Difference between revisions of "2019 AMC 8 Problems/Problem 16"
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The only option that is easily divisible by <math>55</math> is <math>110</math>. Which gives 2 hours of travel. And by the formula <math>\frac{15}{30} + \frac{110}{50} = \frac{5}{2}</math> | The only option that is easily divisible by <math>55</math> is <math>110</math>. Which gives 2 hours of travel. And by the formula <math>\frac{15}{30} + \frac{110}{50} = \frac{5}{2}</math> | ||
| − | And <math>\text{Average Speed}</math> = <math>\frac{\text{Total Distance}{{\text{Total Time}}</math> | + | And <math>\text{Average Speed}</math> = <math>\frac{\text{Total Distance}}{{\text{Total Time}}</math> |
Thus <math>\frac{125}{50} = \frac{5}{2}</math> | Thus <math>\frac{125}{50} = \frac{5}{2}</math> | ||
Revision as of 18:42, 25 December 2019
Problem 16
Qiang drives 
miles at an average speed of 
miles per hour. How many additional miles will he have to drive at 
miles per hour to average 
miles per hour for the entire trip?
Solution 1(answer options)
The only option that is easily divisible by is 
. Which gives 2 hours of travel. And by the formula 
And 
= $\frac{\text{Total Distance}}{{\text{Total Time}}$ (Error compiling LaTeX. Unknown error_msg)
Thus 
Both are equal and thus our answer is 
~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 
Therefore, the total distance is 
and the total time (in hours) is ![\[\frac{15}{30}+\frac{x}{55}=\frac{1}{2}+\frac{x}{55}.\]](http://latex.artofproblemsolving.com/8/f/a/8fa0a8f8d7aee64b52fba8f7ec3c7cc0e5a67416.png)
We can set up the following equation: ![\[\frac{15+x}{\frac{1}{2}+\frac{x}{55}}=50.\]](http://latex.artofproblemsolving.com/8/0/4/80419da878b6f43d2ce0eb55fdb332542584e9b5.png)
Simplifying the equation, we get ![\[15+x=25+\frac{10x}{11}.\]](http://latex.artofproblemsolving.com/7/7/c/77c48a48d74aa538bf0f66c7a122dfea6e814dba.png)
Solving the equation yields 
so our answer is 
.
~twinemma
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. 
