🔬 Model Lab

## Step 1: Problem 1 - Restate the problem
The problem asks us to find the approximate number of rose bushes needed to surround a circular patio with a radius of 12 feet, given that the bushes are spaced about 1 foot apart.

## Step 2: Problem 1 - Calculate the circumference of the circular patio
To find the number of rose bushes needed, we first need to calculate the circumference of the circular patio. The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius. So, $C = 2\pi \times 12 = 24\pi$ feet.

## Step 3: Problem 1 - Approximate the circumference
Now, we approximate the circumference using $\pi \approx 3.14$. Thus, $C \approx 24 \times 3.14 = 75.36$ feet.

## Step 4: Problem 1 - Determine the number of rose bushes needed
Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference. So, we need approximately 75.36 bushes. Rounding to the nearest whole number gives us about 75 bushes.

## Step 5: Problem 1 - Choose the correct answer
Comparing our result to the given choices, we see that 75 is the closest value.

ANSWER 1: D

## Step 6: Problem 2 - Restate the problem
We need to find the smallest possible whole number value of $s$ such that the sides of a triangle have lengths 6.5, 10, and $s$.

## Step 7: Problem 2 - Apply the triangle inequality
For a triangle with sides of lengths $a$, $b$, and $c$, the triangle inequality states that $a + b > c$, $a + c > b$, and $b + c > a$. Applying this to our problem: $6.5 + 10 > s$, $6.5 + s > 10$, and $10 + s > 6.5$. Simplifying these gives $16.5 > s$, $s > 3.5$, and $s > -3.5$. Since $s$ is a whole number and length can't be negative, we consider $s > 3.5$ and $s < 16.5$.

## Step 8: Problem 2 - Determine the smallest whole number value for $s$
The smallest whole number greater than 3.5 is 4.

## Step 9: Problem 2 - Choose the correct answer
Thus, the smallest possible value of $s$ is 4.

ANSWER 2: B

## Step 10: Problem 3 - Restate the problem
We are asked to find the number of possible seating arrangements for Ali, Bonnie, Carlo, and Dianna in a car with four seats, where Bonnie and Carlo are the only ones who can drive.

## Step 11: Problem 3 - Calculate the number of ways to choose the driver
There are 2 choices for the driver's seat (Bonnie or Carlo).

## Step 12: Problem 3 - Calculate the number of arrangements for the remaining seats
After choosing the driver, there are 3 people left to fill the 3 remaining seats. The number of ways to arrange 3 people in 3 seats is $3! = 3 \times 2 \times 1 = 6$.

## Step 13: Problem 3 - Determine the total number of seating arrangements
The total number of seating arrangements is the product of the number of ways to choose the driver and the number of arrangements for the remaining seats: $2 \times 6 = 12$.

## Step 14: Problem 3 - Choose the correct answer
So, there are 12 possible seating arrangements.

ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
The problem asks us to evaluate the expression $1990 - 1980 + 1970 - 1960 + \cdots - 20 + 10$.

## Step 16: Problem 4 - Simplify the given expression
We can simplify the expression by pairing the terms: $(1990 - 1980) + (1970 - 1960) + \cdots + (20 - 10)$. Each pair equals 10.

## Step 17: Problem 4 - Count the number of pairs
The sequence starts at 1990 and ends at 10, decreasing by 10 each time. The number of terms is $(1990 - 10)/10 + 1 = 199$. Since we are pairing them, the number of pairs is $199/2 = 99.5$. However, we actually have 100 terms, so there are 50 pairs (as the terms are paired: 1990 with 1980, 1970 with 1960, and so on until 20 with 10).

## Step 18: Problem 4 - Calculate the sum
The sum is $10 \times 50 = 500$ (since there are 50 pairs, and each pair sums to 10, but we actually directly count the pairs as 10, 20, ..., 1990, which directly gives 50 pairs of 10).

## Step 19: Problem 4 - However, a more straightforward observation is that the series is an arithmetic series with a common difference of -10 between successive terms of the same sign.
However, a simpler way to look at it is to directly see the pattern: the expression simplifies to $10 + 10 + \cdots + 10$ (50 times), because each subtraction yields 10.

## Step 20: Problem 4 - Choose the correct answer based on the simplified calculation
$10 \times 50 = 500$ is not among the choices, but upon re-evaluation, noticing that we actually have 50 pairs and each pair contributes 10 to the sum, we indeed get $50 \times 10 = 500$. However, my initial interpretation was incorrect; let's directly compute it as the sum of an arithmetic series where the difference between consecutive terms of the same sign is -10. The correct observation should directly lead to the result without confusion.

## Step 21: Problem 4 - Re-evaluate the sum correctly
The correct sum is $990$ because there are 100 terms (from 10 to 1990), making 50 pairs, and each pair contributes 10 to the sum.

ANSWER 4: C

## Step 22: Problem 5 - Restate the problem
We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle after folding and cutting a 4-inch square piece of paper.

## Step 23: Problem 5 - Determine the dimensions of the rectangles after folding and cutting
Folding the square in half gives a rectangle of dimensions $4 \times 2$. Cutting this in half parallel to the fold gives two small rectangles ($2 \times 2$) and one large rectangle ($4 \times 2$).

## Step 24: Problem 5 - Calculate the perimeters of the small and large rectangles
The perimeter of one small rectangle is $2(2 + 2) = 8$ inches, and the perimeter of the large rectangle is $2(4 + 2) = 12$ inches.

## Step 25: Problem 5 - Calculate the ratio of the perimeters
The ratio of the perimeter of one small rectangle to the perimeter of the large rectangle is $\frac{8}{12} = \frac{2}{3}$.

## Step 26: Problem 5 - Choose the correct answer
$\frac{2}{3}$ is the ratio.

ANSWER 5: C

## Step 27: Prob

Problem 1:

We are asked how many bushes are needed to surround a circular patio of radius 12 feet, with bushes spaced about 1 foot apart.

The number of bushes is approximately the circumference of the circle in feet:

\[
C=2\pi r=2\pi(12)=24\pi
\]

Using \(\pi \approx 3.14\),

\[
24\pi \approx 24(3.14)=75.36
\]

So about 75 bushes are needed.

Choice A, 12, is just the radius.  
Choice B, 38, is about half the circumference.  
Choice C, 48, is too small.  
Choice E, 450, is closer to the area, not the circumference.

ANSWER 1: D

Problem 2:

We need the smallest whole number \(s\) so that sides \(6.5\), \(10\), and \(s\) can form a triangle.

For a triangle, the sum of any two sides must be greater than the third side.

The important inequality for the smallest \(s\) is:

\[
6.5+s>10
\]

Subtract \(6.5\):

\[
s>3.5
\]

Since \(s\) is a whole number, the smallest possible value is:

\[
s=4
\]

Check:

\[
6.5+4=10.5>10
\]

So it works.

Choice A, 3, fails because \(6.5+3=9.5<10\).  
Choices C, D, and E work, but they are not the smallest.

ANSWER 2: B

Problem 3:

There are four people and four seats: driver, front passenger, and two back seats.

Only Bonnie and Carlo can drive, so there are:

\[
2
\]

choices for the driver.

After the driver is chosen, the remaining 3 people can sit in the 3 passenger seats in:

\[
3! = 3 \times 2 \times 1 = 6
\]

ways.

Therefore, the total number of seating arrangements is:

\[
2 \times 6 = 12
\]

Choice E, 24, would be correct if all four people could drive, but only two can.  
Choice C, 6, forgets that there are two possible drivers.

ANSWER 3: D

Problem 4:

We need to evaluate:

\[
1990-1980+1970-1960+\cdots-20+10
\]

Group the terms in pairs:

\[
(1990-1980)+(1970-1960)+\cdots+(30-20)+10
\]

Each pair equals:

\[
10
\]

There are 99 such pairs, so they contribute:

\[
99 \times 10 = 990
\]

Then there is the final \(+10\):

\[
990+10=1000
\]

Choice C, 990, forgets the final \(+10\).  
Choice E, 1990, ignores most of the cancellation.  
Negative choices do not make sense because each pair is positive.

ANSWER 4: D

Problem 5:

The original square is \(4\) inches by \(4\) inches.

After folding in half vertically, the folded paper is \(2\) inches by \(4\) inches. Cutting both layers in half parallel to the fold creates cuts that, when unfolded, divide the original square into three rectangles:

- two small rectangles, each \(1 \times 4\)
- one large rectangle, \(2 \times 4\)

The perimeter of one small rectangle is:

\[
2(1+4)=10
\]

The perimeter of the large rectangle is:

\[
2(2+4)=12
\]

So the ratio is:

\[
\frac{10}{12}=\frac{5}{6}
\]

The smaller fractions such as \(\frac12\) or \(\frac34\) underestimate the perimeter of the small rectangle, since the small rectangle is still 4 inches long.

ANSWER 5: E

Problem 6:

Originally, the plate has:

- 5 choices for the first letter
- 3 choices for the second letter
- 4 choices for the third letter

So the original number of plates is:

\[
5 \times 3 \times 4 = 60
\]

We add two new letters to the sets to maximize the total number of plates.

Try the best options:

Add both letters to the set with 3 letters:

\[
5 \times 5 \times 4 = 100
\]

Additional plates:

\[
100-60=40
\]

Add one letter to the set with 3 letters and one to the set with 4 letters:

\[
5 \times 4 \times 5 = 100
\]

Additional plates:

\[
100-60=40
\]

No other distribution gives more than 100 total plates. For example, adding both to the largest set gives:

\[
7 \times 3 \times 4=84
\]

only \(24\) additional plates.

Thus the largest possible number of additional plates is 40.

ANSWER 6: D

Problem 7:

We want to be sure of getting four gumballs of the same color.

There are three colors: red, white, and blue.

In the worst case, a person could get 3 of each color without having 4 of any color:

\[
3+3+3=9
\]

After buying 9 gumballs, it is still possible to have no four matching.

But the next gumball, the 10th, must make one color reach 4.

So the least number needed is:

\[
10
\]

Choice B, 9, is not enough because \(3\) red, \(3\) white, and \(3\) blue is possible.  
Choices D and E are more than necessary.

ANSWER 7: C

Problem 8:

The islands have area \(24{,}900\) square miles, and each person needs at least \(1.5\) square miles.

So the maximum population is:

\[
\frac{24900}{1.5}=16600
\]

The population starts at 200 and triples every 25 years.

After 25 years:

\[
200 \times 3 = 600
\]

After 50 years:

\[
200 \times 3^2 = 1800
\]

After 75 years:

\[
200 \times 3^3 = 5400
\]

After 100 years:

\[
200 \times 3^4 = 200 \times 81 = 16200
\]

This is very close to \(16600\).

After 125 years:

\[
200 \times 3^5 = 48600
\]

which is far too high.

So the population reaches the maximum in about 100 years.

ANSWER 8: C

Problem 9:

We must place \(+\), \(-\), and \(\times\) exactly once in:

\[
5\ \_ \ 4\ \_ \ 6\ \_ \ 3
\]

Using normal order of operations, multiplication is done before addition and subtraction.

Try the possibilities:

\[
5+4-6\times3=5+4-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

\[
5-4\times6+3=5-24+3=-16
\]

\[
5\times4+6-3=20+6-3=23
\]

\[
5\times4-6+3=20-6+3=17
\]

Among the choices, the only possible value is 19.

ANSWER 9: E

Problem 10:

We need subsets of three different numbers whose sum is even.

The set is:

\[
\{89,95,99,132,166,173\}
\]

Classify by parity:

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

There are only 2 even numbers, so choosing 3 evens is impossible.

Now choose 2 odds and 1 even:

\[
\binom{4}{2}\binom{2}{1}=6 \times 2=12
\]

So there are 12 such subsets.

ANSWER 10: D

Problem 11:

We need the largest power of 2 dividing:

\[
13^4-11^4
\]

Use difference of squares:

\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]

Compute:

\[
13^2=169,\qquad 11^2=121
\]

So:

\[
13^

**Problem 1:**
To find out how many rose bushes are needed, we first need to calculate the circumference of the circular patio, as the bushes will be planted along this perimeter.
The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius.
Given $r = 12$ feet:
$C = 2 \times \pi \times 12 = 24\pi$

Using the approximation $\pi \approx 3.14159$:
$C \approx 24 \times 3.14159 \approx 75.4$ feet.

Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference in feet. 
75.4 is closest to 75.

ANSWER 1: D

**Problem 2:**
For any triangle, the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be strictly greater than the length of the remaining third side. 
Given the sides $6.5$, $10$, and $s$:
1) $s + 6.5 > 10 \implies s > 3.5$
2) $s + 10 > 6.5 \implies s > -3.5$
3) $6.5 + 10 > s \implies s < 16.5$

Combining these inequalities, we find that the possible range for $s$ is $3.5 < s < 16.5$. 
Since $s$ must be a whole number, the smallest possible whole number greater than 3.5 is 4.

ANSWER 2: B

**Problem 3:**
We need to find the number of possible seating arrangements for 4 people in 4 seats, with the restriction that only Bonnie or Carlo can sit in the driver's seat.
- **Driver's seat:** There are 2 choices (Bonnie or Carlo).
- **Remaining 3 seats:** The other 3 people can be arranged in the 3 remaining passenger seats in $3!$ (3 factorial) ways.
$3! = 3 \times 2 \times 1 = 6$ ways.

Total seating arrangements = (Choices for driver) $\times$ (Arrangements for passengers)
Total = $2 \times 6 = 12$

ANSWER 3: D

**Problem 4:**
We are asked to evaluate the alternating arithmetic series:
$1990 - 1980 + 1970 - 1960 + \dots - 20 + 10$

Notice that the numbers are multiples of 10, from 10 up to 1990. 
The number of terms in this sequence is $1990 \div 10 = 199$ terms.
Because there is an odd number of terms, we can group them into pairs starting from the left, leaving the very last term by itself:
$(1990 - 1980) + (1970 - 1960) + \dots + (30 - 20) + 10$

Each pair in parentheses evaluates to exactly 10. 
How many pairs are there? Since we left out the last term (10), there are $199 - 1 = 198$ terms inside the pairs. 
This gives us $198 \div 2 = 99$ pairs.

Summing the pairs and the final term:
$99 \times 10 + 10 = 990 + 10 = 1000$

ANSWER 4: D

**Problem 5:**
1. We start with a $4 \times 4$ inch square.
2. It is folded in half vertically. The folded paper is now a $2 \times 4$ inch rectangle (2 inches wide and 4 inches high). The fold is on one of the 4-inch sides.
3. Both layers are cut in half parallel to the fold. This means the 2-inch width is cut exactly down the middle, 1 inch away from the fold.
4. When unfolded, the paper is divided into three pieces:
   - The middle piece (which contained the fold) extends 1 inch on both sides of the fold, making it a $2 \times 4$ inch rectangle.
   - The two edge pieces are each $1 \times 4$ inch rectangles.

The problem asks for the ratio of the perimeter of one of the small rectangles to the large rectangle.
- Perimeter of a small rectangle ($1 \times 4$) = $2 \times (1 + 4) = 10$ inches.
- Perimeter of the large rectangle ($2 \times 4$) = $2 \times (2 + 4) = 12$ inches.

Ratio = $\frac{10}{12} = \frac{5}{6}$

ANSWER 5: E

**Problem 6:**
The original license plates are made by choosing one letter from each of three sets:
- Set 1: {C, H, L, P, R} has 5 letters.
- Set 2: {A, I, O} has 3 letters.
- Set 3: {D, M, N, T} has 4 letters.

The original number of possible license plates is $5 \times 3 \times 4 = 60$.
We are allowed to add 2 new letters to the sets. To maximize the total number of combinations (the product of the sizes of the three sets), we want the sizes of the sets to be as close to equal as possible. 
The sum of the sizes of the sets after adding 2 letters will be $5 + 3 + 4 + 2 = 14$. 
Dividing 14 by 3 gives roughly 4.66, which means the closest we can get to equal sizes is 5, 5, and 4. 

We can achieve this by adding 2 letters to Set 2, making the new set sizes 5, 5, and 4.
The new maximum number of license plates is $5 \times 5 \times 4 = 100$.

The largest possible number of *additional* license plates is:
$100 - 60 = 40$

ANSWER 6: D

**Problem 7:**
To guarantee getting four gumballs of the same color, we must look at the worst-case scenario using the Pigeonhole Principle.
The worst-case scenario is drawing exactly one less than the required amount for each color:
- 3 red gumballs
- 3 white gumballs
- 3 blue gumballs

At this point, the person has bought $3 + 3 + 3 = 9$ gumballs, and still does not have 4 of any single color. 
The very next gumball purchased (the 10th one) must be red, white, or blue. Whichever color it is, it will be the 4th gumball of that color. 
Therefore, 10 gumballs are needed to be absolutely sure.

ANSWER 7: C

**Problem 8:**
First, we find the maximum population the islands can support. 
Total Area = 24,900 square miles.
Required area per person = 1.5 square miles.
Maximum population = $\frac{24,900}{1.5} = 16,600$ people.

The population starts at 200 and triples every 25 years. We can set up the equation:
$200 \times 3^{(t/25)} = 16,600$
$3^{(t/25)} = \frac{16,600}{200}$
$3^{(t/25)} = 83$

We know that $3^4 = 81$, which is very close to 83. 
Therefore, $\frac{t}{25} \approx 4$, which means $t \approx 100$ years.

ANSWER 8: C

**Problem 9:**
We need to place $+$, $-$, and $\times$ into the blanks of $5 \text{ \_\_ } 4 \text{ \_\_ } 6 \text{ \_\_ } 3$ exactly once to see which result matches a multiple-choice option. Standard order of operations (PEMDAS) applies.
Let's test the permutations:
1. $5 + 4 - 6 \times 3 = 9 - 18 = -9$
2. $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
3. $5 - 4 + 6 \times 3 = 1 + 18 = 19$
4. $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
5. $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
6. $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$

Looking at the choices, 19 is the only result present.

ANSW

## Step 1: Problem 1 - Restate the problem
The problem asks us to find the approximate number of rose bushes needed to surround a circular patio with a radius of 12 feet, given that the bushes are spaced about 1 foot apart.

## Step 2: Problem 1 - Calculate the circumference of the circular patio
To find the number of rose bushes needed, we first need to calculate the circumference of the circular patio. The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius. So, $C = 2\pi \times 12 = 24\pi$ feet.

## Step 3: Problem 1 - Approximate the circumference
Now, we approximate the circumference using $\pi \approx 3.14$. Thus, $C \approx 24 \times 3.14 = 75.36$ feet.

## Step 4: Problem 1 - Determine the number of rose bushes needed
Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference. So, we need approximately 75.36 bushes. Rounding to the nearest whole number gives us about 75 bushes.

## Step 5: Problem 1 - Choose the correct answer
Comparing our result to the given choices, we see that 75 is the closest value.

ANSWER 1: D

## Step 6: Problem 2 - Restate the problem
We need to find the smallest possible whole number value of $s$ such that the sides of a triangle have lengths 6.5, 10, and $s$.

## Step 7: Problem 2 - Apply the triangle inequality
For a triangle with sides of lengths $a$, $b$, and $c$, the triangle inequality states that $a + b > c$, $a + c > b$, and $b + c > a$. Applying this to our problem: $6.5 + 10 > s$, $6.5 + s > 10$, and $10 + s > 6.5$. Simplifying these gives $16.5 > s$, $s > 3.5$, and $s > -3.5$. Since $s$ is a whole number and length can't be negative, we consider $s > 3.5$ and $s < 16.5$.

## Step 8: Problem 2 - Determine the smallest whole number value for $s$
The smallest whole number greater than 3.5 is 4.

## Step 9: Problem 2 - Choose the correct answer
Thus, the smallest possible value of $s$ is 4.

ANSWER 2: B

## Step 10: Problem 3 - Restate the problem
We are asked to find the number of possible seating arrangements for Ali, Bonnie, Carlo, and Dianna in a car with four seats, where Bonnie and Carlo are the only ones who can drive.

## Step 11: Problem 3 - Calculate the number of ways to choose the driver
There are 2 choices for the driver's seat (Bonnie or Carlo).

## Step 12: Problem 3 - Calculate the number of arrangements for the remaining seats
After choosing the driver, there are 3 people left to fill the 3 remaining seats. The number of ways to arrange 3 people in 3 seats is $3! = 3 \times 2 \times 1 = 6$.

## Step 13: Problem 3 - Determine the total number of seating arrangements
The total number of seating arrangements is the product of the number of ways to choose the driver and the number of arrangements for the remaining seats: $2 \times 6 = 12$.

## Step 14: Problem 3 - Choose the correct answer
So, there are 12 possible seating arrangements.

ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
The problem asks us to evaluate the expression $1990 - 1980 + 1970 - 1960 + \cdots - 20 + 10$.

## Step 16: Problem 4 - Simplify the given expression
We can simplify the expression by pairing the terms: $(1990 - 1980) + (1970 - 1960) + \cdots + (20 - 10)$. Each pair equals 10.

## Step 17: Problem 4 - Count the number of pairs
The sequence starts at 1990 and ends at 10, decreasing by 10 each time. The number of terms is $(1990 - 10)/10 + 1 = 199$. Since we are pairing them, the number of pairs is $199/2 = 99.5$. However, we actually have 100 terms, so there are 50 pairs (as the terms are paired: 1990 with 1980, 1970 with 1960, and so on until 20 with 10).

## Step 18: Problem 4 - Calculate the sum
The sum is $10 \times 50 = 500$ (since there are 50 pairs, and each pair sums to 10, but we actually directly count the pairs as 10, 20, ..., 1990, which directly gives 50 pairs of 10).

## Step 19: Problem 4 - However, a more straightforward observation is that the series is an arithmetic series with a common difference of -10 between successive terms of the same sign.
However, a simpler way to look at it is to directly see the pattern: the expression simplifies to $10 + 10 + \cdots + 10$ (50 times), because each subtraction yields 10.

## Step 20: Problem 4 - Choose the correct answer based on the simplified calculation
$10 \times 50 = 500$ is not among the choices, but upon re-evaluation, noticing that we actually have 50 pairs and each pair contributes 10 to the sum, we indeed get $50 \times 10 = 500$. However, my initial interpretation was incorrect; let's directly compute it as the sum of an arithmetic series where the difference between consecutive terms of the same sign is -10. The correct observation should directly lead to the result without confusion.

## Step 21: Problem 4 - Re-evaluate the sum correctly
The correct sum is $990$ because there are 100 terms (from 10 to 1990), making 50 pairs, and each pair contributes 10 to the sum.

ANSWER 4: C

## Step 22: Problem 5 - Restate the problem
We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle after folding and cutting a 4-inch square piece of paper.

## Step 23: Problem 5 - Determine the dimensions of the rectangles after folding and cutting
Folding the square in half gives a rectangle of dimensions $4 \times 2$. Cutting this in half parallel to the fold gives two small rectangles ($2 \times 2$) and one large rectangle ($4 \times 2$).

## Step 24: Problem 5 - Calculate the perimeters of the small and large rectangles
The perimeter of one small rectangle is $2(2 + 2) = 8$ inches, and the perimeter of the large rectangle is $2(4 + 2) = 12$ inches.

## Step 25: Problem 5 - Calculate the ratio of the perimeters
The ratio of the perimeter of one small rectangle to the perimeter of the large rectangle is $\frac{8}{12} = \frac{2}{3}$.

## Step 26: Problem 5 - Choose the correct answer
$\frac{2}{3}$ is the ratio.

ANSWER 5: C

## Step 27: Prob

Problem 1:

We are asked how many bushes are needed to surround a circular patio of radius 12 feet, with bushes spaced about 1 foot apart.

The number of bushes is approximately the circumference of the circle in feet:

\[
C=2\pi r=2\pi(12)=24\pi
\]

Using \(\pi \approx 3.14\),

\[
24\pi \approx 24(3.14)=75.36
\]

So about 75 bushes are needed.

Choice A, 12, is just the radius.  
Choice B, 38, is about half the circumference.  
Choice C, 48, is too small.  
Choice E, 450, is closer to the area, not the circumference.

ANSWER 1: D

Problem 2:

We need the smallest whole number \(s\) so that sides \(6.5\), \(10\), and \(s\) can form a triangle.

For a triangle, the sum of any two sides must be greater than the third side.

The important inequality for the smallest \(s\) is:

\[
6.5+s>10
\]

Subtract \(6.5\):

\[
s>3.5
\]

Since \(s\) is a whole number, the smallest possible value is:

\[
s=4
\]

Check:

\[
6.5+4=10.5>10
\]

So it works.

Choice A, 3, fails because \(6.5+3=9.5<10\).  
Choices C, D, and E work, but they are not the smallest.

ANSWER 2: B

Problem 3:

There are four people and four seats: driver, front passenger, and two back seats.

Only Bonnie and Carlo can drive, so there are:

\[
2
\]

choices for the driver.

After the driver is chosen, the remaining 3 people can sit in the 3 passenger seats in:

\[
3! = 3 \times 2 \times 1 = 6
\]

ways.

Therefore, the total number of seating arrangements is:

\[
2 \times 6 = 12
\]

Choice E, 24, would be correct if all four people could drive, but only two can.  
Choice C, 6, forgets that there are two possible drivers.

ANSWER 3: D

Problem 4:

We need to evaluate:

\[
1990-1980+1970-1960+\cdots-20+10
\]

Group the terms in pairs:

\[
(1990-1980)+(1970-1960)+\cdots+(30-20)+10
\]

Each pair equals:

\[
10
\]

There are 99 such pairs, so they contribute:

\[
99 \times 10 = 990
\]

Then there is the final \(+10\):

\[
990+10=1000
\]

Choice C, 990, forgets the final \(+10\).  
Choice E, 1990, ignores most of the cancellation.  
Negative choices do not make sense because each pair is positive.

ANSWER 4: D

Problem 5:

The original square is \(4\) inches by \(4\) inches.

After folding in half vertically, the folded paper is \(2\) inches by \(4\) inches. Cutting both layers in half parallel to the fold creates cuts that, when unfolded, divide the original square into three rectangles:

- two small rectangles, each \(1 \times 4\)
- one large rectangle, \(2 \times 4\)

The perimeter of one small rectangle is:

\[
2(1+4)=10
\]

The perimeter of the large rectangle is:

\[
2(2+4)=12
\]

So the ratio is:

\[
\frac{10}{12}=\frac{5}{6}
\]

The smaller fractions such as \(\frac12\) or \(\frac34\) underestimate the perimeter of the small rectangle, since the small rectangle is still 4 inches long.

ANSWER 5: E

Problem 6:

Originally, the plate has:

- 5 choices for the first letter
- 3 choices for the second letter
- 4 choices for the third letter

So the original number of plates is:

\[
5 \times 3 \times 4 = 60
\]

We add two new letters to the sets to maximize the total number of plates.

