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IB Maths Applications & Interpretation · SL and HL · Topic 1

Number and algebra

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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →

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+ Add to test✓ In your test1IBMI-030-00-01Approximation, percentage error and standard form Paper SLEasy5 marks

Light from the Sun takes 499 seconds to reach the Earth. The speed of light is taken to be 3.00×10⁸ m s⁻¹.

(a)

Calculate the distance from the Sun to the Earth, in metres. Give your answer exactly, in the form $a\times10^k$ where $1\le a<10$ and $k\in\mathbb{Z}$.

(2)

(b)

Write down your answer to part (a) correct to two significant figures.

(1)

(c)

The accepted mean Sun–Earth distance is 1.496×10¹¹ m. Calculate the percentage error in your answer to part (a).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for distance = 3.00×10⁸ × 499	✓ 1
A1 1.497×10¹¹ m	✓ 1
Part (b)
A1 1.5×10¹¹ m	✓ 1
Part (c)
M1 for $\left\|\frac{1.497\times10^{11}-1.496\times10^{11}}{1.496\times10^{11}}\right\|\times100$	✓ 1
A1 0.0668 %	✓ 1

Exam tip: In a percentage error, always divide by the exact (accepted) value, not by the approximate value.

+ Add to test✓ In your test2IBMI-030-00-02Approximation, percentage error and standard form Paper SLEasy4 marks

A surveyor estimates by eye that a footbridge is 85 m long. The actual length of the footbridge is 82.4 m.

(a)

Calculate the percentage error in the surveyor's estimate.

(2)

(b)

A second person underestimates the length of the footbridge with a percentage error of exactly 5 %. Find this person's estimate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\left\|\frac{85-82.4}{82.4}\right\|\times100$	✓ 1
A1 3.16 % (accept 3.155…)	✓ 1
Part (b)
M1 for 82.4 × 0.95 (or 82.4 − 0.05 × 82.4)	✓ 1
A1 78.28 m (accept 78.3 m)	✓ 1

Common error: An underestimate means the estimate is below the true value, so multiply the true value by 0.95, not 1.05.

+ Add to test✓ In your test3IBMI-030-00-03Approximation, percentage error and standard form Paper SLEasy6 marks

A rectangular vegetable garden measures 12.47 m by 8.53 m.

(a)

Calculate the area of the garden. Write down your full calculator answer and then give it correct to three significant figures.

(2)

(b)

A gardener estimates the area by first rounding each length to the nearest metre. Find the gardener's estimate.

(2)

(c)

Using the full calculator value of the area as the exact value, calculate the percentage error in the gardener's estimate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 106.3691 m²	✓ 1
A1 106 m²	✓ 1
Part (b)
M1 for 12 × 9	✓ 1
A1 108 m²	✓ 1
Part (c)
M1 for $\left\|\frac{108-106.3691}{106.3691}\right\|\times100$	✓ 1
A1 1.53 %	✓ 1

Exam tip: Use the unrounded area in the percentage error calculation. Using 106 instead gives 1.89%, which is wrong.

+ Add to test✓ In your test4IBMI-030-00-04Approximation, percentage error and standard form Paper SLEasy5 marks

A country has a population of 6.72×10⁷ people and an annual gross domestic product (GDP) of 2.85×10¹² US dollars (USD). Its land area is 2.43×10⁵ km².

(a)

Calculate the GDP per person, giving your answer in USD correct to three significant figures.

(2)

(b)

Write your answer to the previous part in the form $a\times10^k$ where $1\le a<10$ and $k\in\mathbb{Z}$.

(1)

(c)

Calculate the population density of the country in people per km², giving your answer correct to three significant figures in the form $a\times10^k$ where $1\le a<10$ and $k\in\mathbb{Z}$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{2.85\times10^{12}}{6.72\times10^{7}}$	✓ 1
A1 42 400 USD	✓ 1
Part (b)
A1 4.24×10⁴ USD	✓ 1
Part (c)
M1 for $\frac{6.72\times10^{7}}{2.43\times10^{5}}$	✓ 1
A1 2.77×10² people per km²	✓ 1

Common error: Writing 42.4×10³ is not standard form: the number in front must satisfy 1 ≤ a < 10.

+ Add to test✓ In your test5IBMI-030-05-01Approximation, percentage error and standard form Paper SLEasy6 marks

A rectangular solar panel is 1.65 m long and 0.992 m wide. A solar farm has 2400 of these panels.

(a)

Calculate the exact area of one panel.

(1)

(b)

Write your answer to part (a) correct to two significant figures.

(1)

(c)

Using your answer to part (a), find the total area of all 2400 panels. Give your answer in the form $a \times 10^k$, where $1 \le a < 10$ and $k \in \mathbb{Z}$, to three significant figures.

(2)

(d)

A planner uses 1.6 m² for the area of each panel. Calculate the percentage error in this value.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 1.6368 m²	✓ 1
Part (b)
A1 1.6 m²	✓ 1
Part (c)
M1 for 2400 × 1.6368 = 3928.32	✓ 1
A1 3.93×10³ m²	✓ 1
Part (d)
M1 for $\left\|\frac{1.6-1.6368}{1.6368}\right\| \times 100\%$	✓ 1
A1 2.25%	✓ 1

Exam tip: In a percentage error the exact value always goes in the denominator. Here that is 1.6368, not the rounded value 1.6.

+ Add to test✓ In your test6IBMI-030-05-02Approximation, percentage error and standard form Paper SLEasy4 marks

The speed of light in a vacuum is 299 792 458 m s⁻¹.

(a)

Write this speed in the form $a \times 10^k$, where $1 \le a < 10$ and $k \in \mathbb{Z}$, correct to three significant figures.

(2)

(b)

The mean distance from the Sun to Neptune is 4.50×10¹² m. Use the unrounded speed to find the time light takes to travel this distance. Give your answer in seconds in the form $a \times 10^k$, to three significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 for 3.00	✓ 1
A1 for ×10⁸ m s⁻¹	✓ 1
Part (b)
M1 for 4.50×10¹² ÷ 299 792 458	✓ 1
A1 1.50×10⁴ s	✓ 1

Common error: 3.00×10⁸ is correct to three significant figures. Writing 3×10⁸ gives only one significant figure and loses the accuracy mark.

+ Add to test✓ In your test7IBMI-030-05-05Approximation, percentage error and standard form Paper SLEasy4 marks

A type of bacterium has a diameter of 0.000 002 3 m.

(a)

Write this diameter in the form $a \times 10^k$, where $1 \le a < 10$ and $k \in \mathbb{Z}$.

(1)

(b)

Bacteria are placed in a single line, touching end to end, across a slide that is 1.5 cm wide. Find the number of complete bacteria that fit across the slide.

(2)

(c)

Write your answer to the previous part in the form $a \times 10^k$, where $1 \le a < 10$ and $k \in \mathbb{Z}$, correct to three significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 2.3×10⁻⁶ m	✓ 1
Part (b)
M1 converts 1.5 cm to 0.015 m and divides: 0.015 ÷ 2.3×10⁻⁶ = 6521.7…	✓ 1
A1 6521 complete bacteria	✓ 1
Part (c)
A1 6.52×10³	✓ 1

Common error: Convert 1.5 cm to 0.015 m before you divide. If you divide 1.5 by 2.3×10⁻⁶, the answer is 100 times too large.

+ Add to test✓ In your test8IBMI-030-05-11Approximation, percentage error and standard form Paper SLEasy5 marks

The attendance at a football match was 58 437. Each ticket cost $47.85.

(a)

Write the attendance correct to the nearest thousand.

(1)

(b)

Write the attendance correct to three significant figures.

(1)

(c)

Estimate the total ticket revenue by rounding both the attendance and the ticket price to one significant figure.

(1)

(d)

Calculate the percentage error in your estimate from the previous part.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 58 000	✓ 1
Part (b)
A1 58 400	✓ 1
Part (c)
A1 60 000 × 50 = $3 000 000	✓ 1
Part (d)
M1 for actual revenue 58 437 × 47.85 = $2 796 210.45 used in $\frac{\|3\,000\,000-2\,796\,210.45\|}{2\,796\,210.45} \times 100\%$	✓ 1
A1 7.29% (7.288…%)	✓ 1

Common error: Rounding 58 437 to 3 s.f. gives 58 400, not 584. The zeros are placeholders and must be kept.

+ Add to test✓ In your test9IBMI-030-06-r1-01Financial mathematics, loans and annuities Paper SLEasy6 marks

Priya invests €4500 in a savings account that pays a nominal annual interest rate of 3.2%, compounded monthly. She makes no further deposits or withdrawals.

(a)

Find the minimum number of complete months before the value of the investment first exceeds €6000.

(3)

(b)

Find the value of the investment after 6 years. Give your answer to the nearest cent.

(2)

(c)

Hence find the total interest earned in the first 6 years.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 equation or TVM entry: 4500(1 + 0.032/12)^N = 6000, or N unknown, I% = 3.2, PV = −4500, FV = 6000, C/Y = 12	✓ 1
A1 N = 108.02…	✓ 1
A1 109 months (must be rounded up, because after 108 months the value is about €5999.61)	✓ 1
Part (b)
M1 4500(1 + 0.032/12)⁷², or N = 72 entered in TVM	✓ 1
A1 €5451.12	✓ 1
Part (c)
A1 5451.12 − 4500 = €951.12	✓ 1

Common error: A solver value of N = 108.02 does not mean 108 months. The value only exceeds €6000 after a whole extra month of interest, so round up to 109.

+ Add to test✓ In your test10IBMI-030-06-r1-02Financial mathematics, loans and annuities Paper SLEasy5 marks

Tomasz takes out a loan of €18 000 to buy a car. The loan has a nominal annual interest rate of 6.5%, compounded monthly. He repays it in equal monthly instalments at the end of each month over 5 years.

(a)

Find the amount of each monthly instalment. Give your answer to the nearest cent.

(3)

(b)

Find the total interest Tomasz pays over the 5 years.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 N = 60 (or 5 × 12) and I% = 6.5 with C/Y = P/Y = 12	✓ 1
M1 PV = 18000 and FV = 0, with PV and PMT of opposite signs	✓ 1
A1 €352.19	✓ 1
Part (b)
M1 total repaid = 60 × 352.19 = €21 131.40 (accept 21 131.44 from the unrounded instalment)	✓ 1
A1 interest = 21 131.40 − 18 000 = €3131.40 (accept €3131.44)	✓ 1

Exam tip: Write down the values you enter into the financial solver (N, I%, PV, PMT, FV, P/Y, C/Y). The method marks come from these values, even if the final answer is slightly wrong.

+ Add to test✓ In your test11IBMI-030-01-r2-01Financial mathematics, loans and annuities Paper SLEasy6 marks

Priya invests 4500 euros in a savings account that pays a nominal annual interest rate of 3.2%, compounded monthly. She makes no further deposits or withdrawals.

(a)

Find the minimum number of complete months needed for the value of the investment to exceed 6000 euros.

(3)

(b)

Find the interest Priya earns in the first 6 years. Give your answer to the nearest cent.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Sets up $4500\left(1+\frac{0.032}{12}\right)^n > 6000$ or equivalent TVM entry (PV = −4500, FV = 6000, I% = 3.2, C/Y = 12) M1	✓ 1
$n = 108.02...$ A1	✓ 1
109 months A1	✓ 1
Part (b)
$4500\left(1+\frac{0.032}{12}\right)^{72}$ or TVM entry with N = 72 M1	✓ 1
Value = 5451.12 euros A1	✓ 1
Interest = 5451.12 − 4500 = 951.12 euros A1	✓ 1

Common error: Writing 108 months loses the final mark. After 108 months the value is still just under 6000 euros, so the answer must be rounded up to 109 complete months.

+ Add to test✓ In your test12IBMI-030-02-r1-01Arithmetic and geometric sequences in context Paper SLEasy6 marks

A new community theatre is designed so that the front row (row 1) has 18 seats. Each row behind it has 3 more seats than the row in front. The theatre has 25 rows.

(a)

Find the number of seats in row 25.

(2)

(b)

Find the first row that has more than 60 seats.

(2)

(c)

Find the total number of seats in the theatre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses $u_{25}=18+(25-1)\times3$ M1 ; 90 seats A1	✓ 2
Part (b)
Forms $18+3(n-1)>60$ or lists terms. Row 15 has exactly 60 seats M1 ; row 16, with 63 seats A1	✓ 2
Part (c)
Uses $S_{25}=\frac{25}{2}(18+90)$ or an equivalent formula M1 ; 1350 seats A1	✓ 2

Common error: In part (b) the condition is 'more than 60'. Row 15 has exactly 60 seats, so it does not qualify. Always check the boundary term against the strict inequality.

+ Add to test✓ In your test13IBMI-030-02-r1-02Arithmetic and geometric sequences in context Paper SLEasy7 marks

A fitness app was downloaded 2400 times in its first week. The number of downloads in each week is modelled as being 15% more than in the previous week.

(a)

Find the number of downloads in week 8.

(2)

(b)

Find the first week in which the number of downloads is more than 10 000.

(3)

(c)

Find the total number of downloads in the first 12 weeks.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses $2400\times1.15^{7}$ M1 ; 6380 downloads (6384.05…) A1	✓ 2
Part (b)
Forms $2400\times1.15^{n-1}>10000$ or uses a table on a GDC M1 ; gives $n-1>10.2$, or week 11 ≈ 9709 and week 12 ≈ 11 166 A1 ; week 12 A1	✓ 3
Part (c)
Uses $S_{12}=\frac{2400(1.15^{12}-1)}{1.15-1}$ M1 ; 69 600 downloads (69 604.0…) A1	✓ 2

Exam tip: Week 1 is the first term, so week n uses the power n − 1. A quick table of values on a GDC is a reliable way to confirm the first week that passes a threshold.

+ Add to test✓ In your test14IBMI-030-06-r2-01Financial mathematics, loans and annuities Paper SLEasy6 marks

Priya invests 4500 euros (EUR) in a savings account. The account pays a nominal annual interest rate of 3.2%, compounded monthly. She makes no further deposits or withdrawals.

(a)

Find the least number of complete months after which the value of Priya's investment first exceeds EUR 6000.

(3)

(b)

Find the interest Priya has earned after exactly 6 years.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Solve $4500\left(1+\frac{0.032}{12}\right)^{n}=6000$, or TVM with PV = −4500, FV = 6000, I% = 3.2, P/Y = C/Y = 12	✓ 1
n = 108.02…	✓ 1
109 months, since after 108 months the value is still just below EUR 6000	✓ 1
Part (b)
$4500\left(1+\frac{0.032}{12}\right)^{72}$	✓ 1
= EUR 5451.12	✓ 1
Interest = 5451.12 − 4500 = EUR 951.12	✓ 1

Exam tip: When n is not a whole number, decide in context whether to round up or down. Here the value has not yet reached EUR 6000 at month 108, so the answer is 109 months.

+ Add to test✓ In your test15IBMI-030-07-r1-01Arithmetic and geometric sequences in context Paper SLEasy6 marks

A community theatre has seating arranged in 24 rows. The first row has 18 seats, and each row after the first has 3 more seats than the row in front of it.

(a)

Find the number of seats in the 24th row.

(2)

(b)

Find the total number of seats in the theatre.

(2)

(c)

Find the number of the first row that has more than 60 seats.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$u_{24} = 18 + (24-1)(3)$ M1	✓ 1
$= 87$ seats A1	✓ 1
Part (b)
$S_{24} = \frac{24}{2}(18 + 87)$ or $\frac{24}{2}(2\times18 + 23\times3)$ M1	✓ 1
$= 1260$ seats A1	✓ 1
Part (c)
$18 + 3(n-1) > 60$, or noting that row 15 has exactly 60 seats M1	✓ 1
Row 16, which has 63 seats A1	✓ 1

Common error: Row 15 has exactly 60 seats, and that is not more than 60. Check the boundary value against the exact wording of the inequality.

+ Add to test✓ In your test16IBMI-030-01-r1-02Financial mathematics, loans and annuities Paper SLEasy6 marks

Tomás pays £150 into a savings account at the end of every month for 8 years. The account pays a nominal annual interest rate of 3%, compounded monthly.

(a)

Find the amount in the account immediately after the final payment. Give your answer to the nearest penny.

(2)

(b)

Find the total interest Tomás earns over the 8 years.

(2)

(c)

Tomás would like to have £20 000 in the account after 8 years, with the same interest conditions. Find the monthly payment he would need to make.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
TVM with N = 96, I% = 3, PV = 0, PMT = −150, P/Y = C/Y = 12, or $150\times\frac{1.0025^{96}-1}{0.0025}$	✓ 1
£16 252.11	✓ 1
Part (b)
Total paid in: 96 × 150 = £14 400	✓ 1
Interest = £1852.11	✓ 1
Part (c)
TVM with N = 96, I% = 3, PV = 0, FV = 20 000, solving for PMT	✓ 1
£184.59 (accept £184.60, which guarantees the target is reached)	✓ 1

Exam tip: When you use the TVM solver, money paid in and money received must have opposite signs. If PMT and FV have the same sign, the solver gives an error or a meaningless answer.

+ Add to test✓ In your test17IBMI-030-08-r2-01Exponents and logarithms Paper SLEasy5 marks

The sound level, L decibels (dB), of a noise with intensity I W m⁻² is modelled by
$L = 10\log_{10}\left(\frac{I}{I_0}\right)$
where $I_0 = 1.0 \times 10^{-12}$ W m⁻².

(a)

A hairdryer produces a sound of intensity 3.2×10⁻⁵ W m⁻². Find its sound level.

(2)

(b)

A motorbike passing a pedestrian produces a sound level of 88 dB. Find the intensity of this sound.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: substitutes correctly, $L = 10\log_{10}\left(\frac{3.2\times10^{-5}}{1.0\times10^{-12}}\right) = 10\log_{10}(3.2\times10^{7})$	✓ 1
A1: $L = 75.1$ dB	✓ 1
Part (b)
M1: substitutes correctly, $88 = 10\log_{10}\left(\frac{I}{10^{-12}}\right)$	✓ 1
M1: rearranges correctly, $\frac{I}{10^{-12}} = 10^{8.8}$, or solves with GDC	✓ 1
A1: $I = 6.31\times10^{-4}$ W m⁻²	✓ 1

Common error: In part (b), divide 88 by 10 before undoing the logarithm. Writing $10^{88}$ instead of $10^{8.8}$ gives an intensity that is far too large.

+ Add to test✓ In your test18IBMI-030-08-r2-02Exponents and logarithms Paper SLEasy6 marks

After a person drinks a large coffee, the amount of caffeine, C mg, remaining in their body t hours later is modelled by
$C(t) = 200e^{-0.139t}, \quad t \ge 0$.

(a)

Write down the amount of caffeine in the coffee.

(1)

(b)

Find the amount of caffeine remaining after 5 hours.

(2)

(c)

The person can sleep well once the amount of caffeine has fallen below 25 mg. Find the least time after drinking the coffee at which this happens.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: 200 mg	✓ 1
Part (b)
M1: substitutes $t = 5$, $200e^{-0.139\times5}$	✓ 1
A1: 99.8 mg	✓ 1
Part (c)
M1: sets up the equation $200e^{-0.139t} = 25$ (or an inequality)	✓ 1
M1: uses logarithms, $-0.139t = \ln(0.125)$, or solves with GDC	✓ 1
A1: $t = 15.0$ hours	✓ 1

Exam tip: For a 'find the time' question, write the equation you are solving before using the GDC. The equation earns the method mark even if a keying error spoils the final answer.

+ Add to test✓ In your test19IBMI-030-02-r2-01Arithmetic and geometric sequences in context Paper SLEasy6 marks

A new outdoor theatre has 25 rows of seats. The front row (row 1) has 18 seats. Each row after the first has 3 more seats than the row in front of it.

(a)

Find the number of seats in row 25.

(2)

(b)

Find the first row that has more than 60 seats.

(2)

(c)

Find the total number of seats in the theatre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$u_{25} = 18 + (25-1)\times 3$ M1	✓ 1
= 90 seats A1	✓ 1
Part (b)
Sets up $18 + 3(n-1) > 60$, or lists terms using a GDC table M1	✓ 1
$n > 15$, so row 16 (which has 63 seats) A1	✓ 1
Part (c)
$S_{25} = \frac{25}{2}(18 + 90)$ or $\frac{25}{2}(2\times 18 + 24\times 3)$ M1	✓ 1
= 1350 seats A1	✓ 1

Exam tip: When a question asks for the first term to exceed a value, an inequality gives a non-integer or boundary value. Round up to the next whole term number and check that term actually exceeds the value.

+ Add to test✓ In your test20IBMI-030-02-r2-02Arithmetic and geometric sequences in context Paper SLEasy5 marks

A short video is shared 40 times on the first day after it is posted online. The number of shares on each day is modelled as being 15% greater than the number of shares on the previous day.

(a)

Write down the common ratio of the geometric sequence that models the number of shares per day.

(1)

(b)

Find the number of shares on day 10.

