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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.
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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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For annuity due,
C = ₹ 20,000, n = 3, I = 0.1, (1.1)–3 = 0.7513
Therefore, P = `square/0.1 xx [1 - (1 + 0.1)^square]`
= 2,00,000 [1 – 0.7513]
= ₹ `square`