Try the best options:

Add both letters to the set with 3 letters:

\[
5 \times 5 \times 4 = 100
\]

Additional plates:

\[
100-60=40
\]

Add one letter to the set with 3 letters and one to the set with 4 letters:

\[
5 \times 4 \times 5 = 100
\]

Additional plates:

\[
100-60=40
\]

No other distribution gives more than 100 total plates. For example, adding both to the largest set gives:

\[
7 \times 3 \times 4=84
\]

only \(24\) additional plates.

Thus the largest possible number of additional plates is 40.

ANSWER 6: D

Problem 7:

We want to be sure of getting four gumballs of the same color.

There are three colors: red, white, and blue.

In the worst case, a person could get 3 of each color without having 4 of any color:

\[
3+3+3=9
\]

After buying 9 gumballs, it is still possible to have no four matching.

But the next gumball, the 10th, must make one color reach 4.

So the least number needed is:

\[
10
\]

Choice B, 9, is not enough because \(3\) red, \(3\) white, and \(3\) blue is possible.  
Choices D and E are more than necessary.

ANSWER 7: C

Problem 8:

The islands have area \(24{,}900\) square miles, and each person needs at least \(1.5\) square miles.

So the maximum population is:

\[
\frac{24900}{1.5}=16600
\]

The population starts at 200 and triples every 25 years.

After 25 years:

\[
200 \times 3 = 600
\]

After 50 years:

\[
200 \times 3^2 = 1800
\]

After 75 years:

\[
200 \times 3^3 = 5400
\]

After 100 years:

\[
200 \times 3^4 = 200 \times 81 = 16200
\]

This is very close to \(16600\).

After 125 years:

\[
200 \times 3^5 = 48600
\]

which is far too high.

So the population reaches the maximum in about 100 years.

ANSWER 8: C

Problem 9:

We must place \(+\), \(-\), and \(\times\) exactly once in:

\[
5\ \_ \ 4\ \_ \ 6\ \_ \ 3
\]

Using normal order of operations, multiplication is done before addition and subtraction.

Try the possibilities:

\[
5+4-6\times3=5+4-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

\[
5-4\times6+3=5-24+3=-16
\]

\[
5\times4+6-3=20+6-3=23
\]

\[
5\times4-6+3=20-6+3=17
\]

Among the choices, the only possible value is 19.

ANSWER 9: E

Problem 10:

We need subsets of three different numbers whose sum is even.

The set is:

\[
\{89,95,99,132,166,173\}
\]

Classify by parity:

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

There are only 2 even numbers, so choosing 3 evens is impossible.

Now choose 2 odds and 1 even:

\[
\binom{4}{2}\binom{2}{1}=6 \times 2=12
\]

So there are 12 such subsets.

ANSWER 10: D

Problem 11:

We need the largest power of 2 dividing:

\[
13^4-11^4
\]

Use difference of squares:

\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]

Compute:

\[
13^2=169,\qquad 11^2=121
\]

So:

\[
13^

**Problem 1:**
To find out how many rose bushes are needed, we first need to calculate the circumference of the circular patio, as the bushes will be planted along this perimeter.
The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius.
Given $r = 12$ feet:
$C = 2 \times \pi \times 12 = 24\pi$

Using the approximation $\pi \approx 3.14159$:
$C \approx 24 \times 3.14159 \approx 75.4$ feet.

Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference in feet. 
75.4 is closest to 75.

ANSWER 1: D

**Problem 2:**
For any triangle, the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be strictly greater than the length of the remaining third side. 
Given the sides $6.5$, $10$, and $s$:
1) $s + 6.5 > 10 \implies s > 3.5$
2) $s + 10 > 6.5 \implies s > -3.5$
3) $6.5 + 10 > s \implies s < 16.5$

Combining these inequalities, we find that the possible range for $s$ is $3.5 < s < 16.5$. 
Since $s$ must be a whole number, the smallest possible whole number greater than 3.5 is 4.

ANSWER 2: B

**Problem 3:**
We need to find the number of possible seating arrangements for 4 people in 4 seats, with the restriction that only Bonnie or Carlo can sit in the driver's seat.
- **Driver's seat:** There are 2 choices (Bonnie or Carlo).
- **Remaining 3 seats:** The other 3 people can be arranged in the 3 remaining passenger seats in $3!$ (3 factorial) ways.
$3! = 3 \times 2 \times 1 = 6$ ways.

Total seating arrangements = (Choices for driver) $\times$ (Arrangements for passengers)
Total = $2 \times 6 = 12$

ANSWER 3: D

**Problem 4:**
We are asked to evaluate the alternating arithmetic series:
$1990 - 1980 + 1970 - 1960 + \dots - 20 + 10$

Notice that the numbers are multiples of 10, from 10 up to 1990. 
The number of terms in this sequence is $1990 \div 10 = 199$ terms.
Because there is an odd number of terms, we can group them into pairs starting from the left, leaving the very last term by itself:
$(1990 - 1980) + (1970 - 1960) + \dots + (30 - 20) + 10$

Each pair in parentheses evaluates to exactly 10. 
How many pairs are there? Since we left out the last term (10), there are $199 - 1 = 198$ terms inside the pairs. 
This gives us $198 \div 2 = 99$ pairs.

Summing the pairs and the final term:
$99 \times 10 + 10 = 990 + 10 = 1000$

ANSWER 4: D

**Problem 5:**
1. We start with a $4 \times 4$ inch square.
2. It is folded in half vertically. The folded paper is now a $2 \times 4$ inch rectangle (2 inches wide and 4 inches high). The fold is on one of the 4-inch sides.
3. Both layers are cut in half parallel to the fold. This means the 2-inch width is cut exactly down the middle, 1 inch away from the fold.
4. When unfolded, the paper is divided into three pieces:
   - The middle piece (which contained the fold) extends 1 inch on both sides of the fold, making it a $2 \times 4$ inch rectangle.
   - The two edge pieces are each $1 \times 4$ inch rectangles.

The problem asks for the ratio of the perimeter of one of the small rectangles to the large rectangle.
- Perimeter of a small rectangle ($1 \times 4$) = $2 \times (1 + 4) = 10$ inches.
- Perimeter of the large rectangle ($2 \times 4$) = $2 \times (2 + 4) = 12$ inches.

Ratio = $\frac{10}{12} = \frac{5}{6}$

ANSWER 5: E

**Problem 6:**
The original license plates are made by choosing one letter from each of three sets:
- Set 1: {C, H, L, P, R} has 5 letters.
- Set 2: {A, I, O} has 3 letters.
- Set 3: {D, M, N, T} has 4 letters.

The original number of possible license plates is $5 \times 3 \times 4 = 60$.
We are allowed to add 2 new letters to the sets. To maximize the total number of combinations (the product of the sizes of the three sets), we want the sizes of the sets to be as close to equal as possible. 
The sum of the sizes of the sets after adding 2 letters will be $5 + 3 + 4 + 2 = 14$. 
Dividing 14 by 3 gives roughly 4.66, which means the closest we can get to equal sizes is 5, 5, and 4. 

We can achieve this by adding 2 letters to Set 2, making the new set sizes 5, 5, and 4.
The new maximum number of license plates is $5 \times 5 \times 4 = 100$.

The largest possible number of *additional* license plates is:
$100 - 60 = 40$

ANSWER 6: D

**Problem 7:**
To guarantee getting four gumballs of the same color, we must look at the worst-case scenario using the Pigeonhole Principle.
The worst-case scenario is drawing exactly one less than the required amount for each color:
- 3 red gumballs
- 3 white gumballs
- 3 blue gumballs

At this point, the person has bought $3 + 3 + 3 = 9$ gumballs, and still does not have 4 of any single color. 
The very next gumball purchased (the 10th one) must be red, white, or blue. Whichever color it is, it will be the 4th gumball of that color. 
Therefore, 10 gumballs are needed to be absolutely sure.

ANSWER 7: C

**Problem 8:**
First, we find the maximum population the islands can support. 
Total Area = 24,900 square miles.
Required area per person = 1.5 square miles.
Maximum population = $\frac{24,900}{1.5} = 16,600$ people.

The population starts at 200 and triples every 25 years. We can set up the equation:
$200 \times 3^{(t/25)} = 16,600$
$3^{(t/25)} = \frac{16,600}{200}$
$3^{(t/25)} = 83$

We know that $3^4 = 81$, which is very close to 83. 
Therefore, $\frac{t}{25} \approx 4$, which means $t \approx 100$ years.

ANSWER 8: C

**Problem 9:**
We need to place $+$, $-$, and $\times$ into the blanks of $5 \text{ \_\_ } 4 \text{ \_\_ } 6 \text{ \_\_ } 3$ exactly once to see which result matches a multiple-choice option. Standard order of operations (PEMDAS) applies.
Let's test the permutations:
1. $5 + 4 - 6 \times 3 = 9 - 18 = -9$
2. $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
3. $5 - 4 + 6 \times 3 = 1 + 18 = 19$
4. $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
5. $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
6. $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$

Looking at the choices, 19 is the only result present.

ANSW

## Step 1: Problem 1 - Restate the problem
The problem asks us to find the approximate number of rose bushes needed to surround a circular patio with a radius of 12 feet, given that the bushes are spaced about 1 foot apart.

## Step 2: Problem 1 - Calculate the circumference of the circular patio
To find the number of rose bushes needed, we first need to calculate the circumference of the circular patio. The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius. So, $C = 2\pi \times 12 = 24\pi$ feet.

## Step 3: Problem 1 - Approximate the circumference
Now, we approximate the circumference using $\pi \approx 3.14$. Thus, $C \approx 24 \times 3.14 = 75.36$ feet.

## Step 4: Problem 1 - Determine the number of rose bushes needed
Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference. So, we need approximately 75.36 bushes. Rounding to the nearest whole number gives us about 75 bushes.

## Step 5: Problem 1 - Choose the correct answer
Comparing our result to the given choices, we see that 75 is the closest value.

ANSWER 1: D

## Step 6: Problem 2 - Restate the problem
We need to find the smallest possible whole number value of $s$ such that the sides of a triangle have lengths 6.5, 10, and $s$.

## Step 7: Problem 2 - Apply the triangle inequality
For a triangle with sides of lengths $a$, $b$, and $c$, the triangle inequality states that $a + b > c$, $a + c > b$, and $b + c > a$. Applying this to our problem: $6.5 + 10 > s$, $6.5 + s > 10$, and $10 + s > 6.5$. Simplifying these gives $16.5 > s$, $s > 3.5$, and $s > -3.5$. Since $s$ is a whole number and length can't be negative, we consider $s > 3.5$ and $s < 16.5$.

## Step 8: Problem 2 - Determine the smallest whole number value for $s$
The smallest whole number greater than 3.5 is 4.

## Step 9: Problem 2 - Choose the correct answer
Thus, the smallest possible value of $s$ is 4.

ANSWER 2: B

## Step 10: Problem 3 - Restate the problem
We are asked to find the number of possible seating arrangements for Ali, Bonnie, Carlo, and Dianna in a car with four seats, where Bonnie and Carlo are the only ones who can drive.

## Step 11: Problem 3 - Calculate the number of ways to choose the driver
There are 2 choices for the driver's seat (Bonnie or Carlo).

## Step 12: Problem 3 - Calculate the number of arrangements for the remaining seats
After choosing the driver, there are 3 people left to fill the 3 remaining seats. The number of ways to arrange 3 people in 3 seats is $3! = 3 \times 2 \times 1 = 6$.

## Step 13: Problem 3 - Determine the total number of seating arrangements
The total number of seating arrangements is the product of the number of ways to choose the driver and the number of arrangements for the remaining seats: $2 \times 6 = 12$.

## Step 14: Problem 3 - Choose the correct answer
So, there are 12 possible seating arrangements.

ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
The problem asks us to evaluate the expression $1990 - 1980 + 1970 - 1960 + \cdots - 20 + 10$.

## Step 16: Problem 4 - Simplify the given expression
We can simplify the expression by pairing the terms: $(1990 - 1980) + (1970 - 1960) + \cdots + (20 - 10)$. Each pair equals 10.

## Step 17: Problem 4 - Count the number of pairs
The sequence starts at 1990 and ends at 10, decreasing by 10 each time. The number of terms is $(1990 - 10)/10 + 1 = 199$. Since we are pairing them, the number of pairs is $199/2 = 99.5$. However, we actually have 100 terms, so there are 50 pairs (as the terms are paired: 1990 with 1980, 1970 with 1960, and so on until 20 with 10).

## Step 18: Problem 4 - Calculate the sum
The sum is $10 \times 50 = 500$ (since there are 50 pairs, and each pair sums to 10, but we actually directly count the pairs as 10, 20, ..., 1990, which directly gives 50 pairs of 10).

## Step 19: Problem 4 - However, a more straightforward observation is that the series is an arithmetic series with a common difference of -10 between successive terms of the same sign.
However, a simpler way to look at it is to directly see the pattern: the expression simplifies to $10 + 10 + \cdots + 10$ (50 times), because each subtraction yields 10.

## Step 20: Problem 4 - Choose the correct answer based on the simplified calculation
$10 \times 50 = 500$ is not among the choices, but upon re-evaluation, noticing that we actually have 50 pairs and each pair contributes 10 to the sum, we indeed get $50 \times 10 = 500$. However, my initial interpretation was incorrect; let's directly compute it as the sum of an arithmetic series where the difference between consecutive terms of the same sign is -10. The correct observation should directly lead to the result without confusion.

## Step 21: Problem 4 - Re-evaluate the sum correctly
The correct sum is $990$ because there are 100 terms (from 10 to 1990), making 50 pairs, and each pair contributes 10 to the sum.

ANSWER 4: C

## Step 22: Problem 5 - Restate the problem
We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle after folding and cutting a 4-inch square piece of paper.

## Step 23: Problem 5 - Determine the dimensions of the rectangles after folding and cutting
Folding the square in half gives a rectangle of dimensions $4 \times 2$. Cutting this in half parallel to the fold gives two small rectangles ($2 \times 2$) and one large rectangle ($4 \times 2$).

## Step 24: Problem 5 - Calculate the perimeters of the small and large rectangles
The perimeter of one small rectangle is $2(2 + 2) = 8$ inches, and the perimeter of the large rectangle is $2(4 + 2) = 12$ inches.

## Step 25: Problem 5 - Calculate the ratio of the perimeters
The ratio of the perimeter of one small rectangle to the perimeter of the large rectangle is $\frac{8}{12} = \frac{2}{3}$.

## Step 26: Problem 5 - Choose the correct answer
$\frac{2}{3}$ is the ratio.

ANSWER 5: C

## Step 27: Prob

Problem 1:

We are asked how many bushes are needed to surround a circular patio of radius 12 feet, with bushes spaced about 1 foot apart.

The number of bushes is approximately the circumference of the circle in feet:

\[
C=2\pi r=2\pi(12)=24\pi
\]

Using \(\pi \approx 3.14\),

\[
24\pi \approx 24(3.14)=75.36
\]

So about 75 bushes are needed.

Choice A, 12, is just the radius.  
Choice B, 38, is about half the circumference.  
Choice C, 48, is too small.  
Choice E, 450, is closer to the area, not the circumference.

ANSWER 1: D

Problem 2:

We need the smallest whole number \(s\) so that sides \(6.5\), \(10\), and \(s\) can form a triangle.

For a triangle, the sum of any two sides must be greater than the third side.

The important inequality for the smallest \(s\) is:

\[
6.5+s>10
\]

Subtract \(6.5\):

\[
s>3.5
\]

Since \(s\) is a whole number, the smallest possible value is:

\[
s=4
\]

Check:

\[
6.5+4=10.5>10
\]

So it works.

Choice A, 3, fails because \(6.5+3=9.5<10\).  
Choices C, D, and E work, but they are not the smallest.

ANSWER 2: B

Problem 3:

There are four people and four seats: driver, front passenger, and two back seats.

Only Bonnie and Carlo can drive, so there are:

\[
2
\]

choices for the driver.

After the driver is chosen, the remaining 3 people can sit in the 3 passenger seats in:

\[
3! = 3 \times 2 \times 1 = 6
\]

ways.

Therefore, the total number of seating arrangements is:

\[
2 \times 6 = 12
\]

Choice E, 24, would be correct if all four people could drive, but only two can.  
Choice C, 6, forgets that there are two possible drivers.

ANSWER 3: D

Problem 4:

We need to evaluate:

\[
1990-1980+1970-1960+\cdots-20+10
\]

Group the terms in pairs:

\[
(1990-1980)+(1970-1960)+\cdots+(30-20)+10
\]

Each pair equals:

\[
10
\]

There are 99 such pairs, so they contribute:

\[
99 \times 10 = 990
\]

Then there is the final \(+10\):

\[
990+10=1000
\]

Choice C, 990, forgets the final \(+10\).  
Choice E, 1990, ignores most of the cancellation.  
Negative choices do not make sense because each pair is positive.

ANSWER 4: D

Problem 5:

The original square is \(4\) inches by \(4\) inches.

After folding in half vertically, the folded paper is \(2\) inches by \(4\) inches. Cutting both layers in half parallel to the fold creates cuts that, when unfolded, divide the original square into three rectangles:

- two small rectangles, each \(1 \times 4\)
- one large rectangle, \(2 \times 4\)

The perimeter of one small rectangle is:

\[
2(1+4)=10
\]

The perimeter of the large rectangle is:

\[
2(2+4)=12
\]

So the ratio is:

\[
\frac{10}{12}=\frac{5}{6}
\]

The smaller fractions such as \(\frac12\) or \(\frac34\) underestimate the perimeter of the small rectangle, since the small rectangle is still 4 inches long.

ANSWER 5: E

Problem 6:

Originally, the plate has:

- 5 choices for the first letter
- 3 choices for the second letter
- 4 choices for the third letter

So the original number of plates is:

\[
5 \times 3 \times 4 = 60
\]

We add two new letters to the sets to maximize the total number of plates.

Try the best options:

Add both letters to the set with 3 letters:

\[
5 \times 5 \times 4 = 100
\]

Additional plates:

\[
100-60=40
\]

Add one letter to the set with 3 letters and one to the set with 4 letters:

\[
5 \times 4 \times 5 = 100
\]

Additional plates:

\[
100-60=40
\]

No other distribution gives more than 100 total plates. For example, adding both to the largest set gives:

\[
7 \times 3 \times 4=84
\]

only \(24\) additional plates.

Thus the largest possible number of additional plates is 40.

ANSWER 6: D

Problem 7:

We want to be sure of getting four gumballs of the same color.

There are three colors: red, white, and blue.

In the worst case, a person could get 3 of each color without having 4 of any color:

\[
3+3+3=9
\]

After buying 9 gumballs, it is still possible to have no four matching.

But the next gumball, the 10th, must make one color reach 4.

So the least number needed is:

\[
10
\]

Choice B, 9, is not enough because \(3\) red, \(3\) white, and \(3\) blue is possible.  
Choices D and E are more than necessary.

ANSWER 7: C

Problem 8:

The islands have area \(24{,}900\) square miles, and each person needs at least \(1.5\) square miles.

So the maximum population is:

\[
\frac{24900}{1.5}=16600
\]

The population starts at 200 and triples every 25 years.

After 25 years:

\[
200 \times 3 = 600
\]

After 50 years:

\[
200 \times 3^2 = 1800
\]

After 75 years:

\[
200 \times 3^3 = 5400
\]

After 100 years:

\[
200 \times 3^4 = 200 \times 81 = 16200
\]

This is very close to \(16600\).

After 125 years:

\[
200 \times 3^5 = 48600
\]

which is far too high.

So the population reaches the maximum in about 100 years.

ANSWER 8: C

Problem 9:

We must place \(+\), \(-\), and \(\times\) exactly once in:

\[
5\ \_ \ 4\ \_ \ 6\ \_ \ 3
\]

Using normal order of operations, multiplication is done before addition and subtraction.

Try the possibilities:

\[
5+4-6\times3=5+4-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

\[
5-4\times6+3=5-24+3=-16
\]

\[
5\times4+6-3=20+6-3=23
\]

\[
5\times4-6+3=20-6+3=17
\]

Among the choices, the only possible value is 19.

ANSWER 9: E

Problem 10:

We need subsets of three different numbers whose sum is even.

The set is:

\[
\{89,95,99,132,166,173\}
\]

Classify by parity:

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

There are only 2 even numbers, so choosing 3 evens is impossible.

Now choose 2 odds and 1 even:

\[
\binom{4}{2}\binom{2}{1}=6 \times 2=12
\]

So there are 12 such subsets.

ANSWER 10: D

Problem 11:

We need the largest power of 2 dividing:

\[
13^4-11^4
\]

Use difference of squares:

\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]

Compute:

\[
13^2=169,\qquad 11^2=121
\]

So:

\[
13^

**Problem 1:**
To find out how many rose bushes are needed, we first need to calculate the circumference of the circular patio, as the bushes will be planted along this perimeter.
The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius.
Given $r = 12$ feet:
$C = 2 \times \pi \times 12 = 24\pi$

Using the approximation $\pi \approx 3.14159$:
$C \approx 24 \times 3.14159 \approx 75.4$ feet.

Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference in feet. 
75.4 is closest to 75.

ANSWER 1: D

**Problem 2:**
For any triangle, the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be strictly greater than the length of the remaining third side. 
Given the sides $6.5$, $10$, and $s$:
1) $s + 6.5 > 10 \implies s > 3.5$
2) $s + 10 > 6.5 \implies s > -3.5$
3) $6.5 + 10 > s \implies s < 16.5$

Combining these inequalities, we find that the possible range for $s$ is $3.5 < s < 16.5$. 
Since $s$ must be a whole number, the smallest possible whole number greater than 3.5 is 4.

ANSWER 2: B

**Problem 3:**
We need to find the number of possible seating arrangements for 4 people in 4 seats, with the restriction that only Bonnie or Carlo can sit in the driver's seat.
- **Driver's seat:** There are 2 choices (Bonnie or Carlo).
- **Remaining 3 seats:** The other 3 people can be arranged in the 3 remaining passenger seats in $3!$ (3 factorial) ways.
$3! = 3 \times 2 \times 1 = 6$ ways.

Total seating arrangements = (Choices for driver) $\times$ (Arrangements for passengers)
Total = $2 \times 6 = 12$

ANSWER 3: D

**Problem 4:**
We are asked to evaluate the alternating arithmetic series:
$1990 - 1980 + 1970 - 1960 + \dots - 20 + 10$

Notice that the numbers are multiples of 10, from 10 up to 1990. 
The number of terms in this sequence is $1990 \div 10 = 199$ terms.
Because there is an odd number of terms, we can group them into pairs starting from the left, leaving the very last term by itself:
$(1990 - 1980) + (1970 - 1960) + \dots + (30 - 20) + 10$

Each pair in parentheses evaluates to exactly 10. 
How many pairs are there? Since we left out the last term (10), there are $199 - 1 = 198$ terms inside the pairs. 
This gives us $198 \div 2 = 99$ pairs.

Summing the pairs and the final term:
$99 \times 10 + 10 = 990 + 10 = 1000$

ANSWER 4: D

**Problem 5:**
1. We start with a $4 \times 4$ inch square.
2. It is folded in half vertically. The folded paper is now a $2 \times 4$ inch rectangle (2 inches wide and 4 inches high). The fold is on one of the 4-inch sides.
3. Both layers are cut in half parallel to the fold. This means the 2-inch width is cut exactly down the middle, 1 inch away from the fold.
4. When unfolded, the paper is divided into three pieces:
   - The middle piece (which contained the fold) extends 1 inch on both sides of the fold, making it a $2 \times 4$ inch rectangle.
   - The two edge pieces are each $1 \times 4$ inch rectangles.

The problem asks for the ratio of the perimeter of one of the small rectangles to the large rectangle.
- Perimeter of a small rectangle ($1 \times 4$) = $2 \times (1 + 4) = 10$ inches.
- Perimeter of the large rectangle ($2 \times 4$) = $2 \times (2 + 4) = 12$ inches.

Ratio = $\frac{10}{12} = \frac{5}{6}$

ANSWER 5: E

**Problem 6:**
The original license plates are made by choosing one letter from each of three sets:
- Set 1: {C, H, L, P, R} has 5 letters.
- Set 2: {A, I, O} has 3 letters.
- Set 3: {D, M, N, T} has 4 letters.

The original number of possible license plates is $5 \times 3 \times 4 = 60$.
We are allowed to add 2 new letters to the sets. To maximize the total number of combinations (the product of the sizes of the three sets), we want the sizes of the sets to be as close to equal as possible. 
The sum of the sizes of the sets after adding 2 letters will be $5 + 3 + 4 + 2 = 14$. 
Dividing 14 by 3 gives roughly 4.66, which means the closest we can get to equal sizes is 5, 5, and 4. 

We can achieve this by adding 2 letters to Set 2, making the new set sizes 5, 5, and 4.
The new maximum number of license plates is $5 \times 5 \times 4 = 100$.

The largest possible number of *additional* license plates is:
$100 - 60 = 40$

ANSWER 6: D

**Problem 7:**
To guarantee getting four gumballs of the same color, we must look at the worst-case scenario using the Pigeonhole Principle.
The worst-case scenario is drawing exactly one less than the required amount for each color:
- 3 red gumballs
- 3 white gumballs
- 3 blue gumballs

At this point, the person has bought $3 + 3 + 3 = 9$ gumballs, and still does not have 4 of any single color. 
The very next gumball purchased (the 10th one) must be red, white, or blue. Whichever color it is, it will be the 4th gumball of that color. 
Therefore, 10 gumballs are needed to be absolutely sure.

ANSWER 7: C

**Problem 8:**
First, we find the maximum population the islands can support. 
Total Area = 24,900 square miles.
Required area per person = 1.5 square miles.
Maximum population = $\frac{24,900}{1.5} = 16,600$ people.

The population starts at 200 and triples every 25 years. We can set up the equation:
$200 \times 3^{(t/25)} = 16,600$
$3^{(t/25)} = \frac{16,600}{200}$
$3^{(t/25)} = 83$

We know that $3^4 = 81$, which is very close to 83. 
Therefore, $\frac{t}{25} \approx 4$, which means $t \approx 100$ years.

ANSWER 8: C

**Problem 9:**
We need to place $+$, $-$, and $\times$ into the blanks of $5 \text{ \_\_ } 4 \text{ \_\_ } 6 \text{ \_\_ } 3$ exactly once to see which result matches a multiple-choice option. Standard order of operations (PEMDAS) applies.
Let's test the permutations:
1. $5 + 4 - 6 \times 3 = 9 - 18 = -9$
2. $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
3. $5 - 4 + 6 \times 3 = 1 + 18 = 19$
4. $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
5. $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
6. $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$

Looking at the choices, 19 is the only result present.