(2)

(c)

Find the total number of shares over the first 10 days.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$r = 1.15$ A1	✓ 1
Part (b)
$u_{10} = 40 \times 1.15^{9}$ M1	✓ 1
= 140.7… ≈ 141 shares A1	✓ 1
Part (c)
$S_{10} = \frac{40(1.15^{10} - 1)}{1.15 - 1}$ M1	✓ 1
= 812.1… ≈ 812 shares A1	✓ 1

Common error: Using $1.15^{10}$ for the 10th term. The first term is day 1, so only 9 increases have happened by day 10: the power is $n-1$.

+ Add to test✓ In your test21IBMI-030-03-r2-02Exponents and logarithms Paper SLEasy5 marks

The value, in dollars, of a new electric car $t$ years after it is bought is modelled by
$V = 28\,500\times0.84^{t}$, $t \ge 0$.

(a)

Find the value of the car 4 years after it is bought. Give your answer to two decimal places.

(2)

(b)

Find the value of $t$ when the value of the car first falls to \$8000.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: substitutes $t=4$, $28\,500\times0.84^{4}$	✓ 1
A1: \$14 189.33	✓ 1
Part (b)
M1: sets up the equation $28\,500\times0.84^{t} = 8000$	✓ 1
M1: uses logarithms or technology, e.g. $t = \dfrac{\log(8000/28\,500)}{\log 0.84}$, or finds the intersection on a GDC	✓ 1
A1: $t = 7.29$ years	✓ 1

Exam tip: Money answers should be given to two decimal places. Other answers should be given to 3 significant figures unless the question asks for something else. Keep full calculator values in your working and round only at the end.

+ Add to test✓ In your test22IBMI-030-08-r1-01Exponents and logarithms Paper SLEasy6 marks

The acidity of a liquid is measured on the pH scale, where $\text{pH} = -\log_{10} H$ and $H$ is the hydrogen ion concentration in mol dm⁻³.

(a)

A sample of rainwater has $H = 3.2$ ×10⁻⁴ mol dm⁻³. Calculate its pH.

(2)

(b)

A sample of seawater has a pH of 8.1. Find its hydrogen ion concentration.

(2)

(c)

Lemon juice has a pH of 2.3 and tomato juice has a pH of 4.1. Find how many times greater the hydrogen ion concentration of lemon juice is than that of tomato juice.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: substitutes correctly, pH = −log₁₀(3.2×10⁻⁴)	✓ 1
A1: pH = 3.49	✓ 1
Part (b)
M1: rearranges to H = 10^−8.1	✓ 1
A1: H = 7.94×10⁻⁹ mol dm⁻³	✓ 1
Part (c)
M1: forms the ratio 10^−2.3 ÷ 10^−4.1 or 10^1.8	✓ 1
A1: 63.1 times	✓ 1

Common error: A difference of 1.8 on the pH scale is not a ratio of 1.8. Each unit on a log₁₀ scale is a factor of 10, so the ratio is 10¹.⁸.

+ Add to test✓ In your test23IBMI-030-08-r1-02Exponents and logarithms Paper SLEasy6 marks

After drinking a large coffee, the amount of caffeine, $C$ mg, in Dana's bloodstream $t$ hours later is modelled by $C = 200(0.87)^t$, for $t \ge 0$.

(a)

Interpret, in context, the values 200 and 0.87 in the model.

(2)

(b)

Find the amount of caffeine in Dana's bloodstream 5 hours after drinking the coffee.

(2)

(c)

Dana can sleep well once the amount of caffeine falls below 25 mg. Find how long after drinking the coffee this happens.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1: 200 mg is the initial amount of caffeine in the bloodstream	✓ 1
R1: the amount of caffeine decreases by 13% each hour	✓ 1
Part (b)
M1: substitutes t = 5, giving 200(0.87)⁵	✓ 1
A1: 99.7 mg	✓ 1
Part (c)
M1: sets up 200(0.87)^t = 25, for example by using a graph, a solver or logarithms	✓ 1
A1: t = 14.9 hours	✓ 1

Exam tip: When you use your GDC to solve an equation, write the equation you entered first. This earns the method mark even if the final value is misread.

+ Add to test✓ In your test24IBMI-030-03-r1-01Exponents and logarithms Paper SLEasy5 marks

A pharmacologist models the mass of caffeine, C mg, remaining in a person's bloodstream t hours after drinking a large coffee by
$C(t) = 200(0.87)^t$, $t \ge 0$.

(a)

State the percentage by which the mass of caffeine decreases each hour.

(1)

(b)

Find the mass of caffeine remaining 3 hours after the coffee is drunk.

(2)

(c)

Find the time taken for the mass of caffeine to fall to 50 mg.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
13% decrease per hour A1	✓ 1
Part (b)
Substituting t = 3: 200(0.87)³ M1	✓ 1
= 132 mg (131.70…) A1	✓ 1
Part (c)
Setting 200(0.87)^t = 50 and solving, e.g. t = ln 0.25 / ln 0.87 or using GDC M1	✓ 1
t = 9.95 hours (9.9546…) A1	✓ 1

Common error: The multiplier 0.87 means 87% remains each hour, so the decrease is 13%, not 87%.

+ Add to test✓ In your test25IBMI-030-03-r1-02Exponents and logarithms Paper SLEasy6 marks

The sound level, L decibels (dB), of a sound of intensity I W m⁻² is given by
$L = 10\log_{10}\left(\frac{I}{I_0}\right)$, where $I_0 = 1.0 \times 10^{-12}$ W m⁻².

(a)

A vacuum cleaner produces sound of intensity 3.2 ×10⁻⁵ W m⁻². Find its sound level.

(2)

(b)

The sound level at a concert is 95 dB. Find the intensity of the sound.

(2)

(c)

Find the factor by which the intensity of a sound must be multiplied to increase its sound level by 6 dB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
L = 10 log₁₀(3.2×10⁻⁵ / 1.0×10⁻¹²) = 10 log₁₀(3.2×10⁷) M1	✓ 1
L = 75.1 dB (75.05…) A1	✓ 1
Part (b)
I = 1.0×10⁻¹² × 10^9.5 (or equivalent rearrangement) M1	✓ 1
I = 3.16×10⁻³ W m⁻² A1	✓ 1
Part (c)
10 log₁₀(k) = 6, so k = 10^0.6 M1	✓ 1
k = 3.98 A1	✓ 1

Exam tip: An increase in decibels corresponds to multiplying intensity, not adding to it: a rise of ΔL dB multiplies the intensity by 10ΔL/¹⁰.

+ Add to test✓ In your test26IBMI-030-01-r1-01Financial mathematics, loans and annuities Paper SLEasy6 marks

Priya invests $4500 in a savings account that pays a nominal annual interest rate of 3.6%, compounded monthly. She makes no further deposits or withdrawals.

(a)

Find the value of Priya's investment after 6 years. Give your answer to the nearest cent.

(2)

(b)

Find the least number of complete months after which the value of the investment first exceeds $6000.

(3)

(c)

Find the value of the investment at the end of the month found in the previous part. Give your answer to the nearest cent.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Correct substitution: $4500\left(1+\frac{0.036}{12}\right)^{72}$, or TVM with N = 72, I% = 3.6, PV = −4500, PMT = 0, P/Y = C/Y = 12	✓ 1
$5583.16	✓ 1
Part (b)
Sets up $4500(1.003)^n > 6000$ or equivalent, e.g. TVM solving for N with FV = 6000	✓ 1
$n = 96.04...$	✓ 1
97 months, because the value has not reached $6000 after 96 months	✓ 1
Part (c)
$4500(1.003)^{97}$ = $6017.32	✓ 1

Common error: Writing 96 months loses the final mark. Solving gives n = 96.04, so the balance is still below $6000 after 96 months. Round up to the next whole month.

+ Add to test✓ In your test27IBMI-030-03-r2-01Exponents and logarithms Paper SLEasy6 marks

The sound intensity level $L$, in decibels (dB), of a sound of intensity $I$ W m⁻² is modelled by
$L = 10\log_{10}\left(\dfrac{I}{I_0}\right)$, where $I_0 = 1\times10^{-12}$ W m⁻².
At an outdoor music festival, the intensity near the front of the stage is measured as 0.25 W m⁻².

(a)

Calculate the sound intensity level near the front of the stage.

(2)

(b)

Hearing protection is recommended for sound intensity levels above 85 dB. Find the intensity of a sound with a sound intensity level of 85 dB.

(2)

(c)

Find how many times greater the intensity near the front of the stage is than the intensity of a sound at 85 dB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: substitutes correctly, $L = 10\log_{10}\left(\dfrac{0.25}{1\times10^{-12}}\right)$	✓ 1
A1: $L = 114$ dB (113.979…)	✓ 1
Part (b)
M1: $85 = 10\log_{10}\left(\dfrac{I}{10^{-12}}\right) \Rightarrow I = 10^{-12}\times10^{8.5}$	✓ 1
A1: $I = 3.16\times10^{-4}$ W m⁻²	✓ 1
Part (c)
M1: $\dfrac{0.25}{3.162\ldots\times10^{-4}}$ or $10^{(L-85)/10}$ using their $L$	✓ 1
A1: 791 times (790.57…); accept any answer from 790 to 794 that comes from using rounded values, e.g. $10^{2.9} \approx 794$	✓ 1

Common error: A difference of about 29 dB does not mean the intensity is 29 times greater. Each 10 dB is a factor of 10 in intensity, so convert back to intensities before comparing.

+ Add to test✓ In your test28IBMI-030-07-r2-02Arithmetic and geometric sequences in context Paper SLEasy5 marks

The population of a coastal town was 42 000 at the start of 2020. A planning model assumes that the population increases by 2.5% each year. Let $u_1$ be the population at the start of 2020, $u_2$ the population at the start of 2021, and so on.

(a)

Write down the common ratio of this geometric sequence.

(1)

(b)

Find the year in which the population is first predicted to exceed 60 000 at the start of the year.

(2)

(c)

Find the predicted population at the start of 2030.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $r=1.025$	✓ 1
Part (b)
M1: sets up $42000(1.025)^{n-1}>60000$ and solves it, e.g. with a GDC or logarithms, giving $n-1>14.4$	✓ 1
A1: $n=16$, so the start of 2035	✓ 1
Part (c)
M1: $42000\times1.025^{10}$, i.e. the correct power for $u_{11}$	✓ 1
A1: 53764 (accept 53800 or 53763)	✓ 1

Common error: Mixing up $n$ and $n-1$ in the power. 2030 is 10 years after 2020, so the factor 1.025 is applied 10 times, even though 2030 is the 11th term.

+ Add to test✓ In your test29IBMI-030-07-r2-01Arithmetic and geometric sequences in context Paper SLEasy6 marks

A new outdoor amphitheatre is designed with curved rows of seats. The front row (row 1) has 18 seats, and each row after that has 3 more seats than the row in front of it.

(a)

Find the number of seats in row 20.

(2)

(b)

The designers consider a larger version in which the same pattern continues for as many rows as needed. Find the number of the first row that would have more than 100 seats.

(2)

(c)

The amphitheatre that is actually built has exactly 20 rows. Find the total number of seats.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: use of $u_n=u_1+(n-1)d$ with $u_1=18$, $d=3$, $n=20$, giving $18+19\times3$	✓ 1
A1: $u_{20}=75$ seats	✓ 1
Part (b)
M1: sets up $18+3(n-1)>100$, or an equivalent equation or table of values	✓ 1
A1: $n>28.3$, so row 29 (which has 102 seats)	✓ 1
Part (c)
M1: use of $S_n=\frac{n}{2}(2u_1+(n-1)d)$ or $\frac{n}{2}(u_1+u_n)$ with $n=20$	✓ 1
A1: 930 seats	✓ 1

Exam tip: When the answer must be a whole row number, solve the inequality first and then round up. Check your answer by working out the number of seats in the row before it (row 28 has 99 seats).

+ Add to test✓ In your test30IBMI-030-06-r2-02Financial mathematics, loans and annuities Paper SLEasy5 marks

Daniel buys a car costing 21 500 Australian dollars (AUD). He pays a deposit of AUD 3500 and borrows the rest. The loan is at a nominal annual interest rate of 6.5%, compounded monthly. It is repaid with equal monthly payments, made at the end of each month, over 5 years.

(a)

Find Daniel's monthly payment, giving your answer to the nearest cent.

(3)

(b)

Find the total amount of interest Daniel pays over the 5 years.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Loan amount PV = 21 500 − 3500 = AUD 18 000	✓ 1
Correct financial-solver entries: N = 60, I% = 6.5, PV = ±18 000, FV = 0, P/Y = C/Y = 12, solve for PMT. Alternatively, uses the formula $18000 = P\cdot\frac{1-(1+\frac{0.065}{12})^{-60}}{\frac{0.065}{12}}$	✓ 1
Monthly payment = AUD 352.19	✓ 1
Part (b)
Total repaid = 60 × 352.19 = AUD 21 131.40	✓ 1
Interest = 21 131.40 − 18 000 = AUD 3131.40	✓ 1
Also accept AUD 3131.48, which comes from the unrounded payment of 352.1913…. Follow through from the answer to (a).	—

Common error: Using the full price of the car, AUD 21 500, as the loan amount. Only the amount borrowed after the deposit earns interest.

+ Add to test✓ In your test31IBMI-030-01-r2-02Financial mathematics, loans and annuities Paper SLEasy7 marks

A new electric car is bought for 28 000 dollars. Its value depreciates by 15% of its value at the start of each year.

(a)

Find the value of the car after 4 years.

(2)

(b)

Find the number of complete years after which the value of the car first falls below 5000 dollars.

(3)

(c)

A different model of car loses exactly half of its value in 4 years. Assuming a constant annual rate of depreciation, find this rate as a percentage.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$28000 \times 0.85^4$ M1	✓ 1
14 616.18 dollars (accept 14 600 dollars to 3 s.f.) A1	✓ 1
Part (b)
$28000 \times 0.85^n < 5000$ or equivalent, e.g. a table of values or a GDC solver M1	✓ 1
$n = 10.6...$, OR values after 10 and 11 years found as 5512 and 4686 dollars A1	✓ 1
11 years A1	✓ 1
Part (c)
$(1-r)^4 = 0.5$ M1	✓ 1
$r = 1 - 0.5^{0.25} = 0.159$, so 15.9% A1	✓ 1

Exam tip: For depreciation, the multiplier is 1 − r, not 1 + r. If you use a TVM solver, enter the rate as negative, for example I% = −15.

+ Add to test✓ In your test32IBMI-030-07-r1-02Arithmetic and geometric sequences in context Paper SLEasy5 marks

A short video gets 1200 views on the day it is posted (day 1). The number of views each day is 15% more than on the previous day. Assume this pattern continues.

(a)

Find the number of views on day 8.

(2)

(b)

Find the total number of views over the first 14 days.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$u_8 = 1200 \times 1.15^{7}$ M1	✓ 1
$= 3190$ views (3192.02…; accept 3192) A1	✓ 1
Part (b)
Recognises a geometric series with $u_1 = 1200$ and $r = 1.15$ M1	✓ 1
$S_{14} = \frac{1200(1.15^{14}-1)}{1.15-1}$ A1	✓ 1
$= 48600$ views (48605.6…; accept 48606) A1	✓ 1

Exam tip: An increase of 15% gives a common ratio of 1.15, not 0.15. Day 8 is the 8th term, so the power is 7.

+ Add to test✓ In your test33IBMI-030-00-05Approximation, percentage error and standard form Paper SLMedium6 marks

A spherical weather balloon has a radius of 1.236 m.
An engineer quickly estimates its volume using $V=\frac{4}{3}\pi r^3$ with $\pi$ taken as 3 and the radius rounded to 1.2 m.

(a)

Calculate the actual volume of the balloon.

(2)

(b)

Find the engineer's estimate of the volume.

(1)

(c)

Calculate the percentage error in the engineer's estimate.

(2)

(d)

Explain why the engineer's estimate is smaller than the actual volume.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{4}{3}\pi(1.236)^3$	✓ 1
A1 7.91 m³	✓ 1
Part (b)
A1 $\frac{4}{3}\times3\times1.2^3=6.912$ m³	✓ 1
Part (c)
M1 for $\left\|\frac{6.912-7.909...}{7.909...}\right\|\times100$	✓ 1
A1 12.6 %	✓ 1
Part (d)
R1 both approximations (3 for $\pi$ and 1.2 for the radius) are smaller than the true values, so the errors do not cancel and the estimate is too small	✓ 1

Exam tip: Keep the full calculator value of the actual volume when finding the percentage error; rounding to 7.91 first can change the third significant figure.

+ Add to test✓ In your test34IBMI-030-00-06Approximation, percentage error and standard form Paper SLMedium8 marks

A car travels 248 km between two cities in 2 hours 47 minutes.

(a)

Calculate the average speed of the car in km h⁻¹.

(2)

(b)

Convert the average speed you found into m s⁻¹.

(2)

(c)

Before the journey, the driver estimated the journey time by assuming an average speed of 90 km h⁻¹.
Find the driver's estimated journey time in hours. Hence, using unrounded values throughout, calculate the percentage error in this estimate of the journey time.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for converting the time to $2\frac{47}{60}=2.783...$ h and dividing 248 by it	✓ 1
A1 89.1 km h⁻¹	✓ 1
Part (b)
M1 for $\frac{89.10...\times1000}{3600}$	✓ 1
A1 24.8 m s⁻¹	✓ 1
Part (c)
M1 for $\frac{248}{90}$	✓ 1
A1 2.76 h	✓ 1
M1 for $\left\|\frac{2.7555...-2.7833...}{2.7833...}\right\|\times100$	✓ 1
A1 0.998 %	✓ 1

Common error: 2 hours 47 minutes is not 2.47 hours. Divide the minutes by 60 before using the time in a calculation, and keep unrounded values in your calculator for later parts.

+ Add to test✓ In your test35IBMI-030-00-07Approximation, percentage error and standard form Paper SLMedium11 marks

In a simple model, the Earth and Mars orbit the Sun in circles in the same plane. The orbital radius of the Earth is 1.50×10⁸ km and that of Mars is 2.28×10⁸ km. Radio signals travel at 3.00×10⁵ km s⁻¹.
Mars may be modelled as a sphere of diameter 6.79×10³ km.

(a)

Find the minimum distance between the Earth and Mars according to this model. Give your answer in the form $a\times10^k$ where $1\le a<10$ and $k\in\mathbb{Z}$.

(2)

(b)

Calculate the time, in minutes, for a radio signal to travel from the Earth to Mars at this minimum distance.

(2)

(c)

Calculate the time, in minutes, for a radio signal to travel between the planets when they are on opposite sides of the Sun.

(2)

(d)

Calculate the volume of Mars in km³, giving your answer in the form $a\times10^k$ where $1\le a<10$ and $k\in\mathbb{Z}$.

(3)

(e)

The volume of the Earth is 1.08×10¹² km³. Find how many times as large the volume of the Earth is as the volume of Mars.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 2.28×10⁸ − 1.50×10⁸	✓ 1
A1 7.80×10⁷ km	✓ 1
Part (b)
M1 for $\frac{7.80\times10^{7}}{3.00\times10^{5}}=260$ s	✓ 1
A1 $\frac{260}{60}=4.33$ minutes	✓ 1
Part (c)
M1 for a distance of 1.50×10⁸ + 2.28×10⁸ = 3.78×10⁸ km and $\frac{3.78\times10^{8}}{3.00\times10^{5}}=1260$ s	✓ 1
A1 $\frac{1260}{60}=21.0$ minutes	✓ 1
Part (d)
M1 for radius = 3.395×10³ km, seen or implied (for example by the use of $\frac{\pi d^3}{6}$)	✓ 1
M1 for $\frac{4}{3}\pi(3.395\times10^{3})^3$ or $\frac{\pi(6.79\times10^{3})^3}{6}$	✓ 1
A1 1.64×10¹¹ km³	✓ 1
Part (e)
M1 for $\frac{1.08\times10^{12}}{1.639...\times10^{11}}$	✓ 1
A1 6.59 (accept 6.58 to 6.59)	✓ 1

Common error: The volume formula needs the radius, not the diameter. Halve 6.79×10³ before cubing, or use $\frac{\pi d^3}{6}$.

+ Add to test✓ In your test36IBMI-030-00-08Approximation, percentage error and standard form Paper SLMedium7 marks

A rectangular solar panel has length 1.65 m and width 0.99 m, each measured correct to two decimal places.

(a)

Write down the lower and upper bounds of the length of the panel.

(1)

(b)

Calculate the upper bound of the area of the panel.

(2)

(c)

Calculate the lower bound of the area of the panel.

