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The difference between compound interest and simple interest on a sum for 2 years at $ 8 \% $ p.a. is $ ₹ 768 $. Find the sum.A. $ ₹ 110000 $
B. $ ₹ 120000 $
C. $ ₹ 100000 $
D. $ ₹ 170000 $
Given :
The difference between compound interest and simple interest on a sum for 2 years at \( 8 \% \) p.a. is \( ₹ 768 \).
To find :
We have to find the sum.
Solution :
Let the sum be $P$, rate of interest be $r$ and $n$ be the number of years.
Therefore,
Compound interest $= P[(1+\frac{r}{100})^n - 1]$
$= P [(1+\frac{8}{100})^2 - 1]$
$= P [(1+\frac{2}{25})^2 - 1]$
$= P [(\frac{27}{25})^2 - 1]$
$= P [\frac{729}{625} - 1]$
$= P[\frac{729-625}{625}]$
$= \frac{104}{625}P$
Simple interest $= \frac{Pnr}{100}$
$= \frac{P \times 2 \times 8}{100}$
$= \frac{16P}{100}$
$= \frac{4}{25}P$
$CI - SI = 768$
$ \frac{104}{625}P - \frac{4}{25}P = 768$
$ \frac{104}{625}P - \frac{4 \times 25}{25 \times 25}P = 768$
$\frac{104}{625}P -\frac{100}{625}P=768 $
$ \frac{104 - 100}{625}P=768$
$\frac{4}{625}P = 768$
$ P = \frac{768 \times 625}{4}$
$P = 192\times625$
$P=Rs.\ 120000$
Therefore, the sum is ₹120000.