2020 AMC 8 Problems/Problem 11

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

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 $\boxed{E) 24}$

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2020 AMC 8 Problems/Problem 11

Problem 11

Solution