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)   (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







Answer by timofer(107)   (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









East bound, 18 mph
West bound, 23 mph
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