https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=VelaDabant&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T04:53:06ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_16&diff=306082009 AMC 12B Problems/Problem 162009-03-04T00:41:32Z<p>VelaDabant: </p>
<hr />
<div>== Problem ==<br />
Trapezoid <math>ABCD</math> has <math>AD||BC</math>, <math>BD = 1</math>, <math>\angle DBA = 23^{\circ}</math>, and <math>\angle BDC = 46^{\circ}</math>. The ratio <math>BC: AD</math> is <math>9: 5</math>. What is <math>CD</math>?<br />
<br />
<br />
<math>\mathrm{(A)}\ \frac 79\qquad<br />
\mathrm{(B)}\ \frac 45\qquad<br />
\mathrm{(C)}\ \frac {13}{15}\qquad<br />
\mathrm{(D)}\ \frac 89\qquad<br />
\mathrm{(E)}\ \frac {14}{15}</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
Extend <math>\overline {AB}</math> and <math>\overline {DC}</math> to meet at <math>E</math>. Then<br />
<br />
<cmath>\begin{align*}<br />
\angle BED &= 180^{\circ} - \angle EDB - \angle DBE\\<br />
&= 180^{\circ} - 134^{\circ} -23^{\circ} = 23^{\circ}.<br />
\end{align*}</cmath><br />
<br />
Thus <math>\triangle BDE</math> is isosceles with <math>DE = BD</math>. Because <math>\overline {AD} \parallel \overline {BC}</math>, it follows that the triangles <math>BCD</math> and <math>ADE</math> are similar. Therefore<br />
<cmath>\frac 95 = \frac {BC}{AD} = \frac {CD + DE}{DE} = \frac {CD}{BD} + 1 = CD + 1,</cmath><br />
so <math>CD = \boxed{\frac 45}.</math><br />
<br />
=== Solution 2 ===<br />
Let <math>E</math> be the intersection of <math>\overline {BC}</math> and the line through <math>D</math> parallel to <math>\overline {AB}.</math> By constuction <math>BE = AD</math> and <math>\angle BDE = 23^{\circ}</math>; it follows that <math>DE</math> is the bisector of the angle <math>BDC</math>. So by the Angle Bisector Theorem we get<br />
<cmath>CD = \frac {CD}{BD} = \frac {EC}{BE} = \frac {BC - BE}{BE} = \frac {BC}{AD} -1 = \frac 95 - 1 = \boxed{\frac 45}.</cmath><br />
The answer is <math>\mathrm{(B)}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2009|ab=B|num-b=15|num-a=17}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_16&diff=306032009 AMC 12B Problems/Problem 162009-03-03T19:19:44Z<p>VelaDabant: </p>
<hr />
<div>== Problem ==<br />
Trapezoid <math>ABCD</math> has <math>AD||BC</math>, <math>BD = 1</math>, <math>\angle DBA = 23^{\circ}</math>, and <math>\angle BDC = 46^{\circ}</math>. The ratio <math>BC: AD</math> is <math>9: 5</math>. What is <math>CD</math>?<br />
<br />
<br />
<math>\mathrm{(A)}\ \frac 79\qquad<br />
\mathrm{(B)}\ \frac 45\qquad<br />
\mathrm{(C)}\ \frac {13}{15}\qquad<br />
\mathrm{(D)}\ \frac 89\qquad<br />
\mathrm{(E)}\ \frac {14}{15}</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
Extend <math>\overline {AB}</math> and <math>\overline {DC}</math> to meet at <math>E</math>. Then<br />
<br />
<cmath>\begin{align*}<br />
\angle BED &= 180^{\circ} - \angle EDB - \angle DBE\\<br />
&= 180^{\circ} - 134^{\circ} -23^{\circ} = 23^{\circ}.<br />
\end{align*}</cmath><br />
<br />
Thus <math>\triangle BDE</math> is isosceles with <math>DE = BD</math>. Because <math>\overline {AD} \parallel \overline {BC}</math>, it follows that the triangles <math>BCD</math> and <math>ADE</math> are similar. Therefore<br />
<cmath>\frac 95 = \frac {BC}{AD} = \frac {CD + DE}{DE} = \frac {CD}{BD} + 1 = CD + 1,</cmath><br />
so <math>CD = \boxed{\frac 45}.</math><br />
<br />
=== Solution 2 ===<br />
Let <math>E</math> be the intersection of <math>\overline {BC}</math> and the line parallel to <math>\overline {AB}.</math> By constuction <math>BE = AD</math> and <math>\angle BDE = 23^{\circ}</math>; it follows that <math>DE</math> is the bisector of the angle <math>BDC</math>. So by the Angle Bisector Theorem we get<br />
<cmath>CD = \frac {CD}{BD} = \frac {EC}{BE} = \frac {BC - BE}{BE} = \frac {BC}{AD} -1 = \frac 95 - 1 = \boxed{\frac 45}.</cmath><br />
The answer is <math>\mathrm{(B)}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2009|ab=B|num-b=15|num-a=17}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_16&diff=305942009 AMC 12B Problems/Problem 162009-03-03T01:35:15Z<p>VelaDabant: New page: == Problem == Trapezoid <math>ABCD</math> has <math>AD||BC</math>, <math>BD = 1</math>, <math>\angle DBA = 23^{\circ}</math>, and <math>\angle BDC = 46^{\circ}</math>. The ratio <math>BC:...</p>
<hr />
<div>== Problem ==<br />
Trapezoid <math>ABCD</math> has <math>AD||BC</math>, <math>BD = 1</math>, <math>\angle DBA = 23^{\circ}</math>, and <math>\angle BDC = 46^{\circ}</math>. The ratio <math>BC: AD</math> is <math>9: 5</math>. What is <math>CD</math>?<br />
<br />
<br />
<math>\mathrm{(A)}\ \frac 79\qquad<br />
\mathrm{(B)}\ \frac 45\qquad<br />
\mathrm{(C)}\ \frac {13}{15}\qquad<br />
\mathrm{(D)}\ \frac 89\qquad<br />
\mathrm{(E)}\ \frac {14}{15}</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
Extend <math>\overline AB</math> and <math>\overline DC</math> to meet at <math>E</math>. Then<br />
<br />
<cmath>\begin{align*}<br />
\angle BED &= 180^{\circ} - \angle EDB - \angle DBE\\<br />
&= 180^{\circ} - 134^{\circ} -23^{\circ} = 23^{\circ}.<br />
\end{align*}</cmath><br />
<br />
Thus <math>\triangle BDE</math> is isosceles with <math>DE = BD</math>. Because <math>\overline {AD} \parallel \overline {BC}</math>, it follows that the triangles <math>BCD</math> and <math>ADE</math> are similar. Therefore<br />
<cmath>\frac 95 = \frac {BC}{AD} = \frac {CD + DE}{DE} = \frac {CD}{BD} + 1 = CD + 1,</cmath><br />
so <math>CD = \boxed{\frac 45}.</math><br />
<br />
=== Solution 2 ===<br />
Let <math>E</math> be the intersection of <math>\overline {BC}</math> and the line parallel to <math>\overline {AB}.</math> By constuction <math>BE = AD</math> and <math>\angle BDE = 23^{\circ}</math>; it follows that <math>DE</math> is the bisector of the angle <math>BDC</math>. So by the Angle Bisector Theorem we get<br />
<cmath>CD = \frac {CD}{BD} = \frac {EC}{BE} = \frac {BC - BE}{BE} = \frac {BC}{AD} -1 = \frac 95 - 1 = \boxed{\frac 45}.</cmath><br />
The answer is <math>\mathrm{(B)}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2009|ab=B|num-b=15|num-a=17}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_13&diff=305912009 AMC 12B Problems/Problem 132009-03-02T22:36:43Z<p>VelaDabant: New page: == Problem == Triangle <math>ABC</math> has <math>AB = 13</math> and <math>AC = 15</math>, and the altitude to <math>\overline{BC}</math> has length <math>12</math>. What is the sum of the...</p>
<hr />
<div>== Problem ==<br />
Triangle <math>ABC</math> has <math>AB = 13</math> and <math>AC = 15</math>, and the altitude to <math>\overline{BC}</math> has length <math>12</math>. What is the sum of the two possible values of <math>BC</math>?<br />
<br />
<math>\mathrm{(A)}\ 15\qquad<br />
\mathrm{(B)}\ 16\qquad<br />
\mathrm{(C)}\ 17\qquad<br />
\mathrm{(D)}\ 18\qquad<br />
\mathrm{(E)}\ 19</math><br />
<br />
== Solution ==<br />
Let <math>D</math> be the foot of the altitude to <math>\overline BC</math>. Then <math>BD = \sqrt {13^2 - 12^2} = 5</math> and <math>DC = \sqrt {15^2 - 12^2} = 9</math>. Thus <math>BC = BD + BC = 5 + 9 = 14</math> or <math>BC = DC - BD = 9 -5 = 4</math>. The sum of the two possible values is <math>14 + 4 = \boxed{18}</math>. The answer is <math>\mathrm{(D)}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2009|ab=B|num-b=12|num-a=14}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_12&diff=305902009 AMC 12B Problems/Problem 122009-03-02T21:09:58Z<p>VelaDabant: New page: == Problem == The fifth and eighth terms of a geometric sequence of real numbers are <math>7!</math> and <math>8!</math> respectively. What is the first term? <math>\mathrm{(A)}\ 60\qqu...</p>
<hr />
<div>== Problem ==<br />
The fifth and eighth terms of a geometric sequence of real numbers are <math>7!</math> and <math>8!</math> respectively. What is the first term?<br />
<br />
<br />
<math>\mathrm{(A)}\ 60\qquad<br />
\mathrm{(B)}\ 75\qquad<br />
\mathrm{(C)}\ 120\qquad<br />
\mathrm{(D)}\ 225\qquad<br />
\mathrm{(E)}\ 315</math><br />
<br />
== Solution ==<br />
Let the <math>n</math>th term of the series be <math>ar^{n-1}</math>. Because<br />
<cmath>\frac {8!}{7!} = \frac {ar^7}{ar^4} = r^3 = 8,</cmath><br />
it follows that <math>r = 2</math> and the first term is <math>a = \frac {7!}{r^4} = \frac {7!}{16} = \boxed{315}</math>. The answer is <math>\mathrm{(E)}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2009|ab=B|num-b=11|num-a=13}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_9&diff=305882009 AMC 12B Problems/Problem 92009-03-02T20:45:33Z<p>VelaDabant: New page: == Problem == Triangle <math>ABC</math> has vertices <math>A = (3,0)</math>, <math>B = (0,3)</math>, and <math>C</math>, where <math>C</math> is on the line <math>x + y = 7</math>. What i...</p>
<hr />
<div>== Problem ==<br />
Triangle <math>ABC</math> has vertices <math>A = (3,0)</math>, <math>B = (0,3)</math>, and <math>C</math>, where <math>C</math> is on the line <math>x + y = 7</math>. What is the area of <math>\triangle ABC</math>?<br />
<br />
<br />
<math>\mathrm{(A)}\ 6\qquad<br />
\mathrm{(B)}\ 8\qquad<br />
\mathrm{(C)}\ 10\qquad<br />
\mathrm{(D)}\ 12\qquad<br />
\mathrm{(E)}\ 14</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
Because the line <math>x + y = 7</math> is parallel to <math>\overline {AB}</math>, the area of <math>\triangle ABC</math> is independent of the location of <math>C</math> on that line. Therefore it may be assumed that <math>C</math> is <math>(7,0)</math>. In that case the triangle has base <math>AC = 4</math> and altitude <math>3</math>, so its area is <math>\frac 12 \cdot 4 \cdot 3 = \boxed {6}</math>.<br />
<br />
=== Solution 2 ===<br />
The base of the triangle is <math>AB = \sqrt{3^2 + 3^2} = 3\sqrt 2</math>. Its altitude is the distance between the point <math>A</math> and the parallel line <math>x + y = 7</math>, which is<br />
<cmath>\frac {|3 + 0 - 7|}{\sqrt {1^2 + 1^2}} = 2\sqrt 2</cmath><br />
