CAT 2022 Slot 3 — DILR Question 16

Mixed PracticeEasy

Passage / Data

Answer the next 5 questions based on the information given below:

All the first-year students in the computer science (CS) department in a university take both the courses (i) AI and (ii) ML. Students from other departments (non-CS students) can also take one of these two courses, but not both. Students who fail in a course get an F grade; others pass and are awarded A or B or C grades depending on their performance. The following are some additional facts about the number of students who took these two courses this year and the grades they obtained.

1. The numbers of non-CS students who took AI and ML were in the ratio 2 : 5.
2. The number of non-CS students who took either AI or ML was equal to the number of CS students.
3. The numbers of non-CS students who failed in the two courses were the same and their total is equal to the number of CS students who got a C grade in ML.
4. In both the courses, 50% of the students who passed got a B grade. But, while the numbers of students who got A and C grades were the same for AI, they were in the ratio 3 : 2 for ML.
5. No CS student failed in AI, while no non-CS student got an A grade in AI.

6. The numbers of CS students who got A, B and C grades respectively in AI were in the ratio 3 : 5 : 2, while in ML the ratio was 4 : 5 : 2.
7. The ratio of the total number of non-CS students failing in one of the two courses to the number of CS students failing in one of the two courses was 3 : 1.
8. 30 students failed in ML.

How many students took AI?

Answer & solution

A
90
B
60
C
210
270

Solution

âââââââFrom (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.

âââââââ

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’

âââââââ

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.

âââââââ

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

ââââââââââââââ

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

ââââââââââââââ

From (7):

⇒ $\frac{a + a}{30 - a}$ = $\frac{3}{1}$

⇒ 2a = 90 – 3a
⇒ a = 18

ââââââââââââââ

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

ââââââââââââââ

⇒ 210 = 3y + 5y + 2y
⇒ y = 21

ââââââââââââââ

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

ââââââââââââââ

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

ââââââââââââââ

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

ââââââââââââââ

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

⇒ $\frac{72 + c}{36 + 57 - c}$ = $\frac{3}{2}$

⇒ 144 + 2c = 279 – 3c

⇒ 5c = 135

⇒ c = 27

ââââââââââââââ

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

Hence, option (d).