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> | ||
| − | ==Solution== | + | ==Solution 1== |
From Alice and Bob, we know that <math>5 < d < 6.</math> | From Alice and Bob, we know that <math>5 < d < 6.</math> | ||
From Charlie, we know that <math>4 < d.</math> | From Charlie, we know that <math>4 < d.</math> | ||
Revision as of 00:55, 16 February 2021
- 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 
be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of ?
Solution 1
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.
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 before 
and after 
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 .
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. 