Therefore its area is <math>\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}</math>. The answer is <math>\mathrm{(A)}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2009|ab=B|num-b=8|num-a=10}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_20&diff=305792009 AMC 12B Problems/Problem 202009-03-02T01:16:55Z<p>VelaDabant: New page: == Problem == A convex polyhedron <math>Q</math> has vertices <math>V_1,V_2,\ldots,V_n</math>, and <math>100</math> edges. The polyhedron is cut by planes <math>P_1,P_2,\ldots,P_n</math> i...</p>
<hr />
<div>== Problem ==<br />
A convex polyhedron <math>Q</math> has vertices <math>V_1,V_2,\ldots,V_n</math>, and <math>100</math> edges. The polyhedron is cut by planes <math>P_1,P_2,\ldots,P_n</math> in such a way that plane <math>P_k</math> cuts only those edges that meet at vertex <math>V_k</math>. In addition, no two planes intersect inside or on <math>Q</math>. The cuts produce <math>n</math> pyramids and a new polyhedron <math>R</math>. How many edges does <math>R</math> have?<br />
<br />
<math>\mathrm{(A)}\ 200\qquad<br />
\mathrm{(B)}\ 2n\qquad<br />
\mathrm{(C)}\ 300\qquad<br />
\mathrm{(D)}\ 400\qquad<br />
\mathrm{(E)}\ 4n</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
Each edge of <math>Q</math> is cut by two planes, so <math>R</math> has <math>200</math> vertices. Three edges of <math>R</math> meet at each vertex, so <math>R</math> has <math>\frac 12 \cdot 3 \cdot 200 = \boxed {300}</math> edges.<br />
<br />
=== Solution 2 ===<br />
At each vertex, as many new edges are created by this process as there are original edges meeting at that vertex. Thus the total number of new edges is the total number of endpoints of the original edges, which is <math>200</math>. A middle portion of each original edge is also present in <math>R</math>, so <math>R</math> has <math>100 + 200 = \boxed {300}</math> edges.<br />
<br />
=== Solution 3 ===<br />
Euler's Polyhedron Formula applied to <math>Q</math> gives <math>n - 100 + F = 2</math>, where F is the number of faces of <math>Q</math>. Each edge of <math>Q</math> is cut by two planes, so <math>R</math> has <math>200</math> vertices. Each cut by a plane <math>P_k</math> creates an additional face on <math>R</math>, so Euler's Polyhedron Formula applied to <math>R</math> gives <math>200 - E + (F+n) = 2</math>, where <math>E</math> is the number of edges of <math>R</math>. Subtracting the first equation from the second gives <math>300 - E = 0</math>, whence <math>E = \boxed {300}</math>. The answer is <math>\mathrm{(C)}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2009|ab=B|num-b=19|num-a=21}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_10&diff=305782009 AMC 12B Problems/Problem 102009-03-02T00:39:51Z<p>VelaDabant: Redirecting to 2009 AMC 10B Problems/Problem 19</p>
<hr />
<div>#redirect [[2009 AMC 10B Problems/Problem 19]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_19&diff=305772009 AMC 10B Problems/Problem 192009-03-02T00:38:00Z<p>VelaDabant: New page: {{duplicate|2009 AMC 10B #19 and 2009 AMC 12B #10}} == Problem == A particular <math>12</math>-hour digital clock displays the hour and...</p>
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<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #19]] and [[2009 AMC 12B Problems|2009 AMC 12B #10]]}}<br />
<br />
== Problem ==<br />
A particular <math>12</math>-hour digital clock displays the hour and minute of a day. Unfortunately, whenever it is supposed to display a <math>1</math>, it mistakenly displays a <math>9</math>. For example, when it is 1:16 PM the clock incorrectly shows 9:96 PM. What fraction of the day will the clock show the correct time?<br />
<br />
<br />
<math>\mathrm{(A)}\ \frac 12\qquad<br />
\mathrm{(B)}\ \frac 58\qquad<br />
\mathrm{(C)}\ \frac 34\qquad<br />
\mathrm{(D)}\ \frac 56\qquad<br />
\mathrm{(E)}\ \frac {9}{10}</math><br />
<br />
== Solution ==<br />
The clock will display the incorrect time for the entire hours of <math>1, 10, 11</math> and <math>12</math>. So the correct hour is displayed <math>\frac 23</math> of the time. The minutes will not display correctly whenever either the tens digit or the ones digit is a <math>1</math>, so the minutes that will not display correctly are <math>10, 11, 12, \dots, 19</math> and <math>01, 21, 31, 41,</math> and <math>51</math>. This amounts to fifteen of the sixty possible minutes for any given hour. Hence the fraction of the day that the clock shows the correct time is <math>\frac 23 \cdot \left(1 - \frac {15}{60}\right) = \frac 23 \cdot \frac 34 = \boxed{\frac 12}</math>. The answer is <math>\mathrm{(A)}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2009|ab=B|num-b=9|num-a=11}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_14&diff=305762009 AMC 12B Problems/Problem 142009-03-02T00:17:22Z<p>VelaDabant: </p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #17]] and [[2009 AMC 12B Problems|2009 AMC 12B #14]]}}<br />
<br />
== Problem ==<br />
Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from <math>(a,0)</math> to <math>(3,3)</math>, divides the entire region into two regions of equal area. What is <math>a</math>?<br />
<asy><br />
unitsize(1cm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
<br />
fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray);<br />
<br />
xaxis("$x$",-0.5,4,EndArrow(HookHead,4));<br />
yaxis("$y$",-0.5,4,EndArrow(4));<br />
<br />
draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0));<br />
draw((1,0)--(1,2)--(3,2));<br />
draw((2/3,0)--(3,3));<br />
<br />
label("$(a,0)$",(2/3,0),S);<br />
label("$(3,3)$",(3,3),NE);<br />
</asy><math>\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45</math><br />
<br />
== Solution ==<br />
<br />
For <math>a\geq 1.5</math> the shaded area is at most <math>1.5</math>, which is too little. Hence <math>a<1.5</math>, and therefore the point <math>(2,1)</math> is indeed inside the shaded part, as shown in the picture.<br />
<br />
Then the area of the shaded part is one less than the area of the triangle with vertices <math>(a,0)</math>, <math>(3,0)</math>, and <math>(3,3)</math>. Its area is obviously <math>\frac{3(3-a)}2</math>, therefore the area of the shaded part is <math>\frac{7-3a}{2}</math>.<br />
<br />
The entire figure has area <math>5</math>, hence we want the shaded part to have area <math>\frac 52</math>. Solving for <math>a</math>, we get <math>a=\boxed{\frac 23}</math>. The answer is <math>\mathrm{(C)}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2009|ab=B|num-b=16|num-a=18}}<br />
{{AMC12 box|year=2009|ab=B|num-b=13|num-a=15}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_17&diff=305752009 AMC 10B Problems/Problem 172009-03-02T00:11:22Z<p>VelaDabant: Redirecting to 2009 AMC 12B Problems/Problem 14</p>
<hr />
<div>#redirect [[2009 AMC 12B Problems/Problem 14]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_8&diff=305612009 AMC 12B Problems/Problem 82009-03-01T16:15:25Z<p>VelaDabant: Redirecting to 2009 AMC 10B Problems/Problem 15</p>
<hr />
<div>#redirect [[2009 AMC 10B Problems/Problem 15]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_15&diff=305602009 AMC 10B Problems/Problem 152009-03-01T16:13:49Z<p>VelaDabant: New page: {{duplicate|2009 AMC 10B #15 and 2009 AMC 12B #8}} == Problem == When a bucket is two-thirds full of water, the bucket and water weigh ...</p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #15]] and [[2009 AMC 12B Problems|2009 AMC 12B #8]]}}<br />
<br />
== Problem ==<br />
When a bucket is two-thirds full of water, the bucket and water weigh <math>a</math> kilograms. When the bucket is one-half full of water the total weight is <math>b</math> kilograms. In terms of <math>a</math> and <math>b</math>, what is the total weight in kilograms when the bucket is full of water?<br />
<br />
<math>\mathrm{(A)}\ \frac23a + \frac13b\qquad<br />
\mathrm{(B)}\ \frac32a - \frac12b\qquad<br />
\mathrm{(C)}\ \frac32a + b\qquad<br />
\mathrm{(D)}\ \frac32a + 2b\qquad<br />
\mathrm{(E)}\ 3a - 2b</math><br />
<br />
== Solution ==<br />
Let <math>x</math> be the weight of the bucket and let <math>y</math> be the weight of the water in a full bucket. Then we are given that <math>x + \frac 23y = a</math> and <math>x + \frac 12y = b</math>. Hence <math>\frac 16y = a-b</math>, so <math>y = 6a-6b</math>. Thus <math>x = b - \frac 12 (6a-6b) = -3a + 4b</math>. Finally <math>x + y = \boxed {3a-2b}</math>. The answer is <math>\mathrm{(E)}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=14|num-a=16}}<br />
{{AMC12 box|year=2009|ab=B|num-b=7|num-a=9}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_11&diff=305542009 AMC 12B Problems/Problem 112009-03-01T05:05:57Z<p>VelaDabant: Redirecting to 2009 AMC 10B Problems/Problem 14</p>
<hr />
<div>#redirect [[2009 AMC 10B Problems/Problem 14]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_14&diff=305522009 AMC 10B Problems/Problem 142009-03-01T05:03:50Z<p>VelaDabant: New page: {{duplicate|2009 AMC 10B #14 and 2009 AMC 12B #11}} == Problem == On Monday, Millie puts a quart of seeds, <math>25\%</math> of which a...</p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #14]] and [[2009 AMC 12B Problems|2009 AMC 12B #11]]}}<br />
<br />
== Problem ==<br />
On Monday, Millie puts a quart of seeds, <math>25\%</math> of which are millet, into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only <math>25\%</math> of the millet in the feeder, but they eat all of the other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?<br />
<br />
<math>\textbf{(A)}\ \text{Tuesday}\qquad<br />
\textbf{(B)}\ \text{Wednesday}\qquad<br />
\textbf{(C)}\ \text{Thursday}\qquad<br />
\textbf{(D)}\ \text{Friday}\qquad<br />
\textbf{(E)}\ \text{Saturday}</math><br />
<br />
== Solution ==<br />
On Monday, day 1, the birds find <math>\frac 14</math> quart of millet in the feeder. On Tuesday they find<br />
<cmath>\frac 14 + \frac 34 \cdot \frac 14</cmath><br />
quarts of millet. On Wednesday, day 3, they find<br />
<cmath>\frac 14 + \frac 34 \cdot \frac 14 + \left(\frac34\right)^2 \cdot \frac 14</cmath><br />
quarts of millet. The number of quarts of millet they find on day <math>n</math> is<br />
