That is a very simple mathematics question. & J7 d4 Q1 T# R qLet x = Male, y = Female, z = Child ! b$ M1 \5 t0 g* c3 _2 D% otvb now,tvbnow,bttvb6x + 3y + z/2 = 100 Formula 1 * c7 O _/ I% W( v) C5 i' o8 Kx + y + z = 100 Formula 29 Y# ]7 H C, u, A- D
Simplify Formula 2 into z, we get z = 100 - x - y Formula 3, O% |! L% g4 C9 y% f8 j
Sub Formula 3 into Formula 1, we get 6x + 3y + ( 100 - x - y )/2 = 100 Formula 4) m5 x+ ]0 M1 I6 c, o
Formula 4 can change into 11x + 5y = 100 Formula 5 ! K, E! z1 |+ V" w) OIn Formula 5, x can only be 0 - 9, otherwise y will be negative number which we dont want.公仔箱論壇% P1 B+ V& f! g+ D5 s
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.+ }, a! I$ R/ W: f- [
Eventually, the answer is 20 female and 80 children and no male.作者: alan1234 時間: 2007-10-14 12:19 AM 標題: two answers
原帖由 twinsdragon 於 2007-10-13 05:40 PM 發表 / C9 J9 w m$ p- m5 B5 ] S8 Q. f
11x + 5y = 100 Formula 5
" X, X3 r1 l' {- p) E4 ?5.39.217.7611X must be 0 or multiple of 5tvb now,tvbnow,bttvb" x& R9 H! ~; [( Z& b/ [
that meanz x must be 0 or 5作者: manstein 時間: 2007-10-15 08:28 PM