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

Line 22: Line 22:
  
 
~savannahsolver
 
~savannahsolver
 +
 +
==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 18:18, 30 April 2021

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)

Video Solution

https://youtu.be/KdhlT6DR_Do

~savannahsolver

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
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. AMC logo.png