https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ariedel&feedformat=atom AoPS Wiki - User contributions [en] 2021-05-14T17:01:10Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Menelaus%27_Theorem&diff=74000 Menelaus' Theorem 2015-12-27T02:55:43Z <p>Ariedel: Got rid of typo</p> <hr /> <div>{{stub}}<br /> <br /> '''Menelaus' Theorem''' deals with the [[collinearity]] of points on each of the three sides (extended when necessary) of a [[triangle]].<br /> It is named for Menelaus of Alexandria.<br /> == Statement ==<br /> A necessary and sufficient condition for points &lt;math&gt;P, Q, R&lt;/math&gt; on the respective sides &lt;math&gt;BC, CA, AB&lt;/math&gt; (or their extensions) of a triangle &lt;math&gt;ABC&lt;/math&gt; to be [[collinear]] is that<br /> <br /> &lt;center&gt;&lt;math&gt;BP\cdot CQ\cdot AR = PC\cdot QA\cdot RB&lt;/math&gt;&lt;/center&gt;<br /> <br /> where all segments in the formula are [[directed segment]]s.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> defaultpen(fontsize(8));<br /> pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R;<br /> draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75);<br /> draw((7,6)--(6,8)--(4,0));<br /> R=intersectionpoint(A--B,Q--P);<br /> dot(A^^B^^C^^P^^Q^^R);<br /> label(&quot;A&quot;,A,(1,1));label(&quot;B&quot;,B,(-1,0));label(&quot;C&quot;,C,(1,0));label(&quot;P&quot;,P,(0,-1));label(&quot;Q&quot;,Q,(1,0));label(&quot;R&quot;,R,(-1,1));<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> == Proof ==<br /> Draw a line parallel to &lt;math&gt;QP&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; to intersect &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;K&lt;/math&gt;:<br /> &lt;center&gt;&lt;asy&gt;<br /> defaultpen(fontsize(8));<br /> pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R, K=(5.5,0);<br /> draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75);<br /> draw((7,6)--(6,8)--(4,0));<br /> draw(A--K, dashed);<br /> R=intersectionpoint(A--B,Q--P);<br /> dot(A^^B^^C^^P^^Q^^R^^K);<br /> label(&quot;A&quot;,A,(1,1));label(&quot;B&quot;,B,(-1,0));label(&quot;C&quot;,C,(1,0));label(&quot;P&quot;,P,(0,-1));label(&quot;Q&quot;,Q,(1,0));label(&quot;R&quot;,R,(-1,1));<br /> label(&quot;K&quot;,K,(0,-1));<br /> &lt;/asy&gt;&lt;/center&gt;<br /> &lt;math&gt;\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}&lt;/math&gt;<br /> <br /> &lt;math&gt;\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{PC}{PK}&lt;/math&gt;<br /> <br /> Multiplying the two equalities together to eliminate the &lt;math&gt;PK&lt;/math&gt; factor, we get:<br /> <br /> &lt;math&gt;\frac{AR}{RB}\cdot\frac{QC}{QA}=\frac{PC}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{PC}=1&lt;/math&gt;<br /> <br /> ==Proof Using [[Barycentric coordinates]]==<br /> Disclaimer: This proof is not nearly as elegant as the above one. It uses a bash-type approach, as barycentric coordinate proofs tend to be. <br /> <br /> Suppose we give the points &lt;math&gt;P, Q, R&lt;/math&gt; the following coordinates:<br /> <br /> &lt;math&gt;P: (0, P, 1-P)&lt;/math&gt;<br /> <br /> &lt;math&gt;R: (R , 1-R, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;Q: (1-Q ,0 , Q)&lt;/math&gt;<br /> <br /> Note that this says the following:<br /> <br /> &lt;math&gt;\frac{CP}{PB}=\frac{1-P}{P}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{BR}{AR}=\frac{1-R}{R}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{QA}{QC}=\frac{1-Q}{Q}&lt;/math&gt;<br /> <br /> The line through &lt;math&gt;R&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; is given by:<br /> &lt;math&gt;\begin{vmatrix} X &amp; 0 &amp; R \\ Y &amp; P &amp; 1-R\\ Z &amp; 1-P &amp; 0 \end{vmatrix} = 0&lt;/math&gt;<br /> <br /> <br /> which yields, after simplification, <br /> <br /> &lt;cmath&gt;-X\cdot (R-1)(P-1)+Y\cdot R(1-P)-Z\cdot PR = 0&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;Z\cdot PR = -X\cdot (R-1)(P-1)+Y\cdot R(1-P).&lt;/cmath&gt;<br /> <br /> Plugging in the coordinates for &lt;math&gt;Q&lt;/math&gt; yields &lt;math&gt;(Q-1)(R-1)(P-1) = QPR&lt;/math&gt;. From &lt;math&gt;\frac{CP}{PB}=\frac{1-P}{P},&lt;/math&gt; we have &lt;cmath&gt;P=\frac{(1-P)\cdot PB}{CP}.&lt;/cmath&gt; Likewise, &lt;cmath&gt;R=\frac{(1-R)\cdot AR}{BR}&lt;/cmath&gt; and &lt;cmath&gt;Q=\frac{(1-Q)\cdot QC}{QA}.&lt;/cmath&gt;<br /> <br /> <br /> Substituting these values yields &lt;cmath&gt;(Q-1)(R-1)(P-1) = \frac{(1-Q)\cdot QC \cdot (1-P) \cdot PB \cdot (1-R) \cdot AR}{QA\cdot CP\cdot BR}&lt;/cmath&gt; which simplifies to &lt;math&gt;QA\cdot CP \cdot BR = -QC \cdot AR \cdot PB.&lt;/math&gt;<br /> <br /> QED<br /> <br /> == See also ==<br /> * [[Ceva's Theorem]]<br /> * [[Stewart's Theorem]]<br /> <br /> [[Category:Theorems]]<br /> <br /> [[Category:Geometry]]</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_11&diff=73176 2015 AMC 8 Problems/Problem 11 2015-11-25T23:22:15Z <p>Ariedel: /* Solution 2 */ misspelling</p> <hr /> <div>In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read &quot;AMC8&quot;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } \frac{1}{22,050} \qquad<br /> \textbf{(B) } \frac{1}{21,000}\qquad<br /> \textbf{(C) } \frac{1}{10,500}\qquad<br /> \textbf{(D) } \frac{1}{2,100} \qquad<br /> \textbf{(E) } \frac{1}{1,050}<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There is one favorable case, which is the license plate says &quot;AMC8&quot;. We must now find how many total cases there are. There are &lt;math&gt;5&lt;/math&gt; choices for the first letter (since it must be a vowel), &lt;math&gt;21&lt;/math&gt; choices for the second letter (since it must be of 21 consonants), &lt;math&gt;20&lt;/math&gt; choices for the third letter (since it must differ from the second letter), and &lt;math&gt;10&lt;/math&gt; choices for the number. This leads to &lt;math&gt;5 \cdot 21 \cdot 20 \cdot 10=21000&lt;/math&gt; total possible license plates. That means the probability of a license plate saying &quot;AMC8&quot; is &lt;math&gt;\boxed{\textbf{(B) } \frac{1}{21,000}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> The probability of choosing A as the first letter is &lt;math&gt;\dfrac{1}{5}&lt;/math&gt;. The probability of choosing &lt;math&gt;M&lt;/math&gt; next is &lt;math&gt;\dfrac{1}{21}&lt;/math&gt;. The probability of choosing C as the third letter is &lt;math&gt;\dfrac{1}{20}&lt;/math&gt; (since there are 20 other consonants to choose from other then M). The probability of having &lt;math&gt;8&lt;/math&gt; as the last number is &lt;math&gt;\dfrac{1}{10}&lt;/math&gt;. We multiply all these to obtain &lt;math&gt;\dfrac{1}{5}\cdot \dfrac{1}{21} \cdot \dfrac{1}{20}\cdot \dfrac{1}{10}=\dfrac{1}{5\times 21\times 20\times 10}=\dfrac{1}{21\times 100\times 10}=\boxed{\textbf{(B)}~\dfrac{1}{21000}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_2&diff=73171 2015 AMC 8 Problems/Problem 2 2015-11-25T22:51:30Z <p>Ariedel: /* Solution 2 */</p> <hr /> <div>Point &lt;math&gt;O&lt;/math&gt; is the center of the regular octagon &lt;math&gt;ABCDEFGH&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; is the midpoint of the side &lt;math&gt;\overline{AB}.&lt;/math&gt; What fraction of the area of the octagon is shaded?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{11}{32} \quad\textbf{(B) }\frac{3}{8} \quad\textbf{(C) }\frac{13}{32} \quad\textbf{(D) }\frac{7}{16}\quad \textbf{(E) }\frac{15}{32}&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G,H,O,X;<br /> A=dir(45);<br /> B=dir(90);<br /> C=dir(135);<br /> D=dir(180);<br /> E=dir(-135);<br /> F=dir(-90);<br /> G=dir(-45);<br /> H=dir(0);<br /> O=(0,0);<br /> X=midpoint(A--B);<br /> <br /> fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75));<br /> draw(A--B--C--D--E--F--G--H--cycle);<br /> <br /> dot(&quot;$A$&quot;,A,dir(45));<br /> dot(&quot;$B$&quot;,B,dir(90));<br /> dot(&quot;$C$&quot;,C,dir(135));<br /> dot(&quot;$D$&quot;,D,dir(180));<br /> dot(&quot;$E$&quot;,E,dir(-135));<br /> dot(&quot;$F$&quot;,F,dir(-90));<br /> dot(&quot;$G$&quot;,G,dir(-45));<br /> dot(&quot;$H$&quot;,H,dir(0));<br /> dot(&quot;$X$&quot;,X,dir(135/2));<br /> dot(&quot;$O$&quot;,O,dir(0));<br /> draw(E--O--X);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since octagon &lt;math&gt;ABCDEFGH&lt;/math&gt; is a regular octagon, it is split into 8 equal parts, such as triangles &lt;math&gt;\bigtriangleup ABO, \bigtriangleup BCO, \bigtriangleup CDO&lt;/math&gt;, etc. These parts, since they are all equal, are &lt;math&gt;\frac{1}{8}&lt;/math&gt; of the octagon each. The shaded region consists of 3 of these equal parts plus half of another, so the fraction of the octagon that is shaded is &lt;math&gt;\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}=\boxed{\text{(D) }\dfrac{7}{16}}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G,H,O,X,a,b,c,d,e,f,g;<br /> A=dir(45);<br /> B=dir(90);<br /> C=dir(135);<br /> D=dir(180);<br /> E=dir(-135);<br /> F=dir(-90);<br /> G=dir(-45);<br /> H=dir(0);<br /> O=(0,0);<br /> X=midpoint(A--B);<br /> a=midpoint(B--C);<br /> b=midpoint(C--D);<br /> c=midpoint(D--E);<br /> d=midpoint(E--F);<br /> e=midpoint(F--G);<br /> f=midpoint(G--H);<br /> g=midpoint(H--A);<br /> <br /> fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75));<br /> draw(A--B--C--D--E--F--G--H--cycle);<br /> <br /> dot(&quot;$A$&quot;,A,dir(45));<br /> dot(&quot;$B$&quot;,B,dir(90));<br /> dot(&quot;$C$&quot;,C,dir(135));<br /> dot(&quot;$D$&quot;,D,dir(180));<br /> dot(&quot;$E$&quot;,E,dir(-135));<br /> dot(&quot;$F$&quot;,F,dir(-90));<br /> dot(&quot;$G$&quot;,G,dir(-45));<br /> dot(&quot;$H$&quot;,H,dir(0));<br /> dot(&quot;$X$&quot;,X,dir(135/2));<br /> dot(&quot;$O$&quot;,O,dir(0));<br /> draw(E--O--X);<br /> draw(B--F);<br /> draw(A--O);<br /> draw(D--H);<br /> draw(C--G);<br /> draw(a--e);<br /> draw(b--f);<br /> draw(c--g);<br /> draw(d--O);<br /> &lt;/asy&gt;<br /> <br /> The octagon has been divided up into 16 identical triangles (and thus they each have equal area). Since the shaded region occupies 7 out of the 16 total triangles, the answer is &lt;math&gt;\boxed{\textbf{(D)}~\dfrac{7}{16}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_2&diff=73170 2015 AMC 8 Problems/Problem 2 2015-11-25T22:50:56Z <p>Ariedel: </p> <hr /> <div>Point &lt;math&gt;O&lt;/math&gt; is the center of the regular octagon &lt;math&gt;ABCDEFGH&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; is the midpoint of the side &lt;math&gt;\overline{AB}.