Tag: amount of an annuity

Questions Related to amount of an annuity

A person takes a loan on compound interest and returns it in $2$ equal installments . If the rate of interest is $10$% per annum and the yearly installment is Rs $1682$. Find the principal (approximately).

  1. $2920$

  2. $3450$

  3. $2346$

  4. $2275$


Correct Option: A
Explanation:

$P=\cfrac{A}{1+(\cfrac{R}{100})^n}$

$\implies \cfrac{1682}{1+(\cfrac{10}{100})^1}$$+\cfrac{1682}{1+(\cfrac{10}{100})^2}$
$ \implies 1592.09+1390.08=2919.17\approx 2920$
$\implies A=Rs.2920$

A sinking fund is created for the redemption of debentures of Rs 10,000 at the end of 25 years. How much money should be provided out of profits each year for the sinking fund if the investment can earn interest at the rate 4% per annum?

  1. 2408.19

  2. 1408.19

  3. 3408.19

  4. 5408.19


Correct Option: A
Explanation:

Using formula of sinking fund

$M=\cfrac{A}{r}[(1+r)^n-1]$
$\implies 1000= \cfrac{A}{0.04}[(1+0.04)^{25}-1]\implies A=2408.19$

A person takes a loan on compound interest and returns it in $2$ equal installments . If the rate of interest is $10$% per annum and the yearly installment is Rs $1682$. Find the interest charged (approximately) with first installment.

  1. Rs. $202$

  2. Rs. $192$

  3. Rs. $92$

  4. Rs. $292$


Correct Option: D
Explanation:

$P=\cfrac{A}{(1+\cfrac{R}{100})^n}$

$\implies \cfrac{1682}{(1+\cfrac{10}{100})^1}$$+\cfrac{1682}{(1+\cfrac{10}{100})^2}$
$ \implies 1592.09+1390.08=2919.17\approx 2920$
$\implies A _1=2920(1+\cfrac{10}{100})=3212$
$CI=A _1-P=3212-2920=Rs.292$

Which of the following is not an example of annuity certain ?

  1. Car Loan

  2. Daughter's Marriage

  3. House Loan

  4. All of above


Correct Option: B
Explanation:
$\Rightarrow$  $Daughter's\,\,Marriage$ is not an example of annuity certain.
$\Rightarrow$  Annuity certain is annuity that, as a minimum, guarantees a fixed number of payments. It continues over the life of the annuitant, even if he or she lives beyond the number of payments specified in the annuity contract. 
$\Rightarrow$ In case the annuitant dies before exhausting the payments, a named beneficiary continues to receive the remaining number. Also called life annuity certain or life annuity certain and continuous.
$\Rightarrow$  House loan and Car loan are examples of annuity certain.

A man borrowed some money and paid back in 3 equal instalments of Rs 2160 each. What sum did he borrow if the rate of interest charged was 20 % p.a .compounded annually?

  1. 4551.12

  2. 4334.24

  3. 4768.97

  4. None of these


Correct Option: A
Explanation:

We know that,

$A=P(1+\cfrac{r}{100})^n$
$\implies 2160=P(1+\cfrac{20}{100})^2\ \implies P=Rs.4551.12$

A man borrowed some money and paid back in $3$ equal instalments of Rs. $2160$ each. The rate of interest charged was $20\%$ p.a .compounded annually. Find the total interest charged in Rs.(approximately).
  1. $1928$

  2. $1980$

  3. $1930$

  4. $1954$


Correct Option: C
Explanation:

$\Rightarrow$  We have, $A=Rs.2,160,\, n=3\, and\, r=20\%$

$\Rightarrow$  First we have to find Sum borrowed i.e. $V$.
$\Rightarrow$  Using, $V=\dfrac{A}{r}[1-(1+r)^{-n}]$


$\Rightarrow$  $V=\dfrac{2160}{0.2}[1-(1.2)^{-3}]$

$\Rightarrow$  $V=10800[1- \dfrac{1000}{(12)^3}]$

$\Rightarrow$  $V=10800[\dfrac{(1728-1000)}{1728}]$

$\Rightarrow$  $V=\dfrac{10800\times 728}{1728}$

$\Rightarrow$  $V=Rs.4550$.
$\Rightarrow$  Total interest = $(A\times n)-V=2160\times 3-4550=Rs. 1930$

Mr Dev purchased a car paying Rs $90,000$ and promising to pay Rs 5000 every 3 months for the next 10 years. The interest is $6$% p.a. compounded quarterly. What is the cash value (approximately) of the car ?

