🔬 Model Lab

Problem 1:
What is the value of this expression? √ 16√ 81 √ 81√ 16
To solve this problem, we need to simplify the given expression step by step.

First, let's simplify the square roots: √16 = 4 and √81 = 9.

Now, substitute these values into the expression: 
√ 16√ 81 √ 81√ 16 = 4 × 9 9 × 4

Next, simplify the expression: 
4 × 9 9 × 4 = 36 36 = 1

Let's examine the answer choices to rule out the incorrect ones:
A. 4/9 = 0.44 (not equal to 1)
B. 2/3 = 0.67 (not equal to 1)
C. 1 = 1 (matches our result)
D. 3/2 = 1.5 (not equal to 1)
E. 9/4 = 2.25 (not equal to 1)

The only correct answer is C.

ANSWER 1: C

Problem 2:
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
Let's denote the capacity of the tank as x gallons.

The tank is initially one-third full, so it contains x/3 gallons of liquid.

After adding 4 gallons, the tank becomes one-half full, so it contains x/2 gallons of liquid.

We can set up an equation based on this information: 
x/3 + 4 = x/2

To solve for x, first multiply the entire equation by 6 to eliminate the fractions: 
2x + 24 = 3x

Now, subtract 2x from both sides: 
24 = x

Let's examine the answer choices to rule out the incorrect ones:
A. 8: 8/3 + 4 = 6.67, and 8/2 = 4 (not equal)
B. 12: 12/3 + 4 = 8, and 12/2 = 6 (not equal)
C. 20: 20/3 + 4 = 10.67, and 20/2 = 10 (not equal)
D. 24: 24/3 + 4 = 12, and 24/2 = 12 (matches our result)
E. 48: 48/3 + 4 = 20, and 48/2 = 24 (not equal)

The correct answer is D.

ANSWER 2: D

Problem 3:
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 1/3 of all the ninth graders are paired with 2/5 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Let's denote the number of sixth graders as S and the number of ninth graders as N.

We are given that 1/3 of N is paired with 2/5 of S. Since the number of pairs is the same, we can set up the following equation: 
(1/3)N = (2/5)S

To simplify, multiply both sides by 15 to eliminate the fractions: 
5N = 6S

The total number of students is S + N. We need to find the fraction of students who have a buddy, which is (2/5)S + (1/3)N.

Since (1/3)N = (2/5)S, we can simplify this to 2 × (2/5)S or 2 × (1/3)N.

Let's express N in terms of S: 
5N = 6S => N = (6/5)S

The total number of students is S + (6/5)S = (11/5)S.

The number of students with a buddy is 2 × (2/5)S = (4/5)S.

Now, we can find the fraction of students who have a buddy: 
((4/5)S) / ((11/5)S) = 4/11

Let's examine the answer choices to rule out the incorrect ones:
A. 2/15 = 0.133 (not equal to 4/11)
B. 4/11 = 0.364 (matches our result)
C. 11/30 = 0.367 (not equal to 4/11)
D. 3/8 = 0.375 (not equal to 4/11)
E. 11/15 = 0.733 (not equal to 4/11)

The correct answer is B.

ANSWER 3: B

Problem 4:
How many integers between 1000 and 9999 have four distinct digits?
To solve this problem, let's consider the constraints for each digit.

The first digit cannot be 0, so there are 9 possibilities (1-9).

The second digit can be any digit except the one chosen for the first digit, so there are 9 possibilities.

The third digit can be any digit except the two chosen for the first and second digits, so there are 8 possibilities.

The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 possibilities.

Now, we can calculate the total number of integers with four distinct digits: 
9 × 9 × 8 × 7 = 4536

Let's examine the answer choices to rule out the incorrect ones:
A. 3024 (not equal to 4536)
B. 4536 (matches our result)
C. 5040 (not equal to 4536)
D. 6480 (not equal to 4536)
E. 6561 (not equal to 4536)

The correct answer is B.

ANSWER 4: B

Problem 5:
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Compared with Blake, Jenny scored 10 points higher on the first test, 10 points lower on the second, and 20 points higher on each of the third and fourth.
Let's denote Blake's scores on the four tests as B1, B2, B3, and B4.

Blake's average score is 78, so (B1 + B2 + B3 + B4)/4 = 78.

Jenny's scores are B1 + 10, B2 - 10, B3 + 20, and B4 + 20.

Jenny's total score is (B1 + 10) + (B2 - 10) + (B3 + 20) + (B4 + 20) = B1 + B2 + B3 + B4 + 40.

Since B1 + B2 + B3 + B4 = 78 × 4 = 312, Jenny's total score is 312 + 40 = 352.

Jenny's average score is 352/4 = 88.

The difference between Jenny's average and Blake's average is 88 - 78 = 10.

Let's examine the answer choices to rule out the incorrect ones:
A. 10 (matches our result)
B. 15 (not equal to 10)
C. 20 (not equal to 10)
D. 25 (not equal to 10)
E. 40 (not equal to 10)

The correct answer is A.

ANSWER 5: A

Problem 6:
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
Let's denote the cost of one pencil as x cents.

Since the cost is more than a penny, x > 1.

Jamar paid 143 cents, so the number of pencils he bought is 143/x.

Sharona paid 187 cents, so the number of pencils she bought is 187/x.

The difference in the number of pencils bought is (187/x) - (143/x) = (187 - 143)/x = 44/x.

Since x is a factor of both 143 and 187, we need to find the common factors.

The prime factorization of 143 is 11 × 13, and the prime factorization of 187 is 11 × 17.

The common factor is 11, so x = 11.

Now, we can calculate the difference: 
(187/11) - (143/11) = 17 - 13 = 4

Let's examine the answer choices to rule out the incorrect ones:
A. 2 (not equal to 4)
B. 3 (not equal to 4)
C. 4 (matches our result)
D. 5 (not equal to 4)
E. 6 (not equal to 4)

The correct answer is C.

ANSWER 6: C

Problem 7:
An American traveling in Italy wishes to exchange American dollars for Italian lire. If 3000 lire = $1.60, how much lire will 

Problem 1:

We are evaluating the expression

\[
\frac{\sqrt{16\sqrt{81}}}{\sqrt{81\sqrt{16}}}.
\]

First simplify the square roots inside:

\[
\sqrt{81}=9,\qquad \sqrt{16}=4.
\]

So the numerator is

\[
\sqrt{16\sqrt{81}}=\sqrt{16\cdot 9}=\sqrt{144}=12.
\]

The denominator is

\[
\sqrt{81\sqrt{16}}=\sqrt{81\cdot 4}=\sqrt{324}=18.
\]

Thus the value is

\[
\frac{12}{18}=\frac{2}{3}.
\]

Choice C, \(1\), would come from incorrectly canceling the two expressions as if they were the same, but the numbers inside the square roots are different.

ANSWER 1: B

---

Problem 2:

Let the capacity of the tank be \(x\) gallons.

The tank starts one-third full:

\[
\frac{x}{3}
\]

After adding 4 gallons, it is one-half full:

\[
\frac{x}{2}
\]

So the added amount is the difference:

\[
\frac{x}{2}-\frac{x}{3}=4
\]

Find a common denominator:

\[
\frac{3x}{6}-\frac{2x}{6}=4
\]

\[
\frac{x}{6}=4
\]

\[
x=24
\]

So the tank holds 24 gallons.

ANSWER 2: D

---

Problem 3:

Let the number of paired ninth graders be \(k\). Since \(\frac13\) of all ninth graders are paired, the total number of ninth graders is

\[
3k.
\]

The same \(k\) ninth graders are paired with sixth graders, and this represents \(\frac25\) of all sixth graders. So

\[
\frac25(\text{sixth graders})=k.
\]

Thus the total number of sixth graders is

\[
\frac52k.
\]

The number of students who have a buddy is both members of each pair:

\[
2k.
\]

The total number of sixth and ninth graders is

\[
3k+\frac52k=\frac62k+\frac52k=\frac{11}{2}k.
\]

So the fraction who have a buddy is

\[
\frac{2k}{\frac{11}{2}k}=\frac{2k\cdot 2}{11k}=\frac{4}{11}.
\]

A tempting mistake is to add \(\frac13+\frac25=\frac{11}{15}\), but those fractions are of two different groups, not of the total group.

ANSWER 3: B

---

Problem 4:

We need four-digit integers from 1000 to 9999 with four distinct digits.

For the thousands digit, we can choose any digit from 1 to 9:

\[
9 \text{ choices}
\]

For the hundreds digit, we can choose any digit except the thousands digit. This includes 0, so there are

\[
9 \text{ choices}
\]

For the tens digit, we must avoid the first two digits:

\[
8 \text{ choices}
\]

For the ones digit, we must avoid the first three digits:

\[
7 \text{ choices}
\]

Therefore the total number is

\[
9\cdot 9\cdot 8\cdot 7=4536.
\]

Choice D, 6480, would come from allowing too many choices after the first digit.

ANSWER 4: B

---

Problem 5:

Blake took four tests. Jenny’s scores compared with Blake’s were:

- first test: \(+10\)
- second test: \(-10\)
- third test: \(+20\)
- fourth test: \(+20\)

So Jenny’s total score compared with Blake’s total score is

\[
10-10+20+20=40
\]

points higher.

Since there are 4 tests, Jenny’s average is higher by

\[
\frac{40}{4}=10.
\]

So Jenny’s average exceeds Blake’s by 10 points.

ANSWER 5: A

---

Problem 6:

Jamar paid \(\$1.43=143\) cents.

Sharona paid \(\$1.87=187\) cents.

The pencils cost more than 1 cent each, and both totals must be multiples of the pencil price.

So the pencil price must divide both 143 and 187.

Factor:

\[
143=11\cdot 13
\]

\[
187=11\cdot 17
\]

The common factor greater than 1 is 11, so each pencil costs 11 cents.

Jamar bought

\[
143\div 11=13
\]

pencils.

Sharona bought

\[
187\div 11=17
\]

pencils.

So Sharona bought

\[
17-13=4
\]

more pencils.

ANSWER 6: C

---

Problem 7:

We are told

\[
3000 \text{ lire}=\$1.60.
\]

We want the number of lire for \(\$1.00\).

Set up the rate:

\[
\frac{3000}{1.60}
\]

Since

\[
1.60=\frac{16}{10},
\]

we get

\[
3000\div 1.60=3000\div \frac{16}{10}=3000\cdot \frac{10}{16}.
\]

\[
=30000\div 16=1875.
\]

So the traveler receives 1875 lire for \(\$1.00\).

ANSWER 7: D

---

Problem 8:

We need four-digit numbers greater than 1000 using the digits of 2012.

The digits are

\[
2,\ 0,\ 1,\ 2.
\]

There are 4 digits, but the digit 2 is repeated. So the total number of distinct arrangements is

\[
\frac{4!}{2!}=\frac{24}{2}=12.
\]

But some arrangements start with 0, and those are not four-digit numbers.

If 0 is first, we arrange the remaining digits \(2,1,2\):

\[
\frac{3!}{2!}=3.
\]

So the number of valid four-digit numbers is

\[
12-3=9.
\]

All of these are greater than 1000, so the answer is 9.

ANSWER 8: D

---

Problem 9:

There are 108 students, and each eats an average of 2 cookies.

So the total number of cookies needed is

\[
108\cdot 2=216.
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough full pans to make at least 216 cookies:

\[
216\div 15=14.4.
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is

\[
15\cdot 2=30.
\]

Eggs are bought by the half-dozen, and one half-dozen is 6 eggs.

\[
30\div 6=5.
\]

So Walter should buy 5 half-dozens.

Choice E, 15, is the number of pans, not the number of half-dozens of eggs.

ANSWER 9: C

---

Problem 10:

We need to simplify

\[
100\times 19.98\times 1.998\times 1000.
\]

Group the factors conveniently:

\[
(100\times 19.98)(1.998\times 1000).
\]

Now compute each group:

\[
100\times 19.98=1998
\]

and

\[
1.998\times 1000=1998.
\]

So the expression becomes

\[
1998\times 1998=(1998)^2.
\]

ANSWER 10: D

---

Problem 11:

You have \(\$10\), but you plan to give the driver a \(\$2\) tip.

So the actual taxi fare can be

\[
10-2=8
\]

dollars.

The first \(\frac12\) mile costs \(\$2.40\).

Money remaining after the first half-mile:

\[
8.00-2.40=5.60.
\]

Additional distance costs \(\$0.20\) per \(0.1\) mile.

The number of additional \(0.1\)-mile segments is

\[
5.60\div 0.20=28.
\]

So the additional distance is

\[
28\cdot 0.1=2.8
\]

miles.

Total distance:

\[
0.5+2.8=3.3
\]

miles.

ANSWER 11: C

---

Problem 12:

The pizza has diameter 12 inches, so its radius is

\[
6
\]

inches.

Six pepperoni circles fit exactly across the diameter of the pizza. Therefore each pepperoni circle has diameter

\[
12\div 6=2
\]

inches, so each pepperoni has radius

\[
1
\]

inch.

Area of the pizza:

\[
\pi\cdot 6^2=36\p

**Problem 1:**
The garbled text `√ 16√ 81 √ 81√ 16` is a result of extracting the mathematical fraction $\frac{\sqrt{16}}{\sqrt{81}}$ (or $\sqrt{\frac{16}{81}}$) from a document. 

We are asked to evaluate this expression:
$\frac{\sqrt{16}}{\sqrt{81}}$

First, we find the square roots of the perfect squares in the numerator and the denominator:
$\sqrt{16} = 4$
$\sqrt{81} = 9$

Substituting these values back into the fraction, we get:
$\frac{4}{9}$

Looking at the choices, this matches option A.

ANSWER 1: A

**Problem 2:**
Let $C$ be the total capacity of the tank in gallons. 
We are told that the tank is initially one-third full, which is $\frac{1}{3}C$. 
When 4 gallons are added, the tank becomes one-half full, which is $\frac{1}{2}C$. 

We can set up the following equation:
$\frac{1}{3}C + 4 = \frac{1}{2}C$

To solve for $C$, we subtract $\frac{1}{3}C$ from both sides:
$4 = \frac{1}{2}C - \frac{1}{3}C$

Find a common denominator to subtract the fractions (which is 6):
$4 = \frac{3}{6}C - \frac{2}{6}C$
$4 = \frac{1}{6}C$

Multiply both sides by 6:
$C = 24$

The capacity of the tank is 24 gallons.

ANSWER 2: D

**Problem 3:**
Let $N$ be the total number of ninth graders and $S$ be the total number of sixth graders.
The number of ninth graders who have a buddy is $\frac{1}{3}N$.
The number of sixth graders who have a buddy is $\frac{2}{5}S$.

Since each paired ninth grader is matched with exactly one sixth grader, the number of paired ninth graders must equal the number of paired sixth graders:
$\frac{1}{3}N = \frac{2}{5}S$

We can express $N$ in terms of $S$ by multiplying both sides by 3:
$N = \frac{6}{5}S$

The total number of students in the program is:
$N + S = \frac{6}{5}S + S = \frac{11}{5}S$

The total number of students who have a buddy is the sum of the paired ninth graders and paired sixth graders:
$\frac{2}{5}S + \frac{2}{5}S = \frac{4}{5}S$

The fraction of the total number of students who have a buddy is:
$\frac{\text{Total paired}}{\text{Total students}} = \frac{\frac{4}{5}S}{\frac{11}{5}S} = \frac{4}{11}$

ANSWER 3: B

**Problem 4:**
We are looking for the number of 4-digit integers (from 1000 to 9999) where all four digits are distinct.
We can determine this by finding the number of choices for each digit place (thousands, hundreds, tens, units):
1. **Thousands digit:** It cannot be 0 (otherwise it wouldn't be a 4-digit number). So, it can be any digit from 1 to 9. That gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, but it must be different from the thousands digit. That gives us **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, but it must be different from the first two digits. That gives us **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, but it must be different from the first three digits. That gives us **7 choices**.

To find the total number of such integers, we multiply the number of choices together:
$9 \times 9 \times 8 \times 7 = 81 \times 56 = 4536$

ANSWER 4: B

**Problem 5:**
Let's find the total difference between Jenny's points and Blake's points across all four tests.
Compared to Blake, Jenny's score differences were:
Test 1: $+10$
Test 2: $-10$
Test 3: $+20$
Test 4: $+20$

Adding these differences together gives the total amount by which Jenny's overall score exceeded Blake's:
$10 - 10 + 20 + 20 = 40$ points

Since the average is the total score divided by the number of tests (4), Jenny's average exceeds Blake's average by:
$\frac{40}{4} = 10$ points

ANSWER 5: A

**Problem 6:**
Let $p$ be the price of a single pencil in cents. Since both Jamar and Sharona bought the same pencils, $p$ must be a common divisor of the amounts they paid (in cents).
Jamar paid $143$ cents, and Sharona paid $187$ cents.

Let's find the prime factorizations of 143 and 187:
$143 = 11 \times 13$
$187 = 11 \times 17$

The common divisors of 143 and 187 are 1 and 11. Since the problem states that the pencils cost *more than a penny each*, the price of one pencil must be $p = 11$ cents.

Now we can find out how many pencils each bought:
Jamar bought: $\frac{143}{11} = 13$ pencils
Sharona bought: $\frac{187}{11} = 17$ pencils

Sharona bought $17 - 13 = 4$ more pencils than Jamar.

ANSWER 6: C

**Problem 7:**
We are given the exchange rate:
$\$1.60 = 3000 \text{ lire}$

To find out how many lire the traveler will receive for $\$1.00$, we divide the amount of lire by $1.60$:
$\text{Lire for } \$1.00 = \frac{3000}{1.60}$

To make the division easier, we can multiply the numerator and the denominator by 10:
$\frac{30000}{16}$

Now, we simplify the fraction:
$\frac{30000}{16} = \frac{15000}{8} = \frac{7500}{4} = \frac{3750}{2} = 1875$

The traveler will receive 1875 lire for $\$1.00$.

ANSWER 7: D

**Problem 8:**
We need to form 4-digit numbers greater than 1000 using the digits 2, 0, 1, and 2.
For the number to be a valid 4-digit number (and greater than 1000), the first digit (thousands place) cannot be 0. Thus, the first digit must be either 1 or 2.

**Case 1: The first digit is 1.**
The remaining digits to arrange are 0, 2, and 2. 
The number of unique arrangements of these three digits is:
$\frac{3!}{2!} = 3$
These numbers are: 1022, 1202, and 1220.

**Case 2: The first digit is 2.**
The remaining digits to arrange are 0, 1, and 2.
The number of unique arrangements of these three distinct digits is:
$3! = 6$
These numbers are: 2012, 2021, 2102, 2120, 2201, and 2210.

Adding the possibilities from both cases gives a total of $3 + 6 = 9$ numbers.

ANSWER 8: D

**Problem 9:**
First, we calculate the total number of cookies needed for the students:
$108 \text{ students} \times 2 \text{ cookies/student} = 216 \text{ cookies}$

Next, we determine how many pans of cookies are required. Each pan makes 15 cookies, and only full recipes (full pans) are made:
$\frac{216}{15} = 14.4$
Since Walter can only make full pans, he must round up and make 15 pans.

Now, we calculate the total number of eggs needed. The recipe calls fo

Problem 1:
What is the value of this expression? √ 16√ 81 √ 81√ 16
To solve this problem, we need to simplify the given expression step by step.

First, let's simplify the square roots: √16 = 4 and √81 = 9.

Now, substitute these values into the expression: 
√ 16√ 81 √ 81√ 16 = 4 × 9 9 × 4

Next, simplify the expression: 
4 × 9 9 × 4 = 36 36 = 1

Let's examine the answer choices to rule out the incorrect ones:
A. 4/9 = 0.44 (not equal to 1)
B. 2/3 = 0.67 (not equal to 1)
C. 1 = 1 (matches our result)
D. 3/2 = 1.5 (not equal to 1)
E. 9/4 = 2.25 (not equal to 1)

The only correct answer is C.

ANSWER 1: C

Problem 2:
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
Let's denote the capacity of the tank as x gallons.

The tank is initially one-third full, so it contains x/3 gallons of liquid.

After adding 4 gallons, the tank becomes one-half full, so it contains x/2 gallons of liquid.

We can set up an equation based on this information: 
x/3 + 4 = x/2

To solve for x, first multiply the entire equation by 6 to eliminate the fractions: 
2x + 24 = 3x

Now, subtract 2x from both sides: 
24 = x

Let's examine the answer choices to rule out the incorrect ones:
A. 8: 8/3 + 4 = 6.67, and 8/2 = 4 (not equal)
B. 12: 12/3 + 4 = 8, and 12/2 = 6 (not equal)
C. 20: 20/3 + 4 = 10.67, and 20/2 = 10 (not equal)
D. 24: 24/3 + 4 = 12, and 24/2 = 12 (matches our result)
E. 48: 48/3 + 4 = 20, and 48/2 = 24 (not equal)

The correct answer is D.

ANSWER 2: D

Problem 3:
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 1/3 of all the ninth graders are paired with 2/5 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Let's denote the number of sixth graders as S and the number of ninth graders as N.

We are given that 1/3 of N is paired with 2/5 of S. Since the number of pairs is the same, we can set up the following equation: 
(1/3)N = (2/5)S

To simplify, multiply both sides by 15 to eliminate the fractions: 
5N = 6S

The total number of students is S + N. We need to find the fraction of students who have a buddy, which is (2/5)S + (1/3)N.

Since (1/3)N = (2/5)S, we can simplify this to 2 × (2/5)S or 2 × (1/3)N.

Let's express N in terms of S: 
5N = 6S => N = (6/5)S

The total number of students is S + (6/5)S = (11/5)S.

The number of students with a buddy is 2 × (2/5)S = (4/5)S.

Now, we can find the fraction of students who have a buddy: 
((4/5)S) / ((11/5)S) = 4/11

Let's examine the answer choices to rule out the incorrect ones:
A. 2/15 = 0.133 (not equal to 4/11)
B. 4/11 = 0.364 (matches our result)
C. 11/30 = 0.367 (not equal to 4/11)
D. 3/8 = 0.375 (not equal to 4/11)
E. 11/15 = 0.733 (not equal to 4/11)

The correct answer is B.

ANSWER 3: B

Problem 4:
How many integers between 1000 and 9999 have four distinct digits?
To solve this problem, let's consider the constraints for each digit.

The first digit cannot be 0, so there are 9 possibilities (1-9).

The second digit can be any digit except the one chosen for the first digit, so there are 9 possibilities.

The third digit can be any digit except the two chosen for the first and second digits, so there are 8 possibilities.

The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 possibilities.

Now, we can calculate the total number of integers with four distinct digits: 
9 × 9 × 8 × 7 = 4536

Let's examine the answer choices to rule out the incorrect ones:
A. 3024 (not equal to 4536)
B. 4536 (matches our result)
C. 5040 (not equal to 4536)
D. 6480 (not equal to 4536)
E. 6561 (not equal to 4536)

The correct answer is B.

ANSWER 4: B

Problem 5:
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Compared with Blake, Jenny scored 10 points higher on the first test, 10 points lower on the second, and 20 points higher on each of the third and fourth.
Let's denote Blake's scores on the four tests as B1, B2, B3, and B4.

Blake's average score is 78, so (B1 + B2 + B3 + B4)/4 = 78.

Jenny's scores are B1 + 10, B2 - 10, B3 + 20, and B4 + 20.

Jenny's total score is (B1 + 10) + (B2 - 10) + (B3 + 20) + (B4 + 20) = B1 + B2 + B3 + B4 + 40.

Since B1 + B2 + B3 + B4 = 78 × 4 = 312, Jenny's total score is 312 + 40 = 352.

Jenny's average score is 352/4 = 88.

The difference between Jenny's average and Blake's average is 88 - 78 = 10.

Let's examine the answer choices to rule out the incorrect ones:
A. 10 (matches our result)
B. 15 (not equal to 10)
C. 20 (not equal to 10)
D. 25 (not equal to 10)
E. 40 (not equal to 10)

The correct answer is A.

ANSWER 5: A

Problem 6:
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
Let's denote the cost of one pencil as x cents.

Since the cost is more than a penny, x > 1.

Jamar paid 143 cents, so the number of pencils he bought is 143/x.

Sharona paid 187 cents, so the number of pencils she bought is 187/x.

The difference in the number of pencils bought is (187/x) - (143/x) = (187 - 143)/x = 44/x.

Since x is a factor of both 143 and 187, we need to find the common factors.

The prime factorization of 143 is 11 × 13, and the prime factorization of 187 is 11 × 17.

The common factor is 11, so x = 11.

Now, we can calculate the difference: 
(187/11) - (143/11) = 17 - 13 = 4

Let's examine the answer choices to rule out the incorrect ones:
A. 2 (not equal to 4)
B. 3 (not equal to 4)
C. 4 (matches our result)
D. 5 (not equal to 4)
E. 6 (not equal to 4)

The correct answer is C.

