Exercise 3.2 — Class 9 Mathematics Solutions

Solution

1. Write the Universal set $U$ in tabular form.

   The universal set consists of all even positive integers up to $30$:

   $$U = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30\}$$

2. Identify elements of set $A$ (multiples of $6$ in $U$).

   $$A = \{6, 12, 18, 24, 30\}$$

3. Identify elements of set $B$ (multiples of $8$ in $U$).

   $$B = \{8, 16, 24\}$$

Solution

1. State the definition of set intersection.

   The intersection $A \cap B$ is the set containing all elements common to both $A$ and $B$.

2. Evaluate $A \cap B$.

   $$\begin{aligned} A \cap B &= \{6, 12, 18, 24, 30\} \cap \{8, 16, 24\} \\ &= \{24\} \end{aligned}$$

Solution

1. Determine the elements in each region.

   • Intersection $A \cap B$: $\{24\}$
   • Only $A$ ($A \setminus B$): $\{6, 12, 18, 30\}$
   • Only $B$ ($B \setminus A$): $\{8, 16\}$
   • Outside $A \cup B$ in $U$: $\{2, 4, 10, 14, 20, 22, 26, 28\}$

2. 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.

Solution

1. Understand the domain $U$.

   $$U = \{1, 2, 3, \dots, 150\}$$

2. List elements of $G$ (powers of $2$ within $U$).

   $$\begin{aligned} 2^0 &= 1 \\ 2^1 &= 2 \\ 2^2 &= 4 \\ 2^3 &= 8 \\ 2^4 &= 16 \\ 2^5 &= 32 \\ 2^6 &= 64 \\ 2^7 &= 128 \end{aligned}$$

   (Note: $2^8 = 256 > 150$, so it is excluded).

   $$G = \{1, 2, 4, 8, 16, 32, 64, 128\}$$

3. List elements of $H$ (perfect squares within $U$).

   $$\begin{aligned} 1^2 = 1, \quad 2^2 = 4, \quad 3^2 = 9, \quad 4^2 = 16, \quad 5^2 = 25, \quad 6^2 = 36, \\ 7^2 = 49, \quad 8^2 = 64, \quad 9^2 = 81, \quad 10^2 = 100, \quad 11^2 = 121, \quad 12^2 = 144 \end{aligned}$$ $$H = \{1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144\}$$

Solution

1. Combine all unique elements from sets $G$ and $H$.

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

2. Evaluate $G \cup H$.

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

Solution

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

   Common elements are powers of $2$ that are also perfect squares:

   $$\begin{aligned} 1 &= 2^0 = 1^2 \\ 4 &= 2^2 = 2^2 \\ 16 &= 2^4 = 4^2 \\ 64 &= 2^6 = 8^2 \end{aligned}$$

2. Evaluate $G \cap H$.

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

Solution

1. 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 $20$:

   $$Q = \{1, 2, 3, 5, 6, 7, 10, 14, 15\}$$

2. Evaluate $P \cap Q$.

   $$\begin{aligned} P \cap Q &= \{2, 3, 5, 7, 11, 13, 17, 19\} \cap \{1, 2, 3, 5, 6, 7, 10, 14, 15\} \\ &= \{2, 3, 5, 7\} \end{aligned}$$

Solution

1. Combine all elements of $P$ and $Q$.

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

2. Evaluate $P \cup Q$.

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

Solution

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

   • Left Hand Side (L.H.S):

   $$\begin{aligned} \text{L.H.S} &= A \cup B \\ &= \{1, 2, 3, 4, 5\} \cup \{4, 6, 8, 10\} \\ &= \{1, 2, 3, 4, 5, 6, 8, 10\} \quad \text{--- (1)} \end{aligned}$$

   • Right Hand Side (R.H.S):

   $$\begin{aligned} \text{R.H.S} &= B \cup A \\ &= \{4, 6, 8, 10\} \cup \{1, 2, 3, 4, 5\} \\ &= \{1, 2, 3, 4, 5, 6, 8, 10\} \quad \text{--- (2)} \end{aligned}$$

   From (1) and (2), $\text{L.H.S} = \text{R.H.S}$. 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):

   $$\begin{aligned} \text{L.H.S} &= A \cap B \\ &= \{1, 2, 3, 4, 5\} \cap \{4, 6, 8, 10\} \\ &= \{4\} \quad \text{--- (3)} \end{aligned}$$

   • Right Hand Side (R.H.S):

   $$\begin{aligned} \text{R.H.S} &= B \cap A \\ &= \{4, 6, 8, 10\} \cap \{1, 2, 3, 4, 5\} \\ &= \{4\} \quad \text{--- (4)} \end{aligned}$$

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

Solution

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

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

   • L.H.S:

   $$\begin{aligned} \text{L.H.S} &= \mathbb{N} \cup \mathbb{Z} \\ &= \mathbb{Z} \quad \text{--- (1)} \end{aligned}$$

