Content-free logic test answers
February 23, 2007 pm28 10:10 pm
Use this space to submit answers and explanations to this logic test.
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JD2718 
I don’t do content-free very well, but I think the answer to 20 (which has some content) must be (E)?
Well, I seem to have fouled something up — perhaps an assumption I made in question 9. Maybe I’ll give it another try later. Assuming this really can be completed, questions 10 and 16 seem like a healthy place to start.
rdt, I imagine (although obviously I can’t be sure) that the rest of the test will force the answers to questions 19 and 20. Actually, I have a guess about what the answer to 20 will be, but I’ll save it for later (so as not to embarrass myself if I’m wrong ;-) ).
I’ve solved this twice in 15 years, long enough ago that I don’t remember details. The nasty piece of this is, as in sudoku, once you’ve made an error you are fairly well condemned to start from scratch. I agree, btw, with your starting point, and quickly move to #12 (free info!) and then start counting how many of each answer. FWIW, I have 12 answers, about 10,000 possibilities, and am proceeding very slowly, since, if I happen to be good so far, I don’t want to introduce a late, stupid mistake.
Well, I figure I know what a barometer does measure, and its not any of the things listed in the first four options for question 20… Which would make question 20 a dig at standardized tests…
Seems awfully likely, but remember, I didn’t write the test.
I am not a fan of standardized tests, but they do have their place. The profileration of them, and mostly high stakes at that, is driving education in a bad direction. But yeah, looks like the author may have been making a point.
-edit- I should add, I don’t hate them for any personal reason. As a kid, I liked tests, and standardized were the best. I scored high, didn’t need to study. I used to want homework to count the least, then classwork more, then the tests the most (bright, good test-taking, lousy student). I used to argue against take-home tests, since they tended to a) level the playing field and b) raise expectations. Not proud words, I’ll tell you that.
Well, here are the answers I have that I believe to be correct (don’t look if you don’t want spoilers), in numerical order (rather than order of deduction):
6D
8E
10A
12A
13D
15A
16D
17B
all correct so far. We only worked partially in the same order though.
I think I got it. If it’s wrong, I don’t think I want to know :)
1 D
2 A
3 D
4 B
5 E
6 D
7 D
8 E
9 D
10 A
11 B
12 A
13 D
14 B
15 A
16 D
17 B
18 E
19 C
20 A
Dan, I got the same as you except 18.A, 19.B, 20.E. Of course I was assuming 20.E from the start… however my solution has the advantage that none of the answers is C at all.
That’s funny, it seems there are 2 solutions that work.
Also, I thought 19 C made sense – as in, when all else fails, the answer is probably C.
You are both right. I assume that the intention was ABE, but without a definitive answer to 20, both work. Still a good puzzle, though, huh?
Okay, first, 6 and 17 have to be D and B because every other combination results in a contradiction. Likewise, 5 is E, 10 is A, and 16 is D; by the phrasing of 2, 6 is D and 17 is B. To see that 12 is A, either consider 7 and 8, or note that the answer is either A and B and then that B implies C, D, or E. Now 15 is A, so 13 must be D.
At this stage, we have 5 E, 6 D, 10 A, 12 A, 13 D, 15 A, 16 D, 17 B. Also, 1 and 2 are not B or else there is a contradiction. So 3 or 4 is B; from 2, exactly one is B, and from 12, the other must be D, so that 8 is E.
Now, assume that 9 is E. Then 2 is C and 14 is E. But from 8 and 14 we get that there are only two B’s and C’s in total; this is a contradiction since 2 is C, 17 is B, and 3 or 4 is B. The only remaining option for 9 is then D.
We now have 5 E, 6 D, 8 E, 9 D, 10 A, 12 A, 13 D, 15 A, 16 D, 17 B. 7 can’t be E because then 2 would be B; so 2 is A and 7 is D. There are at least two E’s, so 3 is D and 4 is B; then 1 is D. By examination, 11 is B.
We’re left with 14, 18, 19, and 20. There are 7 D’s so far, so 14 can’t be A; from 13, neither is 19. So by 4, 18 and 20 must be A, forcing 19 and 14 to be B.