&lt;/math&gt; What fraction of the area of the octagon is shaded?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{11}{32} \quad\textbf{(B) }\frac{3}{8} \quad\textbf{(C) }\frac{13}{32} \quad\textbf{(D) }\frac{7}{16}\quad \textbf{(E) }\frac{15}{32}&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G,H,O,X;<br /> A=dir(45);<br /> B=dir(90);<br /> C=dir(135);<br /> D=dir(180);<br /> E=dir(-135);<br /> F=dir(-90);<br /> G=dir(-45);<br /> H=dir(0);<br /> O=(0,0);<br /> X=midpoint(A--B);<br /> <br /> fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75));<br /> draw(A--B--C--D--E--F--G--H--cycle);<br /> <br /> dot(&quot;$A$&quot;,A,dir(45));<br /> dot(&quot;$B$&quot;,B,dir(90));<br /> dot(&quot;$C$&quot;,C,dir(135));<br /> dot(&quot;$D$&quot;,D,dir(180));<br /> dot(&quot;$E$&quot;,E,dir(-135));<br /> dot(&quot;$F$&quot;,F,dir(-90));<br /> dot(&quot;$G$&quot;,G,dir(-45));<br /> dot(&quot;$H$&quot;,H,dir(0));<br /> dot(&quot;$X$&quot;,X,dir(135/2));<br /> dot(&quot;$O$&quot;,O,dir(0));<br /> draw(E--O--X);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since octagon &lt;math&gt;ABCDEFGH&lt;/math&gt; is a regular octagon, it is split into 8 equal parts, such as triangles &lt;math&gt;\bigtriangleup ABO, \bigtriangleup BCO, \bigtriangleup CDO&lt;/math&gt;, etc. These parts, since they are all equal, are &lt;math&gt;\frac{1}{8}&lt;/math&gt; of the octagon each. The shaded region consists of 3 of these equal parts plus half of another, so the fraction of the octagon that is shaded is &lt;math&gt;\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}=\boxed{\text{(D) }\dfrac{7}{16}}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> [asy]<br /> pair A,B,C,D,E,F,G,H,O,X,a,b,c,d,e,f,g;<br /> A=dir(45);<br /> B=dir(90);<br /> C=dir(135);<br /> D=dir(180);<br /> E=dir(-135);<br /> F=dir(-90);<br /> G=dir(-45);<br /> H=dir(0);<br /> O=(0,0);<br /> X=midpoint(A--B);<br /> a=midpoint(B--C);<br /> b=midpoint(C--D);<br /> c=midpoint(D--E);<br /> d=midpoint(E--F);<br /> e=midpoint(F--G);<br /> f=midpoint(G--H);<br /> g=midpoint(H--A);<br /> <br /> fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75));<br /> draw(A--B--C--D--E--F--G--H--cycle);<br /> <br /> dot(&quot;&lt;math&gt;A&lt;/math&gt;&quot;,A,dir(45));<br /> dot(&quot;&lt;math&gt;B&lt;/math&gt;&quot;,B,dir(90));<br /> dot(&quot;&lt;math&gt;C&lt;/math&gt;&quot;,C,dir(135));<br /> dot(&quot;&lt;math&gt;D&lt;/math&gt;&quot;,D,dir(180));<br /> dot(&quot;&lt;math&gt;E&lt;/math&gt;&quot;,E,dir(-135));<br /> dot(&quot;&lt;math&gt;F&lt;/math&gt;&quot;,F,dir(-90));<br /> dot(&quot;&lt;math&gt;G&lt;/math&gt;&quot;,G,dir(-45));<br /> dot(&quot;&lt;math&gt;H&lt;/math&gt;&quot;,H,dir(0));<br /> dot(&quot;&lt;math&gt;X&lt;/math&gt;&quot;,X,dir(135/2));<br /> dot(&quot;&lt;math&gt;O&lt;/math&gt;&quot;,O,dir(0));<br /> draw(E--O--X);<br /> draw(B--F);<br /> draw(A--O);<br /> draw(D--H);<br /> draw(C--G);<br /> draw(a--e);<br /> draw(b--f);<br /> draw(c--g);<br /> draw(d--O);<br /> [/asy]<br /> <br /> The octagon has been divided up into 16 identical triangles (and thus they each have equal area). Since the shaded region occupies 7 out of the 16 total triangles, the answer is &lt;math&gt;\boxed{\textbf{(D)}~\dfrac{7}{16}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_12&diff=73167 2015 AMC 8 Problems/Problem 12 2015-11-25T22:47:10Z <p>Ariedel: </p> <hr /> <div>How many pairs of parallel edges, such as &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{GH}&lt;/math&gt; or &lt;math&gt;\overline{EH}&lt;/math&gt; and &lt;math&gt;\overline{FG}&lt;/math&gt;, does a cube have?<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }6 \quad\textbf{(B) }12 \quad\textbf{(C) } 18 \quad\textbf{(D) } 24 \quad \textbf{(E) } 36&lt;/math&gt;<br /> &lt;asy&gt; import three;<br /> currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */<br /> draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle);<br /> draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1));<br /> draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); <br /> draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle);<br /> label(&quot;$D$&quot;,(0,0,0),S);<br /> label(&quot;$A$&quot;,(0,0,1),N);<br /> label(&quot;$H$&quot;,(0,1,0),S);<br /> label(&quot;$E$&quot;,(0,1,1),N);<br /> label(&quot;$C$&quot;,(1,0,0),S);<br /> label(&quot;$B$&quot;,(1,0,1),N);<br /> label(&quot;$G$&quot;,(1,1,0),S);<br /> label(&quot;$F$&quot;,(1,1,1),N);<br /> &lt;/asy&gt;<br /> ==Solution 1==<br /> We first count the number of pairs of parallel lines that are in the same direction as &lt;math&gt;\overline{AB}&lt;/math&gt;. The pairs of parallel lines are &lt;math&gt;\overline{AB}\text{ and }\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{CD}\text{ and }\overline{GH}&lt;/math&gt;, &lt;math&gt;\overline{AB}\text{ and }\overline{CD}&lt;/math&gt;, &lt;math&gt;\overline{EF}\text{ and }\overline{GH}&lt;/math&gt;, &lt;math&gt;\overline{AB}\text{ and }\overline{GH}&lt;/math&gt;, and &lt;math&gt;\overline{CD}\text{ and }\overline{EF}&lt;/math&gt;. These are &lt;math&gt;6&lt;/math&gt; pairs total. We can do the same for the lines in the same direction as &lt;math&gt;\overline{AE}&lt;/math&gt; and &lt;math&gt;\overline{AD}&lt;/math&gt;. This means there are &lt;math&gt;6\cdot 3=\boxed{\textbf{(C) } 18}&lt;/math&gt; total pairs of parallel lines.<br /> <br /> ==Solution 2==<br /> Pick a random edge. Given another edge, the probability that it is parallel to this edge is &lt;math&gt;\frac{3}{12-1}=\frac{3}{11}&lt;/math&gt;. Keep in mind we already used one edge. There are &lt;math&gt;12&lt;/math&gt; edges so &lt;math&gt;\binom{12}{2}=66&lt;/math&gt; pairs. So our answer is &lt;math&gt;\frac{3}{11} \times 66=\boxed{\textbf{(C)}~18}&lt;/math&gt;.<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_16&diff=73166 2015 AMC 8 Problems/Problem 16 2015-11-25T22:44:31Z <p>Ariedel: </p> <hr /> <div>In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If &lt;math&gt;\tfrac{1}{3}&lt;/math&gt; of all the ninth graders are paired with &lt;math&gt;\tfrac{2}{5}&lt;/math&gt; of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } \frac{2}{15} \qquad<br /> \textbf{(B) } \frac{4}{11} \qquad<br /> \textbf{(C) } \frac{11}{30} \qquad<br /> \textbf{(D) } \frac{3}{8} \qquad<br /> \textbf{(E) } \frac{11}{15}<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the number of sixth graders be &lt;math&gt;s&lt;/math&gt;, and the number of ninth graders be &lt;math&gt;n&lt;/math&gt;. Thus, &lt;math&gt;n/3=2s/5&lt;/math&gt;, which simplifies to &lt;math&gt;n=6s/5&lt;/math&gt;. Since we are trying to find the value of &lt;math&gt;\frac{n/3+2s/5}{n+s}&lt;/math&gt;, we can just substitute &lt;math&gt;n&lt;/math&gt; for &lt;math&gt;6s/5&lt;/math&gt; into the equation. We then get a value of &lt;math&gt;\boxed{\textbf{(B)}~\frac{4}{11}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We see that the minimum number of ninth graders is &lt;math&gt;6&lt;/math&gt;, because if there are &lt;math&gt;3&lt;/math&gt; then there is &lt;math&gt;1&lt;/math&gt; ninth grader with a buddy, which would mean &lt;math&gt;2.5&lt;/math&gt; sixth graders with a buddy, and that's impossible. With &lt;math&gt;6&lt;/math&gt; ninth graders, &lt;math&gt;2&lt;/math&gt; of them are in the buddy program, so there &lt;math&gt;\frac{2}{\tfrac{2}{5}}=5&lt;/math&gt; sixth graders total, two of whom have a buddy. Thus, the desired probability is &lt;math&gt;\frac{2+2}{5+6}=\frac{4}{11}\implies\boxed{\textbf{(B) }\frac{4}{11}}&lt;/math&gt;.<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_16&diff=73165 2015 AMC 8 Problems/Problem 16 2015-11-25T22:43:51Z <p>Ariedel: </p> <hr /> <div>In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If &lt;math&gt;\tfrac{1}{3}&lt;/math&gt; of all the ninth graders are paired with &lt;math&gt;\tfrac{2}{5}&lt;/math&gt; of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } \frac{2}{15} \qquad<br /> \textbf{(B) } \frac{4}{11} \qquad<br /> \textbf{(C) } \frac{11}{30} \qquad<br /> \textbf{(D) } \frac{3}{8} \qquad<br /> \textbf{(E) } \frac{11}{15}<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the number of sixth graders be &lt;math&gt;s&lt;/math&gt;, and the number of ninth graders be &lt;math&gt;n&lt;/math&gt;. Thus, &lt;math&gt;n/3=2s/5&lt;/math&gt;, which simplifies to &lt;math&gt;n=6s/5&lt;/math&gt;. Since we are trying to find the value of &lt;math&gt;\frac{n/3+2s/5}{n+s}&lt;/math&gt;, we can just substitute &lt;math&gt;n&lt;/math&gt; for &lt;math&gt;6s/5&lt;/math&gt; into the equation. We then get a value of &lt;math&gt;\boxed{\textbf{(B)}~\frac{4}{11}}&lt;/math&gt;<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_20&diff=73164 2015 AMC 8 Problems/Problem 20 2015-11-25T22:41:39Z <p>Ariedel: </p> <hr /> <div>Ralph went to the store and bought 12 pairs of socks for a total of \$24. Some of the socks he bought cost \$1 a pair, some of the socks he bought cost \$3 a pair, and some of the socks he bought cost \$4 a pair. If he bought at least one pair of each type, how many pairs of \$1 socks did Ralph buy?