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2019 AMC 8 Problems/Problem 16

Revision as of 14:41, 30 June 2020 by Diegojun (talk | contribs) (→‎Solution 1)

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

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

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

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

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

Solution 3

If he travels $15$ miles at a speed of $30$ miles per hour, he travels for 30 min. Average rate is total distance over total time so $(15+d)/(0.5 + t) = 50$, where d is the distance left to travel and t is the time to travel that distance. solve for $d$ to get $d = 10+50t$. you also know that he has to travel $55$ miles per hour for some time, so $d=55t$ plug that in for d to get $55t = 10+50t$ and $t=2$ and since $d=55t$, $d = 2\cdot55 =110$ the answer is $\boxed{\textbf{(D)}\ 110}$.

-goldenn

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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