🔬 Model Lab

Problem 1:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

To find the total number of pages in the book, we need to add the number of pages Isabella read each day.

First, let's calculate the total number of pages she read in the first three days: $36 \times 3 = 108$ pages.
Next, we calculate the total number of pages she read in the next three days: $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.

The total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low since $108 + 132 + 10 = 250$.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 1: B

Problem 2:
What is the smallest possible average of four distinct positive even integers?

To minimize the average of four distinct positive even integers, we need to choose the smallest four distinct positive even integers.

The smallest four distinct positive even integers are 2, 4, 6, and 8.

Let's calculate their average: $\frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$.

Let's examine the choices:
A. 3: Too low since the smallest even number is 2.
B. 4: Too low since the average of 2, 4, 6, and 8 is 5.
C. 5: Matches our calculation.
D. 6: Too high.
E. 7: Too high.

ANSWER 2: C

Problem 3:
Four students take an exam. Three of their scores are 70, 80, and 90. If the average of their four scores is 70, then what is the remaining score?

To find the remaining score, we first need to find the total score of all four students.

The average score is 70, and there are 4 students, so the total score is $70 \times 4 = 280$.

The sum of the three known scores is $70 + 80 + 90 = 240$.

The remaining score is $280 - 240 = 40$.

Let's examine the choices:
A. 40: Matches our calculation.
B. 50: Too high.
C. 55: Too high.
D. 60: Too high.
E. 70: Too high.

ANSWER 3: A

Problem 4:
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

Let's denote the side length of the square as $s$ and the radius of the circle as $r$.

The area of the square is $s^2$, and the area of the circle is $\pi r^2$.

Since they have the same area, we have $s^2 = \pi r^2$.

To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$.

Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$.

Let's examine the choices:
A. $\sqrt{\pi} / 2$: Too low.
B. $\sqrt{\pi}$: Matches our calculation.
C. $\pi$: Too high.
D. $2\pi$: Too high.
E. $\pi^2$: Too high.

ANSWER 4: B

Problem 5:
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Let's denote the smallest of these integers as $n$. Then the six consecutive integers are $n$, $n+1$, $n+2$, $n+3$, $n+4$, and $n+5$.

Their sum is $n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15 = 2013$.

Solving for $n$, we get $6n = 2013 - 15 = 1998$, so $n = \frac{1998}{6} = 333$.

The largest of these six integers is $n + 5 = 333 + 5 = 338$.

Let's examine the choices:
A. 335: Too low.
B. 338: Matches our calculation.
C. 340: Too high.
D. 345: Too high.
E. 350: Too high.

ANSWER 5: B

Problem 6:
On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

A standard checkerboard is an $8 \times 8$ grid, so it has 64 unit squares.

The unit squares that touch the outer edge are those on the first and last rows and columns. There are $8 + 8 + 8 + 8 = 32$ such squares, but the 4 corner squares are counted twice, so we subtract 4 to avoid double-counting: $32 - 4 = 28$.

Thus, there are $64 - 28 = 36$ unit squares that do not touch the outer edge.

The probability that a randomly chosen unit square does not touch the outer edge is $\frac{36}{64} = \frac{9}{16}$.

Let's examine the choices:
A. $\frac{1}{16}$: Too low.
B. $\frac{7}{16}$: Too low.
C. $\frac{1}{2}$: Too low since $\frac{9}{16} > \frac{1}{2}$.
D. $\frac{9}{16}$: Matches our calculation.
E. $\frac{49}{64}$: Too high.

ANSWER 6: D

Problem 7:
What is the sum of the mean, median, and mode of the numbers 2, 3, 0, 3, 1, 4, 0, 3?

First, let's arrange the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.

The mean is the average of the numbers: $\frac{0 + 0 + 1 + 2 + 3 + 3 + 3 + 4}{8} = \frac{16}{8} = 2$.

The median is the average of the 4th and 5th numbers (since there are 8 numbers): $\frac{2 + 3}{2} = 2.5$.

The mode is the number that appears most frequently, which is 3.

The sum of the mean, median, and mode is $2 + 2.5 + 3 = 7.5$.

Let's examine the choices:
A. 6.5: Too low.
B. 7: Too low.
C. 7.5: Matches our calculation.
D. 8.5: Too high.
E. 9: Too high.

ANSWER 7: C

Problem 8:
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

Georgie has 6 choices for the window to enter.

After entering, Georgie has 5 remaining choices for the window to leave (since it must be different from the entry window).

The total number of ways Georgie can enter and leave is $6 \times 5 = 30$.

Let's examine the choices:
A. 12: Too low.
B. 15: Too low.
C. 18: Too low.
D. 30: Matches our calculation.
E. 36: Too high.

ANSWER 8: D

Problem 9:
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?

Let's calculate Jack's total:
1. $90.00 + 6\%$ of $90.00 = 90 + 0.06 \times 90 = 90 + 5.40 = 95.40$.
2. $20\%$ 

Problem 1:

Isabella read the book over 7 days.

- First 3 days:  
  \[
  3 \times 36 = 108
  \]
  pages

- Next 3 days:  
  \[
  3 \times 44 = 132
  \]
  pages

- Last day:  
  \[
  10
  \]
  pages

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 1: B

Problem 2:

We want the smallest possible average of four distinct positive even integers.

The smallest positive even integers are:

\[
2,\ 4,\ 6,\ 8
\]

Their average is:

\[
\frac{2+4+6+8}{4}=\frac{20}{4}=5
\]

We cannot use 0 because the integers must be positive. We also cannot repeat numbers because they must be distinct.

ANSWER 2: C

Problem 3:

The average of 4 scores is 70, so the total of the 4 scores is:

\[
4 \times 70 = 280
\]

The three known scores add to:

\[
70+80+90=240
\]

So the remaining score is:

\[
280-240=40
\]

ANSWER 3: A

Problem 4:

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The area of the square is:

\[
s^2
\]

The area of the circle is:

\[
\pi r^2
\]

They have the same area, so:

\[
s^2=\pi r^2
\]

Taking the square root of both sides:

\[
s=r\sqrt{\pi}
\]

Therefore,

\[
\frac{s}{r}=\sqrt{\pi}
\]

So the ratio of the side length of the square to the radius of the circle is \(\sqrt{\pi}\).

ANSWER 4: B

Problem 5:

Let the six consecutive positive integers be:

\[
n,\ n+1,\ n+2,\ n+3,\ n+4,\ n+5
\]

Their sum is 2013:

\[
n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=2013
\]

Combine like terms:

\[
6n+15=2013
\]

Subtract 15:

\[
6n=1998
\]

Divide by 6:

\[
n=333
\]

So the six integers are:

\[
333,\ 334,\ 335,\ 336,\ 337,\ 338
\]

The largest is 338.

ANSWER 5: B

Problem 6:

A checkerboard has 64 unit squares, so it is an \(8 \times 8\) board.

A square does not touch the outer edge only if it is not in the first row, last row, first column, or last column.

That leaves the interior squares:

\[
6 \times 6 = 36
\]

So the probability is:

\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:

The numbers are:

\[
2,\ 3,\ 0,\ 3,\ 1,\ 4,\ 0,\ 3
\]

First find the mean.

Sum:

\[
2+3+0+3+1+4+0+3=16
\]

There are 8 numbers, so the mean is:

\[
\frac{16}{8}=2
\]

Now sort the numbers:

\[
0,\ 0,\ 1,\ 2,\ 3,\ 3,\ 3,\ 4
\]

There are 8 numbers, so the median is the average of the 4th and 5th numbers:

\[
\frac{2+3}{2}=2.5
\]

The mode is the number that appears most often. The number 3 appears three times, so the mode is:

\[
3
\]

Now add mean, median, and mode:

\[
2+2.5+3=7.5
\]

ANSWER 7: C

Problem 8:

Georgie has 6 choices for the window he enters.

After entering, he must leave by a different window, so he has 5 choices for the window he leaves.

Thus the number of ways is:

\[
6 \times 5 = 30
\]

This is not 15 because entering window A and leaving window B is different from entering window B and leaving window A.

ANSWER 8: D

Problem 9:

Jack’s method:

He first adds 6% sales tax to \$90:

\[
90 \times 1.06 = 95.40
\]

Then he subtracts 20% from that total, meaning he keeps 80%:

\[
95.40 \times 0.80 = 76.32
\]

So Jack’s total is \$76.32.

Jill’s method:

She first subtracts 20% from \$90:

\[
90 \times 0.80 = 72
\]

Then she adds 6% sales tax:

\[
72 \times 1.06 = 76.32
\]

So Jill’s total is also \$76.32.

Therefore:

\[
76.32-76.32=0
\]

ANSWER 9: C

Problem 10:

The expression is:

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}+\frac{55}{10}
\]

Since all fractions have denominator 10, add the numerators:

\[
\frac{1+2+3+4+5+6+7+8+9+55}{10}
\]

Now,

\[
1+2+3+4+5+6+7+8+9=45
\]

So:

\[
\frac{45+55}{10}=\frac{100}{10}=10
\]

ANSWER 10: D

Problem 11:

The code phrase is:

\[
\text{BEST OF LUCK}
\]

The ten letters represent the digits 0 through 9 in order:

\[
B=0,\ E=1,\ S=2,\ T=3,\ O=4,\ F=5,\ L=6,\ U=7,\ C=8,\ K=9
\]

Now decode the word CLUE:

\[
C=8,\quad L=6,\quad U=7,\quad E=1
\]

So CLUE represents:

\[
8671
\]

ANSWER 11: A

Problem 12:

The sunrise was 6:57 AM, and the length of daylight was 10 hours and 24 minutes.

So the sunset time is:

\[
6:57\text{ AM} + 10\text{ hours }24\text{ minutes}
\]

First add 10 hours:

\[
6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}
\]

Now add 24 minutes:

\[
4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}
\]

So the sun really set at 5:21 PM.

ANSWER 12: B

**Problem 1:**
To find the total number of pages in the book, we need to calculate the number of pages Isabella read in each part of the week and add them together.
- For the first 3 days, she read an average of 36 pages per day: 
  3 days × 36 pages/day = 108 pages.
- For the next 3 days, she read an average of 44 pages per day: 
  3 days × 44 pages/day = 132 pages.
- On the last day (the 7th day), she read 10 pages.

Adding these amounts together gives the total number of pages:
108 + 132 + 10 = 250 pages.

ANSWER 1: B

**Problem 2:**
To find the smallest possible average of four distinct positive even integers, we must first select the four smallest distinct positive even integers. 
These integers are 2, 4, 6, and 8.
Next, we find their sum:
2 + 4 + 6 + 8 = 20
Finally, we divide the sum by the number of integers (4) to find the average:
20 / 4 = 5

ANSWER 2: C

**Problem 3:**
Let the unknown remaining score be $x$.
The average of the four scores is given as 70. The formula for the average is the sum of the scores divided by the number of scores:
(70 + 80 + 90 + $x$) / 4 = 70

Multiply both sides by 4 to find the total sum of the scores:
70 + 80 + 90 + $x$ = 280
240 + $x$ = 280

Subtract 240 from both sides:
$x$ = 40
The remaining score is 40.

ANSWER 3: A

**Problem 4:**
Let the side length of the square be $s$ and the radius of the circle be $r$.
The area of the square is $s^2$, and the area of the circle is $\pi r^2$.
We are told that their areas are equal:
$s^2 = \pi r^2$

To find the ratio of the side length to the radius ($s/r$), we take the square root of both sides:
$s = r\sqrt{\pi}$
$s / r = \sqrt{\pi}$

ANSWER 4: B

**Problem 5:**
Let the six consecutive positive integers be $x$, $x+1$, $x+2$, $x+3$, $x+4$, and $x+5$.
Their sum is:
$x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 2013$
$6x + 15 = 2013$

Subtract 15 from both sides:
$6x = 1998$

Divide by 6:
$x = 333$

The problem asks for the largest of these integers, which is $x+5$:
333 + 5 = 338

ANSWER 5: B

**Problem 6:**
A standard checkerboard is an $8 \times 8$ grid, meaning it has a total of 64 unit squares.
The squares that touch the outer edge form the border of the board. If we remove this 1-square-thick border, the remaining inner squares form a smaller $6 \times 6$ grid.
The number of squares that do *not* touch the outer edge is:
$6 \times 6 = 36$

The probability of randomly choosing one of these inner squares is the number of inner squares divided by the total number of squares:
$P = 36 / 64$

Dividing the numerator and the denominator by 4 simplifies the fraction:
$P = 9 / 16$

ANSWER 6: D

**Problem 7:**
First, let's list the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.
There are 8 numbers in total.

- **Mean:** The sum of the numbers divided by the count.
  Sum = 0 + 0 + 1 + 2 + 3 + 3 + 3 + 4 = 16
  Mean = 16 / 8 = 2

- **Median:** The middle value. Since there is an even number of values (8), the median is the average of the 4th and 5th numbers.
  The 4th number is 2, and the 5th number is 3.
  Median = (2 + 3) / 2 = 2.5

- **Mode:** The number that appears most frequently.
  The number 3 appears three times, which is more than any other number.
  Mode = 3

The sum of the mean, median, and mode is:
2 + 2.5 + 3 = 7.5

ANSWER 7: C

**Problem 8:**
Georgie the Ghost has 6 different windows to choose from when entering the house.
Because he must leave by a *different* window, he cannot use the window he entered through. This leaves him with 5 choices for his exit.
By the fundamental counting principle, the total number of ways he can enter and leave is:
6 (entry choices) × 5 (exit choices) = 30 ways.

ANSWER 8: D

**Problem 9:**
Let's calculate the final total for both clerks.
- **Jack's method:** He adds the 6% tax first, making the subtotal $90 \times 1.06$. Then he applies the 20% discount, which means the customer pays 80% of the subtotal.
  Jack's Total = $90 \times 1.06 \times 0.80$

- **Jill's method:** She applies the 20% discount first, making the subtotal $90 \times 0.80$. Then she adds the 6% tax, which means multiplying by 1.06.
  Jill's Total = $90 \times 0.80 \times 1.06$

Because multiplication is commutative (the order in which you multiply numbers does not change the product), Jack's total and Jill's total are mathematically identical.
Therefore, Jack's total minus Jill's total is exactly $0.

ANSWER 9: C

**Problem 10:**
We are asked to find the sum of the following fractions:
$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$

Since all the fractions share a common denominator of 10, we just need to add their numerators:
Sum of numerators = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55

The sum of the digits 1 through 9 is 45.
Total numerator = 45 + 55 = 100

Now divide the total numerator by the denominator:
$\frac{100}{10} = 10$

ANSWER 10: D

**Problem 11:**
The ten-letter code "BEST OF LUCK" maps to the digits 0 through 9 in order:
B = 0
E = 1
S = 2
T = 3
O = 4
F = 5
L = 6
U = 7
C = 8
K = 9

We need to find the 4-digit number for "CLUE":
C = 8
L = 6
U = 7
E = 1

Putting the digits together, "CLUE" represents 8671.

ANSWER 11: A

**Problem 12:**
To find the correct sunset time, we need to add the length of daylight to the sunrise time.
Sunrise: 6:57 AM
Length of daylight: 10 hours and 24 minutes

First, add the 10 hours to the sunrise time:
6:57 AM + 10 hours = 4:57 PM

Next, add the remaining 24 minutes to 4:57 PM:
Adding 3 minutes brings the time to 5:00 PM.
We have 21 minutes left to add (24 - 3 = 21).
5:00 PM + 21 minutes = 5:21 PM.

ANSWER 12: B

Problem 1:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

To find the total number of pages in the book, we need to add the number of pages Isabella read each day.

First, let's calculate the total number of pages she read in the first three days: $36 \times 3 = 108$ pages.
Next, we calculate the total number of pages she read in the next three days: $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.

The total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low since $108 + 132 + 10 = 250$.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 1: B

Problem 2:
What is the smallest possible average of four distinct positive even integers?

To minimize the average of four distinct positive even integers, we need to choose the smallest four distinct positive even integers.

The smallest four distinct positive even integers are 2, 4, 6, and 8.

Let's calculate their average: $\frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$.

Let's examine the choices:
A. 3: Too low since the smallest even number is 2.
B. 4: Too low since the average of 2, 4, 6, and 8 is 5.
C. 5: Matches our calculation.
D. 6: Too high.
E. 7: Too high.

ANSWER 2: C

Problem 3:
Four students take an exam. Three of their scores are 70, 80, and 90. If the average of their four scores is 70, then what is the remaining score?

To find the remaining score, we first need to find the total score of all four students.

The average score is 70, and there are 4 students, so the total score is $70 \times 4 = 280$.

The sum of the three known scores is $70 + 80 + 90 = 240$.

The remaining score is $280 - 240 = 40$.

Let's examine the choices:
A. 40: Matches our calculation.
B. 50: Too high.
C. 55: Too high.
D. 60: Too high.
E. 70: Too high.

ANSWER 3: A

Problem 4:
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

Let's denote the side length of the square as $s$ and the radius of the circle as $r$.

The area of the square is $s^2$, and the area of the circle is $\pi r^2$.

Since they have the same area, we have $s^2 = \pi r^2$.

To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$.

Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$.

Let's examine the choices:
A. $\sqrt{\pi} / 2$: Too low.
B. $\sqrt{\pi}$: Matches our calculation.
C. $\pi$: Too high.
D. $2\pi$: Too high.
E. $\pi^2$: Too high.

ANSWER 4: B

Problem 5:
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Let's denote the smallest of these integers as $n$. Then the six consecutive integers are $n$, $n+1$, $n+2$, $n+3$, $n+4$, and $n+5$.

Their sum is $n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15 = 2013$.

Solving for $n$, we get $6n = 2013 - 15 = 1998$, so $n = \frac{1998}{6} = 333$.

The largest of these six integers is $n + 5 = 333 + 5 = 338$.

Let's examine the choices:
A. 335: Too low.
B. 338: Matches our calculation.
C. 340: Too high.
D. 345: Too high.
E. 350: Too high.

ANSWER 5: B

Problem 6:
On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

A standard checkerboard is an $8 \times 8$ grid, so it has 64 unit squares.

The unit squares that touch the outer edge are those on the first and last rows and columns. There are $8 + 8 + 8 + 8 = 32$ such squares, but the 4 corner squares are counted twice, so we subtract 4 to avoid double-counting: $32 - 4 = 28$.

Thus, there are $64 - 28 = 36$ unit squares that do not touch the outer edge.

The probability that a randomly chosen unit square does not touch the outer edge is $\frac{36}{64} = \frac{9}{16}$.

Let's examine the choices:
A. $\frac{1}{16}$: Too low.
B. $\frac{7}{16}$: Too low.
C. $\frac{1}{2}$: Too low since $\frac{9}{16} > \frac{1}{2}$.
D. $\frac{9}{16}$: Matches our calculation.
E. $\frac{49}{64}$: Too high.

ANSWER 6: D

Problem 7:
What is the sum of the mean, median, and mode of the numbers 2, 3, 0, 3, 1, 4, 0, 3?

First, let's arrange the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.

The mean is the average of the numbers: $\frac{0 + 0 + 1 + 2 + 3 + 3 + 3 + 4}{8} = \frac{16}{8} = 2$.

The median is the average of the 4th and 5th numbers (since there are 8 numbers): $\frac{2 + 3}{2} = 2.5$.

The mode is the number that appears most frequently, which is 3.

The sum of the mean, median, and mode is $2 + 2.5 + 3 = 7.5$.

Let's examine the choices:
A. 6.5: Too low.
B. 7: Too low.
C. 7.5: Matches our calculation.
D. 8.5: Too high.
E. 9: Too high.

ANSWER 7: C

Problem 8:
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

Georgie has 6 choices for the window to enter.

After entering, Georgie has 5 remaining choices for the window to leave (since it must be different from the entry window).

The total number of ways Georgie can enter and leave is $6 \times 5 = 30$.

Let's examine the choices:
A. 12: Too low.
B. 15: Too low.
C. 18: Too low.
D. 30: Matches our calculation.
E. 36: Too high.

ANSWER 8: D

Problem 9:
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?

Let's calculate Jack's total:
1. $90.00 + 6\%$ of $90.00 = 90 + 0.06 \times 90 = 90 + 5.40 = 95.40$.
2. $20\%$ 

Problem 1:

Isabella read the book over 7 days.

- First 3 days:  
  \[
  3 \times 36 = 108
  \]
  pages

- Next 3 days:  
  \[
  3 \times 44 = 132
  \]
  pages

- Last day:  
  \[
  10
  \]
  pages

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 1: B

Problem 2:

We want the smallest possible average of four distinct positive even integers.

The smallest positive even integers are:

\[
2,\ 4,\ 6,\ 8
\]

Their average is:

\[
\frac{2+4+6+8}{4}=\frac{20}{4}=5
\]

We cannot use 0 because the integers must be positive. We also cannot repeat numbers because they must be distinct.

ANSWER 2: C

Problem 3:

The average of 4 scores is 70, so the total of the 4 scores is:

\[
4 \times 70 = 280
\]

The three known scores add to:

\[
70+80+90=240
\]

So the remaining score is:

\[
280-240=40
\]

ANSWER 3: A

Problem 4:

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The area of the square is:

\[
s^2
\]

The area of the circle is:

\[
\pi r^2
\]

They have the same area, so:

\[
s^2=\pi r^2
\]

Taking the square root of both sides:

\[
s=r\sqrt{\pi}
\]

Therefore,

\[
\frac{s}{r}=\sqrt{\pi}
\]

So the ratio of the side length of the square to the radius of the circle is \(\sqrt{\pi}\).

ANSWER 4: B

Problem 5:

Let the six consecutive positive integers be:

\[
n,\ n+1,\ n+2,\ n+3,\ n+4,\ n+5
\]

Their sum is 2013:

\[
n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=2013
\]

Combine like terms:

\[
6n+15=2013
\]

Subtract 15:

\[
6n=1998
\]

Divide by 6:

\[
n=333
\]

So the six integers are:

\[
333,\ 334,\ 335,\ 336,\ 337,\ 338
\]

The largest is 338.

ANSWER 5: B

Problem 6:

A checkerboard has 64 unit squares, so it is an \(8 \times 8\) board.

A square does not touch the outer edge only if it is not in the first row, last row, first column, or last column.

That leaves the interior squares:

\[
6 \times 6 = 36
\]

So the probability is:

\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:

The numbers are:

\[
2,\ 3,\ 0,\ 3,\ 1,\ 4,\ 0,\ 3
\]

First find the mean.

Sum:

\[
2+3+0+3+1+4+0+3=16
\]

There are 8 numbers, so the mean is:

\[
\frac{16}{8}=2
\]

Now sort the numbers:

\[
0,\ 0,\ 1,\ 2,\ 3,\ 3,\ 3,\ 4
\]

There are 8 numbers, so the median is the average of the 4th and 5th numbers:

\[
\frac{2+3}{2}=2.5
\]

The mode is the number that appears most often. The number 3 appears three times, so the mode is:

\[
3
\]

Now add mean, median, and mode:

\[
2+2.5+3=7.5
\]

ANSWER 7: C

Problem 8:

Georgie has 6 choices for the window he enters.

