Hands coincide every 64 minutes: By how much does the watch gain (per day) if successive coincidences occur every 64 minutes?

Introduction / Context:
On a perfect clock, the time between coincidences is T₀ = 720/11 minutes ≈ 65 5/11 min. If the observed interval (in true time) is shorter, 64 minutes, the watch runs fast by a constant scale factor.

Given Data / Assumptions:

• Perfect coincidence interval T₀ = 720/11 min.
• Observed true-time interval T = 64 min.
• Uniform speedup factor r (applies to both hands equally).

Concept / Approach:
With scaling factor r, T = T₀ / r ⇒ r = T₀ / T = (720/11)/64 = 720/704 = 45/44 > 1 (fast). Daily gain = indicated − true over 24 hours = (r − 1) × 24 hours.

Step-by-Step Solution:
1) r = 45/44.2) Daily gain = (45/44 − 1) × 24 h = (1/44) × 24 h = 24/44 h = 6/11 h.3) Convert to minutes: (6/11) × 60 = 360/11 min = 32 8/11 min.

Verification / Alternative check:
Because 64 < 65 5/11, the watch must gain time; 32 8/11 min ≈ 32.73 min/day.

Why Other Options Are Wrong:
37 and 31 minutes are off; “None of the above” is incorrect since an exact value exists.

Common Pitfalls:
Subtracting intervals directly (T₀ − T) without converting to a rate; forgetting both hands scale equally.

Final Answer:
32 8/11 min