Mr. Krishnan deposits ₹ 1,000 per month in a recurring deposit account with State Bank of India for 2 years at 8% p.a. simple interest

Based on above information answer the following :

(i) Find the equivalent principal for 1 month.

(ii) Find the amount of maturity Mr. Krishnan will get at the end of 2 years.

(iii) If the bank revised the rate of interest 6% p.a. from 8% p.a., then by how much the interest paid by the bank will be reduced.

(i) n = 2 years = 24 months, P = ₹ 1,000, r = 8%

The monthly installment deposited by him = ₹ 1,000

So for 1 month, the principal is ₹ 1,000.

Hence, Mr. Krishnan’s equivalent principal for 1 month = ₹ 1,000.

(ii) Given, n = 2 years = 24 months, P = ₹ 1,000, r = 8%

We know that,

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

$\therefore I = ₹ 1000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{8}{100} \\[1em] = ₹ 1000 \times \dfrac{600}{24} \times \dfrac{8}{100} \\[1em] = ₹ 1000 \times 25 \times \dfrac{8}{100} \\[1em] = ₹ 2,000$

Maturity value = Sum deposited + Interest

= ₹ 1,000 × 24 + ₹ 2,000

= ₹ 24,000 + ₹ 2,000

= ₹ 26,000.

Hence, Mr. Krishnan will receive ₹ 26,000 at maturity.

(iii) If the rate is reduced to 6%:

P = ₹ 1000, r = 6% and n = (2 × 12) = 24 months.

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

$\therefore I = ₹ 1000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{6}{100} \\[1em] = ₹ 1000 \times \dfrac{3}{2} \\[1em] = ₹ 1,500.$

Reduction in interest paid to Mr. Krishnan :

= ₹ 2,000 − ₹ 1,500

= ₹ 500.

Hence, the bank will pay ₹ 500 less interest to Mr. Krishnan.