Edexcel International GCSE Maths Upper and Lower Bounds Past Paper Questions and Solutions (4MA1 Higher)

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Edexcel International GCSE Maths A (4MA1) · Higher · Numbers and the Number System

Bounds and Degree of Accuracy: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every bounds and degree of accuracy question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 14 questions worth 46 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.

12 marksHigherJune 2025 · Paper 1HR (4MA1/1HR) · Q2

The length of a ship is 142.8 m, correct to 1 decimal place.

(i) Write down the lower bound of the length of the ship. (1)

(ii) Write down the upper bound of the length of the ship. (1)

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  1. Correct to 1 decimal place means correct to the nearest 0.1 m. The bounds are half of 0.1 (that is 0.05) either side.

Part (i)

  1. 142.8 − 0.05

Answer: 142.75 m

Part (ii)

  1. 142.8 + 0.05

Answer: 142.85 m

24 marksHigherJanuary 2020 · Paper 1HR (4MA1/1HR) · Q11

The weight of a cat is 4.3 kg correct to 2 significant figures.

(a) Write down the upper bound of the weight of the cat. (1)

(b) Write down the lower bound of the weight of the cat. (1)

G = e − f, where e = 17 correct to the nearest integer and f = 9.4 correct to one decimal place.

(c) Work out the upper bound for the value of G. (2)

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Part (a)

  1. 2 significant figures here means to the nearest 0.1 kg, so add half of 0.1.

Answer: 4.35 kg

Part (b)

  1. Subtract half of 0.1.

Answer: 4.25 kg

Part (c)

  1. e is between 16.5 and 17.5; f is between 9.35 and 9.45.
  2. For the biggest difference, take the biggest e and the smallest f: 17.5 − 9.35

Answer: 8.15

33 marksHigherNovember 2021 · Paper 1H (4MA1/1H) · Q13

Here is a triangle XYZ with XZ = 16 cm, angle YXZ = 130° and angle XYZ = 25°. The length XZ and the angles YXZ and XYZ are each given correct to 2 significant figures. Calculate the upper bound for the length YZ. Give your answer correct to one decimal place. Show your working clearly.

XYZ130°25°16 cm
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  1. Sine rule: YZ⁄sin X = XZ⁄sin Y, so YZ = XZ × sin X⁄sin Y
  2. Bounds (2 s.f.): XZ is from 15.5 to 16.5; angle X from 125° to 135°; angle Y from 24.5° to 25.5°.
  3. For the biggest YZ, make the top as large as possible and the bottom as small as possible.
  4. sin X is largest when X = 125° (sin 125° = 0.819… is bigger than sin 135° = 0.707…, because sine decreases after 90°).
  5. sin Y is smallest when Y = 24.5°.
  6. YZ = 16.5 × sin 125°⁄sin 24.5° = 32.59…

Answer: 32.6 cm

42 marksHigherJanuary 2023 · Paper 1HR (4MA1/1HR) · Q14

T = p⁄r. p = 0.51 correct to 2 significant figures. r = 6.3 correct to 2 significant figures. Work out the upper bound for the value of T. Show your working clearly.

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  1. Bounds: p from 0.505 to 0.515; r from 6.25 to 6.35
  2. To make a division as big as possible, use the largest top and the smallest bottom.
  3. Upper bound of T = 0.515⁄6.25 = 0.0824

Answer: 0.0824

53 marksHigherJanuary 2021 · Paper 1H (4MA1/1H) · Q15

a = v − u⁄t. v = 9.6 correct to 1 decimal place, u = 3.8 correct to 1 decimal place, and t = 1.84 correct to 2 decimal places. Calculate the upper bound for the value of a. Give your answer as a decimal correct to 2 decimal places. Show your working clearly.

Show solution
  1. Bounds: v from 9.55 to 9.65; u from 3.75 to 3.85; t from 1.835 to 1.845.
  2. Largest a: largest v, smallest u (it is subtracted), smallest t (we divide by it).
  3. a = 9.65 − 3.75⁄1.835 = 5.9⁄1.835 = 3.215…

Answer: 3.22

63 marksHigherJanuary 2020 · Paper 1H (4MA1/1H) · Q16

P = 2a − c⁄d. a = 58.4 correct to 3 significant figures, c = 20 correct to 2 significant figures and d = 3.6 correct to 2 significant figures. Work out the upper bound for the value of P. Show your working clearly. Give your answer correct to 2 decimal places.

Show solution
  1. Bounds: a from 58.35 to 58.45; c from 19.5 to 20.5; d from 3.55 to 3.65.
  2. For the biggest P: biggest 2a, smallest c (because it is subtracted) and smallest d (because we divide by it).
  3. P = 2 × 58.45 − 19.5⁄3.55 = 97.4⁄3.55 = 27.436…

Answer: 27.44

74 marksHigherJune 2020 · Paper 1H (4MA1/1H) · Q17

A metal block has a mass of 5 kg, correct to the nearest 50 grams. The block has a volume of (1.84 × 10−3) m3, correct to 3 significant figures. Work out the upper bound for the density of the block. Give your answer in kg/m3 correct to 1 decimal place. Show your working clearly.

Show solution
  1. Mass bounds (nearest 50 g = 0.05 kg): 4.975 kg to 5.025 kg
  2. Volume bounds (3 s.f.): 1.835 × 10−3 m3 to 1.845 × 10−3 m3
  3. Density = mass ÷ volume. For the upper bound, use the largest mass and the smallest volume.
  4. 5.025 ÷ (1.835 × 10−3) = 2738.41…

Answer: 2738.4 kg/m3

83 marksHigherJanuary 2022 · Paper 1H (4MA1/1H) · Q18

A = w − x2⁄y. w = 3.45 correct to 2 decimal places, x = 1.9 correct to 1 decimal place and y = 5 correct to the nearest whole number. Work out the lower bound of the value of A. Show your working clearly.

