Quantitative Aptitude: Data Interpretation Questions Set 96

Directions(1-5) :There are five companies and we have given the no. of employee working in different companies. In the table we have also given the percentage of male and female employees of HR and Marketing department.

  1. The ratio of male employee in HR dept. of company P and R together to female employee of Marketing department in company S and T together?
    187:40
    178:35
    153:33
    189:45
    188:43
    Option E
    Required Ratio = (48+140)/(22+21)
    = 188:43

     

  2. Difference between female employees of HR dept. in all companies together (excluding company S) and the female employees of Marketing dept. in all companies together (excluding company Q)?
    115
    121
    139
    183
    147
    Option C
    Required difference
    = (56 + 65 + 70 + 54) – (28 + 35 + 22 + 21)
    = 245 – 106 = 139

     

  3. The ratio of female employee of company Q in HR dept. to male employee of company R in Marketing dept. ?
    13:4
    11:7
    13:6
    18:5
    14:3
    Option A
    Required Ratio = 13 : 4

     

  4. If 60% of the employees of company T in HR department have MBA degree and 40% of the employees of the same company in the Marketing dept. have MBA degree, then how many employees have MBA degree in company T in both dept. together .
    68
    66
    63
    67
    60
    Option B
    Required no. of employees
    = 60/100 * 30/100 * 300 + 40/100 * 10/100 * 300
    = 66

     

  5. Total number of HR employees of company P is what % more then the total no. of marketing employee in company T?
    246.67%
    247.12%
    201.35%
    298.56%
    203.37%
    Option A
    HR employees in company P = 26*4 = 104
    Marketing employee in company T = 10*3 = 30
    Required % = (104 – 30)/30*100 = 246.67%

     

  6. Directions(6-10) : Study the pie-charts given below and answer the following questions.
    Percentage of students studying in various branches of an Engineering college

    Percentage of students interested in various sports of the Engineering college

  7. If 50% Mechanical students and 40% Electrical students are interested in Football then what is their ratio?

    25:22
    23:19
    21:17
    31:21
    13:11
    Option A
    Number of Mechanical students interested in Football
    = 2500 *20/100* 50/100 = 250
    Number of Electrical students interested in Football = 2500 *22/100* 40/100 = 220
    Required ratio = 25 : 22

     

  8. The percentage of students who are interested in other games are same (20%) in all branches. What is the difference between the number of students of Electrical and Mechanical branches who are interested in other games?
    8
    10
    13
    21
    14
    Option B
    Students of Mechanical branch interested in other games
    = 2500*20/100* 20/100 = 100
    Student of Electrical branch interested in other games
    = 2500*22/100* 20/100 = 110
    Difference = (110 – 100) = 10

     

  9. What is the ratio of the number of students who play Volleyball to the number of students who study in Mechanical branch?
    3:5
    5:9
    1:4
    7:3
    5:2
    Option C
    Required ratio = 2500* 5/100 : 2500* 20/100
    = 125 : 500 = 1 : 4

     

  10. If 20% students of Electronics branch fail, and out of these 60% are not interested in sports, then the number of failed Electronics students who are not interested in sports is what per cent of the total number of students who are not interested in sports?
    14%
    17%
    13%
    15%
    12%
    Option E
    Number of failed students of Electronics branch
    = 2500 * 15/100* 20/100 = 75
    Now, failed Electronic students who are not interested in sports = 75*60/100 = 45
    Total number of students of all branches who are not interested in sports = 2500*15/100 = 375
    Required % = 45/375*100 = 12%

     

  11. If 10% of Civil students, 20% of Mechanical students and 12% of Electrical students are not interested in sports then what is the average number of students of these branches who are interested in sports? (Calculate approximate value)
    323
    385
    354
    320
    311
    Option B
    Number of Civil students not interested in sports
    = 2500* 12/100* 10/100 = 30
    Now,
    Number of Civil students interested in sports = 2500 *12/100 = 300 – 30 = 270
    Number of Mechanical students not interested in sports = 2500 *20/100* 20 /100 = 100
    Number of Mechanical students interested in sports = 2500 *20/100 – 100 = 400
    Again,
    Number of Electrical students interested in sports = 2500 *22 /100 – 2500*22 /100*12/100 = 484
    Average number of students of these branches who are interested in sports
    = [270 +400+ 484]/3 = 1154/3 = 384.66 == 385

     


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