A person left home between 2 p.m. and 3 p.m. and returned between 4 p.m. and 5…

Concept: On a clock face, divide the dial into 60 equal minute-spaces (each 6 degrees). At m minutes past hour-mark H, the minute hand sits at position m, and — since the hour hand moves twelve times slower — the hour hand sits at position 5H + m/12. When a watch's two hands trade places between an earlier reading and a later reading, the later reading's hour-hand position must equal the earlier reading's minute-hand position, and the later reading's minute-hand position must equal the earlier reading's hour-hand position. These two conditions give two linear equations in the two elapsed-minute values, and solving them measures the total time that passed.

Apply this to the given watch:

1. Let the person leave x minutes after 2 p.m. and return y minutes after 4 p.m.

2. At departure: minute hand position = x; hour hand position = 10 + x/12 (the 2 p.m. mark sits at 10 minute-spaces).

3. At return: minute hand position = y; hour hand position = 20 + y/12 (the 4 p.m. mark sits at 20 minute-spaces).

4. Since the hands have exchanged places: hour-hand-at-return = minute-hand-at-departure, so 20 + y/12 = x; and minute-hand-at-return = hour-hand-at-departure, so y = 10 + x/12.

5. Substituting the second equation into the first: 20 + (10 + x/12)/12 = x. Multiplying through by 144 gives 2880 + 120 + x = 144x, so 143x = 3000 and x = 3000/143 minutes.

6. Then y = 10 + x/12 = 10 + 250/143 = 1680/143 minutes.

7. Total time elapsed = 120 + y minus x = 120 + 1680/143 minus 3000/143 = 120 minus 1320/143 = 15840/143 = 1440/13 = 110 (10/13) minutes.

Cross-check: substituting x = 3000/143 back gives the departure hour-hand position 10 + x/12 = 1680/143, which indeed equals y (the return minute-hand position); and 20 + y/12 = 3000/143, which indeed equals x (the departure minute-hand position) — confirming the swap is consistent and the elapsed time is 110 (10/13) minutes.