A bicyclist riding against the wind averages 10 miles per hour in traveling from A to B, but with the wind averages 15 miles per hour in returning from B to A. How many miles per hour is his average speed for the trip?
+636 Here is my solution:
Let a = The average speed for the trip
Let d = The one-way distance
Next we Write a time equation: time = distance/speed
Against time + with time = total time
d/15 + d/10 + 2d/a
Then we multiply by 30a
2a(d) + 3a(d) = 30(2d)
2ad + 3ad = 60d
5ad = 60d
divide both sides by d
5a = 60
a = 60/5
a = 12 mph is the average speed for the trip
D/10 = T................(1)
D/15 = T - 1/3.........(2)
When he cycles @ 15mph, that means it is going to take him 10/15 =2/3 of the time of cycling @ 10 mph.
Solve (1) and (2) above and you will get:
D = 10 miles
T = 1 hour
[2*10] / [1+2/3] =20 x 3/5 =12 mph - his average speed.
