https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Andb501&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T14:18:47ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems&diff=747682010 AMC 12B Problems2016-01-23T18:27:19Z<p>Andb501: /* Problem 5 */</p>
<hr />
<div>== Problem 1 ==<br />
Makarla attended two meetings during her <math>9</math>-hour work day. The first meeting took <math>45</math> minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?<br />
<br />
<math>\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35</math><br />
<br />
[[2010 AMC 12B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
A big <math>L</math> is formed as shown. What is its area?<br />
<br />
<center><asy><br />
unitsize(4mm);<br />
defaultpen(linewidth(.8pt));<br />
<br />
draw((0,0)--(5,0)--(5,2)--(2,2)--(2,8)--(0,8)--cycle);<br />
label("8",(0,4),W);<br />
label("5",(5/2,0),S);<br />
label("2",(5,1),E);<br />
label("2",(1,8),N);<br />
</asy></center><br />
<br />
<math>\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 30</math><br />
<br />
[[2010 AMC 12B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
A ticket to a school play cost <math>x</math> dollars, where <math>x</math> is a whole number. A group of 9<sub>th</sub> graders buys tickets costing a total of &#36;<math>48</math>, and a group of 10<sub>th</sub> graders buys tickets costing a total of &#36;<math>64</math>. How many values for <math>x</math> are possible?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math><br />
<br />
[[2010 AMC 12B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A month with <math>31</math> days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?<br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6</math><br />
<br />
[[2010 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Lucky Larry's teacher asked him to substitute numbers for <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> in the expression <math>a-(b-(c-(d+e)))</math> and evaluate the result. Larry ignored the parentheses but added and subtracted correctly and obtained the correct result by coincidence. The numbers Larry substituted for <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> were <math>1</math>, <math>2</math>, <math>3</math>, and <math>4</math>, respectively. What number did Larry substitute for <math>e</math>?<br />
<br />
<math>\textbf{(A)}\ -5 \qquad \textbf{(B)}\ -3 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 5</math><br />
<br />
[[2010 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
At the beginning of the school year, <math>50\%</math> of all students in Mr. Wells' math class answered "Yes" to the question "Do you love math", and <math>50\%</math> answered "No." At the end of the school year, <math>70\%</math> answered "Yes" and <math>30\%</math> answerws "No." Altogether, <math>x\%</math> of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80</math><br />
<br />
[[2010 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
<br />
== Problem 7 ==<br />
Shelby drives her scooter at a speed of <math>30</math> miles per hour if it is not raining, and <math>20</math> miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of <math>16</math> miles in <math>40</math> minutes. How many minutes did she drive in the rain?<br />
<br />
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30</math><br />
<br />
[[2010 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
<br />
<br />
== Problem 8 ==<br />
Every high school in the city of Euclid sent a team of <math>3</math> students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed <math>37</math><sup>th</sup> and <math>64</math><sup>th</sup>, respectively. How many schools are in the city?<br />
<br />
<math>\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26</math><br />
<br />
[[2010 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Let <math>n</math> be the smallest positive integer such that <math>n</math> is divisible by <math>20</math>, <math>n^2</math> is a perfect cube, and <math>n^3</math> is a perfect square. What is the number of digits of <math>n</math>?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7</math><br />
<br />
[[2010 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
The average of the numbers <math>1, 2, 3,\cdots, 98, 99,</math> and <math>x</math> is <math>100x</math>. What is <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{49}{101} \qquad \textbf{(B)}\ \dfrac{50}{101} \qquad \textbf{(C)}\ \dfrac{1}{2} \qquad \textbf{(D)}\ \dfrac{51}{101} \qquad \textbf{(E)}\ \dfrac{50}{99}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
<br />
== Problem 11 ==<br />
A palindrome between <math>1000</math> and <math>10,000</math> is chosen at random. What is the probability that it is divisible by <math>7</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
For what value of <math>x</math> does<br />
<br />
<cmath>\log_{\sqrt{2}}\sqrt{x}+\log_{2}{x}+\log_{4}{x^2}+\log_{8}{x^3}+\log_{16}{x^4}=40?</cmath><br />
<br />
<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 256 \qquad \textbf{(E)}\ 1024</math><br />
<br />
[[2010 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
In <math>\triangle ABC</math>, <math>\cos(2A-B)+\sin(A+B)=2</math> and <math>AB=4</math>. What is <math>BC</math>?<br />
<br />
<math>\textbf{(A)}\ \sqrt{2} \qquad \textbf{(B)}\ \sqrt{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 2\sqrt{2} \qquad \textbf{(E)}\ 2\sqrt{3}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> be positive 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 />
[[2010 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
For how many ordered triples <math>(x,y,z)</math> of nonnegative integers less than <math>20</math> are there exactly two distinct elements in the set <math>\{i^x, (1+i)^y, z\}</math>, where <math>i=\sqrt{-1}</math>?<br />
<br />
<math>\textbf{(A)}\ 149 \qquad \textbf{(B)}\ 205 \qquad \textbf{(C)}\ 215 \qquad \textbf{(D)}\ 225 \qquad \textbf{(E)}\ 235</math><br />
<br />
[[2010 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Positive integers <math>a</math>, <math>b</math>, and <math>c</math> are randomly and independently selected with replacement from the set <math>\{1, 2, 3,\dots, 2010\}</math>. What is the probability that <math>abc + ab + a</math> is divisible by <math>3</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
The entries in a <math>3 \times 3</math> array include all the digits from <math>1</math> through <math>9</math>, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?<br />
<br />
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60</math><br />
<br />
[[2010 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
A frog makes <math>3</math> jumps, each exactly <math>1</math> meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than <math>1</math> meter from its starting position?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than <math>100</math> points. What was the total number of points scored by the two teams in the first half?<br />
<br />
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math><br />
<br />
[[2010 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
A geometric sequence <math>(a_n)</math> has <math>a_1=\sin x</math>, <math>a_2=\cos x</math>, and <math>a_3= \tan x</math> for some real number <math>x</math>. For what value of <math>n</math> does <math>a_n=1+\cos x</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8</math><br />
<br />
[[2010 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Let <math>a > 0</math>, and let <math>P(x)</math> be a polynomial with integer coefficients such that<br />
<br />
<center><br />
<math>P(1) = P(3) = P(5) = P(7) = a</math>, and<br/><br />
<math>P(2) = P(4) = P(6) = P(8) = -a</math>.<br />
</center><br />
<br />
What is the smallest possible value of <math>a</math>?<br />
<br />
<math>\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!</math><br />
<br />
[[2010 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Let <math>ABCD</math> be a cyclic quadrilateral. The side lengths of <math>ABCD</math> are distinct integers less than <math>15</math> such that <math>BC\cdot CD=AB\cdot DA</math>. What is the largest possible value of <math>BD</math>?<br />
<br />
<math>\textbf{(A)}\ \sqrt{\dfrac{325}{2}} \qquad \textbf{(B)}\ \sqrt{185} \qquad \textbf{(C)}\ \sqrt{\dfrac{389}{2}} \qquad \textbf{(D)}\ \sqrt{\dfrac{425}{2}} \qquad \textbf{(E)}\ \sqrt{\dfrac{533}{2}}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Monic quadratic polynomials <math>P(x)</math> and <math>Q(x)</math> have the property that <math>P(Q(x))</math> has zeros at <math>x=-23, -21, -17,</math> and <math>-15</math>, and <math>Q(P(x))</math> has zeros at <math>x=-59,-57,-51</math> and <math>-49</math>. What is the sum of the minimum values of <math>P(x)</math> and <math>Q(x)</math>? <br />
<br />
<math>\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0</math><br />
<br />
[[2010 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
The set of real numbers <math>x</math> for which <br />
<br />
<cmath>\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge1</cmath><br />
<br />
is the union of intervals of the form <math>a<x\le b</math>. What is the sum of the lengths of these intervals?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1003}{335} \qquad \textbf{(B)}\ \dfrac{1004}{335} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \dfrac{403}{134} \qquad \textbf{(E)}\ \dfrac{202}{67}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
For every integer <math>n\ge2</math>, let <math>\text{pow}(n)</math> be the largest power of the largest prime that divides <math>n</math>. For example <math>\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2</math>. What is the largest integer <math>m</math> such that <math>2010^m</math> divides<br />
<br />
<center><br />
<math>\prod_{n=2}^{5300}\text{pow}(n)</math>?<br />
</center><br />
<br />
<br />
<math>\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \qquad \textbf{(E)}\ 78</math><br />
<br />
[[2010 AMC 12B Problems/Problem 25|Solution]]<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems&diff=747672010 AMC 12B Problems2016-01-23T18:26:20Z<p>Andb501: /* Problem 4 */</p>
<hr />
<div>== Problem 1 ==<br />
Makarla attended two meetings during her <math>9</math>-hour work day. The first meeting took <math>45</math> minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?<br />
<br />
<math>\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35</math><br />
<br />
[[2010 AMC 12B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
A big <math>L</math> is formed as shown. What is its area?<br />
