Answer: 142.0

It is possible to place positive integers into the vacant twenty-one squares of the $5 \times 5$ square shown below so that the numbers in each row and column form arithmetic sequences.  Find the number that must occupy the vacant square marked by the asterisk (*).
1988 AIME-6.png

Solution:
Let the coordinates of the square at the bottom left be $(0,0)$ , the square to the right $(1,0)$ , etc.
Label the leftmost column (from bottom to top) $0, a, 2a, 3a, 4a$ and the bottom-most row (from left to right) $0, b, 2b, 3b, 4b$ . Our method will be to use the given numbers to set up equations to solve for $a$ and $b$ , and then calculate $(*)$
$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & & & * & \\ \hline 3a & 74 & & & \\ \hline 2a & & & & 186 \\ \hline a & & 103 & & \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$
We can compute the squares at the intersections of two existing numbers in terms of $a$ and $b$ ; two such equations will give us the values of $a$ and $b$ . On the fourth row from the bottom, the common difference is $74 - 3a$ , so the square at $(2,3)$ has a value of $148 - 3a$ . On the third column from the left, the common difference is $103 - 2b$ , so that square also has a value of $2b + 3(103 - 2b) = 309 - 4b$ . Equating, we get $148 - 3a = 309 - 4b \Longrightarrow 4b - 3a = 161$
Now we compute the square $(2,2)$ . By rows, this value is simply the average of $2a$ and $186$ , so it is equal to $\frac{2a + 186}{2} = a + 93$ . By columns, the common difference is $103 - 2b$ , so our value is $206 - 2b$ . Equating, $a + 93 = 206 - 2b \Longrightarrow a + 2b = 113$
Solving \begin{align*}4b - 3a &= 161\\ a + 2b &= 113 \end{align*}
gives $a = 13$ $b = 50$ . Now it is simple to calculate $(4,3)$ . One way to do it is to see that $(2,2)$ has $206 - 2b = 106$ and $(4,2)$ has $186$ , so $(3,2)$ has $\frac{106 + 186}{2} = 146$ . Now, $(3,0)$ has $3b = 150$ , so $(3,2) = \frac{(3,0) + (3,4)}{2} \Longrightarrow (3,4) = * = \boxed{142}$

Solution:
First, let $a =$ the number to be placed in the first column, fourth row.  Let $b =$ the number to be placed in the second column, fifth row.  We can determine the entire first column and fifth row in terms of $a$ and $b$
$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & & & & \\ \hline 3a & & & & \\ \hline 2a & & & & \\ \hline a & & & & \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$
Next, let $a + b + c =$ the number to be placed in the second column, fourth row.  We can determine the entire second column and fourth row in terms of $a$ $b$ , and $c$
$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & 4a + b + 4c & & & \\ \hline 3a & 3a + b + 3c & & & \\ \hline 2a & 2a + b + 2c & & & \\ \hline a & a + b + c & a + 2b + 2c & a + 3b + 3c & a + 4b + 4c \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$
We have now determined at least two values in each row and column.  We can finish the table without introducing any more variables:
$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & 4a + b + 4c & 4a + 2b + 8c & 4a + 3b + 12c & 4a + 4b + 16c \\ \hline 3a & 3a + b + 3c & 3a + 2b + 6c & 3a + 3b + 9c & 3a + 4b + 12c \\ \hline 2a & 2a + b + 2c & 2a + 2b + 4c & 2a + 3b + 6c & 2a + 4b + 8c \\ \hline a & a + b + c & a + 2b + 2c & a + 3b + 3c & a + 4b + 4c \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$
We now have a system of equations.
Solving, we find that $(a,b,c) = (13,50, - 5)$ .  The number in the square marked by the asterisk is $4a + 3b + 12c = \boxed{142}$

Solution:
We begin with the table that was given to us and add in the following arithmetic progression on the bottom:
$\begin{tabular}[b]{|c|c|c|c|c|}\hline & & & * & \\ \hline & 74 & & & \\ \hline & & & & 186 \\ \hline & & 103 & & \\ \hline 0 & x & 2x & 3x & 4x \\ \hline \end{tabular}$
Since all the rows and columns satisfy an arithmetic progression, we have the following:
$\begin{tabular}[b]{|c|c|c|c|c|}\hline & & 412 - 6x & 392 - 5x & 372 - 4x \\ \hline & 74 & 309 - 4x  & 294 - 3x  & 279 - 2x \\ \hline & & 206 - 2x & 196 - x & 186 \\ \hline & & 103 & 98 + x & 2x + 93 \\ \hline 0 & x & 2x & 3x & 4x \\ \hline \end{tabular}$
We can solve for $x$ in the 2nd row, namely $324 - 5x = 74$ because the arithmetic progression from left to right has difference $x - 15$ . Therefore, we have $x = 50$ , and because the desired asterisk is $392 - 5x$ , the answer is $392 - 250$ $\boxed{142}$


Answer: 3.0

It takes Anna $30$ minutes to walk uphill $1$ km from her home to school, but it takes her only $10$ minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip?
$\mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 3.125\qquad \mathrm{(C) \ } 3.5\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 4.5$

Solution:
Since she walked $1$ km to school and $1$ km back home, her total distance is $1+1=2$ km.
Since she spent $30$ minutes walking to school and $10$ minutes walking back home, her total time is $30+10=40$ minutes = $\frac{40}{60}=\frac{2}{3}$ hours.
Therefore her average speed in km/hr is $\frac{2}{\frac{2}{3}}=\boxed{3}$

Solution: The average speed of two speeds that travel the same distance is the harmonic mean of the speeds, or $\dfrac{2}{\dfrac{1}{x}+\dfrac{1}{y}}=\dfrac{2xy}{x+y}$ (for speeds $x$ and $y$ ). Mary's speed going to school is $2\,\text{km/hr}$ , and her speed coming back is $6\,\text{km/hr}$ . Plugging the numbers in, we get that the average speed is $\dfrac{2\times 6\times 2}{6+2}=\dfrac{24}{8}=\boxed{3}$


Answer: 3.0

Solution:
Since she walked $1$ km to school and $1$ km back home, her total distance is $1+1=2$ km.
Since she spent $30$ minutes walking to school and $10$ minutes walking back home, her total time is $30+10=40$ minutes = $\frac{40}{60}=\frac{2}{3}$ hours.
Therefore her average speed in km/hr is $\frac{2}{\frac{2}{3}}=\boxed{3}$

Solution: The average speed of two speeds that travel the same distance is the harmonic mean of the speeds, or $\dfrac{2}{\dfrac{1}{x}+\dfrac{1}{y}}=\dfrac{2xy}{x+y}$ (for speeds $x$ and $y$ ). Mary's speed going to school is $2\,\text{km/hr}$ , and her speed coming back is $6\,\text{km/hr}$ . Plugging the numbers in, we get that the average speed is $\dfrac{2\times 6\times 2}{6+2}=\dfrac{24}{8}=\boxed{3}$


Answer: 40.0

It takes Clea 60 seconds to walk down an escalator when it is not moving, and 24 seconds when it is moving. How many seconds would it take Clea to ride the escalator down when she is not walking?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$

Solution: She walks at a rate of $x$ units per second to travel a distance $y$ . As $vt=d$ , we find $60x=y$ and $24*(x+k)=y$ , where $k$ is the speed of the escalator. Setting the two equations equal to each other, $60x=24x+24k$ , which means that $k=1.5x$ . Now we divide $60$ by $1.5$ because you add the speed of the escalator but remove the walking, leaving the final answer that it takes to ride the escalator alone as $\boxed{40}$

Solution: We write two equations using distance = rate * time. Let $r$ be the rate she is walking, and $e$ be the speed the escalator moves. WLOG, let the distance of the escalator be $120$ , as the distance is constant. Thus, our $2$ equations are $120 = 60r$ and $120 = 24(r+e)$ . Solving for $e$ , we get $e = 3$ . Thus, it will take Clea $\dfrac{120}{3} = \boxed{40}$ seconds.

Solution: Clea covers $\dfrac{1}{60}$ of the escalator every second. Say the escalator covers $\dfrac{1}{r}$ of the escalator every second. Since Clea and the escalator cover the entire escalator in $24$ seconds, we can use distance $=$ rate $\cdot$ time to get $24\left(\dfrac{1}{60} + \dfrac{1}{r}\right) = 1$ . Solving gives us $r = 40$ , so if Clea were to just stand on the escalator, it would take her $\boxed{40}$ seconds to get down. ~Extremelysupercooldude


Answer: 40.0

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.

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 $\dfrac{1}{\dfrac{1}{40}}=\boxed{40}$

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: 9.0

Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of $10$ miles per hour. Jack walks to the pool at a constant speed of $4$ miles per hour. How many minutes before Jack does Jill arrive?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$

Solution:
Using $d=rt$ , we can set up an equation for when Jill arrives at the swimming pool:
$1=10t$
Solving for $t$ , we get that Jill gets to the pool in $\frac{1}{10}$ of an hour, which is $6$ minutes.  Doing the same for Jack, we get that
Jack arrives at the pool in $\frac{1}{4}$ of an hour, which in turn is $15$ minutes.  Thus, Jill has to wait $15-6=\boxed{9}$


Answer: 71.0

Jack had a bag of $128$ apples.  He sold $25\%$ of them to Jill.  Next he sold $25\%$ of those remaining to June.  Of those apples still in his bag, he gave the shiniest one to his teacher.  How many apples did Jack have then?
$\text{(A)}\ 7 \qquad \text{(B)}\ 63 \qquad \text{(C)}\ 65 \qquad \text{(D)}\ 71 \qquad \text{(E)}\ 111$

Solution:
First he gives $128\times .25 = 32$ apples to Jill, so he has $128-32=96$ apples left.  Then he gives $96\times .25 = 24$ apples to June, so he has $96-24=72$ left.
Finally, he gives one to the teacher, leaving $71\rightarrow \boxed{71}$


Answer: 515.0

Jackie and Phil have two fair coins and a third coin that comes up heads with probability $\frac47$ . Jackie flips the three coins, and then Phil flips the three coins. Let $\frac {m}{n}$ be the probability that Jackie gets the same number of heads as Phil, where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
This can be solved quickly and easily with generating functions
Let $x^n$ represent flipping $n$ heads.
The generating functions for these coins are $(1+x)$ $(1+x)$ ,and $(3+4x)$ in order.
The product is $3+10x+11x^2+4x^3$ . ( $ax^n$ means there are $a$ ways to get $n$ heads, eg there are $10$ ways to get $1$ head, and therefore $2$ tails, here.)
The sum of the coefficients squared (total number of possible outcomes, squared because the event is occurring twice) is $(4 + 11 + 10 + 3)^2 = 28^2 = 784$ and the sum of the squares of each coefficient (the sum of the number of ways that each coefficient can be chosen by the two people) is $4^2 + 11^2 + 10^2 + 3^2=246$ .
The probability is then $\frac{4^2 + 11^2 + 10^2 + 3^2}{28^2}  = \frac{246}{784} = \frac{123}{392}$ .
(Notice the relationship between the addends of the numerator here and the cases in the following solution.)
$123 + 392 = \boxed{515}$


Answer: 1625.0

Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?
$\textbf{(A) } 162,500 \qquad\textbf{(B) } 162.5 \qquad\textbf{(C) }1,625 \qquad\textbf{(D) }1,625,000 \qquad\textbf{(E) } 16,250$

Solution: $6.5$ millimeters is equal to $0.65$ centimeters. $25$ meters is $2500$ centimeters. The answer is $0.65 \times 2500$ , so the answer is $\boxed{1,625}$

Solution: $6.5$ millimeters can be represented as $65 \times 10^{-2}$ centimeters. $25$ meters is $25 \times 10^{2}$ centimeters. Multiplying out these results in $(65 \times 10^{-2}) \times (25 \times 10^{2})$ , which is $65 \times 25$ making the answer $\boxed{1,625}$


Answer: 1625.0

Solution: $6.5$ millimeters is equal to $0.65$ centimeters. $25$ meters is $2500$ centimeters. The answer is $0.65 \times 2500$ , so the answer is $\boxed{1,625}$

Solution: $6.5$ millimeters can be represented as $65 \times 10^{-2}$ centimeters. $25$ meters is $25 \times 10^{2}$ centimeters. Multiplying out these results in $(65 \times 10^{-2}) \times (25 \times 10^{2})$ , which is $65 \times 25$ making the answer $\boxed{1,625}$


Answer: 9.0

Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Solution: After Jamal adds $x$ purple socks, he has $(18+x)$ purple socks and $6+18+12+x=(36+x)$ total socks. This means the probability of drawing a purple sock is $\frac{18+x}{36+x}$ , so we obtain \[\frac{18+x}{36+x}=\frac{3}{5}\] Since $\frac{18+9}{36+9}=\frac{27}{45}=\frac{3}{5}$ , the answer is $\boxed{9}$

Solution: As in Solution 1, we have the equation $\frac{18+x}{36+x}=\frac{3}{5}$ . Cross-multiplying yields $90+5x=108+3x \Rightarrow 2x=18 \Rightarrow x=9$ . Thus, Jamal added $\boxed{9}$ purple socks.

