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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.
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Solution
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
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