Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour?
[asy] import graph; size(8.76cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.58,xmax=10.19,ymin=-4.43,ymax=9.63;  draw((0,0)--(0,8)); draw((0,0)--(8,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); draw((1,0)--(1,8)); draw((2,0)--(2,8)); draw((3,0)--(3,8)); draw((4,0)--(4,8)); draw((5,0)--(5,8)); draw((6,0)--(6,8)); draw((7,0)--(7,8)); label("$1$",(0.95,-0.24),SE*lsf); label("$2$",(1.92,-0.26),SE*lsf); label("$3$",(2.92,-0.31),SE*lsf); label("$4$",(3.93,-0.26),SE*lsf); label("$5$",(4.92,-0.27),SE*lsf); label("$6$",(5.95,-0.29),SE*lsf); label("$7$",(6.94,-0.27),SE*lsf); label("$5$",(-0.49,1.22),SE*lsf); label("$10$",(-0.59,2.23),SE*lsf); label("$15$",(-0.61,3.22),SE*lsf); label("$20$",(-0.61,4.23),SE*lsf); label("$25$",(-0.59,5.22),SE*lsf); label("$30$",(-0.59,6.2),SE*lsf); label("$35$",(-0.56,7.18),SE*lsf); draw((0,0)--(1,1),linewidth(1.6)); draw((1,1)--(2,3),linewidth(1.6)); draw((2,3)--(4,4),linewidth(1.6)); draw((4,4)--(7,7),linewidth(1.6)); label("HOURS",(3.41,-0.85),SE*lsf); label("M",(-1.39,5.32),SE*lsf); label("I",(-1.34,4.93),SE*lsf); label("L",(-1.36,4.51),SE*lsf); label("E",(-1.37,4.11),SE*lsf); label("S",(-1.39,3.7),SE*lsf);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]
$\textbf{(A)}2\qquad\textbf{(B)}2.5\qquad\textbf{(C)}4\qquad\textbf{(D)}4.5\qquad\textbf{(E)}5$

Solution: 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 $\boxed{5}$

Answer: E 5
Problem: The taxi fare in Gotham City is $2.40 for the first $\frac12$ mile and additional mileage charged at the rate $0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles can you ride for $10?
$\textbf{(A) }3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75$

Solution: 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{3.3}

Answer: C 3.3
Problem: The taxi fare in Gotham City is $2.40 for the first $\frac12$ mile and additional mileage charged at the rate $0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles can you ride for $10?
$\textbf{(A) }3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75$

Solution: \begin{array}{|c|c|c|}
Miles & Money & Remark\\
\hline
0 & 8 & \text{2  for  tip} \\
\hline
0.5 & 5.6 & \text{-2.4  for  first  0.5  mi} \\
\hline
3.3 & 0 & \text{-5.6  for  2.8  mi}
\end{array}
Miles travelled = $\boxed{3.3}$

Answer: C 3.3
Problem: 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]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$

Solution: Average the differences between each day. We get $10, -10,\text{ } 20,\text{ } 30,-20$ . We find the average of this list to get $\boxed{6}$

Answer: A 6
Problem: 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]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$

Solution: This solution may take longer to do than the first solution. In total, Asha studied for 400 minutes a week (80 minutes per day) and Sasha studied for 430 minutes a week (86 minutes per day).  86 - 80 = 6.  Therefore, the answer is $\boxed{6}$

Answer: A 6
Problem: Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?
$\textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34$

Solution: 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 $\frac26=\boxed{13}$

Answer: B 13
Problem: Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?
$\textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34$

Solution: If we seat Angie first, there would be only one out of three ways Carlos can sit across from Angie. So the final answer is $\boxed{13}$

Answer: B 13
Problem: Two congruent squares, $ABCD$ and $PQRS$ , have side length $15$ . They overlap to form the $15$ by $25$ rectangle $AQRD$ shown. What percent of the area of rectangle $AQRD$ is shaded?
[asy] filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black); label("D",(0,0),S); label("R",(25,0),S); label("Q",(25,15),N); label("A",(0,15),N); filldraw((10,0)--(15,0)--(15,15)--(10,15)--cycle,mediumgrey,black); label("S",(10,0),S); label("C",(15,0),S); label("B",(15,15),N); label("P",(10,15),N);[/asy]
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 25$

Solution: The overlap length is $2(15)-25=5$ , so the shaded area is $5 \cdot 15 =75$ . The area of the whole shape is $25 \cdot 15 = 375$ . The fraction $\dfrac{75}{375}$ reduces to $\dfrac{1}{5}$ or 20%. Therefore, the answer is $\boxed{20}$

Answer: C 20
Problem: Let $w$ $x$ $y$ , and $z$ be whole numbers. If $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$ , then what does $2w + 3x + 5y + 7z$ equal?
$\textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56$

