AMC 8 · 2010 · #20

Grade 7 countinglogic

fraction-arithmeticlcmset-partitioncomplementary-counting complementary-countingidentify-subproblems ↑ Prerequisites: fraction-arithmeticlcm

📏 Medium solution 💡 4 insights

Problem

In a room, $2/5$ of the people are wearing gloves, and $3/4$ of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and gloves?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: In a room, exactly $\tfrac{2}{5}$ of the people are wearing gloves and exactly $\tfrac{3}{4}$ of the people are wearing hats. Both fractions must come out to whole numbers of people. Over all room sizes that work, what is the smallest possible number of people who are wearing both a hat and gloves?

Givens: Fraction wearing gloves $= \tfrac{2}{5}$ of the total; Fraction wearing hats $= \tfrac{3}{4}$ of the total; Glove count and hat count must each be whole numbers; Answer choices: (A) $3$, (B) $5$, (C) $8$, (D) $15$, (E) $20$

Unknowns: The minimum possible number of people wearing both a hat and gloves

Understand

Plan

Primary tool: #12 Draw a Venn Diagram

Secondary: #7 Identify Subproblems, #11 Try Extreme Cases

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.

Execute — Answer: A

#7 Identify Subproblems 6.NS.B.4 Step 1

Subproblem 1: find the smallest total $T$ that makes both fractions count whole people.
$\tfrac{2}{5}T$ is a whole number when $T$ is a multiple of $5$; $\tfrac{3}{4}T$ is a whole number when $T$ is a multiple of $4$.
The smallest such $T$ is $\text{lcm}(5,4) = 20$.

$$T_{\min} = \text{lcm}(5, 4) = 20$$

💡 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.

#7 Identify Subproblems 5.NF.B.6 Step 2

Subproblem 2: with $T = 20$, count gloves and hats.
Take $\tfrac{2}{5}$ of $20$ for gloves and $\tfrac{3}{4}$ of $20$ for hats.

$$|G| = \tfrac{2}{5} \times 20 = 8, \quad |H| = \tfrac{3}{4} \times 20 = 15$$

💡 "Fraction of a quantity" as multiplication is the Grade 5 fraction-word-problem skill, applied separately to each circle.

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

Draw the Venn diagram.
Inside a $20$-person room, sketch a gloves circle ($8$ people) overlapping a hats circle ($15$ people).
Label the overlap (people wearing both) as $x$, glove-only as $8 - x$, and hat-only as $15 - x$.
Everyone in the room is either in the union of the two circles or in neither, so the four regions (glove-only, both, hat-only, neither) must add to $20$.

$$(8 - x) + x + (15 - x) + (\text{neither}) = 20$$

💡 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.

#11 Try Extreme Cases 7.EE.B.3 Step 4

Apply the extreme-case idea.
To minimize the overlap $x$, push as many people as possible *out* of the overlap into the glove-only and hat-only regions.
The hardest constraint is that all $8 + 15 = 23$ "slots" of gloves-and-hats must be filled by people in a room of only $20$, so at least $23 - 20 = 3$ people are forced to wear both.
Equivalently, set "neither" $= 0$ in the Venn equation and solve for $x$.

$$(8 - x) + x + (15 - x) + 0 = 20 \;\Rightarrow\; 23 - x = 20 \;\Rightarrow\; x = 3$$

💡 "As small as possible" $=$ push the other regions to their extreme; that is exactly Tool #11. Algebraically it is the inclusion-exclusion identity $|G \cap H| = |G| + |H| - |G \cup H|$ minimized by maximizing $|G \cup H|$.

#11 Try Extreme Cases 6.NS.B.4 Step 5

Confirm $x = 3$ on the choice list and rule out the others.
$3$ matches (A); (B) $5$, (C) $8$, (D) $15$, (E) $20$ all exceed the forced minimum we just derived, so none of them is the *smallest* possible overlap.
Larger room sizes (multiples of $20$) scale every region by the same factor and give the same minimum overlap rate, so $T = 20$ already realizes the smallest count.

$$x_{\min} = 3 \;\Rightarrow\; \textbf{(A)}$$

💡 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.

[1] #7 6.NS.B.4 Subproblem 1: find the smallest total $T$ that makes both fractions count whole

[2] #7 5.NF.B.6 Subproblem 2: with $T = 20$, count gloves and hats. Take $\tfrac{2}{5}$ of $20$

[3] #12 7.EE.B.3 Draw the Venn diagram. Inside a $20$-person room, sketch a gloves circle ($8$ pe

[4] #11 7.EE.B.3 Apply the extreme-case idea. To minimize the overlap $x$, push as many people as

[5] #11 6.NS.B.4 Confirm $x = 3$ on the choice list and rule out the others. $3$ matches (A); (B)

Review

Reasonableness: Sanity check the Venn diagram with $x = 3$: glove-only $= 8 - 3 = 5$, both $= 3$, hat-only $= 15 - 3 = 12$, neither $= 0$. Sum $= 5 + 3 + 12 + 0 = 20$. Gloves total: $5 + 3 = 8$ ✓. Hats total: $3 + 12 = 15$ ✓. Trying $x = 2$ would force the union to be $8 + 15 - 2 = 21 > 20$, which is impossible — so $x = 3$ really is the smallest legal value.

Alternative: Tool #6 (Guess and Check) on the answer list: try $x = 3$ first because it is the smallest. The Venn check above succeeds. Try $x = 2$: union would be $21$, more than the $20$-person room — fails. So $3$ is the floor, eliminating (B)-(E). Tool #13 (Convert to Algebra) gives the same answer directly via inclusion-exclusion: $|G \cap H| \geq |G| + |H| - T = 8 + 15 - 20 = 3$, with equality when $|G \cup H| = T$ (no one wears neither). We led with Tool #12 because the Venn picture makes "both means the center" obvious to a younger reader.

CCSS standards used (min grade 7)

6.NS.B.4 Find the greatest common factor and least common multiple (Finding the smallest legal room size $T = \text{lcm}(5, 4) = 20$ so that both $\tfrac{2}{5}T$ and $\tfrac{3}{4}T$ are whole numbers, and checking that larger multiples of $20$ do not produce a smaller overlap.)
5.NF.B.6 Solve real-world problems involving multiplication of fractions (Computing the number of glove-wearers $\tfrac{2}{5} \times 20 = 8$ and hat-wearers $\tfrac{3}{4} \times 20 = 15$ inside the $20$-person room.)
7.EE.B.3 Solve multi-step real-life problems with positive and negative rational numbers in any form (Setting up the Venn equation $(8 - x) + x + (15 - x) + (\text{neither}) = 20$ and solving the inclusion-exclusion inequality $x \geq 8 + 15 - 20 = 3$ to extract the minimum overlap.)

⭐ Once 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 7)

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