Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many hours will it take to fill Steve's pool?
$\textbf{(A)}\  40  \qquad \textbf{(B)}\   42  \qquad \textbf{(C)}\  44   \qquad \textbf{(D)}\  46   \qquad \textbf{(E)}\   48$

Solution: If all four hoses fill $2.5$ gallons a minute, every minute $10$ gallons would be added. Since every hour has $60$ minutes, $600$ gallons of water would be added every hour. $24000/600=\boxed{40}$ hours.

Answer: A 40
Problem: The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad.  Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles.  The width of the railroad track can be ignored.  How many square miles are in the plot of land ACD?
[asy] size(250); defaultpen(linewidth(0.55)); pair A=(-6,0), B=origin, C=(0,6), D=(0,12); pair ac=C+2.828*dir(45), ca=A+2.828*dir(225), ad=D+2.828*dir(A--D), da=A+2.828*dir(D--A), ab=(2.828,0), ba=(-6-2.828, 0);  fill(A--C--D--cycle, gray); draw(ba--ab); draw(ac--ca); draw(ad--da); draw((0,-1)--(0,15)); draw((1/3, -1)--(1/3, 15)); int i; for(i=1; i<15; i=i+1) { draw((-1/10, i)--(13/30, i)); }  label("$A$", A, SE); label("$B$", B, SE); label("$C$", C, SE); label("$D$", D, SE); label("$3$", (1/3,3), E); label("$3$", (1/3,9), E); label("$3$", (-3,0), S); label("Main", (-3,0), N); label(rotate(45)*"Aspen", A--C, SE); label(rotate(63.43494882)*"Brown", A--D, NW); [/asy]
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$

Solution: The area of a triangle is $\frac12 bh$ . If we let $CD$ be the base of the triangle, then the height is $AB$ , and the area is $\frac12 \cdot 3 \cdot 3 = \boxed{4.5}$

Answer: C 4.5
Problem: The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad.  Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles.  The width of the railroad track can be ignored.  How many square miles are in the plot of land ACD?
[asy] size(250); defaultpen(linewidth(0.55)); pair A=(-6,0), B=origin, C=(0,6), D=(0,12); pair ac=C+2.828*dir(45), ca=A+2.828*dir(225), ad=D+2.828*dir(A--D), da=A+2.828*dir(D--A), ab=(2.828,0), ba=(-6-2.828, 0);  fill(A--C--D--cycle, gray); draw(ba--ab); draw(ac--ca); draw(ad--da); draw((0,-1)--(0,15)); draw((1/3, -1)--(1/3, 15)); int i; for(i=1; i<15; i=i+1) { draw((-1/10, i)--(13/30, i)); }  label("$A$", A, SE); label("$B$", B, SE); label("$C$", C, SE); label("$D$", D, SE); label("$3$", (1/3,3), E); label("$3$", (1/3,9), E); label("$3$", (-3,0), S); label("Main", (-3,0), N); label(rotate(45)*"Aspen", A--C, SE); label(rotate(63.43494882)*"Brown", A--D, NW); [/asy]
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$

Solution: We can see that there is a big triangle encasing $ACD$ . The area of that triangle is $\frac12 \cdot 3 \cdot 6 = 9$ . We can easily see that triangle $ABC$ is $\frac12 \cdot 3 \cdot 3$ , which is $4.5$ $9-4.5$ is $4.5$ , so the answer is $\boxed{4.5}$

Answer: C 4.5
Problem: The length of a rectangle is increased by $10\%$ percent and the width is decreased by $10\%$ percent. What percent of the old area is the new area?
$\textbf{(A)}\  90  \qquad \textbf{(B)}\   99  \qquad \textbf{(C)}\   100  \qquad \textbf{(D)}\   101  \qquad \textbf{(E)}\   110$

Solution: In a rectangle with dimensions $10 \times 10$ , the new rectangle would have dimensions $11 \times 9$ . The ratio of the new area to the old area is $99/100 = \boxed{99}$