<cmath>\frac 14 + \frac 34 \cdot \frac 14 + \left(\frac34\right)^2 \cdot \frac 14 + \cdots + \left(\frac 34\right)^{n-1} \cdot \frac 14 = \frac {(\frac 14)(1 - (\frac 34)^n)}{1 - \frac 34} = 1 - \left(\frac 34\right)^n .</cmath><br />
The birds always find <math>\frac 34</math> quart of other seeds, so more than half the seeds are millet if <math>1 - \left(\frac 34\right)^n > \frac 34</math>, that is, when <math>\left(\frac 34\right)^n < \frac 14</math>. Because <math>\left(\frac 34\right)^4 = \frac {81}{256} > \frac 14</math> and <math>\left(\frac 34\right)^5 = \frac {243}{1024} < \frac 14</math>, this will first occur on day <math>5</math> which is <math>\boxed {\text{Friday}}</math>. The answer is <math>\mathrm{(D)}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=13|num-a=15}}<br />
{{AMC12 box|year=2009|ab=B|num-b=10|num-a=12}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_7&diff=305472009 AMC 12B Problems/Problem 72009-02-28T18:54:58Z<p>VelaDabant: Redirecting to 2009 AMC 10B Problems/Problem 8</p>
<hr />
<div>#redirect [[2009 AMC 10B Problems/Problem 8]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_8&diff=305462009 AMC 10B Problems/Problem 82009-02-28T18:52:17Z<p>VelaDabant: New page: {{duplicate|2009 AMC 10B #8 and 2009 AMC 12B #7}} == Problem == In a certain year the price of gasoline rose by <math>20\%</math> durin...</p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #8]] and [[2009 AMC 12B Problems|2009 AMC 12B #7]]}}<br />
<br />
== Problem ==<br />
In a certain year the price of gasoline rose by <math>20\%</math> during January, fell by <math>20\%</math> during February, rose by <math>25\%</math> during March, and fell by <math>x\%</math> during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is <math>x</math><br />
<br />
<math>\mathrm{(A)}\ 12\qquad<br />
\mathrm{(B)}\ 17\qquad<br />
\mathrm{(C)}\ 20\qquad<br />
\mathrm{(D)}\ 25\qquad<br />
\mathrm{(E)}\ 35</math><br />
<br />
== Solution ==<br />
Let <math>p</math> be the price at the beginning of January. The price at the end of March was <math>(1.2)(0.8)(1.25)p = 1.2p.</math> Because the price at the of April was <math>p</math>, the price decreased by <math>0.2p</math> during April, and the percent decrease was<br />
<cmath>x = 100 \cdot \frac{0.2p}{1.2p} = \frac {100}{6} \approx 16.7.</cmath><br />
So to the nearest integer <math>x</math> is <math>\boxed{17}</math>. The answer is <math>\mathrm{(B)}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=7|num-a=9}}<br />
{{AMC12 box|year=2009|ab=B|num-b=6|num-a=8}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_6&diff=305442009 AMC 12B Problems/Problem 62009-02-28T16:31:54Z<p>VelaDabant: Redirecting to 2009 AMC 10B Problems/Problem 7</p>
<hr />
<div>#redirect [[2009 AMC 10B Problems/Problem 7]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_7&diff=305432009 AMC 10B Problems/Problem 72009-02-28T16:30:05Z<p>VelaDabant: New page: {{duplicate|2009 AMC 10B #7 and 2009 AMC 12B #6}} == Problem == By inserting parentheses, it is possible to give the expression <cmath>...</p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #7]] and [[2009 AMC 12B Problems|2009 AMC 12B #6]]}}<br />
<br />
== Problem ==<br />
By inserting parentheses, it is possible to give the expression<br />
<cmath>2\times3 + 4\times5</cmath><br />
several values. How many different values can be obtained?<br />
<br />
<math>\mathrm{(A)}\ 2\qquad<br />
\mathrm{(B)}\ 3\qquad<br />
\mathrm{(C)}\ 4\qquad<br />
\mathrm{(D)}\ 5\qquad<br />
\mathrm{(E)}\ 6</math><br />
<br />
== Solution ==<br />
The three operations can be performed on any of <math>3! = 6</math> orders. However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to check that the values of the four expressions<br />
<cmath>\begin{align*}<br />
(2\times3) + (4\times5) &= 26,\\<br />
(2\times3 + 4)\times5 &= 50,\\<br />
2\times(3 + (4\times5)) &= 46,\\<br />
2\times(3 + 4)\times5 &= 70<br />
\end{align*}</cmath><br />
are in fact all distinct. So the answer is <math>\boxed{4}</math>, which is choice <math>\mathrm{(C)}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=6|num-a=8}}<br />
{{AMC12 box|year=2009|ab=B|num-b=5|num-a=7}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_4&diff=305392009 AMC 12B Problems/Problem 42009-02-28T04:17:18Z<p>VelaDabant: Redirecting to 2009 AMC 10B Problems/Problem 4</p>
<hr />
<div>#redirect [[2009 AMC 10B Problems/Problem 4]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_4&diff=305382009 AMC 10B Problems/Problem 42009-02-28T04:15:44Z<p>VelaDabant: New page: {{duplicate|2009 AMC 10B #4 and 2009 AMC 12B #4}} == Problem == A rectangular yard contains two flower beds in the shape of congruent i...</p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #4]] and [[2009 AMC 12B Problems|2009 AMC 12B #4]]}}<br />
<br />
== Problem ==<br />
A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths <math>15</math> and <math>25</math> meters. What fraction of the yard is occupied by the flower beds?<br />
<br />
<center><asy><br />
unitsize(2mm);<br />
defaultpen(linewidth(.8pt));<br />
<br />
fill((0,0)--(0,5)--(5,5)--cycle,gray);<br />
fill((25,0)--(25,5)--(20,5)--cycle,gray);<br />
draw((0,0)--(0,5)--(25,5)--(25,0)--cycle);<br />
draw((0,0)--(5,5));<br />
draw((20,5)--(25,0));<br />
</asy></center><br />
<br />
<math>\mathrm{(A)}\frac 18\qquad<br />
\mathrm{(B)}\frac 16\qquad<br />
\mathrm{(C)}\frac 15\qquad<br />
\mathrm{(D)}\frac 14\qquad<br />
\mathrm{(E)}\frac 13</math><br />
<br />
== Solution ==<br />
Each triangle has leg length <math>\frac 12 \cdot (25 - 15) = 5</math> meters and area <math>\frac 12 \cdot 5^2 = \frac {25}{2}</math> square meters. Thus the flower beds have a total area of 25 square meters. The entire yard has length 25 m and width 5 m, so its area is 125 square meters. The fraction of the yard occupied by the flower beds is <math>\frac {25}{125} = \boxed{\frac15}</math>. The answer is <math>\mathrm{(C)}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=3|num-a=5}}<br />
{{AMC12 box|year=2009|ab=B|num-b=3|num-a=5}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_18&diff=305372009 AMC 12B Problems/Problem 182009-02-28T03:48:28Z<p>VelaDabant: Redirecting to 2009 AMC 10B Problems/Problem 23</p>
<hr />
<div>#redirect [[2009 AMC 10B Problems/Problem 23]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_17&diff=305362009 AMC 12B Problems/Problem 172009-02-28T03:46:45Z<p>VelaDabant: Redirecting to 2009 AMC 10B Problems/Problem 25</p>
<hr />
<div>#redirect [[2009 AMC 10B Problems/Problem 25]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_5&diff=305352009 AMC 12B Problems/Problem 52009-02-28T03:44:35Z<p>VelaDabant: Redirecting to 2009 AMC 10B Problems/Problem 6</p>
<hr />
<div>#redirect [[2009 AMC 10B Problems/Problem 6]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_3&diff=305342009 AMC 12B Problems/Problem 32009-02-28T03:42:32Z<p>VelaDabant: Redirecting to 2009 AMC 10B Problems/Problem 5</p>
<hr />
<div>#redirect [[2009 AMC 10B Problems/Problem 5]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_1&diff=305332009 AMC 12B Problems/Problem 12009-02-28T03:40:11Z<p>VelaDabant: Redirecting to 2009 AMC 10B Problems/Problem 1</p>
<hr />
<div>#redirect [[2009 AMC 10B Problems/Problem 1]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_2&diff=305322009 AMC 12B Problems/Problem 22009-02-28T03:36:22Z<p>VelaDabant: Redirecting to 2009 AMC 10B Problems/Problem 3</p>
<hr />
<div>#redirect [[2009 AMC 10B Problems/Problem 3]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_6&diff=305272009 AMC 10B Problems/Problem 62009-02-27T22:34:44Z<p>VelaDabant: New page: {{duplicate|2009 AMC 10B #6 and 2009 AMC 12B #5}} == Problem == Kiana has two older twin brothers. The product of their three ages is ...</p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #6]] and [[2009 AMC 12B Problems|2009 AMC 12B #5]]}}<br />
<br />
== Problem ==<br />
Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?<br />
<br />
<math>\mathrm{(A)}\ 10\qquad<br />
\mathrm{(B)}\ 12\qquad<br />
\mathrm{(C)}\ 16\qquad<br />
\mathrm{(D)}\ 18\qquad<br />
\mathrm{(E)}\ 24</math><br />
<br />
== Solution ==<br />
The age of each person is a factor of <math>128 = 2^7</math>. So the twins could be <math>2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8</math> years of age and, consequently Kiana could be 128, 32, 8 or 2 years old, respectively. Because Kiana is younger than her brothers, she must be 2 years old. So the sum of their ages is <math>2 + 8 + 8 = \boxed{18}</math>. The answer is <math>\mathrm{(D)}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=5|num-a=7}}<br />
{{AMC12 box|year=2009|ab=B|num-b=4|num-a=6}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_5&diff=305262009 AMC 10B Problems/Problem 52009-02-27T22:13:33Z<p>VelaDabant: New page: {{duplicate|2009 AMC 10B #5 and 2009 AMC 12B #3}} == Problem == Twenty percent less than 60 is one-third more than what number? <math>...</p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #5]] and [[2009 AMC 12B Problems|2009 AMC 12B #3]]}}<br />
<br />
== Problem ==<br />
Twenty percent less than 60 is one-third more than what number?<br />
<br />
<math>\mathrm{(A)}\ 16\qquad<br />
\mathrm{(B)}\ 30\qquad<br />
\mathrm{(C)}\ 32\qquad<br />
\mathrm{(D)}\ 36\qquad<br />
\mathrm{(E)}\ 48</math><br />
<br />
== Solution ==<br />
Twenty percent less than 60 is <math>\frac 45 \cdot 60 = 48</math>. One-third more than a number ''n'' is <math>\frac 43n</math>. Therefore <math>\frac 43n = 48</math> and the number is <math>\boxed {36}</math>. The answer is <math>\mathrm{(D)}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=4|num-a=6}}<br />
{{AMC12 box|year=2009|ab=B|num-b=2|num-a=4}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_21&diff=305052009 AMC 10B Problems/Problem 212009-02-26T23:42:38Z<p>VelaDabant: New page: == Problem == What is the remainder when <math>3^0 + 3^1 + 3^2 + \cdots + 3^{2009}</math> is divided by 8? <math>\mathrm{(A)}\ 0\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 2\qquad \mathrm{...</p>
<hr />
<div>== Problem ==<br />
