Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$

Solution: We immediately see that $E$ is the only possible amount of seniors, as $10\%$ can only correspond with an answer choice ending with $0$ . Thus the number of seniors is $20$ and the number of juniors is $28-20=8\rightarrow \boxed{8}$ . ~samrocksnature

Answer: C 8
Problem: In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$

Solution: Since there are an equal number of juniors and seniors on the debate team, suppose there are $x$ juniors and $x$ seniors. This number represents $25\% =\frac{1}{4}$ of the juniors and $10\%= \frac{1}{10}$ of the seniors, which tells us that there are $4x$ juniors and $10x$ seniors. There are $28$ juniors and seniors in the program altogether, so we get \[10x+4x=28,\] \[14x=28,\] \[x=2.\] Which means there are $4x=8$ juniors on the debate team, $\boxed{8}$

Answer: C 8
Problem: In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$

Solution: The amount of juniors must be a multiple of $4$ , since exactly $\frac{1}{4}$ of the students are on the debate team. Thus, we can immediately see that $\boxed{8}$

Answer: C 8
Problem: At a math contest, $57$ students are wearing blue shirts, and another $75$ students are wearing yellow shirts. The $132$ students are assigned into $66$ pairs. In exactly $23$ of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?
$\textbf{(A)} ~23 \qquad\textbf{(B)} ~32 \qquad\textbf{(C)} ~37 \qquad\textbf{(D)} ~41 \qquad\textbf{(E)} ~64$

Solution: There are $46$ students paired with a blue partner. The other $11$ students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are $64$ students remaining. Therefore the requested number of pairs is $\tfrac{64}{2}=\boxed{32}$ ~Punxsutawney Phil

Answer: B 32
Problem: The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give $24$ , while the other two multiply to $30$ . What is the sum of the ages of Jonie's four cousins?
$\textbf{(A)} ~21 \qquad\textbf{(B)} ~22 \qquad\textbf{(C)} ~23 \qquad\textbf{(D)} ~24 \qquad\textbf{(E)} ~25$

Solution: First look at the two cousins' ages that multiply to $24$ . Since the ages must be single-digit, the ages must either be $3 \text{ and } 8$ or $4 \text{ and } 6.$
Next, look at the two cousins' ages that multiply to $30$ . Since the ages must be single-digit, the only ages that work are $5 \text{ and } 6.$ Remembering that all the ages must all be distinct, the only solution that works is when the ages are $3, 8$ and $5, 6$
We are required to find the sum of the ages, which is \[3 + 8 + 5 + 6 = \boxed{22}.\]

Answer: B 22
Problem: Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$ , and the afternoon class's mean score is $70$ . The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$ . What is the mean of the scores of all the students?
$\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78$

Solution: Let there be $3x$ students in the morning class and $4x$ students in the afternoon class. The total number of students is $3x + 4x = 7x$ . The average is $\frac{3x\cdot84 + 4x\cdot70}{7x}=76$ . Therefore, the answer is $\boxed{76}$

Answer: C 76
Problem: Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$ , and the afternoon class's mean score is $70$ . The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$ . What is the mean of the scores of all the students?
$\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78$

Solution: Suppose the morning class has $m$ students and the afternoon class has $a$ students. We have the following table: \[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] & \textbf{\# of Students} & \textbf{Mean} & \textbf{Total} \\ \hline & & & \\ [-2.5ex] \textbf{Morning} & m & 84 & 84m \\ \hline & & & \\ [-2.5ex] \textbf{Afternoon} & a & 70 & 70a \end{array}\] We are also given that $\frac ma=\frac34,$ which rearranges as $m=\frac34a.$
The mean of the scores of all the students is \[\frac{84m+70a}{m+a}=\frac{84\left(\frac34a\right)+70a}{\frac34a+a}=\frac{133a}{\frac74a}=133\cdot\frac47=\boxed{76}.\] ~MRENTHUSIASM

Answer: C 76
Problem: Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$ , and the afternoon class's mean score is $70$ . The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$ . What is the mean of the scores of all the students?
$\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78$

Solution: Of the average, $\frac{3}{3+4}=\frac{3}{7}$ of the scores came from the morning class and $\frac{4}{7}$ came from the afternoon class. The average is $\frac{3}{7}\cdot 84+\frac{4}{7}\cdot 70=\boxed{76}.$

