https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Bz1&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T23:26:43ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_8_Problems/Problem_17&diff=803832005 AMC 8 Problems/Problem 172016-09-24T22:32:05Z<p>Bz1: /* Solution */</p>
<hr />
<div>==Problem==<br />
The results of a cross-country team's training run are graphed below. Which student has the greatest average speed?<br />
<asy><br />
for ( int i = 1; i <= 7; ++i )<br />
{<br />
draw((i,0)--(i,6));<br />
}<br />
<br />
for ( int i = 1; i <= 5; ++i )<br />
{<br />
draw((0,i)--(8,i));<br />
}<br />
draw((-0.5,0)--(8,0), linewidth(1));<br />
draw((0,-0.5)--(0,6), linewidth(1));<br />
label("$O$", (0,0), SW);<br />
label(scale(.85)*rotate(90)*"distance", (0, 3), W);<br />
label(scale(.85)*"time", (4, 0), S);<br />
dot((1.25, 4.5));<br />
label(scale(.85)*"Evelyn", (1.25, 4.8), N);<br />
dot((2.5, 2.2));<br />
label(scale(.85)*"Briana", (2.5, 2.2), S);<br />
dot((4.25,5.2));<br />
label(scale(.85)*"Carla", (4.25, 5.2), SE);<br />
dot((5.6, 2.8));<br />
label(scale(.85)*"Debra", (5.6, 2.8), N);<br />
dot((6.8, 1.4));<br />
label(scale(.85)*"Angela", (6.8, 1.4), E);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \text{Angela}\qquad\textbf{(B)}\ \text{Briana}\qquad\textbf{(C)}\ \text{Carla}\qquad\textbf{(D)}\ \text{Debra}\qquad\textbf{(E)}\ \text{Evelyn} </math><br />
<br />
==Solution==<br />
Average speed is distance over time, or the slope of the line through the point and the origin. <math>\boxed{\textbf{(E)}\ \text{Evelyn}}</math> has the steepest line, and runs the greatest distance for the shortest amount of time.<br />
When activated, a prismarine block will be shoot out in the direction pointed. The block will begin to continuously throw Treasure Shards around it in a circle for 30 seconds.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2005|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Bz1https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_8_Problems/Problem_6&diff=803762005 AMC 8 Problems/Problem 62016-09-24T22:21:27Z<p>Bz1: /* See Also */</p>
<hr />
<div>==Problem ==<br />
Suppose <math>d</math> is a digit. For how many values of <math>d</math> is <math>2.00d5 > 2.005</math>?<br />
<br />
<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 10 </math><br />
<br />
==Solution==<br />
We see that <math>2.0055</math> works but <math>2.0045</math> does not. The digit <math>d</math> can be from <math>5</math> through <math>9</math>, which is <math>\boxed{\textbf{(C)}\ 5}</math> values.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2005|num-b=5|num-a=7}}<br />
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png and do not sponsor or endorse this.</div>Bz1https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_8_Problems/Problem_25&diff=803742004 AMC 8 Problems/Problem 252016-09-24T22:17:41Z<p>Bz1: /* Problem */</p>
<hr />
<div>==juan likes tacos==<br />
Two <math>4 \times 4</math> squares intersect at right angles, bisecting their intersecting sides, as shown. The circle's diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?<br />
<br />
<asy><br />
unitsize(6mm);<br />
draw(unitcircle);<br />
filldraw((0,1)--(1,2)--(3,0)--(1,-2)--(0,-1)--(-1,-2)--(-3,0)--(-1,2)--cycle,lightgray,black);<br />
filldraw(unitcircle,white,black);<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 16-4\pi\qquad \textbf{(B)}\ 16-2\pi \qquad \textbf{(C)}\ 28-4\pi \qquad \textbf{(D)}\ 28-2\pi \qquad \textbf{(E)}\ 32-2\pi</math><br />
<br />
==Solution==<br />
If the circle was shaded in, the intersection of the two squares would be a smaller square with half the sidelength, <math>2</math>. The area of this region would be the two larger squares minus the area of the intersection, the smaller square. This is <math>4^2 + 4^2 - 2^2 = 28</math>.<br />
