BPSC 70th CCE Re-exam Prelims General Studies · Question 42 of 142

Correct — B, 10 10/11 minutes past 2. Work in minute-spaces, the sixty divisions on the dial. At exactly 2 o'clock the minute hand is on 12, that is on minute-space 0, and the hour hand is on 2, that is on minute-space 10. The minute hand is therefore ten spaces behind and must gain exactly ten spaces on the hour hand for the two to coincide. Now the rates: in sixty minutes the minute hand travels sixty spaces and the hour hand travels five, so the minute hand GAINS 55 spaces every hour. Gaining one space takes 60/55 = 12/11 minutes, so gaining ten spaces takes 10 × 12/11 = 120/11 minutes, which is 10 and 10/11 minutes. The hands are together at 10 10/11 minutes past 2, roughly 2:10:55. The same result in degrees: the minute hand sweeps 6° per minute and the hour hand 0.5°, a relative rate of 5.5° per minute; at 2:00 the hour hand leads by 60°, so the gap closes in 60 ÷ 5.5 = 120/11 minutes. Generalise it once and you never redo the arithmetic: between H and H+1 o'clock the hands coincide at 60H/11 minutes past H. That gives 5 5/11 past 1, 10 10/11 past 2, 16 4/11 past 3, and so on. Two structural facts follow from the same formula and are worth carrying. The denominator of every coincidence answer is 11 — it comes from 55 = 60 − 5 and can never be anything else, which is the fastest way to reject a printed option. And because the interval between successive coincidences is 720/11 = 65 5/11 minutes, the hands coincide 11 times in twelve hours and 22 times in a day, with no coincidence at all between 11 and 12 o'clock, since the eleventh one falls exactly at 12.

• (a)10 10/43 minutes past 2 — The integer part and the numerator are right and only the denominator has been changed, which makes this the most dangerous option on the page for a candidate who half-remembers the answer. No step of the calculation can produce 43: the relative gain is 55 minute-spaces per hour, so every clock-coincidence answer has 11 in its denominator. Check the denominator first and this option dies immediately.
• (c)38 10/11 minutes past 2 — A hybrid built from the other two — the fraction 10/11 of the correct answer grafted onto the integer 38 of option (d). It corresponds to no event on the dial at all: 38 10/11 is not 60H/11 for any whole H, so it is neither a coincidence time nor a standard right-angle or straight-line time. It exists only to reward pattern-matching on the fraction.
• (d)38 2/11 minutes past 2 — A genuine clock answer moved to the wrong hour. 38 2/11 = 420/11 = 60 × 7 / 11, so it is the moment the hands coincide between SEVEN and eight o'clock, not between two and three. The hands cannot be together thirty-eight minutes past two: by then the minute hand has passed the hour hand and is about twenty-five spaces ahead of it.

Every clock problem reduces to one number: the relative speed of the two hands. The minute hand completes 360° or 60 minute-spaces in an hour; the hour hand completes 30° or 5 minute-spaces in the same hour. The minute hand therefore gains 55 minute-spaces, equivalently 330°, on the hour hand every hour — 5.5° per minute. Once that is fixed, three standard positions follow. The hands coincide when the gain equals the initial lag, are at right angles when they are 15 minute-spaces or 90° apart, and are in a straight line pointing opposite ways when they are 30 minute-spaces or 180° apart. Counting over a full circuit: the hands coincide 11 times in twelve hours and 22 times a day; they are at right angles 22 times in twelve hours and 44 times a day; they are opposite 11 times in twelve hours and 22 times a day. All three counts are one short of the naive answer for the same reason — the hour hand is itself moving, so one crossing per twelve hours is swallowed.

The efficient route through a question like this is elimination rather than computation. First, the denominator: coincidence, right-angle and opposition times between two whole hours all carry 11 in the denominator, so option (a) with 43 is out on inspection. Second, the rough position: between 2 and 3 the hour hand sits a little past the 10-minute mark, and the minute hand reaches it early in the hour, so the answer must have a small integer part — that removes both 38-minute options without any arithmetic. What remains is (b), and the full calculation confirms it. If you do want to compute, use the formula 60H/11 for coincidence and check it against a case you already know: between 12 and 1 the formula gives 0, and the hands are indeed together at twelve o'clock exactly. Reasoning questions on this paper are pure marks — no source can be misremembered and no fact can go out of date — so the discipline is worth building.

• The minute hand gains 55 minute-spaces (330°) on the hour hand every hour, that is 5.5° per minute — the single number every clock question uses.
• The hands coincide at 60H/11 minutes past H o'clock: 5 5/11 past 1, 10 10/11 past 2, 16 4/11 past 3, 38 2/11 past 7, 43 7/11 past 8.
• Successive coincidences are 720/11 = 65 5/11 minutes apart, so the hands coincide 11 times in twelve hours and 22 times in twenty-four.
• There is no coincidence between 11 and 12 o'clock — the eleventh coincidence of the cycle falls exactly at 12:00.
• Between 2 and 3 o'clock the hands are at right angles once, at 27 3/11 minutes past 2, and directly opposite once, at 43 7/11 minutes past 2.

• Assuming the hands meet when the minute hand reaches the 10-minute mark. By the time it gets there the hour hand has already moved on, which is why the answer is 10 10/11 and not 10.
• Accepting an option because its integer part looks right. Option (a) has the correct 10 and the correct numerator 10, and is still wrong on the denominator alone.
• Using 60/55 where 55/60 is needed, or the reverse. Fix the direction by a sanity check: gaining ten spaces must take a little more than ten minutes, not less.

BPSC has asked clocks in both of the standard forms — the counting question, how many times the hands coincide in a day, and this computational one, at what exact time between two given hours. UPSC's basic numeracy block asks the same relative-speed idea through angles instead: how many degrees the minute hand sweeps in a stated interval, or the acute angle between the hands at a stated time.

• practice — not a real PYQ

  At what time between 3 and 4 o'clock will the hands of a clock be together ?

  • (a)15 minutes past 3
  • (b)16 4/11 minutes past 3
  • (c)16 4/13 minutes past 3
  • (d)18 2/11 minutes past 3

  Answer(b) 16 4/11 minutes past 3 — apply 60H/11 with H = 3, giving 180/11 = 16 4/11. It is a little after 15 minutes because the hour hand has itself moved on past the 15-space mark.

• practice — not a real PYQ

  How many times in a day are the hands of a clock at right angles to each other ?

  • (a)24
  • (b)44
  • (c)48
  • (d)22

  Answer(b) 44 — the hands are perpendicular 22 times in every twelve-hour cycle, hence 44 times in twenty-four hours. The count is 44 and not 48 for the same reason coincidences number 22 and not 24: the hour hand is also moving.