
Calculator
During roasting, the mass of moisture in a batch of coffee beans falls by 8% each minute. Initially the beans contain 1.50 kg of moisture. The moisture mass after \(t\) minutes is modelled by \(M = 1.50 \times 0.92^t\) kg.
(a)
Find the mass of moisture remaining after 5 minutes. Give your answer to 3 significant figures.
(2)
(b)
Treating \(t\) as continuous, find the time at which the mass of moisture first equals 0.500 kg. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1.50 \times 0.92^5\)
✓ 1
A1 0.989 kg
✓ 1
Part (b)
M1 \(1.50 \times 0.92^t = 0.500\), so \(0.92^t = \frac{1}{3}\)
✓ 1
M1 \(t = \frac{\ln(1/3)}{\ln 0.92}\) (or equivalent use of logarithms)
✓ 1
A1 \(t = 13.2\) minutes
✓ 1
Exam tip: Isolate the exponential term first, then take logarithms of both sides.
394IBMA3S-1-16r25-051.4 Financial applicationsPaper 2Medium6 marks
Calculator
The yearly maintenance cost of a bridge is €40000 in year 1. Each year the cost is 3.5% higher than in the previous year, so the yearly costs form a geometric sequence.
(a)
Find the maintenance cost in year 12, correct to the nearest euro.
(2)
(b)
Find the total maintenance cost over the first 20 years, correct to the nearest euro.
(2)
(c)
Find the percentage increase in the yearly cost from year 1 to year 12. Give your answer to 3 significant figures.
Common error: The 12th term uses the ratio to the power 11, not 12.
395IBMA3S-1-16r25-081.1 Standard formPaper 2Medium7 marks
Calculator
A radioisotope generator on a space probe produces 120 W of power at launch. The power falls by 0.8% each year, so the power after \(n\) years is \(P = 120 \times 0.992^n\) watts.
(a)
Find the power produced after 10 years, correct to the nearest watt.
(2)
(b)
The instruments on the probe need at least 76 W to operate. Find the number of complete years for which the generator can power them.
(3)
(c)
Find the time, to 3 significant figures, for the power to fall to half of its initial value.
Common error: When a question asks for complete years, round down a time for which the condition still holds rather than rounding to the nearest whole number.
396IBMA3S-1-16r25-091.7 Rational exponents and laws of logarithmsPaper 2Medium7 marks
Calculator
During roasting, the amount of an aroma compound in a batch of coffee is modelled by \(A = 5.0 \times 1.4^t\) micrograms, where \(t\) is the time in minutes after the roast begins.
(a)
Find the amount of the compound after 3 minutes, correct to 3 significant figures.
(2)
(b)
Find the time at which the amount of the compound reaches 40 micrograms. Give your answer to 3 significant figures.
(3)
(c)
Find the time taken for the amount of the compound to double. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(5.0 \times 1.4^3\)
✓ 1
A1 13.7 micrograms
✓ 1
Part (b)
M1 \(5.0 \times 1.4^t = 40\), so \(1.4^t = 8\)
✓ 1
M1 \(t = \frac{\ln 8}{\ln 1.4}\)
✓ 1
A1 6.18 minutes
✓ 1
Part (c)
M1 \(1.4^t = 2\), so \(t = \frac{\ln 2}{\ln 1.4}\)
✓ 1
A1 2.06 minutes
✓ 1
Exam tip: The doubling time of an exponential model does not depend on the starting amount, which is why the 5.0 cancels.
397IBMA3S-1-18r29-071.4 Financial applicationsPaper 2Medium5 marks
Calculator
A swimming club buys equipment for £14 500. Its value falls by 12% each year.
(a)
Find the value of the equipment after 3 years, to the nearest pound.
(2)
(b)
Find the number of complete years after which the value first falls below £5000.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of 14 500 × 0.883
✓ 1
A1 £9881
✓ 1
Part (b)
M1 14 500 × 0.88n < 5000, or a table of values
✓ 1
A1 n > 8.33… (value after 8 years £5215 is too large)
✓ 1
A1 n = 9 years (value after 9 years is £4589)
✓ 1
Exam tip: A fall of 12% each year means multiplying by 0.88 each year, not subtracting 0.12 of the original value.
398IBMA3S-1-18r29-091.4 Financial applicationsPaper 2Medium5 marks
Calculator
Elena has €6000 to invest for 5 years. Account A pays 4.2% per year compounded annually. Account B pays a nominal annual rate of 4.1% compounded monthly. Determine which account gives the greater value after 5 years, and find the difference in value. Give money values to 2 decimal places.
(a)
(5)
Show mark scheme
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Mark
Notes
Part (a)
M1 account A: 6000 × 1.0425
✓ 1
A1 €7370.38
✓ 1
M1 account B: 6000(1 + 0.041/12)60
✓ 1
A1 €7362.58
✓ 1
A1 account A is greater, by €7.80
✓ 1
Common error: A nominal rate compounded monthly is divided by 12 and applied 12 times a year; it is not the same as an annual rate of 4.1%.
399IBMA3S-1-18r29-131.1 Standard formPaper 2Medium5 marks
Calculator
A library has digitised 1.8 × 105 books. Each book has on average 3.2 × 102 pages, and each scanned page uses 2.4 × 106 bytes of storage.
(a)
Find the total storage used. Give your answer in the form a × 10k where 1 ≤ a < 10 and k ∈ ℤ, with a correct to 3 significant figures.
(3)
(b)
The library stores the files on servers that each hold 4.0 × 1013 bytes. Find the least number of servers needed. Use the unrounded value from (a).
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 1.8 × 105 × 3.2 × 102 × 2.4 × 106
✓ 1
A1 13.824 × 1013
✓ 1
A1 1.38 × 1014 bytes
✓ 1
Part (b)
M1 1.3824 × 1014 ÷ 4.0 × 1013 = 3.456
✓ 1
A1 4 servers
✓ 1
Common error: Servers come in whole numbers: round 3.456 up to 4, not to the nearest whole number 3.
400IBMA3S-1-19r29-051.2 Arithmetic sequences and seriesPaper 2Medium7 marks
Calculator
A fashion label launches a scarf. Sales in week 1 are 240 scarves, and in each later week sales are 85% of the sales in the week before.
(a)
Find the total number of scarves the label can expect to sell in the long term.
(2)
(b)
Find the first week in which sales are below 20 scarves.
(3)
(c)
Find the percentage of the long-term total that is sold in the first 10 weeks. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(S_\infty=\frac{240}{1-0.85}\)
✓ 1
A1 1600 scarves
✓ 1
Part (b)
M1 \(240\times0.85^{n-1}<20\)
✓ 1
M1 \(n-1>\frac{\ln(20/240)}{\ln0.85}=15.3\)
✓ 1
A1 week 17
✓ 1
Part (c)
M1 \(\frac{S_{10}}{S_\infty}=1-0.85^{10}\)
✓ 1
A1 80.3%
✓ 1
Exam tip: The nth term is \(u_1r^{n-1}\), not \(u_1r^n\). Test your n by substituting it into the term formula.
401IBMA3S-1-19r29-091.7 Rational exponents and laws of logarithmsPaper 2Medium6 marks
Calculator
The stock of a species in a managed lake is modelled by \(N=N_0e^{-kt}\), where \(N\) is the number of fish, \(t\) is the time in years and \(k>0\). The stock falls from 9600 fish at \(t=0\) to 6100 fish at \(t=5\).
(a)
Find the value of \(k\), giving your answer to 3 significant figures.
(3)
(b)
Find the time for the stock to halve. Give your answer in years to 3 significant figures, using the unrounded value of \(k\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(6100=9600e^{-5k}\)
✓ 1
M1 \(k=\frac{1}{5}\ln\frac{9600}{6100}\)
✓ 1
A1 \(k=0.0907\)
✓ 1
Part (b)
M1 \(\frac12=e^{-kt}\)
✓ 1
M1 \(t=\frac{\ln2}{k}\)
✓ 1
A1 \(t=7.64\) years
✓ 1
Exam tip: Use the unrounded k from your calculator memory in later parts: a rounded k can shift the third significant figure.
402IBMA3S-1-18r31-021.4 Financial applicationsPaper 2Medium5 marks
Calculator
A game studio invests €25 000 in an account that pays a nominal annual interest rate of 3.6%, compounded quarterly.
(a)
Calculate the value of the investment after 5 years. Give your answer to the nearest euro.
(2)
(b)
The studio wants the investment to be worth €40 000 after 8 years. A different account pays interest compounded annually at a rate of \(r\%\) per year. Find the value of \(r\), giving your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(25000\left(1+\frac{3.6}{400}\right)^{20}\) (correct use of the compound interest formula with \(k=4\), \(n=5\))
Exam tip: For compound interest compounded \(k\) times a year, the rate per period is the annual rate divided by \(k\) and the number of periods is \(kn\). Here that is 0.9% per quarter for 20 quarters.
403IBMA3S-1-18r31-081.4 Financial applicationsPaper 2Medium5 marks
Calculator
A gaming PC is bought for €1200. Its value falls by the same percentage each year. After 4 years it is worth €702.
(a)
Find the annual percentage rate of depreciation. Give your answer to 3 significant figures.
(3)
(b)
Find the value of the PC after 6 years. Use the unrounded annual multiplier from (a). Give your answer to the nearest euro.
Exam tip: If a quantity falls by r% each year the yearly multiplier is \(1-\frac{r}{100}\). Keep the unrounded multiplier in your calculator for later parts.
404IBMA3S-1-18r31-111.3 Geometric sequences and seriesPaper 2Medium12 marks
Calculator
A water company pumps water from a borehole. On day 1 it pumps 5000 m3. Each day the volume pumped is 6% less than the volume pumped the day before.
(a)
Find the volume pumped on day 10. Give your answer to the nearest m3.
(2)
(b)
Find the total volume pumped during the first 30 days. Give your answer to 3 significant figures.
(2)
(c)
The company continued pumping for ever at this rate. Find the total volume that would be pumped. Give your answer to 3 significant figures.
(2)
(d)
The aquifer can safely supply 79 000 m3. Find the least number of days after which the total volume pumped exceeds 79 000 m3.
(4)
(e)
Find the volume pumped on the day found in (d). Give your answer to the nearest m3.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(5000\times0.94^{9}\)
✓ 1
A1 2865 m3
✓ 1
Part (b)
M1 \(S_{30}=\frac{5000(1-0.94^{30})}{1-0.94}\)
✓ 1
A1 70 300 m3
✓ 1
Part (c)
M1 \(S_\infty=\frac{5000}{1-0.94}\)
✓ 1
A1 83 300 m3
✓ 1
Part (d)
M1 \(\frac{5000(1-0.94^n)}{0.06}>79000\)
✓ 1
M1 \(0.94^n<0.052\) (using the unrounded sum to infinity \(83\,333.\overline{3}\))
✓ 1
M1 solves using logarithms or a table of values, e.g. \(n>47.8\)
✓ 1
A1 \(n=48\) days
✓ 1
Part (e)
M1 \(5000\times0.94^{47}\) (ECF from (d))
✓ 1
A1 273 m3
✓ 1
Common error: Day \(n\) uses the power \(n-1\) in \(u_n=u_1r^{n-1}\). For day 48 the power is 47, not 48.
405IBMA3S-1-18r31-121.4 Financial applicationsPaper 2Medium11 marks
Calculator
A game studio invests €50 000 in an account that pays a nominal annual interest rate of 4.8%, compounded monthly.
(a)
Find the value of the investment after 3 years. Give your answer to the nearest euro.
(2)
(b)
Show that the effective annual interest rate is 4.91%, correct to 3 significant figures.
(3)
(c)
Find the least whole number of months after which the investment exceeds €80 000.
(3)
(d)
The studio considers another account, with the interest compounded monthly, so that the investment reaches €80 000 after exactly 8 years. Find the nominal annual interest rate needed. Give your answer as a percentage to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(50000\left(1+\frac{4.8}{1200}\right)^{36}\)
✓ 1
A1 €57 728
✓ 1
Part (b)
M1 \(\left(1+\frac{4.8}{1200}\right)^{12}\)
✓ 1
M1 \(=1.04907\ldots\)
✓ 1
A1 effective annual rate \(=4.91\%\) (stated to 3 s.f.)
✓ 1
Part (c)
M1 \(50000(1.004)^{m}>80000\)
✓ 1
M1 \((1.004)^m>1.6\), solved using logarithms or a table, e.g. \(m>117.7\)
Exam tip: When a question says 'least whole number', solve the equation first, then round up to the next integer so that the condition is satisfied.
406IBMA3S-1-17r32-051.3 Geometric sequences and seriesPaper 2Medium7 marks
Calculator
After intense exercise, an athlete's blood lactate concentration is modelled by \(C=1.5\times2^{t/4}\) mmol per litre, where \(t\) is the time in minutes after the start of the effort.
(a)
Find the concentration when \(t=6\). Give your answer to 3 significant figures.
(2)
(b)
Find the value of \(t\) for which \(C=12\).
(2)
(c)
Find the time at which the concentration first reaches 25 mmol per litre. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1.5\times2^{1.5}\)
✓ 1
A1 4.24 mmol per litre
✓ 1
Part (b)
M1 \(2^{t/4}=8=2^{3}\)
✓ 1
A1 \(t=12\) minutes
✓ 1
Part (c)
M1 \(2^{t/4}=\frac{25}{1.5}\)
✓ 1
M1 take logarithms: \(\frac{t}{4}=\frac{\ln(25/1.5)}{\ln 2}\)
✓ 1
A1 \(t=16.2\) minutes
✓ 1
Exam tip: To solve for an exponent that is not a whole number, take logarithms of both sides and use \(\log a^{x}=x\log a\).
407IBMA3S-1-17r32-061.5 Exponents and introduction to logarithmsPaper 2Medium8 marks
Calculator
Astronomers compare the brightness of stars using apparent magnitude. For two stars with magnitudes \(m_1\) and \(m_2\) and received fluxes \(F_1\) and \(F_2\), \(m_2-m_1=2.5\log_{10}\!\left(\frac{F_1}{F_2}\right)\).
(a)
Star P delivers 100 times the flux of star Q. Find \(m_Q-m_P\).
(2)
(b)
Two stars with the same luminosity \(L\) are at distances \(d_A\) and \(d_B\). The flux from a star is \(F=\frac{L}{4\pi d^{2}}\). Show that \(m_B-m_A=5\log_{10}\!\left(\frac{d_B}{d_A}\right)\).
(3)
(c)
Star X has a magnitude 1.2 greater than star Y. Find the value of \(\frac{F_Y}{F_X}\), correct to 3 significant figures.
