Difference between revisions of "2016 AMC 8 Problems/Problem 4"

(Solution)
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<math>\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30</math>
 
<math>\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30</math>
 
==Solution==
 
When Cheenu was a boy, he could run <math>15</math> miles in <math>3</math> hours and <math>30</math> minutes <math>= 3\times60 + 30</math> minutes <math>= 210</math> minutes, thus running <math>\frac{210}{15} = 14</math> minutes per mile. When he is an old man, he can walk <math>10</math> miles in <math>4</math> hours <math>= 4 \times 60</math> minutes <math>= 240</math> minutes, thus walking <math>\frac{240}{10} = 24</math> minutes per mile. Therefore it takes him <math>\boxed{\textbf{(B)}\ 10}</math> minutes longer to walk a mile now compared to when he was a boy.
 
 
 
{{AMC8 box|year=2016|num-b=3|num-a=5}}
 
{{MAA Notice}}
 

Revision as of 18:18, 12 November 2019

When Cheenu was a boy he could run $15$ miles in $3$ hours and $30$ minutes. As an old man he can now walk $10$ miles in $4$ hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?

$\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30$

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