https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_43&feed=atom&action=history 1953 AHSME Problems/Problem 43 - Revision history 2022-01-24T23:55:55Z Revision history for this page on the wiki MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_43&diff=127159&oldid=prev Duck master: Created page and added solution. 2020-06-30T23:35:08Z <p>Created page and added solution.</p> <p><b>New page</b></p><div>If the price of an article is increased by percent &lt;math&gt;p&lt;/math&gt;, then the decrease in percent of sales must not exceed &lt;math&gt;d&lt;/math&gt; in order to yield the same income. The value of &lt;math&gt;d&lt;/math&gt; is:<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{1+p} \qquad \textbf{(B)}\ \frac{1}{1-p} \qquad \textbf{(C)}\ \frac{p}{1+p} \qquad \textbf{(D)}\ \frac{p}{p-1}\qquad \textbf{(E)}\ \frac{1-p}{1+p}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> It turns out that none of the listed solutions are the true solution for the problem as written, so we'll edit it slightly by replacing &quot;percent&quot; with &quot;proportion&quot;. To solve the modified problem, note that the price of the article is &lt;math&gt;1+p&lt;/math&gt; times what it was originally, so that demand must be at least &lt;math&gt;\frac{1}{1+p}&lt;/math&gt; times what it was originally to yield at least the same income. Since &lt;math&gt;1-d = \frac{1}{1+p}&lt;/math&gt;, this implies that &lt;math&gt;d = \frac{p}{1+p}&lt;/math&gt;. Thusly, our answer is &lt;math&gt;\boxed{\text{(C)}}&lt;/math&gt;, and we are done.</div> Duck master