<br />
The diagonal of this smaller square created by connecting the two points of intersection of the squares is the diameter of the circle. This value can be found with Pythagorean or a <math>45^\circ - 45^\circ - 90^\circ</math> circle to be <math>2\sqrt{2}</math>. The radius is half the diameter, <math>\sqrt{2}</math>. The area of the circle is <math>\pi r^2 = \pi (\sqrt{2})^2 = 2\pi</math>.<br />
<br />
The area of the shaded region is the area of the circle minus the area of the two squares which is <math>\boxed{\textbf{(D)}\ 28-2\pi}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2004|num-b=24|after=Last <br /> Question}}<br />
{{MAA Notice}}</div>Bz1https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_8_Problems/Problem_18&diff=803722004 AMC 8 Problems/Problem 182016-09-24T21:30:11Z<p>Bz1: /* Solution */</p>
<hr />
<div>==Problem==<br />
Five friends compete in a dart-throwing contest. Each one has two darts to throw at the same circular target, and each individual's score is the sum of the scores in the target regions that are hit. The scores for the target regions are the whole numbers <math>1</math> through <math>10</math>. Each throw hits the target in a region with a different value. The scores are: Alice <math>16</math> points, Ben <math>4</math> points, Cindy <math>7</math> points, Dave <math>11</math> points, and Ellen <math>17</math> points. Who hits the region worth <math>6</math> points?<br />
<br />
<math>\textbf{(A)}\ \text{Alice}\qquad \textbf{(B)}\ \text{Ben}\qquad \textbf{(C)}\ \text{Cindy}\qquad \textbf{(D)}\ \text{Dave} \qquad \textbf{(E)}\ \text{Ellen}</math><br />
<br />
==Solution==<br />
The only way to get Ben's score is with <math>1+3=4</math>. Cindy's score can be made of <math>3+4</math> or <math>2+5</math>, but since Ben already hit the <math>3</math>, Cindy hit <math>2+5=7</math>. Similar, Dave's darts were in the region <math>4+7=11</math>. Lastly, because there is no <math>7</math> left, <math>\boxed{\textbf{(A)}\ \text{Alice}}</math> must have hit the regions <math>6+10=16</math> and Ellen <math>8+9=17</math>.<br />
<br />
Devenware likes the US<br />
<br />
==See Also==<br />
{{AMC8 box|year=2004|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Bz1https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_8_Problems/Problem_13&diff=803712004 AMC 8 Problems/Problem 132016-09-24T21:29:02Z<p>Bz1: /* Problem */</p>
<hr />
<div>==Problem==<br />
Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true.<br />
I. Bill is the oldest.<br />
II. Amy is not the oldest.<br />
III. Celine is not the youngest.<br />
IIII. Cobra is swag.<br />
Rank the friends from the oldest to the youngest.<br />
<br />
<math>\textbf{(A)}\ \text{Bill, Amy, Celine}\qquad \textbf{(B)}\ \text{Amy, Bill, Celine}\qquad \textbf{(C)}\ \text{Celine, Amy, Bill}\\<br />
\textbf{(D)}\ \text{Celine, Bill, Amy} \qquad \textbf{(E)}\ \text{Amy, Celine, Bill}</math><br />
<br />
==Solution==<br />
If Bill is the oldest, then Amy is not the oldest, and both statements I and II are true, so statement I is not the true one.<br />
<br />
If Amy is not the oldest, and we know Bill cannot be the oldest, then Celine is the oldest. This would mean she is not the youngest, and both statements II and III are true, so statement II is not the true one.<br />
<br />
Therefore, statement III is the true statement, and both I and II are false. From this, Amy is the oldest, Celine is in the middle, and lastly Bill is the youngest. This order is <math>\boxed{\textbf{(E)}\ \text{Amy, Celine, Bill}}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2004|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Bz1https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems/Problem_12&diff=803702003 AMC 8 Problems/Problem 122016-09-24T21:04:54Z<p>Bz1: /* See Also */</p>
