https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Debussy&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T19:38:13ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_8_Problems/Problem_14&diff=882492008 AMC 8 Problems/Problem 142017-11-14T01:10:56Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
Three <math>\text{A's}</math>, three <math>\text{B's}</math>, and three <math>\text{C's}</math> are placed in the nine spaces so that each row and column contain one of each letter. If <math>\text{A}</math> is placed in the upper left corner, how many arrangements are possible?<br />
<asy><br />
size((80));<br />
draw((0,0)--(9,0)--(9,9)--(0,9)--(0,0));<br />
draw((3,0)--(3,9));<br />
draw((6,0)--(6,9));<br />
draw((0,3)--(9,3));<br />
draw((0,6)--(9,6));<br />
label("A", (1.5,7.5));<br />
</asy><br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6 </math><br />
<br />
==Solution==<br />
There are <math>2</math> ways to place the remaining <math>\text{As}</math>, <math>2</math> ways to place the remaining <math>\text{Bs}</math>, and <math>1</math> way to place the remaining <math>\text{Cs}</math> for a total of <math>(2)(2)(1) = \boxed{\textbf{(C)}\ 4}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2008|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_8_Problems&diff=882472008 AMC 8 Problems2017-11-14T01:05:56Z<p>Debussy: /* Problem 21 */</p>
<hr />
<div>==Problem 1==<br />
Susan had 50 dollars to spend at the carnival. She spent 12 dollars on food and twice as much on rides. How many dollars did she have left to spend?<br />
<br />
<math>\textbf{(A)}\ \text{12} \qquad<br />
\textbf{(B)}\ \text{14} \qquad<br />
\textbf{(C)}\ \text{26} \qquad<br />
\textbf{(D)}\ \text{38}\qquad<br />
\textbf{(E)}\ \text{50} </math><br />
<br />
<br />
[[2008 AMC 8 Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
The ten-letter code <math>\text{BEST OF LUCK}</math> represents the ten digits <math>0-9</math>, in order. What 4-digit number is represented by the code word <math>\text{CLUE}</math>?<br />
<br />
<math>\textbf{(A)}\ 8671 \qquad<br />
\textbf{(B)}\ 8672 \qquad<br />
\textbf{(C)}\ 9781 \qquad<br />
\textbf{(D)}\ 9782 \qquad<br />
\textbf{(E)}\ 9872</math><br />
<br />
[[2008 AMC 8 Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
If February is a month that contains Friday the <math>13^{\text{th}}</math>, what day of the week is February 1?<br />
<br />
<math>\textbf{(A)}\ \text{Sunday} \qquad<br />
\textbf{(B)}\ \text{Monday} \qquad<br />
\textbf{(C)}\ \text{Wednesday} \qquad<br />
\textbf{(D)}\ \text{Thursday}\qquad<br />
\textbf{(E)}\ \text{Saturday} </math><br />
<br />
[[2008 AMC 8 Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
In the figure, the outer equilateral triangle has area <math>16</math>, the inner equilateral triangle has area <math>1</math>, and the three trapezoids are congruent. What is the area of one of the trapezoids?<br />
<asy><br />
size((70));<br />
draw((0,0)--(7.5,13)--(15,0)--(0,0));<br />
draw((1.88,3.25)--(9.45,3.25));<br />
draw((11.2,0)--(7.5,6.5));<br />
draw((9.4,9.7)--(5.6,3.25));<br />
</asy><br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7</math><br />
<br />
[[2008 AMC 8 Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
Barney Schwinn notices that the odometer on his bicycle reads <math>1441</math>, a palindrome, because it reads the same forward and backward. After riding <math>4</math> more hours that day and <math>6</math> the next, he notices that the odometer shows another palindrome, <math>1661</math>. What was his average speed in miles per hour?<br />
<br />
<math>\textbf{(A)}\ 15\qquad<br />
\textbf{(B)}\ 16\qquad<br />
\textbf{(C)}\ 18\qquad<br />
\textbf{(D)}\ 20\qquad<br />
\textbf{(E)}\ 22</math><br />
<br />
[[2008 AMC 8 Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
In the figure, what is the ratio of the area of the gray squares to the area of the white squares?<br />
<asy><br />
size((70));<br />
draw((10,0)--(0,10)--(-10,0)--(0,-10)--(10,0));<br />
draw((-2.5,-7.5)--(7.5,2.5));<br />
draw((-5,-5)--(5,5));<br />
draw((-7.5,-2.5)--(2.5,7.5));<br />
draw((-7.5,2.5)--(2.5,-7.5));<br />
draw((-5,5)--(5,-5));<br />
draw((-2.5,7.5)--(7.5,-2.5));<br />
fill((-10,0)--(-7.5,2.5)--(-5,0)--(-7.5,-2.5)--cycle, gray);<br />
fill((-5,0)--(0,5)--(5,0)--(0,-5)--cycle, gray);<br />
fill((5,0)--(7.5,2.5)--(10,0)--(7.5,-2.5)--cycle, gray);<br />
</asy><br />
<math> \textbf{(A)}\ 3:10 \qquad\textbf{(B)}\ 3:8 \qquad\textbf{(C)}\ 3:7 \qquad\textbf{(D)}\ 3:5 \qquad\textbf{(E)}\ 1:1 </math><br />
<br />
[[2008 AMC 8 Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
If <math>\frac{3}{5}=\frac{M}{45}=\frac{60}{N}</math>, what is <math>M+N</math>?<br />
<br />
<math>\textbf{(A)}\ 27\qquad<br />
\textbf{(B)}\ 29 \qquad<br />
\textbf{(C)}\ 45 \qquad<br />
\textbf{(D)}\ 105\qquad<br />
\textbf{(E)}\ 127</math><br />
<br />
[[2008 AMC 8 Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
Candy sales from the Boosters Club from January through April are shown. What were the average sales per month in dollars?<br />
<asy><br />
draw((0,0)--(36,0)--(36,24)--(0,24)--cycle);<br />
draw((0,4)--(36,4));<br />
draw((0,8)--(36,8));<br />
draw((0,12)--(36,12));<br />
draw((0,16)--(36,16));<br />
draw((0,20)--(36,20));<br />
fill((4,0)--(8,0)--(8,20)--(4,20)--cycle, black);<br />
fill((12,0)--(16,0)--(16,12)--(12,12)--cycle, black);<br />
fill((20,0)--(24,0)--(24,8)--(20,8)--cycle, black);<br />
fill((28,0)--(32,0)--(32,24)--(28,24)--cycle, black);<br />
label("120", (0,24), W);<br />
label("80", (0,16), W);<br />
label("40", (0,8), W);<br />
label("Jan", (6,0), S);<br />
label("Feb", (14,0), S);<br />
label("Mar", (22,0), S);<br />
label("Apr", (30,0), S);<br />
</asy><br />
<math> \textbf{(A)}\ 60\qquad\textbf{(B)}\ 70\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 80\qquad\textbf{(E)}\ 85 </math><br />
<br />
[[2008 AMC 8 Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
In <math>2005</math> Tycoon Tammy invested <math>100</math> dollars for two years. During the the first year<br />
her investment suffered a <math>15\%</math> loss, but during the second year the remaining<br />
investment showed a <math>20\%</math> gain. Over the two-year period, what was the change<br />
in Tammy's investment?<br />
<br />
<math>\textbf{(A)}\ 5\%\text{ loss}\qquad<br />
\textbf{(B)}\ 2\%\text{ loss}\qquad<br />
\textbf{(C)}\ 1\%\text{ gain}\qquad<br />
\textbf{(D)}\ 2\% \text{ gain} \qquad<br />
\textbf{(E)}\ 5\%\text{ gain}</math><br />
<br />
[[2008 AMC 8 Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
The average age of the <math>6</math> people in Room A is <math>40</math>. The average age of the <math>4</math> people in Room B is <math>25</math>. If the two groups are combined, what is the average age of all the people?<br />
<br />
<math>\textbf{(A)}\ 32.5 \qquad<br />
\textbf{(B)}\ 33 \qquad<br />
\textbf{(C)}\ 33.5 \qquad<br />
\textbf{(D)}\ 34\qquad<br />
\textbf{(E)}\ 35</math><br />
<br />
[[2008 AMC 8 Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
Each of the <math>39</math> students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and <math>26</math> students have a cat. How many students have both a dog and a cat?<br />
<br />
<math>\textbf{(A)}\ 7\qquad<br />
\textbf{(B)}\ 13\qquad<br />
\textbf{(C)}\ 19\qquad<br />
\textbf{(D)}\ 39\qquad<br />
\textbf{(E)}\ 46</math><br />
<br />
[[2008 AMC 8 Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
A ball is dropped from a height of <math>3</math> meters. On its first bounce it rises to a height of <math>2</math> meters. It keeps falling and bouncing to <math>\frac{2}{3}</math> of the height it reached in the previous bounce. On which bounce will it rise to a height less than <math>0.5</math> meters?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad<br />
\textbf{(B)}\ 4 \qquad<br />
\textbf{(C)}\ 5 \qquad<br />
\textbf{(D)}\ 6 \qquad<br />
\textbf{(E)}\ 7</math><br />
<br />
[[2008 AMC 8 Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than <math>100</math> pounds or more than <math>150</math> pounds. So the boxes are weighed in pairs in every possible way. The results are <math>122</math>, <math>125</math> and <math>127</math> pounds. What is the combined weight in pounds of the three boxes?<br />
<br />
<math>\textbf{(A)}\ 160\qquad<br />
\textbf{(B)}\ 170\qquad<br />
\textbf{(C)}\ 187\qquad<br />
\textbf{(D)}\ 195\qquad<br />
\textbf{(E)}\ 354</math><br />
<br />
[[2008 AMC 8 Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
Three <math>\text{A's}</math>, three <math>\text{B's}</math>, and three <math>\text{C's}</math> are placed in the nine spaces so that each row and column contain one of each letter. If <math>\text{A}</math> is placed in the upper left corner, how many arrangements are possible?<br />
<asy><br />
size((80));<br />
draw((0,0)--(9,0)--(9,9)--(0,9)--(0,0));<br />
draw((3,0)--(3,9));<br />
draw((6,0)--(6,9));<br />
draw((0,3)--(9,3));<br />
draw((0,6)--(9,6));<br />
label("A", (1.5,7.5));<br />
</asy><br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6 </math><br />
<br />
[[2008 AMC 8 Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
In Theresa's first <math>8</math> basketball games, she scored <math>7, 4, 3, 6, 8, 3, 1</math> and <math>5</math> points. In her ninth game, she scored fewer than <math>10</math> points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than <math>10</math> points and her points-per-game average for the <math>10</math> games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?<br />
<br />
<math>\textbf{(A)}\ 35\qquad<br />
\textbf{(B)}\ 40\qquad<br />
\textbf{(C)}\ 48\qquad<br />
\textbf{(D)}\ 56\qquad<br />
\textbf{(E)}\ 72</math><br />
<br />
[[2008 AMC 8 Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units?<br />
<br />
<asy><br />
import three;<br />
defaultpen(linewidth(0.8));<br />
real r=0.5;<br />
currentprojection=orthographic(1,1/2,1/4);<br />
draw(unitcube, white, thick(), nolight);<br />
draw(shift(1,0,0)*unitcube, white, thick(), nolight);<br />
draw(shift(1,-1,0)*unitcube, white, thick(), nolight);<br />
draw(shift(1,0,-1)*unitcube, white, thick(), nolight);<br />
draw(shift(2,0,0)*unitcube, white, thick(), nolight);<br />
draw(shift(1,1,0)*unitcube, white, thick(), nolight);<br />
draw(shift(1,0,1)*unitcube, white, thick(), nolight);</asy><br />
<br />
<math>\textbf{(A)} \:1 : 6 \qquad\textbf{ (B)}\: 7 : 36 \qquad\textbf{(C)}\: 1 : 5 \qquad\textbf{(D)}\: 7 : 30\qquad\textbf{ (E)}\: 6 : 25</math><br />
<br />
[[2008 AMC 8 Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of <math>50</math> units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?<br />
<br />
<math>\textbf{(A)}\ 76\qquad<br />
\textbf{(B)}\ 120\qquad<br />
\textbf{(C)}\ 128\qquad<br />
\textbf{(D)}\ 132\qquad<br />
\textbf{(E)}\ 136</math><br />
<br />
[[2008 AMC 8 Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
Two circles that share the same center have radii <math>10</math> meters and <math>20</math> meters. An aardvark runs along the path shown, starting at <math>A</math> and ending at <math>K</math>. How many meters does the aardvark run?<br />
<asy><br />
size((150));<br />
draw((10,0)..(0,10)..(-10,0)..(0,-10)..cycle);<br />
draw((20,0)..(0,20)..(-20,0)..(0,-20)..cycle);<br />
draw((20,0)--(-20,0));<br />
draw((0,20)--(0,-20));<br />
draw((-2,21.5)..(-15.4, 15.4)..(-22,0), EndArrow);<br />
draw((-18,1)--(-12, 1), EndArrow);<br />
draw((-12,0)..(-8.3,-8.3)..(0,-12), EndArrow);<br />
draw((1,-9)--(1,9), EndArrow);<br />
draw((0,12)..(8.3, 8.3)..(12,0), EndArrow);<br />
draw((12,-1)--(18,-1), EndArrow);<br />
label("$A$", (0,20), N);<br />
label("$K$", (20,0), E);<br />
</asy><br />
<math> \textbf{(A)}\ 10\pi+20\qquad\textbf{(B)}\ 10\pi+30\qquad\textbf{(C)}\ 10\pi+40\qquad\textbf{(D)}\ 20\pi+20\qquad \\ \textbf{(E)}\ 20\pi+40</math><br />
<br />
[[2008 AMC 8 Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
Eight points are spaced around at intervals of one unit around a <math>2 \times 2</math> square, as shown. Two of the <math>8</math> points are chosen at random. What is the probability that the two points are one unit apart?<br />
<asy><br />
size((50));<br />
dot((5,0));<br />
dot((5,5));<br />
dot((0,5));<br />
dot((-5,5));<br />
dot((-5,0));<br />
dot((-5,-5));<br />
dot((0,-5));<br />
dot((5,-5));<br />
</asy><br />
<math> \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{2}{7}\qquad\textbf{(C)}\ \frac{4}{11}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{4}{7} </math><br />
<br />
[[2008 AMC 8 Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and <math>\frac{3}{4}</math> of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?<br />
<br />
<math>\textbf{(A)}\ 12\qquad<br />
\textbf{(B)}\ 17\qquad<br />
\textbf{(C)}\ 24\qquad<br />
\textbf{(D)}\ 27\qquad<br />
\textbf{(E)}\ 36</math><br />
<br />
[[2008 AMC 8 Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
Jerry cuts a wedge from a <math>6</math>-cm cylinder of bologna as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?<br />
<asy><br />
defaultpen(linewidth(0.65));<br />
real d=90-63.43494882;<br />
draw(ellipse((origin), 2, 4));<br />
fill((0,4)--(0,-4)--(-8,-4)--(-8,4)--cycle, white);<br />
draw(ellipse((-4,0), 2, 4));<br />
draw((0,4)--(-4,4));<br />
draw((0,-4)--(-4,-4));<br />
draw(shift(-2,0)*rotate(-d-5)*ellipse(origin, 1.82, 4.56), linetype("10 10"));<br />
draw((-4,4)--(-8,4), dashed);<br />
draw((-4,-4)--(-8,-4), dashed);<br />
draw((-4,4.3)--(-4,5));<br />
draw((0,4.3)--(0,5));<br />
draw((-7,4)--(-7,-4), Arrows(5));<br />
draw((-4,4.7)--(0,4.7), Arrows(5));<br />
label("$8$ cm", (-7,0), W);<br />
label("$6$ cm", (-2,4.7), N);</asy><br />
<br />
<math>\textbf{(A)} 48 \qquad<br />
\textbf{(B)} 75 \qquad<br />
\textbf{(C)}151\qquad<br />
\textbf{(D)}192 \qquad<br />
\textbf{(E)}603</math><br />
<br />
[[2008 AMC 8 Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
For how many positive integer values of <math>n</math> are both <math>\frac{n}{3}</math> and <math>3n</math> three-digit whole numbers?<br />
<br />
<math>\textbf{(A)}\ 12\qquad<br />
\textbf{(B)}\ 21\qquad<br />
\textbf{(C)}\ 27\qquad<br />
\textbf{(D)}\ 33\qquad<br />
\textbf{(E)}\ 34</math><br />
<br />
[[2008 AMC 8 Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
In square <math>ABCE</math>, <math>AF=2FE</math> and <math>CD=2DE</math>. What is the ratio of the area of <math>\triangle BFD</math> to the area of square <math>ABCE</math>?<br />
<asy><br />
size((100));<br />
draw((0,0)--(9,0)--(9,9)--(0,9)--cycle);<br />
draw((3,0)--(9,9)--(0,3)--cycle);<br />
dot((3,0));<br />
dot((0,3));<br />
dot((9,9));<br />
dot((0,0));<br />
dot((9,0));<br />
dot((0,9));<br />
label("$A$", (0,9), NW);<br />
label("$B$", (9,9), NE);<br />
label("$C$", (9,0), SE);<br />
label("$D$", (3,0), S);<br />
label("$E$", (0,0), SW);<br />
label("$F$", (0,3), W);<br />
</asy><br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{2}{9}\qquad\textbf{(C)}\ \frac{5}{18}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{7}{20} </math><br />
