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प्रश्न
Solve the following :
Find the amount a company should set aside at the end of every year if it wants to buy a machine expected to cost ₹1,00,000 at the end of 4 years and interest rate is 5% p. a. compounded annually. [(1.05)4 = 1.21550625]
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उत्तर
Given, A = ₹1,00,000, n = 4 years, r = 5% p.a.
∴ i = `"r"/(100) = (5)/(100)` = 0.05
Since, A = `"C"/"i"[(1 + "i")^"n" - 1]`
∴ 1,00,000 = `"C"/(0.05)[(1 + 0.05)^4 - 1]`
∴ 1,00,000 x 0.05 = C[(1.05)4 – 1]
∴ 5,000 = C(1.21550625 – 1)
∴ 5,000 = C x 0.21550625
∴ C = `(5000)/(0.21550625)`
∴ C = ₹23,201.18
∴ The company should set aside a sum of ₹23,201.18 in order to buy the machine.
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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`
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
