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D01-31

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Bunuel
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D01-31 [#permalink] New post   16 Sep 2014, 00:13
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How many liters of pure alcohol need to be added to a 40-liter solution, which is 10% alcohol by volume, in order to double the alcohol proportion?

A. 4
B. 5
C. 10
D. 20
E. 40
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B
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Bunuel
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Re D01-31 [#permalink] New post   16 Sep 2014, 00:13
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Official Solution:

How many liters of pure alcohol need to be added to a 40-liter solution, which is 10% alcohol by volume, in order to double the alcohol proportion?

A. 4
B. 5
C. 10
D. 20
E. 40


The original 40-liter solution contains 10% alcohol by volume, which amounts to \(0.1*40=4\) liters of alcohol.

Let's say we need to add \(x\) liters of pure alcohol to the 40-liter solution to create a new solution (with a total volume of \(40+x\) liters) that is 20% alcohol by volume.

Now, we can set up an equation to represent the amount of alcohol in the new solution: \(4+x=0.2*(40+x)\). Solving this equation, we find that \(x=5\).

Thus, 5 liters of pure alcohol must be added to the original 40-liter solution to double its alcohol proportion.


Answer: B
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mockingjay
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Re: D01-31 [#permalink] New post   02 Jan 2015, 05:23
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Allegation method:

Alch in Sol(original)......Alch in Sol(added)

10%.........................100% ------> put 100% here because we are adding pure alcohol
...............20%................. -------> we require a solution that has double the alcohol proportion from original one
80%.........................10%

therefore, we get ratio of 8:1 ----> original solution:added solution or pure alcohol

ie, 40 ltrs : 5 ltrs

so, 5 ltrs of pure alcohol must be added
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Sallyzodiac
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Re: D01-31 [#permalink] New post   26 May 2016, 05:09
Bunuel; I tried solving with the following equation:

\(\frac{4+x}{40} = \frac{1}{5}\), where \(\frac{1}{5}\) is the desired (alcohol/total solution) ratio. Solving for \(x\) we get 4. Why doesn't this work?
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Bunuel
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Re: D01-31 [#permalink] New post   26 May 2016, 08:57
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Sallyzodiac wrote:
Bunuel; I tried solving with the following equation:

\(\frac{4+x}{40} = \frac{1}{5}\), where \(\frac{1}{5}\) is the desired (alcohol/total solution) ratio. Solving for \(x\) we get 4. Why doesn't this work?


Because when you add x liters of alcohol, total becomes 40+x liters not 40 liters. It should be (4+x)/(40+x)=1/5, which is the same formula as in my solution above.
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mevicks
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Re: D01-31 [#permalink] New post   06 Jul 2016, 05:32
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MockingJays solution is great! Heres the pictorial representation of the same:
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Untitled.png [ 4.57 KiB | Viewed 66756 times ]

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BrentGMATPrepNow
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Re: D01-31 [#permalink] New post   10 Feb 2020, 21:29
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Bunuel wrote:
How many liters of pure alcohol must be added to a 40-liter solution that is 10% alcohol by volume in order to double the alcohol proportion?

A. 4
B. 5
C. 10
D. 20
E. 40


We start with a 40-liter solution that is 10% alcohol
So, there are 4 liters of pure alcohol in the ORIGINAL 40 liters.
Let x = the number of liters of pure alcohol that we will add to the ORIGINAL mixture.

So, 4 + x = the number of liters in the RESULTING mixture.
Also, since we are adding x liters to the original 40 liters, the RESULTING mixture has a total volume of 40 + x

We want the resulting mixture to be 20% alcohol.
So we can write: (4 + x)/(40 + x) = 20%
In other words: (4 + x)/(40 + x) = 20/100
In other words: (4 + x)/(40 + x) = 1/5
Cross multiply to get: 5(4 + x) = 1(40 + x)
Expand: 20 + 5x = 40 + x
So: 5x = 20 + x
So: 4x = 20
Sol x = 5

Answer: B

Cheers,
Brent
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SJ1295
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Re: D01-31 [#permalink] New post   07 Sep 2020, 14:22
I think this is a high-quality question and I agree with explanation.
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Bunuel
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Re: D01-31 [#permalink] New post   10 Apr 2023, 08:34
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: D01-31 [#permalink]
10 Apr 2023, 08:34
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