document.write( "Question 5300: A retailer spent $48 to purchase a number of mugs. Two of them were broken in the store, but by selling each of the remaining mugs for $3 above the original cost per mug, she made a total profit of $22.\r
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document.write( "If the price for n mugs is $48, how can I express the cost per mug?\r
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document.write( "The number of mugs that are available for sale?\r
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document.write( "Rewrite the equation from above in the standard form of a quadratic equation? \n" );
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Algebra.Com's Answer #2687 by rapaljer(4671)![]() ![]() You can put this solution on YOUR website! This is an interesting problem. Let's start by letting \n" ); document.write( "n= number of mugs purchased by the retailer \n" ); document.write( "n-2 = number of mugs sold (two were broken and not able to be sold!) \n" ); document.write( "C = cost per mug paid by the retailer \n" ); document.write( "C + 3 = retail selling price of the mugs (for which retailer gets a $3 profit)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "To answer your other questions: \n" ); document.write( "If the price for n mugs is $48, then the cost per mug is $ \n" ); document.write( "The number of mugs available for sale = n-2.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "The equation is based upon the revenue from the sale of the mugs: \n" ); document.write( "\"Selling price per mug TIMES number of mugs sold = Revenue = Cost + Profit\" \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Substitute for the C in this equation: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Find the Least Common Denominator for the binomial above: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Multiply both sides by the denominator \"n\": \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Subtract 70n from each side of the equation to set it equal to zero:\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now this is a GREAT place to use Ichudov's new process for solving quadratic equations, but unfortunately I haven't learned it yet. Maybe the Maestro will step in and add his touch to this solution. Anyway, I DO know that this equation DOES factor, since if you calculate \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If you factor it, this comes out to \r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "n cannot be a negative fraction, so \n" ); document.write( "n= 12 mugs were purchased. \n" ); document.write( "n-2 = 10 mugs were sold.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "Add a $3 profit to each mug = $7 for each of the 10 mugs that were sold.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Check: \n" ); document.write( "12 mugs were purchased at $4 each for a total of $48. \n" ); document.write( "10 mugs were sold at $7 each for a total of $70. That gives a profit of $22.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "R^2 at SCC\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |