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Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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Question Stats: 77% (01:56) correct 23% (02:23) wrong based on 3214 sessions

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

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Akgmat85 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

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.

The formula and its application in mixtures are discussed in the following two posts:
http://www.veritasprep.com/blog/2011/03 ... -averages/
http://www.veritasprep.com/blog/2011/04 ... -mixtures/
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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 mixer 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

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

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

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If you enjoy my solutions, you'll love my GMAT prep course. Originally posted by BrentGMATPrepNow on 10 Aug 2017, 11:49.
Last edited by BrentGMATPrepNow on 26 Jul 2020, 13:58, edited 2 times in total.
Intern  Joined: 02 Jul 2015
Posts: 16
Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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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.
##### General Discussion
Manager  Joined: 07 Apr 2015
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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, 07:17.
Last edited by noTh1ng on 25 Aug 2015, 06:34, edited 1 time in total.
Intern  Joined: 24 Aug 2015
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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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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

Hi blendercroix,

She got the 6 because it's the unknown multiplier of the ratio of 3x:7x. That is, $$3x + 7x = 10x$$, and 10x into $$60$$ is $$6$$. Six isn't a value but what helps calculate the ratio. Anyhow, it seems your logic is similar to what she posted, but instead of $$x$$ you made the unknown multiplier a "cup."

Kr,
Mejia

Originally posted by mejia401 on 24 Aug 2015, 16:49.
Last edited by mejia401 on 25 Aug 2015, 02:33, 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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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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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

Take a look at the ratios post: http://www.veritasprep.com/blog/2011/03 ... of-ratios/
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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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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....

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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

let x=liters of 2% solution needed
.02x+.12(60-x)=.05*60
x=42 liters
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Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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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

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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

let x=liters of 2% solution needed
.02x+.12y=.05(x+y)
x/y=7/3
x=7/10*60=42 liters
E
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Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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Quantity of Cheaper / Quantity of Dearer= CP of Dearer – Mean Price / Mean Price – CP of Cheaper

Quantity of Cheaper +Quantity of Dearer=60

CP of Dearer – Mean Price / Mean Price – CP of Cheaper= 12-5/5-2=7/3=7/10*60=42
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Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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.02A +.12B=.05(A+B)
2A+12B=5A+5B
7B=3A
7/3 = A/B
7*6=42
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Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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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

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

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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

EugeneFish has given the pictorial representation of the scale method.

You should check out this post instead: https://www.veritasprep.com/blog/2011/0 ... -averages/

It discusses the scale method, why it works and its pictorial and formula depiction.
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Veritas Prep GMAT Instructor Re: Jackie has two solutions that are 2 percent sulfuric acid   [#permalink] 27 Sep 2017, 02:10

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