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

Answer: A 21
Problem: How many rectangles are in this figure?
[asy] pair A,B,C,D,E,F,G,H,I,J,K,L; A=(0,0); B=(20,0); C=(20,20); D=(0,20); draw(A--B--C--D--cycle); E=(-10,-5); F=(13,-5); G=(13,5); H=(-10,5); draw(E--F--G--H--cycle); I=(10,-20); J=(18,-20); K=(18,13); L=(10,13); draw(I--J--K--L--cycle);[/asy]
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution: 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 $3.$
$3+5+3=\boxed{11}$

Answer: D 11
Problem: Quadrilateral $ABCD$ is a trapezoid, $AD = 15$ $AB = 50$ $BC = 20$ , and the altitude is $12$ . What is the area of the trapezoid?
[asy] pair A,B,C,D; A=(3,20); B=(35,20); C=(47,0); D=(0,0); draw(A--B--C--D--cycle); dot((0,0)); dot((3,20)); dot((35,20)); dot((47,0)); label("A",A,N); label("B",B,N); label("C",C,S); label("D",D,S); draw((19,20)--(19,0)); dot((19,20)); dot((19,0)); draw((19,3)--(22,3)--(22,0)); label("12",(21,10),E); label("50",(19,22),N); label("15",(1,10),W); label("20",(41,12),E);[/asy]
$\textbf{(A) }600\qquad\textbf{(B) }650\qquad\textbf{(C) }700\qquad\textbf{(D) }750\qquad\textbf{(E) }800$

Solution: [asy] unitsize(1.5mm); defaultpen(linewidth(.9pt)+fontsize(10pt)); dotfactor=3;  pair A,B,C,D,X,Y; A=(9,12); B=(59,12); C=(75,0); D=(0,0); X=(9,0); Y=(59,0); draw(A--B--C--D--cycle); draw(A--X); draw(B--Y);  pair[] ps={A,B,C,D,X,Y}; dot(ps);  label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$X$",X,SE); label("$Y$",Y,S); label("$a$",D--X,S); label("$b$",Y--C,S); label("$15$",D--A,NW); label("$50$",B--A,N); label("$20$",B--C,NE); label("$12$",X--A,E); label("$12$",Y--B,W); [/asy]
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
\[12\cdot \frac{(50+75)}{2} = 6(125) = \boxed{750}\]

Answer: D 750
Problem: Students guess that Norb's age is $24, 28, 30, 32, 36, 38, 41, 44, 47$ , and $49$ . Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?
$\textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48$

Solution: If at least half the guesses are too low, then 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 $\boxed{37}$

Answer: C 37
Problem: Students guess that Norb's age is $24, 28, 30, 32, 36, 38, 41, 44, 47$ , and $49$ . Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?
$\textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48$

Solution: Since two guesses are off by one, we know that both $x+1$ and $x-1$ are in the list where $x$ is the age of Norb. Now, we know that $x+1$ and $x-1$ are $28$ and $30$ $30$ and $32$ $36$ and $38$ and $47$ and $49$ . From these values, we know that $x$ must be $29$ $31$ , and $37$ . Since half of the guesses are too low, $24, 28, 30, 32,$ and $36$ are all too low so we can eliminate all numbers in our list lesser than or equal to $36$ . Therefore, our list has only $37$ left so the answer is $\boxed{37}$

Answer: C 37
Problem: What is the tens digit of $7^{2011}$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7$

Solution: 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 $\boxed{4}$

Answer: D 4
Problem: What is the tens digit of $7^{2011}$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7$

Solution: We can use patterns to figure out the answer.
7 to the power of 2 is 49. So, the tens digit is 4.
7 to the power of 3 is 343. So, the tens digit is 4. 
7 to the power of 4 is 2401.So, the tens digit is 0.
7 to the power of 5 is 16807. So, the tens digit is 0.
By now, we can notice the pattern. The tens digit for 7 to the power of 2 is 4, then 4, then 0, then 0. It keeps on going, 2 fours, and then 2 zeros in 4 numbers. If we round up for 2011/4, we get 503. 503 * 4 is 2012. So 7 to the 2012 power has a tens digit of 0, since 2012 is a mutiple of 4, and 7 to the power of 4 has a tens digit of 0. We have to subtract a power from 7 to the 2012 power, so the tens digit goes back from 0 to 4 because if we subtract a power from 7 to the power of 4, we have 7 to the power of 3, which has a tens digit of 4. Hence the answer is $\boxed{4}$

Answer: D 4
Problem: How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108$

Solution: 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 $48+36=\boxed{84}$

Answer: D 84
Problem: How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108$

