document.write( "Question 1044273: A metallurgist has one alloy containing 24% aluminum and another containing 70% aluminum. How many pounds of each alloy must he use to make 44 pounds of a third alloy containing 58% aluminum? \n" ); document.write( "

Algebra.Com's Answer #659542 by ikleyn(52809) $\"\"$ $\"About$

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\n" ); document.write( "A metallurgist has one alloy containing 24% aluminum and another containing 70% aluminum.
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document.write( "Take \"x\" pounds    of the first  alloy containing 24% aluminum.\r\n" );
document.write( "Take (44-x) pounds of the second alloy containing 70% aluminum.\r\n" );
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document.write( "After melting and mixing these alloys you will have 44 pounds of a new alloy.\r\n" );
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document.write( "To provide a 58% content of aluminum in it, the amount \"x\" must satisfies this equation\r\n" );
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document.write( "0.24x + 0.7*(44-x) = 0.58*44,   or\r\n" );
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document.write( "0.24x + 30.8 - 0.7x = 25.52,   or\r\n" );
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document.write( "-0.46x = 25.52 - 30.8,\r\n" );
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document.write( "-0.46x = --5.28,\r\n" );
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document.write( "x =  = 11.478.\r\n" );
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document.write( "So, you need to mix 11.478 pounds of the 24% alloy with 44-11.478 = 32.52 pounds of the 70% alloy.\r\n" );
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document.write( "Answer.  Metallurgist need to mix 11.478 pounds of the 24% alloy with 32.522 pounds of the 70% alloy.\r\n" );
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\n" ); document.write( "\n" ); document.write( "For more mixture word problems on alloys see the lesson\r
\n" ); document.write( "\n" ); document.write( " - Word problems on mixtures for alloys \r
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