Question
On what sum will the difference between compound interest and the simple interest for $2$ years at $7 \frac{1}{2} \%$ be $Rs.22.50 ?$
✓
Answer
$P=x ; t=2$ years $; r=7 \frac{1}{2} \%=\frac{15}{2} \%$
For S.I. :
$I=\frac{P \times r \times t}{100} $
$=\frac{x \times \frac{15}{2} \times 2}{100} $
$ =\frac{3 x}{20}$
For C.I. :
$ \text { C.I. }= P \left(1+\frac{ r }{100}\right)^{ t }- P $
$= x \left(1+\frac{\frac{15}{2}}{100}\right)^2- x$
$ = x \left(1+\frac{15}{2 \times 100}\right)^2- x $
$= x \left(1+\frac{3}{40}\right)^2- x$
$=( x \times 1.075 \times 1.075)- x $
$ =1.155625 x - x$
$ =0.155625 x$
$\text { Given C.I. - S.I. }=\text { Rs } 22.50$
$ \Rightarrow 0.155625 x-\frac{3 x}{20}=\operatorname{Rs} 22.50$
$ \Rightarrow 0.155625 x-0.15 x=\operatorname{Rs} 22.50$
$ \Rightarrow 0.005625 x=\operatorname{Rs} 22.50$
$\Rightarrow x-R s.4,000$
Hence, sum $= Rs.4,000$
For S.I. :
$I=\frac{P \times r \times t}{100} $
$=\frac{x \times \frac{15}{2} \times 2}{100} $
$ =\frac{3 x}{20}$
For C.I. :
$ \text { C.I. }= P \left(1+\frac{ r }{100}\right)^{ t }- P $
$= x \left(1+\frac{\frac{15}{2}}{100}\right)^2- x$
$ = x \left(1+\frac{15}{2 \times 100}\right)^2- x $
$= x \left(1+\frac{3}{40}\right)^2- x$
$=( x \times 1.075 \times 1.075)- x $
$ =1.155625 x - x$
$ =0.155625 x$
$\text { Given C.I. - S.I. }=\text { Rs } 22.50$
$ \Rightarrow 0.155625 x-\frac{3 x}{20}=\operatorname{Rs} 22.50$
$ \Rightarrow 0.155625 x-0.15 x=\operatorname{Rs} 22.50$
$ \Rightarrow 0.005625 x=\operatorname{Rs} 22.50$
$\Rightarrow x-R s.4,000$
Hence, sum $= Rs.4,000$
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