Difference between revisions of "2010 AMC 8 Problems/Problem 8"

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==Solution==
 
==Solution==
coming soon....
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Because they are skating in the same direction, Emily is skating relative to Emerson <math>12-8=4</math> mph. Now we can look at it as if Emerson is standing still and Emily is skating <math>4</math> mph. It takes her
  
answer: <math>\boxed{\textbf{(D) } 15}</math>
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<cmath>\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}</cmath>
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to skate the <math>1/2</math> mile to reach him, and then the same amount of time to be <math>1/2</math> mile ahead of him. This totals to
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<cmath>2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{\textbf{(D)}\ 15}\ \text{minutes}</cmath>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2010|num-b=7|num-a=9}}
 
{{AMC8 box|year=2010|num-b=7|num-a=9}}

Revision as of 17:09, 25 December 2012

Problem

As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction $1/2$ mile in front of her. After she passes him, she can see him in her rear mirror until he is $1/2$ mile behind her. Emily rides at a constant rate of $12$ miles per hour, and Emerson skates at a constant rate of $8$ miles per hour. For how many minutes can Emily see Emerson?

$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16$

Solution

Because they are skating in the same direction, Emily is skating relative to Emerson $12-8=4$ mph. Now we can look at it as if Emerson is standing still and Emily is skating $4$ mph. It takes her

\[\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}\]

to skate the $1/2$ mile to reach him, and then the same amount of time to be $1/2$ mile ahead of him. This totals to

\[2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{\textbf{(D)}\ 15}\ \text{minutes}\]

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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