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Suppose the trains meet at point "A" after time "t".
Let the trains be Train1 and Train2. Suppose Train1 travels a distance "x" in time "t" and Train2 travels a distance "y" in the same time "t". Since the two trains are travelling in opposite directions, the total distance each must travel is x + y.
Train1 has traveled "x" distance, so it has to travel "y" distance to complete the journey. It does this in 1 hour. If its speed is S1, then S1 = x/t. Also S1 = y/1 (because it travels "y" distance in 1 hour). So, x/t = y. So, t = x/y.
Train2 has traveled "y" distance, so it has to travel "x" distance to complete its journey. It travels this distance in 4 hours. If its speed is S2, then S2 = y/t, and, also, S2 = x/4. So, y/t = x/4. So, t = 4y/x.
So, x/y = 4y/x. So, x2 = 4y2.
So, x = 2y. Now, "x" and "y" are distances, so they have to be positive and can't be negative.
Consider the speeds of the trains: S1 = y and S2 = x/4. So, S1 = y and S2 = 2y/4; thus, S2 = y/2. Therefore, S2 = S1/2. This means the speed of Train2 is half the speed of Train1. I hope I got it right.
Some useful links for
- Union Public Service Commission - www.upsc.gov.in
- IIT-Kharagpur - www.iitkgp.ac.in
- Indian Statistical Institute - www.isical.ac.in
- Indian Institute of Technology Madras - www.iitm.ac.in
- Indian Institute of Management, Ahmedabad - www.iimahd.ernet.in
- Indian Institute of Mass Commission - www.iimc.nic.in
- IIT Bombay - www.iitb.ac.in
- Indian School of Mines, Dhanbad - www.ismdhanbad.ac.in
- Birla Institute of Technology, Ranchi - www.bitmesra.ac.in
- Central Institute of Fisheries Nautical and Engineering Training - www.cifnet.nic.in
- Indian Institute of Information Technology, Allahabad (Deemed University) - www.iiita.ac.in
- Central Marine Fisheries Research Institute, Kochi - www.cmfri.com
- Tata Institute of Social Sciences, Mumbai - www.tiss.edu