document.write( "Question 1138859: A retailer planned to buy some computers form a wholesaler for a total of ksh 1,800,000, before the retailer could buy the computers the price per unit was reduced by ksh.4000. this reduction in price enabled the retailer to buy 5 more computers using the same amount of money as originally planned. Determine the number of computers the retailer bought. \n" ); document.write( "

Algebra.Com's Answer #756657 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "Let n be the number of computers the retailer bought (the value under the question).\r\n" );
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document.write( "The price for each single computer then was  .\r\n" );
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document.write( "The planned price (before reducing) was  .\r\n" );
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document.write( "The difference in prices was 4000:\r\n" );
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document.write( "     -  = 4000.\r\n" );
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document.write( "To solve this equation and to find \"n\", first cancel the factor \"1000\" in both sides; then multiply both sides by n*(n-5).\r\n" );
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document.write( "     -  = 4\r\n" );
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document.write( "     1800*n - 1800*(n-5) = 4n*(x-5)\r\n" );
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document.write( "     1800x + 1800n + 9000 = 4n^2 - 20n\r\n" );
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document.write( "      4n^2 - 20n + 9000 = 0\r\n" );
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document.write( "      n^2  -  5n + 2250 = 0\r\n" );
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document.write( "      (n+45)*(n-50) = 0\r\n" );
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document.write( "The two roots are n= -45  and  n= 50,  but only positive value x= 50 is a meaningful solution.\r\n" );
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document.write( "ANSWER.  The retailer bought 50 computers.\r\n" );
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\n" ); document.write( "\n" ); document.write( "To see many other similar solved problems, look into the lesson\r
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