Manisha deposited ₹ 1,000 per month in a recurring deposit account for a period of $2\dfrac{1}{2}$ years. She received ₹ 33,100 at the time of maturity. Find :

(i) the rate of interest

(ii) how much less interest will Manisha receive, if she deposited ₹ 200 less per month at the same rate of interest and for the same time ?

(i) Given,

n = $2\dfrac{1}{2}$ years = 30 months, P = ₹ 1,000

Maturity amount received by Manisha = ₹ 33,100

Total amount deposited = ₹ 1,000 × 30 = ₹ 30,000

Interest received = Maturity value - Sum deposited

= ₹ 33,100 − ₹ 30,000 = ₹ 3,100.

By formula,

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

Substituting values we get :

$\Rightarrow 3100 = 1000 \times \dfrac{30 \times (31)}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 3100 = 1000 \times \dfrac{930}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 3100 = 1000 \times 38.75 \times \dfrac{r}{100} \\[1em] \Rightarrow 3100 = 38750 \times \dfrac{r}{100} \\[1em] \Rightarrow 3100 = 387.5r \\[1em] \Rightarrow r = \dfrac{3100}{387.5} \\[1em] \Rightarrow r = 8\%$

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

(ii) Now, if she deposited ₹ 800 per month.

$I = 800 \times \dfrac{30 \times (31)}{24} \times \dfrac{8}{100} \\[1em] = 800 \times 38.75 \times \dfrac{8}{100} \\[1em] = 800 \times 3.1 \\[1em] = 2,480.$

The difference in the interest she received = ₹ 3,100 − ₹ 2,480

= ₹ 620.

Hence, Manisha will receive ₹ 620 less interest if she deposits ₹ 200 less per month.