Edexcel International GCSE Maths A (Modular) · Higher · Unit 1

Edexcel International GCSE Maths Solutions: 4WM1H/01 May 2025 (Unit 1 Higher)

Full worked solutions to the Pearson Edexcel International GCSE Mathematics A (Modular) Unit 1 Higher Tier exam paper from May 2025 (4WM1H/01, sat Thursday 15 May 2025): 25 questions, 100 marks, 2 hours, calculator allowed. Attempt each question yourself first, then click “Show solution” to check your working step by step.

13 marksHigherMay 2025 · Unit 1H (4WM1H/01)

Without using a calculator, work out 323 × 247

Show each stage of your working. Give your answer as a mixed number in its simplest form.

Show solution

Step 1: convert to improper fractions. 323 = 113 and 247 = 187
Step 2: multiply. 113 × 187 = 19821 = 667 (or cancel the 3 into 18 first: 11 × 67 = 667)
Step 3: convert back to a mixed number. 66 ÷ 7 = 9 remainder 3, so the answer is 937

Exam tip: “without a calculator” means the improper fractions and the multiplication must be written out. The answer alone does not get full marks.

24 marksHigherMay 2025 · Unit 1H (4WM1H/01)

The height of a tree is 17.8 metres correct to one decimal place.

(a) Write down the lower bound for the height of the tree.

The weight of an apple is 145 grams correct to the nearest 5 grams.

(b) Write down the upper bound for the weight of the apple.

A = 4.766 × 8150.399

(c) By rounding each number to one significant figure, work out an estimate for the value of A. Show your working clearly.

Show solution

(a) One decimal place means ± 0.05: 17.8 − 0.05 = 17.75 m
(b) Nearest 5 grams means ± 2.5: 145 + 2.5 = 147.5 g
(c) To 1 significant figure: 4.766 → 5, 815 → 800, 0.399 → 0.4
5 × 8000.4 = 40000.4 = 10 000

34 marksHigherMay 2025 · Unit 1H (4WM1H/01)

Here is a biased spinner. The spinner can land on pink or on green or on red or on yellow or on blue. The table gives information about the probability that, when the spinner is spun, it will land on each colour.

Colour	pink	green	red	yellow	blue
Probability	0.14	0.23	2x	0.18	7x

Jim is going to spin the spinner 500 times. Work out the expected number of times the spinner will land on blue.

Show solution

Step 1: The probabilities add to 1: 0.14 + 0.23 + 0.18 + 2x + 7x = 1 → 0.55 + 9x = 1
Step 2: 9x = 0.45 → x = 0.05
Step 3: P(blue) = 7x = 7 × 0.05 = 0.35
Step 4: Expected number = 0.35 × 500 = 175

43 marksHigherMay 2025 · Unit 1H (4WM1H/01)

The diagram shows triangle ABC and a semicircle.

Diagram not accurately drawn

AB = 15 cm, BC = 20 cm, AC = 25 cm, angle ABC = 90°. The diameter of the semicircle is 8 cm. Work out the area of the region shown shaded in the diagram. Give your answer correct to 3 significant figures.

Show solution

Step 1: area of the triangle. The right angle is at B, so ½ × 15 × 20 = 150 cm²
Step 2: area of the semicircle. Radius = 8 ÷ 2 = 4 cm. ½ × π × 4² = 8π = 25.132… cm²
Step 3: shaded area = triangle − semicircle. 150 − 25.132… = 124.867… = 125 cm² (3 s.f.)

Watch out: 8 cm is the diameter. Using 8 as the radius gives 49.5 cm², which is wrong.

54 marksHigherMay 2025 · Unit 1H (4WM1H/01)

x⁵ × x⁷ = x^m (a) Find the value of m

y⁸ ÷ y³ = yⁿ (b) Find the value of n

Show solution

(a) When multiplying, add the powers: 5 + 7, so m = 12
(b) When dividing, subtract the powers: 8 − 3, so n = 5
(c) Cube each part: 5³ = 125, (a⁴)³ = a¹², (r²)³ = r⁶
Answer: 125a¹²r⁶

63 marksHigherMay 2025 · Unit 1H (4WM1H/01)

The diagram shows a block in the shape of a cuboid.

