Explanation:

- Let the usual speed be $x$ km/h.

- Then, the usual time taken is $\frac{400}{x}$ hours.

- With the increased speed ($x+10$ km/h), time taken becomes $\frac{400}{x+10}$ hours.

- Given that the increased speed reduces travel time by 2 hours, we have:

$$

\frac{400}{x} - \frac{400}{x + 10} = 2

$$

- Solving the equation, cross-multiply and simplify:

$$

400(x + 10) - 400x = 2x(x + 10)

$$

$$

4000 = 2x^2 + 20x

$$

$$

x^2 + 10x - 2000 = 0

$$

- Solve this quadratic equation using the quadratic formula, $\left(\frac{-b\pm \sqrt{b^2-4ac}}{2a}\right)$, where $a=1,b=10,c=-2000$.

- The solution gives two values for $x$, and the relevant one plugged back into $\frac{400}{x}$ gives 10 hours.

- Option 4: 10 is correct.