Question

A sum of money is invested at 10% per annum compounded half yearly. If the difference of amounts at the end of 6 months and 12 months is Rs.189, find the sum of money invested.

Answer 1

Let sum of money invested be ₹ x.

When interest is compounded half-yearly:

A = `P(1 + r/(2 xx 100))^(n xx 2)`

For first `1/2` year:

`A = x zz (1 + 10/(2 xx 100))^(1/2 xx 2)`

`= x xx (1 + 1/20)`

`= x xx 21/20`

= `(21x)/20`

For first 1 year:

`A = x xx (1 + 10/2 xx 100)^(1 xx 2)`

= `x xx (1 + 1/20)^2`

= `x xx (21/20)^2`

= `(441x)/(400)`

Given,

Difference of amounts at end of 6 months and 12 months is ₹ 189.

`=> (441x)/(400) - (21x)/(20) = 189`

`=> (441x - 420x)/(400) = 189`

`=> (21x)/(400) = 189`

`=> x = (189 xx 400)/(21)`

⇒ x = ₹ 3600