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Question
Solve the following :
A man borrowed some money and paid back in 3 equal installments of ₹2,160 each. What amount did he borrow if the rate of interest was 20% per annum compounded annually? Also find the total interest charged. [(1.2)3 = 0.5787]
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Solution
Given, C = ₹2,160, n = 3 years, r = 20% p.a.
∴ i = `"r"/(100) = (2)/(100)` = 0.2
Here, we have to find present value of annuity.
∴ P = `"C"/"i"[1 - (1 + "i")^-"n"]`
= `(2,160)/(0.2)[1 - (1 + 0.2)^-3]`
= 10,800[1 – (1.2)–3]
= 10,800[1 – 0.5787]
= 10,800[0.4213]
∴ P = ₹4,550
The man has paid 3 equal instalments of ₹2,160 each.
∴ Total paid value of instalments
= 3 x 2,160
= ₹6,480
Interest = Total paid value of instalments – Present Value
= 6,480 – 4,550
= ₹1,930.
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For annuity due,
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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
