Mathematics

Class 92025–26 SNC35 Questions SolvedEnglish Medium

Exercise 3.2 — Class 9 Mathematics Solutions

All 35 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Exercise 3.2

13 Questions (22 Sub-parts)

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Q1Question 1

Medium

Consider the universal set $U = {x | x is multiple of 2 and 0 < x \leq 30}$ , $A = {x | x is a multiple of 6}$ and $B = {x | x is a multiple of 8}$

Q1Part (i)

EasyApproved by Miss Alisha • Aug 30, 2026

List all elements of sets $A$ and $B$ in tabular form

Hint

Multiples of $6$ in $U$ : $A = {6, 12, 18, 24, 30}$ ; Multiples of $8$ in $U$ : $B = {8, 16, 24}$ .

Solution

Write the Universal set $U$ in tabular form.

The universal set consists of all even positive integers up to $30$ :
$U = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30}$
Identify elements of set $A$ (multiples of $6$ in $U$ ).
$A = {6, 12, 18, 24, 30}$
Identify elements of set $B$ (multiples of $8$ in $U$ ).
$B = {8, 16, 24}$

Answer

$A = {6, 12, 18, 24, 30}, B = {8, 16, 24}$

Q1Part (ii)

EasyApproved by Miss Alisha • Aug 30, 2026

Find $A \cap B$

Hint

Find common elements of $A$ and $B$ : $A \cap B = {24}$ .

Solution

State the definition of set intersection.

The intersection $A \cap B$ is the set containing all elements common to both $A$ and $B$ .
Evaluate $A \cap B$ .
$\begin{matrix} A \cap B & = {6, 12, 18, 24, 30} \cap {8, 16, 24} \\ = {24} \end{matrix}$

Answer

$A \cap B = {24}$

Q1Part (iii)

MediumApproved by Miss Alisha • Aug 30, 2026

Draw a Venn diagram

Hint

Draw universal set rectangle $U$ with overlapping circles for $A$ and $B$ , with $24$ in the intersection region.

Solution

Determine the elements in each region.
- Intersection $A \cap B$ : ${24}$
- Only $A$ ( $A ∖ B$ ): ${6, 12, 18, 30}$
- Only $B$ ( $B ∖ A$ ): ${8, 16}$
- Outside $A \cup B$ in $U$ : ${2, 4, 10, 14, 20, 22, 26, 28}$
Venn Diagram Representation.

Venn Diagram of Sets A and B in Uvenn diagram
Figure: Set A (multiples of 6), Set B (multiples of 8), with 24 in intersection and remaining even numbers outside.

Answer

$A \cap B = {24}, A ∖ B = {6, 12, 18, 30}, B ∖ A = {8, 16}, (A \cup B)^{'} = {2, 4, 10, 14, 20, 22, 26, 28}$

Q2Question 2

Medium

Let $U = {x | x is an integer and 0 < x \leq 150}$ , $G = {x | x = 2^{m} for integer m and 0 \leq m \leq 12}$ and $H = {x | x is a square}$

Q2Part (i)

EasyApproved by Miss Alisha • Aug 30, 2026

List all elements of sets $G$ and $H$ in tabular form

Hint

$G = {1, 2, 4, 8, 16, 32, 64, 128}$ and $H = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}$ .

Solution

Understand the domain $U$ .
$U = {1, 2, 3, \dots, 150}$
List elements of $G$ (powers of $2$ within $U$ ).
$\begin{matrix} 2^{0} & = 1 \\ 2^{1} & = 2 \\ 2^{2} & = 4 \\ 2^{3} & = 8 \\ 2^{4} & = 16 \\ 2^{5} & = 32 \\ 2^{6} & = 64 \\ 2^{7} & = 128 \end{matrix}$
(Note: $2^{8} = 256 > 150$ , so it is excluded).
$G = {1, 2, 4, 8, 16, 32, 64, 128}$
List elements of $H$ (perfect squares within $U$ ).
$\begin{matrix} 1^{2} = 1, 2^{2} = 4, 3^{2} = 9, 4^{2} = 16, 5^{2} = 25, 6^{2} = 36, \\ 7^{2} = 49, 8^{2} = 64, 9^{2} = 81, 10^{2} = 100, 11^{2} = 121, 12^{2} = 144 \end{matrix}$ $H = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}$

Answer

$G = {1, 2, 4, 8, 16, 32, 64, 128}, H = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}$

Q2Part (ii)

EasyApproved by Miss Alisha • Aug 30, 2026

Find $G \cup H$

Hint

Combine all unique elements of $G$ and $H$ .