What is the remainder when <math>3^0 + 3^1 + 3^2 + \cdots + 3^{2009}</math> is divided by 8?<br />
<br />
<math>\mathrm{(A)}\ 0\qquad<br />
\mathrm{(B)}\ 1\qquad<br />
\mathrm{(C)}\ 2\qquad<br />
\mathrm{(D)}\ 4\qquad<br />
\mathrm{(E)}\ 6</math><br />
<br />
== Solution ==<br />
The sum of any four consecutive powers of 3 is divisible by <math>3^0 + 3^1 + 3^2 +3^3 = 40</math> and hence is divisible by 8. Therefore<br />
: <math>(3^2 + 3^3 + 3^4 + 3^5) + \cdots + (3^{2006} + 3^{2007} + 3^{2008} + 3^{3009})</math><br />
is divisible by 8. So the required remainder is <math>3^0 + 3^1 = \boxed {4}</math>. The answer is <math>\mathrm{(D)}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=20|num-a=22}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_1&diff=304982009 AMC 10B Problems/Problem 12009-02-26T23:00:26Z<p>VelaDabant: </p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #1]] and [[2009 AMC 12B Problems|2009 AMC 12B #1]]}}<br />
<br />
== Problem ==<br />
Each morning of her five-day workweek, Jane bought either a 50-cent muffin or a 75-cent bagel. Her total cost for the week was a whole number of dollars, How many bagels did she buy?<br />
<br />
<math>\mathrm{(A)}\ 1\qquad<br />
\mathrm{(B)}\ 2\qquad<br />
\mathrm{(C)}\ 3\qquad<br />
\mathrm{(D)}\ 4\qquad<br />
\mathrm{(E)}\ 5</math><br />
<br />
== Solution ==<br />
The only combination of five items with total cost a whole number of dollars is 3 muffins and <math>\boxed {2}</math> bagels. The answer is <math>\mathrm{(B)}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|before=First Question|num-a=2}}<br />
{{AMC12 box|year=2009|ab=B|before=First Question|num-a=2}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_3&diff=304972009 AMC 10B Problems/Problem 32009-02-26T22:59:18Z<p>VelaDabant: </p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #3]] and [[2009 AMC 12B Problems|2009 AMC 12B #2]]}}<br />
<br />
== Problem ==<br />
Paula the painter had just enough paint for 30 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms. How many cans of paint did she use for the 25 rooms?<br />
<br />
<math>\mathrm{(A)}\ 10\qquad<br />
\mathrm{(B)}\ 12\qquad<br />
\mathrm{(C)}\ 15\qquad<br />
\mathrm{(D)}\ 18\qquad<br />
\mathrm{(E)}\ 25</math><br />
<br />
== Solution ==<br />
Losing three cans of paint corresponds to being able to paint five fewer rooms. So <math>\frac 35 \cdot 25 = \boxed{15}</math>. The answer is <math>\mathrm{(C)}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=2|num-a=4}}<br />
{{AMC12 box|year=2009|ab=B|num-b=1|num-a=3}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_23&diff=304962009 AMC 10B Problems/Problem 232009-02-26T22:58:21Z<p>VelaDabant: </p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #23]] and [[2009 AMC 12B Problems|2009 AMC 12B #18]]}}<br />
<br />
== Problem ==<br />
Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. Both start from the same line at the same time. At some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?<br />
<br />
<math>\mathrm{(A)}\frac {1}{16}\qquad<br />
\mathrm{(B)}\frac 18\qquad<br />
\mathrm{(C)}\frac {3}{16}\qquad<br />
\mathrm{(D)}\frac 14\qquad<br />
\mathrm{(E)}\frac {5}{16}</math><br />
<br />
== Solution ==<br />
After 10 minutes (600 seconds), Rachel will have completed 6 laps and be 30 seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in 22.5 seconds, she will be in the picture between 18.75 seconds and 41.25 seconds of the tenth minute. After 10 minutes Robert will have completed 7 laps and will be 40 seconds past the starting line. Because Robert runs one-fourth of a lap in 20 seconds, he will be in the picture between 30 and 50 seconds of the tenth minute. Hence both Rachel and Robert will be in the picture if it is taken between 30 and 41.25 seconds of the tenth minute. So the probability that both runners are in the picture is <math>\frac {41.25 - 30} {60} = \boxed{\frac {3}{16}}</math>. The answer is <math>\mathrm{(C)}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2009|ab=B|num-b=17|num-a=19}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_25&diff=304952009 AMC 10B Problems/Problem 252009-02-26T22:56:36Z<p>VelaDabant: </p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #25]] and [[2009 AMC 12B Problems|2009 AMC 12B #17]]}}<br />
<br />
== Problem ==<br />
Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?<br />
<br />
<math>\mathrm{(A)}\frac 18\qquad<br />
\mathrm{(B)}\frac {3}{16}\qquad<br />
\mathrm{(C)}\frac 14\qquad<br />
\mathrm{(D)}\frac 38\qquad<br />
\mathrm{(E)}\frac 12</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
There are two possible stripe orientations for each of the six faces of the cube, so there are <math>2^6 = 64</math> possible stripe combinations. There are three pairs of parallel faces so, if there is an encircling stripe, then the pair of faces that do not contribute uniquely determine the stripe orientation for the remaining faces. In addition, the stripe on each face that does not contribute may be oriented in either of two different ways. So a total of <math>3 \cdot 2 \cdot 2 = 12</math> stripe combinations on the cube result in a continuous stripe around the cube. The required probability is <math>\frac {12}{64} = \boxed{\frac {3}{16}}</math>.<br />
<br />
=== Solution 2 ===<br />
Without loss of generality, orient the cube so that the stripe on the top face goes from front to back. There are two mutually exclusive ways for there to be an encircling stripe: either the front, bottom and back faces are painted to complete an encircling stripe with the top face's stripe or the front, right, back and left faces are painted to form an encircling stripe. The probability of the first case is <math>(\frac 12)^3 = \frac 18</math>, and the probability of the second case is <math>(\frac 12)^4 = \frac {1}{16}</math>. The cases are disjoint, so the probabilities sum <math>\frac 18 + \frac {1}{16} = \boxed {\frac {3}{16}}</math>.<br />
<br />
=== Solution 3 ===<br />
There are three possible orientations of an encircling stripe. For any one of these to appear, the stripes on the four faces through which the continuous stripe is to pass must be properly aligned. The probability of each such stripe alignment is <math>(\frac 12)^4 = \frac {1}{16}</math>. Since there are three such possibilities and they are disjoint, the total probability is <math>3 \cdot \frac {1}{16} = \boxed{\frac {3}{16}}</math>. The answer is <math>\mathrm{(B)}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=24|after=Last question}}<br />
{{AMC12 box|year=2009|ab=B|num-b=16|num-a=18}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_1&diff=304922009 AMC 10B Problems/Problem 12009-02-26T20:02:45Z<p>VelaDabant: </p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #1]] and [[2009 AMC 12B Problems|2009 AMC 12B #1]]}}<br />
<br />
== Problem ==<br />
Each morning of her five-day workweek, Jane bought either a 50-cent muffin or a 75-cent bagel. Her total cost for the week was a whole number of dollars, How many bagels did she buy?<br />
<br />
<math>\mathrm{(A)}\ 1\qquad<br />
\mathrm{(B)}\ 2\qquad<br />
\mathrm{(C)}\ 3\qquad<br />
\mathrm{(D)}\ 4\qquad<br />
\mathrm{(E)}\ 5</math><br />
<br />
== Solution ==<br />
The only combination of five items with total cost a whole number of dollars is 3 muffins and <math>\boxed {2}</math> bagels. The answer is <math>\mathrm{(B)}</math>.<br />
<br />
{{AMC10 box|year=2009|ab=B|before=First Question|num-a=2}}<br />
{{AMC12 box|year=2009|ab=B|before=First Question|num-a=2}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_3&diff=304912009 AMC 10B Problems/Problem 32009-02-26T19:58:07Z<p>VelaDabant: </p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #3]] and [[2009 AMC 12B Problems|2009 AMC 12B #2]]}}<br />
<br />
== Problem ==<br />
Paula the painter had just enough paint for 30 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms. How many cans of paint did she use for the 25 rooms?<br />
<br />
<math>\mathrm{(A)}\ 10\qquad<br />
\mathrm{(B)}\ 12\qquad<br />
\mathrm{(C)}\ 15\qquad<br />
\mathrm{(D)}\ 18\qquad<br />
\mathrm{(E)}\ 25</math><br />
<br />
== Solution ==<br />
Losing three cans of paint corresponds to being able to paint five fewer rooms. So <math>\frac 35 \cdot 25 = \boxed{15}</math>. The answer is <math>\mathrm{(C)}</math>.<br />
<br />
{{AMC10 box|year=2009|ab=B|num-b=2|num-a=4}}<br />
{{AMC12 box|year=2009|ab=B|num-b=1|num-a=3}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_23&diff=304902009 AMC 10B Problems/Problem 232009-02-26T19:55:25Z<p>VelaDabant: </p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #23]] and [[2009 AMC 12B Problems|2009 AMC 12B #18]]}}<br />
<br />
== Problem ==<br />
Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. Both start from the same line at the same time. At some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?<br />
<br />
<math>\mathrm{(A)}\frac {1}{16}\qquad<br />
\mathrm{(B)}\frac 18\qquad<br />
\mathrm{(C)}\frac {3}{16}\qquad<br />
\mathrm{(D)}\frac 14\qquad<br />
\mathrm{(E)}\frac {5}{16}</math><br />
<br />
== Solution ==<br />
After 10 minutes (600 seconds), Rachel will have completed 6 laps and be 30 seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in 22.5 seconds, she will be in the picture between 18.75 seconds and 41.25 seconds of the tenth minute. After 10 minutes Robert will have completed 7 laps and will be 40 seconds past the starting line. Because Robert runs one-fourth of a lap in 20 seconds, he will be in the picture between 30 and 50 seconds of the tenth minute. Hence both Rachel and Robert will be in the picture if it is taken between 30 and 41.25 seconds of the tenth minute. So the probability that both runners are in the picture is <math>\frac {41.25 - 30} {60} = \boxed{\frac {3}{16}}</math>. The answer is <math>\mathrm{(C)}</math>.<br />
<br />
{{AMC10 box|year=2009|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2009|ab=B|num-b=17|num-a=19}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_25&diff=304892009 AMC 10B Problems/Problem 252009-02-26T19:43:57Z<p>VelaDabant: </p>
<hr />