(1)

(d)

The area is calculated using the measured values 1.65 m and 0.99 m. The exact area of the panel could be any value between the bounds found in parts (b) and (c). Using percentage error $= \left|\frac{v_A - v_E}{v_E}\right| \times 100$, where $v_A$ is the calculated area and $v_E$ is the exact area, find the maximum possible percentage error in the calculated area.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 lower bound 1.645 m and upper bound 1.655 m, i.e. 1.645 ≤ length < 1.655	✓ 1
Part (b)
M1 for 1.655 × 0.995	✓ 1
A1 1.65 m² (1.646725)	✓ 1
Part (c)
A1 1.645 × 0.985 = 1.62 m² (1.620325)	✓ 1
Part (d)
M1 for calculated area 1.65 × 0.99 = 1.6335 m²	✓ 1
M1 for substituting the lower bound as $v_E$: $\frac{1.6335-1.620325}{1.620325}\times100$. Also award this mark if both bounds are substituted and the two results are compared.	✓ 1
A1 0.813 % (accept 0.81 %)	✓ 1

Exam tip: The exact value always goes in the denominator. Try both bounds. The lower bound gives 0.813 % and the upper bound gives 0.803 %, so the maximum is 0.813 %.

+ Add to test✓ In your test37IBMI-030-00-09Approximation, percentage error and standard form Paper SLMedium5 marks

A person's heart beats on average 72 times per minute. Assume that a year has 365.25 days and that the person lives for 80 years.

(a)

Calculate the total number of heartbeats in the person's lifetime. Give your answer in the form $a\times10^k$ where $1\le a<10$ and $k\in\mathbb{Z}$.

(3)

(b)

A student estimates the total by using 70 beats per minute and 500 000 minutes per year. Calculate the percentage error in the student's estimate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for minutes in a year 60 × 24 × 365.25 = 525 960	✓ 1
M1 for 72 × 525 960 × 80	✓ 1
A1 3.03×10⁹	✓ 1
Part (b)
M1 for estimate 70 × 500 000 × 80 = 2.8×10⁹ and $\left\|\frac{2.8\times10^9-3.0295...\times10^9}{3.0295...\times10^9}\right\|\times100$	✓ 1
A1 7.58 %	✓ 1

Exam tip: When an answer will be used again, store the unrounded value in your calculator to avoid accumulated rounding errors.

+ Add to test✓ In your test38IBMI-030-00-10Approximation, percentage error and standard form Paper SLMedium6 marks

The dose of a medicine for a child is 0.025 mg per kilogram of body mass. A child has a mass of 18.6 kg.

(a)

Calculate the correct dose for this child in mg. Give your answer in the form $a\times10^k$ where $1\le a<10$ and $k\in\mathbb{Z}$.

(2)

(b)

Write down the dose in grams, in the form $a\times10^k$ where $1\le a<10$ and $k\in\mathbb{Z}$.

(1)

(c)

A nurse rounds the child's mass to 20 kg before calculating the dose. Calculate the percentage error in the dose given.

(2)

(d)

Hospital policy requires the percentage error in any dose to be less than 5 %. State, with a reason, whether the nurse's method meets this policy.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 0.025 × 18.6 = 0.465	✓ 1
A1 4.65×10⁻¹ mg	✓ 1
Part (b)
A1 4.65×10⁻⁴ g (follow through from their answer to part (a))	✓ 1
Part (c)
M1 for a dose of 0.025 × 20 = 0.5 mg substituted into $\left\|\frac{0.5-0.465}{0.465}\right\|\times100$	✓ 1
A1 7.53 % (accept 7.5 %)	✓ 1
Part (d)
R1 No, because 7.53 % > 5 %. Follow through from their part (c) answer if the conclusion is consistent with it	✓ 1

Common error: 1 g = 1000 mg, so converting mg to g means dividing by 10³. The power of ten becomes more negative, not less.

+ Add to test✓ In your test39IBMI-030-05-03Approximation, percentage error and standard form Paper SLMedium6 marks

A surveyor estimates by eye that a radio tower is 48 m tall. A laser measurement then gives the height as 52.6 m, correct to the nearest 0.1 m.

(a)

Taking 52.6 m as the exact height, calculate the percentage error in the surveyor's estimate.

(2)

(b)

Write down the lower bound and the upper bound for the true height of the tower.

(2)

(c)

Find the largest possible percentage error when 52.6 m is used in place of the true height.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{\|48-52.6\|}{52.6} \times 100\%$	✓ 1
A1 8.75%	✓ 1
Part (b)
A1 lower bound 52.55 m	✓ 1
A1 upper bound 52.65 m	✓ 1
Part (c)
M1 for using the lower bound as the true value, $\frac{0.05}{52.55} \times 100\%$	✓ 1
A1 0.0951%	✓ 1

Exam tip: The maximum error is 0.05 m for both bounds. Dividing by the smaller true value, 52.55, gives the larger percentage.

+ Add to test✓ In your test40IBMI-030-05-06Approximation, percentage error and standard form Paper SLMedium7 marks

A car travels 386 km, measured to the nearest km. It uses 27.4 litres of fuel, measured to the nearest 0.1 litre.

(a)

Write down the lower bound and the upper bound for the distance travelled.

(1)

(b)

Calculate the upper bound for the fuel efficiency of the car, in km per litre.

(2)

(c)

Calculate the lower bound for the fuel efficiency of the car, in km per litre.

(2)

(d)

Hence state the fuel efficiency to an appropriate degree of accuracy, giving a reason.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 385.5 km and 386.5 km, both required	✓ 1
Part (b)
M1 for upper distance ÷ lower fuel, 386.5 ÷ 27.35	✓ 1
A1 14.1 km/L (14.131…)	✓ 1
Part (c)
M1 for lower distance ÷ upper fuel, 385.5 ÷ 27.45	✓ 1
A1 14.0 km/L (14.043…)	✓ 1
Part (d)
A1 14 km/L	✓ 1
R1 both bounds round to 14 at two significant figures, but they differ at three significant figures (14.1 and 14.0); award only with the correct value 14	✓ 1

Common error: For the upper bound of a quotient, divide the upper bound of the numerator by the LOWER bound of the denominator. Using both upper bounds does not give the maximum.

+ Add to test✓ In your test41IBMI-030-05-07Approximation, percentage error and standard form Paper SLMedium6 marks

The distance from Earth to Proxima Centauri is 4.02×10¹³ km. A spacecraft travels at a constant speed of 6.1×10⁴ km h⁻¹.

(a)

Find the time, in hours, that the spacecraft takes to reach Proxima Centauri. Give your answer in the form $a \times 10^k$, where $1 \le a < 10$ and $k \in \mathbb{Z}$.

(2)

(b)

Taking one year as 365.25 days, convert your answer to part (a) into years. Give your answer in the form $a \times 10^k$.

(2)

(c)

A student repeats the calculation in part (a) with the distance rounded to 4×10¹³ km. Find the percentage error in the student's time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 4.02×10¹³ ÷ 6.1×10⁴	✓ 1
A1 6.59×10⁸ hours	✓ 1
Part (b)
M1 for dividing by 24 × 365.25	✓ 1
A1 7.52×10⁴ years	✓ 1
Part (c)
M1 for recognising that time is proportional to distance, so the percentage error is $\frac{0.02}{4.02} \times 100\%$, or for comparing the two times directly	✓ 1
A1 0.498%	✓ 1

Exam tip: Time is directly proportional to distance at constant speed. The percentage error in the time is therefore the same as the percentage error in the distance, so you do not need to redo the full calculation.

+ Add to test✓ In your test42IBMI-030-05-09Approximation, percentage error and standard form Paper SLMedium5 marks

A streaming company stores video files that are each 2.35×10⁹ bytes. One of its servers has a capacity of 4.8×10¹⁵ bytes.

(a)

Find the number of complete video files that the server can store.

(2)

(b)

Write the number of complete video files found in the previous part in the form $a \times 10^k$, where $1 \le a < 10$ and $k \in \mathbb{Z}$, correct to three significant figures.

(1)

(c)

An advert claims that the server holds "2 million videos". Taking the number of complete video files as the exact value, calculate the percentage error in this claim.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for (4.8×10¹⁵) ÷ (2.35×10⁹) = 2 042 553.19…	✓ 1
A1 2 042 553 (rounded down to a whole number of complete files)	✓ 1
Part (b)
A1 2.04×10⁶	✓ 1
Part (c)
M1 for $\frac{\|2\,000\,000-2\,042\,553\|}{2\,042\,553} \times 100\%$	✓ 1
A1 2.08%	✓ 1

Common error: When dividing in standard form, divide the front numbers (4.8 ÷ 2.35) and subtract the indices as 15 − 9 = 6. Subtracting them the wrong way round gives 10⁻⁶, which is clearly unrealistic for a number of files. Also remember to round down: a partial file cannot be stored.

+ Add to test✓ In your test43IBMI-030-05-12Approximation, percentage error and standard form Paper SLMedium10 marks

In one year, global carbon dioxide emissions were 3.71×10¹⁰ tonnes. Country A emitted 5.06×10⁹ tonnes and had a population of 3.34×10⁸. The mass of Earth's atmosphere is 5.15×10¹⁸ kg (1 tonne = 1000 kg).

(a)

Calculate the percentage of global emissions produced by Country A.

(2)

(b)

Calculate the emissions per person in Country A, in tonnes.

(2)

(c)

Write the global emissions in kg. Hence find the ratio of the mass of the atmosphere to the mass of one year's global emissions. Give your answer in the form $a \times 10^k$, where $1 \le a < 10$ and $k \in \mathbb{Z}$.

(3)

(d)

A journalist rounds global emissions to 4×10¹⁰ tonnes and Country A's emissions to 5×10⁹ tonnes. The journalist then reports Country A's share as 12.5%. Calculate the percentage error in the journalist's figure, compared with your unrounded answer to part (a).

(2)

(e)

The journalist changed Country A's figure by about 1.2% and the global figure by about 7.8%. Explain why the percentage error in the reported share is larger than either of these changes.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{5.06\times10^9}{3.71\times10^{10}} \times 100$	✓ 1
A1 13.6% (13.638…)	✓ 1
Part (b)
M1 for 5.06×10⁹ ÷ 3.34×10⁸	✓ 1
A1 15.1 tonnes per person (15.149…)	✓ 1
Part (c)
A1 3.71×10¹³ kg	✓ 1
M1 for 5.15×10¹⁸ ÷ 3.71×10¹³	✓ 1
A1 1.39×10⁵	✓ 1
Part (d)
M1 for $\frac{\|12.5-13.638\ldots\|}{13.638\ldots} \times 100\%$	✓ 1
A1 8.35%	✓ 1
Part (e)
R1 the numerator was rounded down and the denominator was rounded up, so both roundings make the quotient smaller and their effects combine rather than cancel	✓ 1

Exam tip: In part (d), use the unrounded value 13.638… from your calculator as the exact value. Using 13.6 gives 8.09%, which is not accurate enough.

+ Add to test✓ In your test44IBMI-030-02-r1-04Arithmetic and geometric sequences in context Paper SLMedium7 marks

A new car costs $28 500. Its value decreases by 12% of its value at the start of 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)

(c)

A different car of the same original price is worth $13 000 after 5 years. Find its annual rate of depreciation, assuming the rate is constant.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses $28500\times0.88^{5}$ M1 ; $15 040.36 A1	✓ 2
Part (b)
Forms $28500\times0.88^{n}<10000$ or uses a table M1 ; $n>8.19$, or value after 8 years ≈ $10 249.58 and after 9 years ≈ $9019.63 A1 ; 9 years A1	✓ 3
Part (c)
Forms $28500(1-r)^{5}=13000$ M1 ; $r=14.5\%$ A1	✓ 2

Common error: Depreciation of 12% means the multiplier is 0.88, not 0.12. In part (c), solve for (1 − r) first and then subtract from 1 to get the rate.

+ Add to test✓ In your test45IBMI-030-06-r2-04Financial mathematics, loans and annuities Paper SLMedium7 marks

Amara needs a mortgage of 320 000 US dollars (USD), repaid with equal monthly payments, made at the end of each month, over 25 years. Two banks make offers.
Bank A: a nominal annual interest rate of 4.8%, compounded monthly, with no fees.
Bank B: a nominal annual interest rate of 4.5%, compounded monthly, with a USD 6500 arrangement fee added to the amount borrowed.

(a)

Find the monthly payment for Bank A.

(2)

(b)

Find the monthly payment for Bank B.

(2)

(c)

Determine which bank offers the lower total cost, and by how much.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
N = 300, I% = 4.8, PV = 320 000, FV = 0, P/Y = C/Y = 12	✓ 1
USD 1833.59	✓ 1
Part (b)
PV = 326 500 with N = 300, I% = 4.5	✓ 1
USD 1814.79	✓ 1
Part (c)
Totals: A = 300 × 1833.59 = USD 550 077.00 and B = 300 × 1814.79 = USD 544 437.00	✓ 1
Difference = USD 5640.00 (accept 5639.19 from unrounded payments)	✓ 1
Bank B costs less in total, even after the fee, because the lower rate saves more than USD 6500 in interest over 25 years	✓ 1

Common error: Leaving out the arrangement fee, or adding it to the total cost without interest. When the fee is added to the loan, interest is charged on it for the full term.

+ Add to test✓ In your test46IBMI-030-07-r1-03Arithmetic and geometric sequences in context Paper SLMedium7 marks

Two athletes follow different training plans.
Amara runs 12 km in week 1 and increases her weekly distance by 2.5 km each week.
Ben runs 10 km in week 1 and increases his weekly distance by 8% each week.

(a)

Find the distance Amara runs in week 10.

(2)

(b)

Find the total distance Ben runs over the first 10 weeks.

(2)

(c)

Find the first week in which Ben's weekly distance is greater than Amara's.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$u_{10} = 12 + 9(2.5)$ M1	✓ 1
$= 34.5$ km A1	✓ 1
Part (b)
$S_{10} = \frac{10(1.08^{10}-1)}{0.08}$ M1	✓ 1
$= 145$ km (144.865…) A1	✓ 1
Part (c)
Sets up $10(1.08)^{n-1} > 12 + 2.5(n-1)$ M1	✓ 1
Uses a table or graph on the GDC, e.g. week 27: Ben 73.96 km, Amara 77 km; week 28: Ben 79.88 km, Amara 79.5 km M1	✓ 1
Week 28 A1	✓ 1

Exam tip: When one side of an inequality is linear and the other exponential, use a GDC table or a graph intersection. Quote the values on either side of the change to support the answer.

+ Add to test✓ In your test47IBMI-030-07-r1-04Arithmetic and geometric sequences in context Paper SLMedium6 marks

A patient takes a 200 mg tablet of a medicine every 24 hours. By the time the next tablet is due, 35% of the medicine present immediately after the previous dose is still in the body.

(a)

Find the amount of medicine in the body immediately after the 3rd tablet.

(2)

(b)

Show that the amount immediately after the $n$th tablet is $\frac{200(1-0.35^{n})}{0.65}$ mg.

(2)

(c)

Find the amount of medicine in the body immediately after the 7th tablet.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$200 + 200(0.35) + 200(0.35)^2$ M1	✓ 1
$= 294.5$ mg (accept 295 mg) A1	✓ 1
Part (b)
The amount is $200 + 200(0.35) + \dots + 200(0.35)^{n-1}$, a geometric series with $u_1 = 200$ and $r = 0.35$ R1	✓ 1
$S_n = \frac{200(1-0.35^{n})}{1-0.35} = \frac{200(1-0.35^{n})}{0.65}$ A1	✓ 1
Part (c)
$\frac{200(1-0.35^{7})}{0.65}$ M1	✓ 1
$= 307$ mg (307.494…) A1	✓ 1

Common error: The first tablet has been in the body the longest, so it has shrunk the most. The newest tablet is the full 200 mg. Do not apply the factor 0.35 to every term.

+ Add to test✓ In your test48IBMI-030-01-r1-03Financial mathematics, loans and annuities Paper SLMedium7 marks

Helena retires with a fund of $250 000. The fund earns a nominal annual interest rate of 4.2%, compounded quarterly. She plans to withdraw an equal amount at the end of every quarter for 20 years, so that the fund is exactly used up.

(a)

Find the amount of each quarterly withdrawal.

(3)

(b)

Find the total interest earned by the fund over the 20 years.

(2)

(c)

Helena decides instead to withdraw $5000 at the end of every quarter. Find the number of full $5000 withdrawals she can make.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
N = 80 with quarterly compounding (P/Y = C/Y = 4)	✓ 1
PV = ±250 000 and FV = 0, with I% = 4.2	✓ 1
$4634.58	✓ 1
Part (b)
Total withdrawn: 80 × 4634.58 = $370 766.40	✓ 1
Interest = 370 766.40 − 250 000 = $120 766.40	✓ 1
Part (c)
TVM with PV = 250 000, PMT = −5000, FV = 0, I% = 4.2, solving for N	✓ 1
N = 71.3, so 71 full withdrawals (followed by a smaller final one)	✓ 1

Exam tip: In context, a non-integer N means the last withdrawal is only part of a full payment. State the number of complete withdrawals, not the decimal.

+ Add to test✓ In your test49IBMI-030-01-r1-04Financial mathematics, loans and annuities Paper SLMedium7 marks

Jonas has €12 000 to invest for 10 years and is comparing two accounts.
Option A: 5.1% per annum, compounded annually.
Option B: 4.95% nominal annual rate, compounded monthly.

(a)

Find the value of the investment after 10 years under Option A.

(2)

(b)

Find the value of the investment after 10 years under Option B.

(2)

(c)

State which option gives the greater return, and by how much.

(1)

(d)

The annual inflation rate is expected to remain at 2.3% for the 10 years. Find the value of the Option A investment after 10 years in terms of today's money. Give your answer to the nearest euro.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$12000(1.051)^{10}$	✓ 1
€19 733.69	✓ 1
Part (b)
$12000\left(1+\frac{0.0495}{12}\right)^{120}$	✓ 1
€19 665.95	✓ 1
Part (c)
Option A, by €67.74	✓ 1
Part (d)
Divides by $1.023^{10}$: $\frac{19733.69}{1.023^{10}}$	✓ 1
€15 720	✓ 1

Common error: A higher compounding frequency does not always give a better return. Compare the final values or the effective annual rates. Here monthly compounding at 4.95% does not make up for the lower nominal rate.

+ Add to test✓ In your test50IBMI-030-08-r2-03Exponents and logarithms Paper SLMedium10 marks

An invasive weed is spreading across the surface of a pond. The area covered, A m², t days after it was first measured is modelled by
$A(t) = 2.4 \times b^{t}$
where $b$ is a constant. Three days after the first measurement, the weed covers 5.9 m².

(a)

Find the value of b.

(2)

(b)

Hence write down the percentage increase in the area covered each day.

(1)

(c)

Find the time it takes for the area covered to double.

(2)

(d)

The surface area of the pond is 500 m². Use the model to find the time at which the weed covers the whole surface.

(3)

(e)

Explain why this model is unlikely to be accurate close to the time found in part (d).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: forms the equation $2.4b^{3} = 5.9$	✓ 1
A1: $b = 1.35$ (1.34963…)	✓ 1
Part (b)
A1: 35.0% (accept 35%)	✓ 1
Part (c)
M1: forms the equation $2.4b^{t} = 4.8$ or $b^{t} = 2$	✓ 1
A1: $t = 2.31$ days	✓ 1
Part (d)
M1: forms the equation $2.4b^{t} = 500$	✓ 1
M1: uses logarithms, $t = \frac{\ln(500/2.4)}{\ln b}$, or solves with GDC	✓ 1
A1: $t = 17.8$ days	✓ 1
Part (e)
R1: the pond surface and resources such as light and nutrients are limited, so exponential growth cannot continue	✓ 1
R1: the rate of spread will slow down as the area covered approaches 500 m², so the model overestimates the area or underestimates the time	✓ 1

Exam tip: Keep the unrounded value of b in your GDC for parts (c) and (d). Using $b = 1.35$ in part (d) gives a time of about 17.8 days here, but early rounding often changes the third significant figure.

+ Add to test✓ In your test51IBMI-030-08-r2-05Exponents and logarithms Paper SLMedium7 marks

A mug of tea is left to cool in a room. Its temperature, T °C, t minutes after it is made is modelled by
$T(t) = 22 + 68e^{-kt}$
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 made.

(1)

(b)

Find the value of k.

(3)

(c)

Find the time it takes for the tea to cool to 30 °C.

(2)

(d)

State the meaning of the value 22 in this context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: 90 °C	✓ 1
Part (b)
M1: substitutes, $55 = 22 + 68e^{-10k}$	✓ 1
M1: isolates the exponential and uses logarithms, $-10k = \ln\left(\frac{33}{68}\right)$, or solves with GDC	✓ 1
A1: $k = 0.0723$	✓ 1
Part (c)
M1: forms the equation $22 + 68e^{-kt} = 30$	✓ 1
A1: $t = 29.6$ minutes	✓ 1
Part (d)
R1: the temperature of the room, which the tea approaches in the long term but does not go below	✓ 1

Common error: Subtract 22 before taking logarithms. Taking logarithms of $22 + 68e^{-kt}$ term by term is invalid, because the logarithm of a sum is not the sum of the logarithms.

