SOLUTION: A bicycle traveling at a constant speed goes 45 miles with the wind turns around and travels an additional 42 miles against the wind. If the speed of the bicycle is 15 miles per ho

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Question 254771: A bicycle traveling at a constant speed goes 45 miles with the wind turns around and travels an additional 42 miles against the wind. If the speed of the bicycle is 15 miles per hour and the total time is 6 hours, find the speed of the wind.
Found 2 solutions by dabanfield, palanisamy:
Answer by dabanfield(803) (Show Source):
You can put this solution on YOUR website!
A bicycle traveling at a constant speed goes 45 miles with the wind turns around and travels an additional 42 miles against the wind. If the speed of the bicycle is 15 miles per hour and the total time is 6 hours, find the speed of the wind.
Remember distance = rate x time or d = r*t
Let w be the speed of the wind and t be the time for the trip with the wind:
The equation for the trip with the wind then looks like:
45 = (w+15)*t or
(1) 45/(w+15) = t
Since the total trip was 6 hours the time for the trip back is 6-t.
The equation for the trip back against the wind looks like:
(2) 42 = (w-15)*(6-t)
Based on (1) we can substitute 45/(w+15) for t in (2) above and solve for w.

Answer by palanisamy(496) (Show Source):
You can put this solution on YOUR website!
Given, the speed of the bicycle = 15 miles per hour
Let the speed of the wind = x miles per hour
A bicycle traveling at a constant speed goes 45 miles with the wind turns around and travels an additional 42 miles against the wind.
Time taken to travel with the wind = distance/speed
= 45/(15+x)
Time taken to travel against the wind = 42/(15-x)
Total time is 45/(15+x) +42/(15-x) = 6
[45(15-x)+42(15+x)]/(15-x)(15+x)= 6
675-45x+630+42x = 6(15+x)(15-x)
-3x+1305 = 6(225-x^2)
0 = 1350-6x^2+3x-1305
6x^2-3x-45=0
2x^2-x-15 = 0
(2x+5)(x-3) = 0
x = 3, -5/2
Since x cannot be negative, we get x = 3
Therefore the speed of the wind = 3 miles per hour