Edexcel IGCSE Maths A (Modular) · Higher · Unit 1

IGCSE Maths Past Paper Solutions: 4WM1H/01 Specimen Paper (Unit 1 Higher)

Full worked solutions to the Pearson Edexcel International GCSE Mathematics A (Modular) Unit 1 Higher Tier specimen paper (4WM1H/01, Sample Assessment Materials, February 2024): 23 questions, 100 marks, 2 hours, calculator allowed. Every final answer below matches the official Pearson mark scheme. Attempt each question yourself first, then click “Show solution” to check your working step by step.

11 markHigherSpecimen 2024 · Unit 1H (4WM1H/01)

The weight of a cake is 2.75 kg, correct to 2 decimal places. Write down the lower bound of the weight of the cake.

Show solution
  1. Correct to 2 decimal places means the true value is within half of 0.01 (= 0.005) either side of 2.75.
    Lower bound = 2.75 − 0.005 = 2.745 kg
23 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

Show that 513 − 267 = 21021

Show solution
  1. Step 1: convert to improper fractions. 513 = 163 and 267 = 207
  2. Step 2: use a common denominator of 21. 163 = 11221 and 207 = 6021
  3. Step 3: subtract. 11221 − 6021 = 5221 = 21021 ✓ (since 52 = 2 × 21 + 10)
Exam tip: this is a “show that” question, so every stage must be written out, including the common-denominator fractions. Writing only the answer scores 0.
34 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

Here is a biased 4-sided spinner. The table gives the probabilities that, when the spinner is spun once, it will land on 1 or it will land on 3.

Number1234
Probability0.260.18

The probability that the spinner will land on 2 is equal to the probability that the spinner will land on 4. Priya is going to spin the spinner 250 times. Work out an estimate for the number of times the spinner will land on a 1 or a 2.

Show solution
  1. Step 1: All probabilities add to 1, so P(2) + P(4) = 1 − (0.26 + 0.18) = 0.56
  2. Step 2: P(2) = P(4), so P(2) = 0.56 ÷ 2 = 0.28
  3. Step 3: P(1 or 2) = 0.26 + 0.28 = 0.54
  4. Step 4: Expected number = 0.54 × 250 = 135
45 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

(a) Expand and simplify (n − 6)(n + 4)

(b) Solve 2x − 3 = 3x − 54  Show clear algebraic working.

Show solution
  1. (a) (n − 6)(n + 4) = n² + 4n − 6n − 24 = n² − 2n − 24
  2. (b) Multiply both sides by 4: 4(2x − 3) = 3x − 5 → 8x − 12 = 3x − 5
    Collect terms: 8x − 3x = −5 + 12 → 5x = 7
    x = 7/5 (= 1.4)
53 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

A plane flies from New York City to Los Angeles. The plane flies a distance of 3980 kilometres in 5 hours 24 minutes. Work out the average speed of the plane. Give your answer in kilometres per hour correct to the nearest whole number.

Show solution
  1. Step 1: convert the time to hours. 24 minutes = 24 ÷ 60 = 0.4 hours, so the time is 5.4 hours.
  2. Step 2: Speed = distance ÷ time = 3980 ÷ 5.4 = 737.03…
  3. To the nearest whole number: 737 km/h
Common mistake: writing 5 hours 24 minutes as 5.24 hours gives 3980 ÷ 5.24 = 759.5, which is wrong. Always divide minutes by 60.
64 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

The diagram shows an 8-sided shape ABCDEFGH.

20 cm ABCD EFGH 5 cm5 cm 12 cm12 cm 28 cm
Diagram not accurately drawn

HG = 28 cm, AH = FG = 12 cm, AB = EF = 5 cm. The height of the shape is 20 cm. CD is parallel to HG. The area of shape ABCDEFGH is 434 cm². Find the length of CD.