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 5 \qquad<br /> \textbf{(C) } 6 \qquad<br /> \textbf{(D) } 7 \qquad<br /> \textbf{(E) } 8<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> So let there be &lt;math&gt;x&lt;/math&gt; pairs of &lt;math&gt;1&lt;/math&gt; dollar socks, &lt;math&gt;y&lt;/math&gt; pairs of &lt;math&gt;3&lt;/math&gt; dollar socks, &lt;math&gt;z&lt;/math&gt; pairs of &lt;math&gt;4&lt;/math&gt; dollar socks.<br /> <br /> We have &lt;math&gt;x+y+z=12&lt;/math&gt;, &lt;math&gt;x+3y+4z=24&lt;/math&gt;, and &lt;math&gt;x,y,z \ge 1&lt;/math&gt;.<br /> <br /> Now we subtract to find &lt;math&gt;2y+3z=12&lt;/math&gt;, and &lt;math&gt;y,z \ge 1&lt;/math&gt;.<br /> It follows that &lt;math&gt;y&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;2y&lt;/math&gt; is a multiple of &lt;math&gt;6&lt;/math&gt;, so since &lt;math&gt;0&lt;2y&lt;12&lt;/math&gt;, we must have &lt;math&gt;2y=6&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;y=3&lt;/math&gt;, and it follows that &lt;math&gt;z=2&lt;/math&gt;. Now &lt;math&gt;x=12-y-z=12-3-2=\boxed{\textbf{(D)}~7}&lt;/math&gt;, as desired.<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_18&diff=73163 2015 AMC 8 Problems/Problem 18 2015-11-25T22:37:19Z <p>Ariedel: </p> <hr /> <div>An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, &lt;math&gt;2,5,8,11,14&lt;/math&gt; is an arithmetic sequence with five terms, in which the first term is &lt;math&gt;2&lt;/math&gt; and the constant added is &lt;math&gt;3&lt;/math&gt;. Each row and each column in this &lt;math&gt;5\times5&lt;/math&gt; array is an arithmetic sequence with five terms. What is the value of &lt;math&gt;X&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(3.85cm);<br /> label(&quot;$X$&quot;,(2.5,2.1),N);<br /> for (int i=0; i&lt;=5; ++i)<br /> draw((i,0)--(i,5), linewidth(.5));<br /> <br /> for (int j=0; j&lt;=5; ++j)<br /> draw((0,j)--(5,j), linewidth(.5));<br /> void draw_num(pair ll_corner, int num) <br /> {<br /> label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt));<br /> }<br /> <br /> draw_num((0,0), 17);<br /> draw_num((4, 0), 81);<br /> <br /> draw_num((0, 4), 1);<br /> <br /> draw_num((4,4), 25);<br /> <br /> <br /> void foo(int x, int y, string n)<br /> {<br /> label(n, (x+0.5,y+0.5), p = fontsize(19pt));<br /> }<br /> <br /> foo(2, 4, &quot; &quot;);<br /> foo(3, 4, &quot; &quot;);<br /> foo(0, 3, &quot; &quot;);<br /> foo(2, 3, &quot; &quot;);<br /> foo(1, 2, &quot; &quot;);<br /> foo(3, 2, &quot; &quot;);<br /> foo(1, 1, &quot; &quot;);<br /> foo(2, 1, &quot; &quot;);<br /> foo(3, 1, &quot; &quot;);<br /> foo(4, 1, &quot; &quot;);<br /> foo(2, 0, &quot; &quot;);<br /> foo(3, 0, &quot; &quot;);<br /> foo(0, 1, &quot; &quot;);<br /> foo(0, 2, &quot; &quot;);<br /> foo(1, 0, &quot; &quot;);<br /> foo(1, 3, &quot; &quot;);<br /> foo(1, 4, &quot; &quot;);<br /> foo(3, 3, &quot; &quot;);<br /> foo(4, 2, &quot; &quot;);<br /> foo(4, 3, &quot; &quot;);<br /> <br /> <br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> We begin filling in the table. The top row has a first term 1 and a fifth term 25, so we have the common difference is &lt;math&gt;\frac{25-1}4=6&lt;/math&gt;. This means we can fill in the first row of the table:<br /> &lt;asy&gt;<br /> size(3.85cm);<br /> label(&quot;$X$&quot;,(2.5,2.1),N);<br /> for (int i=0; i&lt;=5; ++i)<br /> draw((i,0)--(i,5), linewidth(.5));<br /> <br /> for (int j=0; j&lt;=5; ++j)<br /> draw((0,j)--(5,j), linewidth(.5));<br /> void draw_num(pair ll_corner, int num) <br /> {<br /> label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt));<br /> }<br /> <br /> draw_num((0,0), 17);<br /> draw_num((4, 0), 81);<br /> draw_num((1,4), 7);<br /> draw_num((2,4), 13);<br /> draw_num((3,4), 19);<br /> draw_num((0, 4), 1);<br /> <br /> draw_num((4,4), 25);<br /> <br /> <br /> void foo(int x, int y, string n)<br /> {<br /> label(n, (x+0.5,y+0.5), p = fontsize(19pt));<br /> }<br /> <br /> foo(2, 4, &quot; &quot;);<br /> foo(3, 4, &quot; &quot;);<br /> foo(0, 3, &quot; &quot;);<br /> foo(2, 3, &quot; &quot;);<br /> foo(1, 2, &quot; &quot;);<br /> foo(3, 2, &quot; &quot;);<br /> foo(1, 1, &quot; &quot;);<br /> foo(2, 1, &quot; &quot;);<br /> foo(3, 1, &quot; &quot;);<br /> foo(4, 1, &quot; &quot;);<br /> foo(2, 0, &quot; &quot;);<br /> foo(3, 0, &quot; &quot;);<br /> foo(0, 1, &quot; &quot;);<br /> foo(0, 2, &quot; &quot;);<br /> foo(1, 0, &quot; &quot;);<br /> foo(1, 3, &quot; &quot;);<br /> foo(1, 4, &quot; &quot;);<br /> foo(3, 3, &quot; &quot;);<br /> foo(4, 2, &quot; &quot;);<br /> foo(4, 3, &quot; &quot;);<br /> <br /> <br /> &lt;/asy&gt;<br /> <br /> The fifth row has a first term of 17 and a fifth term of 81, so the common difference is &lt;math&gt;\frac{81-17}4=16&lt;/math&gt;. We can fill in the fifth row of the table as shown:<br /> &lt;asy&gt;<br /> size(3.85cm);<br /> label(&quot;$X$&quot;,(2.5,2.1),N);<br /> for (int i=0; i&lt;=5; ++i)<br /> draw((i,0)--(i,5), linewidth(.5));<br /> <br /> for (int j=0; j&lt;=5; ++j)<br /> draw((0,j)--(5,j), linewidth(.5));<br /> void draw_num(pair ll_corner, int num) <br /> {<br /> label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt));<br /> }<br /> <br /> draw_num((0,0), 17);<br /> draw_num((4, 0), 81);<br /> draw_num((1,4), 7);<br /> draw_num((2,4), 13);<br /> draw_num((3,4), 19);<br /> draw_num((4, 4), 25);<br /> draw_num((0, 4), 1);<br /> draw_num((1, 0), 33);<br /> draw_num((2, 0), 49);<br /> draw_num((3, 0), 65);<br /> <br /> <br /> <br /> void foo(int x, int y, string n)<br /> {<br /> label(n, (x+0.5,y+0.5), p = fontsize(19pt));<br /> }<br /> <br /> foo(2, 4, &quot; &quot;);<br /> foo(3, 4, &quot; &quot;);<br /> foo(0, 3, &quot; &quot;);<br /> foo(2, 3, &quot; &quot;);<br /> foo(1, 2, &quot; &quot;);<br /> foo(3, 2, &quot; &quot;);<br /> foo(1, 1, &quot; &quot;);<br /> foo(2, 1, &quot; &quot;);<br /> foo(3, 1, &quot; &quot;);<br /> foo(4, 1, &quot; &quot;);<br /> foo(2, 0, &quot; &quot;);<br /> foo(3, 0, &quot; &quot;);<br /> foo(0, 1, &quot; &quot;);<br /> foo(0, 2, &quot; &quot;);<br /> foo(1, 0, &quot; &quot;);<br /> foo(1, 3, &quot; &quot;);<br /> foo(1, 4, &quot; &quot;);<br /> foo(3, 3, &quot; &quot;);<br /> foo(4, 2, &quot; &quot;);<br /> foo(4, 3, &quot; &quot;);<br /> <br /> <br /> &lt;/asy&gt;<br /> <br /> We must find the third term of the arithmetic sequence with a first term of 13 and a fifth term of 49. The common difference of this sequence is &lt;math&gt;\frac{49-13}4=9&lt;/math&gt;, so the third term is &lt;math&gt;13+2\cdot 9=\boxed{\textbf{(B) }31}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The middle term of the first row is &lt;math&gt;\frac{25+1}{2}=13&lt;/math&gt;, since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is &lt;math&gt;\frac{17+81}{2}=49&lt;/math&gt;. Applying this again for the middle column, the answer is &lt;math&gt;\frac{49+13}{2}=\boxed{\textbf{(B)}~31}&lt;/math&gt;.<br /> ==See Also==<br /> <br /> <br /> {{AMC8 box|year=2015|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_18&diff=73162 2015 AMC 8 Problems/Problem 18 2015-11-25T22:36:40Z <p>Ariedel: </p> <hr /> <div>An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, &lt;math&gt;2,5,8,11,14&lt;/math&gt; is an arithmetic sequence with five terms, in which the first term is &lt;math&gt;2&lt;/math&gt; and the constant added is &lt;math&gt;3&lt;/math&gt;. Each row and each column in this &lt;math&gt;5\times5&lt;/math&gt; array is an arithmetic sequence with five terms. What is the value of &lt;math&gt;X&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(3.85cm);<br /> label(&quot;$X$&quot;,(2.5,2.1),N);<br /> for (int i=0; i&lt;=5; ++i)<br /> draw((i,0)--(i,5), linewidth(.5));<br /> <br /> for (int j=0; j&lt;=5; ++j)<br /> draw((0,j)--(5,j), linewidth(.5));<br /> void draw_num(pair ll_corner, int num) <br /> {<br /> label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt));<br /> }<br /> <br /> draw_num((0,0), 17);<br /> draw_num((4, 0), 81);<br /> <br /> draw_num((0, 4), 1);<br /> <br /> draw_num((4,4), 25);<br /> <br /> <br /> void foo(int x, int y, string n)<br /> {<br /> label(n, (x+0.5,y+0.5), p = fontsize(19pt));<br /> }<br /> <br /> foo(2, 4, &quot; &quot;);<br /> foo(3, 4, &quot; &quot;);<br /> foo(0, 3, &quot; &quot;);<br /> foo(2, 3, &quot; &quot;);<br /> foo(1, 2, &quot; &quot;);<br /> foo(3, 2, &quot; &quot;);<br /> foo(1, 1, &quot; &quot;);<br /> foo(2, 1, &quot; &quot;);<br /> foo(3, 1, &quot; &quot;);<br /> foo(4, 1, &quot; &quot;);<br /> foo(2, 0, &quot; &quot;);<br /> foo(3, 0, &quot; &quot;);<br /> foo(0, 1, &quot; &quot;);<br /> foo(0, 2, &quot; &quot;);<br /> foo(1, 0, &quot; &quot;);<br /> foo(1, 3, &quot; &quot;);<br /> foo(1, 4, &quot; &quot;);<br /> foo(3, 3, &quot; &quot;);<br /> foo(4, 2, &quot; &quot;);<br /> foo(4, 3, &quot; &quot;);<br /> <br /> <br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> We begin filling