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

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

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

To calculate the average speed, simply evaluate 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 $\boxed{\textbf{(D)}\ 110}$ .

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

Solution 4

Let $h$ be the amount of hours Qiang drives after his first 15 miles. Average speed, which we know is $50$ mph, means total distance over total time. For 15 miles at 30 mph, the time taken is $\frac{1}{2}$ hour, so the total time for this trip would be $\frac{1}{2} + h$ hours. For the total distance, 15 miles are traveled in the first part and $55h$ miles in the second. This gives the following equation:

$\dfrac{15+55h}{\frac{1}{2}+h} = 50.$

Cross multiplying, we get that $15 + 55h = 50h + 25$ , and simple algebra gives $h=2$ . In 2 hours traveling at 55 mph, the distance traveled is $\frac{2 \hspace{0.05 in} \text{hours}}{1} \cdot \frac{55 \hspace{0.05 in} \text{miles}}{1 \hspace{0.05 in} \text{hour}} = 2 \cdot 55 \hspace{0.05 in} \text{miles} = 110 \hspace{0.05 in} \text{miles}$ , which is choice $\boxed{\textbf{(D)}\ 110}$ .

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2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2019 AMC 8 Problems/Problem 16

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Problem 16

Solution 1

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