Question
The difference between the compound interest and the simple interest on a certain sum for $2$ years at $6\%$ per annum is $Rs. 90.$ Find the sum.
✓
Answer
Difference between $C.I$ and $S.I. = Rs. 90$
Rate $(R) = 6\%\ p.a$
Period $(n) = 2$ years
Let principal $(P) = Rs. 100$
$\therefore$ Amount on $C.I =\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=100\Big(1+\frac{6}{100}\Big)^2$
$=\text{Rs. }100\times\Big(\frac{53}{50}\Big)^2$
$=\text{Rs. }100\times\frac{53}{50}\times\frac{53}{50}=\text{Rs. }\frac{2809}{25}$
$\therefore\text{C.I}=\text{A}-\text{P}=\text{Rs. }\frac{2809}{25}-100$
$=\text{Rs. }\frac{2809-2500}{25}=\frac{309}{25}$ and
$S.I =\frac{\text{PRT}}{100}=\frac{100\times6\times2}{100}=\text{Rs. }12$
Difference between $C.I$ and $S.I. =\text{Rs. }\frac{309}{25}-12$
$\text{Rs. }\frac{309-300}{25}=\frac{9}{25}$
Now if difference is $\text{Rs. }\frac{9}{25}$
then Principal $= Rs. 100$ and if difference is $Rs. 90,$
then principal$=\frac{100\times25\times90}{9}=\text{Rs. }25000$
Rate $(R) = 6\%\ p.a$
Period $(n) = 2$ years
Let principal $(P) = Rs. 100$
$\therefore$ Amount on $C.I =\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=100\Big(1+\frac{6}{100}\Big)^2$
$=\text{Rs. }100\times\Big(\frac{53}{50}\Big)^2$
$=\text{Rs. }100\times\frac{53}{50}\times\frac{53}{50}=\text{Rs. }\frac{2809}{25}$
$\therefore\text{C.I}=\text{A}-\text{P}=\text{Rs. }\frac{2809}{25}-100$
$=\text{Rs. }\frac{2809-2500}{25}=\frac{309}{25}$ and
$S.I =\frac{\text{PRT}}{100}=\frac{100\times6\times2}{100}=\text{Rs. }12$
Difference between $C.I$ and $S.I. =\text{Rs. }\frac{309}{25}-12$
$\text{Rs. }\frac{309-300}{25}=\frac{9}{25}$
Now if difference is $\text{Rs. }\frac{9}{25}$
then Principal $= Rs. 100$ and if difference is $Rs. 90,$
then principal$=\frac{100\times25\times90}{9}=\text{Rs. }25000$
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