The cost of a car, purchased 2 years ago, depreciates at the rate of 20% every year. If its present worth is ₹315600, find :

(i) its purchase price

(ii) its value after 3 years.

(i) By depreciation formula,

V = $V_0\Big(1 - \dfrac{r}{100}\Big)^n$

Let initial value of car be V₀.

Putting values in formula we get,

$\Rightarrow 315600 = V$0\Big(1 - \dfrac{20}{100}\Big)^2 \\[1em] \Rightarrow 315600 = V0\Big(1 - \dfrac{20}{100}\Big)^2 \\[1em] \Rightarrow 315600 = V0\Big(\dfrac{80}{100}\Big)^2 \\[1em] \Rightarrow 315600 = V0\Big(\dfrac{4}{5}\Big)^2 \\[1em] \Rightarrow 315600 = V0 \times \dfrac{4}{5} \times \dfrac{4}{5} \\[1em] \Rightarrow 315600 = V0 \times \dfrac{16}{25} \\[1em] \Rightarrow V0 = 315600 \times \dfrac{25}{16} \\[1em] \Rightarrow V0 = ₹493125.

Hence, the purchase price of car was ₹493125.

(ii) By depreciation formula,

V = $V_0\Big(1 - \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$V = 315600\Big(1 - \dfrac{20}{100}\Big)^3 \\[1em] = 315600 \times \Big(\dfrac{80}{100}\Big)^3 \\[1em] = 315600 \times \Big(\dfrac{4}{5}\Big)^3 \\[1em] = 315600 \times \Big(\dfrac{64}{125}\Big) \\[1em] = \dfrac{20198400}{125} \\[1em] = ₹161587.20$

Hence, the value of car after 3 years = ₹161587.20