A sum of ₹9600 is invested for 3 years at 10% per annum at compound interest.

(i) What is the sum due at the end of the first year?

(ii) What is the sum due at the end of the second year?

(iii) Find the compound interest earned in 2 years.

(iv) Find the difference between the answers in (ii) and (i) and find the interest on this sum for one year.

(v) Hence, write down the compound interest for the third year.

(i) Principal for first year = ₹9600.

Interest for the first year = ₹ $\dfrac{9600 \times 10 \times 1}{100} = \dfrac{96000}{100}$ = ₹960.

Amount after one year = ₹9600 + ₹960 = ₹10560.

Hence, the amount due at the end of first year = ₹10560.

(ii) Principal for second year = ₹10560.

Interest for the second year = ₹ $\dfrac{10560 \times 10 \times 1}{100} = \dfrac{105600}{100}$ = ₹1056.

Amount after 2 years = ₹10560 + ₹1056 = ₹11616.

Hence, the amount due at the end of second year = ₹11616.

(iii) Compound interest = Final amount - Principal = ₹11616 - ₹9600 = ₹2016.

Hence, the compound interest earned in 2 years = ₹2016.

(iv) Difference between (ii) and (i) = ₹11616 - ₹10560 = ₹1056.

Interest on the above sum for 1 year = $\dfrac{1056 \times 10 \times 1}{100} = \dfrac{10560}{100}$ = ₹105.60

Hence, the difference = ₹1056 and interest earned on it = ₹105.60

(v) Principal for third year = ₹11616.

Interest for the third year = ₹ $\dfrac{11616 \times 10 \times 1}{100} = \dfrac{116160}{100}$ = ₹1161.60

Hence, the compound interest for third year = ₹1161.60