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AMC8_0 | The Blue Bird High School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?
$\textbf{(A) }2\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }12$
| There are $2! = 2$ ways to order the boys on the ends, and there are $3!=6$ ways to order the girls in the middle. We get the answer to be $2 \cdot 6 = \boxed{\textbf{(E) }12}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_4 | 12 |
AMC8_1 | Initially, a spinner points west. Chenille moves it clockwise $2 \dfrac{1}{4}$ revolutions and then counterclockwise $3 \dfrac{3}{4}$ revolutions. In what direction does the spinner point after the two moves?
[asy]size(96); draw(circle((0,0),1),linewidth(1)); draw((0,0.75)--(0,1.25),linewidth(1)); draw((0,-0.75)--(0,-1.25),linewidth(1)); draw((0.75,0)--(1.25,0),linewidth(1)); draw((-0.75,0)--(-1.25,0),linewidth(1)); label("$N$",(0,1.25), N); label("$W$",(-1.25,0), W); label("$E$",(1.25,0), E); label("$S$",(0,-1.25), S); draw((0,0)--(-0.5,0),EndArrow);[/asy]
$\textbf{(A)}\ \text{north} \qquad \textbf{(B)}\ \text{east} \qquad \textbf{(C)}\ \text{south} \qquad \textbf{(D)}\ \text{west} \qquad \textbf{(E)}\ \text{northwest}$
| If the spinner goes clockwise $2 \dfrac{1}{4}$ revolutions and then counterclockwise $3 \dfrac{3}{4}$ revolutions, it ultimately goes counterclockwise $1 \dfrac{1}{2}$ which brings the spinner pointing $\boxed{\textbf{(B)}\ \text{east}}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_4 | east |
AMC8_3 | A three-digit integer contains one of each of the digits $1$ , $3$ , and $5$ . What is the probability that the integer is divisible by $5$ ?
$\textbf{(A)}\ \frac{1}{6} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{2}{3} \qquad \textbf{(E)}\ \frac{5}{6}$
| The three digit numbers are $135,153,351,315,513,531$ . The numbers that end in $5$ are divisible are $5$ , and the probability of choosing those numbers is $\boxed{\textbf{(B)}\ \frac13}$ .
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_13 | $\frac{1}{3}$ |
AMC8_4 | The operation $\otimes$ is defined for all nonzero numbers by $a\otimes b =\frac{a^{2}}{b}$ . Determine $[(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)]$ .
$\text{(A)}\ -\frac{2}{3}\qquad\text{(B)}\ -\frac{1}{4}\qquad\text{(C)}\ 0\qquad\text{(D)}\ \frac{1}{4}\qquad\text{(E)}\ \frac{2}{3}$
| Follow PE(MD)(AS), doing the innermost parentheses first.
$[(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)]$
$[\frac{1^2}{2}\otimes 3]-[1\otimes \frac{2^2}{3}]$
$[\frac{1}{2}\otimes 3]-[1\otimes \frac{4}{3}]$
$[\frac{(\frac{1}{2})^{2}}{3}]-[\frac{1^2}{(\frac{4}{3})}]$
$[\frac{1}{4} \cdot \frac{1}{3}]-[\frac{3}{4}]$
$\frac{1}{12} - \frac{3}{4}$
$\frac{1}{12} - \frac{9}{12}$
$\frac{-8}{12}$
$-\frac{2}{3}$ , which is answer $\boxed{A}$
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_17 | A |
AMC8_5 | A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?
[asy] path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(3,0)*p); draw(shift(3,1)*p); draw(shift(4,0)*p); draw(shift(4,1)*p); draw(shift(7,0)*p); draw(shift(7,1)*p); draw(shift(7,2)*p); draw(shift(8,0)*p); draw(shift(8,1)*p); draw(shift(8,2)*p); draw(shift(9,0)*p); draw(shift(9,1)*p); draw(shift(9,2)*p);[/asy]
$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$
| Solution 1The first square has a sidelength of $1$ , the second square $2$ , and so on. The seventh square has $7$ and is made of $7^2=49$ unit tiles. The sixth square has $6$ and is made of $6^2=36$ unit tiles. The seventh square has $49-36=\boxed{\text{(C)}\ 13}$ more tiles than the sixth square.
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_11 | 13 |
AMC8_6 | How many different four-digit numbers can be formed by rearranging the four digits in $2004$ ?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 81$
| We can solve this problem easily, just by calculating how many choices there are for each of the four digits.
First off, we know there are only $2$ choices for the first digit, because $0$ isn't a valid choice, or the number would a 3-digit number, which is not what we want.
We have $3$ choices for the second digit, since we already used up one of the digits, and $2$ choices for the third, and finally just $1$ choices for the fourth and final one.
$2*3*2*1$ is 12, but there are 2 zeros that have been counted as different numbers, so divide by 2 to get $\boxed{\textbf{(B)}\ 6}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_2 | 6 |
AMC8_7 | 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?
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$
| 50% off the price of half a pound of fish is $3, so 100%, the regular price, of a half pound of fish is $6. If half a pound of fish costs $6, then a whole pound of fish is $\boxed{\textbf{(D)}\ 12}$ dollars.
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_2 | 12 |
AMC8_8 | What is the sum of the distinct prime integer divisors of $2016$ ?
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$
Solution
| Solution 1The prime factorization is $2016=2^5\times3^2\times7$ . Since the problem is only asking us for the distinct prime factors, we have $2,3,7$ . Their desired sum is then $\boxed{\textbf{(B) }12}$ .
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_9 | 12 |
AMC8_9 | Jamar bought some pencils costing more than a penny each at the school bookstore and paid $\textdollar 1.43$ . Sharona bought some of the same pencils and paid $\textdollar 1.87$ . How many more pencils did Sharona buy than Jamar?
$\textbf{(A)}\hspace{.05in}2\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}4\qquad\textbf{(D)}\hspace{.05in}5\qquad\textbf{(E)}\hspace{.05in}6$
| We assume that the price of the pencils remains constant. Convert $\textdollar 1.43$ and $\textdollar 1.87$ to cents. Since the price of the pencils is more than one penny, we can find the price of one pencil (in cents) by taking the greatest common divisor of $143$ and $187$ , which is $11$ . Therefore, Jamar bought $\frac{143}{11} \implies 13$ pencils and Sharona bought $\frac{187}{11} \implies 17$ pencils. Thus, Sharona bought $17-13 = \boxed{\textbf{(C)}\ 4}$ more pencils than Jamar.
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_13 | 4 |
AMC8_10 | Let $w$ , $x$ , $y$ , and $z$ be whole numbers. If $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$ , then what does $2w + 3x + 5y + 7z$ equal?
$\textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56$
| The prime factorization of $588$ is $2^2\cdot3\cdot7^2.$ We can see $w=2, x=1,$ and $z=2.$ Because $5^0=1, y=0.$
\[2w+3x+5y+7z=4+3+0+14=\boxed{\textbf{(A)}\ 21}\]
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_17 | 21 |
AMC8_11 | Problems 8,9 and 10 use the data found in the accompanying paragraph and table:
| France and Spain are European countries. The number of '80s stamps from France is $15$ and the number of '80s stamps from Spain is $9$ . The total number of stamps is $15+9=\boxed{\text{(D)}\ 24}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_8 | 24 |
AMC8_12 | The degree measure of angle $A$ is
[asy] unitsize(12); draw((0,0)--(20,0)--(1,-10)--(9,5)--(18,-8)--cycle); draw(arc((1,-10),(1+19/sqrt(461),-10+10/sqrt(461)),(25/17,-155/17),CCW)); draw(arc((19/3,0),(19/3-8/17,-15/17),(22/3,0),CCW)); 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)); label(rotate(30)*"$40^\circ$",(2,-8.9),ENE); label("$100^\circ$",(21/3,-2/3),SE); label("$110^\circ$",(900/83,-317/83),NNW); label("$A$",(0,0),NW); [/asy]
$\text{(A)}\ 20 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 45$
| Solution 1Angle-chasing using the small triangles:
Use the line below and to the left of the $110^\circ$ angle to find that the rightmost angle in the small lower-left triangle is $180 - 110 = 70^\circ$ .
Then use the small lower-left triangle to find that the remaining angle in that triangle is $180 - 70 - 40 = 70^\circ$ .
Use congruent vertical angles to find that the lower angle in the smallest triangle containing $A$ is also $70^\circ$ .
Next, use line segment $AB$ to find that the other angle in the smallest triangle containing $A$ is $180 - 100 = 80^\circ$ .
The small triangle containing $A$ has a $70^\circ$ angle and an $80^\circ$ angle. The remaining angle must be $180 - 70 - 80 = \boxed{30^\circ, B}$
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_21 | 30 |
AMC8_13 | A circle of radius $2$ is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?
[asy] size(0,50); draw((-1,1)..(-2,2)..(-3,1)..(-2,0)..cycle); dot((-1,1)); dot((-2,2)); dot((-3,1)); dot((-2,0)); draw((1,0){up}..{left}(0,1)); dot((1,0)); dot((0,1)); draw((0,1){right}..{up}(1,2)); dot((1,2)); draw((1,2){down}..{right}(2,1)); dot((2,1)); draw((2,1){left}..{down}(1,0));[/asy]
$\textbf{(A)}\hspace{.05in}\frac{4-\pi}{\pi}\qquad\textbf{(B)}\hspace{.05in}\frac{1}\pi\qquad\textbf{(C)}\hspace{.05in}\frac{\sqrt2}{\pi}\qquad\textbf{(D)}\hspace{.05in}\frac{\pi-1}{\pi}\qquad\textbf{(E)}\hspace{.05in}\frac{3}\pi$
| [asy] dot((0,0),red); dot((0,2),red); dot((2,0),red); dot((2,2),red); draw((0,0)--(0,2)--(2,2)--(2,0)--cycle,red); size(0,50); draw((1,0){up}..{left}(0,1)); dot((1,0)); dot((0,1)); draw((0,1){right}..{up}(1,2)); dot((1,2)); draw((1,2){down}..{right}(2,1)); dot((2,1)); draw((2,1){left}..{down}(1,0));[/asy]
Draw a square around the star figure. The side length of this square is $4$ , because the side length is the diameter of the circle. The square forms $4$ -quarter circles around the star figure. This is the equivalent of one large circle with radius $2$ , meaning that the total area of the quarter circles is $4\pi$ . The area of the square is $16$ . Thus, the area of the star figure is $16 - 4\pi$ . The area of the circle is $4\pi$ . Taking the ratio of the two areas, we find the answer is $\boxed{\textbf{(A)}\ \frac{4-\pi}{\pi}}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_24 | (A) (4-π)/π |
AMC8_14 | The graph shows the constant rate at which Suzanna rides her bike. If she rides a total of a half an hour at the same speed, how many miles would she have ridden?
[asy] import graph; /* this is a label */ Label f; f.p=fontsize(0); xaxis(-0.9,20,Ticks(f, 5.0, 5.0)); yaxis(-0.9,20, Ticks(f, 22.0,5.0)); // real f(real x) { return x; } draw(graph(f,-1,22),black+linewidth(1)); label("1", (-1,5), black); label("2", (-1, 10), black); label("3", (-1, 15), black); label("4", (-1, 20), black); dot((5,5), black+linewidth(5)); dot((10,10), black+linewidth(5)); dot((15, 15), black+linewidth(5)); dot((20,20), black+linewidth(5)); label("MINUTES", (11,-5), S); label(rotate(90)*"MILES", (-5,11), W);[/asy]
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 5.5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 6.5\qquad\textbf{(E)}\ 7$
| Suzanna's speed is $\frac{1}{5}$ . This means she runs $\frac{1}{5} \cdot 30 = \boxed{ \textbf{(C) }6 }$
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_3 | 6 |
AMC8_15 | Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$
| Let the smallest number be $x.$ It follows that the largest number is $4x.$
Since $x,15,$ and $4x$ are equally spaced on a number line, we have
\begin{align*} 4x-15 &= 15-x \\ 5x &= 30 \\ x &= \boxed{\textbf{(C) } 6}. \end{align*}
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6 | 6 |
AMC8_16 | Connie multiplies a number by $2$ and gets $60$ as her answer. However, she should have divided the number by $2$ to get the correct answer. What is the correct answer?
$\textbf{(A)}\ 7.5\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 240$
| If $x$ is the number, then $2x=60$ and $x=30$ . Dividing the number by $2$ yields $\dfrac{30}{2} = \boxed{\textbf{(B)}\ 15}$ .
A second way to do it is to divide the number by $4$ , as you multiplied by $2$ when you were supposed to divide by $2$ . So, $\dfrac{60}{4} = \boxed{\textbf{(B)}\ 15}$ .
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_1 | 15 |
AMC8_17 | A mixture of $30$ liters of paint is $25\%$ red tint, $30\%$ yellow
tint and $45\%$ water. Five liters of yellow tint are added to
the original mixture. What is the percent of yellow tint
in the new mixture?
$\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 35 \qquad \mathrm{(C)}\ 40 \qquad \mathrm{(D)}\ 45 \qquad \mathrm{(E)}\ 50$
| Since $30\%$ of the original $30$ liters of paint was yellow, and $5$ liters of yellow paint were added to make the new mixture, there are $9+5=14$ liters of yellow tint in the new mixture. Since only $5$ liters of paint were added to the original $30$ , there are a total of $35$ liters of paint in the new mixture. This gives $40\%$ of yellow tint in the new mixture, which is $\boxed{\textbf{(C) 40}}$ .
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_17 | 40 |
AMC8_18 | For any positive integer $n$ , define $\boxed{n}$ to be the sum of the positive factors of $n$ .
For example, $\boxed{6} = 1 + 2 + 3 + 6 = 12$ . Find $\boxed{\boxed{11}}$ .
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 30$
| We have
\begin{align*} \boxed{\boxed{11}}&=\boxed{1+11} \\ &=\boxed{12} \\ &=1+2+3+4+6+12 \\ &=28, \end{align*}
from which the answer is $\boxed{\textbf{(D)}\ 28}.$
~Aplus95 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_10 | 28 |
AMC8_19 | 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?
$\textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34$
| If we designate a person to be on a certain side, then all placements of the other people can be considered unique. WLOG, assign Angie to be on the side. There are then $3!=6$ total seating arrangements. If Carlos is across from Angie, there are only $2!=2$ ways to fill the remaining two seats. Then the probability Angie and Carlos are seated opposite each other is $\frac26=\boxed{\textbf{(B)}\ \frac13}$
.
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_12 | 1/3 |
AMC8_20 | From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]
$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$
| Solution 1Choose side "lengths" $a,b,c$ for the triangle, where "length" is how many vertices of the octagon are skipped between vertices of the triangle, starting from the shortest side, and going clockwise, and choosing $a=b$ if the triangle is isosceles: $a+b+c=5$ , where either [ $a\leq b$ and $a < c$ ] or [ $a=b=c$ (but this is impossible in an octagon)].
Options are: $a=0$ with $b,c$ in { 0,5 ; 1,4 ; 2,3 ; 3,2 ; 4,1 }, and $a=1$ with { 1,3 ; 2,2}. $\frac{5}{7}$ of these have a side with length 1, which corresponds to an edge of the octagon. So, our answer is $\boxed{\textbf{(D) } \frac 57}$
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23 | 5/7 |
AMC8_21 | Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah's party?
[asy]/* AMC8 2003 #18 Problem */ pair a=(102,256), b=(68,131), c=(162,101), d=(134,150); pair e=(269,105), f=(359,104), g=(303,12), h=(579,211); pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501); pair m=(282,411), n=(147,451), o=(103,437), p=(31,373); pair q=(419,175), r=(462,209), s=(477,288), t=(443,358); pair oval=(282,303); draw(l--m--n--cycle); draw(p--oval); draw(o--oval); draw(b--d--oval); draw(c--d--e--oval); draw(e--f--g--h--i--j--oval); draw(k--oval); draw(q--oval); draw(s--oval); draw(r--s--t--oval); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h); dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p); dot(q); dot(r); dot(s); dot(t); filldraw(yscale(.5)*Circle((282,606),80),white,black); label(scale(0.75)*"Sarah", oval);[/asy]
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$
| There are $3$ people who are friends with only each other who won't be invited, plus $1$ person who has no friends, and $2$ people who are friends of friends of friends who won’t be invited. So the answer is $\boxed{\textbf{(D)}\ 6}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_18 | 6 |
AMC8_22 | Alina writes the numbers $1, 2, \dots , 9$ on separate cards, one number per card. She wishes to divide the cards into $3$ groups of $3$ cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$
| First, we need to find the sum of each group when split. This is the total sum of all the elements divided by the # of groups. $1 + 2 \cdots + 9 = \frac{9(10)}{2} = 45$ . Then, dividing by $3$ , we have $\frac{45}{3} = 15$ , so each group of $3$ must have a sum of 15. To make the counting easier, we will just see the possible groups 9 can be with. The possible groups 9 can be with 2 distinct numbers are $(9, 2, 4)$ and $(9, 1, 5)$ . Going down both of these avenues, we will repeat the same process for $8$ using the remaining elements in the list. Where there is only 1 set of elements getting the sum of $7$ , $8$ needs in both cases. After $8$ is decided, the remaining 3 elements are forced in a group, yielding us an answer of $\boxed{\textbf{(C)}\ 2}$ as our sets are $(9, 1, 5) (8, 3, 4) (7, 2, 6)$ and $(9, 2, 4) (8, 1, 6) (7, 3 ,5)$ .
~CHECKMATE2021, apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_21 | 2 |
AMC8_23 | A square with area $4$ is inscribed in a square with area $5$ , with each vertex of the smaller square on a side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length $a$ , and the other of length $b$ . What is the value of $ab$ ?
[asy] draw((0,2)--(2,2)--(2,0)--(0,0)--cycle); draw((0,0.3)--(0.3,2)--(2,1.7)--(1.7,0)--cycle); label("$a$",(-0.1,0.15)); label("$b$",(-0.1,1.15));[/asy]
$\textbf{(A)}\hspace{.05in}\frac{1}5\qquad\textbf{(B)}\hspace{.05in}\frac{2}5\qquad\textbf{(C)}\hspace{.05in}\frac{1}2\qquad\textbf{(D)}\hspace{.05in}1\qquad\textbf{(E)}\hspace{.05in}4$
| The total area of the four congruent triangles formed by the squares is $5-4 = 1$ . Therefore, the area of one of these triangles is $\frac{1}{4}$ . The height of one of these triangles is $a$ and the base is $b$ . Using the formula for area of the triangle, we have $\frac{ab}{2} = \frac{1}{4}$ . Multiply by $2$ on both sides to find that the value of $ab$ is $\boxed{\textbf{(C)}\ \frac{1}2}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_25 | 1/2 |
AMC8_24 | Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by $5$ minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?
$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82$
| It is well known that $\text{Distance}=\text{Speed} \cdot \text{Time}$ . In the question, we want distance. From the question, we have that the time is $60$ minutes or $1$ hour. By the equation derived from $\text{Distance}=\text{Speed} \cdot \text{Time}$ , we have $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$ , so the speed is $1$ mile per $x$ minutes. Because we want the distance, we multiply the time and speed together yielding $60\text{ mins}\cdot \frac{1\text{ mile}}{x\text{ mins}}$ . The minutes cancel out, so now we have $\dfrac{60}{x}$ as our distance for the first day. The distance for the following days are:
\[\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}.\]
We know that $x,x+5,x+10,x+15$ are all factors of $60$ , therefore, $x=5$ because the factors have to be in an arithmetic sequence with the common difference being $5$ and $x=5$ is the only solution.
\[\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=12+6+4+3=\boxed{\textbf{(C)}\ 25}.\]
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_23 | 25 |
AMC8_25 | In Theresa's first $8$ basketball games, she scored $7, 4, 3, 6, 8, 3, 1$ and $5$ points. In her ninth game, she scored fewer than $10$ points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than $10$ points and her points-per-game average for the $10$ games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?
$\textbf{(A)}\ 35\qquad \textbf{(B)}\ 40\qquad \textbf{(C)}\ 48\qquad \textbf{(D)}\ 56\qquad \textbf{(E)}\ 72$
| The total number of points from the first $8$ games is $7+4+3+6+8+3+1+5=37$ . We have to make this a multiple of $9$ by scoring less than $10$ points. The smallest multiple of $9$ that is greater than $37$ is $45$ . $45-37=8$ . Now we have to add a number to get a multiple of 10. The next multiple of $10$ is $50$ , which means that we add $5$ . Multiplying these together, you get $8\cdot5 =$ $\boxed{\textbf{(B)}\ 40}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_15 | 40 |
AMC8_26 | Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$
| For each person to have at least one pencil, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use Ball-and-urn to find the number of possibilities is $\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}$ .
Solution by phoenixfire
Minor Edit by Yuvag
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_17 | 10 |
AMC8_27 | Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz Bunny start on the ground, make a sequence of $6$ hops, and end up back on the ground?
(For example, one sequence of hops is up-up-down-down-up-down.)
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 12$
| Looking at the answer choices, you see that you can list them out.
Doing this gets you:
$\mathit{UUDDUD}$
$\mathit{UDUDUD}$
$\mathit{UUUDDD}$
$\mathit{UDUUDD}$
$\mathit{UUDUDD}$
Counting all the paths listed above gets you $\boxed{\textbf{(B)} \ 5}$ .
~ALWAYSRIGHT11
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_13 | 5 |
AMC8_28 | Problems 8,9 and 10 use the data found in the accompanying paragraph and table:
| The price of all the stamps in the '70s together over the total number of stamps is equal to the average price.
\[\frac{(12)(0.06)+(12)(0.06)+(6)(0.04)+(13)(0.05)}{12+12+6+13}\\ = \frac{0.72+0.72+0.24+0.65}{43}\\ = \frac{2.33}{43} \approx \boxed{\text{(E)}\ 5.5\ \text{cents}}\]
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_10 | 5.5 cents |
AMC8_29 | Problems 8, 9 and 10 use the data found in the accompanying paragraph and figures
Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.
$\circ$ Art's cookies are trapezoids.
[asy] size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(5,0)--(5,3)--(2,3)--cycle); draw(rightanglemark((5,3), (5,0), origin)); label("5 in", (2.5,0), S); label("3 in", (5,1.5), E); label("3 in", (3.5,3), N); [/asy]
$\circ$ Roger's cookies are rectangles.
[asy] size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(4,0)--(4,2)--(0,2)--cycle); draw(rightanglemark((4,2), (4,0), origin)); draw(rightanglemark((0,2), origin, (4,0))); label("4 in", (2,0), S); label("2 in", (4,1), E); [/asy]
$\circ$ Paul's cookies are parallelograms.
[asy] size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle); draw((2.5,2)--(2.5,0), dashed); draw(rightanglemark((2.5,2),(2.5,0), origin)); label("3 in", (1.5,0), S); label("2 in", (2.5,1), W); [/asy]
$\circ$ Trisha's cookies are triangles.
[asy] size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(3,4)--cycle); draw(rightanglemark((3,4),(3,0), origin)); label("3 in", (1.5,0), S); label("4 in", (3,2), E); [/asy]
Each friend uses the same amount of dough, and Art makes exactly 12 cookies. Who gets the fewest cookies from one batch of cookie dough?
$\textbf{(A)}\ \text{Art}\qquad\textbf{(B)}\ \text{Roger}\qquad\textbf{(C)}\ \text{Paul}\qquad\textbf{(D)}\ \text{Trisha}\qquad\textbf{(E)}\ \text{There is a tie for fewest.}$
| Out of all the cookies, Art's has an area of $12 \text{ in}^2$ , which was the greatest area out of all the cookies' areas. Roger's cookie had an area of $8 \text{ in} ^2$ , and both Paul and Trisha's cookies had an area of $6 \text{ in}^2$ . This means $\boxed{\textbf{(A)}\ \text{Art}}$ makes the fewest cookies, since his cookie area is the greatest. The answer is not that there is a tie between Paul and Trisha because they can make the most cookies with a given amount of cookie dough, not the least.
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_8 | Art |
AMC8_30 | Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have?
[asy] defaultpen(linewidth(0.6)); pair O=origin, A=(0,1), B=A+1*dir(60), C=(1,1), D=(1,0), E=D+1*dir(-72), F=E+1*dir(-144), G=O+1*dir(-108); draw(O--A--B--C--D--E--F--G--cycle); draw(O--D, dashed); draw(A--C, dashed);[/asy]
$\textbf{(A)}\ 21 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 29$
| Of the six shapes used to create the polygon, the triangle and octagon are adjacent to the others on one side, and the others are adjacent on two sides. In the triangle and octagon $3+8-2(1)=9$ sides are on the outside of the final polygon. In the other shapes $4+5+6+7-4(2) = 14$ sides are on the outside. The resulting polygon has $9+14 = \boxed{\textbf{(B)}\ 23}$ sides.
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_9 | 23 |
AMC8_31 | Triangle $ABC$ is an isosceles triangle with $\overline{AB}=\overline{BC}$ . Point $D$ is the midpoint of both $\overline{BC}$ and $\overline{AE}$ , and $\overline{CE}$ is 11 units long. Triangle $ABD$ is congruent to triangle $ECD$ . What is the length of $\overline{BD}$ ?
[asy] size(100); draw((0,0)--(2,4)--(4,0)--(6,4)--cycle--(4,0),linewidth(1)); label("$A$", (0,0), SW); label("$B$", (2,4), N); label("$C$", (4,0), SE); label("$D$", shift(0.2,0.1)*intersectionpoint((0,0)--(6,4),(2,4)--(4,0)), N); label("$E$", (6,4), NE);[/asy]
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.5\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5.5\qquad\textbf{(E)}\ 6$
| Since triangle $ABD$ is congruent to triangle $ECD$ and $\overline{CE} =11$ , $\overline{AB}=11$ . Since $\overline{AB}=\overline{BC}$ , $\overline{BC}=11$ . Because point $D$ is the midpoint of $\overline{BC}$ , $\overline{BD}=\frac{\overline{BC}}{2}=\frac{11}{2}=\boxed{\textbf{(D)}\ 5.5}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_19 | 5.5 |
AMC8_32 | Two-thirds of the people in a room are seated in three-fourths of the chairs. The rest of the people are standing. If there are $6$ empty chairs, how many people are in the room?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 18\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$
| Working backwards, if $3/4$ of the chairs are taken and $6$ are empty, then there are three times as many taken chairs as empty chairs, or $3 \cdot 6 = 18$ . If $x$ is the number of people in the room and $2/3$ are seated, then $\frac23 x = 18$ and $x = \boxed{(\text{D}) 27}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_20 | 27 |
AMC8_33 | In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term is $4000$ . What is the first term?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10$
| In this solution, we will use trial and error to solve.
$4000$ can be expressed as $200 \times 20$ . We divide $200$ by $20$ and get $10$ , divide $20$ by $10$ and get $2$ , and divide $10$ by $2$ to get $\boxed{\textbf{(D)}\ 5}$ . No one said that they have to be in ascending order!
Solution by ILoveMath31415926535 and clarification edits by apex304
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_22 | 5 |
AMC8_34 | Square $ABCD$ has sides of length 3. Segments $CM$ and $CN$ divide the square's area into three equal parts. How long is segment $CM$ ?
[asy] pair A,B,C,D,M,N; A = (0,0); B = (0,3); C = (3,3); D = (3,0); M = (0,1); N = (1,0); draw(A--B--C--D--cycle); draw(M--C--N); label("$A$",A,SW); label("$M$",M,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,SE); label("$N$",N,S); [/asy]
$\text{(A)}\ \sqrt{10} \qquad \text{(B)}\ \sqrt{12} \qquad \text{(C)}\ \sqrt{13} \qquad \text{(D)}\ \sqrt{14} \qquad \text{(E)}\ \sqrt{15}$
| Since the square has side length $3$ , the area of the entire square is $9$ .
The segments divide the square into 3 equal parts, so the area of each part is $9 \div 3 = 3$ .
Since $\triangle CBM$ has area $3$ and base $CB = 3$ , using the area formula for a triangle:
$A_{tri} = \frac{1}{2}bh$
$3 = \frac{1}{2}3h$
$h = 2$
Thus, height $BM = 2$ .
Since $\triangle CBM$ is a right triangle, $CM = \sqrt{BM^2 + BC^2} = \sqrt{2^2 + 3^2} = \boxed{\text{(C)}\ \sqrt{13}}$ .
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_23 | √13 |
AMC8_35 | 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?
[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(1,1/2,1/4); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(1,-1,0)*unitcube, white, thick(), nolight); draw(shift(1,0,-1)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(1,1,0)*unitcube, white, thick(), nolight); draw(shift(1,0,1)*unitcube, white, thick(), nolight);[/asy]
$\textbf{(A)} \:1 : 6 \qquad\textbf{ (B)}\: 7 : 36 \qquad\textbf{(C)}\: 1 : 5 \qquad\textbf{(D)}\: 7 : 30\qquad\textbf{ (E)}\: 6 : 25$
| The volume is of seven unit cubes which is $7$ . The surface area is the same for each of the protruding cubes which is $5\cdot 6=30$ . The ratio of the volume to the surface area is $\boxed{\textbf{(D)}\: 7 : 30}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_16 | 7 : 30 |
AMC8_36 | The number $N$ is a two-digit number.
• When $N$ is divided by $9$ , the remainder is $1$ .
• When $N$ is divided by $10$ , the remainder is $3$ .
What is the remainder when $N$ is divided by $11$ ?
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$
| From the second bullet point, we know that the second digit must be $3$ , for a number divisible by $10$ ends in zero. Since there is a remainder of $1$ when $N$ is divided by $9$ , the multiple of $9$ must end in a $2$ for it to have the desired remainder $\pmod {10}.$ We now look for this one:
$9(1)=9\\ 9(2)=18\\ 9(3)=27\\ 9(4)=36\\ 9(5)=45\\ 9(6)=54\\ 9(7)=63\\ 9(8)=72$
The number $72+1=73$ satisfies both conditions. We subtract the biggest multiple of $11$ less than $73$ to get the remainder. Thus, $73-11(6)=73-66=\boxed{\textbf{(E) }7}$ .
~CHECKMATE2021
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_5 | 7 |
AMC8_37 | A burger at Ricky C's weighs $120$ grams, of which $30$ grams are filler.
What percent of the burger is not filler?
$\mathrm{(A)}\ 60\% \qquad\mathrm{(B)}\ 65\% \qquad\mathrm{(C)}\ 70\% \qquad\mathrm{(D)}\ 75\% \qquad\mathrm{(E)}\ 90\%$
| There are $30$ grams of filler, so there are $120-30= 90$ grams that aren't filler. $\frac{90}{120}=\frac{3}{4}=\boxed{\mathrm{(D)}\ 75\%}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_3 | 75% |
AMC8_38 | Two $4 \times 4$ squares intersect at right angles, bisecting their intersecting sides, as shown. The circle's diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?
[asy] unitsize(6mm); draw(unitcircle); filldraw((0,1)--(1,2)--(3,0)--(1,-2)--(0,-1)--(-1,-2)--(-3,0)--(-1,2)--cycle,lightgray,black); filldraw(unitcircle,white,black); [/asy]
$\text{(A)}\ 16-4\pi\qquad \text{(B)}\ 16-2\pi \qquad \text{(C)}\ 28-4\pi \qquad \text{(D)}\ 28-2\pi \qquad \text{(E)}\ 32-2\pi$
| If the circle was shaded in, the intersection of the two squares would be a smaller square with half the sidelength, $2$ . The area of this region would be the two larger squares minus the area of the intersection, the smaller square. This is $4^2 + 4^2 - 2^2 = 28$ .
The diagonal of this smaller square created by connecting the two points of intersection of the squares is the diameter of the circle. This value can be found with Pythagorean or a $45^\circ - 45^\circ - 90^\circ$ circle to be $2\sqrt{2}$ . The radius is half the diameter, $\sqrt{2}$ . The area of the circle is $\pi r^2 = \pi (\sqrt{2})^2 = 2\pi$ .
The area of the shaded region is the area of the two squares minus the area of the circle which is $\boxed{\textbf{(D)}\ 28-2\pi}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_25 | 28-2π |
AMC8_39 | All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$
| The $6$ green marbles and yellow marbles form $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total marbles. Now, suppose the total number of marbles is $x$ . We know the number of yellow marbles is $\frac{5}{12}x - 6$ and a positive integer. Therefore, $12$ must divide $x$ . Trying the smallest multiples of $12$ for $x$ , we see that when $x = 12$ , we get there are $-1$ yellow marbles, which is impossible. However when $x = 24$ , there are $\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}$ yellow marbles, which must be the smallest possible.
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_9 | 4 |
AMC8_40 | Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise.
In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$
| Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$ , is a multiple of $12$ . Since this number must be a multiple of $12$ , as stated in the previous sentence, $14$ has a factor $2$ , $k$ must have a factor of $6$ . The smallest number of turns that is a multiple of $6$ is $\boxed{\textbf{(A)}\ 6}$ .
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_20 | 6 |
AMC8_41 | Three dice with faces numbered $1$ through $6$ are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots NOT visible in this view is
[asy] draw((0,0)--(2,0)--(3,1)--(3,7)--(1,7)--(0,6)--cycle); draw((3,7)--(2,6)--(0,6)); draw((3,5)--(2,4)--(0,4)); draw((3,3)--(2,2)--(0,2)); draw((2,0)--(2,6)); dot((1,1)); dot((.5,.5)); dot((1.5,.5)); dot((1.5,1.5)); dot((.5,1.5)); dot((2.5,1.5)); dot((.5,2.5)); dot((1.5,2.5)); dot((1.5,3.5)); dot((.5,3.5)); dot((2.25,2.75)); dot((2.5,3)); dot((2.75,3.25)); dot((2.25,3.75)); dot((2.5,4)); dot((2.75,4.25)); dot((.5,5.5)); dot((1.5,4.5)); dot((2.25,4.75)); dot((2.5,5.5)); dot((2.75,6.25)); dot((1.5,6.5)); [/asy]
$\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 31 \qquad \text{(D)}\ 41 \qquad \text{(E)}\ 53$
| The numbers on one die total $1+2+3+4+5+6 = 21$ , so the numbers
on the three dice total $63$ . Numbers $1, 1, 2, 3, 4, 5, 6$ are visible, and these total $22$ .
This leaves $63 - 22 = \boxed{\text{(D) 41}}$ not seen.
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_8 | 41 |
AMC8_42 | Rectangle $ABCD$ has sides $CD=3$ and $DA=5$ . A circle of radius $1$ is centered at $A$ , a circle of radius $2$ is centered at $B$ , and a circle of radius $3$ is centered at $C$ . Which of the following is closest to the area of the region inside the rectangle but outside all three circles?
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw(Circle((0,0),1)); draw(Circle((0,3),2)); draw(Circle((5,3),3)); label("A",(0.2,0),W); label("B",(0.2,2.8),NW); label("C",(4.8,2.8),NE); label("D",(5,0),SE); label("5",(2.5,0),N); label("3",(5,1.5),E); [/asy]
$\text{(A) }3.5\qquad\text{(B) }4.0\qquad\text{(C) }4.5\qquad\text{(D) }5.0\qquad\text{(E) }5.5$
| The area in the rectangle but outside the circles is the area of the rectangle minus the area of all three of the quarter circles in the rectangle.
The area of the rectangle is $3\cdot5 =15$ . The area of all 3 quarter circles is $\frac{\pi}{4}+\frac{\pi(2)^2}{4}+\frac{\pi(3)^2}{4} = \frac{14\pi}{4} = \frac{7\pi}{2}$ . Therefore the area in the rectangle but outside the circles is $15-\frac{7\pi}{2}$ . $\pi$ is approximately $\dfrac{22}{7},$ and substituting that in will give $15-11=\boxed{\text{(B) }4.0}$
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_20 | 4.0 |
AMC8_43 | 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?
$\textbf{(A)}\ 25 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 33 \qquad \textbf{(D)}\ 40 \qquad \textbf{(E)}\ 45$
| First, find the amount of money one will pay for three sandals without the discount. We have $\textdollar 50\times 3 \text{ sandals} = \textdollar 150$ .
Then, find the amount of money using the discount: $50 + 0.6 \times 50 + \frac{1}{2} \times 50 = \textdollar 105$ .
Finding the percentage yields $\frac{105}{150} = 70 \%$ .
To find the percent saved, we have $100 \% -70 \%= \boxed{\textbf{(B)}\ 30 \%}$
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_12 | 30 |
AMC8_44 | Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]
$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$
| Let the side length of each square $S_k$ be $s_k$ . Then, from the diagram, we can line up the top horizontal lengths of $S_1$ , $S_2$ , and $S_3$ to cover the top side of the large rectangle, so $s_{1}+s_{2}+s_{3}=3322$ . Similarly, the short side of $R_2$ will be $s_1-s_2$ , and lining this up with the left side of $S_3$ to cover the vertical side of the large rectangle gives $s_{1}-s_{2}+s_{3}=2020$ . We subtract the second equation from the first to obtain $2s_{2}=1302$ , and thus $s_{2}=\boxed{\textbf{(A) }651}$ .
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25 | 651 |
AMC8_45 | Rachelle uses 3 pounds of meat to make 8 hamburgers for her family. How many pounds of meat does she need to make 24 hamburgers for a neighbourhood picnic?
$\textbf{(A)}\hspace{.05in}6 \qquad \textbf{(B)}\hspace{.05in}6\dfrac23 \qquad \textbf{(C)}\hspace{.05in}7\dfrac12 \qquad \textbf{(D)}\hspace{.05in}8 \qquad \textbf{(E)}\hspace{.05in}9$
| Since Rachelle uses $3$ pounds of meat to make $8$ hamburgers, she uses $\frac{3}{8}$ pounds of meat to make one hamburger. She'll need 24 times that amount of meat for 24 hamburgers, or $\frac{3}{8} \cdot 24 = \boxed{\textbf{(E)}\ 9}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_1 | 9 |
AMC8_46 | Five friends compete in a dart-throwing contest. Each one has two darts to throw at the same circular target, and each individual's score is the sum of the scores in the target regions that are hit. The scores for the target regions are the whole numbers $1$ through $10$ . Each throw hits the target in a region with a different value. The scores are: Alice $16$ points, Ben $4$ points, Cindy $7$ points, Dave $11$ points, and Ellen $17$ points. Who hits the region worth $6$ points?
$\textbf{(A)}\ \text{Alice}\qquad \textbf{(B)}\ \text{Ben}\qquad \textbf{(C)}\ \text{Cindy}\qquad \textbf{(D)}\ \text{Dave} \qquad \textbf{(E)}\ \text{Ellen}$
| The only way to get Ben's score is with $1+3=4$ . Cindy's score can be made of $3+4$ or $2+5$ , but since Ben already hit the $3$ , Cindy hit $2+5=7$ . Similarly, Dave's darts were in the region $4+7=11$ . Lastly, because there is no $7$ left, $\boxed{\textbf{(A)}\ \text{Alice}}$ must have hit the regions $6+10=16$ and Ellen $8+9=17$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_18 | Alice |
AMC8_47 | The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually $21$ participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?
| On Monday, $20$ people come. On Tuesday, $26$ people come. On Wednesday, $16$ people come. On Thursday, $22$ people come. Finally, on Friday, $16$ people come. $20+26+16+22+16=100$ , so the mean is $20$ . The median is $(16, 16, 20, 22, 26)$ $20$ . The coach figures out that actually $21$ people come on Wednesday. The new mean is $21$ , while the new median is $(16, 20, 21, 22, 26)$ $21$ . Also, the median increases by $1$ because now the median is $21$ instead of $20$ . The median and mean both change, so the answer is $\boxed{\textbf{(B)}}$ .
Another way to compute the change in mean is to notice that the sum increased by $5$ with the correction. So, the average increased by $5/5 = 1$ .
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_10 | B |
AMC8_48 | The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?
$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$
| Solution 1It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes. We know that there is $60$ minutes in a hour. Therefore, there are $11 \cdot 60 = 660$ minutes in 11 hours. Adding the second part(the 5 minutes) we get $660 + 5 = \boxed{\textbf{(C)}\ 665}$ .
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_1 | 665 |
AMC8_49 | A cube has edge length $2$ . Suppose that we glue a cube of edge length $1$ on top of the big cube so that one of its faces rests entirely on the top face of the larger cube. The percent increase in the surface area (sides, top, and bottom) from the original cube to the new solid formed is closest to
[asy] draw((0,0)--(2,0)--(3,1)--(3,3)--(2,2)--(0,2)--cycle); draw((2,0)--(2,2)); draw((0,2)--(1,3)); draw((1,7/3)--(1,10/3)--(2,10/3)--(2,7/3)--cycle); draw((2,7/3)--(5/2,17/6)--(5/2,23/6)--(3/2,23/6)--(1,10/3)); draw((2,10/3)--(5/2,23/6)); draw((3,3)--(5/2,3));[/asy]
$\text{(A)}\ 10\qquad\text{(B)}\ 15\qquad\text{(C)}\ 17\qquad\text{(D)}\ 21\qquad\text{(E)}\ 25$
| The original cube has $6$ faces, each with an area of $2\cdot 2 = 4$ square units. Thus the original figure had a total surface area of $24$ square units.
The new figure has the original surface, with $6$ new faces that each have an area of $1$ square unit, for a total surface area of $6$ additional square units added to it. But $1$ square unit of the top of the bigger cube, and $1$ square unit on the bottom of smaller cube, is not on the surface, and does not count towards the surface area.
The total surface area is therefore $24 + 6 - 1 - 1 = 28$ square units.
The percent increase in surface area is $\frac{SA_{new} - SA_{old}}{SA_{old}}\cdot 100\% = \frac{28-24}{24}\cdot 100\% \approx 16.67\%$ , giving the closest answer as $\boxed{C}$ .
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_22 | C |
AMC8_50 | Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?
$\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}$
| Write the order of the cars as $\square\square\square\square\square$ , where the left end of the row represents the back of the train and the right end represents the front. Call the people $A$ , $D$ , $K$ , $M$ , and $S$ respectively. The first condition gives $M\square\square\square\square$ , so we try $MAS\square\square$ , $M\square AS\square$ , and $M\square\square AS$ . In the first case, as $D$ sat in front of $A$ , we must have $MASDK$ or $MASKD$ , which do not comply with the last condition. In the second case, we obtain $MKASD$ , which works, while the third case is obviously impossible since it results in there being no way for $D$ to sit in front of $A$ . It follows that, with the only possible arrangement being $MKASD$ , the person sitting in the middle car is $\boxed{\textbf{(A) }\text{Aaron}}$ .
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_6 | Aaron |
AMC8_51 | The average age of $5$ people in a room is $30$ years. An $18$ -year-old person leaves
the room. What is the average age of the four remaining people?
$\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36$
| Let $x$ be the average of the remaining $4$ people.
The equation we get is $\frac{4x + 18}{5} = 30$
Simplify,
$4x + 18 = 150$
$4x = 132$
$x = 33$
Therefore, the answer is $\boxed{\textbf{(D)}\ 33}$
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_7 | 33 |
AMC8_52 | The "Middle School Eight" basketball conference has $8$ teams. Every season, each team plays every other conference team twice (home and away), and each team also plays $4$ games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?
$\textbf{(A) }60\qquad\textbf{(B) }88\qquad\textbf{(C) }96\qquad\textbf{(D) }144\qquad \textbf{(E) }160$
| Within the conference, there are 8 teams, so there are $\dbinom{8}{2}=28$ pairings of teams, and each pair must play two games, for a total of $28\cdot 2=56$ games within the conference.
Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of $4\cdot 8 =32$ games outside the conference.
Therefore, the total number of games is $56 + 32 = \boxed{\text{(B) }88}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_16 | 88 |
AMC8_53 | What is the smallest possible average of four distinct positive even integers?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$
| In order to get the smallest possible average, we want the 4 even numbers to be as small as possible. The first 4 positive even numbers are 2, 4, 6, and 8. Their average is $\frac{2+4+6+8}{4}=\frac{20}{4}=\boxed{\text{(C)}\ 5}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_3 | 5 |
AMC8_54 | The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome?
$\text{(A)}\ 0 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 25$
| The palindrome right after 2002 is 2112. The product of the digits of 2112 is $\boxed{\text{(B)}\ 4}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_4 | 4 |
AMC8_55 | There are $81$ grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point $P$ is in the center of the square. Given that point $Q$ is randomly chosen among the other $80$ points, what is the probability that the line $PQ$ is a line of symmetry for the square?
[asy] draw((0,0)--(0,8)); draw((0,8)--(8,8)); draw((8,8)--(8,0)); draw((8,0)--(0,0)); dot((0,0)); dot((0,1)); dot((0,2)); dot((0,3)); dot((0,4)); dot((0,5)); dot((0,6)); dot((0,7)); dot((0,8)); dot((1,0)); dot((1,1)); dot((1,2)); dot((1,3)); dot((1,4)); dot((1,5)); dot((1,6)); dot((1,7)); dot((1,8)); dot((2,0)); dot((2,1)); dot((2,2)); dot((2,3)); dot((2,4)); dot((2,5)); dot((2,6)); dot((2,7)); dot((2,8)); dot((3,0)); dot((3,1)); dot((3,2)); dot((3,3)); dot((3,4)); dot((3,5)); dot((3,6)); dot((3,7)); dot((3,8)); dot((4,0)); dot((4,1)); dot((4,2)); dot((4,3)); dot((4,4)); dot((4,5)); dot((4,6)); dot((4,7)); dot((4,8)); dot((5,0)); dot((5,1)); dot((5,2)); dot((5,3)); dot((5,4)); dot((5,5)); dot((5,6)); dot((5,7)); dot((5,8)); dot((6,0)); dot((6,1)); dot((6,2)); dot((6,3)); dot((6,4)); dot((6,5)); dot((6,6)); dot((6,7)); dot((6,8)); dot((7,0)); dot((7,1)); dot((7,2)); dot((7,3)); dot((7,4)); dot((7,5)); dot((7,6)); dot((7,7)); dot((7,8)); dot((8,0)); dot((8,1)); dot((8,2)); dot((8,3)); dot((8,4)); dot((8,5)); dot((8,6)); dot((8,7)); dot((8,8)); label("P",(4,4),NE); [/asy]
$\textbf{(A) }\frac{1}{5}\qquad\textbf{(B) }\frac{1}{4} \qquad\textbf{(C) }\frac{2}{5} \qquad\textbf{(D) }\frac{9}{20} \qquad\textbf{(E) }\frac{1}{2}$
| [asy] draw((0,0)--(0,8)); draw((0,8)--(8,8)); draw((8,8)--(8,0)); draw((8,0)--(0,0)); dot((0,0)); dot((0,1)); dot((0,2)); dot((0,3)); dot((0,4)); dot((0,5)); dot((0,6)); dot((0,7)); dot((0,8)); dot((1,0)); dot((1,1)); dot((1,2)); dot((1,3)); dot((1,4)); dot((1,5)); dot((1,6)); dot((1,7)); dot((1,8)); dot((2,0)); dot((2,1)); dot((2,2)); dot((2,3)); dot((2,4)); dot((2,5)); dot((2,6)); dot((2,7)); dot((2,8)); dot((3,0)); dot((3,1)); dot((3,2)); dot((3,3)); dot((3,4)); dot((3,5)); dot((3,6)); dot((3,7)); dot((3,8)); dot((4,0)); dot((4,1)); dot((4,2)); dot((4,3)); dot((4,4)); dot((4,5)); dot((4,6)); dot((4,7)); dot((4,8)); dot((5,0)); dot((5,1)); dot((5,2)); dot((5,3)); dot((5,4)); dot((5,5)); dot((5,6)); dot((5,7)); dot((5,8)); dot((6,0)); dot((6,1)); dot((6,2)); dot((6,3)); dot((6,4)); dot((6,5)); dot((6,6)); dot((6,7)); dot((6,8)); dot((7,0)); dot((7,1)); dot((7,2)); dot((7,3)); dot((7,4)); dot((7,5)); dot((7,6)); dot((7,7)); dot((7,8)); dot((8,0)); dot((8,1)); dot((8,2)); dot((8,3)); dot((8,4)); dot((8,5)); dot((8,6)); dot((8,7)); dot((8,8)); label("P",(4,4),NE); draw((0,4)--(3,4)); draw((0,8)--(3,5)); draw((8,8)--(5,5)); draw((4,8)--(4,5)); draw((4,0)--(4,3)); draw((0,0)--(3,3)); draw((8,0)--(5,3)); draw((5,4)--(8,4)); [/asy]
Lines of symmetry go through point $P$ , and there are $8$ directions the lines could go, and there are $4$ dots at each direction. $\frac{4\times8}{80}=\boxed{\textbf{(C)} \frac{2}{5}}$ .
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_6 | 2/5 |
AMC8_56 | A $1\times 2$ rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?
$\textbf{(A)}\ \frac\pi2 \qquad \textbf{(B)}\ \frac{2\pi}3 \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ \frac{4\pi}3 \qquad \textbf{(E)}\ \frac{5\pi}3$
| [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; usepackage("amsmath"); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = 2.392515856236789, xmax = 4.844947145877386, ymin = 6.070697674418619, ymax = 8.062241014799170; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((3.119707204019531,7.403678646934482)--(4.119707204019532,7.403678646934476)--(4.119707204019532,6.903678646934476)--(3.119707204019531,6.903678646934476)--cycle, zzttqq); /* draw figures */ draw((2.912600422832983,6.903678646934476)--(4.326813985206080,6.903678646934476)); draw(shift((3.619707204019532,6.903678646934476))*xscale(0.7071067811865487)*yscale(0.7071067811865487)*arc((0,0),1,0.000000000000000,180.0000000000000)); draw((3.619707204019532,6.903678646934476)--(4.119707204019532,6.903678646934476)); draw((3.619707204019532,6.903678646934476)--(3.119707204019531,6.903678646934476)); draw((3.119707204019531,7.403678646934482)--(4.119707204019532,7.403678646934476), zzttqq); draw((4.119707204019532,7.403678646934476)--(4.119707204019532,6.903678646934476), zzttqq); draw((4.119707204019532,6.903678646934476)--(3.119707204019531,6.903678646934476), zzttqq); draw((3.119707204019531,6.903678646934476)--(3.119707204019531,7.403678646934482), zzttqq); label("$1$",(3.847061310782247,6.924820295983102),SE*labelscalefactor); label("$1$",(4.155729386892184,7.208118393234687),SE*labelscalefactor); draw((3.619707204019532,6.903678646934476)--(4.119707204019532,7.403678646934476)); label("$\sqrt{2}$",(3.711754756871041,7.288456659619466),SE*labelscalefactor); label("$2$",(3.563763213530660,7.563298097251601),SE*labelscalefactor); /* dots and labels */ dot((2.912600422832983,6.903678646934476)); dot((4.326813985206080,6.903678646934476)); dot((3.619707204019532,6.903678646934476)); dot((4.119707204019532,6.903678646934476),blue); dot((3.619707204019532,6.903678646934476)); dot((3.119707204019531,6.903678646934476),blue); dot((3.119707204019531,7.403678646934482),blue); dot((4.119707204019532,7.403678646934476),blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]
A semicircle has symmetry, so the center is exactly at the midpoint of the 2 side on the rectangle, making the radius, by the Pythagorean Theorem, $\sqrt{1^2+1^2}=\sqrt{2}$ . The area is $\frac{2\pi}{2}=\boxed{\textbf{(C)}\ \pi}$ .
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_20 | π |
AMC8_57 | What is the perimeter of trapezoid $ABCD$ ?
[asy]size(3inch, 1.5inch); pair a=(0,0), b=(18,24), c=(68,24), d=(75,0), f=(68,0), e=(18,0); draw(a--b--c--d--cycle); draw(b--e); draw(shift(0,2)*e--shift(2,2)*e--shift(2,0)*e); label("30", (9,12), W); label("50", (43,24), N); label("25", (71.5, 12), E); label("24", (18, 12), E); label("$A$", a, SW); label("$B$", b, N); label("$C$", c, N); label("$D$", d, SE); label("$E$", e, S);[/asy]
$\textbf{(A)}\ 180\qquad\textbf{(B)}\ 188\qquad\textbf{(C)}\ 196\qquad\textbf{(D)}\ 200\qquad\textbf{(E)}\ 204$
| Draw altitudes from $B$ and $C$ to base $AD$ to create a rectangle and two right triangles. The side opposite $BC$ is equal to $50$ . The bases of the right triangles can be found using Pythagorean or special triangles to be $18$ and $7$ . Add it together to get $AD=18+50+7=75$ . The perimeter is $75+30+50+25=\boxed{\textbf{(A)}\ 180}$ .
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_19 | 180 |
AMC8_58 | Amanda Reckonwith draws five circles with radii $1, 2, 3, 4$ and $5$ . Then for each circle she plots the point $(C,A)$ ,
where $C$ is its circumference and $A$ is its area. Which of the
following could be her graph?
$\textbf{(A)}$
[asy] size(75); pair A= (1.5,2) , B= (3,4) , C= (4.5,7) , D= (6,11) , E= (7.5,16) ; draw((0,-1)--(0,16)); draw((-1,0)--(16,0)); dot(A^^B^^C^^D^^E); label("$A$", (0,8), W); label("$C$", (8,0), S);[/asy]
$\textbf{(B)}$
[asy] size(75); pair A= (1.5,9) , B= (3,6) , C= (4.5,6) , D= (6,9) , E= (7.5,15) ; draw((0,-1)--(0,16)); draw((-1,0)--(16,0)); dot(A^^B^^C^^D^^E); label("$A$", (0,8), W); label("$C$", (8,0), S);[/asy]
$\textbf{(C)}$
[asy] size(75); pair A= (1.5,2) , B= (3,6) , C= (4.5,8) , D= (6,6) , E= (7.5,2) ; draw((0,-1)--(0,16)); draw((-1,0)--(16,0)); dot(A^^B^^C^^D^^E); label("$A$", (0,8), W); label("$C$", (8,0), S);[/asy]
$\textbf{(D)}$
[asy] size(75); pair A= (1.5,2) , B= (3,5) , C= (4.5,8) , D= (6,11) , E= (7.5,14) ; draw((0,-1)--(0,16)); draw((-1,0)--(16,0)); dot(A^^B^^C^^D^^E); label("$A$", (0,8), W); label("$C$", (8,0), S);[/asy]
$\textbf{(E)}$
[asy] size(75); pair A= (1.5,15) , B= (3,10) , C= (4.5,6) , D= (6,3) , E= (7.5,1) ; draw((0,-1)--(0,16)); draw((-1,0)--(16,0)); dot(A^^B^^C^^D^^E); label("$A$", (0,8), W); label("$C$", (8,0), S);[/asy]
| The circumference of a circle is obtained by simply multiplying the radius by $2\pi$ . So, the C-coordinate (in this case, it is the x-coordinate) will increase at a steady rate. The area, however, is obtained by squaring the radius and multiplying it by $\pi$ . Since squares do not increase in an evenly spaced arithmetic sequence, the increase in the A-coordinates (aka the y- coordinates) will be much more significant. The answer is $\boxed{\textbf{(A)}},$ [asy] size(75); pair A= (1.5,2) , B= (3,4) , C= (4.5,7) , D= (6,11) , E= (7.5,16) ; draw((0,-1)--(0,16)); draw((-1,0)--(16,0)); dot(A^^B^^C^^D^^E); label("$A$", (0,8), W); label("$C$", (8,0), S);[/asy] .
-RBANDA
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_16 | A |
AMC8_59 | Let $A$ be the area of the triangle with sides of length $25, 25$ , and $30$ . Let $B$ be the area of the triangle with sides of length $25, 25,$ and $40$ . What is the relationship between $A$ and $B$ ?
$\textbf{(A) } A = \dfrac9{16}B \qquad\textbf{(B) } A = \dfrac34B \qquad\textbf{(C) } A=B \qquad \textbf{(D) } A = \dfrac43B \qquad \textbf{(E) }A = \dfrac{16}9B$
| 25-25-30
We can draw the altitude for the side with length 30. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 30 into two segments with length 15. By the Pythagorean Theorem , we have
\[15^2 + x^2 =25^2\]
\[x^2 = 25^2 - 15^2\]
\[x^2 = (25 + 15)(25-15)\]
\[x^2= 40\cdot 10\]
\[x^2= 400\]
\[x = \sqrt{400}\]
\[x= 20\]
Thus we have two 15-20-25 right triangles.
25-25-40
We can draw the altitude for the side with length 40. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 40 into two segments with length 20. From the 25-25-30 case, we know that the other side length is 15, so we have two 15-20-25 right triangles.
Let the area of a 15-20-25 right triangle be $x$ .
\[a = 2x\]
\[b = 2x\]
\[\boxed{\textbf{(C) } A = B}\]
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_16 | A = B |
AMC8_60 | The clock in Sri's car, which is not accurate, gains time at a constant rate. One day as he begins shopping, he notes that his car clock and his watch (which is accurate) both say 12:00 noon. When he is done shopping, his watch says 12:30 and his car clock says 12:35. Later that day, Sri loses his watch. He looks at his car clock and it says 7:00. What is the actual time?
$\textbf{ (A) }5:50\qquad\textbf{(B) }6:00\qquad\textbf{(C) }6:30\qquad\textbf{(D) }6:55\qquad \textbf{(E) }8:10$
| We see that every $35$ minutes the clock passes, the watch passes $30$ minutes. That means that the clock is $\frac{7}{6}$ as fast the watch, so we can set up proportions.
$\dfrac{\text{car clock}}{\text{watch}}=\dfrac{7}{6}=\dfrac{7 \text{ hours}}{x \text{ hours}}$ . Cross-multiplying we get $x=6$ . Now, this is obviously redundant, because we could just eyeball it to see that the watch would have passed $6$ hours. But this method is better when the numbers are a bit more complex, which makes it both easier and reliable. Either way, our answer is $\boxed{\textbf{(B) }6:00}$ .
--BakedPotato66
--Rishi09
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_12 | 6:00 |
AMC8_61 | $650$ students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?
[asy] size(200); defaultpen(linewidth(0.7)); defaultpen(fontsize(8)); draw(origin--(0,250)); int i; for(i=0; i<6; i=i+1) { draw((0,50*i)--(5,50*i)); } filldraw((25,0)--(75,0)--(75,150)--(25,150)--cycle, gray, black); filldraw((75,0)--(125,0)--(125,100)--(75,100)--cycle, gray, black); filldraw((125,0)--(175,0)--(175,150)--(125,150)--cycle, gray, black); filldraw((225,0)--(175,0)--(175,250)--(225,250)--cycle, gray, black); label("$50$", (0,50), W); label("$100$", (0,100), W); label("$150$", (0,150), W); label("$200$", (0,200), W); label("$250$", (0,250), W); label(rotate(90)*"Lasagna", (50,0), S); label(rotate(90)*"Manicotti", (100,0), S); label(rotate(90)*"Ravioli", (150,0), S); label(rotate(90)*"Spaghetti", (200,0), S); label(rotate(90)*"$\mbox{Number of People}$", (-40,140), W); [/asy]
$\mathrm{(A)} \frac{2}{5} \qquad \mathrm{(B)} \frac{1}{2} \qquad \mathrm{(C)} \frac{5}{4} \qquad \mathrm{(D)} \frac{5}{3} \qquad \mathrm{(E)} \frac{5}{2}$
| The answer is $\dfrac{\text{number of students who preferred spaghetti}}{\text{number of students who preferred manicotti}}$
So,
$\frac{250}{100}$
Simplify,
$\frac{5}{2}$
The answer is $\boxed{\textbf{(E)}\ \dfrac{5}{2}}$
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_2 | 5/2 |
AMC8_62 | Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?
$\textbf{(A) } \text{Arn}\qquad\textbf{(B) }\text{Bob}\qquad\textbf{(C) }\text{Cyd}\qquad\textbf{(D) }\text{Dan}\qquad \textbf{(E) }\text{Eve}\qquad$
| The five numbers which cause people to leave the circle are $7, 14, 17, 21,$ and $27.$
The most straightforward way to do this would be to draw out the circle with the people, and cross off people as you count.
Assuming the six people start with $1$ , Arn counts $7$ so he leaves first. Then, Cyd counts $14$ as there are $7$ numbers to be counted from this point. Then, Fon, Bob, and Eve, count, $17,$ $21,$ and $27$ , respectively, so the last one standing is Dan. Hence, the answer would be $\boxed{\textbf{(D) }\text{Dan}}$ .
~Nivaar
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_3 | Dan |
AMC8_63 | What is the largest power of $2$ that is a divisor of $13^4 - 11^4$ ?
$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$
| First, we use difference of squares on $13^4 - 11^4 = (13^2)^2 - (11^2)^2$ to get $13^4 - 11^4 = (13^2 + 11^2)(13^2 - 11^2)$ . Using difference of squares again and simplifying, we get $(169 + 121)(13+11)(13-11) = 290 \cdot 24 \cdot 2 = (2\cdot 8 \cdot 2) \cdot (3 \cdot 145)$ . Realizing that we don't need the right-hand side because it doesn't contain any factor of 2, we see that the greatest power of $2$ that is a divisor $13^4 - 11^4$ is $\boxed{\textbf{(C)}\ 32}$ .
~CHECKMATE2021
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_15 | 32 |
AMC8_64 | Thirteen black and six white hexagonal tiles were used to create the figure below. If a new figure is created by attaching a border of white tiles with the same size and shape as the others, what will be the difference between the total number of white tiles and the total number of black tiles in the new figure?
[asy] defaultpen(linewidth(1)); real x=sqrt(3)/2; path p=rotate(30)*polygon(6); filldraw(p^^shift(0,3)*p^^shift(4x,0)*p^^shift(3x,1.5)*p^^shift(2x,3)*p^^shift(-4x,0)*p^^shift(-3x,1.5)*p^^shift(-2x,3)*p^^shift(3x,-1.5)*p^^shift(-3x,-1.5)*p^^shift(2x,-3)*p^^shift(-2x,-3)*p^^shift(0,-3)*p, black, black); draw(shift(2x,0)*p^^shift(-2x,0)*p^^shift(x,1.5)*p^^shift(-x,1.5)*p^^shift(x,-1.5)*p^^shift(-x,-1.5)*p); [/asy]
$\textbf{(A)}\ 5\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 11\qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$
| The first ring around the middle tile has $6$ tiles, and the second has $12$ . From this pattern, the third ring has $18$ tiles. Of these, $6+18=24$ are white and $1+12=13$ are black, with a difference of $24-13 = \boxed{\textbf{(C)}\ 11}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_15 | 11 |
AMC8_65 | 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?
$\textbf{(A) }4 \qquad\textbf{(B) }5 \qquad\textbf{(C) }6 \qquad\textbf{(D) }7 \qquad\textbf{(E) }9$
| By adding a number from Bag A and a number from Bag B together, the values we can get are $3, 5, 7, 5, 7, 9, 7, 9, 11.$ Therefore the number of different values is $\boxed{\textbf{(B)}\ 5}$ .
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_8 | 5 |
AMC8_66 | Which of the following is the correct order of the fractions $\frac{15}{11},\frac{19}{15},$ and $\frac{17}{13},$ from least to greatest?
$\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}<\frac{17}{13}<\frac{15}{11}$
| Solution 1 (Bashing/Butterfly Method)We take a common denominator:
\[\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.\]
Since $2717<2805<2925$ it follows that the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .
Another approach to this problem is using the properties of one fraction being greater than another, also known as the butterfly method . That is, if
$\frac{a}{b}>\frac{c}{d}$ , then it must be true that $a * d$ is greater than $b * c$ . Using this approach, we can check for at least two distinct pairs of fractions and find out the greater one of those two, logically giving us the expected answer of $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .
-xMidnightFirex
~ dolphin7 - I took your idea and made it an explanation.
- Clearness by doulai1
- Alternate Solution by Nivaar
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_3 | $\frac{19}{15}<\frac{17}{13}<\frac{15}{11}$ |
AMC8_67 | 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?
$\text{(A)}\ 250\qquad\text{(B)}\ 500\qquad\text{(C)}\ 625\qquad\text{(D)}\ 750\qquad\text{(E)}\ 1000$
| Since the volume ratio is equal to the sides ratio cubed, then the ratio of the larger box's volume to the smaller one is 2 cubed.
$2^3=8$
Now multiply 125 (the number of jellybeans that Bert's box can hold) by 8.
$8\cdot125= \boxed{\text{(E)}\ 1000}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_13 | 1000 |
AMC8_68 | In an All-Area track meet, $216$ sprinters enter a $100-$ meter dash competition. The track has $6$ lanes, so only $6$ 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?
$\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72$
| Solution 1From any $n-$ th race, only $\frac{1}{6}$ will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal.
Starting with the first race:
\[\frac{216}{6}=36\]
\[\frac{36}{6}=6\]
\[\frac{6}{6}=1\]
Adding all of the numbers in the second column yields $\boxed{\textbf{(C)}\ 43}$
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_18 | 43 |
AMC8_69 | Two dice are thrown. What is the probability that the product of the two numbers is a multiple of 5?
$\text{(A)}\ \dfrac{1}{36} \qquad \text{(B)}\ \dfrac{1}{18} \qquad \text{(C)}\ \dfrac{1}{6} \qquad \text{(D)}\ \dfrac{11}{36} \qquad \text{(E)}\ \dfrac{1}{3}$
| This is equivalent to asking for the probability that at least one of the numbers is a multiple of $5$ , since if one of the numbers is a multiple of $5$ , then the product with it and another integer is also a multiple of $5$ , and if a number is a multiple of $5$ , then since $5$ is prime, one of the factors must also have a factor of $5$ , and $5$ is the only multiple of $5$ on a die, so one of the numbers rolled must be a $5$ . To find the probability of rolling at least one $5$ , we can find the probability of not rolling a $5$ and subtract that from $1$ , since you either roll a $5$ or not roll a $5$ . The probability of not rolling a $5$ on either dice is $\left(\frac{5}{6} \right) \left(\frac{5}{6} \right)=\frac{25}{36}$ . Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of $5$ , is $1-\frac{25}{36}=\frac{11}{36}, \boxed{\text{D}}$
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_18 | D |
AMC8_70 | Niki usually leaves her cell phone on. If her cell phone is on but
she is not actually using it, the battery will last for $24$ hours. If
she is using it constantly, the battery will last for only $3$ hours.
Since the last recharge, her phone has been on $9$ hours, and during
that time she has used it for $60$ minutes. If she doesn’t use it any
more but leaves the phone on, how many more hours will the battery last?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$
| When not being used, the cell phone uses up $\frac{1}{24}$ of its battery per hour. When being used, the cell phone uses up $\frac{1}{3}$ of its battery per hour. Since Niki's phone has been on for $9$ hours, of those $8$ simply on and $1$ with Niki using her phone, $8(\frac{1}{24}) + 1(\frac{1}{3}) = \frac{2}{3}$ of its battery has been used up. To drain the remaining $\frac{1}{3}$ the phone can last for $\frac{\frac{1}{3}}{\frac{1}{24}}=\boxed{\textbf{(B)}\ 8}$ more hours without being used.
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_12 | 8 |
AMC8_71 | Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive?
$\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25$
| Each cup is filled with $\frac{3}{4} \cdot \frac{1}{5} = \frac{3}{20}$ of the amount of juice in the pitcher, so the percentage is $\frac{3}{20} \cdot 100 = \boxed{\textbf{(C) }15}$ .
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_5 | 15 |
AMC8_72 | How many positive factors does $23,232$ have?
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42$
| We can first find the prime factorization of $23,232$ , which is $2^6\cdot3^1\cdot11^2$ . Now, we add one to our powers and multiply. Therefore, the answer is $(6+1)\cdot(1+1)\cdot(2+1)=7\cdot2\cdot3=\boxed{\textbf{(E) }42}$
Note:
23232 is a large number, so we can look for shortcuts to factor it.
One way to factor it quickly is to use 3 and 11 divisibility rules to observe that $23232 = 3 \cdot 7744 = 3 \cdot 11 \cdot 704 = 3 \cdot 11^2 \cdot 64 = 3^1 \cdot 11^2 \cdot 2^6$ .
Another way is to spot the "32" and compute that $23232 = 32\cdot(101 + 10000/16) = 32\cdot (101+ 5^4) = 32\cdot 726 = 32 \cdot 11 \cdot 66$ .
A third way to factor it is to observe $23232 = 24000 - 768$ . Factoring out the 3 gives us $3(8000 - 256)$ . Since $8000 = 2^6 \cdot 5^3$ and $256 = 2^8$ , we have $2^6 \cdot 3 (5^3 - 2^2) = 2^6 \cdot 3 (125-4) = 2^6 \cdot 3 \cdot 121 = 2^6 \cdot 3 \cdot 11^2$ .
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_18 | 42 |
AMC8_73 | For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$
| We have $5! = 2 \cdot 3 \cdot 4 \cdot 5$ , and $2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!$ . Therefore, the equation becomes $3 \cdot 4 \cdot 10! = 12 \cdot N!$ , and so $12 \cdot 10! = 12 \cdot N!$ . Cancelling the $12$ s, it is clear that $N=\boxed{\textbf{(A) }10}$ .
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_12 | 10 |
AMC8_74 | Kei draws a $6$ -by- $6$ grid. He colors $13$ of the unit squares silver and the remaining squares gold. Kei then folds the grid in half vertically, forming pairs of overlapping unit squares. Let $m$ and $M$ equal the least and greatest possible number of gold-on-gold pairs, respectively. What is the value of $m+M$ ?
[asy] import graph; size(100); pen gridPen = black; void drawSquare(pair p) { draw(box(p, p + (1,1)), gridPen); } int[][] grid = { {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, }; int rows = grid.length; int cols = grid[0].length; for (int i = 0; i < rows; ++i) { for (int j = 0; j < cols; ++j) { if (grid[i][j] == 1) { drawSquare((j, rows - i - 1)); } } } [/asy]
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 16\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$
| First, we note that there are $18$ "pairs" of squares folded on top of each other after the folding. The minimum number of gold pairs occurs when silver squares occupy the maximum number of pairs, and the maximum number of gold pairs occurs when silver squares occupy the minimum number of pairs. The former case occurs when all $13$ silver squares are placed in different pairs, resulting in $18-13=5$ gold pairs. The latter case occurs when the silver squares are paired up as much as possible, resulting in $6$ complete pairs and another square occupying another pair slot. Then there are $18-7=11$ gold pairs. Our answer is $11+5=\boxed{\textbf{(C) }16}$ . ~cxsmi
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_15 | 16 |
AMC8_75 | Isosceles right triangle $ABC$ encloses a semicircle of area $2\pi$ . The circle has its center $O$ on hypotenuse $\overline{AB}$ and is tangent to sides $\overline{AC}$ and $\overline{BC}$ . What is the area of triangle $ABC$ ?
[asy]pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(circle(o, 2)); clip(a--b--c--cycle); draw(a--b--c--cycle); dot(o); label("$C$", c, NW); label("$A$", a, NE); label("$B$", b, SW);[/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 3\pi\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 4\pi$
| First, we notice half a square so first let's create a square. Once we have a square, we will have a full circle. This circle has a diameter of 4 which will be the side of the square. The area would be $4\cdot 4 = 16.$ Divide 16 by 2 to get the original shape and you get $\boxed{8}$
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_23 | 8 |
AMC8_76 | In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid?
[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("$+$",(0,0)); draw(shift(1,0)*box); label("$-$",(1,0)); draw(shift(2,0)*box); label("$+$",(2,0)); draw(shift(3,0)*box); label("$-$",(3,0)); draw(shift(0.5,0.4)*box); label("$-$",(0.5,0.4)); draw(shift(1.5,0.4)*box); label("$-$",(1.5,0.4)); draw(shift(2.5,0.4)*box); label("$-$",(2.5,0.4)); draw(shift(1,0.8)*box); label("$+$",(1,0.8)); draw(shift(2,0.8)*box); label("$+$",(2,0.8)); draw(shift(1.5,1.2)*box); label("$+$",(1.5,1.2)); [/asy]
$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$
| You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns:
+−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++. There are 8 patterns and so the answer is $\boxed{\textbf{(C) } 8}$ .
-NinjaBoi2000
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_19 | 8 |
AMC8_77 | On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?
[mathjax]\textbf{(A) }50\qquad\textbf{(B) }70\qquad\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100[/mathjax]
| Since Anh spends half an hour to drive 10 miles on the coastal road, his speed is [mathjax]r=\dfrac dt=\dfrac{10}{0.5}=20[/mathjax] mph. His speed on the highway then is [mathjax]60[/mathjax] mph. He drives [mathjax]50[/mathjax] miles, so he drives for [mathjax]\dfrac{5}{6}[/mathjax] hours, which is equal to [mathjax]50[/mathjax] minutes (Note that [mathjax]60[/mathjax] miles per hour is the same as [mathjax]1[/mathjax] mile per minute). The total amount of minutes spent on his trip is [mathjax]30+50\implies \boxed{\textbf{(C) }80}[/mathjax].
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_6 | 80 |
AMC8_78 | The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?
$\textbf{(A)}\hspace{.05in}61\qquad\textbf{(B)}\hspace{.05in}122\qquad\textbf{(C)}\hspace{.05in}139\qquad\textbf{(D)}\hspace{.05in}150\qquad\textbf{(E)}\hspace{.05in}161$
| Let the number of two-legged birds be $x$ and the number of four-legged mammals be $y$ . We can now use systems of equations to solve this problem.
Write two equations:
$2x + 4y = 522$
$x + y = 200$
Now multiply the latter equation by $2$ .
$2x + 4y = 522$
$2x + 2y = 400$
By subtracting the second equation from the first equation, we find that $2y = 122 \implies y = 61$ . Since there were $200$ heads, meaning that there were $200$ animals, there were $200 - 61 = \boxed{\textbf{(C)}\ 139}$ two-legged birds.
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_9 | 139 |
AMC8_79 | If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?
$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$
| The sum of the ratios is $10$ . Since the sum of the angles of a triangle is $180^{\circ}$ , the ratio can be scaled up to $54:54:72$ $(3\cdot 18:3\cdot 18:4\cdot 18).$ The numbers in the ratio $54:54:72$ represent the angles of the triangle. The question asks for the largest, so the answer is $\boxed{\textbf{(D) }72}$ .
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_6 | 72 |
AMC8_81 | 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?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$
| We use that fact that $d=rt$ . Let d= distance, r= rate or speed, and t=time. In this case, let $x$ represent the time.
On Monday, he was at a rate of $5 \text{ m.p.h}$ . So, $5x = 2 \text{ miles}\implies x = \frac{2}{5} \text { hours}$ .
For Wednesday, he walked at a rate of $3 \text{ m.p.h}$ . Therefore, $3x = 2 \text{ miles}\implies x = \frac{2}{3} \text { hours}$ .
On Friday, he walked at a rate of $4 \text{ m.p.h}$ . So, $4x = 2 \text{ miles}\implies x=\frac{2}{4}=\frac{1}{2} \text {hours}$ .
Adding up the hours yields $\frac{2}{5} \text { hours}$ + $\frac{2}{3} \text { hours}$ + $\frac{1}{2} \text { hours}$ = $\frac{47}{30} \text { hours}$ .
We now find the amount of time Grandfather would have taken if he walked at $4 \text{ m.p.h}$ per day. Set up the equation, $4x = 2 \text{ miles} \times 3 \text{ days}\implies x = \frac{3}{2} \text { hours}$ .
To find the amount of time saved, subtract the two amounts: $\frac{47}{30} \text { hours}$ - $\frac{3}{2} \text { hours}$ = $\frac{1}{15} \text { hours}$ . To convert this to minutes, we multiply by $60$ .
Thus, the solution to this problem is $\dfrac{1}{15}\times 60=\boxed{\textbf{(D)}\ 4}$
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_11 | 4 |
AMC8_82 | Three-digit powers of $2$ and $5$ are used in this "cross-number" puzzle. What is the only possible digit for the outlined square?
\[\begin{array}{lcl} \textbf{ACROSS} & & \textbf{DOWN} \\ \textbf{2}.~ 2^m & & \textbf{1}.~ 5^n \end{array}\]
[asy] draw((0,-1)--(1,-1)--(1,2)--(0,2)--cycle); draw((0,1)--(3,1)--(3,0)--(0,0)); draw((3,0)--(2,0)--(2,1)--(3,1)--cycle,linewidth(2)); label("$1$",(0,2),SE); label("$2$",(0,1),SE); [/asy]
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$
| The $3$ -digit powers of $5$ are $125$ and $625$ , so space $2$ is filled with a $2$ .
The only $3$ -digit power of $2$ beginning with $2$ is $256$ , so the outlined block is filled with
a $\boxed{\text{(D) 6}}$ .
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_9 | 6 |
AMC8_83 | Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have to read?
$\textbf{(A)}\ 6400\qquad\textbf{(B)}\ 6600\qquad\textbf{(C)}\ 6800\qquad\textbf{(D)}\ 7000\qquad\textbf{(E)}\ 7200$
| The amount of pages Bob, Chandra, and Alice will read is in the ratio 4:6:9. Therefore, Bob, Chandra, and Alice read 160, 240, and 360 pages respectively. They would also be reading for the same amount of time because the ratio of the pages read was based on the time it takes each of them to read a page. Therefore, the amount of seconds each person reads is simply $160 \cdot 45 = \boxed{\textbf{(E)}\ 7200}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_16 | 7200 |
AMC8_84 | Austin and Temple are $50$ miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging $60$ miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged $40$ miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?
$\textbf{(A)}\ 46 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 54$
| The way to Temple took $\frac{50}{60}=\frac56$ hours, and the way back took $\frac{50}{40}=\frac54$ for a total of $\frac56 + \frac54 = \frac{25}{12}$ hours. The trip is $50\cdot2=100$ miles. The average speed is $\frac{100}{25/12} = \boxed{\textbf{(B)}\ 48}$ miles per hour.
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_14 | 48 |
AMC8_85 | In order for Mateen to walk a kilometer (1000m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters?
$\text{(A)}\ 40 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 400 \qquad \text{(D)}\ 500 \qquad \text{(E)}\ 1000$
| The length $L$ of the rectangle is $\frac{1000}{25}=40$ meters. The perimeter $P$ is $\frac{1000}{10}=100$ meters. Since $P_{rect} = 2L + 2W$ , we plug values in to get:
$100 = 2\cdot 40 + 2W$
$100 = 80 + 2W$
$2W = 20$
$W = 10$ meters
Since $A_{rect} = LW$ , the area is $40\cdot 10=400$ square meters or $\boxed{C}$ .
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_16 | C |
AMC8_86 | Candy sales from the Boosters Club from January through April are shown. What were the average sales per month in dollars?
[asy] draw((0,0)--(36,0)--(36,24)--(0,24)--cycle); draw((0,4)--(36,4)); draw((0,8)--(36,8)); draw((0,12)--(36,12)); draw((0,16)--(36,16)); draw((0,20)--(36,20)); fill((4,0)--(8,0)--(8,20)--(4,20)--cycle, black); fill((12,0)--(16,0)--(16,12)--(12,12)--cycle, black); fill((20,0)--(24,0)--(24,8)--(20,8)--cycle, black); fill((28,0)--(32,0)--(32,24)--(28,24)--cycle, black); label("120", (0,24), W); label("80", (0,16), W); label("40", (0,8), W); label("Jan", (6,0), S); label("Feb", (14,0), S); label("Mar", (22,0), S); label("Apr", (30,0), S); [/asy]
$\textbf{(A)}\ 60\qquad\textbf{(B)}\ 70\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 80\qquad\textbf{(E)}\ 85$
| There are a total of $100+60+40+120=320$ dollars of sales spread through $4$ months, for an average of $320/4 = \boxed{\textbf{(D)}\ 80}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_8 | 80 |
AMC8_87 | A small airplane has $4$ rows of seats with $3$ seats in each row. Eight passengers have boarded the plane and are distributed randomly among the seats. A married couple is next to board. What is the probability there will be 2 adjacent seats in the same row for the couple?
$\textbf{(A)} \frac{8}{15}\qquad\textbf{(B)} \frac{32}{55}\qquad\textbf{(C) } \frac{20}{33}\qquad\textbf{(D) } \frac{34}{55}\qquad\textbf{(E) } \frac{8}{11}$
| Suppose the passengers are indistinguishable. There are $\binom{12}{8} = 495$ total ways to distribute the passengers. We proceed with complementary counting, and instead, will count the number of passenger arrangements such that the couple cannot sit anywhere. Consider the partitions of $8$ among the rows of $3$ seats, to make our lives easier, assuming they are non-increasing. We have $(3, 3, 2, 0), (3, 3, 1, 1), (3, 2, 2, 1), (2, 2, 2, 2)$ .
For the first partition, clearly, the couple will always be able to sit in the row with $0$ occupied seats, so we have $0$ cases here.
For the second partition, there are $\frac{4!}{2!2!} = 6$ ways to permute the partition. Now the rows with exactly $1$ passenger must be in the middle, so this case generates $6$ cases.
For the third partition, there are $\frac{4!}{2!} = 12$ ways to permute the partition. For rows with $2$ passengers, there are $\binom{3}{2} = 3$ ways to arrange them in the row so that the couple cannot sit there. The row with $1$ passenger must be in the middle. We obtain $12 \cdot 3^2 = 108$ cases.
For the fourth partition, there is $1$ way to permute the partition. As said before, rows with $2$ passengers can be arranged in $3$ ways, so we obtain $3^4 = 81$ cases.
Collectively, we obtain a total of $6 + 108 + 81 = 195$ cases. The final probability is $1 - \frac{195}{495} = \boxed{\textbf{(C)}~\frac{20}{33}}$ .
~ blueprimes , Wrenmath
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_25 | 20/33 |
AMC8_88 | How many three-digit numbers are divisible by 13?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ 69\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 77$
| Let $k$ be any positive integer so that $13k$ is a multiple of $13$ . For the smallest three-digit number, $13k>100$ and $k>\frac{100}{13} \approx 7.7$ . For the greatest three-digit number, $13k<999$ and $k<\frac{999}{13} \approx 76.8$ . The number $k$ can range from $8$ to $76$ so there are $\boxed{\textbf{(C)}\ 69}$ three-digit numbers.
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_18 | 69 |
AMC8_89 | Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?
$\textbf{(A) }0\quad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$
| Given $n$ games, there must be a total of $n$ wins and $n$ losses. Hence, $4 + 3 + K = 2 + 3 + 3$ where $K$ is Kyler's wins. $K = 1$ , so our final answer is $\boxed{\textbf{(B)}\ 1}.$
~CHECKMATE2021
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_13 | 1 |
AMC8_90 | The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?
[asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$1$",(0,.5)); label("$3$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$5$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]
[asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$2$",(0,.5)); label("$4$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$6$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]
$\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{2}{3}\qquad\textbf{(C)}\ \frac{3}{4}\qquad\textbf{(D)}\ \frac{7}{9}\qquad\textbf{(E)}\ \frac{5}{6}$
| The possible sums are
\[\begin{tabular}{c|ccc} & 1 & 3 & 5 \\ \hline 2 & 3 & 5 & 7 \\ 4 & 5 & 7 & 9 \\ 6 & 7 & 9 & 11 \end{tabular}\]
Only $9$ is not prime, so there are $7$ prime numbers and $9$ total numbers for a probability of $\boxed{\textbf{(D)}\ \frac79}$ .
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_12 | 7/9 |
AMC8_92 | Jorge's teacher asks him to plot all the ordered pairs $(w. l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area 12. What should his graph look like?
$\textbf{(A)}$ [asy] size(75); draw((0,-1)--(0,13)); draw((-1,0)--(13,0)); dot((1,12)); dot((2,6)); dot((3,4)); dot((4,3)); dot((6,2)); dot((12,1)); label("$l$", (0,6), W); label("$w$", (6,0), S);[/asy]
$\textbf{(B)}$ [asy] size(75); draw((0,-1)--(0,13)); draw((-1,0)--(13,0)); dot((1,1)); dot((3,3)); dot((5,5)); dot((7,7)); dot((9,9)); dot((11,11)); label("$l$", (0,6), W); label("$w$", (6,0), S);[/asy]
$\textbf{(C)}$ [asy] size(75); draw((0,-1)--(0,13)); draw((-1,0)--(13,0)); dot((1,11)); dot((3,9)); dot((5,7)); dot((7,5)); dot((9,3)); dot((11,1)); label("$l$", (0,6), W); label("$w$", (6,0), S);[/asy]
$\textbf{(D)}$ [asy] size(75); draw((0,-1)--(0,13)); draw((-1,0)--(13,0)); dot((1,6)); dot((3,6)); dot((5,6)); dot((7,6)); dot((9,6)); dot((11,6)); label("$l$", (0,6), W); label("$w$", (6,0), S);[/asy]
$\textbf{(E)}$ [asy] size(75); draw((0,-1)--(0,13)); draw((-1,0)--(13,0)); dot((6,1)); dot((6,3)); dot((6,5)); dot((6,7)); dot((6,9)); dot((6,11)); label("$l$", (0,6), W); label("$w$", (6,0), S);[/asy]
| The length of the rectangle will relate invertly to the width, specifically using the theorem $l=\frac{12}{w}$ . The only graph that could represent a inverted relationship is $\boxed{\textbf{(A)}}$ . (The rest are linear graphs that represent direct relationships, therefore they are incorrect.)
