document.write( "Question 66883: A candy store sells boxes of candy containing caramels and chocolates. Each box sells for $12.50 and holds 30 pieces of candy (all pieces are the same size). If the caramels cost $0.25 to produce and the chocolates cost $0.45 to produce, how many of each should be in a box to make a profit of $3.00? \n" ); document.write( "
Algebra.Com's Answer #47541 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Let x= number of caramels in the box \n" ); document.write( "And 30-x=number of chocolates in the box\r \n" ); document.write( "\n" ); document.write( "Now we are told that the cost of the caramels ($0.25x) plus the cost of the chocolates ($0.45(30-x)) must equal $12.50-$3.00=$9.50 in order to give us a profit of $3.00. So that gives us:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "$0.25x+$0.45(30-x)=$9.50 multiplying out\r \n" ); document.write( "\n" ); document.write( "$0.25x+$13.5-$0.45x=$9.50 subtract $13.50 from both sides \n" ); document.write( "$0.25x+$13.50-$13.50-$0.45x=$9.50-$13.50 collecting like terms\r \n" ); document.write( "\n" ); document.write( "-$0.20x=-$4\r \n" ); document.write( "\n" ); document.write( "x=20-------------number of caramels in the box\r \n" ); document.write( "\n" ); document.write( "30-x=30-20=10--------------number of chocolates in the box\r \n" ); document.write( "\n" ); document.write( "CK\r \n" ); document.write( "\n" ); document.write( "$0.25(20)+$0.45(10)=$12.50-$3.00 \n" ); document.write( "$5+$4.50=$9.50 \n" ); document.write( "$9.50=$9.50\r \n" ); document.write( "\n" ); document.write( "Hope this helps-----ptaylor\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |