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

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==Solution==
 
==Solution==
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Suppose that his old car runs at <math>x</math> km per liter. Then his new car runs at <math>\frac{3}{2}x</math> km per liter, or <math>x</math> km per <math>\frac{2}{3}</math> of a liter. Let the cost of the old car's fuel be <math>c</math>, so the trip in the old car takes <math>xc</math> dollars, while the trip in the new car takes <math>\frac{2}{3}\cdot\frac{6}{5}xc = \frac{4}{5}xc</math> He saves <math>\frac{\frac{1}{5}xc}{xc} = \boxed{\textbf{(A) } 20%}</math>.
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==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=11|num-a=13}}
 
{{AMC10 box|year=2017|ab=B|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:14, 16 February 2017

Problem

Elmer's new car gives $50\%$ percent better fuel efficiency. However, the new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline the 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

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 $\frac{\frac{1}{5}xc}{xc} = \boxed{\textbf{(A) } 20%}$ (Error compiling LaTeX. Unknown error_msg).

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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

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