Solution

Combine all unique elements from sets $G$ and $H$ .
$\begin{matrix} G & = {1, 2, 4, 8, 16, 32, 64, 128} \\ H & = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144} \end{matrix}$
Evaluate $G \cup H$ .
$\begin{matrix} G \cup H & = {1, 2, 4, 8, 16, 32, 64, 128} \cup {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144} \\ = {1, 2, 4, 8, 9, 16, 25, 32, 36, 49, 64, 81, 100, 121, 128, 144} \end{matrix}$

Answer

$G \cup H = {1, 2, 4, 8, 9, 16, 25, 32, 36, 49, 64, 81, 100, 121, 128, 144}$

Q2Part (iii)

EasyApproved by Miss Alisha • Aug 30, 2026

Find $G \cap H$

Hint

Common elements that are both powers of $2$ and perfect squares: $G \cap H = {1, 4, 16, 64}$ .

Solution

Find the common elements between $G$ and $H$ .

Common elements are powers of $2$ that are also perfect squares:
$\begin{matrix} 1 & = 2^{0} = 1^{2} \\ 4 & = 2^{2} = 2^{2} \\ 16 & = 2^{4} = 4^{2} \\ 64 & = 2^{6} = 8^{2} \end{matrix}$
Evaluate $G \cap H$ .
$\begin{matrix} G \cap H & = {1, 2, 4, 8, 16, 32, 64, 128} \cap {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144} \\ = {1, 4, 16, 64} \end{matrix}$

Answer

$G \cap H = {1, 4, 16, 64}$

Q3Question 3

Medium

Consider the sets $P = {x | x is a prime number and 0 < x < 20}$ and $Q = {x | x is a divisor of 210 and 0 < x < 20}$

Q3Part (i)

EasyApproved by Miss Alisha • Aug 30, 2026

Find $P \cap Q$

Hint

$P = {2, 3, 5, 7, 11, 13, 17, 19}$ , $Q = {1, 2, 3, 5, 6, 7, 10, 14, 15}$ . Then $P \cap Q = {2, 3, 5, 7}$ .

Solution

Write sets $P$ and $Q$ in tabular form.
- $P$ is the set of prime numbers strictly less than $20$ :
P={2,3,5,7,11,13,17,19}
- $Q$ is the set of positive divisors of $210$ strictly less than $20$ :
Q={1,2,3,5,6,7,10,14,15}
Evaluate $P \cap Q$ .
$\begin{matrix} P \cap Q & = {2, 3, 5, 7, 11, 13, 17, 19} \cap {1, 2, 3, 5, 6, 7, 10, 14, 15} \\ = {2, 3, 5, 7} \end{matrix}$

Answer

$P \cap Q = {2, 3, 5, 7}$

Q3Part (ii)

EasyApproved by Miss Alisha • Aug 30, 2026

Find $P \cup Q$

Hint

Combine all elements from $P$ and $Q$ : $P \cup Q = {1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19}$ .

Solution

Combine all elements of $P$ and $Q$ .
$\begin{matrix} P & = {2, 3, 5, 7, 11, 13, 17, 19} \\ Q & = {1, 2, 3, 5, 6, 7, 10, 14, 15} \end{matrix}$
Evaluate $P \cup Q$ .
$\begin{matrix} P \cup Q & = {2, 3, 5, 7, 11, 13, 17, 19} \cup {1, 2, 3, 5, 6, 7, 10, 14, 15} \\ = {1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19} \end{matrix}$

Answer

$P \cup Q = {1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19}$

Q4Question 4

Medium

Verify the commutative properties of union and intersection for the following pairs of sets:

Q4Part (i)

EasyApproved by Miss Alisha • Aug 30, 2026

$A = {1, 2, 3, 4, 5}, B = {4, 6, 8, 10}$

Hint

Verify $A \cup B = B \cup A = {1, 2, 3, 4, 5, 6, 8, 10}$ and $A \cap B = B \cap A = {4}$ .

Solution

(a) Verification of Commutative Property of Union: $A \cup B = B \cup A$

Left Hand Side (L.H.S):

\begin{matrix} L.H.S & = A \cup B \\ = {1, 2, 3, 4, 5} \cup {4, 6, 8, 10} \\ = {1, 2, 3, 4, 5, 6, 8, 10} — (1) \end{matrix}

Right Hand Side (R.H.S):

\begin{matrix} R.H.S & = B \cup A \\ = {4, 6, 8, 10} \cup {1, 2, 3, 4, 5} \\ = {1, 2, 3, 4, 5, 6, 8, 10} — (2) \end{matrix}

From (1) and (2), $L.H.S = R.H.S$ . Hence, $A \cup B = B \cup A$ .

(b) Verification of Commutative Property of Intersection: $A \cap B = B \cap A$

Left Hand Side (L.H.S):

\begin{matrix} L.H.S & = A \cap B \\ = {1, 2, 3, 4, 5} \cap {4, 6, 8, 10} \\ = {4} — (3) \end{matrix}

Right Hand Side (R.H.S):

\begin{matrix} R.H.S & = B \cap A \\ = {4, 6, 8, 10} \cap {1, 2, 3, 4, 5} \\ = {4} — (4) \end{matrix}

From (3) and (4), $L.H.S = R.H.S$ . Hence, $A \cap B = B \cap A$ .

Answer

$A \cup B = B \cup A = {1, 2, 3, 4, 5, 6, 8, 10} and A \cap B = B \cap A = {4}$

Q4Part (ii)

EasyApproved by Miss Alisha • Aug 30, 2026

$ℕ, ℤ$

Hint

Verify $ℕ \cup ℤ = ℤ \cup ℕ = ℤ$ and $ℕ \cap ℤ = ℤ \cap ℕ = ℕ$ .

Solution

Given sets: $ℕ = {1, 2, 3, 4, \dots}$ and $ℤ = {0, \pm 1, \pm 2, \pm 3, \dots}$ . Notice that $ℕ \subset ℤ$ .

(a) Verification of Commutative Property of Union: $ℕ \cup ℤ = ℤ \cup ℕ$

L.H.S:

\begin{matrix} L.H.S & = ℕ \cup ℤ \\ = ℤ — (1) \end{matrix}

R.H.S:

\begin{matrix} R.H.S & = ℤ \cup ℕ \\ = ℤ — (2) \end{matrix}

From (1) and (2), $L.H.S = R.H.S$ . Hence, $ℕ \cup ℤ = ℤ \cup ℕ$ .

(b) Verification of Commutative Property of Intersection: $ℕ \cap ℤ = ℤ \cap ℕ$

L.H.S:

\begin{matrix} L.H.S & = ℕ \cap ℤ \\ = ℕ — (3) \end{matrix}

R.H.S:

\begin{matrix} R.H.S & = ℤ \cap ℕ \\ = ℕ — (4) \end{matrix}

From (3) and (4), $L.H.S = R.H.S$ . Hence, $ℕ \cap ℤ = ℤ \cap ℕ$ .

Answer

$ℕ \cup ℤ = ℤ \cup ℕ = ℤ and ℕ \cap ℤ = ℤ \cap ℕ = ℕ$

Q4Part (iii)

EasyApproved by Miss Alisha • Aug 30, 2026

$A = {x | x \in ℝ \land x \geq 0}, B = ℝ$

Hint

Verify $A \cup B = B \cup A = ℝ$ and $A \cap B = B \cap A = A$ .

Solution

Given sets: $A$ is the set of all non-negative real numbers and $B = ℝ$ is the set of all real numbers. Notice that $A \subset B$ .

(a) Verification of Commutative Property of Union: $A \cup B = B \cup A$

L.H.S:

\begin{matrix} L.H.S & = A \cup B \\ = B = ℝ — (1) \end{matrix}

R.H.S:

\begin{matrix} R.H.S & = B \cup A \\ = B = ℝ — (2) \end{matrix}

From (1) and (2), $L.H.S = R.H.S = ℝ$ .

(b) Verification of Commutative Property of Intersection: $A \cap B = B \cap A$

L.H.S:

\begin{matrix} L.H.S & = A \cap B \\ = A — (3) \end{matrix}

R.H.S:

\begin{matrix} R.H.S & = B \cap A \\ = A — (4) \end{matrix}

From (3) and (4), $L.H.S = R.H.S = A$ .

Answer

$A \cup B = B \cup A = ℝ and A \cap B = B \cap A = A$

Question 5

MediumApproved by Miss Alisha • Aug 30, 2026

Let $U = {a, b, c, d, e, f, g, h, i, j}$ , $A = {a, b, c, d, g, h}$ , $B = {c, d, e, f, j}$ , Verify De Morgan's Laws for these sets. Draw Venn diagram.

Hint

Verify (1) $(A \cup B)^{'} = A^{'} \cap B^{'} = {i}$ and (2) $(A \cap B)^{'} = A^{'} \cup B^{'} = {a, b, e, f, g, h, i, j}$ .

