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AMC8_418 | Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides $5$ times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \tfrac{1}{2}$ feet with each step. How many steps will Bella take by the time she meets Ella?
$\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520$
| Every 10 feet Bella goes, Ella goes 50 feet, which means a total of 60 feet. They need to travel that 60 feet $10560\div60=176$ times to travel the entire 2 miles. Since Bella goes 10 feet 176 times, this means that she travels a total of 1760 feet. And since she walks 2.5 feet each step, $1760\div2.5=\boxed{\textbf{(A) }704}$
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_17 | 704 |
AMC8_419 | Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$ ?
$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$
| To check, if a number is divisible by 19, take its unit digit and multiply it by 2, then add the result to the rest of the number, and repeat this step until the number is reduced to two digits. If the result is divisible by 19, then the original number is also divisible by 19.
After we got 19 eliminated, we can see that the other options have a lot of 1's in them. The divisibility rule for 11 is add alternating digits up, then take the difference of them. The example number works like that. If we add variables, ABCDEF to make number ABCABC, we can see that (A+C+B) - (B+A+C) = 0. Which is divisible by 11, so our answer choice is $\boxed{\textbf{(A)}\ 11}$ .
by: CHECKMATE2021 (edited by CHECKMATE2021)
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_7 | 11 |
AMC8_420 | What is the sum of the two smallest prime factors of $250$ ?
$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 5 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 12$
| The prime factorization of $250$ is $2 \cdot 5^3$ . The smallest two are $2$ and $5$ . $2+5 = \boxed{\text{(C) }7}$ .
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_3 | 7 |
AMC8_421 | Tyler has entered a buffet line in which he chooses one kind of meat, two different vegetables and one dessert. If the order of food items is not important, how many different meals might he choose?
| There are $3$ possibilities for the meat and $4$ possibilites for the dessert, for a total of $4\times3=12$ possibilities for the meat and the dessert. There are $4$ possibilities for the first vegetable and $3$ possibilities for the second, but order doesn't matter, so we overcounted by a factor of $2$ . For example, we counted 'baked beans and corn' and 'corn and baked beans' as $2$ different possibilities, so the total possibilites for the two vegetables is $\frac{4\times3}{2}=6$ , and the total number of possibilites is $12\times6=\boxed{\textbf{(C)}\ 72}.$
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_14 | 72 |
AMC8_422 | Sarika, Dev, and Rajiv are sharing a large block of cheese. They take turns cutting off half of what remains and eating it: first Sarika eats half of the cheese, then Dev eats half of the remaining half, then Rajiv eats half of what remains, then back to Sarika, and so on. They stop when the cheese is too small to see. About what fraction of the original block of cheese does Sarika eat in total?
$\textbf{(A)}\ \frac{4}{7} \qquad \textbf{(B)}\ \frac{3}{5} \qquad \textbf{(C)}\ \frac{2}{3} \qquad \textbf{(D)}\ \frac{3}{4} \qquad \textbf{(E)}\ \frac{7}{8}$
| WLOG, let the amount of total cheese be $1$ . Then Sarika eats $\dfrac{1}{2}$ , Dev eats $\dfrac{1}{4}$ , Rajiv eats $\dfrac{1}{8}$ , Sarika eats $\dfrac{1}{16}$ and so on. After a couple for attempts, we see that Sarika eats cheese in an infinite geometric sequence with first term $\dfrac{1}{2}$ and common ratio of $\dfrac{1}{8}$ . Therefore, we use the infinite geometric sequence formula and get \[\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{8}}=\dfrac{\dfrac{1}{2}}{\dfrac{7}{8}}=\dfrac{4}{7}\] To find how much Sarika eats, we just divide this by our original total and get $\dfrac{\dfrac{4}{7}}{1}=1$ .
Therefore, Sarika eats $\boxed{\textbf{(A) }\frac{4}{7}}$ of the cheese.
~athreyay
~ Edited by Aoum
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_20 | 4/7 |
AMC8_423 | The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5); draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray); draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray); [/asy]
$\textbf{(A) } 10 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$
| We can see that there are 4 whole squares, since the area of each square will be 1, 4 * 1 = 4. Next, there are 12 half squares, and 2 half squares is 1 whole square, so 12/2 = 6 whole squares. The area of this will be 6 * 1 = 6. Finally, we can add the 2 numbers. 4 + 6 = $\boxed{\textbf{(A)} ~10}$ .
~~Brainiacs77~~
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1 | 10 |
AMC8_424 | Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
$\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4}$
| Divide it into $2$ cases:
1) Keiko and Ephriam both get $0$ heads:
This means that they both roll all tails, so there is only $1$ way for this to happen.
2) Keiko and Ephriam both get $1$ head:
For Keiko, there is only $1$ way for this to happen because he is only flipping 1 penny, but for Ephriam, there are 2 ways since there are $2$ choices for when he can flip the head. So, in total there are $2 \cdot 1 = 2$ ways for this case.
Thus, in total there are $3$ ways that work. Since there are $2$ choices for each coin flip (Heads or Tails), there are $2^3 = 8$ total ways of flipping 3 coins.
Thus, since all possible coin flips of 3 coins are equally likely, the probability is $\boxed{(B) \frac38}$ .
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_21 | 3/8 |
AMC8_425 | Four friends do yardwork for their neighbors over the weekend, earning $$15, $20, $25,$ and $$40,$ respectively. They decide to split their earnings equally among themselves. In total, how much will the friend who earned $$40$ give to the others?
$\textbf{(A) }$5 \qquad \textbf{(B) }$10 \qquad \textbf{(C) }$15 \qquad \textbf{(D) }$20 \qquad \textbf{(E) }$25$
| The friends earn $$\left(15+20+25+40\right)=$100$ in total. Since they decided to split their earnings equally, it follows that each person will get $$\left(\frac{100}{4}\right)=$25$ . Since the friend who earned $$40$ will need to leave with $$25$ , he will have to give $$\left(40-25\right)=\boxed{\textbf{(C) }$15}$ to the others.
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_2 | 15 |
AMC8_426 | When the World Wide Web first became popular in the $1990$ s, download speeds reached a maximum of about $56$ kilobits per second. Approximately how many minutes would the download of a $4.2$ -megabyte song have taken at that speed? (Note that there are $8000$ kilobits in a megabyte.)
$\textbf{(A) } 0.6 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 1800 \qquad \textbf{(D) } 7200 \qquad \textbf{(E) } 36000$
| Notice that the number of kilobits in this song is $4.2 \cdot 8000 = 8 \cdot 7 \cdot 6 \cdot 100.$
We must divide this by $56$ in order to find out how many seconds this song would take to download: $\frac{\cancel{8}\cdot\cancel{7}\cdot6\cdot100}{\cancel{56}} = 600.$
Finally, we divide this number by $60$ because this is the number of seconds to get the answer $\frac{600}{60}=\boxed{\textbf{(B) } 10}.$
~wamofan
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_7 | 10 |
AMC8_427 | Each square in a $3 \times 3$ grid is randomly filled with one of the $4$ gray and white tiles shown below on the right.
[asy] size(5.663333333cm); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,gray); draw((1,0)--(1,3)--(2,3)--(2,0),gray); draw((0,1)--(3,1)--(3,2)--(0,2),gray); fill((6,.33)--(7,.33)--(7,1.33)--cycle,mediumgray); draw((6,.33)--(7,.33)--(7,1.33)--(6,1.33)--cycle,gray); fill((6,1.67)--(7,2.67)--(6,2.67)--cycle,mediumgray); draw((6,1.67)--(7,1.67)--(7,2.67)--(6,2.67)--cycle,gray); fill((7.33,.33)--(8.33,.33)--(7.33,1.33)--cycle,mediumgray); draw((7.33,.33)--(8.33,.33)--(8.33,1.33)--(7.33,1.33)--cycle,gray); fill((8.33,1.67)--(8.33,2.67)--(7.33,2.67)--cycle,mediumgray); draw((7.33,1.67)--(8.33,1.67)--(8.33,2.67)--(7.33,2.67)--cycle,gray); [/asy]
What is the probability that the tiling will contain a large gray diamond in one of the smaller $2 \times 2$ grids? Below is an example of such tiling.
[asy] size(2cm); fill((1,0)--(0,1)--(0,2)--(1,1)--cycle,mediumgray); fill((2,0)--(3,1)--(2,2)--(1,1)--cycle,mediumgray); fill((1,2)--(1,3)--(0,3)--cycle,mediumgray); fill((1,2)--(2,2)--(2,3)--cycle,mediumgray); fill((3,2)--(3,3)--(2,3)--cycle,mediumgray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,gray); draw((1,0)--(1,3)--(2,3)--(2,0),gray); draw((0,1)--(3,1)--(3,2)--(0,2),gray); [/asy]
$\textbf{(A) } \frac{1}{1024} \qquad \textbf{(B) } \frac{1}{256} \qquad \textbf{(C) } \frac{1}{64} \qquad \textbf{(D) } \frac{1}{16} \qquad \textbf{(E) } \frac{1}{4}$
| There are $4$ cases that the tiling will contain a large gray diamond in one of the smaller $2 \times 2$ grids, as shown below:
[asy] size(375); fill((1,1)--(2,2)--(1,3)--(0,2)--cycle,mediumgray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,gray); draw((1,0)--(1,3)--(2,3)--(2,0),gray); draw((0,1)--(3,1)--(3,2)--(0,2),gray); fill(shift(7,0)*((1,1)--(2,2)--(1,3)--(0,2)--cycle),mediumgray); draw(shift(6,0)*((0,0)--(3,0)--(3,3)--(0,3)--cycle),gray); draw(shift(6,0)*((1,0)--(1,3)--(2,3)--(2,0)),gray); draw(shift(6,0)*((0,1)--(3,1)--(3,2)--(0,2)),gray); fill(shift(12,-1)*((1,1)--(2,2)--(1,3)--(0,2)--cycle),mediumgray); draw(shift(12,0)*((0,0)--(3,0)--(3,3)--(0,3)--cycle),gray); draw(shift(12,0)*((1,0)--(1,3)--(2,3)--(2,0)),gray); draw(shift(12,0)*((0,1)--(3,1)--(3,2)--(0,2)),gray); fill(shift(19,-1)*((1,1)--(2,2)--(1,3)--(0,2)--cycle),mediumgray); draw(shift(18,0)*((0,0)--(3,0)--(3,3)--(0,3)--cycle),gray); draw(shift(18,0)*((1,0)--(1,3)--(2,3)--(2,0)),gray); draw(shift(18,0)*((0,1)--(3,1)--(3,2)--(0,2)),gray); [/asy]
There are $4^5$ ways to decide the $5$ white squares for each case, and the cases do not have any overlap.
So, the requested probability is \[\frac{4\cdot4^5}{4^9} = \frac{4^6}{4^9} = \frac{1}{4^3} = \boxed{\textbf{(C) } \frac{1}{64}}.\]
~apex304, TaeKim, MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_23 | 1/64 |
AMC8_428 | An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$ . Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. The square in the center is labelled $X$ as shown. What is the value of $X$ ?
$\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42$
[asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy]
| Solution 1We begin filling in the table. The top row has a first term $1$ and a fifth term $25$ , so we have the common difference is $\frac{25-1}4=6$ . This means we can fill in the first row of the table:
[asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((1,4), 7); draw_num((2,4), 13); draw_num((3,4), 19); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy]
The fifth row has a first term of $17$ and a fifth term of $81$ , so the common difference is $\frac{81-17}4=16$ . We can fill in the fifth row of the table as shown:
[asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((1,4), 7); draw_num((2,4), 13); draw_num((3,4), 19); draw_num((4, 4), 25); draw_num((0, 4), 1); draw_num((1, 0), 33); draw_num((2, 0), 49); draw_num((3, 0), 65); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy]
We must find the third term of the arithmetic sequence with a first term of $13$ and a fifth term of $49$ . The common difference of this sequence is $\frac{49-13}4=9$ , so the third term is $13+2\cdot 9=\boxed{\textbf{(B) }31}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_18 | 31 |
AMC8_429 | There are $270$ students at Colfax Middle School, where the ratio of boys to girls is $5 : 4$ . There are $180$ students at Winthrop Middle School, where the ratio of boys to girls is $4 : 5$ . The two schools hold a dance and all students from both schools attend. What fraction of the students at the dance are girls?
$\textbf{(A) } \dfrac7{18} \qquad\textbf{(B) } \dfrac7{15} \qquad\textbf{(C) } \dfrac{22}{45} \qquad\textbf{(D) } \dfrac12 \qquad\textbf{(E) } \dfrac{23}{45}$
| At Colfax Middle School, there are $\frac49 \times 270 = 120$ girls. At Winthrop Middle School, there are $\frac59 \times 180 = 100$ girls. The ratio of girls to the total number of students is $\frac{120+100}{270+180} = \frac{220}{450} = \boxed{\textbf{(C)}\ \frac{22}{45}}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_14 | $\frac{22}{45}$ |
AMC8_430 | Bo, Coe, Flo, Jo, and Moe have different amounts of money. Neither Jo nor Bo has as much money as Flo. Both Bo and Coe have more than Moe. Jo has more than Moe, but less than Bo. Who has the least amount of money?
$\text{(A)}\ \text{Bo}\qquad\text{(B)}\ \text{Coe}\qquad\text{(C)}\ \text{Flo}\qquad\text{(D)}\ \text{Jo}\qquad\text{(E)}\ \text{Moe}$
| Use logic to solve this problem. You don't actually need to use any equations.
Neither Jo nor Bo has as much money as Flo. So Flo clearly does not have the least amount of money. Rule out Flo.
Both Bo and Coe have more than Moe. Rule out Bo and Coe; they clearly do not have the least amount of money.
Jo has more than Moe. Rule out Jo.
The only person who has not been ruled out is Moe. So $\boxed{\text{(E)}}$ is the answer.
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_6 | E |
AMC8_432 | What is the sum of the prime factors of $2010$ ?
$\textbf{(A)}\ 67\qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 77\qquad\textbf{(D)}\ 201\qquad\textbf{(E)}\ 210$
| First, we must find the prime factorization of $2010$ . $2010=2\cdot 3 \cdot 5 \cdot 67$ . We add the factors up to get $\boxed{\textbf{(C)}\ 77}$
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_14 | 77 |
AMC8_433 | Let $a, b$ and $c$ be numbers with $0 < a < b < c$ . Which of the following is
impossible?
$\mathrm{(A)} \ a + c < b \qquad \mathrm{(B)} \ a \cdot b < c \qquad \mathrm{(C)} \ a + b < c \qquad \mathrm{(D)} \ a \cdot c < b \qquad \mathrm{(E)}\frac{b}{c} = a$
| According to the given rules, every number needs to be positive. Since $c$ is always greater than $b$ , adding a positive number ( $a$ ) to $c$ will always make it greater than $b$ .
Therefore, the answer is $\boxed{\textbf{(A)}\ a+c<b}$
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_15 | a+c<b |
AMC8_434 | When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 47 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 49$
| Let the two digits be $a$ and $b$ .
The correct score was $10a+b$ . Clara misinterpreted it as $10b+a$ . The difference between the two is $|9a-9b|$ which factors into $|9(a-b)|$ . Therefore, since the difference is a multiple of 9, the only answer choice that is a multiple of 9 is $\boxed{\textbf{(A)}\ 45}$ .
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_13 | 45 |
AMC8_435 | If $20\%$ of a number is $12$ , what is $30\%$ of the same number?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 30$
| $20\%$ of a number is equal to $\frac{1}{5}$ of that number. Let $n$ =the number
$\frac{1}{5}n = 12$ Multiply both sides by 5
$n = 60$
$30\%$ of $n$ is equal to $\frac{3}{10}n = \frac{3}{10}\cdot60 = 3\cdot6 = \boxed{\textbf{(B)}18}$
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_5 | 18 |
AMC8_436 | Handy Aaron helped a neighbor $1 \frac14$ hours on Monday, $50$ minutes on Tuesday, from 8:20 to 10:45 on Wednesday morning, and a half-hour on Friday. He is paid $\textdollar 3$ per hour. How much did he earn for the week?
$\textbf{(A)}\ \textdollar 8 \qquad \textbf{(B)}\ \textdollar 9 \qquad \textbf{(C)}\ \textdollar 10 \qquad \textbf{(D)}\ \textdollar 12 \qquad \textbf{(E)}\ \textdollar 15$
| Let's convert everything to minutes and add them together. On Monday he worked for $\frac54 \cdot 60 = 75$ minutes. On Tuesday he worked $50$ minutes. On Wednesday he worked for $2$ hours $25$ minutes, or $2(60)+25=145$ minutes. On Friday he worked $\frac{60}{2}=30$ minutes. This adds up to $75+50+145+30=300$ minutes, or $300/60=5$ hours and $5\cdot 3 = \boxed{\textbf{(E)}\ \textdollar 15}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_10 | 15 |
AMC8_437 | Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together?
[asy] draw((-1,0)--(0,0)--(0,1)); draw((0,0)--(0.309, -0.951)); filldraw(arc((0,0), (0,1), (-1,0))--(0,0)--cycle, lightgray); filldraw(arc((0,0), (0.309, -0.951), (0,1))--(0,0)--cycle, gray); draw(arc((0,0), (-1,0), (0.309, -0.951))); label("Colby", (-0.5, 0.5)); label("25\%", (-0.5, 0.3)); label("Alicia", (0.7, 0.2)); label("45\%", (0.7, 0)); label("Brenda", (-0.5, -0.4)); label("30\%", (-0.5, -0.6)); [/asy]
$\textbf{(A) }70 \qquad \textbf{(B) }84 \qquad \textbf{(C) }100 \qquad \textbf{(D) }106 \qquad \textbf{(E) }120$
| Let $x$ be the total amount of votes casted. From the chart, Brenda received $30\%$ of the votes and had $36$ votes. We can express this relationship as $\frac{30}{100}x=36$ . Solving for $x$ , we get $x=\boxed{\textbf{(E)}\ 120}.$
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_2 | 120 |
AMC8_438 | Which of the following numbers is not a perfect square?
$\textbf{(A) }1^{2016}\qquad\textbf{(B) }2^{2017}\qquad\textbf{(C) }3^{2018}\qquad\textbf{(D) }4^{2019}\qquad \textbf{(E) }5^{2020}$
| Our answer must have an odd exponent in order for it to not be a square. Because $4$ is a perfect square, $4^{2019}$ is also a perfect square, so our answer is $\boxed{\textbf{(B) }2^{2017}}$ .
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_7 | 2^2017 |
AMC8_439 | Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?
$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18$
| Say Laila gets a value of $x$ on her first 4 tests, and a value of $y$ on her last test. Thus, $4x+y=82 \cdot 5=410$ .
Because $x$ and $y$ are different, $x$ must be less than $82$ and $y$ must be greater than $82$ . When $x$ decreases by $1$ , $y$ must increase by $4$ to keep the total constant.
The greatest value for $y$ is $98$ (as $y=100$ would make $x$ non-integer). In the range $83\le y\le 98$ , only $4$ values for $y$ result in integer values for $x$ : 86, 90, 94 and 98. Thus, the answer is $\boxed{\textbf{(A) }4}$ .
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_13 | 4 |
AMC8_440 | A collector offers to buy state quarters for 2000% of their face value. At that rate how much will Bryden get for his four state quarters?
$\text{(A)}\ 20\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 200\text{ dollars} \qquad \text{(D)}\ 500\text{ dollars} \qquad \text{(E)}\ 2000\text{ dollars}$
| $2000\%$ is equivalent to $20\times100\%$ . Therefore, $2000\%$ of a number is the same as $20$ times that number. $4$ quarters is $1$ dollar, so Bryden will get $20\times1={20}$ dollars, $\boxed{\text{A}}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_10 | A |
AMC8_441 | The eight-pointed star, shown in the figure below, is a popular quilting pattern. What percent of the entire $4\times4$ grid is covered by the star?
[asy] path x = (0,1)--(1,2)--(2,2)--(1,1)--cycle; path y = reflect((0,0),(4,4)) * x; path z = (1,0)--(2,1)--(3,0)--(3,1)--(2,2)--(1,1); fill(x, gray(0.6)); fill(rotate(90, (2,2)) * x, gray(0.6)); fill(rotate(180, (2,2)) * x, gray(0.6)); fill(rotate(270, (2,2)) * x, gray(0.6)); fill(y, gray(0.8)); fill(rotate(90, (2,2)) * y, gray(0.8)); fill(rotate(180, (2,2)) * y, gray(0.8)); fill(rotate(270, (2,2)) * y, gray(0.8)); draw(z); draw(rotate(90, (2,2)) * z); draw(rotate(180, (2,2)) * z); draw(rotate(270, (2,2)) * z); add(grid(4,4)); [/asy]
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 80$
| Each of the unshaded triangles has base length $2$ and height $1$ , so they all have area $\frac{2 \cdot 1}{2} = 1$ . Each of the unshaded unit squares has area $1$ . The area of the shaded region is equal to the area of the entire grid minus the area of the unshaded region, or $4^2 - 4 \cdot 1 - 4 \cdot 1 = 8$ . The star is then $\frac{8}{16} = \frac{1}{2} = \frac{50}{100}$ , or $\boxed{\textbf{(B)}~50}$ percent of the entire grid.
~cxsmi
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_1 | 50 |
AMC8_442 | A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted $7$ children and $19$ wheels. How many tricycles were there?
$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7$
| Solution 1If all the children were riding bicycles, there would be $2 \times 7=14$ wheels. Each tricycle adds an extra wheel and $19-14=5$ extra wheels are needed, so there are $\boxed{\mathrm{(C)}\ 5}$ tricycles.
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_4 | 5 |
AMC8_443 | Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom, and sides.
[asy] /* AMC8 2002 #22 Problem */ draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1)); draw((1,0)--(1.5,0.5)--(1.5,1.5)); draw((0.5,1.5)--(1,2)--(1.5,2)); draw((1.5,1.5)--(1.5,3.5)--(2,4)--(3,4)--(2.5,3.5)--(2.5,0.5)--(1.5,.5)); draw((1.5,3.5)--(2.5,3.5)); draw((1.5,1.5)--(3.5,1.5)--(3.5,2.5)--(1.5,2.5)); draw((3,4)--(3,3)--(2.5,2.5)); draw((3,3)--(4,3)--(4,2)--(3.5,1.5)); draw((4,3)--(3.5,2.5)); draw((2.5,.5)--(3,1)--(3,1.5));[/asy]
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 36$
| Count the number of sides that are not exposed, where a cube is connected to another cube and subtract it from the total number of faces. There are $5$ places with two adjacent cubes, covering $10$ sides, and $(6)(6)=36$ faces. The exposed surface area is $36-10 = \boxed{\text{(C)}\ 26}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_22 | 26 |
AMC8_444 | Paul owes Paula $35$ cents and has a pocket full of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins that he can use to pay her. What is the difference between the largest and the smallest number of coins he can use to pay her?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$
| The fewest amount of coins that can be used is $2$ (a quarter and a dime). The greatest amount is $7$ , if he only uses nickels. Therefore we have $7-2=\boxed{\textbf{(E)}~5}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_2 | 5 |
AMC8_445 | In square $ABCE$ , $AF=2FE$ and $CD=2DE$ . What is the ratio of the area of $\triangle BFD$ to the area of square $ABCE$ ?
[asy] size((100)); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((3,0)--(9,9)--(0,3)--cycle); dot((3,0)); dot((0,3)); dot((9,9)); dot((0,0)); dot((9,0)); dot((0,9)); label("$A$", (0,9), NW); label("$B$", (9,9), NE); label("$C$", (9,0), SE); label("$D$", (3,0), S); label("$E$", (0,0), SW); label("$F$", (0,3), W); [/asy]
$\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{2}{9}\qquad\textbf{(C)}\ \frac{5}{18}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{7}{20}$
| The area of $\triangle BFD$ is the area of square $ABCE$ subtracted by the the area of the three triangles around it. Arbitrarily assign the side length of the square to be $6$ .
[asy] size((100)); pair A=(0,9), B=(9,9), C=(9,0), D=(3,0), E=(0,0), F=(0,3); pair[] ps={A,B,C,D,E,F}; dot(ps); draw(A--B--C--E--cycle); draw(B--F--D--cycle); label("$A$",A, NW); label("$B$",B, NE); label("$C$",C, SE); label("$D$",D, S); label("$E$",E, SW); label("$F$",F, W); label("$6$",A--B,N); label("$6$",(10,4.5),E); label("$4$",D--C,S); label("$2$",E--D,S); label("$2$",E--F,W); label("$4$",F--A,W); [/asy]
The ratio of the area of $\triangle BFD$ to the area of $ABCE$ is
\[\frac{36-12-12-2}{36} = \frac{10}{36} = \boxed{\textbf{(C)}\ \frac{5}{18}}\]
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_23 | 5/18 |
AMC8_446 | A cup of boiling water ( $212^{\circ}\text{F}$ ) is placed to cool in a room whose temperature remains constant at $68^{\circ}\text{F}$ . Suppose the difference between the water temperature and the room temperature is halved every $5$ minutes. What is the water temperature, in degrees Fahrenheit, after $15$ minutes?
$\textbf{(A) } 77 \qquad \textbf{(B) } 86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 98 \qquad \textbf{(E) } 104$
| Initially, the difference between the water temperature and the room temperature is $212-68=144$ degrees Fahrenheit.
After $5$ minutes, the difference between the temperatures is $144\div2=72$ degrees Fahrenheit.
After $10$ minutes, the difference between the temperatures is $72\div2=36$ degrees Fahrenheit.
After $15$ minutes, the difference between the temperatures is $36\div2=18$ degrees Fahrenheit. At this point, the water temperature is $68+18=\boxed{\textbf{(B) } 86}$ degrees Fahrenheit.
Remark
Alternatively, we can condense the solution above into the following equation: \[68+(212-68)\cdot\left(\frac12\right)^{\tfrac{15}{5}}=86.\]
~MRENTHUSIASM ~MathFun1000
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_9 | 86 |
AMC8_447 | A square piece of paper, 4 inches on a side, is folded in half vertically. Both layers are then cut in half parallel to the fold. Three new rectangles are formed, a large one and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle?
[asy] draw((0,8)--(0,0)--(4,0)--(4,8)--(0,8)--(3.5,8.5)--(3.5,8)); draw((2,-1)--(2,9),dashed); [/asy]
$\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{1}{2} \qquad \text{(C)}\ \dfrac{3}{4} \qquad \text{(D)}\ \dfrac{4}{5} \qquad \text{(E)}\ \dfrac{5}{6}$
| The smaller rectangles each have the same height as the original square, but have $\frac{1}{4}$ the length, since the paper is folded in half and then cut in half the same way. The larger rectangle has the same height as the original square but has $\frac{1}{2}$ the length, since the paper is cut in half after the fold but the fold retains both sides of the larger rectangle. Therefore, the smaller rectangles have dimensions $4 \times 1$ and the larger rectangle has dimensions $4 \times 2$ . The ratio of their perimeters is $\frac{2(4+1)}{2(4+2)}=\frac{5}{6}, \boxed{\text{E}}$
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_16 | E |
AMC8_448 | [] | There are $9$ choices for the first number, since it cannot be $0$ , there are only $9$ choices left for the second number since it must differ from the first, $8$ choices for the third number, since it must differ from the first two, and $7$ choices for the fourth number, since it must differ from all three. This means there are $9 \times 9 \times 8 \times 7=\boxed{\textbf{(B) }4536}$ integers between $1000$ and $9999$ with four distinct digits.
If you're too lazy to do $9 \times 9 \times 8 \times 7= 4536$ , we can notice $9 \times 9 = 81$ and $8 \times 7=56$ The unit digit must be 6. Therefore the answer is the only answer with a a unit digit of 6. $\boxed{B}$ )
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_10 | 4536 |
AMC8_450 | Car M traveled at a constant speed for a given time. This is shown by the dashed line. Car N traveled at twice the speed for the same distance. If Car M and Car N's speed and time are shown as solid line, which graph illustrates this?
[asy] unitsize(12); draw((0,9)--(0,0)--(9,0)); label("time",(4.5,0),S); label("s",(0,7),W); label("p",(0,6),W); label("e",(0,5),W); label("e",(0,4),W); label("d",(0,3),W); label("(A)",(-1,9),NW); draw((0,4)--(4,4),dashed); label("M",(4,4),E); draw((0,8)--(4,8),linewidth(1)); label("N",(4,8),E); draw((15,9)--(15,0)--(24,0)); label("time",(19.5,0),S); label("s",(15,7),W); label("p",(15,6),W); label("e",(15,5),W); label("e",(15,4),W); label("d",(15,3),W); label("(B)",(14,9),NW); draw((15,4)--(19,4),dashed); label("M",(19,4),E); draw((15,8)--(23,8),linewidth(1)); label("N",(23,8),E); draw((30,9)--(30,0)--(39,0)); label("time",(34.5,0),S); label("s",(30,7),W); label("p",(30,6),W); label("e",(30,5),W); label("e",(30,4),W); label("d",(30,3),W); label("(C)",(29,9),NW); draw((30,4)--(34,4),dashed); label("M",(34,4),E); draw((30,2)--(34,2),linewidth(1)); label("N",(34,2),E); draw((0,-6)--(0,-15)--(9,-15)); label("time",(4.5,-15),S); label("s",(0,-8),W); label("p",(0,-9),W); label("e",(0,-10),W); label("e",(0,-11),W); label("d",(0,-12),W); label("(D)",(-1,-6),NW); draw((0,-11)--(4,-11),dashed); label("M",(4,-11),E); draw((0,-7)--(2,-7),linewidth(1)); label("N",(2,-7),E); draw((15,-6)--(15,-15)--(24,-15)); label("time",(19.5,-15),S); label("s",(15,-8),W); label("p",(15,-9),W); label("e",(15,-10),W); label("e",(15,-11),W); label("d",(15,-12),W); label("(E)",(14,-6),NW); draw((15,-11)--(19,-11),dashed); label("M",(19,-11),E); draw((15,-13)--(23,-13),linewidth(1)); label("N",(23,-13),E); [/asy]
| Since car N has twice the speed, it must be twice as high on the speed axis. Also, since cars M and N travel at the same distance but car N has twice the speed, car N must take half the time. Therefore, line N must be half the size of line M. Since the speeds are constant, both lines are horizontal. Reviewing the graphs, we see that the only one satisfying these conditions is graph $\boxed{\text{D}}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_19 | D |
AMC8_451 | Carlos Montado was born on Saturday, November 9, 2002. On what day of the week will Carlos be 706 days old?
$\text{(A)}\ \text{Monday}\qquad\text{(B)}\ \text{Wednesday}\qquad\text{(C)}\ \text{Friday}\qquad\text{(D)}\ \text{Saturday}\qquad\text{(E)}\ \text{Sunday}$
| Days of the week have a cycle that repeats every $7$ days. Thus, after $100$ cycles, or $700$ days, it will be Saturday again. Six more days will make it $\text{Friday} \rightarrow \boxed{C}$
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_5 | C |
AMC8_452 | Fifteen integers $a_1, a_2, a_3, \dots, a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that
\[1 \le a_1 \le 10, \thickspace 13 \le a_2 \le 20, \thickspace \text{ and } \thickspace 241 \le a_{15}\le 250.\]
What is the sum of digits of $a_{14}?$
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12$
| We can find the possible values of the common difference by finding the numbers which satisfy the conditions. To do this, we find the minimum of the last two: $241-20=221$ , and the maximum– $250-13=237$ . There is a difference of $13$ between them, so only $17$ and $18$ work, as $17\cdot13=221$ , so $17$ satisfies $221\leq 13x\leq237$ . The number $18$ is similarly found. $19$ , however, is too much.
Now, we check with the first and last equations using the same method. We know $241-10\leq 14x\leq250-1$ . Therefore, $231\leq 14x\leq249$ . We test both values we just got, and we can realize that $18$ is too large to satisfy this inequality. On the other hand, we can now find that the difference will be $17$ , which satisfies this inequality.
The last step is to find the first term. We know that the first term can only be from $1$ to $3$ since any larger value would render the second inequality invalid. Testing these three, we find that only $a_1=3$ will satisfy all the inequalities. Therefore, $a_{14}=13\cdot17+3=224$ . The sum of the digits is therefore $\boxed{\textbf{(A)}\ 8}$ .
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_25 | 8 |
AMC8_453 | A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are 6.1 cm, 8.2 cm and 9.7 cm. What is the area of the square in square centimeters?
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64$
| The perimeter of the triangle is $6.1+8.2+9.7=24$ cm. A square's perimeter is four times its sidelength, since all its sidelengths are equal. If the square's perimeter is $24$ , the sidelength is $24/4=6$ , and the area is $6^2=\boxed{\textbf{(C)}\ 36}$ .
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_4 | 36 |
AMC8_454 | On a dark and stormy night Snoopy suddenly saw a flash of lightning. Ten seconds later he heard the sound of thunder. The speed of sound is 1088 feet per second and one mile is 5280 feet. Estimate, to the nearest half-mile, how far Snoopy was from the flash of lightning.
$\text{(A)}\ 1 \qquad \text{(B)}\ 1\frac{1}{2} \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 2\frac{1}{2} \qquad \text{(E)}\ 3$
| During the $10$ seconds, the sound traveled $1088\times10=10880$ feet from the lightning to Snoopy. This is equivalent to $\frac{10880}{5280}\approx2$ miles, $\boxed{\text{C}}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_5 | C |
AMC8_455 | A jar contains $5$ different colors of gumdrops. $30\%$ are blue, $20\%$ are brown, $15\%$ are red, $10\%$ are yellow, and other $30$ gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?
$\textbf{(A)}\ 35\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64$
| We do $100-30-20-15-10$ to find the percent of gumdrops that are green. We find that $25\%$ of the gumdrops are green. That means there are $120$ gumdrops. If we replace half of the blue gumdrops with brown gumdrops, then $15\%$ of the jar's gumdrops are brown. $\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{\textbf{(C)}\ 42}$
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_15 | 42 |
AMC8_456 | A straight one-mile stretch of highway, 40 feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at 5 miles per hour, how many hours will it take to cover the one-mile stretch?
