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AMC8_104 | Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?
$\text{(A)}\ 30\qquad\text{(B)}\ 40\qquad\text{(C)}\ 50\qquad\text{(D)}\ 60\qquad\text{(E)}\ 70$
| A pear gives $8/3$ ounces of juice per pear. An orange gives $8/2=4$ ounces of juice per orange. If the pear-orange juice blend used one pear and one orange each, the percentage of pear juice would be
\[\frac{8/3}{8/3+4} \times 100 = \frac{8}{8+12} \times 100 = \boxed{\text{(B)}\ 40}\]
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_24 | 40 |
AMC8_105 | A ship travels from point $A$ to point $B$ along a semicircular path, centered at Island $X$ . Then it travels along a straight path from $B$ to $C$ . Which of these graphs best shows the ship's distance from Island $X$ as it moves along its course?
[asy]size(150); pair X=origin, A=(-5,0), B=(5,0), C=(0,5); draw(Arc(X, 5, 180, 360)^^B--C); dot(X); label("$X$", X, NE); label("$C$", C, N); label("$B$", B, E); label("$A$", A, W); [/asy]
| The distance from Island $\text{X}$ to any point on the semicircle will always be constant. On the graph, this will represent a straight line. The distance between Island $\text{X}$ and line $\text{BC}$ will not be constant though. We can easily prove that the distance between $\text{X}$ and line $\text{BC}$ will represent a curve. As the ship travels from $B$ to $C$ , the distance between the ship and Island $X$ will first decrease until it reaches the point $Y$ so that $\overline{XY}$ is perpendicular to $\overline{BC}$ , and then increase afterwards. Hence the answer choice that fits them all is $\boxed{\text{(B)}}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_24 | B |
AMC8_106 | In $\bigtriangleup ABC$ , $AB=BC=29$ , and $AC=42$ . What is the area of $\bigtriangleup ABC$ ?
$\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701$
| Solution 1We know the semi-perimeter of $\triangle ABC$ is $\frac{29+29+42}{2}=50$ . Next, we use Heron's Formula to find that the area of the triangle is just $\sqrt{50(50-29)^2(50-42)}=\sqrt{50 \cdot 21^2 \cdot 8}=\boxed{\textbf{(B) }420}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_6 | 420 |
AMC8_107 | The twelve-sided figure shown has been drawn on $1 \text{ cm}\times 1 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$ ?
[asy] unitsize(8mm); for (int i=0; i<7; ++i) { draw((i,0)--(i,7),gray); draw((0,i+1)--(7,i+1),gray); } draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); [/asy]
$\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14$
| We count $3 \cdot 3=9$ unit squares in the middle, and $8$ small triangles, which gives 4 rectangles each with an area of $1$ . Thus, the answer is $9+4=\boxed{\textbf{(C) } 13}$ .
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_4 | 13 |
AMC8_108 | The digits 1, 2, 3, 4 and 9 are each used once to form the smallest possible even five-digit number. The digit in the tens place is
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 9$
| Since the number is even, the last digit must be $2$ or $4$ . To make the smallest possible number, the ten-thousands digit must be as small as possible, so the ten-thousands digit is $1$ . Simillarly, the thousands digit has second priority, so it must also be as small as possible once the ten-thousands digit is decided, so the thousands digit is $2$ . Similarly, the hundreds digit needs to be the next smallest number, so it is $3$ . However, for the tens digit, we can't use $4$ , since we already used $2$ and the number must be even, so the units digit must be $4$ and the tens digit is $9, \boxed{\text{E}}$
(The number is $12394$ .)
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_4 | E |
AMC8_109 | In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$ , $BC=4$ , $CD=3$ , and $AD=13$ . What is the area of quadrilateral $ABCD$ ?
[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);[/asy]
$\textbf{(A) }12 \qquad \textbf{(B) }24 \qquad \textbf{(C) }26 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36$
| We first connect point $B$ with point $D$ .
[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); draw((0,0)--(0,5)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);[/asy]
We can see that $\triangle BCD$ is a 3-4-5 right triangle. We can also see that $\triangle BDA$ is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of $\triangle BDA$ is $\frac{5\cdot 12}{2}$ , and the area of $\triangle BCD$ is $\frac{3\cdot 4}{2}$ . Thus, the area of quadrilateral $ABCD$ is $30-6 = \boxed{\textbf{(B)}\ 24}.$
~CHECKMATE2021
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_18 | 24 |
AMC8_110 | Three members of the Euclid Middle School girls' softball team had the following conversation.
Ashley: I just realized that our uniform numbers are all $2$ -digit primes.
Bethany : And the sum of your two uniform numbers is the date of my birthday earlier this month.
Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month.
Ashley: And the sum of your two uniform numbers is today's date.
What number does Caitlin wear?
$\textbf{(A) }11\qquad\textbf{(B) }13\qquad\textbf{(C) }17\qquad\textbf{(D) }19\qquad \textbf{(E) }23$
| The maximum amount of days any given month can have is $31$ , and the smallest two-digit primes are $11, 13,$ and $17$ . There are a few different sums that can be deduced from the following numbers, which are $24, 30,$ and $28$ , all of which represent the three days. Therefore, since Bethany says that the other two people's uniform numbers are earlier, so that means Caitlin and Ashley's numbers must add up to $24$ . Similarly, Caitlin says that the other two people's uniform numbers are later, so the sum must add up to $30$ . This leaves $28$ as today's date. From this, Caitlin was referring to the uniform wearers $13$ and $17$ , telling us that her number is $11$ , giving our solution as $\boxed{(A) 11}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_23 | 11 |
AMC8_111 | How many subsets of two elements can be removed from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ so that the mean (average) of the remaining numbers is 6?
$\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}$
| Solution 1Since there will be $9$ elements after removal, and their mean is $6$ , we know their sum is $54$ . We also know that the sum of the set pre-removal is $66$ . Thus, the sum of the $2$ elements removed is $66-54=12$ . There are only $\boxed{\textbf{(D)}~5}$ subsets of $2$ elements that sum to $12$ : $\{1,11\}, \{2,10\}, \{3, 9\}, \{4, 8\}, \{5, 7\}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_13 | 5 |
AMC8_112 | A semicircle is inscribed in an isosceles triangle with base $16$ and height $15$ so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?
[asy]draw((0,0)--(8,15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,180));[/asy]
$\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}$
| [asy] pair A, B, C, D; A=(0,0); B=(16,0); C=(8,15); D=B/2; draw(A--B--C--cycle); draw(C--D); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$8$",(A+D)/2,S); label("$15$",(C+D)/2,NE); label("$17$",(A+C)/2,W); [/asy]
First, we drop a perpendicular, shown above, to the base of the triangle, cutting the triangle into two congruent right triangles. This triangle is isosceles, which means perpendiculars are medians and vice versa. The base of the resulting right triangle is $8$ for both sides, and the height is $15,$ as given. Using the Pythagorean theorem, we can find the length of the hypotenuse, or $17.$ Using the two legs of the right triangle, we find the area of the right triangle, $60$ . $\frac{60}{17}$ times $2$ results in the radius, which is the height of the right triangle when using the hypotenuse as the base. Hence, the answer is $\boxed{\textbf{(B) }\frac{120}{17}}$ .
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_25 | 120/17 |
AMC8_113 | Triangles $ABC$ , $ADE$ , and $EFG$ are all equilateral. Points $D$ and $G$ are midpoints of $\overline{AC}$ and $\overline{AE}$ , respectively. If $AB = 4$ , what is the perimeter of figure $ABCDEFG$ ?
[asy] pair A,B,C,D,EE,F,G; A = (4,0); B = (0,0); C = (2,2*sqrt(3)); D = (3,sqrt(3)); EE = (5,sqrt(3)); F = (5.5,sqrt(3)/2); G = (4.5,sqrt(3)/2); draw(A--B--C--cycle); draw(D--EE--A); draw(EE--F--G); label("$A$",A,S); label("$B$",B,SW); label("$C$",C,N); label("$D$",D,NE); label("$E$",EE,NE); label("$F$",F,SE); label("$G$",G,SE);[/asy]
$\text{(A)}\ 12\qquad\text{(B)}\ 13\qquad\text{(C)}\ 15\qquad\text{(D)}\ 18\qquad\text{(E)}\ 21$
| The large triangle $ABC$ has sides of length $4$ . The medium triangle has sides of length $2$ . The small triangle has sides of length $1$ . There are $3$ segment sizes, and all segments depicted are one of these lengths.
Starting at $A$ and going clockwise, the perimeter is:
$AB + BC + CD + DE + EF + FG + GA$
$4 + 4 + 2 + 2 + 1 + 1 + 1$
$15$ , thus the answer is $\boxed{C}$
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_15 | C |
AMC8_114 | At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: $1\frac{1}{2}$ cups of flour, $2$ eggs, $3$ tablespoons butter, $\frac{3}{4}$ cups sugar, and $1$ package of chocolate drops. They will make only full recipes, not partial recipes.
Walter can buy eggs by the half-dozen. How many half-dozens should he buy to make enough cookies? (Some eggs and some cookies may be left over.)
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 15$
| If $108$ students eat $2$ cookies on average, there will need to be $108\cdot 2 = 216$ cookies. There are $15$ cookies per pan, meaning there needs to be $\frac{216}{15} = 14.4$ pans. However, since half-recipes are forbidden, we need to round up and make $\lceil \frac{216}{15}\rceil = 15$ pans.
$1$ pan requires $2$ eggs, so $15$ pans require $2\cdot 15 = 30$ eggs. Since there are $6$ eggs in a half dozen, we need $\frac{30}{6} = 5$ half-dozens of eggs, and the answer is $\boxed{C}$
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_17 | C |
AMC8_115 | For the game show Who Wants To Be A Millionaire? , the dollar values of each question are shown in the following table (where K = 1000).
\[\begin{tabular}{rccccccccccccccc} \text{Question} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ \text{Value} & 100 & 200 & 300 & 500 & 1\text{K} & 2\text{K} & 4\text{K} & 8\text{K} & 16\text{K} & 32\text{K} & 64\text{K} & 125\text{K} & 250\text{K} & 500\text{K} & 1000\text{K} \end{tabular}\]
Between which two questions is the percent increase of the value the smallest?
$\text{(A)}\ \text{From 1 to 2} \qquad \text{(B)}\ \text{From 2 to 3} \qquad \text{(C)}\ \text{From 3 to 4} \qquad \text{(D)}\ \text{From 11 to 12} \qquad \text{(E)}\ \text{From 14 to 15}$
| Notice that in two of the increases, the dollar amount doubles. The increases in which this is not true is $2$ to $3$ , $3$ to $4$ , and $11$ to $12$ . We can disregard $11$ to $12$ since that increase is almost $2$ times. The increase from $2$ to $3$ is $\frac{300-200}{200}=\frac{1}{2}$ , so it's only multiplied by a factor of $1.5$ . The increase from $3$ to $4$ is $\frac{500-300}{300}=\frac{2}{3}$ , so it is multiplied by a factor of $1.\bar{6}$ . Therefore, the smallest percent increase is $\text{From 2 to 3}, \boxed{\text{B}}$
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_17 | B |
AMC8_117 | Each principal of Lincoln High School serves exactly one $3$ -year term. What is the maximum number of principals this school could have during an $8$ -year period?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 8$
| If the first year of the $8$ -year period was the final year of a principal's
term, then in the next six years two more principals would serve, and the last year of the
period would be the first year of the fourth principal's term. Therefore, the maximum
number of principals who can serve during an $8$ -year period is $4$ , so the answer is $\boxed{C}$
if the terms are divided $1\ |\ 2\ 3\ 4\ |\ 5\ 6\ 7\ |\ 8$
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_5 | C |
AMC8_118 | The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $11/4$ . To the nearest whole percent, what percent of its games did the team lose?
$\text{(A)}\ 24\% \qquad \text{(B)}\ 27\% \qquad \text{(C)}\ 36\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 73\%$
| The ratio means that for every $11$ games won, $4$ are lost, so the team has won $11x$ games, lost $4x$ games, and played $15x$ games for some positive integer $x$ . The percentage of games lost is just $\dfrac{4x}{15x}\times100=\dfrac{4}{15}\times 100=26.\overline{6}\%\approx\boxed{\text{(B)}\ 27\%}$
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_12 | 27 |
AMC8_119 | A recipe that makes $5$ servings of hot chocolate requires $2$ squares of chocolate, $\frac{1}{4}$ cup sugar, $1$ cup water and $4$ cups milk. Jordan has $5$ squares of chocolate, $2$ cups of sugar, lots of water, and $7$ cups of milk. If he maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate he can make?
$\textbf{(A)}\ 5 \frac18 \qquad \textbf{(B)}\ 6\frac14 \qquad \textbf{(C)}\ 7\frac12 \qquad \textbf{(D)}\ 8 \frac34 \qquad \textbf{(E)}\ 9\frac78$
| Assuming excesses of the other ingredients, the chocolate can make $\frac52 \cdot 5=12.5$ servings, the sugar can make $\frac{2}{1/4} \cdot 5 = 40$ servings, the water can make unlimited servings, and the milk can make $\frac74 \cdot 5 = 8.75$ servings. Limited by the amount of milk, Jordan can make at most $\boxed{\textbf{(D)}\ 8 \frac34}$ servings.
