To determine how many minutes early $Y$ completes the journey compared to $X$, we need to calculate the total time taken by both $X$ and $Y$ for their respective journeys.

Step 1: Calculate the time taken by $X$

• For the first 1 km at 5 km/h: $\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{1}{5} \text{ hours}$
• For the next 2 km at 10 km/h: $\text{Time} = \frac{2}{10} = \frac{1}{5} \text{ hours}$
• For the last 3 km at 4 km/h: $\text{Time} = \frac{3}{4} \text{ hours}$

Total time taken by $X$: $\frac{1}{5} + \frac{1}{5} + \frac{3}{4} = \frac{2}{5} + \frac{3}{4}$ To add these, find a common denominator (20): $\frac{2}{5} = \frac{8}{20}, \quad \frac{3}{4} = \frac{15}{20}$ $\frac{8}{20} + \frac{15}{20} = \frac{23}{20} \text{ hours}$

Step 2: Calculate the time taken by $Y$

• For the first 1 km at 4 km/h: $\text{Time} = \frac{1}{4} \text{ hours}$
• For the next 2 km at 10 km/h: $\text{Time} = \frac{2}{10} = \frac{1}{5} \text{ hours}$
• For the last 3 km at 5 km/h: $\text{Time} = \frac{3}{5} \text{ hours}$

Total time taken by $Y$: $\frac{1}{4} + \frac{1}{5} + \frac{3}{5}$ Convert to a common denominator (20): $\frac{1}{4} = \frac{5}{20}, \quad \frac{1}{5} = \frac{4}{20}, \quad \frac{3}{5} = \frac{12}{20}$ $\frac{5}{20} + \frac{4}{20} + \frac{12}{20} = \frac{21}{20} \text{ hours}$

Step 3: Calculate the difference in time

The difference in time taken by $X$ and $Y$ is: $\frac{23}{20} - \frac{21}{20} = \frac{2}{20} = \frac{1}{10} \text{ hours}$ Convert this to minutes: $\frac{1}{10} \times 60 = 6 \text{ minutes}$

Therefore, $Y$ completes the journey 6 minutes earlier than $X$. The correct option is A.