AMC 8 · 2015 · #15
Easy mode Grade 6Problem
At Euler Middle School, 198 students voted on two different issues.
- 149 students voted YES on the first issue.
- 119 students voted YES on the second issue.
- 29 students voted NO on BOTH issues.
A student is allowed to vote YES on the first issue, the second issue, both, or neither.
How many students voted YES on both issues?
Pick an answer.
AMC 8 2015 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.
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
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) $149$
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
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.
💡 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.
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.
💡 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.
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.
💡 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 $|B|$.
6.EE.B.7 Step 4 Substitute the known values $|A| = 149$, $|B| = 119$, $|A \cup B| = 169$ to compute the overlap.
💡 Solving the one-variable equation $x = a + b - u$ with given whole numbers is a straight Grade 6 substitution.
4.OA.A.3 Set up the Venn diagram. Draw two overlapping circles for $A$ (yes on issue 1) a 4.OA.A.3 Find the union $|A \cup B|$ — students who voted yes on at least one issue. Ever 6.EE.B.7 Apply the two-set inclusion-exclusion identity. Adding $|A|$ and $|B|$ counts ev 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.3Solve 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.7Solve 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 $149 + 119 - 169 = 99$.
⭐ 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 $149 + 119 - 169 = 99$.
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