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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 - Mathematics and Statistics

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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]

योग

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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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Insurance and Annuity - Miscellaneous Exercise 2 [पृष्ठ ३१]

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बालभारती Mathematics and Statistics 2 (Commerce) [English] 12 Standard HSC Maharashtra State Board
अध्याय 2 Insurance and Annuity
Miscellaneous Exercise 2 | Q 4.15 | पृष्ठ ३१

संबंधित प्रश्न

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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`


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`


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