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by naaga » Sun Feb 15, 2009 10:19 pm
1. If 12 ounces of strong solution of vinegar is mixed with 50 ounces of water to form three percent vinegar, what was the original strength of the vinegar solution?

15.5%
20%
18%
12%
19.5%


folks ....how to solve this kind of questions
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by sureshbala » Sun Feb 15, 2009 10:32 pm
Let x% be the strength of vinegar in the original solution.

Given that x%(12)=3%(62)

So x = 15.5

Hence the original solution contains 15.5% of vinegar
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by suddhasil » Sun Feb 15, 2009 10:35 pm
3 percent vinegar = 3 ounces of vinegar in 100 ounces of the solution.

So in 62 ounces (50 water + 12 vinegar) we have (.03 * 62) 1.86 ounces of vinegar.

All the 1.86 ounces are in the vinegar solution.

So the percentage of vinegar in the original solution will be

(1.86/12) * 100 = 15.5%

IMO A
OA?
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thanks

by naaga » Sun Feb 15, 2009 11:03 pm
thanks sureshbala and suddhasil...
the OA is (A)
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