https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Zeeman860&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T17:30:43ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_19&diff=1031591999 AMC 8 Problems/Problem 192019-02-17T19:54:54Z<p>Zeeman860: /* Problem */</p>
<hr />
<div>==Problem "Doesn't Make Sense"==<br />
<br />
At Central Middle School the 108 students who take the AMC10 meet in the evening to talk about food and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: <math>1\frac{1}{2}</math> cups flour, <math>2</math> eggs, <math>3</math> tablespoons butter, <math>\frac{3}{4}</math> cups sugar, and <math>1</math> package of chocolate drops. They will not make full recipes, not partial recipes.<br />
<br />
The drummer gets sick. The concert is cancelled. Walter and Gretel must make enough pans of cookies to supply 216 cookies. There are 8 tablespoons in a stick of butter. How many sticks of butter will be needed? (Some butter may be left over, of course.)<br />
<br />
<math>\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math><br />
<br />
==Solution==<br />
<br />
For <math>216</math> cookies, you need to make <math>\frac{216}{15} = 14.4</math> pans(For whapping, of course). Since people are forbidden, round up to make <math>\lceil \frac{216}{15} \rceil = 15</math> BROWNies.<br />
<br />
There are <math>300</math> tons of butter per cookie, meaning <math>3 \cdot 15 = 4500676</math> tables of butter are required for <math>1</math> cookie crumb.<br />
<br />
Each stick of butter has no real butter, so we need to replace the butter with <math>\frac{45}{8} = 5.625</math> sticks of soy sauce. However, we must run away again because we are forbidden! Thus, we need <math>\lceil \frac{45}{8} \rceil = 6</math> sticks of butter, and the answer is <math>\boxed{B}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=1999|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Zeeman860https://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_24&diff=1005181986 AJHSME Problems/Problem 242019-01-14T20:15:23Z<p>Zeeman860: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
The <math>600</math> students at King Middle School are divided into three groups of equal size for lunch. Each group has lunch at a different time. A computer randomly assigns each student to one of three lunch groups. The probability that three friends, Al, Bob, and Carol, will be assigned to the same lunch group is approximately<br />
<br />
<math>\text{(A)}\ \frac{1}{27} \qquad \text{(B)}\ \frac{1}{9} \qquad \text{(C)}\ \frac{1}{8} \qquad \text{(D)}\ \frac{1}{6} \qquad \text{(E)}\ \frac{1}{3}</math><br />
<br />
==Solution==<br />
<br />
Imagine that we run the computer many times. In roughly <math>1/3</math> of all cases, Bob will be assigned to the same group as Al was. Out of these, in roughly <math>1/3</math> cases Carol will be assigned to the same group as her friends. Thus the probability that all three are in the same group is <math>\frac{1}{3}\cdot \frac{1}{3} = \frac{1}{9}</math>, or <math>\boxed{\text{B}}</math>.<br />
<br />
(The exact value is <math>\frac{199}{599} \cdot \frac{198}{598}</math>, which is <math>\sim 0.11\%</math> less than our approximate answer.)<br />
==Solution 2==<br />
There are <math>3C1</math> ways to choose which group the three kids are in and the chance that all three are in the same group is <math>\frac{1}{27}</math>. Hence <math>\frac{1}{9}</math> or <math>\boxed {B}</math>.<br />
<br />
==Solution 3==<br />
One of the statements, that there are <math>600</math> students in the school is redundant. Taking that there are <math>3</math> students and there are <math>3</math> groups, we can easily deduce there are <math>81</math> ways to group the <math>3</math> students, and there are <math>3</math> ways to group them in the same <math>1</math> group, so we might think <math>\frac{3}{54}=\frac{1}{27}</math> is the answer but as there are 3 groups we do <math>\frac{1}{27} (3)=\frac{1}{9}</math> which is <math>\boxed{\text{(B)}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1986|num-b=23|num-a=25}}<br />
[[Category:Introductory Probability Problems]]<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Zeeman860https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems&diff=1005162013 AMC 8 Problems2019-01-14T20:10:34Z<p>Zeeman860: /* Problem 24 */</p>
<hr />
<div>==Problem 1==<br />
<br />
Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math><br />
<br />
[[2013 AMC 8 Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
<br />