Show solution
  1. Step 1: split the shape into the 28 cm × 12 cm rectangle at the bottom and a trapezium BCDE on top.
    Rectangle area = 28 × 12 = 336 cm²
  2. Step 2: trapezium dimensions. Bottom side BE = 28 − 5 − 5 = 18 cm. Height = 20 − 12 = 8 cm.
    Trapezium area = 434 − 336 = 98 cm²
  3. Step 3: use the trapezium area formula. ½(18 + CD) × 8 = 98 → 4(18 + CD) = 98 → 18 + CD = 24.5
  4. CD = 6.5 cm
74 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

(a) Simplify 8 × (4t)0

x6 ÷ x−5 = xp

(b) Find the value of p

(c) Simplify fully (2k2m4)3

Show solution
  1. (a) Anything to the power 0 equals 1, so (4t)0 = 1 and 8 × 1 = 8
  2. (b) When dividing, subtract the powers: 6 − (−5) = 11, so p = 11
  3. (c) Cube every part: 2³ = 8, (k²)³ = k6, (m4)³ = m12
    Answer: 8k⁶m¹²
83 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

Change a speed of 81 kilometres per hour to a speed in metres per second.

Show solution
  1. Step 1: 81 km = 81 × 1000 = 81 000 m
  2. Step 2: 1 hour = 60 × 60 = 3600 seconds
  3. Step 3: 81 000 ÷ 3600 = 22.5 m/s
Shortcut: to convert km/h to m/s, divide by 3.6 (81 ÷ 3.6 = 22.5).
97 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

(a) Simplify 3a4b5 × 4a7b2

(b) Factorise fully 14x2y4 + 21x3y2

The diagram shows a straight line drawn on a grid.

xy O−112 12345−1

(c) Find an equation of the line.

A different straight line has equation y = 3x − 5

(d) Write down the coordinates of the point at which the line crosses the y-axis.

Show solution
  1. (a) Multiply the numbers and add the powers: 3 × 4 = 12, a4+7 = a11, b5+2 = b7
    Answer: 12a¹¹b⁷
  2. (b) The HCF of 14 and 21 is 7; the lowest powers are x² and y². Take out 7x²y²:
    14x²y4 ÷ 7x²y² = 2y², and 21x³y² ÷ 7x²y² = 3x
    Answer: 7x²y²(2y² + 3x)
  3. (c) The line crosses the y-axis at (0, 4), so c = 4. It passes through (2, 0), so the gradient = (0 − 4) ÷ (2 − 0) = −2.
    Equation: y = −2x + 4
  4. (d) In y = mx + c, the line crosses the y-axis at (0, c): (0, −5)
106 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

The diagram shows a quadrilateral ABCD.

ABCD 6 cm7.5 cm
Diagram not accurately drawn

In the diagram, ABC and DAC are right-angled triangles (right angles at B and A). BC = 6 cm, AC = 7.5 cm. The area of quadrilateral ABCD is 31.5 cm².

(a) Work out the length of AB.

(b) Work out the length of AD.

Show solution
  1. (a) Triangle ABC is right-angled at B, and AC is the hypotenuse. By Pythagoras:
    AB = √(7.5² − 6²) = √(56.25 − 36) = √20.25 = 4.5 cm
  2. (b) Step 1: Area of triangle ABC = ½ × 6 × 4.5 = 13.5 cm²
  3. Step 2: Area of triangle DAC = 31.5 − 13.5 = 18 cm²
  4. Step 3: Triangle DAC is right-angled at A, so AC and AD are the base and height: ½ × 7.5 × AD = 18 → AD = 18 ÷ 3.75 = 4.8 cm
116 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

Tess has a bag containing 11 beads. 4 of the beads are red. 7 of the beads are blue. Tess takes at random a bead from the bag and keeps it. Tess takes at random another bead from the bag and keeps it.

(a) Complete the probability tree diagram.

First beadSecond bead redblue redblueredblue 4/117/11 3/107/10 …………

(b) Work out the probability that both of the beads are red.

(c) Work out the probability that the two beads are of different colours.

Show solution
  1. (a) If the first bead is blue, 10 beads remain: 4 red and 6 blue.
    Missing branches: red = 4/10 blue = 6/10
  2. (b) P(red, red) = 411 × 310 = 12110 = 6/55
  3. (c) “Different colours” means red then blue or blue then red:
    P(R, B) = 411 × 710 = 28110   P(B, R) = 711 × 410 = 28110
    Total = 56110 = 28/55
125 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

(a) Factorise 9x2 − 4y2

(b) Express 78 − x + 34x as a single fraction in its simplest form.

