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another simple one

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by magical cook » Mon Jan 28, 2008 10:08 am
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
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by cris » Mon Jan 28, 2008 10:25 am
Its an easy one but kind of tricky from my point of view...because I just calculated the vinegar content and the alcohol content of the 6 liters of solution and I did not need too. I hate when Gmat gives us data that we dont need (thank god it doesnt happen that often).

Anyway :) D??
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by roy_priya » Mon Jan 28, 2008 3:01 pm
The answer I get is E: 1.1 L. Please post the OA
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Re: another simple one

by camitava » Mon Jan 28, 2008 8:24 pm
magical cook 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
Guys, I have solved the prob like -
Let,
x L of saline solution is added.
So the total solution becomes = 6 + x
Now in the solution, saline will be = 12x/(100 * (6 + x)) L
So
12x/(100 * (6 + x)) = 2/100
or x = 1.2
- So IMO D and agree with Cris!
Ohhh! Forgot to mention, what's the OA, Magical Cook?
Correct me If I am wrong


Regards,

Amitava
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by lalabee » Mon Jan 28, 2008 11:18 pm
i got 1.2 (answer D) is that correct ?

i agree this question is easy but tricky !! :twisted: [/list]
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by Bronson » Thu Jan 31, 2008 2:42 am
Hello guys, can someone explain and post gradually the procedure for solving this question?
Thanks
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by Saffa » Thu Jan 31, 2008 3:40 am
The V and A has nothing to do with the question, so see the question as: How much of the Saline solution must be added to 6 liters of "water" to make a 2% Saline solution?

Let n be the amount of Saline Solution,

thus: 0,12n/(n + 6) = 0,02

i.e: 0,12n = 0,02n + 0,12

i.e: 0,1n = 0,12

i.e.\: n = 1.2
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by Bronson » Fri Feb 01, 2008 1:50 am
Thanks Saffa,
I was wondering all the time why it is not n/(6+n) and I realized that than we would calculate the percentage of the whole saline solution instead the % of salt in the new solution which was required.
It explained 0.12n which was confusing me.
Regards
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