Caselets, Sets and Conditional Data Reasoning Practice Questions for CAT: 276+ Solved Questions with Step-by-Step Solutions

Solve 276+ Caselets, Sets and Conditional Data Reasoning practice 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

A business incubator has 250 startups, classified into two sectors: 'DeepTech' (150 startups) and 'E-Commerce' (100 startups). Each startup is registered for at least one of three government grant schemes: Alpha, Beta, and Gamma.

For the E-Commerce sector:

The total number of registrations across the three schemes is 220 (with Alpha: 90, Beta: 80, Gamma: 50).
The number of startups registered for exactly two schemes is equal to the number registered for all three schemes.

For the DeepTech sector:

The total number of registrations across the three schemes is 230 (with Alpha: 100, Beta: 80, Gamma: 50).

Global Condition:

The total number of startups registered for ALL THREE schemes across both sectors combined is exactly 65.

What is the maximum possible number of DeepTech startups registered for ONLY scheme Alpha?

A.
65
B.
70
C.
75
D.
80

Correct Answer:

Step-by-Step Solution

Key idea: This is a multi-set optimization problem linked by a global constraint. We must use the Conservation of Sums identity ( $S - N = E_{2} + 2 E_{3}$ ) to find the exact parameters of each sector before applying bounding techniques to maximize a specific region.

Step 1: Analyze the E-Commerce sector.

Total startups $N_{E C} = 100$ . Sum of registrations $S_{E C} = 220$ .

Using the identity: $S_{E C} - N_{E C} = E_{2, E C} + 2 E_{3, E C} ⟹ 220 - 100 = 120$ .

We are given $E_{2, E C} = E_{3, E C}$ . Substituting this gives $3 E_{3, E C} = 120 ⟹ E_{3, E C} = 40$ .

Step 2: Use the global condition to find DeepTech parameters.

Total ALL THREE across both sectors = $E_{3, E C} + E_{3, D T} = 65$ .

Since $E_{3, E C} = 40$ , we have $E_{3, D T} = 65 - 40 = 25$ .

For DeepTech, $N_{D T} = 150$ and $S_{D T} = 230$ .

$S_{D T} - N_{D T} = E_{2, D T} + 2 E_{3, D T} ⟹ 230 - 150 = 80$ .

Substituting $E_{3, D T} = 25$ : $E_{2, D T} + 50 = 80 ⟹ E_{2, D T} = 30$ .

The number of DeepTech startups in exactly one scheme is $E_{1, D T} = N_{D T} - E_{2, D T} - E_{3, D T} = 150 - 30 - 25 = 95$ .

Step 3: Maximize DeepTech ONLY Alpha ( $X$ ).

Let $A, B, C$ be the exactly-two regions for DeepTech (Alpha-Beta, Beta-Gamma, Gamma-Alpha).

We know $A + B + C = E_{2, D T} = 30$ .

The total for Alpha is $X + A + C + E_{3, D T} = 100 ⟹ X + A + C = 75$ .

To maximize $X$ , we must minimize $(A + C)$ . Since $A + C = 30 - B$ , we must maximize $B$ .

The total for Gamma is $Z + B + C + E_{3, D T} = 50 ⟹ Z + B + C = 25$ .

Since $Z \geq 0$ and $C \geq 0$ , the maximum possible value for $B$ is 25.

If $B = 25$ , then $C = 0$ and $Z = 0$ .

This gives $A + C = 30 - 25 = 5$ .

Substituting back into the Alpha equation: $X + 5 = 75 ⟹ X = 70$ .

(Checking Beta: $Y + A + B = 80 - 25 = 55 ⟹ Y + 5 + 25 = 55 ⟹ Y = 25$ . Total exactly one: $X + Y + Z = 70 + 25 + 0 = 95$ , which matches $E_{1, D T}$ . All constraints are satisfied).

Answer: 70

Question 2 · Data Interpretation and Logical Reasoning NAT

A tech company has three specialized departments: Cloud (C), Security (S), and AI (A). Every employee is assigned to at least one department. The following data is known about the workforce:

The total number of employees in Cloud is $120$ .
The total number of employees in Security is $100$ .
The total number of employees in AI is $80$ .
The number of employees working in exactly two departments is $x$ .
The number of employees working in all three departments is $y$ .

If it is known that $x + y = 60$ and $x \geq 2 y$ , what is the sum of all possible integer values for the total number of employees in the company?

Correct Answer:

4830

Step-by-Step Solution

Key idea: This is a reverse_engineering problem combining Venn algebra with inequality constraints and arithmetic progression sums.

Step 1: Establish the Sum of Sets and Union relationship.

Sum of sets = $120 + 100 + 80 = 300$ .

Let $N$ be the total number of employees. Since everyone is in at least one department, Union = $N$ .

Formula: $\sum ∣ S e t s ∣ = E_{1} + 2 E_{2} + 3 E_{3}$ .

Also, Union = $E_{1} + E_{2} + E_{3} ⟹ E_{1} = N - E_{2} - E_{3}$ .

Substitute $E_{1}$ : $300 = (N - E_{2} - E_{3}) + 2 E_{2} + 3 E_{3} ⟹ 300 = N + E_{2} + 2 E_{3}$ .

