On what sum will the difference between the simple and compound interest for 3 years at 10% p.a. is ₹232.50?

Let the sum be ₹P.

S.I. = $\dfrac{P \times 10 \times 3}{100} = \dfrac{3P}{10}.$

By formula,

C.I. = $P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] = P\Big[\Big(1 + \dfrac{10}{100}\Big)^3 - 1\Big] \\[1em] = P\Big[\Big(\dfrac{110}{100}\Big)^3 - 1\Big] \\[1em] = P\Big[\Big(\dfrac{11}{10}\Big)^3 - 1\Big] \\[1em] = P\Big[\dfrac{1331}{1000} - 1\Big] \\[1em] = P\Big[\dfrac{1331 - 1000}{1000} \Big] \\[1em] = P \times \dfrac{331}{1000} \\[1em] = \dfrac{331P}{1000}.$

Given, difference between C.I. and S.I. = ₹232.50

$\therefore \dfrac{331P}{1000} - \dfrac{3P}{10} = 232.50 \\[1em] \Rightarrow \dfrac{331P- 300P}{1000} = 232.50 \\[1em] \Rightarrow \dfrac{31P}{1000} = 232.50 \\[1em] \Rightarrow P = \dfrac{232.50 \times 1000}{31} \\[1em] \Rightarrow P = \dfrac{232500}{31} \\[1em] \Rightarrow P = ₹7500.$

Hence, sum = ₹7500.