CAT (Common Admission Test) · Data Interpretation & Logical Puzzles

Venn Diagrams and Set Theory

Set unions, intersections and region counting using Venn diagrams.

Eight concepts on set overlap for CAT DILR. The exam booklet draws circles; you account for regions — peel from the centre, read exactly versus at least before touching a formula, and cross-check every count against the universal total.

CAT (Common Admission Test)
Medium level
8 concepts
17 practice questions

1Fill from the centre

A three-set Venn question never gives you the seven region counts directly. It gives set totals and pairwise overlaps, and every pairwise overlap already includes whoever sits in all three. Start with |A \cap B \cap C|, subtract it from each pairwise figure to get the pairwise-only bands, then subtract those bands from |A|, |B| and |C| for the singles. Skip a layer and you double-count the triple.

The exam booklet draws overlapping circles; you draw a ledger. The order is fixed: triple first, then the three pairwise-only regions, then the three single-set regions, then neither if a universal total is given.

Figure. Seven mutually exclusive Venn regions as separate nodes — no edges. Top row is the triple (fill first); middle row the three pairwise-only bands; bottom row the three singles. Layout y-order is the peel order; connecting disjoint partitions would fake inclusion.

Order of attack

Anchor the tripleWrite |A \cap B \cap C| first — from the stem or from a region already fixed.
Peel pairwise-onlyEach pairwise-only band is the given pairwise total minus the triple: |A \cap B \text{ only}| = |A \cap B| - |A \cap B \cap C|, and likewise for B \cap C and A \cap C.
Peel singlesEach single-set region is the set total minus its two pairwise-only bands minus the triple again.
Close on the universalSum the seven filled regions and subtract from the universal total for neither; the sum should match inclusion–exclusion on |A \cup B \cup C|.

Seven regions and what to subtract
Region	From the stem
A \cap B \cap C	Given, or fixed first
A \cap B only	\|A \cap B\| - \|A \cap B \cap C\|
B \cap C only	\|B \cap C\| - \|A \cap B \cap C\|
A \cap C only	\|A \cap C\| - \|A \cap B \cap C\|
A only	\|A\| - \text{two pairwise-only bands} - \text{triple}
B only	\|B\| - \text{two pairwise-only bands} - \text{triple}
C only	\|C\| - \text{two pairwise-only bands} - \text{triple}

Three-set region peel

In a survey of 100 people: 50 read A, 40 read B, 30 read C; 15 read A and B, 10 read B and C, 12 read A and C, and 5 read all three. How many read at least one publication?

Triple overlap givenA \cap B \cap C = 5
|A \cap B| only = 15 - 5; |B \cap C| only = 10 - 5; |A \cap C| only = 12 - 510, 5, 7
A only = 50 - 10 - 7 - 5; B only = 40 - 10 - 5 - 5; C only = 30 - 5 - 7 - 528, 20, 13
Sum 5 + 10 + 5 + 7 + 28 + 20 + 1388 read at least one

Pro tip. The triple sits inside every pairwise total — that is why you peel it before touching the singles. If you jump straight to |A| - |A \cap B| - |A \cap C|, you subtract the centre twice and the singles come out too small.

Using the same survey: |A \cap B| = 15 and |A \cap B \cap C| = 5. How many read A and B but not C?

The pairwise total 15 counts everyone in both A and B, including the 5 who also read C. The band who read A and B only is 15 - 5 = 10. Answering 15 treats the pairwise total as if it excluded the triple; answering 5 confuses the triple with the whole pairwise region.

2Two-set union

Before three sets, lock the two-set rule: |A \cup B| = |A| + |B| - |A \cap B|. Adding the two set totals counts the overlap twice — once inside each total — so subtract |A \cap B| once. The overlap is the only region shared; everything else in A or B is counted exactly once after the subtraction.

When only two sets appear, there are four regions: A only, B only, both, and neither. Most CAT stems give three of the four numbers and ask for the fourth; write the union formula first, then rearrange for the missing piece.

Figure. Four-region partition of the universal set — A only, overlap, B only, and neither. No inclusion edges: the regions are disjoint; the union formula subtracts the overlap because set totals already count it twice.

Two-set solve

List the four regionsA only, B only, A \cap B, and neither — four buckets that partition the universal set.
Apply the union formula|A \cup B| = |A| + |B| - |A \cap B|. If the stem asks for neither, subtract the union from the universal total.
Peel singles if neededA only = |A| - |A \cap B| and B only = |B| - |A \cap B| when the stem gives set totals and the overlap separately.

