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M07 #8

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M07 #8 [#permalink] New post 18 Sep 2009, 18:13
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%

Weak in mixtures and the OE is not clear.Other methods?
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Re: M07 #8 [#permalink] New post 18 Sep 2009, 18:19
tejal777 wrote:
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%

Weak in mixtures and the OE is not clear.Other methods?


Answer is D.

New volume of the diluted solution = 12+50 = 62 oz.
Total vinegar present in solution = 3% = 62*3/100 = 1.86 oz
Concentration of original solution = 1.86*100/12 = 15.5%
Re: M07 #8   [#permalink] 18 Sep 2009, 18:19
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