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Comparing Quantities Test - 52

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Comparing Quantities Test - 52
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  • Question 1
    1 / -0
    Aman gave Rs.$$9500$$  as a loan to Megh at the rate of $$4\%$$ p.a. compounded yearly. Megh returned the amount after two years. Calculate the interest on the first year's interest.
    Solution
    $$P=9500,R=4,T=2$$ years

    Interest in $$1$$ year$$=\cfrac { P\times R\times 1 }{ 100 } $$$$=\cfrac { 9500\times 4 }{ 100 } $$$$=380$$

    Interest on interest$$=\cfrac { 380\times 4\times 1 }{ 100 } $$$$=15.2$$
  • Question 2
    1 / -0
    The bank offers senior citizens intrest at the rate $$9\%$$ p.a compounded yearly on a fixed deposit. What interest will be received at the end of $$5$$ years if Rs.$$10,000$$ are deposited. Calculate by the method of simple interest.
    Solution
    $$P=10000,R=9,T=5$$ yrs
    Interest in $$1$$st year$$=\cfrac { P\times R\times T }{ 100 } $$
    $$=\cfrac { 10000\times 9\times 1 }{ 100 } $$
    $$=900$$
    $$P'=(900+10000)$$
    Interest in $$2$$nd year$$=\cfrac { 10900\times 9\times 1 }{ 100 } =981$$
    Interest in $$3$$rd year$$=\cfrac { 11881\times 9\times 1 }{ 100 }$$
    $$=1069.29$$
    Interest in $$4$$th year$$=\cfrac { 12950.29\times 9\times 1 }{ 100 }$$
    $$=1165.5261$$
    Interest in $$5$$th year$$=\cfrac { 14115.81\times 9\times 1 }{ 100 }$$
    $$=1270.42$$
    Total interest$$=900+981+1069.29+1165.52+1270.42$$
    $$=5386.23$$

  • Question 3
    1 / -0
    A bank pays interest at the rate of $$8\%$$ per annum compounded yearly. Find the interest by simple interest method if $$Rs.\ 8000$$ was deposited for $$3$$ years.
    Solution
    $$P=Rs.\ 8000,R=8$$%, Time period$$=3$$ years

    Interest in $$1$$st year $$=\cfrac { P\times R\times T }{ 100 } $$
                                    $$=\cfrac { 8000\times 8\times 1 }{ 100 } $$
                                    $$=Rs.\ 640$$

    Principal amount for $$2$$nd year will be,
    $$P'=P+640$$
          $$=Rs.\ 8000+Rs.\ 640$$
          $$=Rs.\ 8640$$

    Interest in $$2$$nd year$$=\cfrac { 8640\times 8 }{ 100 } $$
                                     $$=Rs.\ 691.2$$

    Principal amount for $$2$$nd year will be,
    $$P''=P'+691.2$$
          $$=Rs.\ 8640+Rs.\ 691.2$$
          $$=Rs.\ 9331.2$$

    Interest in $$3$$rd year $$=\cfrac { 9331.2\times 8 }{ 100 } $$
                                     $$=Rs.\ 746.5$$

    So, total interest $$=640+691.2+746.5$$
                                $$=Rs.\ 2077.69$$
                                $$\approx Rs.\ 2077.7$$
  • Question 4
    1 / -0
    A businessman borrowed the Rs.$$45$$ lakh  for two years at the rate $$2.5\%$$ p.a. compounded annually. Calculate the compound interest by simple interest method.
    Solution
    $$\Rightarrow$$  Here, Principal amount for first year  is $$Rs.45,00,000$$ and $$R=2.5\%$$.
    $$\Rightarrow$$  Interest for first year = $$\dfrac{4500000\times 2.5}{100}=Rs.1,12,500$$

    $$\Rightarrow$$  Principal amount for second year = $$Rs.4500000+Rs.112500=Rs.46,12,500$$

    $$\Rightarrow$$  Second year interest = $$\dfrac{4612500\times 2.5}{100}=Rs.1,15,312.5$$

    $$\Rightarrow$$  Total interest for 2 yeras = $$Rs.1,12,500+Rs.1,15,312.5=Rs.2,27,812.5$$
  • Question 5
    1 / -0
    Amar gave Rs.$$50,000 $$ as a loan to Amir at the rate of $$4\%$$ p.a. Amir return the amount after two years. Calculate the interest on the first year's interest.
    Solution
    $$1$$ year interest$$=\cfrac { P\times R\times T }{ 100 } $$
    $$=\cfrac { 50000\times 4\times 1 }{ 100 } =2000$$
    Interest of $$1$$ year interest$$=\cfrac { \left( { P }^{ ' }\times R\times T \right)  }{ 100 } $$
    $$=\cfrac { 2000\times 4\times 1 }{ 100 } $$
    $$=80$$
  • Question 6
    1 / -0
    A man lends Rs. $$12,500$$ at $$12$$% for the first year, at $$15$$% for the second year and at $$18$$% for the third year. If the rates of interest are compounded yearly; find the difference between the C.I. for the first year and the compound interest for the third year.
  • Question 7
    1 / -0
    The shopkeeper gives a discount of $$12$$% on the pair of shoes marked for Rs. $$1425$$, then S.P is:
    Solution
    Discount $$=12$$ %
    MP $$=1425$$
    SP =MP -Discount
    $$=1425-\dfrac{12}{100} \times 1425$$
    $$=1254$$
  • Question 8
    1 / -0
    The cost price of an article is $$\displaystyle \frac {3}{5}$$ times of its selling price. Find the loss or the gain as percent.
    Solution
    Let SP $$=x$$
    So, CP $$=\dfrac{3}{5} x$$
    $$Gain =SP -CP$$ 
    $$=x -\dfrac{3}{5}x$$
    $$=\dfrac{2}{5}x $$
    $$Gain\%=\dfrac{2x}{3x}\times 100 $$
    Profit $$=67$$%



  • Question 9
    1 / -0
    The population of a town $$3$$ years ago was Rs. $$50,000$$.If the population in these three years has increased at the rate of $$10\%,15\%$$ and $$8\%$$ respectively, find the present population.
    Solution
    Let $$P=.50,000$$ and 
    $$R_1=10\%,\\ R_2=15\%,\\ R_3=8\%$$.
    Present population = $$P\times \left (1+\dfrac{R_1}{100}\right)\left (1+\dfrac{R_2}{100}\right)\left (1+\dfrac{R_3}{100}\right)$$
    On substituiting the given values,
    Present population $$=$$ $$50000\times \left (1+\dfrac{10}{100}\right)\times \left (1+\dfrac{15}{100}\right)\times \left (1+\dfrac{8}{100}\right)$$
                                     $$=$$ $$50000\times \dfrac{11}{10}\times \dfrac {23}{20}\times \dfrac{27}{25} =68310$$
    Therefore, present population $$=$$ $$68310$$.
  • 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
    $$C.P$$ of the scooter=Rs.$$18000$$
    Gain=$$\dfrac{100}{9}\%$$

    $$SP=\dfrac{(100+\dfrac{100}{9})}{100}\times 18000$$
           
           $$=\dfrac{100}{9}\times 180=Rs.20000$$
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