Difference between revisions of "2018 AMC 10A Problems/Problem 5"

Revision as of 09:07, 4 January 2022

The following problem is from both the 2018 AMC 12A #4 and 2018 AMC 10A #5, 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 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, which yields the interval $\boxed{\textbf{(D) } (5,6)}$ .

Solution 3 (Illustration)

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

$\mathrm{False}\cap\mathrm{True}=\varnothing.$

$\mathrm{False}\cup\mathrm{True}=[0,\infty).$

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

Video Solutions

https://youtu.be/vO-ELYmgRI8

https://youtu.be/cLJ87xJzcWI

~savannahsolver

https://youtu.be/nNEXCxWLJzc

Education, the Study of Everything

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 8: / Line 8: @@
 ==Solution 2==
-Think of the distances as if they are on a number line. Alice claims that <math>d > 6</math>, Bob says <math>d < 5</math>, while Charlie thinks <math>d < 4</math>. This means that all possible numbers before <math>5</math> and after <math>6</math> are included. But since the three statements are actually false, the distance to the nearest town is one of the numbers not covered, which yields the interval <math>\boxed{\textbf{(D) } (5,6)}</math>.
+Think of the distances as if they are on a number line. Alice claims that <math>d > 6</math>, Bob says <math>d < 5</math>, while Charlie thinks <math>d < 4</math>. This means that all possible numbers less than <math>5</math> and greater than <math>6</math> are included. However, since the three statements are actually false, the distance to the nearest town is one of the numbers not covered, which yields the interval <math>\boxed{\textbf{(D) } (5,6)}</math>.
 ==Solution 3 (Illustration)==

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