Answer: D — $6$

For $X$, the time taken is:

$T_X = \frac{1}{5} + \frac{2}{10} + \frac{3}{4}$

Since $\frac{2}{10} = \frac{1}{5}$, we get:

$T_X = \frac{1}{5} + \frac{1}{5} + \frac{3}{4}$

$T_X = \frac{2}{5} + \frac{3}{4}$

Taking LCM as $20$:

$T_X = \frac{8}{20} + \frac{15}{20} = \frac{23}{20}$

So, $X$ takes $\frac{23}{20}$ hours.

For $Y$, the time taken is:

$T_Y = \frac{1}{4} + \frac{2}{10} + \frac{3}{5}$

Since $\frac{2}{10} = \frac{1}{5}$, we get:

$T_Y = \frac{1}{4} + \frac{1}{5} + \frac{3}{5}$

Taking LCM as $20$:

$T_Y = \frac{5}{20} + \frac{4}{20} + \frac{12}{20} = \frac{21}{20}$

So, $Y$ takes $\frac{21}{20}$ hours.

The time saved by $Y$ is:

$T_X - T_Y = \frac{23}{20} - \frac{21}{20} = \frac{2}{20} = \frac{1}{10}$

Thus, the time saved is $\frac{1}{10}$ hour.

Converting hours into minutes:

$\frac{1}{10} \times 60 = 6$

Therefore, $Y$ completes the journey $6$ minutes earlier than $X$.