Solution

De Morgan's Laws to verify:
1. $(A \cup B)^{'} = A^{'} \cap B^{'}$
2. $(A \cap B)^{'} = A^{'} \cup B^{'}$
Part 1 — Verification of $(A \cup B)^{'} = A^{'} \cap B^{'}$

Compute Left Hand Side $L.H.S = (A \cup B)^{'}$ :
A∪B={a,b,c,d,g,h}∪{c,d,e,f,j}={a,b,c,d,e,f,g,h,j}
Now, taking the complement with respect to $U$ :
(A∪B)′=U∖(A∪B)={a,b,c,d,e,f,g,h,i,j}∖{a,b,c,d,e,f,g,h,j}={i}— (1)
Compute Right Hand Side $R.H.S = A^{'} \cap B^{'}$ :
$\begin{matrix} A^{'} & = U ∖ A = {e, f, i, j} \\ B^{'} & = U ∖ B = {a, b, g, h, i} \\ A^{'} \cap B^{'} & = {e, f, i, j} \cap {a, b, g, h, i} \\ = {i} — (2) \end{matrix}$
From (1) and (2), $L.H.S = R.H.S = {i}$ . Hence, $(A \cup B)^{'} = A^{'} \cap B^{'}$ is verified.

Part 2 — Verification of $(A \cap B)^{'} = A^{'} \cup B^{'}$
Compute Left Hand Side $L.H.S = (A \cap B)^{'}$ :
$\begin{matrix} A \cap B & = {a, b, c, d, g, h} \cap {c, d, e, f, j} \\ = {c, d} \end{matrix}$
Now, taking the complement:
$\begin{matrix} (A \cap B)^{'} & = U ∖ (A \cap B) \\ = {a, b, c, d, e, f, g, h, i, j} ∖ {c, d} \\ = {a, b, e, f, g, h, i, j} — (3) \end{matrix}$
Compute Right Hand Side $R.H.S = A^{'} \cup B^{'}$ :
$\begin{matrix} A^{'} \cup B^{'} & = {e, f, i, j} \cup {a, b, g, h, i} \\ = {a, b, e, f, g, h, i, j} — (4) \end{matrix}$
From (3) and (4), $L.H.S = R.H.S = {a, b, e, f, g, h, i, j}$ . Hence, $(A \cap B)^{'} = A^{'} \cup B^{'}$ is verified.

Part 3 — Venn Diagram Verification

Venn Diagrams: De Morgan's Laws Verificationvenn diagram
The shaded outer region shows (A ∪ B)' = A' ∩ B' = {i}.
The shaded area (everything except intersection {c, d}) represents (A ∩ B)' = A' ∪ B' = {a, b, e, f, g, h, i, j}.

Answer

$(A \cup B)^{'} = A^{'} \cap B^{'} = {i} and (A \cap B)^{'} = A^{'} \cup B^{'} = {a, b, e, f, g, h, i, j}$

Q6Question 6

Easy

If $U = {1, 2, 3, \dots, 20}$ and $A = {1, 3, 5, \dots, 19}$ , verify the following:

Q6Part (i)

EasyApproved by Miss Alisha • Aug 30, 2026

$A \cup A^{'} = U$

Hint

Find $A^{'} = U ∖ A = {2, 4, 6, \dots, 20}$ , then take union with $A$ to get $U$ .

Solution

Find the complement $A^{'}$ .
$\begin{matrix} A^{'} & = U ∖ A \\ = {1, 2, 3, \dots, 20} ∖ {1, 3, 5, \dots, 19} \\ = {2, 4, 6, \dots, 20} \end{matrix}$
Evaluate $A \cup A^{'}$ .
$\begin{matrix} L.H.S & = A \cup A^{'} \\ = {1, 3, 5, \dots, 19} \cup {2, 4, 6, \dots, 20} \\ = {1, 2, 3, 4, \dots, 20} \\ = U \\ = R.H.S \end{matrix}$
Hence proved.

Answer

$A \cup A^{'} = {1, 2, 3, \dots, 20} = U$

Q6Part (ii)

EasyApproved by Miss Alisha • Aug 30, 2026

$A \cap U = A$

Hint

The intersection of any subset $A$ with universal set $U$ is $A$ .

Solution

Evaluate $A \cap U$ .

\begin{matrix} L.H.S & = A \cap U \\ = {1, 3, 5, \dots, 19} \cap {1, 2, 3, \dots, 20} \\ = {1, 3, 5, \dots, 19} \\ = A \\ = R.H.S \end{matrix}

Hence proved.

Answer

$A \cap U = A$

Q6Part (iii)

EasyApproved by Miss Alisha • Aug 30, 2026

$A \cap A^{'} = ϕ$

Hint

A set $A$ and its complement $A^{'}$ share no elements, so $A \cap A^{'} = \emptyset$ .

Solution

Evaluate $A \cap A^{'}$ .

\begin{matrix} L.H.S & = A \cap A^{'} \\ = {1, 3, 5, \dots, 19} \cap {2, 4, 6, \dots, 20} \\ = \emptyset = ϕ \\ = R.H.S \end{matrix}

Hence proved.

