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Onefourth of a solution that was 10% by weight was replaced
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Updated on: 14 Mar 2013, 02:15
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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% 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 x16% and 6% in the last one
this gives me x16/6= ? "stuck here "
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Originally posted by guerrero25 on 13 Mar 2013, 13:52.
Last edited by Bunuel on 14 Mar 2013, 02:15, edited 1 time in total.
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Re: Onefourth of a solution that was 10% by weight was replaced
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13 Mar 2013, 20:52
guerrero25 wrote: Onefourth 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 x16% and 6% in the last one
this gives me x16/6= ? "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/
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Re: Onefourth of a solution that was 10% by weight was replaced
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13 Mar 2013, 22:22
guerrero25 wrote: Onefourth 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.
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Re: Onefourth of a solution that was 10% by weight was replaced
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13 Mar 2013, 18:30
guerrero25 wrote: Onefourth 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 x16% and 6% in the last one
this gives me x16/6= ? "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: Onefourth of a solution that was 10% by weight was replaced
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14 Mar 2013, 02:16
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.
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Re: Onefourth of a solution that was 10% by weight was replaced
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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!



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Re: Onefourth of a solution that was 10% by weight was replaced
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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



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Re: Onefourth of a solution that was 10% by weight was replaced
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21 Oct 2013, 03:41
Vips0000 wrote: guerrero25 wrote: Onefourth 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



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Re: Onefourth of a solution that was 10% by weight was replaced
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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% 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 x16% and 6% in the last one
this gives me x16/6= ? "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) + (x16) = 0 18 + x 16 = 0 X= 34 Hence (A) is the answer Hope it helps Cheers! J



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Re: Onefourth of a solution that was 10% by weight was replaced
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19 Apr 2014, 06:41
Alligation is the KEY
3/4 of 10%1/4 of X%
16%
(X16)%6%
(X16)/6=3/1
X=34%



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Re: Onefourth of a solution that was 10% by weight was replaced
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04 Aug 2014, 22:46
TGC wrote: Alligation is the KEY
3/4 of 10%1/4 of X%
16%
(X16)%6%
(X16)/6=3/1
X=34% Is it called cross method? I think it is the fastest solution
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Re: Onefourth of a solution that was 10% by weight was replaced
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05 Aug 2014, 00:05
vad3tha wrote: TGC wrote: Alligation is the KEY
3/4 of 10%1/4 of X%
16%
(X16)%6%
(X16)/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/
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Onefourth of a solution that was 10% by weight was replaced
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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= (1x)*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 + (1x)*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: Onefourth of a solution that was 10% by weight was replaced
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05 Aug 2014, 03:07
Total Solution ........................ Sugar Concentration 100 ............................................ 10 (Assumed) 25% solution removed 10025 ......................................... \(10\frac{10}{4}\) 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% Answer = A
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Onefourth of a solution that was 10% by weight was replaced
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24 Oct 2014, 06:20
Differential method
1016x problem says that 1/4 was replaced by higher concentration solution, meaning that ratio of 10% sol to X% sol was 3:1 1610=6 > 6*3=18 > 16+18=34% needs to be second solution
Checking
101634 6x=18y x/y=18/6=3:1, it is correct
A



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Onefourth of a solution that was 10% by weight was replaced
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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% 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 x16% and 6% in the last one
this gives me x16/6= ? "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: Onefourth of a solution that was 10% by weight was replaced
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07 May 2015, 03:23
Vips0000 wrote: guerrero25 wrote: Onefourth 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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Onefourth of a solution that was 10% by weight was replaced
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Updated on: 19 Sep 2018, 14:41
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 x=first solution weight y=second solution sugar % 3/4*.1x+1/4*xy=.16x→ .3x+xy=.64x→ y=.34=34% A
Originally posted by gracie on 26 Aug 2015, 16:54.
Last edited by gracie on 19 Sep 2018, 14:41, edited 1 time in total.



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Re: Onefourth of a solution that was 10% by weight was replaced
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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. Answer A



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Re: Onefourth of a solution that was 10% by weight was replaced
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18 Jan 2016, 16:02
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: Onefourth 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 x16% and 6% in the last one
this gives me x16/6= ? "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/




Re: Onefourth of a solution that was 10% by weight was replaced
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