A watch is a minute slow at 1 p.m. on Tuesday and 2 minutes fast at 1 p.m. on Thursday. When did it show the correct time?

Question

What will be the measurement of the angle made by the hour and minute hands of a clock when the time is ‘quarter past 3’?

Answer 1

Treat this as a simple rate problem: the watch's error moves at a steady speed, and we only need to find when that error crosses zero.

Set up the error line: at 1 p.m. Tuesday the error is $-1$ minute (slow), and 48 hours later, at 1 p.m. Thursday, the error is $+2$ minutes (fast). The error rises by $2-(-1)=3$ minutes over 48 hours, so the watch gains at $\frac{3}{48}=\frac{1}{16}$ minute per hour.
Find the zero-crossing: the error needs to rise by 1 minute (from $-1$ to $0$) to hit the correct time, which takes $1 \div \frac{1}{16} = 16$ hours after 1 p.m. Tuesday.
Convert to a clock reading: 16 hours after 1 p.m. Tuesday is 5 a.m. Wednesday (12 hours reaches 1 a.m. Wednesday, and 4 more hours reaches 5 a.m.).

So the watch shows the exact correct time at 5:00 a.m. on Wednesday, option B. \[ \boxed{\text{5:00 a.m. Wednesday}} \]

Answer 2

Treat this as a simple rate problem: the watch's error moves at a steady speed, and we only need to find when that error crosses zero.

Set up the error line: at 1 p.m. Tuesday the error is $-1$ minute (slow), and 48 hours later, at 1 p.m. Thursday, the error is $+2$ minutes (fast). The error rises by $2-(-1)=3$ minutes over 48 hours, so the watch gains at $\frac{3}{48}=\frac{1}{16}$ minute per hour.
Find the zero-crossing: the error needs to rise by 1 minute (from $-1$ to $0$) to hit the correct time, which takes $1 \div \frac{1}{16} = 16$ hours after 1 p.m. Tuesday.
Convert to a clock reading: 16 hours after 1 p.m. Tuesday is 5 a.m. Wednesday (12 hours reaches 1 a.m. Wednesday, and 4 more hours reaches 5 a.m.).

So the watch shows the exact correct time at 5:00 a.m. on Wednesday, option B. \[ \boxed{\text{5:00 a.m. Wednesday}} \]

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