How many times do the hands of a clock coincide in a day ?

Question

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Correct — C, 22. Treat it as a relative-speed problem, which is the only way to get it right without counting on your fingers. The minute hand sweeps 360° in 60 minutes = 6° per minute; the hour hand sweeps 360° in 12 hours = 0.5° per minute. The minute hand therefore gains on the hour hand at 6 − 0.5 = 5.5° per minute. The hands coincide each time that gain completes a full lap of 360°, which takes 360 ÷ 5.5 = 720/11 = 65 5/11 minutes — a little over an hour, not exactly an hour. In one 12-hour cycle of 720 minutes the number of complete laps is 720 ÷ (720/11) = 11, so the hands coincide 11 times in 12 hours and 11 + 11 = 22 times in a full day of 24 hours. Starting from 12:00 the eleven coincidences fall at 12:00, 1:05 5/11, 2:10 10/11, 3:16 4/11, 4:21 9/11, 5:27 3/11, 6:32 8/11, 7:38 2/11, 8:43 7/11, 9:49 1/11 and 10:54 6/11 — and then the next one is 12:00 again. That gap is the whole question: there is no coincidence anywhere inside the 11 o'clock hour, because by 11:00 the minute hand has fallen so far behind that it does not catch up until both hands reach 12. One hour of every twelve has no overlap at all, so the count is 11 per half-day, not 12.

Why the other options are wrong — (a) This is the count you get by assuming one coincidence in every hour and then, mistakenly, applying it only to a 12-hour clock face rather than to a 24-hour day. It fails twice over: the true rate is 11 per 12 hours, not 12, and the question asks about a day, not a clock face.  (b) The classic near-miss. It comes from getting the 11-per-12-hours rate right for one half of the day but slipping back to 12 for the other half, or from counting the 12 o'clock midnight overlap twice. The two halves of the day are identical, so the total must be an even number — 23 cannot be right on parity alone.  (d) The naive answer: one overlap per hour × 24 hours. It ignores the fact that the hour hand is itself moving, so the minute hand needs 65 5/11 minutes — not 60 — to catch it again. Over a day that extra 5 5/11 minutes per lap costs exactly two overlaps.

Concept: Every clock-hands question reduces to one number: the relative angular speed of the two hands, 5.5° per minute (6° per minute for the minute hand minus 0.5° per minute for the hour hand). From that single figure every standard result follows. The hands return to the same relative position every 720/11 = 65 5/11 minutes, so any specified configuration — overlapping at 0°, opposite at 180°, or perpendicular at 90° — recurs on that same cycle. The general working formula for the angle between the hands at H hours and M minutes is |30H − 5.5M| degrees, taking the smaller of that value and 360° minus it. Because 11 cycles fit into 12 hours rather than 12, every 'once an hour' intuition about clocks is off by exactly two events per day.

State PCS 2024 Q121 — Science & Technology PYQ with Answer

State PCS 2024 Q121 — Science & Technology PYQ with Answer