Question 1201012
<font color=black size=3>
Let's draw a vertical number line to represent the locations of airports A, B, and C. 
The order from top to bottom is: C, B, A.
We'll also mark the distances between each adjacent airport.
Each distance is in miles.
{{{
drawing(400, 200, -5,5,-4,4,
line(0,-3-3,0,3+3),
line(0.2,-3,-0.2,-3),

line(0.2,0,-0.2,0),

line(0.2,3,-0.2,3),
locate(-0.5,-2.5,"A"),
locate(-0.5,0.5,"B"),
locate(-0.5,3.5,"C"),

locate(-0.75,-1,"645"),
locate(-0.75,2,"540"),

locate(3,-3,matrix(1,2,Diagram,not)),
locate(3,-3.5,matrix(1,2,to,scale))
)
}}}


Add in the wind blowing from the south (i.e. the wind is blowing toward the north) at 15 mph.
This applies only when the plane is going from A to B.
{{{
drawing(400, 200, -5,5,-4,4,
line(0,-3-3,0,3+3),
line(0.2,-3,-0.2,-3),

line(0.2,0,-0.2,0),

line(0.2,3,-0.2,3),
locate(-0.5,-2.5,"A"),
locate(-0.5,0.5,"B"),
locate(-0.5,3.5,"C"),

locate(-0.75,-1,"645"),
locate(-0.75,2,"540"),

red(line(1,-3,1,3)),
red(line(1,3,0.7,2.4804)),
red(line(1,3,1.3,2.4804)),

red(locate(1.2,0.5,matrix(1,2,wind,speed))),
red(locate(1.2,0,matrix(1,2,15,mph))),

blue(locate(-4.8,3,matrix(1,3,When,the,plane))),
blue(locate(-4.8,2.5,matrix(1,5,goes,from,A,to,B))),

locate(3,-3,matrix(1,2,Diagram,not)),
locate(3,-3.5,matrix(1,2,to,scale))
)
}}}


Since the plane and the wind are both going north (when the plane goes from A to B), the plane's speed is increased.


x = speed of the plane with no wind
x+15 = speed after adding the 15 mph wind boost


distance = rate*time
time = distance/rate
time = (645 miles)/(x+15 mph)
time = 645/(x+15)
This is the time, in hours, it takes to go from A to B.


Side note: 
A headwind slows the plane down (because it attacks the head or front of the plane). 
A tailwind speeds the plane up (because it comes from the tail to push the plane forward).


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When the plane goes from B to C, the wind blows from the north. 
The wind is now aiming south. It moves at 20 mph.
{{{
drawing(400, 200, -5,5,-4,4,
line(0,-3-3,0,3+3),
line(0.2,-3,-0.2,-3),

line(0.2,0,-0.2,0),

line(0.2,3,-0.2,3),
locate(-0.5,-2.5,"A"),
locate(-0.5,0.5,"B"),
locate(-0.5,3.5,"C"),

locate(-0.75,-1,"645"),
locate(-0.75,2,"540"),

red(line(1,3,1,-3)),
red(line(1,-3,1.3,-2.4804)),
red(line(1,-3,0.7,-2.4804)),

red(locate(1.2,0.5,matrix(1,2,wind,speed))),
red(locate(1.2,0,matrix(1,2,20,mph))),

blue(locate(-4.8,3,matrix(1,3,When,the,plane))),
blue(locate(-4.8,2.5,matrix(1,5,goes,from,B,to,C))),

locate(3,-3,matrix(1,2,Diagram,not)),
locate(3,-3.5,matrix(1,2,to,scale))
)
}}}
At this point the plane's speed x is now decreased to x-20 because it's fighting the wind (the plane goes north while the wind is going south).
Note: This assumes the plane's speed without wind (x) is the same as before.



Then,
distance = rate*time
time = distance/rate
time = (540 miles)/(x-20 mph)
time = 540/(x-20)
This is the time, in hours, it takes to go from B to C


------------------------------------------


To summarize what we know so far:<ul><li>time from A to B = 645/(x+15)</li><li>time from B to C = 540/(x-20)</li></ul>Each time value is in hours.


The instructions mention "<font color=blue>each flight required the same time</font>". 
So we'll set each time expression equal to one another. 
Then solve for x.


time from A to B = time from B to C
645/(x+15) = 540/(x-20)
645(x-20) = 540(x+15)
645x-12900 = 540x+8100
645x-540x = 8100+12900
105x = 21000
x = 21000/105
x = <font color=red>200</font>


The plane's speed without wind is <font color=red>200 mph</font>


When going from A to B, the time spent is:
time = 645/(x+15)
time = 645/(200+15)
time = 645/215
time = 3 hours


When going from B to C, the time spent is:
time = 540/(x-20)
time = 540/(200-20)
time = 540/180
time = 3 hours


Both portions of the trip are 3 hours each. This confirms we have the correct answer.


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Answer: The plane's speed without wind is <font color=red>200 mph</font>
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