Sets, Venn Diagrams and Functions Practice Questions for CAT: 38+ Solved Questions with Step-by-Step Solutions

Solve 38+ Sets, Venn Diagrams and Functions practice questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Sets, Venn Diagrams and Functions

Chapter Roadmap

Sets, Venn Diagrams and Functions

A compact Modern Maths chapter where CAT tests overlap counting and set-to-set assignment logic.

⭕

t1. Venn Diagrams and Inclusion-Exclusion

Master overlaps, exactly-one regions, all-three intersections, and max-min range questions.

3 PYQs
higher weight

🔁

t2. Onto Functions and Set Mappings

Count functions where every element in the target set must be hit at least once.

1 PYQ
support topic

By the end: you should be able to convert word data into regions, equations, and valid ranges without double-counting.

Venn Diagrams and Inclusion-Exclusion

Selected Topic

Venn Diagrams and Inclusion-Exclusion

Fix double-counting when groups overlap.

Chapter: Sets, Venn Diagrams and Functions Topic t1 3 direct CAT PYQs

What you will learn here

Draw and fill two-set and three-set Venn regions
Use inclusion-exclusion without double-counting
Handle “none”, “at least one”, and “all three”
Find minimum and maximum possible overlap
Convert CAT word conditions into equations

Sets, Venn Diagrams and Functions: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability MCQ

In a class of $100$ students, every student likes at least one of four sports: A, B, C, and D. $90$ like A, $85$ like B, $80$ like C, and $75$ like D. What is the difference between the maximum and minimum possible number of students who like EXACTLY three of these sports?

A.
70
B.
50
C.
60
D.
80

Correct Answer:

Step-by-Step Solution

Key idea: This is a "multi-set range" question. The signal is four overlapping sets with high totals, asking for the range of an "exactly k" region.

Step 1: Set up the core equations.

Let $x_{k}$ be the number of students in exactly $k$ sets. Since everyone likes at least one, $x_{0} = 0$ .

Total students: $x_{1} + x_{2} + x_{3} + x_{4} = 100$ . (Eq 1)

Sum of sets: $90 + 85 + 80 + 75 = 330$ .

$x_{1} + 2 x_{2} + 3 x_{3} + 4 x_{4} = 330$ . (Eq 2)

Step 2: Find the maximum $x_{3}$ .

Subtract $3 \times$ (Eq 1) from Eq 2:

$(x_{1} + 2 x_{2} + 3 x_{3} + 4 x_{4}) - 3 (x_{1} + x_{2} + x_{3} + x_{4}) = 330 - 300$

$- 2 x_{1} - x_{2} + x_{4} = 30 ⟹ x_{4} = 30 + 2 x_{1} + x_{2}$ .

Substitute $x_{4}$ into Eq 1:

$x_{1} + x_{2} + x_{3} + (30 + 2 x_{1} + x_{2}) = 100$

$x_{3} = 70 - 3 x_{1} - 2 x_{2}$ .

To maximize $x_{3}$ , minimize $x_{1}$ and $x_{2}$ . Let $x_{1} = 0, x_{2} = 0$ .

Max $x_{3} = 70$ . (This is valid as $x_{4} = 30$ , and individual sets can be formed).

Step 3: Find the minimum $x_{3}$ .

From $x_{3} = 70 - 3 x_{1} - 2 x_{2}$ , we minimize $x_{3}$ by maximizing $3 x_{1} + 2 x_{2}$ .

Notice that $3 x_{1} + 2 x_{2}$ represents the total number of "missing" elements (complements) if $x_{0} = 0$ .

The complements of the sets are: $A^{'} = 10, B^{'} = 15, C^{'} = 20, D^{'} = 25$ .

Total missing elements = $10 + 15 + 20 + 25 = 70$ .

Each student in $x_{1}$ misses 3 sets. Each in $x_{2}$ misses 2 sets.

So, $3 x_{1} + 2 x_{2} = 70$ .

Substitute this back into the $x_{3}$ equation:

$x_{3} = 70 - (3 x_{1} + 2 x_{2}) = 70 - 70 = 0$ .

Min $x_{3} = 0$ . (This is valid as we can distribute the 70 missing elements using only 2s and 3s).

