Aptitude - Problems on Trains - Discussion

Hey if we follow the physics concept then relative distance should decrease while crossing each other. Then it should be (27x - 17y)/(x+y). Isn't it?

Lets assume,

The train 1 speed= x m/sec ==> distance traveled by the train 1= 27*x m.
Similarly assume the speed of the train 2 is y m/sec then the train 2 has traveled a distance of 17*y meter.

The total distance (x+y)*23 m.

Therefore,

23(x+y) = 27x+17y.
23x+23y=27x+17y.
==> 4x=6y.
==> x/y=3/2.

Therefore x:y= 3:2. Answer B.

@Arashan has describe very well with formula of speed.

SPEED = DISTANCE/TIME.

DISTANCE/SPEED = TIME.

i.e 27X+17Y/X+Y = 23.

I have solved it in yet simple(crude) way.

Don't worry about their crossing each other in 23 seconds or whatever.

On the face of it, one can make out that he has asked the simplified ration for 27:17.

Simply take them to next higher level so that both of them are divisible by common highest divisor.

i.e. 30/20 = 3/2.

(but recommended not to follow it, as it doesn't apply to every problem?).

Very short form is:

(27-23):(23-17),
4:6,
2:3.

So the ratio is 2:3 or 3:2.

Let speed of train1: x m/s.
Speed of train2: y m/s.

Then relative speed of train1 = (x+y) m/s............(1).

Train are crosses a man means they cross their length.

Length of train1: 27x m.
Length of train2: 17y m.

According to Qst:

Train crosses Each other in 23sec.
Means train1 crosses his length & also the length of train2.

=> Total length crosses by train1 = Length of train1+Length of train2.
=>Total length covered by train1(D)=27x+17y...........(2).

We knows,

Speed(s) = Distance(D)/Time(t)............Eqn(3).

Put Values OF eqn(1) &(2) in eqn(3).

x+y = (27x+27y)/23.
=> 23x+23y = 27x+17y.
=> 27x-23x = 23y-17y.
=> 4x = 6y.

=> x/y = 6/4 = 3/2..........Answer.

Velocity = distance/time.
Let velocities are x, y m/s.

x = d1/27; y = d2/17;
d1 = 27x; d2 = 17y;

Step 2: According to relative speed, since they are running in opposite direction their speeds are added.

x+y = d1+d2/23.

Sub d1 and d2; find answer.

There are 2 trains .
Here is some experiment ,one man is standing on platform.

Trian 1:
=========
we assume X is train 1 speed.
this train cross that man in 27X meters.

Trian 2:
===========
We assume Y is train 2 speed.
This train cross that man in 17Y meters.

We concentrate trains are moving in opposite direction.
So speed is X+Y.

Distance is (train_1_length+trian_2_length) = 27X+17Y.
Time = 23 seconds.

Formule : distance = speed*time.

27X+17Y = (X+Y)*23.

X/Y = 3/2.

Let the speed of the two trains bee x and y meters.
Then a/c to the formula,

Speed = distance/time.
Distance = speed*time.

Distance of one train becomes 17x.
Second become 27y.

a/c to formula distance + distance/speed = time.
17x+27y/x+y = 23.

Now by solving you can get the answer.