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प्रश्न
Fill in the blank :
If payments of an annuity fall due at the end of every period, the series is called annuity __________.
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उत्तर
If payments of an annuity fall due at the end of every period, the series is called annuity immediate.
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संबंधित प्रश्न
Find the amount accumulated after 2 years if a sum of ₹ 24,000 is invested every six months at 12% p.a. compounded half yearly. [Given (1.06)4 = 1.2625]
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
Choose the correct alternative :
Rental payment for an apartment is an example of
______ is a series of constant cash flows over a limited period of time.
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 contingent begins and ends on certain fixed dates.
State whether the following is True or False :
The present value of an annuity is the sum of the present value of all installments.
State whether the following is True or False :
Sinking fund is set aside at the beginning of a business.
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 :
A man borrowed some money and paid back in 3 equal installments of ₹2,160 each. What amount did he borrow if the rate of interest was 20% per annum compounded annually? Also find the total interest charged. [(1.2)3 = 0.5787]
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:
In an ordinary annuity, payments or receipts occur at ______
Multiple choice questions:
In annuity calculations, the interest is usually taken as ______
State whether the following statement is True or False:
The relation between accumulated value ‘A’ and present value ‘P’ is A = P(1+ i)n
State whether the following statement is True or False:
An annuity where payments continue forever is called perpetuity
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 ______
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. [Given (1.03)20 = 1.8061]
A company decides to set aside a certain sum at the end of each year to create a sinking fund, which should amount to ₹ 4 lakhs in 4 years at 10% p.a. Find the amount to be set aside each year?
[Given (1.1)4 = 1.4641]
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`
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
