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AMC8_315 | Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?
$\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140$
| Clearly, for every $5$ seconds, $3$ cars pass. It's more convenient to have everything in seconds: $2$ minutes and $45$ seconds $=2\cdot60 + 45 = 165$ seconds. We then set up a ratio: \[\frac{3}{5}=\frac{x}{165}\]
\[3(165)=5x\]
\[x=3(33)=99\approx\boxed{\textbf{(C)}\ 100}.\]
~megaboy6679 ~MiracleMaths
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_7 | 100 |
AMC8_316 | The numbers from $1$ to $49$ are arranged in a spiral pattern on a square grid, beginning at the center. The first few numbers have been entered into the grid below. Consider the four numbers that will appear in the shaded squares, on the same diagonal as the number $7.$ How many of these four numbers are prime?
[asy] /* Made by MRENTHUSIASM */ size(175); void ds(pair p) { filldraw((0.5,0.5)+p--(-0.5,0.5)+p--(-0.5,-0.5)+p--(0.5,-0.5)+p--cycle,mediumgrey); } ds((0.5,4.5)); ds((1.5,3.5)); ds((3.5,1.5)); ds((4.5,0.5)); add(grid(7,7,grey+linewidth(1.25))); int adj = 1; int curUp = 2; int curLeft = 4; int curDown = 6; label("$1$",(3.5,3.5)); for (int len = 3; len<=3; len+=2) { for (int i=1; i<=len-1; ++i) { label("$"+string(curUp)+"$",(3.5+adj,3.5-adj+i)); label("$"+string(curLeft)+"$",(3.5+adj-i,3.5+adj)); label("$"+string(curDown)+"$",(3.5-adj,3.5+adj-i)); ++curDown; ++curLeft; ++curUp; } ++adj; curUp = len^2 + 1; curLeft = len^2 + len + 2; curDown = len^2 + 2*len + 3; } draw((4,4)--(3,4)--(3,3)--(5,3)--(5,5)--(2,5)--(2,2)--(6,2)--(6,6)--(1,6)--(1,1)--(7,1)--(7,7)--(0,7)--(0,0)--(7,0),linewidth(2)); [/asy]
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
| We fill out the grid, as shown below:
[asy] /* Made by MRENTHUSIASM */ size(175); void ds(pair p) { filldraw((0.5,0.5)+p--(-0.5,0.5)+p--(-0.5,-0.5)+p--(0.5,-0.5)+p--cycle,mediumgrey); } ds((0.5,4.5)); ds((1.5,3.5)); ds((3.5,1.5)); ds((4.5,0.5)); add(grid(7,7,grey+linewidth(1.25))); int adj = 1; int curUp = 2; int curLeft = 4; int curDown = 6; int curRight = 8; label("$1$",(3.5,3.5)); for (int len = 3; len<=7; len+=2) { for (int i=1; i<=len-1; ++i) { label("$"+string(curUp)+"$",(3.5+adj,3.5-adj+i)); label("$"+string(curLeft)+"$",(3.5+adj-i,3.5+adj)); label("$"+string(curDown)+"$",(3.5-adj,3.5+adj-i)); label("$"+string(curRight)+"$",(3.5-adj+i,3.5-adj)); ++curDown; ++curLeft; ++curUp; ++curRight; } ++adj; curUp = len^2 + 1; curLeft = len^2 + len + 2; curDown = len^2 + 2*len + 3; curRight = len^2 + 3*len + 4; } draw((4,4)--(3,4)--(3,3)--(5,3)--(5,5)--(2,5)--(2,2)--(6,2)--(6,6)--(1,6)--(1,1)--(7,1)--(7,7)--(0,7)--(0,0)--(7,0),linewidth(2)); [/asy]
From the four numbers that appear in the shaded squares, $\boxed{\textbf{(D)}\ 3}$ of them are prime: $19,23,$ and $47.$
~MathFun1000, MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_4 | 3 |
AMC8_317 | The region shown below consists of 24 squares, each with side length 1 centimeter. What is the area, in square centimeters, of the largest circle that can fit inside the region, possibly touching the boundaries?
[asy] size(100); void drawSquare(pair p) { draw(box(p, p + (1,1)), black); } int[][] grid = { {0, 0, 0, 0, 0, 0}, {0, 0, 1, 1, 0, 0}, {0, 1, 1, 1, 1, 0}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {0, 1, 1, 1, 1, 0}, {0, 0, 1, 1, 0, 0}, {0, 0, 0, 0, 0, 0} }; int rows = grid.length; int cols = grid[0].length; for (int i = 0; i < rows; ++i) { for (int j = 0; j < cols; ++j) { if (grid[i][j] == 1) { drawSquare((j, rows - i - 1)); } } } [/asy]
$\textbf{(A)}\ 3\pi\qquad \textbf{(B)}\ 4\pi\qquad \textbf{(C)}\ 5\pi\qquad \textbf{(D)}\ 6\pi\qquad \textbf{(E)}\ 8\pi$
| The largest circle that can fit inside the figure has its center in the middle of the figure and will be tangent to the figure in $8$ points. By the Pythagorean Theorem, the distance from the center to one of these $8$ points is $\sqrt{2^2 + 1^2} = \sqrt5$ , so the area of this circle is $\pi \sqrt{5}^2 = \boxed{\textbf{(C)} 5\pi}$ .
~Soupboy0
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_12 | 5π |
AMC8_318 | In rectangle $ABCD$ , $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$ ?
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); label("$4$", (0, 2), W); label("$6$", (3, 0), S);[/asy]
$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$
| Solution 1Using the triangle area formula for triangles: $A = \frac{bh}{2},$ where $A$ is the area, $b$ is the base, and $h$ is the height. This equation gives us $A = \frac{4 \cdot 6}{2} = \frac{24}{2} =\boxed{\textbf{(A) } 12}$ .
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2 | 12 |
AMC8_319 | The digits $1$ , $2$ , $3$ , $4$ , and $5$ are each used once to write a five-digit number $PQRST$ . The three-digit number $PQR$ is divisible by $4$ , the three-digit number $QRS$ is divisible by $5$ , and the three-digit number $RST$ is divisible by $3$ . What is $P$ ?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$
| Solution 1 (Modular Arithmetic)We see that since $QRS$ is divisible by $5$ , $S$ must equal either $0$ or $5$ , but it cannot equal $0$ , so $S=5$ . We notice that since $PQR$ must be even, $R$ must be either $2$ or $4$ . However, when $R=2$ , we see that $T \equiv 2 \pmod{3}$ , which cannot happen because $2$ and $5$ are already used up; so $R=4$ . This gives $T \equiv 3 \pmod{4}$ , meaning $T=3$ . Now, we see that $Q$ could be either $1$ or $2$ , but $14$ is not divisible by $4$ , but $24$ is. This means that $Q=2$ and $P=\boxed{\textbf{(A)}\ 1}$ .
~CHECKMATE2021
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_24 | 1 |
AMC8_320 | Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of $5$ -cent, $10$ -cent, and $25$ -cent stamps, with exactly $20$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $$7.10$ in postage?
(Note: The amount $$7.10$ corresponds to $7$ dollars and $10$ cents. One dollar is worth $100$ cents.)
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 54\qquad \textbf{(E)}\ 55$
| Let's use the most stamps to make $7.10.$ We have $20$ of each stamp, $5$ -cent (nickels), $10$ -cent (dimes), and $25$ -cent (quarters).
If we want the highest number of stamps, we must have the highest number of the smaller value stamps (like the coins above). We can use $20$ nickels and $20$ dimes to bring our total cost to $7.10 - 3.00 = 4.10$ . However, when we try to use quarters, the $25$ cents don’t fit evenly, so we have to give back $15$ cents to make the quarter amount $4.25$ . The most efficient way to do this is to give back a $10$ -cent (dime) stamp and a $5$ -cent (nickel) stamp to have $38$ stamps used so far. Now, we just use $\frac{425}{25} = 17$ quarters to get a grand total of $38 + 17 = \boxed{\textbf{(E)}\ 55}$ .
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, InterstellerApex, mahika99
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_14 | 55 |
AMC8_321 | The number $64$ has the property that it is divisible by its unit digit. How many whole numbers between 10 and 50 have this property?
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$
| Casework by the units digit $u$ will help organize the answer.
$u=0$ gives no solutions, since no real numbers are divisible by $0$
$u=1$ has $4$ solutions, since all numbers are divisible by $1$ .
$u=2$ has $4$ solutions, since every number ending in $2$ is even (ie divisible by $2$ ).
$u=3$ has $1$ solution: $33$ . $\pm 10$ or $\pm 20$ will retain the units digit, but will stop the number from being divisible by $3$ . $\pm 30$ is the smallest multiple of $10$ that will keep the number divisible by $3$ , but those numbers are $3$ and $63$ , which are out of the range of the problem.
$u=4$ has $2$ solutions: $24$ and $44$ . Adding or subtracting $10$ will kill divisibility by $4$ , since $10$ is not divisible by $4$ .
$u=5$ has $4$ solutions: every number ending in $5$ is divisible by $5$ .
$u=6$ has $1$ solution: $36$ . $\pm 10$ or $\pm 20$ will kill divisibility by $3$ , and thus kill divisibility by $6$ .
$u=7$ has no solutions. The first multiples of $7$ that end in $7$ are $7$ and $77$ , but both are outside of the range of this problem.
$u=8$ has $1$ solution: $48$ . $\pm 10, \pm 20, \pm 30$ will all kill divisibility by $8$ since $10, 20,$ and $30$ are not divisible by $8$ .
$u=9$ has no solutions. $9$ and $99$ are the smallest multiples of $9$ that end in $9$ .
Totalling the solutions, we have $0 + 4 + 4 + 1 + 2 + 4 + 1 + 0 + 1 + 0 = 17$ solutions, giving the answer $\boxed{C}$ , which is 17.
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_11 | C |
AMC8_322 | The figure below shows a polygon $ABCDEFGH$ , consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that $AH = EF = 8$ and $GH = 14$ . What is the volume of the prism?
[asy] usepackage("mathptmx"); size(275); defaultpen(linewidth(0.8)); real r = 2, s = 2.5, theta = 14; pair G = (0,0), F = (r,0), C = (r,s), B = (0,s), M = (C+F)/2, I = M + s/2 * dir(-theta); pair N = (B+G)/2, J = N + s/2 * dir(180+theta); pair E = F + r * dir(- 45 - theta/2), D = I+E-F; pair H = J + r * dir(135 + theta/2), A = B+H-J; draw(A--B--C--I--D--E--F--G--J--H--cycle^^rightanglemark(F,I,C)^^rightanglemark(G,J,B)); draw(J--B--G^^C--F--I,linetype ("4 4")); dot("$A$",A,N); dot("$B$",B,1.2*N); dot("$C$",C,N); dot("$D$",D,dir(0)); dot("$E$",E,S); dot("$F$",F,1.5*dir(-100)); dot("$G$",G,S); dot("$H$",H,W); dot("$I$",I,NE); dot("$J$",J,1.5*S); [/asy]
$\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288$
| While imagining the folding, $\overline{AB}$ goes on $\overline{BC},$ $\overline{AH}$ goes on $\overline{CI},$ and $\overline{EF}$ goes on $\overline{FG}.$ So, $BJ=CI=8$ and $FG=BC=8.$ Also, $\overline{HJ}$ becomes an edge parallel to $\overline{FG},$ so that means $HJ=8.$
Since $GH=14,$ then $JG=14-8=6.$ So, the area of $\triangle BJG$ is $\frac{8\cdot6}{2}=24.$ If we let $\triangle BJG$ be the base, then the height is $FG=8.$ So, the volume is $24\cdot8=\boxed{\textbf{(C)} ~192}.$
~aops-g5-gethsemanea2
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_24 | 192 |
AMC8_323 | A bag contains four pieces of paper, each labeled with one of the digits $1$ , $2$ , $3$ or $4$ , with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $3$ ?
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4}$
| The number of ways to form a 3-digit number is $4 \cdot 3 \cdot 2 = 24$ . The combination of digits that give us multiples of 3 are (1,2,3) and (2,3,4), as the integers in the subsets have a sum which is divisible by 3. The number of 3-digit numbers that contain these numbers is $3! + 3! = 12$ .
Therefore, the probability is $\frac{12}{24} = \boxed{\frac{1}{2}}$ .
~abc2142
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_24 | 1/2 |
AMC8_324 | Minh enters the numbers $1$ through $81$ into the cells of a $9 \times 9$ grid in some order. She calculates the product of the numbers in each row and column. What is the least number of rows and columns that could have a product divisible by $3$ ?
$\textbf{(A) } 8\qquad\textbf{(B) } 9\qquad\textbf{(C) } 10\qquad\textbf{(D) } 11\qquad\textbf{(E) } 12$
| Note you can swap/rotate any configuration of rows, such that all the rows and columns that have a product of 3 are in the top left. Hence the points are bounded by a $a \times b$ rectangle. This has $ab$ area and $a+b$ rows and columns divisible by $3$ . We want $ab\ge 27$ and $a+b$ minimized.
If $ab=27$ , we achieve minimum with $a+b=9+3=12$ .
If $ab=28$ ,our best is $a+b=7+4=11$ . Note if $a+b=10$ , $ab=25$ . Because $25<27$ , there is no smaller answer, and we get $\boxed{\textbf{(D)} 11}$ .
- SahanWijetunga
~vockey(minor edits)
~phy6(minor edits)
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_16 | 11 |
AMC8_325 | At Central Middle School the $108$ students who take the AMC8 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 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.
They learn that a big concert is scheduled for the same night and attendance will be down $25\%$ . How many recipes of cookies should they make for their smaller party?
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$
| Solution 1If $108$ students eat $2$ cookies on average, there will need to be $108\cdot 2 = 216$ cookies. But with the smaller attendance, you will only need $100\% - 25\% = 75\%$ of these cookies, or $75\% \cdot 216 = 0.75\cdot 216 = 162$ cookies.
$162$ cookies requires $\frac{162}{15} = 10.8$ batches. However, since half-batches are forbidden, we must round up to get $\left\lceil \frac{162}{15} \right\rceil = 11$ batches, and the correct answer is $\boxed{E}$ .
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_18 | E |
AMC8_327 | In the figure below, choose point $D$ on $\overline{BC}$ so that $\triangle ACD$ and $\triangle ABD$ have equal perimeters. What is the area of $\triangle ABD$ ?
