very difficult problem

Question

An Airplane travels from A to B and back with no wind in 4h and 48 min; same airplane travels from A to B and back, with wind speeds of 100mph, and this time takes 5hrs instead; what is the distance between A and B? 

Answer 1

D is the distance ONE WAY from either A to B or B to A
R is the speed of the plane
t is the time in takes the plane to travel one way either with or against the wind

Note that 48 min = 48/60 = 4/5 hr = 0.8 hr

2D = R * 4.8 ---> D = 2.4R

D = (R+100)*(5-t) <--- with the wind

D = (R-100)(t) <--- against the wind

Substitutes D=2.4R into the second and third equation


2.4R = (R+100)*(5-t)
2.4R = (R-100)(t)



2.4r = 5r + 500 - rt - 100t
2.4r = rt - 100t

Solves first equation for r:
0 = 2.6r + 500 - rt - 100t
100t - 500 = 2.6r - rt
100t - 500 = (2.6-t)r

(100t-500)/(2.6-t) = r

r = -100(t-5)/(t-2.6)



Solves the second equation for r:
2.4r - rt = -100t
r(2.4 - t) = -100t
r(t-2.4) = 100t
r = 100t/(t-2.4)

Sets them equal:

-100(t-5)/(t-2.6) = 100t/(t-2.4)

cross multiplies:

-100(t-5)(t-2.4) = 100t(t-2.6)

-(t-5)(t-2.4) = t(t-2.6)

-(t^2 - 7.4t + 12) = t^2 - 2.6t

-t^2 + 7.4t - 12 = t^2 - 2.6t

0 = 2t^2 - 7.4t - 2.6t + 12

0 = 2t^2 - 10t + 12

0 = t^2 - 5t + 6

0 = (t - 3)(t - 2 )

t-3=0 --> t=3
t-2=0 --> t=2

2 hours with the wind and 3 hours against the wind


Using the ORIGINAL 2nd and 3rd equations:
2(r+100) = 3(r-100)

2r + 200 = 3r - 300
200 + 300 = 3r - 2r
500 = r

The speed of the plane is 500 mph
The round trip distance is 500 * 4.8 = 2400 miles
The distance one way is 1200 miles

With the wind: 1200 miles in 2 hours = 600 mph = 500 + 100 mph <--- YES IT CHECKS
Against the wind: 1200 miles in 3 hours = 400 mph= 500 mph - 100 mph <--- YES IT CHECKS