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_15 | 8 3/4 |
AMC8_120 | Any three vertices of the cube $PQRSTUVW$ , shown in the figure below, can be connected to form a triangle. (For example, vertices $P$ , $Q$ , and $R$ can be connected to form isosceles $\triangle PQR$ .) How many of these triangles are equilateral and contain $P$ as a vertex?
[asy] unitsize(4); pair P,Q,R,S,T,U,V,W; P=(0,30); Q=(30,30); R=(40,40); S=(10,40); T=(10,10); U=(40,10); V=(30,0); W=(0,0); draw(W--V); draw(V--Q); draw(Q--P); draw(P--W); draw(T--U); draw(U--R); draw(R--S); draw(S--T); draw(W--T); draw(P--S); draw(V--U); draw(Q--R); dot(P); dot(Q); dot(R); dot(S); dot(T); dot(U); dot(V); dot(W); label("$P$",P,NW); label("$Q$",Q,NW); label("$R$",R,NE); label("$S$",S,N); label("$T$",T,NE); label("$U$",U,NE); label("$V$",V,SE); label("$W$",W,SW); [/asy]
$\textbf{(A)}0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }6$
| The only equilateral triangles that can be formed are through the diagonals of the faces of the square. From P you have $3$ possible vertices that are possible to form a diagonal through one of the faces. Therefore, there are $3$ possible triangles. So the answer is $\boxed{\textbf{(D) }3}$
~Math645, andliu766, e___
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_20 | 3 |
AMC8_121 | Mr. Ramos gave a test to his class of $20$ students. The dot plot below shows the distribution of test scores.
[asy] //diagram by pog . give me 1,000,000,000 dollars for this diagram size(5cm); defaultpen(0.7); dot((0.5,1)); dot((0.5,1.5)); dot((1.5,1)); dot((1.5,1.5)); dot((2.5,1)); dot((2.5,1.5)); dot((2.5,2)); dot((2.5,2.5)); dot((3.5,1)); dot((3.5,1.5)); dot((3.5,2)); dot((3.5,2.5)); dot((3.5,3)); dot((4.5,1)); dot((4.5,1.5)); dot((5.5,1)); dot((5.5,1.5)); dot((5.5,2)); dot((6.5,1)); dot((7.5,1)); draw((0,0.5)--(8,0.5),linewidth(0.7)); defaultpen(fontsize(10.5pt)); label("$65$", (0.5,-0.1)); label("$70$", (1.5,-0.1)); label("$75$", (2.5,-0.1)); label("$80$", (3.5,-0.1)); label("$85$", (4.5,-0.1)); label("$90$", (5.5,-0.1)); label("$95$", (6.5,-0.1)); label("$100$", (7.5,-0.1)); [/asy]
Later Mr. Ramos discovered that there was a scoring error on one of the questions. He regraded the tests, awarding some of the students $5$ extra points, which increased the median test score to $85$ . What is the minimum number of students who received extra points?
(Note that the median test score equals the average of the $2$ scores in the middle if the $20$ test scores are arranged in increasing order.)
$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~5\qquad\textbf{(E)} ~6\qquad$
| We notice that $13$ students have scores under $85$ currently and only $5$ have scores over $85$ . We find the median of these two numbers, getting:
\[13-5=8\]
\[\frac{8}{2}=4\]
\[13-4=9\]
Thus, we realize that $4$ students must have their score increased by $5$ .
So, the correct answer is $\boxed{\textbf{(C)}4}$ .
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_19 | 4 |
AMC8_122 | A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are there from $P$ to $Q?$ (The figure shows a sample path.)
[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("$P$", (5.5, 0.5)); label("$Q$", (6.5, 7.5)); [/asy]
$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$
| Notice that, to step onto any particular white square, the marker must have come from one of the $1$ or $2$ white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from $P$ to that square is the sum of the number of ways to move from $P$ to each of the white squares immediately beneath it (also called the Waterfall Method). To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated).
[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label("$1$", (5.5, .5)); label("$1$", (4.5, 1.5)); label("$1$", (6.5, 1.5)); label("$1$", (3.5, 2.5)); label("$1$", (7.5, 2.5)); label("$2$", (5.5, 2.5)); label("$1$", (2.5, 3.5)); label("$3$", (6.5, 3.5)); label("$3$", (4.5, 3.5)); label("$4$", (3.5, 4.5)); label("$3$", (7.5, 4.5)); label("$6$", (5.5, 4.5)); label("$10$", (4.5, 5.5)); label("$9$", (6.5, 5.5)); label("$19$", (5.5, 6.5)); label("$9$", (7.5, 6.5)); label("$28$", (6.5, 7.5)); [/asy]
The answer is therefore $\boxed{\textbf{(A) }28}$ .
Note: This is a classic example of a problem involving Pascal's triangle .
~ edited by Dream9, WrenMath and algebraic_algorithmic
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_21 | 28 |
AMC8_123 | To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.
[asy] size(85); for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } } draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy]
Genevieve puts bracing on her large kite in the form of a cross connecting opposite corners of the kite. How many inches of bracing material does she need?
$\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 38 \qquad \text{(E)}\ 39$
| Each diagonal of the large kite is $3$ times the length of the corresponding diagonal of the short kite since it was made with a grid $3$ times as long in both height and width. The diagonals of the small kite are $6$ and $7$ , so the diagonals of the large kite are $3\cdot6=18$ and $3\cdot7=21$ , and the amount of bracing Genevieve needs is the sum of these lengths, which is $39, \boxed{\text{E}}$
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_8 | E |
AMC8_124 | For how many positive integer values of $n$ are both $\frac{n}{3}$ and $3n$ three-digit whole numbers?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 27\qquad \textbf{(D)}\ 33\qquad \textbf{(E)}\ 34$
| Instead of finding n, we find $x=\frac{n}{3}$ . We want $x$ and $9x$ to be three-digit whole numbers. The smallest three-digit whole number is $100$ , so that is our minimum value for $x$ , since if $x \in \mathbb{Z^+}$ , then $9x \in \mathbb{Z^+}$ . The largest three-digit whole number divisible by $9$ is $999$ , so our maximum value for $x$ is $\frac{999}{9}=111$ . There are $12$ whole numbers in the closed set $\left[100,111\right]$ , so the answer is $\boxed{\textbf{(A)}\ 12}$ .
- ColtsFan10
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_22 | 12 |
AMC8_125 | The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43$ dollars. Some of the $30$ sixth graders each bought a pencil, and they paid a total of $1.95$ dollars. How many more sixth graders than seventh graders bought a pencil?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$
| Because the pencil costs a whole number of cents, the cost must be a factor of both $143$ and $195$ . They can be factored into $11\cdot13$ and $3\cdot5\cdot13$ . The common factor cannot be $1$ or there would have to be more than $30$ sixth graders, so the pencil costs $13$ cents. The difference in costs that the sixth and seventh graders paid is $195-143=52$ cents, which is equal to $52/13 = \boxed{\textbf{(D)}\ 4}$ sixth graders.
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_11 | 4 |
AMC8_126 | One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?
[asy] size(75); draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); [/asy]
$\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17$
| Solution 1We can draw a diagram as shown:
[asy] size(75); draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); path arc = arc((2.5,4),1.5,0,90); pair P = intersectionpoint(arc,(0,5)--(5,5)); pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; draw(P--Pp--Ppp--Pppp--cycle); [/asy]
Let us focus on the $4$ big triangles that, together with the inscribed square, fill in the large square. If we zoom in on one of the four triangles, we can see that it is composed of a small unit square in the corner of the large square and two triangles, one smaller than the other. We are going to focus specifically on the smaller of the two triangles. This triangle is similar to the big triangle itself by $\mathrm{AA}$ similarity (because the two sides of a square are parallel. To prove this fact, draw a diagonal of the square and find congruent triangles). Let the shorter leg of the big triangle be $x$ ; then $\dfrac{x}{x-1}=\dfrac{5-x}{1}$ .
\[x=-x^2+6x-5\]
\[x^2-5x+5=0\]
\[x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}\]
\[x=\dfrac{5\pm \sqrt{5}}{2}\]
Thus, $x=\dfrac{5-\sqrt{5}}{2}$ , because by symmetry, $x < \dfrac52$ . Note that the other solution we got, namely, $x=\dfrac{5+\sqrt 5} 2$ , is the length of the segment $5-x$ . $x$ and $5-x$ together sum to $5$ , the side of the length of the large square, and similarly, the sum of the solutions is $5$ . This solution is a result of the symmetry of the problem; if we had set the longer leg of the big triangle to be $x$ , then we would solve the same quadratic to find the same roots, the only difference being that we take the other root.
This means the area of each triangle is $\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}$ .
Thus, the area of the square is $25-(4*\dfrac{5}{2})=15\implies \boxed{\textbf{(C)}}$ .
Don't use this in a contest scenario, and only use it when practicing math skills. :)
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_25 | C |
AMC8_127 | Harry and Terry are each told to calculate $8-(2+5)$ . Harry gets the correct answer. Terry ignores the parentheses and calculates $8-2+5$ . If Harry's answer is $H$ and Terry's answer is $T$ , what is $H-T$ ?
$\textbf{(A) }-10\qquad\textbf{(B) }-6\qquad\textbf{(C) }0\qquad\textbf{(D) }6\qquad \textbf{(E) }10$
| We have $H=8-7=1$ and $T=8-2+5=11$ . Clearly $1-11=-10$
, so our answer is $\boxed{\textbf{(A)}-10}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_1 | -10 |
AMC8_128 | Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$ ?
[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label("A",(-13,0),W); label("B",(0,5),N); label("C",(13,0),E); label("D",(0,-5),S); [/asy]
$\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144$
| [asy] draw((-12,0)--(0,5)); draw((0,5)--(12,0)); draw((12,0)--(0,-5)); draw((0,-5)--(-12,0)); draw((0,0)--(12,0)); draw((0,0)--(0,5)); draw((0,0)--(-12,0)); draw((0,0)--(0,-5)); dot((-12,0)); dot((0,5)); dot((12,0)); dot((0,-5)); label("A",(-12,0),W); label("B",(0,5),N); label("C",(12,0),E); label("D",(0,-5),S); label("E",(0,0),SW); [/asy]
A rhombus has sides of equal length. Because the perimeter of the rhombus is $52$ , each side is $\frac{52}{4}=13$ . In a rhombus, diagonals are perpendicular and bisect each other, which means $\overline{AE}$ = $12$ = $\overline{EC}$ .
Consider one of the
[asy] draw((-12,0)--(0,5)); draw((0,0)--(-12,0)); draw((0,0)--(0,5)); dot((-12,0)); dot((0,5)); label("A",(-12,0),W); label("B",(0,5),N); label("E",(0,0),SE); [/asy]
$\overline{AB}$ = $13$ , and $\overline{AE}$ = $12$ . Using the Pythagorean theorem, we find that $\overline{BE}$ = $5$ .
You know the Pythagorean triple, (5, 12, 13).
Thus the values of the two diagonals are $\overline{AC}$ = $24$ and $\overline{BD}$ = $10$ .
The area of a rhombus is = $\frac{d_1\cdot{d_2}}{2}$ = $\frac{24\cdot{10}}{2}$ = $120$
$\boxed{\textbf{(D)}\ 120}$
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_4 | 120 |
AMC8_129 | At Euclid Middle School the mathematics teachers are Mrs. Germain, Mr. Newton, and Mrs. Young. There are $11$ students in Mrs. Germain's class, $8$ students in Mr. Newton's class, and $9$ students in Mrs. Young's class taking the AMC $8$ this year. How many mathematics students at Euclid Middle School are taking the contest?
$\textbf{(A)}\ 26 \qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30$
| Given that these are the only math teachers at Euclid Middle School and we are told how many from each class are taking the AMC 8, we simply add the three numbers to find the total. $11+8+9=\boxed{\textbf{(C)}\ 28}$
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_1 | 28 |
AMC8_130 | In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?
$\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$
| This isn't finished
to another. This gives equality, as each team wins once and loses once as well. For a win, we have $3$ points, so a team gets $3\times2=6$ points if they each win a game and lose a game. This case brings a total of $18+6=24$ points.
Therefore, we use Case 2 since it brings the greater amount of points, or $\boxed {\textbf {(C) }24}$ .
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_19 | 24 |
AMC8_131 | Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour?
