The difference between the simple interest and the compound interest on a sum of money for 3 years at 10% per annum is ₹ 558. Find the sum.

Given,

n = 3 years

r = 10 %

Let sum of money be ₹ P.

C.I. = A - P

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

Calculating S.I.,

R = 10%

T = 3 years

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{P \times 10 \times 3}{100} \\[1em] = \dfrac{3P}{10}.$

Given,

Difference between S.I. and C.I. = ₹ 558

$\Rightarrow \dfrac{331P}{1000} - \dfrac{3P}{10} = 558 \\[1em] \Rightarrow \dfrac{331P - 300P}{1000} = 558 \\[1em] \Rightarrow \dfrac{31P}{1000} = 558 \\[1em] \Rightarrow P = \dfrac{558 \times 1000}{31} \\[1em] \Rightarrow P = 18 \times 1000 \\[1em] \Rightarrow P = ₹ 18,000.$

Hence, the sum = ₹ 18,000.