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
Answer: E
Source: Official Guide
Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these
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When it comes to mixture questions, it's often useful to sketch the solutions with their components separatedGmat_mission wrote: ↑Fri Jan 08, 2021 4:23 amJackie 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
Answer: E
Source: Official Guide
Since the question asked us to find the number of liters of 2% solution needed, let's let x = number of liters of 2% solution needed
2% of x = 0.02x
So, the initial solution contains, 0.02x liters of pure acid.
Since we want a final total of 60 liters, we need to now add 60-x liters of 12% solution.
12% of (60 - x) = 0.12(60 - x) = 7.2 - 0.12x
To find the volume of pure acid in the resulting solution, we'll add the acid from each solution
Total volume of acid = 0.02x + 7.2 - 0.12x = 7.2 - 0.1x
So, the resulting solution has a total of (7.2 - 0.1x) liters of acid
The NEW solution is 5% PURE acid.
So, we can write: (7.2 - 0.1x)/60 = 5/100
Cross 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
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If you don't want to sketch the solutions, another approach is to just keep track of the acid
Let x = number of liters of 2% solution needed
So, 60 - x = number of liters of 12% solution needed
2% of x = 0.02x
So, 0.02x = the number of liters of PURE acid in the 2% solution
12% of 60 - x = 0.12(60 - x) = 7.2 - 0.12x
So, 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 = 5/100
Cross 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
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Solution:Gmat_mission wrote: ↑Fri Jan 08, 2021 4:23 amJackie 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
Answer: E
Source: Official Guide
We can let the amount of 2% sulfuric acid solution = x and the amount of 12% sulfuric acid solution = y. Thus:
x + y = 60
y = 60 - x
and
0.02x + 0.12y = 0.05(x + y)
2x + 12y = 5x + 5y
7y = 3x
Thus:
7(60 - x) = 3x
420 - 7x = 3x
420 = 10x
42 = x
Alternate Solution:
We will mix x liters of 2% sulfuric acid with (60 – x) liters of 12% sulfuric acid to produce 60 liters of 5% sulfuric acid. We can create an equation from this information and solve for x:
0.02x + 0.12(60 – x) = (0.05)(60)
0.02x + 7.2 – 0.12x = 3
-0.10x = -4.2
x = 42
Answer: E
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