That is a very simple mathematics question.
2 K) P" R4 X o9 l# pLet x = Male, y = Female, z = Child- c% E0 e8 }% O T# c
6x + 3y + z/2 = 100 Formula 1: E+ L% B6 s0 C$ U
x + y + z = 100 Formula 2
+ x6 {% o$ a5 ^" ^2 z8 E) }9 u, x5 USimplify Formula 2 into z, we get z = 100 - x - y Formula 3tvb now,tvbnow,bttvb, q2 [! F# ^" Z( @; Y" o' f$ h* S
Sub Formula 3 into Formula 1, we get 6x + 3y + ( 100 - x - y )/2 = 100 Formula 4tvb now,tvbnow,bttvb, q5 y: V+ @% \+ L
Formula 4 can change into 11x + 5y = 100 Formula 5
: W9 C9 R- Y+ M# V1 pIn Formula 5, x can only be 0 - 9, otherwise y will be negative number which we dont want.公仔箱論壇" w( }- ~$ [5 Z& a% a q: D
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.tvb now,tvbnow,bttvb- o- C7 j _! O' o
Eventually, the answer is 20 female and 80 children and no male. |