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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 - Mathematics and Statistics

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Question

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]

Sum
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Solution

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
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Chapter 2: Insurance and Annuity - Miscellaneous Exercise 2 [Page 31]

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Balbharati Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 2 Insurance and Annuity
Miscellaneous Exercise 2 | Q 4.16 | Page 31

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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%

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n = Number of quarters

= 4 × 1

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⇒ 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


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