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Question
Solve the following :
Find the rate of interest compounded annually if an ordinary annuity of ₹20,000 per year amounts to ₹41,000 in 2 years.
Solution
Given, A = ₹41,000, C = ₹20,000, n = 2 years
We need to find r such that an ordinary annuity of ₹20,000 amounts to ₹41,000 in 2 years.
Since, A = `"C"/"i"[(1 + "i")^"n" - 1]`
41,000 = `"C"/"i"[(1 + "i")^2 - 1]`
41,000 = `(20,000)/"i"[(1 + "i")^2 - 1]`
`(41,000)/(20,000)= (1 + 2"i" + "i"^2 - 1)/"i"`
`(41)/(20) = ("i"^2 + 2"i")/"i"`
`(41)/(20)` = i + 2
`(41)/(20)` – 2 = i
`(41 - 40)/(20)` = i
∴ i = `(1)/(20)` = 0.05
But i = `"r"/(100)`
∴ 0.05 = `"r"/(100)`
∴ r = 5%
The rate of interest is 5%.
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Period, n = 20, r = 5%, (1.025)20 = 1.675
A = `"C"/"I" [(1 + "i")^"n" - 1]`
I = `5/200` = `square` as interest is calculated semi-annually
A = 10,00,000 = `"C"/"I" [(1 + "i")^"n" - 1]`
10,00,000 = `"C"/0.025 [(1 + 0.025)^square - 1]`
= `"C"/0.025 [1.675 - 1]`
10,00,000 = `("C" xx 0.675)/0.025`
C = ₹ `square`