Sorry, in my last paragraph I miscalculated the number of A’s. We need one A and one E. If we’re forced to have 20 E, then we must have 18 A, 19 B, 14 B. Otherwise, we have the following possibilities:
14 B
18 E
19 C/D
20 A
14 C/D
18 E
19 D
20 A
14 B
18 A
19 E
20 B
Alon,
I will check carefully – I didn’t think there were that many solutions.
I got these possibilities for {14,18,19,20}:
14 B 18 E 19 C 20 A
14 C 18 E 19 D 20 A
14 B 18 A 19 E 20 B
14 B 18 A 19 B 20 E
Alon also lists:
(X) 14 B 18 E 19 D 20 A
(Y) 14 D 18 E 19 D 20 A
But (X) can’t work because then there would be 8 Ds (1 3 6 7 9 13 16 and 19), meaning 14 should be C not B.
and (Y) can’t work because 13 is already D, and Q2=A implies that 14 can’t have the same answer as 13.
Thank you. Absolutely.
Wouldn’t 9 be A since 10 is A? Why would it be D?
This test is not valid and impossible to do, here’s why:
10 is A and therefore 9 is A. This makes 13 a paradox, since choosing the correct answer (A), makes it incorrect (since now 13 is an odd numbered question with answer A).
Am I missing something?
9 is D, and thus 13 is also D.
If you make a mistake on this test, it is not possible to fix it. You must start again from the beginning.
I think d’s answer is correct (reply #9)
Dan’s (reply #8) answer is incorrect: If 20=A, there are 2 possible solutions: 14=B and 19=C, or 14=C and 19=D
Without touching the Q20, I can get some answers for Q1 to 13 and 15 to 17 (same as Dan’s), leaving Q14, 18, 19, 20. In this case, Q14=B or C, Q18=A or E, Q19=B, C, D or E, Q20=A, B, C, D or E. Only Q20=E would give an unique solution
Jim lists:
1. 14 B 18 E 19 C 20 A
2. 14 C 18 E 19 D 20 A
3. 14 B 18 A 19 E 20 B
4. 14 B 18 A 19 B 20 E
Q20 couldn’t be A, otherwise there will be 2 possible solutions. As a results, there leaves only two possibilities:
X: 19E, 20B
Y: 19B, 20E
Since Q19 does not provide any information for itself, the only real question is whether 20=B or E. So, what is a barometer? A instrument to measure atmospheric pressure? So none of the answers were correct.
(My previous reply said that only Q20=E gives an unique solution was wrong)
I enjoyed this greatly, thanks for the puzzle!
It would be good to specify that the puzzle has a unique solution where every question has exactly one answer: multiple-choice multiple-answer tests are not too uncommon, and there are options like “all of the above”…
What I especially liked about this problem was that although it wasn’t trivial, it was straightforward; I was never stuck and having to guess, and there wasn’t much backtracking required.
This is great, thank you! I got C E D A as a unique solution, on the grounds that if 14 is B, the solution is not unique, so 14 is C, which then gives C E D A.
#19 doesn’t really answer Austin’s question. Given that 10 is A, then if 9 is A also, then that makes 9 true – fair enough. But whether the answer choice makes the statement true is not enough (see 19 for an example!). There are other constraints. Here, another constraint on 9 is the answer to 2. If 2 isn’t D, then if 10 is A then 9 can’t be A – even though 9 itself would be true if it were. Right?
thanks for all answers
Pro Vision.
Great puzzle. Congrats who made and thanks for posting this.
for (14,18,19,20) I initially though of (B,A,B,A), but I agree with Dave M that
“if 14 is B, the solution is not unique, so 14 is C, which then gives C E D A.”
20 is the only question on this test for which asks for a real-world answer, and I firmly believe that the answer is A. Saying that a barometer tells you something about latitude or longitude is wrong. And when it comes to choosing between A and B, the answer is A because air pressure is a proxy for the weather (cloudy or not, which correlates to temperature not wind) just like the knowledge of math and grammar is a proxy for intelligence on the SAT.
PS the pro vision comment above is spam.
See also http://www.chaos.org.uk/~eddy/craft/srat-Q.html and http://www.maa.org/mathhorizons/Puzzles/Feb05-SRAT/original_puzzle_solution.htm; Dan’s answers (which seem largely undisputed, and which I agree with after the correction made by Dave) don’t match up too well with what is on those two pages.