document.write( "Question 1200052: What amount of pure acid must be added to 500 mL of a 25% acid solution to produce a 50% acid solution?
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Algebra.Com's Answer #834078 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "The response from the other tutor shows a standard formal algebraic solution -- which you should understand, if you are a student learning how to solve problems using algebra.

\n" ); document.write( "If the speed of obtaining the answer is important and any method can be used (as in a timed math competition), here is an informal solution method that is much faster and easier than the formal algebraic method.

\n" ); document.write( "The 50% target concentration is 1/3 of the way from the 25% concentration of the original acid solution to the 100% concentration of the acid that is added. (25 to 50 is a difference of 25; 25 to 100 is a difference of 75; 25/75 = 1/3.)

\n" ); document.write( "That means 1/3 of the mixture should be what is being added.

\n" ); document.write( "So 2/3 of the final mixture is the original 25% acid solution; and that means the amount of 100% acid being added is half the amount of the original 25% acid solution.

\n" ); document.write( "ANSWER: 1/2 of 500 mL = 250 mL

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