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

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~phoenixfire
 
~phoenixfire
  
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==Solution 2==
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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 <math>x.</math> Therefore, the total distance is <math>15+x</math> and the total time (in hours) is <cmath>\frac{15}{30}+\frac{x}{55}=\frac{1}{2}+\frac{x}{55}.</cmath> We can set up the following equation: <cmath>\frac{15+x}{\frac{1}{2}+\frac{x}{55}}=50.</cmath> Simplifying the equation, we get <cmath>15+x=25+\frac{10x}{11}.</cmath> Solving the equation yields <math>x=110,</math> so our answer is <math>\boxed{\textbf{(D)}\ 110}</math>.
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2019|num-b=15|num-a=17}}
 
{{AMC8 box|year=2019|num-b=15|num-a=17}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:07, 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

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 $\boxed{\textbf{(D)}\ 110}$.

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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