Difference between revisions of "2017 AMC 10B Problems/Problem 12"
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==Solution 1== | ==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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+ | ==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>. | ||
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+ | -Lcz | ||
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+ | ==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 <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 | ||
+ | (edited by mydad) | ||
{{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
Contents
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 .
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 .
-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 , and the ratio between the fuel price of these two cars is . Therefore, by multiplying these two ratios, we get that the costs of using these two cars is So the percentage of money saved is .
-Quadraticfunctions (edited by mydad)
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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