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two buses leave towns 604 mi apart at the same time and travel toward eachother. One bus travels 9 mi/h faster than the other one. If they meet in 4 hours, what is the rate of each bus?

User JerodG
by
6.5k points

1 Answer

4 votes

Answer:

s = 71 mph, and s+9 = f = 80 mph

Explanation:

The distances the two busses travel add up to 604 mi.

Letting f be the faster speed and s the slower. Then f = s + 9 (mph).

Then (s + 9)(mph)(4 hr) + (s)(mph)(4 hr) = 604 mi. Solve this for s:

4s + 36 + 4s = 604 mi, or 8s = 568. Finally, s = 71 mph, and s+9 = f = 80 mph.


User Xno
by
6.8k points
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