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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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The future amount, A = ₹ 10,00,000
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`
For an annuity due, C = ₹ 2000, rate = 16% p.a. compounded quarterly for 1 year
∴ Rate of interest per quarter = `square/4` = 4
⇒ r = 4%
⇒ i = `square/100 = 4/100` = 0.04
n = Number of quarters
= 4 × 1
= `square`
⇒ P' = `(C(1 + i))/i [1 - (1 + i)^-n]`
⇒ P' = `(square(1 + square))/0.04 [1 - (square + 0.04)^-square]`
= `(2000(square))/square [1 - (square)^-4]`
= 50,000`(square)`[1 – 0.8548]
= ₹ 7,550.40
