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International GCSE Maths · Pearson Edexcel 4MA1 · Unit 1

Numbers and the number system

335 free, high-quality past-paper-style questions on Numbers and the number system for Edexcel IGCSE Maths, each with a full mark scheme. Covers: Powers, roots and standard form; Integers, fractions, decimals and percentages; Ratio, proportion and degree of accuracy; and more. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.1,771 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

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+ Add to test✓ In your test1IGME-058-01-01Powers, roots and standard form Paper FEasy4 marks

Standard form.

(a)

Write 34 000 000 in standard form.

(1)

(b)

Write 5.2×10⁻⁴ as an ordinary number.

(1)

(c)

Work out (3×10⁵) × (4×10³).
Give your answer in standard form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 3.4×10⁷	✓ 1
Part (b)
B1 for 0.00052	✓ 1
Part (c)
M1 for 12×10⁸ or 1 200 000 000	✓ 1
A1 for 1.2×10⁹	✓ 1

Common error: 12×10⁸ is not in standard form. The first number must be at least 1 and less than 10, so adjust the power of ten.

+ Add to test✓ In your test2IGME-058-01-04Powers, roots and standard form Paper FEasy4 marks

Zero and negative powers.

(a)

Write down the value of 7⁰.

(1)

(b)

Write down the value of 4⁻² as a fraction.

(1)

(c)

\(3^n \times 3^4 = 3^{11}\). Find the value of \(n\).

(1)

(d)

\(2^n = \frac{1}{32}\). Find the value of \(n\).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 1	✓ 1
Part (b)
B1 for \(\frac{1}{16}\)	✓ 1
Part (c)
B1 for \(n = 7\)	✓ 1
Part (d)
B1 for \(n = -5\)	✓ 1

Common error: Any non-zero number to the power 0 is 1, not 0. A negative power gives a reciprocal, not a negative number.

+ Add to test✓ In your test3IGME-058-00-01Integers, fractions, decimals and percentages Paper FEasy5 marks

This question is about the numbers 84 and 126.

(a)

Write 84 as a product of its prime factors.

(2)

(b)

Given that 126 = 2 × 3² × 7, find the highest common factor (HCF) of 84 and 126.

(1)

(c)

Find the lowest common multiple (LCM) of 84 and 126.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a complete method of repeated division or a factor tree, e.g. 84 = 2 × 42 = 2 × 2 × 21	✓ 1
A1 2² × 3 × 7 or 2 × 2 × 3 × 7	✓ 1
Part (b)
B1 42	✓ 1
Part (c)
M1 2² × 3² × 7, or lists of multiples of both numbers with at least one common multiple, or 84 × 126 ÷ 42	✓ 1
A1 252	✓ 1

Exam tip: For the HCF take the lowest power of each common prime; for the LCM take the highest power of every prime that appears.

+ Add to test✓ In your test4IGME-058-00-05Integers, fractions, decimals and percentages Paper FEasy4 marks

At midnight the temperature in a mountain town was −7 °C.

(a)

By midday the temperature had risen by 15 degrees.
Work out the temperature at midday.

(1)

(b)

At midnight the temperature in a second town was −13 °C.
Work out the difference between the midnight temperatures in the two towns.

(1)

(c)

The midnight temperatures in the first town on five nights were
−4 °C, 3 °C, −9 °C, 6 °C, −1 °C
Work out the mean of these temperatures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 8 °C	✓ 1
Part (b)
B1 6 degrees	✓ 1
Part (c)
M1 (−4 + 3 − 9 + 6 − 1) ÷ 5 or −5 ÷ 5	✓ 1
A1 −1 °C	✓ 1

Common error: Losing the negative sign when adding a mix of positive and negative values. Add the positives and negatives separately, then combine.

+ Add to test✓ In your test5IGME-058-02-04Ratio, proportion and degree of accuracy Paper FEasy4 marks

This question is about rounding.

(a)

Write 0.0047368 correct to 3 significant figures.

(1)

(b)

Write 28 463 correct to 2 significant figures.

(1)

(c)

Work out \(\frac{\sqrt{18.7}}{3.2^2-4.85}\)
Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
0.00474	✓ 1
Part (b)
28 000	✓ 1
Part (c)
M1 for 4.324… or 5.39 seen	✓ 1
A1 for 0.802	✓ 1

Common error: Leading zeros are not significant: in 0.0047368 the first significant figure is the 4.

+ Add to test✓ In your test6IGME-058-03-01Set language, notation and Venn diagrams (both tiers) Paper FEasy5 marks

ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
A = {multiples of 3}
B = {factors of 12}

(a)

List the members of the set A ∩ B.

(2)

(b)

Find n(A ∪ B).

(1)

(c)

List the members of the set A′.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A = {3, 6, 9, 12} and B = {1, 2, 3, 4, 6, 12}. B1 for at least two of 3, 6, 12 with no incorrect members . B1 for {3, 6, 12} only .	✓ 2
Part (b)
A ∪ B = {1, 2, 3, 4, 6, 9, 12}, so B1 for 7 .	✓ 1
Part (c)
B1 for at least six correct members with no incorrect members . B1 for {1, 2, 4, 5, 7, 8, 10, 11} .	✓ 2

Common error: n(A ∪ B) asks for the number of members, not a list. Count each shared member once only: 4 + 6 − 3 = 7, not 10.

+ Add to test✓ In your test7IGME-058-03-03Set language, notation and Venn diagrams (both tiers) Paper FEasy4 marks

There are 40 students in a youth club.
18 students play football.
15 students play tennis.
6 students play both football and tennis.

(a)

Find the number of students who play neither football nor tennis.

(2)

(b)

One of the 40 students is chosen at random. Find the probability that this student plays exactly one of the two sports.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 18 − 6 = 12 and 15 − 6 = 9, or for 18 + 15 − 6 = 27 . A1 for 40 − 27 = 13 .	✓ 2
Part (b)
M1 for 12 + 9 = 21 students playing exactly one sport . A1 for \(\frac{21}{40}\) or 0.525 .	✓ 2

Common error: The 18 football players already include the 6 who play both. Fill in the overlap first, then subtract it from each set total.

+ Add to test✓ In your test8IGME-058-03-10Set language, notation and Venn diagrams (both tiers) Paper FEasy4 marks

A Venn diagram shows two overlapping sets, A and B, inside the universal set ξ.
The region in A only contains 2 and 8.
The region in both A and B contains 4.
The region in B only contains 1 and 3.
The region outside both circles contains 5 and 7.

Diagram not to scale

(a)

List the members of A ∪ B.

(1)

(b)

List the members of B′.

(1)

(c)

Find n(A ∩ B′).

(1)

(d)

A member of ξ is chosen at random. Find the probability that it is a member of B.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for {1, 2, 3, 4, 8} .	✓ 1
Part (b)
B1 for {2, 5, 7, 8} .	✓ 1
Part (c)
B1 for 2 .	✓ 1
Part (d)
B1 for \(\frac{3}{7}\) .	✓ 1

Exam tip: B′ includes the members outside both circles, not just the members in A only.

+ Add to test✓ In your test9IGME-058-02-03Ratio, proportion and degree of accuracy Paper FEasy3 marks

Shampoo is sold in two sizes.
Small bottle: 250 ml for £3.20
Large bottle: 400 ml for £4.96

(a)

Which bottle is the better value for money?
You must show your working.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for any valid comparable method, e.g. 3.20 ÷ 2.5 and 4.96 ÷ 4, or 250 ÷ 3.20 and 400 ÷ 4.96, or the cost of a common volume such as 2000 ml	✓ 1
A1 for both correct comparable values, e.g. £1.28 and £1.24 per 100 ml, or 78.1 ml and 80.6 ml per £1, or £25.60 and £24.80 for 2000 ml	✓ 1
A1 for 'large bottle', supported by correct values	✓ 1

Exam tip: Always compare the same quantity, such as cost per 100 ml or amount per pound, and state your conclusion clearly.

+ Add to test✓ In your test10IGME-058-03-02Set language, notation and Venn diagrams (both tiers) Paper FEasy5 marks

ξ = {whole numbers from 1 to 12}
P = {2, 3, 5, 7, 11}
Q = {odd numbers}

(a)

Explain what is meant by 7 ∈ P.

(1)

(b)

Is the statement {3, 5} ⊂ P true or false? Give a reason for your answer.

(1)

(c)

List the members of P ∩ Q.

(1)

(d)

Find n(Q ∩ P′).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
States that 7 is a member (element) of the set P .	✓ 1
Part (b)
'True', with the reason that both 3 and 5 are members of P .	✓ 1
Part (c)
Q = {1, 3, 5, 7, 9, 11}, so P ∩ Q = {3, 5, 7, 11} .	✓ 1
Part (d)
Identifies the members of Q that are not in P as {1, 9} . Answer: 2 .	✓ 2

Exam tip: Read P′ as 'in ξ but not in P'. Q ∩ P′ means the members that are in Q but not in P. n( ) asks how many members there are, not for a list.

+ Add to test✓ In your test11IGME-058-01-02Powers, roots and standard form Paper FEasy4 marks

Write each of the following as a single power of a number.

(a)

\(5^6 \times 5^3\)

(1)

(b)

\(7^{12} \div 7^4\)

(1)

(c)

\(\frac{(2^3)^4}{2^5}\)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for \(5^9\)	✓ 1
Part (b)
B1 for \(7^8\)	✓ 1
Part (c)
B1 for \((2^3)^4 = 2^{12}\) seen or implied	✓ 1
B1 for \(2^7\)	✓ 1

Common error: For a power of a power, such as \((2^3)^4\), multiply the indices to get \(2^{12}\). Do not add them to get \(2^7\). When multiplying powers of the same number, keep the base the same: \(5^6 \times 5^3 = 5^9\), not \(25^9\).

+ Add to test✓ In your test12IGME-058-01-05Powers, roots and standard form Paper FEasy3 marks

The mass of one grain of fine sand is 6.4×10⁻⁷ kg.

(a)

Write 6.4×10⁻⁷ as an ordinary number.

(1)

(b)

A bag contains 2 kg of this sand. Work out the number of grains of sand in the bag.
Give your answer in standard form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 0.00000064	✓ 1
Part (b)
M1 for 2 ÷ (6.4×10⁻⁷) or 3 125 000 seen	✓ 1
A1 for 3.125×10⁶ (accept 3.13×10⁶)	✓ 1

Exam tip: Use brackets or the ×10x key when typing standard form into a calculator, so that 2 is divided by the whole of 6.4×10⁻⁷ and not just by 6.4.

+ Add to test✓ In your test13IGME-058-00-03Integers, fractions, decimals and percentages Paper FEasy5 marks

Percentages in everyday life.

(a)

In a quiz, Priya scored 18 marks out of a possible 75.
Write 18 as a percentage of 75.

(2)

(b)

A jacket normally costs £72. In a sale, the price is reduced by 15%.
Work out the sale price of the jacket.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{18}{75} \times 100\) or equivalent, e.g. 0.24 seen	✓ 1
A1 24%	✓ 1
Part (b)
M1 a method for finding 15% of 72, e.g. 0.15 × 72, or 7.2 + 3.6 (10% + 5%), or 10.8 seen; or 0.85 (85%) seen	✓ 1
M1 72 − 10.8 or 0.85 × 72	✓ 1
A1 £61.20	✓ 1

Exam tip: Multiplying by 0.85 finds the reduced price in one step and avoids forgetting to subtract the discount. Write money answers to 2 decimal places, e.g. £61.20.

+ Add to test✓ In your test14IGME-058-02-01Ratio, proportion and degree of accuracy Paper FEasy3 marks

Priya, Tom and Jess are sharing money.

(a)

Priya has £45 and Tom has £75.
Write the ratio of Priya's money to Tom's money in its simplest form.

(1)

(b)

On another day, Priya, Tom and Jess share £360 in the ratio 2 : 3 : 7.
Work out the largest share.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
3 : 5 (dividing both by the HCF, 15)	✓ 1
Part (b)
M1 for 360 ÷ (2 + 3 + 7) = 30, or for \(\frac{7}{12} \times 360\)	✓ 1
A1 for £210	✓ 1

Exam tip: Find the value of one part by dividing by the total number of parts. Then multiply by the number of parts you need. Check that all the shares add up to the original amount.

+ Add to test✓ In your test15IGME-058-00-04Integers, fractions, decimals and percentages Paper FEasy5 marks

This question is about decimals, fractions and percentages.

(a)

Write 0.0047382 correct to 3 significant figures.

(1)

(b)

Write these numbers in order of size. Start with the smallest.
0.6 \(\frac{5}{8}\) 62% 0.618

(2)

(c)

Write 0.375 as a fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 0.00474	✓ 1
Part (b)
M1 converts \(\frac{5}{8}\) to 0.625 and 62% to 0.62 (or all four to percentages), or a list in which only one pair of adjacent numbers is the wrong way round	✓ 1
A1 0.6, 0.618, 62%, \(\frac{5}{8}\)	✓ 1
Part (c)
M1 \(\frac{375}{1000}\) or a correct partially simplified equivalent, e.g. \(\frac{75}{200}\) or \(\frac{15}{40}\)	✓ 1
A1 \(\frac{3}{8}\)	✓ 1

Common error: Leading zeros are not significant. The first significant figure of 0.0047382 is the 4, so 3 s.f. means 4, 7 and 3 followed by the rounding check on the 8.

+ Add to test✓ In your test16IGME-058-00-02Integers, fractions, decimals and percentages Paper FEasy5 marks

Show all your working in this question. Give each answer in its simplest form.

(a)

Work out \(1\frac{1}{3} \times 2\frac{1}{4}\)

(2)

(b)

Work out \(2\frac{3}{5} - 1\frac{3}{4}\)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 converts both to improper fractions and multiplies, e.g. \(\frac{4}{3} \times \frac{9}{4}\) or \(\frac{36}{12}\)	✓ 1
A1 3	✓ 1
Part (b)
M1 correct first step: converts to improper fractions \(\frac{13}{5}\) and \(\frac{7}{4}\), OR separates whole numbers and fractions, e.g. \(1 + \frac{3}{5} - \frac{3}{4}\)	✓ 1
M1 uses a common denominator, e.g. \(\frac{52}{20} - \frac{35}{20}\) or \(\frac{12}{20} - \frac{15}{20}\)	✓ 1
A1 \(\frac{17}{20}\)	✓ 1

Common error: Do not multiply mixed numbers by multiplying the whole numbers and the fractions separately. For example, \(1 \times 2\) and \(\frac{1}{3} \times \frac{1}{4}\) gives the wrong answer. Convert to improper fractions first.

+ Add to test✓ In your test17IGME-058-03-04Set language, notation and Venn diagrams (both tiers) Paper FEasy5 marks

A and B are two sets in a universal set ξ.

Diagram not to scale

(a)

Use set notation to describe the set of members that are in A but not in B.

(1)

(b)

Use set notation to describe the set of members that are in neither A nor B.

(1)

(c)

You are told that n(ξ) = 30, n(A) = 14, n(B) = 11 and n(A ∩ B) = 5.
Find n(A ∩ B′).

(1)

(d)

Find n((A ∪ B)′).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for A ∩ B′ (accept A \ B or A − B) .	✓ 1
Part (b)
B1 for (A ∪ B)′ or A′ ∩ B′ .	✓ 1
Part (c)
B1 for 14 − 5 = 9 .	✓ 1
Part (d)
M1 for n(A ∪ B) = 14 + 11 − 5 = 20, or 9 + 5 + 6 = 20 . A1 for 30 − 20 = 10 .	✓ 2

Exam tip: Sketch two overlapping circles and put 9, 5 and 6 in the regions. This makes every count quick to check. The 10 members outside both circles must bring the total to 30.

+ Add to test✓ In your test18IGME-058-01-03Powers, roots and standard form Paper FEasy3 marks

Use your calculator to work out
\(\frac{\sqrt{17.6 + 3.2^2}}{2.5^3 - 9.4}\)

(a)

Write down all the figures on your calculator display.

(2)

(b)

Give your answer to part (a) correct to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 5.276… (or √27.84) or 6.225 seen	✓ 1
A1 for 0.8476084… (accept any correct full or truncated display beginning 0.847608)	✓ 1
Part (b)
B1 for 0.848 (ft from their (a))	✓ 1

Exam tip: Write down the value of the numerator and the denominator separately before dividing. This can earn the method mark even if a key is pressed wrongly.

+ Add to test✓ In your test19IGME-058-02-02Ratio, proportion and degree of accuracy Paper FEasy5 marks

Here is a list of ingredients for making pancakes for 6 people.
450 g of flour
3 eggs
240 ml of milk

(a)

Maya has 1 kg of flour, 7 eggs and plenty of milk.
Work out the greatest number of people she can make pancakes for.

(3)

(b)

Work out the amount of flour needed to make pancakes for 10 people.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for using flour as the limit: 1000 ÷ 75 = 13.3… people (or 1000 ÷ 450 × 6), OR showing 13 people needs 975 g flour and 14 people needs 1050 g flour	✓ 1
M1 for using eggs as the limit: 7 ÷ 0.5 = 14 people (or 7 ÷ 3 × 6), OR showing 13 people needs 6.5 eggs (≤ 7)	✓ 1
A1 for 13	✓ 1
Part (b)
M1 for 450 ÷ 6 (= 75) or scale factor 10 ÷ 6	✓ 1
A1 for 750 g	✓ 1

Common error: Do not round 13.3 up to 14. There is not enough flour for 14 people, so round down to 13. Always check every ingredient and use the one that runs out first.

+ Add to test✓ In your test20IGME3-1-07d-01Powers, roots and standard form Paper FEasy6 marks

An electric car has a battery with a capacity of 60 kWh. On a journey, 35% of the battery capacity is used.

(a)

Work out the energy used on the journey. (2)

(2)

(b)

Write the percentage of the battery that is left as a fraction in its simplest form. (2)

(2)

(c)

Over one year the battery capacity falls by 8%. Work out the new capacity of the battery. (2)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.35 × 60 or 35/100 × 60	✓ 1
A1 21 kWh	✓ 1
Part (b)
B1 65% or 100 − 35 seen	✓ 1
B1 \(\frac{13}{20}\)	✓ 1
Part (c)
M1 60 × 0.92 or 60 − 0.08 × 60	✓ 1
A1 55.2 kWh	✓ 1

Exam tip: For a percentage decrease, a single multiplier such as 0.92 is quicker and safer than finding the decrease and subtracting.

+ Add to test✓ In your test21IGME3-1-07d-02Powers, roots and standard form Paper FEasy5 marks

The Sun is 1.5 × 10⁸ km from the Earth. Light travels at 3.0 × 10⁵ km/s.

(a)

Write 4.2 × 10⁻⁵ as an ordinary number. (1)

(1)

(b)

Work out the time, in seconds, for light to travel from the Sun to the Earth. (2)

(2)

(c)

Jupiter is 7.8 × 10⁸ km from the Sun. Work out the time, in seconds, for light to travel from the Sun to Jupiter. Give your answer in standard form. (2)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 0.000042	✓ 1
Part (b)
M1 (1.5 × 10⁸) ÷ (3.0 × 10⁵)	✓ 1
A1 500 s	✓ 1
Part (c)
M1 (7.8 × 10⁸) ÷ (3.0 × 10⁵) or 2600 seen	✓ 1
A1 2.6 × 10³ s	✓ 1

Common error: When dividing numbers in standard form, subtract the powers of 10 (8 − 5 = 3) rather than dividing them.

+ Add to test✓ In your test22IGME3-1-07d-03Powers, roots and standard form Paper FEasy5 marks

A builder mixes cement, sand and gravel in the ratio 1 : 2 : 4 by volume. The builder makes 10.5 m³ of the mixture.

(a)

Work out the volume of sand in the mixture. (2)

(2)

(b)

Cement costs €84 per m³, sand costs €30 per m³ and gravel costs €25 per m³. Work out the total cost of the materials for the 10.5 m³ of mixture. (3)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 10.5 ÷ (1 + 2 + 4) or 10.5 ÷ 7 or 1.5 seen	✓ 1
A1 3 m³	✓ 1
Part (b)
M1 cement 1.5 × 84 (= 126) or gravel 6 × 25 (= 150)	✓ 1
M1 sum of all three costs, e.g. 126 + 90 + 150	✓ 1
A1 €366	✓ 1

Exam tip: Find the value of one part first, then multiply by the number of parts for each item.

+ Add to test✓ In your test23IGME3-1-07d-04Set language, notation and Venn diagrams (both tiers) Paper FEasy5 marks

In a group of 40 students, 25 play football (F) and 18 play tennis (T). 7 students play both sports.

(a)

Work out how many students play neither sport. (2)

(2)

(b)

Describe in words the set F′ ∩ T. (1)

(1)

(c)

One student is chosen at random. Find the probability that this student plays football but not tennis. (2)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 25 + 18 − 7 (= 36) or 18 + 7 + 11 seen	✓ 1
A1 4	✓ 1
Part (b)
B1 students who play tennis but do not play football	✓ 1
Part (c)
M1 (25 − 7) = 18 football only, over 40	✓ 1
A1 \(\frac{9}{20}\) (= 0.45)	✓ 1

Common error: The 25 who play football already include the 7 who play both, so subtract 7 to find football only.

+ Add to test✓ In your test24IGME3-1-07d-05Ratio, proportion and degree of accuracy Paper FEasy4 marks

A runner's time for a 100 m race is recorded as 12.4 seconds, correct to 1 decimal place. The track length is exactly 100 m.

(a)

Write down the error interval for the time, t seconds. (2)

(2)

(b)

Work out the greatest possible average speed of the runner, in m/s, correct to 3 significant figures. (2)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 12.35 ≤ t	✓ 1
B1 t < 12.45	✓ 1
Part (b)
M1 100 ÷ 12.35	✓ 1
A1 8.10 (8.0971...)	✓ 1

Exam tip: The greatest speed over a fixed distance comes from the smallest possible time.

+ Add to test✓ In your test25IGME3-1-11r7-01Integers, fractions, decimals and percentages Paper FEasy4 marks

A rail operator sells season tickets and single fares.

(a)

A season ticket costs £1240. The price increases by 3.5%.
Work out the new price of the season ticket.

(2)

(b)

A single fare is reduced by 12% in a sale. The sale price is £35.20.
Work out the original price of the single fare.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1240 × 1.035 or 1240 × 0.035 = 43.40	✓ 1
A1 £1283.40	✓ 1
Part (b)
M1 35.20 ÷ 0.88 or 35.20 ÷ 88 × 100	✓ 1
A1 £40	✓ 1

Common error: In a reverse percentage problem the sale price is 88% of the original, so do not find 12% of £35.20 and add it on.

+ Add to test✓ In your test26IGME3-1-11r7-02Powers, roots and standard form Paper FEasy4 marks

A sports scientist plans a training session that lasts 1 hour 20 minutes. The first \(\frac{2}{5}\) of the session is a warm-up and the next \(\frac{1}{4}\) is sprint work. The rest of the session is a cool-down.

(a)

Work out the fraction of the session that is the cool-down.

(2)

(b)

Work out the length of the cool-down in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(1-\frac{2}{5}-\frac{1}{4}\) or \(\frac{8}{20}+\frac{5}{20}\)	✓ 1
A1 \(\frac{7}{20}\)	✓ 1
Part (b)
M1 \(\frac{7}{20}\times 80\)	✓ 1
A1 28	✓ 1

Exam tip: Convert the time to a single unit (80 minutes) before taking a fraction of it.

+ Add to test✓ In your test27IGME3-1-11r7-03Powers, roots and standard form Paper FEasy4 marks

Radio signals travel at 3.0 × 10⁵ km/s. A space probe is 1.44 × 10⁸ km from Earth.

(a)

The Moon is 384 400 km from Earth.
Write 384 400 in standard form.

(1)

(b)

Work out the time, in seconds, for a signal to travel from the probe to Earth.
Give your answer in standard form.

(2)

(c)

Write your answer to part (b) in minutes.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3.844 × 10⁵	✓ 1
Part (b)
M1 1.44 × 10⁸ ÷ (3.0 × 10⁵)	✓ 1
A1 4.8 × 10²	✓ 1
Part (c)
B1 8 (ECF from (b))	✓ 1

Common error: When dividing numbers in standard form, divide the numbers and subtract the powers separately; do not subtract the whole numbers.

+ Add to test✓ In your test28IGME3-1-11r7-08Integers, fractions, decimals and percentages Paper FEasy5 marks

Two rail services leave a station at 06:00. Service A then leaves every 12 minutes and service B then leaves every 18 minutes.

(a)

Write 12 and 18 as products of their prime factors.

(2)

(b)

Write down the highest common factor (HCF) of 12 and 18.

(1)

(c)

Work out how many minutes after 06:00 both services next leave together.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 12 = 2² × 3	✓ 1
B1 18 = 2 × 3²	✓ 1
Part (b)
B1 6	✓ 1
Part (c)
M1 LCM = 2² × 3² or listing multiples of 12 and 18	✓ 1
A1 36	✓ 1

Exam tip: The lowest common multiple uses the highest power of each prime from the two factorisations.

+ Add to test✓ In your test29IGME3-1-12r13-01Set language, notation and Venn diagrams (both tiers) Paper FEasy5 marks

A shop surveyed 60 customers. The set ξ is all 60 customers. B is the set of customers who bought bread and M is the set who bought milk. n(B) = 35, n(M) = 28 and n(B ∩ M) = 10.

(a)

Work out how many customers bought bread but not milk.

(1)

(b)

Work out how many customers bought neither bread nor milk.

(2)

(c)

Work out n(B ∪ M′).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 25	✓ 1
Part (b)
M1 n(B ∪ M) = 35 + 28 − 10 (= 53)	✓ 1
A1 7	✓ 1
Part (c)
M1 identifies this set as everyone in B together with those in neither set (35 and 7), or 60 − (28 − 10)	✓ 1
A1 42	✓ 1

Common error: n(B) = 35 already includes the 10 customers in B ∩ M. Adding 35 and 28 without subtracting 10 counts them twice.

+ Add to test✓ In your test30IGME3-1-12r13-02Integers, fractions, decimals and percentages Paper FEasy4 marks

During an outbreak, a clinic recorded 480 new cases in one week and 612 new cases in the next week.

(a)

Work out the percentage increase in weekly cases.

(2)

(b)

In the third week the number of new cases fell by 25% from 612. Work out the number of new cases in the third week.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (612 − 480)/480 × 100	✓ 1
A1 27.5 (%)	✓ 1
Part (b)
M1 0.75 × 612 or 612 − 0.25 × 612	✓ 1
A1 459	✓ 1

Common error: Percentage change is found as a fraction of the original value: divide the change by 480, not by 612.

+ Add to test✓ In your test31IGME3-1-12r13-03Powers, roots and standard form Paper FEasy5 marks

A wind farm and a solar farm supply electricity to a town.

(a)

A wind farm produces 4 500 000 W. Write this in standard form.

(1)

(b)

The solar farm produces 2.4 × 10⁵ W and the wind farm produces 3.6 × 10⁶ W. Work out their total output. Give your answer in standard form.

(2)

(c)

A small turbine produces 7.2 × 10⁴ W. How many of these small turbines would produce 3.6 × 10⁶ W in total?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 4.5 × 10⁶	✓ 1
Part (b)
M1 3 600 000 + 240 000 or 36 × 10⁵ + 2.4 × 10⁵	✓ 1
A1 3.84 × 10⁶ (W)	✓ 1
Part (c)
M1 3.6 × 10⁶ ÷ 7.2 × 10⁴	✓ 1
A1 50	✓ 1

Exam tip: To add numbers in standard form, first write them with the same power of ten, or convert both to ordinary numbers.

+ Add to test✓ In your test32IGME3-1-12r13-04Powers, roots and standard form Paper FEasy4 marks

A shop makes 2.5 kg of trail mix from nuts, raisins and seeds in the ratio 3 : 2 : 5 by mass.

(a)

Work out the mass of raisins in the mix. Give your answer in grams.

(2)

(b)

Nuts cost €12 per kilogram. Work out the cost of the nuts in the mix.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2/10 × 2500 or 2500 ÷ 10 × 2	✓ 1
A1 500 (g)	✓ 1
Part (b)
M1 nuts mass = 3/10 × 2.5 (= 0.75 kg) or 750 g	✓ 1
A1 9 (€)	✓ 1

Exam tip: Add the ratio parts first (3 + 2 + 5 = 10) to find the value of one part, and convert units before using a price per kilogram.

+ Add to test✓ In your test33IGME3-1-13r14-01Ratio, proportion and degree of accuracy Paper FEasy5 marks

A fishery lands 1260 kg of fish in one day. \(\frac{2}{7}\) of the catch is cod. Of the rest of the catch, 35% is haddock and all the remaining fish are mackerel.

(a)

Work out the mass of cod landed.

(2)

(b)

Work out the mass of mackerel landed.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1260 ÷ 7 × 2 or 1260 × \(\frac{2}{7}\)	✓ 1
A1 360 kg	✓ 1
Part (b)
M1 1260 − 360 = 900	✓ 1
M1 900 × 0.65 or 900 − 0.35 × 900	✓ 1
A1 585 kg	✓ 1

Exam tip: Find what is left after each step and say clearly which quantity you are taking the percentage of.

+ Add to test✓ In your test34IGME3-1-13r14-02Powers, roots and standard form Paper FEasy4 marks

A robot's sensor takes one reading every 4.5 × 10⁻³ seconds.

(a)

Work out how many readings the sensor takes in 3.6 × 10² seconds. Give your answer in standard form.

(2)

(b)

A faster sensor takes one reading every 2.5 × 10⁻⁴ seconds. Work out how many times as many readings per second the faster sensor takes as the first sensor.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3.6 × 10² ÷ (4.5 × 10⁻³) or 360 ÷ 0.0045	✓ 1
A1 8 × 10⁴	✓ 1
Part (b)
M1 4.5 × 10⁻³ ÷ (2.5 × 10⁻⁴)	✓ 1
A1 18	✓ 1

Common error: Dividing the powers of ten wrongly: 10⁻³ ÷ 10⁻⁴ is 10¹, not 10⁻¹.

+ Add to test✓ In your test35IGME3-1-13r14-03Ratio, proportion and degree of accuracy Paper FEasy4 marks

A fashion designer mixes red, blue and white dye in the ratio 2 : 3 : 5 to make a shade of mauve.

(a)

The designer makes 8 litres of mauve dye. Work out the volume of blue dye in the mixture.

(2)

(b)

The designer has only 5.5 litres of white dye but plenty of red and blue dye. Work out the greatest total volume of mauve dye that can be made.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 8 ÷ (2 + 3 + 5) × 3 or \(\frac{3}{10}\) × 8	✓ 1
A1 2.4 litres	✓ 1
Part (b)
M1 5.5 ÷ 5 × 10 (or 1.1 litres per part)	✓ 1
A1 11 litres	✓ 1

Exam tip: Find the value of one part first; it makes every later step a simple multiplication.

+ Add to test✓ In your test36IGME3-1-13r14-04Set language, notation and Venn diagrams (both tiers) Paper FEasy4 marks

There are 40 members of a fashion club. Each member was asked whether they knit and whether they sew. Let K be the set of members who knit and S the set of members who sew. 24 members knit, 19 members sew and 7 members do neither.

(a)

Work out the number of members who both knit and sew.

(3)

(b)

Find n(K′ ∩ S).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 − 7 = 33 members do at least one of the two	✓ 1
M1 24 + 19 − 33	✓ 1
A1 10	✓ 1
Part (b)
B1 9	✓ 1

Common error: n(K′ ∩ S) means members who sew but do not knit, not everyone who sews.

+ Add to test✓ In your test37IGME3-1-13r16-04Integers, fractions, decimals and percentages Paper FEasy6 marks

A cargo ship carries 840 containers. Three-eighths of the containers are refrigerated.

(a)

Work out the number of refrigerated containers.

(2)

(b)

Of the containers that are not refrigerated, 40% hold machinery. Work out how many containers hold machinery.

(2)

(c)

Write the number of containers holding machinery as a fraction of all 840 containers. Give your fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3/8 × 840	✓ 1
A1 315	✓ 1
Part (b)
M1 840 − 315 = 525	✓ 1
A1 0.4 × 525 = 210	✓ 1
Part (c)
M1 210/840	✓ 1
A1 1/4	✓ 1

Common error: Take the percentage of the remaining containers, not of the original 840.

+ Add to test✓ In your test38IGME3-1-13r16-08Ratio, proportion and degree of accuracy Paper FEasy4 marks

A ship's hull is painted with a mixture of blue, white and grey paint in the ratio 2 : 5 : 3. A batch contains 40 litres of paint altogether.

(a)

Work out the volume of white paint in the batch.

(2)

(b)

A second batch, in the same ratio, contains 9 litres of grey paint. Work out the volume of blue paint in the second batch.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 ÷ (2 + 5 + 3) = 4	✓ 1
A1 5 × 4 = 20 litres	✓ 1
Part (b)
M1 9 ÷ 3 = 3 litres per part	✓ 1
A1 2 × 3 = 6 litres	✓ 1

Exam tip: Find the value of one part first; every other quantity is then a multiple of it.

+ Add to test✓ In your test39IGME3-1-13r16-10Ratio, proportion and degree of accuracy Paper FEasy6 marks

The height of a volcano is 2847 m, to the nearest metre.

(a)

Write 2847 correct to 2 significant figures.

(1)

(b)

The width of the crater is 380 m, correct to the nearest 10 m. Write down the error interval for the width.

(2)

(c)

Estimate the value of (4.8 × 19.6) ÷ 0.52 by rounding each number to 1 significant figure.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2800	✓ 1
Part (b)
B1 375 ≤ width	✓ 1
B1 width < 385	✓ 1
Part (c)
M1 rounding to 5, 20 and 0.5	✓ 1
M1 (5 × 20) ÷ 0.5	✓ 1
A1 200	✓ 1

Exam tip: In an estimate, show the rounded numbers so that you can earn the method marks even if the arithmetic slips.

+ Add to test✓ In your test40IGME3-1-13r16-14Set language, notation and Venn diagrams (both tiers) Paper FEasy6 marks

Two cargo ships leave a port on regular cycles. The first leaves every 12 days and the second every 18 days. Both leave on 1 March.

(a)

Work out the number of days until both ships next leave on the same day.

(2)

(b)

Write 360 as a product of its prime factors.

(2)

(c)

A ship takes 2¾ hours to cross a channel and a further 1⅚ hours to enter harbour. Work out the total time. Give your answer as a mixed number.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiples of 12 and 18, or 2² × 3 and 2 × 3²	✓ 1
A1 36 days	✓ 1
Part (b)
M1 factor tree or repeated division by primes	✓ 1
A1 2³ × 3² × 5	✓ 1
Part (c)
M1 33/12 + 22/12 (common denominator)	✓ 1
A1 4⁷⁄₁₂ hours	✓ 1

Exam tip: Convert mixed numbers to improper fractions, or add whole numbers and fractions separately, but keep the common denominator.

+ Add to test✓ In your test41IGME3-1-12r19-01Powers, roots and standard form Paper FEasy4 marks

Each turbine on a wind farm produces 3.6 × 10⁶ watts at full output. The farm has 25 turbines.

(a)

Work out the total power of the farm at full output. Give your answer in standard form.

(2)

(b)

A home uses 2.4 × 10³ watts. Work out how many homes one turbine could supply at full output.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3.6 × 25 = 90	✓ 1
A1 9 × 10⁷ W	✓ 1
Part (b)
M1 (3.6 × 10⁶) ÷ (2.4 × 10³)	✓ 1
A1 1500	✓ 1

Exam tip: Check that your standard form answer has a number between 1 and 10 before the power of ten.

+ Add to test✓ In your test42IGME3-1-12r19-02Set language, notation and Venn diagrams (both tiers) Paper FEasy5 marks

A library surveys 60 members. 33 members borrow fiction, 27 members borrow non-fiction and 12 members borrow both fiction and non-fiction.

(a)

Work out the number of members who borrow fiction only, the number who borrow non-fiction only, and the number who borrow neither.

(3)

(b)

One member is chosen at random. Find the probability that this member borrows exactly one of the two types.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 33 − 12 and 27 − 12	✓ 1
A1 fiction only 21 and non-fiction only 15	✓ 1
A1 neither = 60 − (21 + 12 + 15) = 12	✓ 1
Part (b)
M1 (21 + 15) ÷ 60, ECF from (a)	✓ 1
A1 3/5 (= 0.6)	✓ 1

Common error: The number who borrow a type includes those who borrow both, so subtract the overlap before filling in the 'only' regions.

+ Add to test✓ In your test43IGME3-1-12r19-03Set language, notation and Venn diagrams (both tiers) Paper FEasy5 marks

A swimming club has 84 members. The ratio of juniors to seniors is 5 : 2. Juniors pay €14.50 per month and seniors pay €22 per month.

(a)

Work out the number of juniors and the number of seniors.

(2)

(b)

Work out the total fees received in one month.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 84 ÷ 7 = 12	✓ 1
A1 60 juniors and 24 seniors	✓ 1
Part (b)
M1 60 × 14.50 = 870	✓ 1
M1 24 × 22 = 528	✓ 1
A1 €1398	✓ 1

Exam tip: Add the parts of the ratio first to find the value of one part.

+ Add to test✓ In your test44IGME3-1-12r19-04Powers, roots and standard form Paper FEasy5 marks

A library holds 640 books. Three-eighths of the books are children's books. Of the other books, two-fifths are fiction.

