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 66
Place the digits from 1 to 9 in each closed area so that the sum of the digits in each large square is the same.
A farmer buys chicken feed for 47c per tonne, pig feed for $1.23 per tonne and cattle feed for $1.44 per tonne. The feed can only be purchased by the tonne, and part tonnes aren't sold.
How much of each would they have if they bought 100 tonnes for $100?
At a recent painting competition, the results were mislaid. Luckily, I remembered the following:
Emery's rendition of the Constable painting was not in last place. The person who painted the Monet was very successful and took first place. Glen only just managed to avoid last place and came third. Alex beat the person who painted the Taylor, and the person who painted the Van Gogh beat Drew.
Can you determine who painted what, and who won?
Answer # Name Artist
1 Alex Monet
2 Emery Constable
3 Glen Van Gogh
4 Drew Taylor Reasoning
The four entrants were: Alex, Drew, Emery, Glen.
The four artists were: Constable, Monet, Taylor, Van Gogh.
By (2), the Monet came first.
1 Monet
2
3
4
By (3), Glen came third.
1 Monet
2
3 Glen
4
By (1), the Constable can't have been last, so must have been second (as Emery painted it, and not Glen).
1 Monet
2 Constable
3 Glen
4
The Taylor was either third or fourth, either way, by (4) Alex must have won.
1 Alex Monet
2 Constable
3 Glen
4
By (4), the Van Gogh came third, leaving Drew in last place.
1 Alex Monet
2 Constable
3 Glen Van Gogh
4 Drew
Leaving Emery in second place, and the Taylor in last place.
1 Alex Monet
2 Emery Constable
3 Glen Van Gogh
4 Drew Taylor
Note: BrainBashers has a Dark Mode option. For BrainBashers, I'd recommend not using your browser's built-in dark mode, or any dark mode extensions (sometimes you can add an exception for a specific website).
