Hard Multiple Choice Test (1 Viewer)

InteGrand

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SELF-REFERENTIAL APTITUDE TEST, by Jim Propp (jimpropp at gmaildotcom)

The solution to the following puzzle is unique; in some cases the
knowledge that the solution is unique may actually give you a short-cut
to finding the answer to a particular question. (Thanks to Andy Latto
for bringing this subtlety to my attention.)

I should mention that if you don't agree with me about the answer to #20,
you will get a different solution to the puzzle than the one I had in mind.
But I should also mention that if you don't agree with me about the answer
to #20, you are just plain wrong. :)

You may now begin work.


1. The first question whose answer is B is question
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

2. The only two consecutive questions with identical answers are questions
(A) 6 and 7
(B) 7 and 8
(C) 8 and 9
(D) 9 and 10
(E) 10 and 11

3. The number of questions with the answer E is
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

4. The number of questions with the answer A is
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

5. The answer to this question is the same as the answer to question
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

6. The answer to question 17 is
(A) C
(B) D
(C) E
(D) none of the above
(E) all of the above

7. Alphabetically, the answer to this question and the answer to the
following question are
(A) 4 apart
(B) 3 apart
(C) 2 apart
(D) 1 apart
(E) the same

8. The number of questions whose answers are vowels is
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

9. The next question with the same answer as this one is question
(A) 10
(B) 11
(C) 12
(D) 13
(E) 14

10. The answer to question 16 is
(A) D
(B) A
(C) E
(D) B
(E) C

11. The number of questions preceding this one with the answer B is
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

12. The number of questions whose answer is a consonant is
(A) an even number
(B) an odd number
(C) a perfect square
(D) a prime
(E) divisible by 5

13. The only odd-numbered problem with answer A is
(A) 9
(B) 11
(C) 13
(D) 15
(E) 17

14. The number of questions with answer D is
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

15. The answer to question 12 is
(A) A
(B) B
(C) C
(D) D
(E) E

16. The answer to question 10 is
(A) D
(B) C
(C) B
(D) A
(E) E

17. The answer to question 6 is
(A) C
(B) D
(C) E
(D) none of the above
(E) all of the above

18. The number of questions with answer A equals the number of questions
with answer
(A) B
(B) C
(C) D
(D) E
(E) none of the above

19. The answer to this question is:
(A) A
(B) B
(C) C
(D) D
(E) E

20. Standardized test is to intelligence as barometer is to
(A) temperature (only)
(B) wind-velocity (only)
(C) latitude (only)
(D) longitude (only)
(E) temperature, wind-velocity, latitude, and longitude

( to go to the main SRAT page, go to jamespropp.org/srat.html )
~~~
Links to solutions and more: http://faculty.uml.edu/jpropp/srat.html
 

Nailgun

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Sorry for necro lol
That was really fun
I had to start again because I tripped up on q7
I was so scared during the whole thing that I'd make a mistake somewhere and not realize until I got to a contradiction lelele

would recommend
 

leehuan

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Uhh so like I'm spending ages on this cause I'm dumb... here's my logic (only looked at answers as I went along, no solutions):

20 is E

10 and 16 form a pair. Only two of them correspond and they are 10A, 16D

1 is definitely not A. And 1 can't be B either, because otherwise both 1 and 2 are B. Turns out 5 can't be A or B either.

Between 6 and 17, for a similar reason one of them is B and the other is D. But 2 just said that 16 and 17 can't have the same answer no matter what. So 17 is B and 6 must therefore be D

In 12, reject CDE because otherwise too many ambiguities arise when doing 8 and even 12 itself (odds can be primes, perfect square 16 is even). By rejecting 12 CDE, 8 can't be A, B or D. But that means there is definitely an even number of vowels and thus consonants. So 12 must be A

But note that 8 is C or E.

Therefore 15 is A and hence 13 is D

Suddenly no other odd number question can have answer A...

Either 3 or 4 is B, because within Q1-5 at least something is B according to Q1. May be useful later.

2 can't be D or E as 9 and 11 can't be with 10A

If 3 is C, 5 is C. If 4 is D, 5 is D. But simultaneously 5 can be E. Hence to eradicate those solutions 5 must be E and 3 can't be C; 4 can't be D.

Which means 3 can no longer be B either as we now have 2 Es. (But cause 3 is not C, there is another E in this thing somewhere... lol not useful)

Consequences of Q2: 4 can't be E. 14 can't be D. 18 can't be B. 19 can't be E.

Because Q8 is either C or E, I have 6 or 8 vowels. But the only way I can have 6 vowels is if Q3 is C and Q4 is A as 2+4=6. Since Q3 is NOT C, I must have 8 vowels. and thus Q8 is E.

So 7 and 9 are no longer E by Q2.

Because 3 is now D or E, the B in Q1-5 must be Q4

This makes Q1 D

2 is not B so 7 can't be E...

Since 4 is B... We have 8 vowels so 8-5=3. Therefore 3 is D.

Awkward moment 1hr in when I realise I don't even have a C floating around yet...

There are 3 Es on the answer set and we only have 3 Es. So nothing else can be E anymore.

Q2 lost B C D AND E. So Q2 is stuck with A.

So 6 and 7, being the same, makes 7 D

9 is not C as 12 is A.

At this point, 9, 11, 14, 18 and 19 are unanswered. We only have 5 As. But 9, 11, 14 AND 19 have been eliminated from being A. So the last A must be on 18.

Because 5 As exist, Q18 now tells us only 5 Bs exist.

So of the last 4 answers. because the only Bs are 4 and 17, we have 3 Bs left. And the last one is a C or a D.

So far, the Ds are 1, 3, 6, 7, 13 and 16. This means we have either 6, or 7 Ds. Q14 is constrained to A or B. But Q15 is A so Q14 can't be A. So Q14 is a B.

So there is no C...! The last 3 are 2 Bs and a D

If 11's not C, 11 is forced into being B.

So 19 is the B and 9 is the D.

DADBEDDEDABADBADBABE

Ugh, 1hr 33min
 

InteGrand

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Imagine if this something like this came up as a BOS Trial Multiple Choice question set.
 

Nailgun

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This took me almost an entire hour leellee
pls be 4U only kthnx
 

Nailgun

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Although it would be much easier to keep track of 10 questions though lol
 

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