Tag: business mathematics and statistics

Questions Related to business mathematics and statistics

A man borrowed some money and returned it in $3$ equal quarterly installments of Rs $4630.50$ each. Find the interest charged (in Rs) on the sum he borrowed, if the rate of interest was $20\%$ p.a. compounded quarterly?

business mathematics and statistics insurance and annuity amount of an annuity annuities financial mathematics

  1. $1731.50$

  2. $1200$

  3. $1300$

  4. $1251.80$

Show answer
Correct Option: A
Explanation:

Here, we have to find present value $(V)$ of an ordinary annuity certain.

Given, $A=$ Rs $4630.50$, $n= 3$

Also $ r= \dfrac{20}{100} \times \dfrac{1}{4}=0.05 $

$ \therefore V=\dfrac{A}{r} \times [1-(1+r)^{(-n)}] =\dfrac{4630.50}{0.05} \times [1-(1.05)^{(-3)}]=$ Rs. $ \: 12610 $

Thus, the sum borrowed was Rs. $12160$

Now, total money repaid $ = 3 \times 4630.50 =$ Rs. $ \: 13891.50 $

Therefore, interest paid $ =$ Rs, $ \: 13891.50$ $-$  Rs. $ \: 12160 =$ Rs. $ \: 1731.50 $.

Find the amount of an annuity due of Rs $500$ per quarter for $8$ years and $9$ months at $6\%$ compounded quarterly.

business mathematics and statistics insurance and annuity amount of an annuity annuities financial mathematics

  1. Rs $27452.30$

  2. Rs $23137.98$

  3. Rs $13740.86$

  4. Rs $24671.30$

Show answer
Correct Option: B
Explanation:

Number of interest periods, $n = 4 \times 8 +3 = 35$
Each installment, A=Rs $500$
Present value of annuity due,
$v = \frac{A}{r} \times (1+r) \times [1-(1+r)^{-n}]$
= $\frac{500}{0.015} \times 1.015 \times [1-(1.015)^{-35}]$
= $13740.86$
$v = \frac{A}{r} \times (1+r) \times [1-(1+r)^{-n}-1]$
= $\frac{500}{0.015} \times 1.015 \times [1-(1.015)^{-35}-1]$
= $23137.98$

Three equal instalments each of $Rs 200$ were paid at the end of the year for the sum borrowed at $20 \%$ interest compounded annually. Find the sum. 

business mathematics and statistics insurance and annuity amount of an annuity annuities financial mathematics

  1. $ 600$

  2. $421.3$

  3. $ 400$

  4. $ 431.1$

Show answer
Correct Option: B
Explanation:
Each Installment = $\dfrac{P{\cdot}r}{100[1-{\{ \dfrac{100}{100+r} \}}^n]}$
Here Installement$ = 200$
Rate of interest $(r) = 20%$
Number of years$ (n) = 3$
Solving the above equation gives $P = 421.3$

Process of loan repayment by installment payments is classified as

business mathematics and statistics insurance and annuity amount of an annuity annuities financial mathematics

  1. appreciation of loan

  2. amortizing a loan

  3. depreciation a loan

  4. appreciation of investment

Show answer
Correct Option: B
Explanation:

$\Rightarrow$   Process of loan repayment by installment payment is classified as $amortizing\,a\,loan.$

$\Rightarrow$   All repayments of interest-bearing debts by a series of payments, usually in size, made at equal intervals of time is called an amortization. Mortgages and many consumer loans are repaid by this method.
$\Rightarrow$  An amortized loan is a loan with scheduled periodic payments that consist of both principal and interest. An amortized loan payment pays the relevant interest expense for the period before any principal is paid and reduced. 
$\Rightarrow$   This is opposed to loans with interest-only payment features, balloon payment features and even negatively amortizing payment features.

Find the present value (in Rs) of a sequence of annual payments of Rs $10000$ each, the first being made at the end of $5^{th}$ year and the last being made at the end of $12^{th}$ year, if money is worth $6\%$.

business mathematics and statistics insurance and annuity amount of an annuity annuities financial mathematics

  1. $40187.38$

  2. $49087.38$

  3. $40107.38$

  4. $49187.38$

Show answer
Correct Option: D
Explanation:

Here, we have a deferred annuity of $8$ terms(n), deferred for $4$ terms.Each installment, A=Rs $10000$
Rate of interest, $r=$ $6\%$= $0.06$
m= $4$, $m + n=$ $12$
Using the formula, present value
$V = \dfrac{A}{r}$ $\times \dfrac{1}{(1+r)^m}$ - $\frac {1}{(1+r)^{m+n}}$
$V = \dfrac{10000}{0.06} \times [(1.06)^{-4}-(1.06)^{-12}]$$ =49187.38$

Find SI if, Amount =  $Rs \ 1120$ , Rate = $2\dfrac{2}{5}\%$ per year , Time = $5$ years 
business mathematics and statistics insurance and annuity amount of an annuity annuities financial mathematics

  1. $1200$

  2. $201$

  3. $120$

  4. $134.4$

Show answer
Correct Option: D
Explanation:

$ SI= prt$

$p$ is the principal amount on which interest is to be calculated
$r$ is the rate of interest at which the loan is taken
$t$ is the time period for which the simple interest is to be calculated
From the question we know that,
$p=Rs 1120, $ $r =2.4$ percent and $t=5$ years
$\Rightarrow SI = (1120)(2.4)(0.01)(5)= 134.4$

A sum of Rs $2500$ is invested at a rate of $5 \%$ per annum for a term of $5$ years. Find the simple interest received at the end of the term.

business mathematics and statistics insurance and annuity amount of an annuity annuities financial mathematics

  1. $1250$

  2. $625$

  3. $1200$

  4. $1500$

Show answer
Correct Option: B
Explanation:

$\Rightarrow$  Here, $P=Rs.2500,\, R=5\%,\, T=5\,years$.

$\Rightarrow$  $Simple\, Interest=\dfrac{P\times R\times T}{100}$
$\Rightarrow$  $\dfrac{2500 \times 5\times 5}{100}$
$\Rightarrow$  $25\times 25$
$\therefore$  $Simple\, Interest\, =Rs.625$

A $5-$year ordinary annuity has a present value of $\$1,000$.  If the interest rate is $8$ percent, the amount of each annuity payment is closest to which of the following? 

    business mathematics and statistics insurance and annuity amount of an annuity annuities financial mathematics

    1. $ $250.44$

    2. $ $231.91$

    3. $ $181.62$

    4. $ $184.08$

    5. $ $170.44$

    Show answer
    Correct Option: A
    Explanation:

    $\Rightarrow$   We have, $V=\$1000\, r=8\%=0.08$ and $n=5$

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

    $\Rightarrow$  $A=\dfrac{1000\times 0.08}{1-(1.08)^{-5}}$

    $\therefore$     $A=\$250.44$

    A man borrows $Rs \ 1820$ and undertakes to payback with a compound interest of $20 \%$ per annum in a $3$  equal yearly installments at the end of first year, second year and third year. Find the amount of each installments 

    business mathematics and statistics insurance and annuity amount of an annuity annuities financial mathematics

    1. $864$

    2. $800$

    3. $900$

    4. $820$

    Show answer
    Correct Option: A
    Explanation:

    Each Installment = $\dfrac{P{\cdot}r}{100[1-{{ \frac{100}{100+r} }}^n]}$

    here $P = Rs1820$
    $r=0.2$
    $n=3$
    solving gives $Installment = 864Rs$

    Mr. Gupta has been accumulating a fund at $8\%$ effective, which will provide him with an annual income of Rs $30000$ for $15$ years, the first payment being paid on his $60^{th}$ birthday. If he wishes to reduce the number of payments to $10$, find how much annual income (in Rs) will he receive?

    business mathematics and statistics insurance and annuity amount of an annuity annuities financial mathematics

    1. $38268$

    2. $30268$

    3. $38208$

    4. $38068$

    Show answer
    Correct Option: A
    Explanation:

    In first case, we have annuity due of $15$ terms.
    Its present value (as on Mr. Gupta's $60^{th}$ birthday),
    $v = \frac{A}{r} \times (1+r) \times [1-(1+r)^{-n}]$
    = $\frac{3000}{0.08} \times 1.08 \times [1-(1.08)^{-15}]$
    Now if only $10$ payments are to be received, we have annuity due of $10$ terms. If A is the amount of each annual installment,
    $\frac{A}{0.08} \times 1.08 \times [1-(1.08)^{-10}]$
    Thus,
    $\frac{A}{0.08} \times 1.08 \times [1-(1.08)^{-10}]$ = $\frac{3000}{0.08} \times 1.08 \times [1-(1.08)^{-15}]$
    A = $30000 \times \frac{[1-(1.08)^{-15}]}{[1-(1.08)^{-10}]}$
    A = $38268$