<br />
<center><asy><br />
unitsize(4mm);<br />
defaultpen(linewidth(.8pt));<br />
<br />
draw((0,0)--(5,0)--(5,2)--(2,2)--(2,8)--(0,8)--cycle);<br />
label("8",(0,4),W);<br />
label("5",(5/2,0),S);<br />
label("2",(5,1),E);<br />
label("2",(1,8),N);<br />
</asy></center><br />
<br />
<math>\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 30</math><br />
<br />
[[2010 AMC 12B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
A ticket to a school play cost <math>x</math> dollars, where <math>x</math> is a whole number. A group of 9<sub>th</sub> graders buys tickets costing a total of &#36;<math>48</math>, and a group of 10<sub>th</sub> graders buys tickets costing a total of &#36;<math>64</math>. How many values for <math>x</math> are possible?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math><br />
<br />
[[2010 AMC 12B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A month with <math>31</math> days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?<br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6</math><br />
<br />
[[2010 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Lucky Larry's teacher asked him to substitute numbers for <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> in the expression <math>a-(b-(c-(d+e)))</math> and evaluate the result. Larry ignored the parenthese but added and subtracted correctly and obtained the correct result by coincidence. The numbers Larry substituted for <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> were <math>1</math>, <math>2</math>, <math>3</math>, and <math>4</math>, respectively. What number did Larry substitute for <math>e</math>?<br />
<br />
<math>\textbf{(A)}\ -5 \qquad \textbf{(B)}\ -3 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 5</math><br />
<br />
[[2010 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
At the beginning of the school year, <math>50\%</math> of all students in Mr. Wells' math class answered "Yes" to the question "Do you love math", and <math>50\%</math> answered "No." At the end of the school year, <math>70\%</math> answered "Yes" and <math>30\%</math> answerws "No." Altogether, <math>x\%</math> of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80</math><br />
<br />
[[2010 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
<br />
== Problem 7 ==<br />
Shelby drives her scooter at a speed of <math>30</math> miles per hour if it is not raining, and <math>20</math> miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of <math>16</math> miles in <math>40</math> minutes. How many minutes did she drive in the rain?<br />
<br />
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30</math><br />
<br />
[[2010 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
<br />
<br />
== Problem 8 ==<br />
Every high school in the city of Euclid sent a team of <math>3</math> students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed <math>37</math><sup>th</sup> and <math>64</math><sup>th</sup>, respectively. How many schools are in the city?<br />
<br />
<math>\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26</math><br />
<br />
[[2010 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Let <math>n</math> be the smallest positive integer such that <math>n</math> is divisible by <math>20</math>, <math>n^2</math> is a perfect cube, and <math>n^3</math> is a perfect square. What is the number of digits of <math>n</math>?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7</math><br />
<br />
[[2010 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
The average of the numbers <math>1, 2, 3,\cdots, 98, 99,</math> and <math>x</math> is <math>100x</math>. What is <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{49}{101} \qquad \textbf{(B)}\ \dfrac{50}{101} \qquad \textbf{(C)}\ \dfrac{1}{2} \qquad \textbf{(D)}\ \dfrac{51}{101} \qquad \textbf{(E)}\ \dfrac{50}{99}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
<br />
== Problem 11 ==<br />
A palindrome between <math>1000</math> and <math>10,000</math> is chosen at random. What is the probability that it is divisible by <math>7</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
For what value of <math>x</math> does<br />
<br />
<cmath>\log_{\sqrt{2}}\sqrt{x}+\log_{2}{x}+\log_{4}{x^2}+\log_{8}{x^3}+\log_{16}{x^4}=40?</cmath><br />
<br />
<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 256 \qquad \textbf{(E)}\ 1024</math><br />
<br />
[[2010 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
In <math>\triangle ABC</math>, <math>\cos(2A-B)+\sin(A+B)=2</math> and <math>AB=4</math>. What is <math>BC</math>?<br />
<br />
<math>\textbf{(A)}\ \sqrt{2} \qquad \textbf{(B)}\ \sqrt{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 2\sqrt{2} \qquad \textbf{(E)}\ 2\sqrt{3}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> be positive 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 />
[[2010 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
For how many ordered triples <math>(x,y,z)</math> of nonnegative integers less than <math>20</math> are there exactly two distinct elements in the set <math>\{i^x, (1+i)^y, z\}</math>, where <math>i=\sqrt{-1}</math>?<br />
<br />
<math>\textbf{(A)}\ 149 \qquad \textbf{(B)}\ 205 \qquad \textbf{(C)}\ 215 \qquad \textbf{(D)}\ 225 \qquad \textbf{(E)}\ 235</math><br />
<br />
[[2010 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Positive integers <math>a</math>, <math>b</math>, and <math>c</math> are randomly and independently selected with replacement from the set <math>\{1, 2, 3,\dots, 2010\}</math>. What is the probability that <math>abc + ab + a</math> is divisible by <math>3</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
The entries in a <math>3 \times 3</math> array include all the digits from <math>1</math> through <math>9</math>, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?<br />
<br />
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60</math><br />
<br />
[[2010 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
A frog makes <math>3</math> jumps, each exactly <math>1</math> meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than <math>1</math> meter from its starting position?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than <math>100</math> points. What was the total number of points scored by the two teams in the first half?<br />
<br />
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math><br />
<br />
[[2010 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
A geometric sequence <math>(a_n)</math> has <math>a_1=\sin x</math>, <math>a_2=\cos x</math>, and <math>a_3= \tan x</math> for some real number <math>x</math>. For what value of <math>n</math> does <math>a_n=1+\cos x</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8</math><br />
<br />
[[2010 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Let <math>a > 0</math>, and let <math>P(x)</math> be a polynomial with integer coefficients such that<br />
<br />
<center><br />
<math>P(1) = P(3) = P(5) = P(7) = a</math>, and<br/><br />
<math>P(2) = P(4) = P(6) = P(8) = -a</math>.<br />
</center><br />
<br />
What is the smallest possible value of <math>a</math>?<br />
<br />
<math>\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!</math><br />
<br />
[[2010 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Let <math>ABCD</math> be a cyclic quadrilateral. The side lengths of <math>ABCD</math> are distinct integers less than <math>15</math> such that <math>BC\cdot CD=AB\cdot DA</math>. What is the largest possible value of <math>BD</math>?<br />
<br />
<math>\textbf{(A)}\ \sqrt{\dfrac{325}{2}} \qquad \textbf{(B)}\ \sqrt{185} \qquad \textbf{(C)}\ \sqrt{\dfrac{389}{2}} \qquad \textbf{(D)}\ \sqrt{\dfrac{425}{2}} \qquad \textbf{(E)}\ \sqrt{\dfrac{533}{2}}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Monic quadratic polynomials <math>P(x)</math> and <math>Q(x)</math> have the property that <math>P(Q(x))</math> has zeros at <math>x=-23, -21, -17,</math> and <math>-15</math>, and <math>Q(P(x))</math> has zeros at <math>x=-59,-57,-51</math> and <math>-49</math>. What is the sum of the minimum values of <math>P(x)</math> and <math>Q(x)</math>? <br />
<br />
<math>\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0</math><br />
<br />
[[2010 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
The set of real numbers <math>x</math> for which <br />
<br />
<cmath>\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge1</cmath><br />
<br />
is the union of intervals of the form <math>a<x\le b</math>. What is the sum of the lengths of these intervals?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1003}{335} \qquad \textbf{(B)}\ \dfrac{1004}{335} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \dfrac{403}{134} \qquad \textbf{(E)}\ \dfrac{202}{67}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
For every integer <math>n\ge2</math>, let <math>\text{pow}(n)</math> be the largest power of the largest prime that divides <math>n</math>. For example <math>\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2</math>. What is the largest integer <math>m</math> such that <math>2010^m</math> divides<br />
<br />
<center><br />
<math>\prod_{n=2}^{5300}\text{pow}(n)</math>?<br />
</center><br />
<br />
<br />
<math>\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \qquad \textbf{(E)}\ 78</math><br />
<br />
[[2010 AMC 12B Problems/Problem 25|Solution]]<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_20&diff=747032005 AMC 10B Problems/Problem 202016-01-19T02:27:27Z<p>Andb501: /* Solution */</p>
<hr />
<div>== Problem ==<br />
What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, 8 and exactly once?<br />
<br />
<br />
<math> \mathrm{(A)}48000\qquad\mathrm{(B)}49999.5\qquad\mathrm{(C)}53332.8\qquad\mathrm{(D)}55555\qquad\mathrm{(E)}56432.8 </math><br />
<br />
== Solution 1 ==<br />
We first look at how many times each number will appear in each slot. If we fix a number in a slot, then there are <math> 4! = 24 </math> ways to arrange the other numbers, so each number appears in each spot <math> 24 </math> times. Therefore, the sum of all such numbers is <math> 24 \times (1+3+5+7+8) \times (11111) = 24 \times 24 \times 11111 = 6399936 </math> Since there are <math> 5! = 120 </math> such numbers, we divide <math> 6399936 \div 120 </math> to get <math> \boxed{\mathrm{(C)}53332.8} </math><br />
<br />
==Solution 2==<br />