Diagram not accurately drawn

The block has a horizontal base with length 6 m and width x m. The block is placed on a table so that the whole of its horizontal base is in contact with the table. The block exerts a force of 702 newtons on the table. The pressure on the table due to the block is 65 newtons/m². Work out the value of x.

pressure = forcearea

Show solution

Step 1: rearrange the formula. area = forcepressure = 70265 = 10.8 m²
Step 2: the base is a 6 × x rectangle. 6x = 10.8 → x = 1.8

76 marksHigherMay 2025 · Unit 1H (4WM1H/01)

(a) Solve x − 4 = 3 + 2x6 Show clear algebraic working.

(b) (i) Factorise y² − 11y + 30

(ii) Hence solve y² − 11y + 30 = 0

Show solution

(a) Multiply both sides by 6: 6(x − 4) = 3 + 2x → 6x − 24 = 3 + 2x
Collect terms: 4x = 27 → x = 27/4 (= 6.75)
(b)(i) Find two numbers that multiply to +30 and add to −11: −5 and −6.
Answer: (y − 5)(y − 6)
(b)(ii) Set each bracket equal to 0: y = 5 or y = 6

85 marksHigherMay 2025 · Unit 1H (4WM1H/01)

ℰ = {18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}
A = {even numbers} B = {multiples of 3} C = {multiples of 5}

(a) List the members of the set (A ∪ B)′

Sophie writes down the statement B ∩ C = Ø

(b) Explain why Sophie’s statement is wrong.

D is a set such that A ∩ D = {18, 26}. The set D has exactly 5 members.

Show solution

(a) A = {18, 20, 22, 24, 26, 28, 30} and B = {18, 21, 24, 27, 30}.
A ∪ B = {18, 20, 21, 22, 24, 26, 27, 28, 30}
The complement is everything in ℰ that is not in A ∪ B: 19, 23, 25, 29
(b) C = {20, 25, 30}. The number 30 is a multiple of both 3 and 5, so B ∩ C = {30}, which is not empty.
(c) D must contain 18 and 26 and no other even numbers, so its other 3 members must be odd numbers from ℰ (19, 21, 23, 25, 27, 29).
One possible answer: D = {18, 19, 21, 23, 26}

Note: part (c) has many correct answers. Any set with 18, 26 and exactly three different odd numbers from ℰ is right.

93 marksHigherMay 2025 · Unit 1H (4WM1H/01)

7^x = 1 (a) Write down the value of x

8⁻⁴ × 8¹¹8¹² = 8ⁿ (b) Work out the value of n

Show solution

(a) Any non-zero number to the power 0 is 1, so x = 0
(b) Numerator: 8^{−4 + 11} = 8⁷. Then 8⁷ ÷ 8¹² = 8^{7 − 12} = 8⁻⁵
n = −5

105 marksHigherMay 2025 · Unit 1H (4WM1H/01)

Diagram not accurately drawn

ABCD is a trapezium with one line of symmetry. Angle ADC = 60°, AD = 12 cm, DC = 47 cm. Work out the area of the trapezium. Give your answer correct to 3 significant figures. Show your working clearly.

Show solution

Step 1: drop perpendiculars from A and B to DC, making a right-angled triangle at each end.
Step 2: height. h = 12 sin 60° = 10.392… cm
Step 3: horizontal overhang. 12 cos 60° = 6 cm at each end (the trapezium is symmetrical).
Step 4: top side. AB = 47 − 6 − 6 = 35 cm
Step 5: area. ½(35 + 47) × 10.392… = 41 × 10.392… = 426.08… = 426 cm² (3 s.f.)

116 marksHigherMay 2025 · Unit 1H (4WM1H/01)

(a) Expand and simplify 5x(5x + 4)(2x − 3)

(b) Express 94 + y − 75y as a single fraction in its simplest form.

Show solution

(a) Step 1: (5x + 4)(2x − 3) = 10x² − 15x + 8x − 12 = 10x² − 7x − 12
Step 2: 5x(10x² − 7x − 12) = 50x³ − 35x² − 60x
(b) The lowest common denominator of 4 and 5y is 20y.
94 = 45y20y and y − 75y = 4(y − 7)20y = 4y − 2820y
Add: 45y + 4y − 2820y = 49y − 2820y

Equivalent form: the numerator factorises to 7(7y − 4), giving 7(7y − 4)20y. Nothing cancels with 20y, so both forms are fully simplified.

124 marksHigherMay 2025 · Unit 1H (4WM1H/01)

Ricardo is going to play one game of badminton and one game of tennis. He can either win or lose each game. The probability that he will win the game of badminton is 0.7. The probability that he will win the game of tennis is 0.4.

(a) Complete the probability tree diagram.

(b) Work out the probability that Ricardo loses both games.

Show solution

(a) Badminton: win 0.7, lose 1 − 0.7 = 0.3. Tennis (on both pairs of branches): win 0.4, lose 1 − 0.4 = 0.6
(b) Multiply along the “lose, lose” path: 0.3 × 0.6 = 0.18

133 marksHigherMay 2025 · Unit 1H (4WM1H/01)

(a) Complete the table of values for y = x³ + 2x + 3

x	−3	−2	−1	0	1	2	3
y		−9	0	3	6		36

(b) On the grid, draw the graph of y = x³ + 2x + 3 for −3 ≤ x ≤ 3

Show solution

(a) x = −3: (−3)³ + 2(−3) + 3 = −27 − 6 + 3 = −30
x = 2: 2³ + 2(2) + 3 = 8 + 4 + 3 = 15
(b) Plot all seven points, (−3, −30), (−2, −9), (−1, 0), (0, 3), (1, 6), (2, 15) and (3, 36), and join them with a smooth curve.

Exam tip: draw one smooth freehand curve through the points. Joining the points with straight ruler lines loses the mark.

144 marksHigherMay 2025 · Unit 1H (4WM1H/01)

The diagram shows sector OAB of a circle, centre O. OA = OB = 12 cm, angle AOB = 40°.

Diagram not accurately drawn

(a) Work out the area of the sector. Give your answer correct to 3 significant figures.

The diagram shows triangle XYZ. XY = 7 cm, YZ = 9 cm, angle XYZ = 115°.

Diagram not accurately drawn

(b) Work out the area of the triangle. Give your answer correct to 3 significant figures.

Show solution

(a) A sector is a fraction θ360 of the whole circle:
Area = 40360 × π × 12² = 19 × 144π = 16π = 50.265… = 50.3 cm²
(b) Two sides and the angle between them are given, so use Area = ½ab sin C:
½ × 7 × 9 × sin 115° = 31.5 × 0.9063… = 28.548… = 28.5 cm²

153 marksHigherMay 2025 · Unit 1H (4WM1H/01)

Solve the equation 5x² + 9x − 17 = 0. Show your working clearly. Give your solutions correct to 3 significant figures.

Show solution

Step 1: Use the quadratic formula with a = 5, b = 9, c = −17:
x = −9 ± √(9² − 4 × 5 × (−17))2 × 5 = −9 ± √42110
Step 2: √421 = 20.518…
x = −9 + 20.518…10 = 1.1518… or x = −9 − 20.518…10 = −2.9518…
x = 1.15 or x = −2.95

Common mistake: −4 × 5 × (−17) is plus 340, so the number under the root is 81 + 340 = 421, not 81 − 340.

163 marksHigherMay 2025 · Unit 1H (4WM1H/01)

Solve the equation 8^{3y + 4} × 4^3y = 2^5y. Show your working clearly.

Show solution

Step 1: write everything as a power of 2. 8 = 2³ and 4 = 2²
8^3y+4 = 2^3(3y+4) = 2^9y+12 and 4^3y = 2^2(3y) = 2^6y
Step 2: add the powers on the left. 2^9y+12+6y = 2^15y+12
Step 3: equate the powers. 15y + 12 = 5y → 10y = −12 → y = −1.2 (= −6/5)

174 marksHigherMay 2025 · Unit 1H (4WM1H/01)

(√3)⁵ = k√3 where k is an integer.

(a) Find the value of k

(b) Show that 213 − √2 can be written in the form c + √d where c and d are integers. Show each stage of your working clearly.

