That is a very simple mathematics question.& ]# [2 d0 ]4 u, T! f
Let x = Male, y = Female, z = Child5.39.217.761 Q' t8 ^ B( L6 I
6x + 3y + z/2 = 100 Formula 1
! |% U, _* j% C* R5.39.217.76x + y + z = 100 Formula 2
: `: w, }1 ^6 m+ Q& Ltvb now,tvbnow,bttvbSimplify Formula 2 into z, we get z = 100 - x - y Formula 35.39.217.76' |* Y/ p& x1 y* J) S
Sub Formula 3 into Formula 1, we get 6x + 3y + ( 100 - x - y )/2 = 100 Formula 4
) S' L E4 T9 ~8 V( v# H jFormula 4 can change into 11x + 5y = 100 Formula 5( q7 F' T ~! J. W8 b
In Formula 5, x can only be 0 - 9, otherwise y will be negative number which we dont want.* u) V2 Z* P" ~- F
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.
- C% g4 ]- s/ P) [. p$ v, tEventually, the answer is 20 female and 80 children and no male. |