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  1. C is the correct answer to how many questions?

A. 1
B. 2
C. 3
D. 4

  1. Which letter is the correct answer fewer times than any other?

A. D
B. B
C. C
D. A

  1. Which letter is the correct answer more times than any other?

A. D
B. C
C. B
D. A

  1. Which one of these is the only true statement?

A. There are four questions with the answer A.
B. A is the right answer to exactly one question.
C. B is never the right answer.
D. There is an equal amount of A's and C's among the correct answers

5.When is A the correct answer for the first time?

A. On the fifth question
B. On the sixth question
C. On the seventh question
D. On the first question

6.How many occurrences of the most common answer are there?

A. 3
B. 4
C. 5
D. 6

7.Which ones are the only two consecutive questions with the same answer?

A. 2 and 3
B. 7 and 8
C. 1 and 2
D. 4 and 5

8.Which question is the only one with the same answer as this one?

A. 4
B. 2
C. 3
D. 7

Original source: Self-Referential Quiz.

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9
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Answers:

1:B, 2:D, 3:C, 4:B, 5:B, 6:A, 7:D, 8:C

Starting Point:

8 cannot be D, as 7 also being D would contradict the 7/8 relationship. It cannot be A, because question 4 would imply 4 A's, and 8 forces only 2. It cannot be B because 2 answers is a full quarter of the 8 available, so cannot be the single lowest number of occurrences. Therefor 8 is C, and 3 is also C.

This allows us to:

State that answer 1 is B, and that none of the remaining answers can be C. Then, since 2 cannot be B or C as stated earlier, and can't be A because of question 5, 2 must be D. 4 can't be A or D because of 2, or C as previously established, therefor 4 must be B.

Now we are at:

1:B, 2:D, 3:C, 4:B, 5:?, 6:? 7:?, 8:C; and so 7 must be D, which forces 5 to be B and 6 to be A.

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  • $\begingroup$ Aha, you got there first :-) $\endgroup$ – Rand al'Thor Dec 29 '17 at 19:34
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Deduction

Question 8 tells us that one of the four letters is the answer to exactly two questions.

Consider the answer to question 8:

  • If it's D, then the answer to question 7 is D not B, contradiction.
  • If it's A, then the answer to question 4 is A, so there are at least 4 A's, contradiction.
  • If it's B, then the answer to question 2 is B, so B is the rarest answer with at least two occurrences, contradiction.

So we have 8C and therefore 3C. Thus, among the answers, C appears exactly twice and B appears the most times. This also gives us 1B.

Consider the answer to question 7:

  • It can't be B or C, by the last spoilertag above.
  • If it's A, then the answer to question 2 is also C, contradiction.

So we have 7D. By question 5, we know there is at least one answer A, so each letter appears at least once among the answers.

Consider the answer to question 6:

  • If it's D, then the distribution of answers is 0/0/2/6 in some order, and question 2 has no unique answer, contradiction.
  • It can't be C, by the second-to-last spoilertag above.
  • If it's B, then the distribution of answers is 1/1/2/4 or 0/2/2/4. In the first case, question 2 has no unique answer; in the second case, one of the letters never appears among the answers. Contradiction.

So we have 6A. Thus the answers are three B's, two C's, and either two A's and one D or two D's and one A. The only answers we don't know are 2, 4, and 5, two of which must be B. Clearly the answer to question 2 isn't B, so we have 4B and 5B. This tells us the final answer, 2D.

Final answer

1B, 2D, 3C, 4B, 5B, 6A, 7D, 8C.

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  • $\begingroup$ Thank you for your answer! the detailed deduction helped me understand further more! thank you! $\endgroup$ – Quaker Dec 30 '17 at 4:13

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