# ML Aggarwal Solutions Class 9 Mathematics Solutions for Compound Interest Exercise 2.1 in Chapter 2 - Compound Interest

Question 7 Compound Interest Exercise 2.1

Mr. Lalit invested ₹ 75000 at a certain rate of interest, compounded annually for two years. At the end of

the first year it amounts to ₹ 5325. Calculate

(i) the rate of interest.

(ii) the amount at the end of the second year, to the nearest rupee.

It is given that

Investment of Mr. Lalit = ₹ 5000

Period (n) = 2 years

(i) We know that

Amount after one year = ₹ 5325

So the interest for the first year = A – P

Substituting the values

= 5325 – 5000

= ₹ 325

Here

Rate = (SI × 100)/ (P × T)

Substituting the values

= (325 × 100)/ (5000 × 1)

So we get

= 13/2

= 6.5 % p.a.

(ii) We know that

Interest for the second year = (5325 × 13 × 1)/ (100 × 2)

By further calculation

= (213 × 13)/ (4 × 2)

So we get

= 2769/8

= ₹ 346.12

So the amount after second year = 5325 + 346.12

We get

= ₹ 5671.12

= ₹ 5671 (to the nearest rupee)

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