Competition · AMC preparation · step 4 of 4
AMC 8 · 2010 · #20
Grade 7 countinglogicPick an answer.
AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The trigger word both points straight at Tool #12 (Venn Diagram): draw a gloves circle and a hats circle, and the answer lives in the overlap. Tool #7 (Identify Subproblems) splits the work into two clean halves — first pin down the smallest legal room size (an LCM question), then count the overlap inside that room. Tool #11 (Try Extreme Cases) is what "minimum" means here: to make the overlap as small as possible, push the two circles apart as far as the room will allow, i.e. fill the room with "glove-only" and "hat-only" people first and only force the leftover into the overlap.
Find the smallest workable total
Find the smallest room size making both fractions whole: the LCM of 5 and 4 is 20.
Splitting off "what room size is even allowed" first is the Tool #7 move; the LCM is the Grade 6 number-theory tool that answers it.
6.NS.B.4Identify SubproblemsCount gloves and hats
With 20 people, gloves = × 20 = 8 and hats = × 20 = 15.
"Fraction of a quantity" as multiplication is the Grade 5 fraction-word-problem skill, applied separately to each circle.
5.NF.B.6Identify SubproblemsDraw the Venn diagram
Draw a gloves circle (8) and hats circle (15); label the overlap x, so the four regions sum to 20.
Tool #12 turns "both" into a labeled center region and lets us write a single equation that ties all four regions to the room total.
7.EE.B.3Draw A Venn DiagramPush to the extreme case
To minimize the overlap, set "neither" = 0: 23 - x = 20, so x = 3 people must wear both.
"As small as possible" = push the other regions to their extreme; that is exactly Tool #11. Algebraically it is the inclusion-exclusion identity |G ∩ H| = |G| + |H| - |G ∪ H| minimized by maximizing |G ∪ H|.
The number of people wearing both a hat and gloves can be no smaller than the number of glove-wearers plus the number of hat-wearers minus the total number of people in the room.
▸ Why?
Adding the number of glove-wearers to the number of hat-wearers counts everyone who wears both twice, so that sum equals the number of people wearing at least one item plus one extra copy of the overlap.
▸ Why?
The glove-wearers split with no overlap into glove-only people and both-wearers, and the hat-wearers split into hat-only people and both-wearers; adding the two group sizes therefore lays down glove-only once, hat-only once, and the both-wearers twice.
▸ Why?
The people wearing at least one item can never outnumber the room, because together with the people wearing neither they fill exactly the 20-person room and nothing more.
▸ Why?
Rearranging 'glove-wearers plus hat-wearers equals (people wearing at least one item) plus overlap' to isolate the overlap uses subtraction to undo the addition, so the overlap is smallest exactly when the people wearing at least one item is largest, which happens when they fill the whole room.
Confirm against the choices
The forced minimum matches (A); bigger rooms (multiples of 20) scale every region equally, so none beats x = 3.
Checking that doubling the room (T = 40) doubles everything and so doesn't beat T = 20 is the Tool #11 "have I really hit the extreme?" follow-up.
6.NS.B.4Work BackwardsOnce the Venn diagram is drawn, the minimum "both" overlap is just (gloves) + (hats) - (room total) — a Grade 7 multi-step reasoning move on top of Grade 5–6 fractions and LCM.
- Find the smallest workable total
- Count gloves and hats
- Draw the Venn diagram
- Push to the extreme case
- Confirm against the choices
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