AMC 8 · 2015 · #15

Grade 6 arithmeticlogic

set-partitioncomplementary-counting complementary-countingidentify-subproblems ↑ Prerequisites: multi-digit-arithmetic

📏 Short solution 💡 3 insights

Problem

At Euler Middle School, $198$ students voted on two issues in a school referendum with the following results: $149$ voted in favor of the first issue and $119$ voted in favor of the second issue. If there were exactly $29$ students who voted against both issues, how many students voted in favor of both issues?

$\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

149

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Toolkit + CCSS Solution

Understand

Restated: At Euler Middle School, $198$ students voted on two issues. $149$ voted yes on the first issue, $119$ voted yes on the second issue, and exactly $29$ voted no on both. How many students voted yes on both issues?

Givens: Total students who voted = $198$; Yes on issue 1: $|A| = 149$; Yes on issue 2: $|B| = 119$; Voted no on both (in neither A nor B) = $29$; Answer choices: (A) $49$, (B) $70$, (C) $79$, (D) $99$, (E) $149$

Unknowns: The number of students who voted yes on both issues, i.e. $|A \cap B|$

Understand

Plan

Primary tool: #12 Draw a Venn Diagram

Secondary: #7 Identify Subproblems

The trigger words "both" and "against both" (neither) point straight at Tool #12 (Venn diagram): two overlapping circles for issues $A$ and $B$, plus an outside region for the $29$ students who voted no on both. Tool #7 (Identify Subproblems) splits the work into two clean steps — first find how many voted yes on at least one issue (the union), then use that union to pin down the overlap. With those two subproblems in hand the Venn diagram fills in by simple subtraction; no algebra heavier than $x = a + b - u$ is needed.

Execute — Answer: D

#12 Draw a Venn Diagram 4.OA.A.3 Step 1

Set up the Venn diagram.
Draw two overlapping circles for $A$ (yes on issue 1) and $B$ (yes on issue 2), inside a box that holds all $198$ voters.
The $29$ students who voted no on both sit in the box but outside both circles.

$$\text{outside both circles} = 29$$

💡 Putting the "neither" group outside the circles separates them from anyone who voted yes on at least one issue — exactly the partition Tool #12 is built for.

#7 Identify Subproblems 4.OA.A.3 Step 2

Find the union $|A \cup B|$ — students who voted yes on at least one issue.
Everyone except the $29$ "no on both" students must be in one of the circles, so subtract from the total.

$$|A \cup B| = 198 - 29 = 169$$

💡 Splitting the $198$ voters into "in at least one circle" vs "in neither circle" is the Tool #7 subproblems move — solve the easier piece first.

#12 Draw a Venn Diagram 6.EE.B.7 Step 3

Apply the two-set inclusion-exclusion identity.
Adding $|A|$ and $|B|$ counts every "both" student twice, so $|A| + |B|$ overshoots the union by exactly $|A \cap B|$.
Rearranging gives a one-step equation for the overlap.

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

💡 The Venn picture makes the double-count obvious: the lens-shaped intersection is inside both $A$ and $B$, so it is counted once in $|A|$ and again in $|B|$.

#12 Draw a Venn Diagram 6.EE.B.7 Step 4

Substitute the known values $|A| = 149$, $|B| = 119$, $|A \cup B| = 169$ to compute the overlap.

$$|A \cap B| = 149 + 119 - 169 = 268 - 169 = 99 \;\Rightarrow\; \textbf{(D)}$$

💡 Solving the one-variable equation $x = a + b - u$ with given whole numbers is a straight Grade 6 substitution.

[1] #12 4.OA.A.3 Set up the Venn diagram. Draw two overlapping circles for $A$ (yes on issue 1) a

[2] #7 4.OA.A.3 Find the union $|A \cup B|$ — students who voted yes on at least one issue. Ever

[3] #12 6.EE.B.7 Apply the two-set inclusion-exclusion identity. Adding $|A|$ and $|B|$ counts ev

[4] #12 6.EE.B.7 Substitute the known values $|A| = 149$, $|B| = 119$, $|A \cup B| = 169$ to comp

Review

Reasonableness: Sanity-check the four Venn regions with the answer $99$: both $= 99$, only $A = 149 - 99 = 50$, only $B = 119 - 99 = 20$, neither $= 29$. Total $= 99 + 50 + 20 + 29 = 198$. The four pieces partition the $198$ voters exactly, so the answer holds. Also, $|A \cap B| = 99$ is at most $\min(|A|, |B|) = 119$, which a real intersection must satisfy.

Alternative: Tool #3 (Eliminate Possibilities) on the five choices: for each candidate $x$, check whether "only $A$" $= 149 - x$, "only $B$" $= 119 - x$, neither $= 29$, sum to $198$. That requires $149 + 119 - x + 29 = 198$, i.e. $x = 99$. Only choice (D) works; the others either overshoot or undershoot $198$.

CCSS standards used (min grade 6)

4.OA.A.3 Solve multistep word problems with whole numbers using the four operations (Subtracting $198 - 29 = 169$ to find the number of students who voted yes on at least one issue (the union of the two yes-sets).)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ and $px = q$ (Writing the inclusion-exclusion identity $|A \cup B| = |A| + |B| - |A \cap B|$ as a one-variable equation and solving it for $|A \cap B| = 149 + 119 - 169 = 99$.)

⭐ Once the Venn diagram is drawn, this AMC 8 problem only needs Grade 6 "write one equation, solve for the unknown" — the overlap pops out as $149 + 119 - 169 = 99$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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