Aptitude - Problems on Trains - Discussion

Ishita Narula is absolutely right, no more debates on this.

@Ishita Narula is correct. See her explanation.

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.

(x+y)/(s1+s2) = x/s1 = y/s2;

x/s2 = 9;
y/s1 = 16;

Hence s1/s2 = 3/4;

Where <----------x-------->|<--------y------->.

<----------------TOTAL----------------->.

->s1 s2<-.

Let distance between the cities = x.

Let speeds be s1 and s2.

The 2 trains will meet after time = x/(s1+s2)....(1) [using relative velocities].

Distance covered by T1 in the above time = s1*(1) = s1*x/(s1+s2)....(2).

Therefore dist remaining for T1 to destination = x-(2) = s2*x/(s1+s2).

Therefore time required to cover this dist by T1 = s2*x/s1*(s1+s2)....(3).

Similarly time required by T2 to cover its remaining distance = s1*x/s2*(s1+s2)....(4).

Now these times are given to be 9 and 16 respectively.

Dividing (3) and (4) eliminates x to give:

(s2/s1)^2 = 9/16.

Hence, s2/s1 = 3:4.

Hey guys sorry to interrupt.

Even I didn't understood why we used square root, its formula I can understand.

Its clear that S1=>9, S2=>16.

Then how could S1/S2=>3/9.

Its not possible 9/16 = 3/4.

Can any 1 explain it better? Really confused.

Read the question carefully, it is given that after they met at particular time, they take further 9 hrs and 16 hrs to reach THEIR destination. Try to solve that way, then you will get 4:3.

Speed, distance, time.

s1, d, t+9 (s1=d/t+9) ----1.

s2, d, t+16 (s2=d/t+16) ----2.

(s1+s2), d, t (s1+s2=d/t) ----3.

From equation 3.

s1+s2 = d/t.

=> (d/t+9) + (d/t+16) = d/t (from 1&2).

d get cancelled.

Finally by solving we get t = 12.

Now by substituting 't' value in equation 1&2.

We get s1 = d/21.

s2 = d/28.

Therefore ratio s1/s2 = 28/21.

= 4/3.

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.

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.