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Jackie has two solutions that are 2 percent sulfuric acid
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20 Aug 2015, 18:27
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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?
Re: Jackie has two solutions that are 2 percent sulfuric acid
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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?
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.
Jackie has two solutions that are 2 percent sulfuric acid
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Updated on: 26 Jul 2020, 13:58
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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
Answer: E
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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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Re: Jackie has two solutions that are 2 percent sulfuric acid
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24 Aug 2015, 10:19
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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
Jackie has two solutions that are 2 percent sulfuric acid
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Updated on: 25 Aug 2015, 02:33
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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
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.
Re: Jackie has two solutions that are 2 percent sulfuric acid
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24 Aug 2015, 20:01
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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
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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
Jackie has two solutions that are 2 percent sulfuric acid
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16 Jan 2016, 21:36
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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
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14 Feb 2017, 22:21
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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.
Re: Jackie has two solutions that are 2 percent sulfuric acid
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16 Feb 2017, 13:00
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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....
Re: Jackie has two solutions that are 2 percent sulfuric acid
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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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10 Aug 2017, 11:36
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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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Jackie has two solutions that are 2 percent sulfuric acid
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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
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
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.