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

(Solution 1)
m (Solution 2 (variant of Solution 1))
(16 intermediate revisions by 9 users not shown)
Line 38: Line 38:
  
 
==Solution 1==
 
==Solution 1==
 +
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>.
  
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.
+
==Solution 2 (Variant of Solution 1)==
Since speed is distance over time, Naomi's speed is <math>36</math> mph.
+
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>.
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
+
==Video Solution by North America Math Contest Go Go Go==
  
==Solution 2==
+
https://www.youtube.com/watch?v=ND0y051eYm0
  
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>.
+
~North America Math Contest Go Go Go
  
-franzliszt
+
==Video Solution by WhyMath==
 +
https://youtu.be/y__IHWpXprY
  
==Solution 3==
+
~savannahsolver
  
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
+
==Video Solution==
 +
https://youtu.be/xjwDsaRE_Wo
  
==Solution 4==
+
==Video Solution by Interstigation==
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>
+
https://youtu.be/YnwkBZTv5Fw?t=456
-[[User:A_mathemagician|A_MatheMagician]]
 
  
==See also== {{AMC8 box|year=2020|num-b=10|num-a=12}} {{MAA Notice}}
+
~Interstigation
 +
 
 +
==See also== {{AMC8 box|year=2020|num-b=10|num-a=12}}  
 +
{{MAA Notice}}

Revision as of 21:59, 6 August 2022

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 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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png