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Comparing Quantities Test - 42

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Comparing Quantities Test - 42
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  • Question 1
    1 / -0
    A man bought at the cost of 55 plums a rupee and 22 oranges a rupee. He sells 1010 plums and 66 oranges at the selling price of 44 plums a rupee and 33 oranges a rupee. What is his gain or loss?
    Solution
    Cost price of 11 plum =1005=20= \dfrac {100}{5} = 20 paise
    Cost price of 1010 plum =20×10=200= 20 \times 10 = 200 paise
    Cost price of 11 orange =1002=50= \dfrac {100}{2} = 50 paise
    Cost price of 66 oranges =50×6=300= 50 \times 6 = 300 paise
    Selling price per plums =1004=25=\dfrac { 100 }{ 4} = 25 paise
    So, selling price of 1010 plums =25×10=250= 25 \times 10 = 250 paise
    Selling price of 66 oranges =200= 200 paise
    Total cost price =200+300=500= 200 + 300 = 500 paise
    Total selling price =250+200=450= 250 + 200 = 450 paise
    So, loss =500450=50= 500 - 450 = 50 paise.
  • Question 2
    1 / -0
    An article is sold at 10%10\% profit . If its cost price and selling price are 5050 less, the profit would be 5%5\% more. Find the cost price?
    Solution
    let,C.P=xlet,\,C.P=x
    and gain%=10%gain\%=10\%
    S.P=(100+gain100)×C.PS.P=110x100S.P=\left(\dfrac{100+gain}{100}\right)\times C.P\Rightarrow S.P=\dfrac{110x}{100}

    newC.P=C.P50new \,C.P=C.P-50
    newS.P=S.P50new \,S.P=S.P-50
    gain%=5%morethanprevious=15%gain\%=5\%\,more \,than\,previous=15\%
    NewS.P=(100+gain100)×(NewC.P)\Rightarrow New\,S.P=\left(\dfrac{100+gain}{100}\right)\times(New\,C.P)
    (S.P50)=(100+15100)×(C.P50)\Rightarrow (S.P-50)=\left(\dfrac{100+15}{100}\right)\times(C.P-50)
    (110x10050)=115100×(x50)\Rightarrow (\dfrac{110x}{100}\,-\,50)=\dfrac{115}{100}\times (x-50)
    110x10050=115x1001152\Rightarrow \dfrac{110x}{100}\,-50=\dfrac{115x}{100}-\dfrac{115}{2}
    11521002=115x100110x100\Rightarrow \dfrac{115}{2}\,-\,\dfrac{100}{2}=\dfrac{115x}{100}-\dfrac{110x}{100}
    152=5x100\Rightarrow \dfrac{15}{2}=\dfrac{5x}{100}
    x=150\therefore x=150
  • Question 3
    1 / -0
    A and B invest in a business in the ratio of 3:23 : 2. If 55 % of total profit goes to charity and A's share is Rs. 85508550, then total profit is :
    Solution
    Let Total Profit is Rs. 100
    After 5 % charity, profit  = Rs 95
    So, share of A = (95 *3)/5 = Rs. 57
    Now, comparing,
    A share 5757 on profit 100100
    So, A's share 1 on profit =100/57= 100/57
    Thus, A shares Rs. 85508550 on  =(8550100)/57=Rs.15000 (8550 *100)/57 = Rs. 15000
  • Question 4
    1 / -0
    A manufacturer undertakes to supply 20002000 pieces of particular component at Rs. 2525 per piece. According to his estimates, even if 5%5\% fail to pass the quality tests, then he will make a profit of 25%25\%. However, as it turned out, 50%50\% of the components were rejected . what is the loss to the manufacturer ?
    Solution

    Total number of component =2000= 2000
    If 15%15\% component found to be defective

    Then, number of non-defective component =20005%= 2000 - 5\% of 2000=20005100×2000=1900 2000 =2000-\dfrac{5}{100}\times 2000= 1900


    So, selling price of 190190 component =1900×25== 1900 \times 25 = Rs. 4750047500


    This content 25%25\% profit of the manufacturer.