Answer: C 76
Problem: Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$ , and the afternoon class's mean score is $70$ . The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$ . What is the mean of the scores of all the students?
$\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78$

Solution: WLOG, assume there are $3$ students in the morning class and $4$ in the afternoon class. Then the average is $\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{76}.$

Answer: C 76
Problem: In a plane, four circles with radii $1,3,5,$ and $7$ are tangent to line $\ell$ at the same point $A,$ but they may be on either side of $\ell$ . Region $S$ consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region $S$
$\textbf{(A) }24\pi \qquad \textbf{(B) }32\pi \qquad \textbf{(C) }64\pi \qquad \textbf{(D) }65\pi \qquad \textbf{(E) }84\pi$

Solution: Suppose that line $\ell$ is horizontal, and each circle lies either north or south to $\ell.$ We construct the circles one by one:
The diagram below shows one possible configuration of the four circles: [asy] /* diagram made by samrocksnature, edited by MRENTHUSIASM */ pair A=(10,0); pair B=(-10,0); draw(A--B); filldraw(circle((0,7),7),yellow); filldraw(circle((0,-5),5),yellow); filldraw(circle((0,-3),3),white); filldraw(circle((0,-1),1),white); dot((0,0)); label("$A$",(0,0),(0,1.5)); label("$\ell$",(10,0),(1.5,0)); [/asy] Together, the answer is $\pi\cdot7^2+\pi\cdot5^2-\pi\cdot3^2=\boxed{65}.$

Answer: D 65
Problem: Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top? [asy] /* Made by samrocksnature */ add(grid(7,7)); label("$\dots$", (0.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (2.5,0.5)); label("$\dots$", (3.5,0.5)); label("$\dots$", (4.5,0.5)); label("$\dots$", (5.5,0.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (0.5,1.5)); label("$\dots$", (0.5,2.5)); label("$\dots$", (0.5,3.5)); label("$\dots$", (0.5,4.5)); label("$\dots$", (0.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (6.5,1.5)); label("$\dots$", (6.5,2.5)); label("$\dots$", (6.5,3.5)); label("$\dots$", (6.5,4.5)); label("$\dots$", (6.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (1.5,6.5)); label("$\dots$", (2.5,6.5)); label("$\dots$", (3.5,6.5)); label("$\dots$", (4.5,6.5)); label("$\dots$", (5.5,6.5)); label("$\dots$", (6.5,6.5)); label("$17$", (1.5,1.5)); label("$18$", (1.5,2.5)); label("$19$", (1.5,3.5)); label("$20$", (1.5,4.5)); label("$21$", (1.5,5.5)); label("$16$", (2.5,1.5)); label("$5$", (2.5,2.5)); label("$6$", (2.5,3.5)); label("$7$", (2.5,4.5)); label("$22$", (2.5,5.5)); label("$15$", (3.5,1.5)); label("$4$", (3.5,2.5)); label("$1$", (3.5,3.5)); label("$8$", (3.5,4.5)); label("$23$", (3.5,5.5)); label("$14$", (4.5,1.5)); label("$3$", (4.5,2.5)); label("$2$", (4.5,3.5)); label("$9$", (4.5,4.5)); label("$24$", (4.5,5.5)); label("$13$", (5.5,1.5)); label("$12$", (5.5,2.5)); label("$11$", (5.5,3.5)); label("$10$", (5.5,4.5)); label("$25$", (5.5,5.5)); [/asy]
$\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380$

