AMC 10 · 2011 · #8

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

if-then-reasoningcontrapositivelogical-deduction logical-deduction ↑ Prerequisites: if-then-reasoning

📏 Medium solution 💡 2 insights

Problem

At a certain beach if it is at least $80^{\circ} F$ and sunny, then the beach will be crowded. On June 10 the beach was not crowded. What can be concluded about the weather conditions on June 10?

Pick an answer.

(A)

The temperature was cooler than } 80^{\circ} \text{F and it was not sunny.

(B)

The temperature was cooler than } 80^{\circ} \text{F or it was not sunny.

(C)

If the temperature was at least } 80^{\circ} \text{F, then it was sunny.

(D)

If the temperature was cooler than } 80^{\circ} \text{F, then it was sunny.

(E)

If the temperature was cooler than } 80^{\circ} \text{F, then it was not sunny.

AMC 10 2011 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.

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Toolkit + CCSS Solution

Understand

Restated: A beach rule says: if it is at least $80^{\circ}\text{F}$ AND sunny, then the beach is crowded. On June 10 the beach was NOT crowded. Among five statements about that day's weather, decide which one MUST be true.

Givens: The rule: (temperature at least $80^{\circ}\text{F}$) AND (sunny) $\Rightarrow$ crowded; On June 10 the beach was not crowded; Five candidate conclusions: (A) cooler than $80^{\circ}$ AND not sunny; (B) cooler than $80^{\circ}$ OR not sunny; (C) if at least $80^{\circ}$ then sunny; (D) if cooler than $80^{\circ}$ then sunny; (E) if cooler than $80^{\circ}$ then not sunny

Unknowns: Which single statement is forced to be true by the rule together with the fact that the beach was not crowded

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #11 Work Backwards, #3 Eliminate Possibilities

The rule is a one-way if-then whose condition is a compound "hot AND sunny." Tool #11 (Work Backwards) turns the not-crowded fact into the only statement that must travel with the rule — its contrapositive: not crowded $\Rightarrow$ not (hot and sunny). Tool #16 (Change Focus / Count the Complement) then does the load-bearing move: the complement of "hot AND sunny" is not "cool AND cloudy" but "cool OR cloudy" — negating an AND flips it to an OR (De Morgan). With that disjunction in hand, Tool #3 (Eliminate Possibilities) tests the five choices; only the correct negation survives while the AND-trap and the converse/inverse traps fall away.

Execute — Answer: B

#11 Work Backwards 6.EE.B.8 Step 1

Write the rule in symbols

Translate the temperature phrase into an inequality: "at least $80^{\circ}\text{F}$" means $T \ge 80$.
The rule's trigger is a compound condition — it needs BOTH parts at once: $(T \ge 80)$ AND (sunny).
So the rule is $\big[(T \ge 80)\text{ and sunny}\big] \Rightarrow \text{crowded}$.
It only runs this one direction; it never claims a crowd can form only this way.

$$\big[(T \ge 80)\ \text{and}\ \text{sunny}\big] \Rightarrow \text{crowded}$$

💡 "At least $80$" is a boundary you can write as $T \ge 80$, and an AND-condition fires only when every part is true.

#11 Work Backwards 6.EE.B.5 Step 2

Reverse it: the contrapositive

Every one-way if-then carries exactly one partner that must be true alongside it — its contrapositive, formed by reversing the arrow and negating both sides.
We are told the beach was not crowded, which is the negated conclusion.
Reversing the rule gives: not crowded $\Rightarrow$ not (hot and sunny).
This is the only new fact the rule forces once we know there was no crowd.

$$\neg\,\text{crowded} \;\Rightarrow\; \neg\big[(T \ge 80)\ \text{and}\ \text{sunny}\big]$$

💡 If a condition always makes something happen, then that something failing proves the condition was not fully met.

