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.