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Basics of Financial Mathematics Test 9

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Basics of Financial Mathematics Test 9
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  • Question 1
    1 / -0
    Calculate the compound interest for the third year on $$Rs.\ 7,500$$ invested for $$5$$ years at $$10\%$$ per annum.
    Solution
    $$A=P\left(1+\dfrac{R}{100}\right)^T$$
    $$P=Rs.\ 7500, R=10% , T=3$$ years

    $$A=7500\left(1+\dfrac{10}{100}\right)^3$$

    $$A=7500\times \dfrac{110}{100}\times \dfrac{110}{100}\times \dfrac{110}{100}$$

    $$A=\dfrac{75\times 121\times 11}{10}$$

    $$A=Rs. 9982.50$$

    For $$2$$ years
    $$A=7500\left(1+\dfrac{10}{100}\right)^2$$

    $$A=7500\times \dfrac{110}{100}\times \dfrac{110}{100}$$

    $$A=75\times 121$$

    $$A=Rs. 9075$$

    Compound interest for third year = $$Rs.\ 9982.50-9075=907.50  Rs.$$

    Hence, option A.
  • Question 2
    1 / -0
    The value of a machine depreciated by $$10 \%$$ per year during the first two years and $$15 \%$$ per year during the third year. Express the total depreciation of the machine, as per cent, during the three years.
    Solution

    Lets Price of article is $$100$$
    Dep value after $$1$$ year $$ =100-\dfrac{100\times10}{100} $$= $$90$$

    Dep. value after $$2$$ year $$ =90- \cfrac{90\times 10}{100}= 81$$
    Dep. value after $$3$$ years $$=81-\cfrac{81\times 15}{100}= 68.85$$
    Total Dep. $$=100-68.85$$ = $$31.15 \%$$
  • Question 3
    1 / -0
    Calculate the amount and the compound interest on Rs. $$6,000$$ in $$3$$ years at $$5\%$$ per year.
    Solution
    C.A = $$P\left (1+ \frac{R}{100} \right )^{T}$$

    Given $$R=5\%$$ and $$P=6000 ,T=3$$

    So C.A =$$6000\left (1+ \frac{5}{100} \right )^{3}$$

     So C.A =$$6000\left ( \frac{105}{100} \right )\left ( \frac{105}{100} \right )\left ( \frac{105}{100} \right )$$

    C.A $$=6945.75$$ Rs

    C.I $$=6945.75 -6000=945.75$$ Rs
  • Question 4
    1 / -0
    Calculate the amount and the compound interest on Rs. $$16,000$$ in $$3$$ years, when the rates of the interest for successive years are $$10\%$$, $$14\%$$ and $$15\%$$ respectively.
    Solution
    Int-rest amt after one year =$$\dfrac{16000\times 10}{100}= 1600$$Rs

    C.A after one year $$=16000+1700=17600$$ Rs

    Intrest after two year =$$\dfrac{17600\times 14}{100}= 2464$$

    C.A after two years $$=17600+2464=20064$$ Rs

    Intrest after thrree year =$$\dfrac{20064\times 15}{100}= 3009.60$$

    C.A after three years $$=20064+3009.60=23073.60$$ Rs

    intrest after three year $$=23073.60-16000=7073.60$$ Rs
  • Question 5
    1 / -0
    The value of an article which was purchased $$2$$ years ago, depreciates at $$12$$ % per annum. If its present value is Rs .$$ 9680$$, the price at which it was purchased is:
    Solution
    Given, $$A=$$ Rs.$$9680,T=2$$ years,$$R=12\%$$
    $$\because A = P\left (1 - \displaystyle \frac{R}{100} \right)$$
    $$9680 = P \left(1 - \displaystyle \frac{12}{100} \right)^{2}$$
    $$9680=P\left(\dfrac{88}{100}\right)^2$$
    $$P = 9680 \times \displaystyle \frac{25}{22} \times \frac{25}{22}$$
    $$=$$ Rs. $$12,500$$
  • Question 6
    1 / -0
    Mohan lent some amount of money at $$9\%$$ simple interest and an equal amount of money at $$10\%$$ simple interest each for $$2$$ years. If his total interest was Rs. $$760$$, what amount was lent in each case?
    Solution
    Let the amount lent in each case be Rs. $$x$$ 
    $$\Rightarrow \dfrac{x\times9\times2}{100}+\dfrac{x\times10\times2}{100}=760$$
    $$\Rightarrow \dfrac{18x}{100}+\dfrac{20x}{100}=760$$
    $$\Rightarrow\;38x=76000$$
    $$\Rightarrow\;x=2000$$
  • Question 7
    1 / -0
    Arun borrowed a sum of money from Jayant at the rate of $$8\%$$ simple interest for the first four years, $$10\%$$ p.a. for the next $$6$$ years and $$12\%$$ p.a. beyond $$10$$ years. If he pays a total of Rs. $$12160$$ as interest only at the end of $$15$$ years, how much money did he borrow?
  • Question 8
    1 / -0
    A person lent a certain sum of money at $$4\%$$ simple interest and in $$5$$ years, the interest amounted to Rs. $$520$$ less than the sum lent. The sum lent was
    Solution
    Let the sum lent be $$Rs. x$$

    The interest in $$5$$ years $$=\dfrac{x\times 4\times 5}{100}$$

    Given: $$\dfrac{x\times4\times5}{100}=x-520$$

    $$\Rightarrow \dfrac{x}{5}=x-520$$

    $$\Rightarrow x-\dfrac{x}{5}=520$$

    $$\Rightarrow \dfrac{4x}{5}=520$$

    $$\Rightarrow 4x=520\times 5$$

    $$\Rightarrow x=\dfrac{520\times 5}{4}$$

    $$\Rightarrow\;x=$$ Rs. $$650$$

    Therefore, the sum lent is Rs. $$650$$
  • Question 9
    1 / -0
    The cost of a vehicle is Rs. $$1,75,000$$. If its value depreciates at the rate of $$20$$ % per annum, then the total depreciation after $$3$$ years will be:
    Solution
    Value of the vehicle after 3 years
    $$= 1,75,000 \times \left(1 - \displaystyle \frac{20}{100} \right)^{3}$$
    $$= 1,75,000 \times \displaystyle \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5}$$
    = Rs. 89, 600
    $$\therefore$$ Total depreciation = 1, 75, 000 - 89, 600
                                      $$= Rs. 85,400$$
  • Question 10
    1 / -0
    Find the compound interest by using simple interest approach for Rs 5000 for 3 years at 8% per annum compounded annnually. 
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