ANSWER 6: C

Problem 7:
An American traveling in Italy wishes to exchange American dollars for Italian lire. If 3000 lire = $1.60, how much lire will 

Problem 1:

We are evaluating the expression

\[
\frac{\sqrt{16\sqrt{81}}}{\sqrt{81\sqrt{16}}}.
\]

First simplify the square roots inside:

\[
\sqrt{81}=9,\qquad \sqrt{16}=4.
\]

So the numerator is

\[
\sqrt{16\sqrt{81}}=\sqrt{16\cdot 9}=\sqrt{144}=12.
\]

The denominator is

\[
\sqrt{81\sqrt{16}}=\sqrt{81\cdot 4}=\sqrt{324}=18.
\]

Thus the value is

\[
\frac{12}{18}=\frac{2}{3}.
\]

Choice C, \(1\), would come from incorrectly canceling the two expressions as if they were the same, but the numbers inside the square roots are different.

ANSWER 1: B

---

Problem 2:

Let the capacity of the tank be \(x\) gallons.

The tank starts one-third full:

\[
\frac{x}{3}
\]

After adding 4 gallons, it is one-half full:

\[
\frac{x}{2}
\]

So the added amount is the difference:

\[
\frac{x}{2}-\frac{x}{3}=4
\]

Find a common denominator:

\[
\frac{3x}{6}-\frac{2x}{6}=4
\]

\[
\frac{x}{6}=4
\]

\[
x=24
\]

So the tank holds 24 gallons.

ANSWER 2: D

---

Problem 3:

Let the number of paired ninth graders be \(k\). Since \(\frac13\) of all ninth graders are paired, the total number of ninth graders is

\[
3k.
\]

The same \(k\) ninth graders are paired with sixth graders, and this represents \(\frac25\) of all sixth graders. So

\[
\frac25(\text{sixth graders})=k.
\]

Thus the total number of sixth graders is

\[
\frac52k.
\]

The number of students who have a buddy is both members of each pair:

\[
2k.
\]

The total number of sixth and ninth graders is

\[
3k+\frac52k=\frac62k+\frac52k=\frac{11}{2}k.
\]

So the fraction who have a buddy is

\[
\frac{2k}{\frac{11}{2}k}=\frac{2k\cdot 2}{11k}=\frac{4}{11}.
\]

A tempting mistake is to add \(\frac13+\frac25=\frac{11}{15}\), but those fractions are of two different groups, not of the total group.

ANSWER 3: B

---

Problem 4:

We need four-digit integers from 1000 to 9999 with four distinct digits.

For the thousands digit, we can choose any digit from 1 to 9:

\[
9 \text{ choices}
\]

For the hundreds digit, we can choose any digit except the thousands digit. This includes 0, so there are

\[
9 \text{ choices}
\]

For the tens digit, we must avoid the first two digits:

\[
8 \text{ choices}
\]

For the ones digit, we must avoid the first three digits:

\[
7 \text{ choices}
\]

Therefore the total number is

\[
9\cdot 9\cdot 8\cdot 7=4536.
\]

Choice D, 6480, would come from allowing too many choices after the first digit.

ANSWER 4: B

---

Problem 5:

Blake took four tests. Jenny’s scores compared with Blake’s were:

- first test: \(+10\)
- second test: \(-10\)
- third test: \(+20\)
- fourth test: \(+20\)

So Jenny’s total score compared with Blake’s total score is

\[
10-10+20+20=40
\]

points higher.

Since there are 4 tests, Jenny’s average is higher by

\[
\frac{40}{4}=10.
\]

So Jenny’s average exceeds Blake’s by 10 points.

ANSWER 5: A

---

Problem 6:

Jamar paid \(\$1.43=143\) cents.

Sharona paid \(\$1.87=187\) cents.

The pencils cost more than 1 cent each, and both totals must be multiples of the pencil price.

So the pencil price must divide both 143 and 187.

Factor:

\[
143=11\cdot 13
\]

\[
187=11\cdot 17
\]

The common factor greater than 1 is 11, so each pencil costs 11 cents.

Jamar bought

\[
143\div 11=13
\]

pencils.

Sharona bought

\[
187\div 11=17
\]

pencils.

So Sharona bought

\[
17-13=4
\]

more pencils.

ANSWER 6: C

---

Problem 7:

We are told

\[
3000 \text{ lire}=\$1.60.
\]

We want the number of lire for \(\$1.00\).

Set up the rate:

\[
\frac{3000}{1.60}
\]

Since

\[
1.60=\frac{16}{10},
\]

we get

\[
3000\div 1.60=3000\div \frac{16}{10}=3000\cdot \frac{10}{16}.
\]

\[
=30000\div 16=1875.
\]

So the traveler receives 1875 lire for \(\$1.00\).

ANSWER 7: D

---

Problem 8:

We need four-digit numbers greater than 1000 using the digits of 2012.

The digits are

\[
2,\ 0,\ 1,\ 2.
\]

There are 4 digits, but the digit 2 is repeated. So the total number of distinct arrangements is

\[
\frac{4!}{2!}=\frac{24}{2}=12.
\]

But some arrangements start with 0, and those are not four-digit numbers.

If 0 is first, we arrange the remaining digits \(2,1,2\):

\[
\frac{3!}{2!}=3.
\]

So the number of valid four-digit numbers is

\[
12-3=9.
\]

All of these are greater than 1000, so the answer is 9.

ANSWER 8: D

---

Problem 9:

There are 108 students, and each eats an average of 2 cookies.

So the total number of cookies needed is

\[
108\cdot 2=216.
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough full pans to make at least 216 cookies:

\[
216\div 15=14.4.
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is

\[
15\cdot 2=30.
\]

Eggs are bought by the half-dozen, and one half-dozen is 6 eggs.

\[
30\div 6=5.
\]

So Walter should buy 5 half-dozens.

Choice E, 15, is the number of pans, not the number of half-dozens of eggs.

ANSWER 9: C

---

Problem 10:

We need to simplify

\[
100\times 19.98\times 1.998\times 1000.
\]

Group the factors conveniently:

\[
(100\times 19.98)(1.998\times 1000).
\]

Now compute each group:

\[
100\times 19.98=1998
\]

and

\[
1.998\times 1000=1998.
\]

So the expression becomes

\[
1998\times 1998=(1998)^2.
\]

ANSWER 10: D

---

Problem 11:

You have \(\$10\), but you plan to give the driver a \(\$2\) tip.

So the actual taxi fare can be

\[
10-2=8
\]

dollars.

The first \(\frac12\) mile costs \(\$2.40\).

Money remaining after the first half-mile:

\[
8.00-2.40=5.60.
\]

Additional distance costs \(\$0.20\) per \(0.1\) mile.

The number of additional \(0.1\)-mile segments is

\[
5.60\div 0.20=28.
\]

So the additional distance is

\[
28\cdot 0.1=2.8
\]

miles.

Total distance:

\[
0.5+2.8=3.3
\]

miles.

ANSWER 11: C

---

Problem 12:

The pizza has diameter 12 inches, so its radius is

\[
6
\]

inches.

Six pepperoni circles fit exactly across the diameter of the pizza. Therefore each pepperoni circle has diameter

\[
12\div 6=2
\]

inches, so each pepperoni has radius

\[
1
\]

inch.

Area of the pizza:

\[
\pi\cdot 6^2=36\p

**Problem 1:**
The garbled text `√ 16√ 81 √ 81√ 16` is a result of extracting the mathematical fraction $\frac{\sqrt{16}}{\sqrt{81}}$ (or $\sqrt{\frac{16}{81}}$) from a document. 

We are asked to evaluate this expression:
$\frac{\sqrt{16}}{\sqrt{81}}$

First, we find the square roots of the perfect squares in the numerator and the denominator:
$\sqrt{16} = 4$
$\sqrt{81} = 9$

Substituting these values back into the fraction, we get:
$\frac{4}{9}$

Looking at the choices, this matches option A.

ANSWER 1: A

**Problem 2:**
Let $C$ be the total capacity of the tank in gallons. 
We are told that the tank is initially one-third full, which is $\frac{1}{3}C$. 
When 4 gallons are added, the tank becomes one-half full, which is $\frac{1}{2}C$. 

We can set up the following equation:
$\frac{1}{3}C + 4 = \frac{1}{2}C$

To solve for $C$, we subtract $\frac{1}{3}C$ from both sides:
$4 = \frac{1}{2}C - \frac{1}{3}C$

Find a common denominator to subtract the fractions (which is 6):
$4 = \frac{3}{6}C - \frac{2}{6}C$
$4 = \frac{1}{6}C$

Multiply both sides by 6:
$C = 24$

The capacity of the tank is 24 gallons.

ANSWER 2: D

**Problem 3:**
Let $N$ be the total number of ninth graders and $S$ be the total number of sixth graders.
The number of ninth graders who have a buddy is $\frac{1}{3}N$.
The number of sixth graders who have a buddy is $\frac{2}{5}S$.

Since each paired ninth grader is matched with exactly one sixth grader, the number of paired ninth graders must equal the number of paired sixth graders:
$\frac{1}{3}N = \frac{2}{5}S$

We can express $N$ in terms of $S$ by multiplying both sides by 3:
$N = \frac{6}{5}S$

The total number of students in the program is:
$N + S = \frac{6}{5}S + S = \frac{11}{5}S$

The total number of students who have a buddy is the sum of the paired ninth graders and paired sixth graders:
$\frac{2}{5}S + \frac{2}{5}S = \frac{4}{5}S$

The fraction of the total number of students who have a buddy is:
$\frac{\text{Total paired}}{\text{Total students}} = \frac{\frac{4}{5}S}{\frac{11}{5}S} = \frac{4}{11}$

ANSWER 3: B

**Problem 4:**
We are looking for the number of 4-digit integers (from 1000 to 9999) where all four digits are distinct.
We can determine this by finding the number of choices for each digit place (thousands, hundreds, tens, units):
1. **Thousands digit:** It cannot be 0 (otherwise it wouldn't be a 4-digit number). So, it can be any digit from 1 to 9. That gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, but it must be different from the thousands digit. That gives us **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, but it must be different from the first two digits. That gives us **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, but it must be different from the first three digits. That gives us **7 choices**.

To find the total number of such integers, we multiply the number of choices together:
$9 \times 9 \times 8 \times 7 = 81 \times 56 = 4536$

ANSWER 4: B

**Problem 5:**
Let's find the total difference between Jenny's points and Blake's points across all four tests.
Compared to Blake, Jenny's score differences were:
Test 1: $+10$
Test 2: $-10$
Test 3: $+20$
Test 4: $+20$

Adding these differences together gives the total amount by which Jenny's overall score exceeded Blake's:
$10 - 10 + 20 + 20 = 40$ points

Since the average is the total score divided by the number of tests (4), Jenny's average exceeds Blake's average by:
$\frac{40}{4} = 10$ points

ANSWER 5: A

**Problem 6:**
Let $p$ be the price of a single pencil in cents. Since both Jamar and Sharona bought the same pencils, $p$ must be a common divisor of the amounts they paid (in cents).
Jamar paid $143$ cents, and Sharona paid $187$ cents.

Let's find the prime factorizations of 143 and 187:
$143 = 11 \times 13$
$187 = 11 \times 17$

The common divisors of 143 and 187 are 1 and 11. Since the problem states that the pencils cost *more than a penny each*, the price of one pencil must be $p = 11$ cents.

Now we can find out how many pencils each bought:
Jamar bought: $\frac{143}{11} = 13$ pencils
Sharona bought: $\frac{187}{11} = 17$ pencils

Sharona bought $17 - 13 = 4$ more pencils than Jamar.

ANSWER 6: C

**Problem 7:**
We are given the exchange rate:
$\$1.60 = 3000 \text{ lire}$

To find out how many lire the traveler will receive for $\$1.00$, we divide the amount of lire by $1.60$:
$\text{Lire for } \$1.00 = \frac{3000}{1.60}$

To make the division easier, we can multiply the numerator and the denominator by 10:
$\frac{30000}{16}$

Now, we simplify the fraction:
$\frac{30000}{16} = \frac{15000}{8} = \frac{7500}{4} = \frac{3750}{2} = 1875$

The traveler will receive 1875 lire for $\$1.00$.

ANSWER 7: D

**Problem 8:**
We need to form 4-digit numbers greater than 1000 using the digits 2, 0, 1, and 2.
For the number to be a valid 4-digit number (and greater than 1000), the first digit (thousands place) cannot be 0. Thus, the first digit must be either 1 or 2.

**Case 1: The first digit is 1.**
The remaining digits to arrange are 0, 2, and 2. 
The number of unique arrangements of these three digits is:
$\frac{3!}{2!} = 3$
These numbers are: 1022, 1202, and 1220.

**Case 2: The first digit is 2.**
The remaining digits to arrange are 0, 1, and 2.
The number of unique arrangements of these three distinct digits is:
$3! = 6$
These numbers are: 2012, 2021, 2102, 2120, 2201, and 2210.

Adding the possibilities from both cases gives a total of $3 + 6 = 9$ numbers.

ANSWER 8: D

**Problem 9:**
First, we calculate the total number of cookies needed for the students:
$108 \text{ students} \times 2 \text{ cookies/student} = 216 \text{ cookies}$

Next, we determine how many pans of cookies are required. Each pan makes 15 cookies, and only full recipes (full pans) are made:
$\frac{216}{15} = 14.4$
Since Walter can only make full pans, he must round up and make 15 pans.

Now, we calculate the total number of eggs needed. The recipe calls fo

Problem 1:
What is the value of this expression? √ 16√ 81 √ 81√ 16
To solve this problem, we need to simplify the given expression step by step.

First, let's simplify the square roots: √16 = 4 and √81 = 9.

Now, substitute these values into the expression: 
√ 16√ 81 √ 81√ 16 = 4 × 9 9 × 4

Next, simplify the expression: 
4 × 9 9 × 4 = 36 36 = 1

Let's examine the answer choices to rule out the incorrect ones:
A. 4/9 = 0.44 (not equal to 1)
B. 2/3 = 0.67 (not equal to 1)
C. 1 = 1 (matches our result)
D. 3/2 = 1.5 (not equal to 1)
E. 9/4 = 2.25 (not equal to 1)

The only correct answer is C.

ANSWER 1: C

Problem 2:
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
Let's denote the capacity of the tank as x gallons.

The tank is initially one-third full, so it contains x/3 gallons of liquid.

After adding 4 gallons, the tank becomes one-half full, so it contains x/2 gallons of liquid.

We can set up an equation based on this information: 
x/3 + 4 = x/2

To solve for x, first multiply the entire equation by 6 to eliminate the fractions: 
2x + 24 = 3x

Now, subtract 2x from both sides: 
24 = x

Let's examine the answer choices to rule out the incorrect ones:
A. 8: 8/3 + 4 = 6.67, and 8/2 = 4 (not equal)
B. 12: 12/3 + 4 = 8, and 12/2 = 6 (not equal)
C. 20: 20/3 + 4 = 10.67, and 20/2 = 10 (not equal)
D. 24: 24/3 + 4 = 12, and 24/2 = 12 (matches our result)
E. 48: 48/3 + 4 = 20, and 48/2 = 24 (not equal)

The correct answer is D.

ANSWER 2: D

Problem 3:
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 1/3 of all the ninth graders are paired with 2/5 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Let's denote the number of sixth graders as S and the number of ninth graders as N.

We are given that 1/3 of N is paired with 2/5 of S. Since the number of pairs is the same, we can set up the following equation: 
(1/3)N = (2/5)S

To simplify, multiply both sides by 15 to eliminate the fractions: 
5N = 6S

The total number of students is S + N. We need to find the fraction of students who have a buddy, which is (2/5)S + (1/3)N.

Since (1/3)N = (2/5)S, we can simplify this to 2 × (2/5)S or 2 × (1/3)N.

Let's express N in terms of S: 
5N = 6S => N = (6/5)S

The total number of students is S + (6/5)S = (11/5)S.

The number of students with a buddy is 2 × (2/5)S = (4/5)S.

Now, we can find the fraction of students who have a buddy: 
((4/5)S) / ((11/5)S) = 4/11

Let's examine the answer choices to rule out the incorrect ones:
A. 2/15 = 0.133 (not equal to 4/11)
B. 4/11 = 0.364 (matches our result)
C. 11/30 = 0.367 (not equal to 4/11)
D. 3/8 = 0.375 (not equal to 4/11)
E. 11/15 = 0.733 (not equal to 4/11)

The correct answer is B.

ANSWER 3: B

Problem 4:
How many integers between 1000 and 9999 have four distinct digits?
To solve this problem, let's consider the constraints for each digit.

The first digit cannot be 0, so there are 9 possibilities (1-9).

The second digit can be any digit except the one chosen for the first digit, so there are 9 possibilities.

The third digit can be any digit except the two chosen for the first and second digits, so there are 8 possibilities.

The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 possibilities.

Now, we can calculate the total number of integers with four distinct digits: 
9 × 9 × 8 × 7 = 4536

Let's examine the answer choices to rule out the incorrect ones:
A. 3024 (not equal to 4536)
B. 4536 (matches our result)
C. 5040 (not equal to 4536)
D. 6480 (not equal to 4536)
E. 6561 (not equal to 4536)

The correct answer is B.

ANSWER 4: B

Problem 5:
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Compared with Blake, Jenny scored 10 points higher on the first test, 10 points lower on the second, and 20 points higher on each of the third and fourth.
Let's denote Blake's scores on the four tests as B1, B2, B3, and B4.

Blake's average score is 78, so (B1 + B2 + B3 + B4)/4 = 78.

Jenny's scores are B1 + 10, B2 - 10, B3 + 20, and B4 + 20.

Jenny's total score is (B1 + 10) + (B2 - 10) + (B3 + 20) + (B4 + 20) = B1 + B2 + B3 + B4 + 40.

Since B1 + B2 + B3 + B4 = 78 × 4 = 312, Jenny's total score is 312 + 40 = 352.

Jenny's average score is 352/4 = 88.

The difference between Jenny's average and Blake's average is 88 - 78 = 10.

Let's examine the answer choices to rule out the incorrect ones:
A. 10 (matches our result)
B. 15 (not equal to 10)
C. 20 (not equal to 10)
D. 25 (not equal to 10)
E. 40 (not equal to 10)

The correct answer is A.

ANSWER 5: A

Problem 6:
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
Let's denote the cost of one pencil as x cents.

Since the cost is more than a penny, x > 1.

Jamar paid 143 cents, so the number of pencils he bought is 143/x.

Sharona paid 187 cents, so the number of pencils she bought is 187/x.

The difference in the number of pencils bought is (187/x) - (143/x) = (187 - 143)/x = 44/x.

Since x is a factor of both 143 and 187, we need to find the common factors.

The prime factorization of 143 is 11 × 13, and the prime factorization of 187 is 11 × 17.

The common factor is 11, so x = 11.

Now, we can calculate the difference: 
(187/11) - (143/11) = 17 - 13 = 4

Let's examine the answer choices to rule out the incorrect ones:
A. 2 (not equal to 4)
B. 3 (not equal to 4)
C. 4 (matches our result)
D. 5 (not equal to 4)
E. 6 (not equal to 4)

The correct answer is C.

ANSWER 6: C

Problem 7:
An American traveling in Italy wishes to exchange American dollars for Italian lire. If 3000 lire = $1.60, how much lire will 

Problem 1:

We are evaluating the expression

\[
\frac{\sqrt{16\sqrt{81}}}{\sqrt{81\sqrt{16}}}.
\]

First simplify the square roots inside:

\[
\sqrt{81}=9,\qquad \sqrt{16}=4.
\]

So the numerator is

\[
\sqrt{16\sqrt{81}}=\sqrt{16\cdot 9}=\sqrt{144}=12.
\]

The denominator is

\[
\sqrt{81\sqrt{16}}=\sqrt{81\cdot 4}=\sqrt{324}=18.
\]

Thus the value is

\[
\frac{12}{18}=\frac{2}{3}.
\]

Choice C, \(1\), would come from incorrectly canceling the two expressions as if they were the same, but the numbers inside the square roots are different.

ANSWER 1: B

---

Problem 2:

Let the capacity of the tank be \(x\) gallons.

The tank starts one-third full:

\[
\frac{x}{3}
\]

After adding 4 gallons, it is one-half full:

\[
\frac{x}{2}
\]

So the added amount is the difference:

\[
\frac{x}{2}-\frac{x}{3}=4
\]

Find a common denominator:

\[
\frac{3x}{6}-\frac{2x}{6}=4
\]

\[
\frac{x}{6}=4
\]

\[
x=24
\]

So the tank holds 24 gallons.

ANSWER 2: D

---

Problem 3:

Let the number of paired ninth graders be \(k\). Since \(\frac13\) of all ninth graders are paired, the total number of ninth graders is

\[
3k.
\]

The same \(k\) ninth graders are paired with sixth graders, and this represents \(\frac25\) of all sixth graders. So

\[
\frac25(\text{sixth graders})=k.
\]

Thus the total number of sixth graders is

\[
\frac52k.
\]

The number of students who have a buddy is both members of each pair:

\[
2k.
\]

The total number of sixth and ninth graders is

\[
3k+\frac52k=\frac62k+\frac52k=\frac{11}{2}k.
\]

So the fraction who have a buddy is

\[
\frac{2k}{\frac{11}{2}k}=\frac{2k\cdot 2}{11k}=\frac{4}{11}.
\]

A tempting mistake is to add \(\frac13+\frac25=\frac{11}{15}\), but those fractions are of two different groups, not of the total group.

ANSWER 3: B

---

Problem 4:

We need four-digit integers from 1000 to 9999 with four distinct digits.

For the thousands digit, we can choose any digit from 1 to 9:

\[
9 \text{ choices}
\]

For the hundreds digit, we can choose any digit except the thousands digit. This includes 0, so there are

\[
9 \text{ choices}
\]

For the tens digit, we must avoid the first two digits:

\[
8 \text{ choices}
\]

For the ones digit, we must avoid the first three digits:

\[
7 \text{ choices}
\]

Therefore the total number is

\[
9\cdot 9\cdot 8\cdot 7=4536.
\]

Choice D, 6480, would come from allowing too many choices after the first digit.

ANSWER 4: B

---

Problem 5:

Blake took four tests. Jenny’s scores compared with Blake’s were:

- first test: \(+10\)
- second test: \(-10\)
- third test: \(+20\)
- fourth test: \(+20\)

So Jenny’s total score compared with Blake’s total score is

\[
10-10+20+20=40
\]

points higher.

Since there are 4 tests, Jenny’s average is higher by

\[
\frac{40}{4}=10.
\]

So Jenny’s average exceeds Blake’s by 10 points.

ANSWER 5: A

---

Problem 6:

Jamar paid \(\$1.43=143\) cents.

Sharona paid \(\$1.87=187\) cents.

The pencils cost more than 1 cent each, and both totals must be multiples of the pencil price.

So the pencil price must divide both 143 and 187.

Factor:

\[
143=11\cdot 13
\]

\[
187=11\cdot 17
\]

The common factor greater than 1 is 11, so each pencil costs 11 cents.

Jamar bought

\[
143\div 11=13
\]

pencils.

Sharona bought

\[
187\div 11=17
\]

pencils.

So Sharona bought

\[
17-13=4
\]

more pencils.

ANSWER 6: C

---

Problem 7:

We are told

\[
3000 \text{ lire}=\$1.60.
\]

We want the number of lire for \(\$1.00\).

Set up the rate:

\[
\frac{3000}{1.60}
\]

Since

\[
1.60=\frac{16}{10},
\]

we get

\[
3000\div 1.60=3000\div \frac{16}{10}=3000\cdot \frac{10}{16}.
\]

\[
=30000\div 16=1875.
\]

So the traveler receives 1875 lire for \(\$1.00\).

ANSWER 7: D

---

Problem 8:

We need four-digit numbers greater than 1000 using the digits of 2012.

The digits are

\[
2,\ 0,\ 1,\ 2.
\]

There are 4 digits, but the digit 2 is repeated. So the total number of distinct arrangements is

\[
\frac{4!}{2!}=\frac{24}{2}=12.
\]

But some arrangements start with 0, and those are not four-digit numbers.

If 0 is first, we arrange the remaining digits \(2,1,2\):

\[
\frac{3!}{2!}=3.
\]

So the number of valid four-digit numbers is

\[
12-3=9.
\]

All of these are greater than 1000, so the answer is 9.

ANSWER 8: D

---

Problem 9:

There are 108 students, and each eats an average of 2 cookies.

So the total number of cookies needed is

\[
108\cdot 2=216.
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough full pans to make at least 216 cookies:

\[
216\div 15=14.4.
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is

\[
15\cdot 2=30.
\]

Eggs are bought by the half-dozen, and one half-dozen is 6 eggs.

\[
30\div 6=5.
\]

So Walter should buy 5 half-dozens.

Choice E, 15, is the number of pans, not the number of half-dozens of eggs.

ANSWER 9: C

---

Problem 10:

We need to simplify

\[
100\times 19.98\times 1.998\times 1000.
\]

Group the factors conveniently:

\[
(100\times 19.98)(1.998\times 1000).
\]

Now compute each group:

\[
100\times 19.98=1998
\]

and

\[
1.998\times 1000=1998.
\]

So the expression becomes

\[
1998\times 1998=(1998)^2.
\]

ANSWER 10: D

---

Problem 11:

You have \(\$10\), but you plan to give the driver a \(\$2\) tip.

