Aptitude - Problems on Trains - Discussion

Let 1st train be x = 27.
2nd train be y = 17.

Two trains crossed each other=23(x+y).
27x + 17y = 23x + 23y.
27x - 23x = 23y - 17y.
4x = 6y.
x = 6y/4.
x/y = 3/2.
= 3:2.

Let the speed of the two trains be x m/sec and y m/sec.
Then, the length of the first train = 27x metres,
And length of the second train = 17y metres,

Time taken by the train to cross each other is 23 sec.
Time taken by the train to cross each other = Total Distance of both trains/ Total of Both trains.
So, 23 = (27x + 17y) / (x + y).
23x + 23y = 27x + 17y.
6y = 4x.
x/y = 6/4 => 3:2.

1) Train A speed = x
2) Train B speed = y.

Train A time = 27
Train B time = 17.

Time to cross each other trains T= 23.

So,
Train Length of A= 27x
Train Length of B= 17y.

27x + 17y/x+y = 23.
27x + 17y = 23(x+y).
27x + 17y = 23x+23y.
27x - 23x = 23y-17y.
4x = 6y,
x/y = 6/4,
So, x/y = 3/2.

So, the final answer = 3:2.

@All.

If a train T1 takes 27 sec to cross the man which is more than the time taken by train T2 (17 sec) then how can the speed of Train T1 is greater than the speed of train T2?

option 2 is only correct when the length of train T1 is much greater than the length of train T2.

It's simple, speed is inversely proportional to time.
so 27-23 :23-17 {alligation rule}.
=6:4 (time ratio).
So the speed ratio will be 4:6 =2:3.

Here is simple solution for this question.

When we take it as mixture;
23 - 17 = 6
27 - 23 = 4
6/4 i.e 3:2.

Let's assume the distance is constant.

1st train has 27 sec time and X speed
2nd train has 17 sec time and Y speed.

Know that, two trains cross a platform in 23 sec.
The relative seed in opposite directions is (s1+s2).
s1--------x.
s2----------y.

Distance = speed * time.
total distance = Distance 1 + distance 2.
D = d1 + d2
(x+y) 23 = [x * 27] + [y * 17]
23x + 23y = x27 + y17
x/y = 3/2.

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.

Time taken by train 1 = 27s.
Time taken by train 2 = 17s.
Time taken for them to cross each other = 23s

Let the velocity of train 1 be v1 and that of train 2 is v2, and their lengths be l1 and l2 resp.
Net velocity when they cross each other v = v1 + v2.
Thus;
(l1+l2) = (v1+v2)x23---> i
and
l1 = v1 * 27 and l2 = v2 * 17---> ii

By Replacing ii in i.
27v1 + 17v2 = 23(v1+v2)
4v1 = 6v2
v1:v2 = 3:2.

Let the speed of the two trains be x m/sec. and y m/sec.
Then, the length of the first train = 27x metres,
and length of the second train = 17y metres,

Time taken by the train to cross each other is 23 sec.
Time taken by the train to cross each other = Total Distance of both trains/ Total of Both trains.
so, 23 = (27x + 17y) / (x + y).
23x + 23y = 27x + 17y.
6y = 4x.
x/y = 6/4 => 3:2.