Caselets, Sets and Conditional Data Reasoning Previous Year Questions (PYQs) for CAT: 17+ Solved Questions with Step-by-Step Solutions

Solve 17+ Caselets, Sets and Conditional Data Reasoning previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Caselets, Sets and Conditional Data Reasoning

Chapter Journey

This chapter tests your ability to extract hidden constraints from dense paragraphs and organize them logically.

Puzzle Solving and Draw Competitions

Decode random scores, moving averages, and sequential progress.

Operator and Registration Set Classification

Classify entities across overlapping sets and binary conditions.

Subscriber and Course Enrollment Sets

Analyze multi-category enrollments and conditional exclusions.

By the end, you will master translating complex text into clean, solvable grids.

Puzzle Solving and Draw Competitions

Mastering the art of extracting hidden numerical constraints from seemingly random events and sequential tasks.

What you will learn here:

01 Decode random draw scores and bounded sums

02 Handle moving averages and consecutive day links

03 Track sequential puzzle progress and time bottlenecks

04 Optimize logic grids and eliminate impossible cases

Caselets, Sets and Conditional Data Reasoning: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [35 - 39]
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 got A grade in AI?

A.
99
B.
42
C.
84
D.
63

Correct Answer:

Step-by-Step Solution

Key idea: This is a multi-ratio set integration problem. We define a common variable for the CS students and express all other groups in terms of it, using the failure and grade distribution ratios.

Step 1: Define Variables.

Let Total Non-CS = $7 k$ (since Non-CS AI : Non-CS ML = 2 : 5).

Fact 2: CS Students = Total Non-CS = $7 k$ .

Step 2: Analyze Failures.

Fact 8: Total Fail ML = 30.

Fact 3: Non-CS Fail AI = Non-CS Fail ML. Let this be $f$ .

Total Non-CS Fail = $2 f$ .

Fact 3: Total Non-CS Fail = CS Students with C in ML. So, $C S_{C, M L} = 2 f$ .

Fact 7: Ratio of Total Non-CS Fail to Total CS Fail = 3 : 1.

So, Total CS Fail = $2 f /3$ .

For this to be an integer, $f$ must be a multiple of 3. Let $f = 3 m$ .

Then Total Non-CS Fail = $6 m$ , Total CS Fail = $2 m$ .

Non-CS Fail AI = $3 m$ , Non-CS Fail ML = $3 m$ .

$C S_{C, M L} = 6 m$ .

Fact 5: No CS student failed in AI. So CS Fail AI = 0.

Therefore, Total CS Fail = CS Fail ML = $2 m$ .

Step 3: Analyze ML Grades.

CS ML Total = $7 k$ .

CS ML Fail = $2 m$ .

CS ML Pass = $7 k - 2 m$ .

Fact 6: CS ML Grades A:B:C = 4:5:2. Sum of parts = 11.

So CS ML Pass must be divisible by 11. Let $C S_{P a ss, M L} = 11 p$ .

$C S_{A, M L} = 4 p, C S_{B, M L} = 5 p, C S_{C, M L} = 2 p$ .

We know $C S_{C, M L} = 6 m$ .

So $2 p = 6 m ⟹ p = 3 m$ .

$C S_{P a ss, M L} = 11 (3 m) = 33 m$ .

Total CS ML = Pass + Fail = $33 m + 2 m = 35 m$ .

So $7 k = 35 m ⟹ k = 5 m$ .

Step 4: Solve for $m$ .

We know Total Fail ML = 30.

Total Fail ML = Non-CS Fail ML + CS Fail ML = $3 m + 2 m = 5 m$ .

So $5 m = 30 ⟹ m = 6$ .

Then $k = 5 (6) = 30$ .

Step 5: Calculate Total AI A grades.

Total CS = $7 k = 210$ .

CS AI Total = 210. CS AI Fail = 0 (Fact 5).

CS AI Pass = 210.

Fact 6: CS AI Grades A:B:C = 3:5:2. Sum of parts = 10.

1 part = $210/10 = 21$ .

CS AI A = $3 \times 21 = 63$ .

Total Non-CS = 210. Non-CS AI = $(2/7) \times 210 = 60$ .

Non-CS AI Fail = $3 m = 18$ .

Non-CS AI Pass = $60 - 18 = 42$ .

Fact 4: 50% of students who passed got a B grade.

Total AI Pass = CS AI Pass + Non-CS AI Pass = $210 + 42 = 252$ .

