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

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Intern
Joined: 27 Sep 2008
Posts: 14
One-fourth of a solution that was 10% by weight was replaced [#permalink]

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04 Oct 2008, 03:53
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0% (00:00) correct 100% (02:34) wrong based on 0 sessions

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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%

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Manager
Joined: 27 Sep 2008
Posts: 76
Re: Solution Problem [#permalink]

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04 Oct 2008, 04:40
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

Intern
Joined: 27 Sep 2008
Posts: 14
Re: Solution Problem [#permalink]

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04 Oct 2008, 05:24
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your solution assumes that x-16/6 = 3/1
it is possible that x-16/6 = 1/3
Manager
Joined: 27 Sep 2008
Posts: 76
Re: Solution Problem [#permalink]

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04 Oct 2008, 05:26
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).
SVP
Joined: 17 Jun 2008
Posts: 1502
Re: Solution Problem [#permalink]

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05 Oct 2008, 14:28
2/3*(0.1) +1/3*(x) = 0.16

or x = 0.34 and hence 34%.
Intern
Joined: 06 Jul 2008
Posts: 29
Re: Solution Problem [#permalink]

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05 Oct 2008, 15:03
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
Retired Moderator
Joined: 18 Jul 2008
Posts: 920
Re: Solution Problem [#permalink]

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25 Mar 2009, 15:06
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 [ 17.72 KiB | Viewed 2087 times ]
Director
Joined: 01 Apr 2008
Posts: 846
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Re: Solution Problem [#permalink]

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25 Mar 2009, 22:26
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
Manager
Joined: 30 Nov 2008
Posts: 91
Re: Solution Problem [#permalink]

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27 Jun 2009, 08:02
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
Manager
Joined: 28 Jan 2004
Posts: 201
Location: India
Re: Solution Problem [#permalink]

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29 Jun 2009, 22:32
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

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Attachments

Mixtures.doc [28 KiB]
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Re: Solution Problem   [#permalink] 29 Jun 2009, 22:32
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# One-fourth of a solution that was 10% by weight was replaced

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