SOLUTION: two buses leave towns 924mi apart at the same time and travel toward each other. one bus travels 10mi/h slower than the other. if they meet in 6 hours, what is the rate of each bus

Question 949597: two buses leave towns 924mi apart at the same time and travel toward each other. one bus travels 10mi/h slower than the other. if they meet in 6 hours, what is the rate of each bus?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
EDIT: This IS the solution process, and the formula can be evaluated at the end if needed for any final answer.
(Note: One omission adjustment)

Rate of "other bus", r
Rate of "one bus", r-10
t, time when meet
d, distance separation at the start
condition: two busses going toward each other.

___________________rate____________time___________distance
One bus____________r-10____________t______________(r-10)*t
Other bus___________r______________t______________(r*t)
Total_______________________________________________d

Only r is unknown.

------(adjusted)

You can derive .
You can substitute the given values to compute r and r-10.

Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

two buses leave towns 924mi apart at the same time and travel toward each other. one bus travels 10mi/h slower than the other. if they meet in 6 hours, what is the rate of each bus?

Let speed of slower bus, be S
Then speed of faster bus = S + 10
In 6 hours, the slower bus has traveled 6S miles
In 6 hours, the faster bus has traveled 6(S + 10), or 6S + 60 miles
In 6 hours both buses have covered a total of 924 miles
We then get: 6S + 6S + 60 = 924
12S + 60 = 924
12S = 924 - 60
12S = 864
S, or speed of slower bus = , or mph
Speed of faster bus: 72 + 10, or mph
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