Show solution

(a) (√3)⁵ = (√3)⁴ × √3 = 9√3, so k = 9
(b) Step 1: rationalise by multiplying top and bottom by (3 + √2):
21(3 + √2)(3 − √2)(3 + √2)
Step 2: denominator. (3 − √2)(3 + √2) = 9 − 2 = 7
Step 3: simplify. 21(3 + √2)7 = 3(3 + √2) = 9 + 3√2
Step 4: write 3√2 as a single surd. 3√2 = √(9 × 2) = √18
So 213 − √2 = 9 + √18 (c = 9, d = 18) ✓

183 marksHigherMay 2025 · Unit 1H (4WM1H/01)

The histogram shows information about the times, in minutes, that some people were in a shop.

Work out an estimate for the proportion of these people who were in the shop for more than 40 minutes.

Show solution

Step 1: frequency = frequency density × class width for every bar:

Time (min) 0–10 10–15 15–30 30–50 50–60

FD 2.5 4 3.6 0.5 1.7

Frequency 25 20 54 10 17

Total number of people = 25 + 20 + 54 + 10 + 17 = 126
Step 2: people with more than 40 minutes. Half of the 30–50 bar (40–50: 10 × 0.5 = 5) plus the whole 50–60 bar (17): 5 + 17 = 22
Step 3: proportion. 22126 = 11/63 (≈ 0.175)

Time (min)	0–10	10–15	15–30	30–50	50–60
FD	2.5	4	3.6	0.5	1.7
Frequency	25	20	54	10	17

Key idea: the question asks for a proportion, so divide by the total. 22 on its own is not the answer. Any equivalent fraction or decimal (22/126, 0.175) is fine.

193 marksHigherMay 2025 · Unit 1H (4WM1H/01)

Simplify fully 3x² − 3y²5x + 5y ÷ xy² − x²y10xy

Show clear algebraic working.

Show solution

Step 1: factorise everything.
1. 3x² − 3y² = 3(x − y)(x + y)
2. 5x + 5y = 5(x + y)
3. xy² − x²y = xy(y − x) = −xy(x − y)
Step 2: flip the second fraction and multiply. 3(x − y)(x + y)5(x + y) × 10xy−xy(x − y)
Step 3: cancel (x + y), (x − y) and xy: 3 × 105 × (−1) = −6

Watch the sign: y − x = −(x − y). Missing this gives +6, which is wrong. Check: x = 2, y = 1 gives 915 ÷ −220 = −6 ✓

203 marksHigherMay 2025 · Unit 1H (4WM1H/01)

Given that k = m + n and m = 34n, express 7k5 − m in the form a + bn²cn − d where a, b and c are integers and d is prime.

Show solution

Step 1: write k as a single fraction. k = 34n + n = 3 + 4n²4n, so 7k = 7(3 + 4n²)4n
Step 2: write 5 − m as a single fraction. 5 − 34n = 20n − 34n
Step 3: divide. The 4n denominators cancel: 7(3 + 4n²)20n − 3
Step 4: expand the numerator. 21 + 28n²20n − 3 (a = 21, b = 28, c = 20, d = 3, and 3 is prime ✓)

215 marksHigherMay 2025 · Unit 1H (4WM1H/01)

Diagram not accurately drawn

In the diagram, AB = 9.4 cm, AD = 12.8 cm, angle BAD = 72°, angle BDC = 39° and angle BCD = 54°. Work out the length of BC. Give your answer correct to 3 significant figures.

Show solution

Step 1: find BD with the cosine rule in triangle ABD (two sides and the angle between them):
BD² = 9.4² + 12.8² − 2 × 9.4 × 12.8 × cos 72°
= 88.36 + 163.84 − 74.36… = 177.83…
BD = 13.335… cm
Step 2: find BC with the sine rule in triangle BCD. BC is opposite the 39° angle and BD is opposite the 54° angle:
BCsin 39° = 13.335…sin 54° → BC = 13.335… × sin 39°sin 54° = 10.373…
BC = 10.4 cm (3 s.f.)

Accuracy: keep the unrounded value of BD in your calculator for step 2. Rounding BD to 13.3 first gives 10.3, which is the wrong answer.

223 marksHigherMay 2025 · Unit 1H (4WM1H/01)

A box contains 20 counters. 9 of the counters are red, 7 of the counters are yellow, 4 of the counters are green. Alex takes at random three counters from the box. Work out the probability that exactly two of the three counters are the same colour.

