Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Alice needs to replace a light bulb located $10$ centimeters below the ceiling in her kitchen. The ceiling is $2.4$ meters above the floor. Alice is $1.5$ meters tall and can reach $46$ centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?
$\textbf{(A)}\ 32 \qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 38\qquad\textbf{(E)}\ 40$

Solution: Convert everything to the same unit. Since the answer is in centimeters, change meters to centimeters by moving the decimal place two places to the right.
The ceiling is $240$ centimeters above the floor. The combined height of Alice and the light bulb when she reaches for it is $10+150+46=206$ centimeters. That means the stool's height needs to be $240-206=\boxed{34}$

Answer: B 34
Problem: Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99$

Solution: You need $2$ dimes, $1$ nickel, and $4$ pennies for the first $25$ cents. From $26$ cents to $50$ cents, you only need to add $1$ quarter. From $51$ cents to $75$ cents, you also only need to add $1$ quarter. The same for $76$ cents to $99$ cents. Notice that instead of $100$ , it is $99$ . We are left with $3$ quarters, $1$ nickel, $2$ dimes, and $4$ pennies. Thus, the correct answer is $3+2+1+4=\boxed{10}$

Answer: B 10
Problem: As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction $1/2$ mile in front of her. After she passes him, she can see him in her rear mirror until he is $1/2$ mile behind her. Emily rides at a constant rate of $12$ miles per hour, and Emerson skates at a constant rate of $8$ miles per hour. For how many minutes can Emily see Emerson?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16$

Solution: Because they are both moving in the same direction, Emily is riding relative to Emerson $12-8=4$ mph. Now we can look at it as if Emerson is not moving at all [on his skateboard] and Emily is riding at $4$ mph. It takes her
\[\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}\]
to ride the $1/2$ mile to reach him, and then the same amount of time to be $1/2$ mile ahead of him. This totals to
\[2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{15}\]

Answer: D 15
Problem: Ryan got $80\%$ of the problems correct on a $25$ -problem test, $90\%$ on a $40$ -problem test, and $70\%$ on a $10$ -problem test. What percent of all the problems did Ryan answer correctly?
$\textbf{(A)}\ 64 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 84\qquad\textbf{(E)}\ 86$

Solution: Ryan answered $(0.8)(25)=20$ problems correct on the first test, $(0.9)(40)=36$ on the second, and $(0.7)(10)=7$ on the third. This amounts to a total of $20+36+7=63$ problems correct. The total number of problems is $25+40+10=75.$ Therefore, the percentage is $\dfrac{63}{75} = 84\% \rightarrow \boxed{84}$

Answer: D 84
Problem: Six pepperoni circles will exactly fit across the diameter of a $12$ -inch pizza when placed. If a total of $24$ circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?
$\textbf{(A)}\ \frac 12 \qquad\textbf{(B)}\ \frac 23 \qquad\textbf{(C)}\ \frac 34 \qquad\textbf{(D)}\ \frac 56 \qquad\textbf{(E)}\ \frac 78$

Solution: The pepperoni circles' diameter is $2$ , since $\dfrac{12}{6} = 2$ . From that we see that the area of the $24$ circles of pepperoni is $\left ( \frac{2}{2} \right )^2 (24\pi) = 24\pi$ . The large pizza's area is $6^2\pi$ . Therefore, the ratio is $\frac{24\pi}{36\pi} = \boxed{23}$

Answer: B 23
Problem: The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4$ . In feet, how tall is the taller tree?
$\textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112$

Solution: Let the height of the taller tree be $h$ and let the height of the smaller tree be $h-16$ . Since the ratio of the smaller tree to the larger tree is $\frac{3}{4}$ , we have $\frac{h-16}{h}=\frac{3}{4}$ . Solving for $h$ gives us $h=64 \Rightarrow \boxed{64}$

Answer: B 64
Problem: The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4$ . In feet, how tall is the taller tree?
$\textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112$

Solution: To answer this problem, you have to make it so that we have the same proportion as 3:4, but the difference between them is 16. Since the two numbers are consecutive, if we multiply both of them by 16, we would get a difference of 16 between them. So, it would be 48:64 and since we need to find the height of the taller tree, we get $h=64 \Rightarrow \boxed{64}$

Answer: B 64
Problem: Of the $500$ balls in a large bag, $80\%$ are red and the rest are blue. How many of the red balls must be removed so that $75\%$ of the remaining balls are red?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150$

