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Mathematics

A certain sum of money doubles itself at a given rate in 8 years compounded yearly. In how many years will it be four times at the same rate compounded yearly ?

Compound Interest

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Answer

Let rate of interest be r% and sum of money be ₹ P.

By formula, A = P(1+r100)tP\Big(1 + \dfrac{r}{100}\Big)^t

Given,

₹ P becomes twice of itself in 8 years.

∴2P=P(1+r100)8⇒2PP=(1+r100)8⇒2=(1+r100)8.....................(1)\therefore 2P = P\Big(1 + \dfrac{r}{100}\Big)^8 \\[1em] \Rightarrow \dfrac{2P}{P} = \Big(1 + \dfrac{r}{100}\Big)^8 \\[1em] \Rightarrow 2 = \Big(1 + \dfrac{r}{100}\Big)^8 …………………(1)

Let the money become four times in n years.

⇒P(1+r100)n=4P⇒(1+r100)n=4PP⇒(1+r100)n=4⇒(1+r100)n=22⇒(1+r100)n=[(1+r100)8]2 [From (1)]⇒(1+r100)n=(1+r100)16\Rightarrow P\Big(1 + \dfrac{r}{100}\Big)^n = 4P\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \dfrac{4P}{P}\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = 4\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = 2^2\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \Big[\Big(1 + \dfrac{r}{100}\Big)^8\Big]^2 \text{ [From (1)]}\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \Big(1 + \dfrac{r}{100}\Big)^{16}\\[1em]

⇒ n = 16.

Hence, in 16 years money will becomes 4 times of itself.

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