Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts.  Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group.  Each of them has the same number of internet friends.  In how many different ways can this happen?
$\text{(A)}\ 60\qquad\text{(B)}\ 170\qquad\text{(C)}\ 290\qquad\text{(D)}\ 320\qquad\text{(E)}\ 660$

Solution: Note that if $n$ is the number of friends each person has, then $n$ can be any integer from $1$ to $4$ , inclusive.
One person can have at most 4 friends since they cannot be all friends (stated in the problem).
Also note that the cases of $n=1$ and $n=4$ are the same, since a map showing a solution for $n=1$ can correspond one-to-one with a map of a solution for $n=4$ by simply making every pair of friends non-friends and vice versa.  The same can be said of configurations with $n=2$ when compared to configurations of $n=3$ .  Thus, we have two cases to examine, $n=1$ and $n=2$ , and we count each of these combinations twice.
(Note: If you aren’t familiar with one-to-one correspondences, think of it like this: the number of ways to choose 4 friends is equal to number of ways to exclude one friend from your friend group. Hence, since the number of ways to choose 1 friend is the same thing as choosing 1 to not be friends with, $n=1$ and $n=4$ have the same number of ways. Similarly, $n=2$ and $n=3$ have the same number of ways as well. ~peelybonehead)
For $n=1$ , if everyone has exactly one friend, that means there must be $3$ pairs of friends, with no other interconnections.  The first person has $5$ choices for a friend.  There are $4$ people left.  The next person has $3$ choices for a friend.  There are two people left, and these remaining two must be friends.  Thus, there are $15$ configurations with $n=1$
For $n=2$ , there are two possibilities.  The group of $6$ can be split into two groups of $3$ , with each group creating a friendship triangle.  The first person has $\binom{5}{2} = 10$ ways to pick two friends from the other five, while the other three are forced together.  Thus, there are $10$ triangular configurations.
However, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has.  The first person may be seated anywhere on the hexagon without loss of generality .  This person has $\binom{5}{2} = 10$ choices for the two friends on the adjoining vertices.  Each of the three remaining people can be seated "across" from one of the original three people, forming a different configuration.  Thus, there are $10 \cdot 3! = 60$ hexagonal configurations, and in total $70$ configurations for $n=2$
As stated before, $n=3$ has $70$ configurations, and $n=4$ has $15$ configurations.  This gives a total of $(70 + 15)\cdot 2 = 170$ configurations, which is option $\boxed{170}$

Answer: B 170
Problem: Let $a$ $b$ , and $c$ be positive integers with $a\ge$ $b\ge$ $c$ such that $a^2-b^2-c^2+ab=2011$ and $a^2+3b^2+3c^2-3ab-2ac-2bc=-1997$
What is $a$
$\textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253$

Solution: Add the two equations.
$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$
Now, this can be rearranged and factored.
$(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14$
$(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$
$a$ $b$ , and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares.
We see that the only is possibility is $14 = 9 + 4 + 1$
$(a-c)^2 = 9 \Rightarrow a-c = 3$ , since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b = 1$ or $2$ , or whether $b-c = 1$ or $2$
We want to solve for $a$ , so take the two cases and solve them each for an expression in terms of $a$ . Our two cases are $(a, b, c) = (a, a-1, a-3)$ or $(a, a-2, a-3)$ . Plug these values into one of the original equations to see if we can get an integer for $a$
$a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011$ , after some algebra, simplifies to $7a = 2021$ $2021$ is not divisible by $7$ , so $a$ is not an integer.
The other case gives $a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011$ , which simplifies to $8a = 2024$ . Thus, $a = 253$ and the answer is $\boxed{253}$

