A car depreciates at 8% per year. The initial price of the car is £25,000. How much will the car be worth after 4 years? After how many years will the car be worth less than £14,000?
This problem is all about compound interest except this time the value is going down not up. To increase a value by 8% you would multiply it by 1.08 as 8% is 0.08 (1+0.08). To decrease a value however you need to minus the interest from 1 (1-0.08) so in this case we'll be multiplying by 0.92However, you can't just keep taking an extra 8% off of £25,000 with each passing year as the price of the car decreases with the years. So, after the first year: 250000.92=23000but after the second year you need to take the new value of the car: 230000.92=211603rd year: 211600.92=19467.204th year: 19467.20.92=17909.82 (Don't forget to put £ back in for your final answer)This is our answer to the first part of the question but you can imagine this can become quite tiring after calculating for many years at a time. Luckily there's an easier way of calculating the depreciation. Essentially what we've done is this: (250000.92) 1st year((250000.92)0.92) 2nd year - which is the same as (250000.922 )(250000.923) 3rd year(250000.924) 4th yearThere's a very obvious pattern here where the powers are the same as the year you're trying to calculate! So if we wanted to find the value of the car after 10 years we could just use this: 250000.9210=10859.71Now for the second part of the question. Since we know after 4 years is > 14000 and 10 years is <14000 we know the year the car passes this value must be between these two. We can use trial and error to find the year:5th year: 250000.925=16477.046th year: 250000.926=15158.887th year: 250000.927=13946.17Therefore the answer is year 7.
