Question 295596: How many ounces of pure alcohol must be added to 40 ounces of a 70% alcohol solution to produce an 85% alcohol solution? Found 4 solutions by mananth, nerdybill, ikleyn, josgarithmetic:Answer by mananth(16949) (Show Source):
You can put this solution on YOUR website! How many ounces of pure alcohol must be added to 40 ounces of a 70% alcohol solution to produce an 85% alcohol solution?
let quantity of pure alcohol to be added be x ounces
70% alcohol is 40 ounces
the total will be x+40 ounces
x+0.7*40 = (x+40)*0.85
x+2.8 =0.85x +34
0.15x=34-2.8
0.15x=31.2
x=208
You can put this solution on YOUR website! How many ounces of pure alcohol must be added to 40 ounces of a 70% alcohol solution to produce an 85% alcohol solution?.
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Let x = ounces of pure alcohol added
then
x + .70(40) = .85(40+x)
x + 28 = 34 + .85x
.15x + 28 = 34
.15x = 6
x = 6/.15
x = 40 ounces
You can put this solution on YOUR website! .
How many ounces of pure alcohol must be added to 40 ounces of a 70% alcohol solution
to produce an 85% alcohol solution?
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The solution in the post by @mananth is incorrect.
I came to bring a correct solution.
let quantity of pure alcohol to be added be x ounces
70% alcohol is 40 ounces
the total will be x+40 ounces
x + 0.7*40 = (x+40)*0.85
x + 28 = 0.85x + 34
0.15x = 34 - 28
0.15x = 6
x = 6/0.15 = 40.
ANSWER. 40 ounces of pure alcohol must be added.
CHECK for the concentration: = 0.85 ! Precisely correct !
Solved correctly.
Notice that 85% concentration is precisely midpoint between 70% and 100%,
so the answer is obvious and can be guessed MENTALLY.
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