Edexcel GCSE Maths (1MA1), Higher tier · Ratio, proportion and rates of change
Practise Compound interest, growth and decay. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
C1 for Account B, supported by both correct totals (or both amounts of interest, £796.37 and £800)
Worked solution: A: \(5000 \times 1.03^5 = 5796.37\) (to the nearest penny). B: interest \(= 0.032 \times 5000 = 160\) per year, so \(5000 + 5 \times 160 = 5800\). Account B gives £3.63 more.
Question 3Hard4 marks
The value of a car decreases by 15% each year. After 3 years the car is worth £9826.
(a) Leo says, 'The car has lost 3 × 15% = 45% of its value, so its value when new was £9826 ÷ 0.55.' Explain why Leo is wrong.[1]
(b) Work out the value of the car when it was new.[3]
Show the answer and mark scheme
(a)Answer: Each year's 15% is taken from the value at the start of that year, which keeps getting smaller, so the multiplier is \(0.85^3\), not 0.55.
C1 for explaining that the decrease is compound (each 15% is of a smaller value), so the multiplier is \(0.85^3\) (= 0.614125)
Worked solution: The 15% in the second and third years is taken from smaller values, so the total loss is less than 45%: the multiplier is \(0.85^3 = 0.614125\).
(b)Answer: £16 000
M1 for \(0.85^3\) (= 0.614125)
M1 for \(9826 \div 0.85^3\)
A1 for £16 000
Worked solution: \(9826 \div 0.85^3 = 9826 \div 0.614125 = 16\,000\), so the car was worth £16 000 when new.