Show solution
  1. (a) This is a difference of two squares: (3x)² − (2y)²
    Answer: (3x + 2y)(3x − 2y)
  2. (b) The lowest common denominator of 8 and 4x is 8x.
    78 = 7x8x and x + 34x = 2(x + 3)8x
    Subtract: 7x − 2x − 68x = 5x − 68x
Watch the sign: the minus applies to the whole of 2(x + 3), giving −2x − 6, not −2x + 6.
136 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

(a) Expand and simplify (3x − 1)(x + 2)(3x + 1)

(b) Simplify fully (2x58xy2)−2

Show solution
  1. (a) Step 1: Multiply the two brackets that make a difference of two squares: (3x − 1)(3x + 1) = 9x² − 1
  2. Step 2: Multiply by the third bracket: (9x² − 1)(x + 2) = 9x³ + 18x² − x − 2
    Answer: 9x³ + 18x² − x − 2
  3. (b) Step 1: simplify inside the bracket. 2x58xy2 = x44y2
  4. Step 2: the negative power flips the fraction. (x44y2)−2 = (4y2x4)2
  5. Step 3: square. Answer: 16y⁴x⁸
148 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

100 farmers are asked if they have goats (G), sheep (S) or chickens (C) on their farms. Of these farmers: 31 have sheep; 53 have chickens; 6 have goats, sheep and chickens; 11 have sheep and goats; 17 have sheep and chickens; 18 have goats and chickens; 20 do not have any goats, sheep or chickens.

(a) Using this information, complete the Venn diagram to show the number of farmers in each appropriate subset.

ℰ GSC

(b) Find (i) n(G)   (ii) n([G ∪ S]′)   (iii) n(G′ ∩ C)

One of the farmers who has chickens is chosen at random.

(c) Find the probability that this farmer also has goats.

Show solution
ℰ GSC 1359 6 1211 24 20
  1. (a) Work from the centre outwards.
    1. All three: 6
    2. Sheep and goats only: 11 − 6 = 5; sheep and chickens only: 17 − 6 = 11; goats and chickens only: 18 − 6 = 12
    3. Sheep only: 31 − (5 + 6 + 11) = 9
    4. Chickens only: 53 − (12 + 6 + 11) = 24
    5. None: 20
    6. Goats only: 100 − (5 + 9 + 6 + 12 + 11 + 24 + 20) = 100 − 87 = 13
  2. (b)(i) n(G) = 13 + 5 + 6 + 12 = 36
  3. (b)(ii) [G ∪ S]′ is everything outside both G and S: 24 + 20 = 44
  4. (b)(iii) G′ ∩ C is chickens but not goats: 11 + 24 = 35
  5. (c) The farmer is chosen only from the 53 with chickens. Of those, the ones with goats as well: 12 + 6 = 18.
    Probability = 18/53
154 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

Martin and Lucia went on two different car journeys.

For Martin’s journey: distance = 80 km correct to the nearest 5 km; time = 2.7 hours correct to 1 decimal place.
For Lucia’s journey: distance = 33 km correct to 2 significant figures; time = 1 hour correct to the nearest 0.1 hour.

Martin says, “My average speed could have been greater than Lucia’s average speed.” By considering bounds, show that Martin is correct. Show your working clearly.

Show solution
  1. Step 1: bounds for Martin. Distance: 77.5 ≤ d < 82.5. Time: 2.65 ≤ t < 2.75
  2. Step 2: bounds for Lucia. Distance: 32.5 ≤ d < 33.5. Time: 0.95 ≤ t < 1.05
  3. Step 3: Martin’s fastest possible speed uses upper bound distance ÷ lower bound time:
    82.5 ÷ 2.65 = 31.13… km/h
  4. Step 4: Lucia’s slowest possible speed uses lower bound distance ÷ upper bound time:
    32.5 ÷ 1.05 = 30.95… km/h
  5. Conclusion: 31.13… > 30.95…, so Martin’s maximum speed is greater than Lucia’s minimum speed. Martin is correct ✓
Key idea: to make a quotient as large as possible, divide the largest numerator by the smallest denominator; to make it as small as possible, do the opposite.
163 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

The histogram shows information about the total time, m minutes, taken by each child in a school to walk to school every day for one week.

