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

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==Problem==
 
==Problem==
Elmer's new car gives <math>50\%</math> percent better fuel efficiency. However, the new car uses diesel fuel, which is <math>20\%</math> 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?
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Elmer's new car gives <math>50\%</math> percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is <math>20\%</math> 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?
  
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<math>\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%</math>
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==Solution 1==
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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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==Solution 2==
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Because they do not give you a given amount of distance, we'll just make that distance <math>3x</math> miles. Then, we find that the new car will use <math>2*1.2=2.4x</math>. The old car will use <math>3x</math>. Thus the answer is <math>(3-2.4)/3=.6/3=20/100=  \boxed{\textbf{(A)}\ 20\%}</math>.
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-Lcz
  
<math>\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%</math>
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==Solution 3==
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You can find that the ratio of fuel used by the old car and the new car in a same amount of distance is <math>3 : 2</math>, and the ratio between the fuel price of these two cars is <math>5 : 6</math>. Therefore, by multiplying these two ratios, we get that the costs of using these two cars is <cmath>15 : 12 = 5 : 4</cmath>So the percentage of money saved is <math>1 - \frac{4}{5} = \boxed{\textbf{(A)}\ 20\%}</math>.
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-Quadraticfunctions
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(edited by mydad)
  
==Solution==
 
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>.
 
  
==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}}

Latest revision as of 13:26, 26 January 2020

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 $1 - \frac{4}{5} = \boxed{\textbf{(A)}\ 20\%}$.

-Quadraticfunctions (edited by mydad)


2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 10 Problems and Solutions

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