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.

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.

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

Anyone, please explain me.

How to solve the following step?

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

Could anyone tell me please?

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).

I have a small doubt regarding the formula.

In the question, they have said to find the ratio of speed. Not the ratio of time.

So aren't we supposed to modify the formula accordingly? Please someone explain.

Given data:
t1=27 sec t2=17 sec
S=d/t,d=s * t.

d is consider as l,

l1=s1 * t1 =17 s1,l2 = s2* t2 =27 s2.

In question ,it clearly move in a opposite direction,so we have to use this formula->(l1+l2) /s1+s2.

Total time = (17s1+27s2)/s1 + s2,
23(s1+s2)= 17s1+27s2.

Rearrange it,
23s1-17s1=27s2-23s2,
6s1 = 4s2,
s1/s2 = 6/4=3/2.

So,the answer is 3:2

Why is the length of the platform not taken into consideration? Anyone, please explain to me.

Let me explain in detail if anyone still not understood yet.

Please remember the formula for finding speed. Speed = Distance/Time.
Therefore, Distance = Speed x Time.

If a train crosses a man/pole/post/tree in T seconds, and speed of the train is S m/s, then

Then, the Length of the Train = (Speed x Time) = ST metres.

Note: Here, the distance traveled by train in T seconds at the Speed of S m/s is equal to the length of the train.

Now, lets come to the given problem.

Let speed of the first train = X.
Time taken taken by the first train to cross a man = 27 seconds.
Therefore, Length of the first train = Speed x Time = X x 27 = 27X metres.

Let speed of the second train = Y.
Time taken taken by the second train to cross a man = 17 seconds.
Therefore, Length of the second train = 17Y metres.

Important Formula:
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.

Given that, (a + b) / (u + v) = 23 seconds.

Here, a = 27X, b = 17Y and u = X, v = Y.

Therefore, (27X + 17Y)/(X + Y) = 23.

=> 27X + 17Y = 23X + 23Y

=> 4X = 6Y

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

=> X : Y = 3 : 2.

Isn't it like X/Y = 4/6?

How come in the above solution like X/Y 6/4=3:2.
X=4 and Y=6 isn't it?