Difference between revisions of "2020 AMC 8 Problems/Problem 11"
(→Video Solution by WhyMath) |
(→Problem) |
||
Line 1: | Line 1: | ||
− | ==Problem== | + | ==Problem 11== |
After school, Maya and Naomi headed to the beach, <math>6</math> 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? | After school, Maya and Naomi headed to the beach, <math>6</math> 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? | ||
Revision as of 21:40, 1 July 2021
Contents
Problem 11
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 by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=ND0y051eYm0
~North America Math Contest Go Go Go
Video Solution by WhyMath
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=456
~Interstigation
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.