A person travels from his house to his office in 30 minutes but he tak

When going from house to office:
Let the house be the origin (displacement = 0) and the office be at a displacement D (assuming a straight line path for simplicity, though the problem just says “same route” implying the distance is the same).
Time taken ($t_{going}$) = 30 minutes = 0.5 hours.
Displacement while going = D.
Average velocity while going ($v_{going}$) = Displacement / Time = D / 0.5 = 2D. (Taking direction from house to office as positive).

When returning from office to house:
The person travels back to the origin.
Time taken ($t_{returning}$) = 45 minutes = 0.75 hours.
Displacement while returning = -D (from office position D back to house position 0).
Average velocity while returning ($v_{returning}$) = Displacement / Time = -D / 0.75 = -4D/3. (Taking direction from office to house as negative).

The question asks for the ratio of the average velocity while going and the average velocity while returning. The options are positive numbers, implying the ratio of the magnitudes of the average velocities (average speeds for each leg, since distance = |displacement|).
Magnitude of $v_{going}$ = $|2D| = 2D$.
Magnitude of $v_{returning}$ = $|-4D/3| = 4D/3$.

Ratio of magnitudes of average velocities = $|v_{going}| / |v_{returning}|$ = $(2D) / (4D/3)$
$= 2 \times (3 / 4) = 6 / 4 = 3 / 2$.

If the question was strictly about vector velocities, the ratio would be $(2D) / (-4D/3) = -3/2$, which is not an option. The positive options confirm the question is asking for the ratio of speeds or magnitudes of velocities.