Club membership overlap

In a batch of 80 students, 45 play cricket and 35 play football; 12 play both. How many play at least one sport, and how many play neither?

|A \cup B| = 45 + 35 - 1268 play at least one
Neither = 80 - 6812 play neither
Cricket only = 45 - 12; football only = 35 - 1233 and 23

Pro tip. Adding 45 and 35 gives 80, which looks like the whole batch — but 12 were counted in both sports. The union is 68, not 80.

Set |A| = 40, |B| = 25, |A \cap B| = 10. What is |A \cup B|?

|A \cup B| = 40 + 25 - 10 = 55. Answering 65 forgets to subtract the overlap; 75 adds the overlap instead of removing the double count.

3Three-set inclusion–exclusion

When the stem hands you set totals and pairwise overlaps directly, the peel method and the inclusion–exclusion shortcut must agree. The formula is |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|: add the three set totals, subtract each pairwise overlap once, then add the triple back because it was removed three times and should be counted once.

Use this when you need |A \cup B \cup C| fast or as a cross-check after filling regions. If the numbers disagree, a pairwise total was read as a pairwise-only band, or the triple was skipped. Do not call |A|, |B| and |C| the 'singles' — that word is reserved for the A-only / B-only / C-only regions in fill-from-centre.

Figure. Sign ledger as bars: add set totals 120, subtract pairwise 37, add triple 5 to get union 88. Stopping at 83 forgets the centre restore.

Shortcut route

Sum the set totalsAdd |A| + |B| + |C| — every person in the union is counted in at least one set total. These are not the single-set regions.
Subtract pairwise totalsSubtract |A \cap B| + |B \cap C| + |A \cap C| — each pairwise count removes one layer of double counting.
Add the triple onceAdd |A \cap B \cap C| back — the triple was subtracted three times in step 2 but belongs in the union exactly once.

Inclusion–exclusion sign pattern
Term	Sign
\|A\| + \|B\| + \|C\|	add
\|A \cap B\| + \|B \cap C\| + \|A \cap C\|	subtract
\|A \cap B \cap C\|	add back once

Three-set region count

In a survey of 100 people: 50 read A, 40 read B, 30 read C; 15 read A and B, 10 read B and C, 12 read A and C, 5 read all three. How many read at least one?

Set totals: 50 + 40 + 30120
Subtract pairwise: 120 - (15 + 10 + 12)83
Add triple: 83 + 588 read at least one

Pro tip. Follow the +set-totals, −pairwise, +triple pattern exactly. Stopping at 83 forgets to restore the centre; skipping the pairwise step double-counts everyone in two sets.

With |A| = 30, |B| = 20, |A \cap B| = 8, what is |A \cup B| before moving to three sets?

Two-set check: 30 + 20 - 8 = 42. The three-set formula collapses to this when C is empty.

4Exactly versus at least

CAT DILR lives on wording. Exactly two sets means precisely two and not three — the triple-intersection is excluded. At least two sets means two or more — the triple counts. Exactly one means a single set with no overlap. Re-read the adjective before choosing a formula; the numbers in the stem are often identical across trap options.

Pairwise totals always include the triple. Exactly-two counts must peel the triple out of each pairwise band first, or use the closed formula on the sum of pairwise totals. At-least-two counts keep the triple but subtract it only twice from the pairwise sum.

Figure. Exactly two equals pairwise-only bands (22). At least two keeps those bands and adds the triple (27). The adjective alone swings the answer by 5.

Read the adjective

Underline exactly / at least / onlyExactly k forbids membership in a (k+1)th set; at least k includes all higher overlaps.
Map to a region sumExactly two = three pairwise-only bands. At least two = those three bands plus the triple. Exactly one = three single-set regions.
Pick the formula rowUse the concept table — do not reuse an at-least formula when the stem says exactly.

Wording → what counts
Stem says	Regions included
Exactly one set	A only + B only + C only
Exactly two sets	Three pairwise-only bands (triple excluded)
At least two sets	Three pairwise-only bands + triple
All three sets	A \cap B \cap C only

Same data, different adjectives

From the survey where |A \cap B \cap C| = 5 and the pairwise-only bands are 10, 5 and 7: how many read exactly two publications versus at least two?

