Question 1041591
80 gal*0.004 per/gal (0.4%)=0.32 gal of "pure". Let x=amount of 1% or 0.01 that is added.
0.32 gal+.01x=(80+x)(0.006), the final concentration.
0.32 +0.01x=0.480+0.006 x
0.004x=0.160
x=40 gal
The total amount will be 120 gal of 0.006 =0.720 "pure".  One third of that was the more concentrated, and 2/3 was the less concentrated.  And the distance between 0.4 and 1.0 is 1/3 of the way at 0.6.  Thus, one could predict because the final concentration was twice as close to the start, that twice as much of the starting concentration would be in the final mixture.  That is the check.