[asy]draw((0,0)--(4,0)--(0,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,0), ESE); label("$C$", (0, 3), N); label("$3$", (0, 1.5), W); label("$4$", (2, 0), S); label("$5$", (2, 1.5), NE);[/asy]
$\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }\frac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\frac{12}{5}\qquad\textbf{(E) }\frac{5}{2}$
| We know that the perimeters of the two small triangles are $3+CD+AD$ and $4+BD+AD$ . Setting both equal and using $BD+CD = 5$ , we have $BD = 2$ and $CD = 3$ . Now, we simply have to find the area of $\triangle ABD$ . Since $\frac{BD}{CD} = \frac{2}{3}$ , we must have $\frac{[ABD]}{[ACD]} = 2/3$ . Combining this with the fact that $[ABC] = [ABD] + [ACD] = \frac{3\cdot4}{2} = 6$ , we get $[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} \cdot 6 = \boxed{\textbf{(D) } \frac{12}{5}}$ .
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_16 | 12/5 |
AMC8_328 | Of the 36 students in Richelle's class, 12 prefer chocolate pie, 8 prefer apple, and 6 prefer blueberry. Half of the remaining students prefer cherry pie and half prefer lemon. For Richelle's pie graph showing this data, how many degrees should she use for cherry pie?
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 50 \qquad \text{(E)}\ 72$
| There are $36$ students in the class: $12$ prefer chocolate pie, $8$ prefer apple pie, and $6$ prefer blueberry pie. Therefore, $36-12-8-6=10$ students prefer cherry pie or lemon pie. Half of these prefer each, so $5$ students prefer cherry pie. This means that $\frac{5}{36}$ of the students prefer cherry pie, so $\frac{5}{36}$ of the full $360^\circ$ should be used for cherry pie. This is $(\frac{5}{36})(360^\circ)=50^\circ, \boxed{\text{D}}$
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_13 | D |
AMC8_329 | Isabella uses one-foot cubical blocks to build a rectangular fort that is $12$ feet long, $10$ feet wide, and $5$ feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?
[asy]import three; currentprojection=orthographic(-8,15,15); triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; A = (0,0,0); B = (0,10,0); C = (12,10,0); D = (12,0,0); E = (0,0,5); F = (0,10,5); G = (12,10,5); H = (12,0,5); I = (1,1,1); J = (1,9,1); K = (11,9,1); L = (11,1,1); M = (1,1,5); N = (1,9,5); O = (11,9,5); P = (11,1,5); //outside box far draw(surface(A--B--C--D--cycle),white,nolight); draw(A--B--C--D--cycle); draw(surface(E--A--D--H--cycle),white,nolight); draw(E--A--D--H--cycle); draw(surface(D--C--G--H--cycle),white,nolight); draw(D--C--G--H--cycle); //inside box far draw(surface(I--J--K--L--cycle),white,nolight); draw(I--J--K--L--cycle); draw(surface(I--L--P--M--cycle),white,nolight); draw(I--L--P--M--cycle); draw(surface(L--K--O--P--cycle),white,nolight); draw(L--K--O--P--cycle); //inside box near draw(surface(I--J--N--M--cycle),white,nolight); draw(I--J--N--M--cycle); draw(surface(J--K--O--N--cycle),white,nolight); draw(J--K--O--N--cycle); //outside box near draw(surface(A--B--F--E--cycle),white,nolight); draw(A--B--F--E--cycle); draw(surface(B--C--G--F--cycle),white,nolight); draw(B--C--G--F--cycle); //top draw(surface(E--H--P--M--cycle),white,nolight); draw(surface(E--M--N--F--cycle),white,nolight); draw(surface(F--N--O--G--cycle),white,nolight); draw(surface(O--G--H--P--cycle),white,nolight); draw(M--N--O--P--cycle); draw(E--F--G--H--cycle); label("10",(A--B),SE); label("12",(C--B),SW); label("5",(F--B),W);[/asy]
$\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600$
| There are $10 \cdot 12 = 120$ cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are $9 + 11 + 9 + 11 = 40$ cubes. Hence, the answer is $120 + 4 \cdot 40 = \boxed{\textbf{(B)}\ 280}$ .
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_18 | 280 |
AMC8_330 | In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 2, and $m\angle TUS = 60^\circ$ . Arcs $\overarc{TR}$ and $\overarc{SR}$ are each one-sixth of a circle with radius 2. What is the area of the region shown?
[asy]draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("$U$", (2,3.464), N); label("$S$", (1,1.732), W); label("$T$", (3,1.732), E); label("$R$", (2,0), S);[/asy]
$\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$
| [asy]draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("$U$", (2,3.464), N); label("$S$", (1,1.732), W); label("$T$", (3,1.732), E); label("$R$", (2,0), S);[/asy]
In addition to the given diagram, we can draw lines $\overline{SR}$ and $\overline{RT}.$ The area of rhombus $SRTU$ is half the product of its diagonals, which is $\frac{2\sqrt3 \cdot 2}{2}=2\sqrt3$ . However, we have to subtract off the circular segments. The area of those can be found by computing the area of the circle with radius 2, multiplying it by $\frac{1}{6}$ , then finally subtracting the area of an equilateral triangle with a side length 2 from the sector. The sum of the areas of the circular segments is $2(\frac{4 \pi}{6}-\sqrt3).$ The area of rhombus $SRTU$ minus the circular segments is $2\sqrt3-\frac{4 \pi}{3}+2\sqrt3= \boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.$
~PEKKA
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_25 | 4√3 - 4π/3 |
AMC8_331 | What is the area of the shaded pinwheel shown in the $5 \times 5$ grid?
[asy] filldraw((2.5,2.5)--(0,1)--(1,1)--(1,0)--(2.5,2.5)--(4,0)--(4,1)--(5,1)--(2.5,2.5)--(5,4)--(4,4)--(4,5)--(2.5,2.5)--(1,5)--(1,4)--(0,4)--cycle, gray, black); int i; for(i=0; i<6; i=i+1) { draw((i,0)--(i,5)); draw((0,i)--(5,i)); } [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)}\: 6\qquad\textbf{(C)}\: 8\qquad\textbf{(D)}\: 10\qquad\textbf{(E)}\: 12$
| The area of the square around the pinwheel is 25. The area of the pinwheel is equal to $\text{the square } - \text{ the white space.}$ Each of the four triangles have a base of 3 units and a height of 2.5 units, and so their combined area is 15 units squared. Then the unshaded space consists of the four triangles with total area of 15, and there are four white corner squares. Therefore the area of the pinwheel is $25-(15+4)$ which is $\boxed{\textbf{(B) 6}}$
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_23 | 6 |
AMC8_332 | In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12$
| $6$ are blue and green - $b+g=6$
$8$ are red and blue - $r+b=8$
$4$ are red and green - $r+g=4$
We can do trial and error. Let's make blue $5$ . That makes green $1$ and red $3$ because $6-5=1$ and $8-5=3$ . To check this, let's plug $1$ and $3$ into $r+g=4$ , which works. Now count the number of marbles - $5+3+1=9$ . So the answer is $\boxed{\textbf{(C)}\ 9}.$
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_19 | 9 |
AMC8_333 | Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is
[asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4,4)--(0,4)); draw((1,5)--(1,1)--(5,1)); label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,NW); label("$1$",(1,4.5),E); label("$1$",(0.5,5),N); label("$3$",(1,2.5),E); label("$3$",(2.5,1),N); label("$1$",(4,0.5),E); label("$1$",(4.5,1),N); [/asy]
$\text{(A)}\ 7 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$
| The side of the large square is $1 + 3 + 1 = 5$ , so the area of the large square is $5^2 = 25$ .
The area of the middle square is $3^2$ , and the sum of the areas of the two smaller squares is $2 * 1^2 = 2$ .
Thus, the big square minus the three smaller squares is $25 - 9 - 2 = 14$ . This is the area of the two congruent L-shaped regions.
So the area of one L-shaped region is $\frac{14}{2} = 7$ , and the answer is $\boxed{A}$
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_6 | A |
AMC8_334 | Problems 8, 9 and 10 use the data found in the accompanying paragraph and figures
Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.
$\circ$ Art's cookies are trapezoids:
[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(5,0)--(5,3)--(2,3)--cycle); draw(rightanglemark((5,3), (5,0), origin)); label("5 in", (2.5,0), S); label("3 in", (5,1.5), E); label("3 in", (3.5,3), N);[/asy]
$\circ$ Roger's cookies are rectangles:
[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(4,0)--(4,2)--(0,2)--cycle); draw(rightanglemark((4,2), (4,0), origin)); draw(rightanglemark((0,2), origin, (4,0))); label("4 in", (2,0), S); label("2 in", (4,1), E);[/asy]
$\circ$ Paul's cookies are parallelograms:
[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle); draw((2.5,2)--(2.5,0), dashed); draw(rightanglemark((2.5,2),(2.5,0), origin)); label("3 in", (1.5,0), S); label("2 in", (2.5,1), W);[/asy]
$\circ$ Trisha's cookies are triangles:
[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(3,4)--cycle); draw(rightanglemark((3,4),(3,0), origin)); label("3 in", (1.5,0), S); label("4 in", (3,2), E);[/asy]
Each friend uses the same amount of dough, and Art makes exactly $12$ cookies. Art's cookies sell for $60$ cents each. To earn the same amount from a single batch, how much should one of Roger's cookies cost in cents?
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 75\qquad\textbf{(E)}\ 90$
| The area of one of Art's cookies is $3 \cdot 3 + \frac{2 \cdot 3}{2}=9+3=12$ . As he has $12$ cookies in a batch, the amount of dough each person used is $12 \cdot 12=144$ . Roger's cookies have an area of $\frac{144}{2 \cdot 4}=\frac{144}{8}= 18$ cookies in a batch. In total, the amount of money Art will earn is $12 \cdot 60=720$ . Thus, the amount Roger would need to charge per cookie is $\frac{720}{18}=\boxed{\textbf{(C)}\ 40}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_9 | 40 |
AMC8_335 | What time was it $2011$ minutes after midnight on January 1, 2011?
$\textbf{(A)}\ \text{January 1 at 9:31PM}$
$\textbf{(B)}\ \text{January 1 at 11:51PM}$
$\textbf{(C)}\ \text{January 2 at 3:11AM}$
$\textbf{(D)}\ \text{January 2 at 9:31AM}$
$\textbf{(E)}\ \text{January 2 at 6:01PM}$
| There are $60$ minutes in an hour. $2011/60=33\text{r}31,$ or $33$ hours and $31$ minutes. There are $24$ hours in a day, so the time is $9$ hours and $31$ minutes after midnight on January 2, 2011. $\Rightarrow\boxed{\textbf{(D)}\text{ January 2 at 9:31AM}}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_5 | January 2 at 9:31AM |
AMC8_336 | Mindy made three purchases for $\textdollar 1.98$ dollars, $\textdollar 5.04$ dollars, and $\textdollar 9.89$ dollars. What was her total, to the nearest dollar?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$
| The three prices round to $\textdollar 2$ , $\textdollar 5$ , and $\textdollar 10$ , which have a sum of $\boxed{\textbf{(D)}\ 17}$ .
We know that there will not be a rounding error, as the total amount rounded is clearly less than $\textdollar 0.50$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_1 | 17 |
AMC8_337 | A $\textit{tetromino}$ consists of four squares connected along their edges. There are five possible tetromino shapes, $I$ , $O$ , $L$ , $T$ , and $S$ , shown below, which can be rotated or flipped over. Three tetrominoes are used to completely cover a $3\times4$ rectangle. At least one of the tiles is an $S$ tile. What are the other two tiles?
[asy] unitsize(12); add(grid(1,4)); label("I", (0.5,-1)); add(shift((5,0)) * grid(2,2)); label("O", (6,-1)); add(shift((11,0)) * grid(1,3)); add(shift((11,0)) * grid(2,1)); label("L", (12,-1)); add(shift((18,0)) * grid(1,1)); add(shift((17,1)) * grid(3,1)); label("T", (18.5,-1)); add(shift((25,1)) * grid(2,1)); add(shift((24,0)) * grid(2,1)); label("S", (25.5,-1)); add(shift((12,-6)) * grid(4,3)); [/asy]
$\textbf{(A)}I$ and $L\qquad \textbf{(B)} I$ and $T\qquad \textbf{(C)} L$ and $L\qquad \textbf{(D)}L$ and $S\qquad \textbf{(E)}O$ and $T$
| The $3\times4$ rectangle allows for $7$ possible places to put the S piece, with each possible placement having an inverted version. One of the cases looks like this:
[asy] path x = (0,0)--(0,2)--(1,2)--(1,3)--(2,3)--(2,1)--(1,1)--(1,0)--cycle; fill(x, rgb(0,30,0)); add(grid(4,3)); [/asy]
As you can see, there is a hole in the top left corner of the board, which would be impossible to fill using the tetrominos. There are three cases in which a hole isn't created; the S lies flat in the bottom left corner, it lies flat in the top right corner, or it stands upright in the center. All three tilings are shown below.
[asy] path z1 = (2,0)--(3,0)--(3,2)--(2,2)--(2,3)--(1,3)--(1,1)--(2,1)--cycle; path z2 = (0,0)--(2,0)--(2,1)--(1,1)--(1,3)--(0,3)--cycle; path z3 = (2,3)--(4,3)--(4,0)--(3,0)--(3,2)--(2,2)--cycle; fill(z1, rgb(0,30,0)); fill(z2, rgb(127,80,0)); fill(z3, rgb(127,100,0)); add(grid(4,3)); [/asy]
[asy] path y1 = (0,0)--(2,0)--(2,1)--(3,1)--(3,2)--(1,2)--(1,1)--(0,1)--cycle; path y2 = (0,1)--(1,1)--(1,2)--(3,2)--(3,3)--(0,3)--cycle; path y3 = (2,0)--(4,0)--(4,3)--(3,3)--(3,1)--(2,1)--cycle; fill(y1, rgb(0,30,0)); fill(y2, rgb(127,80,0)); fill(y3, rgb(127,100,0)); add(grid(4,3)); [/asy]
[asy] path w1 = (1,1)--(3,1)--(3,2)--(4,2)--(4,3)--(2,3)--(2,2)--(1,2)--cycle; path w2 = (0,0)--(0,3)--(2,3)--(2,2)--(1,2)--(1,0)--cycle; path w3 = (1,0)--(4,0)--(4,2)--(3,2)--(3,1)--(1,1)--cycle; fill(w1, rgb(0,30,0)); fill(w2, rgb(127,80,0)); fill(w3, rgb(127,100,0)); add(grid(4,3)); [/asy]
For each of the inverted cases, the L pieces can be inverted along with the S piece. Because the only cases that fill the rectangle after the S is placed are the ones that use two L pieces, the answer must be $\boxed{\textbf{(C)}~L \ and \ L}$ .