Solution: Since 80 percent of the 500 balls are red, there are 400 red balls. Therefore, there must be 100 blue balls. For the 100 blue balls to be 25% or $\dfrac{1}{4}$ of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is $\boxed{100}$

Answer: D 100
Problem: Of the $500$ balls in a large bag, $80\%$ are red and the rest are blue. How many of the red balls must be removed so that $75\%$ of the remaining balls are red?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150$

Solution: We could also set up a proportion. Since we know there are 400 red balls, we let the amount of red balls removed be $x$ , so $\frac{400-x}{500-x}=\frac{3}{4}$ . Cross-multiplying gives us $1600-4x=1500-3x \implies x=100$ , so our answer is $\boxed{100}$

Answer: D 100
Problem: Of the $500$ balls in a large bag, $80\%$ are red and the rest are blue. How many of the red balls must be removed so that $75\%$ of the remaining balls are red?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150$

Solution: Before any balls are taken out of the bag, there are 400 red balls. Let the number of red balls we take out of the bag be $x$ . Since we want the number of red ball to be 75% of the remaining balls, we can set up the equation: $0.75(500–x)=400–x$ .
In words, 75% of the total remaining balls are equal to the remaining red balls. 
Solving we get $x=100$ , so our answer is $\boxed{100}$ .
~J.L.L

Answer: D 100
Problem: The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is $30\%$ of the perimeter. What is the length of the longest side?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution: Let $n$ $n+1$ , and $n+2$ be the lengths of the sides of the triangle. Then the perimeter of the triangle is $n + (n+1) + (n+2) = 3n+3$ . Using the fact that the length of the smallest side is $30\%$ of the perimeter, it follows that:
$n = 0.3(3n+3) \Rightarrow n = 0.9n+0.9 \Rightarrow 0.1n = 0.9 \Rightarrow n=9$ . The longest side is then $n+2 = 11$ . Thus, answer choice $\boxed{11}$ is correct.

Answer: E 11
Problem: The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is $30\%$ of the perimeter. What is the length of the longest side?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution: Since the length of the shortest side is a whole number and is equal to $\frac{3}{10}$ of the perimeter, it follows that the perimeter must be a multiple of $10$ . Adding the two previous integers to each answer choice, we see that $11+10+9=30$ . Thus, answer choice $\boxed{11}$ is correct.

Answer: E 11
Problem: What is the sum of the prime factors of $2010$
$\textbf{(A)}\ 67\qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 77\qquad\textbf{(D)}\ 201\qquad\textbf{(E)}\ 210$

Solution: First, we must find the prime factorization of $2010$ $2010=2\cdot 3 \cdot 5 \cdot 67$ . We add the factors up to get $\boxed{77}$

Answer: C 77
Problem: A jar contains $5$ different colors of gumdrops. $30\%$ are blue, $20\%$ are brown, $15\%$ are red, $10\%$ are yellow, and other $30$ gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?
$\textbf{(A)}\ 35\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64$

Solution: We do $100-30-20-15-10$ to find the percent of gumdrops that are green. We find that $25\%$ of the gumdrops are green. That means there are $120$ gumdrops. If we replace half of the blue gumdrops with brown gumdrops, then $15\%$ of the jar's gumdrops are brown. $\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{42}$

Answer: C 42
Problem: A decorative window is made up of a rectangle with semicircles at either end. The ratio of $AD$ to $AB$ is $3:2$ . And $AB$ is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircles?
[asy] import graph; size(5cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(8); defaultpen(dps); pen ds=black; real xmin=-4.27,xmax=14.73,ymin=-3.22,ymax=6.8; draw((0,4)--(0,0)); draw((0,0)--(2.5,0)); draw((2.5,0)--(2.5,4)); draw((2.5,4)--(0,4)); draw(shift((1.25,4))*xscale(1.25)*yscale(1.25)*arc((0,0),1,0,180)); draw(shift((1.25,0))*xscale(1.25)*yscale(1.25)*arc((0,0),1,-180,0)); dot((0,0),ds); label("$A$",(-0.26,-0.23),NE*lsf); dot((2.5,0),ds); label("$B$",(2.61,-0.26),NE*lsf); dot((0,4),ds); label("$D$",(-0.26,4.02),NE*lsf); dot((2.5,4),ds); label("$C$",(2.64,3.98),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]
$\textbf{(A)}\ 2:3\qquad\textbf{(B)}\ 3:2\qquad\textbf{(C)}\ 6:\pi\qquad\textbf{(D)}\ 9:\pi\qquad\textbf{(E)}\ 30 :\pi$

