A sum of money doubles itself at compound interest in 15 yrs. In how many yrs it will become eight times?

A 40

B 55

C 45

D 50

E None of these

Correct Answer: Option C

Let us assume that the certain amount of money is `P’

Given that it doubles itself after 15 years (i.e. after 15 years it becomes 2P)

After 15 years, the principal is 2P

So, after another 15 years 2P doubles itself i.e. it becomes 4P (similarly as P becomes 2P in 15 years)

i.e. after 30 years from starting, P becomes 4P.

Let, after 30 years, the principal is 4P.

So, after another 15 years, 4P becomes doubles itself.

i.e. it becomes 8P (similarly as P becomes 2P in 15 years)

Therefore, after 45 years P becomes eight times of itself i.e. become 8P.

Alternative method:

Given that amount doubles itself at compound interest in 15 years

We know that when interest is compounded annually, Amount \(= \;P\;{\left[ {1\; + \frac{R}{{100}}} \right]^n}\)

Where P = Principal, Rate = R% per annum, Time = n years

\(\therefore \;P\;{\left[ {1\; + \frac{R}{{100}}} \right]^{15}}\; = \;2P\)

\(\Rightarrow \;{\left[ {1\; + \frac{R}{{100}}} \right]^{15}}\; = \;2\) …………. (1)

Let us assume that in `x’ years `P’ amounts to `8P’

\(\therefore \;P\;{\left[ {1\; + \frac{R}{{100}}} \right]^x}\; = \;8P\)

\(\Rightarrow \;{\left[ {1\; + \frac{R}{{100}}} \right]^x}\; = \;8\; = \;{2^3}\)

\(\Rightarrow \;{\left[ {1\; + \frac{R}{{100}}} \right]^x}\; = \;{\left\{ {{{\left[ {1\; + \frac{R}{{100}}} \right]}^{15}}} \right\}^3}\) [∵From (1)]

\(\Rightarrow \;{\left[ {1\; + \frac{R}{{100}}} \right]^x}\; = \;{\left[ {1\; + \frac{R}{{100}}} \right]^{45}}\)

⇒ x = 45 [∵If am = an then m = n]

∴ In `45 years’ amount becomes 8 times to the principal at the given rate.