document.write( "Question 1164056: A merchant has coffee worth $20 a pound that she wishes to mix with 90 pounds of coffee worth $80 a pound to get a mixture that can be sold for $40 a pound . How many pounds of the $20 coffee should be used ? \n" ); document.write( "

Algebra.Com's Answer #788352 by greenestamps(13216) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "A formal algebraic setup....

\n" ); document.write( "x pounds of coffee at $20 per pound, plus 90 pounds of coffee at $80 per pound, makes (90+x) pounds of coffee at $40 per pound:

\n" ); document.write( " $\"20%28x%29%2B80%2890%29+=+40%2890%2Bx%29\"$

\n" ); document.write( "Solve using basic algebra.

\n" ); document.write( "Informally, if formal algebra is not required....

\n" ); document.write( "The $40 per pound price of the mixture is \"twice as close to $20\" as it is to $80. (The difference between 20 and 40 is 20; the difference between 40 and 80 is 40. 20 is half of 40, so 40 is twice as close to 20 as it is to 80.)

\n" ); document.write( "Therefore, the mixture must contain twice as much of the $20 per pound coffee as it does the $80 per pound coffee. Since there are 90 pounds of the more expensive coffee, you need 180 pounds of the less expensive coffee.

\n" ); document.write( "ANSWER: 180 pounds

\n" ); document.write( "Obviously, that is the answer you should get if you finish the problem by the formal algebraic method.

\n" ); document.write( " \n" ); document.write( "

\n" );