https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=G1zq&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-13T12:45:00Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems&diff=139882 2003 AMC 8 Problems 2020-12-17T22:33:07Z <p>G1zq: /* Problem 10 */</p> <hr /> <div>==Problem 1==<br /> Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Which of the following numbers has the smallest prime factor?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 55 \qquad\mathrm{(B)}\ 57 \qquad\mathrm{(C)}\ 58 \qquad\mathrm{(D)}\ 59 \qquad\mathrm{(E)}\ 61&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> A burger at Ricky C's weighs 120 grams, of which 30 grams are filler. What percent of the burger is not filler?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 60\% \qquad\mathrm{(B)}\ 65\% \qquad\mathrm{(C)}\ 70\% \qquad\mathrm{(D)}\ 75\% \qquad\mathrm{(E)}\ 90\%&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> If 20% of a number is 12, what is 30% of the same number?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 24 \qquad\mathrm{(E)}\ 30&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Given the areas of the three squares in the figure, what is the area of the interior triangle?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1));<br /> label(&quot;$25$&quot;,(14.5,1),N);<br /> label(&quot;$144$&quot;,(6,-7.5),N);<br /> label(&quot;$169$&quot;,(3.5,7),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Jenny scored 10 points higher than Blake on the first test, 10 points lower than him on the second test, and 20 points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?<br /> <br /> &lt;math&gt; \mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> &lt;math&gt;\textbf{Bake Sale}&lt;/math&gt;<br /> <br /> (Problems 8, 9, and 10 use the data found in the accompanying paragraph and figures)<br /> <br /> Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Art's cookies are trapezoids. <br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(5,0)--(5,3)--(2,3)--cycle);<br /> draw(rightanglemark((5,3), (5,0), origin));<br /> label(&quot;5 in&quot;, (2.5,0), S);<br /> label(&quot;3 in&quot;, (5,1.5), E);<br /> label(&quot;3 in&quot;, (3.5,3), N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Roger's cookies are rectangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(4,0)--(4,2)--(0,2)--cycle);<br /> draw(rightanglemark((4,2), (4,0), origin));<br /> draw(rightanglemark((0,2), origin, (4,0)));<br /> label(&quot;4 in&quot;, (2,0), S);<br /> label(&quot;2 in&quot;, (4,1), E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Paul's cookies are parallelograms.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle);<br /> draw((2.5,2)--(2.5,0), dashed);<br /> draw(rightanglemark((2.5,2),(2.5,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;2 in&quot;, (2.5,1), W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Trisha's cookies are triangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(3,4)--cycle);<br /> draw(rightanglemark((3,4),(3,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;4 in&quot;, (3,2), E);<br /> &lt;/asy&gt;<br /> <br /> Each friend uses the same amount of dough, and Art makes exactly 12 cookies. Who gets the fewest cookies from one batch of cookie dough?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Art}\qquad\textbf{(B)}\ \text{Roger}\qquad\textbf{(C)}\ \text{Paul}\qquad\textbf{(D)}\ \text{Trisha}\qquad\textbf{(E)}\ \text{There is a tie for fewest.} &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Each friend uses the same amount of dough, and Art makes exactly &lt;math&gt;12&lt;/math&gt; cookies. Art's cookies sell for &lt;math&gt;60&lt;/math&gt; cents each. To earn the same amount from a single batch, how much should one of Roger's cookies cost in cents?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 18\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 75\qquad\textbf{(E)}\ 90&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> How many cookies will be in one batch of Trisha's cookies?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> Business is a little slow at Lou's Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday's prices by 10%. Over the weekend, Lou advertises the sale: &quot;Ten percent off the listed price. Sale starts Monday.&quot; How much does a pair of shoes cost on Monday that cost 40 dollars on Thursday?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 39.60\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 40.40\qquad\textbf{(E)}\ 44 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> When a fair six-sided die is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the numbers on the five faces that can be seen is divisible by 6?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{5}{6}\qquad\textbf{(E)}\ 1&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> defaultpen(linewidth(0.8));<br /> real r=0.5;<br /> currentprojection=orthographic(3/4,8/15,7/15);<br /> draw(unitcube, white, thick(), nolight);<br /> draw(shift(1,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(0,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,1)*unitcube, white, thick(), nolight);<br /> draw(shift(1,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,1)*unitcube, white, thick(), nolight);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> In this addition problem, each letter stands for a different digit. <br /> <br /> &lt;math&gt; \setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&amp;T &amp; W &amp; O\\ + &amp;T &amp; W &amp; O\\ \hline F&amp; O &amp; U &amp; R\end{array} &lt;/math&gt;<br /> <br /> If &lt;math&gt;T = 7&lt;/math&gt; and the letter &lt;math&gt;O&lt;/math&gt; represents an even number, what is the only possible value for &lt;math&gt;W&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.8));<br /> path p=unitsquare;<br /> draw(p^^shift(0,1)*p^^shift(1,0)*p);<br /> draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p);<br /> label(&quot;FRONT&quot;, (1,0), S);<br /> label(&quot;SIDE&quot;, (5,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has 4 seats: 1 driver's seat, 1 front passenger seat, and 2 back passenger seats. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?<br /> <br /> &lt;cmath&gt; \begin{array}{c|c|c}\text{Child}&amp;\text{Eye Color}&amp;\text{Hair Color}\\ \hline\text{Benjamin}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Jim}&amp;\text{Brown}&amp;\text{Blonde}\\ \hline\text{Nadeen}&amp;\text{Brown}&amp;\text{Black}\\ \hline\text{Austin}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\text{Tevyn}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Sue}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\end{array} &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Nadeen and Austin}\qquad\textbf{(B)}\ \text{Benjamin and Sue}\qquad\textbf{(C)}\ \text{Benjamin and Austin}\qquad\textbf{(D)}\ \text{Nadeen and Tevyn}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(E)}\ \text{Austin and Sue} &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah's party?<br /> &lt;asy&gt;/* AMC8 2003 #18 Problem */<br /> pair a=(102,256), b=(68,131), c=(162,101), d=(134,150);<br /> pair e=(269,105), f=(359,104), g=(303,12), h=(579,211);<br /> pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501);<br /> pair m=(282,411), n=(147,451), o=(103,437), p=(31,373);<br /> pair q=(419,175), r=(462,209), s=(477,288), t=(443,358);<br /> pair oval=(282,303);<br /> draw(l--m--n--cycle);<br /> draw(p--oval);<br /> draw(o--oval);<br /> draw(b--d--oval);<br /> draw(c--d--e--oval);<br /> draw(e--f--g--h--i--j--oval);<br /> draw(k--oval);<br /> draw(q--oval);<br /> draw(s--oval);<br /> draw(r--s--t--oval);<br /> dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h);<br /> dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p);<br /> dot(q); dot(r); dot(s); dot(t);<br /> filldraw(yscale(.5)*Circle((282,606),80),white,black);<br /> label(scale(0.75)*&quot;Sarah&quot;, oval);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> The area of trapezoid &lt;math&gt; ABCD&lt;/math&gt; is &lt;math&gt;164\text{ cm}^2&lt;/math&gt;. The altitude is 8 cm, &lt;math&gt;AB&lt;/math&gt; is 10 cm, and &lt;math&gt;CD&lt;/math&gt; is 17 cm. What is &lt;math&gt;BC&lt;/math&gt;, in centimeters?<br /> <br /> &lt;asy&gt;/* AMC8 2003 #21 Problem */<br /> size(4inch,2inch);<br /> draw((0,0)--(31,0)--(16,8)--(6,8)--cycle);<br /> draw((11,8)--(11,0), linetype(&quot;8 4&quot;));<br /> draw((11,1)--(12,1)--(12,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$D$&quot;, (31,0), SE);<br /> label(&quot;$B$&quot;, (6,8), NW);<br /> label(&quot;$C$&quot;, (16,8), NE);<br /> label(&quot;10&quot;, (3,5), W);<br /> label(&quot;8&quot;, (11,4), E);<br /> label(&quot;17&quot;, (22.5,5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> The following figures are composed of squares and circles. Which figure has a shaded region with largest area?<br /> &lt;asy&gt;/* AMC8 2003 #22 Problem */<br /> size(3inch, 2inch);<br /> unitsize(1cm);<br /> pen outline = black+linewidth(1);<br /> filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline);<br /> filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline);<br /> filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1));<br /> filldraw(Circle((1,1), 1), white, outline);<br /> filldraw(Circle((3.5,.5), .5), white, outline);<br /> filldraw(Circle((4.5,.5), .5), white, outline);<br /> filldraw(Circle((3.5,1.5), .5), white, outline);<br /> filldraw(Circle((4.5,1.5), .5), white, outline);<br /> filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline);<br /> label(&quot;A&quot;, (1, 2), N);<br /> label(&quot;B&quot;, (4, 2), N);<br /> label(&quot;C&quot;, (7, 2), N);<br /> draw((0,-.5)--(.5,-.5), BeginArrow);<br /> draw((1.5, -.5)--(2, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (1, -.5));<br /> <br /> draw((3,-.5)--(3.5,-.5), BeginArrow);<br /> draw((4.5, -.5)--(5, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (4, -.5));<br /> <br /> draw((6,-.5)--(6.5,-.5), BeginArrow);<br /> draw((7.5, -.5)--(8, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (7, -.5));<br /> <br /> draw((6,1)--(6,-.5), linetype(&quot;4 4&quot;));<br /> draw((8,1)--(8,-.5), linetype(&quot;4 4&quot;));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{A only}\qquad\textbf{(B)}\ \text{B only}\qquad\textbf{(C)}\ \text{C only}\qquad\textbf{(D)}\ \text{both A and B}\qquad\textbf{(E)}\ \text{all are equal}&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> In the pattern below, the cat moves clockwise through the four squares and the mouse moves counterclockwise through the eight exterior segments of the four squares.<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23a.png|800px]]<br /> &lt;/center&gt;<br /> <br /> If the pattern is continued, where would the cat and mouse be after the 247th move?<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23b.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A ship travels from point &lt;math&gt;A&lt;/math&gt; to point &lt;math&gt;B&lt;/math&gt; along a semicircular path, centered at Island &lt;math&gt;X&lt;/math&gt;. Then it travels along a straight path from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt;. Which of these graphs best shows the ship's distance from Island &lt;math&gt;X&lt;/math&gt; as it moves along its course?<br /> <br /> &lt;asy&gt;size(150);<br /> pair X=origin, A=(-5,0), B=(5,0), C=(0,5);<br /> draw(Arc(X, 5, 180, 360)^^B--C);<br /> dot(X);<br /> label(&quot;$X$&quot;, X, NE);<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$B$&quot;, B, E);<br /> label(&quot;$A$&quot;, A, W);<br /> &lt;/asy&gt;<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob24ans.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> In the figure, the area of square &lt;math&gt;WXYZ&lt;/math&gt; is &lt;math&gt;25 \text{ cm}^2&lt;/math&gt;. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = AC&lt;/math&gt;, and when &lt;math&gt;\triangle ABC&lt;/math&gt; is folded over side &lt;math&gt;\overline{BC}&lt;/math&gt;, point &lt;math&gt;A&lt;/math&gt; coincides with &lt;math&gt;O&lt;/math&gt;, the center of square &lt;math&gt;WXYZ&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;, in square centimeters?<br /> <br /> &lt;asy&gt;<br /> defaultpen(fontsize(8));<br /> size(225);<br /> pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5);<br /> draw((-4,0)--Y--X--(-4,10)--cycle);<br /> draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle);<br /> dot(O);<br /> label(&quot;$A$&quot;, A, NW);<br /> label(&quot;$O$&quot;, O, NE);<br /> label(&quot;$B$&quot;, B, SW);<br /> label(&quot;$C$&quot;, C, NW);<br /> label(&quot;$W$&quot;,W , NE);<br /> label(&quot;$X$&quot;, X, N);<br /> label(&quot;$Y$&quot;, Y, S);<br /> label(&quot;$Z$&quot;, Z, SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2003|before=[[2002 AMC 8 Problems|2002 AMC 8]]|after=[[2004 AMC 8 Problems|2004 AMC 8]]}}<br /> * [[AMC 8]]<br /> * [[AMC 8 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems&diff=139881 2003 AMC 8 Problems 2020-12-17T22:32:53Z <p>G1zq: /* Problem 9 */</p> <hr /> <div>==Problem 1==<br /> Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Which of the following numbers has the smallest prime factor?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 55 \qquad\mathrm{(B)}\ 57 \qquad\mathrm{(C)}\ 58 \qquad\mathrm{(D)}\ 59 \qquad\mathrm{(E)}\ 61&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> A burger at Ricky C's weighs 120 grams, of which 30 grams are filler. What percent of the burger is not filler?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 60\% \qquad\mathrm{(B)}\ 65\% \qquad\mathrm{(C)}\ 70\% \qquad\mathrm{(D)}\ 75\% \qquad\mathrm{(E)}\ 90\%&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> If 20% of a number is 12, what is 30% of the same number?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 24 \qquad\mathrm{(E)}\ 30&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Given the areas of the three squares in the figure, what is the area of the interior triangle?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1));<br /> label(&quot;$25$&quot;,(14.5,1),N);<br /> label(&quot;$144$&quot;,(6,-7.5),N);<br /> label(&quot;$169$&quot;,(3.5,7),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Jenny scored 10 points higher than Blake on the first test, 10 points lower than him on the second test, and 20 points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?<br /> <br /> &lt;math&gt; \mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> &lt;math&gt;\textbf{Bake Sale}&lt;/math&gt;<br /> <br /> (Problems 8, 9, and 10 use the data found in the accompanying paragraph and figures)<br /> <br /> Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Art's cookies are trapezoids. <br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(5,0)--(5,3)--(2,3)--cycle);<br /> draw(rightanglemark((5,3), (5,0), origin));<br /> label(&quot;5 in&quot;, (2.5,0), S);<br /> label(&quot;3 in&quot;, (5,1.5), E);<br /> label(&quot;3 in&quot;, (3.5,3), N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Roger's cookies are rectangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(4,0)--(4,2)--(0,2)--cycle);<br /> draw(rightanglemark((4,2), (4,0), origin));<br /> draw(rightanglemark((0,2), origin, (4,0)));<br /> label(&quot;4 in&quot;, (2,0), S);<br /> label(&quot;2 in&quot;, (4,1), E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Paul's cookies are parallelograms.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle);<br /> draw((2.5,2)--(2.5,0), dashed);<br /> draw(rightanglemark((2.5,2),(2.5,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;2 in&quot;, (2.5,1), W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Trisha's cookies are triangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(3,4)--cycle);<br /> draw(rightanglemark((3,4),(3,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;4 in&quot;, (3,2), E);<br /> &lt;/asy&gt;<br /> <br /> Each friend uses the same amount of dough, and Art makes exactly 12 cookies. Who gets the fewest cookies from one batch of cookie dough?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Art}\qquad\textbf{(B)}\ \text{Roger}\qquad\textbf{(C)}\ \text{Paul}\qquad\textbf{(D)}\ \text{Trisha}\qquad\textbf{(E)}\ \text{There is a tie for fewest.} &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Each friend uses the same amount of dough, and Art makes exactly &lt;math&gt;12&lt;/math&gt; cookies. Art's cookies sell for &lt;math&gt;60&lt;/math&gt; cents each. To earn the same amount from a single batch, how much should one of Roger's cookies cost in cents?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 18\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 75\qquad\textbf{(E)}\ 90&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> How many cookies will be in one batch of Trisha's cookies?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> ==Problem 11==<br /> <br /> Business is a little slow at Lou's Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday's prices by 10%. Over the weekend, Lou advertises the sale: &quot;Ten percent off the listed price. Sale starts Monday.&quot; How much does a pair of shoes cost on Monday that cost 40 dollars on Thursday?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 39.60\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 40.40\qquad\textbf{(E)}\ 44 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> When a fair six-sided die is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the numbers on the five faces that can be seen is divisible by 6?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{5}{6}\qquad\textbf{(E)}\ 1&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> defaultpen(linewidth(0.8));<br /> real r=0.5;<br /> currentprojection=orthographic(3/4,8/15,7/15);<br /> draw(unitcube, white, thick(), nolight);<br /> draw(shift(1,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(0,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,1)*unitcube, white, thick(), nolight);<br /> draw(shift(1,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,1)*unitcube, white, thick(), nolight);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> In this addition problem, each letter stands for a different digit. <br /> <br /> &lt;math&gt; \setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&amp;T &amp; W &amp; O\\ + &amp;T &amp; W &amp; O\\ \hline F&amp; O &amp; U &amp; R\end{array} &lt;/math&gt;<br /> <br /> If &lt;math&gt;T = 7&lt;/math&gt; and the letter &lt;math&gt;O&lt;/math&gt; represents an even number, what is the only possible value for &lt;math&gt;W&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.8));<br /> path p=unitsquare;<br /> draw(p^^shift(0,1)*p^^shift(1,0)*p);<br /> draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p);<br /> label(&quot;FRONT&quot;, (1,0), S);<br /> label(&quot;SIDE&quot;, (5,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has 4 seats: 1 driver's seat, 1 front passenger seat, and 2 back passenger seats. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?<br /> <br /> &lt;cmath&gt; \begin{array}{c|c|c}\text{Child}&amp;\text{Eye Color}&amp;\text{Hair Color}\\ \hline\text{Benjamin}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Jim}&amp;\text{Brown}&amp;\text{Blonde}\\ \hline\text{Nadeen}&amp;\text{Brown}&amp;\text{Black}\\ \hline\text{Austin}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\text{Tevyn}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Sue}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\end{array} &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Nadeen and Austin}\qquad\textbf{(B)}\ \text{Benjamin and Sue}\qquad\textbf{(C)}\ \text{Benjamin and Austin}\qquad\textbf{(D)}\ \text{Nadeen and Tevyn}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(E)}\ \text{Austin and Sue} &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah's party?<br /> &lt;asy&gt;/* AMC8 2003 #18 Problem */<br /> pair a=(102,256), b=(68,131), c=(162,101), d=(134,150);<br /> pair e=(269,105), f=(359,104), g=(303,12), h=(579,211);<br /> pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501);<br /> pair m=(282,411), n=(147,451), o=(103,437), p=(31,373);<br /> pair q=(419,175), r=(462,209), s=(477,288), t=(443,358);<br /> pair oval=(282,303);<br /> draw(l--m--n--cycle);<br /> draw(p--oval);<br /> draw(o--oval);<br /> draw(b--d--oval);<br /> draw(c--d--e--oval);<br /> draw(e--f--g--h--i--j--oval);<br /> draw(k--oval);<br /> draw(q--oval);<br /> draw(s--oval);<br /> draw(r--s--t--oval);<br /> dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h);<br /> dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p);<br /> dot(q); dot(r); dot(s); dot(t);<br /> filldraw(yscale(.5)*Circle((282,606),80),white,black);<br /> label(scale(0.75)*&quot;Sarah&quot;, oval);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> The area of trapezoid &lt;math&gt; ABCD&lt;/math&gt; is &lt;math&gt;164\text{ cm}^2&lt;/math&gt;. The altitude is 8 cm, &lt;math&gt;AB&lt;/math&gt; is 10 cm, and &lt;math&gt;CD&lt;/math&gt; is 17 cm. What is &lt;math&gt;BC&lt;/math&gt;, in centimeters?<br /> <br /> &lt;asy&gt;/* AMC8 2003 #21 Problem */<br /> size(4inch,2inch);<br /> draw((0,0)--(31,0)--(16,8)--(6,8)--cycle);<br /> draw((11,8)--(11,0), linetype(&quot;8 4&quot;));<br /> draw((11,1)--(12,1)--(12,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$D$&quot;, (31,0), SE);<br /> label(&quot;$B$&quot;, (6,8), NW);<br /> label(&quot;$C$&quot;, (16,8), NE);<br /> label(&quot;10&quot;, (3,5), W);<br /> label(&quot;8&quot;, (11,4), E);<br /> label(&quot;17&quot;, (22.5,5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> The following figures are composed of squares and circles. Which figure has a shaded region with largest area?<br /> &lt;asy&gt;/* AMC8 2003 #22 Problem */<br /> size(3inch, 2inch);<br /> unitsize(1cm);<br /> pen outline = black+linewidth(1);<br /> filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline);<br /> filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline);<br /> filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1));<br /> filldraw(Circle((1,1), 1), white, outline);<br /> filldraw(Circle((3.5,.5), .5), white, outline);<br /> filldraw(Circle((4.5,.5), .5), white, outline);<br /> filldraw(Circle((3.5,1.5), .5), white, outline);<br /> filldraw(Circle((4.5,1.5), .5), white, outline);<br /> filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline);<br /> label(&quot;A&quot;, (1, 2), N);<br /> label(&quot;B&quot;, (4, 2), N);<br /> label(&quot;C&quot;, (7, 2), N);<br /> draw((0,-.5)--(.5,-.5), BeginArrow);<br /> draw((1.5, -.5)--(2, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (1, -.5));<br /> <br /> draw((3,-.5)--(3.5,-.5), BeginArrow);<br /> draw((4.5, -.5)--(5, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (4, -.5));<br /> <br /> draw((6,-.5)--(6.5,-.5), BeginArrow);<br /> draw((7.5, -.5)--(8, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (7, -.5));<br /> <br /> draw((6,1)--(6,-.5), linetype(&quot;4 4&quot;));<br /> draw((8,1)--(8,-.5), linetype(&quot;4 4&quot;));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{A only}\qquad\textbf{(B)}\ \text{B only}\qquad\textbf{(C)}\ \text{C only}\qquad\textbf{(D)}\ \text{both A and B}\qquad\textbf{(E)}\ \text{all are equal}&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> In the pattern below, the cat moves clockwise through the four squares and the mouse moves counterclockwise through the eight exterior segments of the four squares.<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23a.png|800px]]<br /> &lt;/center&gt;<br /> <br /> If the pattern is continued, where would the cat and mouse be after the 247th move?<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23b.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A ship travels from point &lt;math&gt;A&lt;/math&gt; to point &lt;math&gt;B&lt;/math&gt; along a semicircular path, centered at Island &lt;math&gt;X&lt;/math&gt;. Then it travels along a straight path from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt;. Which of these graphs best shows the ship's distance from Island &lt;math&gt;X&lt;/math&gt; as it moves along its course?<br /> <br /> &lt;asy&gt;size(150);<br /> pair X=origin, A=(-5,0), B=(5,0), C=(0,5);<br /> draw(Arc(X, 5, 180, 360)^^B--C);<br /> dot(X);<br /> label(&quot;$X$&quot;, X, NE);<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$B$&quot;, B, E);<br /> label(&quot;$A$&quot;, A, W);<br /> &lt;/asy&gt;<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob24ans.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> In the figure, the area of square &lt;math&gt;WXYZ&lt;/math&gt; is &lt;math&gt;25 \text{ cm}^2&lt;/math&gt;. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = AC&lt;/math&gt;, and when &lt;math&gt;\triangle ABC&lt;/math&gt; is folded over side &lt;math&gt;\overline{BC}&lt;/math&gt;, point &lt;math&gt;A&lt;/math&gt; coincides with &lt;math&gt;O&lt;/math&gt;, the center of square &lt;math&gt;WXYZ&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;, in square centimeters?<br /> <br /> &lt;asy&gt;<br /> defaultpen(fontsize(8));<br /> size(225);<br /> pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5);<br /> draw((-4,0)--Y--X--(-4,10)--cycle);<br /> draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle);<br /> dot(O);<br /> label(&quot;$A$&quot;, A, NW);<br /> label(&quot;$O$&quot;, O, NE);<br /> label(&quot;$B$&quot;, B, SW);<br /> label(&quot;$C$&quot;, C, NW);<br /> label(&quot;$W$&quot;,W , NE);<br /> label(&quot;$X$&quot;, X, N);<br /> label(&quot;$Y$&quot;, Y, S);<br /> label(&quot;$Z$&quot;, Z, SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2003|before=[[2002 AMC 8 Problems|2002 AMC 8]]|after=[[2004 AMC 8 Problems|2004 AMC 8]]}}<br /> * [[AMC 8]]<br /> * [[AMC 8 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems&diff=139880 2003 AMC 8 Problems 2020-12-17T22:32:27Z <p>G1zq: /* Problem 8 */</p> <hr /> <div>==Problem 1==<br /> Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Which of the following numbers has the smallest prime factor?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 55 \qquad\mathrm{(B)}\ 57 \qquad\mathrm{(C)}\ 58 \qquad\mathrm{(D)}\ 59 \qquad\mathrm{(E)}\ 61&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> A burger at Ricky C's weighs 120 grams, of which 30 grams are filler. What percent of the burger is not filler?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 60\% \qquad\mathrm{(B)}\ 65\% \qquad\mathrm{(C)}\ 70\% \qquad\mathrm{(D)}\ 75\% \qquad\mathrm{(E)}\ 90\%&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> If 20% of a number is 12, what is 30% of the same number?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 24 \qquad\mathrm{(E)}\ 30&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Given the areas of the three squares in the figure, what is the area of the interior triangle?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1));<br /> label(&quot;$25$&quot;,(14.5,1),N);<br /> label(&quot;$144$&quot;,(6,-7.5),N);<br /> label(&quot;$169$&quot;,(3.5,7),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Jenny scored 10 points higher than Blake on the first test, 10 points lower than him on the second test, and 20 points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?<br /> <br /> &lt;math&gt; \mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> &lt;math&gt;\textbf{Bake Sale}&lt;/math&gt;<br /> <br /> (Problems 8, 9, and 10 use the data found in the accompanying paragraph and figures)<br /> <br /> Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Art's cookies are trapezoids. <br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(5,0)--(5,3)--(2,3)--cycle);<br /> draw(rightanglemark((5,3), (5,0), origin));<br /> label(&quot;5 in&quot;, (2.5,0), S);<br /> label(&quot;3 in&quot;, (5,1.5), E);<br /> label(&quot;3 in&quot;, (3.5,3), N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Roger's cookies are rectangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(4,0)--(4,2)--(0,2)--cycle);<br /> draw(rightanglemark((4,2), (4,0), origin));<br /> draw(rightanglemark((0,2), origin, (4,0)));<br /> label(&quot;4 in&quot;, (2,0), S);<br /> label(&quot;2 in&quot;, (4,1), E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Paul's cookies are parallelograms.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle);<br /> draw((2.5,2)--(2.5,0), dashed);<br /> draw(rightanglemark((2.5,2),(2.5,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;2 in&quot;, (2.5,1), W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Trisha's cookies are triangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(3,4)--cycle);<br /> draw(rightanglemark((3,4),(3,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;4 in&quot;, (3,2), E);<br /> &lt;/asy&gt;<br /> <br /> Each friend uses the same amount of dough, and Art makes exactly 12 cookies. Who gets the fewest cookies from one batch of cookie dough?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Art}\qquad\textbf{(B)}\ \text{Roger}\qquad\textbf{(C)}\ \text{Paul}\qquad\textbf{(D)}\ \text{Trisha}\qquad\textbf{(E)}\ \text{There is a tie for fewest.} &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Each friend uses the same amount of dough, and Art makes exactly &lt;math&gt;12&lt;/math&gt; cookies. Art's cookies sell for &lt;math&gt;60&lt;/math&gt; cents each. To earn the same amount from a single batch, how much should one of Roger's cookies cost in cents?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 18\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 75\qquad\textbf{(E)}\ 90&lt;/math&gt;<br /> <br /> ==Problem 10==<br /> How many cookies will be in one batch of Trisha's cookies?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> ==Problem 11==<br /> <br /> Business is a little slow at Lou's Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday's prices by 10%. Over the weekend, Lou advertises the sale: &quot;Ten percent off the listed price. Sale starts Monday.&quot; How much does a pair of shoes cost on Monday that cost 40 dollars on Thursday?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 39.60\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 40.40\qquad\textbf{(E)}\ 44 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> When a fair six-sided die is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the numbers on the five faces that can be seen is divisible by 6?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{5}{6}\qquad\textbf{(E)}\ 1&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> defaultpen(linewidth(0.8));<br /> real r=0.5;<br /> currentprojection=orthographic(3/4,8/15,7/15);<br /> draw(unitcube, white, thick(), nolight);<br /> draw(shift(1,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(0,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,1)*unitcube, white, thick(), nolight);<br /> draw(shift(1,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,1)*unitcube, white, thick(), nolight);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> In this addition problem, each letter stands for a different digit. <br /> <br /> &lt;math&gt; \setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&amp;T &amp; W &amp; O\\ + &amp;T &amp; W &amp; O\\ \hline F&amp; O &amp; U &amp; R\end{array} &lt;/math&gt;<br /> <br /> If &lt;math&gt;T = 7&lt;/math&gt; and the letter &lt;math&gt;O&lt;/math&gt; represents an even number, what is the only possible value for &lt;math&gt;W&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.8));<br /> path p=unitsquare;<br /> draw(p^^shift(0,1)*p^^shift(1,0)*p);<br /> draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p);<br /> label(&quot;FRONT&quot;, (1,0), S);<br /> label(&quot;SIDE&quot;, (5,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has 4 seats: 1 driver's seat, 1 front passenger seat, and 2 back passenger seats. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?<br /> <br /> &lt;cmath&gt; \begin{array}{c|c|c}\text{Child}&amp;\text{Eye Color}&amp;\text{Hair Color}\\ \hline\text{Benjamin}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Jim}&amp;\text{Brown}&amp;\text{Blonde}\\ \hline\text{Nadeen}&amp;\text{Brown}&amp;\text{Black}\\ \hline\text{Austin}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\text{Tevyn}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Sue}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\end{array} &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Nadeen and Austin}\qquad\textbf{(B)}\ \text{Benjamin and Sue}\qquad\textbf{(C)}\ \text{Benjamin and Austin}\qquad\textbf{(D)}\ \text{Nadeen and Tevyn}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(E)}\ \text{Austin and Sue} &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah's party?<br /> &lt;asy&gt;/* AMC8 2003 #18 Problem */<br /> pair a=(102,256), b=(68,131), c=(162,101), d=(134,150);<br /> pair e=(269,105), f=(359,104), g=(303,12), h=(579,211);<br /> pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501);<br /> pair m=(282,411), n=(147,451), o=(103,437), p=(31,373);<br /> pair q=(419,175), r=(462,209), s=(477,288), t=(443,358);<br /> pair oval=(282,303);<br /> draw(l--m--n--cycle);<br /> draw(p--oval);<br /> draw(o--oval);<br /> draw(b--d--oval);<br /> draw(c--d--e--oval);<br /> draw(e--f--g--h--i--j--oval);<br /> draw(k--oval);<br /> draw(q--oval);<br /> draw(s--oval);<br /> draw(r--s--t--oval);<br /> dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h);<br /> dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p);<br /> dot(q); dot(r); dot(s); dot(t);<br /> filldraw(yscale(.5)*Circle((282,606),80),white,black);<br /> label(scale(0.75)*&quot;Sarah&quot;, oval);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> The area of trapezoid &lt;math&gt; ABCD&lt;/math&gt; is &lt;math&gt;164\text{ cm}^2&lt;/math&gt;. The altitude is 8 cm, &lt;math&gt;AB&lt;/math&gt; is 10 cm, and &lt;math&gt;CD&lt;/math&gt; is 17 cm. What is &lt;math&gt;BC&lt;/math&gt;, in centimeters?<br /> <br /> &lt;asy&gt;/* AMC8 2003 #21 Problem */<br /> size(4inch,2inch);<br /> draw((0,0)--(31,0)--(16,8)--(6,8)--cycle);<br /> draw((11,8)--(11,0), linetype(&quot;8 4&quot;));<br /> draw((11,1)--(12,1)--(12,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$D$&quot;, (31,0), SE);<br /> label(&quot;$B$&quot;, (6,8), NW);<br /> label(&quot;$C$&quot;, (16,8), NE);<br /> label(&quot;10&quot;, (3,5), W);<br /> label(&quot;8&quot;, (11,4), E);<br /> label(&quot;17&quot;, (22.5,5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> The following figures are composed of squares and circles. Which figure has a shaded region with largest area?<br /> &lt;asy&gt;/* AMC8 2003 #22 Problem */<br /> size(3inch, 2inch);<br /> unitsize(1cm);<br /> pen outline = black+linewidth(1);<br /> filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline);<br /> filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline);<br /> filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1));<br /> filldraw(Circle((1,1), 1), white, outline);<br /> filldraw(Circle((3.5,.5), .5), white, outline);<br /> filldraw(Circle((4.5,.5), .5), white, outline);<br /> filldraw(Circle((3.5,1.5), .5), white, outline);<br /> filldraw(Circle((4.5,1.5), .5), white, outline);<br /> filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline);<br /> label(&quot;A&quot;, (1, 2), N);<br /> label(&quot;B&quot;, (4, 2), N);<br /> label(&quot;C&quot;, (7, 2), N);<br /> draw((0,-.5)--(.5,-.5), BeginArrow);<br /> draw((1.5, -.5)--(2, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (1, -.5));<br /> <br /> draw((3,-.5)--(3.5,-.5), BeginArrow);<br /> draw((4.5, -.5)--(5, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (4, -.5));<br /> <br /> draw((6,-.5)--(6.5,-.5), BeginArrow);<br /> draw((7.5, -.5)--(8, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (7, -.5));<br /> <br /> draw((6,1)--(6,-.5), linetype(&quot;4 4&quot;));<br /> draw((8,1)--(8,-.5), linetype(&quot;4 4&quot;));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{A only}\qquad\textbf{(B)}\ \text{B only}\qquad\textbf{(C)}\ \text{C only}\qquad\textbf{(D)}\ \text{both A and B}\qquad\textbf{(E)}\ \text{all are equal}&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> In the pattern below, the cat moves clockwise through the four squares and the mouse moves counterclockwise through the eight exterior segments of the four squares.<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23a.png|800px]]<br /> &lt;/center&gt;<br /> <br /> If the pattern is continued, where would the cat and mouse be after the 247th move?<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23b.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A ship travels from point &lt;math&gt;A&lt;/math&gt; to point &lt;math&gt;B&lt;/math&gt; along a semicircular path, centered at Island &lt;math&gt;X&lt;/math&gt;. Then it travels along a straight path from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt;. Which of these graphs best shows the ship's distance from Island &lt;math&gt;X&lt;/math&gt; as it moves along its course?<br /> <br /> &lt;asy&gt;size(150);<br /> pair X=origin, A=(-5,0), B=(5,0), C=(0,5);<br /> draw(Arc(X, 5, 180, 360)^^B--C);<br /> dot(X);<br /> label(&quot;$X$&quot;, X, NE);<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$B$&quot;, B, E);<br /> label(&quot;$A$&quot;, A, W);<br /> &lt;/asy&gt;<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob24ans.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> In the figure, the area of square &lt;math&gt;WXYZ&lt;/math&gt; is &lt;math&gt;25 \text{ cm}^2&lt;/math&gt;. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = AC&lt;/math&gt;, and when &lt;math&gt;\triangle ABC&lt;/math&gt; is folded over side &lt;math&gt;\overline{BC}&lt;/math&gt;, point &lt;math&gt;A&lt;/math&gt; coincides with &lt;math&gt;O&lt;/math&gt;, the center of square &lt;math&gt;WXYZ&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;, in square centimeters?<br /> <br /> &lt;asy&gt;<br /> defaultpen(fontsize(8));<br /> size(225);<br /> pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5);<br /> draw((-4,0)--Y--X--(-4,10)--cycle);<br /> draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle);<br /> dot(O);<br /> label(&quot;$A$&quot;, A, NW);<br /> label(&quot;$O$&quot;, O, NE);<br /> label(&quot;$B$&quot;, B, SW);<br /> label(&quot;$C$&quot;, C, NW);<br /> label(&quot;$W$&quot;,W , NE);<br /> label(&quot;$X$&quot;, X, N);<br /> label(&quot;$Y$&quot;, Y, S);<br /> label(&quot;$Z$&quot;, Z, SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2003|before=[[2002 AMC 8 Problems|2002 AMC 8]]|after=[[2004 AMC 8 Problems|2004 AMC 8]]}}<br /> * [[AMC 8]]<br /> * [[AMC 8 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems&diff=139879 2003 AMC 8 Problems 2020-12-17T22:31:41Z <p>G1zq: /* Problem 10 */</p> <hr /> <div>==Problem 1==<br /> Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Which of the following numbers has the smallest prime factor?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 55 \qquad\mathrm{(B)}\ 57 \qquad\mathrm{(C)}\ 58 \qquad\mathrm{(D)}\ 59 \qquad\mathrm{(E)}\ 61&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> A burger at Ricky C's weighs 120 grams, of which 30 grams are filler. What percent of the burger is not filler?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 60\% \qquad\mathrm{(B)}\ 65\% \qquad\mathrm{(C)}\ 70\% \qquad\mathrm{(D)}\ 75\% \qquad\mathrm{(E)}\ 90\%&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> If 20% of a number is 12, what is 30% of the same number?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 24 \qquad\mathrm{(E)}\ 30&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Given the areas of the three squares in the figure, what is the area of the interior triangle?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1));<br /> label(&quot;$25$&quot;,(14.5,1),N);<br /> label(&quot;$144$&quot;,(6,-7.5),N);<br /> label(&quot;$169$&quot;,(3.5,7),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Jenny scored 10 points higher than Blake on the first test, 10 points lower than him on the second test, and 20 points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?<br /> <br /> &lt;math&gt; \mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> &lt;math&gt;\textbf{Bake Sale}&lt;/math&gt;<br /> <br /> (Problems 8, 9, and 10 use the data found in the accompanying paragraph and figures)<br /> <br /> Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Art's cookies are trapezoids. <br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(5,0)--(5,3)--(2,3)--cycle);<br /> draw(rightanglemark((5,3), (5,0), origin));<br /> label(&quot;5 in&quot;, (2.5,0), S);<br /> label(&quot;3 in&quot;, (5,1.5), E);<br /> label(&quot;3 in&quot;, (3.5,3), N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Roger's cookies are rectangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(4,0)--(4,2)--(0,2)--cycle);<br /> draw(rightanglemark((4,2), (4,0), origin));<br /> draw(rightanglemark((0,2), origin, (4,0)));<br /> label(&quot;4 in&quot;, (2,0), S);<br /> label(&quot;2 in&quot;, (4,1), E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Paul's cookies are parallelograms.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle);<br /> draw((2.5,2)--(2.5,0), dashed);<br /> draw(rightanglemark((2.5,2),(2.5,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;2 in&quot;, (2.5,1), W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Trisha's cookies are triangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(3,4)--cycle);<br /> draw(rightanglemark((3,4),(3,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;4 in&quot;, (3,2), E);<br /> &lt;/asy&gt;<br /> <br /> Each friend uses the same amount of dough, and Art makes exactly 12 cookies. Who gets the fewest cookies from one batch of cookie dough?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Art}\qquad\textbf{(B)}\ \text{Roger}\qquad\textbf{(C)}\ \text{Paul}\qquad\textbf{(D)}\ \text{Trisha}\qquad\textbf{(E)}\ \text{There is a tie for fewest.} &lt;/math&gt;<br /> <br /> ==Problem 9==<br /> Each friend uses the same amount of dough, and Art makes exactly &lt;math&gt;12&lt;/math&gt; cookies. Art's cookies sell for &lt;math&gt;60&lt;/math&gt; cents each. To earn the same amount from a single batch, how much should one of Roger's cookies cost in cents?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 18\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 75\qquad\textbf{(E)}\ 90&lt;/math&gt;<br /> <br /> ==Problem 10==<br /> How many cookies will be in one batch of Trisha's cookies?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> ==Problem 11==<br /> <br /> Business is a little slow at Lou's Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday's prices by 10%. Over the weekend, Lou advertises the sale: &quot;Ten percent off the listed price. Sale starts Monday.&quot; How much does a pair of shoes cost on Monday that cost 40 dollars on Thursday?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 39.60\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 40.40\qquad\textbf{(E)}\ 44 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> When a fair six-sided die is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the numbers on the five faces that can be seen is divisible by 6?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{5}{6}\qquad\textbf{(E)}\ 1&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> defaultpen(linewidth(0.8));<br /> real r=0.5;<br /> currentprojection=orthographic(3/4,8/15,7/15);<br /> draw(unitcube, white, thick(), nolight);<br /> draw(shift(1,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(0,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,1)*unitcube, white, thick(), nolight);<br /> draw(shift(1,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,1)*unitcube, white, thick(), nolight);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> In this addition problem, each letter stands for a different digit. <br /> <br /> &lt;math&gt; \setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&amp;T &amp; W &amp; O\\ + &amp;T &amp; W &amp; O\\ \hline F&amp; O &amp; U &amp; R\end{array} &lt;/math&gt;<br /> <br /> If &lt;math&gt;T = 7&lt;/math&gt; and the letter &lt;math&gt;O&lt;/math&gt; represents an even number, what is the only possible value for &lt;math&gt;W&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.8));<br /> path p=unitsquare;<br /> draw(p^^shift(0,1)*p^^shift(1,0)*p);<br /> draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p);<br /> label(&quot;FRONT&quot;, (1,0), S);<br /> label(&quot;SIDE&quot;, (5,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has 4 seats: 1 driver's seat, 1 front passenger seat, and 2 back passenger seats. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?<br /> <br /> &lt;cmath&gt; \begin{array}{c|c|c}\text{Child}&amp;\text{Eye Color}&amp;\text{Hair Color}\\ \hline\text{Benjamin}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Jim}&amp;\text{Brown}&amp;\text{Blonde}\\ \hline\text{Nadeen}&amp;\text{Brown}&amp;\text{Black}\\ \hline\text{Austin}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\text{Tevyn}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Sue}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\end{array} &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Nadeen and Austin}\qquad\textbf{(B)}\ \text{Benjamin and Sue}\qquad\textbf{(C)}\ \text{Benjamin and Austin}\qquad\textbf{(D)}\ \text{Nadeen and Tevyn}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(E)}\ \text{Austin and Sue} &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah's party?<br /> &lt;asy&gt;/* AMC8 2003 #18 Problem */<br /> pair a=(102,256), b=(68,131), c=(162,101), d=(134,150);<br /> pair e=(269,105), f=(359,104), g=(303,12), h=(579,211);<br /> pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501);<br /> pair m=(282,411), n=(147,451), o=(103,437), p=(31,373);<br /> pair q=(419,175), r=(462,209), s=(477,288), t=(443,358);<br /> pair oval=(282,303);<br /> draw(l--m--n--cycle);<br /> draw(p--oval);<br /> draw(o--oval);<br /> draw(b--d--oval);<br /> draw(c--d--e--oval);<br /> draw(e--f--g--h--i--j--oval);<br /> draw(k--oval);<br /> draw(q--oval);<br /> draw(s--oval);<br /> draw(r--s--t--oval);<br /> dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h);<br /> dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p);<br /> dot(q); dot(r); dot(s); dot(t);<br /> filldraw(yscale(.5)*Circle((282,606),80),white,black);<br /> label(scale(0.75)*&quot;Sarah&quot;, oval);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> The area of trapezoid &lt;math&gt; ABCD&lt;/math&gt; is &lt;math&gt;164\text{ cm}^2&lt;/math&gt;. The altitude is 8 cm, &lt;math&gt;AB&lt;/math&gt; is 10 cm, and &lt;math&gt;CD&lt;/math&gt; is 17 cm. What is &lt;math&gt;BC&lt;/math&gt;, in centimeters?<br /> <br /> &lt;asy&gt;/* AMC8 2003 #21 Problem */<br /> size(4inch,2inch);<br /> draw((0,0)--(31,0)--(16,8)--(6,8)--cycle);<br /> draw((11,8)--(11,0), linetype(&quot;8 4&quot;));<br /> draw((11,1)--(12,1)--(12,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$D$&quot;, (31,0), SE);<br /> label(&quot;$B$&quot;, (6,8), NW);<br /> label(&quot;$C$&quot;, (16,8), NE);<br /> label(&quot;10&quot;, (3,5), W);<br /> label(&quot;8&quot;, (11,4), E);<br /> label(&quot;17&quot;, (22.5,5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> The following figures are composed of squares and circles. Which figure has a shaded region with largest area?<br /> &lt;asy&gt;/* AMC8 2003 #22 Problem */<br /> size(3inch, 2inch);<br /> unitsize(1cm);<br /> pen outline = black+linewidth(1);<br /> filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline);<br /> filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline);<br /> filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1));<br /> filldraw(Circle((1,1), 1), white, outline);<br /> filldraw(Circle((3.5,.5), .5), white, outline);<br /> filldraw(Circle((4.5,.5), .5), white, outline);<br /> filldraw(Circle((3.5,1.5), .5), white, outline);<br /> filldraw(Circle((4.5,1.5), .5), white, outline);<br /> filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline);<br /> label(&quot;A&quot;, (1, 2), N);<br /> label(&quot;B&quot;, (4, 2), N);<br /> label(&quot;C&quot;, (7, 2), N);<br /> draw((0,-.5)--(.5,-.5), BeginArrow);<br /> draw((1.5, -.5)--(2, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (1, -.5));<br /> <br /> draw((3,-.5)--(3.5,-.5), BeginArrow);<br /> draw((4.5, -.5)--(5, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (4, -.5));<br /> <br /> draw((6,-.5)--(6.5,-.5), BeginArrow);<br /> draw((7.5, -.5)--(8, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (7, -.5));<br /> <br /> draw((6,1)--(6,-.5), linetype(&quot;4 4&quot;));<br /> draw((8,1)--(8,-.5), linetype(&quot;4 4&quot;));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{A only}\qquad\textbf{(B)}\ \text{B only}\qquad\textbf{(C)}\ \text{C only}\qquad\textbf{(D)}\ \text{both A and B}\qquad\textbf{(E)}\ \text{all are equal}&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> In the pattern below, the cat moves clockwise through the four squares and the mouse moves counterclockwise through the eight exterior segments of the four squares.<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23a.png|800px]]<br /> &lt;/center&gt;<br /> <br /> If the pattern is continued, where would the cat and mouse be after the 247th move?<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23b.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A ship travels from point &lt;math&gt;A&lt;/math&gt; to point &lt;math&gt;B&lt;/math&gt; along a semicircular path, centered at Island &lt;math&gt;X&lt;/math&gt;. Then it travels along a straight path from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt;. Which of these graphs best shows the ship's distance from Island &lt;math&gt;X&lt;/math&gt; as it moves along its course?