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?

To solve this, let the speed of train A be $v$ and train B be $2v$. Let the length of train A be $L_A$ and train B be $L_B$. The total distance covered to cross a platform is the sum of the train's length and the platform's length. For train A: $T_A = (L_A + 300) / v$. For train B: $T_B = (L_B + 300) / 2v$. Given $T_B = T_A / 3$, we substitute: $(L_B + 300) / 2v = (L_A + 300) / 3v$. Simplifying gives $3(L_B + 300) = 2(L_A + 300)$, which leads to $3L_B + 900 = 2L_A + 600$, or $2L_A - 3L_B = 300$. We are also given $L_A + L_B = 500$. Solving these simultaneous equations: $L_A = 500 - L_B$. Substituting into the first equation: $2(500 - L_B) - 3L_B = 300 \Rightarrow 1000 - 5L_B = 300 \Rightarrow 5L_B = 700 \Rightarrow L_B = 140$ m. Thus, $L_A = 360$ m. The difference is $360 - 140 = 220$ m. Wait, re-calculating: $360 - 140 = 220$ m. Option 1 is 220m. Re-checking the math: $2(360) - 3(140) = 720 - 420 = 300$. Correct. The difference is 220m.