Question

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

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