+ Add to test✓ In your test52IBMI-030-07-r2-04Arithmetic and geometric sequences in context Paper SLMedium7 marks

A small gold mine produces 12 000 kg of ore in its first year of operation. A geologist models the output in each following year as 6% less than the output in the year before.

(a)

Find the output in the 8th year.

(2)

(b)

Find the total output over the first 15 years.

(2)

(c)

The mine's owners claim that, if the model continues indefinitely, the total output will eventually exceed 200 000 kg. Use the model to show that this claim is incorrect.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $12000\times0.94^{7}$	✓ 1
A1: 7780 kg (7781.7)	✓ 1
Part (b)
M1: use of $S_{15}=\frac{12000(1-0.94^{15})}{1-0.94}$	✓ 1
A1: 121 000 kg (120 942)	✓ 1
Part (c)
M1: use of $S_\infty=\frac{u_1}{1-r}=\frac{12000}{0.06}$	✓ 1
A1: $S_\infty=200000$ kg	✓ 1
R1: since $\|r\|=0.94<1$, the partial sums increase towards 200 000 kg but never reach it, so the total can never exceed 200 000 kg and the claim is incorrect	✓ 1

Common error: Using $r=0.06$ instead of $r=0.94$. A decrease of 6% leaves 94% of the previous value, so the common ratio is 0.94.

+ Add to test✓ In your test53IBMI-030-02-r2-03Arithmetic and geometric sequences in context Paper SLMedium7 marks

A ball is dropped from a height of 2.5 m onto a hard floor. After each bounce it rises to 70% of the height from which it last fell.

(a)

Find the maximum height reached by the ball after its 4th bounce.

(2)

(b)

Find the number of the first bounce after which the maximum height is less than 5 cm.

(2)

(c)

Find the total vertical distance travelled by the ball from the moment it is dropped until it hits the floor for the 5th time.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$2.5 \times 0.7^{4}$ M1	✓ 1
= 0.600 m (0.60025 m) A1	✓ 1
Part (b)
Sets up $2.5 \times 0.7^{n} < 0.05$ and solves (logs or GDC), giving $n > 10.97$ M1	✓ 1
11th bounce A1	✓ 1
Part (c)
Sum of first four rebound heights: $\frac{1.75(1 - 0.7^{4})}{1 - 0.7} = 4.43275$ M1	✓ 1
Total $= 2.5 + 2 \times 4.43275$ (each rebound is travelled up and down) M1	✓ 1
= 11.4 m (11.3655 m) A1	✓ 1

Common error: Forgetting that each rebound height is travelled twice (up and down) while the initial drop is travelled only once, or including a fifth rebound that has not happened before the 5th impact.

+ Add to test✓ In your test54IBMI-030-02-r2-04Arithmetic and geometric sequences in context Paper SLMedium7 marks

Priya is offered two job contracts, each lasting 10 years.
Contract A: a starting annual salary of \$42 000, increasing by \$1500 each year.
Contract B: a starting annual salary of \$40 000, increasing by 4.5% each year.

(a)

Find Priya's annual salary in year 8 under each contract.

(3)

(b)

Determine which contract gives Priya the greater total earnings over the 10 years. Justify your answer.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Contract A: $42000 + 7 \times 1500 = \$52\,500.00$ A1	✓ 1
Contract B: $40000 \times 1.045^{7}$ M1	✓ 1
= \$54 434.47 A1	✓ 1
Part (b)
Contract A: $S_{10} = \frac{10}{2}(2\times 42000 + 9\times 1500) = \$487\,500.00$ A1	✓ 1
Contract B: $S_{10} = \frac{40000(1.045^{10} - 1)}{0.045}$ M1	✓ 1
= \$491 528.37 A1	✓ 1
Contract B gives greater total earnings (by about \$4028) R1	✓ 1

Exam tip: For a comparison question, the reasoning mark depends on stating both totals and a clear conclusion in context. Give money answers to 2 decimal places.

+ Add to test✓ In your test55IBMI-030-03-r2-04Exponents and logarithms Paper SLMedium11 marks

A cup of coffee contains 180 mg of caffeine. The amount of caffeine, $C$ mg, remaining in an adult's body $t$ hours after drinking the coffee is modelled by
$C(t) = 180\,e^{-kt}$, $t\ge0$,
where $k$ is a positive constant. The half-life of caffeine in this adult's body is 5.5 hours.

(a)

Show that $k = 0.126$, correct to three significant figures.

(2)

(b)

Find the amount of caffeine remaining 8 hours after the coffee is drunk.

(2)

(c)

Find the time taken for the amount of caffeine to fall to 20 mg.

(3)

(d)

The adult drinks one cup of this coffee at 07:00 and a second, identical cup at 12:00. Assuming the model applies to each cup independently, find the total amount of caffeine in the adult's body at 15:00.

(3)

(e)

State one limitation of the model in this context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: sets up the equation $90 = 180e^{-5.5k}$, or $e^{-5.5k} = 0.5$	✓ 1
A1: $k = \dfrac{\ln 2}{5.5} = 0.126027\ldots$, which gives 0.126	✓ 1
Part (b)
M1: substitutes $t=8$, $180e^{-0.126027\times8}$	✓ 1
A1: 65.7 mg	✓ 1
Part (c)
M1: sets up the equation $180e^{-kt} = 20$	✓ 1
M1: rearranges with logarithms, $t = \dfrac{\ln 9}{k}$, or uses a GDC	✓ 1
A1: $t = 17.4$ hours	✓ 1
Part (d)
M1: uses $t=8$ for the first cup and $t=3$ for the second cup	✓ 1
M1: adds the two amounts, $180e^{-8k} + 180e^{-3k}$ (65.68 + 123.33)	✓ 1
A1: 189 mg	✓ 1
Part (e)
R1: any valid limitation, e.g. it assumes all the caffeine is absorbed instantly when the coffee is drunk; the rate of removal varies between people and with factors such as food or medication; it predicts that caffeine is never completely removed	✓ 1

Exam tip: In a 'show that' question you must show the unrounded value, such as 0.126027…, before rounding it. In later parts, use the unrounded value of k so that rounding errors do not build up.

+ Add to test✓ In your test56IBMI-030-03-r1-04Exponents and logarithms Paper SLMedium7 marks

The populations of two towns, A and B, are modelled t years after the start of 2025 by
$P_A = 12\,000(1.035)^t$ and $P_B = 18\,500(1.012)^t$.

(a)

Find the population of town A after 10 years.

(2)

(b)

Interpret the value 1.012 in the context of the model for town B.

(1)

(c)

Find the value of t at which the two populations are equal.

(3)

(d)

Suggest one reason why the model for town A may not be valid for large values of t.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
12000(1.035)¹⁰ M1	✓ 1
= 16 900 (16 927.2…) A1	✓ 1
Part (b)
The population of town B increases by 1.2% each year A1	✓ 1
Part (c)
12000(1.035)^t = 18500(1.012)^t M1	✓ 1
(1.035/1.012)^t = 18500/12000, so t = ln(1.5417)/ln(1.0227) or solved graphically M1	✓ 1
t = 19.3 years (19.26…) A1	✓ 1
Part (d)
E.g. a constant 3.5% growth rate cannot continue indefinitely because of limited housing, land or resources R1	✓ 1

Common error: Dividing the two growth factors is correct, but subtracting the percentages (3.5% − 1.2%) to form a single factor of 1.023 gives an inaccurate answer.

+ Add to test✓ In your test57IBMI-030-03-r2-03Exponents and logarithms Paper SLMedium7 marks

The acidity of a solution is measured by its pH, defined by
$\text{pH} = -\log_{10}[\text{H}^+]$,
where $[\text{H}^+]$ is the hydrogen ion concentration in mol dm⁻³.

(a)

A sample of lemon juice has $[\text{H}^+] = 6.3\times10^{-3}$ mol dm⁻³. Calculate its pH.

(2)

(b)

A sample of rainwater has a pH of 5.6. Find its hydrogen ion concentration.

(2)

(c)

Over several years, the pH of a mountain lake fell from 6.8 to 6.2. Find the factor by which the hydrogen ion concentration of the lake increased, and interpret your answer in context.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $-\log_{10}(6.3\times10^{-3})$	✓ 1
A1: pH = 2.20	✓ 1
Part (b)
M1: $[\text{H}^+] = 10^{-5.6}$	✓ 1
A1: $2.51\times10^{-6}$ mol dm⁻³	✓ 1
Part (c)
M1: forms the ratio $\dfrac{10^{-6.2}}{10^{-6.8}}$ or $10^{0.6}$	✓ 1
A1: factor = 3.98	✓ 1
R1: the hydrogen ion concentration of the lake water became about 4 times greater, even though the pH fell by only 0.6	✓ 1

Common error: A pH change of 0.6 does not mean the concentration changes by 0.6. On a logarithmic scale, a difference in pH corresponds to multiplying the concentration by 10 raised to that difference.

+ Add to test✓ In your test58IBMI-030-07-r1-05Arithmetic and geometric sequences in context Paper SLMedium13 marks

Priya is offered two jobs.
Offer A: a salary of \$42 000 in year 1, rising by \$1800 each year.
Offer B: a salary of \$38 000 in year 1, rising by 6% each year.

(a)

Find Priya's salary in year 5 under each offer.

(3)

(b)

Find the total amount Priya would earn over the first 10 years under each offer.

(4)

(c)

Priya plans to stay in the job for exactly 10 years. Using your answers to the previous part, state which offer earns her more money, giving a reason.

(1)

(d)

Find the first year in which the annual salary under Offer B is greater than under Offer A.

(2)

(e)

Find the least number of complete years after which the total earned under Offer B is greater than the total earned under Offer A.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Offer A: $42000 + 4(1800) = \$49\,200$ A1	✓ 1
Offer B: $38000 \times 1.06^{4}$ M1	✓ 1
$= \$47\,974.12$ A1	✓ 1
Part (b)
Offer A: $\frac{10}{2}(2\times42000 + 9\times1800)$ M1	✓ 1
$= \$501\,000$ A1	✓ 1
Offer B: $\frac{38000(1.06^{10}-1)}{0.06}$ M1	✓ 1
$= \$500\,870.21$ A1	✓ 1
Part (c)
Offer A, because its 10-year total (\$501 000) is about \$130 more than Offer B's (\$500 870.21) R1	✓ 1
Part (d)
Compares $38000(1.06)^{n-1}$ with $42000 + 1800(n-1)$ by table, graph or solver, e.g. year 6: B \$50 852.57 < A \$51 000; year 7: B \$53 903.73 > A \$52 800 M1	✓ 1
Year 7 A1	✓ 1
Part (e)
Sets up $\frac{38000(1.06^{n}-1)}{0.06} > \frac{n}{2}(84000 + 1800(n-1))$ or equivalent M1	✓ 1
Evaluates both totals for $n = 10$ and $n = 11$: $n=10$: B \$500 870 < A \$501 000; $n=11$: B \$568 922 > A \$561 000 (or uses a graph or solver to find the crossing at $n \approx 10.02$) M1	✓ 1
11 years A1	✓ 1

Exam tip: Check whether the question compares the salary in a single year (terms) or the amount earned so far (sums). The two cross over in different years.

+ Add to test✓ In your test59IBMI-030-06-r2-05Financial mathematics, loans and annuities Paper SLMedium6 marks

Kenji wants to save 40 000 Canadian dollars (CAD) for a house deposit in 6 years. He makes equal deposits at the end of each month into an account that pays a nominal annual interest rate of 3.6%, compounded monthly.

(a)

Find the monthly deposit needed if the account starts with a balance of zero.

(3)

(b)

Kenji's grandmother gives him CAD 5000, which he puts into the account at the start. Find the new monthly deposit needed to reach CAD 40 000 in 6 years.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses 72 monthly periods at a monthly rate of 0.3% (e.g. N = 72, I% = 3.6, P/Y = C/Y = 12)	✓ 1
Correct set-up for end-of-month payments: PV = 0, FV = 40 000, or $\text{PMT}\times\frac{1.003^{72}-1}{0.003}=40000$	✓ 1
Monthly deposit = CAD 498.54	✓ 1
Part (b)
Includes the initial CAD 5000 with the same N and interest rate (e.g. PV = −5000, with PV and PMT both outflows), or finds that 5000 grows to $5000\times1.003^{72}\approx 6203.51$	✓ 1
Correct equation: $5000\times1.003^{72}+\text{PMT}\times\frac{1.003^{72}-1}{0.003}=40000$, or the equivalent TVM entry with FV = 40 000	✓ 1
Monthly deposit = CAD 421.23	✓ 1

Exam tip: In a TVM solver, money you pay into the account (the initial gift and the deposits) takes one sign, and the amount you receive at the end takes the opposite sign. The PMT shown will therefore be negative, so give the deposit as a positive amount.

+ Add to test✓ In your test60IBMI-030-02-r1-05Arithmetic and geometric sequences in context Paper SLMedium11 marks

A ball is dropped from a height of 2.5 m onto a hard floor. After each bounce it rises to 70% of the height it reached before that bounce. Assume the ball moves only vertically.

(a)

Find the height the ball reaches after the 4th bounce.

(2)

(b)

Find the greatest number of bounces for which the ball still rises to a height of at least 1 cm after the bounce.

(3)

(c)

Find the total vertical distance the ball has travelled when it hits the floor for the 5th time.

(3)

(d)

Using the model, find the total vertical distance the ball would travel if it bounced indefinitely.

(2)

(e)

Comment on the validity, in context, of your answer to the previous part.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses $2.5\times0.7^{4}$ M1 ; 0.600 m (0.60025) A1	✓ 2
Part (b)
Forms $2.5\times0.7^{n}\ge0.01$ (or an equivalent with consistent units) M1 ; $n\le15.48$, or bounce 15 gives about 1.19 cm and bounce 16 gives about 0.831 cm A1 ; 15 bounces A1	✓ 3
Part (c)
Recognises that the initial 2.5 m is travelled once and each of the first 4 bounce heights is travelled twice M1 ; uses $2.5+2\times\frac{1.75(1-0.7^{4})}{0.3}$ or $2.5+2(1.75+1.225+0.8575+0.60025)$ M1 ; 11.4 m (11.3655) A1	✓ 3
Part (d)
Uses $2.5+2\times\frac{1.75}{1-0.7}$ M1 ; 14.2 m (14.1667) A1	✓ 2
Part (e)
The model assumes infinitely many bounces, but a real ball comes to rest after a finite number of bounces, for example because of energy losses. The answer is therefore an upper estimate, although the later bounces add very little distance R1	✓ 1

Common error: The first drop is travelled only once. Every height after a bounce is travelled twice, once up and once down. Add the 2.5 m separately rather than doubling it.

+ Add to test✓ In your test61IBMI-030-06-r2-03Financial mathematics, loans and annuities Paper SLMedium10 marks

Marguerite retires with 250 000 Swiss francs (CHF) in an annuity account. The account pays a nominal annual interest rate of 4.2%, compounded quarterly. She plans to withdraw an equal amount at the end of every quarter for 20 years, so that the account is empty after the final withdrawal.

(a)

Find the amount of each quarterly withdrawal.

(3)

(b)

Find the total interest earned by the account over the 20 years.

(2)

(c)

Marguerite instead withdraws CHF 5000 at the end of each quarter. Find the number of full CHF 5000 withdrawals she can make.

(3)

(d)

Find the amount left in the account immediately after the last full CHF 5000 withdrawal.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
N = 80 and I% = 4.2 with P/Y = C/Y = 4, or quarterly rate r = 0.0105	✓ 1
PV = 250 000, FV = 0	✓ 1
Withdrawal = CHF 4634.58	✓ 1
Part (b)
80 × 4634.58 = CHF 370 766.40	✓ 1
Interest = 370 766.40 − 250 000 = CHF 120 766.40 (accept CHF 120 766 from an unrounded withdrawal)	✓ 1
Part (c)
TVM with PV = 250 000, PMT = −5000, FV = 0, I% = 4.2, P/Y = C/Y = 4	✓ 1
N = 71.27…	✓ 1
71 full withdrawals, because the balance is not enough for a 72nd full one	✓ 1
Part (d)
$250000(1.0105)^{71}-5000\times\frac{(1.0105)^{71}-1}{0.0105}$, or TVM with N = 71 solving for FV	✓ 1
= CHF 1344.37	✓ 1

Exam tip: For an annuity paid from a lump sum, the lump sum is PV and the final balance is FV = 0. Give PV and PMT opposite signs on the calculator.

+ Add to test✓ In your test62IBMI-030-01-r2-03Financial mathematics, loans and annuities Paper SLMedium11 marks

Marta takes out a mortgage of 240 000 pounds to buy a flat. The loan is to be repaid over 25 years by equal monthly payments made at the end of each month. The nominal annual interest rate is 4.8%, compounded monthly.

(a)

Find Marta's monthly payment.

(3)

(b)

Find the total interest Marta would pay over the 25 years.

(2)

(c)

Find the amount Marta still owes immediately after her 120th payment.

(3)

(d)

After the 120th payment the interest rate rises to 6.0% per annum, compounded monthly. Marta keeps the original end date of the mortgage. Find her new monthly payment.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
N = 300 and I% = 4.8 with C/Y = 12, or $i = 0.004$ per month M1	✓ 1
PV = 240 000, FV = 0, or $\frac{240000 \times 0.004}{1-1.004^{-300}}$ M1	✓ 1
1375.19 pounds A1	✓ 1
Part (b)
$300 \times 1375.19 - 240000$ M1	✓ 1
172 557 pounds (accept 172 557 to 172 558) A1	✓ 1
Part (c)
Uses N = 120 with PMT = 1375.19 and finds FV, or uses the present value of the remaining 180 payments M1	✓ 1
Correct set-up, e.g. $1375.19 \times \frac{1-1.004^{-180}}{0.004}$ A1	✓ 1
176 213 pounds (accept 176 212 to 176 213) A1	✓ 1
Part (d)
N = 180, I% = 6, PV = 176 213 (their (c)), FV = 0 M1	✓ 1
Correct method, e.g. $\frac{176213 \times 0.005}{1-1.005^{-180}}$ A1	✓ 1
1486.99 pounds (accept 1486.98 to 1486.99) A1	✓ 1

Exam tip: Store the unrounded payment and balance in your calculator and carry them forward. Rounding to the nearest pound in an early part can push a later answer outside the accepted range.

+ Add to test✓ In your test63IBMI-030-03-r1-03Exponents and logarithms Paper SLMedium7 marks

A cup of tea is left on a table in a room. Its temperature, T °C, after t minutes is modelled by
$T(t) = 20 + 75e^{-kt}$, where k is a positive constant.
After 8 minutes the temperature of the tea is 60 °C.

(a)

Write down the initial temperature of the tea.

(1)

(b)

Find the value of k.

(2)

(c)

Find the time taken for the tea to cool to 35 °C.

(2)

(d)

Describe what happens to the temperature of the tea in the long term, justifying your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
T(0) = 20 + 75 = 95 °C A1	✓ 1
Part (b)
20 + 75e^−8k = 60 ⇒ e^−8k = 40/75 (or equivalent, including a correct GDC approach) M1	✓ 1
k = 0.0786 (0.078576…) A1	✓ 1
Part (c)
20 + 75e^−kt = 35 ⇒ e^−kt = 0.2 (or equivalent, including a correct GDC approach) M1	✓ 1
t = 20.5 minutes (20.48…) A1	✓ 1
Part (d)
Since k > 0, as t → ∞, e^−kt → 0 R1	✓ 1
so T decreases and approaches 20 °C (room temperature) from above, never falling below it A1	✓ 1

Exam tip: Store the unrounded value of k in your GDC and use it in part (c). Rounding a value early and then reusing it can introduce errors into later answers.

+ Add to test✓ In your test64IBMI-030-07-r2-03Arithmetic and geometric sequences in context Paper SLMedium7 marks

Lena starts a savings plan. She saves €50 in the first month. In each following month she saves €8 more than in the previous month.

(a)

Find the amount Lena saves in the 12th month.

(2)

(b)

Lena wants her total savings to reach at least €5000. Find the number of months it takes her to reach this target.

(4)

(c)

Find how much her total savings exceed €5000 at the end of that month.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $50+11\times8$	✓ 1
A1: €138	✓ 1
Part (b)
M1: use of the arithmetic series formula $\frac{n}{2}(2\times50+8(n-1))$	✓ 1
M1: sets this $\geq5000$ (or $=5000$), e.g. $4n^2+46n-5000=0$	✓ 1
A1: $n=30.07$ (from GDC or the quadratic formula)	✓ 1
A1: 31 months, since $S_{30}=4980$ euros, which is below the target	✓ 1
Part (c)
A1: $S_{31}=5270$, so she exceeds the target by €270	✓ 1

Exam tip: An answer of $n=30.07$ means the target is reached during month 31, not month 30. State a whole number of months and show why 30 months is not enough.

