Difference between revisions of "2020 AMC 10A Problems/Problem 4"
(→Video Solution) |
MRENTHUSIASM (talk | contribs) m (→Video Solution 1) |
||
| (13 intermediate revisions by 8 users not shown) | |||
| Line 7: | Line 7: | ||
<math> \textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 26 </math> | <math> \textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 26 </math> | ||
| − | ==Solution== | + | ==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 | + | Since the driver travels <math>60</math> miles per hour and each hour she uses <math>2</math> gallons of gasoline, she spends <math>\$4</math> per hour on gas. If she gets <math>\$0.50</math> per mile, then she gets <math>\$30</math> 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 | ||
| − | ==Video Solution== | + | ==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>. | ||
| + | ~sosiaops | ||
| + | |||
| + | ==Video Solution 1== | ||
| + | |||
| + | https://youtu.be/J-Ery8I0yAg | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution 2== | ||
| + | |||
https://youtu.be/WUcbVNy2uv0 | https://youtu.be/WUcbVNy2uv0 | ||
~IceMatrix | ~IceMatrix | ||
| + | ==Video Solution 3== | ||
https://www.youtube.com/watch?v=7-3sl1pSojc | https://www.youtube.com/watch?v=7-3sl1pSojc | ||
| Line 27: | Line 39: | ||
{{AMC10 box|year=2020|ab=A|num-b=3|num-a=5}} | {{AMC10 box|year=2020|ab=A|num-b=3|num-a=5}} | ||
| − | |||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 03:19, 7 October 2022
- 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 miles per hour and each hour she uses gallons of gasoline, she spends 
per hour on gas. If she gets per mile, then she gets 
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
~Education, the Study of Everything
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