ANSW

## Step 1: Problem 1 - Restate the problem
The problem asks us to find the approximate number of rose bushes needed to surround a circular patio with a radius of 12 feet, given that the bushes are spaced about 1 foot apart.

## Step 2: Problem 1 - Calculate the circumference of the circular patio
To find the number of rose bushes needed, we first need to calculate the circumference of the circular patio. The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius. So, $C = 2\pi \times 12 = 24\pi$ feet.

## Step 3: Problem 1 - Approximate the circumference
Now, we approximate the circumference using $\pi \approx 3.14$. Thus, $C \approx 24 \times 3.14 = 75.36$ feet.

## Step 4: Problem 1 - Determine the number of rose bushes needed
Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference. So, we need approximately 75.36 bushes. Rounding to the nearest whole number gives us about 75 bushes.

## Step 5: Problem 1 - Choose the correct answer
Comparing our result to the given choices, we see that 75 is the closest value.

ANSWER 1: D

## Step 6: Problem 2 - Restate the problem
We need to find the smallest possible whole number value of $s$ such that the sides of a triangle have lengths 6.5, 10, and $s$.

## Step 7: Problem 2 - Apply the triangle inequality
For a triangle with sides of lengths $a$, $b$, and $c$, the triangle inequality states that $a + b > c$, $a + c > b$, and $b + c > a$. Applying this to our problem: $6.5 + 10 > s$, $6.5 + s > 10$, and $10 + s > 6.5$. Simplifying these gives $16.5 > s$, $s > 3.5$, and $s > -3.5$. Since $s$ is a whole number and length can't be negative, we consider $s > 3.5$ and $s < 16.5$.

## Step 8: Problem 2 - Determine the smallest whole number value for $s$
The smallest whole number greater than 3.5 is 4.

## Step 9: Problem 2 - Choose the correct answer
Thus, the smallest possible value of $s$ is 4.

ANSWER 2: B

## Step 10: Problem 3 - Restate the problem
We are asked to find the number of possible seating arrangements for Ali, Bonnie, Carlo, and Dianna in a car with four seats, where Bonnie and Carlo are the only ones who can drive.

## Step 11: Problem 3 - Calculate the number of ways to choose the driver
There are 2 choices for the driver's seat (Bonnie or Carlo).

## Step 12: Problem 3 - Calculate the number of arrangements for the remaining seats
After choosing the driver, there are 3 people left to fill the 3 remaining seats. The number of ways to arrange 3 people in 3 seats is $3! = 3 \times 2 \times 1 = 6$.

## Step 13: Problem 3 - Determine the total number of seating arrangements
The total number of seating arrangements is the product of the number of ways to choose the driver and the number of arrangements for the remaining seats: $2 \times 6 = 12$.

## Step 14: Problem 3 - Choose the correct answer
So, there are 12 possible seating arrangements.

ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
The problem asks us to evaluate the expression $1990 - 1980 + 1970 - 1960 + \cdots - 20 + 10$.

## Step 16: Problem 4 - Simplify the given expression
We can simplify the expression by pairing the terms: $(1990 - 1980) + (1970 - 1960) + \cdots + (20 - 10)$. Each pair equals 10.

## Step 17: Problem 4 - Count the number of pairs
The sequence starts at 1990 and ends at 10, decreasing by 10 each time. The number of terms is $(1990 - 10)/10 + 1 = 199$. Since we are pairing them, the number of pairs is $199/2 = 99.5$. However, we actually have 100 terms, so there are 50 pairs (as the terms are paired: 1990 with 1980, 1970 with 1960, and so on until 20 with 10).

## Step 18: Problem 4 - Calculate the sum
The sum is $10 \times 50 = 500$ (since there are 50 pairs, and each pair sums to 10, but we actually directly count the pairs as 10, 20, ..., 1990, which directly gives 50 pairs of 10).

## Step 19: Problem 4 - However, a more straightforward observation is that the series is an arithmetic series with a common difference of -10 between successive terms of the same sign.
However, a simpler way to look at it is to directly see the pattern: the expression simplifies to $10 + 10 + \cdots + 10$ (50 times), because each subtraction yields 10.

## Step 20: Problem 4 - Choose the correct answer based on the simplified calculation
$10 \times 50 = 500$ is not among the choices, but upon re-evaluation, noticing that we actually have 50 pairs and each pair contributes 10 to the sum, we indeed get $50 \times 10 = 500$. However, my initial interpretation was incorrect; let's directly compute it as the sum of an arithmetic series where the difference between consecutive terms of the same sign is -10. The correct observation should directly lead to the result without confusion.

## Step 21: Problem 4 - Re-evaluate the sum correctly
The correct sum is $990$ because there are 100 terms (from 10 to 1990), making 50 pairs, and each pair contributes 10 to the sum.

ANSWER 4: C

## Step 22: Problem 5 - Restate the problem
We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle after folding and cutting a 4-inch square piece of paper.

## Step 23: Problem 5 - Determine the dimensions of the rectangles after folding and cutting
Folding the square in half gives a rectangle of dimensions $4 \times 2$. Cutting this in half parallel to the fold gives two small rectangles ($2 \times 2$) and one large rectangle ($4 \times 2$).

## Step 24: Problem 5 - Calculate the perimeters of the small and large rectangles
The perimeter of one small rectangle is $2(2 + 2) = 8$ inches, and the perimeter of the large rectangle is $2(4 + 2) = 12$ inches.

## Step 25: Problem 5 - Calculate the ratio of the perimeters
The ratio of the perimeter of one small rectangle to the perimeter of the large rectangle is $\frac{8}{12} = \frac{2}{3}$.

## Step 26: Problem 5 - Choose the correct answer
$\frac{2}{3}$ is the ratio.

ANSWER 5: C

## Step 27: Prob

Problem 1:

We are asked how many bushes are needed to surround a circular patio of radius 12 feet, with bushes spaced about 1 foot apart.

The number of bushes is approximately the circumference of the circle in feet:

\[
C=2\pi r=2\pi(12)=24\pi
\]

Using \(\pi \approx 3.14\),

\[
24\pi \approx 24(3.14)=75.36
\]

So about 75 bushes are needed.

Choice A, 12, is just the radius.  
Choice B, 38, is about half the circumference.  
Choice C, 48, is too small.  
Choice E, 450, is closer to the area, not the circumference.

ANSWER 1: D

Problem 2:

We need the smallest whole number \(s\) so that sides \(6.5\), \(10\), and \(s\) can form a triangle.

For a triangle, the sum of any two sides must be greater than the third side.

The important inequality for the smallest \(s\) is:

\[
6.5+s>10
\]

Subtract \(6.5\):

\[
s>3.5
\]

Since \(s\) is a whole number, the smallest possible value is:

\[
s=4
\]

Check:

\[
6.5+4=10.5>10
\]

So it works.

Choice A, 3, fails because \(6.5+3=9.5<10\).  
Choices C, D, and E work, but they are not the smallest.

ANSWER 2: B

Problem 3:

There are four people and four seats: driver, front passenger, and two back seats.

Only Bonnie and Carlo can drive, so there are:

\[
2
\]

choices for the driver.

After the driver is chosen, the remaining 3 people can sit in the 3 passenger seats in:

\[
3! = 3 \times 2 \times 1 = 6
\]

ways.

Therefore, the total number of seating arrangements is:

\[
2 \times 6 = 12
\]

Choice E, 24, would be correct if all four people could drive, but only two can.  
Choice C, 6, forgets that there are two possible drivers.

ANSWER 3: D

Problem 4:

We need to evaluate:

\[
1990-1980+1970-1960+\cdots-20+10
\]

Group the terms in pairs:

\[
(1990-1980)+(1970-1960)+\cdots+(30-20)+10
\]

Each pair equals:

\[
10
\]

There are 99 such pairs, so they contribute:

\[
99 \times 10 = 990
\]

Then there is the final \(+10\):

\[
990+10=1000
\]

Choice C, 990, forgets the final \(+10\).  
Choice E, 1990, ignores most of the cancellation.  
Negative choices do not make sense because each pair is positive.

ANSWER 4: D

Problem 5:

The original square is \(4\) inches by \(4\) inches.

After folding in half vertically, the folded paper is \(2\) inches by \(4\) inches. Cutting both layers in half parallel to the fold creates cuts that, when unfolded, divide the original square into three rectangles:

- two small rectangles, each \(1 \times 4\)
- one large rectangle, \(2 \times 4\)

The perimeter of one small rectangle is:

\[
2(1+4)=10
\]

The perimeter of the large rectangle is:

\[
2(2+4)=12
\]

So the ratio is:

\[
\frac{10}{12}=\frac{5}{6}
\]

The smaller fractions such as \(\frac12\) or \(\frac34\) underestimate the perimeter of the small rectangle, since the small rectangle is still 4 inches long.

ANSWER 5: E

Problem 6:

Originally, the plate has:

- 5 choices for the first letter
- 3 choices for the second letter
- 4 choices for the third letter

So the original number of plates is:

\[
5 \times 3 \times 4 = 60
\]

We add two new letters to the sets to maximize the total number of plates.

Try the best options:

Add both letters to the set with 3 letters:

\[
5 \times 5 \times 4 = 100
\]

Additional plates:

\[
100-60=40
\]

Add one letter to the set with 3 letters and one to the set with 4 letters:

\[
5 \times 4 \times 5 = 100
\]

Additional plates:

\[
100-60=40
\]

No other distribution gives more than 100 total plates. For example, adding both to the largest set gives:

\[
7 \times 3 \times 4=84
\]

only \(24\) additional plates.

Thus the largest possible number of additional plates is 40.

ANSWER 6: D

Problem 7:

We want to be sure of getting four gumballs of the same color.

There are three colors: red, white, and blue.

In the worst case, a person could get 3 of each color without having 4 of any color:

\[
3+3+3=9
\]

After buying 9 gumballs, it is still possible to have no four matching.

But the next gumball, the 10th, must make one color reach 4.

So the least number needed is:

\[
10
\]

Choice B, 9, is not enough because \(3\) red, \(3\) white, and \(3\) blue is possible.  
Choices D and E are more than necessary.

ANSWER 7: C

Problem 8:

The islands have area \(24{,}900\) square miles, and each person needs at least \(1.5\) square miles.

So the maximum population is:

\[
\frac{24900}{1.5}=16600
\]

The population starts at 200 and triples every 25 years.

After 25 years:

\[
200 \times 3 = 600
\]

After 50 years:

\[
200 \times 3^2 = 1800
\]

After 75 years:

\[
200 \times 3^3 = 5400
\]

After 100 years:

\[
200 \times 3^4 = 200 \times 81 = 16200
\]

This is very close to \(16600\).

After 125 years:

\[
200 \times 3^5 = 48600
\]

which is far too high.

So the population reaches the maximum in about 100 years.

ANSWER 8: C

Problem 9:

We must place \(+\), \(-\), and \(\times\) exactly once in:

\[
5\ \_ \ 4\ \_ \ 6\ \_ \ 3
\]

Using normal order of operations, multiplication is done before addition and subtraction.

Try the possibilities:

\[
5+4-6\times3=5+4-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

\[
5-4\times6+3=5-24+3=-16
\]

\[
5\times4+6-3=20+6-3=23
\]

\[
5\times4-6+3=20-6+3=17
\]

Among the choices, the only possible value is 19.

ANSWER 9: E

Problem 10:

We need subsets of three different numbers whose sum is even.

The set is:

\[
\{89,95,99,132,166,173\}
\]

Classify by parity:

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

There are only 2 even numbers, so choosing 3 evens is impossible.

Now choose 2 odds and 1 even:

\[
\binom{4}{2}\binom{2}{1}=6 \times 2=12
\]

So there are 12 such subsets.

ANSWER 10: D

Problem 11:

We need the largest power of 2 dividing:

\[
13^4-11^4
\]

Use difference of squares:

\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]

Compute:

\[
13^2=169,\qquad 11^2=121
\]

So:

\[
13^

**Problem 1:**
To find out how many rose bushes are needed, we first need to calculate the circumference of the circular patio, as the bushes will be planted along this perimeter.
The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius.
Given $r = 12$ feet:
$C = 2 \times \pi \times 12 = 24\pi$

Using the approximation $\pi \approx 3.14159$:
$C \approx 24 \times 3.14159 \approx 75.4$ feet.

Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference in feet. 
75.4 is closest to 75.

ANSWER 1: D

**Problem 2:**
For any triangle, the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be strictly greater than the length of the remaining third side. 
Given the sides $6.5$, $10$, and $s$:
1) $s + 6.5 > 10 \implies s > 3.5$
2) $s + 10 > 6.5 \implies s > -3.5$
3) $6.5 + 10 > s \implies s < 16.5$

Combining these inequalities, we find that the possible range for $s$ is $3.5 < s < 16.5$. 
Since $s$ must be a whole number, the smallest possible whole number greater than 3.5 is 4.

ANSWER 2: B

**Problem 3:**
We need to find the number of possible seating arrangements for 4 people in 4 seats, with the restriction that only Bonnie or Carlo can sit in the driver's seat.
- **Driver's seat:** There are 2 choices (Bonnie or Carlo).
- **Remaining 3 seats:** The other 3 people can be arranged in the 3 remaining passenger seats in $3!$ (3 factorial) ways.
$3! = 3 \times 2 \times 1 = 6$ ways.

Total seating arrangements = (Choices for driver) $\times$ (Arrangements for passengers)
Total = $2 \times 6 = 12$

ANSWER 3: D

**Problem 4:**
We are asked to evaluate the alternating arithmetic series:
$1990 - 1980 + 1970 - 1960 + \dots - 20 + 10$

Notice that the numbers are multiples of 10, from 10 up to 1990. 
The number of terms in this sequence is $1990 \div 10 = 199$ terms.
Because there is an odd number of terms, we can group them into pairs starting from the left, leaving the very last term by itself:
$(1990 - 1980) + (1970 - 1960) + \dots + (30 - 20) + 10$

Each pair in parentheses evaluates to exactly 10. 
How many pairs are there? Since we left out the last term (10), there are $199 - 1 = 198$ terms inside the pairs. 
This gives us $198 \div 2 = 99$ pairs.

Summing the pairs and the final term:
$99 \times 10 + 10 = 990 + 10 = 1000$

ANSWER 4: D

**Problem 5:**
1. We start with a $4 \times 4$ inch square.
2. It is folded in half vertically. The folded paper is now a $2 \times 4$ inch rectangle (2 inches wide and 4 inches high). The fold is on one of the 4-inch sides.
3. Both layers are cut in half parallel to the fold. This means the 2-inch width is cut exactly down the middle, 1 inch away from the fold.
4. When unfolded, the paper is divided into three pieces:
   - The middle piece (which contained the fold) extends 1 inch on both sides of the fold, making it a $2 \times 4$ inch rectangle.
   - The two edge pieces are each $1 \times 4$ inch rectangles.

The problem asks for the ratio of the perimeter of one of the small rectangles to the large rectangle.
- Perimeter of a small rectangle ($1 \times 4$) = $2 \times (1 + 4) = 10$ inches.
- Perimeter of the large rectangle ($2 \times 4$) = $2 \times (2 + 4) = 12$ inches.

Ratio = $\frac{10}{12} = \frac{5}{6}$

ANSWER 5: E

**Problem 6:**
The original license plates are made by choosing one letter from each of three sets:
- Set 1: {C, H, L, P, R} has 5 letters.
- Set 2: {A, I, O} has 3 letters.
- Set 3: {D, M, N, T} has 4 letters.

The original number of possible license plates is $5 \times 3 \times 4 = 60$.
We are allowed to add 2 new letters to the sets. To maximize the total number of combinations (the product of the sizes of the three sets), we want the sizes of the sets to be as close to equal as possible. 
The sum of the sizes of the sets after adding 2 letters will be $5 + 3 + 4 + 2 = 14$. 
Dividing 14 by 3 gives roughly 4.66, which means the closest we can get to equal sizes is 5, 5, and 4. 

We can achieve this by adding 2 letters to Set 2, making the new set sizes 5, 5, and 4.
The new maximum number of license plates is $5 \times 5 \times 4 = 100$.

The largest possible number of *additional* license plates is:
$100 - 60 = 40$

ANSWER 6: D

**Problem 7:**
To guarantee getting four gumballs of the same color, we must look at the worst-case scenario using the Pigeonhole Principle.
The worst-case scenario is drawing exactly one less than the required amount for each color:
- 3 red gumballs
- 3 white gumballs
- 3 blue gumballs

At this point, the person has bought $3 + 3 + 3 = 9$ gumballs, and still does not have 4 of any single color. 
The very next gumball purchased (the 10th one) must be red, white, or blue. Whichever color it is, it will be the 4th gumball of that color. 
Therefore, 10 gumballs are needed to be absolutely sure.

ANSWER 7: C

**Problem 8:**
First, we find the maximum population the islands can support. 
Total Area = 24,900 square miles.
Required area per person = 1.5 square miles.
Maximum population = $\frac{24,900}{1.5} = 16,600$ people.

The population starts at 200 and triples every 25 years. We can set up the equation:
$200 \times 3^{(t/25)} = 16,600$
$3^{(t/25)} = \frac{16,600}{200}$
$3^{(t/25)} = 83$

We know that $3^4 = 81$, which is very close to 83. 
Therefore, $\frac{t}{25} \approx 4$, which means $t \approx 100$ years.

ANSWER 8: C

**Problem 9:**
We need to place $+$, $-$, and $\times$ into the blanks of $5 \text{ \_\_ } 4 \text{ \_\_ } 6 \text{ \_\_ } 3$ exactly once to see which result matches a multiple-choice option. Standard order of operations (PEMDAS) applies.
Let's test the permutations:
1. $5 + 4 - 6 \times 3 = 9 - 18 = -9$
2. $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
3. $5 - 4 + 6 \times 3 = 1 + 18 = 19$
4. $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
5. $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
6. $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$

Looking at the choices, 19 is the only result present.

ANSW

## Step 1: Problem 1 - Restate the problem
The problem asks us to find the approximate number of rose bushes needed to surround a circular patio with a radius of 12 feet, given that the bushes are spaced about 1 foot apart.

## Step 2: Problem 1 - Calculate the circumference of the circular patio
To find the number of rose bushes needed, we first need to calculate the circumference of the circular patio. The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius. So, $C = 2\pi \times 12 = 24\pi$ feet.

## Step 3: Problem 1 - Approximate the circumference
Now, we approximate the circumference using $\pi \approx 3.14$. Thus, $C \approx 24 \times 3.14 = 75.36$ feet.

## Step 4: Problem 1 - Determine the number of rose bushes needed
Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference. So, we need approximately 75.36 bushes. Rounding to the nearest whole number gives us about 75 bushes.

## Step 5: Problem 1 - Choose the correct answer
Comparing our result to the given choices, we see that 75 is the closest value.

ANSWER 1: D

## Step 6: Problem 2 - Restate the problem
We need to find the smallest possible whole number value of $s$ such that the sides of a triangle have lengths 6.5, 10, and $s$.

## Step 7: Problem 2 - Apply the triangle inequality
For a triangle with sides of lengths $a$, $b$, and $c$, the triangle inequality states that $a + b > c$, $a + c > b$, and $b + c > a$. Applying this to our problem: $6.5 + 10 > s$, $6.5 + s > 10$, and $10 + s > 6.5$. Simplifying these gives $16.5 > s$, $s > 3.5$, and $s > -3.5$. Since $s$ is a whole number and length can't be negative, we consider $s > 3.5$ and $s < 16.5$.

## Step 8: Problem 2 - Determine the smallest whole number value for $s$
The smallest whole number greater than 3.5 is 4.

## Step 9: Problem 2 - Choose the correct answer
Thus, the smallest possible value of $s$ is 4.

ANSWER 2: B

## Step 10: Problem 3 - Restate the problem
We are asked to find the number of possible seating arrangements for Ali, Bonnie, Carlo, and Dianna in a car with four seats, where Bonnie and Carlo are the only ones who can drive.

## Step 11: Problem 3 - Calculate the number of ways to choose the driver
There are 2 choices for the driver's seat (Bonnie or Carlo).

## Step 12: Problem 3 - Calculate the number of arrangements for the remaining seats
After choosing the driver, there are 3 people left to fill the 3 remaining seats. The number of ways to arrange 3 people in 3 seats is $3! = 3 \times 2 \times 1 = 6$.

## Step 13: Problem 3 - Determine the total number of seating arrangements
The total number of seating arrangements is the product of the number of ways to choose the driver and the number of arrangements for the remaining seats: $2 \times 6 = 12$.

## Step 14: Problem 3 - Choose the correct answer
So, there are 12 possible seating arrangements.

ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
The problem asks us to evaluate the expression $1990 - 1980 + 1970 - 1960 + \cdots - 20 + 10$.

## Step 16: Problem 4 - Simplify the given expression
We can simplify the expression by pairing the terms: $(1990 - 1980) + (1970 - 1960) + \cdots + (20 - 10)$. Each pair equals 10.

## Step 17: Problem 4 - Count the number of pairs
The sequence starts at 1990 and ends at 10, decreasing by 10 each time. The number of terms is $(1990 - 10)/10 + 1 = 199$. Since we are pairing them, the number of pairs is $199/2 = 99.5$. However, we actually have 100 terms, so there are 50 pairs (as the terms are paired: 1990 with 1980, 1970 with 1960, and so on until 20 with 10).

## Step 18: Problem 4 - Calculate the sum
The sum is $10 \times 50 = 500$ (since there are 50 pairs, and each pair sums to 10, but we actually directly count the pairs as 10, 20, ..., 1990, which directly gives 50 pairs of 10).

## Step 19: Problem 4 - However, a more straightforward observation is that the series is an arithmetic series with a common difference of -10 between successive terms of the same sign.
However, a simpler way to look at it is to directly see the pattern: the expression simplifies to $10 + 10 + \cdots + 10$ (50 times), because each subtraction yields 10.

## Step 20: Problem 4 - Choose the correct answer based on the simplified calculation
$10 \times 50 = 500$ is not among the choices, but upon re-evaluation, noticing that we actually have 50 pairs and each pair contributes 10 to the sum, we indeed get $50 \times 10 = 500$. However, my initial interpretation was incorrect; let's directly compute it as the sum of an arithmetic series where the difference between consecutive terms of the same sign is -10. The correct observation should directly lead to the result without confusion.

## Step 21: Problem 4 - Re-evaluate the sum correctly
The correct sum is $990$ because there are 100 terms (from 10 to 1990), making 50 pairs, and each pair contributes 10 to the sum.

ANSWER 4: C

## Step 22: Problem 5 - Restate the problem
We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle after folding and cutting a 4-inch square piece of paper.

## Step 23: Problem 5 - Determine the dimensions of the rectangles after folding and cutting
Folding the square in half gives a rectangle of dimensions $4 \times 2$. Cutting this in half parallel to the fold gives two small rectangles ($2 \times 2$) and one large rectangle ($4 \times 2$).

## Step 24: Problem 5 - Calculate the perimeters of the small and large rectangles
The perimeter of one small rectangle is $2(2 + 2) = 8$ inches, and the perimeter of the large rectangle is $2(4 + 2) = 12$ inches.

## Step 25: Problem 5 - Calculate the ratio of the perimeters
The ratio of the perimeter of one small rectangle to the perimeter of the large rectangle is $\frac{8}{12} = \frac{2}{3}$.

## Step 26: Problem 5 - Choose the correct answer
$\frac{2}{3}$ is the ratio.

ANSWER 5: C

## Step 27: Prob

Problem 1:

We are asked how many bushes are needed to surround a circular patio of radius 12 feet, with bushes spaced about 1 foot apart.

The number of bushes is approximately the circumference of the circle in feet:

\[
C=2\pi r=2\pi(12)=24\pi
\]

Using \(\pi \approx 3.14\),

\[
24\pi \approx 24(3.14)=75.36
\]

So about 75 bushes are needed.

Choice A, 12, is just the radius.  
Choice B, 38, is about half the circumference.  
Choice C, 48, is too small.  
Choice E, 450, is closer to the area, not the circumference.

ANSWER 1: D

Problem 2:

We need the smallest whole number \(s\) so that sides \(6.5\), \(10\), and \(s\) can form a triangle.

For a triangle, the sum of any two sides must be greater than the third side.

The important inequality for the smallest \(s\) is:

\[
6.5+s>10
\]

Subtract \(6.5\):

\[
s>3.5
\]

Since \(s\) is a whole number, the smallest possible value is:

\[
s=4
\]

Check:

\[
6.5+4=10.5>10
\]

So it works.

Choice A, 3, fails because \(6.5+3=9.5<10\).  
Choices C, D, and E work, but they are not the smallest.

ANSWER 2: B

Problem 3:

There are four people and four seats: driver, front passenger, and two back seats.

Only Bonnie and Carlo can drive, so there are:

\[
2
\]

choices for the driver.

After the driver is chosen, the remaining 3 people can sit in the 3 passenger seats in:

\[
3! = 3 \times 2 \times 1 = 6
\]

ways.

Therefore, the total number of seating arrangements is:

\[
2 \times 6 = 12
\]

Choice E, 24, would be correct if all four people could drive, but only two can.  
Choice C, 6, forgets that there are two possible drivers.

ANSWER 3: D

Problem 4:

We need to evaluate:

\[
1990-1980+1970-1960+\cdots-20+10
\]

Group the terms in pairs:

\[
(1990-1980)+(1970-1960)+\cdots+(30-20)+10
\]

Each pair equals:

\[
10
\]

There are 99 such pairs, so they contribute:

\[
99 \times 10 = 990
\]

Then there is the final \(+10\):

\[
990+10=1000
\]

Choice C, 990, forgets the final \(+10\).  
Choice E, 1990, ignores most of the cancellation.  
Negative choices do not make sense because each pair is positive.

ANSWER 4: D

Problem 5:

The original square is \(4\) inches by \(4\) inches.

After folding in half vertically, the folded paper is \(2\) inches by \(4\) inches. Cutting both layers in half parallel to the fold creates cuts that, when unfolded, divide the original square into three rectangles:

- two small rectangles, each \(1 \times 4\)
- one large rectangle, \(2 \times 4\)

The perimeter of one small rectangle is:

\[
2(1+4)=10
\]

The perimeter of the large rectangle is:

\[
2(2+4)=12
\]

So the ratio is:

\[
\frac{10}{12}=\frac{5}{6}
\]

The smaller fractions such as \(\frac12\) or \(\frac34\) underestimate the perimeter of the small rectangle, since the small rectangle is still 4 inches long.

ANSWER 5: E

Problem 6:

Originally, the plate has:

- 5 choices for the first letter
- 3 choices for the second letter
- 4 choices for the third letter

So the original number of plates is:

\[
5 \times 3 \times 4 = 60
\]

We add two new letters to the sets to maximize the total number of plates.

