Priya bought a car for \(\pounds 12\,500\). Each year, the value of the car depreciates by 18% of its value at the start of that year. Work out the value of...
Question 1 Report
Priya bought a car for \(\pounds 12\,500\). Each year, the value of the car depreciates by 18% of its value at the start of that year.
- Work out the value of the car at the end of the first year. (2)
- Work out the value of the car at the end of three years. Give your answer to the nearest pound. (2)
- After how many complete years will the value of the car first fall below \(\pounds 5000\)? You must show your working. (2)
Answer Details
This question tests repeated percentage decrease (compound depreciation) and finding when a quantity first drops below a threshold.
- After one year, the car retains \(100\% - 18\% = 82\%\) of its value: \(12500 \times 0.82\) [1] \(= \) £10,250 [1].
- Because the depreciation is 18% of the value at the start of each year, the value after 3 years is found by multiplying by 0.82 three times: \(12500 \times 0.82^3\) [1] \(= 12500 \times 0.551368 = \) £6892 to the nearest pound [1].
- Continuing the pattern: after 4 years, \(12500 \times 0.82^4 \approx £5651.52\), still above £5000; after 5 years, \(12500 \times 0.82^5 \approx £4634.25\), which is below £5000 [1]. The value first falls below £5000 after 5 complete years [1].
Depreciation and other repeated percentage changes are modelled by a single multiplier raised to a power (\(0.82^n\)), never by multiplying \(n \times 18\%\) and subtracting it all at once, which would overstate the fall in value.
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