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प्रश्न
Fill in the blank :
The intervening time between payment of two successive installments is called as ___________.
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उत्तर
The intervening time between payment of two successive installments is called as payment 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 the present value of an annuity immediate of ₹36,000 p.a. for 3 years at 9% p.a. compounded annually. [Given (1.09)−3 = 0.7722]
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 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.
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]
A person plans to put ₹400 at the beginning of each year for 2 years in a deposit that gives interest at 2% p.a. compounded annually. Find the amount that will be accumulated at the end of 2 years.
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 :
Amount of money today which is equal to series of payments in future is called
Choose the correct alternative :
Rental payment for an apartment is an example of
Fill in the blank :
The person who receives annuity is called __________.
Fill in the blank :
An annuity where payments continue forever is called __________.
State whether the following is True or False :
Payment of every annuity is called an installment.
Solve the following :
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]
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 :
Find the rate of interest compounded annually if an ordinary annuity of ₹20,000 per year amounts to ₹41,000 in 2 years.
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 :
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]
Solve the following :
After how many years would an annuity due of ₹3,000 p.a. accumulated ₹19,324.80 at 20% p. a. compounded yearly? [Given (1.2)4 = 2.0736]
Multiple choice questions:
Rental payment for an apartment is an example of ______
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:
The future value of an annuity is the accumulated values of all instalments
State whether the following statement is True or False:
Annuity contingent begins and ends on certain fixed dates
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 ______
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`
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
