Mr. Durani bought a plot of land for ₹180000 and a car for ₹320000 at the same time. The value of the plot of land grows uniformly at the rate of 30% p.a., while the value of the car depreciates by 20% in the first year and by 15% p.a. thereafter. If he sells the plot of land as well as the car after 3 years, what will be his profit or loss?

Since, value of land grows uniformly at the rate of 30% p.a. hence, by growth formula,

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

Substituting values we get value of land after 3 years,

$V = ₹180000\Big(1 + \dfrac{30}{100}\Big)^3 \\[1em] = ₹180000\Big(\dfrac{130}{100}\Big)^3 \\[1em] = ₹180000\Big(\dfrac{13}{10}\Big)^3 \\[1em] = ₹180000 \times \dfrac{13}{10} \times \dfrac{13}{10} \times \dfrac{13}{10} \\[1em] = ₹180000 \times \dfrac{2197}{1000} \\[1em] = ₹180 \times 2197 \\[1em] = ₹395460.$

Given, the value of the car depreciates by 20% in the first year and by 15% p.a. thereafter,

Hence, value of car after 3 years = $V = V$0\Big(1 - \dfrac{r1}{100}\Big)\Big(1 - \dfrac{r2}{100}\Big)\Big(1 - \dfrac{r3}{100}\Big)

Substituting values, we get value of car after 3 years,

$V = ₹320000\Big(1 - \dfrac{20}{100}\Big)\Big(1 - \dfrac{15}{100}\Big)\Big(1 - \dfrac{15}{100}\Big) \\[1em] = ₹320000 \times \dfrac{80}{100} \times \dfrac{85}{100} \times \dfrac{85}{100} \\[1em] = ₹(32 \times 4 \times 17 \times 85) \\[1em] = ₹184960.$

Present value of land and car = ₹180000 + ₹320000 = ₹500000.

Total value of land and car after 3 years = ₹395460 + ₹184960 = ₹580420.

Profit = Amount - Initial value = ₹580420 - ₹500000 = ₹80420.

After 3 years profit of Mr. Durani would be ₹80420.