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AMC8_208 | Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$ . In the figure below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of $h$ of $\triangle ABC$ ? (Diagram not drawn to scale.)
[asy] //Diagram by TheMathGuyd size(12cm); real h = 2.5; // height real g=4; //c2c space real s = 0.65; //Xcord of Hline real adj = 0.08; //adjust line diffs pair A,B,C; B=(0,h); C=(1,0); A=-conj(C); pair PONE=(s,h*(1-s)); //Endpoint of Hline ONE pair PTWO=(s+adj,h*(1-s-adj)); //Endpoint of Hline ONE path LONE=PONE--(-conj(PONE)); //Hline ONE path LTWO=PTWO--(-conj(PTWO)); path T=A--B--C--cycle; //Triangle fill (shift(g,0)*(LTWO--B--cycle),mediumgrey); fill (LONE--A--C--cycle,mediumgrey); draw(LONE); draw(T); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); draw(shift(g,0)*LTWO); draw(shift(g,0)*T); label("$A$",shift(g,0)*A,SW); label("$B$",shift(g,0)*B,N); label("$C$",shift(g,0)*C,SE); draw(B--shift(g,0)*B,dashed); draw(C--shift(g,0)*A,dashed); draw((g/2,0)--(g/2,h),dashed); draw((0,h*(1-s))--B,dashed); draw((g,h*(1-s-adj))--(g,0),dashed); label("$5$", midpoint((g,h*(1-s-adj))--(g,0)),UnFill); label("$h$", midpoint((g/2,0)--(g/2,h)),UnFill); label("$11$", midpoint((0,h*(1-s))--B),UnFill); [/asy]
$\textbf{(A) } 14.6 \qquad \textbf{(B) } 14.8 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 15.2 \qquad \textbf{(E) } 15.4$
| First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is $[ABC]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)$ . Similarly, we can find that the area of the gray part in the second triangle is $[ABC]\cdot\left(\tfrac{h-5}{h}\right)^2$ . These areas are equal, so $1-\left(\frac{11}{h}\right)^2=\left(\frac{h-5}{h}\right)^2$ . Simplifying yields $10h=146$ so $h=\boxed{\textbf{(A) }14.6}$ .
~MathFun1000 (~edits apex304)
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_24 | 14.6 |
AMC8_209 | What is the minimum number of small squares that must be colored black so that a line of symmetry lies on the diagonal $\overline{BD}$ of square $ABCD$ ?
[asy] defaultpen(linewidth(1)); for ( int x = 0; x < 5; ++x ) { draw((0,x)--(4,x)); draw((x,0)--(x,4)); } fill((1,0)--(2,0)--(2,1)--(1,1)--cycle); fill((0,3)--(1,3)--(1,4)--(0,4)--cycle); fill((2,3)--(4,3)--(4,4)--(2,4)--cycle); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle); label("$A$", (0, 4), NW); label("$B$", (4, 4), NE); label("$C$", (4, 0), SE); label("$D$", (0, 0), SW); [/asy]
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
| Rotating square $ABCD$ counterclockwise $45^\circ$ so that the line of symmetry $BD$ is a vertical line makes it easier to see that $\boxed{\textbf{(D)}\ 4}$ squares need to be colored to match its corresponding square.
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_3 | 4 |
AMC8_210 | Business is a little slow at Lou's Fine Shoes, so Lou decides to have a
sale. On Friday, Lou increases all of Thursday's prices by $10$ percent. Over the
weekend, Lou advertises the sale: "Ten percent off the listed price. Sale
starts Monday." How much does a pair of shoes cost on Monday that
cost $40$ dollars on Thursday?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 39.60\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 40.40\qquad\textbf{(E)}\ 44$
| On Friday, the shoes would cost $40 \cdot 1.1= 44$ dollars. Then on Monday, the shoes would cost $44- \frac{44}{10}=44-4.4=\boxed{\textbf{(B)}\ 39.60}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_11 | 39.60 |
AMC8_211 | Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, "My house number has two digits, and exactly three of the following four statements about it are true."
(1) It is prime.
(2) It is even.
(3) It is divisible by 7.
(4) One of its digits is 9.
This information allows Malcolm to determine Isabella's house number. What is its units digit?
$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$
| Notice that (1) cannot be true. Otherwise, the number would have to be prime and be either even or divisible by 7. This only happens if the number is 2 or 7, neither of which are two-digit numbers, so we run into a contradiction. Thus, we must have (2), (3), and (4) be true. By (2), the $2$ -digit number is even, and thus, the digit in the tens place must be $9$ . The only even $2$ -digit number starting with $9$ and divisible by $7$ is $98$ , which has a units digit of $\boxed{\textbf{(D)}\ 8}.$
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_8 | 8 |
AMC8_212 | The coordinates of $\triangle ABC$ are $A(5,7)$ , $B(11,7)$ , and $C(3,y)$ , with $y>7$ . The area of $\triangle ABC$ is 12. What is the value of $y$ ?
[asy] draw((3,11)--(11,7)--(5,7)--(3,11)); dot((5,7)); label("$A(5,7)$",(5,7),S); dot((11,7)); label("$B(11,7)$",(11,7),S); dot((3,11)); label("$C(3,y)$",(3,11),NW); [/asy]
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12$
| The triangle has base $6,$ which means its height satisfies
\[\dfrac{6h}{2}=3h=12.\]
This means that $h=4,$ so the answer is $7+4=\boxed{(D) 11}$
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_11 | 11 |
AMC8_213 | Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$
| The least multiple of 6 greater than 23 is 24. So she will need to add $24-23=\boxed{\textbf{(A)}\ 1}$ more model car.
~avamarora
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_1 | 1 |
AMC8_214 | Two cards are dealt from a deck of four red cards labeled $A$ , $B$ , $C$ , $D$ and four green cards labeled $A$ , $B$ , $C$ , $D$ . A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?
$\textbf{(A)}\ \frac{2}{7}\qquad\textbf{(B)}\ \frac{3}{8}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{4}{7}\qquad\textbf{(E)}\ \frac{5}{8}$
| Notice that, no matter which card you choose, there are exactly $4$ cards that either have the same color or letter as it. Since there are $7$ cards left to choose from, the probability is $\boxed{\textbf{(D)}\frac47}$ .
-theepiccarrot7
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_21 | 4/7 |
AMC8_215 | A birdbath is designed to overflow so that it will be self-cleaning. Water flows in at the rate of 20 milliliters per minute and drains at the rate of 18 milliliters per minute. Which one of these graphs shows the volume of water in the birdbath during the filling time and continuing into the overflow time?
[asy] size(450); defaultpen(linewidth(0.8)); path[] p={origin--(8,8)--(14,8), (0,10)--(4,10)--(14,0), origin--(14,14), (0,14)--(14,14), origin--(7,7)--(14,0)}; int i; for(i=0; i<5; i=i+1) { draw(shift(21i,0)*((0,16)--origin--(14,0))); draw(shift(21i,0)*(p[i])); label("Time", (7+21i,0), S); label(rotate(90)*"Volume", (21i,8), W); } label("$A$", (0*21 + 7,-5), S); label("$B$", (1*21 + 7,-5), S); label("$C$", (2*21 + 7,-5), S); label("$D$", (3*21 + 7,-5), S); label("$E$", (4*21 + 7,-5), S); [/asy]
$\text{(A)}\ \text{A} \qquad \text{(B)}\ \text{B} \qquad \text{(C)}\ \text{C} \qquad \text{(D)}\ \text{D} \qquad \text{(E)}\ \text{E}$
| The change in the water volume has a net gain of $20-18=2$ millimeters per minute. The birdbath's volume increases at a constant rate until it reaches its maximum and starts overflowing to keep a constant volume. This is best represented by graph $\boxed{\text{(A)}\ A}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_6 | A |
AMC8_216 | What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?
$\textbf{(A)}\hspace{.05in}3127\qquad\textbf{(B)}\hspace{.05in}3133\qquad\textbf{(C)}\hspace{.05in}3137\qquad\textbf{(D)}\hspace{.05in}3139\qquad\textbf{(E)}\hspace{.05in}3149$
| The problem states that the answer cannot be a perfect square or have prime factors less than $50$ . Therefore, the answer will be the product of at least two different primes greater than $50$ . The two smallest primes greater than $50$ are $53$ and $59$ . Multiplying these two primes, we obtain the number $3127$ , which is also the smallest number on the list of answer choices.
So we are done, and the answer is $\boxed{\textbf{(A)}\ 3127}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_18 | 3127 |
AMC8_217 | Viswam walks half a mile to get to school each day. His route consists of $10$ city blocks of equal length and he takes $1$ minute to walk each block. Today, after walking $5$ blocks, Viswam discovers he has to make a detour, walking $3$ blocks of equal length instead of $1$ block to reach the next corner. From the time he starts his detour, at what speed, in mph, must he walk, in order to get to school at his usual time?
[asy] // Diagram by TheMathGuyd size(13cm); // this is an important stickman to the left of the origin pair C=midpoint((-0.5,0.5)--(-0.6,0.05)); draw((-0.5,0.5)--(-0.6,0.05)); // Head to butt draw((-0.64,0.16)--(-0.7,0.2)--C--(-0.47,0.2)--(-0.4,0.22)); // LH-C-RH draw((-0.6,0.05)--(-0.55,-0.1)--(-0.57,-0.25)); draw((-0.6,0.05)--(-0.68,-0.12)--(-0.8,-0.20)); filldraw(circle((-0.5,0.5),0.1),white,black); int i; real d,s; // gap and side d=0.2; s=1-2*d; for(i=0; i<10; i=i+1) { //dot((i,0), red); //marks to start filldraw((i+d,d)--(i+1-d,d)--(i+1-d,1-d)--(i+d,1-d)--cycle, lightgrey, black); filldraw(conj((i+d,d))--conj((i+1-d,d))--conj((i+1-d,1-d))--conj((i+d,1-d))--cycle,lightgrey,black); } fill((5+d,-d/2)--(6-d,-d/2)--(6-d,d/2)--(5+d,d/2)--cycle,lightred); draw((0,0)--(5,0)--(5,1)--(6,1)--(6,0)--(10.1,0),deepblue+linewidth(1.25)); //Who even noticed label("School", (10,0),E, Draw()); [/asy]
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4.2 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 4.8 \qquad \textbf{(E)}\ 5$
| Note that Viswam walks at a constant speed of $60$ blocks per hour as he takes $1$ minute to walk each block. After walking $5$ blocks, he has taken $5$ minutes and has $5$ minutes remaining to walk $7$ blocks. Therefore, he must walk at a speed of $7 \cdot 60 \div 5 = 84$ blocks per hour to get to school on time, from the time he starts his detour. Since he normally walks $\frac{1}{2}$ mile, which is equal to $10$ blocks, $1$ mile is equal to $20$ blocks. Therefore, he must walk at $84 \div 20 = 4.2$ mph from the time he starts his detour to get to school on time, so the answer is $\boxed{\textbf{(B)}\ 4.2}$ .
~pianoboy (Edits by ILoveMath31415926535 , apex304 and MrThinker)
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_15 | 4.2 |
AMC8_219 | Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers?
[asy] path card=((0,0)--(0,3)--(2,3)--(2,0)--cycle); draw(card, linewidth(1)); draw(shift(2.5,0)*card, linewidth(1)); draw(shift(5,0)*card, linewidth(1)); label("$44$", (1,1.5)); label("$59$", shift(2.5,0)*(1,1.5)); label("$38$", shift(5,0)*(1,1.5));[/asy]
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 17$
| Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number(common sum) would be to add another even number to 44, and a different one to 38. Since there is only one even prime (2), the middle card's hidden number cannot be an odd prime, and so must be even. Therefore, the middle card's hidden number must be 2, so the constant sum is $59+2=61$ . Thus, the first card's hidden number is $61-44=17$ , and the last card's hidden number is $61-38=23$ .
Since the sum of the hidden primes is $2+17+23=42$ , the average of the primes is $\dfrac{42}{3}=\boxed{\textbf{(B)} 14}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_25 | 14 |
AMC8_220 | The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$ ?
$\textbf{(A)}\hspace{.05in}5\qquad\textbf{(B)}\hspace{.05in}6\qquad\textbf{(C)}\hspace{.05in}7\qquad\textbf{(D)}\hspace{.05in}11\qquad\textbf{(E)}\hspace{.05in}12$
| We can eliminate answer choices ${\textbf{(A)}\ 5}$ and ${\textbf{(C)}\ 7}$ , because of the above statement. Now we need to test the remaining answer choices.
Case 1: $x = 6$
Mode: $6$
Median: $6$
Mean: $\frac{37}{7}$
Since the mean does not equal the median or mode, ${\textbf{(B)}\ 6}$ can also be eliminated.
