Question

A sum of money doubles itself at compound interest in 10 years. In how many years will it become eight
times?
​

Answer 1

Given-

• A sum of money doubles itself at compound introduction in 10 years.

Need to find-

• Years it will take to become eight times.

$\huge{ \mathtt{ \purple{Solution-}}}$

(i) Money doubles itself-

$\large{ \implies{2p = p(1+r/100) ^{2} }}$

$\large{ \implies{2 = (1+r/100) ^{2} 15}}$

(ii) Money becomes eight times-

$\large{ \implies{8p =p(1+r/100)^{n} }}$

$\large{ \implies{8 =p(1+r/100)^{n} }}$

$\large{ \implies{ {2}^{2} = p(1 + { \frac{r}{100} )}^{2} }}$

$\large{ \implies{(1 + { \frac{r}{100} }^{15} )}} ^{3} = 1 + \frac{r}{100} ^{n}$

$\large{ \implies{1 + \frac{r}{100} ^{45} = 1 + \frac{r}{100} ^{n} }}$

$\large{ \implies{n = 45}}$

Therefore 45 years is the required time.