<hr />
<div>==Problem==<br />
When a fair six-sided dice is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the faces than can be seen is divisible by <math>6</math>?<br />
<br />
<math> \textbf{(A)}\ 1/3\qquad\textbf{(B)}\ 1/2\qquad\textbf{(C)}\ 2/3\qquad\textbf{(D)}\ 5/6\qquad\textbf{(E)}\ 1 </math><br />
<br />
==Solution==<br />
All the possibilities where <math>6</math> is on any of the five sides is always divisible by six, and <math>1 \times 2 \times 3 \times 4 \times 5</math> is divisible by <math>6</math> since <math>2 \times 3 = 6</math>. So, the answer is <math>\boxed{\textbf{(E)}\ 1}</math> because the outcome is always divisible by <math>6</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2015|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Bz1https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_24&diff=803692002 AMC 8 Problems/Problem 242016-09-24T20:43:31Z<p>Bz1: /* See Also */</p>
<hr />
<div>==Problem==<br />
Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?<br />
<br />
<math> \text{(A)}\ 30\qquad\text{(B)}\ 40\qquad\text{(C)}\ 50\qquad\text{(D)}\ 60\qquad\text{(E)}\ 70 </math><br />
<br />
==Solution==<br />
A pear gives <math>8/3</math> ounces of juice per pear. An orange gives <math>8/2=4</math> ounces of juice per orange. If the pear-orange juice blend used one pear and one orange each, the percentage of pear juice would be<br />
<br />
<cmath>\frac{8/3}{8/3+4} \times 100 = \frac{8}{8+12} \times 100 = \boxed{\text{(B)}\ 40}</cmath><br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=23|num-a=25}}<br />
This problem is copyrighted by the MAA. Thanks for listening.</div>Bz1https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_8&diff=803682002 AMC 8 Problems/Problem 82016-09-24T20:32:50Z<p>Bz1: /* Juan's Old Stamping Grounds */</p>
<hr />
<div>==Juan's Tacos! Only 50 cents at Mexico Grililed Cheese!==<br />
<br />
Problems 8,9 and 10 use the data found in the accompanying paragraph and table:<br />
<br />
<center><br />
Juan organizes the stamps in his collection by country and by the decade in which they were issued. The prices he paid for them at a stamp shop were: Brazil and<br />
France, 6 cents each, Peru 4 cents each, and Spain 5 cents each. (Brazil and Peru are South American countries and France and Spain are in Europe.)<br />
</center><br />
<br />
<asy><br />
/* AMC8 2002 #8, 9, 10 Problem */<br />
size(3inch, 1.5inch);<br />
for ( int y = 0; y &lt;= 5; ++y )<br />
{<br />
draw((0,y)--(18,y));<br />
}<br />
draw((0,0)--(0,5));<br />
draw((6,0)--(6,5));<br />
draw((9,0)--(9,5));<br />
draw((12,0)--(12,5));<br />
draw((15,0)--(15,5));<br />
draw((18,0)--(18,5));<br />
draw(scale(0.8)*"50s", (7.5,4.5));<br />
draw(scale(0.8)*"4", (7.5,3.5));<br />
draw(scale(0.8)*"8", (7.5,2.5));<br />
draw(scale(0.8)*"6", (7.5,1.5));<br />
draw(scale(0.8)*"3", (7.5,0.5));<br />
draw(scale(0.8)*"60s", (10.5,4.5));<br />
draw(scale(0.8)*"7", (10.5,3.5));<br />
draw(scale(0.8)*"4", (10.5,2.5));<br />
draw(scale(0.8)*"4", (10.5,1.5));<br />
draw(scale(0.8)*"9", (10.5,0.5));<br />
draw(scale(0.8)*"70s", (13.5,4.5));<br />
draw(scale(0.8)*"12", (13.5,3.5));<br />
draw(scale(0.8)*"12", (13.5,2.5));<br />
draw(scale(0.8)*"6", (13.5,1.5));<br />
draw(scale(0.8)*"13", (13.5,0.5));<br />
draw(scale(0.8)*"80s", (16.5,4.5));<br />
draw(scale(0.8)*"8", (16.5,3.5));<br />
draw(scale(0.8)*"15", (16.5,2.5));<br />
draw(scale(0.8)*"10", (16.5,1.5));<br />
draw(scale(0.8)*"9", (16.5,0.5));<br />
label(scale(0.8)*"Country", (3,4.5));<br />
label(scale(0.8)*"Brazil", (3,3.5));<br />
label(scale(0.8)*"France", (3,2.5));<br />
