Question
The difference between compound interest for a year payable half$-$yearly and simple interest on a certain sum of money lent out at $10\%$ for a year is $Rs. 15.$ Find the sum of money lent out.
✓
Answer
Let sum of money be $Rs. y$
To calculate $ \text{S.I.}$
$\text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{y \times 10 \times 1}{100}=R s . \frac{y}{10}$
To calculate $ \text{C.I}.($compounded half$-$yearly$)$
$\therefore \text { C.I. }=\mathrm{P}$
$ {\left[\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}-1\right]=y\left[\left(1+\frac{10}{2 x 100}\right)^{1 \times 2}-1\right]}$
$ =y\left[\left(\frac{21}{20}\right)^2-1\right]=\left(\frac{41}{400}\right) y$
Given $:\text {C.I.} - \text{S.I.} = Rs. 15$
$\Rightarrow\left(\frac{41}{400}\right) y-\frac{y}{10}=15$
$ \Rightarrow \frac{y}{400}=15$
$\Rightarrow y=\text { Rs. } 6,000 .$
To calculate $ \text{S.I.}$
$\text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{y \times 10 \times 1}{100}=R s . \frac{y}{10}$
To calculate $ \text{C.I}.($compounded half$-$yearly$)$
$\therefore \text { C.I. }=\mathrm{P}$
$ {\left[\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}-1\right]=y\left[\left(1+\frac{10}{2 x 100}\right)^{1 \times 2}-1\right]}$
$ =y\left[\left(\frac{21}{20}\right)^2-1\right]=\left(\frac{41}{400}\right) y$
Given $:\text {C.I.} - \text{S.I.} = Rs. 15$
$\Rightarrow\left(\frac{41}{400}\right) y-\frac{y}{10}=15$
$ \Rightarrow \frac{y}{400}=15$
$\Rightarrow y=\text { Rs. } 6,000 .$
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