Carcass wrote:
Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required?
A) 18
B) 20
C) 24
D) 36
E) 42
Another approach is to
keep track of the acidLet
x = number of liters of 2% solution needed
So,
60 - x = number of liters of 12% solution needed
2% of
x =
0.02xSo,
0.02x = the number of liters of PURE acid in the 2% solution
12% of
60 - x =
0.12(60 - x) =
7.2 - 0.12xSo,
7.2 - 0.12x = the number of liters of PURE acid in the 12% solution
Now let's COMBINE the two solutions.
Total volume of PURE acid =
0.02x +
7.2 - 0.12x= 7.2 - 0.1x
So, our NEW solution contains 7.2 - 0.1x liters of PURE acid
Also, the NEW solution has a total volume of 60 liters
Since the NEW solution is
5% PURE acid, we can write: (7.2 - 0.1x)/60 =
5/100Cross multiply to get: 100(7.2 - 0.1x) = 5(60)
Expand: 720 - 10x = 300
Add 10 x to both sides: 720 = 300 + 10x
Subtract 300 from both sides: 420 = 10x
Solve: x = 42
Answer: E