Aptitude - Problems on Trains - Discussion

Short method is;

x person 27.
y person 17 add both distances and divide the total distance and multiple the same for both persons.

27+17/23 *2.
37/23 = 1.6 *2.
1.6*2.
Ans : 3.2.

@Poornima.

Thank you for the explanation.

Can the ratio be 2:3? Anyone explain me, please.

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.

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.

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?

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

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.

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.

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

Anyone, please explain me.