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# One-fourth of a solution that was 10% by weight was replaced

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Senior Manager
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One-fourth of a solution that was 10% by weight was replaced [#permalink]

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13 Mar 2013, 13:52
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One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight?

A. 34%
B. 24%
C. 22%
D. 18%
E. 8.5%

[Reveal] Spoiler:
I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach .
many thanks !

I started with 10% ------------ x% in the first row

16% in the second row

and x-16% and 6% in the last one

this gives me x-16/6= ? "stuck here "
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Mar 2013, 02:15, edited 1 time in total.
Edited the question.

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Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]

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13 Mar 2013, 18:30
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guerrero25 wrote:
One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.

34%
24%
22%
18%
8.5%

I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach .
many thanks !

I started with 10% ------------ x% in the first row

16% in the second row

and x-16% and 6% in the last one

this gives me x-16/6= ? "stuck here "

This is the first question I trying to explain here, so please bear with my mistakes.

The best I could think of solving this is using nos.

Lets say we have a 100 ml of solution and it has 10 gm of sugar.

100 ml ---- 10 gm
75 ml ------- 7.5 gm
25 ml ------- 2.5 gm

Once we remove 25 ml, we have only 7.5 gm sugar left to make 16% of 100 ml solution we need 8.5 gm more sugar.
so we need 8.5 gm in 25 ml, which is 8.5 * 4 = 34 gm/ (25 * 4).

Ans A) 34 %

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Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]

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13 Mar 2013, 20:52
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guerrero25 wrote:
One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.

34%
24%
22%
18%
8.5%

I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach .
many thanks !

I started with 10% ------------ x% in the first row

16% in the second row

and x-16% and 6% in the last one

this gives me x-16/6= ? "stuck here "

The allegation formula is this:

w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 is the ratio of weights of the two solutions. Here, since 1/4 of the first solution is replaced, the weight of the first solution is 3/4 and that of the second solution is 1/4 so w1/w2 = 3/1

A2 - Concentration of second solution which we have to find here
Aavg - Concentration of mixture which is 16%
A1 - Concentration of first solution which is 10%