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  1. Bounds: w from 3.445 to 3.455; x from 1.85 to 1.95; y from 4.5 to 5.5.
  2. For the smallest A, use the smallest w and subtract the largest possible x2⁄y (largest x, smallest y).
  3. 1.952⁄4.5 = 3.8025⁄4.5 = 0.845
  4. A = 3.445 − 0.845

Answer: 2.6

94 marksHigherJune 2023 · Paper 1HR (4MA1/1HR) · Q18

Diego builds a fence using fence panels. The total length of the fence is 50 metres, correct to the nearest 5 metres. The length of each fence panel is 1.3 metres, correct to the nearest 10 cm. The cost of each fence panel is £8.65. Diego may only buy complete fence panels. Diego only pays for the number of panels he needs to build the fence. Work out the greatest difference in the possible amounts that Diego could pay to build the fence. Show your working clearly.

Show solution
  1. Fence length bounds (nearest 5 m): 47.5 m to 52.5 m. Panel length bounds (nearest 0.1 m): 1.25 m to 1.35 m.
  2. Most panels: longest fence with shortest panels: 52.5 ÷ 1.25 = 42 panels exactly.
  3. Fewest panels: shortest fence with longest panels: 47.5 ÷ 1.35 = 35.18…, so he needs 36 panels (35 panels would not be long enough).
  4. Difference = 42 − 36 = 6 panels
  5. 6 × £8.65

Answer: £51.90

103 marksHigherJune 2020 · Paper 1HR (4MA1/1HR) · Q19

The diagram shows rectangle ABCD with rectangle EFGH cut out to form the shaded region. AD = 8.3 cm, DC = 7.2 cm, EH = 6.2 cm and HG = 5.3 cm, each correct to one decimal place. Work out the upper bound of the area of the shaded region. Show your working clearly.

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  1. Each measurement can be up to 0.05 cm more or less than given.
  2. For the largest shaded area, use the largest outer rectangle and the smallest hole.
  3. Largest ABCD = 8.35 × 7.25 = 60.5375 cm2
  4. Smallest EFGH = 6.15 × 5.25 = 32.2875 cm2
  5. 60.5375 − 32.2875

Answer: 28.25 cm2

113 marksHigherJanuary 2021 · Paper 1HR (4MA1/1HR) · Q19

k = t⁄a − h. t = 14, a = 7.8 and h = 3.4, each correct to 2 significant figures. Work out the lower bound for the value of k. Show your working clearly.

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  1. Bounds: t from 13.5 to 14.5; a from 7.75 to 7.85; h from 3.35 to 3.45.
  2. For the smallest k: smallest top, largest bottom. Largest a − h = 7.85 − 3.35 = 4.5
  3. k = 13.5 ÷ 4.5

Answer: 3

125 marksHigherJanuary 2023 · Paper 1H (4MA1/1H) · Q19

The acceleration, a, of an object is given by a = v − u⁄t where v = 45.23 correct to 2 decimal places, u = 5.12 correct to 2 decimal places and t = 8.5 correct to 2 significant figures. By considering bounds, work out the value of a to a suitable degree of accuracy. Show your working clearly and give a reason for your answer.

Show solution
  1. Bounds: v from 45.225 to 45.235; u from 5.115 to 5.125; t from 8.45 to 8.55.
  2. Upper bound of a = 45.235 − 5.115⁄8.45 = 40.12⁄8.45 = 4.747…
  3. Lower bound of a = 45.225 − 5.125⁄8.55 = 40.1⁄8.55 = 4.690…
  4. Both bounds round to 4.7 (to 2 significant figures), but they do not agree to 3 significant figures (4.75 and 4.69).

Answer: a = 4.7, because both the upper and lower bounds round to 4.7

133 marksHigherJune 2021 · Paper 1H (4MA1/1H) · Q20

P = t − w⁄y. t = 9.7 correct to 1 decimal place, w = 5.9 correct to 1 decimal place, and y = 3 correct to 1 significant figure. Calculate the upper bound for the value of P. Show your working clearly.

Show solution
  1. Bounds: t is from 9.65 to 9.75, w is from 5.85 to 5.95, and y is from 2.5 to 3.5.
  2. To make P as big as possible, make the top as big as possible and the bottom as small as possible.
  3. Biggest top: biggest t minus smallest w = 9.75 − 5.85 = 3.9
  4. Smallest bottom: y = 2.5
  5. P = 3.9 ÷ 2.5

Answer: 1.56

144 marksHigherJanuary 2022 · Paper 1HR (4MA1/1HR) · Q22

The diagram shows triangle ABC with AB = c cm, AC = b cm, angle CAB = x° and angle ACB = y°. c = 11.5 correct to one decimal place, x = 80 correct to the nearest whole number and y = 75 correct to the nearest whole number. Calculate the upper bound for the value of b. Show your working clearly. Give your answer correct to 3 significant figures.

Show solution
  1. Angle B = 180° − x − y. Sine rule: b⁄sin B = c⁄sin y, so b = c × sin(180° − x − y)⁄sin y = c × sin(x + y)⁄sin y
  2. Bounds: c up to 11.55; x from 79.5 to 80.5; y from 74.5 to 75.5.
  3. x + y is between 154° and 156°. Sine gets smaller as the angle increases beyond 90°, so sin(x + y) is largest when x + y = 154°.
  4. sin y is smallest when y = 74.5°. That fits with x = 79.5°, giving x + y = 154°.
  5. b = 11.55 × sin 154°⁄sin 74.5° = 5.254…

Answer: 5.25 cm

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