document.write( "Question 866200: Two cars leave a city on the same road, one driving 12 mph faster than the other. After 4 hours, the car traveling faster stops for lunch. After 4 hours and 30 minutes, the car traveling slower stops for lunch. Assuming that the person in the faster car is still eating lunch, the cars are now 24 miles apart. How fast is each car driving? \n" ); document.write( "
Algebra.Com's Answer #522116 by stanbon(75887)\"\" \"About 
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Two cars leave a city on the same road, one driving 12 mph faster than the other. After 4 hours, the car traveling faster stops for lunch. After 4 hours and 30 minutes, the car traveling slower stops for lunch. Assuming that the person in the faster car is still eating lunch, the cars are now 24 miles apart. How fast is each car driving?
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\n" ); document.write( "Faster car DATA:
\n" ); document.write( "rate = r+12 mph ; time = 4 hrs ; distance = r*t = 4(r+12) miles
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\n" ); document.write( "Slower car DATA:
\n" ); document.write( "rate = r mph ; time = (9/2) hrs ; distance = r*t = (9/2)r miles
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\n" ); document.write( "Solve for \"r\"
\n" ); document.write( "4(r+12) - (9/2)r = 24
\n" ); document.write( "4r+48 -(9/2)r = 24
\n" ); document.write( "(-1/2)r = -24
\n" ); document.write( "r = 48 mph (rate of the slower car)
\n" ); document.write( "r+12 = 60 mph (rate of the faster car)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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