AMC 10 · 2011 · #6

Grade 4 arithmetic

Archetype Overlap / Complement Counting └ Sub-archetype Two-Set Inclusion-Exclusion: Solve for Missing Region

set-partitionoptimizationprinciple-of-inclusion-exclusion extremal-construction ↑ Prerequisites: set-partition

📏 Short solution 💡 1 insight

Problem

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

Pick an answer.

(A)

(B)

(C)

(D)

(E)

300

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Set $A$ holds $20$ elements and set $B$ holds $15$ elements. Arrange the two sets so their union $A \cup B$ is as small as it can possibly be, and report how many elements that smallest union has.

Givens: $|A| = 20$ (set $A$ has $20$ elements); $|B| = 15$ (set $B$ has $15$ elements); Nothing forces the sets to be separate — they may share elements; Answer choices: (A) $5$, (B) $15$, (C) $20$, (D) $35$, (E) $300$

Unknowns: The smallest possible value of $|A \cup B|$

Understand

Plan

Primary tool: #14 Extreme Principle

Secondary: #12 Draw a Venn Diagram

The question asks for the *smallest possible* union, which is a min/max question — exactly what Tool #14, the Extreme Principle, is for. Instead of trying every arrangement, jump straight to the extreme: shared elements are the only way to shrink a union, so push the overlap as large as it can go. A Venn diagram (Tool #12) makes that extreme easy to see — slide $B$'s circle completely inside $A$'s circle so nothing pokes out. Then only counting is left.

Execute — Answer: C

#14 Extreme Principle 4.OA.A.3 Step 1

Aim for maximum overlap

The union counts every element that is in $A$, or in $B$, or in both — but each element only once.
Shared elements are the only thing that keeps the count from growing.
So to make $A \cup B$ as small as possible, make the overlap $A \cap B$ as large as possible.
That is the extreme case worth testing.

$|A \cup B| = |A| + |B| - |A \cap B|$, so a bigger $|A \cap B|$ means a smaller union.

💡 Overlap is the only discount on a union, so grab the biggest discount you can.

#12 Draw a Venn Diagram 4.OA.A.3 Step 2

Slide B inside A

How much can the overlap be?
$B$ only has $15$ elements, so at most $15$ elements can be shared.
The extreme arrangement is to let all $15$ of $B$'s elements also belong to $A$ — that is, draw $B$'s circle entirely inside $A$'s circle.
Now $B$ contributes no new elements; everything in $B$ was already counted in $A$.

$$B \subseteq A \;\Rightarrow\; A \cap B = B \;\Rightarrow\; |A \cap B| = 15$$

💡 If the small circle sits wholly inside the big circle, the picture of the union is just the big circle.

#14 Extreme Principle 2.NBT.B.5 Step 3

Count the union

With $B$ tucked inside $A$, the union is simply all of $A$.
Check it with the counting formula: add the two sizes and subtract the overlap.
Both routes give the same number, and it matches choice (C).

$$|A \cup B| = 20 + 15 - 15 = 20 \;\Rightarrow\; \textbf{(C)}$$

💡 When the smaller set adds nothing new, the union is exactly the bigger set's size.

[1] #14 4.OA.A.3 The union counts every element that is in $A$, or in $B$, or in both — but each

[2] #12 4.OA.A.3 How much can the overlap be? $B$ only has $15$ elements, so at most $15$ element

[3] #14 2.NBT.B.5 With $B$ tucked inside $A$, the union is simply all of $A$. Check it with the co

Review

Reasonableness: The union must be at least $20$, because $A$'s $20$ elements are always inside it — that rules out (A) $5$ and (B) $15$ as too small. It can be at most $20 + 15 = 35$ when the sets are completely separate, which is choice (D), the *largest* union, not the smallest. So the smallest sits at the bottom of that range, $20$, and (E) $300$ is far outside any possibility. Choice (C) $20$ is the only value that fits.

Alternative: Think about the range instead of one extreme. The union always satisfies $\max(|A|,|B|) \le |A \cup B| \le |A| + |B|$, i.e. $20 \le |A \cup B| \le 35$. The smallest allowed value is the left endpoint, $\max(20,15) = 20$, giving (C) at once.

CCSS standards used (min grade 4)

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Reasoning that overlap is the only way to shrink a union and that the shared part can be at most the size of the smaller set ($15$).)
2.NBT.B.5 Fluently add and subtract within 100 (Computing the union size $20 + 15 - 15 = 20$ once the maximum overlap is fixed.)

⭐ A union is smallest when the sets overlap the most, so the smallest possible union is just the size of the bigger set.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Aim for maximum overlap

Slide B inside A

Count the union

Review

CCSS standards used (min grade 4)

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