Answer

$A \cap A^{'} = \emptyset = ϕ$

Question 7

MediumApproved by Miss Alisha • Aug 30, 2026

In a class of 55 students, 34 like to play cricket and 30 like to play hockey. Also each student likes to play at least one of the two games. How many students like to play both games?

Hint

Use the inclusion-exclusion principle: $n (C \cup H) = n (C) + n (H) - n (C \cap H) ⟹ 55 = 34 + 30 - n (C \cap H) ⟹ n (C \cap H) = 9$ .

Solution

Define the sets and state given values.

Let:
- $C$ = set of students who like to play cricket
- $H$ = set of students who like to play hockey
Given data:
n(C∪H)=55n(C)=34n(H)=30
Let $x = n (C \cap H)$ represent the number of students who like to play both games.
Apply the Principle of Inclusion-Exclusion for two sets.
$n (C \cup H) = n (C) + n (H) - n (C \cap H)$
Substitute and solve for $x$ .
$\begin{matrix} 55 & = 34 + 30 - x \\ 55 & = 64 - x \\ x & = 64 - 55 \\ x & = 9 \end{matrix}$
Regional distribution breakdown:
- Only Cricket: $34 - 9 = 25$
- Only Hockey: $30 - 9 = 21$
- Both Cricket & Hockey: $9$
- Total: $25 + 9 + 21 = 55$
Sports Preference Distribution (Cricket vs Hockey)venn diagram
Figure: 25 like Cricket only, 21 like Hockey only, and 9 like both games.

Answer

$9$ students

Question 8

HardApproved by Miss Alisha • Aug 30, 2026

In a group of 500 employees, 250 can speak Urdu, 150 can speak English, 50 can speak Punjabi, 40 can speak Urdu and English, 30 can speak both English and Punjabi, and 10 can speak Urdu and Punjabi. How many can speak all three languages?

Hint

Use 3-set inclusion-exclusion formula: $n (U \cup E \cup P) = n (U) + n (E) + n (P) - n (U \cap E) - n (E \cap P) - n (U \cap P) + n (U \cap E \cap P) ⟹ 500 = 370 + x ⟹ x = 130$ .

Solution

Define the sets and state given values.

Let:
- $U$ = set of employees who speak Urdu
- $E$ = set of employees who speak English
- $P$ = set of employees who speak Punjabi
Given data:
n(U∪E∪P)=500n(U)=250n(E)=150n(P)=50n(U∩E)=40n(E∩P)=30n(U∩P)=10
Let $x = n (U \cap E \cap P)$ be the number of employees who speak all three languages.
Apply the Principle of Inclusion-Exclusion for three sets.
$n (U \cup E \cup P) = n (U) + n (E) + n (P) - n (U \cap E) - n (E \cap P) - n (U \cap P) + n (U \cap E \cap P)$
Substitute the values and solve for $x$ .
$\begin{matrix} 500 & = 250 + 150 + 50 - 40 - 30 - 10 + x \\ 500 & = 450 - 80 + x \\ 500 & = 370 + x \\ x & = 500 - 370 \\ x & = 130 \end{matrix}$
Conclude.

Therefore, $130$ employees can speak all three languages.

Answer

$130$ employees

Question 9

HardApproved by Miss Alisha • Aug 30, 2026

In sports events, 19 people wear blue shirts, 15 wear green shirts, 3 wear blue and green shirts, 4 wear a cap and blue shirts, and 2 wear a cap and green shirts. The total number of people with either a blue or green shirt or cap is 34. How many people are wearing caps?

Hint

Apply the 3-set formula with $n (B \cap G \cap C) = 0$ : $34 = 19 + 15 + x - 3 - 4 - 2 + 0 ⟹ 34 = 25 + x ⟹ x = 9$ .

Solution

Define the sets and list given data.

Let:
- $B$ = set of people wearing blue shirts
- $G$ = set of people wearing green shirts
- $C$ = set of people wearing caps
Given data:
n(B)=19n(G)=15n(B∩G)=3n(C∩B)=4n(C∩G)=2n(B∪G∪C)=34n(B∩G∩C)=0(no one wears all three items)
Let $x = n (C)$ be the total number of people wearing caps.
Apply the Principle of Inclusion-Exclusion for three sets.
$n (B \cup G \cup C) = n (B) + n (G) + n (C) - n (B \cap G) - n (C \cap B) - n (C \cap G) + n (B \cap G \cap C)$
Substitute the values and solve for $x$ .
$\begin{matrix} 34 & = 19 + 15 + x - 3 - 4 - 2 + 0 \\ 34 & = 34 + x - 9 \\ 34 & = 25 + x \\ x & = 34 - 25 \\ x & = 9 \end{matrix}$
Conclude.

