SOLUTION: How much of a 5% solution was there originally if, after 66 mgs of water evaporated, it turned into a 11% solution? I can do mixture problems when you are adding but for some re

Question 800637: How much of a 5% solution was there originally if, after 66 mgs of water evaporated, it turned into a 11% solution?
I can do mixture problems when you are adding but for some reason I get stuck up on these problems when you have to subtract (i.e. evaporating).
Found 2 solutions by josmiceli, DrBeeee:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = the original mgs of solution
= mgs of what is being dissolved ( I'll say acid )
-------------
The key is that the mgs of acid never changes

Can't find calculator- you can finish & verify

Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
I'm with you! Never did evaporation, but should be the same as adding except for a minus sign lol.
Let s = mgm of solution (never changes)
Let w = initial mgm of water
Initially we have
(1) s/(s+w) = 0.05 or
(2) s = 0.05*(s+w) or
(3) s = 0.05s + 0.05w or
(4) s(1-0.05) = 0.05w or
(5) 0.95s = 0.05w or
(6) s = (0.05/0.95)*w or
(7) s = w/19
With enough experience solving solutions, you may have been able to write equation (6) directly from the problem statement, i.e.
(8) s/w = (5%/95%)
Now that 66 mgms of water has evaporated, the new solution has increased to 11%.
You may readily see how we can write
(8) s = (11/89)*(w-66) for the new solution. If not let's go slower.
We now have
(9) s/(s+w-66) = 0.11 or
(10) s = 0.11s + 0.11*(w-66) or
(11) 0.89s = 0.11*(w-66) or
(12) s = (11/89)*(w-66)
Since the amount of solution, s, does not change we can set (7) equal to (12) to get
(13) w/19 = 11/89w -11*66/89 or
(14) (11/89 - 1/19)*w = 11*66/89 or
(15) (209-89)/(89*19))*w = 11*66/89 or
(16) w = 11*66*19/120 or
(17) w = 114.95
Then
(18) s = w/19 or
(10) s = 6.05
Answer: initially there is 6.05 mgm of solution and 114.95 mgm of water.

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