<br />
[[2008 AMC 8 Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
Ten tiles numbered <math>1</math> through <math>10</math> are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{10}\qquad<br />
\textbf{(B)}\ \frac{1}{6}\qquad<br />
\textbf{(C)}\ \frac{11}{60}\qquad<br />
\textbf{(D)}\ \frac{1}{5}\qquad<br />
\textbf{(E)}\ \frac{7}{30}</math><br />
<br />
[[2008 AMC 8 Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
Mary's winning art design is shown. The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Approximately what percent of the design is black?<br />
<br />
<asy><br />
real d=320;<br />
pair O=origin;<br />
pair P=O+8*dir(d);<br />
pair A0 = origin;<br />
pair A1 = O+1*dir(d);<br />
pair A2 = O+2*dir(d);<br />
pair A3 = O+3*dir(d);<br />
pair A4 = O+4*dir(d);<br />
pair A5 = O+5*dir(d);<br />
filldraw(Circle(A0, 6), white, black);<br />
filldraw(circle(A1, 5), black, black);<br />
filldraw(circle(A2, 4), white, black);<br />
filldraw(circle(A3, 3), black, black);<br />
filldraw(circle(A4, 2), white, black);<br />
filldraw(circle(A5, 1), black, black);<br />
</asy><br />
<math> \textbf{(A)}\ 42\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 48\qquad</math><br />
<br />
[[2008 AMC 8 Problems/Problem 25|Solution]]<br />
<br />
==See Also==<br />
{{AMC8 box|year=2008|before=[[2007 AMC 8 Problems|2007 AMC 8]]|after=[[2009 AMC 8 Problems|2009 AMC 8]]}}<br />
* [[AMC 8]]<br />
* [[AMC 8 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_13&diff=881702010 AMC 8 Problems/Problem 132017-11-11T01:00:52Z<p>Debussy: /* =Problem234523453422 */</p>
<hr />
<div>The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is <math>30\%</math> of the perimeter. What is the length of the longest side?<br />
<br />
<math> \textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math><br />
<br />
==Solution 1==<br />
Let <math>n</math>, <math>n+1</math>, and <math>n+2</math> be the lengths of the sides of the triangle. Then the perimeter of the triangle is <math>n + (n+1) + (n+2) = 3n+3</math>. Using the fact that the length of the smallest side is <math>30\%</math> of the perimeter, it follows that:<br />
<br />
<math>n = 0.3(3n+3) \Rightarrow n = 0.9n+0.9 \Rightarrow 0.1n = 0.9 \Rightarrow n=9</math>. The longest side is then <math>n+2 = 11</math>. Thus, answer choice <math>\boxed{\textbf{(E)}\ 11}</math> is correct.<br />
<br />
==Solution 2==<br />
Since the length of the shortest side is a whole number and is equal to <math>\frac{3}{10}</math> of the perimeter, it follows that the perimeter must be a multiple of <math>10</math>. Adding the two previous integers to each answer choice, we see that <math>11+10+9=30</math>. Thus, answer choice <math>\boxed{\textbf{(E)}\ 11}</math> is correct.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_8&diff=881692010 AMC 8 Problems/Problem 82017-11-11T00:48:51Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction <math>1/2</math> mile in front of her. After she passes him, she can see him in her rear mirror until he is <math>1/2</math> mile behind her. Emily rides at a constant rate of <math>12</math> miles per hour, and Emerson skates at a constant rate of <math>8</math> miles per hour. For how many minutes can Emily see Emerson? <br />
<br />
<math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16 </math><br />
<br />
==Solution==<br />
Because they are both moving in the same direction, Emily is skating relative to Emerson <math>12-8=4</math> mph. Now we can look at it as if Emerson is not moving at all[on his skateboard] and Emily is riding at <math>4</math> mph. It takes her<br />
<br />
<cmath>\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}</cmath><br />
<br />
to skate the <math>1/2</math> mile to reach him, and then the same amount of time to be <math>1/2</math> mile ahead of him. This totals to<br />
<br />
<cmath>2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{\textbf{(D)}\ 15}\ \text{minutes}</cmath><br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems&diff=881632011 AMC 8 Problems2017-11-10T21:04:22Z<p>Debussy: /* Problem 10 */</p>
<hr />
<div>==Problem 1==<br />
Margie bought <math> 3 </math> apples at a cost of <math> 50 </math> cents per apple. She paid with a 5-dollar bill. How much change did Margie recieve?<br />
<br />
<math>\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</math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Karl's rectangular vegetable garden is <math> 20 </math> feet by <math> 45 </math> feet, and Makenna's is <math> 25 </math> feet by <math> 40 </math> feet. Whose garden is larger in area?<br />
<br />
<math>\textbf{(A) }\text{Karl's garden is larger by 100 square feet.}</math><br />
<br />
<math>\textbf{(B) }\text{Karl's garden is larger by 25 square feet.}</math><br />
<br />
<math>\textbf{(C) }\text{The gardens are the same size.}</math> <br />
<br />
<math>\textbf{(D) }\text{Makenna's garden is larger by 25 square feet.}</math><br />
<br />
<math>\textbf{(E) }\text{Makenna's garden is larger by 100 square feet.}</math><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?<br /><br />
<asy><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 />
</asy><br />
<br />
<math> \textbf{(A) }8:17 \qquad\textbf{(B) }25:49 \qquad\textbf{(C) }36:25 \qquad\textbf{(D) }32:17 \qquad\textbf{(E) }36:17</math><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: <cmath>2,0,1,3,0,3,3,1,2.</cmath> Which statement about the mean, median, and mode is true?<br />
<br />
<math>\textbf{(A) }\text{median} < \text{mean} < \text{mode} \qquad \textbf{(B) }\text{mean} < \text{mode} < \text{median} \\ \\ \textbf{(C) }\text{mean} < \text{median} < \text{mode} \qquad \textbf{(D) }\text{median} < \text{mode} < \text{mean} \\ \\ \textbf{(E) }\text{mode} < \text{median} < \text{mean}</math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
What time was it <math>2011</math> minutes after midnight on January 1, 2011?<br />
<br />
<math>\textbf{(A) }\text{January 1 at 9:31 PM}</math><br />
<br />
<math>\textbf{(B) }\text{January 1 at 11:51 PM}</math> <br />
<br />
<math>\textbf{(C) }\text{January 2 at 3:11 AM}</math> <br />
<br />
<math>\textbf{(D) }\text{January 2 at 9:31 AM}</math> <br />
<br />
<math>\textbf{(E) }\text{January 2 at 6:01 PM}</math><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 />
<math> \textbf{(A) }20 \qquad\textbf{(B) }25 \qquad\textbf{(C) }45 \qquad\textbf{(D) }306 \qquad\textbf{(E) }351</math><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 />
<asy><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);</asy><br />
<br />
<math> \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} </math><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 />
<math> \textbf{(A) }4 \qquad\textbf{(B) }5 \qquad\textbf{(C) }6 \qquad\textbf{(D) }7 \qquad\textbf{(E) }9 </math><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 />
<asy><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("$1$",(0.95,-0.24),SE*lsf); label("$2$",(1.92,-0.26),SE*lsf); label("$3$",(2.92,-0.31),SE*lsf); label("$4$",(3.93,-0.26),SE*lsf); label("$5$",(4.92,-0.27),SE*lsf); label("$6$",(5.95,-0.29),SE*lsf); label("$7$",(6.94,-0.27),SE*lsf); label("$5$",(-0.49,1.22),SE*lsf); label("$10$",(-0.59,2.23),SE*lsf); label("$15$",(-0.61,3.22),SE*lsf); label("$20$",(-0.61,4.23),SE*lsf); label("$25$",(-0.59,5.22),SE*lsf); label("$30$",(-0.59,6.2),SE*lsf); label("$35$",(-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("HOURS",(3.41,-0.85),SE*lsf); label("M",(-1.39,5.32),SE*lsf); label("I",(-1.34,4.93),SE*lsf); label("L",(-1.36,4.51),SE*lsf); label("E",(-1.37,4.11),SE*lsf); label("S",(-1.39,3.7),SE*lsf); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<br />
<math> \textbf{(A)}2\qquad\textbf{(B)}2.5\qquad\textbf{(C)}4\qquad\textbf{(D)}4.5\qquad\textbf{(E)}5 </math><br />
<br />
[[2011 AMC 8 Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
The taxi fare in Gotham City is <nowiki>$2.40</nowiki> for the first <math>\frac12</math> mile and additional mileage charged at the rate <nowiki>$0.20</nowiki> for each additional 0.1 mile. You plan to give the driver a <nowiki>$2</nowiki> tip. How many miles can you ride for <nowiki>$10</nowiki>?<br />
<br />
<math> \textbf{(A) } 3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75 </math><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 />
<asy><br />
size(300);<br />
real i;<br />
defaultpen(linewidth(0.8));<br />
draw((0,140)--origin--(220,0));<br />
for(i=1;i<13;i=i+1) {<br />
draw((0,10*i)--(220,10*i));<br />
}<br />
label("$0$",origin,W);<br />
label("$20$",(0,20),W);<br />
label("$40$",(0,40),W);<br />
label("$60$",(0,60),W);<br />
label("$80$",(0,80),W);<br />
label("$100$",(0,100),W);<br />
label("$120$",(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("M",(30,-5),S);<br />
label("Tu",(70,-5),S);<br />
label("W",(110,-5),S);<br />
label("Th",(150,-5),S);<br />
label("F",(190,-5),S);<br />
label("M",(-25,85),W);<br />
label("I",(-27,75),W);<br />
label("N",(-25,65),W);<br />
label("U",(-25,55),W);<br />
label("T",(-25,45),W);<br />
label("E",(-25,35),W);<br />
label("S",(-26,25),W);</asy><br />
<br />
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12 </math><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 />
<math> \textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
Two congruent squares, <math>ABCD</math> and <math>PQRS</math>, have side length <math>15</math>. They overlap to form the <math>15</math> by <math>25</math> rectangle <math>AQRD</math> shown. What percent of the area of rectangle <math>AQRD</math> is shaded? <br />
<asy><br />
filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black);<br />
label("D",(0,0),S);<br />
label("R",(25,0),S);<br />
label("Q",(25,15),N);<br />
label("A",(0,15),N);<br />
filldraw((10,0)--(15,0)--(15,15)--(10,15)--cycle,mediumgrey,black);<br />
label("S",(10,0),S);<br />
label("C",(15,0),S);<br />
label("B",(15,15),N);<br />
label("P",(10,15),N);</asy><br />
<br />
<math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 25 </math><br />
<br />
[[2011 AMC 8 Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
There are <math>270</math> students at Colfax Middle School, where the ratio of boys to girls is <math>5 : 4</math>. There are <math>180</math> students at Winthrop Middle School, where the ratio of boys to girls is <math>4 : 5</math>. 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 />
<math> \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} </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
<br />
How many digits are in the product <math>4^5 \cdot 5^{10}</math>?<br />
<br />
<math> \textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 12 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
Let <math>A</math> be the area of the triangle with sides of length <math>25, 25</math>, and <math>30</math>. Let <math>B</math> be the area of the triangle with sides of length <math>25, 25,</math> and <math>40</math>. What is the relationship between <math>A</math> and <math>B</math>?<br />
<br />
<math> \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 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
Let <math>w</math>, <math>x</math>, <math>y</math>, and <math>z</math> be whole numbers. If <math>2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588</math>, then what does <math>2w + 3x + 5y + 7z</math> equal?<br />
<br />
<math> \textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56 </math><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 />
<math> \textbf{(A) }\dfrac16\qquad\textbf{(B) }\dfrac5{12}\qquad\textbf{(C) }\dfrac12\qquad\textbf{(D) }\dfrac7{12}\qquad\textbf{(E) }\dfrac56 </math><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 />
<asy><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);</asy><br />
<br />
<math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
Quadrilateral <math>ABCD</math> is a trapezoid, <math>AD = 15</math>, <math>AB = 50</math>, <math>BC = 20</math>, and the altitude is <math>12</math>. What is the area of the trapezoid?<br />
<br />
<asy><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("A",A,N);<br />
label("B",B,N);<br />
label("C",C,S);<br />
label("D",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("12",(21,10),E);<br />
label("50",(19,22),N);<br />
label("15",(1,10),W);<br />
label("20",(41,12),E);</asy><br />
<br />
<math> \textbf{(A) }600\qquad\textbf{(B) }650\qquad\textbf{(C) }700\qquad\textbf{(D) }750\qquad\textbf{(E) }800 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
Students guess that Norb's age is <math>24, 28, 30, 32, 36, 38, 41, 44, 47</math>, and <math>49</math>. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?<br />
<br />
<math> \textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
What is the '''tens''' digit of <math>7^{2011}</math>?<br />
<br />
<math> \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7 </math><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 />
<math> \textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108 </math><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 />
<math> \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A circle with radius <math>1</math> 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 />
<asy><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);</asy><br />
<br />
<math> \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 </math><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>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems&diff=881622011 AMC 8 Problems2017-11-10T20:55:27Z<p>Debussy: /* Problem 11 */</p>
<hr />
<div>==Problem 1==<br />
Margie bought <math> 3 </math> apples at a cost of <math> 50 </math> cents per apple. She paid with a 5-dollar bill. How much change did Margie recieve?<br />
<br />
<math>\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</math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Karl's rectangular vegetable garden is <math> 20 </math> feet by <math> 45 </math> feet, and Makenna's is <math> 25 </math> feet by <math> 40 </math> feet. Whose garden is larger in area?<br />
<br />
<math>\textbf{(A) }\text{Karl's garden is larger by 100 square feet.}</math><br />
<br />
<math>\textbf{(B) }\text{Karl's garden is larger by 25 square feet.}</math><br />
<br />
<math>\textbf{(C) }\text{The gardens are the same size.}</math> <br />
<br />
<math>\textbf{(D) }\text{Makenna's garden is larger by 25 square feet.}</math><br />
<br />
<math>\textbf{(E) }\text{Makenna's garden is larger by 100 square feet.}</math><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?<br /><br />
<asy><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 />
</asy><br />
<br />
<math> \textbf{(A) }8:17 \qquad\textbf{(B) }25:49 \qquad\textbf{(C) }36:25 \qquad\textbf{(D) }32:17 \qquad\textbf{(E) }36:17</math><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: <cmath>2,0,1,3,0,3,3,1,2.</cmath> Which statement about the mean, median, and mode is true?<br />
<br />
