<PRE><B>Question 181158</B>
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.

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