Let u=speed of car in m/min
v= speed of man in m/min
t1=usual time car takes to reach factory from station in min.
t2= time spent by engineer in walking in min (to be determined)
t3= time taken by car to reach factory after picking the engineer on the way in min
Now
t2+t3 = 60 + t1 - 10
t1-t3 = t2 - 50
Cosidering distances covered
v . t2 = u (t1 - t3) ( Because car travelled so much distance less that day)
v t2 =u (t2 - 50) ( From above)
v/u = 1 -50/t2
We know that car reached that day 10 mts early. So car saved traveling twice the distance travelled by man by avoiding to and fro driving.
Hence 10 u =2 t2 v (car would have travelled additional 10 min had the engineer not walked)
v/u =5/t2
Substituting v/u from above we get
1 - 50/t2 = 5/t2
t2 -50 = 5
t2 = 5 + 50 = 55 min
I hope it is right! May be there is a shorter way of doing it !