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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, 19:27
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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?

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, 21:48
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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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Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Most Helpful Community Reply Intern Joined: 02 Jul 2015 Posts: 17 Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink] Show Tags 24 Aug 2015, 06:18 13 3 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 Posts: 176 Jackie has two solutions that are 2 percent sulfuric acid [#permalink] Show Tags Updated on: 25 Aug 2015, 07:34 5 $$0,02*x + 0,12 * (60-x) = 0,05 * 60$$ $$0,02*x + 7,2 - 0,12*x = 3$$ $$4,2 = 0,1*x$$ $$x = 0,42$$ = 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. Intern Joined: 24 Aug 2015 Posts: 8 Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink] Show Tags 24 Aug 2015, 11:19 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 Senior Manager Joined: 15 Sep 2011 Posts: 343 Location: United States WE: Corporate Finance (Manufacturing) Jackie has two solutions that are 2 percent sulfuric acid [#permalink] Show Tags Updated on: 25 Aug 2015, 03:33 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, 17:49. Last edited by mejia401 on 25 Aug 2015, 03:33, edited 1 time in total. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8184 Location: Pune, India Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink] Show Tags 24 Aug 2015, 21:01 1 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. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >

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

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24 Aug 2015, 21:04
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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Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 30 Dec 2015 Posts: 91 GPA: 3.92 WE: Engineering (Aerospace and Defense) Jackie has two solutions that are 2 percent sulfuric acid [#permalink] Show Tags 16 Jan 2016, 22:36 5 2 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 _________________ If you analyze enough data, you can predict the future.....its calculating probability, nothing more! Manager Joined: 08 Sep 2015 Posts: 70 Jackie has two solutions that are 2 percent sulfuric acid [#permalink] Show Tags 14 Feb 2017, 23:21 2 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. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12175 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Jackie has two solutions that are 2 percent sulfuric acid [#permalink] Show Tags 16 Feb 2017, 14:00 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.... Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Joined: 07 Dec 2014
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Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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16 Feb 2017, 17:56
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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, 13:01
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Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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10 Aug 2017, 12:36
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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, 12:49
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, 12:14
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, 12:24
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, 12:02
.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, 06:15
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, 03:10
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, 03:10

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