We can first solve for the mean for the digits 1, 3, 5, 7, and 9 since each is 2 away from each other. The mean of the numbers than can be solved using these digits is <math>55555</math>. The total amount of numbers that can be formed using these digits is <math>4! =120</math>. The sum of these numbers is <math>55555(120) = 6666600</math>. Now we can find out the total value that was gained by replacing the 8 with a 9. We can start how be calculating the gain when the 8 was in the ones digit. Since there are <math>4! = 24</math> numbers with the 8 in the ones digit and 1 was gain from each of them, 24 is the number gained. Then, we repeat this with the tens, hundreds, thousands, and ten thousands place, leading to a total of <math>24+240+2400+24000+240000=266664</math> as the total amount that was gained. Subtract this amount from the sum of the digits using the 9 instead of the 8 to get <math>6666600-266664=6399936</math>. Finally, we divide this by 120 to get the average. <math>6399936/120= \boxed{\mathrm{(C)}53332.8} </math><br />
<br />
== See Also ==<br />
{{AMC10 box|year=2005|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_9&diff=745592008 AMC 10B Problems/Problem 92016-01-15T01:36:01Z<p>Andb501: /* Another Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A quadratic equation <math>ax^2 - 2ax + b = 0</math> has two real solutions. What is the average of these two solutions?<br />
<br />
<math>\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ \frac ba\qquad\mathrm{(D)}\ \frac{2b}a\qquad\mathrm{(E)}\ \sqrt{2b-a}</math><br />
<br />
==Solution 1==<br />
<br />
Dividing both sides by <math>a</math>, we get <math>x^2 - 2x + b/a = 0</math>. By Vieta's formulas, the sum of the roots is <math>2</math>, therefore their average is <math>1\Rightarrow \boxed{A}</math>.<br />
<br />
==Solution 2==<br />
<br />
We know that for an equation <math>ax^2 + bx + c = 0</math>, the sum of the roots is <math>-b/a</math>. This means that the sum of the roots for <math>ax^2 - 2ax + b = 0</math> is <math>2a/a</math>, or 2. The average is the sum of the two roots divided by two, so the average is <math>2/2 = 1</math>.<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=B|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_9&diff=745582008 AMC 10B Problems/Problem 92016-01-15T01:35:48Z<p>Andb501: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A quadratic equation <math>ax^2 - 2ax + b = 0</math> has two real solutions. What is the average of these two solutions?<br />
<br />
<math>\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ \frac ba\qquad\mathrm{(D)}\ \frac{2b}a\qquad\mathrm{(E)}\ \sqrt{2b-a}</math><br />
<br />
==Solution 1==<br />
<br />
Dividing both sides by <math>a</math>, we get <math>x^2 - 2x + b/a = 0</math>. By Vieta's formulas, the sum of the roots is <math>2</math>, therefore their average is <math>1\Rightarrow \boxed{A}</math>.<br />
<br />
== Another Solution==<br />
<br />
We know that for an equation <math>ax^2 + bx + c = 0</math>, the sum of the roots is <math>-b/a</math>. This means that the sum of the roots for <math>ax^2 - 2ax + b = 0</math> is <math>2a/a</math>, or 2. The average is the sum of the two roots divided by two, so the average is <math>2/2 = 1</math>.<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=B|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_1&diff=745572008 AMC 10B Problems/Problem 12016-01-15T00:44:12Z<p>Andb501: /* Solution */</p>
<hr />
<div>==Problem==<br />
A basketball player made 5 baskets during a game. Each basket was worth either 2 or 3 points. How many different numbers could represent the total points scored by the player?<br />
<br />
<math>\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6</math><br />
<br />
==Solution==<br />
The number of points could have been 10, 11, 12, 13, 14, or 15. Thus, the answer is <math>\boxed{\mathrm{(E)}\ 6}</math>.<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=B|before=First Question|num-a=2}}<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_7&diff=745562008 AMC 10B Problems/Problem 72016-01-15T00:39:17Z<p>Andb501: /* Solution */</p>
<hr />
<div>==Problem==<br />
An equilateral triangle of side length <math>10</math> is completely filled in by non-overlapping equilateral triangles of side length <math>1</math>. How many small triangles are required?<br />
<br />
<math>\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000</math><br />
<br />
==Solution 1==<br />
'''(C)''' The area of the large triangle is <math>\frac{10^2\sqrt3}{4}</math>, while the area each small triangle is <math>\frac{1^2\sqrt3}{4}</math>. Dividing these two quantities, we get 100, therefore <math>\boxed{100}</math> small triangles can fit in the large one.<br />
<br />
<br />
==Solution 2==<br />
<asy><br />
unitsize(0.5cm);<br />
defaultpen(0.8);<br />
for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); }<br />
for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); }<br />
for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); }<br />
</asy><br />
<br />
The number of triangles is <math>1+3+\dots+19 = \boxed{100}</math>.<br />
<br />
Also, another way to do it is to notice that as you go row by row (from the bottom), the number of triangles decrease by 2 from 19, so we have:<br />
<math>19+17+15...+3+1 = \frac{19+1}{2}\cdot 10 = \boxed{100}</math><br />
<br />
A fourth solution is to notice that the small triangles are similar to the large triangle as they are both equilateral. Therefore, the ratio of their areas is the square of the ratios of their side lengths. Hence the ratio of their areas is <math>(1/10)^2=1/100</math>, so the answer is <math>\boxed{100}</math>.<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=B|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_20&diff=742842003 AMC 10B Problems/Problem 202016-01-03T17:22:37Z<p>Andb501: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #14]] and [[2003 AMC 10B Problems|2003 AMC 10B #20]]}}<br />
<br />
==Problem==<br />
In rectangle <math>ABCD, AB=5</math> and <math>BC=3</math>. Points <math>F</math> and <math>G</math> are on <math>\overline{CD}</math> so that <math>DF=1</math> and <math>GC=2</math>. Lines <math>AF</math> and <math>BG</math> intersect at <math>E</math>. Find the area of <math>\triangle AEB</math>.<br />
<center><asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair A=(0,0), B=(5,0), C=(5,3), D=(0,3);<br />
pair F=(1,3), G=(3,3);<br />
pair E=(5/3,5);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(A--E);<br />
draw(B--E);<br />
<br />
pair[] ps={A,B,C,D,E,F,G}; dot(ps);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
label("$E$",E,N);<br />
label("$F$",F,SE);<br />
label("$G$",G,SW);<br />
label("$1$",midpoint(D--F),N);<br />
label("$2$",midpoint(G--C),N);<br />
label("$5$",midpoint(A--B),S);<br />
label("$3$",midpoint(A--D),W);<br />
</asy></center><br />
<math>\textbf{(A) } 10 \qquad\textbf{(B) } \frac{21}{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } \frac{25}{2} \qquad\textbf{(E) } 15</math><br />
<br />
==Solution 1==<br />
<br />
<math>\triangle EFG \sim \triangle EAB</math> because <math>FG \parallel AB.</math> The ratio of <math>\triangle EFG</math> to <math>\triangle EAB</math> is <math>2:5</math> since <math>AB=5</math> and <math>FG=2</math> from subtraction. If we let <math>h</math> be the height of <math>\triangle EAB,</math><br />
<br />
<cmath>\frac{2}{5} = \frac{h-3}{h}</cmath><br />
<cmath>2h = 5h-15</cmath><br />
<cmath>3h = 15</cmath><br />
<cmath>h = 5</cmath><br />
<br />
The height is <math>5</math> so the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math>.<br />
<br />
==Solution 2==<br />
<br />
We can look at this diagram as if it were a coordinate plane with point <math>A</math> being <math>(0,0)</math>. This means that the equation of the line <math>AE</math> is <math>y=3x</math> and the equation of the line <math>EB</math> is <math>y=\frac{-3}{2}x+\frac{15}{2}</math>. From this we can set of the follow equation to find the <math>x</math> coordinate of point <math>E</math>:<br />
<br />
<cmath>3x=\frac{-3}{2}x+\frac{15}{2}</cmath><br />
<cmath>6x=-3x+15</cmath><br />
<cmath>9x=15</cmath><br />
<cmath>x=\frac{5}{3}</cmath><br />
<br />
We can plug this into one of our original equations to find that the <math>y</math> coordinate is <math>5</math>, meaning the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math><br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2003|ab=B|num-b=13|num-a=15}}<br />
{{AMC10 box|year=2003|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_20&diff=742832003 AMC 10B Problems/Problem 202016-01-03T17:22:14Z<p>Andb501: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #14]] and [[2003 AMC 10B Problems|2003 AMC 10B #20]]}}<br />
<br />
==Problem==<br />
In rectangle <math>ABCD, AB=5</math> and <math>BC=3</math>. Points <math>F</math> and <math>G</math> are on <math>\overline{CD}</math> so that <math>DF=1</math> and <math>GC=2</math>. Lines <math>AF</math> and <math>BG</math> intersect at <math>E</math>. Find the area of <math>\triangle AEB</math>.<br />
<center><asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair A=(0,0), B=(5,0), C=(5,3), D=(0,3);<br />
pair F=(1,3), G=(3,3);<br />
pair E=(5/3,5);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(A--E);<br />
draw(B--E);<br />
<br />
pair[] ps={A,B,C,D,E,F,G}; dot(ps);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
label("$E$",E,N);<br />
label("$F$",F,SE);<br />
label("$G$",G,SW);<br />
label("$1$",midpoint(D--F),N);<br />
label("$2$",midpoint(G--C),N);<br />
label("$5$",midpoint(A--B),S);<br />
label("$3$",midpoint(A--D),W);<br />
</asy></center><br />
<math>\textbf{(A) } 10 \qquad\textbf{(B) } \frac{21}{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } \frac{25}{2} \qquad\textbf{(E) } 15</math><br />
<br />
==Solution 1==<br />
<br />
<math>\triangle EFG \sim \triangle EAB</math> because <math>FG \parallel AB.</math> The ratio of <math>\triangle EFG</math> to <math>\triangle EAB</math> is <math>2:5</math> since <math>AB=5</math> and <math>FG=2</math> from subtraction. If we let <math>h</math> be the height of <math>\triangle EAB,</math><br />
<br />
<cmath>\frac{2}{5} = \frac{h-3}{h}</cmath><br />
<cmath>2h = 5h-15</cmath><br />
<cmath>3h = 15</cmath><br />
<cmath>h = 5</cmath><br />
<br />
The height is <math>5</math> so the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math>.<br />
<br />
==Solution 2==<br />
<br />
We can look at this diagram as if it were a coordinate plane with point <math>A</math> being <math>(0,0)</math>. This means that the equation of the line <math>AE</math> is <math>y=3x</math> and the equation of the line <math>EB</math> is <math>y=\frac{-3}{2}x+\frac{15}{2}</math>. From this we can set of the follow equation to find the <math>x</math> coordinate of point <math>E</math>:<br />
<br />
<cmath>3x=\frac{-3}{2}x+\frac{15}{2}</cmath><br />
