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 1==
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==Solution==
 
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 19:50, 9 February 2018

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

From Alice and Bob, we know that $5 < d < 6.$ From Charlie, we know that $4 < d.$ We take the union of these two intervals to yield $\boxed{\textbf{(D)} (5,6)}.$

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
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