~bubby617
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_11 | L and L |
AMC8_338 | Three $\text{A's}$ , three $\text{B's}$ , and three $\text{C's}$ are placed in the nine spaces so that each row and column contains one of each letter. If $\text{A}$ is placed in the upper left corner, how many arrangements are possible?
[asy] size((80)); draw((0,0)--(9,0)--(9,9)--(0,9)--(0,0)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); draw((0,3)--(9,3)); draw((0,6)--(9,6)); label("A", (1.5,7.5)); [/asy]
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$
| There are $2$ ways to place the remaining $\text{As}$ , $2$ ways to place the remaining $\text{Bs}$ , and $1$ way to place the remaining $\text{Cs}$ for a total of $(2)(2)(1) = \boxed{\textbf{(C)}\ 4}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_14 | 4 |
AMC8_339 | Elisa swims laps in the pool. When she first started, she completed 10 laps in 25 minutes. Now, she can finish 12 laps in 24 minutes. By how many minutes has she improved her lap time?
$\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{3}{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$
| When Elisa started, she finished a lap in $\frac{25}{10}=2.5$ minutes. Now, she finishes a lap is $\frac{24}{12}= 2$ minutes. The difference is $2.5-2=\boxed{\textbf{(A)}\ \frac{1}{2}}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_3 | 1/2 |
AMC8_340 | The base of isosceles $\triangle ABC$ is $24$ and its area is $60$ . What is the length of one
of the congruent sides?
$\mathrm{(A)}\ 5 \qquad \mathrm{(B)}\ 8 \qquad \mathrm{(C)}\ 13 \qquad \mathrm{(D)}\ 14 \qquad \mathrm{(E)}\ 18$
| The area of a triangle is shown by $\frac{1}{2}bh$ . We set the base equal to $24$ , and the area equal to $60$ , and we get the triangle's height, or altitude, to be $5$ . In this isosceles triangle, the height bisects the base, so by using the Pythagorean Theorem, $a^2+b^2=c^2$ , we can solve for one of the legs of the triangle (it will be the hypotenuse, $c$ ).
$a = 12$ , $b = 5$ ,
$c = 13$ .
The answer is $\boxed{\textbf{(C)}\ 13}$ .
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_14 | 13 |
AMC8_341 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$ ?
[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label("$A$",A,N); dot(B); label("$B$", B,SW);dot(C); label("$C$",C,SE); dot(DD); label("$D$",DD,NE); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$
| We use the line-segment ratios to infer area ratios and height ratios.
Areas:
$AD:DC = 1:2 \implies AD:AC = 1:3 \implies [ABD] =\frac{[ABC]}{3} = 120$ .
$BE:BD = 1:2 \text{ (midpoint)} \implies [ABE] = \frac{[ABD]}{2} = \frac{120}{2} = 60$ .
Heights:
Let $h_A$ = height (of altitude) from $\overline{BC}$ to $A$ .
$AD:DC = 1:2 \implies CD:CA = 2:3 \implies \text{height } h_D$ from $\overline{BC}$ to $D$ is $\frac{2}{3}h_A$ .
$BE:BD = 1:2 \text{ (midpoint)} \implies \text{height } h_E$ from $\overline{BC}$ to $E$ is $\frac{1}{2} h_D = \frac{1}{2}(\frac{2}{3} h_A) = \frac{1}{3} h_A$ .
Conclusion:
$\frac{[EBF]} {[ABF]} = \frac{[EBF]} {[EBF] + [ABE]} = \frac{[EBF]} {[EBF]+60}$ , and also $\frac{[EBF]} {[ABF]} = \frac{h_E}{h_A} = \frac{1}{3}$ .
So, $\frac{[EBF]} {[EBF] + 60} = \frac{1}{3}$ , and thus, $[EBF] = \boxed{\textbf{(B) }30}$
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | 30 |
AMC8_342 | The numbers $-2, 4, 6, 9$ and $12$ are rearranged according to these rules:
| From rule 1, the largest number, $12$ , can be second or third. From rule 2, because there are five places, the smallest number $-2$ can either be third or fourth. The median, $6$ can be second, third, or fourth. Because we know the middle three numbers, the first and last numbers are $4$ and $9$ , disregarding their order. Their average is $(4+9)/2 = \boxed{\textbf{(C)}\ 6.5}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_11 | 6.5 |
AMC8_343 | Aaliyah rolls two standard 6-sided dice. She notices that the product of the two numbers rolled is a multiple of $6$ . Which of the following integers cannot be the sum of the two numbers?
$\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9$
| First, figure out all pairs of numbers whose product is 6. Then, using the process of elimination, we can find the following:
$\textbf{(A)}$ is possible: $2\times 3$
$\textbf{(C)}$ is possible: $1\times 6$
$\textbf{(D)}$ is possible: $2\times 6$
The only integer that cannot be the sum is $\boxed{\textbf{(B) } 6}$ .
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_5 | 6 |
AMC8_344 | Consider these two operations:
\begin{align*} a \, \blacklozenge \, b &= a^2 - b^2\\ a \, \bigstar \, b &= (a - b)^2 \end{align*}
What is the output of $(5 \, \blacklozenge \, 3) \, \bigstar \, 6?$
$\textbf{(A) } {-}20 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 100 \qquad \textbf{(E) } 220$
| We can find a general solution to any $((a \, \blacklozenge \, b) \, \bigstar \, c)$ .
\[((a \, \blacklozenge \, b) \, \bigstar \, c)\]
\[=((a^2-b^2) \, \bigstar \, c)\]
\[=(a^2-b^2-c)^2\]
\[=a^4+b^4-(a^2)(b^2)-2(a^2)(c)-(b^2)(a^2)+2(b^2)(c)+c^2\]
\[=5^4+3^4-(5^2)(3^2)-2(5^2)(6)-(3^2)(5^2)+2(3^2)(6)+6^2\]
\[=625+81-225-300-225+108+36\]
\[=\boxed{\textbf{(D) } 100}\]
~megaboy6679
~algebraic_algorithmic "This is very time wasting"
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_2 | 100 |
AMC8_345 | Sergai skated around an ice rink, gliding along different paths. The gray lines in the figures below show four of the paths labeled P, Q, R, and S. What is the sorted order of the four paths from shortest to longest?
$\textbf{(A)}\ P,Q,R,S \qquad \textbf{(B)}\ P,R,S,Q \qquad \textbf{(C)}\ Q,S,P,R \qquad \textbf{(D)}\ R,P,S,Q \qquad \textbf{(E)}\ R,S,P,Q$
| You can measure the lengths of the paths until you find a couple of guaranteed true inferred statements as such:
$Q$ is greater than $S$ ,
$P$ is greater than $R$ ,
and $R$ and $P$ are the smallest two, therefore the order is $R, P, S, Q.$
Thus we get the answer $\boxed{\textbf{(D)}~R, P, S, Q}$ .
- U-King
~TabHawaii (minor formatting edits)
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_6 | R, P, S, Q |
AMC8_346 | The length of a rectangle is increased by $10\%$ percent and the width is decreased by $10\%$ percent. What percent of the old area is the new area?
$\textbf{(A)}\ 90 \qquad \textbf{(B)}\ 99 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 101 \qquad \textbf{(E)}\ 110$
| In a rectangle with dimensions $10 \times 10$ , the new rectangle would have dimensions $11 \times 9$ . The ratio of the new area to the old area is $99/100 = \boxed{\textbf{(B)}\ 99}$ .
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_8 | 99 |
AMC8_347 | Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?
[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1)); draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5)); label("$A$",(-1,1)); label("$B$",(2,1)); label("$1$",(-.4,.4)); label("$2$",(.4,.4)); label("$3$",(.4,-.4)); label("$4$",(-.4,-.4)); label("$1$",(2.6,.4)); label("$2$",(3.4,.4)); label("$3$",(3,-.5)); [/asy]
$\textbf{(A)}\ \frac14\qquad \textbf{(B)}\ \frac13\qquad \textbf{(C)}\ \frac12\qquad \textbf{(D)}\ \frac23\qquad \textbf{(E)}\ \frac34$
| An even number comes from multiplying an even and even, even and odd, or odd and even. Since an odd number only comes from multiplying an odd and odd, there are less cases and it would be easier to find the probability of spinning two odd numbers from $1$ . Multiply the independent probabilities of each spinner getting an odd number together and subtract it from $1$ .
\[1-\frac24 \cdot \frac23 = 1- \frac13 = \boxed{\textbf{(D)}\ \frac23}\]
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_21 | 2/3 |
AMC8_348 | Semicircles $POQ$ and $ROS$ pass through the center $O$ . What is the ratio of the combined areas of the two semicircles to the area of circle $O$ ?
[asy] import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,-1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]
$\textbf{(A)}\ \frac{\sqrt 2}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{\pi}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{\sqrt 2}{2}$
| By the Pythagorean Theorem, the radius of the larger circle turns out to be $\sqrt{1^2 + 1^2} = \sqrt{2}$ . Therefore, the area of the larger circle is $(\sqrt{2})^2\pi = 2\pi$ . Using the coordinate plane given, we find that the radius of each of the two semicircles to be 1. So, the area of the two semicircles is $1^2\pi=\pi$ . Finally, the ratio of the combined areas of the two semicircles to the area of circle $O$ is $\boxed{\textbf{(B)}\ \frac{1}{2}}$ .
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_23 | 1/2 |
AMC8_349 | If $a\otimes b = \dfrac{a + b}{a - b}$ , then $(6\otimes 4)\otimes 3 =$
$\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72$
| $6\otimes4=\frac{6+4}{6-4}=\frac{10}{2}=5$ .
$5\otimes3=\frac{5+3}{5-3}=\frac{8}{2}=\boxed{\textbf{(A)}\ 4}$
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_12 | 4 |
AMC8_350 | Jefferson Middle School has the same number of boys and girls. $\frac{3}{4}$ of the girls and $\frac{2}{3}$
of the boys went on a field trip. What fraction of the students on the field trip were girls?
$\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{9}{17}\qquad\textbf{(C) }\frac{7}{13}\qquad\textbf{(D) }\frac{2}{3}\qquad \textbf{(E) }\frac{14}{15}$
| Let there be $b$ boys and $g$ girls in the school. We see $g=b$ , which means $\frac{3}{4}b+\frac{2}{3}b=\frac{17}{12}b$ kids went on the trip and $\frac{3}{4}b$ kids are girls. So, the answer is $\frac{\frac{3}{4}b}{\frac{17}{12}b}=\frac{9}{17}$ , which is $\boxed{\textbf{(B)} \frac{9}{17}}$ .
~CHECKMATE2021
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_12 | $\frac{9}{17}$ |
AMC8_351 | The two circles pictured have the same center $C$ . Chord $\overline{AD}$ is tangent to the inner circle at $B$ , $AC$ is $10$ , and chord $\overline{AD}$ has length $16$ . What is the area between the two circles?
[asy] unitsize(45); 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((2,0.15)--(1.85,0.15)--(1.85,0)--(2,0)--cycle); draw(circle((2,1),2.24)); draw(circle((2,1),1)); draw((0,0)--(4,0)); draw((0,0)--(2,1)); draw((2,1)--(2,0)); draw((2,1)--(4,0)); dot((0,0),ds); label("$A$", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label("$B$", (1.97,-0.31),NE*lsf); dot((2,1),ds); label("$C$", (1.96,1.09),NE*lsf); dot((4,0),ds); label("$D$", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle); [/asy]
$\textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi$
| Since $\triangle ACD$ is isosceles, $CB$ bisects $AD$ . Thus $AB=BD=8$ . From the Pythagorean Theorem, $CB=6$ . Thus the area between the two circles is
$100\pi - 36\pi=64\pi$ $\boxed{\textbf{(C)}\ 64\pi}$
Note: The length $AC$ is necessary information, as this tells us the radius of the larger circle. The area of the annulus is $\pi(AC^2-BC^2)=\pi AB^2=64\pi$ .
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_19 | 64π |
AMC8_352 | Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$
| [asy] unitsize(8mm); for(int i=0; i<=8; ++i) { draw((0,i)--(8,i)); draw((i,0)--(i,8)); } fill((2,1)--(6,1)--(6,6)--(2,6)--cycle); for(int j=0; j<= 38; ++j) { draw((0,0)--(2,0)--(2,1)--(0,1)--(0,0), black+linewidth(3)); draw((6,6)--(6,8)--(8,8)--(8,6)--(6,6), black+linewidth(3)); }[/asy]
Using combinations , we get that the number of ways to get from Samantha's house to City Park is $\binom31 = \dfrac{3!}{1!2!} = 3$ , and the number of ways to get from City Park to school is $\binom42= \dfrac{4!}{2!2!} = \dfrac{4\cdot 3}{2} = 6$ . Since there's one way to go through City Park (just walking straight through), the number of different ways to go from Samantha's house to City Park to school $3\cdot 6 = \boxed{\textbf{(E)}\ 18}$ .
~Note by Theraccoon: The person who posted this did not include their name.
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_21 | 18 |
AMC8_353 | Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$ ?
[asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,N); label("$E$",E,S); label("$5$",A/2,W); label("$6$",(A+D)/2,N); [/asy]
$\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$
| The area of $\bigtriangleup CDE$ is $\frac{DC\cdot CE}{2}$ . The area of $ABCD$ is $AB\cdot AD=5\cdot 6=30$ , which also must be equal to the area of $\bigtriangleup CDE$ , which, since $DC=5$ , must in turn equal $\frac{5\cdot CE}{2}$ . Through transitivity, then, $\frac{5\cdot CE}{2}=30$ , and $CE=12$ . Then, using the Pythagorean Theorem, you should be able to figure out that $\bigtriangleup CDE$ is a $5-12-13$ triangle, so $DE=\boxed{13}$ , or $\boxed{(B)}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_14 | 13 |
AMC8_354 | Toothpicks are used to make a grid that is $60$ toothpicks long and $32$ toothpicks wide. How many toothpicks are used altogether?
[asy] picture corner; draw(corner,(5,0)--(35,0)); draw(corner,(0,-5)--(0,-35)); for (int i=0; i<3; ++i){for (int j=0; j>-2; --j){if ((i-j)<3){add(corner,(50i,50j));}}} draw((5,-100)--(45,-100)); draw((155,0)--(185,0),dotted+linewidth(2)); draw((105,-50)--(135,-50),dotted+linewidth(2)); draw((100,-55)--(100,-85),dotted+linewidth(2)); draw((55,-100)--(85,-100),dotted+linewidth(2)); draw((50,-105)--(50,-135),dotted+linewidth(2)); draw((0,-105)--(0,-135),dotted+linewidth(2));[/asy]
$\textbf{(A)}\ 1920 \qquad \textbf{(B)}\ 1952 \qquad \textbf{(C)}\ 1980 \qquad \textbf{(D)}\ 2013 \qquad \textbf{(E)}\ 3932$
| There are $61$ vertical columns with a length of $32$ toothpicks, and there are $33$ horizontal rows with a length of $60$ toothpicks, because $32$ and $60$ are the number of intervals. You can verify this by trying a smaller case, i.e. a $3 \times 4$ grid of toothpicks, with $3 \times 3$ and $2 \times 4$ .
Thus, our answer is $61\cdot 32 + 33 \cdot 60 = \boxed{\textbf{(E)}\ 3932}$ .
~Note by Theraccoon: The person who posted this answer did not include their name. Minor edit by ~NXC
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_22 | 3932 |
AMC8_355 | The faces of a cube are painted in six different colors: red $(R)$ , white $(W)$ , green $(G)$ , brown $(B)$ , aqua $(A)$ , and purple $(P)$ . Three views of the cube are shown below. What is the color of the face opposite the aqua face?
[asy] unitsize(2cm); pair x, y, z, trans; int i; x = dir(-5); y = (0.6,0.5); z = (0,1); trans = (2,0); for (i = 0; i <= 2; ++i) { draw(shift(i*trans)*((0,0)--x--(x + y)--(x + y + z)--(y + z)--z--cycle)); draw(shift(i*trans)*((x + z)--x)); draw(shift(i*trans)*((x + z)--(x + y + z))); draw(shift(i*trans)*((x + z)--z)); } label(rotate(-3)*"$R$", (x + z)/2); label(rotate(-5)*slant(0.5)*"$B$", ((x + z) + (y + z))/2); label(rotate(35)*slant(0.5)*"$G$", ((x + z) + (x + y))/2); label(rotate(-3)*"$W$", (x + z)/2 + trans); label(rotate(50)*slant(-1)*"$B$", ((x + z) + (y + z))/2 + trans); label(rotate(35)*slant(0.5)*"$R$", ((x + z) + (x + y))/2 + trans); label(rotate(-3)*"$P$", (x + z)/2 + 2*trans); label(rotate(-5)*slant(0.5)*"$R$", ((x + z) + (y + z))/2 + 2*trans); label(rotate(-85)*slant(-1)*"$G$", ((x + z) + (x + y))/2 + 2*trans); [/asy]
$\textbf{(A)} \text{ red}\qquad\textbf{(B)} \text{ white}\qquad\textbf{(C)} \text{ green}\qquad\textbf{(D)} \text{ brown}\qquad\textbf{(E)} \text{ purple}$
| $B$ is on the top, and $R$ is on the side, and $G$ is on the right side. That means that (image $2$ ) $W$ is on the left side. From the third image, you know that $P$ must be on the bottom since $G$ is sideways. That leaves us with the back, so the back must be $A$ . The front is opposite of the back, so the answer is $\boxed{\textbf{(A)}\ R}$ .
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_12 | R |
AMC8_356 | If February is a month that contains Friday the $13^{\text{th}}$ , what day of the week is February 1?
$\textbf{(A)}\ \text{Sunday} \qquad \textbf{(B)}\ \text{Monday} \qquad \textbf{(C)}\ \text{Wednesday} \qquad \textbf{(D)}\ \text{Thursday}\qquad \textbf{(E)}\ \text{Saturday}$
| We can go backwards by days, but we can also backwards by weeks. If we go backwards by weeks, we see that February 6 is a Friday. If we now go backwards by days, February 1 is a $\boxed{\textbf{(A)}\ \text{Sunday}}$
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_3 | Sunday |
AMC8_357 | The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is
[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW);[/asy]
$\text{(A)}\ 21\qquad\text{(B)}\ 27\qquad\text{(C)}\ 30\qquad\text{(D)}\ 36\qquad\text{(E)}\ 40$
| To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that $ABCD$ can have any dimension. Give the rectangle dimensions of $AB = CD = 12$ and $BC = AD= 6$ , which is the easiest way to avoid fractions. Labelling the right midpoint as $M$ , and the bottom midpoint as $N$ , we know that $DN = NC = 6$ , and $BM = MC = 3$ .
$[\triangle ADN] = \frac{1}{2}\cdot 6\cdot 6 = 18$
$[\triangle MNC] = \frac{1}{2}\cdot 3\cdot 6 = 9$
$[\triangle ABM] = \frac{1}{2}\cdot 12\cdot 3 = 18$
$[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]$
$[\triangle AMN] = 72 - 18 - 9 - 18$
$[\triangle AMN] = 27$ , and the answer is $\boxed{B}$
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_25 | B |
AMC8_358 | In a room, $2/5$ of the people are wearing gloves, and $3/4$ of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$
| Let $x$ be the number of people wearing both a hat and a glove. Since the number of people wearing a hat or a glove must be whole numbers, the number of people in the room must be a multiple of 4 and 5. Since we are trying to find the minimum $x$ , we must use the least common multiple. $lcm(4,5) = 20$ . Thus, we can say that there are $20$ people in the room, all of which are wearing at least a hat or a glove. (Any people wearing neither item would unnecessarily increase the number of people in the room.)
It follows that there are $\frac{2}{5}\cdot 20 = 8$ people wearing gloves and $\frac{3}{4}\cdot 20 = 15$ people wearing hats. Then by applying the Principle of Inclusion and Exclusion (PIE), the total number of people in the room wearing either a hat or a glove or both is $8+15-x = 23-x$ , where $x$ is the number wearing both. Since everyone in the room is wearing at least one item (see above), $23-x = 20$ , and so $x=\boxed{\textbf{(A)}\ 3}$ .
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_20 | 3 |
AMC8_359 | The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4$ . In feet, how tall is the taller tree?
$\textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112$
| Let the height of the taller tree be $h$ and let the height of the smaller tree be $h-16$ . Since the ratio of the smaller tree to the larger tree is $\frac{3}{4}$ , we have $\frac{h-16}{h}=\frac{3}{4}$ . Solving for $h$ gives us $h=64 \Rightarrow \boxed{\textbf{(B)}\ 64}$
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_11 | 64 |
AMC8_360 | On the most recent exam on Prof. Xochi's class,
$5$ students earned a score of at least $95\%$ ,
$13$ students earned a score of at least $90\%$ ,
$27$ students earned a score of at least $85\%$ ,
$50$ students earned a score of at least $80\%$ ,
How many students earned a score of at least $80\%$ and less than $90\%$ ?
$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 22\qquad \textbf{(D)}\ 37\qquad \textbf{(E)}\ 45$
| $50$ people scored at least $80\%$ , and out of these $50$ people, $13$ of them earned a score that was not less than $90\%$ , so the number of people that scored in between at least $80\%$ and less than $90\%$ is $50-13 = \boxed{\text{(D)\ 37}}$ .
~Soupboy0
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_7 | 37 |
AMC8_361 | Given the areas of the three squares in the figure, what is the area of the interior triangle?
[asy] draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1)); label("$25$",(14.5,1),N); label("$144$",(6,-7.5),N); label("$169$",(3.5,7),N); [/asy]
$\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800$
| The sides of the squares are $5, 12$ and $13$ for the square with area $25, 144$ and $169$ , respectively. The legs of the interior triangle are $5$ and $12$ , so the area is $\frac{5 \times 12}{2}=\boxed{\mathrm{(B)}\ 30}$
(note: the ab/2 area method only works because the converse Pythagorean theorem holds for the triple [5, 12, 13]. Therefore, we can find the solution this way because we know the triangle is right ~megaboy6679)
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_6 | 30 |
AMC8_362 | Jean has made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is $8$ feet high while the other peak is $12$ feet high. Each peak forms a $90^\circ$ angle, and the straight sides form a $45^\circ$ angle with the ground. The artwork has an area of $183$ square feet. The sides of the mountain meet at an intersection point near the center of the artwork, $h$ feet above the ground. What is the value of $h?$
[asy] unitsize(.3cm); filldraw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,gray(0.7),linewidth(1)); draw((-1,0)--(-1,8),linewidth(.75)); draw((-1.4,0)--(-.6,0),linewidth(.75)); draw((-1.4,8)--(-.6,8),linewidth(.75)); label("$8$",(-1,4),W); label("$12$",(31,6),E); draw((-1,8)--(8,8),dashed); draw((31,0)--(31,12),linewidth(.75)); draw((30.6,0)--(31.4,0),linewidth(.75)); draw((30.6,12)--(31.4,12),linewidth(.75)); draw((31,12)--(18,12),dashed); label("$45^{\circ}$",(.75,0),NE,fontsize(10pt)); label("$45^{\circ}$",(29.25,0),NW,fontsize(10pt)); draw((8,8)--(7.5,7.5)--(8,7)--(8.5,7.5)--cycle); draw((18,12)--(17.5,11.5)--(18,11)--(18.5,11.5)--cycle); draw((11,5)--(11,0),dashed); label("$h$",(11,2.5),E); [/asy]
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 5\sqrt{2}$
| Extend the "inner part" of the mountain so that the image is two right triangles that overlap in a third right triangle as shown.
[asy] unitsize(.2cm); draw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,linewidth(1)); draw((11,5)--(6, 0)--(16, 0)--cycle,linewidth(0.5)); label("$8\sqrt{2}$",(4,4),NW); label("$12\sqrt{2}$",(24,6),NE); draw((8,8)--(7.5,7.5)--(8,7)--(8.5,7.5)--cycle); draw((18,12)--(17.5,11.5)--(18,11)--(18.5,11.5)--cycle); draw((11,5)--(11,0),dashed); label("$h$",(11,2.5),E); [/asy]
The side length of the largest right triangle is $12\sqrt{2},$ which means its area is $144.$ Similarly, the area of the second largest right triangle is $64$ (the side length is $8\sqrt{2}$ ), and the area of the overlap is $h^2$ (the side length is $h\sqrt{2}$ ). Because the right triangles have a side ratio of 1:1: $\sqrt{2}$ .Thus,
\[144+64-h^2=183,\]
which means that the answer is $\boxed{\mathbf{(B)}\text{ 5}}.$
~BS201
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_24 | 5 |
AMC8_363 | Four squares of side length $4, 7, 9,$ and $10$ are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units?
[asy] size(150); filldraw((0,0)--(10,0)--(10,10)--(0,10)--cycle,gray(0.7),linewidth(1)); filldraw((0,0)--(9,0)--(9,9)--(0,9)--cycle,white,linewidth(1)); filldraw((0,0)--(7,0)--(7,7)--(0,7)--cycle,gray(0.7),linewidth(1)); filldraw((0,0)--(4,0)--(4,4)--(0,4)--cycle,white,linewidth(1)); draw((11,0)--(11,4),linewidth(1)); draw((11,6)--(11,10),linewidth(1)); label("$10$",(11,5),fontsize(14pt)); draw((10.75,0)--(11.25,0),linewidth(1)); draw((10.75,10)--(11.25,10),linewidth(1)); draw((0,11)--(3,11),linewidth(1)); draw((5,11)--(9,11),linewidth(1)); draw((0,11.25)--(0,10.75),linewidth(1)); draw((9,11.25)--(9,10.75),linewidth(1)); label("$9$",(4,11),fontsize(14pt)); draw((-1,0)--(-1,1),linewidth(1)); draw((-1,3)--(-1,7),linewidth(1)); draw((-1.25,0)--(-0.75,0),linewidth(1)); draw((-1.25,7)--(-0.75,7),linewidth(1)); label("$7$",(-1,2),fontsize(14pt)); draw((0,-1)--(1,-1),linewidth(1)); draw((3,-1)--(4,-1),linewidth(1)); draw((0,-1.25)--(0,-.75),linewidth(1)); draw((4,-1.25)--(4,-.75),linewidth(1)); label("$4$",(2,-1),fontsize(14pt)); [/asy]
$\textbf{(A)}\ 42 \qquad \textbf{(B)}\ 45\qquad \textbf{(C)}\ 49\qquad \textbf{(D)}\ 50\qquad \textbf{(E)}\ 52$
| We work inwards. The area of the outer shaded square is the area of the whole square minus the area of the second largest square. The area of the inner shaded region is the area of the third largest square minus the area of the smallest square. The sum of these areas is
\[10^2 - 9^2 + 7^2 - 4^2 = 100 - 81 + 49 - 16 = 19 + 33 = \boxed{\textbf{(E)}\ 52}\]
This problem appears multiple times in various math competitions including the AMC and MATHCOUNTS.
-Benedict (countmath1) ~Nivaar and anabel.disher
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_3 | 52 |
AMC8_364 | At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is $\frac25$ . What fraction of the people in the room are married men?
$\textbf{(A)}\ \frac13\qquad \textbf{(B)}\ \frac38\qquad \textbf{(C)}\ \frac25\qquad \textbf{(D)}\ \frac{5}{12}\qquad \textbf{(E)}\ \frac35$
| Assume arbitrarily (and WLOG) there are $5$ women in the room, of which $5 \cdot \frac25 = 2$ are single and $5-2=3$ are married. Each married woman came with her husband, so there are $3$ married men in the room as well for a total of $5+3=8$ people. The fraction of the people that are married men is $\boxed{\textbf{(B)}\ \frac38}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_22 | 3/8 |
AMC8_365 | Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
$\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152$
| We use Principle of Inclusion-Exclusion. There are $365$ days in the year, and we subtract the days that she gets at least $1$ phone call, which is \[\left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor.\]
To this result we add the number of days where she gets at least $2$ phone calls in a day because we double subtracted these days, which is \[\left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor.\]
We now subtract the number of days where she gets three phone calls, which is $\left \lfloor \frac{365}{60} \right \rfloor.$ Therefore, our answer is \[365 - \left( \left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor \right) + \left( \left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor \right) - \left \lfloor \frac{365}{60} \right \rfloor = 365 - 285+72 - 6 = \boxed{\textbf{(D) }146}.\]
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_24 | 146 |
AMC8_366 | How many whole numbers lie in the interval between $\frac{5}{3}$ and $2\pi$ ?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ \text{infinitely many}$
| The smallest whole number in the interval is $2$ because $5/3$ is more than $1$ but less than $2$ . The largest whole number in the interval is $6$ because $2\pi$ is more than $6$ but less than $7$ . There are five whole numbers in the interval. They are $2$ , $3$ , $4$ , $5$ , and $6$ , so the answer is $\boxed{\text{(D)}\ 5}$ .