   • R.H.S:

   $$\begin{aligned} \text{R.H.S} &= \mathbb{Z} \cup \mathbb{N} \\ &= \mathbb{Z} \quad \text{--- (2)} \end{aligned}$$

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

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

   • L.H.S:

   $$\begin{aligned} \text{L.H.S} &= \mathbb{N} \cap \mathbb{Z} \\ &= \mathbb{N} \quad \text{--- (3)} \end{aligned}$$

   • R.H.S:

   $$\begin{aligned} \text{R.H.S} &= \mathbb{Z} \cap \mathbb{N} \\ &= \mathbb{N} \quad \text{--- (4)} \end{aligned}$$

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

Solution

1. Given sets: $A$ is the set of all non-negative real numbers and $B = \mathbb{R}$ 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:

   $$\begin{aligned} \text{L.H.S} &= A \cup B \\ &= B = \mathbb{R} \quad \text{--- (1)} \end{aligned}$$

   • R.H.S:

   $$\begin{aligned} \text{R.H.S} &= B \cup A \\ &= B = \mathbb{R} \quad \text{--- (2)} \end{aligned}$$

   From (1) and (2), $\text{L.H.S} = \text{R.H.S} = \mathbb{R}$.

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

   • L.H.S:

   $$\begin{aligned} \text{L.H.S} &= A \cap B \\ &= A \quad \text{--- (3)} \end{aligned}$$

   • R.H.S:

   $$\begin{aligned} \text{R.H.S} &= B \cap A \\ &= A \quad \text{--- (4)} \end{aligned}$$

   From (3) and (4), $\text{L.H.S} = \text{R.H.S} = A$.

Solution

1. 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 $\text{L.H.S} = (A \cup B)'$:

   $$\begin{aligned} A \cup B &= \{a, b, c, d, g, h\} \cup \{c, d, e, f, j\} \\ &= \{a, b, c, d, e, f, g, h, j\} \end{aligned}$$

   Now, taking the complement with respect to $U$:

   $$\begin{aligned} (A \cup B)' &= U \setminus (A \cup B) \\ &= \{a, b, c, d, e, f, g, h, i, j\} \setminus \{a, b, c, d, e, f, g, h, j\} \\ &= \{i\} \quad \text{--- (1)} \end{aligned}$$

2. Compute Right Hand Side $\text{R.H.S} = A' \cap B'$:

   $$\begin{aligned} A' &= U \setminus A = \{e, f, i, j\} \\ B' &= U \setminus B = \{a, b, g, h, i\} \\ A' \cap B' &= \{e, f, i, j\} \cap \{a, b, g, h, i\} \\ &= \{i\} \quad \text{--- (2)} \end{aligned}$$

   From (1) and (2), $\text{L.H.S} = \text{R.H.S} = \{i\}$. Hence, $(A \cup B)' = A' \cap B'$ is verified.

   ---

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

1. Compute Left Hand Side $\text{L.H.S} = (A \cap B)'$:

   $$\begin{aligned} A \cap B &= \{a, b, c, d, g, h\} \cap \{c, d, e, f, j\} \\ &= \{c, d\} \end{aligned}$$

   Now, taking the complement:

   $$\begin{aligned} (A \cap B)' &= U \setminus (A \cap B) \\ &= \{a, b, c, d, e, f, g, h, i, j\} \setminus \{c, d\} \\ &= \{a, b, e, f, g, h, i, j\} \quad \text{--- (3)} \end{aligned}$$

2. Compute Right Hand Side $\text{R.H.S} = A' \cup B'$:

   $$\begin{aligned} A' \cup B' &= \{e, f, i, j\} \cup \{a, b, g, h, i\} \\ &= \{a, b, e, f, g, h, i, j\} \quad \text{--- (4)} \end{aligned}$$

   From (3) and (4), $\text{L.H.S} = \text{R.H.S} = \{a, b, e, f, g, h, i, j\}$. 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}.

Solution

1. Find the complement $A'$.

   $$\begin{aligned} A' &= U \setminus A \\ &= \{1, 2, 3, \dots, 20\} \setminus \{1, 3, 5, \dots, 19\} \\ &= \{2, 4, 6, \dots, 20\} \end{aligned}$$

2. Evaluate $A \cup A'$.

   $$\begin{aligned} \text{L.H.S} &= A \cup A' \\ &= \{1, 3, 5, \dots, 19\} \cup \{2, 4, 6, \dots, 20\} \\ &= \{1, 2, 3, 4, \dots, 20\} \\ &= U \\ &= \text{R.H.S} \end{aligned}$$

   Hence proved.

Solution

1. Evaluate $A \cap U$.

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

   Hence proved.

Solution

1. Evaluate $A \cap A'$.

   $$\begin{aligned} \text{L.H.S} &= A \cap A' \\ &= \{1, 3, 5, \dots, 19\} \cap \{2, 4, 6, \dots, 20\} \\ &= \emptyset = \phi \\ &= \text{R.H.S} \end{aligned}$$

   Hence proved.

