<PRE><B>Question 211001</B>
How much solution of Hydrochloric acid at 11% concentration should be mixed
 with 8% concentration to produce a solution of 800 milliliters at 10%?
:
Let x = amt of 11% stuff
Resulting amt is to be 800 ml, therefore
(800-x) = amt of 8% stuff
:
.11x + .08(800-x) = .10(800)
:
.11x + 64 - .08x = 80
:
.11x - .08x = 80 - 64
:
.03x = 16
x = {{{16/.03}}}
x = 533{{{1/3}}} ml of 11% stuff
then
800 - 533{{{1/3}}} = 266{{{2/3}}} of 8% stuff
:
:
Check solution using decimals
.11(533.33) + .08(266.67) = .10(800)
58.67 + 21.33 = 80</PRE>