document.write( "Question 524212: An alloy of silver and gold weighs 15 ounces in air and 14 ounces in
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document.write( "water. Assuming that the silver loses 1/10 of its weight in water and that
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document.write( "gold loses 1/19 of its weight, how many ounces of each metal are in the
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document.write( "alloy? \n" );
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Algebra.Com's Answer #347718 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! An alloy of silver and gold weighs 15 ounces in air and 14 ounces in \n" ); document.write( "water. Assuming that the silver loses 1/10 of its weight in water and that \n" ); document.write( "gold loses 1/19 of its weight, how many ounces of each metal are in the \n" ); document.write( "alloy? \n" ); document.write( ": \n" ); document.write( "Let s = amt of silver \n" ); document.write( "Let g = amt of gold \n" ); document.write( ": \n" ); document.write( "\"An alloy of silver and gold weighs 15 ounces in air\" \n" ); document.write( "s + g = 15 \n" ); document.write( "s = (15-g) \n" ); document.write( ": \n" ); document.write( "And 14 ounces in water. \n" ); document.write( ": Assuming that the silver loses 1/10 of its weight in water and that \n" ); document.write( "gold loses 1/19 of its weight, \n" ); document.write( "therefore \n" ); document.write( " \n" ); document.write( "Multiply by 190, to get rid of the denominators \n" ); document.write( "19(9s) + 10(18g) = 190(14) \n" ); document.write( "171s + 180g = 2660 \n" ); document.write( ": \n" ); document.write( "replace s with (15-g), find g: \n" ); document.write( "171(15-g) + 180g = 2660 \n" ); document.write( "2565 - 171g + 180g = 2660 \n" ); document.write( "-171g + 180g = 2660 - 2565 \n" ); document.write( "9g = 95 \n" ); document.write( "g = \n" ); document.write( "g = 10.556 oz of gold \n" ); document.write( "then \n" ); document.write( "15 - 10.556 = 4.444 oz of silver\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |