Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In $2005$ Tycoon Tammy invested $100$ dollars for two years. During the first year
her investment suffered a $15\%$ loss, but during the second year the remaining
investment showed a $20\%$ gain. Over the two-year period, what was the change
in Tammy's investment?
$\textbf{(A)}\  5\%\text{ loss}\qquad \textbf{(B)}\ 2\%\text{ loss}\qquad \textbf{(C)}\ 1\%\text{ gain}\qquad \textbf{(D)}\ 2\% \text{ gain} \qquad \textbf{(E)}\  5\%\text{ gain}$

Solution: After the $15 \%$ loss, Tammy has $100 \cdot 0.85 = 85$ dollars. After the $20 \%$ gain, she has $85 \cdot 1.2 = 102$ dollars. This is an increase in $2$ dollars from her original $100$ dollars, a $\boxed{2}$

Answer: D 2
Problem: The average age of the $6$ people in Room A is $40$ . The average age of the $4$ people in Room B is $25$ . If the two groups are combined, what is the average age of all the people?
$\textbf{(A)}\ 32.5 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 33.5 \qquad \textbf{(D)}\ 34\qquad \textbf{(E)}\ 35$

Solution: The total of all their ages over the number of people is
\[\frac{6 \cdot 40 + 4\cdot 25}{6+4} = \frac{340}{10} = \boxed{34}.\]

Answer: D 34
Problem: Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ 39\qquad \textbf{(E)}\ 46$

Solution: The union of two sets is equal to the sum of each set minus their intersection. The number of students that have both a dog and a cat is $20+26-39 = \boxed{7}$

Answer: A 7
Problem: Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ 39\qquad \textbf{(E)}\ 46$

Solution: We create a diagram: [asy] draw(circle((0,0),5)); draw(circle((5,0),5)); label("$x$",(2.3,0),S); label("$20$",(7,0),S); label("$26$",(-2,0),S); [/asy]
Let $x$ be the number of students with both a dog and a cat.
Therefore, we have
\[26+20-x = 39\] \[46-x = 39\] \[x = \boxed{7}\]

Answer: A 7
Problem: A ball is dropped from a height of $3$ meters. On its first bounce it rises to a height of $2$ meters. It keeps falling and bouncing to $\frac{2}{3}$ of the height it reached in the previous bounce. On which bounce will it not rise to a height of $0.5$ meters?
$\textbf{(A)}\  3 \qquad \textbf{(B)}\  4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

Solution: Each bounce is $2/3$ times the height of the previous bounce. The first bounce reaches $2$ meters, the second $4/3$ , the third $8/9$ , the fourth $16/27$ , and the fifth $32/81$ . Half of $81$ is $40.5$ , so the ball does not reach the required height on bounce $\boxed{5}$

Answer: C 5
Problem: Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than $100$ pounds or more than $150$ pounds. So the boxes are weighed in pairs in every possible way. The results are $122$ $125$ and $127$ pounds. What is the combined weight in pounds of the three boxes?
$\textbf{(A)}\ 160\qquad \textbf{(B)}\ 170\qquad \textbf{(C)}\ 187\qquad \textbf{(D)}\ 195\qquad \textbf{(E)}\ 354$

Solution: Each box is weighed twice during this, so the combined weight of the three boxes is half the weight of these separate measures:
\[\frac{122+125+127}{2} = \frac{374}{2} = \boxed{187}.\]

Answer: C 187
Problem: Three $\text{A's}$ , three $\text{B's}$ , and three $\text{C's}$ are placed in the nine spaces so that each row and column contains one of each letter. If $\text{A}$ is placed in the upper left corner, how many arrangements are possible?
[asy] size((80)); draw((0,0)--(9,0)--(9,9)--(0,9)--(0,0)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); draw((0,3)--(9,3)); draw((0,6)--(9,6)); label("A", (1.5,7.5)); [/asy]
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution: There are $2$ ways to place the remaining $\text{As}$ $2$ ways to place the remaining $\text{Bs}$ , and $1$ way to place the remaining $\text{Cs}$ for a total of $(2)(2)(1) = \boxed{4}$

