SOLUTION: Two buses leave towns 760mi apart at the same time and travel toward each other. One bus travels
18 mi/h slower than the other. If they meet in
5 hours, what is the rate of eac
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-> SOLUTION: Two buses leave towns 760mi apart at the same time and travel toward each other. One bus travels
18 mi/h slower than the other. If they meet in
5 hours, what is the rate of eac
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Question 1088926: Two buses leave towns 760mi apart at the same time and travel toward each other. One bus travels
18 mi/h slower than the other. If they meet in
5 hours, what is the rate of each bus? Answer by ikleyn(52772) (Show Source):
You can put this solution on YOUR website! .
Two buses leave towns 760 mi apart at the same time and travel toward each other. One bus travels
18 mi/h slower than the other. If they meet in
5 hours, what is the rate of each bus?
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Let x be the rate of the slower bus, in miles per hour.
Then the rate of the faster bus is (x+18) mph, according to the condition.
The slower bus covered 5x miles before the buses meet each other.
The faster bus covered 5*(x+18) miles before they meet each other.
The sum of distances covered by buses is 3x + 5*(x+18).
It is equal exactly 760 miles. It gives you an equation
5x + 5(x+18) = 760.
Simplify and solve for x:
5x + 5x + 90 = 760,
10x = 760 - 90,
10x = 670 ====> x = = 67.
Thus you found the rate of the slower bus. It is 67 miles per hour.
Then the rate of the faster bus is 67 + 18 = 85 mph.
Solved.
It is a typical Travel and Distance problem for two bodies moving toward each other.
