Ashok borrowed ₹ 12000 at some rate per cent compound interest. After a year, he paid back ₹ 4000. If compound interest for the second year be ₹ 920, find :

(i) the rate of interest charged

(ii) the amount of debt at the end of the second year.

(i) Let rate of interest be x%.

For first year :

P = ₹ 12000

R = x%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{12000 \times x \times 1}{100}$ = 120x.

Amount = P + I = ₹ 12000 + ₹ 120x.

Given amount paid back after a year = ₹ 4000

Amount left = ₹ 12000 + ₹ 120x - ₹ 4000 = ₹ 8000 + ₹ 120x

For second year :

P = ₹ 8000 + ₹ 120x

R = x%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{(8000 + 120x) \times x \times 1}{100}$.

Given,

Interest for second year = ₹ 920

$\therefore \dfrac{(8000 + 120x) \times x \times 1}{100} = 920 \\[1em] \Rightarrow (8000 + 120x)x = 92000 \\[1em] \Rightarrow 8000x + 120x^2 = 92000 \\[1em] \Rightarrow 40(200x + 3x^2) = 40(2300) \\[1em] \Rightarrow 3x^2 + 200x = 2300 \\[1em] \Rightarrow 3x^2 + 200x - 2300 = 0 \\[1em] \Rightarrow 3x^2 + 230x - 30x - 2300 = 0 \\[1em] \Rightarrow x(3x + 230) - 10(3x + 230) = 0 \\[1em] \Rightarrow (x - 10)(3x + 230) = 0 \\[1em] \Rightarrow x - 10 = 0 \text{ or } 3x + 230 = 0 \\[1em] \Rightarrow x = 10 \text{ or } 3x = -230 \\[1em] \Rightarrow x = 10 \text{ or } x = -\dfrac{230}{3}.$

Since, rate of interest cannot be negative.

Hence, rate of interest = 10%.

(ii) We know that :

For second year :

P = ₹ 8000 + ₹ 120x = ₹ 8000 + ₹ 120 × 10 = ₹ 9200.

I = ₹ 920

Amount at end of second year = P + I = ₹ 9200 + ₹ 920 = ₹ 10120.

Hence, the amount of debt at end of second year = ₹ 10120.