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

## 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 $\text{Average Speed}$ = $\frac{\text{Total Distance}{{\text{Total Time}}$ (Error compiling LaTeX. ! File ended while scanning use of \frac .)

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 byProblem 15 Followed byProblem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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