Difference between revisions of "2020 AMC 10A Problems/Problem 4"
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==Solution 2 (longer)== | ==Solution 2 (longer)== | ||
The driver is driving for <math>2</math> hours at <math>60</math> miles per hour, she drives <math>120</math> miles. Therefore, she uses <math>\frac{120}{30}=4</math> gallons of gasoline. So, total she has <math>\$0.50\cdot120-\$2.00\cdot4=\$60-\$8=\$52</math>. So, her rate is <math>\frac{52}{2}=\boxed{\textbf{(E)26}}</math> | The driver is driving for <math>2</math> hours at <math>60</math> miles per hour, she drives <math>120</math> miles. Therefore, she uses <math>\frac{120}{30}=4</math> gallons of gasoline. So, total she has <math>\$0.50\cdot120-\$2.00\cdot4=\$60-\$8=\$52</math>. So, her rate is <math>\frac{52}{2}=\boxed{\textbf{(E)26}}</math> | ||
+ | ~sosiaops | ||
==Video Solution 1== | ==Video Solution 1== |
Revision as of 17:28, 17 January 2021
- The following problem is from both the 2020 AMC 12A #3 and 2020 AMC 10A #4, so both problems redirect to this page.
Contents
Problem
A driver travels for hours at miles per hour, during which her car gets miles per gallon of gasoline. She is paid per mile, and her only expense is gasoline at per gallon. What is her net rate of pay, in dollars per hour, after this expense?
Solution 1
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 . ~mathsmiley
Solution 2 (longer)
The driver is driving for hours at miles per hour, she drives miles. Therefore, she uses gallons of gasoline. So, total she has . So, her rate is ~sosiaops
Video Solution 1
Video Solution 2
~IceMatrix
Video Solution 3
https://www.youtube.com/watch?v=7-3sl1pSojc
~bobthefam
~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.