OpenStudy (anonymous):

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 he takes. one day the engineer arrived at the station one hour before his usual time and , without waiting for the car ,started walking for work . on his way he met the car and reached the plant 10 minutes before the usual time. how long did the engineer walk before he met the car ?

5 years ago
OpenStudy (anonymous):

in units it should be 6

5 years ago
OpenStudy (anonymous):

you have to ans ... for how much time he had to walk.

5 years ago
OpenStudy (anonymous):

then w8

5 years ago
OpenStudy (anonymous):

nw see fr d usual routine if you plot along the time horizon scale.at t=0 he is at the station and car is there to pick up

5 years ago
OpenStudy (anonymous):

bt when he arrives 1 hour before..then shift the time horizon

5 years ago
OpenStudy (anonymous):

ans is 55 minutes..

5 years ago
OpenStudy (anonymous):

@shreya pls share how do u find that???????//

5 years ago
OpenStudy (anonymous):

well you can start like this- suppose the train and car reach at 10 am but one day the train arrived at 9 am and the engineer started to walk ... you'll get the answer if u go this way... i m stuck ..i just need some help to get the ans. i have done it before but now m not able to get the ans..

5 years ago
OpenStudy (shubhamsrg):

Lets say the train used to arrive at T . it took 10 mins less, thus the car must have picket the man 5 mins before T,symmetrically. Hence we walked for 60-5 mins = 55 mins

5 years ago
OpenStudy (anonymous):

@shubhamsrg can u explain in detail...

5 years ago
OpenStudy (shubhamsrg):

It was a good question, hmm

5 years ago
OpenStudy (anonymous):

hey why did yu delete it ???

5 years ago
OpenStudy (shubhamsrg):

Understand it like this: lets say his train arrives at 12:00 (i.e. T) On a usual day, car will leave plant at 12:00 - x , will arrive at station at 12:00 reaches work at 12:00 + x on this very special day, man starts walking at 11:00 , car leaves plant at same 12:00 - x car meets man at 12:00 - x + y He reaches to work at 12:00 - x + y + y Note that 12:00 - x+2y = 12:00 + x - 10 => x-y= 5 Thus car meets man at 12:00 - x + y = 12:00 - 5 = 11:55 Hence he walked for 55 mins

5 years ago
OpenStudy (shubhamsrg):

nevermind.

5 years ago
OpenStudy (anonymous):

ah..!! brilliant ..!! amazing yar.. thanks :D .

5 years ago
OpenStudy (shubhamsrg):

Its one of those rare occasions when I did a question! :P Thanks anyways. :)

5 years ago
OpenStudy (anonymous):

bt you can also use shifting the time horizon as we do in our transformation of curves

5 years ago
OpenStudy (anonymous):

@manishsatywali I need an easy explanation..i have no idea about what you're saying

5 years ago
OpenStudy (shubhamsrg):

Firstly I was going all crazy on it with too many variables. I even tried graphing but failed. This one some how worked out.

5 years ago
OpenStudy (anonymous):

k i got it

5 years ago
OpenStudy (anonymous):

u will learn transformation of curves............same concept applies here

5 years ago