Tag: annuities

Questions Related to annuities

Find the future value of an annuity of Rs. $500$ made annually for $7$ years at interest rate of $14\%$ compounded annually. Given that $(1.14)^7=2.5023$.

  1. $5,563.25$

  2. $5,365.35$

  3. $5,365.53$

  4. $5,356.35$


Correct Option: B
Explanation:

Annual payment A$=$Rs. $500$
$n=7$
$i=14\%=0.14$
$A(7, 0.14)=500\left[\displaystyle\frac{(1+1.014)^7-1}{0.14}\right]$
$=$Rs. $5365.35$

A limited company intends to create a depreciation fund to replace at the end of the 25th year assets costing Rs 100000.Calculate the amount (approximately) to be retained out of profits every year if the interest rate is 3%.

  1. 2755

  2. 3245

  3. 5431

  4. 1200


Correct Option: A

Veena is allotted an LIG flat for which she has to make an immediate payment of $ $ 100,000$ and $10$ semi-annual payments of $ $50,000$ each, the first being made at the end of $3$ years. If money is worth $10\%$ per annum compounded half-yearly, find the cash price (in $) of the flat.

  1. $302,509$

  2. $400,509$

  3. $302,009$

  4. $402,509$


Correct Option: D
Explanation:

$\Rightarrow$  Cash price = Down payment + Present value of annuity    --- ( 1 )

$\Rightarrow$  Here, down payment is $\$100,000$, while we have annuity 10 terms i.e. $n$, deferred for $2\dfrac{1}{2}$ years i.e. 5 terms.
$\Rightarrow$  Each installment, $A = \$50,000$
$\Rightarrow$  Rate of interest, $r=10\%\, p.a.$ compounded half-yearly = 0.05
$\Rightarrow$  $m=5$ and $m+n=15$
$\therefore$  Present value of annuity,
$\Rightarrow$  $V=\dfrac{A}{r}\times [\dfrac{1}{(1+r)^m}-\dfrac{1}{(1+r)^{m+n}}]$
$\Rightarrow$  $\dfrac{50,000}{0.05}\times [(1.05)^{-5}-(1.05)^{-15}]$
$\Rightarrow$  $\$ 302509.07$
Now, from ( 1 )
$\Rightarrow$  Cash price of the flat = $\$100,000+\$302,509=\$402,509$

An investor deposits Rs 1000 in a saving institution. Each payment is made at the end of year.If the payment deposited earns 12% interest compounded annually how much amount(approximately) will he receive at the end of 10 years.

  1. $1234$

  2. $2345$

  3. $17548$

  4. $4567$


Correct Option: C
Explanation:

$P=Rs.1000,r=12\%,t=10$

$A=\sum _{ n=1 }^{ 9 }{ P{ (1+\cfrac { r }{ 100 } ) }^{ n }+P } \ =\sum _{ n=1 }^{ 9 }{ 1000{ (1+\cfrac { 12 }{ 100 } ) }^{ n }+1000 } \ =3000\times 16.548\ =Rs.17548$

A machine costing Rs $2$ lacs has effective life of $7$ years and its scrap value is Rs $30000$. What amount (in Rs) should the company put into a sinking fund earning $5\%$ per annum so that it can replace the machine after its useful life? Assume that a new machine will cost Rs $3$ lacs after $7$ years.

  1. $30161.35$

  2. $33101.35$

  3. $33161.35$

  4. $33111.35$


Correct Option: C
Explanation:

$\Rightarrow$  Cost of new machine is $Rs.3\,lacs$

$\Rightarrow$  Scrap value of old machine is $Rs.30000$
$\Rightarrow$  Hence, money required for new machine after 7 years  = $Rs.300000-Rs.30000=Rs.270000$.
$\Rightarrow$  If A is the annual deposit into sinking fund, then we have
$\Rightarrow$  Amount of annuity, $M=Rs.270000$
$\Rightarrow$  Number of periods = $7\, years$
$\Rightarrow$  $r=5\%=0.05$
$\therefore$   $M=\dfrac{A}{r}\times [(1+r)^n - 1]$
$\Rightarrow$  $270000=\dfrac{A}{0.05}\times[(1.05)^7-1]$

$\Rightarrow$  $A=\dfrac{270000\times 0.05}{(1.05)^7 - 1}$

$\therefore$   $A=33161.35$ Rs

A man decides to deposit Rs 3000 at the end of each year in a bank which pays 3% p.a compound interest . If the instalments are allowed to accumulate , what will be the total accumulation at the end of 15 years.

