Question

The difference in simple interest and compound interest on a certain sum of money at $6\frac{2}{3}$ per annum for 3 years is Rs. 46. Determine the sum.

Answer 1

Let sum (P) = Rs. 100
Rate (R) $=6\frac{2}{3}\%=\frac{20}{3}\%$ p.a
Period (n) = 3 years
$\therefore$ S.I. = $=\frac{PRT}{100}=\frac{100\times 20\times 3}{100\times 3}=Rs.20$
Compound interest (C.I) = A - P
$=P{(1+\frac{R}{100})}^n-P$
$=P[{(1+\frac{R}{100})}^n-1]$
$=Rs.100[{(1+\frac{20}{3\times 100})}^3-1]$
$=Rs.100[{\left(\frac{16}{15}\right)}^3-1]$
$=Rs.100[\frac{4096}{3375}-1]$
$=Rs.100\left[\frac{4096-3375}{3375}\right]$
$=Rs.100\times \frac{721}{3375}=Rs.\frac{2884}{135}$
$\therefore$ Difference in C.I. and S.I.
$=Rs.\frac{2884}{135}-20$
$=Rs.\frac{2884-2700}{135}=Rs.\frac{184}{135}$
If difference is Rs. $\frac{184}{135}$, then sum = Rs. 100
if difference is Re. 1, then sum $=Rs.\frac{100\times 135}{184}$
and if difference is Rs. 46, then sum $=Rs.\frac{100\times 135\times 46}{184}=Rs.3375$
$\therefore$ Sum = Rs. 3375