Mi faster than the other. If they meet in 5 hours,
Two buses leave towns 720 miles apart at the same time and travel toward each other. One bus travels 16
what is the rate of each bus?
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Rate of the faster bus:
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Rate of the slower bus:
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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?

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%2F10 = 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.

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Hope this helps :D

5y

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351.23

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