document.write( "Question 1169800: Hi
\n" ); document.write( "John has 36 coins comprising 20c 50c and $1 coins. He has twice as many $1 coins as 20c coins and the value of the 50c coins $4.40 more than the total value of the 20c coins.\r
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Algebra.Com's Answer #794646 by ikleyn(52776) $\"\"$ $\"About$

You can put this solution on YOUR website!
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document.write( "Let x be the number of the 20c coins.\r\n" );
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document.write( "Then the number of the $1 coins is 2x, and the number of the 50c coins is (36-x-2x) = 36-3x.\r\n" );
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document.write( "From the condition, you have this equation\r\n" );
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document.write( "    50*(36-3x) = 20x + 440   cents.\r\n" );
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document.write( "Solve it\r\n" );
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document.write( "    1800 - 150x = 20x + 440\r\n" );
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document.write( "    1800 - 440  = 20x + 150x\r\n" );
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document.write( "       1360     = 170x\r\n" );
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document.write( "         x      =  = 8.\r\n" );
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document.write( "Thus the number of the 20c coins is 8.\r\n" );
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document.write( "Hence, the number of the $1 coins is twice as many, i.e. 16.    ANSWER\r\n" );
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\n" ); document.write( "\n" ); document.write( "The problem is easily solved using one single unknown and one single equation.\r
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