IGCSE Mathematics: Number — Practice Questions

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: Number revision notes

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Section A

1. Write 84 as a product of its prime factors. [2]

2. Find the HCF and LCM of 84 and 126. [3]

Section B

3. Work out, giving your answers as fractions in their lowest terms:

(a) 3/4 + 2/5 [2] (b) 5/6 ÷ 10/9 [3]

4. A shop increases all prices by 8%, then reduces the new prices by 8% in a sale.

(a) Show that the final price is not the same as the original. [3] (b) Calculate the overall percentage change. [2]

5. $3600 is shared between A, B and C in the ratio 2 : 3 : 7.

(a) Calculate each share. [3] (b) C gives one third of his share to A. Calculate the new ratio A : B : C in its simplest form. [3]

6. Simple interest of $270 is earned on $1500 over 3 years. Calculate the annual rate. [3]

7. Calculate, giving your answer in standard form: (6.4 × 10⁻³) ÷ (1.6 × 10²) [3]

8. A rectangle measures 12.4 cm by 7.8 cm, each to 1 decimal place. Calculate the lower bound of its perimeter. [3]

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Section C

9. A price is $84 after a 20% increase. Calculate the original price. [2]

10. Calculate (4 × 10⁵) × (3 × 10³), giving your answer in standard form. [3]

11. Convert a speed of 72 km/h to m/s. [2]

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Answers

1. 84 = 2 × 2 × 3 × 7 [1] = 2² × 3 × 7 [1].

2. 126 = 2 × 3² × 7 [1]. HCF = 2 × 3 × 7 = 42 [1]; LCM = 2² × 3² × 7 = 252 [1].

3. (a) 15/20 + 8/20 [1] = 23/20 (or 1 3/20) [1]. (b) 5/6 × 9/10 [1] = 45/60 [1] = 3/4 [1].

4. (a) Take an original price of $100. After the increase: 100 × 1.08 = $108 [1]. After the reduction: 108 × 0.92 = $99.36 [1], which is less than $100, because the 8% reduction is calculated on the larger amount [1]. (b) Change = −0.64 ÷ 100 [1] = a decrease of 0.64% [1].

5. (a) Total parts = 12 [1]; one part = $300 [1]; A = $600, B = $900, C = $2100 [1]. (b) C gives away $700 [1]; A = 1300, B = 900, C = 1400 [1]; ratio = 1300 : 900 : 1400 = 13 : 9 : 14 [1].

6. I = PRT ÷ 100, so 270 = 1500 × R × 3 ÷ 100 [1]; 270 = 45R [1]; R = 6% per year [1].

7. 6.4 ÷ 1.6 = 4 [1]; 10⁻³ ÷ 10² = 10⁻⁵ [1]; = 4 × 10⁻⁵ [1].

8. Lower bounds are 12.35 cm and 7.75 cm [1]; perimeter = 2(12.35 + 7.75) [1] = 40.2 cm [1].

9. 84 ÷ 1.2 [1] = $70 [1]. Dividing (not subtracting 20%) is essential — 20% of $84 is not the same as 20% of the original price.

10. 4 × 3 = 12 [1]; 10⁵ × 10³ = 10⁸ [1]; 12 × 10⁸ = 1.2 × 10⁹ (renormalised, since 12 lies outside 1 ⩽ a < 10) [1].

11. 72 km/h = 72 000 m/h [1]; ÷ 3600 = 20 m/s [1]. Both the distance unit (km→m) and the time unit (h→s) must be converted — converting only one is the standard error.

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Where marks are usually lost

• Assuming an 8% rise followed by an 8% fall returns to the original price.
• Dividing the total by the number of names rather than by the number of parts.
• Using the upper bound of one dimension and the lower bound of the other.
• Giving a standard form answer with a first factor outside 1 ≤ a < 10.
• Subtracting the percentage directly from the final value in a reverse-percentage question, instead of dividing by the multiplier.

Examiner report insight

• Working left to right instead of following the order of operations (BIDMAS/PENDMAS) – e.g. treating 28 - 8 / 2 as (28 - 8) / 2 rather than 28 - (8 / 2).
• Rounding only the final answer when a question specifically instructs each value to be rounded first (e.g. “correct each number to 1 significant figure, then calculate”) – the instruction applies before the calculation, not after.
• Misreading which digits recur in a recurring decimal before applying the standard “multiply by 10 to the n, subtract” method – treating it as terminating, or misidentifying the repeating block, invalidates the rest of a correct method.
• After multiplying or dividing two numbers in standard form, mishandling the index arithmetic – indices are added when multiplying and subtracted when dividing, separately from renormalising the mantissa back into the range 1 <= a < 10.
• In a compound unit conversion, converting only the unit that catches the eye and forgetting the other – how many parts need converting depends on what’s changing: km/h to m/h only requires converting the distance unit (km to m), since the time unit (h) is unchanged, but km/h to m/s requires converting both the distance (km to m) AND the time (h to s), since both units differ between the two forms.

Source: Cambridge International, 0580 Mathematics Principal Examiner Report, June 2024 series, Papers 11, 12, 13, 21, 23 (verified 2026-09-02).