Data Interpretation Capsules - Edwin Jose, CAT DILR 100 Percentiler - Set 4


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    DIRECTIONS: Study the following table and answer the questions.

    0_1487526586525_di10.png

    (1) Total number of candidates qualified from all the states together in 1997 is approximately what percentage of the total number of candidates qualified from all the states together in 1998?
    (A 72%
    (B) 77%
    (C) 80%
    (D) 83%

    (2) What is the average candidates who appeared from State Q during the given years?
    (A) 8700
    (B) 8760
    (C) 8990
    (D) 8920

    (3) In which of the given years the number of candidates appeared from State P has maximum percentage of qualified candidates?
    (A) 1997
    (B) 1998
    (C) 1999
    (D) 2001

    (4) What is the percentage of candidates qualified from State N for all the years together, over the candidates appeared from State N during all the years together?
    (A) 12.36%
    (B) 12.16%
    (C) 11.47%
    (D) 11.15%

    (5) The percentage of total number of qualified candidates to the total number of appeared candidates among all the five states in 1999 is?
    (A) 11.49%
    (B) 11.84%
    (C) 12.21%
    (D) 12.57%

    Solutions:

    (1) Option C
    Required percentage = ((720 + 840 + 780 + 950 + 870))/((980 + 1050 + 1020 + 1240 + 940)) × 100
    = 4160/5230 × 100
    =80%

    (2) Option C
    Required average = ((8100 + 9500 + 8700 + 9700 + 8950))/5
    = 44950/5
    =8990

    (3) Option D
    The percentages of candidates qualified to candidates appeared from State P during different years are:
    For 1997 780/6400 × 100 = 12.19 %
    For 1998 1020/8800 × 100 = 11.59 %
    For 1999 890/7800 × 100 = 11.41 %
    For 2000 1010/8750 × 100 = 11.54%
    For 2001 1250/9750 × 100 = 12.82 %

    (4) Option D
    Required percentage = ((840 + 1050 + 920 + 980 + 1020))/((7500 + 9200 + 8450 + 9200 + 8800)) × 100
    = 4810/43150 × 100
    =11.15%

    (5) Option B
    Required percentage = ((850 + 920 + 890 + 980 + 1350))/((7400 + 8450 + 7800 + 8700 + 9800))×100
    = 4990/42150 ×100
    = 11.84%

    DIRECTIONS: The following table gives the percentage of marks obtained by seven students in six different subjects in an examination.

    0_1487527102761_di11.png

    (1) What are the average marks obtained by all the seven students in Physics? (rounded off to two digits after decimal)
    (A) 77.26
    (B) 89.14
    (C) 91.37
    (D) 96.11

    (2) The number of students who obtained 60% and above marks in all subjects is?
    (A) 1
    (B) 2
    (C) 3
    (D) NONE

    (3) What was the aggregate of marks obtained by Sajal in all the six subjects?
    (A) 409
    (B) 419
    (C) 429
    (D) 449

    (4) In which subject is the overall percentage the best?
    (A) Maths
    (B) Chemistry
    (C) Physics
    (D) History

    (5) What is the overall percentage of Tarun?
    (A) 52.5%
    (B) 55%
    (C) 60 %
    (D) 63 %

    Solutions:

    (1) OPTION B
    Average marks obtained in Physics by all the seven students
    =1/7 x ([ (90% of 120) + (80% of 120) + (70% of 120) + (80% of 120) + (85% of 120) + (65% of 120) + (50% of 120)])
    = 1/7 x [ (90 + 80 + 70 + 80 + 85 + 65 + 50)% of 120 ]
    = 1/7 x [ 520% of 120 ]
    = 624/7
    = 89.14%

    (2) OPTION B
    From the table it is clear that Sajal and Rohit have 60% or more marks in each of the six subjects.

    (3) OPTION D
    Aggregate marks obtained by Sajal
    = [ (90% of 150) + (60% of 130) + (70% of 120) + (70% of 100) + (90% of 60) + (70% of 40) ]
    = [ 135 + 78 + 84 + 70 + 54 + 28 ]
    = 449.