Solution: We make four cases based off where the multiple of $5$ digit ( $0$ or $5$ ) is. The number has to end with either $5$ or $0$ since it's a multiple of $5$ . In all but the last case, the $5$ and $0$ are used at the end and in another spot which separates the cases.
Case 1: The first digit can't be $0$ , so it must be $5$ . There are $4\cdot3$ to choose the middle two digits. After that, the last digit has to be $0$ , so there are a total of $1\cdot4\cdot3\cdot1=12$ numbers.
Case 2: The second digit can be $0$ or $5$ , leaving $2$ choices. The first and third numbers can be chosen in $4\cdot3$ ways, like last time. The last digit has to be $0$ or $5$ , but not the one we already used. There are a total of $4\cdot2\cdot3\cdot1=24$ numbers.
Case 3: There are the same choices, but the digits $0$ and $5$ are at the last and second-to-last spots. So there are $4\cdot3\cdot2\cdot1=24$ numbers again.
Case 4: There are $4\cdot3\cdot2$ ways to choose the first three numbers. There has to be a $5$ in the number because the largest digit is $5$ . Coincidentally, there are $4\cdot3\cdot2\cdot1$ numbers again.
There are a total of $12 + 24 \cdot 3=\boxed{84}$ numbers.

Answer: D 84
Problem: In how many ways can $10001$ be written as the sum of two primes?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

Solution: 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 $9999.$ However, $9999$ is clearly divisible by $3$ ,  
so the number of ways $10001$ can be written as the sum of two primes is $\boxed{0}$

Answer: A 0
Problem: In how many ways can $10001$ be written as the sum of two primes?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

Solution: First, we noticed that 10001 is equal to 5000+5001, if you subtract n to 5000 and add n to 5001, you always get an even number, even number is never a prime number except 2. We also see that whenever an addend is an odd number, the other addend will be even, so having an odd number as an addend is not possible, other than 9999 and 2, because 2 is a prime. We try 2 and 9999 but we can see 9999 is divisible by 3 and 9 clearly. So the answer is $\boxed{0}$

Answer: A 0
Problem: In how many ways can $10001$ be written as the sum of two primes?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

Solution: It is helpful to know and understand the Goldbach Conjecture - that every even number can be written as the sum of $2$ primes - and also, that the ${\textbf{odd numbers}\ }$ that are the sum of two primes are exactly two more than a prime. This is because to make the sum of two numbers odd, you must have one even and one odd. There is only one even prime, which is two, so the sum will be of the form $2+p$ . Hence, the odd numbers that are the sum of two primes are exactly $2$ more than a prime. Relating to the problem, $10001$ is not $2$ more than a prime, because $10001-2=9999$ and we can easily see that $9999$ is divisible by $3$ . Therefore, $10001$ cannot be written as the sum of two primes, and the answer is $\boxed{0}$ ~mk

Answer: A 0
Problem: A circle with radius $1$ is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?
[asy] filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,gray,black); filldraw(Circle((0,0),1), mediumgray,black); filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);[/asy]
$\textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2$

Solution: 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
\[\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{12}\]

Answer: A 12
Problem: A circle with radius $1$ is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?
[asy] filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,gray,black); filldraw(Circle((0,0),1), mediumgray,black); filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);[/asy]
$\textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2$

Solution: For the ratio of the circle's shaded area to the area between the squares to be $1,$ they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately $\boxed{12}$

Answer: A 12
Problem: A circle with radius $1$ is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?
[asy] filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,gray,black); filldraw(Circle((0,0),1), mediumgray,black); filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);[/asy]
$\textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2$

Solution: Set the side length of the bigger square to be $8$ . 
Then the area of the big square is $8^2 =64$ and 
the area of the small square $(4\sqrt{2})^2  = 32$ . 
The difference is $32$ . 
The area of the circle is $4^2$ times $\pi$ which is $16 \pi$ or about $48$ . 
Knowing the area of the small square is $32$ $48-32$ is $16$ . 
The area of the big square is $64$ . 
So $32/64$ is $1/2$ , or $\boxed{12}$

Answer: A 12
Problem: At Euclid Middle School the mathematics teachers are Mrs. Germain, Mr. Newton, and Mrs. Young. There are $11$ students in Mrs. Germain's class, $8$ students in Mr. Newton's class, and $9$ students in Mrs. Young's class taking the AMC $8$ this year. How many mathematics students at Euclid Middle School are taking the contest?
$\textbf{(A)}\ 26 \qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30$

Solution: 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. $11+8+9=\boxed{28}$

Answer: C 28
Problem: What is the sum of the mean, median, and mode of the numbers $2,3,0,3,1,4,0,3$
$\textbf{(A)}\ 6.5 \qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 7.5\qquad\textbf{(D)}\ 8.5\qquad\textbf{(E)}\ 9$

Solution: 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 $3+2.5+2=\boxed{7.5}$

Answer: C 7.5