SOLUTION: suppose that one cocktail is 25% ethanol and another cocktail is 110% ethanol. How many milliliters of each solution should be mixed to make 10.5 milliliters of an 80% ethanol solu

Question 793912: suppose that one cocktail is 25% ethanol and another cocktail is 110% ethanol. How many milliliters of each solution should be mixed to make 10.5 milliliters of an 80% ethanol solution?
x + y = 10.5
-x -x
y = 10.5 - x
.25x + 1.10y = 10.5(0.8)
.25x + 1.10(10.5 - x) = 10.5(0.8)
.25x + 11.55 - 1.10x = 8.4
-.25x .25x
11.55 - 1.35x = 8.4
-11.55 -11.55
-1.35x = -3.15
/-1.35 /-1.35
x = 2.33Reapeating.
Am I correct on my work? My answer seems wrong to me. If I move my answer into percentage form....it will be 233%?
Please help!!!
Thank you for your time, greatly appreciated!!!!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
suppose that one cocktail is 25% ethanol and another cocktail is 110% ethanol.
How many milliliters of each solution should be mixed to make 10.5 milliliters of an 80% ethanol solution?
:
I think you are on the right track here, but you can't have 110% ethanol, 100%
is all ethenol, lets say it is 100%
:
.25x + 1(10.5-x) = .80(10.5)
.25x - 1x + 10.5 = 8.4
-.75x = 8.4 - 10.5
-.75x = -2.1
x = -2.1/-.75
x = +2.8 milliliters of the 25% ethanol
and
10.5 - 2.8 = 7.7 milliliters of 100% ethanol
;
;
You can confirm this
.25(2.8) + 1(7.7) = .8(10.5)
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