Quantitative Aptitude: Data Interpretation Questions Set 30

Data Interpretation Sets SBI PO 2017, IBPS PO, NIACL, NICL, RBI Grade B, Dena Bank PO PGDBF, BOI, Bank of Baroda and other competitive exams.

Directions (1 – 5): Study the following table carefully and answer the questions that follow:

The table shows the number of students (boys and girls) who participated in different activities from 5 schools – A, B, C, D and E and the individual percentages. (Though some values are missing)

  1. What is the number of students who did not take part in painting from school C?
    A) 529
    B) 543
    C) 481
    D) 427
    E) 465
    View Answer
    Option C
    Solution:

    Students in painting = 26%, so who are not in painting = 100 – 26 = 74%
    So required number = 74% of (330+320) = 481
  2. Students from school A who did not take part in Swimming and Dancing is 147. The number of students who took part in Swimming is 49 more than who took part in Dancing. Find the number of students who took part in swimming from same school.
    A) 187
    B) 196
    C) 173
    D) 205
    E) 224
    View Answer
    Option B
    Solution:

    30% of z = 147 [z – number of students in school A]
    Solve, z = 490
    Let students who took part in swimming is x% and dancing is y%
    So x + y = 70% (100-30) ………………………(1)
    Also x% of 490 = 49 + y% of 490 ………….(2)
    Solve both equations, x = 40%
    So students from school A who took part in Swimming = 40% of 490 = 196
  3. There is a difference of 100 students in number from school C and E. Find the number of students who took part in painting and dancing from school E given that number of girls in school E is less than 400.
    A) 363
    B) 357
    C) 326
    D) 341
    E) 371
    View Answer
    Option D
    Solution:

    Students from C = 330+320 = 650
    So now students from E is 550 or 750. Given that girls in E < 400. So students from E = 550 because boys are 230
    So who took part in painting and dancing = (100-38)% of 550 = 341
  4. The percentage point of students who took part in swimming is 10% greater than that of students who took part in painting from school D. Boys from school B is 6 2/3% less than boys from school D. What is the number of students who did not take part in swimming if total number of girls from school D is 280?
    A) 377
    B) 336
    C) 343
    D) 312
    E) 351
    View Answer
    Option A
    Solution:

    From school D: swimming + painting = 100 – 40 = 60%
    percentage point of students who took part in swimming is 10% greater than that of students who took part in painting from school D. So swimming = 35%, and painting = 25%
    Boys from school B = 280 which is 6 2/3% = 20/3% less than boys from school D
    So (x – 280)/x * 100 = 20/3
    Solve, x = 300 = boys from D
    Girls from school D = 280, so total students = 300+280 = 580
    So number of students who did not take part in swimming = (40+25)% of 580 = 377
  5. Number of girls from school B is 390 less than number of students from school C. If a total of 105 girls took part in painting from school C, then what percent of students from school C who took part in painting are girls?
    A) 59.99%
    B) 55.55%
    C) 57.77%
    D) 51.11%
    E) None of these
    View Answer
    Option B
    Solution:

    Number of girls from B = (330+320) – 390 = 260
    School B:
    Total students who took part in painting = 100 – 65 = 35%
    Let x% are girls from this 35%
    So x% of 35% of (280+260) = 105
    Solve, x = 500/9%

Directions (6 – 10): Study the following table carefully and answer the questions that follow:

The table shows the individual scores of 5 players – A, B, C, D and E in 4 difference matches. (Though some values are missing)
The missing values in column for match 1 are the lowest scores in that match.
Also, the missing values in column for match 3 except the scores of B are the lowest scores in that match.

  1. What is the total score of A and C together?
    A) 196
    B) 204
    C) 200
    D) 191
    E) 210
    View Answer
    Option C
    Solution:

    In Match 1:
    Total of missing value = 150 – (27+36+35) = 52
    Given that the missing value is the lowest in Match 1. Then in match 1 the missing values should be Score of B in match 3 = 121 – (26+49+20) = 26
    Now we know Score of B in Match 3. So
    Match 3:
    Proceed similar as above in Match 1, so both the missing values in Match 3 are 12.
    Score of A in match 4 = 87 – (27+32+12) = 16
    Score of D in match 2 = 93 – (35+12+18) = 28
    Score of E in match 4 = 107 – (26+33+13) = 35
    So total in match 4 = 16 + 20+ 22 + 18 + 35 = 111
    So total in match 2 = 111 + 57 = 168
    So, Score of C in match 2 = 168 – (32+49+28+33) = 26
    So, last total score of C = 36 + 26 + 29 + 22 = 113
  2. What is the difference between the average of best 2 scores of payer E and average of best 3 scores of player D?
    A) 9
    B) 6
    C) 8
    D) 7
    E) 5
    View Answer
    Option D
    Solution:

    The average of best 3 scores of player D = (35+28+18)/3 = 27
    The average of best 2 scores of payer E = (33+35)/2 = 34
    So required difference = 34 – 27 = 7
  3. The total of best 2 scores of A is how much % greater than the total of lowest 2 scores of C?
    A) 25%
    B) 23%
    C) 28%
    D) 30%
    E) 32%
    View Answer
    Option B
    Solution:

    Total of Best 2 scores of A = 27 + 32 = 59
    Total of lowest 2 scores of C = 26 + 22 = 48
    So required % = (59-48)/48 * 100 = 23%
  4. What is the difference between the highest individual score and lowest individual score?
    A) 27
    B) 28
    C) 32
    D) 39
    E) 37
    View Answer
    Option E
    Solution:

    Highest individual score = 49 [B – match 2]
    Lowest individual score = 12 [D – match 3]
    So required difference = 49 – 12 = 37
  5. If one more match is added and it was found that 4 of 5 players secured equal score in match 5 and the 5th one secured 1 score more than those 4, then find the score of the 5th one given that total score in match 5 is 1 1/3 of total score of A in given 4 matches.
    A) 23
    B) 24
    C) 25
    D) 26
    E) 27
    View Answer
    Option B
    Solution:

    Total in match 5 = 4/3 * 87 = 116
    So 4x + (x+1) = 116
    Solve, x = 23
    So of 5th player = 24

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8 Thoughts to “Quantitative Aptitude: Data Interpretation Questions Set 30”

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