Question 12332: A 100% concentrate is to be mixed with having a concentration of 40% to obtain 55 gallons of a mixture with a concentration of 75%. How much of the 100% concentrate will be needed?
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! we have a volume of liquid1 added to a volume of liquid2 to produce a mixture.
Let x = volume of 100% concentrate
Let y = volume of 40% concentrate
so we have
(x gallons of 100%) + (y gallons of 40%) = (55 gallons of 75%)
converting this into Algebra gives:
(x*1) + (y*0.40) = (55*0.75) --> or in fact, keep the percentages as percentages, it doesn't matter.
x + 0.4y = 41.25
OK...so we have 1 equation with 2 unknowns, which we therefore cannot solve. However, we also know that x+y=55. So, y=55-x, hence:
x + 0.4(55-x) = 41.25
x + 22 - 0.4x = 41.25
0.6x = 19.25
--> x = 32.08333 gallons
--> x = 32.08 gallons (to 4 significant figures)
jon.
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