The simple interest on a certain sum for 3 years is ₹1080 and the compound interest on the same sum at the same rate for 2 years is ₹741.60. Find :

(i) the rate of interest

(ii) the principal.

Let the sum be ₹x and rate be r%.

Given, S.I. for 3 years = ₹1080.

$\therefore \dfrac{P \times R \times T}{100} = 1080 \\[1em] \Rightarrow \dfrac{x \times r \times 3}{100} = 1080 \\[1em] \Rightarrow x \times r = \dfrac{1080 \times 100}{3} \\[1em] \Rightarrow x \times r = 36000 …….(i)$

Given, C.I. for 2 years = ₹741.60

$C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\Big(1 + \dfrac{r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\Big(\dfrac{100 + r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\dfrac{(100 + r)^2}{100^2} - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\dfrac{(100 + r)^2 - 100^2}{100^2}\Big] \\[1em] \Rightarrow 741.60 = x\Big[\dfrac{100^2 + r^2 + 200r - 100^2}{100^2}\Big] \\[1em] \Rightarrow 741.60 = x\Big[\dfrac{r^2 + 200r}{100^2}\Big] \\[1em] \Rightarrow 741.60 \times 100^2 = x \times r \times (r + 200) \\[1em] \Rightarrow 7416000 = 36000(r + 200) …….\text{ (Using (i))} \\[1em] \Rightarrow r + 200 = \dfrac{7416000}{36000} \\[1em] \Rightarrow r + 200 = 206 \\[1em] \Rightarrow r = 6\%.$

Hence, the rate of interest = 6%.

(ii) Putting value of r in (i) we get,

$\Rightarrow x \times r = 36000 \\[1em] \Rightarrow 6x = 36000 \\[1em] \Rightarrow x = ₹6000.$

Hence, principal = ₹6000.