5th grade math word problem
posted by tryston on .
Santa was making up a batch of dolls but could not remember exactly how many of each kind he needed. He knew that he needed 57 in total that 27 had to have blue eyes and 29 had to have fair hair. Also some had to have both features and 3 dolls were needed having blue eyes and fair hair but were not able to say " I love math". Also santa needed a total of 34 dolls able to say " I love Math" of whom 17 needed to have fair hair. Every doll had at least 1 of sthe features named and there was one combination of features not asked for at all. Several children had asked for all 3 features in the one doll. How many fairhaired, blue eyed dolls saying "I love math" were needed.

help help help please

13. Here's why:
There are eight possible combinations, which I will refer to as A  H, as follows:
A. Blue, Fair, LoveMath
B. Blue, Fair, NotLoveMath
C. Blue, NotFair, LoveMath
D. Blue, NotFair, NotLoveMath
E. NotBlue, Fair, LoveMath
F. NotBlue, Fair, NotLove Math
G. NotBlue, NotFair, LoveMath
H. NotBlue, NotFair, NotLoveMath
Here are the clues:
1. He knew that he needed 57 in total,
2. that 27 had to have blue eyes
[2a. so 30 do not have blue eyes]
3. and 29 had to have fair hair.
4. His assistant gnome pointed out that some had to have both features
5. and remembered that 3 dolls were needed having blue eyes and fair hair, but which were not able to say 'I Love Math'.
6. he needed a total of 34 dolls able to say ''I Love Math',
7. of whom 17 needed to have fair hair [7a which means that 17 did not have fair hair].
8. "Every doll had at least 1 of the features named";
9. "That there was one combination of features not asked for at all"
10. "Several children had asked for all 3 features in the one doll".
From clue 5 you know B = 3
From clue 8 you know H = 0
From clue 3, A+B+E+F = 29. We already know B = 3, and from clue 7 A+E = 17. So F=9.
From clue 9 you know one of the remaining is 0. Let's guess that C=0 (you can try others, but this seems to work out)
Then from c7a, since C=0 that means G = 17.
From 2a, since E+F+G = 30 and already know F=9 and G=17, so E=4.
From 7, given E=4 then A = 13.
Finally from 2, since A+B+C+D = 30, then D=11.
So:
A = 13 = Blue, Fair, LoveMath 
2588963

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mommy? is that you? no!!!! its santa!