A plane flies from Alphaville to Betaville and then back to Alphaville. When there is no wind, the round trip takes 4 hours and 48 minutes, but when there is a wind blowing from Alphaville to Betaville at 100 miles per hour, the trip takes 5 hours. How many miles is the distance from Alphaville to Betaville?
4hrs, 48 min = 4.8 hrs.
Let D be the distance between the towns, and let R be the normal rate of the plane
So we have this equation [ with no wind ]
2D / R = 4.8
Solving for D we have that
2D = 4.8R
D = 2.4R (1)
And when the wind is blowing, we have this equation
D / (R + 100) + D / ( R - 100) = 5 simplify
[ D(R - 100) + D(R + 100) ] / [ (R + 100)(R - 100) ] = 5
[ 2DR] = 5 [ (R + 100) (R - 100) ] sub (1) for D
2 (2.4R)R = 5 (R^2 - 10000)
4.8R^2 = 5R^2 - 50000
5R^2 - 4.8R^2 - 50000 = 0
.2R^2 - 50000 = 0
.2R^2 = 50000
R^2 = 50000/ .2
R^2 = 50000 * 10/ 2
R^2 = 250,000 take the square root
R = 500
And using (1 )the distance between the towns is (2.4)(500) = 1200 miles