[asy] import graph; size(8.76cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.58,xmax=10.19,ymin=-4.43,ymax=9.63; draw((0,0)--(0,8)); draw((0,0)--(8,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); draw((1,0)--(1,8)); draw((2,0)--(2,8)); draw((3,0)--(3,8)); draw((4,0)--(4,8)); draw((5,0)--(5,8)); draw((6,0)--(6,8)); draw((7,0)--(7,8)); label("$1$",(0.95,-0.24),SE*lsf); label("$2$",(1.92,-0.26),SE*lsf); label("$3$",(2.92,-0.31),SE*lsf); label("$4$",(3.93,-0.26),SE*lsf); label("$5$",(4.92,-0.27),SE*lsf); label("$6$",(5.95,-0.29),SE*lsf); label("$7$",(6.94,-0.27),SE*lsf); label("$5$",(-0.49,1.22),SE*lsf); label("$10$",(-0.59,2.23),SE*lsf); label("$15$",(-0.61,3.22),SE*lsf); label("$20$",(-0.61,4.23),SE*lsf); label("$25$",(-0.59,5.22),SE*lsf); label("$30$",(-0.59,6.2),SE*lsf); label("$35$",(-0.56,7.18),SE*lsf); draw((0,0)--(1,1),linewidth(1.6)); draw((1,1)--(2,3),linewidth(1.6)); draw((2,3)--(4,4),linewidth(1.6)); draw((4,4)--(7,7),linewidth(1.6)); label("HOURS",(3.41,-0.85),SE*lsf); label("M",(-1.39,5.32),SE*lsf); label("I",(-1.34,4.93),SE*lsf); label("L",(-1.36,4.51),SE*lsf); label("E",(-1.37,4.11),SE*lsf); label("S",(-1.39,3.7),SE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]
$\textbf{(A)}2\qquad\textbf{(B)}2.5\qquad\textbf{(C)}4\qquad\textbf{(D)}4.5\qquad\textbf{(E)}5$
| We observe the graph and see that the shape of the graph does not matter. We only want the total time it took Carmen and the total distance she traveled. Based on the graph, Carmen traveled 35 miles for 7 hours. Therefore, her average speed is $\boxed{\textbf{(E)}\ 5}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_9 | 5 |
AMC8_132 | What is the tens digit of $7^{2011}$ ?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7$
| Since we want the tens digit, we can find the last two digits of $7^{2011}$ . We can do this by using modular arithmetic.
\[7^1\equiv 07 \pmod{100}.\]
\[7^2\equiv 49 \pmod{100}.\]
\[7^3\equiv 43 \pmod{100}.\]
\[7^4\equiv 01 \pmod{100}.\]
We can write $7^{2011}$ as $(7^4)^{502}\times 7^3$ . Using this, we can say:
\[7^{2011}\equiv (7^4)^{502}\times 7^3\equiv 7^3\equiv 343\equiv 43\pmod{100}.\]
From the above, we can conclude that the last two digits of $7^{2011}$ are 43. Since they have asked us to find the tens digit, our answer is $\boxed{\textbf{(D)}\ 4}$ .
-Ilovefruits
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_22 | 4 |
AMC8_133 | How many different real numbers $x$ satisfy the equation \[(x^{2}-5)^{2}=16?\]
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$
| We have that $(x^2-5)^2 = 16$ if and only if $x^2-5 = \pm 4$ . If $x^2-5 = 4$ , then $x^2 = 9 \implies x = \pm 3$ , giving 2 solutions. If $x^2-5 = -4$ , then $x^2 = 1 \implies x = \pm 1$ , giving 2 more solutions. All four of these solutions work, so the answer is $\boxed{\textbf{(D) }4}$ . Further, the equation is a quartic in $x$ , so by the Fundamental Theorem of Algebra , there can be at most four real solutions.
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_20 | 4 |
AMC8_134 | In $\bigtriangleup ABC$ , $D$ is a point on side $\overline{AC}$ such that $BD=DC$ and $\angle BCD$ measures $70^\circ$ . What is the degree measure of $\angle ADB$ ?
[asy] size(300); defaultpen(linewidth(0.8)); pair A=(-1,0),C=(1,0),B=dir(40),D=origin; draw(A--B--C--A); draw(D--B); dot("$A$", A, SW); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, S); label("$70^\circ$",C,2*dir(180-35));[/asy]
$\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150$
| Using angle chasing is a good way to solve this problem. $BD = DC$ , so $\angle DBC = \angle DCB = 70$ , because it is an isosceles triangle. Then $\angle CDB = 180-(70+70) = 40$ . Since $\angle ADB$ and $\angle BDC$ are supplementary, $\angle ADB = 180 - 40 = \boxed{\textbf{(D)}~140}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_9 | 140 |
AMC8_135 | When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes. As an old man, he can now walk $10$ miles in $4$ hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?
$\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30$
| When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes $= 3\times60 + 30$ minutes $= 210$ minutes, thus running $\frac{210}{15} = 14$ minutes per mile. Now that he is an old man, he can walk $10$ miles in $4$ hours $= 4 \times 60$ minutes $= 240$ minutes, thus walking $\frac{240}{10} = 24$ minutes per mile. Therefore, it takes him $\boxed{\textbf{(B)}\ 10}$ minutes longer to walk a mile now compared to when he was a boy.
~CHECKMATE2021
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_4 | 10 |
AMC8_136 | Lola, Lolo, Tiya, and Tiyo participated in a ping pong tournament. Each player competed against each of the other three players exactly twice. Shown below are the win-loss records for the players. The numbers $1$ and $0$ represent a win or loss, respectively. For example, Lola won five matches and lost the fourth match. What was Tiyo’s win-loss record?
\[\begin{tabular}{c | c} Player & Result \\ \hline Lola & \texttt{111011}\\ Lolo & \texttt{101010}\\ Tiya & \texttt{010100}\\ Tiyo & \texttt{??????} \end{tabular}\]
$\textbf{(A)}\ \texttt{000101} \qquad \textbf{(B)}\ \texttt{001001} \qquad \textbf{(C)}\ \texttt{010000} \qquad \textbf{(D)}\ \texttt{010101} \qquad \textbf{(E)}\ \texttt{011000}$
| We can calculate the total number of wins ( $1$ 's) by seeing how many matches were players, which is $12$ matches played. Then, we can calculate the # of wins already on the table, which is $5 + 3 + 2 = 10$ , so there are $12 - 10 = 2$ wins left in the mystery player. Now, we will make the key observation that there is only $2$ wins ( $1$ 's) per column as there are $2$ winners and $2$ losers in each round. Strategically looking through the columns counting the $1$ 's and putting our own $2$ $1$ 's when the column isn't already full yields $\boxed{\textbf{(A)}\ \texttt{000101}}$ .
~SohumUttamchandani -edits apex304, lpieleanu
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_8 | 000101 |
AMC8_137 | The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$
| Let the hundreds, tens, and units digits of the original three-digit number be $a$ , $b$ , and $c$ , respectively. We are given that $a=c+2$ . The original three-digit number is equal to $100a+10b+c = 100(c+2)+10b+c = 101c+10b+200$ . The hundreds, tens, and units digits of the reversed three-digit number are $c$ , $b$ , and $a$ , respectively. This number is equal to $100c+10b+a = 100c+10b+(c+2) = 101c+10b+2$ . Subtracting this expression from the expression for the original number, we get $(101c+10b+200) - (101c+10b+2) = 198$ . Thus, the units digit in the final result is $\boxed{\textbf{(E)}\ 8}$
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_22 | 8 |
AMC8_138 | An amusement park has a collection of scale models, with a ratio of $1: 20$ , of buildings and other sights from around the country. The height of the United States Capitol is $289$ feet. What is the height in feet of its duplicate to the nearest whole number?
$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20$
| You can see that since the ratio of real building's heights to the model building's height is $1:20$ . We also know that the U.S Capitol is $289$ feet in real life, so to find the height of the model, we divide by 20. That gives us $14.45$ which rounds to 14. Therefore, to the nearest whole number, the duplicate is $\boxed{\textbf{(A) }14}$ .
~avamarora, Nivaar..
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_1 | 14 |
AMC8_139 | A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased $?$
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$
| Suppose the fraction of discount is $x$ . That means $(1-x)(1+x)=0.84$ ; so, $1-x^{2}=0.84$ , and $(x^{2})=0.16$ , procuring $x=0.4$ . Therefore, the price was increased and decreased by $40$ %, or $\boxed{\textbf{(E)}\ 40}$ .
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_22 | 40 |
AMC8_140 | Each half of this figure is composed of 3 red triangles, 5 blue triangles and 8 white triangles. When the upper half is folded down over the centerline, 2 pairs of red triangles coincide, as do 3 pairs of blue triangles. There are 2 red-white pairs. How many white pairs coincide?
[asy] draw((0,0)--(4,4*sqrt(3))); draw((1,-sqrt(3))--(5,3*sqrt(3))); draw((2,-2*sqrt(3))--(6,2*sqrt(3))); draw((3,-3*sqrt(3))--(7,sqrt(3))); draw((4,-4*sqrt(3))--(8,0)); draw((8,0)--(4,4*sqrt(3))); draw((7,-sqrt(3))--(3,3*sqrt(3))); draw((6,-2*sqrt(3))--(2,2*sqrt(3))); draw((5,-3*sqrt(3))--(1,sqrt(3))); draw((4,-4*sqrt(3))--(0,0)); draw((3,3*sqrt(3))--(5,3*sqrt(3))); draw((2,2*sqrt(3))--(6,2*sqrt(3))); draw((1,sqrt(3))--(7,sqrt(3))); draw((-1,0)--(9,0)); draw((1,-sqrt(3))--(7,-sqrt(3))); draw((2,-2*sqrt(3))--(6,-2*sqrt(3))); draw((3,-3*sqrt(3))--(5,-3*sqrt(3))); [/asy]
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 9$
| Each half has $3$ red triangles, $5$ blue triangles, and $8$ white triangles. There are also $2$ pairs of red triangles, so $2$ red triangles on each side are used, leaving $1$ red triangle, $5$ blue triangles, and $8$ white triangles remaining on each half. Also, there are $3$ pairs of blue triangles, using $3$ blue triangles on each side, so there is $1$ red triangle, $2$ blue triangles, and $8$ white triangles remaining on each half. Also, we have $2$ red-white pairs. This obviously can't use $2$ red triangles on one side, since there is only $1$ on each side, so we must use $1$ red triangle and $1$ white triangle per side, leaving $2$ blue triangles and $7$ white triangles apiece. The remaining blue triangles cannot be matched with other blue triangles since that would mean there were more than $3$ blue pairs, so the remaining blue triangles must be paired with white triangles, yielding $4$ blue-white pairs, one for each of the remaining blue triangles. This uses $2$ blue triangles and $2$ white triangles on each side, leaving $5$ white triangles apiece, which must be paired with each other, so there are $5$ white-white pairs, $\boxed{\text{B}}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_24 | B |
AMC8_141 | What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
$\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660$
| To find either the LCM or the GCF of two numbers, always prime factorize first.
The prime factorization of $180 = 3^2 \times 5 \times 2^2$ .
The prime factorization of $594 = 3^3 \times 11 \times 2$ .
Then, to find the LCM, we have to find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is $3^3, 5, 11, 2^2$ ). Multiply all of these to get 5940.
For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. $3^2 \times 2$ = 18.
Thus the answer = $\frac{5940}{18}$ = $\boxed{\textbf{(C)}\ 330}$ .
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_10 | 330 |
AMC8_142 | The circle shown below on the left has a radius of 1 unit. The region between the circle and the inscribed square is shaded. In the circle shown on the right, one quarter of the region between the circle and the inscribed square is shaded. The shaded regions in the two circles have the same area. What is the radius $R$ , in units, of the circle on the right?
[asy] unitsize(40); real a = 0.707; fill(circle((a,a), 1), grey); fill((0,0)--(0,1.414)--(1.414,1.414)--(1.414,0)--cycle, white); draw((0,0)--(0,1.414)--(1.414,1.414)--(1.414,0)--cycle); draw(circle((a,a), 1)); draw((0.707,0.707)--(1.414,1.414)); dot((0.707,0.707)); label("$1$", (1,1), SE); fill(circle((4+a, a), 2*a), grey); fill(shift((4+a,a)) * ((-2,-2)--(1,-2)--(1,2)--(-2,2)--cycle), white); draw(shift((4+a,a)) * ((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle)); draw(circle((4+a, a), 2*a)); draw((4+a,a)--(5+a,1+a)); dot((4+a,a)); label("$R$", (a+4.5,a+0.5), SE); [/asy]
$\textbf{(A)}\ \sqrt2\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 2\sqrt2\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ 4\sqrt2$
| The area of the shaded region in the circle on the left is the area of the circle minus the area of the square, or $\big(\pi-2)$ . The shaded area in the circle on the right is $\dfrac{1}{4}$ of the area of the circle minus the area of the square, or $\dfrac{\pi R^2-2R^2}{4}$ , which can be factored as $\dfrac{R^2(\pi-2)}{4}$ . Since the shaded areas are equal to each other, we have $\pi-2=\dfrac{R^2(\pi-2)}{4}$ , which simplifies to $R^2=4$ . Taking the square root, we have $R=\boxed{\text{(B)\ 2}}$
~mrtnvlknv
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_18 | 2 |
AMC8_144 | Antonette gets $70 \%$ on a 10-problem test, $80 \%$ on a 20-problem test and $90 \%$ on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is closest to her overall score?
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 77\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 83\qquad\textbf{(E)}\ 87$
| $70 \% \cdot 10=7$
$80 \% \cdot 20=16$
$90 \% \cdot 30=27$
Adding them up gets $7+16+27=50$ . The overall percentage correct would be $\frac{50}{60}=\frac{5}{6}=5 \cdot 16.\overline{6}=83.\overline{3} \approx \boxed{\textbf{(D)}\ 83}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_12 | 83 |
AMC8_145 | Two angles of an isosceles triangle measure $70^\circ$ and $x^\circ$ . What is the sum of the three possible values of $x$ ?
$\textbf{(A)}\ 95 \qquad \textbf{(B)}\ 125 \qquad \textbf{(C)}\ 140 \qquad \textbf{(D)}\ 165 \qquad \textbf{(E)}\ 180$
| There are 3 cases: where $x^\circ$ is a base angle with the $70^\circ$ as the other angle, where $x^\circ$ is a base angle with $70^\circ$ as the vertex angle, and where $x^\circ$ is the vertex angle with $70^\circ$ as a base angle.
Case 1: $x^\circ$ is a base angle with $70^\circ$ as the other angle:
Here, $x=70$ , since base angles are congruent.
