The hands coincide at 10¹⁰⁄₁₁ minutes past 2 — option 4.

Set up the positions. Measure angles clockwise from the 12 mark. At m minutes past 2 :

Set them equal. The hands are together when the two angles are the same :

\(6m=60+0.5m\)

\(5.5m=60\)

\(m=\dfrac{60}{5.5}=\dfrac{120}{11}=10\dfrac{10}{11}\)

The shortcut. The minute hand gains on the hour hand at 5½° a minute. At 2 o’clock exactly, the hour hand is 60° ahead, so the minute hand needs

\(\dfrac{60}{5.5}=\dfrac{120}{11}\ \text{minutes}\)

to catch it. The general rule for the hands coinciding between H and H + 1 o’clock is \(\dfrac{60H}{11}\) minutes past H — here \(120/11\), as found.

A sanity check on the options. The answer must lie a little after 2:10, because at 2:10 the minute hand is exactly on the 2 while the hour hand has already moved five degrees past it. Any answer of about 38 minutes past 2 would put the minute hand near the 8 and the hour hand near the 2 — nowhere near each other. That rules out options 1 and 2 at a glance. Option 3 has the right whole-number part but the wrong fraction: the denominator in every clock problem of this kind is 11, from the 5½° relative speed, never 43.

Hence, the answer is 10¹⁰⁄₁₁ minutes past 2.