<math>\textbf{(A) }\text{median} < \text{mean} < \text{mode} \qquad \textbf{(B) }\text{mean} < \text{mode} < \text{median} \\ \\ \textbf{(C) }\text{mean} < \text{median} < \text{mode} \qquad \textbf{(D) }\text{median} < \text{mode} < \text{mean} \\ \\ \textbf{(E) }\text{mode} < \text{median} < \text{mean}</math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
What time was it <math>2011</math> minutes after midnight on January 1, 2011?<br />
<br />
<math>\textbf{(A) }\text{January 1 at 9:31 PM}</math><br />
<br />
<math>\textbf{(B) }\text{January 1 at 11:51 PM}</math> <br />
<br />
<math>\textbf{(C) }\text{January 2 at 3:11 AM}</math> <br />
<br />
<math>\textbf{(D) }\text{January 2 at 9:31 AM}</math> <br />
<br />
<math>\textbf{(E) }\text{January 2 at 6:01 PM}</math><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 />
<math> \textbf{(A) }20 \qquad\textbf{(B) }25 \qquad\textbf{(C) }45 \qquad\textbf{(D) }306 \qquad\textbf{(E) }351</math><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 />
<asy><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);</asy><br />
<br />
<math> \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} </math><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 />
<math> \textbf{(A) }4 \qquad\textbf{(B) }5 \qquad\textbf{(C) }6 \qquad\textbf{(D) }7 \qquad\textbf{(E) }9 </math><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 />
<asy><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("$1$",(0.95,-0.24),SE*lsf); label("$2$",(1.92,-0.26),SE*lsf); label("$3$",(2.92,-0.31),SE*lsf); label("$4$",(3.93,-0.26),SE*lsf); label("$5$",(4.92,-0.27),SE*lsf); label("$6$",(5.95,-0.29),SE*lsf); label("$7$",(6.94,-0.27),SE*lsf); label("$5$",(-0.49,1.22),SE*lsf); label("$10$",(-0.59,2.23),SE*lsf); label("$15$",(-0.61,3.22),SE*lsf); label("$20$",(-0.61,4.23),SE*lsf); label("$25$",(-0.59,5.22),SE*lsf); label("$30$",(-0.59,6.2),SE*lsf); label("$35$",(-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("HOURS",(3.41,-0.85),SE*lsf); label("M",(-1.39,5.32),SE*lsf); label("I",(-1.34,4.93),SE*lsf); label("L",(-1.36,4.51),SE*lsf); label("E",(-1.37,4.11),SE*lsf); label("S",(-1.39,3.7),SE*lsf); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<br />
<math> \textbf{(A)}2\qquad\textbf{(B)}2.5\qquad\textbf{(C)}4\qquad\textbf{(D)}4.5\qquad\textbf{(E)}5 </math><br />
<br />
[[2011 AMC 8 Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
The taxi fare in Got Ham City is <nowiki>$2.40</nowiki> for the first <math>\frac12</math> mile and additional mileage charged at the rate <nowiki>$0.20</nowiki> for each additional 0.1 mile. You plan to give the driver a <nowiki>$2</nowiki> tip. How many miles can you ride for <nowiki>$10</nowiki>?<br />
<br />
<math> \textbf{(A) } 3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75 </math><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 />
<asy><br />
size(300);<br />
real i;<br />
defaultpen(linewidth(0.8));<br />
draw((0,140)--origin--(220,0));<br />
for(i=1;i<13;i=i+1) {<br />
draw((0,10*i)--(220,10*i));<br />
}<br />
label("$0$",origin,W);<br />
label("$20$",(0,20),W);<br />
label("$40$",(0,40),W);<br />
label("$60$",(0,60),W);<br />
label("$80$",(0,80),W);<br />
label("$100$",(0,100),W);<br />
label("$120$",(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("M",(30,-5),S);<br />
label("Tu",(70,-5),S);<br />
label("W",(110,-5),S);<br />
label("Th",(150,-5),S);<br />
label("F",(190,-5),S);<br />
label("M",(-25,85),W);<br />
label("I",(-27,75),W);<br />
label("N",(-25,65),W);<br />
label("U",(-25,55),W);<br />
label("T",(-25,45),W);<br />
label("E",(-25,35),W);<br />
label("S",(-26,25),W);</asy><br />
<br />
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12 </math><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 />
<math> \textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
Two congruent squares, <math>ABCD</math> and <math>PQRS</math>, have side length <math>15</math>. They overlap to form the <math>15</math> by <math>25</math> rectangle <math>AQRD</math> shown. What percent of the area of rectangle <math>AQRD</math> is shaded? <br />
<asy><br />
filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black);<br />
label("D",(0,0),S);<br />
label("R",(25,0),S);<br />
label("Q",(25,15),N);<br />
label("A",(0,15),N);<br />
filldraw((10,0)--(15,0)--(15,15)--(10,15)--cycle,mediumgrey,black);<br />
label("S",(10,0),S);<br />
label("C",(15,0),S);<br />
label("B",(15,15),N);<br />
label("P",(10,15),N);</asy><br />
<br />
<math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 25 </math><br />
<br />
[[2011 AMC 8 Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
There are <math>270</math> students at Colfax Middle School, where the ratio of boys to girls is <math>5 : 4</math>. There are <math>180</math> students at Winthrop Middle School, where the ratio of boys to girls is <math>4 : 5</math>. 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 />
<math> \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} </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
<br />
How many digits are in the product <math>4^5 \cdot 5^{10}</math>?<br />
<br />
<math> \textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 12 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
Let <math>A</math> be the area of the triangle with sides of length <math>25, 25</math>, and <math>30</math>. Let <math>B</math> be the area of the triangle with sides of length <math>25, 25,</math> and <math>40</math>. What is the relationship between <math>A</math> and <math>B</math>?<br />
<br />
<math> \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 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
Let <math>w</math>, <math>x</math>, <math>y</math>, and <math>z</math> be whole numbers. If <math>2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588</math>, then what does <math>2w + 3x + 5y + 7z</math> equal?<br />
<br />
<math> \textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56 </math><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 />
<math> \textbf{(A) }\dfrac16\qquad\textbf{(B) }\dfrac5{12}\qquad\textbf{(C) }\dfrac12\qquad\textbf{(D) }\dfrac7{12}\qquad\textbf{(E) }\dfrac56 </math><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 />
<asy><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);</asy><br />
<br />
<math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
Quadrilateral <math>ABCD</math> is a trapezoid, <math>AD = 15</math>, <math>AB = 50</math>, <math>BC = 20</math>, and the altitude is <math>12</math>. What is the area of the trapezoid?<br />
<br />
<asy><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("A",A,N);<br />
label("B",B,N);<br />
label("C",C,S);<br />
label("D",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("12",(21,10),E);<br />
label("50",(19,22),N);<br />
label("15",(1,10),W);<br />
label("20",(41,12),E);</asy><br />
<br />
<math> \textbf{(A) }600\qquad\textbf{(B) }650\qquad\textbf{(C) }700\qquad\textbf{(D) }750\qquad\textbf{(E) }800 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
Students guess that Norb's age is <math>24, 28, 30, 32, 36, 38, 41, 44, 47</math>, and <math>49</math>. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?<br />
<br />
<math> \textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
What is the '''tens''' digit of <math>7^{2011}</math>?<br />
<br />
<math> \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7 </math><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 />
<math> \textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108 </math><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 />
<math> \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A circle with radius <math>1</math> 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 />
<asy><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);</asy><br />
<br />
<math> \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 </math><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>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems&diff=881562016 AMC 8 Problems2017-11-10T01:57:39Z<p>Debussy: /* 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 />
<math>\textbf{(A)} 605 \qquad\textbf{(B)} 655\qquad\textbf{(C)} 665\qquad\textbf{(D)} 1005\qquad \textbf{(E)} 1105</math><br />
<br />
==Problem 2==<br />
<br />
In rectangle <math>ABCD</math>, <math>AB=6</math> and <math>AD=8</math>. Point <math>M</math> is the midpoint of <math>\overline{AD}</math>. What is the area of <math>\triangle AMC</math>?<br />
<br />
<math>\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24</math><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 <math>70, 80,</math> and <math>90</math>. If the average of their four scores is <math>70</math>, then what is the remaining score?<br />
<br />
<math>\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70</math><br />
<br />
[[2016 AMC 8 Problems/Problem 3|Solution<br />
]]<br />
<br />
==Problem 4==<br />
<br />
When Cheenu was a boy he could run <math>15</math> miles in <math>3</math> hours and <math>30</math> minutes. As an old man he can now walk <math>10</math> miles in <math>4</math> hours. How many minutes longer does it take for him to travel a mile now compared to when he was a boy?<br />
<br />
<math>\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30</math><br />
<br />
[[2016 AMC 8 Problems/Problem 4|Solution<br />
]]<br />
<br />
==Problem 5==<br />
<br />
The number <math>N</math> is a two-digit number.<br />
<br />
• When <math>N</math> is divided by <math>9</math>, the remainder is <math>1</math>.<br />
<br />
• When <math>N</math> is divided by <math>10</math>, the remainder is <math>3</math>.<br />
<br />
What is the remainder when <math>N</math> is divided by <math>11</math>?<br />
<br />
<br />
<math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7</math><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 />
<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7</math><br />
<asy><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("frequency",(-0.5,8));<br />
label("0", (-1, 0));<br />
label("1", (-1, 1));<br />
label("2", (-1, 2));<br />
label("3", (-1, 3));<br />
label("4", (-1, 4));<br />
label("5", (-1, 5));<br />
label("6", (-1, 6));<br />
label("7", (-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("3", (0.5, -0.5));<br />
label("4", (2.5, -0.5));<br />
label("5", (4.5, -0.5));<br />
label("6", (6.5, -0.5));<br />
label("7", (8.5, -0.5));<br />
label("name length", (4.5, -1));<br />
</asy><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 />
<math>\textbf{(A) }1^{2016}\qquad\textbf{(B) }2^{2017}\qquad\textbf{(C) }3^{2018}\qquad\textbf{(D) }4^{2019}\qquad \textbf{(E) }5^{2020}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 7|Solution<br />
]]<br />
<br />
==Problem 8==<br />
<br />
Find the value of the expression<br />
<cmath>100-98+96-94+92-90+\cdots+8-6+4-2.</cmath><math>\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100</math><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 <math>2016</math>?<br />
<br />
<math>\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63</math><br />
<br />
[[2016 AMC 8 Problems/Problem 9|Solution<br />
]]<br />
<br />
==Problem 10==<br />
<br />
Suppose that <math>a * b</math> means <math>3a-b.</math> What is the value of <math>x</math> if<br />
<cmath>2 * (5 * x)=1</cmath><br />
<math>\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14</math><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 <math>132.</math><br />
<br />
<math>\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12</math><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. 3/4 of the girls and 2/3<br />
of the boys went on a field trip. What fraction of the students were girls?<br />
<math>\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}</math><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 <math>{ - 2, -1, 0, 3, 4, 5}</math> and multiplied together. What is the probability that the product is <math>0</math>?<br />
<br />
<math>\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 13|Solution<br />
]]<br />
<br />
==Problem 14==<br />
<br />
Karl's car uses a gallon of gas every <math>35</math> miles, and his gas tank holds <math>14</math> gallons when it is full. One day, Karl started with a full tank of gas, <br />
drove <math>350</math> miles, bought <math>8</math> 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 />
<math>\textbf{(A)}\mbox{ }525\qquad\textbf{(B)}\mbox{ }560\qquad\textbf{(C)}\mbox{ }595\qquad\textbf{(D)}\mbox{ }665\qquad\textbf{(E)}\mbox{ }735</math><br />
<br />
[[2016 AMC 8 Problems/Problem 14|Solution<br />
]]<br />
<br />
==Problem 15==<br />
<br />
What is the largest power of <math>2</math> that is a divisor of <math>13^4 - 11^4</math>?<br />
<br />
<math>\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128</math><br />
<br />
[[2016 AMC 8 Problems/Problem 15|Solution<br />
]]<br />
<br />
==Problem 16==<br />
<br />
Annie and Bonnie are running laps around a <math>400</math>-meter oval track. They started together, but Annie has pulled ahead because she runs <math>25\%</math> faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?<br />
<br />
<math>\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</math><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 <math>0</math> to <math>9</math>, with repeated digits allowable. If no password may begin with the sequence <math>9,1,1,</math> then how many passwords are possible?<br />
<br />
<math>\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999</math><br />
<br />
[[2016 AMC 8 Problems/Problem 17|Solution<br />
]]<br />
<br />
==Problem 18==<br />
<br />
In an All-Area track meet, <math>216</math> sprinters enter a <math>100-</math>meter dash competition. The track has <math>6</math> lanes, so only <math>6</math> 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 />
<math>\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72</math><br />
<br />
[[2016 AMC 8 Problems/Problem 18|Solution<br />
]]<br />
<br />
==Problem 19==<br />
<br />
The sum of <math>25</math> consecutive even integers is <math>10,000</math>. What is the largest of these <math>25</math> consecutive integers?<br />
<br />
<math>\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424</math><br />
<br />
[[2016 AMC 8 Problems/Problem 19|Solution<br />
]]<br />
<br />
==Problem 20==<br />
<br />
The least common multiple of <math>a</math> and <math>b</math> is <math>12</math>, and the least common multiple of <math>b</math> and <math>c</math> is <math>15</math>. What is the least possible value of the least common multiple of <math>a</math> and <math>c</math>?<br />
<br />
<math>\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180</math><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 />
<math>\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}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 21|Solution<br />
]]<br />
<br />
==Problem 22==<br />
Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA</math>. What is the area of the "bat wings" (shaded area)?<br />
<asy><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("$A$",(3.05,4.2));<br />
label("$B$",(2,4.2));<br />
label("$C$",(1,4.2));<br />
label("$D$",(0,4.2));<br />
label("$E$", (0,-0.2));<br />
label("$F$", (3,-0.2));<br />
</asy><br />
<br />
<math>\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</math><br />
<br />
[[2016 AMC 8 Problems/Problem 22|Solution<br />
]]<br />
<br />
==Problem 23==<br />
<br />
Two congruent circles centered at points <math>A</math> and <math>B</math> each pass through the other circle's center. The line containing both <math>A</math> and <math>B</math> is extended to intersect the circles at points <math>C</math> and <math>D</math>. The circles intersect at two points, one of which is <math>E</math>. What is the degree measure of <math>\angle CED</math>?<br />
<br />
<math>\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150</math><br />
<br />
[[2016 AMC 8 Problems/Problem 23|Solution<br />
]]<br />
<br />
==Problem 24==<br />
<br />
The digits <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math> are each used once to write a five-digit number <math>PQRST</math>. The three-digit number <math>PQR</math> is divisible by <math>4</math>, the three-digit number <math>QRS</math> is divisible by <math>5</math>, and the three-digit number <math>RST</math> is divisible by <math>3</math>. What is <math>P</math>?<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 />
[[2016 AMC 8 Problems/Problem 24|Solution<br />
]]<br />
<br />
==Problem 25== <br />
<br />
A semicircle is inscribed in an isosceles triangle with base <math>16</math> and height <math>15</math> 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 />
<asy>draw((0,0)--(8,15)--(16,0)--(0,0));<br />
draw(arc((8,0),7.0588,0,180));</asy><br />
<br />
<math>\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}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 25|Solution<br />
]]<br />
<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems&diff=879662013 AMC 8 Problems2017-10-25T02:30:41Z<p>Debussy: /* Problem 14 */</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?<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 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 />
<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>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1990_AJHSME_Problems/Problem_4&diff=876331990 AJHSME Problems/Problem 42017-09-29T20:57:53Z<p>Debussy: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
Which of the following could '''not''' be the units digit <nowiki>[ones digit]</nowiki> of the square of a whole number?<br />
<br />
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math><br />
<br />
==Solution==<br />
<br />
We see that <math>1^2=1</math>, <math>2^2=4</math>, <math>5^2=25</math>, and <math>4^2=16</math>, so already we know that either <math>\text{E}</math> is the answer or the problem has some issues.<br />
<br />
For integers, only the units digit affects the units digit of the final result, so we only need to test the squares of the integers from <math>0</math> through <math>9</math> inclusive. Testing shows that <math>8</math> is unachievable, so the answer is <math>\boxed{\text{E}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1990|num-b=3|num-a=5}}<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1990_AJHSME_Problems/Problem_4&diff=876321990 AJHSME Problems/Problem 42017-09-29T20:57:32Z<p>Debussy: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
Which of the following could '''not''' be the units digit <nowiki>[one's digit]</nowiki> of the square of a whole number?<br />
<br />
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math><br />
<br />
==Solution==<br />
<br />
We see that <math>1^2=1</math>, <math>2^2=4</math>, <math>5^2=25</math>, and <math>4^2=16</math>, so already we know that either <math>\text{E}</math> is the answer or the problem has some issues.<br />
<br />
For integers, only the units digit affects the units digit of the final result, so we only need to test the squares of the integers from <math>0</math> through <math>9</math> inclusive. Testing shows that <math>8</math> is unachievable, so the answer is <math>\boxed{\text{E}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1990|num-b=3|num-a=5}}<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1995_AJHSME_Problems/Problem_25&diff=875891995 AJHSME Problems/Problem 252017-09-23T00:09:07Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
<br />
Buses from Dallas to Houston leave every hour on the hour. Buses from Houston to Dallas leave every hour on the half hour. The trip from one city to the other takes <math>5</math> hours. Assuming the buses travel on the same highway, how many Dallas-bound buses does a Houston-bound bus pass in the highway (not in the station)?<br />
<br />
<math> \text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 9\qquad\text{(D)}\ 10\qquad\text{(E)}\ 11 </math><br />
<br />
==Solution==<br />
<br />
Say you are on the Houston-bound bus that left at 12:30 in the afternoon, looking out the window to see how many buses you pass. At 12:45 pm, the Dallas bus that left at 8:00 am is 4:45 away (Note - <math>a:b</math> - <math>a</math> is for hrs. and <math>b</math> is for min.) from Dallas, and therefore 15 minutes from Houston. Your bus is also 15 minutes from Houston, so you are delighted at the first bus you have passed. <br />
<br />
At 1:15 pm, the 9:00 am Dallas bus meets you again, being 4:15 away from Dallas and therefore 45 minutes from Houston. After a while you might notice that the buses meet you in 30 minute intervals after 12:45 pm; indeed, the bus that left an hour later than the most recently passed bus has traveled 30 minutes closer, and so have you, for a total of an hour closer, and when you passed the most recent bus, the bus after it had indeed left an hour later. You decide to predict how many buses you pass by finding all times that are 30 minute intervals from 12:45. They are<br />
12:45<br />
1:15, 1:45<br />
2:15, 2:45<br />
3:15, 3:45<br />
4:15, 4:45<br />
5:15, arrive at 5:30,<br />
Hence, you will pass <math>\boxed{\text{(D)} 10}.</math><br />
<br />
==See Also==<br />
{{AJHSME box|year=1995|num-b=24|after=Last <br /> Problem}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1995_AJHSME_Problems/Problem_25&diff=875881995 AJHSME Problems/Problem 252017-09-23T00:08:33Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
<br />
Buses from Dallas to Houston leave every hour on the hour. Buses from Houston to Dallas leave every hour on the half hour. The trip from one city to the other takes <math>5</math> hours. Assuming the buses travel on the same highway, how many Dallas-bound buses does a Houston-bound bus pass in the highway (not in the station)?<br />
<br />
<math> \text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 9\qquad\text{(D)}\ 10\qquad\text{(E)}\ 11 </math><br />
<br />
==Solution==<br />
<br />
Say you are on the Houston-bound bus that left at 12:30 in the afternoon, looking out the window to see how many buses you pass. At 12:45 pm, the Dallas bus that left at 8:00 am is 4:45 away (Note - <math>a:b</math> - <math>a</math> is for hrs. and <math>b</math> is for min.) from Dallas, and therefore 15 minutes from Houston. Your bus is also 15 minutes from Houston, so you are delighted at the first bus you have passed. <br />
<br />
At 1:15 pm, the 9:00 am Dallas bus meets you again, being 4:15 away from Dallas and therefore 45 minutes from Houston. After a while you might notice that the buses meet you in 30 minute intervals after 12:45 pm; indeed, the bus that left an hour later than the most recently passed bus has traveled 30 minutes closer, and so have you, for a total of an hour closer, and when you passed the most recent bus, the bus after it had indeed left an hour later. You decide to predict how many buses you pass by finding all times that are 30 minute intervals from 12:45. They are<br />
12:45<br />
1:15, 1:45<br />
2:15, 2:45<br />
3:15, 3:45<br />
4:15, 4:45<br />
5:15, arrive at 5:30,<br />
Hence, you will pass <math>\boxed{\text{(D)}10}</math><br />
<br />
==See Also==<br />
{{AJHSME box|year=1995|num-b=24|after=Last <br /> Problem}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1995_AJHSME_Problems/Problem_25&diff=875871995 AJHSME Problems/Problem 252017-09-23T00:08:18Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
<br />
Buses from Dallas to Houston leave every hour on the hour. Buses from Houston to Dallas leave every hour on the half hour. The trip from one city to the other takes <math>5</math> hours. Assuming the buses travel on the same highway, how many Dallas-bound buses does a Houston-bound bus pass in the highway (not in the station)?<br />
<br />
<math> \text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 9\qquad\text{(D)}\ 10\qquad\text{(E)}\ 11 </math><br />
<br />
==Solution==<br />
<br />
Say you are on the Houston-bound bus that left at 12:30 in the afternoon, looking out the window to see how many buses you pass. At 12:45 pm, the Dallas bus that left at 8:00 am is 4:45 away (Note - <math>a:b</math> - <math>a</math> is for hrs. and <math>b</math> is for min.) from Dallas, and therefore 15 minutes from Houston. Your bus is also 15 minutes from Houston, so you are delighted at the first bus you have passed. <br />
<br />
At 1:15 pm, the 9:00 am Dallas bus meets you again, being 4:15 away from Dallas and therefore 45 minutes from Houston. After a while you might notice that the buses meet you in 30 minute intervals after 12:45 pm; indeed, the bus that left an hour later than the most recently passed bus has traveled 30 minutes closer, and so have you, for a total of an hour closer, and when you passed the most recent bus, the bus after it had indeed left an hour later. You decide to predict how many buses you pass by finding all times that are 30 minute intervals from 12:45. They are<br />
12:45<br />
1:15, 1:45<br />
2:15, 2:45<br />
3:15, 3:45<br />
4:15, 4:45<br />
5:15, arrive at 5:30,<br />
Hence, you will pass <math>\boxed{\text{(D}10}</math><br />
<br />
==See Also==<br />
{{AJHSME box|year=1995|num-b=24|after=Last <br /> Problem}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1992_AJHSME_Problems/Problem_12&diff=875861992 AJHSME Problems/Problem 122017-09-22T21:03:48Z<p>Debussy: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
The five tires of a car (four road tires and a full-sized spare) were rotated so that each tire was used the same number of miles during the first <math>30,000</math> miles the car traveled. For how many miles was each tire used?<br />
<br />
<math>\text{(A)}\ 6000 \qquad \text{(B)}\ 7500 \qquad \text{(C)}\ 24,000 \qquad \text{(D)}\ 30,000 \qquad \text{(E)}\ 37,500</math><br />
<br />
== Solution ==<br />
<br />
In the <math> 30,000 </math> miles, four tires were always used at one time, so the total amount of miles the five tires were used in total is <math> 30,000 \times 4=120,000 </math>. Five tires were used and each was used equally, so each tire was used for <math> \frac{120,000}{5}=\boxed{\text{(C)}\ 24,000} </math>.<br />
<br />
==See Also==<br />
{{AJHSME box|year=1992|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1992_AJHSME_Problems/Problem_12&diff=875851992 AJHSME Problems/Problem 122017-09-22T21:03:18Z<p>Debussy: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
The five tires of a car (four road tires and a full-sized spare) were rotated so that each tire was used the same number of miles during the first <math>30,000</math> miles the car traveled. For how many miles was each tire used?<br />
<br />
<math>\text{(A)}\ 6000 \qquad \text{(B)}\ 7500 \qquad \text{(C)}\ 24,000 \qquad \text{(D)}\ 30,000 \qquad \text{(E)}\ 37,500</math><br />
<br />
== Solution ==<br />
<br />
In the <math> 30,000 </math> miles, four tires were always used at one time, so the total amount of miles the five tires were used in total is <math> 30,000 \times 4=120,000 </math>. Five tires were used and each were used equally, so each tire was used for <math> \frac{120,000}{5}=\boxed{\text{(C)}\ 24,000} </math>.<br />
<br />
==See Also==<br />
{{AJHSME box|year=1992|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_19&diff=856171999 AMC 8 Problems/Problem 192017-05-11T22:12:09Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems 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 these 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 only 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. Since fractional pans are forbidden, round up to make <math>\lceil \frac{216}{15} \rceil = 15</math> pans.<br />
<br />
There are <math>3</math> tablespoons of butter per pan, meaning <math>3 \cdot 15 = 45</math> tablespoons of butter are required for <math>15</math> pans.<br />
<br />
Each stick of butter has <math>8</math> tablespoons, so we need <math>\frac{45}{8} = 5.625</math> sticks of butter. However, we must round up again because partial sticks of butter 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>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_19&diff=856161999 AMC 8 Problems/Problem 192017-05-11T22:11:39Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems 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 these 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 only 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. Since fractional pans are forbidden, round up to make <math>\lceil \frac{216}{15} \rceil = 15</math> pans.<br />
<br />
There are <math>3</math> tablespoons of butter per pan, meaning <math>3 \cdot 15 = 45</math> tablespoons of butter is required for <math>15</math> pans.<br />
<br />
Each stick of butter has <math>8</math> tablespoons, so we need <math>\frac{45}{8} = 5.625</math> sticks of butter. However, we must round up again because partial sticks of butter 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>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_19&diff=856151999 AMC 8 Problems/Problem 192017-05-11T22:11:12Z<p>Debussy: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems 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 these 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 only 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> cakes, you need to make <math>\frac{216}{15} = 14.4</math> pans. Since fractional pans are forbidden, round up to make <math>\lceil \frac{216}{15} \rceil = 15</math> pans.<br />
<br />
There are <math>3</math> tablespoons of butter per pan, meaning <math>3 \cdot 15 = 45</math> tablespoons of butter is required for <math>15</math> pans.<br />
<br />
Each stick of butter has <math>8</math> tablespoons, so we need <math>\frac{45}{8} = 5.625</math> sticks of butter. However, we must round up again because partial sticks of butter are forbidden since they will be eaten in one gulp. 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>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_17&diff=856141999 AMC 8 Problems/Problem 172017-05-11T22:09:31Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems 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 of 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 make only full recipes, not partial recipes.<br />