<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #25]] and [[2009 AMC 12B Problems|2009 AMC 12B #17]]}}<br />
<br />
== Problem ==<br />
Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?<br />
<br />
<math>\mathrm{(A)}\frac 18\qquad<br />
\mathrm{(B)}\frac {3}{16}\qquad<br />
\mathrm{(C)}\frac 14\qquad<br />
\mathrm{(D)}\frac 38\qquad<br />
\mathrm{(E)}\frac 12</math><br />
<br />
== Solution ==<br />
<math>\boxed{\frac {3}{16}}</math>. The answer is <math>\mathrm{(B)}</math>.<br />
<br />
{{AMC10 box|year=2009|ab=B|num-b=24|after=Last question}}<br />
{{AMC12 box|year=2009|ab=B|num-b=16|num-a=18}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_25&diff=304882009 AMC 10B Problems/Problem 252009-02-26T19:32:18Z<p>VelaDabant: </p>
<hr />
<div>{{duplicate|[[2009 AMC 12B Problems|2009 AMC 12B #17]] and [[2009 AMC 10B Problems|2009 AMC 10B #25]]}}<br />
<br />
== Problem ==<br />
Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?<br />
<br />
<math>\mathrm{(A)}\frac 18\qquad<br />
\mathrm{(B)}\frac {3}{16}\qquad<br />
\mathrm{(C)}\frac 14\qquad<br />
\mathrm{(D)}\frac 38\qquad<br />
\mathrm{(E)}\frac 12</math><br />
<br />
== Solution ==<br />
<math>\boxed{\frac {3}{16}}</math>. The answer is <math>\mathrm{(B)}</math>.<br />
<br />
{{AMC12 box|year=2009|ab=B|num-b=16|num-a=18}}<br />
{{AMC10 box|year=2009|ab=B|num-b=24|after=Last question}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_25&diff=304872009 AMC 10B Problems/Problem 252009-02-26T19:23:32Z<p>VelaDabant: New page: == Problem == Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random ...</p>
<hr />
<div>== Problem ==<br />
Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?<br />
<br />
<math>\mathrm{(A)}\frac 18\qquad<br />
\mathrm{(B)}\frac {3}{16}\qquad<br />
\mathrm{(C)}\frac 14\qquad<br />
\mathrm{(D)}\frac 38\qquad<br />
\mathrm{(E)}\frac 12</math><br />
<br />
== Solution ==<br />
<math>\boxed{\frac {3}{16}}</math>. The answer is <math>\mathrm{(B)}</math>.<br />
<br />
{{AMC10 box|year=2009|ab=B|num-b=24|after=Last question}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_23&diff=304862009 AMC 10B Problems/Problem 232009-02-26T19:01:13Z<p>VelaDabant: New page: == Problem == Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds....</p>
<hr />
<div>== Problem ==<br />
Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. Both start from the same line at the same time. At some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?<br />
<br />
<math>\mathrm{(A)}\frac {1}{16}\qquad<br />
\mathrm{(B)}\frac 18\qquad<br />
\mathrm{(C)}\frac {3}{16}\qquad<br />
\mathrm{(D)}\frac 14\qquad<br />
\mathrm{(E)}\frac {5}{16}</math><br />
<br />
== Solution ==<br />
After 10 minutes (600 seconds), Rachel will have completed 6 laps and be 30 seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in 22.5 seconds, she will be in the picture between 18.75 seconds and 41.25 seconds of the tenth minute. After 10 minutes Robert will have completed 7 laps and will be 40 seconds past the starting line. Because Robert runs one-fourth of a lap in 20 seconds, he will be in the picture between 30 and 50 seconds of the tenth minute. Hence both Rachel and Robert will be in the picture if it is taken between 30 and 41.25 seconds of the tenth minute. So the probability that both runners are in the picture is <math>\frac {41.25 - 30} {60} = \boxed{\frac {3}{16}}</math>. The answer is <math>\mathrm{(C)}</math>.<br />
<br />
{{AMC10 box|year=2009|ab=B|num-b=22|num-a=24}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_3&diff=304842009 AMC 10B Problems/Problem 32009-02-26T14:54:49Z<p>VelaDabant: New page: == Problem == Paula the painter had just enough paint for 30 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough pai...</p>
<hr />
<div>== Problem ==<br />
Paula the painter had just enough paint for 30 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms. How many cans of paint did she use for the 25 rooms?<br />
<br />
<math>\mathrm{(A)}\ 10\qquad<br />
\mathrm{(B)}\ 12\qquad<br />
\mathrm{(C)}\ 15\qquad<br />
\mathrm{(D)}\ 18\qquad<br />
\mathrm{(E)}\ 25</math><br />
<br />
== Solution ==<br />
Losing three cans of paint corresponds to being able to paint five fewer rooms. So <math>\frac 35 \cdot 25 = \boxed{15}</math>. The answer is <math>\mathrm{(C)}</math>.<br />
<br />
{{AMC10 box|year=2009|ab=B|num-b=2|num-a=4}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_1&diff=304832009 AMC 10B Problems/Problem 12009-02-26T14:36:40Z<p>VelaDabant: New page: == Problem == Each morning of her five-day workweek, Jane bought either a 50-cent muffin or a 75-cent bagel. Her total cost for the week was a whole number of dollars, How many bagels di...</p>
<hr />
<div>== Problem ==<br />
Each morning of her five-day workweek, Jane bought either a 50-cent muffin or a 75-cent bagel. Her total cost for the week was a whole number of dollars, How many bagels did she buy?<br />
<br />
<math>\mathrm{(A)}\ 1\qquad<br />
\mathrm{(B)}\ 2\qquad<br />
\mathrm{(C)}\ 3\qquad<br />
\mathrm{(D)}\ 4\qquad<br />
\mathrm{(E)}\ 5</math><br />
<br />
== Solution ==<br />
The only combination of five items with total cost a whole number of dollars is 3 muffins and 2 bagels. Thus, the answer is <math>\mathrm{(B)}</math>.<br />
<br />
{{AMC10 box|year=2009|ab=B|before=First Question|num-a=2}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems&diff=304142009 AMC 10A Problems2009-02-19T01:15:55Z<p>VelaDabant: /* Problem 7 */</p>
<hr />
<div>== Problem 1 ==<br />
One can holds <math>12</math> ounces of soda. What is the minimum number of cans needed to provide a gallon (128 ounces) of soda?<br />
<br />
<math><br />
\mathrm{(A)}\ 7<br />
\qquad<br />
\mathrm{(B)}\ 8<br />
\qquad<br />
\mathrm{(C)}\ 9<br />
\qquad<br />
\mathrm{(D)}\ 10<br />
\qquad<br />
\mathrm{(E)}\ 11<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes and quarters. Which of the following could ''not'' be the total value of the four coins, in cents?<br />
<br />
<math><br />
\mathrm{(A)}\ 15<br />
\qquad<br />
\mathrm{(B)}\ 25<br />
\qquad<br />
\mathrm{(C)}\ 35<br />
\qquad<br />
\mathrm{(D)}\ 45<br />
\qquad<br />
\mathrm{(E)}\ 55<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
Which of the following is equal to <math>1 + \frac{1}{1+\frac{1}{1+1}}</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{5}{4}<br />
\qquad<br />
\mathrm{(B)}\ \frac{3}{2}<br />
\qquad<br />
\mathrm{(C)}\ \frac{5}{3}<br />
\qquad<br />
\mathrm{(D)}\ 2<br />
\qquad<br />
\mathrm{(E)}\ 3<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Eric plans to compete in a triathlon. He can average <math>2</math> miles per hour in the <math>\frac{1}{4}</math>-mile swim and <math>6</math> miles per hour in the <math>3</math>-mile run. His goal is to finish the triathlon in <math>2</math> hours. To accomplish his goal what must his average speed in miles per hour, be for the <math>15</math>-mile bicycle ride?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{120}{11}<br />
\qquad<br />
\mathrm{(B)}\ 11<br />
\qquad<br />
\mathrm{(C)}\ \frac{56}{5}<br />
\qquad<br />
\mathrm{(D)}\ \frac{45}{4}<br />
\qquad<br />
\mathrm{(E)}\ 12<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
What is the sum of the digits of the square of <math>111,111,111</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 18<br />
\qquad<br />
\mathrm{(B)}\ 27<br />
\qquad<br />
\mathrm{(C)}\ 45<br />
\qquad<br />
\mathrm{(D)}\ 63<br />
\qquad<br />
\mathrm{(E)}\ 81<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
A circle of radius <math>2</math> is inscribed in a semicircle, as shown. The area inside the semicircle but outside the circle is shaded. What fraction of the semicircle's area is shaded? <br />
<br />
<asy><br />
unitsize(6mm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
dotfactor=4;<br />
<br />
filldraw(Arc((0,0),4,0,180)--cycle,gray,black);<br />
filldraw(Circle((0,2),2),white,black);<br />
dot((0,2));<br />
draw((0,2)--((0,2)+2*dir(60)));<br />
label("$2$",midpoint((0,2)--((0,2)+2*dir(60))),SE);<br />
</asy><br />
<br />
<math><br />
\mathrm{(A)}\ \frac{1}{2}<br />
\qquad<br />
\mathrm{(B)}\ \frac{\pi}{6}<br />
\qquad<br />
\mathrm{(C)}\ \frac{2}{\pi}<br />
\qquad<br />
\mathrm{(D)}\ \frac{2}{3}<br />
\qquad<br />
\mathrm{(E)}\ \frac{3}{\pi}<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
A carton contains milk that is <math>2</math>% fat, an amount that is <math>40</math>% less fat than the amount contained in a carton of whole milk. What is the percentage of fat in whole milk?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{12}{5}<br />
\qquad<br />
\mathrm{(B)}\ 3<br />
\qquad<br />
\mathrm{(C)}\ \frac{10}{3}<br />
\qquad<br />
\mathrm{(D)}\ 38<br />
\qquad<br />
\mathrm{(E)}\ 42<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Three Generations of the Wen family are going to the movies, two from each generation. The two members of the youngest generation receive a <math>50</math>% discount as children. The two members of the oldest generation receive a <math>25\%</math> discount as senior citizens. The two members of the middle generation receive no discount. Grandfather Wen, whose senior ticket costs <dollar/><math>6.00</math>, is paying for everyone. How many dollars must he pay?<br />
<br />
<math><br />
\mathrm{(A)}\ 34<br />
\qquad<br />
\mathrm{(B)}\ 36<br />
\qquad<br />
\mathrm{(C)}\ 42<br />
\qquad<br />
\mathrm{(D)}\ 46<br />
\qquad<br />
\mathrm{(E)}\ 48<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
Positive integers <math>a</math>, <math>b</math>, and <math>2009</math>, with <math>a<b<2009</math>, form a geometric sequence with an integer ratio. What is <math>a</math>? <br />
<br />
<math><br />
\mathrm{(A)}\ 7<br />
\qquad<br />
\mathrm{(B)}\ 41<br />
\qquad<br />
\mathrm{(C)}\ 49<br />
\qquad<br />
\mathrm{(D)}\ 289<br />
\qquad<br />
\mathrm{(E)}\ 2009<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
Triangle <math>ABC</math> has a right angle at <math>B</math>. Point <math>D</math> is the foot of the altitude from <math>B</math>, <math>AD=3</math>, and <math>DC=4</math>. What is the area of <math>\triangle ABC</math>?<br />
