Two trains A and B are entering a railway platform from opposite direction. The length of the platform is 300 m. The speed of B is two times the speed of A. The time taken by B to cross the platform is one-third of the time taken by A to cross the platform. If the sum of the lengths of A and B is 500 m, what is the difference in their lengths?

Correct — A, 220 m. To cross a platform a train must travel its own length plus the platform's, so the time is (train length + 300) divided by its speed. Let A's speed be v; then B's is 2v. B's time is one third of A's, so (L_B + 300)/2v = (1/3) x (L_A + 300)/v. The v cancels and cross-multiplying gives 3(L_B + 300) = 2(L_A + 300), that is 2L_A − 3L_B = 300. The second condition is L_A + L_B = 500. Substituting L_A = 500 − L_B gives 1000 − 2L_B − 3L_B = 300, so 5L_B = 700 and L_B = 140 m, leaving L_A = 360 m. The difference is 360 − 140 = 220 m. Check it: A needs (360 + 300)/v = 660/v and B needs (140 + 300)/2v = 220/v, which is a third of A's time.

• (b)250 m — Would need lengths of 375 m and 125 m. Those give A a crossing time of 675/v and B one of 425/2v = 212.5/v, and 212.5 is not a third of 675.
• (c)200 m — Would mean 350 m and 150 m. B's time would be 450/2v = 225/v against A's 650/v, and a third of 650 is 216.7, not 225.
• (d)300 m — Would put the trains at 400 m and 100 m. B would take 400/2v = 200/v while a third of A's 700/v is 233.3/v, so the ratio fails.

Crossing problems are all one formula: the distance a train covers while crossing an object is its own length plus the object's length, and the time is that distance over the relevant speed. When the train passes a pole the object has no length; when it passes a platform, a bridge or another train, the lengths add.

The detail that the trains enter from opposite directions is scenery, not physics — each train crosses the platform on its own, so no relative speed is involved. The real work is that speed and time pull in opposite directions: B is twice as fast, which alone would make its time half of A's, and the stem asks for a third, so B must also be shorter. That intuition makes 140 m against 360 m sound right before any algebra. Two conditions, two unknowns, and the check at the end is worth the ten seconds it costs.

• Time to cross a platform = (length of train + length of platform) ÷ speed of train.
• Here (L_B + 300)/2v = (L_A + 300)/3v, which reduces to 2L_A − 3L_B = 300.
• With L_A + L_B = 500 the lengths are L_A = 360 m and L_B = 140 m.
• The difference in lengths is 220 m.
• Verification: A's time is 660/v and B's is 220/v, exactly one third.

• Treating 'from opposite direction' as a cue to use relative speed.
• Forgetting to add the platform length to the train length.
• Solving one condition and picking the option nearest the result without using the second.

A two-condition crossing item, where one clause in the stem is deliberate scenery.

• practice — not a real PYQ

  A train 200 m long crosses a platform 400 m long in 30 seconds. What is its speed?

  • (a)48 km/h
  • (b)60 km/h
  • (c)72 km/h
  • (d)80 km/h

  Answer(c) 72 km/h — it covers 600 m in 30 s, that is 20 m/s, and 20 x 18/5 = 72 km/h.

• practice — not a real PYQ

  A train crosses a pole in 12 seconds and a 180 m platform in 21 seconds. How long is the train?

  • (a)180 m
  • (b)200 m
  • (c)240 m
  • (d)300 m

  Answer(c) 240 m — the extra 9 seconds cover the 180 m platform, so the speed is 20 m/s and the length is 12 x 20.