<br /> <br /> &lt;asy&gt;size(150);<br /> pair X=origin, A=(-5,0), B=(5,0), C=(0,5);<br /> draw(Arc(X, 5, 180, 360)^^B--C);<br /> dot(X);<br /> label(&quot;$X$&quot;, X, NE);<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$B$&quot;, B, E);<br /> label(&quot;$A$&quot;, A, W);<br /> &lt;/asy&gt;<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob24ans.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> In the figure, the area of square &lt;math&gt;WXYZ&lt;/math&gt; is &lt;math&gt;25 \text{ cm}^2&lt;/math&gt;. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = AC&lt;/math&gt;, and when &lt;math&gt;\triangle ABC&lt;/math&gt; is folded over side &lt;math&gt;\overline{BC}&lt;/math&gt;, point &lt;math&gt;A&lt;/math&gt; coincides with &lt;math&gt;O&lt;/math&gt;, the center of square &lt;math&gt;WXYZ&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;, in square centimeters?<br /> <br /> &lt;asy&gt;<br /> defaultpen(fontsize(8));<br /> size(225);<br /> pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5);<br /> draw((-4,0)--Y--X--(-4,10)--cycle);<br /> draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle);<br /> dot(O);<br /> label(&quot;$A$&quot;, A, NW);<br /> label(&quot;$O$&quot;, O, NE);<br /> label(&quot;$B$&quot;, B, SW);<br /> label(&quot;$C$&quot;, C, NW);<br /> label(&quot;$W$&quot;,W , NE);<br /> label(&quot;$X$&quot;, X, N);<br /> label(&quot;$Y$&quot;, Y, S);<br /> label(&quot;$Z$&quot;, Z, SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2003|before=[[2002 AMC 8 Problems|2002 AMC 8]]|after=[[2004 AMC 8 Problems|2004 AMC 8]]}}<br /> * [[AMC 8]]<br /> * [[AMC 8 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems&diff=139878 2003 AMC 8 Problems 2020-12-17T22:30:42Z <p>G1zq: /* Problem 9 */</p> <hr /> <div>==Problem 1==<br /> Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Which of the following numbers has the smallest prime factor?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 55 \qquad\mathrm{(B)}\ 57 \qquad\mathrm{(C)}\ 58 \qquad\mathrm{(D)}\ 59 \qquad\mathrm{(E)}\ 61&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> A burger at Ricky C's weighs 120 grams, of which 30 grams are filler. What percent of the burger is not filler?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 60\% \qquad\mathrm{(B)}\ 65\% \qquad\mathrm{(C)}\ 70\% \qquad\mathrm{(D)}\ 75\% \qquad\mathrm{(E)}\ 90\%&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> If 20% of a number is 12, what is 30% of the same number?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 24 \qquad\mathrm{(E)}\ 30&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Given the areas of the three squares in the figure, what is the area of the interior triangle?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1));<br /> label(&quot;$25$&quot;,(14.5,1),N);<br /> label(&quot;$144$&quot;,(6,-7.5),N);<br /> label(&quot;$169$&quot;,(3.5,7),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Jenny scored 10 points higher than Blake on the first test, 10 points lower than him on the second test, and 20 points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?<br /> <br /> &lt;math&gt; \mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> &lt;math&gt;\textbf{Bake Sale}&lt;/math&gt;<br /> <br /> (Problems 8, 9, and 10 use the data found in the accompanying paragraph and figures)<br /> <br /> Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Art's cookies are trapezoids. <br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(5,0)--(5,3)--(2,3)--cycle);<br /> draw(rightanglemark((5,3), (5,0), origin));<br /> label(&quot;5 in&quot;, (2.5,0), S);<br /> label(&quot;3 in&quot;, (5,1.5), E);<br /> label(&quot;3 in&quot;, (3.5,3), N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Roger's cookies are rectangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(4,0)--(4,2)--(0,2)--cycle);<br /> draw(rightanglemark((4,2), (4,0), origin));<br /> draw(rightanglemark((0,2), origin, (4,0)));<br /> label(&quot;4 in&quot;, (2,0), S);<br /> label(&quot;2 in&quot;, (4,1), E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Paul's cookies are parallelograms.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle);<br /> draw((2.5,2)--(2.5,0), dashed);<br /> draw(rightanglemark((2.5,2),(2.5,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;2 in&quot;, (2.5,1), W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Trisha's cookies are triangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(3,4)--cycle);<br /> draw(rightanglemark((3,4),(3,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;4 in&quot;, (3,2), E);<br /> &lt;/asy&gt;<br /> <br /> Each friend uses the same amount of dough, and Art makes exactly 12 cookies. Who gets the fewest cookies from one batch of cookie dough?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Art}\qquad\textbf{(B)}\ \text{Roger}\qquad\textbf{(C)}\ \text{Paul}\qquad\textbf{(D)}\ \text{Trisha}\qquad\textbf{(E)}\ \text{There is a tie for fewest.} &lt;/math&gt;<br /> <br /> ==Problem 9==<br /> Each friend uses the same amount of dough, and Art makes exactly &lt;math&gt;12&lt;/math&gt; cookies. Art's cookies sell for &lt;math&gt;60&lt;/math&gt; cents each. To earn the same amount from a single batch, how much should one of Roger's cookies cost in cents?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 18\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 75\qquad\textbf{(E)}\ 90&lt;/math&gt;<br /> <br /> ==Problem 10==<br /> <br /> ==Problem 11==<br /> <br /> Business is a little slow at Lou's Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday's prices by 10%. Over the weekend, Lou advertises the sale: &quot;Ten percent off the listed price. Sale starts Monday.&quot; How much does a pair of shoes cost on Monday that cost 40 dollars on Thursday?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 39.60\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 40.40\qquad\textbf{(E)}\ 44 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> When a fair six-sided die is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the numbers on the five faces that can be seen is divisible by 6?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{5}{6}\qquad\textbf{(E)}\ 1&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> defaultpen(linewidth(0.8));<br /> real r=0.5;<br /> currentprojection=orthographic(3/4,8/15,7/15);<br /> draw(unitcube, white, thick(), nolight);<br /> draw(shift(1,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(0,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,1)*unitcube, white, thick(), nolight);<br /> draw(shift(1,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,1)*unitcube, white, thick(), nolight);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> In this addition problem, each letter stands for a different digit. <br /> <br /> &lt;math&gt; \setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&amp;T &amp; W &amp; O\\ + &amp;T &amp; W &amp; O\\ \hline F&amp; O &amp; U &amp; R\end{array} &lt;/math&gt;<br /> <br /> If &lt;math&gt;T = 7&lt;/math&gt; and the letter &lt;math&gt;O&lt;/math&gt; represents an even number, what is the only possible value for &lt;math&gt;W&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.8));<br /> path p=unitsquare;<br /> draw(p^^shift(0,1)*p^^shift(1,0)*p);<br /> draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p);<br /> label(&quot;FRONT&quot;, (1,0), S);<br /> label(&quot;SIDE&quot;, (5,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has 4 seats: 1 driver's seat, 1 front passenger seat, and 2 back passenger seats. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?<br /> <br /> &lt;cmath&gt; \begin{array}{c|c|c}\text{Child}&amp;\text{Eye Color}&amp;\text{Hair Color}\\ \hline\text{Benjamin}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Jim}&amp;\text{Brown}&amp;\text{Blonde}\\ \hline\text{Nadeen}&amp;\text{Brown}&amp;\text{Black}\\ \hline\text{Austin}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\text{Tevyn}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Sue}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\end{array} &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Nadeen and Austin}\qquad\textbf{(B)}\ \text{Benjamin and Sue}\qquad\textbf{(C)}\ \text{Benjamin and Austin}\qquad\textbf{(D)}\ \text{Nadeen and Tevyn}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(E)}\ \text{Austin and Sue} &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah's party?<br /> &lt;asy&gt;/* AMC8 2003 #18 Problem */<br /> pair a=(102,256), b=(68,131), c=(162,101), d=(134,150);<br /> pair e=(269,105), f=(359,104), g=(303,12), h=(579,211);<br /> pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501);<br /> pair m=(282,411), n=(147,451), o=(103,437), p=(31,373);<br /> pair q=(419,175), r=(462,209), s=(477,288), t=(443,358);<br /> pair oval=(282,303);<br /> draw(l--m--n--cycle);<br /> draw(p--oval);<br /> draw(o--oval);<br /> draw(b--d--oval);<br /> draw(c--d--e--oval);<br /> draw(e--f--g--h--i--j--oval);<br /> draw(k--oval);<br /> draw(q--oval);<br /> draw(s--oval);<br /> draw(r--s--t--oval);<br /> dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h);<br /> dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p);<br /> dot(q); dot(r); dot(s); dot(t);<br /> filldraw(yscale(.5)*Circle((282,606),80),white,black);<br /> label(scale(0.75)*&quot;Sarah&quot;, oval);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> The area of trapezoid &lt;math&gt; ABCD&lt;/math&gt; is &lt;math&gt;164\text{ cm}^2&lt;/math&gt;. The altitude is 8 cm, &lt;math&gt;AB&lt;/math&gt; is 10 cm, and &lt;math&gt;CD&lt;/math&gt; is 17 cm. What is &lt;math&gt;BC&lt;/math&gt;, in centimeters?<br /> <br /> &lt;asy&gt;/* AMC8 2003 #21 Problem */<br /> size(4inch,2inch);<br /> draw((0,0)--(31,0)--(16,8)--(6,8)--cycle);<br /> draw((11,8)--(11,0), linetype(&quot;8 4&quot;));<br /> draw((11,1)--(12,1)--(12,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$D$&quot;, (31,0), SE);<br /> label(&quot;$B$&quot;, (6,8), NW);<br /> label(&quot;$C$&quot;, (16,8), NE);<br /> label(&quot;10&quot;, (3,5), W);<br /> label(&quot;8&quot;, (11,4), E);<br /> label(&quot;17&quot;, (22.5,5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> The following figures are composed of squares and circles. Which figure has a shaded region with largest area?<br /> &lt;asy&gt;/* AMC8 2003 #22 Problem */<br /> size(3inch, 2inch);<br /> unitsize(1cm);<br /> pen outline = black+linewidth(1);<br /> filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline);<br /> filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline);<br /> filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1));<br /> filldraw(Circle((1,1), 1), white, outline);<br /> filldraw(Circle((3.5,.5), .5), white, outline);<br /> filldraw(Circle((4.5,.5), .5), white, outline);<br /> filldraw(Circle((3.5,1.5), .5), white, outline);<br /> filldraw(Circle((4.5,1.5), .5), white, outline);<br /> filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline);<br /> label(&quot;A&quot;, (1, 2), N);<br /> label(&quot;B&quot;, (4, 2), N);<br /> label(&quot;C&quot;, (7, 2), N);<br /> draw((0,-.5)--(.5,-.5), BeginArrow);<br /> draw((1.5, -.5)--(2, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (1, -.5));<br /> <br /> draw((3,-.5)--(3.5,-.5), BeginArrow);<br /> draw((4.5, -.5)--(5, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (4, -.5));<br /> <br /> draw((6,-.5)--(6.5,-.5), BeginArrow);<br /> draw((7.5, -.5)--(8, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (7, -.5));<br /> <br /> draw((6,1)--(6,-.5), linetype(&quot;4 4&quot;));<br /> draw((8,1)--(8,-.5), linetype(&quot;4 4&quot;));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{A only}\qquad\textbf{(B)}\ \text{B only}\qquad\textbf{(C)}\ \text{C only}\qquad\textbf{(D)}\ \text{both A and B}\qquad\textbf{(E)}\ \text{all are equal}&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> In the pattern below, the cat moves clockwise through the four squares and the mouse moves counterclockwise through the eight exterior segments of the four squares.<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23a.png|800px]]<br /> &lt;/center&gt;<br /> <br /> If the pattern is continued, where would the cat and mouse be after the 247th move?<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23b.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A ship travels from point &lt;math&gt;A&lt;/math&gt; to point &lt;math&gt;B&lt;/math&gt; along a semicircular path, centered at Island &lt;math&gt;X&lt;/math&gt;. Then it travels along a straight path from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt;. Which of these graphs best shows the ship's distance from Island &lt;math&gt;X&lt;/math&gt; as it moves along its course?<br /> <br /> &lt;asy&gt;size(150);<br /> pair X=origin, A=(-5,0), B=(5,0), C=(0,5);<br /> draw(Arc(X, 5, 180, 360)^^B--C);<br /> dot(X);<br /> label(&quot;$X$&quot;, X, NE);<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$B$&quot;, B, E);<br /> label(&quot;$A$&quot;, A, W);<br /> &lt;/asy&gt;<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob24ans.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> In the figure, the area of square &lt;math&gt;WXYZ&lt;/math&gt; is &lt;math&gt;25 \text{ cm}^2&lt;/math&gt;. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = AC&lt;/math&gt;, and when &lt;math&gt;\triangle ABC&lt;/math&gt; is folded over side &lt;math&gt;\overline{BC}&lt;/math&gt;, point &lt;math&gt;A&lt;/math&gt; coincides with &lt;math&gt;O&lt;/math&gt;, the center of square &lt;math&gt;WXYZ&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;, in square centimeters?<br /> <br /> &lt;asy&gt;<br /> defaultpen(fontsize(8));<br /> size(225);<br /> pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5);<br /> draw((-4,0)--Y--X--(-4,10)--cycle);<br /> draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle);<br /> dot(O);<br /> label(&quot;$A$&quot;, A, NW);<br /> label(&quot;$O$&quot;, O, NE);<br /> label(&quot;$B$&quot;, B, SW);<br /> label(&quot;$C$&quot;, C, NW);<br /> label(&quot;$W$&quot;,W , NE);<br /> label(&quot;$X$&quot;, X, N);<br /> label(&quot;$Y$&quot;, Y, S);<br /> label(&quot;$Z$&quot;, Z, SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2003|before=[[2002 AMC 8 Problems|2002 AMC 8]]|after=[[2004 AMC 8 Problems|2004 AMC 8]]}}<br /> * [[AMC 8]]<br /> * [[AMC 8 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems&diff=139877 2003 AMC 8 Problems 2020-12-17T22:29:47Z <p>G1zq: Added Problems 9 and 10</p> <hr /> <div>==Problem 1==<br /> Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Which of the following numbers has the smallest prime factor?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 55 \qquad\mathrm{(B)}\ 57 \qquad\mathrm{(C)}\ 58 \qquad\mathrm{(D)}\ 59 \qquad\mathrm{(E)}\ 61&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> A burger at Ricky C's weighs 120 grams, of which 30 grams are filler. What percent of the burger is not filler?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 60\% \qquad\mathrm{(B)}\ 65\% \qquad\mathrm{(C)}\ 70\% \qquad\mathrm{(D)}\ 75\% \qquad\mathrm{(E)}\ 90\%&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> If 20% of a number is 12, what is 30% of the same number?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 24 \qquad\mathrm{(E)}\ 30&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Given the areas of the three squares in the figure, what is the area of the interior triangle?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1));<br /> label(&quot;$25$&quot;,(14.5,1),N);<br /> label(&quot;$144$&quot;,(6,-7.5),N);<br /> label(&quot;$169$&quot;,(3.5,7),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Jenny scored 10 points higher than Blake on the first test, 10 points lower than him on the second test, and 20 points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?<br /> <br /> &lt;math&gt; \mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> &lt;math&gt;\textbf{Bake Sale}&lt;/math&gt;<br /> <br /> (Problems 8, 9, and 10 use the data found in the accompanying paragraph and figures)<br /> <br /> Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Art's cookies are trapezoids. <br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(5,0)--(5,3)--(2,3)--cycle);<br /> draw(rightanglemark((5,3), (5,0), origin));<br /> label(&quot;5 in&quot;, (2.5,0), S);<br /> label(&quot;3 in&quot;, (5,1.5), E);<br /> label(&quot;3 in&quot;, (3.5,3), N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Roger's cookies are rectangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(4,0)--(4,2)--(0,2)--cycle);<br /> draw(rightanglemark((4,2), (4,0), origin));<br /> draw(rightanglemark((0,2), origin, (4,0)));<br /> label(&quot;4 in&quot;, (2,0), S);<br /> label(&quot;2 in&quot;, (4,1), E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Paul's cookies are parallelograms.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle);<br /> draw((2.5,2)--(2.5,0), dashed);<br /> draw(rightanglemark((2.5,2),(2.5,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;2 in&quot;, (2.5,1), W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Trisha's cookies are triangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(3,4)--cycle);<br /> draw(rightanglemark((3,4),(3,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;4 in&quot;, (3,2), E);<br /> &lt;/asy&gt;<br /> <br /> Each friend uses the same amount of dough, and Art makes exactly 12 cookies. Who gets the fewest cookies from one batch of cookie dough?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Art}\qquad\textbf{(B)}\ \text{Roger}\qquad\textbf{(C)}\ \text{Paul}\qquad\textbf{(D)}\ \text{Trisha}\qquad\textbf{(E)}\ \text{There is a tie for fewest.} &lt;/math&gt;<br /> <br /> ==Problem 9==<br /> <br /> ==Problem 10==<br /> <br /> ==Problem 11==<br /> <br /> Business is a little slow at Lou's Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday's prices by 10%. Over the weekend, Lou advertises the sale: &quot;Ten percent off the listed price. Sale starts Monday.&quot; How much does a pair of shoes cost on Monday that cost 40 dollars on Thursday?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 39.60\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 40.40\qquad\textbf{(E)}\ 44 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> When a fair six-sided die is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the numbers on the five faces that can be seen is divisible by 6?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{5}{6}\qquad\textbf{(E)}\ 1&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> defaultpen(linewidth(0.8));<br /> real r=0.5;<br /> currentprojection=orthographic(3/4,8/15,7/15);<br /> draw(unitcube, white, thick(), nolight);<br /> draw(shift(1,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(0,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,1)*unitcube, white, thick(), nolight);<br /> draw(shift(1,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,1)*unitcube, white, thick(), nolight);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> In this addition problem, each letter stands for a different digit. <br /> <br /> &lt;math&gt; \setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&amp;T &amp; W &amp; O\\ + &amp;T &amp; W &amp; O\\ \hline F&amp; O &amp; U &amp; R\end{array} &lt;/math&gt;<br /> <br /> If &lt;math&gt;T = 7&lt;/math&gt; and the letter &lt;math&gt;O&lt;/math&gt; represents an even number, what is the only possible value for &lt;math&gt;W&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.8));<br /> path p=unitsquare;<br /> draw(p^^shift(0,1)*p^^shift(1,0)*p);<br /> draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p);<br /> label(&quot;FRONT&quot;, (1,0), S);<br /> label(&quot;SIDE&quot;, (5,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has 4 seats: 1 driver's seat, 1 front passenger seat, and 2 back passenger seats. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?<br /> <br /> &lt;cmath&gt; \begin{array}{c|c|c}\text{Child}&amp;\text{Eye Color}&amp;\text{Hair Color}\\ \hline\text{Benjamin}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Jim}&amp;\text{Brown}&amp;\text{Blonde}\\ \hline\text{Nadeen}&amp;\text{Brown}&amp;\text{Black}\\ \hline\text{Austin}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\text{Tevyn}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Sue}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\end{array} &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Nadeen and Austin}\qquad\textbf{(B)}\ \text{Benjamin and Sue}\qquad\textbf{(C)}\ \text{Benjamin and Austin}\qquad\textbf{(D)}\ \text{Nadeen and Tevyn}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(E)}\ \text{Austin and Sue} &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah's party?<br /> &lt;asy&gt;/* AMC8 2003 #18 Problem */<br /> pair a=(102,256), b=(68,131), c=(162,101), d=(134,150);<br /> pair e=(269,105), f=(359,104), g=(303,12), h=(579,211);<br /> pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501);<br /> pair m=(282,411), n=(147,451), o=(103,437), p=(31,373);<br /> pair q=(419,175), r=(462,209), s=(477,288), t=(443,358);<br /> pair oval=(282,303);<br /> draw(l--m--n--cycle);<br /> draw(p--oval);<br /> draw(o--oval);<br /> draw(b--d--oval);<br /> draw(c--d--e--oval);<br /> draw(e--f--g--h--i--j--oval);<br /> draw(k--oval);<br /> draw(q--oval);<br /> draw(s--oval);<br /> draw(r--s--t--oval);<br /> dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h);<br /> dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p);<br /> dot(q); dot(r); dot(s); dot(t);<br /> filldraw(yscale(.5)*Circle((282,606),80),white,black);<br /> label(scale(0.75)*&quot;Sarah&quot;, oval);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> The area of trapezoid &lt;math&gt; ABCD&lt;/math&gt; is &lt;math&gt;164\text{ cm}^2&lt;/math&gt;. The altitude is 8 cm, &lt;math&gt;AB&lt;/math&gt; is 10 cm, and &lt;math&gt;CD&lt;/math&gt; is 17 cm. What is &lt;math&gt;BC&lt;/math&gt;, in centimeters?<br /> <br /> &lt;asy&gt;/* AMC8 2003 #21 Problem */<br /> size(4inch,2inch);<br /> draw((0,0)--(31,0)--(16,8)--(6,8)--cycle);<br /> draw((11,8)--(11,0), linetype(&quot;8 4&quot;));<br /> draw((11,1)--(12,1)--(12,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$D$&quot;, (31,0), SE);<br /> label(&quot;$B$&quot;, (6,8), NW);<br /> label(&quot;$C$&quot;, (16,8), NE);<br /> label(&quot;10&quot;, (3,5), W);<br /> label(&quot;8&quot;, (11,4), E);<br /> label(&quot;17&quot;, (22.5,5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> The following figures are composed of squares and circles. Which figure has a shaded region with largest area?<br /> &lt;asy&gt;/* AMC8 2003 #22 Problem */<br /> size(3inch, 2inch);<br /> unitsize(1cm);<br /> pen outline = black+linewidth(1);<br /> filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline);<br /> filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline);<br /> filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1));<br /> filldraw(Circle((1,1), 1), white, outline);<br /> filldraw(Circle((3.5,.5), .5), white, outline);<br /> filldraw(Circle((4.5,.5), .5), white, outline);<br /> filldraw(Circle((3.5,1.5), .5), white, outline);<br /> filldraw(Circle((4.5,1.5), .5), white, outline);<br /> filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline);<br /> label(&quot;A&quot;, (1, 2), N);<br /> label(&quot;B&quot;, (4, 2), N);<br /> label(&quot;C&quot;, (7, 2), N);<br /> draw((0,-.5)--(.5,-.5), BeginArrow);<br /> draw((1.5, -.5)--(2, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (1, -.5));<br /> <br /> draw((3,-.5)--(3.5,-.5), BeginArrow);<br /> draw((4.5, -.5)--(5, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (4, -.5));<br /> <br /> draw((6,-.5)--(6.5,-.5), BeginArrow);<br /> draw((7.5, -.5)--(8, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (7, -.5));<br /> <br /> draw((6,1)--(6,-.5), linetype(&quot;4 4&quot;));<br /> draw((8,1)--(8,-.5), linetype(&quot;4 4&quot;));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{A only}\qquad\textbf{(B)}\ \text{B only}\qquad\textbf{(C)}\ \text{C only}\qquad\textbf{(D)}\ \text{both A and B}\qquad\textbf{(E)}\ \text{all are equal}&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> In the pattern below, the cat moves clockwise through the four squares and the mouse moves counterclockwise through the eight exterior segments of the four squares.<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23a.png|800px]]<br /> &lt;/center&gt;<br /> <br /> If the pattern is continued, where would the cat and mouse be after the 247th move?<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23b.