https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Flyingpenguin&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T19:05:27ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_20&diff=515222013 AMC 12B Problems/Problem 202013-02-23T01:36:28Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>==Problem==<br />
For <math>135^\circ < x < 180^\circ</math>, points <math>P=(\cos x, \cos^2 x), Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)</math> and <math>S =(\tan x, \tan^2 x)</math> are the vertices of a trapezoid. What is <math>\sin(2x)</math>?<br />
<br />
<math> \textbf{(A)}\ 2-2\sqrt{2}\qquad\textbf{(B)}\3\sqrt{3}-6\qquad\textbf{(C)}\ 3\sqrt{2}-5\qquad\textbf{(D}}\ -\frac{3}{4}\qquad\textbf{(E)}\ 1-\sqrt{3}</math><br />
<br />
==Solution==<br />
<br />
Let <math>f,g,h,j</math> be <math>\sin, \cos, \tan, \cot</math> (not respectively). Then we have four points <math>(f,f^2),(g,g^2),(h,h^2),(j,j^2)</math>, and a pair of lines each connecting two points must be parallel (as we are dealing with a trapezoid). WLOG, take the line connecting the first two points and the line connecting the last two points to be parallel, so that <math>\frac{g^2-f^2}{g-f} = \frac{j^2-h^2}{j-h}</math>, or <math>g+f = j+h</math>.<br />
<br />
Now, we must find how to match up <math>\sin, \cos, \tan, \cot</math> to <math>f,g,h,j</math> so that the above equation has a solution. The three possible ways are <math>\sin x+\cos x = \tan x+\cot x</math>, <math>\sin x+\tan x = \cos x+\cot x</math>, and <math>\sin x+\cot x = \cos x+\tan x</math>. On the interval <math>135^\circ < x < 180^\circ</math>, we have <math>0<\sin x<\frac{1}{\sqrt{2}},-1< \cos x<-\frac{1}{\sqrt{2}},-1< \tan x<0,-\infty< \cot x<-1</math>, so the first two of those possible ways do not work because the LHS and the RHS have non-overlapping ranges. Thus, it must be the last possible way, <math>\sin x+\cot x = \cos x+\tan x</math>.<br />
<br />
Now, we perform some algebraic manipulation to find <math>\sin (2x)</math>:<br />
<br />
<cmath><br />
\sin x+\cot x = \cos x+\tan x \\<br />
(\sin x+\cot x)(\sin x \cos x ) = (\cos x+\tan x)(\sin x \cos x ) \\<br />
\sin^2 x \cos x+\cos^2 x = \cos^2 x \sin x+\sin^2 x \\<br />
\sin^2 x \cos x -\cos^2 x \sin x = \sin^2 x - \cos^2 x \\<br />
\sin x \cos x (\sin x -\cos x) = (\sin x - \cos x)(\sin x + \cos x) \\<br />
\sin x \cos x = \sin x + \cos x \\<br />
(\sin x \cos x)^2 = (\sin x + \cos x)^2 \\<br />
(\sin x \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x \\<br />
(\sin x \cos x)^2 = 1 + 2\sin x \cos x \\<br />
(\sin x \cos x)^2 - 2\sin x \cos x - 1 =0<br />
</cmath><br />
<br />
Solve the quadratic to find <math>\sin x \cos x = \frac{2 - 2\sqrt{2}}{2}</math>, so that <math>\sin(2x) = 2 \sin x \cos x = \boxed{\textbf{(A)} \ 2 - 2\sqrt{2}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2013|ab=B|num-b=19|num-a=21}}</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_20&diff=515212013 AMC 12B Problems/Problem 202013-02-23T01:32:24Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>==Problem==<br />
For <math>135^\circ < x < 180^\circ</math>, points <math>P=(\cos x, \cos^2 x), Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)</math> and <math>S =(\tan x, \tan^2 x)</math> are the vertices of a trapezoid. What is <math>\sin(2x)</math>?<br />
<br />
<math> \textbf{(A)}\ 2-2\sqrt{2}\qquad\textbf{(B)}\3\sqrt{3}-6\qquad\textbf{(C)}\ 3\sqrt{2}-5\qquad\textbf{(D}}\ -\frac{3}{4}\qquad\textbf{(E)}\ 1-\sqrt{3}</math><br />
<br />
==Solution==<br />
<br />
Let <math>f,g,h,j</math> be <math>\sin, \cos, \tan, \cot</math> (not respectively). Then we have four points <math>(f,f^2),(g,g^2),(h,h^2),(j,j^2)</math>, and the two lines connecting two pairs of them must be parallel (as we are dealing with a trapezoid). WLOG, we take the line connecting the first two and the line connecting the last two to be parallel, so that <math>\frac{g^2-f^2}{g-f} = \frac{j^2-h^2}{j-h}</math>, or <math>g+f = j+h</math>.<br />
<br />
Now, we must find how to match up <math>\sin, \cos, \tan, \cot</math> to <math>f,g,h,j</math> so that the above equation has a solution. The three possible ways are <math>\sin x+\cos x = \tan x+\cot x</math>, <math>\sin x+\tan x = \cos x+\cot x</math>, and <math>\sin x+\cot x = \cos x+\tan x</math>. On the interval <math>135^\circ < x < 180^\circ</math>, we have <math>0<\sin x<\frac{1}{\sqrt{2}},-1< \cos x<-\frac{1}{\sqrt{2}},-1< \tan x<0,-\infty< \cot x<-1</math>, so the first two of those possible ways do not work because the LHS and the RHS have different ranges. Thus, it must be the last possible way, <math>\sin x+\cot x = \cos x+\tan x</math>.<br />
<br />
Now, we perform some algebraic manipulation to find <math>\sin (2x)</math>:<br />
<br />
<cmath><br />
\sin x+\cot x = \cos x+\tan x \\<br />
(\sin x+\cot x)(\sin x \cos x ) = (\cos x+\tan x)(\sin x \cos x ) \\<br />
\sin^2 x \cos x+\cos^2 x = \cos^2 x \sin x+\sin^2 x \\<br />
\sin^2 x \cos x -\cos^2 x \sin x = \sin^2 x - \cos^2 x \\<br />
\sin x \cos x (\sin x -\cos x) = (\sin x - \cos x)(\sin x + \cos x) \\<br />
\sin x \cos x = \sin x + \cos x \\<br />
(\sin x \cos x)^2 = (\sin x + \cos x)^2 \\<br />
(\sin x \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x \\<br />
(\sin x \cos x)^2 = 1 + 2\sin x \cos x \\<br />
(\sin x \cos x)^2 - 2\sin x \cos x - 1 =0<br />
</cmath><br />
<br />
Solve the quadratic to find <math>\sin x \cos x = \frac{2 - 2\sqrt{2}}{2}</math>, so that <math>\sin(2x) = 2 \sin x \cos x = \boxed{\textbf{(A)} \ 2 - 2\sqrt{2}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2013|ab=B|num-b=19|num-a=21}}</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_16&diff=515192013 AMC 12B Problems/Problem 162013-02-23T00:56:40Z<p>Flyingpenguin: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCDE</math> be an equiangular convex pentagon of perimeter <math>1</math>. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let <math>s</math> be the perimeter of this star. What is the difference between the maximum and the minimum possible values of <math>s</math>.<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\ \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}</math><br />
<br />
==Solution==<br />
<br />
The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure <math>\frac{180^\circ (5-2)}{5}=108^\circ</math>, and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure <math>180^\circ - 108^\circ = 72^\circ</math>. The base angles are equal, so the triangles must be isosceles.<br />
<br />
Let one of the sides of the pentagon have length <math>x_1</math> (and the others <math>x_2, x_3, x_4, x_5</math>). Then, by trigonometry, the non-base sides of the triangle sitting on that side of the pentagon each has length <math>\frac{x_1}{2} \sec 72^\circ</math>, and so the two sides together have length <math>x_1 \sec 72^\circ</math>. To find the perimeter of the star, we sum up the lengths of the non-base sides for each of the five triangles to get <math>(x_1+x_2+x_3+x_4+x_5) \sec 72^\circ = (1) \sec 72^\circ = \sec 72^\circ</math> (because the perimeter of the pentagon is <math>1</math>). The perimeter of the star is constant, so the difference between the maximum and minimum perimeters is <math>\boxed{\textbf{(D)} \ 0}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2013|ab=B|num-b=15|num-a=17}}</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_19&diff=442422008 AMC 12A Problems/Problem 192012-01-15T10:18:10Z<p>Flyingpenguin: </p>
<hr />
<div>==Problem==<br />
In the expansion of<br />
<cmath>\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,</cmath><br />
what is the [[coefficient]] of <math>x^{28}</math>?<br />
<br />
<math>\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405</math><br />
<br />
==Solution== <br />
Let <math>A = \left(1 + x + x^2 + \cdots + x^{14}\right)</math> and <math>B = \left(1 + x + x^2 + \cdots + x^{27}\right)</math>. We are expanding <math>A \cdot A \cdot B</math>. <br />
<br />
Since there are <math>15</math> terms in <math>A</math>, there are <math>15^2 = 225</math> ways to choose one term from each <math>A</math>. The product of the selected terms is <math>x^n</math> for some integer <math>n</math> between <math>0</math> and <math>28</math> inclusive. For each <math>n \neq 0</math>, there is one and only one <math>x^{28 - n}</math> in <math>B</math>. For example, if I choose <math>x^2</math> from <math>A</math> , then there is exactly one power of <math>x</math> in <math>B</math> that I can choose; in this case, it would be <math>x^{24}</math>. Since there is only one way to choose one term from each <math>A</math> to get a product of <math>x^0</math>, there are <math>225 - 1 = 224</math> ways to choose one term from each <math>A</math> and one term from <math>B</math> to get a product of <math>x^{28}</math>. Thus the coefficient of the <math>x^{28}</math> term is <math>224 \Rightarrow C</math>.<br />
<br />
== Solution 2 ==<br />
Let <math>P(x) = \left(1 + x + x^2 + \cdots + x^{14}\right)^2 = a_0 + a_1x + a_2x^2 + \cdots + a_{28}x^{28}</math>. Then the <math>x^{28}</math> term from the product in question <math>\left(1 + x + x^2 + \cdots + x^{27}\right)(a_0 + a_1x + a_2x^2 + \cdots + a_{28}x^{28})</math> is<br />
<br />
<math>1a_{28}x^{28} + xa_{27}x^{27} + x^2a_{26}x^{26} + \cdots + x^{27}a_1x = \left(a_1 + a_2 + \cdots a_{28}\right)x^{28}</math><br />
<br />
So we are trying to find the sum of the coefficients of <math>P(x)</math> minus <math>a_0</math>. Since the constant term <math>a_0</math> in <math>P(x)</math> (when expanded) is <math>1</math>, and the sum of the coefficients of <math>P(x)</math> is <math>P(1)</math>, we find the answer to be<br />
<math>P(1) - a_0<br />
= \left(1 + 1 + 1^2 + \cdots 1^{14}\right)^2 - 1<br />
= 15^2 - 1<br />
= 224 \rightarrow C</math><br />
<br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=A|num-b=18|num-a=20}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_24&diff=441472002 AMC 12A Problems/Problem 242012-01-05T04:22:32Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Find the number of ordered pairs of real numbers <math>(a,b)</math> such that <math>(a+bi)^{2002} = a-bi</math>.<br />
<br />
<math><br />
\text{(A) }1001<br />
\qquad<br />
\text{(B) }1002<br />
\qquad<br />
\text{(C) }2001<br />
\qquad<br />
\text{(D) }2002<br />
\qquad<br />
\text{(E) }2004<br />
</math><br />
<br />
== Solution ==<br />
<br />
Let <math>s=\sqrt{a^2+b^2}</math> be the magnitude of <math>a+bi</math>. Then the magnitude of <math>(a+bi)^{2002}</math> is <math>s^{2002}</math>, while the magnitude of <math>a-bi</math> is <math>s</math>. We get that <math>s^{2002}=s</math>, hence either <math>s=0</math> or <math>s=1</math>.<br />
<br />
For <math>s=0</math> we get a single solution <math>(a,b)=(0,0)</math>.<br />
<br />
Let's now assume that <math>s=1</math>. Multiply both sides by <math>a+bi</math>. The left hand side becomes <math>(a+bi)^{2003}</math>, the right hand side becomes <math>(a-bi)(a+bi)=a^2 + b^2 = 1</math>. Hence the solutions for this case are precisely all the <math>2003</math>rd complex roots of unity, and there are <math>2003</math> of those.<br />
<br />
The total number of solutions is therefore <math>1+2003 = \boxed{2004}</math>.<br />
<br />
<br />
== Solution 2 ==<br />
<br />
As in the other solution, split the problem into when <math>s=0</math> and when <math>s=1</math>. When <math>s=1</math> and <math>a+bi=\cos\theta+\sin\theta</math>, <math>(a+bi)^{2002}= \cos(2002\theta)+i\sin(2002\theta)= \cos\theta-i\sin\theta= \cos(-\theta)+i\sin(-\theta) = a - bi</math>, so we must have <math>2002\theta=-\theta+2\pi k</math> and hence <math>\theta=\frac{2\pi k}{2003}</math>. Since <math>\theta</math> is restricted to <math>[0,2\pi)</math>, <math>k</math> can range from <math>0</math> to <math>2002</math> inclusive, which is <math>2002-0+1=2003</math> values. Thus the total is <math>1+2003 = \boxed{\textbf{(E)}\ 2004}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2002|ab=A|num-b=23|num-a=25}}</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_15&diff=381742011 AIME II Problems/Problem 152011-04-20T00:13:04Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>P(x) = x^2 - 3x - 9</math>. A real number <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}</math> is equal to <math>\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}</math> , where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are positive integers. Find <math>a + b + c + d + e</math>.<br />
