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International GCSE Maths · Pearson Edexcel 4MA1 · Unit 1

Numbers and the number system

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

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+ Add to test✓ In your 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.

(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.

(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 test20IGME-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 test21IGME-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 test22IGME-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 test23IGME-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 test24IGME-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 test25IGME-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 test26IGME-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 test27IGME-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 test28IGME-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

(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 test29IGME-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 test30IGME-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 test31IGME-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 test32IGME-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

(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 test33IGME-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 test34IGME-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 test35IGME-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 test36IGME-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 test37IGME-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 test38IGME-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 test39IGME-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 test40IGME-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 test41IGME-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 test42IGME-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 test43IGME-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 test44IGME-058-04-05Surds Paper HEasy3 marks

A square garden tile has an area of 45 cm².

(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 test45IGME-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 test46IGME-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 test47IGME-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 test48IGME-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 test49IGME-058-04-07Surds Paper HMedium5 marks

A rectangular solar panel has length $(5+\sqrt{3})$ m and width $(5-\sqrt{3})$ m.

(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 test50IGME-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 test51IGME-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 test52IGME-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 test53IGME-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.

(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 test54IGME-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 test55IGME-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 test56IGME-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 test57IGME-058-01-12Powers, roots and standard form 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 test58IGME-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 test59IGME-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 test60IGME-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 test61IGME-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 test62IGME-058-04-12Surds Paper HHard6 marks

In triangle $PQR$, $PQ=(2+\sqrt{3})$ cm, $PR=(2-\sqrt{3})$ cm and angle $QPR=60°$.

(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 test63IGME-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 test64IGME-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 test65IGME-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 test66IGME-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 test67IGME-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 test68IGME-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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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