A Problem in the Engine of the Train

let original speed be 's',distance be 'x' & time be 't' as per schedule.

let the new speed be "s' ",distance is 'x' and time be "t' ".

now the train's speed has decreased by 25% so the new speed is s'=s-(25/100)*s=0.75s .

and the new time t' is t'=t+20.

now divide the new time by old time.

s'/s=(x/t')/(x/t).

or (.75s)/s=(x/(t+20))*(t/x)

or .75=t/(t+20)

or 3/4=t/(t+20)

or 3t+60=4t

or t=60.

$\frac{4}{3}t = t + 20$ $\frac{t}{3} = 20$ $\boxed{t = 60}$

where v=initial velocity, s=displacement and t=initial time required:

v=s/t

t=s/v

25v/100=s/(t+20)

3v(t+20)=4s

3tv+60v=4s

t=(4s-60v)/3v=4s/3v-20

t=4t/3-20

t/3=20

t=60