Case 2: $x = 11$
Mode: $6$
Median: $6$
Mean: $6$
We are done with this problem, because we have found when $x = 11$ , the condition is satisfied. Therefore, the answer is $\boxed{{\textbf{(D)}\ 11}}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_11 | 11 |
AMC8_221 | The average age of the 40 members of a computer science camp is 17 years. There are 20 girls, 15 boys, and 5 adults. If the average age of the girls is 15 and the average age of the boys is 16, what is the average age of the adults?
$\text{(A)}\ 26 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 29 \qquad \text{(E)}\ 30$
| First, find the total amount of the girl's ages and add it to the total amount of the boy's ages. It equals $(20)(15)+(15)(16)=540$ . The total amount of everyone's ages can be found from the average age, $17\cdot40=680$ . Then you do $680-540=140$ to find the sum of the adult's ages. The average age of an adult is divided among the five of them, $140\div5=\boxed{\text{(C)}\ 28}$ .
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_13 | 28 |
AMC8_222 | Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 12 miles per hour. Brian leaves Marquette at 9:00 AM heading for Escanaba on his bike. He bikes at a uniform rate of 16 miles per hour. They both bike on the same 62-mile route between Escanaba and Marquette. At what time in the morning do they meet?
$\textbf{(A)}\ 10: 00\qquad\textbf{(B)}\ 10: 15\qquad\textbf{(C)}\ 10: 30\qquad\textbf{(D)}\ 11: 00\qquad\textbf{(E)}\ 11: 30$
| If Cassie leaves $\frac{1}{2}$ an hour earlier then Brian, when Brian starts, the distance between them will be $62-\frac{12}{2}=56$ . Every hour, they will get $12+16=28$ miles closer. $\frac{56}{28}=2$ , so 2 hours from $9:00$ AM is when they meet, which is $\boxed{\textbf{(D)}\ 11: 00}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_13 | 11:00 |
AMC8_223 | There are four more girls than boys in Ms. Raub's class of $28$ students. What is the ratio of number of girls to the number of boys in her class?
$\textbf{(A) }3 : 4\qquad\textbf{(B) }4 : 3\qquad\textbf{(C) }3 : 2\qquad\textbf{(D) }7 : 4\qquad \textbf{(E) }2 : 1$
| Let $g$ being the number of girls in the class. The number of boys in the class is equal to $g-4$ . Since the total number of students is equal to $28$ , we get $g+g-4=28$ . Solving this equation, we get $g=16$ . There are $16-4=12$ boys in our class, and our answer is $16:12=\boxed{\textbf{(B)}~4:3}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_7 | 4:3 |
AMC8_224 | What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$
| Imagine the clock as a circle. The minute hand will be at the 4 at 20 minutes past the hour. The central angle formed between $4$ and $5$ is $30$ degrees (since it is 1/12 of a full circle, 360). By $4:20$ , the hour hand would have moved $\frac{1}{3}$ way from 4 to 5 since $\frac{20}{60}$ is reducible to $\frac{1}{3}$ . One third of the way from 4 to 5 is one third of 30 degrees, which is 10 degrees past the 4. Recall that the minute hand is at the 4, so the angle between them is $\boxed{(D) 10}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_20 | 10 |
AMC8_225 | In triangle $CAT$ , we have $\angle ACT =\angle ATC$ and $\angle CAT = 36^\circ$ . If $\overline{TR}$ bisects $\angle ATC$ , then $\angle CRT =$
[asy] pair A,C,T,R; C = (0,0); T = (2,0); A = (1,sqrt(5+sqrt(20))); R = (3/2 - sqrt(5)/2,1.175570); draw(C--A--T--cycle); draw(T--R); label("$A$",A,N); label("$T$",T,SE); label("$C$",C,SW); label("$R$",R,NW);[/asy]
$\text{(A)}\ 36^\circ\qquad\text{(B)}\ 54^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 90^\circ\qquad\text{(E)}\ 108^\circ$
| In $\triangle ACT$ , the three angles sum to $180^\circ$ , and $\angle C = \angle T$
$\angle CAT + \angle ATC + \angle ACT = 180$
$36 + \angle ATC + \angle ATC = 180$
$2 \angle ATC = 144$
$\angle ATC = 72$
Since $\angle ATC$ is bisected by $\overline{TR}$ , $\angle RTC = \frac{72}{2} = 36$
Now focusing on the smaller $\triangle RTC$ , the sum of the angles in that triangle is $180^\circ$ , so:
$\angle RTC + \angle TCR + \angle CRT = 180$
$36 + \angle ACT + \angle CRT = 180$
$36 + \angle ATC + \angle CRT = 180$
$36 + 72 + \angle CRT = 180$
$\angle CRT = 72^\circ$ , giving the answer $\boxed{C}$
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_13 | C |
AMC8_226 | One day the Beverage Barn sold $252$ cans of soda to $100$ customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?
$\textbf{(A) }2.5\qquad\textbf{(B) }3.0\qquad\textbf{(C) }3.5\qquad\textbf{(D) }4.0\qquad \textbf{(E) }4.5$
| In order to maximize the median, we need to make the first half of the numbers as small as possible. Since there are $100$ people, the median will be the average of the $50\text{th}$ and $51\text{st}$ largest amount of cans per person. To minimize the first $49$ , they would each have one can. Subtracting these $49$ cans from the $252$ cans gives us $203$ cans left to divide among $51$ people. Taking $\frac{203}{51}$ gives us $3$ and a remainder of $50$ . Seeing this, the largest number of cans the $50$ th person could have is $3$ , which leaves $4$ to the rest of the people. The average of $3$ and $4$ is $3.5$ . Thus our answer is $\boxed{\text{(C) }3.5}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_24 | 3.5 |
AMC8_227 | How many non-congruent triangles have vertices at three of the eight points in the array shown below?
[asy]dot((0,0)); dot((.5,.5)); dot((.5,0)); dot((.0,.5)); dot((1,0)); dot((1,.5)); dot((1.5,0)); dot((1.5,.5));[/asy]
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$
| Assume the base of the triangle is on the bottom four points because a congruent triangle can be made by reflecting the base on the top four points. For a triangle with a base of length $1$ , there are $3$ triangles. For a triangle with a base of length $2$ , there are $3$ triangles. For length $3$ , there are $2$ . In total, the number of non-congruent triangles is $3+3+2=\boxed{\textbf{(D)}\ 8}$ .
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_20 | 8 |
AMC8_228 | Margie's car can go $32$ miles on a gallon of gas, and gas currently costs $ $4$ per gallon. How many miles can Margie drive on $\textdollar 20$ worth of gas?
$\textbf{(A) }64\qquad\textbf{(B) }128\qquad\textbf{(C) }160\qquad\textbf{(D) }320\qquad \textbf{(E) }640$
| Margie can afford $20/4=5$ gallons of gas. She can go $32\cdot5=\boxed{\textbf{(C)}~160}$ miles on this amount of gas.
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_5 | 160 |
AMC8_229 | As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction $1/2$ mile in front of her. After she passes him, she can see him in her rear mirror until he is $1/2$ mile behind her. Emily rides at a constant rate of $12$ miles per hour, and Emerson skates at a constant rate of $8$ miles per hour. For how many minutes can Emily see Emerson?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16$
| Because they are both moving in the same direction, Emily is riding relative to Emerson $12-8=4$ mph. Now we can look at it as if Emerson is not moving at all [on his skateboard] and Emily is riding at $4$ mph. It takes her
\[\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}\]
to ride the $1/2$ mile to reach him, and then the same amount of time to be $1/2$ mile ahead of him. This totals to
\[2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{\textbf{(D)}\ 15}\ \text{minutes}\]
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_8 | 15 |
AMC8_230 | If $n$ is an even positive integer, the $\emph{double factorial}$ notation $n!!$ represents the product of all the even integers from $2$ to $n$ . For example, $8!! = 2 \cdot 4 \cdot 6 \cdot 8$ . What is the units digit of the following sum? \[2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!\]
$\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$
| Notice that once $n>8,$ the units digit of $n!!$ will be $0$ because there will be a factor of $10.$ Thus, we only need to calculate the units digit of \[2!!+4!!+6!!+8!! = 2+8+48+48\cdot8.\] We only care about units digits, so we have $2+8+8+8\cdot8,$ which has the same units digit as $2+8+8+4.$ The answer is $\boxed{\textbf{(B) } 2}.$
~wamofan
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_17 | 2 |
AMC8_231 | Students guess that Norb's age is $24, 28, 30, 32, 36, 38, 41, 44, 47$ , and $49$ . Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?
$\textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48$
| If at least half the guesses are too low, so Norb's age must be greater than $36$ .
If two of the guesses are off by one, then his age is in between two guesses whose difference is $2$ . It could be $31,37,$ or $48,$ but because his age is greater than $36$ it can only be $37$ or $48$ .
Lastly, Norb's age is a prime number so the answer must be $\boxed{\textbf{(C)}\ 37}$ .
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_21 | 37 |
AMC8_232 | Suppose $d$ is a digit. For how many values of $d$ is $2.00d5 > 2.005$ ?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 10$
| We see that $2.0055$ works but $2.0045$ does not. The digit $d$ can be from $5$ through $9$ , which is $\boxed{\textbf{(C)}\ 5}$ values.
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_6 | 5 |
AMC8_233 | An aquarium has a rectangular base that measures $100$ cm by $40$ cm and has a height of $50$ cm. The aquarium is filled with water to a depth of $37$ cm. A rock with volume $1000\text{cm}^3$ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
$\textbf{(A)}\ 0.25\qquad\textbf{(B)}\ 0.5\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 1.25\qquad\textbf{(E)}\ 2.5$
| The water level will rise $1$ cm for every $100 \cdot 40 = 4000\text{cm}^2$ . Since $1000$ is $\frac{1}{4}$ of $4000$ , the water will rise $\frac{1}{4}\cdot1 = \boxed{\textbf{(A)}\ 0.25}$
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_21 | 0.25 |
AMC8_234 | A chess king is said to attack all the squares one step away from it, horizontally, vertically, or diagonally. For instance, a king on the center square of a $3$ x $3$ grid attacks all $8$ other squares, as shown below. Suppose a white king and a black king are placed on different squares of a $3$ x $3$ grid so that they do not attack each other (in other words, not right next to each other). In how many ways can this be done?
[asy] /* AMC8 P17 2024, revised by Teacher David */ unitsize(29pt); import math; add(grid(3,3)); pair [] a = {(0.5,0.5), (0.5, 1.5), (0.5, 2.5), (1.5, 2.5), (2.5,2.5), (2.5,1.5), (2.5,0.5), (1.5,0.5)}; for (int i=0; i<a.length; ++i) { pair x = (1.5,1.5) + 0.4*dir(225-45*i); draw(x -- a[i], arrow=EndArrow()); } label("$K$", (1.5,1.5)); [/asy]
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 27 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 32$
~Diagram by Andrei.martynau
| If you place a king in any of the $4$ corners, the other king will have $5$ spots to go and there are $4$ corners, so $5 \times 4=20$ .
If you place a king in any of the $4$ edges, the other king will have $3$ spots to go and there are $4$ edges so $3 \times 4=12$ .
That gives us $20+12=32$ spots for the other king to go into in total.
So $\boxed{\textbf{(E)} 32}$ is the answer.
~andliu766, captaindiamond868 (minor edits)
~AliceDubbleYou
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_17 | 32 |
AMC8_235 | Starting with some gold coins and some empty treasure chests, I tried to put $9$ gold coins in each treasure chest, but that left $2$ treasure chests empty. So instead I put $6$ gold coins in each treasure chest, but then I had $3$ gold coins left over. How many gold coins did I have?
$\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(D) }63\qquad\textbf{(E) }81$
| We can represent the amount of gold with $g$ and the amount of chests with $c$ . We can use the problem to make the following equations:
\[9c-18 = g\]
\[6c+3 = g\]
We do this because for 9 chests there are 2 empty and if 9 were in each 9 multiplied by 2 is 18 left.
Therefore, $6c+3 = 9c-18.$ This implies that $c = 7.$ We therefore have $g = 45.$ So, our answer is $\boxed{\textbf{(C)}\ 45}$ .
~CHECKMATE2021
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_17 | 45 |
AMC8_236 | Figure 1 is called a "stack map." The numbers tell how many cubes are stacked in each position. Fig. 2 shows these cubes, and Fig. 3 shows the view of the stacked cubes as seen from the front.
Which of the following is the front view for the stack map in Fig. 4?
