SOLUTION: an athlete covered a distance of 57 miles by walking for 2 hours and then riding a bicycle for the remainder of the trip. her rate increased by 10 mph on the bike. if the entire tr

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Question 181158: an athlete covered a distance of 57 miles by walking for 2 hours and then riding a bicycle for the remainder of the trip. her rate increased by 10 mph on the bike. if the entire trip took 5 1/2 hours, what was her riding rate?
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
an athlete covered a distance of 57 miles by walking for 2 hours and then riding a bicycle for the remainder of the trip. her rate increased by 10 mph on the bike. if the entire trip took 5 1/2 hours, what was her riding rate?
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Walking DATA:
time = 2 hrs ; distance = x miles ; rate = x/2 mph
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Bicycle DATA:
time = (7/2) hrs ; distance = (57-x) miles ; rate = d/t = (57-x)/(7/2) mph
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Equation:
bike rate - walk rate = 10 mph
(57-x)/(7/2) - (x/2) = 10
Modify:
(2/7)(57-x) - x/2 = 10
(2/7)*57 - [(2/7)+(1/2)]x = 10
[(2/7)+(1/2)]x = (2/7)*57 + 10
(11/14)x = 184/7
x = (14/11)(184/7)
x = 368/11 = 33.45
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bike rate = (57-x)/(7/2) mph
bike rate = (57-33.45)/(7/2) = 6.73 mph
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Cheers,
Stan H.

Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

Let r be the riding rate, then the walking rate must be r - 10. Let be the riding distance and the walking distance be . And since she walked for 2 hours of a 5.5 hour trip, the riding time must have been 3.5 hours.

Using the basic equation for distance, rate, and time, i.e. , we can describe the walking part of the trip as:

and the riding part of the trip:

But we know that the total distance, which can be expressed as is 57 miles, so:

John