<cmath>6x=-3x+15</cmath><br />
<cmath>9x=15</cmath><br />
<cmath>x=\frac{5}{3}</cmath><br />
<br />
We can plug this into one of our original equations to find that the <math>y</math> coordinate is <math>5</math>, meaning the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2003|ab=B|num-b=13|num-a=15}}<br />
{{AMC10 box|year=2003|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_20&diff=742822003 AMC 10B Problems/Problem 202016-01-03T17:21:18Z<p>Andb501: /* Solution */</p>
<hr />
<div>{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #14]] and [[2003 AMC 10B Problems|2003 AMC 10B #20]]}}<br />
<br />
==Problem==<br />
In rectangle <math>ABCD, AB=5</math> and <math>BC=3</math>. Points <math>F</math> and <math>G</math> are on <math>\overline{CD}</math> so that <math>DF=1</math> and <math>GC=2</math>. Lines <math>AF</math> and <math>BG</math> intersect at <math>E</math>. Find the area of <math>\triangle AEB</math>.<br />
<center><asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair A=(0,0), B=(5,0), C=(5,3), D=(0,3);<br />
pair F=(1,3), G=(3,3);<br />
pair E=(5/3,5);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(A--E);<br />
draw(B--E);<br />
<br />
pair[] ps={A,B,C,D,E,F,G}; dot(ps);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
label("$E$",E,N);<br />
label("$F$",F,SE);<br />
label("$G$",G,SW);<br />
label("$1$",midpoint(D--F),N);<br />
label("$2$",midpoint(G--C),N);<br />
label("$5$",midpoint(A--B),S);<br />
label("$3$",midpoint(A--D),W);<br />
</asy></center><br />
<math>\textbf{(A) } 10 \qquad\textbf{(B) } \frac{21}{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } \frac{25}{2} \qquad\textbf{(E) } 15</math><br />
<br />
==Solution 1==<br />
<br />
<math>\triangle EFG \sim \triangle EAB</math> because <math>FG \parallel AB.</math> The ratio of <math>\triangle EFG</math> to <math>\triangle EAB</math> is <math>2:5</math> since <math>AB=5</math> and <math>FG=2</math> from subtraction. If we let <math>h</math> be the height of <math>\triangle EAB,</math><br />
<br />
<cmath>\frac{2}{5} = \frac{h-3}{h}</cmath><br />
<cmath>2h = 5h-15</cmath><br />
<cmath>3h = 15</cmath><br />
<cmath>h = 5</cmath><br />
<br />
The height is <math>5</math> so the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math>.<br />
<br />
==Solution 2==<br />
<br />
We can look at this diagram as if it were a coordinate plane with point <math>A</math> being <math>(0,0)</math>. This means that the equation of the line <math>AE</math> is <math>y=3x</math> and the equation of the line <math>EB</math> is <math>y=\frac{-3}{2}x+\frac{15}{2}</math>. From this we can set of the follow equation to find the <math>x</math> coordinate of point <math>e</math>:<br />
<br />
<cmath>3x=\frac{-3}{2}x+\frac{15}{2}</cmath><br />
<cmath>6x=-3x+15</cmath><br />
<cmath>9x=15</cmath><br />
<cmath>x=\frac{5}{3}</cmath><br />
<br />
We can plug this into one of our original equations to find that the <math>y</math> coordinate is <math>5</math>, meaning the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2003|ab=B|num-b=13|num-a=15}}<br />
{{AMC10 box|year=2003|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_20&diff=742812003 AMC 10B Problems/Problem 202016-01-03T17:04:15Z<p>Andb501: /* Solution */</p>
<hr />
<div>{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #14]] and [[2003 AMC 10B Problems|2003 AMC 10B #20]]}}<br />
<br />
==Problem==<br />
In rectangle <math>ABCD, AB=5</math> and <math>BC=3</math>. Points <math>F</math> and <math>G</math> are on <math>\overline{CD}</math> so that <math>DF=1</math> and <math>GC=2</math>. Lines <math>AF</math> and <math>BG</math> intersect at <math>E</math>. Find the area of <math>\triangle AEB</math>.<br />
<center><asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair A=(0,0), B=(5,0), C=(5,3), D=(0,3);<br />
pair F=(1,3), G=(3,3);<br />
pair E=(5/3,5);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(A--E);<br />
draw(B--E);<br />
<br />
pair[] ps={A,B,C,D,E,F,G}; dot(ps);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
label("$E$",E,N);<br />
label("$F$",F,SE);<br />
label("$G$",G,SW);<br />
label("$1$",midpoint(D--F),N);<br />
label("$2$",midpoint(G--C),N);<br />
label("$5$",midpoint(A--B),S);<br />
label("$3$",midpoint(A--D),W);<br />
</asy></center><br />
<math>\textbf{(A) } 10 \qquad\textbf{(B) } \frac{21}{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } \frac{25}{2} \qquad\textbf{(E) } 15</math><br />
<br />
==Solution==<br />
<br />
<math>\triangle EFG \sim \triangle EAB</math> because <math>FG \parallel AB.</math> The ratio of <math>\triangle EFG</math> to <math>\triangle EAB</math> is <math>2:5</math> since <math>AB=5</math> and <math>FG=2</math> from subtraction. If we let <math>h</math> be the height of <math>\triangle EAB,</math><br />
<br />
<cmath>\frac{2}{5} = \frac{h-3}{h}</cmath><br />
<cmath>2h = 5h-15</cmath><br />
<cmath>3h = 15</cmath><br />
<cmath>h = 5</cmath><br />
<br />
The height is <math>5</math> so the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2003|ab=B|num-b=13|num-a=15}}<br />
{{AMC10 box|year=2003|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_20&diff=742802003 AMC 10B Problems/Problem 202016-01-03T17:03:45Z<p>Andb501: /* Problem */</p>
<hr />
<div>{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #14]] and [[2003 AMC 10B Problems|2003 AMC 10B #20]]}}<br />
<br />
==Problem==<br />
In rectangle <math>ABCD, AB=5</math> and <math>BC=3</math>. Points <math>F</math> and <math>G</math> are on <math>\overline{CD}</math> so that <math>DF=1</math> and <math>GC=2</math>. Lines <math>AF</math> and <math>BG</math> intersect at <math>E</math>. Find the area of <math>\triangle AEB</math>.<br />
<center><asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair A=(0,0), B=(5,0), C=(5,3), D=(0,3);<br />
pair F=(1,3), G=(3,3);<br />
pair E=(5/3,5);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(A--E);<br />
draw(B--E);<br />
<br />
pair[] ps={A,B,C,D,E,F,G}; dot(ps);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
label("$E$",E,N);<br />
label("$F$",F,SE);<br />
label("$G$",G,SW);<br />
label("$1$",midpoint(D--F),N);<br />
label("$2$",midpoint(G--C),N);<br />
label("$5$",midpoint(A--B),S);<br />
label("$3$",midpoint(A--D),W);<br />
</asy></center><br />
<math>\textbf{(A) } 10 \qquad\textbf{(B) } \frac{21}{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } \frac{25}{2} \qquad\textbf{(E) } 15</math><br />
<br />
==Solution==<br />
<center><asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair A=(0,0), B=(5,0), C=(5,3), D=(0,3);<br />
pair F=(1,3), G=(3,3);<br />
pair E=(5/3,5);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(A--E);<br />
draw(B--E);<br />
<br />
pair[] ps={A,B,C,D,E,F,G}; dot(ps);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
label("$E$",E,N);<br />
label("$F$",F,SE);<br />
label("$G$",G,SW);<br />
label("$1$",midpoint(D--F),N);<br />
label("$2$",midpoint(G--C),N);<br />
label("$5$",midpoint(A--B),S);<br />
label("$3$",midpoint(A--D),W);<br />
</asy></center><br />
<br />
<math>\triangle EFG \sim \triangle EAB</math> because <math>FG \parallel AB.</math> The ratio of <math>\triangle EFG</math> to <math>\triangle EAB</math> is <math>2:5</math> since <math>AB=5</math> and <math>FG=2</math> from subtraction. If we let <math>h</math> be the height of <math>\triangle EAB,</math><br />
<br />
<cmath>\frac{2}{5} = \frac{h-3}{h}</cmath><br />
<cmath>2h = 5h-15</cmath><br />
<cmath>3h = 15</cmath><br />
<cmath>h = 5</cmath><br />
<br />
The height is <math>5</math> so the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2003|ab=B|num-b=13|num-a=15}}<br />
{{AMC10 box|year=2003|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_4&diff=742382002 AMC 12B Problems/Problem 42016-01-02T03:29:32Z<p>Andb501: /* Solution */</p>
<hr />
<div>{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #4]] and [[2002 AMC 10B Problems|2002 AMC 10B #7]]}}<br />
== Problem ==<br />
Let <math>n</math> be a positive [[integer]] such that <math>\frac 12 + \frac 13 + \frac 17 + \frac 1n</math> is an integer. Which of the following statements is '''not ''' true:<br />
<br />
<math>\mathrm{(A)}\ 2\ \text{divides\ }n<br />
\qquad\mathrm{(B)}\ 3\ \text{divides\ }n<br />
\qquad\mathrm{(C)}</math> <math>\ 6\ \text{divides\ }n <br />
\qquad\mathrm{(D)}\ 7\ \text{divides\ }n<br />
\qquad\mathrm{(E)}\ n > 84</math><br />
<br />
== Solution ==<br />
Since <math>\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}</math>,<br />
<br />
<cmath>0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2</cmath><br />
<br />
From which it follows that <math>\frac{41}{42} + \frac 1n = 1</math> and <math>n = 42</math>. The only answer choice that is not true is <math>\boxed{\mathrm{(E)}\ n>84}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2002|ab=B|num-b=6|num-a=8}}<br />
{{AMC12 box|year=2002|ab=B|num-b=3|num-a=5}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_14&diff=742372002 AMC 12B Problems/Problem 142016-01-02T03:26:35Z<p>Andb501: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #14]] and [[2002 AMC 10B Problems|2002 AMC 10B #18]]}}<br />
== Problem ==<br />
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Four distinct [[circle]]s are drawn in a [[plane]]. What is the maximum number of points where at least two of the circles intersect?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude><br />
<br />
<math>\mathrm{(A)}\ 8<br />
\qquad\mathrm{(B)}\ 9<br />
\qquad\mathrm{(C)}\ 10<br />
\qquad\mathrm{(D)}\ 12<br />
\qquad\mathrm{(E)}\ 16</math><br />
<br />
== Solution 1==<br />
For any given pair of circles, they can intersect at most <math>2</math> times. Since there are <math>{4\choose 2} = 6</math> pairs of circles, the maximum number of possible intersections is <math>6 \cdot 2 = 12</math>. We can construct such a situation as below, so the answer is <math>\boxed{\mathrm{(D)}\ 12}</math>.<br />
<br />
[[Image:2002_12B_AMC-14.png]]<br />
<br />
==Solution 2==<br />
Because a pair or circles can intersect at most <math>2</math> times, the first circle can intersect the second at <math>2</math> points, the third can intersect the first two at <math>4</math> points, and the fourth can intersect the first three at <math>6</math> points. This means that our answer is <math>2+4+6=\boxed{\mathrm{(D)}\ 12}.</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2002|ab=B|num-b=17|num-a=19}}<br />
{{AMC12 box|year=2002|ab=B|num-b=13|num-a=15}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_14&diff=742362002 AMC 12B Problems/Problem 142016-01-02T03:25:43Z<p>Andb501: /* Solution */</p>
<hr />
