That is a very simple mathematics question.
. m; A% p) p z: _+ B/ bLet x = Male, y = Female, z = Child% i# U- f+ x/ h# q
6x + 3y + z/2 = 100 Formula 1
; G* h3 R) a4 p, T8 s0 ~5.39.217.76x + y + z = 100 Formula 2
; }3 p) y" _- ?2 D0 {) p" ?1 [" f3 mSimplify Formula 2 into z, we get z = 100 - x - y Formula 3公仔箱論壇& _& E3 _0 e* D% [$ O2 M' D$ x' I
Sub Formula 3 into Formula 1, we get 6x + 3y + ( 100 - x - y )/2 = 100 Formula 4# o: R2 Q' Z7 y2 E
Formula 4 can change into 11x + 5y = 100 Formula 5公仔箱論壇; o( [6 f1 B# b( R2 S6 d6 c. r
In Formula 5, x can only be 0 - 9, otherwise y will be negative number which we dont want.
e- a7 H6 w7 k; L5.39.217.76And 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.
3 L& o0 c2 v$ q5.39.217.76Eventually, the answer is 20 female and 80 children and no male. |
發表於 2007-10-13 05:08 PM
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