Case 2: $x^\circ$ is a base angle with $70^\circ$ as the vertex angle:
Here, the 2 base angles are both $x^\circ$ , so we can use the equation $2x+70=180$ , which simplifies to $x=55$ .
Case 3: $x^\circ$ is the vertex angle with $70^\circ$ as a base angle:
Here, both base angles are $70^\circ$ , since base angles are congruent. Thus, we can use the equation $x+140=180$ , which simplifies to $x=40$ .
Adding up all the cases, we get $70+55+40=165$ , so the answer is $\boxed{\textbf{(D)}\ 165}$ .
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_19 | 165 |
AMC8_146 | In the country of East Westmore, statisticians estimate there is a baby born every $8$ hours and a death every day. To the nearest hundred, how many people are added to the population of East Westmore each year?
$\textbf{(A)}\hspace{.05in}600\qquad\textbf{(B)}\hspace{.05in}700\qquad\textbf{(C)}\hspace{.05in}800\qquad\textbf{(D)}\hspace{.05in}900\qquad\textbf{(E)}\hspace{.05in}1000$
| There are $24\text{ hours}\div8\text{ hours} = 3$ births and one death everyday in East Westmore. Therefore, the population increases by $3$ - $1$ = $2$ people everyday. Thus, there are $2 \times 365 = 730$ people added to the population every year. Rounding, we find the answer is $\boxed{\textbf{(B)}\ 700}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_2 | 700 |
AMC8_147 | A shop advertises everything is "half price in today's sale." In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price?
$\textbf{(A)}\hspace{.05in}10\qquad\textbf{(B)}\hspace{.05in}33\qquad\textbf{(C)}\hspace{.05in}40\qquad\textbf{(D)}\hspace{.05in}60\qquad\textbf{(E)}\hspace{.05in}70$
| Let the original price of an item be $x$ .
First, everything is half-off, so the price is now $\frac{x}{2} = 0.5x$ .
Next, the extra coupon applies 20% off on the sale price , so the price after this discount will be $100\% - 20\% = 80\%$ of what it was before. (Notice how this is not applied to the original price; if it were, the solution would be applying 50% + 20 % = 70% off the original price.)
$80\% \cdot 0.5 x = \frac{4}{5} \cdot 0.5 x = 0.4x$
The price of the item after all discounts have been applied is $0.4x = 40\% \cdot x$ . However, we need to find the percentage off the original price, not the current percentage of the original price. We then subtract $40\% x$ from $100\% x$ (the original price of the item), to find the answer, $\boxed{\textbf{(D)}\ 60}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_8 | 60 |
AMC8_148 | The average cost of a long-distance call in the USA in $1985$ was
$41$ cents per minute, and the average cost of a long-distance
call in the USA in $2005$ was $7$ cents per minute. Find the
approximate percent decrease in the cost per minute of a long-
distance call.
$\mathrm{(A)}\ 7 \qquad\mathrm{(B)}\ 17 \qquad\mathrm{(C)}\ 34 \qquad\mathrm{(D)}\ 41 \qquad\mathrm{(E)}\ 80$
| The percent decrease is (the amount of decrease)/(original amount)
the amount of decrease is $41 - 7 = 34$
so the percent decrease is $\frac{34}{41}$ which is about $\boxed{\textbf{(E)}\ 80\%}$
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_6 | 80% |
AMC8_149 | The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD?
[asy] size(250); defaultpen(linewidth(0.55)); pair A=(-6,0), B=origin, C=(0,6), D=(0,12); pair ac=C+2.828*dir(45), ca=A+2.828*dir(225), ad=D+2.828*dir(A--D), da=A+2.828*dir(D--A), ab=(2.828,0), ba=(-6-2.828, 0); fill(A--C--D--cycle, gray); draw(ba--ab); draw(ac--ca); draw(ad--da); draw((0,-1)--(0,15)); draw((1/3, -1)--(1/3, 15)); int i; for(i=1; i<15; i=i+1) { draw((-1/10, i)--(13/30, i)); } label("$A$", A, SE); label("$B$", B, SE); label("$C$", C, SE); label("$D$", D, SE); label("$3$", (1/3,3), E); label("$3$", (1/3,9), E); label("$3$", (-3,0), S); label("Main", (-3,0), N); label(rotate(45)*"Aspen", A--C, SE); label(rotate(63.43494882)*"Brown", A--D, NW); [/asy]
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$
| The area of a triangle is $\frac12 bh$ . If we let $CD$ be the base of the triangle, then the height is $AB$ , and the area is $\frac12 \cdot 3 \cdot 3 = \boxed{\textbf{(C)}\ 4.5}$ .
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_7 | 4.5 |
AMC8_150 | Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump $5$ pads to the right or $3$ pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located $2023$ pads to the right of her starting position?
$\textbf{(A) } 405 \qquad \textbf{(B) } 407 \qquad \textbf{(C) } 409 \qquad \textbf{(D) } 411 \qquad \textbf{(E) } 413$
| We have $2$ directions going $5$ right or $3$ left. We can assign a variable to each of these directions. We can call going right $1$ direction $\text{X}$ and we can call going $1$ left $\text{Y}$ . We can build an equation of $5\text{X}-3\text{Y}=2023$ , where we have to limit the number of moves we do. We can do this by making more of our moves the $5$ move turn than the $3$ move turn. The first obvious step is to go some amount of moves in the right direction then subtract off in the left direction to land on $2023$ . The least amount of $3$ ’s added to $2023$ to make a multiple of $5$ is $4$ as $2023 + 4(3) = 2035$ . So now, we have solved the problem as we just go $\frac{2035}{5} = 407$ hops right, and just do 4 more hops left. Yielding $407 + 4 = \boxed{\textbf{(D)}\ 411}$ as our answer.
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_18 | 411 |
AMC8_151 | On a map, a $12$ -centimeter length represents $72$ kilometers. How many kilometers does a $17$ -centimeter length represent?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 102\qquad\textbf{(C)}\ 204\qquad\textbf{(D)}\ 864\qquad\textbf{(E)}\ 1224$
| We set up the proportion $\frac{12 \text{cm}}{72 \text{km}}=\frac{17 \text{cm}}{x \text{km}}$ . Thus $x=102 \Rightarrow \boxed{\textbf{(B)}\ 102}$
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_1 | 102 |
AMC8_152 | Points $B$ , $D$ , and $J$ are midpoints of the sides of right triangle $ACG$ . Points $K$ , $E$ , $I$ are midpoints of the sides of triangle $JDG$ , etc. If the dividing and shading process is done 100 times (the first three are shown) and $AC=CG=6$ , then the total area of the shaded triangles is nearest
[asy] draw((0,0)--(6,0)--(6,6)--cycle); draw((3,0)--(3,3)--(6,3)); draw((4.5,3)--(4.5,4.5)--(6,4.5)); draw((5.25,4.5)--(5.25,5.25)--(6,5.25)); fill((3,0)--(6,0)--(6,3)--cycle,black); fill((4.5,3)--(6,3)--(6,4.5)--cycle,black); fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black); label("$A$",(0,0),SW); label("$B$",(3,0),S); label("$C$",(6,0),SE); label("$D$",(6,3),E); label("$E$",(6,4.5),E); label("$F$",(6,5.25),E); label("$G$",(6,6),NE); label("$H$",(5.25,5.25),NW); label("$I$",(4.5,4.5),NW); label("$J$",(3,3),NW); label("$K$",(4.5,3),S); label("$L$",(5.25,4.5),S); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$
| Since $\triangle FGH$ is fairly small relative to the rest of the diagram, we can make an underestimate by using the current diagram. All triangles are right-isosceles triangles.
\[CD = \frac {CG}{2} = 3, DE = \frac{CD}{2} = \frac{3}{2}, EF = \frac{DE}{2} = \frac{3}{4}\]
\[CB = CD = 3, DK = DE = \frac{3}{2}, EL = EF = \frac{3}{4}\]
\[[CBD] = \frac{1}{2}3^2 = \frac{9}{2}\]
\[[DKE] = \frac{1}{2}(\frac{3}{2})^2 = \frac{9}{8}\]
\[[ELF] = \frac{1}{2}(\frac{3}{4})^2 = \frac{9}{32}\]
The sum of the shaded regions is $\frac{9}{2} + \frac{9}{8} + \frac{9}{32} = \frac{189}{32} \approx 5.9$
$5.9$ is an underestimate, as some portion (but not all) of $\triangle FGH$ will be shaded in future iterations.
If you shade all of $\triangle FGH$ , this will add an additional $\frac{9}{32}$ to the area, giving $\frac{198}{32} \approx 6.2$ , which is an overestimate.
Thus, $\boxed{\text{(A)}\ 6}$ is the only answer that is both over the underestimate and under the overestimate.
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_25 | 6 |
AMC8_153 | A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color. The Martian pulls out one sock at a time without looking. How many socks must the Martian remove from the drawer to be certain there will be 5 socks of the same color?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$
| The Martian can pull out $12$ socks, $4$ of each color, without having $5$ of the same kind yet. However, the next one he pulls out must be the fifth of one of the colors so he must remove $\boxed{\textbf{(D)}\ 13}$ socks.
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_16 | 13 |
AMC8_154 | On the AMC 8 contest Billy answers 13 questions correctly, answers 7 questions incorrectly and doesn't answer the last 5. What is his score?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 26$
| As the AMC 8 only rewards 1 point for each correct answer, everything is irrelevant except the number Billy answered correctly, $\boxed{\textbf{(C)}\ 13}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_2 | 13 |
AMC8_155 | Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?
[asy] path cell=((0,0)--(1,0)--(1,1)--(0,1)--cycle); path sw=((0,0)--(1,sqrt(3))); path se=((5,0)--(4,sqrt(3))); draw(cell, linewidth(1)); draw(shift(2,0)*cell, linewidth(1)); draw(shift(4,0)*cell, linewidth(1)); draw(shift(1,3)*cell, linewidth(1)); draw(shift(3,3)*cell, linewidth(1)); draw(shift(2,6)*cell, linewidth(1)); draw(shift(0.45,1.125)*sw, EndArrow); draw(shift(2.45,1.125)*sw, EndArrow); draw(shift(1.45,4.125)*sw, EndArrow); draw(shift(-0.45,1.125)*se, EndArrow); draw(shift(-2.45,1.125)*se, EndArrow); draw(shift(-1.45,4.125)*se, EndArrow); label("$+$", (1.5,1.5)); label("$+$", (3.5,1.5)); label("$+$", (2.5,4.5));[/asy]
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 35$
| If the lower cells contain $A, B$ and $C$ , then the second row will contain
$A + B$ and $B + C$ , and the top cell will contain $A + 2B + C$ . To obtain the
smallest sum, place $1$ in the center cell and $2$ and $3$ in the outer ones. The top
number will be $7$ . For the largest sum, place $9$ in the center cell and $7$ and $8$ in
the outer ones. This top number will be $33$ . The difference is $33 - 7 = \boxed{\textbf{(D)}\ 26 }$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_22 | 26 |
AMC8_156 | What is the degree measure of the smaller angle formed by the hands of a clock at 10 o'clock?
[asy] draw(circle((0,0),2)); dot((0,0)); for(int i = 0; i < 12; ++i) { dot(2*dir(30*i)); } label("$3$",2*dir(0),W); label("$2$",2*dir(30),WSW); label("$1$",2*dir(60),SSW); label("$12$",2*dir(90),S); label("$11$",2*dir(120),SSE); label("$10$",2*dir(150),ESE); label("$9$",2*dir(180),E); label("$8$",2*dir(210),ENE); label("$7$",2*dir(240),NNE); label("$6$",2*dir(270),N); label("$5$",2*dir(300),NNW); label("$4$",2*dir(330),WNW); [/asy]
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 90$
| At $10:00$ , the hour hand will be on the $10$ while the minute hand on the $12$ .
This makes them $\frac{1}{6}$ th of a circle apart, and $\frac{1}{6}\cdot360^{\circ}=\boxed{\textbf{(C) } 60}$ .
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_2 | 60 |
AMC8_157 | The ten-letter code $\text{BEST OF LUCK}$ represents the ten digits $0-9$ , in order. What 4-digit number is represented by the code word $\text{CLUE}$ ?
$\textbf{(A)}\ 8671 \qquad \textbf{(B)}\ 8672 \qquad \textbf{(C)}\ 9781 \qquad \textbf{(D)}\ 9782 \qquad \textbf{(E)}\ 9872$
| We can derive that $C=8$ , $L=6$ , $U=7$ , and $E=1$ . Therefore, the answer is $\boxed{\textbf{(A)}\ 8671}$
~edited by Owencheng
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_2 | 8671 |
AMC8_158 | How many pairs of parallel edges, such as $\overline{AB}$ and $\overline{GH}$ or $\overline{EH}$ and $\overline{FG}$ , does a cube have?
[asy] import three; currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); label("$D$",(0,0,0),S); label("$A$",(0,0,1),N); label("$H$",(0,1,0),S); label("$E$",(0,1,1),N); label("$C$",(1,0,0),S); label("$B$",(1,0,1),N); label("$G$",(1,1,0),S); label("$F$",(1,1,1),N); [/asy]
$\textbf{(A) }6\qquad\textbf{(B) }12\qquad\textbf{(C) }18\qquad\textbf{(D) }24\qquad \textbf{(E) }36$
| Solution 1We first count the number of pairs of parallel lines that are in the same direction as $\overline{AB}$ . The pairs of parallel lines are $\overline{AB}\text{ and }\overline{EF}$ , $\overline{CD}\text{ and }\overline{GH}$ , $\overline{AB}\text{ and }\overline{CD}$ , $\overline{EF}\text{ and }\overline{GH}$ , $\overline{AB}\text{ and }\overline{GH}$ , and $\overline{CD}\text{ and }\overline{EF}$ . These are $6$ pairs total. We can do the same for the lines in the same direction as $\overline{AE}$ and $\overline{AD}$ . This means there are $6\cdot 3=\boxed{\textbf{(C) } 18}$ total pairs of parallel lines.
