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Difference between revisions of "2014 AMC 8 Problems/Problem 17"
 
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<math>\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad \textbf{(E) }12</math>
 
<math>\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad \textbf{(E) }12</math>
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==Solution==
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==Solution 1==
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Note that on a normal day, it takes him <math>1/3</math> hour to get to school. However, today it took <math>\frac{1/2 \text{ mile}}{2 \text{ mph}}=1/4</math> hour to walk the first <math>1/2</math> mile. That means that he has <math>1/3 -1/4 = 1/12</math> hours left to get to school, and <math>1/2</math> mile left to go. Therefore, his speed must be <math>\frac{1/2 \text{ mile}}{1/12 \text { hour}}=\boxed{6 \text{ mph}}</math>, so <math>\boxed{\text{(B) }6}</math> is the answer.
 +
 
 +
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/Il6IcNHKkWk
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
 
 +
== Video Solution ==
 +
https://youtu.be/rQUwNC0gqdg?t=3191
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution==
 +
https://youtu.be/1jW0bwR_DPQ ~savannahsolver
 +
 
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=16|num-a=18}}
 
{{AMC8 box|year=2014|num-b=16|num-a=18}}
 
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:12, 7 May 2024

Problem

George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first $\frac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must George run the last $\frac{1}{2}$ mile in order to arrive just as school begins today?

$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad \textbf{(E) }12$

Solution 1

Note that on a normal day, it takes him $1/3$ hour to get to school. However, today it took $\frac{1/2 \text{ mile}}{2 \text{ mph}}=1/4$ hour to walk the first $1/2$ mile. That means that he has $1/3 -1/4 = 1/12$ hours left to get to school, and $1/2$ mile left to go. Therefore, his speed must be $\frac{1/2 \text{ mile}}{1/12 \text { hour}}=\boxed{6 \text{ mph}}$, so $\boxed{\text{(B) }6}$ is the answer.

Video Solution (CREATIVE THINKING)

https://youtu.be/Il6IcNHKkWk

~Education, the Study of Everything


Video Solution

https://youtu.be/rQUwNC0gqdg?t=3191

~ pi_is_3.14

Video Solution

https://youtu.be/1jW0bwR_DPQ ~savannahsolver

See Also

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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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