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Difficulty: 505-555 Level,   Algebra,   Mixture Problems,                           
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Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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blockman wrote:
Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required?

A) 18
B) 20
C) 24
D) 36
E) 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

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Originally posted by BrentGMATPrepNow on 10 Aug 2017, 12:49.
Last edited by BrentGMATPrepNow on 10 Nov 2020, 07:24, edited 3 times in total.
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Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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\(0,02*x + 0,12 * (60-x) = 0,05 * 60\)
\(0,02*x + 7,2 - 0,12*x = 3\)
\(4,2 = 0,1*x\)
\(x = 0,42\)

= 42%

Originally posted by noTh1ng on 24 Aug 2015, 08:17.
Last edited by noTh1ng on 25 Aug 2015, 07:34, edited 1 time in total.
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Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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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
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Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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blendercroix wrote:
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.
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Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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blendercroix wrote:
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



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

Originally posted by KarishmaB on 24 Aug 2015, 21:04.
Last edited by KarishmaB on 19 Sep 2023, 04:24, edited 1 time in total.
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Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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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
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Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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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.
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Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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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:

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Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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blockman wrote:
Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required?

A) 18
B) 20
C) 24
D) 36
E) 42


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

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Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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.
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Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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adkikani wrote:
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.

Originally posted by KarishmaB on 27 Sep 2017, 03:10.
Last edited by KarishmaB on 19 Sep 2023, 04:25, edited 1 time in total.
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Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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blockman wrote:
Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required?

A) 18
B) 20
C) 24
D) 36
E) 42


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
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Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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!
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Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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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.

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Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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Hi Bunuel math expert! how can I practice more questions on this topic i.e. mixtures ? Thanks
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Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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VQ wrote:
Hi Bunuel math expert! how can I practice more questions on this topic i.e. mixtures ? Thanks


18. Mixture Problems



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink]
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blockman wrote:
Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required?

A) 18
B) 20
C) 24
D) 36
E) 42


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