020406080100 Time (m minutes) Freq.density
Redrawn for illustration: bars are 0–20, 20–30, 30–40, 40–60, 60–75 and 75–100. Use the original paper’s grid for exact readings.

There are no children for whom m > 100. There are 10 children for whom m ≤ 20. Work out an estimate for the number of children for whom 50 < m ≤ 80.

Show solution
  1. Step 1: find the frequency density scale. The 0–20 bar holds 10 children, so its frequency density = 10 ÷ 20 = 0.5. Reading the rest of the graph against this scale gives: 40–60 bar FD = 2.9, 60–75 bar FD = 3.2, 75–100 bar FD = 2.
  2. Step 2: frequency = frequency density × class width for each part of 50 < m ≤ 80:
    1. 50–60 (part of the 40–60 bar): 10 × 2.9 = 29
    2. 60–75: 15 × 3.2 = 48
    3. 75–80 (part of the 75–100 bar): 5 × 2 = 10
  3. Step 3: Total = 29 + 48 + 10 = 87
Alternative method: on the original grid, 25 small squares represent 5 children, so 1 small square = 0.2 children. Counting the small squares between 50 and 80 gives 145 + 240 + 50 = 435 squares, and 435 × 0.2 = 87.
174 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

Express 3 + √8(√2 − 1)2 in the form p + √q where p and q are integers. Show clearly each stage of your working.

Show solution
  1. Step 1: expand the denominator. (√2 − 1)² = 2 − 2√2 + 1 = 3 − 2√2
  2. Step 2: simplify √8. √8 = 2√2, so the numerator is 3 + 2√2. The expression becomes 3 + 2√23 − 2√2
  3. Step 3: rationalise by multiplying top and bottom by (3 + 2√2):
    Denominator: (3 − 2√2)(3 + 2√2) = 9 − 8 = 1
    Numerator: (3 + 2√2)² = 9 + 12√2 + 8 = 17 + 12√2
  4. Step 4: write 12√2 as a single surd. 12√2 = √(144 × 2) = √288
    Answer: 17 + √288 (p = 17, q = 288)
184 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

The diagram shows two identical circles drawn inside a rectangle.

4x cm2x cm
Diagram not accurately drawn

The length of the rectangle is 4x cm and the width of the rectangle is 2x cm. Each circle has a radius of x cm. The circles touch each other and each circle touches three sides of the rectangle. The region inside the rectangle that is outside the circles, shown shaded in the diagram, has a total area of 20 cm². Work out the perimeter of the rectangle. Give your answer correct to 3 significant figures.

Show solution
  1. Step 1: shaded area in terms of x. Rectangle − 2 circles = 4x × 2x − 2πx² = 8x² − 2πx²
  2. Step 2: set equal to 20. x²(8 − 2π) = 20 → x² = 208 − 2π = 11.649… → x = 3.4131…
  3. Step 3: perimeter. 2(4x + 2x) = 12x = 12 × 3.4131… = 40.957…
  4. To 3 significant figures: 41.0 cm
Accuracy: keep the full calculator value of x until the final step. The mark scheme accepts any answer that rounds to 41.
194 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

The diagram shows a triangular prism, ABCDEF, with a rectangular base ABCD.

EDABCF 2.2 cm18°6 cm
Diagram not accurately drawn

AB = 6 cm, DE = 2.2 cm, angle DAE = 18°, angle ADE = 90°. Work out the angle that BE makes with the plane ABCD. Give your answer correct to one decimal place.