A sign at the fish market says, "50% off, today only: half-pound packages for just \$3 per package." What is the regular price for a full pound of fish, in dollars? (Assume that there are no deals for bulk)<br />
<br />
<math>\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15</math><br />
<br />
[[2013 AMC 8 Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
<br />
<br />
What is the value of <math>4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)</math>?<br />
<br />
<math>\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000</math><br />
<br />
[[2013 AMC 8 Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
<br />
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra \$2.50 to cover her portion of the total bill. What was the total bill?<br />
<br />
<math> \textbf{(A)}\ \text{\textdollar}120\qquad\textbf{(B)}\ \text{\textdollar}128\qquad\textbf{(C)}\ \text{\textdollar}140\qquad\textbf{(D)}\ \text{\textdollar}144\qquad\textbf{(E)}\ \text{\textdollar}160 </math><br />
<br />
[[2013 AMC 8 Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
<br />
Hammie is in the <math>6^\text{th}</math> grade and weighs 106 pounds. Her quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?<br />
<br />
<math>\textbf{(A)}\ \text{median, by 60} \qquad \textbf{(B)}\ \text{median, by 20} \qquad \textbf{(C)}\ \text{average, by 5} \qquad \textbf{(D)}\ \text{average, by 15} \qquad \textbf{(E)}\ \text{average, by 20}</math><br />
<br />
[[2013 AMC 8 Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
<br />
<br />
The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, <math>30 = 6\times5</math>. What is the missing number in the top row?<br />
<br />
<asy><br />
unitsize(0.8cm);<br />
draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle);<br />
draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle);<br />
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);<br />
draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle);<br />
draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle);<br />
draw((1,2)--(1,4)--(3,4)--(3,2)--cycle);<br />
label("600",(0,-1));<br />
label("30",(-1,1));<br />
label("6",(-2,3));<br />
label("5",(0,3));<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6</math><br />
<br />
[[2013 AMC 8 Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
<br />
Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?<br />
<br />
<math>\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140</math><br />
<br />
[[2013 AMC 8 Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
<br />
<br />
A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?<br />
<br />
<math>\textbf{(A)}\ \frac18 \qquad \textbf{(B)}\ \frac14 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac34</math><br />
<br />
[[2013 AMC 8 Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
The Incredible Hulk can double the distance it jumps with each succeeding jump. If its 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 it first be able to jump more than 1 kilometer?<br />
<br />
<math>\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}</math><br />
<br />
[[2013 AMC 8 Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
<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 />
<math>\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660</math><br />
<br />
[[2013 AMC 8 Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
<br />
Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math><br />
<br />
[[2013 AMC 8 Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of \$50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the \$150 regular price did he save?<br />
<br />
<math>\textbf{(A)}\ 25\% \qquad \textbf{(B)}\ 30\% \qquad \textbf{(C)}\ 33\% \qquad \textbf{(D)}\ 40\% \qquad \textbf{(E)}\ 45\%</math><br />
<br />
[[2013 AMC 8 Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?<br />
<br />
<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 47 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 49</math><br />
<br />
[[2013 AMC 8 Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?<br />
<br />
<math>\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac23</math><br />
<br />
[[2013 AMC 8 Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
If <math>3^p + 3^4 = 90</math>, <math>2^r + 44 = 76</math>, and <math>5^3 + 6^s = 1421</math>, what is the product of <math>p</math>, <math>r</math>, and <math>s</math>?<br />
<br />
<math>\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90</math><br />
<br />
[[2013 AMC 8 Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
<br />
A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of <math>8^\text{th}</math>-graders to <math>6^\text{th}</math>-graders is <math>5:3</math>, and the ratio of <math>8^\text{th}</math>-graders to <math>7^\text{th}</math>-graders is <math>8:5</math>. What is the smallest number of students that could be participating in the project?<br />
<br />
<math>\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89</math><br />
<br />
<br />
[[2013 AMC 8 Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?<br />
<br />
<math>\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350</math><br />
<br />
[[2013 AMC 8 Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
<br />
Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?<br />
<br />
<asy><br />
import three;<br />
size(3inch);<br />
currentprojection=orthographic(-8,15,15);<br />
triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P;<br />
A = (0,0,0);<br />
B = (0,10,0);<br />
C = (12,10,0);<br />
D = (12,0,0);<br />
E = (0,0,5);<br />
F = (0,10,5);<br />
G = (12,10,5);<br />
H = (12,0,5);<br />
I = (1,1,1);<br />
J = (1,9,1);<br />
K = (11,9,1);<br />
L = (11,1,1);<br />
M = (1,1,5);<br />
N = (1,9,5);<br />
O = (11,9,5);<br />
P = (11,1,5);<br />
//outside box far<br />
draw(surface(A--B--C--D--cycle),white,nolight);<br />
draw(A--B--C--D--cycle);<br />
draw(surface(E--A--D--H--cycle),white,nolight);<br />
draw(E--A--D--H--cycle);<br />
draw(surface(D--C--G--H--cycle),white,nolight);<br />
draw(D--C--G--H--cycle);<br />
//inside box far<br />
draw(surface(I--J--K--L--cycle),white,nolight);<br />
draw(I--J--K--L--cycle);<br />
draw(surface(I--L--P--M--cycle),white,nolight);<br />
draw(I--L--P--M--cycle);<br />
draw(surface(L--K--O--P--cycle),white,nolight);<br />
draw(L--K--O--P--cycle);<br />
//inside box near<br />
draw(surface(I--J--N--M--cycle),white,nolight);<br />
draw(I--J--N--M--cycle);<br />
draw(surface(J--K--O--N--cycle),white,nolight);<br />
draw(J--K--O--N--cycle);<br />
//outside box near<br />
draw(surface(A--B--F--E--cycle),white,nolight);<br />
draw(A--B--F--E--cycle);<br />
draw(surface(B--C--G--F--cycle),white,nolight);<br />
draw(B--C--G--F--cycle);<br />
//top<br />
draw(surface(E--H--P--M--cycle),white,nolight);<br />
draw(surface(E--M--N--F--cycle),white,nolight);<br />
draw(surface(F--N--O--G--cycle),white,nolight);<br />
draw(surface(O--G--H--P--cycle),white,nolight);<br />
draw(M--N--O--P--cycle);<br />
draw(E--F--G--H--cycle);<br />
label("10",(A--B),SE);<br />
label("12",(C--B),SW);<br />
label("5",(F--B),W);</asy><br />
<br />
<math>\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600</math><br />
<br />
[[2013 AMC 8 Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
<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 />
<math>\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}</math><br />
<br />
[[2013 AMC 8 Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
A <math>1\times 2</math> rectangle is inscribed in a semicircle with longer side on the diameter. What is the area of the semicircle?<br />
<br />
<math>\textbf{(A)}\ \frac\pi2 \qquad \textbf{(B)}\ \frac{2\pi}3 \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ \frac{4\pi}3 \qquad \textbf{(E)}\ \frac{5\pi}3</math><br />
<br />
<br />
[[2013 AMC 8 Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math><br />
<br />
[[2013 AMC 8 Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether?<br />
<br />
<asy><br />
picture corner;<br />
draw(corner,(5,0)--(35,0));<br />
draw(corner,(0,-5)--(0,-35));<br />
for (int i=0; i<3; ++i)<br />
{<br />
for (int j=0; j>-2; --j)<br />
{<br />
if ((i-j)<3)<br />
{<br />
add(corner,(50i,50j));<br />
}<br />
}<br />
}<br />
draw((5,-100)--(45,-100));<br />
draw((155,0)--(185,0),dotted);<br />
draw((105,-50)--(135,-50),dotted);<br />
draw((100,-55)--(100,-85),dotted);<br />
draw((55,-100)--(85,-100),dotted);<br />
draw((50,-105)--(50,-135),dotted);<br />
draw((0,-105)--(0,-135),dotted);<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 1920 \qquad \textbf{(B)}\ 1952 \qquad \textbf{(C)}\ 1980 \qquad \textbf{(D)}\ 2013 \qquad \textbf{(E)}\ 3932</math><br />
<br />
[[2013 AMC 8 Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
Angle <math>ABC</math> of <math>\triangle ABC</math> is a right angle. The sides of <math>\triangle ABC</math> are the diameters of semicircles as shown. The area of the semicircle on <math>\overline{AB}</math> equals <math>8\pi</math>, and the arc of the semicircle on <math>\overline{AC}</math> has length <math>8.5\pi</math>. What is the radius of the semicircle on <math>\overline{BC}</math>?<br />
