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

Question

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

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

Bunuel · Answer

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.

skovinsky · Answer

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:

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.

LM · Answer

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

hermit84 · Answer

Using alligation method as shown below answer comes out to be D. Untitled.jpg

gettinit · Answer

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

skovinsky · Answer

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.

gettinit · Answer

Thank you. Sounds like a lot of work just to back solve. 
Better to learn the algebra on these questions.

jullysabat · Answer

Iwould also go for this method....
Very eassy and time consuming...

gracie · Answer

let x=amount of saline solution to be added
.12x=.02(x+6)
x=6/5=1.2 liters

coylahood · Answer

We can use the weighted average method to resolve this problem. According to this method, the weights required when two solutions are mixed together can be expressed as follows:

(Weight of solution 1)/Weight of solution 2 = (concentration of solution 2 - average)/(average - concentration of solution 1).

Applying this concept to the problem at hand we can treat the existing mixture of alcohol and vinegar as a single solution containing no saline (i.e., concentration of saline in this solution is 0). This solution has to be combined with a 12% saline solution and we expect the concentration in the combined solution to be at least 2%.

therefore, w1/w2= (12-2)/(2-0) = 5/1

Therefore every 5 liters of the combined alcohol and vinegar solution require 1 litre of the saline solution. This implies that 6 liters of the combined alcohol and vinegar solution must contain 6/5 = 1.2 liters of the saline solution. So the correct answer is D.

coylahood · Answer

We can use the weighted average method to resolve this problem. According to this method, the weights required when two solutions are mixed together can be expressed as follows:

(Weight of solution 1)/Weight of solution 2 = (concentration of solution 2 - average)/(average - concentration of solution 1).

Applying this concept to the problem at hand we can treat the existing mixture of alcohol and vinegar as a single solution containing no saline (i.e., concentration of saline in this solution is 0). This solution has to be combined with a 12% saline solution and we expect the concentration in the combined solution to be at least 2%.

therefore, w1/w2= (12-2)/(2-0) = 5/1

Therefore every 5 liters of the combined alcohol and vinegar solution require 1 litre of the saline solution. This implies that 6 liters of the combined alcohol and vinegar solution must contain 6/5 = 1.2 liters of the saline solution. So the correct answer is D.

satya2029 · Answer

Resulting mixture=6 litres
minimum added amount =Y
(6+Y)* 0.02=(12/100)Y
Y=1.2 Litres
D:)

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards