F2 Math question

Let s be the number of workers to make sails
and h be the number of workers to make holders.

"there are 10 workers" => s + h = 10 ---------- (1)
"the procedures in making the sail and holders can be finished at the same time"
=> number of sails made per min = number of holders made per min
=> (s/3) = (h/2) ------------------------------------ (2)

From (2) : s = (3/2)h ----------------------------- (3)
Sub. (3) into (1) : (3/2)h + h = 10
_________________3h + 2h = 20
______________________h = 4 --------------- (4)
Sub. (4) into (1) : s + 4 = 10 => s = 6

Let x=the workers can make a sail in 3min

y=the workers can make a holder in 2 min

the 2 equations

x+y=10─(1)《10 workers》

x/3=y/2─(2)《each worker can make( 1/3 )sails in( 1 )min ,〞x 〝workers can make( x/3) s

ails in( x )min ;the same reasons,〞y〝workers can make( y/3) holders in( y)min ;so,〞x/3=

y/2〝→produce the sail and the holder are the 2 main procedures to produce 〞the umberlla

s〝》

(1)→y=10-x

(2)→2x=3y

2x=3(10-x)
2x=30-3x
5x=30
x=6
y=10-6=4

A: 6 workers should make sails and 4 workers should make the holders

to assume that both workers finish in 6mins
and y = # of sail and holder they made
and x = #of workers are making sail
so that 10-x = #of workers are making holder

first equation is 6x/3 = y
second equation is (10-x)6/2 = y

then 6x/3 = (10-x)6/2
2x = (10-x)3
2x = 30-3x
2x+3x = 30
5x = 30
x = 6
so,there have 6workers making sails and (10-x ) = (10-6) = 4workers making holder

2009-02-26 11:58:51 補充：
is it clear working steps?