Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$
[asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,N); label("$E$",E,S); label("$5$",A/2,W); label("$6$",(A+D)/2,N); [/asy]
$\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$

Solution: The area of $\bigtriangleup CDE$ is $\frac{DC\cdot CE}{2}$ . The area of $ABCD$ is $AB\cdot AD=5\cdot 6=30$ , which also must be equal to the area of $\bigtriangleup CDE$ , which, since $DC=5$ , must in turn equal $\frac{5\cdot CE}{2}$ . Through transitivity, then, $\frac{5\cdot CE}{2}=30$ , and $CE=12$ . Then, using the Pythagorean Theorem, you should be able to figure out that $\bigtriangleup CDE$ is a $5-12-13$ triangle, so $DE=\boxed{13}$

Answer: B 13
Problem: Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$
[asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,N); label("$E$",E,S); label("$5$",A/2,W); label("$6$",(A+D)/2,N); [/asy]
$\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$

Solution: The area of the rectangle is $5\times6=30.$ Since the parallel line pairs are identical, $DC=5$ . Let $CE$ be $x$ $\dfrac{5x}{2}=30$ is the area of the right triangle. Solving for $x$ , we get $x=12.$ According to the Pythagorean Theorem, we have a $5-12-13$ triangle. So, the hypotenuse $DE$ has to be $\boxed{13}$

Answer: B 13
Problem: Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$
[asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,N); label("$E$",E,S); label("$5$",A/2,W); label("$6$",(A+D)/2,N); [/asy]
$\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$

Solution: This problem can be solved with the Pythagorean Theorem ( $a^2 + b^2 = c^2$ ).  We know $AB = DC$ , so $DC = 5$ $CE$ is twice the length of $AD$ , so $CE = 12$ $5^2 + 12^2 = c^2$ $5^2 = 25$ $12^2 = 144$ $25 + 144 = 169$ $169$ has a square root of $13$ , so the hypotenuse or $DE$ is $13$ .  The answer is $\boxed{13}$

Answer: B 13
Problem: The "Middle School Eight" basketball conference has $8$ teams. Every season, each team plays every other conference team twice (home and away), and each team also plays $4$ games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?
$\textbf{(A) }60\qquad\textbf{(B) }88\qquad\textbf{(C) }96\qquad\textbf{(D) }144\qquad \textbf{(E) }160$

Solution: Within the conference, there are 8 teams, so there are $\dbinom{8}{2}=28$ pairings of teams, and each pair must play two games, for a total of $28\cdot 2=56$ games within the conference.
Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of $4\cdot 8 =32$ games outside the conference.
Therefore, the total number of games is $56+32 = \boxed{88}$ is our answer.

Answer: B 88
Problem: George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first $\frac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must George run the last $\frac{1}{2}$ mile in order to arrive just as school begins today?
$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad \textbf{(E) }12$

Solution: Note that on a normal day, it takes him $1/3$ hour to get to school. However, today it took $\frac{1/2 \text{ mile}}{2 \text{ mph}}=1/4$ hour to walk the first $1/2$ mile. That means that he has $1/3 -1/4 = 1/12$ hours left to get to school, and $1/2$ mile left to go. Therefore, his speed must be $\frac{1/2 \text{ mile}}{1/12 \text { hour}}=\boxed{6}$ is the answer.

Answer: B 6
Problem: George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first $\frac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must George run the last $\frac{1}{2}$ mile in order to arrive just as school begins today?
$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad \textbf{(E) }12$

Solution: Using the harmonic mean formula, and making the speed he needs to take to get to school for the last half $x$ , we can make the expression:
$\frac{2 * 2x}{2 + x}$
Which simplifies to:
$\frac{4x}{2 + x}$
Then, because we know that since he goes at $3$ mph on a normal day, we can say that it is the harmonic mean of the two rates he goes at today, so we add that to our expression and turn it into something familiar:
$\frac{4x}{2 + x} = 3$
Solving that equation gives us:
$\boxed{6}$

Answer: B 6
Problem: Rectangle $ABCD$ has sides $CD=3$ and $DA=5$ . A circle of radius $1$ is centered at $A$ , a circle of radius $2$ is centered at $B$ , and a circle of radius $3$ is centered at $C$ . Which of the following is closest to the area of the region inside the rectangle but outside all three circles?
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw(Circle((0,0),1)); draw(Circle((0,3),2)); draw(Circle((5,3),3)); label("A",(0.2,0),W); label("B",(0.2,2.8),NW); label("C",(4.8,2.8),NE); label("D",(5,0),SE); label("5",(2.5,0),N); label("3",(5,1.5),E); [/asy]
$\text{(A) }3.5\qquad\text{(B) }4.0\qquad\text{(C) }4.5\qquad\text{(D) }5.0\qquad\text{(E) }5.5$

