SOLUTION: Abigail lives in an off-campus apartment. Some days she rides her bike to campus, other days she
walks. When she rides her bike, she gets to her first classroom building 36 minut
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-> SOLUTION: Abigail lives in an off-campus apartment. Some days she rides her bike to campus, other days she
walks. When she rides her bike, she gets to her first classroom building 36 minut
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Question 1109679: Abigail lives in an off-campus apartment. Some days she rides her bike to campus, other days she
walks. When she rides her bike, she gets to her first classroom building 36 minutes faster than when she
walks. If her average walking speed is 3 mph and her average biking speed is 12 mph, how far is it from
her apartment to the classroom building? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! When she rides her bike, she gets to her first classroom building 36 minutes faster than when she walks.
If her average walking speed is 3 mph and her average biking speed is 12 mph, how far is it from her apartment to the classroom building?
:
let d = dist from home to school
since we are dealing mph, change 36 min: 36/60 = .6 hrs
:
Write a time equation; time = dist/speed
:
Walk time - bike time = .6 hrs - = .6
multiply by 12, cancel the denominators
4d - d = .6(12)
3d = 7.2
d = 7.2/3
d = 2.4 mi to the school
:
:
Check this, find the actual time each way
2.4/3 = .8 hrs
2.4/12 - .2 hrs
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time dif: .6 hrs which is 36 min
