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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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Question Stats:

53% (02:33) correct
47% (01:39) wrong based on 168 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

1

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

4

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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

3

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

2

This post received KUDOS

Expert's post

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. _________________