algebra mixture problem?

Comments

• mitchonlinenl

  June 2021

  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

  June 2021

  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

  June 2021

  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

  June 2021

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

  11x+y=40

  x+y=20

  10x=20

  x=2

  2 gallons cream

  18 gallons milk