you have a 35 min. communte to work. Since you leave early, the trip going to work is easier than the trip home. You can travel to work in the same time that it takes you to make it 28 miles back home. Your average speed coming home is 5 MPH slower than your average speed going to work. What is your average speed going to work?

Question

you have a 35 min. communte to work. Since you leave early, the trip going to work is easier than the trip home.  You can travel to work in the same time that it takes you to make it 28 miles back home. Your average speed coming home is 5 MPH slower than your average speed going to work. What is your average speed going to work?

anonymous · Answer

I got 15.5...?

anonymous · Answer

15.5 MPH

chris · Answer

hmmm.  let's see.  So on the way to work:
1. x * 35min/(60min/hour)   = 28miles

anonymous · Answer

why does that work?

chris · Answer

hmm, or are they not the same length?

anonymous · Answer

They are the same length, I'm sure.

chris · Answer

Hmmm, but in that case it doesn't make as much sense.

anonymous · Answer

It seems like it would be?

chris · Answer

If the trips are the same length (28 miles), and you know how long it takes to get to work (35 min), you would know the average speed

chris · Answer

by distance/time

chris · Answer

I think what the problem is trying to do is give you three unknowns, and enough information for three equations

chris · Answer

three unknowns would be x=speed to work, y = speed back from work, d= distance to work

anonymous · Answer

Gotcha.

anonymous · Answer

I did 1/35 + 1/28 = 1/t(unknown time)

anonymous · Answer

I found the LCD to be 980t.

chris · Answer

ah ok, here we go:
1.  so y*35/60 = 28  (with y being the speed coming back from work)
2.  x = y+5 (speed going to work is five mph faster)
so now just solve for y in the first equation and add 5

chris · Answer

sorry for the confusion heh.  So that means that y=48mph, and x = 53mph