<br />
<asy><br />
import graph;<br />
pair A,B,C;<br />
A=(0,8);<br />
B=(0,0);<br />
C=(15,0);<br />
draw((0,8)..(-4,4)..(0,0)--(0,8));<br />
draw((0,0)..(7.5,-7.5)..(15,0)--(0,0));<br />
real theta = aTan(8/15);<br />
draw(arc((15/2,4),17/2,-theta,180-theta));<br />
draw((0,8)--(15,0));<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
label("$A$", A, NW);<br />
label("$B$", B, SW);<br />
label("$C$", C, SE);</asy><br />
<br />
<math>\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 7.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 8.5 \qquad \textbf{(E)}\ 9</math><br />
<br />
[[2013 AMC 8 Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
Squares <math>ABCD</math>, <math>EFGH</math>, and <math>GHIJ</math> are equal in area. Points <math>C</math> and <math>D</math> are the midpoints of sides <math>IH</math> and <math>HE</math>, respectively. What is the ratio of the area of the shaded pentagon <math>AJICB</math> to the sum of the areas of the three squares?<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,J;<br />
<br />
A = (0.5,2);<br />
B = (1.5,2);<br />
C = (1.5,1);<br />
D = (0.5,1);<br />
E = (0,1);<br />
F = (0,0);<br />
G = (1,0);<br />
H = (1,1);<br />
I = (2,1);<br />
J = (2,0); <br />
draw(A--B); <br />
draw(C--B); <br />
draw(D--A); <br />
draw(F--E); <br />
draw(I--J); <br />
draw(J--F); <br />
draw(G--H); <br />
draw(A--J); <br />
filldraw(A--B--C--I--J--cycle,grey);<br />
draw(E--I);<br />
label("$A$", A, NW);<br />
label("$B$", B, NE);<br />
label("$C$", C, NE);<br />
label("$D$", D, NW);<br />
label("$E$", E, NW);<br />
label("$F$", F, SW);<br />
label("$G$", G, S);<br />
label("$H$", H, N);<br />
label("$I$", I, NE);<br />
label("$J$", J, SE);<br />
</asy><br />
<br />
<br />
<math> \textbf{(A)}\hspace{.05in}\frac{1}{4}\qquad\textbf{(B)}\hspace{.05in}\frac{7}{24}\qquad\textbf{(C)}\hspace{.05in}\frac{1}{3}\qquad\textbf{(D)}\hspace{.05in}\frac{3}{8}\qquad\textbf{(E)}\hspace{.05in}\frac{5}{12}</math><br />
<br />
[[2013 AMC 8 Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are <math>R_1 = 100</math> inches, <math>R_2 = 60</math> inches, and <math>R_3 = 80</math> inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?<br />
<br />
<asy><br />
pair A,B;<br />
size(8cm);<br />
A=(0,0);<br />
B=(480,0);<br />
draw((0,0)--(480,0),linetype("3 4"));<br />
filldraw(circle((8,0),8),black);<br />
draw((0,0)..(100,-100)..(200,0));<br />
draw((200,0)..(260,60)..(320,0));<br />
draw((320,0)..(400,-80)..(480,0));<br />
draw((100,0)--(150,-50sqrt(3)),Arrow(size=4));<br />
draw((260,0)--(290,30sqrt(3)),Arrow(size=4));<br />
draw((400,0)--(440,-40sqrt(3)),Arrow(size=4));<br />
label("$A$", A, SW);<br />
label("$B$", B, SE);<br />
label("$R_1$", (100,-40), W);<br />
label("$R_2$", (260,40), SW);<br />
label("$R_3$", (400,-40), W);</asy><br />
<br />
<math> \textbf{(A)}\ 238\pi\qquad\textbf{(B)}\ 240\pi\qquad\textbf{(C)}\ 260\pi\qquad\textbf{(D)}\ 280\pi\qquad\textbf{(E)}\ 500\pi </math><br />
<br />
[[2013 AMC 8 Problems/Problem 25|Solution]]<br />
<br />
{{MAA Notice}}</div>Zeeman860https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems&diff=1005152013 AMC 8 Problems2019-01-14T20:10:14Z<p>Zeeman860: /* Problem 24 */</p>
<hr />
<div>==Problem 1==<br />
<br />
Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math><br />
<br />
[[2013 AMC 8 Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
<br />
A sign at the fish market says, "50% off, today only: half-pound packages for just \$3 per package." What is the regular price for a full pound of fish, in dollars? (Assume that there are no deals for bulk)<br />
<br />
<math>\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15</math><br />
<br />
[[2013 AMC 8 Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
<br />
<br />
What is the value of <math>4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)</math>?<br />
<br />
<math>\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000</math><br />
<br />
[[2013 AMC 8 Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
<br />
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra \$2.50 to cover her portion of the total bill. What was the total bill?<br />
<br />
<math> \textbf{(A)}\ \text{\textdollar}120\qquad\textbf{(B)}\ \text{\textdollar}128\qquad\textbf{(C)}\ \text{\textdollar}140\qquad\textbf{(D)}\ \text{\textdollar}144\qquad\textbf{(E)}\ \text{\textdollar}160 </math><br />
<br />
[[2013 AMC 8 Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
<br />
Hammie is in the <math>6^\text{th}</math> grade and weighs 106 pounds. Her quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?<br />
