document.write( "Question 155357: A certain metal is 20% tin. How many kilograms of this metal must be mixed with 80 kg of a metal that is 70% tin to get a mental that is 50% tin? \n" ); document.write( "
Algebra.Com's Answer #114409 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Let x=number of kilograms of 20% tin alloy needed\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure tin in the 20% alloy (0.20x) plus the amount of pure tin in the 70% alloy (0.70*80) has to equal the amount of pure tin in the final mixture (0.50(80+x)). So, our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "0.20x+0.70*80=0.50(80+x) get rid of parens and simplify\r
\n" ); document.write( "\n" ); document.write( "0.20x+56=40+0.50x subtract 0.20x and also 40 from each side\r
\n" ); document.write( "\n" ); document.write( "0.20x-0.20x+56-40=40-40+0.50x-0.20x collect like terms\r
\n" ); document.write( "\n" ); document.write( "16=0.30x divide each side by 0.30\r
\n" ); document.write( "\n" ); document.write( "x=53.33 kilograms -----------amount of 20% alloy needed\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "53.33*0.20+0.70*80=0.50*133.33
\n" ); document.write( "10.67+56=66.67
\n" ); document.write( "66.67=66.67\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor
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