<br />
==Solution==<br />
<br />
Table of values of <math>P(x)</math>:<br />
<br />
<math>\begin{array*}<br />
P(5) = 1 \\<br />
P(6) = 9 \\<br />
P(7) = 19 \\<br />
P(8) = 31 \\<br />
P(9) = 45 \\<br />
P(10) = 61 \\<br />
P(11) = 79 \\<br />
P(12) = 99 \\<br />
P(13) = 121 \\<br />
P(14) = 145 \\<br />
P(15) = 171 \\<br />
\end{array*}</math><br />
<br />
In order for <math>\lfloor \sqrt{P(x)} \rfloor = \sqrt{P(\lfloor x \rfloor)}</math> to hold, <math>\sqrt{P(\lfloor x \rfloor)}</math> must be an integer and hence <math>P(\lfloor x \rfloor)</math> must be a perfect square. This limits <math>x</math> to <math>5 < x < 6</math> or <math>6 < x < 7</math> or <math>13 < x < 14</math> since, from the table above, those are the only values of <math>x</math> for which <math>P(\lfloor x \rfloor)</math> is an perfect square. However, in order for <math>\sqrt{P(x)}</math> to be rounded down to <math>P(\lfloor x \rfloor)</math>, <math>P(x)</math> must not be greater than the next perfect square after <math>P(\lfloor x \rfloor)</math> (for the said intervals). Note that in all the cases the next value of <math>P(x)</math> always passes the next perfect square after <math>P(\lfloor x \rfloor)</math>, so in no cases will all values of <math>x</math> in the said intervals work. Now, we consider the three difference cases.<br />
<br />
<br />
Case <math>5 < x < 6</math>:<br />
<br />
<math>P(x)</math> must not be greater than the first perfect square after <math>1</math>, which is <math>4</math>. Since <math>P(x)</math> is increasing for <math>x > 5</math>, we just need to find where <math>P(x) = 4</math> and the values that will work will be <math>5 < x < \text{root}</math>.<br />
<br />
<math>\begin{array*}<br />
x^2 - 3x - 9 = 4 \\<br />
x = \frac{3 + \sqrt{61}}{2}<br />
\end{array*}</math><br />
<br />
So in this case, the only values that will work are <math>5 < x < \frac{3 + \sqrt{61}}{2}</math>.<br />
<br />
Case <math>6 < x < 7</math>:<br />
<br />
<math>P(x)</math> must not be greater than the first perfect square after <math>9</math>, which is <math>16</math>.<br />
<br />
<math>\begin{array*}<br />
x^2 - 3x - 9 = 16 \\<br />
x = \frac{3 + \sqrt{109}}{2}<br />
\end{array*}</math><br />
<br />
So in this case, the only values that will work are <math>6 < x < \frac{3 + \sqrt{109}}{2}</math>.<br />
<br />
Case <math>13 < x < 14</math>:<br />
<br />
<math>P(x)</math> must not be greater than the first perfect square after <math>121</math>, which is <math>144</math>.<br />
<br />
<math>\begin{array*}<br />
x^2 - 3x - 9 = 144 \\<br />
x = \frac{3 + \sqrt{621}}{2}<br />
\end{array*}</math><br />
<br />
So in this case, the only values that will work are <math>13 < x < \frac{3 + \sqrt{621}}{2}</math>.<br />
<br />
Now, we find the length of the working intervals and divide it by the length of the total interval, <math>15 - 5 = 10</math>:<br />
<br />
<math>\begin{array*}<br />
\frac{\left( \frac{3 + \sqrt{61}}{2} - 5 \right) + \left( \frac{3 + \sqrt{109}}{2} - 6 \right) + \left( \frac{3 + \sqrt{621}}{2} - 13 \right)}{10} \\<br />
= \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20}<br />
\end{array*}</math><br />
<br />
So the answer is <math>61 + 109 + 621 + 39 + 20 = \fbox{850}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_15&diff=381722011 AIME II Problems/Problem 152011-04-20T00:08:30Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>P(x) = x^2 - 3x - 9</math>. A real number <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}</math> is equal to <math>\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}</math> , where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are positive integers. Find <math>a + b + c + d + e</math>.<br />
<br />
==Solution==<br />
<br />
Table of values of <math>P(x)</math>:<br />
<br />
<math>\begin{array*}<br />
P(5) = 1 \\<br />
P(6) = 9 \\<br />
P(7) = 19 \\<br />
P(8) = 31 \\<br />
P(9) = 45 \\<br />
P(10) = 61 \\<br />
P(11) = 79 \\<br />
P(12) = 99 \\<br />
P(13) = 121 \\<br />
P(14) = 145 \\<br />
P(15) = 171 \\<br />
\end{array*}</math><br />
<br />
In order for <math>\lfloor \sqrt{P(x)} \rfloor = \sqrt{P(\lfloor x \rfloor)}</math> to hold, <math>\sqrt{P(\lfloor x \rfloor)}</math> must be an integer and hence <math>P(\lfloor x \rfloor)</math> must be a perfect square. This limits <math>x</math> to <math>5 < x < 6</math> or <math>6 < x < 7</math> or <math>13 < x < 14</math> since, from the table above, those are the only values of <math>x</math> for which <math>P(\lfloor x \rfloor)</math> is an perfect square. However, in order for <math>\sqrt{P(x)}</math> to be rounded down to <math>P(\lfloor x \rfloor)</math>, <math>P(x)</math> must not be greater than the next perfect square after <math>P(\lfloor x \rfloor)</math> (for the said intervals). Note that in all the cases the next value of <math>P(x)</math> always passes the next perfect square after <math>P(\lfloor x \rfloor)</math>, so in no cases will all values of <math>x</math> in the said intervals work. Now, we consider the three difference cases.<br />
<br />
<br />
Case <math>1</math>:<br />
<math>5 < x < 6 \\</math>:<br />
<math>P(x)</math> must not be greater than the first perfect square after <math>1</math>, which is <math>4</math>. Since <math>P(x)</math> is increasing for <math>x > 5</math>, we just need to find where <math>P(x) = 4</math> and the values that will work will be <math>5 < x < \text{root}</math>.<br />
<br />
<math>\begin{array*}<br />
x^2 - 3x - 9 = 4 \\<br />
x = \frac{3 + \sqrt{61}}{2}<br />
\end{array*}</math><br />
<br />
So in this case, the only values that will work are <math>5 < x < \frac{3 + \sqrt{61}}{2}</math>.<br />
<br />
Case <math>2</math>:<br />
<math>6 < x < 7 \\</math>:<br />
<math>P(x)</math> must not be greater than the first perfect square after <math>9</math>, which is <math>16</math>.<br />
<br />
<math>\begin{array*}<br />
x^2 - 3x - 9 = 16 \\<br />
x = \frac{3 + \sqrt{109}}{2}<br />
\end{array*}</math><br />
<br />
So in this case, the only values that will work are <math>6 < x < \frac{3 + \sqrt{109}}{2}</math>.<br />
<br />
Case <math>3</math>:<br />
<math>13 < x < 14 \\</math>:<br />
<math>P(x)</math> must not be greater than the first perfect square after <math>121</math>, which is <math>144</math>.<br />
<br />
<math>\begin{array*}<br />
x^2 - 3x - 9 = 144 \\<br />
x = \frac{3 + \sqrt{621}}{2}<br />
\end{array*}</math><br />
<br />
So in this case, the only values that will work are <math>13 < x < \frac{3 + \sqrt{621}}{2}</math>.<br />
<br />
Now, we find the length of the working intervals and divide it by the length of the total interval, <math>15 - 5 = 10</math>:<br />
<br />
<math>\begin{array*}<br />
\frac{(\frac{3 + \sqrt{61}}{2} - 5) + (\frac{3 + \sqrt{109}}{2} - 6) + (\frac{3 + \sqrt{621}}{2} - 13)}{10} \\<br />
= \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39)}{20}<br />
\end{array*}</math><br />
<br />
So the answer is <math>61 + 109 + 621 + 39 + 20 = 850</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_15&diff=381702011 AIME II Problems/Problem 152011-04-19T23:45:31Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>P(x) = x^2 - 3x - 9</math>. A real number <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}</math> is equal to <math>\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}</math> , where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are positive integers. Find <math>a + b + c + d + e</math>.<br />
<br />
==Solution==<br />
<br />
Table of values of P([x]):<br />
<br />
P(5) = 1<br />
P(6) = 9<br />
P(7) = 19<br />
P(8) = 31<br />
P(9) = 45<br />
P(10) = 61<br />
P(11) = 79<br />
P(12) = 99<br />
P(13) = 121<br />
P(14) = 145<br />
P(15) = 171<br />
<br />
In order for [√P(x)] = √P([x]) to hold, √P([x]) must be an integer and hence P([x]) must be a perfect square. This limits x to 5 < x < 6 or 6 < x < 7 or 13 < x < 14 since, from the table above, those are the only values of x for which P([x]) is an integer. However, in order for √P(x) to be rounded down to √P([x]), P(x) must not be greater than the next perfect square after P([x]) (for 5 < x < 6, etc.). Note that in all the cases the next value of P(x) always passes the next perfect square after P([x]), so in no cases will all values of x in the said intervals work. Now, we consider the three difference cases.<br />
<br />
5 < x < 6:<br />
P(x) must not be greater than the first perfect square after 1, which is 4. Since P(x) is increasing for x > 5, we just need to find where P(x) = 4 and the values that will work will be 5 < x < root.<br />
<br />
x^2 - 3x - 9 = 4<br />
x = (3 + √61)/2<br />
<br />
So in this case, the only values that will work are 5 < x < (3 + √61)/2.<br />
<br />
6 < x < 7:<br />
P(x) must not be greater than the first perfect square after 9, which is 16.<br />
<br />
x^2 - 3x - 9 = 16<br />
x = (3 + √109)/2<br />
<br />
So in this case, the only values that will work are 6 < x < (3 + √109)/2.<br />
<br />
13 < x < 14:<br />
P(x) must not be greater than the first perfect square after 121, which is 144.<br />
<br />
x^2 - 3x - 9 = 144<br />
x = (3 + √721)/2<br />
<br />
So in this case, the only values that will work are 13 < x < (3 + √721)/2.<br />
<br />
Now, we find the length of the working intervals and divide it by the length of the total interval, 15 - 5 = 10:<br />
<br />
(((3 + √61)/2 - 5) + ((3 + √109)/2 - 6) + ((3 + √721)/2 - 13))/10<br />
= (√61 + √109 + √721 - 39)/20<br />
<br />
So the answer is 61 + 109 + 721 + 39 + 20 = 950.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_18&diff=369832011 AMC 10A Problems/Problem 182011-02-16T05:45:25Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>== Problem 18 ==<br />
<br />
Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}</math>. What is the area inside Circle <math>C</math> but outside circle <math>A</math> and circle <math>B</math> ?<br />
<br />
== Solution ==<br />
<br />
Draw a rectangle with vertices at the centers of <math>A</math> and <math>B</math> and the intersection of <math>A, C</math> and <math>B, C</math>. Then, we can compute the shades area as the area of half of <math>C</math> plus the area of the rectangle minus the area of the two sectors created by <math>A</math> and <math>B</math>. This is <math>\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{2 \ \mathbf{(C)}}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=369822011 AMC 10A Problems2011-02-16T05:44:09Z<p>Flyingpenguin: /* Problem 18 */</p>
<hr />
<div>== Problem 1 ==<br />
<br />
A cell phone plan costs <math>\</math><math>20</math> each month, plus <math>5</math>¢ per text message sent, plus 10¢ for each minute used over <math>30</math> hours. In January Michelle sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did she have to pay?<br />
<br />
<math> \textbf{(A)}\ </math> <math>24.00 \qquad\textbf{(B)}\ </math> <math>24.50 \qquad\textbf{(C)}\ </math> <math>25.50\qquad\textbf{(D)}\ </math> <math>28.00\qquad\textbf{(E)}\ </math> <math>30.00 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
A small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?<br />
<br />
<math> \textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
Suppose [<math>a</math> <math>b</math>] denotes the average of <math>a</math> and <math>b</math>, and {<math>a</math> <math>b</math> <math>c</math>} denotes the average of <math>a</math>, <math>b</math>, and <math>c</math>. What is {{1 1 0} [0 1] 0}?<br />
<br />
<math> \textbf{(A)}\ \frac{2}{9} \qquad\textbf{(B)}\ \frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
Let <math>X</math> and <math>Y</math> be the following sums of arithmetic sequences: <cmath> \begin{eqnarray*} X &=& 10 + 12 + 14 + \cdots + 100, \\ Y &=& 12 + 14 + 16 + \cdots + 102. \end{eqnarray*} </cmath> What is the value of <math>Y - X</math>?<br />
<br />
<math> \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 </math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of <math>12</math>, <math>15</math>, and <math>10</math> minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students? <br />
<br />
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
Set <math>A </math> has 20 elements, and set <math>B </math> has 15 elements. What is the smallest possible number of elements in <math>A \cup B </math>, the union of <math>A </math> and <math>B </math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Which of the following equations does NOT have a solution?<br />
<br />
<math>\text{(A)}\:(x+7)^2=0</math><br />
<br />
<math>\text{(B)}\:|-3x|+5=0</math><br />
<br />
<math>\text{(C)}\:\sqrt{-x}-2=0</math><br />
<br />
<math>\text{(D)}\:\sqrt{x}-8=0</math><br />
<br />
<math>\text{(E)}\:|-3x|-4=0 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?<br />
<br />
<br />
<math> \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A rectangular region is bounded by the graphs of the equations <math>y=a, y=-b, x=-c,</math> and <math>x=d</math>, where <math>a,b,c,</math> and <math>d</math> are all positive numbers. Which of the following represents the area of this region?<br />
<br />
<math> \textbf{(A)}\ ac+ad+bc+bd\qquad\textbf{(B)}\ ac-ad+bc-bd\qquad\textbf{(C)}\ ac+ad-bc-bd \quad\quad\qquad\textbf{(D)}\ -ac-ad+bc+bd\qquad\textbf{(E)}\ ac-ad-bc+bd </math><br />
<br />
[[2011 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was <math>\</math><math>17.71</math>. What was the cost of a pencil in cents?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
[[2011 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Square <math>EFGH</math> has one vertex on each side of square <math>ABCD</math>. Point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=7\cdot EB</math>. What is the ratio of the area of <math>EFGH</math> to the area of <math>ABCD</math>?<br />
<br />
<math>\text{(A)}\,\frac{49}{64} \qquad\text{(B)}\,\frac{25}{32} \qquad\text{(C)}\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?<br />
<br />
<math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200</math><br />
<br />
[[2011 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br />
<br />
<math>\text{(A)}\,\frac{1}{36} \qquad\text{(B)}\,\frac{1}{12} \qquad\text{(C)}\,\frac{1}{6} \qquad\text{(D)}\,\frac{1}{4} \qquad\text{(E)}\,\frac{5}{18}</math><br />
<br />
[[2011 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?<br />
<br />
<math>\text{(A)}\,140 \qquad\text{(B)}\,240 \qquad\text{(C)}\,440 \qquad\text{(D)}\,640 \qquad\text{(E)}\,840</math><br />
<br />
[[2011 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Which of the following is equal to <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>?<br />
<br />
<math>\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6</math><br />
<br />
[[2011 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
In the eight-term sequence <math>A,B,C,D,E,F,G,H</math>, the value of <math>C</math> is 5 and the sum of any three consecutive terms is 30. What is <math>A+H</math>?<br />
<br />
<math>\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43</math><br />
<br />
[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}</math>. What is the area inside Circle <math>C</math> but outside circle <math>A</math> and circle <math>B</math> ?<br />
<br />
[[2011 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br />
<br />
<math> \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?<br />
<br />
<math> \textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math> \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Seven students count from 1 to 1000 as follows:<br />
<br />
•Alice says all the numbers, except she ships the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br />
<br />
•Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br />
<br />
•Finally, George says the only number that no one else says.<br />
<br />
What number does George say?<br />
<br />
<math> \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Let <math>R</math> be a square region and <math>n\ge4</math> an integer. A point <math>X</math> in the interior of <math>R</math> is called <math>n\text{-}ray</math> partitional if there are <math>n</math> rays emanating from <math>X</math> that divide <math>R</math> into <math>n</math> triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?<br />
<br />
<math>\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500</math><br />
<br />
[[2011 AMC 10A Problems/Problem 25|Solution]]</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=369812011 AMC 10A Problems2011-02-16T05:43:23Z<p>Flyingpenguin: /* Problem 18 */</p>
<hr />
<div>== Problem 1 ==<br />
<br />
A cell phone plan costs <math>\</math><math>20</math> each month, plus <math>5</math>¢ per text message sent, plus 10¢ for each minute used over <math>30</math> hours. In January Michelle sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did she have to pay?<br />
<br />
<math> \textbf{(A)}\ </math> <math>24.00 \qquad\textbf{(B)}\ </math> <math>24.50 \qquad\textbf{(C)}\ </math> <math>25.50\qquad\textbf{(D)}\ </math> <math>28.00\qquad\textbf{(E)}\ </math> <math>30.00 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
A small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?<br />
<br />
<math> \textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
Suppose [<math>a</math> <math>b</math>] denotes the average of <math>a</math> and <math>b</math>, and {<math>a</math> <math>b</math> <math>c</math>} denotes the average of <math>a</math>, <math>b</math>, and <math>c</math>. What is {{1 1 0} [0 1] 0}?<br />
<br />
<math> \textbf{(A)}\ \frac{2}{9} \qquad\textbf{(B)}\ \frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
Let <math>X</math> and <math>Y</math> be the following sums of arithmetic sequences: <cmath> \begin{eqnarray*} X &=& 10 + 12 + 14 + \cdots + 100, \\ Y &=& 12 + 14 + 16 + \cdots + 102. \end{eqnarray*} </cmath> What is the value of <math>Y - X</math>?<br />
<br />
<math> \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 </math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of <math>12</math>, <math>15</math>, and <math>10</math> minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students? <br />
<br />
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
Set <math>A </math> has 20 elements, and set <math>B </math> has 15 elements. What is the smallest possible number of elements in <math>A \cup B </math>, the union of <math>A </math> and <math>B </math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Which of the following equations does NOT have a solution?<br />
<br />
<math>\text{(A)}\:(x+7)^2=0</math><br />
<br />
<math>\text{(B)}\:|-3x|+5=0</math><br />
<br />
<math>\text{(C)}\:\sqrt{-x}-2=0</math><br />
<br />
<math>\text{(D)}\:\sqrt{x}-8=0</math><br />
<br />
<math>\text{(E)}\:|-3x|-4=0 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?<br />
<br />
<br />
<math> \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A rectangular region is bounded by the graphs of the equations <math>y=a, y=-b, x=-c,</math> and <math>x=d</math>, where <math>a,b,c,</math> and <math>d</math> are all positive numbers. Which of the following represents the area of this region?<br />
<br />
<math> \textbf{(A)}\ ac+ad+bc+bd\qquad\textbf{(B)}\ ac-ad+bc-bd\qquad\textbf{(C)}\ ac+ad-bc-bd \quad\quad\qquad\textbf{(D)}\ -ac-ad+bc+bd\qquad\textbf{(E)}\ ac-ad-bc+bd </math><br />
<br />
[[2011 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was <math>\</math><math>17.71</math>. What was the cost of a pencil in cents?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
[[2011 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Square <math>EFGH</math> has one vertex on each side of square <math>ABCD</math>. Point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=7\cdot EB</math>. What is the ratio of the area of <math>EFGH</math> to the area of <math>ABCD</math>?<br />
<br />
<math>\text{(A)}\,\frac{49}{64} \qquad\text{(B)}\,\frac{25}{32} \qquad\text{(C)}\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?<br />
<br />
<math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200</math><br />
<br />
[[2011 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br />
<br />
<math>\text{(A)}\,\frac{1}{36} \qquad\text{(B)}\,\frac{1}{12} \qquad\text{(C)}\,\frac{1}{6} \qquad\text{(D)}\,\frac{1}{4} \qquad\text{(E)}\,\frac{5}{18}</math><br />
<br />
[[2011 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?<br />
<br />
<math>\text{(A)}\,140 \qquad\text{(B)}\,240 \qquad\text{(C)}\,440 \qquad\text{(D)}\,640 \qquad\text{(E)}\,840</math><br />
<br />
[[2011 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Which of the following is equal to <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>?<br />
<br />
<math>\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6</math><br />
<br />
[[2011 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
In the eight-term sequence <math>A,B,C,D,E,F,G,H</math>, the value of <math>C</math> is 5 and the sum of any three consecutive terms is 30. What is <math>A+H</math>?<br />
<br />
<math>\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43</math><br />
<br />
[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}</math>. What is the area inside Circle <math>C</math> but outside circle <math>A</math> and circle <math>B</math> ?<br />
<br />
[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br />
<br />
<math> \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?<br />
<br />
<math> \textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math> \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Seven students count from 1 to 1000 as follows:<br />
<br />
•Alice says all the numbers, except she ships the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br />
<br />
•Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br />
<br />
•Finally, George says the only number that no one else says.<br />
<br />
What number does George say?<br />
<br />
<math> \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Let <math>R</math> be a square region and <math>n\ge4</math> an integer. A point <math>X</math> in the interior of <math>R</math> is called <math>n\text{-}ray</math> partitional if there are <math>n</math> rays emanating from <math>X</math> that divide <math>R</math> into <math>n</math> triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?<br />
<br />
<math>\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500</math><br />
<br />
[[2011 AMC 10A Problems/Problem 25|Solution]]</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=369802011 AMC 10A Problems2011-02-16T05:41:09Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>== Problem 1 ==<br />
<br />
A cell phone plan costs <math>\</math><math>20</math> each month, plus <math>5</math>¢ per text message sent, plus 10¢ for each minute used over <math>30</math> hours. In January Michelle sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did she have to pay?<br />
<br />
<math> \textbf{(A)}\ </math> <math>24.00 \qquad\textbf{(B)}\ </math> <math>24.50 \qquad\textbf{(C)}\ </math> <math>25.50\qquad\textbf{(D)}\ </math> <math>28.00\qquad\textbf{(E)}\ </math> <math>30.00 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
A small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?<br />
<br />
<math> \textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
Suppose [<math>a</math> <math>b</math>] denotes the average of <math>a</math> and <math>b</math>, and {<math>a</math> <math>b</math> <math>c</math>} denotes the average of <math>a</math>, <math>b</math>, and <math>c</math>. What is {{1 1 0} [0 1] 0}?<br />
<br />
<math> \textbf{(A)}\ \frac{2}{9} \qquad\textbf{(B)}\ \frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
Let <math>X</math> and <math>Y</math> be the following sums of arithmetic sequences: <cmath> \begin{eqnarray*} X &=& 10 + 12 + 14 + \cdots + 100, \\ Y &=& 12 + 14 + 16 + \cdots + 102. \end{eqnarray*} </cmath> What is the value of <math>Y - X</math>?<br />
<br />
<math> \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 </math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of <math>12</math>, <math>15</math>, and <math>10</math> minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students? <br />
<br />
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
Set <math>A </math> has 20 elements, and set <math>B </math> has 15 elements. What is the smallest possible number of elements in <math>A \cup B </math>, the union of <math>A </math> and <math>B </math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Which of the following equations does NOT have a solution?<br />
<br />
<math>\text{(A)}\:(x+7)^2=0</math><br />
<br />
<math>\text{(B)}\:|-3x|+5=0</math><br />
<br />
<math>\text{(C)}\:\sqrt{-x}-2=0</math><br />
<br />
<math>\text{(D)}\:\sqrt{x}-8=0</math><br />
<br />
