Question

a sum of money lent at simple interest amounts to rupees 3224 in 2 years and rupees 4160 in 5 years find the sum and the rate of interest

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{We}\mathrm{know},\mathrm{S}.\mathrm{I}.=\frac{\mathrm{p}\times \mathrm{r}\times \mathrm{t}}{100} \\ \mathrm{and}\mathrm{Amount}=\mathrm{P}+\mathrm{S}.\mathrm{I}. \\ \mathrm{Let}\hspace{0.17em}\mathrm{P}=\mathrm{x} \\ \mathrm{Case}1:\mathrm{time}=2\mathrm{years} \\ 3224=\mathrm{x}+\frac{\mathrm{x}\times \mathrm{r}\times 2}{100} \\ 3224-\mathrm{x}=\frac{\mathrm{x}\times \mathrm{r}\times 2}{100} \\ \frac{\left(3224-\mathrm{x}\right)\times 100}{2\mathrm{x}}=\mathrm{r}...\left(1\right) \\ \mathrm{Case}2:\mathrm{time}=5\mathrm{years} \\ 4160=\mathrm{x}+\frac{\mathrm{x}\times \mathrm{r}\times 5}{100} \\ \frac{\left(4160-\mathrm{x}\right)\times 100}{5\mathrm{x}}=\mathrm{r}...\left(2\right) \\ \mathrm{Since}\mathrm{the}\mathrm{rate}\mathrm{is}\mathrm{same}\mathrm{in}\mathrm{both}\mathrm{the}\mathrm{cases}, \\ \frac{\left(4160-\mathrm{x}\right)\times 100}{5\mathrm{x}}=\frac{\left(3224-\mathrm{x}\right)\times 100}{2\mathrm{x}} \\ \frac{\left(4160-\mathrm{x}\right)\times 20}{\mathrm{x}}-\frac{\left(3224-\mathrm{x}\right)\times 50}{\mathrm{x}}=0 \\ \frac{4160\times 20-20\mathrm{x}-3224\times 50+50\mathrm{x}}{\mathrm{x}}=0 \\ \frac{30\mathrm{x}+83200-161200}{\mathrm{x}}=0 \\ 30\mathrm{x}=78000 \\ \mathrm{x}=\frac{78000}{30} \\ \mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{Rs}\mathbf{}\mathbf{2600} \end{gathered}$$

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