<div>{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #14]] and [[2002 AMC 10B Problems|2002 AMC 10B #18]]}}<br />
== Problem ==<br />
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Four distinct [[circle]]s are drawn in a [[plane]]. What is the maximum number of points where at least two of the circles intersect?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude><br />
<br />
<math>\mathrm{(A)}\ 8<br />
\qquad\mathrm{(B)}\ 9<br />
\qquad\mathrm{(C)}\ 10<br />
\qquad\mathrm{(D)}\ 12<br />
\qquad\mathrm{(E)}\ 16</math><br />
<br />
== Solution 1==<br />
For any given pair of circles, they can intersect at most <math>2</math> times. Since there are <math>{4\choose 2} = 6</math> pairs of circles, the maximum number of possible intersections is <math>6 \cdot 2 = 12</math>. We can construct such a situation as below, so the answer is <math>\boxed{\mathrm{(D)}\ 12}</math>.<br />
<br />
[[Image:2002_12B_AMC-14.png]]<br />
<br />
==Solution 2==<br />
Because a pair or circles can intersect at most <math>2</math> times, the first circle can intersect the second at <math>2</math> points, the third can intersect the first two at <math>4</math> points, and the fourth can intersect the first three at <math>6</math> points. This means that our answer is <math>2+4+6=\boxed{\mathrm{(D)}\ 12}</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2002|ab=B|num-b=17|num-a=19}}<br />
{{AMC12 box|year=2002|ab=B|num-b=13|num-a=15}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10B_Problems&diff=733162007 AMC 10B Problems2015-11-29T19:22:53Z<p>Andb501: /* Problem 9 */</p>
<hr />
<div>==Problem 1==<br />
<br />
Isabella's house has <math>3</math> bedrooms. Each bedroom is <math>12</math> feet long, <math>10</math> feet wide, and <math>8</math> feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy <math>60</math> square feet in each bedroom. How many square feet of walls must be painted?<br />
<br />
<math>\textbf{(A) } 678 \qquad\textbf{(B) } 768 \qquad\textbf{(C) } 786 \qquad\textbf{(D) } 867 \qquad\textbf{(E) } 876</math><br />
<br />
[[2007 AMC 10B Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
<br />
Define the operation <math>\star</math> by <math>a \star b = (a+b)b.</math> What is <math>(3 \star 5) - (5 \star 3)?</math><br />
<br />
<math>\textbf{(A) } -16 \qquad\textbf{(B) } -8 \qquad\textbf{(C) } 0 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 16</math><br />
<br />
[[2007 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
<br />
A college student drove his compact car <math>120</math> miles home for the weekend and averaged <math>30</math> miles per gallon. On the return trip the student drove his parents' SUV and averaged only <math>20</math> miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?<br />
<br />
<math>\textbf{(A) } 22 \qquad\textbf{(B) } 24 \qquad\textbf{(C) } 25 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 28</math><br />
<br />
[[2007 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
<br />
The point <math>O</math> is the center of the circle circumscribed about <math>\triangle ABC,</math> with <math>\angle BOC=120^\circ</math> and <math>\angle AOB=140^\circ</math>. What is the degree measure of <math>\angle ABC?</math><br />
<br />
<math>\textbf{(A) } 35 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 50 \qquad\textbf{(E) } 60</math><br />
<br />
[[2007 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
In a certain land, all Arogs are Brafs, all Crups are Brafs, all Dramps are Arogs, and all Crups are Dramps. Which of the following statements is implied by these facts?<br />
<br />
<math>\textbf{(A) } \text{All Dramps are Brafs and are Crups.}\\<br />
\textbf{(B) } \text{All Brafs are Crups and are Dramps.}\\<br />
\textbf{(C) } \text{All Arogs are Crups and are Dramps.}\\<br />
\textbf{(D) } \text{All Crups are Arogs and are Brafs.}\\<br />
\textbf{(E) } \text{All Arogs are Dramps and some Arogs may not be Crumps.}</math><br />
<br />
[[2007 AMC 10B Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
<br />
The 2007 AMC10 will be scored by awarding <math>6</math> points for each correct response, <math>0</math> points for each incorrect response, and <math>1.5</math> points for each problem left unanswered. After looking over the <math>25</math> problems, Sarah has decided to attempt the first <math>22</math> and leave only the last <math>3</math> unanswered. How many of the first <math>22</math> problems must she solve correctly in order to score at least <math>100</math> points?<br />
<br />
<math>\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17</math><br />
<br />
[[2007 AMC 10B Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
<br />
All sides of the convex pentagon <math>ABCDE</math> are of equal length, and <math>\angle A= \angle B = 90^\circ.</math> What is the degree measure of <math>\angle E?</math><br />
<br />
<math>\textbf{(A) } 90 \qquad\textbf{(B) } 108 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 150</math><br />
<br />
[[2007 AMC 10B Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
<br />
On the trip home from the meeting where this AMC10 was constructed, the Contest Chair noted that his airport parking receipt had digits of the form <math>bbcac,</math> where <math>0 \le a < b < c \le 9,</math> and <math>b</math> was the average of <math>a</math> and <math>c.</math> How many different five-digit numbers satisfy all these properties?<br />
<br />
<math>\textbf{(A) } 12 \qquad\textbf{(B) } 16 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 25</math><br />
<br />
[[2007 AMC 10B Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
A cryptographic code is designed as follows. The first time a letter appears in a given message it is replaced by the letter that is <math>1</math> place to its right in the alphabet (assuming that the letter <math>A</math> is one place to the right of the letter <math>Z</math>). The second time this same letter appears in the given message, it is replaced by the letter that is <math>1+2</math> places to the right, the third time it is replaced by the letter that is <math>1+2+3</math> places to the right, and so on. For example, with this code the word "banana" becomes "cbodqg". What letter will replace the last letter <math>s</math> in the message<br />
<cmath>\text{"Lee's sis is a Mississippi miss, Chriss!"?}</cmath><br />
<br />
<math>\textbf{(A) } g \qquad\textbf{(B) } h \qquad\textbf{(C) } o \qquad\textbf{(D) } s \qquad\textbf{(E) } t</math><br />
<br />
[[2007 AMC 10B Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
Two points <math>B</math> and <math>C</math> are in a plane. Let <math>S</math> be the set of all points <math>A</math> in the plane for which <math>\triangle ABC</math> has area <math>1.</math> Which of the following describes <math>S?</math><br />
<br />
<math>\textbf{(A) } \text{two parallel lines} \qquad\textbf{(B) } \text{a parabola} \qquad\textbf{(C) } \text{a circle} \qquad\textbf{(D) } \text{a line segment} \qquad\textbf{(E) } \text{two points}</math><br />
<br />
[[2007 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
<br />
A circle passes through the three vertices of an isosceles triangle that has two sides of length <math>3</math> and a base of length <math>2.</math> What is the area of this circle?<br />
<br />
<math>\textbf{(A) } 2\pi \qquad\textbf{(B) } \frac{5}{2}\pi \qquad\textbf{(C) } \frac{81}{32}\pi \qquad\textbf{(D) } 3\pi \qquad\textbf{(E) } \frac{7}{2}\pi</math><br />
<br />
[[2007 AMC 10B Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
Tom's age is <math>T</math> years, which is also the sum of the ages of his three children. His age <math>N</math> years ago was twice the sum of their ages then. What is <math>T/N?</math><br />
<br />
<math>\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6</math><br />
<br />
[[2007 AMC 10B Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
Two circles of radius <math>2</math> are centered at <math>(2,0)</math> and at <math>(0,2).</math> What is the area of the intersection of the interiors of the two circles?<br />
<br />
<math>\textbf{(A) } \pi -2 \qquad\textbf{(B) } \frac{\pi}{2} \qquad\textbf{(C) } \frac{\pi \sqrt{3}}{3} \qquad\textbf{(D) } 2(\pi -2) \qquad\textbf{(E) } \pi</math><br />
<br />
[[2007 AMC 10B Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
Some boys and girls are having a car wash to raise money for a class trip to China. Initially <math>40\%</math> of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then <math>30\%</math> of the group are girls. How many girls were initially in the group?<br />
<br />
<math>\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 8 \qquad\textbf{(D) } 10 \qquad\textbf{(E) } 12</math><br />
<br />
[[2007 AMC 10B Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
<br />
The angles of quadrilateral <math>ABCD</math> satisfy <math>\angle A=2 \angle B=3 \angle C=4 \angle D.</math> What is the degree measure of <math>\angle A,</math> rounded to the nearest whole number?<br />
<br />
<math>\textbf{(A) } 125 \qquad\textbf{(B) } 144 \qquad\textbf{(C) } 153 \qquad\textbf{(D) } 173 \qquad\textbf{(E) } 180</math><br />
<br />
[[2007 AMC 10B Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
A teacher gave a test to a class in which <math>10\%</math> of the students are juniors and <math>90\%</math> are seniors. The average score on the test was <math>84.</math> The juniors all received the same score, and the average score of the seniors was <math>83.</math> What score did each of the juniors receive on the test?<br />
<br />
<math>\textbf{(A) } 85 \qquad\textbf{(B) } 88 \qquad\textbf{(C) } 93 \qquad\textbf{(D) } 94 \qquad\textbf{(E) } 98</math><br />
<br />
[[2007 AMC 10B Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
Point <math>P</math> is inside equilateral <math>\triangle ABC.</math> Points <math>Q, R,</math> and <math>S</math> are the feet of the perpendiculars from <math>P</math> to <math>\overline{AB}, \overline{BC},</math> and <math>\overline{CA},</math> respectively. Given that <math>PQ=1, PR=2,</math> and <math>PS=3,</math> what is <math>AB?</math><br />
<br />
<math>\textbf{(A) } 4 \qquad\textbf{(B) } 3\sqrt{3} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 4\sqrt{3} \qquad\textbf{(E) } 9</math><br />
<br />
[[2007 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
<br />
A circle of radius <math>1</math> is surrounded by <math>4</math> circles of radius <math>r</math> as shown. What is <math>r</math>?<br />
<br />
<center><asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(.8pt)+fontsize(7pt));<br />
dotfactor=4;<br />
<br />
real r1=1, r2=1+sqrt(2);<br />
