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प्रश्न
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]
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उत्तर
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]
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`
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
