Aptitude - Problems on Trains - Discussion

Short cut:

T2 = 23-17 = 6.
T1 = 27-23 = 4.

S1/S2 = T2/T1.
= 6/4 = 3/2.

@James.

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

4x = 6y.
4x/2 = 6y/2.

2x = 3y.
x/y = 3/2.

Can someone please explain to me how 27x+17y = 23x + 23y & 4x + 6y?

Time x speed = distance.

Let the speeds of two train be x and y.

Then distances will be 27x for 1st train.

17y for 2nd train.

Now when they both cross each other fully. Time is 23 seconds.

Distance will be length of 1st train + length of 2nd train.

Speed will be as we assumed above is x and y (sum of both the speed).

So,

23 = (27x+17y)/x+y.

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

23y-17y = 27x-23x.

6y = 4x.
3y = 2x.

Hence the answer 3:2.

Let we have to assume those to get ratio of that speeds.

Can anyone please tel me how x+y is came.

They asked to find speed of that we have to use speed = length\time only know but how x+y is came.

This is very good method to understand.

This method is easiest method I can understand this sum.

@Vadivel.

There may be possible that the lengths are same but in question given that the crossing time is 23. So we will apply the formula of crossing time (a+b)/(u+v). And for length a & b we will use the man crossing conditions.

Please don't apply anything from your side only use the conditions which are given in question.

Another way be :

Train A and B meet each other at 23 sec. It is the mean/common time of both the trains. Taking it as the reference train A will take 27-23 = 4 sec to reach end point. Similarly train B will take mod|17-23| = 6 sec to reach other end.

Using Formula:

If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then:

(A's speed) : (B's speed) = (under root b : under root a).

A:B = 6:4 =3:2.

Thanks!