So the actual taxi fare can be

\[
10-2=8
\]

dollars.

The first \(\frac12\) mile costs \(\$2.40\).

Money remaining after the first half-mile:

\[
8.00-2.40=5.60.
\]

Additional distance costs \(\$0.20\) per \(0.1\) mile.

The number of additional \(0.1\)-mile segments is

\[
5.60\div 0.20=28.
\]

So the additional distance is

\[
28\cdot 0.1=2.8
\]

miles.

Total distance:

\[
0.5+2.8=3.3
\]

miles.

ANSWER 11: C

---

Problem 12:

The pizza has diameter 12 inches, so its radius is

\[
6
\]

inches.

Six pepperoni circles fit exactly across the diameter of the pizza. Therefore each pepperoni circle has diameter

\[
12\div 6=2
\]

inches, so each pepperoni has radius

\[
1
\]

inch.

Area of the pizza:

\[
\pi\cdot 6^2=36\p

**Problem 1:**
The garbled text `√ 16√ 81 √ 81√ 16` is a result of extracting the mathematical fraction $\frac{\sqrt{16}}{\sqrt{81}}$ (or $\sqrt{\frac{16}{81}}$) from a document. 

We are asked to evaluate this expression:
$\frac{\sqrt{16}}{\sqrt{81}}$

First, we find the square roots of the perfect squares in the numerator and the denominator:
$\sqrt{16} = 4$
$\sqrt{81} = 9$

Substituting these values back into the fraction, we get:
$\frac{4}{9}$

Looking at the choices, this matches option A.

ANSWER 1: A

**Problem 2:**
Let $C$ be the total capacity of the tank in gallons. 
We are told that the tank is initially one-third full, which is $\frac{1}{3}C$. 
When 4 gallons are added, the tank becomes one-half full, which is $\frac{1}{2}C$. 

We can set up the following equation:
$\frac{1}{3}C + 4 = \frac{1}{2}C$

To solve for $C$, we subtract $\frac{1}{3}C$ from both sides:
$4 = \frac{1}{2}C - \frac{1}{3}C$

Find a common denominator to subtract the fractions (which is 6):
$4 = \frac{3}{6}C - \frac{2}{6}C$
$4 = \frac{1}{6}C$

Multiply both sides by 6:
$C = 24$

The capacity of the tank is 24 gallons.

ANSWER 2: D

**Problem 3:**
Let $N$ be the total number of ninth graders and $S$ be the total number of sixth graders.
The number of ninth graders who have a buddy is $\frac{1}{3}N$.
The number of sixth graders who have a buddy is $\frac{2}{5}S$.

Since each paired ninth grader is matched with exactly one sixth grader, the number of paired ninth graders must equal the number of paired sixth graders:
$\frac{1}{3}N = \frac{2}{5}S$

We can express $N$ in terms of $S$ by multiplying both sides by 3:
$N = \frac{6}{5}S$

The total number of students in the program is:
$N + S = \frac{6}{5}S + S = \frac{11}{5}S$

The total number of students who have a buddy is the sum of the paired ninth graders and paired sixth graders:
$\frac{2}{5}S + \frac{2}{5}S = \frac{4}{5}S$

The fraction of the total number of students who have a buddy is:
$\frac{\text{Total paired}}{\text{Total students}} = \frac{\frac{4}{5}S}{\frac{11}{5}S} = \frac{4}{11}$

ANSWER 3: B

**Problem 4:**
We are looking for the number of 4-digit integers (from 1000 to 9999) where all four digits are distinct.
We can determine this by finding the number of choices for each digit place (thousands, hundreds, tens, units):
1. **Thousands digit:** It cannot be 0 (otherwise it wouldn't be a 4-digit number). So, it can be any digit from 1 to 9. That gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, but it must be different from the thousands digit. That gives us **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, but it must be different from the first two digits. That gives us **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, but it must be different from the first three digits. That gives us **7 choices**.

To find the total number of such integers, we multiply the number of choices together:
$9 \times 9 \times 8 \times 7 = 81 \times 56 = 4536$

ANSWER 4: B

**Problem 5:**
Let's find the total difference between Jenny's points and Blake's points across all four tests.
Compared to Blake, Jenny's score differences were:
Test 1: $+10$
Test 2: $-10$
Test 3: $+20$
Test 4: $+20$

Adding these differences together gives the total amount by which Jenny's overall score exceeded Blake's:
$10 - 10 + 20 + 20 = 40$ points

Since the average is the total score divided by the number of tests (4), Jenny's average exceeds Blake's average by:
$\frac{40}{4} = 10$ points

ANSWER 5: A

**Problem 6:**
Let $p$ be the price of a single pencil in cents. Since both Jamar and Sharona bought the same pencils, $p$ must be a common divisor of the amounts they paid (in cents).
Jamar paid $143$ cents, and Sharona paid $187$ cents.

Let's find the prime factorizations of 143 and 187:
$143 = 11 \times 13$
$187 = 11 \times 17$

The common divisors of 143 and 187 are 1 and 11. Since the problem states that the pencils cost *more than a penny each*, the price of one pencil must be $p = 11$ cents.

Now we can find out how many pencils each bought:
Jamar bought: $\frac{143}{11} = 13$ pencils
Sharona bought: $\frac{187}{11} = 17$ pencils

Sharona bought $17 - 13 = 4$ more pencils than Jamar.

ANSWER 6: C

**Problem 7:**
We are given the exchange rate:
$\$1.60 = 3000 \text{ lire}$

To find out how many lire the traveler will receive for $\$1.00$, we divide the amount of lire by $1.60$:
$\text{Lire for } \$1.00 = \frac{3000}{1.60}$

To make the division easier, we can multiply the numerator and the denominator by 10:
$\frac{30000}{16}$

Now, we simplify the fraction:
$\frac{30000}{16} = \frac{15000}{8} = \frac{7500}{4} = \frac{3750}{2} = 1875$

The traveler will receive 1875 lire for $\$1.00$.

ANSWER 7: D

**Problem 8:**
We need to form 4-digit numbers greater than 1000 using the digits 2, 0, 1, and 2.
For the number to be a valid 4-digit number (and greater than 1000), the first digit (thousands place) cannot be 0. Thus, the first digit must be either 1 or 2.

**Case 1: The first digit is 1.**
The remaining digits to arrange are 0, 2, and 2. 
The number of unique arrangements of these three digits is:
$\frac{3!}{2!} = 3$
These numbers are: 1022, 1202, and 1220.

**Case 2: The first digit is 2.**
The remaining digits to arrange are 0, 1, and 2.
The number of unique arrangements of these three distinct digits is:
$3! = 6$
These numbers are: 2012, 2021, 2102, 2120, 2201, and 2210.

Adding the possibilities from both cases gives a total of $3 + 6 = 9$ numbers.

ANSWER 8: D

**Problem 9:**
First, we calculate the total number of cookies needed for the students:
$108 \text{ students} \times 2 \text{ cookies/student} = 216 \text{ cookies}$

Next, we determine how many pans of cookies are required. Each pan makes 15 cookies, and only full recipes (full pans) are made:
$\frac{216}{15} = 14.4$
Since Walter can only make full pans, he must round up and make 15 pans.

Now, we calculate the total number of eggs needed. The recipe calls fo

Problem 1:
What is the value of this expression? √ 16√ 81 √ 81√ 16
To solve this problem, we need to simplify the given expression step by step.

First, let's simplify the square roots: √16 = 4 and √81 = 9.

Now, substitute these values into the expression: 
√ 16√ 81 √ 81√ 16 = 4 × 9 9 × 4

Next, simplify the expression: 
4 × 9 9 × 4 = 36 36 = 1

Let's examine the answer choices to rule out the incorrect ones:
A. 4/9 = 0.44 (not equal to 1)
B. 2/3 = 0.67 (not equal to 1)
C. 1 = 1 (matches our result)
D. 3/2 = 1.5 (not equal to 1)
E. 9/4 = 2.25 (not equal to 1)

The only correct answer is C.

ANSWER 1: C

Problem 2:
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
Let's denote the capacity of the tank as x gallons.

The tank is initially one-third full, so it contains x/3 gallons of liquid.

After adding 4 gallons, the tank becomes one-half full, so it contains x/2 gallons of liquid.

We can set up an equation based on this information: 
x/3 + 4 = x/2

To solve for x, first multiply the entire equation by 6 to eliminate the fractions: 
2x + 24 = 3x

Now, subtract 2x from both sides: 
24 = x

Let's examine the answer choices to rule out the incorrect ones:
A. 8: 8/3 + 4 = 6.67, and 8/2 = 4 (not equal)
B. 12: 12/3 + 4 = 8, and 12/2 = 6 (not equal)
C. 20: 20/3 + 4 = 10.67, and 20/2 = 10 (not equal)
D. 24: 24/3 + 4 = 12, and 24/2 = 12 (matches our result)
E. 48: 48/3 + 4 = 20, and 48/2 = 24 (not equal)

The correct answer is D.

ANSWER 2: D

Problem 3:
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 1/3 of all the ninth graders are paired with 2/5 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Let's denote the number of sixth graders as S and the number of ninth graders as N.

We are given that 1/3 of N is paired with 2/5 of S. Since the number of pairs is the same, we can set up the following equation: 
(1/3)N = (2/5)S

To simplify, multiply both sides by 15 to eliminate the fractions: 
5N = 6S

The total number of students is S + N. We need to find the fraction of students who have a buddy, which is (2/5)S + (1/3)N.

Since (1/3)N = (2/5)S, we can simplify this to 2 × (2/5)S or 2 × (1/3)N.

Let's express N in terms of S: 
5N = 6S => N = (6/5)S

The total number of students is S + (6/5)S = (11/5)S.

The number of students with a buddy is 2 × (2/5)S = (4/5)S.

Now, we can find the fraction of students who have a buddy: 
((4/5)S) / ((11/5)S) = 4/11

Let's examine the answer choices to rule out the incorrect ones:
A. 2/15 = 0.133 (not equal to 4/11)
B. 4/11 = 0.364 (matches our result)
C. 11/30 = 0.367 (not equal to 4/11)
D. 3/8 = 0.375 (not equal to 4/11)
E. 11/15 = 0.733 (not equal to 4/11)

The correct answer is B.

ANSWER 3: B

Problem 4:
How many integers between 1000 and 9999 have four distinct digits?
To solve this problem, let's consider the constraints for each digit.

The first digit cannot be 0, so there are 9 possibilities (1-9).

The second digit can be any digit except the one chosen for the first digit, so there are 9 possibilities.

The third digit can be any digit except the two chosen for the first and second digits, so there are 8 possibilities.

The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 possibilities.

Now, we can calculate the total number of integers with four distinct digits: 
9 × 9 × 8 × 7 = 4536

Let's examine the answer choices to rule out the incorrect ones:
A. 3024 (not equal to 4536)
B. 4536 (matches our result)
C. 5040 (not equal to 4536)
D. 6480 (not equal to 4536)
E. 6561 (not equal to 4536)

The correct answer is B.

ANSWER 4: B

Problem 5:
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Compared with Blake, Jenny scored 10 points higher on the first test, 10 points lower on the second, and 20 points higher on each of the third and fourth.
Let's denote Blake's scores on the four tests as B1, B2, B3, and B4.

Blake's average score is 78, so (B1 + B2 + B3 + B4)/4 = 78.

Jenny's scores are B1 + 10, B2 - 10, B3 + 20, and B4 + 20.

Jenny's total score is (B1 + 10) + (B2 - 10) + (B3 + 20) + (B4 + 20) = B1 + B2 + B3 + B4 + 40.

Since B1 + B2 + B3 + B4 = 78 × 4 = 312, Jenny's total score is 312 + 40 = 352.

Jenny's average score is 352/4 = 88.

The difference between Jenny's average and Blake's average is 88 - 78 = 10.

Let's examine the answer choices to rule out the incorrect ones:
A. 10 (matches our result)
B. 15 (not equal to 10)
C. 20 (not equal to 10)
D. 25 (not equal to 10)
E. 40 (not equal to 10)

The correct answer is A.

ANSWER 5: A

Problem 6:
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
Let's denote the cost of one pencil as x cents.

Since the cost is more than a penny, x > 1.

Jamar paid 143 cents, so the number of pencils he bought is 143/x.

Sharona paid 187 cents, so the number of pencils she bought is 187/x.

The difference in the number of pencils bought is (187/x) - (143/x) = (187 - 143)/x = 44/x.

Since x is a factor of both 143 and 187, we need to find the common factors.

The prime factorization of 143 is 11 × 13, and the prime factorization of 187 is 11 × 17.

The common factor is 11, so x = 11.

Now, we can calculate the difference: 
(187/11) - (143/11) = 17 - 13 = 4

Let's examine the answer choices to rule out the incorrect ones:
A. 2 (not equal to 4)
B. 3 (not equal to 4)
C. 4 (matches our result)
D. 5 (not equal to 4)
E. 6 (not equal to 4)

The correct answer is C.

ANSWER 6: C

Problem 7:
An American traveling in Italy wishes to exchange American dollars for Italian lire. If 3000 lire = $1.60, how much lire will 

Problem 1:

We are evaluating the expression

\[
\frac{\sqrt{16\sqrt{81}}}{\sqrt{81\sqrt{16}}}.
\]

First simplify the square roots inside:

\[
\sqrt{81}=9,\qquad \sqrt{16}=4.
\]

So the numerator is

\[
\sqrt{16\sqrt{81}}=\sqrt{16\cdot 9}=\sqrt{144}=12.
\]

The denominator is

\[
\sqrt{81\sqrt{16}}=\sqrt{81\cdot 4}=\sqrt{324}=18.
\]

Thus the value is

\[
\frac{12}{18}=\frac{2}{3}.
\]

Choice C, \(1\), would come from incorrectly canceling the two expressions as if they were the same, but the numbers inside the square roots are different.

ANSWER 1: B

---

Problem 2:

Let the capacity of the tank be \(x\) gallons.

The tank starts one-third full:

\[
\frac{x}{3}
\]

After adding 4 gallons, it is one-half full:

\[
\frac{x}{2}
\]

So the added amount is the difference:

\[
\frac{x}{2}-\frac{x}{3}=4
\]

Find a common denominator:

\[
\frac{3x}{6}-\frac{2x}{6}=4
\]

\[
\frac{x}{6}=4
\]

\[
x=24
\]

So the tank holds 24 gallons.

ANSWER 2: D

---

Problem 3:

Let the number of paired ninth graders be \(k\). Since \(\frac13\) of all ninth graders are paired, the total number of ninth graders is

\[
3k.
\]

The same \(k\) ninth graders are paired with sixth graders, and this represents \(\frac25\) of all sixth graders. So

\[
\frac25(\text{sixth graders})=k.
\]

Thus the total number of sixth graders is

\[
\frac52k.
\]

The number of students who have a buddy is both members of each pair:

\[
2k.
\]

The total number of sixth and ninth graders is

\[
3k+\frac52k=\frac62k+\frac52k=\frac{11}{2}k.
\]

So the fraction who have a buddy is

\[
\frac{2k}{\frac{11}{2}k}=\frac{2k\cdot 2}{11k}=\frac{4}{11}.
\]

A tempting mistake is to add \(\frac13+\frac25=\frac{11}{15}\), but those fractions are of two different groups, not of the total group.

ANSWER 3: B

---

Problem 4:

We need four-digit integers from 1000 to 9999 with four distinct digits.

For the thousands digit, we can choose any digit from 1 to 9:

\[
9 \text{ choices}
\]

For the hundreds digit, we can choose any digit except the thousands digit. This includes 0, so there are

\[
9 \text{ choices}
\]

For the tens digit, we must avoid the first two digits:

\[
8 \text{ choices}
\]

For the ones digit, we must avoid the first three digits:

\[
7 \text{ choices}
\]

Therefore the total number is

\[
9\cdot 9\cdot 8\cdot 7=4536.
\]

Choice D, 6480, would come from allowing too many choices after the first digit.

ANSWER 4: B

---

Problem 5:

Blake took four tests. Jenny’s scores compared with Blake’s were:

- first test: \(+10\)
- second test: \(-10\)
- third test: \(+20\)
- fourth test: \(+20\)

So Jenny’s total score compared with Blake’s total score is

\[
10-10+20+20=40
\]

points higher.

Since there are 4 tests, Jenny’s average is higher by

\[
\frac{40}{4}=10.
\]

So Jenny’s average exceeds Blake’s by 10 points.

ANSWER 5: A

---

Problem 6:

Jamar paid \(\$1.43=143\) cents.

Sharona paid \(\$1.87=187\) cents.

The pencils cost more than 1 cent each, and both totals must be multiples of the pencil price.

So the pencil price must divide both 143 and 187.

Factor:

\[
143=11\cdot 13
\]

\[
187=11\cdot 17
\]

The common factor greater than 1 is 11, so each pencil costs 11 cents.

Jamar bought

\[
143\div 11=13
\]

pencils.

Sharona bought

\[
187\div 11=17
\]

pencils.

So Sharona bought

\[
17-13=4
\]

more pencils.

ANSWER 6: C

---

Problem 7:

We are told

\[
3000 \text{ lire}=\$1.60.
\]

We want the number of lire for \(\$1.00\).

Set up the rate:

\[
\frac{3000}{1.60}
\]

Since

\[
1.60=\frac{16}{10},
\]

we get

\[
3000\div 1.60=3000\div \frac{16}{10}=3000\cdot \frac{10}{16}.
\]

\[
=30000\div 16=1875.
\]

So the traveler receives 1875 lire for \(\$1.00\).

ANSWER 7: D

---

Problem 8:

We need four-digit numbers greater than 1000 using the digits of 2012.

The digits are

\[
2,\ 0,\ 1,\ 2.
\]

There are 4 digits, but the digit 2 is repeated. So the total number of distinct arrangements is

\[
\frac{4!}{2!}=\frac{24}{2}=12.
\]

But some arrangements start with 0, and those are not four-digit numbers.

If 0 is first, we arrange the remaining digits \(2,1,2\):

\[
\frac{3!}{2!}=3.
\]

So the number of valid four-digit numbers is

\[
12-3=9.
\]

All of these are greater than 1000, so the answer is 9.

ANSWER 8: D

---

Problem 9:

There are 108 students, and each eats an average of 2 cookies.

So the total number of cookies needed is

\[
108\cdot 2=216.
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough full pans to make at least 216 cookies:

\[
216\div 15=14.4.
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is

\[
15\cdot 2=30.
\]

Eggs are bought by the half-dozen, and one half-dozen is 6 eggs.

\[
30\div 6=5.
\]

So Walter should buy 5 half-dozens.

Choice E, 15, is the number of pans, not the number of half-dozens of eggs.

ANSWER 9: C

---

Problem 10:

We need to simplify

\[
100\times 19.98\times 1.998\times 1000.
\]

Group the factors conveniently:

\[
(100\times 19.98)(1.998\times 1000).
\]

Now compute each group:

\[
100\times 19.98=1998
\]

and

\[
1.998\times 1000=1998.
\]

So the expression becomes

\[
1998\times 1998=(1998)^2.
\]

ANSWER 10: D

---

Problem 11:

You have \(\$10\), but you plan to give the driver a \(\$2\) tip.

So the actual taxi fare can be

\[
10-2=8
\]

dollars.

The first \(\frac12\) mile costs \(\$2.40\).

Money remaining after the first half-mile:

\[
8.00-2.40=5.60.
\]

Additional distance costs \(\$0.20\) per \(0.1\) mile.

The number of additional \(0.1\)-mile segments is

\[
5.60\div 0.20=28.
\]

So the additional distance is

\[
28\cdot 0.1=2.8
\]

miles.

Total distance:

\[
0.5+2.8=3.3
\]

miles.

ANSWER 11: C

---

Problem 12:

The pizza has diameter 12 inches, so its radius is

\[
6
\]

inches.

Six pepperoni circles fit exactly across the diameter of the pizza. Therefore each pepperoni circle has diameter

\[
12\div 6=2
\]

inches, so each pepperoni has radius

\[
1
\]

inch.

Area of the pizza:

\[
\pi\cdot 6^2=36\p

**Problem 1:**
The garbled text `√ 16√ 81 √ 81√ 16` is a result of extracting the mathematical fraction $\frac{\sqrt{16}}{\sqrt{81}}$ (or $\sqrt{\frac{16}{81}}$) from a document. 

We are asked to evaluate this expression:
$\frac{\sqrt{16}}{\sqrt{81}}$

First, we find the square roots of the perfect squares in the numerator and the denominator:
$\sqrt{16} = 4$
$\sqrt{81} = 9$

Substituting these values back into the fraction, we get:
$\frac{4}{9}$

Looking at the choices, this matches option A.

ANSWER 1: A

**Problem 2:**
Let $C$ be the total capacity of the tank in gallons. 
We are told that the tank is initially one-third full, which is $\frac{1}{3}C$. 
When 4 gallons are added, the tank becomes one-half full, which is $\frac{1}{2}C$. 

We can set up the following equation:
$\frac{1}{3}C + 4 = \frac{1}{2}C$

To solve for $C$, we subtract $\frac{1}{3}C$ from both sides:
$4 = \frac{1}{2}C - \frac{1}{3}C$

Find a common denominator to subtract the fractions (which is 6):
$4 = \frac{3}{6}C - \frac{2}{6}C$
$4 = \frac{1}{6}C$

Multiply both sides by 6:
$C = 24$

The capacity of the tank is 24 gallons.

ANSWER 2: D

**Problem 3:**
Let $N$ be the total number of ninth graders and $S$ be the total number of sixth graders.
The number of ninth graders who have a buddy is $\frac{1}{3}N$.
The number of sixth graders who have a buddy is $\frac{2}{5}S$.

Since each paired ninth grader is matched with exactly one sixth grader, the number of paired ninth graders must equal the number of paired sixth graders:
$\frac{1}{3}N = \frac{2}{5}S$

We can express $N$ in terms of $S$ by multiplying both sides by 3:
$N = \frac{6}{5}S$

The total number of students in the program is:
$N + S = \frac{6}{5}S + S = \frac{11}{5}S$

The total number of students who have a buddy is the sum of the paired ninth graders and paired sixth graders:
$\frac{2}{5}S + \frac{2}{5}S = \frac{4}{5}S$

The fraction of the total number of students who have a buddy is:
$\frac{\text{Total paired}}{\text{Total students}} = \frac{\frac{4}{5}S}{\frac{11}{5}S} = \frac{4}{11}$

ANSWER 3: B

**Problem 4:**
We are looking for the number of 4-digit integers (from 1000 to 9999) where all four digits are distinct.
We can determine this by finding the number of choices for each digit place (thousands, hundreds, tens, units):
1. **Thousands digit:** It cannot be 0 (otherwise it wouldn't be a 4-digit number). So, it can be any digit from 1 to 9. That gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, but it must be different from the thousands digit. That gives us **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, but it must be different from the first two digits. That gives us **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, but it must be different from the first three digits. That gives us **7 choices**.

To find the total number of such integers, we multiply the number of choices together:
$9 \times 9 \times 8 \times 7 = 81 \times 56 = 4536$

ANSWER 4: B

**Problem 5:**
Let's find the total difference between Jenny's points and Blake's points across all four tests.
Compared to Blake, Jenny's score differences were:
Test 1: $+10$
Test 2: $-10$
Test 3: $+20$
Test 4: $+20$

Adding these differences together gives the total amount by which Jenny's overall score exceeded Blake's:
$10 - 10 + 20 + 20 = 40$ points

Since the average is the total score divided by the number of tests (4), Jenny's average exceeds Blake's average by:
$\frac{40}{4} = 10$ points

ANSWER 5: A

**Problem 6:**
Let $p$ be the price of a single pencil in cents. Since both Jamar and Sharona bought the same pencils, $p$ must be a common divisor of the amounts they paid (in cents).
Jamar paid $143$ cents, and Sharona paid $187$ cents.

Let's find the prime factorizations of 143 and 187:
$143 = 11 \times 13$
$187 = 11 \times 17$

The common divisors of 143 and 187 are 1 and 11. Since the problem states that the pencils cost *more than a penny each*, the price of one pencil must be $p = 11$ cents.

Now we can find out how many pencils each bought:
Jamar bought: $\frac{143}{11} = 13$ pencils
Sharona bought: $\frac{187}{11} = 17$ pencils

Sharona bought $17 - 13 = 4$ more pencils than Jamar.

ANSWER 6: C

**Problem 7:**
We are given the exchange rate:
$\$1.60 = 3000 \text{ lire}$

To find out how many lire the traveler will receive for $\$1.00$, we divide the amount of lire by $1.60$:
$\text{Lire for } \$1.00 = \frac{3000}{1.60}$

To make the division easier, we can multiply the numerator and the denominator by 10:
$\frac{30000}{16}$

Now, we simplify the fraction:
$\frac{30000}{16} = \frac{15000}{8} = \frac{7500}{4} = \frac{3750}{2} = 1875$

The traveler will receive 1875 lire for $\$1.00$.

ANSWER 7: D

**Problem 8:**
We need to form 4-digit numbers greater than 1000 using the digits 2, 0, 1, and 2.
For the number to be a valid 4-digit number (and greater than 1000), the first digit (thousands place) cannot be 0. Thus, the first digit must be either 1 or 2.

