Question 1179655
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The bird capable of the fastest flying speed is the swift. A swift flying with the wind to a favorite feeding spot 
traveled 42 mi in 0.3 h. On returning, now against the wind, the swift was able to travel only 27 mi 
in the same amount of time. What is the rate of the swift in calm air, and what was the rate of the wind?
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Let x be the rate of the swift in calm air, in miles per hour.

Let y be the rate of the wind.


Then  the effective speed of the swift with    the wind is (x+y) mph,

while the effective speed of the swift against the wind is (x-y) mph,



From the problem, the effective rate of the swift with the wind is  the distance divided by the travel time

    {{{42/0.3}}} = 140 miles per hour,


and the effective rate of the plane against the wind is  the distance divided by the travel time

    {{{27/0.3}}} = 90 miles per hour.


So, we have these two equations

    x + y = 140     (1)

    x - y =  90     (2)


To find x, add the equation.  You will get

    2x = 140 + 90 = 230,  x = 230/2 = 115.


Now express y from equation (1) and calculate

    y = 140 - x = 140 - 115 = 25.


At this point the problem is solved completely.


<U>ANSWER</U>.  The speed of the swift in calm air is  115 mph.  The rate of the wind is 25 mph.
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Solved correctly.