Answer: E 253
Problem: Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution: Since $x,y,z$ are all reals located in $[0, n]$ , the number of choices for each one is continuous so we use geometric probability.
WLOG( Without loss of generality ), assume that $n\geq x \geq y \geq z \geq 0$ . Then the set of points $(x,y,z)$ is a tetrahedron, or a triangular pyramid. The point $(x,y,z)$ distributes uniformly in this region. If this is not easy to understand, read Solution II.
The altitude of the tetrahedron is $n$ and the base is an isosceles right triangle with a leg length $n$ . The volume is $V_1=\dfrac{n^3}{6}$ , as shown in the first figure in red.
[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2,-1,2/3);  // three - currentprojection, orthographic draw((1,1,0)--(0,1,0)--(0,0,0),dashed); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1),dashed); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle);  draw((0,0,0)--(1,0,0)--(1,1,0)--(1,1,1), red); draw((1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((1,1,1)--(1,0,0), red); [/asy]
Now we will find the region with points satisfying $|x-y|\geq1$ $|y-z|\geq1$ $|z-x|\geq1$
Since $n\geq x \geq y \geq z \geq 0$ , we have $x-y\geq1$ $y-z\geq1$
The region of points $(x,y,z)$ satisfying the condition is shown in the second figure in black. It is a tetrahedron, too.
[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2, -1, 2/3);  // three - currentprojection,  orthographic draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green); draw((0, 0, 0)--(0, 0, 1), green); draw((0, 1, 0)--(0, 1, 1), dashed+green);  draw((1, 0, 0)--(1, 0, 1), green); draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green);    draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((0,0,0)--(1,0,0)--(1,1,1), red); draw((1,1,1)--(1,1,0)--(1,0.9,0), red);  draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle); draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed); draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed);  [/asy]
The volume of this region is $V_2=\dfrac{(n-2)^3}{6}$
So the probability is $p=\dfrac{V_2}{V_1}=\dfrac{(n-2)^3}{n^3}$
Substituting $n$ with the values in the choices, we find that when $n=10$ $p=\frac{512}{1000}>\frac{1}{2}$ , when $n=9$ $p=\frac{343}{729}<\frac{1}{2}$ . So $n\geq 10$
So the answer is $\boxed{10}$

Answer: D 10
Problem: Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution: Because $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ , which means that $x$ $y$ , and $z$ distributes uniformly and independently in the interval $[0,n]$ . So the point $(x, y, z)$ distributes uniformly in the cubic $0\leqslant x, y, z \leqslant n$ , as shown in the figure below. The volume of this cubic is $V_0=n^3$
Cubic.png
As we want to find the probablity of the incident $A=\big\{ |x-y| \geq 1, |y-z| \geq1, |z-x| \geq 1 \big\}$ , 
we should find the volume of the region of points such that $|x-y|\geq 1$ $|y-z|\geq 1$ $|z-x|\geq 1$ and $0\leq x, y, z \leq n$
Now we will find the region $\big\{ (x,y,z)\ | \ 0\leq x, y, z \leq n, |x-y|\geq 1, |y-z|\geq 1, |z-x|\geq 1 \big\}$
The region can be generated by cutting off 3 slices corresponding to $|x-y|< 1$ $|y-z|< 1$ , and $|z-x|< 1$ , respectively, from the cubic.
After cutting off a slice corresponding to $|x-y|< 1$ , we get two triangular prisms, as shown in the figure.
2.png
In order to observe the object clearly, we rotate the object by the $z$ axis, as shown.
3.png
We can draw the slice corresponding to $|y-z|< 1$ on the object.
4B.png
After cutting off the slice corresponding to $|y-z|< 1$ , we have 4 pieces left.
5.png
After cutting off the slice corresponding to $|z-x|< 1$ , we have 6 congruent triangular prisms.
6B.png
Here we draw all the pictures in colors in order to explain the solution clearly. That does not mean that the students should do it in the examination. They can draw a figure with lines only, as shown below.
7.png
Every triangular pyramid has an altitude $n-2$ and a base of isoceless right triangle with leg length $n-2$ , so the volume is $(n-2)^3/6$ .
Then the volume of the region $\big\{ (x,y,z)\ | \ 0\leqslant x, y, z \leqslant n, |x-y|\geqslant 1, |y-z|\geqslant 1, |z-x|\geqslant 1 \big\}$ is $V_A=6\times(n-2)^3/6$ $(n-2)^3$
So the probability of the incident $A$ is $P(A)=\dfrac{V_A}{V_0}$ $\dfrac{(n-2)^3}{n^3}$
Then we can get the answer the same way as Solution I.
The answer is $\boxed{10}$

