Aptitude - Problems on Trains - Discussion

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.

Let the speed of the train is Xm/s & Ym/s resp.
accordingto the qus.
27x meter is the speed of the first train
17x " " " second "
finally both the the train meet =23(x+y)
27x+17y=23(x+y)
4x=6y
x/y=3/2

@Narayana superb.

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).

23x + 23y = 27x + 17y
23y - 17y = 27x - 23x
6y = 4x
x/y = 6/4 = 3/2
Thus,
1st train's speeds (x) / 2nd train's speed (y) = x/y = 3/2.

Given:
1. time taken by the first train to cross a man (t1) = 27 sec
2. time taken by the second train to cross a man (t2) = 17 sec

3. Time taken by the 1st & 2nd train to cross each other (T)= 23 sec

Find: Ratio
1st train's speeds (x) / 2nd train's speed (y) = x/y = ?

Formula:

speed =
distance(or length)
---------------------
time

Let,
Speed of 1st train to cross a man (x) = distance(d1) / time(t1)
Speed of 2nd train to cross a man (y) = distance(d2) / time(t2)

Thus,
x = d1/27 .................I
y = d2/17 .................II

So, Total speed of the train (S) = speed of 1st train (x) + speed of 2nd train (y)
Therefore, S = x + y

Now, we can say that
Total distance (D) = d1 + d2
D = 27x + 17y............from I & II

Total time (T) = 23 seconds

Total speed of the train (S) = Total distance (D) / Total time (T)

x + y =

27x + 17y
-----------
23

23x + 23y = 27x + 17y
23y - 17y = 27x - 23x
6y = 4x
x/y = 6/4 = 3/2
Thus,
1st train's speeds (x) / 2nd train's speed (y) = x/y = 3/2

Step (1)
Speed= (Distance/Time) meters/sec. Distance= Length here.
From this formula, Length = Speed* Time

So,Length of 1st Train= Speed * Time = 27* x where x=speed
Likewise,length of 2nd Train .= 17* y where y=speed

Step (2):
Note:Imagine one train is stationary. Then the distance traveled by each train would be equal to the length of the other train + its own length.Because a train can cross the other train after the last bogie of the 1st train crosses the last bogie of the other train.
So total length of both trains = 27x+17y.
Now the relative speed of the trains = x+y (since both run in opposite directions).

So Time = Length/speed= 27x+17y/x+y =23(given)
Cross multiply,transpose and simplify. Then you get the answer.

23-17/27-23=6:4 or 3:2 (LINK-UP METHOD)

Thanku Anjan :).

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.

a=27x, b=27y, u=x, v=y.

Lets speed of 1st train = x and of 2nd = y.
when they cross man then.....
so length of 1st = 27x and second 17y.
total length = 27x + 17y
now when they cross each other in opp direction they take 23 sec and we know when 2 object run in opp direction their rel speed is
speed of 1st + speed of 2nd = x+y so now length = time * speed = 23(x+y)

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