Question 684514
A pharmacist wants to prepare a 60liter mixture that is 40% boola using three different kinds of boola.
3 substances x + y + z = 60
:
The first concentration is 15% hula boola, the second is 35 % mula boola, and the third is 55 % gula boola.
.15x; .35y; .55z
:
 Because of the amounts of the boola solutions available, the pharmacist wants to use twice as much of the mula solution as the gula solution.
y = 2z
therefore
x + 2z + z = 60
x + 3z = 60
x = (60-3z)
:
If no chemical reactions among three different boolas take place, how much of each solution should 
:
.15x + .35y + .55z = .40(60)
:
x = (60-3z) and y = 2z
.15(60-3z) + .35(2z) + .55z = 24
9 - .45z + .70z + .55z = 24
.80z = 24 - 9
.80z = 15
z = 15/.8
z = 18.75 liters of g.b
then
x = 60-3(18.57)
x = 60 - 56.25
x = 3.75 liters of h.b
and
y = 2(18.75)
y = 37.5 liters of m.b.
:
:
see if this mess checks out
.15(3.75) + .35(37.5) + .55(18.75) = .40(60)
.5625 + 13.125 + 10.3125 = 24, it does!