document.write( "Question 272631: How many quarts of pure antifreeze must be added to 6 quarts of a 10% solution to obtain a 20% antifreeze soltuion? (Round to the nearest tenth.) \n" ); document.write( "

Algebra.Com's Answer #854113 by greenestamps(13326) $\"\"$ $\"About$

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\n" ); document.write( "While the problem was probably intended as one to be solved by formal algebra, here is quick and easy method for solving any 2-part mixture problem like this. The method is based on the fact that the ratio in which the two ingredients need to be mixed is exactly determined by where the target percentage lies between the two given percentages.

\n" ); document.write( "Without any more words than necessary, here is the solution to this problem using this method.

\n" ); document.write( "(1) The percentages of the two ingredients are 10% and 100%.
\n" ); document.write( "(2) The target percentage, 20%, is \"8 times as close\" to 10% as it is to 100% (the difference between 10% and 20% is 10%; the difference between 20% and 100% is 80%; 80% is 8 times as much as 10%).
\n" ); document.write( "(3) That means the amount of the 10% ingredient must be 8 times the amount of the 100% ingredient.
\n" ); document.write( "(4) The given amount of the 10% ingredient is 6 quarts, so the needed amount of the 100% ingredient is 1/8 of 6 quarts, which is 6/8 = 3/4 quarts.

\n" ); document.write( "ANSWER 3/4 of a quart, or 0.75 quarts

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