Exactly two = 10 + 5 + 722
At least two = 22 + 527

Pro tip. Twenty-two and twenty-seven differ by exactly the triple size. If your exactly-two answer equals the pairwise sum 37, you forgot to subtract the triple three times.

A stem asks how many students take exactly two of three subjects. Which region set belongs in the sum?

The three pairwise-only bands
The three pairwise totals from the stem
The triple overlap only
Pairwise totals plus the triple

Exactly two excludes the triple, so only the pairwise-only bands count. Pairwise totals still contain the triple; adding the triple gives at least two, not exactly two.

5Exactly two sets

Cross-check by peeling: exactly two should equal the sum of the three pairwise-only bands from the fill-from-centre method. If the shortcut and the peel disagree, the triple was mis-read.

Figure. Exactly two = pairwise sum minus 3 times triple = 37 - 15 = 22. Three pairwise totals embed the triple three times; exactly-two wants it zero times.

Closed formula

Sum the three pairwise totalsAdd |A \cap B| + |B \cap C| + |A \cap C| as given — each still carries the triple inside.
Subtract three triplesSubtract 3|A \cap B \cap C| to remove the triple from all three pairwise counts at once.
Cross-check with peelOptionally verify against pairwise-only bands; the two routes must match.

Exactly two sets

Using the same survey data: 50 read A, 40 read B, 30 read C; 15 read A and B, 10 read B and C, 12 read A and C, 5 read all three. How many read exactly two of the three publications?

Sum of pairwise overlaps = 15 + 10 + 1237
Exactly two = 37 - 3 \times 522

Pro tip. Exactly two subtracts three times the triple-overlap from the pairwise sum — do not confuse with at least two, which subtracts only twice.

Pairwise totals sum to 37 and |A \cap B \cap C| = 5. How many read exactly two sets?

37 - 3 \times 5 = 22. Answering 27 uses the at-least-two formula (37 - 2 \times 5); 37 forgets to peel the triple entirely.

6At least two sets

Equivalently: at least two = exactly two + triple. If you already have exactly two from the previous concept, add |A \cap B \cap C| and stop.

Figure. At least two = pairwise sum minus 2 times triple = 37 - 10 = 27. Keep the triple once; net subtract two copies from the pairwise sum.

Two routes

Route A — adjust pairwise sumSum pairwise totals, subtract 2|A \cap B \cap C|.
Route B — add to exactly twoCompute exactly two first, then add the triple.
Sanity checkAt least two must exceed exactly two by exactly |A \cap B \cap C| when the triple is non-zero.

At least two publications

Same survey: pairwise totals 15, 10 and 12; triple overlap 5. How many read at least two publications?

Pairwise sum = 15 + 10 + 1237
At least two = 37 - 2 \times 527
Check: exactly two 22 + 527

Pro tip. The gap between exactly-two and at-least-two answers is always the triple size. If both answers match, you applied the same formula twice.

Exactly two count is 22 and |A \cap B \cap C| = 5. How many read at least two?

At least two includes the triple: 22 + 5 = 27. Answering 22 reports exactly two; 17 subtracts the triple instead of adding it.

7Neither and the universal total

Every Venn item sits inside a universal set — all survey respondents, every student in the batch. Neither equals the universal total minus the union count. Compute the union first (peel or inclusion–exclusion), then subtract from N.

The region sum identity must close: singles + pairwise-only bands + triple + neither = N. If your neither count goes negative, the stem numbers are inconsistent or a pairwise total was mis-read as pairwise-only.

Figure. Neither = N - union = 100 - 88 = 12. The neither band sits outside the circles; a negative neither means inconsistent stem numbers.

Close the books

Find the unionPeel regions or run inclusion–exclusion to get |A \cup B \cup C|.
Subtract from universalNeither = N - |A \cup B \cup C|.
Verify the seven regionsSum all filled regions plus neither; the total must equal N.

Neither in the survey

In the survey of 100 people where 88 read at least one publication, how many read none of A, B or C?

Neither = 100 - 8812 read none
Region sum check: 28 + 20 + 13 + 10 + 5 + 7 + 5 + 12100

Pro tip. Neither is not a Venn region inside the circles — it is everything outside the union. Always subtract the union from N, not the sum of set totals.

Universal total 80; |A \cup B| = 68. How many are in neither set?

Neither = 80 - 68 = 12. Answering 148 adds instead of subtracts; 68 confuses neither with the union count.

