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Solve the following : Find the rate of interest compounded annually if an ordinary annuity of ₹20,000 per year amounts to ₹41,000 in 2 years. - Mathematics and Statistics

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प्रश्न

Solve the following :

Find the rate of interest compounded annually if an ordinary annuity of ₹20,000 per year amounts to ₹41,000 in 2 years.

योग
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उत्तर

Given, A = ₹41,000, C = ₹20,000, n = 2 years
We need to find r such that an ordinary annuity of ₹20,000 amounts to ₹41,000 in 2 years.

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

41,000 = `"C"/"i"[(1 + "i")^2 - 1]`

41,000 = `(20,000)/"i"[(1 + "i")^2 - 1]`

`(41,000)/(20,000)= (1 + 2"i" + "i"^2 - 1)/"i"`

`(41)/(20) = ("i"^2 + 2"i")/"i"`

`(41)/(20)` = i + 2

`(41)/(20)` – 2 = i

`(41 - 40)/(20)` = i

∴ i = `(1)/(20)` = 0.05

But i = `"r"/(100)`

∴ 0.05 = `"r"/(100)`

∴ r = 5%
The rate of interest is 5%.

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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] Standard 12 Maharashtra State Board
अध्याय 2 Insurance and Annuity
Miscellaneous Exercise 2 | Q 4.17 | पृष्ठ ३१

संबंधित प्रश्न

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[Given (1.1)4 = 1.4641]


For annuity due,

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= 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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