A car was purchased for £18 000 on 1st January - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 4

We need to find when the value of the car falls below £1000.

Using the same depreciation formula:

$V = 18000(0.8)^n < 1000$

This can be rearranged to:

(0.8)^n < rac{1000}{18000} = rac{1}{18}

Taking logarithms on both sides:

n imes ext{log}(0.8) < ext{log}rac{1}{18}

Thus,

n > rac{ ext{log}rac{1}{18}}{ ext{log}(0.8)}

Calculating these values:

• ext{log}rac{1}{18} = -1.255 (approximately)
• $ ext{log}(0.8) = -0.097 $$ (approximately)

Therefore,

n > rac{-1.255}{-0.097} \\ n > 12.95

Rounding up, the first whole year when the car value is below £1000 is at $n = 13$.

The cost of the maintenance scheme increases by 12% each year. The cost for the first year is £200. We can find the cost for the 5th year using the formula for geometric progression:

$C = C_1 imes (1 + r)^{n-1}$ Where:

• $C_1 = 200$ (initial cost for the first year)
• $r = 0.12$ (increase rate)
• $n = 5$ (the 5th year)

Substituting the values:

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

Calculating gives:

$C = 200 imes 1.5748 \\ C ext{ (for the 5th year)} ext{ is approximately } £314.07$

Thus, the cost of the scheme for the 5th year is approximately £314.07.

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

The cost for each year can be calculated as follows:

• Year 1: £200
• Year 2: £200 × 1.12 = £224
• Year 3: £224 × 1.12 = £250.88
• Year 4: £250.88 × 1.12 = £280.99 ... continuing this up to Year 15:

To find this quickly, we can use the formula for the sum of a geometric series:

S_n = C_1 rac{1 - r^n}{1 - r}

Where:

• $C_1 = 200$ (initial cost)
• $r = 1.12$ (increase rate)
• $n = 15$ (number of terms)

Calculating gives:

S_{15} = 200 rac{1 - (1.12)^{15}}{1 - 1.12}

Once evaluated:

$S_{15} ext{ is approximately } £7455.91$

Thus, the total cost of the insurance scheme for the first 15 years is approximately £7455.91.