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Difference between revisions of "2018 AMC 10A Problems/Problem 5"
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{{duplicate|[[2018 AMC 12A Problems|2018 AMC 12A #4]] and [[2018 AMC 10A Problems|2018 AMC 10A #5]]}}
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{{duplicate|[[2018 AMC 10A Problems/Problem 5|2018 AMC 10A #5]] and [[2018 AMC 12A Problems/Problem 4|2018 AMC 12A #4]]}}
  
 
==Problem==
 
==Problem==
  
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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 <math>d</math> be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of <math>d</math>?
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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 <math>6</math> miles away," Bob replied, "We are at most <math>5</math> miles away." Charlie then remarked, "Actually the nearest town is at most <math>4</math> miles away." It turned out that none of the three statements were true. Let <math>d</math> be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of <math>d</math>?
  
 
<math>\textbf{(A) }  (0,4)  \qquad        \textbf{(B) }  (4,5)  \qquad    \textbf{(C) }  (4,6)  \qquad  \textbf{(D) }  (5,6)  \qquad  \textbf{(E) }  (5,\infty) </math>
 
<math>\textbf{(A) }  (0,4)  \qquad        \textbf{(B) }  (4,5)  \qquad    \textbf{(C) }  (4,6)  \qquad  \textbf{(D) }  (5,6)  \qquad  \textbf{(E) }  (5,\infty) </math>
  
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==Solution==
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==Solution 1==
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From Alice and Bob, we know that <math>5 < d < 6.</math>
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For each of the false statements, we identify its corresponding true statement. Note that:
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From Charlie, we know that <math>4 < d.</math>
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<ol style="margin-left: 1.5em;">
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We take the intersection of these two intervals to yield <math>\boxed{\textbf{(D) } (5,6)}</math>, because the nearest town is between 5 and 6 miles away.
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  <li><math>\mathrm{False}\cap\mathrm{True}=\varnothing.</math></li><p>
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  <li><math>\mathrm{False}\cup\mathrm{True}=[0,\infty).</math></li><p>
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</ol>
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We construct the following table:
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<cmath>\begin{array}{c||c|c}
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& &  \\ [-2.5ex]
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\textbf{Hiker} & \textbf{False Statement} & \textbf{True Statement} \\ [0.5ex]
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\hline
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& & \\ [-2ex]
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\textbf{Alice} & [6,\infty) & [0,6) \\
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& & \\ [-2.25ex]
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\textbf{Bob} & [0,5] & (5,\infty)  \\
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& & \\ [-2.25ex]
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\textbf{Charlie} & [0,4] & (4,\infty)
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\end{array}</cmath>
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Taking the intersection of the true statements, we have <cmath>[0,6)\cap(5,\infty)\cap(4,\infty)=(5,6)\cap(4,\infty)=\boxed{\textbf{(D) } (5,6)}.</cmath>
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~MRENTHUSIASM
  
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==Video Solutions==
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==Solution 2==
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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. Therefore, the answer is <math>\boxed{\textbf{(D) } (5,6)}</math>.
 +
 
 +
==Video Solution (HOW TO THINK CREATIVELY!)==
 +
https://youtu.be/nNEXCxWLJzc
 +
 
 +
Education, the Study of Everything
 +
 
 +
 
 +
 
 +
 
 +
==Video Solution==
 
https://youtu.be/vO-ELYmgRI8
 
https://youtu.be/vO-ELYmgRI8
  
 +
~IceMatrix
 +
 +
==Video Solution==
 
https://youtu.be/cLJ87xJzcWI
 
https://youtu.be/cLJ87xJzcWI
  
 
~savannahsolver
 
~savannahsolver
−
 
−
https://youtu.be/nNEXCxWLJzc
 
−
 
−
Education, the Study of Everything
 
  
 
== See Also ==
 
== See Also ==

Latest revision as of 13:38, 3 July 2023

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!)

https://youtu.be/nNEXCxWLJzc

Education, the Study of Everything



Video Solution

https://youtu.be/vO-ELYmgRI8

~IceMatrix

Video Solution

https://youtu.be/cLJ87xJzcWI

~savannahsolver

See Also

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. AMC logo.png

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