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A ship travels from point &lt;math&gt;A&lt;/math&gt; to point &lt;math&gt;B&lt;/math&gt; along a semicircular path, centered at Island &lt;math&gt;X&lt;/math&gt;. Then it travels along a straight path from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt;. Which of these graphs best shows the ship's distance from Island &lt;math&gt;X&lt;/math&gt; as it moves along its course?<br /> <br /> &lt;asy&gt;size(150);<br /> pair X=origin, A=(-5,0), B=(5,0), C=(0,5);<br /> draw(Arc(X, 5, 180, 360)^^B--C);<br /> dot(X);<br /> label(&quot;$X$&quot;, X, NE);<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$B$&quot;, B, E);<br /> label(&quot;$A$&quot;, A, W);<br /> &lt;/asy&gt;<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob24ans.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> In the figure, the area of square &lt;math&gt;WXYZ&lt;/math&gt; is &lt;math&gt;25 \text{ cm}^2&lt;/math&gt;. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = AC&lt;/math&gt;, and when &lt;math&gt;\triangle ABC&lt;/math&gt; is folded over side &lt;math&gt;\overline{BC}&lt;/math&gt;, point &lt;math&gt;A&lt;/math&gt; coincides with &lt;math&gt;O&lt;/math&gt;, the center of square &lt;math&gt;WXYZ&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;, in square centimeters?<br /> <br /> &lt;asy&gt;<br /> defaultpen(fontsize(8));<br /> size(225);<br /> pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5);<br /> draw((-4,0)--Y--X--(-4,10)--cycle);<br /> draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle);<br /> dot(O);<br /> label(&quot;$A$&quot;, A, NW);<br /> label(&quot;$O$&quot;, O, NE);<br /> label(&quot;$B$&quot;, B, SW);<br /> label(&quot;$C$&quot;, C, NW);<br /> label(&quot;$W$&quot;,W , NE);<br /> label(&quot;$X$&quot;, X, N);<br /> label(&quot;$Y$&quot;, Y, S);<br /> label(&quot;$Z$&quot;, Z, SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2003|before=[[2002 AMC 8 Problems|2002 AMC 8]]|after=[[2004 AMC 8 Problems|2004 AMC 8]]}}<br /> * [[AMC 8]]<br /> * [[AMC 8 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems&diff=139876 2003 AMC 8 Problems 2020-12-17T22:28:55Z <p>G1zq: /* Problem 8 */</p> <hr /> <div>==Problem 1==<br /> Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Which of the following numbers has the smallest prime factor?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 55 \qquad\mathrm{(B)}\ 57 \qquad\mathrm{(C)}\ 58 \qquad\mathrm{(D)}\ 59 \qquad\mathrm{(E)}\ 61&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> A burger at Ricky C's weighs 120 grams, of which 30 grams are filler. What percent of the burger is not filler?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 60\% \qquad\mathrm{(B)}\ 65\% \qquad\mathrm{(C)}\ 70\% \qquad\mathrm{(D)}\ 75\% \qquad\mathrm{(E)}\ 90\%&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> If 20% of a number is 12, what is 30% of the same number?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 24 \qquad\mathrm{(E)}\ 30&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Given the areas of the three squares in the figure, what is the area of the interior triangle?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1));<br /> label(&quot;$25$&quot;,(14.5,1),N);<br /> label(&quot;$144$&quot;,(6,-7.5),N);<br /> label(&quot;$169$&quot;,(3.5,7),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Jenny scored 10 points higher than Blake on the first test, 10 points lower than him on the second test, and 20 points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?<br /> <br /> &lt;math&gt; \mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> &lt;math&gt;\textbf{Bake Sale}&lt;/math&gt;<br /> <br /> (Problems 8, 9, and 10 use the data found in the accompanying paragraph and figures)<br /> <br /> Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Art's cookies are trapezoids. <br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(5,0)--(5,3)--(2,3)--cycle);<br /> draw(rightanglemark((5,3), (5,0), origin));<br /> label(&quot;5 in&quot;, (2.5,0), S);<br /> label(&quot;3 in&quot;, (5,1.5), E);<br /> label(&quot;3 in&quot;, (3.5,3), N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Roger's cookies are rectangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(4,0)--(4,2)--(0,2)--cycle);<br /> draw(rightanglemark((4,2), (4,0), origin));<br /> draw(rightanglemark((0,2), origin, (4,0)));<br /> label(&quot;4 in&quot;, (2,0), S);<br /> label(&quot;2 in&quot;, (4,1), E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Paul's cookies are parallelograms.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle);<br /> draw((2.5,2)--(2.5,0), dashed);<br /> draw(rightanglemark((2.5,2),(2.5,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;2 in&quot;, (2.5,1), W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Trisha's cookies are triangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(3,4)--cycle);<br /> draw(rightanglemark((3,4),(3,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;4 in&quot;, (3,2), E);<br /> &lt;/asy&gt;<br /> <br /> Each friend uses the same amount of dough, and Art makes exactly 12 cookies. Who gets the fewest cookies from one batch of cookie dough?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Art}\qquad\textbf{(B)}\ \text{Roger}\qquad\textbf{(C)}\ \text{Paul}\qquad\textbf{(D)}\ \text{Trisha}\qquad\textbf{(E)}\ \text{There is a tie for fewest.} &lt;/math&gt;<br /> <br /> ==Problem 11==<br /> <br /> Business is a little slow at Lou's Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday's prices by 10%. Over the weekend, Lou advertises the sale: &quot;Ten percent off the listed price. Sale starts Monday.&quot; How much does a pair of shoes cost on Monday that cost 40 dollars on Thursday?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 39.60\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 40.40\qquad\textbf{(E)}\ 44 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> When a fair six-sided die is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the numbers on the five faces that can be seen is divisible by 6?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{5}{6}\qquad\textbf{(E)}\ 1&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> defaultpen(linewidth(0.8));<br /> real r=0.5;<br /> currentprojection=orthographic(3/4,8/15,7/15);<br /> draw(unitcube, white, thick(), nolight);<br /> draw(shift(1,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(0,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,1)*unitcube, white, thick(), nolight);<br /> draw(shift(1,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,1)*unitcube, white, thick(), nolight);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> In this addition problem, each letter stands for a different digit. <br /> <br /> &lt;math&gt; \setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&amp;T &amp; W &amp; O\\ + &amp;T &amp; W &amp; O\\ \hline F&amp; O &amp; U &amp; R\end{array} &lt;/math&gt;<br /> <br /> If &lt;math&gt;T = 7&lt;/math&gt; and the letter &lt;math&gt;O&lt;/math&gt; represents an even number, what is the only possible value for &lt;math&gt;W&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.8));<br /> path p=unitsquare;<br /> draw(p^^shift(0,1)*p^^shift(1,0)*p);<br /> draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p);<br /> label(&quot;FRONT&quot;, (1,0), S);<br /> label(&quot;SIDE&quot;, (5,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has 4 seats: 1 driver's seat, 1 front passenger seat, and 2 back passenger seats. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?<br /> <br /> &lt;cmath&gt; \begin{array}{c|c|c}\text{Child}&amp;\text{Eye Color}&amp;\text{Hair Color}\\ \hline\text{Benjamin}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Jim}&amp;\text{Brown}&amp;\text{Blonde}\\ \hline\text{Nadeen}&amp;\text{Brown}&amp;\text{Black}\\ \hline\text{Austin}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\text{Tevyn}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Sue}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\end{array} &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Nadeen and Austin}\qquad\textbf{(B)}\ \text{Benjamin and Sue}\qquad\textbf{(C)}\ \text{Benjamin and Austin}\qquad\textbf{(D)}\ \text{Nadeen and Tevyn}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(E)}\ \text{Austin and Sue} &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah's party?<br /> &lt;asy&gt;/* AMC8 2003 #18 Problem */<br /> pair a=(102,256), b=(68,131), c=(162,101), d=(134,150);<br /> pair e=(269,105), f=(359,104), g=(303,12), h=(579,211);<br /> pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501);<br /> pair m=(282,411), n=(147,451), o=(103,437), p=(31,373);<br /> pair q=(419,175), r=(462,209), s=(477,288), t=(443,358);<br /> pair oval=(282,303);<br /> draw(l--m--n--cycle);<br /> draw(p--oval);<br /> draw(o--oval);<br /> draw(b--d--oval);<br /> draw(c--d--e--oval);<br /> draw(e--f--g--h--i--j--oval);<br /> draw(k--oval);<br /> draw(q--oval);<br /> draw(s--oval);<br /> draw(r--s--t--oval);<br /> dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h);<br /> dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p);<br /> dot(q); dot(r); dot(s); dot(t);<br /> filldraw(yscale(.5)*Circle((282,606),80),white,black);<br /> label(scale(0.75)*&quot;Sarah&quot;, oval);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> The area of trapezoid &lt;math&gt; ABCD&lt;/math&gt; is &lt;math&gt;164\text{ cm}^2&lt;/math&gt;. The altitude is 8 cm, &lt;math&gt;AB&lt;/math&gt; is 10 cm, and &lt;math&gt;CD&lt;/math&gt; is 17 cm. What is &lt;math&gt;BC&lt;/math&gt;, in centimeters?<br /> <br /> &lt;asy&gt;/* AMC8 2003 #21 Problem */<br /> size(4inch,2inch);<br /> draw((0,0)--(31,0)--(16,8)--(6,8)--cycle);<br /> draw((11,8)--(11,0), linetype(&quot;8 4&quot;));<br /> draw((11,1)--(12,1)--(12,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$D$&quot;, (31,0), SE);<br /> label(&quot;$B$&quot;, (6,8), NW);<br /> label(&quot;$C$&quot;, (16,8), NE);<br /> label(&quot;10&quot;, (3,5), W);<br /> label(&quot;8&quot;, (11,4), E);<br /> label(&quot;17&quot;, (22.5,5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> The following figures are composed of squares and circles. Which figure has a shaded region with largest area?<br /> &lt;asy&gt;/* AMC8 2003 #22 Problem */<br /> size(3inch, 2inch);<br /> unitsize(1cm);<br /> pen outline = black+linewidth(1);<br /> filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline);<br /> filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline);<br /> filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1));<br /> filldraw(Circle((1,1), 1), white, outline);<br /> filldraw(Circle((3.5,.5), .5), white, outline);<br /> filldraw(Circle((4.5,.5), .5), white, outline);<br /> filldraw(Circle((3.5,1.5), .5), white, outline);<br /> filldraw(Circle((4.5,1.5), .5), white, outline);<br /> filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline);<br /> label(&quot;A&quot;, (1, 2), N);<br /> label(&quot;B&quot;, (4, 2), N);<br /> label(&quot;C&quot;, (7, 2), N);<br /> draw((0,-.5)--(.5,-.5), BeginArrow);<br /> draw((1.5, -.5)--(2, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (1, -.5));<br /> <br /> draw((3,-.5)--(3.5,-.5), BeginArrow);<br /> draw((4.5, -.5)--(5, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (4, -.5));<br /> <br /> draw((6,-.5)--(6.5,-.5), BeginArrow);<br /> draw((7.5, -.5)--(8, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (7, -.5));<br /> <br /> draw((6,1)--(6,-.5), linetype(&quot;4 4&quot;));<br /> draw((8,1)--(8,-.5), linetype(&quot;4 4&quot;));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{A only}\qquad\textbf{(B)}\ \text{B only}\qquad\textbf{(C)}\ \text{C only}\qquad\textbf{(D)}\ \text{both A and B}\qquad\textbf{(E)}\ \text{all are equal}&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> In the pattern below, the cat moves clockwise through the four squares and the mouse moves counterclockwise through the eight exterior segments of the four squares.<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23a.png|800px]]<br /> &lt;/center&gt;<br /> <br /> If the pattern is continued, where would the cat and mouse be after the 247th move?<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23b.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A ship travels from point &lt;math&gt;A&lt;/math&gt; to point &lt;math&gt;B&lt;/math&gt; along a semicircular path, centered at Island &lt;math&gt;X&lt;/math&gt;. Then it travels along a straight path from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt;. Which of these graphs best shows the ship's distance from Island &lt;math&gt;X&lt;/math&gt; as it moves along its course?<br /> <br /> &lt;asy&gt;size(150);<br /> pair X=origin, A=(-5,0), B=(5,0), C=(0,5);<br /> draw(Arc(X, 5, 180, 360)^^B--C);<br /> dot(X);<br /> label(&quot;$X$&quot;, X, NE);<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$B$&quot;, B, E);<br /> label(&quot;$A$&quot;, A, W);<br /> &lt;/asy&gt;<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob24ans.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> In the figure, the area of square &lt;math&gt;WXYZ&lt;/math&gt; is &lt;math&gt;25 \text{ cm}^2&lt;/math&gt;. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = AC&lt;/math&gt;, and when &lt;math&gt;\triangle ABC&lt;/math&gt; is folded over side &lt;math&gt;\overline{BC}&lt;/math&gt;, point &lt;math&gt;A&lt;/math&gt; coincides with &lt;math&gt;O&lt;/math&gt;, the center of square &lt;math&gt;WXYZ&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;, in square centimeters?<br /> <br /> &lt;asy&gt;<br /> defaultpen(fontsize(8));<br /> size(225);<br /> pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5);<br /> draw((-4,0)--Y--X--(-4,10)--cycle);<br /> draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle);<br /> dot(O);<br /> label(&quot;$A$&quot;, A, NW);<br /> label(&quot;$O$&quot;, O, NE);<br /> label(&quot;$B$&quot;, B, SW);<br /> label(&quot;$C$&quot;, C, NW);<br /> label(&quot;$W$&quot;,W , NE);<br /> label(&quot;$X$&quot;, X, N);<br /> label(&quot;$Y$&quot;, Y, S);<br /> label(&quot;$Z$&quot;, Z, SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2003|before=[[2002 AMC 8 Problems|2002 AMC 8]]|after=[[2004 AMC 8 Problems|2004 AMC 8]]}}<br /> * [[AMC 8]]<br /> * [[AMC 8 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems&diff=139875 2003 AMC 8 Problems 2020-12-17T22:27:51Z <p>G1zq: Added Problem 8</p> <hr /> <div>==Problem 1==<br /> Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Which of the following numbers has the smallest prime factor?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 55 \qquad\mathrm{(B)}\ 57 \qquad\mathrm{(C)}\ 58 \qquad\mathrm{(D)}\ 59 \qquad\mathrm{(E)}\ 61&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> A burger at Ricky C's weighs 120 grams, of which 30 grams are filler. What percent of the burger is not filler?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 60\% \qquad\mathrm{(B)}\ 65\% \qquad\mathrm{(C)}\ 70\% \qquad\mathrm{(D)}\ 75\% \qquad\mathrm{(E)}\ 90\%&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> If 20% of a number is 12, what is 30% of the same number?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 24 \qquad\mathrm{(E)}\ 30&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Given the areas of the three squares in the figure, what is the area of the interior triangle?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1));<br /> label(&quot;$25$&quot;,(14.5,1),N);<br /> label(&quot;$144$&quot;,(6,-7.5),N);<br /> label(&quot;$169$&quot;,(3.5,7),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Jenny scored 10 points higher than Blake on the first test, 10 points lower than him on the second test, and 20 points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?<br /> <br /> &lt;math&gt; \mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> &lt;math&gt;\textbf{Bake Sale}&lt;/math&gt;<br /> <br /> Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Art's cookies are trapezoids. <br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(5,0)--(5,3)--(2,3)--cycle);<br /> draw(rightanglemark((5,3), (5,0), origin));<br /> label(&quot;5 in&quot;, (2.5,0), S);<br /> label(&quot;3 in&quot;, (5,1.5), E);<br /> label(&quot;3 in&quot;, (3.5,3), N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Roger's cookies are rectangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(4,0)--(4,2)--(0,2)--cycle);<br /> draw(rightanglemark((4,2), (4,0), origin));<br /> draw(rightanglemark((0,2), origin, (4,0)));<br /> label(&quot;4 in&quot;, (2,0), S);<br /> label(&quot;2 in&quot;, (4,1), E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Paul's cookies are parallelograms.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle);<br /> draw((2.5,2)--(2.5,0), dashed);<br /> draw(rightanglemark((2.5,2),(2.5,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;2 in&quot;, (2.5,1), W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\circ&lt;/math&gt; Trisha's cookies are triangles.<br /> &lt;asy&gt;<br /> size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));<br /> draw(origin--(3,0)--(3,4)--cycle);<br /> draw(rightanglemark((3,4),(3,0), origin));<br /> label(&quot;3 in&quot;, (1.5,0), S);<br /> label(&quot;4 in&quot;, (3,2), E);<br /> &lt;/asy&gt;<br /> <br /> Each friend uses the same amount of dough, and Art makes exactly 12 cookies. Who gets the fewest cookies from one batch of cookie dough?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Art}\qquad\textbf{(B)}\ \text{Roger}\qquad\textbf{(C)}\ \text{Paul}\qquad\textbf{(D)}\ \text{Trisha}\qquad\textbf{(E)}\ \text{There is a tie for fewest.} &lt;/math&gt;<br /> <br /> ==Problem 11==<br /> <br /> Business is a little slow at Lou's Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday's prices by 10%. Over the weekend, Lou advertises the sale: &quot;Ten percent off the listed price. Sale starts Monday.&quot; How much does a pair of shoes cost on Monday that cost 40 dollars on Thursday?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 39.60\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 40.40\qquad\textbf{(E)}\ 44 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> When a fair six-sided die is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the numbers on the five faces that can be seen is divisible by 6?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{5}{6}\qquad\textbf{(E)}\ 1&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> defaultpen(linewidth(0.8));<br /> real r=0.5;<br /> currentprojection=orthographic(3/4,8/15,7/15);<br /> draw(unitcube, white, thick(), nolight);<br /> draw(shift(1,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(0,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,1)*unitcube, white, thick(), nolight);<br /> draw(shift(1,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,1)*unitcube, white, thick(), nolight);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> In this addition problem, each letter stands for a different digit. <br /> <br /> &lt;math&gt; \setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&amp;T &amp; W &amp; O\\ + &amp;T &amp; W &amp; O\\ \hline F&amp; O &amp; U &amp; R\end{array} &lt;/math&gt;<br /> <br /> If &lt;math&gt;T = 7&lt;/math&gt; and the letter &lt;math&gt;O&lt;/math&gt; represents an even number, what is the only possible value for &lt;math&gt;W&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.8));<br /> path p=unitsquare;<br /> draw(p^^shift(0,1)*p^^shift(1,0)*p);<br /> draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p);<br /> label(&quot;FRONT&quot;, (1,0), S);<br /> label(&quot;SIDE&quot;, (5,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has 4 seats: 1 driver's seat, 1 front passenger seat, and 2 back passenger seats. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?<br /> <br /> &lt;cmath&gt; \begin{array}{c|c|c}\text{Child}&amp;\text{Eye Color}&amp;\text{Hair Color}\\ \hline\text{Benjamin}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Jim}&amp;\text{Brown}&amp;\text{Blonde}\\ \hline\text{Nadeen}&amp;\text{Brown}&amp;\text{Black}\\ \hline\text{Austin}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\text{Tevyn}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Sue}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\end{array} &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Nadeen and Austin}\qquad\textbf{(B)}\ \text{Benjamin and Sue}\qquad\textbf{(C)}\ \text{Benjamin and Austin}\qquad\textbf{(D)}\ \text{Nadeen and Tevyn}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(E)}\ \text{Austin and Sue} &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah's party?<br /> &lt;asy&gt;/* AMC8 2003 #18 Problem */<br /> pair a=(102,256), b=(68,131), c=(162,101), d=(134,150);<br /> pair e=(269,105), f=(359,104), g=(303,12), h=(579,211);<br /> pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501);<br /> pair m=(282,411), n=(147,451), o=(103,437), p=(31,373);<br /> pair q=(419,175), r=(462,209), s=(477,288), t=(443,358);<br /> pair oval=(282,303);<br /> draw(l--m--n--cycle);<br /> draw(p--oval);<br /> draw(o--oval);<br /> draw(b--d--oval);<br /> draw(c--d--e--oval);<br /> draw(e--f--g--h--i--j--oval);<br /> draw(k--oval);<br /> draw(q--oval);<br /> draw(s--oval);<br /> draw(r--s--t--oval);<br /> dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h);<br /> dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p);<br /> dot(q); dot(r); dot(s); dot(t);<br /> filldraw(yscale(.5)*Circle((282,606),80),white,black);<br /> label(scale(0.75)*&quot;Sarah&quot;, oval);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> The area of trapezoid &lt;math&gt; ABCD&lt;/math&gt; is &lt;math&gt;164\text{ cm}^2&lt;/math&gt;. The altitude is 8 cm, &lt;math&gt;AB&lt;/math&gt; is 10 cm, and &lt;math&gt;CD&lt;/math&gt; is 17 cm. What is &lt;math&gt;BC&lt;/math&gt;, in centimeters?<br /> <br /> &lt;asy&gt;/* AMC8 2003 #21 Problem */<br /> size(4inch,2inch);<br /> draw((0,0)--(31,0)--(16,8)--(6,8)--cycle);<br /> draw((11,8)--(11,0), linetype(&quot;8 4&quot;));<br /> draw((11,1)--(12,1)--(12,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$D$&quot;, (31,0), SE);<br /> label(&quot;$B$&quot;, (6,8), NW);<br /> label(&quot;$C$&quot;, (16,8), NE);<br /> label(&quot;10&quot;, (3,5), W);<br /> label(&quot;8&quot;, (11,4), E);<br /> label(&quot;17&quot;, (22.5,5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> The following figures are composed of squares and circles. Which figure has a shaded region with largest area?<br /> &lt;asy&gt;/* AMC8 2003 #22 Problem */<br /> size(3inch, 2inch);<br /> unitsize(1cm);<br /> pen outline = black+linewidth(1);<br /> filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline);<br /> filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline);<br /> filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1));<br /> filldraw(Circle((1,1), 1), white, outline);<br /> filldraw(Circle((3.5,.5), .5), white, outline);<br /> filldraw(Circle((4.5,.5), .5), white, outline);<br /> filldraw(Circle((3.5,1.5), .5), white, outline);<br /> filldraw(Circle((4.5,1.5), .5), white, outline);<br /> filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline);<br /> label(&quot;A&quot;, (1, 2), N);<br /> label(&quot;B&quot;, (4, 2), N);<br /> label(&quot;C&quot;, (7, 2), N);<br /> draw((0,-.5)--(.5,-.5), BeginArrow);<br /> draw((1.5, -.5)--(2, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (1, -.5));<br /> <br /> draw((3,-.5)--(3.5,-.5), BeginArrow);<br /> draw((4.5, -.5)--(5, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (4, -.5));<br /> <br /> draw((6,-.5)--(6.5,-.5), BeginArrow);<br /> draw((7.5, -.5)--(8, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (7, -.5));<br /> <br /> draw((6,1)--(6,-.5), linetype(&quot;4 4&quot;));<br /> draw((8,1)--(8,-.5), linetype(&quot;4 4&quot;));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{A only}\qquad\textbf{(B)}\ \text{B only}\qquad\textbf{(C)}\ \text{C only}\qquad\textbf{(D)}\ \text{both A and B}\qquad\textbf{(E)}\ \text{all are equal}&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> In the pattern below, the cat moves clockwise through the four squares and the mouse moves counterclockwise through the eight exterior segments of the four squares.<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23a.png|800px]]<br /> &lt;/center&gt;<br /> <br /> If the pattern is continued, where would the cat and mouse be after the 247th move?<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23b.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A ship travels from point &lt;math&gt;A&lt;/math&gt; to point &lt;math&gt;B&lt;/math&gt; along a semicircular path, centered at Island &lt;math&gt;X&lt;/math&gt;. Then it travels along a straight path from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt;. Which of these graphs best shows the ship's distance from Island &lt;math&gt;X&lt;/math&gt; as it moves along its course?