 # ML Aggarwal Solutions Class 9 Mathematics Solutions for Compound Interest Exercise 2.2 in Chapter 2 - Compound Interest

A man invests ₹ 1200 for two years at compound interest. After one year the money amounts to ₹ 1275.

Find the interest for the second year correct to the nearest rupee.

It is given that

Principal = ₹ 1200

After one year, the amount = ₹ 1275

So the interest for one year = 1275 – 1200 = ₹ 75

We know that

Rate of interest = (SI × 100)/ (P × t)

Substituting the values

= (75 × 100)/ (1200 × 1)

By further calculation

= 75/12

= 25/4

= 6 ¼ % p.a.

Here

Interest for the second year on ₹ 1275 at the rate of 25/4% = Prt/100

Substituting the values

= (1275 × 25 × 1)/ (100 × 4)

By further calculation

= 1275/16

= ₹ 79.70

= ₹ 80

Related Questions

Lido

Courses

Teachers

Book a Demo with us

Syllabus

Maths
CBSE
Maths
ICSE
Science
CBSE

Science
ICSE
English
CBSE
English
ICSE
Coding

Terms & Policies

Selina Question Bank

Maths
Physics
Biology

Allied Question Bank

Chemistry
Connect with us on social media!