Edexcel International GCSE Mathematics A 4MA1: Degree of accuracy and standard form – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 1.8 (Degree of accuracy) and 1.9 (Standard form) of Topic 1 in the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on it. Questions 1–7 use content on both tiers. Questions 8–12 are (Higher): they use the Higher tier only statements on solving problems with bounds and with standard form. A calculator may be used on every 4MA1 paper, but show your working: method marks depend on it. The questions are different from those in the Numbers and the number system practice, so you can use both.

Learn the content first in the study guide and the revision notes. The course hub is Edexcel IGCSE Mathematics and the printable checklist lists every statement.

Questions

1. Round 846 291

(a) to the nearest 1000 [1] (b) to the nearest 100 000 [1]

2. Write

(a) 0.050 784 9 correct to 3 significant figures [1] (b) 0.050 784 9 correct to 2 decimal places [1] (c) 29.996 correct to 2 decimal places [1]

3.

(a) A plank is 47 cm long, correct to the nearest centimetre. Write down the lower bound and the upper bound of its length. [2] (b) A bag of flour has a mass of 0.38 kg, correct to 2 significant figures. Write down the lower bound and the upper bound of the mass. [2] (c) A distance d is 52 000 m, correct to 2 significant figures. Complete the inequality … ≤ d < … [1]

4. Work out an estimate for (7.93 × 0.512) ÷ 0.0198. You must show your working. [3]

5.

(a) Write 63 800 000 in standard form. [1] (b) Write 0.000 905 in standard form. [1] (c) Write 7.14 × 10⁻⁴ as an ordinary number. [1]

6. Write these numbers in order of size. Start with the smallest. [2]

4.2 × 10⁻³ 0.000 39 3.9 × 10⁻² 4.02 × 10⁻⁴

7. Work out each of these. Give your answers in standard form.

(a) (6 × 10⁵) × (7 × 10⁻⁹) [2] (b) (1.8 × 10⁴) ÷ (4.5 × 10⁻³) [2] (c) 3.7 × 10⁵ − 9.4 × 10⁴ [2]

8. (Higher) A metal block has a mass of 640 g, correct to the nearest 10 g. Its volume is 82 cm³, correct to the nearest cm³. Density = mass ÷ volume.

(a) Calculate the upper bound and the lower bound of the density of the block. [4] (b) Use your answers to (a) to write down the density to a suitable degree of accuracy. Give a reason for your answer. [2]

9. (Higher) A rectangular field is 86 m long and 54 m wide. Both measurements are correct to the nearest metre.

(a) Work out the upper bound of the perimeter of the field. [2] (b) Fencing is sold in rolls of exactly 25 m. Work out the least number of rolls that must be bought to be sure of fencing the whole perimeter. [2] (c) Work out the lower bound of the area of the field. [2] (d) The owner says the area is definitely at least 4600 m². Is the owner right? Give a reason. [1]

10. (Higher) A server stores 4.8 × 10¹⁵ bytes of data. The mean size of a file on the server is 1.6 × 10⁶ bytes.

(a) Work out the number of files on the server. Give your answer in standard form. [2] (b) The amount of data stored increases by 35%. Work out the new amount in standard form. [2] (c) 1 terabyte = 10¹² bytes. Write 4.8 × 10¹⁵ bytes in terabytes, as an ordinary number. [1] (d) The original data is copied at 2.5 × 10⁹ bytes per second. Work out how long the copy takes. Give your answer in days, correct to 1 decimal place. [2]

11. (Higher) p = 4 × 10ᵏ, where k is an integer.

(a) Write p² in standard form. Give your answer in terms of k. [2] (b) Write 1/p in standard form. Give your answer in terms of k. [2] (c) p² = 1.6 × 10⁹. Find the value of k. [1]

12. (Higher) A space probe is 4.52 × 10⁸ km from a receiving station, correct to 3 significant figures. A radio signal travels at 3.00 × 10⁵ km per second, correct to 3 significant figures.

Calculate the upper and lower bounds of the time the signal takes to reach the station. Hence write down the time, in seconds, to a suitable degree of accuracy. Give a reason for your answer. [8]

Answers

1. (a) 846 000 [1] (b) 800 000 [1] Examiner insight: An answer like “846” for the nearest 1000 changes the size of the number and scores 0; the place-value zeros are part of the answer.

2. (a) 0.0508 [1] (b) 0.05 [1] (c) 29.99|6 rounds up through 29.99 → 30.00, so 30.00 [1] Examiner insight: In (c) “30” is marked wrong because 2 d.p. requires two digits after the point; in (a) counting the leading zero of 0.05 as a significant figure gives 0.051, which scores 0.

3. (a) Lower bound 46.5 cm [1]; upper bound 47.5 cm [1] (b) Unit is 0.01, half is 0.005 [1]: 0.375 kg and 0.385 kg [1] (c) Unit is 1000, half is 500: 51 500 ≤ d < 52 500 [1] Examiner insight: Bounds written the wrong way round (upper as lower) usually lose the mark for that pair, so label each one clearly.

4. 7.93 → 8, 0.512 → 0.5, 0.0198 → 0.02 [1] (8 × 0.5) ÷ 0.02 = 4 ÷ 0.02 [1] = 200 [1] Examiner insight: The first mark is for showing at least two of the numbers rounded to 1 s.f.; the calculator value 205.06 on its own scores 0 because no estimate is shown.

5. (a) 6.38 × 10⁷ [1] (b) 9.05 × 10⁻⁴ [1] (c) 0.000 714 [1] Examiner insight: 63.8 × 10⁶ has the right value but is not standard form, so it does not earn the mark in (a).