Exam tip: A magnitude difference is a logarithm of a flux ratio, so to find the ratio you must undo the logarithm with a power of 10.
408IBMA3S-1-17r32-071.8 Infinite geometric seriesPaper 2Medium7 marks
Calculator
A tennis ball is dropped from a height of 3.0 m onto a hard court. After each bounce it rises to 0.64 of the height it reached before that bounce. Let \(h_n\) be the height reached after the \(n\)th bounce, so \(h_1=1.92\) m.
(a)
Find the height reached after the 3rd bounce, in metres, to 3 significant figures.
(2)
(b)
Find the exact value of the sum of all the rebound heights \(h_1+h_2+h_3+\cdots\) in metres.
(3)
(c)
Hence find the total distance travelled by the ball, from the moment it is dropped until it comes to rest. Use the exact value from part (b) and give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(h_3=3\times0.64^{3}\) or \(1.92\times0.64^{2}\)
✓ 1
A1 0.786 m
✓ 1
Part (b)
A1 geometric series with \(u_1=1.92\) and \(r=0.64\)
M1 \(3+2\times\frac{16}{3}\) (the ball goes up and down after each bounce)
✓ 1
A1 13.7 m
✓ 1
Common error: After the first drop the ball travels each rebound height twice (up and down), so the total distance is not simply 3 plus the sum of the rebound heights.
409IBMA3S-1-17r32-111.8 Infinite geometric seriesPaper 2Medium8 marks
Calculator
A train begins to brake. In the first second it travels 24 m and in the second second it travels 21.6 m. The distances travelled in successive seconds form a geometric sequence.
(a)
Find the common ratio.
(2)
(b)
Find the distance travelled in the 5th second, correct to 3 significant figures.
(2)
(c)
Find the greatest total distance that the train could travel while braking, assuming the pattern continues indefinitely.
(2)
(d)
Find the total distance travelled in the first 10 seconds of braking, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{21.6}{24}\)
✓ 1
A1 0.9
✓ 1
Part (b)
M1 \(24\times0.9^{4}\)
✓ 1
A1 15.7 m
✓ 1
Part (c)
M1 \(S_\infty=\frac{24}{1-0.9}\)
✓ 1
A1 240 m
✓ 1
Part (d)
M1 \(S_{10}=\frac{24(1-0.9^{10})}{1-0.9}\)
✓ 1
A1 156 m
✓ 1
Exam tip: Use \(S_\infty=\frac{u_1}{1-r}\) only when \(|r|<1\); for a finite number of terms use \(S_n=\frac{u_1(1-r^{n})}{1-r}\).
410IBMA3S-1-18r38-071.1 Standard formPaper 2Medium5 marks
Calculator
A telecoms company buys a satellite dish for €46 000. Its value falls by 18% each year.
(a)
Find the value of the dish after 3 years, to the nearest euro.
(2)
(b)
Find the time, in years, for the value to fall to €10 000. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 46 000 × 0.823
✓ 1
A1 €25 363
✓ 1
Part (b)
M1 46 000 × 0.82t = 10 000
✓ 1
M1 t = ln(10 000/46 000) / ln 0.82
✓ 1
A1 7.69 years
✓ 1
Exam tip: A fall of 18% is a multiplier of 0.82, not 0.18.
411IBMA3S-1-18r38-101.7 Rational exponents and laws of logarithmsPaper 2Medium5 marks
Calculator
The data traffic on a telecoms network is modelled by T = T0ekt, where t is the time in years. The traffic doubles every 2.4 years.
(a)
Find the value of k, to 3 significant figures.
(2)
(b)
Find the time for the traffic to triple. Use the unrounded value of k and give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 2T0 = T0e2.4k, so 2.4k = ln 2
✓ 1
A1 k = 0.289
✓ 1
Part (b)
M1 ekt = 3
✓ 1
M1 t = ln 3 / k = 2.4 ln 3 / ln 2
✓ 1
A1 3.80 years
✓ 1
Exam tip: Use the unrounded k (or the exact form ln 2 / 2.4) in later parts to avoid rounding drift.
412IBMA3S-1-18r41-031.3 Geometric sequences and seriesPaper 2Medium5 marks
Calculator
A renewable-energy company buys a solar inverter for €2400. Its value falls by 12% each year.
(a)
Find the value of the inverter after 5 years. Give your answer to the nearest cent.
(2)
(b)
Find the time, in years, for the value to fall to €1000. Give your answer to 3 significant figures.
Exam tip: A fall of 12% each year means multiplying by 0.88 each year, not subtracting 0.12.
413IBMA3S-1-18r41-051.3 Geometric sequences and seriesPaper 2Medium6 marks
Calculator
A wind farm is being built in stages. In month 1, 4 turbines are installed, in month 2, 7 are installed, in month 3, 10 are installed, and so on. The numbers installed each month form an arithmetic sequence.
Each dot represents one turbine installed. Diagram not to scale
(a)
Find the total number of turbines installed in the first 15 months.
(2)
(b)
The total number of turbines installed first exceeds 1200 in month \(n\). Find the number of turbines installed in month \(n\).
M1 \(\frac{n}{2}\left(8+3(n-1)\right)>1200\) (equation or inequality; accept a table of values)
✓ 1
A1 \(n>27.46\) (or \(S_{27}=1161\) and \(S_{28}=1246\))
✓ 1
A1 \(n=28\)
✓ 1
A1 \(u_{28}=4+27\times3=85\) turbines
✓ 1
Exam tip: For 'first exceeds' questions, solve the equation first and then round up to the next whole number, and check by testing the two nearest values.
414IBMA3S-1-18r41-101.1 Standard formPaper 2Medium7 marks
Calculator
A solar farm operator plans to meet a regional demand of \(1.68\times10^{13}\) W using identical solar farms, each producing \(4.2\times10^{8}\) W. Each farm covers an area of \(8.0\times10^{6}\) m2.
(a)
Find the number of farms needed. Give your answer in the form \(a\times10^k\), where \(1\le a<10\) and \(k\in\mathbb{Z}\).
(2)
(b)
Find the total area covered by these farms, in m2, in the form \(a\times10^k\), where \(1\le a<10\) and \(k\in\mathbb{Z}\).
(2)
(c)
A country has an area of \(9.6\times10^{6}\) km2. Given that 1 km2 = \(10^{6}\) m2, find the area in part (b) as a percentage of the area of the country. Give your answer to 3 significant figures.
Common error: 1 km² is 10⁶ m², not 10³ m². Convert area units by squaring the length conversion factor.
415IBMA3S-1-18r41-131.2 Arithmetic sequences and seriesPaper 2Medium6 marks
Calculator
A homeowner pays £7000 for solar panels. The panels save £540 on electricity bills in year 1. Because the panels degrade, the saving falls by £12 each year, so the yearly savings form an arithmetic sequence.
(a)
Find the total savings over the first 20 years.
(2)
(b)
Find the number of years after which the total savings first exceed the £7000 cost.
M1 \(\frac{n}{2}\left(1080-12(n-1)\right)>7000\) (equation, inequality or table of values)
✓ 1
A1 \(n>15.4\) (or \(S_{15}=6840\) and \(S_{16}=7200\))
✓ 1
A1 \(n=16\)
✓ 1
Part (c)
A1 \(u_{16}=540-15\times12=\text{£}360\)
✓ 1
Exam tip: The common difference is negative here. Substitute −12 carefully and use the whole-number rule: round up the solution of the equation.
416IBMA3S-1-17r42-041.8 Infinite geometric seriesPaper 2Medium7 marks
Calculator
On day 1 of an acclimatisation trek, a mountaineer climbs 600 m. On each later day she climbs 0.8 times the height she climbed on the previous day.
(a)
Find the greatest total height she could climb if the trek continued indefinitely.
(2)
(b)
Find the least number of days after which the total height climbed exceeds 2900 m.
(3)
(c)
After 10 days, find the difference between the total height climbed and your answer to (a). Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(S_\infty=\dfrac{600}{1-0.8}\)
✓ 1
A1 \(3000\) m
✓ 1
Part (b)
M1 \(3000\left(1-0.8^{n}\right)>2900\)
✓ 1
M1 \(0.8^{n}<\dfrac1{30}\), so \(n>\dfrac{\ln(1/30)}{\ln 0.8}=15.24\ldots\) (inequality reverses on dividing by \(\ln 0.8<0\)); trial values \(n=15\) (2894 m) and \(n=16\) (2916 m) are also accepted
Common error: Dividing both sides of an inequality by ln 0.8, a negative number, reverses the inequality sign.
417IBMA3S-1-17r42-071.5 Exponents and introduction to logarithmsPaper 2Medium6 marks
Calculator
In food science, acidity is measured by pH \(=-\log_{10}[\mathrm{H^{+}}]\), where \([\mathrm{H^{+}}]\) is the hydrogen-ion concentration in mol dm−3.
(a)
Lemon juice has \([\mathrm{H^{+}}]=3.6\times10^{-3}\). Find its pH to 3 significant figures.
(2)
(b)
A sample of vinegar has pH 2.9. Find its \([\mathrm{H^{+}}]\) to 3 significant figures.
(2)
(c)
Find how many times greater the hydrogen-ion concentration of the lemon juice is than that of the vinegar. Use the unrounded value from part (b) and give your answer to 3 significant figures.
Common error: A lower pH means a higher concentration of hydrogen ions, so do not compare the pH values directly as if they were concentrations.
418IBMA3S-1-17r42-121.4 Financial applicationsPaper 2Medium10 marks
Calculator
Operator A is a telecoms company with 40 000 subscribers at time \(t=0\), where \(t\) is in years. After 3 years it has 69 120 subscribers. Assume that the number of subscribers grows according to \(S=40\,000\,r^{t}\).
(a)
Show that \(r=1.2\).
(3)
(b)
Find the time for Operator A to reach 200 000 subscribers. Give your answer to 3 significant figures.
(3)
(c)
Operator B has 150 000 subscribers at \(t=0\), and its number of subscribers falls by 8% each year. Find the time at which the two operators have equal numbers of subscribers. Give your answer to 3 significant figures.
M1 \(\left(\dfrac{1.2}{0.92}\right)^{t}=3.75\), so \(t=\dfrac{\ln3.75}{\ln(1.2/0.92)}\)
✓ 1
A1 \(t=4.97\) years (4.9746...)
✓ 1
Common error: A fall of 8% each year is a multiplier of 0.92, not 0.08.
419IBMA3S-1-17r42-131.5 Exponents and introduction to logarithmsPaper 2Medium5 marks
Calculator
Atmospheric pressure \(P\) kPa at height \(h\) metres above sea level is modelled by \(P=101\,\mathrm{e}^{-kh}\), where \(k\) is a positive constant. At a height of 4000 m the pressure is 61.3 kPa.
(a)
Find the value of \(k\) to 3 significant figures.
(3)
(b)
Using the unrounded value of \(k\), find the height at which the pressure is 35.0 kPa. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(61.3=101\,\mathrm{e}^{-4000k}\)
✓ 1
M1 \(-4000k=\ln\dfrac{61.3}{101}\)
✓ 1
A1 \(k=1.25\times10^{-4}\) (1.2484...×10−4)
✓ 1
Part (b)
M1 \(h=\dfrac{\ln(101/35.0)}{k}\)
✓ 1
A1 \(h=8490\) m (8489.4...)
✓ 1
Exam tip: Keep the unrounded value of k in your calculator memory so that later parts are not affected by rounding.
420IBMA3S-1-17r43-041.4 Financial applicationsPaper 2Medium6 marks
Calculator
A Formula 1 team invests €250 000 of sponsorship income in an account that pays a nominal annual interest rate of 3.6%, compounded monthly.
(a)
Find the value of the investment after 5 years. Give your answer to the nearest euro.
(2)
(b)
A different account compounds interest quarterly at a nominal annual rate of \(r\%\). The team wants €250 000 to grow to €320 000 in 6 years. Find the value of \(r\), correct to 3 significant figures.
M1 \(\left(1+\frac{r}{400}\right)^{24}=1.28\) and take the 24th root
✓ 1
A1 \(\frac{r}{400}=0.010339\ldots\)
✓ 1
A1 \(r=4.14\)
✓ 1
(accept a correct finance-solver method)
—
Exam tip: With a finance solver, state the values entered: N, I%, PV, FV, P/Y and C/Y, so method marks can be awarded.
421IBMA3S-1-17r43-071.1 Standard formPaper 2Medium6 marks
Calculator
In a hospital laboratory, a disinfectant is applied to a bacterial culture. The number of bacteria, \(N\), after \(t\) hours is modelled by \(N=4.0\times10^{6}\times0.82^{t}\).
(a)
Find the number of bacteria after 5 hours. Give your answer in the form \(a\times10^{k}\), where \(1\le a<10\), \(a\) to 3 significant figures.
(2)
(b)
Find the time, in hours, for the number of bacteria to fall to 1000. Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(4.0\times10^{6}\times0.82^{5}\)
✓ 1
A1 \(1.48\times10^{6}\)
✓ 1
Part (b)
M1 \(4.0\times10^{6}\times0.82^{t}=1000\)
✓ 1
M1 \(0.82^{t}=2.5\times10^{-4}\)
✓ 1
M1 take logarithms: \(t=\frac{\log(2.5\times10^{-4})}{\log 0.82}\)
✓ 1
A1 \(t=41.8\) hours
✓ 1
Exam tip: To solve an equation with the unknown in an index, isolate the exponential term and take logarithms of both sides.
422IBMA3S-1-17r43-081.1 Standard formPaper 2Medium6 marks
Calculator
A hospital blood bank stores a sample in which the concentration of red blood cells is \(5.2\times10^{6}\) cells per microlitre. A patient has \(5.0\times10^{6}\) microlitres of blood. Each red blood cell has mass \(2.7\times10^{-11}\) g.
(a)
Calculate the number of red blood cells in the patient's blood. Give your answer in the form \(a\times10^{k}\) where \(1\le a<10\) and \(k\in\mathbb{Z}\).
(2)
(b)
Find the total mass of these red blood cells, in grams, in the form \(a\times10^{k}\) where \(1\le a<10\) and \(k\in\mathbb{Z}\).
(2)
(c)
The patient's body mass is \(7.0\times10^{4}\) g. Find the mass of red blood cells as a percentage of body mass, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(5.2\times10^{6}\times5.0\times10^{6}\)
✓ 1
A1 \(2.6\times10^{13}\)
✓ 1
Part (b)
M1 \(2.6\times10^{13}\times2.7\times10^{-11}\)
✓ 1
A1 \(7.02\times10^{2}\) g
✓ 1
Part (c)
M1 \(\frac{702}{7.0\times10^{4}}\times100\)
✓ 1
A1 1.00%
✓ 1
Common error: When multiplying numbers in standard form, add the powers of ten and then check that the leading number is still between 1 and 10.
