Difference between revisions of "2020 AMC 8 Problems/Problem 11"

Revision as of 17:24, 18 November 2020

Problem 11

After school, Maya and Naomi headed to the beach, $6$ 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?

$[asy] unitsize(1.25cm); dotfactor = 10; pen shortdashed=linetype(new real[] {2.7,2.7}); for (int i = 0; i < 6; ++i) { for (int j = 0; j < 6; ++j) { draw((i,0)--(i,6), grey); draw((0,j)--(6,j), grey); } } for (int i = 1; i <= 6; ++i) { draw((-0.1,i)--(0.1,i),linewidth(1.25)); draw((i,-0.1)--(i,0.1),linewidth(1.25)); label(string(5*i), (i,0), 2*S); label(string(i), (0, i), 2*W); } draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25)); label(rotate(90) * "Distance (miles)", (-0.5,3), W); label("Time (minutes)", (3,-0.5), S); dot("Naomi", (2,6), 3*dir(305)); dot((6,6)); label("Maya", (4.45,3.5)); draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35)); draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed); [/asy]$

$\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24$

Solution 1

We use the formula $\text{speed}=\dfrac{\text{distance}}{\text{time}}$ . Naomi's distance is $6$ miles, and her time is $10$ minutes, which is equivalent to $\dfrac{1}{6}$ of an hour. Since speed is distance over time, Naomi's speed is $36$ mph. Using the same process, Maya's speed is $12$ mph. Subtracting those, we get an answer of $\boxed{(\text{E}) 24}$ .

Solution 2

Notice that Naomi travels at a rate of $6$ miles every $10$ minutes or $36$ miles an hour. Maya travels at a rate of $6$ miles every $30$ minutes or $12$ miles an hour. Hence, the answer is $36-12=\textbf{(E) }24$ .

-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 $12$ minutes, their difference in speed is $2\cdot12=\textbf{(E) }24$

@@ Line 48: / Line 48: @@
 -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 <math>12</math> minutes, their difference in speed is <math>2\cdot12=\textbf{(E) }24</math>
 ==See also== {{AMC8 box|year=2020|num-b=10|num-a=12}} {{MAA Notice}}

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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