Q.Find the intersection of each pair of sets of question 1 above.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Set Operations
The idea in plain words
Think of your two favourite groups of friends — the ones who play cricket and the ones who play football. Some friends are in both groups, some in only one, and some in neither. Set operations are simply the mathematical ways to answer questions like: "Who's in at least one team?" or "Who's only in the cricket team?".
The whole secret? Each operation is just a different way of combining or comparing two collections — like sorting your friends into different buckets.
Why this works
Sets are just labelled buckets that hold distinct items. The universal set U is the "whole world" of things we're talking about — say, all your friends. Then each operation picks out a specific bucket:
| Operation | What it asks | Bucket contains |
|---|---|---|
| Union (A∪B) | In either? | Everything from A or B (or both) |
| Intersection (A∩B) | In both? | Only the overlap |
| Difference (A∖B) | In A but not B? | Just the part of A that doesn't touch B |
| Complement (Ac) | Not in A? | Everything outside A (inside U) |
| Symmetric Difference (A△B) | In exactly one? | The two crescent-shaped parts, excluding the overlap |
Step by step
Let's take two concrete sets so you can see each operation in action:
A={1,2,3},B={3,4,5}
Step 1: Union — gather everything from both, but don't repeat anything.
A∪B={1,2,3,4,5}
Step 2: Intersection — only what's common to both.
A∩B={3}
Step 3: Difference (A minus B) — start with A, remove anything that's also in B.
A∖B={1,2}
Step 4: Complement — needs a universal set. Let U={1,2,3,4,5}. Then:
Ac={4,5}
Step 5: Symmetric Difference — combine the two differences:
A△B=(A∖B)∪(B∖A)={1,2}∪{4,5}={1,2,4,5}
A common slip …
Intersection means the elements common to both sets. Using the five pairs of sets from Question 1 of this exercise:
- {1,3,5}∩{1,2,3}={1,3}
- {a,e,i,o,u}∩{a,b,c}={a}
- (multiples of 3) ∩ (naturals less than 6) ={3} …
Using the same five pairs of sets as Question 1 of this exercise, their intersections are: (i) {1,3},
(ii) {a},
(iii) {3},
(iv) ∅,
(v) ∅.
This question reuses the five pairs of sets from Question 1 of the same exercise and asks for the intersection (∩) of each pair -- the elements common to both sets.
(i) X={1,3,5}, Y={1,2,3}. The elements appearing in both are 1 and 3.
X∩Y={1,3}
(ii) A={a,e,i,o,u}, B={a,b,c}. Only a appears in both.
A∩B={a}
(iii) A={x:x is a natural number and a multiple of 3}={3,6,9,12,…}, B={x:x is a natural number less than 6}={1,2,3,4,5}. The only multiple of 3 that is also less than 6 is 3.
A∩B={3}
(iv) A={x:x is a natural number,1<x≤6}={2,3,4,5,6}, B={x:x is a natural number,6<x<10}={7,8,9}. No number can be both ≤6 and >6 at once.
A∩B=∅ …
Method: Direct Element Comparison (Set Intersection)
Concept: The intersection of two sets A and B, written A∩B, is the set of all elements that belong to both A and B.
Steps
- List all elements of the first set clearly.
- List all elements of the second set clearly.
- Compare each element of the first set with the elements of the second set.
- Pick only those elements that appear in both sets.
- Write the result as a set (curly braces, elements separated by commas).
Example (from a typical Question 1)
Let’s assume the pairs from Question 1 were:
- Pair (i): A={1,2,3,4}, B={3,4,5,6}
- Pair (ii): C={a,b,c}, D={b,c,d,e}
For Pair (i):
- Elements of A: 1,2,3,4
- Elements of B: 3,4,5,6
- Common elements: 3 and 4
- Answer: A∩B={3,4}
For Pair (ii): …
Here’s a breakdown of the common mistakes students make when finding the intersection of sets (based on a typical "Question 1" style from NCERT or state board textbooks), along with how to avoid each.
✗ Mistake 1: Confusing Intersection with Union
What students do:
They list all elements from both sets instead of only the common ones.
Example:
A={1,2,3}, B={2,3,4}
Wrong answer: A∩B={1,2,3,4} (this is actually A∪B).
Why it happens:
Students mix up the symbols ∩ (intersection) and ∪ (union).
✓ How to avoid:
- Memorise the shape: ∩ looks like an arch — think “Arch for And” (both sets must have it).
- Always ask: “Is this element in both sets?” If no, leave it out.
✗ Mistake 2: Forgetting to Check All Elements in Both Sets
What students do:
They only check elements from the first set against the second, but miss elements that appear only in the second set.
Example:
X={a,b,c}, Y={b,c,d}
Wrong: X∩Y={b} (missed c).
Why it happens:
Rushing — they stop after scanning the first set.
✓ How to avoid:
- Systematic method:
- List all elements of set 1.
- For each, check if it’s in set 2.
- Then repeat by checking elements of set 2 against set 1 (or just verify your list).
- Use a Venn diagram to visualise overlap.
✗ Mistake 3: Including Duplicate Elements in the Answer
What students do:
They write the same element twice in the intersection set.
Example:
A={1,2,2,3}, B={2,3,4}
Wrong: A∩B={2,2,3}
Why it happens:
They copy duplicates from the original sets without realising sets don’t have repetitions.
✓ How to avoid:
- Remember: Sets never contain duplicates.
- Write the intersection as a set of distinct elements only.
- After listing, check: “Did I write any number twice?”
✗ Mistake 4: Writing the Answer in Wrong Notation
What students do:
They write the answer as a list without curly braces, or use round brackets.
Example:
Wrong: A∩B=2,3 or (2,3)
Why it happens:
Carelessness with set notation.
✓ How to avoid:
- Always use curly braces {} for sets.
- Double-check your final answer format before moving to the next question.
✗ Mistake 5: Assuming Intersection Always Has Elements …
Showing the 12 most recent of 52 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.The shaded region in the given Venn-diagram represents:(a) A ∪ B(b) A ∩ B(c) (A ∪ B)'(d) (A ∩ B)'
›Reveal solutionSolution
The shaded region is everything in the universal set except A and B combined, which is exactly (A∪B)′.
The rectangle is the universal set U, and the two overlapping circles are sets A and B. The description tells us the shading covers the rectangle except the two circles — i.e. every point that lies outside both A and B.
…
- CBSE 2026Set ANNUAL1 markQ.If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 6, 7}, then (A ∪ B)' = ..............
›Reveal solutionSolution
Find A∪B first, then take its complement in U.
Given U={1,2,3,4,5,6,7,8,9}, A={2,4,6,8}, B={2,3,6,7}.
First find A∪B (all elements in A or B or both):
A∪B={2,3,4,6,7,8}
…
- CBSE 2026Set ANNUAL1 markMCQQ.If X={1,3,5} and Y={1,2,3} then X∩Y=?(a) {1,2,3,4,5}(b) {1,2,3,5}(c) {1,3}(d) ϕ
›Reveal solutionSolution
X∩Y consists of elements present in both X and Y, which gives {1,3}.
Given X={1,3,5} and Y={1,2,3}. The intersection X∩Y contains only those elements that belong to BOTH sets.
…
- CBSE 2026Set ANNUAL1 markQ.Write True/False: Sets {2,6,10} and {3,7,11} are disjoint sets.
›Reveal solutionSolution
Sets are disjoint when their intersection is empty; comparing the elements of {2,6,10} and {3,7,11} shows no overlap.
Set A={2,6,10} and set B={3,7,11}.
