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Comparing Quantities Test - 10

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Comparing Quantities Test - 10
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  • Question 1
    1 / -0
    1.8% == .......
    Solution
    1.81.8% = 1.8100=0.018\dfrac{1.8}{100} = 0.018
  • Question 2
    1 / -0
    A shopkeeper sold an article offering a discount of 5% and earned a profit of 23.5%. What would have been the percentage of profit earned if no discount was offered ?
    Solution
    Let the M.P. of the article =Rs.100= Rs.100

    Discount =5%= 5\% S.P.=Rs.95\Rightarrow S.P. = Rs.95

    Profit =23.5%= 23.5\% \Rightarrow C.P. = Rs.(95×100123.5)=Rs.950001235\displaystyle \left ( \frac{95\times 100}{123.5} \right )=Rs.\frac{95000}{1235}

    Had there been no discount, S.P.=Rs100S.P. = Rs100

    Then Profit % =100950001235950001235×100 =\cfrac{100-\dfrac{95000}{1235}}{\cfrac{95000}{1235}}\times 100

    =(123500950001235)950001235×100 =\cfrac{\left ( \dfrac{123500-95000}{1235} \right )}{\cfrac{95000}{1235}}\times 100

    =2850095000×100=30% =\cfrac{28500}{95000}\times 100=30\%
  • Question 3
    1 / -0
    Find the principle if S.IS.I is Rs.31.50Rs. 31.50, time period is 114\displaystyle \frac{1}{4} years and rate percent per annum is 5 14\displaystyle \frac{1}{4}%.
    Solution
    P=100×S.IT×RP = \displaystyle \frac{100 \times S.I}{T \times R}

        =100×31.5054×214=Rs.480= \displaystyle \frac{100 \times 31.50}{\displaystyle \frac{5}{4} \times \frac{21}{4}} = Rs.\, 480
  • Question 4
    1 / -0
    If C.P is Rs. 1750, S.P is Rs. 1925 then the profit percent is
    Solution
    C.P.=Rs1750C. P. = Rs 1750
    S.P.=Rs1925S.P. = Rs 1925
    Profit=19251750=175Profit = 1925 - 1750 = 175
    ProfitProfit % = ProfitC.P.×100\frac{Profit}{C.P.} \times 100
    ProfitProfit % = 1751750×100=10\frac{175}{1750} \times 100 = 10 %
  • Question 5
    1 / -0
    18\frac{1}{8}% means 
    Solution
    18\frac{1}{8} % = 18100=1800\frac{\frac{1}{8}}{100} = \frac{1}{800}
  • Question 6
    1 / -0
    A person spends 331333\frac{1}{3}% of his total income on food.Amount spend on food will be what part of his income
    Solution
    Let the persons income = II
    He spends, 331333 \frac{1}{3} % of his income = 1003\frac{100}{3} %
    Thus, Money spend on food = 1003\frac{100}{3} % of I = 1003100\frac{\frac{100}{3}}{100} of I = I3\frac{I}{3}

    Thus, the man spends 13\frac{1}{3} of his income on food.
  • Question 7
    1 / -0
    80% of a non-leap year =.......= ....... days
    Solution
    A non-leap year has =365= 365 days
    80%80\% of a non-leap year =80100×365= \dfrac{80}{100}\times 365
                                            =292= 292 days
  • Question 8
    1 / -0
    25\dfrac{2}{5} of total students in a class have opted Maths. Find the percentage of students opting Maths
    Solution
    Percentage of students opting for math's =25 ×100= \dfrac{2}{5} \times 100
                                                                           = 40% = 40\% of the students
  • Question 9
    1 / -0
    Express:

    35% as fraction
    Solution
    3535 % = 35100\frac{35}{100} = 720\frac{7}{20}
  • Question 10
    1 / -0
    A man bought a number of clips at 3 for a rupee and an equal number at 2 for a rupee At what price per dozen should he sell them to make a profit of 20%?
    Solution
    Let he bought 1 dozen clips of each kind.
    C.P of 2 dozen=13×12+12×12=10  Rs\frac{1}{3}\times 12+\frac{1}{2}\times 12=10   Rs
    S.P of 2 dozen=120%of10=Rs120100×10=12 Rs120\% of 10 = Rs\dfrac{120}{100}\times 10=12  Rs
    Hence S.P per dozen=6  Rs.
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