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Question
Solve the following :
After how many years would an annuity due of ₹3,000 p.a. accumulated ₹19,324.80 at 20% p. a. compounded yearly? [Given (1.2)4 = 2.0736]
Solution
Given, C = ₹3,000, A' = 19324.80, r = 20% p.a.
∴ i = `"r"/(100) = (20)/(100)` = 0.2
Since, A' = `("C"(1 + "i"))/"i" [(1 + "i")^"n" - 1]`
∴ 19,324.80 = `(3,000(1 + 0.2))/(0.2) [(1 + 0.2)^"n" - 1]`
∴ 19,324.80 = `(3,000 xx 1.2)/(0.2) [(1.2)^"n" - 1]`
∴ 19,324.80 = 3,000 x 6 [(1.2)n – 1]
∴ `(19,324.80)/(18,000)` = (1.2)n – 1
∴ `(19,32,480)/(18,000 xx 100)` = (1.2)n – 1
∴ `(1,07,360)/(1,00,000)` = (1.2)n – 1
∴ 1.0736 = (1.2)n – 1
∴ (1.2)n = 1.0736 + 1
∴ (1.2)n = 2.0736
∴ (1.2)n = (1.2)4 ...[ `Theta` (1.2)4 = 2.0736]
∴ n = 4 years
∴ After 4 years, an annuity due of ₹3,000 p.a. would accumulate to ₹19,324.80 at 20% p.a. compounded annually.
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[Given (1.1)4 = 1.4641]
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`