[asy]size(10cm); pathpen=black; pointpen=black; D(arc((-2,0),1,300,360)); D(arc((0,0),1,0,180)); D(arc((2,0),1,180,360)); D(arc((4,0),1,0,180)); D(arc((6,0),1,180,240)); D((-1.5,-1)--(5.5,-1));[/asy]
Note: 1 mile = 5280 feet
$\textbf{(A) }\frac{\pi}{11}\qquad\textbf{(B) }\frac{\pi}{10}\qquad\textbf{(C) }\frac{\pi}{5}\qquad\textbf{(D) }\frac{2\pi}{5}\qquad\textbf{(E) }\frac{2\pi}{3}$
| Solution 1There are two possible interpretations of the problem: that the road as a whole is $40$ feet wide, or that each lane is $40$ feet wide. Both interpretations will arrive at the same result. However, let us stick with the first interpretation for simplicity. Each lane must then be $20$ feet wide, so Robert must be riding his bike in semicircles with radius $20$ feet and diameter $40$ feet. Since the road is $5280$ feet long, over the whole mile, Robert rides $\frac{5280}{40} =132$ semicircles in total. Were the semicircles full circles, their circumference would be $2\pi\cdot 20=40\pi$ feet; as it is, the circumference of each is half that, or $20\pi$ feet. Therefore, over the stretch of highway, Robert rides a total of $132\cdot 20\pi =2640\pi$ feet, equivalent to $\frac{\pi}{2}$ miles. Robert rides at 5 miles per hour, so divide the $\frac{\pi}{2}$ miles by $5$ mph (because $t = \frac{d}{r}$ and time = distance/rate) to arrive at $\boxed{\textbf{(B) }\frac{\pi}{10}}$ hours.
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_25 | frac{π}{10} |
AMC8_457 | Joe had walked half way from home to school when he realized he was late. He ran the rest of the way to school. He ran 3 times as fast as he walked. Joe took 6 minutes to walk half way to school. How many minutes did it take Joe to get from home to school?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 7.3\qquad\textbf{(C)}\ 7.7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 8.3$
| We can use the equation $d=rt$ where $d$ is the distance, $r$ is the rate, and $t$ is the time. The distances he ran and walked are equal, so $r_rt_r=r_wt_w$ , where $r_r$ is the rate at which he ran, $t_r$ is the time for which he ran, $r_w$ is the rate at which he walked, and $t_w$ is the time for which he walked. Because he runs three times faster than he walks, $r_r=3r_w$ . We want to find the time he ran, $t_r = \frac{r_wt_w}{t_r} = \frac{(r_w)(6)}{3r_w} = 2$ minutes. He traveled for a total of $6+2=\boxed{\textbf{(D)}\ 8}$ minutes.
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_10 | 8 |
AMC8_458 | In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.
[asy] fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray); fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray); label("$8$", (1, 0)); label("$C$", (2, 0)); label("$8$", (3, 0)); label("$8$", (0, 1)); label("$C$", (1, 1)); label("$M$", (2, 1)); label("$C$", (3, 1)); label("$8$", (4, 1)); label("$C$", (0, 2)); label("$M$", (1, 2)); label("$A$", (2, 2)); label("$M$", (3, 2)); label("$C$", (4, 2)); label("$8$", (0, 3)); label("$C$", (1, 3)); label("$M$", (2, 3)); label("$C$", (3, 3)); label("$8$", (4, 3)); label("$8$", (1, 4)); label("$C$", (2, 4)); label("$8$", (3, 4));[/asy]
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36$
| Notice that the upper-most section contains a 3 by 3 square that looks like:
[asy]label("$8$", (1, 2)); label("$C$", (2, 2)); label("$8$", (3, 2)); label("$C$", (1, 1)); label("$M$", (2, 1)); label("$C$", (3, 1)); label("$M$", (1, 0)); label("$A$", (2, 0)); label("$M$", (3, 0));[/asy]
It has 6 paths in which you can spell out AMC8. You will find four identical copies of this square in the figure, so multiply ${6 \cdot 4}$ to get $\boxed{\textbf{(D)}\ 24}$ total paths.
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15 | 24 |
AMC8_459 | Ryan got $80\%$ of the problems correct on a $25$ -problem test, $90\%$ on a $40$ -problem test, and $70\%$ on a $10$ -problem test. What percent of all the problems did Ryan answer correctly?
$\textbf{(A)}\ 64 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 84\qquad\textbf{(E)}\ 86$
| Ryan answered $(0.8)(25)=20$ problems correct on the first test, $(0.9)(40)=36$ on the second, and $(0.7)(10)=7$ on the third. This amounts to a total of $20+36+7=63$ problems correct. The total number of problems is $25+40+10=75.$ Therefore, the percentage is $\dfrac{63}{75} = 84\% \rightarrow \boxed{\textbf{(D)}\ 84}$
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_9 | 84 |
AMC8_460 | The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer (1,000 meters)?
$\textbf{(A)}\ 9^\text{th} \qquad \textbf{(B)}\ 10^\text{th} \qquad \textbf{(C)}\ 11^\text{th} \qquad \textbf{(D)}\ 12^\text{th} \qquad \textbf{(E)}\ 13^\text{th}$
| This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that $2^{10}=1024$ .
However, because the first term is $2^0=1$ and not $2^1=2$ , the solution to the problem is $10-0+1=\boxed{\textbf{(C)}\ 11^{\text{th}}}$
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_9 | 11th |
AMC8_462 | A large square region is paved with $n^2$ gray square tiles, each measuring $s$ inches on a side. A border $d$ inches wide surrounds each tile. The figure below shows the case for $n=3$ . When $n=24$ , the $576$ gray tiles cover $64\%$ of the area of the large square region. What is the ratio $\frac{d}{s}$ for this larger value of $n?$
[asy] draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray); filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray); filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray); filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray); filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray); filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray); filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray); filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray); filldraw((12,12)--(12,9)--(9,9)--(9,12)--cycle, mediumgray); [/asy]
$\textbf{(A) }\frac{6}{25} \qquad \textbf{(B) }\frac{1}{4} \qquad \textbf{(C) }\frac{9}{25} \qquad \textbf{(D) }\frac{7}{16} \qquad \textbf{(E) }\frac{9}{16}$
| The area of the shaded region is $(24s)^2$ . To find the area of the large square, we note that there is a $d$ -inch border between each of the $23$ pairs of consecutive squares, as well as from between the first/last squares and the large square, for a total of $23+2 = 25$ times the length of the border, i.e. $25d$ . Adding this to the total length of the consecutive squares, which is $24s$ , the side length of the large square is $(24s+25d)$ , yielding the equation $\frac{(24s)^2}{(24s+25d)^2}=\frac{64}{100}$ . Taking the square root of both sides (and using the fact that lengths are non-negative) gives $\frac{24s}{24s+25d}=\frac{8}{10} = \frac{4}{5}$ , and cross-multiplying now gives $120s = 96s + 100d \Rightarrow 24s = 100d \Rightarrow \frac{d}{s} = \frac{24}{100} = \boxed{\textbf{(A) }\frac{6}{25}}$ .
Note: Once we obtain $\tfrac{24s}{24s+25d} = \tfrac{4}{5},$ to ease computation, we may take the reciprocal of both sides to yield $\tfrac{24s+25d}{24s} = 1 + \tfrac{25d}{24s} = \tfrac{5}{4},$ so $\tfrac{25d}{24s} = \tfrac{1}{4}.$ Multiplying both sides by $\tfrac{24}{25}$ yields the same answer as before.
~peace09
~Minor Edits by WrenMath
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_24 | 6/25 |
AMC8_463 | In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If $\frac{1}{3}$ of all the ninth graders are paired with $\frac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
$\textbf{(A) } \frac{2}{15} \qquad\textbf{(B) } \frac{4}{11} \qquad\textbf{(C) } \frac{11}{30} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{11}{15}$
| Let the number of sixth graders be $s$ , and the number of ninth graders be $n$ . Thus, $\frac{n}{3}=\frac{2s}{5}$ , which simplifies to $n=\frac{6s}{5}$ . Since we are trying to find the value of $\frac{\frac{n}{3}+\frac{2s}{5}}{n+s}$ , we can just substitute $\frac{6s}{5}$ for $n$ into the equation. We then get a value of $\frac{\frac{\frac{6s}{5}}{3}+\frac{2s}{5}}{\frac{6s}{5}+s} = \frac{\frac{6s+6s}{15}}{\frac{11s}{5}} = \frac{\frac{4s}{5}}{\frac{11s}{5}} = \boxed{\textbf{(B)}~\frac{4}{11}}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_16 | 4/11 |
AMC8_464 | Isabella has $6$ coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every $10$ days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the $6$ dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?
$\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}$
| Let $\text{Day }1$ to $\text{Day\\ }2$ denote a day where one coupon is redeemed and the day when the second coupon is redeemed.
If she starts on a $\text{Monday}$ , she redeems her next coupon on $\text{Thursday}$ .
$\text{Thursday}$ to $\text{Sunday}$ .
Thus, $\textbf{(A)}\ \text{Monday}$ is incorrect.
If she starts on a $\text{Tuesday}$ , she redeems her next coupon on $\text{Friday}$ .
$\text{Friday}$ to $\text{Monday}$ .
$\text{Monday}$ to $\text{Thursday}$ .
$\text{Thursday}$ to $\text{Sunday}$ .
Thus $\textbf{(B)}\ \text{Tuesday}$ is incorrect.
If she starts on a $\text{Wednesday}$ , she redeems her next coupon on $\text{Saturday}$ .
$\text{Saturday}$ to $\text{Tuesday}$ .
$\text{Tuesday}$ to $\text{Friday}$ .
$\text{Friday}$ to $\text{Monday}$ .
$\text{Monday}$ to $\text{Thursday}$ .
And on $\text{Thursday}$ , she redeems her last coupon.
No Sunday occured; thus, $\boxed{\textbf{(C)}\ \text{Wednesday}}$ is correct.
Checking for the other options,
If she starts on a $\text{Thursday}$ , she redeems her next coupon on $\text{Sunday}$ .
Thus, $\textbf{(D)}\ \text{Thursday}$ is incorrect.
If she starts on a $\text{Friday}$ , she redeems her next coupon on $\text{Monday}$ .
$\text{Monday}$ to $\text{Thursday}$ .
$\text{Thursday}$ to $\text{Sunday}$ .
Checking for the other options gave us negative results; thus, the answer is $\boxed{\textbf{(C)}\ \text{Wednesday}}$ .
~KangarooPrecise and ( ??? )
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_14 | Wednesday |
AMC8_465 | A decorative window is made up of a rectangle with semicircles at either end. The ratio of $AD$ to $AB$ is $3:2$ . And $AB$ is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircles?
[asy] import graph; size(5cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(8); defaultpen(dps); pen ds=black; real xmin=-4.27,xmax=14.73,ymin=-3.22,ymax=6.8; draw((0,4)--(0,0)); draw((0,0)--(2.5,0)); draw((2.5,0)--(2.5,4)); draw((2.5,4)--(0,4)); draw(shift((1.25,4))*xscale(1.25)*yscale(1.25)*arc((0,0),1,0,180)); draw(shift((1.25,0))*xscale(1.25)*yscale(1.25)*arc((0,0),1,-180,0)); dot((0,0),ds); label("$A$",(-0.26,-0.23),NE*lsf); dot((2.5,0),ds); label("$B$",(2.61,-0.26),NE*lsf); dot((0,4),ds); label("$D$",(-0.26,4.02),NE*lsf); dot((2.5,4),ds); label("$C$",(2.64,3.98),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]
$\textbf{(A)}\ 2:3\qquad\textbf{(B)}\ 3:2\qquad\textbf{(C)}\ 6:\pi\qquad\textbf{(D)}\ 9:\pi\qquad\textbf{(E)}\ 30 :\pi$
| We can set a proportion:
\[\dfrac{AD}{AB}=\dfrac{3}{2}\]
We substitute $AB$ with 30 and solve for $AD$ .
\[\dfrac{AD}{30}=\dfrac{3}{2}\]
\[AD=45\]
We calculate the combined area of semicircle by putting together semicircle $AB$ and $CD$ to get a circle with radius $15$ . Thus, the area is $225\pi$ . The area of the rectangle is $30\cdot 45=1350$ . We calculate the ratio:
\[\dfrac{1350}{225\pi}=\dfrac{6}{\pi}\Rightarrow\boxed{\textbf{(C)}\ 6:\pi}\]
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_18 | 6:π |
AMC8_466 | Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's overall percentage of correct answers?
