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Solve the following : Find the least number of years for which an annuity of ₹3,000 per annum must run in order that its amount exceeds ₹60,000 at 10% compounded annually. [(1.1)11 = 2.8531, (1.1)12

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प्रश्न

Solve the following :

Find the least number of years for which an annuity of ₹3,000 per annum must run in order that its amount exceeds ₹60,000 at 10% compounded annually. [(1.1)11 = 2.8531, (1.1)12 = 3.1384]

योग
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उत्तर

Given, C = ₹3,000, A = ₹60,000, r = 10% p.a.

∴ i = `"r"/(100) = (10)/(100)` = 0.1

Since, A = `"C"/"i"[(1 + "i")^"n" - 1]`

∴ 60,000 = `(3,000)/(0.1)[(1 + 0.1)^"n" - 1]`

∴ `(60,000 xx 0.1)/(3,000)` = (1.1)n – 1

∴ 2 = (1.1)n – 1
∴ (1.1)n = 2 + 1
∴ (1.1)n = 3
It is given that (1.1)11 = 2.8531 and (1.1)12 = 3.1384
∴ n will be between 11 years and 12 years.
Thus, the least number of years for which an annuity of ₹3,000 per annum must run is 12 years.

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

APPEARS IN

बालभारती Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 2 Insurance and Annuity
Miscellaneous Exercise 2 | Q 4.16 | पृष्ठ ३१

संबंधित प्रश्न

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Payment of every annuity is called an installment.


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State whether the following is True or False :

Annuity contingent begins and ends on certain fixed dates.


State whether the following is True or False :

Sinking fund is set aside at the beginning of a business.


Solve the following :

A shopkeeper insures his shop and godown valued at ₹5,00,000 and ₹10,00,000 respectively for 80 % of their values. If the rate of premium is 8 %, find the total annual premium.


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In an ordinary annuity, payments or receipts occur at ______


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An annuity where payments continue forever is called perpetuity


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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]`

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= 50,000`(square)`[1 – 0.8548]

= ₹ 7,550.40


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