Straight-line (180°) configuration between 9 and 10 o’clock: At what time between 9:00 and 10:00 are the hands in the same straight line but not together (i.e., 180° apart)?

Introduction / Context:
We seek the first time after 9:00 when the hands are 180° apart (a straight line). This is a standard relative-speed problem for clock hands.

Given Data / Assumptions:

• Hour hand speed = 0.5° per minute.
• Minute hand speed = 6° per minute.
• At 9:00, hour hand angle = 270°, minute hand angle = 0°.

Concept / Approach:
Angle difference at t minutes after 9:00 is |(270 + 0.5t) − 6t| = |270 − 5.5t|. For a straight line, set this to 180° and solve for t within 0 < t < 60.

Step-by-Step Solution:
1) |270 − 5.5t| = 180.2) 270 − 5.5t = 180 ⇒ 5.5t = 90 ⇒ t = 90/5.5 = 180/11 minutes.3) The other case 270 − 5.5t = −180 ⇒ t = 900/11 ≈ 81.82 min exceeds the hour; reject.

Verification / Alternative check:
At t = 180/11, the computed difference is exactly 180°.

Why Other Options Are Wrong:
162/11 and 164/11 are distractors that come from mixing the 90° case or rounding; 162/11 past 10 is outside the 9–10 window.

Common Pitfalls:
Confusing 90° (right angle) with 180°; forgetting the hour hand moves.

Final Answer:
180/11 min past 9