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2011 AMC 10A Problem 6

2011 AMC 10A Problem 6AMC 10 Step by Step

Name: Date: September 13, 2026

Sets, Estimation & Miscellaneous · difficulty 1 of 5 · about 1 min · position P1-10

Set $A$ has 20 elements, and set $B$ has 15 elements. What is the smallest possible number of elements in $A \cup B$ , the union of $A$ and $B$ ?

A)
$5$
B)
$15$
C)
$20$
D)
$35$
E)
$300$

Answer

Key insight

The union is smallest when the smaller set sits entirely inside the larger one, so it cannot have fewer than 20 elements and 20 is attainable.

Solution

The union contains all of $A$ , so it has at least $20$ elements.

Exactly $20$ is achievable: let $B$ consist of $15$ of the elements of $A$ . Then $A \cup B = A$ , which has $20$ elements.

The answer is $\boxed{\textbf{(C)}\ 20}$ .

Why this works

By inclusion-exclusion, $|A \cup B| = 20 + 15 - |A \cap B|$ , and the overlap is at most $|B| = 15$ (a set cannot share more elements than it has). The union is minimized when the overlap is maximized, i.e. when $B \subseteq A$ . Minimum and maximum questions about unions come down to the two extreme configurations: nested sets and disjoint sets.

The trap

Subtracting to get 20 - 15 = 5, or treating the sets as disjoint and answering 35 (the maximum, not the minimum).

Common mistakes

Subtracting to get 20 - 15 = 5, or treating the sets as disjoint and answering 35 (the maximum, not the minimum).
Confusing the union with the intersection and answering $15$ .

Techniques

Consider the largest/smallest element or boundary case