A train takes $11$ seconds and $10$ seconds to cross $2$ men who are walking in the same direction of the train at the speed of $6\text{km/hr}$ and $5\text{km/hr}$ respectively. Find the speed of the train.

Let the speed of the train be $V\text{km/hr}$ and the length of the train be $L$ meters.
When the train crosses a person walking in the same direction, the relative speed is subtraction of their speeds.

For the first man:
Speed of the man $=6\text{km/hr}$
Relative speed $=(V-6)\text{km/hr}=(V-6)\times \frac{5}{18}\text{m/s}$
Time taken $=11\text{seconds}$
Distance covered (length of train) is:

$$L=\text{Relative Speed}\times \text{Time}=(V-6)\times \frac{5}{18}\times 11\text{— (Equation 1)}$$

For the second man:
Speed of the man $=5\text{km/hr}$
Relative speed $=(V-5)\text{km/hr}=(V-5)\times \frac{5}{18}\text{m/s}$
Time taken $=10\text{seconds}$
Distance covered (length of train) is:

$$L=(V-5)\times \frac{5}{18}\times 10\text{— (Equation 2)}$$

Since the length of the train ($L$) is constant, we equate Equation 1 and Equation 2:

$$(V-6)\times \frac{5}{18}\times 11=(V-5)\times \frac{5}{18}\times 10$$

Cancel $\frac{5}{18}$ from both sides:

$$(V-6)\times 11=(V-5)\times 10$$

$$11V-66=10V-50$$

$$11V-10V=66-50$$

$$V=16$$

Thus, the speed of the train is $16\text{km/hr}$.