[asy] unitsize(24); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((1,0)--(1,2)); draw((0,1)--(2,1)); draw((5,0)--(7,0)--(7,1)--(20/3,4/3)--(20/3,13/3)--(19/3,14/3)--(16/3,14/3)--(16/3,11/3)--(13/3,11/3)--(13/3,2/3)--cycle); draw((20/3,13/3)--(17/3,13/3)--(17/3,10/3)--(14/3,10/3)--(14/3,1/3)); draw((20/3,10/3)--(17/3,10/3)--(17/3,7/3)--(20/3,7/3)); draw((17/3,7/3)--(14/3,7/3)); draw((7,1)--(6,1)--(6,2)--(5,2)--(5,0)); draw((5,1)--(6,1)--(6,0)); draw((20/3,4/3)--(6,4/3)); draw((17/3,13/3)--(16/3,14/3)); draw((17/3,10/3)--(16/3,11/3)); draw((14/3,10/3)--(13/3,11/3)); draw((5,2)--(13/3,8/3)); draw((5,1)--(13/3,5/3)); draw((6,2)--(17/3,7/3)); draw((9,0)--(11,0)--(11,4)--(10,4)--(10,3)--(9,3)--cycle); draw((11,3)--(10,3)--(10,0)); draw((11,2)--(9,2)); draw((11,1)--(9,1)); draw((13,0)--(16,0)--(16,2)--(13,2)--cycle); draw((13,1)--(16,1)); draw((14,0)--(14,2)); draw((15,0)--(15,2)); label("Figure 1",(1,0),S); label("Figure 2",(17/3,0),S); label("Figure 3",(10,0),S); label("Figure 4",(14.5,0),S); label("$1$",(1.5,.2),N); label("$2$",(.5,.2),N); label("$3$",(.5,1.2),N); label("$4$",(1.5,1.2),N); label("$1$",(13.5,.2),N); label("$3$",(14.5,.2),N); label("$1$",(15.5,.2),N); label("$2$",(13.5,1.2),N); label("$2$",(14.5,1.2),N); label("$4$",(15.5,1.2),N); [/asy]
[asy] unitsize(18); draw((0,0)--(3,0)--(3,2)--(1,2)--(1,4)--(0,4)--cycle); draw((0,3)--(1,3)); draw((0,2)--(1,2)--(1,0)); draw((0,1)--(3,1)); draw((2,0)--(2,2)); draw((5,0)--(8,0)--(8,4)--(7,4)--(7,3)--(6,3)--(6,2)--(5,2)--cycle); draw((8,3)--(7,3)--(7,0)); draw((8,2)--(6,2)--(6,0)); draw((8,1)--(5,1)); draw((10,0)--(12,0)--(12,4)--(11,4)--(11,3)--(10,3)--cycle); draw((12,3)--(11,3)--(11,0)); draw((12,2)--(10,2)); draw((12,1)--(10,1)); draw((14,0)--(17,0)--(17,4)--(16,4)--(16,2)--(14,2)--cycle); draw((17,3)--(16,3)); draw((17,2)--(16,2)--(16,0)); draw((17,1)--(14,1)); draw((15,0)--(15,2)); draw((19,0)--(22,0)--(22,4)--(20,4)--(20,1)--(19,1)--cycle); draw((22,3)--(20,3)); draw((22,2)--(20,2)); draw((22,1)--(20,1)--(20,0)); draw((21,0)--(21,4)); label("(A)",(1.5,0),S); label("(B)",(6.5,0),S); label("(C)",(11,0),S); label("(D)",(15.5,0),S); label("(E)",(20.5,0),S); [/asy]
| The third view is a direct, head-on view of the cubes. Thus, you will only see the highest (or, in these cases, higher) tower in each up-down column. For figure $4$ :
The highest tower in the first up-down column is $2$ in the upper-left box.
The highest tower in the second up-down column is the $3$ in the lower-middle box.
The highest tower in the third up-down column is the $4$ in the upper-right box.
Thus, the head-on view of this tower should have $2$ boxes on the left, $3$ in the middle, and $4$ on the right.
Diagram $\boxed{B}$ shows this description.
As trivia, option C shows the stack from the point of view of an observer on the right, facing towards the left.
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_20 | B |
AMC8_237 | What is the area of the triangle formed by the lines $y=5$ , $y=1+x$ , and $y=1-x$ ?
$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$
| First, we need to find the coordinates where the graphs intersect.
We want the points x and y to be the same. Thus, we set $5=x+1,$ and get $x=4.$ Plugging this into the equation, $y=1-x,$
$y=5$ , and $y=1+x$ intersect at $(4,5)$ , we call this line x.
Doing the same thing, we get $x=-4.$ Thus, $y=5$ . Also,
$y=5$ and $y=1-x$ intersect at $(-4,5)$ , and we call this line y.
It's apparent the only solution to $1-x=1+x$ is $0.$ Thus, $y=1.$
$y=1-x$ and $y=1+x$ intersect at $(0,1)$ , we call this line z.
Using the Shoelace Theorem we get: \[\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}\] $=$ So, our answer is $\boxed{\textbf{(E)}\ 16.}$
We might also see that the lines $y$ and $x$ are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by $-1$ to get the other. As the base is horizontal, this is an isosceles triangle with base 8, as the intersection points have a distance of 8. The height is $5-1=4,$ so $\frac{4\cdot 8}{2} = \boxed{\textbf{(E)} 16.}$
~Edits by WrenMath
Warning: Do not use the distance formula for the base then use Heron's formula. It will take you half of the time you have left!
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_21 | 16 |
AMC8_238 | In a town of $351$ adults, every adult owns a car, motorcycle, or both. If $331$ adults own cars and $45$ adults own motorcycles, how many of the car owners do not own a motorcycle?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 306 \qquad \textbf{(E)}\ 351$
| By PIE , the number of adults who own both cars and motorcycles is $331+45-351=25.$ Out of the $331$ car owners, $25$ of them own motorcycles and $331-25=\boxed{\textbf{(D)}\ 306}$ of them don't.
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_6 | 306 |
AMC8_239 | Big Al, the ape, ate 100 bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many bananas did Big Al eat on May 5?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$
| There are $5$ days from May 1 to May 5. The number of bananas he eats each day is an arithmetic sequence. He eats $n$ bananas on May 5, and $n-4(6)=n-24$ bananas on May 1. The sum of this arithmetic sequence is equal to $100$ .
\begin{align*} \frac{n+n-24}{2} \cdot 5 &= 100\\ n-12&=20\\ n&=\boxed{\textbf{(D)}\ 32} \end{align*}
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_12 | 32 |
AMC8_241 | A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$ . Each team plays a $76$ game schedule. How many games does a team play within its own division?
$\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$
| On one team they play $3N$ games in their division and $4M$ games in the other. This gives $3N+4M=76$ .
Since $M>4$ we start by trying M=5. This doesn't work because $56$ is not divisible by $3$ .
Next, $M=6$ does not work because $52$ is not divisible by $3$ .
We try $M=7$ does work by giving $N=16$ , $~M=7$ and thus $3\times 16=\boxed{\textbf{(B)}~48}$ games in their division.
$M=10$ seems to work, until we realize this gives $N=12$ , but $N>2M$ so this will not work.
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_24 | 48 |
AMC8_242 | You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $ $1.02$ , with at least one coin of each type. How many dimes must you have?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$
| Since you have one coin of each type, $1 + 5 + 10 + 25 = 41$ cents are already determined, leaving you with a total of $102 - 41 = 61$ cents remaining for $5$ coins.
You must have $1$ more penny. If you had more than $1$ penny, you must have at least $6$ pennies to leave a multiple of $5$ for the nickels, dimes, and quarters. But you only have $5$ more coins to assign.
Now you have $61 - 1 = 60$ cents remaining for $4$ coins, which may be nickels, quarters, or dimes. If you have only one more dime, that leaves $50$ cents in $3$ nickels or quarters, which is impossible. If you have two dimes, that leaves $40$ cents for $2$ nickels or quarters, which is again impossible. If you have three dimes, that leaves $30$ cents for $1$ nickel or quarter, which is still impossible. And all four remaining coins being dimes will not be enough.
Therefore, you must have no more dimes to assign, and the $60$ cents in $4$ coins must be divided between the quarters and nickels. We quickly see that $2$ nickels and $2$ quarters work. Thus, the total count is $2$ quarters, $2$ nickels, $1$ penny, plus one more coin of each type that we originally subtracted. Double-checking, that gives a total $2 + 2 + 1 + 4 = 9$ coins, and a total of $2\cdot 25 + 2\cdot 5 + 1 + (1 + 5 + 10 + 25) = 102$ cents.
There is only $1$ dime in that combo, so the answer is $\boxed{A}$ .
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_20 | A |
AMC8_243 | A $3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$ , $1\times4$ , and $1\times1$ , shown below. What is the minimum possible number of $1\times1$ tiles used?
| We can eliminate B, C, and D, because they are not $21$ subtracted by any multiple of $4$ . Finally, we see that there is no way to have A, so the solution is $\boxed{\textbf{(E)\ 5}}$ .
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_7 | 5 |
AMC8_244 | The students in Mrs. Sawyer's class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy E?
[asy] real[] r={6, 8, 4, 2, 5}; int i; for(i=0; i<5; i=i+1) { filldraw((4i,0)--(4i+3,0)--(4i+3,2r[i])--(4i,2r[i])--cycle, black, black); } draw(origin--(19,0)--(19,16)--(0,16)--cycle, linewidth(0.9)); for(i=1; i<8; i=i+1) { draw((0,2i)--(19,2i)); } label("$0$", (0,2*0), W); label("$1$", (0,2*1), W); label("$2$", (0,2*2), W); label("$3$", (0,2*3), W); label("$4$", (0,2*4), W); label("$5$", (0,2*5), W); label("$6$", (0,2*6), W); label("$7$", (0,2*7), W); label("$8$", (0,2*8), W); label("$A$", (0*4+1.5, 0), S); label("$B$", (1*4+1.5, 0), S); label("$C$", (2*4+1.5, 0), S); label("$D$", (3*4+1.5, 0), S); label("$E$", (4*4+1.5, 0), S); label("SWEET TOOTH", (9.5,18), N); label("Kinds of candy", (9.5,-2), S); label(rotate(90)*"Number of students", (-2,8), W);[/asy]
$\text{(A)}\ 5 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$
| From the bar graph, we can see that $5$ students chose candy E. There are $6+8+4+2+5=25$ total students in Mrs. Sawyers class. The percent that chose E is $\frac{5}{25} \cdot 100 = \boxed{\text{(E)}\ 20}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_7 | 20 |
AMC8_245 | The area of polygon $ABCDEF$ is 52 with $AB=8$ , $BC=9$ and $FA=5$ . What is $DE+EF$ ?
[asy] pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4); draw(a--b--c--d--e--f--cycle); draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a); draw(shift(-.25,0)*b--shift(-.25,-.25)*b--shift(0,-.25)*b); draw(shift(-.25,0)*c--shift(-.25,.25)*c--shift(0,.25)*c); draw(shift(.25,0)*d--shift(.25,.25)*d--shift(0,.25)*d); draw(shift(.25,0)*f--shift(.25,.25)*f--shift(0,.25)*f); label("$A$", a, NW); label("$B$", b, NE); label("$C$", c, SE); label("$D$", d, SW); label("$E$", e, SW); label("$F$", f, SW); label("5", (0,6.5), W); label("8", (4,9), N); label("9", (8, 4.5), E); [/asy]
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$
| Notice that $AF + DE = BC$ , so $DE=4$ . Let $O$ be the intersection of the extensions of $AF$ and $DC$ , which makes rectangle $ABCO$ . The area of the polygon is the area of $FEDO$ subtracted from the area of $ABCO$ .
\[\text{Area} = 52 = 8 \cdot 9- EF \cdot 4\]
Solving for the unknown, $EF=5$ , therefore $DE+EF=4+5=\boxed{\textbf{(C)}\ 9}$ .
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_13 | 9 |
AMC8_246 | How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108$
| We can separate this into two cases. If an integer is a multiple of $5,$ the last digit must be either $0$ or $5.$
Case 1: The last digit is $5.$ The leading digit can be $1,2,3,$ or $4.$ Because the second digit can be $0$ but not the leading digit, there are also $4$ choices. The third digit cannot be the leading digit or the second digit, so there are $3$ choices. The number of integers is this case is $4\cdot4\cdot3\cdot1=48.$
Case 2: The last digit is $0.$ Because $5$ is the largest digit, one of the remaining three digits must be $5.$ There are $3$ ways to choose which digit should be $5.$ The remaining digits can be $1,2,3,$ or $4,$ but since they have to be different there are $4\cdot3$ ways to choose. The number of integers in this case is $1\cdot3\cdot4\cdot3=36.$
Therefore, the answer is $48+36=\boxed{\textbf{(D)}\ 84}$ .
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_23 | 84 |
AMC8_247 | Four students take an exam. Three of their scores are $70, 80,$ and $90$ . If the average of their four scores is $70$ , then what is the remaining score?
