AMC8 2010

Given that these are the only math teachers at Euclid Middle School and we are told how many from each class are taking the AMC 8, we simply add the three numbers to find the total. $11+8+9=\boxed{\textbf{(C)}\ 28}$

Answer (D): $5 * 10 = \frac{5 \times 10}{5+10} = \frac{50}{15} = \frac{5 \times 10}{5 \times 3} = \frac{10}{3}.$

The highest price was in Month 1, which was $17. The lowest price was in Month 3, which was $10. 17 is $\frac{17}{10}\cdot100=170\%$ of 10, and is $170-100=70\%$ more than 10. Therefore, the answer is $\boxed{\textbf{(C)}\ 70}$

Putting the numbers in numerical order we get the list $0,0,1,2,3,3,3,4.$ The mode is $3.$ The median is $\frac{2+3}{2}=2.5.$ The average is $\frac{0+0+1+2+3+3+3+4}{8}=\frac{16}{8}=2.$ The sum of all three is $3+2.5+2=\boxed{\textbf{(C)}\ 7.5}$

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 $240-206=\boxed{\textbf{(B)}\ 34}$

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 $\boxed{ \textbf{(E)}\ \text{square} }$.

To make any units digit, you can use 4 pennies and 1 nickel. Using 3 quarters, you can make 75, so you will need 2 more dimes. Therefore, you will need $\boxed{4 + 1 + 2 + 3 = B(10)}$ coins.

Because they are both moving in the same direction, Emily is riding relative to Emerson $12-8=4$ mph. Now we can look at it as if Emerson is not moving at all [on his skateboard] and Emily is riding at $4$ mph. It takes her \[\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}\] to ride the $1/2$ mile to reach him, and then the same amount of time to be $1/2$ mile ahead of him. This totals to \[2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{\textbf{(D)}\ 15}\ \text{minutes}\]

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 $\dfrac{63}{75} = 84\% \rightarrow \boxed{\textbf{(D)}\ 84}$

The pepperoni circles' diameter is $2$, since $\dfrac{12}{6} = 2$. From that we see that the area of the $24$ circles of pepperoni is $\left ( \frac{2}{2} \right )^2 (24\pi) = 24\pi$. The large pizza's area is $6^2\pi$. Therefore, the ratio is $\frac{24\pi}{36\pi} = \boxed{\textbf{(B) }\frac{2}{3}}$

Let the height of the taller tree be $h$ and let the height of the smaller tree be $h-16$. Since the ratio of the smaller tree to the larger tree is $\frac{3}{4}$, we have $\frac{h-16}{h}=\frac{3}{4}$. Solving for $h$ gives us $h=64 \Rightarrow \boxed{\textbf{(B)}\ 64}$

Since 80 percent of the 500 balls are red, there are 400 red balls. Therefore, there must be 100 blue balls. For the 100 blue balls to be 25% or $\dfrac{1}{4}$ of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is $\boxed{\textbf{(D)}\ 100}$.

Let $n$, $n+1$, and $n+2$ be the lengths of the sides of the triangle. Then the perimeter of the triangle is $n + (n+1) + (n+2) = 3n+3$. Using the fact that the length of the smallest side is $30\%$ of the perimeter, it follows that: $n = 0.3(3n+3) \Rightarrow n = 0.9n+0.9 \Rightarrow 0.1n = 0.9 \Rightarrow n=9$. The longest side is then $n+2 = 11$. Thus, answer choice $\boxed{\textbf{(E)}\ 11}$ is correct.

First, we must find the prime factorization of $2010$. $2010=2\cdot 3 \cdot 5 \cdot 67$. We add the factors up to get $\boxed{\textbf{(C)}\ 77}$

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. $\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{\textbf{(C)}\ 42}$

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 $\frac{s}{r}=\sqrt{\pi}\Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}$

We see that half the area of the octagon is $5$. We see that the triangle area is $5-1 = 4$. That means that $\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}$. \[\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}\] Meaning, $\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}$

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: \[\dfrac{1350}{225\pi}=\dfrac{6}{\pi}\Rightarrow\boxed{\textbf{(C)}\ 6:\pi}\]

Since $\triangle ACD$ is isosceles, $CB$ bisects $AD$. Thus $AB=BD=8$. From the Pythagorean Theorem, $CB=6$. Thus the area between the two circles is $100\pi - 36\pi=64\pi$ $\boxed{\textbf{(C)}\ 64\pi}$ Note: The length $AC$ is necessary information, as this tells us the radius of the larger circle. The area of the annulus is $\pi(AC^2-BC^2)=\pi AB^2=64\pi$.