Answer: D 10
Problem: Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution: If $x$ $y$ , and $z$ are separated by at least one, then by subtracting the minimum space between the three variables (which is $2$ ), $x$ $y$ , and $z$ can be chosen randomly in the interval $[0,n-2]$ . The probability is hence $\dfrac{(n-2)^3}{n^3} > \dfrac{1}{2}.$ $\boxed{10}$ is the minimum value in the answer choices that satisfies this inequality.

Answer: D 10
Problem: Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution: Imagine Points $x$ $y$ $z$ as the "starting points" of three "blocks" of real numbers that have length $1$ . We are just trying to find the probability that those three "blocks" do not overlap. To do this we can set each unit of $1$ into $\mu$ equal little increments, and take the limit of the probability as $\mu$ approaches $\infty$ . (This is because there are indefinitely many real numbers in a given interval.) We see that the total number of arrangements for the three blocks of $1$ , without considering the rule that no two blocks shall overlap is $(n\mu + 1)^3$ . We see that the number of ways to arrange the three blocks such that no three of them are overlapping is simply $\dbinom{(n-2)\mu + 3}{3}$ . Taking the limit as $\mu$ approaches infinity, we obtain our closed form to be $\lim_{\mu \to \infty} \frac{\dbinom{(n-2)\mu + 3}{3}}{ (n\mu + 1)^3  }.$ Dividing leading coefficients, we obtain $\lim_{\mu \to \infty} \frac{\dbinom{(n-2)\mu + 3}{3}}{ (n\mu + 1)^3  } =   \frac{\mu(n-2)^3}{\mu n^3} = \frac{(n-2)^3}{n^3}$ . Solving the inequality $\frac{(n-2)^3}{n^3} > \frac{1}{2}$ , we get the least value of $n$ as $n=\boxed{10}$

Answer:  10
Problem: Each third-grade classroom at Pearl Creek Elementary has $18$ students and $2$ pet rabbits. How many more students than rabbits are there in all $4$ of the third-grade classrooms?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80$

Solution: In each class, there are $18-2=16$ more students than rabbits. So for all classrooms, the difference between students and rabbits is $16 \times 4 = \boxed{64}$

Answer: C 64
Problem: A circle of radius 5 is inscribed in a rectangle as shown.  The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?
[asy] draw((0,0)--(0,10)--(20,10)--(20,0)--cycle);  draw(circle((10,5),5));[/asy]
$\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\  125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200$

Solution: Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is $5$ , the diameter is $2\cdot 5 = 10$ .
Since the sides of the rectangle are in a $2:1$ ratio, the longer side has length $2\cdot 10 = 20$ .
Therefore the area is $20\cdot 10 = 200$ or $\boxed{200}$

Answer: E 200
Problem: The point in the $xy$ -plane with coordinates $(1000, 2012)$ is reflected across the line $y=2000$ .  What are the coordinates of the reflected point?
$\textbf{(A)}\ (998,2012)\qquad\textbf{(B)}\ (1000,1988)\qquad\textbf{(C)}\ (1000,2024)\qquad\textbf{(D)}\ (1000,4012)\qquad\textbf{(E)}\ (1012,2012)$

Solution: The line $y = 2000$ is a horizontal line located $12$ units beneath the point $(1000, 2012)$ . When a point is reflected about a horizontal line, only the $y$ - coordinate will change. The $x$ - coordinate remains the same. Since the $y$ -coordinate of the point is $12$ units above the line of reflection, the new $y$ - coordinate  will be $2000 - 12 = 1988$ . Thus, the coordinates of the reflected point are $(1000, 1988)$ $\boxed{1000,1988}$