$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$
| Solution 1Let $r$ be the remaining student's score. We know that the average, 70, is equal to $\frac{70 + 80 + 90 + r}{4}$ . We can use basic algebra to solve for $r$ : \[\frac{70 + 80 + 90 + r}{4} = 70\] \[\frac{240 + r}{4} = 70\] \[240 + r = 280\] \[r = 40\] giving us the answer of $\boxed{\textbf{(A)}\ 40}$ .
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_3 | 40 |
AMC8_248 | On February 13 The Oshkosh Northwester listed the length of daylight as 10 hours and 24 minutes, the sunrise was $6:57\textsc{am}$ , and the sunset as $8:15\textsc{pm}$ . The length of daylight and sunrise were correct, but the sunset was wrong. When did the sun really set?
$\textbf{(A)}\hspace{.05in}5:10\textsc{pm}\quad\textbf{(B)}\hspace{.05in}5:21\textsc{pm}\quad\textbf{(C)}\hspace{.05in}5:41\textsc{pm}\quad\textbf{(D)}\hspace{.05in}5:57\textsc{pm}\quad\textbf{(E)}\hspace{.05in}6:03\textsc{pm}$
| The problem wants us to find the time of sunset and gives us the length of daylight and time of sunrise. So all we have to do is add the length of daylight to the time of sunrise to obtain the answer. Convert 10 hours and 24 minutes into $10:24$ in order to add easier.
Adding, we find that the time of sunset is $6:57\textsc{am} + 10:24 \implies 17:21 \implies \boxed{\textbf{(B)}\ 5:21\textsc{pm}}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_3 | 5:21pm |
AMC8_249 | Two congruent circles centered at points $A$ and $B$ each pass through the other circle's center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D$ . The circles intersect at two points, one of which is $E$ . What is the degree measure of $\angle CED$ ?
$\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150$
| Observe that $\triangle{EAB}$ is equilateral (all are radii of congruent circles). Therefore, $m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}$ . Since $CD$ is a straight line, we conclude that $m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}$ . Since $BE=BD$ (both are radii of the same circle), $\triangle{BED}$ is isosceles, meaning that $m\angle{BED}=m\angle{BDE}=30^{\circ}$ . Similarly, $m\angle{AEC}=m\angle{ACE}=30^{\circ}$ .
Now, $\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}$ . Therefore, the answer is $\boxed{\textbf{(C) }\ 120}$ .
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_23 | 120 |
AMC8_250 | To complete the grid below, each of the digits 1 through 4 must occur once
in each row and once in each column. What number will occupy the lower
right-hand square?
\[\begin{tabular}{|c|c|c|c|}\hline 1 & & 2 &\\ \hline 2 & 3 & &\\ \hline & &&4\\ \hline & &&\\ \hline\end{tabular}\]
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad\textbf{(E)}\ \text{cannot be determined}$
| The number in the first row, last column must be a $3$ due to the fact if a $3$ was in the first row, second column, there would be two threes in that column. By the same reasoning, the number in the second row, last column has to be a $1$ . Therefore the number in the lower right-hand square is $\boxed{\textbf{(B)}\ 2}$ .
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_9 | 2 |
AMC8_251 | An athlete's target heart rate, in beats per minute, is $80\%$ of the theoretical maximum heart rate. The maximum heart rate is found by subtracting the athlete's age, in years, from $220$ . To the nearest whole number, what is the target heart rate of an athlete who is $26$ years old?
$\textbf{(A)}\ 134\qquad\textbf{(B)}\ 155\qquad\textbf{(C)}\ 176\qquad\textbf{(D)}\ 194\qquad\textbf{(E)}\ 243$
| The maximum heart rate is $220-26=194$ beats per minute. The target heart rate is then $0.8*194 \approx \boxed{\textbf{(B)}\ 155}$ beats per minute.
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_7 | 155 |
AMC8_252 | Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
Chandra and Bob, who each have a copy of the book, decide that they can save time by "team reading" the novel. In this scheme, Chandra will read from page 1 to a certain page and Bob will read from the next page through page 760, finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?
$\textbf{(A)}\ 425\qquad\textbf{(B)}\ 444\qquad\textbf{(C)}\ 456\qquad\textbf{(D)}\ 484\qquad\textbf{(E)}\ 506$
| Same as the previous problem, we only use the information we need. Note that it's not just Chandra reads half of it and Bob reads the rest since they have different reading rates. In this case, we set up an equation and solve.
Let $x$ be the number of pages that Chandra reads.
$30x = 45(760-x)$ Distribute the $45$
$30x = 45(760) - 45x$ Add $45x$ to both sides
$75x = 45(760)$ Divide both sides by $15$ to make it easier to solve
$5x = 3(760)$ Divide both sides by $5$
$x = 3(152) = \boxed{\textbf{(C)} 456}$
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_15 | 456 |
AMC8_253 | How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$
| Looking at the values, we notice that $11-7=4$ , $9-5=4$ and $6-2=4$ . This means we are looking for a value that is four less than a multiple of $11$ , $9$ , and $6$ . The least common multiple of these numbers is $11\cdot3^{2}\cdot2=198$ , so the numbers that fulfill this can be written as $198k-4$ , where $k$ is a positive integer. This value is only a three-digit integer when $k$ is $1, 2, 3, 4$ or $5$ , which gives $194, 392, 590, 788,$ and $986$ respectively. Thus, we have $5$ values, so our answer is $\boxed{\textbf{(E) }5}$ .
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_21 | 5 |
AMC8_254 | Of the $500$ balls in a large bag, $80\%$ are red and the rest are blue. How many of the red balls must be removed so that $75\%$ of the remaining balls are red?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150$
| Since 80 percent of the 500 balls are red, there are 400 red balls. Therefore, there must be 100 blue balls. For the 100 blue balls to be 25% or $\dfrac{1}{4}$ of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is $\boxed{\textbf{(D)}\ 100}$ .
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_12 | 100 |
AMC8_255 | In the land of Markovia, there are three cities: \( A \), \( B \), and \( C \). There are 100 people who live in \( A \), 120 who live in \( B \), and 160 who live in \( C \). Everyone works in one of the three cities, and a person may work in the same city where they live. In the figure below, an arrow pointing from one city to another is labeled with the fraction of people living in the first city who work in the second city. (For example, \( \frac{1}{4} \) of the people who live in \( A \) work in \( B \).) How many people work in \( A \)?
[asy] /* AMC8 P17 2025, by NUMANA: BUI VAN HIEU, [email protected], https://husc.edu.vn */ import graph; unitsize(2cm); real r=0.15; pair A, B, C;B = (0,0);C = (2,0);A = (1,sqrt(3)); // Drawing the nodes draw(circle(A,r)); label("$A$", A); draw(circle(B,r)); label("$B$", B); draw(circle(C,r)); label("$C$", C); guide AB=A+r*dir(-135)..{down}B+r*dir(90), BA=B+r*dir(60)..{up}A+r*dir(-105), BC=B+r*dir(0)..(1,-0.2)..C+r*dir(180), CB=C+r*dir(150)..(1,0.3)..B+r*dir(30), CA=C+r*dir(90){up}..A+r*dir(-45), AC=A+r*dir(-75){down}..C+r*dir(120); draw(AB,L=Label("$1/4$", MidPoint, W),Arrow(HookHead)); draw(BA,L=Label("$1/3$", MidPoint, W),Arrow(HookHead)); draw(BC,L=Label("$1/6$", MidPoint, S),Arrow(HookHead)); draw(CB,L=Label("$1/10$", MidPoint, S),Arrow(HookHead)); draw(CA,L=Label("$1/8$", MidPoint, E),Arrow(HookHead)); draw(AC,L=Label("$1/5$", MidPoint, E),Arrow(HookHead)); [/asy]
$\textbf{(A)}\ 55\qquad \textbf{(B)}\ 60\qquad \textbf{(C)}\ 85\qquad \textbf{(D)}\ 115\qquad \textbf{(E)}\ 160$
| $100 (1/4 + 1/5) = 100 \cdot \frac{9}{20} = 45$ people do not work in city $A$ that live in city $A$ , meaning $55$ people that live in city $A$ work in city $A$ . $\frac{1}{3} \cdot 120 = 40$ people that live in $B$ work in $A$ and $\frac{1}{8} \cdot 160 = 20$ people that live in $C$ work in $A$ , so the answer is $55 + 40 + 20 = \boxed{\textbf{(D)}115}$ .
~ alwaysgonnagiveyouup
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_17 | 115 |
AMC8_256 | The figure below shows a large white circle with a number of smaller white and shaded circles in its
interior. What fraction of the interior of the large white circle is shaded?
[asy] // Diagram by TheMathGuyd size(6cm); draw(circle((3,3),3)); filldraw(circle((2,3),2),lightgrey); filldraw(circle((3,3),1),white); filldraw(circle((1,3),1),white); filldraw(circle((5.5,3),0.5),lightgrey); filldraw(circle((4.5,4.5),0.5),lightgrey); filldraw(circle((4.5,1.5),0.5),lightgrey); int i, j; for(i=0; i<7; i=i+1) { draw((0,i)--(6,i), dashed+grey); draw((i,0)--(i,6), dashed+grey); } [/asy]
$\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{11}{36} \qquad \textbf{(C)}\ \frac{1}{3} \qquad \textbf{(D)}\ \frac{19}{36} \qquad \textbf{(E)}\ \frac{5}{9}$
| First, the total area of the radius $3$ circle is simply just $9\cdot \pi$ when using our area of a circle formula.
Now from here, we have to find our shaded area. This can be done by adding the areas of the $\frac{1}{2}$ -radius circles and add; then, take the area of the $1$ radius circles and subtract that from the area of the $2$ radius circle to get our resulting complex shape area. Adding these up, we will get $3\cdot \frac{1}{4} \pi + 4 \pi -\pi - \pi = \frac{3}{4} \pi + 2 \pi = \frac{11\cdot \pi}{4}$ .
So, our answer is $\frac {\frac{11}{4} \pi}{9 \pi} = \boxed{\textbf{(B)}\ \frac{11}{36}}$ .
~apex304
Minor edits by ~NXC
Minor edits by ~Shriyans Chowdhury
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_12 | 11/36 |
AMC8_257 | A whole number larger than $2$ leaves a remainder of $2$ when divided by each of the numbers $3, 4, 5,$ and $6$ . The smallest such number lies between which two numbers?
$\textbf{(A)}\ 40\ \text{and}\ 49 \qquad \textbf{(B)}\ 60 \text{ and } 79 \qquad \textbf{(C)}\ 100\ \text{and}\ 129 \qquad \textbf{(D)}\ 210\ \text{and}\ 249\qquad \textbf{(E)}\ 320\ \text{and}\ 369$
| The smallest number divisible by $3,4,5,$ and $6$ , or their least common multiple, can be found to be $60$ . When $2$ is added to a multiple of number, its remainder when divided by that number is $2$ . The number we are looking for is therefore $62$ , and between $\boxed{\textbf{(B)}\ 60\ \text{and}\ 79}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_19 | 60 and 79 |
AMC8_258 | An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon?
$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3$
| Let the perimeter of the equilateral triangle be $3s$ . The side length of the equilateral triangle would then be $s$ and the sidelength of the hexagon would be $\frac{s}{2}$ .
A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio $1 : 4$ , since the sidelength of the small equilateral triangle is half the side length of the large one. Thus, the area of one of the small equilateral triangles is $1$ . The area of the hexagon is then $1 \times 6 = \boxed{\textbf{(C)}\ 6}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_23 | 6 |
AMC8_259 | Six pepperoni circles will exactly fit across the diameter of a $12$ -inch pizza when placed. If a total of $24$ circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?
$\textbf{(A)}\ \frac 12 \qquad\textbf{(B)}\ \frac 23 \qquad\textbf{(C)}\ \frac 34 \qquad\textbf{(D)}\ \frac 56 \qquad\textbf{(E)}\ \frac 78$
| The pepperoni circles' diameter is $2$ , since $\dfrac{12}{6} = 2$ . From that we see that the area of the $24$ circles of pepperoni is $\left ( \frac{2}{2} \right )^2 (24\pi) = 24\pi$ . The large pizza's area is $6^2\pi$ . Therefore, the ratio is $\frac{24\pi}{36\pi} = \boxed{\textbf{(B) }\frac{2}{3}}$
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_10 | 2/3 |
AMC8_260 | The sales tax rate in Rubenenkoville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?
$\textbf{(A)}\ -\textdollar 1.06\qquad\textbf{(B)}\ -\textdollar 0.53 \qquad\textbf{(C)}\ \textdollar 0\qquad\textbf{(D)}\ \textdollar 0.53\qquad\textbf{(E)}\ \textdollar 1.06$
| The price Jack rings up is $\textdollar{(90.00)(1.06)(0.80)}$ . The price Jill rings up is $\textdollar{(90.00)(0.80)(1.06)}$ . By the commutative property of multiplication, these quantities are the same, and the difference is $\boxed{\textbf{(C)}\ \textdollar 0}$ .
