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Difference between revisions of "2020 AMC 10A Problems/Problem 4"

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==Solution==
 
==Solution==
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Since the driver travels 60 miles per hour and each hour she uses 2 gallons of gasoline, she spends \$4 per hour on gas. If she gets \$0.50 per mile, then she gets \$30 per hour of driving. Subtracting the gas cost, her net rate of pay per hour is <math>\boxed{\textbf{(E)}\ 26}</math>.
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Since the driver travels 60 miles per hour and each hour she uses 2 gallons of gasoline, she spends \$4 per hour on gas. If she gets \$0.50 per mile, then she gets \$30 per hour of driving. Subtracting the gas cost, her net rate of money earned per hour is <math>\boxed{\textbf{(E)}\ 26}</math>.
 
~mathsmiley
 
~mathsmiley
  

Revision as of 15:20, 26 October 2020

The following problem is from both the 2020 AMC 12A #3 and 2020 AMC 10A #4, so both problems redirect to this page.

Problem

A driver travels for $2$ hours at $60$ miles per hour, during which her car gets $30$ miles per gallon of gasoline. She is paid $$0.50$ per mile, and her only expense is gasoline at $$2.00$ per gallon. What is her net rate of pay, in dollars per hour, after this expense?

$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 26$

Solution

Since the driver travels 60 miles per hour and each hour she uses 2 gallons of gasoline, she spends $4 per hour on gas. If she gets $0.50 per mile, then she gets $30 per hour of driving. Subtracting the gas cost, her net rate of money earned per hour is $\boxed{\textbf{(E)}\ 26}$. ~mathsmiley

Video Solution

https://youtu.be/WUcbVNy2uv0

~IceMatrix

https://www.youtube.com/watch?v=7-3sl1pSojc

~bobthefam

https://youtu.be/Dj_DFoZO-xw

~savannahsolver

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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
2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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