Show solution

Method: 1 − P(all three the same) − P(all three different). The counters are not replaced.
Step 1: P(all the same colour).
1. RRR: 920 × 819 × 718 = 5046840
2. YYY: 720 × 619 × 518 = 2106840
3. GGG: 420 × 319 × 218 = 246840
Total = 7386840
Step 2: P(all different colours). One order is 920 × 719 × 418 = 2526840. There are 3! = 6 orders, so the total is 15126840
Step 3: P(exactly two the same) = 1 − 7386840 − 15126840 = 45906840 = 51/76 (≈ 0.671)

Direct method (same answer): add up every “two of one colour, one of another” case. For example, P(two red, one not red) = 3 × 920 × 819 × 1118 = 23766840. Doing the same for yellow (16386840) and green (5766840) gives a total of 45906840 = 5176.

236 marksHigherMay 2025 · Unit 1H (4WM1H/01)

The diagram shows cuboid ABCDEFGH.

Diagram not accurately drawn

The point P lies on CF. AD = 12 cm, DE = 10 cm, CP = 3 cm. The angle of elevation of P from A is 24°. Calculate the size of angle APG. Give your answer correct to the nearest degree. Show your working clearly.

Show solution

P′ is the point on the base directly below P. Angle APG is the angle at P in the shaded triangle APG.

Step 1: the point below P. P is 3 cm along the 10 cm edge, so the point P′ directly below it on the base is 12 cm along and 3 cm across from A.
AP′ = √(12² + 3²) = √153 = 12.369… cm
Step 2: height of the cuboid from the angle of elevation: h = PP′ = √153 × tan 24° = 5.5071… cm
Step 3: the three sides of triangle APG.
1. AP = √153cos 24° = 13.539… cm
2. PG lies in the top face: 12 cm along and 10 − 3 = 7 cm across, so PG = √(12² + 7²) = √193 = 13.892… cm
3. AG is the diagonal of the end face ABGH (10 cm by h): AG = √(10² + 5.5071…²) = 11.416… cm
Step 4: cosine rule for angle APG.
cos APG = AP² + PG² − AG²2 × AP × PG = 183.33… + 193 − 130.33…2 × 13.539… × 13.892… = 246376.20… = 0.6539…
APG = cos⁻¹(0.6539…) = 49.16…° = 49°

Nice check: AP² − AG² = (153 + h²) − (100 + h²) = 53 exactly, so the numerator is 53 + 193 = 246 whatever the height is.

245 marksHigherMay 2025 · Unit 1H (4WM1H/01)

PQRS is a square. PR is a diagonal of the square. P is the point with coordinates (4, 7). R is the point with coordinates (8, −5). Find an equation of the straight line that passes through the points Q and S. Give your answer in the form ay = bx + c where a, b and c are integers.

Show solution

Key fact: the diagonals of a square bisect each other at right angles. So QS passes through the midpoint of PR and is perpendicular to PR.
Step 1: midpoint of PR. (4 + 82, 7 + (−5)2) = (6, 1)
Step 2: gradient of PR. −5 − 78 − 4 = −124 = −3
Step 3: gradient of QS. Perpendicular gradients multiply to −1, so m = 13
Step 4: equation through (6, 1). y − 1 = 13(x − 6) → 3y − 3 = x − 6
3y = x − 3

Check: substitute the midpoint (6, 1): 3(1) = 6 − 3 ✓

255 marksHigherMay 2025 · Unit 1H (4WM1H/01)

(a) Express 8 − 12x − 4x² in the form a − b(x + c)² where a, b and c are constants.

(b) Hence solve 8 − 12x − 4x² = 0. Show your working clearly. Give your solutions in surd form.

Show solution

(a) Step 1: factor −4 out of the x terms. 8 − 12x − 4x² = 8 − 4(x² + 3x)
Step 2: complete the square. x² + 3x = (x + 32)² − 94
Step 3: substitute back. 8 − 4[(x + 32)² − 94] = 8 + 9 − 4(x + 32)²
17 − 4(x + 1.5)² (a = 17, b = 4, c = 32)
(b) 17 − 4(x + 32)² = 0 → (x + 32)² = 174 → x + 32 = ±√172
x = −3 + √172 or −3 − √172

Check: these decimals are 0.5616 and −3.5616, and substituting either into 8 − 12x − 4x² gives 0 ✓. “Hence” means you must use part (a). Do not switch to the quadratic formula.

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