document.write( "Question 628905: Forty ounces of a 35% gold alloy are mixed with 62 ounces of a 20% alloy. Find the concentration of the resulting gold alloy. \n" ); document.write( "
Algebra.Com's Answer #395888 by Maths68(1474)\"\" \"About 
You can put this solution on YOUR website!
Alloy-A
\n" ); document.write( "========
\n" ); document.write( "Amount = 40 ounces
\n" ); document.write( "Concentration = 35%=0.35\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Alloy-B
\n" ); document.write( "========
\n" ); document.write( "Amount = 62 ounces
\n" ); document.write( "Concentration = 20%=0.20\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Resultant Alloy
\n" ); document.write( "===============
\n" ); document.write( "Amount = Alloy-A + Alloy-B = 40+62 = 102 ounces
\n" ); document.write( "Concentration = x\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "(Amount of Alloy-A)(Concentration of Alloy-A)+(Amount of Alloy-B)(Concentration of Alloy-B)= (Amount of Resultant Alloy)(Concentration of Resultant Alloy)
\n" ); document.write( "(40)(0.35)+(62)(0.20)= (102)(x)
\n" ); document.write( "14+12.4=102x
\n" ); document.write( "26.40=102x
\n" ); document.write( "102x=26.04
\n" ); document.write( "102x/102=26.04/102
\n" ); document.write( "x=0.2588\r
\n" ); document.write( "\n" ); document.write( "Concentration of the resultant alloy will be approximately 0.2588 or approximately 26%.
\n" ); document.write( " \n" ); document.write( "
\n" );