Difference between revisions of "2011 AMC 10A Problems/Problem 15"
(→Problem 15) |
m |
||
Line 11: | Line 11: | ||
</math> | </math> | ||
− | == Solution == | + | == Solution 1 == |
We know that <math>\frac{\text{total miles}}{\text{total gas}}=55</math>. Let <math>x</math> be the distance the car traveled during the time it ran on gasoline, then the amount of gas used is <math>0.02x</math>. The total distance traveled is <math>40+x</math>, so we get <math>\frac{40+x}{0.02x}=55</math>. Solving this equation, we get <math>x=400</math>, so the total distance is <math>400 + 40 = \boxed{440 \ \mathbf{(C)}}</math>. | We know that <math>\frac{\text{total miles}}{\text{total gas}}=55</math>. Let <math>x</math> be the distance the car traveled during the time it ran on gasoline, then the amount of gas used is <math>0.02x</math>. The total distance traveled is <math>40+x</math>, so we get <math>\frac{40+x}{0.02x}=55</math>. Solving this equation, we get <math>x=400</math>, so the total distance is <math>400 + 40 = \boxed{440 \ \mathbf{(C)}}</math>. | ||
− | Solution 2 | + | == Solution 2 == |
− | The answer has to be | + | The answer has to be divisible by <math>55</math>, and the only answer that is divisible by 55 is <math>\boxed{440 \ \mathbf{(C)}}</math>. |
== See Also == | == See Also == | ||
{{AMC10 box|year=2011|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2011|ab=A|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:00, 3 December 2016
Contents
Problem 15
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of gallons per mile. On the whole trip he averaged miles per gallon. How long was the trip in miles?
Solution 1
We know that . Let be the distance the car traveled during the time it ran on gasoline, then the amount of gas used is . The total distance traveled is , so we get . Solving this equation, we get , so the total distance is .
Solution 2
The answer has to be divisible by , and the only answer that is divisible by 55 is .
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.