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_3 | 5 |
AMC8_367 | What is the value of $4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)$ ?
$\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000$
| We group the addends inside the parentheses two at a time:
\begin{align*} -1 + 2 - 3 + 4 - 5 + 6 - 7 + \ldots + 1000 &= (-1 + 2) + (-3 + 4) + (-5 + 6) + \ldots + (-999 + 1000) \\ &= \underbrace{1+1+1+\ldots + 1}_{\text{500 1's}} \\ &= 500. \end{align*}
Then the desired answer is $4 \times 500 = \boxed{\textbf{(E)}\ 2000}$ .
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_3 | 2000 |
AMC8_368 | Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?
$\textbf{(A)}\hspace{.05in}76531\qquad\textbf{(B)}\hspace{.05in}86724\qquad\textbf{(C)}\hspace{.05in}87431\qquad\textbf{(D)}\hspace{.05in}96240\qquad\textbf{(E)}\hspace{.05in}97403$
| In order to maximize the sum of the numbers, the numbers must have their digits ordered in decreasing value. There are only two numbers from the answer choices with this property: $76531$ and $87431$ . To determine the answer we will have to use estimation and the first two digits of the numbers.
For $76531$ the number that would maximize the sum would start with $98$ . The first two digits of $76531$ (when rounded) are $77$ . Adding $98$ and $77$ , we find that the first three digits of the sum of the two numbers would be $175$ .
For $87431$ the number that would maximize the sum would start with $96$ . The first two digits of $87431$ (when rounded) are $87$ . Adding $96$ and $87$ , we find that the first three digits of the sum of the two numbers would be $183$ .
From the estimations, we can say that the answer to this problem is $\boxed{\textbf{(C)}\ 87431}$ .
p.s. USE INTUITION, see answer choices before solving any question -litttle_master
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_16 | 87431 |
AMC8_369 | Two circles that share the same center have radii $10$ meters and $20$ meters. An aardvark runs along the path shown, starting at $A$ and ending at $K$ . How many meters does the aardvark run?
[asy] size((150)); draw((10,0)..(0,10)..(-10,0)..(0,-10)..cycle); draw((20,0)..(0,20)..(-20,0)..(0,-20)..cycle); draw((20,0)--(-20,0)); draw((0,20)--(0,-20)); draw((-2,21.5)..(-15.4, 15.4)..(-22,0), EndArrow); draw((-18,1)--(-12, 1), EndArrow); draw((-12,0)..(-8.3,-8.3)..(0,-12), EndArrow); draw((1,-9)--(1,9), EndArrow); draw((0,12)..(8.3, 8.3)..(12,0), EndArrow); draw((12,-1)--(18,-1), EndArrow); label("$A$", (0,20), N); label("$K$", (20,0), E); [/asy]
$\textbf{(A)}\ 10\pi+20\qquad\textbf{(B)}\ 10\pi+30\qquad\textbf{(C)}\ 10\pi+40\qquad\textbf{(D)}\ 20\pi+20\qquad \\ \textbf{(E)}\ 20\pi+40$
| We will deal with this part by part:
Part 1: 1/4 circumference of big circle= $\frac{2\pi r}{4}=\frac{\pi r}{2}=\frac{20\pi}{2}=10\pi$
Part 2: Big radius minus small radius= $20-10=10$
Part 3: 1/4 circumference of small circle= $\frac{\pi r}{2}=\frac{10\pi}{2}=5\pi$
Part 4: Diameter of small circle: $2*10=20$
Part 5: Same as part 3: $5\pi$
Part 6: Same as part 2: $10$
Total: $10\pi + 10 + 5\pi + 20 + 5\pi + 10 = \boxed{E = 20\pi + 40}$
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_18 | 20π + 40 |
AMC8_370 | Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by $B?$
[asy] size(200); dotfactor = 10; pair p1 = (-28,0); pair p2 = (-111,213); draw(p1--p2,linewidth(1)); pair p3 = (-160,0); pair p4 = (-244,213); draw(p3--p4,linewidth(1)); pair p5 = (-316,0); pair p6 = (-67,213); draw(p5--p6,linewidth(1)); pair p7 = (0, 68); pair p8 = (-350,10); draw(p7--p8,linewidth(1)); pair p9 = (0, 150); pair p10 = (-350, 62); draw(p9--p10,linewidth(1)); pair A = intersectionpoint(p1--p2, p5--p6); dot("$A$", A, 2*W); pair B = intersectionpoint(p5--p6, p3--p4); dot("$B$", B, 2*WNW); pair C = intersectionpoint(p7--p8, p5--p6); dot("$C$", C, 1.5*NW); pair D = intersectionpoint(p3--p4, p7--p8); dot("$D$", D, 2*NNE); pair EE = intersectionpoint(p1--p2, p7--p8); dot("$E$", EE, 2*NNE); pair F = intersectionpoint(p1--p2, p9--p10); dot("$F$", F, 2*NNE); [/asy]
$\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5$
| We can form the following expressions for the sum along each line:
\[\begin{dcases}A+B+C\\A+E+F\\C+D+E\\B+D\\B+F\end{dcases}\]
Adding these together, we must have $2A+3B+2C+2D+2E+2F=47$ , i.e. $2(A+B+C+D+E+F)+B=47$ . Since $A,B,C,D,E,F$ are unique integers between $1$ and $6$ , we obtain $A+B+C+D+E+F=1+2+3+4+5+6=21$ (where the order doesn't matter as addition is commutative), so our equation simplifies to $42 + B = 47$ . This means $B = \boxed{\textbf{(E) }5}$ .
~RJ5303707
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_16 | 5 |
AMC8_371 | Extend the square pattern of 8 black and 17 white square tiles by attaching a border of black tiles around the square. What is the ratio of black tiles to white tiles in the extended pattern?
[asy] filldraw((0,0)--(5,0)--(5,5)--(0,5)--cycle,white,black); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle,mediumgray,black); filldraw((2,2)--(3,2)--(3,3)--(2,3)--cycle,white,black); draw((4,0)--(4,5)); draw((3,0)--(3,5)); draw((2,0)--(2,5)); draw((1,0)--(1,5)); draw((0,4)--(5,4)); draw((0,3)--(5,3)); draw((0,2)--(5,2)); draw((0,1)--(5,1)); [/asy]
$\textbf{(A) }8:17 \qquad\textbf{(B) }25:49 \qquad\textbf{(C) }36:25 \qquad\textbf{(D) }32:17 \qquad\textbf{(E) }36:17$
| One way of approaching this is drawing the next circle of boxes around the current square.
[asy] filldraw((-1,-1)--(6,-1)--(6,6)--(-1,6)--cycle,mediumgray,black); filldraw((0,0)--(5,0)--(5,5)--(0,5)--cycle,white,black); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle,mediumgray,black); filldraw((2,2)--(3,2)--(3,3)--(2,3)--cycle,white,black); draw((4,-1)--(4,6)); draw((3,-1)--(3,6)); draw((2,-1)--(2,6)); draw((1,-1)--(1,6)); draw((-1,4)--(6,4)); draw((-1,3)--(6,3)); draw((-1,2)--(6,2)); draw((-1,1)--(6,1)); draw((0,-1)--(0,6)); draw((-1,5)--(6,5)); draw((-1,0)--(6,0)); draw((5,-1)--(5,6)); [/asy]
We can now count the number of black and white tiles; 32 black tiles and 17 white tiles. This means the answer is $\boxed{\textbf{(D) }32:17}$ .
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_3 | 32:17 |
AMC8_372 | Which of the following polygons has the largest area?
[asy] size(330); int i,j,k; for(i=0;i<5; i=i+1) { for(j=0;j<5;j=j+1) { for(k=0;k<5;k=k+1) { dot((6i+j, k)); }}} draw((0,0)--(4,0)--(3,1)--(3,3)--(2,3)--(2,1)--(1,1)--cycle); draw(shift(6,0)*((0,0)--(4,0)--(4,1)--(3,1)--(3,2)--(2,1)--(1,1)--(0,2)--cycle)); draw(shift(12,0)*((0,1)--(1,0)--(3,2)--(3,3)--(1,1)--(1,3)--(0,4)--cycle)); draw(shift(18,0)*((0,1)--(2,1)--(3,0)--(3,3)--(2,2)--(1,3)--(1,2)--(0,2)--cycle)); draw(shift(24,0)*((1,0)--(2,1)--(2,3)--(3,2)--(3,4)--(0,4)--(1,3)--cycle)); label("$A$", (0*6+2, 0), S); label("$B$", (1*6+2, 0), S); label("$C$", (2*6+2, 0), S); label("$D$", (3*6+2, 0), S); label("$E$", (4*6+2, 0), S);[/asy]
$\textbf{(A)}\text{A}\qquad\textbf{(B)}\ \text{B}\qquad\textbf{(C)}\ \text{C}\qquad\textbf{(D)}\ \text{D}\qquad\textbf{(E)}\ \text{E}$
| Each polygon can be partitioned into unit squares and right triangles with sidelength $1$ . Count the number of boxes enclosed by each polygon, with the unit square being $1$ , and the triangle being $.5$ . A has 5, B has 5, C has 5, D has 4.5, and E has 5.5. Therefore, the polygon with the largest area is $\boxed{\textbf{(E)}\ \text{E}}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_15 | E |
AMC8_373 | Circle $X$ has a radius of $\pi$ . Circle $Y$ has a circumference of $8 \pi$ . Circle $Z$ has an area of $9 \pi$ . List the circles in order from smallest to the largest radius.
$\textbf{(A)}\ X, Y, Z\qquad\textbf{(B)}\ Z, X, Y\qquad\textbf{(C)}\ Y, X, Z\qquad\textbf{(D)}\ Z, Y, X\qquad\textbf{(E)}\ X, Z, Y$
| Using the formulas of circles, $C=2 \pi r$ and $A= \pi r^2$ , we find that circle $Y$ has a radius of $4$ and circle $Z$ has a radius of $3$ . Also, circle X has a radius of $\pi$ . Thus, the order from smallest to largest radius is $\boxed{\textbf{(B)}\ Z, X, Y}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_7 | Z, X, Y |
AMC8_374 | Tiles $I, II, III$ and $IV$ are translated so one tile coincides with each of the rectangles $A, B, C$ and $D$ . In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle $C$ ?
[asy] size(400); defaultpen(linewidth(0.8)); path p=origin--(8,0)--(8,6)--(0,6)--cycle; draw(p^^shift(8.5,0)*p^^shift(8.5,10)*p^^shift(0,10)*p); draw(shift(20,2)*p^^shift(28,2)*p^^shift(20,8)*p^^shift(28,8)*p); label("8", (4,6+10), S); label("6", (4+8.5,6+10), S); label("7", (4,6), S); label("2", (4+8.5,6), S); label("I", (4,6+10), N); label("II", (4+8.5,6+10), N); label("III", (4,6), N); label("IV", (4+8.5,6), N); label("3", (0,3+10), E); label("4", (0+8.5,3+10), E); label("1", (0,3), E); label("9", (0+8.5,3), E); label("7", (4,10), N); label("2", (4+8.5,10), N); label("0", (4,0), N); label("6", (4+8.5,0), N); label("9", (8,3+10), W); label("3", (8+8.5,3+10), W); label("5", (8,3), W); label("1", (8+8.5,3), W); label("A", (24,10), N); label("B", (32,10), N); label("C", (24,4), N); label("D", (32,4), N);[/asy]
$\mathrm{(A)}\ I \qquad \mathrm{(B)}\ II \qquad \mathrm{(C)}\ III \qquad \mathrm{(D)}\ IV \qquad \mathrm{(E)}$ cannot be determined
| We first notice that tile III has a $0$ on the bottom and a $5$ on the right side. Since no other tile has a $0$ or a $5$ , Tile III must be in rectangle $D$ . Tile III also has a $1$ on the left, so Tile IV must be in Rectangle $C$ .
The answer is $\boxed{\textbf{(D)}}$
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_11 | D |
AMC8_375 | A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance $d$ traveled by the two animals over time $t$ from start to finish?
| SolutionFirst, the tortoise walks at a constant rate, ruling out $(D)$ . Second, when the hare is resting, the distance will stay the same, ruling out $(E)$ and $(C)$ . Third, the tortoise wins the race, meaning that the non-constant one should go off the graph last, ruling out $(A)$ . The answer $\boxed{\textbf{(B)}}$ is the only one left.
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_5 | B |
AMC8_376 | Tori's mathematics test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry problems. Although she answered 70% of the arithmetic, 40% of the algebra, and 60% of the geometry problems correctly, she did not pass the test because she got less than 60% of the problems right. How many more problems would she have needed to answer correctly to earn a 60% passing grade?
$\text{(A)}\ 1 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 11$
| First, calculate how many of each type of problem she got right:
Arithmetic: $70\% \cdot 10 = 0.70 \cdot 10 = 7$
Algebra: $40\% \cdot 30 = 0.40 \cdot 30 = 12$
Geometry: $60\% \cdot 35 = 0.60 \cdot 35 = 21$
Altogether, Tori answered $7 + 12 + 21 = 40$ questions correct.
To get a $60\%$ on her test overall, she needed to get $60\% \cdot 75 = 0.60 \cdot 75 = 45$ questions right.
Therefore, she needed to answer $45 - 40 = 5$ more questions to pass, so the correct answer is $\boxed{(B) 5}$
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_16 | 5 |
AMC8_377 | Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle $ABCD$ ?
