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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
Find the accumulated (future) value of annuity of ₹ 800 for 3 years at interest rate 8% compounded annually. [Given (1.08)3 = 1.2597]
Find the present value of an ordinary annuity of ₹63,000 p.a. for 4 years at 14% p.a. compounded annually. [Given (1.14)−4 = 0.5921]
Find the number of years for which an annuity of ₹500 is paid at the end of every year, if the accumulated amount works out to be ₹1,655 when interest is compounded annually at 10% p.a.
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]
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.
Fill in the blank :
The payment of each single annuity is called __________.
Fill in the blank :
The intervening time between payment of two successive installments is called as ___________.
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 :
The present value of an annuity is the sum of the present value of all installments.
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 :
A company decides to set aside a certain amount at the end of every year to create a sinking fund that should amount to ₹9,28,200 in 4 years at 10% p.a. Find the amount to be set aside every year. [(1.1)4 = 1.4641]
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:
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:
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 ______
An annuity in which each payment is made at the end of period is called ______
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]
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`
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