Solution: In the diagram below, the red arrows indicate the progression of numbers. In the second row from the top, the greatest number and the least number are $D$ and $E,$ respectively. Note that the numbers in the yellow cells are consecutive odd perfect squares, as we can prove by induction. [asy]  /* Made by MRENTHUSIASM */ size(11.5cm);  for (real i=7.5; i<=14.5; ++i)  { 	fill((i+0.5,i+0.5)--(i-0.5,i+0.5)--(i-0.5,i-0.5)--(i+0.5,i-0.5)--cycle,yellow); }  fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green); fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green);  label("$A$",(14.5,14.5)); label("$B$",(13.5,13.5)); label("$C$",(0.5,14.5)); label("$E$",(1.5,13.5)); label("$D$",(0.5,13.5));  add(grid(15,15,linewidth(1.25)));  draw((1.5,13.5)--(14.5,13.5)--(14.5,0.5)--(0.5,0.5)--(0.5,14.5)--(14.5,14.5),red+linewidth(1.125),EndArrow); [/asy] By observations, we proceed as follows: \begin{alignat*}{6} A=15^2=225, \ B=13^2=169  \quad &\implies \quad &C &= \hspace{1mm}&&A-14\hspace{1mm} &= 211& \\  \quad &\implies \quad &D &= &&C-1  &= 210& \\  \quad &\implies \quad &E &= &&B-12 &= 157&.  \end{alignat*} Therefore, the answer is $D+E=\boxed{367}.$

Answer: A 367
Problem: Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top? [asy] /* Made by samrocksnature */ add(grid(7,7)); label("$\dots$", (0.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (2.5,0.5)); label("$\dots$", (3.5,0.5)); label("$\dots$", (4.5,0.5)); label("$\dots$", (5.5,0.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (0.5,1.5)); label("$\dots$", (0.5,2.5)); label("$\dots$", (0.5,3.5)); label("$\dots$", (0.5,4.5)); label("$\dots$", (0.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (6.5,1.5)); label("$\dots$", (6.5,2.5)); label("$\dots$", (6.5,3.5)); label("$\dots$", (6.5,4.5)); label("$\dots$", (6.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (1.5,6.5)); label("$\dots$", (2.5,6.5)); label("$\dots$", (3.5,6.5)); label("$\dots$", (4.5,6.5)); label("$\dots$", (5.5,6.5)); label("$\dots$", (6.5,6.5)); label("$17$", (1.5,1.5)); label("$18$", (1.5,2.5)); label("$19$", (1.5,3.5)); label("$20$", (1.5,4.5)); label("$21$", (1.5,5.5)); label("$16$", (2.5,1.5)); label("$5$", (2.5,2.5)); label("$6$", (2.5,3.5)); label("$7$", (2.5,4.5)); label("$22$", (2.5,5.5)); label("$15$", (3.5,1.5)); label("$4$", (3.5,2.5)); label("$1$", (3.5,3.5)); label("$8$", (3.5,4.5)); label("$23$", (3.5,5.5)); label("$14$", (4.5,1.5)); label("$3$", (4.5,2.5)); label("$2$", (4.5,3.5)); label("$9$", (4.5,4.5)); label("$24$", (4.5,5.5)); label("$13$", (5.5,1.5)); label("$12$", (5.5,2.5)); label("$11$", (5.5,3.5)); label("$10$", (5.5,4.5)); label("$25$", (5.5,5.5)); [/asy]
$\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380$

Solution: In the diagram below, the red arrows indicate the progression of numbers. In the second row from the top, the greatest number and the least number are $C$ and $G,$ respectively. [asy] /* Made by MRENTHUSIASM */ size(11.5cm);  fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green); fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green);  label("$A$",(14.5,14.5)); label("$B$",(0.5,14.5)); label("$C$",(0.5,13.5)); label("$D$",(0.5,0.5)); label("$E$",(14.5,0.5)); label("$F$",(14.5,13.5)); label("$G$",(1.5,13.5));  add(grid(15,15,linewidth(1.25)));  draw((1.5,13.5)--(14.5,13.5)--(14.5,0.5)--(0.5,0.5)--(0.5,14.5)--(14.5,14.5),red+linewidth(1.125),EndArrow); [/asy] By observations, we proceed as follows: \begin{alignat*}{6} A=15^2=225  \quad &\implies \quad &B &= \hspace{1mm}&&A-14\hspace{1mm} &= 211& \\  \quad &\implies \quad &C &= &&B-1  &= 210& \\  \quad &\implies \quad &D &= &&C-13 &= 197& \\ \quad &\implies \quad &E &= &&D-14 &= 183& \\ \quad &\implies \quad &F &= &&E-13 &= 170& \\ \quad &\implies \quad &G &= &&F-13 &= 157&.  \end{alignat*} Therefore, the answer is $C+G=\boxed{367}.$

