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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 04 Oct 2008, 02:53
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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%
Manager
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Re: Solution Problem [#permalink] New post 04 Oct 2008, 03:40
The answer is (A) = 34%

start with 10% and x% in the first row

then 16% in the second row

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

then you are being told that:

x-16/6 = 3/1 (your started with 4 parts but replaced 1 part).

x-16 = 18

x = 34


:)
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Re: Solution Problem [#permalink] New post 04 Oct 2008, 04:24
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your solution assumes that x-16/6 = 3/1
it is possible that x-16/6 = 1/3
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Re: Solution Problem [#permalink] New post 04 Oct 2008, 04:26
ast wrote:
your solution assumes that x-16/6 = 3/1
it is possible that x-16/6 = 1/3


No - since x is the unknown part meaning its the part that was replaced - given as one part out of four (three where left unchanged).
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Re: Solution Problem [#permalink] New post 05 Oct 2008, 13:28
2/3*(0.1) +1/3*(x) = 0.16

or x = 0.34 and hence 34%.
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Re: Solution Problem [#permalink] New post 05 Oct 2008, 14:03
Lets look at the sugar and Compare them :

Lets say K is the total volume:
Remaining sugar after replacing 1/4th = 3/4 (10K)/100
Addition of 1/4 with x sugar concentration = K/4(x/100)
Final Sugar = k*(16/100)

Equation

3/4 (10K)/100 + K/4(x/100) = k*(16/100)

X = 34.

Ans - A
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Re: Solution Problem [#permalink] New post 25 Mar 2009, 14:06
How do you solve this using the 2D grid method?

This is how I translated the information. Where did I go wrong?

Attachment:
grid.JPG
grid.JPG [ 17.72 KiB | Viewed 1009 times ]
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Re: Solution Problem [#permalink] New post 25 Mar 2009, 21:26
Hi bigfernhead,
Lets apply the method mentioned in my earlier post abt mixtures. The table becomes:
Orig Rem Added Result
Conc 0.10 0.10 x 0.16
Amt 1 0.25 0.25 1

Now, if you have the equation: Orig-Rem+Added=Result, you will get 34%.

bigfernhead wrote:
How do you solve this using the 2D grid method?

This is how I translated the information. Where did I go wrong?

Attachment:
grid.JPG
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Re: Solution Problem [#permalink] New post 27 Jun 2009, 07:02
I swear to God, I love you guys - but you make things so complicated.....

try:

3/4*(10) + 1/4*(x) = 16

30/4 + x/4 = 64/4

x/4 = 34/4

x=34
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Re: Solution Problem [#permalink] New post 29 Jun 2009, 21:32
Refer to the attached document whihc shows the solution in the simplest and the least complicaed way.
I don't know how to post images hence putting it in a document.

Can someone tell me how can we upoad the images ?

Ans 34
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Mixtures.doc [28 KiB]
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Re: Solution Problem   [#permalink] 29 Jun 2009, 21:32
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