<PRE><B>Question 583180</B>
A 55 gallon barrel contains a mixture with a concentration of 40%.
 How much of this mixture must be withdrawn and replaced by 100% concentrate to bring up the mixture up to 75% concentration.
:
Let x = amt of 100% stuff required, and the amt of 40% mixture to be removed
:
.40(55-x) + x = .75(55)
22 - .4x + x = 41.25
.6x = 41.25 - 22
.6x = 19.25
x = 19.25/,6
x = 32.1 mixture removed, then 32.1 gal of 100% added to replace it</PRE>