Show solution
θ EDAB 2.2 cmDB
E sits directly above D, so the angle between BE and the base is angle DBE (θ) in the right-angled triangle EDB.
  1. Step 1: find AD in triangle ADE (right angle at D): tan 18° = 2.2AD → AD = 2.2tan 18° = 6.7709… cm
  2. Step 2: find DB, the diagonal of the rectangular base: DB = √(6.7709…² + 6²) = 9.0468… cm
  3. Step 3: find the angle. E is directly above D, so in triangle EDB (right angle at D): tan θ = 2.29.0468… = 0.2432…
  4. θ = tan⁻¹(0.2432…) = 13.67…° = 13.7°
Key idea: the angle between a line and a plane is measured to the line’s projection on the plane. Here the projection of BE onto ABCD is BD.
204 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

Find the values of a, b and c so that 7 + 12x − 2x2 is written as a − b(x − c)2.

Show solution
  1. Step 1: factor −2 out of the x terms. 7 + 12x − 2x² = 7 − 2(x² − 6x)
  2. Step 2: complete the square inside the bracket. x² − 6x = (x − 3)² − 9
  3. Step 3: substitute back and simplify. 7 − 2[(x − 3)² − 9] = 7 + 18 − 2(x − 3)² = 25 − 2(x − 3)²
  4. a = 25 b = 2 c = 3
Alternative method: expand a − b(x − c)² = a − bc² + 2bcx − bx² and compare coefficients: b = 2, then 2bc = 12 gives c = 3, then a − bc² = 7 gives a = 25.
213 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

Express (20x2 − 36 − 2x − 6) × 14 − x as a single fraction in its simplest form. Show clear algebraic working.

Show solution
  1. Step 1: factorise. x² − 36 = (x − 6)(x + 6), so the common denominator is (x − 6)(x + 6).
  2. Step 2: combine the bracket. 20(x − 6)(x + 6) − 2(x + 6)(x − 6)(x + 6) = 20 − 2x − 12(x − 6)(x + 6) = 8 − 2x(x − 6)(x + 6)
  3. Step 3: factorise the numerator. 8 − 2x = 2(4 − x)
  4. Step 4: multiply by 14 − x and cancel (4 − x): 2(4 − x)(x − 6)(x + 6)(4 − x) = 2x² − 36
223 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

18 × (√27)4n + 66 × 92n + 8 = 3x

Express x in terms of n. Show your working clearly and simplify your expression.

Show solution
  1. Step 1: write everything as a power of 3.
    1. 186 = 3 = 31
    2. √27 = 27½ = 33/2, so (√27)4n+6 = 3(3/2)(4n+6) = 36n+9
    3. 9 = 3², so 92n+8 = 32(2n+8) = 34n+16
  2. Step 2: combine using the index laws. 31 × 36n+934n+16 = 31 + 6n + 9 − 4n − 16 = 32n − 6
  3. So x = 2n − 6 (or 2(n − 3))
Common mistake: writing √27 as 3√3 instead of 33/2. You need a single power of 3 to apply the index laws.
236 marksHigherSpecimen 2024 · Unit 1H (4WM1H/01)

ABCD is a kite. AB = AD and CB = CD. The point B has coordinates (k, 1) where k is a negative constant. The point D has coordinates (8, 7). The straight line L passes through the points B and D and has a gradient of 35.

Find an equation of AC. Give your answer in the form px + qy = r where p, q and r are integers. Show your working clearly.

Show solution
  1. Key fact: in a kite, the diagonal AC is the perpendicular bisector of the other diagonal BD. So AC passes through the midpoint of BD and is perpendicular to it.
  2. Step 1: find k. Gradient of BD = 7 − 18 − k = 35 → 30 = 24 − 3k → k = −2. So B = (−2, 1).
  3. Step 2: midpoint of BD. (−2 + 82, 1 + 72) = (3, 4)
  4. Step 3: gradient of AC. Perpendicular gradients multiply to −1, so m = −53
  5. Step 4: equation through (3, 4). y − 4 = −53(x − 3) → 3y − 12 = −5x + 15
  6. Rearrange: 5x + 3y = 27
Check: substitute the midpoint (3, 4): 5(3) + 3(4) = 15 + 12 = 27 ✓

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