Answer: C 4
Problem: In Theresa's first $8$ basketball games, she scored $7, 4, 3, 6, 8, 3, 1$ and $5$ points. In her ninth game, she scored fewer than $10$ points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than $10$ points and her points-per-game average for the $10$ games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?
$\textbf{(A)}\ 35\qquad \textbf{(B)}\ 40\qquad \textbf{(C)}\ 48\qquad \textbf{(D)}\ 56\qquad \textbf{(E)}\ 72$

Solution: The total number of points from the first $8$ games is $7+4+3+6+8+3+1+5=37$ . We have to make this a multiple of $9$ by scoring less than $10$ points. The closest multiple of $9$ is $45$ $45-37=8$ . Now we have to add a number to get a multiple of 10. The next multiple is $50$ we added $5$ , multiplying these together you get $8\cdot5$ is $40$ . The answer is $\boxed{40}$

Answer: B 40
Problem: Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of $50$ units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?
$\textbf{(A)}\ 76\qquad \textbf{(B)}\ 120\qquad \textbf{(C)}\ 128\qquad \textbf{(D)}\ 132\qquad \textbf{(E)}\ 136$

Solution: A rectangle's area is maximized when its length and width are equivalent, or the two side lengths are closest together in this case with integer lengths. This occurs with the sides $12 \times 13 = 156$ . Likewise, the area is smallest when the side lengths have the greatest difference, which is $1 \times 24 = 24$ . The difference in area is $156-24=\boxed{132}$

Answer: D 132
Problem: Eight points are spaced around at intervals of one unit around a $2 \times 2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the two points are one unit apart? [asy] size((50)); dot((5,0)); dot((5,5)); dot((0,5)); dot((-5,5)); dot((-5,0)); dot((-5,-5)); dot((0,-5)); dot((5,-5)); [/asy] $\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{2}{7}\qquad\textbf{(C)}\ \frac{4}{11}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{4}{7}$

Solution: The two points are one unit apart at $8$ places around the edge of the square. There are $8 \choose 2$ $= 28$ ways to choose two points. The probability is
\[\frac{8}{28} = \boxed{27}\]

Answer: B 27
Problem: Eight points are spaced around at intervals of one unit around a $2 \times 2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the two points are one unit apart? [asy] size((50)); dot((5,0)); dot((5,5)); dot((0,5)); dot((-5,5)); dot((-5,0)); dot((-5,-5)); dot((0,-5)); dot((5,-5)); [/asy] $\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{2}{7}\qquad\textbf{(C)}\ \frac{4}{11}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{4}{7}$

Solution: Arbitrarily pick a point in the grid. Clearly, we see two options for the other point to be placed, so the answer is $\boxed{27}$

Answer: B 27
Problem: The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and $\frac{3}{4}$ of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 17\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$

Solution: Let $b$ be the number of boys and $g$ be the number of girls.
\[\frac23 b = \frac34 g \Rightarrow b = \frac98 g\]
For $g$ and $b$ to be integers, $g$ must cancel out with the denominator, and the smallest possible value is $8$ . This yields $9$ boys. The minimum number of students is $8+9=\boxed{17}$

Answer: B 17
Problem: The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and $\frac{3}{4}$ of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 17\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$

Solution: We know that $\frac23 B = \frac34 G$ or $\frac69 B = \frac68 G$ . So, the ratio of the number of boys to girls is 9:8. The smallest total number of students is $9 + 8 = \boxed{17}$ . ~DY

Answer: B 17
Problem: Jerry cuts a wedge from a $6$ -cm cylinder of bologna as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?
[asy] defaultpen(linewidth(0.65)); real d=90-63.43494882; draw(ellipse((origin), 2, 4)); fill((0,4)--(0,-4)--(-8,-4)--(-8,4)--cycle, white); draw(ellipse((-4,0), 2, 4)); draw((0,4)--(-4,4)); draw((0,-4)--(-4,-4)); draw(shift(-2,0)*rotate(-d-5)*ellipse(origin, 1.82, 4.56), linetype("10 10")); draw((-4,4)--(-8,4), dashed); draw((-4,-4)--(-8,-4), dashed); draw((-4,4.3)--(-4,5)); draw((0,4.3)--(0,5)); draw((-7,4)--(-7,-4), Arrows(5)); draw((-4,4.7)--(0,4.7), Arrows(5)); label("$8$ cm", (-7,0), W); label("$6$ cm", (-2,4.7), N);[/asy]
$\textbf{(A)} \ 48 \qquad \textbf{(B)} \ 75 \qquad \textbf{(C)} \ 151\qquad \textbf{(D)} \ 192 \qquad \textbf{(E)} \ 603$

Solution: The slice is cutting the cylinder into two equal wedges with equal volume. The cylinder's volume is $\pi r^2 h = \pi (4^2)(6) = 96\pi$ . The volume of the wedge is half this which is $48\pi \approx \boxed{151}$

Answer: C 151
Problem: For how many positive integer values of $n$ are both $\frac{n}{3}$ and $3n$ three-digit whole numbers?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 27\qquad \textbf{(D)}\ 33\qquad \textbf{(E)}\ 34$

Solution: Instead of finding n, we find $x=\frac{n}{3}$ . We want $x$ and $9x$ to be three-digit whole numbers. The smallest three-digit whole number is $100$ , so that is our minimum value for $x$ , since if $x \in \mathbb{Z^+}$ , then $9x \in \mathbb{Z^+}$ . The largest three-digit whole number divisible by $9$ is $999$ , so our maximum value for $x$ is $\frac{999}{9}=111$ . There are $12$ whole numbers in the closed set $\left[100,111\right]$ , so the answer is $\boxed{12}$

Answer: A 12
Problem: For how many positive integer values of $n$ are both $\frac{n}{3}$ and $3n$ three-digit whole numbers?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 27\qquad \textbf{(D)}\ 33\qquad \textbf{(E)}\ 34$

Solution: We can set the following inequalities up to satisfy the conditions given by the question, $100 \leq \frac{n}{3} \leq 999$ ,
and $100 \leq 3n \leq 999$ .
Once we simplify these and combine the restrictions, we get the inequality, $300 \leq n \leq 333$ .
Now we have to find all multiples of 3 in this range for $\frac{n}{3}$ to be an integer. We can compute this by setting $\frac{n} {3}=x$ , where $x \in \mathbb{Z^+}$ . Substituting $x$ for $n$ in the previous inequality, we get, $100 \leq x \leq 111$ , and there are $111-100+1$ integers in this range giving us the answer, $\boxed{12}$

Answer: A 12
Problem: For how many positive integer values of $n$ are both $\frac{n}{3}$ and $3n$ three-digit whole numbers?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 27\qquad \textbf{(D)}\ 33\qquad \textbf{(E)}\ 34$

Solution: So we know the largest $3$ digit number is $999$ and the lowest is $100$ . This means $\dfrac{n}{3} \ge 100 \rightarrow n \ge 300$ but $3n \le 999 \rightarrow n \le 333$ . So we have the set ${300, 301, 302, \cdots, 330, 331, 332, 333}$ for $n$ . Now we have to find the multiples of $3$ suitable for $n$ , or else $\dfrac{n}{3}$ will be a decimal. Only numbers ${300, 303, \cdots, 333}$ are counted. We can  divide by $3$ to make the difference $1$ again, getting ${100, 101 \cdots , 111}$ . Due to it being inclusive, we have $111-100+1 =\boxed{12}$

Answer: A 12
Problem: Margie's winning art design is shown. The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Which of the following is closest to the percent of the design that is black?
[asy] real d=320; pair O=origin; pair P=O+8*dir(d); pair A0 = origin; pair A1 = O+1*dir(d); pair A2 = O+2*dir(d); pair A3 = O+3*dir(d); pair A4 = O+4*dir(d); pair A5 = O+5*dir(d); filldraw(Circle(A0, 6), white, black); filldraw(circle(A1, 5), black, black); filldraw(circle(A2, 4), white, black); filldraw(circle(A3, 3), black, black); filldraw(circle(A4, 2), white, black); filldraw(circle(A5, 1), black, black); [/asy]
$\textbf{(A)}\ 42\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 48\qquad$

Solution: Let the smallest circle be 1, the second smallest circle be 2, the third smallest circle be 3, etc. \[\begin{array}{c|cc} \text{circle \#} & \text{radius} & \text{area} \\ \hline 1 & 2 & 4\pi \\ 2 & 4 & 16\pi \\ 3 & 6 & 36\pi \\ 4 & 8 & 64\pi \\ 5 & 10 & 100\pi \\ 6 & 12 & 144\pi \end{array}\]
The entire circle's area is $144\pi$ . The area of the black regions is $(100-64)\pi + (36-16)\pi + 4\pi = 60\pi$ . The percentage of the design that is black is $\frac{60\pi}{144\pi} = \frac{5}{12} \approx \boxed{42}$

Answer: A 42
Problem: Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of $10$ hours per week helping around the house for $6$ weeks. For the first $5$ weeks she helps around the house for $8$ $11$ $7$ $12$ and $10$ hours. How many hours must she work for the final week to earn the tickets?
$\mathrm{(A)}\ 9 \qquad\mathrm{(B)}\ 10 \qquad\mathrm{(C)}\ 11 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 13$

Solution: Let $x$ be the number of hours she must work for the final week. We are looking for the average, so \[\frac{8 + 11 + 7 + 12 + 10 + x}{6} = 10\] Solving gives: \[\frac{48 + x}{6} = 10\]
\[48 + x = 60\]
\[x = 12\]
So, the answer is $\boxed{12}$

Answer: D 12
Problem: Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of $10$ hours per week helping around the house for $6$ weeks. For the first $5$ weeks she helps around the house for $8$ $11$ $7$ $12$ and $10$ hours. How many hours must she work for the final week to earn the tickets?
$\mathrm{(A)}\ 9 \qquad\mathrm{(B)}\ 10 \qquad\mathrm{(C)}\ 11 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 13$

Solution: Use average deviation:
The average is 10 hour per day. If work 8 hours then it is 2 hours short; if work 11 hours then there is 1 hour surplus, the last day need to cancel out the collective deviation from the previous 5 days.
So we got
\[-2+1-3+2+0=-2\]
The last day need to have +2 deviation to cancel out the -2 collective deviation to get 10 as average value, so $\boxed{12}$

Answer: D 12
Problem: A haunted house has six windows. In how many ways can
Georgie the Ghost enter the house by one window and leave
by a different window?
$\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 18 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 36$

Solution: Georgie can enter the haunted house through any of the six windows. Then, he can leave through any of the remaining five windows.
So, Georgie has a total of $6 \cdot 5$ ways he can enter the house by one window and leave
by a different window.
Therefore, we have $\boxed{30}$ ways.

Answer: D 30
Problem: Chandler wants to buy a $500$ dollar mountain bike. For his birthday, his grandparents
send him $50$ dollars, his aunt sends him $35$ dollars and his cousin gives him $15$ dollars. He earns $16$ dollars per week for his paper route. He will use all of his birthday money and all
of the money he earns from his paper route. In how many weeks will he be able
to buy the mountain bike?
$\mathrm{(A)}\ 24 \qquad\mathrm{(B)}\ 25 \qquad\mathrm{(C)}\ 26 \qquad\mathrm{(D)}\ 27 \qquad\mathrm{(E)}\ 28$

Solution: Let $x$ be the number of weeks.
Thus, we have the equation $50 + 35 + 15 + 16x = 500$
Simplify,
$100 + 16x = 500$
$16x = 400$
$x = 25$
The answer is $\boxed{25}$

Answer: B 25
Problem: The average cost of a long-distance call in the USA in $1985$ was $41$ cents per minute, and the average cost of a long-distance
call in the USA in $2005$ was $7$ cents per minute. Find the
approximate percent decrease in the cost per minute of a long-
distance call.
$\mathrm{(A)}\ 7 \qquad\mathrm{(B)}\ 17 \qquad\mathrm{(C)}\ 34 \qquad\mathrm{(D)}\ 41 \qquad\mathrm{(E)}\ 80$

