#### 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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