Therefore, $9$ people are wearing caps.

Answer

$9$ people

Question 10

HardApproved by Miss Alisha • Aug 30, 2026

In a training session, 17 participants have laptops, 11 have tablets, 9 have laptops and tablets, 6 have laptops and books, and 4 have both tablets and books. Four participants have all three items. The total number of participants with laptops, tablets, or books is 35. How many participants have books?

Hint

Use $n (L \cup T \cup B) = n (L) + n (T) + n (B) - n (L \cap T) - n (L \cap B) - n (T \cap B) + n (L \cap T \cap B) ⟹ 35 = 13 + x ⟹ x = 22$ .

Solution

Define the sets and list given data.

Let:
- $L$ = set of participants having laptops
- $T$ = set of participants having tablets
- $B$ = set of participants having books
Given data:
n(L)=17n(T)=11n(L∩T)=9n(L∩B)=6n(T∩B)=4n(L∩T∩B)=4n(L∪T∪B)=35
Let $x = n (B)$ be the total number of participants having books.
Apply the Principle of Inclusion-Exclusion for three sets.
$n (L \cup T \cup B) = n (L) + n (T) + n (B) - n (L \cap T) - n (L \cap B) - n (T \cap B) + n (L \cap T \cap B)$
Substitute the values and solve for $x$ .
$\begin{matrix} 35 & = 17 + 11 + x - 9 - 6 - 4 + 4 \\ 35 & = 28 - 19 + 4 + x \\ 35 & = 9 + 4 + x \\ 35 & = 13 + x \\ x & = 35 - 13 \\ x & = 22 \end{matrix}$
Conclude.

Therefore, $22$ participants have books.

Answer

$22$ participants

Q11Question 11

Hard

A shopping mall has 150 employees labelled 1 to 150, representing the Universal set $U$ . The employees fall into the following categories: • Set A: 40 employees with a salary range of 30k-45k, labelled from 50 to 89. • Set B: 50 employees with a salary range of 50k-80k, labelled from 101 to 150. • Set C: 60 employees with a salary range of 100k-150k, labelled from 1 to 49 and 90 to 100.

Q11Part (a)

MediumApproved by Miss Alisha • Aug 30, 2026

Find $(A^{'} \cup B^{'}) \cap C$

Hint

By De Morgan's Law, $A^{'} \cup B^{'} = (A \cap B)^{'}$ . Since $A \cap B = \emptyset$ , $(A \cap B)^{'} = U$ , so $U \cap C = C$ .

Solution

Write the sets in tabular form.
$\begin{matrix} U & = {1, 2, 3, \dots, 150} \\ A & = {50, 51, 52, \dots, 89} (n (A) = 40) \\ B & = {101, 102, 103, \dots, 150} (n (B) = 50) \\ C & = {1, 2, \dots, 49, 90, 91, \dots, 100} (n (C) = 60) \end{matrix}$
Find $A^{'}$ and $B^{'}$ .
$\begin{matrix} A^{'} & = U ∖ A = {1, 2, \dots, 49, 90, 91, \dots, 150} \\ B^{'} & = U ∖ B = {1, 2, \dots, 100} \end{matrix}$
Find $A^{'} \cup B^{'}$ .
$\begin{matrix} A^{'} \cup B^{'} & = {1, 2, \dots, 49, 90, 91, \dots, 150} \cup {1, 2, \dots, 100} \\ = {1, 2, 3, \dots, 150} = U \end{matrix}$
Evaluate $(A^{'} \cup B^{'}) \cap C$ .
$\begin{matrix} (A^{'} \cup B^{'}) \cap C & = U \cap C \\ = C \\ = {1, 2, \dots, 49, 90, 91, \dots, 100} \end{matrix}$

Answer

${1, 2, \dots, 49, 90, 91, \dots, 100} (which is Set C)$

Q11Part (b)

MediumApproved by Miss Alisha • Aug 30, 2026

Find $n {A \cap (B^{'} \cap C^{'})}$

Hint

$B^{'} \cap C^{'} = (B \cup C)^{'} = A$ , so $A \cap (B^{'} \cap C^{'}) = A \cap A = A$ , giving cardinality $n (A) = 40$ .