Total AI Pass B = $50%$ of $252 = 126$ .

Total AI Pass A + Total AI Pass C = $252 - 126 = 126$ .

Fact 4: Numbers of students who got A and C grades were the same for AI.

So Total AI Pass A = Total AI Pass C = $126/2 = 63$ .

Answer: 63

Question 2 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [35 - 39]
In a coaching class, some students register online, and some others register offline. No student registers both online and offline; hence the total registration number is the sum of online and offline registrations. The following facts and table pertain to these registration numbers for the five months - January to May of 2023. The table shows the minimum, maximum, median registration numbers of these five months, separately for online, offline and total number of registrations. The following additional facts are known.
1. In every month, both online and offline registration numbers were multiples of 10.
2. In January, the number of offline registrations was twice that of online registrations.
3. In April, the number of online registrations was twice that of offline registrations.
4. The number of online registrations in March was the same as the number of offline registrations in February.
5. The number of online registrations was the largest in May.

	Minimum	Maximum	Median
Online	40	100	80
Offline	30	80	50
Total	110	130	120

What best can be concluded about the number of offline registrations in February?

A.
80
B.
50 or 80
C.
30 or 50 or 80
D.
50

Correct Answer:

Step-by-Step Solution

Key idea: This is a constraint satisfaction and bounding question, recognizable by the combination of Min/Max/Median statistics and discrete arithmetic constraints (multiples of 10).

Step 1: Establish fixed points using extreme constraints. Online ( $O$ ) Max = 100. Fact 5 says May $O$ is largest, so May $O = 100$ . Total ( $T$ ) Max = 130. Since $T = O + F$ , May $F \leq 130 - 100 = 30$ . Offline ( $F$ ) Min = 30. Thus, May $F$ MUST be exactly 30. (May $T = 130$ ).

Step 2: Determine January and April. Jan: $F = 2 O$ . Possible pairs $(O, F)$ summing to $[110, 130]$ with $O \geq 40, F \leq 80$ : Only $(40, 80)$ works (Sum 120). Apr: $O = 2 F$ . Possible pairs: Only $(80, 40)$ works (Sum 120).

Step 3: Analyze February and March. We have used $O$ values: 40 (Jan), 80 (Apr), 100 (May). Median $O$ must be 80. The remaining two $O$ values (Feb, Mar) must be such that when sorted with {40, 80, 100}, the median is 80. This requires at least one of them to be $\geq 80$ and the other $\leq 80$ , or both to be 80.

We have used $F$ values: 80 (Jan), 40 (Apr), 30 (May). Median $F$ must be 50. The remaining two $F$ values must make the median of {30, 40, 80, $F_{f e b}$ , $F_{ma r}$ } equal to 50.

Fact 4: Mar $O$ = Feb $F$ . Let this value be $x$ .

Step 4: Test possible values for $x$ (Feb $F$ ).

If $x = 50$ : Feb $F = 50$ , Mar $O = 50$ . To make $O$ median 80 with {40, 50, 80, 100, Feb $O$ }, Feb $O$ must be $\geq 80$ . Since Max $T = 130$ and Feb $F = 50$ , Feb $O$ can be at most 80. Thus Feb $O = 80$ . This gives Feb $T = 130$ .

Now check Mar $F$ . We need $F$ median to be 50 from {30, 40, 50, 80, Mar $F$ }. Mar $F$ must be $\geq 50$ . Also, Mar $T = 50 +$ Mar $F \geq 110 ⟹$ Mar $F \geq 60$ . If Mar $F = 60$ , Mar $T = 110$ . The $T$ values are {110, 120, 120, 130, 130}, median is 120. This perfectly satisfies all conditions!

If $x = 30$ : Mar $O = 30$ , which violates Min $O = 40$ .
If $x = 80$ : Feb $F = 80$ , Mar $O = 80$ . To make $F$ median 50 from {30, 40, 80, 80, Mar $F$ }, Mar $F$ must be $\leq 40$ . If Mar $F = 40$ , Mar $T = 80 + 40 = 120$ . But then $F$ set is {30, 40, 40, 80, 80}, median is 40, not 50. Contradiction.

Thus, Feb $F$ must be exactly 50.

