7
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This is also another puzzle I found on the internet. The link to the first one is here.

  1. The first question with B as the correct answer is:

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

  1. The answer to Question 4 is:

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

  1. The answer to Question 1 is:

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

  1. The number of questions which have D as the correct answer is:

A. 3
B. 2
C. 1
D. 0

  1. The number of questions which have B as the correct answer is:

A. 0
B. 2
C. 3
D. 1

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2
  • $\begingroup$ So many people with the same time! $\endgroup$
    – user41805
    Commented Jun 7, 2015 at 14:21
  • $\begingroup$ I was just writing my explanation as I solved it. $\endgroup$
    – CyanogenCX
    Commented Jun 7, 2015 at 14:29

2 Answers 2

Reset to default
4
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The answer is:

1)C
2)D
3)B
4)C
5)B

Proof:

The answer to question 1 cannot be A (since then it would be B) or B (since then it couldn't be B), so it must be C or D. If it's D, then the answer to question 2 is B, so the answer to question 4 is A, so there are three D's (not answers to questions 2 or 4), so the answer to question 3 must be D, which contradicts question 1. So the answer to question 1 is C, which immediately means the answer to question 3 is B.

Now the answer to question 2 can't be A (since then the answer to question 4 would be D, which is self-contradiction), nor B or C (since then the answer to question 4 would be A or B and there would be at least two D's even though question 5 is the only one that could be D now). So the answer to question 2 is D, which immediately means the answer to question 4 is C.

The total number of B's must now be 1 or 2. If it's 1, then the answer to question 5 is D, so there are two D's, contradicting question 4. So there are two B's and the answer to question 5 is B.

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3
  • 2
    $\begingroup$ Aaaaahhh 14 seconds! $\endgroup$
    – leoll2
    Commented Jun 7, 2015 at 14:17
  • $\begingroup$ @leoll2 You wasted time editing the OP first! ;-) $\endgroup$
    – Rand al'Thor
    Commented Jun 7, 2015 at 14:19
  • $\begingroup$ That's true indeed! $\endgroup$
    – leoll2
    Commented Jun 7, 2015 at 14:56
0
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Answer:

C,D,B,C,B

The explanation is as follows:

Q1 can't be A because it asks which one has the answer as B. Can't be B because then Q1 is already the one with the first answer as B. If Q1 is D, then Q2 links to Q4 saying that 3 questions have the answer as D, and since we already know the answer to Q2 and Q4 none of which are D, then Q1, Q3, and Q5 are D, but Q3 can't be D because according to Q3 and choosing D, the answer to Q1 is A, but we already know that the answer to Q1 isn't A, so this choice fails as well. Therefore the only option left is C.

For Question 3

We know that the correct answer to Q1 is C, and that says that Q3 is the first question with the answer as B, so therefore Q3 is B.

For Question 4, 2, and 5

If we choose A, then there are only 2 questions left which can have the answer as D, but A says there are 3 left, so therefore A is false. Also, we can cross out D because if it is indeed zero, then the answer for this question can't be D. It can't be B because that means Q2 and Q5 have the answer as B because then Q2 says the answer to Q4 is C, but we have chosen B. Therefore only C works, which means Q2 has the answer D. And then the only answer that works for Q5 is B.

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