BPSC 70th CCE Prelims General Studies · Question 121 of 143

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.

• (a)12 — 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)23 — 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)24 — 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.

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.

The trap is that 'the hands meet about once an hour' is almost true, and the question is built on the word 'about'. Do not count the overlaps; count the laps. In 12 hours the minute hand completes 12 revolutions and the hour hand completes 1, so the minute hand laps the hour hand 12 − 1 = 11 times. That subtraction is the whole derivation and it takes two seconds. Then double it for 24 hours: 22. The same lap-counting logic gives the other two standard results the examiner may ask instead — the hands are opposite (180° apart) 11 times per 12 hours, so 22 times a day, and they are at right angles twice per lap, so 22 times per 12 hours and 44 times a day. If you remember only one thing, remember that 11-not-12 is the discriminator, and that a day-based answer must be even.

• Minute hand 6° per minute, hour hand 0.5° per minute; relative gain 5.5° per minute
• Interval between successive coincidences = 360 ÷ 5.5 = 720/11 = 65 5/11 minutes, i.e. 65 minutes 27 3/11 seconds
• 11 coincidences in every 12 hours (the minute hand laps the hour hand 12 − 1 = 11 times), so 22 in 24 hours
• There is no overlap during the 11 o'clock hour in either half of the day — the 11 o'clock and 12 o'clock overlaps merge into the single event at 12:00
• Companion results from the same cycle: hands opposite 22 times a day; hands at right angles 44 times a day; hands in a straight line (0° or 180°) 44 times a day
• Angle between the hands at H:M is |30H − 5.5M| degrees, reduced to the smaller of that and its 360° complement

• Assuming one overlap per hour and answering 24 — the hour hand's own motion stretches the interval to 65 5/11 minutes
• Answering for a 12-hour clock face when the question says 'in a day' (24 hours), or the reverse
• Confusing the three standard counts: coincide 22 a day, opposite 22 a day, at right angles 44 a day

BPSC keeps its reasoning block short and factual — a single-line clock, calendar, coding or direction question with four numerical options and no working required, as here. UPSC has historically asked the same relative-speed machinery in a computational form: 'through how many degrees will the hour hand rotate', 'the acute angle between the hands at 3:25', or the companion count of right angles in a day. Learn the 5.5°-per-minute figure and both styles collapse into one method.

• practice — not a real PYQ

  How many times in a day are the hands of a clock in a straight line but pointing in opposite directions ?

  • (a)44
  • (b)24
  • (c)22
  • (d)11

  Answer(c) 22 — the hands are 180° apart once in every 720/11-minute cycle, i.e. 11 times in 12 hours and 22 times in a day.

• practice — not a real PYQ

  At what time between 4 o'clock and 5 o'clock will the hands of a clock coincide ?

  • (a)4 : 20
  • (b)4 : 21 9/11
  • (c)4 : 22 2/11
  • (d)4 : 24

  Answer(b) 4 : 21 9/11 — at 4:00 the hour hand leads by 120°; the minute hand closes that at 5.5° per minute, taking 120 ÷ 5.5 = 21 9/11 minutes.