Art of Problem Solving
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

2011 AMC 10A Problems/Problem 15: Difference between revisions

AlphaPi17 (talk | contribs)
Happia (talk | contribs)
mNo edit summary
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 divisble by 55, and the only answer that is divisible by 55 is C. so it's C.
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

Problem 15

Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first $40$ miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of $0.02$ gallons per mile. On the whole trip he averaged $55$ miles per gallon. How long was the trip in miles?

$\mathrm{(A)}\ 140 \qquad \mathrm{(B)}\ 240 \qquad \mathrm{(C)}\ 440 \qquad \mathrm{(D)}\ 640 \qquad \mathrm{(E)}\ 840$

Solution 1

We know that $\frac{\text{total miles}}{\text{total gas}}=55$. Let $x$ be the distance the car traveled during the time it ran on gasoline, then the amount of gas used is $0.02x$. The total distance traveled is $40+x$, so we get $\frac{40+x}{0.02x}=55$. Solving this equation, we get $x=400$, so the total distance is $400 + 40 = \boxed{440 \ \mathbf{(C)}}$.

Solution 2

The answer has to be divisible by $55$, and the only answer that is divisible by 55 is $\boxed{440 \ \mathbf{(C)}}$.

See Also

2011 AMC 10A (ProblemsAnswer KeyResources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination