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Question
Multiple choice questions:
If for an immediate annuity r = 10% p.a., P = ₹ 12,679.46 and A = ₹ 18,564, then the amount of each annuity paid is ______
Options
₹ 4,000
₹ 4,500
₹ 3,500
₹ 4,200
Solution
₹ 4,000
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Choose the correct alternative :
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The payment of each single annuity is called __________.
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The intervening time between payment of two successive installments is called as ___________.
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An annuity where payments continue forever is called __________.
State whether the following is True or False :
Payment of every annuity is called an installment.
State whether the following is True or False :
The future value of an annuity is the accumulated values of all installments.
State whether the following is True or False :
Sinking fund is set aside at the beginning of a business.
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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. [(1.03)20 = 1.8061]
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Find the future value after 2 years if an amount of ₹12,000 is invested at the end of every half year at 12% p. a. compounded half yearly. [(1.06)4 = 1.2625]
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Multiple choice questions:
In annuity calculations, the interest is usually taken as ______
State whether the following statement is True or False:
The future value of an annuity is the accumulated values of all instalments
State whether the following statement is True or False:
An annuity where payments continue forever is called perpetuity
In ordinary annuity, payments or receipts occur at ______
If for an immediate annuity r = 10% p.a., P = ₹ 12,679.46 and A = ₹ 18,564, then the amount of each annuity paid is ______
An annuity in which each payment is made at the end of period is called ______
A 35-year old person takes a policy for ₹ 1,00,000 for a period of 20 years. The rate of premium is ₹ 76 and the average rate of bonus is ₹ 7 per thousand p.a. If he dies after paying 10 annual premiums, what amount will his nominee receive?
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`