Aptitude - Problems on Trains - Discussion

Given,

Speed of first train = 50km/hr (after convert it into m/s it becomes 125/9m/s).

Speed of second train = ?

Length of first train = 108m.

Length of second train = 112m.

Now, lets take the speed of second train be x m/s.

So by speed formula i.e.

Speed = distance/time keeping the values in this formula we'll get,

125/9+x = 108+112/6 (here we are adding these value b'coz they are moving in opposite direction).

By taking lcm of 9 on L.H.S it becomes,

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

By cross multiplication.

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

750+54x = 1980.

54x = 1980-750.

54x = 1230.

x = 1230/54.

x = 205/9 m/s.

But now we've to convert it into km/hr so.

(205/9*18/5) km/hr.

After cancellation it become.

Answer: 82 km/hr.

Let me explain in detail.

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). Then no need to confuse yourself by converting the speed to m/s.

Take it as it is, so as to get our answer in km/hr (as the given options are in km/hr). Now just convert the distances 108m & 112m into km/hr.

So add the distances & then convert it.

D = 108+112 = 220m = (220 x 18/5) = 792km/hr.

So now distance is in km/hr.

T = D/S.
6 = 792/(x+50).
6x+300 = 792.

Therefore x = 82km/hr.

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.

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

Distance = (220/1000) km {Want to convert meter into kilometer).

Time = (6/3600) seconds {Second into hours - one hour has 3600 seconds}.

Distance/Time = Speed.

= (220/1000)/(6/3600) = x+50 km {Opposite speed should be added).

Now without converting, follow normally:

220/6*3600 = (50+x) speed (If goes 220 m within 6 seconds, what is the distance per hour or 3600 seconds).

132000 m per hour = (50+x).

So, 132 kilometer = (50+x).

x = 132-50 = 82.

Simple solution method:.

Length of train 1st is = 108m.

Speed of train 1st is = 50km/h => 50 * 5/18 == 125/9 m/sec.

Length of trian - 2nd is = 112m.

Now, let the speed of 2nd tarin is = x m/sec.

Relative speed =x + (125/9).

Relative distance = 108 + 112 = 220 m.

Time = 6 sec.

Now, s = d/t.

=> x + 125/9 = 220/6 (therefore 220/6= 110/3).
=> x = (110/3) - (125/9).
=> x = 205/9 m/sec (205/9) * (18/5).

x = 82 km/h answer.

The answer is 82 km/h is correct. They have shown calculation for relative speed in wrong way.

Speed for 1st train is 50 km/h after conversion it will be 250/18 m/s.
And suppose speed for the second train in x m/s.
Relative speed will be (125 +9x)/9 m/s.

Then,
(125+9x)/9 = 108+112/6.
750+54x = 1980.
54x = 1230.
x = 22.78 m/s.

After conversion to km/h.
22.78x18/5= 82 km/h. Approx.

I believe this how they should have done.

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.

Sorry I make a mistake in my last comment.

D = 108 + 112 = 120.
Converting the speed of 1st Train = 50x 5/18 m/s.
Let the speed of 2nd Train= V m/s.

We get,

220/ (50x 5/18) + V = 6 sec.

=> 220= 6x(250+18V)/18,
=> 220=(250+18V)/3,
=>660= 250+18V,
=>18V=410,
=> V= 410/18 m/s.

Converting the speed of the second Train we Get,
=> (410/18) x (18/5).
=> 82 Kmph.

This is the right answer

Let me explain in details.

First train length, a = 108m.
Its speed, u = 50 km/hr = 50*5/18 m/sec= 125/9 m/sec.

Second train length,b = 112m.
Let, its speed, v = x m/sec.
Each other crosses in 6 sec.

So, we have, t= ((a+b)/(u+v)).
6= ((108+112)/((125/9)+x)).

(125/9)+x = 220/6.
x = (220/6)-(125/9)= 205/9 m/sec.

Converting in km/hr.

(205/9)*(18/5) = 82 km/hr.