in the table. The top row has a first term 1 and a fifth term 25, so we have the common difference is &lt;math&gt;\frac{25-1}4=6&lt;/math&gt;. This means we can fill in the first row of the table:<br /> &lt;asy&gt;<br /> size(3.85cm);<br /> label(&quot;$X$&quot;,(2.5,2.1),N);<br /> for (int i=0; i&lt;=5; ++i)<br /> draw((i,0)--(i,5), linewidth(.5));<br /> <br /> for (int j=0; j&lt;=5; ++j)<br /> draw((0,j)--(5,j), linewidth(.5));<br /> void draw_num(pair ll_corner, int num) <br /> {<br /> label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt));<br /> }<br /> <br /> draw_num((0,0), 17);<br /> draw_num((4, 0), 81);<br /> draw_num((1,4), 7);<br /> draw_num((2,4), 13);<br /> draw_num((3,4), 19);<br /> draw_num((0, 4), 1);<br /> <br /> draw_num((4,4), 25);<br /> <br /> <br /> void foo(int x, int y, string n)<br /> {<br /> label(n, (x+0.5,y+0.5), p = fontsize(19pt));<br /> }<br /> <br /> foo(2, 4, &quot; &quot;);<br /> foo(3, 4, &quot; &quot;);<br /> foo(0, 3, &quot; &quot;);<br /> foo(2, 3, &quot; &quot;);<br /> foo(1, 2, &quot; &quot;);<br /> foo(3, 2, &quot; &quot;);<br /> foo(1, 1, &quot; &quot;);<br /> foo(2, 1, &quot; &quot;);<br /> foo(3, 1, &quot; &quot;);<br /> foo(4, 1, &quot; &quot;);<br /> foo(2, 0, &quot; &quot;);<br /> foo(3, 0, &quot; &quot;);<br /> foo(0, 1, &quot; &quot;);<br /> foo(0, 2, &quot; &quot;);<br /> foo(1, 0, &quot; &quot;);<br /> foo(1, 3, &quot; &quot;);<br /> foo(1, 4, &quot; &quot;);<br /> foo(3, 3, &quot; &quot;);<br /> foo(4, 2, &quot; &quot;);<br /> foo(4, 3, &quot; &quot;);<br /> <br /> <br /> &lt;/asy&gt;<br /> <br /> The fifth row has a first term of 17 and a fifth term of 81, so the common difference is &lt;math&gt;\frac{81-17}4=16&lt;/math&gt;. We can fill in the fifth row of the table as shown:<br /> &lt;asy&gt;<br /> size(3.85cm);<br /> label(&quot;$X$&quot;,(2.5,2.1),N);<br /> for (int i=0; i&lt;=5; ++i)<br /> draw((i,0)--(i,5), linewidth(.5));<br /> <br /> for (int j=0; j&lt;=5; ++j)<br /> draw((0,j)--(5,j), linewidth(.5));<br /> void draw_num(pair ll_corner, int num) <br /> {<br /> label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt));<br /> }<br /> <br /> draw_num((0,0), 17);<br /> draw_num((4, 0), 81);<br /> draw_num((1,4), 7);<br /> draw_num((2,4), 13);<br /> draw_num((3,4), 19);<br /> draw_num((4, 4), 25);<br /> draw_num((0, 4), 1);<br /> draw_num((1, 0), 33);<br /> draw_num((2, 0), 49);<br /> draw_num((3, 0), 65);<br /> <br /> <br /> <br /> void foo(int x, int y, string n)<br /> {<br /> label(n, (x+0.5,y+0.5), p = fontsize(19pt));<br /> }<br /> <br /> foo(2, 4, &quot; &quot;);<br /> foo(3, 4, &quot; &quot;);<br /> foo(0, 3, &quot; &quot;);<br /> foo(2, 3, &quot; &quot;);<br /> foo(1, 2, &quot; &quot;);<br /> foo(3, 2, &quot; &quot;);<br /> foo(1, 1, &quot; &quot;);<br /> foo(2, 1, &quot; &quot;);<br /> foo(3, 1, &quot; &quot;);<br /> foo(4, 1, &quot; &quot;);<br /> foo(2, 0, &quot; &quot;);<br /> foo(3, 0, &quot; &quot;);<br /> foo(0, 1, &quot; &quot;);<br /> foo(0, 2, &quot; &quot;);<br /> foo(1, 0, &quot; &quot;);<br /> foo(1, 3, &quot; &quot;);<br /> foo(1, 4, &quot; &quot;);<br /> foo(3, 3, &quot; &quot;);<br /> foo(4, 2, &quot; &quot;);<br /> foo(4, 3, &quot; &quot;);<br /> <br /> <br /> &lt;/asy&gt;<br /> <br /> We must find the third term of the arithmetic sequence with a first term of 13 and a fifth term of 49. The common difference of this sequence is &lt;math&gt;\frac{49-13}4=9&lt;/math&gt;, so the third term is &lt;math&gt;13+2\cdot 9=\boxed{\textbf{(B) }31}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> The middle term of the first row is &lt;math&gt;\frac{25+1}{2}=13&lt;/math&gt;, since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is &lt;math&gt;\frac{17+81}{2}=49&lt;/math&gt;. Applying this again for the middle column, the answer is &lt;math&gt;\frac{49+13}{2}=\boxed{\textbf{(B)~31}&lt;/math&gt;.<br /> ==See Also==<br /> <br /> <br /> {{AMC8 box|year=2015|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_8&diff=73161 2015 AMC 8 Problems/Problem 8 2015-11-25T22:33:28Z <p>Ariedel: </p> <hr /> <div>What is the smallest whole number larger than the perimeter of any triangle with a side of length &lt;math&gt; 5&lt;/math&gt; and a side of length &lt;math&gt;19&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57&lt;/math&gt;<br /> <br /> ===Solution===<br /> We know from the triangle inequality that the last side, &lt;math&gt;c&lt;/math&gt;, fulfills &lt;math&gt;c&lt;5+19=24&lt;/math&gt;. Adding &lt;math&gt;5+19&lt;/math&gt; to both sides of the inequality, we get &lt;math&gt;c+5+19&lt;48&lt;/math&gt;. However, we know that &lt;math&gt;c+5+19&lt;/math&gt; is the perimeter of our triangle, so we get &lt;math&gt;\boxed{\textbf{(D) }~48}&lt;/math&gt; as our answer.<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_19&diff=73160 2015 AMC 8 Problems/Problem 19 2015-11-25T22:31:20Z <p>Ariedel: /* Solution 3 */</p> <hr /> <div>A triangle with vertices as &lt;math&gt;A=(1,3)&lt;/math&gt;, &lt;math&gt;B=(5,1)&lt;/math&gt;, and &lt;math&gt;C=(4,4)&lt;/math&gt; is plotted on a &lt;math&gt;6\times5&lt;/math&gt; grid. What fraction of the grid is covered by the triangle?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> <br /> draw((1,0)--(1,5),linewidth(.5));<br /> draw((2,0)--(2,5),linewidth(.5));<br /> draw((3,0)--(3,5),linewidth(.5));<br /> draw((4,0)--(4,5),linewidth(.5));<br /> draw((5,0)--(5,5),linewidth(.5));<br /> draw((6,0)--(6,5),linewidth(.5));<br /> draw((0,1)--(6,1),linewidth(.5));<br /> draw((0,2)--(6,2),linewidth(.5));<br /> draw((0,3)--(6,3),linewidth(.5));<br /> draw((0,4)--(6,4),linewidth(.5));<br /> draw((0,5)--(6,5),linewidth(.5)); <br /> draw((0,0)--(0,6),EndArrow);<br /> draw((0,0)--(7,0),EndArrow);<br /> draw((1,3)--(4,4)--(5,1)--cycle);<br /> label(&quot;$y$&quot;,(0,6),W); label(&quot;$x$&quot;,(7,0),S);<br /> label(&quot;$A$&quot;,(1,3),dir(210)); label(&quot;$B$&quot;,(5,1),SE); label(&quot;$C$&quot;,(4,4),dir(100));<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> The area of &lt;math&gt;\triangle ABC&lt;/math&gt; is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is &lt;math&gt;\sqrt{1^2+2^2}=\sqrt{5}&lt;/math&gt;, and its base is &lt;math&gt;\sqrt{2^2+4^2}=\sqrt{20}&lt;/math&gt;. We multiply these and divide by 2 to find the of the triangle is &lt;math&gt;\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5&lt;/math&gt;. Since the grid has an area of &lt;math&gt;30&lt;/math&gt;, the fraction of the grid covered by the triangle is &lt;math&gt;\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Note angle &lt;math&gt;\angle ACB&lt;/math&gt; is right, thus the area is &lt;math&gt;\sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=(10)\times \dfrac{1}{2}=5&lt;/math&gt; thus the fraction of the total is &lt;math&gt;\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> By the shoelace theorem the area of &lt;math&gt;\triangle ABC=|\dfrac{1}{2}(15+4+4-1-20-12)|=|\dfrac{1}{2}(-10)|=5&lt;/math&gt;<br /> This means the fraction of the total area is &lt;math&gt;\boxed{\textbf{(A)}~\dfrac{1}{6}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_19&diff=73159 2015 AMC 8 Problems/Problem 19 2015-11-25T22:30:52Z <p>Ariedel: </p> <hr /> <div>A triangle with vertices as &lt;math&gt;A=(1,3)&lt;/math&gt;, &lt;math&gt;B=(5,1)&lt;/math&gt;, and &lt;math&gt;C=(4,4)&lt;/math&gt; is plotted on a &lt;math&gt;6\times5&lt;/math&gt; grid. What fraction of the grid is covered by the triangle?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> <br /> draw((1,0)--(1,5),linewidth(.5));<br /> draw((2,0)--(2,5),linewidth(.5));<br /> draw((3,0)--(3,5),linewidth(.5));<br /> draw((4,0)--(4,5),linewidth(.5));<br /> draw((5,0)--(5,5),linewidth(.5));<br /> draw((6,0)--(6,5),linewidth(.5));<br /> draw((0,1)--(6,1),linewidth(.5));<br /> draw((0,2)--(6,2),linewidth(.5));<br /> draw((0,3)--(6,3),linewidth(.5));<br /> draw((0,4)--(6,4),linewidth(.5));<br /> draw((0,5)--(6,5),linewidth(.5)); <br /> draw((0,0)--(0,6),EndArrow);<br /> draw((0,0)--(7,0),EndArrow);<br /> draw((1,3)--(4,4)--(5,1)--cycle);<br /> label(&quot;$y$&quot;,(0,6),W); label(&quot;$x$&quot;,(7,0),S);<br /> label(&quot;$A$&quot;,(1,3),dir(210)); label(&quot;$B$&quot;,(5,1),SE); label(&quot;$C$&quot;,(4,4),dir(100));<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> The area of &lt;math&gt;\triangle ABC&lt;/math&gt; is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is &lt;math&gt;\sqrt{1^2+2^2}=\sqrt{5}&lt;/math&gt;, and its base is &lt;math&gt;\sqrt{2^2+4^2}=\sqrt{20}&lt;/math&gt;. We multiply these and divide by 2 to find the of the triangle is &lt;math&gt;\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5&lt;/math&gt;. Since the grid has an area of &lt;math&gt;30&lt;/math&gt;, the fraction of the grid covered by the triangle is &lt;math&gt;\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Note angle &lt;math&gt;\angle ACB&lt;/math&gt; is right, thus the area is &lt;math&gt;\sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=(10)\times \dfrac{1}{2}=5&lt;/math&gt; thus the fraction of the total is &lt;math&gt;\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> By the shoelace theorem the area of &lt;math&gt;\triangle ABC=|\dfrac{1}{2}(15+4+4-1-20-12}|=|\dfrac{1}{2}(-10)|=5&lt;/math&gt;<br /> This means the fraction of the total area is &lt;math&gt;\boxed{\textbf{(A)}~\dfrac{1}{6}}&lt;/math&gt;<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_19&diff=73156 2015 AMC 8 Problems/Problem 19 2015-11-25T22:26:59Z <p>Ariedel: /* Solution 2 */</p> <hr /> <div>A triangle with vertices as &lt;math&gt;A=(1,3)&lt;/math&gt;, &lt;math&gt;B=(5,1)&lt;/math&gt;, and &lt;math&gt;C=(4,4)&lt;/math&gt; is plotted on a &lt;math&gt;6\times5&lt;/math&gt; grid. What fraction of the grid is covered by the triangle?