  1. Rs.$57450$

  2. Rs.$67780$

  3. Rs.$67050$

  4. Rs.$98450$


Correct Option: A
Explanation:

$P=Rs.3000,r=3\%,t=15$
$A=\sum _{ n=1 }^{ 15 }{ P{ (1+\cfrac { r }{ 100 } ) }^{ n } } \ =\sum _{ n=1 }^{ 15 }{ 3000{ (1+\cfrac { 3 }{ 100 } ) }^{ n } } \ =3000\times 19.15\ =Rs.57450$

Find the present value of a sequence of annual payments of Rs 25000 each , the first being made at the  end of 5th year and the last being paid at the end of 12th year, if money is worth 6%.

  1. $122875$

  2. $102875$

  3. $132875$

  4. None of the above


Correct Option: D
Explanation:

We know that formula of present value is 

$V=\cfrac{A}{r}[\cfrac{1}{(1+\cfrac{R}{100})^m}$$-\cfrac{1}{(1+\cfrac{R}{100})^{m+n}}]$
We have  annuity of $8$ terms $(n)$
For $4$ terms $(m)\implies m=4\ \implies m+n=12$
$V=\cfrac{25000}{0.06}[\cfrac{1}{(1+\cfrac{6}{100})^4}$$-\cfrac{1}{(1+\cfrac{6}{100})^{12}}]=Rs.122968.45$

A company borrows Rs 10000 on condition to repay it with compound interest at $5$% p.a . by annual instalments at Rs 1000 each. In how many years will the debt be paid off?

  1. 14.2

  2. 21.7

  3. 12.67

  4. None of these


Correct Option: A
Explanation:
$\Rightarrow$  Company borrow Rs.10000 i.e. $pv=Rs.10000$ and $I=5\%=0.05$. $A$ is also given which is $Rs.1000$
$\Rightarrow$  Present value of annuity regular
$\Rightarrow$  $pv=A\times [\dfrac{(1+I)^n-1}{I\times (1+I)^n}]$

$\Rightarrow$  $10000=1000\times [\dfrac{(1+0.05)^n-1}{0.05\times (1+0.05)^n}]$

$\Rightarrow$  $(1.05)^n-0.5\times (0.5)^n=1$
$\Rightarrow$  $(1.05)^n=2$
Taking log both sides
$\Rightarrow$  $n=\dfrac{log\,2}{log\, 1.05}$
$\therefore$   $n=14.2\, years$

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. If at the end of 5th year , he wants to finish his liability by a single payment , how much should he pay?

  1. 90100

  2. 80100

  3. 34504

  4. 54345


Correct Option: A
Explanation:

$\Rightarrow$  We have $A=Rs.5000,\,I=\dfrac{6}{100}\times \dfrac{1}{4}=0.015$ and $n = 20$

$\Rightarrow$  If at the end of 5th year, i.e., at the time of 20th payment, he wants to finish off the liability, then lump sum payment required is,
$\Rightarrow$  $5000$ + Present value of the remaining 20 installments.
$\Rightarrow$  $5000+V$
$\Rightarrow$  $5000$ + $\dfrac{A}{I}[1-(1+I)^{-n}]$

$\Rightarrow$  $5000+\dfrac{5000}{0.015}[1-(0.015)^{-20}]$    ---- ( 1 )

$\Rightarrow$  Let $x=(1.015)^{-20}$
$\Rightarrow$  $log\,x=-20\,log\,(1.015)$
$\Rightarrow$  $log\,x=-20(0.0064)=-0.128=\bar{1}.8720$
$\Rightarrow$  $x=antilog\,(\bar{1}.8720)=0.7447$
Substitute value of $x$ in ( 1 ),
$\Rightarrow$  $5000+\dfrac{5000}{0.015}(1-0.7447)$

$\Rightarrow$  $5000+\dfrac{5000}{0.015}\times 0.2553$

$\Rightarrow$  $Rs.90100$

Amit buys a house for Rs 500000. The contract is that amit will pay Rs 200000 immediately and the balance in 15 equal instalments with 15 % p.a compound interest . How much has he to pay annually (approximately)?

  1. Rs$51,305$

  2. Rs$54,005$

  3. Rs$51,843$

  4. Rs$91,305$


Correct Option: A
Explanation:

Present value $=Rs.50,000Rs.20,000=Rs.30,000$

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

$\implies 30,000=\cfrac{A}{1+(\cfrac{15}{100})}$$+\cfrac{A}{1+(\cfrac{15}{100})^2}+.....$4
$ \implies A\times 5,847 $

$\implies A=Rs51,305$