<br />
<asy><br />
unitsize(5mm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
dotfactor=4;<br />
<br />
pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21));<br />
pair D=foot(B,A,C);<br />
pair[] ps={B,C,A,D};<br />
<br />
draw(A--B--C--cycle);<br />
draw(B--D);<br />
draw(rightanglemark(B,D,C));<br />
<br />
dot(ps);<br />
label("$A$",A,NW);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NE);<br />
label("$3$",midpoint(A--D),NE);<br />
label("$4$",midpoint(D--C),NE);<br />
</asy><br />
<br />
<math><br />
\mathrm{(A)}\ 4\sqrt3<br />
\qquad<br />
\mathrm{(B)}\ 7\sqrt3<br />
\qquad<br />
\mathrm{(C)}\ 21<br />
\qquad<br />
\mathrm{(D)}\ 14\sqrt3 <br />
\qquad<br />
\mathrm{(E)}\ 42<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
One dimension of a cube is increased by <math>1</math>, another is decreased by <math>1</math>, and the third is left unchanged. The volume of the new rectangular solid is <math>5</math> less than that of the cube. What was the volume of the cube?<br />
<br />
<math><br />
\mathrm{(A)}\ 8<br />
\qquad<br />
\mathrm{(B)}\ 27<br />
\qquad<br />
\mathrm{(C)}\ 64<br />
\qquad<br />
\mathrm{(D)}\ 125<br />
\qquad<br />
\mathrm{(E)}\ 216<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
In quadrilateral <math>ABCD</math>, <math>AB = 5</math>, <math>BC = 17</math>, <math>CD = 5</math>, <math>DA = 9</math>, and <math>BD</math> is an integer. What is <math>BD</math>?<br />
<center><asy><br />
unitsize(4mm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
dotfactor=4;<br />
<br />
pair C=(0,0), B=(17,0);<br />
pair D=intersectionpoints(Circle(C,5),Circle(B,13))[0];<br />
pair A=intersectionpoints(Circle(D,9),Circle(B,5))[0];<br />
pair[] dotted={A,B,C,D};<br />
<br />
draw(D--A--B--C--D--B);<br />
dot(dotted);<br />
label("$D$",D,NW);<br />
label("$C$",C,W);<br />
label("$B$",B,E);<br />
label("$A$",A,NE);<br />
</asy></center><br />
<br />
<math><br />
\mathrm{(A)}\ 11<br />
\qquad<br />
\mathrm{(B)}\ 12<br />
\qquad<br />
\mathrm{(C)}\ 13<br />
\qquad<br />
\mathrm{(D)}\ 14<br />
\qquad<br />
\mathrm{(E)}\ 15<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Suppose that <math>P = 2^m</math> and <math>Q = 3^n</math>. Which of the following is equal to <math>12^{mn}</math> for every pair of integers <math>(m,n)</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ P^2Q<br />
\qquad<br />
\mathrm{(B)}\ P^nQ^m<br />
\qquad<br />
\mathrm{(C)}\ P^nQ^{2m}<br />
\qquad<br />
\mathrm{(D)}\ P^{2m}Q^n<br />
\qquad<br />
\mathrm{(E)}\ P^{2n}Q^m<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
Four congruent rectangles are placed as shown. The area of the outer square is <math>4</math> times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?<br />
<center><asy><br />
unitsize(6mm);<br />
defaultpen(linewidth(.8pt));<br />
<br />
path p=(1,1)--(-2,1)--(-2,2)--(1,2);<br />
draw(p);<br />
draw(rotate(90)*p);<br />
draw(rotate(180)*p);<br />
draw(rotate(270)*p);<br />
</asy></center><br />
<br />
<math><br />
\mathrm{(A)}\ 3<br />
\qquad<br />
\mathrm{(B)}\ \sqrt {10}<br />
\qquad<br />
\mathrm{(C)}\ 2 + \sqrt2<br />
\qquad<br />
\mathrm{(D)}\ 2\sqrt3 <br />
\qquad<br />
\mathrm{(E)}\ 4<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
The figures <math>F_1</math>, <math>F_2</math>, <math>F_3</math>, and <math>F_4</math> shown are the first in a sequence of figures. For <math>n\ge3</math>, <math>F_n</math> is constructed from <math>F_{n - 1}</math> by surrounding it with a square and placing one more diamond on each side of the new square than <math>F_{n - 1}</math> had on each side of its outside square. For example, figure <math>F_3</math> has <math>13</math> diamonds. How many diamonds are there in figure <math>F_{20}</math>?<br />
<center><asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
<br />
path d=(1/2,0)--(0,sqrt(3)/2)--(-1/2,0)--(0,-sqrt(3)/2)--cycle;<br />
marker m=marker(scale(5)*d,Fill);<br />
path f1=(0,0);<br />
path f2=(0,0)--(-1,1)--(1,1)--(1,-1)--(-1,-1);<br />
path[] g2=(-1,1)--(-1,-1)--(0,0)^^(1,-1)--(0,0)--(1,1);<br />
path f3=f2--(-2,-2)--(-2,0)--(-2,2)--(0,2)--(2,2)--(2,0)--(2,-2)--(0,-2);<br />
path[] g3=g2^^(-2,-2)--(0,-2)^^(2,-2)--(1,-1)^^(1,1)--(2,2)^^(-1,1)--(-2,2);<br />
path[] f4=f3^^(-3,-3)--(-3,-1)--(-3,1)--(-3,3)--(-1,3)--(1,3)--(3,3)--<br />
(3,1)--(3,-1)--(3,-3)--(1,-3)--(-1,-3);<br />
path[] g4=g3^^(-2,-2)--(-3,-3)--(-1,-3)^^(3,-3)--(2,-2)^^(2,2)--(3,3)^^<br />
(-2,2)--(-3,3);<br />
<br />
draw(f1,m);<br />
draw(shift(5,0)*f2,m);<br />
draw(shift(5,0)*g2);<br />
draw(shift(12,0)*f3,m);<br />
draw(shift(12,0)*g3);<br />
draw(shift(21,0)*f4,m);<br />
draw(shift(21,0)*g4);<br />
label("$F_1$",(0,-4));<br />
label("$F_2$",(5,-4));<br />
label("$F_3$",(12,-4));<br />
label("$F_4$",(21,-4));<br />
</asy></center><br />
<br />
<math><br />
\mathrm{(A)}\ 401<br />
\qquad<br />
\mathrm{(B)}\ 485<br />
\qquad<br />
\mathrm{(C)}\ 585<br />
\qquad<br />
\mathrm{(D)}\ 626<br />
\qquad<br />
\mathrm{(E)}\ 761<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
Let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be real numbers with <math>|a-b|=2</math>, <math>|b-c|=3</math>, and <math>|c-d|=4</math>. What is the sum of all possible values of <math>|a-d|</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 9<br />
\qquad<br />
\mathrm{(B)}\ 12<br />
\qquad<br />
\mathrm{(C)}\ 15<br />
\qquad<br />
\mathrm{(D)}\ 18<br />
\qquad<br />
\mathrm{(E)}\ 24<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Rectangle <math>ABCD</math> has <math>AB=4</math> and <math>BC=3</math>. Segment <math>EF</math> is constructed through <math>B</math> so that <math>EF</math> is perpendicular to <math>DB</math>, and <math>A</math> and <math>C</math> lie on <math>DE</math> and <math>DF</math>, respectively. What is <math>EF</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 9<br />
\qquad<br />
\mathrm{(B)}\ 10<br />
\qquad<br />
\mathrm{(C)}\ \frac {125}{12}<br />
\qquad<br />
\mathrm{(D)}\ \frac {103}{9}<br />
\qquad<br />
\mathrm{(E)}\ 12<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
At Jefferson Summer Camp, <math>60\%</math> of the children play soccer, <math>30\%</math> of the children swim, and <math>40\%</math> of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer? <br />
<br />
<math><br />
\mathrm{(A)}\ 30\%<br />
\qquad<br />
\mathrm{(B)}\ 40\%<br />
\qquad<br />
\mathrm{(C)}\ 49\%<br />
\qquad<br />
\mathrm{(D)}\ 51\%<br />
\qquad<br />
\mathrm{(E)}\ 70\%<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
Circle <math>A</math> has radius <math>100</math>. Circle <math>B</math> has an integer radius <math>r<100</math> and remains internally tangent to circle <math>A</math> as it rolls once around the circumference of circle <math>A</math>. The two circles have the same points of tangency at the beginning and end of circle <math>B</math>'s trip. How many possible values can <math>r</math> have?<br />
<br />
<math><br />
\mathrm{(A)}\ 4<br />
\qquad<br />
\mathrm{(B)}\ 8<br />
\qquad<br />
\mathrm{(C)}\ 9<br />
\qquad<br />
\mathrm{(D)}\ 50<br />
\qquad<br />
\mathrm{(E)}\ 90<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
Andrea and Lauren are <math>20</math> kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of <math>1</math> kilometer per minute. After <math>5</math> minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?<br />
<br />
<math><br />
\mathrm{(A)}\ 20<br />
\qquad<br />
\mathrm{(B)}\ 30<br />
\qquad<br />
\mathrm{(C)}\ 55<br />
\qquad<br />
\mathrm{(D)}\ 65<br />
\qquad<br />
\mathrm{(E)}\ 80<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle? <br />
<br />
<asy><br />
unitsize(6mm);<br />
defaultpen(linewidth(.8pt));<br />
<br />
draw(Circle((0,0),1+sqrt(2)));<br />
draw(Circle((sqrt(2),0),1));<br />
draw(Circle((0,sqrt(2)),1));<br />
draw(Circle((-sqrt(2),0),1));<br />
draw(Circle((0,-sqrt(2)),1));<br />
</asy><br />
<br />
<math><br />
\mathrm{(A)}\ 3-2\sqrt2<br />
\qquad<br />
\mathrm{(B)}\ 2-\sqrt2<br />
\qquad<br />
\mathrm{(C)}\ 4(3-2\sqrt2)<br />
\qquad<br />
\mathrm{(D)}\ \frac12(3-\sqrt2)<br />
\qquad<br />
\mathrm{(E)}\ 2\sqrt2-2<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
Two cubical dice each have removable numbers <math>1</math> through <math>6</math>. The twelve numbers on the two dice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is <math>7</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{1}{9}<br />
\qquad<br />
\mathrm{(B)}\ \frac{1}{8}<br />
\qquad<br />
\mathrm{(C)}\ \frac{1}{6}<br />
\qquad<br />
\mathrm{(D)}\ \frac{2}{11}<br />
\qquad<br />
\mathrm{(E)}\ \frac{1}{5}<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
Convex quadrilateral <math>ABCD</math> has <math>AB=9</math> and <math>CD=12</math>. Diagonals <math>AC</math> and <math>BD</math> intersect at <math>E</math>, <math>AC=14</math>, and <math>\triangle AED</math> and <math>\triangle BEC</math> have equal areas. What is <math>AE</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{9}{2}<br />
\qquad<br />
\mathrm{(B)}\ \frac{50}{11}<br />
\qquad<br />
\mathrm{(C)}\ \frac{21}{4}<br />
\qquad<br />
\mathrm{(D)}\ \frac{17}{3}<br />
\qquad<br />
\mathrm{(E)}\ 6<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{1}{4}<br />
\qquad<br />
\mathrm{(B)}\ \frac{3}{8}<br />
\qquad<br />
\mathrm{(C)}\ \frac{4}{7}<br />
\qquad<br />
\mathrm{(D)}\ \frac{5}{7}<br />
\qquad<br />
\mathrm{(E)}\ \frac{3}{4}<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
For <math>k > 0</math>, let <math>I_k = 10\ldots 064</math>, where there are <math>k</math> zeros between the <math>1</math> and the <math>6</math>. Let <math>N(k)</math> be the number of factors of <math>2</math> in the prime factorization of <math>I_k</math>. What is the maximum value of <math>N(k)</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 6<br />
\qquad<br />
\mathrm{(B)}\ 7<br />
\qquad<br />
\mathrm{(C)}\ 8<br />
\qquad<br />
\mathrm{(D)}\ 9<br />
\qquad<br />
\mathrm{(E)}\ 10<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 25|Solution]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_7&diff=304132009 AMC 10A Problems/Problem 72009-02-19T01:13:41Z<p>VelaDabant: /* Problem */</p>