After entering, he must leave by a different window, so he has 5 choices for the window he leaves.

Thus the number of ways is:

\[
6 \times 5 = 30
\]

This is not 15 because entering window A and leaving window B is different from entering window B and leaving window A.

ANSWER 8: D

Problem 9:

Jack’s method:

He first adds 6% sales tax to \$90:

\[
90 \times 1.06 = 95.40
\]

Then he subtracts 20% from that total, meaning he keeps 80%:

\[
95.40 \times 0.80 = 76.32
\]

So Jack’s total is \$76.32.

Jill’s method:

She first subtracts 20% from \$90:

\[
90 \times 0.80 = 72
\]

Then she adds 6% sales tax:

\[
72 \times 1.06 = 76.32
\]

So Jill’s total is also \$76.32.

Therefore:

\[
76.32-76.32=0
\]

ANSWER 9: C

Problem 10:

The expression is:

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}+\frac{55}{10}
\]

Since all fractions have denominator 10, add the numerators:

\[
\frac{1+2+3+4+5+6+7+8+9+55}{10}
\]

Now,

\[
1+2+3+4+5+6+7+8+9=45
\]

So:

\[
\frac{45+55}{10}=\frac{100}{10}=10
\]

ANSWER 10: D

Problem 11:

The code phrase is:

\[
\text{BEST OF LUCK}
\]

The ten letters represent the digits 0 through 9 in order:

\[
B=0,\ E=1,\ S=2,\ T=3,\ O=4,\ F=5,\ L=6,\ U=7,\ C=8,\ K=9
\]

Now decode the word CLUE:

\[
C=8,\quad L=6,\quad U=7,\quad E=1
\]

So CLUE represents:

\[
8671
\]

ANSWER 11: A

Problem 12:

The sunrise was 6:57 AM, and the length of daylight was 10 hours and 24 minutes.

So the sunset time is:

\[
6:57\text{ AM} + 10\text{ hours }24\text{ minutes}
\]

First add 10 hours:

\[
6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}
\]

Now add 24 minutes:

\[
4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}
\]

So the sun really set at 5:21 PM.

ANSWER 12: B

**Problem 1:**
To find the total number of pages in the book, we need to calculate the number of pages Isabella read in each part of the week and add them together.
- For the first 3 days, she read an average of 36 pages per day: 
  3 days × 36 pages/day = 108 pages.
- For the next 3 days, she read an average of 44 pages per day: 
  3 days × 44 pages/day = 132 pages.
- On the last day (the 7th day), she read 10 pages.

Adding these amounts together gives the total number of pages:
108 + 132 + 10 = 250 pages.

ANSWER 1: B

**Problem 2:**
To find the smallest possible average of four distinct positive even integers, we must first select the four smallest distinct positive even integers. 
These integers are 2, 4, 6, and 8.
Next, we find their sum:
2 + 4 + 6 + 8 = 20
Finally, we divide the sum by the number of integers (4) to find the average:
20 / 4 = 5

ANSWER 2: C

**Problem 3:**
Let the unknown remaining score be $x$.
The average of the four scores is given as 70. The formula for the average is the sum of the scores divided by the number of scores:
(70 + 80 + 90 + $x$) / 4 = 70

Multiply both sides by 4 to find the total sum of the scores:
70 + 80 + 90 + $x$ = 280
240 + $x$ = 280

Subtract 240 from both sides:
$x$ = 40
The remaining score is 40.

ANSWER 3: A

**Problem 4:**
Let the side length of the square be $s$ and the radius of the circle be $r$.
The area of the square is $s^2$, and the area of the circle is $\pi r^2$.
We are told that their areas are equal:
$s^2 = \pi r^2$

To find the ratio of the side length to the radius ($s/r$), we take the square root of both sides:
$s = r\sqrt{\pi}$
$s / r = \sqrt{\pi}$

ANSWER 4: B

**Problem 5:**
Let the six consecutive positive integers be $x$, $x+1$, $x+2$, $x+3$, $x+4$, and $x+5$.
Their sum is:
$x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 2013$
$6x + 15 = 2013$

Subtract 15 from both sides:
$6x = 1998$

Divide by 6:
$x = 333$

The problem asks for the largest of these integers, which is $x+5$:
333 + 5 = 338

ANSWER 5: B

**Problem 6:**
A standard checkerboard is an $8 \times 8$ grid, meaning it has a total of 64 unit squares.
The squares that touch the outer edge form the border of the board. If we remove this 1-square-thick border, the remaining inner squares form a smaller $6 \times 6$ grid.
The number of squares that do *not* touch the outer edge is:
$6 \times 6 = 36$

The probability of randomly choosing one of these inner squares is the number of inner squares divided by the total number of squares:
$P = 36 / 64$

Dividing the numerator and the denominator by 4 simplifies the fraction:
$P = 9 / 16$

ANSWER 6: D

**Problem 7:**
First, let's list the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.
There are 8 numbers in total.

- **Mean:** The sum of the numbers divided by the count.
  Sum = 0 + 0 + 1 + 2 + 3 + 3 + 3 + 4 = 16
  Mean = 16 / 8 = 2

- **Median:** The middle value. Since there is an even number of values (8), the median is the average of the 4th and 5th numbers.
  The 4th number is 2, and the 5th number is 3.
  Median = (2 + 3) / 2 = 2.5

- **Mode:** The number that appears most frequently.
  The number 3 appears three times, which is more than any other number.
  Mode = 3

The sum of the mean, median, and mode is:
2 + 2.5 + 3 = 7.5

ANSWER 7: C

**Problem 8:**
Georgie the Ghost has 6 different windows to choose from when entering the house.
Because he must leave by a *different* window, he cannot use the window he entered through. This leaves him with 5 choices for his exit.
By the fundamental counting principle, the total number of ways he can enter and leave is:
6 (entry choices) × 5 (exit choices) = 30 ways.

ANSWER 8: D

**Problem 9:**
Let's calculate the final total for both clerks.
- **Jack's method:** He adds the 6% tax first, making the subtotal $90 \times 1.06$. Then he applies the 20% discount, which means the customer pays 80% of the subtotal.
  Jack's Total = $90 \times 1.06 \times 0.80$

- **Jill's method:** She applies the 20% discount first, making the subtotal $90 \times 0.80$. Then she adds the 6% tax, which means multiplying by 1.06.
  Jill's Total = $90 \times 0.80 \times 1.06$

Because multiplication is commutative (the order in which you multiply numbers does not change the product), Jack's total and Jill's total are mathematically identical.
Therefore, Jack's total minus Jill's total is exactly $0.

ANSWER 9: C

**Problem 10:**
We are asked to find the sum of the following fractions:
$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$

Since all the fractions share a common denominator of 10, we just need to add their numerators:
Sum of numerators = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55

The sum of the digits 1 through 9 is 45.
Total numerator = 45 + 55 = 100

Now divide the total numerator by the denominator:
$\frac{100}{10} = 10$

ANSWER 10: D

**Problem 11:**
The ten-letter code "BEST OF LUCK" maps to the digits 0 through 9 in order:
B = 0
E = 1
S = 2
T = 3
O = 4
F = 5
L = 6
U = 7
C = 8
K = 9

We need to find the 4-digit number for "CLUE":
C = 8
L = 6
U = 7
E = 1

Putting the digits together, "CLUE" represents 8671.

ANSWER 11: A

**Problem 12:**
To find the correct sunset time, we need to add the length of daylight to the sunrise time.
Sunrise: 6:57 AM
Length of daylight: 10 hours and 24 minutes

First, add the 10 hours to the sunrise time:
6:57 AM + 10 hours = 4:57 PM

Next, add the remaining 24 minutes to 4:57 PM:
Adding 3 minutes brings the time to 5:00 PM.
We have 21 minutes left to add (24 - 3 = 21).
5:00 PM + 21 minutes = 5:21 PM.

ANSWER 12: B

Problem 1:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

To find the total number of pages in the book, we need to add the number of pages Isabella read each day.

First, let's calculate the total number of pages she read in the first three days: $36 \times 3 = 108$ pages.
Next, we calculate the total number of pages she read in the next three days: $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.

The total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low since $108 + 132 + 10 = 250$.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 1: B

Problem 2:
What is the smallest possible average of four distinct positive even integers?

To minimize the average of four distinct positive even integers, we need to choose the smallest four distinct positive even integers.

The smallest four distinct positive even integers are 2, 4, 6, and 8.

Let's calculate their average: $\frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$.

Let's examine the choices:
A. 3: Too low since the smallest even number is 2.
B. 4: Too low since the average of 2, 4, 6, and 8 is 5.
C. 5: Matches our calculation.
D. 6: Too high.
E. 7: Too high.

ANSWER 2: C

Problem 3:
Four students take an exam. Three of their scores are 70, 80, and 90. If the average of their four scores is 70, then what is the remaining score?

To find the remaining score, we first need to find the total score of all four students.

The average score is 70, and there are 4 students, so the total score is $70 \times 4 = 280$.

The sum of the three known scores is $70 + 80 + 90 = 240$.

The remaining score is $280 - 240 = 40$.

Let's examine the choices:
A. 40: Matches our calculation.
B. 50: Too high.
C. 55: Too high.
D. 60: Too high.
E. 70: Too high.

ANSWER 3: A

Problem 4:
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

Let's denote the side length of the square as $s$ and the radius of the circle as $r$.

The area of the square is $s^2$, and the area of the circle is $\pi r^2$.

Since they have the same area, we have $s^2 = \pi r^2$.

To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$.

Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$.

Let's examine the choices:
A. $\sqrt{\pi} / 2$: Too low.
B. $\sqrt{\pi}$: Matches our calculation.
C. $\pi$: Too high.
D. $2\pi$: Too high.
E. $\pi^2$: Too high.

ANSWER 4: B

Problem 5:
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Let's denote the smallest of these integers as $n$. Then the six consecutive integers are $n$, $n+1$, $n+2$, $n+3$, $n+4$, and $n+5$.

Their sum is $n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15 = 2013$.

Solving for $n$, we get $6n = 2013 - 15 = 1998$, so $n = \frac{1998}{6} = 333$.

The largest of these six integers is $n + 5 = 333 + 5 = 338$.

Let's examine the choices:
A. 335: Too low.
B. 338: Matches our calculation.
C. 340: Too high.
D. 345: Too high.
E. 350: Too high.

ANSWER 5: B

Problem 6:
On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

A standard checkerboard is an $8 \times 8$ grid, so it has 64 unit squares.

The unit squares that touch the outer edge are those on the first and last rows and columns. There are $8 + 8 + 8 + 8 = 32$ such squares, but the 4 corner squares are counted twice, so we subtract 4 to avoid double-counting: $32 - 4 = 28$.

Thus, there are $64 - 28 = 36$ unit squares that do not touch the outer edge.

The probability that a randomly chosen unit square does not touch the outer edge is $\frac{36}{64} = \frac{9}{16}$.

Let's examine the choices:
A. $\frac{1}{16}$: Too low.
B. $\frac{7}{16}$: Too low.
C. $\frac{1}{2}$: Too low since $\frac{9}{16} > \frac{1}{2}$.
D. $\frac{9}{16}$: Matches our calculation.
E. $\frac{49}{64}$: Too high.

ANSWER 6: D

Problem 7:
What is the sum of the mean, median, and mode of the numbers 2, 3, 0, 3, 1, 4, 0, 3?

First, let's arrange the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.

The mean is the average of the numbers: $\frac{0 + 0 + 1 + 2 + 3 + 3 + 3 + 4}{8} = \frac{16}{8} = 2$.

The median is the average of the 4th and 5th numbers (since there are 8 numbers): $\frac{2 + 3}{2} = 2.5$.

The mode is the number that appears most frequently, which is 3.

The sum of the mean, median, and mode is $2 + 2.5 + 3 = 7.5$.

Let's examine the choices:
A. 6.5: Too low.
B. 7: Too low.
C. 7.5: Matches our calculation.
D. 8.5: Too high.
E. 9: Too high.

ANSWER 7: C

Problem 8:
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

Georgie has 6 choices for the window to enter.

After entering, Georgie has 5 remaining choices for the window to leave (since it must be different from the entry window).

The total number of ways Georgie can enter and leave is $6 \times 5 = 30$.

Let's examine the choices:
A. 12: Too low.
B. 15: Too low.
C. 18: Too low.
D. 30: Matches our calculation.
E. 36: Too high.

ANSWER 8: D

Problem 9:
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?

Let's calculate Jack's total:
1. $90.00 + 6\%$ of $90.00 = 90 + 0.06 \times 90 = 90 + 5.40 = 95.40$.
2. $20\%$ 

Problem 1:

Isabella read the book over 7 days.

- First 3 days:  
  \[
  3 \times 36 = 108
  \]
  pages

- Next 3 days:  
  \[
  3 \times 44 = 132
  \]
  pages

- Last day:  
  \[
  10
  \]
  pages

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 1: B

Problem 2:

We want the smallest possible average of four distinct positive even integers.

The smallest positive even integers are:

\[
2,\ 4,\ 6,\ 8
\]

Their average is:

\[
\frac{2+4+6+8}{4}=\frac{20}{4}=5
\]

We cannot use 0 because the integers must be positive. We also cannot repeat numbers because they must be distinct.

ANSWER 2: C

Problem 3:

The average of 4 scores is 70, so the total of the 4 scores is:

\[
4 \times 70 = 280
\]

The three known scores add to:

\[
70+80+90=240
\]

So the remaining score is:

\[
280-240=40
\]

ANSWER 3: A

Problem 4:

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The area of the square is:

\[
s^2
\]

The area of the circle is:

\[
\pi r^2
\]

They have the same area, so:

\[
s^2=\pi r^2
\]

Taking the square root of both sides:

\[
s=r\sqrt{\pi}
\]

Therefore,

\[
\frac{s}{r}=\sqrt{\pi}
\]

So the ratio of the side length of the square to the radius of the circle is \(\sqrt{\pi}\).

ANSWER 4: B

Problem 5:

Let the six consecutive positive integers be:

\[
n,\ n+1,\ n+2,\ n+3,\ n+4,\ n+5
\]

Their sum is 2013:

\[
n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=2013
\]

Combine like terms:

\[
6n+15=2013
\]

Subtract 15:

\[
6n=1998
\]

Divide by 6:

\[
n=333
\]

So the six integers are:

\[
333,\ 334,\ 335,\ 336,\ 337,\ 338
\]

The largest is 338.

ANSWER 5: B

Problem 6:

A checkerboard has 64 unit squares, so it is an \(8 \times 8\) board.

A square does not touch the outer edge only if it is not in the first row, last row, first column, or last column.

That leaves the interior squares:

\[
6 \times 6 = 36
\]

So the probability is:

\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:

The numbers are:

\[
2,\ 3,\ 0,\ 3,\ 1,\ 4,\ 0,\ 3
\]

First find the mean.

Sum:

\[
2+3+0+3+1+4+0+3=16
\]

There are 8 numbers, so the mean is:

\[
\frac{16}{8}=2
\]

Now sort the numbers:

\[
0,\ 0,\ 1,\ 2,\ 3,\ 3,\ 3,\ 4
\]

There are 8 numbers, so the median is the average of the 4th and 5th numbers:

\[
\frac{2+3}{2}=2.5
\]

The mode is the number that appears most often. The number 3 appears three times, so the mode is:

\[
3
\]

Now add mean, median, and mode:

\[
2+2.5+3=7.5
\]

ANSWER 7: C

Problem 8:

Georgie has 6 choices for the window he enters.

After entering, he must leave by a different window, so he has 5 choices for the window he leaves.

Thus the number of ways is:

\[
6 \times 5 = 30
\]

This is not 15 because entering window A and leaving window B is different from entering window B and leaving window A.

ANSWER 8: D

Problem 9:

Jack’s method:

He first adds 6% sales tax to \$90:

\[
90 \times 1.06 = 95.40
\]

Then he subtracts 20% from that total, meaning he keeps 80%:

\[
95.40 \times 0.80 = 76.32
\]

So Jack’s total is \$76.32.

Jill’s method:

She first subtracts 20% from \$90:

\[
90 \times 0.80 = 72
\]

Then she adds 6% sales tax:

\[
72 \times 1.06 = 76.32
\]

So Jill’s total is also \$76.32.

Therefore:

\[
76.32-76.32=0
\]

ANSWER 9: C

Problem 10:

The expression is:

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}+\frac{55}{10}
\]

Since all fractions have denominator 10, add the numerators:

\[
\frac{1+2+3+4+5+6+7+8+9+55}{10}
\]

Now,

\[
1+2+3+4+5+6+7+8+9=45
\]

So:

\[
\frac{45+55}{10}=\frac{100}{10}=10
\]

ANSWER 10: D

Problem 11:

The code phrase is:

\[
\text{BEST OF LUCK}
\]

The ten letters represent the digits 0 through 9 in order:

\[
B=0,\ E=1,\ S=2,\ T=3,\ O=4,\ F=5,\ L=6,\ U=7,\ C=8,\ K=9
\]

Now decode the word CLUE:

\[
C=8,\quad L=6,\quad U=7,\quad E=1
\]

So CLUE represents:

\[
8671
\]

ANSWER 11: A

Problem 12:

The sunrise was 6:57 AM, and the length of daylight was 10 hours and 24 minutes.

So the sunset time is:

\[
6:57\text{ AM} + 10\text{ hours }24\text{ minutes}
\]

First add 10 hours:

\[
6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}
\]

Now add 24 minutes:

\[
4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}
\]

So the sun really set at 5:21 PM.

ANSWER 12: B

**Problem 1:**
To find the total number of pages in the book, we need to calculate the number of pages Isabella read in each part of the week and add them together.
- For the first 3 days, she read an average of 36 pages per day: 
  3 days × 36 pages/day = 108 pages.
- For the next 3 days, she read an average of 44 pages per day: 
  3 days × 44 pages/day = 132 pages.
- On the last day (the 7th day), she read 10 pages.

Adding these amounts together gives the total number of pages:
108 + 132 + 10 = 250 pages.

ANSWER 1: B

**Problem 2:**
To find the smallest possible average of four distinct positive even integers, we must first select the four smallest distinct positive even integers. 
These integers are 2, 4, 6, and 8.
Next, we find their sum:
2 + 4 + 6 + 8 = 20
Finally, we divide the sum by the number of integers (4) to find the average:
20 / 4 = 5

ANSWER 2: C

**Problem 3:**
Let the unknown remaining score be $x$.
The average of the four scores is given as 70. The formula for the average is the sum of the scores divided by the number of scores:
(70 + 80 + 90 + $x$) / 4 = 70

Multiply both sides by 4 to find the total sum of the scores:
70 + 80 + 90 + $x$ = 280
240 + $x$ = 280

Subtract 240 from both sides:
$x$ = 40
The remaining score is 40.

ANSWER 3: A

**Problem 4:**
Let the side length of the square be $s$ and the radius of the circle be $r$.
The area of the square is $s^2$, and the area of the circle is $\pi r^2$.
We are told that their areas are equal:
$s^2 = \pi r^2$

To find the ratio of the side length to the radius ($s/r$), we take the square root of both sides:
$s = r\sqrt{\pi}$
$s / r = \sqrt{\pi}$

ANSWER 4: B

**Problem 5:**
Let the six consecutive positive integers be $x$, $x+1$, $x+2$, $x+3$, $x+4$, and $x+5$.
Their sum is:
$x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 2013$
$6x + 15 = 2013$

Subtract 15 from both sides:
$6x = 1998$

Divide by 6:
$x = 333$

The problem asks for the largest of these integers, which is $x+5$:
333 + 5 = 338

ANSWER 5: B

**Problem 6:**
A standard checkerboard is an $8 \times 8$ grid, meaning it has a total of 64 unit squares.
The squares that touch the outer edge form the border of the board. If we remove this 1-square-thick border, the remaining inner squares form a smaller $6 \times 6$ grid.
The number of squares that do *not* touch the outer edge is:
$6 \times 6 = 36$

The probability of randomly choosing one of these inner squares is the number of inner squares divided by the total number of squares:
$P = 36 / 64$

Dividing the numerator and the denominator by 4 simplifies the fraction:
$P = 9 / 16$

ANSWER 6: D

**Problem 7:**
First, let's list the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.
There are 8 numbers in total.

- **Mean:** The sum of the numbers divided by the count.
  Sum = 0 + 0 + 1 + 2 + 3 + 3 + 3 + 4 = 16
  Mean = 16 / 8 = 2

- **Median:** The middle value. Since there is an even number of values (8), the median is the average of the 4th and 5th numbers.
  The 4th number is 2, and the 5th number is 3.
  Median = (2 + 3) / 2 = 2.5

- **Mode:** The number that appears most frequently.
  The number 3 appears three times, which is more than any other number.
  Mode = 3

The sum of the mean, median, and mode is:
2 + 2.5 + 3 = 7.5

ANSWER 7: C

**Problem 8:**
Georgie the Ghost has 6 different windows to choose from when entering the house.
Because he must leave by a *different* window, he cannot use the window he entered through. This leaves him with 5 choices for his exit.
By the fundamental counting principle, the total number of ways he can enter and leave is:
6 (entry choices) × 5 (exit choices) = 30 ways.

ANSWER 8: D

**Problem 9:**
Let's calculate the final total for both clerks.
- **Jack's method:** He adds the 6% tax first, making the subtotal $90 \times 1.06$. Then he applies the 20% discount, which means the customer pays 80% of the subtotal.
  Jack's Total = $90 \times 1.06 \times 0.80$

- **Jill's method:** She applies the 20% discount first, making the subtotal $90 \times 0.80$. Then she adds the 6% tax, which means multiplying by 1.06.
  Jill's Total = $90 \times 0.80 \times 1.06$

Because multiplication is commutative (the order in which you multiply numbers does not change the product), Jack's total and Jill's total are mathematically identical.
Therefore, Jack's total minus Jill's total is exactly $0.

ANSWER 9: C

**Problem 10:**
We are asked to find the sum of the following fractions:
$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$

Since all the fractions share a common denominator of 10, we just need to add their numerators:
Sum of numerators = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55

The sum of the digits 1 through 9 is 45.
Total numerator = 45 + 55 = 100

Now divide the total numerator by the denominator:
$\frac{100}{10} = 10$

ANSWER 10: D

**Problem 11:**
The ten-letter code "BEST OF LUCK" maps to the digits 0 through 9 in order:
B = 0
E = 1
S = 2
T = 3
O = 4
F = 5
L = 6
U = 7
C = 8
K = 9

We need to find the 4-digit number for "CLUE":
C = 8
L = 6
U = 7
E = 1

Putting the digits together, "CLUE" represents 8671.

ANSWER 11: A

**Problem 12:**
To find the correct sunset time, we need to add the length of daylight to the sunrise time.
Sunrise: 6:57 AM
Length of daylight: 10 hours and 24 minutes

First, add the 10 hours to the sunrise time:
6:57 AM + 10 hours = 4:57 PM

Next, add the remaining 24 minutes to 4:57 PM:
Adding 3 minutes brings the time to 5:00 PM.
We have 21 minutes left to add (24 - 3 = 21).
5:00 PM + 21 minutes = 5:21 PM.