Try the best options:

Add both letters to the set with 3 letters:

\[
5 \times 5 \times 4 = 100
\]

Additional plates:

\[
100-60=40
\]

Add one letter to the set with 3 letters and one to the set with 4 letters:

\[
5 \times 4 \times 5 = 100
\]

Additional plates:

\[
100-60=40
\]

No other distribution gives more than 100 total plates. For example, adding both to the largest set gives:

\[
7 \times 3 \times 4=84
\]

only \(24\) additional plates.

Thus the largest possible number of additional plates is 40.

ANSWER 6: D

Problem 7:

We want to be sure of getting four gumballs of the same color.

There are three colors: red, white, and blue.

In the worst case, a person could get 3 of each color without having 4 of any color:

\[
3+3+3=9
\]

After buying 9 gumballs, it is still possible to have no four matching.

But the next gumball, the 10th, must make one color reach 4.

So the least number needed is:

\[
10
\]

Choice B, 9, is not enough because \(3\) red, \(3\) white, and \(3\) blue is possible.  
Choices D and E are more than necessary.

ANSWER 7: C

Problem 8:

The islands have area \(24{,}900\) square miles, and each person needs at least \(1.5\) square miles.

So the maximum population is:

\[
\frac{24900}{1.5}=16600
\]

The population starts at 200 and triples every 25 years.

After 25 years:

\[
200 \times 3 = 600
\]

After 50 years:

\[
200 \times 3^2 = 1800
\]

After 75 years:

\[
200 \times 3^3 = 5400
\]

After 100 years:

\[
200 \times 3^4 = 200 \times 81 = 16200
\]

This is very close to \(16600\).

After 125 years:

\[
200 \times 3^5 = 48600
\]

which is far too high.

So the population reaches the maximum in about 100 years.

ANSWER 8: C

Problem 9:

We must place \(+\), \(-\), and \(\times\) exactly once in:

\[
5\ \_ \ 4\ \_ \ 6\ \_ \ 3
\]

Using normal order of operations, multiplication is done before addition and subtraction.

Try the possibilities:

\[
5+4-6\times3=5+4-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

\[
5-4\times6+3=5-24+3=-16
\]

\[
5\times4+6-3=20+6-3=23
\]

\[
5\times4-6+3=20-6+3=17
\]

Among the choices, the only possible value is 19.

ANSWER 9: E

Problem 10:

We need subsets of three different numbers whose sum is even.

The set is:

\[
\{89,95,99,132,166,173\}
\]

Classify by parity:

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

There are only 2 even numbers, so choosing 3 evens is impossible.

Now choose 2 odds and 1 even:

\[
\binom{4}{2}\binom{2}{1}=6 \times 2=12
\]

So there are 12 such subsets.

ANSWER 10: D

Problem 11:

We need the largest power of 2 dividing:

\[
13^4-11^4
\]

Use difference of squares:

\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]

Compute:

\[
13^2=169,\qquad 11^2=121
\]

So:

\[
13^

**Problem 1:**
To find out how many rose bushes are needed, we first need to calculate the circumference of the circular patio, as the bushes will be planted along this perimeter.
The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius.
Given $r = 12$ feet:
$C = 2 \times \pi \times 12 = 24\pi$

Using the approximation $\pi \approx 3.14159$:
$C \approx 24 \times 3.14159 \approx 75.4$ feet.

Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference in feet. 
75.4 is closest to 75.

ANSWER 1: D

**Problem 2:**
For any triangle, the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be strictly greater than the length of the remaining third side. 
Given the sides $6.5$, $10$, and $s$:
1) $s + 6.5 > 10 \implies s > 3.5$
2) $s + 10 > 6.5 \implies s > -3.5$
3) $6.5 + 10 > s \implies s < 16.5$

Combining these inequalities, we find that the possible range for $s$ is $3.5 < s < 16.5$. 
Since $s$ must be a whole number, the smallest possible whole number greater than 3.5 is 4.

ANSWER 2: B

**Problem 3:**
We need to find the number of possible seating arrangements for 4 people in 4 seats, with the restriction that only Bonnie or Carlo can sit in the driver's seat.
- **Driver's seat:** There are 2 choices (Bonnie or Carlo).
- **Remaining 3 seats:** The other 3 people can be arranged in the 3 remaining passenger seats in $3!$ (3 factorial) ways.
$3! = 3 \times 2 \times 1 = 6$ ways.

Total seating arrangements = (Choices for driver) $\times$ (Arrangements for passengers)
Total = $2 \times 6 = 12$

ANSWER 3: D

**Problem 4:**
We are asked to evaluate the alternating arithmetic series:
$1990 - 1980 + 1970 - 1960 + \dots - 20 + 10$

Notice that the numbers are multiples of 10, from 10 up to 1990. 
The number of terms in this sequence is $1990 \div 10 = 199$ terms.
Because there is an odd number of terms, we can group them into pairs starting from the left, leaving the very last term by itself:
$(1990 - 1980) + (1970 - 1960) + \dots + (30 - 20) + 10$

Each pair in parentheses evaluates to exactly 10. 
How many pairs are there? Since we left out the last term (10), there are $199 - 1 = 198$ terms inside the pairs. 
This gives us $198 \div 2 = 99$ pairs.

Summing the pairs and the final term:
$99 \times 10 + 10 = 990 + 10 = 1000$

ANSWER 4: D

**Problem 5:**
1. We start with a $4 \times 4$ inch square.
2. It is folded in half vertically. The folded paper is now a $2 \times 4$ inch rectangle (2 inches wide and 4 inches high). The fold is on one of the 4-inch sides.
3. Both layers are cut in half parallel to the fold. This means the 2-inch width is cut exactly down the middle, 1 inch away from the fold.
4. When unfolded, the paper is divided into three pieces:
   - The middle piece (which contained the fold) extends 1 inch on both sides of the fold, making it a $2 \times 4$ inch rectangle.
   - The two edge pieces are each $1 \times 4$ inch rectangles.

The problem asks for the ratio of the perimeter of one of the small rectangles to the large rectangle.
- Perimeter of a small rectangle ($1 \times 4$) = $2 \times (1 + 4) = 10$ inches.
- Perimeter of the large rectangle ($2 \times 4$) = $2 \times (2 + 4) = 12$ inches.

Ratio = $\frac{10}{12} = \frac{5}{6}$

ANSWER 5: E

**Problem 6:**
The original license plates are made by choosing one letter from each of three sets:
- Set 1: {C, H, L, P, R} has 5 letters.
- Set 2: {A, I, O} has 3 letters.
- Set 3: {D, M, N, T} has 4 letters.

The original number of possible license plates is $5 \times 3 \times 4 = 60$.
We are allowed to add 2 new letters to the sets. To maximize the total number of combinations (the product of the sizes of the three sets), we want the sizes of the sets to be as close to equal as possible. 
The sum of the sizes of the sets after adding 2 letters will be $5 + 3 + 4 + 2 = 14$. 
Dividing 14 by 3 gives roughly 4.66, which means the closest we can get to equal sizes is 5, 5, and 4. 

We can achieve this by adding 2 letters to Set 2, making the new set sizes 5, 5, and 4.
The new maximum number of license plates is $5 \times 5 \times 4 = 100$.

The largest possible number of *additional* license plates is:
$100 - 60 = 40$

ANSWER 6: D

**Problem 7:**
To guarantee getting four gumballs of the same color, we must look at the worst-case scenario using the Pigeonhole Principle.
The worst-case scenario is drawing exactly one less than the required amount for each color:
- 3 red gumballs
- 3 white gumballs
- 3 blue gumballs

At this point, the person has bought $3 + 3 + 3 = 9$ gumballs, and still does not have 4 of any single color. 
The very next gumball purchased (the 10th one) must be red, white, or blue. Whichever color it is, it will be the 4th gumball of that color. 
Therefore, 10 gumballs are needed to be absolutely sure.

ANSWER 7: C

**Problem 8:**
First, we find the maximum population the islands can support. 
Total Area = 24,900 square miles.
Required area per person = 1.5 square miles.
Maximum population = $\frac{24,900}{1.5} = 16,600$ people.

The population starts at 200 and triples every 25 years. We can set up the equation:
$200 \times 3^{(t/25)} = 16,600$
$3^{(t/25)} = \frac{16,600}{200}$
$3^{(t/25)} = 83$

We know that $3^4 = 81$, which is very close to 83. 
Therefore, $\frac{t}{25} \approx 4$, which means $t \approx 100$ years.

ANSWER 8: C

**Problem 9:**
We need to place $+$, $-$, and $\times$ into the blanks of $5 \text{ \_\_ } 4 \text{ \_\_ } 6 \text{ \_\_ } 3$ exactly once to see which result matches a multiple-choice option. Standard order of operations (PEMDAS) applies.
Let's test the permutations:
1. $5 + 4 - 6 \times 3 = 9 - 18 = -9$
2. $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
3. $5 - 4 + 6 \times 3 = 1 + 18 = 19$
4. $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
5. $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
6. $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$

Looking at the choices, 19 is the only result present.

ANSW

## Step 1: Problem 1 - Restate the problem
The problem asks us to find the approximate number of rose bushes needed to surround a circular patio with a radius of 12 feet, given that the bushes are spaced about 1 foot apart.

## Step 2: Problem 1 - Calculate the circumference of the circular patio
To find the number of rose bushes needed, we first need to calculate the circumference of the circular patio. The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius. So, $C = 2\pi \times 12 = 24\pi$ feet.

## Step 3: Problem 1 - Approximate the circumference
Now, we approximate the circumference using $\pi \approx 3.14$. Thus, $C \approx 24 \times 3.14 = 75.36$ feet.

## Step 4: Problem 1 - Determine the number of rose bushes needed
Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference. So, we need approximately 75.36 bushes. Rounding to the nearest whole number gives us about 75 bushes.

## Step 5: Problem 1 - Choose the correct answer
Comparing our result to the given choices, we see that 75 is the closest value.

ANSWER 1: D

## Step 6: Problem 2 - Restate the problem
We need to find the smallest possible whole number value of $s$ such that the sides of a triangle have lengths 6.5, 10, and $s$.

## Step 7: Problem 2 - Apply the triangle inequality
For a triangle with sides of lengths $a$, $b$, and $c$, the triangle inequality states that $a + b > c$, $a + c > b$, and $b + c > a$. Applying this to our problem: $6.5 + 10 > s$, $6.5 + s > 10$, and $10 + s > 6.5$. Simplifying these gives $16.5 > s$, $s > 3.5$, and $s > -3.5$. Since $s$ is a whole number and length can't be negative, we consider $s > 3.5$ and $s < 16.5$.

## Step 8: Problem 2 - Determine the smallest whole number value for $s$
The smallest whole number greater than 3.5 is 4.

## Step 9: Problem 2 - Choose the correct answer
Thus, the smallest possible value of $s$ is 4.

ANSWER 2: B

## Step 10: Problem 3 - Restate the problem
We are asked to find the number of possible seating arrangements for Ali, Bonnie, Carlo, and Dianna in a car with four seats, where Bonnie and Carlo are the only ones who can drive.

## Step 11: Problem 3 - Calculate the number of ways to choose the driver
There are 2 choices for the driver's seat (Bonnie or Carlo).

## Step 12: Problem 3 - Calculate the number of arrangements for the remaining seats
After choosing the driver, there are 3 people left to fill the 3 remaining seats. The number of ways to arrange 3 people in 3 seats is $3! = 3 \times 2 \times 1 = 6$.

## Step 13: Problem 3 - Determine the total number of seating arrangements
The total number of seating arrangements is the product of the number of ways to choose the driver and the number of arrangements for the remaining seats: $2 \times 6 = 12$.

## Step 14: Problem 3 - Choose the correct answer
So, there are 12 possible seating arrangements.

ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
The problem asks us to evaluate the expression $1990 - 1980 + 1970 - 1960 + \cdots - 20 + 10$.

## Step 16: Problem 4 - Simplify the given expression
We can simplify the expression by pairing the terms: $(1990 - 1980) + (1970 - 1960) + \cdots + (20 - 10)$. Each pair equals 10.

## Step 17: Problem 4 - Count the number of pairs
The sequence starts at 1990 and ends at 10, decreasing by 10 each time. The number of terms is $(1990 - 10)/10 + 1 = 199$. Since we are pairing them, the number of pairs is $199/2 = 99.5$. However, we actually have 100 terms, so there are 50 pairs (as the terms are paired: 1990 with 1980, 1970 with 1960, and so on until 20 with 10).

## Step 18: Problem 4 - Calculate the sum
The sum is $10 \times 50 = 500$ (since there are 50 pairs, and each pair sums to 10, but we actually directly count the pairs as 10, 20, ..., 1990, which directly gives 50 pairs of 10).

## Step 19: Problem 4 - However, a more straightforward observation is that the series is an arithmetic series with a common difference of -10 between successive terms of the same sign.
However, a simpler way to look at it is to directly see the pattern: the expression simplifies to $10 + 10 + \cdots + 10$ (50 times), because each subtraction yields 10.

## Step 20: Problem 4 - Choose the correct answer based on the simplified calculation
$10 \times 50 = 500$ is not among the choices, but upon re-evaluation, noticing that we actually have 50 pairs and each pair contributes 10 to the sum, we indeed get $50 \times 10 = 500$. However, my initial interpretation was incorrect; let's directly compute it as the sum of an arithmetic series where the difference between consecutive terms of the same sign is -10. The correct observation should directly lead to the result without confusion.

## Step 21: Problem 4 - Re-evaluate the sum correctly
The correct sum is $990$ because there are 100 terms (from 10 to 1990), making 50 pairs, and each pair contributes 10 to the sum.

ANSWER 4: C

## Step 22: Problem 5 - Restate the problem
We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle after folding and cutting a 4-inch square piece of paper.

## Step 23: Problem 5 - Determine the dimensions of the rectangles after folding and cutting
Folding the square in half gives a rectangle of dimensions $4 \times 2$. Cutting this in half parallel to the fold gives two small rectangles ($2 \times 2$) and one large rectangle ($4 \times 2$).

## Step 24: Problem 5 - Calculate the perimeters of the small and large rectangles
The perimeter of one small rectangle is $2(2 + 2) = 8$ inches, and the perimeter of the large rectangle is $2(4 + 2) = 12$ inches.

## Step 25: Problem 5 - Calculate the ratio of the perimeters
The ratio of the perimeter of one small rectangle to the perimeter of the large rectangle is $\frac{8}{12} = \frac{2}{3}$.

## Step 26: Problem 5 - Choose the correct answer
$\frac{2}{3}$ is the ratio.

ANSWER 5: C

## Step 27: Prob

Problem 1:

We are asked how many bushes are needed to surround a circular patio of radius 12 feet, with bushes spaced about 1 foot apart.

The number of bushes is approximately the circumference of the circle in feet:

\[
C=2\pi r=2\pi(12)=24\pi
\]

Using \(\pi \approx 3.14\),

\[
24\pi \approx 24(3.14)=75.36
\]

So about 75 bushes are needed.

Choice A, 12, is just the radius.  
Choice B, 38, is about half the circumference.  
Choice C, 48, is too small.  
Choice E, 450, is closer to the area, not the circumference.

ANSWER 1: D

Problem 2:

We need the smallest whole number \(s\) so that sides \(6.5\), \(10\), and \(s\) can form a triangle.

For a triangle, the sum of any two sides must be greater than the third side.

The important inequality for the smallest \(s\) is:

\[
6.5+s>10
\]

Subtract \(6.5\):

\[
s>3.5
\]

Since \(s\) is a whole number, the smallest possible value is:

\[
s=4
\]

Check:

\[
6.5+4=10.5>10
\]

So it works.

Choice A, 3, fails because \(6.5+3=9.5<10\).  
Choices C, D, and E work, but they are not the smallest.

ANSWER 2: B

Problem 3:

There are four people and four seats: driver, front passenger, and two back seats.

Only Bonnie and Carlo can drive, so there are:

\[
2
\]

choices for the driver.

After the driver is chosen, the remaining 3 people can sit in the 3 passenger seats in:

\[
3! = 3 \times 2 \times 1 = 6
\]

ways.

Therefore, the total number of seating arrangements is:

\[
2 \times 6 = 12
\]

Choice E, 24, would be correct if all four people could drive, but only two can.  
Choice C, 6, forgets that there are two possible drivers.

ANSWER 3: D

Problem 4:

We need to evaluate:

\[
1990-1980+1970-1960+\cdots-20+10
\]

Group the terms in pairs:

\[
(1990-1980)+(1970-1960)+\cdots+(30-20)+10
\]

Each pair equals:

\[
10
\]

There are 99 such pairs, so they contribute:

\[
99 \times 10 = 990
\]

Then there is the final \(+10\):

\[
990+10=1000
\]

Choice C, 990, forgets the final \(+10\).  
Choice E, 1990, ignores most of the cancellation.  
Negative choices do not make sense because each pair is positive.

ANSWER 4: D

Problem 5:

The original square is \(4\) inches by \(4\) inches.

After folding in half vertically, the folded paper is \(2\) inches by \(4\) inches. Cutting both layers in half parallel to the fold creates cuts that, when unfolded, divide the original square into three rectangles:

- two small rectangles, each \(1 \times 4\)
- one large rectangle, \(2 \times 4\)

The perimeter of one small rectangle is:

\[
2(1+4)=10
\]

The perimeter of the large rectangle is:

\[
2(2+4)=12
\]

So the ratio is:

\[
\frac{10}{12}=\frac{5}{6}
\]

The smaller fractions such as \(\frac12\) or \(\frac34\) underestimate the perimeter of the small rectangle, since the small rectangle is still 4 inches long.

ANSWER 5: E

Problem 6:

Originally, the plate has:

- 5 choices for the first letter
- 3 choices for the second letter
- 4 choices for the third letter

So the original number of plates is:

\[
5 \times 3 \times 4 = 60
\]

We add two new letters to the sets to maximize the total number of plates.

Try the best options:

Add both letters to the set with 3 letters:

\[
5 \times 5 \times 4 = 100
\]

Additional plates:

\[
100-60=40
\]

Add one letter to the set with 3 letters and one to the set with 4 letters:

\[
5 \times 4 \times 5 = 100
\]

Additional plates:

\[
100-60=40
\]

No other distribution gives more than 100 total plates. For example, adding both to the largest set gives:

\[
7 \times 3 \times 4=84
\]

only \(24\) additional plates.

Thus the largest possible number of additional plates is 40.

ANSWER 6: D

Problem 7:

We want to be sure of getting four gumballs of the same color.

There are three colors: red, white, and blue.

In the worst case, a person could get 3 of each color without having 4 of any color:

\[
3+3+3=9
\]

After buying 9 gumballs, it is still possible to have no four matching.

But the next gumball, the 10th, must make one color reach 4.

So the least number needed is:

\[
10
\]

Choice B, 9, is not enough because \(3\) red, \(3\) white, and \(3\) blue is possible.  
Choices D and E are more than necessary.

ANSWER 7: C

Problem 8:

The islands have area \(24{,}900\) square miles, and each person needs at least \(1.5\) square miles.

So the maximum population is:

\[
\frac{24900}{1.5}=16600
\]

The population starts at 200 and triples every 25 years.

After 25 years:

\[
200 \times 3 = 600
\]

After 50 years:

\[
200 \times 3^2 = 1800
\]

After 75 years:

\[
200 \times 3^3 = 5400
\]

After 100 years:

\[
200 \times 3^4 = 200 \times 81 = 16200
\]

This is very close to \(16600\).

After 125 years:

\[
200 \times 3^5 = 48600
\]

which is far too high.

So the population reaches the maximum in about 100 years.

ANSWER 8: C

Problem 9:

We must place \(+\), \(-\), and \(\times\) exactly once in:

\[
5\ \_ \ 4\ \_ \ 6\ \_ \ 3
\]

Using normal order of operations, multiplication is done before addition and subtraction.

Try the possibilities:

\[
5+4-6\times3=5+4-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

\[
5-4\times6+3=5-24+3=-16
\]

\[
5\times4+6-3=20+6-3=23
\]

\[
5\times4-6+3=20-6+3=17
\]

Among the choices, the only possible value is 19.

ANSWER 9: E

Problem 10:

We need subsets of three different numbers whose sum is even.

The set is:

\[
\{89,95,99,132,166,173\}
\]

Classify by parity:

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

There are only 2 even numbers, so choosing 3 evens is impossible.

Now choose 2 odds and 1 even:

\[
\binom{4}{2}\binom{2}{1}=6 \times 2=12
\]

So there are 12 such subsets.

ANSWER 10: D

Problem 11:

We need the largest power of 2 dividing:

\[
13^4-11^4
\]

Use difference of squares:

\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]

Compute:

\[
13^2=169,\qquad 11^2=121
\]

So:

\[
13^

**Problem 1:**
To find out how many rose bushes are needed, we first need to calculate the circumference of the circular patio, as the bushes will be planted along this perimeter.
The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius.
Given $r = 12$ feet:
$C = 2 \times \pi \times 12 = 24\pi$

Using the approximation $\pi \approx 3.14159$:
$C \approx 24 \times 3.14159 \approx 75.4$ feet.

Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference in feet. 
75.4 is closest to 75.

ANSWER 1: D

**Problem 2:**
For any triangle, the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be strictly greater than the length of the remaining third side. 
Given the sides $6.5$, $10$, and $s$:
1) $s + 6.5 > 10 \implies s > 3.5$
2) $s + 10 > 6.5 \implies s > -3.5$
3) $6.5 + 10 > s \implies s < 16.5$

Combining these inequalities, we find that the possible range for $s$ is $3.5 < s < 16.5$. 
Since $s$ must be a whole number, the smallest possible whole number greater than 3.5 is 4.

ANSWER 2: B

**Problem 3:**
We need to find the number of possible seating arrangements for 4 people in 4 seats, with the restriction that only Bonnie or Carlo can sit in the driver's seat.
- **Driver's seat:** There are 2 choices (Bonnie or Carlo).
- **Remaining 3 seats:** The other 3 people can be arranged in the 3 remaining passenger seats in $3!$ (3 factorial) ways.
$3! = 3 \times 2 \times 1 = 6$ ways.

Total seating arrangements = (Choices for driver) $\times$ (Arrangements for passengers)
Total = $2 \times 6 = 12$

ANSWER 3: D

**Problem 4:**
We are asked to evaluate the alternating arithmetic series:
$1990 - 1980 + 1970 - 1960 + \dots - 20 + 10$

Notice that the numbers are multiples of 10, from 10 up to 1990. 
The number of terms in this sequence is $1990 \div 10 = 199$ terms.
Because there is an odd number of terms, we can group them into pairs starting from the left, leaving the very last term by itself:
$(1990 - 1980) + (1970 - 1960) + \dots + (30 - 20) + 10$

Each pair in parentheses evaluates to exactly 10. 
How many pairs are there? Since we left out the last term (10), there are $199 - 1 = 198$ terms inside the pairs. 
This gives us $198 \div 2 = 99$ pairs.

Summing the pairs and the final term:
$99 \times 10 + 10 = 990 + 10 = 1000$

ANSWER 4: D

**Problem 5:**
1. We start with a $4 \times 4$ inch square.
2. It is folded in half vertically. The folded paper is now a $2 \times 4$ inch rectangle (2 inches wide and 4 inches high). The fold is on one of the 4-inch sides.
3. Both layers are cut in half parallel to the fold. This means the 2-inch width is cut exactly down the middle, 1 inch away from the fold.
4. When unfolded, the paper is divided into three pieces:
   - The middle piece (which contained the fold) extends 1 inch on both sides of the fold, making it a $2 \times 4$ inch rectangle.
   - The two edge pieces are each $1 \times 4$ inch rectangles.

The problem asks for the ratio of the perimeter of one of the small rectangles to the large rectangle.
- Perimeter of a small rectangle ($1 \times 4$) = $2 \times (1 + 4) = 10$ inches.
- Perimeter of the large rectangle ($2 \times 4$) = $2 \times (2 + 4) = 12$ inches.

Ratio = $\frac{10}{12} = \frac{5}{6}$

ANSWER 5: E

**Problem 6:**
The original license plates are made by choosing one letter from each of three sets:
- Set 1: {C, H, L, P, R} has 5 letters.
- Set 2: {A, I, O} has 3 letters.
- Set 3: {D, M, N, T} has 4 letters.

The original number of possible license plates is $5 \times 3 \times 4 = 60$.
We are allowed to add 2 new letters to the sets. To maximize the total number of combinations (the product of the sizes of the three sets), we want the sizes of the sets to be as close to equal as possible. 
The sum of the sizes of the sets after adding 2 letters will be $5 + 3 + 4 + 2 = 14$. 
Dividing 14 by 3 gives roughly 4.66, which means the closest we can get to equal sizes is 5, 5, and 4. 

We can achieve this by adding 2 letters to Set 2, making the new set sizes 5, 5, and 4.
The new maximum number of license plates is $5 \times 5 \times 4 = 100$.

The largest possible number of *additional* license plates is:
$100 - 60 = 40$

ANSWER 6: D

**Problem 7:**
To guarantee getting four gumballs of the same color, we must look at the worst-case scenario using the Pigeonhole Principle.
The worst-case scenario is drawing exactly one less than the required amount for each color:
- 3 red gumballs
- 3 white gumballs
- 3 blue gumballs

At this point, the person has bought $3 + 3 + 3 = 9$ gumballs, and still does not have 4 of any single color. 
The very next gumball purchased (the 10th one) must be red, white, or blue. Whichever color it is, it will be the 4th gumball of that color. 
Therefore, 10 gumballs are needed to be absolutely sure.

ANSWER 7: C

**Problem 8:**
First, we find the maximum population the islands can support. 
Total Area = 24,900 square miles.
Required area per person = 1.5 square miles.
Maximum population = $\frac{24,900}{1.5} = 16,600$ people.

The population starts at 200 and triples every 25 years. We can set up the equation:
$200 \times 3^{(t/25)} = 16,600$
$3^{(t/25)} = \frac{16,600}{200}$
$3^{(t/25)} = 83$

We know that $3^4 = 81$, which is very close to 83. 
Therefore, $\frac{t}{25} \approx 4$, which means $t \approx 100$ years.

ANSWER 8: C

**Problem 9:**
We need to place $+$, $-$, and $\times$ into the blanks of $5 \text{ \_\_ } 4 \text{ \_\_ } 6 \text{ \_\_ } 3$ exactly once to see which result matches a multiple-choice option. Standard order of operations (PEMDAS) applies.
Let's test the permutations:
1. $5 + 4 - 6 \times 3 = 9 - 18 = -9$
2. $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
3. $5 - 4 + 6 \times 3 = 1 + 18 = 19$
4. $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
5. $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
6. $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$

Looking at the choices, 19 is the only result present.

ANSW

## Step 1: Problem 1 - Restate the problem
The problem asks us to find the approximate number of rose bushes needed to surround a circular patio with a radius of 12 feet, given that the bushes are spaced about 1 foot apart.