423IBMA3S-1-17r43-091.4 Financial applicationsPaper 2Medium5 marks
Calculator
A beekeeper buys hive equipment for £1800. Its value decreases by 12% each year.
(a)
Find the value of the equipment after 5 years, to the nearest pound.
(2)
(b)
A second set of equipment costs £2400 and is worth £1100 after 6 years. Assuming a constant annual percentage depreciation, find that annual rate, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1800\times0.88^{5}\)
✓ 1
A1 £950
✓ 1
Part (b)
M1 \(2400\,r^{6}=1100\) where \(r\) is the annual multiplier
Common error: A multiplier of 0.878 means a depreciation of \(1-0.878\), not 0.878; subtract from 1 at the end.
424IBMA3S-1-19r44-051.7 Rational exponents and laws of logarithmsPaper 2Medium7 marks
Calculator
During a race a Formula 1 tyre loses grip. A grip index is 100 on lap 1 and falls by 1.5% on each following lap, so the indices on successive laps form a geometric sequence.
(a)
Find the grip index on lap 20. Give your answer to 3 significant figures.
(2)
(b)
Find the first lap on which the grip index is below 70.
(3)
(c)
Find the sum of the grip indices over laps 1 to 10. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 identify r = 0.985 and use u20 = 100 × 0.98519
✓ 1
A1 75.0
✓ 1
Part (b)
M1 set up 100 × 0.985n−1 < 70
✓ 1
M1 solve by logarithms, n − 1 > ln 0.7 ÷ ln 0.985 ≈ 23.6, or by trial of values
✓ 1
A1 lap 25
✓ 1
Part (c)
M1 use S10 = 100(1 − 0.98510) ÷ (1 − 0.985)
✓ 1
A1 935
✓ 1
Exam tip: A decrease of 1.5% each step gives a common ratio of 0.985, not 0.015 and not 1.015.
425IBMA3S-1-19r44-061.4 Financial applicationsPaper 2Medium6 marks
Calculator
A beekeeper buys an extractor for €3200. Its value falls by 12% each year, so its value at the end of each year forms a geometric sequence.
(a)
Find the value of the extractor after 4 years. Give your answer to the nearest euro.
(3)
(b)
A second extractor is bought for €3200. Its value after 6 years is €1500. Find the annual percentage rate at which its value falls. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of a ratio of 0.88
✓ 1
M1 3200 × 0.884
✓ 1
A1 €1919
✓ 1
Part (b)
M1 set up 3200(1 − r/100)6 = 1500, or 3200r6 = 1500 for a ratio r
✓ 1
M1 take the sixth root, (1500/3200)1/6 = 0.8814
✓ 1
A1 11.9%
✓ 1
Common error: A fall of 12% per year over 4 years is not a 48% fall; each year's fall is a percentage of the previous year's value.
426IBMA3S-1-19r44-101.3 Geometric sequences and seriesPaper 2Medium6 marks
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A Formula 1 team signs a sponsorship contract. The sponsor pays €2.4 million in year 1, and each year's payment is 4% more than the previous year's payment.
(a)
Find the payment in year 6. Give your answer to the nearest €1000.
(3)
(b)
Find the total amount paid over the first 10 years. Give your answer in millions of euros to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 identify a geometric sequence with r = 1.04
✓ 1
M1 2 400 000 × 1.045
✓ 1
A1 €2 920 000
✓ 1
Part (b)
M1 use S10 = 2.4(1.0410 − 1) ÷ (1.04 − 1)
✓ 1
A1 28.81…
✓ 1
A1 €28.8 million
✓ 1
Exam tip: Year 6 is the sixth term, so the power of the ratio is 5, not 6.
427IBMA3S-1-19r46-051.4 Financial applicationsPaper 2Medium5 marks
Calculator
A data centre operator invests €250 000 in a fund paying 4.2% per annum nominal interest, compounded monthly.
(a)
Find the value of the investment after 5 years, to the nearest euro.
(2)
(b)
Find how many years it takes for the investment to double in value. Give your answer to 3 significant figures.
M1 use of logarithms: \(12n\ln\left(1+\frac{4.2}{1200}\right)=\ln 2\)
✓ 1
A1 \(n=16.5\) years
✓ 1
Exam tip: The doubling time does not depend on the amount invested, so the €250 000 can be replaced by any starting value.
428IBMA3S-1-19r46-061.4 Financial applicationsPaper 2Medium6 marks
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An archaeology department receives a grant of €8000 and invests it in an account paying a nominal annual interest rate of \(r\%\), compounded monthly. After 5 years the account holds €9600.
(a)
Find the value of \(r\), giving your answer to 3 significant figures.
(4)
(b)
Using the unrounded value of \(r\), find the value of the account after 2 years, to the nearest cent.
M1 \(1+\frac{r}{1200}=1.2^{1/60}\) (or use of a GDC solver)
✓ 1
A1 correct unrounded value, \(r=3.6519\ldots\)
✓ 1
A1 \(r=3.65\)
✓ 1
Part (b)
M1 \(8000\left(1+\frac{r}{1200}\right)^{24}\) with unrounded \(r\) (equivalently \(8000\times1.2^{0.4}\))
✓ 1
A1 €8605.23
✓ 1
Common error: Rounding \(r\) to 3.65 before part (b) changes the final cent value, so keep the unrounded value in your calculator.
429IBMA3S-1-19r46-071.4 Financial applicationsPaper 2Medium6 marks
Calculator
A cycling club has a fund of €5000 earning 4% interest per year, compounded annually. The club wants to buy a tandem that now costs €5400, and whose price rises by 2.5% each year.
(a)
Find the value of the fund after 3 years.
(2)
(b)
Find the time, in years to 3 significant figures, at which the value of the fund first equals the price of the tandem.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(5000\times 1.04^3\)
✓ 1
A1 €5624.32
✓ 1
Part (b)
M1 \(5000\times1.04^t=5400\times1.025^t\)
✓ 1
M1 rearrange to \(\left(\frac{1.04}{1.025}\right)^t=1.08\)
✓ 1
M1 take logarithms: \(t=\frac{\ln 1.08}{\ln(1.04/1.025)}\)
✓ 1
A1 \(t=5.30\) years
✓ 1
Exam tip: When two quantities grow at different rates, gather the powers on one side and take the logarithm of a single ratio.
430IBMA3S-1-19r46-101.4 Financial applicationsPaper 2Medium5 marks
Calculator
The amount of carbon-14 remaining in organic material decays according to the model \(N=N_0\left(\frac{1}{2}\right)^{t/5730}\), where \(t\) is the time in years since the organism died and \(N_0\) is the initial amount.
(a)
A wooden artefact is found to contain 62% of the carbon-14 it originally had. Find the age of the artefact, to the nearest year.
(3)
(b)
Find the percentage of the original carbon-14 remaining in a sample that died 10 000 years ago. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.62=\left(\frac{1}{2}\right)^{t/5730}\)
✓ 1
M1 use of logarithms: \(t=5730\times\frac{\ln 0.62}{\ln 0.5}\)
Common error: The exponent is \(t/5730\), so the decay constant is built into the half-life; do not divide the percentage by 5730.
431IBMA3S-1-19r46-111.7 Rational exponents and laws of logarithmsPaper 2Medium6 marks
Calculator
After a cooling failure in a data centre, the temperature of a server room, \(T\) °C, is modelled by \(T=40-22e^{-0.12t}\), where \(t\) is the time in minutes after the failure.
(a)
Write down the temperature at the moment of the failure.
(1)
(b)
Find the temperature 10 minutes after the failure, to 3 significant figures.
(2)
(c)
Find the time at which the temperature reaches 35 °C. Give your answer in minutes to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 18 °C
✓ 1
Part (b)
M1 \(40-22e^{-1.2}\)
✓ 1
A1 33.4 °C
✓ 1
Part (c)
M1 \(35=40-22e^{-0.12t}\) leading to \(e^{-0.12t}=\frac{5}{22}\)
✓ 1
M1 take natural logarithms: \(-0.12t=\ln\frac{5}{22}\)
✓ 1
A1 12.3 minutes
✓ 1
Exam tip: Isolate the exponential term before taking logarithms; taking ln of the whole equation too early does not work.
432IBMA3S-1-17r47-051.2 Arithmetic sequences and seriesPaper 2Medium11 marks
Calculator
A swimmer swims 800 m on day 1 of a training plan. On each later day she swims 4% further than on the previous day, so the daily distances form a geometric sequence.
(a)
Write down the common ratio and find the distance she swims on day 10, to the nearest metre.
(3)
(b)
Show that the total distance swum in the first \(n\) days is \(20000(1.04^n - 1)\) metres.
(2)
(c)
Find the least number of days after which the total distance swum exceeds 20000 m.
(4)
(d)
Find the total distance swum in the first 18 days, to the nearest metre.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 common ratio 1.04
✓ 1
M1 \(800 \times 1.04^{9}\)
✓ 1
A1 1139 m
✓ 1
Part (b)
M1 use of \(S_n = \frac{u_1(r^n - 1)}{r - 1}\) with \(u_1 = 800\), \(r = 1.04\)
M1 use logarithms, \(n > \frac{\ln 2}{\ln 1.04}\) (\(= 17.67\ldots\)), or a table or graph of values
✓ 1
A1 \(n = 18\)
✓ 1
Part (d)
M1 \(20000(1.04^{18} - 1)\)
✓ 1
A1 20516 m
✓ 1
Common error: When solving \(r^n > k\) with logarithms, round the value of \(n\) up to the next whole number, not to the nearest one.
433IBMA3S-1-17r47-061.7 Rational exponents and laws of logarithmsPaper 2Medium6 marks
Calculator
The installed capacity \(C\) megawatts (MW) of a regional wind farm is modelled by \(C = 12 \times 1.15^{t}\), where \(t\) is the time in years since the wind farm began operating.
(a)
Find the capacity after 5 years, correct to three significant figures.
(2)
(b)
Find the time at which the capacity reaches 100 MW. Give your answer in years, correct to three significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(12 \times 1.15^{5}\)
✓ 1
A1 24.1 MW
✓ 1
Part (b)
M1 \(12 \times 1.15^{t} = 100\)
✓ 1
M1 \(1.15^{t} = \frac{100}{12}\)
✓ 1
M1 take logarithms, \(t = \frac{\ln(100/12)}{\ln 1.15}\)
✓ 1
A1 \(t = 15.2\) years
✓ 1
Exam tip: Isolate the exponential term first, then take logarithms of both sides so that the unknown exponent comes down as a multiplier.
434IBMA3S-1-17r47-081.4 Financial applicationsPaper 2Medium5 marks
Calculator
A library's digital archive holds 250 GB of data. Four years later it holds 410 GB. Assume that the archive grows by a constant percentage \(r\%\) each year.
(a)
Find the value of \(r\), correct to three significant figures.
(3)
(b)
Using the unrounded annual growth factor from part (a), predict the size of the archive 10 years after it held 250 GB. Give your answer in GB, correct to three significant figures.
M1 \(250 \times 1.64^{\frac{10}{4}}\) or \(250 \times 1.1316\ldots^{10}\)
✓ 1
A1 861 GB
✓ 1
Exam tip: To undo a power of 4 take a fourth root (power \(\frac{1}{4}\)); dividing by 4 is a common mistake.
435IBMA3S-1-17r47-141.1 Standard formPaper 2Medium7 marks
Calculator
The acidity of swimming pool water is measured by its pH, where \(\text{pH} = -\log_{10}[\text{H}^+]\) and \([\text{H}^+]\) is the concentration of hydrogen ions in mol dm−3.
(a)
A sample of pool water has \([\text{H}^+] = 2.5 \times 10^{-8}\). Find its pH, correct to three significant figures.
(2)
(b)
Another sample has a pH of 7.2. Find \([\text{H}^+]\) for this sample. Give your answer in the form a × 10k where 1 ≤ a < 10, correct to three significant figures.
(2)
(c)
Adding a chemical lowers the pH of the pool from 7.8 to 7.2. Find the factor by which \([\text{H}^+]\) increases, correct to three significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(-\log_{10}(2.5 \times 10^{-8})\)
✓ 1
A1 7.60
✓ 1
Part (b)
M1 \([\text{H}^+] = 10^{-7.2}\)
✓ 1
A1 6.31 × 10−8 mol dm−3
✓ 1
Part (c)
M1 \(\frac{10^{-7.2}}{10^{-7.8}}\)
✓ 1
M1 \(10^{0.6}\)
✓ 1
A1 3.98
✓ 1
Common error: A lower pH means a higher hydrogen ion concentration; the minus sign in the definition reverses the direction.
436AASL1-q00321.3 Geometric sequences and seriesPaper 2Hard14 marks
A graduate is offered two jobs.
Offer A: a salary of €42 000 in the first year, increasing by €1500 each year.
Offer B: a salary of €40 000 in the first year, increasing by 4.5% each year.
(a)
Find the salary in the fifth year under Offer A.
(2)
(b)
Find the salary in the fifth year under Offer B.
(2)
(c)
Find the total amount earned in the first 10 years under each offer.
(4)
(d)
Find the first year in which the annual salary under Offer B is greater than the annual salary under Offer A.
(3)
(e)
Find the least number of complete years after which the total amount earned under Offer B is greater than the total amount earned under Offer A.
437AASL1-q00331.3 Geometric sequences and seriesPaper 2Hard13 marks
During an outbreak of a disease, 40 new cases were recorded on day 1. For days 1 to 10, the number of new cases each day was 25% greater than on the previous day. From day 11 onwards, after restrictions were introduced, the number of new cases each day was 20% less than on the previous day.
In this question, give numbers of cases to the nearest whole number.
(a)
Find the number of new cases recorded on day 10.
(2)
(b)
Find the total number of cases recorded in the first 10 days.
(2)
(c)
Find the first day at the end of which the total number of cases recorded exceeded 1000.
(3)
(d)
Find the first day on which fewer than 20 new cases were recorded.
(3)
(e)(i)
The total number of cases recorded from day 11 to day 20 inclusive.
Write down an expression, using sigma notation, for this total.
(1)
(e)(ii)
Hence find the total number of cases recorded in the first 20 days.
Accept equivalent forms, e.g. \(\sum_{m=1}^{10}40(1.25)^{9}(0.8)^{m}\) or \(\sum_{m=0}^{9}238.42(0.8)^{m}\).
(e)(ii)
\(\dfrac{238.42\left(1-0.8^{10}\right)}{1-0.8}\;(=1064.09)\) added to their (b)
(M1)
\(2394\)
A1
\(2394.2\ldots\) Accept 2390 or 2395 from rounded values.