ANSWER 12: B

Problem 1:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

To find the total number of pages in the book, we need to add the number of pages Isabella read each day.

First, let's calculate the total number of pages she read in the first three days: $36 \times 3 = 108$ pages.
Next, we calculate the total number of pages she read in the next three days: $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.

The total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low since $108 + 132 + 10 = 250$.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 1: B

Problem 2:
What is the smallest possible average of four distinct positive even integers?

To minimize the average of four distinct positive even integers, we need to choose the smallest four distinct positive even integers.

The smallest four distinct positive even integers are 2, 4, 6, and 8.

Let's calculate their average: $\frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$.

Let's examine the choices:
A. 3: Too low since the smallest even number is 2.
B. 4: Too low since the average of 2, 4, 6, and 8 is 5.
C. 5: Matches our calculation.
D. 6: Too high.
E. 7: Too high.

ANSWER 2: C

Problem 3:
Four students take an exam. Three of their scores are 70, 80, and 90. If the average of their four scores is 70, then what is the remaining score?

To find the remaining score, we first need to find the total score of all four students.

The average score is 70, and there are 4 students, so the total score is $70 \times 4 = 280$.

The sum of the three known scores is $70 + 80 + 90 = 240$.

The remaining score is $280 - 240 = 40$.

Let's examine the choices:
A. 40: Matches our calculation.
B. 50: Too high.
C. 55: Too high.
D. 60: Too high.
E. 70: Too high.

ANSWER 3: A

Problem 4:
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

Let's denote the side length of the square as $s$ and the radius of the circle as $r$.

The area of the square is $s^2$, and the area of the circle is $\pi r^2$.

Since they have the same area, we have $s^2 = \pi r^2$.

To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$.

Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$.

Let's examine the choices:
A. $\sqrt{\pi} / 2$: Too low.
B. $\sqrt{\pi}$: Matches our calculation.
C. $\pi$: Too high.
D. $2\pi$: Too high.
E. $\pi^2$: Too high.

ANSWER 4: B

Problem 5:
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Let's denote the smallest of these integers as $n$. Then the six consecutive integers are $n$, $n+1$, $n+2$, $n+3$, $n+4$, and $n+5$.

Their sum is $n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15 = 2013$.

Solving for $n$, we get $6n = 2013 - 15 = 1998$, so $n = \frac{1998}{6} = 333$.

The largest of these six integers is $n + 5 = 333 + 5 = 338$.

Let's examine the choices:
A. 335: Too low.
B. 338: Matches our calculation.
C. 340: Too high.
D. 345: Too high.
E. 350: Too high.

ANSWER 5: B

Problem 6:
On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

A standard checkerboard is an $8 \times 8$ grid, so it has 64 unit squares.

The unit squares that touch the outer edge are those on the first and last rows and columns. There are $8 + 8 + 8 + 8 = 32$ such squares, but the 4 corner squares are counted twice, so we subtract 4 to avoid double-counting: $32 - 4 = 28$.

Thus, there are $64 - 28 = 36$ unit squares that do not touch the outer edge.

The probability that a randomly chosen unit square does not touch the outer edge is $\frac{36}{64} = \frac{9}{16}$.

Let's examine the choices:
A. $\frac{1}{16}$: Too low.
B. $\frac{7}{16}$: Too low.
C. $\frac{1}{2}$: Too low since $\frac{9}{16} > \frac{1}{2}$.
D. $\frac{9}{16}$: Matches our calculation.
E. $\frac{49}{64}$: Too high.

ANSWER 6: D

Problem 7:
What is the sum of the mean, median, and mode of the numbers 2, 3, 0, 3, 1, 4, 0, 3?

First, let's arrange the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.

The mean is the average of the numbers: $\frac{0 + 0 + 1 + 2 + 3 + 3 + 3 + 4}{8} = \frac{16}{8} = 2$.

The median is the average of the 4th and 5th numbers (since there are 8 numbers): $\frac{2 + 3}{2} = 2.5$.

The mode is the number that appears most frequently, which is 3.

The sum of the mean, median, and mode is $2 + 2.5 + 3 = 7.5$.

Let's examine the choices:
A. 6.5: Too low.
B. 7: Too low.
C. 7.5: Matches our calculation.
D. 8.5: Too high.
E. 9: Too high.

ANSWER 7: C

Problem 8:
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

Georgie has 6 choices for the window to enter.

After entering, Georgie has 5 remaining choices for the window to leave (since it must be different from the entry window).

The total number of ways Georgie can enter and leave is $6 \times 5 = 30$.

Let's examine the choices:
A. 12: Too low.
B. 15: Too low.
C. 18: Too low.
D. 30: Matches our calculation.
E. 36: Too high.

ANSWER 8: D

Problem 9:
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?

Let's calculate Jack's total:
1. $90.00 + 6\%$ of $90.00 = 90 + 0.06 \times 90 = 90 + 5.40 = 95.40$.
2. $20\%$ 

Problem 1:

Isabella read the book over 7 days.

- First 3 days:  
  \[
  3 \times 36 = 108
  \]
  pages

- Next 3 days:  
  \[
  3 \times 44 = 132
  \]
  pages

- Last day:  
  \[
  10
  \]
  pages

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 1: B

Problem 2:

We want the smallest possible average of four distinct positive even integers.

The smallest positive even integers are:

\[
2,\ 4,\ 6,\ 8
\]

Their average is:

\[
\frac{2+4+6+8}{4}=\frac{20}{4}=5
\]

We cannot use 0 because the integers must be positive. We also cannot repeat numbers because they must be distinct.

ANSWER 2: C

Problem 3:

The average of 4 scores is 70, so the total of the 4 scores is:

\[
4 \times 70 = 280
\]

The three known scores add to:

\[
70+80+90=240
\]

So the remaining score is:

\[
280-240=40
\]

ANSWER 3: A

Problem 4:

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The area of the square is:

\[
s^2
\]

The area of the circle is:

\[
\pi r^2
\]

They have the same area, so:

\[
s^2=\pi r^2
\]

Taking the square root of both sides:

\[
s=r\sqrt{\pi}
\]

Therefore,

\[
\frac{s}{r}=\sqrt{\pi}
\]

So the ratio of the side length of the square to the radius of the circle is \(\sqrt{\pi}\).

ANSWER 4: B

Problem 5:

Let the six consecutive positive integers be:

\[
n,\ n+1,\ n+2,\ n+3,\ n+4,\ n+5
\]

Their sum is 2013:

\[
n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=2013
\]

Combine like terms:

\[
6n+15=2013
\]

Subtract 15:

\[
6n=1998
\]

Divide by 6:

\[
n=333
\]

So the six integers are:

\[
333,\ 334,\ 335,\ 336,\ 337,\ 338
\]

The largest is 338.

ANSWER 5: B

Problem 6:

A checkerboard has 64 unit squares, so it is an \(8 \times 8\) board.

A square does not touch the outer edge only if it is not in the first row, last row, first column, or last column.

That leaves the interior squares:

\[
6 \times 6 = 36
\]

So the probability is:

\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:

The numbers are:

\[
2,\ 3,\ 0,\ 3,\ 1,\ 4,\ 0,\ 3
\]

First find the mean.

Sum:

\[
2+3+0+3+1+4+0+3=16
\]

There are 8 numbers, so the mean is:

\[
\frac{16}{8}=2
\]

Now sort the numbers:

\[
0,\ 0,\ 1,\ 2,\ 3,\ 3,\ 3,\ 4
\]

There are 8 numbers, so the median is the average of the 4th and 5th numbers:

\[
\frac{2+3}{2}=2.5
\]

The mode is the number that appears most often. The number 3 appears three times, so the mode is:

\[
3
\]

Now add mean, median, and mode:

\[
2+2.5+3=7.5
\]

ANSWER 7: C

Problem 8:

Georgie has 6 choices for the window he enters.

After entering, he must leave by a different window, so he has 5 choices for the window he leaves.

Thus the number of ways is:

\[
6 \times 5 = 30
\]

This is not 15 because entering window A and leaving window B is different from entering window B and leaving window A.

ANSWER 8: D

Problem 9:

Jack’s method:

He first adds 6% sales tax to \$90:

\[
90 \times 1.06 = 95.40
\]

Then he subtracts 20% from that total, meaning he keeps 80%:

\[
95.40 \times 0.80 = 76.32
\]

So Jack’s total is \$76.32.

Jill’s method:

She first subtracts 20% from \$90:

\[
90 \times 0.80 = 72
\]

Then she adds 6% sales tax:

\[
72 \times 1.06 = 76.32
\]

So Jill’s total is also \$76.32.

Therefore:

\[
76.32-76.32=0
\]

ANSWER 9: C

Problem 10:

The expression is:

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}+\frac{55}{10}
\]

Since all fractions have denominator 10, add the numerators:

\[
\frac{1+2+3+4+5+6+7+8+9+55}{10}
\]

Now,

\[
1+2+3+4+5+6+7+8+9=45
\]

So:

\[
\frac{45+55}{10}=\frac{100}{10}=10
\]

ANSWER 10: D

Problem 11:

The code phrase is:

\[
\text{BEST OF LUCK}
\]

The ten letters represent the digits 0 through 9 in order:

\[
B=0,\ E=1,\ S=2,\ T=3,\ O=4,\ F=5,\ L=6,\ U=7,\ C=8,\ K=9
\]

Now decode the word CLUE:

\[
C=8,\quad L=6,\quad U=7,\quad E=1
\]

So CLUE represents:

\[
8671
\]

ANSWER 11: A

Problem 12:

The sunrise was 6:57 AM, and the length of daylight was 10 hours and 24 minutes.

So the sunset time is:

\[
6:57\text{ AM} + 10\text{ hours }24\text{ minutes}
\]

First add 10 hours:

\[
6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}
\]

Now add 24 minutes:

\[
4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}
\]

So the sun really set at 5:21 PM.

ANSWER 12: B

**Problem 1:**
To find the total number of pages in the book, we need to calculate the number of pages Isabella read in each part of the week and add them together.
- For the first 3 days, she read an average of 36 pages per day: 
  3 days × 36 pages/day = 108 pages.
- For the next 3 days, she read an average of 44 pages per day: 
  3 days × 44 pages/day = 132 pages.
- On the last day (the 7th day), she read 10 pages.

Adding these amounts together gives the total number of pages:
108 + 132 + 10 = 250 pages.

ANSWER 1: B

**Problem 2:**
To find the smallest possible average of four distinct positive even integers, we must first select the four smallest distinct positive even integers. 
These integers are 2, 4, 6, and 8.
Next, we find their sum:
2 + 4 + 6 + 8 = 20
Finally, we divide the sum by the number of integers (4) to find the average:
20 / 4 = 5

ANSWER 2: C

**Problem 3:**
Let the unknown remaining score be $x$.
The average of the four scores is given as 70. The formula for the average is the sum of the scores divided by the number of scores:
(70 + 80 + 90 + $x$) / 4 = 70

Multiply both sides by 4 to find the total sum of the scores:
70 + 80 + 90 + $x$ = 280
240 + $x$ = 280

Subtract 240 from both sides:
$x$ = 40
The remaining score is 40.

ANSWER 3: A

**Problem 4:**
Let the side length of the square be $s$ and the radius of the circle be $r$.
The area of the square is $s^2$, and the area of the circle is $\pi r^2$.
We are told that their areas are equal:
$s^2 = \pi r^2$

To find the ratio of the side length to the radius ($s/r$), we take the square root of both sides:
$s = r\sqrt{\pi}$
$s / r = \sqrt{\pi}$

ANSWER 4: B

**Problem 5:**
Let the six consecutive positive integers be $x$, $x+1$, $x+2$, $x+3$, $x+4$, and $x+5$.
Their sum is:
$x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 2013$
$6x + 15 = 2013$

Subtract 15 from both sides:
$6x = 1998$

Divide by 6:
$x = 333$

The problem asks for the largest of these integers, which is $x+5$:
333 + 5 = 338

ANSWER 5: B

**Problem 6:**
A standard checkerboard is an $8 \times 8$ grid, meaning it has a total of 64 unit squares.
The squares that touch the outer edge form the border of the board. If we remove this 1-square-thick border, the remaining inner squares form a smaller $6 \times 6$ grid.
The number of squares that do *not* touch the outer edge is:
$6 \times 6 = 36$

The probability of randomly choosing one of these inner squares is the number of inner squares divided by the total number of squares:
$P = 36 / 64$

Dividing the numerator and the denominator by 4 simplifies the fraction:
$P = 9 / 16$

ANSWER 6: D

**Problem 7:**
First, let's list the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.
There are 8 numbers in total.

- **Mean:** The sum of the numbers divided by the count.
  Sum = 0 + 0 + 1 + 2 + 3 + 3 + 3 + 4 = 16
  Mean = 16 / 8 = 2

- **Median:** The middle value. Since there is an even number of values (8), the median is the average of the 4th and 5th numbers.
  The 4th number is 2, and the 5th number is 3.
  Median = (2 + 3) / 2 = 2.5

- **Mode:** The number that appears most frequently.
  The number 3 appears three times, which is more than any other number.
  Mode = 3

The sum of the mean, median, and mode is:
2 + 2.5 + 3 = 7.5

ANSWER 7: C

**Problem 8:**
Georgie the Ghost has 6 different windows to choose from when entering the house.
Because he must leave by a *different* window, he cannot use the window he entered through. This leaves him with 5 choices for his exit.
By the fundamental counting principle, the total number of ways he can enter and leave is:
6 (entry choices) × 5 (exit choices) = 30 ways.

ANSWER 8: D

**Problem 9:**
Let's calculate the final total for both clerks.
- **Jack's method:** He adds the 6% tax first, making the subtotal $90 \times 1.06$. Then he applies the 20% discount, which means the customer pays 80% of the subtotal.
  Jack's Total = $90 \times 1.06 \times 0.80$

- **Jill's method:** She applies the 20% discount first, making the subtotal $90 \times 0.80$. Then she adds the 6% tax, which means multiplying by 1.06.
  Jill's Total = $90 \times 0.80 \times 1.06$

Because multiplication is commutative (the order in which you multiply numbers does not change the product), Jack's total and Jill's total are mathematically identical.
Therefore, Jack's total minus Jill's total is exactly $0.

ANSWER 9: C

**Problem 10:**
We are asked to find the sum of the following fractions:
$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$

Since all the fractions share a common denominator of 10, we just need to add their numerators:
Sum of numerators = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55

The sum of the digits 1 through 9 is 45.
Total numerator = 45 + 55 = 100

Now divide the total numerator by the denominator:
$\frac{100}{10} = 10$

ANSWER 10: D

**Problem 11:**
The ten-letter code "BEST OF LUCK" maps to the digits 0 through 9 in order:
B = 0
E = 1
S = 2
T = 3
O = 4
F = 5
L = 6
U = 7
C = 8
K = 9

We need to find the 4-digit number for "CLUE":
C = 8
L = 6
U = 7
E = 1

Putting the digits together, "CLUE" represents 8671.

ANSWER 11: A

**Problem 12:**
To find the correct sunset time, we need to add the length of daylight to the sunrise time.
Sunrise: 6:57 AM
Length of daylight: 10 hours and 24 minutes

First, add the 10 hours to the sunrise time:
6:57 AM + 10 hours = 4:57 PM

Next, add the remaining 24 minutes to 4:57 PM:
Adding 3 minutes brings the time to 5:00 PM.
We have 21 minutes left to add (24 - 3 = 21).
5:00 PM + 21 minutes = 5:21 PM.

ANSWER 12: B

Problem 1:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

To find the total number of pages in the book, we need to add the number of pages Isabella read each day.

First, let's calculate the total number of pages she read in the first three days: $36 \times 3 = 108$ pages.
Next, we calculate the total number of pages she read in the next three days: $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.

The total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low since $108 + 132 + 10 = 250$.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 1: B

Problem 2:
What is the smallest possible average of four distinct positive even integers?

To minimize the average of four distinct positive even integers, we need to choose the smallest four distinct positive even integers.

The smallest four distinct positive even integers are 2, 4, 6, and 8.

Let's calculate their average: $\frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$.

Let's examine the choices:
A. 3: Too low since the smallest even number is 2.
B. 4: Too low since the average of 2, 4, 6, and 8 is 5.
C. 5: Matches our calculation.
D. 6: Too high.
E. 7: Too high.

ANSWER 2: C

Problem 3:
Four students take an exam. Three of their scores are 70, 80, and 90. If the average of their four scores is 70, then what is the remaining score?

To find the remaining score, we first need to find the total score of all four students.

The average score is 70, and there are 4 students, so the total score is $70 \times 4 = 280$.

The sum of the three known scores is $70 + 80 + 90 = 240$.

The remaining score is $280 - 240 = 40$.

Let's examine the choices:
A. 40: Matches our calculation.
B. 50: Too high.
C. 55: Too high.
D. 60: Too high.
E. 70: Too high.

ANSWER 3: A

Problem 4:
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

Let's denote the side length of the square as $s$ and the radius of the circle as $r$.

The area of the square is $s^2$, and the area of the circle is $\pi r^2$.

Since they have the same area, we have $s^2 = \pi r^2$.

To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$.

Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$.

Let's examine the choices:
A. $\sqrt{\pi} / 2$: Too low.
B. $\sqrt{\pi}$: Matches our calculation.
C. $\pi$: Too high.
D. $2\pi$: Too high.
E. $\pi^2$: Too high.

ANSWER 4: B

Problem 5:
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Let's denote the smallest of these integers as $n$. Then the six consecutive integers are $n$, $n+1$, $n+2$, $n+3$, $n+4$, and $n+5$.

Their sum is $n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15 = 2013$.

Solving for $n$, we get $6n = 2013 - 15 = 1998$, so $n = \frac{1998}{6} = 333$.

The largest of these six integers is $n + 5 = 333 + 5 = 338$.

Let's examine the choices:
A. 335: Too low.
B. 338: Matches our calculation.
C. 340: Too high.
D. 345: Too high.
E. 350: Too high.

ANSWER 5: B

Problem 6:
On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

A standard checkerboard is an $8 \times 8$ grid, so it has 64 unit squares.

The unit squares that touch the outer edge are those on the first and last rows and columns. There are $8 + 8 + 8 + 8 = 32$ such squares, but the 4 corner squares are counted twice, so we subtract 4 to avoid double-counting: $32 - 4 = 28$.

Thus, there are $64 - 28 = 36$ unit squares that do not touch the outer edge.

The probability that a randomly chosen unit square does not touch the outer edge is $\frac{36}{64} = \frac{9}{16}$.

Let's examine the choices:
A. $\frac{1}{16}$: Too low.
B. $\frac{7}{16}$: Too low.
C. $\frac{1}{2}$: Too low since $\frac{9}{16} > \frac{1}{2}$.
D. $\frac{9}{16}$: Matches our calculation.
E. $\frac{49}{64}$: Too high.

ANSWER 6: D

Problem 7:
What is the sum of the mean, median, and mode of the numbers 2, 3, 0, 3, 1, 4, 0, 3?

First, let's arrange the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.

The mean is the average of the numbers: $\frac{0 + 0 + 1 + 2 + 3 + 3 + 3 + 4}{8} = \frac{16}{8} = 2$.

The median is the average of the 4th and 5th numbers (since there are 8 numbers): $\frac{2 + 3}{2} = 2.5$.

The mode is the number that appears most frequently, which is 3.

The sum of the mean, median, and mode is $2 + 2.5 + 3 = 7.5$.

Let's examine the choices:
A. 6.5: Too low.
B. 7: Too low.
C. 7.5: Matches our calculation.
D. 8.5: Too high.
E. 9: Too high.

ANSWER 7: C

Problem 8:
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

Georgie has 6 choices for the window to enter.

After entering, Georgie has 5 remaining choices for the window to leave (since it must be different from the entry window).

The total number of ways Georgie can enter and leave is $6 \times 5 = 30$.

Let's examine the choices:
A. 12: Too low.
B. 15: Too low.
C. 18: Too low.
D. 30: Matches our calculation.
E. 36: Too high.

ANSWER 8: D

Problem 9:
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?

Let's calculate Jack's total:
1. $90.00 + 6\%$ of $90.00 = 90 + 0.06 \times 90 = 90 + 5.40 = 95.40$.
2. $20\%$ 

Problem 1:

Isabella read the book over 7 days.

- First 3 days:  
  \[
  3 \times 36 = 108
  \]
  pages

- Next 3 days:  
  \[
  3 \times 44 = 132
  \]
  pages

- Last day:  
  \[
  10
  \]
  pages

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 1: B

Problem 2:

We want the smallest possible average of four distinct positive even integers.

The smallest positive even integers are:

\[
2,\ 4,\ 6,\ 8
\]

Their average is:

\[
\frac{2+4+6+8}{4}=\frac{20}{4}=5
\]

We cannot use 0 because the integers must be positive. We also cannot repeat numbers because they must be distinct.

ANSWER 2: C

Problem 3:

The average of 4 scores is 70, so the total of the 4 scores is:

\[
4 \times 70 = 280
\]

The three known scores add to:

\[
70+80+90=240
\]

So the remaining score is:

\[
280-240=40
\]

ANSWER 3: A

Problem 4:

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The area of the square is:

\[
s^2
\]

The area of the circle is:

\[
\pi r^2
\]

They have the same area, so:

\[
s^2=\pi r^2
\]

Taking the square root of both sides:

\[
s=r\sqrt{\pi}
\]

Therefore,

\[
\frac{s}{r}=\sqrt{\pi}
\]

So the ratio of the side length of the square to the radius of the circle is \(\sqrt{\pi}\).

ANSWER 4: B

Problem 5:

Let the six consecutive positive integers be:

\[
n,\ n+1,\ n+2,\ n+3,\ n+4,\ n+5
\]

Their sum is 2013:

\[
n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=2013
\]

Combine like terms:

\[
6n+15=2013
\]

Subtract 15:

\[
6n=1998
\]

Divide by 6:

\[
n=333
\]

So the six integers are:

\[
333,\ 334,\ 335,\ 336,\ 337,\ 338
\]

The largest is 338.

ANSWER 5: B

Problem 6:

A checkerboard has 64 unit squares, so it is an \(8 \times 8\) board.

A square does not touch the outer edge only if it is not in the first row, last row, first column, or last column.

That leaves the interior squares:

\[
6 \times 6 = 36
\]

So the probability is:

\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:

The numbers are:

\[
2,\ 3,\ 0,\ 3,\ 1,\ 4,\ 0,\ 3
\]

First find the mean.

Sum:

\[
2+3+0+3+1+4+0+3=16
\]

There are 8 numbers, so the mean is:

\[
\frac{16}{8}=2
\]

Now sort the numbers:

\[
0,\ 0,\ 1,\ 2,\ 3,\ 3,\ 3,\ 4
\]

There are 8 numbers, so the median is the average of the 4th and 5th numbers:

\[
\frac{2+3}{2}=2.5
\]

The mode is the number that appears most often. The number 3 appears three times, so the mode is:

\[
3
\]

Now add mean, median, and mode:

\[
2+2.5+3=7.5
\]

ANSWER 7: C

Problem 8:

Georgie has 6 choices for the window he enters.

After entering, he must leave by a different window, so he has 5 choices for the window he leaves.