<br />
Walter can buy eggs by the half-dozen. How many half-dozens should he buy to make enough cookies? (Some eggs and some cookies may be left over.)<br />
<br />
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 15</math><br />
<br />
==Solution==<br />
<br />
If <math>108</math> students eat <math>2</math> cookies on average, there will need to be <math>108\cdot 2 = 216</math> cookies. There are <math>15</math> cookies per pan, meaning there needs to be <math>\frac{216}{15} = 14.4</math> pans. However, since half-recipes are forbidden, we need to round up and make <math>\lceil \frac{216}{15}\rceil = 15</math> pans.<br />
<br />
<math>1</math> pan requires <math>2</math> eggs, so <math>15</math> pans require <math>2\cdot 15 = 30</math> eggs. Since there are <math>6</math> eggs in a half dozen, we need <math>\frac{30}{6} = 5</math> half-dozens of eggs, and the answer is <math>\boxed{C}</math><br />
<br />
==See Also==<br />
{{AMC8 box|year=1999|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_17&diff=856131999 AMC 8 Problems/Problem 172017-05-11T22:09:15Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems 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 of 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 make only full recipes, not partial recipes.<br />
<br />
Walter can buy eggs by the half-dozen. How many half-dozens should he buy to make enough cookies? (Some eggs and some cookies may be left over.)<br />
<br />
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 15</math><br />
<br />
==Solution==<br />
<br />
If <math>108</math> students eat <math>2</math> cakes on average, there will need to be <math>108\cdot 2 = 216</math> cookies. There are <math>15</math> cookies per pan, meaning there needs to be <math>\frac{216}{15} = 14.4</math> pans. However, since half-recipes are forbidden, we need to round up and make <math>\lceil \frac{216}{15}\rceil = 15</math> pans.<br />
<br />
<math>1</math> pan requires <math>2</math> eggs, so <math>15</math> pans require <math>2\cdot 15 = 30</math> eggs. Since there are <math>6</math> eggs in a half dozen, we need <math>\frac{30}{6} = 5</math> half-dozens of eggs, and the answer is <math>\boxed{C}</math><br />
<br />
==See Also==<br />
{{AMC8 box|year=1999|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_17&diff=856121999 AMC 8 Problems/Problem 172017-05-11T22:08:41Z<p>Debussy: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems 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 of 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 make only full recipes, not partial recipes.<br />
<br />
Walter can buy eggs by the half-dozen. How many half-dozens should he buy to make enough cookies? (Some eggs and some cookies may be left over.)<br />
<br />
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 15</math><br />
<br />
==Solution==<br />
<br />
If <math>108</math> students eat <math>2</math> cakes on average, there will need to be <math>108\cdot 2 = 216</math> cakes. There are <math>15</math> cakes per pan, meaning there needs to be <math>\frac{216}{15} = 14.4</math> pans. However, since half-recipes are forbidden, we need to round up and make <math>\lceil \frac{216}{15}\rceil = 15</math> pans.<br />
<br />
<math>1</math> pan requires <math>2</math> eggs, so <math>15</math> pans require <math>2\cdot 15 = 30</math> eggs. Since there are <math>6</math> eggs in a half dozen, we need <math>\frac{30}{6} = 5</math> half-dozens of eggs, and the answer is <math>\boxed{C}</math><br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=1999|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_12&diff=854382011 AMC 8 Problems/Problem 122017-04-27T21:53:29Z<p>Debussy: /* Solution 1 */</p>
<hr />
<div>==Problem==<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 />
<math> \textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34 </math><br />
<br />
==Solution 1==<br />
If we designate Angie to be on a certain side, then all placements of the other people can be considered unique. There are then <math>3!=6</math> total seating arrangements. If Carlos is across from Angie, there are only <math>2!=2</math> ways to fill the remaining two seats. Then the probability Angie and Carlos are seated opposite each other is <math>\frac26=\boxed{\textbf{(B)}\ \frac13}</math><br />
<br />
==Solution 2==<br />
If we seat Angie first, there would be only one out of three ways Carlos can sit across from Angie. So the final answer is <math>\boxed{\textbf{(B) }\frac{1}{3}}</math><br />
<br />
==See Also==<br />
{{AMC8 box|year=2011|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=Trigonometry&diff=84742Trigonometry2017-03-17T22:07:37Z<p>Debussy: /* Trigonometry Definitions for non-acute angles */</p>
<hr />
<div>Trigonometry seeks to find the lengths of a [[triangle]]'s sides, given 2 [[angle]]s and a side. Trigonometry is closely related to [[analytic geometry]].<br />
<br />
==Basic definitions==<br />
Usually we call an angle <math>\theta</math>, read "theta", but <math>\theta</math> is just a variable. We could just as well call it <math> a</math>.<br />
<br />
For the following definitions, the "opposite side" is the side opposite of angle <math>\theta</math>, and the "adjacent side" is the side that is part of angle <math>\theta</math>, but is not the hypotenuse. <br />
<br />
i.e. If ABC is a right triangle with right angle C, and angle A = <math>\theta</math>, then BC is the "opposite side", AC is the "adjacent side", and AB is the hypotenuse. <br />
<br />
[[Image:306090triangle.gif]]<br />
<br />
===Sine===<br />
The sine of an angle <math>\theta</math>, abbreviated <math>\sin \theta</math>, is the ratio between the opposite side and the [[hypotenuse]] of a triangle. For instance, in the 30-60-90 triangle above, <math>\sin 30^{\circ}=\frac 12</math>.<br />
<br />
===Cosine===<br />
The cosine of an angle <math>\theta</math>, abbreviated <math>\cos \theta</math>, is the ratio between the adjacent side and the [[hypotenuse]] of a triangle. For instance, in the 30-60-90 triangle above, <math>\cos 30^{\circ} =\frac{\sqrt{3}}{2}</math>.<br />
<br />
===Tangent===<br />
The tangent of an angle <math>\theta</math>, abbreviated <math>\tan \theta</math>, is the ratio between the opposite side and the adjacent side of a triangle. For instance, in the 30-60-90 triangle above, <math>\tan 30^{\circ}=\frac{\sqrt{3}}{3}</math>. (Note that <math> \tan \theta=\frac{\sin\theta}{\cos\theta}</math>.)<br />
<br />
===Cosecant===<br />
The cosecant of an angle <math>\theta</math>, abbreviated <math>\csc \theta</math>, is the ratio between the [[hypotenuse]] and the opposite side of a triangle. For instance, in the 30-60-90 triangle above, <math>\csc 30^{\circ}=2</math>. (Note that <math> \csc \theta=\frac{1}{\sin \theta}</math>.)<br />
<br />
===Secant===<br />
The secant of an angle <math>\theta</math>, abbreviated <math>\sec \theta</math>, is the ratio between the [[hypotenuse]] and the adjacent side of a triangle. For instance, in the 30-60-90 triangle above, <math>\sec 30^{\circ}=\frac{2\sqrt{3}}{3}</math>. (Note that <math> \sec \theta=\frac{1}{\cos \theta}</math>.)<br />
<br />
<br />
===Cotangent===<br />
The cotangent of an angle <math>\theta</math>, abbreviated <math>\cot \theta</math>, is the ratio between the adjacent side and the opposite side of a triangle. For instance, in the 30-60-90 triangle above, <math>\cot 30^{\circ}=\sqrt{3}</math>. (Note that <math> \cot \theta=\frac{\cos\theta}{\sin\theta}</math> or <math> \cot \theta = \frac{1}{\tan \theta}</math>.)<br />
<br />
==Trigonometry Definitions for non-acute angles==<br />
Consider a [[unit circle]] that is centered at the origin. By picking a point on the circle, and dropping a perpendicular line to the x-axis, a right triangle is formed with a [[hypotenuse]] 1 unit long. Letting the angle at the origin be <math>\theta </math> and the coordinates of the point we picked to be <math>(x,y) </math>, we have:<br />
<br />
<cmath>\begin{align*} \sin \theta &= y \\<br />
\cos \theta &= x \\<br />
\tan \theta &= \frac{y}{x} \\<br />
\csc \theta &= \frac{1}{y} \\<br />
\sec \theta &= \frac{1}{x} \\<br />
\cot \theta &= \frac{x}{y} \end{align*}</cmath><br />
<br />
Note that <math>(x,y) </math> is the rectangular coordinates for the point <math> (1,\theta) </math>.<br />
<br />
This is true for all angles, even negative angles and angles greater than 360 degrees. Due to the way trig ratios are defined for non-acute angles, the value of a trig ratio could be positive or negative, or even 0.<br />
<br />
==Trigonometric Identities==<br />
<br />
There are many identities that are based on trigonometric functions.<br />
<br />
===Pythagorean Identities===<br />
<br />
*<math>\sin^2\theta+\cos^2\theta=1</math> <br />
*<math>1+\tan^2\theta=\sec^2\theta</math> <br />
*<math>1+\cot^2\theta=\csc^2\theta</math><br />
<br />
===Double-Angle Identities===<br />
<br />
*<math>\sin 2\theta=2\sin\theta\cos\theta</math><br />
*<math>\cos 2\theta=\cos^2\theta-\sin^2\theta</math><br />
*<math>\tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta}</math><br />
<br />
===Half-Angle Identities===<br />
<br />
*<math>\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}}</math><br />
*<math>\cos\frac{\theta}{2}=\pm\sqrt{\frac{1+\cos\theta}{2}}</math><br />
*<math>\tan\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}</math><br />
<br />
==See also==<br />
* [[Trigonometric identities]]<br />
* [[Trigonometric substitution]]<br />
* [[Geometry]]<br />
[[Category:Trigonometry]]</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_25&diff=810462012 AMC 8 Problems/Problem 252016-11-11T03:07:14Z<p>Debussy: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length <math> a </math>, and the other of length <math> b </math>. What is the value of <math> ab </math>?<br />
<br />
<asy><br />
draw((0,2)--(2,2)--(2,0)--(0,0)--cycle);<br />
draw((0,0.3)--(0.3,2)--(2,1.7)--(1.7,0)--cycle);<br />
label("$a$",(-0.1,0.15));<br />
label("$b$",(-0.1,1.15));</asy><br />
<br />
<math> \textbf{(A)}\hspace{.05in}\frac{1}5\qquad\textbf{(B)}\hspace{.05in}\frac{2}5\qquad\textbf{(C)}\hspace{.05in}\frac{1}2\qquad\textbf{(D)}\hspace{.05in}1\qquad\textbf{(E)}\hspace{.05in}4 </math><br />
<br />
==Solution 1==<br />
The total area of the four congruent triangles formed by the squares is <math>5-4 = 1 </math>. Therefore, the area of one of these triangles is <math> \frac{1}{4} </math>. The height of one of these triangles is <math> a </math> and the base is <math> b </math>. Using the formula for area of the triangle, we have <math> \frac{ab}{2} = \frac{1}{4} </math>. Multiply by <math> 2 </math> on both sides to find that the value of <math> ab </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>.<br />
<br />
==Solution 2==<br />
To solve this problem you could also use algebraic manipulation.<br />
<br />
Since the area of the large square is <math> 5 </math>, the sidelength is <math> \sqrt{5} </math>.<br />
<br />
We then have the equation <math> a + b = \sqrt{5} </math>.<br />
<br />
We also know that the side length of the smaller square is <math> 2 </math>, since its area is <math> 4 </math>. Then, the segment of length <math> a </math> and segment of length <math> b </math> form a right triangle whose hypotenuse would have length <math> 2 </math>. <br />
<br />
So our second equation is <math> \sqrt{{a^2}+{b^2}} = 2 </math>.<br />
<br />
Square both equations.<br />
<br />
<math> a^2 + 2ab + b^2 = 5 </math><br />
<br />
<math> a^2 + b^2 = 4 </math><br />
<br />
Now, subtract, and obtain the equation <math> 2ab = 1 </math>. We can deduce that the value of <math> ab </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2012|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_23&diff=810362010 AMC 8 Problems/Problem 232016-11-10T23:56:38Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
Semicircles <math>POQ</math> and <math>ROS</math> pass through the center <math>O</math>. What is the ratio of the combined areas of the two semicircles to the area of circle <math>O</math>? <br />
<asy><br />
import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,-1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); <br />
dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<math> \textbf{(A)}\ \frac{\sqrt 2}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{\pi}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{\sqrt 2}{2} </math><br />
<br />
==Solution==<br />
By the Pythagorean Theorem, the radius of the larger circle turns out to be <math>1^2 + 1^2 = \sqrt{2}</math>. Therefore, the area of the larger circle is <math>(\sqrt{2})^2\pi = 2\pi </math>. Using the coordinate plane given, we find that the radius of each of the two semicircles to be 1. So, the area of the two semicircles is <math>1^2\pi=\pi</math>. Finally, the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>\boxed{\textbf{(B)}\ \frac{1}{2}}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_23&diff=810352010 AMC 8 Problems/Problem 232016-11-10T23:56:12Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
Semicircles <math>POQ</math> and <math>ROS</math> pass through the center <math>O</math>. What is the ratio of the combined areas of the two semicircles to the area of circle <math>O</math>? <br />
<asy><br />
import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,-1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); <br />
dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<math> \textbf{(A)}\ \frac{\sqrt 2}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{\pi}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{\sqrt 2}{2} </math><br />
<br />
==Solution==<br />
By the Pythagorean Theorem, the radius of the larger circle turns out to be <math>1^2 + 1^2 = \sqrt{2}</math>. Therefore, the area of the larger circle is <math>(\sqrt{2})^2\pi = 2\pi </math>. Using the coordinate plane given, we find that the radius of the two semicircles to be 1. So, the area of the two semicircles is <math>1^2\pi=\pi</math>. Finally, the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>\boxed{\textbf{(B)}\ \frac{1}{2}}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_23&diff=810342010 AMC 8 Problems/Problem 232016-11-10T23:55:14Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
Semicircles <math>POQ</math> and <math>ROS</math> pass through the center <math>O</math>. What is the ratio of the combined areas of the two semicircles to the area of circle <math>O</math>? <br />
<asy><br />
import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,-1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); <br />
dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<math> \textbf{(A)}\ \frac{\sqrt 2}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{\pi}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{\sqrt 2}{2} </math><br />
<br />
==Solution==<br />
By the Pythagorean Theorem, the radius of the larger circle turns out to be <math>1^2 + 1^2 = \sqrt{2}</math>. Therefore, the area of the larger circle is <math>(\sqrt{2})^2\pi = 2\pi </math>. Using the coordinate plane given we find that the radius of the two semicircles to be 1. So, the area of the two semicircles is <math>1^2\pi=\pi</math>. Finally, the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>\boxed{\textbf{(B)}\ \frac{1}{2}}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_23&diff=810332010 AMC 8 Problems/Problem 232016-11-10T23:52:58Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
Semicircles <math>POQ</math> and <math>ROS</math> pass through the center <math>O</math>. What is the ratio of the combined areas of the two semicircles to the area of circle <math>O</math>? <br />
<asy><br />
import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,-1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); <br />
dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<math> \textbf{(A)}\ \frac{\sqrt 2}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{\pi}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{\sqrt 2}{2} </math><br />
<br />
==Solution==<br />
According to the Pythagorean Theorem, the radius of the larger circle is <math>1^2 + 1^2 = \sqrt{2}</math>. Therefore, the area of the larger circle is <math>(\sqrt{2})^2\pi = 2\pi </math>. Using the coordinate plane given we find that the radius of the two semicircles to be 1. So, the area of the two semicircles is <math>1^2\pi=\pi</math>. Finally, the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>\boxed{\textbf{(B)}\ \frac{1}{2}}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_21&diff=810312010 AMC 8 Problems/Problem 212016-11-10T23:32:33Z<p>Debussy: /* Problem */</p>
<hr />
<div>==Problem==<br />
Hui is an avid reader. She bought a copy of the best seller ''Math is Beautiful''. On the first day, Hui read <math>1/5</math> of the pages plus <math>12</math> more, and on the second day she read <math>1/4</math> of the remaining pages plus <math>15</math> pages. On the third day she read <math>1/3</math> of the remaining pages plus <math>18</math> pages. She then realized that there were only <math>62</math> pages left to read, which she read the next day. How many pages are in this book? <br />
<br />
<math> \textbf{(A)}\ 120 \qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 240\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 360 </math><br />
<br />
==Solution==<br />
<br />
Let <math>x</math> be the number of pages in the book. After the first day, Hui had <math>\frac{4x}{5}-12</math> pages left to read. After the second, she had <math>(\frac{3}{4})(\frac{4x}{5}-12)-15 = \frac{3x}{5}-24</math> left. After the third, she had <math>(\frac{2}{3})(\frac{3x}{5}-24)-18=\frac{2x}{5}-34</math> left. This is equivalent to <math>62.</math><br />
<br />
<cmath>\begin{align*} \frac{2x}{5}-34&=62\\<br />
2x - 170 &= 310\\<br />
2x &= 480\\<br />
x &= \boxed{\textbf{(C)}\ 240} \end{align*}</cmath><br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_21&diff=810302010 AMC 8 Problems/Problem 212016-11-10T23:32:21Z<p>Debussy: /* Problem */</p>
<hr />
<div>==Problem==<br />
Hu is an avid reader. She bought a copy of the best seller ''Math is Beautiful''. On the first day, Hui read <math>1/5</math> of the pages plus <math>12</math> more, and on the second day she read <math>1/4</math> of the remaining pages plus <math>15</math> pages. On the third day she read <math>1/3</math> of the remaining pages plus <math>18</math> pages. She then realized that there were only <math>62</math> pages left to read, which she read the next day. How many pages are in this book? <br />
<br />
<math> \textbf{(A)}\ 120 \qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 240\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 360 </math><br />
<br />
==Solution==<br />
<br />
Let <math>x</math> be the number of pages in the book. After the first day, Hui had <math>\frac{4x}{5}-12</math> pages left to read. After the second, she had <math>(\frac{3}{4})(\frac{4x}{5}-12)-15 = \frac{3x}{5}-24</math> left. After the third, she had <math>(\frac{2}{3})(\frac{3x}{5}-24)-18=\frac{2x}{5}-34</math> left. This is equivalent to <math>62.</math><br />
<br />
<cmath>\begin{align*} \frac{2x}{5}-34&=62\\<br />
2x - 170 &= 310\\<br />
2x &= 480\\<br />
x &= \boxed{\textbf{(C)}\ 240} \end{align*}</cmath><br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_8_Problems/Problem_8&diff=809112005 AMC 8 Problems/Problem 82016-11-02T01:18:09Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?<br />
<br />
<math> \textbf{(A)}\ m+3n\qquad\textbf{(B)}\ 3m-n\qquad\textbf{(C)}\ 3m^2 + 3n^2\qquad\textbf{(D)}\ (nm + 3)^2\qquad\textbf{(E)}\ 3mn </math><br />
<br />
==Solution==<br />
Assume [[WLOG]] that <math>m</math> and <math>n</math> are both <math>1</math>. Plugging into each of the choices, we get <math>4, 2, 6, 16,</math> and <math>3</math>. The only odd integer is <math>\boxed{\textbf{(E)}\ 3mn}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2005|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_13&diff=808832002 AMC 8 Problems/Problem 132016-10-31T02:25:49Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
For his birthday, Bert gets a box that holds 125 jellybeans when filled to capacity. A few weeks later, Carrie gets a larger box full of jellybeans. Her box is twice as high, twice as wide and twice as long as Bert's. Approximately, how many jellybeans did Carrie get? <br />
<br />
<math> \text{(A)}\ 250\qquad\text{(B)}\ 500\qquad\text{(C)}\ 625\qquad\text{(D)}\ 750\qquad\text{(E)}\ 1000 </math><br />
<br />
==Solution==<br />
<math>1^3=1<br />
2^3=8<br />
8*125=1000<br />
\boxed{\text{(E)}\ 1000}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_12&diff=808822002 AMC 8 Problems/Problem 122016-10-31T02:24:25Z<p>Debussy: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
A board game spinner is divided into three regions labeled <math>A</math>, <math>B</math> and <math>C</math>. The probability of the arrow stopping on region <math>A</math> is <math>\frac{1}{3}</math> and on region <math>B</math> is <math>\frac{1}{2}</math>. The probability of the arrow stopping on region <math>C</math> is:<br />
<br />
<br />
<math> \text{(A)}\ \frac{1}{12}\qquad\text{(B)}\ \frac{1}{6}\qquad\text{(C)}\ \frac{1}{5}\qquad\text{(D)}\ \frac{1}{3}\qquad\text{(E)}\ \frac{2}{5} </math><br />
<br />
==Solution==<br />
Since the arrow must land in one of the three regions, the sum of the probabilities must be 1. Thus the answer is <math>1-\frac{1}{2}-\frac{1}{3}=\boxed{\text{(B)}\ \frac16}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems&diff=808811999 AMC 8 Problems2016-10-31T01:18:09Z<p>Debussy: /* Problem 6 */</p>
<hr />
<div>==Problem 1==<br />
<br />
<math>(6?3) + 4 - (2 - 1) = 5.</math> To make this statement true, the question mark between the 6 and the 3 should be replaced by<br />
<br />
<math>\text{(A)} \div \qquad \text{(B)}\ \times \qquad \text{(C)} + \qquad \text{(D)}\ - \qquad \text{(E)}\ \text{None of these}</math><br />
<br />
[[1999 AMC 8 Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
What is the degree measure of the smaller angle formed by the hands of a clock at 10 o'clock?<br />
<br />
<asy><br />
draw(circle((0,0),2));<br />
dot((0,0));<br />
for(int i = 0; i < 12; ++i)<br />
{<br />
dot(2*dir(30*i));<br />
}<br />
<br />
label("$3$",2*dir(0),W);<br />
label("$2$",2*dir(30),WSW);<br />
label("$1$",2*dir(60),SSW);<br />
label("$12$",2*dir(90),S);<br />
label("$11$",2*dir(120),SSE);<br />
label("$10$",2*dir(150),ESE);<br />
label("$9$",2*dir(180),E);<br />
label("$8$",2*dir(210),ENE);<br />
label("$7$",2*dir(240),NNE);<br />
label("$6$",2*dir(270),N);<br />
label("$5$",2*dir(300),NNW);<br />
label("$4$",2*dir(330),WNW);<br />
</asy><br />
<br />
<math>\text{(A)}\ 30 \qquad \text{(B)}\ 45 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 90</math><br />
<br />
[[1999 AMC 8 Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
<br />
Which triplet of numbers has a sum NOT equal to 1?<br />
<br />
<math>\text{(A)}\ (1/2,1/3,1/6) \qquad \text{(B)}\ (2,-2,1) \qquad \text{(C)}\ (0.1,0.3,0.6) \qquad \text{(D)}\ (1.1,-2.1,1.0) \qquad \text{(E)}\ (-3/2,-5/2,5)</math><br />
<br />
[[1999 AMC 8 Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
<br />
The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn?<br />
<br />
<asy><br />
for (int a = 0; a < 6; ++a)<br />
{<br />
for (int b = 0; b < 6; ++b)<br />
{<br />
dot((4*a,3*b));<br />
}<br />
}<br />
draw((0,0)--(20,0)--(20,15)--(0,15)--cycle);<br />
draw((0,0)--(16,12));<br />
draw((0,0)--(16,9));<br />
<br />
label(rotate(30)*"Bjorn",(12,6.75),SE);<br />
label(rotate(37)*"Alberto",(11,8.25),NW);<br />
<br />
label("$0$",(0,0),S);<br />
label("$1$",(4,0),S);<br />
label("$2$",(8,0),S);<br />
label("$3$",(12,0),S);<br />
label("$4$",(16,0),S);<br />
label("$5$",(20,0),S);<br />
label("$0$",(0,0),W);<br />
label("$15$",(0,3),W);<br />
label("$30$",(0,6),W);<br />
label("$45$",(0,9),W);<br />
label("$60$",(0,12),W);<br />
label("$75$",(0,15),W);<br />
<br />
label("H",(6,-2),S);<br />
label("O",(8,-2),S);<br />
label("U",(10,-2),S);<br />
label("R",(12,-2),S);<br />
label("S",(14,-2),S);<br />
<br />
label("M",(-4,11),N);<br />
label("I",(-4,9),N);<br />
label("L",(-4,7),N);<br />
label("E",(-4,5),N);<br />
label("S",(-4,3),N);<br />
</asy><br />
<br />
<math>\text{(A)}\ 15 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 25 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35</math><br />
<br />
[[1999 AMC 8 Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
A rectangular garden 50 feet long and 10 feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden?<br />
<br />
<math>\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500</math><br />
<br />
[[1999 AMC 8 Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
<br />
Bo, Coe, Flo, Jo, and Moe have different amounts of money. Neither Jo nor Bo has as much money as Flo. Both Bo and Coe have more than Moe. Jo has more than Moe, but less than Bo. Who has the least amount of money?<br />
<br />
<math>\text{(A)}\ \text{Bo} \qquad \text{(B)}\ \text{Coe} \qquad \text{(C)}\ \text{Flo} \qquad \text{(D)}\ \text{Joe} \qquad \text{(E)}\ \text{Moe}</math><br />
<br />
[[1999 AMC 8 Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
<br />
The third exit on a highway is located at milepost 40 and the tenth exit is at milepost 160. There is a service center on the highway located three-fourths of the way from the third exit to the tenth exit. At what milepost would you expect to find this service center?<br />
<br />
<math>\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 110 \qquad \text{(D)}\ 120 \qquad \text{(E)}\ 130</math><br />
<br />
[[1999 AMC 8 Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
<br />
Six squares are colored, front and back, (R = red, B = blue, O = orange, Y = yellow, G = green, and W = white). They are hinged together as shown, then folded to form a cube. The face opposite the white face is<br />
<br />
<asy><br />
draw((0,2)--(1,2)--(1,1)--(2,1)--(2,0)--(3,0)--(3,1)--(4,1)--(4,2)--(2,2)--(2,3)--(0,3)--cycle);<br />
draw((1,3)--(1,2)--(2,2)--(2,1)--(3,1)--(3,2));<br />
label("R",(.5,2.3),N);<br />
label("B",(1.5,2.3),N);<br />
label("G",(1.5,1.3),N);<br />
label("Y",(2.5,1.3),N);<br />
label("W",(2.5,.3),N);<br />
label("O",(3.5,1.3),N);<br />
</asy><br />
<br />
<math>\text{(A)}\ \text{B} \qquad \text{(B)}\ \text{G} \qquad \text{(C)}\ \text{O} \qquad \text{(D)}\ \text{R} \qquad \text{(E)}\ \text{Y}</math><br />
<br />
[[1999 AMC 8 Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
Three flower beds overlap as shown. Bed A has 500 plants, bed B has 450 plants, and bed C has 350 plants. Beds A and B share 50 plants, while beds A and C share 100. The total number of plants is<br />
<br />
<asy><br />
draw((0,0)--(3,0)--(3,1)--(0,1)--cycle);<br />
draw(circle((.3,-.1),.7));<br />
draw(circle((2.8,-.2),.8));<br />
label("A",(1.3,.5),N);<br />
label("B",(3.1,-.2),S);<br />
label("C",(.6,-.2),S);<br />
</asy><br />
<br />
<math>\text{(A)}\ 850 \qquad \text{(B)}\ 1000 \qquad \text{(C)}\ 1150 \qquad \text{(D)}\ 1300 \qquad \text{(E)}\ 1450</math><br />
<br />
[[1999 AMC 8 Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
A complete cycle of a traffic light takes 60 seconds. During each cycle the light is green for 25 seconds, yellow for 5 seconds, and red for 30 seconds. At a randomly chosen time, what is the probability that the light will NOT be green?<br />
<br />
<math>\text{(A)}\ \frac{1}{4} \qquad \text{(B)}\ \frac{1}{3} \qquad \text{(C)}\ \frac{5}{12} \qquad \text{(D)}\ \frac{1}{2} \qquad \text{(E)}\ \frac{7}{12}</math><br />
<br />
[[1999 AMC 8 Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
<br />
Each of the five numbers 1, 4, 7, 10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is<br />
<br />
<asy><br />
draw((0,0)--(3,0)--(3,1)--(0,1)--cycle);<br />
draw((1,-1)--(2,-1)--(2,2)--(1,2)--cycle);<br />
</asy><br />
<br />
<math>\text{(A)}\ 20 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 30</math><br />
<br />
[[1999 AMC 8 Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is <math>11/4</math>. To the nearest whole percent, what percent of its games did the team lose?<br />
<br />
<math>\text{(A)}\ 24 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 45 \qquad \text{(E)}\ 73</math><br />
<br />
[[1999 AMC 8 Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
The average age of the 40 members of a computer science camp is 17 years. There are 20 girls, 15 boys, and 5 adults. If the average age of the girls is 15 and the average age of the boys is 16, what is the average age of the adults?<br />
<br />
<math>\text{(A)}\ 26 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 29 \qquad \text{(E)}\ 30</math><br />
<br />
[[1999 AMC 8 Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
In trapezoid <math>ABCD</math>, the sides <math>AB</math> and <math>CD</math> are equal. The perimeter of <math>ABCD</math> is<br />
<br />
<asy><br />
draw((0,0)--(4,3)--(12,3)--(16,0)--cycle);<br />
draw((4,3)--(4,0),dashed);<br />
draw((3.2,0)--(3.2,.8)--(4,.8));<br />
<br />
label("$A$",(0,0),SW);<br />
label("$B$",(4,3),NW);<br />
label("$C$",(12,3),NE);<br />
label("$D$",(16,0),SE);<br />
label("$8$",(8,3),N);<br />
label("$16$",(8,0),S);<br />
label("$3$",(4,1.5),E);<br />
</asy><br />
<br />
<math>\text{(A)}\ 27 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 34 \qquad \text{(E)}\ 48</math><br />
<br />
[[1999 AMC 8 Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
<br />
Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}. <br />
<br />
When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters?<br />
<br />
<math>\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60</math><br />