Solution: The area in the rectangle but outside the circles is the area of the rectangle minus the area of all three of the quarter circles in the rectangle.
The area of the rectangle is $3\cdot5 =15$ . The area of all 3 quarter circles is $\frac{\pi}{4}+\frac{\pi(2)^2}{4}+\frac{\pi(3)^2}{4} = \frac{14\pi}{4} = \frac{7\pi}{2}$ . Therefore the area in the rectangle but outside the circles is $15-\frac{7\pi}{2}$ $\pi$ is approximately $\dfrac{22}{7},$ and substituting that in will give $15-11=\boxed{4.0}$

Answer: B 4.0
Problem: The $7$ -digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ are each multiples of $3$ . Which of the following could be the value of $C$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8$

Solution: The number $\mod{3}$ is congruent to sum of a number's digits $\mod{3}$ is congruent to the number $\mod{3}$ $74A52B1 \pmod{3}$ must be congruent to 0, since it is divisible by 3. Therefore, $7+4+A+5+2+B+1 \pmod{3}$ is also congruent to 0. $7+4+5+2+1 \equiv 1 \pmod{3}$ , so $A+B\equiv 2 \pmod{3}$ . As we know, $326AB4C\equiv 0 \pmod{3}$ , so $3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}$ , and therefore $A+B+C\equiv 0 \pmod{3}$ . We can substitute 2 for $A+B$ , so $2+C\equiv 0 \pmod{3}$ , and therefore $C\equiv 1\pmod{3}$ . This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is $\boxed{1}$

Answer: A 1
Problem: The $7$ -digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ are each multiples of $3$ . Which of the following could be the value of $C$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8$

Solution: Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. $7 + 4 + 5 + 2 + 1 = 19$ . To be a multiple of $3$ $A + B$ has to be either $2$ or $5$ or $8$ ... and so on. We add up the numerical digits in the second number; $3 + 2 + 6 + 4 = 15$ . We then add two of the selected values, $5$ to $15$ , to get $20$ . We then see that C = $1, 4$ or $7, 10$ ... and so on, otherwise the number will not be divisible by three. We then add $8$ to $15$ , to get $23$ , which shows us that C = $1$ or $4$ or $7$ ... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be $1, 4,$ and $7$ . However, in the answer choices, there is no $7$ or $4$ or anything greater than $7$ , but there is a $1$ , so $\boxed{1}$ is our answer.

Answer: A 1
Problem: Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the greatest number of additional cars she must buy in order to be able to arrange all her cars this way?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 533$

Solution: The least multiple of 6 greater than 23 is 24. So she will need to add $24-23=\boxed{1}$ more model car.
~avamarora

Answer: A 1
Problem: Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the greatest number of additional cars she must buy in order to be able to arrange all her cars this way?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 533$

Solution: 6 x 4 = 24, which is 1 more than 23. So, the answer is $\boxed{1}$

Answer: A 1
Problem: A sign at the fish market says, "50 $\%$ off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars?
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$

Solution: 50% off the price of half a pound of fish is $3, so 100%, the regular price, of a half pound of fish is $6. If half a pound of fish costs $6, then a whole pound of fish is $\boxed{12}$ dollars.

Answer: D 12
Problem: A sign at the fish market says, "50 $\%$ off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars?
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$

Solution: Suppose a full pound at normal price costs $x$ dollars. Then, with the 50% off deal, the full pound would cost $x/2$ dollars. A half pound of this would cost $x/4$ dollars (including the 50% off). 50% off half a pound is 3, so we form the equation $x/4=3\Rightarrow x=\boxed{12}$

Answer: D 12
Problem: What is the value of $4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)$
$\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000$

Solution: We group the addends inside the parentheses two at a time: \begin{align*} -1 + 2 - 3 + 4 - 5 + 6 - 7 + \ldots + 1000 &= (-1 + 2) + (-3 + 4) + (-5 + 6) + \ldots + (-999 + 1000) \\ &= \underbrace{1+1+1+\ldots + 1}_{\text{500 1's}} \\ &= 500. \end{align*} Then the desired answer is $4 \times 500 = \boxed{2000}$

