Here is a snippet of section A of the curious multiple-choice entrance exam into the exclusive BrainBashers puzzle club.
Q1. The first question with B as the correct answer is:
A. Q1
B. Q4
C. Q3
D. Q2
Q2. The answer to Q4 is:
A. D
B. A
C. B
D. C
Q3. The answer to Q1 is:
A. D
B. C
C. B
D. A
Q4. The number of questions that have D as the correct answer is:
A. 3
B. 2
C. 1
D. 0
Q5. The number of questions that have B as the correct answer is:
Answers
Q1. C (the first question with B as the correct answer is Q3)
Q2. D (the answer to Question 4 is C)
Q3. B (the answer to Question 1 is C)
Q4. C (the number of questions that have D as the correct answer is one)
Q5. B (the number of questions that have B as the correct answer is two)
Reasoning
A complicated, and sometimes, confusing answer!
Can Q1 be A? No, because this would tell us that Q1 was the first question with B as the answer, which would be a contradiction (as we've just assumed Q1 was A).
Can Q1 be B? No, because this would tell us that Q4 was the first question with B as the answer, which would be a contradiction (as we've just assumed it was Q1).
Can Q1 be C? Possibly, because Q3 points back to Q1 correctly, and is logically consistent.
Can Q1 be D? No, because:
Q2 would be B Q4 would be A, which means that we have three questions with D which means that Q3 would have to be D, and this would tell us that Q1 is A, which would be a contradiction (as we've just assumed that Q1 is D).
Therefore, Q1 is C, which means that Q3 is B.
We can ignore Q2 for a moment, as it asks us about Q4, and look at Q4 first.
Looking at Q4 (how many questions have D as the answer) it can't be D (zero), as this would contradict itself.
It can't be A (three) as we only have two other questions without an answer.
If Q4 was B, then the remaining questions (Q2 and Q5) would both have to be D, and:
Q5 being D would mean that only one question was B, which would be a contradiction (as Q3 is already B, and we've just assumed that Q4 is also B).
So Q4 must be C, which means that Q2 is D.
Looking at Q5, it can't be A (as Q3 is B), it can't be D (as Q4 tells us that we only have one question with D as the answer, Q2). It can't be C as we don't have three questions that are B. So Q5 is B (the two questions are Q3 and Q5).
??
Puzzle 131
Using all of the letters A to Z, each once only, complete these common words. -e-er--eue--oma--p-a-e--erso----k-am-on-ouse-a--ur----igent
Answer -e-er = fever (FV)--eue = queue (QU)--o = zoo (ZO)ma- = mat (T)-p-a-e- = speaker (SEKR)-erso- = person (PN)---k-am-on = backgammon (BACGM)-ouse = house (H)-a- = wax (WX)-ur- = jury (JY)---igent = diligent (DIL)
This puzzle isn't guaranteed to have a unique answer, can you find another?
??
Puzzle 132
Below you can find 8 words that have had their beginnings and endings removed. In each case, the same two letters can be found at the beginning and the end. For example REspiRE. --qui----yli----tat----gib----liv----rmi----ifi----alg--