Answers: (a) 298 · (b) 1330 · (c) day 9 · (d) day 23 · (e)(ii) 2394
438AASL1-q00421.4 Financial applicationsPaper 2Hard6 marks
Bank X offers an interest rate of 4.1% per year, compounded annually. Bank Y offers a nominal annual interest rate of 4.0%, compounded monthly.
(a)
Determine which bank gives the greater value after 5 years for an investment of €10 000. Justify your answer by finding both values to the nearest cent.
(3)
(b)
Bank Y decides to change its nominal annual rate, still compounded monthly, so that its investments grow at exactly the same rate as those at Bank X. Find the new nominal annual rate. Give your answer correct to three significant figures.
(3)
Show mark scheme
Working
Mark
Notes
(a)
Bank X: \(10\,000(1.041)^{5}=\)€12 225.13
A1
Bank Y: \(10\,000\left(1+\dfrac{4}{1200}\right)^{60}=\)€12 209.97
A1
Bank X gives the greater value
R1
Must follow from two values.
(b)
\(\left(1+\dfrac{r}{1200}\right)^{12}=1.041\)
(M1)
Or TVM with any equal starting value.
\(r=4.0249\ldots\)
(A1)
\(4.02\%\)
A1
Answers: (a) Bank X (€12 225.13 versus €12 209.97) · (b) 4.02%
439AASL1-q00431.4 Financial applicationsPaper 2Hard8 marks
Throughout this question, the annual rate of inflation is 2.4%. The real value of an amount after \(n\) years, in terms of today’s money, is found by dividing its value after \(n\) years by \(1.024^{n}\).
(a)
Omar keeps €1500 in cash for 10 years. Find the real value of this cash after 10 years, in terms of today’s money.
(2)
(b)
Find the number of complete years after which the real value of Omar’s €1500 is first less than €1200.
(3)
(c)
Omar instead invests €20 000 for 10 years at a nominal annual rate of \(r\%\), compounded monthly. Find the least value of \(r\), correct to two decimal places, for which the real value of the investment after 10 years is at least €22 000.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{1500}{1.024^{10}}\)
(M1)
€1183.29
A1
(b)
\(\dfrac{1500}{1.024^{n}}\lt 1200\)
(M1)
\(n\gt 9.41\); or after 9 years €1211.69, after 10 years €1183.29
Accept use of FV \(=22\,000\times1.024^{10}=27\,888.31\) in TVM.
\(r=3.3293\ldots\)
(A1)
\(r=3.33\)
A1
Award A0 for 3.32 (gives a real value of €21 979.46, which is too small). The approximate method of subtracting the inflation rate from \(r\) (not the method given in the stem) gives \(3.35\): award (M1)(A0)A0.
Answers: (a) €1183.29 · (b) 10 years · (c) \(r=3.33\)
440AASL1-q00441.4 Financial applicationsPaper 2Hard13 marks
At the start of each year, Priya deposits $2000 into a savings account that pays interest at 3% per year, compounded annually. She makes no withdrawals. Let \(V_{n}\) be the value of the account at the end of year \(n\), immediately after interest has been added.
Timeline not to scale
(a)
Show that \(V_{2}=4181.80\).
(2)
(b)
Show that \(V_{n}=\dfrac{206\,000}{3}\left(1.03^{n}-1\right)\).
(3)
(c)
Find the value of \(V_{10}\).
(2)
(d)
Find the least value of \(n\) for which \(V_{n}\gt 50\,000\).
(3)
(e)
Instead, Priya could deposit a single amount of $20 000 at the start of year 1 in the same account. Find how much greater the value of the account would be at the end of year 10 under this option.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(V_{1}=2000\times1.03=2060\), then \(V_{2}=(2060+2000)\times1.03\)
441IBMA2S-044-04-111.9 The binomial theoremPaper 2Hard14 marks
Calculator
The expansion of \((1+ax)^n\), where \(n\) is a positive integer and \(a\) is a constant, begins \(1+12x+63x^2+\dots\)
(a)
Write down the first three terms of the expansion of \((1+ax)^n\) in terms of \(a\), \(n\) and \(x\).
(2)
(b)
Find the value of \(a\) and the value of \(n\).
(6)
(c)
Find the coefficient of \(x^3\).
(2)
(d)
A streaming subscription costs $12.00 per month. The price rises by 1.5% each year. The monthly price after 8 years is \(12(1.015)^8\). Write \(1.015\) in the form \(1+ax\), using your value of \(a\). Then use the first four terms of the expansion to estimate the monthly price after 8 years. Give your answer to the nearest cent.
(3)
(e)
Explain why your estimate is very close to the true price.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(1+nax\) A1 \(+\frac{n(n-1)}{2}a^2x^2\)
✓ 2
Part (b)
M1 \(na=12\) M1 \(\frac{n(n-1)}{2}a^2=63\) M1 substituting \(na=12\) to eliminate a variable: \(12a(n-1)=126\) A1 \(a(n-1)=10.5\), so \(12-a=10.5\) A1 \(a=1.5\) A1 \(n=8\)
M1 \(1+1.5x=1.015\) gives \(x=0.01\) M1 \(1+12(0.01)+63(0.01)^2+189(0.01)^3=1.126489\) A1 \(12\times1.126489=13.5179\), so $13.52
✓ 3
Part (e)
R1 the omitted terms contain \(x^4=10^{-8}\) and higher powers of \(x\). Their total is only about \(3.6\times10^{-6}\), which changes the price by much less than one cent
✓ 1
Exam tip: Two unknowns need two equations. Substitute \(na=12\) into the \(x^2\) equation instead of solving a cubic; this keeps the algebra short.
442IBMA2S-044-06-111.5 Exponents and introduction to logarithmsPaper 2Hard12 marks
Calculator
The sound level L decibels (dB) of a sound with intensity I W m−2 is given by L = 10 log10\(\left(\frac{I}{I_0}\right)\), where I0 = 1.0×10−12 W m−2.
(a)
Near a jet engine, the intensity is 1.0×102 W m−2. Find the sound level.
(2)
(b)
A pair of headphones produces a sound level of 98 dB. Find the intensity of this sound, giving your answer in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ, with a correct to 3 significant figures.
(3)
(c)
Show that increasing the sound level by 10 dB multiplies the intensity by 10.
(2)
(d)
A safety guideline allows 8 hours of listening at 85 dB. The allowed time halves for every 3 dB increase above 85 dB. Find the allowed listening time at 98 dB, in minutes.
(3)
(e)
Find how many times more intense a 98 dB sound is than an 85 dB sound.
Common error: In part (d), (98 − 85) ÷ 3 = 4.33… is not a whole number, so use the fractional power 13/3. Rounding it to 4 halvings gives 30 minutes, which is wrong.
443IBMA2S-044-06-121.2 Arithmetic sequences and seriesPaper 2Hard13 marks
Calculator
In 1971 a computer processor contained 2.3×103 transistors. A model assumes that the number of transistors on a leading processor doubles every 2 years. Let Tn be the number of transistors predicted in the year 1971 + 2n, where n ≥ 0.
(a)
Write down an expression for Tn in terms of n, and justify your answer.
(2)
(b)
Find the number of transistors predicted for 2021. Give your answer in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ, with a correct to 3 significant figures.
(2)
(c)
A processor released in 2021 actually contained 5.7×1010 transistors. Find the percentage error of the model's prediction.
(2)
(d)
Find the first year in which the model predicts more than 1.0×1012 transistors.
(4)
(e)
A factory makes 1.5×106 of the 2021 processors in its first year of production, and production increases by 12% each year after that. Each processor contains 5.7×1010 transistors. Find the total number of transistors in all the processors made in the first 10 years. Give your answer in the form a×10k, where 1 ≤ a < 10 and k ∈ ℤ, with a correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the number doubles every 2 years, so the values form a geometric sequence with common ratio 2 A1 Tn = 2.3×103×2n
M1 setting up 2.3×103×2n > 1.0×1012 A1 n > 28.7 R1 n must be an integer, so n = 29 A1 year 1971 + 58 = 2029
✓ 4
Part (e)
M1 using a geometric series with r = 1.12: S10 = 1.5×106(1.1210 − 1) ÷ 0.12 A1 2.63×107 processors A1 multiplying by 5.7×1010 gives 1.50×1018 transistors
✓ 3
Exam tip: n counts 2-year steps, not years. Always convert back: n = 25 means the year 2021, and n = 29 means the year 2029.
444IBMA2S-044-01-081.3 Geometric sequences and seriesPaper 2Hard14 marks
Calculator
An engineer models the power output \(P\) kW of a wind turbine at wind speed \(v\) m s−1 by \(P=kv^{n}\), where \(k\) and \(n\) are constants. Tests show that \(P=24\) when \(v=6\), and \(P=81\) when \(v=9\).
(a)
Show that \(\ln P=\ln k+n\ln v\).
(2)
(b)
Find the value of \(n\).
(4)
(c)
Find the value of \(k\).
(2)
(d)
Find the wind speed needed for the turbine to produce 150 kW.
(3)
(e)
The wind speed increases by 20%. Find the percentage increase in power output.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 takes logs of both sides and uses the product law: \(\ln P=\ln k+\ln v^{n}\)
✓ 1
A1 uses the power law to obtain \(\ln P=\ln k+n\ln v\) (AG)
✓ 1
Part (b)
M1 eliminates \(k\), either by dividing the equations, \(\left(\frac{9}{6}\right)^{n}=\frac{81}{24}\), or by subtracting the log forms, \(n(\ln 9-\ln 6)=\ln 81-\ln 24\)
✓ 1
A1 a correct simplified equation, e.g. \(1.5^{n}=3.375\) or \(n\ln 1.5=\ln 3.375\)
✓ 1
M1 rearranges using logs: \(n=\dfrac{\ln 3.375}{\ln 1.5}\) (or equivalent)
Exam tip: To remove the unknown constant \(k\), divide one equation by the other or subtract their log forms. Either way you are left with a single equation in \(n\), which you can solve with logs.
445IBMA2S-044-00-101.8 Infinite geometric seriesPaper 2Hard11 marks
Calculator
A 50 m length of cable is wound onto a drum. The first turn has radius 4.0 cm. Each new turn sits on top of the previous turn, so its radius is 0.25 cm greater. Model each turn as a circle, so the lengths of successive turns form an arithmetic sequence.
(a)
Find the length of the first turn and the common difference of the sequence, giving exact answers in terms of π.
(2)
(b)
Find the total length of cable in the first 40 turns, giving your answer in metres.
(3)
(c)
Find the number of complete turns the 50 m cable makes on the drum.
(3)
(d)
Find the length of cable left over after the last complete turn.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(u_1=2\pi\times4=8\pi\) cm
✓ 1
A1 \(d=2\pi\times0.25=0.5\pi\) cm
✓ 1
Part (b)
M1 \(S_{40}=\frac{40}{2}(16\pi+39\times0.5\pi)\)
✓ 1
A1 \(710\pi\) cm
✓ 1
A1 22.3 m
✓ 1
Part (c)
M1 \(\frac{n}{2}(16\pi+0.5\pi(n-1))\le5000\)
✓ 1
M1 solves e.g. \(n^2+31n-\frac{20000}{\pi}\le0\) to get \(n\le65.8\)
Common error: The common difference is the increase in circumference, \(2\pi\times0.25\), not the increase in radius, 0.25 cm. Also use one unit throughout: 50 m = 5000 cm.
446IBMA2S-044-02-101.3 Geometric sequences and seriesPaper 2Hard14 marks
Calculator
Two streaming services launch in the same month. Service A gets 5000 new sign-ups in month 1, and the number of new sign-ups increases by 400 each month after that. Service B gets 2000 new sign-ups in month 1, and the number of new sign-ups increases by 9% each month after that.
(a)
Find the number of new sign-ups in month 12 for each service.
(3)
(b)
Find the first month in which Service B gets more new sign-ups than Service A.
(3)
(c)
Find the total number of sign-ups for Service B in the first 12 months.
(2)
(d)
Find the first month at the end of which the total number of sign-ups for Service B is greater than the total for Service A.
(4)
(e)
Find by how much the total for Service B exceeds the total for Service A at the end of that month.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 Service A: \(5000+11\times400=9400\)
✓ 1
M1 Service B: \(2000\times1.09^{11}\)
✓ 1
A1 5160 (5160.85…)
✓ 1
Part (b)
M1 \(2000\times1.09^{n-1}>5000+400(n-1)\)
✓ 1
M1 uses a GDC table or graph, e.g. month 23: 13317 < 13800 and month 24: 14516 > 14200
✓ 1
A1 month 24
✓ 1
Part (c)
M1 \(\frac{2000(1.09^{12}-1)}{0.09}\)
✓ 1
A1 40300 (40281.4…)
✓ 1
Part (d)
M1 \(S_A=\frac{n}{2}(10000+400(n-1))\) or \(200n^2+4800n\)
✓ 1
M1 \(S_B=\frac{2000(1.09^{n}-1)}{0.09}\)
✓ 1
M1 compares values, e.g. \(n=34\): 393965 < 394400 and \(n=35\): 431422 > 413000
✓ 1
A1 month 35
✓ 1
Part (e)
M1 \(431421.5-413000\)
✓ 1
A1 18400 (18421.5…); accept the unrounded value
✓ 1
Common error: Comparing the sign-ups in a single month uses the terms of the sequences. Comparing the running totals uses the sums of the series. The two comparisons give different months, so check which one the question asks for.
447IBMA2S-044-05-121.7 Rational exponents and laws of logarithmsPaper 2Hard12 marks
Calculator
The capacitor in a camera flash charges so that its voltage \(t\) seconds after switching on is \(V=300\left(1-e^{-t/\tau}\right)\) volts, where \(\tau\) is a positive constant. After 2.5 seconds the voltage is 200 V.
(a)
Show that \(\tau=\frac{2.5}{\ln 3}\).
(3)
(b)
The flash is ready when the voltage reaches 285 V. Find the time taken for the flash to be ready.
(3)
(c)
When the flash fires, the voltage falls according to \(V=285e^{-kt}\) and reaches 10 V after 0.004 seconds. Find \(k\).
(3)
(d)
An improved design uses a different value of \(\tau\) so that the flash is ready after exactly 4 seconds. Find the voltage 2.5 seconds after switching on for the improved design.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets \(300(1-e^{-2.5/\tau})=200\)
✓ 1
M1 \(e^{-2.5/\tau}=\frac{1}{3}\)
✓ 1
A1 \(\frac{2.5}{\tau}=\ln3\), so \(\tau=\frac{2.5}{\ln3}\) (AG)
Common error: Setting \(e^{-t/\tau}=0.95\) instead of \(1-e^{-t/\tau}=0.95\) gives the time for the voltage to reach 5% of its maximum, not 95%.