+ Add to test✓ In your test65IBMI-030-08-r1-03Exponents and logarithms Paper SLMedium7 marks

The sound level, $L$ decibels (dB), of a noise with intensity $I$ W m⁻² is given by $L = 10\log_{10}\left(\frac{I}{I_0}\right)$, where $I_0 = 1.0$×10⁻¹² W m⁻².
A workshop machine produces sound with intensity 4.5×10⁻⁵ W m⁻² at the operator's position.

(a)

Calculate the sound level of the machine at the operator's position.

(2)

(b)

Ear protection must be worn if the sound level reaches 95 dB. Find the intensity that corresponds to 95 dB.

(2)

(c)

A second, identical machine is switched on, so the intensity at the operator's position doubles. Find the increase in the sound level.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: substitutes correctly, L = 10 log₁₀(4.5×10⁻⁵ ÷ 1.0×10⁻¹²) = 10 log₁₀(4.5×10⁷)	✓ 1
A1: L = 76.5 dB (76.53…)	✓ 1
Part (b)
M1: rearranges to I = 1.0×10⁻¹² × 10^9.5	✓ 1
A1: I = 3.16×10⁻³ W m⁻²	✓ 1
Part (c)
M1: uses the doubled intensity 9.0×10⁻⁵ W m⁻² in the formula, OR writes the increase as 10 log₁₀(2I/I₀) − 10 log₁₀(I/I₀)	✓ 1
A1: new level is 79.5 dB (79.54…), OR simplifies the increase to 10 log₁₀2	✓ 1
A1: increase of 3.01 dB (accept 3.0 dB from subtracting the rounded levels)	✓ 1

Common error: Doubling the intensity does not double the decibel level. On a logarithmic scale it adds 10 log₁₀2 ≈ 3.01 dB. Keep unrounded values in your calculator when you subtract two levels.

+ Add to test✓ In your test66IBMI-030-02-r2-05Arithmetic and geometric sequences in context Paper SLMedium11 marks

Marco is training for a long-distance event. He runs 3.2 km on day 1 of his training plan and increases his daily distance by 0.4 km each day.
His friend Lena runs 2.5 km on day 1 and increases her daily distance by 6% each day.

(a)

Find the distance Marco runs on day 15.

(2)

(b)

Find the total distance Marco runs in the first 15 days.

(2)

(c)

Find the first day on which Lena runs further than Marco.

(3)

(d)

Marco's coach sets him a target of running a total of at least 200 km. Find the minimum number of days needed to reach this target.

(3)

(e)

Comment on the suitability of Lena's model for predicting her daily distance after many months of training.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$u_{15} = 3.2 + 14 \times 0.4$ M1	✓ 1
= 8.8 km A1	✓ 1
Part (b)
$S_{15} = \frac{15}{2}(3.2 + 8.8)$ M1	✓ 1
= 90 km A1	✓ 1
Part (c)
Marco: $3.2 + 0.4(n-1)$; Lena: $2.5 \times 1.06^{n-1}$ M1	✓ 1
Compares values using a GDC table or graph, e.g. day 32: Marco 15.6 km, Lena 15.2 km; day 33: Marco 16.0 km, Lena 16.1 km M1	✓ 1
Day 33 A1	✓ 1
Part (d)
$\frac{n}{2}(2\times 3.2 + 0.4(n-1)) \geq 200$ M1	✓ 1
$0.2n^2 + 3n - 200 = 0$ solved, giving $n = 25$ (or $-40$) A1	✓ 1
25 days. The total after 25 days is exactly 200 km, so the target is met. The negative root is rejected. A1	✓ 1
Part (e)
Not suitable: exponential growth is unbounded. For example, by day 60 the model predicts about 78 km per day, which is unrealistic for a runner. R1	✓ 1

Exam tip: When you compare an arithmetic model with a geometric one, a GDC table of both sequences is the most reliable method. Quote the values on both sides of the change to show where it happens.

+ Add to test✓ In your test67IBMI-030-06-r1-05Financial mathematics, loans and annuities Paper SLMedium10 marks

A small bakery needs to borrow £40 000 for new ovens. It is offered two loans. Each loan is repaid over 6 years with payments at the end of each period.
Option A: nominal annual rate 7.2%, compounded monthly, with monthly payments.
Option B: nominal annual rate 6.9%, compounded quarterly, with quarterly payments. A one-off arrangement fee of £500 is paid when the loan starts.

(a)

Find the monthly payment for Option A.

(2)

(b)

Find the quarterly payment for Option B.

(2)

(c)

Calculate the effective annual interest rate for each option, giving your answers as percentages to 3 significant figures.

(2)

(d)

Taking the fee into account, determine which option costs the bakery less in total, and by how much.

(3)

(e)

Explain why Option B costs more in total even though its effective annual rate is lower.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 N = 72, I% = 7.2, PV = 40 000, FV = 0, C/Y = P/Y = 12	✓ 1
A1 £685.81	✓ 1
Part (b)
M1 N = 24, I% = 6.9, PV = 40 000, FV = 0, C/Y = P/Y = 4	✓ 1
A1 £2049.54	✓ 1
Part (c)
A1 Option A: (1 + 0.072/12)¹² − 1 = 7.44%	✓ 1
A1 Option B: (1 + 0.069/4)⁴ − 1 = 7.08%	✓ 1
Part (d)
M1 total = payment × number of payments, with the £500 fee added for B	✓ 1
A1 A: 72 × 685.81 = £49 378.32; B: 24 × 2049.54 + 500 = £49 688.96	✓ 1
A1 Option A is cheaper, by £310.64 (accept £310.69 from unrounded payments, or £311)	✓ 1
Part (e)
R1 Option B charges about £190 less in interest, but the £500 fee more than cancels this saving	✓ 1

Exam tip: A lower effective annual rate does not always mean a cheaper loan. Fees paid up front are part of the true cost, so include them before you choose.

+ Add to test✓ In your test68IBMI-030-01-r1-05Financial mathematics, loans and annuities Paper SLMedium7 marks

The Okafor family take out a $320 000 mortgage to buy a house. The loan is repaid with equal monthly payments over 25 years, at a nominal annual interest rate of 4.8%, compounded monthly.

(a)

Find the monthly payment.

(3)

(b)

Find the amount still owed immediately after the 60th payment.

(2)

(c)

Find the amount of interest paid during the first 5 years.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
N = 300 and I% = 4.8 with P/Y = C/Y = 12	✓ 1
PV = 320 000, FV = 0, solving for PMT	✓ 1
$1833.59	✓ 1
Part (b)
TVM with N = 60, PV = 320 000, PMT = −1833.59, solving for FV, or the present value of the remaining 240 payments	✓ 1
$282 544 (accept answers from $282 543 to $282 545, from rounded or unrounded PMT)	✓ 1
Part (c)
Principal repaid = 320 000 − 282 544 = $37 456, and total paid = 60 × 1833.59 = $110 015.40	✓ 1
Interest = $72 559.40 (accept answers from $72 558 to $72 561, follow through from their (b))	✓ 1

Exam tip: Interest paid over a period equals the total payments minus the reduction in the balance, not minus the full loan. In the first 5 years about two-thirds of each payment is interest.

+ Add to test✓ In your test69IBMI-030-01-r2-05Financial mathematics, loans and annuities Paper SLMedium12 marks

Lena is saving for a house deposit. She compares two savings accounts.
Account A: 5.1% per annum, compounded annually.
Account B: 5.0% per annum, compounded monthly.

(a)

By finding the effective annual interest rate of Account B, determine which account gives the better return. Justify your answer.

(4)

(b)

Lena chooses Account B. She deposits 350 euros at the end of every month for 5 years. Find the value of her savings at the end of the 5 years.

(3)

(c)

Inflation is expected to be 2.4% per year over the 5 years. Find the value of Lena's savings at the end of the 5 years in terms of today's money.

(2)

(d)

Lena wants exactly 30 000 euros in Account B after 5 years of equal monthly deposits, each made at the end of a month. Find the monthly deposit she needs to make.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\left(1+\frac{0.05}{12}\right)^{12} - 1$ or equivalent GDC method M1	✓ 1
5.116% (accept 5.12%; a value rounded to 5.1% does not allow a valid comparison) A1	✓ 1
Compares this explicitly with the 5.1% effective rate of Account A, e.g. 5.116% > 5.1% R1	✓ 1
Account B gives the better return A1	✓ 1
Part (b)
Attempts a TVM calculation or the annuity formula with 60 monthly periods M1	✓ 1
Correct values: either N = 60, I% = 5, C/Y = P/Y = 12, PMT = ±350, PV = 0, or $350 \times \frac{(1+0.05/12)^{60}-1}{0.05/12}$ A1	✓ 1
23 802.13 euros (accept 23 800) A1	✓ 1
Part (c)
$\frac{23802.13}{1.024^5}$ M1	✓ 1
21 140.54 euros (accept 21 100) A1	✓ 1
Part (d)
Attempts to solve for the payment with 60 monthly periods M1	✓ 1
Correct values: either N = 60, I% = 5, C/Y = P/Y = 12, FV = 30 000, PV = 0, or $\frac{30000 \times 0.05/12}{(1+0.05/12)^{60}-1}$ A1	✓ 1
441.14 euros A1	✓ 1

Exam tip: When a question asks you to justify a choice, state both effective rates to enough decimal places to tell them apart and compare them explicitly. Naming the better account without a comparison does not earn the reasoning mark.

+ Add to test✓ In your test70IBMI-030-00-11Approximation, percentage error and standard form Paper SLHard13 marks

A city has a population of 2.4×10⁶, correct to two significant figures. Each person uses on average 145 litres of water per day, correct to three significant figures. 1 m³ = 1000 litres.
The city's reservoir holds exactly 4.2×10⁷ m³ of water.

(a)

Using the given values, calculate the total daily water use of the city in litres and in m³. Give both answers in the form $a\times10^k$ where $1\le a<10$ and $k\in\mathbb{Z}$.

(3)

(b)

Write down the lower and upper bounds of the population and of the daily use per person. Hence find the upper bound of the city's daily water use in m³.

(4)

(c)

Assuming no rainfall and no other supply, find the number of complete days for which the reservoir is guaranteed to supply the city.

(2)

(d)

A planner calculates the supply time using the daily water use in m³ found in the first part of this question. Calculate the percentage error of the planner's value compared with the guaranteed supply time of 117.8... days found in the previous part. Comment on the planner's method in context.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 2.4×10⁶ × 145	✓ 1
A1 3.48×10⁸ litres	✓ 1
A1 3.48×10⁵ m³	✓ 1
Part (b)
A1 population: 2.35×10⁶ and 2.45×10⁶	✓ 1
A1 use per person: 144.5 and 145.5 litres	✓ 1
M1 for 2.45×10⁶ × 145.5 ÷ 1000	✓ 1
A1 3.56×10⁵ m³ (3.56475×10⁵)	✓ 1
Part (c)
M1 for $\frac{4.2\times10^{7}}{3.56475\times10^{5}}=117.8...$	✓ 1
A1 117 complete days	✓ 1
Part (d)
M1 for $\frac{4.2\times10^{7}}{3.48\times10^{5}}=120.7$ days	✓ 1
M1 for $\left\|\frac{120.69...-117.82...}{117.82...}\right\|\times100$	✓ 1
A1 2.44 %	✓ 1
R1 the planner overestimates how long the water will last, so plans based on it could leave the city short; the worst-case (upper bound) usage should be used	✓ 1

Exam tip: For a guaranteed number of days, divide by the upper bound of usage and round down: 117.8 days gives only 117 complete days.

+ Add to test✓ In your test71IBMI-030-05-10Approximation, percentage error and standard form Paper SLHard6 marks

A courier estimates the mass of a parcel as 4.2 kg. The percentage error in this estimate is 12.5%.

(a)

Find the two possible values of the actual mass of the parcel.

(3)

(b)

The actual mass, rounded to the nearest kilogram, is 5 kg. State the actual mass of the parcel, giving a reason.

(2)

(c)

The courier charges $3.15 per kg of actual mass. Calculate the charge for this parcel.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for an equation such as $\frac{\|4.2-A\|}{A}=0.125$	✓ 1
A1 A = 4.8 kg	✓ 1
A1 A = 3.73 kg (3.7333…)	✓ 1
Part (b)
A1 4.8 kg	✓ 1
R1 because 3.73 rounds to 4 kg, not 5 kg, while 4.8 rounds to 5 kg	✓ 1
Part (c)
A1 $15.12	✓ 1

Exam tip: The percentage error is measured relative to the actual value, which is unknown. Do not just work out 4.2 × 1.125 and 4.2 × 0.875. Instead, solve 4.2 = A(1 ± 0.125).

+ Add to test✓ In your test72IBMI-030-06-r1-06Financial mathematics, loans and annuities Paper SLHard12 marks

Aisha borrows £30 000 to pay for a postgraduate course. She repays the loan with payments of £438.26 at the end of each month for 7 years. Interest is charged at a nominal annual rate of r%, compounded monthly.

(a)

Show that r = 6.00, correct to 3 significant figures.

(2)

(b)

Using r = 6, find the amount Aisha still owes immediately after her 36th payment. Give your answer to the nearest pound.

(3)

(c)

Immediately after her 36th payment, Aisha pays a lump sum of £7500 off the loan. She continues paying £438.26 per month, with a smaller final payment if needed.
Find the number of further payments she makes, and hence state how many months earlier the loan is repaid than originally planned.

(4)

(d)

Instead, Aisha decides to keep the original end date after paying the £7500 lump sum. Find her new monthly payment.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 TVM entry: N = 84, PV = 30 000, PMT = −438.26, FV = 0, C/Y = P/Y = 12, solving for I%	✓ 1
A1 I% = 6.0004… so r = 6.00	✓ 1
Part (b)
M1 N = 36 solving for FV, or the present value of the 48 remaining payments	✓ 1
M1 correct values entered: PV = 30 000, PMT = −438.26, I% = 6	✓ 1
A1 £18 661	✓ 1
Part (c)
A1 new balance = 18 661 − 7500 = £11 161	✓ 1
M1 PV = 11 161, PMT = −438.26, I% = 6, FV = 0, solving for N	✓ 1
A1 N = 27.3…, so 28 further payments (27 full payments and one smaller payment)	✓ 1
A1 48 − 28 = 20 months earlier	✓ 1
Part (d)
M1 PV = 11 161 (their balance after the lump sum), N = 48	✓ 1
M1 I% = 6, FV = 0, solving for PMT	✓ 1
A1 £262.12	✓ 1

Common error: A non-integer N, such as 27.3, means a final part payment. That still counts as a payment, so the loan takes 28 months, not 27.

+ Add to test✓ In your test73IBMI-030-02-r1-06Arithmetic and geometric sequences in context Paper SLHard12 marks

A patient takes a 250 mg dose of a medication every 8 hours. By the time the next dose is due, 60% of the medication in the body has been eliminated, so 40% remains. Let $A_n$ be the amount of medication in the body, in mg, immediately after the nth dose.

(a)

Show that $A_3=390$ mg.

(2)

(b)

Find $A_6$.

(2)

(c)

The medication is effective only once the amount immediately after a dose is at least 400 mg. Find how many hours after the first dose this first happens.

(3)

(d)

Find the long-term amount immediately after a dose, according to the model.

(2)

(e)

Instead, the doctor gives a larger first dose of L mg, followed by 250 mg every 8 hours. Find the value of L for which the amount immediately after every dose is the same.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$A_3=250+250(0.4)+250(0.4)^2$ or $A_2=350$, $A_3=0.4\times350+250$ M1 ; $250+100+40=390$ A1	✓ 2
Part (b)
Uses $\frac{250(1-0.4^{6})}{1-0.4}$ M1 ; 415 mg (414.96) A1	✓ 2
Part (c)
Forms $\frac{250(1-0.4^{n})}{0.6}\ge400$ or tests values: $A_3=390$, $A_4=406$ M1 ; $n=4$ A1 ; 24 hours after the first dose A1	✓ 3
Part (d)
Uses the sum to infinity $\frac{250}{1-0.4}$ M1 ; 417 mg (416.67) A1	✓ 2
Part (e)
Forms $0.4L+250=L$ M1 ; rearranges to $0.6L=250$ M1 ; $L=417$ mg (416.67) A1	✓ 3

Exam tip: In part (c) the answer needed is a time, not a number of doses. The 4th dose is taken after 3 intervals of 8 hours, so it is 24 hours after the first dose, not 32 hours.

+ Add to test✓ In your test74IBMI-030-07-r1-06Arithmetic and geometric sequences in context Paper SLHard13 marks

A fish farm has 8000 fish in a lake at the start. At the end of each year, 10% of the fish in the lake are harvested, and then 500 young fish are added. Let $F_n$ be the number of fish after $n$ years, so $F_0 = 8000$ and $F_n = 0.9F_{n-1} + 500$. Assume no fish die or are born naturally.

(a)

Find $F_1$ and $F_2$.

(2)

(b)

Show that $F_n = 8000(0.9)^n + 500\left(1 + 0.9 + 0.9^2 + \dots + 0.9^{n-1}\right)$, and hence that $F_n = 5000 + 3000(0.9)^n$.

(3)

(c)

Find the number of fish in the lake after 10 years, to the nearest whole fish.

(2)

(d)

Find the least number of years after which there are fewer than 5200 fish in the lake.

(3)

(e)

Describe what happens to the number of fish in the long term. The farmer wants the long-term population to be 6000 fish, still harvesting 10% each year. Find the number of young fish that should be added each year.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$F_1 = 0.9(8000) + 500 = 7700$ A1	✓ 1
$F_2 = 0.9(7700) + 500 = 7430$ A1	✓ 1
Part (b)
Expands the rule, e.g. $F_2 = 0.9^2(8000) + 500(0.9) + 500$, and generalises the pattern R1	✓ 1
Sums the geometric series: $500 \times \frac{1-0.9^{n}}{1-0.9} = 5000(1-0.9^{n})$ M1	✓ 1
$F_n = 8000(0.9)^n + 5000 - 5000(0.9)^n = 5000 + 3000(0.9)^n$ A1	✓ 1
Part (c)
$5000 + 3000(0.9)^{10}$ M1	✓ 1
$= 6046$ fish A1	✓ 1
Part (d)
$5000 + 3000(0.9)^n < 5200 \Rightarrow 0.9^{n} < \frac{1}{15}$ M1	✓ 1
$n > 25.7$ (by logarithms or GDC) A1	✓ 1
26 years A1	✓ 1
Part (e)
$0.9^{n} \to 0$, so the population decreases towards 5000 fish (the harvest of 10% then balances the 500 added) R1	✓ 1
In the long term, $10\%$ of the population equals the number added: $0.1 \times 6000 = J$, or $\frac{J}{0.1} = 6000$ M1	✓ 1
$J = 600$ fish per year A1	✓ 1

Exam tip: A 'show that' answer must reach the printed result through visible algebra. Write the geometric sum formula and simplify it step by step; do not just check a value of $n$.

+ Add to test✓ In your test75IBMI-030-02-r2-06Arithmetic and geometric sequences in context Paper SLHard14 marks

Two factories are reducing their carbon dioxide emissions.
Factory A emits 48 000 tonnes in year 1, and its annual emission decreases by 8% each year.
Factory B emits 40 000 tonnes in year 1, and its annual emission decreases by 1500 tonnes each year.

(a)

Find the annual emission of Factory A in year 5.

(2)

(b)

Find the total emission of Factory A over the first 10 years.

(2)

(c)

Find the first year in which the annual emission of Factory A is less than that of Factory B.

(3)

(d)

Find the number of years needed for the total emission of Factory A to first exceed 450 000 tonnes.

(3)

(e)

Find the total emission of Factory B over the first 20 years, and explain why this model cannot be used to find the total emission of Factory B over the first 30 years.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$48000 \times 0.92^{4}$ M1	✓ 1
= 34 400 tonnes (34 386.9) A1	✓ 1
Part (b)
$S_{10} = \frac{48000(1 - 0.92^{10})}{1 - 0.92}$ M1	✓ 1
= 339 000 tonnes (339 367) A1	✓ 1
Part (c)
Sets up $48000 \times 0.92^{n-1} < 40000 - 1500(n-1)$ M1	✓ 1
Year 5: A = 34 387, B = 34 000; year 6: A = 31 636, B = 32 500 A1	✓ 1
Year 6 A1	✓ 1
Part (d)
$\frac{48000(1 - 0.92^{n})}{0.08} > 450000$ M1	✓ 1
$0.92^{n} < 0.25$, giving $n > 16.6$ A1	✓ 1
17 years A1	✓ 1
Part (e)
$S_{20} = \frac{20}{2}(2\times 40000 - 19\times 1500)$ M1	✓ 1
= 515 000 tonnes A1	✓ 1
$40000 - 1500(n-1) < 0$ when $n > 27.7$ M1	✓ 1
So the model predicts negative annual emissions from year 28, which is impossible R1	✓ 1

Common error: Giving 16.6 years as the answer. The total only exceeds 450 000 tonnes once a whole year is complete, so the result must be rounded up to 17 years.

