Difference between revisions of "2019 AMC 8 Problems/Problem 16"
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==Solution 2== | ==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 <math>x.</math> Therefore, the total distance is <math>15+x</math> and the total time (in hours) is <cmath>\frac{15}{30}+\frac{x}{55}=\frac{1}{2}+\frac{x}{55}.</cmath> We can set up the following equation: <cmath>\frac{15+x}{\frac{1}{2}+\frac{x}{55}}=50.</cmath> Simplifying the equation, we get <cmath>15+x=25+\frac{10x}{11}.</cmath> Solving the equation yields <math>x=110,</math> so our answer is <math>\boxed{\textbf{(D)}\ 110}</math>. | 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 <math>x.</math> Therefore, the total distance is <math>15+x</math> and the total time (in hours) is <cmath>\frac{15}{30}+\frac{x}{55}=\frac{1}{2}+\frac{x}{55}.</cmath> We can set up the following equation: <cmath>\frac{15+x}{\frac{1}{2}+\frac{x}{55}}=50.</cmath> Simplifying the equation, we get <cmath>15+x=25+\frac{10x}{11}.</cmath> Solving the equation yields <math>x=110,</math> so our answer is <math>\boxed{\textbf{(D)}\ 110}</math>. | ||
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+ | ~twinemma | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2019|num-b=15|num-a=17}} | {{AMC8 box|year=2019|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:08, 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
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
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.