document.write( "Question 1156877: Hi\r
\n" ); document.write( "\n" ); document.write( "An alloy contains copper and silver. The amount of silver exceeds copper by 50 grams. After 25 percent of the original amount of silver was added to the alloy,the new alloy contains 60 percent silver.\r
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Algebra.Com's Answer #779663 by ikleyn(52781) $\"\"$ $\"About$

You can put this solution on YOUR website!
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document.write( "Let x = original mass of silver in the alloy (in grams); y = mass of of copper.\r\n" );
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document.write( "After adding 25% of the original mass of silver, there are x+0.25x = 1.25x grams silver in the alloy.\r\n" );
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document.write( "From the condition, you have these two equations\r\n" );
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document.write( "    x - y = 50   grams         (1)   (the difference of the original amounts)\r\n" );
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document.write( "     = 0.6             (2)   (silver concentration of 60% after adding)\r\n" );
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document.write( "Simplify equation (2)\r\n" );
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document.write( "    1.25x = 0.6*(1.25x + y)\r\n" );
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document.write( "    1.25x = 0.75x + 0.6y.\r\n" );
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document.write( "    0.5x = 0.6y                (3)\r\n" );
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document.write( "From equation (1), express x = 50 + y and substitute it into equation (3).\r\n" );
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document.write( "    0.5*(50 + y) = 0.6y\r\n" );
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document.write( "    25 + 0.5y    = 0.6y \r\n" );
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document.write( "    25           = 0.6y - 0.5y = 0.1y\r\n" );
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document.write( "    y = 25/0.1 = 250.\r\n" );
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document.write( "Then from equation (1),  x = 250 + 50 = 300.\r\n" );
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document.write( "ANSWER.  Originally, the alloy was 300 gram solver and 250 grams copper.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved.\r
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\n" ); document.write( "\n" ); document.write( "You may check that my solution is correct.\r
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