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> | ||
==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 
mile to school. He leaves home at the same time each day, walks at a steady speed of 
miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first 
mile at a speed of only 
miles per hour. At how many miles per hour must George run the last mile in order to arrive just as school begins today?
Solution
Note that on a normal day, it takes him 
hour to get to school. However, today it took 
hour to walk the first 1/2 mile.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