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> <br /> draw((1,0)--(1,5),linewidth(.5));<br /> draw((2,0)--(2,5),linewidth(.5));<br /> draw((3,0)--(3,5),linewidth(.5));<br /> draw((4,0)--(4,5),linewidth(.5));<br /> draw((5,0)--(5,5),linewidth(.5));<br /> draw((6,0)--(6,5),linewidth(.5));<br /> draw((0,1)--(6,1),linewidth(.5));<br /> draw((0,2)--(6,2),linewidth(.5));<br /> draw((0,3)--(6,3),linewidth(.5));<br /> draw((0,4)--(6,4),linewidth(.5));<br /> draw((0,5)--(6,5),linewidth(.5)); <br /> draw((0,0)--(0,6),EndArrow);<br /> draw((0,0)--(7,0),EndArrow);<br /> draw((1,3)--(4,4)--(5,1)--cycle);<br /> label(&quot;$y$&quot;,(0,6),W); label(&quot;$x$&quot;,(7,0),S);<br /> label(&quot;$A$&quot;,(1,3),dir(210)); label(&quot;$B$&quot;,(5,1),SE); label(&quot;$C$&quot;,(4,4),dir(100));<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> The area of &lt;math&gt;\triangle ABC&lt;/math&gt; is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is &lt;math&gt;\sqrt{1^2+2^2}=\sqrt{5}&lt;/math&gt;, and its base is &lt;math&gt;\sqrt{2^2+4^2}=\sqrt{20}&lt;/math&gt;. We multiply these and divide by 2 to find the of the triangle is &lt;math&gt;\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5&lt;/math&gt;. Since the grid has an area of &lt;math&gt;30&lt;/math&gt;, the fraction of the grid covered by the triangle is &lt;math&gt;\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Note angle &lt;math&gt;\angle ACB&lt;/math&gt; is right, thus the area is &lt;math&gt;\sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=(10)\times \dfrac{1}{2}=5&lt;/math&gt; thus the fraction of the total is &lt;math&gt;\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_19&diff=73155 2015 AMC 8 Problems/Problem 19 2015-11-25T22:26:29Z <p>Ariedel: </p> <hr /> <div>A triangle with vertices as &lt;math&gt;A=(1,3)&lt;/math&gt;, &lt;math&gt;B=(5,1)&lt;/math&gt;, and &lt;math&gt;C=(4,4)&lt;/math&gt; is plotted on a &lt;math&gt;6\times5&lt;/math&gt; grid. What fraction of the grid is covered by the triangle?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> <br /> draw((1,0)--(1,5),linewidth(.5));<br /> draw((2,0)--(2,5),linewidth(.5));<br /> draw((3,0)--(3,5),linewidth(.5));<br /> draw((4,0)--(4,5),linewidth(.5));<br /> draw((5,0)--(5,5),linewidth(.5));<br /> draw((6,0)--(6,5),linewidth(.5));<br /> draw((0,1)--(6,1),linewidth(.5));<br /> draw((0,2)--(6,2),linewidth(.5));<br /> draw((0,3)--(6,3),linewidth(.5));<br /> draw((0,4)--(6,4),linewidth(.5));<br /> draw((0,5)--(6,5),linewidth(.5)); <br /> draw((0,0)--(0,6),EndArrow);<br /> draw((0,0)--(7,0),EndArrow);<br /> draw((1,3)--(4,4)--(5,1)--cycle);<br /> label(&quot;$y$&quot;,(0,6),W); label(&quot;$x$&quot;,(7,0),S);<br /> label(&quot;$A$&quot;,(1,3),dir(210)); label(&quot;$B$&quot;,(5,1),SE); label(&quot;$C$&quot;,(4,4),dir(100));<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> The area of &lt;math&gt;\triangle ABC&lt;/math&gt; is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is &lt;math&gt;\sqrt{1^2+2^2}=\sqrt{5}&lt;/math&gt;, and its base is &lt;math&gt;\sqrt{2^2+4^2}=\sqrt{20}&lt;/math&gt;. We multiply these and divide by 2 to find the of the triangle is &lt;math&gt;\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5&lt;/math&gt;. Since the grid has an area of &lt;math&gt;30&lt;/math&gt;, the fraction of the grid covered by the triangle is &lt;math&gt;\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Note angle &lt;math&gt;\angle ACB&lt;/math&gt; is right this the area is &lt;math&gt;\sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=(10)\times \dfrac{1}{2}=5&lt;/math&gt; thus the fraction of the total is &lt;math&gt;\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}&lt;/math&gt;<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_11&diff=73153 2015 AMC 8 Problems/Problem 11 2015-11-25T22:21:43Z <p>Ariedel: /* Solution 2 */</p> <hr /> <div>In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read &quot;AMC8&quot;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } \frac{1}{22,050} \qquad<br /> \textbf{(B) } \frac{1}{21,000}\qquad<br /> \textbf{(C) } \frac{1}{10,500}\qquad<br /> \textbf{(D) } \frac{1}{2,100} \qquad<br /> \textbf{(E) } \frac{1}{1,050}<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There is one favorable case, which is the license plate says &quot;AMC8&quot;. We must now find how many total cases there are. There are &lt;math&gt;5&lt;/math&gt; choices for the first letter (since it must be a vowel), &lt;math&gt;21&lt;/math&gt; choices for the second letter (since it must be of 21 consonants), &lt;math&gt;20&lt;/math&gt; choices for the third letter (since it must differ from the second letter), and &lt;math&gt;10&lt;/math&gt; choices for the number. This leads to &lt;math&gt;5 \cdot 21 \cdot 20 \cdot 10=21000&lt;/math&gt; total possible license plates. That means the probability of a license plate saying &quot;AMC8&quot; is &lt;math&gt;\boxed{\textbf{(B) } \frac{1}{21,000}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> The probability of choosing A as the first letter is &lt;math&gt;\dfrac{1}{5}&lt;/math&gt;. The probability of choosing &lt;math&gt;M&lt;/math&gt; next is &lt;math&gt;\dfrac{1}{21}&lt;/math&gt;. The probability of choosing C as the third letter is &lt;math&gt;\dfrac{1}{20}&lt;/math&gt; (since there are 20 other consonants to choos from other then M). The probability of having &lt;math&gt;8&lt;/math&gt; as the last number is &lt;math&gt;\dfrac{1}{10}&lt;/math&gt;. We multiply all these to obtain &lt;math&gt;\dfrac{1}{5}\cdot \dfrac{1}{21} \cdot \dfrac{1}{20}\cdot \dfrac{1}{10}=\dfrac{1}{5\times 21\times 20\times 10}=\dfrac{1}{21\times 100\times 10}=\boxed{\textbf{(B)}~\dfrac{1}{21000}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_11&diff=73152 2015 AMC 8 Problems/Problem 11 2015-11-25T22:20:25Z <p>Ariedel: /* Solution 2 */</p> <hr /> <div>In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read &quot;AMC8&quot;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } \frac{1}{22,050} \qquad<br /> \textbf{(B) } \frac{1}{21,000}\qquad<br /> \textbf{(C) } \frac{1}{10,500}\qquad<br /> \textbf{(D) } \frac{1}{2,100} \qquad<br /> \textbf{(E) } \frac{1}{1,050}<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There is one favorable case, which is the license plate says &quot;AMC8&quot;. We must now find how many total cases there are. There are &lt;math&gt;5&lt;/math&gt; choices for the first letter (since it must be a vowel), &lt;math&gt;21&lt;/math&gt; choices for the second letter (since it must be of 21 consonants), &lt;math&gt;20&lt;/math&gt; choices for the third letter (since it must differ from the second letter), and &lt;math&gt;10&lt;/math&gt; choices for the number. This leads to &lt;math&gt;5 \cdot 21 \cdot 20 \cdot 10=21000&lt;/math&gt; total possible license plates. That means the probability of a license plate saying &quot;AMC8&quot; is &lt;math&gt;\boxed{\textbf{(B) } \frac{1}{21,000}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> The probability of choosing A as the first letter is &lt;math&gt;\dfrac{1}{5}&lt;/math&gt;. The probability of choosing &lt;math&gt;M&lt;/math&gt; next is &lt;math&gt;\dfrac{1}{21}&lt;/math&gt;. The probability of choosing C as the third letter is &lt;math&gt;\dfrac{1}{20}&lt;/math&gt; (since there are 20 other consonants to choos from other then M). The probability of having &lt;math&gt;8&lt;/math&gt; as the last number is &lt;math&gt;\dfrac{1}{10}&lt;/math&gt;. We multiply all these to obtain &lt;math&gt;\dfrac{1}{5}\cdot \dfrac{1}{21} \cdot \dfrac{1}{29}\cdot \dfrac{1}{10}=\dfrac{1}{5}\times 21\times 20\times 10}=\dfrac{1}{21\times 100\times 10}=\boxed{\textbf{(B)}~\dfrac{1}{21000}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_11&diff=73151 2015 AMC 8 Problems/Problem 11 2015-11-25T22:20:09Z <p>Ariedel: /* Solution 2 */</p> <hr /> <div>In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read &quot;AMC8&quot;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } \frac{1}{22,050} \qquad<br /> \textbf{(B) } \frac{1}{21,000}\qquad<br /> \textbf{(C) } \frac{1}{10,500}\qquad<br /> \textbf{(D) } \frac{1}{2,100} \qquad<br /> \textbf{(E) } \frac{1}{1,050}<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There is one favorable case, which is the license plate says &quot;AMC8&quot;. We must now find how many total cases there are. There are &lt;math&gt;5&lt;/math&gt; choices for the first letter (since it must be a vowel), &lt;math&gt;21&lt;/math&gt; choices for the second letter (since it must be of 21 consonants), &lt;math&gt;20&lt;/math&gt; choices for the third letter (since it must differ from the second letter), and &lt;math&gt;10&lt;/math&gt; choices for the number. This leads to &lt;math&gt;5 \cdot 21 \cdot 20 \cdot 10=21000&lt;/math&gt; total possible license plates. That means the probability of a license plate saying &quot;AMC8&quot; is &lt;math&gt;\boxed{\textbf{(B) } \frac{1}{21,000}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> The probability of choosing A as the first letter is &lt;math&gt;\dfrac{1}{5}&lt;/math&gt;. The probability of choosing &lt;math&gt;M&lt;/math&gt; next is &lt;math&gt;\dfrac{1}{21}&lt;/math&gt;. The probability of choosing C as the third letter is &lt;math&gt;\dfrac{1}{20}&lt;/math&gt; (since there are 20 other consonants to choos from other then M). The probability of having &lt;math&gt;8&lt;/math&gt; as the last number is &lt;math&gt;\dfrac{1}{10}&lt;/math&gt;. We multiply all these to obtain &lt;math&gt;\dfrac{1}{5}\cdot \dfrac{1}{21} \cdot \dfrac{1}{29}\cdot \dfrac{1}{10}=\dfrac{1}{5}\times 21\times 20\times 10}=\dfrac{1}{21\times 100\times 10}=\boxed{\textbf{(B)}~\dfrac{1}{21000}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_11&diff=73150 2015 AMC 8 Problems/Problem 11 2015-11-25T22:19:30Z <p>Ariedel: </p> <hr /> <div>In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read &quot;AMC8&quot;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } \frac{1}{22,050} \qquad<br /> \textbf{(B) } \frac{1}{21,000}\qquad<br /> \textbf{(C) } \frac{1}{10,500}\qquad<br /> \textbf{(D) } \frac{1}{2,100} \qquad<br /> \textbf{(E) } \frac{1}{1,050}<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There is one favorable case, which is the license plate says &quot;AMC8&quot;. We must now find how many total cases there are. There are &lt;math&gt;5&lt;/math&gt; choices for the first letter (since it must be a vowel), &lt;math&gt;21&lt;/math&gt; choices for the second letter (since it must be of 21 consonants), &lt;math&gt;20&lt;/math&gt; choices for the third letter (since it must differ from the second letter), and &lt;math&gt;10&lt;/math&gt; choices for the number. This leads to &lt;math&gt;5 \cdot 21 \cdot 20 \cdot 10=21000&lt;/math&gt; total possible license plates. That means the probability of a license plate saying &quot;AMC8&quot; is &lt;math&gt;\boxed{\textbf{(B) } \frac{1}{21,000}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> The probability of choosing A as the first letter is &lt;math&gt;\dfrac{1}{5}&lt;/math&gt;. The probability of choosing &lt;math&gt;M&lt;/math&gt; next is &lt;math&gt;\dfrac{1}{21}&lt;/math&gt;. The probability of choosing C as the third letter is &lt;math&gt;\dfrac{1}{20}&lt;/math&gt; (since there are 20 other consonants to choos from other then M). The probability of having &lt;math&gt;8&lt;/math&gt; as the last number is &lt;math&gt;\dfrac{1}{10}&lt;/math&gt;. We multiply all these to obtain &lt;math&gt;\dfrac{1}{5}\cdot \dfrac{1}{21} \cdot \dfrac{1}{29}\cdot \dfrac{1}{10}=\dfrac{1}{5\times 21\times 20\times 10}=\dfrac{1}{21\times 100\times 10}=\boxed{\textbf{(B)}~\dfrac{1}{21000}&lt;/math&gt;<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_22&diff=73149 2015 AMC 8 Problems/Problem 22 2015-11-25T22:09:30Z <p>Ariedel: </p> <hr /> <div>On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 21 \qquad<br /> \textbf{(B) } 30 \qquad<br /> \textbf{(C) } 60 \qquad<br /> \textbf{(D) } 90 \qquad<br /> \textbf{(E) } 1080<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> The given information means that the number of students in the match has &lt;math&gt;12&lt;/math&gt; factors. The least integer with &lt;math&gt;12&lt;/math&gt; factors is &lt;math&gt;2^2\cdot3\cdot5=60\implies\boxed{\textbf{(C) }~60}&lt;/math&gt;.<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=73147 2015 AMC 8 Problems/Problem 17 2015-11-25T22:08:11Z <p>Ariedel: /* Solution 3 */</p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> So &lt;math&gt;\frac{d}{v}=\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{d}{v+18}=\frac{1}{5}&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;d=\frac{1}{5}v+3.6=\frac{1}{3}v&lt;/math&gt;, which gives &lt;math&gt;v=27&lt;/math&gt;, which then gives &lt;math&gt;d=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> &lt;math&gt;d = rt&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt; is obviously constant<br /> <br /> &lt;math&gt;\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{r}{3} = \frac{r}{5} + \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2r}{15} = \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;10r = 270&lt;/math&gt; so &lt;math&gt;r = 27&lt;/math&gt;, plug into the first one and it's &lt;math&gt;\boxed{\textbf{(D)}~9}&lt;/math&gt; miles to school<br /> <br /> ===Solution 3===<br /> We set up an equation in terms of &lt;math&gt;d&lt;/math&gt; the distance and &lt;math&gt;x&lt;/math&gt; the speed In miles per hour. We have &lt;math&gt;d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)&lt;/math&gt;<br /> &lt;cmath&gt;d=(5)(x)=(3)(x+18)&lt;/cmath&gt;<br /> &lt;cmath&gt;5x=3x+54&lt;/cmath&gt;<br /> &lt;cmath&gt;2x=54&lt;/cmath&gt;<br /> &lt;cmath&gt;x=27&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;d=\dfrac{27}{3}=\boxed{\textbf{(D)},~9}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=73146 2015 AMC 8 Problems/Problem 17 2015-11-25T22:07:46Z <p>Ariedel: /* Solution 3 */</p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> So &lt;math&gt;\frac{d}{v}=\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{d}{v+18}=\frac{1}{5}&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;d=\frac{1}{5}v+3.6=\frac{1}{3}v&lt;/math&gt;, which gives &lt;math&gt;v=27&lt;/math&gt;, which then gives &lt;math&gt;d=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> &lt;math&gt;d = rt&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt; is obviously constant<br /> <br /> &lt;math&gt;\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{r}{3} = \frac{r}{5} + \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2r}{15} = \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;10r = 270&lt;/math&gt; so &lt;math&gt;r = 27&lt;/math&gt;, plug into the first one and it's &lt;math&gt;\boxed{\textbf{(D)}~9}&lt;/math&gt; miles to school<br /> <br /> ===Solution 3===<br /> We set up an equation in terms of &lt;math&gt;d&lt;/math&gt; the distance and &lt;math&gt;x&lt;/math&gt; the speed In miles per hour. We have &lt;math&gt;\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)&lt;/math&gt;<br /> &lt;cmath&gt;d=(5)(x)=(3)(x+18)&lt;/cmath&gt;<br /> &lt;cmath&gt;5x=3x+54&lt;/cmath&gt;<br /> &lt;cmath&gt;2x=54&lt;/cmath&gt;<br /> &lt;cmath&gt;x=27&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;d=\dfrac{27}{3}=\boxed{\textbf{(D)},~9}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=73145 2015 AMC 8 Problems/Problem 17 2015-11-25T22:06:58Z <p>Ariedel: </p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> So &lt;math&gt;\frac{d}{v}=\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{d}{v+18}=\frac{1}{5}&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;d=\frac{1}{5}v+3.6=\frac{1}{3}v&lt;/math&gt;, which gives &lt;math&gt;v=27&lt;/math&gt;, which then gives &lt;math&gt;d=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> &lt;math&gt;d = rt&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt; is obviously constant<br /> <br /> &lt;math&gt;\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{r}{3} = \frac{r}{5} + \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2r}{15} = \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;10r = 270&lt;/math&gt; so &lt;math&gt;r = 27&lt;/math&gt;, plug into the first one and it's &lt;math&gt;\boxed{\textbf{(D)}~9}&lt;/math&gt; miles to school<br /> <br /> ===Solution 3===<br /> We set up an equation in terms of &lt;math&gt;d&lt;/math&gt; the distance and &lt;math&gt;x&lt;/math&gt; the speed In miles per hour. We have &lt;math&gt;\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)&lt;/math&gt;<br /> &lt;math&gt;d=(5)(x)=(3)(x+18)&lt;/math&gt;<br /> &lt;math&gt;5x=3x+54&lt;/math&gt;<br /> &lt;math&gt;2x=54&lt;/math&gt;<br /> &lt;math&gt;x=27&lt;/math&gt;<br /> <br /> So &lt;math&gt;d=\dfrac{27}{3}=\boxed{\textbf{(D)},~9}&lt;/math&gt;<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=73144 2015 AMC 8 Problems/Problem 17 2015-11-25T22:06:03Z <p>Ariedel: /* Solution 2 */</p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> So &lt;math&gt;\frac{d}{v}=\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{d}{v+18}=\frac{1}{5}&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;d=\frac{1}{5}v+3.6=\frac{1}{3}v&lt;/math&gt;, which gives &lt;math&gt;v=27&lt;/math&gt;, which then gives &lt;math&gt;d=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> &lt;math&gt;d = rt&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt; is obviously constant<br /> <br /> &lt;math&gt;\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{r}{3} = \frac{r}{5} + \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2r}{15} = \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;10r = 270&lt;/math&gt; so &lt;math&gt;r = 27&lt;/math&gt;, plug into the first one and it's &lt;math&gt;\boxed{\textbf{(D)}~9}&lt;/math&gt; miles to school<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=73143 2015 AMC 8 Problems/Problem 17 2015-11-25T22:05:46Z <p>Ariedel: </p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> So &lt;math&gt;\frac{d}{v}=\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{d}{v+18}=\frac{1}{5}&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;d=\frac{1}{5}v+3.6=\frac{1}{3}v&lt;/math&gt;, which gives &lt;math&gt;v=27&lt;/math&gt;, which then gives &lt;math&gt;d=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> &lt;math&gt;d = rt&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt; is obviously constant<br /> <br /> &lt;math&gt;\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{r}{3} = \frac{r}{5} + \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2r}{15} = \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;10r = 270&lt;/math&gt; so &lt;math&gt;r = 27&lt;/math&gt;, plug into the first one and it's &lt;math&gt;\boxed{\textbf{B}~9}&lt;/math&gt; miles to school<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=73142 2015 AMC 8 Problems/Problem 17 2015-11-25T22:04:40Z <p>Ariedel: </p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> So &lt;math&gt;\frac{d}{v}=\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{d}{v+18}=\frac{1}{5}&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;d=\frac{1}{5}v+3.6=\frac{1}{3}v&lt;/math&gt;, which gives &lt;math&gt;v=27&lt;/math&gt;, which then gives &lt;math&gt;d=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_9&diff=73140 2015 AMC 8 Problems/Problem 9 2015-11-25T21:57:16Z <p>Ariedel: </p> <hr /> <div>On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working &lt;math&gt;20&lt;/math&gt; days?<br /> <br /> &lt;math&gt;\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> The sum of &lt;math&gt;1,3,5, ........ 39&lt;/math&gt; is &lt;math&gt;\dfrac{(1 + 39)(20)}{2}= \boxed{\textbf{(D)}~400}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> The sum is just the sum of the first 20 odd integers, which is &lt;math&gt;20^2=\boxed{\textbf{(D)}~400}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_9&diff=73139 2015 AMC 8 Problems/Problem 9 2015-11-25T21:56:23Z <p>Ariedel: </p> <hr /> <div>On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working &lt;math&gt;20&lt;/math&gt; days?<br /> <br /> &lt;math&gt;\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> The sum of &lt;math&gt;1,3,5, ........ 39&lt;/math&gt; is &lt;math&gt;\dfrac{(1 + 39)(20)}{2}= \boxed{\textbf{(D)}~400}&lt;/math&gt;<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_7&diff=73133 2015 AMC 8 Problems/Problem 7 2015-11-25T21:39:12Z <p>Ariedel: </p> <hr /> <div>Each of two boxes contains three chips numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{2}{9}\qquad\textbf{(C) }\frac{4}{9}\qquad\textbf{(D) }\frac{1}{2}\qquad \textbf{(E) }\frac{5}{9}&lt;/math&gt;<br /> <br /> ===Solution===<br /> We can instead find the probability that their product is odd, and subtract this from &lt;math&gt;1&lt;/math&gt;. In order to get an odd product, we have to draw an odd number from each box. We have a &lt;math&gt;\frac{2}{3}&lt;/math&gt; probability of drawing an odd number from one box, so there is a &lt;math&gt;{\frac{2}{3}}^2=\frac{4}{9}&lt;/math&gt; of having an odd product. Thus, there is a &lt;math&gt;1-\frac{4}{9}=\frac{5}{9}&lt;/math&gt; probability of having an even product. We get our answer to be &lt;math&gt;\boxed{\textbf{(E) }\frac{5}{9}}&lt;/math&gt;.<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_6&diff=73129 2015 AMC 8 Problems/Problem 6 2015-11-25T21:35:16Z <p>Ariedel: </p> <hr /> <div>In &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, &lt;math&gt;AB=BC=29&lt;/math&gt;, and &lt;math&gt;AC=42&lt;/math&gt;. What is the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> We know the semi-perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{29+29+42}{2}=50&lt;/math&gt;. Next, we use heron's formula to find that the area of the triangle is just &lt;math&gt;\sqrt{50(50-29)^2(50-42)}=\sqrt{50 \cdot 21^2 \cdot 8}=\textbf{(B) }420.&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> Splitting the isosceles triangle in half, we get a right triangle with hypotenuse 29 and leg 21. By Pythagorean Theorem the height is 20 so the area is<br /> &lt;math&gt;\dfrac{(20)(42)}{2} = \boxed{\textbf{(B)}~420}&lt;/math&gt;.<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_6&diff=73125 2015 AMC 8 Problems/Problem 6 2015-11-25T21:33:38Z <p>Ariedel: </p> <hr /> <div>In &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, &lt;math&gt;AB=BC=29&lt;/math&gt;, and &lt;math&gt;AC=42&lt;/math&gt;. What is the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> We know the semi-perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{29+29+42}{2}=50&lt;/math&gt;. Next, we use heron's formula to find that the area of the triangle is just &lt;math&gt;\sqrt{50(50-29)^2(50-42)}=\sqrt{50 \cdot 21^2 \cdot 8}=\textbf{(B) }420.&lt;/math&gt;<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_4&diff=73122 2015 AMC 8 Problems/Problem 4 2015-11-25T21:32:46Z <p>Ariedel: </p> <hr /> <div>The Centerville Middle School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }12&lt;/math&gt;<br /> <br /> ===Solution===<br /> There are &lt;math&gt;2&lt;/math&gt; ways to order the boys on the end, and there are &lt;math&gt;3!=6&lt;/math&gt; ways to order the girls in the middle. We get the answer to be &lt;math&gt;2 \cdot 6 = \textbf{(E) }12&lt;/math&gt;.<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_14&diff=73120 2015 AMC 8 Problems/Problem 14 2015-11-25T21:31:27Z <p>Ariedel: /* Solution 2 */</p> <hr /> <div>Which of the following integers cannot be written as the sum of four consecutive odd integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> Let our &lt;math&gt;4&lt;/math&gt; numbers be &lt;math&gt;n, n+2, n+4, n+6&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is odd. Then our sum is &lt;math&gt;4n+12&lt;/math&gt;. The only answer choice that cannot be written as &lt;math&gt;4n+12&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is odd, is &lt;math&gt;\boxed{\textbf{(D)}\text{ 100}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> If the four consecutive odd integers are &lt;math&gt;2n-3,~ 2n-1, ~2n+1&lt;/math&gt; and &lt;math&gt;2n+3&lt;/math&gt; then the sum is &lt;math&gt;8n&lt;/math&gt;. All the integers are divisible by &lt;math&gt;8&lt;/math&gt; except &lt;math&gt;100&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(D)}~100}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_14&diff=73117 2015 AMC 8 Problems/Problem 14 2015-11-25T21:31:12Z <p>Ariedel: </p> <hr /> <div>Which of the following integers cannot be written as the sum of four consecutive odd integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> Let our &lt;math&gt;4&lt;/math&gt; numbers be &lt;math&gt;n, n+2, n+4, n+6&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is odd. Then our sum is &lt;math&gt;4n+12&lt;/math&gt;. The only answer choice that cannot be written as &lt;math&gt;4n+12&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is odd, is &lt;math&gt;\boxed{\textbf{(D)}\text{ 100}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> If the four consecutive odd integers are &lt;math&gt;2n-3,~ 2n-1, ~2n+1&lt;/math&gt; and &lt;math&gt;2n+3&lt;/math&gt; then the sum is &lt;math&gt;8n&lt;/math&gt;. All the integers are divisible by &lt;math&gt;8&lt;/math&gt; except &lt;math&gt;100&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{D}~100}&lt;/math&gt;.<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_14&diff=73113 2015 AMC 8 Problems/Problem 14 2015-11-25T21:27:13Z <p>Ariedel: </p> <hr /> <div>Which of the following integers cannot be written as the sum of four consecutive odd integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}&lt;/math&gt;<br /> <br /> ===Solution===<br /> Let our &lt;math&gt;4&lt;/math&gt; numbers be &lt;math&gt;n, n+2, n+4, n+6&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is odd. Then our sum is &lt;math&gt;4n+12&lt;/math&gt;. The only answer choice that cannot be written as &lt;math&gt;4n+12&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is odd, is &lt;math&gt;\boxed{\textbf{(D)}\text{ 100}}&lt;/math&gt;.<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_3&diff=73107 2015 AMC 8 Problems/Problem 3 2015-11-25T21:25:08Z <p>Ariedel: </p> <hr /> <div>Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of &lt;math&gt;10&lt;/math&gt; miles per hour. Jack walks to the pool at a constant speed of &lt;math&gt;4&lt;/math&gt; miles per hour. How many minutes before Jack does Jill arrive?<br /> <br /> &lt;math&gt;\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> ===Solution===<br /> Jill arrives in &lt;math&gt;\dfrac{1}{10}&lt;/math&gt; of an hour, which is &lt;math&gt;6&lt;/math&gt; minutes. Jack arrives in &lt;math&gt;\dfrac{1}{4}&lt;/math&gt; of an hour which is &lt;math&gt;15&lt;/math&gt; minutes. Thus, the time difference is &lt;math&gt;\boxed{\textbf{(D)}~9}&lt;/math&gt; minutes.<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_23&diff=73100 2015 AMC 8 Problems/Problem 23 2015-11-25T21:22:27Z <p>Ariedel: </p> <hr /> <div>Tom has twelve slips of paper which he wants to put into five cups labeled &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;. The numbers on the papers are &lt;math&gt;2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,&lt;/math&gt; and &lt;math&gt;4.5&lt;/math&gt;. If a slip with &lt;math&gt;2&lt;/math&gt; goes into cup &lt;math&gt;E&lt;/math&gt; and a slip with &lt;math&gt;3&lt;/math&gt; goes into cup &lt;math&gt;B&lt;/math&gt;, then the slip with &lt;math&gt;3.5&lt;/math&gt; must go into what cup?