AAPL wrote:How many gallons of milk that is 10 percent butter-fat must be added to 8 gallons of milk that is 35 percent butter-fat to obtain milk that is 20 percent butter-fat?

A. 6
B. 12
C. 14
D. 16
E. 28

One option is to focus on the butter fat

We begin with 8 gallons of milk that is 35 percent butter-fat
So, the volume of butter fat = 35% of 8
= (0.35)(8)
= 2.8 gallons of butter fat (to begin)

Let x = number of gallons of milk that is 10 percent butter-fat to be added to the initial 8 gallons
The volume of butter fat = 10% of x
= 0.1x gallons of butter fat (to be added)

So, in the RESULTING mixture, we have:
Total volume of butter fat = 2.8 + 0.1x
TOTAL VOLUME of mixture = 8 + x

We want the RESULTING mixture to be 20 percent butter-fat
In other words, we want: (Total volume of butter fat)/(TOTAL VOLUME of mixture) = 20%
We can now write: (2.8 + 0.1x)/(8 + x) = 20/100
Cross multiply to get: 100(2.8 + 0.1x) = 20(8 + x)
Expand: 280 + 10x = 160 + 20x
Rearrange: 120 = 10x
Solve: x = 12

Answer: B

Cheers,
Brent