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M15-26

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M15-26  [#permalink]

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New post 16 Sep 2014, 00:56
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A
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Question Stats:

66% (02:31) correct 34% (02:58) wrong based on 76 sessions

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Two solutions of acid were mixed to obtain 10 liters of a new solution. Before they were mixed, the first solution contained 0.8 liters of acid while the second contained 0.6 liters of acid. If the percentage of acid in the first solution was twice that in the second, what was the volume of the first solution?

A. 3.0 liters
B. 3.2 liters
C. 3.6 liters
D. 4.0 liters
E. 4.2 liters

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Re M15-26  [#permalink]

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New post 16 Sep 2014, 00:56
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Official Solution:

Two solutions of acid were mixed to obtain 10 liters of a new solution. Before they were mixed, the first solution contained 0.8 liters of acid while the second contained 0.6 liters of acid. If the percentage of acid in the first solution was twice that in the second, what was the volume of the first solution?

A. 3.0 liters
B. 3.2 liters
C. 3.6 liters
D. 4.0 liters
E. 4.2 liters

Denote \(x\) as the volume of the first solution. Using the data from the stem we can build an equation \(\frac{0.8}{x} = \frac{0.6}{10 - x}*2\). Solving it we get \(x = 4\).

Answer: D
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Re: M15-26  [#permalink]

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New post 18 Jun 2015, 09:06
Bunuel wrote:
Two solutions of acid were mixed to obtain 10 liters of a new solution. Before they were mixed, the first solution contained 0.8 liters of acid while the second contained 0.6 liters of acid. If the percentage of acid in the first solution was twice that in the second, what was the volume of the first solution?

A. 3.0 liters
B. 3.2 liters
C. 3.6 liters
D. 4.0 liters
E. 4.2 liters


Hi Bunuel

I found this question on the mobile GMAT APP. Is there a way to bookmark, so I can see it on my account too? If i bookmark within the app, it does not pop up here on the web.

Thanks
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Re: M15-26  [#permalink]

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New post 01 Feb 2017, 07:08
Proceed by substituting the options.... The answer comes out to be option D.
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Re: M15-26  [#permalink]

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New post 25 Jun 2017, 00:48
I used an another method which works.

Solution A1 (0.8 litres)
Solution A2 (0.6 litres)

A1+A2=10

(0.8/A1)=2*(0.6/A2)

Therefore A2=1.5A1

A1+1.5A1 =10

A1= 10/2.5 = 4 litres.
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Re M15-26  [#permalink]

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New post 06 Sep 2018, 09:13
I think this is a high-quality question and I agree with explanation. I used a different approach: Solicit your opinion on it.

First solution has 0.8 lt Acid.
We also know the concentration of acid is twice that of the second solution.
So, if sample 1 had same concentration then sample1 would contain 0.4 lt acid.
Now we got a ratio between volumes between the two different solutions.
0.4:0.6 which is 2:3
We can say there was 4 liter of solution 1 in 10 liter of mixed solution.

Let me know your opinion on it.
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M15-26  [#permalink]

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New post 03 Jan 2019, 20:09
Looks like a strange question but is actually doable.

From Q stem, I ascertained the follows:
2x:x is the ratio of the concentration of acid in sol 1 to that in sol 2
Then I set 2x=.5 and x=.25

This gave me .8/1.6=.5 and .6/2.4=.25
I noticed something interesting here, 1.6+2.4=4 and 4*2.5=10.
So when I multiplied 1.6 by 2.5 and voila, I got 4.

@Bunnel chetan2u, is this approach correct? Also, how can I tell if a question is worth spending more than 2 mins.

Many many thanks in advance!!!
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Re: M15-26  [#permalink]

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New post 25 Aug 2019, 09:37
Sol A(acid) = .8
Sol B(acid) = .6 * 2 --> .12 (it's stated that it's twice the percent of acid A to total vol A)

Acid A/B = .8/.12 = 2/3 ratio of Sol A to Sol B.

If total solution is volume is 10 then Sol A is 4 liters.
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Re: M15-26  [#permalink]

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New post 09 Oct 2019, 07:42
Bunuel How would we know that both solutions must sum to 10? Can the 10L solution not be made from much larger solutions?
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New post 09 Oct 2019, 07:47
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Re: M15-26  [#permalink]

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New post 09 Oct 2019, 09:24
Bunuel wrote:
ghnlrug wrote:
Bunuel How would we know that both solutions must sum to 10? Can the 10L solution not be made from much larger solutions?


Two solutions of acid were mixed to obtain 10 liters of a new solution


but can those solutions not come from two larger solutions? I can obtain a 10L mixture from 2 mixtures that are 100L each.
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Re: M15-26   [#permalink] 09 Oct 2019, 09:24
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