Answer: A 367
Problem: Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top? [asy] /* Made by samrocksnature */ add(grid(7,7)); label("$\dots$", (0.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (2.5,0.5)); label("$\dots$", (3.5,0.5)); label("$\dots$", (4.5,0.5)); label("$\dots$", (5.5,0.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (0.5,1.5)); label("$\dots$", (0.5,2.5)); label("$\dots$", (0.5,3.5)); label("$\dots$", (0.5,4.5)); label("$\dots$", (0.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (6.5,1.5)); label("$\dots$", (6.5,2.5)); label("$\dots$", (6.5,3.5)); label("$\dots$", (6.5,4.5)); label("$\dots$", (6.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (1.5,6.5)); label("$\dots$", (2.5,6.5)); label("$\dots$", (3.5,6.5)); label("$\dots$", (4.5,6.5)); label("$\dots$", (5.5,6.5)); label("$\dots$", (6.5,6.5)); label("$17$", (1.5,1.5)); label("$18$", (1.5,2.5)); label("$19$", (1.5,3.5)); label("$20$", (1.5,4.5)); label("$21$", (1.5,5.5)); label("$16$", (2.5,1.5)); label("$5$", (2.5,2.5)); label("$6$", (2.5,3.5)); label("$7$", (2.5,4.5)); label("$22$", (2.5,5.5)); label("$15$", (3.5,1.5)); label("$4$", (3.5,2.5)); label("$1$", (3.5,3.5)); label("$8$", (3.5,4.5)); label("$23$", (3.5,5.5)); label("$14$", (4.5,1.5)); label("$3$", (4.5,2.5)); label("$2$", (4.5,3.5)); label("$9$", (4.5,4.5)); label("$24$", (4.5,5.5)); label("$13$", (5.5,1.5)); label("$12$", (5.5,2.5)); label("$11$", (5.5,3.5)); label("$10$", (5.5,4.5)); label("$25$", (5.5,5.5)); [/asy]
$\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380$

Solution: From the full diagram below, the answer is $210+157=\boxed{367} [/asy] This solution is extremely time-consuming and error-prone. Please do not try it in a real competition unless you have no other options.

Answer: A 367
Problem: The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9$

Solution: The final image of $P$ is $(-6,3)$ . We know the reflection rule for reflecting over $y=-x$ is $(x,y) \rightarrow (-y, -x)$ . So before the reflection and after rotation the point is $(-3,6)$
By definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and $(1,5)$ . The first slope is $\frac{5-6}{1-(-3)} = \frac{-1}{4}$ . This means the slope of $P$ and $(1,5)$ is $4$
Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from $(3,-6)$ to $(1,5)$ it follows we shall only use the slope once to travel from $(1,5)$ to $P$
Therefore point $P$ is located at $(1+1, 5+4) = (2,9)$ . The answer is $9-2 = 7 = \boxed{7}$

Answer: D 7
Problem: The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9$

Solution: Let us reconstruct that coordinate plane as the complex plane. Then, the point $P(a, b)$ becomes $a+b\cdot{i}$ . 
A $90^\circ$ rotation around the point $(1, 5)$ can be done by translating the point $(1, 5)$ to the origin, rotating around the origin by $90^\circ$ , and then translating the origin back to the point $(1, 5)$ \[a+b\cdot{i}  \implies (a-1)+(b-5)\cdot{i} \implies ((a-1)+(b-5)\cdot{i})\cdot{i} = 5-b+(a-1)i \implies 5+1-b+(a-1+5)i = 6-b+(a+4)i.\] By basis reflection rules, the reflection of $(-6, 3)$ about the line $y = -x$ is $(-3, 6)$ . 
Hence, we have \[6-b+(a+4)i = -3+6i \implies b=9, a=2,\] from which $b-a = 9-2 = \boxed{7}$

