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One-fourth of a solution that was 10% by weight was replaced [#permalink]
13 Mar 2013, 12:52
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Difficulty:
75% (hard)
Question Stats:
57% (02:44) correct
43% (01:55) wrong based on 373 sessions
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?
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
Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]
13 Mar 2013, 17:30
4
This post received KUDOS
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).
Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]
13 Mar 2013, 19:52
8
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Expert's post
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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%
Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]
13 Mar 2013, 21:22
5
This post received KUDOS
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%
Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]
14 Mar 2013, 01:16
4
This post received KUDOS
Expert's post
2
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.
Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]
17 Sep 2013, 04:38
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
Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]
21 Oct 2013, 02:41
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%
Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]
16 Dec 2013, 15:05
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?
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
Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________
One-fourth of a solution that was 10% by weight was replaced [#permalink]
04 Aug 2014, 23:50
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. _________________
One-fourth of a solution that was 10% by weight was replaced [#permalink]
24 Oct 2014, 05:20
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
One-fourth of a solution that was 10% by weight was replaced [#permalink]
25 Oct 2014, 05:45
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?
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
Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]
07 May 2015, 02:23
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!
Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]
18 Jan 2016, 04:52
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%.
Re: One-fourth of a solution that was 10% by weight was replaced [#permalink]
18 Jan 2016, 15:02
1
This post received KUDOS
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%
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