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.
How do you change the sign of x and y?

Anyone please explain to me.

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.

u=27s
v=17s

Cross = 23s.

l1 = 27x.
l2 = 17y.

23 = l1 + l2/x + y
23 = 27x + 17y/x + y.
23x + 23y = 27x + 17y.
4x - 6y = 0.
4x = 6y.
x/y = 6/4.
x/y = 3/2.

Hence, option (b).

Well, I have something different, 1st train is 27 sec so take this time as a total as 100% minus the 2nd train which is 17 sec so 62. 96 % is the percent of 1st train slower than the second train or vice versa.

So, 62.

96 is a 3 ratio to 100 so the remaining 38.04 is 2.
So, it's 3:2 am I right?

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.

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.

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.

But why we can't use the formula of B:A?

Anyone, please explain me.

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.