SOLUTION: A bike race consists of two segments whose total length is 90 km. The first segment is covered at 10 kph and takes 2 hours longer to complete than the second segment, which is cov

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Question 96972: A bike race consists of two segments whose total length is 90 km. The first segment is covered at 10 kph and takes 2 hours longer to complete than the second segment, which is covered at 25 kph. How long is each segment?
I have tried setting it up as t/25 = (t+2)/10 but for some reason I can't figure out the algebra.
Thanks!
Found 2 solutions by Earlsdon, mathslover:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
From your equation, it appears that you assumed that the two segments were the same distance, but there is nothing in the problem to support this assumption.
So, back to basics:
d%5B1%5D+=+r%5B1%5Dt%5B1%5D and...
d%5B2%5D+=+r%5B2%5Dt%5B2%5D
We know that:
r%5B1%5D+=+10 and...
r%5B2%5D+=+25
t%5B1%5D+=+t%5B2%5D+%2B+2 and we also know that:
d%5B1%5D%2Bd%5B2%5D+=+90 So we can write:
d%5B1%5D+=+10t%5B1%5D
d%5B2%5D+=+25t%5B2%5D aagh...to many variables! Let's replace t%5B1%5D with t%5B2%5D%2B2 and replace d%5B2%5D with 90-d%5B1%5D, so now we have:
d%5B1%5D+=+10%28t%5B2%5D%2B2%29 and
90-d%5B1%5D+=+25t%5B2%5D Solve this for d%5B1%5D
d%5B1%5D+=+90-25t%5B2%5D...but, of course, d%5B1%5D+=+d%5B1%5D, so...
10%28t%5B2%5D%2B2%29+=+90-25t%5B2%5D Good, now we have one variable t%5B2%5D, so let's solve for it.
10t%5B2%5D%2B20+=+90-25t%5B2%5D
35t%5B2%5D%2B20+=+90
35t%5B2%5D+=+70 and...
t%5B2%5D+=+2hrs. Now we can find d%5B1%5D and d%5B2%5D
d%5B1%5D+=+10%28t%5B2%5D%2B2%29
d%5B1%5D+=+10%282%2B2%29
d%5B1%5D+=+40km and...
d%5B2%5D+=+90-d%5B1%5D
d%5B2%5D+=+90-40
d%5B2%5D+=+50km
Check:
Total distance is d%5B1%5D%2Bd%5B2%5D+=+40%2B50 = 90km

Answer by mathslover(157) About Me  (Show Source):
You can put this solution on YOUR website!
Let the first segment be x km
therefore the second segment is 90- x km
time taken to cover the first segment (t1) = x/10
time taken to cover the second segment (t2) = (90-x)/25
given that the time taken to cover the first segment is 2 hours more
t1 = t2 + 2
substituting the values of t1 andf t2 from above
x/10 = (90-x)/25 +2
Multipying by 50 on both sides
5x = 180 -2x + 100
7x =280
x=40
therefore the other segment is 90 -40 = 50
hence the segments is (40, 50)