That is a very simple mathematics question.
- G0 ]6 \4 } m' `0 k ^5.39.217.76Let x = Male, y = Female, z = Childtvb now,tvbnow,bttvb- D5 v- [* M! ]+ d2 T3 m1 ?. `
6x + 3y + z/2 = 100 Formula 1
. x- m- v4 ]* I; O( ftvb now,tvbnow,bttvbx + y + z = 100 Formula 25.39.217.76' p, h; Y$ z1 I
Simplify Formula 2 into z, we get z = 100 - x - y Formula 35.39.217.76& g7 L. o7 w/ z9 c9 p% M
Sub Formula 3 into Formula 1, we get 6x + 3y + ( 100 - x - y )/2 = 100 Formula 4 I K& o, O2 G; e9 g
Formula 4 can change into 11x + 5y = 100 Formula 53 @2 G& X. O# ~
In Formula 5, x can only be 0 - 9, otherwise y will be negative number which we dont want.
5 `$ x* V2 \/ {( g* Q+ T3 d; _5.39.217.76And between 0 and 9, only 0 can give me a single number of y. So, x has to be 0 and y have to be 20. Then put x and y back to formula 1, we get z = 80.tvb now,tvbnow,bttvb7 E& |/ N" n' Y
Eventually, the answer is 20 female and 80 children and no male. |
發表於 2007-10-13 05:08 PM
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