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PS Weighted Averages

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by logitech » Fri Jan 16, 2009 12:43 am
Now it is time to practice the Weighted Averages concept:

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?
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by bluementor » Fri Jan 16, 2009 1:20 am
12*(x/100) + 50*(0/100) = 62*(3/100)
12x = 62*3
4x = 62
x=15.5

Answer 15.5%

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Re: PS Weighted Averages

by logitech » Fri Jan 16, 2009 1:51 am
(12) X---------------3------0 (50)

So lets compare the distances:

X-3/3

Lets compare the WEIGHTS

50/12

SO

X-3/3 = 50/12

x=15.5
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by 4meonly » Fri Jan 16, 2009 4:00 am
bluementor wrote:12*(x/100) + 50*(0/100) = 62*(3/100)
12x = 62*3
4x = 62
x=15.5

Answer 15.5%

-BM-
I went through the % of water (x)

(12*x/100) + 50 = 0.97(50+12)
x=84.5%
therefore, concentration of vinegar is 15.5%
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by DavoodBeater » Fri Jan 16, 2009 7:46 am
I like your way Logitech, Great.
Thanks
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by missrochelle » Tue Aug 24, 2010 10:14 am
Given the wording on this problem - I mistook it for a volume problem rather than weighted average. I thought the NEW solution would be 12oz of vinegar, plus 50 oz of water...and tried to calculate 12 is 3% of what number ....

Any guidance here on how to identify this problem as a mixture/weighted average not volume?

Also - I still don't get the solution - step by step can someone explain if we are assuming the old mixture to be 12% vinegar and what % water?? I tried to set it up as a chart but don't know what to fill in.
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