algebra question?

Question

a man drives from home to work at a speed of 50 miles per hours, the return trip from work to home id traveled at the more leisurely pace of 30 miles per hour, whta is the man's average speed for the trip
is it 40

Answer 1

Let distance from home to work = D miles
Time to work = D/50 hours
Time for return = D/30 hours
Total time = D/50 + D/30 = 3D/150 + 5D/150
Total time = 8D/150
Total distance = 2D
Average speed = 2D/(8D / 150)
= 1/4 x 150
= 37.5 m.p.h.

Answer 2

Remember that average speed is a function of time. If the man traveled for x amount of *time* at 50mph, and then for the same (x) amount of time at 30mph, *then* his average speed would be the average of the two speeds, i.e. 40mph. [(50x + 30x) / 2x = 80x / 2x = 80 / 2 = 40]

However, the problems states that the man travelled the same *distance* at two different speeds. In other words, he spent more time travelling at 30mph than he spent travelling at 50mph. If you consider it that way, it becomes obvious that his average speed was less than 40mph. To get the actual number, we need to divide the distance he traveled by the amount of time he spent travelling:
d = distance between work and home, so total distance traveled = 2d
Time spent going to work = d / 50
Time spent going home = d / 30
Total time = 3d/150 + 5d/150 = 8d/150 = 4d/75
Total distance/total time = 2d / (4d/75) = 75 * 2d/4d = 75 / 2 = 37.5

Answer 3

yeap. 40

Answer 4

if it's the exact same distance then yes

Answer 5

40mph