document.write( "Question 903383: A train, after travelling for 3 hours, was detained 40 minutes. Then it proceeded at 6/5 its former rate and arrived 10 minutes late. If the delay had occurred at a point 15 kilometers farther from its destination, the train would have been only 5 minutes late. Find the rate of the train and the total distance traveled. \n" ); document.write( "
Algebra.Com's Answer #548149 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
let s = the normal speed of the train
\n" ); document.write( "let d = total distance of the trip
\n" ); document.write( ":
\n" ); document.write( "A train, after travelling for 3 hours, was detained 40 minutes.
\n" ); document.write( "3s = train distance up to this point
\n" ); document.write( "then
\n" ); document.write( "\"d%2Fs\" = normal time for the trip
\n" ); document.write( ":
\n" ); document.write( " Then it proceeded at 6/5 its former rate and arrived 10 minutes late.
\n" ); document.write( "faster speed = 1.2s
\n" ); document.write( "at this speed it cut half an hour off the normal travel time (40-10)
\n" ); document.write( "\"d%2Fs\" - (3+\"%28%28d-3s%29%29%2F%28%281.2s%29%29\") = .5 hrs
\n" ); document.write( "\"d%2Fs\" - 3 - \"%28%28d-3s%29%29%2F%28%281.2s%29%29\" = .5 hrs
\n" ); document.write( "add 3 to both sides
\n" ); document.write( "\"d%2Fs\" - \"%28%28d-3s%29%29%2F%28%281.2s%29%29\" = 3.5
\n" ); document.write( "multiply by 1.2s, resulting in
\n" ); document.write( "1.2d - d + 3s = 3.5(1.2s)
\n" ); document.write( ".2d = 4.2s - 3s
\n" ); document.write( ".2d = 1.2s
\n" ); document.write( "d = 1.2s/.2
\n" ); document.write( "d = 6s
\n" ); document.write( "then
\n" ); document.write( "3s = halfway point, where the 1st delay occurred
\n" ); document.write( ":
\n" ); document.write( " If the delay had occurred at a point 15 kilometers farther from its destination, the train would have been only 5 minutes late.
\n" ); document.write( "This time the train gained 40-5 = 35 minutes over the original undelayed time
\n" ); document.write( "35 min = 7/12 hrs
\n" ); document.write( ":
\n" ); document.write( "the distance traveled at the faster speed this time; (3s+15)
\n" ); document.write( "The distance traveled at the normal speed: (3s-15)
\n" ); document.write( ":
\n" ); document.write( "time at normal speed + time at faster speed = normal time - 35 min
\n" ); document.write( "\"%283s-15%29%2Fs\" + \"%283s%2B15%29%2F1.2s\" = \"6s%2Fs\" - \"7%2F12\"
\n" ); document.write( "multiply equation by 12s
\n" ); document.write( "12(3s-15) + 10(3s+15) = 12(6s) - 7s
\n" ); document.write( "36s - 180 + 30s + 150 = 72s - 7s
\n" ); document.write( "66s - 30 = 65s
\n" ); document.write( "66s - 65s = 30
\n" ); document.write( "s = 30 mph is the normal speed of the train
\n" ); document.write( "Find the distance
\n" ); document.write( "6(30) = 180 mi is the distance
\n" ); document.write( " \r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );