Question
The difference between simple interest and compound interest compounded annually on a certain sum is $Rs.448$ for $2$ years at $8$ percent per annum. Find the sum.
✓
Answer
Since, $C.I.$
$=A - P$
$\text { C.I. }=\left(1+\frac{8}{100}\right)^2-P$
$=P\left(\frac{108}{100}\right)^2-P$
$=\frac{11664 P }{10000}- P$
$=\frac{11664 P -10000 P }{10000}$
$=\frac{1664 P }{10000}$
$\text { S.L. }=\frac{ P \times 8 \times 2}{100}$
$=\frac{16 P }{100} $
Now, $C.I. - S.I = Rs. 448$
$\Rightarrow \frac{1664 P }{10000}-\frac{16 P }{100}=\text { Rs. } 448$
$\Rightarrow \frac{1664 P -1600 P }{10000}= Rs .448$
$\Rightarrow 64 P = Rs .4480000$
$\Rightarrow P = Rs {70000}$
Hence, the sum is $Rs. 70000 .$
$=A - P$
$\text { C.I. }=\left(1+\frac{8}{100}\right)^2-P$
$=P\left(\frac{108}{100}\right)^2-P$
$=\frac{11664 P }{10000}- P$
$=\frac{11664 P -10000 P }{10000}$
$=\frac{1664 P }{10000}$
$\text { S.L. }=\frac{ P \times 8 \times 2}{100}$
$=\frac{16 P }{100} $
Now, $C.I. - S.I = Rs. 448$
$\Rightarrow \frac{1664 P }{10000}-\frac{16 P }{100}=\text { Rs. } 448$
$\Rightarrow \frac{1664 P -1600 P }{10000}= Rs .448$
$\Rightarrow 64 P = Rs .4480000$
$\Rightarrow P = Rs {70000}$
Hence, the sum is $Rs. 70000 .$
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