IGCSE Mathematics: Number (Extended) — Practice Questions (Cambridge 0580)

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 — Cambridge International holds copyright in its own papers. Use these alongside the official past papers available from your board.

Each question practises a skill from the Extended (Supplement) content of the syllabus. After each answer there is a common mistake to watch out for.

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Questions

1. (Extended) A used car is bought for $18 500. Its value falls by 12% each year. Calculate its value after 4 years, giving your answer to the nearest dollar. [3]

2. (Extended) The population of a town is 42 000. It increases by 3.5% each year.

(a) Calculate the population after 6 years. Give your answer correct to 3 significant figures.

(b) Find the number of complete years after which the population first exceeds 60 000. [4]

3. (Extended) A machine is bought for $24 000. After 5 years its value is $15 000. The value falls by the same percentage each year. Calculate this annual percentage decrease. [3]

4. (Extended) Simplify √72 + √50 − √18, giving your answer in the form k√2. [2]

5. (Extended) Rationalise the denominator and simplify 14 / (3 + √2). [3]

6. (Extended) Expand and simplify (2 + √3)(5 − 2√3). [2]

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Answers

1. (Extended) Multiplier for a 12% decrease = 0.88 [1]. Value = 18 500 × 0.88⁴ [1] = 11 094.36… so the value is $11 094 [1].

Common mistake: Subtracting 12% of the original price four times (simple decrease) gives $9620, which is wrong: each year’s 12% is taken from the previous year’s value.

2. (Extended) (a) 42 000 × 1.035⁶ [1] = 51 628.7… = 51 600 (3 s.f.) [1].

(b) After 10 years: 42 000 × 1.035¹⁰ = 59 245 (not yet above 60 000); after 11 years: 42 000 × 1.035¹¹ = 61 319 [1]. So the population first exceeds 60 000 after 11 years [1].

Common mistake: Using 0.35 or 1.35 as the multiplier. A 3.5% increase means multiplying by 1.035 each year.

3. (Extended) 24 000 × m⁵ = 15 000, so m⁵ = 15 000 ÷ 24 000 = 0.625 [1]. m = 0.625^(1/5) = 0.9103 [1]. Annual decrease = 1 − 0.9103 = 0.0897, so 8.97% (3 s.f.) [1].

Common mistake: Dividing the total fall of 37.5% by 5 to get 7.5% per year. That treats the decrease as simple, not compound; take the fifth root of the overall multiplier instead.

4. (Extended) √72 = 6√2, √50 = 5√2, √18 = 3√2 [1]. 6√2 + 5√2 − 3√2 = 8√2 [1].

Common mistake: Writing √72 + √50 − √18 as √(72 + 50 − 18) = √104. Surds can only be added or subtracted once they are written as multiples of the same root.

5. (Extended) Multiply numerator and denominator by (3 − √2) [1]. Denominator: (3 + √2)(3 − √2) = 9 − 2 = 7 [1]. 14(3 − √2) ÷ 7 = 2(3 − √2) = 6 − 2√2 [1].

Common mistake: Multiplying by (3 + √2) or by √2 alone. Use the conjugate (change the sign between the terms) so the surd disappears from the denominator.

6. (Extended) 2 × 5 − 4√3 + 5√3 − 2 × 3 [1] = 10 − 6 + √3 = 4 + √3 [1].

Common mistake: Treating √3 × √3 as 9 or as √9 left unsimplified. √3 × √3 = 3, so −2√3 × √3 = −6.

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Where marks are usually lost

• Using simple (linear) change instead of a repeated multiplier for growth and decay.
• Writing the wrong multiplier, such as 0.12 instead of 0.88 for a 12% decrease.
• Stopping at the first year that looks close in “how many years” questions instead of checking the years either side.
• Adding surds with different roots, or not simplifying each surd fully first.
• Forgetting to use the conjugate when rationalising a denominator of the form a + √b.