The compound interest accrued on an amount of Rs. 25500 at the end of 3 years is Rs. 8440.50. What amount at the same rate in the same period ?

1Rs. 4650
2Rs. 5650
3Rs. 6650
4Rs. 7650

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let the rate be R % p .a . \\ then, \\ 25500 {(1 + \frac{R}{100})}^{3} \\ = 25500 + 8440.50 \\ = 33940.50 \\ \Rightarrow {(1 + \frac{R}{100})}^{3} = \frac{33940.50}{25500} = \frac{1331}{1000} = {(\frac{11}{10})}^{3} \\ \Rightarrow 1 + \frac{R}{100} = \frac{11}{10} \\ \Rightarrow \frac{R}{100} = \frac{1}{10} \\ \Rightarrow R = 10 % \\ S . I . = R s . (\frac{25500 \times 10 \times 3}{100}) \\ = R s .7650 \end{aligned}

$\eqalign{ & {\text{Let the rate be R}}\% {\text{ p}}{\text{.a}}{\text{. }} \cr & {\text{then,}} \cr & {\text{25500}}{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} \cr & = 25500 + 8440.50 \cr & = 33940.50 \cr & \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} = \frac{{33940.50}}{{25500}} = \frac{{1331}}{{1000}} = {\left( {\frac{{11}}{{10}}} \right)^3} \cr & \Rightarrow 1 + \frac{{\text{R}}}{{100}} = \frac{{11}}{{10}} \cr & \Rightarrow \frac{{\text{R}}}{{100}} = \frac{1}{{10}} \cr & \Rightarrow {\text{R}} = 10\,\% \cr & S.I. = {\text{R}}s.\left( {\frac{{25500 \times 10 \times 3}}{{100}}} \right) \cr & = {\text{R}}s.7650 \cr}$