<PRE><B>Question 257068</B>
How many gallons of pure alcohol should be mixed with 20 gal of a 15% alcohol solution to obtain a mixture that is 25% alcohol?
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Let x = amt of pure alcohol required
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A per cent equation
.15(20) + 1x = .25(x+20)
3 + 1x = .25x + 5
1x - .25x = 5 - 3
.75x = 2
x = {{{2/.75}}}
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x = 2{{{2/3}}} gal of pure alcohol required
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Marin Caswell needs 10% hydrocholoric acid for a chemistry experiment.
 How much 5% acid should mix with 60 mL of 20% acid to get a 10% solution? 
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Let x = amt of 5% mixture required
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.05x + .20(60) = .10(x+60)
.05x + 12 = .10x + 6
12 - 6 = .10x - .05x
6 = .05x
x = {{{6/.05}}}
x = 120 ml of 5% stuff
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Prove this
.05(120) + .20(60) = .10(120 + 60)
6 + 12 = .10(180)
18 = 18
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You can prove the 1st one, like we did this one. </PRE>