Difference between revisions of "2019 AMC 8 Problems/Problem 16"

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<math>\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135</math>
 
<math>\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135</math>
  
==Solution 1==
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==Solution 1(answer options)==
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The only option that is easily divisible by <math>55</math> is <math>110</math>. Which gives 2 hours of travel. And by the formula <math>\frac{15}{30} + \frac{110}{50} = \frac{5}{2}</math>
  
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And <math>Average Speed</math> = <math>\frac{Total Distance}{Total Time}</math>
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Thus <math>\frac{125}{50} = \frac{5}{2}</math>
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Both are equal and thus our answer <math>\boxed{\textbf{(D)}\ 110}</math>
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~phoenixfire
  
 
==See Also==
 
==See Also==

Revision as of 13:39, 20 November 2019

Problem 16

Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?

$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$

Solution 1(answer options)

The only option that is easily divisible by $55$ is $110$. Which gives 2 hours of travel. And by the formula $\frac{15}{30} + \frac{110}{50} = \frac{5}{2}$

And $Average Speed$ = $\frac{Total Distance}{Total Time}$

Thus $\frac{125}{50} = \frac{5}{2}$

Both are equal and thus our answer $\boxed{\textbf{(D)}\ 110}$

~phoenixfire

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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