$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$
| Let the number of questions that they solved alone be $x$ . Let the percentage of problems they correctly solve together be $a$ %.
As given, \[\frac{80x}{100} + \frac{ax}{100} = \frac{2 \cdot 88x}{100}\] .
Hence, $a = 96$ .
Zoe got $\frac{90x}{100} + \frac{ax}{100} = \frac{186x}{100}$ problems right out of $2x$ . Therefore, Zoe got $\frac{\frac{186x}{100}}{2x} = \frac{93}{100} = \boxed{\textbf{(C) } 93}$ percent of the problems correct.
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14 | 93 |
AMC8_467 | The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$ . Altitude $\overline{XC}$ bisects $\overline{YZ}$ . The area (in square inches) of the shaded region is
[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), N); label(scale(0.8)*"$Y$", (0,0), W); label(scale(0.8)*"$Z$", (10,0), E); label(scale(0.8)*"$A$", (2.5,2.2), W); label(scale(0.8)*"$B$", (7.5,2.2), E); label(scale(0.8)*"$C$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white);[/asy]
$\textbf{(A)}\ 1\frac{1}2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac{1}2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 3\frac{1}2$
| We know the area of triangle $XYZ$ is $8$ square inches. The area of a triangle can also be represented as $\frac{bh}{2}$ or in this problem $\frac{XC\cdot YZ}{2}$ . By solving, we have \[\frac{XC\cdot YZ}{2} = 8,\] \[XC\cdot YZ = 16.\]
With SAS congruence, triangles $XCY$ and $XCZ$ are congruent. Hence, triangle $XCY = \frac{8}{2} = 4$ . (Let's say point $D$ is the intersection between line segments $XC$ and $AB$ .) We can find the area of the trapezoid $ADCY$ by subtracting the area of triangle $XAD$ from $4$ .
We find the area of triangle $XAD$ by the $\frac{bh}{2}$ formula- $\frac{XD\cdot AD}{2} = \frac{\frac{XC}{2}\cdot AD}{2}$ . $AD$ is $\frac{1}{4}$ of $YZ$ from solution 1. The area of $XAD$ is \[\frac{\frac{XC}{2}\cdot \frac{YZ}{4}}{2} = \frac{16}{16} = 1\] .
Therefore, the area of the shaded area- trapezoid $ADCY$ has area $4-1 = \boxed{\text{(D)}\ 3}$ .
- sarah07
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_20 | 3 |
AMC8_468 | Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?
$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac23$
| The favorable responses are either they both show a green bean or they both show a red bean. The probability that both show a green bean is $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$ . The probability that both show a red bean is $\frac{1}{2}\cdot\frac{2}{4}=\frac{1}{4}$ . Therefore the probability is $\frac{1}{4}+\frac{1}{8}=\boxed{\textbf{(C)}\ \frac{3}{8}}$
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_14 | 3/8 |
AMC8_469 | Margie's winning art design is shown. The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Which of the following is closest to the percent of the design that is black?
[asy] real d=320; pair O=origin; pair P=O+8*dir(d); pair A0 = origin; pair A1 = O+1*dir(d); pair A2 = O+2*dir(d); pair A3 = O+3*dir(d); pair A4 = O+4*dir(d); pair A5 = O+5*dir(d); filldraw(Circle(A0, 6), white, black); filldraw(circle(A1, 5), black, black); filldraw(circle(A2, 4), white, black); filldraw(circle(A3, 3), black, black); filldraw(circle(A4, 2), white, black); filldraw(circle(A5, 1), black, black); [/asy]
$\textbf{(A)}\ 42\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 48\qquad$
| Let the smallest circle be 1, the second smallest circle be 2, the third smallest circle be 3, etc.
\[\begin{array}{c|cc} \text{circle \#} & \text{radius} & \text{area} \\ \hline 1 & 2 & 4\pi \\ 2 & 4 & 16\pi \\ 3 & 6 & 36\pi \\ 4 & 8 & 64\pi \\ 5 & 10 & 100\pi \\ 6 & 12 & 144\pi \end{array}\]
The entire circle's area is $144\pi$ . The area of the black regions is $(100-64)\pi + (36-16)\pi + 4\pi = 60\pi$ . The percentage of the design that is black is $\frac{60\pi}{144\pi} = \frac{5}{12} \approx \boxed{\textbf{(A)}\ 42}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_25 | 42 |
AMC8_470 | Points $R$ , $S$ and $T$ are vertices of an equilateral triangle, and points $X$ , $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be
drawn using any three of these six points as vertices?
[asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3)); dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z); label("$S$",SS,N); label("$R$",R,SW); label("$T$",T,SE); label("$X$",X,S); label("$Y$",Y,NW); label("$Z$",Z,NE); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20$
| There are $6$ points in the figure, and $3$ of them are needed to form a triangle, so there are ${6\choose{3}} =20$ possible triplets of the $6$ points. However, some of these created congruent triangles, and some don't even make triangles at all.
Case 1: Triangles congruent to There is obviously only $1$ of these: $\triangle RST$ itself.
Case 2: Triangles congruent to There are $4$ of these: $\triangle SYZ, \triangle RXY, \triangle TXZ,$ and $\triangle XYZ$ .
Case 3: Triangles congruent to There are $6$ of these: $\triangle RSX, \triangle TSX, \triangle STY, \triangle RTY, \triangle RSZ,$ and $\triangle RTZ$ .
Case 4: Triangles congruent to There are again $6$ of these: $\triangle SYX, \triangle SZX, \triangle TYZ, \triangle TYX, \triangle RXZ,$ and $\triangle RYZ$ .
However, if we add these up, we accounted for only $1+4+6+6=17$ of the $20$ possible triplets. We see that the remaining triplets don't even form triangles; they are $SYR, RXT,$ and $TZS$ . Adding these $3$ into the total yields for all of the possible triplets, so we see that there are only $4$ possible non-congruent, non-degenerate triangles, $\boxed{(D)}$
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_23 | D |
AMC8_471 | Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}.
When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters?
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60$
| Solution 1There are currently $5$ choices for the first letter, $3$ choices for the second letter, and $4$ choices for the third letter, for a total of $5 \cdot 3 \cdot 4 = 60$ license plates.
Adding $2$ letters to the start gives $7\cdot 3 \cdot 4 = 84$ plates.
Adding $2$ letters to the middle gives $5 \cdot 5 \cdot 4 = 100$ plates.
Adding $2$ letters to the end gives $5 \cdot 3 \cdot 6 = 90$ plates.
Adding a letter to the start and middle gives $6 \cdot 4 \cdot 4 = 96$ plates.
Adding a letter to the start and end gives $6 \cdot 3 \cdot 5 = 90$ plates.
Adding a letter to the middle and end gives $5 \cdot 4 \cdot 5 = 100$ plates.
You can get at most $100$ license plates total, giving an additional $100 - 60 = 40$ plates, making the answer $\boxed {D}$
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_15 | D |
AMC8_472 | In how many ways can the letters in $\textbf{BEEKEEPER}$ be rearranged so that two or more $\textbf{E}$ s do not appear together?
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 24 \qquad \textbf{(E) } 120$
| All valid arrangements of the letters must be of the form \[\textbf{E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E}.\]
The problem is equivalent to counting the arrangements of $\textbf{B},\textbf{K},\textbf{P},$ and $\textbf{R}$ into the four blanks, in which there are $4!=\boxed{\textbf{(D) } 24}$ ways.
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_14 | 24 |
AMC8_473 | Onkon wants to cover his room's floor with his favourite red carpet. How many square yards of red carpet are required to cover a rectangular floor that is $12$ feet long and $9$ feet wide? (There are 3 feet in a yard.)
$\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972$
| We need $12\cdot9$ square feet of carpet to cover the floor. Since there are $9$ square feet in a square yard, we divide this by $9$ to get $\bold{\boxed{\textbf{(A) }12}}$ square yards.
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_1 | 12 |
AMC8_474 | For any positive integer $M$ , the notation $M!$ denotes the product of the integers $1$ through
$M$ . What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$ ?
$\textbf{(A) }23 \qquad \textbf{(B) }24 \qquad \textbf{(C) }25 \qquad \textbf{(D) }26 \qquad \textbf{(E) }27$
| Factoring out $98!+99!+100!$ , we have $98! (1+99+99*100)$ , which is $98! (10000)$ . Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors of $5$ . The $19$ is because of all the multiples of $5$ .The $3$ is because of all the multiples of $25$ . Now, $10,000$ has $4$ factors of $5$ , so there are a total of $22 + 4 = \boxed{\textbf{(D)}\ 26}$ factors of $5$ .
~CHECKMATE2021
Note: Do you know what formula this uses? Most AMC 8 test takers won't know it.
It's Legendre's Formula.
https://artofproblemsolving.com/wiki/index.php/Legendre%27s_Formula?srsltid=AfmBOoovTeAI0nS9kUtxSrX27kDSKO89Fw2xxDngnBwnXcKW5eql12H5
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_19 | 26 |
AMC8_475 | How many distinct triangles can be drawn using three of the dots below as vertices?
[asy]dot(origin^^(1,0)^^(2,0)^^(0,1)^^(1,1)^^(2,1));[/asy]
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$
| The number of ways to choose three points to make a triangle is $\binom 63 = 20$ . However, two* of these are a straight line so we subtract $2$ to get $\boxed{\textbf{(C)}\ 18}$ .
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_21 | 18 |
AMC8_476 | Alice needs to replace a light bulb located $10$ centimeters below the ceiling in her kitchen. The ceiling is $2.4$ meters above the floor. Alice is $1.5$ meters tall and can reach $46$ centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?
$\textbf{(A)}\ 32 \qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 38\qquad\textbf{(E)}\ 40$
| Convert everything to the same unit. Since the answer is in centimeters, change meters to centimeters by moving the decimal place two places to the right.
The ceiling is $240$ centimeters above the floor. The combined height of Alice and the light bulb when she reaches for it is $10+150+46=206$ centimeters. That means the stool's height needs to be $240-206=\boxed{\textbf{(B)}\ 34}$
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_5 | 34 |
AMC8_477 | Annie and Bonnie are running laps around a $400$ -meter oval track. They started together, but Annie has pulled ahead, because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
$\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25$
| Solution 1Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken. This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run $\boxed{\textbf{(D)}\ 5 }$ laps.
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_16 | 5 |
AMC8_478 | Which of the following figures has the greatest number of lines of symmetry?
$\textbf{(A)}\ \text{equilateral triangle}$
$\textbf{(B)}\ \text{non-square rhombus}$
$\textbf{(C)}\ \text{non-square rectangle}$
$\textbf{(D)}\ \text{isosceles trapezoid}$
$\textbf{(E)}\ \text{square}$
| An equilateral triangle has $3$ lines of symmetry.
A non-square rhombus has $2$ lines of symmetry.
A non-square rectangle has $2$ lines of symmetry.
An isosceles trapezoid has $1$ line of symmetry.
A square has $4$ lines of symmetry.
Therefore, the answer is $\boxed{ \textbf{(E)}\ \text{square} }$ .
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_6 | square |
AMC8_479 | The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and $\frac{3}{4}$ of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 17\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$
| Let $b$ be the number of boys and $g$ be the number of girls.
\[\frac23 b = \frac34 g \Rightarrow b = \frac98 g\]
For $g$ and $b$ to be integers, $g$ must cancel out with the denominator, and the smallest possible value is $8$ . This yields $9$ boys. The minimum number of students is $8+9=\boxed{\textbf{(B)}\ 17}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_20 | 17 |
AMC8_480 | A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is $\frac{1}{2}$ foot from the top face. The second cut is $\frac{1}{3}$ foot below the first cut, and the third cut is $\frac{1}{17}$ foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet?
[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label("A", (1,0,3/4), W); label("B", (1,0,1/3), W); label("C", (1,0,1/6-d/4), W); label("D", (1,0,d/2), W); label("1/2", (1,1,3/4), E); label("1/3", (1,1,1/3), E); label("1/17", (0,1,1/6-d/4), E);[/asy]
[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label("D", (1/2, 1, 0), SE); label("A", (1+1/2, 1, 0), SE); label("B", (2+1/2, 1, 0), SE); label("C", (3+1/2, 1, 0), SE);[/asy]
$\textbf{(A)}\:6\qquad\textbf{(B)}\:7\qquad\textbf{(C)}\:\frac{419}{51}\qquad\textbf{(D)}\:\frac{158}{17}\qquad\textbf{(E)}\:11$
| The areas of the tops of $A$ , $B$ , $C$ , and $D$ in the figure formed has sum $1+1+1+1 = 4$ as do the bottoms. Thus, the total so far is $8$ . Now, one of the sides has an area of one, since it combines all of the heights of $A$ , $B$ , $C$ , and $D$ , which is $1$ . The other side is also the same. Thus the total area now is $10$ . From the front, the surface area is half, because if you looked at it straight from the front it would look exactly like a side of $A$ , with a surface area of half. From the back, it is the same thing. Thus, the total surface area is $10+\frac{1}{2}+\frac{1}{2}= 11$ , or $\boxed{\textbf{(E)}\:11 }$ .
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_25 | 11 |
AMC8_481 | On a twenty-question test, each correct answer is worth 5 points, each unanswered question is worth 1 point and each incorrect answer is worth 0 points. Which of the following scores is NOT possible?
$\text{(A)}\ 90 \qquad \text{(B)}\ 91 \qquad \text{(C)}\ 92 \qquad \text{(D)}\ 95 \qquad \text{(E)}\ 97$
| The highest possible score is if you get every answer right, to get $5(20)=100$ . The second highest possible score is if you get $19$ questions right and leave the remaining one blank, to get a $5(19)+1(1)=96$ . Therefore, no score between $96$ and $100$ , exclusive, is possible, so $97$ is not possible, $\boxed{\text{E}}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_22 | E |
AMC8_482 | In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of $1$ square unit, then what is the area of the shaded region, in square units?
[asy] size(4cm); filldraw(scale(2)*unitcircle,gray,black); filldraw(shift(-1,0)*unitcircle,white,black); filldraw(shift(1,0)*unitcircle,white,black); [/asy]
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}$
| Let the radius of the large circle be $R$ . Then, the radius of the smaller circles are $\frac R2$ . The areas of the circles are directly proportional to the square of the radii, so the ratio of the area of the small circle to the large one is $\frac 14$ ( $\frac {R^2}{4}$ is $\frac 14$ of $R^2$ .) This means the combined area of the 2 smaller circles is half of the larger circle, and therefore the shaded region is equal to the combined area of the 2 smaller circles, which is $\boxed{\textbf{(D) } 1}$ .
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_15 | 1 |
AMC8_483 | How many whole numbers between 1 and 1000 do not contain the digit 1?
$\textbf{(A)}\ 512 \qquad \textbf{(B)}\ 648 \qquad \textbf{(C)}\ 720 \qquad \textbf{(D)}\ 728 \qquad \textbf{(E)}\ 800$
| Note that this is the same as finding how many numbers with up to three digits do not contain 1.
Since there are 10 total possible digits, and only one of them is not allowed (1), each place value has its choice of 9 digits, for a total of $9*9*9=729$ such numbers. However, we overcounted by one; 0 is not between 1 and 1000, so there are $\boxed{\textbf{(D)}\ 728}$ numbers.
Note: As stated above, 020 is equivalent to 20, but 000 is not between 1 and 1000.
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_22 | 728 |
AMC8_484 | Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers?
$\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 28$
| Note that the sum of the first two numbers is $21\cdot2=42,$ the sum of the middle two numbers is $26\cdot2=52,$ and the sum of the last two numbers is $30\cdot2=60.$
It follows that the sum of the four numbers is $42+60=102,$ so the sum of the first and last numbers is $102-52=50.$ Therefore, the average of the first and last numbers is $50\div2=\boxed{\textbf{(B) } 25}.$
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_16 | 25 |
AMC8_485 | Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test?
$\textbf{(A)}\hspace{.05in}90\qquad\textbf{(B)}\hspace{.05in}92\qquad\textbf{(C)}\hspace{.05in}95\qquad\textbf{(D)}\hspace{.05in}96\qquad\textbf{(E)}\hspace{.05in}97$
| Isabella wants an average grade of $95$ on her 4 tests; this also means that she wants the sum of her test scores to be at least $95 \times 4 = 380$ (if she goes over this number, she'll be over her goal!). She's already taken two tests, which sum to $97+91 = 188$ , which means she needs $192$ more points to achieve her desired average. In order to minimize the score on the third test, we assume that Isabella will receive all $100$ points on the fourth test. Therefore, the lowest Isabella could have scored on the third test would be $192-100 = \boxed{\textbf{(B)}\ 92}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_7 | 92 |
AMC8_486 | In the figure, the outer equilateral triangle has area $16$ , the inner equilateral triangle has area $1$ , and the three trapezoids are congruent. What is the area of one of the trapezoids?
[asy] size((70)); draw((0,0)--(7.5,13)--(15,0)--(0,0)); draw((1.88,3.25)--(9.45,3.25)); draw((11.2,0)--(7.5,6.5)); draw((9.4,9.7)--(5.6,3.25)); [/asy]
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$
| The area outside the small triangle but inside the large triangle is $16-1=15$ . This is equally distributed between the three trapezoids. Each trapezoid has an area of $15/3 = \boxed{\textbf{(C)}\ 5}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_4 | 5 |
AMC8_487 | Barney Schwinn notices that the odometer on his bicycle reads $1441$ , a palindrome, because it reads the same forward and backward. After riding $4$ more hours that day and $6$ the next, he notices that the odometer shows another palindrome, $1661$ . What was his average speed in miles per hour?
$\textbf{(A)}\ 15\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 20\qquad \textbf{(E)}\ 22$
| Barney travels $1661-1441=220$ miles in $4+6=10$ hours for an average of $220/10=\boxed{\textbf{(E)}\ 22}$ miles per hour.
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_5 | 22 |
AMC8_488 | In the pattern below, the cat moves clockwise through the four squares, and the mouse moves counterclockwise through the eight exterior segments of the four squares.
| Break this problem into two parts: where the cat will be after the $247^{\text{th}}$ move, and where the mouse will be.
The cat has four possible positions in 1 cycle which are repeated every four moves. $247$ has a remainder of $3$ when divided by $4$ . This corresponds to the position the cat has after the 3rd move, which is the bottom right corner.
Similarly, the mouse has eight possible positions in 1 cycle that repeat every eight moves. $247$ has a remainder of $7$ when divided by $8$ . This corresponds to the position the mouse has after the 7th move which is bottom left corner.
The only arrangement with the mouse in that position and the cat in the bottom right square is $\boxed{\textbf{(A)}}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_23 | A |
AMC8_489 | On Monday, Taye has $$2$ . Every day, he either gains $$3$ or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, $3$ days later?
$\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7$
| How many dollar values could be on the first day? Only $2$ dollars. The second day, you can either add $3$ dollars, or double, so you can have $5$ dollars, or $4$ . For each of these values, you have $2$ values for each. For $5$ dollars, you have $10$ dollars or $8$ , and for $4$ dollars, you have $8$ dollars or $7$ dollars. Now, you have $2$ values for each of these. For $10$ dollars, you have $13$ dollars or $20$ , for $8$ dollars, you have $16$ dollars or $11$ , for $8$ dollars, you have $16$ dollars or $11$ , and for $7$ dollars, you have $14$ dollars or $10$ .
On the final day, there are 11, 11, 16, and 16 repeating, leaving you with $8-2 = \boxed{\textbf{(D)\ 6}}$ different values.
~ cxsmi (minor formatting edits)
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_8 | 6 |
AMC8_490 | Lucius is counting backward by $7$ s. His first three numbers are $100$ , $93$ , and $86$ . What is his $10$ th number?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 37 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 47$
| We plug $a=100, d=-7$ and $n=10$ into the formula $a+d(n-1)$ for the $n$ th term of an arithmetic sequence whose first term is $a$ and common difference is $d$ to get $100-7(10-1) = \boxed{\text{(B)\ 37}}$ .
~Soupboy0
| https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_4 | 37 |
AMC8_491 | Margie bought $3$ apples at a cost of $50$ cents per apple. She paid with a 5-dollar bill. How much change did Margie receive?
$\textbf{(A)}\ \textdollar 1.50 \qquad \textbf{(B)}\ \textdollar 2.00 \qquad \textbf{(C)}\ \textdollar 2.50 \qquad \textbf{(D)}\ \textdollar 3.00 \qquad \textbf{(E)}\ \textdollar 3.50$
| $50$ cents is equivalent to $\textdollar 0.50.$ Then the three apples cost $3 \times \textdollar 0.50 = \textdollar 1.50.$ The change Margie receives is $\textdollar 5.00 - \textdollar 1.50 = \boxed{\textbf{(E)}\ \textdollar 3.50}$
-Mathloveraldoxxx5223
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_1 | 3.50 |
AMC8_492 | Let $R$ be a set of nine distinct integers. Six of the elements are $2$ , $3$ , $4$ , $6$ , $9$ , and $14$ . What is the number of possible values of the median of $R$ ?
$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}8$
| First, we find that the minimum value of the median of $R$ will be $3$ .
We then experiment with sequences of numbers to determine other possible medians.
Median: $3$
Sequence: $-2, -1, 0, 2, 3, 4, 6, 9, 14$
Median: $4$
Sequence: $-1, 0, 2, 3, 4, 6, 9, 10, 14$
Median: $5$
Sequence: $0, 2, 3, 4, 5, 6, 9, 10, 14$
Median: $6$
Sequence: $0, 2, 3, 4, 6, 9, 10, 14, 15$
Median: $7$
Sequence: $2, 3, 4, 6, 7, 8, 9, 10, 14$
Median: $8$
Sequence: $2, 3, 4, 6, 8, 9, 10, 14, 15$
Median: $9$
Sequence: $2, 3, 4, 6, 9, 14, 15, 16, 17$
Any number greater than $9$ also cannot be a median of set $R$ .
Therefore, the answer is $\boxed{\textbf{(D)}\ 7}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_22 | 7 |
AMC8_493 | What is the last digit of: \[222{,}222-22{,}222-2{,}222-222-22-2?\]
$\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 8\qquad\textbf{(E) } 10$
| We can rewrite the expression as $222,222-(22,222+2,222+222+22+2)$ . We note that the units digit of $22,222+2,222+222+22+2$ is $0$ because all the units digits of the five numbers are $2$ and $5\cdot2=10$ , which has a units digit of $0$ . Now, we have something with a units digit of $0$ subtracted from $222,222$ , and so the units digit of this expression is $\boxed{\textbf{(B) } 2}$ .
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_1 | 2 |
AMC8_494 | What is the product of $\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdots\times\frac{2006}{2005}$ ?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1002\qquad\textbf{(C)}\ 1003\qquad\textbf{(D)}\ 2005\qquad\textbf{(E)}\ 2006$
| The numerator in each fraction cancels out with the denominator of the next fraction. There are only two numbers that didn't cancel: $\frac{2006}{2}=\boxed{\textbf{(C)}\ 1003}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_9 | 1003 |
AMC8_495 | A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?
