Question

A sum of money invested at compound interest amounts to rs. 4624 in 2 years and to rs. 4913 in 3 years. the sum of money is​

Answer 1

Answer :-

Let the sum of money be x and rate of interest be R.

We know that,

$\implies\sf A = x \times \Big(1 + \dfrac{R}{100}\Big)^n$

For 2 years :-

$\implies\sf 4624 = x \times \Big(1 + \dfrac{R}{100}\Big)^2 \: \: \: \: \: \: -i$

For 3 years :-

$\implies\sf 4913 = x \times \Big(1 + \dfrac{R}{100}\Big)^3 \: \: \: \: \: \: -ii$

Dividing equation i by ii :-

$\implies\sf \dfrac{4913}{4624} = \dfrac{x \times \Big(1 + \dfrac{R}{100}\Big)^3 }{x \times \Big(1 + \dfrac{R}{100}\Big)^2}$

$\implies\sf \dfrac{17\times 17\times 17}{17 \times 17 \times 16} = 1 + \dfrac{R}{100}$

$\implies\sf \dfrac{17}{16} = \dfrac{100+R}{100}$

$\implies\sf 100 \times 17 = 16 ( 100 + R )$

$\implies\sf 1700 = 1600 + 16 R$

$\implies\sf 16 R = 100$

$\implies\sf R = \dfrac{100}{16}$

$\implies\sf R = 6.25$

Substituting the value in equation i :-

$\implies\sf 4624 = x \times \Big(1 + \dfrac{R}{100}\Big)^2$

$\implies\sf 4624 = x \times \Big( 1 + \dfrac{6.25}{100}\Big)^2$

$\implies\sf 4624 = x \times \Big( \dfrac{17}{16} \Big)^2$

$\implies\sf x = \dfrac{4624}{\Big( \dfrac{17}{16} \Big)^2}$

$\implies\sf x = 4096$

Sum of money = Rs. 4096