Q.If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find

Concept: Set Intersection and Union

Intersection $A \cap B$ picks elements common to both sets; union $A \cup B$ combines all elements from both (no repetition). Apply these operations step by step.

Given:

$A = \{3, 5, 7, 9, 11\}$, $B = \{7, 9, 11, 13\}$, $C = \{11, 13, 15\}$, $D = \{15, 17\}$

(i) $A \cap B$: common elements in $A$ and $B$ are $7, 9, 11$.

(ii) $B \cap C$: common elements in $B$ and $C$ are $11, 13$.

(iii) $A \cap C \cap D$: first $A \cap C = \{11\}$, then $\{11\} \cap D = \emptyset$ (no overlap).

(iv) $A \cap C$: only $11$ appears in both.

(v) $B \cap D$: no common elements.

(vi) $A \cap (B \cup C)$: $B \cup C = \{7, 9, 11, 13, 15\}$, so $A \cap \{7, 9, 11, 13, 15\} = \{7, 9, 11\}$.

(vii) $A \cap D$: no overlap.

(viii) $A \cap (B \cup D)$: $B \cup D = \{7, 9, 11, 13, 15, 17\}$, so $A \cap \{7, 9, 11, 13, 15, 17\} = \{7, 9, 11\}$. …

The intersection of two sets is the collection of elements they share. Think of it as the overlap in a Venn diagram: if an element appears in both sets, it belongs to their intersection; otherwise, it doesn't. The union, by contrast, gathers everything from both sets without duplication.

The strategy is straightforward: for each intersection, scan through the sets and pick out elements that appear in all the sets being intersected. For unions, collect all distinct elements from the sets involved, then proceed with any further operations.

Let me work through each part:

(i) $A \cap B$

We need elements that appear in both $A = \{3, 5, 7, 9, 11\}$ and $B = \{7, 9, 11, 13\}$.

Checking each element of $A$:

• $3 \in A$ but $3 \notin B$
• $5 \in A$ but $5 \notin B$
• $7 \in A$ and $7 \in B$ ✓
• $9 \in A$ and $9 \in B$ ✓
• $11 \in A$ and $11 \in B$ ✓

So $A \cap B = \{7, 9, 11\}$.

(ii) $B \cap C$

Elements common to $B = \{7, 9, 11, 13\}$ and $C = \{11, 13, 15\}$:

• $7 \in B$ but $7 \notin C$
• $9 \in B$ but $9 \notin C$
• $11 \in B$ and $11 \in C$ ✓
• $13 \in B$ and $13 \in C$ ✓

So $B \cap C = \{11, 13\}$.

(iii) $A \cap C \cap D$

We need elements present in all three sets: $A = \{3, 5, 7, 9, 11\}$, $C = \{11, 13, 15\}$, and $D = \{15, 17\}$.

First, $A \cap C$: only $11$ appears in both $A$ and $C$, so $A \cap C = \{11\}$.

Now intersect with $D = \{15, 17\}$: does $11 \in D$? No.

So $A \cap C \cap D = \emptyset$ (the empty set).

(iv) $A \cap C$

Elements common to $A = \{3, 5, 7, 9, 11\}$ and $C = \{11, 13, 15\}$:

Only $11$ appears in both.

So $A \cap C = \{11\}$.

(v) $B \cap D$

Elements common to $B = \{7, 9, 11, 13\}$ and $D = \{15, 17\}$:

No element of $B$ appears in $D$.

So $B \cap D = \emptyset$.

(vi) $A \cap (B \cup C)$

First find $B \cup C$, the union of $B = \{7, 9, 11, 13\}$ and $C = \{11, 13, 15\}$:

$B \cup C = \{7, 9, 11, 13, 15\}$.

Now intersect with $A = \{3, 5, 7, 9, 11\}$:

• $3 \in A$ but $3 \notin (B \cup C)$
• $5 \in A$ but $5 \notin (B \cup C)$
• $7 \in A$ and $7 \in (B \cup C)$ ✓
• $9 \in A$ and $9 \in (B \cup C)$ ✓
• $11 \in A$ and $11 \in (B \cup C)$ ✓

So $A \cap (B \cup C) = \{7, 9, 11\}$.

(vii) $A \cap D$

Elements common to $A = \{3, 5, 7, 9, 11\}$ and $D = \{15, 17\}$:

No overlap.

So $A \cap D = \emptyset$.

(viii) $A \cap (B \cup D)$

First find $B \cup D$, the union of $B = \{7, 9, 11, 13\}$ and $D = \{15, 17\}$:

$B \cup D = \{7, 9, 11, 13, 15, 17\}$.

Now intersect with $A = \{3, 5, 7, 9, 11\}$:

The common elements are $7, 9, 11$.