Solution

Find $B^{'} \cap C^{'}$ .
$\begin{matrix} B^{'} & = U ∖ B = {1, 2, 3, \dots, 100} \\ C^{'} & = U ∖ C = {50, 51, 52, \dots, 89, 101, 102, \dots, 150} \end{matrix}$
Taking intersection:
$\begin{matrix} B^{'} \cap C^{'} & = {1, 2, \dots, 100} \cap {50, 51, \dots, 89, 101, \dots, 150} \\ = {50, 51, 52, \dots, 89} \\ = A \end{matrix}$
Evaluate $A \cap (B^{'} \cap C^{'})$ .
$\begin{matrix} A \cap (B^{'} \cap C^{'}) & = A \cap A \\ = A \\ = {50, 51, 52, \dots, 89} \end{matrix}$
Find the cardinality $n {A \cap (B^{'} \cap C^{'})}$ .
$\begin{matrix} n {A \cap (B^{'} \cap C^{'})} & = n (A) \\ = 40 \end{matrix}$

Answer

$40$

Q12Question 12

Hard

In a secondary school 125 students participate in at least one of the following sports: cricket, football, or hockey. • 60 students play cricket. • 70 students play football. • 40 students play hockey. • 25 students play both cricket and football. • 15 students play both football and hockey. • 10 students play both cricket and hockey.

Q12Part (a)

MediumApproved by Miss Alisha • Aug 30, 2026

How many students play all three sports?

Hint

Use formula: $125 = 60 + 70 + 40 - 25 - 15 - 10 + x ⟹ 125 = 120 + x ⟹ x = 5$ .

Solution

Define the sets and list given values.

Let:
- $C$ = set of students who play cricket
- $F$ = set of students who play football
- $H$ = set of students who play hockey
Given data:
n(C∪F∪H)=125n(C)=60n(F)=70n(H)=40n(C∩F)=25n(F∩H)=15n(C∩H)=10
Let $x = n (C \cap F \cap H)$ be the number of students who play all three sports.
Apply the Principle of Inclusion-Exclusion for three sets.
$n (C \cup F \cup H) = n (C) + n (F) + n (H) - n (C \cap F) - n (F \cap H) - n (C \cap H) + n (C \cap F \cap H)$
Substitute the values and solve for $x$ .
$\begin{matrix} 125 & = 60 + 70 + 40 - 25 - 15 - 10 + x \\ 125 & = 170 - 50 + x \\ 125 & = 120 + x \\ x & = 125 - 120 \\ x & = 5 \end{matrix}$
Conclude.

Therefore, $5$ students play all three sports.

Answer

$5$ students

Q12Part (b)

MediumApproved by Miss Alisha • Aug 30, 2026

Draw a Venn diagram showing the distribution of sports participation in all the games.

Hint

Draw 3 intersecting circles for Cricket, Football, and Hockey with center value $5$ and verified sum $125$ .

Solution

Calculate values for each individual disjoint region.
- All three sports (Center):
n(C∩F∩H)=5
- Cricket and Football only:
n(C∩F)−5=25−5=20
- Football and Hockey only:
n(F∩H)−5=15−5=10
- Cricket and Hockey only:
n(C∩H)−5=10−5=5
- Cricket only:
n(C)−(20+5+5)=60−30=30
- Football only:
n(F)−(20+5+10)=70−35=35
- Hockey only:
n(H)−(5+5+10)=40−20=20
Verify the total sum:
$Total = 30 + 35 + 20 + 20 + 10 + 5 + 5 = 125$
Rendered Venn Diagram:

3-Set Venn Diagram: Sports Participation Distributionvenn diagram
Figure: Distribution of 125 students across Cricket, Football, and Hockey with 5 in the center.

Answer

$Only Cricket = 30, Only Football = 35, Only Hockey = 20, All Three = 5$

Q13Question 13

Hard

A survey was conducted in which 130 people were asked about their favourite foods. The survey results showed the following information: • 40 people said they liked nihari • 65 people said they liked biryani • 50 people said they liked korma • 20 people said they liked nihari and biryani • 35 people said they liked biryani and korma • 27 people said they liked nihari and korma • 12 people said they liked all three foods nihari, biryani, and korma

Q13Part (a)

MediumApproved by Miss Alisha • Aug 30, 2026

At least how many people like nihari, biryani or korma?

Hint

Find $n (N \cup B \cup K) = 40 + 65 + 50 - 20 - 35 - 27 + 12 = 85$ .

Solution

Define the sets and list given data.

Let:
- $N$ = set of people who like nihari
- $B$ = set of people who like biryani
- $K$ = set of people who like korma
Given:
n(U)=130n(N)=40n(B)=65n(K)=50n(N∩B)=20n(B∩K)=35n(N∩K)=27n(N∩B∩K)=12
Apply the Principle of Inclusion-Exclusion for three sets.
$n (N \cup B \cup K) = n (N) + n (B) + n (K) - n (N \cap B) - n (B \cap K) - n (N \cap K) + n (N \cap B \cap K)$
Substitute the values and calculate.
$\begin{matrix} n (N \cup B \cup K) & = 40 + 65 + 50 - 20 - 35 - 27 + 12 \\ = 155 - 82 + 12 \\ = 167 - 82 \\ = 85 \end{matrix}$
Conclude.

