On what sum of money will the difference between simple interest and compound interest for 2 years at 5% per annum be equal to Rs. 63 ?

1Rs. 24600
2Rs. 24800
3Rs. 25200
4Rs. 25500

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Rate of interest  =  5 % per annum \\ Time  =  2 year \\ Accroding to question, \\ P [{(1 + \frac{r}{100})}^{n} - 1] - \frac{P \times r \times t}{100} = 63 \\ P [{(1 + \frac{5}{100})}^{2} - 1] - \frac{P \times 5 \times 2}{100} = 63 \\ \Rightarrow P [{(1 + \frac{5}{100})}^{2} - 1] - \frac{10 P}{100} = 63 \\ \Rightarrow P [{(\frac{105}{100})}^{2} - 1] - \frac{10 P}{100} = 63 \\ \Rightarrow P (\frac{11025 - 10000}{10000}) - \frac{10 P}{100} = 63 \\ \Rightarrow \frac{1025 P}{10000} - \frac{10 P}{100} = 63 \\ \Rightarrow \frac{1025 P - 1000 P}{10000} = 63 \\ \Rightarrow 65 P = R s .630000 \\ \Rightarrow P = \frac{630000}{25} \\ \Rightarrow P = R s 25200 \\ Hence, sum Rs . 25200 \end{aligned}

$\eqalign{ & {\text{Rate of interest = 5}}\% {\text{ per annum}} \cr & {\text{Time = 2 year}} \cr & {\text{Accroding to question,}} \cr & P\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^n} - 1} \right] - \frac{{P \times r \times t}}{{100}}{\text{ = 63}} \cr & P\left[ {{{\left( {1 + \frac{5}{{100}}} \right)}^2} - 1} \right] - \frac{{P \times 5 \times 2}}{{100}}{\text{ = 63}} \cr & \Rightarrow P\left[ {{{\left( {1 + \frac{5}{{100}}} \right)}^2} - 1} \right] - \frac{{10P}}{{100}}{\text{ = 63}} \cr & \Rightarrow P\left[ {{{\left( {\frac{{105}}{{100}}} \right)}^2} - 1} \right] - \frac{{10P}}{{100}}{\text{ = 63}} \cr & \Rightarrow P\left( {\frac{{11025 - 10000}}{{10000}}} \right) - \frac{{10P}}{{100}} = 63 \cr & \Rightarrow \frac{{1025P}}{{10000}} - \frac{{10P}}{{100}} = 63 \cr & \Rightarrow \frac{{1025P - 1000P}}{{10000}} = 63 \cr & \Rightarrow 65P = Rs.630000 \cr & \Rightarrow P = \frac{{630000}}{{25}} \cr & \Rightarrow P = Rs25200 \cr & {\text{Hence}},\,{\text{sum Rs}}{\text{. 25200}} \cr}$