Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 3 -SweetMango77)
Line 50: Line 50:
  
 
==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}}
−
 
−
==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>
 

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_11&oldid=137659"