Answer: B 1000,1988
Problem: When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over.  When Paul does the same with his marbles, he has 3 marbles left over.  Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag.  How many marbles will be leftover?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: In total, there were $3+4=7$ marbles left from both Ringo and Paul.We know that $7 \equiv 1 \pmod{6}$ . This means that there would be $1$ marble leftover, or $\boxed{1}$

Answer: A 1
Problem: When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over.  When Paul does the same with his marbles, he has 3 marbles left over.  Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag.  How many marbles will be leftover?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: Let $r$ be the number of marbles Ringo has and let $p$ be the number of marbles Paul has. we have the following equations: \[r \equiv 4 \mod{6}\] \[p \equiv 3 \mod{6}\] Adding these equations we get: \[p + r \equiv 7 \mod{6}\] We know that $7 \equiv 1 \mod{6}$ so therefore: \[p + r \equiv 7 \equiv 1 \mod{6} \implies p + r \equiv 1 \mod{6}\] Thus when Ringo and Paul pool their marbles, they will have $\boxed{1}$ marble left over.

Answer: A 1
Problem: Anna enjoys dinner at a restaurant in Washington, D.C., where the sales tax on meals is 10%. She leaves a 15% tip on the price of her meal before the sales tax is added, and the tax is calculated on the pre-tip amount. She spends a total of 27.50 dollars for dinner. What is the cost of her dinner without tax or tip in dollars?
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

Solution: Let $x$ be the cost of her dinner.
$27.50=x+\frac{1}{10}*x+\frac{3}{20}*x$
$27+\frac{1}{2}=\frac{5}{4}*x$
$\frac{55}{2}=\frac{5}{4}x$
$\frac{55}{2}*\frac{4}{5}=x$
$x=22$
$\boxed{22}$

Answer: D 22
Problem: For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54$

Solution: Let $x$ be the number of acorns that both animals had.
So by the info in the problem:
$\frac{x}{3}=\left( \frac{x}{4} \right)+4$
Subtracting $\frac{x}{4}$ from both sides leaves
$\frac{x}{12}=4$
$\boxed{48}$

Answer:  48
Problem: What is the sum of all integer solutions to $1<(x-2)^2<25$
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25$

Solution: $(x-2)^2$ = perfect square.
1 < perfect square < 25
Perfect square can equal: 4, 9, or 16
Solve for $x$
$(x-2)^2=4$
$x=4,0$
and
$(x-2)^2=9$
$x=5,-1$
and
$(x-2)^2=16$
$x=6,-2$
The sum of all integer solutions is
$4+5+6+0+(-1)+(-2)=\boxed{12}$

Answer: B 12
Problem: Two integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of the previous four integers the sum is 57. What is the minimum number of odd integers among the 6 integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: Out of the first two integers, it's possible for both to be even: for example, $10 + 16 = 26.$ But the next two integers, when added, increase the sum by $15,$ which is odd, so one of them must be odd and the other must be even: for example, $3 + 12 = 15.$ Finally, the next two integers increase the sum by $16,$ which is even, so we can have both be even: for example, $2 + 14 = 16.$ Therefore, $\boxed{1}$ is the minimum number of integers that must be odd.

