Aptitude - Problems on Trains - Discussion

According to me :

1. Let speed of train from Howrah to Patna train be x kmph
2. Let speed of train from patna to howrah be y kmph.
3. Now since they take 9 hours and 16 hours respectively to cover remaining distance that means the y kmph train covers 9x distance before meeting at a common point and x kmph train covers 16y distance before meeting at the common point.
4. Now since speed is directly proportional to distance.
x/y=16y/9x
5. 9x^2 = 16y^2
6 Therefore, x/y=4/3

The above explanations are not correct. For example,

2 trains at station A&B. It starts at the same time at 8 AM in the morning. Train A speed is 100km/hr&train B speed is 150km/hr. Then they will meet after 4 hrs. If the total distance is 1000km.

After meeting, B takes 9hrs (TOTAL 13hrs) to complete its journey and A takes 16hrs (TOTAL 20HRS) to complete its journey. So the speeds of both trains decreasing to 50 km/hr &76.92km/hr. So equations 2&3 are not correct.

Let speed of train from Howrah to Patna = u.
And speed of train from Patna to Howrah= v.

Now let they meet at a distance x from Howrah and y from Patna after time t.

So t= (x/u ) = (y/ v)
=> x/y = u/v ----------------------------(1)

Now we are given that y/u = 9 --------(A)
and x/v =16 ----------------(B)

Divide B by A. We will get, (x/y) * (u/v ) = 16/9.

From equation (1), replace x/y by u/v,
= > (u/v)2 = 16/9
= > u/v = 4/3.

Suppose the total distance is x,

Time taken by trains to meet = t

if speeds be s1 and s2, and they meet after time t,
=> s1*t + s2*t =x (eqn 1)

Train A takes total t+9 hours to travel x with speed s1,

=> s1(t+9)=x (eqn 2)

for train B, total time t+16

=> s2(t+16)=x (eqn 3)


9*s1= t*s2 (equate eqn 1 and 2)

t*s1= 16*s2 (equate eqn 1 and 3)

divide,

9/t= t/16 => t*t= 9*16

substitute t,

=> s1/s2= sqre-root (16/9)

Suppose after t time they will meet.
Let the speed of the train A is a and speed of train B is b.

after time t total distance covered by
train A is a*t and train B is b*t

Now for train A, b*t distance is traveled by 9 hrs with speed a.
So, b*t/a = 9 -----(1)
Similarly for train B, a*t distance is covered by 16 hrs with speed b.
So, a*t/b = 16 -----(2)

Now (2) / (1):
a^2/b^2 = 16/9
or, a:b = 4:3

The Best One:

Suppose after t time they will meet.
Let the speed of the train A is a and speed of train B is b.

After time t total distance covered by,
Train A is a*t and train B is b*t.

Now for train A, b*t distance is traveled by 9 hrs with speed a.
So, b*t/a = 9 -----(1).

Similarly for train B, a*t distance is covered by 16 hrs with speed b.
So, a*t/b = 16 -----(2).

Now (2) / (1):
a^2/b^2 = 16/9.
or, a:b = 4:3.

Friends best answer is here.

Let train from Howrah to Patna (thp) speed is a kmph.

Let train from Patna to Howrah (tph) speed is b kmph.

Total distance from h to p = 9a+16b.

Train starts simultaneously, hence when both train meets then both spend same time while travel some distance.

Thp<------------16b----><-------9a----->tph.

Time through thp = Time through tph.

16b/a = 9a/b.

Hence, a/b = 4/3.

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 the speed of first train be x and that of second one be y.

After they meet first train travels distance=9x and second train travels distance=16y, Because distance=speed *time.

Now before they meet second train covers the distance 9x with velocity y. And first train covers the distance 16y with velocity x.

Since they start simultaneously so 9x/y = 16y/x.

Therefore x^2/y^2 = 16/9 or x/y = 4/3.

Let us take both trains to meet after time t.

And the speed of train A and B as x and y.

So now whatever distance covered by train A in time t is covered by train B in 16 hours.
So x*t = y*16 --> eq(1)

Similarly whatever distance covered by train B in time t is covered by train A in 9 hours.
so y*t = x*4 --> eq(2).

From eq(1) and eq(2),
We get (x^2 / y^2) =( 16/4).