$\textbf{(A)}\ \frac{\sqrt{\pi}}{2}\qquad\textbf{(B)}\ \sqrt{\pi}\qquad\textbf{(C)}\ \pi\qquad\textbf{(D)}\ 2\pi\qquad\textbf{(E)}\ \pi^{2}$
| Let the side length of the square be $s$ , and let the radius of the circle be $r$ . Thus we have $s^2=r^2\pi$ . Dividing each side by $r^2$ , we get $\frac{s^2}{r^2}=\pi$ . Since $\left(\frac{s}{r}\right)^2=\frac{s^2}{r^2}$ , we have $\frac{s}{r}=\sqrt{\pi}\Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}$
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_16 | √π |
AMC8_496 | Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has $4$ seats: $1$ Driver seat, $1$ front passenger seat, and $2$ back passenger seats. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 24$
| There are only $2$ people who can go in the driver's seat--Bonnie and Carlo. Any of the $3$ remaining people can go in the front passenger seat. There are $2$ people who can go in the first back passenger seat, and the remaining person must go in the last seat. Thus, there are $2\cdot3\cdot2$ or $12$ ways. The answer is then $\boxed{\textbf{(D)}\ 12}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_16 | 12 |
AMC8_497 | Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of $10$ miles per hour. Jack walks to the pool at a constant speed of $4$ miles per hour. How many minutes before Jack does Jill arrive?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$
| Using $d=rt$ , we can set up an equation for when Jill arrives at the swimming pool:
$1=10t$
Solving for $t$ , we get that Jill gets to the pool in $\frac{1}{10}$ of an hour, which is $6$ minutes. Doing the same for Jack, we get that
Jack arrives at the pool in $\frac{1}{4}$ of an hour, which in turn is $15$ minutes. Thus, Jill has to wait $15-6=\boxed{\textbf{(D)}~9}$
minutes for Jack to arrive at the pool.
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_3 | 9 |
AMC8_498 | The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is
$\text{(A)}\ 19 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 40$
| Since there is an odd number of terms, the median is the number in the middle, specifically, the third largest number is $18$ , and there are $2$ numbers less than $18$ and $2$ numbers greater than $18$ . The sum of these integers is $5(15)=75$ , since the mean is $15$ . To make the largest possible number with a given sum, the other numbers must be as small as possible. The two numbers less than $18$ must be positive and distinct, so the smallest possible numbers for these are $1$ and $2$ . The number right after $18$ also needs to be as small as possible, so it must be $19$ . This means that the remaining number, the maximum possible value for a number in the set, is $75-1-2-18-19=35, \boxed{\text{(D) 35}}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_21 | 35 |
AMC8_499 | Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textdollar\underline{1} \underline{A} \underline{2}$ . What is the missing digit $A$ of this $3$ -digit number?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4$
| Since all the eleven members paid the same amount, that means that the total must be divisible by $11$ . We can do some trial-and-error to get $A=3$ , so our answer is $\boxed{\textbf{(D)}~3}$
~SparklyFlowers
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_8 | 3 |
AMC8_500 | A group of frogs (called an army) is living in a tree. A frog turns green when in the shade and turns yellow
when in the sun. Initially, the ratio of green to yellow frogs was $3 : 1$ . Then $3$ green frogs moved to the
sunny side and $5$ yellow frogs moved to the shady side. Now the ratio is $4 : 1$ . What is the difference
between the number of green frogs and the number of yellow frogs now?
$\textbf{(A) } 10\qquad\textbf{(B) } 12\qquad\textbf{(C) } 16\qquad\textbf{(D) } 20\qquad\textbf{(E) } 24$
| Let the initial number of green frogs be $g$ and the initial number of yellow frogs be $y$ . Since the ratio of the number of green frogs to yellow frogs is initially $3 : 1$ , $g = 3y$ . Now, $3$ green frogs move to the sunny side and $5$ yellow frogs move to the shade side, thus the new number of green frogs is $g + 2$ and the new number of yellow frogs is $y - 2$ . We are given that $\frac{g + 2}{y - 2} = \frac{4}{1}$ , so $g + 2 = 4y - 8$ , since $g = 3y$ , we have $3y + 2 = 4y - 8$ , so $y = 10$ and $g = 30$ . Thus the answer is $(g + 2) - (y - 2) = 32 - 8 = \boxed{(E) \hspace{1 mm} 24}.$
-anonchalantdreadhead
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_21 | 24 |
AMC8_501 | The graph shows the number of minutes studied by both Asha (black bar) and Sasha (grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha?
[asy] size(300); real i; defaultpen(linewidth(0.8)); draw((0,140)--origin--(220,0)); for(i=1;i<13;i=i+1) { draw((0,10*i)--(220,10*i)); } label("$0$",origin,W); label("$20$",(0,20),W); label("$40$",(0,40),W); label("$60$",(0,60),W); label("$80$",(0,80),W); label("$100$",(0,100),W); label("$120$",(0,120),W); path MonD=(20,0)--(20,60)--(30,60)--(30,0)--cycle,MonL=(30,0)--(30,70)--(40,70)--(40,0)--cycle,TuesD=(60,0)--(60,90)--(70,90)--(70,0)--cycle,TuesL=(70,0)--(70,80)--(80,80)--(80,0)--cycle,WedD=(100,0)--(100,100)--(110,100)--(110,0)--cycle,WedL=(110,0)--(110,120)--(120,120)--(120,0)--cycle,ThurD=(140,0)--(140,80)--(150,80)--(150,0)--cycle,ThurL=(150,0)--(150,110)--(160,110)--(160,0)--cycle,FriD=(180,0)--(180,70)--(190,70)--(190,0)--cycle,FriL=(190,0)--(190,50)--(200,50)--(200,0)--cycle; fill(MonD,grey); fill(MonL,lightgrey); fill(TuesD,grey); fill(TuesL,lightgrey); fill(WedD,grey); fill(WedL,lightgrey); fill(ThurD,grey); fill(ThurL,lightgrey); fill(FriD,grey); fill(FriL,lightgrey); draw(MonD^^MonL^^TuesD^^TuesL^^WedD^^WedL^^ThurD^^ThurL^^FriD^^FriL); label("M",(30,-5),S); label("Tu",(70,-5),S); label("W",(110,-5),S); label("Th",(150,-5),S); label("F",(190,-5),S); label("M",(-25,85),W); label("I",(-27,75),W); label("N",(-25,65),W); label("U",(-25,55),W); label("T",(-25,45),W); label("E",(-25,35),W); label("S",(-26,25),W);[/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$
| Average the differences between each day. We get $10, -10,\text{ } 20,\text{ } 30,-20$ . We find the average of this list to get $\boxed{\textbf{(A)}\ 6}$ . ( In case you were wondering, the way to calculate the average is $\frac{(10+(-10)+20+30+(-20))}{5} = \frac{ 30}{5} = 6$ . So the answer is indeed $\boxed{\textbf{(A)}\ 6}$ )
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_11 | 6 |
AMC8_502 | An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it
is an odd integer whose digits are all distinct?
(A) $\dfrac{14}{75}$ (B) $\dfrac{56}{225}$ (C) $\dfrac{107}{400}$ (D) $\dfrac{7}{25}$ (E) $\dfrac{9}{25}$
| There are $5$ options for the last digit as the integer must be odd. The first digit now has $8$ options left (it can't be $0$ or the same as the last digit). The second digit also has $8$ options left (it can't be the same as the first or last digit). Finally, the third digit has $7$ options (it can't be the same as the three digits that are already chosen).
Since there are $9,000$ total integers, our answer is \[\frac{8 \cdot 8 \cdot 7 \cdot 5}{9000} = \boxed{\textbf{(B)}\ \frac{56}{225}}.\]
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_20 | 56/225 |
AMC8_503 | Hammie is in $6^\text{th}$ grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?
$\textbf{(A)}\ \text{median, by 60} \qquad \textbf{(B)}\ \text{median, by 20} \qquad \textbf{(C)}\ \text{average, by 5}\qquad \textbf{(D)}\ \text{average, by 15} \qquad \textbf{(E)}\ \text{average, by 20}$
| Listing the elements from least to greatest, we have $(5, 5, 6, 8, 106)$ , we see that the median weight is 6 pounds.
The average weight of the five kids is $\frac{5+5+6+8+106}{5} = \frac{130}{5} = 26$ .
Hence, \[26-6=\boxed{\textbf{(E)}\ \text{average, by 20}}.\]
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_5 | 20 |
AMC8_504 | NASA’s Perseverance Rover was launched on July $30,$ $2020.$ After traveling $292{,}526{,}838$ miles, it landed on Mars in Jezero Crater about $6.5$ months later. Which of the following is closest to the Rover’s average interplanetary speed in miles per hour?
$\textbf{(A)}\ 6{,}000 \qquad \textbf{(B)}\ 12{,}000 \qquad \textbf{(C)}\ 60{,}000 \qquad \textbf{(D)}\ 120{,}000 \qquad \textbf{(E)}\ 600{,}000$
| Note that $6.5$ months is approximately $6.5\cdot30\cdot24$ hours. Therefore, the speed (in miles per hour) is \[\frac{292{,}526{,}838}{6.5\cdot30\cdot24} \approx \frac{300{,}000{,}000}{6.5\cdot30\cdot24} = \frac{10{,}000{,}000}{6.5\cdot24} \approx \frac{10{,}000{,}000}{6.4\cdot25} = \frac{10{,}000{,}000}{160} = 62500 \approx \boxed{\textbf{(C)}\ 60{,}000}.\]
As the answer choices are far apart from each other, we can ensure that the approximation is correct.
~apex304, SohumUttamchandani, MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_11 | 60,000 |
AMC8_505 | Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ 39\qquad \textbf{(E)}\ 46$
| The union of two sets is equal to the sum of each set minus their intersection. The number of students that have both a dog and a cat is $20+26-39 = \boxed{\textbf{(A)}\ 7}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_11 | 7 |
AMC8_506 | Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?
$\textbf{(A)}\ \text{\textdollar}120\qquad\textbf{(B)}\ \text{\textdollar}128\qquad\textbf{(C)}\ \text{\textdollar}140\qquad\textbf{(D)}\ \text{\textdollar}144\qquad\textbf{(E)}\ \text{\textdollar}160$
| Since Judi's 7 friends had to pay $2.50 extra each to cover the total amount that Judi should have paid, we multiply $2.50\cdot7=17.50$ is the bill Judi would have paid if she had money. Hence, to calculate the total amount, we multiply $17.50\cdot8=\boxed{\textbf{(C) } 140}$ to find the total the 8 friends paid.
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_4 | 140 |
AMC8_507 | Suppose that $a * b$ means $3a-b.$ What is the value of $x$ if
\[2 * (5 * x)=1\]
$\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14$
| Let us plug in $(5 * x)=1$ into $3a-b$ . Thus it would be $3(5)-x$ . Now we have $2*(15-x)=1$ . Plugging $2*(15-x)$ into $3a-b$ , we have $6-15+x=1$ . Solving for $x$ we have \[-9+x=1\] \[x=\boxed{\textbf{(D)} \, 10}\]
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_10 | 10 |
AMC8_508 | Alice has $24$ apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?
$\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$
| Note: This solution uses the non-negative version for stars and bars. A solution using the positive version of stars is similar (first removing an apple from each person instead of 2).
This method uses the counting method of stars and bars (non-negative version). Since each person must have at least $2$ apples, we can remove $2*3$ apples from the total that need to be sorted. With the remaining $18$ apples, we can use stars and bars to determine the number of possibilities. Assume there are $18$ stars in a row, and $2$ bars, which will be placed to separate the stars into groups of $3$ . In total, there are $18$ spaces for stars $+ 2$ spaces for bars, for a total of $20$ spaces. We can now do $20 \choose 2$ . This is because if we choose distinct $2$ spots for the bars to be placed, each combo of $3$ groups will be different, and all apples will add up to $18$ . We can also do this because the apples are indistinguishable. $20 \choose 2$ is $190$ , therefore the answer is $\boxed{\textbf{(C) }190}$ .
