# math

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Mary has a solution that is 60% alcohol and another that is 20% alcohol. How much of each should she use to make 100 milliliters of a solution that is 52% alcohol?

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let the volume of 60% solution used be x
then the volume of the 20% has to be 100-x

solve:

.6x +.2(100-x) = .52(100)

• math -

where did you get 100-x???

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If the both of us together have 100 cents, and I have x, then you have 100-x

Mary wants to end up with 100 ml
if x ml are of the 60% type, then
100-x ml must be the 40% kind.

notice (x) + (100-x) = 100

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and that doesnt answer the question, the question asks how much of each solution she should use

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its 20% not 40% percent

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if the 60% one is x, then 100-x = 100-60 which equals 40, not 20

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Emmet, look how I defined x.
Once you find x you have the amount of 60% solution used, plug x into 100-x and you have the 20% solution.

Do you know how to solve the equation?

(the 40% is a typo, look at my equation, I used 20%)

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