Solution

1. 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:

   $$\begin{aligned} n(C \cup H) &= 55 \\ n(C) &= 34 \\ n(H) &= 30 \end{aligned}$$

   Let $x = n(C \cap H)$ represent the number of students who like to play both games.

2. Apply the Principle of Inclusion-Exclusion for two sets.

   $$n(C \cup H) = n(C) + n(H) - n(C \cap H)$$

3. Substitute and solve for $x$.

   $$\begin{aligned} 55 &= 34 + 30 - x \\ 55 &= 64 - x \\ x &= 64 - 55 \\ x &= 9 \end{aligned}$$

4. 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.

Solution

1. 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:

   $$\begin{aligned} n(U \cup E \cup P) &= 500 \\ n(U) &= 250 \\ n(E) &= 150 \\ n(P) &= 50 \\ n(U \cap E) &= 40 \\ n(E \cap P) &= 30 \\ n(U \cap P) &= 10 \end{aligned}$$

   Let $x = n(U \cap E \cap P)$ be the number of employees who speak all three languages.

2. Apply the Principle of Inclusion-Exclusion for three sets.

   $$n(U \cup E \cup P) = n(U) + n(E) + n(P) - n(U \cap E) - n(E \cap P) - n(U \cap P) + n(U \cap E \cap P)$$

3. Substitute the values and solve for $x$.

   $$\begin{aligned} 500 &= 250 + 150 + 50 - 40 - 30 - 10 + x \\ 500 &= 450 - 80 + x \\ 500 &= 370 + x \\ x &= 500 - 370 \\ x &= 130 \end{aligned}$$

4. Conclude.

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

Solution

1. 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:

   $$\begin{aligned} n(B) &= 19 \\ n(G) &= 15 \\ n(B \cap G) &= 3 \\ n(C \cap B) &= 4 \\ n(C \cap G) &= 2 \\ n(B \cup G \cup C) &= 34 \\ n(B \cap G \cap C) &= 0 \quad \text{(no one wears all three items)} \end{aligned}$$

   Let $x = n(C)$ be the total number of people wearing caps.

2. Apply the Principle of Inclusion-Exclusion for three sets.

   $$n(B \cup G \cup C) = n(B) + n(G) + n(C) - n(B \cap G) - n(C \cap B) - n(C \cap G) + n(B \cap G \cap C)$$

3. Substitute the values and solve for $x$.

   $$\begin{aligned} 34 &= 19 + 15 + x - 3 - 4 - 2 + 0 \\ 34 &= 34 + x - 9 \\ 34 &= 25 + x \\ x &= 34 - 25 \\ x &= 9 \end{aligned}$$

4. Conclude.

   Therefore, $9$ people are wearing caps.

Solution

1. 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:

   $$\begin{aligned} n(L) &= 17 \\ n(T) &= 11 \\ n(L \cap T) &= 9 \\ n(L \cap B) &= 6 \\ n(T \cap B) &= 4 \\ n(L \cap T \cap B) &= 4 \\ n(L \cup T \cup B) &= 35 \end{aligned}$$

   Let $x = n(B)$ be the total number of participants having books.

2. Apply the Principle of Inclusion-Exclusion for three sets.

   $$n(L \cup T \cup B) = n(L) + n(T) + n(B) - n(L \cap T) - n(L \cap B) - n(T \cap B) + n(L \cap T \cap B)$$

3. Substitute the values and solve for $x$.

   $$\begin{aligned} 35 &= 17 + 11 + x - 9 - 6 - 4 + 4 \\ 35 &= 28 - 19 + 4 + x \\ 35 &= 9 + 4 + x \\ 35 &= 13 + x \\ x &= 35 - 13 \\ x &= 22 \end{aligned}$$

4. Conclude.

   Therefore, $22$ participants have books.

Solution

1. Write the sets in tabular form.

   $$\begin{aligned} U &= \{1, 2, 3, \dots, 150\} \\ A &= \{50, 51, 52, \dots, 89\} \quad (n(A) = 40) \\ B &= \{101, 102, 103, \dots, 150\} \quad (n(B) = 50) \\ C &= \{1, 2, \dots, 49, 90, 91, \dots, 100\} \quad (n(C) = 60) \end{aligned}$$

2. Find $A'$ and $B'$.

   $$\begin{aligned} A' &= U \setminus A = \{1, 2, \dots, 49, 90, 91, \dots, 150\} \\ B' &= U \setminus B = \{1, 2, \dots, 100\} \end{aligned}$$

3. Find $A' \cup B'$.

   $$\begin{aligned} A' \cup B' &= \{1, 2, \dots, 49, 90, 91, \dots, 150\} \cup \{1, 2, \dots, 100\} \\ &= \{1, 2, 3, \dots, 150\} = U \end{aligned}$$

4. Evaluate $(A' \cup B') \cap C$.

   $$\begin{aligned} (A' \cup B') \cap C &= U \cap C \\ &= C \\ &= \{1, 2, \dots, 49, 90, 91, \dots, 100\} \end{aligned}$$