<hr />
<div>== Problem ==<br />
A carton contains milk that is <math>2</math>% fat, an amount that is <math>40</math>% less fat than the amount contained in a carton of whole milk. What is the percentage of fat in whole milk?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{12}{5}<br />
\qquad<br />
\mathrm{(B)}\ 3<br />
\qquad<br />
\mathrm{(C)}\ \frac{10}{3}<br />
\qquad<br />
\mathrm{(D)}\ 38<br />
\qquad<br />
\mathrm{(E)}\ 42<br />
</math><br />
<br />
==Solution==<br />
<math>\longrightarrow \fbox{C}</math><br />
<br />
==See also==<br />
{{AMC10 box|year=2009|ab=A|num-b=6|num-a=8}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_7&diff=304122009 AMC 10A Problems/Problem 72009-02-19T01:11:34Z<p>VelaDabant: New page: == Problem == A carton contains milk that is <math>2</math>% fat, an amount that is <math>40</math>% less fat than the amount contained in a carton of whole milk. What is the percentage of...</p>
<hr />
<div>== Problem ==<br />
A carton contains milk that is <math>2</math>% fat, an amount that is <math>40</math>% less fat than the amount contained in a carton of whole milk. What is the percentage of fat in whole milk?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{12}{5}<br />
\qquad<br />
\mathrm{(B)}\ \frac{10}{3}<br />
\qquad<br />
\mathrm{(C)}\ 9<br />
\qquad<br />
\mathrm{(D)}\ 38<br />
\qquad<br />
\mathrm{(E)}\ 42<br />
</math><br />
<br />
==Solution==<br />
<math>\longrightarrow \fbox{C}</math><br />
<br />
==See also==<br />
{{AMC10 box|year=2009|ab=A|num-b=6|num-a=8}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_6&diff=304112009 AMC 10A Problems/Problem 62009-02-19T01:07:19Z<p>VelaDabant: New page: == Problem == A circle of radius <math>2</math> is inscribed in a semicircle, as shown. The area inside the semicircle but outside the circle is shaded. What fraction of the semicircle's ...</p>
<hr />
<div>== Problem ==<br />
<br />
A circle of radius <math>2</math> is inscribed in a semicircle, as shown. The area inside the semicircle but outside the circle is shaded. What fraction of the semicircle's area is shaded? <br />
<br />
<asy><br />
unitsize(6mm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
dotfactor=4;<br />
<br />
filldraw(Arc((0,0),4,0,180)--cycle,gray,black);<br />
filldraw(Circle((0,2),2),white,black);<br />
dot((0,2));<br />
draw((0,2)--((0,2)+2*dir(60)));<br />
label("$2$",midpoint((0,2)--((0,2)+2*dir(60))),SE);<br />
</asy><br />
<br />
<math><br />
\mathrm{(A)}\ \frac{1}{2}<br />
\qquad<br />
\mathrm{(B)}\ \frac{\pi}{6}<br />
\qquad<br />
\mathrm{(C)}\ \frac{2}{\pi}<br />
\qquad<br />
\mathrm{(D)}\ \frac{2}{3}<br />
\qquad<br />
\mathrm{(E)}\ \frac{3}{\pi}<br />
</math><br />
<br />
==Solution==<br />
<math>\longrightarrow \fbox{A}</math><br />
<br />
==See also==<br />
{{AMC10 box|year=2009|ab=A|num-b=5|num-a=7}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_4&diff=304102009 AMC 10A Problems/Problem 42009-02-19T01:00:36Z<p>VelaDabant: New page: ==Problem== Eric plans to compete in a triathlon. He can average <math>2</math> miles per hour in the <math>\frac{1}{4}</math>-mile swim and <math>6</math> miles per hour in the <math>3</m...</p>
<hr />
<div>==Problem==<br />
Eric plans to compete in a triathlon. He can average <math>2</math> miles per hour in the <math>\frac{1}{4}</math>-mile swim and <math>6</math> miles per hour in the <math>3</math>-mile run. His goal is to finish the triathlon in <math>2</math> hours. To accomplish his goal what must his average speed in miles per hour, be for the <math>15</math>-mile bicycle ride?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{120}{11}<br />
\qquad<br />
\mathrm{(B)}\ 11<br />
\qquad<br />
\mathrm{(C)}\ \frac{56}{5}<br />
\qquad<br />
\mathrm{(D)}\ \frac{45}{4}<br />
\qquad<br />
\mathrm{(E)}\ 12<br />
</math><br />
<br />
==Solution==<br />
<math>\longrightarrow \fbox{A}</math><br />
<br />
==See also==<br />
{{AMC10 box|year=2009|ab=A|num-b=3|num-a=5}}</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=AMC_12_Problems_and_Solutions&diff=30409AMC 12 Problems and Solutions2009-02-19T00:49:50Z<p>VelaDabant: </p>
<hr />
<div>[[AMC 12]] problems and solutions by test:<br />
<br />
* [[1951 AHSME]]<br />
* [[1952 AHSME]]<br />
* [[1953 AHSME]]<br />
* [[1954 AHSME]]<br />
* [[1955 AHSME]]<br />
* [[1956 AHSME]]<br />
* [[1957 AHSME]]<br />
* [[1958 AHSME]]<br />
* [[1959 AHSME]]<br />
* [[1960 AHSME]]<br />
* [[1961 AHSME]]<br />
* [[1962 AHSME]]<br />
* [[1963 AHSME]]<br />
* [[1964 AHSME]]<br />
* [[1965 AHSME]]<br />
* [[1966 AHSME]]<br />
* [[1970 AHSME]]<br />
* [[2000 AMC 12]]<br />
* [[2001 AMC 12]]<br />
* [[2002 AMC 12A]]<br />
* [[2002 AMC 12B]]<br />
* [[2003 AMC 12A]]<br />
* [[2003 AMC 12B]]<br />
* [[2004 AMC 12A]]<br />
* [[2004 AMC 12B]]<br />
* [[2005 AMC 12A]]<br />
* [[2005 AMC 12B]]<br />
* [[2006 AMC 12A]]<br />
* [[2006 AMC 12B]]<br />
* [[2007 AMC 12A]]<br />
* [[2007 AMC 12B]]<br />
* [[2008 AMC 12A]]<br />
* [[2008 AMC 12B]]<br />
* [[2009 AMC 12A]]<br />
<br />
== Resources ==<br />
* [[American Mathematics Competitions]]<br />
* [[AMC Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
==Other==<br />
* [[AMC 12 Problems and Solutions/Tasks | AMC 12-related tasks which need completion]]<br />
<br />
[[Category:Math Contest Problems]]</div>VelaDabanthttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems&diff=304082009 AMC 10A Problems2009-02-19T00:47:31Z<p>VelaDabant: </p>
<hr />
<div>== Problem 1 ==<br />
One can holds <math>12</math> ounces of soda. What is the minimum number of cans needed to provide a gallon (128 ounces) of soda?<br />
<br />
<math><br />
\mathrm{(A)}\ 7<br />
\qquad<br />
\mathrm{(B)}\ 8<br />
\qquad<br />
\mathrm{(C)}\ 9<br />
\qquad<br />
\mathrm{(D)}\ 10<br />
\qquad<br />
\mathrm{(E)}\ 11<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes and quarters. Which of the following could ''not'' be the total value of the four coins, in cents?<br />
<br />
<math><br />
\mathrm{(A)}\ 15<br />
\qquad<br />
\mathrm{(B)}\ 25<br />
\qquad<br />
\mathrm{(C)}\ 35<br />
\qquad<br />
\mathrm{(D)}\ 45<br />
\qquad<br />
\mathrm{(E)}\ 55<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
Which of the following is equal to <math>1 + \frac{1}{1+\frac{1}{1+1}}</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{5}{4}<br />
\qquad<br />
\mathrm{(B)}\ \frac{3}{2}<br />
\qquad<br />
\mathrm{(C)}\ \frac{5}{3}<br />
\qquad<br />
\mathrm{(D)}\ 2<br />
\qquad<br />
\mathrm{(E)}\ 3<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Eric plans to compete in a triathlon. He can average <math>2</math> miles per hour in the <math>\frac{1}{4}</math>-mile swim and <math>6</math> miles per hour in the <math>3</math>-mile run. His goal is to finish the triathlon in <math>2</math> hours. To accomplish his goal what must his average speed in miles per hour, be for the <math>15</math>-mile bicycle ride?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{120}{11}<br />
\qquad<br />
\mathrm{(B)}\ 11<br />
\qquad<br />
\mathrm{(C)}\ \frac{56}{5}<br />
\qquad<br />
\mathrm{(D)}\ \frac{45}{4}<br />
\qquad<br />
\mathrm{(E)}\ 12<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
What is the sum of the digits of the square of <math>111,111,111</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 18<br />
\qquad<br />
\mathrm{(B)}\ 27<br />
\qquad<br />
\mathrm{(C)}\ 45<br />
\qquad<br />
\mathrm{(D)}\ 63<br />
\qquad<br />
\mathrm{(E)}\ 81<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
A circle of radius <math>2</math> is inscribed in a semicircle, as shown. The area inside the semicircle but outside the circle is shaded. What fraction of the semicircle's area is shaded? <br />
<br />
<asy><br />
unitsize(6mm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
dotfactor=4;<br />
<br />
filldraw(Arc((0,0),4,0,180)--cycle,gray,black);<br />
filldraw(Circle((0,2),2),white,black);<br />
dot((0,2));<br />
draw((0,2)--((0,2)+2*dir(60)));<br />
label("$2$",midpoint((0,2)--((0,2)+2*dir(60))),SE);<br />
</asy><br />
<br />
<math><br />
\mathrm{(A)}\ \frac{1}{2}<br />
\qquad<br />
\mathrm{(B)}\ \frac{\pi}{6}<br />
\qquad<br />
\mathrm{(C)}\ \frac{2}{\pi}<br />
\qquad<br />
\mathrm{(D)}\ \frac{2}{3}<br />
\qquad<br />
\mathrm{(E)}\ \frac{3}{\pi}<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
A carton contains milk that is <math>2</math>% fat, an amount that is <math>40</math>% less fat than the amount contained in a carton of whole milk. What is the percentage of fat in whole milk?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{12}{5}<br />
\qquad<br />
\mathrm{(B)}\ \frac{10}{3}<br />
\qquad<br />
\mathrm{(C)}\ 9<br />
\qquad<br />
\mathrm{(D)}\ 38<br />
\qquad<br />
\mathrm{(E)}\ 42<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Three Generations of the Wen family are going to the movies, two from each generation. The two members of the youngest generation receive a <math>50</math>% discount as children. The two members of the oldest generation receive a <math>25\%</math> discount as senior citizens. The two members of the middle generation receive no discount. Grandfather Wen, whose senior ticket costs <dollar/><math>6.00</math>, is paying for everyone. How many dollars must he pay?<br />
<br />
<math><br />
\mathrm{(A)}\ 34<br />
\qquad<br />
\mathrm{(B)}\ 36<br />
\qquad<br />
\mathrm{(C)}\ 42<br />
\qquad<br />
\mathrm{(D)}\ 46<br />
\qquad<br />
\mathrm{(E)}\ 48<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
Positive integers <math>a</math>, <math>b</math>, and <math>2009</math>, with <math>a<b<2009</math>, form a geometric sequence with an integer ratio. What is <math>a</math>? <br />
<br />
<math><br />
\mathrm{(A)}\ 7<br />
\qquad<br />
\mathrm{(B)}\ 41<br />
\qquad<br />
\mathrm{(C)}\ 49<br />
\qquad<br />
\mathrm{(D)}\ 289<br />
\qquad<br />
\mathrm{(E)}\ 2009<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
Triangle <math>ABC</math> has a right angle at <math>B</math>. Point <math>D</math> is the foot of the altitude from <math>B</math>, <math>AD=3</math>, and <math>DC=4</math>. What is the area of <math>\triangle ABC</math>?<br />
