CAT 2022 Slot 3 · DILR · Set 3: Grades in AI and ML Courses
The full set and all 5 questions (Q11–Q15) from CAT 2022 Slot 3, with answers and solutions.
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.
Question 11 · Caselet DI
- A60
- B210
- C90
- D270
Answer & solution
Answer: Option D
Solving the set
Step 1 – Let non-CS students in AI and ML be 2k and 5k (fact 1). The number of CS students equals 7k (fact 2), and every CS student takes both courses.
Step 2 – AI. No CS student fails AI and CS grades are A : B : C = 3 : 5 : 2, so CS AI = 10m (A 3m, B 5m, C 2m). No non-CS student gets A in AI, so all AI A grades are 3m; AI C = AI A = 3m ⇒ non-CS C = m; AI B = A + C = 6m ⇒ non-CS B = m.
Step 3 – Failures. Non-CS failures are f in AI and f in ML; 2f = CS C grades in ML (fact 3). CS ML grades are A : B : C = 4 : 5 : 2, so CS C in ML = 2f ⇒ CS ML: A 4f, B 5f, C 2f (passed 11f). Non-CS failures (2f) : CS failures = 3 : 1 ⇒ CS failures = 2f/3 (all in ML). ML failures f + 2f/3 = 30 ⇒ f = 18. CS ML failures = 12; CS passed ML = 11 × 18 = 198 (A 72, B 90, C 36).
Step 4 – CS students = 198 + 12 = 210 = 10m ⇒ m = 21; 7k = 210 ⇒ k = 30. Non-CS AI = 60 (B 21, C 21, F 18 ✓); non-CS ML = 150 (F 18, passed 132).
Step 5 – ML passed = 198 + 132 = 330: B = 165, A : C = 3 : 2 ⇒ A = 99, C = 66. Non-CS ML: A 27, B 75, C 30.
Group | Course | A | B | C | F | Total |
CS | AI | 63 | 105 | 42 | 0 | 210 |
Non-CS | AI | 0 | 21 | 21 | 18 | 60 |
All | AI | 63 | 126 | 63 | 18 | 270 |
CS | ML | 72 | 90 | 36 | 12 | 210 |
Non-CS | ML | 27 | 75 | 30 | 18 | 150 |
All | ML | 99 | 165 | 66 | 30 | 360 |
This question
From the grade table, AI was taken by 210 CS + 60 non-CS = 270 students.
Hence, the answer is option 4.
Question 12 · Caselet DI
Answer & solution
Answer: 12
Solving the set
Step 1 – Let non-CS students in AI and ML be 2k and 5k (fact 1). The number of CS students equals 7k (fact 2), and every CS student takes both courses.
Step 2 – AI. No CS student fails AI and CS grades are A : B : C = 3 : 5 : 2, so CS AI = 10m (A 3m, B 5m, C 2m). No non-CS student gets A in AI, so all AI A grades are 3m; AI C = AI A = 3m ⇒ non-CS C = m; AI B = A + C = 6m ⇒ non-CS B = m.
Step 3 – Failures. Non-CS failures are f in AI and f in ML; 2f = CS C grades in ML (fact 3). CS ML grades are A : B : C = 4 : 5 : 2, so CS C in ML = 2f ⇒ CS ML: A 4f, B 5f, C 2f (passed 11f). Non-CS failures (2f) : CS failures = 3 : 1 ⇒ CS failures = 2f/3 (all in ML). ML failures f + 2f/3 = 30 ⇒ f = 18. CS ML failures = 12; CS passed ML = 11 × 18 = 198 (A 72, B 90, C 36).
Step 4 – CS students = 198 + 12 = 210 = 10m ⇒ m = 21; 7k = 210 ⇒ k = 30. Non-CS AI = 60 (B 21, C 21, F 18 ✓); non-CS ML = 150 (F 18, passed 132).
Step 5 – ML passed = 198 + 132 = 330: B = 165, A : C = 3 : 2 ⇒ A = 99, C = 66. Non-CS ML: A 27, B 75, C 30.
Group | Course | A | B | C | F | Total |
CS | AI | 63 | 105 | 42 | 0 | 210 |
Non-CS | AI | 0 | 21 | 21 | 18 | 60 |
All | AI | 63 | 126 | 63 | 18 | 270 |
CS | ML | 72 | 90 | 36 | 12 | 210 |
Non-CS | ML | 27 | 75 | 30 | 18 | 150 |
All | ML | 99 | 165 | 66 | 30 | 360 |
This question
From Step 3, 12 CS students failed in ML.
Hence, the answer is 12.
Question 13 · Caselet DI
Answer & solution
Answer: 27
Solving the set
Step 1 – Let non-CS students in AI and ML be 2k and 5k (fact 1). The number of CS students equals 7k (fact 2), and every CS student takes both courses.
Step 2 – AI. No CS student fails AI and CS grades are A : B : C = 3 : 5 : 2, so CS AI = 10m (A 3m, B 5m, C 2m). No non-CS student gets A in AI, so all AI A grades are 3m; AI C = AI A = 3m ⇒ non-CS C = m; AI B = A + C = 6m ⇒ non-CS B = m.
Step 3 – Failures. Non-CS failures are f in AI and f in ML; 2f = CS C grades in ML (fact 3). CS ML grades are A : B : C = 4 : 5 : 2, so CS C in ML = 2f ⇒ CS ML: A 4f, B 5f, C 2f (passed 11f). Non-CS failures (2f) : CS failures = 3 : 1 ⇒ CS failures = 2f/3 (all in ML). ML failures f + 2f/3 = 30 ⇒ f = 18. CS ML failures = 12; CS passed ML = 11 × 18 = 198 (A 72, B 90, C 36).
