Difference between revisions of "2019 AMC 8 Problems/Problem 16"
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<math>\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135</math> | <math>\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135</math> | ||
| − | ==Solution 1== | + | ==Solution 1(answer options)== |
| + | 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>Average Speed</math> = <math>\frac{Total Distance}{Total Time}</math> | ||
| + | |||
| + | Thus <math>\frac{125}{50} = \frac{5}{2}</math> | ||
| + | |||
| + | Both are equal and thus our answer <math>\boxed{\textbf{(D)}\ 110}</math> | ||
| + | |||
| + | ~phoenixfire | ||
==See Also== | ==See Also== | ||
Revision as of 13:39, 20 November 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 
= 
Thus 
Both are equal and thus our answer 
~phoenixfire
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. 