(a)

Work out the number of children's books.

(2)

(b)

Work out the number of fiction books that are not children's books.

(2)

(c)

Write the number in (b) as a fraction of all the books. Give your answer in its simplest form.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 640 ÷ 8 × 3	✓ 1
A1 240	✓ 1
Part (b)
M1 400 × 2/5	✓ 1
A1 160	✓ 1
Part (c)
A1 160/640 = 1/4 (accept 0.25)	✓ 1

Common error: 'Of the other books' means a fraction of the remainder, not of the original total.

+ Add to test✓ In your test45IGME3-1-11r25-01Integers, fractions, decimals and percentages Paper FEasy4 marks

A mountaineering team sets up base camp at an altitude of 3 200 m, where the temperature is 4 °C. The summit is at an altitude of 6 400 m. The temperature falls by 6.5 °C for every 1 000 m climbed.

(a)

Work out the difference in altitude between base camp and the summit.

(1)

(b)

Work out the temperature at the summit.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3 200 m	✓ 1
Part (b)
M1 3.2 × 6.5 or 20.8 (fall in temperature)	✓ 1
M1 4 − 20.8	✓ 1
A1 −16.8 °C	✓ 1

Common error: When a temperature falls below zero, subtract the total fall from the starting value and keep the negative sign in the answer.

+ Add to test✓ In your test46IGME3-1-11r25-02Ratio, proportion and degree of accuracy Paper FEasy3 marks

A food scientist makes a sauce from tomato purée, water and oil in the ratio 5 : 3 : 2 by mass. The total mass of the sauce is 450 g.

(a)

Work out the mass of tomato purée in the sauce.

(2)

(b)

Write the mass of oil as a percentage of the total mass of the sauce.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 450 ÷ (5 + 3 + 2) = 45	✓ 1
A1 225 g	✓ 1
Part (b)
B1 20%	✓ 1

Exam tip: Add the parts of a ratio first to find the value of one part, then multiply by the number of parts you need.

+ Add to test✓ In your test47IGME3-1-11r25-11Powers, roots and standard form Paper FEasy4 marks

A food laboratory measures the mass of a vitamin in one tablet as 0.03951 g.

(a)

Write 0.03951 correct to 2 significant figures.

(1)

(b)

Write 0.03951 in standard form.

(1)

(c)

There are 250 tablets in a bottle. Use the value 0.03951 g to work out the total mass of the vitamin in the bottle. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 0.040	✓ 1
Part (b)
B1 3.951 × 10⁻²	✓ 1
Part (c)
M1 250 × 0.03951 (= 9.8775)	✓ 1
A1 9.88 g	✓ 1

Common error: A zero written after the 4 in 0.040 is a significant figure and must be kept; leading zeros are not significant.

+ Add to test✓ In your test48IGME3-1-12r25-01Powers, roots and standard form Paper FEasy3 marks

A wind farm has 12 turbines. The total power output of the farm is 3.6 × 10⁷ W. Every turbine produces the same power.

(a)

Work out the power produced by one turbine. Give your answer in standard form.

(2)

(b)

Write your answer to part (a) as an ordinary number.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (3.6 × 10⁷) ÷ 12	✓ 1
A1 3 × 10⁶ W	✓ 1
Part (b)
B1 3 000 000 (ECF from (a))	✓ 1

Exam tip: Standard form is a × 10ⁿ with 1 ≤ a < 10, so 3 × 10⁶ is already in standard form.

+ Add to test✓ In your test49IGME3-1-12r25-02Ratio, proportion and degree of accuracy Paper FEasy4 marks

A swimming club has junior, intermediate and senior swimmers in the ratio 2 : 3 : 5. There are 80 swimmers in the club.

(a)

Work out the number of senior swimmers.

(2)

(b)

Write the number of junior swimmers as a fraction of all the swimmers. Give your answer in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 80 ÷ (2 + 3 + 5) = 8	✓ 1
A1 40	✓ 1
Part (b)
M1 16 ⁄ 80 or 2 ⁄ 10	✓ 1
A1 ¹⁄₅	✓ 1

Common error: Divide by the total number of parts (here 10), not by the number of groups (here 3).

+ Add to test✓ In your test50IGME3-1-12r25-12Integers, fractions, decimals and percentages Paper FEasy4 marks

Two library reading groups meet regularly. Group A meets every 12 days and group B meets every 18 days. Both groups meet today.

(a)

Work out the number of days until the groups next meet on the same day.

(2)

(b)

Write 36 as a product of its prime factors.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 lists multiples of 12 and 18, or uses prime factors, to find a common multiple	✓ 1
A1 36	✓ 1
Part (b)
M1 correct method, for example a factor tree or repeated division	✓ 1
A1 2² × 3²	✓ 1

Exam tip: Use the lowest common multiple when events repeat and you need the first time they coincide.

+ Add to test✓ In your test51IGME3-1-13r28-01Powers, roots and standard form Paper FEasy4 marks

A space probe travels 4.2 × 10⁹ km at an average speed of 1.4 × 10⁵ km/h.

(a)

Work out the time taken. Give your answer in hours in standard form.

(2)

(b)

Convert your answer to (a) into days.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4.2 × 10⁹ ÷ 1.4 × 10⁵	✓ 1
A1 3 × 10⁴ hours	✓ 1
Part (b)
M1 3 × 10⁴ ÷ 24	✓ 1
A1 1250 days	✓ 1

Exam tip: Write standard form as A × 10ⁿ with 1 ≤ A < 10 and check your calculator display matches your written answer.

+ Add to test✓ In your test52IGME3-1-13r28-02Powers, roots and standard form Paper FEasy4 marks

Green coffee beans lose 18% of their mass during roasting.

(a)

A roaster puts 25 kg of green beans into the drum. Work out the mass of the roasted beans.

(2)

(b)

After roasting a different batch, the mass of roasted beans is 12.30 kg. Work out the mass of green beans in this batch.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 25 × 0.82 or 25 − 0.18 × 25	✓ 1
A1 20.5 kg	✓ 1
Part (b)
M1 12.30 ÷ 0.82	✓ 1
A1 15 kg	✓ 1

Common error: In a reverse percentage question, do not find 18% of the final mass; the percentage is of the original amount.

+ Add to test✓ In your test53IGME3-1-13r28-13Ratio, proportion and degree of accuracy Paper FEasy4 marks

A café brews coffee using coffee and water in the ratio 1 : 15 by mass. Assume 1 litre of water has a mass of 1 kg.

(a)

Work out the mass of coffee needed for 1.2 kg of water. Give your answer in grams.

(2)

(b)

The café uses 2.4 kg of coffee in one day. Work out the volume of water used, in litres.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1200 ÷ 15	✓ 1
A1 80 g	✓ 1
Part (b)
M1 2.4 × 15	✓ 1
A1 36 litres	✓ 1

Exam tip: Check units before you divide: convert 1.2 kg to 1200 g first.

+ Add to test✓ In your test54IGME3-1-13r29-01Powers, roots and standard form Paper FEasy5 marks

A railway line between two cities is 360 km long. A freight train completes \(\frac{3}{8}\) of the line before its first stop.

(a)

Work out the distance the train has travelled at the first stop.

(2)

(b)

Write down the fraction of the line still to travel.

(1)

(c)

Between the first stop and the second stop the train covers \(\frac{2}{3}\) of the distance still to travel. Work out the distance between the first and second stops.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{3}{8}\times 360\)	✓ 1
A1 135 km	✓ 1
Part (b)
B1 \(\frac{5}{8}\)	✓ 1
Part (c)
M1 \(\frac{2}{3}\times(360-135)\) or \(\frac{2}{3}\times 225\)	✓ 1
A1 150 km	✓ 1

Exam tip: In fraction-of problems, find what is left before taking the next fraction; the second fraction applies to the remainder, not the whole line.

+ Add to test✓ In your test55IGME3-1-13r29-02Powers, roots and standard form Paper FEasy5 marks

Light from the Sun reaches Earth after travelling a distance of \(1.5\times10^{11}\) m. Light travels at \(3.0\times10^{8}\) m/s.

(a)

Write \(4.2\times10^{5}\) as an ordinary number.

(1)

(b)

Work out the time, in seconds, taken for light to travel from the Sun to Earth.

(2)

(c)

Convert your answer to part (b) into minutes. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 420 000	✓ 1
Part (b)
M1 \(\frac{1.5\times10^{11}}{3.0\times10^{8}}\)	✓ 1
A1 500 s	✓ 1
Part (c)
M1 \(500\div60\)	✓ 1
A1 8.33 minutes	✓ 1

Common error: Dividing by 100 instead of 60 when converting seconds to minutes; there are 60 seconds in a minute.

+ Add to test✓ In your test56IGME3-1-13r29-03Ratio, proportion and degree of accuracy Paper FEasy4 marks

A sports drink is made by mixing concentrate and water in the ratio 1 : 7 (concentrate : water). An athlete makes 600 ml of the drink.

(a)

Work out how many millilitres of concentrate are needed.

(2)

(b)

The concentrate is sold in 500 ml bottles costing £4.80 each. Work out the cost of the concentrate used to make the 600 ml of drink.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(600\div(1+7)\)	✓ 1
A1 75 ml	✓ 1
Part (b)
M1 \(\frac{75}{500}\times4.80\)	✓ 1
A1 £0.72	✓ 1

Exam tip: Add the parts of the ratio first (1 + 7 = 8) to find the size of one part.

+ Add to test✓ In your test57IGME3-1-13r29-04Set language, notation and Venn diagrams (both tiers) Paper FEasy4 marks

40 athletes attend a training camp. Of these, 22 take part in running sessions and 17 take part in swimming sessions. 9 athletes take part in both. Every athlete takes part in running, swimming, both or neither.
\(\xi\) = all 40 athletes, \(R\) = athletes who run, \(S\) = athletes who swim.

(a)

Work out the number of athletes who run but do not swim.

(1)

(b)

Work out the number of athletes who take part in neither running nor swimming.

(2)

(c)

One athlete is chosen at random from the 40. Find the probability that this athlete swims but does not run.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 13	✓ 1
Part (b)
M1 \(40-(13+9+8)\) or \(40-(22+17-9)\)	✓ 1
A1 10	✓ 1
Part (c)
B1 \(\frac{8}{40}=\frac{1}{5}\) (or 0.2)	✓ 1

Exam tip: The number in the overlap is counted once in each circle total, so subtract it when you work out how many are in at least one set.

+ Add to test✓ In your test58IGME3-1-14r33-01Powers, roots and standard form Paper FEasy5 marks

A communications satellite orbits at a height of 36 000 000 m above a ground station. A radio signal travels at 3 × 10⁸ m/s.

(a)

Write 36 000 000 in standard form.

(1)

(b)

Work out the time, in seconds, for the signal to travel from the ground station up to the satellite.

(2)

(c)

The signal travels up to the satellite and straight back down again. Write the total time in standard form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3.6 × 10⁷	✓ 1
Part (b)
M1 3.6 × 10⁷ ÷ (3 × 10⁸)	✓ 1
A1 0.12 s	✓ 1
Part (c)
M1 2 × 0.12 (ECF from (b))	✓ 1
A1 2.4 × 10⁻¹ s	✓ 1

Common error: Check that the first number in standard form is at least 1 and less than 10; 0.24 × 10⁰ is not standard form.

+ Add to test✓ In your test59IGME3-1-14r33-02Powers, roots and standard form Paper FEasy5 marks

A mountaineering team packs a food mixture of rice, lentils and nuts in the ratio 5 : 3 : 2 by mass. The team eats 4.5 kg of the mixture each day.

(a)

Work out the mass of lentils in 4.5 kg of the mixture.

(2)

(b)

The expedition lasts 12 days. Work out the total mass of nuts the team needs.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4.5 ÷ (5 + 3 + 2) × 3 or 3/10 × 4.5	✓ 1
A1 1.35 kg	✓ 1
Part (b)
M1 4.5 × 12 = 54 kg of mixture	✓ 1
M1 2/10 × 54	✓ 1
A1 10.8 kg	✓ 1

Exam tip: Find the total number of parts first, then the mass of one part, before multiplying.

+ Add to test✓ In your test60IGME3-1-14r33-04Set language, notation and Venn diagrams (both tiers) Paper FEasy5 marks

A mountaineering club has 40 members. Of these, 26 members climb, 19 members ski and 8 members do neither.
Let ξ = all members of the club, C = members who climb and S = members who ski.

(a)

Work out the number of members who both climb and ski.

(3)

(b)

One member is chosen at random. Find the probability that this member climbs but does not ski.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 − 8 = 32 members do at least one activity	✓ 1
M1 26 + 19 − 32	✓ 1
A1 13 members	✓ 1
Part (b)
M1 26 − 13 = 13 climb only (ECF from (a))	✓ 1
A1 13/40 (or 0.325)	✓ 1

Exam tip: Use n(C ∪ S) = n(C) + n(S) − n(C ∩ S) to avoid counting the overlap twice.

+ Add to test✓ In your test61IGME3-1-14r33-05Ratio, proportion and degree of accuracy Paper FEasy4 marks

A baker has 2¹⁄₂ kg of flour. Each batch of rolls uses ³⁄₄ kg of flour.

(a)

Work out 2¹⁄₂ ÷ ³⁄₄. Give your answer as a fraction.

(2)

(b)

The baker makes as many complete batches as possible. Work out the mass of flour left over.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5/2 × 4/3	✓ 1
A1 10/3 (or 3¹⁄₃)	✓ 1
Part (b)
M1 3 complete batches use 3 × 0.75 = 2.25 kg	✓ 1
A1 0.25 kg (or 1/4 kg)	✓ 1

Common error: Dividing by a fraction means multiplying by its reciprocal; change the mixed number to an improper fraction first.

+ Add to test✓ In your test62IGME3-1-14r34-01Integers, fractions, decimals and percentages Paper FEasy4 marks

A library has 8400 books. 35% of the books are fiction.

(a)

Work out the number of fiction books.

(2)

(b)

1428 of the books are for children. Work out 1428 as a percentage of all the books.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.35 × 8400 or 35/100 × 8400	✓ 1
A1 2940	✓ 1
Part (b)
M1 1428 ÷ 8400 (× 100)	✓ 1
A1 17%	✓ 1

Exam tip: To find a percentage of a total, divide the part by the whole and multiply by 100.

+ Add to test✓ In your test63IGME3-1-14r34-02Powers, roots and standard form Paper FEasy3 marks

A wind farm has 15 turbines. Each turbine produces 2.4 × 10⁶ watts.

(a)

Write 0.00045 in standard form.

(1)

(b)

Work out the total power produced by the 15 turbines. Give your answer in standard form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 4.5 × 10⁻⁴	✓ 1
Part (b)
M1 15 × 2.4 × 10⁶ = 36 000 000	✓ 1
A1 3.6 × 10⁷ watts	✓ 1

Common error: 36 × 10⁶ is not in standard form; the number part must be between 1 and 10.

+ Add to test✓ In your test64IGME3-1-14r34-03Set language, notation and Venn diagrams (both tiers) Paper FEasy4 marks

A swimming club has 40 members. 24 members swim front crawl (C) and 18 swim butterfly (B). 5 members swim neither stroke.

(a)

Work out the number of members who swim both front crawl and butterfly.

(3)

(b)

Find n(B ∩ C′).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 − 5 = 35 swim at least one stroke	✓ 1
M1 24 + 18 − 35	✓ 1
A1 7	✓ 1
Part (b)
B1 11	✓ 1

Common error: C′ means “not C”, so B ∩ C′ is butterfly swimmers who do not swim front crawl.

+ Add to test✓ In your test65IGME3-1-14r34-04Powers, roots and standard form Paper FEasy5 marks

Fractions.

(a)

Work out 2 3/4 + 1 2/3. Give your answer as a mixed number.

(3)

(b)

Work out 3/4 × 2 2/3.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 convert to a common denominator, e.g. 33/12 + 20/12	✓ 1
M1 53/12	✓ 1
A1 4 5/12	✓ 1
Part (b)
M1 3/4 × 8/3	✓ 1
A1 2	✓ 1

Exam tip: Convert mixed numbers to improper fractions before multiplying.

+ Add to test✓ In your test66IGME3-1-14r39-01Powers, roots and standard form Paper FEasy5 marks

A volcano crater has a volume of 4.5 × 10⁷ m³.

(a)

Write 7.5 × 10⁴ as an ordinary number.

(1)

(b)

Ash falls at a rate of 6 × 10⁵ tonnes per day. Work out the total mass, in tonnes, that falls in 4 days. Give your answer in standard form.

(2)

(c)

Lava flows into the crater at 3 × 10³ m³ per hour. Work out the number of hours needed to fill the crater. Give your answer in standard form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 75 000	✓ 1
Part (b)
M1 6 × 10⁵ × 4 (or 2 400 000)	✓ 1
A1 2.4 × 10⁶	✓ 1
Part (c)
M1 4.5 × 10⁷ ÷ (3 × 10³)	✓ 1
A1 1.5 × 10⁴ hours	✓ 1

Common error: When multiplying in standard form, check that the first number is still between 1 and 10 before you finish.

+ Add to test✓ In your test67IGME3-1-14r39-02Integers, fractions, decimals and percentages Paper FEasy6 marks

A theatre has 480 seats. 35% of the seats are in the balcony.

(a)

Work out the number of seats in the balcony.

(2)

(b)

On Friday, 396 seats were sold. Work out the percentage of seats sold.

(2)

(c)

A ticket normally costs £28. In a sale the price is reduced by 15%. Work out the sale price.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.35 × 480 (or 10% = 48 used)	✓ 1
A1 168	✓ 1
Part (b)
M1 396 ÷ 480 × 100	✓ 1
A1 82.5%	✓ 1
Part (c)
M1 28 × 0.85 (or 28 − 4.20)	✓ 1
A1 £23.80	✓ 1

Exam tip: A percentage decrease of 15% is a multiplier of 0.85, which gives the new price in one step.

+ Add to test✓ In your test68IGME3-1-14r39-03Ratio, proportion and degree of accuracy Paper FEasy5 marks

A cargo ship carries red, blue and green containers in the ratio 5 : 3 : 2. The ship carries 2400 containers altogether.

(a)

Write down the ratio of red containers to green containers in its simplest form.

(1)

(b)

Work out the number of blue containers.

(2)

(c)

20% of the red containers are refrigerated. Work out the number of refrigerated containers.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 5 : 2	✓ 1
Part (b)
M1 2400 ÷ (5 + 3 + 2) × 3	✓ 1
A1 720	✓ 1
Part (c)
M1 1200 × 0.2 (red containers = 1200)	✓ 1
A1 240	✓ 1

Common error: Divide by the sum of all the parts of the ratio, not by the number of parts you want.

+ Add to test✓ In your test69IGME3-1-14r39-04Set language, notation and Venn diagrams (both tiers) Paper FEasy6 marks

A drama club has 40 members. Let A be the set of members who act and C be the set of members who work on the stage crew. 24 members act, 18 work on the stage crew and 6 members do neither.

(a)

Work out the number of members who both act and work on the stage crew.

(3)

(b)

Work out n(A ∩ C′).

(1)

(c)

One of the members who act is chosen at random. Find the probability that this member also works on the stage crew.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 − 6 = 34 members in at least one set	✓ 1
M1 24 + 18 − 34	✓ 1
A1 8	✓ 1
Part (b)
B1 16 (ECF from (a))	✓ 1
Part (c)
M1 8 ÷ 24 (ECF from (a))	✓ 1
A1 1/3 (or equivalent)	✓ 1

Common error: C′ means 'not in C': A ∩ C′ counts members who act and are not on the stage crew.

+ Add to test✓ In your test70IGME3-1-14r40-01Integers, fractions, decimals and percentages Paper FEasy4 marks

A doctor prescribes a dose of 80 mg of a drug.

(a)

The pharmacist reduces the dose by 15%. Work out the reduced dose.

(2)

(b)

Later the dose is increased from 68 mg to 85 mg. Work out the percentage increase.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.85 × 80 or 80 − 0.15 × 80	✓ 1
A1 68 mg	✓ 1
Part (b)
M1 (85 − 68) ÷ 68 × 100	✓ 1
A1 25%	✓ 1

Common error: A percentage change is always found as a fraction of the original value, here 68, not the new value.

+ Add to test✓ In your test71IGME3-1-14r40-02Ratio, proportion and degree of accuracy Paper FEasy3 marks

In a video game, Mia collects gold coins and gems in the ratio 5 : 3. She collects 240 items in total.

(a)

Write the fraction of her items that are gems. Give your answer in its simplest form.

(1)

(b)

Work out the number of gold coins Mia collects.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3/8	✓ 1
Part (b)
M1 240 ÷ (5 + 3) (= 30)	✓ 1
A1 150	✓ 1

Exam tip: To share in a ratio, add the parts first; each part is then total ÷ sum of parts.

+ Add to test✓ In your test72IGME3-1-14r40-03Powers, roots and standard form Paper FEasy3 marks

A reservoir holds 4.2 × 10⁷ litres of water. A town uses 3 × 10⁵ litres of water each day.

(a)

Write 4.2 × 10⁷ as an ordinary number.

(1)

(b)

Work out how many days the water in the reservoir would last if no more water enters it.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 42 000 000	✓ 1
Part (b)
M1 (4.2 × 10⁷) ÷ (3 × 10⁵)	✓ 1
A1 140 days	✓ 1

Exam tip: To divide numbers in standard form, divide the number parts and use the index laws on the powers of 10 separately.

+ Add to test✓ In your test73IGME3-1-14r40-04Set language, notation and Venn diagrams (both tiers) Paper FEasy4 marks

There are 40 members in a gaming club. 24 members play strategy games (S), 18 members play racing games (R) and 7 members play neither type of game. Some members play both.

(a)

Work out the number of members who play both strategy games and racing games.

(3)

(b)

Work out the number of members who play strategy games but not racing games.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 − 7 = 33 members play at least one type	✓ 1
M1 24 + 18 − 33	✓ 1
A1 9	✓ 1
Part (b)
B1 15 (ECF from (a))	✓ 1

Common error: The numbers 24 and 18 each include the members who play both, so adding them double counts the overlap.

+ Add to test✓ In your test74IGME3-1-14r41-01Powers, roots and standard form Paper FEasy4 marks

A mountain summit is 8.85 × 10³ m above sea level and base camp is 5.3 × 10³ m above sea level.

(a)

Work out the height of the summit above base camp. Give your answer in standard form.

(2)

(b)

A climber's oxygen cylinder holds 3.2 × 10³ litres and she uses 4 × 10¹ litres per hour. For how many hours will the cylinder last?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 8850 − 5300 (or 8.85 − 5.3)	✓ 1
A1 3.55 × 10³	✓ 1
Part (b)
M1 3200 ÷ 40	✓ 1
A1 80	✓ 1

Exam tip: Convert standard-form numbers to ordinary numbers if that helps with the arithmetic, then convert back at the end.

+ Add to test✓ In your test75IGME3-1-14r41-02Ratio, proportion and degree of accuracy Paper FEasy4 marks

A cereal bar of mass 250 g contains 18% sugar by mass.

(a)

Work out the mass of sugar in the bar.

(2)

(b)

A new recipe reduces the mass of sugar by 20%. Work out the mass of sugar in the new recipe.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.18 × 250	✓ 1
A1 45 g	✓ 1
Part (b)
M1 0.8 × 45 (or 45 − 9)	✓ 1
A1 36 g	✓ 1

Common error: Reducing by 20% means multiplying by 0.8, not subtracting 20 from the amount.

+ Add to test✓ In your test76IGME3-1-14r41-03Set language, notation and Venn diagrams (both tiers) Paper FEasy4 marks

In a taste test of 40 people, 22 liked salty snacks (S), 17 liked sweet snacks (W) and 6 liked both. Everyone took part in the test.

(a)

Work out the number of people who liked neither type of snack.

(2)

(b)

One person is chosen at random. Find the probability that they liked sweet snacks only.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 22 + 17 − 6 = 33	✓ 1
A1 40 − 33 = 7	✓ 1
Part (b)
M1 17 − 6 = 11	✓ 1
A1 ¹¹⁄₄₀ (= 0.275)	✓ 1

Common error: The 6 people who liked both are counted inside both 22 and 17, so subtract them once when finding the union.

+ Add to test✓ In your test77IGME3-1-14r41-04Ratio, proportion and degree of accuracy Paper FEasy5 marks

Sam and Priya share a monthly phone bill of €72 in the ratio 5 : 3.

(a)

Work out how much Sam pays.

(2)

(b)

Next month the total bill increases by 15%. They share the new bill in the same ratio. Work out how much Priya pays next month.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 72 ÷ 8 (= 9)	✓ 1
A1 €45	✓ 1
Part (b)
M1 72 × 1.15 (= 82.80)	✓ 1
M1 82.80 × ³⁄₈	✓ 1
A1 €31.05	✓ 1

Exam tip: Add the ratio parts first to find the number of equal shares, then multiply to find each person's amount.

+ Add to test✓ In your test78IGME3-1-14r42-02Integers, fractions, decimals and percentages Paper FEasy5 marks

A cycling shop has a sale.

(a)

A bicycle has a normal price of £560. In the sale the price is reduced by 15%.
Work out the sale price.

(3)

(b)

A helmet costs £36. The price increases by 12.5%.
Work out the new price.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.15 × 560 = 84	✓ 1
M1 560 − 84	✓ 1
A1 £476	✓ 1
Part (b)
M1 36 × 1.125, or 36 + 0.125 × 36	✓ 1
A1 £40.50	✓ 1

Exam tip: A single multiplier does the job in one step: reduce by 15% means × 0.85, increase by 12.5% means × 1.125.

+ Add to test✓ In your test79IGME3-1-14r42-06Ratio, proportion and degree of accuracy Paper FEasy5 marks

A cyclist rides 42 km in 1 hour 45 minutes.

(a)

Work out the cyclist's average speed in km/h.

(2)

(b)

The cyclist rides a further 90 km at the same average speed.
Work out the time this takes. Give your answer in hours and minutes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 42 ÷ 1.75	✓ 1
A1 24 km/h	✓ 1
Part (b)
M1 90 ÷ 24	✓ 1
A1 3.75 hours	✓ 1
A1 3 hours 45 minutes	✓ 1

Common error: 1 hour 45 minutes is 1.75 hours, not 1.45 hours.

+ Add to test✓ In your test80IGME3-1-14r42-09Set language, notation and Venn diagrams (both tiers) Paper FEasy5 marks

The layers of an excavation are numbered 1 to 15.
ξ = {1, 2, 3, …, 15}
P = {prime numbers in ξ}
Q = {odd numbers in ξ}

(a)

List the elements of P ∩ Q′.

(1)

(b)

Find n(P ∪ Q).

(2)

(c)

List the elements of P′ ∩ Q.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 {2}	✓ 1
Part (b)
M1 lists or counts the 8 odd numbers plus the prime 2	✓ 1
A1 9	✓ 1
Part (c)
M1 at least 2 of 1, 9, 15 and no more than 1 extra	✓ 1
A1 {1, 9, 15}	✓ 1

Common error: 1 is not a prime number, so it belongs in P′.

+ Add to test✓ In your test81IGME3-1-14r42-13Integers, fractions, decimals and percentages Paper FEasy5 marks

Two backup programs run in a data centre. Program A runs every 18 minutes. Program B runs every 24 minutes. Both programs run together at 09:00.

(a)

Write 18 and 24 as products of their prime factors.

(2)

(b)

Work out the lowest common multiple of 18 and 24.

(2)

(c)

Work out the next time at which both programs run together.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 18 = 2 × 3²	✓ 1
B1 24 = 2³ × 3	✓ 1
Part (b)
M1 2³ × 3², or lists multiples of both numbers	✓ 1
A1 72	✓ 1
Part (c)
B1 10:12 (ECF from (b))	✓ 1

Common error: For the LCM take the highest power of each prime; for the HCF take the lowest power of the primes they share.

+ Add to test✓ In your test82IGME3-1-14r43-02Integers, fractions, decimals and percentages Paper FEasy5 marks

In a video game, a player starts a level with 240 coins. The player spends ¼ of the coins on armour. The player then spends ⅖ of the coins that remain on potions.

(a)

Work out how many coins the player spends on armour.

(1)

(b)

Work out how many coins the player spends on potions.

(2)

(c)

Work out the fraction of the original 240 coins that the player has left. Give your answer in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 60	✓ 1
Part (b)
M1 (240 − 60) × ⅖ (ECF from (a))	✓ 1
A1 72	✓ 1
Part (c)
M1 240 − 60 − 72 = 108 (ECF)	✓ 1
A1 \(\frac{9}{20}\)	✓ 1

Exam tip: 'Of the coins that remain' means use the amount left after the first purchase, not the original 240.

+ Add to test✓ In your test83IGME3-1-14r43-03Powers, roots and standard form Paper FEasy4 marks

A reservoir holds 4.2 × 10⁷ litres of water. A small town uses 3 × 10⁵ litres of water each day.

(a)

Work out how many days the water in the reservoir would last if the town were the only user.

(2)

(b)

A second reservoir holds 1.5 × 10⁸ litres. Work out the total volume of water in the two reservoirs. Give your answer in standard form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4.2 × 10⁷ ÷ 3 × 10⁵ (as a division of the two numbers)	✓ 1
A1 140 days	✓ 1
Part (b)
M1 4.2 × 10⁷ + 1.5 × 10⁸, or both written with the same power of 10	✓ 1
A1 1.92 × 10⁸ litres	✓ 1

Common error: Adding the powers of ten when adding two numbers in standard form. Write both with the same power of ten first.

+ Add to test✓ In your test84IGME3-1-14r43-14Integers, fractions, decimals and percentages Paper FEasy4 marks

A household's water bill has a standing charge of £24.60 plus £1.85 for every cubic metre of water used. In one quarter the household uses 62 m³.

(a)

Work out the bill before any discount.

(2)

(b)

The household receives a 4% discount for paying by direct debit. Work out the amount to pay. Give your answer to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 24.60 + 1.85 × 62	✓ 1
A1 £139.30	✓ 1
Part (b)
M1 139.30 × 0.96 (ECF from (a))	✓ 1
A1 £133.73	✓ 1

Exam tip: A 4% discount can be found in one step by multiplying by 0.96.

+ Add to test✓ In your test85IGME3-1-12r47-01Set language, notation and Venn diagrams (both tiers) Paper FEasy4 marks

A marine biology club has 40 students. F is the set of students who study fish and C is the set of students who study corals. n(F) = 25, n(C) = 18 and n(F ∩ C) = 7.

(a)

Work out the number of students who study only corals.

(1)

(b)

Work out the number of students who study neither fish nor corals.

(2)

(c)

Write down n(F′).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 18 − 7 = 11	✓ 1
Part (b)
M1 25 + 11 = 36 (students in at least one set)	✓ 1
A1 4	✓ 1
Part (c)
B1 15	✓ 1

Common error: The 7 students in the overlap are counted in both totals, so subtract them once when finding the union.

+ Add to test✓ In your test86IGME3-1-12r47-02Integers, fractions, decimals and percentages Paper FEasy7 marks

An architecture firm has a budget of €48000 for a small building. 35% of the budget is spent on materials and 3/8 of the budget is spent on labour.

(a)

Work out how much is spent on materials.

(2)

(b)

Work out how much is spent on labour.

(2)

(c)

Write the amount of the budget that remains as a percentage of the whole budget.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.35 × 48000	✓ 1
A1 16800	✓ 1
Part (b)
M1 3/8 × 48000	✓ 1
A1 18000	✓ 1
Part (c)
M1 48000 − 16800 − 18000 = 13200	✓ 1
M1 13200 ÷ 48000 × 100	✓ 1
A1 27.5	✓ 1

Exam tip: Find the remaining amount in euros first, then convert it to a percentage of the original budget.

+ Add to test✓ In your test87IGME3-1-12r47-03Powers, roots and standard form Paper FEasy4 marks

Diatoms are tiny marine algae. One type of diatom has length 3.2 × 10⁻⁵ m.

(a)

Write 0.000072 in standard form.

(1)

(b)

A chain is made of 2.5 × 10⁴ of these diatoms placed end to end. Work out the length of the chain in metres. Give your answer in standard form.

(2)

(c)

Write the length of the chain in centimetres, as a decimal number.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 7.2 × 10⁻⁵	✓ 1
Part (b)
M1 2.5 × 10⁴ × 3.2 × 10⁻⁵	✓ 1
A1 8 × 10⁻¹	✓ 1
Part (c)
B1 80	✓ 1

Common error: When multiplying numbers in standard form, multiply the front numbers and add the powers of 10; do not add the front numbers.

+ Add to test✓ In your test88IGME3-1-12r47-04Integers, fractions, decimals and percentages Paper FEasy5 marks

A distribution centre has 840 kg of cargo to share between three warehouses in the ratio 3 : 5 : 6.

(a)

Work out the mass of cargo given to the warehouse with the smallest share.

(2)

(b)

A lorry uses 14 litres of fuel for every 100 km. Fuel costs €1.62 per litre. Work out the cost of fuel for a journey of 350 km.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 840 ÷ (3 + 5 + 6) = 60	✓ 1
A1 180	✓ 1
Part (b)
M1 14 × 3.5 = 49 litres	✓ 1
M1 49 × 1.62	✓ 1
A1 79.38	✓ 1

Exam tip: Add the parts of a ratio first to find the value of one part.

+ Add to test✓ In your test89IGME3-1-13r48-01Ratio, proportion and degree of accuracy Paper FEasy5 marks

A sports club runs a training session lasting 1 hour 45 minutes. One third of the session is a warm-up. Three fifths of the time that remains after the warm-up is spent on skills drills. The rest of the session is a practice match.

(a)

Work out the length of the warm-up, in minutes.

(1)

(b)

Work out the number of minutes spent on skills drills.

(2)

(c)

Work out the number of minutes spent on the practice match.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 35	✓ 1
Part (b)
M1 (105 − 35) × 3/5 or 70 × 3/5	✓ 1
A1 42	✓ 1
Part (c)
M1 105 − 35 − 42 or 70 − 42	✓ 1
A1 28 (ECF from earlier parts)	✓ 1

Exam tip: Convert 1 hour 45 minutes to minutes before taking fractions of it, and remember the second fraction applies only to what remains.

+ Add to test✓ In your test90IGME3-1-13r48-02Powers, roots and standard form Paper FEasy5 marks

At its closest approach, Mars is 54 600 000 km from Earth. Radio signals travel at 3 × 10⁵ km/s.

(a)

Write 54 600 000 in standard form.

(1)

(b)

Work out the time, in seconds, for a radio signal to travel from Earth to Mars at this closest approach.

(2)

(c)

Write your answer to part (b) in minutes. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 5.46 × 10⁷	✓ 1
Part (b)
M1 5.46 × 10⁷ ÷ (3 × 10⁵)	✓ 1
A1 182	✓ 1
Part (c)
M1 182 ÷ 60	✓ 1
A1 3.03	✓ 1

Common error: Dividing the powers of ten wrongly: 10⁷ ÷ 10⁵ = 10², not 10¹² or 10³⁵.

+ Add to test✓ In your test91IGME3-1-13r48-03Ratio, proportion and degree of accuracy Paper FEasy6 marks

A train has 360 seats. The ratio of first-class seats to standard-class seats is 1 : 8.

(a)

Work out the number of first-class seats.

(2)

(b)

Three eighths of the standard-class seats are reserved. Work out the number of reserved standard-class seats.

(2)

(c)

Each reserved seat costs £12.50 to reserve. Work out the total paid for all the reserved standard-class seats.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ (1 + 8) or 360 ÷ 9	✓ 1
A1 40	✓ 1
Part (b)
M1 (360 − 40) × 3/8 or 320 × 3/8	✓ 1
A1 120	✓ 1
Part (c)
M1 120 × 12.5 (ECF from (b))	✓ 1
A1 1500 or £1500.00	✓ 1

Exam tip: Add the parts of a ratio first to find the value of one part, then multiply.

+ Add to test✓ In your test92IGME3-1-13r48-08Integers, fractions, decimals and percentages Paper FEasy4 marks

Train A leaves a station every 12 minutes. Train B leaves the same station every 18 minutes. Both trains leave together at 07:00.

(a)

Write 18 as a product of its prime factors.

(2)

(b)

Work out the next time that both trains leave together.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 divide by primes, factor tree or 2 × 9 seen	✓ 1
A1 2 × 3²	✓ 1
Part (b)
M1 attempt at LCM of 12 and 18 or multiples of both listed	✓ 1
A1 07:36	✓ 1

Exam tip: Write each number as a product of primes; the lowest common multiple uses the highest power of each prime that appears.

+ Add to test✓ In your test93IGME3-1-14r48-01Integers, fractions, decimals and percentages Paper FEasy4 marks

Two robots start together at the start line of a circular test track. Robot A completes one lap every 24 seconds. Robot B completes one lap every 36 seconds.

(a)

Work out the number of seconds until the robots are next at the start line together.

(2)

(b)

The robots run for 12 minutes. Work out the number of times, after the start, that they are together at the start line.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at LCM of 24 and 36 or multiples of both listed	✓ 1
A1 72	✓ 1
Part (b)
M1 12 × 60 = 720 seconds and 720 ÷ 72	✓ 1
A1 10 (ECF from (a))	✓ 1

Common error: Convert 12 minutes to seconds before dividing by a time given in seconds.