448IBMA2S-044-03-111.4 Financial applicationsPaper 2Hard11 marks
Calculator
An athlete invests her prize money of 20 000 dollars. For the first k years it earns 4% per year, compounded annually. After k years the whole amount is moved to a second account earning 6% per year, compounded annually, until the end of year 10. Here k is an integer with 0 ≤ k ≤ 10.
(a)
Find the value of the investment at the end of year 10 when k = 3.
(2)
(b)
Show that the value, V dollars, at the end of year 10 is \(V = 20000 \times 1.06^{10} \times \left(\frac{1.04}{1.06}\right)^k\).
(2)
(c)
The athlete wants at least 32 000 dollars at the end of year 10. Find the largest value of k for which this is possible.
(4)
(d)
For the value of k found in part (c), find how much more the athlete has at the end of year 10 than if she had kept all the money at 4% for the full 10 years.
M1 writes V = 20 000 × 1.04k × 1.0610−k A1 uses 1.0610−k = 1.0610 ÷ 1.06k to obtain the given result
✓ 2
Part (c)
M1 sets up 20 000 × 1.0610 × (1.04/1.06)k ≥ 32 000 A1 boundary value k = 5.92 (5.916…), OR evaluates V(5) = 32 563 and V(6) = 31 948 R1 since 1.04/1.06 < 1, V decreases as k increases, so the condition holds for every k up to the boundary and fails beyond it A1 k = 5
Common error: Dividing by ln(1.04/1.06), which is negative, reverses the inequality. Give a reason why the largest k is 5 rather than 6.
449IBMA2S-044-03-121.3 Geometric sequences and seriesPaper 2Hard11 marks
Calculator
A satellite operator values a communications transponder using reducing-balance depreciation at a rate of r per year, where 0 < r < 1. The original value of the transponder is V0 dollars. After 3 years the transponder has lost exactly 40% of its original value.
(a)
Show that r = 0.157, correct to 3 significant figures.
(3)
(b)
Write down an expression, in terms of V0, r and n, for the amount of depreciation during year n.
(2)
(c)
Find the minimum number of complete years after which the transponder is worth less than 10% of its original value.
(3)
(d)
The depreciation during year 4 is $28 000. Find V0.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets up (1 − r)3 = 0.6 M1 1 − r = 0.61/3 = 0.84343… A1 r = 0.156567… ≈ 0.157
✓ 3
Part (b)
M1 difference between the value at the start and the end of year n: V0(1 − r)n−1 − V0(1 − r)n A1 \(V_0 r(1-r)^{n-1}\)
✓ 2
Part (c)
M1 sets up (1 − r)n < 0.1 using the unrounded value of r A1 n > 13.5 A1 14 years
Exam tip: In part (d), replace (1 − r)³ with exactly 0.6 from the given information. This avoids rounding errors from using r = 0.157.
450IBMA3S-1-16r11-081.4 Financial applicationsPaper 2Hard12 marks
Calculator
A wind farm's annual energy output falls as its turbines age. In year 1 the output is 240 GWh and in each later year the output is 96% of the output in the previous year, so the yearly outputs form a geometric sequence.
(a)
Find the output in year 10, to the nearest GWh.
(2)
(b)
Find the total output over the first 10 years, to the nearest GWh.
(2)
(c)
Show that if the wind farm could operate indefinitely, the total output would be 6000 GWh.
(2)
(d)
Find the least number of years \(n\) for which the total output exceeds 5000 GWh.
(4)
(e)
Find the output in year 44, correct to 4 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_{10}=240\times0.96^{9}\)
✓ 1
A1 166 GWh
✓ 1
Part (b)
M1 \(S_{10}=\frac{240(1-0.96^{10})}{1-0.96}\)
✓ 1
A1 2011 GWh
✓ 1
Part (c)
M1 \(S_\infty=\frac{240}{1-0.96}\)
✓ 1
A1 \(=6000\) GWh
✓ 1
Part (d)
M1 \(6000(1-0.96^n)>5000\)
✓ 1
M1 \(0.96^n<\frac16\) and take logarithms, reversing the inequality because \(\ln0.96<0\)
✓ 1
A1 \(n>43.9\)
✓ 1
A1 \(n=44\)
✓ 1
Part (e)
M1 \(240\times0.96^{43}\)
✓ 1
A1 41.48 GWh
✓ 1
Common error: When dividing both sides of an inequality by a negative number, such as \(\ln 0.96\), the inequality sign reverses.
451IBMA3S-1-16r11-131.2 Arithmetic sequences and seriesPaper 2Hard14 marks
Calculator
A library display has shelves 1, 2, 3, ... The numbers of books on the shelves form an arithmetic sequence with first term \(a\) and common difference \(d\), where \(d\ne0\). The numbers of books on shelves 1, 3 and 7 are the first three terms of a geometric sequence.
(a)
Show that \(a=2d\).
(4)
(b)
Find the common ratio of the geometric sequence.
(2)
(c)
Shelf 1 holds 4 books. Find the common difference and the number of books on shelf 12.
(2)
(d)
Find the total number of books on the first 12 shelves.
(2)
(e)
The display can hold at most 400 books. Find the greatest number of complete shelves that can be used, and the total number of books on them.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{a+2d}{a}=\frac{a+6d}{a+2d}\)
✓ 1
A1 \((a+2d)^2=a(a+6d)\)
✓ 1
A1 \(4d^2=2ad\)
✓ 1
R1 divide by \(2d\) since \(d\ne0\), giving \(a=2d\)
✓ 1
Part (b)
M1 \(r=\frac{a+2d}{a}=\frac{4d}{2d}\)
✓ 1
A1 \(r=2\)
✓ 1
Part (c)
A1 \(d=2\)
✓ 1
A1 \(u_{12}=4+11\times2=26\)
✓ 1
Part (d)
M1 \(S_{12}=\frac{12}{2}(2\times4+11\times2)\)
✓ 1
A1 180
✓ 1
Part (e)
M1 \(S_n=\frac n2(8+2(n-1))=n(n+3)\)
✓ 1
M1 \(n(n+3)\le400\)
✓ 1
A1 \(n=18\)
✓ 1
A1 \(S_{18}=18\times21=378\)
✓ 1
Common error: Shelves 1, 3 and 7 are terms \(a\), \(a+2d\) and \(a+6d\). Using \(a+3d\) and \(a+7d\) is a common slip with term numbers.
452IBMA3S-1-16r13-131.7 Rational exponents and laws of logarithmsPaper 2Hard11 marks
Calculator
Kleiber's law models the daily energy need \(R\) kcal of a farm animal of mass \(m\) kg by \(R=70\,m^{3/4}\).
(a)
Find \(R\) for a cow of mass 450 kg. Give your answer to 3 significant figures.
(2)
(b)
A calf needs 3000 kcal per day. Find its mass, to 3 significant figures.
(3)
(c)
The mass of an animal is multiplied by \(k\). Show that \(R\) is multiplied by \(k^{3/4}\), and hence find the percentage increase in \(R\) when the mass is doubled. Give your answer to 3 significant figures.
(3)
(d)
Given that \(\log_{10}R=a+b\log_{10}m\), find the exact value of \(b\) and the value of \(a\) to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(70\times450^{3/4}\)
✓ 1
A1 6840 (6839.2...)
✓ 1
Part (b)
M1 \(m^{3/4}=\frac{3000}{70}\)
✓ 1
M1 raise both sides to the power \(\frac43\)
✓ 1
A1 150 kg (149.98...)
✓ 1
Part (c)
M1 new \(R=70(km)^{3/4}=k^{3/4}\times70m^{3/4}\)
✓ 1
A1 for \(k=2\), \(2^{3/4}=1.68\) (1.6818...)
✓ 1
A1 68.2% increase
✓ 1
Part (d)
M1 take \(\log_{10}\) of both sides: \(\log_{10}R=\log_{10}70+\frac34\log_{10}m\)
✓ 1
A1 \(b=\frac34\)
✓ 1
A1 \(a=\log_{10}70=1.85\)
✓ 1
Common error: A fractional power of a product splits as \((km)^{3/4}=k^{3/4}m^{3/4}\); do not multiply the exponent by \(k\).
453IBMA3S-1-16r13-141.2 Arithmetic sequences and seriesPaper 2Hard10 marks
Calculator
A composer writes a piece in which the numbers of notes in successive bars form an arithmetic sequence. Bar 4 has 15 notes and the first 10 bars contain 195 notes in total.
(a)
Find the number of notes in bar 1 and the common difference.
(4)
(b)
Find the number of notes in bar 30.
(2)
(c)
The whole piece contains 1995 notes. Find the number of bars in the piece.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(a+3d=15\)
✓ 1
M1 \(\frac{10}{2}(2a+9d)=195\)
✓ 1
M1 solve simultaneously
✓ 1
A1 \(a=6\), \(d=3\)
✓ 1
Part (b)
M1 \(u_{30}=6+29\times3\) (ECF from (a))
✓ 1
A1 93
✓ 1
Part (c)
M1 \(\frac n2\big(12+3(n-1)\big)=1995\)
✓ 1
A1 \(n^2+3n-1330=0\)
✓ 1
A1 \(n=35\)
✓ 1
R1 reject \(n=-38\) because \(n\) is a positive integer
✓ 1
Exam tip: Write each given fact as an equation using \(u_n\) or \(S_n\) before solving the pair together.
454IBMA3S-1-17r13-131.7 Rational exponents and laws of logarithmsPaper 2Hard12 marks
Calculator
A laboratory culture contains \(N_n = 5\times 4^n\) bacteria \(n\) hours after the start of an experiment. Let \(u_n = \ln N_n\).
(a)
Show that \(u_1, u_2, u_3, \ldots\) is an arithmetic sequence.
(3)
(b)
Find \(u_1\) in the form \(\ln k\), where \(k\) is an integer.
(1)
(c)
Show that \(u_1 + u_2 + \cdots + u_{10} = a\ln 2 + b\ln 5\) where \(a, b\) are integers, and find \(a\) and \(b\).
(3)
(d)
Find the least whole number of hours \(n\) for which \(N_n > 1\,000\,000\).
(3)
(e)
Find the value of \(n\) for which \(u_n = \ln 5 + 14\ln 2\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_n = \ln 5 + n\ln 4\) using \(\ln(ab) = \ln a + \ln b\) and \(\ln a^b = b\ln a\)
✓ 1
M1 find \(u_{n+1} - u_n\)
✓ 1
R1 \(u_{n+1}-u_n = \ln 4\), independent of \(n\), so the sequence is arithmetic
M1 use \(\ln 20 = 2\ln 2 + \ln 5\) and \(\ln 4 = 2\ln 2\)
✓ 1
A1 \(a = 110\), \(b = 10\)
✓ 1
Part (d)
M1 \(5\times 4^n > 10^6\), i.e. \(4^n > 200\,000\)
✓ 1
M1 solve, e.g. \(n > \frac{\ln 200000}{\ln 4} = 8.80\) or compare \(N_8 = 327\,680\) and \(N_9 = 1\,310\,720\)
✓ 1
A1 \(n = 9\)
✓ 1
Part (e)
M1 \(n\ln 4 = 14\ln 2\), so \(2n\ln 2 = 14\ln 2\)
✓ 1
A1 \(n = 7\)
✓ 1
Exam tip: Rewrite \(\ln 4\) as \(2\ln 2\) so that every term uses the same logarithms. Then you can compare coefficients.
455IBMA3S-1-17r13-141.8 Infinite geometric seriesPaper 2Hard13 marks
Calculator
A hospital patient is given a 200 mg dose of a drug at the start of each day. After 24 hours, 30% of the drug present is still in the body. Let \(A_n\) be the amount of drug in the body, in mg, immediately after the \(n\)th dose, so \(A_1 = 200\).
(a)
Show that \(A_3 = 278\).
(2)
(b)
Show that \(A_n = \dfrac{2000}{7}\left(1 - 0.3^n\right)\).
(3)
(c)
Find the value approached by \(A_n\) as \(n\) becomes very large. Give your answer to 3 significant figures.
(2)
(d)
Amounts above 280 mg are unsafe. Find the amount in the body immediately after the first dose at which this happens.
(4)
(e)
Show that \(A_n\) never exceeds \(\dfrac{2000}{7}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(A_2 = 200 + 0.3\times 200 = 260\)
✓ 1
A1 \(A_3 = 200 + 0.3\times 260 = 278\)
✓ 1
Part (b)
M1 \(A_n = 200 + 200(0.3) + \cdots + 200(0.3)^{n-1}\), a geometric series with \(r=0.3\)
✓ 1
M1 \(S_n = \frac{200(1-0.3^n)}{1-0.3}\)
✓ 1
A1 \(= \frac{2000}{7}(1-0.3^n)\) (AG)
✓ 1
Part (c)
M1 \(0.3^n\to 0\), or \(S_\infty = \frac{200}{1-0.3}\)
✓ 1
A1 \(\frac{2000}{7} \approx 286\) mg
✓ 1
Part (d)
M1 attempt to solve \(\frac{2000}{7}(1-0.3^n) > 280\), i.e. \(0.3^n < 0.02\), or evaluate successive terms
✓ 1
A1 \(A_3 = 278 < 280\)
✓ 1
A1 \(n = 4\)
✓ 1
A1 \(A_4 = 283.4\) mg
✓ 1
Part (e)
R1 \(0.3^n > 0\) for all \(n\), so \(1 - 0.3^n < 1\)
✓ 1
R1 hence \(A_n < \frac{2000}{7}\) for every \(n\)
✓ 1
Common error: Treating \(A_n\) as a single geometric term. Here \(A_n\) is a sum of \(n\) terms, so use the sum formula.
456IBMA3S-1-16r14-101.3 Geometric sequences and seriesPaper 2Hard13 marks
Calculator
A hospital patient is given a 500 mg tablet of a drug every 8 hours. After 8 hours, 30% of the drug present in the body remains. Let \(A_n\) be the total amount of the drug in the body, in mg, immediately after the \(n\)th tablet is taken.
(a)
Show that \(A_3 = 695\).
(2)
(b)
Show that \(A_n = \frac{500(1 - 0.3^n)}{0.7}\).
(3)
(c)
Find the value that \(A_n\) approaches as \(n\) becomes very large. Give your answer to the nearest mg.
(2)
(d)
The hospital requires that \(A_n\) must not exceed 700 mg. Find the number of the first tablet after which \(A_n > 700\).
