Difference between revisions of "2015 AMC 8 Problems/Problem 3"

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<math>\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10</math>
 
<math>\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10</math>
  
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===Solution===
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Jill arrives in <math>\dfrac{1}{10}</math> of an hour, which is <math>6</math> minutes. Jack arrives in <math>\dfrac{1}{4}</math> of an hour which is <math>15</math> minutes. Thus, the time difference is <math>\boxed{\textbf{(D)}~9}</math> minutes.
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=2015|num-b=2|num-a=4}}
 
{{AMC8 box|year=2015|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:25, 25 November 2015

Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of $10$ miles per hour. Jack walks to the pool at a constant speed of $4$ miles per hour. How many minutes before Jack does Jill arrive?

$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$

Solution

Jill arrives in $\dfrac{1}{10}$ of an hour, which is $6$ minutes. Jack arrives in $\dfrac{1}{4}$ of an hour which is $15$ minutes. Thus, the time difference is $\boxed{\textbf{(D)}~9}$ minutes.

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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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