Difference between revisions of "2017 AMC 10B Problems/Problem 12"
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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>. | + | 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== | ==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 12:18, 16 February 2017
Problem
Elmer's new car gives percent better fuel efficiency. However, the new car uses diesel fuel, which is 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?
Solution
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 .
See Also
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 |
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