8Max and min overlap

CAT often asks for the largest or smallest possible overlap given set totals and a universal cap. Maximum |A \cap B| is \min(|A|, |B|) — pack one set entirely inside the other. When a universal cap is given, minimum overlap is the sum of set totals minus that cap, floored at zero; without a cap, the minimum is zero.

For three sets, maximise the triple by min-ing set totals; minimise it by pushing singles to the edge of the universal bound. Always check feasibility — a negative minimum means zero is the floor.

Figure. Batch of 100: max intersection = min(60,50) = 50; min intersection = max(0, 60+50-100) = 10. Min forces the union onto the cap.

Push to extremes

Maximise overlapMake the smaller set a subset of the larger: max |A \cap B| = \min(|A|, |B|).
Minimise overlap with a capSpread sets apart until the union hits N: min |A \cap B| = \max(0, |A| + |B| - N).
Check union boundNeither cannot go negative; if |A| + |B| - N is negative, the minimum overlap is zero.

Extreme overlap bounds (two sets, universal N)
Target	Formula
Maximum \|A \cap B\|	\min(\|A\|, \|B\|)
Minimum \|A \cap B\|	\max(0, \|A\| + \|B\| - N)
Maximum \|A \cup B\|	\min(N, \|A\| + \|B\|)
Minimum \|A \cup B\|	\max(\|A\|, \|B\|)

Overlap bounds in a batch

In a batch of 100 students, 60 study maths and 50 study physics. What are the maximum and minimum possible values of |M \cap P|?

Maximum: \min(60, 50)50 (all physics students also take maths)
Minimum: \max(0, 60 + 50 - 100)10 (union hits the batch cap)
Union at minimum overlap: 60 + 50 - 10100 — no room for neither

Pro tip. When the minimum overlap forces the union to equal N, neither is zero — every student is in at least one set.

|A| = 40, |B| = 35, universal total 100. What is the minimum possible |A \cap B|?

|A| + |B| - N = 40 + 35 - 100 = -25, floored to 0. The sets can be disjoint with 25 in neither. Answering 25 treats -25 as the answer without the floor.

Notes

Region Counting - Fill from the Centre: For two or three sets, start with the innermost (all-sets) intersection and work outward, subtracting overlaps so each region is counted exactly once.
Inclusion-Exclusion: Totals across overlapping sets require adding singles, subtracting pairwise overlaps, and re-adding the triple overlap; skipping a term double-counts.
Exactly vs At Least: 'Exactly two' excludes the triple-overlap; 'at least two' includes it. Distinguish these carefully in wording.
Maximisation/Minimisation: CAT often asks the max or min possible in a region given constraints; push overlaps to extremes within the allowed totals to bound the answer.
Common trap: Confusing 'exactly two sets' with 'at least two sets' - the triple-intersection is counted in the latter but not the former.

Formulas

|A\cup B\cup C| = |A|+|B|+|C| - |A\cap B| - |B\cap C| - |C\cap A| + |A\cap B\cap C|.
Exactly two sets = (\Sigma\,\text{pairwise}) - 3\,|A\cap B\cap C|.
Exactly one set = \Sigma|A| - 2(\Sigma\,\text{pairwise}) + 3|A\cap B\cap C|.
At least two = (\Sigma\,\text{pairwise}) - 2|A\cap B\cap C|.
Neither = universal -\,|A\cup B\cup C|.

Exam traps & shortcuts

Always fill the central triple-overlap first, then pairwise-only regions, then singles.
Re-read whether the question says 'exactly' or 'at least' before choosing the formula.
For max/min region questions, push overlaps to their extreme allowed values.
Cross-check region sum against the given universal total using inclusion-exclusion.

Reference tables

Closed forms for three-set wording
Asked count	Formula
Exactly one set	\Sigma\|A\| - 2(\Sigma\,\text{pairwise}) + 3\|A \cap B \cap C\|
Exactly two sets	(\Sigma\,\text{pairwise}) - 3\|A \cap B \cap C\|
At least two sets	(\Sigma\,\text{pairwise}) - 2\|A \cap B \cap C\|
Neither	N - \|A \cup B \cup C\|

Recap

Read only this the night before.

Peel order: Triple first, then three pairwise-only bands, then singles. Every pairwise total already hides the triple inside it.
Exactly vs at least: Exactly two peels the triple three times from the pairwise sum; at least two peels it twice. The gap between the two answers is always the triple size.
Close the books: Neither = universal minus union. Max overlap = min of the set totals; min overlap = max(0, sum of set totals − universal cap).

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