SOLUTION: Two cyclists leave a city at the same time, one going east and the other going west. The westbound cyclist bikes 5 mph faster than the eastbound cyclist.After 6 hours, they are 246

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Question 1208999: Two cyclists leave a city at the same time, one going east and the other going west. The westbound cyclist bikes 5 mph faster than the eastbound cyclist.After 6 hours, they are 246 miles apart. How fast is each cyclist riding?
Found 2 solutions by josgarithmetic, timofer:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
Many example exercises fit the form of this one.
Two cyclists leave a city at the same time, one going east and the other going west. The westbound cyclist
bikes k mph faster than the eastbound cyclist. After t hours, they are d miles apart. How fast is each
cyclist riding?

One unknown variable, x, the speed for the eastward cyclist
All other numbers are given.
                 SPEED       TIME       DISTANCE
EASTWARD           x           t         xt         

WESTWARD           x+k         t        t(x+k)

TOTAL                                     d

highlight_green%28xt%2Bt%28x%2Bk%29=d%29

xt%2Btx%2Btk=d
tx%2Btx%2Btk=d
2tx=d-tk
highlight%28x=%28d-tk%29%2F%282t%29%29

Answer by timofer(107) About Me  (Show Source):
You can put this solution on YOUR website!
They go in opposite directions so the sum of the distances each travels by 6 hours
must be 246 miles.

eastbound + westbound = 246

6s%2B6%28s%2B5%29=246
6s%2B6s%2B30=246
12s%2B30=246
12s=246-30
12s=216
s=216%2F12
s=18

East bound, 18 mph
West bound, 23 mph