Sets, Venn Diagrams and Functions Short Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Sets, Venn Diagrams and Functions short notes for CAT: 29 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Why Venn Diagrams Exist

A Venn diagram is a “count once” machine

When groups overlap, a person can belong to more than one group. Venn diagrams split the universe into non-overlapping regions.

The goal is not drawing circles beautifully. The goal is counting every person exactly once.

Two-Set Inclusion-Exclusion

The first overlap formula

When two groups overlap, adding their totals counts the common part twice.

n (A \cup B) = n (A) + n (B) - n (A \cap B)

At least one:

n (A \cup B)

Both:

n (A \cap B)

Neither: total universe minus

n (A \cup B)

Two-Set Max-Min Overlap

How small or large can “both” be?

For two sets $A$ and $B$ inside a universe of size $N$ :

Minimum overlap

max (0, n (A) + n (B) - N)

Forced overlap when totals exceed the universe.

Maximum overlap

min (n (A), n (B))

The smaller set can fit inside the larger one.

CAT often hides this inside percentages or gender-wise data.

PYQ: Swimming and Running with Boys and Girls

CAT PYQ PATTERN

Minimum and maximum students who chose both

Let girls be $G$ . Then boys are

250 - G

Girls are between

44%

and

60%

250

, so

110 \leq G \leq 150

Swimming:

S = 0.5 (250 - G) + 0.8 G = 125 + 0.3 G

Running:

R = 0.7 (250 - G) + 0.6 G = 175 - 0.1 G

Since every student chose swimming or running or both:

n (S \cap R) = S + R - 250 = 50 + 0.2 G

For all percentage counts to be integers,

G

can be

110, 120, 130, 140, 150

.
Therefore both ranges from:

50 + 0.2 (110) = 72 to 50 + 0.2 (150) = 80

Answer: minimum

72

, maximum

80

Sets, Venn Diagrams and Functions: Solved Questions with Step-by-Step Explanations (2 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 $47$ .

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Sets, Venn Diagrams and Functions Short Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Sets, Venn Diagrams and Functions short notes for CAT: 29 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

A question from this chapter

Question 1

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?

Question 2

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?

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