<br /> <br /> &lt;asy&gt;size(150);<br /> pair X=origin, A=(-5,0), B=(5,0), C=(0,5);<br /> draw(Arc(X, 5, 180, 360)^^B--C);<br /> dot(X);<br /> label(&quot;$X$&quot;, X, NE);<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$B$&quot;, B, E);<br /> label(&quot;$A$&quot;, A, W);<br /> &lt;/asy&gt;<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob24ans.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> In the figure, the area of square &lt;math&gt;WXYZ&lt;/math&gt; is &lt;math&gt;25 \text{ cm}^2&lt;/math&gt;. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = AC&lt;/math&gt;, and when &lt;math&gt;\triangle ABC&lt;/math&gt; is folded over side &lt;math&gt;\overline{BC}&lt;/math&gt;, point &lt;math&gt;A&lt;/math&gt; coincides with &lt;math&gt;O&lt;/math&gt;, the center of square &lt;math&gt;WXYZ&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;, in square centimeters?<br /> <br /> &lt;asy&gt;<br /> defaultpen(fontsize(8));<br /> size(225);<br /> pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5);<br /> draw((-4,0)--Y--X--(-4,10)--cycle);<br /> draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle);<br /> dot(O);<br /> label(&quot;$A$&quot;, A, NW);<br /> label(&quot;$O$&quot;, O, NE);<br /> label(&quot;$B$&quot;, B, SW);<br /> label(&quot;$C$&quot;, C, NW);<br /> label(&quot;$W$&quot;,W , NE);<br /> label(&quot;$X$&quot;, X, N);<br /> label(&quot;$Y$&quot;, Y, S);<br /> label(&quot;$Z$&quot;, Z, SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2003|before=[[2002 AMC 8 Problems|2002 AMC 8]]|after=[[2004 AMC 8 Problems|2004 AMC 8]]}}<br /> * [[AMC 8]]<br /> * [[AMC 8 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=User:G1zq&diff=115661 User:G1zq 2020-01-27T09:46:45Z <p>G1zq: /* About Me */</p> <hr /> <div>===About Me===<br /> Hello! I'm Spottedshade. I am a seventh grader who is obsessed with Nintendo. Feel free to PM me about anything.</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_13&diff=114899 2014 AMC 8 Problems/Problem 13 2020-01-17T07:07:14Z <p>G1zq: /* Solution */</p> <hr /> <div>==Problem==<br /> If &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; are integers and &lt;math&gt;n^2+m^2&lt;/math&gt; is even, which of the following is impossible?<br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; are even &lt;math&gt;\qquad\textbf{(B) }&lt;/math&gt; &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; are odd &lt;math&gt;\qquad\textbf{(C) }&lt;/math&gt; &lt;math&gt;n+m&lt;/math&gt; is even &lt;math&gt;\qquad\textbf{(D) }&lt;/math&gt; &lt;math&gt;n+m&lt;/math&gt; is odd &lt;math&gt;\qquad \textbf{(E) }&lt;/math&gt; none of these are impossible<br /> ==Solution==<br /> Since &lt;math&gt;n^2+m^2&lt;/math&gt; is even, either both &lt;math&gt;n^2&lt;/math&gt; and &lt;math&gt;m^2&lt;/math&gt; are even, or they are both odd. Therefore, &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; are either both even or both odd, since the square of an even number is even and the square of an odd number is odd. As a result, &lt;math&gt;n+m&lt;/math&gt; must be even. The answer, then, is &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2014|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_12&diff=114898 2014 AMC 8 Problems/Problem 12 2020-01-17T06:58:43Z <p>G1zq: /* Problem */</p> <hr /> <div>==Problem==<br /> A magazine printed photos of three celebrities along with three photos of the celebrities as babies. The baby pictures did not identify the celebrities. Readers were asked to match each celebrity with the correct baby pictures. What is the probability that a reader guessing at random will match all three correctly? <br /> <br /> &lt;math&gt; \textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{1}{6}\qquad\textbf{(C) }\frac{1}{4}\qquad\textbf{(D) }\frac{1}{3}\qquad\textbf{(E) }\frac{1}{2} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let's call the celebrities A, B, and C. <br /> There is a &lt;math&gt;\frac{1}{3}&lt;/math&gt; chance that celebrity A's picture will be selected, and a &lt;math&gt;\frac{1}{3}&lt;/math&gt; chance that his baby picture will be selected. That means there are two celebrities left. There is now a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance that celebrity B's picture will be selected, and another &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance that his baby picture will be selected. This leaves a &lt;math&gt;\frac{1}{1}&lt;/math&gt; chance for the last celebrity, so the total probability is &lt;math&gt;\frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{36}&lt;/math&gt;. However, the order of the celebrities doesn't matter, so the final probability will be &lt;math&gt; 3!\cdot \frac{1}{36}=\boxed{\text{(B) }\dfrac{1}{6}}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> There is a &lt;math&gt;\frac{1}{3}&lt;/math&gt; chance that the reader will choose the correct baby picture for the first person. Next, the second person gives a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance, and the last person leaves only 1 choice. Thus, the probability is &lt;math&gt;\dfrac{1}{3\cdot 2}=\boxed{\text{(B) }\dfrac{1}{6}}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> There are &lt;math&gt;3!&lt;/math&gt; ways assign the pictures to each of the celebrities. There is one favorable outcome where all of them are matched correctly, so the answer is &lt;math&gt;\boxed{\textbf{(B)}~\frac{1}{6}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2014|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_25&diff=104599 2009 AMC 8 Problems/Problem 25 2019-03-17T00:18:15Z <p>G1zq: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is &lt;math&gt;\frac{1}{2}&lt;/math&gt; foot from the top face. The second cut is &lt;math&gt;\frac{1}{3}&lt;/math&gt; foot below the first cut, and the third cut is &lt;math&gt;\frac{1}{17}&lt;/math&gt; foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet?<br /> &lt;asy&gt;<br /> import three;<br /> real d=11/102;<br /> defaultpen(fontsize(8));<br /> defaultpen(linewidth(0.8));<br /> currentprojection=orthographic(1,8/15,7/15);<br /> draw(unitcube, white, thick(), nolight);<br /> void f(real x) {<br /> draw((0,1,x)--(1,1,x)--(1,0,x));<br /> }<br /> f(d);<br /> f(1/6);<br /> f(1/2);<br /> label(&quot;A&quot;, (1,0,3/4), W);<br /> label(&quot;B&quot;, (1,0,1/3), W);<br /> label(&quot;C&quot;, (1,0,1/6-d/4), W);<br /> label(&quot;D&quot;, (1,0,d/2), W);<br /> label(&quot;1/2&quot;, (1,1,3/4), E);<br /> label(&quot;1/3&quot;, (1,1,1/3), E);<br /> label(&quot;1/17&quot;, (0,1,1/6-d/4), E);&lt;/asy&gt;<br /> <br /> &lt;asy&gt;<br /> import three;<br /> real d=11/102;<br /> defaultpen(fontsize(8));<br /> defaultpen(linewidth(0.8));<br /> currentprojection=orthographic(2,8/15,7/15);<br /> int t=0;<br /> void f(real x) {<br /> path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle;<br /> path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle;<br /> path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle;<br /> draw(surface(r), white, nolight);<br /> draw(surface(f), white, nolight);<br /> draw(surface(u), white, nolight);<br /> draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x));<br /> t=t+1;<br /> }<br /> f(d);<br /> f(1/2);<br /> f(1/3);<br /> f(1/17);<br /> label(&quot;D&quot;, (1/2, 1, 0), SE);<br /> label(&quot;A&quot;, (1+1/2, 1, 0), SE);<br /> label(&quot;B&quot;, (2+1/2, 1, 0), SE);<br /> label(&quot;C&quot;, (3+1/2, 1, 0), SE);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\:6\qquad\textbf{(B)}\:7\qquad\textbf{(C)}\:\frac{419}{51}\qquad\textbf{(D)}\:\frac{158}{17}\qquad\textbf{(E)}\:11 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> The tops of &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt; in the figure formed has sum &lt;math&gt; 1+1+1+1 = 4 &lt;/math&gt; as do the bottoms. Thus, the total so far is &lt;math&gt;8&lt;/math&gt;. Now, one of the sides has area one, since it combines all of the heights of &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;, which is &lt;math&gt;1&lt;/math&gt;. The other side also satisfies this. Thus the total area now is &lt;math&gt;10&lt;/math&gt;. From the front, the surface area is half, because if you looked at it straight from the front it would look exactly like &lt;math&gt;A&lt;/math&gt;, with surface area half. From the back it is the same thing. Thus, the total is &lt;math&gt; 10+\frac{1}{2}+\frac{1}{2}= 11 &lt;/math&gt;, or &lt;math&gt; \boxed{\textbf{(E)}\:11 } &lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> <br /> The top parts and the bottom parts sum to 8. The sides add on another 2. Therefore, the only logical answer would be &lt;math&gt;\boxed{\textbf{(E)}\:11 }&lt;/math&gt; since it is the only answer greater than 10.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2009|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_18&diff=104162 2006 AMC 8 Problems/Problem 18 2019-03-08T20:11:58Z <p>G1zq: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> A cube with 3-inch edges is made using 27 cubes with 1-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white? <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{9}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{4}{9}\qquad\textbf{(D)}\ \frac{5}{9}\qquad\textbf{(E)}\ \frac{19}{27} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The surface area of the cube is &lt;math&gt;6(3)(3)=54&lt;/math&gt;. Each of the eight black cubes has 3 faces on the outside, making &lt;math&gt;3(8)=24&lt;/math&gt; black faces. Therefore there are &lt;math&gt;54-24=30&lt;/math&gt; white faces. To find the ratio, we evaluate &lt;math&gt;\frac{30}{54}= \boxed{\textbf{(D)}\ \frac{5}{9}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> We can notice that each face is the same, so each face is &lt;math&gt;\boxed{\textbf{(D)}\ \frac{5}{9}}&lt;/math&gt; white.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2006|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_18&diff=104161 2006 AMC 8 Problems/Problem 18 2019-03-08T20:11:48Z <p>G1zq: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> A cube with 3-inch edges is made using 27 cubes with 1-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white? <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{9}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{4}{9}\qquad\textbf{(D)}\ \frac{5}{9}\qquad\textbf{(E)}\ \frac{19}{27} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The surface area of the cube is &lt;math&gt;6(3)(3)=54&lt;/math&gt;. Each of the eight black cubes has 3 faces on the outside, making &lt;math&gt;3(8)=24&lt;/math&gt; black faces. Therefore there are &lt;math&gt;54-24=30&lt;/math&gt; white faces. To find the ratio, we evaluate &lt;math&gt;\frac{30}{54}= \boxed{\textbf{(D)}\ \frac{5}{9}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> We can notice that each face is the same, so each face is &lt;math&gt;boxed{\textbf{(D)}\ \frac{5}{9}}&lt;/math&gt; white.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2006|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_18&diff=104160 2006 AMC 8 Problems/Problem 18 2019-03-08T20:11:32Z <p>G1zq: /* Solution */</p> <hr /> <div>==Problem==<br /> A cube with 3-inch edges is made using 27 cubes with 1-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white? <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{9}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{4}{9}\qquad\textbf{(D)}\ \frac{5}{9}\qquad\textbf{(E)}\ \frac{19}{27} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The surface area of the cube is &lt;math&gt;6(3)(3)=54&lt;/math&gt;. Each of the eight black cubes has 3 faces on the outside, making &lt;math&gt;3(8)=24&lt;/math&gt; black faces. Therefore there are &lt;math&gt;54-24=30&lt;/math&gt; white faces. To find the ratio, we evaluate &lt;math&gt;\frac{30}{54}= \boxed{\textbf{(D)}\ \frac{5}{9}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> We can notice that each face is the same, so each face is boxed{\textbf{(D)}\ \frac{5}{9}}white.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2006|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=User_talk:G1zq&diff=100797 User talk:G1zq 2019-01-23T23:38:02Z <p>G1zq: Created page with &quot;===My Talk Page=== Please, feel free to talk here or to discuss!&quot;</p> <hr /> <div>===My Talk Page===<br /> <br /> Please, feel free to talk here or to discuss!</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=User:G1zq&diff=100796 User:G1zq 2019-01-23T23:36:24Z <p>G1zq: /* About Me */</p> <hr /> <div>===About Me===<br /> Hello! I'm Spottedshade. I am a fifth grader who adores math, novels, and most importantly, SUCCULENTS!!! I LOOVE succulents. I don't know why, but I just do. Please feel free to call me Spottedshade, g1zq(though Spottedshade is preferred), Spotty, Succulentspot, or Rainy! The first 3 are the most common.</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=User:G1zq&diff=100795 User:G1zq 2019-01-23T23:36:07Z <p>G1zq: </p> <hr /> <div>===About Me===<br /> Hello! I'm Spottedshade. I am a fifth grader who adores math, novels, and most importantly, SUCCULENTS!!! I LOOVE succulents. I don't know why, but I just do. Please feel free to call me Spottedshade, g[b]1[/b]zq(though Spottedshade is preferred), Spotty, Succulentspot, or Rainy! The first 3 are the most common.</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_23&diff=100794 2009 AMC 8 Problems/Problem 23 2019-01-23T23:31:49Z <p>G1zq: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought &lt;math&gt;400&lt;/math&gt; jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class? <br /> <br /> &lt;math&gt; \textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34 &lt;/math&gt;<br /> <br /> ==Solution==<br /> If there are &lt;math&gt;x&lt;/math&gt; girls, then there are &lt;math&gt;x+2&lt;/math&gt; boys. She gave each girl &lt;math&gt;x&lt;/math&gt; jellybeans and each boy &lt;math&gt;x+2&lt;/math&gt; jellybeans, for a total of &lt;math&gt;x^2 + (x+2)^2&lt;/math&gt; jellybeans. She gave away &lt;math&gt;400-6=394&lt;/math&gt; jellybeans.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> x^2+(x+2)^2 &amp;= 394\\<br /> x^2+x^2+4x+4 &amp;= 394\\<br /> 2x^2 + 4x - 390 &amp;= 0\\<br /> x^2 + 2x - 195 &amp;= 0\\<br /> (x+15)(x-13) &amp;=0<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;x=-15,13&lt;/math&gt;, and there cannot be a negative amount of students, there are &lt;math&gt;13&lt;/math&gt; girls, &lt;math&gt;13+2=15&lt;/math&gt; boys, and &lt;math&gt;13+15=\boxed{\textbf{(B)}\ 28}&lt;/math&gt; students.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2009|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_25&diff=100087 2007 AMC 8 Problems/Problem 25 2019-01-04T22:19:12Z <p>G1zq: /* Solution */</p> <hr /> <div>===Problem===<br /> On the dart board shown in the figure below, the outer circle has radius &lt;math&gt;6&lt;/math&gt; and the inner circle has radius &lt;math&gt;3&lt;/math&gt;. Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> draw(circle((0,0),6));<br /> draw(circle((0,0),3));<br /> draw((0,0)--(0,6));<br /> draw((0,0)--(rotate(120)*(0,6)));<br /> draw((0,0)--(rotate(-120)*(0,6)));<br /> label('1',(rotate(60)*(0,3/2)));<br /> label('2',(rotate(-60)*(0,3/2)));<br /> label('2',(0,-3/2));<br /> label('2',(rotate(60)*(0,9/2)));<br /> label('1',(rotate(-60)*(0,9/2)));<br /> label('1',(0,-9/2));<br /> &lt;/asy&gt;&lt;/center&gt;<br /> &lt;math&gt;\mathrm{(A)} \frac{17}{36} \qquad \mathrm{(B)} \frac{35}{72} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{37}{72} \qquad \mathrm{(E)} \frac{19}{36}&lt;/math&gt;<br /> <br /> ===Solution===<br /> To get an odd sum, we must add an even number and an odd number. So we have a little casework to do. Before we do that, we also have to figure out some relative areas. You could either find the absolute areas (as is done below for completeness), or apply some proportional reasoning and symmetry to realize the relative areas (which is really all that matters).<br /> <br /> <br /> To find the areas of the sections, notice that the three smaller sections trisect a circle with radius &lt;math&gt;3&lt;/math&gt;. The area of this entire circle is &lt;math&gt;9\pi&lt;/math&gt;. The area of each smaller section then must be &lt;math&gt;\frac{9\pi}{3}&lt;/math&gt; or &lt;math&gt;3\pi&lt;/math&gt;. The larger sections trisect a &quot;ring&quot; which is the difference of two circles, one with radius &lt;math&gt;3&lt;/math&gt;, the other radius &lt;math&gt;6&lt;/math&gt;. So, the area of the ring (''annulus'') is &lt;math&gt;36\pi - 9\pi&lt;/math&gt; or &lt;math&gt;27\pi&lt;/math&gt;. The area of each larger section must be &lt;math&gt;\frac{27\pi}{3}&lt;/math&gt; or &lt;math&gt;9\pi&lt;/math&gt;. Note that the area of the whole circle is &lt;math&gt;36\pi&lt;/math&gt;.<br /> <br /> <br /> One smaller section and two larger sections contain an odd number (that is, 1). So the probability of throwing an odd number is &lt;math&gt;3\pi + (2 \cdot 9\pi) = 21\pi&lt;/math&gt;. Since the area of the whole circle is &lt;math&gt;36\pi&lt;/math&gt;, the probability of getting an odd is &lt;math&gt;\frac{21\pi}{36\pi} = \frac{21}{36} = \frac{7}{12}&lt;/math&gt;.<br /> <br /> Since the remaining sections contain even numbers (that is, 2), the probability of throwing an even is the complement, or &lt;math&gt;1 - \frac{7}{12} = \frac{5}{12}&lt;/math&gt;.<br /> <br /> <br /> Now, the two cases: You could either get an odd then an even, or an even then an odd.<br /> <br /> <br /> Case 1: Odd then even<br /> Multiply the probabilities to get &lt;math&gt;\frac{7}{12} \cdot \frac{5}{12} = \frac{35}{144}&lt;/math&gt;.<br /> <br /> <br /> Case 2: Even then odd<br /> Multiply the probabilities to get &lt;math&gt;\frac{5}{12} \cdot \frac{7}{12} = \frac{35}{144}&lt;/math&gt;. Notice that this is the same.<br /> <br /> <br /> Thus, the total probability of an odd sum is &lt;math&gt;\frac{35}{144} \cdot \frac{2}{1}&lt;/math&gt; = &lt;math&gt;\boxed{B = \frac{35}{72}}&lt;/math&gt;.<br /> <br /> <br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_16&diff=99622 2014 AMC 8 Problems/Problem 16 2018-12-21T01:20:28Z <p>G1zq: /* Solution */</p> <hr /> <div>==Problem==<br /> The &quot;Middle School Eight&quot; basketball conference has &lt;math&gt;8&lt;/math&gt; teams. Every season, each team plays every other conference team twice (home and away), and each team also plays &lt;math&gt;4&lt;/math&gt; games against non-conference opponents. What is the total number of games in a season involving the &quot;Middle School Eight&quot; teams?<br /> <br /> &lt;math&gt;\textbf{(A) }60\qquad\textbf{(B) }88\qquad\textbf{(C) }96\qquad\textbf{(D) }144\qquad \textbf{(E) }160&lt;/math&gt;<br /> ==Solution==<br /> <br /> Within the conference, there are 8 teams, so there are &lt;math&gt;\dbinom{8}{2}=28&lt;/math&gt; pairings of teams, and each pair must play two games, for a total of &lt;math&gt;28\cdot 2=56&lt;/math&gt; games within the conference.<br /> <br /> Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of &lt;math&gt;4\cdot 8 =32&lt;/math&gt; games outside the conference.<br /> <br /> Therefore, the total number of games is &lt;math&gt;56+32 = \boxed{88}&lt;/math&gt;, so &lt;math&gt;\boxed{\text{(B)}}&lt;/math&gt; is our answer.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2014|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=Simon%27s_Favorite_Factoring_Trick&diff=94201 Simon's Favorite Factoring Trick 2018-04-24T21:40:04Z <p>G1zq: /* About */</p> <hr /> <div><br /> ==About==<br /> '''Dr. Simon's Favorite Factoring Trick''' (abbreviated '''SFFT''') is a special factorization first popularized by [[AoPS]] user [[user:ComplexZeta | Simon Rubinstein-Salzedo]].<br /> <br /> ==The General Statement==<br /> The general statement of SFFT is: &lt;math&gt;{xy}+{xk}+{jy}+{jk}=(x+j)(y+k)&lt;/math&gt;. Two special common cases are: &lt;math&gt;xy + x + y + 1 = (x+1)(y+1)&lt;/math&gt; and &lt;math&gt;xy - x - y +1 = (x-1)(y-1)&lt;/math&gt;.<br /> <br /> The act of adding &lt;math&gt;{jk}&lt;/math&gt; to &lt;math&gt;{xy}+{xk}+{jy}&lt;/math&gt; in order to be able to factor it could be called &quot;completing the rectangle&quot; in analogy to the more familiar &quot;completing the square.&quot;<br /> <br /> == Applications ==<br /> This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are variables and &lt;math&gt;j,k&lt;/math&gt; are known constants. Also, it is typically necessary to add the &lt;math&gt;jk&lt;/math&gt; term to both sides to perform the factorization.<br /> <br /> == Problems ==<br /> ===Introductory===<br /> *Two different [[prime number]]s between &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;18&lt;/math&gt; are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 } &lt;/math&gt;<br /> <br /> ([[2000 AMC 12/Problem 6|Source]])<br /> <br /> ===Intermediate===<br /> *&lt;math&gt;m, n&lt;/math&gt; are integers such that &lt;math&gt;m^2 + 3m^2n^2 = 30n^2 + 517&lt;/math&gt;. Find &lt;math&gt;3m^2n^2&lt;/math&gt;.<br /> <br /> ([[1987 AIME Problems/Problem 5|Source]])<br /> <br /> *The integer &lt;math&gt;N&lt;/math&gt; is positive. There are exactly &lt;math&gt;2005&lt;/math&gt; pairs &lt;math&gt;(x, y)&lt;/math&gt; of positive integers satisfying:<br /> <br /> &lt;cmath&gt;\frac 1x +\frac 1y = \frac 1N&lt;/cmath&gt;<br /> <br /> Prove that &lt;math&gt;N&lt;/math&gt; is a perfect square. (British Mathematical Olympiad Round 2, 2005)<br /> <br /> == See Also ==<br /> * [[Algebra]]<br /> * [[Factoring]]<br /> <br /> [[Category:Elementary algebra]]<br /> [[Category:Theorems]]</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems&diff=93994 2011 AMC 8 Problems 2018-04-17T01:11:00Z <p>G1zq: /* Problem 2 */</p> <hr /> <div>==Problem 1==<br /> Margie bought &lt;math&gt; 3 &lt;/math&gt; apples at a cost of &lt;math&gt; 50 &lt;/math&gt; cents per apple. She paid with a 5-dollar bill. How much change did Margie recieve?<br /> <br /> &lt;math&gt;\textbf{(A) }\ \textdollar 1.50 \qquad \textbf{(B) }\ \textdollar 2.00 \qquad \textbf{(C) }\ \textdollar 2.50 \qquad \textbf{(D) }\ \textdollar 3.00 \qquad \textbf{(E) }\ \textdollar 3.50&lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Karl's rectangular vegetable garden is &lt;math&gt; 20 &lt;/math&gt; feet by &lt;math&gt; 45 &lt;/math&gt; feet, and Makenna's is &lt;math&gt; 25 &lt;/math&gt; feet by &lt;math&gt; 40 &lt;/math&gt; feet. Which of the following statements are true?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Karl's garden is larger by 100 square feet.}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(B) }\text{Karl's garden is larger by 25 square feet.}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(C) }\text{The gardens are the same size.}&lt;/math&gt; <br /> <br /> &lt;math&gt;\textbf{(D) }\text{Makenna's garden is larger by 25 square feet.}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(E) }\text{Makenna's garden is larger by 100 square feet.}&lt;/math&gt;<br /> <br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Extend the square pattern of 8 black and 17 white square tiles by attaching a border of black tiles around the square. What is the ratio of black tiles to white tiles in the extended pattern?&lt;br /&gt;<br /> &lt;asy&gt;<br /> filldraw((0,0)--(5,0)--(5,5)--(0,5)--cycle,white,black);<br /> filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle,mediumgray,black);<br /> filldraw((2,2)--(3,2)--(3,3)--(2,3)--cycle,white,black);<br /> draw((4,0)--(4,5));<br /> draw((3,0)--(3,5));<br /> draw((2,0)--(2,5));<br /> draw((1,0)--(1,5));<br /> draw((0,4)--(5,4));<br /> draw((0,3)--(5,3));<br /> draw((0,2)--(5,2));<br /> draw((0,1)--(5,1));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A) }8:17 \qquad\textbf{(B) }25:49 \qquad\textbf{(C) }36:25 \qquad\textbf{(D) }32:17 \qquad\textbf{(E) }36:17&lt;/math&gt;<br /> <br /> [[2011 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Here is a list of the numbers of fish that Tyler caught in nine outings last summer: &lt;cmath&gt;2,0,1,3,0,3,3,1,2.&lt;/cmath&gt; Which statement about the mean, median, and mode is true?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{median} &lt; \text{mean} &lt; \text{mode} \qquad \textbf{(B) }\text{mean} &lt; \text{mode} &lt; \text{median} \\ \\ \textbf{(C) }\text{mean} &lt; \text{median} &lt; \text{mode} \qquad \textbf{(D) }\text{median} &lt; \text{mode} &lt; \text{mean} \\ \\ \textbf{(E) }\text{mode} &lt; \text{median} &lt; \text{mean}&lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> What time was it &lt;math&gt;2011&lt;/math&gt; minutes after midnight on January 1, 2011?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{January 1 at 9:31 PM}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(B) }\text{January 1 at 11:51 PM}&lt;/math&gt; <br /> <br /> &lt;math&gt;\textbf{(C) }\text{January 2 at 3:11 AM}&lt;/math&gt; <br /> <br /> &lt;math&gt;\textbf{(D) }\text{January 2 at 9:31 AM}&lt;/math&gt; <br /> <br /> &lt;math&gt;\textbf{(E) }\text{January 2 at 6:01 PM}&lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> In a town of 351 adults, every adult owns a car, motorcycle, or both. If 331 adults own cars and 45 adults own motorcycles, how many of the car owners do not own a motorcycle?