Aptitude - Problems on Trains - Discussion

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.

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.

Still not getting Why square root is used ?

In this question we need to calculate the ratio only after the trains have met each other. For overall distance the ration will be a:b. For distance after each trains met will be in square roots.

Let the time taken before they meet be 't' hours.
After meeting 1st train takes 9 hours and 2nd takes 16 hours.

dist= speed*time, but here speed is relative before they meet,

Therefore, d=(s1+s2)t -----(1).
d=s1.t+s2.t.

After they meet,

d= s1(t+9) ----------(2) for train-1.
d= s1.t+s1.9.
d= s2(t+16)----------(3) for train-2.
d= s2.t+s2.16.

Equate (1) and (2),
s2.t = s1.9.
s1/s2= t/9 ---------------(4).

Equate (1) and (3),
s1.t = s2.16.
s1/s2=16/t----------------(5).

Divide (4) by (5),
1=(t/9)/(16/t).
16/t = t/9.
t*t=16/9.
Taking square root on both side,
t=4/3.

Therefore, speeds are in ratio 4:3.