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_12 | 18 |
AMC8_159 | The five pieces shown below can be arranged to form four of the five figures shown in the choices. Which figure cannot be formed?
[asy] defaultpen(linewidth(0.6)); size(80); real r=0.5, s=1.5; path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(s,r)*p); draw(shift(s,-r)*p); draw(shift(2s,2r)*p); draw(shift(2s,0)*p); draw(shift(2s,-2r)*p); draw(shift(3s,3r)*p); draw(shift(3s,-3r)*p); draw(shift(3s,r)*p); draw(shift(3s,-r)*p); draw(shift(4s,-4r)*p); draw(shift(4s,-2r)*p); draw(shift(4s,0)*p); draw(shift(4s,2r)*p); draw(shift(4s,4r)*p); [/asy]
[asy] size(350); defaultpen(linewidth(0.6)); path p=origin--(1,0)--(1,1)--(0,1)--cycle; pair[] a={(0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (1,1), (1,2), (2,0), (2,1), (3,0), (3,1), (3,2), (3,3), (3,4)}; pair[] b={(5,3), (5,4), (6,2), (6,3), (6,4), (7,1), (7,2), (7,3), (7,4), (8,0), (8,1), (8,2), (9,0), (9,1), (9,2)}; pair[] c={(11,0), (11,1), (11,2), (11,3), (11,4), (12,1), (12,2), (12,3), (12,4), (13,2), (13,3), (13,4), (14,3), (14,4), (15,4)}; pair[] d={(17,0), (17,1), (17,2), (17,3), (17,4), (18,0), (18,1), (18,2), (18,3), (18,4), (19,0), (19,1), (19,2), (19,3), (19,4)}; pair[] e={(21,4), (22,1), (22,2), (22,3), (22,4), (23,0), (23,1), (23,2), (23,3), (23,4), (24,1), (24,2), (24,3), (24,4), (25,4)}; int i; for(int i=0; i<15; i=i+1) { draw(shift(a[i])*p); draw(shift(b[i])*p); draw(shift(c[i])*p); draw(shift(d[i])*p); draw(shift(e[i])*p); } [/asy]
\[\textbf{(A)}\qquad\qquad\qquad\textbf{(B)}\quad\qquad\qquad\textbf{(C)}\:\qquad\qquad\qquad\textbf{(D)}\quad\qquad\qquad\textbf{(E)}\]
| The answer is $\boxed{\textbf{(B)}}$ because the longest piece cannot fit into the figure.
Note that the five pieces can be arranged to form the figures in $\textbf{(A)},\textbf{(C)},\textbf{(D)},$ and $\textbf{(E)},$ as shown below:
[asy] defaultpen(linewidth(0.6)); size(80); real r=0.5, s=1.5; path p=origin--(1,0)--(1,1)--(0,1)--cycle; fill(p,red); fill(shift(s,r)*p,yellow); fill(shift(s,-r)*p,yellow); fill(shift(2s,2r)*p,green); fill(shift(2s,0)*p,green); fill(shift(2s,-2r)*p,green); fill(shift(3s,3r)*p,cyan); fill(shift(3s,-3r)*p,cyan); fill(shift(3s,r)*p,cyan); fill(shift(3s,-r)*p,cyan); fill(shift(4s,-4r)*p,pink); fill(shift(4s,-2r)*p,pink); fill(shift(4s,0)*p,pink); fill(shift(4s,2r)*p,pink); fill(shift(4s,4r)*p,pink); draw(p); draw(shift(s,r)*p); draw(shift(s,-r)*p); draw(shift(2s,2r)*p); draw(shift(2s,0)*p); draw(shift(2s,-2r)*p); draw(shift(3s,3r)*p); draw(shift(3s,-3r)*p); draw(shift(3s,r)*p); draw(shift(3s,-r)*p); draw(shift(4s,-4r)*p); draw(shift(4s,-2r)*p); draw(shift(4s,0)*p); draw(shift(4s,2r)*p); draw(shift(4s,4r)*p); [/asy]
[asy] size(350); defaultpen(linewidth(0.6)); path p=origin--(1,0)--(1,1)--(0,1)--cycle; pair[] a={(0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (1,1), (1,2), (2,0), (2,1), (3,0), (3,1), (3,2), (3,3), (3,4)}; pair[] b={(5,3), (5,4), (6,2), (6,3), (6,4), (7,1), (7,2), (7,3), (7,4), (8,0), (8,1), (8,2), (9,0), (9,1), (9,2)}; pair[] c={(11,0), (11,1), (11,2), (11,3), (11,4), (12,1), (12,2), (12,3), (12,4), (13,2), (13,3), (13,4), (14,3), (14,4), (15,4)}; pair[] d={(17,0), (17,1), (17,2), (17,3), (17,4), (18,0), (18,1), (18,2), (18,3), (18,4), (19,0), (19,1), (19,2), (19,3), (19,4)}; pair[] e={(21,4), (22,1), (22,2), (22,3), (22,4), (23,0), (23,1), (23,2), (23,3), (23,4), (24,1), (24,2), (24,3), (24,4), (25,4)}; fill((0,0)--(0,5)--(1,5)--(1,0)--cycle,pink); fill((1,0)--(1,3)--(2,3)--(2,0)--cycle,green); fill((2,0)--(2,2)--(3,2)--(3,0)--cycle,yellow); fill((3,4)--(3,5)--(4,5)--(4,4)--cycle,red); fill((3,0)--(3,4)--(4,4)--(4,0)--cycle,cyan); fill((11,0)--(11,5)--(12,5)--(12,0)--cycle,pink); fill((12,1)--(12,5)--(13,5)--(13,1)--cycle,cyan); fill((13,2)--(13,5)--(14,5)--(14,2)--cycle,green); fill((14,3)--(14,5)--(15,5)--(15,3)--cycle,yellow); fill((15,4)--(15,5)--(16,5)--(16,4)--cycle,red); fill((17,0)--(17,5)--(18,5)--(18,0)--cycle,pink); fill((18,0)--(18,4)--(19,4)--(19,0)--cycle,cyan); fill((18,4)--(18,5)--(19,5)--(19,4)--cycle,red); fill((19,0)--(19,3)--(20,3)--(20,0)--cycle,green); fill((19,3)--(19,5)--(20,5)--(20,3)--cycle,yellow); fill((21,4)--(21,5)--(22,5)--(22,4)--cycle,red); fill((22,1)--(22,5)--(23,5)--(23,1)--cycle,cyan); fill((23,0)--(23,5)--(24,5)--(24,0)--cycle,pink); fill((24,1)--(24,4)--(25,4)--(25,1)--cycle,green); fill((24,4)--(24,5)--(26,5)--(26,4)--cycle,yellow); int i; for(int i=0; i<15; i=i+1) { draw(shift(a[i])*p); draw(shift(b[i])*p); draw(shift(c[i])*p); draw(shift(d[i])*p); draw(shift(e[i])*p); } [/asy]
\[\textbf{(A)}\qquad\qquad\qquad\textbf{(B)}\quad\qquad\qquad\textbf{(C)}\qquad\qquad\qquad\textbf{(D)}\quad\qquad\qquad\textbf{(E)}\]
~Basketball8 ~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_4 | B |
AMC8_160 | There are $20$ cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all $20$ cities?
[asy] // made by SirCalcsALot size(300); pen shortdashed=linetype(new real[] {6,6}); for (int i = 2000; i < 9000; i = i + 2000) { draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey); label(string(i), (0,i), W); } for (int i = 500; i < 9300; i=i+500) { draw((0,i)--(150,i),linewidth(1.25)); if (i % 2000 == 0) { draw((0,i)--(250,i),linewidth(1.25)); } } int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750}; int data_length = 20; int r = 550; for (int i = 0; i < data_length; ++i) { fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey); draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)); } draw((0,4750)--(11450,4750),shortdashed); label("Cities", (11450*0.5,0), S); label(rotate(90)*"Population", (0,9000*0.5), 10*W); // axis draw((0,0)--(0,9300), linewidth(1.25)); draw((0,0)--(11550,0), linewidth(1.25)); [/asy]
$\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000$
| We can see that the dotted line is exactly halfway between $4{,}500$ and $5{,}000$ , so it is at $4{,}750$ . As this is the average population of all $20$ cities, the total population is simply $4{,}750 \cdot 20 = \boxed{\textbf{(D) }95{,}000}$ .
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_14 | 95,000 |
AMC8_161 | The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$
| The arithmetic mean of these numbers is $\frac{\frac{2013}{3}}{2}=\frac{671}{2}=335.5$ . Therefore the numbers are $333$ , $334$ , $335$ , $336$ , $337$ , $338$ , so the answer is $\boxed{\textbf{(B)}\ 338}$ .
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_17 | 338 |
AMC8_162 | Right isosceles triangles are constructed on the sides of a 3-4-5 right triangle, as shown. A capital letter represents the area of each triangle. Which one of the following is true?
[asy] /* AMC8 2002 #16 Problem */ draw((0,0)--(4,0)--(4,3)--cycle); draw((4,3)--(-4,4)--(0,0)); draw((-0.15,0.1)--(0,0.25)--(.15,0.1)); draw((0,0)--(4,-4)--(4,0)); draw((4,0.2)--(3.8,0.2)--(3.8,-0.2)--(4,-0.2)); draw((4,0)--(7,3)--(4,3)); draw((4,2.8)--(4.2,2.8)--(4.2,3)); label(scale(0.8)*"$Z$", (0, 3), S); label(scale(0.8)*"$Y$", (3,-2)); label(scale(0.8)*"$X$", (5.5, 2.5)); label(scale(0.8)*"$W$", (2.6,1)); label(scale(0.65)*"5", (2,2)); label(scale(0.65)*"4", (2.3,-0.4)); label(scale(0.65)*"3", (4.3,1.5));[/asy]
$\textbf{(A)}\ X+Z=W+Y\qquad\textbf{(B)}\ W+X=Z\qquad\textbf{(C)}\ 3X+4Y=5Z\qquad$
$\textbf{(D)}\ X+W=\frac{1}{2}(Y+Z)\qquad\textbf{(E)}\ X+Y=Z$
| The area of a right triangle can be found by using the legs of triangle as the base and height. In the three isosceles triangles, the length of their second leg is the same as the side that is connected to the $3-4-5$ triangle.
\begin{align*} W&=(3)(4)/2 = 6\\ X&=(3)(3)/2=4.5\\ Y&=(4)(4)/2 = 8\\ Z&=(5)(5)/2 = 12.5 \end{align*}
Plugging into the answer choices, the only that works is $\boxed{\textbf{(E)}\ X+Y=Z}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_16 | X+Y=Z |
AMC8_163 | Three concentric circles centered at $O$ have radii of $1$ , $2$ , and $3$ . Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angles $BOC$ , as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of $\angle{BOC}$ in degrees?
[asy] size(150); import graph; draw(circle((0,0),3)); real radius = 3; real angleStart = -54; // starting angle of the sector real angleEnd = 54; // ending angle of the sector label("$O$",(0,0),W); pair O = (0, 0); filldraw(arc(O, radius, angleStart, angleEnd)--O--cycle, gray); filldraw(circle((0,0),2),gray); filldraw(circle((0,0),1),white); draw((1.763,2.427)--(0,0)--(1.763,-2.427)); label("$B$",(1.763,2.427),NE); label("$C$",(1.763,-2.427),SE); [/asy]
$\textbf{(A) } 108\qquad\textbf{(B) } 120\qquad\textbf{(C) } 135\qquad\textbf{(D) } 144\qquad\textbf{(E) } 150$
| Let $x=\angle{BOC}$ .
We see that the shaded region is the inner ring plus a sector $x^\circ$ of the outer ring. Using the formula for the area of a circle ( $A = \pi r^2$ ), we find that the area of $x$ is $\left( 4 \pi - \pi \right)+\frac{x}{360} \left( 9 \pi - 4 \pi \right)$ . This simplifies to $3 \pi + \frac{x}{360}(5 \pi)$ .
The unshaded portion is comprised of the smallest circle plus the sector $(360-x)^\circ$ of the outer ring, which evaluates to $\pi + \frac{360-x}{360}(5 \pi)$ .
We are told these are equal. Therefore, $3 \pi + \frac{x}{360}(5 \pi) = \pi + \frac{360-x}{360}(5 \pi)$ . Solving for $x$ reveals $x=\boxed{\textbf{(A) } 108}$ .
~MrThinker
~Dash11 (Edited coefficient of pi. Credit goes to MrThinker for the solution and explanation.)
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_18 | 108 |
AMC8_164 | Each of the even numbers $2, 4, 6, \ldots, 50$ is divided by $7$ . The remainders are recorded. Which histogram displays the number of times each remainder occurs?
[asy] /*By Reda_mandymath*/ unitsize(15); void histogram(pair p, string _str, int[] n) { /* p is shift transformation, _str is choice string, n[] is the array of number of remainders, _pen is the pen style of block, a is the width of block, b is the width of gap _scale is the font scale of labels*/ pen _pen; real a = 0.8; real b = 0.3; real _scale = 0.8; draw(shift(p) * ((0, 0) -- (9, 0) -- (9, 5) -- (0, 5) -- cycle)); label(scale(_scale) * rotate(90) * "Count", (-0.4, 2.5)+p); label(scale(_scale) * "Remainder", (4.5, -1)+p); for (int i = 0; i <= 6; ++i) { if (n[i] == 3) { _pen = mediumgray; } else { _pen = heavygray; } fill(shift(p) * ((a*(i+1) + b*i, 0) -- (a*(i+1) + b*i, n[i]) -- (a*(i+2) + b*i, n[i]) -- (a*(i+2) + b*i, 0) -- cycle), _pen); label(scale(_scale) * string(i), shift(p) * (a*(i+1.5) + b*i, 0), S); label(scale(_scale) * string(n[i]), shift(p) * (a*(i+1.5) + b*i, n[i]), N); } label(_str, shift(p) * (-0.4, 6)); } histogram((0, 0), "$\textbf{(A)}$", new int[] {3, 4, 4, 3, 4, 3, 4}); histogram((12, 0), "$\textbf{(B)}$", new int[] {3, 4, 4, 4, 3, 3, 4}); histogram((24, 0), "$\textbf{(C)}$", new int[] {3, 4, 4, 4, 4, 3, 3}); histogram((0, -8), "$\textbf{(D)}$", new int[] {4, 3, 4, 3, 4, 3, 4}); histogram((12, -8), "$\textbf{(E)}$", new int[] {4, 4, 3, 4, 3, 4, 3}); [/asy]
| Let's take the numbers 2 through 14 (evens). The remainders will be 2, 4, 6, 1, 3, 5, and 0. This sequence keeps repeating itself over and over. We can take floor(50/14) = 3, so after the number 42, every remainder has been achieved 3 times. However, since 44, 46, 48, and 50 are left, the remainders of those will be 2, 4, 6, and 1 respectively. The only histogram in which those 4 numbers are set at 4 is histogram $\boxed{\textbf{(A)}}$ .
~Sigmacuber
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_13 | A |
AMC8_165 | The sum of two prime numbers is $85$ . What is the product of these two prime numbers?
$\textbf{(A) }85\qquad\textbf{(B) }91\qquad\textbf{(C) }115\qquad\textbf{(D) }133\qquad \textbf{(E) }166$
| Since the two prime numbers sum to an odd number, one of them must be even. The only even prime number is $2$ . The other prime number is $85-2=83$ , and the product of these two numbers is $83\cdot2=\boxed{\textbf{(E)}~166}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_4 | 166 |
AMC8_166 | A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 2 feet long or 1 foot long (no blocks may be cut). The vertical joins in the blocks must be staggered as shown, and the wall must be even on the ends. What is the smallest number of blocks needed to build this wall?
[asy] draw((0,0)--(6,0)--(6,1)--(5,1)--(5,2)--(0,2)--cycle); draw((0,1)--(5,1)); draw((1,1)--(1,2)); draw((3,1)--(3,2)); draw((2,0)--(2,1)); draw((4,0)--(4,1)); [/asy]
$\text{(A)}\ 344\qquad\text{(B)}\ 347\qquad\text{(C)}\ 350\qquad\text{(D)}\ 353\qquad\text{(E)}\ 356$
| Since the bricks are $1$ foot high, there will be $7$ rows. To minimize the number of blocks used, rows $1, 3, 5,$ and $7$ will look like the bottom row of the picture, which takes $\frac{100}{2} = 50$ bricks to construct. Rows $2, 4,$ and $6$ will look like the upper row pictured, which has $49$ 2-foot bricks in the middle, and $2$ 1-foot bricks on each end for a total of $51$ bricks.
Four rows of $50$ bricks and three rows of $51$ bricks totals $4\cdot 50 + 3\cdot 51 = 200 + 153 = 353$ bricks, giving the answer $\boxed{D}.$
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_12 | D |
AMC8_167 | In $\triangle ABC,$ a point $E$ is on $\overline{AB}$ with $AE=1$ and $EB=2.$ Point $D$ is on $\overline{AC}$ so that $\overline{DE} \parallel \overline{BC}$ and point $F$ is on $\overline{BC}$ so that $\overline{EF} \parallel \overline{AC}.$ What is the ratio of the area of $CDEF$ to the area of $\triangle ABC?$
[asy] size(7cm); pair A,B,C,DD,EE,FF; A = (0,0); B = (3,0); C = (0.5,2.5); EE = (1,0); DD = intersectionpoint(A--C,EE--EE+(C-B)); FF = intersectionpoint(B--C,EE--EE+(C-A)); draw(A--B--C--A--DD--EE--FF,black+1bp); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",DD,W); label("$E$",EE,S); label("$F$",FF,NE); label("$1$",(A+EE)/2,S); label("$2$",(EE+B)/2,S); [/asy]
$\textbf{(A) } \frac{4}{9} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{5}{9} \qquad \textbf{(D) } \frac{3}{5} \qquad \textbf{(E) } \frac{2}{3}$
| By similar triangles, we have $[ADE] = \frac{1}{9}[ABC]$ . Similarly, we see that [mathjax][BEF] = \dfrac{4}{9}[ABC][/mathjax]. Using this information, we get \[[ACFE] = \frac{5}{9}[ABC].\] Then, since $[ADE] = \frac{1}{9}[ABC]$ , it follows that the [mathjax][CDEF] = \dfrac{4}{9}[ABC][/mathjax]. Thus, the answer would be $\boxed{\textbf{(A) } \frac{4}{9}}$ .
Sidenote: [mathjax][ABC][/mathjax] denotes the area of triangle [mathjax]ABC[/mathjax]. Similarly, [mathjax][ABCD][/mathjax] denotes the area of figure [mathjax]ABCD[/mathjax].
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_20 | 4/9 |
AMC8_168 | Three circular arcs of radius $5$ units bound the region shown. Arcs $AB$ and $AD$ are quarter-circles, and arc $BCD$ is a semicircle. What is the area, in square units, of the region?
[asy] pair A,B,C,D; A = (0,0); B = (-5,5); C = (0,10); D = (5,5); draw(arc((-5,0),A,B,CCW)); draw(arc((0,5),B,D,CW)); draw(arc((5,0),D,A,CCW)); label("$A$",A,S); label("$B$",B,W); label("$C$",C,N); label("$D$",D,E);[/asy]
$\text{(A)}\ 25\qquad\text{(B)}\ 10+5\pi\qquad\text{(C)}\ 50\qquad\text{(D)}\ 50+5\pi\qquad\text{(E)}\ 25\pi$
| Solution 1Draw two squares: one that has opposing corners at $A$ and $B$ , and one that has opposing corners at $A$ and $D$ . These squares share side $\overline{AO}$ , where $O$ is the center of the large semicircle.
These two squares have a total area of $2 \cdot 5^2$ , but have two quarter circle "bites" of radius $5$ that must be removed. Thus, the bottom part of the figure has area
$2\cdot 25 - 2 \cdot \frac{1}{4}\pi \cdot 5^2$
$50 - \frac{25\pi}{2}$
This is the area of the part of the figure underneath $\overline{BD}$ . The part of the figure over $\overline{BD}$ is just a semicircle with radius $5$ , which has area of $\frac{1}{2}\pi\cdot 5^2 = \frac{25\pi}{2}$
Adding the two areas gives a total area of $\boxed{(c)50}$
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_19 | 50 |
AMC8_169 | In $2005$ Tycoon Tammy invested $100$ dollars for two years. During the first year
her investment suffered a $15\%$ loss, but during the second year the remaining
investment showed a $20\%$ gain. Over the two-year period, what was the change
in Tammy's investment?
$\textbf{(A)}\ 5\%\text{ loss}\qquad \textbf{(B)}\ 2\%\text{ loss}\qquad \textbf{(C)}\ 1\%\text{ gain}\qquad \textbf{(D)}\ 2\% \text{ gain} \qquad \textbf{(E)}\ 5\%\text{ gain}$
| After the $15 \%$ loss, Tammy has $100 \cdot 0.85 = 85$ dollars. After the $20 \%$ gain, she has $85 \cdot 1.2 = 102$ dollars. This is an increase in $2$ dollars from her original $100$ dollars, a $\boxed{\textbf{(D)}\ 2 \%\ \text{gain}}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_9 | 2% gain |
AMC8_170 | Wind chill is a measure of how cold people feel when exposed to wind outside. A good estimate for wind chill can be found using this calculation
\[(\text{wind chill}) = (\text{air temperature}) - 0.7 \times (\text{wind speed}),\]
where temperature is measured in degrees Fahrenheit $(^{\circ}\text{F})$ and the wind speed is measured in miles per hour (mph). Suppose the air temperature is $36^{\circ}\text{F}$ and the wind speed is $18$ mph. Which of the following is closest to the approximate wind chill?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 35$
| By substitution, we have
\begin{align*} (\text{wind chill}) &= 36 - 0.7 \times 18 \\ &= 36 - 12.6 \\ &= 23.4 \\ &\approx \boxed{\textbf{(B)}\ 23}. \end{align*}
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_3 | 23 |
AMC8_171 | Soda is sold in packs of 6, 12 and 24 cans. What is the minimum number of packs needed to buy exactly 90 cans of soda?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 15$
| Start by buying the largest packs first. After three $24$ -packs, $90-3(24)=18$ cans are left. After one $12$ -pack, $18-12=6$ cans are left. Then buy one more $6$ -pack. The total number of packs is $3+1+1=\boxed{\textbf{(B)}\ 5}$ .
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_5 | 5 |
AMC8_172 | Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces
[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight); [/asy]
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$
| This is the number cubes that are adjacent to another cube on exactly two sides. The bottom corner cubes are connected on three sides, and the top corner cubes are connected on one. The number we are looking for is the number of middle cubes, which is $\boxed{\textbf{(B)}\ 6}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_13 | 6 |
AMC8_173 | A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$
| A number is divisible by $15$ precisely if it is divisible by $3$ and $5$ . The latter means the last digit must be either $5$ or $0$ , and the former means the sum of the digits must be divisible by $3$ . If the last digit is $0$ , the first digit would be $0$ (because the digits alternate), which is not possible. Hence the last digit must be $5$ , and the number is of the form $5\square 5\square 5$ . If the unknown digit is $x$ , we deduce $5+x+5+x+5 \equiv 0 \pmod{3} \Rightarrow 2x \equiv 0 \pmod{3}$ . We know $2^{-1}$ exists modulo $3$ because 2 is relatively prime to 3, so we conclude that $x$ (i.e. the second and fourth digit of the number) must be a multiple of $3$ . It can be $0$ , $3$ , $6$ , or $9$ , so there are $\boxed{\textbf{(B) }4}$ options: $50505$ , $53535$ , $56565$ , and $59595$ .
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_19 | 4 |
AMC8_174 | The table below shows the ancient Egyptian hieroglyphs that were used to represent different numbers.
For example, the number $32$ was represented by the hieroglyphs $\cap \cap \cap ||$ . What number is represented by the following combination of hieroglyphs?
$\textbf{(A)}\ 1,423 \qquad \textbf{(B)}\ 10,423 \qquad \textbf{(C)}\ 14,023 \qquad \textbf{(D)}\ 14,203 \qquad \textbf{(E)}\ 14,230$
| The first hieroglyph is worth $10,000$ , the next 4 are worth $100 \cdot 4 = 400$ , the next $2$ are worth $10 \cdot 2 = 20$ , and the last $3$ are worth $1 \cdot 3 = 3$ . Therefore, the answer is $10,000 + 400 + 20 + 3 = \boxed{\textbf{(B)}\ 10,423}$
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_2 | 10,423 |
AMC8_175 | A unit hexagram is composed of a regular hexagon of side length $1$ and its $6$
equilateral triangular extensions, as shown in the diagram. What is the ratio of
the area of the extensions to the area of the original hexagon?
[asy] defaultpen(linewidth(0.7)); draw(polygon(3)); pair D=origin+1*dir(270), E=origin+1*dir(150), F=1*dir(30); draw(D--E--F--cycle); [/asy]
$\mathrm{(A)}\ 1:1 \qquad \mathrm{(B)}\ 6:5 \qquad \mathrm{(C)}\ 3:2 \qquad \mathrm{(D)}\ 2:1 \qquad \mathrm{(E)}\ 3:1$
| The six equilateral triangular extensions fit perfectly into the hexagon meaning the answer is $\boxed{\textbf{(A) }1:1}$
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_12 | 1:1 |
AMC8_176 | What is the smallest whole number larger than the perimeter of any triangle with a side of length $5$ and a side of length $19$ ?
$\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57$
| We know from the Triangle Inequality that the last side, $s$ , fulfills $s<5+19=24$ . Adding $5+19$ to both sides of the inequality, we get $s+5+19<48$ , and because $s+5+19$ is the perimeter of our triangle, $\boxed{\textbf{(D)}\ 48}$ is our answer.
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_8 | 48 |
AMC8_177 | The Little Twelve Basketball Conference has two divisions, with six teams in each division. Each team plays each of the other teams in its own division twice and every team in the other division once. How many conference games are scheduled?