Thus the number of ways is:

\[
6 \times 5 = 30
\]

This is not 15 because entering window A and leaving window B is different from entering window B and leaving window A.

ANSWER 8: D

Problem 9:

Jack’s method:

He first adds 6% sales tax to \$90:

\[
90 \times 1.06 = 95.40
\]

Then he subtracts 20% from that total, meaning he keeps 80%:

\[
95.40 \times 0.80 = 76.32
\]

So Jack’s total is \$76.32.

Jill’s method:

She first subtracts 20% from \$90:

\[
90 \times 0.80 = 72
\]

Then she adds 6% sales tax:

\[
72 \times 1.06 = 76.32
\]

So Jill’s total is also \$76.32.

Therefore:

\[
76.32-76.32=0
\]

ANSWER 9: C

Problem 10:

The expression is:

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}+\frac{55}{10}
\]

Since all fractions have denominator 10, add the numerators:

\[
\frac{1+2+3+4+5+6+7+8+9+55}{10}
\]

Now,

\[
1+2+3+4+5+6+7+8+9=45
\]

So:

\[
\frac{45+55}{10}=\frac{100}{10}=10
\]

ANSWER 10: D

Problem 11:

The code phrase is:

\[
\text{BEST OF LUCK}
\]

The ten letters represent the digits 0 through 9 in order:

\[
B=0,\ E=1,\ S=2,\ T=3,\ O=4,\ F=5,\ L=6,\ U=7,\ C=8,\ K=9
\]

Now decode the word CLUE:

\[
C=8,\quad L=6,\quad U=7,\quad E=1
\]

So CLUE represents:

\[
8671
\]

ANSWER 11: A

Problem 12:

The sunrise was 6:57 AM, and the length of daylight was 10 hours and 24 minutes.

So the sunset time is:

\[
6:57\text{ AM} + 10\text{ hours }24\text{ minutes}
\]

First add 10 hours:

\[
6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}
\]

Now add 24 minutes:

\[
4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}
\]

So the sun really set at 5:21 PM.

ANSWER 12: B

**Problem 1:**
To find the total number of pages in the book, we need to calculate the number of pages Isabella read in each part of the week and add them together.
- For the first 3 days, she read an average of 36 pages per day: 
  3 days × 36 pages/day = 108 pages.
- For the next 3 days, she read an average of 44 pages per day: 
  3 days × 44 pages/day = 132 pages.
- On the last day (the 7th day), she read 10 pages.

Adding these amounts together gives the total number of pages:
108 + 132 + 10 = 250 pages.

ANSWER 1: B

**Problem 2:**
To find the smallest possible average of four distinct positive even integers, we must first select the four smallest distinct positive even integers. 
These integers are 2, 4, 6, and 8.
Next, we find their sum:
2 + 4 + 6 + 8 = 20
Finally, we divide the sum by the number of integers (4) to find the average:
20 / 4 = 5

ANSWER 2: C

**Problem 3:**
Let the unknown remaining score be $x$.
The average of the four scores is given as 70. The formula for the average is the sum of the scores divided by the number of scores:
(70 + 80 + 90 + $x$) / 4 = 70

Multiply both sides by 4 to find the total sum of the scores:
70 + 80 + 90 + $x$ = 280
240 + $x$ = 280

Subtract 240 from both sides:
$x$ = 40
The remaining score is 40.

ANSWER 3: A

**Problem 4:**
Let the side length of the square be $s$ and the radius of the circle be $r$.
The area of the square is $s^2$, and the area of the circle is $\pi r^2$.
We are told that their areas are equal:
$s^2 = \pi r^2$

To find the ratio of the side length to the radius ($s/r$), we take the square root of both sides:
$s = r\sqrt{\pi}$
$s / r = \sqrt{\pi}$

ANSWER 4: B

**Problem 5:**
Let the six consecutive positive integers be $x$, $x+1$, $x+2$, $x+3$, $x+4$, and $x+5$.
Their sum is:
$x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 2013$
$6x + 15 = 2013$

Subtract 15 from both sides:
$6x = 1998$

Divide by 6:
$x = 333$

The problem asks for the largest of these integers, which is $x+5$:
333 + 5 = 338

ANSWER 5: B

**Problem 6:**
A standard checkerboard is an $8 \times 8$ grid, meaning it has a total of 64 unit squares.
The squares that touch the outer edge form the border of the board. If we remove this 1-square-thick border, the remaining inner squares form a smaller $6 \times 6$ grid.
The number of squares that do *not* touch the outer edge is:
$6 \times 6 = 36$

The probability of randomly choosing one of these inner squares is the number of inner squares divided by the total number of squares:
$P = 36 / 64$

Dividing the numerator and the denominator by 4 simplifies the fraction:
$P = 9 / 16$

ANSWER 6: D

**Problem 7:**
First, let's list the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.
There are 8 numbers in total.

- **Mean:** The sum of the numbers divided by the count.
  Sum = 0 + 0 + 1 + 2 + 3 + 3 + 3 + 4 = 16
  Mean = 16 / 8 = 2

- **Median:** The middle value. Since there is an even number of values (8), the median is the average of the 4th and 5th numbers.
  The 4th number is 2, and the 5th number is 3.
  Median = (2 + 3) / 2 = 2.5

- **Mode:** The number that appears most frequently.
  The number 3 appears three times, which is more than any other number.
  Mode = 3

The sum of the mean, median, and mode is:
2 + 2.5 + 3 = 7.5

ANSWER 7: C

**Problem 8:**
Georgie the Ghost has 6 different windows to choose from when entering the house.
Because he must leave by a *different* window, he cannot use the window he entered through. This leaves him with 5 choices for his exit.
By the fundamental counting principle, the total number of ways he can enter and leave is:
6 (entry choices) × 5 (exit choices) = 30 ways.

ANSWER 8: D

**Problem 9:**
Let's calculate the final total for both clerks.
- **Jack's method:** He adds the 6% tax first, making the subtotal $90 \times 1.06$. Then he applies the 20% discount, which means the customer pays 80% of the subtotal.
  Jack's Total = $90 \times 1.06 \times 0.80$

- **Jill's method:** She applies the 20% discount first, making the subtotal $90 \times 0.80$. Then she adds the 6% tax, which means multiplying by 1.06.
  Jill's Total = $90 \times 0.80 \times 1.06$

Because multiplication is commutative (the order in which you multiply numbers does not change the product), Jack's total and Jill's total are mathematically identical.
Therefore, Jack's total minus Jill's total is exactly $0.

ANSWER 9: C

**Problem 10:**
We are asked to find the sum of the following fractions:
$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$

Since all the fractions share a common denominator of 10, we just need to add their numerators:
Sum of numerators = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55

The sum of the digits 1 through 9 is 45.
Total numerator = 45 + 55 = 100

Now divide the total numerator by the denominator:
$\frac{100}{10} = 10$

ANSWER 10: D

**Problem 11:**
The ten-letter code "BEST OF LUCK" maps to the digits 0 through 9 in order:
B = 0
E = 1
S = 2
T = 3
O = 4
F = 5
L = 6
U = 7
C = 8
K = 9

We need to find the 4-digit number for "CLUE":
C = 8
L = 6
U = 7
E = 1

Putting the digits together, "CLUE" represents 8671.

ANSWER 11: A

**Problem 12:**
To find the correct sunset time, we need to add the length of daylight to the sunrise time.
Sunrise: 6:57 AM
Length of daylight: 10 hours and 24 minutes

First, add the 10 hours to the sunrise time:
6:57 AM + 10 hours = 4:57 PM

Next, add the remaining 24 minutes to 4:57 PM:
Adding 3 minutes brings the time to 5:00 PM.
We have 21 minutes left to add (24 - 3 = 21).
5:00 PM + 21 minutes = 5:21 PM.

ANSWER 12: B

Problem 1:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

To find the total number of pages in the book, we need to add the number of pages Isabella read each day.

First, let's calculate the total number of pages she read in the first three days: $36 \times 3 = 108$ pages.
Next, we calculate the total number of pages she read in the next three days: $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.

The total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low since $108 + 132 + 10 = 250$.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 1: B

Problem 2:
What is the smallest possible average of four distinct positive even integers?

To minimize the average of four distinct positive even integers, we need to choose the smallest four distinct positive even integers.

The smallest four distinct positive even integers are 2, 4, 6, and 8.

Let's calculate their average: $\frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$.

Let's examine the choices:
A. 3: Too low since the smallest even number is 2.
B. 4: Too low since the average of 2, 4, 6, and 8 is 5.
C. 5: Matches our calculation.
D. 6: Too high.
E. 7: Too high.

ANSWER 2: C

Problem 3:
Four students take an exam. Three of their scores are 70, 80, and 90. If the average of their four scores is 70, then what is the remaining score?

To find the remaining score, we first need to find the total score of all four students.

The average score is 70, and there are 4 students, so the total score is $70 \times 4 = 280$.

The sum of the three known scores is $70 + 80 + 90 = 240$.

The remaining score is $280 - 240 = 40$.

Let's examine the choices:
A. 40: Matches our calculation.
B. 50: Too high.
C. 55: Too high.
D. 60: Too high.
E. 70: Too high.

ANSWER 3: A

Problem 4:
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

Let's denote the side length of the square as $s$ and the radius of the circle as $r$.

The area of the square is $s^2$, and the area of the circle is $\pi r^2$.

Since they have the same area, we have $s^2 = \pi r^2$.

To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$.

Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$.

Let's examine the choices:
A. $\sqrt{\pi} / 2$: Too low.
B. $\sqrt{\pi}$: Matches our calculation.
C. $\pi$: Too high.
D. $2\pi$: Too high.
E. $\pi^2$: Too high.

ANSWER 4: B

Problem 5:
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Let's denote the smallest of these integers as $n$. Then the six consecutive integers are $n$, $n+1$, $n+2$, $n+3$, $n+4$, and $n+5$.

Their sum is $n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15 = 2013$.

Solving for $n$, we get $6n = 2013 - 15 = 1998$, so $n = \frac{1998}{6} = 333$.

The largest of these six integers is $n + 5 = 333 + 5 = 338$.

Let's examine the choices:
A. 335: Too low.
B. 338: Matches our calculation.
C. 340: Too high.
D. 345: Too high.
E. 350: Too high.

ANSWER 5: B

Problem 6:
On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

A standard checkerboard is an $8 \times 8$ grid, so it has 64 unit squares.

The unit squares that touch the outer edge are those on the first and last rows and columns. There are $8 + 8 + 8 + 8 = 32$ such squares, but the 4 corner squares are counted twice, so we subtract 4 to avoid double-counting: $32 - 4 = 28$.

Thus, there are $64 - 28 = 36$ unit squares that do not touch the outer edge.

The probability that a randomly chosen unit square does not touch the outer edge is $\frac{36}{64} = \frac{9}{16}$.

Let's examine the choices:
A. $\frac{1}{16}$: Too low.
B. $\frac{7}{16}$: Too low.
C. $\frac{1}{2}$: Too low since $\frac{9}{16} > \frac{1}{2}$.
D. $\frac{9}{16}$: Matches our calculation.
E. $\frac{49}{64}$: Too high.

ANSWER 6: D

Problem 7:
What is the sum of the mean, median, and mode of the numbers 2, 3, 0, 3, 1, 4, 0, 3?

First, let's arrange the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.

The mean is the average of the numbers: $\frac{0 + 0 + 1 + 2 + 3 + 3 + 3 + 4}{8} = \frac{16}{8} = 2$.

The median is the average of the 4th and 5th numbers (since there are 8 numbers): $\frac{2 + 3}{2} = 2.5$.

The mode is the number that appears most frequently, which is 3.

The sum of the mean, median, and mode is $2 + 2.5 + 3 = 7.5$.

Let's examine the choices:
A. 6.5: Too low.
B. 7: Too low.
C. 7.5: Matches our calculation.
D. 8.5: Too high.
E. 9: Too high.

ANSWER 7: C

Problem 8:
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

Georgie has 6 choices for the window to enter.

After entering, Georgie has 5 remaining choices for the window to leave (since it must be different from the entry window).

The total number of ways Georgie can enter and leave is $6 \times 5 = 30$.

Let's examine the choices:
A. 12: Too low.
B. 15: Too low.
C. 18: Too low.
D. 30: Matches our calculation.
E. 36: Too high.

ANSWER 8: D

Problem 9:
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?

Let's calculate Jack's total:
1. $90.00 + 6\%$ of $90.00 = 90 + 0.06 \times 90 = 90 + 5.40 = 95.40$.
2. $20\%$ 

Problem 1:

Isabella read the book over 7 days.

- First 3 days:  
  \[
  3 \times 36 = 108
  \]
  pages

- Next 3 days:  
  \[
  3 \times 44 = 132
  \]
  pages

- Last day:  
  \[
  10
  \]
  pages

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 1: B

Problem 2:

We want the smallest possible average of four distinct positive even integers.

The smallest positive even integers are:

\[
2,\ 4,\ 6,\ 8
\]

Their average is:

\[
\frac{2+4+6+8}{4}=\frac{20}{4}=5
\]

We cannot use 0 because the integers must be positive. We also cannot repeat numbers because they must be distinct.

ANSWER 2: C

Problem 3:

The average of 4 scores is 70, so the total of the 4 scores is:

\[
4 \times 70 = 280
\]

The three known scores add to:

\[
70+80+90=240
\]

So the remaining score is:

\[
280-240=40
\]

ANSWER 3: A

Problem 4:

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The area of the square is:

\[
s^2
\]

The area of the circle is:

\[
\pi r^2
\]

They have the same area, so:

\[
s^2=\pi r^2
\]

Taking the square root of both sides:

\[
s=r\sqrt{\pi}
\]

Therefore,

\[
\frac{s}{r}=\sqrt{\pi}
\]

So the ratio of the side length of the square to the radius of the circle is \(\sqrt{\pi}\).

ANSWER 4: B

Problem 5:

Let the six consecutive positive integers be:

\[
n,\ n+1,\ n+2,\ n+3,\ n+4,\ n+5
\]

Their sum is 2013:

\[
n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=2013
\]

Combine like terms:

\[
6n+15=2013
\]

Subtract 15:

\[
6n=1998
\]

Divide by 6:

\[
n=333
\]

So the six integers are:

\[
333,\ 334,\ 335,\ 336,\ 337,\ 338
\]

The largest is 338.

ANSWER 5: B

Problem 6:

A checkerboard has 64 unit squares, so it is an \(8 \times 8\) board.

A square does not touch the outer edge only if it is not in the first row, last row, first column, or last column.

That leaves the interior squares:

\[
6 \times 6 = 36
\]

So the probability is:

\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:

The numbers are:

\[
2,\ 3,\ 0,\ 3,\ 1,\ 4,\ 0,\ 3
\]

First find the mean.

Sum:

\[
2+3+0+3+1+4+0+3=16
\]

There are 8 numbers, so the mean is:

\[
\frac{16}{8}=2
\]

Now sort the numbers:

\[
0,\ 0,\ 1,\ 2,\ 3,\ 3,\ 3,\ 4
\]

There are 8 numbers, so the median is the average of the 4th and 5th numbers:

\[
\frac{2+3}{2}=2.5
\]

The mode is the number that appears most often. The number 3 appears three times, so the mode is:

\[
3
\]

Now add mean, median, and mode:

\[
2+2.5+3=7.5
\]

ANSWER 7: C

Problem 8:

Georgie has 6 choices for the window he enters.

After entering, he must leave by a different window, so he has 5 choices for the window he leaves.

Thus the number of ways is:

\[
6 \times 5 = 30
\]

This is not 15 because entering window A and leaving window B is different from entering window B and leaving window A.

ANSWER 8: D

Problem 9:

Jack’s method:

He first adds 6% sales tax to \$90:

\[
90 \times 1.06 = 95.40
\]

Then he subtracts 20% from that total, meaning he keeps 80%:

\[
95.40 \times 0.80 = 76.32
\]

So Jack’s total is \$76.32.

Jill’s method:

She first subtracts 20% from \$90:

\[
90 \times 0.80 = 72
\]

Then she adds 6% sales tax:

\[
72 \times 1.06 = 76.32
\]

So Jill’s total is also \$76.32.

Therefore:

\[
76.32-76.32=0
\]

ANSWER 9: C

Problem 10:

The expression is:

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}+\frac{55}{10}
\]

Since all fractions have denominator 10, add the numerators:

\[
\frac{1+2+3+4+5+6+7+8+9+55}{10}
\]

Now,

\[
1+2+3+4+5+6+7+8+9=45
\]

So:

\[
\frac{45+55}{10}=\frac{100}{10}=10
\]

ANSWER 10: D

Problem 11:

The code phrase is:

\[
\text{BEST OF LUCK}
\]

The ten letters represent the digits 0 through 9 in order:

\[
B=0,\ E=1,\ S=2,\ T=3,\ O=4,\ F=5,\ L=6,\ U=7,\ C=8,\ K=9
\]

Now decode the word CLUE:

\[
C=8,\quad L=6,\quad U=7,\quad E=1
\]

So CLUE represents:

\[
8671
\]

ANSWER 11: A

Problem 12:

The sunrise was 6:57 AM, and the length of daylight was 10 hours and 24 minutes.

So the sunset time is:

\[
6:57\text{ AM} + 10\text{ hours }24\text{ minutes}
\]

First add 10 hours:

\[
6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}
\]

Now add 24 minutes:

\[
4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}
\]

So the sun really set at 5:21 PM.

ANSWER 12: B

**Problem 1:**
To find the total number of pages in the book, we need to calculate the number of pages Isabella read in each part of the week and add them together.
- For the first 3 days, she read an average of 36 pages per day: 
  3 days × 36 pages/day = 108 pages.
- For the next 3 days, she read an average of 44 pages per day: 
  3 days × 44 pages/day = 132 pages.
- On the last day (the 7th day), she read 10 pages.

Adding these amounts together gives the total number of pages:
108 + 132 + 10 = 250 pages.

ANSWER 1: B

**Problem 2:**
To find the smallest possible average of four distinct positive even integers, we must first select the four smallest distinct positive even integers. 
These integers are 2, 4, 6, and 8.
Next, we find their sum:
2 + 4 + 6 + 8 = 20
Finally, we divide the sum by the number of integers (4) to find the average:
20 / 4 = 5

ANSWER 2: C

**Problem 3:**
Let the unknown remaining score be $x$.
The average of the four scores is given as 70. The formula for the average is the sum of the scores divided by the number of scores:
(70 + 80 + 90 + $x$) / 4 = 70

Multiply both sides by 4 to find the total sum of the scores:
70 + 80 + 90 + $x$ = 280
240 + $x$ = 280

Subtract 240 from both sides:
$x$ = 40
The remaining score is 40.

ANSWER 3: A

**Problem 4:**
Let the side length of the square be $s$ and the radius of the circle be $r$.
The area of the square is $s^2$, and the area of the circle is $\pi r^2$.
We are told that their areas are equal:
$s^2 = \pi r^2$

To find the ratio of the side length to the radius ($s/r$), we take the square root of both sides:
$s = r\sqrt{\pi}$
$s / r = \sqrt{\pi}$

ANSWER 4: B

**Problem 5:**
Let the six consecutive positive integers be $x$, $x+1$, $x+2$, $x+3$, $x+4$, and $x+5$.
Their sum is:
$x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 2013$
$6x + 15 = 2013$

Subtract 15 from both sides:
$6x = 1998$

Divide by 6:
$x = 333$

The problem asks for the largest of these integers, which is $x+5$:
333 + 5 = 338

ANSWER 5: B

**Problem 6:**
A standard checkerboard is an $8 \times 8$ grid, meaning it has a total of 64 unit squares.
The squares that touch the outer edge form the border of the board. If we remove this 1-square-thick border, the remaining inner squares form a smaller $6 \times 6$ grid.
The number of squares that do *not* touch the outer edge is:
$6 \times 6 = 36$

The probability of randomly choosing one of these inner squares is the number of inner squares divided by the total number of squares:
$P = 36 / 64$

Dividing the numerator and the denominator by 4 simplifies the fraction:
$P = 9 / 16$

ANSWER 6: D

**Problem 7:**
First, let's list the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.
There are 8 numbers in total.

- **Mean:** The sum of the numbers divided by the count.
  Sum = 0 + 0 + 1 + 2 + 3 + 3 + 3 + 4 = 16
  Mean = 16 / 8 = 2

- **Median:** The middle value. Since there is an even number of values (8), the median is the average of the 4th and 5th numbers.
  The 4th number is 2, and the 5th number is 3.
  Median = (2 + 3) / 2 = 2.5

- **Mode:** The number that appears most frequently.
  The number 3 appears three times, which is more than any other number.
  Mode = 3

The sum of the mean, median, and mode is:
2 + 2.5 + 3 = 7.5

ANSWER 7: C

**Problem 8:**
Georgie the Ghost has 6 different windows to choose from when entering the house.
Because he must leave by a *different* window, he cannot use the window he entered through. This leaves him with 5 choices for his exit.
By the fundamental counting principle, the total number of ways he can enter and leave is:
6 (entry choices) × 5 (exit choices) = 30 ways.

ANSWER 8: D

**Problem 9:**
Let's calculate the final total for both clerks.
- **Jack's method:** He adds the 6% tax first, making the subtotal $90 \times 1.06$. Then he applies the 20% discount, which means the customer pays 80% of the subtotal.
  Jack's Total = $90 \times 1.06 \times 0.80$

- **Jill's method:** She applies the 20% discount first, making the subtotal $90 \times 0.80$. Then she adds the 6% tax, which means multiplying by 1.06.
  Jill's Total = $90 \times 0.80 \times 1.06$

Because multiplication is commutative (the order in which you multiply numbers does not change the product), Jack's total and Jill's total are mathematically identical.
Therefore, Jack's total minus Jill's total is exactly $0.

ANSWER 9: C

**Problem 10:**
We are asked to find the sum of the following fractions:
$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$

Since all the fractions share a common denominator of 10, we just need to add their numerators:
Sum of numerators = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55

The sum of the digits 1 through 9 is 45.
Total numerator = 45 + 55 = 100

Now divide the total numerator by the denominator:
$\frac{100}{10} = 10$

ANSWER 10: D

**Problem 11:**
The ten-letter code "BEST OF LUCK" maps to the digits 0 through 9 in order:
B = 0
E = 1
S = 2
T = 3
O = 4
F = 5
L = 6
U = 7
C = 8
K = 9

We need to find the 4-digit number for "CLUE":
C = 8
L = 6
U = 7
E = 1

Putting the digits together, "CLUE" represents 8671.

ANSWER 11: A

**Problem 12:**
To find the correct sunset time, we need to add the length of daylight to the sunrise time.
Sunrise: 6:57 AM
Length of daylight: 10 hours and 24 minutes

First, add the 10 hours to the sunrise time:
6:57 AM + 10 hours = 4:57 PM

Next, add the remaining 24 minutes to 4:57 PM:
Adding 3 minutes brings the time to 5:00 PM.
We have 21 minutes left to add (24 - 3 = 21).
5:00 PM + 21 minutes = 5:21 PM.

ANSWER 12: B

Problem 1:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

To find the total number of pages in the book, we need to add the number of pages Isabella read each day.

First, let's calculate the total number of pages she read in the first three days: $36 \times 3 = 108$ pages.
Next, we calculate the total number of pages she read in the next three days: $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.

The total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low since $108 + 132 + 10 = 250$.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 1: B

Problem 2:
What is the smallest possible average of four distinct positive even integers?

To minimize the average of four distinct positive even integers, we need to choose the smallest four distinct positive even integers.

The smallest four distinct positive even integers are 2, 4, 6, and 8.

Let's calculate their average: $\frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$.

Let's examine the choices:
A. 3: Too low since the smallest even number is 2.
B. 4: Too low since the average of 2, 4, 6, and 8 is 5.
C. 5: Matches our calculation.
D. 6: Too high.
E. 7: Too high.

ANSWER 2: C

Problem 3:
Four students take an exam. Three of their scores are 70, 80, and 90. If the average of their four scores is 70, then what is the remaining score?

To find the remaining score, we first need to find the total score of all four students.

The average score is 70, and there are 4 students, so the total score is $70 \times 4 = 280$.

The sum of the three known scores is $70 + 80 + 90 = 240$.

The remaining score is $280 - 240 = 40$.

Let's examine the choices:
A. 40: Matches our calculation.
B. 50: Too high.
C. 55: Too high.
D. 60: Too high.
E. 70: Too high.

ANSWER 3: A

Problem 4:
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

Let's denote the side length of the square as $s$ and the radius of the circle as $r$.

The area of the square is $s^2$, and the area of the circle is $\pi r^2$.

Since they have the same area, we have $s^2 = \pi r^2$.

To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$.

Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$.

Let's examine the choices:
A. $\sqrt{\pi} / 2$: Too low.
B. $\sqrt{\pi}$: Matches our calculation.
C. $\pi$: Too high.
D. $2\pi$: Too high.
E. $\pi^2$: Too high.

ANSWER 4: B

Problem 5:
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Let's denote the smallest of these integers as $n$. Then the six consecutive integers are $n$, $n+1$, $n+2$, $n+3$, $n+4$, and $n+5$.

Their sum is $n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15 = 2013$.

Solving for $n$, we get $6n = 2013 - 15 = 1998$, so $n = \frac{1998}{6} = 333$.

The largest of these six integers is $n + 5 = 333 + 5 = 338$.

Let's examine the choices:
A. 335: Too low.
B. 338: Matches our calculation.
C. 340: Too high.
D. 345: Too high.
E. 350: Too high.

ANSWER 5: B

Problem 6:
On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

A standard checkerboard is an $8 \times 8$ grid, so it has 64 unit squares.

The unit squares that touch the outer edge are those on the first and last rows and columns. There are $8 + 8 + 8 + 8 = 32$ such squares, but the 4 corner squares are counted twice, so we subtract 4 to avoid double-counting: $32 - 4 = 28$.

Thus, there are $64 - 28 = 36$ unit squares that do not touch the outer edge.

The probability that a randomly chosen unit square does not touch the outer edge is $\frac{36}{64} = \frac{9}{16}$.

Let's examine the choices:
A. $\frac{1}{16}$: Too low.
B. $\frac{7}{16}$: Too low.
C. $\frac{1}{2}$: Too low since $\frac{9}{16} > \frac{1}{2}$.
D. $\frac{9}{16}$: Matches our calculation.
E. $\frac{49}{64}$: Too high.

ANSWER 6: D

Problem 7:
What is the sum of the mean, median, and mode of the numbers 2, 3, 0, 3, 1, 4, 0, 3?

First, let's arrange the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.

The mean is the average of the numbers: $\frac{0 + 0 + 1 + 2 + 3 + 3 + 3 + 4}{8} = \frac{16}{8} = 2$.

The median is the average of the 4th and 5th numbers (since there are 8 numbers): $\frac{2 + 3}{2} = 2.5$.

The mode is the number that appears most frequently, which is 3.

The sum of the mean, median, and mode is $2 + 2.5 + 3 = 7.5$.

Let's examine the choices:
A. 6.5: Too low.
B. 7: Too low.
C. 7.5: Matches our calculation.
D. 8.5: Too high.
E. 9: Too high.

ANSWER 7: C

Problem 8:
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

Georgie has 6 choices for the window to enter.

After entering, Georgie has 5 remaining choices for the window to leave (since it must be different from the entry window).

The total number of ways Georgie can enter and leave is $6 \times 5 = 30$.

Let's examine the choices:
A. 12: Too low.
B. 15: Too low.
C. 18: Too low.
D. 30: Matches our calculation.
E. 36: Too high.

ANSWER 8: D

Problem 9:
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?

Let's calculate Jack's total:
1. $90.00 + 6\%$ of $90.00 = 90 + 0.06 \times 90 = 90 + 5.40 = 95.40$.
2. $20\%$ 

Problem 1:

Isabella read the book over 7 days.

- First 3 days:  
  \[
  3 \times 36 = 108
  \]
  pages

- Next 3 days:  
  \[
  3 \times 44 = 132
  \]
  pages

- Last day:  
  \[
  10
  \]
  pages

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 1: B

Problem 2:

We want the smallest possible average of four distinct positive even integers.

The smallest positive even integers are:

\[
2,\ 4,\ 6,\ 8
\]

Their average is:

\[
\frac{2+4+6+8}{4}=\frac{20}{4}=5
\]

We cannot use 0 because the integers must be positive. We also cannot repeat numbers because they must be distinct.

ANSWER 2: C

Problem 3:

The average of 4 scores is 70, so the total of the 4 scores is:

\[
4 \times 70 = 280
\]

The three known scores add to:

\[
70+80+90=240
\]

So the remaining score is:

\[
280-240=40
\]

ANSWER 3: A

Problem 4:

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The area of the square is:

\[
s^2
\]

The area of the circle is:

\[
\pi r^2
\]

They have the same area, so:

\[
s^2=\pi r^2
\]

Taking the square root of both sides:

\[
s=r\sqrt{\pi}
\]

Therefore,

\[
\frac{s}{r}=\sqrt{\pi}
\]

So the ratio of the side length of the square to the radius of the circle is \(\sqrt{\pi}\).

ANSWER 4: B

Problem 5:

Let the six consecutive positive integers be:

\[
n,\ n+1,\ n+2,\ n+3,\ n+4,\ n+5
\]

Their sum is 2013:

\[
n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=2013
\]

Combine like terms:

\[
6n+15=2013
\]

Subtract 15:

\[
6n=1998
\]

Divide by 6:

\[
n=333
\]

So the six integers are:

\[
333,\ 334,\ 335,\ 336,\ 337,\ 338
\]

The largest is 338.

ANSWER 5: B

Problem 6:

A checkerboard has 64 unit squares, so it is an \(8 \times 8\) board.

A square does not touch the outer edge only if it is not in the first row, last row, first column, or last column.

That leaves the interior squares:

\[
6 \times 6 = 36
\]

So the probability is:

\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:

The numbers are:

\[
2,\ 3,\ 0,\ 3,\ 1,\ 4,\ 0,\ 3
\]

First find the mean.

Sum:

\[
2+3+0+3+1+4+0+3=16
\]

There are 8 numbers, so the mean is:

\[
\frac{16}{8}=2
\]

Now sort the numbers:

\[
0,\ 0,\ 1,\ 2,\ 3,\ 3,\ 3,\ 4
\]

There are 8 numbers, so the median is the average of the 4th and 5th numbers:

\[
\frac{2+3}{2}=2.5
\]

The mode is the number that appears most often. The number 3 appears three times, so the mode is:

\[
3
\]

Now add mean, median, and mode:

\[
2+2.5+3=7.5
\]

ANSWER 7: C

Problem 8:

Georgie has 6 choices for the window he enters.

After entering, he must leave by a different window, so he has 5 choices for the window he leaves.

Thus the number of ways is:

\[
6 \times 5 = 30
\]

This is not 15 because entering window A and leaving window B is different from entering window B and leaving window A.

ANSWER 8: D

Problem 9:

Jack’s method:

He first adds 6% sales tax to \$90:

\[
90 \times 1.06 = 95.40
\]

Then he subtracts 20% from that total, meaning he keeps 80%:

\[
95.40 \times 0.80 = 76.32
\]

So Jack’s total is \$76.32.

Jill’s method:

She first subtracts 20% from \$90:

\[
90 \times 0.80 = 72
\]

Then she adds 6% sales tax:

\[
72 \times 1.06 = 76.32
\]

So Jill’s total is also \$76.32.

Therefore:

\[
76.32-76.32=0
\]

ANSWER 9: C

Problem 10:

The expression is:

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}+\frac{55}{10}
\]

Since all fractions have denominator 10, add the numerators:

\[
\frac{1+2+3+4+5+6+7+8+9+55}{10}
\]

Now,

\[
1+2+3+4+5+6+7+8+9=45
\]

So:

\[
\frac{45+55}{10}=\frac{100}{10}=10
\]

ANSWER 10: D

Problem 11:

The code phrase is:

\[
\text{BEST OF LUCK}
\]

The ten letters represent the digits 0 through 9 in order:

\[
B=0,\ E=1,\ S=2,\ T=3,\ O=4,\ F=5,\ L=6,\ U=7,\ C=8,\ K=9
\]

Now decode the word CLUE:

\[
C=8,\quad L=6,\quad U=7,\quad E=1
\]

So CLUE represents:

\[
8671
\]

ANSWER 11: A

Problem 12:

The sunrise was 6:57 AM, and the length of daylight was 10 hours and 24 minutes.

So the sunset time is:

\[
6:57\text{ AM} + 10\text{ hours }24\text{ minutes}
\]

First add 10 hours:

\[
6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}
\]

Now add 24 minutes:

\[
4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}
\]

So the sun really set at 5:21 PM.

ANSWER 12: B

**Problem 1:**
To find the total number of pages in the book, we need to calculate the number of pages Isabella read in each part of the week and add them together.
- For the first 3 days, she read an average of 36 pages per day: 
  3 days × 36 pages/day = 108 pages.
- For the next 3 days, she read an average of 44 pages per day: 
  3 days × 44 pages/day = 132 pages.
- On the last day (the 7th day), she read 10 pages.

Adding these amounts together gives the total number of pages:
108 + 132 + 10 = 250 pages.

ANSWER 1: B

**Problem 2:**
To find the smallest possible average of four distinct positive even integers, we must first select the four smallest distinct positive even integers. 
These integers are 2, 4, 6, and 8.
Next, we find their sum:
2 + 4 + 6 + 8 = 20
Finally, we divide the sum by the number of integers (4) to find the average:
20 / 4 = 5

ANSWER 2: C

**Problem 3:**
Let the unknown remaining score be $x$.
The average of the four scores is given as 70. The formula for the average is the sum of the scores divided by the number of scores:
(70 + 80 + 90 + $x$) / 4 = 70

Multiply both sides by 4 to find the total sum of the scores:
70 + 80 + 90 + $x$ = 280
240 + $x$ = 280

Subtract 240 from both sides:
$x$ = 40
The remaining score is 40.

ANSWER 3: A

**Problem 4:**
Let the side length of the square be $s$ and the radius of the circle be $r$.
The area of the square is $s^2$, and the area of the circle is $\pi r^2$.
We are told that their areas are equal:
$s^2 = \pi r^2$

To find the ratio of the side length to the radius ($s/r$), we take the square root of both sides:
$s = r\sqrt{\pi}$
$s / r = \sqrt{\pi}$

ANSWER 4: B

**Problem 5:**
Let the six consecutive positive integers be $x$, $x+1$, $x+2$, $x+3$, $x+4$, and $x+5$.
Their sum is:
$x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 2013$
$6x + 15 = 2013$

Subtract 15 from both sides:
$6x = 1998$

Divide by 6:
$x = 333$

The problem asks for the largest of these integers, which is $x+5$:
333 + 5 = 338

ANSWER 5: B

**Problem 6:**
A standard checkerboard is an $8 \times 8$ grid, meaning it has a total of 64 unit squares.
The squares that touch the outer edge form the border of the board. If we remove this 1-square-thick border, the remaining inner squares form a smaller $6 \times 6$ grid.
The number of squares that do *not* touch the outer edge is:
$6 \times 6 = 36$

The probability of randomly choosing one of these inner squares is the number of inner squares divided by the total number of squares:
$P = 36 / 64$

Dividing the numerator and the denominator by 4 simplifies the fraction:
$P = 9 / 16$

ANSWER 6: D

**Problem 7:**
First, let's list the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.
There are 8 numbers in total.

- **Mean:** The sum of the numbers divided by the count.
  Sum = 0 + 0 + 1 + 2 + 3 + 3 + 3 + 4 = 16
  Mean = 16 / 8 = 2

- **Median:** The middle value. Since there is an even number of values (8), the median is the average of the 4th and 5th numbers.
  The 4th number is 2, and the 5th number is 3.
  Median = (2 + 3) / 2 = 2.5

- **Mode:** The number that appears most frequently.
  The number 3 appears three times, which is more than any other number.
  Mode = 3

The sum of the mean, median, and mode is:
2 + 2.5 + 3 = 7.5

ANSWER 7: C

**Problem 8:**
Georgie the Ghost has 6 different windows to choose from when entering the house.
Because he must leave by a *different* window, he cannot use the window he entered through. This leaves him with 5 choices for his exit.
By the fundamental counting principle, the total number of ways he can enter and leave is:
6 (entry choices) × 5 (exit choices) = 30 ways.

ANSWER 8: D

**Problem 9:**
Let's calculate the final total for both clerks.
- **Jack's method:** He adds the 6% tax first, making the subtotal $90 \times 1.06$. Then he applies the 20% discount, which means the customer pays 80% of the subtotal.
  Jack's Total = $90 \times 1.06 \times 0.80$

- **Jill's method:** She applies the 20% discount first, making the subtotal $90 \times 0.80$. Then she adds the 6% tax, which means multiplying by 1.06.
  Jill's Total = $90 \times 0.80 \times 1.06$

Because multiplication is commutative (the order in which you multiply numbers does not change the product), Jack's total and Jill's total are mathematically identical.
Therefore, Jack's total minus Jill's total is exactly $0.

ANSWER 9: C

**Problem 10:**
We are asked to find the sum of the following fractions:
$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$

Since all the fractions share a common denominator of 10, we just need to add their numerators:
Sum of numerators = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55

The sum of the digits 1 through 9 is 45.
Total numerator = 45 + 55 = 100

Now divide the total numerator by the denominator:
$\frac{100}{10} = 10$

ANSWER 10: D

**Problem 11:**
The ten-letter code "BEST OF LUCK" maps to the digits 0 through 9 in order:
B = 0
E = 1
S = 2
T = 3
O = 4
F = 5
L = 6
U = 7
C = 8
K = 9

We need to find the 4-digit number for "CLUE":
C = 8
L = 6
U = 7
E = 1

Putting the digits together, "CLUE" represents 8671.

ANSWER 11: A

**Problem 12:**
To find the correct sunset time, we need to add the length of daylight to the sunrise time.
Sunrise: 6:57 AM
Length of daylight: 10 hours and 24 minutes

First, add the 10 hours to the sunrise time:
6:57 AM + 10 hours = 4:57 PM

Next, add the remaining 24 minutes to 4:57 PM:
Adding 3 minutes brings the time to 5:00 PM.
We have 21 minutes left to add (24 - 3 = 21).
5:00 PM + 21 minutes = 5:21 PM.

ANSWER 12: B

Problem 1:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

To find the total number of pages in the book, we need to add the number of pages Isabella read each day.

First, let's calculate the total number of pages she read in the first three days: $36 \times 3 = 108$ pages.
Next, we calculate the total number of pages she read in the next three days: $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.

The total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low since $108 + 132 + 10 = 250$.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 1: B

Problem 2:
What is the smallest possible average of four distinct positive even integers?

To minimize the average of four distinct positive even integers, we need to choose the smallest four distinct positive even integers.

The smallest four distinct positive even integers are 2, 4, 6, and 8.

Let's calculate their average: $\frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$.

Let's examine the choices:
A. 3: Too low since the smallest even number is 2.
B. 4: Too low since the average of 2, 4, 6, and 8 is 5.
C. 5: Matches our calculation.
D. 6: Too high.
E. 7: Too high.

ANSWER 2: C

Problem 3:
Four students take an exam. Three of their scores are 70, 80, and 90. If the average of their four scores is 70, then what is the remaining score?

To find the remaining score, we first need to find the total score of all four students.

The average score is 70, and there are 4 students, so the total score is $70 \times 4 = 280$.

The sum of the three known scores is $70 + 80 + 90 = 240$.

The remaining score is $280 - 240 = 40$.

Let's examine the choices:
A. 40: Matches our calculation.
B. 50: Too high.
C. 55: Too high.
D. 60: Too high.
E. 70: Too high.

ANSWER 3: A

Problem 4:
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

Let's denote the side length of the square as $s$ and the radius of the circle as $r$.

The area of the square is $s^2$, and the area of the circle is $\pi r^2$.

Since they have the same area, we have $s^2 = \pi r^2$.

To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$.

Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$.

Let's examine the choices:
A. $\sqrt{\pi} / 2$: Too low.
B. $\sqrt{\pi}$: Matches our calculation.
C. $\pi$: Too high.
D. $2\pi$: Too high.
E. $\pi^2$: Too high.

ANSWER 4: B

Problem 5:
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Let's denote the smallest of these integers as $n$. Then the six consecutive integers are $n$, $n+1$, $n+2$, $n+3$, $n+4$, and $n+5$.

Their sum is $n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15 = 2013$.

Solving for $n$, we get $6n = 2013 - 15 = 1998$, so $n = \frac{1998}{6} = 333$.

The largest of these six integers is $n + 5 = 333 + 5 = 338$.

Let's examine the choices:
A. 335: Too low.
B. 338: Matches our calculation.
C. 340: Too high.
D. 345: Too high.
E. 350: Too high.

ANSWER 5: B

Problem 6:
On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

A standard checkerboard is an $8 \times 8$ grid, so it has 64 unit squares.

The unit squares that touch the outer edge are those on the first and last rows and columns. There are $8 + 8 + 8 + 8 = 32$ such squares, but the 4 corner squares are counted twice, so we subtract 4 to avoid double-counting: $32 - 4 = 28$.

Thus, there are $64 - 28 = 36$ unit squares that do not touch the outer edge.

The probability that a randomly chosen unit square does not touch the outer edge is $\frac{36}{64} = \frac{9}{16}$.

Let's examine the choices:
A. $\frac{1}{16}$: Too low.
B. $\frac{7}{16}$: Too low.
C. $\frac{1}{2}$: Too low since $\frac{9}{16} > \frac{1}{2}$.
D. $\frac{9}{16}$: Matches our calculation.
E. $\frac{49}{64}$: Too high.

ANSWER 6: D

Problem 7:
What is the sum of the mean, median, and mode of the numbers 2, 3, 0, 3, 1, 4, 0, 3?

First, let's arrange the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.

The mean is the average of the numbers: $\frac{0 + 0 + 1 + 2 + 3 + 3 + 3 + 4}{8} = \frac{16}{8} = 2$.

The median is the average of the 4th and 5th numbers (since there are 8 numbers): $\frac{2 + 3}{2} = 2.5$.

The mode is the number that appears most frequently, which is 3.

The sum of the mean, median, and mode is $2 + 2.5 + 3 = 7.5$.

Let's examine the choices:
A. 6.5: Too low.
B. 7: Too low.
C. 7.5: Matches our calculation.
D. 8.5: Too high.
E. 9: Too high.

ANSWER 7: C

Problem 8:
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

Georgie has 6 choices for the window to enter.

After entering, Georgie has 5 remaining choices for the window to leave (since it must be different from the entry window).

The total number of ways Georgie can enter and leave is $6 \times 5 = 30$.

Let's examine the choices:
A. 12: Too low.
B. 15: Too low.
C. 18: Too low.
D. 30: Matches our calculation.
E. 36: Too high.

ANSWER 8: D

Problem 9:
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?

Let's calculate Jack's total:
1. $90.00 + 6\%$ of $90.00 = 90 + 0.06 \times 90 = 90 + 5.40 = 95.40$.
2. $20\%$ 

Problem 1:

Isabella read the book over 7 days.

- First 3 days:  
  \[
  3 \times 36 = 108
  \]
  pages

- Next 3 days:  
  \[
  3 \times 44 = 132
  \]
  pages

- Last day:  
  \[
  10
  \]
  pages

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 1: B

Problem 2:

We want the smallest possible average of four distinct positive even integers.

The smallest positive even integers are:

\[
2,\ 4,\ 6,\ 8
\]

Their average is:

\[
\frac{2+4+6+8}{4}=\frac{20}{4}=5
\]

We cannot use 0 because the integers must be positive. We also cannot repeat numbers because they must be distinct.

ANSWER 2: C

Problem 3:

The average of 4 scores is 70, so the total of the 4 scores is:

\[
4 \times 70 = 280
\]

The three known scores add to:

\[
70+80+90=240
\]

So the remaining score is:

\[
280-240=40
\]

ANSWER 3: A

Problem 4:

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The area of the square is:

\[
s^2
\]

The area of the circle is:

\[
\pi r^2
\]

They have the same area, so:

\[
s^2=\pi r^2
\]

Taking the square root of both sides:

\[
s=r\sqrt{\pi}
\]

Therefore,

\[
\frac{s}{r}=\sqrt{\pi}
\]

So the ratio of the side length of the square to the radius of the circle is \(\sqrt{\pi}\).

ANSWER 4: B

Problem 5:

Let the six consecutive positive integers be:

\[
n,\ n+1,\ n+2,\ n+3,\ n+4,\ n+5
\]

Their sum is 2013:

\[
n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=2013
\]

Combine like terms:

\[
6n+15=2013
\]

Subtract 15:

\[
6n=1998
\]

Divide by 6:

\[
n=333
\]

So the six integers are:

\[
333,\ 334,\ 335,\ 336,\ 337,\ 338
\]

The largest is 338.

ANSWER 5: B

Problem 6:

A checkerboard has 64 unit squares, so it is an \(8 \times 8\) board.

A square does not touch the outer edge only if it is not in the first row, last row, first column, or last column.

That leaves the interior squares:

\[
6 \times 6 = 36
\]

So the probability is:

\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:

The numbers are:

\[
2,\ 3,\ 0,\ 3,\ 1,\ 4,\ 0,\ 3
\]

First find the mean.

Sum:

\[
2+3+0+3+1+4+0+3=16
\]

There are 8 numbers, so the mean is:

\[
\frac{16}{8}=2
\]

Now sort the numbers:

\[
0,\ 0,\ 1,\ 2,\ 3,\ 3,\ 3,\ 4
\]

There are 8 numbers, so the median is the average of the 4th and 5th numbers:

\[
\frac{2+3}{2}=2.5
\]

The mode is the number that appears most often. The number 3 appears three times, so the mode is:

\[
3
\]

Now add mean, median, and mode:

\[
2+2.5+3=7.5
\]

ANSWER 7: C

Problem 8:

Georgie has 6 choices for the window he enters.

After entering, he must leave by a different window, so he has 5 choices for the window he leaves.

Thus the number of ways is:

\[
6 \times 5 = 30
\]

This is not 15 because entering window A and leaving window B is different from entering window B and leaving window A.

ANSWER 8: D

Problem 9:

Jack’s method:

He first adds 6% sales tax to \$90:

\[
90 \times 1.06 = 95.40
\]

Then he subtracts 20% from that total, meaning he keeps 80%:

\[
95.40 \times 0.80 = 76.32
\]

So Jack’s total is \$76.32.

Jill’s method:

She first subtracts 20% from \$90:

\[
90 \times 0.80 = 72
\]

Then she adds 6% sales tax:

\[
72 \times 1.06 = 76.32
\]

So Jill’s total is also \$76.32.

Therefore:

\[
76.32-76.32=0
\]

ANSWER 9: C

Problem 10:

The expression is:

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}+\frac{55}{10}
\]

Since all fractions have denominator 10, add the numerators:

\[
\frac{1+2+3+4+5+6+7+8+9+55}{10}
\]

Now,

\[
1+2+3+4+5+6+7+8+9=45
\]

So:

\[
\frac{45+55}{10}=\frac{100}{10}=10
\]

ANSWER 10: D

Problem 11:

The code phrase is:

\[
\text{BEST OF LUCK}
\]

The ten letters represent the digits 0 through 9 in order:

\[
B=0,\ E=1,\ S=2,\ T=3,\ O=4,\ F=5,\ L=6,\ U=7,\ C=8,\ K=9
\]

Now decode the word CLUE:

\[
C=8,\quad L=6,\quad U=7,\quad E=1
\]

So CLUE represents:

\[
8671
\]

ANSWER 11: A

Problem 12:

The sunrise was 6:57 AM, and the length of daylight was 10 hours and 24 minutes.

So the sunset time is:

\[
6:57\text{ AM} + 10\text{ hours }24\text{ minutes}
\]

First add 10 hours:

\[
6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}
\]

Now add 24 minutes:

\[
4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}
\]

So the sun really set at 5:21 PM.

ANSWER 12: B

**Problem 1:**
To find the total number of pages in the book, we need to calculate the number of pages Isabella read in each part of the week and add them together.
- For the first 3 days, she read an average of 36 pages per day: 
  3 days × 36 pages/day = 108 pages.
- For the next 3 days, she read an average of 44 pages per day: 
  3 days × 44 pages/day = 132 pages.
- On the last day (the 7th day), she read 10 pages.

Adding these amounts together gives the total number of pages:
108 + 132 + 10 = 250 pages.

ANSWER 1: B

**Problem 2:**
To find the smallest possible average of four distinct positive even integers, we must first select the four smallest distinct positive even integers. 
These integers are 2, 4, 6, and 8.
Next, we find their sum:
2 + 4 + 6 + 8 = 20
Finally, we divide the sum by the number of integers (4) to find the average:
20 / 4 = 5

ANSWER 2: C

**Problem 3:**
Let the unknown remaining score be $x$.
The average of the four scores is given as 70. The formula for the average is the sum of the scores divided by the number of scores:
(70 + 80 + 90 + $x$) / 4 = 70

Multiply both sides by 4 to find the total sum of the scores:
70 + 80 + 90 + $x$ = 280
240 + $x$ = 280

Subtract 240 from both sides:
$x$ = 40
The remaining score is 40.

ANSWER 3: A

**Problem 4:**
Let the side length of the square be $s$ and the radius of the circle be $r$.
The area of the square is $s^2$, and the area of the circle is $\pi r^2$.
We are told that their areas are equal:
$s^2 = \pi r^2$

To find the ratio of the side length to the radius ($s/r$), we take the square root of both sides:
$s = r\sqrt{\pi}$
$s / r = \sqrt{\pi}$

ANSWER 4: B

**Problem 5:**
Let the six consecutive positive integers be $x$, $x+1$, $x+2$, $x+3$, $x+4$, and $x+5$.
Their sum is:
$x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 2013$
$6x + 15 = 2013$

Subtract 15 from both sides:
$6x = 1998$

Divide by 6:
$x = 333$

The problem asks for the largest of these integers, which is $x+5$:
333 + 5 = 338

ANSWER 5: B

**Problem 6:**
A standard checkerboard is an $8 \times 8$ grid, meaning it has a total of 64 unit squares.
The squares that touch the outer edge form the border of the board. If we remove this 1-square-thick border, the remaining inner squares form a smaller $6 \times 6$ grid.
The number of squares that do *not* touch the outer edge is:
$6 \times 6 = 36$

The probability of randomly choosing one of these inner squares is the number of inner squares divided by the total number of squares:
$P = 36 / 64$

Dividing the numerator and the denominator by 4 simplifies the fraction:
$P = 9 / 16$

ANSWER 6: D

**Problem 7:**
First, let's list the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.
There are 8 numbers in total.

- **Mean:** The sum of the numbers divided by the count.
  Sum = 0 + 0 + 1 + 2 + 3 + 3 + 3 + 4 = 16
  Mean = 16 / 8 = 2

- **Median:** The middle value. Since there is an even number of values (8), the median is the average of the 4th and 5th numbers.
  The 4th number is 2, and the 5th number is 3.
  Median = (2 + 3) / 2 = 2.5

- **Mode:** The number that appears most frequently.
  The number 3 appears three times, which is more than any other number.
  Mode = 3

The sum of the mean, median, and mode is:
2 + 2.5 + 3 = 7.5

ANSWER 7: C

**Problem 8:**
Georgie the Ghost has 6 different windows to choose from when entering the house.
Because he must leave by a *different* window, he cannot use the window he entered through. This leaves him with 5 choices for his exit.
By the fundamental counting principle, the total number of ways he can enter and leave is:
6 (entry choices) × 5 (exit choices) = 30 ways.

ANSWER 8: D

**Problem 9:**
Let's calculate the final total for both clerks.
- **Jack's method:** He adds the 6% tax first, making the subtotal $90 \times 1.06$. Then he applies the 20% discount, which means the customer pays 80% of the subtotal.
  Jack's Total = $90 \times 1.06 \times 0.80$

- **Jill's method:** She applies the 20% discount first, making the subtotal $90 \times 0.80$. Then she adds the 6% tax, which means multiplying by 1.06.
  Jill's Total = $90 \times 0.80 \times 1.06$

Because multiplication is commutative (the order in which you multiply numbers does not change the product), Jack's total and Jill's total are mathematically identical.
Therefore, Jack's total minus Jill's total is exactly $0.

ANSWER 9: C

**Problem 10:**
We are asked to find the sum of the following fractions:
$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$

Since all the fractions share a common denominator of 10, we just need to add their numerators:
Sum of numerators = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55

The sum of the digits 1 through 9 is 45.
Total numerator = 45 + 55 = 100

Now divide the total numerator by the denominator:
$\frac{100}{10} = 10$

ANSWER 10: D

**Problem 11:**
The ten-letter code "BEST OF LUCK" maps to the digits 0 through 9 in order:
B = 0
E = 1
S = 2
T = 3
O = 4
F = 5
L = 6
U = 7
C = 8
K = 9

We need to find the 4-digit number for "CLUE":
C = 8
L = 6
U = 7
E = 1

Putting the digits together, "CLUE" represents 8671.

ANSWER 11: A

**Problem 12:**
To find the correct sunset time, we need to add the length of daylight to the sunrise time.
Sunrise: 6:57 AM
Length of daylight: 10 hours and 24 minutes

First, add the 10 hours to the sunrise time:
6:57 AM + 10 hours = 4:57 PM

Next, add the remaining 24 minutes to 4:57 PM:
Adding 3 minutes brings the time to 5:00 PM.
We have 21 minutes left to add (24 - 3 = 21).
5:00 PM + 21 minutes = 5:21 PM.

ANSWER 12: B

Problem 1:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

To find the total number of pages in the book, we need to add the number of pages Isabella read each day.

First, let's calculate the total number of pages she read in the first three days: $36 \times 3 = 108$ pages.
Next, we calculate the total number of pages she read in the next three days: $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.

The total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low since $108 + 132 + 10 = 250$.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 1: B

Problem 2:
What is the smallest possible average of four distinct positive even integers?

To minimize the average of four distinct positive even integers, we need to choose the smallest four distinct positive even integers.

The smallest four distinct positive even integers are 2, 4, 6, and 8.

Let's calculate their average: $\frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$.

Let's examine the choices:
A. 3: Too low since the smallest even number is 2.
B. 4: Too low since the average of 2, 4, 6, and 8 is 5.
C. 5: Matches our calculation.
D. 6: Too high.
E. 7: Too high.

ANSWER 2: C

Problem 3:
Four students take an exam. Three of their scores are 70, 80, and 90. If the average of their four scores is 70, then what is the remaining score?

To find the remaining score, we first need to find the total score of all four students.

The average score is 70, and there are 4 students, so the total score is $70 \times 4 = 280$.

The sum of the three known scores is $70 + 80 + 90 = 240$.

The remaining score is $280 - 240 = 40$.

Let's examine the choices:
A. 40: Matches our calculation.
B. 50: Too high.
C. 55: Too high.
D. 60: Too high.
E. 70: Too high.

ANSWER 3: A

Problem 4:
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

Let's denote the side length of the square as $s$ and the radius of the circle as $r$.

The area of the square is $s^2$, and the area of the circle is $\pi r^2$.

Since they have the same area, we have $s^2 = \pi r^2$.

To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$.

Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$.

Let's examine the choices:
A. $\sqrt{\pi} / 2$: Too low.
B. $\sqrt{\pi}$: Matches our calculation.
C. $\pi$: Too high.
D. $2\pi$: Too high.
E. $\pi^2$: Too high.

ANSWER 4: B

Problem 5:
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Let's denote the smallest of these integers as $n$. Then the six consecutive integers are $n$, $n+1$, $n+2$, $n+3$, $n+4$, and $n+5$.

Their sum is $n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15 = 2013$.

Solving for $n$, we get $6n = 2013 - 15 = 1998$, so $n = \frac{1998}{6} = 333$.

The largest of these six integers is $n + 5 = 333 + 5 = 338$.

Let's examine the choices:
A. 335: Too low.
B. 338: Matches our calculation.
C. 340: Too high.
D. 345: Too high.
E. 350: Too high.

ANSWER 5: B

Problem 6:
On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

A standard checkerboard is an $8 \times 8$ grid, so it has 64 unit squares.

The unit squares that touch the outer edge are those on the first and last rows and columns. There are $8 + 8 + 8 + 8 = 32$ such squares, but the 4 corner squares are counted twice, so we subtract 4 to avoid double-counting: $32 - 4 = 28$.

Thus, there are $64 - 28 = 36$ unit squares that do not touch the outer edge.

The probability that a randomly chosen unit square does not touch the outer edge is $\frac{36}{64} = \frac{9}{16}$.

Let's examine the choices:
A. $\frac{1}{16}$: Too low.
B. $\frac{7}{16}$: Too low.
C. $\frac{1}{2}$: Too low since $\frac{9}{16} > \frac{1}{2}$.
D. $\frac{9}{16}$: Matches our calculation.
E. $\frac{49}{64}$: Too high.

ANSWER 6: D

Problem 7:
What is the sum of the mean, median, and mode of the numbers 2, 3, 0, 3, 1, 4, 0, 3?

First, let's arrange the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.

The mean is the average of the numbers: $\frac{0 + 0 + 1 + 2 + 3 + 3 + 3 + 4}{8} = \frac{16}{8} = 2$.

The median is the average of the 4th and 5th numbers (since there are 8 numbers): $\frac{2 + 3}{2} = 2.5$.

The mode is the number that appears most frequently, which is 3.

The sum of the mean, median, and mode is $2 + 2.5 + 3 = 7.5$.

Let's examine the choices:
A. 6.5: Too low.
B. 7: Too low.
C. 7.5: Matches our calculation.
D. 8.5: Too high.
E. 9: Too high.

ANSWER 7: C

Problem 8:
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

Georgie has 6 choices for the window to enter.

After entering, Georgie has 5 remaining choices for the window to leave (since it must be different from the entry window).

The total number of ways Georgie can enter and leave is $6 \times 5 = 30$.

Let's examine the choices:
A. 12: Too low.
B. 15: Too low.
C. 18: Too low.
D. 30: Matches our calculation.
E. 36: Too high.

ANSWER 8: D

Problem 9:
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?

Let's calculate Jack's total:
1. $90.00 + 6\%$ of $90.00 = 90 + 0.06 \times 90 = 90 + 5.40 = 95.40$.
2. $20\%$ 

Problem 1:

Isabella read the book over 7 days.

- First 3 days:  
  \[
  3 \times 36 = 108
  \]
  pages

- Next 3 days:  
  \[
  3 \times 44 = 132
  \]
  pages

- Last day:  
  \[
  10
  \]
  pages

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 1: B

Problem 2:

We want the smallest possible average of four distinct positive even integers.

The smallest positive even integers are:

\[
2,\ 4,\ 6,\ 8
\]

Their average is:

\[
\frac{2+4+6+8}{4}=\frac{20}{4}=5
\]

We cannot use 0 because the integers must be positive. We also cannot repeat numbers because they must be distinct.

ANSWER 2: C

Problem 3:

The average of 4 scores is 70, so the total of the 4 scores is:

\[
4 \times 70 = 280
\]

The three known scores add to:

\[
70+80+90=240
\]

So the remaining score is:

\[
280-240=40
\]

ANSWER 3: A

Problem 4:

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The area of the square is:

\[
s^2
\]

The area of the circle is:

\[
\pi r^2
\]

They have the same area, so:

\[
s^2=\pi r^2
\]

Taking the square root of both sides:

\[
s=r\sqrt{\pi}
\]

Therefore,

\[
\frac{s}{r}=\sqrt{\pi}
\]

So the ratio of the side length of the square to the radius of the circle is \(\sqrt{\pi}\).

ANSWER 4: B

Problem 5:

Let the six consecutive positive integers be:

\[
n,\ n+1,\ n+2,\ n+3,\ n+4,\ n+5
\]

Their sum is 2013:

\[
n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=2013
\]

Combine like terms:

\[
6n+15=2013
\]

Subtract 15:

\[
6n=1998
\]

Divide by 6:

\[
n=333
\]

So the six integers are:

\[
333,\ 334,\ 335,\ 336,\ 337,\ 338
\]

The largest is 338.

ANSWER 5: B

Problem 6:

A checkerboard has 64 unit squares, so it is an \(8 \times 8\) board.

A square does not touch the outer edge only if it is not in the first row, last row, first column, or last column.

That leaves the interior squares:

\[
6 \times 6 = 36
\]

So the probability is:

\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:

The numbers are:

\[
2,\ 3,\ 0,\ 3,\ 1,\ 4,\ 0,\ 3
\]

First find the mean.

Sum:

\[
2+3+0+3+1+4+0+3=16
\]

There are 8 numbers, so the mean is:

\[
\frac{16}{8}=2
\]

Now sort the numbers:

\[
0,\ 0,\ 1,\ 2,\ 3,\ 3,\ 3,\ 4
\]

There are 8 numbers, so the median is the average of the 4th and 5th numbers:

\[
\frac{2+3}{2}=2.5
\]

The mode is the number that appears most often. The number 3 appears three times, so the mode is:

\[
3
\]

Now add mean, median, and mode:

\[
2+2.5+3=7.5
\]

ANSWER 7: C

Problem 8:

Georgie has 6 choices for the window he enters.

After entering, he must leave by a different window, so he has 5 choices for the window he leaves.

Thus the number of ways is:

\[
6 \times 5 = 30
\]

This is not 15 because entering window A and leaving window B is different from entering window B and leaving window A.

ANSWER 8: D

Problem 9:

Jack’s method:

He first adds 6% sales tax to \$90:

\[
90 \times 1.06 = 95.40
\]

Then he subtracts 20% from that total, meaning he keeps 80%:

\[
95.40 \times 0.80 = 76.32
\]

So Jack’s total is \$76.32.

Jill’s method:

She first subtracts 20% from \$90:

\[
90 \times 0.80 = 72
\]

Then she adds 6% sales tax:

\[
72 \times 1.06 = 76.32
\]

So Jill’s total is also \$76.32.

Therefore:

\[
76.32-76.32=0
\]

ANSWER 9: C

Problem 10:

The expression is:

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}+\frac{55}{10}
\]

Since all fractions have denominator 10, add the numerators:

\[
\frac{1+2+3+4+5+6+7+8+9+55}{10}
\]

Now,

\[
1+2+3+4+5+6+7+8+9=45
\]

So:

\[
\frac{45+55}{10}=\frac{100}{10}=10
\]

ANSWER 10: D

Problem 11:

The code phrase is:

\[
\text{BEST OF LUCK}
\]

The ten letters represent the digits 0 through 9 in order:

\[
B=0,\ E=1,\ S=2,\ T=3,\ O=4,\ F=5,\ L=6,\ U=7,\ C=8,\ K=9
\]

Now decode the word CLUE:

\[
C=8,\quad L=6,\quad U=7,\quad E=1
\]

So CLUE represents:

\[
8671
\]

ANSWER 11: A

Problem 12:

The sunrise was 6:57 AM, and the length of daylight was 10 hours and 24 minutes.

So the sunset time is:

\[
6:57\text{ AM} + 10\text{ hours }24\text{ minutes}
\]

First add 10 hours:

\[
6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}
\]

Now add 24 minutes:

\[
4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}
\]

So the sun really set at 5:21 PM.

ANSWER 12: B

**Problem 1:**
To find the total number of pages in the book, we need to calculate the number of pages Isabella read in each part of the week and add them together.
- For the first 3 days, she read an average of 36 pages per day: 
  3 days × 36 pages/day = 108 pages.
- For the next 3 days, she read an average of 44 pages per day: 
  3 days × 44 pages/day = 132 pages.
- On the last day (the 7th day), she read 10 pages.

Adding these amounts together gives the total number of pages:
108 + 132 + 10 = 250 pages.

ANSWER 1: B

**Problem 2:**
To find the smallest possible average of four distinct positive even integers, we must first select the four smallest distinct positive even integers. 
These integers are 2, 4, 6, and 8.
Next, we find their sum:
2 + 4 + 6 + 8 = 20
Finally, we divide the sum by the number of integers (4) to find the average:
20 / 4 = 5

ANSWER 2: C

**Problem 3:**
Let the unknown remaining score be $x$.
The average of the four scores is given as 70. The formula for the average is the sum of the scores divided by the number of scores:
(70 + 80 + 90 + $x$) / 4 = 70

Multiply both sides by 4 to find the total sum of the scores:
70 + 80 + 90 + $x$ = 280
240 + $x$ = 280

Subtract 240 from both sides:
$x$ = 40
The remaining score is 40.

ANSWER 3: A

**Problem 4:**
Let the side length of the square be $s$ and the radius of the circle be $r$.
The area of the square is $s^2$, and the area of the circle is $\pi r^2$.
We are told that their areas are equal:
$s^2 = \pi r^2$

To find the ratio of the side length to the radius ($s/r$), we take the square root of both sides:
$s = r\sqrt{\pi}$
$s / r = \sqrt{\pi}$

ANSWER 4: B

**Problem 5:**
Let the six consecutive positive integers be $x$, $x+1$, $x+2$, $x+3$, $x+4$, and $x+5$.
Their sum is:
$x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 2013$
$6x + 15 = 2013$

Subtract 15 from both sides:
$6x = 1998$

Divide by 6:
$x = 333$

The problem asks for the largest of these integers, which is $x+5$:
333 + 5 = 338

ANSWER 5: B

**Problem 6:**
A standard checkerboard is an $8 \times 8$ grid, meaning it has a total of 64 unit squares.
The squares that touch the outer edge form the border of the board. If we remove this 1-square-thick border, the remaining inner squares form a smaller $6 \times 6$ grid.
The number of squares that do *not* touch the outer edge is:
$6 \times 6 = 36$

The probability of randomly choosing one of these inner squares is the number of inner squares divided by the total number of squares:
$P = 36 / 64$

Dividing the numerator and the denominator by 4 simplifies the fraction:
$P = 9 / 16$

ANSWER 6: D

**Problem 7:**
First, let's list the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.
There are 8 numbers in total.

- **Mean:** The sum of the numbers divided by the count.
  Sum = 0 + 0 + 1 + 2 + 3 + 3 + 3 + 4 = 16
  Mean = 16 / 8 = 2

- **Median:** The middle value. Since there is an even number of values (8), the median is the average of the 4th and 5th numbers.
  The 4th number is 2, and the 5th number is 3.
  Median = (2 + 3) / 2 = 2.5

- **Mode:** The number that appears most frequently.
  The number 3 appears three times, which is more than any other number.
  Mode = 3

The sum of the mean, median, and mode is:
2 + 2.5 + 3 = 7.5

ANSWER 7: C

**Problem 8:**
Georgie the Ghost has 6 different windows to choose from when entering the house.
Because he must leave by a *different* window, he cannot use the window he entered through. This leaves him with 5 choices for his exit.
By the fundamental counting principle, the total number of ways he can enter and leave is:
6 (entry choices) × 5 (exit choices) = 30 ways.

ANSWER 8: D

**Problem 9:**
Let's calculate the final total for both clerks.
- **Jack's method:** He adds the 6% tax first, making the subtotal $90 \times 1.06$. Then he applies the 20% discount, which means the customer pays 80% of the subtotal.
  Jack's Total = $90 \times 1.06 \times 0.80$

- **Jill's method:** She applies the 20% discount first, making the subtotal $90 \times 0.80$. Then she adds the 6% tax, which means multiplying by 1.06.
  Jill's Total = $90 \times 0.80 \times 1.06$

Because multiplication is commutative (the order in which you multiply numbers does not change the product), Jack's total and Jill's total are mathematically identical.
Therefore, Jack's total minus Jill's total is exactly $0.

ANSWER 9: C

**Problem 10:**
We are asked to find the sum of the following fractions:
$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$

Since all the fractions share a common denominator of 10, we just need to add their numerators:
Sum of numerators = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55

The sum of the digits 1 through 9 is 45.
Total numerator = 45 + 55 = 100

Now divide the total numerator by the denominator:
$\frac{100}{10} = 10$

ANSWER 10: D

**Problem 11:**
The ten-letter code "BEST OF LUCK" maps to the digits 0 through 9 in order:
B = 0
E = 1
S = 2
T = 3
O = 4
F = 5
L = 6
U = 7
C = 8
K = 9

We need to find the 4-digit number for "CLUE":
C = 8
L = 6
U = 7
E = 1

Putting the digits together, "CLUE" represents 8671.

ANSWER 11: A

**Problem 12:**
To find the correct sunset time, we need to add the length of daylight to the sunrise time.
Sunrise: 6:57 AM
Length of daylight: 10 hours and 24 minutes

First, add the 10 hours to the sunrise time:
6:57 AM + 10 hours = 4:57 PM

Next, add the remaining 24 minutes to 4:57 PM:
Adding 3 minutes brings the time to 5:00 PM.
We have 21 minutes left to add (24 - 3 = 21).
5:00 PM + 21 minutes = 5:21 PM.

ANSWER 12: B

Problem 1:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

To find the total number of pages in the book, we need to add the number of pages Isabella read each day.

First, let's calculate the total number of pages she read in the first three days: $36 \times 3 = 108$ pages.
Next, we calculate the total number of pages she read in the next three days: $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.

The total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low since $108 + 132 + 10 = 250$.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 1: B

Problem 2:
What is the smallest possible average of four distinct positive even integers?

To minimize the average of four distinct positive even integers, we need to choose the smallest four distinct positive even integers.

The smallest four distinct positive even integers are 2, 4, 6, and 8.

Let's calculate their average: $\frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$.

Let's examine the choices:
A. 3: Too low since the smallest even number is 2.
B. 4: Too low since the average of 2, 4, 6, and 8 is 5.
C. 5: Matches our calculation.
D. 6: Too high.
E. 7: Too high.

ANSWER 2: C

Problem 3:
Four students take an exam. Three of their scores are 70, 80, and 90. If the average of their four scores is 70, then what is the remaining score?

To find the remaining score, we first need to find the total score of all four students.

The average score is 70, and there are 4 students, so the total score is $70 \times 4 = 280$.

The sum of the three known scores is $70 + 80 + 90 = 240$.

The remaining score is $280 - 240 = 40$.

Let's examine the choices:
A. 40: Matches our calculation.
B. 50: Too high.
C. 55: Too high.
D. 60: Too high.
E. 70: Too high.

ANSWER 3: A

Problem 4:
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

Let's denote the side length of the square as $s$ and the radius of the circle as $r$.

The area of the square is $s^2$, and the area of the circle is $\pi r^2$.

Since they have the same area, we have $s^2 = \pi r^2$.

To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$.

Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$.

Let's examine the choices:
A. $\sqrt{\pi} / 2$: Too low.
B. $\sqrt{\pi}$: Matches our calculation.
C. $\pi$: Too high.
D. $2\pi$: Too high.
E. $\pi^2$: Too high.

ANSWER 4: B

Problem 5:
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Let's denote the smallest of these integers as $n$. Then the six consecutive integers are $n$, $n+1$, $n+2$, $n+3$, $n+4$, and $n+5$.

Their sum is $n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15 = 2013$.

Solving for $n$, we get $6n = 2013 - 15 = 1998$, so $n = \frac{1998}{6} = 333$.

The largest of these six integers is $n + 5 = 333 + 5 = 338$.

Let's examine the choices:
A. 335: Too low.
B. 338: Matches our calculation.
C. 340: Too high.
D. 345: Too high.
E. 350: Too high.

ANSWER 5: B

Problem 6:
On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

A standard checkerboard is an $8 \times 8$ grid, so it has 64 unit squares.

The unit squares that touch the outer edge are those on the first and last rows and columns. There are $8 + 8 + 8 + 8 = 32$ such squares, but the 4 corner squares are counted twice, so we subtract 4 to avoid double-counting: $32 - 4 = 28$.

Thus, there are $64 - 28 = 36$ unit squares that do not touch the outer edge.

The probability that a randomly chosen unit square does not touch the outer edge is $\frac{36}{64} = \frac{9}{16}$.

Let's examine the choices:
A. $\frac{1}{16}$: Too low.
B. $\frac{7}{16}$: Too low.
C. $\frac{1}{2}$: Too low since $\frac{9}{16} > \frac{1}{2}$.
D. $\frac{9}{16}$: Matches our calculation.
E. $\frac{49}{64}$: Too high.

ANSWER 6: D

Problem 7:
What is the sum of the mean, median, and mode of the numbers 2, 3, 0, 3, 1, 4, 0, 3?

First, let's arrange the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.

The mean is the average of the numbers: $\frac{0 + 0 + 1 + 2 + 3 + 3 + 3 + 4}{8} = \frac{16}{8} = 2$.

The median is the average of the 4th and 5th numbers (since there are 8 numbers): $\frac{2 + 3}{2} = 2.5$.

The mode is the number that appears most frequently, which is 3.

The sum of the mean, median, and mode is $2 + 2.5 + 3 = 7.5$.

Let's examine the choices:
A. 6.5: Too low.
B. 7: Too low.
C. 7.5: Matches our calculation.
D. 8.5: Too high.
E. 9: Too high.

ANSWER 7: C

Problem 8:
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

Georgie has 6 choices for the window to enter.

After entering, Georgie has 5 remaining choices for the window to leave (since it must be different from the entry window).

The total number of ways Georgie can enter and leave is $6 \times 5 = 30$.

Let's examine the choices:
A. 12: Too low.
B. 15: Too low.
C. 18: Too low.
D. 30: Matches our calculation.
E. 36: Too high.

ANSWER 8: D

Problem 9:
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?

Let's calculate Jack's total:
1. $90.00 + 6\%$ of $90.00 = 90 + 0.06 \times 90 = 90 + 5.40 = 95.40$.
2. $20\%$ 

Problem 1:

Isabella read the book over 7 days.

- First 3 days:  
  \[
  3 \times 36 = 108
  \]
  pages

- Next 3 days:  
  \[
  3 \times 44 = 132
  \]
  pages

- Last day:  
  \[
  10
  \]
  pages

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 1: B

Problem 2:

We want the smallest possible average of four distinct positive even integers.

The smallest positive even integers are:

\[
2,\ 4,\ 6,\ 8
\]

Their average is:

\[
\frac{2+4+6+8}{4}=\frac{20}{4}=5
\]

We cannot use 0 because the integers must be positive. We also cannot repeat numbers because they must be distinct.

ANSWER 2: C

Problem 3:

The average of 4 scores is 70, so the total of the 4 scores is:

\[
4 \times 70 = 280
\]

The three known scores add to:

\[
70+80+90=240
\]

So the remaining score is:

\[
280-240=40
\]

ANSWER 3: A

Problem 4:

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The area of the square is:

\[
s^2
\]

The area of the circle is:

\[
\pi r^2
\]

They have the same area, so:

\[
s^2=\pi r^2
\]

Taking the square root of both sides:

\[
s=r\sqrt{\pi}
\]

Therefore,

\[
\frac{s}{r}=\sqrt{\pi}
\]

So the ratio of the side length of the square to the radius of the circle is \(\sqrt{\pi}\).

ANSWER 4: B

Problem 5:

Let the six consecutive positive integers be:

\[
n,\ n+1,\ n+2,\ n+3,\ n+4,\ n+5
\]

Their sum is 2013:

\[
n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=2013
\]

Combine like terms:

\[
6n+15=2013
\]

Subtract 15:

\[
6n=1998
\]

Divide by 6:

\[
n=333
\]

So the six integers are:

\[
333,\ 334,\ 335,\ 336,\ 337,\ 338
\]

The largest is 338.

ANSWER 5: B

Problem 6:

A checkerboard has 64 unit squares, so it is an \(8 \times 8\) board.

A square does not touch the outer edge only if it is not in the first row, last row, first column, or last column.

That leaves the interior squares:

\[
6 \times 6 = 36
\]

So the probability is:

\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:

The numbers are:

\[
2,\ 3,\ 0,\ 3,\ 1,\ 4,\ 0,\ 3
\]

First find the mean.

Sum:

\[
2+3+0+3+1+4+0+3=16
\]

There are 8 numbers, so the mean is:

\[
\frac{16}{8}=2
\]

Now sort the numbers:

\[
0,\ 0,\ 1,\ 2,\ 3,\ 3,\ 3,\ 4
\]

There are 8 numbers, so the median is the average of the 4th and 5th numbers:

\[
\frac{2+3}{2}=2.5
\]

The mode is the number that appears most often. The number 3 appears three times, so the mode is:

\[
3
\]

Now add mean, median, and mode:

\[
2+2.5+3=7.5
\]

ANSWER 7: C

Problem 8:

Georgie has 6 choices for the window he enters.

After entering, he must leave by a different window, so he has 5 choices for the window he leaves.

Thus the number of ways is:

\[
6 \times 5 = 30
\]

This is not 15 because entering window A and leaving window B is different from entering window B and leaving window A.

ANSWER 8: D

Problem 9:

Jack’s method:

He first adds 6% sales tax to \$90:

\[
90 \times 1.06 = 95.40
\]

Then he subtracts 20% from that total, meaning he keeps 80%:

\[
95.40 \times 0.80 = 76.32
\]

So Jack’s total is \$76.32.

Jill’s method:

She first subtracts 20% from \$90:

\[
90 \times 0.80 = 72
\]

Then she adds 6% sales tax:

\[
72 \times 1.06 = 76.32
\]

So Jill’s total is also \$76.32.

Therefore:

\[
76.32-76.32=0
\]

ANSWER 9: C

Problem 10:

The expression is:

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}+\frac{55}{10}
\]

Since all fractions have denominator 10, add the numerators:

\[
\frac{1+2+3+4+5+6+7+8+9+55}{10}
\]

Now,

\[
1+2+3+4+5+6+7+8+9=45
\]

So:

\[
\frac{45+55}{10}=\frac{100}{10}=10
\]

ANSWER 10: D

Problem 11:

The code phrase is:

\[
\text{BEST OF LUCK}
\]

The ten letters represent the digits 0 through 9 in order:

\[
B=0,\ E=1,\ S=2,\ T=3,\ O=4,\ F=5,\ L=6,\ U=7,\ C=8,\ K=9
\]

Now decode the word CLUE:

\[
C=8,\quad L=6,\quad U=7,\quad E=1
\]

So CLUE represents:

\[
8671
\]

ANSWER 11: A

Problem 12:

The sunrise was 6:57 AM, and the length of daylight was 10 hours and 24 minutes.

So the sunset time is:

\[
6:57\text{ AM} + 10\text{ hours }24\text{ minutes}
\]

First add 10 hours:

\[
6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}
\]

Now add 24 minutes:

\[
4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}
\]

So the sun really set at 5:21 PM.

ANSWER 12: B

**Problem 1:**
To find the total number of pages in the book, we need to calculate the number of pages Isabella read in each part of the week and add them together.
- For the first 3 days, she read an average of 36 pages per day: 
  3 days × 36 pages/day = 108 pages.
- For the next 3 days, she read an average of 44 pages per day: 
  3 days × 44 pages/day = 132 pages.
- On the last day (the 7th day), she read 10 pages.

Adding these amounts together gives the total number of pages:
108 + 132 + 10 = 250 pages.

ANSWER 1: B

**Problem 2:**
To find the smallest possible average of four distinct positive even integers, we must first select the four smallest distinct positive even integers. 
These integers are 2, 4, 6, and 8.
Next, we find their sum:
2 + 4 + 6 + 8 = 20
Finally, we divide the sum by the number of integers (4) to find the average:
20 / 4 = 5

ANSWER 2: C

**Problem 3:**
Let the unknown remaining score be $x$.
The average of the four scores is given as 70. The formula for the average is the sum of the scores divided by the number of scores:
(70 + 80 + 90 + $x$) / 4 = 70

Multiply both sides by 4 to find the total sum of the scores:
70 + 80 + 90 + $x$ = 280
240 + $x$ = 280

Subtract 240 from both sides:
$x$ = 40
The remaining score is 40.

ANSWER 3: A

**Problem 4:**
Let the side length of the square be $s$ and the radius of the circle be $r$.
The area of the square is $s^2$, and the area of the circle is $\pi r^2$.
We are told that their areas are equal:
$s^2 = \pi r^2$

To find the ratio of the side length to the radius ($s/r$), we take the square root of both sides:
$s = r\sqrt{\pi}$
$s / r = \sqrt{\pi}$

ANSWER 4: B

**Problem 5:**
Let the six consecutive positive integers be $x$, $x+1$, $x+2$, $x+3$, $x+4$, and $x+5$.
Their sum is:
$x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 2013$
$6x + 15 = 2013$

Subtract 15 from both sides:
$6x = 1998$

Divide by 6:
$x = 333$

The problem asks for the largest of these integers, which is $x+5$:
333 + 5 = 338

ANSWER 5: B

**Problem 6:**
A standard checkerboard is an $8 \times 8$ grid, meaning it has a total of 64 unit squares.
The squares that touch the outer edge form the border of the board. If we remove this 1-square-thick border, the remaining inner squares form a smaller $6 \times 6$ grid.
The number of squares that do *not* touch the outer edge is:
$6 \times 6 = 36$

The probability of randomly choosing one of these inner squares is the number of inner squares divided by the total number of squares:
$P = 36 / 64$

Dividing the numerator and the denominator by 4 simplifies the fraction:
$P = 9 / 16$

ANSWER 6: D

**Problem 7:**
First, let's list the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.
There are 8 numbers in total.

- **Mean:** The sum of the numbers divided by the count.
  Sum = 0 + 0 + 1 + 2 + 3 + 3 + 3 + 4 = 16
  Mean = 16 / 8 = 2

- **Median:** The middle value. Since there is an even number of values (8), the median is the average of the 4th and 5th numbers.
  The 4th number is 2, and the 5th number is 3.
  Median = (2 + 3) / 2 = 2.5

- **Mode:** The number that appears most frequently.
  The number 3 appears three times, which is more than any other number.
  Mode = 3

The sum of the mean, median, and mode is:
2 + 2.5 + 3 = 7.5

ANSWER 7: C

**Problem 8:**
Georgie the Ghost has 6 different windows to choose from when entering the house.
Because he must leave by a *different* window, he cannot use the window he entered through. This leaves him with 5 choices for his exit.
By the fundamental counting principle, the total number of ways he can enter and leave is:
6 (entry choices) × 5 (exit choices) = 30 ways.

ANSWER 8: D

**Problem 9:**
Let's calculate the final total for both clerks.
- **Jack's method:** He adds the 6% tax first, making the subtotal $90 \times 1.06$. Then he applies the 20% discount, which means the customer pays 80% of the subtotal.
  Jack's Total = $90 \times 1.06 \times 0.80$

- **Jill's method:** She applies the 20% discount first, making the subtotal $90 \times 0.80$. Then she adds the 6% tax, which means multiplying by 1.06.
  Jill's Total = $90 \times 0.80 \times 1.06$

Because multiplication is commutative (the order in which you multiply numbers does not change the product), Jack's total and Jill's total are mathematically identical.
Therefore, Jack's total minus Jill's total is exactly $0.

ANSWER 9: C

**Problem 10:**
We are asked to find the sum of the following fractions:
$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$

Since all the fractions share a common denominator of 10, we just need to add their numerators:
Sum of numerators = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55

The sum of the digits 1 through 9 is 45.
Total numerator = 45 + 55 = 100

Now divide the total numerator by the denominator:
$\frac{100}{10} = 10$

ANSWER 10: D

**Problem 11:**
The ten-letter code "BEST OF LUCK" maps to the digits 0 through 9 in order:
B = 0
E = 1
S = 2
T = 3
O = 4
F = 5
L = 6
U = 7
C = 8
K = 9

We need to find the 4-digit number for "CLUE":
C = 8
L = 6
U = 7
E = 1

Putting the digits together, "CLUE" represents 8671.

ANSWER 11: A

**Problem 12:**
To find the correct sunset time, we need to add the length of daylight to the sunrise time.
Sunrise: 6:57 AM
Length of daylight: 10 hours and 24 minutes

First, add the 10 hours to the sunrise time:
6:57 AM + 10 hours = 4:57 PM

Next, add the remaining 24 minutes to 4:57 PM:
Adding 3 minutes brings the time to 5:00 PM.
We have 21 minutes left to add (24 - 3 = 21).
5:00 PM + 21 minutes = 5:21 PM.

ANSWER 12: B

Problem 1:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

To find the total number of pages in the book, we need to add the number of pages Isabella read each day.

First, let's calculate the total number of pages she read in the first three days: $36 \times 3 = 108$ pages.
Next, we calculate the total number of pages she read in the next three days: $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.

The total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low since $108 + 132 + 10 = 250$.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 1: B

Problem 2:
What is the smallest possible average of four distinct positive even integers?

To minimize the average of four distinct positive even integers, we need to choose the smallest four distinct positive even integers.

The smallest four distinct positive even integers are 2, 4, 6, and 8.

Let's calculate their average: $\frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$.

Let's examine the choices:
A. 3: Too low since the smallest even number is 2.
B. 4: Too low since the average of 2, 4, 6, and 8 is 5.
C. 5: Matches our calculation.
D. 6: Too high.
E. 7: Too high.

ANSWER 2: C

Problem 3:
Four students take an exam. Three of their scores are 70, 80, and 90. If the average of their four scores is 70, then what is the remaining score?

To find the remaining score, we first need to find the total score of all four students.

The average score is 70, and there are 4 students, so the total score is $70 \times 4 = 280$.

The sum of the three known scores is $70 + 80 + 90 = 240$.

The remaining score is $280 - 240 = 40$.

Let's examine the choices:
A. 40: Matches our calculation.
B. 50: Too high.
C. 55: Too high.
D. 60: Too high.
E. 70: Too high.

ANSWER 3: A

Problem 4:
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

Let's denote the side length of the square as $s$ and the radius of the circle as $r$.

The area of the square is $s^2$, and the area of the circle is $\pi r^2$.

Since they have the same area, we have $s^2 = \pi r^2$.

To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$.

Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$.

Let's examine the choices:
A. $\sqrt{\pi} / 2$: Too low.
B. $\sqrt{\pi}$: Matches our calculation.
C. $\pi$: Too high.
D. $2\pi$: Too high.
E. $\pi^2$: Too high.

ANSWER 4: B

Problem 5:
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Let's denote the smallest of these integers as $n$. Then the six consecutive integers are $n$, $n+1$, $n+2$, $n+3$, $n+4$, and $n+5$.

Their sum is $n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15 = 2013$.

Solving for $n$, we get $6n = 2013 - 15 = 1998$, so $n = \frac{1998}{6} = 333$.

The largest of these six integers is $n + 5 = 333 + 5 = 338$.

Let's examine the choices:
A. 335: Too low.
B. 338: Matches our calculation.
C. 340: Too high.
D. 345: Too high.
E. 350: Too high.

ANSWER 5: B

Problem 6:
On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

A standard checkerboard is an $8 \times 8$ grid, so it has 64 unit squares.

The unit squares that touch the outer edge are those on the first and last rows and columns. There are $8 + 8 + 8 + 8 = 32$ such squares, but the 4 corner squares are counted twice, so we subtract 4 to avoid double-counting: $32 - 4 = 28$.

Thus, there are $64 - 28 = 36$ unit squares that do not touch the outer edge.

The probability that a randomly chosen unit square does not touch the outer edge is $\frac{36}{64} = \frac{9}{16}$.

Let's examine the choices:
A. $\frac{1}{16}$: Too low.
B. $\frac{7}{16}$: Too low.
C. $\frac{1}{2}$: Too low since $\frac{9}{16} > \frac{1}{2}$.
D. $\frac{9}{16}$: Matches our calculation.
E. $\frac{49}{64}$: Too high.

ANSWER 6: D

Problem 7:
What is the sum of the mean, median, and mode of the numbers 2, 3, 0, 3, 1, 4, 0, 3?

First, let's arrange the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.

The mean is the average of the numbers: $\frac{0 + 0 + 1 + 2 + 3 + 3 + 3 + 4}{8} = \frac{16}{8} = 2$.

The median is the average of the 4th and 5th numbers (since there are 8 numbers): $\frac{2 + 3}{2} = 2.5$.

The mode is the number that appears most frequently, which is 3.

The sum of the mean, median, and mode is $2 + 2.5 + 3 = 7.5$.

Let's examine the choices:
A. 6.5: Too low.
B. 7: Too low.
C. 7.5: Matches our calculation.
D. 8.5: Too high.
E. 9: Too high.

ANSWER 7: C

Problem 8:
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

Georgie has 6 choices for the window to enter.

After entering, Georgie has 5 remaining choices for the window to leave (since it must be different from the entry window).

The total number of ways Georgie can enter and leave is $6 \times 5 = 30$.

Let's examine the choices:
A. 12: Too low.
B. 15: Too low.
C. 18: Too low.
D. 30: Matches our calculation.
E. 36: Too high.

ANSWER 8: D

Problem 9:
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?

Let's calculate Jack's total:
1. $90.00 + 6\%$ of $90.00 = 90 + 0.06 \times 90 = 90 + 5.40 = 95.40$.
2. $20\%$ 

Problem 1:

Isabella read the book over 7 days.

- First 3 days:  
  \[
  3 \times 36 = 108
  \]
  pages

- Next 3 days:  
  \[
  3 \times 44 = 132
  \]
  pages

- Last day:  
  \[
  10
  \]
  pages

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 1: B

Problem 2:

We want the smallest possible average of four distinct positive even integers.

The smallest positive even integers are:

\[
2,\ 4,\ 6,\ 8
\]

Their average is:

\[
\frac{2+4+6+8}{4}=\frac{20}{4}=5
\]

We cannot use 0 because the integers must be positive. We also cannot repeat numbers because they must be distinct.

ANSWER 2: C

Problem 3:

The average of 4 scores is 70, so the total of the 4 scores is:

\[
4 \times 70 = 280
\]

The three known scores add to:

\[
70+80+90=240
\]

So the remaining score is:

\[
280-240=40
\]

ANSWER 3: A

Problem 4:

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The area of the square is:

\[
s^2
\]

The area of the circle is:

\[
\pi r^2
\]

They have the same area, so:

\[
s^2=\pi r^2
\]

Taking the square root of both sides:

\[
s=r\sqrt{\pi}
\]

Therefore,

\[
\frac{s}{r}=\sqrt{\pi}
\]

So the ratio of the side length of the square to the radius of the circle is \(\sqrt{\pi}\).

ANSWER 4: B

Problem 5:

Let the six consecutive positive integers be:

\[
n,\ n+1,\ n+2,\ n+3,\ n+4,\ n+5
\]

Their sum is 2013:

\[
n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=2013
\]

Combine like terms:

\[
6n+15=2013
\]

Subtract 15:

\[
6n=1998
\]

Divide by 6:

\[
n=333
\]

So the six integers are:

\[
333,\ 334,\ 335,\ 336,\ 337,\ 338
\]

The largest is 338.

ANSWER 5: B

Problem 6:

A checkerboard has 64 unit squares, so it is an \(8 \times 8\) board.

A square does not touch the outer edge only if it is not in the first row, last row, first column, or last column.

That leaves the interior squares:

\[
6 \times 6 = 36
\]

So the probability is:

\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:

The numbers are:

\[
2,\ 3,\ 0,\ 3,\ 1,\ 4,\ 0,\ 3
\]

First find the mean.

Sum:

\[
2+3+0+3+1+4+0+3=16
\]

There are 8 numbers, so the mean is:

\[
\frac{16}{8}=2
\]

Now sort the numbers:

\[
0,\ 0,\ 1,\ 2,\ 3,\ 3,\ 3,\ 4
\]

There are 8 numbers, so the median is the average of the 4th and 5th numbers:

\[
\frac{2+3}{2}=2.5
\]

The mode is the number that appears most often. The number 3 appears three times, so the mode is:

\[
3
\]

Now add mean, median, and mode:

\[
2+2.5+3=7.5
\]

ANSWER 7: C

Problem 8:

Georgie has 6 choices for the window he enters.

After entering, he must leave by a different window, so he has 5 choices for the window he leaves.