**Case 1: The first digit is 1.**
The remaining digits to arrange are 0, 2, and 2. 
The number of unique arrangements of these three digits is:
$\frac{3!}{2!} = 3$
These numbers are: 1022, 1202, and 1220.

**Case 2: The first digit is 2.**
The remaining digits to arrange are 0, 1, and 2.
The number of unique arrangements of these three distinct digits is:
$3! = 6$
These numbers are: 2012, 2021, 2102, 2120, 2201, and 2210.

Adding the possibilities from both cases gives a total of $3 + 6 = 9$ numbers.

ANSWER 8: D

**Problem 9:**
First, we calculate the total number of cookies needed for the students:
$108 \text{ students} \times 2 \text{ cookies/student} = 216 \text{ cookies}$

Next, we determine how many pans of cookies are required. Each pan makes 15 cookies, and only full recipes (full pans) are made:
$\frac{216}{15} = 14.4$
Since Walter can only make full pans, he must round up and make 15 pans.

Now, we calculate the total number of eggs needed. The recipe calls fo

Problem 1:
What is the value of this expression? √ 16√ 81 √ 81√ 16
To solve this problem, we need to simplify the given expression step by step.

First, let's simplify the square roots: √16 = 4 and √81 = 9.

Now, substitute these values into the expression: 
√ 16√ 81 √ 81√ 16 = 4 × 9 9 × 4

Next, simplify the expression: 
4 × 9 9 × 4 = 36 36 = 1

Let's examine the answer choices to rule out the incorrect ones:
A. 4/9 = 0.44 (not equal to 1)
B. 2/3 = 0.67 (not equal to 1)
C. 1 = 1 (matches our result)
D. 3/2 = 1.5 (not equal to 1)
E. 9/4 = 2.25 (not equal to 1)

The only correct answer is C.

ANSWER 1: C

Problem 2:
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
Let's denote the capacity of the tank as x gallons.

The tank is initially one-third full, so it contains x/3 gallons of liquid.

After adding 4 gallons, the tank becomes one-half full, so it contains x/2 gallons of liquid.

We can set up an equation based on this information: 
x/3 + 4 = x/2

To solve for x, first multiply the entire equation by 6 to eliminate the fractions: 
2x + 24 = 3x

Now, subtract 2x from both sides: 
24 = x

Let's examine the answer choices to rule out the incorrect ones:
A. 8: 8/3 + 4 = 6.67, and 8/2 = 4 (not equal)
B. 12: 12/3 + 4 = 8, and 12/2 = 6 (not equal)
C. 20: 20/3 + 4 = 10.67, and 20/2 = 10 (not equal)
D. 24: 24/3 + 4 = 12, and 24/2 = 12 (matches our result)
E. 48: 48/3 + 4 = 20, and 48/2 = 24 (not equal)

The correct answer is D.

ANSWER 2: D

Problem 3:
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 1/3 of all the ninth graders are paired with 2/5 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Let's denote the number of sixth graders as S and the number of ninth graders as N.

We are given that 1/3 of N is paired with 2/5 of S. Since the number of pairs is the same, we can set up the following equation: 
(1/3)N = (2/5)S

To simplify, multiply both sides by 15 to eliminate the fractions: 
5N = 6S

The total number of students is S + N. We need to find the fraction of students who have a buddy, which is (2/5)S + (1/3)N.

Since (1/3)N = (2/5)S, we can simplify this to 2 × (2/5)S or 2 × (1/3)N.

Let's express N in terms of S: 
5N = 6S => N = (6/5)S

The total number of students is S + (6/5)S = (11/5)S.

The number of students with a buddy is 2 × (2/5)S = (4/5)S.

Now, we can find the fraction of students who have a buddy: 
((4/5)S) / ((11/5)S) = 4/11

Let's examine the answer choices to rule out the incorrect ones:
A. 2/15 = 0.133 (not equal to 4/11)
B. 4/11 = 0.364 (matches our result)
C. 11/30 = 0.367 (not equal to 4/11)
D. 3/8 = 0.375 (not equal to 4/11)
E. 11/15 = 0.733 (not equal to 4/11)

The correct answer is B.

ANSWER 3: B

Problem 4:
How many integers between 1000 and 9999 have four distinct digits?
To solve this problem, let's consider the constraints for each digit.

The first digit cannot be 0, so there are 9 possibilities (1-9).

The second digit can be any digit except the one chosen for the first digit, so there are 9 possibilities.

The third digit can be any digit except the two chosen for the first and second digits, so there are 8 possibilities.

The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 possibilities.

Now, we can calculate the total number of integers with four distinct digits: 
9 × 9 × 8 × 7 = 4536

Let's examine the answer choices to rule out the incorrect ones:
A. 3024 (not equal to 4536)
B. 4536 (matches our result)
C. 5040 (not equal to 4536)
D. 6480 (not equal to 4536)
E. 6561 (not equal to 4536)

The correct answer is B.

ANSWER 4: B

Problem 5:
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Compared with Blake, Jenny scored 10 points higher on the first test, 10 points lower on the second, and 20 points higher on each of the third and fourth.
Let's denote Blake's scores on the four tests as B1, B2, B3, and B4.

Blake's average score is 78, so (B1 + B2 + B3 + B4)/4 = 78.

Jenny's scores are B1 + 10, B2 - 10, B3 + 20, and B4 + 20.

Jenny's total score is (B1 + 10) + (B2 - 10) + (B3 + 20) + (B4 + 20) = B1 + B2 + B3 + B4 + 40.

Since B1 + B2 + B3 + B4 = 78 × 4 = 312, Jenny's total score is 312 + 40 = 352.

Jenny's average score is 352/4 = 88.

The difference between Jenny's average and Blake's average is 88 - 78 = 10.

Let's examine the answer choices to rule out the incorrect ones:
A. 10 (matches our result)
B. 15 (not equal to 10)
C. 20 (not equal to 10)
D. 25 (not equal to 10)
E. 40 (not equal to 10)

The correct answer is A.

ANSWER 5: A

Problem 6:
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
Let's denote the cost of one pencil as x cents.

Since the cost is more than a penny, x > 1.

Jamar paid 143 cents, so the number of pencils he bought is 143/x.

Sharona paid 187 cents, so the number of pencils she bought is 187/x.

The difference in the number of pencils bought is (187/x) - (143/x) = (187 - 143)/x = 44/x.

Since x is a factor of both 143 and 187, we need to find the common factors.

The prime factorization of 143 is 11 × 13, and the prime factorization of 187 is 11 × 17.

The common factor is 11, so x = 11.

Now, we can calculate the difference: 
(187/11) - (143/11) = 17 - 13 = 4

Let's examine the answer choices to rule out the incorrect ones:
A. 2 (not equal to 4)
B. 3 (not equal to 4)
C. 4 (matches our result)
D. 5 (not equal to 4)
E. 6 (not equal to 4)

The correct answer is C.

ANSWER 6: C

Problem 7:
An American traveling in Italy wishes to exchange American dollars for Italian lire. If 3000 lire = $1.60, how much lire will 

Problem 1:

We are evaluating the expression

\[
\frac{\sqrt{16\sqrt{81}}}{\sqrt{81\sqrt{16}}}.
\]

First simplify the square roots inside:

\[
\sqrt{81}=9,\qquad \sqrt{16}=4.
\]

So the numerator is

\[
\sqrt{16\sqrt{81}}=\sqrt{16\cdot 9}=\sqrt{144}=12.
\]

The denominator is

\[
\sqrt{81\sqrt{16}}=\sqrt{81\cdot 4}=\sqrt{324}=18.
\]

Thus the value is

\[
\frac{12}{18}=\frac{2}{3}.
\]

Choice C, \(1\), would come from incorrectly canceling the two expressions as if they were the same, but the numbers inside the square roots are different.

ANSWER 1: B

---

Problem 2:

Let the capacity of the tank be \(x\) gallons.

The tank starts one-third full:

\[
\frac{x}{3}
\]

After adding 4 gallons, it is one-half full:

\[
\frac{x}{2}
\]

So the added amount is the difference:

\[
\frac{x}{2}-\frac{x}{3}=4
\]

Find a common denominator:

\[
\frac{3x}{6}-\frac{2x}{6}=4
\]

\[
\frac{x}{6}=4
\]

\[
x=24
\]

So the tank holds 24 gallons.

ANSWER 2: D

---

Problem 3:

Let the number of paired ninth graders be \(k\). Since \(\frac13\) of all ninth graders are paired, the total number of ninth graders is

\[
3k.
\]

The same \(k\) ninth graders are paired with sixth graders, and this represents \(\frac25\) of all sixth graders. So

\[
\frac25(\text{sixth graders})=k.
\]

Thus the total number of sixth graders is

\[
\frac52k.
\]

The number of students who have a buddy is both members of each pair:

\[
2k.
\]

The total number of sixth and ninth graders is

\[
3k+\frac52k=\frac62k+\frac52k=\frac{11}{2}k.
\]

So the fraction who have a buddy is

\[
\frac{2k}{\frac{11}{2}k}=\frac{2k\cdot 2}{11k}=\frac{4}{11}.
\]

A tempting mistake is to add \(\frac13+\frac25=\frac{11}{15}\), but those fractions are of two different groups, not of the total group.

ANSWER 3: B

---

Problem 4:

We need four-digit integers from 1000 to 9999 with four distinct digits.

For the thousands digit, we can choose any digit from 1 to 9:

\[
9 \text{ choices}
\]

For the hundreds digit, we can choose any digit except the thousands digit. This includes 0, so there are

\[
9 \text{ choices}
\]

For the tens digit, we must avoid the first two digits:

\[
8 \text{ choices}
\]

For the ones digit, we must avoid the first three digits:

\[
7 \text{ choices}
\]

Therefore the total number is

\[
9\cdot 9\cdot 8\cdot 7=4536.
\]

Choice D, 6480, would come from allowing too many choices after the first digit.

ANSWER 4: B

---

Problem 5:

Blake took four tests. Jenny’s scores compared with Blake’s were:

- first test: \(+10\)
- second test: \(-10\)
- third test: \(+20\)
- fourth test: \(+20\)

So Jenny’s total score compared with Blake’s total score is

\[
10-10+20+20=40
\]

points higher.

Since there are 4 tests, Jenny’s average is higher by

\[
\frac{40}{4}=10.
\]

So Jenny’s average exceeds Blake’s by 10 points.

ANSWER 5: A

---

Problem 6:

Jamar paid \(\$1.43=143\) cents.

Sharona paid \(\$1.87=187\) cents.

The pencils cost more than 1 cent each, and both totals must be multiples of the pencil price.

So the pencil price must divide both 143 and 187.

Factor:

\[
143=11\cdot 13
\]

\[
187=11\cdot 17
\]

The common factor greater than 1 is 11, so each pencil costs 11 cents.

Jamar bought

\[
143\div 11=13
\]

pencils.

Sharona bought

\[
187\div 11=17
\]

pencils.

So Sharona bought

\[
17-13=4
\]

more pencils.

ANSWER 6: C

---

Problem 7:

We are told

\[
3000 \text{ lire}=\$1.60.
\]

We want the number of lire for \(\$1.00\).

Set up the rate:

\[
\frac{3000}{1.60}
\]

Since

\[
1.60=\frac{16}{10},
\]

we get

\[
3000\div 1.60=3000\div \frac{16}{10}=3000\cdot \frac{10}{16}.
\]

\[
=30000\div 16=1875.
\]

So the traveler receives 1875 lire for \(\$1.00\).

ANSWER 7: D

---

Problem 8:

We need four-digit numbers greater than 1000 using the digits of 2012.

The digits are

\[
2,\ 0,\ 1,\ 2.
\]

There are 4 digits, but the digit 2 is repeated. So the total number of distinct arrangements is

\[
\frac{4!}{2!}=\frac{24}{2}=12.
\]

But some arrangements start with 0, and those are not four-digit numbers.

If 0 is first, we arrange the remaining digits \(2,1,2\):

\[
\frac{3!}{2!}=3.
\]

So the number of valid four-digit numbers is

\[
12-3=9.
\]

All of these are greater than 1000, so the answer is 9.

ANSWER 8: D

---

Problem 9:

There are 108 students, and each eats an average of 2 cookies.

So the total number of cookies needed is

\[
108\cdot 2=216.
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough full pans to make at least 216 cookies:

\[
216\div 15=14.4.
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is

\[
15\cdot 2=30.
\]

Eggs are bought by the half-dozen, and one half-dozen is 6 eggs.

\[
30\div 6=5.
\]

So Walter should buy 5 half-dozens.

Choice E, 15, is the number of pans, not the number of half-dozens of eggs.

ANSWER 9: C

---

Problem 10:

We need to simplify

\[
100\times 19.98\times 1.998\times 1000.
\]

Group the factors conveniently:

\[
(100\times 19.98)(1.998\times 1000).
\]

Now compute each group:

\[
100\times 19.98=1998
\]

and

\[
1.998\times 1000=1998.
\]

So the expression becomes

\[
1998\times 1998=(1998)^2.
\]

ANSWER 10: D

---

Problem 11:

You have \(\$10\), but you plan to give the driver a \(\$2\) tip.

So the actual taxi fare can be

\[
10-2=8
\]

dollars.

The first \(\frac12\) mile costs \(\$2.40\).

Money remaining after the first half-mile:

\[
8.00-2.40=5.60.
\]

Additional distance costs \(\$0.20\) per \(0.1\) mile.

The number of additional \(0.1\)-mile segments is

\[
5.60\div 0.20=28.
\]

So the additional distance is

\[
28\cdot 0.1=2.8
\]

miles.

Total distance:

\[
0.5+2.8=3.3
\]

miles.

ANSWER 11: C

---

Problem 12:

The pizza has diameter 12 inches, so its radius is

\[
6
\]

inches.

Six pepperoni circles fit exactly across the diameter of the pizza. Therefore each pepperoni circle has diameter

\[
12\div 6=2
\]

inches, so each pepperoni has radius

\[
1
\]

inch.

Area of the pizza:

\[
\pi\cdot 6^2=36\p

**Problem 1:**
The garbled text `√ 16√ 81 √ 81√ 16` is a result of extracting the mathematical fraction $\frac{\sqrt{16}}{\sqrt{81}}$ (or $\sqrt{\frac{16}{81}}$) from a document. 

We are asked to evaluate this expression:
$\frac{\sqrt{16}}{\sqrt{81}}$

First, we find the square roots of the perfect squares in the numerator and the denominator:
$\sqrt{16} = 4$
$\sqrt{81} = 9$

Substituting these values back into the fraction, we get:
$\frac{4}{9}$

Looking at the choices, this matches option A.

ANSWER 1: A

**Problem 2:**
Let $C$ be the total capacity of the tank in gallons. 
We are told that the tank is initially one-third full, which is $\frac{1}{3}C$. 
When 4 gallons are added, the tank becomes one-half full, which is $\frac{1}{2}C$. 

We can set up the following equation:
$\frac{1}{3}C + 4 = \frac{1}{2}C$

To solve for $C$, we subtract $\frac{1}{3}C$ from both sides:
$4 = \frac{1}{2}C - \frac{1}{3}C$

Find a common denominator to subtract the fractions (which is 6):
$4 = \frac{3}{6}C - \frac{2}{6}C$
$4 = \frac{1}{6}C$

Multiply both sides by 6:
$C = 24$

The capacity of the tank is 24 gallons.

ANSWER 2: D

**Problem 3:**
Let $N$ be the total number of ninth graders and $S$ be the total number of sixth graders.
The number of ninth graders who have a buddy is $\frac{1}{3}N$.
The number of sixth graders who have a buddy is $\frac{2}{5}S$.

Since each paired ninth grader is matched with exactly one sixth grader, the number of paired ninth graders must equal the number of paired sixth graders:
$\frac{1}{3}N = \frac{2}{5}S$

We can express $N$ in terms of $S$ by multiplying both sides by 3:
$N = \frac{6}{5}S$

The total number of students in the program is:
$N + S = \frac{6}{5}S + S = \frac{11}{5}S$

The total number of students who have a buddy is the sum of the paired ninth graders and paired sixth graders:
$\frac{2}{5}S + \frac{2}{5}S = \frac{4}{5}S$

The fraction of the total number of students who have a buddy is:
$\frac{\text{Total paired}}{\text{Total students}} = \frac{\frac{4}{5}S}{\frac{11}{5}S} = \frac{4}{11}$

ANSWER 3: B

**Problem 4:**
We are looking for the number of 4-digit integers (from 1000 to 9999) where all four digits are distinct.
We can determine this by finding the number of choices for each digit place (thousands, hundreds, tens, units):
1. **Thousands digit:** It cannot be 0 (otherwise it wouldn't be a 4-digit number). So, it can be any digit from 1 to 9. That gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, but it must be different from the thousands digit. That gives us **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, but it must be different from the first two digits. That gives us **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, but it must be different from the first three digits. That gives us **7 choices**.

To find the total number of such integers, we multiply the number of choices together:
$9 \times 9 \times 8 \times 7 = 81 \times 56 = 4536$

ANSWER 4: B

**Problem 5:**
Let's find the total difference between Jenny's points and Blake's points across all four tests.
Compared to Blake, Jenny's score differences were:
Test 1: $+10$
Test 2: $-10$
Test 3: $+20$
Test 4: $+20$

Adding these differences together gives the total amount by which Jenny's overall score exceeded Blake's:
$10 - 10 + 20 + 20 = 40$ points

Since the average is the total score divided by the number of tests (4), Jenny's average exceeds Blake's average by:
$\frac{40}{4} = 10$ points

ANSWER 5: A

**Problem 6:**
Let $p$ be the price of a single pencil in cents. Since both Jamar and Sharona bought the same pencils, $p$ must be a common divisor of the amounts they paid (in cents).
Jamar paid $143$ cents, and Sharona paid $187$ cents.

Let's find the prime factorizations of 143 and 187:
$143 = 11 \times 13$
$187 = 11 \times 17$

The common divisors of 143 and 187 are 1 and 11. Since the problem states that the pencils cost *more than a penny each*, the price of one pencil must be $p = 11$ cents.

Now we can find out how many pencils each bought:
Jamar bought: $\frac{143}{11} = 13$ pencils
Sharona bought: $\frac{187}{11} = 17$ pencils

Sharona bought $17 - 13 = 4$ more pencils than Jamar.

ANSWER 6: C

**Problem 7:**
We are given the exchange rate:
$\$1.60 = 3000 \text{ lire}$

To find out how many lire the traveler will receive for $\$1.00$, we divide the amount of lire by $1.60$:
$\text{Lire for } \$1.00 = \frac{3000}{1.60}$

To make the division easier, we can multiply the numerator and the denominator by 10:
$\frac{30000}{16}$

Now, we simplify the fraction:
$\frac{30000}{16} = \frac{15000}{8} = \frac{7500}{4} = \frac{3750}{2} = 1875$

The traveler will receive 1875 lire for $\$1.00$.

ANSWER 7: D

**Problem 8:**
We need to form 4-digit numbers greater than 1000 using the digits 2, 0, 1, and 2.
For the number to be a valid 4-digit number (and greater than 1000), the first digit (thousands place) cannot be 0. Thus, the first digit must be either 1 or 2.

**Case 1: The first digit is 1.**
The remaining digits to arrange are 0, 2, and 2. 
The number of unique arrangements of these three digits is:
$\frac{3!}{2!} = 3$
These numbers are: 1022, 1202, and 1220.

**Case 2: The first digit is 2.**
The remaining digits to arrange are 0, 1, and 2.
The number of unique arrangements of these three distinct digits is:
$3! = 6$
These numbers are: 2012, 2021, 2102, 2120, 2201, and 2210.

Adding the possibilities from both cases gives a total of $3 + 6 = 9$ numbers.

ANSWER 8: D

**Problem 9:**
First, we calculate the total number of cookies needed for the students:
$108 \text{ students} \times 2 \text{ cookies/student} = 216 \text{ cookies}$

Next, we determine how many pans of cookies are required. Each pan makes 15 cookies, and only full recipes (full pans) are made:
$\frac{216}{15} = 14.4$
Since Walter can only make full pans, he must round up and make 15 pans.

Now, we calculate the total number of eggs needed. The recipe calls fo

Problem 1:
What is the value of this expression? √ 16√ 81 √ 81√ 16
To solve this problem, we need to simplify the given expression step by step.

First, let's simplify the square roots: √16 = 4 and √81 = 9.

Now, substitute these values into the expression: 
√ 16√ 81 √ 81√ 16 = 4 × 9 9 × 4

Next, simplify the expression: 
4 × 9 9 × 4 = 36 36 = 1

Let's examine the answer choices to rule out the incorrect ones:
A. 4/9 = 0.44 (not equal to 1)
B. 2/3 = 0.67 (not equal to 1)
C. 1 = 1 (matches our result)
D. 3/2 = 1.5 (not equal to 1)
E. 9/4 = 2.25 (not equal to 1)

The only correct answer is C.

ANSWER 1: C

Problem 2:
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
Let's denote the capacity of the tank as x gallons.

The tank is initially one-third full, so it contains x/3 gallons of liquid.

After adding 4 gallons, the tank becomes one-half full, so it contains x/2 gallons of liquid.

We can set up an equation based on this information: 
x/3 + 4 = x/2

To solve for x, first multiply the entire equation by 6 to eliminate the fractions: 
2x + 24 = 3x

Now, subtract 2x from both sides: 
24 = x

Let's examine the answer choices to rule out the incorrect ones:
A. 8: 8/3 + 4 = 6.67, and 8/2 = 4 (not equal)
B. 12: 12/3 + 4 = 8, and 12/2 = 6 (not equal)
C. 20: 20/3 + 4 = 10.67, and 20/2 = 10 (not equal)
D. 24: 24/3 + 4 = 12, and 24/2 = 12 (matches our result)
E. 48: 48/3 + 4 = 20, and 48/2 = 24 (not equal)

The correct answer is D.

ANSWER 2: D

Problem 3:
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 1/3 of all the ninth graders are paired with 2/5 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Let's denote the number of sixth graders as S and the number of ninth graders as N.

We are given that 1/3 of N is paired with 2/5 of S. Since the number of pairs is the same, we can set up the following equation: 
(1/3)N = (2/5)S

To simplify, multiply both sides by 15 to eliminate the fractions: 
5N = 6S

The total number of students is S + N. We need to find the fraction of students who have a buddy, which is (2/5)S + (1/3)N.

Since (1/3)N = (2/5)S, we can simplify this to 2 × (2/5)S or 2 × (1/3)N.

Let's express N in terms of S: 
5N = 6S => N = (6/5)S

The total number of students is S + (6/5)S = (11/5)S.

The number of students with a buddy is 2 × (2/5)S = (4/5)S.

Now, we can find the fraction of students who have a buddy: 
((4/5)S) / ((11/5)S) = 4/11

Let's examine the answer choices to rule out the incorrect ones:
A. 2/15 = 0.133 (not equal to 4/11)
B. 4/11 = 0.364 (matches our result)
C. 11/30 = 0.367 (not equal to 4/11)
D. 3/8 = 0.375 (not equal to 4/11)
E. 11/15 = 0.733 (not equal to 4/11)

The correct answer is B.

ANSWER 3: B

Problem 4:
How many integers between 1000 and 9999 have four distinct digits?
To solve this problem, let's consider the constraints for each digit.

The first digit cannot be 0, so there are 9 possibilities (1-9).

The second digit can be any digit except the one chosen for the first digit, so there are 9 possibilities.

The third digit can be any digit except the two chosen for the first and second digits, so there are 8 possibilities.

The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 possibilities.

Now, we can calculate the total number of integers with four distinct digits: 
9 × 9 × 8 × 7 = 4536

Let's examine the answer choices to rule out the incorrect ones:
A. 3024 (not equal to 4536)
B. 4536 (matches our result)
C. 5040 (not equal to 4536)
D. 6480 (not equal to 4536)
E. 6561 (not equal to 4536)

The correct answer is B.

ANSWER 4: B

Problem 5:
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Compared with Blake, Jenny scored 10 points higher on the first test, 10 points lower on the second, and 20 points higher on each of the third and fourth.
Let's denote Blake's scores on the four tests as B1, B2, B3, and B4.

Blake's average score is 78, so (B1 + B2 + B3 + B4)/4 = 78.

Jenny's scores are B1 + 10, B2 - 10, B3 + 20, and B4 + 20.

Jenny's total score is (B1 + 10) + (B2 - 10) + (B3 + 20) + (B4 + 20) = B1 + B2 + B3 + B4 + 40.

Since B1 + B2 + B3 + B4 = 78 × 4 = 312, Jenny's total score is 312 + 40 = 352.

Jenny's average score is 352/4 = 88.

The difference between Jenny's average and Blake's average is 88 - 78 = 10.

Let's examine the answer choices to rule out the incorrect ones:
A. 10 (matches our result)
B. 15 (not equal to 10)
C. 20 (not equal to 10)
D. 25 (not equal to 10)
E. 40 (not equal to 10)

The correct answer is A.

ANSWER 5: A

Problem 6:
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
Let's denote the cost of one pencil as x cents.