    (4) OPTION A
    We shall find the overall percentage (for all the seven students) with respect to each subject. The overall percentage for any subject is equal to the average of percentages obtained by all the seven students since the maximum marks for any subject is the same for all the students.
    Therefore, overall percentage for:
    Maths = 1/7 x (90 + 100 + 90 + 80 + 80 + 70 + 65)
    =1/7 x (575)
    = 82.14%.
    Chemistry = 1/7 x (50 + 80 + 60 + 65 + 65 + 75 + 35)
    =1/7 x (430)
    = 61.43%.
    Physics = 1/7 x (90 + 80 + 70 + 80 + 85 + 65 +50)
    =1/7 x (520)
    = 74.29%.
    Geography = 1/7 x (60 + 40 + 70 + 80 + 95 + 85 + 77)
    =1/7 x (507)
    = 72.43%.
    History = 1/7 x (70 + 80 + 90 + 60 + 50 + 40 + 80)
    = 1/7 x (470)
    = 67.14%.
    Comp. Science = 1/7 x (80 + 70 + 70 + 60 + 90 + 60 + 80)
    =1/7 x (510)
    = 72.86%.

    (5) OPTION C
    Aggregate marks obtained by Tarun
    = [ (65% of 150) + (35% of 130) + (50% of 120) + ((77% of 100) + (80% of 60) + (80% of 40)]
    = [ 97.5 + 45.5 + 60 + 77 + 48 + 32]
    = 360.
    The maximum marks (of all the six subjects)
    = (150 + 130 + 120 + 100 + 60 + 40)
    = 600.

    DIRECTIONS: Study the following table and answer the questions.

    0_1487527533078_di12.png

    Classification of 100 Students Based on the Marks Obtained by them in Physics and Chemistry in an Examination.

    (1) What is the different between the number of students passed with 30 as cut-off marks in Chemistry and those passed with 30 as cut-off marks in aggregate?
    (A) 3
    (B) 4
    (C) 5
    (D) 6

    (2) If at least 60% marks in Physics are required for pursuing higher studies in Physics, how many students will be eligible to pursue higher studies in Physics?
    (A) 27
    (B) 32
    (C) 34
    (D) 41

    (3) The percentage of number of students getting at least 60% marks in Chemistry over those getting at least 40% marks in aggregate, is approximately?
    (A) 21
    (B) 27
    (C) 29
    (D) 31

    (4) The number of students scoring less than 40% marks in aggregate is?
    (A) 13
    (B) 19
    (C) 20
    (D) 27

    (5) If it is known that at least 23 students were eligible for a Symposium on Chemistry, then the minimum qualifying marks in Chemistry for eligibility to Symposium would lie in the range?
    (A) 40-45
    (B) 30-40
    (C) 20-30
    (D) Below 20

    Solutions:

    (1) Option D
    Required difference
    = (No. of students scoring 30 and above marks in Chemistry) - (Number of students scoring 30 and above marks in aggregate)
    = 27 - 21
    = 6.

    (2) We have 60% of 50 = 60/100 X 50
    Required number = No. of students scoring 30 and above marks in Physics
    = 32

    (3) Option C
    Number of students getting at least 60% marks in Chemistry
    = Number of students getting 30 and above marks in Chemistry
    = 21
    Number of students getting at least 40% marks in aggregate
    = Number of students getting 20 and above marks in aggregate
    = 73.
    Required percentage = 21/73 X 100
    = 28.77

    (4) Option D
    We have 40% of 50 = 20
    Required number
    = Number of students scoring less than 20 marks in aggregate
    = 100 - Number of students scoring 20 and above marks in aggregate
    = 100 - 73
    = 27.

    (5) Option C
    Since 66 students get 20 and above marks in Chemistry and out of these 21 students get 30 and above marks, therefore to select top 35 students in Chemistry, the qualifying marks should lie in the range 20-30.


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