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Basics of Financial Mathematics Test 15

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Basics of Financial Mathematics Test 15
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  • Question 1
    1 / -0
    A sewing machine was purchased $$2$$ years ago. Its value decreases by $$10\%$$ every year. Its present value is $$Rs.\ 19,083.60$$. For how much was the machine purchased?
    Solution
    Given, 
    Let, principal $$=P$$
    Rate of depreciation $$R=10\%$$
    Present price $$A=Rs.\ 19,083.60$$
    Time $$T=2$$ years
    So, $$n=2$$

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

    $$\Rightarrow 19,083.60= P \times \left( 1-\dfrac { 10 }{ 100 }  \right) ^{ 2 } $$

    $$\Rightarrow 19,083.60= P \times \left( 1-\dfrac { 1 }{ 10 }  \right) ^{ 2 } $$

    $$\Rightarrow 19,083.60= P \times \left(\dfrac { 9 }{ 10 }  \right) ^{ 2 } $$

    $$\Rightarrow 19,083.60= P \times \left(\dfrac { 81 }{ 100 }  \right)  $$

    $$\Rightarrow P= 19,083.60\times \left(\dfrac { 100 }{ 81 }  \right)  $$

    $$\Rightarrow P= 23,560 $$

    Hence, the machine was purchased for $$Rs.\ 23,560$$ .

  • Question 2
    1 / -0
    An article can be sold for 20000 cash or for 12000 down payment and 4 equal monthly instalments of 2200 each. Find the interest paid.
    Solution
    With the downpayment option, total amount paid for article $$ = Rs  12000 + 4(2200) = Rs 20800 $$

    But actually SP of article $$ = Rs 20000 $$

    So, interest paid $$ = Rs  20800 - 20000 = Rs  800 $$
  • Question 3
    1 / -0
    A sum of money invested at simple interest amounts to 2480 at the end of four years and 4080 at the end of eight years. Find the principal.
    Solution
    Simple Interest $$ SI = \dfrac {PNR}{100} $$
    Also Amount $$ A = P + SI $$
    So, for four years,
    $$ 2480 - P = \dfrac {P \times 4 \times R}{100}  $$  -- (1)

    And, for eight years,
    $$ 4080 - P = \dfrac {P \times 8 \times R}{100}  $$  -- (2)

    Subtracting equation (1) from equation (2), we get
    $$ => 1600 = \dfrac {4PR}{100} $$
    $$ => PR = 40,000 $$

    Substituting it in equation (1), we get
    $$ 2480 - P = \dfrac { 4 \times 40,000}{100}  $$
    $$ => P = Rs$$  $$880 $$
  • Question 4
    1 / -0
    A sum of money at simple interest amounts to  800 in 2 years and to 1200 in 6 years. The sum is
    Solution
    Simple Interest $$ SI = \frac {PNR}{100} $$

    Given,
    $$ Amount  A = 800 $$
    $$ => P + SI = 800 $$
    $$ => P + \frac {P \times 2 \times R}{100} = 800 $$
    $$ => P(100+2R) = 80,000 $$  -- (2)

    And $$ Amount  A = 1200 $$
    $$ => P + SI = 1200 $$
    $$ => P + \frac {P \times 6\times R}{100} = 1200 $$
    $$ => P(100+6R) = 120000 $$   - (2)

    Dividing eqn 2 by 1, we get

    $$ \frac {100+6R}{100+2R} = \frac {12}{8} $$
    $$ => R = \frac {100}{6} $$ %

    Substituting in eqn 2, we get
    $$ P = Rs 600 $$
  • Question 5
    1 / -0
    A man had $$\text{Rs. } 2,000$$. He lent a part of this at $$5\%$$ interest and the rest at $$4\%$$ interest. The total interest he received in one year was $$\text{Rs. } 92$$. The money he lent at $$5\%$$ interest was
    Solution
    Let the money lent at the rate of $$5\%$$ be $$x.$$ So, the money lent at $$4\%$$ will be $$2000-x.$$

    It is given that the total interest obtained in $$1$$ year is $$\text{Rs. } 92.$$ So,
    $$\begin{aligned}{}\frac{{x \times 5 \times 1}}{{100}} + \frac{{\left( {2000 - x} \right) \times 4 \times 1}}{{100}} &= 92\\5x + 4\left( {2000 - x} \right) &= 9200\\5x + 8000 - 4x& = 9200\\x &=\text{Rs. } 1200\end{aligned}$$

