Question 340310: Two vessels contain mixtures of alcohol and water. In one there is three times as much as alcohol as water; in other, five times as much as water as alcohol. How much much be drawn off from each to fill a third vessel which holds 7 gallons, in order that its contents may be half alcohol and half water.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Two vessels contain mixtures of alcohol and water.
In one there is three times as much as alcohol as water;
3/4 alcohol
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in other, five times as much as water as alcohol.
1/6 alcohol
How much much be drawn off from each to fill a third vessel which holds 7 gallons, in order that its contents may be half alcohol and half water.
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Quantity: x + y = 7 gallons
Alcohol::::0.75x+0.1667y = 0.50*7
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Multiply 1st by 75
Multiply 2nd by 100
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75x + 75y = 75*7
75x + 16.67y = 50*7
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Substract 2nd from 1st and solve for "y":
58.33y = 25*7
y = 3 gallons (amt to draw from the 2nd vessel)
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x = 4 gallons (amt to draw from the 1st vessel)
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Cheers,
Stan H.
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