[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$
| Solution 1We can see that there are $2$ rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is $5$ , so the bigger side is $10$ , if we do $5 \cdot 2 = 10$ . Now we get the sides of the big rectangle being $15$ and $10$ , so the area is $\boxed{\textbf{(E)}\ 150}$ . ~avamarora
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_2 | 150 |
AMC8_378 | Betty drives a truck to deliver packages in a neighborhood whose street map is shown below. Betty starts at the factory (labled $F$ ) and drives to location $A$ , then $B$ , then $C$ , before returning to $F$ . What is the shortest distance, in blocks, she can drive to complete the route?
[asy] unitsize(20); add(grid(8,6)); path w = circle((0,0),0.4); fill(w, white); draw(w); label("$B$",(0,0)); fill(shift((2,4)) * w, white); draw(shift((2,4)) * w); label("$C$",(2,4)); fill(shift((7,3)) * w, white); draw(shift((7,3)) * w); label("$A$",(7,3)); fill(shift((6,5)) * w, white); draw(shift((6,5)) * w); label("$F$",(6,5)); draw((6,-0.2)--(7,-0.2), EndArrow(3)); draw((7,-0.2)--(6,-0.2), EndArrow(3)); draw(shift(6.5, -0.48) * scale(0.03) * texpath("1 block")); draw((8.2,1)--(8.2,2), EndArrow(3)); draw((8.2,2)--(8.2,1), EndArrow(3)); draw(shift(8.88, 1.5) * scale(0.03) * texpath("1 block")); [/asy]
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26\qquad \textbf{(E)}\ 28$
| Each shortest possible path from $A$ to $B$ follows the edges of the rectangle. The following path outlines a path of $\boxed{\textbf{(C)}\ 24}$ units:
[asy] unitsize(20); add(grid(8,6)); draw((6,5)--(7,5)--(7,0)--(0,0)--(0,4)--(2,4)--(2,5)--cycle,green); path w = circle((0,0),0.4); fill(w, white); draw(w); label("$B$",(0,0)); fill(shift((2,4)) * w, white); draw(shift((2,4)) * w); label("$C$",(2,4)); fill(shift((7,3)) * w, white); draw(shift((7,3)) * w); label("$A$",(7,3)); fill(shift((6,5)) * w, white); draw(shift((6,5)) * w); label("$F$",(6,5)); [/asy]
~ zhenghua
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_5 | 24 |
AMC8_379 | A cube with $3$ -inch edges is to be constructed from $27$ smaller cubes with $1$ -inch edges. Twenty-one of the cubes are colored red and $6$ are colored white. If the $3$ -inch cube is constructed to have the smallest possible white surface area showing, what fraction of the surface area is white?
$\textbf{(A) }\frac{5}{54}\qquad\textbf{(B) }\frac{1}{9}\qquad\textbf{(C) }\frac{5}{27}\qquad\textbf{(D) }\frac{2}{9}\qquad\textbf{(E) }\frac{1}{3}$
| For the least possible surface area that is white, we should have 1 cube in the center, and the other 5 with only 1 face exposed. This gives 5 square inches of white, surface area. Since the cube has a surface area of 54 square inches, our answer is $\boxed{\textbf{(A) }\frac{5}{54}}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_19 | 5/54 |
AMC8_380 | Aunt Anna is $42$ years old. Caitlin is $5$ years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?
$\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 21\qquad\mathrm{(E)}\ 37$
| If Brianna is half as old as Aunt Anna, then Brianna is $\frac{42}{2}$ years old, or $21$ years old.
If Caitlin is $5$ years younger than Brianna, she is $21-5$ years old, or $16$ .
So, the answer is $\boxed{B}$
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_1 | B |
AMC8_381 | Problems 8,9 and 10 use the data found in the accompanying paragraph and table:
| Brazil 50s and 60s total 11 stamps with each 6 cents, Peru 50s and 60s total 10 stamps with each 4 cents. So total $11*0.06+10*0.04 = \boxed{\text{(B)}\ $1.06}$
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_9 | 1.06 |
AMC8_382 | Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?
$\text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2}$
| Since Ott gets equal amounts of money from each friend, we can say that he gets $x$ dollars from each friend. This means that Moe has $5x$ dollars, Loki has $4x$ dollars, and Nick has $3x$ dollars. The total amount is $12x$ dollars, and since Ott gets $3x$ dollars total, $\frac{3x}{12x}= \frac{3}{12} = \boxed{\text{(B)}\ \frac14}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_25 | 1/4 |
AMC8_383 | Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first?
[asy] size(7cm); draw((-8,27)--(72,27)); draw((16,0)--(16,35)); draw((40,0)--(40,35)); label("12", (28,3)); draw((25,6.5)--(25,12)--(31,12)--(31,6.5)--cycle); draw((25,5.5)--(31,5.5)); label("18", (56,3)); draw((53,6.5)--(53,12)--(59,12)--(59,6.5)--cycle); draw((53,5.5)--(59,5.5)); draw((53,5.5)--(59,5.5)); label("20", (28,18)); label("15", (28,24)); draw((25,21)--(31,21)); label("10", (56,18)); label("10", (56,24)); draw((53,21)--(59,21)); label("First Half", (28,31)); label("Second Half", (56,31)); label("Candace", (2.35,6)); label("Steph", (0,21)); [/asy]
$\textbf{(A) } 7\qquad\textbf{(B) } 8\qquad\textbf{(C) } 9\qquad\textbf{(D) } 10\qquad\textbf{(E) } 11$
| Let $x$ be the number of shots that Candace made in the first half, and let $y$ be the number of shots Candace made in the second half. Since Candace and Steph took the same number of attempts, with an equal percentage of baskets scored, we have $x+y=10+15=25.$ In addition, we have the following inequalities: \[\frac{x}{12}<\frac{15}{20} \implies x<9,\] and \[\frac{y}{18}<\frac{10}{10} \implies y<18.\] Pairing this up with $x+y=25$ we see the only possible solution is $(x,y)=(8,17),$ for an answer of $17-8 = \boxed{\textbf{(C) } 9}.$
~wamofan
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_21 | 9 |
AMC8_384 | How many two-digit numbers have digits whose sum is a perfect square?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19$
| There is $1$ integer whose digits sum to $1$ : $10$ .
There are $4$ integers whose digits sum to $4$ : $13, 22, 31, \text{and } 40$ .
There are $9$ integers whose digits sum to $9$ : $18, 27, 36, 45, 54, 63, 72, 81, \text{and } 90$ .
There are $3$ integers whose digits sum to $16$ : $79, 88, \text{and } 97$ .
Two digits cannot sum to $25$ or any greater square since the greatest sum of digits of a two-digit number is $9 + 9 = 18$ .
Thus, the answer is $1 + 4 + 9 + 3 = \boxed{\textbf{(C)} 17}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_11 | 17 |
AMC8_385 | Which of the following numbers has the smallest prime factor?
$\mathrm{(A)}\ 55 \qquad\mathrm{(B)}\ 57 \qquad\mathrm{(C)}\ 58 \qquad\mathrm{(D)}\ 59 \qquad\mathrm{(E)}\ 61$
| The smallest prime factor is $2$ , and since $58$ is the only multiple of $2$ , the answer is $\boxed{\mathrm{(C)}\ 58}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_2 | 58 |
AMC8_386 | What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$ ?
$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$
| Rearranging the terms, we get $(1-2)+(3-4)+(5-6)+...(2017-2018)+2019$ , and our answer is $-1009+2019=\boxed{\textbf{(E) }1010}$ .
If you were stuck on this problem, refer to AOPS arithmetic lessons.
~Nivaar
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_5 | 1010 |
AMC8_387 | Find the number of two-digit positive integers whose digits total $7$ .
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$
| The numbers are $16, 25, 34, 43, 52, 61, 70$ which gives us a total of $\boxed{\textbf{(B)}\ 7}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_8 | 7 |
AMC8_388 | In the given figure, hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$ . What is the area of $\triangle KBC$ ?
[asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw((-8,0)--(-4,-6*sqrt(2))); draw((4,6*sqrt(2))--(8,0)); draw((8,0)--(4,-6*sqrt(2))); draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle); draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("$I$",(-4,8+6*sqrt(2)),dir(100)); label("$J$",(4,8+6*sqrt(2)),dir(80)); label("$A$",(-4,6*sqrt(2)),dir(280)); label("$B$",(4,6*sqrt(2)),dir(250)); label("$C$",(8,0),W); label("$D$",(4,-6*sqrt(2)),NW); label("$E$",(-4,-6*sqrt(2)),NE); label("$F$",(-8,0),E); draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle); label("$K$",(4+4*sqrt(3),4+6*sqrt(2)),E); draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed); label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S); label("$G$",(-8-6*sqrt(2),-4),W); label("$32$",(-10,-8),N); label("$18$",(0,6*sqrt(2)+2),N); [/asy]
$\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32$ .
| Clearly, since $\overline{FE}$ is a side of a square with area $32$ , $\overline{FE} = \sqrt{32} = 4 \sqrt{2}$ . Now, since $\overline{FE} = \overline{BC}$ , we have $\overline{BC} = 4 \sqrt{2}$ .
We know that $\overline{JB}$ is a side of a square with area $18$ , so $\overline{JB} = \sqrt{18} = 3 \sqrt{2}$ . Since $\Delta JBK$ is equilateral, $\overline{BK} = \overline{JB} = 3 \sqrt{2}$ .
Lastly, $\Delta KBC$ is a right triangle. We see that $\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360^{\circ} \rightarrow 90^{\circ} + 120^{\circ} + \angle CBK + 60^{\circ} = 360^{\circ} \rightarrow \angle CBK = 90^{\circ}$ , so $\Delta KBC$ is a right triangle with legs $3 \sqrt{2}$ and $4 \sqrt{2}$ . Now, its area is $\dfrac{3 \sqrt{2} \cdot 4 \sqrt{2}}{2} = \dfrac{24}{2} = \boxed{\textbf{(C)}~12}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_21 | 12 |
AMC8_389 | Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially bolded . What percent of the total area is partially bolded?
[asy] import graph; size(7.01cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.42,xmax=14.59,ymin=-10.08,ymax=5.26; pair A=(0,0), B=(4,0), C=(0,4), D=(4,4), F=(2,0), G=(3,0), H=(1,4), I=(2,4), J=(3,4), K=(0,-2), L=(4,-2), M=(0,-6), O=(0,-4), P=(4,-4), Q=(2,-2), R=(2,-6), T=(6,4), U=(10,0), V=(10,4), Z=(10,2), A_1=(8,4), B_1=(8,0), C_1=(6,-2), D_1=(10,-2), E_1=(6,-6), F_1=(10,-6), G_1=(6,-4), H_1=(10,-4), I_1=(8,-2), J_1=(8,-6), K_1=(8,-4); draw(C--H--(1,0)--A--cycle,linewidth(1.6)); draw(M--O--Q--R--cycle,linewidth(1.6)); draw(A_1--V--Z--cycle,linewidth(1.6)); draw(G_1--K_1--J_1--E_1--cycle,linewidth(1.6)); draw(C--D); draw(D--B); draw(B--A); draw(A--C); draw(H--(1,0)); draw(I--F); draw(J--G); draw(C--H,linewidth(1.6)); draw(H--(1,0),linewidth(1.6)); draw((1,0)--A,linewidth(1.6)); draw(A--C,linewidth(1.6)); draw(K--L); draw((4,-6)--L); draw((4,-6)--M); draw(M--K); draw(O--P); draw(Q--R); draw(O--Q); draw(M--O,linewidth(1.6)); draw(O--Q,linewidth(1.6)); draw(Q--R,linewidth(1.6)); draw(R--M,linewidth(1.6)); draw(T--V); draw(V--U); draw(U--(6,0)); draw((6,0)--T); draw((6,2)--Z); draw(A_1--B_1); draw(A_1--Z); draw(A_1--V,linewidth(1.6)); draw(V--Z,linewidth(1.6)); draw(Z--A_1,linewidth(1.6)); draw(C_1--D_1); draw(D_1--F_1); draw(F_1--E_1); draw(E_1--C_1); draw(G_1--H_1); draw(I_1--J_1); draw(G_1--K_1,linewidth(1.6)); draw(K_1--J_1,linewidth(1.6)); draw(J_1--E_1,linewidth(1.6)); draw(E_1--G_1,linewidth(1.6)); dot(A,linewidth(1pt)+ds); dot(B,linewidth(1pt)+ds); dot(C,linewidth(1pt)+ds); dot(D,linewidth(1pt)+ds); dot((1,0),linewidth(1pt)+ds); dot(F,linewidth(1pt)+ds); dot(G,linewidth(1pt)+ds); dot(H,linewidth(1pt)+ds); dot(I,linewidth(1pt)+ds); dot(J,linewidth(1pt)+ds); dot(K,linewidth(1pt)+ds); dot(L,linewidth(1pt)+ds); dot(M,linewidth(1pt)+ds); dot((4,-6),linewidth(1pt)+ds); dot(O,linewidth(1pt)+ds); dot(P,linewidth(1pt)+ds); dot(Q,linewidth(1pt)+ds); dot(R,linewidth(1pt)+ds); dot((6,0),linewidth(1pt)+ds); dot(T,linewidth(1pt)+ds); dot(U,linewidth(1pt)+ds); dot(V,linewidth(1pt)+ds); dot((6,2),linewidth(1pt)+ds); dot(Z,linewidth(1pt)+ds); dot(A_1,linewidth(1pt)+ds); dot(B_1,linewidth(1pt)+ds); dot(C_1,linewidth(1pt)+ds); dot(D_1,linewidth(1pt)+ds); dot(E_1,linewidth(1pt)+ds); dot(F_1,linewidth(1pt)+ds); dot(G_1,linewidth(1pt)+ds); dot(H_1,linewidth(1pt)+ds); dot(I_1,linewidth(1pt)+ds); dot(J_1,linewidth(1pt)+ds); dot(K_1,linewidth(1pt)+ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]
$\textbf{(A) }12\frac{1}{2} \qquad\textbf{(B) }20 \qquad\textbf{(C) }25 \qquad\textbf{(D) }33\frac{1}{3} \qquad\textbf{(E) }37\frac{1}{2}$
| Assume that the area of each square is $1$ . Then, the area of the bolded region in the top left square is $\dfrac{1}{4}$ . The area of the top right bolded region is $\dfrac{1}{8}$ . The area of the bottom left bolded region is $\dfrac{3}{8}$ . And the area of the bottom right bolded region is $\dfrac{1}{4}$ . Add the four fractions: $\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{3}{8} + \dfrac{1}{4} = 1$ . The four squares together have an area of $4$ , so the percentage bolded is $\dfrac{1}{4} \cdot 100 = \boxed{\textbf{(C)}\ 25}$ .
