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

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One-fourth of a solution that was 10% by weight was replaced [#permalink] New post 13 Mar 2013, 12:52
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53% (02:36) correct 47% (01:38) wrong based on 189 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?

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, 01:15, 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] New post 13 Mar 2013, 17:30
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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] New post 13 Mar 2013, 19:52
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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/
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Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] New post 13 Mar 2013, 21:22
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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 !


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: One-fourth of a solution that was 10% by weight was replaced [#permalink] New post 14 Mar 2013, 01:16
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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%

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: One-fourth of a solution that was 10% by weight was replaced [#permalink] New post 14 Mar 2013, 08:51
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

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

10*3x/4 + p*1x/4 = 16*x
=> 30 + p = 64
=> p =
A
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Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] New post 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%

ans A it is.


thanks m8 that really helped. clear and easy
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Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] New post 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?

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!
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Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] New post 19 Apr 2014, 05:41
Alligation is the KEY

3/4 of 10%------------------------------1/4 of X%

-------------------16%----------------------------


(X-16)%------------------------------------------6%

(X-16)/6=3/1

X=34%
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Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] New post 04 Aug 2014, 21:46
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 :)
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Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] New post 04 Aug 2014, 23:05
Expert's post
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/
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One-fourth of a solution that was 10% by weight was replaced [#permalink] New post 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.
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Re: One-fourth of a solution that was 10% by weight was replaced [#permalink] New post 05 Aug 2014, 02:07
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%

Answer = A
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One-fourth of a solution that was 10% by weight was replaced [#permalink] New post 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

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

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 %
One-fourth of a solution that was 10% by weight was replaced   [#permalink] 25 Oct 2014, 05:45
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