Aptitude - Problems on Trains - Discussion

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.

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.

Let the trains be named as A & B.

And the total distance b/w Howrah to Pune be d.
WHEN BOTH THE TRAINS MEET,

Distance covered by train A, d1 = s1*t.........1.
Distance covered by train B, d2 = s2*t.........2.

Divide equation 1 by 2,
d1/d2 = s1/s2.........3.

WHEN BOTH TRAINS HAVE CROSSED AND ARE MOVING TO THEIR RESPECTIVE STATIONS,

Distance covered by train A, d2=s1*9..........4.
Distance covered by train B, d1=s2*16..........5.

Divide equation 5 by 4.
d1/d2=s2*16/s1*9.........6.

Solving equation 3 & 6.
We get,

s1/s2 = s2*16/s1*9.
s1*s1/s2*s2 = 16*9.

s1/s2 = 4*3. Answer.

Howrah A--> B<--Patna

|--Va*t----->|<---Vb*t--|

|<--Tb = 16hr--|--Ta=9hr->|

Va = (Vb*t)/9 --- Eq1

Vb = (Va*t)/16 --- Eq2

Eq1/Eq2

(Va/vb) = (Vb/Va)*(16/9)

(Va:Vb) = sqrt(16/9) = 4:3.

Let the total distance b/w two station = x.

Let after t time they will meet and their speed is a and b respectively, then
distance travel by trains,

at +bt = x ... (eq1).

After meeting they reach their destinations after 9hr and 16 hrs,

So, a(t+9) = x ... (eq2).
And b(t+16) = x ... (eq3).

Now, by equating the eq1 and eq2.
at+bt = a(t+9).

a/b = t/9... (eq4).

By, eq1 and eq3.
a/b = 16/t ... (eq5).


by, eq4 and eq5.

t*t = 9*16.
t = 3*4.
Put the value of t in eq5.

a/b = 4/3 answer.

Just think like they all traveled equal time before meeting.
Take that time as T so.

A traveled to the point of meeting in T time while the same was traveled by B in 9 hours after meeting. similar situation for B.

So Considering the ratio of their speed remains constant.
T/16=9/T, This will give value of T as 12.

Now calculate time by each and thus Ratio.

@Abhishek.

Dude it is mentioned after they meet. At this point both trains are travelling with different speeds so they did not meet at mid point.

Hope you are clear.

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.

@Bhargav. Buddy see here we have to calculate the ratio and the ratio of two quantities will be unit less, so here we don't need any units. I hope you will get this explanation.

Two things strikes out here:

1. Just memorize the formula and save your time.

2. By logic, the answer has to be 4:3 coz this is the only option with speed of train A more than the train B, and from the question, we know train A is faster than train B.