Determine the amounts of 20% and 50% salt solutions that should be mixed to obtain 300 gallons of 41% salt solution.

1 answer:

3 0

Assume that solution with 20% and 50% salt is x and y gallons respectively.

Therefore,

x+y = 300 => y=300-x

Additionally
20x+50y = 300*41 = 12300

Substituting for y

20x+50(300-x) = 12300 => 20x+15000-50x = 123000 => -30x = -2700

x= 90 gallons
y= 300-x = 300 - 90 =210 gallons

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