<br /> <br /> &lt;math&gt; \textbf{(A) }20 \qquad\textbf{(B) }25 \qquad\textbf{(C) }45 \qquad\textbf{(D) }306 \qquad\textbf{(E) }351&lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially bolded. What percent of the total area is partially bolded?<br /> <br /> &lt;asy&gt;<br /> import graph; size(7.01cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.42,xmax=14.59,ymin=-10.08,ymax=5.26; <br /> pair A=(0,0), B=(4,0), C=(0,4), D=(4,4), F=(2,0), G=(3,0), H=(1,4), I=(2,4), J=(3,4), K=(0,-2), L=(4,-2), M=(0,-6), O=(0,-4), P=(4,-4), Q=(2,-2), R=(2,-6), T=(6,4), U=(10,0), V=(10,4), Z=(10,2), A_1=(8,4), B_1=(8,0), C_1=(6,-2), D_1=(10,-2), E_1=(6,-6), F_1=(10,-6), G_1=(6,-4), H_1=(10,-4), I_1=(8,-2), J_1=(8,-6), K_1=(8,-4); <br /> draw(C--H--(1,0)--A--cycle,linewidth(1.6)); draw(M--O--Q--R--cycle,linewidth(1.6)); draw(A_1--V--Z--cycle,linewidth(1.6)); draw(G_1--K_1--J_1--E_1--cycle,linewidth(1.6)); <br /> draw(C--D); draw(D--B); draw(B--A); draw(A--C); draw(H--(1,0)); draw(I--F); draw(J--G); draw(C--H,linewidth(1.6)); draw(H--(1,0),linewidth(1.6)); draw((1,0)--A,linewidth(1.6)); draw(A--C,linewidth(1.6)); draw(K--L); draw((4,-6)--L); draw((4,-6)--M); draw(M--K); draw(O--P); draw(Q--R); draw(O--Q); draw(M--O,linewidth(1.6)); draw(O--Q,linewidth(1.6)); draw(Q--R,linewidth(1.6)); draw(R--M,linewidth(1.6)); draw(T--V); draw(V--U); draw(U--(6,0)); draw((6,0)--T); draw((6,2)--Z); draw(A_1--B_1); draw(A_1--Z); draw(A_1--V,linewidth(1.6)); draw(V--Z,linewidth(1.6)); draw(Z--A_1,linewidth(1.6)); draw(C_1--D_1); draw(D_1--F_1); draw(F_1--E_1); draw(E_1--C_1); draw(G_1--H_1); draw(I_1--J_1); draw(G_1--K_1,linewidth(1.6)); draw(K_1--J_1,linewidth(1.6)); draw(J_1--E_1,linewidth(1.6)); draw(E_1--G_1,linewidth(1.6)); <br /> dot(A,linewidth(1pt)+ds); dot(B,linewidth(1pt)+ds); dot(C,linewidth(1pt)+ds); dot(D,linewidth(1pt)+ds); dot((1,0),linewidth(1pt)+ds); dot(F,linewidth(1pt)+ds); dot(G,linewidth(1pt)+ds); dot(H,linewidth(1pt)+ds); dot(I,linewidth(1pt)+ds); dot(J,linewidth(1pt)+ds); dot(K,linewidth(1pt)+ds); dot(L,linewidth(1pt)+ds); dot(M,linewidth(1pt)+ds); dot((4,-6),linewidth(1pt)+ds); dot(O,linewidth(1pt)+ds); dot(P,linewidth(1pt)+ds); dot(Q,linewidth(1pt)+ds); dot(R,linewidth(1pt)+ds); dot((6,0),linewidth(1pt)+ds); dot(T,linewidth(1pt)+ds); dot(U,linewidth(1pt)+ds); dot(V,linewidth(1pt)+ds); dot((6,2),linewidth(1pt)+ds); dot(Z,linewidth(1pt)+ds); dot(A_1,linewidth(1pt)+ds); dot(B_1,linewidth(1pt)+ds); dot(C_1,linewidth(1pt)+ds); dot(D_1,linewidth(1pt)+ds); dot(E_1,linewidth(1pt)+ds); dot(F_1,linewidth(1pt)+ds); dot(G_1,linewidth(1pt)+ds); dot(H_1,linewidth(1pt)+ds); dot(I_1,linewidth(1pt)+ds); dot(J_1,linewidth(1pt)+ds); dot(K_1,linewidth(1pt)+ds); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}12\frac{1}{2}\qquad\textbf{(B)}20\qquad\textbf{(C)}25\qquad\textbf{(D)}33\frac{1}{3}\qquad\textbf{(E)}37\frac{1}{2} &lt;/math&gt;<br /> <br /> [[2011 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?<br /> <br /> &lt;math&gt; \textbf{(A) }4 \qquad\textbf{(B) }5 \qquad\textbf{(C) }6 \qquad\textbf{(D) }7 \qquad\textbf{(E) }9 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour?<br /> &lt;asy&gt;<br /> import graph; size(8.76cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.58,xmax=10.19,ymin=-4.43,ymax=9.63; <br /> draw((0,0)--(0,8)); draw((0,0)--(8,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); draw((1,0)--(1,8)); draw((2,0)--(2,8)); draw((3,0)--(3,8)); draw((4,0)--(4,8)); draw((5,0)--(5,8)); draw((6,0)--(6,8)); draw((7,0)--(7,8)); label(&quot;1$&quot;,(0.95,-0.24),SE*lsf); label(&quot;$2$&quot;,(1.92,-0.26),SE*lsf); label(&quot;$3$&quot;,(2.92,-0.31),SE*lsf); label(&quot;$4$&quot;,(3.93,-0.26),SE*lsf); label(&quot;$5$&quot;,(4.92,-0.27),SE*lsf); label(&quot;$6$&quot;,(5.95,-0.29),SE*lsf); label(&quot;$7$&quot;,(6.94,-0.27),SE*lsf); label(&quot;$5$&quot;,(-0.49,1.22),SE*lsf); label(&quot;$10$&quot;,(-0.59,2.23),SE*lsf); label(&quot;$15$&quot;,(-0.61,3.22),SE*lsf); label(&quot;$20$&quot;,(-0.61,4.23),SE*lsf); label(&quot;$25$&quot;,(-0.59,5.22),SE*lsf); label(&quot;$30$&quot;,(-0.59,6.2),SE*lsf); label(&quot;$35$&quot;,(-0.56,7.18),SE*lsf); draw((0,0)--(1,1),linewidth(1.6)); draw((1,1)--(2,3),linewidth(1.6)); draw((2,3)--(4,4),linewidth(1.6)); draw((4,4)--(7,7),linewidth(1.6)); label(&quot;HOURS&quot;,(3.41,-0.85),SE*lsf); label(&quot;M&quot;,(-1.39,5.32),SE*lsf); label(&quot;I&quot;,(-1.34,4.93),SE*lsf); label(&quot;L&quot;,(-1.36,4.51),SE*lsf); label(&quot;E&quot;,(-1.37,4.11),SE*lsf); label(&quot;S&quot;,(-1.39,3.7),SE*lsf); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}2\qquad\textbf{(B)}2.5\qquad\textbf{(C)}4\qquad\textbf{(D)}4.5\qquad\textbf{(E)}5 &lt;/math&gt;<br /> <br /> [[2011 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> The taxi fare in Gotham City is &lt;nowiki&gt;$2.40&lt;/nowiki&gt; for the first &lt;math&gt;\frac12&lt;/math&gt; mile and additional mileage charged at the rate &lt;nowiki&gt;$0.20&lt;/nowiki&gt; for each additional 0.1 mile. You plan to give the driver a &lt;nowiki&gt;$2&lt;/nowiki&gt; tip. How many miles can you ride for &lt;nowiki&gt;$10&lt;/nowiki&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) } 3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> The graph shows the number of minutes studied by both Asha (black bar) and Sasha(grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha?<br /> &lt;asy&gt;<br /> size(300);<br /> real i;<br /> defaultpen(linewidth(0.8));<br /> draw((0,140)--origin--(220,0));<br /> for(i=1;i&lt;13;i=i+1) {<br /> draw((0,10*i)--(220,10*i));<br /> }<br /> label(&quot;$0$&quot;,origin,W);<br /> label(&quot;$20$&quot;,(0,20),W);<br /> label(&quot;$40$&quot;,(0,40),W);<br /> label(&quot;$60$&quot;,(0,60),W);<br /> label(&quot;$80$&quot;,(0,80),W);<br /> label(&quot;$100$&quot;,(0,100),W);<br /> label(&quot;$120$&quot;,(0,120),W);<br /> path MonD=(20,0)--(20,60)--(30,60)--(30,0)--cycle,MonL=(30,0)--(30,70)--(40,70)--(40,0)--cycle,TuesD=(60,0)--(60,90)--(70,90)--(70,0)--cycle,TuesL=(70,0)--(70,80)--(80,80)--(80,0)--cycle,WedD=(100,0)--(100,100)--(110,100)--(110,0)--cycle,WedL=(110,0)--(110,120)--(120,120)--(120,0)--cycle,ThurD=(140,0)--(140,80)--(150,80)--(150,0)--cycle,ThurL=(150,0)--(150,110)--(160,110)--(160,0)--cycle,FriD=(180,0)--(180,70)--(190,70)--(190,0)--cycle,FriL=(190,0)--(190,50)--(200,50)--(200,0)--cycle;<br /> fill(MonD,grey);<br /> fill(MonL,lightgrey);<br /> fill(TuesD,grey);<br /> fill(TuesL,lightgrey);<br /> fill(WedD,grey);<br /> fill(WedL,lightgrey);<br /> fill(ThurD,grey);<br /> fill(ThurL,lightgrey);<br /> fill(FriD,grey);<br /> fill(FriL,lightgrey);<br /> draw(MonD^^MonL^^TuesD^^TuesL^^WedD^^WedL^^ThurD^^ThurL^^FriD^^FriL);<br /> label(&quot;M&quot;,(30,-5),S);<br /> label(&quot;Tu&quot;,(70,-5),S);<br /> label(&quot;W&quot;,(110,-5),S);<br /> label(&quot;Th&quot;,(150,-5),S);<br /> label(&quot;F&quot;,(190,-5),S);<br /> label(&quot;M&quot;,(-25,85),W);<br /> label(&quot;I&quot;,(-27,75),W);<br /> label(&quot;N&quot;,(-25,65),W);<br /> label(&quot;U&quot;,(-25,55),W);<br /> label(&quot;T&quot;,(-25,45),W);<br /> label(&quot;E&quot;,(-25,35),W);<br /> label(&quot;S&quot;,(-26,25),W);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?<br /> <br /> &lt;math&gt; \textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Two congruent squares, &lt;math&gt;ABCD&lt;/math&gt; and &lt;math&gt;PQRS&lt;/math&gt;, have side length &lt;math&gt;15&lt;/math&gt;. They overlap to form the &lt;math&gt;15&lt;/math&gt; by &lt;math&gt;25&lt;/math&gt; rectangle &lt;math&gt;AQRD&lt;/math&gt; shown. What percent of the area of rectangle &lt;math&gt;AQRD&lt;/math&gt; is shaded? <br /> &lt;asy&gt;<br /> filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black);<br /> label(&quot;D&quot;,(0,0),S);<br /> label(&quot;R&quot;,(25,0),S);<br /> label(&quot;Q&quot;,(25,15),N);<br /> label(&quot;A&quot;,(0,15),N);<br /> filldraw((10,0)--(15,0)--(15,15)--(10,15)--cycle,mediumgrey,black);<br /> label(&quot;S&quot;,(10,0),S);<br /> label(&quot;C&quot;,(15,0),S);<br /> label(&quot;B&quot;,(15,15),N);<br /> label(&quot;P&quot;,(10,15),N);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 25 &lt;/math&gt;<br /> <br /> [[2011 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> There are &lt;math&gt;270&lt;/math&gt; students at Colfax Middle School, where the ratio of boys to girls is &lt;math&gt;5 : 4&lt;/math&gt;. There are &lt;math&gt;180&lt;/math&gt; students at Winthrop Middle School, where the ratio of boys to girls is &lt;math&gt;4 : 5&lt;/math&gt;. The two schools hold a dance and all students from both schools attend. What fraction of the students at the dance are girls?<br /> <br /> &lt;math&gt; \textbf{(A) } \dfrac7{18} \qquad\textbf{(B) } \dfrac7{15} \qquad\textbf{(C) } \dfrac{22}{45} \qquad\textbf{(D) } \dfrac12 \qquad\textbf{(E) } \dfrac{23}{45} &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> How many digits are in the product &lt;math&gt;4^5 \cdot 5^{10}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 12 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Let &lt;math&gt;A&lt;/math&gt; be the area of the triangle with sides of length &lt;math&gt;25, 25&lt;/math&gt;, and &lt;math&gt;30&lt;/math&gt;. Let &lt;math&gt;B&lt;/math&gt; be the area of the triangle with sides of length &lt;math&gt;25, 25,&lt;/math&gt; and &lt;math&gt;40&lt;/math&gt;. What is the relationship between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) } A = \dfrac9{16}B \qquad\textbf{(B) } A = \dfrac34B \qquad\textbf{(C) } A=B \qquad\textbf{(D) } A = \dfrac43B \\ \\ \textbf{(E) }A = \dfrac{16}9B &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Let &lt;math&gt;w&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; be whole numbers. If &lt;math&gt;2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588&lt;/math&gt;, then what does &lt;math&gt;2w + 3x + 5y + 7z&lt;/math&gt; equal?<br /> <br /> &lt;math&gt; \textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> A fair 6-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?<br /> <br /> &lt;math&gt; \textbf{(A) }\dfrac16\qquad\textbf{(B) }\dfrac5{12}\qquad\textbf{(C) }\dfrac12\qquad\textbf{(D) }\dfrac7{12}\qquad\textbf{(E) }\dfrac56 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> How many rectangles are in this figure?<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G,H,I,J,K,L;<br /> A=(0,0);<br /> B=(20,0);<br /> C=(20,20);<br /> D=(0,20);<br /> draw(A--B--C--D--cycle);<br /> E=(-10,-5);<br /> F=(13,-5);<br /> G=(13,5);<br /> H=(-10,5);<br /> draw(E--F--G--H--cycle);<br /> I=(10,-20);<br /> J=(18,-20);<br /> K=(18,13);<br /> L=(10,13);<br /> draw(I--J--K--L--cycle);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is a trapezoid, &lt;math&gt;AD = 15&lt;/math&gt;, &lt;math&gt;AB = 50&lt;/math&gt;, &lt;math&gt;BC = 20&lt;/math&gt;, and the altitude is &lt;math&gt;12&lt;/math&gt;. What is the area of the trapezoid?<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D;<br /> A=(3,20);<br /> B=(35,20);<br /> C=(47,0);<br /> D=(0,0);<br /> draw(A--B--C--D--cycle);<br /> dot((0,0));<br /> dot((3,20));<br /> dot((35,20));<br /> dot((47,0));<br /> label(&quot;A&quot;,A,N);<br /> label(&quot;B&quot;,B,N);<br /> label(&quot;C&quot;,C,S);<br /> label(&quot;D&quot;,D,S);<br /> draw((19,20)--(19,0));<br /> dot((19,20));<br /> dot((19,0));<br /> draw((19,3)--(22,3)--(22,0));<br /> label(&quot;12&quot;,(21,10),E);<br /> label(&quot;50&quot;,(19,22),N);<br /> label(&quot;15&quot;,(1,10),W);<br /> label(&quot;20&quot;,(41,12),E);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A) }600\qquad\textbf{(B) }650\qquad\textbf{(C) }700\qquad\textbf{(D) }750\qquad\textbf{(E) }800 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> Students guess that Norb's age is &lt;math&gt;24, 28, 30, 32, 36, 38, 41, 44, 47&lt;/math&gt;, and &lt;math&gt;49&lt;/math&gt;. Norb says, &quot;At least half of you guessed too low, two of you are off by one, and my age is a prime number.&quot; How old is Norb?<br /> <br /> &lt;math&gt; \textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> What is the '''tens''' digit of &lt;math&gt;7^{2011}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?<br /> <br /> &lt;math&gt; \textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> In how many ways can 10001 be written as the sum of two primes?<br /> <br /> &lt;math&gt; \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> A circle with radius &lt;math&gt;1&lt;/math&gt; is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?<br /> <br /> &lt;asy&gt;<br /> filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,mediumgray,black);<br /> filldraw(Circle((0,0),1), mediumgray,black);<br /> filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2011|before=[[2010 AMC 8 Problems|2010 AMC 8]]|after=[[2012 AMC 8 Problems|2012 AMC 8]]}}<br /> * [[AMC 8]]<br /> * [[AMC 8 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_9&diff=93572 2012 AMC 8 Problems/Problem 9 2018-03-28T21:03:19Z <p>G1zq: /* Solution */</p> <hr /> <div>==Problem==<br /> The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?<br /> <br /> &lt;math&gt; \textbf{(A)}\hspace{.05in}61\qquad\textbf{(B)}\hspace{.05in}122\qquad\textbf{(C)}\hspace{.05in}139\qquad\textbf{(D)}\hspace{.05in}150\qquad\textbf{(E)}\hspace{.05in}161 &lt;/math&gt;<br /> <br /> ==Solution 1: Algebra==<br /> Let the number of two-legged birds be &lt;math&gt;x&lt;/math&gt; and the number of four-legged mammals be &lt;math&gt;y&lt;/math&gt;. We can now use systems of equations to solve this problem.<br /> <br /> Write two equations:<br /> <br /> &lt;math&gt; 2x + 4y = 522 &lt;/math&gt;<br /> <br /> &lt;math&gt; x + y = 200 &lt;/math&gt;<br /> <br /> Now multiply the latter equation by &lt;math&gt; 2 &lt;/math&gt;.<br /> <br /> &lt;math&gt; 2x + 4y = 522 &lt;/math&gt;<br /> <br /> &lt;math&gt; 2x + 2y = 400 &lt;/math&gt;<br /> <br /> By subtracting the second equation from the first equation, we find that &lt;math&gt; 2y = 122 \implies y = 61 &lt;/math&gt;. Since there were &lt;math&gt; 200 &lt;/math&gt; heads, meaning that there were &lt;math&gt; 200 &lt;/math&gt; animals, there were &lt;math&gt; 200 - 61 = \boxed{\textbf{(C)}\ 139} &lt;/math&gt; two-legged birds.<br /> <br /> ==Solution 2: Cheating the System==<br /> First, we &quot;assume&quot; there are 200 two-legged birds only, and 0 four-legged mammals. Of course, this poses a problem, as then there would only be &lt;math&gt;200\cdot2=400&lt;/math&gt; legs.<br /> <br /> Now we have to do some swapping--for every two-legged bird we swap for a four-legged mammal, we gain 2 legs. For example, if we swapped one bird for one mammal, giving 199 birds and 1 mammal, there would be &lt;math&gt;400 + 1(2) = 402&lt;/math&gt; legs. If we swapped two birds for two mammals, there would be &lt;math&gt;400 + 2(2) = 404&lt;/math&gt; legs. If we swapped 50 birds for 50 mammals, there would be &lt;math&gt;400 + 50(2) = 500&lt;/math&gt; legs.<br /> <br /> Notice that we must gain &lt;math&gt;522-400 = 122&lt;/math&gt; legs. This means we must swap out &lt;math&gt;122\div2 = 61&lt;/math&gt; birds. Therefore, there must be &lt;math&gt;200-61 = \boxed{\textbf{(C)}\ 139}&lt;/math&gt; birds. Checking our work, we find that &lt;math&gt;139\cdot2 + 61 \cdot 4 = 522&lt;/math&gt;, and we are correct.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2012|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_19&diff=93335 2013 AMC 8 Problems/Problem 19 2018-03-20T19:38:29Z <p>G1zq: /* Problem */</p> <hr /> <div>==Problem==<br /> Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Hannah, Cassie, Bridget} \qquad \textbf{(B)}\ \text{Hannah, Bridget, Cassie} \\ \qquad \textbf{(C)}\ \text{Cassie, Bridget, Hannah} \qquad \textbf{(D)}\ \text{Cassie, Hannah, Bridget} \\ \qquad \textbf{(E)}\ \text{Bridget, Cassie, Hannah}&lt;/math&gt;<br /> <br /> ==Solution==<br /> If Hannah did better than Cassie, there would be no way she could know for sure that she didn't get the lowest score in the class. Therefore, Hannah did worse than Cassie. Similarly, if Hannah did worse than Bridget, there is no way Bridget could have known that she didn't get the highest in the class. Therefore, Hannah did better than Bridget, so our order is &lt;math&gt;\boxed{\textbf{(D) Cassie, Hannah, Bridget}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2013|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_10&diff=93334 2013 AMC 8 Problems/Problem 10 2018-03-20T18:57:03Z <p>G1zq: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> To find either the LCM or the GCF of two numbers, always prime factorize first.<br /> <br /> The prime factorization of &lt;math&gt;180 = 3^2 \times 5 \times 2^2&lt;/math&gt;.<br /> <br /> The prime factorization of &lt;math&gt;594 = 3^3 \times 11 \times 2&lt;/math&gt;.<br /> <br /> Then, to find the LCM, we have to find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is &lt;math&gt;3^3, 5, 11, 2^2&lt;/math&gt;). Multiply all of these to get 5940. <br /> <br /> For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. &lt;math&gt;3^2 \times 2&lt;/math&gt; = 18.<br /> <br /> Thus the answer = &lt;math&gt;\frac{5940}{18}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(C)}\ 330}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We start off with a similar approach as the original solution. From the prime factorizations, the GCF is &lt;math&gt;18&lt;/math&gt;.<br /> <br /> It is a well known fact that &lt;math&gt;\gcd(m,n)\times \operatorname{lcm}(m,n)=|mn|&lt;/math&gt;. So we have, &lt;math&gt;18\times \operatorname{lcm} (180,594)=594\times 180&lt;/math&gt;.<br /> <br /> Dividing by &lt;math&gt;18&lt;/math&gt; yields &lt;math&gt;\operatorname{lcm} (180,594)=594\times 10=5940&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf}(180,594)}=\frac{5940}{18}=\boxed{\textbf{(C)}\ 330}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> From Solution 1, <br /> the prime factorization of &lt;math&gt;180 = 2^2 \cdot 3^2 \cdot 5&lt;/math&gt;.<br /> The prime factorization of &lt;math&gt;594 = 2 \cdot 3^3 \cdot 11&lt;/math&gt;.<br /> Hence, &lt;math&gt;\operatorname{lcm} (180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11&lt;/math&gt;, and &lt;math&gt;\operatorname{gcf} (180,594) = 2 \cdot 3^2&lt;/math&gt;.<br /> Therefore, &lt;math&gt;\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf} (180,594)} = \frac{2^2 \cdot 3^3 \cdot 5 \cdot 11}{2 \cdot 3^2} = 2 \cdot 3 \cdot 5 \cdot 11 \cdot = 330 \Longrightarrow \boxed{\textbf{(C)}\ 330}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2013|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_10&diff=93333 2013 AMC 8 Problems/Problem 10 2018-03-20T18:56:41Z <p>G1zq: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> To find either the LCM or the GCF of two numbers, always prime factorize first.<br /> <br /> The prime factorization of &lt;math&gt;180 = 3^2 \times 5 \times 2^2&lt;/math&gt;.<br /> <br /> The prime factorization of &lt;math&gt;594 = 3^3 \times 11 \times 2&lt;/math&gt;.<br /> <br /> Then, to find the LCM, we have to find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is &lt;math&gt;3^3, 5, 11, 2^2&lt;/math&gt;). Multiply all of these to get 5940. <br /> <br /> For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. &lt;math&gt;3^2 \times 2&lt;/math&gt; = 18.<br /> <br /> Thus the answer = &lt;math&gt;\frac{5940}{18}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(C)}\ 330}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We start off with a similar approach as the original solution. From the prime factorizations, the GCF is &lt;math&gt;18&lt;/math&gt;.<br /> <br /> It is a well known fact that &lt;math&gt;\gcd(m,n)\times \operatorname{lcm}(m,n)=|mn|&lt;/math&gt;. So we have, &lt;math&gt;18\times \operatorname{lcm} (180,594)=594\times 180&lt;/math&gt;.<br /> <br /> Dividing by &lt;math&gt;18&lt;/math&gt; yields &lt;math&gt;\operatorname{lcm} (180,594)=594\times 10=5940&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf}(180,594)}=\frac{5940}{18}=\boxed{\textbf{(C)}\ 330}&lt;/math&gt;.<br /> <br /> ==Another Similar Solution==<br /> From Solution 1, <br /> the prime factorization of &lt;math&gt;180 = 2^2 \cdot 3^2 \cdot 5&lt;/math&gt;.<br /> The prime factorization of &lt;math&gt;594 = 2 \cdot 3^3 \cdot 11&lt;/math&gt;.<br /> Hence, &lt;math&gt;\operatorname{lcm} (180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11&lt;/math&gt;, and &lt;math&gt;\operatorname{gcf} (180,594) = 2 \cdot 3^2&lt;/math&gt;.<br /> Therefore, &lt;math&gt;\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf} (180,594)} = \frac{2^2 \cdot 3^3 \cdot 5 \cdot 11}{2 \cdot 3^2} = 2 \cdot 3 \cdot 5 \cdot 11 \cdot = 330 \Longrightarrow \boxed{\textbf{(C)}\ 330}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2013|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_9&diff=93327 2013 AMC 8 Problems/Problem 9 2018-03-19T21:52:47Z <p>G1zq: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> The Inedible Bulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9^\text{th} \qquad \textbf{(B)}\ 10^\text{th} \qquad \textbf{(C)}\ 11^\text{th} \qquad \textbf{(D)}\ 12^\text{th} \qquad \textbf{(E)}\ 13^\text{th}&lt;/math&gt;<br /> <br /> ==Solution==<br /> This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that &lt;math&gt;2^{10}=1024&lt;/math&gt;.<br /> <br /> However, because the first term is &lt;math&gt;2^0=1&lt;/math&gt; and not &lt;math&gt;2^1=2&lt;/math&gt;, the solution to the problem is &lt;math&gt;10-0+1=\boxed{\textbf{(C)}\ 11^{\text{th}}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We can also solve this problem by listing out how far the Hulk jumps on each jump: on the first jump, he goes 1 meter, the second jump 2 meters, and so on. Listing out these numbers, we get:<br /> <br /> &lt;math&gt;1, 2, 4, 8, 16, 32, 64, 128, 256, 512, \boxed{\textbf{1024}}&lt;/math&gt;<br /> <br /> On the 11th jump, the Hulk jumps 1024 meters &gt; 1000 meters (1 kilometer), so our answer is the 11th jump, or &lt;math&gt;\boxed{\textbf{(C)}}.&lt;/math&gt;<br /> <br /> <br /> ==Solution 3==<br /> The smallest power of two that is larger or equal to 1000 is &lt;math&gt;2^{10}&lt;/math&gt;, or &lt;math&gt;1024.&lt;/math&gt; Thus, the answer is &lt;math&gt;\boxed{\textbf{(C)}}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2013|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_9&diff=93326 2013 AMC 8 Problems/Problem 9 2018-03-19T21:52:18Z <p>G1zq: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The Inedible Bulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9^\text{th} \qquad \textbf{(B)}\ 10^\text{th} \qquad \textbf{(C)}\ 11^\text{th} \qquad \textbf{(D)}\ 12^\text{th} \qquad \textbf{(E)}\ 13^\text{th}&lt;/math&gt;<br /> <br /> ==Solution==<br /> This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that &lt;math&gt;2^{10}=1024&lt;/math&gt;.<br /> <br /> However, because the first term is &lt;math&gt;2^0=1&lt;/math&gt; and not &lt;math&gt;2^1=2&lt;/math&gt;, the solution to the problem is &lt;math&gt;10-0+1=\boxed{\textbf{(C)}\ 11^{\text{th}}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We can also solve this problem by listing out how far the Hulk jumps on each jump: on the first jump, he goes 1 meter, the second jump 2 meters, and so on. Listing out these numbers, we get:<br /> <br /> &lt;math&gt;1, 2, 4, 8, 16, 32, 64, 128, 256, 512, \boxed{\textbf{1024}}&lt;/math&gt;<br /> <br /> On the 11th jump, the Hulk jumps 1024 meters &gt; 1000 meters (1 kilometer), so our answer is the 11th jump, or &lt;math&gt;\boxed{\textbf{(C)}}.&lt;/math&gt;<br /> <br /> <br /> ==Solution 3==<br /> The smallest power of two that is larger or equal to 1000 is &lt;math&gt;2^{10}&lt;/math&gt;, or &lt;math&gt;1024.&lt;/math&gt; Thus, the answer is &lt;math&gt;\boxed{\textbf{(B)}}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2013|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_25&diff=93313 2014 AMC 8 Problems/Problem 25 2018-03-18T19:37:27Z <p>G1zq: /* Solution */</p> <hr /> <div>==Problem==<br /> A straight one-mile stretch of highway, &lt;math&gt;40&lt;/math&gt; feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at &lt;math&gt;5&lt;/math&gt; miles per hour, how many hours will it take to cover the one-mile stretch?