Jackie has two solutions that are 2 percent sulfuric acid

Use weighted average:

2% and 12% solutions mix to give 5% solution.

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (12 - 5)/(5 - 2) = 7/3

You need 7 parts of 2% solution and 3 parts of 12% solution to get 10 parts of 5% solution.
If total 5% solution is actually 60 litres, you need 7*6 = 42 litres of 2% solution and 3*6 = 18 litres of 12% solution.

Answer (E)

Weighted Avg and Mixtures Basics are discussed here:
https://anaprep.com/arithmetic-weighted-averages/
https://anaprep.com/arithmetic-mixtures/
and these videos:
https://www.youtube.com/watch?v=_GOAU7moZ2Q
https://www.youtube.com/watch?v=VdBl9Hw0HBg

let a=amount of 2% acid and b= amount of 12% acid.

Now, The equation translates to,
0.02a + .12b = .05(a+b)
but a+b= 60

therefore .02a + .12b = .05(60)
=> 2a + 12b = 300

but b=60-a

therefore
2a+ 12(60-a) = 300
=> 10a = 420
hence a = 42.

When it comes to mixture questions, it's often useful to sketch the solutions with their components separated

Since the question asked us to find the number of liters of 2% solution needed, let's let x = number of liters of 2% solution needed
2% of x = 0.02x
So, the initial solution contains, 0.02x liters of pure acid.



Since we want a final total of 60 liters, we need to now add 60-x liters of 12% solution.
12% of (60 - x) = 0.12(60 - x) = 7.2 - 0.12x


To find the volume of pure acid in the resulting solution, we'll add the acid from each solution
Total volume of acid = 0.02x + 7.2 - 0.12x = 7.2 - 0.1x

So, the resulting solution has a total of (7.2 - 0.1x) liters of acid

The NEW solution is 5% PURE acid.
So, we can write: (7.2 - 0.1x)/60 = 5/100
Cross multiply to get: 100(7.2 - 0.1x) = 5(60)
Expand: 720 - 10x = 300
Add 10 x to both sides: 720 = 300 + 10x
Subtract 300 from both sides: 420 = 10x
Solve: x = 42

Answer: E

----------------------------------------

If you don't want to sketch the solutions, another approach is to just keep track of the acid
Let x = number of liters of 2% solution needed
So, 60 - x = number of liters of 12% solution needed

2% of x = 0.02x
So, 0.02x = the number of liters of PURE acid in the 2% solution

12% of 60 - x = 0.12(60 - x) = 7.2 - 0.12x
So, 7.2 - 0.12x = the number of liters of PURE acid in the 12% solution

Now let's COMBINE the two solutions.
Total volume of PURE acid = 0.02x + 7.2 - 0.12x
= 7.2 - 0.1x
So, our NEW solution contains 7.2 - 0.1x liters of PURE acid
Also, the NEW solution has a total volume of 60 liters

Since the NEW solution is 5% PURE acid, we can write: (7.2 - 0.1x)/60 = 5/100
Cross multiply to get: 100(7.2 - 0.1x) = 5(60)
Expand: 720 - 10x = 300
Add 10 x to both sides: 720 = 300 + 10x
Subtract 300 from both sides: 420 = 10x
Solve: x = 42

Answer: E

RELATED VIDEO

\(0,02*x + 0,12 * (60-x) = 0,05 * 60\)
\(0,02*x + 7,2 - 0,12*x = 3\)
\(4,2 = 0,1*x\)
\(x = 0,42\)

= 42%

Alright I'm so frustrated but after reading the article she posted. I try my best to answer using my way of thinking:

7 cups + 3 cups = 10 cups..so you need 10 cups of 5% solution to produce 60 liters. There is 10 cups and there is 60 liters. We have 60, we have 10 so the other number is missing is 6 therefore 6*7 cups = 42 liters of solution

Yes, you are right. Look, for every 10 cups/liters/ml/gallons/units etc of 5% solution, you need 7 cups/liters/ml/gallons/units etc of 2% solution and 3 cups/liters/ml/gallons/units etc of 12% solution.

So for each 10 litres of 5% solution, you need 7 litres of 2% and 3 litres of 12%.
Then how much will you need for 60 litres of 5% solution?

You will need 6 times the original quantity so you will need 7*6 = 42 litres of 2% solution and 3*6 = 18 litres of 12% solution.

