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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
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]
A lady plans to save for her daughter’s marriage. She wishes to accumulate a sum of ₹ 4,64,100 at the end of 4 years. What amount should she invest every year if she gets an interest of 10% p.a. compounded annually? [Given (1.1)4 = 1.4641]
A person wants to create a fund of ₹ 6,96,150 after 4 years at the time of his retirement. He decides to invest a fixed amount at the end of every year in a bank that offers him interest of 10% p.a. compounded annually. What amount should he invest every year? [Given (1.1)4 = 1.4641]
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 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 :
The person who receives annuity is called __________.
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 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 :
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 :
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]
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:
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:
A sinking fund is a fund established by financial organization
State whether the following statement is True or False:
The future value of an annuity is the accumulated values of all instalments
If payments of an annuity fall due at the beginning of every period, the series is called annuity ______
The intervening time between payment of two successive installments is called as ______
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]
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`
