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Bunuel
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M15-26 [#permalink] New post  15 Sep 2014, 23:56
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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] New post  15 Sep 2014, 23:56
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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

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] 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] New post  24 Jun 2017, 23:48
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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] 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] 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.

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Re: M15-26 [#permalink] New post  25 Aug 2019, 08:37
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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] New post  09 Oct 2019, 06:42
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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Re: M15-26 [#permalink] New post  09 Oct 2019, 06:47
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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
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Re: M15-26 [#permalink] New post  09 Oct 2019, 08:24
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] New post  19 Nov 2019, 07:41
Here what I did, I am getting an Incorrect answer, can someone tell me why ?
F + S = 10
0.8F = 2*0.6S

After substitution I am getting F = 6, and Second SOlution as 4.

Can someone tell me how EXPERT
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M15-26 [#permalink] New post  19 Nov 2019, 09:17
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hero_with_1000_faces wrote:
Here what I did, I am getting an Incorrect answer, can someone tell me why ?
F + S = 10
0.8F = 2*0.6S

After substitution I am getting F = 6, and Second SOlution as 4.

Can someone tell me how EXPERT


What is the logic behind the highlighted portion? The solution above give the correct version: \(\frac{0.8}{F} = \frac{0.6}{S}*2\) (the percentage of acid in the first solution (0.8/F) was (=) twice (2*) that in the second (0.6/S)).
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Re: M15-26 [#permalink] New post  19 Nov 2019, 21:19
Bunuel wrote:
hero_with_1000_faces wrote:
Here what I did, I am getting an Incorrect answer, can someone tell me why ?
F + S = 10
0.8F = 2*0.6S

After substitution I am getting F = 6, and Second SOlution as 4.

Can someone tell me how EXPERT


What is the logic behind the highlighted portion? The solution above give the correct version: \(\frac{0.8}{F} = \frac{0.6}{S}*2\) (the percentage of acid in the first solution (0.8/F) was (=) twice (2*) that in the second (0.6/S)).



Thanks for your reply,

The question states, if the percentage of acid in the first solution is twice....

I think I have realized my mistake, I assumed if percentage is twice than the actual value of acid will also be twice, I think this was my mistake.
am I right ?
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Re: M15-26 [#permalink] New post  19 Nov 2019, 21:53
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hero_with_1000_faces wrote:
Bunuel wrote:
hero_with_1000_faces wrote:
Here what I did, I am getting an Incorrect answer, can someone tell me why ?
F + S = 10
0.8F = 2*0.6S

After substitution I am getting F = 6, and Second SOlution as 4.

Can someone tell me how EXPERT


What is the logic behind the highlighted portion? The solution above give the correct version: \(\frac{0.8}{F} = \frac{0.6}{S}*2\) (the percentage of acid in the first solution (0.8/F) was (=) twice (2*) that in the second (0.6/S)).



Thanks for your reply,

The question states, if the percentage of acid in the first solution is twice....

I think I have realized my mistake, I assumed if percentage is twice than the actual value of acid will also be twice, I think this was my mistake.
am I right ?


What is "value of acid"? You multiply 0.8 (acid in liters in the first solution) by F (first solution in liters). This does not make sense.
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Re: M15-26 [#permalink] New post  19 Nov 2019, 23:23
Sorry for my hasty writing, I wanted to write.

I assumed that "the percentage of First solution is twice the Percentage of Second solution" is same as "the actual value of acid in First solution will also be twice that of second solution", I think this was my mistake.
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Re: M15-26 [#permalink] New post  18 May 2020, 23:54
what do we mean by two solutions of acid?? does that solution has only acid or it's like acid + water??

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Re: M15-26 [#permalink] New post  18 May 2020, 23:56
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nikitamaheshwari wrote:
what do we mean by two solutions of acid?? does that solution has only acid or it's like acid + water??

Bunuel


An acid solution = acid + water.
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M15-26 [#permalink] New post  04 Dec 2020, 01:50
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


@VeritasKarishma/@Bunuel,

Is it possible to attempt the question with weightage average (esp. scale method)? If yes, can you please explain it?

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