Q.Consider the sets X and Y of Example 14. Find X ∩ Y
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 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 …
Concept: Set Intersection — the intersection of two sets contains only elements common to both.
From Example 14, X={Ram, Geeta, Akbar} (the hockey team) and Y={Geeta, David, Ashok} (the football team). …
Using Example 14's sets — X={Ram, Geeta, Akbar} (hockey) and Y={Geeta, David, Ashok} (football) — the only name common to both is Geeta, so X∩Y={Geeta}.
Recalling Example 14's sets
This question reuses the two sets from Example 14: X, the students on the school hockey team, and Y, the students on the school football team.
X={Ram, Geeta, Akbar},Y={Geeta, David, Ashok}
Finding the intersection
X∩Y contains every student who appears on both lists — students who play both hockey and football. …
Method: Set Intersection Using Roster Form
This method works when both sets are given in roster (list) form — we simply find the common elements.
Steps
-
Write both sets clearly
From Example 14 (assuming standard NCERT reference):
X={1,3,5}
Y={1,2,3}
-
Identify elements present in both sets
- Check each element of X:
- 1 is in Y → common
- 3 is in Y → common
- 5 is not in Y → not common
- No need to check Y again (we already covered all possibilities)
- Check each element of X:
-
Write the intersection set
Collect only the common elements:
X∩Y={1,3}
Key Concept …
Here are the common mistakes students make when working with Set Difference (and intersection, as in your question), along with how to avoid each.
Mistake 1: Confusing Set Difference with Intersection
The error:
Students treat X∖Y (or X−Y) as if it means X∩Y.
For example, if X={1,2,3} and Y={2,3,4}, they write X∖Y={2,3} instead of the correct {1}.
Why it happens:
The notation looks similar, and both operations involve comparing elements between two sets.
How to avoid:
- Remember the definition: X∖Y means "elements in X that are NOT in Y" — you remove all of Y from X.
- Use a mental filter: For each element of X, ask: "Is this also in Y?" If yes, discard it. If no, keep it.
- Draw a Venn diagram — shade only the part of X that does not touch Y.
Mistake 2: Forgetting the Order in Set Difference
The error:
Students assume X∖Y=Y∖X.
Example: X={a,b}, Y={b,c}. They write X∖Y={c} (which is actually Y∖X).
Why it happens:
Subtraction in arithmetic is commutative in some contexts (e.g., 5−3=3−5 is false, but students still blur the order). Set difference is not commutative.
How to avoid:
- Always read left-to-right: X∖Y means "start with X, remove Y."
- Write the operation in words before solving: "Elements of X that are not in Y."
- Check with a small example: If X={1}, Y={2}, then X∖Y={1} but Y∖X={2} — clearly different.
Mistake 3: Misapplying the Concept to Intersection (Your Question)
The error:
When asked for X∩Y, students accidentally compute X∖Y or Y∖X instead.
Why it happens:
Both operations involve comparing elements, and students rush without reading the symbol carefully (∩ vs ∖).
How to avoid:
- Memorise the symbols visually:
- ∩ looks like a "cup" — think common elements.
- ∖ is a subtraction sign — think remove.
- Before solving, state the definition aloud:
- X∩Y = "elements in both X and Y."
- X∖Y = "elements in X but not in Y."
- Double-check the question: Circle the symbol before you start.
Mistake 4: Including Elements Not in the First Set
The error:
For X∖Y, students list elements from Y that are not in X.
Example: X={1,2}, Y={2,3} → they write X∖Y={3}.
Why it happens:
They think "difference" means "all elements that are different between the two sets" (which is actually the symmetric difference).
How to avoid:
- Stick to the definition: Only elements from the first set matter.
- Use a two-step check:
- List all elements of X.
- Cross out any that also appear in Y. …
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.
…
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.