Difference between revisions of "2018 AMC 10A Problems/Problem 5"
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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>? | + | {{duplicate|[[2018 AMC 10A Problems/Problem 5|2018 AMC 10A #5]] and [[2018 AMC 12A Problems/Problem 4|2018 AMC 12A #4]]}} |
+ | |||
+ | ==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 <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> | ||
+ | |||
+ | ==Solution 1== | ||
+ | For each of the false statements, we identify its corresponding true statement. Note that: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li><math>\mathrm{False}\cap\mathrm{True}=\varnothing.</math></li><p> | ||
+ | <li><math>\mathrm{False}\cup\mathrm{True}=[0,\infty).</math></li><p> | ||
+ | </ol> | ||
+ | We construct the following table: | ||
+ | <cmath>\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}</cmath> | ||
+ | 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> | ||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==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 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 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/cLJ87xJzcWI | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2018|ab=A|num-b=4|num-a=6}} | {{AMC10 box|year=2018|ab=A|num-b=4|num-a=6}} | ||
+ | {{AMC12 box|year=2018|ab=A|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Logic Problems]] |
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.
Contents
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 miles away," Bob replied, "We are at most miles away." Charlie then remarked, "Actually the nearest town is at most miles away." It turned out that none of the three statements were true. Let be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of ?
Solution 1
For each of the false statements, we identify its corresponding true statement. Note that:
We construct the following table: Taking the intersection of the true statements, we have ~MRENTHUSIASM
Solution 2
Think of the distances as if they are on a number line. Alice claims that , Bob says , while Charlie thinks . This means that all possible numbers less than and greater than 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 .
Video Solution (HOW TO THINK CREATIVELY!)
Education, the Study of Everything
Video Solution
~IceMatrix
Video Solution
~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.