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Find the rate of interest compounded annually if an annuity immediate at ₹20,000 per year amounts to ₹2,60,000 in 3 years. - Mathematics and Statistics

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प्रश्न

Find the rate of interest compounded annually if an annuity immediate at ₹20,000 per year amounts to ₹2,60,000 in 3 years.

योग
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उत्तर

Given, C = ₹20,000, A = ₹2,60,000, n = 3 years.

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

∴ 2,60,000 = `(20,000)/"i"[(1 + "i")^3 - 1]`

∴ `(2,60,000)/(20,000) = (1)/"i" [1 + 3"i" + 3"i"^2 + "i"^3 - 1]`

∴ 13 = `(3"i" + 3"i"^2 + "i"^3)/"i"`

∴ 13 = 3 +  3i + i2
∴ i2 + 3i – 10 = 0
∴ i2 + 5i – 2i – 10 = 0
∴ i(i + 5) – 2(i + 5) = 0
∴ (i + 5)(i – 2) = 0
∴ i = – 5 or i = 2
But, i cannot be negative.
∴ i = 2
∴ `"r"/(100)` = 2
∴ r = 200% p.a.
∴ The rate of interest is 200% p.a.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Insurance and Annuity - Exercise 2.2 [पृष्ठ २८]

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बालभारती Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 2 Insurance and Annuity
Exercise 2.2 | Q 1.09 | पृष्ठ २८

संबंधित प्रश्न

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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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