Answer: B 99
Problem: Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have?
[asy] defaultpen(linewidth(0.6)); pair O=origin, A=(0,1), B=A+1*dir(60), C=(1,1), D=(1,0), E=D+1*dir(-72), F=E+1*dir(-144), G=O+1*dir(-108); draw(O--A--B--C--D--E--F--G--cycle); draw(O--D, dashed); draw(A--C, dashed);[/asy]
$\textbf{(A)}\ 21 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 29$

Solution: Of the six shapes used to create the polygon, the triangle and octagon are adjacent to the others on one side, and the others are adjacent on two sides. In the triangle and octagon $3+8-2(1)=9$ sides are on the outside of the final polygon. In the other shapes $4+5+6+7-4(2) = 14$ sides are on the outside. The resulting polygon has $9+14 = \boxed{23}$ sides.

Answer: B 23
Problem: Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have?
[asy] defaultpen(linewidth(0.6)); pair O=origin, A=(0,1), B=A+1*dir(60), C=(1,1), D=(1,0), E=D+1*dir(-72), F=E+1*dir(-144), G=O+1*dir(-108); draw(O--A--B--C--D--E--F--G--cycle); draw(O--D, dashed); draw(A--C, dashed);[/asy]
$\textbf{(A)}\ 21 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 29$

Solution: We can quickly see a pattern if we draw out the other shapes. Every shape will have two of its sides taken out except the triangle and octagon. We can then make the expression $2+2+3+4+5+7$ which is $\boxed{23}$

Answer: B 23
Problem: The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43$ dollars. Some of the $30$ sixth graders each bought a pencil, and they paid a total of $1.95$ dollars. How many more sixth graders than seventh graders bought a pencil?
$\textbf{(A)}\  1  \qquad \textbf{(B)}\   2  \qquad \textbf{(C)}\   3  \qquad \textbf{(D)}\   4  \qquad \textbf{(E)}\   5$

Solution: Because the pencil costs a whole number of cents, the cost must be a factor of both $143$ and $195$ . They can be factored into $11\cdot13$ and $3\cdot5\cdot13$ . The common factor cannot be $1$ or there would have to be more than $30$ sixth graders, so the pencil costs $13$ cents. The difference in costs that the sixth and seventh graders paid is $195-143=52$ cents, which is equal to $52/13 = \boxed{4}$ sixth graders.

Answer: D 4
Problem: The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?
[asy] unitsize(30);  draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$1$",(0,.5)); label("$3$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$5$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy] [asy] unitsize(30);  draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$2$",(0,.5)); label("$4$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$6$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]
$\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{2}{3}\qquad\textbf{(C)}\ \frac{3}{4}\qquad\textbf{(D)}\ \frac{7}{9}\qquad\textbf{(E)}\ \frac{5}{6}$

Solution: The possible sums are \[\begin{tabular}{c|ccc} & 1 & 3 & 5 \\ \hline 2 & 3 & 5 & 7 \\ 4 & 5 & 7 & 9 \\ 6 & 7 & 9 & 11 \end{tabular}\]
Only $9$ is not prime, so there are $7$ prime numbers and $9$ total numbers for a probability of $\boxed{79}$

Answer: D 79
Problem: A three-digit integer contains one of each of the digits $1$ $3$ , and $5$ . What is the probability that the integer is divisible by $5$
$\textbf{(A)}\  \frac{1}{6}  \qquad \textbf{(B)}\   \frac{1}{3}  \qquad \textbf{(C)}\   \frac{1}{2}  \qquad \textbf{(D)}\  \frac{2}{3}   \qquad \textbf{(E)}\   \frac{5}{6}$

Solution: The three digit numbers are $135,153,351,315,513,531$ . The numbers that end in $5$ are divisible are $5$ , and the probability of choosing those numbers is $\boxed{13}$

