Mr. Ahuja deposited ₹ 500 per month in an R.D. account for a period of 3 years. He received ₹ 20,220 at the time of maturity. Find :

(i) rate of interest.

(ii) how much more interest Mr. Ahuja will receive, if he had deposited ₹ 100 more every month.

Given,

n = 3 years = 36 months, P = ₹ 500

Let r be the rate of interest.

Maturity amount = ₹ 20,220

Total amount deposited = 500 × 36 = ₹ 18,000.

Interest received = Maturity amount - Amount deposited

= ₹ 20,220 − ₹ 18,000

= ₹ 2,220.

By formula,

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

Substituting values we get :

$\Rightarrow 2220 = 500 \times \dfrac{36 \times (37)}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 2220 = 500 \times \dfrac{1332}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 2220 = 500 \times 55.5 \times \dfrac{r}{100} \\[1em] \Rightarrow 2220 = 27750 \times \dfrac{r}{100} \\[1em] \Rightarrow 2220 = 277.5r \\[1em] \Rightarrow r = \dfrac{2220}{277.5}\\[1em] \Rightarrow r = 8\%$

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

(ii) If Mr. Ahuja had deposited ₹ 100 more per month then the monthly deposit would have been ₹ 600.

By formula,

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

Substituting values we get :

$\Rightarrow I = 600 \times \dfrac{36 \times (37)}{24} \times \dfrac{8}{100} \\[1em] \Rightarrow I = 600 \times 55.5 \times \dfrac{8}{100} \\[1em] \Rightarrow I = 600 \times 4.44 \\[1em] \Rightarrow I = ₹ 2,664.$

The difference in the interest Mr. Ahuja received

= ₹ 2,664 − ₹ 2,220

= ₹ 444.

Hence, Mr. Ahuja would receive ₹ 444 more interest if he deposited ₹ 100 more per month.