Aptitude - Problems on Trains - Discussion

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.

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