(4)
(e)
The dose is changed to \(D\) mg per tablet. Find the value of \(D\) for which \(A_n\) approaches exactly 700 mg.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(A_2 = 500 + 0.3 \times 500 = 650\)
✓ 1
A1 \(A_3 = 500 + 0.3 \times 650 = 695\)
✓ 1
Part (b)
M1 \(A_n = 500 + 500(0.3) + \ldots + 500(0.3)^{n-1}\), a geometric series with \(u_1 = 500\) and \(r = 0.3\)
Exam tip: When solving an inequality with a base less than 1, take logarithms of both sides carefully: dividing by ln 0.3 (a negative number) reverses the inequality sign.
457IBMA3S-1-16r14-131.9 The binomial theoremPaper 2Hard6 marks
Calculator
In the expansion of \((3 + x)^n\), where \(n \in \mathbb{N}\), \(n \geq 3\), the coefficient of \(x^2\) is equal to the coefficient of \(x^3\).
(a)
Find the value of \(n\).
(4)
(b)
Find the value of this coefficient.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 coefficients \(\binom{n}{2}3^{n-2}\) and \(\binom{n}{3}3^{n-3}\)
✓ 1
M1 equate the two coefficients
✓ 1
M1 simplify, e.g. \(\frac{n-2}{3} = 3\)
✓ 1
A1 \(n = 11\)
✓ 1
Part (b)
M1 \(\binom{11}{3} \times 3^8\) or \(\binom{11}{2} \times 3^9\)
✓ 1
A1 1 082 565
✓ 1
Common error: Powers of 3 differ between the two terms: write them as 3ⁿ⁻² and 3ⁿ⁻³ and cancel the common factor carefully.
458IBMA3S-1-16r15-131.5 Exponents and introduction to logarithmsPaper 2Hard12 marks
Calculator
A homeowner borrows £20000 at 5% interest per year. Interest is added at the end of each year and then a payment of £3000 is made. Let \(B_n\) be the balance owed immediately after the \(n\)th payment.
Timeline not to scale
(a)
Find \(B_1\) and \(B_2\).
(2)
(b)
Show that \(B_n=60000-40000\times1.05^{n}\).
(4)
(c)
Find the smallest number of payments needed to clear the loan.
(3)
(d)
The 9th payment is smaller than £3000 and clears the loan exactly. Find the amount of the 9th payment. Use the unrounded value of \(B_8\) and give your answer to the nearest penny.
M1 \(60000-40000\times1.05^{n}\le0\), so \(1.05^{n}\ge1.5\)
✓ 1
M1 attempt to solve using logarithms or technology
✓ 1
A1 \(n>8.31\), so 9 payments
✓ 1
Part (d)
M1 \(B_8=60000-40000\times1.05^{8}\)
✓ 1
A1 \(B_8=901.78\)
✓ 1
A1 final payment \(901.78\times1.05=\text{£}946.87\)
✓ 1
Common error: Interest is added before each payment, so the final payment is the remaining balance plus one year's interest.
459IBMA3S-1-16r15-141.8 Infinite geometric seriesPaper 2Hard11 marks
Calculator
During a braking test a train travels 120 m in the first second after the brakes are applied. In each later second it travels 0.9 times the distance it travelled in the previous second. Assume the pattern continues indefinitely.
(a)
Find the distances travelled in the second and third seconds.
(2)
(b)
Find the total distance the train travels before it stops.
(2)
(c)
Find the smallest whole number of seconds after which the train has travelled at least 90% of this total distance.
(4)
(d)
A signal is 1100 m from the point where the brakes are applied. Determine whether the train stops before it reaches the signal and, if it does not, find how far past the signal it stops.
M1 \(1200(1-0.9^{n})\ge1080\), so \(0.9^{n}\le0.1\)
✓ 1
A1 \(n\ge21.85\dots\)
✓ 1
A1 22 s
✓ 1
Part (d)
R1 total distance 1200 m is greater than 1100 m, so the train does not stop before the signal
✓ 1
M1 \(1200-1100\)
✓ 1
A1 100 m past the signal
✓ 1
Exam tip: A sum to infinity gives the limit of the total; compare it with the given distance.
460IBMA3S-1-17r15-111.4 Financial applicationsPaper 2Hard8 marks
Calculator
Maya saves for a race trip. At the start of each month she deposits €200 into an account that pays interest of 0.4% per month, added at the end of each month. The first deposit is made at the start of month 1.
(a)
Find the value of the first deposit at the end of month 24, to the nearest cent.
(2)
(b)
Show that the total value of the account at the end of month 24 is €5047.52, to the nearest cent.
(3)
(c)
Maya instead wants the account to be worth exactly €10 000 at the end of month 36. Find the monthly deposit, to the nearest euro.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(200\times1.004^{24}\)
✓ 1
A1 €220.11
✓ 1
Part (b)
M1 values of deposits form a geometric sequence \(200\times1.004^{24},200\times1.004^{23},\ldots,200\times1.004\)
✓ 1
M1 \(S_{24}=\frac{200\times1.004(1.004^{24}-1)}{1.004-1}\) or equivalent
Exam tip: When deposits are made at the start of a period, the first deposit earns interest for the full number of periods, so the series starts at \(1.004^{n}\).
461IBMA3S-1-17r15-141.5 Exponents and introduction to logarithmsPaper 2Hard12 marks
Calculator
The brightness of a star of apparent magnitude \(m\) is \(b=k\times10^{-0.4m}\), where \(k\) is a constant. A star of smaller magnitude is brighter. The magnitude of Sirius is \(-1.46\) and the magnitude of Polaris is \(1.98\).
(a)
Show that a star of magnitude 1 is 100 times as bright as a star of magnitude 6.
(2)
(b)
Find how many times as bright Sirius is as Polaris, to 3 significant figures.
(3)
(c)
A star X is 250 times as bright as Polaris. Find the magnitude of X, to 3 significant figures.
(4)
(d)
The brightnesses of stars of magnitude 1, 2, 3, 4, 5, 6 form a geometric sequence. Find the combined brightness of one star of each of these magnitudes as a multiple of the brightness of the magnitude 6 star, to 3 significant figures.
M1 ratio between consecutive terms in order 6 to 1 is \(10^{0.4}\)
✓ 1
M1 \(\frac{10^{2.4}-1}{10^{0.4}-1}\)
✓ 1
A1 \(165\) (from \(165.48\ldots\))
✓ 1
Common error: Brighter stars have smaller magnitudes: the exponent is \(0.4\times\)(magnitude of the fainter star minus magnitude of the brighter one).
462IBMA3S-1-17r16-131.4 Financial applicationsPaper 2Hard7 marks
Calculator
A library invests €20 000 in an account paying a nominal annual interest rate of r%, compounded monthly. After exactly 3 years the account holds €22 400.
(a)
Find the value of r, correct to 3 significant figures.
(4)
(b)
Using the unrounded value of r, find the smallest whole number of months after which the account first exceeds €30 000.
Exam tip: Store the unrounded value of r in your calculator and use it in the next part; rounding early can change later digits.
463IBMA3S-1-16r19-101.4 Financial applicationsPaper 2Hard11 marks
Calculator
A swimming club deposits €500 at the start of each year into an account that pays 4% interest per year, compounded annually. The first deposit is made at the start of year 1. No money is withdrawn.
Timeline not to scale
(a)
Show that the first deposit is worth €740.12 at the end of year 10.
(2)
(b)
Find the total value of all ten deposits at the end of year 10. Give your answer to the nearest cent.
(4)
(c)
The club deposits money at the start of each year until the total value at the end of year \(n\) first exceeds €10 000. Find the value of \(n\).
(3)
(d)
Find the total interest earned by the ten deposits at the end of year 10. Use the unrounded value from (b), and give your answer to the nearest cent.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(500\times1.04^{10}\)
✓ 1
A1 \(740.12\) (to 2 decimal places, must show unrounded value or \(740.122\ldots\))
✓ 1
Part (b)
M1 recognising the values \(500\times1.04^k\) for \(k=1,\ldots,10\) form a geometric series
✓ 1
A1 first term \(520\) and common ratio \(1.04\) (or equivalent)
✓ 1
M1 use of \(S_{10}=\frac{520(1.04^{10}-1)}{1.04-1}\)
✓ 1
A1 €6243.18
✓ 1
Part (c)
M1 \(\frac{520(1.04^{n}-1)}{0.04}>10000\)
✓ 1
M1 attempt to solve, e.g. by logarithms: \(1.04^n>1.769\), so \(n>14.5\)
✓ 1
A1 \(n=15\)
✓ 1
Part (d)
M1 subtract total deposited, \(10\times500=5000\)
✓ 1
A1 €1243.18 (ECF from (b))
✓ 1
Common error: The deposit made at the start of year 1 earns 10 years of interest and the last deposit earns only 1 year, so the series begins at \(500\times1.04\).
464IBMA3S-1-16r19-121.7 Rational exponents and laws of logarithmsPaper 2Hard10 marks
Calculator
The power output \(P\) kW of a wind turbine at wind speed \(v\) m/s is modelled by \(P=kv^{n}\), where \(k\) and \(n\) are constants. When \(v=4\), \(P=32\). When \(v=10\), \(P=500\).
(a)
Show that \(n=3\).
(3)
(b)
Find the value of \(k\).
(2)
(c)
Rearrange the model to give \(v=aP^{b}\), where \(a\) and \(b\) are constants. Give \(a\) to 3 significant figures.
(3)
(d)
Hence find the wind speed needed for an output of 108 kW.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writing \(32=k\times4^n\) and \(500=k\times10^n\)
✓ 1
M1 dividing the equations: \(\frac{500}{32}=\left(\frac{10}{4}\right)^n\)
✓ 1
A1 \(n=\frac{\ln15.625}{\ln2.5}=3\) (AG)
✓ 1
Part (b)
M1 substituting into \(32=k\times4^3\)
✓ 1
A1 \(k=0.5\)
✓ 1
Part (c)
M1 \(v^3=\frac{P}{0.5}=2P\)
✓ 1
M1 taking cube roots, e.g. \(v=(2P)^{\frac13}\)
✓ 1
A1 \(a=2^{\frac13}=1.26\), \(b=\frac13\)
✓ 1
Part (d)
M1 substituting \(P=108\) into \(v=(2P)^{\frac13}\)
✓ 1
A1 \(v=6\) m/s
✓ 1
Exam tip: Dividing two equations of the form \(y=kx^n\) eliminates \(k\) and leaves a single equation for \(n\).
465IBMA3S-1-16r20-121.3 Geometric sequences and seriesPaper 2Hard10 marks
Calculator
The number of new cases per day in an epidemic is modelled by \(C(t)=A\,b^{t}\), where \(t\) is the number of days after monitoring begins, \(A>0\) and \(b>0\). On day \(t=0\) there are 50 new cases and on day \(t=6\) there are 800 new cases.
(a)
Show that \(b=2^{\frac{2}{3}}\).
(3)
(b)
Find the number of new cases on day \(t=10\). Give your answer to the nearest whole number.
(2)
(c)
Find the first whole day on which the number of new cases exceeds 20 000.
(3)
(d)
The model can be written as \(C(t)=50e^{kt}\). Find the value of \(k\) to 3 significant figures.
M1 \(50\times 2^{\frac{2t}{3}}=20\,000\), so \(2^{\frac{2t}{3}}=400\)
✓ 1
M1 takes logarithms: \(t=\frac{3\ln 400}{2\ln 2}=12.97\) (or uses a solver)
✓ 1
A1 day 13 (since \(C(12)=12\,800\) is below 20 000)
✓ 1
Part (d)
M1 \(e^{k}=2^{\frac23}\) so \(k=\frac{2}{3}\ln 2\)
✓ 1
A1 \(k=0.462\)
✓ 1
Common error: After solving for \(t\) in a question about whole days, do not round to the nearest value automatically; check which whole day first satisfies the condition.
466IBMA3S-1-16r20-131.9 The binomial theoremPaper 2Hard12 marks
Calculator
The output of a solar panel falls by 2% each year, so after \(t\) years its output is \(0.98^{t}\) times its original output. Let \(f(x)=(1+ax)^n\), where \(n\) is a positive integer. The expansion of \(f(x)\) begins \(1-12x+60x^2+\dots\)
(a)
Find the value of \(n\) and the value of \(a\).
(5)
(b)
Find the coefficient of \(x^3\) in the expansion of \(f(x)\).
(2)
(c)
By choosing a suitable value of \(x\) and using the first four terms of the expansion, estimate the fraction of its original output that the panel produces after 6 years.
(2)
(d)
Find the value of \(t\) for which the panel's output has fallen to 70% of its original output. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(na=-12\)
✓ 1
M1 \(\binom{n}{2}a^2=60\)
✓ 1
M1 eliminates \(a\) (for example \(a=-\frac{12}{n}\) gives \(\frac{n(n-1)}{2}\times\frac{144}{n^2}=60\))
M1 \(t=\frac{\ln 0.7}{\ln 0.98}\) (or uses a solver)
✓ 1
A1 \(t=17.7\)
✓ 1
Exam tip: When a binomial expansion is used to estimate a power such as \(0.98^6\), write the base as \(1+ax\) and decide the value of \(x\) before substituting.
467IBMA3S-1-17r20-121.2 Arithmetic sequences and seriesPaper 2Hard14 marks
Calculator
A rail crew inspects track in a network. The lengths inspected each week form a geometric sequence. In week 3 the crew inspects 36 km and in week 6 it inspects 15.1875 km.
(a)
Find the common ratio.
(3)
(b)
Find the length inspected in week 1.
(2)
(c)
Find the sum to infinity of the sequence.
(2)
(d)
The network is 255 km long. Find the least number of weeks needed for the total length inspected to exceed 255 km.
(5)
(e)
Find the length inspected in week 20, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_6=u_3r^3\), so \(r^3=\dfrac{15.1875}{36}\)
✓ 1
A1 \(r^3=0.421875\)
✓ 1
A1 \(r=0.75\)
✓ 1
Part (b)
M1 \(u_1=\dfrac{36}{0.75^2}\)
✓ 1
A1 64 km
✓ 1
Part (c)
M1 \(\dfrac{64}{1-0.75}\)
✓ 1
A1 256 km
✓ 1
Part (d)
M1 \(\dfrac{64(1-0.75^n)}{0.25}>255\)
✓ 1
M1 \(1-0.75^n>\dfrac{255}{256}\), so \(0.75^n<\dfrac{1}{256}\)
✓ 1
M1 take logarithms
✓ 1
A1 \(n>\dfrac{\ln(1/256)}{\ln0.75}=19.3\), with the inequality reversed because \(\ln0.75<0\)
✓ 1
A1 \(n=20\)
✓ 1
Part (e)
M1 \(64\times0.75^{19}\)
✓ 1
A1 0.271 km
✓ 1
Common error: Dividing both sides of an inequality by a negative number, such as \(\ln0.75\), reverses the inequality sign.