So, $N = 300 - E_{2} - 2 E_{3}$ .

Step 2: Substitute the given constraints.

We are given $E_{2} = x$ and $E_{3} = y$ .

$N = 300 - x - 2 y$ .

We are also given $x + y = 60 ⟹ x = 60 - y$ .

Substitute $x$ : $N = 300 - (60 - y) - 2 y = 240 - y$ .

Step 3: Determine the valid range for $y$ .

Constraint 1: $x \geq 2 y ⟹ (60 - y) \geq 2 y ⟹ 60 \geq 3 y ⟹ y \leq 20$ .

Constraint 2: $y \geq 0$ (Region counts cannot be negative).

Constraint 3: $x \geq 0 ⟹ 60 - y \geq 0 ⟹ y \leq 60$ .

Combining these, the valid integer range for $y$ is $0 \leq y \leq 20$ .

Step 4: Calculate the sum of all possible values for $N$ .

Since $N = 240 - y$ , as $y$ ranges from $0$ to $20$ , $N$ ranges from $240$ down to $220$ .

This is an arithmetic progression with $21$ terms: $240, 239, 238, \dots, 220$ .

Sum = $\frac{Number of terms}{2} \times (First term + Last term)$

Sum = $\frac{21}{2} \times (240 + 220) = \frac{21}{2} \times 460 = 21 \times 230 = 4830$ .

Answer: 4830

Question 3 · Data Interpretation and Logical Reasoning NAT

Four analysts — P, Q, R, and S — process data batches over five consecutive days. Each day, every analyst processes an integer number of batches between 10 and 50. The daily total is the sum of all four analysts' outputs that day. The "Rolling Average" for Day

k

(

k > 1

) is defined as the average of the Daily Totals of Day

k - 1

, Day

k

, and Day

k + 1

. For Day 2 and Day 4, this 3-day rolling average is provided below. Note that Day 1 and Day 5 do not have rolling averages. Table 1: Rolling Averages

Day 2	Day 4
120	135

Additional Constraints: 1. The Daily Total strictly increased each day from Day 1 to Day 5. 2. On any given day, no two analysts processed the same number of batches. 3. Analyst P always processed more batches than Analyst Q but fewer than Analyst R on every single day. 4. The sum of all batches processed across all five days by all four analysts was exactly 680. 5. On Day 3, Analyst S processed exactly 20 batches. What is the maximum possible number of batches Analyst R could have processed on Day 3?

Correct Answer:

Step-by-Step Solution

Key idea: This is a bounded sum reconstruction problem combined with ranking constraints. We must first deduce the exact Daily Totals using the Rolling Average definition and the monotonicity constraint, then maximize one individual's score subject to strict ordering.

Step 1: Define variables and equations.

Let $T_{k}$ be the Daily Total on Day $k$ .

Rolling Average Day 2: $\frac{T _{1} + T _{2} + T _{3}}{3} = 120 \Rightarrow T_{1} + T_{2} + T_{3} = 360$ .

Rolling Average Day 4: $\frac{T _{3} + T _{4} + T _{5}}{3} = 135 \Rightarrow T_{3} + T_{4} + T_{5} = 405$ .

Grand Total: $\sum_{k = 1}^{5} T_{k} = 680$ .

Step 2: Solve for $T_{3}$ .

Sum of the two rolling sums: $(T_{1} + T_{2} + T_{3}) + (T_{3} + T_{4} + T_{5}) = 360 + 405 = 765$ .

This equals $(\sum T_{k}) + T_{3} = 680 + T_{3}$ .

Therefore, $680 + T_{3} = 765 \Rightarrow T_{3} = 85$ .

Step 3: Apply constraints to Day 3.

We know $T_{3} = 85$ . On Day 3, let scores be $P_{3}, Q_{3}, R_{3}, S_{3}$ .

Given $S_{3} = 20$ , so $P_{3} + Q_{3} + R_{3} = 85 - 20 = 65$ .

Ranking constraint: $Q_{3} < P_{3} < R_{3}$ . All are distinct integers.

Bounds: $10 \leq Q_{3}, P_{3}, R_{3} \leq 50$ .

Step 4: Maximize $R_{3}$ .

To maximize $R_{3}$ , we must minimize $P_{3} + Q_{3}$ subject to $Q_{3} < P_{3} < R_{3}$ and lower bound 10.

Minimum distinct integers for $Q_{3}, P_{3}$ are 10 and 11.

If $Q_{3} = 10$ and $P_{3} = 11$ , then $R_{3} = 65 - (10 + 11) = 44$ .

Check validity: Is $11 < 44$ ? Yes. Is $44 \leq 50$ ? Yes.

Thus, max $R_{3} = 44$ .

Answer: 44

Question 4 · Data Interpretation and Logical Reasoning MCQ

A company has 200 employees, split into two departments: Tech (120 employees) and HR (80 employees). They are surveyed on their adoption of 3 new software benefits: B1, B2, and B3.

For the Tech department:

50% opted for B1.
40% opted for B2.
30% opted for B3.
The number of Tech employees who opted for EXACTLY TWO benefits is equal to the number who opted for ALL THREE.