Answer: B 13
Problem: A three-digit integer contains one of each of the digits $1$ $3$ , and $5$ . What is the probability that the integer is divisible by $5$
$\textbf{(A)}\  \frac{1}{6}  \qquad \textbf{(B)}\   \frac{1}{3}  \qquad \textbf{(C)}\   \frac{1}{2}  \qquad \textbf{(D)}\  \frac{2}{3}   \qquad \textbf{(E)}\   \frac{5}{6}$

Solution: The number is divisible by 5 if and only if the number ends in $5$ (also $0$ , but that case can be ignored, as none of the digits are $0$ )
If we randomly arrange the three digits, the probability of the last digit being $5$ is $\boxed{13}$

Answer: B 13
Problem: Austin and Temple are $50$ miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging $60$ miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged $40$ miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?
$\textbf{(A)}\  46  \qquad \textbf{(B)}\   48  \qquad \textbf{(C)}\   50  \qquad \textbf{(D)}\   52  \qquad \textbf{(E)}\   54$

Solution: The way to Temple took $\frac{50}{60}=\frac56$ hours, and the way back took $\frac{50}{40}=\frac54$ for a total of $\frac56 + \frac54 = \frac{25}{12}$ hours. The trip is $50\cdot2=100$ miles. The average speed is $\frac{100}{25/12} = \boxed{48}$ miles per hour.

Answer: B 48
Problem: Austin and Temple are $50$ miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging $60$ miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged $40$ miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?
$\textbf{(A)}\  46  \qquad \textbf{(B)}\   48  \qquad \textbf{(C)}\   50  \qquad \textbf{(D)}\   52  \qquad \textbf{(E)}\   54$

Solution: We calculate the harmonic mean of Austin and Temple. 
Plugging in, we have: $\frac{2ab}{a+b} = \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = \boxed{48}$ miles per hour.

Answer: B 48
Problem: Austin and Temple are $50$ miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging $60$ miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged $40$ miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?
$\textbf{(A)}\  46  \qquad \textbf{(B)}\   48  \qquad \textbf{(C)}\   50  \qquad \textbf{(D)}\   52  \qquad \textbf{(E)}\   54$

Solution: The 50 miles part of this question is redundant. We find that the average speed is skewed towards the lower speed, since you spend more time in the lower speed. So, we take the LCM of 40 and 60, getting 120. So, we do (40 + 40 + 40 + 60 + 60)/5, calculation the ratio of the parts and taking the mean. Solving, we get 240/5  = $\boxed{48} \qquad$ -themathgood

Answer: B 48
Problem: How many $3$ -digit positive integers have digits whose product equals $24$
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 24$

Solution: With the digits listed from least to greatest, the $3$ -digit integers are $138,146,226,234$ $226$ can be arranged in $\frac{3!}{2!} = 3$ ways, and the other three can be arranged in $3!=6$ ways. There are $3+6(3) = \boxed{21}$ $3$ -digit positive integers.

Answer: D 21
Problem: The diagram represents a $7$ -foot-by- $7$ -foot floor that is tiled with $1$ -square-foot black tiles and white tiles. Notice that the corners have white tiles. If a $15$ -foot-by- $15$ -foot floor is to be tiled in the same manner, how many white tiles will be needed?
[asy]unitsize(10); draw((0,0)--(7,0)--(7,7)--(0,7)--cycle); draw((1,7)--(1,0)); draw((6,7)--(6,0)); draw((5,7)--(5,0)); draw((4,7)--(4,0)); draw((3,7)--(3,0)); draw((2,7)--(2,0)); draw((0,1)--(7,1)); draw((0,2)--(7,2)); draw((0,3)--(7,3)); draw((0,4)--(7,4)); draw((0,5)--(7,5)); draw((0,6)--(7,6)); fill((1,0)--(2,0)--(2,7)--(1,7)--cycle,black); fill((3,0)--(4,0)--(4,7)--(3,7)--cycle,black); fill((5,0)--(6,0)--(6,7)--(5,7)--cycle,black); fill((0,5)--(0,6)--(7,6)--(7,5)--cycle,black); fill((0,3)--(0,4)--(7,4)--(7,3)--cycle,black); fill((0,1)--(0,2)--(7,2)--(7,1)--cycle,black);[/asy]
$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 57 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 96 \qquad \textbf{(E)}\ 126$

