Difference between revisions of "2014 AMC 8 Problems/Problem 17"

(Solution)
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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>
 
==Solution==
 
==Solution==
Note that
+
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 1/2 mile.
  
 
==See Also==
 
==See Also==

Revision as of 21:55, 27 November 2014

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

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.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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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