Difference between revisions of "2020 AMC 8 Problems/Problem 11"
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==Solution 1== | ==Solution 1== | ||
− | Naomi travels <math>6</math> miles in a time of <math>10</math> minutes, which is equivalent to <math>\dfrac{1}{6}</math> of an hour. Since <math>\text{speed} = \frac{\text{distance}}{\text{time}}</math>, her speed is <math>\frac{6}{\left(\frac{1}{6}\right)} = 36</math> mph. By a similar calculation, Maya's speed is <math>12</math> mph, so the answer is <math>36-12 = \boxed{\textbf{(E) }24}</math> | + | Naomi travels <math>6</math> miles in a time of <math>10</math> minutes, which is equivalent to <math>\dfrac{1}{6}</math> of an hour. Since <math>\text{speed} = \frac{\text{distance}}{\text{time}}</math>, her speed is <math>\frac{6}{\left(\frac{1}{6}\right)} = 36</math> mph. By a similar calculation, Maya's speed is <math>12</math> mph, so the answer is <math>36-12 = \boxed{\textbf{(E) }24}</math>. |
==Solution 2 (variant of Solution 1)== | ==Solution 2 (variant of Solution 1)== |
Revision as of 12:52, 30 December 2020
Problem
After school, Maya and Naomi headed to the beach, miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?
Solution 1
Naomi travels miles in a time of minutes, which is equivalent to of an hour. Since , her speed is mph. By a similar calculation, Maya's speed is mph, so the answer is .
Solution 2 (variant of Solution 1)
Naomi's speed of miles in minutes is equivalent to miles per hour, while Maya's speed of miles in minutes (i.e. half an hour) is equivalent to miles per hour. The difference is consequently .
Video Solution
https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.