Solution: $6$ green socks and $12$ orange socks together should be $100\%-60\% = 40\%$ of the new total number of socks, so that new total must be $\frac{6+12}{0.4}= 45$ . Therefore, $45-6-18-12=\boxed{9}$ purple socks were added.


Answer: 13.0

Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)} 16$

Solution:
$0.8$ MB file can either be on its own disk, or share it with a $0.4$ MB. Clearly it is better to pick the second possibility. Thus we will have $3$ disks, each with one $0.8$ MB file and one $0.4$ MB file.
We are left with $12$ files of $0.7$ MB each, and $12$ files of $0.4$ MB each. Their total size is $12\cdot (0.7 + 0.4) = 13.2$ MB. The total capacity of $9$ disks is $9\cdot 1.44 = 12.96$ MB, hence we need at least $10$ more disks. And we can easily verify that $10$ disks are indeed enough: six of them will carry two $0.7$ MB files each, and four will carry three $0.4$ MB files each.
Thus our answer is $3+10 = \boxed{13}$


Answer: 13.0

Solution:
Similarly to Solution 1, we see that there must be $3$ disks to account for the $0.8$ MB file. Secondly, since there are $[30-(3+12)]-3 = 12$ files(for both 0.4 MB and 0.7 MB) left, it is easy to see that the optimal way to place the files would be $3$ files per disks for the 0.4 MB files and hence would require 4 disks.
We are left with $12$ files( $0.7$ MB), where the optimal number of files per disks is $2$ , so the optimal number of disks for this type of file would be $6$ disks. Therefore, the answer is $3+4+6=\boxed{13}$


Answer: 13.0

Solution:
$0.8$ MB file can either be on its own disk, or share it with a $0.4$ MB. Clearly it is better to pick the second possibility. Thus we will have $3$ disks, each with one $0.8$ MB file and one $0.4$ MB file.
We are left with $12$ files of $0.7$ MB each, and $12$ files of $0.4$ MB each. Their total size is $12\cdot (0.7 + 0.4) = 13.2$ MB. The total capacity of $9$ disks is $9\cdot 1.44 = 12.96$ MB, hence we need at least $10$ more disks. And we can easily verify that $10$ disks are indeed enough: six of them will carry two $0.7$ MB files each, and four will carry three $0.4$ MB files each.
Thus our answer is $3+10 = \boxed{13}$


Answer: 13.0

Solution:
Similarly to Solution 1, we see that there must be $3$ disks to account for the $0.8$ MB file. Secondly, since there are $[30-(3+12)]-3 = 12$ files(for both 0.4 MB and 0.7 MB) left, it is easy to see that the optimal way to place the files would be $3$ files per disks for the 0.4 MB files and hence would require 4 disks.
We are left with $12$ files( $0.7$ MB), where the optimal number of files per disks is $2$ , so the optimal number of disks for this type of file would be $6$ disks. Therefore, the answer is $3+4+6=\boxed{13}$


Answer: 4.0

Jamar bought some pencils costing more than a penny each at the school bookstore and paid $\textdollar 1.43$ . Sharona bought some of the same pencils and paid $\textdollar 1.87$ . How many more pencils did Sharona buy than Jamar?
$\textbf{(A)}\hspace{.05in}2\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}4\qquad\textbf{(D)}\hspace{.05in}5\qquad\textbf{(E)}\hspace{.05in}6$

Solution: We assume that the price of the pencils remains constant. Convert $\textdollar 1.43$ and $\textdollar 1.87$ to cents.  Since the price of the pencils is more than one penny, we can find the price of one pencil (in cents) by taking the greatest common divisor of $143$ and $187$ , which is $11$ . Therefore, Jamar bought $\frac{143}{11} \implies 13$ pencils and Sharona bought $\frac{187}{11} \implies 17$ pencils. Thus, Sharona bought $17-13 = \boxed{4}$ more pencils than Jamar.

Solution: We find the difference between $1.43$ and $1.87$ is $1.87-1.43 = 0.44$ , which is the extra cost of Sharona's pencils than Jamar's pencils. Because the difference between the amounts of the pencils they bought must be divided evenly by $0.44$ , looking into the answers, $2$ or $4$ is possibly correct. It gives us the price of each pencil should be $0.44/2=0.22$ or $0.44/4=0.11$ , respectively. Then we find only $0.11$ can be divided evenly by $1.43$ and $1.87$ . So the answer is $\boxed{4}$

Solution: You can quickly tell, if you have memorized your times tables, that $187$ and $143$ are both multiples of $11$ . Additionally, the difference is 44 cents. That means that there are $\boxed{4}$ more pencils purchased by Sharona because $11*4 = 44$


Answer: 26.0

Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum?
$\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26$

Solution: On a cube, there are $12$ edges, $8$ corners, and $6$ faces. Adding them up gets $12+8+6= \boxed{26}$


Answer: 25.0

Jane is 25 years old. Dick is older than Jane. In $n$ years, where $n$ is a positive integer, Dick's age and Jane's age will both be two-digit numbers and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let $d$ be Dick's present age. How many ordered pairs of positive integers $(d,n)$ are possible?

Solution:
Let Jane's age $n$ years from now be $10a+b$ , and let Dick's age be $10b+a$ . If $10b+a>10a+b$ , then $b>a$ . The possible pairs of $a,b$ are:
That makes 36. But $10a+b>25$ , so we subtract all the extraneous pairs: $(1,2), (1,3), (2,3), (1,4), (2,4), (1,5), (2,5), (1,6), (1,7), (1,8),$ and $(1,9)$ $36-11=\boxed{025}$

Solution: Start by assuming that $n < 5$ (essentially, Jane is in the 20s when their ages are 'reverses' of each other). Then we get the pairs \[(61,1),(70,2),(79,3),(88,4).\] Repeating this for the 30s gives \[(34,9),(43,10),(52,11),(61,12),(70,13),(79,14).\] From here, it's pretty clear that every decade we go up we get $(d,n+11)$ as a pair. Since both ages must always be two-digit numbers, we can show that each decade after the 30s, we get 1 fewer option. Therefore, our answer is $4+6+5+\dots+2+1=4+21=\boxed{025}.$


Answer: 85.0

Jar A contains four liters of a solution that is 45% acid.  Jar B contains five liters of a solution that is 48% acid.  Jar C contains one liter of a solution that is $k\%$ acid.  From jar C, $\frac{m}{n}$ liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B.  At the end both jar A and jar B contain solutions that are 50% acid.  Given that $m$ and $n$ are relatively prime positive integers, find $k + m + n$

Solution:
Jar A contains $\frac{11}{5}$ liters of water, and $\frac{9}{5}$ liters of acid; jar B contains $\frac{13}{5}$ liters of water and $\frac{12}{5}$ liters of acid.
The gap between the amount of water and acid in the first jar, $\frac{2}{5}$ , is double that of the gap in the second jar, $\frac{1}{5}$ . Therefore, we must add twice as much of jar C into the jar $A$ over jar $B$ . So, we must add $\frac{2}{3}$ of jar C into jar $A$ , so $m = 2, n=3$
Since jar C contains $1$ liter of solution, we are adding $\frac{2}{3}$ of a liter of solution to jar $A$ . In order to close the gap between water and acid, there must be $\frac{2}{5}$ more liters of acid than liters of water in these $\frac{2}{3}$ liters of solution. So, in the $\frac{2}{3}$ liters of solution, there are $\frac{2}{15}$ liters of water, and $\frac{8}{15}$ liters of acid.  So, 80% of the $\frac{2}{3}$ sample is acid, so overall, in jar C, 80% of the sample is acid.
Therefore, our answer is $80 + 2 + 3 = \boxed{85}$

Solution: There are $\frac{45}{100}(4)=\frac{9}{5}$ L of acid in Jar A. There are $\frac{48}{100}(5)=\frac{12}{5}$ L of acid in Jar B. And there are $\frac{k}{100}$ L of acid in Jar C. After transferring the solutions from jar C, there will be $4+\frac{m}{n}$ L of solution in Jar A and $\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}$ L of acid in Jar A. $6-\frac{m}{n}$ L of solution in Jar B and $\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}$ of acid in Jar B. Since the solutions are 50% acid, we can multiply the amount of acid for each jar by 2, then equate them to the amount of solution. \[\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}\] \[\frac{24}{5}-\frac{km}{50n}+\frac{k}{50}=6-\frac{m}{n}\] Add the equations to get \[\frac{42}{5}+\frac{k}{50}=10\] Solving gives $k=80$ If we substitute back in the original equation we get $\frac{m}{n}=\frac{2}{3}$ so $3m=2n$ . Since $m$ and $n$ are relatively prime, $m=2$ and $n=3$ . Thus $k+m+n=80+2+3=\boxed{085}$

Solution:
One might cleverly change the content of both Jars.
Since the end result of both Jars are $50\%$ acid, we can turn Jar A into a 1 gallon liquid with $50\%-4(5\%) = 30\%$ acid
and Jar B into 1 gallon liquid with $50\%-5(2\%) =40\%$ acid.
Now, since Jar A and Jar B contain the same amount of liquid, twice as much liquid will be pour into Jar A than Jar B, so $\dfrac{2}{3}$ of Jar C will be pour into Jar A.
Thus, $m=2$ and $n=3$
$\dfrac{30\% + \frac{2}{3} \cdot k\%}{\frac{5}{3}} = 50\%$
Solving for $k$ yields $k=80$
So the answer is $80+2+3 = \boxed{085}$

Solution:
One may first combine all three jars in to a single container.  That container will have $10$ liters of liquid, and it should be $50\%$ acidic.  Thus there must be $5$ liters of acid.
Jar A contained $45\% \cdot 4L$ , or $1.8L$ of acid, and jar B $48\% \cdot 5L$ or $2.4L$ .  Solving for the amount of acid in jar C, $k = (5 - 2.4 - 1.8) = .8$ , or $80\%$
Once one knowss that the jar C is $80\%$ acid, use solution 1 to figure out m and n for $k+m+n=80+2+3=\boxed{085}$


Answer: 5.0

Jean has made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is $8$ feet high while the other peak is $12$ feet high. Each peak forms a $90^\circ$ angle, and the straight sides form a $45^\circ$ angle with the ground. The artwork has an area of $183$ square feet. The sides of the mountain meet at an intersection point near the center of the artwork, $h$ feet above the ground. What is the value of $h?$
[asy] unitsize(.3cm); filldraw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,gray(0.7),linewidth(1)); draw((-1,0)--(-1,8),linewidth(.75)); draw((-1.4,0)--(-.6,0),linewidth(.75)); draw((-1.4,8)--(-.6,8),linewidth(.75)); label("$8$",(-1,4),W); label("$12$",(31,6),E); draw((-1,8)--(8,8),dashed); draw((31,0)--(31,12),linewidth(.75)); draw((30.6,0)--(31.4,0),linewidth(.75)); draw((30.6,12)--(31.4,12),linewidth(.75)); draw((31,12)--(18,12),dashed); label("$45^{\circ}$",(.75,0),NE,fontsize(10pt)); label("$45^{\circ}$",(29.25,0),NW,fontsize(10pt)); draw((8,8)--(7.5,7.5)--(8,7)--(8.5,7.5)--cycle); draw((18,12)--(17.5,11.5)--(18,11)--(18.5,11.5)--cycle); draw((11,5)--(11,0),dashed); label("$h$",(11,2.5),E); [/asy]
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 5\sqrt{2} \qquad \textbf{(E)}\ 6$