label(scale(0.8)*"Peru", (3,1.5));<br />
label(scale(0.8)*"Spain", (3,0.5));<br />
label(scale(0.9)*"Juan's Stamp Collection", (9,0), S);<br />
label(scale(0.9)*"Number of Stamps by Decade", (9,5), N);</asy><br />
<br />
==Problem==<br />
<br />
How many of his European stamps were issued in the '80s? <br />
<br />
<math>\text{(A)}\ 9 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 42</math><br />
<br />
==Solution==<br />
Franc and Spain are European countries. The number of '80s stamps from France is <math>15</math> and the number of '80s stamps from Spain is <math>9</math>. The total number of stamps is <math>15+9=\boxed{\text{(D)}\ 24}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Bz1https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_1&diff=735142015 AMC 8 Problems/Problem 12015-12-03T21:41:23Z<p>Bz1: /* Solution */</p>
<hr />
<div>How many square yards of carpet are required to cover a rectangular floor that is <math>12</math> feet long and <math>9</math> feet wide? (There are 3 feet in a yard.)<br />
<br />
<math>\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972</math><br />
<br />
==Solution==<br />
<br />
First, we multiply <math>12\cdot9</math> to get that you need <math>108</math> square feet of carpet you need to cover. Since there are <math>9</math> square feet in a square yard, you divide <math>108</math> by <math>9</math> to get <math>12</math> square yards, so our answer is 12.<br />
<br />
==Solution 2==<br />
Since there are <math>3</math> feet in a yard, we divide <math>9</math> by <math>3</math> to get <math>3</math>, and <math>12</math> by <math>3</math> to get <math>4</math>. To find the area of the carpet, we then multiply these two values together to get <math>\boxed{\textbf{(A)}~12}</math><br />
<br />
==See Also==<br />
{{AMC8 box|year=2015|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Bz1https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems&diff=732012015 AMC 8 Problems2015-11-26T13:37:39Z<p>Bz1: /* Problem 1 */</p>
<hr />
<div>==Problem 1==<br />
<br />
How many square yards of carpet are required to cover a rectangular floor that is <math>12</math> feet long and <math>9</math> feet wide? (There are [b]3 feet[/b] in a yard.)<br />
<br />
<math>\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972</math><br />
<br />
[[2015 AMC 8 Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
<br />
Point <math>O</math> is the center of the regular octagon <math>ABCDEFGH</math>, and <math>X</math> is the midpoint of the side <math>\overline{AB}.</math> What fraction of the area of the octagon is shaded?<br />
<br />
<math>\textbf{(A) }\frac{11}{32} \quad\textbf{(B) }\frac{3}{8} \quad\textbf{(C) }\frac{13}{32} \quad\textbf{(D) }\frac{7}{16}\quad \textbf{(E) }\frac{15}{32}</math><br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,O,X;<br />
A=dir(45);<br />
B=dir(90);<br />
C=dir(135);<br />
D=dir(180);<br />
E=dir(-135);<br />
F=dir(-90);<br />
G=dir(-45);<br />
H=dir(0);<br />
O=(0,0);<br />
X=midpoint(A--B);<br />
<br />
fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75));<br />
draw(A--B--C--D--E--F--G--H--cycle);<br />
<br />
dot("$A$",A,dir(45));<br />
dot("$B$",B,dir(90));<br />
dot("$C$",C,dir(135));<br />
dot("$D$",D,dir(180));<br />
dot("$E$",E,dir(-135));<br />
dot("$F$",F,dir(-90));<br />
dot("$G$",G,dir(-45));<br />
dot("$H$",H,dir(0));<br />
dot("$X$",X,dir(135/2));<br />
dot("$O$",O,dir(0));<br />
draw(E--O--X);<br />
</asy><br />
[[2015 AMC 8 Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
<br />
Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of <math>10</math> miles per hour. Jack walks to the pool at a constant speed of <math>4</math> miles per hour. How many minutes before Jack does Jill arrive?<br />