<math>\text{(E)}\:|-3x|-4=0 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?<br />
<br />
<br />
<math> \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A rectangular region is bounded by the graphs of the equations <math>y=a, y=-b, x=-c,</math> and <math>x=d</math>, where <math>a,b,c,</math> and <math>d</math> are all positive numbers. Which of the following represents the area of this region?<br />
<br />
<math> \textbf{(A)}\ ac+ad+bc+bd\qquad\textbf{(B)}\ ac-ad+bc-bd\qquad\textbf{(C)}\ ac+ad-bc-bd \quad\quad\qquad\textbf{(D)}\ -ac-ad+bc+bd\qquad\textbf{(E)}\ ac-ad-bc+bd </math><br />
<br />
[[2011 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was <math>\</math><math>17.71</math>. What was the cost of a pencil in cents?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
[[2011 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Square <math>EFGH</math> has one vertex on each side of square <math>ABCD</math>. Point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=7\cdot EB</math>. What is the ratio of the area of <math>EFGH</math> to the area of <math>ABCD</math>?<br />
<br />
<math>\text{(A)}\,\frac{49}{64} \qquad\text{(B)}\,\frac{25}{32} \qquad\text{(C)}\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?<br />
<br />
<math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200</math><br />
<br />
[[2011 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br />
<br />
<math>\text{(A)}\,\frac{1}{36} \qquad\text{(B)}\,\frac{1}{12} \qquad\text{(C)}\,\frac{1}{6} \qquad\text{(D)}\,\frac{1}{4} \qquad\text{(E)}\,\frac{5}{18}</math><br />
<br />
[[2011 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?<br />
<br />
<math>\text{(A)}\,140 \qquad\text{(B)}\,240 \qquad\text{(C)}\,440 \qquad\text{(D)}\,640 \qquad\text{(E)}\,840</math><br />
<br />
[[2011 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Which of the following is equal to <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>?<br />
<br />
<math>\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6</math><br />
<br />
[[2011 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
In the eight-term sequence <math>A,B,C,D,E,F,G,H</math>, the value of <math>C</math> is 5 and the sum of any three consecutive terms is 30. What is <math>A+H</math>?<br />
<br />
<math>\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43</math><br />
<br />
[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}</math>. What is the area inside Circle <math>C</math> but outside circle <math>A</math> and circle <math>B</math> ?<br />
<br />
== Problem 19 ==<br />
<br />
In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br />
<br />
<math> \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?<br />
<br />
<math> \textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math> \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Seven students count from 1 to 1000 as follows:<br />
<br />
•Alice says all the numbers, except she ships the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br />
<br />
•Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br />
<br />
•Finally, George says the only number that no one else says.<br />
<br />
What number does George say?<br />
<br />
<math> \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Let <math>R</math> be a square region and <math>n\ge4</math> an integer. A point <math>X</math> in the interior of <math>R</math> is called <math>n\text{-}ray</math> partitional if there are <math>n</math> rays emanating from <math>X</math> that divide <math>R</math> into <math>n</math> triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?<br />
<br />
<math>\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500</math><br />
<br />
[[2011 AMC 10A Problems/Problem 25|Solution]]</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=369792011 AMC 10A Problems2011-02-16T05:40:44Z<p>Flyingpenguin: /* Problem 18 */</p>
<hr />
<div>== Problem 1 ==<br />
<br />
A cell phone plan costs <math>\</math><math>20</math> each month, plus <math>5</math>¢ per text message sent, plus 10¢ for each minute used over <math>30</math> hours. In January Michelle sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did she have to pay?<br />
<br />
<math> \textbf{(A)}\ </math> <math>24.00 \qquad\textbf{(B)}\ </math> <math>24.50 \qquad\textbf{(C)}\ </math> <math>25.50\qquad\textbf{(D)}\ </math> <math>28.00\qquad\textbf{(E)}\ </math> <math>30.00 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
A small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?<br />
<br />
<math> \textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
Suppose [<math>a</math> <math>b</math>] denotes the average of <math>a</math> and <math>b</math>, and {<math>a</math> <math>b</math> <math>c</math>} denotes the average of <math>a</math>, <math>b</math>, and <math>c</math>. What is {{1 1 0} [0 1] 0}?<br />
<br />
<math> \textbf{(A)}\ \frac{2}{9} \qquad\textbf{(B)}\ \frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
Let <math>X</math> and <math>Y</math> be the following sums of arithmetic sequences: <cmath> \begin{eqnarray*} X &=& 10 + 12 + 14 + \cdots + 100, \\ Y &=& 12 + 14 + 16 + \cdots + 102. \end{eqnarray*} </cmath> What is the value of <math>Y - X</math>?<br />
<br />
<math> \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 </math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of <math>12</math>, <math>15</math>, and <math>10</math> minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students? <br />
<br />
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
Set <math>A </math> has 20 elements, and set <math>B </math> has 15 elements. What is the smallest possible number of elements in <math>A \cup B </math>, the union of <math>A </math> and <math>B </math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Which of the following equations does NOT have a solution?<br />
<br />
<math>\text{(A)}\:(x+7)^2=0</math><br />
<br />
<math>\text{(B)}\:|-3x|+5=0</math><br />
<br />
<math>\text{(C)}\:\sqrt{-x}-2=0</math><br />
<br />
<math>\text{(D)}\:\sqrt{x}-8=0</math><br />
<br />
<math>\text{(E)}\:|-3x|-4=0 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?<br />
<br />
<br />
<math> \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A rectangular region is bounded by the graphs of the equations <math>y=a, y=-b, x=-c,</math> and <math>x=d</math>, where <math>a,b,c,</math> and <math>d</math> are all positive numbers. Which of the following represents the area of this region?<br />
<br />
<math> \textbf{(A)}\ ac+ad+bc+bd\qquad\textbf{(B)}\ ac-ad+bc-bd\qquad\textbf{(C)}\ ac+ad-bc-bd \quad\quad\qquad\textbf{(D)}\ -ac-ad+bc+bd\qquad\textbf{(E)}\ ac-ad-bc+bd </math><br />
<br />
[[2011 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was <math>\</math><math>17.71</math>. What was the cost of a pencil in cents?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
[[2011 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Square <math>EFGH</math> has one vertex on each side of square <math>ABCD</math>. Point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=7\cdot EB</math>. What is the ratio of the area of <math>EFGH</math> to the area of <math>ABCD</math>?<br />
<br />
<math>\text{(A)}\,\frac{49}{64} \qquad\text{(B)}\,\frac{25}{32} \qquad\text{(C)}\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?<br />
<br />
<math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200</math><br />
<br />
[[2011 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br />
<br />
<math>\text{(A)}\,\frac{1}{36} \qquad\text{(B)}\,\frac{1}{12} \qquad\text{(C)}\,\frac{1}{6} \qquad\text{(D)}\,\frac{1}{4} \qquad\text{(E)}\,\frac{5}{18}</math><br />
<br />
[[2011 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?<br />
<br />
<math>\text{(A)}\,140 \qquad\text{(B)}\,240 \qquad\text{(C)}\,440 \qquad\text{(D)}\,640 \qquad\text{(E)}\,840</math><br />
<br />
[[2011 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Which of the following is equal to <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>?<br />
<br />
<math>\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6</math><br />
<br />
[[2011 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
In the eight-term sequence <math>A,B,C,D,E,F,G,H</math>, the value of <math>C</math> is 5 and the sum of any three consecutive terms is 30. What is <math>A+H</math>?<br />
<br />
<math>\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43</math><br />
<br />
[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}</math>. What is the area inside Circle <math>C</math> but outside circle <math>A</math> and circle <math>B</math> ?<br />
<br />
== Solution ==<br />
<br />
Draw a rectangle with vertices at the centers of <math>A</math> and <math>B</math> and the intersection of <math>A, C</math> and <math>B, C</math>. Then, we can compute the shades area as the area of half of <math>C</math> plus the area of the rectangle minus the area of the two sectors created by <math>A</math> and <math>B</math>. This is <math>\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{2 \ \mathbf{(C)}}</math>.<br />
<br />
== Problem 19 ==<br />
<br />
In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br />
<br />
<math> \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?<br />
<br />
<math> \textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math> \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Seven students count from 1 to 1000 as follows:<br />
<br />
•Alice says all the numbers, except she ships the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br />
<br />
•Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br />
<br />
•Finally, George says the only number that no one else says.<br />
<br />
What number does George say?<br />
<br />
<math> \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Let <math>R</math> be a square region and <math>n\ge4</math> an integer. A point <math>X</math> in the interior of <math>R</math> is called <math>n\text{-}ray</math> partitional if there are <math>n</math> rays emanating from <math>X</math> that divide <math>R</math> into <math>n</math> triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?<br />
<br />
<math>\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500</math><br />
<br />
[[2011 AMC 10A Problems/Problem 25|Solution]]</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_18&diff=369782011 AMC 10A Problems/Problem 182011-02-16T05:39:21Z<p>Flyingpenguin: Created page with '== Solution == Draw a rectangle with vertices at the centers of <math>A</math> and <math>B</math> and the intersection of <math>A, C</math> and <math>B, C</math>. Then, we can c…'</p>
<hr />
<div>== Solution ==<br />
<br />
Draw a rectangle with vertices at the centers of <math>A</math> and <math>B</math> and the intersection of <math>A, C</math> and <math>B, C</math>. Then, we can compute the shades area as the area of half of <math>C</math> plus the area of the rectangle minus the area of the two sectors created by <math>A</math> and <math>B</math>. This is <math>\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{2 \ \mathbf{(C)}}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=369772011 AMC 10A Problems2011-02-16T05:38:15Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>== Problem 1 ==<br />
<br />
A cell phone plan costs <math>\</math><math>20</math> each month, plus <math>5</math>¢ per text message sent, plus 10¢ for each minute used over <math>30</math> hours. In January Michelle sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did she have to pay?<br />
<br />
<math> \textbf{(A)}\ </math> <math>24.00 \qquad\textbf{(B)}\ </math> <math>24.50 \qquad\textbf{(C)}\ </math> <math>25.50\qquad\textbf{(D)}\ </math> <math>28.00\qquad\textbf{(E)}\ </math> <math>30.00 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
A small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?<br />
<br />
<math> \textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
Suppose [<math>a</math> <math>b</math>] denotes the average of <math>a</math> and <math>b</math>, and {<math>a</math> <math>b</math> <math>c</math>} denotes the average of <math>a</math>, <math>b</math>, and <math>c</math>. What is {{1 1 0} [0 1] 0}?<br />
<br />
<math> \textbf{(A)}\ \frac{2}{9} \qquad\textbf{(B)}\ \frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
Let <math>X</math> and <math>Y</math> be the following sums of arithmetic sequences: <cmath> \begin{eqnarray*} X &=& 10 + 12 + 14 + \cdots + 100, \\ Y &=& 12 + 14 + 16 + \cdots + 102. \end{eqnarray*} </cmath> What is the value of <math>Y - X</math>?<br />