Solution: Extend the "inner part" of the mountain so that the image is two right triangles that overlap in a third right triangle as shown.  The side length of the largest right triangle is $12\sqrt{2},$ which means its area is $144.$ Similarly, the area of the second largest right triangle is $64$ (the side length is $8\sqrt{2}$ ), and the area of the overlap is $h^2$ (the side length is $h\sqrt{2}$ ). Thus, \[144+64-h^2=183,\] which means that the answer is $\boxed{5}.$

Solution: You can measure $h$ with a ruler(rulers are allowed on the AMC 8), and see that $h$ is closest to $\boxed{5}.$


Answer: 116.0

Jen enters a lottery by picking $4$ distinct numbers from $S=\{1,2,3,\cdots,9,10\}.$ $4$ numbers are randomly chosen from $S.$ She wins a prize if at least two of her numbers were $2$ of the randomly chosen numbers, and wins the grand prize if all four of her numbers were the randomly chosen numbers. The probability of her winning the grand prize given that she won a prize is $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
This is a conditional probability problem. Bayes' Theorem states that \[P(A|B)=\dfrac{P(B|A)\cdot P(A)}{P(B)}\]
Let us calculate the probability of winning a prize. We do this through casework: how many of Jen's drawn numbers match the lottery's drawn numbers?
To win a prize, Jen must draw at least $2$ numbers identical to the lottery. Thus, our cases are drawing $2$ $3$ , or $4$ numbers identical.
Let us first calculate the number of ways to draw exactly $2$ identical numbers to the lottery. Let Jen choose the numbers $a$ $b$ $c$ , and $d$ ; we have $\dbinom42$ ways to choose which $2$ of these $4$ numbers are identical to the lottery. We have now determined $2$ of the $4$ numbers drawn in the lottery; since the other $2$ numbers Jen chose can not be chosen by the lottery, the lottery now has $10-2-2=6$ numbers to choose the last $2$ numbers from. Thus, this case is $\dbinom62$ , so this case yields $\dbinom42\dbinom62=6\cdot15=90$ possibilities.
Next, let us calculate the number of ways to draw exactly $3$ identical numbers to the lottery. Again, let Jen choose $a$ $b$ $c$ , and $d$ . This time, we have $\dbinom43$ ways to choose the identical numbers and again $6$ numbers left for the lottery to choose from; however, since $3$ of the lottery's numbers have already been determined, the lottery only needs to choose $1$ more number, so this is $\dbinom61$ . This case yields $\dbinom43\dbinom61=4\cdot6=24$
Finally, let us calculate the number of ways to all $4$ numbers matching. There is actually just one way for this to happen.
In total, we have $90+24+1=115$ ways to win a prize. The lottery has $\dbinom{10}4=210$ possible combinations to draw, so the probability of winning a prize is $\dfrac{115}{210}$ . There is actually no need to simplify it or even evaluate $\dbinom{10}4$ or actually even know that it has to be $\dbinom{10}4$ ; it suffices to call it $a$ or some other variable, as it will cancel out later. However, let us just go through with this. The probability of winning a prize is $\dfrac{115}{210}$ . Note that the probability of winning a grand prize is just matching all $4$ numbers, which we already calculated to have $1$ possibility and thus have probability $\dfrac1{210}$ . Thus, our answer is $\dfrac{\frac1{210}}{\frac{115}{210}}=\dfrac1{115}$ . Therefore, our answer is $1+115=\boxed{116}$

Solution:
For getting all $4$ right, there is only $1$ way.
For getting $3$ right, there is $\dbinom43$ multiplied by $\dbinom61$ $24$ ways.
For getting $2$ right, there is $\dbinom42$ multiplied by $\dbinom62$ $90$ ways.
$\frac{1}{1+24+90}$ $\frac{1}{115}$
Therefore, the answer is $1+115 = \boxed{116}$


Answer: 29.0

Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$ . Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$ . The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
$B-J \ne 0$ because $B \ne J$ , so the probability that $B-J < 0$ is $\frac{1}{2}$ by symmetry.
The probability that $B-J = 1$ is $\frac{19}{20 \times 19} = \frac{1}{20}$ because there are 19 pairs: $(B,J) = (2,1), \ldots, (20,19)$
The probability that $B-J \ge 2$ is $1-\frac{1}{2}-\frac{1}{20} = \frac{9}{20} \implies \boxed{029}$

Solution: By symmetry, the desired probability is equal to the probability that $J - B$ is at most $-2$ , which is $\frac{1-P}{2}$ where $P$ is the probability that $B$ and $J$ differ by $1$ (no zero, because the two numbers are distinct). There are $20 \cdot 19 = 380$ total possible combinations of $B$ and $J$ , and $1 + 18 \cdot 2 + 1 = 38$ ones that form $P$ , so $P = \frac{38}{380} = \frac{1}{10}$ . Therefore the answer is $\frac{9}{20} \rightarrow \boxed{029}$

Solution:
This problem is essentially asking how many ways there are to choose $2$ distinct elements from a $20$ element set such that no $2$ elements are adjacent. Using the well-known formula $\dbinom{n-k+1}{k}$ , there are $\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171$ ways. Dividing $171$ by $380$ , our desired probability is $\frac{171}{380} = \frac{9}{20}$ . Thus, our answer is $9+20=\boxed{029}$ .
-Fidgetboss_4000

Solution: Create a grid using graph paper, with $20$ columns for the values of $J$ from $1$ to $20$ and $20$ rows for the values of $B$ from $1$ to $20$ . Since $B$ cannot equal $J$ , we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since $B - J$ must be at least $2$ , we can mark the line where $B - J = 2$ . Now we sum the number of squares that are on this line and below it. We get $171$ . Then we find the number of total squares, which is $400 - 20 = 380$ . Finally, we take the ratio $\frac{171}{380}$ , which simplifies to $\frac{9}{20}$ . Our answer is $9+20=\boxed{029}$

Solution: We can see that if $B$ chooses $20$ $J$ can choose from $1$ through $18$ such that $B-J\geq 2$ . If $B$ chooses $19$ $J$ has choices $1$ $17$ . By continuing this pattern, $B$ will choose $3$ and $J$ will have $1$ option. Summing up the total, we get $18+17+\cdots+1$ as the total number of solutions. The total amount of choices is $20\times19$ (B and J must choose different numbers), so the probability is $\frac{18\cdot19\div2}{20\cdot19}=\frac{9}{20}$ . Therefore, the answer is $9+20=\boxed{029}$

Solution: Similar to solution 4, we can go through the possible values of $J$ to find all the values of $B$ that makes $B-J\geq 2$ . If $J$ chooses $1$ , then $B$ can choose anything from $3$ to $20$ . If $J$ chooses $2$ , then $B$ can choose anything from $4$ to $20$ . By continuing this pattern, we can see that there is $18+17+\cdots+1$ possible solutions. The amount of solutions is, therefore, $\frac{18\cdot19}{2}=171$ . Now, because $B$ and $J$ must be different, we have $20\times19=380$ possible choices, so the probability is $\frac{171}{380}=\frac{9}{20}$ . Therefore, the final answer is $9+20=\boxed{029}$


Answer: 163.0

Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let $t\,$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t\,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Solution: Consider the unit cicle of radius 50. Assume that they start at points $(-50,100)$ and $(-50,-100).$ Then at time $t$ , they end up at points $(-50+t,100)$ and $(-50+3t,-100).$ The equation of the line connecting these points and the equation of the circle are \begin{align}y&=-\frac{100}{t}x+200-\frac{5000}{t}\\50^2&=x^2+y^2\end{align}. When they see each other again, the line connecting the two points will be tangent to the circle at the point $(x,y).$ Since the radius is perpendicular to the tangent we get \[-\frac{x}{y}=-\frac{100}{t}\] or $xt=100y.$ Now substitute \[y= \frac{xt}{100}\] into $(2)$ and get \[x=\frac{5000}{\sqrt{100^2+t^2}}.\] Now substitute this and \[y=\frac{xt}{100}\] into $(1)$ and solve for $t$ to get \[t=\frac{160}{3}.\] Finally, the sum of the numerator and denominator is $160+3=\boxed{163}.$

Solution:
Let $A$ and $B$ be Kenny's initial and final points respectively and define $C$ and $D$ similarly for Jenny. Let $O$ be the center of the building. Also, let $X$ be the intersection of $AC$ and $BD$ . Finaly, let $P$ and $Q$ be the points of tangency of circle $O$ to $AC$ and $BD$ respectively.
From the problem statement, $AB=3t$ , and $CD=t$ . Since $\Delta ABX \sim \Delta CDX$ $CX=AC\cdot\left(\frac{CD}{AB-CD}\right)=200\cdot\left(\frac{t}{3t-t}\right)=100$
Since $PC=100$ $PX=200$ . So, $\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}$
Since circle $O$ is tangent to $BX$ and $AX$ $OX$ is the angle bisector of $\angle BXA$
Thus, $\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}$
Therefore, $t = CD = CX\cdot\tan(\angle BXA) = 100 \cdot \frac{8}{15} = \frac{160}{3}$ , and the answer is $\boxed{163}$

Solution:
Let $t$ be the time they walk.  Then $CD=t$ and $AB=3t$
Draw a line from point $O$ to $Q$ such that $OQ$ is perpendicular to $BD$ .  Further, draw a line passing through points $O$ and $P$ , so $OP$ is parallel to $AB$ and $CD$ and is midway between those two lines.  Then $PR=\dfrac{AB+CD}{2}=\dfrac{3t+t}{2}=2t$ .  Draw another line passing through point $D$ and parallel to $AC$ , and call the point of intersection of this line with $AB$ as $S$ .  Then $SB=AB-AS=3t-t=2t$
We see that $m\angle SBD=m\angle ORQ$ since they are corresponding angles, and thus by angle-angle similarity, $\triangle QOR\sim\triangle SDB$
Then \begin{align*} \dfrac{OQ}{DS}=\dfrac{RO}{BD}&\implies\dfrac{50}{200}=\dfrac{RO}{\sqrt{200^2+4t^2}}\\ &\implies RO=\dfrac{1}{4}\left(\sqrt{200^2+4t^2}\right)\\ &\implies RO=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right) \end{align*}
And we obtain \begin{align*} PR-OP&=RO\\ 2t-50&=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right)\\ 4t-100&=\sqrt{100^2+t^2}\\ (4t-100)^2&=\left(\sqrt{100^2+t^2}\right)^2\\ 16t^2-800t+100^2&=t^2+100^2\\ 15t^2&=800t\\ t&=\dfrac{800}{15} \end{align*}
so we have $t=\frac{160}{3}$ , and our answer is thus $160+3=\boxed{163}$


Answer: 9.0

Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$

Solution:
For starters, we identify d as distance and v as velocity (speed)
Writing the equation gives us: $\frac{d}{v}=\frac{1}{3}$ and $\frac{d}{v+18}=\frac{1}{5}$
This gives $d=\frac{1}{5}v+3.6=\frac{1}{3}v$ , which gives $v=27$ , which then gives $d=\boxed{9}$

Solution:
$d = rt$ $d=\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)$
$\frac{r}{3} = \frac{r}{5} + \frac{18}{5}$
$10r = 270$ so $r = 27$ , plug into the first one and it's $\boxed{9}$ miles to school.

