SOLUTION: I have been struggling with this word problem and I don't know where to start: What amount of a 50% acid solution must be mixed with a 25% solution to produce 500 mL of a 45% solut

Question 990833: I have been struggling with this word problem and I don't know where to start: What amount of a 50% acid solution must be mixed with a 25% solution to produce 500 mL of a 45% solution?
If you could just help me get started that would be greatly appreciated.
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Do it like this one:
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One alloy contains 80% gold (by weight)and another alloy contains 55% gold.how many grams of each alloy should be combined to make 40g of an alloy that contains 70% gold.
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e = amount of 80%
f = amount of 55%
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e + f = 40 (total alloy)
80e + 55f = 70*40 (total gold)
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Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
What amount of a 50% acid solution must be mixed with a 25% solution to produce 500 mL of a 45% solution?
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Let x be the volume of a 50% acid solution (in mL).
Then the volume of the pure acid in this solution is 0.5x mL.

Next, the volume of a 25% solution is (500-x) mL, and it contains 0.25*(500-x) of the pure acid.

Now we have 0.5x + 0.25*(500-x) of the pure acid in the mixt, and it is 45% of 500 mL, i.e 0.45*500 = 225 mL.

Thus we have an equation

0.5x + 0.25(500-x) = 225.

Solve it:

0.5x + 125 - 0.25x = 225,

0.25x = 225 - 125 = 100,

x = = 400.

Answer. 400 mL of 50% acid solution should be mixed with (500-400) = 100 mL of 25% acid solution.

Check: 0.5*400 + 0.25*100 = 225 mL of pure acid.

             = 0.45.

The answer is correct.

You may want to look into the lesson Mixture problems in this site where many of similar problems are considered and solved.

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