# Selina Solutions Class 9 Mathematics Solutions for Exercise 2(B) in Chapter 2 - Chapter 2- Compound Interest [Without Using Formula

Question 2 Exercise 2(B)

A man lends ₹ 12,500 at 12% for the first year, at 15% for the second year and at 18% for the third year. If the rates of interest are compounded yearly; find the difference between C.I of the first year and the compound interest for the third year.

For 1st year

P = Rs. 12500

R = 12%

R = 1 year

I=\frac{12500 \times 12 \times 1}{100}=R s, 1500

A = 12500 + 1500 = Rs. 14000

For 2nd year

P = Rs. 1400

R = 15%

T = 1 year

I=\frac{14000 \times 15 \times 1}{100}=R s, 2898I=\frac{14000 \times 15 \times 1}{100}=R s. 2898

A = 1400 + 2100 = Rs. 16100

For 3rd year

P = Rs. 16100

R = 18%

T = 1 year

I=\frac{16100 \times 18 \times 1}{100}=R s .2898

A = 16100 + 2898 = Rs. 3998

Difference between the compound interest of the third year and first year

= Rs. 2893 - Rs. 1500

= Rs. 1398

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