Question 1179659
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Flying with the wind, a plane flew 1,120 mi in 4 h. 
Against the wind, the plane required 7 h to fly the same distance. 
Find the rate of the plane in calm air and the rate of the wind.
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Let u be the rate of the plane at no wind (in miles per hour)
and v be the rate of the wind (in the same units).


Then the effective rate  of the plane with   the wind is u + v
and  the effective rate of the plane against the wind is u - v.


From the problem, the effective rate of the plane with the wind is the distance of 1120 miles 
divided by the time of 4 hours  {{1120/4}}} = 280 mph.


                  The effective rate of the plane against the wind is the distance of 1120 miles 
divided by the time of 7 hours  {{{1120/7}}} = 160 mph.


So, we have two equations to find 'u' and 'v'

    u + v = 280,    (1)

    u - v = 160.    (2)


To solve, add equations (1) and (2).  The terms 'v' and '-v' will cancel each other, and you will get

    2u = 280 + 160 = 440  --->   u = 440/2 = 220.

Now from equation (1)

     v = 280 - u = 280 - 220 = 60.


<U>ANSWER</U>.  The rate of the plane in still air is 220 mph.  The rate of the wind is 60 mph km/h.
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Solved.