Two cars run to a place at the speeds of $45\text{km/h}$ and $55\text{km/h}$, respectively. If the second car takes $40$ minutes less than the first for the journey, then what is the length of the journey (in $\text{km}$)?

Let the total distance of the journey be $D\text{km}$.
Time taken by the first car:

$$T_1=\frac{D}{45}\text{hours}$$

Time taken by the second car:

$$T_2=\frac{D}{55}\text{hours}$$

The difference between the time taken by both cars is $40$ minutes. Converting minutes into hours:

$$40\text{minutes}=\frac{40}{60}\text{hours}=\frac{2}{3}\text{hours}$$

According to the question:

$$T_1-T_2=\frac{2}{3}$$

$$\frac{D}{45}-\frac{D}{55}=\frac{2}{3}$$

Taking the LCM of $45$ and $55$, which is $495$:

$$\frac{11D-9D}{495}=\frac{2}{3}$$

$$\frac{2D}{495}=\frac{2}{3}$$

$$2D=\frac{2\times 495}{3}$$

$$2D=2\times 165$$

$$D=165\text{km}$$

Therefore, the length of the journey is $165\text{km}$.