Carcass wrote:
How many gallons of milk that is 10 percent butter-fat must be added to 8 gallons of milk that is 35 percent butterfat to obtain milk that is 20 percent butterfat?
a) 6
b) 12
c) 14
d) 16
e) 28
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITIONGRE - Math BookOne 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 +
0.1xTOTAL 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