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M15-26

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M15-26  [#permalink]

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New post 15 Sep 2014, 23:56
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  55% (hard)

Question Stats:

65% (02:00) correct 35% (02:09) wrong based on 89 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 15 Sep 2014, 23:56
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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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Collection of Questions:
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Re: M15-26  [#permalink]

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New post 18 Jun 2015, 08:06
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, 06:08
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 24 Jun 2017, 23:48
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, 08:13
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, 19:09
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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M15-26   [#permalink] 03 Jan 2019, 19:09
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