Solution X, which is 50% alcohol, is combined with solution

[swerve]

Solution X, which is 50% alcohol, is combined with solution Y, which is 30% alcohol, to form 16 liters of a new solution that is 35% alcohol. How much of solution Y is used?

A. 4 liters
B. 6 liters
C. 8 liters
D. 10 liters
E. 12 liters

The OA is E.

Source: Veritas Prep

[ceilidh.erickson]

This is a WEIGHTED AVERAGE question.

With any mixture, the resulting solution will be closer in composition to whichever component we added more of. In other words, if we add more of a 50% solution, the resulting mixture will be closer to 50%, and if we add more of the 30% solution, the resulting mixture will be closer to 30%. If we add exactly half of each, the result will be 40%. (It could never be less than 30% or more than 50%, if those were the only two solutions added).

Since in this case the resulting mixture of 35% is closer to 30% than 50%, there must be more of solution Y in the overall mixture; Y must account for more than half of the 16 liters. Right away, we can eliminate anything 8 or less: A, B, and C.

In a weighted average, the ratio of PART : PART will be in the same proportion as the ratio of the differences from the original amounts (30 and 50) to the actual (35). 50 - 35 = 15 and 35 - 30 = 5. Thus, the ratio is 15 : 5, or 3 : 1.

If the ratio of Y to X is 3 to 1, then Y is 3/4 of the total mixture. 3/4 of 16 is 12.

The answer is E.

[fskilnik@GMATH]

Solution X, which is 50% alcohol, is combined with solution Y, which is 30% alcohol, to form 16 liters of a new solution that is 35% alcohol. How much of solution Y is used?

A. 4 liters
B. 6 liters
C. 8 liters
D. 10 liters
E. 12 liters

Let´s do it in the beautiful "Indian style", that is, using alligation... the diagram follows:



? = y

Based on it, we have: y/16 = (50-35)/(50-30) = 3/4 , therefore y = (3/4) 16 = 12.

The above follows the notations and rationale taught in the GMATH method.

[GMATGuruNY]

Solution X, which is 50% alcohol, is combined with solution Y, which is 30% alcohol, to form 16 liters of a new solution that is 35% alcohol. How much of solution Y is used?

A. 4 liters
B. 6 liters
C. 8 liters
D. 10 liters
E. 12 liters

We can PLUG IN THE ANSWERS, which represent the amount of Y.
Since the percentage for the mixture (35%) is closer to Y's percentage (30%) than to X's percentage (50%), Y must constitute MORE THAN 1/2 of the 16-liter mixture.
Thus, the amount of Y must be equal to more than 8 liters.
Eliminate A, B and C.
When the correct answer is plugged in, the mixture will be 35% alcohol.

D: Y= 10 liters, implying that X = 6 liters
Since Y is 30% alcohol, the amount of alcohol in 10 liters of Y = (3/10)(10) = 3.
Since X is 50% alcohol, the amount of alcohol in 6 liters of X = (1/2)(6) = 3.
Percentage of alcohol in the 16-liter mixture = alcohol/total = (3+3)/(16) = 6/16 = 3/8 = 37.5%.
Eliminate D.

The correct answer is E.

[Brent@GMATPrepNow]

Solution X, which is 50% alcohol, is combined with solution Y, which is 30% alcohol, to form 16 liters of a new solution that is 35% alcohol. How much of solution Y is used?

A. 4 liters
B. 6 liters
C. 8 liters
D. 10 liters
E. 12 liters

The OA is E.

Source: Veritas Prep

When solving mixture questions, I find it useful to sketch the solutions with the ingredients SEPARATED:


Since we want to determine the volume of solution Y needed, let's...
...let y = volume (in liters) of solution Y needed
This means 16 - y = volume (in liters) of solution X needed (since the combined volume of both amounts is 16 liters)

So, we get:


Now let's determine the volume of alcohol in each container.

Solution Y is 30% alcohol. We have y liters of solution Y.
So, the volume of alcohol = 0.3y

Solution X is 50% alcohol. We have 16 - y liters of solution X.
So, the volume of alcohol = 0.5(16 - y) = 8 - 0.5y

The combined solution is 35% alcohol. There are 16 liters of this solution.
So, the volume of alcohol = 0.35(16) = 5.6

So, our sketch looks like this:


At this point, we can focus on the volume of alcohol in each container.
We know that: (volume of alcohol in 1st container) + (volume of alcohol in 2nd container) = volume of alcohol in combined solution.
In other words: 0.3y + (8 - 0.5y) = 5.6
Simplify: 8 - 0.2y = 5.6
Subtract 8 from both sides to get: -0.2y = -2.4
Solve: y = 12

Answer: E

[Jeff@TargetTestPrep]

Solution X, which is 50% alcohol, is combined with solution Y, which is 30% alcohol, to form 16 liters of a new solution that is 35% alcohol. How much of solution Y is used?

A. 4 liters
B. 6 liters
C. 8 liters
D. 10 liters
E. 12 liters

We start with x liters of a solution that is 50% alcohol. We add to it y liters of a solution that is 30% alcohol. The result is (x + y) liters of a solution that is 35% alcohol. We can create the equations:

0.5x + 0.3y = 0.35(x + y)

50x + 30y = 35x + 35y

15x = 5y

3x = y

and

x + y = 16

Substituting, we have:

x + 3x = 16

4x = 16

x = 4, so y = 12

Answer: E