Let $x$ be the number of people wearing both a hat and a glove. Since the number of people wearing a hat or a glove must be whole numbers, the number of people in the room must be a multiple of 4 and 5. Since we are trying to find the minimum $x$, we must use the least common multiple. $lcm(4,5) = 20$. Thus, we can say that there are $20$ people in the room, all of which are wearing at least a hat or a glove. (Any people wearing neither item would unnecessarily increase the number of people in the room.) It follows that there are $\frac{2}{5}\cdot 20 = 8$ people wearing gloves and $\frac{3}{4}\cdot 20 = 15$ people wearing hats. Then by applying the Principle of Inclusion and Exclusion (PIE), the total number of people in the room wearing either a hat or a glove or both is $8+15-x = 23-x$, where $x$ is the number wearing both. Since everyone in the room is wearing at least one item (see above), $23-x = 20$, and so $x=\boxed{\textbf{(A)}\ 3}$.

Let $x$ be the number of pages in the book. After the first day, Hui had $\frac{4x}{5}-12$ pages left to read. After the second, she had $\left(\frac{3}{4}\right)\left(\frac{4x}{5}-12\right)-15 = \frac{3x}{5}-24$ left. After the third, she had $\left(\frac{2}{3}\right)\left(\frac{3x}{5}-24\right)-18=\frac{2x}{5}-34$ left. This is equivalent to $62.$ \begin{align*} \frac{2x}{5}-34&=62\\ 2x - 170 &= 310\\ 2x &= 480\\ x &= \boxed{\textbf{(C)}\ 240} \end{align*}

Let the hundreds, tens, and units digits of the original three-digit number be $a$, $b$, and $c$, respectively. We are given that $a=c+2$. The original three-digit number is equal to $100a+10b+c = 100(c+2)+10b+c = 101c+10b+200$. The hundreds, tens, and units digits of the reversed three-digit number are $c$, $b$, and $a$, respectively. This number is equal to $100c+10b+a = 100c+10b+(c+2) = 101c+10b+2$. Subtracting this expression from the expression for the original number, we get $(101c+10b+200) - (101c+10b+2) = 198$. Thus, the units digit in the final result is $\boxed{\textbf{(E)}\ 8}$

By the Pythagorean Theorem, the radius of the larger circle turns out to be $\sqrt{1^2 + 1^2} = \sqrt{2}$. Therefore, the area of the larger circle is $(\sqrt{2})^2\pi = 2\pi$. Using the coordinate plane given, we find that the radius of each of the two semicircles to be 1. So, the area of the two semicircles is $1^2\pi=\pi$. Finally, the ratio of the combined areas of the two semicircles to the area of circle $O$ is $\boxed{\textbf{(B)}\ \frac{1}{2}}$.

Since all of the exponents are multiples of four, we can simplify the problem by taking the fourth root of each number. Evaluating we get $10^2=100$, $5^3=125$, and $2^6=64$. Since $64<100<125$, it follows that $\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}$ is the correct answer.

A dynamics programming approach is quick and easy. The number of ways to climb one stair is $1$. There are $2$ ways to climb two stairs: $1$,$1$ or $2$. For 3 stairs, there are $4$ ways: ($1$,$1$,$1$) ($1$,$2$) ($2$,$1$) ($3$) For four stairs, consider what step they came from to land on the fourth stair. They could have hopped straight from the 1st, done a double from #2, or used a single step from #3. The ways to get to each of these steps are $1+2+4=7$ ways to get to step 4. The pattern can then be extended: $4$ steps: $1+2+4=7$ ways. $5$ steps: $2+4+7=13$ ways. $6$ steps: $4+7+13=24$ ways. Thus, there are $\boxed{\textbf{(E) } 24}$ ways to get to step $6.$