Reminder: Commutative states abc = bac= acb =..... and is the same in any order. This only works for addition and multiplication.
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_11 | 0 |
AMC8_261 | How many digits are in the product $4^5 \cdot 5^{10}$ ?
$\textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 12$
| \[4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = 10^{10}.\]
That is one $1$ followed by ten $0$ 's, which is $\boxed{\textbf{(D)}\ 11}$ digits.
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_15 | 11 |
AMC8_262 | Tess runs counterclockwise around rectangular block $JKLM$ . She lives at corner $J$ . Which graph could represent her straight-line distance from home?
[asy] unitsize(5mm); pair J=(-3,2); pair K=(-3,-2); pair L=(3,-2); pair M=(3,2); draw(J--K--L--M--cycle); label("$J$",J,NW); label("$K$",K,SW); label("$L$",L,SE); label("$M$",M,NE); [/asy]
$\textbf{(A)}$
[asy] size(80);defaultpen(linewidth(0.8)); //A draw((16,0)--origin--(0,16)); draw(origin--(15,15)); label("time", (8,0), S); label(rotate(90)*"distance", (0,8), W); [/asy]
$\textbf{(B)}$
[asy] size(80);defaultpen(linewidth(0.8)); //B draw((16,0)--origin--(0,16)); draw((0,6)--(1,6)--(1,12)--(2,12)--(2,11)--(3,11)--(3,1)--(12,1)--(12,0)); label("time", (8,0), S); label(rotate(90)*"distance", (0,8), W); [/asy]
$\textbf{(C)}$
[asy] size(80);defaultpen(linewidth(0.8)); //C draw((16,0)--origin--(0,16)); draw(origin--(2.7,8)--(3,9)^^(11,9)--(14,0)); draw(Arc((4,9), 1, 0, 180)); draw(Arc((10,9), 1, 0, 180)); draw(Arc((7,9), 2, 180,360)); label("time", (8,0), S); label(rotate(90)*"distance", (0,8), W); [/asy]
$\textbf{(D)}$
[asy] size(80);defaultpen(linewidth(0.8)); //D draw((16,0)--origin--(0,16)); draw(origin--(2,6)--(7,14)--(10,12)--(14,0)); label("time", (8,0), S); label(rotate(90)*"distance", (0,8), W); [/asy]
$\textbf{(E)}$
[asy] size(80);defaultpen(linewidth(0.8)); //E draw((16,0)--origin--(0,16)); draw(origin--(3,6)--(7,6)--(10,12)--(14,12)); label("time", (8,0), S); label(rotate(90)*"distance", (0,8), W); [/asy]
| For her distance to be represented as a constant horizontal line, Tess would have to be running in a circular shape with her home as the center. Since she is running around a rectangle, this is not possible, ruling out $B$ and $E$ with straight lines. Because $JL$ is the diagonal of the rectangle, and $L$ is at the middle distance around the perimeter, her maximum distance should be in the middle of her journey. The maximum in $A$ is at the end, and $C$ has two maximums, ruling both out. Thus the answer is $\boxed{\textbf{(D)}}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_23 | D |
AMC8_263 | The letter T is formed by placing two $2 \times 4$ inch rectangles next to each other, as shown. What is the perimeter of the T, in inches?
[asy] size(150); draw((0,6)--(4,6)--(4,4)--(3,4)--(3,0)--(1,0)--(1,4)--(0,4)--cycle, linewidth(1));[/asy]
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$
| If the two rectangles were seperate, the perimeter would be $2(2(2+4)=24$ . It easy to see that their connection erases 2 from each of the rectangles, so the final perimeter is $24-2 \times 2 = \boxed{\textbf{(C)}\ 20}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_6 | 20 |
AMC8_264 | In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$
| We can try to guess and check to find the answer. If she got five right, her score would be $(5*5)-(5*2)=15$ . If she got six right her score would be $(6*5)-(2*4)=22$ . That's close, but it's still not right! If she got 7 right, her score would be $(7*5)-(2*3)=29$ . Thus, our answer is $\boxed{\text{(C)}\ 7}$ . ~avamarora
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_17 | 7 |
AMC8_265 | A haunted house has six windows. In how many ways can
Georgie the Ghost enter the house by one window and leave
by a different window?
$\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 18 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 36$
| Georgie can enter the haunted house through any of the six windows. Then, he can leave through any of the remaining five windows.
So, Georgie has a total of $6 \cdot 5$ ways he can enter the house by one window and leave
by a different window.
Therefore, we have $\boxed{\textbf{(D)}\ 30}$ ways.
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_4 | 30 |
AMC8_266 | Points $A, B, C$ and $D$ are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?
[asy]size(100); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle,linewidth(1)); draw((0,1)--(1,2)--(2,1)--(1,0)--cycle); label("$A$", (1,2), N); label("$B$", (2,1), E); label("$C$", (1,0), S); label("$D$", (0,1), W);[/asy]
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$
| Solution 1Drawing segments $AC$ and $BD$ , the number of triangles outside square $ABCD$ is the same as the number of triangles inside the square. Thus areas must be equal so the area of $ABCD$ is half the area of the larger square which is $\frac{60}{2}=\boxed{\textbf{(D)}\ 30 }$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_5 | 30 |
AMC8_267 | Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many hours will it take to fill Steve's pool?
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 42 \qquad \textbf{(C)}\ 44 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 48$
| Each of the four hoses hose fills $24,000/4 = 6,000$ gallons of water. At the rate it goes at it will take $6,000/2.5 = 2400$ minutes, or $\boxed{\textbf{(A)}\ 40}$ hours.
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_6 | 40 |
AMC8_268 | Buffalo Shuffle-o is a card game in which all the cards are distributed evenly among all players at the start of the game. When Annika and $3$ of her friends play Buffalo Shuffle-o , each player is dealt $15$ cards. Suppose $2$ more friends join the next game. How many cards will be dealt to each player?
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12$
| We start with Annika and $3$ of her friends playing, meaning that there are $4$ players. This must mean that there is a total of $4 \cdot 15 = 60$ cards. If $2$ more players joined, there would be $6$ players, and since the cards need to be split evenly, this would mean that each player gets $\frac{60}{6}=\boxed{\text{(C)\ 10}}$ cards.
~shreyan.chethan
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_3 | 10 |
AMC8_269 | A bus takes $2$ minutes to drive from one stop to the next, and waits $1$ minute at each stop to let passengers board. Zia takes $5$ minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwise she will start walking toward the next stop. Suppose the bus and Zia start at the same time toward the library, with the bus $3$ stops behind. After how many minutes will Zia board the bus?
| Initially, suppose that the bus is at Stop $0$ (starting point) and Zia is at Stop $3.$
We construct the following table of $5$ -minute intervals:
\[\begin{array}{c||c|c} & & \\ [-2.5ex] \textbf{Time} & \textbf{Bus's Location} & \textbf{Zia's Location} \\ [0.5ex] \hline & & \\ [-2ex] \boldsymbol{5} \ \textbf{Minutes} & \text{Stop} \ 2 \ \text{(Waiting)} & \text{Stop} \ 4 \\ \boldsymbol{10} \ \textbf{Minutes} & \text{Midpoint of Stops} \ 3 \ \text{and} \ 4 & \text{Stop} \ 5 \\ \boldsymbol{15} \ \textbf{Minutes} & \text{Stop} \ 5 \ \text{(Leaving)} & \text{Stop} \ 6 \end{array}\]
Note that Zia will wait for the bus after $15$ minutes, and the bus will arrive $2$ minutes later.
Therefore, the answer is $15+2=\boxed{\textbf{(A) } 17}.$
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_22 | 17 |
AMC8_272 | What is the correct ordering of the three numbers $\frac{5}{19}$ , $\frac{7}{21}$ , and $\frac{9}{23}$ , in increasing order?
$\textbf{(A)}\hspace{.05in}\frac{9}{23}<\frac{7}{21}<\frac{5}{19}\quad\textbf{(B)}\hspace{.05in}\frac{5}{19}<\frac{7}{21}<\frac{9}{23}\quad\textbf{(C)}\hspace{.05in}\frac{9}{23}<\frac{5}{19}<\frac{7}{21}$
$\textbf{(D)}\hspace{.05in}\frac{5}{19}<\frac{9}{23}<\frac{7}{21}\quad\textbf{(E)}\hspace{.05in}\frac{7}{21}<\frac{5}{19}<\frac{9}{23}$
| The value of $\frac{7}{21}$ is $\frac{1}{3}$ . Now we give all the fractions a common denominator.
$\frac{5}{19} \implies \frac{345}{1311}$
$\frac{1}{3} \implies \frac{437}{1311}$
$\frac{9}{23} \implies \frac{513}{1311}$
Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_20 | (B) 5/19 < 7/21 < 9/23 |
AMC8_273 | Which of the following values is largest?
$\textbf{(A) }2+0+1+7\qquad\textbf{(B) }2 \times 0 +1+7\qquad\textbf{(C) }2+0 \times 1 + 7\qquad\textbf{(D) }2+0+1 \times 7\qquad\textbf{(E) }2 \times 0 \times 1 \times 7$
| We will compute each expression.
A) $2 + 0 + 1 + 7 = 10$
B) $2 \times 0 + 1 + 7 = 8$
C) $2 + 0 \times 1 + 7 = 9$
D) $2 + 0 + 1 \times 7 = 9$
E) $2 \times 0 \times 1 \times 7 = 0$
Ordering these, we get $10, 8, 9, 9, 0$ . Out of these, $10$ is the largest number and option $(A)$ adds up to $10$ . Therefore, the answer is $\boxed{\textbf{(A) } 2+0+1+7}$ .
- SBose
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_1 | A |
AMC8_274 | On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?
$\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401$
| Solution 1First, we recognize that the number of widgets Janabel sells each day forms an arithmetic sequence. On day 1, she sells 1 widget, on day 2, she sells 3 widgets, on day 3, she sells 5 widgets, and so on. This sequence can be described by the first term \( a = 1 \) and the common difference \( d = 2 \).
Step 1: Find the number of widgets sold on day 20
The formula for the \( n \)-th term of an arithmetic sequence is:
\[a_n = a + (n-1)d\]
For \( n = 20 \):
\[a_{20} = 1 + (20-1) \cdot 2 = 1 + 38 = 39\]
So, Janabel sells 39 widgets on day 20.
Step 2: Calculate the total number of widgets sold in 20 days}
The sum \( S_n \) of the first \( n \) terms of an arithmetic sequence is given by:
\[S_n = \frac{n}{2} \cdot (a + a_n)\]
For \( n = 20 \):
\[S_{20} = \frac{20}{2} \cdot (1 + 39) = 10 \cdot 40 = 400\]
Therefore, the total number of widgets sold after 20 days is:
$(40\cdot10)=\boxed{\textbf{(D)}~400}$
~ GeometryMystery
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_9 | 400 |
AMC8_275 | Karl's rectangular vegetable garden is $20$ feet by $45$ feet, and Makenna's is $25$ feet by $40$ feet. Whose garden is larger in area?
$\textbf{(A)}\ \text{Karl's garden is larger by 100 square feet.}$
$\textbf{(B)}\ \text{Karl's garden is larger by 25 square feet.}$
$\textbf{(C)}\ \text{The gardens are the same size.}$
$\textbf{(D)}\ \text{Makenna's garden is larger by 25 square feet.}$
$\textbf{(E)}\ \text{Makenna's garden is larger by 100 square feet.}$
| The area of a rectangle is given by the formula length times width. Karl's garden is $20 \times 45 = 900$ square feet and Makenna's garden is $25 \times 40 = 1000$ square feet. Since $1000 > 900,$ Makenna's garden is larger by $1000-900=100$ square feet. $\Rightarrow \boxed{ \textbf{(E)}\ \text{Makenna's garden is larger by 100 square feet.} }$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_2 | Makenna's garden is larger by 100 square feet. |
AMC8_276 | How many rectangles are in this figure?
[asy] pair A,B,C,D,E,F,G,H,I,J,K,L; A=(0,0); B=(20,0); C=(20,20); D=(0,20); draw(A--B--C--D--cycle); E=(-10,-5); F=(13,-5); G=(13,5); H=(-10,5); draw(E--F--G--H--cycle); I=(10,-20); J=(18,-20); K=(18,13); L=(10,13); draw(I--J--K--L--cycle);[/asy]
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$
| The figure can be divided into $7$ sections. The number of rectangles with just one section is $3.$ The number of rectangles with two sections is $5.$ There are none with only three sections. The number of rectangles with four sections is $3.$
$3+5+3=\boxed{\textbf{(D)}\ 11}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_19 | 11 |
AMC8_277 | A company sells detergent in three different sized boxes: small (S), medium (M) and large (L). The medium size costs 50% more than the small size and contains 20% less detergent than the large size. The large size contains twice as much detergent as the small size and costs 30% more than the medium size. Rank the three sizes from best to worst buy.
$\text{(A)}\, SML \qquad \text{(B)}\, LMS \qquad \text{(C)}\, MSL \qquad \text{(D)}\, LSM \qquad \text{(E)}\, MLS$
| Suppose the small size costs $\textdollar 1$ and has $5$ oz. The medium size then costs $\textdollar 1.50$ and has $8$ oz. The large size costs $\textdollar 1.95$ and has 10 oz. The small, medium, and large size cost respectively, $0.200, 0.188, 0.195$ dollars per oz. The sizes from best to worst buy are $\boxed{\textbf{(E)}\ \text{MLS}}$ .
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_22 | MLS |
AMC8_278 | A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?
[asy] pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(a--d--b--c--cycle); draw(circle(o, 2.5)); [/asy]
$\textbf{(A)}\ \frac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \frac{1+\sqrt{2}}{2} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}$
| Solution 1Let the region within the circle and square be $a$ . In other words, it is the area inside the circle $\textbf{and}$ the square. Let $r$ be the radius. We know that the area of the circle minus $a$ is equal to the area of the square, minus $a$ .
We get:
$\pi r^2 -a=4-a$
$r^2=\frac{4}{\pi}$
$r=\frac{2}{\sqrt{\pi}}$
So the answer is $\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}$ .
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_25 | $\frac{2}{\sqrt{\pi}}$ |
AMC8_279 | When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.
[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label("$N$",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label("if $N$ is even",(9.25,1.25),N); label("if $N$ is odd",(9.25,-1.25),N); label("$\frac N2$",(12,1.25)); label("$3N+1$",(12.6,-1.25)); [/asy]
For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$ \[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\] When the same $6$ -step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$ \[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\]
$\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$
| We start with final output of $1$ and work backward, taking cares to consider all possible inputs that could have resulted in any particular output. This produces following set of possibilities each stage:
\[\{1\}\rightarrow\{2\}\rightarrow\{4\}\rightarrow\{1,8\}\rightarrow\{2,16\}\rightarrow\{4,5,32\}\rightarrow\{1,8,10,64\}\]
where, for example, $2$ must come from $4$ (as there is no integer $n$ satisfying $3n+1=2$ ), but $16$ could come from $32$ or $5$ (as $\frac{32}{2} = 3 \cdot 5 + 1 = 16$ , and $32$ is even while $5$ is odd). By construction, last set in this sequence contains all the numbers which will lead to number $1$ to end of the $6$ -step process, and sum is $1+8+10+64=\boxed{\textbf{(E) }83}$ .
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_22 | 83 |
AMC8_280 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$
| Solution 1For starters, we identify d as distance and v as velocity (speed)
Writing the equation gives us: $\frac{d}{v}=\frac{1}{3}$ and $\frac{d}{v+18}=\frac{1}{5}$ .
This gives $d=\frac{1}{5}v+3.6=\frac{1}{3}v$ , which gives $v=27$ , which then gives $d=\boxed{\textbf{(D)}~9}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | 9 |
AMC8_281 | What is the value of this expression in decimal form?
\[\frac{44}{11} + \frac{110}{44} + \frac{44}{1100}\]
$\textbf{(A) } 6.4\qquad\textbf{(B) } 6.504\qquad\textbf{(C) } 6.54\qquad\textbf{(D) } 6.9\qquad\textbf{(E) } 6.94$
| We see that $\frac{44}{11}$ is $4$ ;
$\frac{110}{44}$ simplifies to $\frac{5}{2}$ , which is $2.5$ ;
and $\frac{44}{1100}$ simplifies to $\frac{1}{25}$ , which is $0.04$ ;
$4+2.5+0.04$ reveals \[\frac{44}{11} + \frac{110}{44} + \frac{44}{1100}\] is $\boxed{\text{(C)\ 6.54}}$ .
~
~Minor Formatting by GreenPlanet2050
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_2 | 6.54 |
AMC8_282 | At Euler Middle School, $198$ students voted on two issues in a school referendum with the following results: $149$ voted in favor of the first issue and $119$ voted in favor of the second issue. If there were exactly $29$ students who voted against both issues, how many students voted in favor of both issues?
$\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149$
| We can see that this is a Venn Diagram Problem.
First, we analyze the information given. There are $198$ students. Let's use A as the first issue and B as the second issue.
$149$ students were for A, and $119$ students were for B. There were also $29$ students against both A and B.
Solving this without a Venn Diagram, we subtract $29$ away from the total, $198$ . Out of the remaining $169$ , we have $149$ people for A and
$119$ people for B. We add this up to get $268$ . Since that is more than what we need, we subtract $169$ from $268$ to get
$\boxed{\textbf{(D)}~99}$ .
[asy] defaultpen(linewidth(0.7)); draw(Circle(origin, 5)); draw(Circle((5,0), 5)); label("$A$", (0,5), N); label("$B$", (5,5), N); label("$99$", (2.5, -0.5), N); label("$50$", (-2.5,-0.5), N); label("$20$", (7.5, -0.5), N); [/asy]
Note: One could use the Principle of Inclusion-Exclusion in a similar way to achieve the same result.
~ cxsmi (note)
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_15 | 99 |
AMC8_283 | Which of the following integers cannot be written as the sum of four consecutive odd integers?
$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$
| Solution 1Let our $4$ numbers be $n, n+2, n+4, n+6$ , where $n$ is odd. Then, our sum is $4n+12$ . The only answer choice that cannot be written as $4n+12$ , where $n$ is odd, is $\boxed{\textbf{(D)}\text{ 100}}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_14 | 100 |
AMC8_284 | Andy and Bethany have a rectangular array of numbers with $40$ rows and $75$ columns. Andy adds the numbers in each row. The average of his $40$ sums is $A$ . Bethany adds the numbers in each column. The average of her $75$ sums is $B$ . What is the value of $\frac{A}{B}$ ?
$\textbf{(A)}\ \frac{64}{225} \qquad \textbf{(B)}\ \frac{8}{15} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{15}{8} \qquad \textbf{(E)}\ \frac{225}{64}$
| Note that $40A=75B=\text{sum of the numbers in the array}$ . This means that $\frac{A}{B}=\boxed{\text{(D) } \frac{15}{8}}$ using basic algebraic manipulation.
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_21 | 15/8 |
AMC8_285 | What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$ ?
$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$
| Directly calculating:
We evaluate both the top and bottom: $\frac{40320}{36}$ . This simplifies to $\boxed{\textbf{(B)}\ 1120}$ .
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_5 | 1120 |
AMC8_286 | How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$
| Find the least common multiple of $15, 20, 25$ by turning the numbers into their prime factorization. \[15 = 3 * 5, 20 = 2^2 * 5, 25 = 5^2\] Gather all necessary multiples
$3, 2^2, 5^2$ when multiplied gets $300$ . The multiples of $300 - 300, 600, 900, 1200, 1500, 1800, 2100$ . The number of multiples between 1000 and 2000 is $\boxed{\textbf{(C)}\ 3}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_19 | 3 |
AMC8_288 | Granny Smith has $63. Elberta has $2 more than Anjou and Anjou has one-third as much as Granny Smith. How many dollars does Elberta have?
$\text{(A)}\ 17 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 23$
| Since Anjou has $\frac{1}{3}$ the amount of money as Granny Smith and Granny Smith has $ $63$ , Anjou has $\frac{1}{3}\times63=21$ dollars. Elberta has $ $2$ more than this, so she has $ $23$ , or $\boxed{\text{E}}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_3 | E |
AMC8_289 | A circle and two distinct lines are drawn on a sheet of paper. What is the largest possible number of points of intersection of these figures?
$\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad {(C)}\ 4 \qquad {(D)}\ 5 \qquad {(E)}\ 6$
| The two lines can both intersect the circle twice, and can intersect each other once, so $2+2+1= \boxed {\text {(D)}\ 5}.$
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_1 | 5 |
AMC8_290 | Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?
[asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); for (int i=0; i<7; ++i) { A[i] = rotate(60*i)*(1,0);} path hex = A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--cycle; fill(p,grey1); draw(scale(1.25)*hex,black+linewidth(1.25)); pair S = 6A[0]+2A[1]; fill(shift(S)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(S+2*A[i])*p,grey2);} draw(shift(S)*scale(3.25)*hex,black+linewidth(1.25)); pair T = 16A[0]+4A[1]; fill(shift(T)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(T+2*A[i])*p,grey2); fill(shift(T+4*A[i])*p,grey1); fill(shift(T+2*A[i]+2*A[i+1])*p,grey1); } draw(shift(T)*scale(5.25)*hex,black+linewidth(1.25)); [/asy]
$\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49$
| Looking at the rows of each hexagon, we see that the first hexagon has $1$ dot, the second has $2+3+2$ dots, and the third has $3+4+5+4+3$ dots. Given the way the hexagons are constructed, it is clear that this pattern continues. Hence, the fourth hexagon has $4+5+6+7+6+5+4=\boxed{\textbf{(B) }37}$ dots.
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_4 | 37 |
AMC8_291 | Jerry cuts a wedge from a $6$ -cm cylinder of bologna as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?
[asy] defaultpen(linewidth(0.65)); real d=90-63.43494882; draw(ellipse((origin), 2, 4)); fill((0,4)--(0,-4)--(-8,-4)--(-8,4)--cycle, white); draw(ellipse((-4,0), 2, 4)); draw((0,4)--(-4,4)); draw((0,-4)--(-4,-4)); draw(shift(-2,0)*rotate(-d-5)*ellipse(origin, 1.82, 4.56), linetype("10 10")); draw((-4,4)--(-8,4), dashed); draw((-4,-4)--(-8,-4), dashed); draw((-4,4.3)--(-4,5)); draw((0,4.3)--(0,5)); draw((-7,4)--(-7,-4), Arrows(5)); draw((-4,4.7)--(0,4.7), Arrows(5)); label("$8$ cm", (-7,0), W); label("$6$ cm", (-2,4.7), N);[/asy]
$\textbf{(A)} \ 48 \qquad \textbf{(B)} \ 75 \qquad \textbf{(C)} \ 151\qquad \textbf{(D)} \ 192 \qquad \textbf{(E)} \ 603$
| The slice is cutting the cylinder into two equal wedges with equal volume. The cylinder's volume is $\pi r^2 h = \pi (4^2)(6) = 96\pi$ . The volume of the wedge is half this which is $48\pi \approx \boxed{\textbf{(C)}\ 151}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_21 | 151 |
AMC8_292 | Angle $ABC$ of $\triangle ABC$ is a right angle. The sides of $\triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $\overline{AB}$ equals $8\pi$ , and the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$ . What is the radius of the semicircle on $\overline{BC}$ ?
[asy] import graph; pair A,B,C; A=(0,8); B=(0,0); C=(15,0); draw((0,8)..(-4,4)..(0,0)--(0,8)); draw((0,0)..(7.5,-7.5)..(15,0)--(0,0)); real theta = aTan(8/15); draw(arc((15/2,4),17/2,-theta,180-theta)); draw((0,8)--(15,0)); dot(A); dot(C); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE);[/asy]
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 7.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 8.5 \qquad \textbf{(E)}\ 9$
| If the semicircle on $\overline{AB}$ were a full circle, the area would be $16\pi$ .
$\pi r^2=16 \pi \Rightarrow r^2=16 \Rightarrow r=+4$ , therefore the diameter of the first circle is $8$ .
The arc of the largest semicircle is $8.5 \pi$ , so if it were a full circle, the circumference would be $17 \pi$ . So the $\text{diameter}=17$ .
By the Pythagorean theorem, the other side has length $15$ , so the radius is $\boxed{\textbf{(B)}\ 7.5}$
~Edited by Theraccoon to correct typos.
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_23 | 7.5 |
AMC8_293 | A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5$
| Solution 1The counting numbers that leave a remainder of $4$ when divided by $6$ are
$4, 10, 16, 22, 28, 34, \cdots$ The counting numbers that leave a remainder of $3$ when
divided by $5$ are $3,8,13,18,23,28,33, \cdots$ So $28$ is the smallest possible number
of coins that meets both conditions. Because 28 is divisible by 7, there are $\boxed{\textbf{(A)}\ 0}$ coins left
when they are divided among seven people.