Take a look at the ratios video here: https://youtu.be/5ODENGG5dvc

For a question like this always have 2 equations, one for concentration and one for volume:

Let X = vol of 2% sol
Y = vol of 12% sol

Since both solutions are mixed to make 60liters of 5% solution
X + Y = 60 ----1
Also,
equation for concentration
2X + 12Y = 5 (60)
2X + 12Y = 300 ------2

Solve equation 1 & 2 simultaneously
Y = 18
X = 60-18 = 42

2% 12%

5%

7% 3%

=> we need 7 parts of the 2% solution and 3 parts of the 12% solution => total 10 parts => 60/10 = 6 liters
Therefore, 7 * 6 liters = 42 liters of 2% solution is needed.

Hi All,

This is essentially a Weighted Average question, but it can be solved in a variety of different ways. Since the answer choices are "spread out" numbers, there's actually a great 'brute force' approach that you can use to logically answer this question without doing that much math.

We're told to mix a 2% acid solution with a 12% acid solution and end up with 60 LITERS of 5% acid solution. We're asked how many liters (of the 60) would be the 2% solution.

IF....
we had 1 liter of each solution, then the acidity of the mixture would be (2% + 12%)/2 = 7%.... this is clearly too high (it's supposed to be 5%), so we need MORE of the 2% mixture.

IF....
we had 2 liters of the 2% solution and 1 liter of the 12% solution, then the acidity of the mixture would be (2% + 2% + 12%)/3 = 16/3% = 5 1/3%.... this is also clearly too high (it's supposed to be 5%), so we need even MORE of the 2% mixture. In this example, it's worth noting that 2/3 of the mixture is the 2% solution.

Since we need even MORE of that solution, we need MORE than 2/3 of the total to be that 2% mixture. There's only one answer that's more than 2/3 of 60....

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Let x be the number of liters of 2% solution in the mixture
Since there are 60 liters in total, 60 - x will equal number of liters of 12% solution in the mixture

Now apply the formula:
5 = (x/60)(2) + [(60-x)/60](12)
Multiply both sides by 60 to get: 300 = 2x + (60-x)(12)
Expand: 300 = 2x + 720 - 12x
Rearrange: -420 = -10x
Solve: x = 42

Answer: E

RELATED VIDEO

VeritasPrepKarishma

Can you please explain solution posted by EugeneFish?
I went through your blog here
but all I am able to proceed beyond assigning two points (weights) on line segment and invert distance from either of them.
say w1/w2 = d2/d1


Thanks in advance.

EugeneFish has given the pictorial representation of the scale method.

You should check out this video: https://youtu.be/_GOAU7moZ2Q

It discusses the scale method, why it works and its pictorial and formula depiction.

We can let the amount of 2% sulfuric acid solution = x and the amount of 12% sulfuric acid solution = y. Thus:

x + y = 60

y = 60 - x

and

0.02x + 0.12y = 0.05(x + y)

2x + 12y = 5x + 5y

7y = 3x

Thus:

7(60 - x) = 3x

420 - 7x = 3x

420 = 10x

42 = x

Alternate Solution:

We will mix x liters of 2% sulfuric acid with (60 – x) liters of 12% sulfuric acid to produce 60 liters of 5% sulfuric acid. We can create an equation from this information and solve for x:

0.02x + 0.12(60 – x) = (0.05)(60)

0.02x + 7.2 – 0.12x = 3

-0.10x = -4.2

x = 42

Answer: E
_________________

Argh this is really frustrating me. I don't know If its just because i'm going wrong with the math or what but i get:

0.02x + 0.12y
---------------- = 3 (5% of 60 is 3)
x+y


This becomes 0.02x +0.12y = 3x + 3y which becomes 2.98x = 2.88y

What am I doing wrong here?

Thanks!

Hi calappa1234,

You made a small math mistake (trying to combine 2 steps into 1 step). Here's what the initial equation should look like:

(.02X + .12Y)/(X + Y) = .05

The prompt tells us that (X + Y) = 60, so you could 'substitute' that into the denominator if you like. However - if you're going to approach this algebraically - based on the wording of the prompt, you would likely find it easier to combine 'like' terms and simplify:

.02X + .12Y = .05X + .05Y
.07Y = .03X
7Y = 3X
7/3 = X/Y

Thus, the amount of X is a bit more than DOUBLE the amount of Y. Since the total of X+Y is 60, that means X = a little more than 40 and Y = a little less than 20.

GMAT assassins aren't born, they're made,
Rich

Hi Bunuel math expert! how can I practice more questions on this topic i.e. mixtures ? Thanks

18. Mixture Problems

• Theory
  Weighted Average and Mixture Problems on the GMAT
  Tips and Tricks: Mixtures
  
  • Questions
    DS Questions
    PS Questions

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
_________________

Answer: Option E

Video solution by GMATinsight



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