Question

Kim’s wristwatch always shows the correct time, including ‘am’ and ‘pm’. Jim’s watch is identical to Kim’s watch in all aspects except its pace, which is slower than the pace of Kim’s watch. At 12 noon on January 1st, Jim sets his watch to the correct time, but an hour later, it shows 12:57 pm. At 12 noon on the next June 1st, Jim resets his watch to the correct time. On how many instances between, and including 12 noon on the two dates mentioned, do Jim’s and Kim’s watches show the exact same t

Accepted Answer

D. Jim's watch loses 3 minutes every hour $\implies$ loses 72 minutes per day. To show exact same time (including AM/PM), Jim's watch must lose 24 hours = 1440 minutes. Time required to lose 24 hours $= 1440 / 72 = 20$ days. Days between Jan 1 (noon) and June 1 (noon) in non-leap year $= 31 + 28 + 31 + 30 + 31 = 151$ days. Number of 20-day intervals $= 151 / 20 = 7$ full intervals. Matches occur at $t = 0, 20, 40, 60, 80, 100, 120, 140$ days, plus reset at $t = 151$ days (June 1 noon). Total instances = $8 + 1 = 9$ times.

Clocks Meeting Instances

Explanation

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