Aptitude - Problems on Trains - Discussion

Lets assume that:

1. They meet at time t.

2. total distance between trains before they start is d.

3. when they meet, one of the train having speed "v1" has traveled "d1" distance. Hence other train having speed "v2" must have "d-d1" distance traveled.

Explanation:

Using the formula , distance=speed * time.

d1=v1*t ......(1) for train 1.

d-d1=v2*t ......(2) for train 2.

By using (1)and (2).
v1/v2=d1/(d-d1) .......(3).

Now , second train takes t+16 hours to cover "d" distance and other train takes t+9 hours for same distance.

So,

d=v1*(t+9) ...........(4).

d=v2*(t+16) ...........(5).

For result (4),

d=t*v1+9*v1.
Replacing t*v1 by result (1).

Then we get
d=d1+9v1 ................(6).

For result (5),
d=t*v2+16*v2.

Replacing t*v2 by result (2).
Then we get,
d=d-d1+16*v2 ................(7).

From (6) and (7) we can write,
9v1/16v2 = (d-d1)/(d1) ...............(8).

From (8) and (3) we write that,
9v1/16v2=v2/v1.

Hence:
v1*v1/v2*v2=16/9.
v1/v2=sqrt(16/9).
v1/v2=4/3.

Distance say x.
So times 9 and 16.

Speed ratio,

x/9:x/16.

16:9.

Answer is 16:9 but there is no answer in option.

So square root booth side,

Then 4:3.

The distance cover by both train = x.

The speed of both trains are s1 and s2.

Then,

s1=x/9 --- (1).
s2=x/16 --- (2).

Then,

s1/s2= x/9/x/16.
s1/s2= 16/9.
s1:s2= 16:9.

Two trains start simultaneously,one traveling at vh the other at VP.

These trains must travel a distance of length l.

They meet at a point x from Howrah that is l-x from Patna. Since they both started at the same time it means they took equal amount of time t to cover the distances.

x = vh*t.
l-x = vp*t.

x/(l-x) = vh/vp.

Now it takes the Patna bound train 9 hours to cover l-x.
It takes the Howrah bound train 16 hours to cover x.

x = 16vp.
l-x = 9vh.

16vp/9vh = vh/vp.
vh2/vp2 = 16/9.

Perform the square root.

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

Suppose total distance is x, and trains are T1 and T2;
Suppose speeds are S1 and S2;
T1 and T2 meet at time t;
.'. total distance is S1.t+S2.t=x;--->(eqn 1)

But according to question T1 and T2 covers the distance 'x' at (t+9) and (t+16) hr respectively.
.'. for T1, S1.(t+9)=x;--->(eqn 2)
And for T2, S2.(t+16)=x;--->(eqn 3)

Now from eqn1 and eqn2;
S2.t=S1.9;
=>S1/S2=t/9;--->(eqn 4)
And from eqn1 and eqn3;
S1.t=S2.16;
=>S1/S2=16/t;--->(eqn 5)

Now multiply eqn4 by S1/S2 from both sides we get
(S1/S2)^2=(t/9).(S1/S2);
=>(S1/S2)^2=(t/9).(16/t); [using eqn5]
=>S1/S2=sq root of (16/9);
=>S1/S2=4/3;

How one decide that we have to take 9:16 or 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

I understand what you did there abhishek. There is just one Problem. The distance "X" which you are taking for both trains, how do you know it is the same. i.e. They are both travelling at different speeds, so they definitely won't meet at a mid point so there cannot be a single X for both train's distances, right?

Good job @ ISHITHA NARULA