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# Jackie has two solutions that are 2 percent sulfuric acid

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Intern
Joined: 02 Aug 2015
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Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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20 Aug 2015, 18:27
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25% (medium)

Question Stats:

78% (01:27) correct 22% (01:50) wrong based on 2165 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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20 Aug 2015, 20:48
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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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Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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24 Aug 2015, 05:18
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5
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
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Joined: 07 Apr 2015
Posts: 164
Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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Updated on: 25 Aug 2015, 06:34
6
$$0,02*x + 0,12 * (60-x) = 0,05 * 60$$
$$0,02*x + 7,2 - 0,12*x = 3$$
$$4,2 = 0,1*x$$
$$x = 0,42$$

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

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24 Aug 2015, 10:19
2
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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Updated on: 25 Aug 2015, 02:33
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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24 Aug 2015, 20:01
2
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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24 Aug 2015, 20:04
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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16 Jan 2016, 21:36
5
3
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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14 Feb 2017, 22:21
4
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 = 3 liters
Therefore, 7 * 3 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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16 Feb 2017, 13:00
3
1
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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16 Feb 2017, 16:56
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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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22 May 2017, 12:01
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Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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10 Aug 2017, 11:36
2
Top Contributor
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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Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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10 Aug 2017, 11:49
Top Contributor
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

Another approach is to 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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Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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11 Aug 2017, 11:14
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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11 Aug 2017, 11:24
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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20 Sep 2017, 11:02
.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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26 Sep 2017, 05:15
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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27 Sep 2017, 02:10
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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Re: Jackie has two solutions that are 2 percent sulfuric acid &nbs [#permalink] 27 Sep 2017, 02:10

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