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

A man borrows ₹ 6000 at 5% compound interest. If he repays ₹ 1200 at the end of each year, find the

amount outstanding at the beginning of the third year.

It is given that

Principal = ₹ 6000

Rate of interest = 5% p.a.

We know that

Interest for the first year = Prt/100

Substituting the values

= (6000 × 5 × 1)/ 100

= ₹ 300

So the amount after one year = 6000 + 300 = ₹ 6300

Principal for the second year = ₹ 6300

Amount paid = ₹ 1200

So the balance = 6300 – 1200 = ₹ 5100

Here

Interest for the second year = (5100 × 5 × 1)/ 100 = ₹ 255

Amount for the second year = 5100 + 255 = ₹ 5355

Amount paid = ₹ 1200

So the balance = 5355 – 1200 = ₹ 4155

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