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_23 | 0 |
AMC8_294 | A lemming sits at a corner of a square with side length $10$ meters. The lemming runs $6.2$ meters along a diagonal toward the opposite corner. It stops, makes a $90^{\circ}$ right turn and runs $2$ more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6.2 \qquad \textbf{(E)}\ 7$
| The shortest segments would be perpendicular to the square. The lemming went $x$ meters horizontally and $y$ meters vertically. No matter how much it went, the lemming would have been $x$ and $y$ meters from the sides and $10-x$ and $10-y$ meters from the remaining two. To find the average, add the lengths of the four segments and divide by four: $\frac {\cancel{x}+10-\cancel{x}+\cancel{y}+10-\cancel{y}}{4} =$ $\boxed{\textbf{(C)}\ 5}$ .
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_22 | 5 |
AMC8_295 | Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned $6$ years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is $30$ years. How many years older than Bella is Anna?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } ~5$
| Five years ago, Bella was $6$ years old, and the kitten was $0$ years old.
Today, Bella is $11$ years old, and the kitten is $5$ years old. It follows that Anna is $30-11-5=14$ years old.
Therefore, Anna is $14-11=\boxed{\textbf{(C) } 3}$ years older than Bella.
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_5 | 3 |
AMC8_297 | Two $600$ mL pitchers contain orange juice. One pitcher is $1/3$ full and the other pitcher is $2/5$ full. Water is added to fill each pitcher completely, then both pitchers are poured into one large container. What fraction of the mixture in the large container is orange juice?
$\textbf{(A)}\ \frac18 \qquad \textbf{(B)}\ \frac{3}{16} \qquad \textbf{(C)}\ \frac{11}{30} \qquad \textbf{(D)}\ \frac{11}{19}\qquad \textbf{(E)}\ \frac{11}{15}$
| The first pitcher contains $600 \cdot \frac13 = 200$ mL of orange juice. The second pitcher contains $600 \cdot \frac25 = 240$ mL of orange juice. In the large pitcher, there is a total of $200+240=440$ mL of orange juice and $600+600=1200$ mL of fluids, giving a fraction of $\frac{440}{1200} = \boxed{\textbf{(C)}\ \frac{11}{30}}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_16 | 11/30 |
AMC8_298 | Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$
| After Jamal adds $x$ purple socks, he has $(18+x)$ purple socks and $6+18+12+x=(36+x)$ total socks. This means the probability of drawing a purple sock is $\frac{18+x}{36+x}$ , so we obtain \[\frac{18+x}{36+x}=\frac{3}{5}\] Since $\frac{18+9}{36+9}=\frac{27}{45}=\frac{3}{5}$ , the answer is $\boxed{\textbf{(B) }9}$ .
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_13 | 9 |
AMC8_299 | An ATM password at Fred's Bank is composed of four digits from $0$ to $9$ , with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?
$\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999$
| Solution 1For the first three digits, there are $10^3-1=999$ combinations since $911$ is not allowed. For the final digit, any of the $10$ numbers are allowed. $999 \cdot 10 = 9990 \rightarrow \boxed{\textbf{(D)}\ 9990}$ .
~CHECKMATE2021
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_17 | 9990 |
AMC8_300 | Henry the donkey has a very long piece of pasta. He takes a number of bites of pasta, each time eating $3$ inches of pasta from the middle of one piece. In the end, he has $10$ pieces of pasta whose total length is $17$ inches. How long, in inches, was the piece of pasta he started with?
$\textbf{(A) } 34\qquad\textbf{(B) } 38\qquad\textbf{(C) } 41\qquad\textbf{(D) } 44\qquad\textbf{(E) } 47$
| Let's say that the first strand of pasta he had is x. The first time he takes a bite of this strand will make it 2 pieces. This is x - 3. The second time he takes the bites of EACH strand will become (x - 3) * 3(2). The 2 is for the two bites he took, 1 for each strand. This equation is showing that for every bite that he is taking, he is taking off 3 inches. This is being showed in the 3(2). He took 2 bites, each taking off 3 inches. Now, the problem is stating that their are 10 pieces of pasta and the TOTAL length is 17 inches. We have to plug in the total number of bites to the equation (x - 3) * 3( ). To find this out, we can see that for the number of strands there are, he took 1 less bite than that number. For example, if there are 2 pieces, he took 1 bite to get those. Since there are 10 pieces, he took 9 bites. The equation then becomes, (x - 3) * 3(9) = 17. To solve this, x = 44. Therefore, the answer is $\boxed{\textbf{(D) } 44}$ .
~~Brainiacs77~~
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_11 | 44 |
AMC8_301 | A rectangle, with sides parallel to the $x$ -axis and $y$ -axis, has opposite vertices located at $(15, 3)$ and $(16, 5)$ . A line is drawn through points $A(0, 0)$ and $B(3, 1)$ . Another line is drawn through points $C(0, 10)$ and $D(2, 9)$ . How many points on the rectangle lie on at least one of the two lines?
[asy] usepackage("mathptmx"); size(9cm); draw((0,-.5)--(0,11),EndArrow(size=.15cm)); draw((1,0)--(1,11),mediumgray); draw((2,0)--(2,11),mediumgray); draw((3,0)--(3,11),mediumgray); draw((4,0)--(4,11),mediumgray); draw((5,0)--(5,11),mediumgray); draw((6,0)--(6,11),mediumgray); draw((7,0)--(7,11),mediumgray); draw((8,0)--(8,11),mediumgray); draw((9,0)--(9,11),mediumgray); draw((10,0)--(10,11),mediumgray); draw((11,0)--(11,11),mediumgray); draw((12,0)--(12,11),mediumgray); draw((13,0)--(13,11),mediumgray); draw((14,0)--(14,11),mediumgray); draw((15,0)--(15,11),mediumgray); draw((16,0)--(16,11),mediumgray); draw((-.5,0)--(17,0),EndArrow(size=.15cm)); draw((0,1)--(17,1),mediumgray); draw((0,2)--(17,2),mediumgray); draw((0,3)--(17,3),mediumgray); draw((0,4)--(17,4),mediumgray); draw((0,5)--(17,5),mediumgray); draw((0,6)--(17,6),mediumgray); draw((0,7)--(17,7),mediumgray); draw((0,8)--(17,8),mediumgray); draw((0,9)--(17,9),mediumgray); draw((0,10)--(17,10),mediumgray); draw((-.13,1)--(.13,1)); draw((-.13,2)--(.13,2)); draw((-.13,3)--(.13,3)); draw((-.13,4)--(.13,4)); draw((-.13,5)--(.13,5)); draw((-.13,6)--(.13,6)); draw((-.13,7)--(.13,7)); draw((-.13,8)--(.13,8)); draw((-.13,9)--(.13,9)); draw((-.13,10)--(.13,10)); draw((1,-.13)--(1,.13)); draw((2,-.13)--(2,.13)); draw((3,-.13)--(3,.13)); draw((4,-.13)--(4,.13)); draw((5,-.13)--(5,.13)); draw((6,-.13)--(6,.13)); draw((7,-.13)--(7,.13)); draw((8,-.13)--(8,.13)); draw((9,-.13)--(9,.13)); draw((10,-.13)--(10,.13)); draw((11,-.13)--(11,.13)); draw((12,-.13)--(12,.13)); draw((13,-.13)--(13,.13)); draw((14,-.13)--(14,.13)); draw((15,-.13)--(15,.13)); draw((16,-.13)--(16,.13)); label(scale(.7)*"$1$", (1,-.13), S); label(scale(.7)*"$2$", (2,-.13), S); label(scale(.7)*"$3$", (3,-.13), S); label(scale(.7)*"$4$", (4,-.13), S); label(scale(.7)*"$5$", (5,-.13), S); label(scale(.7)*"$6$", (6,-.13), S); label(scale(.7)*"$7$", (7,-.13), S); label(scale(.7)*"$8$", (8,-.13), S); label(scale(.7)*"$9$", (9,-.13), S); label(scale(.7)*"$10$", (10,-.13), S); label(scale(.7)*"$11$", (11,-.13), S); label(scale(.7)*"$12$", (12,-.13), S); label(scale(.7)*"$13$", (13,-.13), S); label(scale(.7)*"$14$", (14,-.13), S); label(scale(.7)*"$15$", (15,-.13), S); label(scale(.7)*"$16$", (16,-.13), S); label(scale(.7)*"$1$", (-.13,1), W); label(scale(.7)*"$2$", (-.13,2), W); label(scale(.7)*"$3$", (-.13,3), W); label(scale(.7)*"$4$", (-.13,4), W); label(scale(.7)*"$5$", (-.13,5), W); label(scale(.7)*"$6$", (-.13,6), W); label(scale(.7)*"$7$", (-.13,7), W); label(scale(.7)*"$8$", (-.13,8), W); label(scale(.7)*"$9$", (-.13,9), W); label(scale(.7)*"$10$", (-.13,10), W); dot((0,0),linewidth(4)); label(scale(.75)*"$A$", (0,0), NE); dot((3,1),linewidth(4)); label(scale(.75)*"$B$", (3,1), NE); dot((0,10),linewidth(4)); label(scale(.75)*"$C$", (0,10), NE); dot((2,9),linewidth(4)); label(scale(.75)*"$D$", (2,9), NE); draw((15,3)--(16,3)--(16,5)--(15,5)--cycle,linewidth(1.125)); dot((15,3),linewidth(4)); dot((16,3),linewidth(4)); dot((16,5),linewidth(4)); dot((15,5),linewidth(4)); [/asy]
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
| If we extend the lines, we have the following diagram:
[asy] usepackage("mathptmx"); size(9cm); draw((0,-.5)--(0,11),EndArrow(size=.15cm)); draw((1,0)--(1,11),mediumgray); draw((2,0)--(2,11),mediumgray); draw((3,0)--(3,11),mediumgray); draw((4,0)--(4,11),mediumgray); draw((5,0)--(5,11),mediumgray); draw((6,0)--(6,11),mediumgray); draw((7,0)--(7,11),mediumgray); draw((8,0)--(8,11),mediumgray); draw((9,0)--(9,11),mediumgray); draw((10,0)--(10,11),mediumgray); draw((11,0)--(11,11),mediumgray); draw((12,0)--(12,11),mediumgray); draw((13,0)--(13,11),mediumgray); draw((14,0)--(14,11),mediumgray); draw((15,0)--(15,11),mediumgray); draw((16,0)--(16,11),mediumgray); draw((-.5,0)--(17,0),EndArrow(size=.15cm)); draw((0,1)--(17,1),mediumgray); draw((0,2)--(17,2),mediumgray); draw((0,3)--(17,3),mediumgray); draw((0,4)--(17,4),mediumgray); draw((0,5)--(17,5),mediumgray); draw((0,6)--(17,6),mediumgray); draw((0,7)--(17,7),mediumgray); draw((0,8)--(17,8),mediumgray); draw((0,9)--(17,9),mediumgray); draw((0,10)--(17,10),mediumgray); draw((-.13,1)--(.13,1)); draw((-.13,2)--(.13,2)); draw((-.13,3)--(.13,3)); draw((-.13,4)--(.13,4)); draw((-.13,5)--(.13,5)); draw((-.13,6)--(.13,6)); draw((-.13,7)--(.13,7)); draw((-.13,8)--(.13,8)); draw((-.13,9)--(.13,9)); draw((-.13,10)--(.13,10)); draw((1,-.13)--(1,.13)); draw((2,-.13)--(2,.13)); draw((3,-.13)--(3,.13)); draw((4,-.13)--(4,.13)); draw((5,-.13)--(5,.13)); draw((6,-.13)--(6,.13)); draw((7,-.13)--(7,.13)); draw((8,-.13)--(8,.13)); draw((9,-.13)--(9,.13)); draw((10,-.13)--(10,.13)); draw((11,-.13)--(11,.13)); draw((12,-.13)--(12,.13)); draw((13,-.13)--(13,.13)); draw((14,-.13)--(14,.13)); draw((15,-.13)--(15,.13)); draw((16,-.13)--(16,.13)); label(scale(.7)*"$1$", (1,-.13), S); label(scale(.7)*"$2$", (2,-.13), S); label(scale(.7)*"$3$", (3,-.13), S); label(scale(.7)*"$4$", (4,-.13), S); label(scale(.7)*"$5$", (5,-.13), S); label(scale(.7)*"$6$", (6,-.13), S); label(scale(.7)*"$7$", (7,-.13), S); label(scale(.7)*"$8$", (8,-.13), S); label(scale(.7)*"$9$", (9,-.13), S); label(scale(.7)*"$10$", (10,-.13), S); label(scale(.7)*"$11$", (11,-.13), S); label(scale(.7)*"$12$", (12,-.13), S); label(scale(.7)*"$13$", (13,-.13), S); label(scale(.7)*"$14$", (14,-.13), S); label(scale(.7)*"$15$", (15,-.13), S); label(scale(.7)*"$16$", (16,-.13), S); label(scale(.7)*"$1$", (-.13,1), W); label(scale(.7)*"$2$", (-.13,2), W); label(scale(.7)*"$3$", (-.13,3), W); label(scale(.7)*"$4$", (-.13,4), W); label(scale(.7)*"$5$", (-.13,5), W); label(scale(.7)*"$6$", (-.13,6), W); label(scale(.7)*"$7$", (-.13,7), W); label(scale(.7)*"$8$", (-.13,8), W); label(scale(.7)*"$9$", (-.13,9), W); label(scale(.7)*"$10$", (-.13,10), W); draw((0,10)--(17,1.5),blue); draw((0,0)--(17,17/3),blue); dot((0,0),linewidth(4)); label(scale(.75)*"$A$", (0,0), NE); dot((3,1),linewidth(4)); label(scale(.75)*"$B$", (3,1), NE); dot((0,10),linewidth(4)); label(scale(.75)*"$C$", (0,10), NE); dot((2,9),linewidth(4)); label(scale(.75)*"$D$", (2,9), NE); draw((15,3)--(16,3)--(16,5)--(15,5)--cycle,linewidth(1.125)); dot((15,3),linewidth(4)); dot((16,3),linewidth(4)); dot((16,5),linewidth(4)); dot((15,5),linewidth(4)); [/asy]
Therefore, we see that the answer is $\boxed{\textbf{(B)}\ 1}.$
~MrThinker
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_7 | 1 |
AMC8_302 | In a far-off land three fish can be traded for two loaves of bread and a loaf of bread can be traded for four bags of rice. How many bags of rice is one fish worth?
