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CAT 2022 Slot 3 DILR · Set 1 of 4Data InterpretationHard DifficultyQ1–Q5 · 3 MCQ + 2 TITA

CAT 2022 Slot 3 DILR Set 1: Tables (5 Questions) with Solutions

The Tables set from the official CAT 2022 Slot 3 DILR paper: the caselet, all 5 questions, the official answer key and a step-by-step solution for each. 100% free, no sign-up.

The Set: Read the Caselet First

CAT 2022 Slot 3 DILR · Questions 1–5
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.

In the exam this set competed with 3 others for 40 minutes. Budget about 10 minutes if you attempt it.

Q1. Tables

MCQ · +3 / −1
How many students took AI?
  1. A) 210
  2. B) 90
  3. C) 60
  4. D) 270
Correct Answer
D) 270
Step-by-step solution
📌 Core Concept
Set up the non-CS students with a ratio variable, then use the failure facts to fix the actual numbers.
🔢 Step-by-Step Solution
1
Let non-CS students in AI and ML be 2k2k and 5k5k. Each non-CS student takes only one course, so non-CS total =7k=7k. Fact 2 gives CS students =7k=7k.
2
Let ff be the number of non-CS students failing in each course. Fact 3 gives total non-CS failures =2f=2f, which equals CS students with a C in ML.
3
Fact 5 says no CS student failed in AI, so all CS failures are in ML. Fact 7 gives 2f:CS failures=3:12f : \text{CS failures} = 3:1, so CS failures =2f3=\frac{2f}{3}.
4
Fact 8: ML failures =f+2f3=30= f + \frac{2f}{3} = 30, so f=18f=18. CS students failing ML =12=12, and CS students with C in ML =36=36.
5
Fact 6: CS ML grades are in the ratio 4:5:24:5:2 among passers, so C is 211\frac{2}{11} of passers. Then passers =36×112=198=36\times\frac{11}{2}=198.
6
CS students =198+12=210=198+12=210, so 7k=2107k=210 and k=30k=30.
7
Non-CS in AI =2k=60=2k=60. Students taking AI =210+60=270=210+60=270.
Quick Check
CS =210=210 equals non-CS =60+150=210=60+150=210, as Fact 2 requires.
Correct Answer: D

Q2. Tables

TITA · +3 / 0
How many CS students failed in ML?
Correct Answer
12
Step-by-step solution
📌 Core Concept
CS students fail only in ML (Fact 5), so the failure ratio in Fact 7 links CS failures to non-CS failures. The total of 30 ML failures then fixes the numbers.
🔢 Step-by-Step Solution
1
Let ff be the number of non-CS students failing in each course (Fact 3). Total non-CS failures =2f=2f.
2
Fact 7 gives 2f:CS failures=3:12f:\text{CS failures}=3:1, so CS failures =2f3=\frac{2f}{3}.
3
Fact 5 says no CS student failed AI, so all CS failures are in ML.
4
ML failures are non-CS failures plus CS failures: f+2f3=30f+\frac{2f}{3}=30. This gives 5f3=30\frac{5f}{3}=30, so f=18f=18.
5
CS students failing ML =2×183=12=\frac{2\times18}{3}=12.
Quick Check
18+12=3018+12=30, which matches Fact 8.
Correct Answer: 12

Q3. Tables

TITA · +3 / 0
How many non-CS students got A grade in ML?
Correct Answer
27
Step-by-step solution
📌 Core Concept
Total A grades in ML come from the ratio of A to C among all passers. Subtracting the CS share leaves the non-CS A grades.
🔢 Step-by-Step Solution
1
Let non-CS students in AI and ML be 2k2k and 5k5k. CS students =7k=7k (Fact 2). Let ff be the number of non-CS students failing in each course.
2
Facts 3, 5, 7 and 8 give f+2f3=30f+\frac{2f}{3}=30, so f=18f=18. CS failures in ML =12=12, and CS students with C in ML =2f=36=2f=36.
3
Fact 6 gives CS ML grades in the ratio 4:5:24:5:2. C is 211\frac{2}{11} of CS passers, so CS passers =198=198. This gives CS A =72=72 and CS B =90=90. CS students =198+12=210=198+12=210, so k=30k=30.
4
Non-CS ML students =150=150. Passers =150−18=132=150-18=132.
5
Total ML passers =198+132=330=198+132=330. Fact 4 says 50% got B, so B =165=165, and A + C =165=165.
6
A:C =3:2=3:2 in ML, so total A =99=99.
7
Non-CS A in ML =99−72=27=99-72=27.
Quick Check
Total C =66=66, so non-CS C =30=30. Then 27+30+75=13227+30+75=132, matching the non-CS passers.
Correct Answer: 27

Q4. Tables

MCQ · +3 / −1
How many students got A grade in AI?
  1. A) 84
  2. B) 42
  3. C) 63
  4. D) 99
Correct Answer
C) 63
Step-by-step solution
📌 Core Concept
In AI, the A and C counts are equal and B is half of the passers. So A is one quarter of all AI passers.
🔢 Step-by-Step Solution
1
Let non-CS students in AI and ML be 2k2k and 5k5k. CS students =7k=7k. Let ff be the number of non-CS students failing in each course.
2
Facts 3, 5, 7 and 8 give f=18f=18. CS students with C in ML =36=36.
3
CS ML grades are in the ratio 4:5:24:5:2, so CS ML passers =36×112=198=36\times\frac{11}{2}=198. CS students =198+12=210=198+12=210, so k=30k=30.
4
Non-CS students in AI =60=60. With 18 failing, non-CS AI passers =42=42.
5
No CS student failed AI, so CS AI passers =210=210. Total AI passers =210+42=252=210+42=252.
6
Fact 4: B is 50% of passers, so A + C =126=126. A and C are equal in AI, so A =63=63.
Quick Check
CS AI grades in the ratio 3:5:23:5:2 give A =63=63, so non-CS A =0=0, as Fact 5 states.
Correct Answer: C

Q5. Tables

MCQ · +3 / −1
How many non-CS students got B grade in ML?
  1. A) 165
  2. B) 25
  3. C) 90
  4. D) 75
Correct Answer
D) 75
Step-by-step solution
📌 Core Concept
In ML, 50% of all passers got B. Subtracting the CS B grades from this total leaves the non-CS B grades.
🔢 Step-by-Step Solution
1
Let non-CS students in AI and ML be 2k2k and 5k5k. CS students =7k=7k. Let ff be the number of non-CS students failing in each course.
2
Facts 3, 5, 7 and 8 give f=18f=18. CS failures in ML =12=12, and CS students with C in ML =36=36.
3
CS ML grades are in the ratio 4:5:24:5:2, so CS passers =36×112=198=36\times\frac{11}{2}=198 and CS B =511×198=90=\frac{5}{11}\times198=90.
4
CS students =198+12=210=7k=198+12=210=7k, so k=30k=30. Non-CS students in ML =150=150, and non-CS ML passers =150−18=132=150-18=132.
5
Total ML passers =198+132=330=198+132=330. By Fact 4, B =165=165.
6
Non-CS B in ML =165−90=75=165-90=75.
Why Not the Others
165 is the total B in ML, including CS students.
25 and 90 do not follow from the data. 90 is the CS B count, not the non-CS count.
Correct Answer: D
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CAT 2022 Slot 3 DILR at a Glance

Sets in this section4
Questions in this section20
Section time40 minutes
Convenor IIMIIM Bangalore

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