algebra mixture problem?

anonymous

"Cream is approximately 22% butterfat. How many gallons of cream must be mixed w/ milk testing at 2% butterfat to get 20 gallons of milk containing 4% butterfat?"

help? thank you

mitchonlinenl

c = amount of cream

m = amount of milk

20 gallons =>

c + m = 20

or m = 20 - c

20 gallons of 4% fat corresponds to 0.8 gallons of fat

=> 0.22c + 0.02m = 0.8

with m = 20 - c:

=> 0.22c + 0.02(20-c) =0.8

=> 0.22c + 0.4 - 0.02c = 0.8

=> 0.2c=0.4

=> c = 2

you need 2 gallons of cream

1ofselbyx27s

let x = gal of cream and y = gal of milk

then 22/100 x + 2/100 y = 4/100 (20)

or 22X + 2Y = 80 and 11 X + Y = 40 (1)

also x + y = 20 (2)

subtract (2) from (1)

10x = 20 or x = 2 therefore y = 18

we need 2 gal at 22% and 18 gal at 2% to get 20 gal at 4%

computerguy103

Create an equation:

x*0.22 + y*0.02 = 20*0.04

x = gallons of cream

y = gallons of milk

Now, since x + y = 20, we can replace y with 20 - x:

x*0.22 + (20 - x)*0.02 = 20*0.04

Now simplify and solve it:

11x + 20 - x = 40

10x = 20

x = 20/10

x = 2

y = 20 - x = 18

You should mix 2 gallons of cream with 18 gallons of 2% milk.

bob-h

.22x+.02y=.04*20=.8

11x+y=40

x+y=20

10x=20

x=2

2 gallons cream

18 gallons milk