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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]
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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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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`