Solution: We set up an equation in terms of $d$ the distance and $x$ the speed In miles per hour. We have $d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)$ , giving \[(5)(x)=(3)(x+18)\] \[5x=3x+54\] \[2x=54\] \[x=27\] Hence, $d=\dfrac{27}{3}=\boxed{9}$

Solution: Since it takes $\frac{3}{5}$ of the original time for him to get to school when there is no traffic, the speed must be $\frac{5}{3}$ of the speed in no traffic or $\frac{2}{3}$ more. Letting $x$ be the rate and we know that $\frac{5}{3}x = x + 18$ , so we have $\frac{2x}{3} = 18$ miles per hour. Solving for $x$ gives us $27$ miles per hour. Because $20$ minutes is a third of an hour, the distance would then be $9$ miles $\boxed{9}$

Solution: When driving in rush hour traffic, he drives $20$ minutes for one distance ( $1d$ ) to the school. It means he drives $60$ minutes for $3$ distances ( $3d$ ) to the school. When driving in no traffic hours, he drives $12$ minutes for one distance ( $1d$ ) to the school. It means he drives $60$ minutes for $5$ distances ( $5d$ ) to the school. Subtracting these two situations, it gives us $5d-3d = 18 = 2d$ , then $d=\frac{18}{2}=9$ . So the distance to the school would be $\boxed{9}$ miles. ----LarryFlora

Solution:
(Ratios) $\textbf{requires edits}$
In rush hour traffic, we can create this ratio for each distance we are going to try supposing the distance is $d$
$d$ $20$ minutes
In no traffic, we can do the same:
$d$ $12$ minutes
We want the ratio to be the distance to $60$ minutes:
$d$ $20$ minutes = $3$ d : $60$ minutes
$d$ $12$ minutes = $5$ d : $60$ minutes
This shows the speed in miles per hour. This means that the speed in the 2nd ratio has to be 18 faster than the first ratio. We get
$3d+18 = 5d \cdot d=9$ , or $\boxed{9}$


Answer: 30.0

Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?
$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 40\%\qquad\textbf{(C)}\ 50\%\qquad\textbf{(D)}\ 60\%\qquad\textbf{(E)}\ 70\%$

Solution: Let $j$ represent how far Jerry walked, and $s$ represent how far Silvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, $j = 2$ Since Silvia walked the diagonal, she walked the hypotenuse of a $45$ $45$ $90$ triangle with leg length $1$ . Thus, $s = \sqrt{2} = 1.414...$ We can then take $\frac{j-s}{j} \approx \frac{2 - 1.4}{2}=0.3 \implies \boxed{30}$


Answer: 30.0

Solution: Let $j$ represent how far Jerry walked, and $s$ represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, $j = 2$ Since Silvia walked the diagonal, she walked the hypotenuse of a $45$ $45$ $90$ triangle with leg length $1$ . Thus, $s = \sqrt{2} = 1.414...$ We can then take $\frac{j-s}{j} \approx \frac{2 - 1.4}{2} = 0.3 = 30\%$ $\boxed{30}$


Answer: 151.0

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: 151.0

Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$ (For example, he succeeds if his sequence of tosses is $HTHHHHHH.$
$\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313$

Solution:
For $6$ to $8$ heads, we are guaranteed to hit $4$ heads, so the sum here is $\binom{8}{2}+\binom{8}{1}+\binom{8}{0}=28+8+1=37$
For $4$ heads, you have to hit the $4$ heads at the start so there's only one way, $1$
For $5$ heads, we either start off with $4$ heads, which gives us $_4\text{C}_1=4$ ways to arrange the other flips, or we start off with five heads and one tail, which has $6$ ways minus the $2$ overlapping cases, $\text{HHHHHTTT}$ and $\text{HHHHTHTT}$ . Total ways: $8$
Then we sum to get $46$ . There are a total of $2^8=256$ possible sequences of $8$ coin flips, so the probability is $\frac{46}{256}=\frac{23}{128}$ . Summing, we get $23+128=\boxed{151}$

Solution:
Reaching 4 will require either 4, 6, or 8 flips. Therefore we can split into 3 cases:
(Case 1): The first four flips are heads. Then, the last four flips can be anything so $2^4=16$ possibilities work.
(Case 2): It takes 6 flips to reach 4. There must be one tail in the first four flips so we don't repeat case 1. The tail can be in one of 4 positions. The next two flips must be heads. The last two flips can be anything so $2^2=4$ flips work. $4*4=16$
(Case 3): It takes 8 flips to reach 4. We can split this case into 2 sub-cases. There can either be 1 or 2 tails in the first 4 flips.
(1 tail in first four flips). In this case, the first tail can be in 4 positions. The second tail can be in either the 5th or 6th position so we don't repeat case 2. Thus, there are $4*2=8$ possibilities.
(2 tails in first four flips). In this case, the tails can be in $\binom{4}{2}=6$ positions.
Adding these cases up and taking the total out of $2^8=256$ yields $\frac{16+16+8+6}{256}=\frac{46}{256}=\frac{23}{128}$ . This means the answer is $23+128=\boxed{151}$

Solution: Notice every $2$ flips, there is a $\dfrac{1}{4}$ chance to go $2$ steps left $(L),$ $\dfrac{1}{2}$ chance to stay put $(P),$ and $\dfrac{1}{4}$ chance to move $2$ steps right $(R).$ We have $4$ choices for how to get to $4:$ \[RR-1\text{ arrangement}-\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{16}\] \[PRR-2\text{ arrangements}-2\cdot\dfrac{1}{2}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{16}\] \[LRRR-2\text{ arrangements}-2\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{128}\] \[PPRR-3\text{ arrangements}-3\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{3}{64}\] Finally, our answer will be $\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{128}+\dfrac{3}{64}=\dfrac{23}{128}\implies a+b=\boxed{151}.$


Answer: 3.0

Jim starts with a positive integer $n$ and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with $n = 55$ , then his sequence contains $5$ numbers:
\[\begin{array}{ccccc} {}&{}&{}&{}&55\\ 55&-&7^2&=&6\\ 6&-&2^2&=&2\\ 2&-&1^2&=&1\\ 1&-&1^2&=&0\\ \end{array}\]
Let $N$ be the smallest number for which Jim’s sequence has $8$ numbers. What is the units digit of $N$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 3 \qquad \mathrm{(C)}\ 5 \qquad \mathrm{(D)}\ 7 \qquad \mathrm{(E)}\ 9$

Solution:
We can find the answer by working backwards. We begin with $1-1^2=0$ on the bottom row, then the $1$ goes to the right of the equal's sign in the row above. We find the smallest value $x$ for which $x-1^2=1$ and $x>1^2$ , which is $x=2$
We repeat the same procedure except with $x-1^2=1$ for the next row and $x-1^2=2$ for the row after that. However, at the fourth row, we see that solving $x-1^2=3$ yields $x=4$ , in which case it would be incorrect since $1^2=1$ is not the greatest perfect square less than or equal to $x$ . So we make it a $2^2$ and solve $x-2^2=3$ . We continue on using this same method where we increase the perfect square until $x$ can be made bigger than it. When we repeat this until we have $8$ rows, we get:
\[\begin{array}{ccccc}{}&{}&{}&{}&7223\\ 7223&-&84^{2}&=&167\\ 167&-&12^{2}&=&23\\ 23&-&4^{2}&=&7\\ 7&-&2^{2}&=&3\\ 3&-&1^{2}&=&2\\ 2&-&1^{2}&=&1\\ 1&-&1^{2}&=&0\\ \end{array}\]
Hence the solution is the last digit of $7223$ , which is $\boxed{3}$

Solution:
Notice that to get the previous term, we must add the smallest square number, (let's call it $n^2$ ) such that the sum is less than $(n+1)^2$ . Otherwise, instead of subtracting $n^2$ from the previous term, we're subtracting a greater square number.
Remember that $(x+1)^2 = x^2 + 2x + 1$ . Recall that to find the previous term, we must add a square number such that it is less than the next square number. $a + n^2 < (n+1)^2$ . For this to be true, $a < 2n + 1$ . What that means is that given a term $a$ , we can find the previous term by adding $n^2$ where $n > \frac {a-1}{2}$
For example, to find the term that precedes $167$ , we know that $n>166/2 = 83$ . Therefore, $n=84$ and the previous term is $167 + 84^2 = 7223$ . The last digit of $7223$ is $3 \Rightarrow \boxed{3}$


Answer: 8.0

Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$

Solution: We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc.  Thus, after nine "turns", $1+2+3+4+5+6+7+8+9=45$ numbers have been said.  In the tenth turn, the eighth number will be the 53rd number said, because $53-45=8$ .  Since we are starting from 1 every turn, the 53rd number said will be $\boxed{8}$

Solution: We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc.  We notice that the number of numbers is $1 + 2 + 3 + 4 ...$ every time we finish a "turn" we notice the sum of these would be the largest number $\frac{n(n+1)}{2}$ under 53, we can easily see that if we double this it's $n^2 + n \simeq 106$ , and we immediately note that 10 is too high, but 9 is perfect, meaning that at 9, 45 numbers have been said so far, $\frac{9(9+1)}{2} = 45$ and $53 - 45 = \boxed{8}$

Solution: Let $T(n)$ denote the $n$ th triangle number. Then, observe that the $T(n)$ th number said is $n$ . It follows that the $55$ th number is $10$ (as $55 = T(10)$ ). Thus, the $53$ rd number is $10 - 2 = 8$ , which is answer choice $\boxed{8}$


Answer: 8.0

Solution: We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc.  Thus, after nine "turns", $1+2+3+4+5+6+7+8+9=45$ numbers have been said.  In the tenth turn, the eighth number will be the 53rd number said, because $53-45=8$ .  Since we are starting from 1 every turn, the 53rd number said will be $\boxed{8}$

Solution: We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc.  We notice that the number of numbers is $1 + 2 + 3 + 4 ...$ every time we finish a "turn" we notice the sum of these would be the largest number $\frac{n(n+1)}{2}$ under 53, we can easily see that if we double this it's $n^2 + n \simeq 106$ , and we immediately note that 10 is too high, but 9 is perfect, meaning that at 9, 45 numbers have been said so far, $\frac{9(9+1)}{2} = 45$ and $53 - 45 = \boxed{8}$

Solution: Let $T(n)$ denote the $n$ th triangle number. Then, observe that the $T(n)$ th number said is $n$ . It follows that the $55$ th number is $10$ (as $55 = T(10)$ ). Thus, the $53$ rd number is $10 - 2 = 8$ , which is answer choice $\boxed{8}$


Answer: 8.0

Joe had walked half way from home to school when he realized he was late. He ran the rest of the way to school. He ran 3 times as fast as he walked. Joe took 6 minutes to walk half way to school. How many minutes did it take Joe to get from home to school?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 7.3\qquad\textbf{(C)}\ 7.7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 8.3$

Solution: We can use the equation $d=rt$ where $d$ is the distance, $r$ is the rate, and $t$ is the time. The distances he ran and walked are equal, so $r_rt_r=r_wt_w$ , where $r_r$ is the rate at which he ran, $t_r$ is the time for which he ran, $r_w$ is the rate at which he walked, and $t_w$ is the time for which he walked. Because he runs three times faster than he walks, $r_r=3r_w$ . We want to find the time he ran, $t_r = \frac{r_wt_w}{t_r} = \frac{(r_w)(6)}{3r_w} = 2$ minutes. He traveled for a total of $6+2=\boxed{8}$ minutes.


