Advertisements
Advertisements
प्रश्न
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]
Advertisements
उत्तर
Given C = ₹1,000, n = 3 years, r = 10% p.a.
∴ i = `"r"/(100) = (10)/(100)` = 0.1
Now, A = `("C"(1 + "i"))/"i"[(1 + "i")^"n" - 1]`
∴ A = `(1,000(1 + 0.1))/(0.1)[(1 + 0.1)^3 - 1]`
= `(1,000(1.1))/(0.1)[(1.1)^3 - 1]`
= (1,000)(11)[1.331 – 1]
= 11,000(0.331)
∴ A = 3,641
∴ Accumulated value of annuity due is ₹3,641.
APPEARS IN
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 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]
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]
Choose the correct alternative :
You get payments of ₹8,000 at the beginning of each year for five years at 6%, what is the value of this annuity?
In an ordinary annuity, payments or receipts occur at ______.
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 person who receives annuity is called __________.
Fill in the blank :
The intervening time between payment of two successive installments is called as ___________.
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 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 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 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 :
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 :
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 ______
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:
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 ______
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`
