document.write( "Question 634535: A train after traveling 150 km meets with an accident and then proceeds with 3/5 of its former speed and arrives its destination 8 hours late. Had the accident occurred 360 km further, it would have reached the destination 4 hours late. What is the total distance traveled by the train? \n" ); document.write( "
Algebra.Com's Answer #399849 by ankor@dixie-net.com(22740)\"\" \"About 
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A train after traveling 150 km meets with an accident and then proceeds
\n" ); document.write( " with 3/5 of its former speed and arrives its destination 8 hours late.
\n" ); document.write( " Had the accident occurred 360 km further, it would have reached the
\n" ); document.write( "destination 4 hours late.
\n" ); document.write( "What is the total distance traveled by the train?
\n" ); document.write( ":
\n" ); document.write( "let s = the normal speed of the train
\n" ); document.write( "then
\n" ); document.write( ".6s = the speed after the accident (3/5 the normal speed)
\n" ); document.write( ":
\n" ); document.write( "let d = total distance
\n" ); document.write( "Then
\n" ); document.write( "(d-150)km = distance traveled at the slower speed, first scenario
\n" ); document.write( "and
\n" ); document.write( "(d - (150+360)) = distance traveled at the slower speed (360 km further)
\n" ); document.write( "(d - 510) km
\n" ); document.write( ":
\n" ); document.write( "Write a time equation for the 1st scenario
\n" ); document.write( "slower time - normal time = 8 hrs
\n" ); document.write( "(\"150%2Fs\" + \"%28d-150%29%2F.6s\") - \"d%2Fs\" = 8
\n" ); document.write( "Multiply by .6s to clear the denominators
\n" ); document.write( "(.6(150) + (d-150)) - .6d = .6s(8)
\n" ); document.write( "90 + d - 150 - .6d = 4.8s
\n" ); document.write( ".4d - 60 = 4.8s
\n" ); document.write( ".4d - 4.8s = 60
\n" ); document.write( ":
\n" ); document.write( "Same with the 2nd scenario, (accident 360 km further down the road)
\n" ); document.write( "(\"510%2Fs\" + \"%28d-510%29%2F.6s\") - \"d%2Fs\" = 4
\n" ); document.write( "Multiply by .6s to clear the denominators
\n" ); document.write( "(.6(510) + (d-510)) - .6d = .6s(4)
\n" ); document.write( "306 + d - 510 - .6d = 2.4s
\n" ); document.write( ".4d - 204 = 2.4s
\n" ); document.write( ".4d - 2.4s = 204
\n" ); document.write( ":
\n" ); document.write( "Find the speed using elimination with these two equations
\n" ); document.write( ".4d - 2.4s = 204
\n" ); document.write( ".4d - 4.8s = 60
\n" ); document.write( "------------------subtraction eliminates d find s
\n" ); document.write( "0 + 2.4s = 144
\n" ); document.write( "s = 144/2.4
\n" ); document.write( "s = 60 km/hr is the normal speed
\n" ); document.write( "then
\n" ); document.write( ".6(60) = 36 km/hr is the \"after accident speed\"
\n" ); document.write( ":
\n" ); document.write( "Find the distance using the 1st scenario equation, replace s with 60
\n" ); document.write( ".4d - 4.8(60) = 60
\n" ); document.write( ".4d - 288 = 60
\n" ); document.write( ".4d = 60 + 288
\n" ); document.write( ".4d = 348
\n" ); document.write( "d = 348/.4
\n" ); document.write( "d = 870 km is the total distance
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check solution using the 2nd scenario
\n" ); document.write( "\"510%2F60\" + \"%28870-510%29%2F36\" - \"870%2F60\" =
\n" ); document.write( "8.5 + 10 - 14.5 = 4 hrs, confirms our solution of d=870km
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