Thus the number of ways is:

\[
6 \times 5 = 30
\]

This is not 15 because entering window A and leaving window B is different from entering window B and leaving window A.

ANSWER 8: D

Problem 9:

Jack’s method:

He first adds 6% sales tax to \$90:

\[
90 \times 1.06 = 95.40
\]

Then he subtracts 20% from that total, meaning he keeps 80%:

\[
95.40 \times 0.80 = 76.32
\]

So Jack’s total is \$76.32.

Jill’s method:

She first subtracts 20% from \$90:

\[
90 \times 0.80 = 72
\]

Then she adds 6% sales tax:

\[
72 \times 1.06 = 76.32
\]

So Jill’s total is also \$76.32.

Therefore:

\[
76.32-76.32=0
\]

ANSWER 9: C

Problem 10:

The expression is:

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}+\frac{55}{10}
\]

Since all fractions have denominator 10, add the numerators:

\[
\frac{1+2+3+4+5+6+7+8+9+55}{10}
\]

Now,

\[
1+2+3+4+5+6+7+8+9=45
\]

So:

\[
\frac{45+55}{10}=\frac{100}{10}=10
\]

ANSWER 10: D

Problem 11:

The code phrase is:

\[
\text{BEST OF LUCK}
\]

The ten letters represent the digits 0 through 9 in order:

\[
B=0,\ E=1,\ S=2,\ T=3,\ O=4,\ F=5,\ L=6,\ U=7,\ C=8,\ K=9
\]

Now decode the word CLUE:

\[
C=8,\quad L=6,\quad U=7,\quad E=1
\]

So CLUE represents:

\[
8671
\]

ANSWER 11: A

Problem 12:

The sunrise was 6:57 AM, and the length of daylight was 10 hours and 24 minutes.

So the sunset time is:

\[
6:57\text{ AM} + 10\text{ hours }24\text{ minutes}
\]

First add 10 hours:

\[
6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}
\]

Now add 24 minutes:

\[
4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}
\]

So the sun really set at 5:21 PM.

ANSWER 12: B

**Problem 1:**
To find the total number of pages in the book, we need to calculate the number of pages Isabella read in each part of the week and add them together.
- For the first 3 days, she read an average of 36 pages per day: 
  3 days × 36 pages/day = 108 pages.
- For the next 3 days, she read an average of 44 pages per day: 
  3 days × 44 pages/day = 132 pages.
- On the last day (the 7th day), she read 10 pages.

Adding these amounts together gives the total number of pages:
108 + 132 + 10 = 250 pages.

ANSWER 1: B

**Problem 2:**
To find the smallest possible average of four distinct positive even integers, we must first select the four smallest distinct positive even integers. 
These integers are 2, 4, 6, and 8.
Next, we find their sum:
2 + 4 + 6 + 8 = 20
Finally, we divide the sum by the number of integers (4) to find the average:
20 / 4 = 5

ANSWER 2: C

**Problem 3:**
Let the unknown remaining score be $x$.
The average of the four scores is given as 70. The formula for the average is the sum of the scores divided by the number of scores:
(70 + 80 + 90 + $x$) / 4 = 70

Multiply both sides by 4 to find the total sum of the scores:
70 + 80 + 90 + $x$ = 280
240 + $x$ = 280

Subtract 240 from both sides:
$x$ = 40
The remaining score is 40.

ANSWER 3: A

**Problem 4:**
Let the side length of the square be $s$ and the radius of the circle be $r$.
The area of the square is $s^2$, and the area of the circle is $\pi r^2$.
We are told that their areas are equal:
$s^2 = \pi r^2$

To find the ratio of the side length to the radius ($s/r$), we take the square root of both sides:
$s = r\sqrt{\pi}$
$s / r = \sqrt{\pi}$

ANSWER 4: B

**Problem 5:**
Let the six consecutive positive integers be $x$, $x+1$, $x+2$, $x+3$, $x+4$, and $x+5$.
Their sum is:
$x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 2013$
$6x + 15 = 2013$

Subtract 15 from both sides:
$6x = 1998$

Divide by 6:
$x = 333$

The problem asks for the largest of these integers, which is $x+5$:
333 + 5 = 338

ANSWER 5: B

**Problem 6:**
A standard checkerboard is an $8 \times 8$ grid, meaning it has a total of 64 unit squares.
The squares that touch the outer edge form the border of the board. If we remove this 1-square-thick border, the remaining inner squares form a smaller $6 \times 6$ grid.
The number of squares that do *not* touch the outer edge is:
$6 \times 6 = 36$

The probability of randomly choosing one of these inner squares is the number of inner squares divided by the total number of squares:
$P = 36 / 64$

Dividing the numerator and the denominator by 4 simplifies the fraction:
$P = 9 / 16$

ANSWER 6: D

**Problem 7:**
First, let's list the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.
There are 8 numbers in total.

- **Mean:** The sum of the numbers divided by the count.
  Sum = 0 + 0 + 1 + 2 + 3 + 3 + 3 + 4 = 16
  Mean = 16 / 8 = 2

- **Median:** The middle value. Since there is an even number of values (8), the median is the average of the 4th and 5th numbers.
  The 4th number is 2, and the 5th number is 3.
  Median = (2 + 3) / 2 = 2.5

- **Mode:** The number that appears most frequently.
  The number 3 appears three times, which is more than any other number.
  Mode = 3

The sum of the mean, median, and mode is:
2 + 2.5 + 3 = 7.5

ANSWER 7: C

**Problem 8:**
Georgie the Ghost has 6 different windows to choose from when entering the house.
Because he must leave by a *different* window, he cannot use the window he entered through. This leaves him with 5 choices for his exit.
By the fundamental counting principle, the total number of ways he can enter and leave is:
6 (entry choices) × 5 (exit choices) = 30 ways.

ANSWER 8: D

**Problem 9:**
Let's calculate the final total for both clerks.
- **Jack's method:** He adds the 6% tax first, making the subtotal $90 \times 1.06$. Then he applies the 20% discount, which means the customer pays 80% of the subtotal.
  Jack's Total = $90 \times 1.06 \times 0.80$

- **Jill's method:** She applies the 20% discount first, making the subtotal $90 \times 0.80$. Then she adds the 6% tax, which means multiplying by 1.06.
  Jill's Total = $90 \times 0.80 \times 1.06$

Because multiplication is commutative (the order in which you multiply numbers does not change the product), Jack's total and Jill's total are mathematically identical.
Therefore, Jack's total minus Jill's total is exactly $0.

ANSWER 9: C

**Problem 10:**
We are asked to find the sum of the following fractions:
$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$

Since all the fractions share a common denominator of 10, we just need to add their numerators:
Sum of numerators = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55

The sum of the digits 1 through 9 is 45.
Total numerator = 45 + 55 = 100

Now divide the total numerator by the denominator:
$\frac{100}{10} = 10$

ANSWER 10: D

**Problem 11:**
The ten-letter code "BEST OF LUCK" maps to the digits 0 through 9 in order:
B = 0
E = 1
S = 2
T = 3
O = 4
F = 5
L = 6
U = 7
C = 8
K = 9

We need to find the 4-digit number for "CLUE":
C = 8
L = 6
U = 7
E = 1

Putting the digits together, "CLUE" represents 8671.

ANSWER 11: A

**Problem 12:**
To find the correct sunset time, we need to add the length of daylight to the sunrise time.
Sunrise: 6:57 AM
Length of daylight: 10 hours and 24 minutes

First, add the 10 hours to the sunrise time:
6:57 AM + 10 hours = 4:57 PM

Next, add the remaining 24 minutes to 4:57 PM:
Adding 3 minutes brings the time to 5:00 PM.
We have 21 minutes left to add (24 - 3 = 21).
5:00 PM + 21 minutes = 5:21 PM.

ANSWER 12: B

Problem 1:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

To find the total number of pages in the book, we need to add the number of pages Isabella read each day.

First, let's calculate the total number of pages she read in the first three days: $36 \times 3 = 108$ pages.
Next, we calculate the total number of pages she read in the next three days: $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.

The total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low since $108 + 132 + 10 = 250$.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 1: B

Problem 2:
What is the smallest possible average of four distinct positive even integers?

To minimize the average of four distinct positive even integers, we need to choose the smallest four distinct positive even integers.

The smallest four distinct positive even integers are 2, 4, 6, and 8.

Let's calculate their average: $\frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$.

Let's examine the choices:
A. 3: Too low since the smallest even number is 2.
B. 4: Too low since the average of 2, 4, 6, and 8 is 5.
C. 5: Matches our calculation.
D. 6: Too high.
E. 7: Too high.

ANSWER 2: C

Problem 3:
Four students take an exam. Three of their scores are 70, 80, and 90. If the average of their four scores is 70, then what is the remaining score?

To find the remaining score, we first need to find the total score of all four students.

The average score is 70, and there are 4 students, so the total score is $70 \times 4 = 280$.

The sum of the three known scores is $70 + 80 + 90 = 240$.

The remaining score is $280 - 240 = 40$.

Let's examine the choices:
A. 40: Matches our calculation.
B. 50: Too high.
C. 55: Too high.
D. 60: Too high.
E. 70: Too high.

ANSWER 3: A

Problem 4:
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

Let's denote the side length of the square as $s$ and the radius of the circle as $r$.

The area of the square is $s^2$, and the area of the circle is $\pi r^2$.

Since they have the same area, we have $s^2 = \pi r^2$.

To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$.

Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$.

Let's examine the choices:
A. $\sqrt{\pi} / 2$: Too low.
B. $\sqrt{\pi}$: Matches our calculation.
C. $\pi$: Too high.
D. $2\pi$: Too high.
E. $\pi^2$: Too high.

ANSWER 4: B

Problem 5:
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Let's denote the smallest of these integers as $n$. Then the six consecutive integers are $n$, $n+1$, $n+2$, $n+3$, $n+4$, and $n+5$.

Their sum is $n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15 = 2013$.

Solving for $n$, we get $6n = 2013 - 15 = 1998$, so $n = \frac{1998}{6} = 333$.

The largest of these six integers is $n + 5 = 333 + 5 = 338$.

Let's examine the choices:
A. 335: Too low.
B. 338: Matches our calculation.
C. 340: Too high.
D. 345: Too high.
E. 350: Too high.

ANSWER 5: B

Problem 6:
On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

A standard checkerboard is an $8 \times 8$ grid, so it has 64 unit squares.

The unit squares that touch the outer edge are those on the first and last rows and columns. There are $8 + 8 + 8 + 8 = 32$ such squares, but the 4 corner squares are counted twice, so we subtract 4 to avoid double-counting: $32 - 4 = 28$.

Thus, there are $64 - 28 = 36$ unit squares that do not touch the outer edge.

The probability that a randomly chosen unit square does not touch the outer edge is $\frac{36}{64} = \frac{9}{16}$.

Let's examine the choices:
A. $\frac{1}{16}$: Too low.
B. $\frac{7}{16}$: Too low.
C. $\frac{1}{2}$: Too low since $\frac{9}{16} > \frac{1}{2}$.
D. $\frac{9}{16}$: Matches our calculation.
E. $\frac{49}{64}$: Too high.

ANSWER 6: D

Problem 7:
What is the sum of the mean, median, and mode of the numbers 2, 3, 0, 3, 1, 4, 0, 3?

First, let's arrange the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.

The mean is the average of the numbers: $\frac{0 + 0 + 1 + 2 + 3 + 3 + 3 + 4}{8} = \frac{16}{8} = 2$.

The median is the average of the 4th and 5th numbers (since there are 8 numbers): $\frac{2 + 3}{2} = 2.5$.

The mode is the number that appears most frequently, which is 3.

The sum of the mean, median, and mode is $2 + 2.5 + 3 = 7.5$.

Let's examine the choices:
A. 6.5: Too low.
B. 7: Too low.
C. 7.5: Matches our calculation.
D. 8.5: Too high.
E. 9: Too high.

ANSWER 7: C

Problem 8:
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

Georgie has 6 choices for the window to enter.

After entering, Georgie has 5 remaining choices for the window to leave (since it must be different from the entry window).

The total number of ways Georgie can enter and leave is $6 \times 5 = 30$.

Let's examine the choices:
A. 12: Too low.
B. 15: Too low.
C. 18: Too low.
D. 30: Matches our calculation.
E. 36: Too high.

ANSWER 8: D

Problem 9:
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?

Let's calculate Jack's total:
1. $90.00 + 6\%$ of $90.00 = 90 + 0.06 \times 90 = 90 + 5.40 = 95.40$.
2. $20\%$ 

Problem 1:

Isabella read the book over 7 days.

- First 3 days:  
  \[
  3 \times 36 = 108
  \]
  pages

- Next 3 days:  
  \[
  3 \times 44 = 132
  \]
  pages

- Last day:  
  \[
  10
  \]
  pages

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 1: B

Problem 2:

We want the smallest possible average of four distinct positive even integers.

The smallest positive even integers are:

\[
2,\ 4,\ 6,\ 8
\]

Their average is:

\[
\frac{2+4+6+8}{4}=\frac{20}{4}=5
\]

We cannot use 0 because the integers must be positive. We also cannot repeat numbers because they must be distinct.

ANSWER 2: C

Problem 3:

The average of 4 scores is 70, so the total of the 4 scores is:

\[
4 \times 70 = 280
\]

The three known scores add to:

\[
70+80+90=240
\]

So the remaining score is:

\[
280-240=40
\]

ANSWER 3: A

Problem 4:

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The area of the square is:

\[
s^2
\]

The area of the circle is:

\[
\pi r^2
\]

They have the same area, so:

\[
s^2=\pi r^2
\]

Taking the square root of both sides:

\[
s=r\sqrt{\pi}
\]

Therefore,

\[
\frac{s}{r}=\sqrt{\pi}
\]

So the ratio of the side length of the square to the radius of the circle is \(\sqrt{\pi}\).

ANSWER 4: B

Problem 5:

Let the six consecutive positive integers be:

\[
n,\ n+1,\ n+2,\ n+3,\ n+4,\ n+5
\]

Their sum is 2013:

\[
n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=2013
\]

Combine like terms:

\[
6n+15=2013
\]

Subtract 15:

\[
6n=1998
\]

Divide by 6:

\[
n=333
\]

So the six integers are:

\[
333,\ 334,\ 335,\ 336,\ 337,\ 338
\]

The largest is 338.

ANSWER 5: B

Problem 6:

A checkerboard has 64 unit squares, so it is an \(8 \times 8\) board.

A square does not touch the outer edge only if it is not in the first row, last row, first column, or last column.

That leaves the interior squares:

\[
6 \times 6 = 36
\]

So the probability is:

\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:

The numbers are:

\[
2,\ 3,\ 0,\ 3,\ 1,\ 4,\ 0,\ 3
\]

First find the mean.

Sum:

\[
2+3+0+3+1+4+0+3=16
\]

There are 8 numbers, so the mean is:

\[
\frac{16}{8}=2
\]

Now sort the numbers:

\[
0,\ 0,\ 1,\ 2,\ 3,\ 3,\ 3,\ 4
\]

There are 8 numbers, so the median is the average of the 4th and 5th numbers:

\[
\frac{2+3}{2}=2.5
\]

The mode is the number that appears most often. The number 3 appears three times, so the mode is:

\[
3
\]

Now add mean, median, and mode:

\[
2+2.5+3=7.5
\]

ANSWER 7: C

Problem 8:

Georgie has 6 choices for the window he enters.

After entering, he must leave by a different window, so he has 5 choices for the window he leaves.

Thus the number of ways is:

\[
6 \times 5 = 30
\]

This is not 15 because entering window A and leaving window B is different from entering window B and leaving window A.

ANSWER 8: D

Problem 9:

Jack’s method:

He first adds 6% sales tax to \$90:

\[
90 \times 1.06 = 95.40
\]

Then he subtracts 20% from that total, meaning he keeps 80%:

\[
95.40 \times 0.80 = 76.32
\]

So Jack’s total is \$76.32.

Jill’s method:

She first subtracts 20% from \$90:

\[
90 \times 0.80 = 72
\]

Then she adds 6% sales tax:

\[
72 \times 1.06 = 76.32
\]

So Jill’s total is also \$76.32.

Therefore:

\[
76.32-76.32=0
\]

ANSWER 9: C

Problem 10:

The expression is:

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}+\frac{55}{10}
\]

Since all fractions have denominator 10, add the numerators:

\[
\frac{1+2+3+4+5+6+7+8+9+55}{10}
\]

Now,

\[
1+2+3+4+5+6+7+8+9=45
\]

So:

\[
\frac{45+55}{10}=\frac{100}{10}=10
\]

ANSWER 10: D

Problem 11:

The code phrase is:

\[
\text{BEST OF LUCK}
\]

The ten letters represent the digits 0 through 9 in order:

\[
B=0,\ E=1,\ S=2,\ T=3,\ O=4,\ F=5,\ L=6,\ U=7,\ C=8,\ K=9
\]

Now decode the word CLUE:

\[
C=8,\quad L=6,\quad U=7,\quad E=1
\]

So CLUE represents:

\[
8671
\]

ANSWER 11: A

Problem 12:

The sunrise was 6:57 AM, and the length of daylight was 10 hours and 24 minutes.

So the sunset time is:

\[
6:57\text{ AM} + 10\text{ hours }24\text{ minutes}
\]

First add 10 hours:

\[
6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}
\]

Now add 24 minutes:

\[
4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}
\]

So the sun really set at 5:21 PM.

ANSWER 12: B

**Problem 1:**
To find the total number of pages in the book, we need to calculate the number of pages Isabella read in each part of the week and add them together.
- For the first 3 days, she read an average of 36 pages per day: 
  3 days × 36 pages/day = 108 pages.
- For the next 3 days, she read an average of 44 pages per day: 
  3 days × 44 pages/day = 132 pages.
- On the last day (the 7th day), she read 10 pages.

Adding these amounts together gives the total number of pages:
108 + 132 + 10 = 250 pages.

ANSWER 1: B

**Problem 2:**
To find the smallest possible average of four distinct positive even integers, we must first select the four smallest distinct positive even integers. 
These integers are 2, 4, 6, and 8.
Next, we find their sum:
2 + 4 + 6 + 8 = 20
Finally, we divide the sum by the number of integers (4) to find the average:
20 / 4 = 5

ANSWER 2: C

**Problem 3:**
Let the unknown remaining score be $x$.
The average of the four scores is given as 70. The formula for the average is the sum of the scores divided by the number of scores:
(70 + 80 + 90 + $x$) / 4 = 70

Multiply both sides by 4 to find the total sum of the scores:
70 + 80 + 90 + $x$ = 280
240 + $x$ = 280

Subtract 240 from both sides:
$x$ = 40
The remaining score is 40.

ANSWER 3: A

**Problem 4:**
Let the side length of the square be $s$ and the radius of the circle be $r$.
The area of the square is $s^2$, and the area of the circle is $\pi r^2$.
We are told that their areas are equal:
$s^2 = \pi r^2$

To find the ratio of the side length to the radius ($s/r$), we take the square root of both sides:
$s = r\sqrt{\pi}$
$s / r = \sqrt{\pi}$

ANSWER 4: B

**Problem 5:**
Let the six consecutive positive integers be $x$, $x+1$, $x+2$, $x+3$, $x+4$, and $x+5$.
Their sum is:
$x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 2013$
$6x + 15 = 2013$

Subtract 15 from both sides:
$6x = 1998$

Divide by 6:
$x = 333$

The problem asks for the largest of these integers, which is $x+5$:
333 + 5 = 338

ANSWER 5: B

**Problem 6:**
A standard checkerboard is an $8 \times 8$ grid, meaning it has a total of 64 unit squares.
The squares that touch the outer edge form the border of the board. If we remove this 1-square-thick border, the remaining inner squares form a smaller $6 \times 6$ grid.
The number of squares that do *not* touch the outer edge is:
$6 \times 6 = 36$

The probability of randomly choosing one of these inner squares is the number of inner squares divided by the total number of squares:
$P = 36 / 64$

Dividing the numerator and the denominator by 4 simplifies the fraction:
$P = 9 / 16$

ANSWER 6: D

**Problem 7:**
First, let's list the numbers in ascending order: 0, 0, 1, 2, 3, 3, 3, 4.
There are 8 numbers in total.

- **Mean:** The sum of the numbers divided by the count.
  Sum = 0 + 0 + 1 + 2 + 3 + 3 + 3 + 4 = 16
  Mean = 16 / 8 = 2

- **Median:** The middle value. Since there is an even number of values (8), the median is the average of the 4th and 5th numbers.
  The 4th number is 2, and the 5th number is 3.
  Median = (2 + 3) / 2 = 2.5

- **Mode:** The number that appears most frequently.
  The number 3 appears three times, which is more than any other number.
  Mode = 3

The sum of the mean, median, and mode is:
2 + 2.5 + 3 = 7.5

ANSWER 7: C

**Problem 8:**
Georgie the Ghost has 6 different windows to choose from when entering the house.
Because he must leave by a *different* window, he cannot use the window he entered through. This leaves him with 5 choices for his exit.
By the fundamental counting principle, the total number of ways he can enter and leave is:
6 (entry choices) × 5 (exit choices) = 30 ways.

ANSWER 8: D

**Problem 9:**
Let's calculate the final total for both clerks.
- **Jack's method:** He adds the 6% tax first, making the subtotal $90 \times 1.06$. Then he applies the 20% discount, which means the customer pays 80% of the subtotal.
  Jack's Total = $90 \times 1.06 \times 0.80$

- **Jill's method:** She applies the 20% discount first, making the subtotal $90 \times 0.80$. Then she adds the 6% tax, which means multiplying by 1.06.
  Jill's Total = $90 \times 0.80 \times 1.06$

Because multiplication is commutative (the order in which you multiply numbers does not change the product), Jack's total and Jill's total are mathematically identical.
Therefore, Jack's total minus Jill's total is exactly $0.

ANSWER 9: C

**Problem 10:**
We are asked to find the sum of the following fractions:
$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$

Since all the fractions share a common denominator of 10, we just need to add their numerators:
Sum of numerators = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55

The sum of the digits 1 through 9 is 45.
Total numerator = 45 + 55 = 100

Now divide the total numerator by the denominator:
$\frac{100}{10} = 10$

ANSWER 10: D

**Problem 11:**
The ten-letter code "BEST OF LUCK" maps to the digits 0 through 9 in order:
B = 0
E = 1
S = 2
T = 3
O = 4
F = 5
L = 6
U = 7
C = 8
K = 9

We need to find the 4-digit number for "CLUE":
C = 8
L = 6
U = 7
E = 1

Putting the digits together, "CLUE" represents 8671.

ANSWER 11: A

**Problem 12:**
To find the correct sunset time, we need to add the length of daylight to the sunrise time.
Sunrise: 6:57 AM
Length of daylight: 10 hours and 24 minutes

First, add the 10 hours to the sunrise time:
6:57 AM + 10 hours = 4:57 PM

Next, add the remaining 24 minutes to 4:57 PM:
Adding 3 minutes brings the time to 5:00 PM.
We have 21 minutes left to add (24 - 3 = 21).
5:00 PM + 21 minutes = 5:21 PM.

ANSWER 12: B