Question

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?

• A. 1:00 a.m. on Wednesday
• B. 5:00 a.m. on Wednesday
• C. 1:00 p.m. on Wednesday
• D. 5:00 p.m. on Wednesday

Hint:

Find how fast the watch gains per hour, then see when it makes up its initial 1-minute lag.

Answer 1

The Correct Option is B

Step 1: Find the total error and the time interval.
At 1 p.m. Tuesday the watch is 1 minute slow, so its error is \(-1\) minute compared to the correct time. At 1 p.m. Thursday, 48 hours later, the watch is 2 minutes fast, so its error is \(+2\) minutes. Over these 48 hours the error has changed from \(-1\) minute to \(+2\) minutes, a total swing of \(2 - (-1) = 3\) minutes.

Step 2: Work out the rate of gain.
The watch gains 3 minutes every 48 hours at a steady rate (the standard assumption in such problems), so it gains \(\frac{3}{48} = \frac{1}{16}\) minute every hour.

Step 3: Find when the error becomes zero.
Starting from \(-1\) minute at 1 p.m. Tuesday, the watch needs to gain exactly 1 minute to read the correct time. Time needed \(= \dfrac{1}{1/16} = 16\) hours. Adding 16 hours to 1 p.m. Tuesday: 12 hours takes us to 1 a.m. Wednesday, and 4 more hours takes us to 5 a.m. Wednesday.

Final Answer:
The watch shows the correct time at 5:00 a.m. on Wednesday. \[ \boxed{\text{5:00 a.m. on Wednesday}} \]