Find the sum on which the difference between the simple interest and the compound interest at the rate of 8% per annum compounded annually be ₹ 64 in 2 years.

Let sum be ₹ x.

For S.I. :

P = ₹ x

R = 8%

T = 2 years

S.I. = $\dfrac{P \times R \times T}{100} = \dfrac{x \times 8 \times 2}{100} = \dfrac{4x}{25}$.

For C.I. :

For first year :

P = ₹ x

R = 8%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{x \times 8 \times 1}{100} = \dfrac{2x}{25}$.

Amount = P + I = $x + \dfrac{2x}{25} = \dfrac{27x}{25}$.

For second year :

P = ₹ $\dfrac{27x}{25}$

R = 8%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{\dfrac{27x}{25} \times 8 \times 1}{100} = \dfrac{54x}{625}$.

Amount = P + I = $\dfrac{27x}{25} + \dfrac{54x}{625} = \dfrac{675x + 54x}{625} = \dfrac{729x}{625}$.

C.I. = Final amount - Initial Principal

$= \dfrac{729x}{625} - x \\[1em] = \dfrac{729x - 625x}{625} \\[1em] = \dfrac{104x}{625}.$

Given, difference between S.I. and C.I. = ₹ 64

$\Rightarrow \dfrac{104x}{625} - \dfrac{4x}{25} = 64 \\[1em] \Rightarrow \dfrac{104x - 100x}{625} = 64 \\[1em] \Rightarrow \dfrac{4x}{625} = 64 \\[1em] \Rightarrow x = \dfrac{64}{4} \times 625 \\[1em] \Rightarrow x = 16 \times 625 = 10000.$

Hence, required sum = ₹ 10000.