A sum of money is invested at compound interest payable annually. The interest in two successive years is ₹ 225 and ₹ 240. Find:
(i) the rate of interest
(ii) the original sum
(iii) the interest earned in the third year.
Solution:
It is given that
Interest for the first year = ₹ 225
Interest for the second year = ₹ 240
So the difference = 240 – 225 = ₹ 15
Here ₹ 15 is the interest on ₹ 225 for 1 year
(i) Rate = (SI × 100)/ (P × t)
Substituting the values
= (15 × 100)/ (225 × 1)
So we get
= 20/3
= 6 2/3% p.a.
(ii) We know that
Sum = (SI × 100)/ (R × t)
Substituting the values
= (225 × 100)/ (20/3 × 1)
It can be written as
= (225 × 100 × 3)/ (20 × 1)
So we get
= 225 × 15
= ₹ 3375
(iii) Here
Amount after second year = 225 + 240 + 3375 = ₹ 3840
So the interest for the third year = Prt/100
Substituting the values
= (3840 × 20 × 1)/ (100 × 3)
= ₹ 256
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