Quantitative Aptitude: Data Interpretation Set 6 (Missing DI)

Directions (1 – 5):
The table shows the number of males (M) and females (F) in Five Companies A, B, C, D and E during Six years.
Some values are missing in the table. To answer each question, refer to data in table and respective question.

Company/
Year
A B C D E
M F M F M F M F M F
2011       202   102 142   12  
2012     112   105     136   20
2013   182 126 164   150   140 38  
2014 122 106   118 104   100     85
2015     124     105   100 100  
2016 138   142 128 186   124     125
  1. If the average number of female workers in all the given companies in 2013 is 156 and the difference between average number of female workers in all the given companies in 2013 and average number of male workers in all the given companies in 2016 is ten, then what is the difference between the male workers in Company D in 2016 and female workers in company E in 2013?
    A) 8
    B) 6
    C) 5
    D) 4
    E) 2
    View Answer
    Option D
    Solution:

    Average number of female workers in all the given companies in 2013 = 156
    Total female workers = 156 * 5 = 780
    So the female workers in Company E in 2013 = 780 – 644 = 136
    Average number of male workers in all the given companies in 2016 = 156-10 = 146
    So total male workers = 146 * 5 = 730
    The male workers in Company E in 2016 = 730 – 590 = 140
    Difference = 140 – 136 = 4
  2. If the total female workers in Company B is 830 and the average female workers in company C is 100 for the years 2012, 2013, 2014 and 2015, then the total number of female workers in Company B in 2012, 2013, 2014 and 2015 together is what percent more than the total number of female workers in Company C in the same years together?
    A) 40%
    B) 16%
    C) 25%
    D) 18%
    E) 20%
    View Answer
    Option C
    Solution:

    Average female workers in company C (2012, 2013, 2014 and 2015) = 100
    Total female workers in company C (2012, 2013, 2014 and 2015) = 100*4 = 400
    Total female workers in Company B = 830
    Total female workers in company B (2012, 2013, 2014 and 2015) = 830 – (202+128) = 500
    Required % = (500 – 400)/400 * 100 = 25%
  3. If the average of total number of workers (male and female) in companies A, B and C together in 2013 is 220 and in the year 2014, the sum of male workers of company B and female workers of company C is 100, then what is the ratio of the total number of workers (male and female) in companies A, B and C together in 2013 to the total number of workers (male and female) in the same companies together in 2014?
    A) 4:3
    B) 6:5
    C) 2:5
    D) 8:5
    E) 2:3
    View Answer
    Option B
    Solution:

    Total number of workers (male and female) in companies A, B and C together in 2013 = 660
    Total number of workers (male and female) in companies A, B and C together in 2014 = 100 + 122 + 106 + 104 +118 = 550
    Ratio = 660 : 550 = 6:5
  4. Total number of workers (male and female) in company D in the year 2011 is 280 and then decreased by 20 for every year. Total number of workers (male and female) in company E in the year 2011 is 20 and then doubled for every year. The total number of female workers in company E during all the given years together is what percent (approximately) of the total number of male workers in company D during all the given years together?
    A) 40%
    B) 56%
    C) 65%
    D) 68%
    E) 70%
    View Answer
    Option E
    Solution:

    For the year 2012, males = 260 -136 =124
    For the year 2013, males = 240 – 140 =100
    For the year 2015, males = 200 – 100 =100
    Total number of male workers = 142 + 124 + 120 + 100 + 100 + 124 = 710
    For the year 2011, females = 20 – 12 =8
    For the year 2013, females = 80 – 38 =42
    For the year 2015, females = 320 – 100 = 220
    Total number of female workers = 8 + 20 + 42 + 85 + 220 + 125 = 500
    Required % = (500/710)* 100 = 70.42%
  5. Total number of workers (male and female) in company D in the year 2011 is 280 and then decreased by 20 for every year. Total number of workers (male and female) in company E in the year 2011 is 20 and then doubled for every year. The total number of female workers in company E during all the given years together is what percent (approximately) less than the total number of male workers in company D during all the given years together?
    A) 14%
    B) 16%
    C) 25%
    D) 29%
    E) 18%
    View Answer
    Option D
    Solution:

    For the year 2012 = 260 -136 =124
    For the year 2013 = 220 – 120 =100
    For the year 2015 = 200 – 100 =100
    Total number of male workers = 142 + 124 + 120 + 100 + 100 + 124 = 710
    For the year 2011 = 20 – 12 =8
    For the year 2013 = 80 – 38 =42
    For the year 2015 = 320 – 100 = 220
    Total number of female workers = 8 + 20 + 42 + 85 + 220 + 145 = 500
    Required % = (710 – 500/710)* 100 = 29.57%

Directions (6 – 10):
The table shows the number of marks secured by each student in different subjects. The marks are out of total 150 marks in each subject.
Some values are missing in the table. To answer each question, refer to data in table and respective question.