+ Add to test✓ In your test94IGME3-1-14r48-02Integers, fractions, decimals and percentages Paper FEasy6 marks

A jacket costs €85 in a fashion shop. In a sale, the price is reduced by 30%.

(a)

Work out the sale price of the jacket.

(2)

(b)

In the last week of the sale, the sale price is reduced by a further 10%. Work out the new price of the jacket.

(2)

(c)

Maya says: “The two reductions of 30% and 10% mean the jacket is 40% cheaper than the original price.” Is Maya correct? You must show your working.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 85 × 0.7 or 85 − 0.3 × 85	✓ 1
A1 59.50	✓ 1
Part (b)
M1 59.50 × 0.9	✓ 1
A1 53.55	✓ 1
Part (c)
M1 53.55 ÷ 85 = 0.63 or 85 − 53.55 = 31.45	✓ 1
A1 No, the total reduction is 37%	✓ 1

Common error: Successive percentage changes cannot be added, because the second change applies to a different amount.

+ Add to test✓ In your test95IGME3-1-14r48-03Ratio, proportion and degree of accuracy Paper FEasy5 marks

On one trip, a fishing boat catches cod, haddock and mackerel in the ratio 5 : 3 : 2. The total mass of the catch is 1800 kg.

(a)

Work out the mass of haddock caught.

(2)

(b)

The mackerel is sold for €2.40 per kilogram. Work out the amount of money received for the mackerel.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1800 ÷ (5 + 3 + 2) × 3	✓ 1
A1 540	✓ 1
Part (b)
M1 1800 ÷ 10 × 2 or 360 kg of mackerel	✓ 1
M1 360 × 2.40	✓ 1
A1 864 or €864.00	✓ 1

Exam tip: Add the ratio parts to find the total number of parts, divide the total mass by this, then multiply by the share you need.

+ Add to test✓ In your test96IGME3-1-14r48-05Set language, notation and Venn diagrams (both tiers) Paper FEasy6 marks

The robots in a warehouse are numbered 1 to 20, so ξ = {whole numbers from 1 to 20}. A is the set of robots whose numbers are multiples of 3. B is the set of robots whose numbers are even.

(a)

List the members of A ∩ B.

(2)

(b)

Work out n(A ∪ B).

(2)

(c)

A robot is chosen at random. Find the probability that its number is in neither A nor B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 at least two of 6, 12, 18 listed	✓ 1
B1 {6, 12, 18} and no others	✓ 1
Part (b)
M1 6 + 10 − 3 or listing all members	✓ 1
A1 13	✓ 1
Part (c)
M1 (20 − 13)/20 or 7/20 (ECF from (b))	✓ 1
A1 7/20 or 0.35	✓ 1

Exam tip: The union A ∪ B contains numbers in either set; check which numbers appear in both so that they are not counted twice.

+ Add to test✓ In your test97IGME-058-01-08Powers, roots and standard form Paper FMedium4 marks

In one year the population of the world was 8.1×10⁹.
The land area of the world is 1.49×10⁸ km².

(a)

Work out the mean number of people per km² of land.
Give your answer to 3 significant figures.

(2)

(b)

In the same year the population of China was 1.41×10⁹.
Work out the population of China as a percentage of the world population.
Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for (8.1×10⁹) ÷ (1.49×10⁸)	✓ 1
A1 for 54.4	✓ 1
Part (b)
M1 for (1.41×10⁹) ÷ (8.1×10⁹) × 100	✓ 1
A1 for 17.4%	✓ 1

Common error: Dividing the world population by China's population gives the reciprocal ratio. The part goes on top: China ÷ world.

+ Add to test✓ In your test98IGME-058-00-06Integers, fractions, decimals and percentages Paper FMedium5 marks

A shop increases the price of a laptop by 20%. The new price is £546.

(a)

Work out the price of the laptop before the increase.

(3)

(b)

Later, the shop reduces the £546 price by 20%.
Leon says the laptop is now back to its original price.
Show that Leon is wrong.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 recognises £546 represents 120% or uses multiplier 1.2	✓ 1
M1 546 ÷ 1.2	✓ 1
A1 £455	✓ 1
Part (b)
M1 546 × 0.8 or 546 − 0.2 × 546	✓ 1
A1 £436.80, which is not equal to £455, so Leon is wrong	✓ 1

Common error: Finding 20% of £546 and subtracting it to get the original price is wrong: the 20% was calculated on the original price, not on £546.

+ Add to test✓ In your test99IGME-058-00-10Integers, fractions, decimals and percentages Paper FMedium5 marks

This question is about the number 180.

(a)

Write 180 as a product of its prime factors. Give your answer in index form.

(2)

(b)

k is the smallest positive integer such that 180k is a square number.
Find k.

(1)

(c)

n is the smallest positive integer such that 180n is a cube number.
Find n.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a complete method, e.g. factor tree with at least two correct steps	✓ 1
A1 2² × 3² × 5	✓ 1
Part (b)
B1 5	✓ 1
Part (c)
M1 recognises each prime needs power 3, e.g. 2³ × 3³ × 5³ or n = 2 × 3 × 5²	✓ 1
A1 150	✓ 1

Exam tip: A number is a perfect square when every prime power in its factorisation is even, and a perfect cube when every power is a multiple of 3.

+ Add to test✓ In your test100IGME-058-00-11Integers, fractions, decimals and percentages Paper FMedium5 marks

A supermarket sells rice in two pack sizes.
Pack A: 500 g for £3.60
Pack B: 750 g, normally £5.80, now with 15% off

(a)

Work out the sale price of Pack B.

(2)

(b)

Which pack is better value? You must show your working.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5.80 × 0.85 or 5.80 − 0.15 × 5.80	✓ 1
A1 £4.93	✓ 1
Part (b)
M1 price per unit for one pack, e.g. Pack A 3.60 ÷ 5 = £0.72 per 100 g	✓ 1
M1 price per same unit for the other pack, e.g. Pack B 4.93 ÷ 7.5 = £0.657 per 100 g	✓ 1
A1 Pack B with correct comparable figures	✓ 1

Exam tip: Compare like with like: work out the cost of the same mass (or the mass per £1) for both packs, and state the conclusion clearly.

+ Add to test✓ In your test101IGME-058-02-07Ratio, proportion and degree of accuracy Paper FMedium4 marks

\(a : b = 3 : 4\) and \(b : c = 6 : 5\)

(a)

Find \(a : b : c\)
Give your answer in its simplest form.

(2)

(b)

\(a + b + c = 496\)
Find the value of \(c\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for making b the same in both, e.g. 9 : 12 and 12 : 10	✓ 1
A1 for 9 : 12 : 10	✓ 1
Part (b)
M1 for 496 ÷ 31 (= 16)	✓ 1
A1 for 160	✓ 1

Common error: Writing 3 : 4 : 5 or 3 : 6 : 5 is wrong – the two values of b must be made equal (use the LCM 12) before combining.

+ Add to test✓ In your test102IGME-058-02-12Ratio, proportion and degree of accuracy Paper FMedium4 marks

A walking map has a scale of 1 : 25 000

(a)

Two villages are 6.4 cm apart on the map.
Work out the real distance between the villages in kilometres.

(2)

(b)

A footpath is 3.15 km long.
Work out the length of the footpath on the map in centimetres.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 6.4 × 25 000 (= 160 000 cm)	✓ 1
A1 for 1.6 km	✓ 1
Part (b)
M1 for 315 000 ÷ 25 000 or 3.15 × 100 000 ÷ 25 000	✓ 1
A1 for 12.6 cm	✓ 1

Common error: There are 100 000 cm in 1 km; converting with 1000 or 100 alone gives an answer out by a power of ten.

+ Add to test✓ In your test103IGME-058-03-06Set language, notation and Venn diagrams (both tiers) Paper FMedium6 marks

In a survey of 50 people, A is the set of people who own a cat and B is the set of people who own a dog.
n(A) = 28, n(B) = 31 and n((A ∪ B)′) = 5.

(a)

Find n(A ∩ B).

(3)

(b)

Find n(A′ ∩ B).

(1)

(c)

One of the 50 people is chosen at random. Find the probability that this person owns exactly one of these two types of pet.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for n(A ∪ B) = 50 − 5 = 45 . M1 for 28 + 31 − n(A ∩ B) = 45, or equivalent . A1 for 14 .	✓ 3
Part (b)
B1 for 31 − 14 = 17 .	✓ 1
Part (c)
M1 for (28 − 14) + 17 = 31 . A1 for \(\frac{31}{50}\) or 0.62 .	✓ 2

Exam tip: When the totals of A and B add to more than the number inside the circles, the excess is the overlap.

+ Add to test✓ In your test104IGME-058-03-09Set language, notation and Venn diagrams (both tiers) Paper FMedium5 marks

ξ = {integers}
A = {x : −3 < x ≤ 4}
B = {x : x² < 10}

(a)

List the members of A.

(1)

(b)

List the members of A ∩ B.

(2)

(c)

Find n(A ∪ B).

(1)

(d)

Is B ⊂ A? Give a reason for your answer.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for {−2, −1, 0, 1, 2, 3, 4} .	✓ 1
Part (b)
M1 for B = {−3, −2, −1, 0, 1, 2, 3} . A1 for {−2, −1, 0, 1, 2, 3} .	✓ 2
Part (c)
B1 for 8, from the members −3 to 4 inclusive .	✓ 1
Part (d)
B1 for 'no', because −3 ∈ B but −3 ∉ A .	✓ 1

Common error: x² < 10 has negative solutions too. (−3)² = 9, so −3 is in B.

+ Add to test✓ In your test105IGME-058-03-14Set language, notation and Venn diagrams (both tiers) Paper FMedium5 marks

A Venn diagram shows two overlapping sets, A and B, inside the universal set ξ, where n(ξ) = 40.
The number of members in each region is:
A only: 2x + 1
A ∩ B: x
B only: 3x − 2
outside both circles: 5

Diagram not to scale

(a)

Find the value of x.

(3)

(b)

Find n(A′).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for (2x + 1) + x + (3x − 2) + 5 = 40 . M1 for 6x + 4 = 40 . A1 for x = 6 .	✓ 3
Part (b)
M1 for (3x − 2) + 5, or for 40 − (2x + 1) − x . A1 for 21 .	✓ 2

Exam tip: Check your value of x: the regions are 13, 6, 16 and 5, which add up to 40.

+ Add to test✓ In your test106IGME-058-03-05Set language, notation and Venn diagrams (both tiers) Paper FMedium6 marks

ξ = {integers from 1 to 20 inclusive}
E = {even numbers}
S = {square numbers}
P = {prime numbers}

(a)

List the members of S ∩ E.

(1)

(b)

Explain why P ∩ S = ∅.

(1)

(c)

Find n(E ∪ P).

(2)

(d)

List the members of (E ∪ S)′ ∩ P′.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for {4, 16} .	✓ 1
Part (b)
B1 for explaining that no square number in ξ is prime. The squares are 1, 4, 9 and 16. 1 is not prime, and each of the others has more than two factors .	✓ 1
Part (c)
M1 for n(E) = 10 and n(P) = 8, with 2 identified as being in both sets, or for a correct list of the 17 members . A1 for 10 + 8 − 1 = 17 .	✓ 2
Part (d)
M1 for finding (E ∪ S)′ = {3, 5, 7, 11, 13, 15, 17, 19}, the odd numbers that are not square, or for a correct method that removes the primes from this set . A1 for {15} only .	✓ 2

Common error: 1 is not a prime number, and 2 is both even and prime. Count 2 only once when you find the number of members in a union.

+ Add to test✓ In your test107IGME-058-02-08Ratio, proportion and degree of accuracy Paper FMedium5 marks

6 identical machines can complete an order in 15 hours.
All machines work at the same rate.

(a)

How long would 9 of these machines take to complete the same order?

(2)

(b)

On another day, the 6 machines start the order. After 3 hours, 3 more machines join them.
Work out the total time taken to complete the order from the start.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 6 × 15 = 90 machine-hours, or 90 ÷ 9	✓ 1
A1 for 10 hours	✓ 1
Part (b)
M1 for 90 − 6 × 3 = 72 machine-hours remaining	✓ 1
M1 for 72 ÷ 9 = 8 hours	✓ 1
A1 for 3 + 8 = 11 hours	✓ 1

Exam tip: For inverse proportion, work out the total amount of work (machines × hours), which stays constant. If the question asks for the time from the start, add on the hours already worked.

+ Add to test✓ In your test108IGME-058-01-10Powers, roots and standard form Paper FMedium6 marks

Calculating with standard form.

(a)

Work out (4.6×10⁵) + (7.3×10⁴).
Give your answer in standard form.

(2)

(b)

Write these numbers in order of size. Start with the smallest.
3.2×10⁻², 0.0034, 31×10⁻³, 3.3×10⁻³

(2)

(c)

Work out (8.4×10⁻³) ÷ (2.1×10⁵).
Give your answer in standard form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 460 000 + 73 000 or 4.6×10⁵ + 0.73×10⁵ or 533 000	✓ 1
A1 for 5.33×10⁵	✓ 1
Part (b)
B1 for writing all four numbers in a common form, e.g. 0.032, 0.0034, 0.031, 0.0033, or 3.2×10⁻², 3.4×10⁻³, 3.1×10⁻², 3.3×10⁻³	✓ 1
B1 for 3.3×10⁻³, 0.0034, 31×10⁻³, 3.2×10⁻²	✓ 1
Part (c)
M1 for 8.4 ÷ 2.1 = 4 or power of ten −3 − 5 = −8	✓ 1
A1 for 4×10⁻⁸	✓ 1

Common error: When adding in standard form you cannot simply add the front numbers and keep one power. Convert both numbers to the same power of ten, or to ordinary numbers, first.

+ Add to test✓ In your test109IGME-058-03-07Set language, notation and Venn diagrams (both tiers) Paper FMedium6 marks

60 people were asked which of tea (T), coffee (C) and juice (J) they drank yesterday. The results are shown in a Venn diagram with three overlapping circles. The regions of the diagram contain these numbers:
T only: x
C only: 2x
J only: 3
T and C but not J: 7
T and J but not C: 5
C and J but not T: 6
all three: 4
outside all circles: 8

Diagram not to scale

(a)

Find the value of x.

(3)

(b)

Find n(T).

(1)

(c)

Find n(T′ ∩ C′).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for forming an equation for the total, e.g. x + 2x + 3 + 7 + 5 + 6 + 4 + 8 = 60 or 3x + 33 = 60 . M1 for 3x = 27 or equivalent rearrangement . A1 for x = 9 .	✓ 3
Part (b)
B1 for 9 + 7 + 5 + 4 = 25 (follow through their x + 16) .	✓ 1
Part (c)
M1 for identifying the regions as J only and outside all circles . A1 for 3 + 8 = 11 .	✓ 2

Common error: T′ ∩ C′ includes the people outside all three circles. Remember to add the 8, not just the 3 in J only.

+ Add to test✓ In your test110IGME-058-02-06Ratio, proportion and degree of accuracy Paper FMedium5 marks

Ali and Ben share some money in the ratio 5 : 3
Ali gets £84 more than Ben.

(a)

Work out the total amount of money they share.

(3)

(b)

Ben gives \(\frac{1}{4}\) of his share to Chloe.
Find the ratio of the amounts Ali : Ben : Chloe now have.
Give your ratio in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 5 − 3 = 2 parts and 84 ÷ 2 = 42 (1 part = £42)	✓ 1
M1 for 42 × 8, or for Ali = 210 and Ben = 126	✓ 1
A1 for £336	✓ 1
Part (b)
M1 for a correct unsimplified ratio, e.g. 210 : 94.5 : 31.5 or 5 : 2.25 : 0.75 or 420 : 189 : 63	✓ 1
A1 for 20 : 9 : 3	✓ 1

Exam tip: The £84 difference matches the difference in the number of parts (5 − 3 = 2), not either person's share.

+ Add to test✓ In your test111IGME-058-00-08Integers, fractions, decimals and percentages Paper FMedium5 marks

A roll of ribbon is \(12\frac{3}{4}\) m long.
Maya cuts as many pieces of length \(\frac{5}{8}\) m as she can from the roll.

(a)

How many pieces does Maya cut? Show your working clearly.

(3)

(b)

Work out the length of ribbon left over. Give your answer as a fraction of a metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{51}{4} \div \frac{5}{8}\)	✓ 1
M1 \(\frac{51}{4} \times \frac{8}{5} = \frac{102}{5}\) or 20.4	✓ 1
A1 20	✓ 1
Part (b)
M1 \(\frac{51}{4} - 20 \times \frac{5}{8}\) or \(\frac{2}{5} \times \frac{5}{8}\)	✓ 1
A1 \(\frac{1}{4}\) m	✓ 1

Exam tip: The number of whole pieces must be rounded down, even if the decimal part is greater than 0.5.

+ Add to test✓ In your test112IGME-058-00-07Integers, fractions, decimals and percentages Paper FMedium5 marks

Kofi invests £3500 in a savings account. The account pays 3.2% compound interest per year.

(a)

Work out the value of Kofi's investment at the end of 4 years.
Give your answer to the nearest penny.

(3)

(b)

Work out the overall percentage increase in the value of the investment over the 4 years.
Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for a correct first step: multiplier 1.032 seen, or 3500 × 0.032 (= 112), or 3500 × 1.032 (= 3612)	✓ 1
M1 for a complete method: 3500 × 1.032⁴, or a correct year-by-year method repeated for all 4 years	✓ 1
A1 £3969.97	✓ 1
Part (b)
M1 (1.032⁴ − 1) × 100, or (3969.97 − 3500) ÷ 3500 × 100, or 3969.97 ÷ 3500 × 100 − 100	✓ 1
A1 13.4%	✓ 1

Common error: Multiplying 3.2% by 4 to get 12.8% treats the interest as simple interest. With compound interest, each year's interest is also earned on the interest already added.

+ Add to test✓ In your test113IGME-058-01-06Powers, roots and standard form Paper FMedium5 marks

The distance from the Earth to the Sun is 1.5×10⁸ km.
Light travels at 3.0×10⁵ km per second.

(a)

Work out the time, in minutes, that light takes to travel from the Sun to the Earth.
Give your answer to 3 significant figures.

(3)

(b)

The distance from Neptune to the Sun is 4.5×10⁹ km.
How many times further from the Sun is Neptune than the Earth?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for (1.5×10⁸) ÷ (3.0×10⁵), or 500 seen, or an equivalent method such as (1.5×10⁸) ÷ (3.0×10⁵ × 60)	✓ 1
M1 for converting seconds to minutes, e.g. 500 ÷ 60 (implied by a correct answer)	✓ 1
A1 for 8.33 minutes (accept 8.333…)	✓ 1
Part (b)
M1 for (4.5×10⁹) ÷ (1.5×10⁸)	✓ 1
A1 for 30	✓ 1

Common error: Dividing gives 500 seconds. Forgetting to convert to minutes loses the final mark, so always check the unit asked for.

+ Add to test✓ In your test114IGME3-1-07d-07Powers, roots and standard form Paper FMedium5 marks

A water tank is \(\frac{3}{8}\) full.

(a)

Work out \(2\frac{1}{4} \div \frac{3}{8}\). (2)

(2)

(b)

A pump adds water at a rate of \(\frac{1}{12}\) of the tank's capacity each hour. Work out how long it takes to fill the rest of the tank. (3)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{9}{4} \times \frac{8}{3}\)	✓ 1
A1 6	✓ 1
Part (b)
B1 \(1 - \frac{3}{8} = \frac{5}{8}\) remaining	✓ 1
M1 \(\frac{5}{8} \div \frac{1}{12}\)	✓ 1
A1 7.5 hours (7 hours 30 minutes)	✓ 1

Exam tip: Convert mixed numbers to improper fractions before dividing, then multiply by the reciprocal of the second fraction.

+ Add to test✓ In your test115IGME3-1-07d-08Ratio, proportion and degree of accuracy Paper FMedium7 marks

A pharmacy sells a liquid medicine in two sizes: a 150 ml bottle for €4.20 and a 400 ml bottle for €10.80.

(a)

Which bottle is better value for money? You must show your working. (3)

(3)

(b)

A patient takes 7.5 ml three times a day for 8 days. Work out the total volume taken. (2)

(2)

(c)

The patient buys two 150 ml bottles. Work out the volume of medicine left over after the 8 days. (2)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4.20 ÷ 150 or 10.80 ÷ 400 or cost per 100 ml found	✓ 1
A1 0.028 and 0.027 per ml (or €2.80 and €2.70 per 100 ml)	✓ 1
B1 the 400 ml bottle, with correct comparison	✓ 1
Part (b)
M1 7.5 × 3 × 8	✓ 1
A1 180 ml	✓ 1
Part (c)
M1 2 × 150 − 180	✓ 1
A1 120 ml	✓ 1

Exam tip: To compare prices, convert both to the same unit rate, such as cost per ml.

+ Add to test✓ In your test116IGME3-1-07d-09Set language, notation and Venn diagrams (both tiers) Paper FMedium5 marks

A survey of 50 households found that 28 have solar panels (S), 21 have a heat pump (H) and 9 have neither.

(a)

Work out how many households have both solar panels and a heat pump. (3)

(3)

(b)

A household with solar panels is chosen at random. Find the probability that it also has a heat pump. (2)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 50 − 9 = 41 have at least one	✓ 1
M1 28 + 21 − 41	✓ 1
A1 8	✓ 1
Part (b)
M1 8 out of 28	✓ 1
A1 \(\frac{2}{7}\) (= 0.285714...)	✓ 1

Common error: The total of 50 includes the 9 households in neither set, so remove them before using n(S ∪ H) = n(S) + n(H) − n(S ∩ H).

+ Add to test✓ In your test117IGME3-1-11r7-04Powers, roots and standard form Paper FMedium5 marks

A freight train carries coal, grain and timber in the ratio 5 : 3 : 2. The total mass of the load is 4500 tonnes.

(a)

Work out the mass of grain.

(2)

(b)

At a depot, 18% of the coal is unloaded. Work out the mass of coal still on the train.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4500 ÷ (5 + 3 + 2) × 3 or 4500 ÷ 10 = 450	✓ 1
A1 1350 tonnes	✓ 1
Part (b)
B1 mass of coal = 2250 tonnes	✓ 1
M1 2250 × 0.82 or 2250 − 0.18 × 2250 (= 405)	✓ 1
A1 1845 tonnes	✓ 1

Exam tip: In a ratio share, first find the value of one part by dividing by the total number of parts.

+ Add to test✓ In your test118IGME3-1-11r7-05Set language, notation and Venn diagrams (both tiers) Paper FMedium6 marks

60 athletes take part in a study. Let ξ = {all 60 athletes}, S = {athletes who do strength training} and E = {athletes who do endurance training}.
n(S) = 38, n(E) = 29 and 12 athletes do neither type of training.

(a)

Work out the number of athletes who do both types of training.

(3)

(b)

Write down the value of n((S ∪ E)′).

(1)

(c)

One athlete is chosen at random. Find the probability that this athlete does endurance training only.
Give your answer as a fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 38 + 29 − x + 12 = 60 or 38 − x + x + 29 − x + 12 = 60	✓ 1
M1 79 − x = 60 solved for x	✓ 1
A1 19	✓ 1
Part (b)
B1 12	✓ 1
Part (c)
M1 (29 − 19) ÷ 60 or \(\frac{10}{60}\)	✓ 1
A1 \(\frac{1}{6}\) (= 0.1667...)	✓ 1

Common error: A common error is to place 38 and 29 in the two circles of a Venn diagram without removing the overlap, which counts the athletes who do both twice.

+ Add to test✓ In your test119IGME3-1-11r7-09Powers, roots and standard form Paper FMedium4 marks

The mass of the Earth is 5.97 × 10²⁴ kg. The mass of the Moon is 7.35 × 10²² kg.

(a)

Work out the total mass of the Earth and the Moon.
Give your answer in standard form correct to 3 significant figures.

(2)

(b)

How many times greater is the mass of the Earth than the mass of the Moon?
Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 writing both masses with the same power of 10, for example 597 × 10²² + 7.35 × 10²²	✓ 1
A1 6.04 × 10²⁴ kg (6.0435 × 10²⁴)	✓ 1
Part (b)
M1 5.97 × 10²⁴ ÷ (7.35 × 10²²)	✓ 1
A1 81.2	✓ 1

Common error: Powers of ten must match before adding or subtracting; adding 5.97 and 7.35 directly gives a value that is wrong by a factor of 100 for the Moon.

+ Add to test✓ In your test120IGME3-1-11r7-10Ratio, proportion and degree of accuracy Paper FMedium5 marks

A shop sells two sizes of protein powder. Tub A holds 900 g and costs £27.90. Tub B holds 1.5 kg and costs £45.00.

(a)

Which tub is better value for money? You must show your working.

(3)

(b)

One serving is 30 g. Work out the cost, in pence, of one serving from Tub B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 27.90 ÷ 900 or 2790 ÷ 900 (cost per gram of A)	✓ 1
M1 45.00 ÷ 1500 or 4500 ÷ 1500 (cost per gram of B)	✓ 1
A1 3.1p per g for A and 3p per g for B, so Tub B is better value	✓ 1
Part (b)
M1 3 × 30 or 45 ÷ 1500 × 30 × 100	✓ 1
A1 90	✓ 1

Exam tip: Convert 1.5 kg to 1500 g before comparing; the two tubs must be compared in the same unit.

+ Add to test✓ In your test121IGME3-1-12r13-05Integers, fractions, decimals and percentages Paper FMedium6 marks

In one month a village used 4800 kWh of electricity. Solar supplied 3/8 of this, wind supplied 1/4 and hydro supplied the rest.

(a)

What fraction of the electricity came from hydro?

(2)

(b)

How much electricity, in kWh, came from wind?

(2)

(c)

The wind electricity was 2 1/2 times the wind electricity in the same month last year. How much did wind supply last year?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1 − 3/8 − 1/4 or 3/8 + 2/8 = 5/8	✓ 1
A1 3/8	✓ 1
Part (b)
M1 1/4 × 4800	✓ 1
A1 1200 (kWh)	✓ 1
Part (c)
M1 1200 ÷ 2.5 (ECF from (b))	✓ 1
A1 480 (kWh)	✓ 1

Common error: Use a common denominator when adding or subtracting fractions: 3/8 + 1/4 is 3/8 + 2/8, not 4/12.

+ Add to test✓ In your test122IGME3-1-12r13-07Powers, roots and standard form Paper FMedium5 marks

A laboratory culture starts with 500 bacteria. The number of bacteria doubles every 20 minutes.

(a)

Work out the number of bacteria after 2 hours.

(2)

(b)

Write your answer to (a) in standard form.

(1)

(c)

After n doublings the number of bacteria is 500 × 2ⁿ. Find n when the number of bacteria is 16 000.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 six doublings: 500 × 2⁶	✓ 1
A1 32 000	✓ 1
Part (b)
B1 3.2 × 10⁴ (ECF from (a))	✓ 1
Part (c)
M1 2ⁿ = 32	✓ 1
A1 n = 5	✓ 1

Common error: Two hours is 120 minutes, which is 6 doublings of 20 minutes. Doubling six times means multiplying by 2⁶, not by 2 × 6.

+ Add to test✓ In your test123IGME3-1-12r13-08Integers, fractions, decimals and percentages Paper FMedium6 marks

A shop reduces all its prices by 20% in a sale.

(a)

The sale price of a jacket is €84. Work out its original price.

(2)

(b)

A coat has an original price of €150. On the last day of the sale, the sale price is reduced by a further 10%. Work out the price of the coat on the last day.

(2)

(c)

Work out the single percentage reduction on the original price that is equivalent to the two reductions of 20% then 10%.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 84 ÷ 0.8	✓ 1
A1 105 (€)	✓ 1
Part (b)
M1 150 × 0.8 (= 120)	✓ 1
A1 108 (€)	✓ 1
Part (c)
M1 0.8 × 0.9 = 0.72 or 108/150	✓ 1
A1 28 (%)	✓ 1

Common error: A sale price is 80% of the original, so divide by 0.8 to reverse it; do not add 20% to the sale price.

+ Add to test✓ In your test124IGME3-1-12r13-10Set language, notation and Venn diagrams (both tiers) Paper FMedium7 marks

During an outbreak, ξ = {days 1 to 14}. A = {days with more than 100 new cases} = {3, 4, 5, 6, 7, 8, 9}. B = {days with schools closed} = {6, 7, 8, 9, 10, 11, 12}. C = {days with a mass testing event} = {4, 8, 12}.

(a)

List the elements of A ∩ B.

(1)

(b)

Work out n(A′ ∩ B′).

(2)

(c)

List the elements of A ∩ B ∩ C. Is C a subset of A? Give a reason.

(2)

(d)

Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B) to work out n(A ∪ B).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 {6, 7, 8, 9}	✓ 1
Part (b)
M1 lists days in neither set: 1, 2, 13, 14	✓ 1
A1 4	✓ 1
Part (c)
B1 {8}	✓ 1
B1 No, because 12 is in C but not in A	✓ 1
Part (d)
M1 7 + 7 − 4	✓ 1
A1 10	✓ 1

Exam tip: A′ means everything in ξ that is not in A. For ∩ look for elements in both sets; for ∪ look for elements in either set.

+ Add to test✓ In your test125IGME3-1-13r14-05Set language, notation and Venn diagrams (both tiers) Paper FMedium5 marks

A fish farm buys feed pellets at €0.82 per kg. It uses 48.7 kg of pellets each day for 29 days.

(a)

Estimate the total cost of the pellets by rounding each number to 1 significant figure.

(2)

(b)

Work out the total cost of the pellets. Give your answer in euros to 2 decimal places.

(2)

(c)

Use your answers to say whether your estimate is an over-estimate or an under-estimate. Give a reason.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 50 × 0.8 × 30	✓ 1
A1 €1200	✓ 1
Part (b)
M1 48.7 × 0.82 × 29	✓ 1
A1 €1158.09	✓ 1
Part (c)
B1 over-estimate because 1200 is greater than 1158.09	✓ 1

Exam tip: When estimating, round every number to 1 significant figure first, then calculate with the rounded values.

+ Add to test✓ In your test126IGME3-1-13r14-12Powers, roots and standard form Paper FMedium6 marks

Two robots flash their lights at the same moment at the start of a display. Robot A flashes every 12 seconds and robot B flashes every 18 seconds.

(a)

Write 12 and 18 as products of prime factors.

(2)

(b)

Find the lowest common multiple of 12 and 18.

(2)

(c)

The start is the 1st time that both robots flash together. How many minutes after the start do they flash together for the 6th time?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 12 = 2 × 2 × 3	✓ 1
B1 18 = 2 × 3 × 3	✓ 1
Part (b)
M1 2 × 2 × 3 × 3 or list of multiples of both	✓ 1
A1 36	✓ 1
Part (c)
M1 5 × 36 = 180 seconds	✓ 1
A1 3 minutes	✓ 1

Common error: The 6th simultaneous flash is 5 intervals after the start, not 6, because the start counts as the 1st.

+ Add to test✓ In your test127IGME3-1-13r14-13Ratio, proportion and degree of accuracy Paper FMedium5 marks

5 identical robots take 12 hours to paint a fleet of fishing boats. All the robots work at the same steady rate.

(a)

Work out how long 8 robots would take to paint the fleet. Give your answer in hours and minutes.

(2)

(b)

Work out the number of robots needed to paint the fleet in 4 hours.

(2)

(c)

State one assumption you made in your calculations.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5 × 12 ÷ 8	✓ 1
A1 7 hours 30 minutes	✓ 1
Part (b)
M1 5 × 12 ÷ 4	✓ 1
A1 15	✓ 1
Part (c)
B1 each robot works at the same rate / the robots do not get in each other's way	✓ 1

Exam tip: More workers means less time, so multiply robots by hours to get a constant total, then divide.

+ Add to test✓ In your test128IGME3-1-13r14-14Powers, roots and standard form Paper FMedium6 marks

A fish tank has a capacity of 4.8 m³ and is \(\frac{5}{8}\) full of water.

(a)

Work out the volume of water in the tank.

(2)

(b)

\(\frac{3}{10}\) of the water in the tank is drained out. Work out the volume of water left in the tank.

(2)

(c)

Write the volume of water left as a fraction of the capacity of the tank. Give your fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4.8 × \(\frac{5}{8}\)	✓ 1
A1 3 m³	✓ 1
Part (b)
M1 3 × (1 − \(\frac{3}{10}\)) or 3 − 0.9	✓ 1
A1 2.1 m³	✓ 1
Part (c)
M1 2.1 ÷ 4.8 or \(\frac{21}{48}\)	✓ 1
A1 \(\frac{7}{16}\) (0.4375)	✓ 1

Common error: \(\frac{3}{10}\) is a fraction of the water in the tank at that moment, not of the full capacity.

+ Add to test✓ In your test129IGME3-1-13r16-05Integers, fractions, decimals and percentages Paper FMedium7 marks

A theatre reduces the price of a ticket by 15% for a matinee performance. The reduced price is £34.00.

(a)

Work out the original price of the ticket.

(3)

(b)

The theatre's average audience is 1500 people. The audience increases by 20% in each of two consecutive years. Work out the average audience after the two years.

(2)

(c)

Work out the overall percentage increase in the audience over the two years. Use 2160 from (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 reduced price is 85% (0.85) of the original	✓ 1
M1 34.00 ÷ 0.85	✓ 1
A1 £40.00	✓ 1
Part (b)
M1 1500 × 1.2²	✓ 1
A1 2160	✓ 1
Part (c)
M1 (2160 − 1500) ÷ 1500 × 100	✓ 1
A1 44%	✓ 1

Common error: Two increases of 20% do not make a 40% increase; the second increase is applied to the already larger amount.

+ Add to test✓ In your test130IGME3-1-13r16-06Powers, roots and standard form Paper FMedium5 marks

A volcano emits 4.5 × 10⁶ kg of ash each day for 12 days.

(a)

Write 0.000 047 in standard form.

(1)

(b)

Work out the total mass of ash emitted. Give your answer in standard form.

(2)

(c)

One cubic metre of the ash has a mass of 1.8 × 10³ kg. Work out the volume of ash emitted. Give your answer in standard form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 4.7 × 10⁻⁵	✓ 1
Part (b)
M1 4.5 × 10⁶ × 12	✓ 1
A1 5.4 × 10⁷ kg	✓ 1
Part (c)
M1 5.4 × 10⁷ ÷ 1.8 × 10³	✓ 1
A1 3.0 × 10⁴ m³	✓ 1

Exam tip: Keep the answer in the form A × 10ⁿ with 1 ≤ A < 10; 30 × 10³ would lose the final mark.

+ Add to test✓ In your test131IGME3-1-13r16-09Ratio, proportion and degree of accuracy Paper FMedium6 marks

Eight dock workers take 15 hours to unload a ship. Assume every worker unloads at the same constant rate.

(a)

How long would 12 workers take to unload the same ship?

(2)

(b)

How many workers are needed to unload the ship in 4 hours?

(2)

(c)

On another day, bad weather reduces the rate of work of each worker by 20%. How long do 8 workers take to unload the ship? Give your answer in hours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 8 × 15 = 120 worker-hours	✓ 1
A1 120 ÷ 12 = 10 hours	✓ 1
Part (b)
M1 120 ÷ 4	✓ 1
A1 30 workers	✓ 1
Part (c)
M1 each worker does 0.8 of the usual work, so time = 15 ÷ 0.8	✓ 1
A1 18.75 hours (accept 18 hours 45 minutes)	✓ 1

Common error: More workers means less time (inverse proportion), so multiply the worker count and time to get the constant.

+ Add to test✓ In your test132IGME3-1-13r16-12Set language, notation and Venn diagrams (both tiers) Paper FMedium5 marks

A theatre society has 40 members. 25 members act, 20 members work on lighting, and 6 members do neither. Some members do both.

(a)

Work out how many members do both acting and lighting.

(3)

(b)

A member is chosen at random. Work out the probability that the member works on lighting but does not act.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 − 6 = 34 members do at least one	✓ 1
M1 25 + 20 − x = 34	✓ 1
A1 x = 11	✓ 1
Part (b)
M1 20 − 11 = 9	✓ 1
A1 9/40 (accept 0.225)	✓ 1

Exam tip: Fill in the overlap of a Venn diagram first; the 'only' regions are then found by subtraction.

+ Add to test✓ In your test133IGME3-1-12r19-05Integers, fractions, decimals and percentages Paper FMedium4 marks

Two wind turbines are inspected at regular intervals. Turbine A is inspected every 360 hours and turbine B every 420 hours. Both were inspected together at time 0.

(a)

Write 360 as a product of its prime factors.

(2)

(b)

Given that 420 = 2² × 3 × 5 × 7, work out after how many hours both turbines will next be inspected together.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 prime factor tree or repeated division by primes	✓ 1
A1 2³ × 3² × 5	✓ 1
Part (b)
M1 lowest common multiple using highest powers of each prime: 2³ × 3² × 5 × 7	✓ 1
A1 2520 hours	✓ 1

Exam tip: For a lowest common multiple, take the highest power of every prime that appears in either number.

+ Add to test✓ In your test134IGME3-1-12r19-06Integers, fractions, decimals and percentages Paper FMedium6 marks

The price of a single swim at a pool rose from €6.40 to €7.36. Later, as a promotion, the new price was reduced by 25%.

(a)

Work out the percentage increase from €6.40 to €7.36.

(2)

(b)

Work out the promotional price.

