Question
Find the difference betlween the compound interest compounded yearly and half-yearly for the following:
$Rs.15,000$ for $1 \frac{1}{2}$ years at $12 \%$ p.a.
$Rs.15,000$ for $1 \frac{1}{2}$ years at $12 \%$ p.a.
✓
Answer
$P=\operatorname{Rs} 15,000 ; t=1 \frac{1}{2}$ years
When compounded yearly : $r=12 \%$ p.a.
$A=P\left(1+\frac{r}{100}\right)^n$
$A=\operatorname{Rs} 15000\left(1+\frac{12}{100}\right)\left(1+\frac{12}{100}\right)^{\frac{1}{2}}$
$=\text { Rs } 15000 \times 1.12 \times\left(1+\frac{1}{2} \times \frac{12}{100}\right)$
$=\text { Rs } 15,000 \times 1.12 \times 1.06$
$=\text { Rs } 17,808$
$\text { C.I. }=A-P$
$=\text { Rs }(17,808-15,000)$
$=\text { Rs } 2808 $
When compounded half-yearly :
$ A=P\left(1+\frac{r}{100}\right)^n$
$A=\operatorname{Rs} 15000\left(1+\frac{6}{100}\right)^3$
$=\operatorname{Rs} 15,000 \times 1.06 \times 1.06 \times 1.06$
$=\operatorname{Rs} 17,865.24$
$\text { C.I. }=A-P$
$=\text { Rs }(17,865.24-15,000)$
$=\text { Rs } 2,865.24 $
Hence the difference in the interest=Rs $(2,865.24-2,808)$
$ =\operatorname{Rs} 57.24 $
When compounded yearly : $r=12 \%$ p.a.
$A=P\left(1+\frac{r}{100}\right)^n$
$A=\operatorname{Rs} 15000\left(1+\frac{12}{100}\right)\left(1+\frac{12}{100}\right)^{\frac{1}{2}}$
$=\text { Rs } 15000 \times 1.12 \times\left(1+\frac{1}{2} \times \frac{12}{100}\right)$
$=\text { Rs } 15,000 \times 1.12 \times 1.06$
$=\text { Rs } 17,808$
$\text { C.I. }=A-P$
$=\text { Rs }(17,808-15,000)$
$=\text { Rs } 2808 $
When compounded half-yearly :
$ A=P\left(1+\frac{r}{100}\right)^n$
$A=\operatorname{Rs} 15000\left(1+\frac{6}{100}\right)^3$
$=\operatorname{Rs} 15,000 \times 1.06 \times 1.06 \times 1.06$
$=\operatorname{Rs} 17,865.24$
$\text { C.I. }=A-P$
$=\text { Rs }(17,865.24-15,000)$
$=\text { Rs } 2,865.24 $
Hence the difference in the interest=Rs $(2,865.24-2,808)$
$ =\operatorname{Rs} 57.24 $
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