Since the cost is more than a penny, x > 1.

Jamar paid 143 cents, so the number of pencils he bought is 143/x.

Sharona paid 187 cents, so the number of pencils she bought is 187/x.

The difference in the number of pencils bought is (187/x) - (143/x) = (187 - 143)/x = 44/x.

Since x is a factor of both 143 and 187, we need to find the common factors.

The prime factorization of 143 is 11 × 13, and the prime factorization of 187 is 11 × 17.

The common factor is 11, so x = 11.

Now, we can calculate the difference: 
(187/11) - (143/11) = 17 - 13 = 4

Let's examine the answer choices to rule out the incorrect ones:
A. 2 (not equal to 4)
B. 3 (not equal to 4)
C. 4 (matches our result)
D. 5 (not equal to 4)
E. 6 (not equal to 4)

The correct answer is C.

ANSWER 6: C

Problem 7:
An American traveling in Italy wishes to exchange American dollars for Italian lire. If 3000 lire = $1.60, how much lire will 

Problem 1:

We are evaluating the expression

\[
\frac{\sqrt{16\sqrt{81}}}{\sqrt{81\sqrt{16}}}.
\]

First simplify the square roots inside:

\[
\sqrt{81}=9,\qquad \sqrt{16}=4.
\]

So the numerator is

\[
\sqrt{16\sqrt{81}}=\sqrt{16\cdot 9}=\sqrt{144}=12.
\]

The denominator is

\[
\sqrt{81\sqrt{16}}=\sqrt{81\cdot 4}=\sqrt{324}=18.
\]

Thus the value is

\[
\frac{12}{18}=\frac{2}{3}.
\]

Choice C, \(1\), would come from incorrectly canceling the two expressions as if they were the same, but the numbers inside the square roots are different.

ANSWER 1: B

---

Problem 2:

Let the capacity of the tank be \(x\) gallons.

The tank starts one-third full:

\[
\frac{x}{3}
\]

After adding 4 gallons, it is one-half full:

\[
\frac{x}{2}
\]

So the added amount is the difference:

\[
\frac{x}{2}-\frac{x}{3}=4
\]

Find a common denominator:

\[
\frac{3x}{6}-\frac{2x}{6}=4
\]

\[
\frac{x}{6}=4
\]

\[
x=24
\]

So the tank holds 24 gallons.

ANSWER 2: D

---

Problem 3:

Let the number of paired ninth graders be \(k\). Since \(\frac13\) of all ninth graders are paired, the total number of ninth graders is

\[
3k.
\]

The same \(k\) ninth graders are paired with sixth graders, and this represents \(\frac25\) of all sixth graders. So

\[
\frac25(\text{sixth graders})=k.
\]

Thus the total number of sixth graders is

\[
\frac52k.
\]

The number of students who have a buddy is both members of each pair:

\[
2k.
\]

The total number of sixth and ninth graders is

\[
3k+\frac52k=\frac62k+\frac52k=\frac{11}{2}k.
\]

So the fraction who have a buddy is

\[
\frac{2k}{\frac{11}{2}k}=\frac{2k\cdot 2}{11k}=\frac{4}{11}.
\]

A tempting mistake is to add \(\frac13+\frac25=\frac{11}{15}\), but those fractions are of two different groups, not of the total group.

ANSWER 3: B

---

Problem 4:

We need four-digit integers from 1000 to 9999 with four distinct digits.

For the thousands digit, we can choose any digit from 1 to 9:

\[
9 \text{ choices}
\]

For the hundreds digit, we can choose any digit except the thousands digit. This includes 0, so there are

\[
9 \text{ choices}
\]

For the tens digit, we must avoid the first two digits:

\[
8 \text{ choices}
\]

For the ones digit, we must avoid the first three digits:

\[
7 \text{ choices}
\]

Therefore the total number is

\[
9\cdot 9\cdot 8\cdot 7=4536.
\]

Choice D, 6480, would come from allowing too many choices after the first digit.

ANSWER 4: B

---

Problem 5:

Blake took four tests. Jenny’s scores compared with Blake’s were:

- first test: \(+10\)
- second test: \(-10\)
- third test: \(+20\)
- fourth test: \(+20\)

So Jenny’s total score compared with Blake’s total score is

\[
10-10+20+20=40
\]

points higher.

Since there are 4 tests, Jenny’s average is higher by

\[
\frac{40}{4}=10.
\]

So Jenny’s average exceeds Blake’s by 10 points.

ANSWER 5: A

---

Problem 6:

Jamar paid \(\$1.43=143\) cents.

Sharona paid \(\$1.87=187\) cents.

The pencils cost more than 1 cent each, and both totals must be multiples of the pencil price.

So the pencil price must divide both 143 and 187.

Factor:

\[
143=11\cdot 13
\]

\[
187=11\cdot 17
\]

The common factor greater than 1 is 11, so each pencil costs 11 cents.

Jamar bought

\[
143\div 11=13
\]

pencils.

Sharona bought

\[
187\div 11=17
\]

pencils.

So Sharona bought

\[
17-13=4
\]

more pencils.

ANSWER 6: C

---

Problem 7:

We are told

\[
3000 \text{ lire}=\$1.60.
\]

We want the number of lire for \(\$1.00\).

Set up the rate:

\[
\frac{3000}{1.60}
\]

Since

\[
1.60=\frac{16}{10},
\]

we get

\[
3000\div 1.60=3000\div \frac{16}{10}=3000\cdot \frac{10}{16}.
\]

\[
=30000\div 16=1875.
\]

So the traveler receives 1875 lire for \(\$1.00\).

ANSWER 7: D

---

Problem 8:

We need four-digit numbers greater than 1000 using the digits of 2012.

The digits are

\[
2,\ 0,\ 1,\ 2.
\]

There are 4 digits, but the digit 2 is repeated. So the total number of distinct arrangements is

\[
\frac{4!}{2!}=\frac{24}{2}=12.
\]

But some arrangements start with 0, and those are not four-digit numbers.

If 0 is first, we arrange the remaining digits \(2,1,2\):

\[
\frac{3!}{2!}=3.
\]

So the number of valid four-digit numbers is

\[
12-3=9.
\]

All of these are greater than 1000, so the answer is 9.

ANSWER 8: D

---

Problem 9:

There are 108 students, and each eats an average of 2 cookies.

So the total number of cookies needed is

\[
108\cdot 2=216.
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough full pans to make at least 216 cookies:

\[
216\div 15=14.4.
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is

\[
15\cdot 2=30.
\]

Eggs are bought by the half-dozen, and one half-dozen is 6 eggs.

\[
30\div 6=5.
\]

So Walter should buy 5 half-dozens.

Choice E, 15, is the number of pans, not the number of half-dozens of eggs.

ANSWER 9: C

---

Problem 10:

We need to simplify

\[
100\times 19.98\times 1.998\times 1000.
\]

Group the factors conveniently:

\[
(100\times 19.98)(1.998\times 1000).
\]

Now compute each group:

\[
100\times 19.98=1998
\]

and

\[
1.998\times 1000=1998.
\]

So the expression becomes

\[
1998\times 1998=(1998)^2.
\]

ANSWER 10: D

---

Problem 11:

You have \(\$10\), but you plan to give the driver a \(\$2\) tip.

So the actual taxi fare can be

\[
10-2=8
\]

dollars.

The first \(\frac12\) mile costs \(\$2.40\).

Money remaining after the first half-mile:

\[
8.00-2.40=5.60.
\]

Additional distance costs \(\$0.20\) per \(0.1\) mile.

The number of additional \(0.1\)-mile segments is

\[
5.60\div 0.20=28.
\]

So the additional distance is

\[
28\cdot 0.1=2.8
\]

miles.

Total distance:

\[
0.5+2.8=3.3
\]

miles.

ANSWER 11: C

---

Problem 12:

The pizza has diameter 12 inches, so its radius is

\[
6
\]

inches.

Six pepperoni circles fit exactly across the diameter of the pizza. Therefore each pepperoni circle has diameter

\[
12\div 6=2
\]

inches, so each pepperoni has radius

\[
1
\]

inch.

Area of the pizza:

\[
\pi\cdot 6^2=36\p

**Problem 1:**
The garbled text `√ 16√ 81 √ 81√ 16` is a result of extracting the mathematical fraction $\frac{\sqrt{16}}{\sqrt{81}}$ (or $\sqrt{\frac{16}{81}}$) from a document. 

We are asked to evaluate this expression:
$\frac{\sqrt{16}}{\sqrt{81}}$

First, we find the square roots of the perfect squares in the numerator and the denominator:
$\sqrt{16} = 4$
$\sqrt{81} = 9$

Substituting these values back into the fraction, we get:
$\frac{4}{9}$

Looking at the choices, this matches option A.

ANSWER 1: A

**Problem 2:**
Let $C$ be the total capacity of the tank in gallons. 
We are told that the tank is initially one-third full, which is $\frac{1}{3}C$. 
When 4 gallons are added, the tank becomes one-half full, which is $\frac{1}{2}C$. 

We can set up the following equation:
$\frac{1}{3}C + 4 = \frac{1}{2}C$

To solve for $C$, we subtract $\frac{1}{3}C$ from both sides:
$4 = \frac{1}{2}C - \frac{1}{3}C$

Find a common denominator to subtract the fractions (which is 6):
$4 = \frac{3}{6}C - \frac{2}{6}C$
$4 = \frac{1}{6}C$

Multiply both sides by 6:
$C = 24$

The capacity of the tank is 24 gallons.

ANSWER 2: D

**Problem 3:**
Let $N$ be the total number of ninth graders and $S$ be the total number of sixth graders.
The number of ninth graders who have a buddy is $\frac{1}{3}N$.
The number of sixth graders who have a buddy is $\frac{2}{5}S$.

Since each paired ninth grader is matched with exactly one sixth grader, the number of paired ninth graders must equal the number of paired sixth graders:
$\frac{1}{3}N = \frac{2}{5}S$

We can express $N$ in terms of $S$ by multiplying both sides by 3:
$N = \frac{6}{5}S$

The total number of students in the program is:
$N + S = \frac{6}{5}S + S = \frac{11}{5}S$

The total number of students who have a buddy is the sum of the paired ninth graders and paired sixth graders:
$\frac{2}{5}S + \frac{2}{5}S = \frac{4}{5}S$

The fraction of the total number of students who have a buddy is:
$\frac{\text{Total paired}}{\text{Total students}} = \frac{\frac{4}{5}S}{\frac{11}{5}S} = \frac{4}{11}$

ANSWER 3: B

**Problem 4:**
We are looking for the number of 4-digit integers (from 1000 to 9999) where all four digits are distinct.
We can determine this by finding the number of choices for each digit place (thousands, hundreds, tens, units):
1. **Thousands digit:** It cannot be 0 (otherwise it wouldn't be a 4-digit number). So, it can be any digit from 1 to 9. That gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, but it must be different from the thousands digit. That gives us **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, but it must be different from the first two digits. That gives us **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, but it must be different from the first three digits. That gives us **7 choices**.

To find the total number of such integers, we multiply the number of choices together:
$9 \times 9 \times 8 \times 7 = 81 \times 56 = 4536$

ANSWER 4: B

**Problem 5:**
Let's find the total difference between Jenny's points and Blake's points across all four tests.
Compared to Blake, Jenny's score differences were:
Test 1: $+10$
Test 2: $-10$
Test 3: $+20$
Test 4: $+20$

Adding these differences together gives the total amount by which Jenny's overall score exceeded Blake's:
$10 - 10 + 20 + 20 = 40$ points

Since the average is the total score divided by the number of tests (4), Jenny's average exceeds Blake's average by:
$\frac{40}{4} = 10$ points

ANSWER 5: A

**Problem 6:**
Let $p$ be the price of a single pencil in cents. Since both Jamar and Sharona bought the same pencils, $p$ must be a common divisor of the amounts they paid (in cents).
Jamar paid $143$ cents, and Sharona paid $187$ cents.

Let's find the prime factorizations of 143 and 187:
$143 = 11 \times 13$
$187 = 11 \times 17$

The common divisors of 143 and 187 are 1 and 11. Since the problem states that the pencils cost *more than a penny each*, the price of one pencil must be $p = 11$ cents.

Now we can find out how many pencils each bought:
Jamar bought: $\frac{143}{11} = 13$ pencils
Sharona bought: $\frac{187}{11} = 17$ pencils

Sharona bought $17 - 13 = 4$ more pencils than Jamar.

ANSWER 6: C

**Problem 7:**
We are given the exchange rate:
$\$1.60 = 3000 \text{ lire}$

To find out how many lire the traveler will receive for $\$1.00$, we divide the amount of lire by $1.60$:
$\text{Lire for } \$1.00 = \frac{3000}{1.60}$

To make the division easier, we can multiply the numerator and the denominator by 10:
$\frac{30000}{16}$

Now, we simplify the fraction:
$\frac{30000}{16} = \frac{15000}{8} = \frac{7500}{4} = \frac{3750}{2} = 1875$

The traveler will receive 1875 lire for $\$1.00$.

ANSWER 7: D

**Problem 8:**
We need to form 4-digit numbers greater than 1000 using the digits 2, 0, 1, and 2.
For the number to be a valid 4-digit number (and greater than 1000), the first digit (thousands place) cannot be 0. Thus, the first digit must be either 1 or 2.

**Case 1: The first digit is 1.**
The remaining digits to arrange are 0, 2, and 2. 
The number of unique arrangements of these three digits is:
$\frac{3!}{2!} = 3$
These numbers are: 1022, 1202, and 1220.

**Case 2: The first digit is 2.**
The remaining digits to arrange are 0, 1, and 2.
The number of unique arrangements of these three distinct digits is:
$3! = 6$
These numbers are: 2012, 2021, 2102, 2120, 2201, and 2210.

Adding the possibilities from both cases gives a total of $3 + 6 = 9$ numbers.

ANSWER 8: D

**Problem 9:**
First, we calculate the total number of cookies needed for the students:
$108 \text{ students} \times 2 \text{ cookies/student} = 216 \text{ cookies}$

Next, we determine how many pans of cookies are required. Each pan makes 15 cookies, and only full recipes (full pans) are made:
$\frac{216}{15} = 14.4$
Since Walter can only make full pans, he must round up and make 15 pans.

Now, we calculate the total number of eggs needed. The recipe calls fo

Problem 1:
What is the value of this expression? √ 16√ 81 √ 81√ 16
To solve this problem, we need to simplify the given expression step by step.

First, let's simplify the square roots: √16 = 4 and √81 = 9.

Now, substitute these values into the expression: 
√ 16√ 81 √ 81√ 16 = 4 × 9 9 × 4

Next, simplify the expression: 
4 × 9 9 × 4 = 36 36 = 1

Let's examine the answer choices to rule out the incorrect ones:
A. 4/9 = 0.44 (not equal to 1)
B. 2/3 = 0.67 (not equal to 1)
C. 1 = 1 (matches our result)
D. 3/2 = 1.5 (not equal to 1)
E. 9/4 = 2.25 (not equal to 1)

The only correct answer is C.

ANSWER 1: C

Problem 2:
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
Let's denote the capacity of the tank as x gallons.

The tank is initially one-third full, so it contains x/3 gallons of liquid.

After adding 4 gallons, the tank becomes one-half full, so it contains x/2 gallons of liquid.

We can set up an equation based on this information: 
x/3 + 4 = x/2

To solve for x, first multiply the entire equation by 6 to eliminate the fractions: 
2x + 24 = 3x

Now, subtract 2x from both sides: 
24 = x

Let's examine the answer choices to rule out the incorrect ones:
A. 8: 8/3 + 4 = 6.67, and 8/2 = 4 (not equal)
B. 12: 12/3 + 4 = 8, and 12/2 = 6 (not equal)
C. 20: 20/3 + 4 = 10.67, and 20/2 = 10 (not equal)
D. 24: 24/3 + 4 = 12, and 24/2 = 12 (matches our result)
E. 48: 48/3 + 4 = 20, and 48/2 = 24 (not equal)

The correct answer is D.

ANSWER 2: D

Problem 3:
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 1/3 of all the ninth graders are paired with 2/5 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Let's denote the number of sixth graders as S and the number of ninth graders as N.

We are given that 1/3 of N is paired with 2/5 of S. Since the number of pairs is the same, we can set up the following equation: 
(1/3)N = (2/5)S

To simplify, multiply both sides by 15 to eliminate the fractions: 
5N = 6S

The total number of students is S + N. We need to find the fraction of students who have a buddy, which is (2/5)S + (1/3)N.

Since (1/3)N = (2/5)S, we can simplify this to 2 × (2/5)S or 2 × (1/3)N.

Let's express N in terms of S: 
5N = 6S => N = (6/5)S

The total number of students is S + (6/5)S = (11/5)S.

The number of students with a buddy is 2 × (2/5)S = (4/5)S.

Now, we can find the fraction of students who have a buddy: 
((4/5)S) / ((11/5)S) = 4/11

Let's examine the answer choices to rule out the incorrect ones:
A. 2/15 = 0.133 (not equal to 4/11)
B. 4/11 = 0.364 (matches our result)
C. 11/30 = 0.367 (not equal to 4/11)
D. 3/8 = 0.375 (not equal to 4/11)
E. 11/15 = 0.733 (not equal to 4/11)

The correct answer is B.

ANSWER 3: B

Problem 4:
How many integers between 1000 and 9999 have four distinct digits?
To solve this problem, let's consider the constraints for each digit.

The first digit cannot be 0, so there are 9 possibilities (1-9).

The second digit can be any digit except the one chosen for the first digit, so there are 9 possibilities.

The third digit can be any digit except the two chosen for the first and second digits, so there are 8 possibilities.

The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 possibilities.

Now, we can calculate the total number of integers with four distinct digits: 
9 × 9 × 8 × 7 = 4536

Let's examine the answer choices to rule out the incorrect ones:
A. 3024 (not equal to 4536)
B. 4536 (matches our result)
C. 5040 (not equal to 4536)
D. 6480 (not equal to 4536)
E. 6561 (not equal to 4536)

The correct answer is B.

ANSWER 4: B

Problem 5:
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Compared with Blake, Jenny scored 10 points higher on the first test, 10 points lower on the second, and 20 points higher on each of the third and fourth.
Let's denote Blake's scores on the four tests as B1, B2, B3, and B4.

Blake's average score is 78, so (B1 + B2 + B3 + B4)/4 = 78.

Jenny's scores are B1 + 10, B2 - 10, B3 + 20, and B4 + 20.

Jenny's total score is (B1 + 10) + (B2 - 10) + (B3 + 20) + (B4 + 20) = B1 + B2 + B3 + B4 + 40.

Since B1 + B2 + B3 + B4 = 78 × 4 = 312, Jenny's total score is 312 + 40 = 352.

Jenny's average score is 352/4 = 88.

The difference between Jenny's average and Blake's average is 88 - 78 = 10.

Let's examine the answer choices to rule out the incorrect ones:
A. 10 (matches our result)
B. 15 (not equal to 10)
C. 20 (not equal to 10)
D. 25 (not equal to 10)
E. 40 (not equal to 10)

The correct answer is A.

ANSWER 5: A

Problem 6:
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
Let's denote the cost of one pencil as x cents.

Since the cost is more than a penny, x > 1.

Jamar paid 143 cents, so the number of pencils he bought is 143/x.

Sharona paid 187 cents, so the number of pencils she bought is 187/x.

The difference in the number of pencils bought is (187/x) - (143/x) = (187 - 143)/x = 44/x.

Since x is a factor of both 143 and 187, we need to find the common factors.

The prime factorization of 143 is 11 × 13, and the prime factorization of 187 is 11 × 17.

The common factor is 11, so x = 11.

Now, we can calculate the difference: 
(187/11) - (143/11) = 17 - 13 = 4

Let's examine the answer choices to rule out the incorrect ones:
A. 2 (not equal to 4)
B. 3 (not equal to 4)
C. 4 (matches our result)
D. 5 (not equal to 4)
E. 6 (not equal to 4)

The correct answer is C.

ANSWER 6: C

Problem 7:
An American traveling in Italy wishes to exchange American dollars for Italian lire. If 3000 lire = $1.60, how much lire will 

Problem 1:

We are evaluating the expression

\[
\frac{\sqrt{16\sqrt{81}}}{\sqrt{81\sqrt{16}}}.
\]

First simplify the square roots inside:

\[
\sqrt{81}=9,\qquad \sqrt{16}=4.
\]

So the numerator is

\[
\sqrt{16\sqrt{81}}=\sqrt{16\cdot 9}=\sqrt{144}=12.
\]

The denominator is

\[
\sqrt{81\sqrt{16}}=\sqrt{81\cdot 4}=\sqrt{324}=18.
\]

Thus the value is

\[
\frac{12}{18}=\frac{2}{3}.
\]

Choice C, \(1\), would come from incorrectly canceling the two expressions as if they were the same, but the numbers inside the square roots are different.

ANSWER 1: B

---

Problem 2:

Let the capacity of the tank be \(x\) gallons.

The tank starts one-third full:

\[
\frac{x}{3}
\]

After adding 4 gallons, it is one-half full:

\[
\frac{x}{2}
\]

So the added amount is the difference:

\[
\frac{x}{2}-\frac{x}{3}=4
\]

Find a common denominator:

\[
\frac{3x}{6}-\frac{2x}{6}=4
\]

\[
\frac{x}{6}=4
\]

\[
x=24
\]

So the tank holds 24 gallons.

ANSWER 2: D

---

Problem 3:

Let the number of paired ninth graders be \(k\). Since \(\frac13\) of all ninth graders are paired, the total number of ninth graders is

\[
3k.
\]

The same \(k\) ninth graders are paired with sixth graders, and this represents \(\frac25\) of all sixth graders. So

\[
\frac25(\text{sixth graders})=k.
\]

Thus the total number of sixth graders is

\[
\frac52k.
\]

The number of students who have a buddy is both members of each pair:

\[
2k.
\]

The total number of sixth and ninth graders is

\[
3k+\frac52k=\frac62k+\frac52k=\frac{11}{2}k.
\]

So the fraction who have a buddy is

\[
\frac{2k}{\frac{11}{2}k}=\frac{2k\cdot 2}{11k}=\frac{4}{11}.
\]

A tempting mistake is to add \(\frac13+\frac25=\frac{11}{15}\), but those fractions are of two different groups, not of the total group.

ANSWER 3: B

---

Problem 4:

We need four-digit integers from 1000 to 9999 with four distinct digits.

For the thousands digit, we can choose any digit from 1 to 9:

\[
9 \text{ choices}
\]

For the hundreds digit, we can choose any digit except the thousands digit. This includes 0, so there are

\[
9 \text{ choices}
\]

For the tens digit, we must avoid the first two digits:

\[
8 \text{ choices}
\]

For the ones digit, we must avoid the first three digits:

\[
7 \text{ choices}
\]

Therefore the total number is

\[
9\cdot 9\cdot 8\cdot 7=4536.
\]

Choice D, 6480, would come from allowing too many choices after the first digit.

ANSWER 4: B

---

Problem 5:

Blake took four tests. Jenny’s scores compared with Blake’s were:

- first test: \(+10\)
- second test: \(-10\)
- third test: \(+20\)
- fourth test: \(+20\)

So Jenny’s total score compared with Blake’s total score is

\[
10-10+20+20=40
\]

points higher.

Since there are 4 tests, Jenny’s average is higher by

\[
\frac{40}{4}=10.
\]

So Jenny’s average exceeds Blake’s by 10 points.

ANSWER 5: A

---

Problem 6:

Jamar paid \(\$1.43=143\) cents.

Sharona paid \(\$1.87=187\) cents.

The pencils cost more than 1 cent each, and both totals must be multiples of the pencil price.

So the pencil price must divide both 143 and 187.

Factor:

\[
143=11\cdot 13
\]

\[
187=11\cdot 17
\]

The common factor greater than 1 is 11, so each pencil costs 11 cents.

Jamar bought

\[
143\div 11=13
\]

pencils.

Sharona bought

\[
187\div 11=17
\]

pencils.

So Sharona bought

\[
17-13=4
\]

more pencils.

ANSWER 6: C

---

Problem 7:

We are told

\[
3000 \text{ lire}=\$1.60.
\]

We want the number of lire for \(\$1.00\).

Set up the rate:

\[
\frac{3000}{1.60}
\]

Since

\[
1.60=\frac{16}{10},
\]

we get

\[
3000\div 1.60=3000\div \frac{16}{10}=3000\cdot \frac{10}{16}.
\]

\[
=30000\div 16=1875.
\]

So the traveler receives 1875 lire for \(\$1.00\).

ANSWER 7: D

---

Problem 8:

We need four-digit numbers greater than 1000 using the digits of 2012.

The digits are

\[
2,\ 0,\ 1,\ 2.
\]

There are 4 digits, but the digit 2 is repeated. So the total number of distinct arrangements is

\[
\frac{4!}{2!}=\frac{24}{2}=12.
\]

But some arrangements start with 0, and those are not four-digit numbers.

If 0 is first, we arrange the remaining digits \(2,1,2\):

\[
\frac{3!}{2!}=3.
\]

So the number of valid four-digit numbers is

\[
12-3=9.
\]

All of these are greater than 1000, so the answer is 9.

