Quantitative Aptitude: Data Interpretation Questions Set 131

Directions(1-5):Answer the questions based on the following information.
The graph given below shows the total number of units produced by five companies P,Q,R,S and T and percentage of units sold by respective company in 2010.

  1. What is the difference between the number of units which remain unsold by company S and T?
    5000
    5200
    4600
    4000
    4800
    Option C
    Required Difference = 6600 -2000 = 4600

     

  2. Number of units sold by company R is how much percent more or less than the number of units sold by company S?
    14.2%
    17.5%
    19.7%
    22.2%
    10.9%
    Option B
    Percentage = [(6000 – 4950)/6000 ]*100 = 17.5%

     

  3. What is the average number of unsold units of companies P,R and T together?
    4200
    3215
    3500
    4750
    3550
    Option D
    Average = (3600+4050+6600)/3 = 4750

     

  4. Company R has three types of unsold products (P,Q,R), out of which 60% were of the type P while type Q and R were in the ratio 5:4 resp. Find the number of type Q products of company R which remain unsold.
    985
    900
    850
    800
    788
    Option B
    Total number of products of type Q and type R were unsold = 0.40*4050 = 1620
    Number of type Q which remain unsold = 5/9*1620 = 900

     

  5. What is the total number of units sold by company P and Q together in 2010?
    7500
    5400
    7000
    6400
    6200
    Option D
    Total number of units = 2400+4000 = 6400

     

  6. Directions(6-10): Answer the questions based on the given information.
    The graph given below shows the number of students who play Cricket and Badminton from five different schools. One student play one game.

  7. Find the average number of students who play Badminton in school A,B and D.
    200
    180
    190
    240
    210
    Option D
    Required Average = (260+180+280)/3 = 240

     

  8. What is the difference between the average number of students who play Cricket in school A and B together and the average number of students who play Badminton in school C and D?
    20
    25
    32
    30
    45
    Option D
    Difference = (240+280)/2 – (210+250)/2 = 30

     

  9. If the ratio of the number of boys to girls who play Cricket in school D is 2:3, Find the number of boys who play Cricket in school D?
    80
    100
    90
    85
    96
    Option E
    Number = (240/5)*2 = 96

     

  10. Find the ratio of the number of students who play Cricket in school C to the number of students who play Badminton in B.
    4:5
    5:6
    7:5
    2:3
    8:9
    Option E
    Ratio = 160:180 = 8:9

     

  11. The number of students who play Cricket in E is how much percent less than the number of students who play Badminton in A and C together?
    20%
    30%
    40%
    50%
    60%
    Option C
    Difference = (260+240)- 300 = 200
    Percentage = 200/500*100 = 40%

     


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