Two cars travel from city A to city B at a speed of $42\text{km/hr}$ and $60\text{km/hr}$ respectively. If one car takes 2 hours lesser time than the other car for the journey, then the distance between City A and City B is:

Let the distance between City A and City B be $d\text{km}$.The time taken by the slower car ($42\text{km/hr}$) is $\frac{d}{42}$ hours.The time taken by the faster car ($60\text{km/hr}$) is $\frac{d}{60}$ hours.According to the question, the difference between their travel times is 2 hours:
$$\frac{d}{42}-\frac{d}{60}=2$$
Find the LCM of 42 and 60, which is 420:
$$\frac{10d-7d}{420}=2$$
$$\frac{3d}{420}=2$$
$$\frac{d}{140}=2$$
$$d=2\times 140=280\text{km}$$
Therefore, the distance between City A and City B is $280\text{km}$.