Step 4: Calculate the difference.

Max $x_{3}$ - Min $x_{3} = 70 - 0 = 70$ .

Answer: 70.

Question 2 · Quantitative Ability MCQ

In a class of $100$ students, exactly $5$ students like none of three genres: Fiction, Non-fiction and Poetry. $70$ students like Fiction, $65$ like Non-fiction and $55$ like Poetry. What is the difference between the maximum and the minimum possible number of students who like all three genres?

A.
45
B.
47
C.
48
D.
55

Correct Answer:

Step-by-Step Solution

Key idea: this is a three-set max-min question with a FIXED union, recognisable because the number who like none is pinned at exactly 5, so the union is exactly 95 rather than flexible.

Step 1: Write the region equations. Let $E_{1}, E_{2}, t$ be the counts of students in exactly one, exactly two and all three genres. The union is $95$ , so $E_{1} + E_{2} + t = 95$ .

Step 2: Add the set sizes. Each exactly-two student is counted twice and each all-three student thrice, so $E_{1} + 2 E_{2} + 3 t = 70 + 65 + 55 = 190$ .

Step 3: Subtract the first equation from the second: $E_{2} + 2 t = 95$ .

Step 4: Maximise $t$ . Since $E_{2} \geq 0$ , $2 t \leq 95$ , so $t \leq 47.5$ , giving $t_{ma x} = 47$ (integer). Construction: $E_{2} = 1$ , $E_{1} = 47$ ; put the single exactly-two person in Fiction-Poetry, then Fiction-only $= 22$ , Non-fiction-only $= 18$ , Poetry-only $= 7$ , which sums to $47$ . Valid.

Step 5: Minimise $t$ . Try $t = 0$ : then $E_{2} = 95$ and $E_{1} = 0$ . Solve the pair splits $x_{F N} + x_{N P} + x_{F P} = 95$ against the set sizes: $x_{F P} = 40$ , $x_{F N} = 30$ ... checking: Fiction needs $x_{F N} + x_{F P} = 70$ , Non-fiction $x_{F N} + x_{N P} = 65$ , Poetry $x_{N P} + x_{F P} = 55$ ; solving gives $x_{F N} = 40$ , $x_{N P} = 25$ , $x_{F P} = 30$ , all non-negative. Valid, so $t_{min} = 0$ .

Step 6: Difference $= 47 - 0 = 47$ .

Answer: B.

Common trap: taking the maximum as the smallest set ( $55$ ). That bound only applies when the union can stretch to absorb it; here the union is fixed at $95$ , and the surplus equation caps $t$ at $47$ .

Question 3 · Quantitative Ability MCQ

In a survey of $100$ people, $70$ like Coffee, $80$ like Tea, and $60$ like Lemonade. Some people may like none of these three drinks. What is the maximum possible number of people who like EXACTLY two of these drinks?

A.
80
B.
90
C.
85
D.
95

Correct Answer:

Step-by-Step Solution

Key idea: This is a "three-set max-min with flexible union" question. The signal is the request to maximize the "exactly two" region while allowing the "none" region to vary.

Step 1: Set up the exact region equations.

Let $x, y, z, w$ be the number of people in exactly 1, exactly 2, exactly 3, and none of the sets.

$x + y + z + w = 100$

Sum of sets: $x + 2 y + 3 z = 70 + 80 + 60 = 210$ .

Step 2: Express $y$ in terms of the other variables.

From the first equation, $x = 100 - y - z - w$ .

Substitute this into the sum of sets equation:

$(100 - y - z - w) + 2 y + 3 z = 210$

$100 + y + 2 z - w = 210 ⟹ y = 110 - 2 z + w$ .

Step 3: Maximize $y$ by bounding $z$ and $w$ .

To maximize $y$ , we want to maximize $w$ and minimize $z$ .

However, we must ensure $x \geq 0$ .

Substitute $y$ back into the expression for $x$ :

$x = 100 - (110 - 2 z + w) - z - w = - 10 + z - 2 w$ .

For $x \geq 0$ , we need $z - 2 w \geq 10 ⟹ z \geq 10 + 2 w$ .

Step 4: Find the absolute maximum.

