Question

The train was moving with uniform speed after covering a distance of 30km some defect develop with the engine of the train and for this reason its speed reduced to 4 by 5 of its original speed. The train reaches is destination late by 45 minutes in case the defects had happened after covering 18 km more the train would have reached just 36 minute late find the speed of the train at the start and the distance of Germany

Answer 1

Speed u kmph distance traveled = S1 = 30km
time taken t1 = 30 / u hrs
New speed = v = 4/5 u
let the additional distance traveled = S2 km
time taken for taveling S2 = t2 = S2/ (4/5 u) = 5 S2 / 4 u hrs
Let Expected arrival time T = total time to trvel S1 and S2

t1 + t2 = 30 / u + 5 S2 / 4 u hrs = T + 45/60
=> 120 + 5 S2 = 4 u T + 3 u ======== equation 1

If defect hapened later.... the total time to tavel S1+S2 distance is
(S1 + 18) / u + (S2 - 18) / (4/5 u) = 48/ u + 5 (S2 -18)/4u
= ( 192 + 5 S2 - 90) / 4u = (102 + 5S2)/4 u
This is equal to T + 36/60 hours = T + 0.6
=> 102 + 5 S2 = 4 u ( T + 0.6) = 4 u T + 2.4 u === = equation 2
Subtract equation 2 from eqation 1 we get
18 = 0.6 u => u = 30 kmph
Equation 1 :: 120 + 5 S2 = 120 T + 90 => 6 + S2 = 24 T ---- eq 3
=> S2 = 24 T - 6

Total distance = S1 + S2 = 30 + 24 T - 6 = 24 (1 + T) km - eq 4