SOLUTION: Deni walks to work and back on the same path. She walks on level ground at 4 MPH, uphill at 3 1/3 MPH, and downhill at 5 MPH. To get to work, she walks first on level ground and th

Question 556604: Deni walks to work and back on the same path. She walks on level ground at 4 MPH, uphill at 3 1/3 MPH, and downhill at 5 MPH. To get to work, she walks first on level ground and then up a hill. If her total time for the round trip is 2 1/2 hours, how many miles does she travel to work?
This problem has been stressing me out because i know that i have to set something = 2 1/2 hours but i have only delt with two other speeds, not 6 or 3 depending on how you look at it.
Thank you for the help
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Deni walks to work and back on the same path.
She walks on level ground at 4 MPH, uphill at 3 1/3 MPH, and downhill at 5 MPH.
To get to work, she walks first on level ground and then up a hill.
If her total time for the round trip is 2 1/2 hours, how many miles does she travel to work?
:
Let x = level ground dist
Let y = hill distance
:
We can write a time equation for the round trip: time = dist/speed:
+ + + = 2.5
+ + = 2.5
multiply thru by 4*3.33*5 = 66.66 to clear the denominators, cancel we have:
16.665(2x) + 20y + 13.333y = 66.66(2.5)
33.333x + 20y + 13.333y = 166.66
33.333x + 33.333y = 166.66
33.333y = -33.333x + 166.66
divide thru by 33.333 and we have a simple equaiton
y = -x + 5
From this equation we can find an x,y combination that will make sense
there are more than one set of solutions to this
Let x = 2 mi on the level
y = -2 + 5
then
y = 3 mi on the hill
therefore:
2 + 3 + 3 + 2 = 10 miles to work
:
see if this checks out time-wise
+ + =
.5 + .9 + .6 + .5 = 2.5 hrs
:
It's interesting that any of the positive solutions will add up 10 miles which is what they want
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