Question 1116795
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Let r = the rate of the slower car.
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>>the rate of the faster car exceeded the rate of the slower 
car by 6 kilometers per hour.<<
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So r+6 = the rate of the faster car
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>>traveling toward each other<<
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So their approach rate is the sum of their rates = r+r+6 = 2r+6
From 11AM to 2:30PM is 3.5 hours.
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>>from towns 315 kilometers apart<<
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So their original distance apart went from 315 km to 0 km distance 
apart in 3.5 hours, so

using Distance = Rate&#8729;Time, with Distance = 315 km, 
Approach Rate = 2r+6, and Time = 3.5 hours

315 = (2r+6)(3.5)
 
Solve that and get r = 42 km/hr for the slower car, and
r+6 = 42+6 = 48 km/hr for the faster car.

Edwin</pre>