Answer: 2.0

Joe has a collection of $23$ coins, consisting of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins. He has $3$ more $10$ -cent coins than $5$ -cent coins, and the total value of his collection is $320$ cents. How many more $25$ -cent coins does Joe have than $5$ -cent coins?
$\textbf{(A) }   0   \qquad        \textbf{(B) }   1   \qquad    \textbf{(C) }   2   \qquad   \textbf{(D) }  3  \qquad  \textbf{(E) }   4$

Solution: Let $x$ be the number of $5$ -cent coins that Joe has. Therefore, he must have $(x+3) \ 10$ -cent coins and $(23-(x+3)-x) \ 25$ -cent coins. Since the total value of his collection is $320$ cents, we can write \begin{align*} 5x + 10(x+3) + 25(23-(x+3)-x) &= 320 \\ 5x + 10x + 30 + 500 - 50x &= 320 \\ 35x &= 210 \\ x &= 6. \end{align*} Joe has six $5$ -cent coins, nine $10$ -cent coins, and eight $25$ -cent coins. Thus, our answer is $8-6 = \boxed{2}.$

Solution:
Let the number of $5$ -cent coins be $x,$ the number of $10$ -cent coins be $x+3,$ and the number of $25$ -cent coins be $y.$
Set up the following two equations with the information given in the problem: \[5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290\] \[x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20\]
From there, multiply the second equation by $25$ to get \[50x+25y=500.\]
Subtract the first equation from the multiplied second equation to get $35x=210,$ or $x=6.$
Substitute $6$ in for $x$ into one of the equations to get $y=8.$
Finally, the answer is $8-6=\boxed{2}.$

Solution:
Let $n,d,$ and $q$ be the numbers of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins in Joe's collection, respectively. We are given that \begin{align*} n+d+q&=23, &(1) \\ 5n+10d+25q&=320, &(2) \\ d&=n+3. &(3) \end{align*} Substituting $(3)$ into each of $(1)$ and $(2)$ and then simplifying, we have \begin{align*} 2n+q&=20, \hspace{17.5mm} &(1\star) \\ 3n+5q&=58. &(2\star) \end{align*} Subtracting $(2\star)$ from $5\cdot(1\star)$ gives $7n=42,$ from which $n=6.$ Substituting this into either $(1\star)$ or $(2\star)$ produces $q=8.$
Finally, the answer is $q-n=\boxed{2}.$


Answer: 11.0

Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$

Solution:
Suppose that Chloe is $c$ years old today, so Joey is $c+1$ years old today. After $n$ years, Chloe and Zoe will be $n+c$ and $n+1$ years old, respectively. We are given that \[\frac{n+c}{n+1}=1+\frac{c-1}{n+1}\] is an integer for $9$ nonnegative integers $n.$ It follows that $c-1$ has $9$ positive divisors. The prime factorization of $c-1$ is either $p^8$ or $p^2q^2.$ Since $c-1<100,$ the only possibility is $c-1=2^2\cdot3^2=36,$ from which $c=37.$ We conclude that Joey is $c+1=38$ years old today.
Suppose that Joey's age is a multiple of Zoe's age after $k$ years, in which Joey and Zoe will be $k+38$ and $k+1$ years old, respectively. We are given that \[\frac{k+38}{k+1}=1+\frac{37}{k+1}\] is an integer for some positive integer $k.$ It follows that $37$ is divisible by $k+1,$ so the only possibility is $k=36.$ We conclude that Joey will be $k+38=74$ years old then, from which the answer is $7+4=\boxed{11}.$

Solution:
Let Joey's age be $j$ , Chloe's age be $c$ , and we know that Zoe's age is $1$
We know that there must be $9$ values $k\in\mathbb{Z}$ such that $c+k=a(1+k)$ where $a$ is an integer.
Therefore, $c-1+(1+k)=a(1+k)$ and $c-1=(1+k)(a-1)$ . Therefore, we know that, as there are $9$ solutions for $k$ , there must be $9$ solutions for $c-1$ . We know that this must be a perfect square. Testing perfect squares, we see that $c-1=36$ , so $c=37$ . Therefore, $j=38$ . Now, since $j-1=37$ , by similar logic, $37=(1+k)(a-1)$ , so $k=36$ and Joey will be $38+36=74$ and the sum of the digits is $\boxed{11}$

Solution:
Here's a different way of stating Solution 2:
If a number is a multiple of both Chloe's age and Zoe's age, then it is a multiple of their difference. Since the difference between their ages does not change, then that means the difference between their ages has $9$ factors. Therefore, the difference between Chloe and Zoe's age is $36$ , so Chloe is $37$ , and Joey is $38$ . The common factor that will divide both of their ages is $37$ , so Joey will be $74$ . The answer is $7 + 4 =  \boxed{11}$

Solution:
Similar approach to above, just explained less concisely and more in terms of the problem (less algebraic).
Let $C+n$ denote Chloe's age, $J+n$ denote Joey's age, and $Z+n$ denote Zoe's age, where $n$ is the number of years from now. We are told that $C+n$ is a multiple of $Z+n$ exactly nine times. Because $Z+n$ is $1$ at $n=0$ and will increase until greater than $C-Z$ , it will hit every natural number less than $C-Z$ , including every factor of $C-Z$ . For $C+n$ to be an integral multiple of $Z+n$ , the difference $C-Z$ must also be a multiple of $Z$ , which happens if $Z$ is a factor of $C-Z$ . Therefore, $C-Z$ has nine factors. The smallest number that has nine positive factors is $2^23^2=36$ . (We want it to be small so that Joey will not have reached three digits of age before his age is a multiple of Zoe's.) We also know $Z=1$ and $J=C+1$ . Thus, \begin{align*} C-Z&=36, \\ J-Z&=37. \end{align*} By our above logic, the next time $J-Z$ is a multiple of $Z+n$ will occur when $Z+n$ is a factor of $J-Z$ . Because $37$ is prime, the next time this happens is at $Z+n=37$ , when $J+n=74$ . The answer is $7+4=\boxed{11}$


Answer: 11.0

Solution:
Suppose that Chloe is $c$ years old today, so Joey is $c+1$ years old today. After $n$ years, Chloe and Zoe will be $n+c$ and $n+1$ years old, respectively. We are given that \[\frac{n+c}{n+1}=1+\frac{c-1}{n+1}\] is an integer for $9$ nonnegative integers $n.$ It follows that $c-1$ has $9$ positive divisors. The prime factorization of $c-1$ is either $p^8$ or $p^2q^2.$ Since $c-1<100,$ the only possibility is $c-1=2^2\cdot3^2=36,$ from which $c=37.$ We conclude that Joey is $c+1=38$ years old today.
Suppose that Joey's age is a multiple of Zoe's age after $k$ years, in which Joey and Zoe will be $k+38$ and $k+1$ years old, respectively. We are given that \[\frac{k+38}{k+1}=1+\frac{37}{k+1}\] is an integer for some positive integer $k.$ It follows that $37$ is divisible by $k+1,$ so the only possibility is $k=36.$ We conclude that Joey will be $k+38=74$ years old then, from which the answer is $7+4=\boxed{11}.$

Solution:
Let Joey's age be $j$ , Chloe's age be $c$ , and we know that Zoe's age is $1$
We know that there must be $9$ values $k\in\mathbb{Z}$ such that $c+k=a(1+k)$ where $a$ is an integer.
Therefore, $c-1+(1+k)=a(1+k)$ and $c-1=(1+k)(a-1)$ . Therefore, we know that, as there are $9$ solutions for $k$ , there must be $9$ solutions for $c-1$ . We know that this must be a perfect square. Testing perfect squares, we see that $c-1=36$ , so $c=37$ . Therefore, $j=38$ . Now, since $j-1=37$ , by similar logic, $37=(1+k)(a-1)$ , so $k=36$ and Joey will be $38+36=74$ and the sum of the digits is $\boxed{11}$

Solution:
Here's a different way of stating Solution 2:
If a number is a multiple of both Chloe's age and Zoe's age, then it is a multiple of their difference. Since the difference between their ages does not change, then that means the difference between their ages has $9$ factors. Therefore, the difference between Chloe and Zoe's age is $36$ , so Chloe is $37$ , and Joey is $38$ . The common factor that will divide both of their ages is $37$ , so Joey will be $74$ . The answer is $7 + 4 =  \boxed{11}$

Solution:
Similar approach to above, just explained less concisely and more in terms of the problem (less algebraic).
Let $C+n$ denote Chloe's age, $J+n$ denote Joey's age, and $Z+n$ denote Zoe's age, where $n$ is the number of years from now. We are told that $C+n$ is a multiple of $Z+n$ exactly nine times. Because $Z+n$ is $1$ at $n=0$ and will increase until greater than $C-Z$ , it will hit every natural number less than $C-Z$ , including every factor of $C-Z$ . For $C+n$ to be an integral multiple of $Z+n$ , the difference $C-Z$ must also be a multiple of $Z$ , which happens if $Z$ is a factor of $C-Z$ . Therefore, $C-Z$ has nine factors. The smallest number that has nine positive factors is $2^23^2=36$ . (We want it to be small so that Joey will not have reached three digits of age before his age is a multiple of Zoe's.) We also know $Z=1$ and $J=C+1$ . Thus, \begin{align*} C-Z&=36, \\ J-Z&=37. \end{align*} By our above logic, the next time $J-Z$ is a multiple of $Z+n$ will occur when $Z+n$ is a factor of $J-Z$ . Because $37$ is prime, the next time this happens is at $Z+n=37$ , when $J+n=74$ . The answer is $7+4=\boxed{11}$


Answer: 11.0

Joey and his five brothers are ages $3$ $5$ $7$ $9$ $11$ , and $13$ .  One afternoon two of his brothers whose ages sum to $16$ went to the movies, two brothers younger than $10$ went to play baseball, and Joey and the $5$ -year-old stayed home.  How old is Joey?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7  \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$

Solution: Because the $5$ -year-old stayed home, we know that the $11$ -year-old did not go to the movies, as the $5$ -year-old did not and $11+5=16$ .  Also, the $11$ -year-old could not have gone to play baseball, as he is older than $10$ .  Thus, the $11$ -year-old must have stayed home, so Joey is $\boxed{11}$

Solution: There are only $4$ kids who are under $10$ but since the $5$ -year old stayed home, the only possible ages who went to play baseball are the brothers who are $3,7,9$ , either $13+3$ or $7+9$ is $16$ but since we need $2$ kids to go to baseball who are under $10$ $13,3$ must have been the pair to go to the movies and $9,7$ must have went to baseball, so only the $11$ -year old is left, which is answer choice $\boxed{11}$


Answer: 56.0

Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\textbf{(E) } 64$

Solution:
We can simplify the problem first, then apply reasoning to the original problem. Let's say that there are $8$ coins. Shaded coins flip heads, and blank coins flip tails. So, after the first flip;

Then, after the second (new heads in blue);

And after the third (new head in green);

So in total, $7$ of the $8$ coins resulted in heads. Now we have the ratio of $\frac{7}{8}$ of the total coins will end up heads. Therefore, we have $\frac{7}{8}\cdot64=\boxed{56}$

Solution: Every time the coins are flipped, half of them are expected to turn up heads. The expected number of heads on the first flip is $32$ , on the second flip is $16$ , and on the third flip, it is $8$ . Adding these gives $\boxed{56}$

Solution: Every time the coins are flipped, each of them has a $1/2$ probability of being tails. Doing this $3$ times, $1/8$ of them will be tails. $64-64*1/8=$ $\boxed{56}$


Answer: 56.0

Solution:
(Similar to solution 2)
Notice how:
$E(\text{Heads on 1st flip, 2nd flip, 3rd flip}) = E(\text{Heads on 1st flip}) + E(\text{Heads on 2nd flip}) + E(\text{Heads on 3rd flip})$
The expected number of heads for the first flip is simply $64 \cdot \frac{1}{2}$ , since each coin has a 1 in 2 chance of being heads. Then, we are left with $64 - 32 = 32$ coins. Then, half of these coins will be heads again, which leaves us with $32 - 16 = 16$ coins. Then, half of these coins will be heads again, which leaves us with $16 - 8 = 8$ coins.
Hence, the expected number of heads is simply:
$32 + 16 + 8 = \boxed{56} \implies D$


