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Mixture problem

by gdshamain » Thu May 22, 2014 8:27 am
How many liters of a solution that is
10% alcohol by volume must be added
to 2 liters of a solution that is 50%
alcohol by volume to create a solution
that is 15% alcohol by volume?
A) 10
B) 12
C) 14
D) 16
E) 18
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by GMATGuruNY » Thu May 22, 2014 8:36 am
gdshamain wrote:How many liters of a solution that is
10% alcohol by volume must be added
to 2 liters of a solution that is 50%
alcohol by volume to create a solution
that is 15% alcohol by volume?
A) 10
B) 12
C) 14
D) 16
E) 18
Alligation approach:

Let X = the 10% solution and Y = the 50% solution.

Step 1: Plot the 3 percentages on a number line, with the percentages for X and Y on the ends and the percentage for the mixture in the middle.
X 10------------15------------50 Y

Step 2: Calculate the distances between the percentages.
X 10-----5-----15-----35-----50 Y

Step 3: Determine the ratio in the mixture.
The required ratio of X to Y is equal to the RECIPROCAL of the distances in red.
X:Y = 35:5 = 7:1.

Since X:Y = 7:1 = 14:2, 14 liters of X must be added to 2 liters of Y.

The correct answer is C.
Last edited by GMATGuruNY on Sun Jun 15, 2014 2:02 am, edited 1 time in total.
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by gdshamain » Thu May 22, 2014 8:43 am
Thanks.. The correct answer is C and not D?
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by GMATGuruNY » Thu May 22, 2014 8:54 am
gdshamain wrote:Thanks.. The correct answer is C and not D?
Yes.
An alternate approach would be to PLUG IN THE ANSWERS, which represent the amount of 10% solution that must be added to 2 liters of 50% solution.

Answer choice C: 14 liters
Amount of alcohol in 14 liters of 10% solution = 10% of 14 = 1.4 liters.
Amount of alcohol in 2 liters of 50% solution = 50% of 2 = 1 liter.
(total alcohol)/(total volume) = (1.4 + 1)/(14 + 2) = (2.4)/16 = 24/160 = 3/20 = 15/100 = 15%.
Success!

The correct answer is C.
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by [email protected] » Thu May 22, 2014 8:02 pm
Hi gdshamain,

Mixture questions can also be solved algebraically:

X = liters of 10% alcohol

Since we're mixing two types of solution (one that's 10% alcohol and one that 50% alcohol), we have to keep track of that information when we use the average formula.

[.1(X) + .5(2)] / (X + 2) = .15

Simplify....

.1X + 1 = .15X + .3

.7 = .05X

Multiply both sides by 20....

14 = X

Final Answer: C

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Mixture problem

by Brent@GMATPrepNow » Sat May 24, 2014 9:19 am
gdshamain wrote:How many liters of a solution that is 10% alcohol by volume must be added to 2 liters of a solution that is 50% alcohol by volume to create a solution that is 15% alcohol by volume?
A) 10
B) 12
C) 14
D) 16
E) 18
When solving mixture questions, I find it useful to sketch the solutions with the ingredients SEPARATED.

Start with 2 liters of solution that is 50% alcohol:
Image
When we draw this with the ingredients separated, we see we have 1 liter of alcohol in the mixture.


Next, we'll let x = the number of liters of 10% solution we need to add.
Since 10% of the mixture is alcohol, we know that 0.1x = the volume of alcohol in this solution:
Image

At this point, we can add the two solutions to get the following volumes:
Image

Since the resulting solution is 15% alcohol (i.e., 15/100 of the solution is alcohol), we can write the following equation:
(1 + 0.1x)/(2 + x) = 15/100
Simplify to get: (1 + 0.1x)/(2 + x) = 3/20
Cross multiply to get: 20(1 + 0.1x) = 3(2 + x)
Expand: 20 + 2x = 6 + 3x
Rearrange/solve: 14 = x

Answer: C

Cheers,
Brent

Here are some additional mixture questions to practice with:
https://www.beatthegmat.com/liters-of-mi ... 71387.html
https://www.beatthegmat.com/percentage-m ... 68631.html
https://www.beatthegmat.com/rodrick-mixe ... 70387.html
https://www.beatthegmat.com/mixure-probl ... 61767.html
https://www.beatthegmat.com/mixture-rati ... 91643.html
Last edited by Brent@GMATPrepNow on Sat Apr 28, 2018 7:05 am, edited 1 time in total.
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by Brent@GMATPrepNow » Sun May 25, 2014 4:02 am
gdshamain wrote:How many liters of a solution that is 10% alcohol by volume must be added
to 2 liters of a solution that is 50% alcohol by volume to create a solution that is 15% alcohol by volume?
A) 10
B) 12
C) 14
D) 16
E) 18
Here's another approach:

IMPORTANT CONCEPT: If EQUAL VOLUMES of 2 solutions are combined, the concentration of the resulting mixture will be the AVERAGE of the 2 solutions.
Example: If 3 liters of a 10% alcohol solution are combined with 3 liters of a 30% alcohol solution, the resulting solution will be 20% alcohol.
Likewise, if 11 liters of 6% alcohol solution are combined with 11 liters of a 7% alcohol solution, the resulting solution will be 6.5% alcohol.

Okay, now onto the solution.....

Begin with 2 liters of a solution that is 50% alcohol.
Add 2 liters of the solution that is 10% alcohol.
The result is 4 liters of solution that is 30% alcohol [since (50+10)/2 = 30]
We want a 15% solution, so we need to keep diluting the mixture.

Next, add 4 liters of the solution that is 10% alcohol.
The result is 8 liters of solution that is 20% alcohol [since (30+10)/2 = 20]
We want a 15% solution, so we need to keep diluting the mixture.

Next, add 8 liters of the solution that is 10% alcohol.
The result is 16 liters of solution that is 15% alcohol [since (20+10)/2 = 15]
We now have our 15% solution.

In TOTAL, the amount of 10% alcohol solution required = 2 + 4 + 8 = [spoiler]14 = C[/spoiler]

Cheers,
Brent
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