Answer: D 7
Problem: The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9$

Solution: Using the same method as in Solution 1, we can obtain that the point before the reflection is $(-3,6)$ . 
If we let the original point be $(x, y)$ , then we can use that the starting point is $(1,5)$ to obtain two vectors $\langle -4,1 \rangle$ and $\langle x-1, y-5 \rangle$ .
We know that two vectors are perpendicular if their dot product is equal to $0$ , and that both points are the same distance ( $\sqrt {17}$ ) from $(1,5)$
Therefore, we can write two equations using these vectors: $(x-1)^2 + (y-5)^2 = 17$ (from distance and pythagorean theorem) and $-4x+y-1 = 0$ (from dot product)
Solving, we simplify the second equation to $y=4x+1$ , and plug it into the first equation.
We obtain $(x-1)^2 + (4x-4)^2 = 17$ . We can simplify this to the quadratic $17x^2-34x=0$ . 
When we factor out $17x$ , we find that $x = 2$ or $x = 0$ . However, $x$ cannot equal $0$ .
Therefore, $x = 2$ , and plugging this into the second equation gives us that $y = 9$ . 
Since the point is $(9, 2)$ , we compute $9-2 = \boxed{7}$

Answer: D 7
Problem: The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9$

Solution: Using the same method as in Solution 1 reflecting $(-6,3)$ about the line $y = -x$ gives us $(-3,6).$
Let the original point be $\langle x,y \rangle.$ From point $(1,5),$ we form the vectors $\langle -4,1 \rangle$ and $\langle x-1, y-5 \rangle$ that extend out from the initial point. If they are perpendicular, we know that their dot product has to equal zero. Therefore, \[\langle -4,1 \rangle \cdot \langle x-1, y-5 \rangle = 0 \implies -4x+y-1= 0.\] Now, we have to do some guess and check from the multiple choices. Let $y - x = A$ where $A$ is one of the answer choices. Then, $A -3x = 1.$ By intuition and logical reasoning we deduce that $A$ must be $1 \pmod 3$ so that brings our potential answers down to $\text{\textbf{(A)}}$ and $\text{\textbf{(C)}}.$ If $A = 1$ from $\text{\textbf{(A)}},$ then $x = 0,$ which we can quickly rule out since we know thar $P$ rotated counterclockwise not clockwise. Hence, $\boxed{7}$ is the answer.

Answer: D 7
Problem: An inverted cone with base radius $12  \mathrm{cm}$ and height $18  \mathrm{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has radius of $24  \mathrm{cm}$ . What is the height in centimeters of the water in the cylinder?
$\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6$

Solution: The volume of a cone is $\frac{1}{3} \cdot\pi \cdot r^2 \cdot h$ where $r$ is the base radius and $h$ is the height. The water completely fills up the cone so the volume of the water is $\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi$
The volume of a cylinder is $\pi \cdot r^2 \cdot h$ so the volume of the water in the cylinder would be $24\cdot24\cdot\pi\cdot h$
We can equate these two expressions because the water volume stays the same like this $24\cdot24\cdot\pi\cdot h = 6\cdot144\pi$ . We get $4h = 6$ and $h=\frac{6}{4}$
So the answer is $\boxed{1.5}.$

Answer: A 1.5
Problem: An inverted cone with base radius $12  \mathrm{cm}$ and height $18  \mathrm{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has radius of $24  \mathrm{cm}$ . What is the height in centimeters of the water in the cylinder?
$\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6$

Solution: The water completely fills up the cone. For now, assume the radius of both cone and cylinder are the same. Then the cone has $\frac{1}{3}$ of the volume of the cylinder, and so the height is divided by $3$ . Then, from the problem statement, the radius is doubled, meaning the area of the base is quadrupled (since $2^2 = 4$ ).
Therefore, the height is divided by $3$ and divided by $4$ , which is $18 \div 3 \div 4 = \boxed{1.5}.$

Answer: A 1.5
Problem: Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?
$\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64$

Solution: Let the side lengths of the rectangular pan be $m$ and $n$ . It follows that $(m-2)(n-2) = \frac{mn}{2}$ , since half of the brownie pieces are in the interior. This gives $2(m-2)(n-2) = mn \iff mn - 4m - 4n + 8 = 0$ . Adding 8 to both sides and applying Simon's Favorite Factoring Trick , we obtain $(m-4)(n-4) = 8$ . Since $m$ and $n$ are both positive, we obtain $(m, n) = (5, 12), (6, 8)$ (up to ordering). By inspection, $5 \cdot 12 = \boxed{60}$ maximizes the number of brownies.

Answer: D 60