# Difference between revisions of "2017 AMC 10B Problems/Problem 12"

## Problem

Elmer's new car gives $50\%$ percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip? $\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%$

## Solution 1

Suppose that his old car runs at $x$ km per liter. Then his new car runs at $\frac{3}{2}x$ km per liter, or $x$ km per $\frac{2}{3}$ of a liter. Let the cost of the old car's fuel be $c$, so the trip in the old car takes $xc$ dollars, while the trip in the new car takes $\frac{2}{3}\cdot\frac{6}{5}xc = \frac{4}{5}xc$. He saves $\frac{\frac{1}{5}xc}{xc} = \boxed{\textbf{(A)}\ 20\%}$.

## Solution 2

Because they do not give you a given amount of distance, we'll just make that distance $3x$ miles. Then, we find that the new car will use $2*1.2=2.4x$. The old car will use $3x$. Thus the answer is $(3-2.4)/3=.6/3=20/100= \boxed{\textbf{(A)}\ 20\%}$.

-Lcz

## Solution 3

You can find that the ratio of fuel used by the old car and the new car in a same amount of distance is $3 : 2$, and the ratio between the fuel price of these two cars is $5 : 6$. Therefore, by multiplying these two ratios, we get that the costs of using these two cars is $$15 : 12 = 5 : 4$$So the percentage of money saved is $1 - \frac{4}{5} = \boxed{\textbf{(A)}\ 20\%}$.

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