CAT 2022 Slot 3 DILR Set 1 Q1–Q5
A DILR caselet from the Data Interpretation & Logical Reasoning (DILR) section of CAT 2022 Slot 3, with 5 questions (Venn & Set Puzzles). Answers are CAT's official key.
The set
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.
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.
Question 1
How many students took AI?
- A60
- B270
- C90
- D210
Answer: (B) 270 Venn & Set Puzzles MCQ
Solution
- Let CS students \(=n\); by fact 2 non-CS students \(=n\) too, split 2:5 as \(2k\) (AI) and \(5k\) (ML) with \(7k=n\).
- AI, CS (no fails): A, B, C \(=3a,5a,2a\) with \(n=10a\). Non-CS get no A, so A = C in AI forces non-CS C \(=a\), and B = half the passes gives non-CS B \(=a\).
- Non-CS fail \(f\) in each course; fact 3 gives \(2f=\) CS C-grades in ML \(=2b\) (CS ML grades \(4b:5b:2b\)), so \(f=b\). Fact 7: \(2f:g=3:1\) where \(g\) = CS fails in ML. With \(f+g=30\): \(f=18,\ g=12,\ b=18\).
- So CS ML passes \(=11b=198\), \(n=210\), \(a=21\), \(k=30\): non-CS AI \(=60\) (check: \(2a+f=42+18\)), non-CS ML \(=150\) with 132 passes.
- AI students \(=210+60=270\).
Question 2
How many CS students failed in ML?
TITA: type in the answer. There are no options and no negative marking.
Answer: 12 Venn & Set Puzzles TITA
Solution
- Let CS students \(=n\); by fact 2 non-CS students \(=n\) too, split 2:5 as \(2k\) (AI) and \(5k\) (ML) with \(7k=n\).
- AI, CS (no fails): A, B, C \(=3a,5a,2a\) with \(n=10a\). Non-CS get no A, so A = C in AI forces non-CS C \(=a\), and B = half the passes gives non-CS B \(=a\).
- Non-CS fail \(f\) in each course; fact 3 gives \(2f=\) CS C-grades in ML \(=2b\) (CS ML grades \(4b:5b:2b\)), so \(f=b\). Fact 7: \(2f:g=3:1\) where \(g\) = CS fails in ML. With \(f+g=30\): \(f=18,\ g=12,\ b=18\).
- CS students who failed ML \(=g=12\).
Question 3
How many non-CS students got A grade in ML?
TITA: type in the answer. There are no options and no negative marking.
Answer: 27 Venn & Set Puzzles TITA
Solution
- Let CS students \(=n\); by fact 2 non-CS students \(=n\) too, split 2:5 as \(2k\) (AI) and \(5k\) (ML) with \(7k=n\).
- AI, CS (no fails): A, B, C \(=3a,5a,2a\) with \(n=10a\). Non-CS get no A, so A = C in AI forces non-CS C \(=a\), and B = half the passes gives non-CS B \(=a\).
- Non-CS fail \(f\) in each course; fact 3 gives \(2f=\) CS C-grades in ML \(=2b\) (CS ML grades \(4b:5b:2b\)), so \(f=b\). Fact 7: \(2f:g=3:1\) where \(g\) = CS fails in ML. With \(f+g=30\): \(f=18,\ g=12,\ b=18\).
- So CS ML passes \(=11b=198\), \(n=210\), \(a=21\), \(k=30\): non-CS AI \(=60\) (check: \(2a+f=42+18\)), non-CS ML \(=150\) with 132 passes.
- ML passes \(=198+132=330\): A 99, B 165, C 66 (3:5:2 split of 50% B). CS has A 72, B 90, C 36, so non-CS has A 27, B 75, C 30.
- Non-CS students with A in ML \(=99-72=27\).
Question 4
How many students got A grade in AI?
- A42
- B99
- C84
- D63
Answer: (D) 63 Venn & Set Puzzles MCQ
Solution
- Let CS students \(=n\); by fact 2 non-CS students \(=n\) too, split 2:5 as \(2k\) (AI) and \(5k\) (ML) with \(7k=n\).
- AI, CS (no fails): A, B, C \(=3a,5a,2a\) with \(n=10a\). Non-CS get no A, so A = C in AI forces non-CS C \(=a\), and B = half the passes gives non-CS B \(=a\).
- Non-CS fail \(f\) in each course; fact 3 gives \(2f=\) CS C-grades in ML \(=2b\) (CS ML grades \(4b:5b:2b\)), so \(f=b\). Fact 7: \(2f:g=3:1\) where \(g\) = CS fails in ML. With \(f+g=30\): \(f=18,\ g=12,\ b=18\).
- So CS ML passes \(=11b=198\), \(n=210\), \(a=21\), \(k=30\): non-CS AI \(=60\) (check: \(2a+f=42+18\)), non-CS ML \(=150\) with 132 passes.
- A grades in AI are all CS: \(3a=63\).
Question 5
How many non-CS students got B grade in ML?
- A165
- B90
- C25
- D75
Answer: (D) 75 Venn & Set Puzzles MCQ
Solution
- Let CS students \(=n\); by fact 2 non-CS students \(=n\) too, split 2:5 as \(2k\) (AI) and \(5k\) (ML) with \(7k=n\).
- AI, CS (no fails): A, B, C \(=3a,5a,2a\) with \(n=10a\). Non-CS get no A, so A = C in AI forces non-CS C \(=a\), and B = half the passes gives non-CS B \(=a\).
- Non-CS fail \(f\) in each course; fact 3 gives \(2f=\) CS C-grades in ML \(=2b\) (CS ML grades \(4b:5b:2b\)), so \(f=b\). Fact 7: \(2f:g=3:1\) where \(g\) = CS fails in ML. With \(f+g=30\): \(f=18,\ g=12,\ b=18\).
- So CS ML passes \(=11b=198\), \(n=210\), \(a=21\), \(k=30\): non-CS AI \(=60\) (check: \(2a+f=42+18\)), non-CS ML \(=150\) with 132 passes.
- ML passes \(=198+132=330\): A 99, B 165, C 66 (3:5:2 split of 50% B). CS has A 72, B 90, C 36, so non-CS has A 27, B 75, C 30.
- Non-CS B grades in ML \(=165-90=75\).