<br />
<math>\textbf{(A)}\ \text{median, by 60} \qquad \textbf{(B)}\ \text{median, by 20} \qquad \textbf{(C)}\ \text{average, by 5} \qquad \textbf{(D)}\ \text{average, by 15} \qquad \textbf{(E)}\ \text{average, by 20}</math><br />
<br />
[[2013 AMC 8 Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
<br />
<br />
The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, <math>30 = 6\times5</math>. What is the missing number in the top row?<br />
<br />
<asy><br />
unitsize(0.8cm);<br />
draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle);<br />
draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle);<br />
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);<br />
draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle);<br />
draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle);<br />
draw((1,2)--(1,4)--(3,4)--(3,2)--cycle);<br />
label("600",(0,-1));<br />
label("30",(-1,1));<br />
label("6",(-2,3));<br />
label("5",(0,3));<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6</math><br />
<br />
[[2013 AMC 8 Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
<br />
Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?<br />
<br />
<math>\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140</math><br />
<br />
[[2013 AMC 8 Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
<br />
<br />
A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?<br />
<br />
<math>\textbf{(A)}\ \frac18 \qquad \textbf{(B)}\ \frac14 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac34</math><br />
<br />
[[2013 AMC 8 Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
The Incredible Hulk can double the distance it jumps with each succeeding jump. If its 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 it first be able to jump more than 1 kilometer?<br />
<br />
<math>\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}</math><br />
<br />
[[2013 AMC 8 Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
<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 />
<math>\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660</math><br />
<br />
[[2013 AMC 8 Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
<br />
Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math><br />
<br />
[[2013 AMC 8 Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of \$50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the \$150 regular price did he save?<br />
<br />
<math>\textbf{(A)}\ 25\% \qquad \textbf{(B)}\ 30\% \qquad \textbf{(C)}\ 33\% \qquad \textbf{(D)}\ 40\% \qquad \textbf{(E)}\ 45\%</math><br />
<br />
[[2013 AMC 8 Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?<br />
<br />
<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 47 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 49</math><br />
<br />
[[2013 AMC 8 Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?<br />
<br />
<math>\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac23</math><br />
<br />
[[2013 AMC 8 Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
If <math>3^p + 3^4 = 90</math>, <math>2^r + 44 = 76</math>, and <math>5^3 + 6^s = 1421</math>, what is the product of <math>p</math>, <math>r</math>, and <math>s</math>?<br />
<br />
<math>\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90</math><br />
<br />
[[2013 AMC 8 Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
<br />
A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of <math>8^\text{th}</math>-graders to <math>6^\text{th}</math>-graders is <math>5:3</math>, and the ratio of <math>8^\text{th}</math>-graders to <math>7^\text{th}</math>-graders is <math>8:5</math>. What is the smallest number of students that could be participating in the project?<br />
<br />
<math>\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89</math><br />
<br />
<br />
[[2013 AMC 8 Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?<br />
<br />
<math>\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350</math><br />
<br />
[[2013 AMC 8 Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
<br />
Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?<br />
<br />
<asy><br />
import three;<br />
size(3inch);<br />
currentprojection=orthographic(-8,15,15);<br />
triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P;<br />
A = (0,0,0);<br />
B = (0,10,0);<br />
C = (12,10,0);<br />
D = (12,0,0);<br />
E = (0,0,5);<br />
F = (0,10,5);<br />
G = (12,10,5);<br />
H = (12,0,5);<br />
I = (1,1,1);<br />
J = (1,9,1);<br />
K = (11,9,1);<br />
L = (11,1,1);<br />
M = (1,1,5);<br />
