A sum of money doubles itself at a certain rate of compound interest in 12 years when the interest is compounded annually. In how many years will it become eight time of itself?

Correct option is A

Given:

Amount = 2 × Principal
Time = 12 years
Solution:
A certain sum at C.I. becomes n times in t years then,
n times → t years
$$n^m$$​ times → (m × t) years
A sum of money placed at C.I. doubles itself in 4 years then,
2 times → 12 years
8 times( $$2^3$$​ times) = (12 × 3) years = 36 years
∴ A sum of money will amount to its 8 times in  36 years.
Alternate Method
Formula Used:
​$$\text{Amount} = \text{Principal} \left(1 + \frac{rate}{100}\right)^{\text{time}}$$​​
Solution:
Here, A = 2P, t = 12 years

​$$A = P \times \left[1 + \frac{r}{100}\right]^t \\2P = P \times \left[1 + \frac{r}{100}\right]^{12} \\2 = \left[1 + \frac{r}{100}\right]^{12}$$ .........  (i)

If sum becomes 8 times in n years,

Then, A = 8P
​$$8P = P [1+\frac{r}{100}]^{t}\\8 = [1+\frac{r}{100}]^{t}\\2^3 = [1+\frac{r}{100}]^{t}\\$$      

​$$2 = [1+\frac{r}{100}]^{t/3}\\$$  ........... (ii) 

from equation (i) and (ii)

$$\left[1 + \frac{r}{100}\right]^{12}$$  = $$[1+\frac{r}{100}]^{t/3}\\$$​​

comparing both sides base  

​$$\frac{t}{3} = 12$$ 

t = 36 ​​

∴ A sum of money will amount to its 8 times in 36 years.