(2)

(c)

Work out the overall percentage decrease from the original price of €6.40 to the promotional price.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 increase is 0.96, and 0.96 ÷ 6.40	✓ 1
A1 15%	✓ 1
Part (b)
M1 7.36 × 0.75	✓ 1
A1 €5.52	✓ 1
Part (c)
M1 (6.40 − 5.52) ÷ 6.40 × 100, ECF from (b)	✓ 1
A1 13.75% decrease	✓ 1

Common error: A 15% rise followed by a 25% fall does not cancel out, because each percentage is taken of a different amount.

+ Add to test✓ In your test135IGME3-1-12r19-07Set language, notation and Venn diagrams (both tiers) Paper FMedium5 marks

40 children attend a swimming lesson. 22 swim front crawl (C), 21 swim breaststroke (B) and 6 swim neither stroke.

(a)

Work out the number of children who swim both strokes, n(C ∩ B).

(3)

(b)

One child is chosen at random. Find the probability that the child swims exactly one of the two strokes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 − 6 = 34 swim at least one stroke	✓ 1
M1 22 + 21 − 34	✓ 1
A1 9	✓ 1
Part (b)
M1 (13 + 12) ÷ 40, ECF from (a)	✓ 1
A1 5/8 (= 0.625)	✓ 1

Common error: Adding n(C) and n(B) counts the overlap twice, so it must be subtracted once.

+ Add to test✓ In your test136IGME3-1-12r19-11Ratio, proportion and degree of accuracy Paper FMedium7 marks

Two libraries each hold 5000 books. In library A the ratio of fiction to non-fiction books is 3 : 7. In library B the ratio of fiction to non-fiction books is 2 : 3.

(a)

Work out the number of fiction books in library A.

(2)

(b)

Work out the number of fiction books in library B.

(2)

(c)

Work out the percentage of all the books in both libraries that are fiction.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5000 ÷ 10 × 3	✓ 1
A1 1500	✓ 1
Part (b)
M1 5000 ÷ 5 × 2	✓ 1
A1 2000	✓ 1
Part (c)
M1 1500 + 2000 = 3500 fiction books	✓ 1
M1 3500 ÷ 10 000 × 100, ECF from (a) and (b)	✓ 1
A1 35%	✓ 1

Common error: The combined percentage is found from the combined totals; do not average the two separate percentages unless the totals are equal.

+ Add to test✓ In your test137IGME3-1-12r19-12Powers, roots and standard form Paper FMedium6 marks

A wind farm has a maximum power output of 1.5 × 10⁸ W. Use 1 year = 3.15 × 10⁷ s. A typical home uses 1.26 × 10¹⁰ J of energy in a year.

(a)

Work out the energy, in joules, the farm would produce in one year if it ran at maximum output all the time. Give your answer in standard form.

(2)

(b)

Work out how many homes this energy could supply for a year.

(2)

(c)

In practice the farm averages 40% of its maximum output. Work out how many homes it can supply in a year.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (1.5 × 10⁸) × (3.15 × 10⁷)	✓ 1
A1 4.725 × 10¹⁵ J	✓ 1
Part (b)
M1 (4.725 × 10¹⁵) ÷ (1.26 × 10¹⁰), ECF from (a)	✓ 1
A1 3.75 × 10⁵ (375 000)	✓ 1
Part (c)
M1 0.4 × 3.75 × 10⁵, ECF from (b)	✓ 1
A1 1.5 × 10⁵ (150 000)	✓ 1

Exam tip: Write standard form numbers with brackets in your calculator, for example (1.5×10^8)×(3.15×10^7).

+ Add to test✓ In your test138IGME3-1-11r25-04Set language, notation and Venn diagrams (both tiers) Paper FMedium4 marks

40 climbers each take at least one of two training courses, or neither. 24 climbers take the rope course (R), 18 climbers take the ice course (I) and 7 climbers take neither course.
ξ = {all 40 climbers}

(a)

Work out the number of climbers who take both courses.

(3)

(b)

Find n(R ∩ I′).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 − 7 = 33 climbers take at least one course	✓ 1
M1 24 + 18 − 33	✓ 1
A1 9	✓ 1
Part (b)
B1 15 (ECF from (a): 24 − their 9)	✓ 1

Common error: n(R ∩ I′) counts only the people in R who are not in I, so the overlap must be taken away from n(R).

+ Add to test✓ In your test139IGME3-1-11r25-05Integers, fractions, decimals and percentages Paper FMedium6 marks

In a sale, the price of a pack of energy gels is reduced by 12%. The sale price is £3.52.

(a)

Work out the original price of the pack.

(3)

(b)

The shop later raises the price from the sale price of £3.52 back to the original price. Work out the percentage increase. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 100% − 12% = 88% or 0.88	✓ 1
M1 3.52 ÷ 0.88	✓ 1
A1 £4.00	✓ 1
Part (b)
M1 4.00 − 3.52 = 0.48	✓ 1
M1 0.48 ÷ 3.52 × 100 (ECF from (a))	✓ 1
A1 13.6%	✓ 1

Common error: A 12% reduction is not undone by a 12% increase, because the increase is a percentage of the smaller, sale price.

+ Add to test✓ In your test140IGME3-1-11r25-12Ratio, proportion and degree of accuracy Paper FMedium5 marks

A baker has 3³⁄₄ kg of flour. She uses ²⁄₅ of the flour to make bread.

(a)

Work out the mass of flour she uses. Give your answer as a mixed number or a decimal.

(2)

(b)

She divides the remaining flour into portions. Each portion has mass ³⁄₈ kg. Work out the number of portions.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ¹⁵⁄₄ × ²⁄₅	✓ 1
A1 1¹⁄₂ kg or 1.5 kg	✓ 1
Part (b)
M1 ¹⁵⁄₄ − their ³⁄₂ = ⁹⁄₄	✓ 1
M1 ⁹⁄₄ ÷ ³⁄₈	✓ 1
A1 6	✓ 1

Common error: To divide by a fraction, multiply by its reciprocal: ⁹⁄₄ × ⁸⁄₃. Convert the mixed number to an improper fraction before doing any arithmetic.

+ Add to test✓ In your test141IGME3-1-11r25-13Set language, notation and Venn diagrams (both tiers) Paper FMedium5 marks

ξ = {1, 2, 3, ..., 20}, the numbers of 20 phone masts.
A = {masts with 4G} = {multiples of 3}
B = {masts with 5G} = {even numbers}

(a)

List the members of A ∩ B.

(2)

(b)

Find n(A ∪ B′).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 at least two of 6, 12, 18 and no incorrect numbers	✓ 1
B1 6, 12, 18	✓ 1
Part (b)
M1 B′ is the 10 odd numbers	✓ 1
M1 adds the members of A that are not odd: 6, 12, 18	✓ 1
A1 13	✓ 1

Common error: B′ means everything in ξ that is not in B. In a union, count each member once only.

+ Add to test✓ In your test142IGME3-1-12r25-04Set language, notation and Venn diagrams (both tiers) Paper FMedium5 marks

A library surveys 60 members. 35 members borrowed fiction books (F), 28 members borrowed non-fiction books (N) and 12 members borrowed both.
ξ = {all 60 members}

(a)

Work out the number of members who borrowed neither fiction nor non-fiction.

(3)

(b)

Find n(F ∩ N′).

(1)

(c)

Find n((F ∩ N)′).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 35 + 28 − 12 = 51	✓ 1
M1 60 − 51	✓ 1
A1 9	✓ 1
Part (b)
B1 23	✓ 1
Part (c)
B1 48	✓ 1

Common error: Do not add 35 and 28 without taking away the 12 members counted twice.

+ Add to test✓ In your test143IGME3-1-12r25-05Integers, fractions, decimals and percentages Paper FMedium5 marks

The power output of a wind farm increases from 48 MW to 54 MW.

(a)

Work out the percentage increase.

(2)

(b)

A second wind farm increases its output by 30% to 84.5 MW. Work out its output before the increase.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (54 − 48) ÷ 48 × 100	✓ 1
A1 12.5%	✓ 1
Part (b)
M1 100% + 30% = 130% or 1.3	✓ 1
M1 84.5 ÷ 1.3	✓ 1
A1 65 MW	✓ 1

Common error: In a reverse percentage problem, divide by the multiplier. Taking 30% off 84.5 MW does not give the original output.

+ Add to test✓ In your test144IGME3-1-12r25-10Set language, notation and Venn diagrams (both tiers) Paper FMedium4 marks

A wind farm has 38 turbines. Each turbine produces 2.96 MW of power.

(a)

By rounding each number to 1 significant figure, estimate the total power produced by the farm.

(2)

(b)

Without calculating the exact value, state whether your estimate is an overestimate or an underestimate. Give a reason.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 × 3	✓ 1
A1 120 MW	✓ 1
Part (b)
B1 overestimate	✓ 1
B1 both numbers were rounded up	✓ 1

Exam tip: When estimating, round to 1 significant figure first and show the rounded values in your working.

+ Add to test✓ In your test145IGME3-1-12r25-11Integers, fractions, decimals and percentages Paper FMedium4 marks

A swimming pool charges £6.40 for each swim. An annual pass costs £275 and allows unlimited swims. Mia swims 52 times in a year.

(a)

Work out how much Mia would pay if she paid for each swim separately.

(1)

(b)

Work out the percentage saving if Mia buys the annual pass instead. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 £332.80	✓ 1
Part (b)
M1 332.80 − 275 = 57.80	✓ 1
M1 57.80 ÷ 332.80 × 100	✓ 1
A1 17.4%	✓ 1

Common error: A percentage saving is found as a fraction of the original (pay-per-swim) cost, not of the new cost.

+ Add to test✓ In your test146IGME3-1-13r28-03Integers, fractions, decimals and percentages Paper FMedium6 marks

A steel bridge deck is repainted over several days. On day 1, ³⁄₈ of the deck is painted. On day 2, ²⁄₅ of the part still unpainted is painted.

(a)

Work out the fraction of the whole deck painted on day 2. Give your answer in its simplest form.

(2)

(b)

Work out the fraction of the whole deck still unpainted after day 2.

(2)

(c)

The area still unpainted after day 2 is 270 m². Work out the total area of the deck.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ²⁄₅ × ⁵⁄₈ (remaining after day 1 is ⁵⁄₈)	✓ 1
A1 ¹⁄₄	✓ 1
Part (b)
M1 1 − ³⁄₈ − ¹⁄₄ (or ⁵⁄₈ − ¹⁄₄)	✓ 1
A1 ³⁄₈	✓ 1
Part (c)
M1 270 ÷ ³⁄₈ (ECF from (b))	✓ 1
A1 720 m²	✓ 1

Exam tip: For a fraction of a remainder, first find the remainder as a fraction of the whole, then multiply.

+ Add to test✓ In your test147IGME3-1-13r28-06Set language, notation and Venn diagrams (both tiers) Paper FMedium5 marks

Of 60 students at an engineering summer school, 28 study structures (S), 22 study materials (M) and 9 study both. The universal set ξ is the set of all 60 students.

(a)

Work out n(S ∪ M).

(2)

(b)

Work out n((S ∪ M)′).

(1)

(c)

A student who studies materials is chosen at random. Find the probability that this student also studies structures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 28 + 22 − 9	✓ 1
A1 41	✓ 1
Part (b)
B1 60 − 41 = 19 (ECF from (a))	✓ 1
Part (c)
M1 9 out of 22 (the 22 students in M form the sample space)	✓ 1
A1 ⁹⁄₂₂	✓ 1

Common error: For a probability given a condition, the denominator is the size of the condition set, not the whole universal set.

+ Add to test✓ In your test148IGME3-1-13r28-09Integers, fractions, decimals and percentages Paper FMedium6 marks

A coffee roaster packs 180 bags of Colombian beans and 300 bags of Kenyan beans into identical boxes. Each box has the same number of Colombian bags and the same number of Kenyan bags, with no bags left over. The roaster wants to use as many boxes as possible.

(a)

Write 180 as a product of its prime factors.

(2)

(b)

Find the highest common factor (HCF) of 180 and 300.

(2)

(c)

Work out the total number of bags in each box.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 repeated division by primes (e.g. 180 = 2 × 90 = …)	✓ 1
A1 2² × 3² × 5	✓ 1
Part (b)
M1 300 = 2² × 3 × 5² or a correct list of common factors	✓ 1
A1 60	✓ 1
Part (c)
M1 180 ÷ 60 and 300 ÷ 60 (ECF from (b))	✓ 1
A1 3 + 5 = 8 bags	✓ 1

Exam tip: The HCF is found from the lowest power of each common prime factor.

+ Add to test✓ In your test149IGME3-1-13r28-11Powers, roots and standard form Paper FMedium5 marks

A bridge uses 3.4 × 10⁵ kg of steel in its deck and 8.6 × 10⁴ kg of steel in its cables.

(a)

Work out the total mass of steel. Give your answer in standard form.

(2)

(b)

The steel costs £0.80 per kg. Work out the total cost of the steel. Give your answer in standard form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 340 000 + 86 000 (or converting to a common power of 10)	✓ 1
A1 4.26 × 10⁵ kg	✓ 1
Part (b)
M1 4.26 × 10⁵ × 0.8 (ECF from (a))	✓ 1
A1 340 800	✓ 1
A1 3.408 × 10⁵ (£)	✓ 1

Common error: When adding numbers in standard form, write them with the same power of 10 first; adding the powers is wrong.

+ Add to test✓ In your test150IGME3-1-13r29-05Integers, fractions, decimals and percentages Paper FMedium6 marks

(a)

Work out the percentage increase in the monthly membership fee.

(2)

(b)

Work out the sale price of the running shoes.

(2)

(c)

Work out the total interest earned on the investment.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{38.40-32}{32}\times100\) or \(\frac{6.4}{32}\)	✓ 1
A1 20%	✓ 1
Part (b)
M1 \(85\times0.88\) or \(85-0.12\times85\)	✓ 1
A1 £74.80	✓ 1
Part (c)
M1 \(2500\times0.03\times4\)	✓ 1
A1 £300	✓ 1

Common error: Percentage change is found as change divided by the original value, not the new value.

+ Add to test✓ In your test151IGME3-1-13r29-06Integers, fractions, decimals and percentages Paper FMedium4 marks

Two satellites pass over a ground station at regular intervals. Satellite A passes every 24 minutes and satellite B passes every 36 minutes. They pass the station together at 09:00.

(a)

Write 360 as a product of its prime factors.

(2)

(b)

Work out the number of minutes after 09:00 until both satellites next pass the station together.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 factor tree or repeated division by primes	✓ 1
A1 \(2^{3}\times3^{2}\times5\)	✓ 1
Part (b)
M1 attempt at the lowest common multiple of 24 and 36 (listing multiples or prime factors \(2^{3}\times3\) and \(2^{2}\times3^{2}\))	✓ 1
A1 72 minutes	✓ 1

Exam tip: Show the prime factors or a list of multiples; a correct method with an arithmetic slip still earns the method mark.

+ Add to test✓ In your test152IGME3-1-13r29-07Ratio, proportion and degree of accuracy Paper FMedium5 marks

In one carriage of a train, the ratio of seated passengers to standing passengers is 5 : 2. There are 36 more seated passengers than standing passengers.

(a)

Work out the number of seated passengers.

(3)

(b)

Six standing passengers then take empty seats. Work out the fraction of all passengers who are now standing. Give your answer in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(5x-2x=36\) or 3 parts = 36	✓ 1
M1 one part = 12	✓ 1
A1 60	✓ 1
Part (b)
M1 \(\frac{24-6}{84}\) or \(\frac{18}{84}\)	✓ 1
A1 \(\frac{3}{14}\)	✓ 1

Common error: The 36 is a difference, not a total, so it equals 3 parts, not 7 parts.

+ Add to test✓ In your test153IGME3-1-13r29-11Set language, notation and Venn diagrams (both tiers) Paper FMedium4 marks

A team of astronomers numbers its observation nights from 1 to 20.
\(\xi=\{1,2,3,\dots,20\}\)
\(A\) = {even numbers}
\(B\) = {multiples of 3}
\(C\) = {prime numbers}

(a)

List the members of \(A\cap B\).

(1)

(b)

List the members of \(A\cap C\).

(1)

(c)

Work out \(n\big((A\cup B)'\big)\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 {6, 12, 18}	✓ 1
Part (b)
B1 {2}	✓ 1
Part (c)
M1 \(20-(10+6-3)\) or listing the 13 members of \(A\cup B\)	✓ 1
A1 7	✓ 1

Common error: Adding \(n(A)\) and \(n(B)\) counts the overlap twice; subtract \(n(A\cap B)\) once.

+ Add to test✓ In your test154IGME3-1-14r33-03Ratio, proportion and degree of accuracy Paper FMedium6 marks

A food scientist tests a cheese of mass 450 g that contains 28% fat by mass. During ripening the cheese loses 12% of its mass as water. No fat is lost.

(a)

Work out the mass of fat in the cheese before ripening.

(2)

(b)

Work out the mass of the cheese after ripening.

(2)

(c)

Work out the percentage of fat in the ripened cheese. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.28 × 450	✓ 1
A1 126 g	✓ 1
Part (b)
M1 0.88 × 450 (or 450 − 0.12 × 450)	✓ 1
A1 396 g	✓ 1
Part (c)
M1 126 ÷ 396 × 100 (ECF)	✓ 1
A1 31.8%	✓ 1

Common error: The fat percentage changes because the total mass changes, even though the mass of fat stays the same.

+ Add to test✓ In your test155IGME3-1-14r33-08Set language, notation and Venn diagrams (both tiers) Paper FMedium6 marks

A mobile network has channels numbered 1 to 20.
ξ = {channels numbered 1 to 20}
A = {channels whose number is a multiple of 3}
B = {channels whose number is even}

(a)

List the members of A ∩ B.

(2)

(b)

Work out n((A ∪ B)′).

(2)

(c)

A channel is chosen at random from ξ. Find the probability that it is in B but not in A.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least two of 6, 12, 18 listed, with no incorrect members	✓ 1
A1 6, 12, 18	✓ 1
Part (b)
M1 n(A ∪ B) = 6 + 10 − 3 = 13 (or listing the 7 channels outside both sets)	✓ 1
A1 7	✓ 1
Part (c)
M1 10 − 3 = 7 channels in B only	✓ 1
A1 7/20 (or 0.35)	✓ 1

Exam tip: (A ∪ B)′ means everything outside both sets, so subtract the union from the total.

+ Add to test✓ In your test156IGME3-1-14r33-11Powers, roots and standard form Paper FMedium6 marks

A data cable transmits 2.4 × 10⁹ bits every second. A video file contains 7.2 × 10¹¹ bits. There are 86 400 seconds in one day.

(a)

Work out the time, in seconds, to send the file. Give your answer in standard form.

(2)

(b)

Work out the number of bits the cable can transmit in one day. Give your answer in standard form correct to 3 significant figures.

(2)

(c)

Work out the number of these video files the cable could send in one day if it runs without a break.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 7.2 × 10¹¹ ÷ (2.4 × 10⁹)	✓ 1
A1 3 × 10² s	✓ 1
Part (b)
M1 2.4 × 10⁹ × 86 400	✓ 1
A1 2.07 × 10¹⁴	✓ 1
Part (c)
M1 86 400 ÷ 300 (or 2.0736 × 10¹⁴ ÷ 7.2 × 10¹¹, using the unrounded value)	✓ 1
A1 288	✓ 1

Exam tip: Use the unrounded value from earlier working when a later part needs it, and round only the final answer.

+ Add to test✓ In your test157IGME3-1-14r34-05Ratio, proportion and degree of accuracy Paper FMedium4 marks

A swim school has 140 swimmers. The ratio of beginners to improvers to advanced swimmers is 5 : 3 : 2.

(a)

20 more improvers join the school. Write the ratio of beginners to improvers in its simplest form.

(2)

(b)

Before the 20 improvers joined, what percentage of the 140 swimmers were advanced?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 beginners 70 and improvers 42 + 20 = 62	✓ 1
A1 35 : 31	✓ 1
Part (b)
M1 2/10 × 140 = 28 advanced, or 2/10 as a percentage	✓ 1
A1 20%	✓ 1

Common error: Share the total using the sum of the ratio parts (10), not by using 140 ÷ 3 or 140 ÷ 5.

+ Add to test✓ In your test158IGME3-1-14r34-06Powers, roots and standard form Paper FMedium5 marks

A wind farm’s yearly output increased by 12% to 33.6 GWh.

(a)

Work out the output before the increase.

(3)

(b)

The next year the output falls from 33.6 GWh to 28.5 GWh. Work out the percentage decrease. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 33.6 = 1.12 × original	✓ 1
M1 33.6 ÷ 1.12	✓ 1
A1 30 GWh	✓ 1
Part (b)
M1 (33.6 − 28.5) / 33.6 × 100	✓ 1
A1 15.2%	✓ 1

Common error: Do not take 12% of 33.6 and subtract it; the 12% was calculated on the earlier, smaller value.

+ Add to test✓ In your test159IGME3-1-14r34-07Powers, roots and standard form Paper FMedium4 marks

A library receives 31 boxes of books. Each box has a mass of 19.6 kg.

(a)

By rounding each number to 1 significant figure, estimate the total mass of the boxes.

(2)

(b)

Work out the actual total mass, and state whether your estimate is an underestimate or an overestimate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 30 × 20	✓ 1
A1 600 kg	✓ 1
Part (b)
M1 31 × 19.6 = 607.6 kg	✓ 1
A1 underestimate (600 < 607.6)	✓ 1

Exam tip: Show the rounded values you use; a correct estimate with no working may earn no marks.

+ Add to test✓ In your test160IGME3-1-14r34-12Powers, roots and standard form Paper FMedium4 marks

Standard form.

(a)

A library has 1.2 × 10⁵ books. Each book has an average thickness of 2.5 × 10⁻² m. Work out the total length of shelf needed. Give your answer in standard form.

(2)

(b)

A wind farm produces 4.8 × 10⁸ kWh in a year, shared equally between 2.4 × 10⁴ homes. Work out the energy for each home. Give your answer in standard form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.2 × 10⁵ × 2.5 × 10⁻² = 3000	✓ 1
A1 3 × 10³ m	✓ 1
Part (b)
M1 4.8 × 10⁸ ÷ 2.4 × 10⁴ = 20 000	✓ 1
A1 2 × 10⁴ kWh	✓ 1

Common error: Writing 3000 or 20 000 alone does not meet the standard form instruction.

+ Add to test✓ In your test161IGME3-1-14r34-13Set language, notation and Venn diagrams (both tiers) Paper FMedium5 marks

60 people at a library were asked what they borrowed. 34 borrowed fiction (F), 27 borrowed non-fiction (N) and 9 borrowed both.

(a)

Work out the number of people who borrowed neither fiction nor non-fiction.

(3)

(b)

Find n(F ∪ N′).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 34 + 27 − 9 = 52 borrowed at least one type	✓ 1
M1 60 − 52	✓ 1
A1 8	✓ 1
Part (b)
M1 people in F together with those not in N (34 + 8)	✓ 1
A1 42	✓ 1

Exam tip: Work out each region of the Venn diagram first, then read off the set you need.

+ Add to test✓ In your test162IGME3-1-14r39-05Surds Paper FMedium7 marks

A ship carries a cargo of mass m kg.

(a)

The mass of the cargo is 4350 kg, correct to the nearest 50 kg. Write down the error interval for m.

(2)

(b)

By rounding each number to 1 significant figure, estimate the value of (48.2 × 19.7) ÷ 0.52.

(3)

(c)

Without calculating the exact value, state whether your estimate is bigger or smaller than the exact value. Give a reason.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 4325 ≤ m	✓ 1
B1 m < 4375	✓ 1
Part (b)
M1 any two of 50, 20, 0.5 used	✓ 1
M1 50 × 20 ÷ 0.5	✓ 1
A1 2000	✓ 1
Part (c)
B1 bigger (an overestimate)	✓ 1
B1 both numerator values were rounded up and the denominator was rounded down	✓ 1

Exam tip: Write the rounded numbers you use in an estimation question: method marks depend on seeing them.

+ Add to test✓ In your test163IGME3-1-14r39-07Integers, fractions, decimals and percentages Paper FMedium6 marks

A theatre offers a student ticket at a 20% discount.

(a)

After the discount, a student ticket costs £36. Work out the original price of the ticket.

(3)

(b)

The theatre's income in one month rose from £48 000 to £55 200. Work out the percentage increase.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 36 ÷ 0.8 (100% − 20% = 80%)	✓ 1
M1 36 ÷ 80 × 100 (or equivalent)	✓ 1
A1 £45	✓ 1
Part (b)
M1 55 200 − 48 000 = 7200	✓ 1
M1 7200 ÷ 48 000 × 100	✓ 1
A1 15%	✓ 1

Common error: For a reverse percentage, the given price is 80% of the original, so divide by 0.8; multiplying by 1.2 is wrong.

+ Add to test✓ In your test164IGME3-1-14r39-10Powers, roots and standard form Paper FMedium7 marks

A theatre company spends 2/5 of its budget on costumes and 1/4 of its budget on sets. The budget is £12 000.

(a)

Work out the fraction of the budget that is left.

(3)

(b)

Work out the amount of money left.

(2)

(c)

3/7 of the money left is spent on lighting. Work out the amount spent on lighting.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common denominator of 20 (8/20 and 5/20)	✓ 1
M1 1 − 8/20 − 5/20	✓ 1
A1 7/20	✓ 1
Part (b)
M1 12 000 × 7/20 (ECF from (a))	✓ 1
A1 £4200	✓ 1
Part (c)
M1 3/7 × 4200 (ECF from (b))	✓ 1
A1 £1800	✓ 1

Common error: Add or subtract fractions only after giving them a common denominator; never add the denominators.

+ Add to test✓ In your test165IGME3-1-14r39-14Integers, fractions, decimals and percentages Paper FMedium7 marks

Three spotlights in a theatre flash at regular intervals. Spotlight P flashes every 12 seconds, spotlight Q every 18 seconds and spotlight R every 15 seconds. All three flash together at the start of a show.

(a)

Write 180 as a product of its prime factors.

(2)

(b)

Find the lowest common multiple of 12 and 18.

(2)

(c)

After how many seconds will all three spotlights next flash together?

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 correct method such as a factor tree or repeated division by primes	✓ 1
A1 2² × 3² × 5	✓ 1
Part (b)
M1 listing multiples or prime factorisation (12 = 2² × 3, 18 = 2 × 3²)	✓ 1
A1 36	✓ 1
Part (c)
M1 prime factors of 15 = 3 × 5 considered with 2² × 3²	✓ 1
M1 2² × 3² × 5 (or listing multiples of 12, 18 and 15)	✓ 1
A1 180 seconds	✓ 1

Common error: The lowest common multiple takes the highest power of each prime; the highest common factor takes the lowest.

+ Add to test✓ In your test166IGME3-1-14r40-05Integers, fractions, decimals and percentages Paper FMedium5 marks

A games console has a sale price of £280 after a discount of 20% was applied to its original price.

(a)

Work out the original price of the console.

(3)

(b)

In a later sale the original price is reduced by 35%. Work out the new sale price.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 £280 represents 80% (100% − 20%) of the original price	✓ 1
M1 280 ÷ 0.8	✓ 1
A1 £350	✓ 1
Part (b)
M1 350 × 0.65 (ECF from (a))	✓ 1
A1 £227.50	✓ 1

Common error: To undo a 20% discount, divide by 0.8; adding 20% of the sale price to the sale price gives the wrong original price.

+ Add to test✓ In your test167IGME3-1-14r40-06Powers, roots and standard form Paper FMedium6 marks

A rainwater tank is 5/8 full. It contains 400 litres of water.

(a)

Work out the capacity of the tank.

(2)

(b)

3/10 of the water in the tank is then used. Work out the volume of water that is left.

(2)

(c)

Write the volume left as a fraction of the capacity of the tank. Give your answer in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 400 ÷ 5 × 8 (or 400 ÷ 0.625)	✓ 1
A1 640 litres	✓ 1
Part (b)
M1 400 × (1 − 3/10) or 400 × 7/10	✓ 1
A1 280 litres	✓ 1
Part (c)
M1 280/640 (ECF from (a) and (b))	✓ 1
A1 7/16	✓ 1

Exam tip: If 5/8 of a quantity is known, divide by the numerator to find 1/8, then multiply by the denominator to find the whole.

+ Add to test✓ In your test168IGME3-1-14r40-08Integers, fractions, decimals and percentages Paper FMedium5 marks

Two pumps at a treatment plant both start a cycle at 08:00. Pump A starts a new cycle every 12 minutes. Pump B starts a new cycle every 18 minutes.

(a)

Write 12 and 18 each as a product of prime factors.

(2)

(b)

Find the lowest common multiple (LCM) of 12 and 18.

(2)

(c)

Work out the next time after 08:00 at which both pumps start a new cycle together.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 12 = 2² × 3	✓ 1
B1 18 = 2 × 3²	✓ 1
Part (b)
M1 2² × 3² or lists multiples of both numbers	✓ 1
A1 36	✓ 1
Part (c)
B1 08:36	✓ 1

Exam tip: To find the LCM from prime factors, take the highest power of each prime that appears in either number.

+ Add to test✓ In your test169IGME3-1-14r40-10Powers, roots and standard form Paper FMedium4 marks

Each tablet in a box contains 2.5 × 10⁻⁴ g of an active ingredient. A box holds 84 tablets.

(a)

Work out the total mass of active ingredient in one box. Give your answer in standard form.

(2)

(b)

A patient needs 1.5 g of the active ingredient. Work out the number of tablets that supply exactly this amount.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 84 × 2.5 × 10⁻⁴ (= 0.021)	✓ 1
A1 2.1 × 10⁻² g	✓ 1
Part (b)
M1 1.5 ÷ (2.5 × 10⁻⁴)	✓ 1
A1 6000	✓ 1

Exam tip: Check your calculator display carefully: 2.5 × 10⁻⁴ must be entered with brackets or the EXP key so the power applies to 10 only.

+ Add to test✓ In your test170IGME3-1-14r40-11Set language, notation and Venn diagrams (both tiers) Paper FMedium6 marks

A game uses a fair 20-sided die numbered 1 to 20 and the die is rolled once. ξ = {1, 2, 3, ..., 20}, A is the set of multiples of 3 and B is the set of factors of 24.

(a)

List the members of A ∩ B.

(2)

(b)

Work out n((A ∪ B)′).

(2)

(c)

Work out the probability that the number rolled is in B but not in A. Give your answer as a fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 at least two of 3, 6, 12 and no incorrect values	✓ 1
B1 {3, 6, 12}	✓ 1
Part (b)
M1 lists or counts A ∪ B, e.g. 6 + 7 − 3 = 10 (A has 6 members, B has 7)	✓ 1
A1 10	✓ 1
Part (c)
B1 4 members: 1, 2, 4, 8	✓ 1
A1 1/5	✓ 1

Common error: Factors of 24 greater than 20 (24 itself) are not in ξ, so they must not be counted in B.

+ Add to test✓ In your test171IGME3-1-14r41-05Powers, roots and standard form Paper FMedium4 marks

A radio signal travels at 3 × 10⁸ m/s. A satellite is 3.6 × 10⁷ m above a ground station.

(a)

Work out the time taken for a signal to travel from the ground station to the satellite. Give your answer in standard form.

(2)

(b)

Work out the distance the signal travels in 2.5 × 10⁻³ s. Give your answer in standard form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3.6 × 10⁷ ÷ 3 × 10⁸	✓ 1
A1 1.2 × 10⁻¹ s	✓ 1
Part (b)
M1 3 × 10⁸ × 2.5 × 10⁻³	✓ 1
A1 7.5 × 10⁵ m (= 750000 m)	✓ 1

Common error: When multiplying or dividing numbers in standard form, deal with the powers of 10 using the index laws and check the answer is still between 1 and 10 before the power.

+ Add to test✓ In your test172IGME3-1-14r41-06Ratio, proportion and degree of accuracy Paper FMedium5 marks

A cake recipe for 12 cakes uses 300 g of flour and 180 g of sugar.

(a)

Work out the mass of flour needed for 30 cakes.

(2)

(b)

A baker has 495 g of sugar. Work out the greatest number of cakes she can make using this recipe.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 300 ÷ 12 = 25 g per cake (or × 2.5)	✓ 1
A1 750 g	✓ 1
Part (b)
M1 180 ÷ 12 = 15 g of sugar per cake	✓ 1
M1 495 ÷ 15	✓ 1
A1 33	✓ 1

Exam tip: Find the amount for one item first (the unitary method), then scale up.

+ Add to test✓ In your test173IGME3-1-14r41-07Ratio, proportion and degree of accuracy Paper FMedium5 marks

A climber walks ²⁄₅ of the length of a route on day 1. On day 2 she walks ¹⁄₃ of the remaining distance.

(a)

What fraction of the whole route is still to be walked after day 2?

(3)

(b)

The remaining distance is 7.2 km. Work out the total length of the route.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 remaining after day 1 is ³⁄₅	✓ 1
M1 ¹⁄₃ × ³⁄₅ = ¹⁄₅ walked on day 2, or ²⁄₃ × ³⁄₅	✓ 1
A1 ²⁄₅	✓ 1
Part (b)
M1 7.2 ÷ ²⁄₅ (or 7.2 ÷ 2 × 5)	✓ 1
A1 18 km	✓ 1

Common error: Day 2 uses a fraction of what remains, not a fraction of the whole route.

+ Add to test✓ In your test174IGME3-1-14r41-08Set language, notation and Venn diagrams (both tiers) Paper FMedium5 marks

A telecoms company has 120 customers. A is the set of customers with fibre broadband and B is the set of customers with a mobile contract. n(A) = 70, n(B) = 55 and 15 customers have neither.

(a)

Work out n(A ∩ B).

(3)

(b)

A customer is chosen at random from those with a mobile contract. Find the probability that this customer also has fibre broadband.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 n(A ∪ B) = 120 − 15 = 105	✓ 1
M1 70 + 55 − 105	✓ 1
A1 20	✓ 1
Part (b)
M1 ²⁰⁄₅₅ using ECF from (a)	✓ 1
A1 ⁴⁄₁₁ (= 0.363636…)	✓ 1

Exam tip: A probability 'given' a group uses only that group as the total, here the 55 people with mobile contracts.

+ Add to test✓ In your test175IGME3-1-14r42-01Powers, roots and standard form Paper FMedium7 marks

A dig site yields 240 pottery fragments. \(\frac{3}{8}\) of the fragments are red clay. The rest are grey clay.
40% of the grey-clay fragments are decorated.

(a)

Work out the number of red-clay fragments.

(2)

(b)

Work out the number of decorated grey-clay fragments.

(3)

(c)

Write the number of decorated grey-clay fragments as a fraction of all 240 fragments. Give your fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3 ÷ 8 × 240, or 240 ÷ 8 × 3	✓ 1
A1 90	✓ 1
Part (b)
M1 240 − 90 = 150 grey-clay fragments	✓ 1
M1 0.4 × 150	✓ 1
A1 60 (ECF from (a))	✓ 1
Part (c)
M1 60 ÷ 240	✓ 1
A1 1/4 (= 0.25)	✓ 1

Exam tip: Find the number of fragments in the remaining group first, then take the percentage of that group, not of the total.

+ Add to test✓ In your test176IGME3-1-14r42-04Powers, roots and standard form Paper FMedium6 marks

A data centre has 2 × 10⁴ identical hard drives. Each drive stores 4.5 × 10¹² bytes.

(a)

Work out the total number of bytes the drives can store. Give your answer in standard form.

(2)

(b)

The average file is 6 × 10⁵ bytes.
Work out how many average files can be stored. Give your answer in standard form.

(2)

(c)

The data centre uses 7.2 × 10⁵ kWh of energy each day.
Work out the energy used each hour. Give your answer in standard form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × 4.5 = 9 and 10⁴ × 10¹² = 10¹⁶	✓ 1
A1 9 × 10¹⁶	✓ 1
Part (b)
M1 9 × 10¹⁶ ÷ 6 × 10⁵ written as (9 ÷ 6) × 10¹⁶⁻⁵	✓ 1
A1 1.5 × 10¹¹ (ECF from (a))	✓ 1
Part (c)
M1 7.2 × 10⁵ ÷ 24	✓ 1
A1 3 × 10⁴ kWh (30000)	✓ 1

Common error: Divide the number parts and subtract the powers of 10 separately, then check that the number part is between 1 and 10.

+ Add to test✓ In your test177IGME3-1-14r42-05Ratio, proportion and degree of accuracy Paper FMedium5 marks

An archaeology team has diggers, surveyors and cataloguers in the ratio 5 : 3 : 2.
There are 40 people in the team.

(a)

Work out the number of surveyors.

(2)

(b)

6 more cataloguers join the team. Nobody else joins or leaves.
Write the new ratio of diggers : surveyors : cataloguers. Give your ratio in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 ÷ (5 + 3 + 2) = 4	✓ 1
A1 12	✓ 1
Part (b)
M1 diggers 20 and cataloguers 8 found	✓ 1
M1 20 : 12 : 14	✓ 1
A1 10 : 6 : 7	✓ 1

Exam tip: Convert the ratio to actual numbers first; only simplify at the end.

+ Add to test✓ In your test178IGME3-1-14r42-08Powers, roots and standard form Paper FMedium5 marks

A data centre has 60 servers. Some servers have solid-state drives (SSD) and some have graphics processors (GPU).
38 servers have an SSD. 25 servers have a GPU. 8 servers have neither an SSD nor a GPU.