468IBMA3S-1-17r20-131.9 The binomial theoremPaper 2Hard10 marks
Calculator
Consider the expansion of \((1+2x)^n\) where \(n\) is a positive integer, \(n\ge3\). The coefficient of \(x^3\) is six times the coefficient of \(x^2\).
(a)
Find the value of \(n\).
(5)
(b)
Write down the coefficient of \(x^3\) in the expansion when \(n=11\).
(2)
(c)
By using the first four terms of the expansion of \((1+2x)^{11}\) with \(x=0.01\), estimate the value of \(1.02^{11}\). Give your answer to 6 significant figures.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 coefficient of \(x^2\) is \(\binom{n}{2}2^2\)
✓ 1
M1 coefficient of \(x^3\) is \(\binom{n}{3}2^3\)
✓ 1
M1 \(8\binom{n}{3}=6\times4\binom{n}{2}\)
✓ 1
M1 simplify, for example \(\dfrac{4n(n-1)(n-2)}{3}=12n(n-1)\), so \(n-2=9\)
Exam tip: Write the coefficient condition as an equation in \(n\) and cancel common factors such as \(n(n-1)\) before solving.
469IBMA3S-1-17r22-131.3 Geometric sequences and seriesPaper 2Hard10 marks
Calculator
A patient takes a tablet containing \(D\) mg of a drug every 12 hours. Just before each later tablet, 60% of the drug present immediately after the previous tablet remains in the body. Let \(A_n\) be the amount of the drug in the body, in mg, immediately after the \(n\)th tablet.
(a)
Show that \(A_n=\frac{D(1-0.6^n)}{0.4}\).
(3)
(b)
The amount in the body must never exceed 600 mg. Find the largest possible value of \(D\).
(2)
(c)
Now let \(D=200\). Find the smallest \(n\) for which \(A_n\) exceeds 450 mg.
(3)
(d)
Find \(A_n\) for this value of \(n\). Give your answer to 2 decimal places.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(A_n=D+0.6D+0.6^2D+\dots+0.6^{n-1}D\) (geometric, first term \(D\), ratio 0.6)
✓ 1
M1 \(S_n=\frac{D(1-0.6^n)}{1-0.6}\)
✓ 1
A1 \(A_n=\frac{D(1-0.6^n)}{0.4}\)
✓ 1
Part (b)
M1 limiting amount \(\frac{D}{1-0.6}\le600\)
✓ 1
A1 \(D=240\)
✓ 1
Part (c)
M1 \(500(1-0.6^n)>450\)
✓ 1
M1 \(0.6^n<0.1\), so \(n>\frac{\ln0.1}{\ln0.6}=4.51\) (inequality reversed on dividing by \(\ln0.6<0\))
✓ 1
A1 \(n=5\)
✓ 1
Part (d)
M1 \(\frac{200(1-0.6^5)}{0.4}\)
✓ 1
A1 461.12 mg
✓ 1
Common error: Dividing both sides of an inequality by ln 0.6, which is negative, reverses the inequality sign.
470IBMA3S-1-16r25-111.2 Arithmetic sequences and seriesPaper 2Hard12 marks
Calculator
A space probe uses an ion thruster for a series of burns. The first burn uses 18.0 kg of fuel. Each later burn uses 85% of the fuel used in the previous burn. The probe carries 130 kg of fuel.
(a)
Find the mass of fuel used in the 4th burn, correct to 3 significant figures.
(2)
(b)
Find the total mass of fuel used in the first 8 burns, correct to 3 significant figures.
(2)
(c)
Show that the probe can never use more than 130 kg of fuel, however many burns are made.
(3)
(d)
Find the least number of burns needed for the total fuel used to exceed 105 kg.
(3)
(e)
Find the mass of fuel remaining after these 13 burns, correct to 3 significant figures. Use the unrounded total for 13 burns.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(18 \times 0.85^{3}\)
✓ 1
A1 11.1 kg
✓ 1
Part (b)
M1 \(S_8 = \frac{18(1 - 0.85^8)}{1 - 0.85}\)
✓ 1
A1 87.3 kg
✓ 1
Part (c)
M1 \(S_\infty = \frac{18}{1 - 0.85}\) with \(|r| < 1\)
✓ 1
A1 120 kg
✓ 1
R1 the total used is always less than 120 kg, which is less than 130 kg
✓ 1
Part (d)
M1 \(\frac{18(1 - 0.85^n)}{0.15} > 105\), so \(0.85^n < 0.125\)
✓ 1
M1 \(n > \frac{\ln 0.125}{\ln 0.85}\)
✓ 1
A1 \(n > 12.8\) (3 s.f.), so \(n = 13\)
✓ 1
Part (e)
M1 \(130 - \frac{18(1 - 0.85^{13})}{0.15}\)
✓ 1
A1 24.5 kg
✓ 1
Exam tip: For a sum to infinity, state that the common ratio satisfies |r| < 1 before using the formula.
471IBMA3S-1-16r25-121.8 Infinite geometric seriesPaper 2Hard10 marks
Calculator
A geometric sequence has first term 4 and sum to infinity 6.
(a)
Find the common ratio.
(2)
(b)
Find the fourth term of the sequence.
(1)
(c)
Show that the sum of the first \(n\) terms is \(S_n = 6\left(1 - \left(\frac{1}{3}\right)^n\right)\).
(2)
(d)
Find the least value of \(n\) for which \(6 - S_n < 0.001\).
Exam tip: Write 6 − S_n as a single power of the ratio, then use logarithms to solve the inequality.
472IBMA3S-1-16r25-131.8 Infinite geometric seriesPaper 2Hard10 marks
Calculator
After a heavy lorry crosses a footbridge, the bridge deck oscillates. The first oscillation has amplitude 15.0 cm and each later oscillation has an amplitude that is 80% of the previous one. Each oscillation takes 1.4 s.
(a)
Find the amplitude of the 5th oscillation, correct to 3 significant figures.
(2)
(b)
Find the sum of the amplitudes of the first 10 oscillations, correct to 3 significant figures.
(2)
(c)
Find the sum of the amplitudes of all the oscillations.
(2)
(d)
Find the number of the first oscillation whose amplitude is less than 1.00 cm.
(3)
(e)
Find the time, in seconds, from the start of the first oscillation to the end of the oscillation found in part (d).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(15 \times 0.8^{4}\)
✓ 1
A1 6.14 cm
✓ 1
Part (b)
M1 \(S_{10} = \frac{15(1 - 0.8^{10})}{1 - 0.8}\)
✓ 1
A1 66.9 cm
✓ 1
Part (c)
M1 \(S_\infty = \frac{15}{1 - 0.8}\)
✓ 1
A1 75 cm
✓ 1
Part (d)
M1 \(15 \times 0.8^{n-1} < 1\), so \(0.8^{n-1} < \frac{1}{15}\)
Common error: If the amplitude of oscillation n is 15 × 0.8^(n − 1), the power is one less than the oscillation number.
473IBMA3S-1-18r29-141.3 Geometric sequences and seriesPaper 2Hard10 marks
Calculator
In its first year a wind farm produces 40 GWh of electricity. Because of ageing, the output in each later year is 95% of the output in the previous year.
(a)
Find the output in year 6, giving your answer to 3 significant figures.
(2)
(b)
Find the total output over the first 20 years, giving your answer to 3 significant figures.
(3)
(c)
Find the total output that the farm would produce if it continued indefinitely.
(2)
(d)
Find the least number of years after which the total output exceeds 90% of the answer to part (c).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of 40 × 0.955
✓ 1
A1 31.0 GWh
✓ 1
Part (b)
M1 use of S20 = 40(1 − 0.9520)/(1 − 0.95)
✓ 1
A1 correct substitution, giving 513.21…
✓ 1
A1 513 GWh
✓ 1
Part (c)
M1 use of S∞ = 40/(1 − 0.95)
✓ 1
A1 800 GWh
✓ 1
Part (d)
M1 800(1 − 0.95n) > 720
✓ 1
M1 0.95n < 0.1 and solve using logarithms or a table
✓ 1
A1 n = 45 (since n > 44.9…)
✓ 1
Exam tip: To solve an inequality of the form rⁿ < c with 0 < r < 1, take logarithms and remember that dividing by ln r (negative) reverses the inequality.
474IBMA3S-1-19r29-121.8 Infinite geometric seriesPaper 2Hard9 marks
Calculator
A ribbon cutter makes cuts of lengths forming the geometric series \(5+5\left(\frac{2x-3}{4}\right)+5\left(\frac{2x-3}{4}\right)^2+\ldots\) metres, where \(x\) is a real constant.
(a)
Find the set of values of \(x\) for which the series converges.
(3)
(b)
The sum to infinity is 8 metres. Find the value of \(x\).
(3)
(c)
For this value of \(x\), find the total length of the first 6 cuts. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 convergence requires \(|r|<1\) with \(r=\frac{2x-3}{4}\)
✓ 1
M1 \(-4<2x-3<4\)
✓ 1
A1 \(-\frac12<x<\frac72\)
✓ 1
Part (b)
M1 \(\frac{5}{1-r}=8\)
✓ 1
A1 \(r=\frac38\)
✓ 1
A1 \(\frac{2x-3}{4}=\frac38\) so \(x=2.25\)
✓ 1
Part (c)
M1 \(S_6=\frac{5\left(1-r^6\right)}{1-r}\)
✓ 1
M1 substitution \(r=\frac38\)
✓ 1
A1 7.98 m
✓ 1
Exam tip: State the convergence condition \(|r|<1\) explicitly and check that any x you find lies inside the interval before using the sum to infinity.
475IBMA3S-1-19r29-141.3 Geometric sequences and seriesPaper 2Hard7 marks
Calculator
A fishery has 3000 tonnes of stock at the start of year 0. During each year the stock falls by 8% through natural losses and then 120 tonnes are added by restocking. Let \(S_n\) be the stock, in tonnes, at the end of year \(n\), so \(S_0=3000\).
(a)
Find \(S_1\) and \(S_2\).
(2)
(b)
Show that \(S_n=1500+1500\times0.92^n\).
(3)
(c)
Find the stock at the end of year 20, to the nearest tonne.
Exam tip: For a recurrence of the form \(S_{n+1}=rS_n+c\), unroll the first few terms: the repeated \(c\) terms form a geometric series you can sum.
476IBMA3S-1-18r31-101.3 Geometric sequences and seriesPaper 2Hard14 marks
Calculator
A patient takes a 150 mg tablet of a drug every 8 hours. In the 8 hours between tablets, 60% of the drug in the bloodstream is eliminated, so 40% remains when the next tablet is taken. Each tablet is absorbed immediately. Let \(A_n\) be the amount of drug, in mg, in the bloodstream immediately after the \(n\)th tablet is taken, so \(A_1=150\).
(a)
Find \(A_3\).
(2)
(b)
Explain why \(A_n\) is the sum of the first \(n\) terms of a geometric sequence. State the first term and the common ratio.
(2)
(c)
Hence show that \(A_n=250\left(1-0.4^{n}\right)\).
(2)
(d)
Find the least number of tablets after which the amount of drug immediately after a tablet exceeds 249.5 mg.
(4)
(e)
Find the limit of \(A_n\) as \(n\) becomes very large, and state what this value means for the patient.
(2)
(f)
The dose is changed to \(D\) mg per tablet, with the same elimination rate. The long-term amount immediately after a tablet must not exceed 400 mg. Find the greatest value of \(D\).
R1 the \(k\)th previous tablet contributes \(150\times0.4^{k}\), so \(A_n=150+150(0.4)+150(0.4)^2+\dots+150(0.4)^{n-1}\)
✓ 1
A1 first term 150, common ratio 0.4
✓ 1
Part (c)
M1 \(A_n=\frac{150(1-0.4^n)}{1-0.4}\)
✓ 1
A1 \(A_n=250(1-0.4^n)\), shown with the simplification \(\frac{150}{0.6}=250\)
✓ 1
Part (d)
M1 \(250(1-0.4^n)>249.5\)
✓ 1
M1 \(0.4^n<0.002\)
✓ 1
M1 solves using logarithms or a table of values, e.g. \(n>6.78\)
✓ 1
A1 \(n=7\)
✓ 1
Part (e)
A1 limit is 250 mg (since \(0.4^n\to0\))
✓ 1
R1 the largest amount in the bloodstream settles close to 250 mg, immediately after each tablet in the long term
✓ 1
Part (f)
M1 \(\frac{D}{1-0.4}\le400\)
✓ 1
A1 \(D=240\) mg
✓ 1
Exam tip: For a repeated-dose problem, write the first few totals as sums such as \(150+150(0.4)+\dots\) to see the geometric structure before using the formula.
477IBMA3S-1-18r31-141.7 Rational exponents and laws of logarithmsPaper 2Hard13 marks
Calculator
A water treatment plant removes impurities from 800 g of contaminated water in repeated passes through a filter. Each pass removes 70% of the impurities still present at the start of that pass.
(a)
Find the mass of impurities remaining after the second pass.
(2)
(b)
Show that the masses removed in successive passes form a geometric sequence. State the first term and the common ratio.
(3)
(c)
Find the total mass of impurities removed during the first 4 passes. Give your answer to 3 significant figures.
(3)
(d)
At least 99.9% of the original impurities must be removed. Find the least number of passes needed.
(3)
(e)
Find the mass of impurities remaining after the number of passes found in (d). Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(800\times0.3^2\) (or \(800-560-168\))
✓ 1
A1 72 g
✓ 1
Part (b)
M1 mass removed in pass 1 is \(0.7\times800=560\) g and in pass 2 is \(0.7\times240=168\) g
✓ 1
M1 ratio \(\frac{168}{560}=0.3\) (and shows that pass \(k\) removes \(560\times0.3^{k-1}\))
✓ 1
A1 first term 560 g, common ratio 0.3
✓ 1
Part (c)
M1 \(S_4=\frac{560(1-0.3^4)}{1-0.3}\)
✓ 1
M1 \(=560\times\frac{0.9919}{0.7}\)
✓ 1
A1 794 g
✓ 1
Part (d)
M1 remaining mass less than \(0.001\times800=0.8\) g, so \(800\times0.3^n<0.8\)
✓ 1
M1 solves using logarithms or a table of values, e.g. \(n>5.74\)
✓ 1
A1 6 passes
✓ 1
Part (e)
M1 \(800\times0.3^{6}\) (ECF from (d))
✓ 1
A1 0.583 g
✓ 1
Exam tip: Two routes give the same remaining mass: \(800\times0.3^n\), or \(800\) minus the sum of the masses removed. Using one to check the other catches arithmetic slips.
