document.write( "Question 526667: a passenger train leaves the station 3 hours after a freight train left the same station. the freight train travels 30 mph slower than the passenger train. Find the rate of each if the passenger train catches up with the freight traing in 4 hours \n" ); document.write( "
Algebra.Com's Answer #348629 by ankor@dixie-net.com(22740)\"\" \"About 
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a passenger train leaves the station 3 hours after a freight train left the same station.
\n" ); document.write( " the freight train travels 30 mph slower than the passenger train.
\n" ); document.write( " Find the rate of each if the passenger train catches up with the freight train in 4 hours
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\n" ); document.write( "Let s = speed of the freight
\n" ); document.write( "then
\n" ); document.write( "(s+30) = speed of the pass train
\n" ); document.write( ":
\n" ); document.write( "From the given information we know the freight traveled 7 hrs, the pass 4 hrs
\n" ); document.write( ":
\n" ); document.write( "When the pass catches the freight, they will have traveled the same distance.
\n" ); document.write( "Write dist equation, dist = speed * time
\n" ); document.write( ":
\n" ); document.write( "Pass dist = freight dist
\n" ); document.write( "4(s+30) = 7s
\n" ); document.write( "4s + 120 = 7s
\n" ); document.write( "120 = 7s - 4s
\n" ); document.write( "120 = 3s
\n" ); document.write( "s + 120/3
\n" ); document.write( "s = 40 mph is the speed of the freight
\n" ); document.write( "then obviously,
\n" ); document.write( "40 + 30 = 70 mph is the pass speed
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\n" ); document.write( "Check this by finding the dist each traveled
\n" ); document.write( "4(70) = 280 mi
\n" ); document.write( "7(40) = 280 mi \n" ); document.write( "
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