SOLUTION: A girl walks up a mountain at 6 mph and arrives at 1pm, she walks another day up the same mountain at 10mph, but arrives at 11am. How fast would she have to walk to arrive there at
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: A girl walks up a mountain at 6 mph and arrives at 1pm, she walks another day up the same mountain at 10mph, but arrives at 11am. How fast would she have to walk to arrive there at
Log On
Question 265431: A girl walks up a mountain at 6 mph and arrives at 1pm, she walks another day up the same mountain at 10mph, but arrives at 11am. How fast would she have to walk to arrive there at 12 noon? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A girl walks up a mountain at 6 mph and arrives at 1pm, she walks another day up the same mountain at 10mph, but arrives at 11am.
How fast would she have to walk to arrive there at 12 noon?
:
Let r = rate she would have to walk to arrive at noon
:
let d = the distance she walks up the mountain each time
:
Write two time equations
r mph time = 6 mph time - 1 hr = - 1
:
r mph time = 10 mph time + 1 hr = + 1
:
therefore we can find d this way + 1 = - 1
Multiply by 30 to eliminate the denominators
3d + 30 = 5d - 30
30 + 30 = 5d - 3d
60 = 2d
d = 30 miles
:
Use the first time equation to find r when d = 30 = - 1 = 5 - 1 = 4
4r = 30
r =
r = 7.5 mph, the walking speed to arrive at noon
:
:
Check solution by finding the times at each speed
30/7.5 = 4 hr (arr at noon)
30/6 = 5 hr (arr at 1 pm)
30/10 = 3 hr (arr at 11 am)