Step 4 – CS students = 198 + 12 = 210 = 10m ⇒ m = 21; 7k = 210 ⇒ k = 30. Non-CS AI = 60 (B 21, C 21, F 18 ✓); non-CS ML = 150 (F 18, passed 132).
Step 5 – ML passed = 198 + 132 = 330: B = 165, A : C = 3 : 2 ⇒ A = 99, C = 66. Non-CS ML: A 27, B 75, C 30.
Group | Course | A | B | C | F | Total |
CS | AI | 63 | 105 | 42 | 0 | 210 |
Non-CS | AI | 0 | 21 | 21 | 18 | 60 |
All | AI | 63 | 126 | 63 | 18 | 270 |
CS | ML | 72 | 90 | 36 | 12 | 210 |
Non-CS | ML | 27 | 75 | 30 | 18 | 150 |
All | ML | 99 | 165 | 66 | 30 | 360 |
This question
From Step 5, non-CS students with A in ML = 99 − 72 = 27.
Hence, the answer is 27.
Question 14 · Caselet DI
- A99
- B42
- C84
- D63
Answer & solution
Answer: Option D
Solving the set
Step 1 – Let non-CS students in AI and ML be 2k and 5k (fact 1). The number of CS students equals 7k (fact 2), and every CS student takes both courses.
Step 2 – AI. No CS student fails AI and CS grades are A : B : C = 3 : 5 : 2, so CS AI = 10m (A 3m, B 5m, C 2m). No non-CS student gets A in AI, so all AI A grades are 3m; AI C = AI A = 3m ⇒ non-CS C = m; AI B = A + C = 6m ⇒ non-CS B = m.
Step 3 – Failures. Non-CS failures are f in AI and f in ML; 2f = CS C grades in ML (fact 3). CS ML grades are A : B : C = 4 : 5 : 2, so CS C in ML = 2f ⇒ CS ML: A 4f, B 5f, C 2f (passed 11f). Non-CS failures (2f) : CS failures = 3 : 1 ⇒ CS failures = 2f/3 (all in ML). ML failures f + 2f/3 = 30 ⇒ f = 18. CS ML failures = 12; CS passed ML = 11 × 18 = 198 (A 72, B 90, C 36).
Step 4 – CS students = 198 + 12 = 210 = 10m ⇒ m = 21; 7k = 210 ⇒ k = 30. Non-CS AI = 60 (B 21, C 21, F 18 ✓); non-CS ML = 150 (F 18, passed 132).
Step 5 – ML passed = 198 + 132 = 330: B = 165, A : C = 3 : 2 ⇒ A = 99, C = 66. Non-CS ML: A 27, B 75, C 30.
Group | Course | A | B | C | F | Total |
CS | AI | 63 | 105 | 42 | 0 | 210 |
Non-CS | AI | 0 | 21 | 21 | 18 | 60 |
All | AI | 63 | 126 | 63 | 18 | 270 |
CS | ML | 72 | 90 | 36 | 12 | 210 |
Non-CS | ML | 27 | 75 | 30 | 18 | 150 |
All | ML | 99 | 165 | 66 | 30 | 360 |
This question
AI A grades are all CS: 3m = 63.
Hence, the answer is option 4.
Question 15 · Caselet DI
- A165
- B75
- C25
- D90
Answer & solution
Answer: Option B
Solving the set
Step 1 – Let non-CS students in AI and ML be 2k and 5k (fact 1). The number of CS students equals 7k (fact 2), and every CS student takes both courses.
Step 2 – AI. No CS student fails AI and CS grades are A : B : C = 3 : 5 : 2, so CS AI = 10m (A 3m, B 5m, C 2m). No non-CS student gets A in AI, so all AI A grades are 3m; AI C = AI A = 3m ⇒ non-CS C = m; AI B = A + C = 6m ⇒ non-CS B = m.
Step 3 – Failures. Non-CS failures are f in AI and f in ML; 2f = CS C grades in ML (fact 3). CS ML grades are A : B : C = 4 : 5 : 2, so CS C in ML = 2f ⇒ CS ML: A 4f, B 5f, C 2f (passed 11f). Non-CS failures (2f) : CS failures = 3 : 1 ⇒ CS failures = 2f/3 (all in ML). ML failures f + 2f/3 = 30 ⇒ f = 18. CS ML failures = 12; CS passed ML = 11 × 18 = 198 (A 72, B 90, C 36).
Step 4 – CS students = 198 + 12 = 210 = 10m ⇒ m = 21; 7k = 210 ⇒ k = 30. Non-CS AI = 60 (B 21, C 21, F 18 ✓); non-CS ML = 150 (F 18, passed 132).
Step 5 – ML passed = 198 + 132 = 330: B = 165, A : C = 3 : 2 ⇒ A = 99, C = 66. Non-CS ML: A 27, B 75, C 30.
Group | Course | A | B | C | F | Total |
CS | AI | 63 | 105 | 42 | 0 | 210 |
Non-CS | AI | 0 | 21 | 21 | 18 | 60 |
All | AI | 63 | 126 | 63 | 18 | 270 |
CS | ML | 72 | 90 | 36 | 12 | 210 |
Non-CS | ML | 27 | 75 | 30 | 18 | 150 |
All | ML | 99 | 165 | 66 | 30 | 360 |
This question
Non-CS B grades in ML = 165 − 90 = 75.
Hence, the answer is option 2.
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