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_7 | 25 |
AMC8_390 | Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. What fraction of the pizza did Peter eat?
$\textbf{(A)}~\frac{1}{24}\qquad\textbf{(B)}~\frac{1}{12}\qquad\textbf{(C)}~\frac{1}{8}\qquad\textbf{(D)}~\frac{1}{6}\qquad\textbf{(E)}~\frac{1}{4}$
| Peter ate $1 + \frac{1}{2} = \frac{3}{2}$ slices. The pizza has $12$ slices total. Taking the ratio of the amount of slices Peter ate to the amount of slices in the pizza, we find that Peter ate $\dfrac{\frac{3}{2}\text{ slices}}{12\text{ slices}} = \boxed{\textbf{(C)}\ \frac{1}{8}}$ of the pizza.
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_4 | 1/8 |
AMC8_391 | Find the value of the expression
\[100-98+96-94+92-90+\cdots+8-6+4-2.\] $\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$
| Solution 1We can group each subtracting pair together:
\[(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).\]
After subtracting, we have:
\[2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).\]
There are $50$ even numbers, therefore there are $\dfrac{50}{2}=25$ even pairs. Therefore the sum is $2 \cdot 25=\boxed{\textbf{(C) }50}$
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_8 | 50 |
AMC8_392 | Points $A$ , $B$ , $C$ and $D$ have these coordinates: $A(3,2)$ , $B(3,-2)$ , $C(-3,-2)$ and $D(-3, 0)$ . The area of quadrilateral $ABCD$ is
[asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } } draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) { draw((i,-1/3)--(i,1/3),linewidth(0.5)); draw((-1/3,i)--(1/3,i),linewidth(0.5)); } [/asy]
$\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24$
| [asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } } draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) { draw((i,-1/3)--(i,1/3),linewidth(0.5)); draw((-1/3,i)--(1/3,i),linewidth(0.5)); } { draw((3,2)--(3,-2)--(-3,-2)--(-3,0)--cycle,linewidth(1)); } label("$A$",(3,2),NE); label("$B$",(3,-2),SE); label("$C$",(-3,-2),SW); label("$D$",(-3,0),NW); [/asy]
This quadrilateral is a trapezoid, because $AB\parallel CD$ but $BC$ is not parallel to $AD$ . The area of a trapezoid is the product of its height and its median, where the median is the average of the side lengths of the bases. The two bases are $AB$ and $CD$ , which have lengths $2$ and $4$ , respectively, so the length of the median is $\frac{2+4}{2}=3$ . $CB$ is perpendicular to the bases, so it is the height, and has length $6$ . Therefore, the area of the trapezoid is $(3)(6)=18, \boxed{\text{C}}$
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_11 | C |
AMC8_393 | A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^\text{th}$ -graders to $6^\text{th}$ -graders is $5:3$ , and the the ratio of $8^\text{th}$ -graders to $7^\text{th}$ -graders is $8:5$ . What is the smallest number of students that could be participating in the project?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89$
| Solution 1: AlgebraWe multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together:
$5:3 = 5(8):3(8) = 40:24$
$8:5 = 8(5):5(5) = 40:25$
Therefore, the ratio of 8th graders to 7th graders to 6th graders is $40:25:24$ . Since the ratio is in lowest terms, the smallest number of students participating in the project is $40+25+24 = \boxed{\textbf{(E)}\ 89}$ .
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_16 | 89 |
AMC8_394 | On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?
[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); 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)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]
$\textbf{(A)}\ \frac{1}{16}\qquad\textbf{(B)}\ \frac{7}{16}\qquad\textbf{(C)}\ \frac{1}2\qquad\textbf{(D)}\ \frac{9}{16}\qquad\textbf{(E)}\ \frac{49}{64}$
| There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\frac{36}{64} = \boxed{\textbf{(D)}\ \frac{9}{16}}$ .
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_10 | 9/16 |
AMC8_395 | Harold tosses a coin four times. The probability that he gets at least as many heads as tails is
$\text{(A)}\ \frac{5}{16}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{5}{8}\qquad\text{(E)}\ \frac{11}{16}$
| Case 1: There are two heads and two tails. There are $\binom{4}{2} = 6$ ways to choose which two tosses are heads, and the other two must be tails.
Case 2: There are three heads, one tail. There are $\binom{4}{1} = 4$ ways to choose which of the four tosses is a tail.
Case 3: There are four heads and no tails. This can only happen $1$ way.
There are a total of $2^4=16$ possible configurations, giving a probability of $\frac{6+4+1}{16} = \boxed{\text{(E)}\ \frac{11}{16}}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_21 | 11/16 |
AMC8_396 | In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is , $X$ in centimeters?
[asy] pair A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R; A=(4,0); B=(7,0); C=(7,4); D=(8,4); E=(8,5); F=(10,5); G=(10,7); H=(7,7); I=(7,8); J=(5,8); K=(5,7); L=(4,7); M=(4,6); N=(0,6); O=(0,5); P=(2,5); Q=(2,3); R=(4,3); draw(A--B--C--D--E--F--G--H--I--J--K--L--M--N--O--P--Q--R--cycle); label("$X$",(3.4,1.5)); label("6",(7.6,1.5)); label("1",(7.6,3.5)); label("1",(8.4,4.6)); label("2",(9.4,4.6)); label("2",(10.4,6)); label("3",(8.4,7.4)); label("1",(7.5,7.8)); label("2",(6,8.5)); label("1",(4.7,7.8)); label("1",(4.3,7.5)); label("1",(3.5,6.5)); label("4",(1.8,6.5)); label("1",(-0.5,5.5)); label("2",(0.8,4.5)); label("2",(1.5,3.8)); label("2",(2.8,2.6));[/asy]
$\textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}2\qquad\textbf{(C)}\hspace{.05in}3\qquad\textbf{(D)}\hspace{.05in}4\qquad\textbf{(E)}\hspace{.05in}5$
|
$1 + 1 + 1 + 2 + X = 1 + 2 + 1 + 6\\ 5 + X = 10\\ X = 5$
Thus, the answer is $\boxed{\textbf{(E)}\ 5}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_5 | 5 |
AMC8_397 | In the figure, the area of square $WXYZ$ is $25 \text{ cm}^2$ . The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In $\triangle ABC$ , $AB = AC$ , and when $\triangle ABC$ is folded over side $\overline{BC}$ , point $A$ coincides with $O$ , the center of square $WXYZ$ . What is the area of $\triangle ABC$ , in square centimeters?
[asy] defaultpen(fontsize(8)); size(225); pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5); draw((-4,0)--Y--X--(-4,10)--cycle); draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle); dot(O); label("$A$", A, NW); label("$O$", O, NE); label("$B$", B, SW); label("$C$", C, NW); label("$W$",W , NE); label("$X$", X, N); label("$Y$", Y, S); label("$Z$", Z, SE); [/asy]
$\textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2$
| We see that $XY = 5$ , the vertical distance between $B$ and $X$ is $1$ , and the vertical distance between $C$ and $Y$ is $1$ . Therefore, $BC = 5 - 1 - 1 = 3$ . We are given that the length of the altitude of $\triangle ABC$ is equal to the distance between $\overline{BC}$ and $O$ , which is $1 + 1 + \frac{5}{2} = \frac{9}{2}$ . So the area of $\triangle ABC$ is $\frac{1}{2}\left(3\cdot \frac{9}{2}\right)$ , which is $\boxed{\textbf{(C)} \ \frac{27}{4}}$ .
~sidkris
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_25 | 27/4 |
AMC8_398 | What is the value of the product
\[\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?\]
$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(D) }7\qquad\textbf{(E) }8$
| By adding up the numbers in each of the $6$ parentheses, we get:
$\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}$ .
Using telescoping, most of the terms cancel out diagonally. We are left with $\frac{7}{1}$ which is equivalent to $7$ . Thus, the answer would be $\boxed{\textbf{(D) }7}$ .
Also, you can also make the fractions $\frac{7!}{6!}$ , which also yields $\boxed{\textbf{(D)}7}$
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_2 | 7 |
AMC8_399 | A complete cycle of a traffic light takes 60 seconds. During each cycle the light is green for 25 seconds, yellow for 5 seconds, and red for 30 seconds. At a randomly chosen time, what is the probability that the light will NOT be green?
$\text{(A)}\ \frac{1}{4} \qquad \text{(B)}\ \frac{1}{3} \qquad \text{(C)}\ \frac{5}{12} \qquad \text{(D)}\ \frac{1}{2} \qquad \text{(E)}\ \frac{7}{12}$
| Solution 1\[\frac{\text{time not green}}{\text{total time}} = \frac{R + Y}{R + Y + G} = \frac{35}{60} = \boxed{\text{(E)}\ \frac{7}{12}}\]
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_10 | 7/12 |
AMC8_400 | An equilateral triangle is placed inside a larger equilateral triangle so that the region between them can be divided into three congruent trapezoids, as shown below. The side length of the inner triangle is $\frac23$ the side length of the larger triangle. What is the ratio of the area of one trapezoid to the area of the inner triangle?
[asy] // Diagram by TheMathGuyd pair A,B,C; A=(0,1); B=(sqrt(3)/2,-1/2); C=-conj(B); fill(2B--3B--3C--2C--cycle,grey); dot(3A); dot(3B); dot(3C); dot(2A); dot(2B); dot(2C); draw(2A--2B--2C--cycle); draw(3A--3B--3C--cycle); draw(2A--3A); draw(2B--3B); draw(2C--3C); [/asy]
$\textbf{(A) } 1 : 3 \qquad \textbf{(B) } 3 : 8 \qquad \textbf{(C) } 5 : 12 \qquad \textbf{(D) } 7 : 16 \qquad \textbf{(E) } 4 : 9$
| All equilateral triangles are similar. For the outer equilateral triangle to the inner equilateral triangle, since their side-length ratio is $\frac32,$ their area ratio is $\left(\frac32\right)^2=\frac94.$ It follows that the area ratio of three trapezoids to the inner equilateral triangle is $\frac94-1=\frac54,$ so the area ratio of one trapezoid to the inner equilateral triangle is \[\frac54\cdot\frac13=\frac{5}{12}=\boxed{\textbf{(C) } 5 : 12}.\]
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_19 | 5 : 12 |
AMC8_401 | Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$
| Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$ , respectively. If we ignore the constraint that $S$ and $T$ cannot be next to each other, we get a total of $4!=24$ ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that $S$ and $T$ can be next to each other. If we place $S$ and $T$ next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. $ST\square\square, \square ST\square, \square\square ST$ ). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $TS\square\square, \square TS\square, \square\square TS$ ). Thus there are 6 ways of placing $S$ and $T$ directly next to each other. Next, notice that for each of these placements, we have two open slots for placing $A$ and $B$ . Specifically, we can place $A$ in the first open slot and $B$ in the second open slot or switch their order and place $B$ in the first open slot and $A$ in the second open slot. This gives us a total of $6\times 2=12$ ways to place $S$ and $T$ next to each other. Subtracting this from the total number of arrangements gives us $24-12=12$ total arrangements $\implies\boxed{\textbf{(C) }12}$ .
We can also solve this problem directly by looking at the number of ways that we can place $S$ and $T$ such that they are not directly next to each other. Observe that there are three ways to place $S$ and $T$ (in that order) into the four slots so they are not next to each other (i.e. $S\square T\square, \square S\square T, S\square\square T$ ). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $T\square S\square, \square T\square S, T\square\square S$ ). Thus there are 6 ways of placing $S$ and $T$ so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing $A$ and $B$ . Specifically, we can place $A$ in the first open slot and $B$ in the second open slot or switch their order and place $B$ in the first open slot and $A$ in the second open slot. This gives us a total of $6\times 2=12$ ways to place $S$ and $T$ such that they are not next to each other $\implies\boxed{\textbf{(C) }12}$ .
~ junaidmansuri
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10 | 12 |
AMC8_402 | After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?
[asy] unitsize(1.25cm); dotfactor = 10; pen shortdashed=linetype(new real[] {2.7,2.7}); for (int i = 0; i < 6; ++i) { for (int j = 0; j < 6; ++j) { draw((i,0)--(i,6), grey); draw((0,j)--(6,j), grey); } } for (int i = 1; i <= 6; ++i) { draw((-0.1,i)--(0.1,i),linewidth(1.25)); draw((i,-0.1)--(i,0.1),linewidth(1.25)); label(string(5*i), (i,0), 2*S); label(string(i), (0, i), 2*W); } draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25)); label(rotate(90) * "Distance (miles)", (-0.5,3), W); label("Time (minutes)", (3,-0.5), S); dot("Naomi", (2,6), 3*dir(305)); dot((6,6)); label("Maya", (4.45,3.5)); draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35)); draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed); [/asy]
$\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24$
| Naomi travels $6$ miles in a time of $10$ minutes, which is equivalent to $\dfrac{1}{6}$ of an hour. Since $\text{speed} = \frac{\text{distance}}{\text{time}}$ , her speed is $\frac{6}{\left(\frac{1}{6}\right)} = 36$ mph. By a similar calculation, Maya's speed is $12$ mph, so the answer is $36-12 = \boxed{\textbf{(E) }24}$ .
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_11 | 24 |
AMC8_403 | Marla has a large white cube that has an edge of 10 feet. She also has enough green paint to cover 300 square feet. Marla uses all the paint to create a white square centered on each face, surrounded by a green border. What is the area of one of the white squares, in square feet?
$\textbf{(A)}\hspace{.05in}5\sqrt2\qquad\textbf{(B)}\hspace{.05in}10\qquad\textbf{(C)}\hspace{.05in}10\sqrt2\qquad\textbf{(D)}\hspace{.05in}50\qquad\textbf{(E)}\hspace{.05in}50\sqrt2$
| If Marla evenly distributes her $300$ square feet of paint between the 6 faces, each face will get $300\div6 = 50$ square feet of paint. The surface area of one of the faces of the cube is $10^2 = 100$ square feet. Therefore, there will be $100-50 = \boxed{\textbf{(D)}\ 50}$ square feet of white on each side.