Substitute the minimum bound for $z$ into the equation for $y$ :

$y \leq 110 - 2 (10 + 2 w) + w = 110 - 20 - 4 w + w = 90 - 3 w$ .

To maximize $y$ , we must minimize $w$ . Let $w = 0$ .

Then $z \geq 10$ , so the minimum $z$ is $10$ .

This gives a maximum $y = 90 - 0 = 90$ .

(This configuration is valid: $x = 0, y = 90, z = 10, w = 0$ ).

Answer: 90.

Question 4 · Quantitative Ability NAT

In a survey of $100$ people, $60$ like Tea, $70$ like Coffee, and $80$ like Milk. The survey report states "some people may like none of these". What is the difference between the maximum and minimum possible number of people who like all three drinks?

Correct Answer:

Step-by-Step Solution

Key idea: this is a three-set max-min question where the "none" condition is given as a possibility rather than a fixed number. This means the union of the three sets can vary, which affects the minimum intersection bound.

Step 1: Find the maximum of the triple intersection.

The triple intersection can never exceed the smallest individual set.

Max $∣ T \cap C \cap M ∣ \leq min (60, 70, 80) = 60$ .

This is achievable: let all $60$ Tea likers also like Coffee and Milk. The remaining $10$ Coffee likers and $20$ Milk likers can be distinct, using $90$ people total. The remaining $10$ people like none. This satisfies "some may like none". So Max = $60$ .

Step 2: Find the minimum of the triple intersection.

The bounding formula for three sets is:

$∣ T \cap C \cap M ∣ \geq ∣ T ∣ + ∣ C ∣ + ∣ M ∣ - 2 \times ∣ T \cup C \cup M ∣$ .

To MINIMIZE the intersection, we must MAXIMIZE the union $∣ T \cup C \cup M ∣$ .

Since "some may like none", the union can be at most the total population, $100$ .

Min $∣ T \cap C \cap M ∣ \geq 60 + 70 + 80 - 2 (100) = 210 - 200 = 10$ .

So Min = $10$ .

Step 3: Compute the difference.

Difference = Max - Min = $60 - 10 = 50$ .

Answer: $50$ .

Common trap: assuming "some may like none" means the union MUST be strictly less than $100$ , or using the formula with union $= 100$ for the maximum instead of the minimum.

Question 5 · Quantitative Ability MCQ

In a class of $120$ students, every student likes at least one of three subjects: Physics, Mathematics, and Chemistry. $60$ students like Physics, $80$ like Mathematics, and $50$ like Chemistry. If at least $10$ students like all three subjects, which of the following CANNOT be the number of students who like EXACTLY two subjects?

A.
48
B.
32
C.
24
D.
15

Correct Answer:

Step-by-Step Solution

Key idea: This is an "algebraic Venn parity" question. The signal is the sum of sets and a condition on the "exactly three" region, asking for an impossible value of "exactly two".

Step 1: Set up the exact region equations.

Let $x, y, z$ be the number of students who like exactly 1, 2, and 3 subjects respectively.

Since everyone likes at least one, $x + y + z = 120$ .

The sum of the individual set sizes is $60 + 80 + 50 = 190$ .

This gives the equation: $x + 2 y + 3 z = 190$ .

Step 2: Eliminate $x$ to link $y$ and $z$ .

Subtract the first equation from the second:

$(x + 2 y + 3 z) - (x + y + z) = 190 - 120$

$y + 2 z = 70 ⟹ y = 70 - 2 z$ .

Step 3: Apply the constraints.

We are given $z \geq 10$ .

Since $z$ is an integer, $2 z$ is always an even number.

Therefore, $y = 70 - even = even$ .

The number of students who like exactly two subjects ( $y$ ) MUST be an even number.

Step 4: Check the options.

48, 32, and 24 are all even numbers (and fall within the valid range $0 \leq y \leq 50$ ).

15 is an odd number, which is mathematically impossible given the parity constraint.

Answer: 15.

Sets, Venn Diagrams and Functions Practice Questions for CAT: 38+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Sets, Venn Diagrams and Functions

Sets, Venn Diagrams and Functions

Venn Diagrams and Inclusion-Exclusion

Venn Diagrams and Inclusion-Exclusion

Sets, Venn Diagrams and Functions: 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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