478IBMA3S-1-17r32-121.2 Arithmetic sequences and seriesPaper 2Hard8 marks
Calculator
A coach plans a tapering programme for an athlete. The training loads (in arbitrary units) in successive weeks form an infinite geometric sequence with common ratio \(r\), where \(0<r<1\). The load in week 2 is 12 and the total load over all weeks is 50.
(a)
Show that \(25r^{2}-25r+6=0\).
(3)
(b)
Hence find the two possible values of \(r\).
(2)
(c)
The coach chooses the plan with the larger load in week 1. Find the total load in the first 5 weeks of this plan, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u_1=\frac{12}{r}\)
✓ 1
M1 \(\frac{u_1}{1-r}=50\) so \(\frac{12}{r(1-r)}=50\)
✓ 1
A1 \(12=50r-50r^{2}\) leading to \(25r^{2}-25r+6=0\) (AG)
✓ 1
Part (b)
M1 attempt to solve the quadratic, e.g. \((5r-2)(5r-3)=0\)
✓ 1
A1 \(r=0.4\) and \(r=0.6\)
✓ 1
Part (c)
A1 larger \(u_1\) is \(\frac{12}{0.4}=30\) so \(r=0.4\)
✓ 1
M1 \(S_5=\frac{30(1-0.4^{5})}{1-0.4}\)
✓ 1
A1 49.5
✓ 1
Exam tip: When two values of \(r\) satisfy \(|r|<1\), test each against the other conditions in the question before choosing.
479IBMA3S-1-17r32-141.7 Rational exponents and laws of logarithmsPaper 2Hard8 marks
Calculator
Two positive real numbers \(x\) and \(y\) satisfy the equations \(x+y=20\) and \(\log_{3}x-\log_{3}y=2\).
(a)
Show that \(x=9y\).
(3)
(b)
Find the values of \(x\) and \(y\).
(3)
(c)
Find the value of \(\log_{3}x+\log_{3}y\), correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\log_{3}\!\left(\frac{x}{y}\right)=2\)
✓ 1
M1 \(\frac{x}{y}=3^{2}\)
✓ 1
A1 \(x=9y\) (AG)
✓ 1
Part (b)
M1 substitute: \(9y+y=20\)
✓ 1
A1 \(y=2\)
✓ 1
A1 \(x=18\)
✓ 1
Part (c)
M1 \(\log_{3}(xy)=\log_{3}36\) and change of base \(\frac{\ln36}{\ln3}\)
✓ 1
A1 3.26
✓ 1
Exam tip: Subtracting logarithms of the same base turns into a quotient: \(\log_a x-\log_a y=\log_a\frac{x}{y}\), which removes the logarithms in one step.
480IBMA3S-1-18r38-141.4 Financial applicationsPaper 2Hard8 marks
Calculator
An investor places a sum €P in a savings plan. For the first 4 years the plan pays 3.1% per year, compounded annually. For the next 3 years it pays a nominal annual rate of 2.4%, compounded quarterly. After the 7 years the plan is worth €9800.
Timeline not to scale
(a)
Find the single number by which the investment is multiplied over the 7 years. Give your answer to 4 significant figures.
(3)
(b)
Find P, to the nearest euro. Use the unrounded value of the multiplier from (a).
(2)
(c)
The investor could instead have used an account paying a single fixed rate, compounded annually, for all 7 years, with the same P and the same final value. Find that annual rate as a percentage to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 1.0314 and (1 + 0.024/4)12
✓ 1
A1 quarterly rate 0.6% over 12 quarters
✓ 1
A1 1.214
✓ 1
Part (b)
M1 P = 9800/(1.0314 × 1.00612)
✓ 1
A1 €8 073
✓ 1
Part (c)
M1 (1 + r/100)7 = multiplier from (a) (unrounded)
✓ 1
M1 1 + r/100 = (1.0314 × 1.00612)1/7
✓ 1
A1 r = 2.81
✓ 1
Exam tip: Keep unrounded intermediate values in your calculator memory when later parts build on them.
481IBMA3S-1-18r41-081.4 Financial applicationsPaper 2Hard10 marks
Calculator
The installed capacity of a regional solar network is modelled by \(P=120\times1.18^{t}\), where \(P\) is in megawatts (MW) and \(t\) is the time in years since the start of 2020. A second network has installed capacity modelled by \(Q=80\times1.32^{t}\) MW.
(a)
Interpret the value 1.18 in the model for \(P\).
(1)
(b)
Find the capacity of the first network at \(t=5\). Give your answer to 3 significant figures.
(2)
(c)
Find the time taken for the capacity of the first network to double. Give your answer in years, to 3 significant figures.
(3)
(d)
Find the value of \(t\) at which the two networks have equal capacity. Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the capacity of the first network grows by 18% each year (increases by a factor of 1.18 each year)
✓ 1
Part (b)
M1 \(120\times1.18^{5}\)
✓ 1
A1 275 MW
✓ 1
Part (c)
M1 \(1.18^{t}=2\)
✓ 1
M1 \(t=\frac{\ln2}{\ln1.18}\)
✓ 1
A1 4.19 years
✓ 1
Part (d)
M1 \(80\times1.32^{t}=120\times1.18^{t}\)
✓ 1
M1 \(\left(\frac{1.32}{1.18}\right)^{t}=1.5\)
✓ 1
M1 \(t=\frac{\ln1.5}{\ln(1.32/1.18)}\)
✓ 1
A1 \(t=3.62\)
✓ 1
Common error: When both sides have the form a×bt, divide to collect the powers into a single exponential before taking logarithms.
482IBMA3S-1-18r41-121.4 Financial applicationsPaper 2Hard10 marks
Calculator
A retailer buys a refrigeration unit for €3200. Its value falls by 15% in each of the first two years, and then by 10% each year after that.
Timeline not to scale
(a)
Find the value of the unit after 2 years.
(2)
(b)
Find the smallest whole number of years after which the value of the unit is less than €1050.
(4)
(c)
The value after 10 years is €995.24. Find the constant annual percentage rate of depreciation that would give this same value after 10 years. Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(3200\times0.85^{2}\)
✓ 1
A1 €2312
✓ 1
Part (b)
M1 value after \(n\) years is \(2312\times0.9^{\,n-2}\), set \(<1050\)
✓ 1
M1 \(0.9^{\,n-2}<0.4541\ldots\), so \(n-2>\frac{\ln(1050/2312)}{\ln0.9}\)
Common error: A change of rate part-way through means two separate powers: apply the first rate for its years, then start a new power with the second rate for the remaining years.
483IBMA3S-1-17r42-091.8 Infinite geometric seriesPaper 2Hard10 marks
Calculator
A signal sent along a fibre produces echoes whose amplitudes, in mV, form an infinite geometric sequence with first term \(a\) and common ratio \(r\), where \(0<r<1\). The sum of all the echo amplitudes is 12 mV and the sum of the first three echo amplitudes is 10.5 mV.
(a)
Show that \(r^{3}=0.125\).
(3)
(b)
Find the value of \(r\) and the value of \(a\).
(2)
(c)
After the \(n\)th echo, the amplitudes of all later echoes have a total below 0.001 mV. Find the least value of \(n\).
(3)
(d)
Find the amplitude of the 8th echo.
(2)
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Marking point
Mark
Notes
Part (a)
M1 \(\dfrac{a}{1-r}=12\)
✓ 1
M1 \(\dfrac{a\left(1-r^{3}\right)}{1-r}=10.5\), so \(12\left(1-r^{3}\right)=10.5\)
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A1 \(r^{3}=0.125\) (AG)
✓ 1
Part (b)
M1 \(r=\sqrt[3]{0.125}=0.5\)
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A1 \(a=12(1-0.5)=6\)
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Part (c)
M1 the total of later echoes is \(S_\infty-S_n=12\times0.5^{n}\)
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M1 \(12\times0.5^{n}<0.001\), so \(n>\dfrac{\log 12000}{\log 2}=13.55\ldots\)
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A1 \(n=14\)
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Part (d)
M1 \(u_{8}=6\times0.5^{7}\)
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A1 \(0.046875\) mV
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Common error: The tail of a series after n terms is the sum to infinity minus S_n, not the (n+1)th term on its own.
484IBMA3S-1-17r42-141.9 The binomial theoremPaper 2Hard10 marks
Calculator
In the expansion of \((1+ax)^{n}\), where \(a>0\) and \(n\) is a positive integer, the coefficient of \(x\) is 12 and the coefficient of \(x^{2}\) is 60. This expansion is used to model the growth of a data-transfer rate.
(a)
Show that \(na=12\) and \(n(n-1)a^{2}=120\).
(2)
(b)
Find the values of \(n\) and \(a\).
(3)
(c)
Find the coefficient of \(x^{3}\) in the expansion.
(2)
(d)
Use the first four terms of the expansion to estimate the value of \(1.02^{6}\). Give your answer to 6 significant figures.
(3)
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Marking point
Mark
Notes
Part (a)
M1 coefficient of \(x\) is \(na\), so \(na=12\)
✓ 1
A1 coefficient of \(x^{2}\) is \(\binom n2a^{2}=60\), so \(n(n-1)a^{2}=120\) (AG)
✓ 1
Part (b)
M1 \(n(n-1)a^{2}=(na)^{2}-a(na)=144-12a=120\)
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A1 \(a=2\)
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A1 \(n=6\)
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Part (c)
M1 \(\binom63\times2^{3}\)
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A1 \(160\)
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Part (d)
M1 uses \(x=0.01\) in \(1+12x+60x^{2}+160x^{3}\)
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A1 \(1+0.12+0.006+0.00016\)
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A1 \(1.12616\)
✓ 1
Exam tip: To estimate a number such as 1.02^6, write it as (1 + 2x)^6 with x = 0.01 so that the expansion applies.
485IBMA3S-1-17r43-121.4 Financial applicationsPaper 2Hard8 marks
Calculator
A racing team places £8000 in Account B, which pays a nominal annual interest rate of 4.8%, compounded quarterly.
(a)
Find the value of Account B after 3 years, to the nearest pound.
(2)
(b)
Account A pays interest of \(r\%\) once per year (compounded annually). Find the value of \(r\) for which Account A gives the same value as Account B after 3 years. Give your answer to 3 significant figures.
(3)
(c)
Find how long, in years, it takes for Account B to reach £12 500. Give your answer to 3 significant figures.
(3)
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Marking point
Mark
Notes
Part (a)
M1 \(8000\left(1+\frac{0.048}{4}\right)^{12}\)
✓ 1
A1 £9231
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Part (b)
M1 \((1+\frac{r}{100})^{3}=1.012^{12}\)
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A1 \(1+\frac{r}{100}=1.012^{4}=1.04887\ldots\)
✓ 1
A1 \(r=4.89\)
✓ 1
Part (c)
M1 \(8000(1.012)^{4n}=12500\)
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M1 \(4n=\frac{\log(1.5625)}{\log(1.012)}\)
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A1 \(n=9.35\) years
✓ 1
Exam tip: Compare accounts through the effective annual rate: the multiplier over one year is \((1+\frac{i}{k})^{k}\).
486IBMA3S-1-19r44-091.3 Geometric sequences and seriesPaper 2Hard7 marks
Calculator
During a nectar flow, the mass of honey (in kg) collected by a hive on successive days forms a geometric sequence with a positive common ratio. The mass collected on day 2 is 24 kg and the mass collected on day 4 is 54 kg.
(a)
Find the common ratio and the mass collected on day 1.
(3)
(b)
The total mass collected over the first \(n\) days first exceeds 2000 kg on day \(N\). Find \(N\), and hence find the total mass collected over the first \(N\) days, to the nearest kg.
(4)
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Notes
Part (a)
M1 use u4 ÷ u2 = r2, so r2 = 54/24
✓ 1
A1 r = 1.5 (reject −1.5)
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A1 u1 = 24 ÷ 1.5 = 16 kg
✓ 1
Part (b)
M1 use Sn = 16(1.5n − 1) ÷ 0.5 and set up Sn > 2000
✓ 1
M1 solve 1.5n > 63.5 using logarithms or trial (n > 10.2)
✓ 1
A1 N = 11
✓ 1
A1 S11 = 2736 kg
✓ 1
Common error: Taking only the positive ratio is justified by the stem; always say why the negative root is rejected rather than ignoring it.
487IBMA3S-1-19r46-131.4 Financial applicationsPaper 2Hard9 marks
Calculator
A data centre operator places €20 000 in a reserve fund. For the first 4 years the fund earns 3% per year, compounded annually. Then the rate changes to a nominal annual rate of \(r\%\), compounded monthly, for a further 3 years, after which the fund is worth €25 000.
Timeline not to scale
(a)
Find the value of the fund after the first 4 years, to the nearest cent.
(2)
(b)
Find the value of \(r\), to 3 significant figures. (You may use either the unrounded value or the answer to (a).)
(4)
(c)
Find the single annual rate of interest, compounded annually for all 7 years, that would turn €20 000 into €25 000. Give your answer as a percentage to 3 significant figures.
Exam tip: Keep each stage's result in your calculator memory, then carry the unrounded value into the next stage.
488IBMA3S-1-17r47-111.4 Financial applicationsPaper 2Hard8 marks
Calculator
A library trust invests €D at the start of each year for 20 years. The fund earns 3% compound interest per year, added at the end of each year. The trust wants the fund to be worth exactly €21 000 at the end of the 20th year, immediately after the final interest payment.
(a)
Write down, in terms of D, the value at the end of the 20th year of (i) the deposit made at the start of year 1, (ii) the deposit made at the start of year 20.
(2)
(b)
Show that the total value of the fund at the end of year 20 is \(\frac{1.03D(1.03^{20} - 1)}{0.03}\).
(3)
(c)
Find the value of D, correct to the nearest cent.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 (i) \(D \times 1.03^{20}\)
✓ 1
A1 (ii) \(D \times 1.03\)
✓ 1
Part (b)
M1 recognise the values form a geometric series with first term \(1.03D\) and common ratio 1.03
✓ 1
M1 use of \(S_{20} = \frac{u_1(r^{20} - 1)}{r - 1}\)
✓ 1
A1 \(\frac{1.03D(1.03^{20} - 1)}{0.03}\) (AG)
✓ 1
Part (c)
M1 \(\frac{1.03D(1.03^{20} - 1)}{0.03} = 21000\)
✓ 1
M1 rearrange for D, e.g. \(D = \frac{21000 \times 0.03}{1.03(1.03^{20} - 1)}\)
✓ 1
A1 €758.77
✓ 1
Common error: The first deposit earns interest for 20 years and the last for only 1 year, so the series starts at \(1.03D\), not \(D\).
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