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ML Aggarwal Solutions Class 9 Mathematics Solutions for Compound Interest Exercise 2.2 in Chapter 2 - Compound Interest
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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.
Answer:
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
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