A sum of ₹ 12,500 is deposited for $1\dfrac{1}{2}$ years, compounded half yearly. It amounts to ₹ 13,000 at the end of first half year. Find:

(i) The rate of interest

(ii) The final amount. Give your answer correct to the nearest rupee.

(i) Given,

P = ₹ 12,500

n = 1 half year

Amount = ₹ 13,000

When rate of interest is compounded half-yearly :

By formula,

$A = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{2 \times n}$

For first half year:

$\Rightarrow 13000 = 12500\Big(1 + \dfrac{R}{2 \times 100}\Big)^{0.5 \times 2} \\[1em] \Rightarrow \dfrac{13000}{12500}= \Big(1 + \dfrac{R}{200}\Big)^1 \\[1em] \Rightarrow 1.04 - 1 = \Big(\dfrac{R}{200}\Big) \\[1em] \Rightarrow 0.04 = \Big(\dfrac{R}{200}\Big) \\[1em] \Rightarrow R = 0.04 \times 200 \\[1em] \Rightarrow R = 8\%$

Hence, Rate of interest = 8% p.a .

(ii) Given,

P = ₹ 12,500

n = 1.5 years

R = 8%

Let's calculate compound interest:

When rate of interest is compounded half-yearly :

By formula,

$A = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{2 \times n}$

$\Rightarrow A = 12500 \Big(1 + \dfrac{8}{2 \times 100}\Big)^{2 \times 1.5} \\[1em] \Rightarrow A = 12500 \Big(\dfrac{200 + 8}{200}\Big)^3 \\[1em] \Rightarrow A = 12500 \Big(\dfrac{208}{200}\Big)^3 \\[1em] \Rightarrow A = 12500 \Big(\dfrac{26}{25}\Big)^3 \\[1em] \Rightarrow A = 12500 \times \dfrac{17576}{15625} \\[1em] \Rightarrow A = ₹ 14,060.80 \approx ₹ 14,061$

Hence, compound interest = ₹ 14,061.