<br /> <br /> Note: &lt;math&gt;1&lt;/math&gt; mile = &lt;math&gt;5280&lt;/math&gt; feet<br /> <br /> &lt;asy&gt; size(10cm); pathpen=black; pointpen=black; D(arc((-2,0),1,300,360)); D(arc((0,0),1,0,180)); D(arc((2,0),1,180,360)); D(arc((4,0),1,0,180)); D(arc((6,0),1,180,240)); D((-1.5,1)--(5.5,1)); D((-1.5,0)--(5.5,0),dashed); D((-1.5,-1)--(5.5,-1)); &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A) }\frac{\pi}{11}\qquad\textbf{(B) }\frac{\pi}{10}\qquad\textbf{(C) }\frac{\pi}{5}\qquad\textbf{(D) }\frac{2\pi}{5}\qquad\textbf{(E) }\frac{2\pi}{3} &lt;/math&gt;<br /> <br /> ==Solution==<br /> There are two possible interpretations of the problem: that the road as a whole is &lt;math&gt;40&lt;/math&gt; feet wide, or that each lane is &lt;math&gt;40&lt;/math&gt; feet wide. Both interpretations will arrive at the same result. However, let us stick with the first interpretation for simplicity. Each lane must then be &lt;math&gt;20&lt;/math&gt; feet wide, so Robert must be riding his bike in semicircles with radius &lt;math&gt;20&lt;/math&gt; feet and diameter &lt;math&gt;40&lt;/math&gt; feet. Since the road is &lt;math&gt;5280&lt;/math&gt; feet long, over the whole mile, Robert rides &lt;math&gt;\frac{5280}{40} =132&lt;/math&gt; semicircles in total. Were the semicircles full circles, their circumference would be &lt;math&gt;2\pi\cdot 20=40\pi&lt;/math&gt; feet; as it is, the circumference of each is half that, or &lt;math&gt;20\pi&lt;/math&gt; feet. Therefore, over the stretch of highway, Robert rides a total of &lt;math&gt;132\cdot 20\pi =2640\pi&lt;/math&gt; feet, equivalent to &lt;math&gt;\frac{\pi}{2}&lt;/math&gt; miles. Robert rides at 5 miles per hour, so divide the &lt;math&gt;\frac{\pi}{2}&lt;/math&gt; miles by &lt;math&gt;5&lt;/math&gt; mph (because t = d/r time = distance/rate) to arrive at &lt;math&gt;\boxed{\textbf{(B) }\frac{\pi}{10}}&lt;/math&gt; hours.</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_21&diff=93312 2014 AMC 8 Problems/Problem 21 2018-03-18T19:27:59Z <p>G1zq: /* Solution */</p> <hr /> <div>==Problem==<br /> The &lt;math&gt;7&lt;/math&gt;-digit numbers &lt;math&gt;\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}&lt;/math&gt; and &lt;math&gt;\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}&lt;/math&gt; are each multiples of &lt;math&gt;3&lt;/math&gt;. Which of the following could be the value of &lt;math&gt;C&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8&lt;/math&gt;<br /> ==Solution==<br /> The sum of a number's digits &lt;math&gt;\mod{3}&lt;/math&gt; is congruent to the number &lt;math&gt;\pmod{3}&lt;/math&gt;. &lt;math&gt;74A52B1 \mod{3}&lt;/math&gt; must be congruent to 0, since it is divisible by 3. Therefore, &lt;math&gt;7+4+A+5+2+B+1 \mod{3}&lt;/math&gt; is also congruent to 0. &lt;math&gt;7+4+5+2+1 \equiv 1 \pmod{3}&lt;/math&gt;, so &lt;math&gt;A+B\equiv 2 \pmod{3}&lt;/math&gt;. As we know, &lt;math&gt;326AB4C\equiv 0 \pmod{3}&lt;/math&gt;, so &lt;math&gt;3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}&lt;/math&gt;, and therefore &lt;math&gt;A+B+C\equiv 0 \pmod{3}&lt;/math&gt;. We can substitute 2 for &lt;math&gt;A+B&lt;/math&gt;, so &lt;math&gt;2+C\equiv 0 \pmod{3}&lt;/math&gt;, and therefore &lt;math&gt;C\equiv 1\pmod{3}&lt;/math&gt;. This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is &lt;math&gt;\boxed{\textbf{(A) }1}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2014|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=91591 Gmaas 2018-02-16T21:31:57Z <p>G1zq: /* Known Facts About Gmaas */</p> <hr /> <div>=== Known Facts About Gmaas ===<br /> - Gmaas is a quadrillionaire. EDIT: Gmaas is an infinity-aire<br /> <br /> -Gmaas really knows that roblox is awful and does not play it seriously, thank god our lord is sane<br /> <br /> -In 2003, Gmaas used calculas to force his reign over AoPS.<br /> <br /> -&quot;Actually, my name is spelled GMAAS&quot;<br /> <br /> -Gmass owns all AoPS staff including Richard Rusczyk<br /> <br /> - Gmass's true number of lives left is unknown; however, Gmass recently confirmed that he had at least a million left. Why does Gmass have so many more lives than other cats? The power of Gmass.<br /> <br /> - sseraj once spelled gmaas as gmass on accident in Introduction to Geometry (1532).<br /> <br /> -Gmaas actively plays Roblox, and is a globally ranked professional gamer: https://www.roblox.com/users/29708533/profile<br /> <br /> -Gmaas has beaten Chuck Norris in a fight...twice<br /> <br /> -Gmass is a South Korean, North Korean, Palestinian, Israeli, U.S., and Soviet citizen at the same time. EDIT: Gmaas has now been found out to be a citizen of every country in the world. Gmaas seems to enjoy the country of AOPS best, however.<br /> <br /> -&quot;i am sand&quot; destroyed Gmass in FTW<br /> <br /> - sseraj posted a picture of gmaas with a game controller in Introduction to Geometry (1532).<br /> <br /> -Gmaas plays roblox mobile edition and likes Minecraft, Candy Crush, and Club Penguin Rewritten<br /> <br /> -Gmaas is Roy Moore's horse in the shape of a cat<br /> <br /> -Gmaas is a known roblox/club penguin rewritten player and is a legend at it. He has over &lt;math&gt;289547987693&lt;/math&gt; robux and &lt;math&gt;190348&lt;/math&gt; in CPR.<br /> <br /> -This is all hypothetical EDIT: This is all factual <br /> <br /> -Gmass's real name is Princess. He has a sibling named Rusty<br /> (Warrior cats reference)<br /> <br /> - He is capable of salmon powers, according to PunSpark (ask him)<br /> <br /> The Gmaas told Richard Rusczyk to make AoPS<br /> <br /> -The Gmaas is everything. Yes, you are part of the Gmaas-Dw789<br /> <br /> -The Gmaas knows every dimension up to 9999999999999999999999999999999999999999999999999th dimension<br /> <br /> -Certain theories provide evidence that he IS darth plagueis the wise<br /> <br /> -Gmaas is &quot;TIRED OF PEOPLE ADDING TO HIS PAGE!!&quot; (Maas 45).<br /> <br /> -Gmaas has multiple accounts; some of them are pifinity, Lord_Baltimore, Spacehead1AU, squareman, and Electro3.0<br /> <br /> -Gmaas has a penguin servant named sm24136 who runs GMAASINC. The penguin may or may not be dead. <br /> <br /> -Gmass owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown.<br /> <br /> - Gmass made this page<br /> <br /> - Gmaas is king of the first men, the anduls<br /> <br /> -Gmass is a well known professor at MEOWston Academy<br /> <br /> -Gmass is a Tuna addict, along with other, more potent fish such as Salmon and Trout<br /> <br /> - Gmass won the reward of being cutest and fattest cat ever--he surpassed grumpy cat<br /> <br /> -Last sighting 1571-stretch-algebra-a 12/6/17<br /> <br /> - owner of sseraj, not pet<br /> <br /> - embodiment of life and universe and beyond <br /> <br /> - Watches memes<br /> <br /> -After Death became the GOD OF HYPERDEATH and obtained over 9000 souls <br /> <br /> -Gmass's real name is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso [STOP RICK ROLLING. (Source)]<br /> <br /> -Gmass is a certified Slytherin and probably the cutest cat ever.<br /> <br /> -Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom <br /> <br /> -Gmaas has superpowers that allow him to overcome the horrors of Mr. Toilet (while he was locked in the bathroom)<br /> <br /> - Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br /> <br /> -Gmaas once conquered the moon and imprinted his face on it until asteroids came.<br /> <br /> -Gmaas is a supreme overlord who must be given &lt;math&gt;10^{1000000000000000000000^{1000000000000000000000}}&lt;/math&gt; minecraft DIAMONDS<br /> <br /> - gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard, plus black holes.<br /> <br /> - Gmaas is 5space's favorite animal. <br /> <br /> - He lives with sseraj. <br /> <br /> -Gmaas is my favorite pokemon<br /> <br /> -Gmaas dislikes number theory but enjoys geometry.<br /> <br /> - Gmaas is cool<br /> <br /> - He is often overfed (with probability &lt;math&gt;\frac{3972}{7891}&lt;/math&gt;), or malnourished (with probability &lt;math&gt;\frac{3919}{7891}&lt;/math&gt;) by sseraj.<br /> <br /> - He has &lt;cmath&gt;\sum_{k=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAS&lt;/cmath&gt; supercars, excluding the Purrari and the 138838383 Teslas. <br /> <br /> - He is an employee of AoPS.<br /> <br /> - He is a gmaas with yellow fur and white hypnotizing eyes.<br /> <br /> - He was born with a tail that is a completely different color from the rest of his fur.<br /> <br /> - His stare is very hypnotizing and effective at getting table scraps.<br /> <br /> - He sometimes appears several minutes before certain classes start as an admin. <br /> <br /> - He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br /> <br /> - It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br /> <br /> - Actually, he is a cat. He said so. And science also says so.<br /> <br /> - He is distant relative of Mathcat1234.<br /> <br /> - He is very famous now, and mods always talk about him before class starts.<br /> <br /> - His favorite food is AoPS textbooks because they help him digest problems.<br /> <br /> - Gmaas tends to reside in sseraj's fridge.<br /> <br /> - Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br /> <br /> - The fur of Gmaas can protect him from the harsh conditions of a freezer.<br /> <br /> - Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br /> <br /> - Gmaas is a sage omniscient cat.<br /> <br /> - He is looking for suitable places other than sseraj's fridge to live in.<br /> <br /> - Places where gmaas sightings have happened: <br /> ~The Royal Scoop ice cream store in Bonita Beach Florida<br /> <br /> ~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br /> <br /> ~Alligator Swamp A 1072 <br /> <br /> ~Alligator Swamp B 1073<br /> <br /> ~Prealgebra A (1488)<br /> <br /> ~Introduction to Algebra A (1170)<br /> <br /> ~Introduction to Algebra B (1529)<br /> <br /> ~Welcome to Panda Town Gate 1076<br /> <br /> ~Welcome to Gmaas Town Gate 1221<br /> <br /> ~Welcome to Gmaas Town Gate 1125<br /> <br /> ~33°01'17.4&quot;N 117°05'40.1&quot;W (Rancho Bernardo Road, San Diego, CA)<br /> <br /> ~The other side of the ice in Antarctica<br /> <br /> ~Feisty Alligator Swamp 1115<br /> <br /> ~Introduction to Geometry 1221 (Taught by sseraj)<br /> <br /> ~Introduction to Counting and Probability 1142 <br /> <br /> ~Feisty-ish Alligator Swamp 1115 (AGAIN)<br /> <br /> ~Intermediate Counting and Probability 1137<br /> <br /> ~Intermediate Counting and Probability 1207<br /> <br /> ~Posting student surveys<br /> <br /> ~USF Castle Walls - Elven Tribe 1203<br /> <br /> ~Dark Lord's Hut 1210<br /> <br /> ~AMC 10 Problem Series 1200<br /> <br /> ~Intermediate Number Theory 1138<br /> <br /> ~Intermediate Number Theory 1476<br /> <br /> ~Introduction To Number Theory 1204. Date:7/27/16.<br /> <br /> ~Algebra B 1112<br /> <br /> ~Intermediate Algebra 1561 7:17 PM 12/11/16<br /> <br /> ~Nowhere Else, Tasmania<br /> <br /> ~Earth Dimension C-137<br /> <br /> <br /> <br /> - These have all been designated as the most glorious sections of Aopsland now (Especially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&amp;B).<br /> <br /> - Gmaas has also been sighted in Olympiad Geometry 1148.<br /> <br /> - Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br /> <br /> - Gmaas once snuck into sseraj's email so he could give pianoman24 an extension in Introduction to Number Theory 1204. This was 1204 minutes after his sighting on 7/27/16.<br /> <br /> - Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br /> <br /> EDIT: Nobody has yet seen him atop a tribal base yet.<br /> <br /> - Gmaas are often under the disguise of a penguin or cat. Look out for them.<br /> <br /> EDIT: Gmaas rarely disguises himself as a penguin.<br /> <br /> - He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br /> <br /> EDIT: He IS an AoPS site admin.<br /> <br /> - If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br /> <br /> - Is this the real life? Is this just fantasy? No. This is Gmaas, the legend.<br /> <br /> -Aha!! An impostor!! <br /> http://www.salford.ac.uk/environment-life-sciences/research/applied-archaeology/greater-manchester-archaeological-advisory-service<br /> (look at the acronym).<br /> <br /> - Gmaas might have been viewing (with a &lt;math&gt;\frac{99999}{100000}&lt;/math&gt; chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br /> <br /> - EDIT: Gmaas is a he.<br /> <br /> -Gmaas is love, Gmaas is life<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mew.<br /> <br /> - Gmaas is on the list of &quot;Elusive Creatures.&quot; If you have questions or want the full list, contact moab33.<br /> <br /> - Gmaas can be summoned using the &lt;math&gt;\tan(90)&lt;/math&gt; ritual. Draw a pentagram and write the numerical value of &lt;math&gt;\tan(90)&lt;/math&gt; in the middle, and he will be summoned.<br /> <br /> - EDIT: The above fact is incorrect. math101010 has done this and commented with screenshot proof at the below link, and Gmaas was not summoned.<br /> https://artofproblemsolving.com/community/c287916h1291232<br /> <br /> - EDIT EDIT: The above 'proof' is non-conclusive. math101010 had only put an approximation.<br /> <br /> - Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br /> <br /> - EDIT: That has never happened and thus it does not contain the singularity of a black hole. <br /> <br /> - Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br /> <br /> -Despite common belief, Harry Potter did not defeat Lord Voldemort. Gmaas did.<br /> <br /> - The original owner of Gmaas is Gmaas.<br /> <br /> - Gmaas was not the fourth Peverell brother, but he ascended into a higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br /> <br /> - It is suspected that Gmaas may be ordering his cyber hairballs to take the forums, along with microbots.<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mu.<br /> <br /> - Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond.<br /> EDIT: that is a lie. he IS THE Illuminati. april fools(it's actually July, I do April fools year round)<br /> <br /> - It has been wondered if Gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br /> <br /> - Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br /> <br /> - It has been confirmed that gmaas uses gmewal as his email service<br /> <br /> - Gmaas enjoys wearing gmean shorts<br /> <br /> - Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br /> <br /> - Gmaas is well known behind his stage name, Michael Stevens (also known as Vsauce XD), or his page name, Purrshanks.<br /> <br /> - Gmaas rekt sseraj at 12:54 June 4, 2016 UTC time zone. And then the Doctor chased him.<br /> <br /> - Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br /> <br /> - Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br /> <br /> - In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both Gmaas and the DL (USF code name of the Dark Lord) love BBC. How Gmaas gave him a TV may be lost to history. And it has been lost.<br /> <br /> - The TV has been noticed to be invincible. Many USF weapons, even volcano rings, have tried (and failed) to destroy it. The last time it was seen was on a Kobold display case outside of a mine. The display case was crushed, and a report showed a spy running off with a non-crushed TV.<br /> <br /> -The reason why Dacammel left the USF is that gmaas entrusted his TV to him, and not wanting to be discovered by LF, Cobra, or Z9, dacammel chose to leave the USF, but is regretting it, as snakes keep spawning from the TV.<br /> <br /> - EDIT: The above fact is somewhat irrelevant.<br /> <br /> - Gmaas is a Super Duper Uper Cat Time Lord. He has &lt;math&gt;57843504&lt;/math&gt; regenerations and has used &lt;math&gt;3&lt;/math&gt;. &lt;cmath&gt;9\cdot12\cdot2\cdot267794=57843504&lt;/cmath&gt;. <br /> <br /> -Gmaas highly enjoys destroying squeaky toys until he finds the squeaky part, then destroys the squeaky part.<br /> <br /> - Gmaas loves to eat turnips. At &lt;math&gt;\frac{13}{32}&lt;/math&gt; of the sites he was spotted at, he was seen with a turnip.<br /> <br /> -Gmaas has a secret hidden garden full of turnips under sseraj's house.<br /> <br /> - Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br /> <br /> -Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br /> <br /> -Gmaas is in alliance with the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything in sight? Nobody knows. (Except him.)<br /> <br /> -Gmaas lives in Gallifrey and in Gotham City (he has sleepovers with Batman).<br /> <br /> -Gmaas is an excellent driver.<br /> <br /> -The native location of Gmaas is the twilight zone.<br /> <br /> -Donald Trump once sang &quot;All Hail the Chief&quot; to Gmaas, 3 days after being sworn in as US President.<br /> <br /> - Gmaas likes to talk with rrusczyk from time to time.<br /> <br /> - Gmaas can shoot fire from his paws.<br /> <br /> - Gmaas is the reason why the USF has the longest thread on AoPS.<br /> <br /> - He (or she) is an avid watcher of the popular T.V. show &quot;Bernie Sanders and the Gauntlet of DOOM&quot;<br /> <br /> - sseraj, in 1521 Introduction to Number Theory, posted an image of Gmaas after saying &quot;Who wants to see 5space?&quot; at around 5:16 PM Mountain Time, noting Gmaas was &quot;also 5space&quot;<br /> <br /> -EDIT: he also did it in Introduction to Algebra A once<br /> <br /> - Gmaas is now my HD background on my Mac.<br /> <br /> - In 1521 Into to Number Theory, sseraj posted an image of a 5space Gmaas fusion. (First sighting) <br /> <br /> - Also confirmed that Gmaas doesn't like ketchup because it was the only food left the photo.<br /> <br /> - In 1447 Intro to Geometry, sseraj posted a picture of Gmaas with a rubik's cube suggesting that Gmaas's has an average solve time of &lt;math&gt;-GMAAS&lt;/math&gt; seconds.<br /> <br /> -Gmass beat Superman in a fight with ease<br /> <br /> -Gmass was an admin of Roblox<br /> <br /> -Gmass traveled around the world, paying so much &lt;math&gt;MONEY&lt;/math&gt; just to eat :D<br /> <br /> -Gmaas is a confirmed Apex predator and should not be approached, unless in a domestic form.<br /> Summary:<br /> <br /> -When Gmaas subtracts &lt;math&gt;0.\overline{99}&lt;/math&gt; from &lt;math&gt;1&lt;/math&gt;, the difference is greater than &lt;math&gt;0&lt;/math&gt;.<br /> <br /> -Gmaas was shown to have fallen on Wed Aug 23 2017: https://ibb.co/bNrtmk https://ibb.co/jzUDmk<br /> <br /> -Gmaas died on August ,24, 2017, but fortunately IceParrot revived him after about 2 mins of being dead.<br /> <br /> -The results of the revival are top secret, and nobody knows what happened.<br /> <br /> -sseraj, in 1496 Prealgebra 2, said that Gmaas is Santacat.<br /> <br /> -sseraj likes to post a picture of gmaas in every class he passes by.<br /> <br /> -sseraj posted a picture of gmaas as an Ewok, suggesting he resides on the moon of Endor. Unfortunately, the moon of Ender is also uninhabitable ever since the wreckage of the Death Star changed the climate there. It is thought gmass is now wandering space in search for a home.<br /> EDIT: What evidence is there Endor was affected? Other Ewoks still live there.<br /> <br /> -Gmaas is the lord of the pokemans<br /> <br /> -Gmass can communicate with, and sometimes control any other cats, however this is very rare, as cats normally have a very strong will<br /> <br /> -Picture of Gmass http://i.imgur.com/PP9xi.png<br /> <br /> -Known by Mike Miller<br /> <br /> -Gmaas got mad at sseraj once, so he locked him in his own freezer<br /> <br /> -Then, sseraj decided to eat all of Gmaas's hidden turnips in the freezer as punishment<br /> <br /> -Gmaas is an obviously omnipotent cat.<br /> <br /> -ehawk11 met him<br /> <br /> -sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop'.<br /> <br /> -sseraj has posted pictures of gmass in '&quot;intro to algebra&quot;, before class started, with the title, &quot;caption contest&quot; anyone who posted a caption mysteriously vanished in the middle of the night. <br /> EDIT: This has happened many times, including in Introduction to Geometry 1533, among other active classes. The person writing this (Poeshuman) did participate, and did not disappear. (You could argue Gmass is typing this through his/her account...)<br /> <br /> - gmass has once slept in your bed and made it wet<br /> <br /> -It is rumored that rrusczyk is actually Gmaas in disguise<br /> <br /> -Gmaas is suspected to be a Mewtwo in disguise<br /> <br /> -Gmaas is a cat but has characteristics of every other animal on Earth.<br /> <br /> -Gmass is the ruler of the universe and has been known to be the creator of the species &quot;Gmassians&quot;.<br /> <br /> -There is a rumor that Gmaas is starting a poll<br /> <br /> -Gmaas is a rumored past ThunderClan cat who ran away, founded GmaasClan, then became a kittypet.<br /> <br /> -There is a rumored sport called &quot;Gmaas Hunting&quot; where people try to successfully capture gmaas in the wild with video/camera/eyes. Strangely, no one has been able to do this, and those that have have mysteriously disappeared into the night. Nobody knows why. The person who is writing this(g1zq) has tried Gmaas Hunting, but has never been successful.<br /> <br /> === Gmaas photos ===<br /> http://cdn.artofproblemsolving.com/images/f/f/8/ff8efef3a0d2eb51254634e54bec215b948a1bba.jpg<br /> <br /> === gmaas in Popular Culture ===<br /> <br /> - [s]Currently, is being written (by themoocow) about the adventures of gmaas. It is aptly titled, &quot;The Adventures of gmaas&quot;.[/s]Sorry, this was a rick roll troll.<br /> <br /> - BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,&quot;What are gwali?&quot; the customary answer &quot;This is gwali&quot; is returned. Scientist 5space is now looking into it.<br /> <br /> - Sullymath and themoocow are also writing a book about Gmaas<br /> <br /> -Ornx the mad god is actually gmass wearing a suit of armor. This explains why he is never truly killed<br /> <br /> - Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br /> <br /> - Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br /> <br /> - Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles<br /> <br /> - gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br /> <br /> -oadaegan is also writing a story about him. He is continuing the book that was started by JpusheenS. when he is done he will post it here<br /> <br /> -Gmaas is a time traveler from 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br /> <br /> -No one knows if Gmass is a Mr. mime in a cat skin, the other way around, or just a downright combination of both.<br /> <br /> -In it, it mentions these four links as things Gmass is having trouble (specifically technical difficulties). What could it mean? Links:<br /> https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br /> https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br /> https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br /> https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br /> <br /> <br /> - Another possible Gmaas sighting [https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project]<br /> <br /> -&lt;math&gt;Another&lt;/math&gt; sighting? [https://www.radarbox24.com/data/flights/GMAAS]<br /> <br /> -Gmaas has been sighted several times on the Global Announcements forum<br /> <br /> -Gmaas uses the following transportation: [img]http://cdn.artofproblemsolving.com/images/3/6/8/368da4e615ea3476355ee3388b39f30a48b8dd48.jpg[/img]</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_14&diff=90278 2014 AMC 8 Problems/Problem 14 2018-02-05T23:06:44Z <p>G1zq: /* Solution */</p> <hr /> <div>==Problem==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; and right triangle &lt;math&gt;DCE&lt;/math&gt; have the same area. They are joined to form a trapezoid, as shown. What is &lt;math&gt;DE&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(linewidth(0.8));<br /> pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0);<br /> draw(A--B--E--D--cycle^^C--D);<br /> draw(rightanglemark(D,C,E,30));<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,S);<br /> label(&quot;$D$&quot;,D,N);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$5$&quot;,A/2,W);<br /> label(&quot;$6$&quot;,(A+D)/2,N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16 &lt;/math&gt;<br /> ==Solution==<br /> The area of &lt;math&gt;\bigtriangleup CDE&lt;/math&gt; is &lt;math&gt;\frac{DC\cdot CE}{2}&lt;/math&gt;. The area of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;AB\cdot AD=5\cdot 6=30&lt;/math&gt;, which also must be equal to the area of &lt;math&gt;\bigtriangleup CDE&lt;/math&gt;, which, since &lt;math&gt;DC=5&lt;/math&gt;, must in turn equal &lt;math&gt;\frac{5\cdot CE}{2}&lt;/math&gt;. Through transitivity, then, &lt;math&gt;\frac{5\cdot CE}{2}=30&lt;/math&gt;, and &lt;math&gt;CE=12&lt;/math&gt;. Then, using the Pythagorean Theorem, you should be able to figure out that &lt;math&gt;\bigtriangleup CDE&lt;/math&gt; is a &lt;math&gt;5-12-13&lt;/math&gt; triangle, so &lt;math&gt;DE=\boxed{13}&lt;/math&gt;, or &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> The area of the rectangle is &lt;math&gt;5\times6=30.