Aptitude - Problems on Trains - Discussion

Its very easy (no need any formula).

Train A speed = X m/s.
Train B speed = Y m/s.

For train A (Unitary method/Tri rashi method).

1sec------Xm (because train A speed =X m/s)
27sec----?.

So,length of train A = 27X ----------------> P.

Same For train B.
So,length of train B = 17Y -----------------> Q.

Train running in opposite so,
Relative speed = (X+Y) m/s ---------->R.

*Train cross each other in 23 sec.
So total distance traavel in 23 sec = 27X+17Y.
(From equations P and Q)

1sec ------------------> (X+Y)m (From eq R).
23sec ----------------(27X+17Y)m.

So, 23*(X+Y) = 27X+17Y.
And we get the answer.

Hai every body,
The speed of one train = x, Speed of another train = y.
We know that speed = length/time || length = speed*time.

So we find the length of the first train = x(speed)*27(time).

The length of the second train = y(speed)*17(time).

Time taken by both train to cross(total time) = 23.

Total length = (length of the first train)+(length of the second train) = (x*27)+(y*17).

Total speed = (speed of first train) + (speed of the second train) = x+y.
Total time = total length/total speed.

23 = [(x*27)+(y*17)]/(x+y).
27x+17y = 23(x+y).
27x+17y = 23x+23y ||27x-23x = 23y-17y || 4x = 6y ||x/y = 3/2.

Hello there if anyone did't understood the concept behind it let me explain in my way.
So, everyone know when two bodies are moving towards each other or having an velocity in opposite direction the velocity is add up (e.g you are going on car at 20 m/s and from front the other car is coming at 10 m/s so you will cross each other at 30 m/s) V1+V2.

Now same concept applies here,

S1+S2 = V1T1+V2T2 (S=VT).
S1+S2 / V1+V2 = T1+T2.
AND T1+T2 is 23.
So S1+S2/V1+V2 = 23.

NOW for S1= 27V1 and S2 = 17V2.
Put values in eq and you will get the ratio,
27V1+17V2/V1+V2 = 23.
V1/V2 = 3/2.

Thanks.

Let the speeds of the two trains be x m/sec and y m/sec respectively.

Then, length of the first train = 27x metres,(length=speed*time..so length=x*27=27x metres)

and length of the second train = 17y metres.(length=speed*time..so length=y*17=17y metres)

If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then:

The time taken by the trains to cross each other = (a + b)/u + v)sec.
By using above formula we get d solution for this.
Here it is given that they cross each other in 23 seconds.
So 23 = (27x+17y) / (x+y).

Let the speeds of the two trains be x m/sec and y m/sec respectively.

Then, length of the first train = 27x metres,(length=speed*time..so length=x*27=27x metres)

and length of the second train = 17y metres.(length=speed*time..so length=y*17=17y metres)

If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then:

Relative speed of the trains =x+y(since they are moving in opp direction)

Total distance covered = L1+L2
So time taken to cross each other = total distance /relative speed

Hence 23=(27x+17y)/(x+y)

Given data:

l1=27 m/s.
l2=17m/s.

Now, we can find speed
s=l/t .

But speed is not given consider speeds x and y.

Now easy to find lengths,
l1=t*s =27x
l2=t*s =17y

If two trains travelling in opposite directions then we add two lengths(l1 and l2) and x and y speed m/s so that we able to find ratios.

Now the time taken by train to cross each other is given i.e 23 seconds.

Now formula forms like this :

( l1 + l2)
---------- = 23.
( x + y )

( 27x + 17 y)=( 23x + 23y ).

(27x - 23x ) = ( 23y - 17y ).

(4x)=(6y).

x 6 3
-=-=-
y 4 2

Another way be :

Train A and B meet each other at 23 sec. It is the mean/common time of both the trains. Taking it as the reference train A will take 27-23 = 4 sec to reach end point. Similarly train B will take mod|17-23| = 6 sec to reach other end.

Using Formula:

If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then:

(A's speed) : (B's speed) = (under root b : under root a).

A:B = 6:4 =3:2.

Thanks!

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.

Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:.

Its given as Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds.

So it should be taken as 27x and 17x seconds only, I don't know why it is given as 27x and 17y meters and also denoted as length of train. Please any one explain me?

Speed=distance*time or length*time
Speed of the first train=X*27
Let X=distance of first train
Let Y=distance of second train
Speed of the second train=y*17
If train moves in opposite direction means total speed is addition of two train speed
27X+17Y

Both trains meet in 23 second by crossing a man
1'st train speed=23X
2nd train speed=23Y
Total speed=23X+23Y

So first train speed =(27-23)*X
2nd=(23-17)*Y

Equating both speed 6X=4Y

X/Y=6/4=3/2.