<br />
<math>\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10</math><br />
<br />
[[2015 AMC 8 Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
<br />
The Centerville Middle School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?<br />
<br />
<math>\textbf{(A) }2\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }12</math><br />
<br />
[[2015 AMC 8 Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
Billy's basketball team scored the following points over the course of the first 11 games of the season: <cmath>42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73</cmath> If his team scores 40 in the 12th game, which of the following statistics will show an increase?<br />
<br />
<math>\textbf{(A) } \text{range} \qquad \textbf{(B) } \text{median} \qquad \textbf{(C) } \text{mean} \qquad \textbf{(D) } \text{mode} \qquad \textbf{(E) } \text{mid-range}</math><br />
<br />
[[2015 AMC 8 Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
<br />
In <math>\bigtriangleup ABC</math>, <math>AB=BC=29</math>, and <math>AC=42</math>. What is the area of <math>\bigtriangleup ABC</math>?<br />
<br />
<math>\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701</math><br />
<br />
[[2015 AMC 8 Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
<br />
Each of two boxes contains three chips numbered <math>1</math>, <math>2</math>, <math>3</math>. 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 />
<math>\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}</math><br />
<br />
[[2015 AMC 8 Problems/Problem 7|Solution]]<br />
<br />
==Problem 8== <br />
<br />
What is the smallest whole number larger than the perimeter of any triangle with a side of length <math>5</math> and a side of length <math>19</math>?<br />
<br />
<math>\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57</math><br />
<br />
[[2015 AMC 8 Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working <math>20</math> days?<br />
<br />
<math>\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401</math><br />
<br />
[[2015 AMC 8 Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
How many integers between <math>1000</math> and <math>9999</math> have four distinct digits?<br />
<br />
<math>\textbf{(A) }3024\qquad\textbf{(B) }4536\qquad\textbf{(C) }5040\qquad\textbf{(D) }6480\qquad \textbf{(E) }6561</math><br />
<br />
[[2015 AMC 8 Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
<br />
In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read "AMC8"?<br />
<br />
<math>\textbf{(A) } \frac{1}{22,050} \qquad \textbf{(B) } \frac{1}{21,000}\qquad \textbf{(C) } \frac{1}{10,500}\qquad \textbf{(D) } \frac{1}{2,100} \qquad \textbf{(E) } \frac{1}{1,050}</math><br />
<br />
[[2015 AMC 8 Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
How many pairs of parallel edges, such as <math>\overline{AB}</math> and <math>\overline{GH}</math> or <math>\overline{EH}</math> and <math>\overline{FG}</math>, does a cube have?<br />
<br />
<br />
<math>\textbf{(A) }6 \quad\textbf{(B) }12 \quad\textbf{(C) } 18 \quad\textbf{(D) } 24 \quad \textbf{(E) } 36</math> <asy> import three; currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); label("$D$",(0,0,0),S); label("$A$",(0,0,1),N); label("$H$",(0,1,0),S); label("$E$",(0,1,1),N); label("$C$",(1,0,0),S); label("$B$",(1,0,1),N); label("$G$",(1,1,0),S); label("$F$",(1,1,1),N); </asy><br />
<br />
[[2015 AMC 8 Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
How many subsets of two elements can be removed from the set <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}</math> so that the mean (average) of the remaining numbers is <math>6</math>?<br />
<br />
<math>\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}</math><br />
