Question: How many students took AI?
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.
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’
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.
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
From (8): 30 students failed in ML hence CS students failing in ML = 30 – a.
From (7):
⇒ a + a 30 - a = 3 1
⇒ 2a = 90 – 3a
⇒ a = 18
⇒ 7x = 72 + 90 + 36 + 12 = 210
⇒ x = 30
⇒ 210 = 3y + 5y + 2y
⇒ y = 21
⇒ 60 = 0 + (non-CS student in AI getting B grade) + 21 + 18
∴ non-CS student in AI getting B grade = 60 – 39 = 21
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
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
From (4): the numbers of students who got A and C grades were in the ratio 3 : 2 for ML.
⇒ 72 + c 36 + 57 - c = 3 2
⇒ 144 + 2c = 279 – 3c
⇒ 5c = 135
⇒ c = 27
∴ Number of students taking AI = 210 (CS students) + 60 (non-CS students) = 270.
Hence, option (d).