Answer: E 2000
Problem: Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?
$\textbf{(A)}\ \text{\textdollar}120\qquad\textbf{(B)}\ \text{\textdollar}128\qquad\textbf{(C)}\ \text{\textdollar}140\qquad\textbf{(D)}\ \text{\textdollar}144\qquad\textbf{(E)}\ \text{\textdollar}160$

Solution: Since Judi's 7 friends had to pay $2.50 extra each to cover the total amount that Judi should have paid, we multiply $2.50\cdot7=17.50$ is the bill Judi would have paid if she had money. Hence, to calculate the total amount, we multiply $17.50\cdot8=\boxed{140}$ to find the total the 8 friends paid.

Answer: C 140
Problem: Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?
$\textbf{(A)}\ \text{\textdollar}120\qquad\textbf{(B)}\ \text{\textdollar}128\qquad\textbf{(C)}\ \text{\textdollar}140\qquad\textbf{(D)}\ \text{\textdollar}144\qquad\textbf{(E)}\ \text{\textdollar}160$

Solution: Let $m$ be the total bill the 8 friends split up; each person was supposed to pay $$\frac{m}{8}$ . Since Judi forgot her money, we have to add $$\frac{m}{8}+2.50$ to get the amount of money each of the $7$ friends have to pay. After multipling by 7, we have $7(\frac{m}{8}+2.50)=\frac{7m}{8} + 17.5$ . This is equal to $m$ because both sides represent the total bill. Solving for $m$ \[\frac{7m}{8} + 17.5=m\] \[\frac{m}{8}=17.5\] \[m=\boxed{140}.\]

Answer: C 140
Problem: Hammie is in $6^\text{th}$ grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?
$\textbf{(A)}\ \text{median, by 60} \qquad \textbf{(B)}\ \text{median, by 20} \qquad \textbf{(C)}\ \text{average, by 5}\qquad \textbf{(D)}\ \text{average, by 15} \qquad \textbf{(E)}\ \text{average, by 20}$

Solution: Listing the elements from least to greatest, we have $(5, 5, 6, 8, 106)$ , we see that the median weight is 6 pounds.
The average weight of the five kids is $\frac{5+5+6+8+106}{5} = \frac{130}{5} = 26$
Hence, \[26-6=\boxed{20}.\]

Answer: E 20
Problem: The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, $30 = 6\times5$ . What is the missing number in the top row?
[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); [/asy]
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$

Solution: Let the value in the empty box in the middle row be $x$ , and the value in the empty box in the top row be $y$ $y$ is the answer we're looking for.
[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); label("$x$",(1,1)); label("$y$",(2,3)); [/asy]
From the diagram, $600 = 30x$ , making $x = 20$
[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); label("20",(1,1)); label("$y$",(2,3)); [/asy]
It follows that $20 = 5y$ , so $y = \boxed{4}$

Answer: C 4
Problem: The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, $30 = 6\times5$ . What is the missing number in the top row?
[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); [/asy]
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$

Solution: Another way to do this problem is to realize what makes up the bottommost number. This method doesn't work quite as well for this problem, but in a larger tree, it might be faster. Again, let the value in the empty box in the middle row be $x$ , and the value in the empty box in the top row be $y$ $y$ is the answer we're looking for.
[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); label("$x$",(1,1)); label("$y$",(2,3)); [/asy]
We can write some equations:
$600 = 30x\\ 30 = 6\cdot 5\\ x = 5y$
Now we can substitute into the first equation using the two others:
$600 = (6\cdot5)(5y)\\ 600= 6\cdot5\cdot5\cdot y\\ 600=6\cdot25\cdot y\\ 600 = 150y\\ \boxed{4} = y$

Answer: C 4
Problem: Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?
$\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140$

Solution: Clearly, for every $5$ seconds, $3$ cars pass. It's more convenient to have everything in seconds: $2$ minutes and $45$ seconds $=2\cdot60 + 45 = 165$ seconds. We then set up a ratio: \[\frac{3}{5}=\frac{x}{165}\] \[3(165)=5x\] \[x=3(33)=99\approx\boxed{100}.\]

Answer: C 100
Problem: Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?
$\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140$

Solution: $2$ minutes and $45$ seconds is equal to $120+45=165\text{ seconds}$
Since $6$ cars pass at around $10$ seconds, there are about $\left \lfloor{\dfrac{165}{10}}\right \rfloor =16$ groups of $6$ cars. There are about $16\cdot6=96\text{ cars}$ , so the closest answer choice is $\boxed{100}$