+ Add to test✓ In your test76IBMI-030-08-r1-06Exponents and logarithms Paper SLHard14 marks

A cup of black coffee is left on a table in a room. Its temperature, $T$ °C, $t$ minutes after it is poured is modelled by $T = 21 + 69e^{-kt}$, where $k$ is a positive constant.
Five minutes after it is poured, the temperature of the coffee is 78 °C.

(a)

Write down the temperature of the coffee when it is poured.

(1)

(b)

Find the value of $k$.

(3)

(c)

Find the temperature of the coffee 20 minutes after it is poured.

(2)

(d)

The coffee is pleasant to drink once its temperature falls to 60 °C. Find how long after pouring this happens.

(3)

(e)

State what happens to the temperature of the coffee in the long term, and interpret this value in context.

(2)

(f)

At the same moment, a cup of coffee with milk is poured at 75 °C. Its temperature is modelled by $T = 21 + 54e^{-0.031t}$. Using your value of $k$, find the time at which the two cups of coffee are at the same temperature.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: 90 °C	✓ 1
Part (b)
M1: substitutes t = 5 and T = 78: 78 = 21 + 69e^−5k	✓ 1
M1: rearranges to e^−5k = 57/69 and takes ln, or uses a GDC solver	✓ 1
A1: k = 0.0382 (0.038211…)	✓ 1
Part (c)
M1: substitutes t = 20 with their k	✓ 1
A1: T = 53.1 °C	✓ 1
Part (d)
M1: sets up 21 + 69e^−kt = 60	✓ 1
M1: rearranges to e^−kt = 39/69 and takes ln, or uses a GDC solver	✓ 1
A1: t = 14.9 minutes	✓ 1
Part (e)
A1: T approaches 21 °C, because e^−kt → 0 as t increases	✓ 1
R1: 21 °C is the temperature of the room, so the coffee cools to room temperature	✓ 1
Part (f)
M1: sets the models equal: 69e^−kt = 54e^−0.031t	✓ 1
M1: takes ln or uses a GDC to obtain (k − 0.031)t = ln(69/54)	✓ 1
A1: t = 34.0 minutes	✓ 1

Exam tip: Store the unrounded value of k in your GDC and use it in later parts. If you use a rounded value, accuracy marks can be lost through rounding errors.

+ Add to test✓ In your test77IBMI-030-07-r2-06Arithmetic and geometric sequences in context Paper SLHard14 marks

A lake is stocked with 5000 fish. Each year the fish population grows by 8%, and then at the end of the year 600 fish are removed by licensed anglers. Let $P_n$ be the number of fish at the end of year $n$, after the fish have been removed, with $P_0=5000$.

(a)

Show that $P_1=4800$.

(2)

(b)

Find $P_3$.

(2)

(c)

Show that $P_n=5000(1.08)^n-600\left(1+1.08+1.08^2+\dots+1.08^{n-1}\right)$, and hence that $P_n=7500-2500(1.08)^n$.

(4)

(d)

Using the formula from part (c), find the year in which the fish population would be wiped out.

(3)

(e)

The lake authority wants the population never to fall below 5000. Find the largest number of fish, $R$, that can be removed each year, and justify your answer.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $5000\times1.08-600$	✓ 1
A1: $5400-600=4800$ (AG)	✓ 1
Part (b)
M1: repeated use of $P_{n+1}=1.08P_n-600$, e.g. $P_2=4584$	✓ 1
A1: $P_3=4350.72$, so about 4351 fish (accept 4350.72, 4351 or 4350)	✓ 1
Part (c)
M1: expands at least two steps, e.g. $P_2=5000(1.08)^2-600(1.08)-600$	✓ 1
A1: generalises correctly to the given expression	✓ 1
M1: uses the sum of a geometric series, $\frac{1.08^n-1}{0.08}$, so that 600 times the bracket becomes $7500(1.08^n-1)$	✓ 1
A1: $5000(1.08)^n-7500(1.08)^n+7500=7500-2500(1.08)^n$ (AG)	✓ 1
Part (d)
M1: sets $7500-2500(1.08)^n=0$, i.e. $1.08^n=3$	✓ 1
A1: $n=\frac{\ln3}{\ln1.08}=14.3$	✓ 1
A1: during year 15, since $P_{14}\approx157>0$ and $P_{15}\approx-430<0$	✓ 1
Part (e)
M1: requires the yearly growth to be at least the number removed, $0.08\times5000\geq R$ (or $P_1=5400-R\geq5000$)	✓ 1
A1: $R=400$	✓ 1
R1: with $R=400$, each year's growth of 400 exactly replaces the fish removed, so the population stays at 5000. Any larger $R$ makes $P_1<5000$, and the population then decreases every year	✓ 1

Common error: Treating the removal as a percentage, or subtracting the 600 fish before applying the 8% growth. Follow the order stated in the context: first grow, then remove.

+ Add to test✓ In your test78IBMI-030-03-r1-06Exponents and logarithms Paper SLHard14 marks

A patient is given doses of a medicine. After a single dose of D mg, the amount of medicine remaining in the body t hours later is modelled by $A(t) = De^{-kt}$. The half-life of the medicine is 6 hours.

(a)

Show that k ≈ 0.116.

(2)

(b)

A single dose of 400 mg is given. Find the amount remaining after 10 hours.

(2)

(c)

Instead, the patient receives a 400 mg dose every 6 hours. Find the amount of medicine in the body immediately after the 4th dose.

(2)

(d)

Show that the amount immediately after the nth dose is $800(1 - 0.5^n)$ mg, and state the long-term value this amount approaches.

(2)

(e)

For safety, the amount immediately after any dose must not exceed 780 mg. Find the maximum number of doses that can be given on this schedule.

(3)

(f)

A doctor proposes giving 400 mg doses every T hours indefinitely, so that the amount immediately after a dose never exceeds 600 mg. Find the minimum value of T.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
e^−6k = 0.5 M1	✓ 1
k = ln 2 / 6 = 0.11552… ≈ 0.116 A1	✓ 1
Part (b)
400e^−10k (= 400 × 2^−10/6) M1	✓ 1
= 126 mg (125.99…); accept 125 mg from k = 0.116 A1	✓ 1
Part (c)
400 + 200 + 100 + 50, since each earlier dose halves every 6-hour interval M1	✓ 1
= 750 mg A1	✓ 1
Part (d)
Geometric series with u₁ = 400 and r = 0.5 (the fraction left after one half-life): S_n = 400(1 − 0.5ⁿ)/(1 − 0.5) = 800(1 − 0.5ⁿ) M1	✓ 1
As n → ∞, 0.5ⁿ → 0, so the amount approaches 800 mg A1	✓ 1
Part (e)
800(1 − 0.5ⁿ) ≤ 780 M1	✓ 1
0.5ⁿ ≥ 0.025 ⇒ n ≤ 5.32 M1	✓ 1
Maximum of 5 doses A1	✓ 1
Part (f)
The amount after each dose increases towards the sum to infinity 400/(1 − e^−kT), with r = e^−kT M1	✓ 1
400/(1 − e^−kT) ≤ 600 ⇒ e^−kT ≤ 1/3 M1	✓ 1
Minimum T = 6 ln 3 / ln 2 = 9.51 hours; accept 9.47 hours from k = 0.116 A1	✓ 1

Exam tip: When doses repeat at equal intervals, the amounts form a geometric series whose common ratio is the fraction remaining over one interval; the sum to infinity gives the long-term level.

+ Add to test✓ In your test79IBMI-030-01-r1-06Financial mathematics, loans and annuities Paper SLHard14 marks

Ravi is 40 years old. He wants to retire at 65 and then receive $2500 at the end of every month for 25 years. His pension fund earns a nominal annual interest rate of 4.5%, compounded monthly, both before and after retirement. He already has $40 000 in the fund. He will add an equal deposit at the end of every month for the 25 years until he retires.

(a)

Find the amount the fund must contain when Ravi retires.

(3)

(b)

Find the monthly deposit Ravi must make until he retires.

(4)

(c)

Find the total amount Ravi pays into the fund, including the initial $40 000. Hence find the total interest earned over the 50 years.

(3)

(d)

Inflation is expected to average 2% per year. Find the value, in today's money, of a $2500 payment made 25 years from now.

(2)

(e)

Comment on whether Ravi's plan meets his needs, referring to your answer to the previous part.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Recognises a present value of an annuity with PMT = 2500	✓ 1
N = 300, I% = 4.5, FV = 0, P/Y = C/Y = 12	✓ 1
$449 775.80	✓ 1
Part (b)
Value of the existing $40 000 after 300 months: $40000(1.00375)^{300}$	✓ 1
= $122 949.70	✓ 1
TVM with N = 300, PV = −40 000, FV = 449 775.80, solving for PMT; or the remaining $326 826.10 equated to the future value of the deposits	✓ 1
$591.01	✓ 1
Part (c)
Total paid in = 300 × 591.01 + 40 000	✓ 1
= $217 303	✓ 1
Total received = 300 × 2500 = $750 000, so interest ≈ $532 697	✓ 1
Part (d)
$\frac{2500}{1.02^{25}}$	✓ 1
$1523.83	✓ 1
Part (e)
A $2500 payment at 65 buys only about as much as $1524 does today	✓ 1
Payments later in retirement are worth even less in real terms, so Ravi may need payments that rise with inflation, or higher deposits	✓ 1

Common error: The $40 000 already in the fund also earns interest for 25 years. Leaving out its growth makes the required monthly deposit far too large.

+ Add to test✓ In your test80IBMI-030-04-r2-01Complex numbers and matrices Paper HLEasy6 marks

In an alternating-current circuit, impedances are modelled as complex numbers measured in ohms (Ω). Two components have impedances $Z_1 = 3 + 4i$ and $Z_2 = 2 - 5i$.
When components are connected in series, the total impedance is $Z_S = Z_1 + Z_2$. When they are connected in parallel, the total impedance $Z_P$ satisfies $\frac{1}{Z_P} = \frac{1}{Z_1} + \frac{1}{Z_2}$.

(a)

Write down $Z_S$ in the form $a + bi$.

(1)

(b)

Show that $Z_P = \frac{Z_1 Z_2}{Z_1 + Z_2}$ and hence find $Z_P$ in the form $a + bi$.

(3)

(c)

Find the magnitude of the parallel impedance, $|Z_P|$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$Z_S = 5 - i$ Ω	✓ 1
Part (b)
Combines the fractions: $\frac{1}{Z_P} = \frac{Z_1+Z_2}{Z_1Z_2}$, so $Z_P = \frac{Z_1Z_2}{Z_1+Z_2}$	✓ 1
Substitutes: $Z_P = \frac{26 - 7i}{5 - i}$	✓ 1
$Z_P = 5.27 - 0.346i$ Ω	✓ 1
Part (c)
Uses $\sqrt{a^2+b^2}$, e.g. $\sqrt{5.269^2 + 0.3462^2}$	✓ 1
$\|Z_P\| = 5.28$ Ω	✓ 1

Exam tip: Your GDC can divide complex numbers directly. Still write the unsimplified fraction first, because that working step earns a method mark even if you mistype a value.

+ Add to test✓ In your test81IBMI-030-04-r2-02Complex numbers and matrices Paper HLEasy6 marks

A café sells coffee, tea and cake. The numbers sold on Monday, Tuesday and Wednesday of one week are given by the matrix
$Q = \begin{pmatrix} 12 & 8 & 15 \\ 10 & 14 & 9 \\ 20 & 11 & 13 \end{pmatrix}$
where the rows are Monday, Tuesday and Wednesday, and the columns are coffee, tea and cake. In that week the prices, in dollars, are $P = \begin{pmatrix} 2.80 \\ 2.20 \\ 3.50 \end{pmatrix}$.

(a)

Find $QP$ and state what its entries represent.

(2)

(b)

The following week the café sells exactly the same quantities but changes its prices. The revenues on Monday, Tuesday and Wednesday are \$116.00, \$101.00 and \$139.50 respectively. By forming a matrix equation, find the new price of a cake.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$QP = \begin{pmatrix} 103.70 \\ 90.30 \\ 125.70 \end{pmatrix}$	✓ 1
These are the total revenues, in dollars, on Monday, Tuesday and Wednesday	✓ 1
Part (b)
Writes $Q\begin{pmatrix} c \\ t \\ k \end{pmatrix} = \begin{pmatrix} 116 \\ 101 \\ 139.5 \end{pmatrix}$	✓ 1
Uses the inverse: $\begin{pmatrix} c \\ t \\ k \end{pmatrix} = Q^{-1}\begin{pmatrix} 116 \\ 101 \\ 139.5 \end{pmatrix}$	✓ 1
Obtains $c = 3.00$, $t = 2.50$	✓ 1
Cake costs \$4.00	✓ 1

Common error: Pre-multiply by the inverse: the answer is $Q^{-1}R$, not $RQ^{-1}$. The product $RQ^{-1}$ is not even defined for a 3×1 column matrix $R$.

+ Add to test✓ In your test82IBMI-030-04-r1-01Complex numbers and matrices Paper HLEasy5 marks

An engineer models two components in an alternating-current circuit using complex impedances, measured in ohms (Ω):
$Z_1 = 4 + 3i$ and $Z_2 = 2 - 6i$.
When the components are connected in series, the total impedance is $Z_S = Z_1 + Z_2$. When they are connected in parallel, the total impedance $Z_P$ satisfies $\frac{1}{Z_P} = \frac{1}{Z_1} + \frac{1}{Z_2}$.

(a)

Find $Z_S$, giving your answer in the form $a + bi$.

(1)

(b)

Find $Z_P$, giving your answer in the form $a + bi$.

(3)

(c)

The magnitude of the impedance is $|Z_P|$. Find $|Z_P|$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $Z_S = 6 - 3i$ Ω	✓ 1
Part (b)
M1 uses $Z_P = \frac{Z_1 Z_2}{Z_1 + Z_2}$ or adds the reciprocals $\frac{1}{Z_1}+\frac{1}{Z_2}$	✓ 1
A1 $Z_1 Z_2 = 26 - 18i$ (or equivalent correct intermediate value)	✓ 1
A1 $Z_P = 4.67 - 0.667i$ Ω (exact: $\frac{14}{3} - \frac{2}{3}i$)	✓ 1
Part (c)
A1 $\|Z_P\| = \sqrt{4.667^2 + 0.667^2} = 4.71$ Ω	✓ 1

Common error: Writing $Z_P = \frac{1}{Z_1} + \frac{1}{Z_2}$ and forgetting to take the reciprocal at the end. Using $\frac{Z_1Z_2}{Z_1+Z_2}$ avoids this, and a GDC in complex mode handles the division directly.

+ Add to test✓ In your test83IBMI-030-04-r1-02Complex numbers and matrices Paper HLEasy5 marks

A graphic designer uses the transformation with matrix $T = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}$ to distort a company logo drawn on a coordinate grid, where units are centimetres. A point with position vector $\begin{pmatrix} x \\ y \end{pmatrix}$ is mapped to $T\begin{pmatrix} x \\ y \end{pmatrix}$.

(a)

Find the image of the point $(3, -2)$.

(1)

(b)

A point on the original logo is mapped to $(5, 8)$. Find the coordinates of this original point.

(2)

(c)

The original logo has an area of 12 cm². Find the area of the transformed logo.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $(4, -9)$	✓ 1
Part (b)
M1 uses $T^{-1}\begin{pmatrix}5\\8\end{pmatrix}$ with $T^{-1} = \frac{1}{7}\begin{pmatrix}3 & -1\\1 & 2\end{pmatrix}$, or solves $2x + y = 5$, $-x + 3y = 8$	✓ 1
A1 $(1, 3)$	✓ 1
Part (c)
M1 $\det T = 2\times3 - 1\times(-1) = 7$	✓ 1
A1 area $= 7 \times 12 = 84$ cm²	✓ 1

Exam tip: The absolute value of the determinant is the area scale factor. If a determinant is negative, the orientation is reversed but the area is still multiplied by its absolute value.

+ Add to test✓ In your test84IBMI-030-04-r2-06Complex numbers and matrices Paper HLMedium7 marks

A population of birds is split into juveniles $J$ and adults $D$, counted each spring. The model is
$\begin{pmatrix} J_{n+1} \\ D_{n+1} \end{pmatrix} = M\begin{pmatrix} J_n \\ D_n \end{pmatrix}$, where $M = \begin{pmatrix} 0 & 1.5 \\ 0.6 & 0.4 \end{pmatrix}$.
Initially there are 100 juveniles and 50 adults.

(a)

Find the total population after 5 years, to the nearest bird.

(2)

(b)

Find the eigenvalues of $M$.

(3)

(c)

Hence state the long-term annual percentage growth rate of the population, and interpret this in context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses $M^5\begin{pmatrix} 100 \\ 50 \end{pmatrix} = \begin{pmatrix} 165.63 \\ 136.69 \end{pmatrix}$	✓ 1
Total = 302 birds	✓ 1
Part (b)
$\det(M - \lambda I) = -\lambda(0.4 - \lambda) - 0.9$	✓ 1
$\lambda^2 - 0.4\lambda - 0.9 = 0$	✓ 1
$\lambda = 1.17$ or $\lambda = -0.770$	✓ 1
Part (c)
The dominant eigenvalue is 1.1695, so the growth rate is 17.0% per year	✓ 1
In the long term the population increases by about 17.0% each year, assuming the survival and birth rates stay constant	✓ 1

Exam tip: The long-term growth factor is the eigenvalue with the largest absolute value. The negative eigenvalue only causes oscillations, and these die away because $|-0.770| < 1.17$.

+ Add to test✓ In your test85IBMI-030-04-r1-03Complex numbers and matrices Paper HLMedium12 marks

A city bike-share scheme has 300 bikes and three docking stations, A, B and C. Each bike is used once per day and is left at a station overnight. Data suggest the following daily behaviour.
• Of bikes starting at A: 60% return to A, 30% go to B, 10% go to C.
• Of bikes starting at B: 20% go to A, 70% return to B, 10% go to C.
• Of bikes starting at C: 30% go to A, 20% go to B, 50% return to C.
On Monday morning there are 100 bikes at each station.

(a)

Write down the transition matrix $T$, where the entry in row $i$, column $j$ is the probability that a bike starting the day at station $j$ ends the day at station $i$.

(2)

(b)

Find the expected number of bikes at each station after three days.

(3)

(c)

Find the long-term expected number of bikes at each station.

(5)

(d)

Each station has 120 docking points. With reference to your answer to part (c), comment on whether this is suitable in the long term.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A2 $T = \begin{pmatrix}0.6 & 0.2 & 0.3\\0.3 & 0.7 & 0.2\\0.1 & 0.1 & 0.5\end{pmatrix}$	✓ 2
(Award A1 only if the matrix is transposed or one column is incorrect.)	—
Part (b)
M1 calculates $T^3\begin{pmatrix}100\\100\\100\end{pmatrix}$	✓ 1
A1 A: 110 (110.2), B: 137 (136.6)	✓ 1
A1 C: 53 (53.2)	✓ 1
Part (c)
M1 sets up $T\mathbf{v} = \mathbf{v}$, or finds the eigenvector for eigenvalue 1, or evaluates $T^n$ for large $n$	✓ 1
M1 uses the condition $a + b + c = 300$ (or proportions summing to 1)	✓ 1
A1 A: 108 ($300\times\frac{13}{36}$)	✓ 1
A1 C: 50 ($300\times\frac{6}{36}$)	✓ 1
A1 B: 142 ($300\times\frac{17}{36}$)	✓ 1
Part (d)
R1 in the long term about 142 bikes are expected at B, which is more than its 120 docking points	✓ 1
A1 so B needs more docks (or bikes must be moved from B), while C has spare capacity (about 50 bikes for 120 docks)	✓ 1

Exam tip: Check that every column of your transition matrix sums to 1 before doing any calculation. A transposed matrix gives completely different predictions, and every later part will be wrong.

+ Add to test✓ In your test86IBMI-030-04-r1-04Complex numbers and matrices Paper HLMedium6 marks

A bakery sells croissants at x euros each, muffins at y euros each and scones at z euros each. Three office orders were placed in one morning.
• Order 1: 3 croissants, 2 muffins and 4 scones cost €19.70.
• Order 2: 2 croissants, 5 muffins and 1 scone cost €16.25.
• Order 3: 4 croissants, 1 muffin and 3 scones cost €18.05.

(a)

Write this information as a matrix equation of the form $A\mathbf{x} = \mathbf{b}$.

(2)

(b)

Hence find the price of each item.

(2)

(c)

The bakery offers a 10% discount on a box containing 6 of each item. Find the cost of this box.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a 3×3 coefficient matrix multiplied by $\begin{pmatrix}x\\y\\z\end{pmatrix}$	✓ 1
A1 $\begin{pmatrix}3&2&4\\2&5&1\\4&1&3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}19.70\\16.25\\18.05\end{pmatrix}$	✓ 1
Part (b)
M1 uses $\mathbf{x} = A^{-1}\mathbf{b}$ (for example, with a GDC)	✓ 1
A1 croissant €2.40, muffin €1.85, scone €2.20	✓ 1
Part (c)
M1 $0.9 \times 6 \times (2.40 + 1.85 + 2.20)$	✓ 1
A1 €34.83	✓ 1

Exam tip: Write money answers to exactly 2 decimal places, for example €34.83 and not €34.8.

