Aptitude - Problems on Trains - Discussion

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.

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.

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.

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.

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

@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.

Thanks all for explaining this.

Thanks for the explaining the answer @Ishita Narula.

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.

I didn't understand, please anyone explain me to get it.