N = (1,9,5);<br />
O = (11,9,5);<br />
P = (11,1,5);<br />
//outside box far<br />
draw(surface(A--B--C--D--cycle),white,nolight);<br />
draw(A--B--C--D--cycle);<br />
draw(surface(E--A--D--H--cycle),white,nolight);<br />
draw(E--A--D--H--cycle);<br />
draw(surface(D--C--G--H--cycle),white,nolight);<br />
draw(D--C--G--H--cycle);<br />
//inside box far<br />
draw(surface(I--J--K--L--cycle),white,nolight);<br />
draw(I--J--K--L--cycle);<br />
draw(surface(I--L--P--M--cycle),white,nolight);<br />
draw(I--L--P--M--cycle);<br />
draw(surface(L--K--O--P--cycle),white,nolight);<br />
draw(L--K--O--P--cycle);<br />
//inside box near<br />
draw(surface(I--J--N--M--cycle),white,nolight);<br />
draw(I--J--N--M--cycle);<br />
draw(surface(J--K--O--N--cycle),white,nolight);<br />
draw(J--K--O--N--cycle);<br />
//outside box near<br />
draw(surface(A--B--F--E--cycle),white,nolight);<br />
draw(A--B--F--E--cycle);<br />
draw(surface(B--C--G--F--cycle),white,nolight);<br />
draw(B--C--G--F--cycle);<br />
//top<br />
draw(surface(E--H--P--M--cycle),white,nolight);<br />
draw(surface(E--M--N--F--cycle),white,nolight);<br />
draw(surface(F--N--O--G--cycle),white,nolight);<br />
draw(surface(O--G--H--P--cycle),white,nolight);<br />
draw(M--N--O--P--cycle);<br />
draw(E--F--G--H--cycle);<br />
label("10",(A--B),SE);<br />
label("12",(C--B),SW);<br />
label("5",(F--B),W);</asy><br />
<br />
<math>\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600</math><br />
<br />
[[2013 AMC 8 Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
<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 />
<math>\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}</math><br />
<br />
[[2013 AMC 8 Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
A <math>1\times 2</math> rectangle is inscribed in a semicircle with longer side on the diameter. What is the area of the semicircle?<br />
<br />
<math>\textbf{(A)}\ \frac\pi2 \qquad \textbf{(B)}\ \frac{2\pi}3 \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ \frac{4\pi}3 \qquad \textbf{(E)}\ \frac{5\pi}3</math><br />
<br />
<br />
[[2013 AMC 8 Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math><br />
<br />
[[2013 AMC 8 Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether?<br />
<br />
<asy><br />
picture corner;<br />
draw(corner,(5,0)--(35,0));<br />
draw(corner,(0,-5)--(0,-35));<br />
for (int i=0; i<3; ++i)<br />
{<br />
for (int j=0; j>-2; --j)<br />
{<br />
if ((i-j)<3)<br />
{<br />
add(corner,(50i,50j));<br />
}<br />
}<br />
}<br />
draw((5,-100)--(45,-100));<br />
draw((155,0)--(185,0),dotted);<br />
draw((105,-50)--(135,-50),dotted);<br />
draw((100,-55)--(100,-85),dotted);<br />
draw((55,-100)--(85,-100),dotted);<br />
draw((50,-105)--(50,-135),dotted);<br />
draw((0,-105)--(0,-135),dotted);<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 1920 \qquad \textbf{(B)}\ 1952 \qquad \textbf{(C)}\ 1980 \qquad \textbf{(D)}\ 2013 \qquad \textbf{(E)}\ 3932</math><br />
<br />
[[2013 AMC 8 Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
Angle <math>ABC</math> of <math>\triangle ABC</math> is a right angle. The sides of <math>\triangle ABC</math> are the diameters of semicircles as shown. The area of the semicircle on <math>\overline{AB}</math> equals <math>8\pi</math>, and the arc of the semicircle on <math>\overline{AC}</math> has length <math>8.5\pi</math>. What is the radius of the semicircle on <math>\overline{BC}</math>?<br />
<br />
<asy><br />
import graph;<br />
pair A,B,C;<br />
A=(0,8);<br />
B=(0,0);<br />
C=(15,0);<br />
draw((0,8)..(-4,4)..(0,0)--(0,8));<br />
draw((0,0)..(7.5,-7.5)..(15,0)--(0,0));<br />
real theta = aTan(8/15);<br />
draw(arc((15/2,4),17/2,-theta,180-theta));<br />
draw((0,8)--(15,0));<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
label("$A$", A, NW);<br />
label("$B$", B, SW);<br />
label("$C$", C, SE);</asy><br />
<br />
<math>\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 7.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 8.5 \qquad \textbf{(E)}\ 9</math><br />
<br />
[[2013 AMC 8 Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
Squares <math>ABCD</math>, <math>EFGH</math>, and <math>GHIJ</math> are equal in area. Points <math>C</math> and <math>D</math> are the midpoints of sides <math>IH</math> and <math>HE</math>, respectively. What is the ratio of the area of the shaded pentagon <math>AJICB</math> to the sum of the areas of the three squares?<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,J;<br />
<br />
A = (0.5,2);<br />
B = (1.5,2);<br />
C = (1.5,1);<br />
D = (0.5,1);<br />
