That is a very simple mathematics question.* r1 {6 {8 j& F0 Y, b- Q7 K
Let x = Male, y = Female, z = Child
; h: A4 a- b/ ]" I6x + 3y + z/2 = 100 Formula 1tvb now,tvbnow,bttvb& n) {7 S/ Z! `, ?1 U
x + y + z = 100 Formula 2
# W% @* O& k M0 x8 L2 J: w" `os.tvboxnow.comSimplify Formula 2 into z, we get z = 100 - x - y Formula 3
1 S% D* w+ U$ b$ v8 |+ NSub Formula 3 into Formula 1, we get 6x + 3y + ( 100 - x - y )/2 = 100 Formula 4
3 M* i& T k# V6 h8 _* b& eFormula 4 can change into 11x + 5y = 100 Formula 5
' z& ~/ F8 D% I) t+ D2 Mos.tvboxnow.comIn Formula 5, x can only be 0 - 9, otherwise y will be negative number which we dont want.
5 g% V* b! m( Ztvb now,tvbnow,bttvbAnd 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.os.tvboxnow.com' P1 z$ Z* b. l; x9 I, ^: @) K7 x
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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