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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?

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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.

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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

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 = 3 liters Therefore, 7 * 3 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

Re: Jackie has two solutions that are 2 percent sulfuric acid
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10 Aug 2017, 11:49

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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

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

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.