3/1 = (A2 - 16)/(16 - 10)
A2 = 34%

To find out how you get this formula, check: http://www.veritasprep.com/blog/2011/03 ... -averages/
For more on mixtures and replacement, check:
http://www.veritasprep.com/blog/2011/04 ... -mixtures/
http://www.veritasprep.com/blog/2012/01 ... -mixtures/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17802 [7], given: 235 Director Status: Done with formalities.. and back.. Joined: 15 Sep 2012 Posts: 635 Kudos [?]: 664 [6], given: 23 Location: India Concentration: Strategy, General Management Schools: Olin - Wash U - Class of 2015 WE: Information Technology (Computer Software) Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] ### Show Tags 13 Mar 2013, 22:22 6 This post received KUDOS 1 This post was BOOKMARKED guerrero25 wrote: One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight. 34% 24% 22% 18% 8.5% I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach . many thanks ! Instead of using complex calculations and remembering formulae, why dont u directly get to weighted average. 3 parts of 10% + 1 part of x (unknown) % = 4 parts of 16% => x% = 64%-30% = 34% ans A it is. _________________ Lets Kudos!!! Black Friday Debrief Kudos [?]: 664 [6], given: 23 Math Expert Joined: 02 Sep 2009 Posts: 42259 Kudos [?]: 132712 [5], given: 12335 Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] ### Show Tags 14 Mar 2013, 02:16 5 This post received KUDOS Expert's post 3 This post was BOOKMARKED One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight? A. 34% B. 24% C. 22% D. 18% E. 8.5% This is a weighted average question. Say the second solution (which was 1/4 th of total) was x% sugar, then 3/4*0.1+1/4*x=1*0.16 --> x=0.34. Alternately you can consider total solution to be 100 liters and in this case you'll have: 75*0.1+25*x=100*0.16 --> x=0.34. Answer: A. Hope it helps. _________________ Kudos [?]: 132712 [5], given: 12335 Senior Manager Joined: 10 Apr 2012 Posts: 277 Kudos [?]: 1181 [0], given: 325 Location: United States Concentration: Technology, Other GPA: 2.44 WE: Project Management (Telecommunications) Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] ### Show Tags 14 Mar 2013, 09:51 Thank you all for helping me understand the problem . Now I am getting a good grip on the concepts .The members /experts here 're truly amazing ! Kudos all Regards! Kudos [?]: 1181 [0], given: 325 Intern Joined: 10 Sep 2013 Posts: 17 Kudos [?]: 8 [0], given: 0 Location: India Concentration: Strategy GMAT 1: 700 Q50 V34 GPA: 3.8 WE: General Management (Energy and Utilities) Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] ### Show Tags 17 Sep 2013, 05:38 One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight? A. 34% B. 24% C. 22% D. 18% E. 8.5% let the volume of the original solution be X and p be the percentage of sugar in the solution which is replaced. then 10*3x/4 + p*1x/4 = 16*x => 30 + p = 64 => p = A Kudos [?]: 8 [0], given: 0 Intern Joined: 06 Feb 2012 Posts: 16 Kudos [?]: 52 [0], given: 7 Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] ### Show Tags 21 Oct 2013, 03:41 Vips0000 wrote: guerrero25 wrote: One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight. 34% 24% 22% 18% 8.5% I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach . many thanks ! Instead of using complex calculations and remembering formulae, why dont u directly get to weighted average. 3 parts of 10% + 1 part of x (unknown) % = 4 parts of 16% => x% = 64%-30% = 34% ans A it is. thanks m8 that really helped. clear and easy Kudos [?]: 52 [0], given: 7 Current Student Joined: 06 Sep 2013 Posts: 1972 Kudos [?]: 741 [0], given: 355 Concentration: Finance Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] ### Show Tags 16 Dec 2013, 16:05 guerrero25 wrote: One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight? A. 34% B. 24% C. 22% D. 18% E. 8.5% [Reveal] Spoiler: I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach . many thanks ! I started with 10% ------------ x% in the first row 16% in the second row and x-16% and 6% in the last one this gives me x-16/6= ? "stuck here " By differentials too So basically we need to find the % of sugar by weight of the second solution. We have this for the first solution 10% and we have the ratio of both and that the final solution was 16%. So then since only 25% (1/4) is being replace the unknown percentage has a weight of 1 while the original amount remains with a weight of 3 (Thus the ratio 3:1). So by applying the differences between the result of 16% and each point value we can solve for ‘x’, in this case the percentage of sugar in the second solution 3(-6) + (x-16) = 0 -18 + x -16 = 0 X= 34 Hence (A) is the answer Hope it helps Cheers! J Kudos [?]: 741 [0], given: 355 Director Joined: 03 Aug 2012 Posts: 899 Kudos [?]: 911 [1], given: 322 Concentration: General Management, General Management GMAT 1: 630 Q47 V29 GMAT 2: 680 Q50 V32 GPA: 3.7 WE: Information Technology (Investment Banking) Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] ### Show Tags 19 Apr 2014, 06:41 1 This post received KUDOS Alligation is the KEY 3/4 of 10%------------------------------1/4 of X% -------------------16%---------------------------- (X-16)%------------------------------------------6% (X-16)/6=3/1 X=34% _________________ Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________ Kudos [?]: 911 [1], given: 322 Manager Joined: 22 Feb 2009 Posts: 210 Kudos [?]: 180 [0], given: 148 Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] ### Show Tags 04 Aug 2014, 22:46 TGC wrote: Alligation is the KEY 3/4 of 10%------------------------------1/4 of X% -------------------16%---------------------------- (X-16)%------------------------------------------6% (X-16)/6=3/1 X=34% Is it called cross method? I think it is the fastest solution _________________ ......................................................................... +1 Kudos please, if you like my post Kudos [?]: 180 [0], given: 148 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7736 Kudos [?]: 17802 [0], given: 235 Location: Pune, India Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] ### Show Tags 05 Aug 2014, 00:05 vad3tha wrote: TGC wrote: Alligation is the KEY 3/4 of 10%------------------------------1/4 of X% -------------------16%---------------------------- (X-16)%------------------------------------------6% (X-16)/6=3/1 X=34% Is it called cross method? I think it is the fastest solution This is called Alligation or the scale method (a variant). Read more about this method here: http://www.veritasprep.com/blog/2011/03 ... -averages/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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One-fourth of a solution that was 10% by weight was replaced [#permalink]

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05 Aug 2014, 00:50
guerrero25 wrote:
One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight?