For the HR department:

75% opted for B1.
50% opted for B2.
25% opted for B3.
The number of HR employees who opted for AT LEAST ONE benefit is exactly 70.

Global Condition:

The total number of employees (across both departments combined) who opted for ALL THREE benefits is exactly 25.

What is the maximum possible number of Tech employees who opted for ONLY benefit B1?

A.
48
B.
50
C.
52
D.
54

Correct Answer:

Step-by-Step Solution

Key idea: This problem combines the Percentage Base Trap with hidden population bounds. We must convert percentages to absolute numbers, use the "At Least One $\leq$ Total Population" inequality to bound $E_{3}$ , and then apply minimize-to-maximize logic on the Venn regions.

Step 1: Convert percentages to absolute sums for Tech.

$N_{T} = 120$ .

$∣ B 1∣ = 50% of 120 = 60$ .

$∣ B 2∣ = 40% of 120 = 48$ .

$∣ B 3∣ = 30% of 120 = 36$ .

Sum $S_{T} = 60 + 48 + 36 = 144$ .

We are given $E_{2 T} = E_{3 T}$ .

Using $S_{T} - A_{T} = E_{2 T} + 2 E_{3 T}$ , where $A_{T}$ is the number of Tech employees in at least one benefit.

$144 - A_{T} = 3 E_{3 T}$ .

Since $A_{T} \leq N_{T} ⟹ A_{T} \leq 120$ .

Therefore, $3 E_{3 T} = 144 - A_{T} \geq 144 - 120 = 24 ⟹ E_{3 T} \geq 8$ .

Step 2: Analyze HR to find the global link.

$N_{H} = 80$ .

$∣ B 1∣ = 60, ∣ B 2∣ = 40, ∣ B 3∣ = 20 ⟹ S_{H} = 120$ .

We are given $A_{H} = 70$ .

$S_{H} - A_{H} = E_{2 H} + 2 E_{3 H} ⟹ 120 - 70 = 50 ⟹ E_{2 H} + 2 E_{3 H} = 50$ .

The global condition states $E_{3 T} + E_{3 H} = 25$ .

To maximize Tech ONLY B1, we must maximize $E_{1 T}$ . Since $E_{1 T} = A_{T} - 2 E_{3 T} = 144 - 5 E_{3 T}$ , we must MINIMIZE $E_{3 T}$ .

To minimize $E_{3 T}$ , we must MAXIMIZE $E_{3 H}$ .

From HR, $2 E_{3 H} \leq 50 ⟹ E_{3 H} \leq 25$ .

However, we also know $E_{3 T} \geq 8 ⟹ 25 - E_{3 H} \geq 8 ⟹ E_{3 H} \leq 17$ .

So the maximum valid $E_{3 H} = 17$ , which forces the minimum $E_{3 T} = 8$ .

Step 3: Maximize Tech ONLY B1 ( $X$ ).

With $E_{3 T} = 8$ , we have $E_{2 T} = 8$ .

$A_{T} = 144 - 3 (8) = 120$ .

$E_{1 T} = 120 - 8 - 8 = 104$ .

Let $a, b, c$ be the exactly-two regions. $a + b + c = 8$ .

The total for B1 is $X + a + c + 8 = 60 ⟹ X + a + c = 52$ .

To maximize $X$ , minimize $a + c$ . Since $a + c = 8 - b$ , we must maximize $b$ .

The total for B3 is $Z + b + c + 8 = 36 ⟹ Z + b + c = 28$ .

Since $Z \geq 0$ and $c \geq 0$ , and $a + b + c = 8$ , the absolute maximum for $b$ is 8.

If $b = 8$ , then $a = 0, c = 0$ .

This gives $a + c = 0$ , so $X = 52 - 0 = 52$ .

(Checking other regions: $Y = 40 - 8 = 32$ , $Z = 28 - 8 = 20$ . Sum $X + Y + Z = 52 + 32 + 20 = 104 = E_{1 T}$ . All valid).

Answer: 52

Question 5 · Data Interpretation and Logical Reasoning MCQ

When approaching a dense caselet with multiple hidden constraints in the exam, what is the recommended first step to organize the data?

A.
Start calculating the final answer immediately
B.
Guess the values based on the options provided
C.
Extract hidden constraints and organize them logically into a grid
D.
Memorize the entire paragraph before reading the questions

Correct Answer:

Step-by-Step Solution

Key idea: This is a framework question about exam execution. Dense caselets require structured organization before any calculation.

Step 1: Understand the nature of dense caselets. They contain hidden constraints scattered across the text.

Step 2: Identify the best approach. Jumping to calculations or guessing leads to errors. Memorizing wastes time.

Step 3: The correct approach is to extract constraints and build a clean, logical grid. This visual representation makes hidden connections obvious.

Answer: C

Caselets, Sets and Conditional Data Reasoning Practice Questions for CAT: 276+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Caselets, Sets and Conditional Data Reasoning

Chapter Journey

Puzzle Solving and Draw Competitions

Puzzle Solving and Draw Competitions

What you will learn here:

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

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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