Aptitude - Problems on Trains - Discussion

Ishita Narula is correct

d1/d2=x1/x2;
d1=16x1;
d2=9x2;
9/16=(x1)2/(x2)2;
x1/x2=3/4;

Good job @ ISHITHA NARULA

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?

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

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

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;

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

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.

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.