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)![]() ![]() 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( " \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( " \n" ); document.write( " \n" ); document.write( "add 3 to both sides \n" ); document.write( " \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( " \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( " |