&lt;/math&gt; Since the parallel line pairs are identical, &lt;math&gt;DC=5&lt;/math&gt;. Let &lt;math&gt;CE&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;\dfrac{5x}{2}=30&lt;/math&gt; is the area of the right triangle. Solving for &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;x=12.&lt;/math&gt; According to the Pythagorean Theorem, we have a 5-12-13 triangle. So, the hypotenuse &lt;math&gt;DE&lt;/math&gt; has to be &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2014|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=User:G1zq&diff=90246 User:G1zq 2018-02-04T21:19:59Z <p>G1zq: /* g1zq */</p> <hr /> <div>===g1zq===<br /> Hi! I'm a fun loving AoPSer who likes math, warrior cats, pokemon, harry potter, video games, and more! Feel free to friend me ;)</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=90147 2015 AMC 8 Problems/Problem 17 2018-02-02T20:08:11Z <p>G1zq: /* Solution 1 */</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 /> [INCOMPLETE]<br /> <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> G1zq https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_15&diff=90146 2015 AMC 8 Problems/Problem 15 2018-02-02T19:58:51Z <p>G1zq: /* Solution 1 */</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 /> ==Solution 1==<br /> We can see that this is a Venn Diagram Problem.[SOMEBODY DRAW IT PLEASE] <br /> <br /> First, we analyze the information given. There are &lt;math&gt;198&lt;/math&gt; students. Let's use A as the first issue and B as the second issue. <br /> <br /> &lt;math&gt;149&lt;/math&gt; students were for the A, and &lt;math&gt;119&lt;/math&gt; students were for B. There were also &lt;math&gt;29&lt;/math&gt; students against both A and B. <br /> <br /> Solving this without a Venn Diagram, we subtract &lt;math&gt;29&lt;/math&gt; away from the total, &lt;math&gt;198&lt;/math&gt;. Out of the remaining &lt;math&gt;169&lt;/math&gt; , we have &lt;math&gt;149&lt;/math&gt; people for A and <br /> <br /> &lt;math&gt;119&lt;/math&gt; people for B. We add this up to get &lt;math&gt;268&lt;/math&gt; . Since that is more than what we need, we subtract &lt;math&gt;169&lt;/math&gt; from &lt;math&gt;268&lt;/math&gt; to get <br /> <br /> &lt;math&gt;\boxed{\textbf{(D)}~99}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> There are 198 people. We know that 29 people voted against both the first issue and the second issue. That leaves us with 169 people that voted for at least one of them. If 119 people voted for both of them, then that would leave 20 people out of the vote, because 149 is less than 169 people. 169-149 is 20, so to make it even, we have to take 20 away from the 119 people, which leaves us with &lt;math&gt;\boxed{\textbf{(D)}~99}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_7&diff=90145 2015 AMC 8 Problems/Problem 7 2018-02-02T16:33:53Z <p>G1zq: /* Solution 3 */</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 calculate 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;\left ( \frac{2}{3} \right )^2=\frac{4}{9}&lt;/math&gt; probability of having an odd product. Thus, there is a &lt;math&gt;1-\frac{4}{9}=\boxed{\textbf{(E)}~\frac{5}{9}}&lt;/math&gt; probability of having an even product.<br /> <br /> ===Solution 2===<br /> You can also make this problem into a spinner problem. You have the first spinner with &lt;math&gt;3&lt;/math&gt; equally divided <br /> <br /> sections, &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3.&lt;/math&gt; You make a second spinner that is identical to the first, with &lt;math&gt;3&lt;/math&gt; equal sections of <br /> <br /> &lt;math&gt;1&lt;/math&gt;,&lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt;. If the first spinner lands on &lt;math&gt;1&lt;/math&gt;, to be even, it must land on two. You write down the first <br /> <br /> combination of numbers &lt;math&gt;(1,2)&lt;/math&gt;. Next, if the spinner lands on &lt;math&gt;2&lt;/math&gt;, it can land on any number on the second <br /> <br /> spinner. We now have the combinations of &lt;math&gt;(1,2) ,(2,1), (2,2), (2,3)&lt;/math&gt;. Finally, if the first spinner ends on &lt;math&gt;3&lt;/math&gt;, we <br /> <br /> have &lt;math&gt;(3,2).&lt;/math&gt; Since there are &lt;math&gt;3*3=9&lt;/math&gt; possible combinations, and we have &lt;math&gt;5&lt;/math&gt; evens, the final answer is <br /> <br /> &lt;math&gt;\boxed{\textbf{(E) }\frac{5}{9}}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We can also list out the numbers. Hat A has chips &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt;, and Hat B also has chips &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt;. Chip &lt;math&gt;1&lt;/math&gt;(from Hat A) <br /> <br /> could be with 3 partners from Hat B. This is also the same for chips &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; from Hat A. &lt;math&gt;3+3+3=9&lt;/math&gt; total sums. Chip &lt;math&gt;1&lt;/math&gt; could be <br /> added with 2 other chips to make an even sum, just like chip &lt;math&gt;3&lt;/math&gt;. Chip &lt;math&gt;2&lt;/math&gt; can only add with 1 chip. &lt;math&gt;2+2+1=5&lt;/math&gt;. The answer is &lt;math&gt;\boxed{\textbf{(E) }\frac{5}{9}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_7&diff=90144 2015 AMC 8 Problems/Problem 7 2018-02-02T16:33:08Z <p>G1zq: </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 calculate 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;\left ( \frac{2}{3} \right )^2=\frac{4}{9}&lt;/math&gt; probability of having an odd product. Thus, there is a &lt;math&gt;1-\frac{4}{9}=\boxed{\textbf{(E)}~\frac{5}{9}}&lt;/math&gt; probability of having an even product.<br /> <br /> ===Solution 2===<br /> You can also make this problem into a spinner problem. You have the first spinner with &lt;math&gt;3&lt;/math&gt; equally divided <br /> <br /> sections, &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3.&lt;/math&gt; You make a second spinner that is identical to the first, with &lt;math&gt;3&lt;/math&gt; equal sections of <br /> <br /> &lt;math&gt;1&lt;/math&gt;,&lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt;. If the first spinner lands on &lt;math&gt;1&lt;/math&gt;, to be even, it must land on two. You write down the first <br /> <br /> combination of numbers &lt;math&gt;(1,2)&lt;/math&gt;. Next, if the spinner lands on &lt;math&gt;2&lt;/math&gt;, it can land on any number on the second <br /> <br /> spinner. We now have the combinations of &lt;math&gt;(1,2) ,(2,1), (2,2), (2,3)&lt;/math&gt;. Finally, if the first spinner ends on &lt;math&gt;3&lt;/math&gt;, we <br /> <br /> have &lt;math&gt;(3,2).&lt;/math&gt; Since there are &lt;math&gt;3*3=9&lt;/math&gt; possible combinations, and we have &lt;math&gt;5&lt;/math&gt; evens, the final answer is <br /> <br /> &lt;math&gt;\boxed{\textbf{(E) }\frac{5}{9}}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We can also list out the numbers. Hat A has chips &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt;, and Hat B also has chips &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt;. Chip &lt;math&gt;1&lt;/math&gt;(from Hat A) could be with 3 partners from Hat B. This is also the same for chips &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; from Hat A. &lt;math&gt;3+3+3=9&lt;/math&gt; total sums. Chip &lt;math&gt;1&lt;/math&gt; could be added with 2 other chips to make an even sum, just like chip &lt;math&gt;3&lt;/math&gt;. Chip &lt;math&gt;2&lt;/math&gt; can only add with 1 chip. &lt;math&gt;2+2+1=5&lt;/math&gt;. The answer is &lt;math&gt;\boxed{\textbf{(E) }\frac{5}{9}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_23&diff=90126 2016 AMC 8 Problems/Problem 23 2018-02-01T20:23:05Z <p>G1zq: /* Solution 1 */</p> <hr /> <div>Two congruent circles centered at points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; each pass through the other circle's center. The line containing both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is extended to intersect the circles at points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. The circles intersect at two points, one of which is &lt;math&gt;E&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle CED&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Drawing the diagram[SOMEONE DRAW IT PLEASE], we see that &lt;math&gt;\triangle EAB&lt;/math&gt; is equilateral as each side is the radius of one of the two circles. Therefore, &lt;math&gt;\overarc{EB}=m\angle EAB-60^\circ&lt;/math&gt;. Therefore, since it is an inscribed angle, &lt;math&gt;m\angle ECB=\frac{60^\circ}{2}=30^\circ&lt;/math&gt;. So, in &lt;math&gt;\triangle ECD&lt;/math&gt;, &lt;math&gt;m\angle ECB=m\angle EDA=30^\circ&lt;/math&gt;, and &lt;math&gt;m\angle CED=180^\circ-30^\circ-30^\circ=120^\circ&lt;/math&gt;. Our answer is &lt;math&gt;\boxed{\textbf{(C) }\ 120}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> As in Solution 1, observe that &lt;math&gt;\triangle{EAB}&lt;/math&gt; is equilateral. Therefore, &lt;math&gt;m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}&lt;/math&gt;. Since &lt;math&gt;CD&lt;/math&gt; is a straight line, we conclude that &lt;math&gt;m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}&lt;/math&gt;. Since &lt;math&gt;BE=BD&lt;/math&gt; (both are radii of the same circle), &lt;math&gt;\triangle{BED}&lt;/math&gt; is isosceles, meaning that &lt;math&gt;m\angle{BED}=m\angle{BDE}=30^{\circ}&lt;/math&gt;. Similarly, &lt;math&gt;m\angle{AEC}=m\angle{ACE}=30^{\circ}&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(C) }\ 120}&lt;/math&gt;.<br /> <br /> <br /> <br /> {{AMC8 box|year=2016|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems&diff=90107 2016 AMC 8 Problems 2018-01-31T20:33:55Z <p>G1zq: /* Problem 12 */</p> <hr /> <div><br /> <br /> ==Problem 1==<br /> <br /> The longest professional tennis match lasted a total of 11 hours and 5 minutes. How many minutes was that?<br /> <br /> &lt;math&gt;\textbf{(A)} 605 \qquad\textbf{(B)} 655\qquad\textbf{(C)} 665\qquad\textbf{(D)} 1005\qquad \textbf{(E)} 1105&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 1|Solution<br /> ]]<br /> <br /> ==Problem 2==<br /> <br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;AB=6&lt;/math&gt; and &lt;math&gt;AD=8&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{AD}&lt;/math&gt;. What is the area of &lt;math&gt;\triangle AMC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 2|Solution<br /> ]]<br /> <br /> ==Problem 3==<br /> <br /> Four students take an exam. Three of their scores are &lt;math&gt;70, 80,&lt;/math&gt; and &lt;math&gt;90&lt;/math&gt;. If the average of their four scores is &lt;math&gt;70&lt;/math&gt;, then what is the remaining score?<br /> <br /> &lt;math&gt;\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 3|Solution<br /> ]]<br /> <br /> ==Problem 4==<br /> <br /> When Cheenu was a boy he could run &lt;math&gt;15&lt;/math&gt; miles in &lt;math&gt;3&lt;/math&gt; hours and &lt;math&gt;30&lt;/math&gt; minutes. As an old man he can now walk &lt;math&gt;10&lt;/math&gt; miles in &lt;math&gt;4&lt;/math&gt; hours. How many minutes longer does it take for him to travel a mile now compared to when he was a boy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 4|Solution<br /> ]]<br /> <br /> ==Problem 5==<br /> <br /> The number &lt;math&gt;N&lt;/math&gt; is a two-digit number.<br /> <br /> • When &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;9&lt;/math&gt;, the remainder is &lt;math&gt;1&lt;/math&gt;.<br /> <br /> • When &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;10&lt;/math&gt;, the remainder is &lt;math&gt;3&lt;/math&gt;.<br /> <br /> What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;11&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 5|Solution<br /> ]]<br /> <br /> ==Problem 6==<br /> The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?<br /> &lt;math&gt;\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7&lt;/math&gt;<br /> &lt;asy&gt;<br /> unitsize(0.9cm);<br /> draw((-0.5,0)--(10,0), linewidth(1.5));<br /> draw((-0.5,1)--(10,1));<br /> draw((-0.5,2)--(10,2));<br /> draw((-0.5,3)--(10,3));<br /> draw((-0.5,4)--(10,4));<br /> draw((-0.5,5)--(10,5));<br /> draw((-0.5,6)--(10,6));<br /> draw((-0.5,7)--(10,7));<br /> label(&quot;frequency&quot;,(-0.5,8));<br /> label(&quot;0&quot;, (-1, 0));<br /> label(&quot;1&quot;, (-1, 1));<br /> label(&quot;2&quot;, (-1, 2));<br /> label(&quot;3&quot;, (-1, 3));<br /> label(&quot;4&quot;, (-1, 4));<br /> label(&quot;5&quot;, (-1, 5));<br /> label(&quot;6&quot;, (-1, 6));<br /> label(&quot;7&quot;, (-1, 7));<br /> filldraw((0,0)--(0,7)--(1,7)--(1,0)--cycle, black);<br /> filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle, black);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, black);<br /> filldraw((6,0)--(6,4)--(7,4)--(7,0)--cycle, black);<br /> filldraw((8,0)--(8,4)--(9,4)--(9,0)--cycle, black);<br /> label(&quot;3&quot;, (0.5, -0.5));<br /> label(&quot;4&quot;, (2.5, -0.5));<br /> label(&quot;5&quot;, (4.5, -0.5));<br /> label(&quot;6&quot;, (6.5, -0.5));<br /> label(&quot;7&quot;, (8.5, -0.5));<br /> label(&quot;name length&quot;, (4.5, -1));<br /> &lt;/asy&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 6|Solution<br /> ]]<br /> <br /> <br /> ==Problem 7==<br /> <br /> Which of the following numbers is not a perfect square?<br /> <br /> &lt;math&gt;\textbf{(A) }1^{2016}\qquad\textbf{(B) }2^{2017}\qquad\textbf{(C) }3^{2018}\qquad\textbf{(D) }4^{2019}\qquad \textbf{(E) }5^{2020}&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 7|Solution<br /> ]]<br /> <br /> ==Problem 8==<br /> <br /> Find the value of the expression<br /> &lt;cmath&gt;100-98+96-94+92-90+\cdots+8-6+4-2.&lt;/cmath&gt;&lt;math&gt;\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 8|Solution<br /> ]]<br /> <br /> ==Problem 9==<br /> <br /> What is the sum of the distinct prime integer divisors of &lt;math&gt;2016&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 9|Solution<br /> ]]<br /> <br /> ==Problem 10==<br /> <br /> Suppose that &lt;math&gt;a * b&lt;/math&gt; means &lt;math&gt;3a-b.&lt;/math&gt; What is the value of &lt;math&gt;x&lt;/math&gt; if<br /> &lt;cmath&gt;2 * (5 * x)=1&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 10|Solution<br /> ]]<br /> <br /> ==Problem 11==<br /> <br /> Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is &lt;math&gt;132.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 11|Solution<br /> ]]<br /> <br /> ==Problem 12==<br /> <br /> Jefferson Middle School has the same number of boys and girls. &lt;math&gt;\frac{3}{4}&lt;/math&gt; of the girls and &lt;math&gt;\frac{2}{3}&lt;/math&gt;<br /> of the boys went on a field trip. What fraction of the students were girls?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{9}{17}\qquad\textbf{(C) }\frac{7}{13}\qquad\textbf{(D) }\frac{2}{3}\qquad \textbf{(E) }\frac{14}{15}&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 12|Solution<br /> ]]<br /> <br /> ==Problem 13==<br /> <br /> Two different numbers are randomly selected from the set &lt;math&gt;{ - 2, -1, 0, 3, 4, 5}&lt;/math&gt; and multiplied together. What is the probability that the product is &lt;math&gt;0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 13|Solution<br /> ]]<br /> <br /> ==Problem 14==<br /> <br /> Karl's car uses a gallon of gas every &lt;math&gt;35&lt;/math&gt; miles, and his gas tank holds &lt;math&gt;14&lt;/math&gt; gallons when it is full. One day, Karl started with a full tank of gas, <br /> drove &lt;math&gt;350&lt;/math&gt; miles, bought &lt;math&gt;8&lt;/math&gt; gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day? <br /> <br /> &lt;math&gt;\textbf{(A)}\mbox{ }525\qquad\textbf{(B)}\mbox{ }560\qquad\textbf{(C)}\mbox{ }595\qquad\textbf{(D)}\mbox{ }665\qquad\textbf{(E)}\mbox{ }735&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 14|Solution<br /> ]]<br /> <br /> ==Problem 15==<br /> <br /> What is the largest power of &lt;math&gt;2&lt;/math&gt; that is a divisor of &lt;math&gt;13^4 - 11^4&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 15|Solution<br /> ]]<br /> <br /> ==Problem 16==<br /> <br /> Annie and Bonnie are running laps around a &lt;math&gt;400&lt;/math&gt;-meter oval track. They started together, but Annie has pulled ahead because she runs &lt;math&gt;25\%&lt;/math&gt; faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?<br /> <br /> &lt;math&gt;\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 16|Solution<br /> ]]<br /> <br /> ==Problem 17==<br /> <br /> An ATM password at Fred's Bank is composed of four digits from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;9&lt;/math&gt;, with repeated digits allowable. If no password may begin with the sequence &lt;math&gt;9,1,1,&lt;/math&gt; then how many passwords are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 17|Solution<br /> ]]<br /> <br /> ==Problem 18==<br /> <br /> In an All-Area track meet, &lt;math&gt;216&lt;/math&gt; sprinters enter a &lt;math&gt;100-&lt;/math&gt;meter dash competition. The track has &lt;math&gt;6&lt;/math&gt; lanes, so only &lt;math&gt;6&lt;/math&gt; sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?<br /> <br /> &lt;math&gt;\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 18|Solution<br /> ]]<br /> <br /> ==Problem 19==<br /> <br /> The sum of &lt;math&gt;25&lt;/math&gt; consecutive even integers is &lt;math&gt;10,000&lt;/math&gt;. What is the largest of these &lt;math&gt;25&lt;/math&gt; consecutive integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 19|Solution<br /> ]]<br /> <br /> ==Problem 20==<br /> <br /> The least common multiple of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; is &lt;math&gt;12&lt;/math&gt;, and the least common multiple of &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; is &lt;math&gt;15&lt;/math&gt;. What is the least possible value of the least common multiple of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 20|Solution<br /> ]]<br /> <br /> ==Problem 21==<br /> <br /> A top hat contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?<br /> <br /> &lt;math&gt;\textbf{(A) }\dfrac{3}{10}\qquad\textbf{(B) }\dfrac{2}{5}\qquad\textbf{(C) }\dfrac{1}{2}\qquad\textbf{(D) }\dfrac{3}{5}\qquad \textbf{(E) }\dfrac{7}{10}&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 21|Solution<br /> ]]<br /> <br /> ==Problem 22==<br /> Rectangle &lt;math&gt;DEFA&lt;/math&gt; below is a &lt;math&gt;3 \times 4&lt;/math&gt; rectangle with &lt;math&gt;DC=CB=BA&lt;/math&gt;. What is the area of the &quot;bat wings&quot; (shaded area)?<br /> &lt;asy&gt;<br /> draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));<br /> draw((3,0)--(1,4)--(0,0));<br /> fill((0,0)--(1,4)--(1.5,3)--cycle, black);<br /> fill((3,0)--(2,4)--(1.5,3)--cycle, black);<br /> label(&quot;$A$&quot;,(3.05,4.2));<br /> label(&quot;$B$&quot;,(2,4.2));<br /> label(&quot;$C$&quot;,(1,4.2));<br /> label(&quot;$D$&quot;,(0,4.2));<br /> label(&quot;$E$&quot;, (0,-0.2));<br /> label(&quot;$F$&quot;, (3,-0.2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }4&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 22|Solution<br /> ]]<br /> <br /> ==Problem 23==<br /> <br /> Two congruent circles centered at points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; each pass through the other circle's center. The line containing both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is extended to intersect the circles at points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. The circles intersect at two points, one of which is &lt;math&gt;E&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle CED&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 23|Solution<br /> ]]<br /> <br /> ==Problem 24==<br /> <br /> The digits &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; are each used once to write a five-digit number &lt;math&gt;PQRST&lt;/math&gt;. The three-digit number &lt;math&gt;PQR&lt;/math&gt; is divisible by &lt;math&gt;4&lt;/math&gt;, the three-digit number &lt;math&gt;QRS&lt;/math&gt; is divisible by &lt;math&gt;5&lt;/math&gt;, and the three-digit number &lt;math&gt;RST&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt;. What is &lt;math&gt;P&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 24|Solution<br /> ]]<br /> <br /> ==Problem 25== <br /> <br /> A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 25|Solution<br /> ]]<br /> <br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_25&diff=90087 2017 AMC 8 Problems/Problem 25 2018-01-31T00:20:07Z <p>G1zq: /* Solution */</p> <hr /> <div>==Problem 25==<br /> <br /> In the figure shown, &lt;math&gt;\overline{US}&lt;/math&gt; and &lt;math&gt;\overline{UT}&lt;/math&gt; are line segments each of length 2, and &lt;math&gt;m\angle TUS = 60^\circ&lt;/math&gt;. Arcs &lt;math&gt;\overarc{TR}&lt;/math&gt; and &lt;math&gt;\overarc{SR}&lt;/math&gt; are each one-sixth of a circle with radius 2. What is the area of the region shown? <br /> <br /> &lt;asy&gt;draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label(&quot;$U$&quot;, (2,3.464), N); label(&quot;$S$&quot;, (1,1.732), W); label(&quot;$T$&quot;, (3,1.732), E); label(&quot;$R$&quot;, (2,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let the centers of the circles containing arcs &lt;math&gt;\overarc{SR}&lt;/math&gt; and &lt;math&gt;\overarc{TR}&lt;/math&gt; be &lt;math&gt;S'&lt;/math&gt; and &lt;math&gt;T'&lt;/math&gt;, respectively. Extend &lt;math&gt;\overline{US}&lt;/math&gt; and &lt;math&gt;\overline{UT}&lt;/math&gt; to &lt;math&gt;S'&lt;/math&gt; and &lt;math&gt;T'&lt;/math&gt;, and connect point &lt;math&gt;S'&lt;/math&gt; with point &lt;math&gt;T'&lt;/math&gt;. <br /> &lt;asy&gt;draw((1,1.732)--(2,3.464)--(3,1.732));<br /> draw((1,1.732)--(0,0)--(4,0)--(3,1.732));<br /> draw(arc((0,0),(2,0),(1,1.732)));<br /> draw(arc((4,0),(3,1.732),(2,0)));<br /> label(&quot;$U$&quot;, (2,3.464), N);<br /> label(&quot;$S$&quot;, (1,1.732), W);<br /> label(&quot;$T$&quot;, (3,1.732), E);<br /> label(&quot;$R$&quot;, (2,0), S);<br /> label(&quot;$S'$&quot;, (0,0), W);<br /> label(&quot;$T'$&quot;, (4,0), E);&lt;/asy&gt;<br /> We can clearly see that &lt;math&gt;\triangle US'T'&lt;/math&gt; is an equilateral triangle, because two of its angles are &lt;math&gt;60^\circ&lt;/math&gt;, which is the degree measure of &lt;math&gt;\frac{1}{6}&lt;/math&gt; a circle. The area of the figure is equal to the area of equilateral triangle &lt;math&gt;\triangle US'T'&lt;/math&gt; minus the combined area of the &lt;math&gt;2&lt;/math&gt; sectors of the circles. Using the area formula for an equilateral triangle, &lt;math&gt;\frac{\sqrt{3}}{4} \cdot a,&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is the side length of the equilateral triangle, the area of &lt;math&gt;\triangle US'T'&lt;/math&gt; is &lt;math&gt;\frac{\sqrt 3}{4} \cdot 4^2 = 4\sqrt 3.&lt;/math&gt; The combined area of the &lt;math&gt;2&lt;/math&gt; sectors is &lt;math&gt;2&lt;/math&gt; times one sixth &lt;math&gt;\pi \cdot r^2&lt;/math&gt;, which is &lt;math&gt;2 \cdot \frac 16 \cdot \pi \cdot 2^2 = \frac{4\pi}{3}.&lt;/math&gt; Our final answer is then &lt;math&gt;\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=24|after=Last Problem}}<br /> <br /> {{MAA Notice}}</div> G1zq https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_18&diff=90085 2017 AMC 8 Problems/Problem 18 2018-01-30T22:38:19Z <p>G1zq: /* Solution */</p> <hr /> <div>==Problem 18==<br /> In the non-convex quadrilateral &lt;math&gt;ABCD&lt;/math&gt; shown below, &lt;math&gt;\angle BCD&lt;/math&gt; is a right angle, &lt;math&gt;AB=12&lt;/math&gt;, &lt;math&gt;BC=4&lt;/math&gt;, &lt;math&gt;CD=3&lt;/math&gt;, and &lt;math&gt;AD=13&lt;/math&gt;. &lt;asy&gt;draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label(&quot;$B$&quot;, (0, 0), SW); label(&quot;$A$&quot;, (12, 0), ESE); label(&quot;$C$&quot;, (2.4, 3.6), SE); label(&quot;$D$&quot;, (0, 5), N);&lt;/asy&gt; What is the area of quadrilateral &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }12\qquad\textbf{(B) }24\qquad\textbf{(C) }26\qquad\textbf{(D) }30\qquad\textbf{(E) }36&lt;/math&gt;<br /> <br /> ==Solution==<br /> We first connect point &lt;math&gt;B&lt;/math&gt; with point &lt;math&gt;D&lt;/math&gt;. <br /> <br /> &lt;asy&gt;draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); draw((0,0)--(0,5)); label(&quot;$B$&quot;, (0, 0), SW); label(&quot;$A$&quot;, (12, 0), ESE); label(&quot;$C$&quot;, (2.4, 3.6), SE); label(&quot;$D\$&quot;, (0, 5), N);&lt;/asy&gt; <br /> <br /> We can see that &lt;math&gt;\triangle BCD&lt;/math&gt; is a 3-4-5 right triangle. We can also see that &lt;math&gt;\triangle BDA&lt;/math&gt; is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of &lt;math&gt;\triangle BDA&lt;/math&gt; is &lt;math&gt;\frac{5\cdot 12}{2}&lt;/math&gt;, and the area of the smaller 3-4-5 triangle is &lt;math&gt;\frac{3\cdot 4}{2}&lt;/math&gt;. Thus, the area of quadrialteral &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;30-6 = \boxed{\textbf{(B)}\ 24}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=17|num-a=19}}<br /> <br /> {{MAA Notice}}</div> G1zq