https://artofproblemsolving.com/wiki/index.php?title=1957_AHSME_Problems/Problem_6&feed=atom&action=history
1957 AHSME Problems/Problem 6 - Revision history
2024-03-29T13:31:49Z
Revision history for this page on the wiki
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https://artofproblemsolving.com/wiki/index.php?title=1957_AHSME_Problems/Problem_6&diff=106640&oldid=prev
Someonenumber011 at 05:55, 19 June 2019
2019-06-19T05:55:17Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 05:55, 19 June 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
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<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">==Problem==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">An open box is constructed by starting with a rectangular sheet of metal <math>10</math> in. by <math>14</math> in. and cutting a square of side <math>x</math> inches from each corner. The resulting projections are folded up and the seams welded. The volume of the resulting box is:</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"><math>\textbf{(A)}\ 140x - 48x^2 + 4x^3 \qquad  \textbf{(B)}\ 140x + 48x^2 + 4x^3\qquad \\ \textbf{(C)}\ 140x+24x^2+x^3\qquad \textbf{(D)}\ 140x-24x^2+x^3\qquad \textbf{(E)}\ \text{none of these}</math></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">==Solution==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>The resulting metal piece looks something like this where the white parts are squares of length <math>x</math>:</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>The resulting metal piece looks something like this where the white parts are squares of length <math>x</math>:</div></td></tr>
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Someonenumber011
https://artofproblemsolving.com/wiki/index.php?title=1957_AHSME_Problems/Problem_6&diff=106639&oldid=prev
Someonenumber011 at 05:05, 19 June 2019
2019-06-19T05:05:14Z
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<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 05:05, 19 June 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l12" >Line 12:</td>
<td colspan="2" class="diff-lineno">Line 12:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>From here, try to visualize the rectangular prism coming together and realize the height is <math>x</math>, the length is <math>14-2x</math>, and the width is <math>10-2x</math>. Therefore, the volume is <math>x(14-2x)(10-2x)=x(4x^2-48x+40)=</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>From here, try to visualize the rectangular prism coming together and realize the height is <math>x</math>, the length is <math>14-2x</math>, and the width is <math>10-2x</math>. Therefore, the volume is <math>x(14-2x)(10-2x)=x(4x^2-48x+40)=</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\boxed{\textbf{(A) } 140x - 48x^2 + 4x^3}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\boxed{\textbf{(A) } 140x - 48x^2 + 4x^3}</math>.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">{{AHSME box|year=1957|ab=B|num-b=5|num-a=7}}</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">{{MAA Notice}}</ins></div></td></tr>
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Someonenumber011
https://artofproblemsolving.com/wiki/index.php?title=1957_AHSME_Problems/Problem_6&diff=106625&oldid=prev
Someonenumber011: Created page with "The resulting metal piece looks something like this where the white parts are squares of length <math>x</math>: <asy> fill((0,4)--(4,4)--(4,0)--(6,0)--(6,4)--(10,4)--(10,10)-..."
2019-06-19T04:36:41Z
<p>Created page with "The resulting metal piece looks something like this where the white parts are squares of length <math>x</math>: <asy> fill((0,4)--(4,4)--(4,0)--(6,0)--(6,4)--(10,4)--(10,10)-..."</p>
<p><b>New page</b></p><div>The resulting metal piece looks something like this where the white parts are squares of length <math>x</math>:<br />
<br />
<asy><br />
fill((0,4)--(4,4)--(4,0)--(6,0)--(6,4)--(10,4)--(10,10)--(6,10)--(6,14)--(4,14)--(4,10)--(0,10)--cycle,grey);<br />
draw((0,0)--(14-4,0)--(10,14)--(0,14)--cycle);<br />
draw((0,0)--(0,4)--(4,4)--(4,0)--cycle);<br />
draw((10-4,0)--(10,0)--(10,4)--(6,4)--cycle);<br />
draw((0,14)--(4,14)--(4,10)--(0,10)--cycle);<br />
draw((6,14)--(6,10)--(10,10)--(10,14)--cycle);<br />
</asy><br />
<br />
From here, try to visualize the rectangular prism coming together and realize the height is <math>x</math>, the length is <math>14-2x</math>, and the width is <math>10-2x</math>. Therefore, the volume is <math>x(14-2x)(10-2x)=x(4x^2-48x+40)=<br />
\boxed{\textbf{(A) } 140x - 48x^2 + 4x^3}</math>.</div>
Someonenumber011