Set theory in aptitude is mostly about counting — how many people like tea, how many like coffee, how many like both? The Venn diagram is our best friend here. Once we draw it, the answer practically jumps out.
In simple language, a set is just a collection of things. A set of students who passed math, a set of people who drink tea, a set of numbers divisible by 3 — anything grouped together.
Basic Set Operations
Inclusion-Exclusion Principle — The Core Formula
This is the formula that solves 90% of set theory problems in exams.
Why do we subtract the intersection? Because when we add n(A) and n(B), the people in both sets get counted twice. Subtracting the intersection fixes that double-counting. For three sets, it’s the same idea but we have to be more careful — we subtract the pairwise overlaps, but then the triple overlap gets subtracted too many times, so we add it back once.
Venn Diagram for 2 Sets
The trick with Venn diagrams: always fill in the intersection first, then work outward. If we fill from outside in, numbers don’t add up correctly.
Example 1: Basic 2-set problem
In a class of 60 students, 35 like cricket, 28 like football, and 10 like both. How many like neither?
n(A ∪ B) = n(A) + n(B) - n(A ∩ B) = 35 + 28 - 10 = 53
Neither = Total - n(A ∪ B) = 60 - 53 = 7 students
Example 2: Finding the intersection
In a group of 100 people, 72 read newspaper A, 54 read newspaper B. How many read both? (Everyone reads at least one.)
Since everyone reads at least one: n(A ∪ B) = 100.
100 = 72 + 54 - n(A ∩ B) n(A ∩ B) = 126 - 100 = 26 people
Venn Diagram for 3 Sets
The 3-Set Filling Strategy
When solving a 3-set problem, fill in this order:
- Center first: n(A ∩ B ∩ C) = g
- Pairwise intersections: subtract center from each. E.g., “A and B only” = n(A ∩ B) - g
- Individual only: subtract all overlaps. E.g., “Only A” = n(A) - d - e - g
- Add everything to verify it matches the total.
Example 3: Three-set problem
In a survey of 200 students: 120 like tea, 80 like coffee, 60 like juice. 40 like tea and coffee, 30 like coffee and juice, 25 like tea and juice, and 15 like all three. How many like none of the three?
Using inclusion-exclusion for 3 sets:
n(T ∪ C ∪ J) = 120 + 80 + 60 - 40 - 30 - 25 + 15 = 180
None = 200 - 180 = 20 students
Let’s also find each region:
- All three: 15
- Tea and coffee only: 40 - 15 = 25
- Coffee and juice only: 30 - 15 = 15
- Tea and juice only: 25 - 15 = 10
- Only tea: 120 - 25 - 10 - 15 = 70
- Only coffee: 80 - 25 - 15 - 15 = 25
- Only juice: 60 - 10 - 15 - 15 = 20
Check: 70 + 25 + 20 + 25 + 15 + 10 + 15 + 20 (none) = 200. It works!
Maxima-Minima Problems
These are the trickiest set theory questions. They ask things like “what is the maximum number who could like all three?” or “at least how many like both?”
The Key Insight
Maximum of A ∩ B = min(n(A), n(B)) — the overlap can’t be more than the smaller set.
Minimum of A ∩ B = n(A) + n(B) - n(U) if this is positive, else 0 — this is when we spread things out as much as possible.
Example 4: Min-max problem
In a class of 40, 25 passed Math and 30 passed English. Find the minimum and maximum number who passed both.
Maximum: min(25, 30) = 25 (all Math-passers also passed English)
Minimum: 25 + 30 - 40 = 15 (spread out as much as possible, 15 must overlap)
Example 5: “At least” problem with 3 sets
100 people were surveyed. 60 read Hindu, 50 read Times, 45 read Express. What is the minimum number reading at least two newspapers?
Maximum reading only one paper = minimize overlaps. Each person reading only one paper “uses” 1 from the count. Total sum = 60 + 50 + 45 = 155. Since there are only 100 people, at least 155 - 100 = 55 “extra” readings come from overlaps. Each double-reader accounts for 1 extra, each triple-reader accounts for 2 extras.
To minimize “at least two” readers, maximize triple-readers (each triple-reader absorbs 2 extras). So we need:
- If all extras are triple-readers: 55/2 = 27.5, which isn’t integer. So 27 triple + 1 double = 55 extras. Minimum reading at least two = 27 + 1 = 28.
Wait, let’s think more carefully. Let d = number reading exactly 2, t = number reading exactly 3. The extras are d + 2t = 55. We want to minimize d + t (people reading at least 2). If t = 27, then d = 55 - 54 = 1, so d + t = 28. If t = 0, d = 55, so d + t = 55.
Minimum reading at least two papers = 28.
Common Exam Variations
- Basic 2-set: find union, intersection, or neither.
- 3-set problems: fill the Venn diagram completely.
- “How many like exactly one?” — sum of the three “only” regions.
- “How many like at least two?” — sum of pairwise + triple.
- Max-min problems: find bounds on intersections.
- Complement questions: “How many do NOT like X?”
Important Identities
- n(A only) = n(A) - n(A ∩ B)
- n(exactly one of A, B) = n(A) + n(B) - 2n(A ∩ B)
- n(at least one) = n(A ∪ B) = n(U) - n(neither)
- For 3 sets: n(exactly one) = n(A) + n(B) + n(C) - 2[n(A∩B) + n(B∩C) + n(A∩C)] + 3n(A∩B∩C)
- For 3 sets: n(exactly two) = n(A∩B) + n(B∩C) + n(A∩C) - 3n(A∩B∩C)
Practice Problems
Problem 1: In a group of 80 people, 50 speak Hindi, 40 speak English, and 20 speak both. How many speak neither?
Problem 2: In a class of 150 students: 90 play cricket, 70 play football, 50 play hockey, 30 play both cricket and football, 25 play both football and hockey, 20 play both cricket and hockey, and 10 play all three. How many play none?
Problem 3: In a group of 50, everyone likes at least one of two subjects — Math or Science. 40 like Math and 35 like Science. How many like exactly one subject?
Answers
Problem 1: n(H ∪ E) = 50 + 40 - 20 = 70. Neither = 80 - 70 = 10 people.
Problem 2: n(C ∪ F ∪ H) = 90 + 70 + 50 - 30 - 25 - 20 + 10 = 145. None = 150 - 145 = 5 students.
Problem 3: n(M ∪ S) = 50 (everyone likes at least one). So 50 = 40 + 35 - n(M ∩ S) → n(M ∩ S) = 25. Exactly one subject = n(M ∪ S) - n(M ∩ S) = 50 - 25 = 25 students. (Or: only Math = 40 - 25 = 15, only Science = 35 - 25 = 10, exactly one = 15 + 10 = 25.)