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Question
Solve the following :
Some machinery is expected to cost 25% more over its present cost of ₹6,96,000 after 20 years. The scrap value of the machinery will realize ₹1,50,000. What amount should be set aside at the end of every year at 5% p.a. compound interest for 20 years to replace the machinery? [Given (1.05)20= 2.653]
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Solution
Since, the machinery is expected to cost 25% more over its present cost i.e., 6,96,000,
∴ Expected value of machinery
= Present cost + 25% of present cost
= `6,96,000 + 25/100 xx 6,96,000`
= 6,96,000 + 1,74,000
= ₹8,70,000
After 20 years, scrap value of the machinery is ₹ 1,50,000.
∴ Accumulated value of machinery = Expected value of machinery – Scrap value of machinery
= 8,70,000 – 1,50,000
= ₹7,20,000
∴ A = ₹ 7,20,000
Also, r = 5% p.a., n = 20 years,
i = `"r"/(100) = (5)/(100)` = 0.05
Since, A = `"C"/"i"[(1 + "i")^"n"` - 1]
∴ 7,20,000 = `"C"/(0.05)[(1 + 0.05)^20 - 1]`
∴ 7,20,000 x 0.05 - C[(1.05)20 – 1]
∴ 36,000 = C (2.653 – 1)
∴ 36,000 = C x 1.653
∴ C = `(36,000)/(1.653)`
∴ C = ₹21,778.58
∴ Sum of ₹21,778.58 should be set aside at the end of each year.
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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`
