A train covers a certain distance at a uniform speed. If the train had been 6 km/hr faster, it would have taken four hours less than the scheduled time. And if the train were slower by 6 km/hr, the train would have taken six hours more than the scheduled time. Find the length of the journey in kilometers.
+130514
Equating times we have this system
D/ ( R + 6) = D/R - 4
D/ (R - 6) = D/R + 6 simplify
D / (R + 6) = (D - 4R) / R
D/ ( R -6) = ( D + 6R) / R
DR = (R +6) ( D - 4R)
DR = (R - 6) ( D + 6R)
DR = DR + 6D - 4R^2 - 24R
DR = DR - 6D + 6R^2 -36R
0 = -4R^2 + 6D - 24R
0 = 6R^2 - 6D -36R add these
0 = 2R^2 + 60R
2R ( R - 30) = 0
Setting the second factor to 0 and solving for R produces the normal rate of 30 Km/ hr
And
-4 (30)^2 + 6D - 24 (30) = 0
-3600 + 6D - 720 = 0
6D = 3600 + 720
6D = 4320
D = 4360 / 6 = 720 km = distance of the journey
