document.write( "Question 292296: i know the basic setup of the problem, but i don't know how to simplify the 4 hours and 30 minutes, i always come up with long decimal answers. please help.\r
\n" ); document.write( "\n" ); document.write( "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 #211167 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r\r
\n" ); document.write( "\n" ); document.write( "let r=rate of one car
\n" ); document.write( "and r+12=rate of the other car
\n" ); document.write( "After 4 hours the car travelling faster has travelled 4(r+12)mi
\n" ); document.write( "After 4.5 hours the car travelling slower has travelled 4.5r mi
\n" ); document.write( "Now we are told that the difference between these distances is 24 mi, so:\r
\n" ); document.write( "\n" ); document.write( "4(r+12)-4.5r=24
\n" ); document.write( "4r+48-4.5r=24 subtract 48 from each side
\n" ); document.write( "4r-4.5r=24-48
\n" ); document.write( "-0.5r=-24
\n" ); document.write( "r=48 mph----------------rate of slower car
\n" ); document.write( "r+12=48+12=60 mph---rate of faster car
\n" ); document.write( "CK
\n" ); document.write( "48*4.5=216 mi
\n" ); document.write( "60*4=240 mi
\n" ); document.write( "240-216=24 mi\r
\n" ); document.write( "\n" ); document.write( "hope this helps--ptaylor\r
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