Clocks & Calendars MCQs for UPSC Prelims

Correct answer: B - Wednesday

All three meet again after LCM(5, 24, 9) days. Prime factors: 5, 24 = 2^3 x 3, 9 = 3^2, so LCM = 8 x 9 x 5 = 360 days. The day of the week advances by 360 mod 7; since 357 = 51 x 7, the remainder is 3. Three days after Sunday is Wednesday, option (b). Option (a) Monday would need remainder 1, option (c) Thursday remainder 4, and option (d) Sunday remainder 0 - 360 is not a multiple of 7, so they cannot meet on a Sunday.

Correct answer: D - None of the above statements is correct.

One watch loses 2 min/day, the other gains 2 min/day, so the gap between the two widens by 4 minutes each day. For the two watches to display the same time again on a 24-hour clock, the accumulated gap must equal one full 24-hour cycle = 24 x 60 = 1440 minutes. Days needed = 1440 / 4 = 360 days. After 30, 90 and 120 days the gaps are only 120, 360 and 480 minutes, none equal to 1440, so the watches do not coincide on those days. Hence none of the options is correct, (d).

Correct answer: D - Between 7:05 hours and 7:10 hours

At 7:00 the hour hand is at 210 degrees and the minute hand at 0 degrees, so the angle is 210 degrees (not 180). At m minutes past 7, the hour hand is at 210 + 0.5m degrees and the minute hand at 6m degrees. Setting 210 + 0.5m - 6m = 180 gives 5.5m = 30, so m = 60/11, approximately 5 minutes 27 seconds. This instant lies between 7:05 and 7:10, option (d). Option (a) is wrong because at 7:00 the angle is 210 degrees. At 7:05 the angle is 210 + 2.5 - 30 = 182.5 degrees, still more than 180, so option (c) is wrong and the hands have not yet reached 180 anywhere in 7:00 to 7:05, ruling out option (b).

Correct answer: C - Both 1 and 2

Statement 1: after 3 o'clock the minute hand catches the hour hand when 6t = 90 + 0.5t, so t = 180/11 = 16 and 4/11 minutes, i.e. about 3:16:22, which lies between 3:16 p.m. and 3:17 p.m. Correct. Statement 2: the second hand laps the minute hand once every 60/59 minutes; the coincidence in the 59th minute occurs at 58 x 60/59 = 58.98 minutes, i.e. about 4:58:59, which lies between 4:58 p.m. and 4:59 p.m. Correct. Both statements hold, so option (c); options (a), (b) and (d) each deny a coincidence that demonstrably occurs in the stated window.

Correct answer: A - 5:36 p.m.

In 24 real hours the faulty clock shows 24 h 10 min = 1450 min, so for every 1440 real minutes the clock displays 1450 minutes. The clock-shown time from 8:00 a.m. Monday to 6:00 p.m. Wednesday is 2 days + 10 hours = 58 hours = 3480 displayed minutes. Real minutes = 3480 * (1440/1450) = 3480 * 1440 / 1450 = 3456 min = 57 h 36 min. Adding 57 h 36 min to 8:00 a.m. Monday: 57 h 36 min = 2 days 9 h 36 min; 8:00 a.m. Monday + 2 days = 8:00 a.m. Wednesday, + 9 h 36 min = 5:36 p.m. Wednesday. Answer (a) 5:36 p.m.

Correct answer: A - 3

All five groups meet together on days that are common multiples of their meeting intervals 2, 3, 4, 5 and 6. The LCM of 2, 3, 4, 5 and 6 is 60, so they all meet on every 60th day. Within 180 days this happens on day 60, day 120 and day 180, which is 3 times.

Correct answer: C - 12th July is a Sunday if the year is a leap year.

Count days from 12 January (a Sunday) to the target date and take the remainder modulo 7. Days remaining in January after the 12th = 31 - 12 = 19. Then Feb + Mar + Apr + May + Jun. To 12 July (leap year, Feb = 29): 19 + 29 + 31 + 30 + 31 + 30 + 12 = 182 days. 182 / 7 = 26 remainder 0, so 12 July is the same weekday as 12 January, a Sunday. So option (c) is correct (and option (d) is false). To 15 July (non-leap, Feb = 28): 19 + 28 + 31 + 30 + 31 + 30 + 15 = 184 days; 184 mod 7 = 2, giving Tuesday, not Sunday, so (b) is wrong. To 15 July (leap): 185 mod 7 = 3, giving Wednesday, not Sunday, so (a) is wrong. Only (c) holds.

Correct answer: D - 5 minutes fast

Let the outbound driving time be t. The return route is 1.25 times as long at twice the speed, so return time = (1.25/2) t = 0.625 t. Total elapsed real time from leaving home (14:30) to returning home (16:00) is 90 minutes, including the 25-minute stay: t + 25 + 0.625 t = 90, giving 1.625 t = 65, t = 40 minutes. So the man really reached the village at 14:30 + 40 = 15:10, but the village clock read 15:15, meaning the village clock is 5 minutes fast. (Return check: 0.625*40 = 25 min, 15:10 + 25 stay + 25 drive = 16:00.)