AMC 10 · 2011 · #6

Easy mode Grade 4

Problem

Anna has a list of $20$ favorite songs. Ben has a list of $15$ favorite songs. Put both lists together and count each song only once. What is the smallest number of different songs the combined list could have?

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.

View mode:

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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