pair A=(0,0), B=(1+sqrt(2),1+sqrt(2)), C=(-1-sqrt(2),1+sqrt(2)), D=(-1-sqrt(2),-1-sqrt(2)), E=(1+sqrt(2),-1-sqrt(2));<br />
pair A1=(1,0), B1=(2+2sqrt(2),1+sqrt(2)), C1=(0,1+sqrt(2)), D1=(0,-1-sqrt(2)), E1=(2+2sqrt(2),-1-sqrt(2));<br />
path circleA=Circle(A,r1); path circleB=Circle(B,r2); path circleC=Circle(C,r2); path circleD=Circle(D,r2); path circleE=Circle(E,r2);<br />
draw(circleA); draw(circleB); draw(circleC); draw(circleD); draw(circleE);<br />
draw(A--A1); draw(B--B1); draw(C--C1); draw(D--D1); draw(E--E1);<br />
<br />
label("$1$",midpoint(A--A1),N);<br />
label("$r$",midpoint(B--B1),N);<br />
label("$r$",midpoint(C--C1),N);<br />
label("$r$",midpoint(D--D1),N);<br />
label("$r$",midpoint(E--E1),N);<br />
</asy></center><br />
<br />
<math>\textbf{(A) } \sqrt{2} \qquad\textbf{(B) } 1+\sqrt{2} \qquad\textbf{(C) } \sqrt{6} \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 2+\sqrt{2}</math><br />
<br />
[[2007 AMC 10B Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
<br />
The wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by <math>4,</math> and the second number is divided by <math>5.</math> The first remainder designates a column, and the second remainder designates a row on the checkerboard shown. What is the probability that the pair of numbers designates a shaded square?<br />
<br />
<center><asy><br />
unitsize(5mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
real r=2;<br />
pair O=(0,0);<br />
pair A=(0,2), A1=(0,-2); draw(A--A1);<br />
pair B=(sqrt(3),1), B1=(-sqrt(3),-1); draw(B--B1);<br />
pair C=(sqrt(3),-1), C1=(-sqrt(3),1); draw(C--C1);<br />
path circleO=Circle(O,r);<br />
draw(circleO);<br />
pair[] ps={O}; dot(ps);<br />
label("$6$",(-0.6,1));<br />
label("$1$",(0.6,1));<br />
label("$2$",(0.6,-1));<br />
label("$9$",(-0.6,-1));<br />
label("$7$",(1.2,0));<br />
label("$3$",(-1.2,0));<br />
<br />
label("$pointer$",(-4,0));<br />
draw((-5.5,0.5)--(-5.5,-0.5)--(-3,-0.5)--(-2.5,0)--(-3,0.5)--cycle);<br />
<br />
fill((4,0)--(4,1)--(5,1)--(5,0)--cycle,gray);<br />
fill((6,2)--(6,1)--(5,1)--(5,2)--cycle,gray);<br />
fill((6,0)--(6,-1)--(5,-1)--(5,0)--cycle,gray);<br />
fill((6,0)--(6,1)--(7,1)--(7,0)--cycle,gray);<br />
fill((4,-1)--(5,-1)--(5,-2)--(4,-2)--cycle,gray);<br />
fill((6,-1)--(7,-1)--(7,-2)--(6,-2)--cycle,gray);<br />
draw((4,2)--(7,2)--(7,-2)--(4,-2)--cycle);<br />
draw((4,1)--(7,1)); draw((4,0)--(7,0)); draw((4,-1)--(7,-1));<br />
draw((5,2)--(5,-2)); draw((6,2)--(6,-2));<br />
label("$1$",midpoint((4,-1)--(4,-2)),W);<br />
label("$2$",midpoint((4,0)--(4,-1)),W);<br />
label("$3$",midpoint((4,1)--(4,0)),W);<br />
label("$4$",midpoint((4,2)--(4,1)),W);<br />
label("$1$",midpoint((4,-2)--(5,-2)),S);<br />
label("$2$",midpoint((5,-2)--(6,-2)),S);<br />
label("$3$",midpoint((7,-2)--(6,-2)),S);<br />
</asy></center><br />
<br />
<math>\textbf{(A) } \frac{1}{3} \qquad\textbf{(B) } \frac{4}{9} \qquad\textbf{(C) } \frac{1}{2} \qquad\textbf{(D) } \frac{5}{9} \qquad\textbf{(E) } \frac{2}{3}</math><br />
<br />
[[2007 AMC 10B Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
A set of <math>25</math> square blocks is arranged into a <math>5 \times 5</math> square. How many different combinations of <math>3</math> blocks can be selected from that set so that no two are in the same row or column?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 125 \qquad\textbf{(C) } 600 \qquad\textbf{(D) } 2300 \qquad\textbf{(E) } 3600</math><br />
<br />
[[2007 AMC 10B Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
Right <math>\triangle ABC</math> has <math>AB=3, BC=4,</math> and <math>AC=5.</math> Square <math>XYZW</math> is inscribed in <math>\triangle ABC</math> with <math>X</math> and <math>Y</math> on <math>\overline{AC}, W</math> on <math>\overline{AB},</math> and <math>Z</math> on <math>\overline{BC}.</math> What is the side length of the square?<br />
<br />
<math>\textbf{(A) } \frac{3}{2} \qquad\textbf{(B) } \frac{60}{37} \qquad\textbf{(C) } \frac{12}{7} \qquad\textbf{(D) } \frac{23}{13} \qquad\textbf{(E) 2} </math><br />
<br />
[[2007 AMC 10B Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
A player chooses one of the numbers <math>1</math> through <math>4</math>. After the choice has been made, two regular four-sided (tetrahedral) dice are rolled, with the sides of the dice numbered <math>1</math> through <math>4.</math> If the number chosen appears on the bottom of exactly one die after it has been rolled, then the player wins <math>1</math> dollar. If the number chosen appears on the bottom of both of the dice, then the player wins <math>2</math> dollars. If the number chosen does not appear on the bottom of either of the dice, the player loses <math>1</math> dollar. What is the expected return to the player, in dollars, for one roll of the dice?<br />
<br />
<math>\textbf{(A) } -\frac{1}{8} \qquad\textbf{(B) } -\frac{1}{16} \qquad\textbf{(C) } 0 \qquad\textbf{(D) } \frac{1}{16} \qquad\textbf{(E) } \frac{1}{8}</math><br />
<br />
[[2007 AMC 10B Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
A pyramid with a square base is cut by a plane that is parallel to its base and <math>2</math> units from the base. The surface area of the smaller pyramid that is cut from the top is half the surface area of the original pyramid. What is the altitude of the original pyramid?<br />
<br />
<math>\textbf{(A) } 2 \qquad\textbf{(B) } 2+\sqrt{2} \qquad\textbf{(C) } 1+2\sqrt{2} \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 4+2\sqrt{2}</math><br />
<br />
[[2007 AMC 10B Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
Let <math>n</math> denote the smallest positive integer that is divisible by both <math>4</math> and <math>9,</math> and whose base-<math>10</math> representation consists of only <math>4</math>'s and <math>9</math>'s, with at least one of each. What are the last four digits of <math>n?</math><br />
<br />
<math>\textbf{(A) } 4444 \qquad\textbf{(B) } 4494 \qquad\textbf{(C) } 4944 \qquad\textbf{(D) } 9444 \qquad\textbf{(E) } 9944</math><br />
<br />
[[2007 AMC 10B Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
<br />
How many pairs of positive integers <math>(a,b)</math> are there such that <math>a</math> and <math>b</math> have no common factors greater than <math>1</math> and<br />
<cmath>\frac{a}{b} + \frac{14b}{9a}</cmath><br />
is an integer?<br />
<br />
<math>\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } \text{infinitely many}</math><br />
<br />
[[2007 AMC 10B Problems/Problem 25|Solution]]<br />
==See also==<br />
{{AMC10 box|year=2007|ab=B|before=[[2007 AMC 10A Problems]]|after=[[2008 AMC 10A Problems]]}}<br />
*[[AMC 10 Problems and Solutions]]<br />
*[[AMC Problems and Solutions]]<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems&diff=733152006 AMC 10B Problems2015-11-29T19:16:03Z<p>Andb501: /* Problem 16 */</p>
<hr />
<div>== Problem 1 ==<br />
What is <math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} </math> ?<br />
<br />
<math> \mathrm{(A) \ } -2006\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } 2006 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
For real numbers <math>x</math> and <math>y</math>, define <math> x \mathop{\spadesuit} y = (x+y)(x-y) </math>. What is <math> 3 \mathop{\spadesuit} (4 \mathop{\spadesuit} 5) </math>?<br />
<br />
<math> \mathrm{(A) \ } -72\qquad \mathrm{(B) \ } -27\qquad \mathrm{(C) \ } -24\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 72 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score? <br />
<br />
<math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 17\qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 24 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Circles of diameter 1 inch and 3 inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area? <br />
<br />
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
A <math> 2 \times 3 </math> rectangle and a <math> 3 \times 4 </math> rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square? <br />
<br />
<math> \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 25\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 49\qquad \mathrm{(E) \ } 64 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
A region is bounded by semicircular arcs constructed on the side of a square whose sides measure <math> \frac{2}{\pi} </math>, as shown. What is the perimeter of this region? <br />
<br />
<asy><br />
unitsize(1cm);<br />
defaultpen(.8);<br />
<br />
filldraw( circle( (0,1), 1 ), lightgray, black );<br />
filldraw( circle( (0,-1), 1 ), lightgray, black );<br />
filldraw( circle( (1,0), 1 ), lightgray, black );<br />
filldraw( circle( (-1,0), 1 ), lightgray, black );<br />
filldraw( (-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle, lightgray, black );<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } \frac{4}{\pi}\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } \frac{8}{\pi}\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } \frac{16}{\pi} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Which of the following is equivalent to <math> \sqrt{\frac{x}{1-\frac{x-1}{x}}} </math> when <math> x < 0 </math>?<br />
<br />
<math> \mathrm{(A) \ } -x\qquad \mathrm{(B) \ } x\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } \sqrt{\frac{x}{2}}\qquad \mathrm{(E) \ } x\sqrt{-1} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle? <br />
<br />
<asy><br />
unitsize(1cm);<br />
defaultpen(.8);<br />
<br />
draw( (-sqrt(5),0) -- (sqrt(5),0), dashed );<br />
draw( (-1,0)--(-1,2)--(1,2)--(1,0)--cycle );<br />
draw( arc( (0,0), sqrt(5), 0, 180 ) );<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 20\pi\qquad \mathrm{(B) \ } 25\pi\qquad \mathrm{(C) \ } 30\pi\qquad \mathrm{(D) \ } 40\pi\qquad \mathrm{(E) \ } 50\pi </math><br />
<br />
[[2006 AMC 10B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade? <br />
<br />
<math> \mathrm{(A) \ } 129\qquad \mathrm{(B) \ } 137\qquad \mathrm{(C) \ } 174\qquad \mathrm{(D) \ } 233\qquad \mathrm{(E) \ } 411 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? <br />
<br />
<math> \mathrm{(A) \ } 43\qquad \mathrm{(B) \ } 44\qquad \mathrm{(C) \ } 45\qquad \mathrm{(D) \ } 46\qquad \mathrm{(E) \ } 47 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
What is the tens digit in the sum <math> 7!+8!+9!+...+2006!</math><br />
<br />
<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 9 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
The lines <math> x=\frac{1}{4}y+a </math> and <math> y=\frac{1}{4}x+b </math> intersect at the point <math> (1,2) </math>. What is <math> a+b </math>?<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } \frac{3}{4}\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } \frac{9}{4} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Joe and JoAnn each bought 12 ounces of coffee in a 16 ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee? <br />