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_21 | 50 |
AMC8_404 | Two different numbers are randomly selected from the set $\{ - 2, -1, 0, 3, 4, 5\}$ and multiplied together. What is the probability that the product is $0$ ?
$\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}$
| Solution 11. Identify the total number of ways to select two different numbers from the set:
The set has 6 elements. The number of ways to choose 2 different numbers from 6 is given by the combination formula: $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$ .
2. Identify the favorable outcomes:
For the product to be zero, one of the chosen numbers must be zero. The set contains one zero (0). To have a product of zero, we need to choose 0 and any other number from the remaining five numbers $-2, -1, 3, 4, 5$ .
The number of ways to choose 0 and one other number from the remaining five is 5.
3. Calculate the probability:
The probability is the number of favorable outcomes divided by the total number of outcomes: $\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{15} = \frac{1}{3}$ .
Thus, the probability that the product is $0$ is $\boxed{\textbf{(D)} \ \frac{1}{3}}.$
~GeometryMystery
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_13 | 1/3 |
AMC8_405 | In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read "AMC8"?
$\textbf{(A) } \frac{1}{22,050} \qquad \textbf{(B) } \frac{1}{21,000}\qquad \textbf{(C) } \frac{1}{10,500}\qquad \textbf{(D) } \frac{1}{2,100} \qquad \textbf{(E) } \frac{1}{1,050}$
| Solution 1There is one favorable case, which is the license plate says "AMC8." We must now find how many total cases there are. There are $5$ choices for the first letter (since it must be a vowel), $21$ choices for the second letter (since it must be of $21$ consonants), $20$ choices for the third letter (since it must differ from the second letter), and $10$ choices for the digit. This leads to $5 \cdot 21 \cdot 20 \cdot 10=21,000$ total possible license plates. Therefore, the probability of a license plate saying "AMC8" is $\boxed{\textbf{(B) } \frac{1}{21,000}}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_11 | 1/21,000 |
AMC8_406 | If $\angle A = 20^\circ$ and $\angle AFG =\angle AGF$ , then $\angle B+\angle D =$
[asy] pair A,B,C,D,EE,F,G; A = (0,0); B = (9,4); C = (21,0); D = (13,-12); EE = (4,-16); F = (13/2,-6); G = (8,0); draw(A--C--EE--B--D--cycle); label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,WSW); label("$G$",G,NW);[/asy]
$\text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\circ\qquad\text{(E)}\ 90^\circ$
| As a strategy, think of how $\angle B + \angle D$ would be determined, particularly without determining either of the angles individually, since it may not be possible to determine $\angle B$ or $\angle D$ alone. If you see $\triangle BFD$ , then you can see that the problem is solved quickly after determining $\angle BFD$ .
But start with $\triangle AGF$ , since that's where most of our information is. Looking at $\triangle AGF$ , since $\angle AFG = \angle AGF$ , and $\angle A = 20$ , we can write:
$\angle A + \angle AFG + \angle AGF = 180$
$20 + 2\angle AFG = 180$
$\angle AFG = 80$
By noting that $\angle AFG$ and $\angle GFD$ make a straight line, we know
$\angle AFG + \angle GFD = 180$
$80 + \angle GFD = 180$
$\angle GFD = 100$
Ignoring all other parts of the figure and looking only at $\triangle BFD$ , you see that $\angle B + \angle D + \angle BFD = 180$ . But $\angle BFD$ is the same as $\angle GFD$ . Therefore:
$\angle B + \angle D + \angle GFD = 180$
$\angle B + \angle D + 100 = 180$
$\angle B + \angle D = 80^\circ$ , and the answer is thus $\boxed{D}$
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_24 | D |
AMC8_407 | Akash's birthday cake is in the form of a $4 \times 4 \times 4$ inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into $64$ smaller cubes, each measuring $1 \times 1 \times 1$ inch, as shown below. How many of the small pieces will have icing on exactly two sides?
[asy] /* Created by SirCalcsALot and sonone Code modfied from https://artofproblemsolving.com/community/c3114h2152994_the_old__aops_logo_with_asymptote */ import three; currentprojection=orthographic(1.75,7,2); //++++ edit colors, names are self-explainatory ++++ //pen top=rgb(27/255, 135/255, 212/255); //pen right=rgb(254/255,245/255,182/255); //pen left=rgb(153/255,200/255,99/255); pen top = rgb(170/255, 170/255, 170/255); pen left = rgb(81/255, 81/255, 81/255); pen right = rgb(165/255, 165/255, 165/255); pen edges=black; int max_side = 4; //+++++++++++++++++++++++++++++++++++++++ path3 leftface=(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle; path3 rightface=(0,1,0)--(1,1,0)--(1,1,1)--(0,1,1)--cycle; path3 topface=(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle; for(int i=0; i<max_side; ++i){ for(int j=0; j<max_side; ++j){ draw(shift(i,j,-1)*surface(topface),top); draw(shift(i,j,-1)*topface,edges); draw(shift(i,-1,j)*surface(rightface),right); draw(shift(i,-1,j)*rightface,edges); draw(shift(-1,j,i)*surface(leftface),left); draw(shift(-1,j,i)*leftface,edges); } } picture CUBE; draw(CUBE,surface(leftface),left,nolight); draw(CUBE,surface(rightface),right,nolight); draw(CUBE,surface(topface),top,nolight); draw(CUBE,topface,edges); draw(CUBE,leftface,edges); draw(CUBE,rightface,edges); // begin made by SirCalcsALot int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}}; for (int i = 0; i < max_side; ++i) { for (int j = 0; j < max_side; ++j) { for (int k = 0; k < min(heights[i][j], max_side); ++k) { add(shift(i,j,k)*CUBE); } } } [/asy]
$\textbf{(A) }12 \qquad \textbf{(B) }16 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24$
| Notice that, for a small cube which does not form part of the bottom face, it will have exactly $2$ faces with icing on them only if it is one of the $2$ center cubes of an edge of the larger cube. There are $12-4 = 8$ such edges (as we exclude the $4$ edges of the bottom face), so this case yields $2 \cdot 8 = 16$ small cubes. As for the bottom face, we can see that only the $4$ corner cubes have exactly $2$ faces with icing, so the total is $16+4 = \boxed{\textbf{(D) }20}$ .
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_9 | 20 |
AMC8_408 | A number $N$ is inserted into the list $2$ , $6$ , $7$ , $7$ , $28$ . The mean is now twice as great as the median. What is $N$ ?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 20\qquad \textbf{(D)}\ 28\qquad \textbf{(E)}\ 34$
| The median of the list is $7$ , so the mean of the new list will be $7 \cdot 2 = 14$ . Since there are $6$ numbers in the new list, the sum of the $6$ numbers will be $14 \cdot 6 = 84$ . Therefore, $2+6+7+7+28+N = 84 \implies N = \boxed{\text{(E)\ 34}}$
~Soupboy0
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_14 | 34 |
AMC8_409 | The first AMC $8$ was given in $1985$ and it has been given annually since that time. Samantha turned $12$ years old the year that she took the seventh AMC $8$ . In what year was Samantha born?
$\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983$
| The seventh AMC 8 would have been given in $1991$ . If Samantha was 12 then, that means she was born 12 years ago, so she was born in $1991-12=1979$ .
Our answer is $\boxed{(\text{A})1979}$
corrections made by DrDominic
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_10 | 1979 |
AMC8_410 | Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$
| Solution 1So, let there be $x$ pairs of $$1$ socks, $y$ pairs of $$3$ socks, and $z$ pairs of $$4$ socks.
We have $x+y+z=12$ , $x+3y+4z=24$ , and $x,y,z \ge 1$ .
Now, we subtract to find $2y+3z=12$ , and $y,z \ge 1$ .
It follows that $2y$ is a multiple of $3$ and $3z$ is a multiple of $3$ . Since sum of 2 multiples of 3 = multiple of 3, so we must have $2y=6$ .
Therefore, $y=3$ , and it follows that $z=2$ . Now, $x=12-y-z=12-3-2=\boxed{\textbf{(D)}~7}$ , as desired.
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_20 | 7 |
AMC8_411 | Jeff rotates spinners $P$ , $Q$ and $R$ and adds the resulting numbers. What is the probability that his sum is an odd number?
[asy] size(200); path circle=circle((0,0),2); path r=(0,0)--(0,2); draw(circle,linewidth(1)); draw(shift(5,0)*circle,linewidth(1)); draw(shift(10,0)*circle,linewidth(1)); draw(r,linewidth(1)); draw(rotate(120)*r,linewidth(1)); draw(rotate(240)*r,linewidth(1)); draw(shift(5,0)*r,linewidth(1)); draw(shift(5,0)*rotate(90)*r,linewidth(1)); draw(shift(5,0)*rotate(180)*r,linewidth(1)); draw(shift(5,0)*rotate(270)*r,linewidth(1)); draw(shift(10,0)*r,linewidth(1)); draw(shift(10,0)*rotate(60)*r,linewidth(1)); draw(shift(10,0)*rotate(120)*r,linewidth(1)); draw(shift(10,0)*rotate(180)*r,linewidth(1)); draw(shift(10,0)*rotate(240)*r,linewidth(1)); draw(shift(10,0)*rotate(300)*r,linewidth(1)); label("$P$", (-2,2)); label("$Q$", shift(5,0)*(-2,2)); label("$R$", shift(10,0)*(-2,2)); label("$1$", (-1,sqrt(2)/2)); label("$2$", (1,sqrt(2)/2)); label("$3$", (0,-1)); label("$2$", shift(5,0)*(-sqrt(2)/2,sqrt(2)/2)); label("$4$", shift(5,0)*(sqrt(2)/2,sqrt(2)/2)); label("$6$", shift(5,0)*(sqrt(2)/2,-sqrt(2)/2)); label("$8$", shift(5,0)*(-sqrt(2)/2,-sqrt(2)/2)); label("$1$", shift(10,0)*(-0.5,1)); label("$3$", shift(10,0)*(0.5,1)); label("$5$", shift(10,0)*(1,0)); label("$7$", shift(10,0)*(0.5,-1)); label("$9$", shift(10,0)*(-0.5,-1)); label("$11$", shift(10,0)*(-1,0)); [/asy]
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4}$
| In order for Jeff to have an odd number sum, the numbers must either be Odd + Odd + Odd or Even + Even + Odd. We easily notice that we cannot obtain Odd + Odd + Odd because spinner $Q$ contains only even numbers. Therefore we must work with Even + Even + Odd and spinner $Q$ will give us one of our even numbers. We also see that spinner $R$ only contains odd, so spinner $R$ must give us our odd number. We still need one even number from spinner $P$ . There is only 1 even number: $2$ . Since spinning the required numbers are automatic on the other spinners, we only have to find the probability of spinning a $2$ in spinner $P$ , which is clearly $\boxed{\textbf{(B)}\ \frac{1}{3}}$
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_17 | 1/3 |
AMC8_412 | What is the value of \[\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdots\frac{18}{20}\cdot\frac{19}{21}\cdot\frac{20}{22}?\]
$\textbf{(A) } \frac{1}{462} \qquad \textbf{(B) } \frac{1}{231} \qquad \textbf{(C) } \frac{1}{132} \qquad \textbf{(D) } \frac{2}{213} \qquad \textbf{(E) } \frac{1}{22}$
| Note that common factors (from $3$ to $20,$ inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes \[\frac{1}{\cancel{3}}\cdot\frac{2}{\cancel{4}}\cdot\frac{\cancel{3}}{\cancel{5}}\cdots\frac{\cancel{18}}{\cancel{20}}\cdot\frac{\cancel{19}}{21}\cdot\frac{\cancel{20}}{22}=\frac{1\cdot2}{21\cdot22}=\frac{1}{21\cdot11}=\boxed{\textbf{(B) } \frac{1}{231}}.\]
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_8 | 1/231 |
AMC8_413 | What is the minimum possible product of three different numbers of the set $\{-8,-6,-4,0,3,5,7\}$ ?
$\text{(A)}\ -336 \qquad \text{(B)}\ -280 \qquad \text{(C)}\ -210 \qquad \text{(D)}\ -192 \qquad \text{(E)}\ 0$
| The only way to get a negative product using three numbers is to multiply one negative number and two positives or three negatives. Only two reasonable choices
exist: $(-8)\times(-6)\times(-4) = (-8)\times(24) = -192$ and $(-8)\times5\times7 = (-8)\times35 = -280$ .
The latter is smaller, so $\boxed{\text{(B)\ -280}}$ .
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_7 | -280 |
AMC8_415 | There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?
$\textbf{(A)}\ 5724 \qquad \textbf{(B)}\ 7245 \qquad \textbf{(C)}\ 7254 \qquad \textbf{(D)}\ 7425 \qquad \textbf{(E)}\ 7542$
| There are only 5 options for the problem so we can just try them. It is easy since that we only need try to use $2$ , $3$ to divide them. Even $5$ will leads to a solution start with $1$ which we don't need.
$5724/2=2862$ , $5724/3=1908$ . $7245/3=2415$ . $7254/2=3627$ , $7254/3=2418$ . $7425/3=2475$ . The answer is $\boxed{\textbf{(D)}}$ . You can obtain the answer in only 6 calculations.
~ M4ST3R0FM4TH
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_25 | D |
AMC8_417 | A classroom has a row of 35 coat hooks. Paulina likes coats to be equally spaced, so that there is the same number of empty hooks before the first coat, after the last coat, and between every coat and the next one. Suppose there is at least 1 coat and at least 1 empty hook. How many different numbers of coats can satisfy Paulina's pattern?
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$
| Suppose there are $c$ coats on the rack. Notice that there are $c+1$ "gaps" formed by these coats, each of which must have the same number of empty spaces (say, $k$ ). Then the values $k$ and $c$ must satisfy $c+k(c+1)=35 \implies kc+k+c=35$ . We now use Simon's Favorite Factoring Trick as follows: \[kc+k+c=35\] \[\implies kc+k+c+1=36\] \[\implies (k+1)(c+1)=36\] Our only restrictions now are that $k>0 \implies k+1 > 1$ and $c>0 \implies c+1>1$ . Other than that, each factor pair of $36$ produces a valid solution $(k,c)$ , which in turn uniquely determines an arrangement. Since $36$ has $9$ factors, our answer is $9-2=\boxed{\textbf{(D)}~7}$ . ~cxsmi
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_22 | 7 |
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