SOLUTION: At 11am two cars started traveling toward each other from towns 315kilometers apart. They passed each other at 2:30pm. If the rate of the faster car exceeded the rate of the slo

Question 1116795: At 11am two cars started traveling toward each other from towns 315kilometers
apart. They passed each other at 2:30pm. If the rate of the faster car exceeded
the rate of the slower car by 6kilometers per hour, find the rate, in kilometers
per hour, of the faster car.
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!

Let r = the rate of the slower car.

>>the rate of the faster car exceeded the rate of the slower
car by 6 kilometers per hour.<<

So r+6 = the rate of the faster car

>>traveling toward each other<<

So their approach rate is the sum of their rates = r+r+6 = 2r+6
From 11AM to 2:30PM is 3.5 hours.

>>from towns 315 kilometers apart<<

So their original distance apart went from 315 km to 0 km distance 
apart in 3.5 hours, so

using Distance = Rate∙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

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