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

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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>
 
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>\text{Average Speed}</math> = <math>\frac{\text{Total Distance}}{{\text{Total Time}}</math>
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And <math>\text{Average Speed}</math> = <math>\frac{Total Distance}{Total Time}</math>
 
   
 
   
 
Thus <math>\frac{125}{50} = \frac{5}{2}</math>
 
Thus <math>\frac{125}{50} = \frac{5}{2}</math>

Revision as of 17:42, 25 December 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 $\text{Average Speed}$ = $\frac{Total Distance}{Total Time}$

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

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

~phoenixfire

Solution 2

Note that the average speed is simply the total distance over the total time. Let the number of additional miles he has to drive be $x.$ Therefore, the total distance is $15+x$ and the total time (in hours) is \[\frac{15}{30}+\frac{x}{55}=\frac{1}{2}+\frac{x}{55}.\] We can set up the following equation: \[\frac{15+x}{\frac{1}{2}+\frac{x}{55}}=50.\] Simplifying the equation, we get \[15+x=25+\frac{10x}{11}.\] Solving the equation yields $x=110,$ so our answer is $\boxed{\textbf{(D)}\ 110}$.

~twinemma

See Also

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 AJHSME/AMC 8 Problems and Solutions

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