Question

Brian buys a computer for £2100.
It depreciates at a rate of 1% per year.
How much will it be worth in 6 years?
Give your answer to the nearest penny where appropriate.

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Price\:of\:computer\:after\:6\:year=\pounds1977.1}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Price \: of \: computer = \pounds 2100 \\ \\ \tt: \implies Depreciate \: rate\% = 1\% \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Price \: of \: computer \: after \: 6 \: year =?$

• According to given question :

$\bold{As \: we \: know \: that : } \\ \tt: \implies 1st \: year \: depreciate \: price = 2100 - 2100 \: of \: 1\% \\ \\ \tt: \implies 1st \: year \: depreciate \: price =2100 - 2100 \times \frac{1}{100} \\ \\ \tt: \implies 1st \: year \: depreciate \: price =2100 - 21 \\ \\ \tt: \implies 1st \: year \: depreciate \: price = \pounds 2079 \\ \\ \bold{For \: 2nd \: year : } \\ \tt: \implies 2nd \: year \: depreciate \: price =2079 - 1\% \: of \: 2079 \\ \\ \tt: \implies 2nd \: year \: depreciate \: price = 2079 - \frac{1}{100} \times 2079 \\ \\ \tt: \implies 2nd\: year \: depreciate \: price =2079 - 20.79 \\ \\ \tt: \implies 2nd\: year \: depreciate \: price = \pounds 2058.21 \\ \\ \bold{Similarly : } \\ \tt: \implies 3rd\: year \: depreciate \: price =2058.21 - 1\% \: of \: 2058.21$$\\ \tt: \implies3rd \: year \: depreciate \: price =2058.21 - 20.5821 \\ \\ \tt: \implies 3rd \: year \: depreciate \: price =\pounds 2037.6279 \\ \\ \bold{For \: 4th \: year : } \\ \tt: \implies 4th \: year \: depreciate \: price = 2037.6279 - 20.376279 \\ \\ \tt: \implies 4th \: year \: depreciate \: price =\pounds 2017.251621 \\ \\ \bold{For \: 5th \: year : } \\ \tt: \implies5th \: year \: depreciate \: price =2017.251621 - 20.17251621 \\ \\ \tt: \implies 5th \: year \: depreciate \: price =\pounds 1997.07910479 \\ \\ \bold{For \: 6th \: year : }\\ \tt: \implies 6th \: year \: depreciate \: price = 1997.07910479 - 19.9707910479 \\ \\ \tt: \implies 6th\: year \: depreciate \: price =1977.1083077421 \\ \\ \green{\tt: \implies 6th\: year \: depreciate \: price = \pounds 1977.1}$

Answer 2

Given

• Brian buys a computer for £2100.

• It depreciates at a rate of 1% per year

To Find

• price after 6 years

Solution

price after 6 years = P ( 1 - R/100 )⁶

→ price after 6 years = £2100 ( 1 - 1/100 )⁶

→ price after 6 years = £2100 ( 99/100 )⁶

→ price after 6 years = £2100 ( 0.99 )⁶

→ price after 6 years = £2100 × 0.941

→ price after 6 years = £1976.2

price after 6 years is £1976.2