Difference between revisions of "2018 AMC 10A Problems/Problem 5"
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<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== | ==Solution== | ||
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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. | 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. | ||
| − | ==Video | + | ==Video Solutions== |
https://youtu.be/vO-ELYmgRI8 | https://youtu.be/vO-ELYmgRI8 | ||
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~savannahsolver | ~savannahsolver | ||
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| + | https://youtu.be/nNEXCxWLJzc | ||
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| + | Education, the Study of Everything | ||
== See Also == | == See Also == | ||
Revision as of 17:53, 29 November 2020
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 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 
be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of ?
Solution
From Alice and Bob, we know that 
From Charlie, we know that 
We take the intersection of these two intervals to yield 
, because the nearest town is between 5 and 6 miles away.
Video Solutions
~savannahsolver
Education, the Study of Everything
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. 
