Difference between revisions of "2020 AMC 8 Problems/Problem 11"
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<math>\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24</math> | <math>\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24</math> | ||
| − | ==Solution== | + | ==Solution 1== |
We use the formula <math>\text{speed}=\dfrac{\text{distance}}{\text{time}}</math>. Naomi's distance is <math>6</math> miles, and her time is <math>10</math> minutes, which is equivalent to <math>\dfrac{1}{6}</math> of an hour. | We use the formula <math>\text{speed}=\dfrac{\text{distance}}{\text{time}}</math>. Naomi's distance is <math>6</math> miles, and her time is <math>10</math> minutes, which is equivalent to <math>\dfrac{1}{6}</math> of an hour. | ||
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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | Notice that Naomi travels at a rate of <math>6</math> miles every <math>10</math> minutes or <math>36</math> miles an hour. Maya travels at a rate of <math>6</math> miles every <math>30</math> minutes or <math>12</math> miles an hour. Hence, the answer is <math>36-12=\textbf{(E) }24</math>. | ||
| + | |||
| + | -franzliszt | ||
==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 13:55, 18 November 2020
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?
![[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
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. 