…
- CBSE 2025Set ANNUAL1 markMCQQ.A={1,2,3},B={2,3,7}⇒A∪B=(a) {1,2,3}(b) {1,3,7}(c) {1,2,3,7}(d) {1,2,7}
›Reveal solutionSolution
A∪B={1,2,3,7}: the union lists every element that is in A or in B (or both), each written once.
For sets A and B, the union is A∪B={x:x∈A or x∈B} — combine both sets and remove duplicate entries.
…
- CBSE 2025Set ANNUAL1 markMCQQ.X={3,5,7},Y={2,3,5}⇒X∩Y=(a) {3,2}(b) {3,7}(c) {5,7}(d) {3,5}
›Reveal solutionSolution
X∩Y={3,5}: the intersection keeps only elements that belong to both sets.
For sets X and Y, X∩Y={x:x∈X and x∈Y}.
…
- CBSE 2025Set ANNUAL1 markMCQQ.X={1,2},Y={2,3,5},Z={4,6}⇒X∪Y∪Z=(a) {1,2,3,5,6}(b) {2,3,4,5,6}(c) {1,2,3,4,5,6}(d) {1,2}
›Reveal solutionSolution
X∪Y∪Z={1,2,3,4,5,6}: list every element appearing in at least one of the three sets, once each.
Given X={1,2}, Y={2,3,5}, Z={4,6}.
…
- CBSE 2025Set ANNUAL1 markMCQQ.X={1,2,3,6},Y={4,5,6},Z={4,2,3,6}⇒(X∪Y)∩Z=(a) {2,3,4,6}(b) {1,5}(c) {1,2,5}(d) {1,2,3,5}
›Reveal solutionSolution
(X∪Y)∩Z={2,3,4,6}, found by first taking the union, then intersecting with Z.
Given X={1,2,3,6}, Y={4,5,6}, Z={4,2,3,6}.
Step 1: X∪Y={1,2,3,4,5,6} (combine both, 6 counted once).
…
- CBSE 2025Set ANNUAL1 markMCQQ.X={a,b,c,d},Y={c,a,r},Z={r,o,b}⇒(X∩Y)∪Z=(a) {a,b,c,o,r}(b) {c,a,r,b}(c) {r,o,b,c}(d) ϕ
›Reveal solutionSolution
(X∩Y)∪Z={a,b,c,o,r}, found by first taking the intersection, then the union with Z.
Given X={a,b,c,d}, Y={c,a,r}, Z={r,o,b}.
…
- CBSE 2025Set ANNUAL1 markMCQQ.A={x:x−2=0},B={x:2x=6}⇒A∪B=(a) {2,6}(b) {−2,6}(c) {2,3}(d) {2,−3}
›Reveal solutionSolution
A∪B={2,3}.
A={x:x−2=0}={2}. B={x:2x=6}={3}.
…
- CBSE 2025Set ANNUAL1 markMCQQ.A={x:x2+5x+6=0},B={x:x2+8x+15=0}⇒(a) A⊂B(b) B⊂A(c) A=B(d) A∩B={−3}
›Reveal solutionSolution
A={−2,−3}, B={−3,−5}, and their only common element is −3, so A∩B={−3}.
A={x:x2+5x+6=0}: factorising, (x+2)(x+3)=0⇒x=−2,−3, so A={−2,−3}.
B={x:x2+8x+15=0}: factorising, (x+3)(x+5)=0⇒x=−3,−5, so B={−3,−5}.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Given U={1,2,…,15}, A={1,2,3,5,15}, B={2,4,6,8,10,12,14}, C={2,3,5,7,11,13}. A′=(a) {4,6,7,8,9,10,11,12,13,14}(b) {4,6,8,10,12,14}(c) {8,10,12,14}(d) ϕ
›Reveal solutionSolution
A′=U−A={4,6,7,8,9,10,11,12,13,14}.
Given U={1,2,…,15} and A={1,2,3,5,15}. The complement A′=U−A consists of every element of U not in A.
…
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