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Question
Find the number of years for which an annuity of ₹500 is paid at the end of every year, if the accumulated amount works out to be ₹1,655 when interest is compounded annually at 10% p.a.
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Solution
Given, C = ₹500, A = ₹1,655, r = 10% p.a.
∴ i = `"r"/(100) = (10)/(100)` = 0.1
Now, A = `"C"/"i"[(1 + "i")^"n" - 1]`
∴ 1,655 = `(500)/(0.1)[(1 + 0.1)^"n" - 1]`
∴ `((1,655)(0.1))/(500)` = (1.1)n – 1
∴ 0.331= (1.1)n – 1
∴ (1.1)n = 1 + 0.331
∴ (1.1)n = 1.331
∴ (1.1)n = (1.1)3
∴ n = 3
∴ The annuity is paid for 3 years.
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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`
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