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_10 | A |
AMC8_93 | Kaleana shows her test score to Quay, Marty and Shana, but the others keep theirs hidden. Quay thinks, "At least two of us have the same score." Marty thinks, "I didn't get the lowest score." Shana thinks, "I didn't get the highest score." List the scores from lowest to highest for Marty (M), Quay (Q) and Shana (S).
$\text{(A)}\ \text{S,Q,M} \qquad \text{(B)}\ \text{Q,M,S} \qquad \text{(C)}\ \text{Q,S,M} \qquad \text{(D)}\ \text{M,S,Q} \qquad \text{(E)}\ \text{S,M,Q}$
| Since the only other score Quay knows is Kaleana's, and he knows that two of them have the same score, Quay and Kaleana must have the same score, and $K=Q$ . Marty knows that he didn't get the lowest score, and the only other score he knows is Kaleana's, so Marty must know that Kaleana must have a lower score than him, and $M>K$ . Finally, Shana knows that she didn't get the highest score, and the only other score she knows is Kaleana's, so Shana must know that Kaleana must have a higher score than her, and $S<K$ . Putting these together and substituting $Q$ for $K$ , we have $S<Q<M$ , and from least to greatest this is $\text{S, Q, M}$ , therefore the answer is $\boxed{\text{A}}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_20 | A |
AMC8_94 | In the BIG N, a middle school football conference, each team plays every other team exactly once. If a total of 21 conference games were played during the 2012 season, how many teams were members of the BIG N conference?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}10$
| This problem is very similar to a handshake problem. We use the formula $\frac{n(n-1)}{2}$ to usually find the number of games played (or handshakes). Now we have to use the formula in reverse.
So we have the equation $\frac{n(n-1)}{2} = 21$ . Solving, we find that the number of teams in the BIG N conference is $\boxed{\textbf{(B)}\ 7}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_14 | 7 |
AMC8_95 | A ball is dropped from a height of $3$ meters. On its first bounce it rises to a height of $2$ meters. It keeps falling and bouncing to $\frac{2}{3}$ of the height it reached in the previous bounce. On which bounce will it not rise to a height of $0.5$ meters?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$
| Each bounce is $2/3$ times the height of the previous bounce. The first bounce reaches $2$ meters, the second $4/3$ , the third $8/9$ , the fourth $16/27$ , and the fifth $32/81$ . Half of $81$ is $40.5$ , so the ball does not reach the required height on bounce $\boxed{\textbf{(C)}\ 5}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_12 | 5 |
AMC8_96 | The letters $\text{P}, \text{Q},$ and $\text{R}$ are entered into a $20\times20$ table according to the pattern shown below. How many $\text{P}$ s, $\text{Q}$ s, and $\text{R}$ s will appear in the completed table?
[asy] /* Made by MRENTHUSIASM, Edited by Kante314 */ usepackage("mathdots"); size(5cm); draw((0,0)--(6,0),linewidth(1.5)+mediumgray); draw((0,1)--(6,1),linewidth(1.5)+mediumgray); draw((0,2)--(6,2),linewidth(1.5)+mediumgray); draw((0,3)--(6,3),linewidth(1.5)+mediumgray); draw((0,4)--(6,4),linewidth(1.5)+mediumgray); draw((0,5)--(6,5),linewidth(1.5)+mediumgray); draw((0,0)--(0,6),linewidth(1.5)+mediumgray); draw((1,0)--(1,6),linewidth(1.5)+mediumgray); draw((2,0)--(2,6),linewidth(1.5)+mediumgray); draw((3,0)--(3,6),linewidth(1.5)+mediumgray); draw((4,0)--(4,6),linewidth(1.5)+mediumgray); draw((5,0)--(5,6),linewidth(1.5)+mediumgray); label(scale(.9)*"\textsf{P}", (.5,.5)); label(scale(.9)*"\textsf{Q}", (.5,1.5)); label(scale(.9)*"\textsf{R}", (.5,2.5)); label(scale(.9)*"\textsf{P}", (.5,3.5)); label(scale(.9)*"\textsf{Q}", (.5,4.5)); label("$\vdots$", (.5,5.6)); label(scale(.9)*"\textsf{Q}", (1.5,.5)); label(scale(.9)*"\textsf{R}", (1.5,1.5)); label(scale(.9)*"\textsf{P}", (1.5,2.5)); label(scale(.9)*"\textsf{Q}", (1.5,3.5)); label(scale(.9)*"\textsf{R}", (1.5,4.5)); label("$\vdots$", (1.5,5.6)); label(scale(.9)*"\textsf{R}", (2.5,.5)); label(scale(.9)*"\textsf{P}", (2.5,1.5)); label(scale(.9)*"\textsf{Q}", (2.5,2.5)); label(scale(.9)*"\textsf{R}", (2.5,3.5)); label(scale(.9)*"\textsf{P}", (2.5,4.5)); label("$\vdots$", (2.5,5.6)); label(scale(.9)*"\textsf{P}", (3.5,.5)); label(scale(.9)*"\textsf{Q}", (3.5,1.5)); label(scale(.9)*"\textsf{R}", (3.5,2.5)); label(scale(.9)*"\textsf{P}", (3.5,3.5)); label(scale(.9)*"\textsf{Q}", (3.5,4.5)); label("$\vdots$", (3.5,5.6)); label(scale(.9)*"\textsf{Q}", (4.5,.5)); label(scale(.9)*"\textsf{R}", (4.5,1.5)); label(scale(.9)*"\textsf{P}", (4.5,2.5)); label(scale(.9)*"\textsf{Q}", (4.5,3.5)); label(scale(.9)*"\textsf{R}", (4.5,4.5)); label("$\vdots$", (4.5,5.6)); label(scale(.9)*"$\dots$", (5.5,.5)); label(scale(.9)*"$\dots$", (5.5,1.5)); label(scale(.9)*"$\dots$", (5.5,2.5)); label(scale(.9)*"$\dots$", (5.5,3.5)); label(scale(.9)*"$\dots$", (5.5,4.5)); label(scale(.9)*"$\iddots$", (5.5,5.6)); [/asy]
$\textbf{(A)}~132\text{ Ps, }134\text{ Qs, }134\text{ Rs}$
$\textbf{(B)}~133\text{ Ps, }133\text{ Qs, }134\text{ Rs}$
$\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}$
$\textbf{(D)}~134\text{ Ps, }132\text{ Qs, }134\text{ Rs}$
$\textbf{(E)}~134\text{ Ps, }133\text{ Qs, }133\text{ Rs}$
| In our $5\times5$ grid, there are $8,9$ and $8$ of the letters $\text{P}, \text{Q},$ and $\text{R}$ , respectively, and in a $2\times2$ grid, there are $1,2$ and $1$ of the letters $\text{P}, \text{Q},$ and $\text{R}$ , respectively. We see that in both grids, there are $x, x+1,$ and $x$ of the $\text{P}, \text{Q},$ and $\text{R}$ , respectively. This is because in any $n\times n$ grid with $n\equiv2\pmod3$ , there are $x, x+1,$ and $x$ of the $\text{P}, \text{Q},$ and $\text{R}$ , respectively. We can see that the only answer choice which satisfies this condition is $\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.$
~CoOlPoTaToEs, apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, Nivaar
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_16 | C |
AMC8_97 | Ningli looks at the $6$ pairs of numbers directly across from each other on a clock. She takes the average of each pair of numbers. What is the average of the resulting $6$ numbers?
[asy] unitsize(1cm); draw(circle((0,0),2)); for(int i = 1; i <= 12; ++i) { draw(1.9*dir(90-i*30)-- 2*dir(90-i*30));//,linewidth(1pt) label("$"+string(i)+"$",2.3*dir(90-i*30)); } draw(2*dir(-150)--2*dir(30),dashed); [/asy]
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9.5 \qquad \textbf{(E)}\ 12$
| Our answer is \[\frac{\frac{1+7}{2} + \frac{2+8}{2} + \cdots + \frac{6+12}{2}}{6} = \frac{\frac{1}{2}((1+7)+(2+8)+\cdots+(6+12))}{6} = \frac{1+2+3+4+5+6+7+8+9+10+11+12}{2 \cdot 6} = \frac{\frac{12 \cdot 13}{2}}{2 \cdot 6} = \frac{78}{12} =\boxed{\textbf{(B)}~6.5}\]
~Sigmacuber
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_9 | 6.5 |
AMC8_99 | The average age of the $6$ people in Room A is $40$ . The average age of the $4$ people in Room B is $25$ . If the two groups are combined, what is the average age of all the people?
$\textbf{(A)}\ 32.5 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 33.5 \qquad \textbf{(D)}\ 34\qquad \textbf{(E)}\ 35$
| The total of all their ages over the number of people is
\[\frac{6 \cdot 40 + 4\cdot 25}{6+4} = \frac{340}{10} = \boxed{\textbf{(D)}\ 34}.\]
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_10 | 34 |
AMC8_100 | Jordan owns 15 pairs of sneakers. Three fifths of the pairs are red and the rest are white. Two thirds of the pairs are high-top and the rest are low-top. The red high-top sneakers make up a fraction of the collection. What is the least possible value of this fraction?
$\textbf{(A) } 0\qquad\textbf{(B) } \dfrac{1}{5} \qquad\textbf{(C) } \dfrac{4}{15} \qquad\textbf{(D) } \dfrac{1}{3} \qquad\textbf{(E) } \dfrac{2}{5}$
| Jordan has $10$ high top sneakers, and $6$ white sneakers. We would want as many white high-top sneakers as possible, so we set $6$ high-top sneakers to be white. Then, we have $10-6=4$ red high-top sneakers, so the answer is $\boxed{\dfrac{4}{15}}.$
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_19 | 4/15 |
AMC8_101 | The results of a cross-country team's training run are graphed below. Which student has the greatest average speed?
[asy] for ( int i = 1; i <= 7; ++i ) { draw((i,0)--(i,6)); } for ( int i = 1; i <= 5; ++i ) { draw((0,i)--(8,i)); } draw((-0.5,0)--(8,0), linewidth(1)); draw((0,-0.5)--(0,6), linewidth(1)); label("$O$", (0,0), SW); label(scale(.85)*rotate(90)*"distance", (0, 3), W); label(scale(.85)*"time", (4, 0), S); dot((1.25, 4.5)); label(scale(.85)*"Evelyn", (1.25, 4.8), N); dot((2.5, 2.2)); label(scale(.85)*"Briana", (2.5, 2.2), S); dot((4.25,5.2)); label(scale(.85)*"Carla", (4.25, 5.2), SE); dot((5.6, 2.8)); label(scale(.85)*"Debra", (5.6, 2.8), N); dot((6.8, 1.4)); label(scale(.85)*"Angela", (6.8, 1.4), E); [/asy]
$\textbf{(A)}\ \text{Angela}\qquad\textbf{(B)}\ \text{Briana}\qquad\textbf{(C)}\ \text{Carla}\qquad\textbf{(D)}\ \text{Debra}\qquad\textbf{(E)}\ \text{Evelyn}$
| Average speed is distance over time, or the slope of the line through the point and the origin. $\boxed{\textbf{(E)}\ \text{Evelyn}}$ has the steepest line, and runs the greatest distance for the shortest amount of time.
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_17 | Evelyn |
AMC8_102 | On a beach $50$ people are wearing sunglasses and $35$ people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is also wearing sunglasses is $\frac{2}{5}$ . If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?
$\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}$
| The number of people wearing caps and sunglasses is
$\frac{2}{5}\cdot35=14$ . So then, 14 people out of the 50 people wearing sunglasses also have caps.
$\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}$
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_15 | 7/25 |
AMC8_103 | Which triplet of numbers has a sum NOT equal to 1?
$\text{(A)}\ (1/2,1/3,1/6) \qquad \text{(B)}\ (2,-2,1) \qquad \text{(C)}\ (0.1,0.3,0.6) \qquad \text{(D)}\ (1.1,-2.1,1.0) \qquad \text{(E)}\ (-3/2,-5/2,5)$
| By adding each triplet, we can see that $\boxed{(D)}$ gives us $0$ , not $1$ , as our sum.
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_3 | D |
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