(a)

Work out the number of servers that have both an SSD and a GPU.

(3)

(b)

Write the number of servers with an SSD but no GPU as a fraction of all 60 servers. Give your fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 60 − 8 = 52 servers have at least one	✓ 1
M1 38 + 25 − 52	✓ 1
A1 11	✓ 1
Part (b)
M1 (38 − 11) ÷ 60 = 27/60	✓ 1
A1 9/20 (= 0.45) (ECF from (a))	✓ 1

Common error: Adding 38 and 25 counts the servers with both twice; subtract the overlap once.

+ Add to test✓ In your test179IGME3-1-14r42-14Ratio, proportion and degree of accuracy Paper FMedium8 marks

A cyclist rides for 2 \(\frac{2}{3}\) hours in the morning and 1 \(\frac{3}{4}\) hours in the afternoon.

(a)

Work out the total time she rides. Give your answer as a mixed number.

(3)

(b)

She planned to ride for 6 \(\frac{1}{2}\) hours in total.
Work out the planned time that remains. Give your answer as a mixed number.

(2)

(c)

In the remaining time she has 50 km still to ride.
Work out the average speed she needs, in km/h.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 converts to improper fractions or a common denominator, 8/3 and 7/4	✓ 1
M1 32/12 + 21/12 = 53/12	✓ 1
A1 4 5/12 hours	✓ 1
Part (b)
M1 6 1/2 − 4 5/12 = 78/12 − 53/12	✓ 1
A1 2 1/12 hours (ECF from (a))	✓ 1
Part (c)
M1 speed = 50 ÷ 2 1/12	✓ 1
M1 50 ÷ 25/12 = 50 × 12/25	✓ 1
A1 24 km/h	✓ 1

Common error: Turn mixed numbers into improper fractions before dividing; dividing by 25/12 means multiplying by 12/25.

+ Add to test✓ In your test180IGME3-1-14r43-01Powers, roots and standard form Paper FMedium5 marks

A pharmacist prepares a medicine for a patient. The dose is 0.35 mg for every kilogram of the patient's body mass. The patient has a body mass of 64 kg. The medicine is supplied in tablets of 7.5 mg. Tablets cannot be split.

(a)

Work out the dose in mg.

(2)

(b)

Work out the least number of tablets that gives at least the full dose.

(2)

(c)

Work out, in mg, how much more than the dose the patient receives.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.35 × 64	✓ 1
A1 22.4 mg	✓ 1
Part (b)
M1 22.4 ÷ 7.5 (= 2.98...) (ECF from (a))	✓ 1
A1 3 tablets	✓ 1
Part (c)
A1 0.1 mg (ECF from (a) and (b))	✓ 1

Common error: Rounding down when a minimum is required. Always check that the total mass given is at least the dose.

+ Add to test✓ In your test181IGME3-1-14r43-04Ratio, proportion and degree of accuracy Paper FMedium4 marks

A laboratory technician mixes concentrate and water in the ratio 2 : 7 by volume to make 450 ml of a solution.

(a)

Work out the volume of concentrate in the solution.

(2)

(b)

A further 150 ml of water is added to the solution. Work out the percentage of the new mixture that is concentrate. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 450 ÷ (2 + 7) = 50	✓ 1
A1 100 ml	✓ 1
Part (b)
M1 100 ÷ (450 + 150) × 100 (ECF from (a))	✓ 1
A1 16.7%	✓ 1

Common error: Forgetting that the total volume has increased when the extra water is added.

+ Add to test✓ In your test182IGME3-1-14r43-05Integers, fractions, decimals and percentages Paper FMedium7 marks

In a sale, the price of a games console is reduced by 15%. The sale price is £212.50.

(a)

Work out the original price of the games console.

(3)

(b)

After the sale, the shop increases the sale price by 8%. Work out the new price.

(2)

(c)

Work out the overall percentage decrease from the original price to the new price.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sale price is 85% of the original, or 100 − 15 = 85	✓ 1
M1 212.50 ÷ 0.85	✓ 1
A1 £250	✓ 1
Part (b)
M1 212.50 × 1.08	✓ 1
A1 £229.50	✓ 1
Part (c)
M1 (250 − 229.50) ÷ 250 × 100 (ECF)	✓ 1
A1 8.2%	✓ 1

Common error: Finding 15% of £212.50 and adding it on. A reverse percentage problem needs division by the multiplier.

+ Add to test✓ In your test183IGME3-1-14r43-09Set language, notation and Venn diagrams (both tiers) Paper FMedium5 marks

An online gaming club has 60 members. 38 members play a puzzle game, 29 members play an action game, and 9 members play neither game.

(a)

Work out the number of members who play both games.

(3)

(b)

One member is chosen at random. Work out the probability that this member plays the puzzle game only. Give your answer as a fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 60 − 9 = 51 play at least one game	✓ 1
M1 38 + 29 − 51	✓ 1
A1 16	✓ 1
Part (b)
M1 (38 − 16) ÷ 60 (ECF from (a))	✓ 1
A1 \(\frac{11}{30}\)	✓ 1

Exam tip: Work from the outside of a Venn diagram inward: find those in neither set first, then use the total to find the overlap.

+ Add to test✓ In your test184IGME3-1-14r43-11Powers, roots and standard form Paper FMedium5 marks

A water engineer needs to estimate the value of (6.8 × 207) ÷ 0.48.

(a)

Write down the three rounded numbers and use them to work out an estimate.

(3)

(b)

Use a calculator to work out the exact value of (6.8 × 207) ÷ 0.48.

(1)

(c)

State whether your estimate in (a) is an underestimate or an overestimate. Give a reason.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 7, 200 and 0.5	✓ 1
M1 7 × 200 ÷ 0.5	✓ 1
A1 2800	✓ 1
Part (b)
B1 2932.5	✓ 1
Part (c)
A1 underestimate, because 2800 < 2932.5	✓ 1

Exam tip: Dividing by a number rounded up makes the answer smaller, so rounding in different directions can partly cancel out.

+ Add to test✓ In your test185IGME3-1-12r47-05Integers, fractions, decimals and percentages Paper FMedium6 marks

Two lamp strips on a building facade are controlled by timers. Strip P flashes every 36 seconds and strip Q flashes every 90 seconds. They flash together at 8:00 pm.

(a)

Write 90 as a product of its prime factors.

(2)

(b)

Find the highest common factor (HCF) of 36 and 90.

(2)

(c)

Find the number of seconds from 8:00 pm until the strips next flash together.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 correct method, e.g. a factor tree or division by primes	✓ 1
A1 2 × 3² × 5	✓ 1
Part (b)
M1 36 = 2² × 3², common factors 2 × 3²	✓ 1
A1 18	✓ 1
Part (c)
M1 lowest common multiple, e.g. 2² × 3² × 5, or listing multiples	✓ 1
A1 180	✓ 1

Exam tip: Use the lowest common multiple, not the HCF, when two repeating events must coincide.

+ Add to test✓ In your test186IGME3-1-12r47-06Set language, notation and Venn diagrams (both tiers) Paper FMedium6 marks

50 sailors were asked about the equipment they use. 28 use sonar (S), 22 use GPS (G), x use both, and 6 use neither.

(a)

Find the value of x.

(3)

(b)

One sailor is chosen at random. Find the probability that this sailor uses sonar only.

(2)

(c)

Write down n(S′ ∩ G).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (28 − x) + x + (22 − x) + 6 = 50	✓ 1
M1 56 − x = 50	✓ 1
A1 x = 6	✓ 1
Part (b)
M1 22/50	✓ 1
A1 11/25 (or 0.44)	✓ 1
Part (c)
B1 16	✓ 1

Common error: The region for 'sonar only' is 28 − x, not 28.

+ Add to test✓ In your test187IGME3-1-12r47-13Powers, roots and standard form Paper FMedium6 marks

A cargo ship carries 1.8 × 10⁴ containers. Each container has a mass of 2.5 × 10⁴ kg. The ship has a mass of 6 × 10⁷ kg when empty.

(a)

Work out the total mass of the containers. Give your answer in standard form.

(2)

(b)

Work out the total mass of the loaded ship. Give your answer in standard form.

(2)

(c)

Work out the mass of the containers as a percentage of the mass of the loaded ship. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.8 × 2.5 = 4.5 (or 1.8 × 10⁴ × 2.5 × 10⁴)	✓ 1
A1 4.5 × 10⁸ kg	✓ 1
Part (b)
M1 4.5 × 10⁸ + 0.6 × 10⁸ (same power of 10)	✓ 1
A1 5.1 × 10⁸ kg	✓ 1
Part (c)
M1 4.5 × 10⁸ ÷ 5.1 × 10⁸ × 100	✓ 1
A1 88.235... = 88.2	✓ 1

Exam tip: Write both numbers with the same power of 10 before adding them.

+ Add to test✓ In your test188IGME3-1-13r48-04Integers, fractions, decimals and percentages Paper FMedium5 marks

The price of a monthly rail pass increased by 12%. The price after the increase is £43.68.

(a)

Work out the price of the pass before the increase.

(3)

(b)

The original price of £39.00 is increased by 12% each year for 3 years. Work out the price after the third increase. Give your answer to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 100% + 12% = 112% or multiplier 1.12	✓ 1
M1 43.68 ÷ 1.12	✓ 1
A1 39 or £39.00	✓ 1
Part (b)
M1 39 × 1.12³	✓ 1
A1 54.79 or £54.79	✓ 1

Common error: To undo a percentage increase, divide by the multiplier; subtracting 12% of the new price gives the wrong answer.

+ Add to test✓ In your test189IGME3-1-13r48-05Set language, notation and Venn diagrams (both tiers) Paper FMedium6 marks

60 athletes at a sports academy were asked about two events. S is the set of athletes who take part in sprinting and J is the set of athletes who take part in jumping.
n(S) = 36, n(J) = 28 and 9 athletes take part in neither event.

(a)

Work out n(S ∩ J).

(3)

(b)

Work out n(S ∩ J′).

(1)

(c)

One of the 60 athletes is chosen at random. Find the probability that this athlete takes part in jumping only. Give your answer as a fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 60 − 9 = 51 athletes in S ∪ J	✓ 1
M1 36 + 28 − 51	✓ 1
A1 13	✓ 1
Part (b)
B1 23 (ECF from (a))	✓ 1
Part (c)
M1 15/60 or 28 − 13 = 15 seen	✓ 1
A1 1/4	✓ 1

Exam tip: n(A ∪ B) = n(A) + n(B) − n(A ∩ B); subtract the intersection once because it has been counted twice.

+ Add to test✓ In your test190IGME3-1-13r48-06Ratio, proportion and degree of accuracy Paper FMedium5 marks

A runner's time for one 400 m lap is 52.4 seconds, correct to the nearest 0.1 second.

(a)

Write down the lower bound and the upper bound of the runner's time.

(2)

(b)

The track is exactly 400 m long. Using the time of 52.4 seconds, work out the runner's average speed. Give your answer in m/s correct to 3 significant figures.

(2)

(c)

Write down the runner's time to 2 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 52.35	✓ 1
B1 52.45	✓ 1
Part (b)
M1 400 ÷ 52.4	✓ 1
A1 7.63	✓ 1
Part (c)
B1 52	✓ 1

Common error: The upper bound of a value given to the nearest 0.1 is 52.45, not 52.49 or 52.5.

+ Add to test✓ In your test191IGME3-1-13r48-07Powers, roots and standard form Paper FMedium4 marks

The mass of the Earth is 5.97 × 10²⁴ kg. The mass of Jupiter is 1.90 × 10²⁷ kg.

(a)

Work out the total mass of the Earth and Jupiter. Give your answer in standard form correct to 3 significant figures.

(2)

(b)

Work out how many times greater the mass of Jupiter is than the mass of the Earth. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.90 × 10²⁷ + 5.97 × 10²⁴ evaluated as 1.90597 × 10²⁷	✓ 1
A1 1.91 × 10²⁷	✓ 1
Part (b)
M1 1.90 × 10²⁷ ÷ 5.97 × 10²⁴ with brackets used correctly	✓ 1
A1 318	✓ 1

Exam tip: Use brackets when entering standard form divisions on a calculator: (1.90 × 10^27) ÷ (5.97 × 10^24).

+ Add to test✓ In your test192IGME3-1-14r48-04Powers, roots and standard form Paper FMedium5 marks

A medical nanobot is 3.5 × 10⁻⁶ m long. Each nanobot has a mass of 4 × 10⁻⁹ g.

(a)

Write 0.000 003 5 in standard form.

(1)

(b)

The nanobots are placed end to end in a straight line 2.1 mm long. Work out the number of nanobots in the line. 1 mm = 1 × 10⁻³ m.

(2)

(c)

Work out the total mass, in grams, of the nanobots in part (b). Give your answer in standard form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3.5 × 10⁻⁶	✓ 1
Part (b)
M1 2.1 × 10⁻³ ÷ (3.5 × 10⁻⁶)	✓ 1
A1 600	✓ 1
Part (c)
M1 600 × 4 × 10⁻⁹ (ECF from (b))	✓ 1
A1 2.4 × 10⁻⁶	✓ 1

Common error: When converting millimetres to metres, 2.1 mm is 2.1 × 10⁻³ m, not 2.1 × 10³ m.

+ Add to test✓ In your test193IGME3-1-14r48-06Set language, notation and Venn diagrams (both tiers) Paper FMedium4 marks

A fashion designer buys 19.6 m of fabric at €7.92 per metre.

(a)

Estimate the total cost of the fabric by rounding each number to 1 significant figure. You must show your working.

(2)

(b)

Work out the exact cost of the fabric.

(1)

(c)

Without calculating, explain whether your estimate in part (a) is bigger or smaller than the exact cost.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rounds to 20 and 8	✓ 1
A1 20 × 8 = 160	✓ 1
Part (b)
B1 155.23 or €155.232	✓ 1
Part (c)
B1 bigger, because both numbers were rounded up	✓ 1

Exam tip: An estimate is found by rounding each number to 1 significant figure; show the rounded numbers as part of the working.

+ Add to test✓ In your test194IGME3-1-14r48-07Powers, roots and standard form Paper FMedium4 marks

A robot's sensor records 3⁴ readings in each cycle. The robot carries out 3² cycles in a test.

(a)

Write the total number of readings in the test as a single power of 3.

(2)

(b)

Work out the total number of readings in the test.

(1)

(c)

The robot stores the readings in a square grid with the same number of rows as columns. Work out the number of readings in one row.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3⁴ × 3² or adds the indices 4 + 2	✓ 1
A1 3⁶	✓ 1
Part (b)
B1 729	✓ 1
Part (c)
B1 27	✓ 1

Common error: When multiplying powers of the same base, add the indices; 3⁴ × 3² is not 9⁶.

+ Add to test✓ In your test195IGME-058-00-13Integers, fractions, decimals and percentages Paper FHard5 marks

Amira receives her monthly pay.
She spends \(\frac{1}{4}\) of it on rent.
She spends \(\frac{2}{3}\) of what is left on bills.
She saves 40% of the money remaining after rent and bills.
She spends the rest, £270, on food.

(a)

What fraction of her pay is left after rent and bills?

(2)

(b)

Work out Amira's monthly pay.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{1}{3} \times \frac{3}{4}\), or \(\frac{3}{4} - \frac{2}{3} \times \frac{3}{4}\) (i.e. \(\frac{3}{4} - \frac{1}{2}\))	✓ 1
A1 \(\frac{1}{4}\)	✓ 1
Part (b)
M1 food is 60% of the remainder: \(0.6 \times \frac{1}{4} = \frac{3}{20}\) of pay, or remainder = 270 ÷ 0.6 = £450	✓ 1
M1 270 ÷ \(\frac{3}{20}\), or 450 × 4	✓ 1
A1 £1800	✓ 1

Common error: Taking \(\frac{2}{3}\) of the whole pay instead of \(\frac{2}{3}\) of what is left after rent. Each fraction here applies to the amount remaining at that stage.

+ Add to test✓ In your test196IGME-058-02-09Ratio, proportion and degree of accuracy Paper FHard5 marks

In a chess club, the ratio of boys to girls is 4 : 5
12 more boys join the club and no one leaves.
The ratio of boys to girls is now 8 : 7

(a)

Work out the number of people in the club before the 12 boys joined.
Show clear algebraic working.

(4)

(b)

How many more girls must now join so that the numbers of boys and girls are equal?

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for boys = 4x, girls = 5x	✓ 1
M1 for \(\frac{4x+12}{5x}=\frac{8}{7}\) or 7(4x + 12) = 8(5x)	✓ 1
M1 for 28x + 84 = 40x leading to x = 7	✓ 1
A1 for 63	✓ 1
Part (b)
B1 for 5 (boys 4(7) + 12 = 40, girls 5(7) = 35), or follow through: (4x + 12) − 5x using their value of x from (a)	✓ 1

Common error: Only the number of boys changes. The girls stay at 5x, so do not add 12 to both parts of the ratio.

+ Add to test✓ In your test197IGME-058-03-13Set language, notation and Venn diagrams (both tiers) Paper FHard6 marks

ξ = {positive integers less than 30}
A = {multiples of 4}
B = {multiples of 6}
C = {factors of 48}

(a)

Describe the members of A ∩ B in words.

(1)

(b)

Find n(A ∪ B).

(2)

(c)

Find n(C ∩ (A ∪ B)′).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 'multiples of 12' (less than 30), or equivalent such as 'common multiples of 4 and 6' .	✓ 1
Part (b)
M1 for n(A) = 7, n(B) = 4 and n(A ∩ B) = 2 (i.e. 12 and 24), OR for a correct list A ∪ B = {4, 6, 8, 12, 16, 18, 20, 24, 28} . A1 for 9 .	✓ 2
Part (c)
M1 for C = {1, 2, 3, 4, 6, 8, 12, 16, 24}, with 48 excluded because it is not in ξ . M1 for identifying 4, 6, 8, 12, 16 and 24 as the members of C that are in A ∪ B, OR for a correct list of (A ∪ B)′ = {1, 2, 3, 5, 7, 9, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 25, 26, 27, 29} . A1 for 3, from {1, 2, 3} .	✓ 3

Common error: Sets can only contain members of ξ. 48 is a factor of 48, but it is not less than 30, so it is not in C.

+ Add to test✓ In your test198IGME3-1-11r7-14Powers, roots and standard form Paper FHard7 marks

A runner plans a long run. She runs \(\frac{2}{5}\) of the total distance and then stops for a drink. She then runs \(\frac{3}{8}\) of the distance that is left and stops for a second drink. She still has 9 km to run.

(a)

Work out the fraction of the total distance she has run at the second drink.

(3)

(b)

Work out the total distance of the run.

(2)

(c)

She runs the whole distance at an average speed of 9.6 km/h.
Work out the time she takes, in hours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 distance left after the first drink = \(\frac{3}{5}\)	✓ 1
M1 \(\frac{3}{8}\times\frac{3}{5}=\frac{9}{40}\)	✓ 1
A1 \(\frac{2}{5}+\frac{9}{40}=\frac{5}{8}\)	✓ 1
Part (b)
M1 9 ÷ \(\frac{3}{8}\) or 9 ÷ (1 − \(\frac{5}{8}\))	✓ 1
A1 24 km	✓ 1
Part (c)
M1 24 ÷ 9.6 (ECF from (b))	✓ 1
A1 2.5 hours (2 hours 30 minutes)	✓ 1

Exam tip: Draw a bar or write a fraction of what remains: \(\frac{3}{8}\) of the distance left is not \(\frac{3}{8}\) of the whole run.

+ Add to test✓ In your test199IGME3-1-12r13-14Powers, roots and standard form Paper FHard7 marks

The number of people infected in an outbreak is 2500 at the start. It is predicted to increase by 12% each week.

(a)

Work out the predicted number infected after 3 weeks. Give your answer to the nearest whole number.

(2)

(b)

Find the number of whole weeks after which the number infected first exceeds 10 000.

(3)

(c)

Use the unrounded value after 13 weeks to write the predicted number infected in standard form, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2500 × 1.12³	✓ 1
A1 3512	✓ 1
Part (b)
M1 tests 1.12ⁿ against 4 or 2500 × 1.12ⁿ against 10 000	✓ 1
M1 after 12 weeks 2500 × 1.12¹² = 9739.9… (below 10 000) and after 13 weeks 10 908.7… (above 10 000)	✓ 1
A1 13 (weeks)	✓ 1
Part (c)
M1 2500 × 1.12¹³ = 10 908.7…	✓ 1
A1 1.09 × 10⁴ (= 10 900)	✓ 1

Common error: Repeated percentage increases are calculated with a multiplier raised to a power (1.12ⁿ), not by adding 12% of the starting amount each time.

+ Add to test✓ In your test200IGME3-1-14r33-12Ratio, proportion and degree of accuracy Paper FHard5 marks

A mountain expedition stores oxygen cylinders at base camp and at camp 2 in the ratio 7 : 3. There are 60 cylinders in total. The team then moves n cylinders from base camp to camp 2. After the move the ratio of cylinders at base camp to cylinders at camp 2 is 3 : 2.

(a)

Work out the number of cylinders at each camp before the move.

(2)

(b)

Find the value of n.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 60 ÷ 10 = 6 per part	✓ 1
A1 42 at base camp and 18 at camp 2	✓ 1
Part (b)
M1 (42 − n)/(18 + n) = 3/2 or 2(42 − n) = 3(18 + n) (ECF from (a))	✓ 1
M1 84 − 2n = 54 + 3n	✓ 1
A1 n = 6	✓ 1

Exam tip: Check your answer: after moving 6 cylinders the camps hold 36 and 24, which is the ratio 3 : 2.

+ Add to test✓ In your test201IGME3-1-12r47-07Set language, notation and Venn diagrams (both tiers) Paper FHard6 marks

ξ = {integers from 1 to 30 inclusive}. P = {prime numbers} and M = {multiples of 3}.

(a)

List the members of P ∩ M.

(1)

(b)

Find n(P ∪ M).

(2)

(c)

A number is chosen at random from the members of M′. Find the probability that it is prime.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 {3}	✓ 1
Part (b)
M1 10 + 10 − 1 (or listing both sets without repeats)	✓ 1
A1 19	✓ 1
Part (c)
M1 n(M′) = 30 − 10 = 20	✓ 1
M1 primes that are not multiples of 3: 9	✓ 1
A1 9/20 (or 0.45)	✓ 1

Common error: Prime numbers up to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29; 1 is not prime.

+ Add to test✓ In your test202IGME3-1-12r47-14Ratio, proportion and degree of accuracy Paper FHard6 marks

A tank holds only cod and haddock, in the ratio 3 : 5. Then 12 cod are added to the tank and 8 haddock are removed. The tank now has the same number of cod as haddock.

(a)

Find the value of one part of the original ratio.

(3)

(b)

Work out the original total number of fish in the tank.

(1)

(c)

Work out the percentage increase in the number of cod.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 original numbers 3k and 5k	✓ 1
M1 3k + 12 = 5k − 8	✓ 1
A1 k = 10	✓ 1
Part (b)
B1 8 × 10 = 80	✓ 1
Part (c)
M1 12 ÷ 30 × 100	✓ 1
A1 40	✓ 1

Exam tip: Represent the unknown amounts as multiples of one part, then form an equation from the new condition.

+ Add to test✓ In your test203IGME-058-04-01Surds Paper HEasy4 marks

Do not use the surd function on your calculator for this question. Show your working.

(a)

Simplify \(\sqrt{72}\), giving your answer in the form \(a\sqrt{2}\) where \(a\) is an integer.

(2)

(b)

Simplify \(\sqrt{50}+\sqrt{18}\), giving your answer in the form \(k\sqrt{2}\) where \(k\) is an integer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\sqrt{36}\times\sqrt{2}\) or a partial simplification such as \(3\sqrt{8}\) or \(2\sqrt{18}\) ; A1 \(6\sqrt{2}\)	✓ 2
Part (b)
M1 for \(5\sqrt{2}\) or \(3\sqrt{2}\) seen ; A1 \(8\sqrt{2}\)	✓ 2

Exam tip: Look for the largest square factor (4, 9, 16, 25, 36, …) of the number under the root. Using a smaller one, such as \(\sqrt{72}=2\sqrt{18}\), means you must simplify again.

+ Add to test✓ In your test204IGME-058-04-02Surds Paper HEasy4 marks

Expand and simplify each expression. Show your working.

(a)

\((3+\sqrt{5})(3-\sqrt{5})\)

(2)

(b)

\((2+\sqrt{3})^2\), giving your answer in the form \(a+b\sqrt{3}\) where \(a\) and \(b\) are integers.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for four terms with at least three correct, e.g. \(9-3\sqrt{5}+3\sqrt{5}-5\) ; A1 \(4\)	✓ 2
Part (b)
M1 for \(4+2\sqrt{3}+2\sqrt{3}+3\) or equivalent with at least three terms correct ; A1 \(7+4\sqrt{3}\)	✓ 2

Common error: \((2+\sqrt{3})^2\) is not \(4+3\). Write the bracket out twice and expand all four terms so the middle terms are not lost.

+ Add to test✓ In your test205IGME-058-04-03Surds Paper HEasy4 marks

Rationalise the denominator of each fraction, giving your answer in its simplest form.

(a)

\(\frac{6}{\sqrt{3}}\)

(2)

(b)

\(\frac{15}{2\sqrt{5}}\)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for multiplying by \(\frac{\sqrt{3}}{\sqrt{3}}\) to get \(\frac{6\sqrt{3}}{3}\) ; A1 \(2\sqrt{3}\)	✓ 2
Part (b)
M1 for \(\frac{15\sqrt{5}}{2\times5}\) or \(\frac{15\sqrt{5}}{10}\) ; A1 \(\frac{3\sqrt{5}}{2}\)	✓ 2

Exam tip: Multiply top and bottom by the surd only, not by the whole denominator: in \(\frac{15}{2\sqrt{5}}\) multiply by \(\frac{\sqrt{5}}{\sqrt{5}}\), giving a denominator of \(2\times5=10\).

+ Add to test✓ In your test206IGME-058-04-04Surds Paper HEasy3 marks

Show your working in each part.

(a)

Write \(\sqrt{12}\times\sqrt{6}\) in the form \(a\sqrt{2}\) where \(a\) is an integer.

(2)

(b)

Work out the value of \(\frac{\sqrt{300}}{\sqrt{3}}\).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\sqrt{72}\) or \(2\sqrt{3}\times\sqrt{6}=2\sqrt{18}\) ; A1 \(6\sqrt{2}\)	✓ 2
Part (b)
B1 \(10\)	✓ 1

Exam tip: Use \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\) and \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) to combine the surds first, then simplify.

+ Add to test✓ In your test207IGME-058-04-05Surds Paper HEasy3 marks

A square garden tile has an area of 45 cm².

Diagram not to scale

(a)

Find the side length of the tile. Give your answer in the form \(a\sqrt{b}\) cm, where \(a\) and \(b\) are integers and \(b\) is prime.

(2)

(b)

Find the perimeter of the tile in the form \(k\sqrt{5}\) cm.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\sqrt{45}=\sqrt{9}\times\sqrt{5}\) or \(\sqrt{9\times5}\) ; A1 \(3\sqrt{5}\) cm	✓ 2
Part (b)
B1 \(12\sqrt{5}\) cm (follow through 4 × their answer to (a) if it is in the form \(a\sqrt{5}\))	✓ 1

Common error: Do not give a decimal such as 6.71 cm when a surd form is asked for. The exact form is needed for the mark.

+ Add to test✓ In your test208IGME3-1-13r28-07Surds Paper HEasy4 marks

A rectangular solar panel on a probe has length (3 + √5) m and width (3 − √5) m.

(a)

Show that the area of the panel is 4 m².

(2)

(b)

Work out the perimeter of the panel.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (3 + √5)(3 − √5) expanded: 9 − 3√5 + 3√5 − 5	✓ 1
A1 9 − 5 = 4 m² with no errors seen	✓ 1
Part (b)
M1 2 × ((3 + √5) + (3 − √5))	✓ 1
A1 12 m	✓ 1

Exam tip: Expanding a difference of two squares gives no surd term, so the answer here is rational.

+ Add to test✓ In your test209IGME-058-01-09Powers, roots and standard form Paper HMedium5 marks

Solving equations with powers.

(a)

Solve \(2^{x+3} = 8^x\)

(3)

(b)

Solve \(9^y = \frac{1}{27}\)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for writing 8 as \(2^3\), giving \(2^{3x}\)	✓ 1
M1 for \(x + 3 = 3x\)	✓ 1
A1 for \(x = 1.5\)	✓ 1
Part (b)
M1 for \(3^{2y} = 3^{-3}\)	✓ 1
A1 for \(y = -\frac{3}{2}\)	✓ 1

Exam tip: Write both sides as powers of the same base before comparing indices; 9 and 27 are both powers of 3.

+ Add to test✓ In your test210IGME-058-01-11Powers, roots and standard form Paper HMedium4 marks

Simplifying with fractional indices.

(a)

Simplify fully \(\left(27x^6y^{-3}\right)^{\frac{2}{3}}\)

(3)

(b)

Hence find the value of \(\left(27x^6y^{-3}\right)^{\frac{2}{3}}\) when \(x = 2\) and \(y = 3\).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 9	✓ 1
B1 for \(x^4\)	✓ 1
B1 for \(y^{-2}\) or \(\frac{1}{y^2}\), giving \(\frac{9x^4}{y^2}\)	✓ 1
Part (b)
B1 for 16	✓ 1

Common error: The index \(\frac{2}{3}\) applies to the number too: \(27^{\frac{2}{3}} = 9\), not 18.

+ Add to test✓ In your test211IGME-058-00-09Integers, fractions, decimals and percentages Paper HMedium5 marks

In this question, use an algebraic method for each part.

(a)

Show that \(0.\dot{2}\dot{7} = \frac{3}{11}\)

(2)

(b)

Write \(0.4\dot{3}\dot{6}\) as a fraction in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x = 0.2727… and 100x = 27.2727…	✓ 1
A1 99x = 27 so x = \(\frac{27}{99} = \frac{3}{11}\)	✓ 1
Part (b)
M1 two recurring decimals whose difference is a terminating decimal, e.g. 10x = 4.3636… and 1000x = 436.3636…	✓ 1
M1 990x = 432 or \(\frac{432}{990}\)	✓ 1
A1 \(\frac{24}{55}\)	✓ 1

Exam tip: Choose two multiples of x so that the recurring parts line up exactly after the decimal point; subtracting then removes the recurring part.

+ Add to test✓ In your test212IGME-058-04-06Surds Paper HMedium6 marks

Show your working clearly in each part.

(a)

Rationalise the denominator and simplify \(\frac{4}{3-\sqrt{5}}\).

(3)

(b)

Show that \(\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{3+\sqrt{5}}{2}\).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for multiplying numerator and denominator by \(3+\sqrt{5}\) ; M1 for denominator \(9-5=4\) ; A1 \(3+\sqrt{5}\)	✓ 3
Part (b)
M1 for multiplying numerator and denominator by \(\sqrt{5}+1\) ; M1 for numerator \(5+2\sqrt{5}+1=6+2\sqrt{5}\) and denominator \(4\) ; A1 for \(\frac{6+2\sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}\) fully shown	✓ 3

Exam tip: For a denominator \(a-\sqrt{b}\), multiply by \(a+\sqrt{b}\). The difference of two squares removes the surd: \((a-\sqrt{b})(a+\sqrt{b})=a^2-b\).

+ Add to test✓ In your test213IGME-058-04-07Surds Paper HMedium5 marks

A rectangular solar panel has length \((5+\sqrt{3})\) m and width \((5-\sqrt{3})\) m.

Diagram not to scale

(a)

Show that the area of the panel is an integer number of square metres, and state its value.

(2)

(b)

Find the length of a diagonal of the panel. Give your answer in the form \(a\sqrt{b}\) m, where \(a\) and \(b\) are integers and \(b\) is as small as possible.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(25-5\sqrt{3}+5\sqrt{3}-3\) ; A1 \(22\) m²	✓ 2
Part (b)
M1 for \((5+\sqrt{3})^2+(5-\sqrt{3})^2\) ; M1 for \(28+10\sqrt{3}\) and \(28-10\sqrt{3}\), giving \(56\) ; A1 \(2\sqrt{14}\) m	✓ 3

Common error: \(\sqrt{56}\) is not fully simplified: \(56=4\times14\), so \(\sqrt{56}=2\sqrt{14}\).

+ Add to test✓ In your test214IGME-058-04-08Surds Paper HMedium4 marks

Show your working in each part.

(a)

Write \(\sqrt{8}+\sqrt{32}-\sqrt{18}\) in the form \(k\sqrt{2}\) where \(k\) is an integer.

(2)

(b)

Given that \(\sqrt{8}+\sqrt{32}-\sqrt{18}=\sqrt{n}\), find the value of \(n\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for at least two of \(2\sqrt{2}\), \(4\sqrt{2}\), \(3\sqrt{2}\) ; A1 \(3\sqrt{2}\)	✓ 2
Part (b)
M1 for \(3\sqrt{2}=\sqrt{9\times2}\) or \((3\sqrt{2})^2\) ; A1 \(18\)	✓ 2

Common error: \(\sqrt{8}+\sqrt{32}-\sqrt{18}\) is not \(\sqrt{8+32-18}=\sqrt{22}\). Surds can only be added once they are written as multiples of the same root.

+ Add to test✓ In your test215IGME-058-04-10Surds Paper HMedium4 marks

Solve the equation below.

(a)

Solve \(\sqrt{3}\,x-2=x+4\). Give your answer in the form \(a+b\sqrt{3}\) where \(a\) and \(b\) are integers.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\sqrt{3}\,x-x=6\) ; M1 for \(x(\sqrt{3}-1)=6\) so \(x=\frac{6}{\sqrt{3}-1}\) ; M1 for multiplying by \(\frac{\sqrt{3}+1}{\sqrt{3}+1}\) to get \(\frac{6(\sqrt{3}+1)}{2}\) ; A1 \(3+3\sqrt{3}\)	✓ 4

Exam tip: Collect the \(x\) terms on one side and factorise out \(x\). Dividing then gives a surd denominator, which must be rationalised to reach the required form.

+ Add to test✓ In your test216IGME-058-04-11Surds Paper HMedium4 marks

Show your working in each part.

(a)

Write \((\sqrt{3})^5\) in the form \(a\sqrt{3}\) where \(a\) is an integer.

(1)

(b)

Write \(\frac{1}{\sqrt{3}}+\sqrt{27}\) in the form \(\frac{k\sqrt{3}}{3}\) where \(k\) is an integer.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(9\sqrt{3}\)	✓ 1
Part (b)
M1 for \(\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\) ; M1 for \(\sqrt{27}=3\sqrt{3}\) or \(\frac{9\sqrt{3}}{3}\) ; A1 \(\frac{10\sqrt{3}}{3}\)	✓ 3

Exam tip: Rationalise first, then write both terms over the same denominator of 3 before adding.

+ Add to test✓ In your test217IGME-058-02-10Ratio, proportion and degree of accuracy Paper HMedium4 marks

A rectangle has length 12.4 cm correct to 1 decimal place.
Its width is 7 cm correct to the nearest centimetre.

Diagram not to scale

(a)

Write down the error interval for the width, w cm.

(2)

(b)

Work out the upper bound for the area of the rectangle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 6.5 ≤ w	✓ 1
B1 for w < 7.5	✓ 1
Part (b)
M1 for 12.45 × 7.5	✓ 1
A1 for 93.375 cm²	✓ 1

Common error: The upper bound of 12.4 to 1 decimal place is 12.45, not 12.5 – halve the degree of accuracy.

+ Add to test✓ In your test218IGME-058-01-07Powers, roots and standard form Paper HMedium5 marks

Work out the exact value of each of the following. Do not use a calculator.

(a)

Find the value of \(81^{\frac{1}{4}}\)

(1)

(b)

Find the value of \(27^{-\frac{2}{3}}\)
Give your answer as a fraction.

(2)

(c)

Find the value of \(\left(\frac{16}{25}\right)^{-\frac{3}{2}}\)
Give your answer as a fraction.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 3	✓ 1
Part (b)
M1 for \(\frac{1}{27^{2/3}}\) seen, or \(27^{\frac{2}{3}} = 9\) seen, or \((\sqrt[3]{27})^2\) seen	✓ 1
A1 for \(\frac{1}{9}\)	✓ 1
Part (c)
M1 for \(\left(\frac{25}{16}\right)^{\frac{3}{2}}\) or \(\left(\frac{5}{4}\right)^3\) seen, or \(\left(\frac{16}{25}\right)^{\frac{3}{2}} = \frac{64}{125}\) seen	✓ 1
A1 for \(\frac{125}{64}\)	✓ 1

Exam tip: Deal with each part of the index in turn. The negative sign flips the fraction, the denominator gives the root and the numerator gives the power. Taking the root first keeps the numbers small.

+ Add to test✓ In your test219IGME-058-03-08Set language, notation and Venn diagrams (both tiers) Paper HMedium7 marks

A gym has 120 members.
70 members use the swimming pool (S).
55 members use the weights room (W).
12 members use neither.
x members use both.

(a)

Find the value of x.

(2)

(b)

A member who uses the weights room is chosen at random. Find the probability that this member also uses the swimming pool.

