8.1 Overlapping Sets & Venn Diagrams

GMAT Focus Quantitative Reasoning is 21 Problem Solving questions in 45 minutes with no calculator. Official Guide Math Review 3.4 lists sets among the tested tools. Overlapping-set items are not abstract set theory: they are classification word problems. A finite population is sorted by one, two, or three attributes, and you must recover a missing cell without double-counting. The algebra is elementary. The difficulty is linguistic — both, only, at least one, and neither name different regions of the same diagram.

Two-Set Formula and Four Regions

When a population of size $T$ is classified by two overlapping attributes $A$ and $B$, every member falls into exactly one of four mutually exclusive regions: only $A$, only $B$, both, or neither.

The union identity is

$|A \cup B| = |A| + |B| - |A \cap B|$

The intersection is subtracted once because it was already counted inside both $|A|$ and $|B|$. Everyone outside the union is the complement:

$\text{Neither} = T - |A \cup B| = T - |A| - |B| + |A \cap B|$

Packed as a single accounting equation:

$T = |A| + |B| - |A \cap B| + \text{Neither}$

Compute the exclusive pieces immediately:

• Only $A = |A| - |A \cap B|$
• Only $B = |B| - |A \cap B|$
• Exactly one $= |A| + |B| - 2|A \cap B|$
• At least one $= |A \cup B| = T - \text{Neither}$

A Venn diagram is a bookkeeping sketch, not a picture to admire. On a no-calculator exam, label each exclusive region, convert the given English into equations, and solve. The most common translation error is treating $|A|$ as if it already meant only A. The total for group $A$ includes the overlap.

Set-Phrasing Translation Table

The Double-Matrix Method

When each attribute is binary and exhaustive — every person is in $A$ or not in $A$, and in $B$ or not in $B$ — a $2 \times 2$ yes/no table is faster than circles. Rows are $A$ versus not-$A$. Columns are $B$ versus not-$B$. Each row sums to a marginal total, each column sums to a marginal total, and the four interior cells are mutually exclusive and complete.

Because every person occupies exactly one cell, any three independent facts determine the fourth. If the prompt gives percents rather than headcounts, assign a convenient total of $100$ or the least common multiple of the denominators, fill known cells, and subtract. Typical GMAT matrix pairs include domestic versus international crossed with full-time versus part-time, and online versus in-person crossed with pass versus fail.

Three-Set Overlap

Three attributes $A$, $B$, and $C$ partition the universe into eight exclusive regions: only $A$, only $B$, only $C$, only $A&B$, only $B&C$, only $C&A$, all three, and neither.

Standard inclusion-exclusion uses pairwise totals that already include the triple:

$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |C \cap A| + |A \cap B \cap C|$

$\text{Neither} = T - |A \cup B \cup C|$

When the prompt instead gives exactly two and exactly three counts, switch identities. In the raw sum $|A| + |B| + |C|$, a person in exactly one set is counted once, a person in exactly two sets is counted twice, and a person in all three is counted three times:

$|A| + |B| + |C| = (\text{Exactly 1}) + 2(\text{Exactly 2}) + 3(\text{Exactly 3})$

Combined with $T = (\text{Exactly 1}) + (\text{Exactly 2}) + (\text{Exactly 3}) + \text{Neither}$, this rearranges to

$T = |A| + |B| + |C| - (\text{Exactly 2}) - 2(\text{Exactly 3}) + \text{Neither}$

The factor of $2$ on the triple is the correction that reduces a triple-counted person from three down to one. Subtracting the triple a third time would erase the person from the total.

Pairwise totals and exclusive pairwise regions are not interchangeable:

• Only $A&B$ (exactly those two) $= |A \cap B| - |A \cap B \cap C|$
• Exactly two $= |A \cap B| + |B \cap C| + |C \cap A| - 3|A \cap B \cap C|$

Extreme Overlap Bounds

If the item asks for the greatest or least possible intersection rather than a unique value:

1. Maximum $|A \cap B| = \min(|A|, |B|)$ — the smaller group sits entirely inside the larger one.
2. Minimum $|A \cap B| = \max(0, |A| + |B| - T)$ — overlap is forced once $A$ and $B$ together exceed the universe.

These bounds are the set-theory version of the discrete-optimization questions in Section 8.4.

Worked Example: Two-Set Regions

A cohort of $40$ candidates includes $22$ who studied algebra, $18$ who studied statistics, and $8$ who studied both. How many studied neither, and how many studied only algebra?

Union $= 22 + 18 - 8 = 32$, so neither $= 40 - 32 = 8$. Only algebra $= 22 - 8 = 14$. Exactly one subject $= 14 + 10 = 24$.

Worked Example: Three-Set Pairwise Totals

A club of $120$ members has $60$ tennis players, $50$ golfers, and $40$ squash players. Twenty play tennis and golf, $15$ play tennis and squash, $10$ play golf and squash, and $5$ play all three (those pairwise figures include the triple). How many play none of the three sports?

Union $= 60 + 50 + 40 - 20 - 15 - 10 + 5 = 110$. None $= 120 - 110 = 10$. Exclusive check: only tennis-and-golf $= 15$, only tennis-and-squash $= 10$, only golf-and-squash $= 5$, all three $= 5$, only tennis $= 30$, only golf $= 25$, only squash $= 20$. Those seven occupied regions sum to $110$; adding $10$ neither recovers $120$.

Worked Example: Forced Minimum Overlap

Among $80$ employees, $50$ drink coffee and $45$ drink tea. The greatest possible number who drink both is $\min(50, 45) = 45$. The least possible number who drink both is $\max(0, 50 + 45 - 80) = 15$.