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# m07 Q8

Author Message
Intern
Joined: 13 Jan 2010
Posts: 23
Followers: 1

Kudos [?]: 6 [0], given: 10

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02 Aug 2010, 05:56
Can someone help explain the solution? I seem to be having problems with mixtures. Would you have any recommendation on study material that I can use for the same?

If 12 ounces of a strong vinegar solution are diluted with 50 ounces of water to form a three-percent vinegar solution, what was the concentration of the original solution?

19.3%
17%
16.67%
15.5%
12.5%
Intern
Joined: 23 Jun 2010
Posts: 36
Followers: 0

Kudos [?]: 28 [1] , given: 5

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02 Aug 2010, 16:39
1
KUDOS
I would approach the question in the following way:
1) The question reads 12oz of strong vinegar solution (we do not know what is the composition of the solution) is added to 50oz of water.
2) Thus we get 12oz + 50oz = 62oz of total solution.
3) The question asks what was the original concentration of the solution.
4) Now the amount of vinegar in the solution remains unchanged since we only added 50oz of water to dilute it to 3%. We use this information to calculate the total amount of vinegar currently present in the solution. i.e. 3% of 62oz = 1.8~
5) From step 3 we know the amount of vinegar is unchanged. That means there is 1.8~oz vinegar in 12oz.
6) Solving for "How many in 100oz" will give us the concentration. That is, (100 * 10)/(12 * 18) = 15.5%
Hope that helped.
(Don't forget to give Kudos if it did! )
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Re: m07 Q8   [#permalink] 02 Aug 2010, 16:39
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# m07 Q8

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