Difference between revisions of "2020 AMC 8 Problems/Problem 11"
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Notice that the difference between Maya's and Naomi's arrival time is 4 units on the graph, or twice as slow as Naomi. Since Naomi's time to go is <math>12</math> minutes, their difference in speed is <math>2\cdot12=\boxed{\textbf{(E) }24}</math> | Notice that the difference between Maya's and Naomi's arrival time is 4 units on the graph, or twice as slow as Naomi. Since Naomi's time to go is <math>12</math> minutes, their difference in speed is <math>2\cdot12=\boxed{\textbf{(E) }24}</math> | ||
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==See also== {{AMC8 box|year=2020|num-b=10|num-a=12}} {{MAA Notice}} | ==See also== {{AMC8 box|year=2020|num-b=10|num-a=12}} {{MAA Notice}} | ||
Revision as of 18:07, 18 November 2020
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?
![[asy] unitsize(1.25cm); dotfactor = 10; pen shortdashed=linetype(new real[] {2.7,2.7}); for (int i = 0; i < 6; ++i) { for (int j = 0; j < 6; ++j) { draw((i,0)--(i,6), grey); draw((0,j)--(6,j), grey); } } for (int i = 1; i <= 6; ++i) { draw((-0.1,i)--(0.1,i),linewidth(1.25)); draw((i,-0.1)--(i,0.1),linewidth(1.25)); label(string(5*i), (i,0), 2*S); label(string(i), (0, i), 2*W); } draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25)); label(rotate(90) * "Distance (miles)", (-0.5,3), W); label("Time (minutes)", (3,-0.5), S); dot("Naomi", (2,6), 3*dir(305)); dot((6,6)); label("Maya", (4.45,3.5)); draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35)); draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed); [/asy]](http://latex.artofproblemsolving.com/1/0/1/1012d4ebea522bd45aaa55a5d54f98b4db812507.png)
Solution 1
We use the formula 
. Naomi's distance is miles, and her time is 
minutes, which is equivalent to 
of an hour.
Since speed is distance over time, Naomi's speed is 
mph.
Using the same process, Maya's speed is 
mph. Subtracting those, we get an answer of 
.
Solution 2
Notice that Naomi travels at a rate of miles every minutes or miles an hour. Maya travels at a rate of miles every 
minutes or miles an hour. Hence, the answer is 
.
-franzliszt
Solution 3 -SweetMango77
Notice that the difference between Maya's and Naomi's arrival time is 4 units on the graph, or twice as slow as Naomi. Since Naomi's time to go is minutes, their difference in speed is 
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. 
