AMC8 2022

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 are 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 = $\boxed{\textbf{(A)} ~10}$.

We can find a general solution to any $((a \, \blacklozenge \, b) \, \bigstar \, c)$. \[((a \, \blacklozenge \, b) \, \bigstar \, c)\] \[=((a^2-b^2) \, \bigstar \, c)\] \[=(a^2-b^2-c)^2\] \[=a^4+b^4-(a^2)(b^2)-2(a^2)(c)-(b^2)(a^2)+2(b^2)(c)+c^2\] \[=5^4+3^4-(5^2)(3^2)-2(5^2)(6)-(3^2)(5^2)+2(3^2)(6)+6^2\] \[=625+81-225-300-225+108+36\] \[=\boxed{\textbf{(D) } 100}\] To time wasting

The positive divisors of $100$ are \[1,2,4,5,10,20,25,50,100.\] It is clear that $10\leq c\leq50,$ so we apply casework to $c:$ Together, the numbers $a,b,$ and $c$ can be chosen in $\boxed{\textbf{(E) } 4}$ ways.

When M is first reflected over the line $q,$ we obtain the following diagram: When M is then reflected over the line $p,$ we obtain the following diagram: Therefore, the answer is $\boxed{\textbf{(E)}}.$

Five years ago, Bella was $6$ years old, and the kitten was $0$ years old. Today, Bella is $11$ years old, and the kitten is $5$ years old. It follows that Anna is $30-11-5=14$ years old. Therefore, Anna is $14-11=\boxed{\textbf{(C) } 3}$ years older than Bella.

Let the smallest number be $x.$ It follows that the largest number is $4x.$ Since $x,15,$ and $4x$ are equally spaced on a number line, we have \begin{align*} 4x-15 &= 15-x \\ 5x &= 30 \\ x &= \boxed{\textbf{(C) } 6}. \end{align*}

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 $\frac{600}{60}=\boxed{\textbf{(B) } 10}.$

Note that common factors (from $3$ to $20,$ inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes \[\frac{1}{\cancel{3}}\cdot\frac{2}{\cancel{4}}\cdot\frac{\cancel{3}}{\cancel{5}}\cdots\frac{\cancel{18}}{\cancel{20}}\cdot\frac{\cancel{19}}{21}\cdot\frac{\cancel{20}}{22}=\frac{1\cdot2}{21\cdot22}=\frac{1}{21\cdot11}=\boxed{\textbf{(B) } \frac{1}{231}}.\]

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: \[68+(212-68)\cdot\left(\frac12\right)^{\tfrac{15}{5}}=86.\]

Note that: Therefore, the answer is $\boxed{\textbf{(E)}}.$

Let's say that the first strand of pasta he had is x. The first time he takes a bite of this strand will make it 2 pieces. This is x - 3. The second time he takes the bites of EACH strand will become (x - 3) * 3(2). The 2 is for the two bites he took, 1 for each strand. This equation is showing that for every bite that he is taking, he is taking off 3 inches. This is being showed in the 3(2). He took 2 bites, each taking off 3 inches. Now, the problem is stating that their are 10 pieces of pasta and the TOTAL length is 17 inches. We have to plug in the total number of bites to the equation (x - 3) * 3( ). To find this out, we can see that for the number of strands there are, he took 1 less bite than that number. For example, if there are 2 pieces, he took 1 bite to get those. Since there are 10 pieces, he took 9 bites. The equation then becomes, (x - 3) * 3(9) = 17. To solve this, x = 44. Therefore, the answer is $\boxed{\textbf{(D) } 44}$.

First, we calculate that there are a total of $4\cdot4=16$ possibilities. Now, we list all of two-digit perfect squares. $64$ and $81$ are the only ones that can be made using the spinner. Consequently, there is a $\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}$ probability that the number formed by the two spinners is a perfect square.

Let $m$ and $n$ be positive integers such that $m>n$ and $m+n=28.$ It follows that $m=2n+d$ for some positive integer $d.$ We wish to find the number of possible values for $d.$ By substitution, we have $(2n+d)+n=28,$ from which $d=28-3n.$ Note that $n=1,2,3,\ldots,9$ each generate a positive integer for $d,$ so there are $\boxed{\textbf{(D) } 9}$ possible values for $d.$

All valid arrangements of the letters must be of the form \[ \mathbf{E\underline{\quad}E\underline{\quad}E\underline{\quad}E\underline{\quad}E}. \] The problem is equivalent to counting the arrangements of $\mathbf{B},\ \mathbf{K},\ \mathbf{P}$, and $\mathbf{R}$ into the four blanks, in which there are $4! = \boxed{\textbf{(D)}\ 24}$ ways.

By the answer choices, we can disregard the points that do not have integer weights. As a result, we obtain the following diagram: We then proceed in the same way that we had done in Solution 1. Following the steps, we figure out the blue dot that yields the lowest slope, along with passing the origin. We then can look at the x-axis(in this situation, the weight) and figure out it has $\boxed{\textbf{(C) } 3}$ ounces.

