Quiz About This Quiz

Answer: 144

Since many of these answers depend on each other, I'm just going to step through a solution independent of the order of the questions, using notation of A..J for the answers to questions 1..10 (Q1..Q10). The first thing to notice is that Q1 boils down to the formula: A + everything else = A; hence the sum of everything else is zero.

Answer: 2

Next, look at Q4. Since it's a count that includes itself, D must be at least one. Since it's included in the zero-sum mentioned above, that means something else must be negative, which gives us our first answer - E is False. This narrows down B a little - there are 4 True-False questions, plus a possibility that J is True, so B can now only be from 0-4.

Answer: True

Skip ahead a moment to Q8. It tells us that A is a square, and that it must be greater than zero (otherwise H would be undefined; hence not an integer). This also tells us that the average F must also be positive. Since we're getting into averages, I should also mention at this point that there are either 5 or 6 numerical answers, depending on what J is.

Answer: 2

Since F is an integer, A must be a multiple of the number of numerical answers, which is either 5 or 6. I'll call that number N. We know A is both a square and a multiple of N, so for some positive integer K, (KN)*(KN) = A. We also get H = (KN) or -(KN), and F = KN*K. Therefore the sum of all known numerics except A, is KN*K+B+D+H, which must equal zero...

Answer: False

Now we already know F is positive, B is at least zero, and D is at least one, so if J is not numeric, then H must be able to offset at least F+1 in order to get the zero sum; hence it would have to be negative. Plugging into the formula above, we get KN*K+1-KN=0, or KN(K-1)+1=0, or KN(K-1)=-1. We know K is greater than 0, and N is 5 or 6, so this cannot be a {solution;} hence H cannot bring the sum to zero.

Answer: 24

Since H alone cannot bring the zero-sum down to zero, J must be numeric, and furthermore, must be negative; hence N=6. We still don't know if H is negative, but now we have enough information to get C ... since A=F*6, and A is greater than 0, A must be at least 36 (smallest square divisible by 6), making it higher than any other answer could possibly be; hence C is True.

Answer: False

We can narrow down B further now - we've shown J cannot be True, and C is True, so B is at least 1 and at most 3. Now to look at D - there are 6 numerics, and at least one is negative (J), so D is at most 5. However, A is at least 36, and so F and H are each at least 6 (if H is positive), so D can't match any of {them;} this takes D down to two possibilities: 1 or 2.

Answer: -12

Since the only possible number D can match is B, and B must be in 1-3, we can break this down into cases. If D is 1, B must be greater than one in order not to match. If D is 2, then B must also be 2 because it must match. In both cases, B is at least two, so it cannot be less than D; hence G is False. Since two True-False questions have been False, B must be 2, so the remaining T-F (I) must be True.

Answer: True

Q9 gives us the formula F=(B-D)-HD (or F=(D-B)-HD, but as we'll see, it doesn't matter). We know D is 1 or 2, so first try D=1: F=1-H (or F=-1-H), but going back we know for some positive K, F=K(KN), and H=KN or -KN, so F=KH or -KH. Substituting, we get (-)KH=(-)1-H, or H((-)K+1)=(-)1, which would require H=-1 or 1, contradicting the fact that H is a multiple of 6.

Answer: -16

By elimination, D=2, which reduces Q9 to F=-2H. Since F=(-)KH with K positive, we finally have K=2, which quickly gives us A=(KN)*(KN)=144. FN=A yields F=24; and F=-2H gives us H=-12. Finally, with all the other numbers in place, we simply solve the original zero-sum equation: B+D+F+H+J=0, or 2+2+24-12+J=0. J=-16.