document.write( "Question 1161205: A merchant has coffee worth $50 a pound that she wishes to mix with 70 pounds of coffee worth $80 a pound to get a mixture that can be sold for $70 a pound. How many pounds of the $50 coffee should be used? \n" ); document.write( "

Algebra.Com's Answer #784697 by greenestamps(13203) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "Here is a typical traditional algebraic setup for solving the problem.

\n" ); document.write( "x pounds of coffee worth $50 per pound, plus 70 pounds of coffee worth $80 per pound, equals (70+x) pounds of coffee worth $70 per pound:

\n" ); document.write( " $\"50%28x%29%2B80%2870%29+=+70%2870%2Bx%29\"$

\n" ); document.write( "The equation is easily solved using basic algebra.

\n" ); document.write( "Here is an alternative method that can be used if a formal algebraic solution is not required. For this particular problem, the amount of work required for both methods is comparable; but for many mixture problems like this, this alternative method can be MUCH faster and easier.

\n" ); document.write( "The per-pound price of the mixture ($70) is 2/3 of the way from the per-pound price of the lower priced coffee ($50) to the per-pound price of the higher priced coffee ($80).

\n" ); document.write( "That means 2/3 of the mixture must be the higher priced coffee. In other words, the more expensive coffee and the less expensive coffee should be mixed in the ratio 2:1.

\n" ); document.write( "Since she is using 70 pounds of the more expensive coffee, she should use half as much -- 35 pounds -- of the less expensive coffee.

\n" ); document.write( "Of course that is the answer you should get using the formal algebraic method shown above.

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

\n" );