AMC8 2011

50 cents is equivalent to \$0.50. Then the three apples cost $3 \times \$0.50 = \$1.50.$ The change Margie receives is \$5.00 - \$1.50 = $\boxed{\textbf{(E)}\ \$3.50}$

The area of a rectangle is given by the formula length times width. Karl's garden is $20 \times 45 = 900$ square feet and Makenna's garden is $25 \times 40 = 1000$ square feet. Since $1000 > 900,$ Makenna's garden is larger by $1000-900=100$ square feet. $\Rightarrow \boxed{ \textbf{(E)}\ \text{Makenna's garden is larger by 100 square feet.} }$

One way of approaching this is drawing the next circle of boxes around the current square. We can now count the number of black and white tiles; 32 black tiles and 17 white tiles. This means the answer is $\boxed{\textbf{(D) }32:17}$.

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 $\boxed{\textbf{(C)}\ \text{mean} < \text{median} < \text{mode}}$

There are $60$ minutes in an hour. $2011/60=33\text{r}31,$ or $33$ hours and $31$ minutes. There are $24$ hours in a day, so the time is $9$ hours and $31$ minutes after midnight on January 2, 2011. $\Rightarrow\boxed{\textbf{(D)}\text{ January 2 at 9:31AM}}$

By PIE, the number of adults who own both cars and motorcycles is $331+45-351=25.$ Out of the $331$ car owners, $25$ of them own motorcycles and $331-25=\boxed{\textbf{(D)}\ 306}$ of them don't.

Assume that the area of each square is $1$. Then, the area of the bolded region in the top left square is $\dfrac{1}{4}$. The area of the top right bolded region is $\dfrac{1}{8}$. The area of the bottom left bolded region is $\dfrac{3}{8}$. And the area of the bottom right bolded region is $\dfrac{1}{4}$. Add the four fractions: $\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{3}{8} + \dfrac{1}{4} = 1$. The four squares together have an area of $4$, so the percentage bolded is $\dfrac{1}{4} \cdot 100 = \boxed{\textbf{(C)}\ 25}$.

By adding a number from Bag A and a number from Bag B together, the values we can get are $3, 5, 7, 5, 7, 9, 7, 9, 11.$ Therefore the number of different values is $\boxed{\textbf{(B)}\ 5}$.

We observe the graph and see that the shape of the graph does not matter. We only want the total time it took Carmen and the total distance she traveled. Based on the graph, Carmen traveled 35 miles for 7 hours. Therefore, her average speed is $\boxed{\textbf{(E)}\ 5}$

Let $x$ be the number of miles you ride. The number of miles you ride after the first half mile is $x-0.5.$ We can write this equation: \begin{align*} 10 &= 2.4 + 0.2 \times \frac{x-0.5}{0.1} + 2\\ 5.6 &= 2(x-0.5)\\ 2.8 &= x-0.5\\ x &= \boxed{\textbf{(C)}\ 3.3}\end{align*}

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 $\boxed{\textbf{(A)}\ 6}$)

If we designate a person to be on a certain side, then all placements of the other people can be considered unique. WLOG, assign Angie to be on the side. There are then $3!=6$ total seating arrangements. If Carlos is across from Angie, there are only $2!=2$ ways to fill the remaining two seats. Then the probability Angie and Carlos are seated opposite each other is $\frac26=\boxed{\textbf{(B)}\ \frac13}$ .

The length of BP is 5. the ratio of the areas is $\dfrac{5\cdot 15}{25\cdot 15}=\dfrac{5}{25}=20\%$

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 $\frac{120+100}{270+180} = \frac{220}{450} = \boxed{\textbf{(C)}\ \frac{22}{45}}$

\[4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = 10^{10}.\] That is one $1$ followed by ten $0$'s, which is $\boxed{\textbf{(D)}\ 11}$ digits.

25-25-30 We can draw the altitude for the side with length 30. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 30 into two segments with length 15. By the Pythagorean Theorem, we have \[15^2 + x^2 =25^2\] \[x^2 = 25^2 - 15^2\] \[x^2 = (25 + 15)(25-15)\] \[x^2= 40\cdot 10\] \[x^2= 400\] \[x = \sqrt{400}\] \[x= 20\] Thus we have two 15-20-25 right triangles. 25-25-40 We can draw the altitude for the side with length 40. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 40 into two segments with length 20. From the 25-25-30 case, we know that the other side length is 15, so we have two 15-20-25 right triangles. Let the area of a 15-20-25 right triangle be $x$. \[a = 2x\] \[b = 2x\] \[\boxed{\textbf{(C) } A = B}\]