$\textbf{(A)}\ 80\qquad\textbf{(B)}\ 96\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 192$
| Within each division, there are $\binom {6}{2} = 15$ pairings, and each of these games happens twice. The same goes for the other division so that there are $4(15)=60$ games within their own divisions. The number of games between the two divisions is $(6)(6)=36$ . Together there are $60+36=\boxed{\textbf{(B)}\ 96}$ conference games.
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_14 | 96 |
AMC8_178 | Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can,
and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how
many items will they buy?
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$
| We know that the sandwiches cost $4.50$ dollars. Guessing will bring us to multiplying $4.50$ by 6, which gives us $27.00$ . Since they can spend $30.00$ they have $3$ dollars left. Since soft drinks cost $1.00$ dollar each, they can buy 3 soft drinks, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 soft drinks, they bought a total of $9$ items. Therefore, the answer is $\boxed{\textbf{(D) }9}$ .
- SBose
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_1 | 9 |
AMC8_179 | How many positive integers can fill the blank in the sentence below?
“One positive integer is _____ more than twice another, and the sum of the two numbers is $28$ .”
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$
| Let $m$ and $n$ be positive integers such that $m>n$ and $m+n=28.$ It follows that $m=2n+d$ for some positive integer $d.$ We wish to find the number of possible values for $d.$
By substitution, we have $(2n+d)+n=28,$ from which $d=28-3n.$ Note that $n=1,2,3,\ldots,9$ each generate a positive integer for $d,$ so there are $\boxed{\textbf{(D) } 9}$ possible values for $d.$
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_13 | 9 |
AMC8_180 | The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?
\[\begin{array}{c|c|c}\text{Child}&\text{Eye Color}&\text{Hair Color}\\ \hline\text{Benjamin}&\text{Blue}&\text{Black}\\ \hline\text{Jim}&\text{Brown}&\text{Blonde}\\ \hline\text{Nadeen}&\text{Brown}&\text{Black}\\ \hline\text{Austin}&\text{Blue}&\text{Blonde}\\ \hline\text{Tevyn}&\text{Blue}&\text{Black}\\ \hline\text{Sue}&\text{Blue}&\text{Blonde}\\ \hline\end{array}\]
$\textbf{(A)}\ \text{Nadeen and Austin}\qquad\textbf{(B)}\ \text{Benjamin and Sue}\qquad\textbf{(C)}\ \text{Benjamin and Austin}\qquad\textbf{(D)}\ \text{Nadeen and Tevyn}$
$\textbf{(E)}\ \text{Austin and Sue}$
| Jim has brown eyes and blonde hair. If you look for anybody who has brown eyes or blonde hair, you find that Nadeen, Austin, and Sue are Jim's possible siblings. However, the children have at least one common characteristics. Since Austin and Sue both have blonde hair, Nadeen is ruled out and therefore $\boxed{\textbf{(E)}\ \text{Austin and Sue}}$ are his siblings. You can also see that in the hair color column, there are three black haired people and three blond haired people. Since Jim has blond hair, all his siblings must be the other two with blond hair.
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_17 | Austin and Sue |
AMC8_181 | A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?
[asy] defaultpen(linewidth(0.8)); path p=unitsquare; draw(p^^shift(0,1)*p^^shift(1,0)*p); draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p); label("FRONT", (1,0), S); label("SIDE", (5,0), S); [/asy]
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$
| In order to minimize the amount of cubes needed, we must match up as many squares of our given figures with each other to make different sides of the same cube. One example of the solution with $\boxed{\textbf{(B)}\ 4}$ cubes. Notice the corner cube cannot be removed for a figure of 3 cubes because at least one face of a cube must be touching another face.
[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,-1,0)*unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(1,0,1)*unitcube, white, thick(), nolight);[/asy]
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_15 | 4 |
AMC8_182 | The harmonic mean of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1, 2, and 4?
$\textbf{(A) }\frac{3}{7}\qquad\textbf{(B) }\frac{7}{12}\qquad\textbf{(C) }\frac{12}{7}\qquad\textbf{(D) }\frac{7}{4}\qquad \textbf{(E) }\frac{7}{3}$
| The sum of the reciprocals is $\frac{1}{1} + \frac{1}{2} + \frac{1}{4}= \frac{7}{4}$ . Their average is $\frac{7}{12}$ . Taking the reciprocal of this gives $\boxed{\textbf{(C) }\frac{12}{7}}$ .
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_10 | 12/7 |
AMC8_183 | The $7$ -digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ are each multiples of $3$ . Which of the following could be the value of $C$ ?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8$
| Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. $7 + 4 + 5 + 2 + 1 = 19$ . To be a multiple of $3$ , $A + B$ has to be either $2$ or $5$ or $8$ ... and so on. We add up the numerical digits in the second number; $3 + 2 + 6 + 4 = 15$ . We then add two of the selected values, $5$ to $15$ , to get $20$ . We then see that C = $1, 4$ or $7, 10$ ... and so on, otherwise the number will not be divisible by three. We then add $8$ to $15$ , to get $23$ , which shows us that C = $1$ or $4$ or $7$ ... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be $1, 4,$ and $7$ . However, in the answer choices, there is no $7$ or $4$ or anything greater than $7$ , but there is a $1$ , so $\boxed{\textbf{(A) }1}$ is our answer.
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_21 | 1 |
AMC8_186 | George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first $\frac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must George run the last $\frac{1}{2}$ mile in order to arrive just as school begins today?
$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad \textbf{(E) }12$
| Note that on a normal day, it takes him $1/3$ hour to get to school. However, today it took $\frac{1/2 \text{ mile}}{2 \text{ mph}}=1/4$ hour to walk the first $1/2$ mile. That means that he has $1/3 -1/4 = 1/12$ hours left to get to school, and $1/2$ mile left to go. Therefore, his speed must be $\frac{1/2 \text{ mile}}{1/12 \text { hour}}=\boxed{6 \text{ mph}}$ , so $\boxed{\text{(B) }6}$ is the answer.
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_17 | 6 |
AMC8_187 | Billy's basketball team scored the following points over the course of the first $11$ games of the season. If his team scores $40$ in the $12^{th}$ game, which of the following statistics will show an increase?
\[42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73\]
$\textbf{(A) } \text{range} \qquad \textbf{(B) } \text{median} \qquad \textbf{(C) } \text{mean} \qquad \textbf{(D) } \text{mode} \qquad \textbf{(E) } \text{mid-range}$
| Solution 1When they score a $40$ on the next game, the range increases from $73-42=31$ to $73-40=33$ . This means the $\boxed{\textbf{(A) } \text{range}}$ increased.
Note: The range is defined to be the difference of the largest element and the smallest element of a set.
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_5 | range |
AMC8_188 | How many $3$ -digit positive integers have digits whose product equals $24$ ?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 24$
| With the digits listed from least to greatest, the $3$ -digit integers are $138,146,226,234$ . $226$ can be arranged in $\frac{3!}{2!} = 3$ ways, and the other three can be arranged in $3!=6$ ways. There are $3+6(3) = \boxed{\textbf{(D)}\ 21}$ $3$ -digit positive integers.
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_16 | 21 |
AMC8_189 | In the figure below, $ABCD$ is a rectangle with sides of length $AB = 5$ inches and $AD$ = 3 inches. Rectangle $ABCD$ is rotated $90^\circ$ clockwise around the midpoint of side $DC$ to give a second rectangle. What is the total area, in square inches, covered by the two overlapping rectangles?
$\textbf{(A)}\ 21 \qquad \textbf{(B)}\ 22.25 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 23.75 \qquad \textbf{(E)}\ 25$
| The area of each rectangle is $5 \cdot 3 = 15$ . Then the sum of the areas of the two regions is the sum of the areas of the two rectangles, minus the area of their overlap. To find the area of the overlap, we note that the region of overlap is a square, each of whose sides have length $2.5$ (as they are formed by the midpoint of one of the long sides and a vertex). Then the answer is $15+15-2.5^2=\boxed{\textbf{(D)}~23.75}$ .
~ cxsmi
~ Edited by Aoum
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_10 | 23.75 |
AMC8_190 | Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); label("$B$",(6,6),NE); label("$C$",(6,0),SE); label("$D$",(0,0),SW); label("$E$",(3,0),S); label("$F$",(4,2),E); [/asy]
$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$
| We can use analytic geometry for this problem.
Let us start by giving $D$ the coordinate $(0,0)$ , $A$ the coordinate $(0,1)$ , and so forth. $\overline{AC}$ and $\overline{EB}$ can be represented by the equations $y=-x+1$ and $y=2x-1$ , respectively. Solving for their intersection gives point $F$ coordinates $\left(\frac{2}{3},\frac{1}{3}\right)$ .
Now, we can see that $\triangle$ $EFC$ ’s area is simply $\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}$ or $\frac{1}{12}$ . This means that pentagon $ABCEF$ ’s area is $\frac{1}{2}+\frac{1}{12}=\frac{7}{12}$ of the entire square, and it follows that quadrilateral $AFED$ ’s area is $\frac{5}{12}$ of the square.
The area of the square is then $\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B) } 108}$ .
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_22 | 108 |
AMC8_191 | A $\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$ -by- $3$ grid. Shown below is a sample configuration with three $\triangle$ s in a line.
[asy] //diagram size(5cm); defaultpen(linewidth(1.5)); real r = 0.37; path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle; draw((0,0)--(0,3)--(3,3)--(3,0)--cycle); draw((0,1)--(3,1)--(3,2)--(0,2)--cycle); draw((1,0)--(1,3)--(2,3)--(2,0)--cycle); draw(circle((3/2,5/2),1/3)); draw(circle((5/2,1/2),1/3)); draw(circle((3/2,3/2),1/3)); draw(shift(0.5,0.38) * equi); draw(shift(1.5,0.38) * equi); draw(shift(0.5,1.38) * equi); draw(shift(2.5,1.38) * equi); draw(shift(0.5,2.38) * equi); draw(shift(2.5,2.38) * equi); [/asy]
How many configurations will have three $\triangle$ s in a line and three $\bigcirc$ s in a line?
$\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96$
| Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end.
We take casework:
Case 1: 3 lines :
In this case, the lines would need to be $2$ of one shape and $1$ of another, so there are $\frac{3!}{2} = 3$ ways to arrange the lines and $2$ ways to pick which shape has only one line. In total, this is $3\cdot 2 = 6.$
Case 2: 2 lines :
In this case, the lines would be one line of triangles, one line of circles, and the last one can be anything that includes both shapes. There are $3! = 6$ ways to arrange the lines and $2^3-2 = 6$ ways to choose the last line. (We subtract $2$ from the last line because one arrangement of the last line is all triangles and the other arrangement of the last line is all circles, which causes Case 2 to overlap with Case 1 and further complicating the solution.) In total, this is $6\cdot 6 = 36.$
Finally, we add and multiply: $2(36+6)=2(42)=\boxed{\textbf{(D) }84}$ .
~wamofan
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_23 | 84 |
AMC8_192 | Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$
| The area of trapezoid $CBFE$ is $\frac{1+3}2\cdot 4=8$ . Next, we find the height of each triangle to calculate their area. The two non-colored isosceles triangles are similar, and are in a $3:1$ ratio by AA similarity (alternate interior and vertical angles) so the height of the larger is $3,$ while the height of the smaller one is $1.$ Thus, their areas are $\frac12$ and $\frac92$ . Subtracting these areas from the trapezoid, we get $8-\frac12-\frac92 =\boxed3$ . Therefore, the answer to this problem is $\boxed{\textbf{(C) }3}$ .
~23orimy412uc3478
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_22 | 3 |
AMC8_193 | The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn?
[asy] for (int a = 0; a < 6; ++a) { for (int b = 0; b < 6; ++b) { dot((4*a,3*b)); } } draw((0,0)--(20,0)--(20,15)--(0,15)--cycle); draw((0,0)--(16,12)); draw((0,0)--(16,9)); label(rotate(30)*"Bjorn",(12,6.75),SE); label(rotate(37)*"Alberto",(11,8.25),NW); label("$0$",(0,0),S); label("$1$",(4,0),S); label("$2$",(8,0),S); label("$3$",(12,0),S); label("$4$",(16,0),S); label("$5$",(20,0),S); label("$0$",(0,0),W); label("$15$",(0,3),W); label("$30$",(0,6),W); label("$45$",(0,9),W); label("$60$",(0,12),W); label("$75$",(0,15),W); label("H",(6,-2),S); label("O",(8,-2),S); label("U",(10,-2),S); label("R",(12,-2),S); label("S",(14,-2),S); label("M",(-4,11),N); label("I",(-4,9),N); label("L",(-4,7),N); label("E",(-4,5),N); label("S",(-4,3),N); [/asy]
$\text{(A)}\ 15 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 25 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$
| After 4 hours, we see that Bjorn biked 45 miles, and Alberto biked 60. Thus the answer is $60-45=15$ $\boxed{\text{(A)}}$ .
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_4 | 15 |
AMC8_194 | Casey's shop class is making a golf trophy. He has to paint 300 dimples on a golf ball. If it takes him 2 seconds to paint one dimple, how many minutes will it take for him to finish
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$
| It will take him $300\cdot2=600$ seconds to paint all the dimples.
This is equivalent to $\frac{600}{60}=10$ minutes $\Rightarrow \boxed{D}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_1 | D |
AMC8_195 | The diagram shows an octagon consisting of $10$ unit squares. The portion below $\overline{PQ}$ is a unit square and a triangle with base $5$ . If $\overline{PQ}$ bisects the area of the octagon, what is the ratio $\dfrac{XQ}{QY}$ ?
