Question

Find the compound interest, correct to the nearest rupee, on Rs. $2,400$ for $2\frac{1}{2}$ years at $5$ per cent per annum.

Answer 1

We first calculate the C.I. for first $2$ years and then the S.I. next $\frac{1}{2}$ year.
For ,
$N=2years$
$R=5$ %
$=>P=Rs2,400$

And on interest being compounded we have,
$Amount=P{(1+\frac{R}{100})}^N=2,400\times {(1+\frac{5}{100})}^2=2,400\times {1.05}^2=Rs2,646$

So, $C.I$ for first $2$ years $=.=A-P=Rs2,646=Rs2,400=Rs246$

This, $Rs2,646$ is the principal for the next $\frac{1}{2}$ year.
So, $S.I.=\frac{PNR}{100}=\frac{2,646\times \frac{1}{2}\times 5}{100}=Rs66.15$

Thus, total interest $=Rs246+Rs66.15=Rs312.15$
To the nearest rupee , it is $Rs312$