The prime factorization of $588$ is $2^2\cdot3\cdot7^2.$ We can see $w=2, x=1,$ and $z=2.$ Because $5^0=1, y=0.$ \[2w+3x+5y+7z=4+3+0+14=\boxed{\textbf{(A)}\ 21}\]

There are $6\cdot6=36$ ways to roll the two dice, and 6 of them result in two of the same number. Out of the remaining $36-6=30$ ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are $30/2=15$ ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is \[\frac{15+6}{36}=\frac{21}{36}=\boxed{\textbf{(D)}\ \frac{7}{12}}\]

The figure can be divided into $7$ sections. The number of rectangles with just one section is $3.$ The number of rectangles with two sections is $5.$ There are none with only three sections. The number of rectangles with four sections is $3.$ $3+5+3=\boxed{\textbf{(D)}\ 11}$

If you draw altitudes from $A$ and $B$ to $CD,$ the trapezoid will be divided into two right triangles and a rectangle. You can find the values of $a$ and $b$ with the Pythagorean theorem. \[a=\sqrt{15^2-12^2}=\sqrt{81}=9\] \[b=\sqrt{20^2-12^2}=\sqrt{256}=16\] $ABYX$ is a rectangle so $XY=AB=50.$ \[CD=a+XY+b=9+50+16=75\] The area of the trapezoid is \[12\cdot \frac{(50+75)}{2} = 6(125) = \boxed{\textbf{(D)}\ 750}\]

At least half the guesses are too low, so Norb's age must be greater than $36$. If two of the guesses are off by one, then his age is in between two guesses whose difference is $2$. It could be $31,37,$ or $48,$ but because his age is greater than $36$ it can only be $37$ or $48$. Lastly, Norb's age is a prime number so the answer must be $\boxed{\textbf{(C)}\ 37}$.

Since we want the tens digit, we can find the last two digits of $7^{2011}$. We can do this by using modular arithmetic. \[7^1\equiv 07 \pmod{100}.\] \[7^2\equiv 49 \pmod{100}.\] \[7^3\equiv 43 \pmod{100}.\] \[7^4\equiv 01 \pmod{100}.\] We can write $7^{2011}$ as $(7^4)^{502}\times 7^3$. Using this, we can say: \[7^{2011}\equiv (7^4)^{502}\times 7^3\equiv 7^3\equiv 343\equiv 43\pmod{100}.\] From the above, we can conclude that the last two digits of $7^{2011}$ are 43. Since they have asked us to find the tens digit, our answer is $\boxed{\textbf{(D)}\ 4}$.

We can separate this into two cases. If an integer is a multiple of $5,$ the last digit must be either $0$ or $5.$ Case 1: The last digit is $5.$ The leading digit can be $1,2,3,$ or $4.$ Because the second digit can be $0$ but not the leading digit, there are also $4$ choices. The third digit cannot be the leading digit or the second digit, so there are $3$ choices. The number of integers is this case is $4\cdot4\cdot3\cdot1=48.$ Case 2: The last digit is $0.$ Because $5$ is the largest digit, one of the remaining three digits must be $5.$ There are $3$ ways to choose which digit should be $5.$ The remaining digits can be $1,2,3,$ or $4,$ but since they have to be different there are $4\cdot3$ ways to choose. The number of integers in this case is $1\cdot3\cdot4\cdot3=36.$ Therefore, the answer is $48+36=\boxed{\textbf{(D)}\ 84}$.

For the sum of two numbers to be odd, one must be odd and the other must be even, because all odd numbers are of the form $2n+1$ where n is an integer, and all even numbers are of the form $2m$ where m is an integer. \[2n + 1 + 2m = 2m + 2n + 1 = 2(m+n) + 1\] and $m+n$ is an integer because $m$ and $n$ are both integers. The only even prime number is $2,$ so our only combination could be $2$ and $9999.$ But, $9999$ is clearly divisible by $3$, so the number of ways $10001$ can be written as the sum of two primes is $\boxed{\textbf{(A)}\ 0}$.

The area of the smaller square is one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: $2 \cdot 2 \cdot \frac{1}{2}=2.$ The circle's shaded area is the area of the smaller square subtracted from the area of the circle: $\pi - 2.$ If you draw the diagonals of the smaller square, you will see that the larger square is split $4$ congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: $2.$ Approximating $\pi$ to $3.14,$ the ratio of the circle's shaded area to the area between the two squares is about \[\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{\textbf{(A)}\ \frac12}\].