## 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 |