Aptitude - Problems on Trains - Discussion

Discussion Forum : Problems on Trains - General Questions (Q.No. 31)
31.
Two, trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:
2 : 3
4 : 3
6 : 7
9 : 16
Answer: Option
Explanation:

Let us name the trains as A and B. Then,

(A's speed) : (B's speed) = b : a = 16 : 9 = 4 : 3.

Discussion:
109 comments Page 6 of 11.

Sk.vali said:   1 decade ago
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.

Rohit said:   1 decade ago
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.

Pratham said:   1 decade ago
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.

Lite said:   1 decade ago
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.

Harshit said:   1 decade ago
(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<-.

Chunky Kumar said:   10 years ago
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.

DEVVRAT KUMAR CHHOKER said:   10 years ago
@Ishita Narula is correct. See her explanation.

Rahul said:   9 years ago
Ishita Narula is absolutely right, no more debates on this.

Shubham said:   9 years ago
Let Total Distance = D.

If they take t time to cross each other then.

A * t + B * t = D.

Total time taken;

A is t + 9 hrs.
B is t + 6 hrs.

Thus.

A * (t + 9) = D.
B * (t + 16) = D.

A * t + B * t = A * (t + 9).
A * t +B * t = B * (t + 16).

Take t common and divide above two equations to get the answer.

Mohan said:   9 years ago
@Ishita Narula.

Thanks for your clear explanation.


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