At what time between 3 o’clock and 4 o’clock will the hour and minute hands meet (be exactly together)?

Introduction / Context:
Clock-angle problems rely on relative speed. The minute hand moves 6 degrees per minute; the hour hand moves 0.5 degrees per minute. They meet when the minute hand catches up the initial angular lead of the hour hand after the hour mark.

Given Data / Assumptions:

• At exactly 3:00, hour hand at 90°; minute hand at 0°.
• Relative speed = 6 − 0.5 = 5.5 degrees per minute.
• Required: time t such that 5.5 t = 90.

Concept / Approach:
Solve t = 90 / 5.5 minutes past 3. Express as an exact fraction to avoid rounding errors; simplify to an improper fraction for neatness.

Step-by-Step Solution:

Verification / Alternative check:
General formula: meeting after h o’clock is t = (60/11)h minutes; for h = 3, t = 180/11 minutes. This corroborates the direct relative-speed method.

Why Other Options Are Wrong:
164/11, 143/11, 132/11 are smaller times and do not satisfy 5.5 t = 90; they correspond to other angles/conditions.

Common Pitfalls:
Using 65/11 (the “between 1 and 2” overlap time) or forgetting that the hour hand also moves during t minutes.

Final Answer:
180/11 minutes past 3.