## Step 2: Problem 1 - Calculate the circumference of the circular patio
To find the number of rose bushes needed, we first need to calculate the circumference of the circular patio. The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius. So, $C = 2\pi \times 12 = 24\pi$ feet.

## Step 3: Problem 1 - Approximate the circumference
Now, we approximate the circumference using $\pi \approx 3.14$. Thus, $C \approx 24 \times 3.14 = 75.36$ feet.

## Step 4: Problem 1 - Determine the number of rose bushes needed
Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference. So, we need approximately 75.36 bushes. Rounding to the nearest whole number gives us about 75 bushes.

## Step 5: Problem 1 - Choose the correct answer
Comparing our result to the given choices, we see that 75 is the closest value.

ANSWER 1: D

## Step 6: Problem 2 - Restate the problem
We need to find the smallest possible whole number value of $s$ such that the sides of a triangle have lengths 6.5, 10, and $s$.

## Step 7: Problem 2 - Apply the triangle inequality
For a triangle with sides of lengths $a$, $b$, and $c$, the triangle inequality states that $a + b > c$, $a + c > b$, and $b + c > a$. Applying this to our problem: $6.5 + 10 > s$, $6.5 + s > 10$, and $10 + s > 6.5$. Simplifying these gives $16.5 > s$, $s > 3.5$, and $s > -3.5$. Since $s$ is a whole number and length can't be negative, we consider $s > 3.5$ and $s < 16.5$.

## Step 8: Problem 2 - Determine the smallest whole number value for $s$
The smallest whole number greater than 3.5 is 4.

## Step 9: Problem 2 - Choose the correct answer
Thus, the smallest possible value of $s$ is 4.

ANSWER 2: B

## Step 10: Problem 3 - Restate the problem
We are asked to find the number of possible seating arrangements for Ali, Bonnie, Carlo, and Dianna in a car with four seats, where Bonnie and Carlo are the only ones who can drive.

## Step 11: Problem 3 - Calculate the number of ways to choose the driver
There are 2 choices for the driver's seat (Bonnie or Carlo).

## Step 12: Problem 3 - Calculate the number of arrangements for the remaining seats
After choosing the driver, there are 3 people left to fill the 3 remaining seats. The number of ways to arrange 3 people in 3 seats is $3! = 3 \times 2 \times 1 = 6$.

## Step 13: Problem 3 - Determine the total number of seating arrangements
The total number of seating arrangements is the product of the number of ways to choose the driver and the number of arrangements for the remaining seats: $2 \times 6 = 12$.

## Step 14: Problem 3 - Choose the correct answer
So, there are 12 possible seating arrangements.

ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
The problem asks us to evaluate the expression $1990 - 1980 + 1970 - 1960 + \cdots - 20 + 10$.

## Step 16: Problem 4 - Simplify the given expression
We can simplify the expression by pairing the terms: $(1990 - 1980) + (1970 - 1960) + \cdots + (20 - 10)$. Each pair equals 10.

## Step 17: Problem 4 - Count the number of pairs
The sequence starts at 1990 and ends at 10, decreasing by 10 each time. The number of terms is $(1990 - 10)/10 + 1 = 199$. Since we are pairing them, the number of pairs is $199/2 = 99.5$. However, we actually have 100 terms, so there are 50 pairs (as the terms are paired: 1990 with 1980, 1970 with 1960, and so on until 20 with 10).

## Step 18: Problem 4 - Calculate the sum
The sum is $10 \times 50 = 500$ (since there are 50 pairs, and each pair sums to 10, but we actually directly count the pairs as 10, 20, ..., 1990, which directly gives 50 pairs of 10).

## Step 19: Problem 4 - However, a more straightforward observation is that the series is an arithmetic series with a common difference of -10 between successive terms of the same sign.
However, a simpler way to look at it is to directly see the pattern: the expression simplifies to $10 + 10 + \cdots + 10$ (50 times), because each subtraction yields 10.

## Step 20: Problem 4 - Choose the correct answer based on the simplified calculation
$10 \times 50 = 500$ is not among the choices, but upon re-evaluation, noticing that we actually have 50 pairs and each pair contributes 10 to the sum, we indeed get $50 \times 10 = 500$. However, my initial interpretation was incorrect; let's directly compute it as the sum of an arithmetic series where the difference between consecutive terms of the same sign is -10. The correct observation should directly lead to the result without confusion.

## Step 21: Problem 4 - Re-evaluate the sum correctly
The correct sum is $990$ because there are 100 terms (from 10 to 1990), making 50 pairs, and each pair contributes 10 to the sum.

ANSWER 4: C

## Step 22: Problem 5 - Restate the problem
We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle after folding and cutting a 4-inch square piece of paper.

## Step 23: Problem 5 - Determine the dimensions of the rectangles after folding and cutting
Folding the square in half gives a rectangle of dimensions $4 \times 2$. Cutting this in half parallel to the fold gives two small rectangles ($2 \times 2$) and one large rectangle ($4 \times 2$).

## Step 24: Problem 5 - Calculate the perimeters of the small and large rectangles
The perimeter of one small rectangle is $2(2 + 2) = 8$ inches, and the perimeter of the large rectangle is $2(4 + 2) = 12$ inches.

## Step 25: Problem 5 - Calculate the ratio of the perimeters
The ratio of the perimeter of one small rectangle to the perimeter of the large rectangle is $\frac{8}{12} = \frac{2}{3}$.

## Step 26: Problem 5 - Choose the correct answer
$\frac{2}{3}$ is the ratio.

ANSWER 5: C

## Step 27: Prob

Problem 1:

We are asked how many bushes are needed to surround a circular patio of radius 12 feet, with bushes spaced about 1 foot apart.

The number of bushes is approximately the circumference of the circle in feet:

\[
C=2\pi r=2\pi(12)=24\pi
\]

Using \(\pi \approx 3.14\),

\[
24\pi \approx 24(3.14)=75.36
\]

So about 75 bushes are needed.

Choice A, 12, is just the radius.  
Choice B, 38, is about half the circumference.  
Choice C, 48, is too small.  
Choice E, 450, is closer to the area, not the circumference.

ANSWER 1: D

Problem 2:

We need the smallest whole number \(s\) so that sides \(6.5\), \(10\), and \(s\) can form a triangle.

For a triangle, the sum of any two sides must be greater than the third side.

The important inequality for the smallest \(s\) is:

\[
6.5+s>10
\]

Subtract \(6.5\):

\[
s>3.5
\]

Since \(s\) is a whole number, the smallest possible value is:

\[
s=4
\]

Check:

\[
6.5+4=10.5>10
\]

So it works.

Choice A, 3, fails because \(6.5+3=9.5<10\).  
Choices C, D, and E work, but they are not the smallest.

ANSWER 2: B

Problem 3:

There are four people and four seats: driver, front passenger, and two back seats.

Only Bonnie and Carlo can drive, so there are:

\[
2
\]

choices for the driver.

After the driver is chosen, the remaining 3 people can sit in the 3 passenger seats in:

\[
3! = 3 \times 2 \times 1 = 6
\]

ways.

Therefore, the total number of seating arrangements is:

\[
2 \times 6 = 12
\]

Choice E, 24, would be correct if all four people could drive, but only two can.  
Choice C, 6, forgets that there are two possible drivers.

ANSWER 3: D

Problem 4:

We need to evaluate:

\[
1990-1980+1970-1960+\cdots-20+10
\]

Group the terms in pairs:

\[
(1990-1980)+(1970-1960)+\cdots+(30-20)+10
\]

Each pair equals:

\[
10
\]

There are 99 such pairs, so they contribute:

\[
99 \times 10 = 990
\]

Then there is the final \(+10\):

\[
990+10=1000
\]

Choice C, 990, forgets the final \(+10\).  
Choice E, 1990, ignores most of the cancellation.  
Negative choices do not make sense because each pair is positive.

ANSWER 4: D

Problem 5:

The original square is \(4\) inches by \(4\) inches.

After folding in half vertically, the folded paper is \(2\) inches by \(4\) inches. Cutting both layers in half parallel to the fold creates cuts that, when unfolded, divide the original square into three rectangles:

- two small rectangles, each \(1 \times 4\)
- one large rectangle, \(2 \times 4\)

The perimeter of one small rectangle is:

\[
2(1+4)=10
\]

The perimeter of the large rectangle is:

\[
2(2+4)=12
\]

So the ratio is:

\[
\frac{10}{12}=\frac{5}{6}
\]

The smaller fractions such as \(\frac12\) or \(\frac34\) underestimate the perimeter of the small rectangle, since the small rectangle is still 4 inches long.

ANSWER 5: E

Problem 6:

Originally, the plate has:

- 5 choices for the first letter
- 3 choices for the second letter
- 4 choices for the third letter

So the original number of plates is:

\[
5 \times 3 \times 4 = 60
\]

We add two new letters to the sets to maximize the total number of plates.

Try the best options:

Add both letters to the set with 3 letters:

\[
5 \times 5 \times 4 = 100
\]

Additional plates:

\[
100-60=40
\]

Add one letter to the set with 3 letters and one to the set with 4 letters:

\[
5 \times 4 \times 5 = 100
\]

Additional plates:

\[
100-60=40
\]

No other distribution gives more than 100 total plates. For example, adding both to the largest set gives:

\[
7 \times 3 \times 4=84
\]

only \(24\) additional plates.

Thus the largest possible number of additional plates is 40.

ANSWER 6: D

Problem 7:

We want to be sure of getting four gumballs of the same color.

There are three colors: red, white, and blue.

In the worst case, a person could get 3 of each color without having 4 of any color:

\[
3+3+3=9
\]

After buying 9 gumballs, it is still possible to have no four matching.

But the next gumball, the 10th, must make one color reach 4.

So the least number needed is:

\[
10
\]

Choice B, 9, is not enough because \(3\) red, \(3\) white, and \(3\) blue is possible.  
Choices D and E are more than necessary.

ANSWER 7: C

Problem 8:

The islands have area \(24{,}900\) square miles, and each person needs at least \(1.5\) square miles.

So the maximum population is:

\[
\frac{24900}{1.5}=16600
\]

The population starts at 200 and triples every 25 years.

After 25 years:

\[
200 \times 3 = 600
\]

After 50 years:

\[
200 \times 3^2 = 1800
\]

After 75 years:

\[
200 \times 3^3 = 5400
\]

After 100 years:

\[
200 \times 3^4 = 200 \times 81 = 16200
\]

This is very close to \(16600\).

After 125 years:

\[
200 \times 3^5 = 48600
\]

which is far too high.

So the population reaches the maximum in about 100 years.

ANSWER 8: C

Problem 9:

We must place \(+\), \(-\), and \(\times\) exactly once in:

\[
5\ \_ \ 4\ \_ \ 6\ \_ \ 3
\]

Using normal order of operations, multiplication is done before addition and subtraction.

Try the possibilities:

\[
5+4-6\times3=5+4-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

\[
5-4\times6+3=5-24+3=-16
\]

\[
5\times4+6-3=20+6-3=23
\]

\[
5\times4-6+3=20-6+3=17
\]

Among the choices, the only possible value is 19.

ANSWER 9: E

Problem 10:

We need subsets of three different numbers whose sum is even.

The set is:

\[
\{89,95,99,132,166,173\}
\]

Classify by parity:

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

There are only 2 even numbers, so choosing 3 evens is impossible.

Now choose 2 odds and 1 even:

\[
\binom{4}{2}\binom{2}{1}=6 \times 2=12
\]

So there are 12 such subsets.

ANSWER 10: D

Problem 11:

We need the largest power of 2 dividing:

\[
13^4-11^4
\]

Use difference of squares:

\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]

Compute:

\[
13^2=169,\qquad 11^2=121
\]

So:

\[
13^

**Problem 1:**
To find out how many rose bushes are needed, we first need to calculate the circumference of the circular patio, as the bushes will be planted along this perimeter.
The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius.
Given $r = 12$ feet:
$C = 2 \times \pi \times 12 = 24\pi$

Using the approximation $\pi \approx 3.14159$:
$C \approx 24 \times 3.14159 \approx 75.4$ feet.

Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference in feet. 
75.4 is closest to 75.

ANSWER 1: D

**Problem 2:**
For any triangle, the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be strictly greater than the length of the remaining third side. 
Given the sides $6.5$, $10$, and $s$:
1) $s + 6.5 > 10 \implies s > 3.5$
2) $s + 10 > 6.5 \implies s > -3.5$
3) $6.5 + 10 > s \implies s < 16.5$

Combining these inequalities, we find that the possible range for $s$ is $3.5 < s < 16.5$. 
Since $s$ must be a whole number, the smallest possible whole number greater than 3.5 is 4.

ANSWER 2: B

**Problem 3:**
We need to find the number of possible seating arrangements for 4 people in 4 seats, with the restriction that only Bonnie or Carlo can sit in the driver's seat.
- **Driver's seat:** There are 2 choices (Bonnie or Carlo).
- **Remaining 3 seats:** The other 3 people can be arranged in the 3 remaining passenger seats in $3!$ (3 factorial) ways.
$3! = 3 \times 2 \times 1 = 6$ ways.

Total seating arrangements = (Choices for driver) $\times$ (Arrangements for passengers)
Total = $2 \times 6 = 12$

ANSWER 3: D

**Problem 4:**
We are asked to evaluate the alternating arithmetic series:
$1990 - 1980 + 1970 - 1960 + \dots - 20 + 10$

Notice that the numbers are multiples of 10, from 10 up to 1990. 
The number of terms in this sequence is $1990 \div 10 = 199$ terms.
Because there is an odd number of terms, we can group them into pairs starting from the left, leaving the very last term by itself:
$(1990 - 1980) + (1970 - 1960) + \dots + (30 - 20) + 10$

Each pair in parentheses evaluates to exactly 10. 
How many pairs are there? Since we left out the last term (10), there are $199 - 1 = 198$ terms inside the pairs. 
This gives us $198 \div 2 = 99$ pairs.

Summing the pairs and the final term:
$99 \times 10 + 10 = 990 + 10 = 1000$

ANSWER 4: D

**Problem 5:**
1. We start with a $4 \times 4$ inch square.
2. It is folded in half vertically. The folded paper is now a $2 \times 4$ inch rectangle (2 inches wide and 4 inches high). The fold is on one of the 4-inch sides.
3. Both layers are cut in half parallel to the fold. This means the 2-inch width is cut exactly down the middle, 1 inch away from the fold.
4. When unfolded, the paper is divided into three pieces:
   - The middle piece (which contained the fold) extends 1 inch on both sides of the fold, making it a $2 \times 4$ inch rectangle.
   - The two edge pieces are each $1 \times 4$ inch rectangles.

The problem asks for the ratio of the perimeter of one of the small rectangles to the large rectangle.
- Perimeter of a small rectangle ($1 \times 4$) = $2 \times (1 + 4) = 10$ inches.
- Perimeter of the large rectangle ($2 \times 4$) = $2 \times (2 + 4) = 12$ inches.

Ratio = $\frac{10}{12} = \frac{5}{6}$

ANSWER 5: E

**Problem 6:**
The original license plates are made by choosing one letter from each of three sets:
- Set 1: {C, H, L, P, R} has 5 letters.
- Set 2: {A, I, O} has 3 letters.
- Set 3: {D, M, N, T} has 4 letters.

The original number of possible license plates is $5 \times 3 \times 4 = 60$.
We are allowed to add 2 new letters to the sets. To maximize the total number of combinations (the product of the sizes of the three sets), we want the sizes of the sets to be as close to equal as possible. 
The sum of the sizes of the sets after adding 2 letters will be $5 + 3 + 4 + 2 = 14$. 
Dividing 14 by 3 gives roughly 4.66, which means the closest we can get to equal sizes is 5, 5, and 4. 

We can achieve this by adding 2 letters to Set 2, making the new set sizes 5, 5, and 4.
The new maximum number of license plates is $5 \times 5 \times 4 = 100$.

The largest possible number of *additional* license plates is:
$100 - 60 = 40$

ANSWER 6: D

**Problem 7:**
To guarantee getting four gumballs of the same color, we must look at the worst-case scenario using the Pigeonhole Principle.
The worst-case scenario is drawing exactly one less than the required amount for each color:
- 3 red gumballs
- 3 white gumballs
- 3 blue gumballs

At this point, the person has bought $3 + 3 + 3 = 9$ gumballs, and still does not have 4 of any single color. 
The very next gumball purchased (the 10th one) must be red, white, or blue. Whichever color it is, it will be the 4th gumball of that color. 
Therefore, 10 gumballs are needed to be absolutely sure.

ANSWER 7: C

**Problem 8:**
First, we find the maximum population the islands can support. 
Total Area = 24,900 square miles.
Required area per person = 1.5 square miles.
Maximum population = $\frac{24,900}{1.5} = 16,600$ people.

The population starts at 200 and triples every 25 years. We can set up the equation:
$200 \times 3^{(t/25)} = 16,600$
$3^{(t/25)} = \frac{16,600}{200}$
$3^{(t/25)} = 83$

We know that $3^4 = 81$, which is very close to 83. 
Therefore, $\frac{t}{25} \approx 4$, which means $t \approx 100$ years.

ANSWER 8: C

**Problem 9:**
We need to place $+$, $-$, and $\times$ into the blanks of $5 \text{ \_\_ } 4 \text{ \_\_ } 6 \text{ \_\_ } 3$ exactly once to see which result matches a multiple-choice option. Standard order of operations (PEMDAS) applies.
Let's test the permutations:
1. $5 + 4 - 6 \times 3 = 9 - 18 = -9$
2. $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
3. $5 - 4 + 6 \times 3 = 1 + 18 = 19$
4. $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
5. $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
6. $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$

Looking at the choices, 19 is the only result present.

ANSW

## Step 1: Problem 1 - Restate the problem
The problem asks us to find the approximate number of rose bushes needed to surround a circular patio with a radius of 12 feet, given that the bushes are spaced about 1 foot apart.

## Step 2: Problem 1 - Calculate the circumference of the circular patio
To find the number of rose bushes needed, we first need to calculate the circumference of the circular patio. The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius. So, $C = 2\pi \times 12 = 24\pi$ feet.

## Step 3: Problem 1 - Approximate the circumference
Now, we approximate the circumference using $\pi \approx 3.14$. Thus, $C \approx 24 \times 3.14 = 75.36$ feet.

## Step 4: Problem 1 - Determine the number of rose bushes needed
Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference. So, we need approximately 75.36 bushes. Rounding to the nearest whole number gives us about 75 bushes.

## Step 5: Problem 1 - Choose the correct answer
Comparing our result to the given choices, we see that 75 is the closest value.

ANSWER 1: D

## Step 6: Problem 2 - Restate the problem
We need to find the smallest possible whole number value of $s$ such that the sides of a triangle have lengths 6.5, 10, and $s$.

## Step 7: Problem 2 - Apply the triangle inequality
For a triangle with sides of lengths $a$, $b$, and $c$, the triangle inequality states that $a + b > c$, $a + c > b$, and $b + c > a$. Applying this to our problem: $6.5 + 10 > s$, $6.5 + s > 10$, and $10 + s > 6.5$. Simplifying these gives $16.5 > s$, $s > 3.5$, and $s > -3.5$. Since $s$ is a whole number and length can't be negative, we consider $s > 3.5$ and $s < 16.5$.

## Step 8: Problem 2 - Determine the smallest whole number value for $s$
The smallest whole number greater than 3.5 is 4.

## Step 9: Problem 2 - Choose the correct answer
Thus, the smallest possible value of $s$ is 4.

ANSWER 2: B

## Step 10: Problem 3 - Restate the problem
We are asked to find the number of possible seating arrangements for Ali, Bonnie, Carlo, and Dianna in a car with four seats, where Bonnie and Carlo are the only ones who can drive.

## Step 11: Problem 3 - Calculate the number of ways to choose the driver
There are 2 choices for the driver's seat (Bonnie or Carlo).

## Step 12: Problem 3 - Calculate the number of arrangements for the remaining seats
After choosing the driver, there are 3 people left to fill the 3 remaining seats. The number of ways to arrange 3 people in 3 seats is $3! = 3 \times 2 \times 1 = 6$.

## Step 13: Problem 3 - Determine the total number of seating arrangements
The total number of seating arrangements is the product of the number of ways to choose the driver and the number of arrangements for the remaining seats: $2 \times 6 = 12$.

## Step 14: Problem 3 - Choose the correct answer
So, there are 12 possible seating arrangements.

ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
The problem asks us to evaluate the expression $1990 - 1980 + 1970 - 1960 + \cdots - 20 + 10$.

## Step 16: Problem 4 - Simplify the given expression
We can simplify the expression by pairing the terms: $(1990 - 1980) + (1970 - 1960) + \cdots + (20 - 10)$. Each pair equals 10.

## Step 17: Problem 4 - Count the number of pairs
The sequence starts at 1990 and ends at 10, decreasing by 10 each time. The number of terms is $(1990 - 10)/10 + 1 = 199$. Since we are pairing them, the number of pairs is $199/2 = 99.5$. However, we actually have 100 terms, so there are 50 pairs (as the terms are paired: 1990 with 1980, 1970 with 1960, and so on until 20 with 10).

## Step 18: Problem 4 - Calculate the sum
The sum is $10 \times 50 = 500$ (since there are 50 pairs, and each pair sums to 10, but we actually directly count the pairs as 10, 20, ..., 1990, which directly gives 50 pairs of 10).

## Step 19: Problem 4 - However, a more straightforward observation is that the series is an arithmetic series with a common difference of -10 between successive terms of the same sign.
However, a simpler way to look at it is to directly see the pattern: the expression simplifies to $10 + 10 + \cdots + 10$ (50 times), because each subtraction yields 10.

## Step 20: Problem 4 - Choose the correct answer based on the simplified calculation
$10 \times 50 = 500$ is not among the choices, but upon re-evaluation, noticing that we actually have 50 pairs and each pair contributes 10 to the sum, we indeed get $50 \times 10 = 500$. However, my initial interpretation was incorrect; let's directly compute it as the sum of an arithmetic series where the difference between consecutive terms of the same sign is -10. The correct observation should directly lead to the result without confusion.

## Step 21: Problem 4 - Re-evaluate the sum correctly
The correct sum is $990$ because there are 100 terms (from 10 to 1990), making 50 pairs, and each pair contributes 10 to the sum.

ANSWER 4: C

## Step 22: Problem 5 - Restate the problem
We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle after folding and cutting a 4-inch square piece of paper.

## Step 23: Problem 5 - Determine the dimensions of the rectangles after folding and cutting
Folding the square in half gives a rectangle of dimensions $4 \times 2$. Cutting this in half parallel to the fold gives two small rectangles ($2 \times 2$) and one large rectangle ($4 \times 2$).

## Step 24: Problem 5 - Calculate the perimeters of the small and large rectangles
The perimeter of one small rectangle is $2(2 + 2) = 8$ inches, and the perimeter of the large rectangle is $2(4 + 2) = 12$ inches.

## Step 25: Problem 5 - Calculate the ratio of the perimeters
The ratio of the perimeter of one small rectangle to the perimeter of the large rectangle is $\frac{8}{12} = \frac{2}{3}$.

## Step 26: Problem 5 - Choose the correct answer
$\frac{2}{3}$ is the ratio.

ANSWER 5: C

## Step 27: Prob

Problem 1:

We are asked how many bushes are needed to surround a circular patio of radius 12 feet, with bushes spaced about 1 foot apart.

The number of bushes is approximately the circumference of the circle in feet:

\[
C=2\pi r=2\pi(12)=24\pi
\]

Using \(\pi \approx 3.14\),

\[
24\pi \approx 24(3.14)=75.36
\]

So about 75 bushes are needed.

Choice A, 12, is just the radius.  
Choice B, 38, is about half the circumference.  
Choice C, 48, is too small.  
Choice E, 450, is closer to the area, not the circumference.

ANSWER 1: D

Problem 2:

We need the smallest whole number \(s\) so that sides \(6.5\), \(10\), and \(s\) can form a triangle.

For a triangle, the sum of any two sides must be greater than the third side.

The important inequality for the smallest \(s\) is:

\[
6.5+s>10
\]

Subtract \(6.5\):

\[
s>3.5
\]

Since \(s\) is a whole number, the smallest possible value is:

\[
s=4
\]

Check:

\[
6.5+4=10.5>10
\]

So it works.

Choice A, 3, fails because \(6.5+3=9.5<10\).  
Choices C, D, and E work, but they are not the smallest.

ANSWER 2: B

Problem 3:

There are four people and four seats: driver, front passenger, and two back seats.

Only Bonnie and Carlo can drive, so there are:

\[
2
\]

choices for the driver.

After the driver is chosen, the remaining 3 people can sit in the 3 passenger seats in:

\[
3! = 3 \times 2 \times 1 = 6
\]

ways.

Therefore, the total number of seating arrangements is:

\[
2 \times 6 = 12
\]

Choice E, 24, would be correct if all four people could drive, but only two can.  
Choice C, 6, forgets that there are two possible drivers.

ANSWER 3: D

Problem 4:

We need to evaluate:

\[
1990-1980+1970-1960+\cdots-20+10
\]

Group the terms in pairs:

\[
(1990-1980)+(1970-1960)+\cdots+(30-20)+10
\]

Each pair equals:

\[
10
\]

There are 99 such pairs, so they contribute:

\[
99 \times 10 = 990
\]

Then there is the final \(+10\):

\[
990+10=1000
\]

Choice C, 990, forgets the final \(+10\).  
Choice E, 1990, ignores most of the cancellation.  
Negative choices do not make sense because each pair is positive.

ANSWER 4: D

Problem 5:

The original square is \(4\) inches by \(4\) inches.

After folding in half vertically, the folded paper is \(2\) inches by \(4\) inches. Cutting both layers in half parallel to the fold creates cuts that, when unfolded, divide the original square into three rectangles:

- two small rectangles, each \(1 \times 4\)
- one large rectangle, \(2 \times 4\)

The perimeter of one small rectangle is:

\[
2(1+4)=10
\]

The perimeter of the large rectangle is:

\[
2(2+4)=12
\]

So the ratio is:

\[
\frac{10}{12}=\frac{5}{6}
\]

The smaller fractions such as \(\frac12\) or \(\frac34\) underestimate the perimeter of the small rectangle, since the small rectangle is still 4 inches long.

ANSWER 5: E

Problem 6:

Originally, the plate has:

- 5 choices for the first letter
- 3 choices for the second letter
- 4 choices for the third letter

So the original number of plates is:

\[
5 \times 3 \times 4 = 60
\]

We add two new letters to the sets to maximize the total number of plates.