Answer: D

Question 3 · Data Interpretation and Logical Reasoning NAT

Common Description: Instructions [35 - 39]
Three participants - Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants’ scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.
Table 1: 2-day averages for Days 2 through 5

Day 2	Day 3	Day 4	Day 5
15	15.5	16	17

Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.
Table 2: Ranks of participants on each day

	Day 1	Day 2	Day 3	Day 4	Day 5
Akhil	1	2	2	3	3
Bimal	2	3	2	1	1
Chatur	3	1	1	2	2

The following information is also known.
1. Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.
2. The total score on Day 3 is the same as the total score on Day 4.
3. Bimal’s scores are the same on Day 1 and Day 3. If Akhil attains a total score of 24, then what is the total score of Bimal?

Correct Answer:

Step-by-Step Solution

Key idea: This requires using a new global constraint (Akhil's total score) to resolve the ambiguous states left from the initial deduction, leveraging invariance.

Step 1: Recall the fixed daily totals and scores from the set's common data.

$T_{1} = 15, T_{2} = 15, T_{3} = 16, T_{4} = 16, T_{5} = 18$ .

$C = {3, 9, 6, 6, 6}$ .

$A_{3} = 5, B_{3} = 5$ .

$A_{4} = 3, B_{4} = 7$ .

$B_{1} = 5, A_{1} = 7$ .

Step 2: Identify the ambiguous days.

Day 2: $T_{2} = 15, C_{2} = 9 ⟹ A_{2} + B_{2} = 6$ . Ranks $C > A > B ⟹ 9 > A_{2} > B_{2}$ . Possible $(A_{2}, B_{2})$ : $(5, 1)$ or $(4, 2)$ .

Day 5: $T_{5} = 18, C_{5} = 6 ⟹ A_{5} + B_{5} = 12$ . Ranks $B > C > A ⟹ B_{5} > 6 > A_{5}$ . Possible $(B_{5}, A_{5})$ : $(7, 5)$ or $(8, 4)$ .

Step 3: Apply Akhil's total score constraint.

$A_{t o t a l} = A_{1} + A_{2} + A_{3} + A_{4} + A_{5} = 24$ .

$7 + A_{2} + 5 + 3 + A_{5} = 24 ⟹ 15 + A_{2} + A_{5} = 24 ⟹ A_{2} + A_{5} = 9$ .

Testing the pairs:

If $A_{2} = 5$ , then $A_{5} = 4$ . This requires $B_{2} = 1$ and $B_{5} = 8$ . Both are valid pairs.

If $A_{2} = 4$ , then $A_{5} = 5$ . This requires $B_{2} = 2$ and $B_{5} = 7$ . Both are valid pairs.

Step 4: Calculate Bimal's total score.

Notice that in both valid scenarios, $B_{2} + B_{5} = 1 + 8 = 9$ or $2 + 7 = 9$ .

$B_{t o t a l} = B_{1} + B_{2} + B_{3} + B_{4} + B_{5} = 5 + B_{2} + 5 + 7 + B_{5} = 17 + (B_{2} + B_{5}) = 17 + 9 = 26$ .

Answer: 26

Question 4 · Data Interpretation and Logical Reasoning MCQ

A.
165
B.
75
C.
25
D.
90

Correct Answer:

Step-by-Step Solution

Key idea: This is a 'Multi-Ratio Set Integration' problem. It requires extracting a specific subgroup value after solving for the base variables using a chain of ratio and constraint clues.

Step 1: Define variables from the facts.

Let Non-CS AI = $2 x$ , Non-CS ML = $5 x$ . Total Non-CS = $7 x$ .

Fact 2: Total CS = Total Non-CS = $7 x$ .

Fact 5 & 7: No CS student failed AI. Let Non-CS fail AI = $f$ , Non-CS fail ML = $f$ . Total Non-CS fail = $2 f$ .

Let CS fail ML = $c_{f}$ . Total CS fail = $c_{f}$ .

Fact 7: Total Non-CS fail / Total CS fail = 3/1 $⟹ 2 f / c_{f} = 3 ⟹ c_{f} = 2 f /3$ .

Step 2: Use the ML failure total.

Fact 8: Total Fail ML = 30.

Total Fail ML = Non-CS fail ML + CS fail ML = $f + 2 f /3 = 5 f /3$ .

So, $5 f /3 = 30 ⟹ 5 f = 90 ⟹ f = 18$ .

Thus, Non-CS fail ML = 18, and CS fail ML = $2 (18) /3 = 12$ .

Step 3: Find CS ML pass grades.

Fact 3: Total Non-CS fail ( $2 f = 36$ ) = CS students who got C in ML. So CS C in ML = 36.

Fact 6: CS ML pass grades A:B:C = 4:5:2.

Since 2 parts = 36, 1 part = 18.