    Hence, the money lent at $$5\%$$ rate is $$\text{Rs. } 1200.$$
  • Question 6
    1 / -0
    A sum was put at SI at a certain rate for 8 years Had it been put at 4% higher rate it would have fetched Rs. 80 more Find the sum___
    Solution
    $$\displaystyle \frac{P\times \left ( R+4 \right )\times 8}{100}-\frac{P\times 8\times R}{100}=80$$
    $$\displaystyle \frac{2PR+8P}{25}-\frac{8PR}{100}=80$$
    $$\displaystyle \frac{8PR+32P-8PR}{100}=80$$
    $$\displaystyle \frac{32P}{100}=80$$
    P = $$\displaystyle \frac{80\times 100}{32}$$
    = Rs. 250
  • Question 7
    1 / -0
    Arjun wants to invest $$Rs. 15,000$$ in two types of bonds. He earns $$12\%$$ in the first type and $$15\%$$ in the second. His investment in $$15\%$$ bond, if he has a total earning of $$Rs. 1,950$$, is
    Solution
    Let Arjun's investment in $$12\%$$ bond be $$x$$ and in $$15\%$$ bond be $$(15000-x)$$
    total earnings is $$1950$$
    earning in $$12\%$$ bond is $$=\dfrac { 12 }{ 100 } \times x$$

    earning in $$15\%$$ bond is $$=\dfrac { 15 }{ 100 } \times \left( 15000-x \right) $$

    As per problem,total earning$$=1950$$

    $$\Rightarrow \left( \dfrac { 12 }{ 100 } \times x \right) +\dfrac { 15 }{ 100 } \times \left( 15000-x \right) =1950$$
     
    $$\Rightarrow \dfrac { 12x }{ 100 } +\dfrac { 225000-15x }{ 100 } =1950$$

    $$\Rightarrow \dfrac { 12x+225000-15x }{ 100 } =1950$$
     
    $$\Rightarrow -3x+225000=195000$$
     $$\Rightarrow -3x=-30000$$
    $$\Rightarrow x=10000$$
    Therefore his investment in $$15\%$$ bond is $$(15000-10000)=Rs5000$$

  • Question 8
    1 / -0
    Ram had Rs 2 lakh part of which he lent at 15% p.a. and rest at 12% p.a. Yearly interest on both the parts was Rs 27,600 How much did he lend at 15%?
    Solution
    Given 200000 Rs lend @ 15% and 12% recd Rs 27600
    So average rate of interest 
    $$\frac{27600}{200000}\times 100=13.8$$ %
    Let x and y  amount lent @ 15% and 12%
    We have x:y =(13.8-12):(15-13.8)=2:3
    So money lend at @ 15% =$$\frac{3}{5}\times 200000=120000$$ Rs

  • Question 9
    1 / -0
    The compound interest on $$Rs. 10,000$$ in $$\displaystyle2\frac{1}{2}$$ years at $$4\%$$ per annum is
    Solution
    Given principal$$=Rs.10,000$$
    Time$$=2.5$$ years
    Rate of interest$$=4\%$$
    Compound interest for first year$$=\dfrac{PTR}{100}=Rs.\dfrac { 10000\times 1\times 4 }{ 100 } =Rs.400$$
    Amount at the end of first year$$=Rs.(10000+400)=Rs.10400$$

    Amount for first year $$=$$ Principal for second year$$=Rs.10400$$
    Compound interest for second year$$=\dfrac{PTR}{100}=Rs.\dfrac { 10400\times 1\times 4 }{ 100 } =Rs.416$$

    For next $$6$$ months,
    Principal $$=Rs.(10400+416)=Rs.10816$$
    Time$$=6\ months=\dfrac{1}{2}\ years$$
    Compound interest for next six months will be$$=\dfrac{PTR}{100}=Rs.\dfrac { 10816\times 1\times 4 }{ 100\times 2 } =Rs.216.32$$
    Total compound interest $$=Rs.(400+416+216.32)=Rs.1032.32$$

  • Question 10
    1 / -0
    Ajay had purchased a second hand scooter for Rs. $$18000$$ and spent Rs $$1800$$ for repairs. After $$1$$ year he wanted to sell the scooter. At what price should he sell it to gain $$\cfrac{100}{9}$$ $$\%$$ , if $$\cfrac {100}{11}$$ $$\%$$ is to be deducted at the end of every year on account of depreciation?
    Solution
    Cost price $$=$$ Rs. $$18000$$
    cost on repairing $$=$$ Rs. $$1800$$
    So, Total cost $$= 18000 + 1800 = 19800$$
    Depreciation $$= \dfrac {100}{11} \%= 9.09 \%$$
    Gain $$= \dfrac {100}{9 } \%= 11.11 \%$$
    After depreciation, the cost price would be  $$= 19800 - 9.09\%$$ of $$19800 =$$ Rs. $$18000$$
    S.P. to gain $$11.11 \%  = 18000 + 11.11 \%$$ of $$18000 = 18000 + 1999.8 =$$ Rs. $$20000$$.
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