 Student/Subject English Maths Physics Chemistry Biology Statistics
(150) (150) (150) (150) (150) (150)
Ravi ___ 140 ___ 120 ___ 140
Pavi ___ ___ 115 126 140 ___
Suresh 80 ___ 64 100 ___ 120
Ramesh ___ 150 120 ___ ___ 130
Geeta 88 ___ ___ 112 ___ 142
Sita 104 132 138 ___ 132 ___
  1. If the average marks of Pavi and Suresh is 105 and the marks scored by Pavi & Suresh in Maths is 60 less than that of marks scored by Ramesh & Suresh in Biology. Find the respective ratio of the total marks obtained by Pavi and Suresh in Maths and that scored by Suresh and Ramesh in Biology?
    A) 1 : 3
    B) 3 : 5
    C) 7 : 5
    D) 7 : 9
    E) None of these
    View Answer
    Option D
    Solution:

    Average marks of Pavi and Suresh = 105
    Marks of Pavi and Suresh = 210
    Marks of Suresh and Ramesh = 270
    Ratio = 210 : 270 = 7 : 9
  2. Total marks obtained by six students in Maths is 700 and average marks of Pavi and Suresh in Maths is 105. If 40% of mark is required to pass each subject then what is the difference between the minimum pass marks and marks scored by Geeta?
    A) 3
    B) 5
    C) 8
    D) 7
    E) None of these
    Solution:
    View Answer
    Option C
    Maths marks of Geeta = 700 – (140 + 150 + 132 + 210) = 68
    Minimum pass marks = 40/100 * 150 = 60
    Difference = 8
  3. If the average marks of Ravi in English, Physics and Biology is 120 and the sum of the marks obtained by Ravi in Maths, Chemistry and Statistics is same as the marks obtained by Ramesh in English, Physics and Chemistry then the total marks scored by Ravi in all subjects together is what % of total marks obtained by Ramesh in all the subjects together?
    A) 65%
    B) 95%
    C) 85%
    D) 75%
    E) None of these
    View Answer
    Option B
    Solution:

    Average marks of Ravi in English, Physics and Biology = 120
    Total marks of Ravi in English, Physics and Biology = 360
    Marks obtained by Ravi in Maths, Chemistry and Statistics = 400
    Total marks scored by Ravi in all subjects together = 360 + 400 = 760
    Total marks scored by Ramesh in all subjects together = 400 + 400 = 800
    Required % = 760/800 * 100 = 95%
  4. If Marks scored by Sita in Chemstry and Statistics is 30 less than the marks scored by herself in Maths then find the overall % of marks obtained by Sita in Maths, Chemistry and Statistics?
    A) 63%
    B) 55%
    C) 84%
    D) 62%
    E) 52%
    View Answer
    Option E
    Solution:

    Marks obtained by Sita in Maths, Chemistry and Statistics = 102 + 132 = 234
    % of marks obtained by Sita in Maths, Chemistry and Statistics = 234/450 * 100 = 52%
  5. Pavi’s marks in Statistics equals to 50% of the marks scored by herself in Biology while Sita’s Marks in Statistics is 10 less than the marks scored by herself in Physics. Then what is the average marks scored by students in Statistics?
    A) 112.3
    B) 121.7
    C) 184.5
    D) 162.4
    E) 152.4
    View Answer
    Option B
    Solution:

    Pavi’s marks in Statistics = 50% of 140 = 70
    Sita’s Marks in Statistics = 138 – 10 = 128
    Average = ( 140 + 70 + 120 + 130 + 142 + 128)/6

 

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One Thought to “Quantitative Aptitude: Data Interpretation Set 6 (Missing DI)”

  1. Dhmn Hinty

    1st question me 2013 female sum 636 aayga

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