A. 34%
B. 24%
C. 22%
D. 18%
E. 8.5%

Original - sugar=0.1*s1, other = 0.9*s1
Removed - sugar= -0.1 * s1/4, other= - 0.9*s1/4
Added - sugar= x * s1/4, other= (1-x)*s1/4
Resultant sugar= 0.16*s1, other=0.84*s1

(0.1*s1 - 0.1*s1/4 + x*s1/4) / (0.9*s1 - 0.9*s1/4 + (1-x)*s1/4) = 16/84
x=0.34 or percent sugar by weight is 34%

The advantage with this approach is that you can use it for any mixture problem.
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Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]

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05 Aug 2014, 03:07
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Total Solution ........................ Sugar Concentration

100 ............................................ 10
(Assumed)

25% solution removed

100-25 ......................................... $$10-\frac{10}{4}$$

75 ................................................... $$\frac{30}{4}$$

Same quantity has been replaced. Say the new solution sugar weight = x%

75 + 25 ......................................... $$\frac{30}{4} + \frac{25x}{100}$$

Given that the resultant solution has sugar concentration of 16%

So equating the above

$$\frac{30}{4} + \frac{25x}{100} = 16$$

x = 34%

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One-fourth of a solution that was 10% by weight was replaced [#permalink]

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24 Oct 2014, 06:20
Differential method

10------16-------x
problem says that 1/4 was replaced by higher concentration solution, meaning that ratio of 10% sol to X% sol was 3:1
16-10=6 --> 6*3=18 ---> 16+18=34% needs to be second solution

Checking

10-----16-----------------34
6x=18y
x/y=18/6=3:1, it is correct

A

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One-fourth of a solution that was 10% by weight was replaced [#permalink]

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25 Oct 2014, 06:45
guerrero25 wrote:
One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight?

A. 34%
B. 24%
C. 22%
D. 18%
E. 8.5%

[Reveal] Spoiler:
I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach .
many thanks !

I started with 10% ------------ x% in the first row

16% in the second row

and x-16% and 6% in the last one

this gives me x-16/6= ? "stuck here "

[img]10% x%
16%
3/4 1/4
3 1
[/img]

The difference 6 obtained by usual cross subtraction is 1 part. So 3 parts is 18 . To get 18, 16 must be subtracted from X. So x is the addition of 16 plus 18

THEREFORE X = 18+16 = 34 %

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Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]

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07 May 2015, 03:23
Vips0000 wrote:
guerrero25 wrote:
One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.
34%
24%
22%
18%
8.5%

I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach .
many thanks !

Instead of using complex calculations and remembering formulae, why dont u directly get to weighted average.

3 parts of 10% + 1 part of x (unknown) % = 4 parts of 16%
=> x% = 64%-30% = 34%

ans A it is.

Thank you!
Your explanation is crisp and clear...kudos to you!

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One-fourth of a solution that was 10% by weight was replaced [#permalink]

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26 Aug 2015, 16:54
let x=sugar % of solution 2
.1-.1/4+x/4=.16
x=.34=34%

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Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]

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18 Jan 2016, 05:52
This question could be solved with logic and simple calculation.

If we use equal amount of solution with different concentrations, then the resulting concentration is the Mean of concentrations

So, if we apply the same concept above, solution 1 has 10 % and solution 2 need to be 31% to make the resulting concentration 16% ( if both with same amount solutions, for example 75 ml)

To have same resulting concentration 16%, we need to decrease the amount of solution 2 and increase its concentration above 31%.

only 1 answer fits.

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Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]

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18 Jan 2016, 16:02
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Thank you SO MUCH VeritasPrepKarishma !!! You made this topic look so easy in your blogs below. I am getting all the mixture questions correct now with the help of your blog.

VeritasPrepKarishma wrote:
guerrero25 wrote:
One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.

34%
24%
22%
18%
8.5%

I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach .
many thanks !

I started with 10% ------------ x% in the first row

16% in the second row

and x-16% and 6% in the last one

this gives me x-16/6= ? "stuck here "

The allegation formula is this:

w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 is the ratio of weights of the two solutions. Here, since 1/4 of the first solution is replaced, the weight of the first solution is 3/4 and that of the second solution is 1/4 so w1/w2 = 3/1

A2 - Concentration of second solution which we have to find here
Aavg - Concentration of mixture which is 16%
A1 - Concentration of first solution which is 10%

3/1 = (A2 - 16)/(16 - 10)
A2 = 34%

To find out how you get this formula, check: http://www.veritasprep.com/blog/2011/03 ... -averages/
For more on mixtures and replacement, check:
http://www.veritasprep.com/blog/2011/04 ... -mixtures/
http://www.veritasprep.com/blog/2012/01 ... -mixtures/

Kudos [?]: 15 [1], given: 6

Re: One-fourth of a solution that was 10% by weight was replaced   [#permalink] 18 Jan 2016, 16:02

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