SOLUTION: A pilot can fly a plane at 125 mph in calm air. A recent trip of 300 mi flying with the wind and 300 mi returning against the wind took 5 h. Find the rate of the wind.

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Question 1116842: A pilot can fly a plane at 125 mph in calm air. A recent trip of 300 mi flying with the wind and 300 mi returning against the wind took 5 h. Find the rate of the wind.
Found 2 solutions by ikleyn, solver91311:
Answer by ikleyn(52903) About Me  (Show Source):
You can put this solution on YOUR website!

The "time" equation is


300%2F%28125%2Bw%29 + 300%2F%28125-w%29 = 5    hours.    (1)


Solve for w, the rate of wind.


Answer.  Rate of wind is 25 miles per hour.

I solved it mentally   (since the answer is OBVIOUS).   //   The formal solution is as follows. 

    Multiply both sides of (1) by (125+w)*(125-w). You will get


    300*(125-w) + 300*(125+w) = 5*(125-w)*(125+w)

    60*(125-w) + 60*(125+w) = (125-w)*(125+w)

    2*60*125 = 125%5E2+-+w%5E2

    w%5E2 = 125%5E2-2%2A60%2A125 = 125*(125-120) = 125*5 = 5%5E4.

    w = 5%5E2 = 25.


    Check.  300%2F%28125%2B25%29 + 300%2F%28125-25%29 = 300%2F100 + 300%2F150 = 3 + 2 = 5  hours.   ! Correct !

Solved.

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It is a typical "tailwind and headwind" word problem.

See the lessons
    - Wind and Current problems
    - Wind and Current problems solvable by quadratic equations
    - Selected problems from the archive on a plane flying with and against the wind
in this site.


Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the section "Word problems",  the topic "Travel and Distance problems".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.


Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Basic formula is

For the outbound trip (with the wind), the speed of the aircraft is where is the rate of speed of the wind.

For the return trip (against the wind), the speed of the aircraft is

Let the time of the outbound trip be , then the time of the return trip must be whatever remains of the total trip time of 5 hours, or .

Since the two legs of the trip are equal in distance, namely 300 miles, we have two equations in and :



and



Although we are asked for the speed of the wind, it will be a much neater calculation to solve for first.





Add the two equations:



Combine the RHS fractions over the common denominator









So is either 2 or 3. However, the way we set up the problem with representing the time of the 'with the wind' leg of the trip, must be the smaller of the two values. Hence:



and



So, since





and



Checking the answer in the return trip equation:







Checks

John

My calculator said it, I believe it, that settles it