<br />
[[2015 AMC 8 Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
Which of the following integers cannot be written as the sum of four consecutive odd integers?<br />
<br />
<math>\textbf{(A)}\text{ 16}\qquad\textbf{(B)}\text{ 40}\qquad\textbf{(C)}\text{ 72}\qquad\textbf{(D)}\text{ 100}\qquad\textbf{(E)}\text{ 200}</math><br />
<br />
[[2015 AMC 8 Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
<br />
At Euler Middle School, <math>198</math> students voted on two issues in a school referendum with the following results: <math>149</math> voted in favor of the first issue and <math>119</math> voted in favor of the second issue. If there were exactly <math>29</math> students who voted against both issues, how many students voted in favor of both issues?<br />
<br />
<math>\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149</math><br />
<br />
[[2015 AMC 8 Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If <math>\tfrac{1}{3}</math> of all the ninth graders are paired with <math>\tfrac{2}{5}</math> of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?<br />
<br />
<math><br />
\textbf{(A) } \frac{2}{15} \qquad<br />
\textbf{(B) } \frac{4}{11} \qquad<br />
\textbf{(C) } \frac{11}{30} \qquad<br />
\textbf{(D) } \frac{3}{8} \qquad<br />
\textbf{(E) } \frac{11}{15}<br />
</math><br />
<br />
[[2015 AMC 8 Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
Jeremy's father drives him to school in rush hour traffic in <math>20</math> minutes. One day there is no traffic, so his father can drive him <math>18</math> miles per hour faster and gets him to school in <math>12</math> minutes. How far in miles is it to school?<br />
<br />
<math><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 />
</math><br />
<br />
[[2015 AMC 8 Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
<br />
An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, <math>2,5,8,11,14</math> is an arithmetic sequence with five terms, in which the first term is <math>2</math> and the constant added is <math>3</math>. Each row and each column in this <math>5\times5</math> array is an arithmetic sequence with five terms. What is the value of <math>X</math>?<br />
<br />
<math>\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42</math><br />
<br />
<asy> size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); </asy><br />
<br />
[[2015 AMC 8 Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
<br />
A triangle with vertices as <math>A=(1,3)</math>, <math>B=(5,1)</math>, and <math>C=(4,4)</math> is plotted on a <math>6\times5</math> grid. What fraction of the grid is covered by the triangle?<br />
<br />
<asy><br />
<br />
draw((1,0)--(1,5),linewidth(.5));<br />
draw((2,0)--(2,5),linewidth(.5));<br />
draw((3,0)--(3,5),linewidth(.5));<br />
draw((4,0)--(4,5),linewidth(.5));<br />
draw((5,0)--(5,5),linewidth(.5));<br />
draw((6,0)--(6,5),linewidth(.5));<br />
draw((0,1)--(6,1),linewidth(.5));<br />
draw((0,2)--(6,2),linewidth(.5));<br />
draw((0,3)--(6,3),linewidth(.5));<br />
draw((0,4)--(6,4),linewidth(.5));<br />
draw((0,5)--(6,5),linewidth(.5)); <br />
draw((0,0)--(0,6),EndArrow);<br />
draw((0,0)--(7,0),EndArrow);<br />
draw((1,3)--(4,4)--(5,1)--cycle);<br />
label("$y$",(0,6),W); label("$x$",(7,0),S);<br />
label("$A$",(1,3),dir(210)); label("$B$",(5,1),SE); label("$C$",(4,4),dir(100));<br />
</asy><br />
<br />
<math>\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}</math><br />
<br />
[[2015 AMC 8 Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
Ralph went to the store and bought 12 pairs of socks for a total of \$24. Some of the socks he bought cost \$1 a pair, some of the socks he bought cost \$3 a pair, and some of the socks he bought cost \$4 a pair. If he bought at least one pair of each type, how many pairs of \$1 socks did Ralph buy?<br />
