Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these

Gmat_mission wrote: ↑

Fri Jan 08, 2021 4:23 am

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

When it comes to mixture questions, it's often useful to sketch the solutions with their components separated

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

----------------------------------------

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

Gmat_mission wrote: ↑

Fri Jan 08, 2021 4:23 am

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

Solution:

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