Venn diagram sets in CAT DILR: regions, formulas and overlap bounds

Venn questions reward one habit above all: fill the diagram region by region, from the centre outward, and never answer from the totals alone. A three-set Venn diagram has seven regions inside the circles and one outside. Once all eight numbers are on paper, every question in the set is a lookup.

CAT uses Venn logic in two ways. Some sets give you enough data to fill every region, and test whether you can translate words like "only" and "exactly two". Others leave the regions open and ask for the minimum or maximum possible overlap. This guide covers both, with each example solved step by step. The overlap bounds were confirmed by an exhaustive search over every possible split.

The eight regions

For three sets A, B and C, name the regions this way:

Every phrase in a question maps to a group of these regions.

The formulas

Two sets: n(A ∪ B) = n(A) + n(B) − n(A ∩ B).

Three sets: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(C ∩ A) + n(A ∩ B ∩ C).

Two more are worth knowing, because CAT data is often given as counts rather than intersections:

• Sum of the three set sizes = (exactly one) + 2 × (exactly two) + 3 × (all three).
• Exactly two = (sum of the three pairwise intersections) − 3 × (all three). At least two = that sum − 2 × (all three).

Example 1: two sets

Of 100 students, 60 like tea, 50 like coffee and 20 like both.

• Tea or coffee: 60 + 50 − 20 = 90.
• Neither: 100 − 90 = 10.
• Tea only: 60 − 20 = 40.

Example 2: three sets from intersections

n(A) = 40, n(B) = 35, n(C) = 30; n(A ∩ B) = 12, n(B ∩ C) = 10, n(C ∩ A) = 8; n(A ∩ B ∩ C) = 5.

Fill from the centre outward.

• All three: 5.
• A and B only: 12 − 5 = 7. B and C only: 10 − 5 = 5. C and A only: 8 − 5 = 3. Exactly two = 7 + 5 + 3 = 15, which matches 30 − 3 × 5.
• Only A: 40 − 7 − 3 − 5 = 25. Only B: 35 − 7 − 5 − 5 = 18. Only C: 30 − 5 − 3 − 5 = 17. Exactly one = 60.
• Union = 60 + 15 + 5 = 80, which matches the formula: 105 − 30 + 5 = 80.

Two routes to the same total is your check.

Worked set: a survey filled region by region

A coaching centre surveyed 120 aspirants about three mock-test series, P, Q and R.

• 10 aspirants use none of the three.
• 70 use P, 50 use Q and 40 use R.
• 10 use all three, and 30 use exactly two.
• 22 use both P and Q, and 18 use both Q and R.

Step 1: the centre. All three = 10.

Step 2: the "exactly two" regions. "Both P and Q" includes the 10 who use all three, so P and Q only = 22 − 10 = 12. Likewise Q and R only = 18 − 10 = 8. The three exactly-two regions total 30, so P and R only = 30 − 12 − 8 = 10.

Step 3: the "only" regions.

• Only P = 70 − 12 − 10 − 10 = 38.
• Only Q = 50 − 12 − 8 − 10 = 20.
• Only R = 40 − 8 − 10 − 10 = 12.

Step 4: check. Inside the circles: 38 + 20 + 12 + 12 + 8 + 10 + 10 = 110. Add the 10 who use none: 120. The data is consistent.

Q1. How many use exactly one series? 38 + 20 + 12 = 70. Check with the formula: 160 = E1 + 2 × 30 + 3 × 10, so E1 = 70.

Q2. How many use P but not R? Only P + P and Q only = 38 + 12 = 50.

Q3. How many use at most one series? Exactly one + none = 70 + 10 = 80.

Q4 (TITA). What percentage of Q's users use no other series? 20 ÷ 50 = 40.

Overlap bounds: minimum and maximum

In a class of 100, 70 passed Maths, 75 passed English and 80 passed Science.

Minimum who passed all three. Count failures instead. 30 failed Maths, 25 failed English, 20 failed Science: at most 75 students failed something. So at least 100 − 75 = 25 passed all three. It is achievable when no student failed two subjects.

In general, with N people, minimum in all three = n(A) + n(B) + n(C) − 2N, if that is positive. Here 225 − 200 = 25.

Maximum who passed all three. It cannot exceed the smallest set: 70, achievable if every Maths passer also passed the other two.

Maximum who passed none. At most 100 − 80 = 20, achievable if the Maths and English passers are all among the 80 Science passers.

Minimum who passed at least two subjects. Let k be that number. Each of these students accounts for at most 3 passes, and each of the other 100 − k for at most 1. The 225 passes need 225 ≤ 3k + (100 − k), so 2k ≥ 125 and k ≥ 62.5. The minimum is 63. One split that achieves it: 62 passed all three, 1 passed English and Science only, and 8, 12 and 17 passed only Maths, only English and only Science.

Common mistakes

• Subtracting a pairwise intersection twice. When filling "only A", subtract the "only two" regions and the centre once each, not the raw pairwise totals.
• Ignoring "none". The union is the total minus those in none of the sets.
• Stating a bound without an example. A minimum is proven only when you show a split that reaches it.
• Mixing "exactly two" and "at least two". The difference is the centre.

Practice set

1. In a group of 60, 35 read a newspaper, 28 read a magazine and 12 read neither. How many read both?
2. In the survey set, how many aspirants use exactly one of Q and R, whatever they do about P?
3. In a group of 50, 40 speak Hindi, 35 speak English and 30 speak Marathi. What is the minimum number who speak all three?

Answers:

1. 15. 48 read at least one, so both = 35 + 28 − 48 = 15.
2. 54. Q but not R: only Q + P and Q only = 20 + 12 = 32. R but not Q: only R + P and R only = 12 + 10 = 22. Total 54.
3. 5. Non-speakers: 10, 15 and 20, so at most 45 miss at least one language, and at least 50 − 45 = 5 speak all three. Formula: 105 − 2 × 50 = 5.

What to do next

• Draw the three-circle diagram with eight labelled regions until you can do it in ten seconds.
• On every Venn set, fill from the centre outward and check the total two ways.
• Practise five minimum-overlap questions using the complement method.
• See set theory in QA for the same formulas in quant questions, and maxima and minima in LR for more bound-then-build reasoning.