# The difference between compound interest and simple interest on a sum for 2 years at $8 \%$ p.a. is $√¢‚Äö¬π 768$. Find the sum.A. $√¢‚Äö¬π 110000$B. $√¢‚Äö¬π 120000$C. $√¢‚Äö¬π 100000$D. $√¢‚Äö¬π 170000$

Given :

The difference between compound interest and simple interest on a sum for 2 years at $8 \%$ p.a. is $‚Çπ 768$.

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 ‚Çπ120000.‚Ää

Updated on: 10-Oct-2022

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