A man invests ₹ 10,000 for 3 years at a certain rate of interest, compounded annually. At the end of one year, it amounts to ₹ 11,200. Calculate :

(i) the rate of interest per annum;

(ii) the interest accrued in the second year;

(iii) the amount at the end of the third year.

(i) Given,

P = ₹ 10,000

T = 3 year

Amount at the end of first year = ₹ 11,200

Interest in the first year = Amount - Principal

= ₹ 11,200 - ₹ 10,000 = ₹ 1,200.

So, for 1 year interest equals to ₹ 1,200 on ₹ 10,000. Let rate of interest be R%. Substituting values we get :

$\Rightarrow I = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow 1200 = \dfrac{10000 \times R \times 1}{100} \\[1em] \Rightarrow 1200 = 100 \times R \\[1em] \Rightarrow R = \dfrac{1200}{100} \\[1em] \Rightarrow R = 12\%.$

Hence, the rate of interest per annum = 12% p.a.

(ii) Given,

For second year :

P = ₹ 11,200

T = 1 year

R = 12%

Interest accrued in the second year,

I = $\dfrac{P \times R \times T}{100}$

$= \dfrac{11200 \times 12 \times 1}{100}$

= ₹ 1,344.

Hence, the interest accrued in the second year = ₹ 1,344.

(iii) For third year,

P = ₹ 11,200 + ₹ 1,344 = ₹ 12,544

I = $\dfrac{P \times R \times T}{100}$

$= \dfrac{12544 \times 12 \times 1}{100}$

= ₹ 1,505.28

Amount at the end of the third year = P + I = ₹ 12,544 + ₹ 1,505.28 = ₹ 14,049.28

Hence, the amount at the end of the third year = ₹ 14,049.28.