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BTG flashcards problem

This topic has 2 expert replies and 0 member replies
sparkles3144 Master | Next Rank: 500 Posts Default Avatar
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Posted:
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Upvotes:
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BTG flashcards problem

Post Sun Jun 15, 2014 1:09 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?

Answer 14

I disagree with the answer.
Can someone show steps?

Thanks!

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GMAT/MBA Expert

Post Sun Jun 15, 2014 5:06 am
Quote:
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:

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:


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


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:
http://www.beatthegmat.com/liters-of-mixture-x-in-the-50-mixture-t271387.html
http://www.beatthegmat.com/percentage-mixture-t268631.html
http://www.beatthegmat.com/rodrick-mixes-a-martini-that-has-a-volume-of-n-ounces-havi-t270387.html
http://www.beatthegmat.com/mixure-problem-quite-confusing-t261767.html
http://www.beatthegmat.com/mixture-ratio-problem-2-t191643.html

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GMAT/MBA Expert

Post Sun Jun 15, 2014 2:02 am
Quote:
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
We can use ALLIGATION -- a great way to handle MIXTURE PROBLEMS.

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.

An alternate approach is 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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GMATGuruNY@gmail.com
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For more information, please email me at GMATGuruNY@gmail.com.

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