ANSWER 8: D

---

Problem 9:

There are 108 students, and each eats an average of 2 cookies.

So the total number of cookies needed is

\[
108\cdot 2=216.
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough full pans to make at least 216 cookies:

\[
216\div 15=14.4.
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is

\[
15\cdot 2=30.
\]

Eggs are bought by the half-dozen, and one half-dozen is 6 eggs.

\[
30\div 6=5.
\]

So Walter should buy 5 half-dozens.

Choice E, 15, is the number of pans, not the number of half-dozens of eggs.

ANSWER 9: C

---

Problem 10:

We need to simplify

\[
100\times 19.98\times 1.998\times 1000.
\]

Group the factors conveniently:

\[
(100\times 19.98)(1.998\times 1000).
\]

Now compute each group:

\[
100\times 19.98=1998
\]

and

\[
1.998\times 1000=1998.
\]

So the expression becomes

\[
1998\times 1998=(1998)^2.
\]

ANSWER 10: D

---

Problem 11:

You have \(\$10\), but you plan to give the driver a \(\$2\) tip.

So the actual taxi fare can be

\[
10-2=8
\]

dollars.

The first \(\frac12\) mile costs \(\$2.40\).

Money remaining after the first half-mile:

\[
8.00-2.40=5.60.
\]

Additional distance costs \(\$0.20\) per \(0.1\) mile.

The number of additional \(0.1\)-mile segments is

\[
5.60\div 0.20=28.
\]

So the additional distance is

\[
28\cdot 0.1=2.8
\]

miles.

Total distance:

\[
0.5+2.8=3.3
\]

miles.

ANSWER 11: C

---

Problem 12:

The pizza has diameter 12 inches, so its radius is

\[
6
\]

inches.

Six pepperoni circles fit exactly across the diameter of the pizza. Therefore each pepperoni circle has diameter

\[
12\div 6=2
\]

inches, so each pepperoni has radius

\[
1
\]

inch.

Area of the pizza:

\[
\pi\cdot 6^2=36\p

**Problem 1:**
The garbled text `√ 16√ 81 √ 81√ 16` is a result of extracting the mathematical fraction $\frac{\sqrt{16}}{\sqrt{81}}$ (or $\sqrt{\frac{16}{81}}$) from a document. 

We are asked to evaluate this expression:
$\frac{\sqrt{16}}{\sqrt{81}}$

First, we find the square roots of the perfect squares in the numerator and the denominator:
$\sqrt{16} = 4$
$\sqrt{81} = 9$

Substituting these values back into the fraction, we get:
$\frac{4}{9}$

Looking at the choices, this matches option A.

ANSWER 1: A

**Problem 2:**
Let $C$ be the total capacity of the tank in gallons. 
We are told that the tank is initially one-third full, which is $\frac{1}{3}C$. 
When 4 gallons are added, the tank becomes one-half full, which is $\frac{1}{2}C$. 

We can set up the following equation:
$\frac{1}{3}C + 4 = \frac{1}{2}C$

To solve for $C$, we subtract $\frac{1}{3}C$ from both sides:
$4 = \frac{1}{2}C - \frac{1}{3}C$

Find a common denominator to subtract the fractions (which is 6):
$4 = \frac{3}{6}C - \frac{2}{6}C$
$4 = \frac{1}{6}C$

Multiply both sides by 6:
$C = 24$

The capacity of the tank is 24 gallons.

ANSWER 2: D

**Problem 3:**
Let $N$ be the total number of ninth graders and $S$ be the total number of sixth graders.
The number of ninth graders who have a buddy is $\frac{1}{3}N$.
The number of sixth graders who have a buddy is $\frac{2}{5}S$.

Since each paired ninth grader is matched with exactly one sixth grader, the number of paired ninth graders must equal the number of paired sixth graders:
$\frac{1}{3}N = \frac{2}{5}S$

We can express $N$ in terms of $S$ by multiplying both sides by 3:
$N = \frac{6}{5}S$

The total number of students in the program is:
$N + S = \frac{6}{5}S + S = \frac{11}{5}S$

The total number of students who have a buddy is the sum of the paired ninth graders and paired sixth graders:
$\frac{2}{5}S + \frac{2}{5}S = \frac{4}{5}S$

The fraction of the total number of students who have a buddy is:
$\frac{\text{Total paired}}{\text{Total students}} = \frac{\frac{4}{5}S}{\frac{11}{5}S} = \frac{4}{11}$

ANSWER 3: B

**Problem 4:**
We are looking for the number of 4-digit integers (from 1000 to 9999) where all four digits are distinct.
We can determine this by finding the number of choices for each digit place (thousands, hundreds, tens, units):
1. **Thousands digit:** It cannot be 0 (otherwise it wouldn't be a 4-digit number). So, it can be any digit from 1 to 9. That gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, but it must be different from the thousands digit. That gives us **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, but it must be different from the first two digits. That gives us **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, but it must be different from the first three digits. That gives us **7 choices**.

To find the total number of such integers, we multiply the number of choices together:
$9 \times 9 \times 8 \times 7 = 81 \times 56 = 4536$

ANSWER 4: B

**Problem 5:**
Let's find the total difference between Jenny's points and Blake's points across all four tests.
Compared to Blake, Jenny's score differences were:
Test 1: $+10$
Test 2: $-10$
Test 3: $+20$
Test 4: $+20$

Adding these differences together gives the total amount by which Jenny's overall score exceeded Blake's:
$10 - 10 + 20 + 20 = 40$ points

Since the average is the total score divided by the number of tests (4), Jenny's average exceeds Blake's average by:
$\frac{40}{4} = 10$ points

ANSWER 5: A

**Problem 6:**
Let $p$ be the price of a single pencil in cents. Since both Jamar and Sharona bought the same pencils, $p$ must be a common divisor of the amounts they paid (in cents).
Jamar paid $143$ cents, and Sharona paid $187$ cents.

Let's find the prime factorizations of 143 and 187:
$143 = 11 \times 13$
$187 = 11 \times 17$

The common divisors of 143 and 187 are 1 and 11. Since the problem states that the pencils cost *more than a penny each*, the price of one pencil must be $p = 11$ cents.

Now we can find out how many pencils each bought:
Jamar bought: $\frac{143}{11} = 13$ pencils
Sharona bought: $\frac{187}{11} = 17$ pencils

Sharona bought $17 - 13 = 4$ more pencils than Jamar.

ANSWER 6: C

**Problem 7:**
We are given the exchange rate:
$\$1.60 = 3000 \text{ lire}$

To find out how many lire the traveler will receive for $\$1.00$, we divide the amount of lire by $1.60$:
$\text{Lire for } \$1.00 = \frac{3000}{1.60}$

To make the division easier, we can multiply the numerator and the denominator by 10:
$\frac{30000}{16}$

Now, we simplify the fraction:
$\frac{30000}{16} = \frac{15000}{8} = \frac{7500}{4} = \frac{3750}{2} = 1875$

The traveler will receive 1875 lire for $\$1.00$.

ANSWER 7: D

**Problem 8:**
We need to form 4-digit numbers greater than 1000 using the digits 2, 0, 1, and 2.
For the number to be a valid 4-digit number (and greater than 1000), the first digit (thousands place) cannot be 0. Thus, the first digit must be either 1 or 2.

**Case 1: The first digit is 1.**
The remaining digits to arrange are 0, 2, and 2. 
The number of unique arrangements of these three digits is:
$\frac{3!}{2!} = 3$
These numbers are: 1022, 1202, and 1220.

**Case 2: The first digit is 2.**
The remaining digits to arrange are 0, 1, and 2.
The number of unique arrangements of these three distinct digits is:
$3! = 6$
These numbers are: 2012, 2021, 2102, 2120, 2201, and 2210.

Adding the possibilities from both cases gives a total of $3 + 6 = 9$ numbers.

ANSWER 8: D

**Problem 9:**
First, we calculate the total number of cookies needed for the students:
$108 \text{ students} \times 2 \text{ cookies/student} = 216 \text{ cookies}$

Next, we determine how many pans of cookies are required. Each pan makes 15 cookies, and only full recipes (full pans) are made:
$\frac{216}{15} = 14.4$
Since Walter can only make full pans, he must round up and make 15 pans.

Now, we calculate the total number of eggs needed. The recipe calls fo

Problem 1:
What is the value of this expression? √ 16√ 81 √ 81√ 16
To solve this problem, we need to simplify the given expression step by step.

First, let's simplify the square roots: √16 = 4 and √81 = 9.

Now, substitute these values into the expression: 
√ 16√ 81 √ 81√ 16 = 4 × 9 9 × 4

Next, simplify the expression: 
4 × 9 9 × 4 = 36 36 = 1

Let's examine the answer choices to rule out the incorrect ones:
A. 4/9 = 0.44 (not equal to 1)
B. 2/3 = 0.67 (not equal to 1)
C. 1 = 1 (matches our result)
D. 3/2 = 1.5 (not equal to 1)
E. 9/4 = 2.25 (not equal to 1)

The only correct answer is C.

ANSWER 1: C

Problem 2:
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
Let's denote the capacity of the tank as x gallons.

The tank is initially one-third full, so it contains x/3 gallons of liquid.

After adding 4 gallons, the tank becomes one-half full, so it contains x/2 gallons of liquid.

We can set up an equation based on this information: 
x/3 + 4 = x/2

To solve for x, first multiply the entire equation by 6 to eliminate the fractions: 
2x + 24 = 3x

Now, subtract 2x from both sides: 
24 = x

Let's examine the answer choices to rule out the incorrect ones:
A. 8: 8/3 + 4 = 6.67, and 8/2 = 4 (not equal)
B. 12: 12/3 + 4 = 8, and 12/2 = 6 (not equal)
C. 20: 20/3 + 4 = 10.67, and 20/2 = 10 (not equal)
D. 24: 24/3 + 4 = 12, and 24/2 = 12 (matches our result)
E. 48: 48/3 + 4 = 20, and 48/2 = 24 (not equal)

The correct answer is D.

ANSWER 2: D

Problem 3:
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 1/3 of all the ninth graders are paired with 2/5 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Let's denote the number of sixth graders as S and the number of ninth graders as N.

We are given that 1/3 of N is paired with 2/5 of S. Since the number of pairs is the same, we can set up the following equation: 
(1/3)N = (2/5)S

To simplify, multiply both sides by 15 to eliminate the fractions: 
5N = 6S

The total number of students is S + N. We need to find the fraction of students who have a buddy, which is (2/5)S + (1/3)N.

Since (1/3)N = (2/5)S, we can simplify this to 2 × (2/5)S or 2 × (1/3)N.

Let's express N in terms of S: 
5N = 6S => N = (6/5)S

The total number of students is S + (6/5)S = (11/5)S.

The number of students with a buddy is 2 × (2/5)S = (4/5)S.

Now, we can find the fraction of students who have a buddy: 
((4/5)S) / ((11/5)S) = 4/11

Let's examine the answer choices to rule out the incorrect ones:
A. 2/15 = 0.133 (not equal to 4/11)
B. 4/11 = 0.364 (matches our result)
C. 11/30 = 0.367 (not equal to 4/11)
D. 3/8 = 0.375 (not equal to 4/11)
E. 11/15 = 0.733 (not equal to 4/11)

The correct answer is B.

ANSWER 3: B

Problem 4:
How many integers between 1000 and 9999 have four distinct digits?
To solve this problem, let's consider the constraints for each digit.

The first digit cannot be 0, so there are 9 possibilities (1-9).

The second digit can be any digit except the one chosen for the first digit, so there are 9 possibilities.

The third digit can be any digit except the two chosen for the first and second digits, so there are 8 possibilities.

The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 possibilities.

Now, we can calculate the total number of integers with four distinct digits: 
9 × 9 × 8 × 7 = 4536

Let's examine the answer choices to rule out the incorrect ones:
A. 3024 (not equal to 4536)
B. 4536 (matches our result)
C. 5040 (not equal to 4536)
D. 6480 (not equal to 4536)
E. 6561 (not equal to 4536)

The correct answer is B.

ANSWER 4: B

Problem 5:
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Compared with Blake, Jenny scored 10 points higher on the first test, 10 points lower on the second, and 20 points higher on each of the third and fourth.
Let's denote Blake's scores on the four tests as B1, B2, B3, and B4.

Blake's average score is 78, so (B1 + B2 + B3 + B4)/4 = 78.

Jenny's scores are B1 + 10, B2 - 10, B3 + 20, and B4 + 20.

Jenny's total score is (B1 + 10) + (B2 - 10) + (B3 + 20) + (B4 + 20) = B1 + B2 + B3 + B4 + 40.

Since B1 + B2 + B3 + B4 = 78 × 4 = 312, Jenny's total score is 312 + 40 = 352.

Jenny's average score is 352/4 = 88.

The difference between Jenny's average and Blake's average is 88 - 78 = 10.

Let's examine the answer choices to rule out the incorrect ones:
A. 10 (matches our result)
B. 15 (not equal to 10)
C. 20 (not equal to 10)
D. 25 (not equal to 10)
E. 40 (not equal to 10)

The correct answer is A.

ANSWER 5: A

Problem 6:
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
Let's denote the cost of one pencil as x cents.

Since the cost is more than a penny, x > 1.

Jamar paid 143 cents, so the number of pencils he bought is 143/x.

Sharona paid 187 cents, so the number of pencils she bought is 187/x.

The difference in the number of pencils bought is (187/x) - (143/x) = (187 - 143)/x = 44/x.

Since x is a factor of both 143 and 187, we need to find the common factors.

The prime factorization of 143 is 11 × 13, and the prime factorization of 187 is 11 × 17.

The common factor is 11, so x = 11.

Now, we can calculate the difference: 
(187/11) - (143/11) = 17 - 13 = 4

Let's examine the answer choices to rule out the incorrect ones:
A. 2 (not equal to 4)
B. 3 (not equal to 4)
C. 4 (matches our result)
D. 5 (not equal to 4)
E. 6 (not equal to 4)

The correct answer is C.

ANSWER 6: C

Problem 7:
An American traveling in Italy wishes to exchange American dollars for Italian lire. If 3000 lire = $1.60, how much lire will 

Problem 1:

We are evaluating the expression

\[
\frac{\sqrt{16\sqrt{81}}}{\sqrt{81\sqrt{16}}}.
\]

First simplify the square roots inside:

\[
\sqrt{81}=9,\qquad \sqrt{16}=4.
\]

So the numerator is

\[
\sqrt{16\sqrt{81}}=\sqrt{16\cdot 9}=\sqrt{144}=12.
\]

The denominator is

\[
\sqrt{81\sqrt{16}}=\sqrt{81\cdot 4}=\sqrt{324}=18.
\]

Thus the value is

\[
\frac{12}{18}=\frac{2}{3}.
\]

Choice C, \(1\), would come from incorrectly canceling the two expressions as if they were the same, but the numbers inside the square roots are different.

ANSWER 1: B

---

Problem 2:

Let the capacity of the tank be \(x\) gallons.

The tank starts one-third full:

\[
\frac{x}{3}
\]

After adding 4 gallons, it is one-half full:

\[
\frac{x}{2}
\]

So the added amount is the difference:

\[
\frac{x}{2}-\frac{x}{3}=4
\]

Find a common denominator:

\[
\frac{3x}{6}-\frac{2x}{6}=4
\]

\[
\frac{x}{6}=4
\]

\[
x=24
\]

So the tank holds 24 gallons.

ANSWER 2: D

---

Problem 3:

Let the number of paired ninth graders be \(k\). Since \(\frac13\) of all ninth graders are paired, the total number of ninth graders is

\[
3k.
\]

The same \(k\) ninth graders are paired with sixth graders, and this represents \(\frac25\) of all sixth graders. So

\[
\frac25(\text{sixth graders})=k.
\]

Thus the total number of sixth graders is

\[
\frac52k.
\]

The number of students who have a buddy is both members of each pair:

\[
2k.
\]

The total number of sixth and ninth graders is

\[
3k+\frac52k=\frac62k+\frac52k=\frac{11}{2}k.
\]

So the fraction who have a buddy is

\[
\frac{2k}{\frac{11}{2}k}=\frac{2k\cdot 2}{11k}=\frac{4}{11}.
\]

A tempting mistake is to add \(\frac13+\frac25=\frac{11}{15}\), but those fractions are of two different groups, not of the total group.

ANSWER 3: B

---

Problem 4:

We need four-digit integers from 1000 to 9999 with four distinct digits.

For the thousands digit, we can choose any digit from 1 to 9:

\[
9 \text{ choices}
\]

For the hundreds digit, we can choose any digit except the thousands digit. This includes 0, so there are

\[
9 \text{ choices}
\]

For the tens digit, we must avoid the first two digits:

\[
8 \text{ choices}
\]

For the ones digit, we must avoid the first three digits:

\[
7 \text{ choices}
\]

Therefore the total number is

\[
9\cdot 9\cdot 8\cdot 7=4536.
\]

Choice D, 6480, would come from allowing too many choices after the first digit.

ANSWER 4: B

---

Problem 5:

Blake took four tests. Jenny’s scores compared with Blake’s were:

- first test: \(+10\)
- second test: \(-10\)
- third test: \(+20\)
- fourth test: \(+20\)

So Jenny’s total score compared with Blake’s total score is

\[
10-10+20+20=40
\]

points higher.

Since there are 4 tests, Jenny’s average is higher by

\[
\frac{40}{4}=10.
\]

So Jenny’s average exceeds Blake’s by 10 points.

ANSWER 5: A

---

Problem 6:

Jamar paid \(\$1.43=143\) cents.

Sharona paid \(\$1.87=187\) cents.

The pencils cost more than 1 cent each, and both totals must be multiples of the pencil price.

So the pencil price must divide both 143 and 187.

Factor:

\[
143=11\cdot 13
\]

\[
187=11\cdot 17
\]

The common factor greater than 1 is 11, so each pencil costs 11 cents.

Jamar bought

\[
143\div 11=13
\]

pencils.

Sharona bought

\[
187\div 11=17
\]

pencils.

So Sharona bought

\[
17-13=4
\]

more pencils.

ANSWER 6: C

---

Problem 7:

We are told

\[
3000 \text{ lire}=\$1.60.
\]

We want the number of lire for \(\$1.00\).

Set up the rate:

\[
\frac{3000}{1.60}
\]

Since

\[
1.60=\frac{16}{10},
\]

we get

\[
3000\div 1.60=3000\div \frac{16}{10}=3000\cdot \frac{10}{16}.
\]

\[
=30000\div 16=1875.
\]

So the traveler receives 1875 lire for \(\$1.00\).

ANSWER 7: D

---

Problem 8:

We need four-digit numbers greater than 1000 using the digits of 2012.

The digits are

\[
2,\ 0,\ 1,\ 2.
\]

There are 4 digits, but the digit 2 is repeated. So the total number of distinct arrangements is

\[
\frac{4!}{2!}=\frac{24}{2}=12.
\]

But some arrangements start with 0, and those are not four-digit numbers.

If 0 is first, we arrange the remaining digits \(2,1,2\):

\[
\frac{3!}{2!}=3.
\]

So the number of valid four-digit numbers is

\[
12-3=9.
\]

All of these are greater than 1000, so the answer is 9.

ANSWER 8: D

---

Problem 9:

There are 108 students, and each eats an average of 2 cookies.

So the total number of cookies needed is

\[
108\cdot 2=216.
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough full pans to make at least 216 cookies:

\[
216\div 15=14.4.
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is

\[
15\cdot 2=30.
\]

Eggs are bought by the half-dozen, and one half-dozen is 6 eggs.

\[
30\div 6=5.
\]

So Walter should buy 5 half-dozens.

Choice E, 15, is the number of pans, not the number of half-dozens of eggs.

ANSWER 9: C

---

Problem 10:

We need to simplify

\[
100\times 19.98\times 1.998\times 1000.
\]

Group the factors conveniently:

\[
(100\times 19.98)(1.998\times 1000).
\]

Now compute each group:

\[
100\times 19.98=1998
\]

and

\[
1.998\times 1000=1998.
\]

So the expression becomes

\[
1998\times 1998=(1998)^2.
\]

ANSWER 10: D

---

Problem 11:

You have \(\$10\), but you plan to give the driver a \(\$2\) tip.

So the actual taxi fare can be

\[
10-2=8
\]

dollars.

The first \(\frac12\) mile costs \(\$2.40\).

Money remaining after the first half-mile:

\[
8.00-2.40=5.60.
\]

Additional distance costs \(\$0.20\) per \(0.1\) mile.

The number of additional \(0.1\)-mile segments is

\[
5.60\div 0.20=28.
\]

So the additional distance is

\[
28\cdot 0.1=2.8
\]

miles.

Total distance:

\[
0.5+2.8=3.3
\]

miles.

ANSWER 11: C

---

Problem 12:

The pizza has diameter 12 inches, so its radius is

\[
6
\]

inches.

Six pepperoni circles fit exactly across the diameter of the pizza. Therefore each pepperoni circle has diameter

\[
12\div 6=2
\]

inches, so each pepperoni has radius

\[
1
\]

inch.

Area of the pizza:

\[
\pi\cdot 6^2=36\p

**Problem 1:**
The garbled text `√ 16√ 81 √ 81√ 16` is a result of extracting the mathematical fraction $\frac{\sqrt{16}}{\sqrt{81}}$ (or $\sqrt{\frac{16}{81}}$) from a document. 

We are asked to evaluate this expression:
$\frac{\sqrt{16}}{\sqrt{81}}$

First, we find the square roots of the perfect squares in the numerator and the denominator:
$\sqrt{16} = 4$
$\sqrt{81} = 9$

Substituting these values back into the fraction, we get:
$\frac{4}{9}$

Looking at the choices, this matches option A.

ANSWER 1: A

**Problem 2:**
Let $C$ be the total capacity of the tank in gallons. 
We are told that the tank is initially one-third full, which is $\frac{1}{3}C$. 
When 4 gallons are added, the tank becomes one-half full, which is $\frac{1}{2}C$. 

We can set up the following equation:
$\frac{1}{3}C + 4 = \frac{1}{2}C$

To solve for $C$, we subtract $\frac{1}{3}C$ from both sides:
$4 = \frac{1}{2}C - \frac{1}{3}C$

Find a common denominator to subtract the fractions (which is 6):
$4 = \frac{3}{6}C - \frac{2}{6}C$
$4 = \frac{1}{6}C$

Multiply both sides by 6:
$C = 24$

The capacity of the tank is 24 gallons.

ANSWER 2: D

**Problem 3:**
Let $N$ be the total number of ninth graders and $S$ be the total number of sixth graders.
The number of ninth graders who have a buddy is $\frac{1}{3}N$.
The number of sixth graders who have a buddy is $\frac{2}{5}S$.

Since each paired ninth grader is matched with exactly one sixth grader, the number of paired ninth graders must equal the number of paired sixth graders:
$\frac{1}{3}N = \frac{2}{5}S$

We can express $N$ in terms of $S$ by multiplying both sides by 3:
$N = \frac{6}{5}S$

The total number of students in the program is:
$N + S = \frac{6}{5}S + S = \frac{11}{5}S$

The total number of students who have a buddy is the sum of the paired ninth graders and paired sixth graders:
$\frac{2}{5}S + \frac{2}{5}S = \frac{4}{5}S$

The fraction of the total number of students who have a buddy is:
$\frac{\text{Total paired}}{\text{Total students}} = \frac{\frac{4}{5}S}{\frac{11}{5}S} = \frac{4}{11}$

ANSWER 3: B

**Problem 4:**
We are looking for the number of 4-digit integers (from 1000 to 9999) where all four digits are distinct.
We can determine this by finding the number of choices for each digit place (thousands, hundreds, tens, units):
1. **Thousands digit:** It cannot be 0 (otherwise it wouldn't be a 4-digit number). So, it can be any digit from 1 to 9. That gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, but it must be different from the thousands digit. That gives us **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, but it must be different from the first two digits. That gives us **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, but it must be different from the first three digits. That gives us **7 choices**.

To find the total number of such integers, we multiply the number of choices together:
$9 \times 9 \times 8 \times 7 = 81 \times 56 = 4536$

ANSWER 4: B

**Problem 5:**
Let's find the total difference between Jenny's points and Blake's points across all four tests.
Compared to Blake, Jenny's score differences were:
Test 1: $+10$
Test 2: $-10$
Test 3: $+20$
Test 4: $+20$

Adding these differences together gives the total amount by which Jenny's overall score exceeded Blake's:
$10 - 10 + 20 + 20 = 40$ points

Since the average is the total score divided by the number of tests (4), Jenny's average exceeds Blake's average by:
$\frac{40}{4} = 10$ points

ANSWER 5: A

**Problem 6:**
Let $p$ be the price of a single pencil in cents. Since both Jamar and Sharona bought the same pencils, $p$ must be a common divisor of the amounts they paid (in cents).
Jamar paid $143$ cents, and Sharona paid $187$ cents.

Let's find the prime factorizations of 143 and 187:
$143 = 11 \times 13$
$187 = 11 \times 17$

The common divisors of 143 and 187 are 1 and 11. Since the problem states that the pencils cost *more than a penny each*, the price of one pencil must be $p = 11$ cents.

