SOLUTION: a tank contains 20 gallons of a mixture of alcohol and water that is 40% alcohol by volume. how much of the mixture should be removed and replaced by an equal volume of water so th
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Question 215928: a tank contains 20 gallons of a mixture of alcohol and water that is 40% alcohol by volume. how much of the mixture should be removed and replaced by an equal volume of water so that the resulting solution will be 25% alcohol by volume?
can you make a representation and solution to this?
i tried making this equation: .40(20)-.40(x)+.1(20)=.25(20-x=20) but this doesn't work out when i check it... Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! a tank contains 20 gallons of a mixture of alcohol and water that is 40% alcohol by volume.
how much of the mixture should be removed and replaced by an equal volume of
water so that the resulting solution will be 25% alcohol by volume?
:
Let x = amt of water to be added,
and
Let x = amt of mixture to be removed
:
:
Water is a 0% solution
:
The equation:
.40(20-x) + .0(x) = .25(20)
or just
.40(20-x) = .25(20)
:
8 - .4x = 5
:
-.4x = 5 - 8
;
-.4x = -3
x =
x = +7.5 gal mixture removed and 7.5 gal water added
;
:
Check in an "amt of alcohol" equation
.4(20-7.5) = .25(20)
.4(12.5) = 5
5 = 5
