Question

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

times?

a. 20

b. 30

c. 40

d. 35
​

Answer 1

Question :-

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

Answer :-

• 30 years

Given :-

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

To find :-

• In how many years will it become eigh times?

Solution :-

Let ,

• Sum of money = P

• Rate of interest = R

• Time = T

$\implies \large \sf \: A = P {\bigg(1 + \dfrac{R}{100} \bigg)}^{10}$

$\\ \\ \large\implies \sf2 \cancel {P} = {\cancel {P} \bigg(1 + \frac{R}{100} \bigg)}^{10}$

$\\ \\ \large \implies \sf2 = \bigg(1 + \frac{R}{100} \bigg)^{10}$

In year it will become eight,

$\large \implies \sf 8 \cancel {P}= {\cancel {P} \bigg(1 + \dfrac{R}{100} \bigg)}^{T}$

$\\ \\ \large \implies \sf8 = \bigg(1 + \dfrac{R}{100} \bigg)^{ T}$

$\\ \\ \large \implies \sf( {2)}^{3} = \bigg(1 + \dfrac{R}{100} \bigg)^{T}$

$\\ \\ \large \implies \sf {\Bigg[\bigg(1+\dfrac{R}{100} \bigg)^{10} \Bigg] }^{3} = \bigg(1 + \dfrac{R}{100} \bigg)^{7}$

$\\ \\ \large \implies \sf \bigg(1 + \dfrac{R}{100} \bigg)^{30} = \bigg(1 + \dfrac{R}{100} \bigg)^{7}$

$\\ \\ \large \implies \boxed { \sf \: T = 30 \: years}$

☞So in 30 years it will become eight