Therefore, $85$ people like at least one of the foods (nihari, biryani, or korma).

Answer

$85$ people

Q13Part (b)

EasyApproved by Miss Alisha • Aug 30, 2026

How many people did not like nihari, biryani, or korma?

Hint

$n ((N \cup B \cup K)^{'}) = 130 - 85 = 45$ people.

Solution

State the relation with the universal set.

The number of people who do not like any of the three foods is the cardinality of the complement of $N \cup B \cup K$ :
$n ((N \cup B \cup K)^{'}) = n (U) - n (N \cup B \cup K)$
Substitute values and evaluate.
$\begin{matrix} n ((N \cup B \cup K)^{'}) & = 130 - 85 \\ = 45 \end{matrix}$
Conclude.

Therefore, $45$ people did not like nihari, biryani, or korma.

Answer

$45$ people

Q13Part (c)

MediumApproved by Miss Alisha • Aug 30, 2026

How many people like only one of the following foods: nihari, biryani, or korma?

Hint

Calculate only Nihari ( $5$ ) + only Biryani ( $22$ ) + only Korma ( $0$ ) = $27$ .

Solution

Calculate the number of people who like only Nihari.
$\begin{matrix} n (Only N) & = n (N) - n (N \cap B) - n (N \cap K) + n (N \cap B \cap K) \\ = 40 - 20 - 27 + 12 \\ = 52 - 47 \\ = 5 \end{matrix}$
Calculate the number of people who like only Biryani.
$\begin{matrix} n (Only B) & = n (B) - n (N \cap B) - n (B \cap K) + n (N \cap B \cap K) \\ = 65 - 20 - 35 + 12 \\ = 77 - 55 \\ = 22 \end{matrix}$
Calculate the number of people who like only Korma.
$\begin{matrix} n (Only K) & = n (K) - n (N \cap K) - n (B \cap K) + n (N \cap B \cap K) \\ = 50 - 27 - 35 + 12 \\ = 62 - 62 \\ = 0 \end{matrix}$
Sum the people who like only one food.
$\begin{matrix} Total & = n (Only N) + n (Only B) + n (Only K) \\ = 5 + 22 + 0 \\ = 27 \end{matrix}$

Answer

$27$ people

Q13Part (d)

MediumApproved by Miss Alisha • Aug 30, 2026

Draw a Venn diagram.

Hint

Draw a 3-set Venn diagram with $12$ in the center, single regions $5, 22, 0$ , and $45$ outside the union.

Solution

Determine the elements in each region of the Venn diagram.
- All three foods: $12$
- Nihari and Biryani only: $20 - 12 = 8$
- Biryani and Korma only: $35 - 12 = 23$
- Nihari and Korma only: $27 - 12 = 15$
- Only Nihari: $5$
- Only Biryani: $22$
- Only Korma: $0$
- Outside the three sets (neither food): $130 - 85 = 45$
Verify the sum:
$\begin{matrix} Total surveyed & = (5 + 22 + 0) + (8 + 23 + 15) + 12 + 45 \\ = 27 + 46 + 12 + 45 \\ = 130 \end{matrix}$
Rendered Venn Diagram:

Favourite Foods Survey Distributionvenn diagram
Figure: Distribution of 130 surveyed individuals across Nihari, Biryani, and Korma with 45 outside the union.

Answer

$Only N = 5, Only B = 22, Only K = 0, Center = 12, Outside = 45$

The exercises and question numbering reproduced on these pages are from Mathematics for Class 9 (National Curriculum of Pakistan 2023), published by the Punjab Curriculum and Textbook Board (PCTB), Lahore, authored by Muhammad Akhtar Shirani, Madiha Mahmood, and Ghulam Murtaza. PCTB holds the copyright in the original textbook. PrepSure is not affiliated with, endorsed by, or sponsored by PCTB. The worked solutions, explanations, hints and method notes are PrepSure's own original work, written and reviewed by our team. If you hold rights in this material and believe anything here exceeds fair use, write to us and we will take it down.

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Exercise 3.2

Part 1 — Verification of (A∪B)′=A′∩B′

Part 2 — Verification of (A∩B)′=A′∪B′

Part 3 — Venn Diagram Verification

Part 1 — Verification of $(A \cup B)^{'} = A^{'} \cap B^{'}$

Part 2 — Verification of $(A \cap B)^{'} = A^{'} \cup B^{'}$