+ Add to test✓ In your test87IBMI-030-08-r2-04Exponents and logarithms Paper HLMedium7 marks

Two bacterial cultures are grown in a laboratory. The number of bacteria in each culture t hours after the start of an experiment is modelled by
Culture A: $N_A = 800 \times 2^{t/5}$
Culture B: $N_B = 50 \times 3^{t/4}$

(a)

Use the laws of logarithms to show that the two cultures contain the same number of bacteria when
$t = \dfrac{\ln 16}{\frac{1}{4}\ln 3 - \frac{1}{5}\ln 2}$.

(4)

(b)

Find the value of t.

(1)

(c)

Find the number of bacteria in each culture at this time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: sets $N_A = N_B$ and rearranges, $16 \times 2^{t/5} = 3^{t/4}$	✓ 1
M1: takes natural logarithms of both sides	✓ 1
M1: uses the product and power laws, $\ln 16 + \frac{t}{5}\ln 2 = \frac{t}{4}\ln 3$	✓ 1
A1: collects terms in t, $\ln 16 = t\left(\frac{1}{4}\ln 3 - \frac{1}{5}\ln 2\right)$, leading to the given answer	✓ 1
Part (b)
A1: $t = 20.4$ hours (20.383…)	✓ 1
Part (c)
M1: substitutes the unrounded value of t into either model, $800 \times 2^{20.383.../5}$	✓ 1
A1: 13 500 bacteria	✓ 1

Common error: In a 'show that' question the logarithm laws must be written out step by step. Jumping straight to the given expression, or only checking it with the GDC, does not earn the method marks.

+ Add to test✓ In your test88IBMI-030-03-r2-05Exponents and logarithms Paper HLMedium7 marks

Astronomers describe the apparent brightness of a star using its magnitude $m$. This is defined by
$m = -2.5\log_{10}\left(\dfrac{F}{F_0}\right)$,
where $F$ is the light flux received from the star and $F_0$ is a fixed reference flux. A brighter star has a smaller magnitude.

(a)

Two stars have magnitudes $m_1$ and $m_2$ and fluxes $F_1$ and $F_2$. Use the laws of logarithms to show that $m_1 - m_2 = 2.5\log_{10}\left(\dfrac{F_2}{F_1}\right)$.

(2)

(b)

Star A has magnitude 1.25 and star B has magnitude 4.85. Find $\dfrac{F_A}{F_B}$.

(2)

(c)

A binary system consists of two stars that appear as a single point of light. Each star on its own has magnitude 3.40. Find the magnitude of the combined light from the system.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $m_1 - m_2 = -2.5\left[\log_{10}F_1 - \log_{10}F_0 - \log_{10}F_2 + \log_{10}F_0\right]$, using $\log(a/b) = \log a - \log b$	✓ 1
A1: $= -2.5\log_{10}\left(\dfrac{F_1}{F_2}\right) = 2.5\log_{10}\left(\dfrac{F_2}{F_1}\right)$, using $-\log x = \log(1/x)$	✓ 1
Part (b)
M1: $4.85 - 1.25 = 2.5\log_{10}\left(\dfrac{F_A}{F_B}\right)$, so $\dfrac{F_A}{F_B} = 10^{3.6/2.5}$	✓ 1
A1: 27.5	✓ 1
Part (c)
M1: the combined flux is $2F$, where $F$ is the flux of one star	✓ 1
M1: $m_{\text{comb}} = 3.40 - 2.5\log_{10}2$	✓ 1
A1: 2.65	✓ 1

Common error: The magnitudes of the two stars do not add, so the answer is not 6.80. It is the fluxes that add. A larger flux gives a smaller magnitude, so the combined magnitude must be less than 3.40.

+ Add to test✓ In your test89IBMI-030-08-r1-05Exponents and logarithms Paper HLMedium8 marks

A biologist believes that the resting metabolic rate, $P$ watts, of a mammal is related to its mass, $m$ kg, by a model of the form $P = a m^b$, where $a$ and $b$ are constants.

(a)

Show that if $P = a m^b$, then $\ln P = b \ln m + \ln a$.

(2)

(b)

Using data from 12 species, the biologist finds that the regression line of $\ln P$ on $\ln m$ is $\ln P = 0.742 \ln m + 1.25$. Find the value of $a$ and the value of $b$.

(3)

(c)

Use the model to estimate the resting metabolic rate of a 60 kg mammal.

(2)

(d)

The heaviest species in the data set has a mass of 450 kg. Comment on the reliability of using the model to estimate the metabolic rate of an elephant of mass 6000 kg.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: takes ln of both sides: ln P = ln(a m^b)	✓ 1
A1: uses ln(xy) = ln x + ln y and ln(m^b) = b ln m to obtain ln P = b ln m + ln a	✓ 1
Part (b)
A1: b = 0.742	✓ 1
M1: ln a = 1.25, so a = e^1.25	✓ 1
A1: a = 3.49	✓ 1
Part (c)
M1: substitutes m = 60, e.g. 3.49×60^0.742 or ln P = 0.742 ln 60 + 1.25	✓ 1
A1: P = 72.8 W	✓ 1
Part (d)
R1: unreliable, because 6000 kg is well outside the range of the data, so this is extrapolation	✓ 1

Common error: The intercept of the ln–ln line is ln a, not a. You must exponentiate it to find a.

+ Add to test✓ In your test90IBMI-030-03-r1-05Exponents and logarithms Paper HLMedium8 marks

A biologist collects data on the body mass, m kg, and resting metabolic rate, R watts, of mammals with masses between 0.02 kg and 100 kg. A log–log graph of the data is approximately linear, and the regression line of ln R on ln m is
$\ln R = 0.74\ln m + 1.25$.

(a)

Show that the relationship can be written in the form $R = am^b$, and find the values of a and b.

(3)

(b)

Estimate the resting metabolic rate of a mammal of mass 65 kg.

(2)

(c)

Find the factor by which body mass must be multiplied for the resting metabolic rate to double.

(2)

(d)

Comment on the reliability of using the model to estimate the metabolic rate of an elephant of mass 5000 kg.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R = e^{0.74 ln m + 1.25} = e^1.25 × m^0.74 M1	✓ 1
a = 3.49 (3.4903…) A1	✓ 1
b = 0.74 A1	✓ 1
Part (b)
R = 3.4903 × 65^0.74 (or ln R = 0.74 ln 65 + 1.25 = 4.339) M1	✓ 1
R = 76.6 W A1	✓ 1
Part (c)
2R = a(km)^0.74 ⇒ k^0.74 = 2 M1	✓ 1
k = 2^1/0.74 = 2.55 A1	✓ 1
Part (d)
Unreliable, since 5000 kg is far outside the data range of 0.02–100 kg, so this is extrapolation R1	✓ 1

Exam tip: In a log–log model the gradient of the line becomes the power b and the intercept c becomes the coefficient a = ec (or 10c if base-10 logs are used).

+ Add to test✓ In your test91IBMI-030-04-r2-05Complex numbers and matrices Paper HLMedium11 marks

A drone's horizontal position, relative to its landing pad at the origin, is modelled by a complex number $z$, with distances in metres. The drone starts at $z_0 = 40$. At each step of a landing manoeuvre, its position is multiplied by $w = 0.9e^{i\pi/6}$, so that $z_n = 40w^n$.

(a)

Write $w$ in the form $a + bi$.

(2)

(b)

Find $z_4$, giving your answer in the form $a + bi$.

(3)

(c)

Find the first step at which the drone is less than 10 m from the landing pad.

(3)

(d)

For the step found in the previous part, find the drone's distance from the pad and the argument of its position in the interval $-\pi < \theta \le \pi$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$0.9\cos\frac{\pi}{6} + 0.9i\sin\frac{\pi}{6}$	✓ 1
$w = 0.779 + 0.45i$	✓ 1
Part (b)
$z_4 = 40(0.9)^4 e^{i4\pi/6}$	✓ 1
$= 26.244e^{i2\pi/3}$	✓ 1
$z_4 = -13.1 + 22.7i$	✓ 1
Part (c)
Uses $\|z_n\| = 40(0.9)^n$	✓ 1
Solves $40(0.9)^n < 10$, giving $n > 13.2$	✓ 1
Step 14	✓ 1
Part (d)
Distance $= 40(0.9)^{14} = 9.15$ m	✓ 1
Argument $= \frac{14\pi}{6} = \frac{7\pi}{3}$	✓ 1
Reduces by $2\pi$: $\frac{\pi}{3}$ (1.05 rad)	✓ 1

Exam tip: Multiplying by $re^{i\theta}$ scales the distance by $r$ and rotates the point by $\theta$. Work with modulus and argument separately rather than expanding powers in Cartesian form.

+ Add to test✓ In your test92IBMI-030-04-r1-05Complex numbers and matrices Paper HLMedium7 marks

Two alternating voltages in a sound-system circuit are given by
$V_1 = 12\sin(100\pi t)$ and $V_2 = 8\sin\left(100\pi t + \frac{\pi}{3}\right)$,
where $V_1$ and $V_2$ are in volts and $t$ is the time in seconds. All angles are in radians. Each voltage can be represented by a complex number (phasor) $Ae^{i\phi}$, where $A$ is the amplitude and $\phi$ is the phase. So $V_1$ is represented by $12$ and $V_2$ by $8e^{i\pi/3}$.

(a)

Write $8e^{i\pi/3}$ in the form $a + bi$.

(2)

(b)

The total voltage is $V = V_1 + V_2$. By adding the phasors, write $V$ in the form $R\sin(100\pi t + \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$.

(3)

(c)

Find the total voltage when $t = 0.002$ s.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $8\cos\frac{\pi}{3} + 8i\sin\frac{\pi}{3}$	✓ 1
A1 $4 + 4\sqrt{3}\,i$ (≈ $4 + 6.93i$)	✓ 1
Part (b)
M1 adds the phasors: $12 + 4 + 4\sqrt{3}\,i = 16 + 4\sqrt{3}\,i$	✓ 1
A1 $R = \sqrt{16^2 + 48} = \sqrt{304} = 4\sqrt{19} \approx 17.4$ V	✓ 1
A1 $\alpha = \arctan\frac{4\sqrt{3}}{16} \approx 0.409$ rad, so $V = 17.4\sin(100\pi t + 0.409)$	✓ 1
Part (c)
M1 substitutes $t = 0.002$ into the expression for $V$ found in the previous part, or into $V_1 + V_2$	✓ 1
A1 $V = 15.0$ V	✓ 1

Common error: Adding the amplitudes (12 + 8 = 20 V) ignores the phase difference. Add the phasors as complex numbers, and keep your GDC in radian mode.

+ Add to test✓ In your test93IBMI-030-00-12Approximation, percentage error and standard form Paper HLHard12 marks

The pH of a solution is defined by $\text{pH}=-\log_{10}c$, where $c$ is the hydrogen ion concentration in mol L⁻¹.

(a)

Lemon juice has $c=6.3\times10^{-3}$ mol L⁻¹. Calculate its pH.

(2)

(b)

Seawater has a pH of 8.1. Find its hydrogen ion concentration, giving your answer in the form $a\times10^k$ where $1\le a<10$ and $k\in\mathbb{Z}$.

(2)

(c)

Find how many times greater the hydrogen ion concentration of lemon juice is than that of seawater. Give your answer in the form $a\times10^k$ where $1\le a<10$ and $k\in\mathbb{Z}$.

(2)

(d)

The hydrogen ion concentration of a solution is multiplied by a factor $k$, where $k>0$. Show that the pH decreases by $\log_{10}k$.

(2)

(e)

A student measures the pH of a sample as 4.6 when its true value is 4.5. Calculate the percentage error in the measured pH and the percentage error in the hydrogen ion concentration deduced from it. Comment on your answers.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $-\log_{10}(6.3\times10^{-3})$	✓ 1
A1 2.20	✓ 1
Part (b)
M1 for $c=10^{-8.1}$	✓ 1
A1 7.94×10⁻⁹ mol L⁻¹	✓ 1
Part (c)
M1 for $\frac{6.3\times10^{-3}}{7.943...\times10^{-9}}$ or $10^{8.1-2.20}$	✓ 1
A1 7.93×10⁵; accept 7.94×10⁵, which comes from using the rounded pH	✓ 1
Part (d)
M1 for $-\log_{10}(kc)=-(\log_{10}k+\log_{10}c)$, using $\log(ab)=\log a+\log b$	✓ 1
A1 $=\text{pH}-\log_{10}k$, so the pH decreases by $\log_{10}k$ AG. If \(0	✓ 1
Part (e)
A1 percentage error in pH: $\frac{4.6-4.5}{4.5}\times100=2.22\%$	✓ 1
M1 for the concentrations $10^{-4.6}=2.51\times10^{-5}$ and $10^{-4.5}=3.16\times10^{-5}$	✓ 1
A1 percentage error in concentration: $\left\|\frac{10^{-4.6}-10^{-4.5}}{10^{-4.5}}\right\|\times100=20.6\%$	✓ 1
R1 a small percentage error in pH produces a much larger percentage error in concentration because the scale is logarithmic	✓ 1

Common error: A pH change of 0.1 is not a 0.1% or 10% change in concentration. It multiplies the concentration by 10⁰.¹, which is about 1.26.

+ Add to test✓ In your test94IBMI-030-04-r2-03Complex numbers and matrices Paper HLHard15 marks

A car-sharing company has 200 cars, which are parked overnight either at the Airport (A) or in the City (C). Each day:
• 70% of cars that start the day at A end the day at A; the rest end at C.
• 80% of cars that start the day at C end the day at C; the rest end at A.
Initially there are 120 cars at A and 80 cars at C. Let $s_n = \begin{pmatrix} A_n \\ C_n \end{pmatrix}$ be the numbers of cars at each location after $n$ days, so that $s_{n} = T^n s_0$.

(a)

Write down the transition matrix $T$.

(2)

(b)

Find the number of cars at each location after 3 days.

(2)

(c)

Find the eigenvalues of $T$ and a corresponding eigenvector for each.

(4)

(d)

Hence show that $A_n = 80 + 40(0.5)^n$.

(3)

(e)

Find the least number of days after which there are fewer than 81 cars at A.

(2)

(f)

State the long-term number of cars in the City, justifying your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$T = \begin{pmatrix} 0.7 & 0.2 \\ 0.3 & 0.8 \end{pmatrix}$	✓ 2
(one mark for each correct column)	—
Part (b)
Uses $T^3\begin{pmatrix} 120 \\ 80 \end{pmatrix}$	✓ 1
85 cars at A and 115 cars at C	✓ 1
Part (c)
Characteristic equation $\lambda^2 - 1.5\lambda + 0.5 = 0$	✓ 1
$\lambda = 1$ and $\lambda = 0.5$	✓ 1
For $\lambda = 1$: eigenvector $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ (or any multiple)	✓ 1
For $\lambda = 0.5$: eigenvector $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ (or any multiple)	✓ 1
Part (d)
Writes $s_0 = a\begin{pmatrix} 2 \\ 3 \end{pmatrix} + b\begin{pmatrix} 1 \\ -1 \end{pmatrix}$, giving $2a + b = 120$, $3a - b = 80$	✓ 1
$a = 40$, $b = 40$	✓ 1
$s_n = 40\begin{pmatrix} 2 \\ 3 \end{pmatrix} + 40(0.5)^n\begin{pmatrix} 1 \\ -1 \end{pmatrix}$, so $A_n = 80 + 40(0.5)^n$	✓ 1
Part (e)
Solves $80 + 40(0.5)^n < 81$, giving $n > 5.32$	✓ 1
$n = 6$ days	✓ 1
Part (f)
As $n \to \infty$, $(0.5)^n \to 0$, so $s_n \to \begin{pmatrix} 80 \\ 120 \end{pmatrix}$	✓ 1
In the long term there are 120 cars in the City (and 80 at the Airport)	✓ 1

Exam tip: The eigenvector for $\lambda = 1$ gives the steady-state proportions. Scaling $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ to total 200 is a quick check on your long-term answer.

+ Add to test✓ In your test95IBMI-030-03-r2-06Exponents and logarithms Paper HLHard13 marks

The populations of two towns, $t$ years after the start of 2025, are modelled by
Town A: $P_A = 24\,000\times1.032^{t}$
Town B: $P_B = 41\,000\times0.987^{t}$

(a)

Using logarithms, find the value of $t$ when the two towns have equal populations.

(4)

(b)

The model for town A can be written as $P_A = 24\,000\,e^{kt}$. Find the value of $k$.

(2)

(c)

Find the time taken for the population of town A to double.

(2)

(d)

A third town, C, had a population of 15 000 at the start of 2025. Its population is modelled as doubling every 18 years. Find the value of $t$ when the population of town C equals the population of town A, and comment on the reliability of your answer.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: sets up the equation $24\,000\times1.032^{t} = 41\,000\times0.987^{t}$	✓ 1
M1: rearranges to $\left(\dfrac{1.032}{0.987}\right)^{t} = \dfrac{41}{24}$, or takes logs of both sides	✓ 1
M1: $t = \dfrac{\ln(41/24)}{\ln(1.032/0.987)}$	✓ 1
A1: $t = 12.0$ years	✓ 1
Part (b)
M1: $e^{k} = 1.032$	✓ 1
A1: $k = \ln1.032 = 0.0315$	✓ 1
Part (c)
M1: $1.032^{t} = 2 \Rightarrow t = \dfrac{\ln2}{\ln1.032}$	✓ 1
A1: 22.0 years	✓ 1
Part (d)
M1: writes the model $P_C = 15\,000\times2^{t/18}$	✓ 1
M1: sets up the equation $15\,000\times2^{t/18} = 24\,000\times1.032^{t}$ and takes logarithms	✓ 1
M1: $t\left(\dfrac{\ln2}{18} - \ln1.032\right) = \ln1.6$	✓ 1
A1: $t = 67.1$ years	✓ 1
R1: unreliable, because this is a long-term extrapolation; constant growth rates are unlikely to continue for about 67 years	✓ 1

Exam tip: When the unknown appears in an exponent on both sides, take logarithms of both sides and collect the terms in t before dividing. A GDC intersection is a good way to check your answer.

+ Add to test✓ In your test96IBMI-030-04-r1-06Complex numbers and matrices Paper HLHard14 marks

A wildlife team models a deer population in two stages: juveniles $J_n$ and adults $A_n$, where $n$ is the number of years after the start of the study.
• Each year, every adult produces on average 1.8 juveniles that survive to the next count. No juveniles remain juveniles for more than one year.
• 20% of juveniles survive and become adults the next year.
• 90% of adults survive to the next year.
Initially $J_0 = 200$ and $A_0 = 300$.

(a)

Find the matrix $M$ such that $\begin{pmatrix}J_{n+1}\\A_{n+1}\end{pmatrix} = M\begin{pmatrix}J_n\\A_n\end{pmatrix}$.

(2)

(b)

Find the eigenvalues of $M$ and a corresponding eigenvector for each.

(4)

(c)

Hence show that $A_n = \frac{800}{3}(1.2)^n + \frac{100}{3}(-0.3)^n$.

(4)

(d)

Describe the long-term behaviour of the population, with reference to the eigenvalues.

(2)

(e)

Find the number of adults predicted after 10 years.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 top row $(0 \;\; 1.8)$	✓ 1
A1 bottom row $(0.2 \;\; 0.9)$, so $M = \begin{pmatrix}0 & 1.8\\0.2 & 0.9\end{pmatrix}$	✓ 1
Part (b)
M1 characteristic equation $\lambda^2 - 0.9\lambda - 0.36 = 0$	✓ 1
A1 $\lambda = 1.2$ and $\lambda = -0.3$	✓ 1
A1 for $\lambda = 1.2$: eigenvector $\begin{pmatrix}3\\2\end{pmatrix}$ (or any non-zero multiple)	✓ 1
A1 for $\lambda = -0.3$: eigenvector $\begin{pmatrix}-6\\1\end{pmatrix}$ (or any non-zero multiple)	✓ 1
Part (c)
M1 writes $\begin{pmatrix}200\\300\end{pmatrix} = p\begin{pmatrix}3\\2\end{pmatrix} + q\begin{pmatrix}-6\\1\end{pmatrix}$, or uses $M^n = PD^nP^{-1}$	✓ 1
A1 $p = \frac{400}{3}$	✓ 1
A1 $q = \frac{100}{3}$	✓ 1
A1 $A_n = 2p(1.2)^n + q(-0.3)^n = \frac{800}{3}(1.2)^n + \frac{100}{3}(-0.3)^n$ (AG)	✓ 1
Part (d)
R1 $(-0.3)^n \to 0$, so the term with $\lambda = 1.2$ dominates for large $n$	✓ 1
A1 the population grows by about 20% per year, with juveniles and adults tending to the ratio $3:2$	✓ 1
Part (e)
M1 substitutes $n = 10$ into the expression for $A_n$, or calculates $M^{10}\begin{pmatrix}200\\300\end{pmatrix}$	✓ 1
A1 $A_{10} = 1651$ adults (accept 1650)	✓ 1

Exam tip: Check your general formula against a value worked out directly. Here $A_1 = 0.2(200) + 0.9(300) = 310$, and the formula gives $320 - 10 = 310$, which confirms the coefficients.