Answer: A 1
Problem: How many ordered pairs of positive integers $(M,N)$ satisfy the equation $\frac{M}{6}=\frac{6}{N}?$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution: Cross-multiplying gives $MN=36.$ We write $36$ as a product of two positive integers: \begin{align*} 36 &= 1\cdot36 \\ &= 2\cdot18 \\ &= 3\cdot12 \\ &= 4\cdot9 \\ &= 6\cdot6. \end{align*} The products $1\cdot36, 2\cdot18, 3\cdot12,$ and $4\cdot9$ each produce $2$ ordered pairs $(M,N),$ as we can switch the order of the factors. The product $6\cdot6$ produces $1$ ordered pair $(M,N).$ Together, we have $4\cdot2+1=\boxed{9}$ ordered pairs $(M,N).$

Answer: D 9
Problem: How many ordered pairs of positive integers $(M,N)$ satisfy the equation $\frac{M}{6}=\frac{6}{N}?$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution: Cross-multiplying gives $MN=36.$ From the prime factorization \[36=2^2\cdot3^2,\] we conclude that $36$ has $(2+1)(2+1)=9$ positive divisors. There are $9$ values of $M,$ and each value generates $1$ ordered pair $(M,N).$ So, there are $\boxed{9}$ ordered pairs $(M,N)$ in total.

Answer: D 9
Problem: A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
$\textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304$

Solution: Desserts must be chosen for $7$ days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.
There are $3$ choices for dessert on Saturday: pie, ice cream, or pudding, as there must be cake on Friday and the same dessert may not be served two days in a row. Likewise, there are $3$ choices for dessert on Thursday. Once dessert for Thursday is selected, there are $3$ choices for dessert on Wednesday, once Wednesday's dessert is selected there are $3$ choices for dessert on Tuesday, etc. Thus, there are $3$ choices for dessert for each of $6$ days, so the total number of possible dessert menus is $3^6$ , or $\boxed{729}$

Answer: A 729
Problem: A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
$\textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304$

Solution: There are $4 \cdot 3^6$ ways for the desserts to be chosen. By symmetry, any of the desserts that are chosen on Friday share $\frac{1}{4}$ of the total arrangements. Therefore our answer is $\frac{4\cdot3^6}{4} = 3^6 = \boxed{729}.$

Answer:  729
Problem: It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$

Solution: Let $s$ be the speed of the escalator and $c$ be the speed of Clea. Using $d = v t$ , the first statement can be translated to the equation $d = 60c$ . The second statement can be translated to $d = 24(c+s)$ . Since the same distance is being covered in each scenario, we can set the two equations equal and solve for $s$ . We find that $s = \dfrac{3c}{2}$ . The problem asks for the time it takes her to ride down the escalator when she just stands on it. Since $t = \dfrac{d}{s}$ and $d = 60c$ , we have $t = \dfrac{60c}{\dfrac{3c}{2}} = 40$ seconds. Answer choice $\boxed{40}$ is correct.

Answer: B 40
Problem: It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$

Solution: Let $s$ be the speed of the escalator and $c$ be the speed of Clea. Then without loss of generality , assume that the length of the escalator be 1. Then $c=\dfrac{1}{60}$ and $c+s=\dfrac{1}{24}$ , so $s=\dfrac{1}{24}-\dfrac{1}{60}=\dfrac{1}{40}$ . Thus the time it takes for Clea to ride down the operating escalator when she just stands on it is $\dfrac{1}{\dfrac{1}{40}}=\boxed{40}$

Answer: B 40
Problem: It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$

Solution: WLOG, let the length of the escalator be $120$ ft. So, Clea's rate is $120/60=2$ feet per second (fps $*$ ), and the escalator and Clea's rate is $120/24=5$ fps. So, the escalator's rate is $5-2=3$ fps. Therefore, $120/3=\boxed{40}$

Answer: B 40
Problem: In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end of the tournament?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution: The total number of games (and wins) in the tournament is $\frac{6 \times 5}{2}= 15$ . A six-way tie is impossible as this would imply each team has 2.5 wins, so the maximum number of tied teams is five. Here's a chart of 15 games where five teams each have 3 wins:
The "X's" are for when it is where a team is set against itself, which cannot happen. The chart says that Team 6 has lost all of its matches, which means that each of the other teams won against it. Then, alternating Wins and Losses were tried. It shows that it is possible for 5 teams to tie for the same amount of wins, which in this case is 3 wins.
Thus, the answer is $\boxed{5}$

Answer: D 5