<br />
<math> \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 </math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of <math>12</math>, <math>15</math>, and <math>10</math> minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students? <br />
<br />
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
Set <math>A </math> has 20 elements, and set <math>B </math> has 15 elements. What is the smallest possible number of elements in <math>A \cup B </math>, the union of <math>A </math> and <math>B </math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Which of the following equations does NOT have a solution?<br />
<br />
<math>\text{(A)}\:(x+7)^2=0</math><br />
<br />
<math>\text{(B)}\:|-3x|+5=0</math><br />
<br />
<math>\text{(C)}\:\sqrt{-x}-2=0</math><br />
<br />
<math>\text{(D)}\:\sqrt{x}-8=0</math><br />
<br />
<math>\text{(E)}\:|-3x|-4=0 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?<br />
<br />
<br />
<math> \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A rectangular region is bounded by the graphs of the equations <math>y=a, y=-b, x=-c,</math> and <math>x=d</math>, where <math>a,b,c,</math> and <math>d</math> are all positive numbers. Which of the following represents the area of this region?<br />
<br />
<math> \textbf{(A)}\ ac+ad+bc+bd\qquad\textbf{(B)}\ ac-ad+bc-bd\qquad\textbf{(C)}\ ac+ad-bc-bd \quad\quad\qquad\textbf{(D)}\ -ac-ad+bc+bd\qquad\textbf{(E)}\ ac-ad-bc+bd </math><br />
<br />
[[2011 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was <math>\</math><math>17.71</math>. What was the cost of a pencil in cents?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
[[2011 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Square <math>EFGH</math> has one vertex on each side of square <math>ABCD</math>. Point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=7\cdot EB</math>. What is the ratio of the area of <math>EFGH</math> to the area of <math>ABCD</math>?<br />
<br />
<math>\text{(A)}\,\frac{49}{64} \qquad\text{(B)}\,\frac{25}{32} \qquad\text{(C)}\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?<br />
<br />
<math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200</math><br />
<br />
[[2011 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br />
<br />
<math>\text{(A)}\,\frac{1}{36} \qquad\text{(B)}\,\frac{1}{12} \qquad\text{(C)}\,\frac{1}{6} \qquad\text{(D)}\,\frac{1}{4} \qquad\text{(E)}\,\frac{5}{18}</math><br />
<br />
[[2011 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?<br />
<br />
<math>\text{(A)}\,140 \qquad\text{(B)}\,240 \qquad\text{(C)}\,440 \qquad\text{(D)}\,640 \qquad\text{(E)}\,840</math><br />
<br />
[[2011 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Which of the following is equal to <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>?<br />
<br />
<math>\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6</math><br />
<br />
[[2011 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
In the eight-term sequence <math>A,B,C,D,E,F,G,H</math>, the value of <math>C</math> is 5 and the sum of any three consecutive terms is 30. What is <math>A+H</math>?<br />
<br />
<math>\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43</math><br />
<br />
[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}</math>. What is the area inside Circle <math>C</math> but outside circle <math>A</math> and circle <math>B</math> ?<br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br />
<br />
<math> \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?<br />
<br />
<math> \textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math> \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Seven students count from 1 to 1000 as follows:<br />
<br />
•Alice says all the numbers, except she ships the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br />
<br />
•Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br />
<br />
•Finally, George says the only number that no one else says.<br />
<br />
What number does George say?<br />
<br />
<math> \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Let <math>R</math> be a square region and <math>n\ge4</math> an integer. A point <math>X</math> in the interior of <math>R</math> is called <math>n\text{-}ray</math> partitional if there are <math>n</math> rays emanating from <math>X</math> that divide <math>R</math> into <math>n</math> triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?<br />
<br />
<math>\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500</math><br />
<br />
[[2011 AMC 10A Problems/Problem 25|Solution]]</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=369762011 AMC 10A Problems2011-02-16T05:37:19Z<p>Flyingpenguin: /* Problem 18 */</p>
<hr />
<div>== Problem 1 ==<br />
<br />
A cell phone plan costs <math>\</math><math>20</math> each month, plus <math>5</math>¢ per text message sent, plus 10¢ for each minute used over <math>30</math> hours. In January Michelle sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did she have to pay?<br />
<br />
<math> \textbf{(A)}\ </math> <math>24.00 \qquad\textbf{(B)}\ </math> <math>24.50 \qquad\textbf{(C)}\ </math> <math>25.50\qquad\textbf{(D)}\ </math> <math>28.00\qquad\textbf{(E)}\ </math> <math>30.00 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
A small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?<br />
<br />
<math> \textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
Suppose [<math>a</math> <math>b</math>] denotes the average of <math>a</math> and <math>b</math>, and {<math>a</math> <math>b</math> <math>c</math>} denotes the average of <math>a</math>, <math>b</math>, and <math>c</math>. What is {{1 1 0} [0 1] 0}?<br />
<br />
<math> \textbf{(A)}\ \frac{2}{9} \qquad\textbf{(B)}\ \frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
Let <math>X</math> and <math>Y</math> be the following sums of arithmetic sequences: <cmath> \begin{eqnarray*} X &=& 10 + 12 + 14 + \cdots + 100, \\ Y &=& 12 + 14 + 16 + \cdots + 102. \end{eqnarray*} </cmath> What is the value of <math>Y - X</math>?<br />
<br />
<math> \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 </math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of <math>12</math>, <math>15</math>, and <math>10</math> minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students? <br />
<br />
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
Set <math>A </math> has 20 elements, and set <math>B </math> has 15 elements. What is the smallest possible number of elements in <math>A \cup B </math>, the union of <math>A </math> and <math>B </math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Which of the following equations does NOT have a solution?<br />
<br />
<math>\text{(A)}\:(x+7)^2=0</math><br />
<br />
<math>\text{(B)}\:|-3x|+5=0</math><br />
<br />
<math>\text{(C)}\:\sqrt{-x}-2=0</math><br />
<br />
<math>\text{(D)}\:\sqrt{x}-8=0</math><br />
<br />
<math>\text{(E)}\:|-3x|-4=0 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?<br />
<br />
<br />
<math> \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A rectangular region is bounded by the graphs of the equations <math>y=a, y=-b, x=-c,</math> and <math>x=d</math>, where <math>a,b,c,</math> and <math>d</math> are all positive numbers. Which of the following represents the area of this region?<br />
<br />
<math> \textbf{(A)}\ ac+ad+bc+bd\qquad\textbf{(B)}\ ac-ad+bc-bd\qquad\textbf{(C)}\ ac+ad-bc-bd \quad\quad\qquad\textbf{(D)}\ -ac-ad+bc+bd\qquad\textbf{(E)}\ ac-ad-bc+bd </math><br />
<br />
[[2011 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was <math>\</math><math>17.71</math>. What was the cost of a pencil in cents?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
[[2011 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Square <math>EFGH</math> has one vertex on each side of square <math>ABCD</math>. Point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=7\cdot EB</math>. What is the ratio of the area of <math>EFGH</math> to the area of <math>ABCD</math>?<br />
<br />
<math>\text{(A)}\,\frac{49}{64} \qquad\text{(B)}\,\frac{25}{32} \qquad\text{(C)}\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?<br />
<br />
<math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200</math><br />
<br />
[[2011 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br />
<br />
<math>\text{(A)}\,\frac{1}{36} \qquad\text{(B)}\,\frac{1}{12} \qquad\text{(C)}\,\frac{1}{6} \qquad\text{(D)}\,\frac{1}{4} \qquad\text{(E)}\,\frac{5}{18}</math><br />
<br />
[[2011 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?<br />
<br />
<math>\text{(A)}\,140 \qquad\text{(B)}\,240 \qquad\text{(C)}\,440 \qquad\text{(D)}\,640 \qquad\text{(E)}\,840</math><br />
<br />
[[2011 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Which of the following is equal to <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>?<br />
<br />
<math>\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6</math><br />
<br />
[[2011 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
In the eight-term sequence <math>A,B,C,D,E,F,G,H</math>, the value of <math>C</math> is 5 and the sum of any three consecutive terms is 30. What is <math>A+H</math>?<br />
<br />
<math>\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43</math><br />
<br />
[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}</math>. What is the area inside Circle <math>C</math> but outside circle <math>A</math> and circle <math>B</math> ?<br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Solution ==<br />
<br />
Draw a rectangle with vertices at the centers of <math>A</math> and <math>B</math> and the intersection of <math>A, C</math> and <math>B, C</math>. Then, we can compute the shades area as the area of half of <math>C</math> plus the area of the rectangle minus the area of the two sectors created by <math>A</math> and <math>B</math>. This is <math>\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{2 \ \mathbf{(C)}}</math>.<br />
<br />
== Problem 19 ==<br />
<br />
In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br />
<br />
<math> \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?<br />
<br />
<math> \textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math> \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Seven students count from 1 to 1000 as follows:<br />
<br />
•Alice says all the numbers, except she ships the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br />
<br />
•Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br />
<br />
•Finally, George says the only number that no one else says.<br />
<br />
What number does George say?<br />
<br />
<math> \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Let <math>R</math> be a square region and <math>n\ge4</math> an integer. A point <math>X</math> in the interior of <math>R</math> is called <math>n\text{-}ray</math> partitional if there are <math>n</math> rays emanating from <math>X</math> that divide <math>R</math> into <math>n</math> triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?<br />
<br />
<math>\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500</math><br />
<br />
[[2011 AMC 10A Problems/Problem 25|Solution]]</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_22&diff=369752011 AMC 10A Problems/Problem 222011-02-16T05:30:59Z<p>Flyingpenguin: </p>
<hr />
<div>== Problem 22 ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math> \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 </math><br />
<br />
== Solution ==<br />
<br />
Let vertex <math>A</math> be any vertex, then vertex <math>B</math> be one of the diagonal vertices to <math>A</math>, <math>C</math> be one of the diagonal vertices to <math>B</math>, and so on. We consider cases for this problem.<br />
<br />
In the case that <math>C</math> has the same color as <math>A</math>, <math>D</math> has a different color than <math>A</math> and so <math>E</math> has a different color than <math>A</math> and <math>D</math>. In this case, <math>A</math> has <math>6</math> choices, <math>B</math> has <math>5</math> choices (any color but the color of <math>A</math>), <math>C</math> has <math>1</math> choice, <math>D</math> has <math>5</math> choices, and <math>E</math> has <math>4</math> choices, resulting in a possible of <math>6 \cdot 5 \cdot 1 \cdot 5 \cdot 4 = 600</math> combinations.<br />
