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 crossing a person walking in the same direction, relative speed is subtracted.

Case 1: Crossing the first man (speed $6\text{km/hr}$) in $11$ seconds.

$$\text{Relative speed}=(v-6)\text{km/hr}=(v-6)\times \frac{5}{18}\text{m/s}$$

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

Case 2: Crossing the second man (speed $5\text{km/hr}$) in $10$ seconds.

$$\text{Relative speed}=(v-5)\text{km/hr}=(v-5)\times \frac{5}{18}\text{m/s}$$

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

Equating the length $L$ from Equation 1 and Equation 2:

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

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

$$11v-66=10v-50$$

$$11v-10v=66-50$$

$$v=16\text{km/hr}$$