Total CS ML Pass = (4 + 5 + 2) parts = 11 parts = $11 \times 18 = 198$ .

Total CS ML = CS ML Pass + CS ML Fail = 198 + 12 = 210.

Step 4: Find Total ML and Total Pass ML.

Since Total CS = 210, Total Non-CS = 210 (Fact 2).

Non-CS ML = $5 x$ . Since Total Non-CS = $7 x = 210$ , $x = 30$ .

So Non-CS ML = $5 \times 30 = 150$ .

Total ML = Non-CS ML + CS ML = 150 + 210 = 360.

Total Pass ML = Total ML - Total Fail ML = 360 - 30 = 330.

Step 5: Calculate Non-CS B grades in ML.

Fact 4: 50% of students who passed got a B grade.

Total B in ML = 50% of 330 = 165.

CS B in ML = 5 parts = $5 \times 18 = 90$ .

Non-CS B in ML = Total B in ML - CS B in ML = 165 - 90 = 75.

Answer: 75

Question 5 · Data Interpretation and Logical Reasoning NAT

Common Description: Anu, Bijay, Chetan, Deepak, Eshan, and Faruq are six friends. Each of them uses a mobile number from exactly one of the two mobile operators - Xitel and Yocel. During the last month, the six friends made several calls to each other. Each call was made by one of these six friends to another. The table below summarizes the number of minutes of calls that each of the six made to (outgoing minutes) and received from (incoming minutes) these friends, grouped by the operators. Some of the entries are missing.

Friend	Operator	Outgoing minutes to Operator Xitel	Outgoing minutes to Operator Yocel	Incoming minutes from Operator Xitel	Incoming minutes from Operator Yocel
Anu	Xitel	100		50	225
Bijay	Xitel		200		125
Chetan	Yocel	50	175	250	150
Deepak	Yocel	100	150	275	100
Eshan	Yocel		100	100	375
Faruq	Yocel	0		100	150

It is known that the duration of calls from Faruq to Eshan was 200 minutes. Also, there were no calls from:
i. Bijay to Eshan,
ii. Chetan to Anu and Chetan to Deepak,
iii. Deepak to Bijay and Deepak to Faruq,
iv. Eshan to Chetan and Eshan to Deepak. What was the total duration of calls (in minutes) made by Anu to friends having mobile numbers from Operator Yocel?

Correct Answer:

150

Step-by-Step Solution

Key idea: This is a flow balance problem in a closed network. The total outgoing minutes to a specific operator group must equal the total incoming minutes from that same operator group.

Step 1: Establish the Balance Equation for Operator Yocel.

Total Incoming to Yocel ( $I_{Y}$ ) = 225 (Anu) + 125 (Bijay) + 150 (Chetan) + 100 (Deepak) + 375 (Eshan) + 150 (Faruq) = 1125.

Total Outgoing to Yocel ( $O_{Y}$ ) = $A_{o u t Y}$ + 200 (Bijay) + 175 (Chetan) + 150 (Deepak) + 100 (Eshan) + $F_{o u t Y}$ = $A_{o u t Y} + F_{o u t Y} + 625$ .

Equating $O_{Y} = I_{Y}$ : $A_{o u t Y} + F_{o u t Y} + 625 = 1125 ⟹ A_{o u t Y} + F_{o u t Y} = 500$ .

Step 2: Determine Faruq's Outgoing to Yocel ( $F_{o u t Y}$ ).

Faruq calls Chetan, Deepak, and Eshan. We are given Call(F $\to$ E) = 200.

Deepak receives 100 mins from Yocel. Constraints state no calls from Chetan or Eshan to Deepak. Thus, Deepak's entire 100 mins from Yocel must come from Faruq. So, Call(F $\to$ D) = 100.

Eshan receives 375 mins from Yocel. Eshan's outgoing to Yocel is 100. Constraints state no calls from Eshan to Chetan or Deepak, so Eshan's entire 100 mins outgoing to Yocel must be to Faruq. Call(E $\to$ F) = 100.

Faruq receives 150 mins from Yocel. Since Call(D $\to$ F) = 0, Faruq receives from Chetan and Eshan. Call(C $\to$ F) + Call(E $\to$ F) = 150 $⟹$ Call(C $\to$ F) + 100 = 150 $⟹$ Call(C $\to$ F) = 50.

Chetan's outgoing to Yocel is 175. Chetan calls Deepak (0), Eshan, and Faruq (50). So Call(C $\to$ E) = 175 - 50 = 125.

