Question 30787: How many gallons of 51% alcohol solution should be added to some of a 57% alcohol solution to make a 438 gal of a 55% solution? Thank you. Found 2 solutions by checkley71, Fermat:Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website! (.51X)+(438-X).57=(438).55
.51X+249.66-.57X=240.9
-.06X=-8.76 OR X=8.76/.06 OR 146 GAL OF .51 % & 438-146=292 GAL OF 57%
You can put this solution on YOUR website! Let there be X gall of 51% alcohol
Let there be Y gall of 57% alcohol
If a solution of alcohol, X gall, has 51% concentration, then 51% of that volume is pure alcohol and the remainder, 49%, is pure water.
51% alcohol: 0.51*X gall of pure alcohol and 0.49*X gall of pure water
57% alcohol: 0.57*Y gall of pure alcohol and 0.43*Y gall of pure water
adding the two mixtures together,
55% alcohol: (0.51*X + 0.57*Y) gall of pure alcohol and (0.49*X + 0.43*Y) gall of pure water.
Concentration of mixture = quantity of alcohol over quantity of mixture (water plus alcohol)
55% = (0.51X + 0.57Y) gall over (X + Y) galls
0.55 = (0.51X + 0.57Y)/(X + Y)
0.55(X+Y) = 0.51X + 0.57Y
substituting for X+Y = 438,
0.55*438 = 0.51X + 0.57Y
240.9 = 0.51X + 0.57Y
substituting for X = 438 - Y
240.9 = 0.51(438 - Y) + 0.57Y
240.9 = 223.38 - 0.51Y + 0.57Y
240.9 - 223.38 = -0.51Y + 0.57Y
17.52 = 0.06Y
Y = 17.52/0.06
Y = 192 gall
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X = 146 gall
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Ans: need 146 gall of 51% alcohol
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