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{Total Distance}{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 19:21, 26 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 =
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 We can set up the following equation: Simplifying the equation, we get Solving the equation yields so our answer is .
~twinemma
Solution 3
If he travels miles at a speed of miles per hour, he travels for 30 min. Average rate is total distance over total time so , where d is the distance left to travel and t is the time to tragvel that distance. solve for to get . you also know that he has to tral miles per hour for some time, so plug that in for d to get and and since , the answer is .
-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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