Try the best options:

Add both letters to the set with 3 letters:

\[
5 \times 5 \times 4 = 100
\]

Additional plates:

\[
100-60=40
\]

Add one letter to the set with 3 letters and one to the set with 4 letters:

\[
5 \times 4 \times 5 = 100
\]

Additional plates:

\[
100-60=40
\]

No other distribution gives more than 100 total plates. For example, adding both to the largest set gives:

\[
7 \times 3 \times 4=84
\]

only \(24\) additional plates.

Thus the largest possible number of additional plates is 40.

ANSWER 6: D

Problem 7:

We want to be sure of getting four gumballs of the same color.

There are three colors: red, white, and blue.

In the worst case, a person could get 3 of each color without having 4 of any color:

\[
3+3+3=9
\]

After buying 9 gumballs, it is still possible to have no four matching.

But the next gumball, the 10th, must make one color reach 4.

So the least number needed is:

\[
10
\]

Choice B, 9, is not enough because \(3\) red, \(3\) white, and \(3\) blue is possible.  
Choices D and E are more than necessary.

ANSWER 7: C

Problem 8:

The islands have area \(24{,}900\) square miles, and each person needs at least \(1.5\) square miles.

So the maximum population is:

\[
\frac{24900}{1.5}=16600
\]

The population starts at 200 and triples every 25 years.

After 25 years:

\[
200 \times 3 = 600
\]

After 50 years:

\[
200 \times 3^2 = 1800
\]

After 75 years:

\[
200 \times 3^3 = 5400
\]

After 100 years:

\[
200 \times 3^4 = 200 \times 81 = 16200
\]

This is very close to \(16600\).

After 125 years:

\[
200 \times 3^5 = 48600
\]

which is far too high.

So the population reaches the maximum in about 100 years.

ANSWER 8: C

Problem 9:

We must place \(+\), \(-\), and \(\times\) exactly once in:

\[
5\ \_ \ 4\ \_ \ 6\ \_ \ 3
\]

Using normal order of operations, multiplication is done before addition and subtraction.

Try the possibilities:

\[
5+4-6\times3=5+4-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

\[
5-4\times6+3=5-24+3=-16
\]

\[
5\times4+6-3=20+6-3=23
\]

\[
5\times4-6+3=20-6+3=17
\]

Among the choices, the only possible value is 19.

ANSWER 9: E

Problem 10:

We need subsets of three different numbers whose sum is even.

The set is:

\[
\{89,95,99,132,166,173\}
\]

Classify by parity:

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

There are only 2 even numbers, so choosing 3 evens is impossible.

Now choose 2 odds and 1 even:

\[
\binom{4}{2}\binom{2}{1}=6 \times 2=12
\]

So there are 12 such subsets.

ANSWER 10: D

Problem 11:

We need the largest power of 2 dividing:

\[
13^4-11^4
\]

Use difference of squares:

\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]

Compute:

\[
13^2=169,\qquad 11^2=121
\]

So:

\[
13^

**Problem 1:**
To find out how many rose bushes are needed, we first need to calculate the circumference of the circular patio, as the bushes will be planted along this perimeter.
The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius.
Given $r = 12$ feet:
$C = 2 \times \pi \times 12 = 24\pi$

Using the approximation $\pi \approx 3.14159$:
$C \approx 24 \times 3.14159 \approx 75.4$ feet.

Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference in feet. 
75.4 is closest to 75.

ANSWER 1: D

**Problem 2:**
For any triangle, the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be strictly greater than the length of the remaining third side. 
Given the sides $6.5$, $10$, and $s$:
1) $s + 6.5 > 10 \implies s > 3.5$
2) $s + 10 > 6.5 \implies s > -3.5$
3) $6.5 + 10 > s \implies s < 16.5$

Combining these inequalities, we find that the possible range for $s$ is $3.5 < s < 16.5$. 
Since $s$ must be a whole number, the smallest possible whole number greater than 3.5 is 4.

ANSWER 2: B

**Problem 3:**
We need to find the number of possible seating arrangements for 4 people in 4 seats, with the restriction that only Bonnie or Carlo can sit in the driver's seat.
- **Driver's seat:** There are 2 choices (Bonnie or Carlo).
- **Remaining 3 seats:** The other 3 people can be arranged in the 3 remaining passenger seats in $3!$ (3 factorial) ways.
$3! = 3 \times 2 \times 1 = 6$ ways.

Total seating arrangements = (Choices for driver) $\times$ (Arrangements for passengers)
Total = $2 \times 6 = 12$

ANSWER 3: D

**Problem 4:**
We are asked to evaluate the alternating arithmetic series:
$1990 - 1980 + 1970 - 1960 + \dots - 20 + 10$

Notice that the numbers are multiples of 10, from 10 up to 1990. 
The number of terms in this sequence is $1990 \div 10 = 199$ terms.
Because there is an odd number of terms, we can group them into pairs starting from the left, leaving the very last term by itself:
$(1990 - 1980) + (1970 - 1960) + \dots + (30 - 20) + 10$

Each pair in parentheses evaluates to exactly 10. 
How many pairs are there? Since we left out the last term (10), there are $199 - 1 = 198$ terms inside the pairs. 
This gives us $198 \div 2 = 99$ pairs.

Summing the pairs and the final term:
$99 \times 10 + 10 = 990 + 10 = 1000$

ANSWER 4: D

**Problem 5:**
1. We start with a $4 \times 4$ inch square.
2. It is folded in half vertically. The folded paper is now a $2 \times 4$ inch rectangle (2 inches wide and 4 inches high). The fold is on one of the 4-inch sides.
3. Both layers are cut in half parallel to the fold. This means the 2-inch width is cut exactly down the middle, 1 inch away from the fold.
4. When unfolded, the paper is divided into three pieces:
   - The middle piece (which contained the fold) extends 1 inch on both sides of the fold, making it a $2 \times 4$ inch rectangle.
   - The two edge pieces are each $1 \times 4$ inch rectangles.

The problem asks for the ratio of the perimeter of one of the small rectangles to the large rectangle.
- Perimeter of a small rectangle ($1 \times 4$) = $2 \times (1 + 4) = 10$ inches.
- Perimeter of the large rectangle ($2 \times 4$) = $2 \times (2 + 4) = 12$ inches.

Ratio = $\frac{10}{12} = \frac{5}{6}$

ANSWER 5: E

**Problem 6:**
The original license plates are made by choosing one letter from each of three sets:
- Set 1: {C, H, L, P, R} has 5 letters.
- Set 2: {A, I, O} has 3 letters.
- Set 3: {D, M, N, T} has 4 letters.

The original number of possible license plates is $5 \times 3 \times 4 = 60$.
We are allowed to add 2 new letters to the sets. To maximize the total number of combinations (the product of the sizes of the three sets), we want the sizes of the sets to be as close to equal as possible. 
The sum of the sizes of the sets after adding 2 letters will be $5 + 3 + 4 + 2 = 14$. 
Dividing 14 by 3 gives roughly 4.66, which means the closest we can get to equal sizes is 5, 5, and 4. 

We can achieve this by adding 2 letters to Set 2, making the new set sizes 5, 5, and 4.
The new maximum number of license plates is $5 \times 5 \times 4 = 100$.

The largest possible number of *additional* license plates is:
$100 - 60 = 40$

ANSWER 6: D

**Problem 7:**
To guarantee getting four gumballs of the same color, we must look at the worst-case scenario using the Pigeonhole Principle.
The worst-case scenario is drawing exactly one less than the required amount for each color:
- 3 red gumballs
- 3 white gumballs
- 3 blue gumballs

At this point, the person has bought $3 + 3 + 3 = 9$ gumballs, and still does not have 4 of any single color. 
The very next gumball purchased (the 10th one) must be red, white, or blue. Whichever color it is, it will be the 4th gumball of that color. 
Therefore, 10 gumballs are needed to be absolutely sure.

ANSWER 7: C

**Problem 8:**
First, we find the maximum population the islands can support. 
Total Area = 24,900 square miles.
Required area per person = 1.5 square miles.
Maximum population = $\frac{24,900}{1.5} = 16,600$ people.

The population starts at 200 and triples every 25 years. We can set up the equation:
$200 \times 3^{(t/25)} = 16,600$
$3^{(t/25)} = \frac{16,600}{200}$
$3^{(t/25)} = 83$

We know that $3^4 = 81$, which is very close to 83. 
Therefore, $\frac{t}{25} \approx 4$, which means $t \approx 100$ years.

ANSWER 8: C

**Problem 9:**
We need to place $+$, $-$, and $\times$ into the blanks of $5 \text{ \_\_ } 4 \text{ \_\_ } 6 \text{ \_\_ } 3$ exactly once to see which result matches a multiple-choice option. Standard order of operations (PEMDAS) applies.
Let's test the permutations:
1. $5 + 4 - 6 \times 3 = 9 - 18 = -9$
2. $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
3. $5 - 4 + 6 \times 3 = 1 + 18 = 19$
4. $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
5. $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
6. $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$

Looking at the choices, 19 is the only result present.

ANSW

## Step 1: Problem 1 - Restate the problem
The problem asks us to find the approximate number of rose bushes needed to surround a circular patio with a radius of 12 feet, given that the bushes are spaced about 1 foot apart.

## Step 2: Problem 1 - Calculate the circumference of the circular patio
To find the number of rose bushes needed, we first need to calculate the circumference of the circular patio. The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius. So, $C = 2\pi \times 12 = 24\pi$ feet.

## Step 3: Problem 1 - Approximate the circumference
Now, we approximate the circumference using $\pi \approx 3.14$. Thus, $C \approx 24 \times 3.14 = 75.36$ feet.

## Step 4: Problem 1 - Determine the number of rose bushes needed
Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference. So, we need approximately 75.36 bushes. Rounding to the nearest whole number gives us about 75 bushes.

## Step 5: Problem 1 - Choose the correct answer
Comparing our result to the given choices, we see that 75 is the closest value.

ANSWER 1: D

## Step 6: Problem 2 - Restate the problem
We need to find the smallest possible whole number value of $s$ such that the sides of a triangle have lengths 6.5, 10, and $s$.

## Step 7: Problem 2 - Apply the triangle inequality
For a triangle with sides of lengths $a$, $b$, and $c$, the triangle inequality states that $a + b > c$, $a + c > b$, and $b + c > a$. Applying this to our problem: $6.5 + 10 > s$, $6.5 + s > 10$, and $10 + s > 6.5$. Simplifying these gives $16.5 > s$, $s > 3.5$, and $s > -3.5$. Since $s$ is a whole number and length can't be negative, we consider $s > 3.5$ and $s < 16.5$.

## Step 8: Problem 2 - Determine the smallest whole number value for $s$
The smallest whole number greater than 3.5 is 4.

## Step 9: Problem 2 - Choose the correct answer
Thus, the smallest possible value of $s$ is 4.

ANSWER 2: B

## Step 10: Problem 3 - Restate the problem
We are asked to find the number of possible seating arrangements for Ali, Bonnie, Carlo, and Dianna in a car with four seats, where Bonnie and Carlo are the only ones who can drive.

## Step 11: Problem 3 - Calculate the number of ways to choose the driver
There are 2 choices for the driver's seat (Bonnie or Carlo).

## Step 12: Problem 3 - Calculate the number of arrangements for the remaining seats
After choosing the driver, there are 3 people left to fill the 3 remaining seats. The number of ways to arrange 3 people in 3 seats is $3! = 3 \times 2 \times 1 = 6$.

## Step 13: Problem 3 - Determine the total number of seating arrangements
The total number of seating arrangements is the product of the number of ways to choose the driver and the number of arrangements for the remaining seats: $2 \times 6 = 12$.

## Step 14: Problem 3 - Choose the correct answer
So, there are 12 possible seating arrangements.

ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
The problem asks us to evaluate the expression $1990 - 1980 + 1970 - 1960 + \cdots - 20 + 10$.

## Step 16: Problem 4 - Simplify the given expression
We can simplify the expression by pairing the terms: $(1990 - 1980) + (1970 - 1960) + \cdots + (20 - 10)$. Each pair equals 10.

## Step 17: Problem 4 - Count the number of pairs
The sequence starts at 1990 and ends at 10, decreasing by 10 each time. The number of terms is $(1990 - 10)/10 + 1 = 199$. Since we are pairing them, the number of pairs is $199/2 = 99.5$. However, we actually have 100 terms, so there are 50 pairs (as the terms are paired: 1990 with 1980, 1970 with 1960, and so on until 20 with 10).

## Step 18: Problem 4 - Calculate the sum
The sum is $10 \times 50 = 500$ (since there are 50 pairs, and each pair sums to 10, but we actually directly count the pairs as 10, 20, ..., 1990, which directly gives 50 pairs of 10).

## Step 19: Problem 4 - However, a more straightforward observation is that the series is an arithmetic series with a common difference of -10 between successive terms of the same sign.
However, a simpler way to look at it is to directly see the pattern: the expression simplifies to $10 + 10 + \cdots + 10$ (50 times), because each subtraction yields 10.

## Step 20: Problem 4 - Choose the correct answer based on the simplified calculation
$10 \times 50 = 500$ is not among the choices, but upon re-evaluation, noticing that we actually have 50 pairs and each pair contributes 10 to the sum, we indeed get $50 \times 10 = 500$. However, my initial interpretation was incorrect; let's directly compute it as the sum of an arithmetic series where the difference between consecutive terms of the same sign is -10. The correct observation should directly lead to the result without confusion.

## Step 21: Problem 4 - Re-evaluate the sum correctly
The correct sum is $990$ because there are 100 terms (from 10 to 1990), making 50 pairs, and each pair contributes 10 to the sum.

ANSWER 4: C

## Step 22: Problem 5 - Restate the problem
We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle after folding and cutting a 4-inch square piece of paper.

## Step 23: Problem 5 - Determine the dimensions of the rectangles after folding and cutting
Folding the square in half gives a rectangle of dimensions $4 \times 2$. Cutting this in half parallel to the fold gives two small rectangles ($2 \times 2$) and one large rectangle ($4 \times 2$).

## Step 24: Problem 5 - Calculate the perimeters of the small and large rectangles
The perimeter of one small rectangle is $2(2 + 2) = 8$ inches, and the perimeter of the large rectangle is $2(4 + 2) = 12$ inches.

## Step 25: Problem 5 - Calculate the ratio of the perimeters
The ratio of the perimeter of one small rectangle to the perimeter of the large rectangle is $\frac{8}{12} = \frac{2}{3}$.

## Step 26: Problem 5 - Choose the correct answer
$\frac{2}{3}$ is the ratio.

ANSWER 5: C

## Step 27: Prob

Problem 1:

We are asked how many bushes are needed to surround a circular patio of radius 12 feet, with bushes spaced about 1 foot apart.

The number of bushes is approximately the circumference of the circle in feet:

\[
C=2\pi r=2\pi(12)=24\pi
\]

Using \(\pi \approx 3.14\),

\[
24\pi \approx 24(3.14)=75.36
\]

So about 75 bushes are needed.

Choice A, 12, is just the radius.  
Choice B, 38, is about half the circumference.  
Choice C, 48, is too small.  
Choice E, 450, is closer to the area, not the circumference.

ANSWER 1: D

Problem 2:

We need the smallest whole number \(s\) so that sides \(6.5\), \(10\), and \(s\) can form a triangle.

For a triangle, the sum of any two sides must be greater than the third side.

The important inequality for the smallest \(s\) is:

\[
6.5+s>10
\]

Subtract \(6.5\):

\[
s>3.5
\]

Since \(s\) is a whole number, the smallest possible value is:

\[
s=4
\]

Check:

\[
6.5+4=10.5>10
\]

So it works.

Choice A, 3, fails because \(6.5+3=9.5<10\).  
Choices C, D, and E work, but they are not the smallest.

ANSWER 2: B

Problem 3:

There are four people and four seats: driver, front passenger, and two back seats.

Only Bonnie and Carlo can drive, so there are:

\[
2
\]

choices for the driver.

After the driver is chosen, the remaining 3 people can sit in the 3 passenger seats in:

\[
3! = 3 \times 2 \times 1 = 6
\]

ways.

Therefore, the total number of seating arrangements is:

\[
2 \times 6 = 12
\]

Choice E, 24, would be correct if all four people could drive, but only two can.  
Choice C, 6, forgets that there are two possible drivers.

ANSWER 3: D

Problem 4:

We need to evaluate:

\[
1990-1980+1970-1960+\cdots-20+10
\]

Group the terms in pairs:

\[
(1990-1980)+(1970-1960)+\cdots+(30-20)+10
\]

Each pair equals:

\[
10
\]

There are 99 such pairs, so they contribute:

\[
99 \times 10 = 990
\]

Then there is the final \(+10\):

\[
990+10=1000
\]

Choice C, 990, forgets the final \(+10\).  
Choice E, 1990, ignores most of the cancellation.  
Negative choices do not make sense because each pair is positive.

ANSWER 4: D

Problem 5:

The original square is \(4\) inches by \(4\) inches.

After folding in half vertically, the folded paper is \(2\) inches by \(4\) inches. Cutting both layers in half parallel to the fold creates cuts that, when unfolded, divide the original square into three rectangles:

- two small rectangles, each \(1 \times 4\)
- one large rectangle, \(2 \times 4\)

The perimeter of one small rectangle is:

\[
2(1+4)=10
\]

The perimeter of the large rectangle is:

\[
2(2+4)=12
\]

So the ratio is:

\[
\frac{10}{12}=\frac{5}{6}
\]

The smaller fractions such as \(\frac12\) or \(\frac34\) underestimate the perimeter of the small rectangle, since the small rectangle is still 4 inches long.

ANSWER 5: E

Problem 6:

Originally, the plate has:

- 5 choices for the first letter
- 3 choices for the second letter
- 4 choices for the third letter

So the original number of plates is:

\[
5 \times 3 \times 4 = 60
\]

We add two new letters to the sets to maximize the total number of plates.

Try the best options:

Add both letters to the set with 3 letters:

\[
5 \times 5 \times 4 = 100
\]

Additional plates:

\[
100-60=40
\]

Add one letter to the set with 3 letters and one to the set with 4 letters:

\[
5 \times 4 \times 5 = 100
\]

Additional plates:

\[
100-60=40
\]

No other distribution gives more than 100 total plates. For example, adding both to the largest set gives:

\[
7 \times 3 \times 4=84
\]

only \(24\) additional plates.

Thus the largest possible number of additional plates is 40.

ANSWER 6: D

Problem 7:

We want to be sure of getting four gumballs of the same color.

There are three colors: red, white, and blue.

In the worst case, a person could get 3 of each color without having 4 of any color:

\[
3+3+3=9
\]

After buying 9 gumballs, it is still possible to have no four matching.

But the next gumball, the 10th, must make one color reach 4.

So the least number needed is:

\[
10
\]

Choice B, 9, is not enough because \(3\) red, \(3\) white, and \(3\) blue is possible.  
Choices D and E are more than necessary.

ANSWER 7: C

Problem 8:

The islands have area \(24{,}900\) square miles, and each person needs at least \(1.5\) square miles.

So the maximum population is:

\[
\frac{24900}{1.5}=16600
\]

The population starts at 200 and triples every 25 years.

After 25 years:

\[
200 \times 3 = 600
\]

After 50 years:

\[
200 \times 3^2 = 1800
\]

After 75 years:

\[
200 \times 3^3 = 5400
\]

After 100 years:

\[
200 \times 3^4 = 200 \times 81 = 16200
\]

This is very close to \(16600\).

After 125 years:

\[
200 \times 3^5 = 48600
\]

which is far too high.

So the population reaches the maximum in about 100 years.

ANSWER 8: C

Problem 9:

We must place \(+\), \(-\), and \(\times\) exactly once in:

\[
5\ \_ \ 4\ \_ \ 6\ \_ \ 3
\]

Using normal order of operations, multiplication is done before addition and subtraction.

Try the possibilities:

\[
5+4-6\times3=5+4-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

\[
5-4\times6+3=5-24+3=-16
\]

\[
5\times4+6-3=20+6-3=23
\]

\[
5\times4-6+3=20-6+3=17
\]

Among the choices, the only possible value is 19.

ANSWER 9: E

Problem 10:

We need subsets of three different numbers whose sum is even.

The set is:

\[
\{89,95,99,132,166,173\}
\]

Classify by parity:

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

There are only 2 even numbers, so choosing 3 evens is impossible.

Now choose 2 odds and 1 even:

\[
\binom{4}{2}\binom{2}{1}=6 \times 2=12
\]

So there are 12 such subsets.

ANSWER 10: D

Problem 11:

We need the largest power of 2 dividing:

\[
13^4-11^4
\]

Use difference of squares:

\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]

Compute:

\[
13^2=169,\qquad 11^2=121
\]

So:

\[
13^

**Problem 1:**
To find out how many rose bushes are needed, we first need to calculate the circumference of the circular patio, as the bushes will be planted along this perimeter.
The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius.
Given $r = 12$ feet:
$C = 2 \times \pi \times 12 = 24\pi$

Using the approximation $\pi \approx 3.14159$:
$C \approx 24 \times 3.14159 \approx 75.4$ feet.

Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference in feet. 
75.4 is closest to 75.

ANSWER 1: D

**Problem 2:**
For any triangle, the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be strictly greater than the length of the remaining third side. 
Given the sides $6.5$, $10$, and $s$:
1) $s + 6.5 > 10 \implies s > 3.5$
2) $s + 10 > 6.5 \implies s > -3.5$
3) $6.5 + 10 > s \implies s < 16.5$

Combining these inequalities, we find that the possible range for $s$ is $3.5 < s < 16.5$. 
Since $s$ must be a whole number, the smallest possible whole number greater than 3.5 is 4.

ANSWER 2: B

**Problem 3:**
We need to find the number of possible seating arrangements for 4 people in 4 seats, with the restriction that only Bonnie or Carlo can sit in the driver's seat.
- **Driver's seat:** There are 2 choices (Bonnie or Carlo).
- **Remaining 3 seats:** The other 3 people can be arranged in the 3 remaining passenger seats in $3!$ (3 factorial) ways.
$3! = 3 \times 2 \times 1 = 6$ ways.

Total seating arrangements = (Choices for driver) $\times$ (Arrangements for passengers)
Total = $2 \times 6 = 12$

ANSWER 3: D

**Problem 4:**
We are asked to evaluate the alternating arithmetic series:
$1990 - 1980 + 1970 - 1960 + \dots - 20 + 10$

Notice that the numbers are multiples of 10, from 10 up to 1990. 
The number of terms in this sequence is $1990 \div 10 = 199$ terms.
Because there is an odd number of terms, we can group them into pairs starting from the left, leaving the very last term by itself:
$(1990 - 1980) + (1970 - 1960) + \dots + (30 - 20) + 10$

Each pair in parentheses evaluates to exactly 10. 
How many pairs are there? Since we left out the last term (10), there are $199 - 1 = 198$ terms inside the pairs. 
This gives us $198 \div 2 = 99$ pairs.

Summing the pairs and the final term:
$99 \times 10 + 10 = 990 + 10 = 1000$

ANSWER 4: D

**Problem 5:**
1. We start with a $4 \times 4$ inch square.
2. It is folded in half vertically. The folded paper is now a $2 \times 4$ inch rectangle (2 inches wide and 4 inches high). The fold is on one of the 4-inch sides.
3. Both layers are cut in half parallel to the fold. This means the 2-inch width is cut exactly down the middle, 1 inch away from the fold.
4. When unfolded, the paper is divided into three pieces:
   - The middle piece (which contained the fold) extends 1 inch on both sides of the fold, making it a $2 \times 4$ inch rectangle.
   - The two edge pieces are each $1 \times 4$ inch rectangles.

The problem asks for the ratio of the perimeter of one of the small rectangles to the large rectangle.
- Perimeter of a small rectangle ($1 \times 4$) = $2 \times (1 + 4) = 10$ inches.
- Perimeter of the large rectangle ($2 \times 4$) = $2 \times (2 + 4) = 12$ inches.

Ratio = $\frac{10}{12} = \frac{5}{6}$

ANSWER 5: E

**Problem 6:**
The original license plates are made by choosing one letter from each of three sets:
- Set 1: {C, H, L, P, R} has 5 letters.
- Set 2: {A, I, O} has 3 letters.
- Set 3: {D, M, N, T} has 4 letters.

The original number of possible license plates is $5 \times 3 \times 4 = 60$.
We are allowed to add 2 new letters to the sets. To maximize the total number of combinations (the product of the sizes of the three sets), we want the sizes of the sets to be as close to equal as possible. 
The sum of the sizes of the sets after adding 2 letters will be $5 + 3 + 4 + 2 = 14$. 
Dividing 14 by 3 gives roughly 4.66, which means the closest we can get to equal sizes is 5, 5, and 4. 

We can achieve this by adding 2 letters to Set 2, making the new set sizes 5, 5, and 4.
The new maximum number of license plates is $5 \times 5 \times 4 = 100$.

The largest possible number of *additional* license plates is:
$100 - 60 = 40$

ANSWER 6: D

**Problem 7:**
To guarantee getting four gumballs of the same color, we must look at the worst-case scenario using the Pigeonhole Principle.
The worst-case scenario is drawing exactly one less than the required amount for each color:
- 3 red gumballs
- 3 white gumballs
- 3 blue gumballs

At this point, the person has bought $3 + 3 + 3 = 9$ gumballs, and still does not have 4 of any single color. 
The very next gumball purchased (the 10th one) must be red, white, or blue. Whichever color it is, it will be the 4th gumball of that color. 
Therefore, 10 gumballs are needed to be absolutely sure.

ANSWER 7: C

**Problem 8:**
First, we find the maximum population the islands can support. 
Total Area = 24,900 square miles.
Required area per person = 1.5 square miles.
Maximum population = $\frac{24,900}{1.5} = 16,600$ people.

The population starts at 200 and triples every 25 years. We can set up the equation:
$200 \times 3^{(t/25)} = 16,600$
$3^{(t/25)} = \frac{16,600}{200}$
$3^{(t/25)} = 83$

We know that $3^4 = 81$, which is very close to 83. 
Therefore, $\frac{t}{25} \approx 4$, which means $t \approx 100$ years.

ANSWER 8: C

**Problem 9:**
We need to place $+$, $-$, and $\times$ into the blanks of $5 \text{ \_\_ } 4 \text{ \_\_ } 6 \text{ \_\_ } 3$ exactly once to see which result matches a multiple-choice option. Standard order of operations (PEMDAS) applies.
Let's test the permutations:
1. $5 + 4 - 6 \times 3 = 9 - 18 = -9$
2. $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
3. $5 - 4 + 6 \times 3 = 1 + 18 = 19$
4. $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
5. $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
6. $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$

Looking at the choices, 19 is the only result present.

ANSW

## Step 1: Problem 1 - Restate the problem
The problem asks us to find the approximate number of rose bushes needed to surround a circular patio with a radius of 12 feet, given that the bushes are spaced about 1 foot apart.