6. Convert to the same form, e.g. 0.000 39 = 3.9 × 10⁻⁴ and 4.02 × 10⁻⁴ = 0.000 402 [1] Order: 0.000 39, 4.02 × 10⁻⁴, 4.2 × 10⁻³, 3.9 × 10⁻² [1] Examiner insight: One mark is usually available for three numbers in the correct order, so converting all four to one form first protects at least the method mark.

7. (a) 6 × 7 = 42 and 10⁵ × 10⁻⁹ = 10⁻⁴, giving 42 × 10⁻⁴ [1] = 4.2 × 10⁻³ [1] (b) 1.8 ÷ 4.5 = 0.4 and 10⁴ ÷ 10⁻³ = 10⁷, giving 0.4 × 10⁷ [1] = 4 × 10⁶ [1] (c) 370 000 − 94 000 [1] = 276 000 = 2.76 × 10⁵ [1] Examiner insight: The method mark is often earned by an unadjusted answer such as 42 × 10⁻⁴, but the accuracy mark needs it in proper standard form.

8. (a) Mass: 635 ≤ m < 645; volume: 81.5 ≤ V < 82.5 [1] Upper bound = 645 ÷ 81.5 [1] = 7.914 g/cm³ (4 s.f.) [1] Lower bound = 635 ÷ 82.5 = 7.697 g/cm³ (4 s.f.) [1] (b) To 2 s.f. the bounds give 7.9 and 7.7, which differ; to 1 s.f. both give 8 [1]. Density = 8 g/cm³ (1 s.f.) [1] Examiner insight: Using 645 ÷ 82.5 for the upper bound is a standard wrong pairing and loses the method mark; the reason in (b) must refer to both bounds rounding to the same value.

9. (a) Upper bounds 86.5 m and 54.5 m [1]; 2 × (86.5 + 54.5) = 282 m [1] (b) 282 ÷ 25 = 11.28 [1], so 12 rolls [1] (c) Lower bounds 85.5 m and 53.5 m; 85.5 × 53.5 [1] = 4574.25 m² [1] (d) No: the lower bound of the area, 4574.25 m², is less than 4600 m², so the area could be under 4600 m² [1] Examiner insight: In (b) using the measured perimeter, 280 m, gives 11.2 and still 12 rolls, but no method mark for bounds is earned; in (d) “No” with no comparison scores 0.

10. (a) (4.8 × 10¹⁵) ÷ (1.6 × 10⁶) [1] = 3 × 10⁹ files [1] (b) 4.8 × 10¹⁵ × 1.35 [1] = 6.48 × 10¹⁵ bytes [1] (c) (4.8 × 10¹⁵) ÷ 10¹² = 4800 terabytes [1] (d) (4.8 × 10¹⁵) ÷ (2.5 × 10⁹) = 1.92 × 10⁶ seconds; ÷ (60 × 60 × 24) [1] = 22.2 days [1] Examiner insight: In (d) the time-conversion step must be visible; dividing by 3600 only (giving hours) and calling it days loses the accuracy mark.

11. (a) p² = 16 × 10²ᵏ [1] = 1.6 × 10²ᵏ⁺¹ [1] (b) 1/p = (1/4) × 10⁻ᵏ = 0.25 × 10⁻ᵏ [1] = 2.5 × 10⁻ᵏ⁻¹ [1] (c) 2k + 1 = 9, so k = 4 [1] Examiner insight: 16 × 10²ᵏ earns the method mark only; the answer mark needs the front number between 1 and 10 with the power adjusted to 2k + 1.

12. Distance bounds: 4.515 × 10⁸ ≤ d < 4.525 × 10⁸ km [1] Speed bounds: 2.995 × 10⁵ ≤ v < 3.005 × 10⁵ km/s [1] Upper bound of time = upper distance ÷ lower speed = (4.525 × 10⁸) ÷ (2.995 × 10⁵) [1] = 1510.85 s [1] Lower bound of time = lower distance ÷ upper speed = (4.515 × 10⁸) ÷ (3.005 × 10⁵) [1] = 1502.50 s [1] To 3 s.f. the bounds give 1510 and 1500, which differ; to 2 s.f. both give 1500 [1] Time = 1500 s (1.5 × 10³ s) to 2 s.f. [1] Examiner insight: The final mark needs both bounds shown rounded to the chosen accuracy; giving 1510 s because it is the “more accurate” bound scores 0 for the conclusion.

Where marks are usually lost

• Dropping place-value zeros when rounding large numbers (846 instead of 846 000).
• Counting the zero in 0.05… as a significant figure.
• Writing 30 instead of 30.00 when 2 d.p. is asked for.
• Estimating with values rounded to the nearest whole number rather than 1 s.f.
• Leaving answers such as 42 × 10⁻⁴ or 0.4 × 10⁷ unadjusted.
• Subtracting 9.4 from 3.7 as if the powers matched.
• (Higher) Pairing upper ÷ upper for the upper bound of a quotient.
• (Higher) Using the measured value, not the upper bound, when the question says “to be sure”.
• (Higher) Stating a “suitable degree of accuracy” without showing both bounds rounded.

Next steps

• Recap with the revision notes.
• Reread the worked examples in the study guide.
• Try more Topic 1 questions in the Numbers and the number system practice.
• Visit the Edexcel IGCSE Mathematics hub and the printable checklist.
• Try one of our free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, first examination June 2018. Published by Pearson Education Limited.