<PRE><B>Question 189860</B>
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Let&#39;s start with a piece of advice that is not only good for algebra word problems, but for life in general.  Begin with the end in mind.


The end result is the number of shares sold, so let &lt;i&gt;&lt;b&gt;x&lt;/b&gt;&lt;/i&gt; represent the number of shares sold.  And we know that the number of shares purchased is 5 more than that, so &lt;i&gt;&lt;b&gt;x&lt;/b&gt;&lt;/i&gt; + 5.  We also need a variable to represent the initial price, let&#39;s use &lt;i&gt;&lt;b&gt;p&lt;/b&gt;&lt;/i&gt;.  We don&#39;t need a separate variable for the sell price because we are told the price increase was $5, so the sell price can be expressed as &lt;i&gt;&lt;b&gt;p&lt;/b&gt;&lt;/i&gt; + 5.


We know that the original share price is equal to the amount invested ($250)divided by the number of shares purchased:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  p = \frac{250}{x + 5} ]


Furthermore, the sell price is equal to $50 more than the amount invested, or $300 divided by the number of shares sold, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  p + 5 = \frac{300}{x} ]


Which can be manipulated thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  p = \frac{300}{x} - \frac{5x}{x} = \frac{300 -5x}{x}]


Now we have two expressions in terms of &lt;i&gt;&lt;b&gt;x&lt;/b&gt;&lt;/i&gt; that are each equal to &lt;i&gt;&lt;b&gt;p&lt;/b&gt;&lt;/i&gt;, so set them equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{250}{x + 5} = \frac{300 -5x}{x}]


From here it is only a matter of cross-multiplying and solving the resulting quadratic.  Remember to exclude the negative root because selling a negative number of shares is a logical absurdity.  The positive root will be the desired answer to the question.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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