2018 AMC 10A Problem 5

P1-10~1 minPrint

Absolute Value & Inequalities Logic Puzzles Set up the equation/formula and compute; no special trick needed

Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, "We are at least $6$ miles away," Bob replied, "We are at most $5$ miles away." Charlie then remarked, "Actually the nearest town is at most $4$ miles away." It turned out that none of the three statements were true. Let $d$ be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of $d$ ?

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Key insight

Every statement is false, so flip each one: less than 6, more than 5, more than 4. The numbers that pass all three lie strictly between 5 and 6.

Solution

All three people are wrong, so the true distance $d$ is whatever makes each statement false.

Alice said "at least $6$ ", meaning $d \ge 6$ . That is false, so $d < 6$ . (If $d$ were exactly $6$ , she would be right, so $6$ is left out.)
Bob said "at most $5$ ", meaning $d \le 5$ . That is false, so $d > 5$ .
Charlie said "at most $4$ ", meaning $d \le 4$ . That is false, so $d > 4$ .
Combine. Any number bigger than $5$ is automatically bigger than $4$ , so Charlie's fact adds nothing new. We need $d > 5$ and $d < 6$ together: $5 < d < 6$ .

In interval notation, $5 < d < 6$ is written $(5, 6)$ .

The answer is $\boxed{\textbf{(D)}\ (5,6)}$ .

Why this works

The opposite of "at least $6$ " is "less than $6$ ", and the opposite of "at most $5$ " is "more than $5$ ". The edge number switches sides: it was allowed in the statement, so it is not allowed in the opposite. After flipping, sketch the three conditions on a number line and keep only the stretch that satisfies all of them.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Translate each statement, then negate it because all three are false.

Alice: $d \ge 6$ is false, so $d < 6$ .
Bob: $d \le 5$ is false, so $d > 5$ .
Charlie: $d \le 4$ is false, so $d > 4$ .

All three must hold. The conditions $d > 5$ and $d > 4$ together are just $d > 5$ , and combined with $d < 6$ we get $5 < d < 6$ .

The answer is $\boxed{\textbf{(D)}\ (5,6)}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Intermediate Counting & Probability, Chapter 2: Sets and Logic (§2.4)
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)
Intermediate Algebra, Chapter 12: Inequalities (§12.3, §12.4)