That is a very simple mathematics question.
5 L1 j3 `3 ~5 G$ T4 n; F h: m2 c! bLet x = Male, y = Female, z = Child+ Z K! V) |/ G" }' v
6x + 3y + z/2 = 100 Formula 1tvb now,tvbnow,bttvb: ]8 N, V- g* t# W* ?7 z2 m/ Y
x + y + z = 100 Formula 2tvb now,tvbnow,bttvb0 h8 A' n0 v. e9 D$ o+ y
Simplify Formula 2 into z, we get z = 100 - x - y Formula 35.39.217.76( v4 B( r6 f* a( x+ X* P. y( m9 ]' D
Sub Formula 3 into Formula 1, we get 6x + 3y + ( 100 - x - y )/2 = 100 Formula 4
, s6 R% b1 K9 k4 c* B% T. M& t公仔箱論壇Formula 4 can change into 11x + 5y = 100 Formula 5, V$ t) S2 q; J9 j1 U- v
In Formula 5, x can only be 0 - 9, otherwise y will be negative number which we dont want.
- [) W4 [0 u. x/ ~' ]8 n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.* e: J- b6 ^% ]8 ?# `- K6 L
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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