<PRE><B>Question 239139</B>
Mixture A is 7.5% acid. To raise the concentration of acid to 10%, some pure acid will be added to the mixture.
 How many liters of mixture A and how many liters of pure acid are needed to end up with 200 liters of 10% solution?
:
Let x = amt of pure acid required, 
It says the total has to be 200 liters, therefore
(200-x) = 7.5% stuff (A)
:
1.0x + .075(200-x) = .10(200)
:
x + 15 - .075x = 20
:
x - .075x = 20 - 15
:
.925x = 5
x = {{{5/.925}}}
x = 5.4 liters of pure acid require
then
200 - 5.4 = 194.6 liters of 7.5% stuff (A)
;
:
Check solution in original equaiton
5.4 + .075(194.6) = .10(200)
5.4 + 14.6 = 20
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