<br />
In the case that <math>C</math> has a different color than <math>A</math> and <math>D</math> has a different color than <math>A</math>, <math>A</math> has <math>6</math> choices, <math>B</math> has <math>5</math> choices, <math>C</math> has <math>4</math> choices (since <math>A</math> and <math>B</math> necessarily have different colors), <math>D</math> has <math>4</math> choices, and <math>E</math> has <math>4</math> choices, resulting in a possible of <math>6 \cdot 5 \cdot 4 \cdot 4 \cdot 4 = 1920</math> combinations.<br />
<br />
In the case that <math>C</math> has a different color than <math>A</math> and <math>D</math> has the same color as <math>A</math>, <math>A</math> has <math>6</math> choices, <math>B</math> has <math>5</math> choices, <math>C</math> has <math>4</math> choices, <math>D</math> has <math>1</math> choice, and <math>E</math> has <math>5</math> choices, resulting in a possible of <math>6 \cdot 5 \cdot 4 \cdot 1 \cdot 5 = 600</math> combinations.<br />
<br />
Adding all those combinations up, we get <math>600+1920+600=\boxed{3120 \ \mathbf{(C)}}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_15&diff=369742011 AMC 10A Problems/Problem 152011-02-16T04:39:10Z<p>Flyingpenguin: </p>
<hr />
<div>Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?<br />
<br />
'''(A) 140 (B) 240 (C) 440 (D) 640 (E) 840'''<br />
<br />
== Solution ==<br />
<br />
We know that <math>\frac{\text{total miles}}{\text{total gas}}=55</math>. Let <math>x</math> be the distance the car traveled during the time it ran on gasoline, then the amount of gas used is <math>0.02x</math>. The total distance traveled is <math>40+x</math>, so we get <math>\frac{40+x}{0.02x}=55</math>. Solving this equation, we get <math>x=400</math>, so the total distance is <math>400 + 40 = \boxed{440 \ \mathbf{(C)}}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_12&diff=369732011 AMC 10A Problems/Problem 122011-02-16T04:13:39Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>==Problem 12==<br />
The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
== Solution ==<br />
<br />
Suppose there were <math>x</math> three-point shots, <math>y</math> two-point shots, and <math>z</math> one-point shots. Then we get the following system of equations:<br />
<cmath>\begin{align}<br />
3x=2y\\ z=y+1\\ 3x+2y+z=61<br />
\end{align}</cmath><br />
<br />
The value we are looking for is <math>z</math>, which is easily found to be <math>z=\boxed{13 \ \mathbf{(A)}}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_14&diff=369722011 AMC 10A Problems/Problem 142011-02-16T04:13:02Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>==Problem 14==<br />
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br />
<br />
<math>\text{(A)}\,\frac{1}{36} \qquad\text{(B)}\,\frac{1}{12} \qquad\text{(C)}\,\frac{1}{6} \qquad\text{(D)}\,\frac{1}{4} \qquad\text{(E)}\,\frac{5}{18}</math><br />
<br />
== Solution ==<br />
<br />
We want the area, <math>\pi r^2</math>, to be less than the circumference, <math>2 \pi r</math>:<br />
<br />
<cmath>\begin{align*}<br />
\pi r^2 &< 2 \pi r \\<br />
r &< 2<br />
\end{align*}</cmath><br />
<br />
If <math>r<2</math> then the dice must show <math>(1,1),(1,2),(2,1)</math> which are <math>3</math> choices out of a total possible of <math>6 \times 6 =36</math>, so the probability is <math>\frac{3}{36}=\boxed{\frac{1}{12} \ \mathbf{(B)}}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_14&diff=369712011 AMC 10A Problems/Problem 142011-02-16T04:12:13Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>==Problem 14==<br />
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br />
<br />
<math>\text{(A)}\,\frac{1}{36} \qquad\text{(B)}\,\frac{1}{12} \qquad\text{(C)}\,\frac{1}{6} \qquad\text{(D)}\,\frac{1}{4} \qquad\text{(E)}\,\frac{5}{18}</math><br />
<br />
== Solution ==<br />
<br />
We want the area, <math>\pi r^2</math>, to be less than the circumference, <math>2 \pi r</math>:<br />
<br />
<cmath>\begin{align*}<br />
\pi r^2 &< 2 \pi r \\<br />
r &< 2<br />
\end{align*}</cmath><br />
<br />
If <math>r<2</math> then the dice must show <math>(1,1),(1,2),(2,1)</math> which are <math>3</math> choices out of a total possible of <math>6 \times 6 =36</math>, so the probability is <math>3/36=1/12</math></div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_12&diff=369702011 AMC 10A Problems/Problem 122011-02-16T04:11:35Z<p>Flyingpenguin: </p>
<hr />
<div>==Problem 12==<br />
The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
== Solution ==<br />
<br />
Suppose there were <math>x</math> three-point shots, <math>y</math> two-point shots, and <math>z</math> one-point shots. Then we get the following system of equations:<br />
<cmath>\begin{array}<br />
3x=2y\\ z=y+1\\ 3x+2y+z=61<br />
\end{array}</cmath><br />
<br />
The value we are looking for is <math>z</math>, which is easily found to be <math>z=\boxed{13 \ \mathbf{(A)}}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_11&diff=369692011 AMC 10A Problems/Problem 112011-02-16T04:06:02Z<p>Flyingpenguin: </p>
<hr />
<div>==Problem 11==<br />
Square <math>EFGH</math> has one vertex on each side of square <math>ABCD</math>. Point <math>E</math> is on <math>AB</math> with <math>AE=7\cdot EB</math>. What is the ratio of the area of <math>EFGH</math> to the area of <math>ABCD</math>?<br />
<br />
<math>\text{(A)}\,\frac{49}{64} \qquad\text{(B)}\,\frac{25}{32} \qquad\text{(C)}\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4} </math><br />
<br />
== Solution ==<br />
<br />
Let <math>8s</math> be the length of the sides of square <math>ABCD</math>, then the length of one of the sides of square <math>EFGH</math> is <math>\sqrt{(7s)^2+s^2}=\sqrt{50s^2}</math>, and hence the ratio in the areas is <math>\frac{\sqrt{50s^2}^2}{(8s)^2}=\frac{50}{64} = \boxed{\frac{25}{32} \ \mathbf{(B)}}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_10&diff=369682011 AMC 10A Problems/Problem 102011-02-16T03:59:03Z<p>Flyingpenguin: </p>
<hr />
<div>==Problem 10==<br />
A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was <math>\</math><math>17.71</math>. What was the cost of a pencil in cents?<br />
<br />
<math>\text{(A)}\,7 \qquad\text{(B)}\,11 \qquad\text{(C)}\,17 \qquad\text{(D)}\,23 \qquad\text{(E)}\,77</math><br />
<br />
== Solution ==<br />
<br />
Let <math>x>15</math> be the number of students at the bookstore, <math>p>1</math> be the number of pencils each student bought, and <math>c>p</math> be the price of one pencil. We see that <math>xcp = 1771 = 7 \cdot 11 \cdot 23</math>. Then we must have <math>x=23, c=11, p=7</math> by the original conditions and the answer is <math>c=\boxed{11 \ \mathbf{(B)}}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_4&diff=369672011 AMC 10A Problems/Problem 42011-02-16T03:51:28Z<p>Flyingpenguin: </p>
<hr />
<div>Let X and Y be the following sums of arithmetic sequences: <br />
<br />
<cmath> \begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end[eqnarray*} </cmath><br />
<br />
What is the value of Y - X? <br />
<br />
<math> \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 </math><br />
<br />
== <math>\mathbf{Solution}</math> ==<br />
We see that both sequences have equal numbers of terms, so reformat the sequence to look like: <br />
<br />
<cmath>\begin{align*}<br />
Y = \ &12 + 14 + \cdots + 100 + 102\\<br />
X = 10 \ + \ &12 + 14 + \cdots + 100\\<br />
\end{align*}</cmath><br />
From here it is obvious that <math>Y - X = 102 - 10 = 92</math></div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_21&diff=369662011 AMC 10A Problems/Problem 212011-02-16T03:45:25Z<p>Flyingpenguin: </p>
<hr />
<div>== Problem 21 ==<br />
Two counterfeit coins of equal weight are mixed with <math>8</math> identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the <math>10</math> coins. A second pair is selected at random without replacement from the remaining <math>8</math> coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all <math>4</math> selected coins are genuine?<br />
<br />
<math> \textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16} </math><br />
<br />
== Solution ==<br />
<br />
Note that we are trying to find the conditional probability <math>P(A \vert B) = \frac{P(A \cap B)}{P(B)}</math> where <math>A</math> is the <math>4</math> coins being genuine and <math>B</math> is the sum of the weight of the coins being equal. The only possibilities for <math>B</math> are (g is abbreviated for genuine and c is abbreviated for counterfeit, and the pairs are the first and last two in the quadruple) <math>(g,g,g,g),(g,c,g,c),(g,c,c,g),(c,g,g,c),(c,g,c,g)</math>. We see that <math>A \cap B</math> happens with probability <math>\frac{8}{10} \times \frac{7}{9} \times \frac{6}{8} \times \frac{5}{7} = \frac{1}{3}</math>, and <math>B</math> happens with probability <math>\frac{1}{3}+4 \times \left( \frac{8}{10} \times \frac{2}{9} \times \frac{7}{8}\times \frac{1}{7}\right) = \frac{19}{45}</math>, hence <math>P(A \vert B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{3}}{\frac{19}{45}}=\boxed{\frac{15}{19} \ \mathbf{(D)}}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_17&diff=369562011 AMC 10A Problems/Problem 172011-02-14T22:55:25Z<p>Flyingpenguin: </p>
<hr />
<div>==Problem 17==<br />
In the eight-term sequence <math>A,B,C,D,E,F,G,H</math>, the value of <math>C</math> is 5 and the sum of any three consecutive terms is 30. What is <math>A+H</math>?<br />
<br />
<math>\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43</math><br />
<br />
== Solution ==<br />
<br />
We consider the sum <math>A+B+C+D+E+F+G+H</math> and use the fact that <math>C=5</math>, and hence <math>A+B=25</math>.<br />
<br />
<cmath>\begin{align*}<br />
&A+B+C+D+E+F+G+H=A+(B+C+D)+(E+F+G)+H=A+30+30+H=A+H+60\\<br />
&A+B+C+D+E+F+G+H=(A+B)+(C+D+E)+(F+G+H)=25+30+30=85<br />
\end{align*}</cmath><br />
<br />
Equating the two values we get for the sum, we get the answer <math>A+H+60=85</math> <math>\Longrightarrow</math> <math>A+H=\boxed{25 \ \mathbf{(C)}}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_14&diff=369542011 AMC 10A Problems/Problem 142011-02-14T19:45:45Z<p>Flyingpenguin: </p>
<hr />
<div>==Problem 14==<br />
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br />
<br />
<math>\text{(A)}\,\frac{1}{36} \qquad\text{(B)}\,\frac{1}{12} \qquad\text{(C)}\,\frac{1}{6} \qquad\text{(D)}\,\frac{1}{4} \qquad\text{(E)}\,\frac{5}{18}</math><br />
<br />
== Solution ==<br />
<br />
We want the area, <math>\pi r^2</math>, to be less than the circumference, <math>2 \pi r</math>:<br />
<br />
<cmath>\begin{align*}<br />
\pi r^2 &< 2 \pi r<br />
r &< 2<br />
\end{align*}</cmath><br />
<br />
If <math>r<2</math> then the dice must show <math>(1,1),(1,2),(2,1)</math> which are <math>3</math> choices out of a total possible of <math>6 \times 6 =36</math>, so the probability is <math>3/36=1/12</math></div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_13&diff=369532011 AMC 10A Problems/Problem 132011-02-14T19:10:37Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>==Problem 13==<br />
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?<br />
<br />
<math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200</math><br />
<br />
== Solution ==<br />
<br />
We split up into cases of the hundreds digits being <math>2</math> or <math>5</math>. If the hundred digits is <math>2</math>, then the units digits must be <math>8</math> in order for the number to be even and then there are <math>4</math> remaining choices (<math>1,5,7,9</math>) for the tens digit, giving <math>1 \times 1 \times 4=4</math> possibilities. Similarly, there are <math>1 \times 2 \times 4=8</math> possibilities for the <math>5</math> case, giving a total of <math>\boxed{4+8=12 \ \mathbf{(A)}}</math> possibilities.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_13&diff=369522011 AMC 10A Problems/Problem 132011-02-14T19:08:52Z<p>Flyingpenguin: </p>
<hr />
<div>==Problem 13==<br />
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?<br />
<br />
<math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200</math><br />
<br />
== Solution ==<br />
<br />
We split up into cases of the hundreds digits being <math>2</math> or <math>5</math>. If the hundred digits is <math>2</math>, then the units digits must be <math>8</math> in order for the number to be even and then there are <math>4</math> remaining choices (<math>1,5,7,9</math>) for the tens digit, giving <math>1 \times 1 \times 4=4</math> possibilities. Similarly, there are <math>1 \times 2 \times 4=8</math> possibilities for the <math>5</math> case, giving a total of \boxed{12 \ \mathbf{(A)}} possibilities.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_6&diff=369512011 AMC 10A Problems/Problem 62011-02-14T19:02:27Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>==Problem 6==<br />
Set <math>A</math> has <math>20</math> elements, and set <math>B</math> has <math>15</math> elements. What is the smallest possible number of elements in <math>A \cup B</math>?<br />
<br />
<math> \textbf{(A)}5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 </math><br />
<br />
== Solution ==<br />
<br />
<math>A \cup B</math> will be smallest if <math>B</math> is completely contained in <math>A</math>, in which case all the elements in <math>B</math> would be counted for in <math>A</math>. So the total would be the number of elements in <math>A</math>, which is <math>\boxed{20 \ \mathbf{(C)}}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_6&diff=369502011 AMC 10A Problems/Problem 62011-02-14T19:02:13Z<p>Flyingpenguin: </p>
<hr />
<div>==Problem 6==<br />
Set <math>A</math> has <math>20</math> elements, and set <math>B</math> has <math>15</math> elements. What is the smallest possible number of elements in <math>A \cup B</math>?<br />
<br />
<math> \textbf{(A)}5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 </math><br />
<br />
== Solution ==<br />
<br />
<math>A \cup B</math> will be smallest if <math>B</math> is completely contained in <math>A</math>, in which case all the elements in <math>B</math> would be counted for in <math>A</math>. So the total would be the number of elements in <math>A</math>, which is <math>\boxed{20 \mathbf{(C)}}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_16&diff=369372011 AMC 10A Problems/Problem 162011-02-14T04:13:56Z<p>Flyingpenguin: </p>
<hr />
<div>==Problem 16==<br />
Which of the following is equal to <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>?<br />
<br />
<math>\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6</math><br />
<br />
== Solution ==<br />
<br />
We find the answer by squaring, then square rooting the expression.<br />
<br />
<cmath>\begin{align*}<br />
&\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\\ = \ &\sqrt{\left(\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\right)^2}\\ = \ &\sqrt{9-6\sqrt{2}+2\sqrt{(9-6\sqrt{2})(9+6\sqrt{2})}+9+6\sqrt{2}}\\ = \ &\sqrt{18+2\sqrt{(9-6\sqrt{2})(9+6\sqrt{2})}}\\ = \ &\sqrt{18+2\sqrt{9^2-(6\sqrt{2})^2}}\\ = \ &\sqrt{18+2\sqrt{81-72}}\\ = \ &\sqrt{18+2\sqrt{9}}\\ = \ &\sqrt{18+6}\\= \ &\sqrt{24}\\ = \ &\boxed{2\sqrt{6} \ \mathbf{(B)}}<br />
\end{align*}</cmath></div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_20&diff=368632011 AMC 10A Problems/Problem 202011-02-11T23:55:25Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>== Problem 20 ==<br />
Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} </math><br />
<br />
== Solution ==<br />
<br />
Fix a point <math>A</math> from which we draw a clockwise chord. In order for the clockwise chord from another point <math>B</math> to intersect that of point <math>A</math>, <math>A</math> and <math>B</math> must be no more than <math>r</math> units apart. By drawing the circle, we quickly see that <math>B</math> can be on <math>\frac{120}{360}=\boxed{\frac{1}{3} \ \textbf{(D)}}</math> of the perimeter of the circle.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_20&diff=368622011 AMC 10A Problems/Problem 202011-02-11T23:55:05Z<p>Flyingpenguin: </p>
<hr />
<div>== Problem 20 ==<br />
Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} </math><br />
<br />
== Solution ==<br />
<br />
Fix a point <math>A</math> from which you draw a clockwise chord. In order for the clockwise chord from another point <math>B</math> to intersect that of point <math>A</math>, <math>A</math> and <math>B</math> must be no more than <math>r</math> units apart. By drawing the circle, we quickly see that <math>B</math> can be on <math>\frac{120}{360}=\boxed{\frac{1}{3} \ \textbf{(D)}}</math> of the perimeter of the circle.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=367312011 AMC 10A Problems2011-02-10T03:36:59Z<p>Flyingpenguin: /* Problem 1 */</p>
<hr />
<div>== Problem 1 ==<br />
A cell phone plan costs <math>\</math><math>20</math> each month, plus <math>5</math>¢ per text message sent, plus 10¢ for each minute used over <math>30</math> hours. In January Michelle sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did she have to pay?<br />
<br />
<math> \textbf{(A)}\ </math> <math>24.00 \qquad\textbf{(B)}\ </math> <math>24.50 \qquad\textbf{(C)}\ </math> <math>25.50\qquad\textbf{(D)}\ </math> <math>28.00\qquad\textbf{(E)}\ </math> <math>30.00 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
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[[2011 AMC 10A Problems/Problem 2|Solution]]<br />
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== Problem 3 ==<br />
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[[2011 AMC 10A Problems/Problem 3|Solution]]<br />
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== Problem 4 ==<br />
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[[2011 AMC 10A Problems/Problem 4|Solution]]<br />
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== Problem 5 ==<br />
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[[2011 AMC 10A Problems/Problem 5|Solution]]<br />
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== Problem 6 ==<br />
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[[2011 AMC 10A Problems/Problem 6|Solution]]<br />
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== Problem 7 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 7|Solution]]<br />
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== Problem 8 ==<br />
[[2011 AMC 10A Problems/Problem 8|Solution]]<br />
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== Problem 9 ==<br />
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[[2011 AMC 10A Problems/Problem 9|Solution]]<br />
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== Problem 10 ==<br />
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[[2011 AMC 10A Problems/Problem 10|Solution]]<br />
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== Problem 11 ==<br />
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[[2011 AMC 10A Problems/Problem 11|Solution]]<br />
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== Problem 12 ==<br />
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[[2011 AMC 10A Problems/Problem 12|Solution]]<br />
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== Problem 13 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 13|Solution]]<br />
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== Problem 14 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 25|Solution]]</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=367252011 AMC 10A Problems2011-02-10T03:29:40Z<p>Flyingpenguin: /* Problem 1 */</p>
<hr />
<div>== Problem 1 ==<br />
A cell phone plan costs <math>\</math><math>20</math> each month, plus <math>5</math>¢ per text message sent, plus 10¢ for each minute used over <math>30</math> hours. In January Michelle sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did she have to pay?<br />
<br />
<math> \textbf{(A)}\ </math> <math>24.00\qquad\textbf{(B)}\ </math> <math>24.50\qquad\textbf{(C)}\ </math> <math>25.50\qquad\textbf{(D)}\ </math> <math>28.00\qquad\textbf{(E)}\ </math> <math>30.00 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 4|Solution]]<br />
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== Problem 5 ==<br />
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[[2011 AMC 10A Problems/Problem 5|Solution]]<br />
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== Problem 6 ==<br />
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[[2011 AMC 10A Problems/Problem 6|Solution]]<br />
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== Problem 7 ==<br />
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[[2011 AMC 10A Problems/Problem 7|Solution]]<br />
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== Problem 8 ==<br />
[[2011 AMC 10A Problems/Problem 8|Solution]]<br />
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== Problem 9 ==<br />
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[[2011 AMC 10A Problems/Problem 9|Solution]]<br />
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== Problem 10 ==<br />
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[[2011 AMC 10A Problems/Problem 10|Solution]]<br />
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== Problem 11 ==<br />
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[[2011 AMC 10A Problems/Problem 11|Solution]]<br />
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== Problem 12 ==<br />
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[[2011 AMC 10A Problems/Problem 12|Solution]]<br />
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== Problem 13 ==<br />
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[[2011 AMC 10A Problems/Problem 13|Solution]]<br />
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== Problem 14 ==<br />
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[[2011 AMC 10A Problems/Problem 14|Solution]]<br />
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== Problem 15 ==<br />
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[[2011 AMC 10A Problems/Problem 15|Solution]]<br />
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== Problem 16 ==<br />
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[[2011 AMC 10A Problems/Problem 16|Solution]]<br />
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== Problem 17 ==<br />
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[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
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== Problem 18 ==<br />
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[[2011 AMC 10A Problems/Problem 18|Solution]]<br />
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== Problem 19 ==<br />
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[[2011 AMC 10A Problems/Problem 19|Solution]]<br />
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== Problem 20 ==<br />
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[[2011 AMC 10A Problems/Problem 20|Solution]]<br />
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== Problem 21 ==<br />
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[[2011 AMC 10A Problems/Problem 21|Solution]]<br />
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== Problem 22 ==<br />
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[[2011 AMC 10A Problems/Problem 22|Solution]]<br />
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== Problem 23 ==<br />
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[[2011 AMC 10A Problems/Problem 23|Solution]]<br />
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== Problem 24 ==<br />
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[[2011 AMC 10A Problems/Problem 24|Solution]]<br />
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== Problem 25 ==<br />
<br />
[[2011 AMC 10A Problems/Problem 25|Solution]]</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_24&diff=366192009 AMC 10A Problems/Problem 242011-02-08T04:38:45Z<p>Flyingpenguin: /* Solution 2 */</p>
<hr />
<div>== Problem ==<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 />
== Solution 1 ==<br />
<br />
We will try to use symmetry as much as possible.<br />
<br />
Pick the first vertex <math>A</math>, its choice clearly does not influence anything.<br />
<br />
Pick the second vertex <math>B</math>. With probability <math>3/7</math> vertices <math>A</math> and <math>B</math> have a common edge, with probability <math>3/7</math> they are in opposite corners of the same face, and with probability <math>1/7</math> they are in opposite corners of the cube. We will handle each of the cases separately.<br />
<br />
In the first case, there are <math>2</math> faces that contain the edge <math>AB</math>. In each of these faces there are <math>2</math> other vertices. If one of these <math>4</math> vertices is the third vertex <math>C</math>, the entire triangle <math>ABC</math> will be on a face. On the other hand, if <math>C</math> is one of the two remaining vertices, the triangle will contain points inside the cube. Hence in this case the probability of choosing a good <math>C</math> is <math>2/6 = 1/3</math>.<br />
