Two equal sums were lent at 5% and 6% per annum compound interest for 2 years. If the difference in the compound interest was ₹422, find :

(i) the equal sums

(ii) compound interest for each sum.

(i) Let the sum be ₹P.

C.I. = $P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]$

C.I. when sum is lent at 5% for 2 years,

$C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] = P\Big[\Big(1 + \dfrac{5}{100}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(1 + \dfrac{1}{20}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(\dfrac{21}{20}\Big)^2 - 1\Big] \\[1em] = P\Big[\dfrac{441}{400} - 1\Big] \\[1em] = P\Big[\dfrac{441 - 400}{400} \Big] \\[1em] = P \times \dfrac{41}{400} \\[1em] = \dfrac{41P}{400}.$

C.I. when sum is lent at 6% for 2 years,

$C.I. = P\Big[\Big(1 + \dfrac{6}{100}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(1 + \dfrac{3}{50}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(\dfrac{53}{50}\Big)^2 - 1\Big] \\[1em] = P\Big[\dfrac{2809}{2500} - 1\Big] \\[1em] = P \times \dfrac{2809 - 2500}{2500} \\[1em] = \dfrac{309P}{2500}.$

Given, difference in C.I. = ₹422.

$\therefore \dfrac{309P}{2500} - \dfrac{41P}{400} = 422 \\[1em] \Rightarrow \dfrac{309P \times 4 - 41P \times 25}{10000} = 422 \\[1em] \Rightarrow \dfrac{1236P - 1025P}{10000} = 422 \\[1em] \Rightarrow \dfrac{211P}{10000} = 422 \\[1em] \Rightarrow P = \dfrac{422 \times 10000}{211} \\[1em] \Rightarrow P = ₹20000.$

Hence, equal sum = ₹20000.

(ii) C.I. when sum is lent at 5% for 2 years = $\dfrac{41P}{400}.$

$C.I. = \dfrac{41 \times 20000}{400} \\[1em] = 41 \times 50 \\[1em] = ₹2050.$

C.I. when sum is lent at 6% for 2 years = $\dfrac{309P}{2500}.$

$C.I. = \dfrac{309 \times 20000}{2500} \\[1em] = 309 \times 8 \\[1em] = ₹2472.$

Hence, C.I. = ₹2050 when sum is lent at 5% for 2 years and C.I. = ₹2472 when sum is lent at 6% for 2 years.