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A car was purchased for £18 000 on 1st January - Edexcel - A-Level Maths Pure - Question 7 - 2010 - Paper 4
Question 7
A car was purchased for £18 000 on 1st January. On 1st January each following year, the value of the car is 80% of its value on 1st January in the previous year. (a... show full transcript
Worked Solution & Example Answer:A car was purchased for £18 000 on 1st January - Edexcel - A-Level Maths Pure - Question 7 - 2010 - Paper 4
Step 1
Show that the value of the car exactly 3 years after it was purchased is £9216.
Answer
To find the value of the car after 3 years, we can use the formula for exponential decay:
t = P(1 - r)^n
where:
- t is the value after n years,
- P is the initial value (£18,000),
- r is the depreciation rate (0.20), and
- n is the number of years (3).
Hence,
t = 18000(0.8)^3 = 18000 imes 0.512 = 9216.
Thus, the value of the car after 3 years is indeed £9216.
Step 2
Find the value of n.
Answer
We are tasked with finding the number of years n when the car's value first goes below £1000.
Using the same formula:
t = 18000(0.8)^n < 1000.
Rearranging gives:
0.8^n < rac{1000}{18000} = rac{1}{18}.
Taking the logarithm of both sides:
a)
log(0.8^n) < log( rac{1}{18}),
b)
n imes log(0.8) < log( rac{1}{18}),
c)
n > rac{log( rac{1}{18})}{log(0.8)}.
Calculating values,
log(0.8) is approximately -0.0969 and log(1/18) is approximately -1.255.
Thus,
n > rac{-1.255}{-0.0969} ≈ 12.95,
which means n must be 13 years.
Step 3
Find the cost of the scheme for the 5th year, giving your answer to the nearest penny.
Answer
To find the cost in the 5th year, we note that the cost increases by 12% each year.
If the cost for the 1st year is £200, then for subsequent years:
Cost for year 2:
£200 × 1.12 = £224,
Cost for year 3:
£224 × 1.12 = £250.88,
Cost for year 4:
£250.88 × 1.12 = £281.00,
Cost for year 5:
£281.00 × 1.12 = £314.72.
Thus, the cost of the scheme for the 5th year, rounded to the nearest penny, is £314.72.
Step 4
Find the total cost of the insurance scheme for the first 15 years.
Answer
To calculate the total cost over 15 years, we sum the costs for each year using the formula:
Cost for each year is given by:
t_n = 200 imes 1.12^{(n-1)}.
Thus, we calculate:
Total Cost = 200 + 224 + 250.88 + ... + (200 imes 1.12^{14}).
This is a geometric series with a first term a = 200 and a common ratio r = 1.12. The number of terms n = 15.
The sum S_n of the series is:
t_n = a rac{(1 - r^n)}{(1 - r)}
Calculating:
t_{15} = 200 rac{(1 - (1.12)^{15})}{(1 - 1.12)}
d ightarrow paved ext{[approximately £7455.94]}.
Adding them gives:
total ext{(to nearest pound)} ightarrow £7460.