I ll choose B all the way for all the questions that I might have to guess.

I’ll leave it blank and hope that the Scantron gives me the mark!

you should mark two then partially erase them and let scantron try to figure it out. worse case scenario is that a grader has to look at it right?

Maybe this strategy makes sense: Tally up each answer letter have already chosen, then choose the least used letter for the unknowns. Assuming the answers are completely random, each letter has a 1 in 3 chance of being picked. In the long run, each letter should be right a third of the time. In this case, the sample size is small (60 questions) so it probably wont be anywhere near this. Nonetheless, I think there should be a slight statistical advantage to picking the least represented letter. Obviously try and adjust the choices for the letters you know are incorrect. My stats knowledge is fairly limited so maybe a math guru might want to chime in on this.

Eliminate the obvious one, then flip a coin Btw, I’ve done scantron research in the past. It reads the darkest mark, you can fill in all 3 answers and it will select the darkest of the 3. No 2 pencils aren’t required anymore (except in old machines), they were needed because marks were read by light shinning through. No.2 led is the right balance of dark and reflective due to the graphite so that it gave the most accurate readings. Newer machines will read pen, pencil, smeared poop, doesn’t matter… the darkest mark wins.

with all 3 marked, when they do the manual checking, they wont;t make it count

I go with F for F’ed

i depend on intuition lol

Anyone else starts to double guess their answer choice when they notice they have 4 or 5 consecutive A’s or B’s ?

I go with F for F’ed - Guille +1

hey mr moose, tell me did you know, your antlers are gone? You must have kids.

ill go with B …

C because it is the biggest breast cup they offer!

Since you’ll probably end with a lot of time remaining, I would count up all the A, B, and C. I know each answer is independent, so having all A as answer is just as likely as all B, etc. But I’d like to think that having 40 A, 40 B, and 40 C is more likely. I also like to think that clusters are more likely to be broken up by opposites. If A and B have been switching off in the last 3-4 answers, then I choose C. It becomes a game of roulette and can be mildly more entertaining than sitting around. Again, I know every answer choice is independent, but I’ve never taken a test where a single number came up more than 7 times in a row.

one of the state is incorrect and one statement is correct