Aptitude - Problems on Trains - Discussion

Total distance of the 2 trains are 220 (108+112) :first don't change anything. Just change the speed 50 km/hr into m/sec.

If we change speed 50 ino m/sec then it will be 125/9. The equation becomes.

(125/9) +x=220/6.

We get x=410/18 m/sec.

Now convert to km/hr we get x= (410/18) (18/5).

X=82.

We have got v (kmph) if it had been v * 5/18 then we would have gotten speed in m/s.

Hey @Prity and @Prem
See
(1) S km/hr = S.5/18 m/sec
(2) S m/sec = S.18/5 km/hr

That's why
50.5/18 -X.5/18 = 220/6 m/sec
250/18 -X.5/18 = 220/6 m/sec
- X.5/18 = 220/6 . 18 /250 m/sec

Now all are in m/ sec but now the game will be changed
Suppose if I move 5/18 from left side to right side so according to second rule m/sec. 18/5 =km/hr

-X = 220/6. 18 /250. (5/18) the value will be in km/hr
After all the calculation you will get the same answer in km/hr.

I hope you got the logic behind it.

I agree with you. @Prity.

Length of the trains=108+112=220m=(220/1000)=(11/50)km

Let the speed of second train be xkm/hr
therefore,relative speed=(50+x)km/hr(since it is opposite direction the speeds should be added)
time=6sec=(6/3600)=(1/600)hr
then solve using the formula time=(distance/speed)
i.e (11/50)/(50+x)=(1/600)
after solving x will be 82km/hr

Relative speed =220/6 * 18/5=132.
Second trains speed =132-50 = 80.

Thank you @Das.

Thank you @Soumitra Das.

(Total distance covered) ÷ (net speed) = time.
Mtrs/(mtr/sec) = sec.
220/(50+x) * 5/18=6,
So x=82.

But the unit is now mtr/sec because all we have converted into mtrs and seconds.

First take 220/6=36.66 m/s.

Now convert into km/h so 132km/h.
and minus 50km/h from 132 - 50 = 82km/h.
Thanks.