Question

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

1. the rate of interest
2. 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?

Answer 1

(i) Given,

n = `2 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 xx (n(n + 1))/(2 xx 12) xx r/100`

Substituting values we get:

⇒ 3100 = `1000 xx (30 xx (31))/24 xx r/100`

⇒ 3100 = `1000 xx 930/24 xx r/100`

⇒ 3100 = `1000 xx 38.75 xx r/100`

⇒ 3100 = `38750 xx r/100`

⇒ 3100 = 387.5r

⇒ r = `3100/387.5`

⇒ r = 8

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

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

∴ I = `P xx (n(n + 1))/(2 xx 12) xx r/100`

I = `800 xx (30 xx (31))/24 xx 8/100`

= `800 xx 38.75 xx 8/100`

= 800 = 3.1

= 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.