A watch is 1 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?

The watch is 1 minute slow (or 1 minute behind the correct time) at 1PM on Tuesday and 2 minutes fast (or 2 minutes ahead of the correct time) at 1PM on Thursday. The correct clock has traversed $$24\times 2= 48 $$ hours, or $$48*60=2880$$ minutes, while the incorrect clock has travelled $$2880+3=2883$$ minutes in the same time.

The speeds of the correct clock and the incorrect clock, respectively, are $$1$$ minute per minute and $$\dfrac{2883}{2880}$$ minutes per minute. The initial distance between them was $$1$$ minute, therefore, they'll next meet after:

$$\dfrac{1}{\frac{2883}{2880}-1} = \dfrac{2880}{3}$$ minutes, or $$960$$ minutes, which is equal to 16 hours. This will be 5:00 AM on Wednesday. The correct answer is option D.