(2)

(c)

Two different members of the gym are chosen at random. Find the probability that both of these members use neither the swimming pool nor the weights room.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 70 + 55 − x + 12 = 120, or for 70 + 55 − (120 − 12) . A1 for x = 17 .	✓ 2
Part (b)
M1 for a denominator of 55 . A1 for \(\frac{17}{55}\) .	✓ 2
Part (c)
M1 for \(\frac{12}{120}\) . M1 for \(\frac{12}{120} \times \frac{11}{119}\) . A1 for \(\frac{11}{1190}\) or 0.00924 .	✓ 3

Common error: 'Given they use the weights room' restricts you to the W circle, so divide by 55, not by 120. When two different members are chosen, the second choice is made from the 119 people who are left, and only 11 of them use neither facility.

+ Add to test✓ In your test220IGME-058-04-09Surds Paper HMedium5 marks

Give each answer in the form \(a+b\sqrt{3}\) where \(a\) and \(b\) are integers.

(a)

Expand and simplify \((2\sqrt{3}-1)(\sqrt{3}+4)\).

(3)

(b)

Expand and simplify \((2\sqrt{3}-1)^2\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Four terms from the expansion, with at least three correct ; fully correct expansion \(6+8\sqrt{3}-\sqrt{3}-4\) ; \(2+7\sqrt{3}\)	✓ 3
Part (b)
Correct expansion \(12-2\sqrt{3}-2\sqrt{3}+1\) or equivalent ; \(13-4\sqrt{3}\)	✓ 2

Common error: \((2\sqrt{3})^2=4\times3=12\), not \(2\times3=6\). Square both the number and the surd.

+ Add to test✓ In your test221IGME3-1-07d-06Integers, fractions, decimals and percentages Paper HMedium5 marks

A solar panel installation costs €2550 after a discount of 15%.

(a)

Work out the original price of the installation. (2)

(2)

(b)

The recorded value of the installation falls by 12% each year. Starting from the original price of €3000, work out its value after 3 years. Give your answer to the nearest cent. (3)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2550 ÷ 0.85	✓ 1
A1 €3000	✓ 1
Part (b)
M1 3000 × 0.88	✓ 1
M1 3000 × 0.88³ (= 2044.416)	✓ 1
A1 €2044.42	✓ 1

Common error: Do not divide by 0.15 for a reverse percentage: the sale price represents 85% of the original, so divide by 0.85.

+ Add to test✓ In your test222IGME3-1-07d-12Surds Paper HMedium5 marks

A rectangular panel has length (3 + √5) m and width (3 − √5) m.

(a)

Show that the area of the panel is 4 m². (2)

(2)

(b)

A strip of length (6 + √20) m is measured in lengths of √5 m. Show that the number of lengths is \(2 + \frac{6}{5}\sqrt{5}\). (3)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (3 + √5)(3 − √5) = 9 − 3√5 + 3√5 − 5	✓ 1
A1 4 shown	✓ 1
Part (b)
B1 √20 = 2√5	✓ 1
M1 \(\frac{6 + 2\sqrt{5}}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}\)	✓ 1
A1 \(\frac{6\sqrt{5} + 10}{5} = 2 + \frac{6}{5}\sqrt{5}\) (= 4.68 to 3 s.f.)	✓ 1

Exam tip: Simplify surds such as √20 first, then rationalise by multiplying the top and bottom by the surd in the denominator.

+ Add to test✓ In your test223IGME3-1-07d-13Ratio, proportion and degree of accuracy Paper HMedium7 marks

A solar panel has a length of 1.65 m and a width of 0.99 m. Each measurement is correct to the nearest centimetre.

(a)

Write down the lower bound and the upper bound of the length. (2)

(2)

(b)

Work out the upper bound of the area of the panel. (2)

(2)

(c)

Work out the lower bound of the area of the panel. (2)

(2)

(d)

Use your answers to explain why the area can be given as 1.6 m² correct to 2 significant figures. (1)

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 1.645	✓ 1
B1 1.655	✓ 1
Part (b)
M1 1.655 × 0.995	✓ 1
A1 1.646725 m²	✓ 1
Part (c)
M1 1.645 × 0.985	✓ 1
A1 1.620325 m²	✓ 1
Part (d)
B1 both bounds round to 1.6 to 2 significant figures	✓ 1

Exam tip: For an area, use the lower bounds of both lengths for the lower bound and the upper bounds of both for the upper bound.

+ Add to test✓ In your test224IGME3-1-11r7-06Ratio, proportion and degree of accuracy Paper HMedium5 marks

A lap of a running track is 400 m, correct to the nearest 10 m. A runner completes one lap in 52 s, correct to the nearest second.

(a)

Write down the lower bound and the upper bound of the length of the lap.

(2)

(b)

Work out the upper bound of the runner's average speed.
Give your answer in m/s correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 lower bound 395 m	✓ 1
B1 upper bound 405 m	✓ 1
Part (b)
B1 lower bound of time = 51.5 s	✓ 1
M1 405 ÷ 51.5	✓ 1
A1 7.86 m/s	✓ 1

Common error: For the greatest speed, divide the largest distance by the smallest time. Using two upper bounds gives a value that is too small.

+ Add to test✓ In your test225IGME3-1-11r7-07Surds Paper HMedium5 marks

A rectangular solar panel on a satellite has length (5 + √3) m and width (3 − √3) m.

Diagram not to scale

(a)

Show that the area of the panel is (12 − 2√3) m².

(2)

(b)

Work out the length of the panel divided by its width.
Give your answer in the form \(\frac{a+b\sqrt{3}}{c}\) where a, b and c are integers.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expanding: 15 − 5√3 + 3√3 − 3	✓ 1
A1 12 − 2√3	✓ 1
Part (b)
M1 multiplying numerator and denominator by (3 + √3)	✓ 1
A1 numerator 18 + 8√3 and denominator 6	✓ 1
A1 \(\frac{9+4\sqrt{3}}{3}\) (≈ 5.309)	✓ 1

Exam tip: To rationalise a denominator of the form a − √b, multiply top and bottom by a + √b so that the denominator becomes the integer a² − b.

+ Add to test✓ In your test226IGME3-1-11r7-13Powers, roots and standard form Paper HMedium6 marks

A rail line goes √18 km due east from a station to a signal box and then √32 km due north to a junction. The two sections of track are at right angles.

Diagram not to scale

(a)

Write √18 and √32 in the form a√2, where a is an integer.

(2)

(b)

Show that the direct distance from the station to the junction is 5√2 km.

(2)

(c)

Work out how much shorter the direct distance is than the distance along the track.
Give your answer in the form a√2.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 √18 = 3√2	✓ 1
B1 √32 = 4√2	✓ 1
Part (b)
M1 18 + 32 = 50 (Pythagoras)	✓ 1
A1 √50 = 5√2	✓ 1
Part (c)
M1 7√2 − 5√2 (ECF from (a))	✓ 1
A1 2√2 km (≈ 2.828 km)	✓ 1

Common error: A common error is √18 + √32 = √50. Surds can be added only after each has been simplified to the same surd, here 3√2 + 4√2 = 7√2.

+ Add to test✓ In your test227IGME3-1-12r13-06Ratio, proportion and degree of accuracy Paper HMedium4 marks

A solar collector is a rectangle whose length is 2.4 m and width is 1.2 m. Each measurement is correct to the nearest 0.1 m.

(a)

Write down the lower bound and the upper bound of the length.

(2)

(b)

Work out the upper bound of the area of the collector. Give your answer exactly.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 lower bound 2.35 m	✓ 1
B1 upper bound 2.45 m	✓ 1
Part (b)
B1 upper bounds 2.45 and 1.25 used	✓ 1
A1 3.0625 (m²)	✓ 1

Common error: For the greatest possible area of a rectangle, multiply the two upper bounds; do not mix a lower bound with an upper bound.

+ Add to test✓ In your test228IGME3-1-12r13-09Surds Paper HMedium6 marks

A rectangular solar array has length (4 + √3) m and width (4 − √3) m.

Diagram not to scale

(a)

Show that the area of the array is 13 m².

(2)

(b)

Work out the perimeter of the array.

(2)

(c)

Rationalise the denominator of 12/√3. Give your answer in the form a√3, where a is an integer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expands (4 + √3)(4 − √3) = 16 − 4√3 + 4√3 − 3	✓ 1
A1 13	✓ 1
Part (b)
M1 2(4 + √3 + 4 − √3)	✓ 1
A1 16 (m)	✓ 1
Part (c)
M1 multiplies numerator and denominator by √3 (12√3/3)	✓ 1
A1 4√3 (≈ 6.928203)	✓ 1

Exam tip: Brackets of the form (a + √b)(a − √b) always give a rational result, a² − b.

+ Add to test✓ In your test229IGME3-1-12r13-11Ratio, proportion and degree of accuracy Paper HMedium5 marks

In a sample of people, the proportion who are infected is 0.2777… as a decimal, where the digit 7 recurs forever.

(a)

Show that this recurring decimal is equal to 5/18.

(3)

(b)

The sample contains 1260 people. Work out how many are infected.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 lets x = 0.2777…, so 10x = 2.777… and 100x = 27.777…	✓ 1
M1 subtracts: 100x − 10x = 27.777… − 2.777… so 90x = 25	✓ 1
A1 x = 25/90 = 5/18	✓ 1
Part (b)
M1 1260 × 5/18	✓ 1
A1 350	✓ 1

Exam tip: Multiply by a power of ten chosen so that the recurring parts line up exactly, then subtract to remove them.

+ Add to test✓ In your test230IGME3-1-13r14-06Integers, fractions, decimals and percentages Paper HMedium5 marks

A fashion shop reduces all its prices by 15% in a sale.

(a)

In the sale a jacket costs £63.75. Work out the price of the jacket before the sale.

(2)

(b)

The shop's sewing machine cost £1450 when new. Its value falls by 12% each year. Work out its value after 3 years. Give your answer to the nearest penny.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 63.75 ÷ 0.85	✓ 1
A1 £75	✓ 1
Part (b)
M1 multiplier 0.88 used	✓ 1
M1 1450 × 0.88³	✓ 1
A1 £988.13	✓ 1

Common error: A 15% reduction means the sale price is 85% of the original, so divide by 0.85. Multiplying the sale price by 1.15 gives the wrong answer.

+ Add to test✓ In your test231IGME3-1-13r14-07Surds Paper HMedium5 marks

A rectangular panel of fabric has length (5 − √3) metres and width (2 + √3) metres.

(a)

Show that the area of the panel is (7 + 3√3) m².

(2)

(b)

A length of ribbon is \(\frac{21}{3+\sqrt{2}}\) metres. Write this length in the form a + b√2, where a and b are integers.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expansion with at least 3 correct terms, e.g. 10 + 5√3 − 2√3 − 3	✓ 1
A1 collects terms to reach 7 + 3√3 with 10 − 3 = 7 and 5√3 − 2√3 = 3√3 seen	✓ 1
Part (b)
M1 multiply numerator and denominator by (3 − √2)	✓ 1
A1 denominator simplified to 7 (numerator 63 − 21√2)	✓ 1
A1 9 − 3√2 (≈ 4.7574)	✓ 1

Exam tip: To rationalise a denominator of the form a + √b, multiply top and bottom by a − √b so the surd disappears from the denominator.

+ Add to test✓ In your test232IGME3-1-13r14-11Powers, roots and standard form Paper HMedium5 marks

A robot moves in a straight line in two stages. In the first stage it travels √48 m and in the second stage it travels √27 m. Its speed throughout is √3 m/s.

(a)

Write the total distance travelled in the form k√3, where k is an integer.

(3)

(b)

Work out the total time taken for the robot to travel this distance.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √48 = 4√3	✓ 1
M1 √27 = 3√3	✓ 1
A1 7√3 m	✓ 1
Part (b)
M1 7√3 ÷ √3 (ECF from (a))	✓ 1
A1 7 s	✓ 1

Exam tip: Write each surd as a multiple of the same simplest surd before adding.

+ Add to test✓ In your test233IGME3-1-13r16-01Surds Paper HMedium4 marks

A surveyor on a volcano measures two distances along a crater rim: √48 m and √27 m. Show your working.

(a)

Write √48 in the form a√3, where a is an integer.

(2)

(b)

Hence, or otherwise, write √48 + √27 in the form b√3, where b is an integer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √48 = √16 × √3	✓ 1
A1 4√3	✓ 1
Part (b)
B1 √27 = 3√3	✓ 1
A1 7√3	✓ 1

Exam tip: Look for the largest square number that divides the number under the root before simplifying.

+ Add to test✓ In your test234IGME3-1-13r16-02Surds Paper HMedium6 marks

The rectangular stage of a small theatre has length (4 + √5) m and width (4 − √5) m.

Diagram not to scale

(a)

Show that the area of the stage is 11 m².

(2)

(b)

Work out the perimeter of the stage.

(1)

(c)

Work out the length of a diagonal of the stage. Give your answer in metres to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (4 + √5)(4 − √5) = 16 − 4√5 + 4√5 − 5 (all four terms)	✓ 1
A1 16 − 5 = 11	✓ 1
Part (b)
B1 2 × [(4 + √5) + (4 − √5)] = 16 m	✓ 1
Part (c)
M1 diagonal² = (4 + √5)² + (4 − √5)²	✓ 1
A1 21 + 8√5 + 21 − 8√5 = 42	✓ 1
A1 √42 = 6.48 m	✓ 1

Common error: When expanding (4 + √5)², do not forget the middle term 8√5; it only cancels once both squares are added.

+ Add to test✓ In your test235IGME3-1-13r16-07Powers, roots and standard form Paper HMedium6 marks

Indices and roots.

(a)

Simplify (16x⁸)^3/4.

(2)

(b)

Work out the value of 27^−2/3.

(2)

(c)

Solve 4^x = 32.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 16^3/4 = 8 or (x⁸)^3/4 = x⁶	✓ 1
A1 8x⁶	✓ 1
Part (b)
M1 27^1/3 = 3 (or reciprocal of 27^2/3)	✓ 1
A1 1/9	✓ 1
Part (c)
M1 2^2x = 2⁵	✓ 1
A1 x = 2.5	✓ 1

Exam tip: Writing both sides with the same base turns an exponential equation into a simple linear one.

+ Add to test✓ In your test236IGME3-1-12r19-08Surds Paper HMedium5 marks

Give exact answers, without using decimals, in this question.

(a)

Write √48 in the form a√3.

(2)

(b)

Show that (√48 + √12) ÷ √3 = 6.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √16 × √3	✓ 1
A1 4√3	✓ 1
Part (b)
M1 √12 = 2√3	✓ 1
M1 4√3 + 2√3 = 6√3	✓ 1
A1 6√3 ÷ √3 = 6	✓ 1

Exam tip: Look for the largest square factor when simplifying a surd.

+ Add to test✓ In your test237IGME3-1-12r19-09Surds Paper HMedium7 marks

In this question, show all of your working with surds.

(a)

Show that 12/√3 = 4√3.

(2)

(b)

Expand and simplify (3 + √5)(3 − √5).

(2)

(c)

Use your answer to (b) to show that 6/(3 + √5) + 6/(3 − √5) = 9.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiplies numerator and denominator by √3	✓ 1
A1 12√3 ÷ 3 = 4√3	✓ 1
Part (b)
M1 9 − 3√5 + 3√5 − 5	✓ 1
A1 4	✓ 1
Part (c)
M1 common denominator (3 + √5)(3 − √5) = 4	✓ 1
M1 numerator 6(3 − √5) + 6(3 + √5) = 36	✓ 1
A1 36 ÷ 4 = 9	✓ 1

Common error: Rationalising uses the conjugate: the middle terms cancel, leaving a rational denominator.

+ Add to test✓ In your test238IGME3-1-11r25-06Integers, fractions, decimals and percentages Paper HMedium6 marks

A telecoms company has 85 000 subscribers. The number of subscribers increases by 4% each year.

(a)

Work out the number of subscribers after 3 years. Give your answer correct to the nearest hundred.

(3)

(b)

Find the least whole number of years after which the company has more than 100 000 subscribers. You must show your working.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.04³ or 1.124864	✓ 1
M1 85 000 × 1.04³ (= 95613.44)	✓ 1
A1 95 600	✓ 1
Part (b)
M1 evaluates 85 000 × 1.04ⁿ for at least one value of n	✓ 1
A1 n = 4 gives 99 438 (less than 100 000) and n = 5 gives 103 415 (more than 100 000)	✓ 1
A1 5 years	✓ 1

Exam tip: For repeated percentage change, use a multiplier raised to a power rather than adding the percentage each year.

+ Add to test✓ In your test239IGME3-1-11r25-07Surds Paper HMedium6 marks

A rectangular antenna panel has length (4 + √2) cm and width (4 − √2) cm.

Diagram not to scale

(a)

Show that the area of the panel is 14 cm².

(2)

(b)

Work out the perimeter of the panel.

(2)

(c)

A cable has diameter 12 ÷ √3 mm. Write this in the form a√3, where a is an integer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (4 + √2)(4 − √2) expanded, for example 16 − 4√2 + 4√2 − 2	✓ 1
A1 16 − 2 = 14 (shown)	✓ 1
Part (b)
M1 2 × ((4 + √2) + (4 − √2))	✓ 1
A1 16 cm	✓ 1
Part (c)
M1 multiplies numerator and denominator by √3	✓ 1
A1 4√3	✓ 1

Exam tip: Terms such as (a + √b)(a − √b) multiply to a whole number, because the surd terms cancel.

+ Add to test✓ In your test240IGME3-1-11r25-10Ratio, proportion and degree of accuracy Paper HMedium6 marks

A climber abseils down a rope. The length of the rope is 60 m, correct to the nearest metre. The climber takes 24 seconds, correct to the nearest second.

(a)

Write down the lower bound and the upper bound of the length of the rope.

(2)

(b)

Write down the lower bound of the time taken.

(1)

(c)

Work out the upper bound of the average speed of the climber. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 59.5 m	✓ 1
B1 60.5 m	✓ 1
Part (b)
B1 23.5 s	✓ 1
Part (c)
M1 upper bound of length ÷ lower bound of time	✓ 1
A1 60.5 ÷ 23.5 = 2.5744...	✓ 1
A1 2.57 m/s	✓ 1

Exam tip: To make a speed as large as possible, divide the largest possible distance by the smallest possible time.

+ Add to test✓ In your test241IGME3-1-12r25-06Ratio, proportion and degree of accuracy Paper HMedium7 marks

The length of a swimming pool is 25.0 m, correct to 3 significant figures. A swimmer completes 4 lengths in 62.4 s, correct to 3 significant figures.

(a)

Write down the lower bound and the upper bound of the length of the pool.

(2)

(b)

Work out the lower bound of the swimmer's average speed. Give your answer to 3 decimal places.

(3)

(c)

The upper bound of the average speed is 1.607 m/s, to 3 decimal places. Write down the average speed to a suitable degree of accuracy. Justify your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 24.95 m	✓ 1
B1 25.05 m	✓ 1
Part (b)
B1 upper bound of the time is 62.45 s	✓ 1
M1 (4 × 24.95) ÷ 62.45 = 99.8 ÷ 62.45	✓ 1
A1 1.598 m/s	✓ 1
Part (c)
B1 both bounds round to 1.6 to 2 significant figures	✓ 1
B1 1.6 m/s	✓ 1

Exam tip: Check whether the upper and lower bounds round to the same value; the degree of accuracy you can quote is the one on which they agree.

+ Add to test✓ In your test242IGME3-1-12r25-07Powers, roots and standard form Paper HMedium7 marks

A wind farm has 2⁵ turbines at each of 2³ sites.

(a)

Write the total number of turbines as a single power of 2, and then as an ordinary number.

(2)

(b)

Work out the value of 16^3/4.

(2)

(c)

Solve 4^x = ¹⁄₃₂.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2⁸	✓ 1
B1 256	✓ 1
Part (b)
M1 (16^1/4)³ or 2³	✓ 1
A1 8	✓ 1
Part (c)
M1 writes both sides as powers of 2: 2^2x = 2⁻⁵	✓ 1
M1 2x = −5	✓ 1
A1 x = −⁵⁄₂ or −2.5	✓ 1

Common error: For a fractional index such as 3/4, the denominator is the root and the numerator is the power. A negative power gives a reciprocal, not a negative number.

+ Add to test✓ In your test243IGME3-1-12r25-13Powers, roots and standard form Paper HMedium6 marks

A wind turbine is bought for £2 400 000. Its value falls by 8% each year.

(a)

Work out the value of the turbine after 5 years. Give your answer correct to 3 significant figures.

(3)

(b)

Find the number of whole years after which the turbine is first worth less than half of its original value. You must show your working.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 100% − 8% = 92% or 0.92	✓ 1
M1 2 400 000 × 0.92⁵ (= 1 581 796)	✓ 1
A1 £1 580 000	✓ 1
Part (b)
M1 half of the original value is £1 200 000 and evaluates 2 400 000 × 0.92ⁿ for at least one value of n	✓ 1
A1 n = 8 gives 1 231 725 (more than £1 200 000) and n = 9 gives 1 133 187 (less than £1 200 000)	✓ 1
A1 9 years	✓ 1

Exam tip: For repeated percentage decrease, the multiplier is 1 − (percentage as a decimal) raised to the number of years.

+ Add to test✓ In your test244IGME3-1-13r28-08Surds Paper HMedium5 marks

A bridge designer works with cable lengths written using surds.

(a)

Show that ⁶⁄_{3 − √3} = 3 + √3

(3)

(b)

Hence write ⁶⁄_{3 − √3} + √12 in the form a + b√3, where a and b are integers.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiply numerator and denominator by (3 + √3)	✓ 1
M1 denominator (3 − √3)(3 + √3) = 9 − 3 = 6	✓ 1
A1 ^{6(3 + √3)}⁄₆ = 3 + √3 shown	✓ 1
Part (b)
M1 √12 = 2√3	✓ 1
A1 3 + 3√3	✓ 1

Exam tip: To rationalise a denominator of the form a − √b, multiply top and bottom by a + √b.

+ Add to test✓ In your test245IGME3-1-13r28-10Powers, roots and standard form Paper HMedium5 marks

Work through the following index questions.

(a)

Work out the value of 27^−2/3. Give your answer as a fraction.

(2)

(b)

Solve 4^x = ¹⁄₃₂

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cube root of 27 is 3 (or reciprocal for the negative power)	✓ 1
A1 ¹⁄₉	✓ 1
Part (b)
M1 write 4^x as 2^2x	✓ 1
M1 ¹⁄₃₂ = 2⁻⁵ so 2x = −5	✓ 1
A1 x = −2.5	✓ 1

Exam tip: A negative power means a reciprocal; a fractional power means a root, so 27⁻²/³ needs both ideas.

+ Add to test✓ In your test246IGME3-1-13r28-14Set language, notation and Venn diagrams (both tiers) Paper HMedium8 marks

The universal set ξ is the set of integers from 1 to 30 inclusive. A is the set of multiples of 3. B is the set of factors of 24.

(a)

List the elements of A ∩ B.

(2)

(b)

Work out n((A ∪ B)′).

(2)

(c)

A number is chosen at random from ξ. Find the probability that it is in B but not in A.

(2)

(d)

A number is chosen at random from A. Find the probability that it is also in B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 factors of 24: 1, 2, 3, 4, 6, 8, 12, 24 (at least 5 correct)	✓ 1
A1 {3, 6, 12, 24}	✓ 1
Part (b)
M1 30 − (10 + 8 − 4)	✓ 1
A1 16	✓ 1
Part (c)
M1 4 numbers (1, 2, 4, 8) out of 30	✓ 1
A1 ²⁄₁₅	✓ 1
Part (d)
M1 4 out of 10	✓ 1
A1 ²⁄₅	✓ 1

Common error: Count elements of a union as n(A) + n(B) − n(A ∩ B); adding n(A) and n(B) alone counts the overlap twice.

+ Add to test✓ In your test247IGME3-1-13r29-08Set language, notation and Venn diagrams (both tiers) Paper HMedium4 marks

A runner increases her weekly training distance by 15%. Her new weekly distance is 69 km. A sports club has 8000 members. The number of members falls by 10% each year for 3 years.

(a)

Work out her original weekly distance.

(2)

(b)

Work out the number of members of the club after 3 years.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(69\div1.15\)	✓ 1
A1 60 km	✓ 1
Part (b)
M1 \(8000\times0.9^{3}\)	✓ 1
A1 5832	✓ 1

Common error: In a reverse percentage problem, dividing by the multiplier is correct; taking 15% of 69 and subtracting gives the wrong answer.

+ Add to test✓ In your test248IGME3-1-13r29-09Surds Paper HMedium6 marks

This question is about surds.

(a)

Show that \(\sqrt{48}+\sqrt{12}=6\sqrt{3}\).

(2)

(b)

Rationalise the denominator of \(\dfrac{14}{\sqrt{7}}\). Give your answer in the form \(a\sqrt{7}\).

(2)

(c)

A rectangular railway signal board measures \((5+\sqrt{2})\) m by \((5-\sqrt{2})\) m. Work out its area. Give your answer as an integer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{48}=4\sqrt{3}\) or \(\sqrt{12}=2\sqrt{3}\)	✓ 1
A1 \(4\sqrt{3}+2\sqrt{3}=6\sqrt{3}\)	✓ 1
Part (b)
M1 multiply numerator and denominator by \(\sqrt{7}\)	✓ 1
A1 \(2\sqrt{7}\)	✓ 1
Part (c)
M1 \((5+\sqrt{2})(5-\sqrt{2})\) expanded or difference of two squares	✓ 1
A1 \(25-2=23\) m\(^{2}\)	✓ 1

Exam tip: Pairs of brackets of the form (a + b)(a − b) lose their surd terms; write the expansion out to show it.

+ Add to test✓ In your test249IGME3-1-14r33-06Powers, roots and standard form Paper HMedium6 marks

Use the laws of indices to answer the following.

(a)

Work out the value of 27^2/3.

(2)

(b)

Work out the value of 16^−3/4.

(2)

(c)

Solve 4^x = 32.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (³√27)² or 3²	✓ 1
A1 9	✓ 1
Part (b)
M1 1 ÷ 16^3/4 or (⁴√16)⁻³ = 2⁻³	✓ 1
A1 1/8 (or 0.125)	✓ 1
Part (c)
M1 write both sides as powers of 2: 2^2x = 2⁵	✓ 1
A1 x = 5/2 (or 2.5)	✓ 1

Exam tip: Take the root first, then the power; it keeps the numbers small.

+ Add to test✓ In your test250IGME3-1-14r33-09Surds Paper HMedium6 marks

Surds are written in exact form.

(a)

Write √75 in the form a√3, where a is an integer.

(1)

(b)

Expand and simplify (2 + √3)(5 − √3).

(2)

(c)

Show that 12/(3 − √3) can be written as 6 + 2√3.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 5√3	✓ 1
Part (b)
M1 10 − 2√3 + 5√3 − 3 (at least three terms correct)	✓ 1
A1 7 + 3√3	✓ 1
Part (c)
M1 multiply numerator and denominator by (3 + √3)	✓ 1
M1 denominator becomes 9 − 3 = 6 and numerator 12(3 + √3)	✓ 1
A1 12(3 + √3)/6 = 2(3 + √3) = 6 + 2√3	✓ 1

Common error: To rationalise a denominator of the form a − √b, multiply by a + √b (the conjugate), not by √b.

+ Add to test✓ In your test251IGME3-1-14r33-13Surds Paper HMedium5 marks

A tent panel is a right-angled triangle. The two shorter sides have lengths (√5 + 1) m and (√5 − 1) m.

Diagram not to scale

(a)

Show that the area of the panel is 2 m².

(2)

(b)

Find the length of the longest side of the panel. Give your answer in the form a√b, where a and b are integers and b is as small as possible.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (√5 + 1)(√5 − 1)	✓ 1
A1 ½ × (5 − 1) = 2	✓ 1
Part (b)
M1 (√5 + 1)² + (√5 − 1)² = (6 + 2√5) + (6 − 2√5)	✓ 1
A1 12	✓ 1
A1 √12 = 2√3 m	✓ 1

Common error: Squaring a bracket with a surd gives three terms; the middle terms here cancel because the signs differ.

+ Add to test✓ In your test252IGME3-1-14r34-09Surds Paper HMedium4 marks

Surds.

(a)

Write √72 in the form a√2, where a is an integer.

(2)

(b)

Simplify √50 + √18.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √36 × √2	✓ 1
A1 6√2	✓ 1
Part (b)
M1 5√2 + 3√2	✓ 1
A1 8√2	✓ 1

Exam tip: Look for the largest square factor first: 72 = 36 × 2 is quicker than 4 × 18.

+ Add to test✓ In your test253IGME3-1-14r34-10Surds Paper HMedium5 marks

Rationalising denominators.

(a)

Rationalise the denominator of 12/√6. Give your answer in its simplest form.

(2)

(b)

Show that 10/(3 − √5) can be written as (15 + 5√5)/2.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiply numerator and denominator by √6	✓ 1
A1 2√6	✓ 1
Part (b)
M1 multiply numerator and denominator by (3 + √5)	✓ 1
M1 denominator (3 − √5)(3 + √5) = 9 − 5 = 4	✓ 1
A1 10(3 + √5)/4 = (15 + 5√5)/2 shown	✓ 1

Common error: To rationalise a denominator of the form a − √b, multiply by a + √b, not by √b.

+ Add to test✓ In your test254IGME3-1-14r39-06Ratio, proportion and degree of accuracy Paper HMedium6 marks

A ship sails a distance of 132 nautical miles, correct to the nearest nautical mile, at an average speed of 15 knots, correct to the nearest knot. (1 knot is 1 nautical mile per hour.)

(a)

Write down the upper bound of the distance and the lower bound of the speed.

(2)

(b)

Work out the upper bound for the time taken. Give your answer in hours to 3 significant figures.

(2)

(c)

Work out the lower bound for the time taken. Give your answer in hours to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 132.5	✓ 1
B1 14.5	✓ 1
Part (b)
M1 132.5 ÷ 14.5	✓ 1
A1 9.14	✓ 1
Part (c)
M1 131.5 ÷ 15.5	✓ 1
A1 8.48	✓ 1

Common error: For the largest time, divide the largest distance by the smallest speed; mixing two upper bounds gives the wrong value.

+ Add to test✓ In your test255IGME3-1-14r39-12Powers, roots and standard form Paper HMedium8 marks

An eruption throws out 3.5 × 10⁸ m³ of ash with density 1.2 × 10³ kg/m³.

(a)

Work out the mass of the ash, in kg. Give your answer in standard form.

(2)

(b)

A cargo ship has mass 7 × 10⁷ kg. Work out how many ships of this mass have the same total mass as the ash.

(2)

(c)

A second eruption produces 8.4 × 10¹⁰ kg of ash. Work out the total mass of ash from the two eruptions. Give your answer in standard form.

(2)

(d)

Work out the percentage of the total mass that came from the second eruption. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3.5 × 10⁸ × 1.2 × 10³	✓ 1
A1 4.2 × 10¹¹ kg	✓ 1
Part (b)
M1 4.2 × 10¹¹ ÷ (7 × 10⁷) (ECF from (a))	✓ 1
A1 6 × 10³	✓ 1
Part (c)
M1 4.2 × 10¹¹ + 0.84 × 10¹¹ (same power of 10)	✓ 1
A1 5.04 × 10¹¹ kg	✓ 1
Part (d)
M1 8.4 × 10¹⁰ ÷ 5.04 × 10¹¹ × 100 (ECF from (c))	✓ 1
A1 16.7%	✓ 1

Exam tip: Before adding or subtracting numbers in standard form, write them with the same power of 10.

+ Add to test✓ In your test256IGME3-1-14r40-07Surds Paper HMedium6 marks

Two square game boards have areas of 75 cm² and 12 cm².

Diagram not to scale

(a)

Show that the sum of the side lengths of the two boards is 7√3 cm.

(3)

(b)

Rationalise the denominator and simplify 6 / (3 − √3).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √75 = √25 × √3	✓ 1
M1 √12 = √4 × √3	✓ 1
A1 5√3 + 2√3 = 7√3	✓ 1
Part (b)
M1 multiplies numerator and denominator by (3 + √3)	✓ 1
A1 denominator 9 − 3 = 6 (numerator 18 + 6√3)	✓ 1
A1 3 + √3	✓ 1

Exam tip: To rationalise a denominator of the form a − √b, multiply top and bottom by a + √b; the denominator becomes a whole number.

+ Add to test✓ In your test257IGME3-1-14r40-09Ratio, proportion and degree of accuracy Paper HMedium6 marks

A rectangular water tank has a base of length 12.5 m and width 8.0 m. Each measurement is correct to 1 decimal place.

Diagram not to scale

(a)

Write down the lower bound and the upper bound of the width of the tank.

(2)

(b)

Work out the lower bound of the area of the base of the tank.

(2)

(c)

Work out the upper bound of the area of the base of the tank.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 lower bound 7.95 m	✓ 1
B1 upper bound 8.05 m	✓ 1
Part (b)
M1 12.45 × 7.95	✓ 1
A1 98.9775 m² (accept 98.98)	✓ 1
Part (c)
M1 12.55 × 8.05	✓ 1
A1 101.0275 m² (accept 101.03)	✓ 1

Common error: For a measurement given to 1 decimal place, the error is up to 0.05 either way, not 0.1.

+ Add to test✓ In your test258IGME3-1-14r41-09Ratio, proportion and degree of accuracy Paper HMedium6 marks

A climber walks d = 1.8 km, measured to the nearest 0.1 km, in a time t = 2.5 hours, measured to the nearest 0.1 hour. Average speed = distance ÷ time.

(a)

Write down the lower bound and upper bound of d, and of t.

(2)

(b)

Calculate the upper bound for the average speed. Give your answer to 3 significant figures.

(2)

(c)

Calculate the lower bound for the average speed. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 1.75 ≤ d < 1.85	✓ 1
B1 2.45 ≤ t < 2.55	✓ 1
Part (b)
M1 1.85 ÷ 2.45	✓ 1
A1 0.755 km/h	✓ 1
Part (c)
M1 1.75 ÷ 2.55	✓ 1
A1 0.686 km/h	✓ 1

Common error: For the upper bound of a quotient, divide the upper bound of the numerator by the lower bound of the denominator.

+ Add to test✓ In your test259IGME3-1-14r41-10Integers, fractions, decimals and percentages Paper HMedium5 marks

The proportion of sugar in a food mixture is the recurring decimal \(0.4\dot{7}\) (0.4777…).

(a)

Show that \(0.4\dot{7}\) can be written as ⁴³⁄₉₀.

(3)

(b)

Write \(0.\dot{2}\dot{7}\) (0.272727…) as a fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 let x = 0.4777…, so 10x = 4.777… and 100x = 47.777…	✓ 1
M1 100x − 10x = 43, i.e. 90x = 43	✓ 1
A1 x = ⁴³⁄₉₀	✓ 1
Part (b)
M1 ²⁷⁄₉₉ (or 100x − x = 27)	✓ 1
A1 ³⁄₁₁ (= 0.272727…)	✓ 1

Exam tip: Multiply by a power of 10 that lines up the repeating digits, then subtract so that the recurring part cancels.

+ Add to test✓ In your test260IGME3-1-14r41-11Powers, roots and standard form Paper HMedium5 marks

Evaluate and solve the following, showing your working.

(a)

Evaluate \(125^{-\frac{2}{3}}\).

(2)

(b)

Solve \(4^{x} = \frac{1}{32}\).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cube root of 125 is 5, or reciprocal taken	✓ 1
A1 ¹⁄₂₅	✓ 1
Part (b)
M1 write both sides as powers of 2: 2^2x = 2⁻⁵	✓ 1
M1 2x = −5	✓ 1
A1 x = −⁵⁄₂ (or −2.5)	✓ 1

Common error: A negative index means a reciprocal, not a negative answer.

+ Add to test✓ In your test261IGME3-1-14r42-03Integers, fractions, decimals and percentages Paper HMedium6 marks

The number of members in a cycling club fell by 8% over one year. There are now 207 members.

(a)

Work out the number of members the club had before the fall.

(3)

(b)

The club expects its membership to grow by 6% each year for the next 3 years, starting from the 207 members it has now.
Work out the expected number of members after 3 years. Give your answer to the nearest whole number.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 207 = 0.92 × original	✓ 1
M1 207 ÷ 0.92	✓ 1
A1 225	✓ 1
Part (b)
M1 multiplier 1.06 used three times, 1.06³	✓ 1
M1 207 × 1.06³ = 246.54…	✓ 1
A1 247	✓ 1

Common error: For a reverse percentage, divide by the multiplier (0.92); do not add 8% of the new amount.

+ Add to test✓ In your test262IGME3-1-14r42-10Surds Paper HMedium7 marks

This question is about surds.

Diagram not to scale

(a)

Write √72 in the form a√2, where a is an integer.

(2)

(b)

A plan for a cycle track is a rectangle of length (3 + √5) km and width (4 − √5) km.
Show that the area of the rectangle is (7 + √5) km².

(3)

(c)

Rationalise the denominator of 10 ÷ √5. Give your answer in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √36 × √2	✓ 1
A1 6√2	✓ 1
Part (b)
M1 expands to 12 − 3√5 + 4√5 − (√5)²	✓ 1
M1 uses √5 × √5 = 5 and collects the √5 terms	✓ 1
A1 12 − 5 + √5 = 7 + √5	✓ 1
Part (c)
M1 multiplies top and bottom by √5	✓ 1
A1 2√5 (≈ 4.47)	✓ 1

Common error: (√5)² is 5, so the product −(√5)(√5) contributes −5, not −√10 or −25.