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } A \qquad<br /> \textbf{(B) } B \qquad<br /> \textbf{(C) } C \qquad<br /> \textbf{(D) } D \qquad<br /> \textbf{(E) } E<br /> &lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_22&diff=73099 2015 AMC 8 Problems/Problem 22 2015-11-25T21:22:04Z <p>Ariedel: </p> <hr /> <div>On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 21 \qquad<br /> \textbf{(B) } 30 \qquad<br /> \textbf{(C) } 60 \qquad<br /> \textbf{(D) } 90 \qquad<br /> \textbf{(E) } 1080<br /> &lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_21&diff=73097 2015 AMC 8 Problems/Problem 21 2015-11-25T21:21:33Z <p>Ariedel: </p> <hr /> <div>In the given figure hexagon &lt;math&gt;ABCDEF&lt;/math&gt; is equiangular, &lt;math&gt;ABJI&lt;/math&gt; and &lt;math&gt;FEHG&lt;/math&gt; are squares with areas &lt;math&gt;18&lt;/math&gt; and &lt;math&gt;32&lt;/math&gt; respectively, &lt;math&gt;\triangle JBK&lt;/math&gt; is equilateral and &lt;math&gt;FE=BC&lt;/math&gt;. What is the area of &lt;math&gt;\triangle KBC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32&lt;/math&gt;.<br /> &lt;asy&gt;<br /> draw((-4,6*sqrt(2))--(4,6*sqrt(2)));<br /> draw((-4,-6*sqrt(2))--(4,-6*sqrt(2)));<br /> draw((-8,0)--(-4,6*sqrt(2)));<br /> draw((-8,0)--(-4,-6*sqrt(2)));<br /> draw((4,6*sqrt(2))--(8,0));<br /> draw((8,0)--(4,-6*sqrt(2)));<br /> draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle);<br /> draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle);<br /> label(&quot;$I$&quot;,(-4,8+6*sqrt(2)),dir(100)); label(&quot;$J$&quot;,(4,8+6*sqrt(2)),dir(80));<br /> label(&quot;$A$&quot;,(-4,6*sqrt(2)),dir(280)); label(&quot;$B$&quot;,(4,6*sqrt(2)),dir(250));<br /> label(&quot;$C$&quot;,(8,0),W); label(&quot;$D$&quot;,(4,-6*sqrt(2)),NW); label(&quot;$E$&quot;,(-4,-6*sqrt(2)),NE); label(&quot;$F$&quot;,(-8,0),E);<br /> draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle);<br /> label(&quot;$K$&quot;,(4+4*sqrt(3),4+6*sqrt(2)),E);<br /> draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed);<br /> label(&quot;$H$&quot;,(-4-6*sqrt(2),-4-6*sqrt(2)),S);<br /> label(&quot;$G$&quot;,(-8-6*sqrt(2),-4),W);<br /> label(&quot;$32$&quot;,(-10,-8),N);<br /> label(&quot;$18$&quot;,(0,6*sqrt(2)+2),N);<br /> &lt;/asy&gt;<br /> <br /> == Solution ==<br /> <br /> Clearly, since &lt;math&gt;\overline{FE}&lt;/math&gt; is a side of a square with area &lt;math&gt;32&lt;/math&gt;, &lt;math&gt;\overline{FE} = \sqrt{32} = 4 \sqrt{2}&lt;/math&gt;. Now, since &lt;math&gt;\overline{FE} = \overline{BC}&lt;/math&gt;, we have &lt;math&gt;\overline{BC} = 4 \sqrt{2}&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;\overline{AB}&lt;/math&gt; is a side of a square with area &lt;math&gt;18&lt;/math&gt;, so &lt;math&gt;\overline{AB} = \sqrt{18} = 3 \sqrt{2}&lt;/math&gt;. Since &lt;math&gt;\Delta JBK&lt;/math&gt; is equilateral, &lt;math&gt;\overline{BK} = 3 \sqrt{2}&lt;/math&gt;.<br /> <br /> Lastly, &lt;math&gt;\Delta KBC&lt;/math&gt; is a right triangle. We see that &lt;math&gt;\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360 \rightarrow 90 + 120 + \angle CBK + 60 = 360 \rightarrow \angle CBK = 90&lt;/math&gt;, so &lt;math&gt;\Delta KBC&lt;/math&gt; is a right triangle with legs &lt;math&gt;3 \sqrt{2}&lt;/math&gt; and &lt;math&gt;4 \sqrt{2}&lt;/math&gt;. Now, its area is &lt;math&gt;\dfrac{3 \sqrt{2} \cdot 4 \sqrt{2}}{2} = \dfrac{24}{2} = 12&lt;/math&gt;, and our answer is &lt;math&gt;\boxed{\text{C}}&lt;/math&gt;.<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_20&diff=73096 2015 AMC 8 Problems/Problem 20 2015-11-25T21:20:28Z <p>Ariedel: </p> <hr /> <div>Ralph went to the store and bought 12 pairs of socks for a total of \$24. Some of the socks he bought cost \$1 a pair, some of the socks he bought cost \$3 a pair, and some of the socks he bought cost \$4 a pair. If he bought at least one pair of each type, how many pairs of \$1 socks did Ralph buy?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 5 \qquad<br /> \textbf{(C) } 6 \qquad<br /> \textbf{(D) } 7 \qquad<br /> \textbf{(E) } 8<br /> &lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_18&diff=73094 2015 AMC 8 Problems/Problem 18 2015-11-25T21:19:47Z <p>Ariedel: </p> <hr /> <div>An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, &lt;math&gt;2,5,8,11,14&lt;/math&gt; is an arithmetic sequence with five terms, in which the first term is &lt;math&gt;2&lt;/math&gt; and the constant added is &lt;math&gt;3&lt;/math&gt;. Each row and each column in this &lt;math&gt;5\times5&lt;/math&gt; array is an arithmetic sequence with five terms. What is the value of &lt;math&gt;X&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(3.85cm);<br /> label(&quot;$X$&quot;,(2.5,2.1),N);<br /> for (int i=0; i&lt;=5; ++i)<br /> draw((i,0)--(i,5), linewidth(.5));<br /> <br /> for (int j=0; j&lt;=5; ++j)<br /> draw((0,j)--(5,j), linewidth(.5));<br /> void draw_num(pair ll_corner, int num) <br /> {<br /> label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt));<br /> }<br /> <br /> draw_num((0,0), 17);<br /> draw_num((4, 0), 81);<br /> <br /> draw_num((0, 4), 1);<br /> <br /> draw_num((4,4), 25);<br /> <br /> <br /> void foo(int x, int y, string n)<br /> {<br /> label(n, (x+0.5,y+0.5), p = fontsize(19pt));<br /> }<br /> <br /> foo(2, 4, &quot; &quot;);<br /> foo(3, 4, &quot; &quot;);<br /> foo(0, 3, &quot; &quot;);<br /> foo(2, 3, &quot; &quot;);<br /> foo(1, 2, &quot; &quot;);<br /> foo(3, 2, &quot; &quot;);<br /> foo(1, 1, &quot; &quot;);<br /> foo(2, 1, &quot; &quot;);<br /> foo(3, 1, &quot; &quot;);<br /> foo(4, 1, &quot; &quot;);<br /> foo(2, 0, &quot; &quot;);<br /> foo(3, 0, &quot; &quot;);<br /> foo(0, 1, &quot; &quot;);<br /> foo(0, 2, &quot; &quot;);<br /> foo(1, 0, &quot; &quot;);<br /> foo(1, 3, &quot; &quot;);<br /> foo(1, 4, &quot; &quot;);<br /> foo(3, 3, &quot; &quot;);<br /> foo(4, 2, &quot; &quot;);<br /> foo(4, 3, &quot; &quot;);<br /> <br /> <br /> &lt;/asy&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=73093 2015 AMC 8 Problems/Problem 17 2015-11-25T21:19:19Z <p>Ariedel: </p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_16&diff=73092 2015 AMC 8 Problems/Problem 16 2015-11-25T21:18:51Z <p>Ariedel: </p> <hr /> <div>In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If &lt;math&gt;\tfrac{1}{3}&lt;/math&gt; of all the ninth graders are paired with &lt;math&gt;\tfrac{2}{5}&lt;/math&gt; of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } \frac{2}{15} \qquad<br /> \textbf{(B) } \frac{4}{11} \qquad<br /> \textbf{(C) } \frac{11}{30} \qquad<br /> \textbf{(D) } \frac{3}{8} \qquad<br /> \textbf{(E) } \frac{11}{15}<br /> &lt;/math&gt;<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_15&diff=73090 2015 AMC 8 Problems/Problem 15 2015-11-25T21:18:25Z <p>Ariedel: </p> <hr /> <div>At Euler Middle School, &lt;math&gt;198&lt;/math&gt; students voted on two issues in a school referendum with the following results: &lt;math&gt;149&lt;/math&gt; voted in favor of the first issue and &lt;math&gt;119&lt;/math&gt; voted in favor of the second issue. If there were exactly &lt;math&gt;29&lt;/math&gt; students who voted against both issues, how many students voted in favor of both issues?<br /> <br /> &lt;math&gt;\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_14&diff=73088 2015 AMC 8 Problems/Problem 14 2015-11-25T21:17:50Z <p>Ariedel: </p> <hr /> <div>Which of the following integers cannot be written as the sum of four consecutive odd integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_13&diff=73087 2015 AMC 8 Problems/Problem 13 2015-11-25T21:17:21Z <p>Ariedel: </p> <hr /> <div>How many subsets of two elements can be removed from the set &lt;math&gt;\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}&lt;/math&gt; so that the mean (average) of the remaining numbers is 6?<br /> <br /> &lt;math&gt;\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Since there will be &lt;math&gt;9&lt;/math&gt; elements after removal, and their mean is &lt;math&gt;6&lt;/math&gt;, we know their sum is &lt;math&gt;54&lt;/math&gt;. We also know that the sum of the set pre-removal is &lt;math&gt;66&lt;/math&gt;. Thus, the sum of the &lt;math&gt;2&lt;/math&gt; elements removed is &lt;math&gt;66-54=12&lt;/math&gt;. There are only &lt;math&gt;5&lt;/math&gt; subsets of &lt;math&gt;2&lt;/math&gt; elements that sum to &lt;math&gt;12&lt;/math&gt;: &lt;math&gt;\{1,11\}, \{2,10\}, \{3, 9\}, \{4, 8\}, \{5, 7\}&lt;/math&gt;. Therefore, our answer is &lt;math&gt;\textbf{(D)}\text{ 5}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=73086 2015 AMC 8 Problems/Problem 17 2015-11-25T21:16:16Z <p>Ariedel: </p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|after=18}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_15&diff=73085 2015 AMC 8 Problems/Problem 15 2015-11-25T21:14:37Z <p>Ariedel: </p> <hr /> <div>At Euler Middle School, &lt;math&gt;198&lt;/math&gt; students voted on two issues in a school referendum with the following results: &lt;math&gt;149&lt;/math&gt; voted in favor of the first issue and &lt;math&gt;119&lt;/math&gt; voted in favor of the second issue. If there were exactly &lt;math&gt;29&lt;/math&gt; students who voted against both issues, how many students voted in favor of both issues?<br /> <br /> &lt;math&gt;\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=14|after=16}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_14&diff=73084 2015 AMC 8 Problems/Problem 14 2015-11-25T21:14:06Z <p>Ariedel: </p> <hr /> <div>Which of the following integers cannot be written as the sum of four consecutive odd integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=13|after=15}}<br /> {{MAA Notice}}</div> Ariedel https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_14&diff=73083 2015 AMC 8 Problems/Problem 14 2015-11-25T21:13:33Z <p>Ariedel: </p> <hr /> <div>Which of the following integers cannot be written as the sum of four consecutive odd integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Ariedel