<br />
<math> \mathrm{(A) \ } \frac{6}{7}\qquad \mathrm{(B) \ } \frac{13}{14}\qquad \mathrm{(C) \ }1 \qquad \mathrm{(D) \ } \frac{14}{13}\qquad \mathrm{(E) \ } \frac{7}{6} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Let <math>a</math> and <math>b</math> be the roots of the equation <math> x^2-mx+2=0 </math>. Suppose that <math> a+(1/b) </math> and <math> b+(1/a) </math> are the roots of the equation <math> x^2-px+q=0 </math>. What is <math>q</math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{5}{2}\qquad \mathrm{(B) \ } \frac{7}{2}\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } \frac{9}{2}\qquad \mathrm{(E) \ } 8 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Rhombus <math>ABCD</math> is similar to rhombus <math>BFDE</math>. The area of rhombus <math>ABCD</math> is <math>24</math> and <math> \angle BAD = 60^\circ </math>. What is the area of rhombus <math>BFDE</math>? <br />
<br />
<asy><br />
unitsize(3cm);<br />
defaultpen(.8);<br />
<br />
pair A=(0,0), B=(1,0), D=dir(60), C=B+D;<br />
<br />
draw(A--B--C--D--cycle);<br />
pair Ep = intersectionpoint( B -- (B+10*dir(150)), D -- (D+10*dir(270)) );<br />
pair F = intersectionpoint( B -- (B+10*dir(90)), D -- (D+10*dir(330)) );<br />
<br />
draw(B--Ep--D--F--cycle);<br />
<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
label("$E$",Ep,SW);<br />
label("$F$",F,NE);<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 6\qquad \mathrm{(B) \ } 4\sqrt{3}\qquad \mathrm{(C) \ } 8\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 6\sqrt{3} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Leap Day, February 29, 2004, occurred on a Sunday. On what day of the week will Leap Day, February 29, 2020, occur? <br />
<br />
<math> \mathrm{(A) \ } \textrm{Tuesday} \qquad \mathrm{(B) \ } \textrm{Wednesday} \qquad \mathrm{(C) \ } \textrm{Thursday} \qquad \mathrm{(D) \ } \textrm{Friday} \qquad \mathrm{(E) \ } \textrm{Saturday} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same? <br />
<br />
<math> \mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{5}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
Let <math> a_1 , a_2 , ... </math> be a sequence for which<br />
<br />
<math> a_1=2 </math> , <math> a_2=3 </math>, and <math>a_n=\frac{a_{n-1}}{a_{n-2}} </math> for each positive integer <math> n \ge 3 </math>. <br />
<br />
What is <math> a_{2006} </math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{3}{2}\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } 3 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
A circle of radius <math>2</math> is centered at <math>O</math>. Square <math>OABC</math> has side length <math>1</math>. Sides <math>AB</math> and <math>CB</math> are extended past <math>B</math> to meet the circle at <math>D</math> and <math>E</math>, respectively. What is the area of the shaded region in the figure, which is bounded by <math>BD</math>, <math>BE</math>, and the minor arc connecting <math>D</math> and <math>E</math>?<br />
<br />
<asy><br />
unitsize(1.5cm);<br />
defaultpen(.8);<br />
<br />
draw( circle( (0,0), 2 ) );<br />
draw( (-2,0) -- (2,0) );<br />
draw( (0,-2) -- (0,2) );<br />
<br />
pair D = intersectionpoint( circle( (0,0), 2 ), (1,0) -- (1,2) );<br />
pair Ep = intersectionpoint( circle( (0,0), 2 ), (0,1) -- (2,1) );<br />
draw( (1,0) -- D );<br />
draw( (0,1) -- Ep );<br />
<br />
filldraw( (1,1) -- arc( (0,0),Ep,D ) -- cycle, mediumgray, black );<br />
<br />
label("$O$",(0,0),SW);<br />
label("$A$",(1,0),S);<br />
label("$C$",(0,1),W);<br />
label("$B$",(1,1),SW);<br />
label("$D$",D,N);<br />
label("$E$",Ep,E);<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } \frac{\pi}{3}+1-\sqrt{3}\qquad \mathrm{(B) \ } \frac{\pi}{2}(2-\sqrt{3})\qquad \mathrm{(C) \ } \pi(2-\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}+\frac{\sqrt{3}+1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}-1+\sqrt{3} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
In rectangle <math>ABCD</math>, we have <math>A=(6,-22)</math>, <math>B=(2006,178)</math>, <math>D=(8,y)</math>, for some integer <math>y</math>. What is the area of rectangle <math>ABCD</math>?<br />
<br />
<math> \mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
For a particular peculiar pair of dice, the probabilities of rolling <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math>, on each die are in the ratio <math>1:2:3:4:5:6</math>. What is the probability of rolling a total of <math>7</math> on the two dice? <br />
<br />
<math> \mathrm{(A) \ } \frac{4}{63}\qquad \mathrm{(B) \ } \frac{1}{8}\qquad \mathrm{(C) \ } \frac{8}{63}\qquad \mathrm{(D) \ } \frac{1}{6}\qquad \mathrm{(E) \ } \frac{2}{7} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Elmo makes <math>N</math> sandwiches for a fundraiser. For each sandwich he uses <math>B</math> globs of peanut butter at <math>4</math>¢ per glob and <math>J</math> blobs of jam at <math>5</math>¢ per blob. The cost of the peanut butter and jam to make all the sandwiches is \$<math>2.53</math>. Assume that <math>B</math>, <math>J</math>, and <math>N</math> are positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches?<br />
<br />
<math> \mathrm{(A) \ } 1.05\qquad \mathrm{(B) \ } 1.25\qquad \mathrm{(C) \ } 1.45\qquad \mathrm{(D) \ } 1.65\qquad \mathrm{(E) \ } 1.85 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral?<br />
<br />
<asy><br />
unitsize(1.5cm);<br />
defaultpen(.8);<br />
<br />
pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br />
pair F = intersectionpoint( A--D, B--Ep );<br />
<br />
draw( A -- B -- C -- cycle );<br />
draw( A -- D );<br />
draw( B -- Ep );<br />
filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br />
<br />
label("$7$",(1.25,0.2));<br />
label("$7$",(2.2,0.45));<br />
label("$3$",(0.45,0.35));<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
Circles with centers <math>O</math> and <math>P</math> have radii <math>2</math> and <math>4</math>, respectively, and are externally tangent. Points <math>A</math> and <math>B</math> on the circle with center <math>O</math> and points <math>C</math> and <math>D</math> on the circle with center <math>P</math> are such that <math>AD</math> and <math>BC</math> are common external tangents to the circles. What is the area of the concave hexagon <math>AOBCPD</math>?<br />
<br />
<asy><br />
unitsize(.7cm);<br />
defaultpen(.8);<br />
<br />
pair O = (0,0), P = (6,0), Q = (-6,0);<br />
pair A = intersectionpoint( arc( (-3,0), (0,0), (-6,0) ), circle( O, 2 ) );<br />
pair B = (A.x, -A.y );<br />
pair D = Q + 2*(A-Q);<br />
pair C = Q + 2*(B-Q);<br />
<br />
draw( circle(O,2) );<br />
draw( circle(P,4) );<br />
draw( (Q + 0.8*(A-Q)) -- ( Q + 2.3*(A-Q) ) );<br />
draw( (Q + 0.8*(B-Q)) -- ( Q + 2.3*(B-Q) ) );<br />
draw( A -- O -- B );<br />
draw( C -- P -- D );<br />
draw( O -- P );<br />
<br />
label("$O$",O,W);<br />
label("$P$",P,E);<br />
<br />
label("$A$",A,NNW);<br />
label("$B$",B,SSW);<br />
<br />
label("$D$",D,NNW);<br />
label("$C$",C,SSW);<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 24\sqrt{3}\qquad \mathrm{(E) \ } 32\sqrt{2} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is <b><i>not</i></b> the age of one of Mr. Jones's children? <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 />
[[2006 AMC 10B Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC10 box|year=2006|ab=A|before=[[2006 AMC 10A Problems]]|after=[[2007 AMC 10A Problems]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[2006 AMC 10B]]<br />
* [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=143 2006 AMC B Math Jam Transcript]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems&diff=733142006 AMC 10B Problems2015-11-29T19:14:59Z<p>Andb501: /* Problem 7 */</p>
<hr />
<div>== Problem 1 ==<br />
What is <math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} </math> ?<br />
<br />
<math> \mathrm{(A) \ } -2006\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } 2006 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
For real numbers <math>x</math> and <math>y</math>, define <math> x \mathop{\spadesuit} y = (x+y)(x-y) </math>. What is <math> 3 \mathop{\spadesuit} (4 \mathop{\spadesuit} 5) </math>?<br />
<br />
<math> \mathrm{(A) \ } -72\qquad \mathrm{(B) \ } -27\qquad \mathrm{(C) \ } -24\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 72 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score? <br />
<br />
<math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 17\qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 24 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Circles of diameter 1 inch and 3 inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area? <br />
<br />
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
A <math> 2 \times 3 </math> rectangle and a <math> 3 \times 4 </math> rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square? <br />
<br />
<math> \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 25\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 49\qquad \mathrm{(E) \ } 64 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
A region is bounded by semicircular arcs constructed on the side of a square whose sides measure <math> \frac{2}{\pi} </math>, as shown. What is the perimeter of this region? <br />
<br />
<asy><br />
unitsize(1cm);<br />
defaultpen(.8);<br />
<br />
filldraw( circle( (0,1), 1 ), lightgray, black );<br />
filldraw( circle( (0,-1), 1 ), lightgray, black );<br />
filldraw( circle( (1,0), 1 ), lightgray, black );<br />
filldraw( circle( (-1,0), 1 ), lightgray, black );<br />
filldraw( (-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle, lightgray, black );<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } \frac{4}{\pi}\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } \frac{8}{\pi}\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } \frac{16}{\pi} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Which of the following is equivalent to <math> \sqrt{\frac{x}{1-\frac{x-1}{x}}} </math> when <math> x < 0 </math>?<br />
<br />
<math> \mathrm{(A) \ } -x\qquad \mathrm{(B) \ } x\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } \sqrt{\frac{x}{2}}\qquad \mathrm{(E) \ } x\sqrt{-1} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle? <br />
<br />
<asy><br />
unitsize(1cm);<br />
defaultpen(.8);<br />
<br />
draw( (-sqrt(5),0) -- (sqrt(5),0), dashed );<br />
draw( (-1,0)--(-1,2)--(1,2)--(1,0)--cycle );<br />
