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Basics of Financial Mathematics Test 39

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Basics of Financial Mathematics Test 39
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  • Question 1
    1 / -0
    If meena gives an interest of Rs. $$45$$ for one year at $$9\%$$ rate p.a. What is the has borrowed ?
    Solution
    Let the amount, Meena has borrowed $$=P=$$ Principal
    Interest of one year $$=Rs. 45$$
    Rate, $$r=9$$% per annum
    Time, $$t=1$$ year

    Simple Interest $$=\cfrac{P\times r\times t}{100}$$

    $$45=\cfrac{P\times 9\times 1}{100}$$

    $$9P=4500$$

    $$P=Rs. 500$$
  • Question 2
    1 / -0
    What would be the $$C.I$$ obtained on an amount of $$12500$$ at the rate of $$12$$ $$p.c.p.a.$$ after $$2$$ years? 
    Solution
    The  formula of for  compound interest is 
    $$C.I. = P\left( 1+\cfrac{r}{100} \right)^{n}-P$$
    $$=12,500 \left( 1+\cfrac{12}{100} \right)^{2}-12500$$
    $$= 12500\times \frac{112}{100}\times \frac{112}{100}-12500$$
    $$=3180$$
  • Question 3
    1 / -0
    What would be the C.I. obtained on an amount of Rs. $$12000$$ at the rate of $$9$$ p.c.p.a for $$3$$ years?
    Solution
    Given:
    $$P=12000$$
    $$R=9\%$$
    $$F=n=3\ year$$

    $$A=P \left (1+\dfrac {R}{100}\right)^n$$

    $$=12000 \left (1+\dfrac {9}{100}\right)^3$$

    $$=12000 \left (\dfrac {109}{100}\right)^3$$

    $$=12000\times \dfrac {109}{100}\times \dfrac {109}{100}\times \dfrac {109}{100}$$

    $$A=15540.348$$

    C.I.$$=A- P$$

    C.I.$$=15540.348-12000$$

    C.I$$=$$ Rs. $$3540.348$$

    Hence option (B) is the correct option
  • Question 4
    1 / -0
    The population of a town in $$2009$$ was $$125000$$. It increase $$10\%$$ per year. What is the population after $$3$$ years?
    Solution
    Taking $$P$$ as the population of the town in $$2009$$, $$n$$ is the number of years after which the population is to be calculated and $$R$$ is the rate at which the population increases, then:
    Population after $$3$$ years $$= P \left(1 + \dfrac{R}{100}\right)^{n} $$

    $$= 12500 \left(1 + \dfrac{10}{100}\right)^{3} $$

    $$= 12500 \left(1 + \dfrac{1}{10}\right)^{3} $$

    $$= 12500 \left( \dfrac{11}{10}\right)^{3} $$

    $$ = 12500 \times \dfrac{1331}{1000} $$

    $$ = 16,637.5$$ 
  • Question 5
    1 / -0
    Decrease $$300$$ by $$24$$%
    Solution
    $$24$$percent of $$300$$ is $$\dfrac{24}{100}\times 300=72$$
    Decrease 300 by $$24$$ percent$$=300-72=228$$
  • Question 6
    1 / -0
    The population of Mumbai increase $$5\%$$ per annum, its population was $$1000000$$ in $$2004$$. What was it's population after $$3$$ year?
    Solution
    Population after $$3$$ years $$= P\left(1 + \dfrac{r}{100}\right)^{n}$$

    $$= 1000000\left(1 + \dfrac{5}{100}\right)^{3}$$

    $$= 1000000\left(1 + \dfrac{1}{20}\right)^{3}$$

    $$= 1000000\left( \dfrac{21}{20}\right)^{3}$$

    $$= 1000000 \times \dfrac{9261}{8000} $$
    $$= 1157625$$

  • Question 7
    1 / -0
    A steno-typist deposits Rs. $$10,000$$ in a bank for a period of $$3$$ years at $$12$$% per annum compound interest. The interest accrued to her after maturity will be  
  • Question 8
    1 / -0
    Calculate the amount and compound interest on 
    (i) INR $$10800$$ for $$3$$ years at $$12\dfrac{1}{2}\% $$   per annum compounded annually . 
    Solution
    The amount after $$n$$ years compounded annually is given as,
    $$A=P\left(1+\dfrac{r}{100}\right)^n$$
    Where,
    $$P=$$Principal
    $$r=$$ rate of interest
    According to the given condition,

    Amount After three years $$=10800\left({1}+{\dfrac{12.5}{100}}\right)^3$$ 

                                          $$=10800\left (\dfrac{112.5}{100}\right)^3\\$$
                                           $$=15377.34$$

    compound interest $$=15377.34-10800$$
                                   $$=Rs\ 4577.34$$
  • Question 9
    1 / -0
    What is CI on INR $$7500$$ for $$4$$ years if the rate of interest is $$10\%$$ p.a. For the first $$2$$ years and $$20\%$$p.a. for the next $$2$$ years?
    Solution
    Given,
    CI $$=P\left[\left(1+\dfrac{r}{100}\right)^t-1\right]$$

    CI $$=CI_1+CI_2$$

    $$CI_1$$(For the first two years)

    $$=7500\left[\left(1+\dfrac{10}{100}\right)^2-1\right]$$

    $$=7500[(1.1)^2-1]$$

    $$=7500\times 0.21=1575$$

    For $$CI_2$$, 
    $$P$$ will be $$7500+1575$$$$=9075$$

    $$CI_2=9075\left[\left(1+\dfrac{20}{100}\right)^2-1\right]$$

    $$=9075[(1.2)^2-1]$$

    $$=9075\times 0.44$$

    $$=3979.8$$

    $$CI=CI_1+CI_2$$

    $$=1575+3979.8$$

    $$=5554.8$$.
  • Question 10
    1 / -0
    The sum on which the compound interest for second year at $$10 \% \,p.a.$$ is $$Rs. \,132$$ is given by
    Solution

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