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Find the amount of an ordinary annuity if a payment of ₹ 500 is made at the end of every quarter for 5 years at the rate of 12% per annum compounded quarterly. [Given (1.03)20 = 1.8061] - Mathematics and Statistics

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प्रश्न

Find the amount of an ordinary annuity if a payment of ₹ 500 is made at the end of every quarter for 5 years at the rate of 12% per annum compounded quarterly. [Given (1.03)20 = 1.8061]

योग
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उत्तर

Given, C = ₹ 500
Amount is invested at the end of every quarter.

∴ It is an immediate annuity.

Rate of interest is 12% p.a.

∴ r = `(12)/(4)`% = 3% per quarter

∴ i = `"r"/(100) = (3)/(100)`  0..03

The period is of 5 years and payment is made on quarterly basis.

∴ n = 5 x 4 = 20

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

= `(500)/(0.03)[(1 + 0.03)^20 - 1]`

= `(500)/(0.03)[(1.03)^20 - 1]`

= `(500)/(0.03)(1.8061 - 1)`

= `(500)/(0.03) xx (0.8061)`

= `(403.05)/(0.03)`

= `(40305)/(3)`

= ₹ 13,435

∴ Amount of ordinary annuity is ₹ 13,435.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2.2: Insurance and Annuity - Q.4

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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`


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


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