On the road again
Today is my last long drive for a while. I’ll try to post a Sunday Sample tonight, or maybe tomorrow. But for now, here’s a problem to chew on.
In a multiple choice test, one question was illegible, but the following choice of answers was clearly printed. Which must be the right answer?
- All of the below
- None of the below
- All of the above
- One of the above
- None of the above
- None of the above
Option 5: None of the above
Comment by ashwin | August 24, 2008 |
I got the same answer, but I’m willing to show my work…
Option 6 (none of the above) being true would mean options 1-5 are false, but since then options 1-4 are false, option 5 (none of the above) would be true, forming a contradiction. Therefore, Option 6 is false.
Since option 6 is false, option 1 (all of the below) is false, meaning option 3 (all of the above) is false.
That leaves options 2, 4, and 5. If option 2 is true (none of the below), that would imply that 4 (one of the above) is false (by 2) and 4 is true (since only 2 above 4 is true), a contradiction. Therefore, 2 is false. Since 1, 2, and 3 are all false, 4 (one of the above) must therefore be false.
What’s left is 5 (none of the above). Since 1, 2, 3, and 4 are known to be false, that means 5 is true. 6 is also known to be false.
Therefore, the only consistent answer is Option 5: None of the above.
Comment by blaisepascal | August 24, 2008 |
Option 5, because most multiple choice tests use two similar answers, where one is a “distractor”. The two most similar answer options are 5 and 6, but as stated above, 6 contradicts itself. No need to even consider options 1-4 in that context (though blaisepascal’s reasoning is sound)
Comment by Bobby Isosceles | August 24, 2008 |
Also option 5, by a much easier argument:
We assume that there is a unique true option. It cannot be option 4, because then one of options 1, 2, and 3 would be true, contradicting uniqueness. So 4 is false; so 1, 2, and 3 are all false; so 5 is true.
Comment by Chad Groft | August 25, 2008 |
Why would you ever make such an assumption? It’s not even valid for real multiple-choice tests, as is shown by the ubiquitous “all of the above” response.
Comment by John Armstrong | August 25, 2008 |
Chad wrote:
John wrote:
Because in posing your puzzle, you said: “Which must be the right answer?”
Comment by John Baez | August 25, 2008 |
Similar language (“fill in the bubble corresponding to the correct answer”) is in the instructions to the SAT, and it also includes “all of the above” answers. I doubt I’d ever have met Chad if he took the definite article that literally.
Comment by John Armstrong | August 25, 2008 |
With a test-designer this devious, we can’t be sure the illegible question wasn’t “Which of the following is false?”…
Comment by Sridhar | August 25, 2008 |
That’s a good point, Sridhar. What does the problem look like in that case?
Comment by John Armstrong | August 25, 2008 |
[…] might recognize a lot of the other physics blogs featured there. I also found this gem while I was browsing around their website, so I thought it was interesting enough to send you guys […]
Pingback by Blog aggregators? « Shores of the Dirac Sea | December 21, 2008 |
Apologies in advance for the proof…
If we take this all as Boolean logic the rules boil down to:
(. == AND, + == OR, ! == NOT, 1,2,3,4,5,6 => A,B,C,D,E,F)
A: B.C.D.E.F
B: !C.!D.!E.!F
C: A.B
D: A!B!C + B!A!C + C!A!B
E: !A.!B.!C.!D
F: !A.!B.!C.!D.!E
So if we start with rule C (#3)
C = BCDEF . !C!D!E!F –Substitution
= 0 –Complementation
Thus, C must be false.
And by subsitution A(#1) must also be false (A = B.0.D.E.F = 0)
Furthermore, D(#4) is as follows:
D = 0.!B.!0 + B.!0.!0 + 0.!0.!B
= 0 + B + 0
D = B
Next E(#5):
E = !A.!B.!C.!D
= !0.!B.!0.!B –Substitution
= 1.!B.1.!B –Idempotence
= !B
Then F(#6):
F = !A.!B.!C.!D.!E
= 1.!B.1.!B.B
= B.!B –Complementation, Idempotence
= 0
Thus F must be false.
Finally, B(#2):
B = !C.!D.!E.!F
= 1 .!B.!!B.1 –Complementation, Double Negation
= 0
Thus B must be false.
And now we have all the pieces to put the rest together:
A = 0
B = 0
C = 0
D = B = 0
E = !B = 1
F = 0
And thus in the negative (by duality):
A = 1
B = 1
C = 1
D = 1
E = 0 –E Must be false.
F = 1
Therefore, E is the only correct answer.
Q.E.D.
Comment by Nik Hodgkinson | December 21, 2008 |
@Sridhar, @John Armstrong: The question cannot be “Which of the following is false?”:
1. Only one is false and all the others are true. (Given.)
2. Either 5 or 6 is true. (From 1.)
3. All “of the above” (1-4) are false. (From 2.)
4. 1 contradicts 3: 1-4 are false, but only one can be false.
Therefore, that question is impossible.
Comment by Tim McCormack | December 21, 2008 |
[…] Wow, people are loving my zero-knowledge test. It got 1,743 views yesterday, thanks to someone redditing it. […]
Pingback by Symmetric Tensors « The Unapologetic Mathematician | December 22, 2008 |
Try this one then:
http://surfacedepth.blogspot.com/2008/12/time-to-test-your-wit.html
Comment by creating2009 | December 23, 2008 |
I chose option 5, and knew I was right, without having to proof anything for anyone.
Ahh, the “high ground” I keep hearing about. Yep, the view is nice.
Comment by dziban | January 9, 2009 |