Difference between revisions of "2017 AMC 10B Problems/Problem 12"
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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> | <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> | ||
| − | ==Solution== | + | ==Solution 1== |
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>. | 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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{{AMC10 box|year=2017|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2017|ab=B|num-b=11|num-a=13}} | ||
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| + | ==Solution 2== | ||
| + | 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>. | ||
Revision as of 11:26, 12 February 2019
Problem
Elmer's new car gives 
percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is 
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?
Solution 1
Suppose that his old car runs at 
km per liter. Then his new car runs at 
km per liter, or km per 
of a liter. Let the cost of the old car's fuel be 
, so the trip in the old car takes 
dollars, while the trip in the new car takes 
. He saves 
.
| 2017 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
Solution 2
Because they do not give you a given amount of distance, we'll just make that distance 
miles. Then, we find that the new car will use 
. The old car will use . Thus the answer is 
.