  1. 238467

  2. 235467

  3. 228467

  4. None of these


Correct Option: A
Explanation:

$\Rightarrow$  We have, $A=Rs.5000,\, n=40$ and $I=\dfrac{6}{100}\times \dfrac{1}{4}=0.015$

$\Rightarrow$  $V=\dfrac{A}{I}[1-(1+I)^{-n}]$

$\Rightarrow$  $V=\dfrac{5000}{0.015}[1-(1+0.015)^{-40}]$      ---   ( 1 )

$\Rightarrow$  Let $x=(1.015)^{-40}$
$\Rightarrow$  $log\, x=-40(0.0064)$
$\Rightarrow$  $log\, x=-0.256=(\bar{1}.7440)$
$\Rightarrow$   $x=antilog (\bar{1}.7440)$
$\Rightarrow$   $x=0.5546$
     Substitute this value in ( 1 ),
$\Rightarrow$  $V=\dfrac{5000}{0.015}(1-0.5546)$

$\Rightarrow$  $V=\dfrac{5000}{0.015}\times 0.4454=Rs.148466.67$

$\therefore$   Total cash value of the car = $Rs.90000+Rs.148466.67$
$\Rightarrow$  Total cash value of car = $Rs.2,38,467$ approx.

Find the amount(approximately) of an annuity immediate of Rs 2000 per annum for 20 years. The rate of interest is $\dfrac{27}{2}$ % per annum.

  1. 171703

  2. 19000

  3. 21345

  4. 43251


Correct Option: A
Explanation:
$n=20,r=(27/2)\%,P=Rs.2000$
$A=P(1+\cfrac{r}{100})^n$
$\implies A=2000[(1+\cfrac{(27/2)}{100})^{20}$$+(1+\cfrac{(27/2)}{100})^{19}......$$+(1+\cfrac{(27/2)}{100})^{1}]\\A=2000\times 85.85\\ A=Rs.171700\approx 171703$

Mr Assem decides to deposit Rs 5000 at the end of year in a bank which pays compound interest the rate of 5% per annum. What will be his total accumulation (approximately) at the end of 15 years?

  1. Rs.$140092$

  2. Rs.$907892$

  3. Rs.$100892$

  4. Rs.$107892$


Correct Option: D
Explanation:
$A=P(1+\cfrac{r}{100})^n$
$\implies A=5000[(1+\cfrac{5}{100})$$+(1+\cfrac{5}{100})^{14}......$$+(1+\cfrac{5}{100})^{1}+5000]\\=5000\times 21.5718\\ Rs.107892$

Mr Dev purchased a car paying Rs $90,000$ and promised to pay Rs 5000 every 3 months for the next 10 years. The interest is $6$% p.a. compounded quarterly. If Mr Dev misses first 6 payments , how much should he pay at the time of 7th payment (approximately) to bring himself up to date.

  1. 36333

  2. 46532

  3. 56734

  4. 60322


Correct Option: A
Explanation:

$\Rightarrow$  We have $A=Rs.5000,\, n=7$  $I=\dfrac{6}{100}\times \dfrac{1}{4}=0.015$

$\Rightarrow$  At the time of 7th payment, equivalent amount of first 6 missed payment has also to be paid. Thus total payment to be made at the end of 7th period is the amount of annuity of 7 terms. Hence, amount required required to be paid is,
$\Rightarrow$  $M=\dfrac{A}{I}[(1+I)^{n}-1]$


$\Rightarrow$  $M=\dfrac{5000}{0.015}[(1.015)^{7}-1]$           -----  ( 1 )
$\Rightarrow$   Let $x=(1.015)^{7}$
$\Rightarrow$   $log\,x=7\,log\,(1.015)$
$\Rightarrow$   $log\,x=7\,(0.0064)=0.0448$
$\Rightarrow$   $x=antilog\,(0.0448)$
$\Rightarrow$   $x=1.109$
Substitute value of $x$ in ( 1 ) we get,
$\Rightarrow$  $M=\dfrac{5000}{0.015}(1.109-1)$

$\Rightarrow$  $M=\dfrac{5000}{0.015}\times 0.109$

$\Rightarrow$  $M=\dfrac{5000\times 1000}{15}\times \dfrac{109}{1000}$

$\Rightarrow$  $M=Rs.36333.33\approx Rs.36333$