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by sanju09 » Sat Feb 04, 2012 4:47 am
To make an alcohol solution, pure alcohol is mixed in water. How many liters of solution X, which is 30 percent alcohol, must be mixed with 10 liters of solution Y, which is k percent alcohol, to obtain solution Z, which is 20 percent alcohol?
(1) k is an odd multiple of 7.
(2) The solution Z so obtained contains 18.4 liters of water contents.



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by WhiteWNNoise » Sat Feb 04, 2012 1:14 pm
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by sanju09 » Mon Feb 06, 2012 1:47 am
WhiteWNNoise wrote:Image

I can't see why statement (1) is not sufficient. See, if x = 20 - k, the only odd multiple of 7 in place of k, which can sustain x positive is 7, and this determines everything we want.

[spoiler]I suggest D as the answer to this DS.[/spoiler]
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by neelgandham » Mon Feb 06, 2012 5:33 am
To make an alcohol solution, pure alcohol is mixed in water. How many liters of solution X, which is 30 percent alcohol, must be mixed with 10 liters of solution Y, which is k percent alcohol, to obtain solution Z, which is 20 percent alcohol?

If x is the number of liters of solution x that is mixed, then, using the scale method(please refer to the attachment)

(0.2-(k/100))/x = (0.3-0.2)/10 => k+x =20

So, the question can be rephrased to If k+x =20, then what is the value of x?
(1)k is an odd multiple of 7.
Alcohol content in Y < Alcohol content of Z < Alcohol content of X
i.e. k<20<30. Since k is an odd multiple of 7, the value of k should be 7. If k = 7, x = 13
(2)The solution Z so obtained contains 18.4 liters of water contents.
Since Z is 80% water:
z = (100/0.8)*18.4 = 23. Since y=10, and z =23, x=13.
SUFFICIENT.
Option D
[Edit: Corrected]
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by GMATGuruNY » Mon Feb 06, 2012 6:10 am
sanju09 wrote:To make an alcohol solution, pure alcohol is mixed in water. How many liters of solution X, which is 30 percent alcohol, must be mixed with 10 liters of solution Y, which is k percent alcohol, to obtain solution Z, which is 20 percent alcohol?
(1) k is an odd multiple of 7.
(2) The solution Z so obtained contains 18.4 liters of water contents.

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Using alligation:

The proportion needed of each element in the mixture is equal to the positive distance between the OTHER two percentages.

Proportion needed of X = percentage of alcohol in Z - percentage of alcohol in Y = 20-k.
Proportion needed of Y = percentage of alcohol in X - percentage of alcohol in Z = 30-20 = 10.

X:Y = (20-k)/10

Since Y = 10 liters:
X:10 = (20-k)/10
X = 20-k.

Statement 1: k is an odd multiple of 7.
Since X = 20-k, and X>0, we know that k=7 and that X=13.
SUFFICIENT.

Statement 2: 18.4 liters of water in Z.
Since Z is 80% water:
18.4 = .8Z
Z = 23.
Since Y=10, Z=23, and X+Y=Z, we know that X=13.
SUFFICIENT.

The correct answer is D.
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by Rastis » Tue Feb 07, 2012 5:25 am
Can someone provide a way to decipher these type of mixing questions? I always have an extremely hard time even setting this problem up. Also Mitch, can you better explain your "alligation"? It's very hard to interpret what you're trying to say.
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by GMATGuruNY » Tue Feb 07, 2012 7:47 am
Rastis wrote:Can someone provide a way to decipher these type of mixing questions? I always have an extremely hard time even setting this problem up. Also Mitch, can you better explain your "alligation"? It's very hard to interpret what you're trying to say.
To illustrate:

X = 10% alcohol
Y = 30% alcohol
Z = percentage of alcohol in a mixture of X and Y

Alligation uses the RELATIVE DISTANCES to determine the proportion needed of X and Y.

If Z is EQUIDISTANT from X and Y, EQUAL amounts of X and Y are needed.
If Z = 20% -- HALFWAY between X and Y -- EQUAL amounts of X and Y will yield the desired percentage.
In other words, X:Y = 1:1.
If we mix 100 liters of X and 100 liters of Y, the percentage of alcohol in Z = alcohol/total = (10+30)/200 = 1/5 = 20%.

The CLOSER Z gets to Y, the MORE we need of Y, the LESS we need of X.
If Z = 25%, which is CLOSER to Y:
The distance between Z and Y tells us the proportion needed of X: 30-25 = 5.
The distance between Z and X tells us the proportion needed of Y: 25-10 = 15.
X:Y = 5:15 = 1:3.
Since the desired percentage is CLOSER to Y, we need LESS of X and MORE of Y.
To confirm that the ratio is correct:
If we mix 100 liters of X and 300 liters of Y, the percentage of alcohol in Z = alcohol/total = (10 + 3*30)/400 = 1/4 = 25%.

