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== |
| − | 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, | + | 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>. |
==Solution 3 (Illustration)== | ==Solution 3 (Illustration)== | ||
Revision as of 09:09, 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 
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
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 
.
Solution 3 (Illustration)
For each of the false statements, we identify its corresponding true statement. Note that:
We construct the following table:
![\[\begin{array}{c||c|c} & & \\ [-2.5ex] \textbf{Hiker} & \textbf{False Statement} & \textbf{True Statement} \\ [0.5ex] \hline & & \\ [-2ex] \textbf{Alice} & [6,\infty) & [0,6) \\ & & \\ [-2.25ex] \textbf{Bob} & [0,5] & (5,\infty) \\ & & \\ [-2.25ex] \textbf{Charlie} & [0,4] & (4,\infty) \end{array}\]](http://latex.artofproblemsolving.com/1/6/4/1645d4ff8df04a7bdc85c3c64aac7453f53711a4.png)
Taking the intersection of the true statements, we have ![\[[0,6)\cap(5,\infty)\cap(4,\infty)=(5,6)\cap(4,\infty)=\boxed{\textbf{(D) } (5,6)}.\]](http://latex.artofproblemsolving.com/3/7/e/37e8ff50fd0dbecfbb52f8742566d5fd1241f893.png)
~MRENTHUSIASM
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. 