draw( arc( (0,0), sqrt(5), 0, 180 ) );<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 20\pi\qquad \mathrm{(B) \ } 25\pi\qquad \mathrm{(C) \ } 30\pi\qquad \mathrm{(D) \ } 40\pi\qquad \mathrm{(E) \ } 50\pi </math><br />
<br />
[[2006 AMC 10B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade? <br />
<br />
<math> \mathrm{(A) \ } 129\qquad \mathrm{(B) \ } 137\qquad \mathrm{(C) \ } 174\qquad \mathrm{(D) \ } 233\qquad \mathrm{(E) \ } 411 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? <br />
<br />
<math> \mathrm{(A) \ } 43\qquad \mathrm{(B) \ } 44\qquad \mathrm{(C) \ } 45\qquad \mathrm{(D) \ } 46\qquad \mathrm{(E) \ } 47 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
What is the tens digit in the sum <math> 7!+8!+9!+...+2006!</math><br />
<br />
<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 9 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
The lines <math> x=\frac{1}{4}y+a </math> and <math> y=\frac{1}{4}x+b </math> intersect at the point <math> (1,2) </math>. What is <math> a+b </math>?<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } \frac{3}{4}\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } \frac{9}{4} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Joe and JoAnn each bought 12 ounces of coffee in a 16 ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee? <br />
<br />
<math> \mathrm{(A) \ } \frac{6}{7}\qquad \mathrm{(B) \ } \frac{13}{14}\qquad \mathrm{(C) \ }1 \qquad \mathrm{(D) \ } \frac{14}{13}\qquad \mathrm{(E) \ } \frac{7}{6} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Let <math>a</math> and <math>b</math> be the roots of the equation <math> x^2-mx+2=0 </math>. Suppose that <math> a+(1/b) </math> and <math> b+(1/a) </math> are the roots of the equation <math> x^2-px+q=0 </math>. What is <math>q</math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{5}{2}\qquad \mathrm{(B) \ } \frac{7}{2}\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } \frac{9}{2}\qquad \mathrm{(E) \ } 8 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Rhombus <math>ABCD</math> is similar to rhombus <math>BFDE</math>. The area of rhombus <math>ABCD</math> is <math>24</math> and <math> \angle BAD = 60^\circ </math>. What is the area of rhombus <math>BFDE</math>? <br />
<br />
<asy><br />
unitsize(3cm);<br />
defaultpen(.8);<br />
<br />
pair A=(0,0), B=(1,0), D=dir(60), C=B+D;<br />
<br />
draw(A--B--C--D--cycle);<br />
pair Ep = intersectionpoint( B -- (B+10*dir(150)), D -- (D+10*dir(270)) );<br />
pair F = intersectionpoint( B -- (B+10*dir(90)), D -- (D+10*dir(330)) );<br />
<br />
draw(B--Ep--D--F--cycle);<br />
<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
label("$E$",Ep,SW);<br />
label("$F$",F,NE);<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 6\qquad \mathrm{(B) \ } 4\sqrt{3}\qquad \mathrm{(C) \ } 8\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 6\sqrt{3} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Leap Day, February 29, 2004, occured on a Sunday. On what day of the week will Leap Day, February 29, 2020, occur? <br />
<br />
<math> \mathrm{(A) \ } \textrm{Tuesday} \qquad \mathrm{(B) \ } \textrm{Wednesday} \qquad \mathrm{(C) \ } \textrm{Thursday} \qquad \mathrm{(D) \ } \textrm{Friday} \qquad \mathrm{(E) \ } \textrm{Saturday} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same? <br />
<br />
<math> \mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{5}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
Let <math> a_1 , a_2 , ... </math> be a sequence for which<br />
<br />
<math> a_1=2 </math> , <math> a_2=3 </math>, and <math>a_n=\frac{a_{n-1}}{a_{n-2}} </math> for each positive integer <math> n \ge 3 </math>. <br />
<br />
What is <math> a_{2006} </math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{3}{2}\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } 3 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
A circle of radius <math>2</math> is centered at <math>O</math>. Square <math>OABC</math> has side length <math>1</math>. Sides <math>AB</math> and <math>CB</math> are extended past <math>B</math> to meet the circle at <math>D</math> and <math>E</math>, respectively. What is the area of the shaded region in the figure, which is bounded by <math>BD</math>, <math>BE</math>, and the minor arc connecting <math>D</math> and <math>E</math>?<br />
<br />
<asy><br />
unitsize(1.5cm);<br />
defaultpen(.8);<br />
<br />
draw( circle( (0,0), 2 ) );<br />
draw( (-2,0) -- (2,0) );<br />
draw( (0,-2) -- (0,2) );<br />
<br />
pair D = intersectionpoint( circle( (0,0), 2 ), (1,0) -- (1,2) );<br />
pair Ep = intersectionpoint( circle( (0,0), 2 ), (0,1) -- (2,1) );<br />
draw( (1,0) -- D );<br />
draw( (0,1) -- Ep );<br />
<br />
filldraw( (1,1) -- arc( (0,0),Ep,D ) -- cycle, mediumgray, black );<br />
<br />
label("$O$",(0,0),SW);<br />
label("$A$",(1,0),S);<br />
label("$C$",(0,1),W);<br />
label("$B$",(1,1),SW);<br />
label("$D$",D,N);<br />
label("$E$",Ep,E);<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } \frac{\pi}{3}+1-\sqrt{3}\qquad \mathrm{(B) \ } \frac{\pi}{2}(2-\sqrt{3})\qquad \mathrm{(C) \ } \pi(2-\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}+\frac{\sqrt{3}+1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}-1+\sqrt{3} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
In rectangle <math>ABCD</math>, we have <math>A=(6,-22)</math>, <math>B=(2006,178)</math>, <math>D=(8,y)</math>, for some integer <math>y</math>. What is the area of rectangle <math>ABCD</math>?<br />
<br />
<math> \mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
For a particular peculiar pair of dice, the probabilities of rolling <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math>, on each die are in the ratio <math>1:2:3:4:5:6</math>. What is the probability of rolling a total of <math>7</math> on the two dice? <br />
<br />
<math> \mathrm{(A) \ } \frac{4}{63}\qquad \mathrm{(B) \ } \frac{1}{8}\qquad \mathrm{(C) \ } \frac{8}{63}\qquad \mathrm{(D) \ } \frac{1}{6}\qquad \mathrm{(E) \ } \frac{2}{7} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Elmo makes <math>N</math> sandwiches for a fundraiser. For each sandwich he uses <math>B</math> globs of peanut butter at <math>4</math>¢ per glob and <math>J</math> blobs of jam at <math>5</math>¢ per blob. The cost of the peanut butter and jam to make all the sandwiches is \$<math>2.53</math>. Assume that <math>B</math>, <math>J</math>, and <math>N</math> are positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches?<br />
<br />
<math> \mathrm{(A) \ } 1.05\qquad \mathrm{(B) \ } 1.25\qquad \mathrm{(C) \ } 1.45\qquad \mathrm{(D) \ } 1.65\qquad \mathrm{(E) \ } 1.85 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral?<br />
<br />
<asy><br />
unitsize(1.5cm);<br />
defaultpen(.8);<br />
<br />
pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br />
pair F = intersectionpoint( A--D, B--Ep );<br />
<br />
draw( A -- B -- C -- cycle );<br />
draw( A -- D );<br />
draw( B -- Ep );<br />
filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br />
<br />
label("$7$",(1.25,0.2));<br />
label("$7$",(2.2,0.45));<br />
label("$3$",(0.45,0.35));<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
Circles with centers <math>O</math> and <math>P</math> have radii <math>2</math> and <math>4</math>, respectively, and are externally tangent. Points <math>A</math> and <math>B</math> on the circle with center <math>O</math> and points <math>C</math> and <math>D</math> on the circle with center <math>P</math> are such that <math>AD</math> and <math>BC</math> are common external tangents to the circles. What is the area of the concave hexagon <math>AOBCPD</math>?<br />
<br />
<asy><br />
unitsize(.7cm);<br />
defaultpen(.8);<br />
<br />
pair O = (0,0), P = (6,0), Q = (-6,0);<br />
pair A = intersectionpoint( arc( (-3,0), (0,0), (-6,0) ), circle( O, 2 ) );<br />
pair B = (A.x, -A.y );<br />
pair D = Q + 2*(A-Q);<br />
pair C = Q + 2*(B-Q);<br />
<br />
draw( circle(O,2) );<br />
draw( circle(P,4) );<br />
draw( (Q + 0.8*(A-Q)) -- ( Q + 2.3*(A-Q) ) );<br />
draw( (Q + 0.8*(B-Q)) -- ( Q + 2.3*(B-Q) ) );<br />
draw( A -- O -- B );<br />
draw( C -- P -- D );<br />
draw( O -- P );<br />
<br />
label("$O$",O,W);<br />
label("$P$",P,E);<br />
<br />
label("$A$",A,NNW);<br />
label("$B$",B,SSW);<br />
<br />
label("$D$",D,NNW);<br />
label("$C$",C,SSW);<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 24\sqrt{3}\qquad \mathrm{(E) \ } 32\sqrt{2} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is <b><i>not</i></b> the age of one of Mr. Jones's children? <br />
<br />
<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math><br />
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[[2006 AMC 10B Problems/Problem 25|Solution]]<br />
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== See also ==<br />
{{AMC10 box|year=2006|ab=A|before=[[2006 AMC 10A Problems]]|after=[[2007 AMC 10A Problems]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[2006 AMC 10B]]<br />
* [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=143 2006 AMC B Math Jam Transcript]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Andb501https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_16&diff=732632002 AMC 10A Problems/Problem 162015-11-27T20:12:21Z<p>Andb501: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5</math>. What is <math>a + b + c + d</math>?<br />
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<math>\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5</math><br />
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==Solution 1==<br />
Let <math>x=a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5</math>. Since one of the sums involves a, b, c, and d, it makes sense to consider 4x. We have <math>4x=(a+1)+(b+2)+(c+3)+(d+4)=a+b+c+d+10=4(a+b+c+d)+20</math>. Rearranging, we have <math>3(a+b+c+d)=-10</math>, so <math>a+b+c+d=\frac{-10}{3}</math>. Thus, our answer is <math>\boxed{\text{(B)}\ -10/3}</math>.<br />
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==Solution 2==<br />
Take <br />
<math>a + 1 = a + b + c + d + 5</math>.<br />
Now we can clearly see:<br />
<math>-4 = b + c + d</math>.<br />
Continuing this same method with <math>b + 2, c + 3</math>, and <math>d + 4</math> we get:<br />
<math> -4 = b + c + d</math>, <br />
<math> -3 = a + c + d</math>,<br />
<math> -2 = a + b + d</math>, and<br />
<math> -1 = a + b + c</math>,<br />
Adding, we see <math> -10 = 3a + 3b + 3c + 3d</math>. Therefore, <math>a + b + c + d = \frac{-10}{3}</math>.<br />
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==See Also==<br />
{{AMC10 box|year=2002|ab=A|num-b=15|num-a=17}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Andb501