The CLOSER Z gets to X, the MORE we need of X, the LESS we need of Y.
If Z = 15%, which is closer to X:
The distance between Z and Y tells us the proportion needed of X: 30-15 = 15.
The distance between Z and X tells us the proportion needed of Y: 15-10 =5.
X:Y = 15:5 = 3:1.
Since the desired percentage is CLOSER to X, we need MORE of X and LESS of Y.
To confirm that the ratio is correct:
If we mix 300 liters of X and 100 liters of Y, the percentage of alcohol in Z = alcohol/total = (3*10 + 30)/400 = 3/20 = 15%.

With this explanation in mind, please revisit my solution above.

Here are some other alligation problems:

https://www.beatthegmat.com/allegation-t69552.html

https://www.beatthegmat.com/mixture-prob ... 89508.html

https://www.beatthegmat.com/mixture-with ... 82022.html

https://www.beatthegmat.com/problem-from ... 49413.html

https://www.beatthegmat.com/a-ps-from-gm ... tml#353227
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by Rastis » Tue Feb 07, 2012 7:51 am
Think I'm more confused than ever. No way would I be able to do that on my own. Thanks anyway!
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by sanju09 » Wed Feb 08, 2012 4:11 am
Rastis wrote:Think I'm more confused than ever. No way would I be able to do that on my own. Thanks anyway!

Rely on basic algebra...

If x liters of solution X, which is 30 percent alcohol, (which means 0.3 x liters of alcohol and 0.7 x liters of water), must be mixed with 10 liters of solution Y, which is k percent alcohol, [which means (k/100) × 10 liters of alcohol and {(100 - k)/100} × 10 liters of water], to obtain (x + 10) liters of solution Z, which is 20 percent alcohol, [which means 0.2 × (x + 10) liters of alcohol and 0.8 × (x + 10) liters of water], then

0.2 × (x + 10) = 0.3 x + (k/100) × 10, and 0.8 × (x + 10) = 0.7 x + {(100 - k)/100} × 10

Either relation yields the same following result, or x + k = 20; and we have to answer x uniquely, any how.

(1) k is an odd multiple of 7. Only k = 7 fits in the loop in order to keep x positive; this gives us x = 13. Sufficient.

(2) The solution Z so obtained contains 18.4 liters of water contents. We know that the solution Z contains 0.8 × (x + 10) liters of water, hence 0.8 × (x + 10) = 18.4 or x = 13. Sufficient.


Alligation Rule is the key to such questions, and very simple to understand. Even then if it sounds like some Rocket Science to any of us, we can still do the difference without mentioning it, algebra rocks everywhere in math and logics.
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by GMATGuruNY » Wed Feb 08, 2012 5:12 am
Rastis wrote:Think I'm more confused than ever. No way would I be able to do that on my own. Thanks anyway!
DS weighted average/mixture problems typically involve 3 KEY FACTS:

FACT 1: The percentage/average in each ingredient.
FACT 2: The ratio of the ingredients.
FACT 3: The percentage/average in the mixture.

If we know FACT 1 (the percentage/average in each ingredient) and FACT 2 (the ratio of the ingredients), we can determine FACT 3 (the percentage/average in the mixture).

If we know FACT 1 (the percentage/average in each ingredient) and FACT 3 (the percentage/average in the mixture), we can determine FACT 2 (the ratio of the ingredients).

Bearing this in mind:
To make an alcohol solution, pure alcohol is mixed in water. How many liters of solution X, which is 30 percent alcohol, must be mixed with 10 liters of solution Y, which is k percent alcohol, to obtain solution Z, which is 20 percent alcohol?

(1) k is an odd multiple of 7.

(2) The solution Z so obtained contains 18.4 liters of water contents.
The question stem gives us FACT 3 (the percentage in the mixture):
Z = 20%.

Statement 1: k=7.
Thus, we have FACT 1 (the percentage in each ingredient):
X=30% and Y=7%.
Thus, we can determine FACT 2 (the ratio of the ingredients).
With the ratio of X to Y and the value of Y (10 liters), we can determine the value of X.
SUFFICIENT.

Statement 2: Z is 18.4 liters of water.
Since Z is 80% water, we can determine the value of Z.
With the value of Z and the value of Y (10 liters), we can determine the value of X.
SUFFICIENT.

The correct answer is D.

While my initial post showed all the math, the reasoning expressed here is what I actually USED to determine the correct answer. No actual math is needed if we understand how weighted averages work.

For many DS questions, we shouldn't do much calculating: as soon as we recognize that we have sufficient -- or insufficient -- information, we should choose the appropriate answer (or eliminate inappropriate answers) and move on.
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by Rastis » Wed Feb 08, 2012 5:43 am
Thanks, Mitch. That explanation is a LOT clearer. Also, I have never, ever heard of allegation until this post.
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