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(C) Let the principal be $P$ and the rate of interest be $R\%$. Simple Interest $(SI)$ for $3$ years $= Rs. 240$. Therefore,$SI$ for $1$ year $= \frac

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$1600$

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(C) Let the principal be $P$ and the rate of interest be $R\%$. Simple Interest $(SI)$ for $3$ years $= Rs. 240$. Therefore,$SI$ for $1$ year $= \frac{240}{3} = Rs. 80$. $SI$ for $2$ years $= 80 	imes 2 = Rs. 160$. We know that $SI = \frac{P 	imes R 	imes T}{100}$,so for $2$ years: $160 = \frac{P 	imes R 	imes 2}{100} \Rightarrow PR = 8000$. Compound Interest $(CI)$ for $2$ years $= Rs. 164$. The difference between $CI$ and $SI$ for $2$ years is given by $CI - SI = P \left( \frac{R}{100} ight)^2$. $164 - 160 = P \left( \frac{R}{100} ight)^2 \Rightarrow 4 = P \left( \frac{R}{100} ight)^2$. Since $PR = 8000$,we have $R = \frac{8000}{P}$. Substituting $R$ in the equation: $4 = P \left( \frac{8000/P}{100} ight)^2 = P \left( \frac{80}{P} ight)^2 = P 	imes \frac{6400}{P^2} = \frac{6400}{P}$. $P = \frac{6400}{4} = 1600$. Thus,the sum is $Rs. 1600$.

If the compound interest on a certain sum of money for $2$ years is $Rs. 164$ whereas the simple interest for $3$ years at the same rate is $Rs. 240$ on the same sum,find the sum (in $Rs.$).

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