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Mathematics

Mr. Goyal bought a car for ₹ 6,25,000. Its value depreciates at 20% p.a. How many years it would take for the price of the car to go down by ₹ 3,69,000?

Compound Interest

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Answer

Given,

Present value of machine (V) = ₹ 6,25,000

r = 20%

Depreciated value = ₹ 6,25,000 - ₹ 3,69,000 = ₹ 2,56,000

By formula,

Value of machine after n years = ₹ V×(1−r100)nV \times \Big(1 - \dfrac{r}{100}\Big)^n

Substituting the values in formula,

⇒256000=625000×(1−20100)n⇒(1−20100)n=256000625000⇒(1−15)n=256625⇒(45)n=(45)4⇒n=4\Rightarrow 256000 = 625000 \times \Big(1 - \dfrac{20}{100}\Big)^n\\[1em] \Rightarrow \Big(1 - \dfrac{20}{100}\Big)^n=\dfrac{256000}{625000} \\[1em] \Rightarrow \Big(1 - \dfrac{1}{5}\Big)^n=\dfrac{256}{625} \\[1em] \Rightarrow \Big(\dfrac{4}{5}\Big)^n=\Big(\dfrac{4}{5}\Big)^4 \\[1em] \Rightarrow n = 4

Hence, It will take 4 years for the price of the car to depreciate.

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