Wednesday

August 20, 2014

August 20, 2014

Posted by **Reiny** on Wednesday, September 16, 2009 at 8:45am.

let the number of men be x

the number of women be y

and the number of children be (100-x-y)

then 3x + 2y + (1/2)(100-x-y) = 100

6x + 4y + 100 - x - y = 200

5x + 3y = 100

this is a linear relations where one obvious solution is (20,0)

the 'slope' is -5/3, that is,

for every decrease of 3 in the x's we can increase y by 5 and we have another solution.

so other solutions are

(17,5)

(14,10)

(11,15) etc.

of course we have to check if the number of children is still valid.

e.g. for (11,15)

we would have 11 men, 15 women and 100-11-15 or 74 children

check: 11(3) + 15(2) + (1/2)(74) = 100 YEAH!

keep my pattern going for the rest of the results.

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