    Let the total manufacturing price of the component =x = x


    SP=47500 = 47500


    x+25%\Rightarrow x + 25\% of x=47500 x = 47500


    100x+25x100=47500\Rightarrow \dfrac{ 100x + 25x }{100 }= 47500


    125x=4750000\Rightarrow 125 x = 4750000


     x=\Rightarrow  x = Rs. 3800038000


    If 5050 component i.e. 10001000 components are defective. then


    Total collection =25×1000=25000= 25 \times 1000 = 25000


    Loss =3800025000= = 38000 - 25000 = Rs. 1300013000

  • Question 5
    1 / -0
    The profit after selling a pair of trousers for Rs. 863863 is the same as the loss incurred after selling the same pair of trousers for Rs. 631631. What is the cost price of the pair of trousers?
    Solution
    Let the C.P. be xx.
    Then, Profit =863x= 863 - x
    Loss =x631= x - 631
    Given, profit == Loss
    863x=x631\Rightarrow 863 - x = x - 631
    2x=863+631=1494\Rightarrow 2x = 863+631 = 1494
    x=14942=747\Rightarrow x = \dfrac {1494}{2} = 747
    Cost Price == Rs. 747747
  • Question 6
    1 / -0
    A man sold 1818 toys for Rs. 1680016800, gaining thereby the cost price of 33 toy. find the cost price of a toy.
    Solution
    Let the cost of one toy be x x
    Then, cost of 1818 toy =18x= 18x
    Gain =3x = 3x
    SP of 1818 toys = = Rs. 1680016800
    Gain == S.P. - C.P.
    3x=1680018x\Rightarrow 3x = 16800 - 18x
    21x=16800\Rightarrow 21x = 16800
    x=1680021=\Rightarrow x = \dfrac{16800}{21}= Rs. 800 800
    Therefore, cost price of a toy is Rs. 800800.
  • Question 7
    1 / -0
    A sells a bicycle to BB at a profit of 20%20\%. BB sells it to CC at a profit of 25%25\%. If CC pays Rs. 225225 for it, the cost price of the bicycle for AA is
    Solution
    Let the C.P for A=A= Rs. x x 
    Profit =20%=20\%
    Then S.P  for AA =120100×x==\dfrac{120}{100}\times x= Rs. 12x10\dfrac{12x}{10}
    S.P. for A=A= C.P. for BB == Rs. 12x10\dfrac{12x}{10}
    Profit =25%=25\%
    then S.P for BB =25100×12x10==\dfrac{25}{100}\times \dfrac{12x}{10}= Rs. 3x2\dfrac{3x}{2}
    S.P. for B=B = C.P. for C=C= Rs. 3x2\dfrac{3x}{2}
    According to the question, 
    C.P for C=C= Rs. 225225
    3x2=225\therefore \dfrac{3x}{2}=225
    x=225×23=\Rightarrow x=\dfrac{225\times 2}{3}= Rs. 150150
    Hence, cost of the bicycle for AA is Rs. 150150.
  • Question 8
    1 / -0
    The prices of two articles are in the ratio of 3:43 : 4. If the price of the first article be increased by 1010% and that of the second by Rs 44, the original ratio remains the same. The original price of second article is ?
    Solution
    Let price of two articles are 3x3x and 4x4x
    After increment price become 3.3x3.3x and (4x+4)(4x + 4)
    Now, according to question,
    3.3x(4x+4)=34\dfrac{3.3x}{(4x + 4)} = \dfrac{3}{4}
    13.2x=12x+113.2x = 12x + 1
    13.2x12x=1213.2x - 12x = 12
    1.2x=121.2x = 12
    x=10x = 10
    So, original price of second article =4x=4×10== 4x = 4 \times10 = Rs 4040.
  • Question 9
    1 / -0
    Jacob bought a scooter for a certain sum of money. He spent 10%10\% of the cost on repairs and sold the scooter for a profit of Rs. 11001100. How much did he spend on repairs if he made a profit of 20%20\%.
    Solution
    Total profit == Rs. 11001100
    Percent profit =20= 20
    20% =1100\Rightarrow 20\%  = 1100
    So, 100%=5500100\%=5500
    Rs. 55005500 would be the cost price for Jacob.
    He spends 10%10\% on repairing. so, the cost price was Rs. 50005000
    Thus, he spends Rs. 500500 on repairing.
  • Question 10
    1 / -0
    A shopkeeper bought 2 dozen of apples at RS. 58 and 4 dozen of oranges at Rs. 48. He sold the oranges to ram at 5% loss and apples to shyam at 15 % gain. What is overall loss or gain percentage in the transaction?
    Solution
    CP of 2 dozen of apples=Rs. 58\text{CP of 2 dozen of apples}=Rs.\ 58
    Loss=5%Loss=5\%
    SP=580.05×58=Rs. 55.1SP=58-0.05\times 58=Rs.\ 55.1

    CP of 4 dozen of oranges=Rs. 48\text{CP of 4 dozen of oranges}=Rs.\ 48
    Gain=15%Gain=15\%
    SP=48+0.15×48=Rs. 55.2SP=48+0.15\times 48=Rs.\ 55.2

    Total cost price =48+58=Rs. 106=48+58=Rs.\ 106
    Total selling price =55.1+55.2=Rs. 110.3=55.1+55.2=Rs.\ 110.3
    Hence, SP>CPSP>CP.
    So, profit =110.3106=Rs. 4.3=110.3-106=Rs.\ 4.3
    P(%)=4.3106×100=4.05%P(\%)=\dfrac{4.3}{106}\times 100=4.05\%
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