SOLUTION: Two vessels contains mixtures of alcohol and water. In one there is three times as much alcohol as water; in the other, five times as much water as alcohol. How much must be drawn
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Question 344970: Two vessels contains mixtures of alcohol and water. In one there is three times as much alcohol as water; in the other, five times as much water as alcohol. How much must be drawn off from each to fill a third vessel which holds 7 gallons, in order that its contents may be half water and half alcohol.
Thanks for the answer! Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Two vessels contains mixtures of alcohol and water.
In one there is three times as much alcohol as water; in the other,
five times as much water as alcohol.
How much must be drawn off from each to fill a third vessel which holds 7
gallons, in order that its contents may be half water and half alcohol.
:
Let x = amt drawn from the 1st vessel
then
(7-x) = amt drawn from the 2nd vessel
:
Write an equation based on the percent alcohol
In the 1st vessel 3 times as much alcohol as water = 75 % alcohol
In the 2nd vessel 5 times as much water as alcohol = 16.7 % alcohol
:
.75x + .167(7-x) = .50(7)
:
.75x + 1.167 - .167x = 3.5
:
.750x - .167x = 3.5 - 1.167
:
.583x = 2.333
x =
x = 4 gal from the 1st vessel
then
3 gal from the 2nd vessel
:
:
Check in original equation
.75(4) + .167(3) = .50(7)
3 + .5 = 3.5