<br />
<asy><br />
unitsize(5mm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
dotfactor=4;<br />
<br />
pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21));<br />
pair D=foot(B,A,C);<br />
pair[] ps={B,C,A,D};<br />
<br />
draw(A--B--C--cycle);<br />
draw(B--D);<br />
draw(rightanglemark(B,D,C));<br />
<br />
dot(ps);<br />
label("$A$",A,NW);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NE);<br />
label("$3$",midpoint(A--D),NE);<br />
label("$4$",midpoint(D--C),NE);<br />
</asy><br />
<br />
<math><br />
\mathrm{(A)}\ 4\sqrt3<br />
\qquad<br />
\mathrm{(B)}\ 7\sqrt3<br />
\qquad<br />
\mathrm{(C)}\ 21<br />
\qquad<br />
\mathrm{(D)}\ 14\sqrt3 <br />
\qquad<br />
\mathrm{(E)}\ 42<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
One dimension of a cube is increased by <math>1</math>, another is decreased by <math>1</math>, and the third is left unchanged. The volume of the new rectangular solid is <math>5</math> less than that of the cube. What was the volume of the cube?<br />
<br />
<math><br />
\mathrm{(A)}\ 8<br />
\qquad<br />
\mathrm{(B)}\ 27<br />
\qquad<br />
\mathrm{(C)}\ 64<br />
\qquad<br />
\mathrm{(D)}\ 125<br />
\qquad<br />
\mathrm{(E)}\ 216<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
In quadrilateral <math>ABCD</math>, <math>AB = 5</math>, <math>BC = 17</math>, <math>CD = 5</math>, <math>DA = 9</math>, and <math>BD</math> is an integer. What is <math>BD</math>?<br />
<center><asy><br />
unitsize(4mm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
dotfactor=4;<br />
<br />
pair C=(0,0), B=(17,0);<br />
pair D=intersectionpoints(Circle(C,5),Circle(B,13))[0];<br />
pair A=intersectionpoints(Circle(D,9),Circle(B,5))[0];<br />
pair[] dotted={A,B,C,D};<br />
<br />
draw(D--A--B--C--D--B);<br />
dot(dotted);<br />
label("$D$",D,NW);<br />
label("$C$",C,W);<br />
label("$B$",B,E);<br />
label("$A$",A,NE);<br />
</asy></center><br />
<br />
<math><br />
\mathrm{(A)}\ 11<br />
\qquad<br />
\mathrm{(B)}\ 12<br />
\qquad<br />
\mathrm{(C)}\ 13<br />
\qquad<br />
\mathrm{(D)}\ 14<br />
\qquad<br />
\mathrm{(E)}\ 15<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Suppose that <math>P = 2^m</math> and <math>Q = 3^n</math>. Which of the following is equal to <math>12^{mn}</math> for every pair of integers <math>(m,n)</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ P^2Q<br />
\qquad<br />
\mathrm{(B)}\ P^nQ^m<br />
\qquad<br />
\mathrm{(C)}\ P^nQ^{2m}<br />
\qquad<br />
\mathrm{(D)}\ P^{2m}Q^n<br />
\qquad<br />
\mathrm{(E)}\ P^{2n}Q^m<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
Four congruent rectangles are placed as shown. The area of the outer square is <math>4</math> times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?<br />
<center><asy><br />
unitsize(6mm);<br />
defaultpen(linewidth(.8pt));<br />
<br />
path p=(1,1)--(-2,1)--(-2,2)--(1,2);<br />
draw(p);<br />
draw(rotate(90)*p);<br />
draw(rotate(180)*p);<br />
draw(rotate(270)*p);<br />
</asy></center><br />
<br />
<math><br />
\mathrm{(A)}\ 3<br />
\qquad<br />
\mathrm{(B)}\ \sqrt {10}<br />
\qquad<br />
\mathrm{(C)}\ 2 + \sqrt2<br />
\qquad<br />
\mathrm{(D)}\ 2\sqrt3 <br />
\qquad<br />
\mathrm{(E)}\ 4<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
The figures <math>F_1</math>, <math>F_2</math>, <math>F_3</math>, and <math>F_4</math> shown are the first in a sequence of figures. For <math>n\ge3</math>, <math>F_n</math> is constructed from <math>F_{n - 1}</math> by surrounding it with a square and placing one more diamond on each side of the new square than <math>F_{n - 1}</math> had on each side of its outside square. For example, figure <math>F_3</math> has <math>13</math> diamonds. How many diamonds are there in figure <math>F_{20}</math>?<br />
<center><asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
<br />
path d=(1/2,0)--(0,sqrt(3)/2)--(-1/2,0)--(0,-sqrt(3)/2)--cycle;<br />
marker m=marker(scale(5)*d,Fill);<br />
path f1=(0,0);<br />
path f2=(0,0)--(-1,1)--(1,1)--(1,-1)--(-1,-1);<br />
path[] g2=(-1,1)--(-1,-1)--(0,0)^^(1,-1)--(0,0)--(1,1);<br />
path f3=f2--(-2,-2)--(-2,0)--(-2,2)--(0,2)--(2,2)--(2,0)--(2,-2)--(0,-2);<br />
path[] g3=g2^^(-2,-2)--(0,-2)^^(2,-2)--(1,-1)^^(1,1)--(2,2)^^(-1,1)--(-2,2);<br />
path[] f4=f3^^(-3,-3)--(-3,-1)--(-3,1)--(-3,3)--(-1,3)--(1,3)--(3,3)--<br />
(3,1)--(3,-1)--(3,-3)--(1,-3)--(-1,-3);<br />
path[] g4=g3^^(-2,-2)--(-3,-3)--(-1,-3)^^(3,-3)--(2,-2)^^(2,2)--(3,3)^^<br />
(-2,2)--(-3,3);<br />
<br />
draw(f1,m);<br />
draw(shift(5,0)*f2,m);<br />
draw(shift(5,0)*g2);<br />
draw(shift(12,0)*f3,m);<br />
draw(shift(12,0)*g3);<br />
draw(shift(21,0)*f4,m);<br />
draw(shift(21,0)*g4);<br />
label("$F_1$",(0,-4));<br />
label("$F_2$",(5,-4));<br />
label("$F_3$",(12,-4));<br />
label("$F_4$",(21,-4));<br />
</asy></center><br />
<br />
<math><br />
\mathrm{(A)}\ 401<br />
\qquad<br />
\mathrm{(B)}\ 485<br />
\qquad<br />
\mathrm{(C)}\ 585<br />
\qquad<br />
\mathrm{(D)}\ 626<br />
\qquad<br />
\mathrm{(E)}\ 761<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
Let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be real numbers with <math>|a-b|=2</math>, <math>|b-c|=3</math>, and <math>|c-d|=4</math>. What is the sum of all possible values of <math>|a-d|</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 9<br />
\qquad<br />
\mathrm{(B)}\ 12<br />
\qquad<br />
\mathrm{(C)}\ 15<br />
\qquad<br />
\mathrm{(D)}\ 18<br />
\qquad<br />
\mathrm{(E)}\ 24<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Rectangle <math>ABCD</math> has <math>AB=4</math> and <math>BC=3</math>. Segment <math>EF</math> is constructed through <math>B</math> so that <math>EF</math> is perpendicular to <math>DB</math>, and <math>A</math> and <math>C</math> lie on <math>DE</math> and <math>DF</math>, respectively. What is <math>EF</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 9<br />
\qquad<br />
\mathrm{(B)}\ 10<br />
\qquad<br />
\mathrm{(C)}\ \frac {125}{12}<br />
\qquad<br />
\mathrm{(D)}\ \frac {103}{9}<br />
\qquad<br />
\mathrm{(E)}\ 12<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
At Jefferson Summer Camp, <math>60\%</math> of the children play soccer, <math>30\%</math> of the children swim, and <math>40\%</math> of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer? <br />
<br />
<math><br />
\mathrm{(A)}\ 30\%<br />
\qquad<br />
\mathrm{(B)}\ 40\%<br />
\qquad<br />
\mathrm{(C)}\ 49\%<br />
\qquad<br />
\mathrm{(D)}\ 51\%<br />
\qquad<br />
\mathrm{(E)}\ 70\%<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
Circle <math>A</math> has radius <math>100</math>. Circle <math>B</math> has an integer radius <math>r<100</math> and remains internally tangent to circle <math>A</math> as it rolls once around the circumference of circle <math>A</math>. The two circles have the same points of tangency at the beginning and end of circle <math>B</math>'s trip. How many possible values can <math>r</math> have?<br />
<br />
<math><br />
\mathrm{(A)}\ 4<br />
\qquad<br />
\mathrm{(B)}\ 8<br />
\qquad<br />
\mathrm{(C)}\ 9<br />
\qquad<br />
\mathrm{(D)}\ 50<br />
\qquad<br />
\mathrm{(E)}\ 90<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
Andrea and Lauren are <math>20</math> kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of <math>1</math> kilometer per minute. After <math>5</math> minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?<br />
<br />
<math><br />
\mathrm{(A)}\ 20<br />
\qquad<br />
\mathrm{(B)}\ 30<br />
\qquad<br />
\mathrm{(C)}\ 55<br />
\qquad<br />
\mathrm{(D)}\ 65<br />
\qquad<br />
\mathrm{(E)}\ 80<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle? <br />
<br />
<asy><br />
unitsize(6mm);<br />
defaultpen(linewidth(.8pt));<br />
<br />
draw(Circle((0,0),1+sqrt(2)));<br />
draw(Circle((sqrt(2),0),1));<br />
draw(Circle((0,sqrt(2)),1));<br />
draw(Circle((-sqrt(2),0),1));<br />
draw(Circle((0,-sqrt(2)),1));<br />
</asy><br />
<br />
<math><br />
\mathrm{(A)}\ 3-2\sqrt2<br />
\qquad<br />
\mathrm{(B)}\ 2-\sqrt2<br />
\qquad<br />
\mathrm{(C)}\ 4(3-2\sqrt2)<br />
\qquad<br />
\mathrm{(D)}\ \frac12(3-\sqrt2)<br />
\qquad<br />
\mathrm{(E)}\ 2\sqrt2-2<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
Two cubical dice each have removable numbers <math>1</math> through <math>6</math>. The twelve numbers on the two dice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is <math>7</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{1}{9}<br />
\qquad<br />
\mathrm{(B)}\ \frac{1}{8}<br />
\qquad<br />
\mathrm{(C)}\ \frac{1}{6}<br />
\qquad<br />
\mathrm{(D)}\ \frac{2}{11}<br />
\qquad<br />
\mathrm{(E)}\ \frac{1}{5}<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
Convex quadrilateral <math>ABCD</math> has <math>AB=9</math> and <math>CD=12</math>. Diagonals <math>AC</math> and <math>BD</math> intersect at <math>E</math>, <math>AC=14</math>, and <math>\triangle AED</math> and <math>\triangle BEC</math> have equal areas. What is <math>AE</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{9}{2}<br />
\qquad<br />
\mathrm{(B)}\ \frac{50}{11}<br />
\qquad<br />
\mathrm{(C)}\ \frac{21}{4}<br />
\qquad<br />
\mathrm{(D)}\ \frac{17}{3}<br />
\qquad<br />
\mathrm{(E)}\ 6<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{1}{4}<br />
\qquad<br />
\mathrm{(B)}\ \frac{3}{8}<br />
\qquad<br />
\mathrm{(C)}\ \frac{4}{7}<br />
\qquad<br />
\mathrm{(D)}\ \frac{5}{7}<br />
\qquad<br />
\mathrm{(E)}\ \frac{3}{4}<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
For <math>k > 0</math>, let <math>I_k = 10\ldots 064</math>, where there are <math>k</math> zeros between the <math>1</math> and the <math>6</math>. Let <math>N(k)</math> be the number of factors of <math>2</math> in the prime factorization of <math>I_k</math>. What is the maximum value of <math>N(k)</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 6<br />
\qquad<br />
\mathrm{(B)}\ 7<br />
\qquad<br />
\mathrm{(C)}\ 8<br />
\qquad<br />
\mathrm{(D)}\ 9<br />
\qquad<br />
\mathrm{(E)}\ 10<br />
</math><br />
<br />
[[2009 AMC 10A Problems/Problem 25|Solution]]</div>VelaDabant