That is a very simple mathematics question. : r: Z% o2 f0 v; f& RLet x = Male, y = Female, z = Child! j% S& G* E& b4 ~. s; x5 m
6x + 3y + z/2 = 100 Formula 1 , m' g9 J6 s, S) \5.39.217.76x + y + z = 100 Formula 2tvb now,tvbnow,bttvb; K, O( t+ D% I' H+ u: ?
Simplify Formula 2 into z, we get z = 100 - x - y Formula 3 D/ U" T4 \% z, P8 ^tvb now,tvbnow,bttvbSub Formula 3 into Formula 1, we get 6x + 3y + ( 100 - x - y )/2 = 100 Formula 4 ( k& u& Y; t6 y6 e5.39.217.76Formula 4 can change into 11x + 5y = 100 Formula 5 9 Y' s2 u) X/ s% _In Formula 5, x can only be 0 - 9, otherwise y will be negative number which we dont want. 9 O$ [: b7 V& p8 [& O/ c1 m5 ^5 t$ e% pAnd 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 = 80.8 v9 ~9 P; k; l% i I" B
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
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11X must be 0 or multiple of 5公仔箱論壇3 u& F8 B) s0 L" [! S7 z3 L
that meanz x must be 0 or 5作者: manstein 時間: 2007-10-15 08:28 PM
