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

(Removed a duplicate solution and an incorrect solution, and improved grammar and clarity)
(→‎Video Solution by Math-X (First understand the problem!!!))
 
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==Problem==
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==Problem 11==
 
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>.
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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 <math>36-12 = \boxed{\textbf{(E) }24}</math>.
  
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==Solution 2 (variant of Solution 1)==
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==Solution 2 (Variant of Solution 1)==
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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>.
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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 consequently <math>36-12=\boxed{\textbf{(E) }24}</math>.
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==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=bHNrBwwUCMI
 +
 
 +
~NiuniuMaths
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=MfI4I15mLBXNvGMm&t=1375
 +
 
 +
~Math-X
 +
 
 +
==Video Solution (CLEVER MANIPULATIONS!!!)==
 +
https://youtu.be/gQrRdpw_tu8
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by North America Math Contest Go Go Go==
 +
 
 +
https://www.youtube.com/watch?v=ND0y051eYm0
 +
 
 +
~North America Math Contest Go Go Go
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/y__IHWpXprY
 +
 
 +
~savannahsolver
  
 
==Video Solution==
 
==Video Solution==
−
https://youtu.be/y__IHWpXprY
+
https://youtu.be/xjwDsaRE_Wo
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/YnwkBZTv5Fw?t=456
 +
 
 +
~Interstigation
 +
 
 +
 
 +
 
  
 
==See also== {{AMC8 box|year=2020|num-b=10|num-a=12}}  
 
==See also== {{AMC8 box|year=2020|num-b=10|num-a=12}}  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:34, 26 January 2024

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

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 $36-12 = \boxed{\textbf{(E) }24}$.

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 consequently $36-12=\boxed{\textbf{(E) }24}$.

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=bHNrBwwUCMI

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=MfI4I15mLBXNvGMm&t=1375

~Math-X

Video Solution (CLEVER MANIPULATIONS!!!)

https://youtu.be/gQrRdpw_tu8

~Education, the Study of Everything

Video Solution by North America Math Contest Go Go Go

https://www.youtube.com/watch?v=ND0y051eYm0

~North America Math Contest Go Go Go

Video Solution by WhyMath

https://youtu.be/y__IHWpXprY

~savannahsolver

Video Solution

https://youtu.be/xjwDsaRE_Wo

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=456

~Interstigation



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

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