Solution: In a $1$ -foot-by- $1$ -foot floor, there is $1$ white tile. In a $3$ -by- $3$ , there are $4$ . Continuing on, you can deduce the $n^{th}$ positive odd integer floor has $n^2$ white tiles. $15$ is the $8^{th}$ odd integer, so there are $\boxed{64}$ white tiles.

Answer: C 64
Problem: Two angles of an isosceles triangle measure $70^\circ$ and $x^\circ$ . What is the sum of the three possible values of $x$
$\textbf{(A)}\   95    \qquad \textbf{(B)}\    125   \qquad \textbf{(C)}\    140   \qquad \textbf{(D)}\    165   \qquad \textbf{(E)}\     180$

Solution: There are 3 cases: where $x^\circ$ is a base angle with the $70^\circ$ as the other angle, where $x^\circ$ is a base angle with $70^\circ$ as the vertex angle, and where $x^\circ$ is the vertex angle with $70^\circ$ as a base angle.
Case 1: $x^\circ$ is a base angle with the $70^\circ$ as the other angle:
Here, $x=70$ , since base angles are congruent.
Case 2: $x^\circ$ is a base angle with $70^\circ$ as the vertex angle:
Here, the 2 base angles are both $x^\circ$ , so we can use the equation $2x+70=180$ , which simplifies to $x=55$
Case 3: $x^\circ$ is the vertex angle with $70^\circ$ as a base angle:
Here, both base angles are $70^\circ$ , since base angles are congruent. Thus, we can use the equation $x+140=180$ , which simplifies to $x=40$
Adding up all the cases, we get $70+55+40=165$ , so the answer is $\boxed{165}$

Answer: D 165
Problem: How many non-congruent triangles have vertices at three of the eight points in the array shown below?
[asy]dot((0,0)); dot((.5,.5)); dot((.5,0)); dot((.0,.5)); dot((1,0)); dot((1,.5)); dot((1.5,0)); dot((1.5,.5));[/asy]
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

Solution: Assume the base of the triangle is on the bottom four points because a congruent triangle can be made by reflecting the base on the top four points. For a triangle with a base of length $1$ , there are $3$ triangles. For a triangle with a base of length $2$ , there are $3$ triangles. For length $3$ , there are $2$ . In total, the number of non-congruent triangles is $3+3+2=\boxed{8}$

Answer: D 8
Problem: How many whole numbers between 1 and 1000 do not contain the digit 1?
$\textbf{(A)}\ 512 \qquad \textbf{(B)}\ 648 \qquad \textbf{(C)}\ 720 \qquad \textbf{(D)}\ 728 \qquad \textbf{(E)}\ 800$

Solution: 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 $\boxed{728}$ numbers.

Answer: D 728
Problem: 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?
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$

Solution: 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 $13+15=\boxed{28}$ students.

Answer: B 28
Problem: The letters $A$ $B$ $C$ and $D$ represent digits. If $\begin{tabular}{ccc}&A&B\\ +&C&A\\ \hline &D&A\end{tabular}$ and $\begin{tabular}{ccc}&A&B\\ -&C&A\\ \hline &&A\end{tabular}$ ,what digit does $D$ represent?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution: Because $B+A=A$ $B$ must be $0$ .
Next, because $B-A=A\implies0-A=A,$ we get $A=5$ as the "0" mentioned above is actually 10 in this case.
Now we can rewrite $\begin{tabular}{ccc}&A&0\\ +&C&A\\ \hline &D&A\end{tabular}$ as $\begin{tabular}{ccc}&5&0\\ +&C&5\\ \hline &D&5\end{tabular}$ . Therefore, $D=5+C$
Finally, $A-1-C=0\implies{A=C+1}\implies{C=4}$ , So we have $D=5+C\implies{D=5+4}=\boxed{9}$

