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Difference between revisions of "2019 AMC 8 Problems/Problem 16"
 
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~Education, the Study of Everything
 
~Education, the Study of Everything
  
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==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)==
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==Video Solution by The Power of Logic(1 to 25 Full Solution)==
 
https://youtu.be/Xm4ZGND9WoY
 
https://youtu.be/Xm4ZGND9WoY
  

Latest revision as of 09:32, 9 November 2024

Problem

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 (by looking at the answer choices)

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



~TaeKim

Video Solution

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/IgpayYB48C4?si=8YldGqXbPzZfeA-z&t=4756

~Math-X


https://www.youtube.com/watch?v=OC1KdFeZFeE

Associated Video

https://youtu.be/5K1AgeZ8rUQ

- happytwin

https://www.youtube.com/watch?v=0rcDe2bDRug ~David

Video Solution

Solution detailing how to solve the problem:

https://www.youtube.com/watch?v=sEZ0sM-d1FA&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=17

Video Solution

https://youtu.be/aFsC5awOWBk

- Soo, DRMS, NM

Video Solution

https://youtu.be/btmFN_C1zSg

~savannahsolver

Video Solution (The Fastest Way)

https://www.youtube.com/watch?v=IOhkO3c3c2A&ab_channel=SaxStreak002

~SaxStreak

Video Solution (MOST EFFICIENT+ CREATIVE THINKING!!!(BTW WE LOVE THIS ONE))

https://youtu.be/JiTos1fFtUA

~Education, the Study of Everything

Video Solution by The Power of Logic(1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

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

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

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