Question

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.

Answer 1

since there is no number for \(m\) lets pick one

Answer 2

okay :)

Answer 3

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

Answer 4

are you sure the variables are both \(m\) ?

Answer 5

yes it states both variables are m

Answer 6

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

Answer 7

so clearly something is wrong with this problem

Answer 8

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

Answer 9

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

Answer 10

\(m\) ounces, which is \(m\%\) salt means you have
\[m\times \frac{m}{100}\] amount of salt

Answer 11

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 \(m\)

Answer 12

did i lose you yet?

Answer 13

i hope so because i made a typo
it should be
\[\frac{m^2}{100}+x=\frac{2m}{100}(m+x)\]

Answer 14

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 :)

Answer 15

then solve for \(x\) which should not be too hard

Answer 16

\[\frac{m^2}{100}+x=\frac{2m^2}{100}+\frac{m}{50}x\] etc

Answer 17

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?

Answer 18

oh did you reduce the 2 and 100 to 1 and 50, then multiply by x? that's what it appears you did...

Answer 19

yes

Answer 20

subtract \(\frac{m^2}{100}\) from both sides next

Answer 21

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?

Answer 22

makes no difference

Answer 23

okay so what do i do with the -m(sq.)/100 ? (since i subtracted like you said)

Answer 24

or should i not have that?

Answer 25

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?

Answer 26

\[\frac{m^2}{100}+x=\frac{2m^2}{100}+\frac{m}{50}x\]
\[x=\frac{m^2}{100}+\frac{m}{50}x\]
\[x-\frac{m}{50}x=\frac{m^2}{100}\]
\[x(1-\frac{m}{50})=\frac{m^2}{100}\] then divide both sides by \(1-\frac{m}{50}\)

Answer 27

so you'd be left with x= m(sq.)/100 + -50m(sq)/100m