Q.Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′
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
Students often confuse difference (A∖B) with complement (Ac). Remember: difference is relative to another set, complement is relative to the whole universe. If U={1,2,3,4,5} and B={3,4,5}, then A∖B depends on what A is, but Bc is always {1,2}.
Another trap: symmetric difference is not the same as union. Union includes the overlap; symmetric difference kicks it out.
Takeaway
Every set operation is just a precise way to answer "which elements go where?" — learn the picture first, then the notation writes itself.
Set Operations — covering union, intersection, difference, and complement — is a foundational topic in the CBSE Class 11 Mathematics chapter on Sets, and Venn-diagram-based formula questions on this exact idea are a recurring feature in NCERT exercises and school exams. Students searching for "set operations class 11 maths" or "union and intersection formula with examples" will find this same definition-formula-example structure useful for board exam preparation and quick revision.
Concept: Set Membership — the complement A′ contains every element of the universal set U that is not in A.
Step 1: List all elements of U: {1,2,3,4,5,6,7,8,9,10}.
Step 2: Remove every element that belongs to A={1,3,5,7,9}.
Step 3: The remaining elements are {2,4,6,8,10}.
The complement A′ is {2,4,6,8,10}.
The complement of a set is everything in the universal set that is not in the original set. Here, A′={2,4,6,8,10}.
The idea of a set complement is simple but powerful: it’s the “other half” of the universe. If U is the universal set (the collection of all elements we care about), then A′ (read “A complement” or “A prime”) contains every element of U that is not in A.
Think of it like a Venn diagram: U is the entire rectangle, A is one circle inside it. A′ is everything outside that circle but still inside the rectangle. So to find A′, you just scan through U and pick out the numbers that are missing from A.
Let’s do it step by step.
-
List the universal set.
U={1,2,3,4,5,6,7,8,9,10}
-
List the given set A.
A={1,3,5,7,9}
-
Identify what’s missing.
Compare each element of U against A:
- 1 is in A → skip
- 2 is not in A → include
- 3 is in A → skip
- 4 is not in A → include
- 5 is in A → skip
- 6 is not in A → include
- 7 is in A → skip
- 8 is not in A → include
- 9 is in A → skip
- 10 is not in A → include
-
Collect the complement.
The elements we kept are 2,4,6,8,10. So:
A′={2,4,6,8,10}
A common mistake is to forget that the complement is defined only with respect to the given universal set U. If U were different, A′ would change. Here, U is clearly given, so we stick to it.
Notice that A contains all the odd numbers from U, and A′ contains all the even numbers. That’s a neat pattern: complements often reveal a natural partition of the universal set.
The complement of A is {2,4,6,8,10}.
Method: Set Difference (Subtraction) Method
The set difference A−B (also written A∖B) contains all elements that are in A but not in B.
Steps for A−B
-
List all elements of A
A={1,2,3,4,5,6}
-
Remove any element that also belongs to B
B={2,4,6,8}
Common elements: 2,4,6
-
Keep only the elements from A that are not in B
Remaining: 1,3,5
-
Write the result
A−B={1,3,5}
Steps for B−A
-
List all elements of B
B={2,4,6,8}
-
Remove any element that also belongs to A
A={1,2,3,4,5,6}
Common elements: 2,4,6
-
Keep only the elements from B that are not in A
Remaining: 8
-
Write the result
B−A={8}
Key Concept Check
- A−B is not the same as B−A — set difference is not commutative.
- An element belongs to A−B if and only if: x∈A and x∈/B
🧠 The Concept: Complement of a Set
Given a universal set U and a set A, the complement A′ (also written Ac or A) is:
All elements in U that are NOT in A.
So for:
- U={1,2,3,4,5,6,7,8,9,10}
- A={1,3,5,7,9}
The correct answer is:
A′={2,4,6,8,10}
✗ Common Mistake #1: Forgetting the Universal Set
What students do:
They list A′ as "everything except A" — but they don't check what U actually contains. They might write {2,4,6,8} (missing 10) or include numbers like 11 or 12.
Why it happens:
They treat "complement" as "opposite of A in all numbers" instead of "opposite of A within U."
✓ How to avoid:
Always write U down before starting. Then cross out every element of A from U. Whatever remains is A′.
✗ Common Mistake #2: Including Elements of A in A′
What students do:
They write A′={1,2,4,6,8,10} — accidentally keeping 1 (which is in A).
Why it happens:
Rushing or misreading the question. They might think "complement" means "add the missing ones" instead of "remove all of A."
✓ How to avoid:
After writing your answer, check every element: is it in A? If yes, it cannot be in A′. Double-check by scanning A again.
✗ Common Mistake #3: Confusing Complement with Difference
What students do:
They write A′=U−A correctly in concept, but then compute U−A as {2,4,6,8} (forgetting 10) or {2,4,6,8,9} (keeping 9).
Why it happens:
They don't systematically subtract — they guess or skip elements.
✓ How to avoid:
Use a systematic method:
- List U in order.
- Circle or tick elements that are in A.
- Write down only the unticked ones.
✗ Common Mistake #4: Writing A′ as a Sentence Instead of a Set
What students do:
They write "all even numbers" or "numbers not in A" instead of {2,4,6,8,10}.
Why it happens:
They think describing the pattern is enough.
✓ How to avoid:
In exams, always write the explicit set using curly braces {}. A description is not a valid answer unless the question asks for one.
✓ Quick Checklist to Avoid All Mistakes
| Step | Action |
|---|---|
| 1 | Write U clearly. |
| 2 | Write A clearly. |
| 3 | Cross out every element of A from U. |
| 4 | List the remaining elements in {}. |
| 5 | Verify: no element of A is in your answer. |
| 6 | Verify: every element of your answer is in U. |
Final correct answer:
A′={2,4,6,8,10}
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.
A point lies in (A∪B)′ exactly when it is not in A∪B, i.e. not in A and not in B (by De Morgan's law, (A∪B)′=A′∩B′). That is precisely the description of the shaded region.
✓Final answerThe shaded region represents (A∪B)′ — option (d).
- 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}
The complement is everything in U not in A∪B:
(A∪B)′=U−(A∪B)={1,5,9}
✓Final answer(A∪B)′={1,5,9}.
- 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.
Check each element of X: is 1∈Y? Yes. Is 3∈Y? Yes. Is 5∈Y? No.
So X∩Y={1,3}.
✓Final answerX∩Y={1,3}, which is option (c).
- 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}.
Comparing every element of A against B: 2∈/B, 6∈/B, 10∈/B. None of A's elements are in B, so A∩B=∅.
By definition, sets with empty intersection are disjoint sets.
✓Final answerTrue.
- 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.
Here A={1,2,3}, B={2,3,7}. Writing all elements of A then adding any elements of B not already listed: 1,2,3 (from A), then 7 (from B, since 2 and 3 are already present).
So A∪B={1,2,3,7}.
✓Final answerThe correct option is (c) {1,2,3,7}.
- 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}.
Here X={3,5,7} and Y={2,3,5}. Checking each element of X against Y: 3∈Y (yes), 5∈Y (yes), 7∈Y (no). So X∩Y={3,5}.
✓Final answerThe correct option is (d) {3,5}.
- 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}.
First take X∪Y={1,2,3,5} (2 is common, written once). Then union with Z: {1,2,3,5}∪{4,6}={1,2,3,4,5,6}, since Z shares no elements with the earlier union.
✓Final answerThe correct option is (c) {1,2,3,4,5,6}.
- 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).
Step 2: (X∪Y)∩Z keeps only elements also in Z={2,3,4,6}. Checking each element of X∪Y against Z: 1∈/Z, 2∈Z, 3∈Z, 4∈Z, 5∈/Z, 6∈Z. So the result is {2,3,4,6}.
✓Final answerThe correct option is (a) {2,3,4,6}.
- 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}.
Step 1: X∩Y keeps elements common to both: a∈Y, c∈Y, so X∩Y={a,c} (b and d are not in Y; r is not in X).
Step 2: (X∩Y)∪Z={a,c}∪{r,o,b}={a,b,c,o,r}.
✓Final answerThe correct option is (a) {a,b,c,o,r}.
- 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}.
A∪B={2}∪{3}={2,3}.
✓Final answerThe correct option is (c) {2,3}.
- 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}.
Neither A⊂B nor B⊂A nor A=B holds (each has an element the other lacks), but both contain −3, so A∩B={−3}.
✓Final answerThe correct option is (d) A∩B={−3}.
- 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.
Removing 1,2,3,5,15 from U leaves {4,6,7,8,9,10,11,12,13,14}.
✓Final answerThe correct option is (a) {4,6,7,8,9,10,11,12,13,14}.
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