+ Add to test✓ In your test263IGME3-1-14r43-06Integers, fractions, decimals and percentages Paper HMedium6 marks

A water pump costs €4800 when new. Its value falls by 12% of its value at the start of each year.

(a)

Work out how many complete years it takes until the value of the pump is first below €2000.

(3)

(b)

Work out the value of the pump after 3 years. Give your answer to the nearest cent.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiplier 0.88 used, with 4800 × 0.88ⁿ evaluated for at least two consecutive values of n	✓ 1
A1 value after 6 years ≈ €2229, which is above €2000	✓ 1
A1 7 years (value ≈ €1962)	✓ 1
Part (b)
M1 multiplier 0.88 seen	✓ 1
M1 4800 × 0.88³	✓ 1
A1 €3271.07	✓ 1

Exam tip: A fall of 12% each year is a repeated multiplication by 0.88, not a fall of 12% of the original value every year.

+ Add to test✓ In your test264IGME3-1-14r43-13Powers, roots and standard form Paper HMedium6 marks

Show your working in each part.

(a)

Work out the value of \(64^{\frac{2}{3}}\).

(2)

(b)

Work out the value of \(\left(\frac{9}{16}\right)^{-\frac{3}{2}}\). Give your answer as a fraction.

(2)

(c)

Solve \(8^{x} = \frac{1}{32}\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cube root of 64 is 4	✓ 1
A1 16	✓ 1
Part (b)
M1 reciprocal \(\frac{16}{9}\) and square root \(\frac{4}{3}\)	✓ 1
A1 \(\frac{64}{27}\)	✓ 1
Part (c)
M1 \(2^{3x} = 2^{-5}\)	✓ 1
A1 \(x = -\frac{5}{3}\)	✓ 1

Common error: Taking the root and power in the wrong sense: for am/ⁿ, the denominator n is the root and the numerator m is the power.

+ Add to test✓ In your test265IGME3-1-12r47-08Ratio, proportion and degree of accuracy Paper HMedium8 marks

An architect gives the length of a rectangular pool as 7.4 m and the width as 3.2 m. Each measurement is correct to 1 decimal place.

Diagram not to scale

(a)

Write down the error interval for the length, using inequalities.

(2)

(b)

Work out the lower bound for the area of the pool.

(2)

(c)

Work out the upper bound for the area of the pool.

(2)

(d)

Rhys says, "The area is 20 m² correct to 1 significant figure." Is Rhys correct? Give a reason.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 7.35 ≤ length	✓ 1
B1 length < 7.45	✓ 1
Part (b)
M1 7.35 × 3.15	✓ 1
A1 23.1525	✓ 1
Part (c)
M1 7.45 × 3.25	✓ 1
A1 24.2125	✓ 1
Part (d)
B1 Yes, because both bounds (23.1525 and 24.2125) round to 20 to 1 significant figure	✓ 1
B1 so the area is 20 m² to 1 significant figure whatever the exact measurements	✓ 1

Common error: The upper bound of a value rounded to 1 d.p. is 0.05 above, and values at the upper bound are not included in the error interval.

+ Add to test✓ In your test266IGME3-1-12r47-09Integers, fractions, decimals and percentages Paper HMedium7 marks

A boat costs €18000 and loses 12% of its value each year. A ferry fare increased by 15% to €27.60.

(a)

Work out the fare before the increase.

(2)

(b)

Find the least number of whole years after which the value of the boat is below €9000.

(3)

(c)

Work out the value of the boat after 3 years. Give your answer to the nearest euro.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 27.60 ÷ 1.15	✓ 1
A1 24	✓ 1
Part (b)
M1 correct method, e.g. 18000 × 0.88⁵ = 9499.17 (still above 9000)	✓ 1
M1 18000 × 0.88⁶ = 8359.27	✓ 1
A1 6	✓ 1
Part (c)
M1 18000 × 0.88³	✓ 1
A1 12266 (from 12266.496)	✓ 1

Exam tip: A 12% fall each year means multiplying by 0.88 repeatedly, not subtracting 12% of the original each time.

+ Add to test✓ In your test267IGME3-1-12r47-10Surds Paper HMedium6 marks

Surds are used in an architect's plans. All lengths are in exact form.

(a)

A square tile has area 72 cm². Write the side length of the tile in the form a√2, where a is an integer.

(2)

(b)

A rectangular window has sides (4 + √2) m and (4 − √2) m. Show that its area is a whole number of square metres.

(2)

(c)

A beam has height 18/√6 m. Rationalise the denominator and simplify.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √72 = √36 × √2	✓ 1
A1 6√2	✓ 1
Part (b)
M1 16 − 2 or full expansion with the surd terms cancelling	✓ 1
A1 14	✓ 1
Part (c)
M1 multiplies top and bottom by √6	✓ 1
A1 3√6 (= 7.3485...)	✓ 1

Common error: (4 + √2)(4 − √2) has no middle terms: they cancel, leaving 16 − 2.

+ Add to test✓ In your test268IGME3-1-12r47-12Surds Paper HMedium6 marks

A triangular glass panel has a right angle. The two shorter sides have lengths 3√2 cm and √8 cm.

Diagram not to scale

(a)

Work out the area of the panel.

(2)

(b)

Show that the longest side has length √26 cm.

(2)

(c)

Write the perimeter in the form a√2 + √26, where a is an integer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1/2 × 3√2 × √8	✓ 1
A1 6	✓ 1
Part (b)
M1 (3√2)² + (√8)² = 18 + 8	✓ 1
A1 √26	✓ 1
Part (c)
M1 √8 = 2√2	✓ 1
A1 5√2 + √26 (= 12.1701...)	✓ 1

Common error: √8 must be simplified to 2√2 before adding it to other multiples of √2.

+ Add to test✓ In your test269IGME3-1-13r48-09Surds Paper HMedium5 marks

A rectangular solar panel has width \((3+\sqrt{2})\) m and length \((5-2\sqrt{2})\) m.

(a)

Show that the area of the panel is \((11-\sqrt{2})\) m².

(3)

(b)

Find the perimeter of the panel. Give your answer in the form \(a-b\sqrt{2}\) where \(a\) and \(b\) are integers.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expand \((3+\sqrt{2})(5-2\sqrt{2})\) with four terms \(15-6\sqrt{2}+5\sqrt{2}-2\sqrt{2}\sqrt{2}\)	✓ 1
M1 uses \(\sqrt{2}\times\sqrt{2}=2\) so \(-4\)	✓ 1
A1 \(15-4-\sqrt{2}=11-\sqrt{2}\)	✓ 1
Part (b)
M1 \(2[(3+\sqrt{2})+(5-2\sqrt{2})]\)	✓ 1
A1 \(16-2\sqrt{2}\)	✓ 1

Common error: In \((3+\sqrt{2})(5-2\sqrt{2})\), the last term is \(+\sqrt{2}\times(-2\sqrt{2})=-4\); many forget the 2 from \(\sqrt{2}\times\sqrt{2}\).

+ Add to test✓ In your test270IGME3-1-13r48-11Integers, fractions, decimals and percentages Paper HMedium5 marks

A footballer's shooting accuracy over a season is recorded as the recurring decimal 0.272727… (the digits 27 repeat forever). Let \(x = 0.272727\ldots\)

(a)

Show that \(x=\dfrac{3}{11}\).

(2)

(b)

Another player's accuracy is 0.1444… (the digit 4 repeats forever). Write 0.1444… as a fraction in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(100x=27.2727\ldots\) and subtracts to get \(99x=27\)	✓ 1
A1 \(x=\dfrac{27}{99}=\dfrac{3}{11}\)	✓ 1
Part (b)
M1 \(10y=1.444\ldots\) and \(100y=14.444\ldots\)	✓ 1
M1 subtracts to get \(90y=13\)	✓ 1
A1 \(\dfrac{13}{90}\)	✓ 1

Exam tip: Choose the multiplier so that the repeating parts line up exactly and cancel on subtraction.

+ Add to test✓ In your test271IGME3-1-13r48-12Powers, roots and standard form Paper HMedium6 marks

In astronomy, indices are used to handle very large numbers. A light year is 9.46 × 10¹⁵ m.

(a)

Work out the value of \(27^{\frac{2}{3}}\).

(2)

(b)

Work out the value of \(8^{-\frac{4}{3}}\). Give your answer as a fraction.

(2)

(c)

A galaxy is 2.4 × 10²² m from Earth. Work out this distance in light years. Give your answer in standard form correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \((\sqrt[3]{27})^2\) or \(3^2\)	✓ 1
A1 9	✓ 1
Part (b)
M1 \(\dfrac{1}{8^{4/3}}\) or \(2^{-4}\)	✓ 1
A1 \(\dfrac{1}{16}\)	✓ 1
Part (c)
M1 2.4 × 10²² ÷ 9.46 × 10¹⁵ with brackets used correctly	✓ 1
A1 2.54 × 10⁶	✓ 1

Common error: A negative index means a reciprocal, not a negative answer: \(8^{-4/3}\) is positive.

+ Add to test✓ In your test272IGME3-1-14r48-09Surds Paper HMedium4 marks

A robot moves \(\sqrt{12}\) m east and then \(\sqrt{27}\) m north.

Diagram not to scale

(a)

Show that the total distance the robot moves is \(5\sqrt{3}\) m.

(2)

(b)

Find the straight-line distance of the robot from its starting point. Give your answer as a surd.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{12}=2\sqrt{3}\) or \(\sqrt{27}=3\sqrt{3}\)	✓ 1
A1 \(2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)	✓ 1
Part (b)
M1 \(12+27=39\)	✓ 1
A1 \(\sqrt{39}\)	✓ 1

Common error: The two distances are at right angles, so add the squares under Pythagoras (\(12+27\)); adding the lengths gives the distance travelled, not the straight-line distance.

+ Add to test✓ In your test273IGME3-1-14r48-13Ratio, proportion and degree of accuracy Paper HMedium5 marks

The time, \(t\) hours, taken by a team of \(n\) identical robots to harvest a field is inversely proportional to \(n\). A team of 6 robots takes 14 hours.

(a)

Show that \(t=\dfrac{84}{n}\).

(2)

(b)

Work out the time taken by 8 robots.

(1)

(c)

Work out the number of robots needed to harvest the field in 4 hours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(t=\dfrac{k}{n}\) and substitutes \(t=14\), \(n=6\)	✓ 1
A1 \(k=84\) so \(t=\dfrac{84}{n}\)	✓ 1
Part (b)
B1 10.5 hours	✓ 1
Part (c)
M1 \(4=\dfrac{84}{n}\)	✓ 1
A1 21	✓ 1

Common error: Inverse proportion means more robots give less time; check that your answer makes sense.

+ Add to test✓ In your test274IGME-058-01-12Surds Paper HHard5 marks

Surds.

(a)

Write \(\sqrt{12} + \sqrt{27}\) in the form \(a\sqrt{3}\), where \(a\) is an integer.

(2)

(b)

Rationalise the denominator of \(\frac{14}{3 - \sqrt{2}}\)
Give your answer in the form \(p + q\sqrt{2}\), where \(p\) and \(q\) are integers.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\sqrt{12} = 2\sqrt{3}\) or \(\sqrt{27} = 3\sqrt{3}\)	✓ 1
A1 for \(5\sqrt{3}\)	✓ 1
Part (b)
M1 for multiplying numerator and denominator by \(3 + \sqrt{2}\)	✓ 1
M1 for denominator \(9 - 2 = 7\)	✓ 1
A1 for \(6 + 2\sqrt{2}\)	✓ 1

Exam tip: Multiply by the conjugate \(3 + \sqrt{2}\); the difference of two squares removes the surd from the denominator.

+ Add to test✓ In your test275IGME-058-01-13Powers, roots and standard form Paper HHard5 marks

A culture starts with 4.0×10³ bacteria.
The number of bacteria doubles every 20 minutes.
The mass of one bacterium is 9.5×10⁻¹³ g.

(a)

Work out the number of bacteria after 10 hours.
Give your answer in standard form correct to 3 significant figures.

(3)

(b)

Work out the total mass, in grams, of the bacteria after 10 hours.
Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 30 doublings (600 ÷ 20)	✓ 1
M1 for 4.0×10³ × 2³⁰	✓ 1
A1 for 4.29×10¹²	✓ 1
Part (b)
M1 for their (a) × 9.5×10⁻¹³	✓ 1
A1 for 4.08 g	✓ 1

Common error: Doubling 30 times means multiplying by 2³⁰, not by 2 × 30 = 60.

+ Add to test✓ In your test276IGME-058-01-14Powers, roots and standard form Paper HHard5 marks

Powers of 2.

(a)

Show that \(4^{3x} \times 8^{2-x}\) can be written as \(2^{3x+6}\)

(2)

(b)

Hence solve \(4^{3x} \times 8^{2-x} = \frac{\sqrt{2}}{16}\)
Give your answer as a fraction.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(2^{6x}\) or \(2^{6-3x}\)	✓ 1
A1 for \(2^{6x + 6 - 3x} = 2^{3x+6}\) with full working	✓ 1
Part (b)
M1 for \(\frac{\sqrt{2}}{16} = 2^{\frac{1}{2} - 4} = 2^{-\frac{7}{2}}\)	✓ 1
M1 for \(3x + 6 = -\frac{7}{2}\)	✓ 1
A1 for \(x = -\frac{19}{6}\)	✓ 1

Exam tip: Write \(\sqrt{2}\) as \(2^{\frac{1}{2}}\) and \(\frac{1}{16}\) as \(2^{-4}\), then add the indices to get a single power of 2.

+ Add to test✓ In your test277IGME-058-00-12Integers, fractions, decimals and percentages Paper HHard5 marks

The number of birds in a wetland reserve decreased by 6% each year for 3 years.
In the fourth year, the number increased by x%.
At the end of the fourth year, the number of birds was the same as at the start of the first year.

(a)

Work out the overall percentage decrease in the number of birds over the first 3 years. Give your answer correct to 3 significant figures.

(2)

(b)

Find the value of x. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.94³ or 0.830584	✓ 1
A1 16.9%	✓ 1
Part (b)
M1 0.94³ × m = 1 or m = 1 ÷ 0.94³	✓ 1
M1 m = 1.20397…	✓ 1
A1 20.4	✓ 1

Common error: Assuming an 18% increase reverses three 6% decreases. Percentage changes combine by multiplying multipliers, not by adding percentages.

+ Add to test✓ In your test278IGME-058-00-14Integers, fractions, decimals and percentages Paper HHard5 marks

A car was bought new for £18 500.
Its value depreciated by the same percentage each year.
After 4 years it was worth £11 840.

(a)

Work out the annual percentage rate of depreciation. Give your answer correct to 3 significant figures.

(3)

(b)

The value continues to depreciate at the same rate.
Work out the value of the car 2 years after it was worth £11 840.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 11 840 ÷ 18 500 = 0.64	✓ 1
M1 \(\sqrt[4]{0.64}\) = 0.8944…	✓ 1
A1 10.6%	✓ 1
Part (b)
M1 11 840 × 0.8944…² or 11 840 × 0.8	✓ 1
A1 £9472	✓ 1

Common error: Dividing the total percentage loss (36%) by 4 to get 9% per year ignores compounding. Take the fourth root of the overall multiplier.

+ Add to test✓ In your test279IGME-058-04-12Surds Paper HHard6 marks

In triangle \(PQR\), \(PQ=(2+\sqrt{3})\) cm, \(PR=(2-\sqrt{3})\) cm and angle \(QPR=60°\).

Diagram not to scale

(a)

Find the exact area of triangle \(PQR\).

(3)

(b)

Find the exact length of \(QR\).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{1}{2}(2+\sqrt{3})(2-\sqrt{3})\sin60°\) ; M1 for \((2+\sqrt{3})(2-\sqrt{3})=1\) ; A1 \(\frac{\sqrt{3}}{4}\) cm²	✓ 3
Part (b)
M1 for \(QR^2=(2+\sqrt{3})^2+(2-\sqrt{3})^2-2(2+\sqrt{3})(2-\sqrt{3})\cos60°\) ; M1 for \((7+4\sqrt{3})+(7-4\sqrt{3})-1=13\) ; A1 \(\sqrt{13}\) cm	✓ 3

Exam tip: Use the exact values \(\sin60°=\frac{\sqrt{3}}{2}\) and \(\cos60°=\frac{1}{2}\); rounding them would lose the exact answer.

+ Add to test✓ In your test280IGME-058-04-13Surds Paper HHard6 marks

\(p\) and \(q\) are positive integers such that \((p+q\sqrt{2})^2=19+6\sqrt{2}\).

(a)

Find the value of \(p\) and the value of \(q\).

(4)

(b)

Hence write \(\frac{1}{\sqrt{19+6\sqrt{2}}}\) in the form \(\frac{a\sqrt{2}+b}{c}\) where \(a\), \(b\) and \(c\) are integers.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(p^2+2pq\sqrt{2}+2q^2\) ; M1 for \(p^2+2q^2=19\) and \(2pq=6\) ; M1 for testing integer pairs with \(pq=3\), e.g. \(p=3,q=1\) gives 11 ; A1 \(p=1\), \(q=3\)	✓ 4
Part (b)
M1 for \(\frac{1}{1+3\sqrt{2}}\times\frac{3\sqrt{2}-1}{3\sqrt{2}-1}\) ; A1 \(\frac{3\sqrt{2}-1}{17}\)	✓ 2

Exam tip: Compare the parts with no surd and the parts with \(\sqrt{2}\) separately; this gives two equations in \(p\) and \(q\).

+ Add to test✓ In your test281IGME-058-02-13Ratio, proportion and degree of accuracy Paper HHard4 marks

A car travels a distance of 145 km, correct to the nearest 5 km.
The journey takes 1.8 hours, correct to the nearest 0.1 hour.

(a)

Write down the upper bound for the distance.

(1)

(b)

Work out the lower bound for the average speed of the car.
Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 147.5 km	✓ 1
Part (b)
B1 for 142.5 and 1.85 seen	✓ 1
M1 for 142.5 ÷ 1.85	✓ 1
A1 for 77.0 km/h	✓ 1

Exam tip: For the lower bound of a quotient, divide the lower bound of the numerator by the upper bound of the denominator.

+ Add to test✓ In your test282IGME-058-02-14Ratio, proportion and degree of accuracy Paper HHard6 marks

The brightness, B, of a lamp measured by a sensor is inversely proportional to the square of the distance, d metres, between the sensor and the lamp.
When d = 5, B = 36

(a)

Find the value of B when d = 7.5

(3)

(b)

The distance d is increased by 20%.
Work out the percentage decrease in B.
Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(B=\frac{k}{d^2}\)	✓ 1
M1 for k = 36 × 25 = 900	✓ 1
A1 for 16	✓ 1
Part (b)
M1 for new d = 1.2d	✓ 1
M1 for new B = \(\frac{1}{1.2^2}\)B = 0.694…B	✓ 1
A1 for 30.6%	✓ 1

Common error: A 20% increase in d does not give a 20% decrease in B: because of the square, B is multiplied by 1 ÷ 1.44.

+ Add to test✓ In your test283IGME-058-03-12Set language, notation and Venn diagrams (both tiers) Paper HHard5 marks

A and B are sets in a universal set ξ with n(ξ) = 90.
n(A) : n(B) = 3 : 2
n(A ∩ B) = 15
n((A ∪ B)′) = 15

(a)

Find n(A).

(3)

(b)

A member of ξ is chosen at random. Find the probability that it is in B but not in A.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for using n(A) = 3k and n(B) = 2k . M1 for 3k + 2k − 15 + 15 = 90 . A1 for k = 18, so n(A) = 54 .	✓ 3
Part (b)
M1 for n(B ∩ A′) = 36 − 15 = 21 . A1 for \(\frac{21}{90} = \frac{7}{30}\) .	✓ 2

Exam tip: Write the ratio with a multiplier k, then use n(ξ) = n(A) + n(B) − n(A ∩ B) + n((A ∪ B)′).

+ Add to test✓ In your test284IGME-058-04-14Surds Paper HHard5 marks

In this question you must show all your working. Answers given without supporting working may not earn full marks.

(a)

Show that \(\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\).

(2)

(b)

Hence, or otherwise, find the exact value of \(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{2+\sqrt{3}}\).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for multiplying numerator and denominator by \(\sqrt{2}-1\) ; A1 for showing the denominator expansion \((\sqrt{2}+1)(\sqrt{2}-1)=2-\sqrt{2}+\sqrt{2}-1=1\), leading to \(\sqrt{2}-1\) with no errors	✓ 2
Part (b)
M1 for multiplying numerator and denominator of at least one of the other two fractions by its conjugate, e.g. \(\sqrt{3}-\sqrt{2}\) or \(2-\sqrt{3}\) ; A1 for both \(\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\sqrt{2}\) and \(\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\) ; A1 for \((\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(2-\sqrt{3})=1\), with the cancelling surds shown	✓ 3

Exam tip: When each denominator is a sum of square roots of consecutive whole numbers, rationalising gives a denominator of 1. Write the three results out in full so that the surds visibly cancel in pairs.

+ Add to test✓ In your test285IGME-058-03-11Set language, notation and Venn diagrams (both tiers) Paper HHard8 marks

80 students were asked which of French (F), German (G) and Spanish (S) they study.
5 students study all three languages.
12 students study both French and German, 9 students study both German and Spanish and 14 students study both French and Spanish. Each of these three numbers includes the 5 students who study all three languages.
n(F) = 40, n(G) = 26 and n(S) = 33.

(a)

Find the number of students who study none of these languages.

(4)

(b)

A student is chosen at random. Find the probability that this student studies exactly two of the languages.

(2)

(c)

A student who studies Spanish is chosen at random. Find the probability that this student studies only one language.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for the exactly-two regions 12 − 5 = 7, 9 − 5 = 4 and 14 − 5 = 9 . M1 for the single regions: F only 40 − 7 − 9 − 5 = 19, G only 26 − 7 − 4 − 5 = 10, S only 33 − 9 − 4 − 5 = 15 . M1 for a total of 19 + 10 + 15 + 7 + 4 + 9 + 5 = 69 . A1 for 80 − 69 = 11 .	✓ 4
Part (b)
M1 for 7 + 4 + 9 = 20 . A1 for \(\frac{20}{80} = \frac{1}{4}\) .	✓ 2
Part (c)
M1 for \(\frac{15}{33}\), or for a denominator of 33 . A1 for \(\frac{5}{11}\) .	✓ 2

Common error: The 12 who study both French and German include the 5 who study all three. Fill the centre of the Venn diagram first and work outwards, subtracting at each step.

+ Add to test✓ In your test286IGME3-1-07d-10Powers, roots and standard form Paper HHard7 marks

Alloy A is made of copper and tin in the ratio 3 : 2 by mass. Alloy B is made of copper and tin in the ratio 1 : 4 by mass. A metalworker melts together 5 kg of alloy A and 10 kg of alloy B.

(a)

Find the ratio of the mass of copper to the mass of tin in the mixture. Give your answer in the form 1 : n. (4)

(4)

(b)

The metalworker adds t kg of pure tin so that 20% of the new alloy is copper. Work out the value of t. (3)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 copper in A = 3 kg or tin in A = 2 kg	✓ 1
M1 copper in B = 2 kg or tin in B = 8 kg	✓ 1
A1 copper 5 kg and tin 10 kg	✓ 1
A1 1 : 2	✓ 1
Part (b)
M1 \(\frac{5}{15 + t} = 0.2\)	✓ 1
M1 15 + t = 25	✓ 1
A1 10	✓ 1

Exam tip: Convert each ratio into actual masses before combining. Adding the ratios directly gives the wrong result when the total masses differ.

+ Add to test✓ In your test287IGME3-1-07d-11Powers, roots and standard form Paper HHard7 marks

One bacterium has a mass of 1.2 × 10⁻¹² g. A culture contains 3.6 × 10⁹ bacteria.

(a)

Work out the total mass of the bacteria in the culture. Give your answer in grams in standard form. (2)

(2)

(b)

Work out how many bacteria have a total mass of 0.6 g. Give your answer in standard form. (2)

(2)

(c)

The number of bacteria in the culture doubles every 30 minutes. Starting with 3.6 × 10⁹ bacteria, find the least whole number of hours after which there are at least 5 × 10¹¹ bacteria. (3)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.2 × 3.6 (= 4.32) with powers combined	✓ 1
A1 4.32 × 10⁻³ g	✓ 1
Part (b)
M1 0.6 ÷ (1.2 × 10⁻¹²)	✓ 1
A1 5 × 10¹¹	✓ 1
Part (c)
M1 5 × 10¹¹ ÷ 3.6 × 10⁹ ≈ 139, or powers of 2 tried (2⁷ = 128 gives 4.6 × 10¹¹, which is too small)	✓ 1
B1 2⁸ = 256 needed, which is 8 doublings	✓ 1
A1 4 hours	✓ 1

Common error: When multiplying numbers in standard form, multiply the front numbers and add the powers separately, then check the result is still in the form a × 10ⁿ with 1 ≤ a < 10.

+ Add to test✓ In your test288IGME3-1-07d-14Powers, roots and standard form Paper HHard7 marks

This question is about recurring decimals and powers.

(a)

Write the recurring decimal 0.272727... (the digits 27 repeat for ever) as a fraction in its simplest form. (2)

(2)

(b)

Work out the value of \(\left(\frac{27}{8}\right)^{-\frac{2}{3}}\). (2)

(2)

(c)

Solve \(2^{3x} = \frac{1}{4\sqrt{2}}\). (3)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 100x − x = 27.2727... − 0.2727... or 99x = 27	✓ 1
A1 \(\frac{3}{11}\)	✓ 1
Part (b)
M1 \(\left(\frac{8}{27}\right)^{\frac{2}{3}}\) or \(\left(\frac{2}{3}\right)^{2}\)	✓ 1
A1 \(\frac{4}{9}\)	✓ 1
Part (c)
M1 4√2 = 2² × 2^1/2 = 2^5/2	✓ 1
M1 3x = −\(\frac{5}{2}\)	✓ 1
A1 x = −\(\frac{5}{6}\) (= −0.833333...)	✓ 1

Common error: A negative power means a reciprocal, and a fractional power means a root: do both before squaring or cubing.

+ Add to test✓ In your test289IGME3-1-11r7-11Integers, fractions, decimals and percentages Paper HHard6 marks

The number of passengers using a rail line was 5.0 million in 2022. In a forecast, the number falls by 6% each year.

(a)

Work out the forecast number of passengers, in millions, at the end of two years of falls.

(2)

(b)

A different forecast has the number of passengers falling by the same percentage each year, from 5.0 million to 4.0 million over 3 years.
Work out this yearly percentage decrease. Give your answer correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5.0 × 0.94²	✓ 1
A1 4.418 million	✓ 1
Part (b)
M1 5.0 × k³ = 4.0 or k³ = 0.8	✓ 1
M1 k = ∛0.8 = 0.9283...	✓ 1
A1 decrease = 1 − 0.9283 = 0.0717	✓ 1
A1 7.17%	✓ 1

Common error: A constant percentage fall over several years uses a repeated multiplier; dividing the total fall of 20% by 3 gives 6.67%, which is wrong.

+ Add to test✓ In your test290IGME3-1-11r7-12Ratio, proportion and degree of accuracy Paper HHard9 marks

Bottle A contains juice and water in the ratio 2 : 3 by volume. Bottle B contains juice and water in the ratio 3 : 7 by volume.

(a)

Work out the volume of juice and the volume of water in 600 ml of the mixture from bottle A.

(2)

(b)

500 ml from bottle A is mixed with 500 ml from bottle B.
Write the ratio of juice to water in the new mixture in the form a : b, where a and b are integers with no common factor.

(3)

(c)

600 ml from bottle A is mixed with v ml from bottle B. The ratio of juice to water in the new mixture is 1 : 2.
Work out the value of v.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 600 ÷ 5 × 2 or 600 ÷ 5 = 120	✓ 1
A1 juice 240 ml and water 360 ml	✓ 1
Part (b)
M1 juice from A = 200 and juice from B = 150 (water 300 and 350)	✓ 1
M1 totals 350 and 650	✓ 1
A1 7 : 13	✓ 1
Part (c)
B1 juice = 240 + 0.3v and water = 360 + 0.7v	✓ 1
M1 2(240 + 0.3v) = 360 + 0.7v	✓ 1
M1 480 + 0.6v = 360 + 0.7v, so 0.1v = 120	✓ 1
A1 1200	✓ 1

Exam tip: When mixing, add the quantities of each ingredient separately, then form the ratio of the totals.

+ Add to test✓ In your test291IGME3-1-12r13-12Surds Paper HHard6 marks

A rectangular roof panel has area 6 m² and width (3 − √3) m.

Diagram not to scale

(a)

Show that the length of the panel is (3 + √3) m.

(3)

(b)

Show that the diagonal of the panel is 2√6 m.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length = 6/(3 − √3), multiplying numerator and denominator by (3 + √3)	✓ 1
M1 denominator (3 − √3)(3 + √3) = 9 − 3 = 6	✓ 1
A1 6(3 + √3)/6 = 3 + √3	✓ 1
Part (b)
M1 diagonal² = (3 + √3)² + (3 − √3)²	✓ 1
A1 (12 + 6√3) + (12 − 6√3) = 24	✓ 1
A1 diagonal = √24 = 2√6	✓ 1

Common error: Squaring a bracket with a surd needs three terms: (3 + √3)² = 9 + 6√3 + 3, not 9 + 3.

+ Add to test✓ In your test292IGME3-1-12r13-13Set language, notation and Venn diagrams (both tiers) Paper HHard5 marks

The numbers of staff in store A and store B are in the ratio 5 : 3. When 4 members of staff move from store A to store B, the ratio of the numbers of staff in A to B becomes 3 : 2.

(a)

Work out the number of staff that store A had originally.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 staff numbers 5k and 3k	✓ 1
M1 (5k − 4)/(3k + 4) = 3/2	✓ 1
M1 cross-multiplies and solves: 10k − 8 = 9k + 12	✓ 1
A1 k = 20	✓ 1
A1 100	✓ 1

Exam tip: Represent a ratio with a multiplier, such as 5k : 3k, and then apply the change to the actual quantities before forming the new ratio.

+ Add to test✓ In your test293IGME3-1-13r14-08Ratio, proportion and degree of accuracy Paper HHard7 marks

The length and width of a rectangular fish pond are measured as 12.4 m and 7.8 m. Each measurement is correct to 1 decimal place. There are 1560 fish in the pond.

Diagram not to scale

(a)

Write down the error interval for the length of the pond, using inequality signs.

(2)

(b)

Work out the upper bound for the area of the pond.

(2)

(c)

Show that, whatever the true dimensions of the pond, there are more than 15 fish per square metre.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 12.35 ≤ length	✓ 1
B1 length < 12.45	✓ 1
Part (b)
M1 12.45 × 7.85	✓ 1
A1 97.7325 m²	✓ 1
Part (c)
M1 divide 1560 by the upper bound of the area, 97.7325	✓ 1
A1 15.96… fish per m² (15.96)	✓ 1
C1 15.96 is greater than 15, and this is the smallest possible density, so the density is always more than 15	✓ 1

Common error: The smallest possible density uses the largest possible area, because the area is the denominator.

+ Add to test✓ In your test294IGME3-1-13r14-09Set language, notation and Venn diagrams (both tiers) Paper HHard7 marks

A robotics club has 50 members. Some have built a line-following robot (set L) and some have written their own program (set P). The numbers of members in each region are: built a robot only, 2x; did both, x; wrote a program only, x + 6; did neither, 4.

Diagram not to scale

(a)

Find the value of x.

(3)

(b)

Find n(L ∪ P′).

(2)

(c)

One member who wrote their own program is chosen at random. Find the probability that this member has also built a line-following robot.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sum of the four regions equated to 50	✓ 1
A1 4x + 10 = 50	✓ 1
A1 x = 10	✓ 1
Part (b)
M1 20 + 10 + 4 or 50 − 16	✓ 1
A1 34	✓ 1
Part (c)
M1 10 ÷ (10 + 16)	✓ 1
A1 \(\frac{5}{13}\) (≈ 0.3846)	✓ 1

Common error: P′ is everyone outside P. In n(L ∪ P′) the 'neither' region is included as well as L.

+ Add to test✓ In your test295IGME3-1-13r14-10Powers, roots and standard form Paper HHard5 marks

The battery meter on a robot shows the charge as the recurring decimal \(0.4\dot{1}\), which means 0.41111…

(a)

Show that \(0.4\dot{1} = \frac{37}{90}\).

(3)

(b)

A gear train in the robot has an overall factor of \(\left(\frac{81}{16}\right)^{\frac{3}{4}}\). Work out the exact value of this factor.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 let x = 0.4111… and form 10x = 4.111… and 100x = 41.111…	✓ 1
M1 subtract to get 90x = 37	✓ 1
A1 x = \(\frac{37}{90}\)	✓ 1
Part (b)
M1 fourth root of \(\frac{81}{16}\) is \(\frac{3}{2}\)	✓ 1
A1 \(\frac{27}{8}\) (3.375)	✓ 1

Exam tip: For a fractional index, take the root first and then the power; it keeps the numbers small.

+ Add to test✓ In your test296IGME3-1-13r16-03Surds Paper HHard7 marks

A rectangular banner has area (7 + 3√5) cm² and length (2 + √5) cm. All lengths are in centimetres.

(a)

Show that the width of the banner is (7 + 3√5)/(2 + √5) cm.

(1)

(b)

Rationalise the denominator to show that the width can be written in the form a + b√5, where a and b are integers. Find a and b.

(4)

(c)

Hence write down the perimeter of the banner in the form p + q√5. Use the exact width from (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 width = area ÷ length, giving (7 + 3√5)/(2 + √5)	✓ 1
Part (b)
M1 multiply numerator and denominator by (2 − √5)	✓ 1
A1 numerator (7 + 3√5)(2 − √5) = 14 − 7√5 + 6√5 − 15 = −1 − √5	✓ 1
A1 denominator (2 + √5)(2 − √5) = 4 − 5 = −1	✓ 1
A1 width = 1 + √5, so a = 1 and b = 1	✓ 1
Part (c)
M1 2[(2 + √5) + (1 + √5)]	✓ 1
A1 6 + 4√5	✓ 1

Exam tip: To rationalise a denominator of the form a + √b, multiply top and bottom by a − √b.

+ Add to test✓ In your test297IGME3-1-13r16-11Ratio, proportion and degree of accuracy Paper HHard8 marks

The rectangular floor of a theatre stage has length 12.4 m and width 8.6 m. Both measurements are correct to 1 decimal place.

Diagram not to scale

(a)

Write down the error interval for the length.

(1)

(b)

Work out the lower bound for the perimeter of the floor.

(2)

(c)

Ellis says that the area of the floor is 107 m², correct to 3 significant figures. Use upper and lower bounds to show that this cannot be justified.

(3)

(d)

The floor is covered with flooring costing £12.50 per m². Work out the greatest possible cost. Use the unrounded upper bound of the area, and give your answer to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 12.35 ≤ length < 12.45	✓ 1
Part (b)
M1 2 × (12.35 + 8.55)	✓ 1
A1 41.8 m	✓ 1
Part (c)
M1 upper bound 12.45 × 8.65 and lower bound 12.35 × 8.55	✓ 1
A1 107.6925 and 105.5925	✓ 1
A1 these round to 108 and 106 (3 s.f.), which are different, so 107 cannot be justified	✓ 1
Part (d)
M1 12.5 × 107.6925	✓ 1
A1 £1346.16 (from 1346.15625)	✓ 1

Common error: For a product, combine the two upper bounds for the upper bound, and the two lower bounds for the lower bound.

+ Add to test✓ In your test298IGME3-1-13r16-13Set language, notation and Venn diagrams (both tiers) Paper HHard8 marks

The universal set is ξ = {integers x : 1 ≤ x ≤ 30}. P is the set of prime numbers in ξ, Q is the set of multiples of 5 in ξ, and R is the set of square numbers in ξ.

(a)

List the elements of P ∩ Q.

(1)

(b)

Find n((P ∪ Q)′).

(2)

(c)

A number is chosen at random from ξ. Given that the number is not a multiple of 5, find the probability that it is a square number.

(3)

(d)

Find n(P′ ∩ Q′ ∩ R′).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 {5}	✓ 1
Part (b)
M1 n(P) = 10, n(Q) = 6, n(P ∩ Q) = 1 so n(P ∪ Q) = 15	✓ 1
A1 30 − 15 = 15	✓ 1
Part (c)
M1 24 numbers are not multiples of 5	✓ 1
M1 squares not in Q are 1, 4, 9, 16, so 4 numbers	✓ 1
A1 4/24 = 1/6	✓ 1
Part (d)
M1 n(P ∪ Q ∪ R) = 15 + 4 = 19	✓ 1
A1 30 − 19 = 11	✓ 1

Common error: 25 is both a square and a multiple of 5, so it must not be counted again when the squares are added to P ∪ Q.

+ Add to test✓ In your test299IGME3-1-12r19-10Ratio, proportion and degree of accuracy Paper HHard8 marks

A swimming pool is 50 m long, measured to the nearest metre. A swimmer completes 4 lengths in 180 seconds, measured to the nearest second. Average speed = total distance ÷ total time.

(a)

Calculate the upper bound for the average speed. Give your answer to 4 significant figures.

(3)

(b)

Calculate the lower bound for the average speed. Give your answer to 4 significant figures.

