A certain sum amounts to ₹5292 in 2 years and to ₹5556.60 in 3 years at compound interest. Find the rate and the sum.

Let the rate of interest be r% per annum.

$A = P\Big(1 + \dfrac{r}{100}\Big)^n.$

As sum amounts to ₹5292 in 2 years, substituting value we get,

$\therefore 5292 = P\Big(1 + \dfrac{r}{100}\Big)^2$ …..(Eq. 1)

As sum amounts to ₹5556.60 in 3 years, substituting value we get,

$\therefore 5556.60 = P\Big(1 + \dfrac{r}{100}\Big)^3$ …..(Eq. 2)

Dividing Eq. 2 by Eq. 1 we get,

$\Rightarrow 1 + \dfrac{r}{100} = \dfrac{5556.60}{5292} \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{555660}{529200} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{555660}{529200} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{26460}{529200} \\[1em] \Rightarrow r = \dfrac{26460}{529200} \times 100 \\[1em] \Rightarrow r = \dfrac{26460}{5292} \\[1em] \Rightarrow r = 5\%.$

Substituting value of r in Eq. 1 we get,

$\Rightarrow 5292 = P\Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] \Rightarrow 5292 = P\Big(1 + \dfrac{1}{20}\Big)^2 \\[1em] \Rightarrow 5292 = P\Big(\dfrac{21}{20}\Big)^2 \\[1em] \Rightarrow 5292 = P \times \dfrac{441}{400} \\[1em] \Rightarrow P = 5292 \times \dfrac{400}{441} \\[1em] \Rightarrow P = 12 \times 400 \\[1em] \Rightarrow P = ₹4800.$

Hence, sum = ₹4800 and rate = 5%.