<br />
[[1999 AMC 8 Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
Tori's mathematics test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry problems. Although she answered 70% of the arithmetic, 40% of the algebra, and 60% of the geometry problems correctly, she did not pass the test because she got less than 60% of the problems right. How many more problems would she have needed to answer correctly to earn a 60% passing grade?<br />
<br />
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 11</math><br />
<br />
[[1999 AMC 8 Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
Problems 17, 18, and 19 refer to the following:<br />
<br />
At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems 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 these 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 make only full recipes, not partial recipes.<br />
<br />
Walter can buy eggs by the half-dozen. How many half-dozens should he buy to make enough cookies? (Some eggs and some cookies may be left over.)<br />
<br />
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 15</math><br />
<br />
[[1999 AMC 8 Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
<br />
Problems 17, 18, and 19 refer to the following:<br />
<br />
At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems 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 these 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 make only full recipes, not partial recipes.<br />
<br />
They learn that a big concert is scheduled for the same night and attendance will be down 25%. How many recipes of cookies should they make for their smaller party?<br />
<br />
<math>\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11</math><br />
<br />
[[1999 AMC 8 Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
Problems 17, 18, and 19 refer to the following:<br />
<br />
At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems 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 these 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 make only 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 />
[[1999 AMC 8 Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
Figure 1 is called a "stack map." The numbers tell how many cubes are stacked in each position. Fig. 2 shows these cubes, and Fig. 3 shows the view of the stacked cubes as seen from the front.<br />
<br />
Which of the following is the front view for the stack map in Fig. 4?<br />
<br />
<asy><br />
unitsize(24);<br />
<br />
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);<br />
draw((1,0)--(1,2));<br />
draw((0,1)--(2,1));<br />
<br />
draw((5,0)--(7,0)--(7,1)--(20/3,4/3)--(20/3,13/3)--(19/3,14/3)--(16/3,14/3)--(16/3,11/3)--(13/3,11/3)--(13/3,2/3)--cycle);<br />
draw((20/3,13/3)--(17/3,13/3)--(17/3,10/3)--(14/3,10/3)--(14/3,1/3));<br />
draw((20/3,10/3)--(17/3,10/3)--(17/3,7/3)--(20/3,7/3));<br />
draw((17/3,7/3)--(14/3,7/3));<br />
draw((7,1)--(6,1)--(6,2)--(5,2)--(5,0));<br />
draw((5,1)--(6,1)--(6,0));<br />
draw((20/3,4/3)--(6,4/3));<br />
draw((17/3,13/3)--(16/3,14/3));<br />
draw((17/3,10/3)--(16/3,11/3));<br />
draw((14/3,10/3)--(13/3,11/3));<br />
draw((5,2)--(13/3,8/3));<br />
draw((5,1)--(13/3,5/3));<br />
draw((6,2)--(17/3,7/3));<br />
<br />
draw((9,0)--(11,0)--(11,4)--(10,4)--(10,3)--(9,3)--cycle);<br />
draw((11,3)--(10,3)--(10,0));<br />
draw((11,2)--(9,2));<br />
draw((11,1)--(9,1));<br />
<br />
draw((13,0)--(16,0)--(16,2)--(13,2)--cycle);<br />
draw((13,1)--(16,1));<br />
draw((14,0)--(14,2));<br />
draw((15,0)--(15,2));<br />
<br />
label("Figure 1",(1,0),S);<br />
label("Figure 2",(17/3,0),S);<br />
label("Figure 3",(10,0),S);<br />
label("Figure 4",(14.5,0),S);<br />
<br />
label("$1$",(1.5,.2),N);<br />
label("$2$",(.5,.2),N);<br />
label("$3$",(.5,1.2),N);<br />
label("$4$",(1.5,1.2),N);<br />
<br />
label("$1$",(13.5,.2),N);<br />
label("$3$",(14.5,.2),N);<br />
label("$1$",(15.5,.2),N);<br />
label("$2$",(13.5,1.2),N);<br />
label("$2$",(14.5,1.2),N);<br />
label("$4$",(15.5,1.2),N);<br />
</asy><br />
<br />
<br /> <br /><br />
<br />
<asy><br />
unitsize(18);<br />
draw((0,0)--(3,0)--(3,2)--(1,2)--(1,4)--(0,4)--cycle);<br />
draw((0,3)--(1,3));<br />
draw((0,2)--(1,2)--(1,0));<br />
draw((0,1)--(3,1));<br />
draw((2,0)--(2,2));<br />
<br />
draw((5,0)--(8,0)--(8,4)--(7,4)--(7,3)--(6,3)--(6,2)--(5,2)--cycle);<br />
draw((8,3)--(7,3)--(7,0));<br />
draw((8,2)--(6,2)--(6,0));<br />
draw((8,1)--(5,1));<br />
<br />
draw((10,0)--(12,0)--(12,4)--(11,4)--(11,3)--(10,3)--cycle);<br />
draw((12,3)--(11,3)--(11,0));<br />
draw((12,2)--(10,2));<br />
draw((12,1)--(10,1));<br />
<br />
draw((14,0)--(17,0)--(17,4)--(16,4)--(16,2)--(14,2)--cycle);<br />
draw((17,3)--(16,3));<br />
draw((17,2)--(16,2)--(16,0));<br />
draw((17,1)--(14,1));<br />
draw((15,0)--(15,2));<br />
<br />
draw((19,0)--(22,0)--(22,4)--(20,4)--(20,1)--(19,1)--cycle);<br />
draw((22,3)--(20,3));<br />
draw((22,2)--(20,2));<br />
draw((22,1)--(20,1)--(20,0));<br />
draw((21,0)--(21,4));<br />
<br />
label("(A)",(1.5,0),S);<br />
label("(B)",(6.5,0),S);<br />
label("(C)",(11,0),S);<br />
label("(D)",(15.5,0),S);<br />
label("(E)",(20.5,0),S);<br />
</asy><br />
<br />
[[1999 AMC 8 Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
The degree measure of angle <math>A</math> is<br />
<br />
<asy><br />
unitsize(12);<br />
draw((0,0)--(20,0)--(1,-10)--(9,5)--(18,-8)--cycle);<br />
draw(arc((1,-10),(1+19/sqrt(461),-10+10/sqrt(461)),(25/17,-155/17),CCW));<br />
draw(arc((19/3,0),(19/3-8/17,-15/17),(22/3,0),CCW));<br />
draw(arc((900/83,-400/83),(900/83+19/sqrt(461),-400/83+10/sqrt(461)),(900/83 - 9/sqrt(97),-400/83 + 4/sqrt(97)),CCW));<br />
label(rotate(30)*"$40^\circ$",(2,-8.9),ENE);<br />
label("$100^\circ$",(21/3,-2/3),SE);<br />
label("$110^\circ$",(900/83,-317/83),NNW);<br />
label("$A$",(0,0),NW);<br />
</asy><br />
<br />
<math>\text{(A)}\ 20 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 45</math><br />
<br />
[[1999 AMC 8 Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
In a far-off land three fish can be traded for two loaves of bread and a loaf of bread can be traded for four bags of rice. How many bags of rice is one fish worth?<br />
<br />
<math>\text{(A)}\ \frac{3}{8} \qquad \text{(B)}\ \frac{1}{2} \qquad \text{(C)}\ \frac{3}{4} \qquad \text{(D)}\ 2\frac{2}{3} \qquad \text{(E)}\ 3\frac{1}{3}</math><br />
<br />
[[1999 AMC 8 Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
Square <math>ABCD</math> has sides of length 3. Segments <math>CM</math> and <math>CN</math> divide the square's area into three equal parts. How long is segment <math>CM</math>?<br />
<br />
<asy><br />
pair A,B,C,D,M,N;<br />
A = (0,0);<br />
B = (0,3);<br />
C = (3,3);<br />
D = (3,0);<br />
M = (0,1);<br />
N = (1,0);<br />
draw(A--B--C--D--cycle);<br />
draw(M--C--N);<br />
label("$A$",A,SW);<br />
label("$M$",M,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,SE);<br />
label("$N$",N,S);<br />
</asy><br />
<br />
<math>\text{(A)}\ \sqrt{10} \qquad \text{(B)}\ \sqrt{12} \qquad \text{(C)}\ \sqrt{13} \qquad \text{(D)}\ \sqrt{14} \qquad \text{(E)}\ \sqrt{15}</math><br />
<br />
[[1999 AMC 8 Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
When <math>1999^{2000}</math> is divided by <math>5</math>, the remainder is <br />
<br />
<math>\text{(A)}\ 4 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 0</math><br />
<br />
[[1999 AMC 8 Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
<br />
Points <math>B</math>, <math>D</math>, and <math>J</math> are midpoints of the sides of right triangle <math>ACG</math>. Points <math>K</math>, <math>E</math>, <math>I</math> are midpoints of the sides of triangle <math>JDG</math>, etc. If the dividing and shading process is done 100 times (the first three are shown) and <math>AC=CG=6</math>, then the total area of the shaded triangles is nearest<br />
<br />
<asy><br />
draw((0,0)--(6,0)--(6,6)--cycle);<br />
draw((3,0)--(3,3)--(6,3));<br />
draw((4.5,3)--(4.5,4.5)--(6,4.5));<br />
draw((5.25,4.5)--(5.25,5.25)--(6,5.25));<br />
fill((3,0)--(6,0)--(6,3)--cycle,black);<br />
fill((4.5,3)--(6,3)--(6,4.5)--cycle,black);<br />
fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black);<br />
<br />
label("$A$",(0,0),SW);<br />
label("$B$",(3,0),S);<br />
label("$C$",(6,0),SE);<br />
label("$D$",(6,3),E);<br />
label("$E$",(6,4.5),E);<br />
label("$F$",(6,5.25),E);<br />
label("$G$",(6,6),NE);<br />
label("$H$",(5.25,5.25),NW);<br />
label("$I$",(4.5,4.5),NW);<br />
label("$J$",(3,3),NW);<br />
label("$K$",(4.5,3),S);<br />
label("$L$",(5.25,4.5),S);<br />
</asy><br />
<br />
<math>\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10</math><br />
<br />
[[1999 AMC 8 Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC8 box|year=1999|before=[[1998 AJHSME Problems|1998 AJHSME]]|after=[[2000 AMC 8 Problems|2000 AMC 8]]}}<br />
* [[AMC 8]]<br />
* [[AMC 8 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
<br />
<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_6&diff=808801999 AMC 8 Problems/Problem 62016-10-31T01:17:11Z<p>Debussy: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
Bo, Coe, Flo, Jo, and Moe have different amounts of money. Neither Jo nor Bo has as much money as Flo. Both Bo and Coe have more than Moe. Jo has more than Moe, but less than Bo. Who has the least amount of money? <br />
<br />
<math> \text{(A)}\ \text{Bo}\qquad\text{(B)}\ \text{Coe}\qquad\text{(C)}\ \text{Flo}\qquad\text{(D)}\ \text{Jo}\qquad\text{(E)}\ \text{Moe} </math><br />
<br />
==Solution==<br />
<br />
Use logic to solve this problem. You don't actually need to use any equations.<br />
<br />
Neither Jo nor Bo has as much money as Flo. So Flo clearly does not have the least amount of money. Rule out Flo.<br />
<br />
Both Bo and Coe have more than Moe. Rule out Bo and Coe; they clearly do not have the least amount of money.<br />
<br />
Jo has more than Moe. Rule out Jo. <br />
<br />
The only person who has not been ruled out is Moe. So <math>\boxed{\text{(E)}}</math> is the answer.<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=1999|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=Pick%27s_Theorem&diff=80662Pick's Theorem2016-10-17T01:08:56Z<p>Debussy: /* Proof */</p>
<hr />
<div>'''Pick's Theorem''' expresses the [[area]] of a [[polygon]], all of whose [[vertex | vertices]] are [[lattice points]] in a [[coordinate plane]], in terms of the number of lattice points inside the polygon and the number of lattice points on the sides of the polygon. The formula is:<br />
<br />
<math>A = I + \frac{B}{2} - 1</math><br />
<br />
where <math>I</math> is the number of lattice points in the interior and <math>B</math> being the number of lattice points on the boundary.<br />
It is similar to the [[Shoelace Theorem]], and although it is less powerful, it is a good tool to have in solving problems.<br />
<br />
<asy><br />
size(150);<br />
defaultpen(linewidth(0.8));<br />
for (int i = 1; i <= 5; i=i+1) {<br />
for (int j = 1; j <= 5; j=j+1) {<br />
dot((i,j));<br />
}<br />
}<br />
draw((1,1)--(1,3)--(3,4)--(2,5)--(5,5)--(2,2)--(4,1)--cycle);</asy><br />
<br />
== Proof ==<br />
{Outline: Show that any triangle on the lattice points with no point in its interior or on its edges has an area of <math>\frac{1}{2}</math>. Then triangulate the shape and apply Euler's Polyhedron formula for graphs to obtain the desired result.}<br />
<br />
== Usage ==<br />
<br />
[[Category:Geometry]]<br />
<br />
[[Category:Theorems]]</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=Pick%27s_Theorem&diff=80661Pick's Theorem2016-10-17T01:08:43Z<p>Debussy: /* Proof */</p>
<hr />
<div>'''Pick's Theorem''' expresses the [[area]] of a [[polygon]], all of whose [[vertex | vertices]] are [[lattice points]] in a [[coordinate plane]], in terms of the number of lattice points inside the polygon and the number of lattice points on the sides of the polygon. The formula is:<br />
<br />
<math>A = I + \frac{B}{2} - 1</math><br />
<br />
where <math>I</math> is the number of lattice points in the interior and <math>B</math> being the number of lattice points on the boundary.<br />
It is similar to the [[Shoelace Theorem]], and although it is less powerful, it is a good tool to have in solving problems.<br />
<br />
<asy><br />
size(150);<br />
defaultpen(linewidth(0.8));<br />
for (int i = 1; i <= 5; i=i+1) {<br />
for (int j = 1; j <= 5; j=j+1) {<br />
dot((i,j));<br />
}<br />
}<br />
draw((1,1)--(1,3)--(3,4)--(2,5)--(5,5)--(2,2)--(4,1)--cycle);</asy><br />
<br />
== Proof ==<br />
{Outline: show that any triangle on the lattice points with no point in its interior or on its edges has an area of <math>\frac{1}{2}</math>. Then triangulate the shape and apply Euler's Polyhedron formula for graphs to obtain the desired result.}<br />
<br />
== Usage ==<br />
<br />
[[Category:Geometry]]<br />
<br />
[[Category:Theorems]]</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=Pick%27s_Theorem&diff=80660Pick's Theorem2016-10-17T01:08:12Z<p>Debussy: </p>
<hr />
<div>'''Pick's Theorem''' expresses the [[area]] of a [[polygon]], all of whose [[vertex | vertices]] are [[lattice points]] in a [[coordinate plane]], in terms of the number of lattice points inside the polygon and the number of lattice points on the sides of the polygon. The formula is:<br />
<br />
<math>A = I + \frac{B}{2} - 1</math><br />
<br />
where <math>I</math> is the number of lattice points in the interior and <math>B</math> being the number of lattice points on the boundary.<br />
It is similar to the [[Shoelace Theorem]], and although it is less powerful, it is a good tool to have in solving problems.<br />
<br />
<asy><br />
size(150);<br />
defaultpen(linewidth(0.8));<br />
for (int i = 1; i <= 5; i=i+1) {<br />
for (int j = 1; j <= 5; j=j+1) {<br />
dot((i,j));<br />
}<br />
}<br />
draw((1,1)--(1,3)--(3,4)--(2,5)--(5,5)--(2,2)--(4,1)--cycle);</asy><br />
<br />
== Proof ==<br />
{Outline: show that any triangle on the lattice with no point in its interior or on its edges has an area of <math>\frac{1}{2}</math>. Then triangulate the shape and apply Euler's Polyhedron formula for graphs to obtain the desired result.}<br />
<br />
== Usage ==<br />
<br />
[[Category:Geometry]]<br />
<br />
[[Category:Theorems]]</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_10&diff=806592015 AMC 8 Problems/Problem 102016-10-17T00:43:09Z<p>Debussy: /* Solution 1 */</p>
<hr />
<div>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 />
==Solution 1==<br />
The question can be rephrased to "How many four-digit positive integers have four distinct digits?",since numbers between <math>1000</math> and <math>9999</math> are four-digit integers. There are <math>9</math> choices for the first number, since it cannot be <math>0</math>, There are only <math>9</math> choices left for the second number since it must differ from the first, <math>8</math> choices for the third number, since it must differ from the first two, and <math>7</math> choices for the fourth number, since it must differ from all three. This means there are <math>9 \times 9 \times 8 \times 7=\boxed{\textbf{(B) }4536}</math> integers between <math>1000</math> and <math>9999</math> with four distinct digits.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2015|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_2&diff=805592014 AMC 8 Problems/Problem 22016-10-09T16:11:42Z<p>Debussy: </p>
