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John has 3 solutions: a 12% saline solution, a 8% vinegar

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John has 3 solutions: a 12% saline solution, a 8% vinegar [#permalink] New post 14 Dec 2010, 16:19
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John has 3 solutions: a 12% saline solution, a 8% vinegar solution, and a 15% alcohol solution. He mixes 3 liters of the alcohol solution with 3 liters of the vinegar solution. What is the minimum amount of the saline solution he must add if the resulting mixture must be at least 2% saline solution?

(A) 7.1 liters
(B) 3 liters
(C) 2.4 liters
(D) 1.2 liters
(E) 1.1 liters
[Reveal] Spoiler: OA

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Re: Mixture problem [#permalink] New post 14 Dec 2010, 16:36
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rxs0005 wrote:
John has 3 solutions: a 12% saline solution, a 8% vinegar solution, and a 15% alcohol solution. He mixes 3 liters of the alcohol solution with 3 liters of the vinegar solution. What is the minimum amount of the saline solution he must add if the resulting mixture must be at least 2% saline solution?

(A) 7.1 liters
(B) 3 liters
(C) 2.4 liters
(D) 1.2 liters
(E) 1.1 liters


So we have 3+3=6 liters of some 0% saline solution (info about 8% vinegar solution, and a 15% alcohol solution is just to confuse us). We want to add x liters of 12% saline solution so that in resulting solution there will be at least 2% of salt (this "at least" part is also to confuse us).

Now, as all salt comes form x liters of 12% saline solution, then amount of salt in 6+x liters and x liters must be equal: 0.12x=0.02(6+x) --> x=1.2.

Answer: D.
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Re: Mixture problem [#permalink] New post 14 Dec 2010, 16:46
rxs0005 wrote:
John has 3 solutions: a 12% saline solution, a 8% vinegar solution, and a 15% alcohol solution. He mixes 3 liters of the alcohol solution with 3 liters of the vinegar solution. What is the minimum amount of the saline solution he must add if the resulting mixture must be at least 2% saline solution?

(A) 7.1 liters
(B) 3 liters
(C) 2.4 liters
(D) 1.2 liters
(E) 1.1 liters


Hi,

I'm not sure if this is a home-made question or has just been reproduced incorrectly, but there are two errors; one is a minor grammar issue, but the other radically changes the answer to the question.

First, the minor grammar isssue: "a 8% vinegar solution" should be "an 8% vinegar solution".

Second, the game-changer: the question should probably read "if the resulting mixture must contain at least 2% saline?"; the current wording ("2% saline solution") really doesn't make sense. There are also other ways the question could have been reworded to fix the problem.

Assuming that our rewording is correct, let's break down the question.

The first, and possibly most important, thing to note is that the question seems much much harder than it actually is. Since we want our final solution to be at least 2% saline, we don't care what concentration we have of the other solutes - all we care about is the amount of saline vs the total amount of solution. So, even though the final mixture will contain vinegar, alcohol and water, we can just lump all 3 in the "non-saline" category.

The second thing to note is that since we want at least 2% saline, the correct answer will reflect a solution that's exactly 2%.

So, paraphrasing and simplifying the question:

Quote:
We have 6 litres of a non-saline solution. How many litres of a 12% saline solution must be added so that the final mixture is 2% saline?


Now that we've got a simple question, we can set up our solution using the percent formula:

% = part/whole * 100%

x = amount of saline solution added
part = amount of saline = .12x
whole = total amount of solution = 6 + x

2% = .12x/(6 + x) * 100%

cancel out the "%" on both sides:

2 = .12x/(6 + x) * 100
2(6 + x) = .12x * 100
12 + 2x = 12x
12 = 10x
12/10 = x
1.2 = x

choose D!

* * *

Backsolving would also have been an excellent approach. Use common sense to eliminate A and B (each would give us way too much saline in the final solution), then plug in D; if D gives too much saline, then E is correct; if D gives too little saline, then C is correct; if D gives exactly 2% saline, then D is correct.
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Re: Mixture problem [#permalink] New post 14 Dec 2010, 20:58
saline: 12% ( total of S liters)
vinegar: 8% ( total of V liters)
alcohol: 15% ( total of A liters)


Vinegar = 3 liters has (24/100) of vinegar
Alcohol = 3 liters has (45/100) of alcohol

Total = 6 liters

x liters of saline is added. It means salinity is (12x/100)


total = (6+x)

(12x/100) = (2/100)*(6+x)
x = 6/5 liters = 1.2liters
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Re: Mixture problem [#permalink] New post 18 Dec 2010, 11:16
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Using alligation method as shown below
answer comes out to be D.
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Re: Mixture problem [#permalink] New post 18 Dec 2010, 12:28
good explanations. I tired using backsolving but took too much time. Slovinsky how would you go about solving using this method. Do you start with either B or D and then look at the 6 liters plus the additional liters of saline added to see if you find 2%? Thanks
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Re: Mixture problem [#permalink] New post 18 Dec 2010, 13:05
gettinit wrote:
good explanations. I tired using backsolving but took too much time. Slovinsky how would you go about solving using this method. Do you start with either B or D and then look at the 6 liters plus the additional liters of saline added to see if you find 2%? Thanks


As I noted, we can quickly eliminate A and B because those will clearly give us too much saline. So, I'd start with the middle of the 3 remaining choices - D.

If we assume that we added 1.2l of saline, we calculate:

12% of 1.2l = .144l of saline

1.2l + 6l = 7.2l total solution

% = part/whole * 100%, so

% = (.144/7.2) * 100% = (14.4/7.2)% = 2%

D gives us 2%, which is exactly what we want - therefore D is correct!

If D had given us less than 2%, then we need to add more and would choose C; if D had given us more than 2% then we need to add less and would choose E.
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Re: Mixture problem [#permalink] New post 19 Dec 2010, 23:51
Thank you. Sounds like a lot of work just to back solve.
Better to learn the algebra on these questions.
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Re: Mixture problem [#permalink] New post 08 Jan 2011, 17:46
hermit84 wrote:
Using alligation method as shown below
answer comes out to be D.
Attachment:
Untitled.jpg



Iwould also go for this method....
Very eassy and time consuming...
Re: Mixture problem   [#permalink] 08 Jan 2011, 17:46
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