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 $50\div2=\boxed{\textbf{(B) } 25}.$

Notice that once $n>8,$ the units digit of $n!!$ will be $0$ because there will be a factor of $10.$ Thus, we only need to calculate the units digit of \[2!!+4!!+6!!+8!! = 2+8+48+48\cdot8.\] We only care about units digits, so we have $2+8+8+8\cdot8,$ which has the same units digit as $2+8+8+4.$ The answer is $\boxed{\textbf{(B) } 2}.$

The midpoints of the four sides of every rectangle are the vertices of a rhombus whose area is half the area of the rectangle: Note that the diagonals of the rhombus have the same lengths as the sides of the rectangle. Let $A=(-3,0), B=(2,0), C=(5,4),$ and $D=(0,4).$ Note that $A,B,C,$ and $D$ are the vertices of a rhombus whose diagonals have lengths $AC=4\sqrt{5}$ and $BD=2\sqrt{5}.$ It follows that the dimensions of the rectangle are $4\sqrt{5}$ and $2\sqrt{5},$ so the area of the rectangle is $4\sqrt{5}\cdot2\sqrt{5}=\boxed{\textbf{(C) } 40}.$

We notice that $13$ students have scores under $85$ currently, and only $5$ have scores over $85$. We find the median of these two numbers, getting: \[13-5=8\] \[\frac{8}{2}=4\] \[13-4=9\] Thus, we realize that $4$ students must have their score increased by $5$. So, the correct answer is $\boxed{\textbf{(C)}4}$.

The sum of the numbers in each row is $12$. Consider the second row. In order for the sum of the numbers in this row to equal $12$, the two shaded numbers must add up to $13$: If two numbers add up to $13$, one of them must be at least $7$: If both shaded numbers are no more than $6$, their sum can be at most $12$. Therefore, for $x$ to be larger than the three missing numbers, $x$ must be at least $8$. We can construct a working scenario where $x=8$: So, our answer is $\boxed{\textbf{(D) } 8}$.

Let $x$ be the number of shots that Candace made in the first half, and let $y$ be the number of shots Candace made in the second half. Since Candace and Steph took the same number of attempts, with an equal percentage of baskets scored, we have $x+y=10+15=25.$ In addition, we have the following inequalities: \[\frac{x}{12}<\frac{15}{20} \implies x<9,\] and \[\frac{y}{18}<\frac{10}{10} \implies y<18.\] Pairing this up with $x+y=25$ we see the only possible solution is $(x,y)=(8,17),$ for an answer of $17-8 = \boxed{\textbf{(C) } 9}.$

Initially, suppose that the bus is at Stop $0$ (starting point) and Zia is at Stop $3.$ We construct the following table of $5$-minute intervals: Note that Zia will wait for the bus after $15$ minutes, and the bus will arrive $2$ minutes later. Therefore, the answer is $15+2=\boxed{\textbf{(A) } 17}.$

We will only consider cases where the three identical symbols are the same column, but at the end we shall double our answer as the same holds true for rows. There are $3$ ways to choose a column with all $\bigcirc$'s and $2$ ways to choose a column with all $\triangle$'s. The third column can be filled in $2^3=8$ ways. Therefore, we have a total of $3\cdot2\cdot8=48$ cases. However, we overcounted the cases with $2$ complete columns of with one symbol and $1$ complete column with another symbol. This happens in $2\cdot3=6$ cases. $48-6=42$. However, we have to remember to double our answer, giving us $\boxed{\textbf{(D) }84}$ ways to complete the grid.

While imagining the folding, $\overline{AB}$ goes on $\overline{BC},$ $\overline{AH}$ goes on $\overline{CI},$ and $\overline{EF}$ goes on $\overline{FG}.$ So, $BJ=CI=8$ and $FG=BC=8.$ Also, $\overline{HJ}$ becomes an edge parallel to $\overline{FG},$ so that means $HJ=8.$ Since $GH=14,$ then $JG=14-8=6.$ So, the area of $\triangle BJG$ is $\frac{8\cdot6}{2}=24.$ If we let $\triangle BJG$ be the base, then the height is $FG=8.$ So, the volume is $24\cdot8=\boxed{\textbf{(C)} ~192}.$

Let $A$ denote the leaf where the cricket starts and $B$ denote one of the other $3$ leaves. Note that: - If the cricket is at $A,$ then the probability that it hops to $B$ next is $1.$ - If the cricket is at $B,$ then the probability that it hops to $A$ next is $\frac13.$ - If the cricket is at $B,$ then the probability that it hops to $B$ next is $\frac23.$ We apply casework to the possible paths of the cricket: 1. $A \rightarrow B \rightarrow A \rightarrow B \rightarrow A$ The probability for this case is $1\cdot\frac13\cdot1\cdot\frac13=\frac19.$ 2. $A \rightarrow B \rightarrow B \rightarrow B \rightarrow A$ The probability for this case is $1\cdot\frac23\cdot\frac23\cdot\frac13=\frac{4}{27}.$ Together, the probability that the cricket returns to $A$ after $4$ hops is $\frac19+\frac{4}{27}=\boxed{\textbf{(E) }\frac{7}{27}}.$