- Sat Sep 22, 2018 6:53 am#788
A train, an hour after starting, meets with an accident which detains it an hour, after which it proceeds at 3/5 of its former rate and arrives three hour after the time; but had the accident happened 50 miles farther on yhe line, it would have arrived one and one-half hour sooner. Find the length of the journey.
50/3x - 50/5x = 1.5
1/3x - 1/5x = 15/500
2/15x = 3/100
x = 200/45 = 40/9
5x = 200/9
3x = 120/9
1*200/9 + 0*1 + 120(t-2+3)/9 = 200t/9
200/9 + 120t/9 + 120/9 = 200t/9
80t/9 = 320/9
t = 4
D = 4*200/9 = 800/9 miles
50/3x - 50/5x = 1.5
1/3x - 1/5x = 15/500
2/15x = 3/100
x = 200/45 = 40/9
5x = 200/9
3x = 120/9
1*200/9 + 0*1 + 120(t-2+3)/9 = 200t/9
200/9 + 120t/9 + 120/9 = 200t/9
80t/9 = 320/9
t = 4
D = 4*200/9 = 800/9 miles