# 2018 AMC 10A Problems/Problem 5

The following problem is from both the 2018 AMC 10A #5 and 2018 AMC 12A #4, so both problems redirect to this page.

## Problem

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 $d$?

$\textbf{(A) } (0,4) \qquad \textbf{(B) } (4,5) \qquad \textbf{(C) } (4,6) \qquad \textbf{(D) } (5,6) \qquad \textbf{(E) } (5,\infty)$

## Solution 1

For each of the false statements, we identify its corresponding true statement. Note that:

1. $\mathrm{False}\cap\mathrm{True}=\varnothing.$
2. $\mathrm{False}\cup\mathrm{True}=[0,\infty).$

We construct the following table: $$\begin{array}{c||c|c} & & \\ [-2.5ex] \textbf{Hiker} & \textbf{False Statement} & \textbf{True Statement} \\ [0.5ex] \hline & & \\ [-2ex] \textbf{Alice} & [6,\infty) & [0,6) \\ & & \\ [-2.25ex] \textbf{Bob} & [0,5] & (5,\infty) \\ & & \\ [-2.25ex] \textbf{Charlie} & [0,4] & (4,\infty) \end{array}$$ Taking the intersection of the true statements, we have $$[0,6)\cap(5,\infty)\cap(4,\infty)=(5,6)\cap(4,\infty)=\boxed{\textbf{(D) } (5,6)}.$$ ~MRENTHUSIASM

## Solution 2

Think of the distances as if they are on a number line. Alice claims that $d > 6$, Bob says $d < 5$, while Charlie thinks $d < 4$. This means that all possible numbers less than $5$ and greater than $6$ are included. However, since the three statements are actually false, the distance to the nearest town is one of the numbers not covered. Therefore, the answer is $\boxed{\textbf{(D) } (5,6)}$.

## Video Solution (HOW TO THINK CREATIVELY!)

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