(3)

(c)

Use your answers to state the average speed to a suitable degree of accuracy. Give a reason for your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 upper bound for distance 4 × 50.5 = 202 m	✓ 1
M1 lower bound for time 179.5 s	✓ 1
A1 1.125 m/s	✓ 1
Part (b)
M1 lower bound for distance 4 × 49.5 = 198 m	✓ 1
M1 upper bound for time 180.5 s	✓ 1
A1 1.097 m/s	✓ 1
Part (c)
M1 both bounds, 1.097 and 1.125, round to 1.1 to 2 significant figures	✓ 1
A1 1.1 m/s	✓ 1

Exam tip: For a quotient, the upper bound uses the upper bound of the numerator and the lower bound of the denominator.

+ Add to test✓ In your test300IGME3-1-12r19-13Powers, roots and standard form Paper HHard6 marks

Evaluate and solve the following, showing your working.

(a)

Work out the value of 27^2/3.

(1)

(b)

Work out the value of 16^−3/4.

(2)

(c)

Solve 8^x+1 = 4^2x−3.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 9	✓ 1
Part (b)
M1 16^1/4 = 2 and 2³ = 8, or equivalent	✓ 1
A1 1/8	✓ 1
Part (c)
M1 writes both sides as powers of 2: 2^3(x+1) = 2^2(2x−3)	✓ 1
M1 3x + 3 = 4x − 6	✓ 1
A1 x = 9	✓ 1

Common error: A negative index gives a reciprocal; the fractional part gives the root first. Do not make the answer negative.

+ Add to test✓ In your test301IGME3-1-12r19-14Set language, notation and Venn diagrams (both tiers) Paper HHard7 marks

50 swimmers are asked which of two events they enter: butterfly (A) or backstroke (B). The number of swimmers entering only butterfly is 2x − 3, both events is x, only backstroke is x + 4, and neither event is 2x + 1.

Diagram not to scale

(a)

Work out the value of x.

(3)

(b)

Work out n(A), the number of swimmers entering butterfly.

(1)

(c)

Two different swimmers are chosen at random. Find the probability that both enter only butterfly. Give your answer as a fraction in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (2x − 3) + x + (x + 4) + (2x + 1) = 50	✓ 1
A1 6x + 2 = 50	✓ 1
A1 x = 8	✓ 1
Part (b)
A1 13 + 8 = 21	✓ 1
Part (c)
M1 13/50	✓ 1
M1 × 12/49 (second choice without replacement)	✓ 1
A1 78/1225 (≈ 0.0637)	✓ 1

Common error: When choosing without replacement, reduce both the numerator and the denominator for the second choice.

+ Add to test✓ In your test302IGME3-1-11r25-08Surds Paper HHard7 marks

A cable car travels a distance of (5 + √2) km in a time of (3 − √2) hours.

(a)

Show that the average speed of the cable car can be written as (a + b√2) ÷ c km/h, where a, b and c are integers. Find a, b and c.

(4)

(b)

Given that (x + √2)² = 11 + k√2, where x > 0, find the value of x and the value of k.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (5 + √2) ÷ (3 − √2) and multiplies numerator and denominator by (3 + √2)	✓ 1
A1 numerator 15 + 5√2 + 3√2 + 2 = 17 + 8√2	✓ 1
B1 denominator 9 − 2 = 7	✓ 1
A1 a = 17, b = 8, c = 7	✓ 1
Part (b)
M1 expands to x² + 2x√2 + 2	✓ 1
A1 x² + 2 = 11 so x = 3	✓ 1
A1 k = 2x = 6	✓ 1

Common error: To remove a surd from the denominator of (a − √b), multiply by (a + √b), not by (a − √b); the sign in the middle must change.

+ Add to test✓ In your test303IGME3-1-11r25-09Surds Paper HHard6 marks

A cheese is made in the shape of a sphere. The mass, M kg, of a cheese is directly proportional to the cube of its radius, r cm. A cheese with radius 10 cm has mass 4 kg.

(a)

Find a formula for M in terms of r.

(2)

(b)

Work out the mass of a cheese with radius 15 cm.

(2)

(c)

A cheese has mass 0.5 kg. Work out its radius.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 M = kr³ and 4 = k × 10³	✓ 1
A1 M = 0.004r³	✓ 1
Part (b)
M1 0.004 × 15³ (ECF from (a))	✓ 1
A1 13.5 kg	✓ 1
Part (c)
M1 0.5 = 0.004r³ leading to r³ = 125	✓ 1
A1 5 cm	✓ 1

Exam tip: "Directly proportional to the cube" means M = kr³. Find k from the known pair of values before using the formula.

+ Add to test✓ In your test304IGME3-1-11r25-14Powers, roots and standard form Paper HHard8 marks

A food scientist measures the fraction of sugar in a syrup as the recurring decimal 0.47777..., where only the digit 7 recurs.

(a)

Show that 0.47777... = ⁴³⁄₉₀.

(3)

(b)

Work out the mass of sugar in a 360 g sample of the syrup.

(2)

(c)

A second syrup has a sugar fraction of ⁷⁄₃₀. The sugar fraction of the first syrup is n times the sugar fraction of the second syrup. Find n, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 lets x = 0.47777... and writes 10x = 4.7777... and 100x = 47.7777...	✓ 1
M1 subtracts to get 90x = 43	✓ 1
A1 x = ⁴³⁄₉₀ (shown)	✓ 1
Part (b)
M1 ⁴³⁄₉₀ × 360	✓ 1
A1 172 g	✓ 1
Part (c)
M1 ⁴³⁄₉₀ ÷ ⁷⁄₃₀	✓ 1
A1 ⁴³⁄₂₁ = 2.047...	✓ 1
A1 2.05	✓ 1

Exam tip: To convert a recurring decimal, multiply by powers of 10 so that the recurring parts line up, then subtract to cancel them.

+ Add to test✓ In your test305IGME3-1-12r25-08Surds Paper HHard7 marks

A training area in a swimming pool is a rectangle of length √75 m and width √12 m.

(a)

Write √75 and √12 in the form a√3, where a is an integer.

(2)

(b)

Work out the area of the training area.

(2)

(c)

Show that (√75 + √12) ÷ (√75 − √12) = 7/3.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 √75 = 5√3	✓ 1
B1 √12 = 2√3	✓ 1
Part (b)
M1 5√3 × 2√3 (ECF from (a))	✓ 1
A1 30 m²	✓ 1
Part (c)
M1 (5√3 + 2√3) ÷ (5√3 − 2√3)	✓ 1
M1 7√3 ÷ 3√3	✓ 1
A1 7/3 (shown)	✓ 1

Exam tip: Simplify each surd first by taking out the largest square factor (75 = 25 × 3, 12 = 4 × 3); like surds can then be added, subtracted and divided.

+ Add to test✓ In your test306IGME3-1-12r25-09Set language, notation and Venn diagrams (both tiers) Paper HHard5 marks

A library has fiction and non-fiction books in the ratio 7 : 3. The library removes 40 fiction books and adds 120 non-fiction books. The ratio of fiction to non-fiction books is then 3 : 2.

(a)

Work out the number of fiction books the library had originally.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 7k fiction and 3k non-fiction books originally	✓ 1
M1 (7k − 40) ⁄ (3k + 120) = 3 ⁄ 2, or 2(7k − 40) = 3(3k + 120)	✓ 1
M1 14k − 80 = 9k + 360 leading to 5k = 440	✓ 1
A1 k = 88	✓ 1
A1 616	✓ 1

Exam tip: Write the original quantities as multiples of a single unknown, then apply each change before using the new ratio.

+ Add to test✓ In your test307IGME3-1-12r25-14Set language, notation and Venn diagrams (both tiers) Paper HHard7 marks

90 library members were asked which types of book they borrowed: fiction (F), non-fiction (N), both or neither.
ξ = {all 90 members}
The number who borrowed only fiction is 3x.
The number who borrowed both is x.
The number who borrowed only non-fiction is x + 8.
12 members borrowed neither.

Diagram not to scale

(a)

Show that 5x + 20 = 90.

(2)

(b)

Solve 5x + 20 = 90.

(2)

(c)

Work out the number of members who borrowed only non-fiction.

(1)

(d)

A member who borrowed fiction is chosen at random. Find the probability that this member also borrowed non-fiction.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3x + x + (x + 8) + 12	✓ 1
A1 5x + 20 = 90 (shown)	✓ 1
Part (b)
M1 5x = 70	✓ 1
A1 x = 14	✓ 1
Part (c)
B1 22 (ECF from (b))	✓ 1
Part (d)
M1 14 ⁄ (42 + 14) or 14 ⁄ 56	✓ 1
A1 1/4	✓ 1

Common error: When the member is chosen from those who borrowed fiction, the denominator is n(F) = 56, not the total of 90.

+ Add to test✓ In your test308IGME3-1-13r28-05Ratio, proportion and degree of accuracy Paper HHard8 marks

A probe lander falls through 130 m in 6.2 seconds. The distance is correct to the nearest 10 m. The time is correct to 2 significant figures.

(a)

Write down the lower bound and the upper bound of the time.

(2)

(b)

Calculate the upper bound of the average speed. Give your answer correct to 3 significant figures.

(3)

(c)

Use the lower bound of the speed and your answer to (b) to find the average speed to an appropriate degree of accuracy. You must show your working.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 lower bound 6.15 s	✓ 1
B1 upper bound 6.25 s	✓ 1
Part (b)
M1 upper bound of distance ÷ lower bound of time	✓ 1
A1 135 ÷ 6.15 = 21.95…	✓ 1
A1 22.0 m/s	✓ 1
Part (c)
M1 lower bound of speed = 125 ÷ 6.25 = 20	✓ 1
M1 compare: the lower bound 20 and the upper bound 21.95… both round to 20 to 1 significant figure	✓ 1
A1 20 m/s	✓ 1

Common error: For the greatest speed, use the greatest distance and the least time; mixing upper bounds for both gives the wrong value.

+ Add to test✓ In your test309IGME3-1-13r28-12Integers, fractions, decimals and percentages Paper HHard5 marks

A coffee machine costs £4800 when new. Its value decreases by 15% of its value at the start of each year.

(a)

Work out the value of the machine after 3 years. Give your answer to the nearest penny.

(2)

(b)

Find the number of complete years after which the value first falls below £2000. You must show your working.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4800 × 0.85³	✓ 1
A1 2947.80 (£)	✓ 1
Part (b)
M1 evidence of trial: 4800 × 0.85ⁿ for at least one value of n	✓ 1
A1 after 5 years £2129.79 (above £2000) and after 6 years £1810.32 (below £2000)	✓ 1
A1 6 years	✓ 1

Exam tip: For a repeated percentage decrease, use a multiplier raised to a power rather than subtracting 15% each time.

+ Add to test✓ In your test310IGME3-1-13r29-12Ratio, proportion and degree of accuracy Paper HHard5 marks

A training squad has sprinters and distance runners in the ratio 3 : 5. Later, 4 more sprinters join the squad and 2 distance runners leave. The number of sprinters is then equal to the number of distance runners.

(a)

Write down the fraction of the original squad who were sprinters.

(1)

(b)

Work out the number of athletes in the original squad.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(\frac{3}{8}\)	✓ 1
Part (b)
M1 let the original numbers be \(3x\) and \(5x\)	✓ 1
M1 \(3x+4=5x-2\)	✓ 1
A1 \(x=3\)	✓ 1
A1 24	✓ 1

Exam tip: Introduce a letter for one part of the ratio, then form an equation from the new situation.

+ Add to test✓ In your test311IGME3-1-13r29-14Surds Paper HHard7 marks

A rectangular railway platform has length \((3+\sqrt{2})\) m and area \((8+5\sqrt{2})\) m\(^{2}\).

Diagram not to scale

(a)

Show that the width of the platform is \((2+\sqrt{2})\) m.

(4)

(b)

Work out the perimeter of the platform. Give your answer in the form \(a+b\sqrt{2}\).

(2)

(c)

Write your answer to part (b) as a decimal, correct to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 width = \(\dfrac{8+5\sqrt{2}}{3+\sqrt{2}}\)	✓ 1
M1 multiply numerator and denominator by \(3-\sqrt{2}\)	✓ 1
A1 numerator \(24-8\sqrt{2}+15\sqrt{2}-10=14+7\sqrt{2}\) and denominator \(9-2=7\)	✓ 1
A1 \(\dfrac{14+7\sqrt{2}}{7}=2+\sqrt{2}\)	✓ 1
Part (b)
M1 \(2\big[(3+\sqrt{2})+(2+\sqrt{2})\big]\)	✓ 1
A1 \(10+4\sqrt{2}\) m	✓ 1
Part (c)
A1 15.7 m	✓ 1

Exam tip: To rationalise a denominator of the form \(a+\sqrt{b}\), multiply top and bottom by \(a-\sqrt{b}\).

+ Add to test✓ In your test312IGME3-1-14r33-07Ratio, proportion and degree of accuracy Paper HHard8 marks

A walker covers a distance of 6.4 km, measured to the nearest 0.1 km, in a time of 1.5 hours, measured to the nearest 0.1 hour. Her average speed is distance ÷ time.

(a)

Write down the lower bound and the upper bound of the distance.

(2)

(b)

Calculate the upper bound for her average speed. Give your answer to 3 significant figures.

(2)

(c)

Calculate the lower bound for her average speed. Give your answer to 3 significant figures.

(2)

(d)

Use your answers to write her average speed to a suitable degree of accuracy. Give a reason.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 lower bound 6.35 km	✓ 1
B1 upper bound 6.45 km	✓ 1
Part (b)
M1 upper bound of distance ÷ lower bound of time = 6.45 ÷ 1.45	✓ 1
A1 4.45 km/h	✓ 1
Part (c)
M1 6.35 ÷ 1.55	✓ 1
A1 4.10 km/h	✓ 1
Part (d)
M1 both bounds (4.45 and 4.10) round to the same value to 1 significant figure	✓ 1
A1 4 km/h, because the bounds both round to 4 to 1 s.f.	✓ 1

Common error: For the upper bound of a quotient use the upper bound on top and the lower bound underneath.

+ Add to test✓ In your test313IGME3-1-14r33-14Set language, notation and Venn diagrams (both tiers) Paper HHard6 marks

In a food-science taste test, 100 people are each asked whether they like a sweet sample (S) and a sour sample (T). Let x be the number of people who like both samples. The number who like only the sweet sample is x + 10. The number who like only the sour sample is 2x − 4. 18 people like neither sample.

Diagram not to scale

(a)

Show that 4x + 24 = 100.

(2)

(b)

Solve the equation to find x.

(1)

(c)

One person who likes the sweet sample is chosen at random. Find the probability that this person also likes the sour sample.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (x + 10) + x + (2x − 4) + 18 = 100	✓ 1
A1 4x + 24 = 100	✓ 1
Part (b)
A1 x = 19	✓ 1
Part (c)
M1 number who like sweet = 29 + 19 = 48 (ECF from (b))	✓ 1
M1 19 as the numerator	✓ 1
A1 19/48	✓ 1

Exam tip: When the person is chosen from a group, such as those who like sweet, the denominator is the size of that group, not 100.

+ Add to test✓ In your test314IGME3-1-14r34-08Ratio, proportion and degree of accuracy Paper HHard8 marks

The length of a swimming pool is 25.0 m, correct to 3 significant figures. A swimmer swims 20 lengths of the pool in 612 seconds, correct to the nearest second.

(a)

Write down the upper bound of the length of the pool.

(1)

(b)

Work out the upper bound of the swimmer’s average speed. Give your answer correct to 3 significant figures.

(3)

(c)

Work out the lower bound of the average speed. Give your answer correct to 3 significant figures.

(2)

(d)

Use your answers to (b) and (c) to give the average speed to a suitable degree of accuracy. Give a reason.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 25.05 m	✓ 1
Part (b)
M1 upper bound of distance 20 × 25.05 = 501	✓ 1
M1 divide by the lower bound of time, 611.5	✓ 1
A1 0.819 m/s	✓ 1
Part (c)
M1 499 ÷ 612.5 (lower bound of distance ÷ upper bound of time)	✓ 1
A1 0.815 m/s	✓ 1
Part (d)
M1 bounds 0.819 and 0.815 round to different values (0.82 and 0.81) at 2 significant figures	✓ 1
A1 0.8 m/s, the only value both bounds share to 1 significant figure	✓ 1

Common error: For a speed, use the upper bound of distance with the lower bound of time to get the upper bound of speed.

+ Add to test✓ In your test315IGME3-1-14r34-11Powers, roots and standard form Paper HHard5 marks

A right-angled triangular pennant flies above a swimming pool. Its two shorter sides measure (2 + √3) m and (2 − √3) m.

Diagram not to scale

(a)

Find the area of the pennant. Give your answer as an exact fraction.

(2)

(b)

Show that the longest side of the pennant is √14 m.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (2 + √3)(2 − √3)	✓ 1
A1 ½ m²	✓ 1
Part (b)
M1 (2 + √3)² + (2 − √3)²	✓ 1
M1 (7 + 4√3) + (7 − 4√3)	✓ 1
A1 14, so the longest side is √14 m	✓ 1

Exam tip: Expand each bracket separately and watch the √3 terms cancel when you add.

+ Add to test✓ In your test316IGME3-1-14r34-14Integers, fractions, decimals and percentages Paper HHard6 marks

A wind farm produces 480 GWh in its first year. Its yearly output then falls by 2.5% each year.

(a)

Work out the output in the fifth year. Give your answer correct to 3 significant figures.

(3)

(b)

In which year of operation does the output first fall below 400 GWh? You must show your working.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiplier 0.975	✓ 1
M1 480 × 0.975⁴	✓ 1
A1 434 GWh	✓ 1
Part (b)
M1 480 × 0.975⁷ = 402.1 (year 8) or equivalent trial	✓ 1
M1 480 × 0.975⁸ = 392.0 (year 9) or equivalent trial	✓ 1
A1 year 9	✓ 1

Common error: The fifth year needs four reductions, not five, because the first year is the starting value.

+ Add to test✓ In your test317IGME3-1-14r39-11Integers, fractions, decimals and percentages Paper HHard6 marks

A volcanologist records the ratio of two tremor frequencies as the recurring decimal 0.272727… (the digits 27 repeat).

(a)

Show that 0.272727… = 3/11.

(3)

(b)

A second ratio is 0.416666… (only the 6 repeats). Write this recurring decimal as a fraction in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 let x = 0.272727… and 100x = 27.2727…	✓ 1
M1 subtract: 99x = 27	✓ 1
A1 x = 27/99 = 3/11	✓ 1
Part (b)
M1 100x = 41.666… and 1000x = 416.666…	✓ 1
M1 subtract: 900x = 375	✓ 1
A1 5/12	✓ 1

Common error: Choose the multipliers so that the recurring parts line up exactly; when only one digit repeats, you need two different powers of 10 that differ by one place.

+ Add to test✓ In your test318IGME3-1-14r39-13Ratio, proportion and degree of accuracy Paper HHard7 marks

Three ships A, B and C carry cargo. The mass carried by A to the mass carried by B is in the ratio 3 : 5. The mass carried by B to the mass carried by C is in the ratio 2 : 3. Together the three ships carry 12 400 tonnes.

(a)

Show that the ratio of the masses carried by A : B : C is 6 : 10 : 15.

(2)

(b)

Work out the mass carried by ship C.

(2)

(c)

A mass of x tonnes is moved from ship C to ship A so that they then carry equal masses. Work out x.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 make the B parts equal (3 : 5 = 6 : 10 and 2 : 3 = 10 : 15)	✓ 1
A1 6 : 10 : 15	✓ 1
Part (b)
M1 12 400 ÷ (6 + 10 + 15) × 15	✓ 1
A1 6000 tonnes	✓ 1
Part (c)
M1 A carries 2400 tonnes (12 400 ÷ 31 × 6) (ECF)	✓ 1
M1 2400 + x = 6000 − x (ECF from (b))	✓ 1
A1 1800	✓ 1

Common error: To combine two ratios, make the shared quantity (here B) the same number in both before joining them.

+ Add to test✓ In your test319IGME3-1-14r40-13Ratio, proportion and degree of accuracy Paper HHard5 marks

The members of a gaming guild are experienced players and new players in the ratio 7 : 3. Then 8 experienced players leave the guild and 4 new players join. The ratio of experienced players to new players is now 5 : 4.

(a)

Work out the number of experienced players in the guild at the start. You must show your working.

(4)

(b)

Work out the total number of players in the guild at the start.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 starts with 7k experienced and 3k new players	✓ 1
M1 (7k − 8) : (3k + 4) = 5 : 4, giving 4(7k − 8) = 5(3k + 4)	✓ 1
M1 28k − 32 = 15k + 20, so 13k = 52	✓ 1
A1 k = 4, so 28 experienced players	✓ 1
Part (b)
B1 40 (ECF from (a))	✓ 1

Exam tip: Represent the starting numbers with a single letter multiplier (7k and 3k), apply the changes, then use the new ratio to form an equation.

+ Add to test✓ In your test320IGME3-1-14r40-14Surds Paper HHard6 marks

A ramp in a video game level is modelled as a right-angled triangle. The two perpendicular sides have lengths (2 + √3) m and (2 − √3) m.

Diagram not to scale

(a)

Show that the area of the triangle is exactly ½ m².

(3)

(b)

Work out the length of the hypotenuse of the triangle. Give your answer as a surd in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (2 + √3)(2 − √3)	✓ 1
M1 expands to 4 − 2√3 + 2√3 − 3 (or difference of two squares 4 − 3)	✓ 1
A1 ½ × 1 = ½	✓ 1
Part (b)
M1 (2 + √3)² + (2 − √3)²	✓ 1
M1 expands to (7 + 4√3) + (7 − 4√3) = 14	✓ 1
A1 √14 m	✓ 1

Exam tip: Expanding (a + √b)² gives a middle term 2a√b; opposite signs in the two brackets mean these terms cancel when added.

+ Add to test✓ In your test321IGME3-1-14r41-13Integers, fractions, decimals and percentages Paper HHard6 marks

A phone is sold in a sale for €396 after a reduction of 12%.

(a)

Work out the original price of the phone.

(2)

(b)

The trade-in value of the phone falls by 18% each year from its original price of €450. Work out the trade-in value after 3 years, to the nearest cent.

(2)

(c)

Work out the overall percentage fall in the trade-in value over the 3 years. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 396 ÷ 0.88	✓ 1
A1 €450	✓ 1
Part (b)
M1 450 × 0.82³	✓ 1
A1 €248.12	✓ 1
Part (c)
M1 (1 − 0.82³) × 100, or (450 − 248.12) ÷ 450 × 100	✓ 1
A1 44.9%	✓ 1

Common error: A reverse percentage problem needs a division by the multiplier, not an increase by the same percentage of the reduced price.

+ Add to test✓ In your test322IGME3-1-14r41-14Surds Paper HHard7 marks

Use surds and algebra to answer the following.

(a)

Expand and simplify (2 + √3)(5 − √3).

(2)

(b)

Find positive integers a and b such that (a + b√3)² = 28 + 10√3.

(4)

(c)

Hence write √(28 + 10√3) in the form p + q√3.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 10 − 2√3 + 5√3 − 3	✓ 1
A1 7 + 3√3	✓ 1
Part (b)
M1 expand to a² + 3b² + 2ab√3	✓ 1
M1 equate: a² + 3b² = 28 and 2ab = 10	✓ 1
M1 use ab = 5 to test (a, b) = (5, 1) or (1, 5)	✓ 1
A1 a = 5, b = 1	✓ 1
Part (c)
5 + √3 (= 6.7320508…)	✓ 1

Exam tip: Compare the rational parts and the surd parts separately when two expressions with surds are equal.

+ Add to test✓ In your test323IGME3-1-14r42-07Ratio, proportion and degree of accuracy Paper HHard8 marks

A cyclist rides a distance of 42 km, measured correct to the nearest kilometre.
The ride takes 1 hour 45 minutes, measured correct to the nearest 5 minutes.

(a)

Write down the lower bound and the upper bound for the distance, and for the time in minutes.

(2)

(b)

Work out the upper bound for the cyclist's average speed in km/h. Give your answer correct to 3 significant figures.

(3)

(c)

Work out the lower bound for the average speed. Hence write the average speed to a suitable degree of accuracy, giving a reason.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 distance: 41.5 ≤ d < 42.5	✓ 1
B1 time: 102.5 ≤ t < 107.5 minutes	✓ 1
Part (b)
M1 largest distance ÷ smallest time, 42.5 ÷ 102.5	✓ 1
M1 converts minutes to hours, e.g. × 60	✓ 1
A1 24.9 km/h (24.878…)	✓ 1
Part (c)
M1 smallest distance ÷ largest time, 41.5 × 60 ÷ 107.5	✓ 1
A1 23.2 km/h (23.16…)	✓ 1
A1 20 km/h to 1 significant figure, because both bounds round to 20	✓ 1

Exam tip: For a speed (distance ÷ time), the upper bound uses the upper bound of distance and the lower bound of time.

+ Add to test✓ In your test324IGME3-1-14r42-11Surds Paper HHard8 marks

A rectangular trench has length (2 + √3) m and area (10 + 6√3) m².

Diagram not to scale

(a)

Show that the width of the trench is (2 + 2√3) m.

(3)

(b)

Work out the perimeter of the trench. Give your answer in the form a + b√3.

(2)

(c)

Show that the square of the length of a diagonal of the trench is (23 + 12√3) m².

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 width = (10 + 6√3) ÷ (2 + √3)	✓ 1
M1 multiplies top and bottom by (2 − √3)	✓ 1
A1 (20 − 10√3 + 12√3 − 18) ÷ (4 − 3) = 2 + 2√3	✓ 1
Part (b)
M1 2 × ((2 + √3) + (2 + 2√3))	✓ 1
A1 8 + 6√3	✓ 1
Part (c)
M1 diagonal² = (2 + √3)² + (2 + 2√3)²	✓ 1
M1 expands (2 + √3)² = 7 + 4√3 or (2 + 2√3)² = 16 + 8√3	✓ 1
A1 7 + 4√3 + 16 + 8√3 = 23 + 12√3 (≈ 43.8)	✓ 1

Exam tip: To divide by a surd expression such as (2 + √3), multiply top and bottom by its conjugate (2 − √3).

+ Add to test✓ In your test325IGME3-1-14r42-12Integers, fractions, decimals and percentages Paper HHard7 marks

This question is about recurring decimals.

(a)

Show that 0.272727… (the digits 27 repeat forever) = 3/11.

(3)

(b)

Write 0.416666… (the digit 6 repeats forever) as a fraction in its simplest form.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 lets x = 0.2727… and 100x = 27.2727…	✓ 1
M1 100x − x = 27, so 99x = 27	✓ 1
A1 x = 27/99 = 3/11	✓ 1
Part (b)
M1 forms two multiples with the same recurring part, e.g. 1000x = 416.666… and 100x = 41.666…	✓ 1
M1 subtracts: 900x = 375	✓ 1
A1 x = 375/900, or an equivalent fraction	✓ 1
A1 5/12 (≈ 0.417)	✓ 1

Exam tip: Choose the multiples of x so that the recurring digits line up exactly, then subtract to remove them.

+ Add to test✓ In your test326IGME3-1-14r43-07Surds Paper HHard7 marks

Work out each part exactly.

(a)

Simplify \(\sqrt{72} - \sqrt{18} + \sqrt{8}\). Give your answer in the form \(a\sqrt{2}\).

(2)

(b)

Express \(\dfrac{7 + \sqrt{3}}{1 + \sqrt{3}}\) in the form \(a + b\sqrt{3}\), where \(a\) and \(b\) are integers.

(3)

(c)

In a game, a rectangular hitbox has area 40 cm² and width \(\sqrt{10}\) cm. Work out its length. Give your answer as a surd in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 any two of \(6\sqrt{2}\), \(3\sqrt{2}\), \(2\sqrt{2}\) seen	✓ 1
A1 \(5\sqrt{2}\)	✓ 1
Part (b)
M1 multiplies numerator and denominator by \(1 - \sqrt{3}\)	✓ 1
M1 numerator \(4 - 6\sqrt{3}\) and denominator \(-2\)	✓ 1
A1 \(-2 + 3\sqrt{3}\)	✓ 1
Part (c)
M1 \(\dfrac{40}{\sqrt{10}}\) rationalised, or \(\dfrac{40\sqrt{10}}{10}\)	✓ 1
A1 \(4\sqrt{10}\) cm	✓ 1

Exam tip: Rationalise by multiplying top and bottom by the conjugate; the denominator becomes a difference of two squares.

+ Add to test✓ In your test327IGME3-1-14r43-08Ratio, proportion and degree of accuracy Paper HHard5 marks

A rectangular water tank has a base of length 2.4 m and width 1.5 m, each correct to 1 decimal place. The depth of the tank is 1.2 m, correct to 2 significant figures. 1 m³ = 1000 litres.

(a)

Write down the upper bound for the area of the base of the tank.

(2)

(b)

Show that the tank is certain to hold at least 3900 litres when full.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 upper bounds 2.45 and 1.55	✓ 1
A1 3.7975 m²	✓ 1
Part (b)
M1 lower bounds 2.35, 1.45 and 1.15 multiplied	✓ 1
A1 volume 3.918625 m³ = 3918.6... litres	✓ 1
A1 3918.6 > 3900, so the tank must hold at least 3900 litres	✓ 1

Common error: Using a mixture of upper and lower bounds. For the smallest possible volume, use the lower bound of every dimension.

+ Add to test✓ In your test328IGME3-1-14r43-12Integers, fractions, decimals and percentages Paper HHard5 marks

In a game, a drop rate is displayed as the recurring decimal 0.2777... in which the digit 7 repeats forever. A second rate is 0.181818... in which the digits 18 repeat forever.

(a)

Show that 0.2777... = \(\frac{5}{18}\).

(3)

(b)

Write 0.181818... as a fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 lets x = 0.2777... and forms 10x = 2.777... and 100x = 27.777...	✓ 1
M1 subtracts: 100x − 10x = 25, so 90x = 25	✓ 1
A1 x = \(\frac{25}{90} = \frac{5}{18}\)	✓ 1
Part (b)
M1 100x − x = 18, so 99x = 18	✓ 1
A1 \(\frac{2}{11}\)	✓ 1

Exam tip: Multiply by 10 for one repeating digit and by 100 for two, so that the repeating tails line up and cancel when you subtract.

+ Add to test✓ In your test329IGME3-1-12r47-11Surds Paper HHard6 marks

Two steel cables have lengths (5 + √3) m and (5 − √3) m.

(a)

Show that (5 + √3) ÷ (5 − √3) can be written as (14 + 5√3)/11.

(3)

(b)

Show that (5 + √3)² + (5 − √3)² is an integer.

(2)

(c)

Work out the ratio of the longer length to the shorter length as a decimal, correct to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiplies top and bottom by (5 + √3)	✓ 1
M1 numerator 25 + 10√3 + 3 = 28 + 10√3 and denominator 25 − 3 = 22	✓ 1
A1 divides by 2 to reach (14 + 5√3)/11	✓ 1
Part (b)
M1 expands both brackets, e.g. 28 + 10√3 and 28 − 10√3	✓ 1
A1 56	✓ 1
Part (c)
B1 2.06 (from 2.0600...)	✓ 1

Exam tip: To rationalise a denominator of the form a − √b, multiply by a + √b: the product is a whole number.

+ Add to test✓ In your test330IGME3-1-13r48-10Surds Paper HHard6 marks

A spacecraft travels a distance of \(\dfrac{18}{\sqrt{6}}\) million km on one leg of a journey.

(a)

Show that this distance can be written as \(3\sqrt{6}\) million km.

(2)

(b)

Write \(\dfrac{5+\sqrt{3}}{5-\sqrt{3}}\) in the form \(a+b\sqrt{3}\), where \(a\) and \(b\) are rational numbers.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiplies numerator and denominator by \(\sqrt{6}\)	✓ 1
A1 \(\dfrac{18\sqrt{6}}{6}=3\sqrt{6}\)	✓ 1
Part (b)
M1 multiplies numerator and denominator by \(5+\sqrt{3}\)	✓ 1
M1 denominator \(25-3=22\)	✓ 1
M1 numerator \(25+10\sqrt{3}+3\) or \(28+10\sqrt{3}\)	✓ 1
A1 \(\dfrac{14}{11}+\dfrac{5}{11}\sqrt{3}\)	✓ 1

Exam tip: To rationalise a denominator of the form \(p-\sqrt{q}\), multiply top and bottom by \(p+\sqrt{q}\) so the surds disappear in the denominator.

+ Add to test✓ In your test331IGME3-1-13r48-13Ratio, proportion and degree of accuracy Paper HHard7 marks

A train travels 84 km in 55 minutes. The distance is correct to the nearest kilometre and the time is correct to the nearest minute.

(a)

Write down the lower bound and the upper bound for the distance, and the lower bound and the upper bound for the time.

(2)

(b)

Work out the upper bound of the average speed of the train in km/h. Give your answer correct to 3 significant figures.

(3)

(c)

Work out the lower bound of the average speed of the train in km/h. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 distance 83.5 km and 84.5 km	✓ 1
B1 time 54.5 minutes and 55.5 minutes	✓ 1
Part (b)
M1 upper bound of distance ÷ lower bound of time, 84.5 ÷ 54.5	✓ 1
M1 × 60 to convert minutes to hours	✓ 1
A1 93.0	✓ 1
Part (c)
M1 83.5 ÷ 55.5 × 60	✓ 1
A1 90.3	✓ 1

Exam tip: For the largest speed, use the largest distance and the smallest time; for the smallest speed, use the smallest distance and the largest time.

+ Add to test✓ In your test332IGME3-1-13r48-14Set language, notation and Venn diagrams (both tiers) Paper HHard6 marks

50 triathletes were asked whether they train in swimming (S) or cycling (C). Let \(x\) be the number who train in both. The number who train in swimming only is \(x+5\). The number who train in cycling only is \(2x-1\). 6 triathletes train in neither.

Diagram not to scale

(a)

Show that \(4x+10=50\).

(2)

(b)

Solve \(4x+10=50\) and find the number of triathletes who train in cycling.

(2)

(c)

One of the triathletes who trains in cycling is chosen at random. Find the probability that this triathlete also trains in swimming.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sum of regions \((x+5)+x+(2x-1)+6\)	✓ 1
A1 \(=4x+10\) and equates to 50	✓ 1
Part (b)
M1 \(x=10\)	✓ 1
A1 number in cycling \(=x+(2x-1)=29\)	✓ 1
Part (c)
M1 10/29 or \(\dfrac{n(S\cap C)}{n(C)}\) (ECF from (b))	✓ 1
A1 \(\dfrac{10}{29}\)	✓ 1

Common error: Conditional probability uses the smaller group as the denominator: only the 29 cyclists, not all 50 triathletes.

+ Add to test✓ In your test333IGME3-1-14r48-11Ratio, proportion and degree of accuracy Paper HHard6 marks

A rectangular piece of fabric has length 2.4 m, correct to 1 decimal place, and width 1.35 m, correct to 2 decimal places.

(a)

Write down the upper bound of the length and the lower bound of the width.

(2)

(b)

Work out the upper bound of the area of the fabric.

(2)

(c)

Work out the lower bound of the area of the fabric.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2.45	✓ 1
B1 1.345	✓ 1
Part (b)
M1 2.45 × 1.355	✓ 1
A1 3.31975	✓ 1
Part (c)
M1 2.35 × 1.345	✓ 1
A1 3.16075	✓ 1

Common error: Upper bounds are used for both length and width when finding the upper bound of an area.

+ Add to test✓ In your test334IGME3-1-14r48-12Powers, roots and standard form Paper HHard5 marks

A tank contains cod and haddock in the ratio 3 : 5. Then 12 cod are added and 4 haddock are removed. After this, the numbers of cod and haddock are equal.

(a)

Let the original numbers be \(3k\) cod and \(5k\) haddock. Show that \(3k+12=5k-4\).

(2)

(b)

Solve the equation and work out the original total number of fish in the tank.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 new numbers \(3k+12\) and \(5k-4\)	✓ 1
A1 equates them	✓ 1
Part (b)
M1 \(2k=16\)	✓ 1
A1 \(k=8\)	✓ 1
A1 \(8\times 3+8\times 5=64\)	✓ 1

Exam tip: Write ratios as multiples such as 3k and 5k, then translate the words of the question into an equation.

+ Add to test✓ In your test335IGME3-1-14r48-14Surds Paper HHard5 marks

A rectangular fashion banner has area \(8\) cm² and width \((\sqrt{5}-1)\) cm.

Diagram not to scale

(a)

Show that the length of the banner is \((2+2\sqrt{5})\) cm.

(3)

(b)

Work out the perimeter of the banner. Give your answer in the form \(a+b\sqrt{5}\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length \(=\dfrac{8}{\sqrt{5}-1}\)	✓ 1
M1 multiplies top and bottom by \(\sqrt{5}+1\)	✓ 1
A1 \(\dfrac{8(\sqrt{5}+1)}{4}=2+2\sqrt{5}\)	✓ 1
Part (b)
M1 \(2[(2+2\sqrt{5})+(\sqrt{5}-1)]\)	✓ 1
A1 \(2+6\sqrt{5}\)	✓ 1

Exam tip: To rationalise \(\dfrac{8}{\sqrt{5}-1}\), multiply by \(\dfrac{\sqrt{5}+1}{\sqrt{5}+1}\) to use the difference of two squares.