Aptitude - Problems on Trains - Discussion

Let the trains name as 1 and 2.
Let the time taken to meet be t.

Train1with speed S1 took 9hrs which took time "t" for train2 .(S1*9 = S2* t)
Similarly, train2 with speed S2 took 16hrs which took time "t" for train1(S1*t = S2*16).
Equating t from both equations we get the answer.

Thanks for the explaining the answer @Ishita Narula.

Thanks all for explaining this.

@All.

LOGICAL WAY:

if train1 reaches in 9hrs and train 2 reaches in 16hrs.
That means the speed of train 1 > speed of train 2.
Now cross check the answers. Only option B has a train1 speed greater than train2. i.e 4:3.

I think answer for this question will be 3:4.

Let's say.

Train A travels 'x' is the distance from Patna with speed 'V1' till the point of meet and
train B travels 'y' is the distance from Patna with speed 'V2' till the point of the meet.

the time taken for them to meet 't' = x/V1= y/V2 ---- (1)

Now Train A will travel 'y' distance that was previously travelled by train 'B' and vice-versa.

As per given info;
y/V1 = 9 and x/V2 = 16.

Taking ratio:
y/V1*V2/x = 9/16.

but from (1) -----> y/x = V1/V2,

=> ( V1/V2)^2 = 9/16.
=> V1:V2 = 3:4.

As per your answer 4:3.

Take A= 40 kmph,
and B= 30 kmph,
So now distance travelled by A = 40*9= 360km,
Distance by B = 30*16 = 480 km.

Is these two distance are same.

Two get the ratio of the speed of trains simply flip the ratio of the time(speed and time are inversely proportional to each other). i.e, 16:9.

Otherwise to get your answer 4:3 simply replace 9 with 12 in question.
16:12 ==== 4:3.

Here we need to find the ratio of the speed of the trains for which trains have travelled.

Distance D is same for both. So the ratio of speed,
=> S1:S2,
=> D/a:D/b,
=>1/a:1/b
=>b:a.

The initial answers were lengthy.

Now since both are departing at same time,
1 step d1/d2=v1*t/v2*t=v1/v2,
2nd step d1/d2=v2*16/v1*9,
And on equating both we get the required solution.

Howrah(A) ------------------------- Patna(B).

Let, distance between Howrah and Patna be x; S1 = speed of 1st train from A to B; S2 = speed of 2nd train from B to A.

Let, after t hours both trains meet each other.

So, distance covered by 1st train in t hours = S1 * t = distance covered by 2nd train after he meets the 1st train and goes towards station A.

So, S1 * t = S2 * 16 Or, S1/S2 = 16/t--> (I)

Again, distance covered by 2nd train in t hours = S2 * t = distance covered by 1st train after he meets the 2nd train and goes towards station B.

So, S2 * t = S1 * 9 Or, S1/S2 = t/9--> (II)
From equations (I) and (II), we get,
16/t = t/9
=> t^2 = 9*16 = 144
=>t = 12 hours ( as t is time hence -12 is neglected).

Hence, ratio of speeds,
S1/S2 = 12/9 = 4/3 (putting t=12 in equation(II)).
Or, S1 : S2 = 4 : 3.