Question

12. A sum of money is invested at compound interest payable . the interest in two successive years is ₹225 and ₹240. Find
a) the rate of interest
b) the original sum
the interest earned in the third year​

Answer 1

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

Answer 2

Answer:

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

=

3

20

=6

3

2

% 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

Step-by-step explanation:

hope helps