Question 27140
Chas has some 12% acid solution and some 32% acid solution. Determine how much
of each he needs in order to get 40ml of a 20% acid solution. 
I tried setting up the problem like this: .12(x)+ .32(x)=.2(40) but i didn't
get the answer right.
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Your mistake is in letting the same letter x represent two different
quantities.  You must let x represent only one of the quantities, and then
either y or 40-x will represent the other.  Never let the same letter 
represent two different quantities in the same problem. There are four ways 
to solve it.
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1. One unknown letting x = no of ml. of weaker solution and 40-x = no of ml. of stronger solution.
`
That equation would be .12(x) + .32(40-x) = .2(40)
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2. One unknown letting x = no of ml. of stronger solution and 40-x = no of ml. of weaker solution.
`
That equation would be .12(40-x) + .32(x) = .2(40)
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3. Two unknowns letting x = no of ml. of weaker solution and y = no of ml. of stronger solution.
`
This would be a system of two equations and two unknowns:

   x +    y =    40
.12x + .32y = .2(40)
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4. Two unknowns letting x = no of ml. of stronger solution and y = no of ml.
of weaker solution.

This would be a system of two equations and two unknowns:

   x +    y =    40
.32x + .12y = .2(40)
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For any of these four ways, the answer is 24 ml of the weaker solution
and 16 ml of the stronger solution.

Edwin
AnlytcPhil@aol.com