<br />
In the second case, the triangle <math>ABC</math> will not intersect the cube iff point <math>C</math> is one of the two points on the side that contains <math>AB</math>. Hence the probability of <math>ABC</math> intersecting the inside of the cube is <math>2/3</math>.<br />
<br />
In the third case, already the diagonal <math>AB</math> contains points inside the cube, hence this case will be good regardless of the choice of <math>C</math>.<br />
<br />
Summing up all cases, the resulting probability is:<br />
<cmath><br />
\frac 37\cdot\frac 13 + \frac 37\cdot \frac 23 + \frac 17\cdot 1 = \boxed{\frac 47}<br />
</cmath><br />
<br />
== Solution 2 ==<br />
<br />
There are <math>\binom{8}{3}=56</math> ways to pick three vertices from eight total vertices; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability.<br />
<br />
There are four ways to choose three points from the vertices of a single face. Since there are six faces, <math>4 \times 6 = 24</math>.<br />
<br />
Thus, the probability of what we don't want is <math>\frac{24}{56} = \frac{3}{7}</math>. Using complementary, <br />
<br />
<cmath><br />
1- \frac 37 = \boxed{\frac 47}<br />
</cmath><br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2009|ab=A|num-b=23|num-a=25}}</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_24&diff=366182009 AMC 10A Problems/Problem 242011-02-08T04:36:31Z<p>Flyingpenguin: /* Solution 2 */</p>
<hr />
<div>== Problem ==<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 />
== Solution 1 ==<br />
<br />
We will try to use symmetry as much as possible.<br />
<br />
Pick the first vertex <math>A</math>, its choice clearly does not influence anything.<br />
<br />
Pick the second vertex <math>B</math>. With probability <math>3/7</math> vertices <math>A</math> and <math>B</math> have a common edge, with probability <math>3/7</math> they are in opposite corners of the same face, and with probability <math>1/7</math> they are in opposite corners of the cube. We will handle each of the cases separately.<br />
<br />
In the first case, there are <math>2</math> faces that contain the edge <math>AB</math>. In each of these faces there are <math>2</math> other vertices. If one of these <math>4</math> vertices is the third vertex <math>C</math>, the entire triangle <math>ABC</math> will be on a face. On the other hand, if <math>C</math> is one of the two remaining vertices, the triangle will contain points inside the cube. Hence in this case the probability of choosing a good <math>C</math> is <math>2/6 = 1/3</math>.<br />
<br />
In the second case, the triangle <math>ABC</math> will not intersect the cube iff point <math>C</math> is one of the two points on the side that contains <math>AB</math>. Hence the probability of <math>ABC</math> intersecting the inside of the cube is <math>2/3</math>.<br />
<br />
In the third case, already the diagonal <math>AB</math> contains points inside the cube, hence this case will be good regardless of the choice of <math>C</math>.<br />
<br />
Summing up all cases, the resulting probability is:<br />
<cmath><br />
\frac 37\cdot\frac 13 + \frac 37\cdot \frac 23 + \frac 17\cdot 1 = \boxed{\frac 47}<br />
</cmath><br />
<br />
== Solution 2 ==<br />
<br />
There are <math>\binom{8}{3}=56</math> ways to pick three vertices from <math>8</math>; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability.<br />
<br />
There are four ways to choose three points from the vertices of a single face. Since there are six faces, <math>4 \times 6 = 24</math>.<br />
<br />
Thus, the probability of what we don't want is <math>\frac{24}{56} = \frac{3}{7}</math>. Using complementary, <br />
<br />
<cmath><br />
1- \frac 37 = \boxed{\frac 47}<br />
</cmath><br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2009|ab=A|num-b=23|num-a=25}}</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_8&diff=365372010 AMC 12A Problems/Problem 82011-02-02T02:52:38Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Triangle <math>ABC</math> has <math>AB=2 \cdot AC</math>. Let <math>D</math> and <math>E</math> be on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, such that <math>\angle BAE = \angle ACD</math>. Let <math>F</math> be the intersection of segments <math>AE</math> and <math>CD</math>, and suppose that <math>\triangle CFE</math> is equilateral. What is <math>\angle ACB</math>?<br />
<br />
<math>\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ</math><br />
<br />
== Solution ==<br />
Let <math>\angle BAE = \angle ACD = x</math>.<br />
<br />
<cmath>\begin{align*}\angle BCD &= \angle AEC = 60^\circ\\<br />
\angle EAC + \angle FCA + \angle ECF + \angle AEC &= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\<br />
\angle EAC &= 60^\circ - x\\<br />
\angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}</cmath><br />
<br />
Since <math>\frac{AC}{AB} = \frac{1}{2}</math>, triangle <math>ABC</math> is a <math>30-60-90</math> triangle, so <math>\angle BCA = \boxed{90^\circ\ \textbf{(C)}}</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_25&diff=365092010 AMC 10A Problems/Problem 252011-01-29T23:22:44Z<p>Flyingpenguin: </p>
<hr />
<div>We can find the answer by working backwards. We begin with <math>1-1^2=0</math> on the bottom row, then the <math>1</math> goes to the right of the equal's sign in the row above. We find the smallest value <math>x</math> for which <math>x-1^2=1</math> and <math>x>1^2</math>, which is <math>x=2</math>.<br />
<br />
We repeat the same procedure except with <math>x-1^2=1</math> for the next row and <math>x-1^2=2</math> for the row after that. However, at the fourth row, we see that solving <math>x-1^2=3</math> yields <math>x=4</math>, in which case it would be incorrect since <math>1^2=1</math> is not the greatest perfect square less than or equal to <math>x</math> . So we make it a <math>2^2</math> and solve <math>x-2^2=3</math>. We continue on using this same method where we increase the perfect square until <math>x</math> can be made bigger than it. When we repeat this until we have <math>8</math> rows, we get:<br />
<br />
<cmath> \begin{array}{ccccc}{}&{}&{}&{}&7223\\ 7223&-&84^{2}&=&167\\ 167&-&12^{2}&=&23\\ 23&-&4^{2}&=&7\\ 7&-&2^{2}&=&3\\ 3&-&1^{2}&=&2\\ 2&-&1^{2}&=&1\\ 1&-&1^{2}&=&0\\ \end{array} </cmath><br />
<br />
Hence the solution is the last digit of <math>7223</math>, which is <math>\boxed{\textbf{(B)}\ 3}</math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems/Problem_18&diff=365022002 AMC 10B Problems/Problem 182011-01-29T07:25:30Z<p>Flyingpenguin: /* Problem */</p>
<hr />
<div>== Problem ==<br />
<br />
Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?<br />
<br />
<math>\textbf{(A) } 8\qquad \textbf{(B) } 9\qquad \textbf{(C) } 10\qquad \textbf{(D) } 12\qquad \textbf{(E) } 16</math><br />
<br />
== Solution ==<br />
We know that <math>2</math> distinct circles can intersect at no more than <math>2</math> points. Thus <math>4</math> circles can intersect at <math>2 \times 4= \boxed{\textbf{(D)}\ 8}</math> points total.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems/Problem_18&diff=365012002 AMC 10B Problems/Problem 182011-01-29T07:24:49Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?<br />
<br />
<math>\textbf{(A) } 8\qquad \textbf{(B) } 9\qquad \textbf{(C) } 10\qquad \textbf{(D) } 12\qquad \textbf{(E) } 16\</math><br />
<br />
== Solution ==<br />
We know that <math>2</math> distinct circles can intersect at no more than <math>2</math> points. Thus <math>4</math> circles can intersect at <math>2 \times 4= \boxed{\textbf{(D)}\ 8}</math> points total.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems/Problem_25&diff=365002002 AMC 10B Problems/Problem 252011-01-29T07:18:30Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>== Problem ==<br />
When 15 is appended to a list of integers, the mean is increased by 2. When 1 is appended to the enlarged list, the mean of the enlarged list is decreased by 1. How many integers were in the original list?<br />
<br />
<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math><br />
<br />
== Solution ==<br />
Let <math>x</math> be the sum of the integers and <math>y</math> be the number of elements in the list. Then we get the equations <math>\frac{x+15}{y+1}=\frac{x}{y}+2</math> and <math>\frac{x+15+1}{y+1+1}=\frac{x+16}{y+2}=\frac{x}{y}+2-1=\frac{x}{y}+1</math>. With a little work, the solution is found to be <math>y= \boxed{\textbf{(A)}\ 4} </math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems/Problem_25&diff=364992002 AMC 10B Problems/Problem 252011-01-29T07:18:14Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>== Problem ==<br />
When 15 is appended to a list of integers, the mean is increased by 2. When 1 is appended to the enlarged list, the mean of the enlarged list is decreased by 1. How many integers were in the original list?<br />
<br />
<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math><br />
<br />
== Solution ==<br />
Let <math>x</math> be the sum of the list of integers and <math>y</math> be the number of elements in the list. Then we get the equations <math>\frac{x+15}{y+1}=\frac{x}{y}+2</math> and <math>\frac{x+15+1}{y+1+1}=\frac{x+16}{y+2}=\frac{x}{y}+2-1=\frac{x}{y}+1</math>. With a little work, the solution is found to be <math>y= \boxed{\textbf{(A)}\ 4} </math>.</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_14&diff=363572010 AMC 12B Problems/Problem 142011-01-18T05:43:04Z<p>Flyingpenguin: /* Solution */</p>
<hr />
<div>== Problem 14 ==<br />
Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> be postive integers with <math>a+b+c+d+e=2010</math> and let <math>M</math> be the largest of the sum <math>a+b</math>, <math>b+c</math>, <math>c+d</math> and <math>d+e</math>. What is the smallest possible value of <math>M</math>?<br />
<br />
<math>\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804</math><br />
<br />
== Solution ==<br />
We want to try make <math>a+b</math>, <math>b+c</math>, <math>c+d</math>, and <math>d+e</math> as close as possible so that <math>M</math>, the maximum of these, if smallest.<br />
<br />
Notice that <math>2010=670+670+670</math>. In order to express <math>2010</math> as a sum of <math>5</math> numbers, we must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers as close as possible): <math>2010=670+1+670+1+668</math> or <math>2010=670+1+669+1+669</math>. We see that in both cases, the value of <math>M</math> is <math>671</math>, so the answer is <math>671</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}}</div>Flyingpenguinhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_25&diff=363562010 AMC 10A Problems/Problem 252011-01-18T04:02:24Z<p>Flyingpenguin: Created page with 'We can find the answer by working backwards. We begin with <math>1-1^2=0</math> on the bottom row, then the <math>1</math> goes to the right of the equal's sign in the row above.…'</p>
<hr />
<div>We can find the answer by working backwards. We begin with <math>1-1^2=0</math> on the bottom row, then the <math>1</math> goes to the right of the equal's sign in the row above. We find the smallest value <math>x</math> for which <math>x-1^2=1</math> (and that does not violate the fact that <math>1^2</math> must be the greatest perfect square less than <math>x</math>), which is <math>2</math>.<br />
<br />
We continue with the same way with the next row, but at the fourth row, we see that solving <math>x-1^2=3</math> yields <math>x=4</math>, but in that case we would have to make the <math>1^2</math> a <math>2^2</math> since <math>4 \geq 2^2</math>. So we make it a <math>2^2</math> and solve <math>x-2^2=3</math>. We continue on using this same method where we move the perfect square up by <math>1</math> each time until we get the <math>x</math> to be a value greater than the RHS plus the perfect square number. When we repeat this until we have <math>8</math> rows, we get:<br />
<br />
<cmath> \begin{align*} 7223\\ 7223-84^2 &= 167\\ 167-12^2 &= 23\\ 23-4^2 &= 7\\ 7-2^2 &= 3\\ 3-1^2 &= 2\\ 2-1^2 &= 1\\ 1-1^2 &= 0\end{align*} </cmath><br />
<br />
Hence the solution is the last digit of <math>7223</math>, which is <math>3</math>.</div>Flyingpenguin