anonymous
  • anonymous
This probably is super simple and I'm just overthinking but here is the question: A chemist has "m" ounces of a salt-water solution that is "m" percent salt. How much salt must be added to the solution to be 2m percent salt.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
since there is no number for \(m\) lets pick one
anonymous
  • anonymous
okay :)
anonymous
  • anonymous
suppose she has 100 ounces of solution that is 100% salt then you can't solve it, since there is no way for it to be 200% salt

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anonymous
  • anonymous
are you sure the variables are both \(m\) ?
anonymous
  • anonymous
yes it states both variables are m
anonymous
  • anonymous
suppose she has 50 ounces that are 50% salt then there is still no way to solve it, because there is no way to make it 100% salt
anonymous
  • anonymous
so clearly something is wrong with this problem
anonymous
  • anonymous
maybe we could do it if \(m=10\) 10 ounces, 10% salt, and you want to make it 20% salt out of the ten ounces you have \(.1\times 10=1\) ounce of salt and 9 of water, add \(x\) ounces of salt and you have \(1+x\) ounces of salt and \(10+x\) ounces all together, you want \(1+x=.2(10+x)\) which you can solve
anonymous
  • anonymous
we can try it with \(m\) and see what happens, maybe it will be clear why it doesn't work with \(m=50\) or above
anonymous
  • anonymous
\(m\) ounces, which is \(m\%\) salt means you have \[m\times \frac{m}{100}\] amount of salt
anonymous
  • anonymous
you add \(x\) ounces of salt to it, you will have \[\frac{m^2}{100}+x\] amount of salt, and \(m+x\) ounces all together you want \[\frac{m^2}{100}+x=\frac{2x}{100}(m+x)\] and you have to solve this for \(x\) in terms of \(m\)
anonymous
  • anonymous
did i lose you yet?
anonymous
  • anonymous
i hope so because i made a typo it should be \[\frac{m^2}{100}+x=\frac{2m}{100}(m+x)\]
anonymous
  • anonymous
okay i understand where you get the m(squared) over 100. oh well i think you fixed up what was confusing with your original so one second :)
anonymous
  • anonymous
then solve for \(x\) which should not be too hard
anonymous
  • anonymous
\[\frac{m^2}{100}+x=\frac{2m^2}{100}+\frac{m}{50}x\] etc
anonymous
  • anonymous
okay so i was gonna ask what to do with the 2m/100 (m+z) and I figured distributing would be correct but what did you do there?
anonymous
  • anonymous
oh did you reduce the 2 and 100 to 1 and 50, then multiply by x? that's what it appears you did...
anonymous
  • anonymous
yes
anonymous
  • anonymous
subtract \(\frac{m^2}{100}\) from both sides next
anonymous
  • anonymous
and a question for writing it down: should the x be right next to the m or it being off to the side a bit doesn't change that?
anonymous
  • anonymous
makes no difference
anonymous
  • anonymous
okay so what do i do with the -m(sq.)/100 ? (since i subtracted like you said)
anonymous
  • anonymous
or should i not have that?
anonymous
  • anonymous
should i add a +m(sq.)/100 to get it on the other side since i have a -m(sq.)/100 adding to the x on the left side?
anonymous
  • anonymous
\[\frac{m^2}{100}+x=\frac{2m^2}{100}+\frac{m}{50}x\] \[x=\frac{m^2}{100}+\frac{m}{50}x\] \[x-\frac{m}{50}x=\frac{m^2}{100}\] \[x(1-\frac{m}{50})=\frac{m^2}{100}\] then divide both sides by \(1-\frac{m}{50}\)
anonymous
  • anonymous
so you'd be left with x= m(sq.)/100 + -50m(sq)/100m

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