$\text{(A)}\ \frac{3}{8} \qquad \text{(B)}\ \frac{1}{2} \qquad \text{(C)}\ \frac{3}{4} \qquad \text{(D)}\ 2\frac{2}{3} \qquad \text{(E)}\ 3\frac{1}{3}$
| Let $f$ represent one fish, $l$ a loaf of bread, and $r$ a bag of rice. Then:
$3f=2l$ ,
$l=4r$
Substituting $l$ from the second equation back into the first gives us $3f=8r$ . So each fish is worth $\frac{8}{3}$ bags of rice, or $2 \frac{2}{3}\Rightarrow \boxed{D}$ .
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_22 | D |
AMC8_303 | Two integers are inserted into the list $3, 3, 8, 11, 28$ to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?
$\textbf{(A) } 56 \qquad \textbf{(B) } 57 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 61$
| To double the range, we must find the current range, which is $28 - 3 = 25$ , to then double to: $2(25) = 50$ . Since we do not want to change the median, we need to get a value less than $8$ (as $8$ would change the mode) for the smaller, making $53$ fixed for the larger. Anything less than $3$ is not beneficial to the optimization because you want to get the largest range without changing the mode. So, taking our optimal values of $7$ and $53$ , we have an answer of $7 + 53 = \boxed{\textbf{(D)}\ 60}$ .
~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower, CHECKMATE2021
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_20 | 60 |
AMC8_305 | Ms. Hamilton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament.
Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10$
| There are $\binom{4}{3}$ ways to choose three starters. Thus the answer is $\boxed{\textbf{(B)}\ 4}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_4 | 4 |
AMC8_306 | The digits $2,0,2,$ and $3$ are placed in the expression below, one digit per box. What is the maximum possible value of the expression?
[asy] // Diagram by TheMathGuyd. I can compress this later size(5cm); real w=2.2; pair O,I,J; O=(0,0);I=(1,0);J=(0,1); path bsqb = O--I; path bsqr = I--I+J; path bsqt = I+J--J; path bsql = J--O; path lsqb = shift((1.2,0.75))*scale(0.5)*bsqb; path lsqr = shift((1.2,0.75))*scale(0.5)*bsqr; path lsqt = shift((1.2,0.75))*scale(0.5)*bsqt; path lsql = shift((1.2,0.75))*scale(0.5)*bsql; draw(bsqb,dashed); draw(bsqr,dashed); draw(bsqt,dashed); draw(bsql,dashed); draw(lsqb,dashed); draw(lsqr,dashed); draw(lsqt,dashed); draw(lsql,dashed); label(scale(3)*"$\times$",(w,1/3)); draw(shift(1.3w,0)*bsqb,dashed); draw(shift(1.3w,0)*bsqr,dashed); draw(shift(1.3w,0)*bsqt,dashed); draw(shift(1.3w,0)*bsql,dashed); draw(shift(1.3w,0)*lsqb,dashed); draw(shift(1.3w,0)*lsqr,dashed); draw(shift(1.3w,0)*lsqt,dashed); draw(shift(1.3w,0)*lsql,dashed); [/asy]
$\textbf{(A) }0 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }16 \qquad \textbf{(E) }18$
| First, let us consider the case where $0$ is a base: This would result in the entire expression being $0.$ Contrastingly, if $0$ is an exponent, we will get a value greater than $0.$ $3^2\times2^0=9$ is greater than $2^3\times2^0=8$ and $2^2\times3^0=4.$ Therefore, the answer is $\boxed{\textbf{(C) }9}.$
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_6 | 9 |
AMC8_307 | Blake and Jenny each took four $100$ -point tests. Blake averaged $78$ on the four tests. Jenny scored $10$ points higher than Blake on the first test, $10$ points lower than him on the second test, and $20$ points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?
$\mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40$
| Solution 1Blake scored a total of $4 \times 78=312$ points. Jenny scored $10-10+20+20=40$ points higher than Blake, so her average is $\frac{312+40}{4}=88$ .
the difference is $88-78=\boxed{\mathrm{(A)}\ 10}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_7 | 10 |
AMC8_308 | I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$
| Let the numbers be $x$ and $y$ . Then we have $x+y=11$ and $xy=24$ . Solving for $x$ in the first equation yields $x=11-y$ , and substituting this into the second equation gives $(11-y)(y)=24$ . Simplifying this gives $-y^2+11y=24$ , or $y^2-11y+24=0$ . This factors as $(y-3)(y-8)=0$ , so $y=3$ or $y=8$ , and the corresponding $x$ values are $x=8$ and $x=3$ . These are essentially the same answer: one number is $3$ and one number is $8$ , so the largest number is $8, \boxed{\text{D}}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_2 | 8 |
AMC8_309 | Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?
$\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880$
| To solve this, treat the two Arabic books as one unit and the four Spanish books as another unit. Along with the three German books, you now have five units to arrange.
You can arrange these five units in $5!$ ways. Within the Arabic block, the two books can be arranged in $2!$ ways, and within the Spanish block, the four books can be arranged in $4!$ ways.
Multiplying all these possibilities gives the total number of arrangements which is $\boxed{\textbf{(C) }5760}.$ ways to arrange the nine books while keeping the Arabic and Spanish books together.
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_16 | 5760 |
AMC8_310 | How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$ )
$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$
| Since $2020 = 2^2 \cdot 5 \cdot 101$ , we can simply list its factors: \[1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.\] There are $12$ factors; only $1, 2, 4, 5, 101$ don't have over $3$ factors, so the remaining $12-5 = \boxed{\textbf{(B) }7}$ factors have more than $3$ factors.
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_17 | 7 |
AMC8_311 | Karl's car uses a gallon of gas every $35$ miles, and his gas tank holds $14$ gallons when it is full. One day, Karl started with a full tank of gas, drove $350$ miles, bought $8$ gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day?
$\textbf{(A) }525\qquad\textbf{(B) }560\qquad\textbf{(C) }595\qquad\textbf{(D) }665\qquad \textbf{(E) }735$
| Since he uses a gallon of gas every $35$ miles, he had used $\frac{350}{35} = 10$ gallons after $350$ miles. Therefore, after the first leg of his trip he had $14 - 10 = 4$ gallons of gas left. Then, he bought $8$ gallons of gas, which brought him up to $12$ gallons of gas in his gas tank. When he arrived, he had $\frac{1}{2} \cdot 14 = 7$ gallons of gas. So he used $5$ gallons of gas on the second leg of his trip. Therefore, the second part of his trip covered $5 \cdot 35 = 175$ miles. Adding this to the $350$ miles, we see that he drove $350 + 175 = \boxed{\textbf{(A)} \, 525}$ miles.
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_14 | 525 |
AMC8_312 | A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 4 games, Ines won 3 games, Janet won 2 games, Kendra won 2 games and Lara won 2 games, how many games did Monica win?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$
| Since there are 6 players, a total of $\frac{6(6-1)}{2}=15$ games are played. So far, $4+3+2+2+2=13$ games finished (one person won from each game), so Monica needs to win $15-13 = \boxed{\textbf{(C)}\ 2}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_20 | 2 |
AMC8_313 | The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?
[asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.5,6)--(10,6)); draw((-0.5,7)--(10,7)); label("frequency",(-0.5,8)); label("0", (-1, 0)); label("1", (-1, 1)); label("2", (-1, 2)); label("3", (-1, 3)); label("4", (-1, 4)); label("5", (-1, 5)); label("6", (-1, 6)); label("7", (-1, 7)); filldraw((0,0)--(0,7)--(1,7)--(1,0)--cycle, black); filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle, black); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, black); filldraw((6,0)--(6,4)--(7,4)--(7,0)--cycle, black); filldraw((8,0)--(8,4)--(9,4)--(9,0)--cycle, black); label("3", (0.5, -0.5)); label("4", (2.5, -0.5)); label("5", (4.5, -0.5)); label("6", (6.5, -0.5)); label("7", (8.5, -0.5)); label("name length", (4.5, -1)); [/asy]
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7$
| We first notice that the median name will be the $(19+1)/2=10^{\mbox{th}}$ name. The $10^{\mbox{th}}$ name is $\boxed{\textbf{(B)}\ 4}$ .
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_6 | 4 |
AMC8_314 | The letter M in the figure below is first reflected over the line $q$ and then reflected over the line $p$ . What is the resulting image?
[asy] // pog diagram usepackage("newtxtext"); size(3cm); draw((-1,0)--(1,0)); draw((0,-1)--(0,1)); label("$\textbf{\textsf{M}}$",(0.25,0.6)); draw((-0.8,-0.8)--(0.8,0.8),linewidth(1.1)); label("$p$", (-1,0),NE); label("$q$", (-0.75,-0.75), N*1.5); [/asy]
[asy] // pog diagram usepackage("newtxtext"); size(12.5cm); draw((-1,0)--(1,0)); draw((0,-1)--(0,1)); label(rotate(90)*"$\textbf{\textsf{M}}$",(0.6,-0.25)); draw((-0.8,-0.8)--(0.8,0.8),linewidth(1.1)); label("$\textbf{(A)}$",(-1,1),W); draw((2,0)--(4,0)); draw((3,-1)--(3,1)); label(rotate(270)*"$\textbf{\textsf{M}}$",(2.8,0.7)); draw((2.2,-0.8)--(3.8,0.8),linewidth(1.1)); label("$\textbf{(B)}$",(2,1),W); draw((5,0)--(7,0)); draw((6,-1)--(6,1)); label(rotate(90)*"$\textbf{\textsf{M}}$",(5.4,0.2)); draw((5.2,-0.8)--(6.8,0.8),linewidth(1.1)); label("$\textbf{(C)}$",(5,1),W); draw((-1,-2.5)--(1,-2.5)); draw((0,-3.5)--(0,-1.5)); label(rotate(180)*"$\textbf{\textsf{M}}$",(-0.25,-3.1)); draw((-0.8,-3.3)--(0.8,-1.7),linewidth(1.1)); label("$\textbf{(D)}$",(-1,-1.5),W); draw((2,-2.5)--(4,-2.5)); draw((3,-3.5)--(3,-1.5)); label(rotate(270)*"$\textbf{\textsf{M}}$",(3.6,-2.75)); draw((2.2,-3.3)--(3.8,-1.7),linewidth(1.1)); label("$\textbf{(E)}$",(2,-1.5),W); [/asy]
| When M is first reflected over the line $q,$ we obtain the following diagram:
[asy] /* Made by MRENTHUSIASM */ usepackage("newtxtext"); size(3cm); draw((-1,0)--(1,0)); draw((0,-1)--(0,1)); label(rotate(270)*"$\textbf{\textsf{M}}$",(0.6,0.25)); draw((-0.8,-0.8)--(0.8,0.8),linewidth(1.1)); [/asy]
When M is then reflected over the line $p,$ we obtain the following diagram:
[asy] /* Made by MRENTHUSIASM */ usepackage("newtxtext"); size(3cm); draw((-1,0)--(1,0)); draw((0,-1)--(0,1)); label(rotate(270)*"$\textbf{\textsf{M}}$",(0.6,-0.25)); draw((-0.8,-0.8)--(0.8,0.8),linewidth(1.1)); [/asy]
Therefore, the answer is $\boxed{\textbf{(E)}}.$
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_4 | E |
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