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2018 AMC 10A Problem 5

2018 AMC 10A Problem 5AMC 10 Step by Step

Name: Date: September 13, 2026

Absolute Value & Inequalities · difficulty 1 of 5 · about 1 min · position P1-10

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 $d$ ?

A)
$(0,4)$
B)
$(4,5)$
C)
$(4,6)$
D)
$(5,6)$
E)
$(5,\infty)$

Answer

Key insight

Negate each claim: d < 6, d > 5, d > 4; the intersection is the open interval (5, 6).

Solution

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

Why this works

The negation of "at least $a$ " is "strictly less than $a$ ," and the negation of "at most $b$ " is "strictly greater than $b$ "; the endpoints flip from included to excluded. After negating, the set of possible values is the intersection of the three half-lines, which a quick number-line sketch shows is the open interval between the strongest lower bound and the only upper bound.

The trap

Forgetting to negate and intersecting the original statements, or turning 'at least 6 is false' into d ≤ 6 instead of d < 6.

Common mistakes

Forgetting to negate and intersecting the original statements, or turning 'at least 6 is false' into d ≤ 6 instead of d < 6.
Using Charlie's bound $d > 4$ instead of Bob's stronger bound $d > 5$ , giving $(4,6)$ , choice (C).

Techniques

Set up the equation/formula and compute; no special trick needed