Question 1088926
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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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<pre>
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 = {{{670/10}}} = 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.
</pre>

Solved.



It is a typical Travel and Distance problem for two bodies moving toward each other.


For more samples of similar solved problems see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Travel-and-Distance-problems-for-two-bodies-moving-toward-each-other.lesson>Travel and Distance problems for two bodies moving in opposite directions</A>

in this site.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this textbook under the section "<U>Word problems</U>", &nbsp;the topic "<U>Travel and Distance problems</U>".