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:
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.
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.
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.
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<-.
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.
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.
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.
Thanks for your clear explanation.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers