“We – A, B and C – each have some children.
(i) A has at least one girl and twice as many boys as girls.
(ii) B has at least one girl and three times as many boys as girls.
(iii) C has at least one girl and three more boys than girls.
(iv) When I tell you the number of children we have – a number less than 25 – you will know how many children I have, but not how many children each of the others has. Altogether we have……..”
Who is the speaker and how many children does the speaker have?
(In reply to
My answer by Jim McCrank)
C cannot have 5 children. C must have 1+n girls and 1+n+3 boys for a minimum of 5 children. 1+n+1+n+3 = 5+2n. n = 0,1,2,...
I believe the answer is either a total of 16 or 19 children. A is the speaker and has 1 girl (16 total children) or 2 girls (19 total children). B and C have either 1 and 3 or 2 and 1 girls respectivly.
16 - A(1+2) + B(1+3) + C(3+6) = 3+4+9 = 16
16 - A(1+2) + B(2+6) + C(1+4) = 3+8+5 = 16
19 - A(2+4) + B(1+3) + C(3+6) = 6+4+9 = 19
19 - A(2+4) + B(2+6) + C(1+4) = 6+8+5 = 19
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Posted by Tony
on 2003-01-28 09:09:16 |
re: My answer
