please empty your brain below 
Yesterday's answers
6) JIGSAW 7) Larry 8) APRIL EIGHTEENTH, SEPTEMBER TWENTYTHIRD 9) 32 10) 140 
Question 7 contradicts itself, it isn't possible for each boy simultaneously to have a different number of answers correct AND for for them to range from all right to all wrong.
Two got some answers right, one got them all wrong and one got them all right. 
Do you not consider zero a number?

@0729 commenter.
Q7 is entirely consistent: three question, four possible scores (0,1,2,3) We are told one boy got all the answers right, so one set of answers is right. Try each set in turn and we get the following scores: Barry all correct: 3, 0, 2, 2 Garry all correct: 0, 3, 1, 0 Harry all correct: 2, 1, 3, 1 Larry all correct: 2, 0, 1, 3 So Larry is the correct answer as it is the only set that results in each boy having a different score. Q10 wasn't difficult, was it? If all the games are drawn that's 760 points. 1000760 = 240 so there must be 240 threepointers. The balance is 140. 
I miscalculated Q10 by assigning 1 point for lost matches (instead of 0); but I really found only 28 squares in Q9: 1 BIG square, 9 medium squares, 12 small squares, 4 squares of 2x2 medium squares, and finally 2 with bisectors make up 1.5x1.5 medium squares. So where are the remaining 4?

IMHO the correct answer to Q10 is 70. This makes it 310 games won and 70 games drawn.
310*3 + 70*1 = 1000 
D'oh... Found my mistake...

08:11 commentator and timbo  clearly you have access to 'invisible' rules that I'm unaware of.
I repeat the wording 'EACH boy had a different number of answers CORRECT'  so how can any of them score zero if they all got at least one answer correct? The wording should be 'Each of the four boys got a different score in the quiz, they ranged from all correct answers to all wrong answers, which boy answered all of the questions correctly?' By the way  why is it four boys, as opposed to girls, or children? 
@Patrickov: There are 13 2x2 squares: the nine obvious ones, plus four less obvious. Dividing the big square notionally into 6x6 and counting from the left and top as (1,1), the 2x2 squares have as their top left corners
(1,1) (2,1)* (3,1) (5,1) (1,2)* (1,3) (3,3) (4,3)* (5,3) (3,4)* (5,1) (5,3) (5,5) I suspect you have overlooked the four I have starred 
@Still anon
If there were only three children, it would have been possible for them all to get at least one correct. But with four, there have to be four different numbers. The number of answers Garry got correct was zero. Why were they boys? Probably because the question setter couldn't think of four girls' names that rhyme. Or, maybe, that it is implausible that any group of girls would do so badly on such a test(!) 
@Noname: Damn it's hard!

Blast.
0 
still anon seems to believe that 'EACH boy had a different number of answers CORRECT' implies 'they all got at least one answer correct'. The special rule to which the rest of us have access is, as an earlier commentor said, 'zero is a number'.
If DG had provided us with dozens of puzzles involving schoolchildren, and all those children had been declared male, there might have been a case to answer sexismwise. But one single question is not really a convincing indication of anything. 
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