The simple interest on a certain sum for 3 years is ₹225 and the compound interest on the same sum at the same rate for 2 years is ₹153. Find the rate of interest and principal.

Let principal be ₹P and rate of interest be R% p.a.

According to the first condition of the question,

S.I. on ₹P for 3 years = ₹225.

$\therefore \dfrac{P \times R \times T}{100} = 225 \\[1em] \Rightarrow \dfrac{P \times R \times 3}{100} = 225 \\[1em] \Rightarrow P \times R = \dfrac{225 \times 100}{3} \\[1em] \Rightarrow P \times R = 7500 \\[1em] \Rightarrow P = \dfrac{7500}{R} …….(i)$

According to the second condition of the question,

C.I. on ₹P for 2 years at R% p.a. = ₹153.

$\therefore P\Big[\Big(1 + \dfrac{R}{100}\Big)^n - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\Big(1 + \dfrac{R}{100}\Big)^2 - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\Big(\dfrac{100 + R}{100}\Big)^2 - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\dfrac{(100 + R)^2}{100^2} - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\dfrac{(100 + R)^2 - 100^2}{100^2}\Big] = 153 \\[1em] \Rightarrow P\Big[\dfrac{100^2 + R^2 + 200R - 100^2}{100^2}\Big] = 153 \\[1em] \Rightarrow P\Big[\dfrac{R^2 + 200R}{100^2}\Big] = 153 \\[1em]$

Using value of P from Eq 1 above:

$\Rightarrow \dfrac{7500}{R} \times \dfrac{R(R + 200)}{100^2} = 153 \\[1em] \Rightarrow \dfrac{7500(R + 200)}{100^2} = 153 \\[1em] \Rightarrow R + 200 = \dfrac{153 \times 100 \times 100}{7500} \\[1em] \Rightarrow R + 200 = \dfrac{15300}{75} \\[1em] \Rightarrow R + 200 = 204 \\[1em] \Rightarrow R = 204 - 200 = 4\%.$

Substituting value of R in (i) we get,

$P = \dfrac{7500}{4} = ₹1875.$

Hence, rate of interest = 4% and sum = ₹1875.