Now we can find out how many pencils each bought:
Jamar bought: $\frac{143}{11} = 13$ pencils
Sharona bought: $\frac{187}{11} = 17$ pencils

Sharona bought $17 - 13 = 4$ more pencils than Jamar.

ANSWER 6: C

**Problem 7:**
We are given the exchange rate:
$\$1.60 = 3000 \text{ lire}$

To find out how many lire the traveler will receive for $\$1.00$, we divide the amount of lire by $1.60$:
$\text{Lire for } \$1.00 = \frac{3000}{1.60}$

To make the division easier, we can multiply the numerator and the denominator by 10:
$\frac{30000}{16}$

Now, we simplify the fraction:
$\frac{30000}{16} = \frac{15000}{8} = \frac{7500}{4} = \frac{3750}{2} = 1875$

The traveler will receive 1875 lire for $\$1.00$.

ANSWER 7: D

**Problem 8:**
We need to form 4-digit numbers greater than 1000 using the digits 2, 0, 1, and 2.
For the number to be a valid 4-digit number (and greater than 1000), the first digit (thousands place) cannot be 0. Thus, the first digit must be either 1 or 2.

**Case 1: The first digit is 1.**
The remaining digits to arrange are 0, 2, and 2. 
The number of unique arrangements of these three digits is:
$\frac{3!}{2!} = 3$
These numbers are: 1022, 1202, and 1220.

**Case 2: The first digit is 2.**
The remaining digits to arrange are 0, 1, and 2.
The number of unique arrangements of these three distinct digits is:
$3! = 6$
These numbers are: 2012, 2021, 2102, 2120, 2201, and 2210.

Adding the possibilities from both cases gives a total of $3 + 6 = 9$ numbers.

ANSWER 8: D

**Problem 9:**
First, we calculate the total number of cookies needed for the students:
$108 \text{ students} \times 2 \text{ cookies/student} = 216 \text{ cookies}$

Next, we determine how many pans of cookies are required. Each pan makes 15 cookies, and only full recipes (full pans) are made:
$\frac{216}{15} = 14.4$
Since Walter can only make full pans, he must round up and make 15 pans.

Now, we calculate the total number of eggs needed. The recipe calls fo

Problem 1:
What is the value of this expression? √ 16√ 81 √ 81√ 16
To solve this problem, we need to simplify the given expression step by step.

First, let's simplify the square roots: √16 = 4 and √81 = 9.

Now, substitute these values into the expression: 
√ 16√ 81 √ 81√ 16 = 4 × 9 9 × 4

Next, simplify the expression: 
4 × 9 9 × 4 = 36 36 = 1

Let's examine the answer choices to rule out the incorrect ones:
A. 4/9 = 0.44 (not equal to 1)
B. 2/3 = 0.67 (not equal to 1)
C. 1 = 1 (matches our result)
D. 3/2 = 1.5 (not equal to 1)
E. 9/4 = 2.25 (not equal to 1)

The only correct answer is C.

ANSWER 1: C

Problem 2:
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
Let's denote the capacity of the tank as x gallons.

The tank is initially one-third full, so it contains x/3 gallons of liquid.

After adding 4 gallons, the tank becomes one-half full, so it contains x/2 gallons of liquid.

We can set up an equation based on this information: 
x/3 + 4 = x/2

To solve for x, first multiply the entire equation by 6 to eliminate the fractions: 
2x + 24 = 3x

Now, subtract 2x from both sides: 
24 = x

Let's examine the answer choices to rule out the incorrect ones:
A. 8: 8/3 + 4 = 6.67, and 8/2 = 4 (not equal)
B. 12: 12/3 + 4 = 8, and 12/2 = 6 (not equal)
C. 20: 20/3 + 4 = 10.67, and 20/2 = 10 (not equal)
D. 24: 24/3 + 4 = 12, and 24/2 = 12 (matches our result)
E. 48: 48/3 + 4 = 20, and 48/2 = 24 (not equal)

The correct answer is D.

ANSWER 2: D

Problem 3:
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 1/3 of all the ninth graders are paired with 2/5 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Let's denote the number of sixth graders as S and the number of ninth graders as N.

We are given that 1/3 of N is paired with 2/5 of S. Since the number of pairs is the same, we can set up the following equation: 
(1/3)N = (2/5)S

To simplify, multiply both sides by 15 to eliminate the fractions: 
5N = 6S

The total number of students is S + N. We need to find the fraction of students who have a buddy, which is (2/5)S + (1/3)N.

Since (1/3)N = (2/5)S, we can simplify this to 2 × (2/5)S or 2 × (1/3)N.

Let's express N in terms of S: 
5N = 6S => N = (6/5)S

The total number of students is S + (6/5)S = (11/5)S.

The number of students with a buddy is 2 × (2/5)S = (4/5)S.

Now, we can find the fraction of students who have a buddy: 
((4/5)S) / ((11/5)S) = 4/11

Let's examine the answer choices to rule out the incorrect ones:
A. 2/15 = 0.133 (not equal to 4/11)
B. 4/11 = 0.364 (matches our result)
C. 11/30 = 0.367 (not equal to 4/11)
D. 3/8 = 0.375 (not equal to 4/11)
E. 11/15 = 0.733 (not equal to 4/11)

The correct answer is B.

ANSWER 3: B

Problem 4:
How many integers between 1000 and 9999 have four distinct digits?
To solve this problem, let's consider the constraints for each digit.

The first digit cannot be 0, so there are 9 possibilities (1-9).

The second digit can be any digit except the one chosen for the first digit, so there are 9 possibilities.

The third digit can be any digit except the two chosen for the first and second digits, so there are 8 possibilities.

The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 possibilities.

Now, we can calculate the total number of integers with four distinct digits: 
9 × 9 × 8 × 7 = 4536

Let's examine the answer choices to rule out the incorrect ones:
A. 3024 (not equal to 4536)
B. 4536 (matches our result)
C. 5040 (not equal to 4536)
D. 6480 (not equal to 4536)
E. 6561 (not equal to 4536)

The correct answer is B.

ANSWER 4: B

Problem 5:
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Compared with Blake, Jenny scored 10 points higher on the first test, 10 points lower on the second, and 20 points higher on each of the third and fourth.
Let's denote Blake's scores on the four tests as B1, B2, B3, and B4.

Blake's average score is 78, so (B1 + B2 + B3 + B4)/4 = 78.

Jenny's scores are B1 + 10, B2 - 10, B3 + 20, and B4 + 20.

Jenny's total score is (B1 + 10) + (B2 - 10) + (B3 + 20) + (B4 + 20) = B1 + B2 + B3 + B4 + 40.

Since B1 + B2 + B3 + B4 = 78 × 4 = 312, Jenny's total score is 312 + 40 = 352.

Jenny's average score is 352/4 = 88.

The difference between Jenny's average and Blake's average is 88 - 78 = 10.

Let's examine the answer choices to rule out the incorrect ones:
A. 10 (matches our result)
B. 15 (not equal to 10)
C. 20 (not equal to 10)
D. 25 (not equal to 10)
E. 40 (not equal to 10)

The correct answer is A.

ANSWER 5: A

Problem 6:
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
Let's denote the cost of one pencil as x cents.

Since the cost is more than a penny, x > 1.

Jamar paid 143 cents, so the number of pencils he bought is 143/x.

Sharona paid 187 cents, so the number of pencils she bought is 187/x.

The difference in the number of pencils bought is (187/x) - (143/x) = (187 - 143)/x = 44/x.

Since x is a factor of both 143 and 187, we need to find the common factors.

The prime factorization of 143 is 11 × 13, and the prime factorization of 187 is 11 × 17.

The common factor is 11, so x = 11.

Now, we can calculate the difference: 
(187/11) - (143/11) = 17 - 13 = 4

Let's examine the answer choices to rule out the incorrect ones:
A. 2 (not equal to 4)
B. 3 (not equal to 4)
C. 4 (matches our result)
D. 5 (not equal to 4)
E. 6 (not equal to 4)

The correct answer is C.

ANSWER 6: C

Problem 7:
An American traveling in Italy wishes to exchange American dollars for Italian lire. If 3000 lire = $1.60, how much lire will 

Problem 1:

We are evaluating the expression

\[
\frac{\sqrt{16\sqrt{81}}}{\sqrt{81\sqrt{16}}}.
\]

First simplify the square roots inside:

\[
\sqrt{81}=9,\qquad \sqrt{16}=4.
\]

So the numerator is

\[
\sqrt{16\sqrt{81}}=\sqrt{16\cdot 9}=\sqrt{144}=12.
\]

The denominator is

\[
\sqrt{81\sqrt{16}}=\sqrt{81\cdot 4}=\sqrt{324}=18.
\]

Thus the value is

\[
\frac{12}{18}=\frac{2}{3}.
\]

Choice C, \(1\), would come from incorrectly canceling the two expressions as if they were the same, but the numbers inside the square roots are different.

ANSWER 1: B

---

Problem 2:

Let the capacity of the tank be \(x\) gallons.

The tank starts one-third full:

\[
\frac{x}{3}
\]

After adding 4 gallons, it is one-half full:

\[
\frac{x}{2}
\]

So the added amount is the difference:

\[
\frac{x}{2}-\frac{x}{3}=4
\]

Find a common denominator:

\[
\frac{3x}{6}-\frac{2x}{6}=4
\]

\[
\frac{x}{6}=4
\]

\[
x=24
\]

So the tank holds 24 gallons.

ANSWER 2: D

---

Problem 3:

Let the number of paired ninth graders be \(k\). Since \(\frac13\) of all ninth graders are paired, the total number of ninth graders is

\[
3k.
\]

The same \(k\) ninth graders are paired with sixth graders, and this represents \(\frac25\) of all sixth graders. So

\[
\frac25(\text{sixth graders})=k.
\]

Thus the total number of sixth graders is

\[
\frac52k.
\]

The number of students who have a buddy is both members of each pair:

\[
2k.
\]

The total number of sixth and ninth graders is

\[
3k+\frac52k=\frac62k+\frac52k=\frac{11}{2}k.
\]

So the fraction who have a buddy is

\[
\frac{2k}{\frac{11}{2}k}=\frac{2k\cdot 2}{11k}=\frac{4}{11}.
\]

A tempting mistake is to add \(\frac13+\frac25=\frac{11}{15}\), but those fractions are of two different groups, not of the total group.

ANSWER 3: B

---

Problem 4:

We need four-digit integers from 1000 to 9999 with four distinct digits.

For the thousands digit, we can choose any digit from 1 to 9:

\[
9 \text{ choices}
\]

For the hundreds digit, we can choose any digit except the thousands digit. This includes 0, so there are

\[
9 \text{ choices}
\]

For the tens digit, we must avoid the first two digits:

\[
8 \text{ choices}
\]

For the ones digit, we must avoid the first three digits:

\[
7 \text{ choices}
\]

Therefore the total number is

\[
9\cdot 9\cdot 8\cdot 7=4536.
\]

Choice D, 6480, would come from allowing too many choices after the first digit.

ANSWER 4: B

---

Problem 5:

Blake took four tests. Jenny’s scores compared with Blake’s were:

- first test: \(+10\)
- second test: \(-10\)
- third test: \(+20\)
- fourth test: \(+20\)

So Jenny’s total score compared with Blake’s total score is

\[
10-10+20+20=40
\]

points higher.

Since there are 4 tests, Jenny’s average is higher by

\[
\frac{40}{4}=10.
\]

So Jenny’s average exceeds Blake’s by 10 points.

ANSWER 5: A

---

Problem 6:

Jamar paid \(\$1.43=143\) cents.

Sharona paid \(\$1.87=187\) cents.

The pencils cost more than 1 cent each, and both totals must be multiples of the pencil price.

So the pencil price must divide both 143 and 187.

Factor:

\[
143=11\cdot 13
\]

\[
187=11\cdot 17
\]

The common factor greater than 1 is 11, so each pencil costs 11 cents.

Jamar bought

\[
143\div 11=13
\]

pencils.

Sharona bought

\[
187\div 11=17
\]

pencils.

So Sharona bought

\[
17-13=4
\]

more pencils.

ANSWER 6: C

---

Problem 7:

We are told

\[
3000 \text{ lire}=\$1.60.
\]

We want the number of lire for \(\$1.00\).

Set up the rate:

\[
\frac{3000}{1.60}
\]

Since

\[
1.60=\frac{16}{10},
\]

we get

\[
3000\div 1.60=3000\div \frac{16}{10}=3000\cdot \frac{10}{16}.
\]

\[
=30000\div 16=1875.
\]

So the traveler receives 1875 lire for \(\$1.00\).

ANSWER 7: D

---

Problem 8:

We need four-digit numbers greater than 1000 using the digits of 2012.

The digits are

\[
2,\ 0,\ 1,\ 2.
\]

There are 4 digits, but the digit 2 is repeated. So the total number of distinct arrangements is

\[
\frac{4!}{2!}=\frac{24}{2}=12.
\]

But some arrangements start with 0, and those are not four-digit numbers.

If 0 is first, we arrange the remaining digits \(2,1,2\):

\[
\frac{3!}{2!}=3.
\]

So the number of valid four-digit numbers is

\[
12-3=9.
\]

All of these are greater than 1000, so the answer is 9.

ANSWER 8: D

---

Problem 9:

There are 108 students, and each eats an average of 2 cookies.

So the total number of cookies needed is

\[
108\cdot 2=216.
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough full pans to make at least 216 cookies:

\[
216\div 15=14.4.
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is

\[
15\cdot 2=30.
\]

Eggs are bought by the half-dozen, and one half-dozen is 6 eggs.

\[
30\div 6=5.
\]

So Walter should buy 5 half-dozens.

Choice E, 15, is the number of pans, not the number of half-dozens of eggs.

ANSWER 9: C

---

Problem 10:

We need to simplify

\[
100\times 19.98\times 1.998\times 1000.
\]

Group the factors conveniently:

\[
(100\times 19.98)(1.998\times 1000).
\]

Now compute each group:

\[
100\times 19.98=1998
\]

and

\[
1.998\times 1000=1998.
\]

So the expression becomes

\[
1998\times 1998=(1998)^2.
\]

ANSWER 10: D

---

Problem 11:

You have \(\$10\), but you plan to give the driver a \(\$2\) tip.

So the actual taxi fare can be

\[
10-2=8
\]

dollars.

The first \(\frac12\) mile costs \(\$2.40\).

Money remaining after the first half-mile:

\[
8.00-2.40=5.60.
\]

Additional distance costs \(\$0.20\) per \(0.1\) mile.

The number of additional \(0.1\)-mile segments is

\[
5.60\div 0.20=28.
\]

So the additional distance is

\[
28\cdot 0.1=2.8
\]

miles.

Total distance:

\[
0.5+2.8=3.3
\]

miles.

ANSWER 11: C

---

Problem 12:

The pizza has diameter 12 inches, so its radius is

\[
6
\]

inches.

Six pepperoni circles fit exactly across the diameter of the pizza. Therefore each pepperoni circle has diameter

\[
12\div 6=2
\]

inches, so each pepperoni has radius

\[
1
\]

inch.

Area of the pizza:

\[
\pi\cdot 6^2=36\p

**Problem 1:**
The garbled text `√ 16√ 81 √ 81√ 16` is a result of extracting the mathematical fraction $\frac{\sqrt{16}}{\sqrt{81}}$ (or $\sqrt{\frac{16}{81}}$) from a document. 

We are asked to evaluate this expression:
$\frac{\sqrt{16}}{\sqrt{81}}$

First, we find the square roots of the perfect squares in the numerator and the denominator:
$\sqrt{16} = 4$
$\sqrt{81} = 9$

Substituting these values back into the fraction, we get:
$\frac{4}{9}$

Looking at the choices, this matches option A.

ANSWER 1: A

**Problem 2:**
Let $C$ be the total capacity of the tank in gallons. 
We are told that the tank is initially one-third full, which is $\frac{1}{3}C$. 
When 4 gallons are added, the tank becomes one-half full, which is $\frac{1}{2}C$. 

We can set up the following equation:
$\frac{1}{3}C + 4 = \frac{1}{2}C$

To solve for $C$, we subtract $\frac{1}{3}C$ from both sides:
$4 = \frac{1}{2}C - \frac{1}{3}C$

Find a common denominator to subtract the fractions (which is 6):
$4 = \frac{3}{6}C - \frac{2}{6}C$
$4 = \frac{1}{6}C$

Multiply both sides by 6:
$C = 24$

The capacity of the tank is 24 gallons.

ANSWER 2: D

**Problem 3:**
Let $N$ be the total number of ninth graders and $S$ be the total number of sixth graders.
The number of ninth graders who have a buddy is $\frac{1}{3}N$.
The number of sixth graders who have a buddy is $\frac{2}{5}S$.

Since each paired ninth grader is matched with exactly one sixth grader, the number of paired ninth graders must equal the number of paired sixth graders:
$\frac{1}{3}N = \frac{2}{5}S$

We can express $N$ in terms of $S$ by multiplying both sides by 3:
$N = \frac{6}{5}S$

The total number of students in the program is:
$N + S = \frac{6}{5}S + S = \frac{11}{5}S$

The total number of students who have a buddy is the sum of the paired ninth graders and paired sixth graders:
$\frac{2}{5}S + \frac{2}{5}S = \frac{4}{5}S$

The fraction of the total number of students who have a buddy is:
$\frac{\text{Total paired}}{\text{Total students}} = \frac{\frac{4}{5}S}{\frac{11}{5}S} = \frac{4}{11}$

ANSWER 3: B

**Problem 4:**
We are looking for the number of 4-digit integers (from 1000 to 9999) where all four digits are distinct.
We can determine this by finding the number of choices for each digit place (thousands, hundreds, tens, units):
1. **Thousands digit:** It cannot be 0 (otherwise it wouldn't be a 4-digit number). So, it can be any digit from 1 to 9. That gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, but it must be different from the thousands digit. That gives us **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, but it must be different from the first two digits. That gives us **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, but it must be different from the first three digits. That gives us **7 choices**.

To find the total number of such integers, we multiply the number of choices together:
$9 \times 9 \times 8 \times 7 = 81 \times 56 = 4536$

ANSWER 4: B

**Problem 5:**
Let's find the total difference between Jenny's points and Blake's points across all four tests.
Compared to Blake, Jenny's score differences were:
Test 1: $+10$
Test 2: $-10$
Test 3: $+20$
Test 4: $+20$

Adding these differences together gives the total amount by which Jenny's overall score exceeded Blake's:
$10 - 10 + 20 + 20 = 40$ points

Since the average is the total score divided by the number of tests (4), Jenny's average exceeds Blake's average by:
$\frac{40}{4} = 10$ points

ANSWER 5: A

**Problem 6:**
Let $p$ be the price of a single pencil in cents. Since both Jamar and Sharona bought the same pencils, $p$ must be a common divisor of the amounts they paid (in cents).
Jamar paid $143$ cents, and Sharona paid $187$ cents.

Let's find the prime factorizations of 143 and 187:
$143 = 11 \times 13$
$187 = 11 \times 17$

The common divisors of 143 and 187 are 1 and 11. Since the problem states that the pencils cost *more than a penny each*, the price of one pencil must be $p = 11$ cents.

Now we can find out how many pencils each bought:
Jamar bought: $\frac{143}{11} = 13$ pencils
Sharona bought: $\frac{187}{11} = 17$ pencils

Sharona bought $17 - 13 = 4$ more pencils than Jamar.

ANSWER 6: C

**Problem 7:**
We are given the exchange rate:
$\$1.60 = 3000 \text{ lire}$

To find out how many lire the traveler will receive for $\$1.00$, we divide the amount of lire by $1.60$:
$\text{Lire for } \$1.00 = \frac{3000}{1.60}$

To make the division easier, we can multiply the numerator and the denominator by 10:
$\frac{30000}{16}$

Now, we simplify the fraction:
$\frac{30000}{16} = \frac{15000}{8} = \frac{7500}{4} = \frac{3750}{2} = 1875$

The traveler will receive 1875 lire for $\$1.00$.

ANSWER 7: D

**Problem 8:**
We need to form 4-digit numbers greater than 1000 using the digits 2, 0, 1, and 2.
For the number to be a valid 4-digit number (and greater than 1000), the first digit (thousands place) cannot be 0. Thus, the first digit must be either 1 or 2.

**Case 1: The first digit is 1.**
The remaining digits to arrange are 0, 2, and 2. 
The number of unique arrangements of these three digits is:
$\frac{3!}{2!} = 3$
These numbers are: 1022, 1202, and 1220.

**Case 2: The first digit is 2.**
The remaining digits to arrange are 0, 1, and 2.
The number of unique arrangements of these three distinct digits is:
$3! = 6$
These numbers are: 2012, 2021, 2102, 2120, 2201, and 2210.

Adding the possibilities from both cases gives a total of $3 + 6 = 9$ numbers.

ANSWER 8: D

**Problem 9:**
First, we calculate the total number of cookies needed for the students:
$108 \text{ students} \times 2 \text{ cookies/student} = 216 \text{ cookies}$

Next, we determine how many pans of cookies are required. Each pan makes 15 cookies, and only full recipes (full pans) are made:
$\frac{216}{15} = 14.4$
Since Walter can only make full pans, he must round up and make 15 pans.

Now, we calculate the total number of eggs needed. The recipe calls fo

Problem 1:
What is the value of this expression? √ 16√ 81 √ 81√ 16
To solve this problem, we need to simplify the given expression step by step.

First, let's simplify the square roots: √16 = 4 and √81 = 9.

Now, substitute these values into the expression: 
√ 16√ 81 √ 81√ 16 = 4 × 9 9 × 4

Next, simplify the expression: 
4 × 9 9 × 4 = 36 36 = 1

Let's examine the answer choices to rule out the incorrect ones:
A. 4/9 = 0.44 (not equal to 1)
B. 2/3 = 0.67 (not equal to 1)
C. 1 = 1 (matches our result)
D. 3/2 = 1.5 (not equal to 1)
E. 9/4 = 2.25 (not equal to 1)

The only correct answer is C.

ANSWER 1: C

Problem 2:
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
Let's denote the capacity of the tank as x gallons.

The tank is initially one-third full, so it contains x/3 gallons of liquid.

After adding 4 gallons, the tank becomes one-half full, so it contains x/2 gallons of liquid.

We can set up an equation based on this information: 
x/3 + 4 = x/2

To solve for x, first multiply the entire equation by 6 to eliminate the fractions: 
2x + 24 = 3x

Now, subtract 2x from both sides: 
24 = x

Let's examine the answer choices to rule out the incorrect ones:
A. 8: 8/3 + 4 = 6.67, and 8/2 = 4 (not equal)
B. 12: 12/3 + 4 = 8, and 12/2 = 6 (not equal)
C. 20: 20/3 + 4 = 10.67, and 20/2 = 10 (not equal)
D. 24: 24/3 + 4 = 12, and 24/2 = 12 (matches our result)
E. 48: 48/3 + 4 = 20, and 48/2 = 24 (not equal)

The correct answer is D.

ANSWER 2: D

Problem 3:
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 1/3 of all the ninth graders are paired with 2/5 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Let's denote the number of sixth graders as S and the number of ninth graders as N.

We are given that 1/3 of N is paired with 2/5 of S. Since the number of pairs is the same, we can set up the following equation: 
(1/3)N = (2/5)S

To simplify, multiply both sides by 15 to eliminate the fractions: 
5N = 6S

The total number of students is S + N. We need to find the fraction of students who have a buddy, which is (2/5)S + (1/3)N.

Since (1/3)N = (2/5)S, we can simplify this to 2 × (2/5)S or 2 × (1/3)N.

Let's express N in terms of S: 
5N = 6S => N = (6/5)S

The total number of students is S + (6/5)S = (11/5)S.

The number of students with a buddy is 2 × (2/5)S = (4/5)S.

Now, we can find the fraction of students who have a buddy: 
((4/5)S) / ((11/5)S) = 4/11

Let's examine the answer choices to rule out the incorrect ones:
A. 2/15 = 0.133 (not equal to 4/11)
B. 4/11 = 0.364 (matches our result)
C. 11/30 = 0.367 (not equal to 4/11)
D. 3/8 = 0.375 (not equal to 4/11)
E. 11/15 = 0.733 (not equal to 4/11)

The correct answer is B.

ANSWER 3: B

Problem 4:
How many integers between 1000 and 9999 have four distinct digits?
To solve this problem, let's consider the constraints for each digit.

The first digit cannot be 0, so there are 9 possibilities (1-9).

The second digit can be any digit except the one chosen for the first digit, so there are 9 possibilities.

The third digit can be any digit except the two chosen for the first and second digits, so there are 8 possibilities.

The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 possibilities.

Now, we can calculate the total number of integers with four distinct digits: 
9 × 9 × 8 × 7 = 4536

Let's examine the answer choices to rule out the incorrect ones:
A. 3024 (not equal to 4536)
B. 4536 (matches our result)
C. 5040 (not equal to 4536)
D. 6480 (not equal to 4536)
E. 6561 (not equal to 4536)

The correct answer is B.