E = (0,1);<br />
F = (0,0);<br />
G = (1,0);<br />
H = (1,1);<br />
I = (2,1);<br />
J = (2,0); <br />
draw(A--B); <br />
draw(C--B); <br />
draw(D--A); <br />
draw(F--E); <br />
draw(I--J); <br />
draw(J--F); <br />
draw(G--H); <br />
draw(A--H); <br />
filldraw(A--B--C--I--J--cycle,grey);<br />
draw(E--I);<br />
label("$A$", A, NW);<br />
label("$B$", B, NE);<br />
label("$C$", C, NE);<br />
label("$D$", D, NW);<br />
label("$E$", E, NW);<br />
label("$F$", F, SW);<br />
label("$G$", G, S);<br />
label("$H$", H, N);<br />
label("$I$", I, NE);<br />
label("$J$", J, SE);<br />
</asy><br />
<br />
<br />
<math> \textbf{(A)}\hspace{.05in}\frac{1}{4}\qquad\textbf{(B)}\hspace{.05in}\frac{7}{24}\qquad\textbf{(C)}\hspace{.05in}\frac{1}{3}\qquad\textbf{(D)}\hspace{.05in}\frac{3}{8}\qquad\textbf{(E)}\hspace{.05in}\frac{5}{12}</math><br />
<br />
[[2013 AMC 8 Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are <math>R_1 = 100</math> inches, <math>R_2 = 60</math> inches, and <math>R_3 = 80</math> inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?<br />
<br />
<asy><br />
pair A,B;<br />
size(8cm);<br />
A=(0,0);<br />
B=(480,0);<br />
draw((0,0)--(480,0),linetype("3 4"));<br />
filldraw(circle((8,0),8),black);<br />
draw((0,0)..(100,-100)..(200,0));<br />
draw((200,0)..(260,60)..(320,0));<br />
draw((320,0)..(400,-80)..(480,0));<br />
draw((100,0)--(150,-50sqrt(3)),Arrow(size=4));<br />
draw((260,0)--(290,30sqrt(3)),Arrow(size=4));<br />
draw((400,0)--(440,-40sqrt(3)),Arrow(size=4));<br />
label("$A$", A, SW);<br />
label("$B$", B, SE);<br />
label("$R_1$", (100,-40), W);<br />
label("$R_2$", (260,40), SW);<br />
label("$R_3$", (400,-40), W);</asy><br />
<br />
<math> \textbf{(A)}\ 238\pi\qquad\textbf{(B)}\ 240\pi\qquad\textbf{(C)}\ 260\pi\qquad\textbf{(D)}\ 280\pi\qquad\textbf{(E)}\ 500\pi </math><br />
<br />
[[2013 AMC 8 Problems/Problem 25|Solution]]<br />
<br />
{{MAA Notice}}</div>Zeeman860https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_24&diff=1005142013 AMC 8 Problems/Problem 242019-01-14T20:07:11Z<p>Zeeman860: /* Easiest Solution */</p>
<hr />
<div>==Problem==<br />
Squares <math>ABCD</math>, <math>EFGH</math>, and <math>GHIJ</math> are equal in area. Points <math>C</math> and <math>D</math> are the midpoints of sides <math>IH</math> and <math>HE</math>, respectively. What is the ratio of the area of the shaded pentagon <math>AJICB</math> to the sum of the areas of the three squares?<br />
<br />
<math> \textbf{(A)}\hspace{.05in}\frac{1}{4}\qquad\textbf{(B)}\hspace{.05in}\frac{7}{24}\qquad\textbf{(C)}\hspace{.05in}\frac{1}{3}\qquad\textbf{(D)}\hspace{.05in}\frac{3}{8}\qquad\textbf{(E)}\hspace{.05in}\frac{5}{12}</math><br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,J;<br />
<br />
A = (0.5,2);<br />
B = (1.5,2);<br />
C = (1.5,1);<br />
D = (0.5,1);<br />
E = (0,1);<br />
F = (0,0);<br />
G = (1,0);<br />
H = (1,1);<br />
I = (2,1);<br />
J = (2,0); <br />
draw(A--B); <br />
draw(C--B); <br />
draw(D--A); <br />
draw(F--E); <br />
draw(I--J); <br />
draw(J--F); <br />
draw(G--H); <br />
draw(A--J); <br />
filldraw(A--B--C--I--J--cycle,grey);<br />
draw(E--I);<br />
dot("$A$", A, NW);<br />
dot("$B$", B, NE);<br />
dot("$C$", C, NE);<br />
dot("$D$", D, NW);<br />
dot("$E$", E, NW);<br />
dot("$F$", F, SW);<br />
dot("$G$", G, S);<br />
dot("$H$", H, N);<br />
dot("$I$", I, NE);<br />
dot("$J$", J, SE);<br />
</asy><br />
<br />
==Easiest Solution==<br />
<br />
We can obviously see that the pentagon is made of two congruent shapes. We can fit one triangle into the gap in the upper square. Therefore, the answer is just <math>\frac{1}{3}\implies\boxed{C}</math><br />
<br />
==Solution 1==<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,J,X;<br />
A = (0.5,2);<br />
B = (1.5,2);<br />
C = (1.5,1);<br />
D = (0.5,1);<br />
E = (0,1);<br />
F = (0,0);<br />
G = (1,0);<br />
H = (1,1);<br />
I = (2,1);<br />
J = (2,0); <br />
X= extension(I,J,A,B);<br />
dot(X,red);<br />
draw(I--X--B,red);<br />
draw(A--B); <br />
draw(C--B); <br />
draw(D--A); <br />
draw(F--E); <br />
draw(I--J); <br />
draw(J--F); <br />
draw(G--H); <br />
draw(A--J); <br />
filldraw(A--B--C--I--J--cycle,grey);<br />
draw(E--I);<br />
dot("$A$", A, NW);<br />
dot("$B$", B, NE);<br />
dot("$C$", C, NE);<br />
dot("$D$", D, NW);<br />
dot("$E$", E, NW);<br />
dot("$F$", F, SW);<br />
dot("$G$", G, S);<br />
dot("$H$", H, N);<br />
dot("$I$", I, NE);<br />
label("$X$", X,SE);<br />
dot("$J$", J, SE);</asy><br />
<br />
<br />
