Numeracy Higher November 2023 Unit 2 Q5
5. A car was bought in 1973 for £2500.
In the first year, this car depreciated by 23% of its value.
In each of the following 39 years, it depreciated by 4% of its value in the previous year.
The car then started to increase in value.
In each of the next 10 years, it increased by 14% of its value in the previous year.
Calculate the value of the car after these 50 years.
You must show all your working. [4]
The value of the car after 50 years is £ ……………
| Scheme | Marks | Comments |
|---|---|---|
| 2500 × (1 – 0.23) × (1 – 0.04)\(^{39}\) × (1 + 0.14)\(^{10}\) or 2500 × 0.77 × 0.96\(^{39}\) × 1.14\(^{10}\) or equivalent | M3 | May be seen in stages M2 for a product with any 3 correct terms OR M1 for a product with any 2 correct terms |
| (£) 1452(.30) | A1 | CAO, ignore premature rounding in working provided answer is (£) 1452.(…), allow rounded to (£)1450 from correct working |
Note
2500 × 0.77 = (£) 1925
2500 × 0.96\(^{39}\) = (£) 508.766….
2500 × 1.14\(^{10}\) = (£) 9268.053…
0.77 × 0.96\(^{39}\) = 0.15669…
0.77 × 1.14\(^{10}\) = 2.854…
0.96\(^{39}\) × 1.14\(^{10}\) = 0.7544…
2500 × 0.77 × 0.96\(^{39}\) = (£) 391.7498…
2500 × 0.77 × 1.14\(^{10}\) = (£) 7136.401…
2500 × 0.96\(^{39}\) × 1.14\(^{10}\) = (£)1886.108…
0.77 × 0.96\(^{39}\) × 1.14\(^{10}\) = 0.5809…
Values may differ by rounding at individual stages