<br />
<math>\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8</math><br />
<br />
[[2015 AMC 8 Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
In the given figure hexagon <math>ABCDEF</math> is equiangular, <math>ABJI</math> and <math>FEHG</math> are squares with areas <math>18</math> and <math>32</math> respectively, <math>\triangle JBK</math> is equilateral and <math>FE=BC</math>. What is the area of <math>\triangle KBC</math>?<br />
<br />
<math>\textbf{(A) }6\sqrt{2}\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }9\sqrt{2}\qquad\textbf{(E) }32</math><br />
<br />
<asy><br />
<br />
draw((-4,6*sqrt(2))--(4,6*sqrt(2)));<br />
draw((-4,-6*sqrt(2))--(4,-6*sqrt(2)));<br />
draw((-8,0)--(-4,6*sqrt(2)));<br />
draw((-8,0)--(-4,-6*sqrt(2)));<br />
draw((4,6*sqrt(2))--(8,0));<br />
draw((8,0)--(4,-6*sqrt(2)));<br />
draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle);<br />
draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle);<br />
label("$I$",(-4,8+6*sqrt(2)),dir(100)); label("$J$",(4,8+6*sqrt(2)),dir(80));<br />
label("$A$",(-4,6*sqrt(2)),dir(280)); label("$B$",(4,6*sqrt(2)),dir(250));<br />
label("$C$",(8,0),W); label("$D$",(4,-6*sqrt(2)),NW); label("$E$",(-4,-6*sqrt(2)),NE); label("$F$",(-8,0),E);<br />
draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle);<br />
label("$K$",(4+4*sqrt(3),4+6*sqrt(2)),E);<br />
draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed);<br />
label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S);<br />
label("$G$",(-8-6*sqrt(2),-4),W);<br />
label("$32$",(-10,-8),N);<br />
label("$18$",(0,6*sqrt(2)+2),N);<br />
<br />
</asy><br />
<br />
[[2015 AMC 8 Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?<br />
<br />
<math>\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080</math><br />
<br />
[[2015 AMC 8 Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
Tom has twelve slips of paper which he wants to put into five cups labeled <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math>. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from <math>A</math> to <math>E</math>. The numbers on the papers are <math>2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,</math> and <math>4.5</math>. If a slip with 2 goes into cup <math>E</math> and a slip with 3 goes into cup <math>B</math>, then the slip with 3.5 must go into what cup?<br />
<br />
<math><br />
\textbf{(A) } A \qquad<br />
\textbf{(B) } B \qquad<br />
\textbf{(C) } C \qquad<br />
\textbf{(D) } D \qquad<br />
\textbf{(E) } E<br />
</math><br />
<br />
[[2015 AMC 8 Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
A baseball league consists of two four-team divisions. Each team plays every other team in its division <math>N</math> games. Each team plays every team in the other division <math>M</math> games with <math>N>2M</math> and <math>M>4</math>. Each team plays a 76 game schedule. How many games does a team play within its own division?<br />
<br />
<math>\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72</math><br />
<br />
[[2015 AMC 8 Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
<br />
One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br />
<br />
<asy><br />
<br />
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br />
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br />
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br />
filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br />
filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br />
<br />
</asy><br />
<br />
<math> \textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17</math><br />
<br />
[[2015 AMC 8 Problems/Problem 25|Solution]]</div>Bz1