If a sum on compound interest becomes 4 times in 3 years, then with the same interest rate, the sum will become 64 times in:

Correct option is D

Given:

A sum of money becomes 4 times itself in 3 years at compound interest.

-We need to find the time it takes for the same sum to become 64 times itself at the same interest rate.

Formula Used:

Compound Interest Formula:

​$$A = P \left(1 + \frac{r}{100}\right)^n$$​​

where:

A = Amount after n years,

P = Principal,

r = Annual interest rate,

n = Time in years.

Solution:

The amount becomes 4 times in 3 years:

​$$4P = P \left(1 + \frac{r}{100}\right)^3 \\ \ \\\left(1 + \frac{r}{100}\right)^3 = 4$$​​

Now, it became 64 times;

​$$64P = P \left(1 + \frac{r}{100}\right)^n \\ \ \\\left(1 + \frac{r}{100}\right)^n = 64 \\ \ \\\left(1 + \frac{r}{100}\right)^n = (4)^3 \\ \ \\\left(1 + \frac{r}{100}\right)^n = \left[\left(1 + \frac{r}{100}\right)^3)\right]^3 \\ \ \\ \left(1 + \frac{r}{100}\right)^n = \left(1 + \frac{r}{100}\right)^9$$​​

Base is equal, so

n = 9 years

Alternate Solution:

If the amount becomes 4 times in 3 years, then the growth follows:

4 = 2²(since 4 is 2 squared)

Time for 64 Times Growth:

64 = 4³, so the growth factor of 64 corresponds to 3 periods of 3 years each:

3$$\times$$​ 3 = 9 years