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प्रश्न
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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उत्तर
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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[Given (1.1)4 = 1.4641]
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`
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∴ 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