Answer: 49.0

Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\tfrac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
For the purposes of this problem, we will use miles and minutes as our units; thus, the bikers travel at speeds of $\dfrac{1}{3}$ mi/min.
Let $d$ be the length of the trains, $r_1$ be the speed of train 1 (the faster train), and $r_2$ be the speed of train 2.
Consider the problem from the bikers' moving frame of reference. In order to pass Jon, the first train has to cover a distance equal to its own length, at a rate of $r_1 - \dfrac{1}{3}$ . Similarly, the second train has to cover a distance equal to its own length, at a rate of $r_2 + \dfrac{1}{3}$ . Since the times are equal and $d = rt$ , we have that $\dfrac{d}{r_1 - \dfrac{1}{3}} = \dfrac{d}{r_2 + \dfrac{1}{3}}$ . Solving for $r_1$ in terms of $r_2$ , we get that $r_1 = r_2 + \dfrac{2}{3}$
Now, let's examine the times it takes the trains to pass Steve. This time, we augment train 1's speed by $\dfrac{1}{3}$ , and decrease train 2's speed by $\dfrac{1}{3}$ . Thus, we have that $\dfrac{d}{r_2 - \dfrac{1}{3}} = 10\dfrac{d}{r_1 + \dfrac{1}{3}}$
Multiplying this out and simplifying, we get that $r_1 = 10r_2 - \dfrac{11}{3}$ . Since we now have 2 expressions for $r_1$ in terms of $r_2$ , we can set them equal to each other:
$r_2 + \dfrac{2}{3} = 10r_2 - \dfrac{11}{3}$ . Solving for $r_2$ , we get that $r_2 = \dfrac{13}{27}$ . Since we know that it took train 2 1 minute to pass Jon, we know that $1 = \dfrac{d}{r_2 + \dfrac{1}{3}}$ . Plugging in $\dfrac{13}{27}$ for $r_2$ and solving for $d$ , we get that $d = \dfrac{22}{27}$ , and our answer is $27 + 22 = \boxed{049}$

Solution:
Using a similar approach to Solution 1, let the speed of the east bound train be $a$ and the speed of the west bound train be $b$
So $a-20=b+20$ and $a+20=10(b-20)$
From the first equation, $a=b+40$ . Substituting into the second equation, \[b+60=10b-200\] \[260=9b\] \[b=\frac{260}{9}\text{ mph}\] This means that \[a=\frac{260}{9}+40=\frac{620}{9}\text{ mph}\] Checking, we get that the common difference in Jon's speed and trains' speeds is $\frac{440}{9}$ and the differences for Steve is $\frac{800}{9}$ and $\frac{80}{9}$
This question assumes the trains' lengths in MILES: \[\frac{440}{9}\cdot \frac{1}{60}=\frac{440}{540}=\frac{22}{27}\text{ miles}\] Adding up, we get $22+27=\boxed{049}$

Solution:
Let the length of the trains be $L$ , let the rate of the westward train be $W_{R}$ , ;et the rate of the eastward train be $E_{R}$ , and let the time it takes for the eastward train to pass Steve be $E_{T}$
We have that
$L=(\frac{1}{60})(W_{R}+20)$
$L=(\frac{1}{60})(E_{R}-20)$
Adding both of the equations together, we get that
$2L=\frac{W_{R}}{60}+\frac{E_{R}}{60}\implies 120L=W_{R}+E_{R}$
Now, from the second part of the problem, we acquire that
$L=(E_{T})(E_{R}+20)$
$L=(10E_{T})(W_{R}-20)$
Dividing the second equation by the first, we get that... $1=\frac{10(W_{R}-20)}{E_{R}+20}\implies E_{R}+20=10W_{R}-200\implies E_{R}+220=10W_{R}\implies E_{R}=10W_{R}-220$
Now, substituting into the $120L=W_{R}+E_{R}$
$120L=W_{R}+(10W_{R}-220)\implies 120L= 11W_{R}-220\implies W_{R}=\frac{120L+220}{11}$
Finally, plugging this back into our very first equation..
$L=(\frac{1}{60})((\frac{120L+220}{11})+20)\implies 660L=120L+440\implies 540L=440\implies L=\frac{22}{27}$
Hence, the answer is $22+27=\boxed{049}$


Answer: 95.0

Josanna's test scores to date are $90, 80, 70, 60,$ and $85$ . Her goal is to raise here test average at least $3$ points with her next test. What is the minimum test score she would need to accomplish this goal?
$\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 82 \qquad\textbf{(C)}\ 85 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 95$

Solution: The average of her current scores is $77$ . To raise it $3$ points, she needs an average of $80$ , and so after her $6$ tests, a sum of $480$ . Her current sum is $385$ , so she needs a $480 - 385 = \boxed{95}$


Answer: 95.0

Josanna's test scores to date are $90, 80, 70, 60,$ and $85.$ Her goal is to raise her test average at least $3$ points with her next test. What is the minimum test score she would need to accomplish this goal?
$\textbf{(A)}\ 80 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 85 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 95$

Solution:
Take the average of her current test scores, which is \[\frac{90+80+70+60+85}{5} = \frac{385}{5} = 77\]
This means that she wants her test average after the sixth test to be $80.$ Let $x$ be the score that Josanna receives on her sixth test. Thus, our equation is
\[\frac{90+80+70+60+85+x}{6} = 80\]
\[385+x = 480\]
\[x = \boxed{95}\]


Answer: 14.0

Jose is $4$ years younger than Zack.  Zack is $3$ years older than Inez.  Inez is $15$ years old.  How old is Jose?
$\text{(A)}\ 8 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 14 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 22$

Solution:
Zack is $3$ years older than Inez, who is $15$ .  Therefore, Zack is $15 + 3 = 18$ years old.
Jose is $4$ years younger than Zack, who is $18$ .  Therefore, Jose is $18 - 4 = 14$ years old, and the answer is $\boxed{14}$


Answer: 5.0

Josh and Mike live $13$ miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?
$\textbf{(A) } 4\qquad \textbf{(B) } 5\qquad \textbf{(C) } 6\qquad \textbf{(D) } 7\qquad \textbf{(E) } 8$

Solution:
Let $m$ be the distance in miles that Mike rode.
Since Josh rode for twice the length of time as Mike and at four-fifths of Mike's rate, he rode $2\cdot\frac{4}{5}\cdot m = \frac{8}{5}m$ miles.
Since their combined distance was $13$ miles,
$\frac{8}{5}m + m = 13$
$\frac{13}{5}m = 13$
$m = \boxed{5}$


Answer: 64.0

Josh writes the numbers $1,2,3,\dots,99,100$ . He marks out $1$ , skips the next number $(2)$ , marks out $3$ , and continues skipping and marking out the next number to the end of the list. Then he goes back to the start of his list, marks out the first remaining number $(2)$ , skips the next number $(4)$ , marks out $6$ , skips $8$ , marks out $10$ , and so on to the end. Josh continues in this manner until only one number remains. What is that number?
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 56 \qquad \textbf{(D)}\ 64 \qquad \textbf{(E)}\ 96$

Solution:
Following the pattern, you are crossing out...
Time 1: Every non-multiple of $2$
Time 2: Every non-multiple of $4$
Time 3: Every non-multiple of $8$
Following this pattern, you are left with every multiple of $64$ which is only $\boxed{64}$


Answer: 17.0

Joy has $30$ thin rods, one each of every integer length from $1 \text{ cm}$ through $30 \text{ cm}$ . She places the rods with lengths $3 \text{ cm}$ $7 \text{ cm}$ , and $15 \text{cm}$ on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 17 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 19  \qquad\textbf{(E)}\ 20$

Solution: The quadrilateral cannot be a straight line. Thus, the fourth side must be longer than $15 - (3 + 7) = 5$ and shorter than $15 + 3 + 7 = 25$ . This means Joy can use the $19$ possible integer rod lengths that fall into $[6, 24]$ . However, she has already used the rods of length $7$ cm and $15$ cm so the answer is $19 - 2 = 17$ $\boxed{17}$


Answer: 17.0

Joy has $30$ thin rods, one each of every integer length from $1$ cm through $30$ cm. She places the rods with lengths $3$ cm, $7$ cm, and $15$ cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20$

Solution: The triangle inequality generalizes to all polygons, so $x < 3+7+15$ and $15<x+3+7$ yields $5<x<25$ . Now, we know that there are $19$ numbers between $5$ and $25$ exclusive, but we must subtract $2$ to account for the 2 lengths already used that are between those numbers, which gives $19-2=\boxed{17}$


Answer: 8.0

Joyce made $12$ of her first $30$ shots in the first three games of this basketball game, so her seasonal shooting average was $40\%$ .  In her next game, she took $10$ shots and raised her seasonal shooting average to $50\%$ .  How many of these $10$ shots did she make?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution:
After the fourth game, she took $40$ shots, $50\%$ of which she made, so she made $40\times .5=20$ shots.  Twelve of them were made in the first three games, so in the last game she made $20-12=8$ shots.
$\boxed{8}$


Answer: 5650.0

Julie is preparing a speech for her class.  Her speech must last between one-half hour and three-quarters of an hour.  The ideal rate of speech is 150 words per minute.  If Julie speaks at the ideal rate, which of the following number of words would be an appropriate length for her speech?
$\text{(A)}\ 2250 \qquad \text{(B)}\ 3000 \qquad \text{(C)}\ 4200 \qquad \text{(D)}\ 4350 \qquad \text{(E)}\ 5650$

Solution: Since there are $60$ minutes in an hour, we need a number of words between $(\frac{1}{2}\times 60)\times150=4500$ and $(\frac{3}{4}\times60)\times150=6750$ .  The only such choice is $\boxed{5650}$


Answer: 2.5

Karl bought five folders from Pay-A-Lot at a cost of $\textdollar 2.50$ each.
Pay-A-Lot had a 20%-off sale the following day. How much could
Karl have saved on the purchase by waiting a day?
$\textbf{(A)}\ \textdollar 1.00  \qquad\textbf{(B)}\ \textdollar 2.00 \qquad\textbf{(C)}\ \textdollar 2.50\qquad\textbf{(D)}\ \textdollar 2.75 \qquad\textbf{(E)}\ \textdollar 5.00$

Solution: Karl paid $5 \cdot 2.50 = \textdollar 12.50$ $20 \%$ of this cost that he saved is $12.50 \cdot .2 = \boxed{2.50}$

Solution: Each folder can also be $5/2$ dollars, and $20\%$ can be shown as $(1/5)$ . We can multiply $(5/2) \cdot (1/5) = (1/2)$ $(1/2)$ is also $50$ cents or the amount of money that is saved after the $20\%$ discount. So each folder is $2.50-0.5 = \textdollar2$ .Since Karl bought 5 folders all of the folders after the discount is $(5)(2) = 10$ , and the money bought before the discount is $(5)(2.50) = \textdollar12.50$ . To find the money Karl saves all we have to do is subtract $12.50 - 10 = 2.50$ . Thus the answer is $\boxed{2.50}$

Solution: Having a $20\%$ off sale is equivalent to a "Buy 4 get 1 free" sale. Thus, he would have saved the price of 1 folder, which is $\boxed{2.50}$


Answer: 525.0

Karl's car uses a gallon of gas every $35$ miles, and his gas tank holds $14$ gallons when it is full. One day, Karl started with a full tank of gas, drove $350$ miles, bought $8$ gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day?
$\textbf{(A) }525\qquad\textbf{(B) }560\qquad\textbf{(C) }595\qquad\textbf{(D) }665\qquad \textbf{(E) }735$

Solution: Since he uses a gallon of gas every $35$ miles, he had used $\frac{350}{35} = 10$ gallons after $350$ miles. Therefore, after the first leg of his trip he had $14 - 10 = 4$ gallons of gas left. Then, he bought $8$ gallons of gas, which brought him up to $12$ gallons of gas in his gas tank. When he arrived, he had $\frac{1}{2} \cdot 14 = 7$ gallons of gas. So he used $5$ gallons of gas on the second leg of his trip. Therefore,                                               the second part of his trip covered $5 \cdot 35 = 175$ miles. Adding this to the $350$ miles, we see that he drove $350 + 175 = \boxed{525}$ miles.


Answer: 100.0

Karl's rectangular vegetable garden is $20$ feet by $45$ feet, and Makenna's is $25$ feet by $40$ feet. Whose garden is larger in area?
$\textbf{(A)}\ \text{Karl's garden is larger by 100 square feet.}$
$\textbf{(B)}\ \text{Karl's garden is larger by 25 square feet.}$
$\textbf{(C)}\ \text{The gardens are the same size.}$
$\textbf{(D)}\ \text{Makenna's garden is larger by 25 square feet.}$
$\textbf{(E)}\ \text{Makenna's garden is larger by 100 square feet.}$

Solution: The area of a rectangle is given by the formula length times width. Karl's garden is $20 \times 45 = 900$ square feet and Makenna's garden is $25 \times 40 = 1000$ square feet. Since $1000 > 900,$ Makenna's garden is larger by $1000-900=100$ square feet. $\Rightarrow \boxed{100}$


Answer: 90.0

Kate bakes a $20$ -inch by $18$ -inch pan of cornbread. The cornbread is cut into pieces that measure $2$ inches by $2$ inches. How many pieces of cornbread does the pan contain?
$\textbf{(A) } 90 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 180 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 360$

Solution: The area of the pan is $20\cdot18=360$ . Since the area of each piece is $2\cdot2=4$ , there are $\frac{360}{4} = \boxed{90}$ pieces.

