SOLUTION: HELP!!! I have worked this problem and can't get it. Problem: Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 10 miles with the wind, and

Algebra.Com
Question 37145: HELP!!! I have worked this problem and can't get it.
Problem:
Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 10 miles with the wind, and in the same amount of time runs 4 miles against the wind. What is the rate of the wind?
Found 3 solutions by fractalier, mbarugel, AnlytcPhil:
Answer by fractalier(6550)   (Show Source): You can put this solution on YOUR website!
Okay. The governing equation for this kind of problem is RT = D. Rate times time equals distance. His rate (with no wind) is given as 5 mph. Then we set up two equations, one with the wind and one against:
(R + W)T = D
(R - W)T = D
now substitute in what you know
(5 + W)T = 10
(5 - W)T = 4
If we solve the second one for T
T = 4 / (5 - W)
and then plug it in to the first equation, we get
(5 + W)( 4 / (5 - W)) = 10
Solving this we get W = 15/7 or about 2.143 mph, the wind speed.
Answer by mbarugel(146)   (Show Source): You can put this solution on YOUR website!
Hello!
We must use the following relationship in order to answer this:
If an object is moving at a speed of V miles per hour, then the time (in hours) it takes to travel a distance D (in miles) is D/V.
Now, we know that Jim can run at 5 mph without wind. Let's call X to the speed of the wind. So Jim's speed is 5 + X mph when he goes with the wind, and 5 - X mph when he goes against it.
Using the aforementioned relationship, and the fact that he takes the same time to make 10 miles with the wind and 4 miles against the wind, we get the following equation:

So we just isolate X:



So the speed of the wind is 2.1428... mph.

I hope this helps!
Get more answers at Online Math Answers.com!
Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
Problem: 
Jim can run 5 miles per hour on level ground on a still day. 
One windy day, he runs 10 miles with the wind, and in the same 
amount of time runs 4 miles against the wind. What is the rate 
of the wind?

Make this chart

               DISTANCE      RATE       TIME
With wind                                      
Against wind 

>>...he runs 10 miles with the wind, and in the same amount of time 
runs 4 miles against the wind...<<

Fill in the two distances                                  

               DISTANCE      RATE       TIME
With wind         10                            
Against wind       4                           

Let the rate of the wind be x miles per hour. 

So when running with the wind, his rate is increased by x mph.
So we add x to his rate of 5mph and get 5+x mph, so fill that
rate in: 

               DISTANCE      RATE       TIME
With wind         10         5+x      
Against wind       4        

When running against the wind, his rate is decreased by x mph.
So we subtract x from his rate of 5mph and get 5-x mph, so fill 
that rate in: 

               DISTANCE      RATE       TIME
With wind         10         5+x      
Against wind       4         5-x

Now use TIME = DISTANCE/RATE to fill in the two times:


               DISTANCE      RATE       TIME
With wind         10         5+x      10/(5+x)
Against wind       4         5-x       4/(5-x)


>>>...in the same amount of time...<<

This says the two times are equal:

               10      4
             ————— = —————
              5+x     5-x  

Multiply thru by LCD = (5+x)(5-x)

           10(5-x) = 4(5+x)

          50 - 10x = 20 + 4x

              -14x = -30

                 x = (-30)/(-14)

                 x = 15/7 or 2 1/7 mph

Checking:

When he runs with the wind he runs 5 + 2 1/7 mph or 7 1/7 mph 
or 50/7 mph for 10 miles.  Since T = D/R, his time is 10/(50/7)
= 7/5 hours

When he runs against the wind he runs 5 - 2 1/7 mph or 2 6/7 mph
or 20/7 mph for 4 miles.  Since T = D/R, his time is 4/(20/7)
= 7/5 hours

So the times are equal, thus the answer is correct.

Edwin
AnlytcPhil@aol.com
RELATED QUESTIONS

Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 10... (answered by Paul)
Solve the problem and show the work. Jim can run 5 miles per hour on level ground on a... (answered by ptaylor)
Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 15... (answered by josmiceli)
jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 10... (answered by Paul)
Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 13... (answered by ankor@dixie-net.com)
Jim can run 5 miles per hour on a level ground on a still day. One windy day, he runs 11... (answered by scott8148,mananth)
Jim can run 5 miles per hour on level ground on a still day. One windy day he runs 15... (answered by josmiceli)
Solve the problem. Jan can run 5 miles per hour on level ground on a still day. One... (answered by checkley71)
I attempted to try this problem for myself, yet I still got the wrong answer. So can you... (answered by rothauserc,MathTherapy)