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

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==Problem 11==
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==Problem==
 
After school, Maya and Naomi headed to the beach, <math>6</math> 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?
 
After school, Maya and Naomi headed to the beach, <math>6</math> 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?
  
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==Solution 1==
 
==Solution 1==
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Naomi travels <math>6</math> miles in a time of <math>10</math> minutes, which is equivalent to <math>\dfrac{1}{6}</math> of an hour. Since <math>\text{speed} = \frac{\text{distance}}{\text{time}}</math>, her speed is <math>\frac{6}{\left(\frac{1}{6}\right)} = 36</math> mph. By a similar calculation, Maya's speed is <math>12</math> mph, so the answer is of <math>36-12 = \boxed{\textbf{(E) }24}</math>.
  
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.
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==Solution 2 (variant of Solution 1)==
Since speed is distance over time, Naomi's speed is <math>36</math> mph.
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Naomi's speed of <math>6</math> miles in <math>10</math> minutes is equivalent to <math>6 \cdot 6 = 36</math> miles per hour, while Maya's speed of <math>6</math> miles in <math>30</math> minutes (i.e. half an hour) is equivalent to <math>6 \cdot 2 = 12</math> miles per hour. The difference is <math>36-12=\boxed{\textbf{(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>.
 
 
 
-wuwang2002
 
 
 
==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
 
 
 
==Solution 3==
 
 
 
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=\boxed{\textbf{(E) }24}</math>~SweetMango77
 
 
 
==Solution 4==
 
We see that Naomi travels <math>6</math> miles in <math>\dfrac{1}{6}</math> of an hour. Thus, his speed is <math>36</math> mph. Maya's speed is <math>12</math> mph. The difference is <math>36 - 12 = \boxed{(\text{E}) 24}.</math>
 
-[[User:A_mathemagician|A_MatheMagician]]
 
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/y__IHWpXprY
 
https://youtu.be/y__IHWpXprY
  
~savannahsolver
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==See also== {{AMC8 box|year=2020|num-b=10|num-a=12}}  
 
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{{MAA Notice}}
==See also== {{AMC8 box|year=2020|num-b=10|num-a=12}} {{MAA Notice}}
 

Revision as of 08:43, 20 November 2020

Problem

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

Naomi travels $6$ miles in a time of $10$ minutes, which is equivalent to $\dfrac{1}{6}$ of an hour. Since $\text{speed} = \frac{\text{distance}}{\text{time}}$, her speed is $\frac{6}{\left(\frac{1}{6}\right)} = 36$ mph. By a similar calculation, Maya's speed is $12$ mph, so the answer is of $36-12 = \boxed{\textbf{(E) }24}$.

Solution 2 (variant of Solution 1)

Naomi's speed of $6$ miles in $10$ minutes is equivalent to $6 \cdot 6 = 36$ miles per hour, while Maya's speed of $6$ miles in $30$ minutes (i.e. half an hour) is equivalent to $6 \cdot 2 = 12$ miles per hour. The difference is $36-12=\boxed{\textbf{(E) }24}$.

Video Solution

https://youtu.be/y__IHWpXprY

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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

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