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Practice Questions

Edexcel IGCSE Mathematics: Numbers and the Number System — Practice Questions

Original exam-style practice questions with full worked answers on fractions, ratio, percentages, standard form and bounds.

Subject
Mathematics
Level
IGCSE
Topic
Numbers and the number system
Updated

Aligned to Pearson Edexcel IGCSE Mathematics (4MA1), Specification Issue 2, November 2017. Official specification .

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These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Numbers and the Number System revision notes


Section A

1. Write 0.375 as a fraction in its simplest form. [2]

2. Divide £450 in the ratio 4 : 5. [3]

Section B

3. Calculate, giving answers in standard form:

(a) (3.2 × 10⁵) × (2.5 × 10⁻³) [2] (b) (8.4 × 10⁷) ÷ (4.0 × 10²) [2]

4. A population increases from 48 000 to 55 200.

(a) Calculate the percentage increase. [3] (b) If it continues to grow at this rate, calculate the population after a further 3 years. [3]

5. The value of a car depreciates by 18% each year. It is bought for $24 000.

(a) Calculate its value after 3 years. [3] (b) Calculate the total percentage loss over the 3 years. [2]

6. x = 6.2 and y = 3.5, both correct to 1 decimal place.

(a) Write the upper and lower bounds of x and y. [2] (b) Calculate the lower bound of x ÷ y. [3]

7. Express 0.2̇7̇ as a fraction in its simplest form. [3]


Answers

1. 375/1000 [1] = 3/8 [1].

2. Total parts = 9 [1]; one part = £50 [1]; £200 and £250 [1].

3. (a) 3.2 × 2.5 = 8; 10⁵ × 10⁻³ = 10² [1]; = 8 × 10² [1]. (b) 8.4 ÷ 4.0 = 2.1; 10⁷ ÷ 10² = 10⁵ [1]; = 2.1 × 10⁵ [1].

4. (a) Increase = 7200 [1]; (7200 ÷ 48 000) × 100 [1] = 15% [1]. (b) 55 200 × 1.15³ [1] [1] = 83 952 (to the nearest whole number) [1].

5. (a) 24 000 × 0.82³ [1] [1] = $13 233 (to the nearest dollar) [1]. (b) Loss = 24 000 − 13 233 = 10 767 [1]; (10 767 ÷ 24 000) × 100 = 44.9% [1].

6. (a) x: 6.15 to 6.25 [1]; y: 3.45 to 3.55 [1]. (b) For the lowest quotient use the smallest numerator and the largest denominator [1]: 6.15 ÷ 3.55 [1] = 1.732 (to 3 d.p.) [1].

7. Let x = 0.272727… Then 100x = 27.2727… [1]; 99x = 27 [1]; x = 27/99 = 3/11 [1].


Section C — additional questions

8. x = 6.2 and y = 3.5 (as in question 6, both correct to 1 decimal place). Calculate the upper bound of x ÷ y. [3]

9. A shop increases all its prices by 8%, then reduces the new prices by 8% in a subsequent sale. Explain, with a calculation, whether the final price is the same as the original price. [4]

10. Simplify (2.4 × 10⁶) ÷ (6.0 × 10⁻²), giving your answer in standard form. [2]

Answers to Section C

8. For the highest quotient, use the largest numerator and the smallest denominator [1]: 6.25 ÷ 3.45 [1] = 1.812 (to 3 d.p.) [1].

9. No, the final price is not the same as the original [1]. Taking an original price of $100: after an 8% increase, the price is $100 × 1.08 = $108 [1]; after an 8% decrease on this new price, the price is $108 × 0.92 = $99.36 [1], which is less than the original $100, because the second 8% is calculated on the larger, increased price rather than on the original amount [1].

10. 2.4 ÷ 6.0 = 0.4; 10⁶ ÷ 10⁻² = 10⁸ [1]; 0.4 × 10⁸ is not in correct standard form, so it must be rewritten as 4 × 10⁷ [1].

A note on percentage increase followed by percentage decrease

Question 9 illustrates a common misconception worth revising directly: applying the same percentage as an increase and then a decrease does not return a value to its original amount, because the two percentages are calculated on different base values. An 8% increase is calculated on the original price, but the following 8% decrease is calculated on the new, already-increased price, which is a larger number — so the absolute amount subtracted in the decrease is larger than the absolute amount added in the increase, leaving the final value below the starting point. This same reasoning explains why a percentage decrease followed by an equal percentage increase also fails to return to the original value, and recognising this pattern helps avoid an intuitive but incorrect assumption that equal and opposite percentage changes cancel out.

A note on choosing bounds for division

Questions 6(b) and 8 both test the same underlying rule for finding the bounds of a quotient, and it is worth stating precisely because it differs from the rule for a sum or product: to find the lowest possible value of x ÷ y, pair the smallest possible numerator with the largest possible denominator, and to find the highest possible value, pair the largest possible numerator with the smallest possible denominator. This is because dividing by a larger number produces a smaller result and dividing by a smaller number produces a larger result, the opposite relationship to what governs a sum or product of two rounded measurements, which is why bounds questions on division consistently catch out candidates who apply the same “both upper” or “both lower” shortcut that works for addition and multiplication.

Where marks are usually lost

  • Dividing by the number of parts rather than finding one part first.
  • Using 0.18 rather than 0.82 as the depreciation multiplier.
  • Using the lower bound of both numbers when dividing.
  • Not simplifying the fraction fully.

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