SOLUTION: This is a science and medicine. A bicyclist rode into the country for 5h. In returning , her speed was 5mi/h faster and the trip took 4h. What was her speed each way? Would the a

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Question 111934: This is a science and medicine.
A bicyclist rode into the country for 5h. In returning , her speed was 5mi/h faster and the trip took 4h. What was her speed each way?
Would the answer be 20mi/h and 25mi/h?
Thank you,
Barb Neely
Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website!
Your answer is correct. Her speed was 20 mph going into the country and 25 mph during her
return ride.
.
You can check this as follows: On her way into the country she rode at 20 mph for 5 hours
so she covered 100 miles. On her return trip she rode at 25 mph and in 4 hours she covered
25 times 4 = 100 miles. Her two rides covered the same distance ... so the speeds
are correct.
.
This comes from recognizing that the two distances covered (distance out into the country and
distance on the return trip) must be the same. So you can use the distance formula of:
.
Distance = speed times time
.
and in equation form this is
.
D = S * T
.
for each leg of the trip. Call R the speed on the outbound trip. Then R + 5 must be the
speed on the return trip. This means that the distance on the outbound trip is:
.
D = R * 5 = 5R
.
where 5 is the time in hours on the outbound leg. Then the return distance is:
.
D = (R + 5)*4 = 4R + 20
.
in which R + 5 is the speed on the return trip and 4 is the number of hours spent at
that speed.
.
Since the two distances are equal, you can set them equal in equation form to get:
.
5R = 4R + 20
.
Subtract 4R from both sides of this equation and it reduces to:
.
R = 20
.
And 20 is the speed in mph for the outbound ride. Therefore 25 mph must be the speed for
the return leg because is is 5 mph faster than the speed on the outbound leg.
.
Hope this helps.