The compound interest on a sum of money for 2 years is ₹1331.20 and the simple interest on the same sum for the same period at the same rate is ₹1280. Find the sum and the rate of interest per annum.

Let the sum be ₹x and rate be r%.

Given, S.I. for 2 years = ₹1280.

$\therefore \dfrac{P \times R \times T}{100} = 1280 \\[1em] \Rightarrow \dfrac{x \times r \times 2}{100} = 1280 \\[1em] \Rightarrow x \times r = \dfrac{1280 \times 100}{2} \\[1em] \Rightarrow x \times r = 64000 …….(i)$

Given, C.I. for 2 years = ₹1331.20

$C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\Big(1 + \dfrac{r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\Big(\dfrac{100 + r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\dfrac{(100 + r)^2}{100^2} - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\dfrac{(100 + r)^2 - 100^2}{100^2}\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\dfrac{100^2 + r^2 + 200r - 100^2}{100^2}\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\dfrac{r^2 + 200r}{100^2}\Big] \\[1em] \Rightarrow 1331.20 \times 100^2 = x \times r \times (r + 200) \\[1em] \Rightarrow 13312000 = 64000(r + 200) …….\text{ (Using (i))} \\[1em] \Rightarrow r + 200 = \dfrac{13312000}{64000} \\[1em] \Rightarrow r + 200 = 208 \\[1em] \Rightarrow r = 8\%.$

Putting value of r in Eq. (i) we get,

$\Rightarrow x \times 8 = 64000 \\[1em] \Rightarrow x = \dfrac{64000}{8} \\[1em] \Rightarrow x = ₹8000.$

Hence, sum = ₹8000 and rate = 8%.