Solution: By dividing each of the dimensions by $2$ , we get a $10\times9$ grid that makes $\boxed{90}$ pieces.


Answer: 90.0

Solution: The area of the pan is $20\cdot18=360$ . Since the area of each piece is $2\cdot2=4$ , there are $\frac{360}{4} = \boxed{90}$ pieces.

Solution: By dividing each of the dimensions by $2$ , we get a $10\times9$ grid that makes $\boxed{90}$ pieces.


Answer: 7.0

Kate rode her bicycle for 30 minutes at a speed of 16 mph, then walked for 90 minutes at a speed of 4 mph. What was her overall average speed in miles per hour?
$\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 9\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$

Solution: \[16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14\] \[\frac{14}2=7\Rightarrow\boxed{7}\]


Answer: 157.0

Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
We have $2+4\cdot 2$ cases total.
The two are all red and all green. Then, you have 4 of one, 1 of other. 3 of one, 2 of other. 2 of one, 3 of other. 1 of one, 4 of other. Then flip the order, so times two.
Obviously the denominator is $10\cdot 9\cdot 8\cdot 7\cdot 6$ , since we are choosing a card without replacement.
Then, we have for the numerator for the two of all red and green: \[5\cdot 4\cdot 3\cdot 2\cdot 1.\]
For the 4 and 1, we have: \[5\cdot 4\cdot 3\cdot 2\cdot 5.\]
For the 3 and 2, we have: \[5\cdot 4\cdot 3\cdot 5\cdot 4.\]
For the 2 and 3, we have: \[5\cdot 4\cdot 5\cdot 4\cdot 3.\]
For the 1 and 4, we have: \[5\cdot 5\cdot 4\cdot 3\cdot 2.\]
Adding up and remembering to double all of them, since they can be reversed and the 5's can be red or green, we get, after simplifying: $\dfrac{31}{126}$
Thus the answer is $31 + 126 = \boxed{157}$ .
-gorefeebuddie

Solution:
Our probability will be $\dfrac{\text{number of "happy" configurations of cards}}{\text{total number of configurations of cards}}.$
First of all, we have $10$ choices for the first card, $9$ choices for the second card, $8$ choices for the third card, $7$ choices for the fourth card, and $6$ choices for the last card. This gives a total of $10\cdot 9\cdot 8\cdot 7\cdot 6$ possible ways for five cards to be chosen.
Finding the number of configurations that make Kathy happy is a more difficult task, however, and we will resort to casework to do it.
First, let's look at the appearances of the "happy configurations" that Kathy likes. Based on the premise of the problem, we can realize that there are ten cases for the appearance of the configurations: \[\text{RRRRR},\] \[\text{GGGGG},\] \[\text{RRRRG},\] \[\text{GGGGR},\] \[\text{RRRGG},\] \[\text{GGGRR},\] \[\text{RRGGG},\] \[\text{GGRRR},\] \[\text{RGGGG},\] \[\text{GRRRR}.\] But this doesn't mean there are 10 "happy configurations" in total-- remember that we've been treating these cards as distinguishable, so we must continue to do so.
To lighten the load of 10 cases on the human brain, we can note that in the eyes of what we will soon do, $RRRRR$ and $GGGGG$ are effectively equivalent, and therefore may be treated in the same case. We will have to multiply by 2 at the end, though.
Similarly, we can equate $RRRRG,$ $GGGGR,$ $RGGGG,$ and $GRRRR,$ as well as $RRRGG,$ $GGGRR,$ $RRGGG,$ and $GGRRR,$ so that we just have three cases. We can approach each of these cases with constructive counting.
Case 1: $RRRRR$ -type.
For this case, there are $5$ available choices for the first card, $4$ available choices for the second card, $3$ for the third card, $2$ for the fourth card, and $1$ for the last card. This leads to a total of $5\cdot 4\cdot 3\cdot 2\cdot 1=120$ configurations for this case. There are $2$ cases of this type.
Case 2: $RRRRG$ -type.
For this case, there are $5$ available choices for the first card, $4$ available choices for the second card, $3$ for the third card, $2$ for the fourth card, and $5$ choices for the last card (not $1$ , because we're doing a new color). This leads to a total of $5\cdot 4\cdot 3\cdot 2\cdot 5=600$ configurations for this case. There are $4$ cases of this type.
Case 3: $RRRGG$ -type.
For this case, there are $5$ available choices for the first card, $4$ available choices for the second card, $3$ for the third card, $5$ for the fourth card, and $4$ choices for the last card. This leads to a total of $5\cdot 4\cdot 3\cdot 5\cdot 4=1200$ configurations for this case. There are $4$ cases of this type.
Adding the cases up gives $2\cdot 120+4\cdot 600+4\cdot 1200=7440$ "happy" configurations in total.
This means that the probability that Kathy is happy will be $\dfrac{7440}{10\cdot 9\cdot 8\cdot 7\cdot 6},$ which simplifies to $\dfrac{31}{126}.$
So the answer is $31+126=\boxed{157}$

Solution:
As the problem states, some examples of valid are $RRGGG$ $GGGGR$ , and $RRRRR$ . Let's use each of these as more general cases.
Let $RRGGG$ be the case when there are 2 adjacents of one color, and 3 adjacents of the other color. This yields $4$ combinations ( $RRGGG$ $GGRRR$ $RRRGG$ , and $GGGRR$ ). The probability of each of these is equal, equating to $\frac{5}{10}\cdot \frac{4}{9}\cdot \frac{5}{8}\cdot \frac{4}{7}\cdot \frac{3}{6}=\frac{5}{126}$ , and since there are $4$ combinations, the probability of this case is $4\cdot \frac{5}{126}=\frac{10}{63}$
Next case is $GGGGR$ . Let this be when there are 4 adjacents of one color, and 1 individual color. Once again, this yields $4$ combinations ( $GGGGR$ $RRRRG$ $RGGGG$ , and $GRRRR$ ). The probability of each is the same, equating to $\frac{5}{10}\cdot \frac{4}{9}\cdot \frac{3}{8}\cdot \frac{2}{7}\cdot \frac{5}{6}=\frac{5}{252}$ , and since there are $4$ combinations, the probability of this case is $4\cdot \frac{5}{252}=\frac{5}{63}$
The final case is $RRRRR$ , in which there is just an adjacent block of $5$ colors. There are only $2$ combinations this time, each equating to the probability $\frac{5}{10}\cdot \frac{4}{9}\cdot \frac{3}{8}\cdot \frac{2}{7}\cdot \frac{1}{6}=\frac{1}{252}$ , and since there are $2$ combinations, the probability of this case is $2\cdot \frac{1}{252}=\frac{1}{126}$
Thus, the total probability is $\frac{10}{63}+\frac{5}{63}+\frac{1}{126}=\frac{31}{126} \implies m+n=\boxed{157}$

Solution: Kathy will draw the 10 cards in some order, but the restriction of having all greens in a row and all reds in a row only applies to the first 5 cards drawn. The total number of ways the 10 cards can be drawn is simply 10 choose 5 which is 252. Now we just count the number of possible successful configurations of the ten cards. The first 5 cards can start either be $GRRRR$ $GGRRR$ $GGGRR$ $GGGGR$ $GGGGG$ or the same thing except starting with a red. The number of ways to order $GRRRR$ is the number of ways to order the last 5 cards, which is 5C1. Doing all of the other cases, the total is $(\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\binom{5}{5})*2 = 62$ $\frac{62}{252} = \frac{31}{126},$ so the solution is $31 + 126 = \boxed{157}$

Solution:
Assume without loss of generality that the first card laid out is red. Then the arrangements that satisfy Kathy’s requirements are RRRRR, RRRRG, RRRGG, RRGGG, and RGGGG. The probability that Kathy will lay out one of these
arrangements is \[\frac49\cdot\frac38\cdot\frac27\cdot\frac16\] \[\frac49\cdot\frac38\cdot\frac27\cdot\frac56\] \[\frac49\cdot\frac38\cdot\frac57\cdot\frac46\] \[\frac49\cdot\frac58\cdot\frac47\cdot\frac36\] \[+\frac59\cdot\frac48\cdot\frac37\cdot\frac26\] \[\overline{..........................}\] \[\frac{31}{126}\] The requested sum is $31+126=\boxed{157}.$


Answer: 38.0

Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
$\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4}$

Solution:
Divide it into $2$ cases:
1) Keiko and Ephriam both get $0$ heads:
This means that they both roll all tails, so there is only $1$ way for this to happen.
2) Keiko and Ephriam both get $1$ head:
For Keiko, there is only $1$ way for this to happen because he is only flipping 1 penny, but for Ephriam, there are 2 ways since there are $2$ choices for when he can flip the head. So, in total there are $2 \cdot 1 = 2$ ways for this case.
Thus, in total there are $3$ ways that work. Since there are $2$ choices for each coin flip (Heads or Tails), there are $2^3 = 8$ total ways of flipping 3 coins.
Thus, since all possible coin flips of 3 coins are equally likely, the probability is $\boxed{38}$


Answer: 3.0

Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of $6$ meters, and it takes her $36$ seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?
$\textbf{(A)}\ \frac{\pi}{3} \qquad\textbf{(B)}\ \frac{2\pi}{3} \qquad\textbf{(C)}\ \pi \qquad\textbf{(D)}\ \frac{4\pi}{3} \qquad\textbf{(E)}\ \frac{5\pi}{3}$

Solution:
Let $s$ be Keiko's speed in meters per second, $a$ be the length of the straight parts of the track, $b$ be the radius of the smaller circles, and $b+6$ be the radius of the larger circles. The length of the inner edge will be $2a+2b \pi$ and the length of the outer edge will be $2a+2\pi (b+6).$ Since it takes $36$ seconds longer for Keiko to walk on the outer edge,
\begin{align*} \frac{2a+2b \pi}{s} + 36 &= \frac{2a+2\pi (b+6)}{s}\\ 2a+2b\pi +36s &= 2a+2b\pi +12\pi\\ 36s&=12\pi\\ s&=\boxed{3}

Solution:
It is basically the same as Solution 1 except you can completely disregard the straight edges of the track since it will take Keiko the same time to walk that length for both inner and outer circles. Instead, focus on the circular part. If the diameter of the smaller circle were $r,$ then the length of the smaller circle would be $r \pi$ and the length of the larger circle would be $(r+12) \pi.$ Since it still takes $36$ seconds longer,
\begin{align*} \frac{r \pi}{s} + 36 &= \frac{(r+12)\pi}{s}\\ r \pi + 36s &= r \pi + 12 \pi\\ s&=\boxed{3}


Answer: 3.0

Keiko walks once around a track at the same constant speed every day.  The sides of the track are straight, and the ends are semicircles.  The track has a width of $6$ meters, and it takes her $36$ seconds longer to walk around the outside edge of the track than around the inside edge.  What is Keiko's speed in meters per second?
$\textbf{(A)}\ \frac{\pi}{3} \qquad \textbf{(B)}\ \frac{2\pi}{3} \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\  \frac{4\pi}{3} \qquad \textbf{(E)}\  \frac{5\pi}{3}$

Solution:
To find Keiko's speed, all we need to find is the difference between the distance around the inside edge of the track and the distance around the outside edge of the track, and divide it by the difference in the time it takes her for each distance.  We are given the difference in time, so all we need to find is the difference between the distances.
The track is divided into lengths and curves.  The lengths of the track will exhibit no difference in distance between the inside and outside edges, so we only need to concern ourselves with the curves.
The curves of the track are semicircles, but since there are two of them, we can consider both of the at the same time by treating them as a single circle.  We need to find the difference in the circumferences of the inside and outside edges of the circle.
The formula for the circumference of a circle is $C = 2 * \pi * r$ where $r$ is the radius of the circle.
Let's define the circumference of the inside circle as $C_1$ and the circumference of the outside circle as $C_2$
If the radius of the inside circle ( $r_1$ ) is $n$ , then given the thickness of the track is 6 meters, the radius of the outside circle ( $r_2$ ) is $n + 6$
Using this, the difference in the circumferences is:
$C_2 - C_1 = 2 * \pi * (r_2 - r_1) = 2 * \pi * (n + 6 - n) = 2 * \pi * 6 = 12\pi$
$12\pi$ is the difference between the inside and outside lengths of the track.  Divided by the time differential, we get:
$12\pi \div 36 = \boxed{3}$


Answer: 18.0

Kiana has two older twin brothers.  The product of their three ages is 128.  What is the sum of their three ages?
$\mathrm{(A)}\ 10\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 16\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 24$

Solution: The age of each person is a factor of $128 = 2^7$ .  So the twins could be $2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8$ years of age and, consequently Kiana could be $128$ $32$ $8$ , or $2$ years old, respectively.  Because Kiana is younger than her brothers, she must be $2$ years old. So the sum of their ages is $2 + 8 + 8 = \boxed{18}$


Answer: 18.0

Solution: The age of each person is a factor of $128 = 2^7$ .  So the twins could be $2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8$ years of age and, consequently Kiana could be $128$ $32$ $8$ , or $2$ years old, respectively.  Because Kiana is younger than her brothers, she must be $2$ years old. So the sum of their ages is $2 + 8 + 8 = \boxed{18}$


Answer: 50.0

King Middle School has $1200$ students.  Each student takes $5$ classes a day.  Each teacher teaches $4$ classes.  Each class has $30$ students and $1$ teacher.  How many teachers are there at King Middle School?
$\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 45 \qquad \text{(E)}\ 50$

Solution:
If each student has $5$ classes, and there are $1200$ students, then they have a total of $5\times 1200=6000$ classes among them.
Each class has $30$ students, so there must be $\frac{6000}{30}=200$ classes. Each class has $1$ teacher, so the teachers have a total of $200$ classes among them.
Each teacher teaches $4$ classes, so if there are $t$ teachers, they have $4t$ classes among them. This was found to be $200$ , so \[4t=200\Rightarrow t=50\]
This is answer choice $\boxed{50}$


Answer: 25.0

Kymbrea's comic book collection currently has $30$ comic books in it, and she is adding to her collection at the rate of $2$ comic books per month. LaShawn's collection currently has $10$ comic books in it, and he is adding to his collection at the rate of $6$ comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25$

Solution: Kymbrea has $30$ comic books initially and every month, she adds two. This can be represented as $30 + 2x$ where x is the number of months elapsed. LaShawn's collection, similarly, is $10 + 6x$ . To find when LaShawn will have twice the number of comic books as Kymbrea, we solve for x with the equation $2(2x + 30) = 6x + 10$ and get $x = \boxed{25}$


Answer: 415.0

Label one disk " $1$ ", two disks " $2$ ", three disks " $3$ $, ...,$ fifty disks " $50$ ". Put these $1+2+3+ \cdots+50=1275$ labeled disks in a box. Disks are then drawn from the box at random without replacement. The minimum number of disks that must be drawn to guarantee drawing at least ten disks with the same label is
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 51 \qquad\textbf{(C)}\ 415 \qquad\textbf{(D)}\ 451 \qquad\textbf{(E)}\ 501$

Solution:
We can solve this problem by thinking of the worst case scenario, essentially an adaptation of the Pigeon-hole principle.
We can start by picking up all the disks numbered 1 to 9 since even if we have all those disks we won't have 10 of any one disk. This gives us 45 disks.
From disks numbered from 10 to 50, we can pick up at most 9 disks to prevent picking up 10. There are 50-10+1 = 41 different numbers from 10 to 50. We pick up 9 from each number, therefore, we multiply $41 \cdot 9 = 369$ . In total, the maximum number we can pick up without picking up 10 of the same kind is $369+45=414$ . We need one more disk to guarantee a complete set of 10. Therefore, the answer is $\boxed{415}$


Answer: 4.0

Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?
$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18$

Solution:
Say Laila gets a value of $x$ on her first 4 tests, and a value of $y$ on her last test. Thus, $4x+y=410.$
The value $y$ has to be greater than $82$ , because otherwise she would receive the same score on her last test. Additionally, the greatest value for $y$ is $98$ (as $y=100$ would make $x$ as a decimal), so therefore, the greatest value $x$ can be is $98$ . As a result, only $4$ numbers work, $86, 90, 94$ and $98$ . Thus, the answer is $\boxed{4}$

Solution: The average point is $82$ leads us to suppose that Laila got all $82$ points for the tests. We know that Laila got the same points in the first four tests and they are all lower than the last test. Let the first four tests is $81$ points, then the last tests should be $86$ points to keep the average point is fixed. The possible points are $86$ $90$ $94$ $98$ . The answer is $\boxed{4}$


Answer: 22.0

Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was $3$ points higher than Zelda's average on all the quizzes she took during the first semester. Yolanda's average on all the quizzes she took during the second semester was $18$ points higher than her average for the first semester and was again $3$ points higher than Zelda's average on all the quizzes Zelda took during her second semester. Which one of the following statements cannot possibly be true?
$\textbf{(A)}$ Yolanda's quiz average for the academic year was $22$ points higher than Zelda's.
$\textbf{(B)}$ Zelda's quiz average for the academic year was higher than Yolanda's.
$\textbf{(C)}$ Yolanda's quiz average for the academic year was $3$ points higher than Zelda's.
$\textbf{(D)}$ Zelda's quiz average for the academic year equaled Yolanda's.
$\textbf{(E)}$ If Zelda had scored $3$ points higher on each quiz she took, then she would have had the same average for the academic year as Yolanda.

Solution:
Denote by $A_i$ the average of person with initial $A$ in semester $i \in \left\{1, 2 \right\}$ Thus, $Y_1 = Z_1 + 3$ $Y_2 = Y_1 + 18$ $Y_2 = Z_2 + 3$
Denote by $A_{12}$ the average of person with initial $A$ in the full year.
Thus, $Y_{12}$ can be any number in $\left( Y_1 , Y_2 \right)$ and $Z_{12}$ can be any number in $\left( Z_1 , Z_2 \right)$
Therefore, the impossible solution is $\boxed{22}$

Solution:
We can use process of elimination by finding possible solutions to answer choices. Let $y_1$ and $y_2$ be the number of quizzes Yolanda took in the first and second semesters, respectively. Define $z_1$ and $z_2$ similarly for Zelda.
Answer choice B is satisfied by $(y_1,y_2,z_1,z_2) = (289,1,1,289)$
Answer choice C and E are both satisfied by $(y_1,y_2,z_1,z_2) = (17,17,17,17)$
Answer choice D is satisfied by $(y_1,y_2,z_1,z_2) = (7,5,5,7)$
Therefore the answer is $\boxed{22}$

Solution:
Let Yolanda's average for semester $1$ be $y_1$ , the number of quizzes Yolanda took in semester $1$ be $n_1$ , Zelda's average for semester $1$ be $z_1$ , the number of quizzes Zelda took in semester $1$ be $k_1$ , Yolanda's average for semester $2$ be $y_2$ , the number of quizzes Yolanda took in semester $2$ be $n_2$ , Zelda's average for semester $2$ be $z_2$ , the number of quizzes Zelda took in semester $2$ be $k_2$ , Yolanda's average for the entire year be $y$ , Zelda's average for the entire year be $z$
From the problem we know that
\[y_1 = z_1 + 3, \quad y_2 = z_2 + 3\]
\[y_2 = y_1 + 18, \quad z_2 = z_1 + 18\]
\[y = \frac{ y_1 n_1 + y_2 n_2 }{ n_1 + n_2} = \frac{ y_1 n_1 + (y_1 + 18) n_2 }{ n_1 + n_2} = y_1 + \frac{18 y_2 }{ n_1 + n_2}\]
\[z = \frac{ z_1 k_1 + z_2 k_2 }{ k_1 + k_2} = \frac{ z_1 k_1 + (z_1 + 18) k_2 }{ k_1 + k_2} = z_1 = \frac{ 18 k_2 }{ k_1 + k_2}\]
\[y - z = y_1 + \frac{18 y_2 }{ n_1 + n_2} - z_1  - \frac{ 18 k_2 }{ k_1 + k_2} = y_1 - z_1 + 18 \left( \frac{y_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2} \right) = 3 + 18 \left( \frac{y_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2} \right)\]
$\frac{y_2 }{ n_1 + n_2}$ at most is $1$ $\frac{k_2 }{ k_1 + k_2}$ is at least $0$ , meaning that $\frac{y_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2}$ is at most $1$
Therefore, \[y - z \le 3 + 18 = 21\]
Hence, $\boxed{22}$ is not possible.

Solution:
Denote $y_1$ and $y_2$ as the quiz averages of Yolanda in the $1$ st and $2$ nd semesters, respectively. Similarly, denote $z_1$ and $z_2$ as the quiz averages of Zelda in the $1$ st and $2$ nd semesters.
We have $y_1 = z_1 + 3$ , so $y_1 - 3 = z_1$ . We also know that $y_2 = 18 +y_1 = 3 + z_2$ , implying $z_2 = 15 + y_1$
The average quiz scores for both students must lie between the averages of each semester, i.e $\mathrm{Avg_{y}}\in[y_1, 18+y_1],$ and $\mathrm{Avg_{z}}\in[y_1-3, y_1+15].$ Since $z_2>z_1$ $y_2 > y_1$ , and $y_1>z_1$ , we have $\min\{\mathrm{Avg}\} = \min\{\mathrm{Avg}_z\} = z_1$ and $\max\{\mathrm{Avg}\} = \max\{\mathrm{Avg}_y\} = y_2$ . Therefore the maximum difference between the two yearly averages is
\[|y_2 - z_1| = |(y_1 + 18)-(y_1-3)| = 21 < 22.\]
Therefore, $\boxed{22}$ is not possible.


Answer: 32.0

Last summer $100$ students attended basketball camp.  Of those attending, $52$ were boys and $48$ were girls.  Also, $40$ students were from Jonas Middle School and $60$ were from Clay Middle School.  Twenty of the girls were from Jonas Middle School.  How many of the boys were from Clay Middle School?
$\text{(A)}\ 20 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 52$

Solution:
Make a table with the information given.
\[\begin{tabular}{c|ccc} & \text{Jonas} & \text{Clay} & \text{total} \\ \hline \text{boys} & & & 52 \\ \text{girls} & 20 & & 48 \\ \text{total} & 40 & 60 & 100  \end{tabular}\]
Because the first two columns must add up to the third column, and the same with the rows, the rest of the empty boxes can be filled in.
\[\begin{tabular}{c|ccc} & \text{Jonas} & \text{Clay} & \text{total} \\ \hline \text{boys} & 20 & 32 & 52 \\ \text{girls} & 20 & 28 & 48 \\ \text{total} & 40 & 60 & 100  \end{tabular}\]
The number of boys from Clay is $\boxed{32}$


Answer: 40.0

Last summer $30\%$ of the birds living on Town Lake were geese, $25\%$ were swans, $10\%$ were herons, and $35\%$ were ducks. What percent of the birds that were not swans were geese?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 60$

Solution: To simplify the problem, WLOG, let us say that there were a total of $100$ birds. The number of birds that are not swans is $75$ . The number of geese is $30$ . Therefore the percentage is just $\frac{30}{75} \times 100 = 40 \Rightarrow \boxed{40}$


Answer: 40.0

Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60$

Solution: 75% of the total birds were not swans. Out of that 75%, there was $30\% / 75\% = \boxed{40}$ of the birds that were not swans that were geese.

Solution: WLOG, suppose there were 100 birds in total living on Town Lake, then 30 were geese, 25 were swans, 10 were herons, and 35 were ducks. $100-25 = 75$ of the birds are not swans and 30 of these are geese, so the answer is $\frac{30}{75} \times 100 = \boxed{40}$