Simple interest on a certain sum of money for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 percent per annum by ₹ 228. Find the sum.

Let sum of money be ₹ x.

For S.I. :

P = ₹ x

Time (T) = 4 years

Rate of interest (R) = 4%

By formula,

S.I. = $\dfrac{P \times R \times T}{100}$

Substituting values we get :

$\Rightarrow S.I. = \dfrac{x \times 4 \times 4}{100} \\[1em] = \dfrac{4x}{25}.$

For C.I. :

P = ₹ x

Rate of interest (r) = 5%

Time (n) = 3 years

By formula,

$\Rightarrow C.I. = A - P \\[1em] \Rightarrow C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = x \times \Big(1 + \dfrac{5}{100}\Big)^3 - x \\[1em] = x \times \Big(\dfrac{105}{100}\Big)^3 - x \\[1em] = x \times \Big(\dfrac{21}{20}\Big)^3 - x \\[1em] = \dfrac{9261x}{8000} - x \\[1em] = \dfrac{9261x - 8000x}{8000} \\[1em] = \dfrac{1261x}{8000}.$

Given,

Simple interest on ₹ x for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 percent per annum by ₹ 228.

$\therefore \dfrac{4x}{25} - \dfrac{1261x}{8000} = 228 \\[1em] \Rightarrow \dfrac{1280x - 1261x}{8000} = 228 \\[1em] \Rightarrow \dfrac{19x}{8000} = 228 \\[1em] \Rightarrow x = \dfrac{228 \times 8000}{19} \\[1em] \Rightarrow x = ₹ 96000.$

Hence, sum of money = ₹ 96000.