SOLUTION: Hi I have a problem that is tricky, a bicyclist and a hiker leave the same place at the sametime and travel in the same direction. The bicyclist travels three times as fast as the
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Question 180134: Hi I have a problem that is tricky, a bicyclist and a hiker leave the same place at the sametime and travel in the same direction. The bicyclist travels three times as fast as the hiker. At the end of 3 hrs they are 24 miles apart. How fast does the hiker travel.
I tried looking at all angles but could not get a suitable ,if you could provide a suitable formula using x + y. Thanks This is a uniform motion problem from a study unit book for an on line course from penn foster its algebra part 3. 2469C-1 Found 2 solutions by Alan3354, stanbon:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! a bicyclist and a hiker leave the same place at the sametime and travel in the same direction. The bicyclist travels three times as fast as the hiker. At the end of 3 hrs they are 24 miles apart. How fast does the hiker travel.
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The hiker moves at r mph, the cyclist at 3r mph.
After 3 hours, the hiker has travelled 3r miles, the cyclist 9r miles.
9r - 3r = 24 miles
6r = 24
r = 4mph, the hiker's speed.
You can put this solution on YOUR website! a bicyclist and a hiker leave the same place at the sametime and travel in the same direction. The bicyclist travels three times as fast as the hiker. At the end of 3 hrs they are 24 miles apart. How fast does the hiker travel.
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Hiker DATA:
rate = x mph ; time = 3 hrs. ; distance = rt = 3x miles
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Bicyclist DATA:
rate = 3x mph ; time = 3 hrs. ; distance = rt = 9x miles
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Equation:
Since they are going in the same direction:
3x-x = 24 miles
2x = 24
x = 12 mph (hiker's rate; extraordinary)
3x = 36 mph (bicyclist's rate; also extraordinary)
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Cheers,
Stan H.
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