log_2 of 60*90 is about 12.4. This shows that a naive scheme would expect to get about 12.4 + (23 - 12.4)/2 = 17.7 questions right. I would be very skeptical about any scheme that purports to guarantee better than this. A reasonable approach would be trade the possibility of doing better than this with the possibility of doing worse.
In fact, I can guarantee 17 correct questions, with only 11 bits which has the same floor for expected value (11 + (12/2) = 17):
Divide the first 18 questions into 6 groups of three.
Use the first 6 bits of time to indicate the majority in each group of three, giving you 12 guaranteed answers. The last 5 questions can be encoded directly.
It seems like there should be room for improvement, as this does very well on each group in the 1/4 time that each is uniform. I also worry that the average here is 12 + 1.5 + 5 > 17.7.
Perhaps overlapping the groups could help, which starts to look like an LDPC code, but with max, rather than parity. The difficulty is that overlapping bits can no longer guarantee more, because the previous ones could have already done so.
In fact, I can guarantee 17 correct questions, with only 11 bits which has the same floor for expected value (11 + (12/2) = 17):
Divide the first 18 questions into 6 groups of three. Use the first 6 bits of time to indicate the majority in each group of three, giving you 12 guaranteed answers. The last 5 questions can be encoded directly.
It seems like there should be room for improvement, as this does very well on each group in the 1/4 time that each is uniform. I also worry that the average here is 12 + 1.5 + 5 > 17.7.
Perhaps overlapping the groups could help, which starts to look like an LDPC code, but with max, rather than parity. The difficulty is that overlapping bits can no longer guarantee more, because the previous ones could have already done so.