Difference between revisions of "2020 AMC 8 Problems/Problem 11"
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Since speed is distance over time, Naomi's speed is <math>36</math> mph. | Since speed is distance over time, Naomi's speed is <math>36</math> mph. | ||
Using the same process, Maya's speed is <math>12</math> mph. Subtracting those, we get an answer of <math>\boxed{(\text{E}) 24}</math>. | Using the same process, Maya's speed is <math>12</math> mph. Subtracting those, we get an answer of <math>\boxed{(\text{E}) 24}</math>. | ||
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| + | ==See also== {{AMC8 box|year=2020|before=First problem|num-a=2}} {{MAA Notice}} | ||
Revision as of 00:10, 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
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 
.
See also
| 2020 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by First problem |
Followed by Problem 2 | |
| 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. 