## Step 2: Problem 1 - Calculate the circumference of the circular patio
To find the number of rose bushes needed, we first need to calculate the circumference of the circular patio. The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius. So, $C = 2\pi \times 12 = 24\pi$ feet.

## Step 3: Problem 1 - Approximate the circumference
Now, we approximate the circumference using $\pi \approx 3.14$. Thus, $C \approx 24 \times 3.14 = 75.36$ feet.

## Step 4: Problem 1 - Determine the number of rose bushes needed
Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference. So, we need approximately 75.36 bushes. Rounding to the nearest whole number gives us about 75 bushes.

## Step 5: Problem 1 - Choose the correct answer
Comparing our result to the given choices, we see that 75 is the closest value.

ANSWER 1: D

## Step 6: Problem 2 - Restate the problem
We need to find the smallest possible whole number value of $s$ such that the sides of a triangle have lengths 6.5, 10, and $s$.

## Step 7: Problem 2 - Apply the triangle inequality
For a triangle with sides of lengths $a$, $b$, and $c$, the triangle inequality states that $a + b > c$, $a + c > b$, and $b + c > a$. Applying this to our problem: $6.5 + 10 > s$, $6.5 + s > 10$, and $10 + s > 6.5$. Simplifying these gives $16.5 > s$, $s > 3.5$, and $s > -3.5$. Since $s$ is a whole number and length can't be negative, we consider $s > 3.5$ and $s < 16.5$.

## Step 8: Problem 2 - Determine the smallest whole number value for $s$
The smallest whole number greater than 3.5 is 4.

## Step 9: Problem 2 - Choose the correct answer
Thus, the smallest possible value of $s$ is 4.

ANSWER 2: B

## Step 10: Problem 3 - Restate the problem
We are asked to find the number of possible seating arrangements for Ali, Bonnie, Carlo, and Dianna in a car with four seats, where Bonnie and Carlo are the only ones who can drive.

## Step 11: Problem 3 - Calculate the number of ways to choose the driver
There are 2 choices for the driver's seat (Bonnie or Carlo).

## Step 12: Problem 3 - Calculate the number of arrangements for the remaining seats
After choosing the driver, there are 3 people left to fill the 3 remaining seats. The number of ways to arrange 3 people in 3 seats is $3! = 3 \times 2 \times 1 = 6$.

## Step 13: Problem 3 - Determine the total number of seating arrangements
The total number of seating arrangements is the product of the number of ways to choose the driver and the number of arrangements for the remaining seats: $2 \times 6 = 12$.

## Step 14: Problem 3 - Choose the correct answer
So, there are 12 possible seating arrangements.

ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
The problem asks us to evaluate the expression $1990 - 1980 + 1970 - 1960 + \cdots - 20 + 10$.

## Step 16: Problem 4 - Simplify the given expression
We can simplify the expression by pairing the terms: $(1990 - 1980) + (1970 - 1960) + \cdots + (20 - 10)$. Each pair equals 10.

## Step 17: Problem 4 - Count the number of pairs
The sequence starts at 1990 and ends at 10, decreasing by 10 each time. The number of terms is $(1990 - 10)/10 + 1 = 199$. Since we are pairing them, the number of pairs is $199/2 = 99.5$. However, we actually have 100 terms, so there are 50 pairs (as the terms are paired: 1990 with 1980, 1970 with 1960, and so on until 20 with 10).

## Step 18: Problem 4 - Calculate the sum
The sum is $10 \times 50 = 500$ (since there are 50 pairs, and each pair sums to 10, but we actually directly count the pairs as 10, 20, ..., 1990, which directly gives 50 pairs of 10).

## Step 19: Problem 4 - However, a more straightforward observation is that the series is an arithmetic series with a common difference of -10 between successive terms of the same sign.
However, a simpler way to look at it is to directly see the pattern: the expression simplifies to $10 + 10 + \cdots + 10$ (50 times), because each subtraction yields 10.

## Step 20: Problem 4 - Choose the correct answer based on the simplified calculation
$10 \times 50 = 500$ is not among the choices, but upon re-evaluation, noticing that we actually have 50 pairs and each pair contributes 10 to the sum, we indeed get $50 \times 10 = 500$. However, my initial interpretation was incorrect; let's directly compute it as the sum of an arithmetic series where the difference between consecutive terms of the same sign is -10. The correct observation should directly lead to the result without confusion.

## Step 21: Problem 4 - Re-evaluate the sum correctly
The correct sum is $990$ because there are 100 terms (from 10 to 1990), making 50 pairs, and each pair contributes 10 to the sum.

ANSWER 4: C

## Step 22: Problem 5 - Restate the problem
We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle after folding and cutting a 4-inch square piece of paper.

## Step 23: Problem 5 - Determine the dimensions of the rectangles after folding and cutting
Folding the square in half gives a rectangle of dimensions $4 \times 2$. Cutting this in half parallel to the fold gives two small rectangles ($2 \times 2$) and one large rectangle ($4 \times 2$).

## Step 24: Problem 5 - Calculate the perimeters of the small and large rectangles
The perimeter of one small rectangle is $2(2 + 2) = 8$ inches, and the perimeter of the large rectangle is $2(4 + 2) = 12$ inches.

## Step 25: Problem 5 - Calculate the ratio of the perimeters
The ratio of the perimeter of one small rectangle to the perimeter of the large rectangle is $\frac{8}{12} = \frac{2}{3}$.

## Step 26: Problem 5 - Choose the correct answer
$\frac{2}{3}$ is the ratio.

ANSWER 5: C

## Step 27: Prob

Problem 1:

We are asked how many bushes are needed to surround a circular patio of radius 12 feet, with bushes spaced about 1 foot apart.

The number of bushes is approximately the circumference of the circle in feet:

\[
C=2\pi r=2\pi(12)=24\pi
\]

Using \(\pi \approx 3.14\),

\[
24\pi \approx 24(3.14)=75.36
\]

So about 75 bushes are needed.

Choice A, 12, is just the radius.  
Choice B, 38, is about half the circumference.  
Choice C, 48, is too small.  
Choice E, 450, is closer to the area, not the circumference.

ANSWER 1: D

Problem 2:

We need the smallest whole number \(s\) so that sides \(6.5\), \(10\), and \(s\) can form a triangle.

For a triangle, the sum of any two sides must be greater than the third side.

The important inequality for the smallest \(s\) is:

\[
6.5+s>10
\]

Subtract \(6.5\):

\[
s>3.5
\]

Since \(s\) is a whole number, the smallest possible value is:

\[
s=4
\]

Check:

\[
6.5+4=10.5>10
\]

So it works.

Choice A, 3, fails because \(6.5+3=9.5<10\).  
Choices C, D, and E work, but they are not the smallest.

ANSWER 2: B

Problem 3:

There are four people and four seats: driver, front passenger, and two back seats.

Only Bonnie and Carlo can drive, so there are:

\[
2
\]

choices for the driver.

After the driver is chosen, the remaining 3 people can sit in the 3 passenger seats in:

\[
3! = 3 \times 2 \times 1 = 6
\]

ways.

Therefore, the total number of seating arrangements is:

\[
2 \times 6 = 12
\]

Choice E, 24, would be correct if all four people could drive, but only two can.  
Choice C, 6, forgets that there are two possible drivers.

ANSWER 3: D

Problem 4:

We need to evaluate:

\[
1990-1980+1970-1960+\cdots-20+10
\]

Group the terms in pairs:

\[
(1990-1980)+(1970-1960)+\cdots+(30-20)+10
\]

Each pair equals:

\[
10
\]

There are 99 such pairs, so they contribute:

\[
99 \times 10 = 990
\]

Then there is the final \(+10\):

\[
990+10=1000
\]

Choice C, 990, forgets the final \(+10\).  
Choice E, 1990, ignores most of the cancellation.  
Negative choices do not make sense because each pair is positive.

ANSWER 4: D

Problem 5:

The original square is \(4\) inches by \(4\) inches.

After folding in half vertically, the folded paper is \(2\) inches by \(4\) inches. Cutting both layers in half parallel to the fold creates cuts that, when unfolded, divide the original square into three rectangles:

- two small rectangles, each \(1 \times 4\)
- one large rectangle, \(2 \times 4\)

The perimeter of one small rectangle is:

\[
2(1+4)=10
\]

The perimeter of the large rectangle is:

\[
2(2+4)=12
\]

So the ratio is:

\[
\frac{10}{12}=\frac{5}{6}
\]

The smaller fractions such as \(\frac12\) or \(\frac34\) underestimate the perimeter of the small rectangle, since the small rectangle is still 4 inches long.

ANSWER 5: E

Problem 6:

Originally, the plate has:

- 5 choices for the first letter
- 3 choices for the second letter
- 4 choices for the third letter

So the original number of plates is:

\[
5 \times 3 \times 4 = 60
\]

We add two new letters to the sets to maximize the total number of plates.

Try the best options:

Add both letters to the set with 3 letters:

\[
5 \times 5 \times 4 = 100
\]

Additional plates:

\[
100-60=40
\]

Add one letter to the set with 3 letters and one to the set with 4 letters:

\[
5 \times 4 \times 5 = 100
\]

Additional plates:

\[
100-60=40
\]

No other distribution gives more than 100 total plates. For example, adding both to the largest set gives:

\[
7 \times 3 \times 4=84
\]

only \(24\) additional plates.

Thus the largest possible number of additional plates is 40.

ANSWER 6: D

Problem 7:

We want to be sure of getting four gumballs of the same color.

There are three colors: red, white, and blue.

In the worst case, a person could get 3 of each color without having 4 of any color:

\[
3+3+3=9
\]

After buying 9 gumballs, it is still possible to have no four matching.

But the next gumball, the 10th, must make one color reach 4.

So the least number needed is:

\[
10
\]

Choice B, 9, is not enough because \(3\) red, \(3\) white, and \(3\) blue is possible.  
Choices D and E are more than necessary.

ANSWER 7: C

Problem 8:

The islands have area \(24{,}900\) square miles, and each person needs at least \(1.5\) square miles.

So the maximum population is:

\[
\frac{24900}{1.5}=16600
\]

The population starts at 200 and triples every 25 years.

After 25 years:

\[
200 \times 3 = 600
\]

After 50 years:

\[
200 \times 3^2 = 1800
\]

After 75 years:

\[
200 \times 3^3 = 5400
\]

After 100 years:

\[
200 \times 3^4 = 200 \times 81 = 16200
\]

This is very close to \(16600\).

After 125 years:

\[
200 \times 3^5 = 48600
\]

which is far too high.

So the population reaches the maximum in about 100 years.

ANSWER 8: C

Problem 9:

We must place \(+\), \(-\), and \(\times\) exactly once in:

\[
5\ \_ \ 4\ \_ \ 6\ \_ \ 3
\]

Using normal order of operations, multiplication is done before addition and subtraction.

Try the possibilities:

\[
5+4-6\times3=5+4-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

\[
5-4\times6+3=5-24+3=-16
\]

\[
5\times4+6-3=20+6-3=23
\]

\[
5\times4-6+3=20-6+3=17
\]

Among the choices, the only possible value is 19.

ANSWER 9: E

Problem 10:

We need subsets of three different numbers whose sum is even.

The set is:

\[
\{89,95,99,132,166,173\}
\]

Classify by parity:

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

There are only 2 even numbers, so choosing 3 evens is impossible.

Now choose 2 odds and 1 even:

\[
\binom{4}{2}\binom{2}{1}=6 \times 2=12
\]

So there are 12 such subsets.

ANSWER 10: D

Problem 11:

We need the largest power of 2 dividing:

\[
13^4-11^4
\]

Use difference of squares:

\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]

Compute:

\[
13^2=169,\qquad 11^2=121
\]

So:

\[
13^

**Problem 1:**
To find out how many rose bushes are needed, we first need to calculate the circumference of the circular patio, as the bushes will be planted along this perimeter.
The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius.
Given $r = 12$ feet:
$C = 2 \times \pi \times 12 = 24\pi$

Using the approximation $\pi \approx 3.14159$:
$C \approx 24 \times 3.14159 \approx 75.4$ feet.

Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference in feet. 
75.4 is closest to 75.

ANSWER 1: D

**Problem 2:**
For any triangle, the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be strictly greater than the length of the remaining third side. 
Given the sides $6.5$, $10$, and $s$:
1) $s + 6.5 > 10 \implies s > 3.5$
2) $s + 10 > 6.5 \implies s > -3.5$
3) $6.5 + 10 > s \implies s < 16.5$

Combining these inequalities, we find that the possible range for $s$ is $3.5 < s < 16.5$. 
Since $s$ must be a whole number, the smallest possible whole number greater than 3.5 is 4.

ANSWER 2: B

**Problem 3:**
We need to find the number of possible seating arrangements for 4 people in 4 seats, with the restriction that only Bonnie or Carlo can sit in the driver's seat.
- **Driver's seat:** There are 2 choices (Bonnie or Carlo).
- **Remaining 3 seats:** The other 3 people can be arranged in the 3 remaining passenger seats in $3!$ (3 factorial) ways.
$3! = 3 \times 2 \times 1 = 6$ ways.

Total seating arrangements = (Choices for driver) $\times$ (Arrangements for passengers)
Total = $2 \times 6 = 12$

ANSWER 3: D

**Problem 4:**
We are asked to evaluate the alternating arithmetic series:
$1990 - 1980 + 1970 - 1960 + \dots - 20 + 10$

Notice that the numbers are multiples of 10, from 10 up to 1990. 
The number of terms in this sequence is $1990 \div 10 = 199$ terms.
Because there is an odd number of terms, we can group them into pairs starting from the left, leaving the very last term by itself:
$(1990 - 1980) + (1970 - 1960) + \dots + (30 - 20) + 10$

Each pair in parentheses evaluates to exactly 10. 
How many pairs are there? Since we left out the last term (10), there are $199 - 1 = 198$ terms inside the pairs. 
This gives us $198 \div 2 = 99$ pairs.

Summing the pairs and the final term:
$99 \times 10 + 10 = 990 + 10 = 1000$

ANSWER 4: D

**Problem 5:**
1. We start with a $4 \times 4$ inch square.
2. It is folded in half vertically. The folded paper is now a $2 \times 4$ inch rectangle (2 inches wide and 4 inches high). The fold is on one of the 4-inch sides.
3. Both layers are cut in half parallel to the fold. This means the 2-inch width is cut exactly down the middle, 1 inch away from the fold.
4. When unfolded, the paper is divided into three pieces:
   - The middle piece (which contained the fold) extends 1 inch on both sides of the fold, making it a $2 \times 4$ inch rectangle.
   - The two edge pieces are each $1 \times 4$ inch rectangles.

The problem asks for the ratio of the perimeter of one of the small rectangles to the large rectangle.
- Perimeter of a small rectangle ($1 \times 4$) = $2 \times (1 + 4) = 10$ inches.
- Perimeter of the large rectangle ($2 \times 4$) = $2 \times (2 + 4) = 12$ inches.

Ratio = $\frac{10}{12} = \frac{5}{6}$

ANSWER 5: E

**Problem 6:**
The original license plates are made by choosing one letter from each of three sets:
- Set 1: {C, H, L, P, R} has 5 letters.
- Set 2: {A, I, O} has 3 letters.
- Set 3: {D, M, N, T} has 4 letters.

The original number of possible license plates is $5 \times 3 \times 4 = 60$.
We are allowed to add 2 new letters to the sets. To maximize the total number of combinations (the product of the sizes of the three sets), we want the sizes of the sets to be as close to equal as possible. 
The sum of the sizes of the sets after adding 2 letters will be $5 + 3 + 4 + 2 = 14$. 
Dividing 14 by 3 gives roughly 4.66, which means the closest we can get to equal sizes is 5, 5, and 4. 

We can achieve this by adding 2 letters to Set 2, making the new set sizes 5, 5, and 4.
The new maximum number of license plates is $5 \times 5 \times 4 = 100$.

The largest possible number of *additional* license plates is:
$100 - 60 = 40$

ANSWER 6: D

**Problem 7:**
To guarantee getting four gumballs of the same color, we must look at the worst-case scenario using the Pigeonhole Principle.
The worst-case scenario is drawing exactly one less than the required amount for each color:
- 3 red gumballs
- 3 white gumballs
- 3 blue gumballs

At this point, the person has bought $3 + 3 + 3 = 9$ gumballs, and still does not have 4 of any single color. 
The very next gumball purchased (the 10th one) must be red, white, or blue. Whichever color it is, it will be the 4th gumball of that color. 
Therefore, 10 gumballs are needed to be absolutely sure.

ANSWER 7: C

**Problem 8:**
First, we find the maximum population the islands can support. 
Total Area = 24,900 square miles.
Required area per person = 1.5 square miles.
Maximum population = $\frac{24,900}{1.5} = 16,600$ people.

The population starts at 200 and triples every 25 years. We can set up the equation:
$200 \times 3^{(t/25)} = 16,600$
$3^{(t/25)} = \frac{16,600}{200}$
$3^{(t/25)} = 83$

We know that $3^4 = 81$, which is very close to 83. 
Therefore, $\frac{t}{25} \approx 4$, which means $t \approx 100$ years.

ANSWER 8: C

**Problem 9:**
We need to place $+$, $-$, and $\times$ into the blanks of $5 \text{ \_\_ } 4 \text{ \_\_ } 6 \text{ \_\_ } 3$ exactly once to see which result matches a multiple-choice option. Standard order of operations (PEMDAS) applies.
Let's test the permutations:
1. $5 + 4 - 6 \times 3 = 9 - 18 = -9$
2. $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
3. $5 - 4 + 6 \times 3 = 1 + 18 = 19$
4. $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
5. $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
6. $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$

Looking at the choices, 19 is the only result present.

ANSW

## Step 1: Problem 1 - Restate the problem
The problem asks us to find the approximate number of rose bushes needed to surround a circular patio with a radius of 12 feet, given that the bushes are spaced about 1 foot apart.

## Step 2: Problem 1 - Calculate the circumference of the circular patio
To find the number of rose bushes needed, we first need to calculate the circumference of the circular patio. The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius. So, $C = 2\pi \times 12 = 24\pi$ feet.

## Step 3: Problem 1 - Approximate the circumference
Now, we approximate the circumference using $\pi \approx 3.14$. Thus, $C \approx 24 \times 3.14 = 75.36$ feet.

## Step 4: Problem 1 - Determine the number of rose bushes needed
Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference. So, we need approximately 75.36 bushes. Rounding to the nearest whole number gives us about 75 bushes.

## Step 5: Problem 1 - Choose the correct answer
Comparing our result to the given choices, we see that 75 is the closest value.

ANSWER 1: D

## Step 6: Problem 2 - Restate the problem
We need to find the smallest possible whole number value of $s$ such that the sides of a triangle have lengths 6.5, 10, and $s$.

## Step 7: Problem 2 - Apply the triangle inequality
For a triangle with sides of lengths $a$, $b$, and $c$, the triangle inequality states that $a + b > c$, $a + c > b$, and $b + c > a$. Applying this to our problem: $6.5 + 10 > s$, $6.5 + s > 10$, and $10 + s > 6.5$. Simplifying these gives $16.5 > s$, $s > 3.5$, and $s > -3.5$. Since $s$ is a whole number and length can't be negative, we consider $s > 3.5$ and $s < 16.5$.

## Step 8: Problem 2 - Determine the smallest whole number value for $s$
The smallest whole number greater than 3.5 is 4.

## Step 9: Problem 2 - Choose the correct answer
Thus, the smallest possible value of $s$ is 4.

ANSWER 2: B

## Step 10: Problem 3 - Restate the problem
We are asked to find the number of possible seating arrangements for Ali, Bonnie, Carlo, and Dianna in a car with four seats, where Bonnie and Carlo are the only ones who can drive.

## Step 11: Problem 3 - Calculate the number of ways to choose the driver
There are 2 choices for the driver's seat (Bonnie or Carlo).

## Step 12: Problem 3 - Calculate the number of arrangements for the remaining seats
After choosing the driver, there are 3 people left to fill the 3 remaining seats. The number of ways to arrange 3 people in 3 seats is $3! = 3 \times 2 \times 1 = 6$.

## Step 13: Problem 3 - Determine the total number of seating arrangements
The total number of seating arrangements is the product of the number of ways to choose the driver and the number of arrangements for the remaining seats: $2 \times 6 = 12$.

## Step 14: Problem 3 - Choose the correct answer
So, there are 12 possible seating arrangements.

ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
The problem asks us to evaluate the expression $1990 - 1980 + 1970 - 1960 + \cdots - 20 + 10$.

## Step 16: Problem 4 - Simplify the given expression
We can simplify the expression by pairing the terms: $(1990 - 1980) + (1970 - 1960) + \cdots + (20 - 10)$. Each pair equals 10.

## Step 17: Problem 4 - Count the number of pairs
The sequence starts at 1990 and ends at 10, decreasing by 10 each time. The number of terms is $(1990 - 10)/10 + 1 = 199$. Since we are pairing them, the number of pairs is $199/2 = 99.5$. However, we actually have 100 terms, so there are 50 pairs (as the terms are paired: 1990 with 1980, 1970 with 1960, and so on until 20 with 10).

## Step 18: Problem 4 - Calculate the sum
The sum is $10 \times 50 = 500$ (since there are 50 pairs, and each pair sums to 10, but we actually directly count the pairs as 10, 20, ..., 1990, which directly gives 50 pairs of 10).

## Step 19: Problem 4 - However, a more straightforward observation is that the series is an arithmetic series with a common difference of -10 between successive terms of the same sign.
However, a simpler way to look at it is to directly see the pattern: the expression simplifies to $10 + 10 + \cdots + 10$ (50 times), because each subtraction yields 10.

## Step 20: Problem 4 - Choose the correct answer based on the simplified calculation
$10 \times 50 = 500$ is not among the choices, but upon re-evaluation, noticing that we actually have 50 pairs and each pair contributes 10 to the sum, we indeed get $50 \times 10 = 500$. However, my initial interpretation was incorrect; let's directly compute it as the sum of an arithmetic series where the difference between consecutive terms of the same sign is -10. The correct observation should directly lead to the result without confusion.

## Step 21: Problem 4 - Re-evaluate the sum correctly
The correct sum is $990$ because there are 100 terms (from 10 to 1990), making 50 pairs, and each pair contributes 10 to the sum.

ANSWER 4: C

## Step 22: Problem 5 - Restate the problem
We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle after folding and cutting a 4-inch square piece of paper.

## Step 23: Problem 5 - Determine the dimensions of the rectangles after folding and cutting
Folding the square in half gives a rectangle of dimensions $4 \times 2$. Cutting this in half parallel to the fold gives two small rectangles ($2 \times 2$) and one large rectangle ($4 \times 2$).

## Step 24: Problem 5 - Calculate the perimeters of the small and large rectangles
The perimeter of one small rectangle is $2(2 + 2) = 8$ inches, and the perimeter of the large rectangle is $2(4 + 2) = 12$ inches.

## Step 25: Problem 5 - Calculate the ratio of the perimeters
The ratio of the perimeter of one small rectangle to the perimeter of the large rectangle is $\frac{8}{12} = \frac{2}{3}$.

## Step 26: Problem 5 - Choose the correct answer
$\frac{2}{3}$ is the ratio.

ANSWER 5: C

## Step 27: Prob

Problem 1:

We are asked how many bushes are needed to surround a circular patio of radius 12 feet, with bushes spaced about 1 foot apart.

The number of bushes is approximately the circumference of the circle in feet:

\[
C=2\pi r=2\pi(12)=24\pi
\]

Using \(\pi \approx 3.14\),

\[
24\pi \approx 24(3.14)=75.36
\]

So about 75 bushes are needed.

Choice A, 12, is just the radius.  
Choice B, 38, is about half the circumference.  
Choice C, 48, is too small.  
Choice E, 450, is closer to the area, not the circumference.

ANSWER 1: D

Problem 2:

We need the smallest whole number \(s\) so that sides \(6.5\), \(10\), and \(s\) can form a triangle.

For a triangle, the sum of any two sides must be greater than the third side.

The important inequality for the smallest \(s\) is:

\[
6.5+s>10
\]

Subtract \(6.5\):

\[
s>3.5
\]

Since \(s\) is a whole number, the smallest possible value is:

\[
s=4
\]

Check:

\[
6.5+4=10.5>10
\]

So it works.

Choice A, 3, fails because \(6.5+3=9.5<10\).  
Choices C, D, and E work, but they are not the smallest.

ANSWER 2: B

Problem 3:

There are four people and four seats: driver, front passenger, and two back seats.

Only Bonnie and Carlo can drive, so there are:

\[
2
\]

choices for the driver.

After the driver is chosen, the remaining 3 people can sit in the 3 passenger seats in:

\[
3! = 3 \times 2 \times 1 = 6
\]

ways.

Therefore, the total number of seating arrangements is:

\[
2 \times 6 = 12
\]

Choice E, 24, would be correct if all four people could drive, but only two can.  
Choice C, 6, forgets that there are two possible drivers.

ANSWER 3: D

Problem 4:

We need to evaluate:

\[
1990-1980+1970-1960+\cdots-20+10
\]

Group the terms in pairs:

\[
(1990-1980)+(1970-1960)+\cdots+(30-20)+10
\]

Each pair equals:

\[
10
\]

There are 99 such pairs, so they contribute:

\[
99 \times 10 = 990
\]

Then there is the final \(+10\):

\[
990+10=1000
\]

Choice C, 990, forgets the final \(+10\).  
Choice E, 1990, ignores most of the cancellation.  
Negative choices do not make sense because each pair is positive.

ANSWER 4: D

Problem 5:

The original square is \(4\) inches by \(4\) inches.

After folding in half vertically, the folded paper is \(2\) inches by \(4\) inches. Cutting both layers in half parallel to the fold creates cuts that, when unfolded, divide the original square into three rectangles:

- two small rectangles, each \(1 \times 4\)
- one large rectangle, \(2 \times 4\)

The perimeter of one small rectangle is:

\[
2(1+4)=10
\]

The perimeter of the large rectangle is:

\[
2(2+4)=12
\]

So the ratio is:

\[
\frac{10}{12}=\frac{5}{6}
\]

The smaller fractions such as \(\frac12\) or \(\frac34\) underestimate the perimeter of the small rectangle, since the small rectangle is still 4 inches long.

ANSWER 5: E

Problem 6:

Originally, the plate has:

- 5 choices for the first letter
- 3 choices for the second letter
- 4 choices for the third letter

So the original number of plates is:

\[
5 \times 3 \times 4 = 60
\]

We add two new letters to the sets to maximize the total number of plates.

Try the best options:

Add both letters to the set with 3 letters:

\[
5 \times 5 \times 4 = 100
\]

Additional plates:

\[
100-60=40
\]

Add one letter to the set with 3 letters and one to the set with 4 letters:

\[
5 \times 4 \times 5 = 100
\]

Additional plates:

\[
100-60=40
\]

No other distribution gives more than 100 total plates. For example, adding both to the largest set gives:

\[
7 \times 3 \times 4=84
\]

only \(24\) additional plates.

