sigh

Let's start with the easy ones. All answers to question 10 except A imply answers to question 16 which imply a contradictory answer to question 10. Therefore 10 is A and 16 is D. Similarly, on question 17, answers A and C imply answers to 6 which generate a contradiction, and answer E is impossible. Since "16 and 17" is not an option in question 2, 17 cannot be D. Therefore 17 is B, and 6 is D.

Consider question 1. Answer A is a contradiction because it implies the answer to 1 is B. Answer B is a contradiction because the first question with answer B would be 1. If the answer is E, then 5 must be B, making 2 also B, in which case 5 is not the first B, again a contradiction. So the answer must be C or D. Therefore at least one of questions 3 and 4 is B.

Now consider question 7. If the answer to 7 is A, B, or E, the answer to 8 can only be E. If the answer to 7 is C, 8 is either A or E. If the answer to 7 is D, 8 is either C or E. So the only possible answers for 8 are A, C, and E. This tells us that the total number of vowel answers is an even number.

Put these two facts together: The total number of As and Es is an even number, and either the number of Es must be 1 or the number of As 5. In either case, the number of As and the number of Es must both be odd numbers. Furthermore, we cannot choose identical answers for questions 3 and 4 due to "3 and 4" not being an option for question 2. The only possibilities are 1 E and 7 As, or 3 Es and 5 As. In either case, the total number of vowel answers is 8. Thus the answer to 8 is E.

If there are 8 vowel answers, there are 12 consonant answers. The only correct answer to question 12 is A. This immediately implies 15 is A, which in turn implies 13 is D. Based on question 13, we can eliminate A as an answer to all odd-numbered questions except 15.

Now to question 5. We could argue that the answer is E because otherwise there would be multiple correct answers, but this will not be necessary. We have already eliminated A as an answer for 1, B as an answer for 2, and C as an answer for 3. The answer cannot be D since "5 and 6" is not an option in question 2. Therefore the answer is E.

Now that we have at least two E answers, we can eliminate B as an answer for question 3. So the answer for 3 is D; there are exactly 3 Es. Since the number of Es and As sum to 8, the answer to 4 is B; there are 5 As. It immediately follows that the answer to 1 is D.

Next, question 9.

- The answer cannot be A because it was already eliminated by question 13.

- The answer cannot be B. If the answer was B, then there would be at least two Bs before question 11, question 9 itself and question 4. But an answer of B for 9 would also imply an answer of B for 11, indicating that only one B precedes question 11. This is a contradiction.

- The answer cannot be C since that implies 12 is C and we know it's A.

- The answer cannot be E. We already know that the answers to 5 and 8 are E. If 9 is also E, it implies that 14 is E as well, meaning there are at least 4 Es. But we know there are 3 Es. This is a contradiction.

Therefore the answer to 9 is D.Back to question 2. B has already been eliminated. We see now that the answers to 8 and 9 are different, as are the answers to 9 and 10. Questions 10 and 11 have different answers since the answer to 10 is A, which has been eliminated as an answer to 11. Thus the answer to 2 is A. We can immediately conclude that the answer to question 7 is the same as the answer to question 6, D.

Question 20 is a jab at standardized testing. Barometers measure air pressure, not any of the listed options. The intended relationship must be that standardized testing does not measure intelligence. Therefore E is the correct answer, because all options are not what a barometer measures.

Now to question 18. We know there are 5 As. We have yet to use C as an answer, and there are only 3 remaining unanswered questions, eliminating answer B. We have already used 7 Ds, eliminating answer C. We know there are exactly 3 Es, eliminating answer D. E is eliminated because there are only 3 Es, and we just used our third E on question 20. Therefore the answer is A; there are the same number of As as Bs.

Damn that's impressive, needed to read the first couple of posts before being able to work out the rest of the answers by myself. How long did it take you to figure out those first few conclusions and did you do any writing to set it up? Or were you able to just think it up in your head and work from there.

>>66876

I think the way it was intended to be solved was to do each question before the standardized testing one and reach the logical conclusion that it must be E since every other possibility is ruled out at that point and you'll need a single E to satisfy one of the few remaining conditions for questions that haven't been met. Recognizing the snarkiness of the author and getting that correct right off the bat though is probably the simplest solution to the point of the questions.>>66913

I was hoping it would be like that, but if you don't assume the last question is E, the solution isn't unique. The simplest alternate solution is switching 19 and 20. In addition, there are three solutions with 18=E, 20=A: 14=B, 19=C; 14=C, 19=D; and 14=D, 19=D.