~goofytaipan91
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_25 | 190 |
AMC8_509 | A triangle with vertices as $A=(1,3)$ , $B=(5,1)$ , and $C=(4,4)$ is plotted on a $6\times5$ grid. What fraction of the grid is covered by the triangle?
$\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}$
[asy] draw((1,0)--(1,5),linewidth(.5)); draw((2,0)--(2,5),linewidth(.5)); draw((3,0)--(3,5),linewidth(.5)); draw((4,0)--(4,5),linewidth(.5)); draw((5,0)--(5,5),linewidth(.5)); draw((6,0)--(6,5),linewidth(.5)); draw((0,1)--(6,1),linewidth(.5)); draw((0,2)--(6,2),linewidth(.5)); draw((0,3)--(6,3),linewidth(.5)); draw((0,4)--(6,4),linewidth(.5)); draw((0,5)--(6,5),linewidth(.5)); draw((0,0)--(0,6),EndArrow); draw((0,0)--(7,0),EndArrow); draw((1,3)--(4,4)--(5,1)--cycle); label("$y$",(0,6),W); label("$x$",(7,0),S); label("$A$",(1,3),dir(210)); label("$B$",(5,1),SE); label("$C$",(4,4),dir(100)); [/asy]
| Solution 1The area of $\triangle ABC$ is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is $\sqrt{1^2+2^2}=\sqrt{5}$ , and its base is $\sqrt{2^2+4^2}=\sqrt{20}$ . We multiply these and divide by $2$ to find the area of the triangle is $\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5$ . Since the grid has an area of $30$ , the fraction of the grid covered by the triangle is $\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_19 | 1/6 |
AMC8_510 | What is the units digit of $13^{2012}$ ?
$\textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}5\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}9$
| The problem wants us to find the units digit of $13^{2012}$ , therefore, we can eliminate the tens digit of $13$ , because the tens digit will not affect the final result. So our new expression is $3^{2012}$ . Now we need to look for a pattern in the units digit.
$3^1 \implies 3$
$3^2 \implies 9$
$3^3 \implies 7$
$3^4 \implies 1$
$3^5 \implies 3$
We observe that there is a pattern for the units digit which recurs every four powers of three. Using this pattern, we can subtract 1 from 2012 and divide by 4. The remainder is the power of three that we are looking for, plus one. $2011$ divided by $4$ leaves a remainder of $3$ , so the answer is the units digit of $3^{3+1}$ , or $3^4$ . Thus, we find that the units digit of $13^{2012}$ is
$\boxed{{\textbf{(A)}\ 1}}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_12 | 1 |
AMC8_511 | A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures $8$ inches high and $10$ inches wide. What is the area of the border, in square inches?
$\textbf{(A)}\hspace{.05in}36\qquad\textbf{(B)}\hspace{.05in}40\qquad\textbf{(C)}\hspace{.05in}64\qquad\textbf{(D)}\hspace{.05in}72\qquad\textbf{(E)}\hspace{.05in}88$
| In order to find the area of the frame, we need to subtract the area of the photograph from the area of the photograph and the frame together. The area of the photograph is $8 \times 10 = 80$ square inches. The height of the whole frame (including the photograph) would be $8+2+2 = 12$ , and the width of the whole frame, $10+2+2 = 14$ . Therefore, the area of the whole figure would be $12 \times 14 = 168$ square inches. Subtracting the area of the photograph from the area of both the frame and photograph, we find the answer to be $168-80 = \boxed{\textbf{(E)}\ 88}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_6 | 88 |
AMC8_512 | A rectangular garden 60 feet long and 20 feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden?
$\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500$
| We need the same perimeter as a $60$ by $20$ rectangle, but the greatest area we can get. right now the perimeter is $160$ . To get the greatest area while keeping a perimeter of $160$ , the sides should all be $40$ . that means an area of $1600$ . Right now, the area is $20 \times 60$ which is $1200$ . $1600-1200=400$ which is $\boxed{D}$ .
| https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_5 | D |
AMC8_513 | A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are $R_1 = 100$ inches, $R_2 = 60$ inches, and $R_3 = 80$ inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?
[asy] pair A,B; size(8cm); A=(0,0); B=(480,0); draw((0,0)--(480,0),linetype("3 4")); filldraw(circle((8,0),8),black); draw((0,0)..(100,-100)..(200,0)); draw((200,0)..(260,60)..(320,0)); draw((320,0)..(400,-80)..(480,0)); draw((100,0)--(150,-50sqrt(3)),Arrow(size=4)); draw((260,0)--(290,30sqrt(3)),Arrow(size=4)); draw((400,0)--(440,-40sqrt(3)),Arrow(size=4)); label("$A$", A, SW); label("$B$", B, SE); label("$R_1$", (100,-40), W); label("$R_2$", (260,40), SW); label("$R_3$", (400,-40), W);[/asy]
$\textbf{(A)}\ 238\pi \qquad \textbf{(B)}\ 240\pi \qquad \textbf{(C)}\ 260\pi \qquad \textbf{(D)}\ 280\pi \qquad \textbf{(E)}\ 500\pi$
| The total length of all of the arcs is $100\pi +80\pi +60\pi=240\pi$ . Since we want the path from the center, the actual distance will be subtracted by $2\pi$ because it's already half the circumference through semicircle A, which needs to go half the circumference extra through semicircle B, and it's already half the circumference through semicircle C, and the circumference is $4\pi$ Therefore, the answer is $240\pi-2\pi=\boxed{\textbf{(A)}\ 238\pi}$ .
~ PowerQualimit
Video Solution:
https://www.youtube.com/watch?v=zZGuBFyiQrk by WhyMath
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_25 | 238π |
AMC8_514 | Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99$
| You need $2$ dimes, $1$ nickel, and $4$ pennies for the first $25$ cents. From $26$ cents to $50$ cents, you only need to add $1$ quarter. From $51$ cents to $75$ cents, you also only need to add $1$ quarter. The same for $76$ cents to $99$ cents. Notice that instead of $100$ , it is $99$ . We are left with $3$ quarters, $1$ nickel, $2$ dimes, and $4$ pennies. Thus, the correct answer is $3+2+1+4=\boxed{\textbf{(B)}\ 10}$ .
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_7 | 10 |
AMC8_515 | Squares $ABCD$ , $EFGH$ , and $GHIJ$ are equal in area. Points $C$ and $D$ are the midpoints of sides $IH$ and $HE$ , respectively. What is the ratio of the area of the shaded pentagon $AJICB$ to the sum of the areas of the three squares?
$\textbf{(A)}\hspace{.05in}\frac{1}{4}\qquad\textbf{(B)}\hspace{.05in}\frac{7}{24}\qquad\textbf{(C)}\hspace{.05in}\frac{1}{3}\qquad\textbf{(D)}\hspace{.05in}\frac{3}{8}\qquad\textbf{(E)}\hspace{.05in}\frac{5}{12}$
[asy] pair A,B,C,D,E,F,G,H,I,J; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, NE); dot("$D$", D, NW); dot("$E$", E, NW); dot("$F$", F, SW); dot("$G$", G, S); dot("$H$", H, N); dot("$I$", I, NE); dot("$J$", J, SE); [/asy]
| We notice that ABCX is a trapezoid with the bases AB and CX. Assuming AB is 1, we find that CX is 0.25 since it is half of CH which is 0.5. Using the area of the trapezoid formula, we calculate the area of ABCX to be 0.625 and XCIJ to be 0.375. The combined areas equal 1. Therefore, the ratio of the area of the hexagon to the three squares is 1:3 because the area of the three squares is 3. The answer is $\boxed{\textbf{(C)}\ \frac {1}{3}}$ -~TheNerdwhoIsNerdy. (I don't know if this is correct, pls check).
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_24 | 1/3 |
AMC8_516 | On the last day of school, Mrs. Wonderful gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought $400$ jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$
| If there are $x$ girls, then there are $x+2$ boys. She gave each girl $x$ jellybeans and each boy $x+2$ jellybeans, for a total of $x^2 + (x+2)^2$ jellybeans. She gave away $400-6=394$ jellybeans.
\begin{align*} x^2+(x+2)^2 &= 394\\ x^2+x^2+4x+4 &= 394\\ 2x^2 + 4x &= 390\\ x^2 + 2x &= 195\\ \end{align*}
From here, we can see that $x = 13$ as $13^2 + 26 = 195$ , so there are $13$ girls, $13+2=15$ boys, and $13+15=\boxed{\textbf{(B)}\ 28}$ students.
Note by FrankensteinQuixoteCabin: The user who made this solution did not mention their name.
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_23 | 28 |
AMC8_517 | Here is a list of the numbers of fish that Tyler caught in nine outings last summer: \[2,0,1,3,0,3,3,1,2.\] Which statement about the mean, median, and mode is true?
$\textbf{(A)}\ \text{median} < \text{mean} < \text{mode} \qquad \textbf{(B)}\ \text{mean} < \text{mode} < \text{median} \\ \\ \textbf{(C)}\ \text{mean} < \text{median} < \text{mode} \qquad \textbf{(D)}\ \text{median} < \text{mode} < \text{mean} \\ \\ \textbf{(E)}\ \text{mode} < \text{median} < \text{mean}$
| First, put the numbers in increasing order.
\[0,0,1,1,2,2,3,3,3\]
The mean is $\frac{0+0+1+1+2+2+3+3+3}{9} = \frac{15}{9},$ the median is $2,$ and the mode is $3.$ Because, $\frac{15}{9} < 2 < 3,$ the answer is $\boxed{\textbf{(C)}\ \text{mean} < \text{median} < \text{mode}}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_4 | mean < median < mode |
AMC8_518 | How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12$
| For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of $2012$ , since all of the valid 4-digit number will always be greater than $1000$ . The best way to solve this problem is by using casework.
There can be only two leading digits, namely $1$ or $2$ .
When the leading digit is $1$ , you can make $\frac{3!}{2!1!} \implies 3$ such numbers.
When the leading digit is $2$ , you can make $3! \implies 6$ such numbers.
Summing the amounts of numbers, we find that there are $\boxed{\textbf{(D)}\ 9}$ such numbers.
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_10 | 9 |
AMC8_519 | Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likely?
(A) all 4 are boys
(B) all 4 are girls
(C) 2 are girls and 2 are boys
(D) 3 are of one gender and 1 is of the other gender
(E) all of these outcomes are equally likely
| We'll just start by breaking cases down. The probability of A occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$ . The probability of B occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$ .
The probability of C occurring is $\dbinom{4}{2}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}$ , because we need to choose 2 of the 4 slots to be girls.
For D, there are two possible cases, 3 girls and 1 boy or 3 boys and 1 girl. The probability of the first case is $\dbinom{4}{1}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{4}$ because we need to choose 1 of the 4 slots to be a boy. However, the second case has the same probability because we are choosing 1 of the 4 children to be a girl, so the total probability is $\frac{1}{4} \cdot 2 = \frac{1}{2}$ .
So out of the four fractions, D is the largest. So our answer is $\boxed{\text{(D) 3 are of one gender and 1 is of the other gender}}.$
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_18 | 3 are of one gender and 1 is of the other gender |
AMC8_520 | A $2$ -digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the units digit of the number?
$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad \textbf{(E) }9$
| We can think of the number as $10a+b$ , where a is the tens digit and b is the unit digit. Since the number is equal to the product of the digits ( $ab$ ) plus the sum of the digits ( $a+b$ ), we can say that $10a+b=ab+a+b$ . We can simplify this to $10a=ab+a$ , which factors to $(10)a=(b+1)a$ . Dividing by $a$ , we have that $b+1=10$ . Therefore, the units digit, $b$ , is $\boxed{\textbf{(E) }9}$
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_22 | 9 |
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