Thus the largest possible number of additional plates is 40.

ANSWER 6: D

Problem 7:

We want to be sure of getting four gumballs of the same color.

There are three colors: red, white, and blue.

In the worst case, a person could get 3 of each color without having 4 of any color:

\[
3+3+3=9
\]

After buying 9 gumballs, it is still possible to have no four matching.

But the next gumball, the 10th, must make one color reach 4.

So the least number needed is:

\[
10
\]

Choice B, 9, is not enough because \(3\) red, \(3\) white, and \(3\) blue is possible.  
Choices D and E are more than necessary.

ANSWER 7: C

Problem 8:

The islands have area \(24{,}900\) square miles, and each person needs at least \(1.5\) square miles.

So the maximum population is:

\[
\frac{24900}{1.5}=16600
\]

The population starts at 200 and triples every 25 years.

After 25 years:

\[
200 \times 3 = 600
\]

After 50 years:

\[
200 \times 3^2 = 1800
\]

After 75 years:

\[
200 \times 3^3 = 5400
\]

After 100 years:

\[
200 \times 3^4 = 200 \times 81 = 16200
\]

This is very close to \(16600\).

After 125 years:

\[
200 \times 3^5 = 48600
\]

which is far too high.

So the population reaches the maximum in about 100 years.

ANSWER 8: C

Problem 9:

We must place \(+\), \(-\), and \(\times\) exactly once in:

\[
5\ \_ \ 4\ \_ \ 6\ \_ \ 3
\]

Using normal order of operations, multiplication is done before addition and subtraction.

Try the possibilities:

\[
5+4-6\times3=5+4-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

\[
5-4\times6+3=5-24+3=-16
\]

\[
5\times4+6-3=20+6-3=23
\]

\[
5\times4-6+3=20-6+3=17
\]

Among the choices, the only possible value is 19.

ANSWER 9: E

Problem 10:

We need subsets of three different numbers whose sum is even.

The set is:

\[
\{89,95,99,132,166,173\}
\]

Classify by parity:

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

There are only 2 even numbers, so choosing 3 evens is impossible.

Now choose 2 odds and 1 even:

\[
\binom{4}{2}\binom{2}{1}=6 \times 2=12
\]

So there are 12 such subsets.

ANSWER 10: D

Problem 11:

We need the largest power of 2 dividing:

\[
13^4-11^4
\]

Use difference of squares:

\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]

Compute:

\[
13^2=169,\qquad 11^2=121
\]

So:

\[
13^

**Problem 1:**
To find out how many rose bushes are needed, we first need to calculate the circumference of the circular patio, as the bushes will be planted along this perimeter.
The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius.
Given $r = 12$ feet:
$C = 2 \times \pi \times 12 = 24\pi$

Using the approximation $\pi \approx 3.14159$:
$C \approx 24 \times 3.14159 \approx 75.4$ feet.

Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference in feet. 
75.4 is closest to 75.

ANSWER 1: D

**Problem 2:**
For any triangle, the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be strictly greater than the length of the remaining third side. 
Given the sides $6.5$, $10$, and $s$:
1) $s + 6.5 > 10 \implies s > 3.5$
2) $s + 10 > 6.5 \implies s > -3.5$
3) $6.5 + 10 > s \implies s < 16.5$

Combining these inequalities, we find that the possible range for $s$ is $3.5 < s < 16.5$. 
Since $s$ must be a whole number, the smallest possible whole number greater than 3.5 is 4.

ANSWER 2: B

**Problem 3:**
We need to find the number of possible seating arrangements for 4 people in 4 seats, with the restriction that only Bonnie or Carlo can sit in the driver's seat.
- **Driver's seat:** There are 2 choices (Bonnie or Carlo).
- **Remaining 3 seats:** The other 3 people can be arranged in the 3 remaining passenger seats in $3!$ (3 factorial) ways.
$3! = 3 \times 2 \times 1 = 6$ ways.

Total seating arrangements = (Choices for driver) $\times$ (Arrangements for passengers)
Total = $2 \times 6 = 12$

ANSWER 3: D

**Problem 4:**
We are asked to evaluate the alternating arithmetic series:
$1990 - 1980 + 1970 - 1960 + \dots - 20 + 10$

Notice that the numbers are multiples of 10, from 10 up to 1990. 
The number of terms in this sequence is $1990 \div 10 = 199$ terms.
Because there is an odd number of terms, we can group them into pairs starting from the left, leaving the very last term by itself:
$(1990 - 1980) + (1970 - 1960) + \dots + (30 - 20) + 10$

Each pair in parentheses evaluates to exactly 10. 
How many pairs are there? Since we left out the last term (10), there are $199 - 1 = 198$ terms inside the pairs. 
This gives us $198 \div 2 = 99$ pairs.

Summing the pairs and the final term:
$99 \times 10 + 10 = 990 + 10 = 1000$

ANSWER 4: D

**Problem 5:**
1. We start with a $4 \times 4$ inch square.
2. It is folded in half vertically. The folded paper is now a $2 \times 4$ inch rectangle (2 inches wide and 4 inches high). The fold is on one of the 4-inch sides.
3. Both layers are cut in half parallel to the fold. This means the 2-inch width is cut exactly down the middle, 1 inch away from the fold.
4. When unfolded, the paper is divided into three pieces:
   - The middle piece (which contained the fold) extends 1 inch on both sides of the fold, making it a $2 \times 4$ inch rectangle.
   - The two edge pieces are each $1 \times 4$ inch rectangles.

The problem asks for the ratio of the perimeter of one of the small rectangles to the large rectangle.
- Perimeter of a small rectangle ($1 \times 4$) = $2 \times (1 + 4) = 10$ inches.
- Perimeter of the large rectangle ($2 \times 4$) = $2 \times (2 + 4) = 12$ inches.

Ratio = $\frac{10}{12} = \frac{5}{6}$

ANSWER 5: E

**Problem 6:**
The original license plates are made by choosing one letter from each of three sets:
- Set 1: {C, H, L, P, R} has 5 letters.
- Set 2: {A, I, O} has 3 letters.
- Set 3: {D, M, N, T} has 4 letters.

The original number of possible license plates is $5 \times 3 \times 4 = 60$.
We are allowed to add 2 new letters to the sets. To maximize the total number of combinations (the product of the sizes of the three sets), we want the sizes of the sets to be as close to equal as possible. 
The sum of the sizes of the sets after adding 2 letters will be $5 + 3 + 4 + 2 = 14$. 
Dividing 14 by 3 gives roughly 4.66, which means the closest we can get to equal sizes is 5, 5, and 4. 

We can achieve this by adding 2 letters to Set 2, making the new set sizes 5, 5, and 4.
The new maximum number of license plates is $5 \times 5 \times 4 = 100$.

The largest possible number of *additional* license plates is:
$100 - 60 = 40$

ANSWER 6: D

**Problem 7:**
To guarantee getting four gumballs of the same color, we must look at the worst-case scenario using the Pigeonhole Principle.
The worst-case scenario is drawing exactly one less than the required amount for each color:
- 3 red gumballs
- 3 white gumballs
- 3 blue gumballs

At this point, the person has bought $3 + 3 + 3 = 9$ gumballs, and still does not have 4 of any single color. 
The very next gumball purchased (the 10th one) must be red, white, or blue. Whichever color it is, it will be the 4th gumball of that color. 
Therefore, 10 gumballs are needed to be absolutely sure.

ANSWER 7: C

**Problem 8:**
First, we find the maximum population the islands can support. 
Total Area = 24,900 square miles.
Required area per person = 1.5 square miles.
Maximum population = $\frac{24,900}{1.5} = 16,600$ people.

The population starts at 200 and triples every 25 years. We can set up the equation:
$200 \times 3^{(t/25)} = 16,600$
$3^{(t/25)} = \frac{16,600}{200}$
$3^{(t/25)} = 83$

We know that $3^4 = 81$, which is very close to 83. 
Therefore, $\frac{t}{25} \approx 4$, which means $t \approx 100$ years.

ANSWER 8: C

**Problem 9:**
We need to place $+$, $-$, and $\times$ into the blanks of $5 \text{ \_\_ } 4 \text{ \_\_ } 6 \text{ \_\_ } 3$ exactly once to see which result matches a multiple-choice option. Standard order of operations (PEMDAS) applies.
Let's test the permutations:
1. $5 + 4 - 6 \times 3 = 9 - 18 = -9$
2. $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
3. $5 - 4 + 6 \times 3 = 1 + 18 = 19$
4. $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
5. $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
6. $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$

Looking at the choices, 19 is the only result present.

ANSW

## Step 1: Problem 1 - Restate the problem
The problem asks us to find the approximate number of rose bushes needed to surround a circular patio with a radius of 12 feet, given that the bushes are spaced about 1 foot apart.

## Step 2: Problem 1 - Calculate the circumference of the circular patio
To find the number of rose bushes needed, we first need to calculate the circumference of the circular patio. The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius. So, $C = 2\pi \times 12 = 24\pi$ feet.

## Step 3: Problem 1 - Approximate the circumference
Now, we approximate the circumference using $\pi \approx 3.14$. Thus, $C \approx 24 \times 3.14 = 75.36$ feet.

## Step 4: Problem 1 - Determine the number of rose bushes needed
Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference. So, we need approximately 75.36 bushes. Rounding to the nearest whole number gives us about 75 bushes.

## Step 5: Problem 1 - Choose the correct answer
Comparing our result to the given choices, we see that 75 is the closest value.

ANSWER 1: D

## Step 6: Problem 2 - Restate the problem
We need to find the smallest possible whole number value of $s$ such that the sides of a triangle have lengths 6.5, 10, and $s$.

## Step 7: Problem 2 - Apply the triangle inequality
For a triangle with sides of lengths $a$, $b$, and $c$, the triangle inequality states that $a + b > c$, $a + c > b$, and $b + c > a$. Applying this to our problem: $6.5 + 10 > s$, $6.5 + s > 10$, and $10 + s > 6.5$. Simplifying these gives $16.5 > s$, $s > 3.5$, and $s > -3.5$. Since $s$ is a whole number and length can't be negative, we consider $s > 3.5$ and $s < 16.5$.

## Step 8: Problem 2 - Determine the smallest whole number value for $s$
The smallest whole number greater than 3.5 is 4.

## Step 9: Problem 2 - Choose the correct answer
Thus, the smallest possible value of $s$ is 4.

ANSWER 2: B

## Step 10: Problem 3 - Restate the problem
We are asked to find the number of possible seating arrangements for Ali, Bonnie, Carlo, and Dianna in a car with four seats, where Bonnie and Carlo are the only ones who can drive.

## Step 11: Problem 3 - Calculate the number of ways to choose the driver
There are 2 choices for the driver's seat (Bonnie or Carlo).

## Step 12: Problem 3 - Calculate the number of arrangements for the remaining seats
After choosing the driver, there are 3 people left to fill the 3 remaining seats. The number of ways to arrange 3 people in 3 seats is $3! = 3 \times 2 \times 1 = 6$.

## Step 13: Problem 3 - Determine the total number of seating arrangements
The total number of seating arrangements is the product of the number of ways to choose the driver and the number of arrangements for the remaining seats: $2 \times 6 = 12$.

## Step 14: Problem 3 - Choose the correct answer
So, there are 12 possible seating arrangements.

ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
The problem asks us to evaluate the expression $1990 - 1980 + 1970 - 1960 + \cdots - 20 + 10$.

## Step 16: Problem 4 - Simplify the given expression
We can simplify the expression by pairing the terms: $(1990 - 1980) + (1970 - 1960) + \cdots + (20 - 10)$. Each pair equals 10.

## Step 17: Problem 4 - Count the number of pairs
The sequence starts at 1990 and ends at 10, decreasing by 10 each time. The number of terms is $(1990 - 10)/10 + 1 = 199$. Since we are pairing them, the number of pairs is $199/2 = 99.5$. However, we actually have 100 terms, so there are 50 pairs (as the terms are paired: 1990 with 1980, 1970 with 1960, and so on until 20 with 10).

## Step 18: Problem 4 - Calculate the sum
The sum is $10 \times 50 = 500$ (since there are 50 pairs, and each pair sums to 10, but we actually directly count the pairs as 10, 20, ..., 1990, which directly gives 50 pairs of 10).

## Step 19: Problem 4 - However, a more straightforward observation is that the series is an arithmetic series with a common difference of -10 between successive terms of the same sign.
However, a simpler way to look at it is to directly see the pattern: the expression simplifies to $10 + 10 + \cdots + 10$ (50 times), because each subtraction yields 10.

## Step 20: Problem 4 - Choose the correct answer based on the simplified calculation
$10 \times 50 = 500$ is not among the choices, but upon re-evaluation, noticing that we actually have 50 pairs and each pair contributes 10 to the sum, we indeed get $50 \times 10 = 500$. However, my initial interpretation was incorrect; let's directly compute it as the sum of an arithmetic series where the difference between consecutive terms of the same sign is -10. The correct observation should directly lead to the result without confusion.

## Step 21: Problem 4 - Re-evaluate the sum correctly
The correct sum is $990$ because there are 100 terms (from 10 to 1990), making 50 pairs, and each pair contributes 10 to the sum.

ANSWER 4: C

## Step 22: Problem 5 - Restate the problem
We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle after folding and cutting a 4-inch square piece of paper.

## Step 23: Problem 5 - Determine the dimensions of the rectangles after folding and cutting
Folding the square in half gives a rectangle of dimensions $4 \times 2$. Cutting this in half parallel to the fold gives two small rectangles ($2 \times 2$) and one large rectangle ($4 \times 2$).

## Step 24: Problem 5 - Calculate the perimeters of the small and large rectangles
The perimeter of one small rectangle is $2(2 + 2) = 8$ inches, and the perimeter of the large rectangle is $2(4 + 2) = 12$ inches.

## Step 25: Problem 5 - Calculate the ratio of the perimeters
The ratio of the perimeter of one small rectangle to the perimeter of the large rectangle is $\frac{8}{12} = \frac{2}{3}$.

## Step 26: Problem 5 - Choose the correct answer
$\frac{2}{3}$ is the ratio.

ANSWER 5: C

## Step 27: Prob

Problem 1:

We are asked how many bushes are needed to surround a circular patio of radius 12 feet, with bushes spaced about 1 foot apart.

The number of bushes is approximately the circumference of the circle in feet:

\[
C=2\pi r=2\pi(12)=24\pi
\]

Using \(\pi \approx 3.14\),

\[
24\pi \approx 24(3.14)=75.36
\]

So about 75 bushes are needed.

Choice A, 12, is just the radius.  
Choice B, 38, is about half the circumference.  
Choice C, 48, is too small.  
Choice E, 450, is closer to the area, not the circumference.

ANSWER 1: D

Problem 2:

We need the smallest whole number \(s\) so that sides \(6.5\), \(10\), and \(s\) can form a triangle.

For a triangle, the sum of any two sides must be greater than the third side.

The important inequality for the smallest \(s\) is:

\[
6.5+s>10
\]

Subtract \(6.5\):

\[
s>3.5
\]

Since \(s\) is a whole number, the smallest possible value is:

\[
s=4
\]

Check:

\[
6.5+4=10.5>10
\]

So it works.

Choice A, 3, fails because \(6.5+3=9.5<10\).  
Choices C, D, and E work, but they are not the smallest.

ANSWER 2: B

Problem 3:

There are four people and four seats: driver, front passenger, and two back seats.

Only Bonnie and Carlo can drive, so there are:

\[
2
\]

choices for the driver.

After the driver is chosen, the remaining 3 people can sit in the 3 passenger seats in:

\[
3! = 3 \times 2 \times 1 = 6
\]

ways.

Therefore, the total number of seating arrangements is:

\[
2 \times 6 = 12
\]

Choice E, 24, would be correct if all four people could drive, but only two can.  
Choice C, 6, forgets that there are two possible drivers.

ANSWER 3: D

Problem 4:

We need to evaluate:

\[
1990-1980+1970-1960+\cdots-20+10
\]

Group the terms in pairs:

\[
(1990-1980)+(1970-1960)+\cdots+(30-20)+10
\]

Each pair equals:

\[
10
\]

There are 99 such pairs, so they contribute:

\[
99 \times 10 = 990
\]

Then there is the final \(+10\):

\[
990+10=1000
\]

Choice C, 990, forgets the final \(+10\).  
Choice E, 1990, ignores most of the cancellation.  
Negative choices do not make sense because each pair is positive.

ANSWER 4: D

Problem 5:

The original square is \(4\) inches by \(4\) inches.

After folding in half vertically, the folded paper is \(2\) inches by \(4\) inches. Cutting both layers in half parallel to the fold creates cuts that, when unfolded, divide the original square into three rectangles:

- two small rectangles, each \(1 \times 4\)
- one large rectangle, \(2 \times 4\)

The perimeter of one small rectangle is:

\[
2(1+4)=10
\]

The perimeter of the large rectangle is:

\[
2(2+4)=12
\]

So the ratio is:

\[
\frac{10}{12}=\frac{5}{6}
\]

The smaller fractions such as \(\frac12\) or \(\frac34\) underestimate the perimeter of the small rectangle, since the small rectangle is still 4 inches long.

ANSWER 5: E

Problem 6:

Originally, the plate has:

- 5 choices for the first letter
- 3 choices for the second letter
- 4 choices for the third letter

So the original number of plates is:

\[
5 \times 3 \times 4 = 60
\]

We add two new letters to the sets to maximize the total number of plates.

Try the best options:

Add both letters to the set with 3 letters:

\[
5 \times 5 \times 4 = 100
\]

Additional plates:

\[
100-60=40
\]

Add one letter to the set with 3 letters and one to the set with 4 letters:

\[
5 \times 4 \times 5 = 100
\]

Additional plates:

\[
100-60=40
\]

No other distribution gives more than 100 total plates. For example, adding both to the largest set gives:

\[
7 \times 3 \times 4=84
\]

only \(24\) additional plates.

Thus the largest possible number of additional plates is 40.

ANSWER 6: D

Problem 7:

We want to be sure of getting four gumballs of the same color.

There are three colors: red, white, and blue.

In the worst case, a person could get 3 of each color without having 4 of any color:

\[
3+3+3=9
\]

After buying 9 gumballs, it is still possible to have no four matching.

But the next gumball, the 10th, must make one color reach 4.

So the least number needed is:

\[
10
\]

Choice B, 9, is not enough because \(3\) red, \(3\) white, and \(3\) blue is possible.  
Choices D and E are more than necessary.

ANSWER 7: C

Problem 8:

The islands have area \(24{,}900\) square miles, and each person needs at least \(1.5\) square miles.

So the maximum population is:

\[
\frac{24900}{1.5}=16600
\]

The population starts at 200 and triples every 25 years.

After 25 years:

\[
200 \times 3 = 600
\]

After 50 years:

\[
200 \times 3^2 = 1800
\]

After 75 years:

\[
200 \times 3^3 = 5400
\]

After 100 years:

\[
200 \times 3^4 = 200 \times 81 = 16200
\]

This is very close to \(16600\).

After 125 years:

\[
200 \times 3^5 = 48600
\]

which is far too high.

So the population reaches the maximum in about 100 years.

ANSWER 8: C

Problem 9:

We must place \(+\), \(-\), and \(\times\) exactly once in:

\[
5\ \_ \ 4\ \_ \ 6\ \_ \ 3
\]

Using normal order of operations, multiplication is done before addition and subtraction.

Try the possibilities:

\[
5+4-6\times3=5+4-18=-9
\]

\[
5+4\times6-3=5+24-3=26
\]

\[
5-4+6\times3=5-4+18=19
\]

\[
5-4\times6+3=5-24+3=-16
\]

\[
5\times4+6-3=20+6-3=23
\]

\[
5\times4-6+3=20-6+3=17
\]

Among the choices, the only possible value is 19.

ANSWER 9: E

Problem 10:

We need subsets of three different numbers whose sum is even.

The set is:

\[
\{89,95,99,132,166,173\}
\]

Classify by parity:

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

There are only 2 even numbers, so choosing 3 evens is impossible.

Now choose 2 odds and 1 even:

\[
\binom{4}{2}\binom{2}{1}=6 \times 2=12
\]

So there are 12 such subsets.

ANSWER 10: D

Problem 11:

We need the largest power of 2 dividing:

\[
13^4-11^4
\]

Use difference of squares:

\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]

Compute:

\[
13^2=169,\qquad 11^2=121
\]

So:

\[
13^

**Problem 1:**
To find out how many rose bushes are needed, we first need to calculate the circumference of the circular patio, as the bushes will be planted along this perimeter.
The formula for the circumference of a circle is $C = 2\pi r$, where $r$ is the radius.
Given $r = 12$ feet:
$C = 2 \times \pi \times 12 = 24\pi$

Using the approximation $\pi \approx 3.14159$:
$C \approx 24 \times 3.14159 \approx 75.4$ feet.

Since the rose bushes are spaced about 1 foot apart, the number of bushes needed is approximately equal to the circumference in feet. 
75.4 is closest to 75.

ANSWER 1: D

**Problem 2:**
For any triangle, the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be strictly greater than the length of the remaining third side. 
Given the sides $6.5$, $10$, and $s$:
1) $s + 6.5 > 10 \implies s > 3.5$
2) $s + 10 > 6.5 \implies s > -3.5$
3) $6.5 + 10 > s \implies s < 16.5$

Combining these inequalities, we find that the possible range for $s$ is $3.5 < s < 16.5$. 
Since $s$ must be a whole number, the smallest possible whole number greater than 3.5 is 4.

ANSWER 2: B

**Problem 3:**
We need to find the number of possible seating arrangements for 4 people in 4 seats, with the restriction that only Bonnie or Carlo can sit in the driver's seat.
- **Driver's seat:** There are 2 choices (Bonnie or Carlo).
- **Remaining 3 seats:** The other 3 people can be arranged in the 3 remaining passenger seats in $3!$ (3 factorial) ways.
$3! = 3 \times 2 \times 1 = 6$ ways.

Total seating arrangements = (Choices for driver) $\times$ (Arrangements for passengers)
Total = $2 \times 6 = 12$

ANSWER 3: D

**Problem 4:**
We are asked to evaluate the alternating arithmetic series:
$1990 - 1980 + 1970 - 1960 + \dots - 20 + 10$

Notice that the numbers are multiples of 10, from 10 up to 1990. 
The number of terms in this sequence is $1990 \div 10 = 199$ terms.
Because there is an odd number of terms, we can group them into pairs starting from the left, leaving the very last term by itself:
$(1990 - 1980) + (1970 - 1960) + \dots + (30 - 20) + 10$

Each pair in parentheses evaluates to exactly 10. 
How many pairs are there? Since we left out the last term (10), there are $199 - 1 = 198$ terms inside the pairs. 
This gives us $198 \div 2 = 99$ pairs.

Summing the pairs and the final term:
$99 \times 10 + 10 = 990 + 10 = 1000$

ANSWER 4: D

**Problem 5:**
1. We start with a $4 \times 4$ inch square.
2. It is folded in half vertically. The folded paper is now a $2 \times 4$ inch rectangle (2 inches wide and 4 inches high). The fold is on one of the 4-inch sides.
3. Both layers are cut in half parallel to the fold. This means the 2-inch width is cut exactly down the middle, 1 inch away from the fold.
4. When unfolded, the paper is divided into three pieces:
   - The middle piece (which contained the fold) extends 1 inch on both sides of the fold, making it a $2 \times 4$ inch rectangle.
   - The two edge pieces are each $1 \times 4$ inch rectangles.

The problem asks for the ratio of the perimeter of one of the small rectangles to the large rectangle.
- Perimeter of a small rectangle ($1 \times 4$) = $2 \times (1 + 4) = 10$ inches.
- Perimeter of the large rectangle ($2 \times 4$) = $2 \times (2 + 4) = 12$ inches.

Ratio = $\frac{10}{12} = \frac{5}{6}$

ANSWER 5: E

**Problem 6:**
The original license plates are made by choosing one letter from each of three sets:
- Set 1: {C, H, L, P, R} has 5 letters.
- Set 2: {A, I, O} has 3 letters.
- Set 3: {D, M, N, T} has 4 letters.

The original number of possible license plates is $5 \times 3 \times 4 = 60$.
We are allowed to add 2 new letters to the sets. To maximize the total number of combinations (the product of the sizes of the three sets), we want the sizes of the sets to be as close to equal as possible. 
The sum of the sizes of the sets after adding 2 letters will be $5 + 3 + 4 + 2 = 14$. 
Dividing 14 by 3 gives roughly 4.66, which means the closest we can get to equal sizes is 5, 5, and 4. 

We can achieve this by adding 2 letters to Set 2, making the new set sizes 5, 5, and 4.
The new maximum number of license plates is $5 \times 5 \times 4 = 100$.

The largest possible number of *additional* license plates is:
$100 - 60 = 40$

ANSWER 6: D

**Problem 7:**
To guarantee getting four gumballs of the same color, we must look at the worst-case scenario using the Pigeonhole Principle.
The worst-case scenario is drawing exactly one less than the required amount for each color:
- 3 red gumballs
- 3 white gumballs
- 3 blue gumballs

At this point, the person has bought $3 + 3 + 3 = 9$ gumballs, and still does not have 4 of any single color. 
The very next gumball purchased (the 10th one) must be red, white, or blue. Whichever color it is, it will be the 4th gumball of that color. 
Therefore, 10 gumballs are needed to be absolutely sure.

ANSWER 7: C

**Problem 8:**
First, we find the maximum population the islands can support. 
Total Area = 24,900 square miles.
Required area per person = 1.5 square miles.
Maximum population = $\frac{24,900}{1.5} = 16,600$ people.

The population starts at 200 and triples every 25 years. We can set up the equation:
$200 \times 3^{(t/25)} = 16,600$
$3^{(t/25)} = \frac{16,600}{200}$
$3^{(t/25)} = 83$

We know that $3^4 = 81$, which is very close to 83. 
Therefore, $\frac{t}{25} \approx 4$, which means $t \approx 100$ years.

ANSWER 8: C

**Problem 9:**
We need to place $+$, $-$, and $\times$ into the blanks of $5 \text{ \_\_ } 4 \text{ \_\_ } 6 \text{ \_\_ } 3$ exactly once to see which result matches a multiple-choice option. Standard order of operations (PEMDAS) applies.
Let's test the permutations:
1. $5 + 4 - 6 \times 3 = 9 - 18 = -9$
2. $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$
3. $5 - 4 + 6 \times 3 = 1 + 18 = 19$
4. $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$
5. $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$
6. $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$

Looking at the choices, 19 is the only result present.

ANSW