twinsdragon

That is a very simple mathematics question.
3 b( [! ?! `/ u) l/ y% m公仔箱論壇Let x = Male, y = Female, z = Child: [1 F. _4 W$ j
6x + 3y + z/2 = 100   Formula 13 d; L9 r# ]: @. h! \
x + y + z = 100    Formula 2
% ~! ~# `( }' Qtvb now,tvbnow,bttvbSimplify Formula 2 into z, we get z = 100 - x - y Formula 3
, `' r: L: l% b, j& ztvb now,tvbnow,bttvbSub Formula 3 into Formula 1, we get 6x + 3y + ( 100 - x - y )/2 = 100 Formula 4tvb now,tvbnow,bttvb1 V$ n1 v* ?1 Y3 v) @+ |
Formula 4 can change into 11x + 5y = 100 Formula 5, g) l* c: \0 C" O2 L+ W( K
In Formula 5, x can only be 0 - 9, otherwise y will be negative number which we dont want.
9 U  p" E) f; q3 s! c9 W$ }# c2 Y公仔箱論壇And 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., u) {$ T$ {/ g  G9 A5 {
Eventually, the answer is 20 female and 80 children and no male.