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Basics of Financial Mathematics Test 10

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Basics of Financial Mathematics Test 10
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  • Question 1
    1 / -0
    A sum of $$\text{Rs.}\ 24000$$ is borrowed for $$\displaystyle 1\frac{1}{2}$$years at the rate of interest $$10\%$$ per annum compounded semi-annually. What is the compound interest ( $$x$$ )?
    Solution
    Principal amount, $$P=\text{Rs.}\ 24000$$, 
    Time $$t=$$ $$\displaystyle 1\frac{1}{2}$$ years $$= 3$$ half years, 
    Rate of interest, $$R = 10\%$$ p.a. $$= 5\%$$ per half year
    Using the formula for calculation of amount when principal is compounded semi-anually:
    $$A=P\left(1+\dfrac{R}{100}\right)^t$$
    $$\displaystyle A=24000\left ( 1+\frac{5}{100} \right )^{3}$$

         $$=24000\times \left ( \dfrac{21}{20} \right )^{3}$$

         $$\displaystyle =24000\times \dfrac{9261}{8000}=$$ Rs. $$27783$$
    $$\displaystyle $$ C.I. $$=\text{ Rs.}\ (27783 -24000) =\text{ Rs.}\ 3783$$
    $$\displaystyle \therefore$$ C.I. $$(x)$$ lies between $$\text{Rs.}\ 3000$$ and $$\text{Rs.}\ 4000$$.
  • Question 2
    1 / -0
    A man borrowed Rs. $$5000$$ at $$10\%$$ per annum compound interest. At the end of each year he has repaid Rs. $$1500$$. The amount of money he still woes after the third year is 
    Solution
    Since the man repays Rs. $$1500$$ at the end of each year
    $$\displaystyle \therefore$$ Amount he owes at the end of $$1^{st}$$ year
    $$\displaystyle =5000\left ( 1+\frac{10}{100} \right )-1500=5500-1500=Rs.4000$$
    Amount he owes at the end of $$2^{nd}$$ year
    $$\displaystyle =4000\left ( 1+\frac{10}{100} \right )-1500=4400-1500=Rs.2900$$
    Amount he owes at the end of $$3^{rd}$$ year
    $$\displaystyle =2900\left ( 1+\frac{10}{100} \right )-1500=3190-1500=Rs.1690$$
  • Question 3
    1 / -0
    If the amount is $$\displaystyle 2\frac{1}{4}$$ times of the sum after $$2$$ years, then the rate of compound interest must be 
    Solution
    Let the sum be Rs. $$x$$. 
    Then amount $$=$$ Rs. $$\displaystyle \frac{9x}{4},n=2,r= ?$$
    $$\displaystyle \therefore \frac{9x}{4}=x\left ( 1+\frac{r}{100} \right )^{2}$$
    $$\Rightarrow \dfrac{9}{4}=\left ( 1+\dfrac{r}{100} \right )^{2}$$
    $$\Rightarrow \left ( \dfrac{3}{2} \right )^{2}=\left ( 1+\dfrac{r}{100} \right )^{2}$$
    $$\displaystyle \Rightarrow \frac{r}{100}=\frac{3}{2}-1=\frac{1}{2}$$
    $$\Rightarrow r=50\%$$ p.a.
  • Question 4
    1 / -0
    What is the compound interest on an amount of Rs. $$4800$$ at the rate of $$6$$ percent p.a. at the end of $$2$$ years?
    Solution
    Given, $$P = $$ $$Rs.$$ $$4800, r = 6\%$$ $$p.a.$$, $$n = 2$$