+ Add to test✓ In your test97IBMI-030-08-r2-06Exponents and logarithms Paper HLHard14 marks

A biologist models the resting metabolic rate, R watts, of a mammal of mass M kg by
$R = aM^{b}$
where $a$ and $b$ are positive constants.

(a)

Show that this model gives a linear relationship between $\log_{10} R$ and $\log_{10} M$.

(2)

(b)

For data from a range of mammals, the regression line of $y = \log_{10} R$ on $x = \log_{10} M$ is $y = 0.74x + 0.53$. Find the value of b and the value of a.

(3)

(c)

Use the model to estimate the resting metabolic rate of a 65 kg mammal.

(2)

(d)

Use the model to estimate the mass of a mammal with a resting metabolic rate of 500 W.

(3)

(e)

Show that, according to the model, doubling the mass of a mammal multiplies its resting metabolic rate by a constant factor, and find this factor.

(2)

(f)

Interpret your answer to the previous part in terms of the energy used per kilogram of body mass.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: takes logarithms of both sides, $\log_{10} R = \log_{10}(aM^{b})$	✓ 1
A1: $\log_{10} R = b\log_{10} M + \log_{10} a$, which has the form $y = mx + c$ with $y = \log_{10} R$ and $x = \log_{10} M$, so the relationship is linear	✓ 1
Part (b)
A1: $b = 0.74$	✓ 1
M1: $\log_{10} a = 0.53$	✓ 1
A1: $a = 10^{0.53} = 3.39$	✓ 1
Part (c)
M1: substitutes, $R = 10^{0.53} \times 65^{0.74}$	✓ 1
A1: $R = 74.4$ W	✓ 1
Part (d)
M1: forms the equation $500 = 10^{0.53}M^{0.74}$, or $\log_{10} 500 = 0.74\log_{10} M + 0.53$	✓ 1
M1: rearranges correctly, $\log_{10} M = \frac{\log_{10} 500 - 0.53}{0.74}$, or solves with GDC	✓ 1
A1: $M = 853$ kg	✓ 1
Part (e)
M1: forms the ratio $\frac{a(2M)^{b}}{aM^{b}} = 2^{b}$, which is independent of M	✓ 1
A1: factor $= 2^{0.74} = 1.67$	✓ 1
Part (f)
R1: the rate is multiplied by less than 2 when the mass doubles, so the rate per kilogram, $\frac{R}{M} = aM^{-0.26}$, decreases as M increases	✓ 1
R1: larger mammals therefore use less energy per kilogram of body mass than smaller mammals	✓ 1

Exam tip: In the line $y = 0.74x + 0.53$, the intercept is $\log_{10} a$, not a itself. Undo the logarithm, $a = 10^{0.53}$, before you substitute into the power model.

Marking point	Mark	Notes
Part (a)
M1 for \(\left\|\frac{85-82.4}{82.4}\right\|\times100\)	✓ 1
A1 3.16 % (accept 3.155…)	✓ 1
Part (b)
M1 for 82.4 × 0.95 (or 82.4 − 0.05 × 82.4)	✓ 1
A1 78.28 m (accept 78.3 m)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{2.85\times10^{12}}{6.72\times10^{7}}\)	✓ 1
A1 42 400 USD	✓ 1
Part (b)
A1 4.24×10⁴ USD	✓ 1
Part (c)
M1 for \(\frac{6.72\times10^{7}}{2.43\times10^{5}}\)	✓ 1
A1 2.77×10² people per km²	✓ 1

Marking point	Mark	Notes
Part (a)
Sets up \(4500\left(1+\frac{0.032}{12}\right)^n > 6000\) or equivalent TVM entry (PV = −4500, FV = 6000, I% = 3.2, C/Y = 12) M1	✓ 1
\(n = 108.02...\) A1	✓ 1
109 months A1	✓ 1
Part (b)
\(4500\left(1+\frac{0.032}{12}\right)^{72}\) or TVM entry with N = 72 M1	✓ 1
Value = 5451.12 euros A1	✓ 1
Interest = 5451.12 − 4500 = 951.12 euros A1	✓ 1

Marking point	Mark	Notes
Part (a)
Uses \(u_{25}=18+(25-1)\times3\) M1 ; 90 seats A1	✓ 2
Part (b)
Forms \(18+3(n-1)>60\) or lists terms. Row 15 has exactly 60 seats M1 ; row 16, with 63 seats A1	✓ 2
Part (c)
Uses \(S_{25}=\frac{25}{2}(18+90)\) or an equivalent formula M1 ; 1350 seats A1	✓ 2

Marking point	Mark	Notes
Part (a)
Uses \(2400\times1.15^{7}\) M1 ; 6380 downloads (6384.05…) A1	✓ 2
Part (b)
Forms \(2400\times1.15^{n-1}>10000\) or uses a table on a GDC M1 ; gives \(n-1>10.2\), or week 11 ≈ 9709 and week 12 ≈ 11 166 A1 ; week 12 A1	✓ 3
Part (c)
Uses \(S_{12}=\frac{2400(1.15^{12}-1)}{1.15-1}\) M1 ; 69 600 downloads (69 604.0…) A1	✓ 2

Marking point	Mark	Notes
Part (a)
Solve \(4500\left(1+\frac{0.032}{12}\right)^{n}=6000\), or TVM with PV = −4500, FV = 6000, I% = 3.2, P/Y = C/Y = 12	✓ 1
n = 108.02…	✓ 1
109 months, since after 108 months the value is still just below EUR 6000	✓ 1
Part (b)
\(4500\left(1+\frac{0.032}{12}\right)^{72}\)	✓ 1
= EUR 5451.12	✓ 1
Interest = 5451.12 − 4500 = EUR 951.12	✓ 1

Marking point	Mark	Notes
Part (a)
\(u_{24} = 18 + (24-1)(3)\) M1	✓ 1
\(= 87\) seats A1	✓ 1
Part (b)
\(S_{24} = \frac{24}{2}(18 + 87)\) or \(\frac{24}{2}(2\times18 + 23\times3)\) M1	✓ 1
\(= 1260\) seats A1	✓ 1
Part (c)
\(18 + 3(n-1) > 60\), or noting that row 15 has exactly 60 seats M1	✓ 1
Row 16, which has 63 seats A1	✓ 1

Marking point	Mark	Notes
Part (a)
M1: substitutes correctly, \(L = 10\log_{10}\left(\frac{3.2\times10^{-5}}{1.0\times10^{-12}}\right) = 10\log_{10}(3.2\times10^{7})\)	✓ 1
A1: \(L = 75.1\) dB	✓ 1
Part (b)
M1: substitutes correctly, \(88 = 10\log_{10}\left(\frac{I}{10^{-12}}\right)\)	✓ 1
M1: rearranges correctly, \(\frac{I}{10^{-12}} = 10^{8.8}\), or solves with GDC	✓ 1
A1: \(I = 6.31\times10^{-4}\) W m⁻²	✓ 1

Marking point	Mark	Notes
Part (a)
\(u_{25} = 18 + (25-1)\times 3\) M1	✓ 1
= 90 seats A1	✓ 1
Part (b)
Sets up \(18 + 3(n-1) > 60\), or lists terms using a GDC table M1	✓ 1
\(n > 15\), so row 16 (which has 63 seats) A1	✓ 1
Part (c)
\(S_{25} = \frac{25}{2}(18 + 90)\) or \(\frac{25}{2}(2\times 18 + 24\times 3)\) M1	✓ 1
= 1350 seats A1	✓ 1

Marking point	Mark	Notes
Part (a)
\(r = 1.15\) A1	✓ 1
Part (b)
\(u_{10} = 40 \times 1.15^{9}\) M1	✓ 1
= 140.7… ≈ 141 shares A1	✓ 1
Part (c)
\(S_{10} = \frac{40(1.15^{10} - 1)}{1.15 - 1}\) M1	✓ 1
= 812.1… ≈ 812 shares A1	✓ 1

Marking point	Mark	Notes
Part (a)
M1: substitutes \(t=4\), \(28\,500\times0.84^{4}\)	✓ 1
A1: \$14 189.33	✓ 1
Part (b)
M1: sets up the equation \(28\,500\times0.84^{t} = 8000\)	✓ 1
M1: uses logarithms or technology, e.g. \(t = \dfrac{\log(8000/28\,500)}{\log 0.84}\), or finds the intersection on a GDC	✓ 1
A1: \(t = 7.29\) years	✓ 1

Marking point	Mark	Notes
Part (a)
Correct substitution: \(4500\left(1+\frac{0.036}{12}\right)^{72}\), or TVM with N = 72, I% = 3.6, PV = −4500, PMT = 0, P/Y = C/Y = 12	✓ 1
$5583.16	✓ 1
Part (b)
Sets up \(4500(1.003)^n > 6000\) or equivalent, e.g. TVM solving for N with FV = 6000	✓ 1
\(n = 96.04...\)	✓ 1
97 months, because the value has not reached $6000 after 96 months	✓ 1
Part (c)
\(4500(1.003)^{97}\) = $6017.32	✓ 1

Marking point	Mark	Notes
Part (a)
A1: \(r=1.025\)	✓ 1
Part (b)
M1: sets up \(42000(1.025)^{n-1}>60000\) and solves it, e.g. with a GDC or logarithms, giving \(n-1>14.4\)	✓ 1
A1: \(n=16\), so the start of 2035	✓ 1
Part (c)
M1: \(42000\times1.025^{10}\), i.e. the correct power for \(u_{11}\)	✓ 1
A1: 53764 (accept 53800 or 53763)	✓ 1

Marking point	Mark	Notes
Part (a)
M1: use of \(u_n=u_1+(n-1)d\) with \(u_1=18\), \(d=3\), \(n=20\), giving \(18+19\times3\)	✓ 1
A1: \(u_{20}=75\) seats	✓ 1
Part (b)
M1: sets up \(18+3(n-1)>100\), or an equivalent equation or table of values	✓ 1
A1: \(n>28.3\), so row 29 (which has 102 seats)	✓ 1
Part (c)
M1: use of \(S_n=\frac{n}{2}(2u_1+(n-1)d)\) or \(\frac{n}{2}(u_1+u_n)\) with \(n=20\)	✓ 1
A1: 930 seats	✓ 1

Marking point	Mark	Notes
Part (a)
\(28000 \times 0.85^4\) M1	✓ 1
14 616.18 dollars (accept 14 600 dollars to 3 s.f.) A1	✓ 1
Part (b)
\(28000 \times 0.85^n < 5000\) or equivalent, e.g. a table of values or a GDC solver M1	✓ 1
\(n = 10.6...\), OR values after 10 and 11 years found as 5512 and 4686 dollars A1	✓ 1
11 years A1	✓ 1
Part (c)
\((1-r)^4 = 0.5\) M1	✓ 1
\(r = 1 - 0.5^{0.25} = 0.159\), so 15.9% A1	✓ 1

Marking point	Mark	Notes
Part (a)
\(u_8 = 1200 \times 1.15^{7}\) M1	✓ 1
\(= 3190\) views (3192.02…; accept 3192) A1	✓ 1
Part (b)
Recognises a geometric series with \(u_1 = 1200\) and \(r = 1.15\) M1	✓ 1
\(S_{14} = \frac{1200(1.15^{14}-1)}{1.15-1}\) A1	✓ 1
\(= 48600\) views (48605.6…; accept 48606) A1	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{4}{3}\pi(1.236)^3\)	✓ 1
A1 7.91 m³	✓ 1
Part (b)
A1 \(\frac{4}{3}\times3\times1.2^3=6.912\) m³	✓ 1
Part (c)
M1 for \(\left\|\frac{6.912-7.909...}{7.909...}\right\|\times100\)	✓ 1
A1 12.6 %	✓ 1
Part (d)
R1 both approximations (3 for \(\pi\) and 1.2 for the radius) are smaller than the true values, so the errors do not cancel and the estimate is too small	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for converting the time to \(2\frac{47}{60}=2.783...\) h and dividing 248 by it	✓ 1
A1 89.1 km h⁻¹	✓ 1
Part (b)
M1 for \(\frac{89.10...\times1000}{3600}\)	✓ 1
A1 24.8 m s⁻¹	✓ 1
Part (c)
M1 for \(\frac{248}{90}\)	✓ 1
A1 2.76 h	✓ 1
M1 for \(\left\|\frac{2.7555...-2.7833...}{2.7833...}\right\|\times100\)	✓ 1
A1 0.998 %	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{\|48-52.6\|}{52.6} \times 100\%\)	✓ 1
A1 8.75%	✓ 1
Part (b)
A1 lower bound 52.55 m	✓ 1
A1 upper bound 52.65 m	✓ 1
Part (c)
M1 for using the lower bound as the true value, \(\frac{0.05}{52.55} \times 100\%\)	✓ 1
A1 0.0951%	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{5.06\times10^9}{3.71\times10^{10}} \times 100\)	✓ 1
A1 13.6% (13.638…)	✓ 1
Part (b)
M1 for 5.06×10⁹ ÷ 3.34×10⁸	✓ 1
A1 15.1 tonnes per person (15.149…)	✓ 1
Part (c)
A1 3.71×10¹³ kg	✓ 1
M1 for 5.15×10¹⁸ ÷ 3.71×10¹³	✓ 1
A1 1.39×10⁵	✓ 1
Part (d)
M1 for \(\frac{\|12.5-13.638\ldots\|}{13.638\ldots} \times 100\%\)	✓ 1
A1 8.35%	✓ 1
Part (e)
R1 the numerator was rounded down and the denominator was rounded up, so both roundings make the quotient smaller and their effects combine rather than cancel	✓ 1

Marking point	Mark	Notes
Part (a)
Uses \(28500\times0.88^{5}\) M1 ; $15 040.36 A1	✓ 2
Part (b)
Forms \(28500\times0.88^{n}<10000\) or uses a table M1 ; \(n>8.19\), or value after 8 years ≈ $10 249.58 and after 9 years ≈ $9019.63 A1 ; 9 years A1	✓ 3
Part (c)
Forms \(28500(1-r)^{5}=13000\) M1 ; \(r=14.5\%\) A1	✓ 2

Marking point	Mark	Notes
Part (a)
\(u_{10} = 12 + 9(2.5)\) M1	✓ 1
\(= 34.5\) km A1	✓ 1
Part (b)
\(S_{10} = \frac{10(1.08^{10}-1)}{0.08}\) M1	✓ 1
\(= 145\) km (144.865…) A1	✓ 1
Part (c)
Sets up \(10(1.08)^{n-1} > 12 + 2.5(n-1)\) M1	✓ 1
Uses a table or graph on the GDC, e.g. week 27: Ben 73.96 km, Amara 77 km; week 28: Ben 79.88 km, Amara 79.5 km M1	✓ 1
Week 28 A1	✓ 1

Marking point	Mark	Notes
Part (a)
\(200 + 200(0.35) + 200(0.35)^2\) M1	✓ 1
\(= 294.5\) mg (accept 295 mg) A1	✓ 1
Part (b)
The amount is \(200 + 200(0.35) + \dots + 200(0.35)^{n-1}\), a geometric series with \(u_1 = 200\) and \(r = 0.35\) R1	✓ 1
\(S_n = \frac{200(1-0.35^{n})}{1-0.35} = \frac{200(1-0.35^{n})}{0.65}\) A1	✓ 1
Part (c)
\(\frac{200(1-0.35^{7})}{0.65}\) M1	✓ 1
\(= 307\) mg (307.494…) A1	✓ 1

Marking point	Mark	Notes
Part (a)
\(12000(1.051)^{10}\)	✓ 1
€19 733.69	✓ 1
Part (b)
\(12000\left(1+\frac{0.0495}{12}\right)^{120}\)	✓ 1
€19 665.95	✓ 1
Part (c)
Option A, by €67.74	✓ 1
Part (d)
Divides by \(1.023^{10}\): \(\frac{19733.69}{1.023^{10}}\)	✓ 1
€15 720	✓ 1

Marking point	Mark	Notes
Part (a)
M1: forms the equation \(2.4b^{3} = 5.9\)	✓ 1
A1: \(b = 1.35\) (1.34963…)	✓ 1
Part (b)
A1: 35.0% (accept 35%)	✓ 1
Part (c)
M1: forms the equation \(2.4b^{t} = 4.8\) or \(b^{t} = 2\)	✓ 1
A1: \(t = 2.31\) days	✓ 1
Part (d)
M1: forms the equation \(2.4b^{t} = 500\)	✓ 1
M1: uses logarithms, \(t = \frac{\ln(500/2.4)}{\ln b}\), or solves with GDC	✓ 1
A1: \(t = 17.8\) days	✓ 1
Part (e)
R1: the pond surface and resources such as light and nutrients are limited, so exponential growth cannot continue	✓ 1
R1: the rate of spread will slow down as the area covered approaches 500 m², so the model overestimates the area or underestimates the time	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(12000\times0.94^{7}\)	✓ 1
A1: 7780 kg (7781.7)	✓ 1
Part (b)
M1: use of \(S_{15}=\frac{12000(1-0.94^{15})}{1-0.94}\)	✓ 1
A1: 121 000 kg (120 942)	✓ 1
Part (c)
M1: use of \(S_\infty=\frac{u_1}{1-r}=\frac{12000}{0.06}\)	✓ 1
A1: \(S_\infty=200000\) kg	✓ 1
R1: since \(\|r\|=0.94<1\), the partial sums increase towards 200 000 kg but never reach it, so the total can never exceed 200 000 kg and the claim is incorrect	✓ 1

Marking point	Mark	Notes
Part (a)
\(2.5 \times 0.7^{4}\) M1	✓ 1
= 0.600 m (0.60025 m) A1	✓ 1
Part (b)
Sets up \(2.5 \times 0.7^{n} < 0.05\) and solves (logs or GDC), giving \(n > 10.97\) M1	✓ 1
11th bounce A1	✓ 1
Part (c)
Sum of first four rebound heights: \(\frac{1.75(1 - 0.7^{4})}{1 - 0.7} = 4.43275\) M1	✓ 1
Total \(= 2.5 + 2 \times 4.43275\) (each rebound is travelled up and down) M1	✓ 1
= 11.4 m (11.3655 m) A1	✓ 1

Marking point	Mark	Notes
Part (a)
Contract A: \(42000 + 7 \times 1500 = \\)52\,500.00$ A1	✓ 1
Contract B: $40000 \times 1.045^{7}$ M1	✓ 1
= \$54 434.47 A1	✓ 1
Part (b)
Contract A: \(S_{10} = \frac{10}{2}(2\times 42000 + 9\times 1500) = \\)487\,500.00$ A1	✓ 1
Contract B: $S_{10} = \frac{40000(1.045^{10} - 1)}{0.045}$ M1	✓ 1
= \$491 528.37 A1	✓ 1
Contract B gives greater total earnings (by about \$4028) R1	✓ 1

Marking point	Mark	Notes
Part (a)
M1: sets up the equation \(90 = 180e^{-5.5k}\), or \(e^{-5.5k} = 0.5\)	✓ 1
A1: \(k = \dfrac{\ln 2}{5.5} = 0.126027\ldots\), which gives 0.126	✓ 1
Part (b)
M1: substitutes \(t=8\), \(180e^{-0.126027\times8}\)	✓ 1
A1: 65.7 mg	✓ 1
Part (c)
M1: sets up the equation \(180e^{-kt} = 20\)	✓ 1
M1: rearranges with logarithms, \(t = \dfrac{\ln 9}{k}\), or uses a GDC	✓ 1
A1: \(t = 17.4\) hours	✓ 1
Part (d)
M1: uses \(t=8\) for the first cup and \(t=3\) for the second cup	✓ 1
M1: adds the two amounts, \(180e^{-8k} + 180e^{-3k}\) (65.68 + 123.33)	✓ 1
A1: 189 mg	✓ 1
Part (e)
R1: any valid limitation, e.g. it assumes all the caffeine is absorbed instantly when the coffee is drunk; the rate of removal varies between people and with factors such as food or medication; it predicts that caffeine is never completely removed	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(-\log_{10}(6.3\times10^{-3})\)	✓ 1
A1: pH = 2.20	✓ 1
Part (b)
M1: \([\text{H}^+] = 10^{-5.6}\)	✓ 1
A1: \(2.51\times10^{-6}\) mol dm⁻³	✓ 1
Part (c)
M1: forms the ratio \(\dfrac{10^{-6.2}}{10^{-6.8}}\) or \(10^{0.6}\)	✓ 1
A1: factor = 3.98	✓ 1
R1: the hydrogen ion concentration of the lake water became about 4 times greater, even though the pH fell by only 0.6	✓ 1