Solution: The percent decrease is (the amount of decrease)/(original amount)
the amount of decrease is $41 - 7 = 34$
so the percent decrease is $\frac{34}{41}$ which is about $\boxed{80}$

Answer: E 80
Problem: The average age of $5$ people in a room is $30$ years. An $18$ -year-old person leaves
the room. What is the average age of the four remaining people?
$\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36$

Solution: Let $x$ be the average of the remaining $4$ people.
The equation we get is $\frac{4x + 18}{5} = 30$
Simplify,
$4x + 18 = 150$
$4x = 132$
$x = 33$
Therefore, the answer is $\boxed{33}$

Answer: D 33
Problem: The average age of $5$ people in a room is $30$ years. An $18$ -year-old person leaves
the room. What is the average age of the four remaining people?
$\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36$

Solution: Since an $18$ year old left from a group of people averaging $30$ , The remaining people must total $30 - 18 = 12$ years older than $30$ . Therefore, the average is $\dfrac{12}{4} = 3$ years over $30$ . Giving us $\boxed{33}$

Answer: D 33
Problem: The average age of $5$ people in a room is $30$ years. An $18$ -year-old person leaves
the room. What is the average age of the four remaining people?
$\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36$

Solution: The total ages would be $30*5=150$ . Then, if one $18$ year old leaves, we subtract $18$ from $150$ and get $132$ . Then, we divide $132$ by $4$ to get the new average, $\boxed{33}$

Answer: D 33
Problem: In trapezoid $ABCD$ $\overline{AD}$ is perpendicular to $\overline{DC}$ $AD = AB = 3$ , and $DC = 6$ . In addition, $E$ is on $\overline{DC}$ , and $\overline{BE}$ is parallel to $\overline{AD}$ . Find the area of $\triangle BEC$ [asy] defaultpen(linewidth(0.7)); pair A=(0,3), B=(3,3), C=(6,0), D=origin, E=(3,0); draw(E--B--C--D--A--B); draw(rightanglemark(A, D, C)); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, SE); label("$D$", D, SW); label("$E$", E, NW); label("$3$", A--D, W); label("$3$", A--B, N); label("$6$", E, S); [/asy]
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 18$

Solution: Clearly, $ABED$ is a square with side-length $3.$ By segment subtraction, we have $EC = DC - DE = 6 - 3 = 3.$
The area of $\triangle BEC$ is \[\frac12\cdot EC\cdot BE = \frac12\cdot3\cdot3 = \boxed{4.5}.\] ~Aplus95 (Solution)

Answer: B 4.5
Problem: In trapezoid $ABCD$ $\overline{AD}$ is perpendicular to $\overline{DC}$ $AD = AB = 3$ , and $DC = 6$ . In addition, $E$ is on $\overline{DC}$ , and $\overline{BE}$ is parallel to $\overline{AD}$ . Find the area of $\triangle BEC$ [asy] defaultpen(linewidth(0.7)); pair A=(0,3), B=(3,3), C=(6,0), D=origin, E=(3,0); draw(E--B--C--D--A--B); draw(rightanglemark(A, D, C)); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, SE); label("$D$", D, SW); label("$E$", E, NW); label("$3$", A--D, W); label("$3$", A--B, N); label("$6$", E, S); [/asy]
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 18$

Solution: Clearly, $ABED$ is a square with side-length $3.$
Let the brackets denote areas. We apply area subtraction to find the area of $\triangle BEC:$ \begin{align*} [BEC]&=[ABCD]-[ABED] \\ &=\frac{AB+CD}{2}\cdot AD - AB^2 \\ &=\frac{3+6}{2}\cdot 3 - 3^2 \\ &=\boxed{4.5} ~MRENTHUSIASM