Answer: C 100
Problem: The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer (1,000 meters)?
$\textbf{(A)}\ 9^\text{th} \qquad \textbf{(B)}\ 10^\text{th} \qquad \textbf{(C)}\ 11^\text{th} \qquad \textbf{(D)}\ 12^\text{th} \qquad \textbf{(E)}\ 13^\text{th}$

Solution: This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that $2^{10}=1024$
However, because the first term is $2^0=1$ and not $2^1=2$ , the solution to the problem is $10-0+1=\boxed{11}$

Answer: C 11
Problem: What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
$\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660$

Solution: To find either the LCM or the GCF of two numbers, always prime factorize first.
The prime factorization of $180 = 3^2 \times  5 \times 2^2$
The prime factorization of $594 = 3^3 \times  11 \times 2$
Then, to find the LCM, we have to find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is $3^3, 5, 11, 2^2$ ). Multiply all of these to get 5940.
For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. $3^2 \times 2$ = 18.
Thus the answer = $\frac{5940}{18}$ $\boxed{330}$

Answer: C 330
Problem: What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
$\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660$

Solution: We start off with a similar approach as the original solution. From the prime factorizations, the GCF is $18$
It is a well known fact that $\gcd(m,n)\times \operatorname{lcm}(m,n)=|mn|$ . So we have, $18\times \operatorname{lcm} (180,594)=594\times 180$
Dividing by $18$ yields $\operatorname{lcm} (180,594)=594\times 10=5940$
Therefore, $\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf}(180,594)}=\frac{5940}{18}=\boxed{330}$

Answer: C 330
Problem: What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
$\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660$

Solution: From Solution 1, 
the prime factorization of $180 = 2^2 \cdot 3^2 \cdot 5$ .
The prime factorization of $594 = 2 \cdot 3^3 \cdot 11$ .
Hence, $\operatorname{lcm} (180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11$ , and $\operatorname{gcf} (180,594) = 2 \cdot 3^2$ .
Therefore, $\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf} (180,594)} = \frac{2^2 \cdot 3^3 \cdot 5 \cdot 11}{2 \cdot 3^2} = 2 \cdot 3 \cdot 5 \cdot 11 = 330 \Longrightarrow \boxed{330}$

Answer: C 330
Problem: Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution: We use that fact that $d=rt$ . Let d= distance, r= rate or speed, and t=time. In this case, let $x$ represent the time.
On Monday, he was at a rate of $5 \text{ m.p.h}$ . So, $5x = 2 \text{ miles}\implies x = \frac{2}{5} \text { hours}$
For Wednesday, he walked at a rate of $3 \text{ m.p.h}$ . Therefore, $3x = 2 \text{ miles}\implies x = \frac{2}{3}  \text { hours}$
On Friday, he walked at a rate of $4 \text{ m.p.h}$ . So, $4x = 2 \text{ miles}\implies x=\frac{2}{4}=\frac{1}{2}  \text {hours}$
Adding up the hours yields $\frac{2}{5}  \text { hours}$ $\frac{2}{3}  \text { hours}$ $\frac{1}{2}  \text { hours}$ $\frac{47}{30}  \text { hours}$
We now find the amount of time Grandfather would have taken if he walked at $4 \text{ m.p.h}$ per day. Set up the equation, $4x = 2 \text{ miles} \times  3 \text{ days}\implies x = \frac{3}{2}  \text { hours}$
To find the amount of time saved, subtract the two amounts: $\frac{47}{30}  \text { hours}$ $\frac{3}{2}  \text { hours}$ $\frac{1}{15}  \text { hours}$ . To convert this to minutes, we multiply by $60$
Thus, the solution to this problem is $\dfrac{1}{15}\times 60=\boxed{4}$

Answer: D 4
Problem: At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $150 regular price did he save?
$\textbf{(A)}\ 25 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 33 \qquad \textbf{(D)}\ 40 \qquad \textbf{(E)}\ 45$

Solution: First, find the amount of money one will pay for three sandals without the discount. We have $\textdollar 50\times 3 \text{ sandals} = \textdollar 150$
Then, find the amount of money using the discount: $50 + 0.6 \times 50 + \frac{1}{2} \times 50 = \textdollar 105$
Finding the percentage yields $\frac{105}{150} = 70 \%$
To find the percent saved, we have $100 \% -70 \%= \boxed{30}$

Answer: B 30