Discussion

Explanation:

​​​​​​​From (1) and (2): Let the number of non-CS students who took AI and ML are 2x and 5x respectively.
⇒ Number of CS students = 2x + 5x = 7x.

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From (3): Let the number of non-CS students failing in two courses is ‘a’ each and CS students who got a C in ML = ‘2a’

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From (6): Let number of CS students  who got A, B and C grades respectively in AI are 3y, 5y, 2y.
While number of CS students  who got A, B and C grades respectively in ML are 4a, 5a, 2a.
From (5): No CS student failed in AI, while no non-CS student got an A grade in AI.

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From (4): The numbers of students who got A and C grades were the same for AI
⇒ 3y + 0 = 2y + non-CS students in AI with C grade
⇒ non-CS students in AI with C grade = 3y – 2y = y

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From (8): 30 students failed in ML hence CS students failing in ML = 30 – a.

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From (7):

⇒ a+a30-a = 31

⇒ 2a = 90 – 3a
⇒ a = 18

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⇒ 7x = 72 + 90 + 36 + 12 = 210
⇒ x = 30

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⇒ 210 = 3y + 5y + 2y
⇒ y = 21

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⇒ 60 = 0 + (non-CS student in AI getting B grade) + 21 + 18
∴ non-CS student in AI getting B grade = 60 – 39 = 21

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From (4): In both the courses, 50% of the students who passed got a B grade
Number of students passing in ML = 72 + 90 + 36 + (150 - 18) = 330
⇒ Number of students receiving B grade in ML = 330/2 = 165.
⇒ non-CS students with B grade in ML = 165 – 90 = 75

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Let number of non-CS students with A grade in ML = c, hence number of non-CS students with C grade in ML = 150 – 75 – 18 – c = 57 – c

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From (4): the numbers of students who got A and C grades were in the ratio 3 : 2 for ML.

⇒ 72+c36+57-c = 32

⇒ 144 + 2c = 279 – 3c

⇒ 5c = 135

⇒ c = 27

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∴ Number of students taking AI = 210 (CS students) + 60 (non-CS students) = 270.

Hence, option (d).

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