So $A \cap (B \cup D) = \{7, 9, 11\}$.

(ix) $(A \cap B) \cap (B \cup C)$

From part (i), $A \cap B = \{7, 9, 11\}$. …

Set Difference & Intersection — Solution Method

Method Name: Venn Diagram / Direct Listing Method

(For small finite sets, we directly list common elements.)

Steps

1. Write each set clearly — list all elements.
2. For intersection ($\cap$): pick elements that appear in all given sets.
3. For union ($\cup$): combine all elements from the sets, removing duplicates.
4. Follow brackets — solve inside parentheses first.
5. Write final answer as a set inside curly braces $\{ \}$.

Given Sets

• $A = \{3, 5, 7, 9, 11\}$
• $B = \{7, 9, 11, 13\}$
• $C = \{11, 13, 15\}$
• $D = \{15, 17\}$

Solutions

(i) $A \cap B$

Common in both: $7, 9, 11$

Answer: $\{7, 9, 11\}$

(ii) $B \cap C$

Common: $11, 13$

Answer: $\{11, 13\}$

(iii) $A \cap C \cap D$

• $A \cap C = \{11\}$
• Then $\{11\} \cap D = \emptyset$ (no common element) Answer: $\emptyset$ (empty set)

(iv) $A \cap C$

Common: $11$

Answer: $\{11\}$

(v) $B \cap D$

$B = \{7,9,11,13\}$, $D = \{15,17\}$ — no common element

Answer: $\emptyset$

(vi) $A \cap (B \cup C)$

• First, $B \cup C = \{7,9,11,13,15\}$
• Then $A \cap \{7,9,11,13,15\} = \{7,9,11\}$ Answer: $\{7, 9, 11\}$

(vii) $A \cap D$

$A = \{3,5,7,9,11\}$, $D = \{15,17\}$ — no common

Answer: $\emptyset$

(viii) $A \cap (B \cup D)$ …

This is a classic exercise from NCERT Class 11 (Chapter 1: Sets). Let's go concept-first, then list the common mistakes.

🧠 Concept Recap: Set Difference vs Intersection

First, note: this question is about intersection ($\cap$), not set difference ($\setminus$).

• Intersection $A \cap B$ = elements common to both A and B.
• Set difference $A \setminus B$ = elements in A but not in B.

Since the question asks for $\cap$, we'll focus on mistakes in intersection problems — but the same vigilance applies to difference.

✗ Common Mistakes & How to Avoid Them

1. ✗ Forgetting that intersection is commutative

Mistake: Thinking $A \cap B$ is different from $B \cap A$ in result.

Why it happens: Students confuse $\cap$ with $\setminus$ (set difference).

✓ Fix: Remember:

Always check: "What is common to both?" — order doesn't matter.

2. ✗ Incorrectly handling three-set intersections like $A \cap C \cap D$

Mistake: Taking intersection of only two sets, or including elements not in all three.

Example: For $A \cap C \cap D$, a student might write $\{11, 13\}$ (only $A \cap C$) instead of checking D too.

✓ Fix:

• Step 1: Find $A \cap C = \{11\}$
• Step 2: Intersect that with D: $\{11\} \cap \{15, 17\} = \emptyset$ Result: $\boxed{\emptyset}$ (empty set)

3. ✗ Mistaking $\cup$ for $\cap$ inside compound expressions

Mistake: In $A \cap (B \cup C)$, a student might take $A \cap B$ and $A \cap C$ separately but then union them instead of intersecting with the union.

✓ Fix:

• First find $B \cup C = \{7, 9, 11, 13, 15\}$
• Then intersect with A: $\{3,5,7,9,11\} \cap \{7,9,11,13,15\} = \{7,9,11\}$ Result: $\boxed{\{7,9,11\}}$

4. ✗ Forgetting the empty set $\emptyset$

Mistake: Writing "no elements" or leaving blank instead of $\emptyset$ or $\{\}$.

Example: $A \cap D = \emptyset$, but students write nothing or "0".

✓ Fix: Always write $\emptyset$ or $\{\}$ for empty intersection. In exams, both are accepted — but be explicit.

5. ✗ Order of operations error in $(A \cap B) \cap (B \cup C)$

Mistake: Doing $(A \cap B)$ and $(B \cup C)$ correctly, but then taking union of the two results instead of intersection.

✓ Fix:

• $A \cap B = \{7,9,11\}$
• $B \cup C = \{7,9,11,13,15\}$
• Intersection: $\{7,9,11\} \cap \{7,9,11,13,15\} = \{7,9,11\}$ Result: $\boxed{\{7,9,11\}}$

6. ✗ Listing elements not in the universal set (if given)

Mistake: Including an element that doesn't belong to any given set. …