Question 150850
A mixture containing 6% salt is to be mixed with 2 ounces of a mixture which is 15% salt, in order to obtain a solution which is 12% salt. How much of the first solution must be used? 
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How do you solve a problem "Interactively"?
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Anyway.
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Let x = amt of 1st solution required
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.06x + .15(2) = .12(x+2)
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.06x + .3 = .12x + .24
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.3 - .24 = .12x - .06x
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.06 = .06x
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x = 1 oz of the 1st solution required
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Check solution:
.06(1) + .15(2) = .12(1+2)
.06 + .30 = .36