document.write( "Question 1198580: Paul bought a number of shares of stock for a total of $3000.Three months later the stock had increased in value by $5 per share and he sold all but 50 shares and regained his original investment of $3000.How many shares did he sell? \n" ); document.write( "

Algebra.Com's Answer #832191 by ikleyn(52797) $\"\"$ $\"About$

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\n" ); document.write( "Paul bought a number of shares of stock for a total of $3000.
\n" ); document.write( "Three months later the stock had increased in value by $5 per share and he sold all
\n" ); document.write( "but 50 shares and regained his original investment of $3000. How many shares did he sell?
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document.write( "Let n be the number of shares Paul sold.\r\n" );
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document.write( "Then the original number of shares he bought was (n+50).\r\n" );
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document.write( "The buying price per share was   dollars.\r\n" );
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document.write( "The selling price per share was  dollars.\r\n" );
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document.write( "The difference of these prices was 5 dollars, according to the problem,\r\n" );
document.write( "which gives this \"price\" equation\r\n" );
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document.write( "     -  = 5  dollars.     (1)\r\n" );
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document.write( "To solve this equation, multiply both sides by n*(n+50), then simplify step by step\r\n" );
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document.write( "    3000*(n+50) - 3000*n = 5n*(n+50)\r\n" );
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document.write( "    3000n + 150000 - 3000n = 5n^2 + 250n\r\n" );
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document.write( "    5n^2 + 250n - 150000 = 0\r\n" );
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document.write( "     n^2 +  50n -  30000 = 0\r\n" );
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document.write( "     (n+200)*(n-150) = 0\r\n" );
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document.write( "The equation has two roots, -200 and 150.  \r\n" );
document.write( "Of these two roots, we discard the negative one and accept the positive value n= 150.\r\n" );
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document.write( "CHECK.  To check, substitute n= 150 into equation (1). You will get\r\n" );
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document.write( "         -  = 20 - 15 = 5,  which is precisely correct.\r\n" );
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document.write( "ANSWER.  Paul sold 150 shares.\r\n" );
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