Competition · AMC preparation · step 4 of 4
AMC 8 · 2015 · #15
Grade 6 arithmeticlogicPick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Set up the Venn diagram
Draw circles A and B inside a box holding all 198 voters; the 29 who voted no on both sit 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.3Draw A Venn DiagramFind the union
Everyone except those 29 is in at least one circle, so the union is 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.
4.OA.A.3Identify SubproblemsWrite the inclusion-exclusion rule
By inclusion-exclusion, |A| + |B| counts the overlap twice, so it exceeds the union by exactly |A ∩ 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|.
Adding circle A's total to circle B's total and then subtracting the count in at least one circle leaves exactly the count in both circles: |A ∩ B| = |A| + |B| - |A ∪ B|.
▸ Why?
Adding circle A's total to circle B's total counts each student who is in both circles twice, because such a student sits inside circle A and also inside circle B.
▸ Why?
Circle A's total splits with no gaps or overlaps into the only-A students and the both students, and those parts add back to the total, so circle A's count contains the both students once.
▸ Why?
Circle B's total splits the same way into the only-B students and the both students, so circle B's count also contains the both students once.
▸ Why?
The count in at least one circle is the only-A, only-B, and both regions joined with no gaps or overlaps, so that union count already contains the both students exactly once.
▸ Why?
So the two totals added together equal the union count plus one extra copy of the both students, and subtracting the union count removes that copy, leaving only the both students.
▸ Why?
Subtraction undoes addition, so taking the union count back off cancels the matching copy hidden inside the sum and leaves just the leftover both students.
Substitute to get the overlap
Substitute: |A ∩ B| = 149 + 119 - 169 = 99, choice (D).
Solving the one-variable equation x = a + b - u with given whole numbers is a straight Grade 6 substitution.
6.EE.B.7Draw A Venn DiagramOnce 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.
- Set up the Venn diagram
- Find the union
- Write the inclusion-exclusion rule
- Substitute to get the overlap
A parent dashboard for the family lives at sensimlab.com.