SOLUTION: 4/7 of the school's faculty are women; 4/5 of the male faculty are married; and 9 of the male faculty are unmarried. How many faculty members are there?
Question 178921: 4/7 of the school's faculty are women; 4/5 of the male faculty are married; and 9 of the male faculty are unmarried. How many faculty members are there? Found 2 solutions by ptaylor, ankor@dixie-net.com:Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=number of faculty members
Then (4/7)x=number of women
And (3/7)x=number of males
Now we are told that:
(4/5)*(3/7)x=married males
9=unmarried males
married males + unmarried males=total males
(4/5)*(3/7)x +9=(3/7)x or
(12/35)x+9=(3/7)x multiply each term by 35
12x+315=15x subtract 12x from each side
12x-12x+315=15x-12x collect like terms
3x=315 divide each side by 3
x=105-----------number of faculty members
CK
(3/7)(105)=45
(4/5)*45=36
36+9=45
45=45
You can put this solution on YOUR website! 4/7 of the school's faculty are women; 4/5 of the male faculty are married; and 9 of the male faculty are unmarried. How many faculty members are there?
:
let x = no. members
:
if of the males are married, then are unmarried and = 9
therefore total males = 5*9 = 45
:
x = x + 45
multiply equation by 7 to get rid of the denominator (what was I thinking?!)
7x = 7*x + 7(45)
cancel denominators
7x = 4x + 315
:
7x - 4x = 315
:
3x = 315
x =
x = 105 members
:
:
Check solution: married males: 45-9 = 36 105 + 36 + 9 =
60 = 36 + 9 = 105