Answer: E 9
Problem: 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]
$\textbf{(A)}\:6\qquad\textbf{(B)}\:7\qquad\textbf{(C)}\:\frac{419}{51}\qquad\textbf{(D)}\:\frac{158}{17}\qquad\textbf{(E)}\:11$

Solution: 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 $\boxed{11}$

Answer: E 11
Problem: 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]
$\textbf{(A)}\:6\qquad\textbf{(B)}\:7\qquad\textbf{(C)}\:\frac{419}{51}\qquad\textbf{(D)}\:\frac{158}{17}\qquad\textbf{(E)}\:11$

Solution: The top parts and the bottom parts sum to 8. The sides add on another 2. Therefore, the only logical answer would be $\boxed{11}$ since it is the only answer greater than 10.

Answer: E 11
Problem: Susan had 50 dollars to spend at the carnival. She spent 12 dollars on food and twice as much on rides. How many dollars did she have left to spend?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 38 \qquad \textbf{(E)}\ 50$

Solution: If Susan spent 12 dollars, then twice that much on rides, then she spent $12+12 \times 2=36$ dollars in total. We subtract $36$ from $50$ to get $\boxed{14}$

Answer: B 14
Problem: The ten-letter code $\text{BEST OF LUCK}$ represents the ten digits $0-9$ , in order. What 4-digit number is represented by the code word $\text{CLUE}$
$\textbf{(A)}\ 8671 \qquad \textbf{(B)}\ 8672 \qquad \textbf{(C)}\ 9781 \qquad \textbf{(D)}\ 9782 \qquad \textbf{(E)}\ 9872$

Solution: We can derive that $C=8$ $L=6$ $U=7$ , and $E=1$ . Therefore, the answer is $\boxed{8671}$ ~edited by Owencheng

Answer: A 8671
Problem: 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?
$\textbf{(A)}\ 15\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 20\qquad \textbf{(E)}\ 22$

Solution: Barney travels $1661-1441=220$ miles in $4+6=10$ hours for an average of $220/10=\boxed{22}$ miles per hour.

Answer: E 22
Problem: If $\frac{3}{5}=\frac{M}{45}=\frac{60}{N}$ , what is $M+N$
$\textbf{(A)}\ 27\qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 105\qquad \textbf{(E)}\ 127$

Solution: Separate into two equations $\frac35 = \frac{M}{45}$ and $\frac35 = \frac{60}{N}$ and solve for the unknowns. $M=27$ and $N=100$ , therefore $M+N=\boxed{127}$

Answer: E 127
Problem: Candy sales from the Boosters Club from January through April are shown. What were the average sales per month in dollars?
[asy] draw((0,0)--(36,0)--(36,24)--(0,24)--cycle); draw((0,4)--(36,4)); draw((0,8)--(36,8)); draw((0,12)--(36,12)); draw((0,16)--(36,16)); draw((0,20)--(36,20)); fill((4,0)--(8,0)--(8,20)--(4,20)--cycle, black); fill((12,0)--(16,0)--(16,12)--(12,12)--cycle, black); fill((20,0)--(24,0)--(24,8)--(20,8)--cycle, black); fill((28,0)--(32,0)--(32,24)--(28,24)--cycle, black); label("120", (0,24), W); label("80", (0,16), W); label("40", (0,8), W); label("Jan", (6,0), S); label("Feb", (14,0), S); label("Mar", (22,0), S); label("Apr", (30,0), S); [/asy]
$\textbf{(A)}\ 60\qquad\textbf{(B)}\ 70\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 80\qquad\textbf{(E)}\ 85$

Solution: There are a total of $100+60+40+120=320$ dollars of sales spread through $4$ months, for an average of $320/4 = \boxed{80}$

Answer: D 80