At what time between 2 and 3 O'Clock will the hands of a clock be together ?

Question

UnlockIAS · Accepted Answer

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.

Why the other options are wrong — (a) 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) 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) 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.

Concept: 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.

State PCS 2025 Q42 — Science & Technology PYQ with Answer

State PCS 2025 Q42 — Science & Technology PYQ with Answer