Solution: We can set a proportion:
\[\dfrac{AD}{AB}=\dfrac{3}{2}\]
We substitute $AB$ with 30 and solve for $AD$
\[\dfrac{AD}{30}=\dfrac{3}{2}\]
\[AD=45\]
We calculate the combined area of semicircle by putting together semicircle $AB$ and $CD$ to get a circle with radius $15$ . Thus, the area is $225\pi$ . The area of the rectangle is $30\cdot 45=1350$ . We calculate the ratio:
\[\dfrac{1350}{225\pi}=\dfrac{6}{\pi}\Rightarrow\boxed{6}\]

Answer: C 6
Problem: The two circles pictured have the same center $C$ . Chord $\overline{AD}$ is tangent to the inner circle at $B$ $AC$ is $10$ , and chord $\overline{AD}$ has length $16$ . What is the area between the two circles?
[asy] unitsize(45); import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); draw((2,0.15)--(1.85,0.15)--(1.85,0)--(2,0)--cycle); draw(circle((2,1),2.24)); draw(circle((2,1),1)); draw((0,0)--(4,0)); draw((0,0)--(2,1)); draw((2,1)--(2,0)); draw((2,1)--(4,0)); dot((0,0),ds); label("$A$", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label("$B$", (1.97,-0.31),NE*lsf); dot((2,1),ds); label("$C$", (1.96,1.09),NE*lsf); dot((4,0),ds); label("$D$", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle); [/asy]
$\textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi$

Solution: Since $\triangle ACD$ is isosceles, $CB$ bisects $AD$ . Thus $AB=BD=8$ . From the Pythagorean Theorem, $CB=6$ . Thus the area between the two circles is $100\pi - 36\pi=64\pi$ $\boxed{64}$

Answer: C 64
Problem: In a room, $2/5$ of the people are wearing gloves, and $3/4$ of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$

Solution: Let $x$ be the number of people wearing both a hat and a glove. Since the number of people wearing a hat or a glove must be whole numbers, the number of people in the room must be a multiple of (4,5), but we do lcm of 4,5 = 20. Since we are trying to find the minimum $x$ , we must use the smallest possible value for the number of people in the room. Similarly, we can assume that there are no people present who are wearing neither of the two items since this would unnecessarily increase the number of people in the room. Thus, we can say that there are $20$ people in the room, all of which are wearing at least a hat or a glove.
It follows that there are $\frac{2}{5}\cdot 20 = 8$ people wearing gloves and $\frac{3}{4}\cdot 20 = 15$ people wearing hats. Then by applying the Principle of Inclusion Exclusion (PIE), the total number of people in the room wearing either a hat or a glove or both is $8+15-x = 23-x$ . Since we know that this equals $20$ , it follows that $23-x = 20$ , which implies that $x=3$ . Thus, $\boxed{3}$ is the correct answer.

Answer: A 3
Problem: Hui is an avid reader. She bought a copy of the best seller Math is Beautiful . On the first day, Hui read $1/5$ of the pages plus $12$ more, and on the second day she read $1/4$ of the remaining pages plus $15$ pages. On the third day she read $1/3$ of the remaining pages plus $18$ pages. She then realized that there were only $62$ pages left to read, which she read the next day. How many pages are in this book?
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 240\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 360$

Solution: Let $x$ be the number of pages in the book. After the first day, Hui had $\frac{4x}{5}-12$ pages left to read. After the second, she had $\left(\frac{3}{4}\right)\left(\frac{4x}{5}-12\right)-15 = \frac{3x}{5}-24$ left. After the third, she had $\left(\frac{2}{3}\right)\left(\frac{3x}{5}-24\right)-18=\frac{2x}{5}-34$ left. This is equivalent to $62.$
\begin{align*} \frac{2x}{5}-34&=62\\ 2x - 170 &= 310\\ 2x &= 480\\ x &= \boxed{240}

Answer: C 240
Problem: Hui is an avid reader. She bought a copy of the best seller Math is Beautiful . On the first day, Hui read $1/5$ of the pages plus $12$ more, and on the second day she read $1/4$ of the remaining pages plus $15$ pages. On the third day she read $1/3$ of the remaining pages plus $18$ pages. She then realized that there were only $62$ pages left to read, which she read the next day. How many pages are in this book?
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 240\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 360$

Solution: If Hui has $62$ pages left to read at the end of Day 3, then on Day 3, before she read those extra 18 pages, she had $80$ pages left to read. $80$ pages is $\frac{2}{3}$ of the pages she had left at the end of Day 2. So at the end of Day 2, she had $80 \cdot \frac{3}{2} = 120$ pages left to read. 
Similarly, we can work backwards on Day 2 to find that at the end of Day 1 Hui had $180$ pages left to read. Again working backwards on Day 1 we find that Hui originally had $\boxed{240}$ pages to read.

Answer: C 240
Problem: The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$

Solution: Let the hundreds, tens, and units digits of the original three-digit number be $a$ $b$ , and $c$ , respectively. We are given that $a=c+2$ . The original three-digit number is equal to $100a+10b+c = 100(c+2)+10b+c = 101c+10b+200$ . The hundreds, tens, and units digits of the reversed three-digit number are $c$ $b$ , and $a$ , respectively. This number is equal to $100c+10b+a = 100c+10b+(c+2) = 101c+10b+2$ . Subtracting this expression from the expression for the original number, we get $(101c+10b+200) - (101c+10b+2) = 198$ . Thus, the units digit in the final result is $\boxed{8}$

Answer: E 8
Problem: The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$

Solution: The result must hold for any three-digit number with hundreds digit being $2$ more than the units digit. $301$ is such a number. Evaluating, we get $301-103=198$ . Thus, the units digit in the final result is $\boxed{8}$

Answer: E 8
Problem: The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$

Solution: Set the units digit of the original number as $x$ . Thus, its hundreds digit is $x+2$ . After the digits are reversed, the hundreds digit of the original number will be the units digit of the new number. Since $x-(x+2) = -2$ , we can do regrouping and "borrow" $1$ from the tens digit and bring it to the units digit as a $10$ . Therefore, the units digit will end up as $-2 + 10 = \boxed{8}$

Answer: E 8
Problem: Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$ $2$ , or $3$ at a time. For example, Jo could climb $3$ , then $1$ , then $2$ . In how many ways can Jo climb the stairs?
$\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

Solution: A dynamics programming approach is quick and easy. The number of ways to climb one stair is $1$ . There are $2$ ways to climb two stairs: $1$ $1$ or $2$ . For 3 stairs, there are $4$ ways: 
( $1$ $1$ $1$ )
( $1$ $2$ )
( $2$ $1$ )
( $3$
For four stairs, consider what step they came from to land on the fourth stair. They could have hopped straight from the 1st, done a double from #2, or used a single step from #3. The ways to get to each of these steps are $1+2+4=7$ ways to get to step 4. The pattern can then be extended: $4$ steps: $1+2+4=7$ ways. $5$ steps: $2+4+7=13$ ways. $6$ steps: $4+7+13=24$ ways.
Thus, there are $\boxed{24}$ ways to get to step $6.$

Answer: E 24
Problem: Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$ $2$ , or $3$ at a time. For example, Jo could climb $3$ , then $1$ , then $2$ . In how many ways can Jo climb the stairs?
$\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

Solution: Complementary counting is also possible. Considering the six steps, Jo has to land on the last step, so there are $2^5=32$ subsets (hit steps) of the other five steps. After that, subtract the number of ways to climb the steps while taking a leap of $4$ $5$ , or $6$ . The eight possible ways for this is ( $4$ $1$ $1$ ), ( $4$ $2$ ), ( $1$ $4$ $1$ ), ( $1$ $1$ $4$ ), ( $2$ $4$ ), ( $1$ $5$ ), ( $5$ $1$ ), and ( $6$
Altogether this makes for $32-8= \boxed{24}$ valid ways for Jo to get to step 6.

Answer: E 24
Problem: Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$ $2$ , or $3$ at a time. For example, Jo could climb $3$ , then $1$ , then $2$ . In how many ways can Jo climb the stairs?
$\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

Solution: We can set up a recursion to solve this problem. Suppose $f(n)$ represents the number of valid ways to get to the $n$ th step. $f(0)=1$ because there is 1 way for Jo to get to the " $0$ th" step (i.e. the ground). There is $1$ way to get to the first step (a $1$ -step), so $f(1)=1$ . There are $2$ ways to get to the second step (two $1$ -steps or one $2$ -step). Thus, $f(2) = 2$ . In general, $f(n) = f(n-1) + f(n-2) + f(n-3)$ . This is because from the $n-3$ th step, Jo can take a $3$ -step to get to the $n$ th step, from the $n-2$ th step, Jo can take a $2$ -step to get to the $n$ th step, and from the $n-1$ th step, Jo can take a $1$ -step to get to the $n$ th step. We now iteratively calculate values of $f(n)$
$f(0) = 1$
$f(1) = 1$
$f(2) = 2$
$f(3) = f(2) + f(1) + f(0) = 2 + 1 + 1 = 4$
$f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
$f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
$f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = \boxed{24}$

Answer: E 24
Problem: On average, for every 4 sports cars sold at the local dealership, 7 sedans are sold. The dealership predicts that it will sell 28 sports cars next month. How many sedans does it expect to sell?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 112$

Solution: This means the ratio is $4:7$ . If the ratio now is $28:x$ , then that means $x= \boxed{49}$

Answer: D 49
Problem: The graph shows the constant rate at which Suzanna rides her bike. If she rides a total of a half an hour at the same speed, how many miles would she have ridden?
[asy] import graph; /* this is a label */ Label f;  f.p=fontsize(0); xaxis(-0.9,20,Ticks(f, 5.0, 5.0));  yaxis(-0.9,20, Ticks(f, 22.0,5.0)); // real f(real x)  {  return x; }  draw(graph(f,-1,22),black+linewidth(1)); label("1", (-1,5), black);  label("2", (-1, 10), black); label("3", (-1, 15), black); label("4", (-1, 20), black); dot((5,5), black+linewidth(5)); dot((10,10), black+linewidth(5)); dot((15, 15), black+linewidth(5)); dot((20,20), black+linewidth(5)); label("MINUTES", (11,-5), S); label(rotate(90)*"MILES", (-5,11), W);[/asy]
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 5.5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 6.5\qquad\textbf{(E)}\ 7$

Solution: Suzanna's speed is $\frac{1}{5}$ . This means she runs $\frac{1}{5} \cdot 30 = \boxed{6}$

Answer: C 6
Problem: The graph shows the constant rate at which Suzanna rides her bike. If she rides a total of a half an hour at the same speed, how many miles would she have ridden?
[asy] import graph; /* this is a label */ Label f;  f.p=fontsize(0); xaxis(-0.9,20,Ticks(f, 5.0, 5.0));  yaxis(-0.9,20, Ticks(f, 22.0,5.0)); // real f(real x)  {  return x; }  draw(graph(f,-1,22),black+linewidth(1)); label("1", (-1,5), black);  label("2", (-1, 10), black); label("3", (-1, 15), black); label("4", (-1, 20), black); dot((5,5), black+linewidth(5)); dot((10,10), black+linewidth(5)); dot((15, 15), black+linewidth(5)); dot((20,20), black+linewidth(5)); label("MINUTES", (11,-5), S); label(rotate(90)*"MILES", (-5,11), W);[/asy]
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 5.5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 6.5\qquad\textbf{(E)}\ 7$

Solution: From the graph, we can see that every $5$ minutes Suzanna goes, her distance increases by $1$ . Since half an hour is $10$ minutes away, she would go $2$ more miles. $4+2$ is $6$ , so the answer is $\boxed{6}$

Answer: C 6
Problem: A sequence of numbers starts with $1$ $2$ , and $3$ . The fourth number of the sequence is the sum of the previous three numbers in the sequence: $1+2+3=6$ . In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?
$\textbf{(A)}\  11  \qquad \textbf{(B)}\   20  \qquad \textbf{(C)}\   37  \qquad \textbf{(D)}\   68  \qquad \textbf{(E)}\    99$

Solution: List them out, adding the three previous numbers to get the next number,
\[1,2,3,6,11,20,37,\boxed{68}\]

Answer: D 68
Problem: Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many hours will it take to fill Steve's pool?
$\textbf{(A)}\  40  \qquad \textbf{(B)}\   42  \qquad \textbf{(C)}\  44   \qquad \textbf{(D)}\  46   \qquad \textbf{(E)}\   48$

Solution: Each of the four hoses hose fills $24,000/4 = 6,000$ gallons of water. At the rate it goes at it will take $6,000/2.5 = 2400$ minutes, or $\boxed{40}$ hours.

Answer: A 40