Question
Find the difference between the compound interest and the simple interest in $3$ years on $Rs.15,000$ at $8\%$ p.a. compounded yearly.
✓
Answer
Here, $P=\operatorname{Rs} 15,000 ; r=8 \% ; t=3$ years
For simple interest :
$ \text { S.I. }=\frac{ P \times r \times t }{100} $
$ \text { S.I. }=\operatorname{Rs} \frac{15000 \times 8 \times 3}{100} $
$ \text { S.I. }=\operatorname{Rs} 3600 $
For compound interest:
$A=P\left(1+\frac{r}{100}\right)^n $
$ A=\text { Rs } 15000\left(1+\frac{8}{100}\right)^3 $
$ A=\text { Rs } 15000 \times \frac{108}{100} \times \frac{108}{100} \times \frac{108}{100}$
$A=\text { Rs } 18895.68$
$\text { C.I. }=A-P$
$ \text { C.I. }=\text { Rs }(18,895.68-15,000)$
$ \text { C.I. }=\text { Rs3, 895.68 } $
The difference in the compound interest and the simple interest $=$ Rs $(3,895.683 .600)= Rs.295.68$
For simple interest :
$ \text { S.I. }=\frac{ P \times r \times t }{100} $
$ \text { S.I. }=\operatorname{Rs} \frac{15000 \times 8 \times 3}{100} $
$ \text { S.I. }=\operatorname{Rs} 3600 $
For compound interest:
$A=P\left(1+\frac{r}{100}\right)^n $
$ A=\text { Rs } 15000\left(1+\frac{8}{100}\right)^3 $
$ A=\text { Rs } 15000 \times \frac{108}{100} \times \frac{108}{100} \times \frac{108}{100}$
$A=\text { Rs } 18895.68$
$\text { C.I. }=A-P$
$ \text { C.I. }=\text { Rs }(18,895.68-15,000)$
$ \text { C.I. }=\text { Rs3, 895.68 } $
The difference in the compound interest and the simple interest $=$ Rs $(3,895.683 .600)= Rs.295.68$
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