Question

The difference between the compound interest and the simple interest on a certain sum for 2 years at 6% per annum is Rs 90. Find the sum.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{Rs}Pbethe\mathrm{sum}. \\ \mathrm{Then}\mathrm{SI}=\left(\frac{P\times 2\times 6}{100}\right)=\mathrm{Rs}.\frac{12P}{100}=\mathrm{Rs}.\frac{3P}{25} \\ \mathrm{Also},\mathrm{CI}=\left\{P\times {\left(1+\frac{6}{100}\right)}^2-P\right\} \\ =\mathrm{Rs}.\left\{P\times {\left(\frac{100+6}{100}\right)}^2-P\right\} \\ =\mathrm{Rs}.\left\{P\times {\left(\frac{53}{50}\right)}^2-P\right\} \\ =\mathrm{Rs}.\left\{\left(\frac{2809P}{2500}\right)-P\right\} \\ =\mathrm{Rs}.\left\{\frac{2809P-2500P}{2500}\right\}=\mathrm{Rs}.\frac{309P}{2500} \\ Now,\left(\mathrm{CI}-\mathrm{SI}\right)=\mathrm{Rs}.\left(\frac{309P}{2500}-\frac{3P}{25}\right) \\ =\mathrm{Rs}.\left(\frac{309P-300P}{2500}\right) \\ =\mathrm{Rs}.\frac{9P}{2500} \\ \mathrm{Now},\mathrm{Rs}.90=\frac{9P}{2500} \\ \mathit{\Rightarrow }P=\left(\frac{90\times 2500}{9}\right)=\mathrm{Rs}.25000 \\ \mathrm{Hence},\mathrm{the}required\mathrm{sum}\mathrm{is}\mathrm{Rs}.25000. \end{gathered}$$