SOLUTION: Jim can run 5 miles per hour on level ground on a still day. One windy day he runs 15 miles with the wind and in the same amount of time runs 4 miles against the wind. What is the

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Jim can run 5 miles per hour on level ground on a still day. One windy day he runs 15 miles with the wind and in the same amount of time runs 4 miles against the wind. What is the       Log On

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Question 846234: Jim can run 5 miles per hour on level ground on a still day. One windy day he runs 15 miles with the wind and in the same amount of time runs 4 miles against the wind. What is the rate of the wind?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +t+ = his time for running both
with the wind and against the wind
Let +w+ = the speed of the wind
--------------------------------
With the wind:
(1) +15+=+%28+5+%2B+w+%29%2At+
Against the wind:
(2) +4+=+%28+5+-+w+%29%2At+
---------------------
(1) +15+=+5t+%2B+w%2At+
and
(2) +4+=+5t+-+w%2At+
---------------------
Add the equations:
(1) +15+=+5t+%2B+w%2At+
(2) +4+=+5t+-+w%2At+
+19+=+10t+
+t+=+1.9+
and
(2) +4+=+%28+5+-+w+%29%2A1.9+
(2) +4+=+9.5+-+1.9w+
(2) +1.9w+=+5.5+
(2) +w+=+2.895+
check:
(1) +15+=+%28+5+%2B+w+%29%2At+
(1) +15+=+%28+5+%2B+2.895+%29%2A1.9+
(1) +15+=+7.895%2A1.9+
(1) +15+=+15.0005+
and
(2) +4+=+%28+5+-+w+%29%2At+
(2) +4+=+%28+5+-+2.895+%29%2A1.9+
(2) +4+=+2.105%2A1.9+
(2) +4+=+3.9995+
close enough
The wind speed is 2.895 mi/hr