Answer is C
Let length of train be x and speed of the train be y
Case 1 – Person A → S = $2*\frac{5}{18}m/sec$ t = 9 sec
Case 2 – Person B → S = $4*\frac{5}{18}m/sec$ t = 10 sec
Dist in Case 1 → $\frac{x}{y-\frac{10}{18}}=9$ → x = 9y – 5 -----(1)
Dist in Case 2 → $\frac{x}{y-\frac{10}{9}}=10$ → x = 10y – $\frac{100}{9}$ -----(2)
On solving (1) and (2), we will get x = 50 and y = $\frac{55}{9}$