document.write( "Question 513472: AN alloy containing 40% gold and an alloy containing 70% gold are to be mixed to produce 50 pounds of an alloy containing 60% gold. How much of each alloy is needed? \n" ); document.write( "
Algebra.Com's Answer #343040 by Maths68(1474)![]() ![]() You can put this solution on YOUR website! Alloy A \n" ); document.write( "Amount = x lb \n" ); document.write( "Concentration =40% =0.4 \n" ); document.write( "================================ \n" ); document.write( "Alloy B \n" ); document.write( "Amount = 50-x lb \n" ); document.write( "Concentration =70% = 0.70 \n" ); document.write( "============================== \n" ); document.write( "Final Alloy \n" ); document.write( "Amount =50 lb \n" ); document.write( "Concentration =60%=0.6 \n" ); document.write( "================================= \n" ); document.write( "[Amount Alloy A * Concentration A] + [Amount Alloy B * Concentration of B] = Amount of Final Alloy * Concentration of Final Alloy \n" ); document.write( "================================== \n" ); document.write( "(x)(0.4)+(50-x)(0.7)=(50)(0.6) \n" ); document.write( "0.4x+35-0.7x=30 \n" ); document.write( "0.4x-0.7x=30-35 \n" ); document.write( "-0.3x=-5 \n" ); document.write( "-0.3x/-0.3=-5/-0.3 \n" ); document.write( "x=16.66 \n" ); document.write( "=========================== \n" ); document.write( "Alloy A \n" ); document.write( "Amount = 16.66 lb \n" ); document.write( "Concentration =40% =0.4 \n" ); document.write( "================================ \n" ); document.write( "Alloy B \n" ); document.write( "Amount = 50-16.66=33.33 lb \n" ); document.write( "Concentration =70% = 0.70 \n" ); document.write( "============================== \n" ); document.write( "==================================================== \n" ); document.write( "A 16.66 pounds alloy containing 40% gold and a 33.33 pounds alloy containing 70% gold are to be mixed to produce 50 pounds of an alloy containing 60% gold \n" ); document.write( " |