A car was purchased for £18 000 on 1st January - Edexcel - A-Level Maths Pure - Question 7 - 2010 - Paper 4

To find the value of $n$ when the car's value drops below £1000, we set up the inequality:

$18000 imes (0.8)^n < 1000$

Solving for $n$:

1. Divide both sides by 18000: $(0.8)^n < \frac{1000}{18000}$ $(0.8)^n < \frac{1}{18}$

2. Taking the logarithm of both sides: $n \log(0.8) < \log(\frac{1}{18})$

3. Therefore, $n > \frac{\log(\frac{1}{18})}{\log(0.8)}$

After computing: $n \approx 12.97$

Thus, rounding up gives us $n = 13$ years.

The cost for the first year is £200, and each subsequent year increases by 12%:

The cost for the nth year can be found using:

$C_n = C_1 (1 + r)^{n-1}$

where:

• $C_n$ is the cost in year n,
• $C_1$ is the cost in the first year (£200), and
• $r$ is the rate of increase (0.12).

Calculating for the 5th year:

$C_5 = 200 (1 + 0.12)^{5-1} = 200 (1.12)^4$

Evaluating gives:

$C_5 \approx 200 \times 1.5748 \approx 314.76$

Thus, the cost for the 5th year is £314.76.

To find the total cost for the first 15 years, we need to sum the costs from year 1 to year 15:

Using the formula:

$S_n = C_1 \frac{(1 + r)^n - 1}{r}$

where:

• $S_n$ is the total cost over n years,
• $C_1$ is the cost in the first year (£200),
• $r$ is the growth factor (0.12), and
• $n$ is the total number of years (15).

Calculating:

$S_{15} = 200 \frac{(1.12)^{15} - 1}{0.12}$

After computing: $S_{15} \approx 200 \frac{5.848 - 1}{0.12} \approx 200 \times 39.5667 \approx 7913.34$

Thus, the total cost for the first 15 years is approximately £7913.34.