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प्रश्न
In ordinary annuity, payments or receipts occur at ______
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उत्तर
end of each period
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संबंधित प्रश्न
A person invested ₹ 5,000 every year in finance company that offered him interest compounded at 10% p.a., what is the amount accumulated after 4 years? [Given (1.1)4 = 1.4641]
Find accumulated value after 1 year of an annuity immediate in which ₹ 10,000 is invested every quarter at 16% p.a. compounded quarterly. [Given (1.04)4 = 1.1699]
Find the rate of interest compounded annually if an annuity immediate at ₹20,000 per year amounts to ₹2,60,000 in 3 years.
Find the accumulated value of annuity due of ₹1,000 p.a. for 3 years at 10% p.a. compounded annually. [Given (1.1)3 = 1.331]
An annuity immediate is to be paid for some years at 12% p.a. The present value of the annuity is ₹ 10,000 and the accumulated value is ₹ 20,000. Find the amount of each annuity payment
For an annuity immediate paid for 3 years with interest compounded at 10% p.a., the present value is ₹24,000. What will be the accumulated value after 3 years? [Given (1.1)3 = 1.331]
A person sets up a sinking fund in order to have ₹ 1,00,000 after 10 years. What amount should be deposited bi-annually in the account that pays him 5% p.a. compounded semi-annually? [Given (1.025)20 = 1.675]
Fill in the blank :
If payments of an annuity fall due at the beginning of every period, the series is called annuity __________.
State whether the following is True or False :
Payment of every annuity is called an installment.
State whether the following is True or False:
Annuity certain begins on a fixed date and ends when an event happens.
State whether the following is True or False :
The future value of an annuity is the accumulated values of all installments.
Solve the following :
A shopkeeper insures his shop and godown valued at ₹5,00,000 and ₹10,00,000 respectively for 80 % of their values. If the rate of premium is 8 %, find the total annual premium.
Solve the following :
Find the amount a company should set aside at the end of every year if it wants to buy a machine expected to cost ₹1,00,000 at the end of 4 years and interest rate is 5% p. a. compounded annually. [(1.05)4 = 1.21550625]
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]
Solve the following :
A person purchases a television by paying ₹20,000 in cash and promising to pay ₹1,000 at end of every month for the next 2 years. If money is worth 12% p. a. converted monthly, find the cash price of the television. [(1.01)–24 = 0.7875]
Solve the following :
Find the present value of an annuity immediate of ₹20,000 per annum for 3 years at 10% p.a. compounded annually. [(1.1)–3 = 0.7513]
Solve the following :
Some machinery is expected to cost 25% more over its present cost of ₹6,96,000 after 20 years. The scrap value of the machinery will realize ₹1,50,000. What amount should be set aside at the end of every year at 5% p.a. compound interest for 20 years to replace the machinery? [Given (1.05)20= 2.653]
Multiple choice questions:
Rental payment for an apartment is an example of ______
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 ______
State whether the following statement is True or False:
Annuity contingent begins and ends on certain fixed dates
The present value of an immediate annuity for 4 years at 10% p.a. compounded annually is ₹ 23,400. It’s accumulated value after 4 years would be ₹ ______
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 ______
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`
The future amount, A = ₹ 10,00,000
Period, n = 20, r = 5%, (1.025)20 = 1.675
A = `"C"/"I" [(1 + "i")^"n" - 1]`
I = `5/200` = `square` as interest is calculated semi-annually
A = 10,00,000 = `"C"/"I" [(1 + "i")^"n" - 1]`
10,00,000 = `"C"/0.025 [(1 + 0.025)^square - 1]`
= `"C"/0.025 [1.675 - 1]`
10,00,000 = `("C" xx 0.675)/0.025`
C = ₹ `square`
