11something less than 50 mins would be ans :P [98]
An engineer works at a plant out of town. A car is sent for him from the plant everyday that arrives at the railway station at the same time as the train on which he comes. One day he arrives at the station one hour before and, without waiting for the car, started walking to work. On his way he met the car and reached the plant 10 minutes before the usual time. How long did the engineer walked before he met the car? Assume all motions with uniform speed.
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41 Answers
1assumption is not allowed !! kaymant is the answer in real numbers or in terms of variables !!
1How long did the engineer walked before he met the car? // does it mean we have to calculate the time he walked or the distance he walked !!
1we ve to calculate d time yaar.............also see d title of d thread.........
1arrey............im not assuming..................i arrived at tht expression.......if u want i can post d solution also............im just verifying it taking real life situations............
and kaymant sir asked for time..........
1ok.........
let d speed of d car be v1....and d distance be x....and speed of d man be v2
for d car to reach at d time of d arrival of d train it shud start at a time t1 b4 its arrival.....so v1.t1=x
now.... d train cums 1 hour b4..... now let the man and d car meet at a distance y from d station
so y=v2.t ==>1
and also x-y=v1.t3 ==>2
now..... 60+t1=t+t3+10
50+t1-t3=t
now eqn 2 and 1...
v1.t1=v1.(50+t1-t)+v2.t
so...solving...
t=50v1/(v1-v2)...........
66the "plane" in the question is a typo.. and I am asking the time he walked for and it will be a numerical answer.. the data given are sufficient to find the time.
1Let the speed of the person = u
and speed of the car = v
normal time =t
time for which he walked = t1
time he went on car = t2
distance he walked= s1
As the distance is same
s=vt=ut1+vt2
t1+t2= t-10
So, vt= ut1+v[t-10-t1]
t1=10v/(u-v)
1/t1=u/10v - 1/10
The car could travel s1 in 60-t1 minutes
60-t1=ut1/v
Hence, u/v= (60/t1) -1
Which gives
1/t1 = (6/t1)- 1/10
Therefore t1=50
ie. the man walks for 50 minutes before getting the car