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

Latest revision as of 00:16, 4 December 2023

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 for the 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)

Solution 4

Assume WLOG that Elmer's old car's range is $100$ miles. So, Elmer's new car's range is $100 \times 1.5 = 150$ miles. Also, assume that the gas Elmer's old car uses is $$10$ , which means that diesel will cost $$12$ . Now we can deduce that Elmer's old car uses $10 \div 100 = $0.10$ per mile, and Elmer's new car uses $12 \div 150 = $0.08$ per mile. Therefore, Elmer's new car saves $\boxed{\textbf{(A) }20\%}$ more money than his old car.

~MrThinker

Video Solution 1

https://youtu.be/KdhlT6DR_Do

~savannahsolver

Video Solution 2

https://www.youtube.com/watch?v=a6dbFrDbo1w

~Math4All999

@@ Line 13: / Line 13: @@
 ==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>.
+You can find that the ratio of fuel used by the old car and the new car for the 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>.
 -Quadraticfunctions
 (edited by mydad)
-==Video Solution==
+==Solution 4==
+Assume WLOG that Elmer's old car's range is <math>100</math> miles. So, Elmer's new car's range is <math>100 \times 1.5 = 150</math> miles. Also, assume that the gas Elmer's old car uses is <math>\$10</math>, which means that diesel will cost <math>\$12</math>. Now we can deduce that Elmer's old car uses <math>10 \div 100 = \$0.10</math> per mile, and Elmer's new car uses <math>12 \div 150 = \$0.08</math> per mile. Therefore, Elmer's new car saves <math>\boxed{\textbf{(A) }20\%}</math> more money than his old car.
+~MrThinker
+==Video Solution 1==
 https://youtu.be/KdhlT6DR_Do
 ~savannahsolver
+==Video Solution 2==
+https://www.youtube.com/watch?v=a6dbFrDbo1w
+~Math4All999
 ==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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