Eshan's incoming from Yocel is 375. This comes from Chetan, Deepak, and Faruq. Call(C $\to$ E) + Call(D $\to$ E) + Call(F $\to$ E) = 375 $⟹$ 125 + Call(D $\to$ E) + 200 = 375 $⟹$ Call(D $\to$ E) = 50.

Deepak's outgoing to Yocel is 150. Deepak calls Chetan, Eshan, and Faruq (0). So Call(D $\to$ C) + Call(D $\to$ E) = 150 $⟹$ Call(D $\to$ C) + 50 = 150 $⟹$ Call(D $\to$ C) = 100.

Chetan's incoming from Yocel is 150. This comes from Deepak, Eshan (0), and Faruq. Call(D $\to$ C) + Call(F $\to$ C) = 150 $⟹$ 100 + Call(F $\to$ C) = 150 $⟹$ Call(F $\to$ C) = 50.

Now sum up Faruq's outgoing to Yocel: $F_{o u t Y}$ = Call(F $\to$ C) + Call(F $\to$ D) + Call(F $\to$ E) = 50 + 100 + 200 = 350.

Step 3: Solve for Anu's Outgoing to Yocel ( $A_{o u t Y}$ ).

$A_{o u t Y} + 350 = 500 ⟹ A_{o u t Y} = 150$ .

Answer: 150

A question from this chapter

Question 1

Common Description: Instructions [35 - 39] 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 got A grade in AI?

Question 2

Common Description: Instructions [35 - 39] In a coaching class, some students register online, and some others register offline. No student registers both online and offline; hence the total registration number is the sum of online and offline registrations. The following facts and table pertain to these registration numbers for the five months - January to May of 2023. The table shows the minimum, maximum, median registration numbers of these five months, separately for online, offline and total number of registrations. The following additional facts are known. 1. In every month, both online and offline registration numbers were multiples of 10. 2. In January, the number of offline registrations was twice that of online registrations. 3. In April, the number of online registrations was twice that of offline registrations. 4. The number of online registrations in March was the same as the number of offline registrations in February. 5. The number of online registrations was the largest in May. Minimum Maximum Median Online 4010080 Offline 308050 Total 110130120 What best can be concluded about the number of offline registrations in February?

Question 3

Common Description: Instructions [35 - 39] Three participants - Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants’ scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5. Table 1: 2-day averages for Days 2 through 5 Day 2 Day 3 Day 4 Day 5 15 15.5 1617 Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2. Table 2: Ranks of participants on each day Day 1 Day 2 Day 3 Day 4 Day 5 Akhil 12233 Bimal 23211 Chatur 31122 The following information is also known. 1. Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4. 2. The total score on Day 3 is the same as the total score on Day 4. 3. Bimal’s scores are the same on Day 1 and Day 3. If Akhil attains a total score of 24, then what is the total score of Bimal?

Question 4

Common Description: Instructions [35 - 39] 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 non-CS students got B grade in ML?

Question 5

Common Description: Anu, Bijay, Chetan, Deepak, Eshan, and Faruq are six friends. Each of them uses a mobile number from exactly one of the two mobile operators - Xitel and Yocel. During the last month, the six friends made several calls to each other. Each call was made by one of these six friends to another. The table below summarizes the number of minutes of calls that each of the six made to (outgoing minutes) and received from (incoming minutes) these friends, grouped by the operators. Some of the entries are missing. Friend Operator Outgoing minutes to Operator Xitel Outgoing minutes to Operator Yocel Incoming minutes from Operator Xitel Incoming minutes from Operator Yocel Anu Xitel 10050225 Bijay Xitel 200125 Chetan Yocel 50175250150 Deepak Yocel 100150275100 Eshan Yocel 100100375 Faruq Yocel 0100150 It is known that the duration of calls from Faruq to Eshan was 200 minutes. Also, there were no calls from: i. Bijay to Eshan, ii. Chetan to Anu and Chetan to Deepak, iii. Deepak to Bijay and Deepak to Faruq, iv. Eshan to Chetan and Eshan to Deepak. What was the total duration of calls (in minutes) made by Anu to friends having mobile numbers from Operator Yocel?

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Caselets, Sets and Conditional Data Reasoning Previous Year Questions (PYQs) for CAT: 17+ Solved Questions with Step-by-Step Solutions

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Caselets, Sets and Conditional Data Reasoning: Solved Questions with Step-by-Step Explanations (5 Problems)

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