<hr />
<div>==Problem==<br />
Paul owes Paula <math>35</math> cents and has a pocket full of <math>5</math>-cent coins, <math>10</math>-cent coins, and <math>25</math>-cent coins that he can use to pay her. What is the difference between the largest and the smallest number of coins he can use to pay her?<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 />
==Solution==<br />
The fewest amount of coins that can be used is 2 (a quarter and a dime). The greatest amount is 7, if he only uses nickels. Therefore we have <math>7-2=\boxed{\textbf{(E)}~5}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2014|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_19&diff=805482011 AMC 8 Problems/Problem 192016-10-08T17:57:30Z<p>Debussy: /* Solution */</p>
<hr />
<div>==Problem==<br />
How many rectangles are in this figure?<br />
<br />
<asy><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);</asy><br />
<br />
<math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 </math><br />
<br />
==Solution==<br />
The figure can be divided into <math>7</math> sections. The number of rectangles with just one section is <math>3.</math> The number of rectangles with two sections is <math>5.</math> There are none with only three sections. The number of rectangles with four sections is <math>3+5+3=\boxed{\textbf{(D)}\ 11}</math><br />
<br />
==See Also==<br />
{{AMC8 box|year=2011|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems&diff=805472011 AMC 8 Problems2016-10-08T17:49:06Z<p>Debussy: /* Problem 17 */</p>
<hr />
<div>==Problem 1==<br />
Margie bought <math> 3 </math> apples at a cost of <math> 50 </math> cents per apple. She paid with a 5-dollar bill. How much change did Margie recieve?<br />
<br />
<math>\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</math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Karl's rectangular vegetable garden is <math> 20 </math> feet by <math> 45 </math> feet, and Makenna's is <math> 25 </math> feet by <math> 40 </math> feet. Whose garden is larger in area?<br />
<br />
<math>\textbf{(A) }\text{Karl's garden is larger by 100 square feet.}</math><br />
<br />
<math>\textbf{(B) }\text{Karl's garden is larger by 25 square feet.}</math><br />
<br />
<math>\textbf{(C) }\text{The gardens are the same size.}</math> <br />
<br />
<math>\textbf{(D) }\text{Makenna's garden is larger by 25 square feet.}</math><br />
<br />
<math>\textbf{(E) }\text{Makenna's garden is larger by 100 square feet.}</math><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?<br /><br />
<asy><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 />
</asy><br />
<br />
<math> \textbf{(A) }8:17 \qquad\textbf{(B) }25:49 \qquad\textbf{(C) }36:25 \qquad\textbf{(D) }32:17 \qquad\textbf{(E) }36:17</math><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: <cmath>2,0,1,3,0,3,3,1,2.</cmath> Which statement about the mean, median, and mode is true?<br />
<br />
<math>\textbf{(A) }\text{median} < \text{mean} < \text{mode} \qquad \textbf{(B) }\text{mean} < \text{mode} < \text{median} \\ \\ \textbf{(C) }\text{mean} < \text{median} < \text{mode} \qquad \textbf{(D) }\text{median} < \text{mode} < \text{mean} \\ \\ \textbf{(E) }\text{mode} < \text{median} < \text{mean}</math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
What time was it <math>2011</math> minutes after midnight on January 1, 2011?<br />
<br />
<math>\textbf{(A) }\text{January 1 at 9:31 PM}</math><br />
<br />
<math>\textbf{(B) }\text{January 1 at 11:51 PM}</math> <br />
<br />
<math>\textbf{(C) }\text{January 2 at 3:11 AM}</math> <br />
<br />
<math>\textbf{(D) }\text{January 2 at 9:31 AM}</math> <br />
<br />
<math>\textbf{(E) }\text{January 2 at 6:01 PM}</math><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 />
<math> \textbf{(A) }20 \qquad\textbf{(B) }25 \qquad\textbf{(C) }45 \qquad\textbf{(D) }306 \qquad\textbf{(E) }351</math><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 />
<asy><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);</asy><br />
<br />
<math> \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} </math><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 />
<math> \textbf{(A) }4 \qquad\textbf{(B) }5 \qquad\textbf{(C) }6 \qquad\textbf{(D) }7 \qquad\textbf{(E) }9 </math><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 />
<asy><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("$1$",(0.95,-0.24),SE*lsf); label("$2$",(1.92,-0.26),SE*lsf); label("$3$",(2.92,-0.31),SE*lsf); label("$4$",(3.93,-0.26),SE*lsf); label("$5$",(4.92,-0.27),SE*lsf); label("$6$",(5.95,-0.29),SE*lsf); label("$7$",(6.94,-0.27),SE*lsf); label("$5$",(-0.49,1.22),SE*lsf); label("$10$",(-0.59,2.23),SE*lsf); label("$15$",(-0.61,3.22),SE*lsf); label("$20$",(-0.61,4.23),SE*lsf); label("$25$",(-0.59,5.22),SE*lsf); label("$30$",(-0.59,6.2),SE*lsf); label("$35$",(-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("HOURS",(3.41,-0.85),SE*lsf); label("M",(-1.39,5.32),SE*lsf); label("I",(-1.34,4.93),SE*lsf); label("L",(-1.36,4.51),SE*lsf); label("E",(-1.37,4.11),SE*lsf); label("S",(-1.39,3.7),SE*lsf); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<br />
<math> \textbf{(A)}2\qquad\textbf{(B)}2.5\qquad\textbf{(C)}4\qquad\textbf{(D)}4.5\qquad\textbf{(E)}5 </math><br />
<br />
[[2011 AMC 8 Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
The taxi fare in Gotham City is <nowiki>$2.40</nowiki> for the first <math>\frac12</math> mile and additional mileage charged at the rate <nowiki>$0.20</nowiki> for each additional 0.1 mile. You plan to give the driver a <nowiki>$2</nowiki> tip. How many miles can you ride for <nowiki>$10</nowiki>?<br />
<br />
<math> \textbf{(A) } 3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75 </math><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 />
<asy><br />
size(300);<br />
real i;<br />
defaultpen(linewidth(0.8));<br />
draw((0,140)--origin--(220,0));<br />
for(i=1;i<13;i=i+1) {<br />
draw((0,10*i)--(220,10*i));<br />
}<br />
label("$0$",origin,W);<br />
label("$20$",(0,20),W);<br />
label("$40$",(0,40),W);<br />
label("$60$",(0,60),W);<br />
label("$80$",(0,80),W);<br />
label("$100$",(0,100),W);<br />
label("$120$",(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("M",(30,-5),S);<br />
label("Tu",(70,-5),S);<br />
label("W",(110,-5),S);<br />
label("Th",(150,-5),S);<br />
label("F",(190,-5),S);<br />
label("M",(-25,85),W);<br />
label("I",(-27,75),W);<br />
label("N",(-25,65),W);<br />
label("U",(-25,55),W);<br />
label("T",(-25,45),W);<br />
label("E",(-25,35),W);<br />
label("S",(-26,25),W);</asy><br />
<br />
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12 </math><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 />
<math> \textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
Two congruent squares, <math>ABCD</math> and <math>PQRS</math>, have side length <math>15</math>. They overlap to form the <math>15</math> by <math>25</math> rectangle <math>AQRD</math> shown. What percent of the area of rectangle <math>AQRD</math> is shaded? <br />
<asy><br />
filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black);<br />
label("D",(0,0),S);<br />
label("R",(25,0),S);<br />
label("Q",(25,15),N);<br />
label("A",(0,15),N);<br />
filldraw((10,0)--(15,0)--(15,15)--(10,15)--cycle,mediumgrey,black);<br />
label("S",(10,0),S);<br />
label("C",(15,0),S);<br />
label("B",(15,15),N);<br />
label("P",(10,15),N);</asy><br />
<br />
<math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 25 </math><br />
<br />
[[2011 AMC 8 Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
There are <math>270</math> students at Colfax Middle School, where the ratio of boys to girls is <math>5 : 4</math>. There are <math>180</math> students at Winthrop Middle School, where the ratio of boys to girls is <math>4 : 5</math>. 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 />
<math> \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} </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
<br />
How many digits are in the product <math>4^5 \cdot 5^{10}</math>?<br />
<br />
<math> \textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 12 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
Let <math>A</math> be the area of the triangle with sides of length <math>25, 25</math>, and <math>30</math>. Let <math>B</math> be the area of the triangle with sides of length <math>25, 25,</math> and <math>40</math>. What is the relationship between <math>A</math> and <math>B</math>?<br />
<br />
<math> \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 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
Let <math>w</math>, <math>x</math>, <math>y</math>, and <math>z</math> be whole numbers. If <math>2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588</math>, then what does <math>2w + 3x + 5y + 7z</math> equal?<br />
<br />
<math> \textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56 </math><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 />
<math> \textbf{(A) }\dfrac16\qquad\textbf{(B) }\dfrac5{12}\qquad\textbf{(C) }\dfrac12\qquad\textbf{(D) }\dfrac7{12}\qquad\textbf{(E) }\dfrac56 </math><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 />
<asy><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);</asy><br />
<br />
<math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
Quadrilateral <math>ABCD</math> is a trapezoid, <math>AD = 15</math>, <math>AB = 50</math>, <math>BC = 20</math>, and the altitude is <math>12</math>. What is the area of the trapezoid?<br />
<br />
<asy><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("A",A,N);<br />
label("B",B,N);<br />
label("C",C,S);<br />
label("D",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("12",(21,10),E);<br />
label("50",(19,22),N);<br />
label("15",(1,10),W);<br />
label("20",(41,12),E);</asy><br />
<br />
<math> \textbf{(A) }600\qquad\textbf{(B) }650\qquad\textbf{(C) }700\qquad\textbf{(D) }750\qquad\textbf{(E) }800 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
Students guess that Norb's age is <math>24, 28, 30, 32, 36, 38, 41, 44, 47</math>, and <math>49</math>. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?<br />
<br />
<math> \textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
What is the '''tens''' digit of <math>7^{2011}</math>?<br />
<br />
<math> \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7 </math><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 />
<math> \textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108 </math><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 />
<math> \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4 </math><br />
<br />
<br />
[[2011 AMC 8 Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A circle with radius <math>1</math> 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 />
<asy><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);</asy><br />
<br />
<math> \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 </math><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>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_8&diff=805462010 AMC 8 Problems/Problem 82016-10-08T16:08:47Z<p>Debussy: </p>
<hr />
<div>==Problem==<br />
As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction <math>1/2</math> mile in front of her. After she passes him, she can see him in her rear mirror until he is <math>1/2</math> mile behind her. Emily rides at a constant rate of <math>12</math> miles per hour, and Emerson skates at a constant rate of <math>8</math> miles per hour. For how many minutes can Emily see Emerson? <br />
<br />
<math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16 </math><br />
<br />
==Solution==<br />
Because they are skating in the same direction, Emily is skating relative to Emerson <math>12-8=4</math> mph. Now we can look at it as if Emerson is not moving at all[on his skateboard] and Emily is riding at <math>4</math> mph. It takes her<br />
<br />
<cmath>\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}</cmath><br />
<br />
to skate the <math>1/2</math> mile to reach him, and then the same amount of time to be <math>1/2</math> mile ahead of him. This totals to<br />
<br />
<cmath>2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{\textbf{(D)}\ 15}\ \text{minutes}</cmath><br />
<br />
Last edited on Nov 11, 2014, 10:23 am.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_8&diff=805452010 AMC 8 Problems/Problem 82016-10-08T16:07:32Z<p>Debussy: </p>
<hr />
<div>==Problem==<br />
As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction <math>1/2</math> mile in front of her. After she passes him, she can see him in her rear mirror until he is <math>1/2</math> mile behind her. Emily rides at a constant rate of <math>12</math> miles per hour, and Emerson skates at a constant rate of <math>8</math> miles per hour. For how many minutes can Emily see Emerson? <br />
<br />
<math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16 </math><br />
<br />
==Solution==<br />
Because they are skating in the same direction, Emily is skating relative to Emerson <math>12-8=4</math> mph. Now we can look at it as if Emerson is not moving at all and Emily is riding at <math>4</math> mph. It takes her<br />
<br />
<cmath>\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}</cmath><br />
<br />
to skate the <math>1/2</math> mile to reach him, and then the same amount of time to be <math>1/2</math> mile ahead of him. This totals to<br />
<br />
<cmath>2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{\textbf{(D)}\ 15}\ \text{minutes}</cmath><br />
<br />
Last edited on Nov 11, 2014, 10:23 am.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Debussyhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_1&diff=801942010 AMC 8 Problems/Problem 12016-09-04T03:21:12Z<p>Debussy: </p>
<hr />
<div>==Problem==<br />
At Euclid Middle School the mathematics teachers are Miss Germain, Mr. Newton, and Mrs. Young. There are <math>11</math> students in Mrs. Germain's class, <math>8</math> students in Mr. Newton's class, and <math>9</math> students in Mrs. Young's class taking the AMC 8 this year. How many mathematics students at Euclid Middle School are taking the contest? <br />
<br />
<math> \textbf{(A)}\ 26 \qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30 </math><br />
<br />
==Solution==<br />
Given that these are the only math teachers at Euclid Middle School and we are told how many from each class are taking the AMC 8, we simply add the three numbers to find the total. <math>11+8+9=\boxed{\textbf{(C)}\ 28}</math><br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Debussy