    $$\displaystyle \therefore C.I.=A-P=P\left ( 1+\frac{r}{100} \right )^{n}-P$$

    $$\displaystyle =4800\left ( 1+\frac{6}{100} \right )^{2}-4800$$

    $$\displaystyle =4800\times \dfrac{106\times 106}{100\times 100}-4800$$

    $$= 5393.28 - 4800 =$$ $$Rs.$$ $$593.28$$
  • Question 5
    1 / -0
    An amount of Rs. $$x$$ at compound interest at $$20\%$$ per annum for $$3$$ years becomes $$y$$. What is $$y : x$$?
    Solution
    Given, $$P= Rs. x$$, A =Rs. y$$, $$r = 20\%$$ p.a., $$n = 3$$ years
    $$\displaystyle \therefore y=x\left ( 1+\frac{20}{100} \right )^{3}$$
    $$\Rightarrow y=x\left ( \dfrac{6}{5} \right )^{3}$$
    $$\Rightarrow \dfrac{y}{x}=\dfrac{216}{125}$$
    Therefore, $$y:x$$ is $$216:125$$.
  • Question 6
    1 / -0
    The value of a machine depreciates at the rate of $$10\%$$ every year. it was purchased $$3$$ years ago . If its present value is Rs. $$8748$$, its purchase price was
    Solution
    We know that
    Final price $$=$$ initial price$${ \left( 1+\frac { rate }{ 100 }  \right)  }^{ time }$$
    Here the final price $$=$$ Rs. $$8784$$, time $$=3$$ yrs, rate $$=-10\% $$p.a.
    The rate is negative since the price is depriciating.
    Let the initial price $$=$$ Rs. $$x$$.
    $$\therefore  x\times { \left( 1-\dfrac { 10 }{ 100 }  \right)  }^{ 3 }=8748$$
    $$ \Rightarrow x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } =8748$$
    $$ \Rightarrow x=8748\times \dfrac { 10 }{ 9 } \times \dfrac { 10 }{ 9 } \times \dfrac { 10 }{ 9 } =$$ Rs. $$12000$$
    So, the price of the machine $$3$$ years back $$=$$ Rs. $$12000$$.
    Ans- Option C.
  • Question 7
    1 / -0
    Find the compound interest on Rs. $$24000$$ at $$15\%$$ per annum for $$\displaystyle 2\frac{1}{3}$$ years,.
    Solution
    Given, $$P=$$ Rs. $$24000, R=15\%$$ p.a., $$T =2\dfrac {1}{3}$$ years
    $$\displaystyle \therefore A=24000\left ( 1+\frac{15}{100} \right )^{2}\left ( 1+\frac{\frac{1}{3}\times 15}{100} \right )=24000\left ( 1+\frac{3}{20} \right )^{2}\left ( 1+\frac{1}{20} \right )$$
    $$\displaystyle =24000\left ( \frac{23}{20} \right )^{2}\times \frac{21}{20}=24000\times \frac{23\times 23\times 21}{8000}=$$ Rs. $$33327$$
    $$\displaystyle \therefore$$ C.I. $$=$$ Rs. $$33327-$$ Rs. $$24000=$$ Rs. $$9327$$
  • Question 8
    1 / -0
    A sum of money at compound interest amount to three times of its self in three years. in how many years will it be nine times of its self?
    Solution
    Given,  $$3P=P$$ $$\displaystyle \left ( 1+\frac{r}{100} \right )^{3}\Rightarrow \left ( 1+\frac{r}{100} \right )^{3}=3$$ ............(i)
    Let $$t$$ be the time in years in which the sum will be nine times of itself. 
    Then $$\displaystyle 9P=P\left ( 1+\frac{r}{100} \right )^{t}$$
    $$\Rightarrow \left ( 1+\dfrac{r}{100} \right )^{t}=9=3^{2}$$
    $$\displaystyle \Rightarrow \left ( 1+\frac{r}{100} \right )^{t}=\left ( \left ( 1+\frac{r}{100} \right )^{3} \right )^{2}$$ ....(From(i))
    $$\displaystyle \Rightarrow \left ( 1+\frac{r}{100} \right )^{t}=\left ( 1+\frac{r}{100} \right )^{60}$$
    $$\Rightarrow t=6$$ years
  • Question 9
    1 / -0
    In how many rears will a sum of Rs. $$800$$ at $$10 \%$$  per annum compound interest, compounded semiannually becomes Rs. $$926.10$$?
    Solution
    Given, $$P =$$ Rs. $$800, r = 10\%$$ p.a. $$= 5 \%$$ per half year, $$A =$$ Rs. $$926.10$$, Time $$= 2n$$
    $$\displaystyle \therefore 926.10=800\left ( 1+\frac{5}{100} \right )^{2n}$$
    $$\displaystyle \Rightarrow \frac{9261}{8000}=\left ( 1+\frac{1}{20} \right )^{2n}$$
    $$\Rightarrow \left ( \dfrac{21}{20} \right )^{3}=\left ( \dfrac{21}{20} \right )^{2n}$$
    $$\displaystyle \Rightarrow 2n=3$$
    $$\Rightarrow n=\dfrac{3}{2}=1\dfrac{1}{2}$$ years
  • Question 10
    1 / -0
    The current birth rate per thousand is $$32$$, whereas the corresponding death rate is $$11$$ per thousand. The net growth rate in terms of population increase in per cent is given by, 
    Solution

    Given, birth rate per $$1000$$ is $$32$$ and corresponding death rate is $$11$$ per thousand.
    therefore, net growth on $$1000=(32-11)=21$$
    Hence, net growth on $$100=\dfrac  {21}{1000}\times 100=2.1\%$$
    Therefore, net growth rate in terms of population increase in per cent is $$2.1\%$$.
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