Aptitude - Problems on Trains - Discussion

It's (220/6*18/5) - 50 = 82km/h.

Can we convert 50 km/hr into m/s and add the speed of other trains as they are in opposite directions?

Can anyone explain me?

Here is the simplest form that you can understand.
Since the two train travel in opp directions, add both speeds. the speed of one train is given, let the speed of another train be x.
Speed=x+50 km/hr.
The relative distance = 108 + 112 = 220 m.
Time = 6 seconds.

Speed = distance/time.
x+50 = 220/6 m/s
x+50 = 110/3 *18/5(to convert m/s to km/hr ,since the answer required in km/hr)
x+50 = 132,
x = 82 km/hr.

@Ambika.

Good explanation. Thanks.

First convert km/hr in m/secs,
x= 410/18 m/secs.
for km/hr.

41O/18 * 18/5 = 82km/hr.

6 = 220 then,
36 = (x+50)*10.
After cross multiple;
6*22 = X + 50.
X = 82.

@All.

First, we are assumed as X in km/hr, but we are applying the formula convert into M/S it is applicable for RELATIVE SPEED=(X+50).

It is not applicable for Assuming X, so the final answer is in KM/H.

The answer 82 is in m/s. We want to convert it into km/hr, because in first step we convert speed into m/s by multiplying 5/18.

First train speed 50sec.

Second train speed?
S = l1 + l2/t1 + t2.
Relative speed = x + 50s.
X + 50s = 108+112/6+t2.
X + 50s =792/6.
6x + 300=792.
6x = 792-300.
6x = 492.
X = 492/6.
X = 82.

Given :
(length of 1st train) L1 = 108 m.
(length of 2nd train) L2 = 112 m.
(speed of 1st train) S1 = 50 KMPH = 50* 5/18 = 250/18.

(time taken to cross each other)T = 6 s.

Solution :
T = (L1 + L2 ) / (S1 + S2).

6 = 220 / ( 250/18 + S2).
(taking the right side denominator to left).

6( 250/18 + S2) = 220.
(multiplying 6 inside).

250/3 + 6 S2 = 220.
6 S2 = 220 - 250/3.
6 S2 = (660-250)/3
S2 = 410/18
( to convert this to KMPH - multiply by 18/5 )

S2 = 410/18 * 18/5
S2 = 82 KMPH