Question 653234
a 3% salt solution is mixed with a 9% salt solution.
 how many grams of the 9% salt solution were used to make 150 g of a 4% salt solution?
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We convert the per cent to the decimal equiv.
In this case we write a amt of salt equation, the amt of salt of the two
solutions, equal amt in the resulting solution
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Let x = grams of 9% solution required
the result will be 150 grams, therefore
(150-x) = grams of 3% solution
:
.09x + .03(150-x) = .04(150)
.09x + 4.5 - .03x = 6
.09x - .03x = 6 - 4.5
.06x = 1.5
x = 1.5/.06
x = 25 grams of 9% solution required, which is what they asked 
then we would like to know
150 - 25 = 125 grams of 3% solution
:
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We can confirm this solution by finding the actual amt of salt in each solution,
see if it adds up to the amt of salt in the resulting solution
.09(25) + .03(125) = .04(150)
2.25 + 3.75 = 6; proves we have good answer
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This is a typical mixture equation, can be applied to most mixture problems.
Always a good idea, when you get the answer, to check it as we did here.
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