[asy] import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); draw((0,0)--(6,0),linewidth(1.2pt)); draw((0,0)--(0,1),linewidth(1.2pt)); draw((0,1)--(1,1),linewidth(1.2pt)); draw((1,1)--(1,2),linewidth(1.2pt)); draw((1,2)--(5,2),linewidth(1.2pt)); draw((5,2)--(5,1),linewidth(1.2pt)); draw((5,1)--(6,1),linewidth(1.2pt)); draw((6,1)--(6,0),linewidth(1.2pt)); draw((1,1)--(5,1),linewidth(1.2pt)); draw((1,1)--(1,0),linewidth(1.2pt)); draw((2,2)--(2,0),linewidth(1.2pt)); draw((3,2)--(3,0),linewidth(1.2pt)); draw((4,2)--(4,0),linewidth(1.2pt)); draw((5,1)--(5,0),linewidth(1.2pt)); draw((0,0)--(5,1.5),linewidth(1.2pt)); dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); dot((0,1),ds); dot((1,1),ds); dot((1,2),ds); dot((5,2),ds); label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); dot((4,2),ds); dot((5,1.5),ds); label("$Q$", (5.14,1.51),NE*lsf); clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle); [/asy]
$\textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4}$
| We see that half the area of the octagon is $5$ . We see that the triangle area is $5-1 = 4$ . That means that $\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}$ .
\[\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}\]
Meaning, $\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}$
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_17 | 2/3 |
AMC8_196 | A square with integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?
$\text{(A)}\hspace{.05in}3\qquad\text{(B)}\hspace{.05in}4\qquad\text{(C)}\hspace{.05in}5\qquad\text{(D)}\hspace{.05in}6\qquad\text{(E)}\hspace{.05in}7$
| The first answer choice ${\textbf{(A)}\ 3}$ , can be eliminated since there must be $10$ squares with integer side lengths. We then test the next smallest sidelength which is $4$ . The square with area $16$ can be partitioned into $8$ squares with area $1$ and two squares with area $4$ , which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is $\boxed{\textbf{(B)}\ 4}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_17 | 4 |
AMC8_197 | The circumference of the circle with center $O$ is divided into $12$ equal arcs, marked the letters $A$ through $L$ as seen below. What is the number of degrees in the sum of the angles $x$ and $y$ ?
[asy] size(230); defaultpen(linewidth(0.65)); pair O=origin; pair[] circum = new pair[12]; string[] let = {"$A$","$B$","$C$","$D$","$E$","$F$","$G$","$H$","$I$","$J$","$K$","$L$"}; draw(unitcircle); for(int i=0;i<=11;i=i+1) { circum[i]=dir(120-30*i); dot(circum[i],linewidth(2.5)); label(let[i],circum[i],2*dir(circum[i])); } draw(O--circum[4]--circum[0]--circum[6]--circum[8]--cycle); label("$x$",circum[0],2.75*(dir(circum[0]--circum[4])+dir(circum[0]--circum[6]))); label("$y$",circum[6],1.75*(dir(circum[6]--circum[0])+dir(circum[6]--circum[8]))); label("$O$",O,dir(60)); [/asy]
$\textbf{(A) }75\qquad\textbf{(B) }80\qquad\textbf{(C) }90\qquad\textbf{(D) }120\qquad\textbf{(E) }150$
| The measure of an inscribed angle is half the measure of its corresponding central angle. Since each unit arc is $\frac{1}{12}$ of the circle's circumference, each unit central angle measures $\frac{360}{12}^{\circ}=30^{\circ}$ . From this, $\angle EOG = 60^{\circ}$ , so $x = 30^{\circ}$ . Also, $\angle AOI = 120^{\circ}$ , so $y = 60^{\circ}$ . The number of degrees in the sum of both angles is $30 + 60 = \boxed{(C)\ 90}.$
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_15 | 90 |
AMC8_198 | The following figures are composed of squares and circles. Which figure has a shaded region with largest area?
[asy]/* AMC8 2003 #22 Problem */ size(3inch, 2inch); unitsize(1cm); pen outline = black+linewidth(1); filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline); filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline); filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1)); filldraw(Circle((1,1), 1), white, outline); filldraw(Circle((3.5,.5), .5), white, outline); filldraw(Circle((4.5,.5), .5), white, outline); filldraw(Circle((3.5,1.5), .5), white, outline); filldraw(Circle((4.5,1.5), .5), white, outline); filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline); label("A", (1, 2), N); label("B", (4, 2), N); label("C", (7, 2), N); draw((0,-.5)--(.5,-.5), BeginArrow); draw((1.5, -.5)--(2, -.5), EndArrow); label("2 cm", (1, -.5)); draw((3,-.5)--(3.5,-.5), BeginArrow); draw((4.5, -.5)--(5, -.5), EndArrow); label("2 cm", (4, -.5)); draw((6,-.5)--(6.5,-.5), BeginArrow); draw((7.5, -.5)--(8, -.5), EndArrow); label("2 cm", (7, -.5)); draw((6,1)--(6,-.5), linetype("4 4")); draw((8,1)--(8,-.5), linetype("4 4"));[/asy]
$\textbf{(A)}\ \text{A only}\qquad\textbf{(B)}\ \text{B only}\qquad\textbf{(C)}\ \text{C only}\qquad\textbf{(D)}\ \text{both A and B}\qquad\textbf{(E)}\ \text{all are equal}$
| First we have to find the area of the shaded region in each of the figures. In figure $\textbf{A}$ the area of the shaded region is the area of the circle subtracted from the area of the square. That is $2^2-1^2 \pi=4-\pi$ . In figure $\textbf{B}$ the area of the shaded region is the sum of the areas of the 4 circles subtracted from the area of the square. That is $2^2-4 \left( \left( \frac{1}{2} \right)^2 \pi \right)=4-4 \left(\frac{\pi}{4} \right)=4-\pi$ . In figure $\textbf{C}$ the area of the shaded region is the area of the square subtracted from the area of the circle. The diameter of the circle and the diagonal of the square are equal to 2. We can easily find the area of the square using the area formula $\frac{d_1 d_2}{2}$ . So the area of the shaded region is $1^2 \pi-\frac{2\cdot{2}}{2}=\pi-2$ . Clearly the largest area that we found among the three shaded regions is $\pi-2$ . Thus the answer is $\boxed{C}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_22 | C |
AMC8_199 | What is the value of the product
\[\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?\]
$\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50$
| We rewrite: \[\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{100}{99}\]
The middle terms cancel, leaving us with
\[\left(\frac{1\cdot100}{2\cdot99}\right)= \boxed{\textbf{(B)}\frac{50}{99}}\]
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_17 | 50/99 |
AMC8_200 | The taxi fare in Gotham City is $2.40 for the first $\frac12$ mile and additional mileage charged at the rate $0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles can you ride for $10?
$\textbf{(A) }3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75$
| Let $x$ be the number of miles you ride. The number of miles you ride after the first half mile is $x-0.5.$ We can write this equation:
\begin{align*} 10 &= 2.4 + 0.2 \times \frac{x-0.5}{0.1} + 2\\ 5.6 &= 2(x-0.5)\\ 2.8 &= x-0.5\\ x &= \boxed{\textbf{(C)}\ 3.3}\end{align*}
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_10 | 3.3 |
AMC8_201 | In trapezoid $ABCD$ , angles $B$ and $C$ measure $60^\circ$ and $AB = DC$ . The side lengths are all positive integers, and the perimeter of $ABCD$ is 30 units. How many non-congruent trapezoids satisfy all of these conditions?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
| Let $a$ be the length of the shorter base, and let $b$ be the length of the longer one. Note that these two parameters, along with the angle measures and the fact that the trapezoid is isosceles, uniquely determine a trapezoid. We drop perpendiculars down from the endpoints of the top base. Then the length from the foot of this perpendicular to either vertex will be half the difference between the lengths of the two bases, or $\frac{b-a}{2}$ . Now, since we have a 30-60-90 triangle and this side length corresponds to the "30" part, the length of the hypotenuse (one of the legs) is $2 \cdot \frac{b-a}{2} = b-a$ . Then the perimeter of the trapezoid is $2(b-a)+a+b=3b-a=30$ . The only other stipulation for this trapezoid to be valid is that $b>a$ (which was our assumption). We can now easily count the valid pairs $(a,b)$ , yielding $(3,11),(6,12),(9,13),(12,14)$ . It is clear that proceeding further would cause $a \geq b$ , so we have $\boxed{\textbf{(E)}~4}$ valid trapezoids.
~cxsmi
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_24 | 4 |
AMC8_202 | Let the letters $F$ , $L$ , $Y$ , $B$ , $U$ , $G$ represent distinct digits. Suppose $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}$ is the greatest number that satisfies the equation
\[8\cdot\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G}.\]
What is the value of $\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G}$ ?
$\textbf{(A)}\ 1089 \qquad \textbf{(B)}\ 1098 \qquad \textbf{(C)}\ 1107 \qquad \textbf{(D)}\ 1116 \qquad \textbf{(E)}\ 1125$
| The highest that $FLYFLY$ can be would have to be $124124$ , and it cannot be higher than that, because then it would be $125125$ , and $125125$ multiplied by 8 is $1001000$ , and then it would exceed the $6$ - digit limit set on $BUGBUG$ .
So, if we start at $124124\cdot8$ , we get $992992$ , which would be wrong because both $B \& U$ would be $9$ , and the numbers cannot be repeated between different letters.
If we move on to the next highest, $123123$ , and multiply by $8$ , we get $984984$ . All the digits are different, so $FLY+BUG$ would be $123+984$ , which is $1107$ . So, the answer is $\boxed{\textbf{(C)}1107}$ .
- Akhil Ravuri of John Adams Middle School
- Aryan Varshney of John Adams Middle School (minor edits... props to Akhil for the main/full answer :D)
~ cxsmi (minor formatting edits)
~ Alice of Evergreen Middle School
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_15 | 1107 |
AMC8_203 | Eight points are spaced around at intervals of one unit around a $2 \times 2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the two points are one unit apart?
[asy] size((50)); dot((5,0)); dot((5,5)); dot((0,5)); dot((-5,5)); dot((-5,0)); dot((-5,-5)); dot((0,-5)); dot((5,-5)); [/asy]
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{2}{7}\qquad\textbf{(C)}\ \frac{4}{11}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{4}{7}$
| The two points are one unit apart at $8$ places around the edge of the square. There are $8 \choose 2$ $= 28$ ways to choose two points. The probability is
\[\frac{8}{28} = \boxed{\textbf{(B)}\ \frac27}\]
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_19 | 2/7 |
AMC8_204 | To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.
[asy] for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } } draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy]
The large kite is covered with gold foil. The foil is cut from a rectangular piece that just covers the entire grid. How many square inches of waste material are cut off from the four corners?
$\textbf{(A)}\ 63 \qquad \textbf{(B)}\ 72 \qquad \textbf{(C)}\ 180 \qquad \textbf{(D)}\ 189 \qquad \textbf{(E)}\ 264$
| The large grid has dimensions three times that of the small grid, so its dimensions are $3(6)\times3(7)$ , or $18\times21$ , so the area is $(18)(21)=378$ . The area of the kite is half of the area of the rectangle as you can see, so the area of the waste material is also half the area of the rectangle. Thus, the area of the waste material is $378/2=\boxed{189\textbf{ (D)}}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_9 | 189 |
AMC8_205 | The average of the five numbers in a list is $54$ . The average of the first two
numbers is $48$ . What is the average of the last three numbers?
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 57 \qquad \textbf{(D)}\ 58 \qquad \textbf{(E)}\ 59$
| Let the $5$ numbers be $a, b, c, d$ , and $e$ . Thus $\frac{a+b+c+d+e}{5}=54$ and $a+b+c+d+e=270$ . Since $\frac{a+b}{2}=48$ , $a+b=96$ . Substituting back into our original equation, we have $96+c+d+e=270$ and $c+d+e=174$ . Dividing by $3$ gives the average of $\boxed{\textbf{(D)}\ 58}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_9 | 58 |
AMC8_206 | Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than $100$ pounds or more than $150$ pounds. So the boxes are weighed in pairs in every possible way. The results are $122$ , $125$ and $127$ pounds. What is the combined weight in pounds of the three boxes?
$\textbf{(A)}\ 160\qquad \textbf{(B)}\ 170\qquad \textbf{(C)}\ 187\qquad \textbf{(D)}\ 195\qquad \textbf{(E)}\ 354$
| Each box is weighed twice during this, so the combined weight of the three boxes is half the weight of these separate measures:
\[\frac{122+125+127}{2} = \frac{374}{2} = \boxed{\textbf{(C)}\ 187}.\]
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_13 | 187 |
AMC8_207 | In this addition problem, each letter stands for a different digit.
$\setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&T & W & O\\ +&T & W & O\\ \hline F& O & U & R\end{array}$
If T = 7 and the letter O represents an even number, what is the only possible value for W?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$
| Since both T's are 7, then O has to equal 4, because 7 + 7 = 14. Then, F has to equal 1. To get R, we do 4 + 4 (since O = 4) to get R = 8. The value for W then has to be a number less than 5, otherwise it will change the value of O, and can't be a number that has already been used, like 4 or 1. The only other possibilities are 2 and 3. 2 doesn't work because it makes U = 4, which is what O already equals. So, the only possible value of W is 3 $\boxed{\text{D}}$
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_14 | D |
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