Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets ₹ 1,200 as interest at the time of maturity, find :

(i) the monthly instalment

(ii) the amount of maturity

(iii) If Mohan decreases his monthly installment by 20%, how much less interest will he get at the same rate of interest and for the same time ?

(i) Let monthly installment be ₹ x.

So,

P = ₹ x, r = 6% and n = (2 × 12) = 24 months.

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

$\therefore I = ₹ x \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{6}{100} \\[1em] = ₹ x \times 25 \times \dfrac{6}{100} \\[1em] = ₹ \dfrac{150x}{100} \\[1em] = ₹ x \times \dfrac{3}{2} \\[1em] = ₹ \dfrac{3x}{2}$

Given, I = ₹ 1200.

$\therefore \dfrac{3x}{2} = 1200 \\[1em] \Rightarrow x = \dfrac{2400}{3} = 800.$

Hence, monthly installment = ₹ 800.

(ii) Maturity value = Sum deposited + Interest

= ₹ 800 × 24 + ₹ 1,200

= ₹ 19,200 + ₹ 1,200

= ₹ 20,400.

Hence, maturity value = ₹ 20,400.

(iii) Given,

Monthly installment is reduced by 20%.

Thus, new monthly installment = ₹ 800 - 20% of ₹ 800

= ₹ 800 - ₹ 160 = ₹ 640.

P = ₹ 640, r = 6% and n = (2 × 12) = 24 months

By formula,

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\therefore I = ₹ 640 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{6}{100} \\[1em] = ₹ 640 \times 25 \times \dfrac{6}{100} \\[1em] = ₹ \dfrac{96000}{100} \\[1em] = ₹ 960.$

Reduction in interest = ₹ 1,200 - ₹ 960 = ₹ 240.

Hence, Mohan will get ₹ 240 less as interest.