Higher June 2025 Paper 2R Q12
12 Osman buys a car for $16 000
The car depreciates at a rate of 12% each year for the first 2 years.
In the third year, the car depreciates at a rate of \(x\)%
At the end of 3 years, the value of the car is $11 461.12
Work out the value of \(x\)
(3)
| Scheme | Marks |
|---|---|
\(16\,000 \times \left(1 - \dfrac{12}{100}\right)^2\) (= 12 390.4) or 16 000 × 0.7744 (= 12 390.4) oe or \(16\,000 \times \left(1 - \dfrac{12}{100}\right)\) (= 14 080) and \(\text{``}{14\,080}\text{''} \times \left(1 - \dfrac{12}{100}\right)\) (=12 390.4) oe or \(\left(1 - \dfrac{12}{100}\right)^2\) (= 0.7744) and \(\dfrac{11461.12}{16000}\left(= \dfrac{4477}{6250} = 0.71632\right)\) | M1 |
eg \(\dfrac{\text{``}{12390.4}\text{''} - 11461.12}{\text{``}{12390.4}\text{''}}\;(\times 100)\;(= 0.075)\) or \(1 - \dfrac{11461.12}{\text{``}{12390.4}\text{''}}\;(\times 100)\) or 1 – 0.925 (= 0.075) or \(\dfrac{11461.12}{\text{``}{12390.4}\text{''}}\;(\times 100)\;(= 0.925)\) or \(\dfrac{\text{``}{0.71632}\text{''}}{\text{``}{0.7744}\text{''}}\;(\times 100)\;(= 0.925)\) | M1 |
| Correct answer only scores full marks (unless from obviously incorrect working) Answer: 7.5 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for a method to find the value of the car after two years
or
a method to find the overall percentage multiplier after two years and the overall percentage multiplier for the three years
May be seen embedded, eg in an equation
Do not allow (1 – 12%) unless processed correctly
M1: for a complete method to find the value of the decimal equivalent of \(x\)%
or
for a complete method to find the percentage multiplier for the third year
May be seen embedded, eg within a correct equation rearranged to one of these equivalent forms
A1: oe
SCB2 for answer –7.5