SOLUTION: You have 2 drums of milk, one contains 16% butterfat, the other contains 48% butterfat. You need 40 gallons of milk at 30% butterfat. How many gallons of each do you have to mix?
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Question 287930: You have 2 drums of milk, one contains 16% butterfat, the other contains 48% butterfat. You need 40 gallons of milk at 30% butterfat. How many gallons of each do you have to mix?
Thank you for your time.
Regards,
Beth Found 2 solutions by Grinnell, ankor@dixie-net.com:Answer by Grinnell(63) (Show Source):
You can put this solution on YOUR website! Let x be number of gallons at 16%
let 40-x be number of gallons at 48%
You want to end up with 40 gallons at 30%
Sooo, off to the races!
x(.16) + (40-x)(.48)=40(.30)
(let's multiply by 100 to remove the decimal)
16x+48(40-x)=(40)30
16x+1920-48x=1200 (do the math)
16x+1920-48x=1200
-32x=-720
x=-720/-32
x=22.5 (I said 720/32=27.5---I caught my simple dividing mistake because I always double
check.)
Sooo 22.5 gallons of 16% butterfat and
17.5 gallons of 48% butterfat must be mixed to yield 40 gallons of 30% butterfat! I have to run. Always check your work. Plug in the numbers and see if they work!!!! Later...(I divided wrong at first--I said 4-4=1 (DUuu)---This threw the whole answer off! I caught my mistake because I re-checked AND double checked. Be careful in the simple arithmetic. WE ALL MAKE MISTAKES!!!!!!!
You can put this solution on YOUR website! You have 2 drums of milk, one contains 16% butterfat, the other contains 48% butterfat.
You need 40 gallons of milk at 30% butterfat.
How many gallons of each do you have to mix?
:
Let x = amt of 48% milk
the result is to be 40 gal, therefore:
(40-x) = amt of 16% milk
:
At typical mixture equation using decimal equiv of percent:
.48x + .16(40-x) = .30(40)
:
.48x + 6.4 - .16x = 12
:
.48x - .16x = 12 - 6.4
:
.32x = 5.6
x =
x = 17.5 gal of 48% milk
and
40 - 17.5 = 22.5 gal of 16% milk
;
:
Check solution in original equation
.48(17.5) + .16(22.5) = .30(40)
8.4 + 3.6 = 12; confirms our solutions
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