A certain sum amounts to ₹798.60 after 3 years and ₹878.46 after 4 years. Find the interest 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 ₹798.60 in 3 years substituting value we get,

$\therefore 798.60 = P\Big(1 + \dfrac{r}{100}\Big)^3$ …..(Eq. 1)

As sum amounts to ₹878.46 in 4 years substituting value we get,

$\therefore 878.46 = P\Big(1 + \dfrac{r}{100}\Big)^4$ …..(Eq. 2)

Dividing Eq. 2 by Eq. 1 we get,

$\Rightarrow 1 + \dfrac{r}{100} = \dfrac{878.46}{798.60} \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{87846}{79860} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{87846}{79860} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{87846 - 79860}{79860} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{7986}{79860} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1}{10} \\[1em] \Rightarrow r = \dfrac{100}{10} \\[1em] \Rightarrow r = 10\%.$

Substituting value of r in Eq. 1 we get,

$\Rightarrow 798.60 = P\Big(1 + \dfrac{10}{100}\Big)^3 \\[1em] \Rightarrow 798.60 = P\Big(1 + \dfrac{1}{10}\Big)^3 \\[1em] \Rightarrow 798.60 = P\Big(\dfrac{11}{10}\Big)^3 \\[1em] \Rightarrow 798.60 = P \times \dfrac{1331}{1000} \\[1em] \Rightarrow P = 798.60 \times \dfrac{1000}{1331} \\[1em] \Rightarrow P = \dfrac{798600}{1331} \\[1em] \Rightarrow P = ₹600.$

Hence, sum = ₹600 and rate = 10%.