Answer: B 4.5
Problem: To complete the grid below, each of the digits 1 through 4 must occur once
in each row and once in each column. What number will occupy the lower
right-hand square?
\[\begin{tabular}{|c|c|c|c|}\hline 1 & & 2 &\\ \hline 2 & 3 & &\\ \hline & &&4\\ \hline & &&\\ \hline\end{tabular}\]
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad\textbf{(E)}\ \text{cannot be determined}$

Solution: The number in the first row, last column must be a $3$ due to the fact if a $3$ was in the first row, second column, there would be two threes in that column. By the same reasoning, the number in the second row, last column has to be a $1$ . Therefore the number in the lower right-hand square is $\boxed{2}$

Answer: B 2
Problem: To complete the grid below, each of the digits 1 through 4 must occur once
in each row and once in each column. What number will occupy the lower
right-hand square?
\[\begin{tabular}{|c|c|c|c|}\hline 1 & & 2 &\\ \hline 2 & 3 & &\\ \hline & &&4\\ \hline & &&\\ \hline\end{tabular}\]
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad\textbf{(E)}\ \text{cannot be determined}$

Solution: Note how the first and second row already contain a $2$ . Since the third row, last column already has a $4$ , the only possible place a $2$ could be in is the bottom right square. Thus our answer is $\boxed{2}$

Answer: B 2
Problem: Sets $A$ and $B$ , shown in the Venn diagram, have the same number of elements.
Their union has $2007$ elements and their intersection has $1001$ elements. Find
the number of elements in $A$
[asy] defaultpen(linewidth(0.7)); draw(Circle(origin, 5)); draw(Circle((5,0), 5)); label("$A$", (0,5), N); label("$B$", (5,5), N); label("$1001$", (2.5, -0.5), N);[/asy]
$\mathrm{(A)}\ 503 \qquad \mathrm{(B)}\ 1006 \qquad \mathrm{(C)}\ 1504 \qquad \mathrm{(D)}\ 1507 \qquad \mathrm{(E)}\ 1510$

Solution: Let $x$ be the number of elements in $A$ and $B$ which is equal.
Then we could form equation $2x-1001 = 2007$
$2x = 3008$
$x = 1504$
The answer is $\boxed{1504}$

Answer: C 1504
Problem: Sets $A$ and $B$ , shown in the Venn diagram, have the same number of elements.
Their union has $2007$ elements and their intersection has $1001$ elements. Find
the number of elements in $A$
[asy] defaultpen(linewidth(0.7)); draw(Circle(origin, 5)); draw(Circle((5,0), 5)); label("$A$", (0,5), N); label("$B$", (5,5), N); label("$1001$", (2.5, -0.5), N);[/asy]
$\mathrm{(A)}\ 503 \qquad \mathrm{(B)}\ 1006 \qquad \mathrm{(C)}\ 1504 \qquad \mathrm{(D)}\ 1507 \qquad \mathrm{(E)}\ 1510$

Solution: Let $x$ be the number of elements in $A$ not including the intersection. $2007-1001=1006$ total elements excluding the intersection. Since we know that $A=B$ , we can find that $x=\frac{1006}2=503$ . Now we need to add the intersection. $503+1001=\boxed{1504}$

Answer: C 1504
Problem: The base of isosceles $\triangle ABC$ is $24$ and its area is $60$ . What is the length of one
of the congruent sides?
$\mathrm{(A)}\ 5 \qquad \mathrm{(B)}\ 8 \qquad \mathrm{(C)}\ 13 \qquad \mathrm{(D)}\ 14 \qquad \mathrm{(E)}\ 18$

Solution: The area of a triangle is shown by $\frac{1}{2}bh$ . We set the base equal to $24$ , and the area equal to $60$ , and we get the height, or altitude, of the triangle to be $5$ . In this isosceles triangle, the height bisects the base, so by using the Pythagorean Theorem, $a^2+b^2=c^2$ , we can solve for one of the legs of the triangle (it will be the the hypotenuse, $c$ ). $a = 12$ $b = 5$ $c = 13$ .
The answer is $\boxed{13}$

Answer: C 13