Current Affairs PDF Sales

Aptitude Questions – Data Interpretation Set 83

AffairsCloud YouTube Channel - Click Here

Hello Aspirants.

Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample From Data Interpretation based on latest pattern that is important for SBI, IBPS, RBI, SSC , NICL, LIC and other exams. We have included Some questions that are repeatedly asked in exams.

I.Study the following graph carefully to answer the given questions

DI Set 83

  1. If the ratio of number of boys & girls are in the ratio 3:2 in N and P, find the average number of boys in N and P together ?
    1.377.15
    2.379.25
    3.320.75
    4.397.5
    5.None of these
    Answer – 4.397.5
    Explanation :
    N:675 =>Boys = 675*3/5 = 405
    P:650 =>Boys = 650*3/5 = 390
    Avg = 405+390/2 = 397.5

  2. Find the total no of failed students in M, O and Q together ?
    1.247
    2.330
    3.341
    4.430
    5.None of these
    Answer – 3.341
    Explanation :
    M =768*25%= 192
    O= 460*15% = 69
    Q = 320*25% = 80
    Total = 341

  3. What will be the ratio of number of boys passed in N to number of girls failed in N, if the number of girls and boys in the ratio 16:11 ?
    1.121:24
    2.7:11
    3.103:529
    4.21:37
    5.None of these
    Answer – 1.121:24
    Explanation :
    675 => 400:275 => 16:11 = G:B
    G:400*12% =48
    B:275*88% =242
    242:48 = 121: 24

  4. Find the difference between the number of passed students in Q to number of failed students in M ?
    1.48
    2.51
    3.27
    4.-32
    5.None of these
    Answer – 1.48
    Explanation :
    Q passed => 75% of 320 = 240
    M failed => 192
    Difference = 240-192 = 48

  5. Which centre has the second least number of failed student ?
    1.M
    2.P
    3.O
    4.Q
    5.N
    Answer – 4.Q
    Explanation :
    M=192
    N=81
    O=69
    P=130
    Q=80

II.Study the following table carefully to answer the given questions

Classification of 50 Students based on the marks obtained by them in Quants & Reasoning in Prelim exam

Subject                                     Mark Out of 100
80 and above 70 and above 50 and above 30 and above 0 and above
Quants 23 28 37 43 50
Reasoning 12 32 35 39 50
Average(Aggregate) 15 22 38 40 50

  1. The number of students scoring less than 30% marks in aggregate is?
    1.15
    2.5
    3.20
    4.10
    5.None of these
    Answer – 4.10
    Explanation :
    We have 30% of 100 = (30/100)x100 = 30
    Therefore Required number
    = Number of students scoring less than 30 marks in aggreagate
    = 50 – Number of students scoring 30 and above marks in aggregate
    = 50 – 40 = 10.

  2. If it is known that at least 20 students were eligible for main exam, then the minimum qualifying marks in Reasoning for eligibility to attend the main exam(Top 20 students only can attend main exam )?
    1.70-80
    2.above 80
    3.70-60
    4.50-70
    5.None of these
    Answer – 1.70-80
    Explanation :
    12 students secured above 80 then remaining 8 student will scored in the range 70-80

  3. The percentage of number of students scored 70% marks in Quants is what percentage of number of students scored 30% marks in aggregate ?
    1.60%
    2.70%
    3.50%
    4.75%
    5.None of these
    Answer –2.70%
    Explanation :
    Number of students scored 70% marks or 70 in Quants  = 28
    Number of students scored 30% marks or 30 in aggregate= 40
    % = 28/40 * 100 = 70%

  4. If at least 50% marks in aggregate are required for interview, how many students will be eligible to attend interview?
    1.50
    2.40
    3.20
    4.38
    5.None of these
    Answer – 4.38
    Explanation :
    50% of 100 = 50*100/100 = 50
    No of students scored 50 aggregate = 38

  5. What is the different between the number of students passed with 80 as cut-off marks in Quants and 30 as cut-off marks in aggregate together to no of students passed with 70 as cut-off marks in aggregate and 50 as cut off in Reasoning?
    1.11
    2.5
    3.8
    4.6
    5.None of these
    Answer – 4.6
    Explanation :
    Difference = (23+40 ) – ( 22+35) =6