#16 Change Focus / Count the Complement 6.EE.B.5 Step 3

Negate the AND — it becomes an OR

Now unpack "not (hot and sunny)." To break a two-part AND-condition you only need ONE part to fail, so its negation is a two-part OR: (not hot) OR (not sunny).
Negating $T \ge 80$ gives $T < 80$, i.e.
cooler than $80^{\circ}\text{F}$.
So "not (hot and sunny)" reads: cooler than $80^{\circ}\text{F}$ OR not sunny.
The connector must be OR, not AND — the day could be cool and still sunny, or hot and still cloudy.

$$\neg\big[(T \ge 80)\ \text{and}\ \text{sunny}\big] \;=\; (T < 80)\ \text{or}\ (\text{not sunny})$$

💡 Wrecking a "both must hold" rule takes just one broken part, so "not both" opens up to "either one."

#3 Eliminate Possibilities 6.EE.B.5 Step 4

Test the five choices

Compare each statement to the forced fact "cooler than $80^{\circ}$ OR not sunny." (A) uses AND — too strong, since a cool-but-sunny day is not crowded yet fails "not sunny." (C) if hot then sunny — not forced; a hot cloudy day can be uncrowded.
(D) if cool then sunny and (E) if cool then not sunny both invent a link the rule never states, so a cool sunny day breaks (E) and a cool cloudy day breaks (D).
Only (B), "cooler than $80^{\circ}\text{F}$ OR not sunny," matches the negation exactly, so the answer is (B).

$$\text{(A) AND ✗, (C) ✗, (D) ✗, (E) ✗, (B) OR ✓} \;\Rightarrow\; \textbf{(B)}$$

💡 Only the exact De Morgan negation is guaranteed; the AND-version and the extra if-then links are the classic traps.

[1] #11 6.EE.B.8 Translate the temperature phrase into an inequality: "at least $80^{\circ}\text{

[2] #11 6.EE.B.5 Every one-way if-then carries exactly one partner that must be true alongside it

[3] #16 6.EE.B.5 Now unpack "not (hot and sunny)." To break a two-part AND-condition you only nee

[4] #3 6.EE.B.5 Compare each statement to the forced fact "cooler than $80^{\circ}$ OR not sunny

Review

Reasonableness: Stress-test with concrete uncrowded days. A cool AND sunny day is not hot-and-sunny, so it can be uncrowded — it satisfies (B) via "cooler than $80^{\circ}$," but breaks (A) and (E), which demand "not sunny." A hot AND cloudy day can be uncrowded too — it satisfies (B) via "not sunny," but breaks (C). A cool AND cloudy day breaks (D). In every uncrowded case at least one part of (B) holds, while each other choice fails some case. This confirms (B).

Alternative: Reason purely by counterexample instead of formal logic. For each of (A), (C), (D), (E), picture a specific uncrowded day — cool-and-sunny, hot-and-cloudy, or cool-and-cloudy — that makes that statement false while keeping the beach uncrowded. Every one of them has such a counterexample. Only (B) has none, because the beach could be crowded only when it is both hot and sunny; so being uncrowded forces at least one of "cool" or "not sunny." The statement with no counterexample is the answer: (B).

CCSS standards used (min grade 6)

6.EE.B.8 Write an inequality of the form x > c or x < c and graph on a number line (Translating "at least $80^{\circ}\text{F}$" into $T \ge 80$ and its negation "cooler than $80^{\circ}\text{F}$" into $T < 80$.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Truth-testing each if-then choice against the rule — forming the contrapositive, applying De Morgan to negate the AND-condition into an OR, and checking which statement stays true in every uncrowded case.)

⭐ To break a "both A and B" rule you only need ONE part to fail, so "not (hot and sunny)" means cool OR cloudy — joined by OR, never AND.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Write the rule in symbols

Reverse it: the contrapositive

Negate the AND — it becomes an OR

Test the five choices

Review

CCSS standards used (min grade 6)

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