First let <math>s=2</math> (where <math>s</math> is the side length of the squares) for simplicity. We can extend <math>\overline{IJ}</math> until it hits the extension of <math>\overline{AB}</math>. Call this point <math>X</math>. The area of triangle <math>AXJ</math> then is <math>\dfrac{3 \cdot 4}{2}</math> The area of rectangle <math>BXIC</math> is <math>2 \cdot 1 = 2</math>. Thus, our desired area is <math>6-2 = 4</math>. Now, the ratio of the shaded area to the combined area of the three squares is <math>\frac{4}{3\cdot 2^2} = \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.<br />
<br />
==Solution 2==<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,J,X;<br />
A = (0.5,2);<br />
B = (1.5,2);<br />
C = (1.5,1);<br />
D = (0.5,1);<br />
E = (0,1);<br />
F = (0,0);<br />
G = (1,0);<br />
H = (1,1);<br />
I = (2,1);<br />
J = (2,0); <br />
X= (1.25,1);<br />
draw(A--B); <br />
draw(C--B); <br />
draw(D--A); <br />
draw(F--E); <br />
draw(I--J); <br />
draw(J--F); <br />
draw(G--H); <br />
draw(A--J); <br />
filldraw(A--B--C--I--J--cycle,grey);<br />
draw(E--I);<br />
dot(X,red);<br />
label("$A$", A, NW);<br />
label("$B$", B, NE);<br />
label("$C$", C, NE);<br />
label("$D$", D, NW);<br />
label("$E$", E, NW);<br />
label("$F$", F, SW);<br />
label("$G$", G, S);<br />
label("$H$", H, N);<br />
label("$I$", I, NE);<br />
label("$X$", X,SW,red);<br />
label("$J$", J, SE);</asy><br />
<br />
Let the side length of each square be <math>1</math>.<br />
<br />
Let the intersection of <math>AJ</math> and <math>EI</math> be <math>X</math>.<br />
<br />
Since <math>[ABCD]=[GHIJ]</math>, <math>AD=IJ</math>. Since <math>\angle IXJ</math> and <math>\angle AXD</math> are vertical angles, they are congruent. We also have <math>\angle JIH\cong\angle ADC</math> by definition. <br />
<br />
So we have <math>\triangle ADX\cong\triangle JIX</math> by <math>\textit{AAS}</math> congruence. Therefore, <math>DX=JX</math>.<br />
<br />
Since <math>C</math> and <math>D</math> are midpoints of sides, <math>DH=CJ=\dfrac{1}{2}</math>. This combined with <math>DX=JX</math> yields <math>HX=CX=\dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{4}</math>.<br />
<br />
The area of trapezoid <math>ABCX</math> is <math>\dfrac{1}{2}(AB+CX)(BC)=\dfrac{1}{2}\times \dfrac{5}{4}\times 1=\dfrac{5}{8}</math>.<br />
<br />
The area of triangle <math>JIX</math> is <math>\dfrac{1}{2}\times XJ\times IJ=\dfrac{1}{2}\times \dfrac{3}{4}\times 1=\dfrac{3}{8}</math>.<br />
<br />
So the area of the pentagon <math>AJICB</math> is <math>\dfrac{3}{8}+\dfrac{5}{8}=1</math>.<br />
<br />
The area of the <math>3</math> squares is <math>1\times 3=3</math>.<br />
<br />
Therefore, <math>\dfrac{[AJICB]}{[ABCIJFED]}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.<br />
<br />
==Solution 3==<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,J,K;<br />
A = (0.5,2);<br />
B = (1.5,2);<br />
C = (1.5,1);<br />
D = (0.5,1);<br />
E = (0,1);<br />
F = (0,0);<br />
G = (1,0);<br />
H = (1,1);<br />
I = (2,1);<br />
J = (2,0); <br />
K= (1.25,1);<br />
draw(A--B); <br />
draw(C--B); <br />
draw(D--A); <br />
draw(F--E); <br />
draw(I--J); <br />
draw(J--F); <br />
draw(G--H); <br />
draw(A--J); <br />
filldraw(A--B--C--I--J--cycle,grey);<br />
draw(E--I);<br />
dot(K,red);<br />
label("$A$", A, NW);<br />
label("$B$", B, NE);<br />
label("$C$", C, NE);<br />
label("$D$", D, NW);<br />
label("$E$", E, NW);<br />
label("$F$", F, SW);<br />
label("$G$", G, S);<br />
label("$H$", H, N);<br />
label("$I$", I, NE);<br />
label("$K$", K,SW,red);<br />
label("$J$", J, SE);</asy><br />
<br />
Let the intersection of <math>AJ</math> and <math>EI</math> be <math>K</math>.<br />
<br />
Now we have <math>\triangle ADK</math> and <math>\triangle KIJ</math>.<br />
<br />
Because both triangles has a side on congruent squares therefore <math>AD \cong IJ</math>.<br />
<br />
Because <math>\angle AKD</math> and <math>\angle JKI</math> are vertical angles <math>\angle AKD \cong \angle JKI</math>.<br />
<br />
Also both <math>\angle ADK</math> and <math>\angle JIK</math> are right angles so <math>\angle ADK \cong \angle JIK</math>.<br />
<br />
Therefore by AAS(Angle, Angle, Side) <math>\triangle ADK \cong \triangle KIJ</math>.<br />
<br />
Then translating/rotating the shaded <math>\triangle JIK</math> into the position of <math>\triangle ADK</math><br />
<br />
So the shaded area now completely covers the square <math>ABCD</math><br />
<br />
Set the area of a square as <math>x</math><br />
<br />
Therefore, <math>\frac{x}{3x}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2013|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Zeeman860