A train travels at a certain average speed for a distance of 63
km and then travels a distance of 72 km at an average speed of
6 km/h more than its original speed. If it takes 3 hours to complete
total journey, what is the original speed?

The correct option is 42 km/h.

Let's solve the problem step-by-step to understand why this is the correct original speed of the train.

Let the original average speed of the train be $x$ km/h.

The distance traveled at the original speed is 63 km.
Using the formula for time:

$\text{Time}=\frac{\text{Distance}}{\text{Speed}}$

The time taken to cover the first part of the journey is:

$t_1=\frac{63}{x}\text{hours}$

Next, the train travels a distance of 72 km at an average speed of 6 km/h more than its original speed.
So, the new speed is:

$(x+6)\text{km/h}$

The time taken to cover this second part of the journey is:

$t_2=\frac{72}{x+6}\text{hours}$

The total time taken to complete the entire journey is given as 3 hours. Therefore, we can set up the equation:

$t_1+t_2=3$

$\frac{63}{x}+\frac{72}{x+6}=3$

To simplify, we can divide the entire equation by 3:

$\frac{21}{x}+\frac{24}{x+6}=1$

Now, let's test the correct option, $x=42$ km/h, to verify it satisfies this equation:

$\text{LHS}=\frac{21}{42}+\frac{24}{42+6}$

$\text{LHS}=\frac{1}{2}+\frac{24}{48}$

$\text{LHS}=\frac{1}{2}+\frac{1}{2}=1$

Since the LHS equals the RHS, the original speed of the train is indeed 42 km/h.