Imagine two clocks, one that runs normally and one that runs at 12 times normal speed, so that the hour hand of the fast clock always matches the minute hand of the normal one. Our short-handed clock will give an ambiguous reading every time these two imaginary clocks show the same time, provided we omit cases where the two hands coincide.

During the period from just after noon to midnight, the minute hand of the fast clock revolves 144 times and the hour hand of the normal clock revolves once. So the two clocks will show the same time 143 times. During that period the two hands of the accurate clock will coincide 11 times, so between noon and midnight we’ll be unable to tell the time 143 – 11 = 132 times.

The same reasoning obtains for the period from midnight to noon, so the answer is 264.

From “A Clock Puzzle,” by Andy Latto, in Puzzler’s Tribute, edited by David Wolfe and Tom Rodgers, 2002.