ANSWER 4: B

Problem 5:
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Compared with Blake, Jenny scored 10 points higher on the first test, 10 points lower on the second, and 20 points higher on each of the third and fourth.
Let's denote Blake's scores on the four tests as B1, B2, B3, and B4.

Blake's average score is 78, so (B1 + B2 + B3 + B4)/4 = 78.

Jenny's scores are B1 + 10, B2 - 10, B3 + 20, and B4 + 20.

Jenny's total score is (B1 + 10) + (B2 - 10) + (B3 + 20) + (B4 + 20) = B1 + B2 + B3 + B4 + 40.

Since B1 + B2 + B3 + B4 = 78 × 4 = 312, Jenny's total score is 312 + 40 = 352.

Jenny's average score is 352/4 = 88.

The difference between Jenny's average and Blake's average is 88 - 78 = 10.

Let's examine the answer choices to rule out the incorrect ones:
A. 10 (matches our result)
B. 15 (not equal to 10)
C. 20 (not equal to 10)
D. 25 (not equal to 10)
E. 40 (not equal to 10)

The correct answer is A.

ANSWER 5: A

Problem 6:
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
Let's denote the cost of one pencil as x cents.

Since the cost is more than a penny, x > 1.

Jamar paid 143 cents, so the number of pencils he bought is 143/x.

Sharona paid 187 cents, so the number of pencils she bought is 187/x.

The difference in the number of pencils bought is (187/x) - (143/x) = (187 - 143)/x = 44/x.

Since x is a factor of both 143 and 187, we need to find the common factors.

The prime factorization of 143 is 11 × 13, and the prime factorization of 187 is 11 × 17.

The common factor is 11, so x = 11.

Now, we can calculate the difference: 
(187/11) - (143/11) = 17 - 13 = 4

Let's examine the answer choices to rule out the incorrect ones:
A. 2 (not equal to 4)
B. 3 (not equal to 4)
C. 4 (matches our result)
D. 5 (not equal to 4)
E. 6 (not equal to 4)

The correct answer is C.

ANSWER 6: C

Problem 7:
An American traveling in Italy wishes to exchange American dollars for Italian lire. If 3000 lire = $1.60, how much lire will 

Problem 1:

We are evaluating the expression

\[
\frac{\sqrt{16\sqrt{81}}}{\sqrt{81\sqrt{16}}}.
\]

First simplify the square roots inside:

\[
\sqrt{81}=9,\qquad \sqrt{16}=4.
\]

So the numerator is

\[
\sqrt{16\sqrt{81}}=\sqrt{16\cdot 9}=\sqrt{144}=12.
\]

The denominator is

\[
\sqrt{81\sqrt{16}}=\sqrt{81\cdot 4}=\sqrt{324}=18.
\]

Thus the value is

\[
\frac{12}{18}=\frac{2}{3}.
\]

Choice C, \(1\), would come from incorrectly canceling the two expressions as if they were the same, but the numbers inside the square roots are different.

ANSWER 1: B

---

Problem 2:

Let the capacity of the tank be \(x\) gallons.

The tank starts one-third full:

\[
\frac{x}{3}
\]

After adding 4 gallons, it is one-half full:

\[
\frac{x}{2}
\]

So the added amount is the difference:

\[
\frac{x}{2}-\frac{x}{3}=4
\]

Find a common denominator:

\[
\frac{3x}{6}-\frac{2x}{6}=4
\]

\[
\frac{x}{6}=4
\]

\[
x=24
\]

So the tank holds 24 gallons.

ANSWER 2: D

---

Problem 3:

Let the number of paired ninth graders be \(k\). Since \(\frac13\) of all ninth graders are paired, the total number of ninth graders is

\[
3k.
\]

The same \(k\) ninth graders are paired with sixth graders, and this represents \(\frac25\) of all sixth graders. So

\[
\frac25(\text{sixth graders})=k.
\]

Thus the total number of sixth graders is

\[
\frac52k.
\]

The number of students who have a buddy is both members of each pair:

\[
2k.
\]

The total number of sixth and ninth graders is

\[
3k+\frac52k=\frac62k+\frac52k=\frac{11}{2}k.
\]

So the fraction who have a buddy is

\[
\frac{2k}{\frac{11}{2}k}=\frac{2k\cdot 2}{11k}=\frac{4}{11}.
\]

A tempting mistake is to add \(\frac13+\frac25=\frac{11}{15}\), but those fractions are of two different groups, not of the total group.

ANSWER 3: B

---

Problem 4:

We need four-digit integers from 1000 to 9999 with four distinct digits.

For the thousands digit, we can choose any digit from 1 to 9:

\[
9 \text{ choices}
\]

For the hundreds digit, we can choose any digit except the thousands digit. This includes 0, so there are

\[
9 \text{ choices}
\]

For the tens digit, we must avoid the first two digits:

\[
8 \text{ choices}
\]

For the ones digit, we must avoid the first three digits:

\[
7 \text{ choices}
\]

Therefore the total number is

\[
9\cdot 9\cdot 8\cdot 7=4536.
\]

Choice D, 6480, would come from allowing too many choices after the first digit.

ANSWER 4: B

---

Problem 5:

Blake took four tests. Jenny’s scores compared with Blake’s were:

- first test: \(+10\)
- second test: \(-10\)
- third test: \(+20\)
- fourth test: \(+20\)

So Jenny’s total score compared with Blake’s total score is

\[
10-10+20+20=40
\]

points higher.

Since there are 4 tests, Jenny’s average is higher by

\[
\frac{40}{4}=10.
\]

So Jenny’s average exceeds Blake’s by 10 points.

ANSWER 5: A

---

Problem 6:

Jamar paid \(\$1.43=143\) cents.

Sharona paid \(\$1.87=187\) cents.

The pencils cost more than 1 cent each, and both totals must be multiples of the pencil price.

So the pencil price must divide both 143 and 187.

Factor:

\[
143=11\cdot 13
\]

\[
187=11\cdot 17
\]

The common factor greater than 1 is 11, so each pencil costs 11 cents.

Jamar bought

\[
143\div 11=13
\]

pencils.

Sharona bought

\[
187\div 11=17
\]

pencils.

So Sharona bought

\[
17-13=4
\]

more pencils.

ANSWER 6: C

---

Problem 7:

We are told

\[
3000 \text{ lire}=\$1.60.
\]

We want the number of lire for \(\$1.00\).

Set up the rate:

\[
\frac{3000}{1.60}
\]

Since

\[
1.60=\frac{16}{10},
\]

we get

\[
3000\div 1.60=3000\div \frac{16}{10}=3000\cdot \frac{10}{16}.
\]

\[
=30000\div 16=1875.
\]

So the traveler receives 1875 lire for \(\$1.00\).

ANSWER 7: D

---

Problem 8:

We need four-digit numbers greater than 1000 using the digits of 2012.

The digits are

\[
2,\ 0,\ 1,\ 2.
\]

There are 4 digits, but the digit 2 is repeated. So the total number of distinct arrangements is

\[
\frac{4!}{2!}=\frac{24}{2}=12.
\]

But some arrangements start with 0, and those are not four-digit numbers.

If 0 is first, we arrange the remaining digits \(2,1,2\):

\[
\frac{3!}{2!}=3.
\]

So the number of valid four-digit numbers is

\[
12-3=9.
\]

All of these are greater than 1000, so the answer is 9.

ANSWER 8: D

---

Problem 9:

There are 108 students, and each eats an average of 2 cookies.

So the total number of cookies needed is

\[
108\cdot 2=216.
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough full pans to make at least 216 cookies:

\[
216\div 15=14.4.
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is

\[
15\cdot 2=30.
\]

Eggs are bought by the half-dozen, and one half-dozen is 6 eggs.

\[
30\div 6=5.
\]

So Walter should buy 5 half-dozens.

Choice E, 15, is the number of pans, not the number of half-dozens of eggs.

ANSWER 9: C

---

Problem 10:

We need to simplify

\[
100\times 19.98\times 1.998\times 1000.
\]

Group the factors conveniently:

\[
(100\times 19.98)(1.998\times 1000).
\]

Now compute each group:

\[
100\times 19.98=1998
\]

and

\[
1.998\times 1000=1998.
\]

So the expression becomes

\[
1998\times 1998=(1998)^2.
\]

ANSWER 10: D

---

Problem 11:

You have \(\$10\), but you plan to give the driver a \(\$2\) tip.

So the actual taxi fare can be

\[
10-2=8
\]

dollars.

The first \(\frac12\) mile costs \(\$2.40\).

Money remaining after the first half-mile:

\[
8.00-2.40=5.60.
\]

Additional distance costs \(\$0.20\) per \(0.1\) mile.

The number of additional \(0.1\)-mile segments is

\[
5.60\div 0.20=28.
\]

So the additional distance is

\[
28\cdot 0.1=2.8
\]

miles.

Total distance:

\[
0.5+2.8=3.3
\]

miles.

ANSWER 11: C

---

Problem 12:

The pizza has diameter 12 inches, so its radius is

\[
6
\]

inches.

Six pepperoni circles fit exactly across the diameter of the pizza. Therefore each pepperoni circle has diameter

\[
12\div 6=2
\]

inches, so each pepperoni has radius

\[
1
\]

inch.

Area of the pizza:

\[
\pi\cdot 6^2=36\p

**Problem 1:**
The garbled text `√ 16√ 81 √ 81√ 16` is a result of extracting the mathematical fraction $\frac{\sqrt{16}}{\sqrt{81}}$ (or $\sqrt{\frac{16}{81}}$) from a document. 

We are asked to evaluate this expression:
$\frac{\sqrt{16}}{\sqrt{81}}$

First, we find the square roots of the perfect squares in the numerator and the denominator:
$\sqrt{16} = 4$
$\sqrt{81} = 9$

Substituting these values back into the fraction, we get:
$\frac{4}{9}$

Looking at the choices, this matches option A.

ANSWER 1: A

**Problem 2:**
Let $C$ be the total capacity of the tank in gallons. 
We are told that the tank is initially one-third full, which is $\frac{1}{3}C$. 
When 4 gallons are added, the tank becomes one-half full, which is $\frac{1}{2}C$. 

We can set up the following equation:
$\frac{1}{3}C + 4 = \frac{1}{2}C$

To solve for $C$, we subtract $\frac{1}{3}C$ from both sides:
$4 = \frac{1}{2}C - \frac{1}{3}C$

Find a common denominator to subtract the fractions (which is 6):
$4 = \frac{3}{6}C - \frac{2}{6}C$
$4 = \frac{1}{6}C$

Multiply both sides by 6:
$C = 24$

The capacity of the tank is 24 gallons.

ANSWER 2: D

**Problem 3:**
Let $N$ be the total number of ninth graders and $S$ be the total number of sixth graders.
The number of ninth graders who have a buddy is $\frac{1}{3}N$.
The number of sixth graders who have a buddy is $\frac{2}{5}S$.

Since each paired ninth grader is matched with exactly one sixth grader, the number of paired ninth graders must equal the number of paired sixth graders:
$\frac{1}{3}N = \frac{2}{5}S$

We can express $N$ in terms of $S$ by multiplying both sides by 3:
$N = \frac{6}{5}S$

The total number of students in the program is:
$N + S = \frac{6}{5}S + S = \frac{11}{5}S$

The total number of students who have a buddy is the sum of the paired ninth graders and paired sixth graders:
$\frac{2}{5}S + \frac{2}{5}S = \frac{4}{5}S$

The fraction of the total number of students who have a buddy is:
$\frac{\text{Total paired}}{\text{Total students}} = \frac{\frac{4}{5}S}{\frac{11}{5}S} = \frac{4}{11}$

ANSWER 3: B

**Problem 4:**
We are looking for the number of 4-digit integers (from 1000 to 9999) where all four digits are distinct.
We can determine this by finding the number of choices for each digit place (thousands, hundreds, tens, units):
1. **Thousands digit:** It cannot be 0 (otherwise it wouldn't be a 4-digit number). So, it can be any digit from 1 to 9. That gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, but it must be different from the thousands digit. That gives us **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, but it must be different from the first two digits. That gives us **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, but it must be different from the first three digits. That gives us **7 choices**.

To find the total number of such integers, we multiply the number of choices together:
$9 \times 9 \times 8 \times 7 = 81 \times 56 = 4536$

ANSWER 4: B

**Problem 5:**
Let's find the total difference between Jenny's points and Blake's points across all four tests.
Compared to Blake, Jenny's score differences were:
Test 1: $+10$
Test 2: $-10$
Test 3: $+20$
Test 4: $+20$

Adding these differences together gives the total amount by which Jenny's overall score exceeded Blake's:
$10 - 10 + 20 + 20 = 40$ points

Since the average is the total score divided by the number of tests (4), Jenny's average exceeds Blake's average by:
$\frac{40}{4} = 10$ points

ANSWER 5: A

**Problem 6:**
Let $p$ be the price of a single pencil in cents. Since both Jamar and Sharona bought the same pencils, $p$ must be a common divisor of the amounts they paid (in cents).
Jamar paid $143$ cents, and Sharona paid $187$ cents.

Let's find the prime factorizations of 143 and 187:
$143 = 11 \times 13$
$187 = 11 \times 17$

The common divisors of 143 and 187 are 1 and 11. Since the problem states that the pencils cost *more than a penny each*, the price of one pencil must be $p = 11$ cents.

Now we can find out how many pencils each bought:
Jamar bought: $\frac{143}{11} = 13$ pencils
Sharona bought: $\frac{187}{11} = 17$ pencils

Sharona bought $17 - 13 = 4$ more pencils than Jamar.

ANSWER 6: C

**Problem 7:**
We are given the exchange rate:
$\$1.60 = 3000 \text{ lire}$

To find out how many lire the traveler will receive for $\$1.00$, we divide the amount of lire by $1.60$:
$\text{Lire for } \$1.00 = \frac{3000}{1.60}$

To make the division easier, we can multiply the numerator and the denominator by 10:
$\frac{30000}{16}$

Now, we simplify the fraction:
$\frac{30000}{16} = \frac{15000}{8} = \frac{7500}{4} = \frac{3750}{2} = 1875$

The traveler will receive 1875 lire for $\$1.00$.

ANSWER 7: D

**Problem 8:**
We need to form 4-digit numbers greater than 1000 using the digits 2, 0, 1, and 2.
For the number to be a valid 4-digit number (and greater than 1000), the first digit (thousands place) cannot be 0. Thus, the first digit must be either 1 or 2.

**Case 1: The first digit is 1.**
The remaining digits to arrange are 0, 2, and 2. 
The number of unique arrangements of these three digits is:
$\frac{3!}{2!} = 3$
These numbers are: 1022, 1202, and 1220.

**Case 2: The first digit is 2.**
The remaining digits to arrange are 0, 1, and 2.
The number of unique arrangements of these three distinct digits is:
$3! = 6$
These numbers are: 2012, 2021, 2102, 2120, 2201, and 2210.

Adding the possibilities from both cases gives a total of $3 + 6 = 9$ numbers.

ANSWER 8: D

**Problem 9:**
First, we calculate the total number of cookies needed for the students:
$108 \text{ students} \times 2 \text{ cookies/student} = 216 \text{ cookies}$

Next, we determine how many pans of cookies are required. Each pan makes 15 cookies, and only full recipes (full pans) are made:
$\frac{216}{15} = 14.4$
Since Walter can only make full pans, he must round up and make 15 pans.

Now, we calculate the total number of eggs needed. The recipe calls fo

Problem 1:
What is the value of this expression? √ 16√ 81 √ 81√ 16
To solve this problem, we need to simplify the given expression step by step.

First, let's simplify the square roots: √16 = 4 and √81 = 9.

Now, substitute these values into the expression: 
√ 16√ 81 √ 81√ 16 = 4 × 9 9 × 4

Next, simplify the expression: 
4 × 9 9 × 4 = 36 36 = 1

Let's examine the answer choices to rule out the incorrect ones:
A. 4/9 = 0.44 (not equal to 1)
B. 2/3 = 0.67 (not equal to 1)
C. 1 = 1 (matches our result)
D. 3/2 = 1.5 (not equal to 1)
E. 9/4 = 2.25 (not equal to 1)

The only correct answer is C.

ANSWER 1: C

Problem 2:
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
Let's denote the capacity of the tank as x gallons.

The tank is initially one-third full, so it contains x/3 gallons of liquid.

After adding 4 gallons, the tank becomes one-half full, so it contains x/2 gallons of liquid.

We can set up an equation based on this information: 
x/3 + 4 = x/2

To solve for x, first multiply the entire equation by 6 to eliminate the fractions: 
2x + 24 = 3x

Now, subtract 2x from both sides: 
24 = x

Let's examine the answer choices to rule out the incorrect ones:
A. 8: 8/3 + 4 = 6.67, and 8/2 = 4 (not equal)
B. 12: 12/3 + 4 = 8, and 12/2 = 6 (not equal)
C. 20: 20/3 + 4 = 10.67, and 20/2 = 10 (not equal)
D. 24: 24/3 + 4 = 12, and 24/2 = 12 (matches our result)
E. 48: 48/3 + 4 = 20, and 48/2 = 24 (not equal)

The correct answer is D.

ANSWER 2: D

Problem 3:
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 1/3 of all the ninth graders are paired with 2/5 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Let's denote the number of sixth graders as S and the number of ninth graders as N.

We are given that 1/3 of N is paired with 2/5 of S. Since the number of pairs is the same, we can set up the following equation: 
(1/3)N = (2/5)S

To simplify, multiply both sides by 15 to eliminate the fractions: 
5N = 6S

The total number of students is S + N. We need to find the fraction of students who have a buddy, which is (2/5)S + (1/3)N.

Since (1/3)N = (2/5)S, we can simplify this to 2 × (2/5)S or 2 × (1/3)N.

Let's express N in terms of S: 
5N = 6S => N = (6/5)S

The total number of students is S + (6/5)S = (11/5)S.

The number of students with a buddy is 2 × (2/5)S = (4/5)S.

Now, we can find the fraction of students who have a buddy: 
((4/5)S) / ((11/5)S) = 4/11

Let's examine the answer choices to rule out the incorrect ones:
A. 2/15 = 0.133 (not equal to 4/11)
B. 4/11 = 0.364 (matches our result)
C. 11/30 = 0.367 (not equal to 4/11)
D. 3/8 = 0.375 (not equal to 4/11)
E. 11/15 = 0.733 (not equal to 4/11)

The correct answer is B.

ANSWER 3: B

Problem 4:
How many integers between 1000 and 9999 have four distinct digits?
To solve this problem, let's consider the constraints for each digit.

The first digit cannot be 0, so there are 9 possibilities (1-9).

The second digit can be any digit except the one chosen for the first digit, so there are 9 possibilities.

The third digit can be any digit except the two chosen for the first and second digits, so there are 8 possibilities.

The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 possibilities.

Now, we can calculate the total number of integers with four distinct digits: 
9 × 9 × 8 × 7 = 4536

Let's examine the answer choices to rule out the incorrect ones:
A. 3024 (not equal to 4536)
B. 4536 (matches our result)
C. 5040 (not equal to 4536)
D. 6480 (not equal to 4536)
E. 6561 (not equal to 4536)

The correct answer is B.

ANSWER 4: B

Problem 5:
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Compared with Blake, Jenny scored 10 points higher on the first test, 10 points lower on the second, and 20 points higher on each of the third and fourth.
Let's denote Blake's scores on the four tests as B1, B2, B3, and B4.

Blake's average score is 78, so (B1 + B2 + B3 + B4)/4 = 78.

Jenny's scores are B1 + 10, B2 - 10, B3 + 20, and B4 + 20.

Jenny's total score is (B1 + 10) + (B2 - 10) + (B3 + 20) + (B4 + 20) = B1 + B2 + B3 + B4 + 40.

Since B1 + B2 + B3 + B4 = 78 × 4 = 312, Jenny's total score is 312 + 40 = 352.

Jenny's average score is 352/4 = 88.

The difference between Jenny's average and Blake's average is 88 - 78 = 10.

Let's examine the answer choices to rule out the incorrect ones:
A. 10 (matches our result)
B. 15 (not equal to 10)
C. 20 (not equal to 10)
D. 25 (not equal to 10)
E. 40 (not equal to 10)

The correct answer is A.

ANSWER 5: A

Problem 6:
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
Let's denote the cost of one pencil as x cents.

Since the cost is more than a penny, x > 1.

Jamar paid 143 cents, so the number of pencils he bought is 143/x.

Sharona paid 187 cents, so the number of pencils she bought is 187/x.

The difference in the number of pencils bought is (187/x) - (143/x) = (187 - 143)/x = 44/x.

Since x is a factor of both 143 and 187, we need to find the common factors.

The prime factorization of 143 is 11 × 13, and the prime factorization of 187 is 11 × 17.

The common factor is 11, so x = 11.

Now, we can calculate the difference: 
(187/11) - (143/11) = 17 - 13 = 4

Let's examine the answer choices to rule out the incorrect ones:
A. 2 (not equal to 4)
B. 3 (not equal to 4)
C. 4 (matches our result)
D. 5 (not equal to 4)
E. 6 (not equal to 4)

The correct answer is C.

ANSWER 6: C

Problem 7:
An American traveling in Italy wishes to exchange American dollars for Italian lire. If 3000 lire = $1.60, how much lire will 

Problem 1:

We are evaluating the expression

\[
\frac{\sqrt{16\sqrt{81}}}{\sqrt{81\sqrt{16}}}.
\]

First simplify the square roots inside:

\[
\sqrt{81}=9,\qquad \sqrt{16}=4.
\]

So the numerator is

\[
\sqrt{16\sqrt{81}}=\sqrt{16\cdot 9}=\sqrt{144}=12.
\]

The denominator is

\[
\sqrt{81\sqrt{16}}=\sqrt{81\cdot 4}=\sqrt{324}=18.
\]

Thus the value is

\[
\frac{12}{18}=\frac{2}{3}.
\]

Choice C, \(1\), would come from incorrectly canceling the two expressions as if they were the same, but the numbers inside the square roots are different.

ANSWER 1: B

---

Problem 2:

Let the capacity of the tank be \(x\) gallons.

The tank starts one-third full:

\[
\frac{x}{3}
\]

After adding 4 gallons, it is one-half full:

\[
\frac{x}{2}
\]

So the added amount is the difference:

\[
\frac{x}{2}-\frac{x}{3}=4
\]

Find a common denominator:

\[
\frac{3x}{6}-\frac{2x}{6}=4
\]

\[
\frac{x}{6}=4
\]

\[
x=24
\]

So the tank holds 24 gallons.

ANSWER 2: D

---

Problem 3:

Let the number of paired ninth graders be \(k\). Since \(\frac13\) of all ninth graders are paired, the total number of ninth graders is

\[
3k.
\]

The same \(k\) ninth graders are paired with sixth graders, and this represents \(\frac25\) of all sixth graders. So

\[
\frac25(\text{sixth graders})=k.
\]

Thus the total number of sixth graders is

\[
\frac52k.
\]

The number of students who have a buddy is both members of each pair:

\[
2k.
\]

The total number of sixth and ninth graders is

\[
3k+\frac52k=\frac62k+\frac52k=\frac{11}{2}k.
\]

So the fraction who have a buddy is

\[
\frac{2k}{\frac{11}{2}k}=\frac{2k\cdot 2}{11k}=\frac{4}{11}.
\]

A tempting mistake is to add \(\frac13+\frac25=\frac{11}{15}\), but those fractions are of two different groups, not of the total group.

ANSWER 3: B

---

Problem 4:

We need four-digit integers from 1000 to 9999 with four distinct digits.

For the thousands digit, we can choose any digit from 1 to 9:

\[
9 \text{ choices}
\]

For the hundreds digit, we can choose any digit except the thousands digit. This includes 0, so there are

\[
9 \text{ choices}
\]

For the tens digit, we must avoid the first two digits:

\[
8 \text{ choices}
\]

For the ones digit, we must avoid the first three digits:

\[
7 \text{ choices}
\]

Therefore the total number is

\[
9\cdot 9\cdot 8\cdot 7=4536.
\]

Choice D, 6480, would come from allowing too many choices after the first digit.

ANSWER 4: B

---

Problem 5:

Blake took four tests. Jenny’s scores compared with Blake’s were:

- first test: \(+10\)
- second test: \(-10\)
- third test: \(+20\)
- fourth test: \(+20\)

So Jenny’s total score compared with Blake’s total score is

\[
10-10+20+20=40
\]

points higher.

Since there are 4 tests, Jenny’s average is higher by

\[
\frac{40}{4}=10.
\]

So Jenny’s average exceeds Blake’s by 10 points.

ANSWER 5: A

---

Problem 6:

Jamar paid \(\$1.43=143\) cents.

Sharona paid \(\$1.87=187\) cents.

The pencils cost more than 1 cent each, and both totals must be multiples of the pencil price.

So the pencil price must divide both 143 and 187.

Factor:

\[
143=11\cdot 13
\]

\[
187=11\cdot 17
\]

The common factor greater than 1 is 11, so each pencil costs 11 cents.

Jamar bought

\[
143\div 11=13
\]

pencils.

Sharona bought

\[
187\div 11=17
\]

pencils.

So Sharona bought

\[
17-13=4
\]

more pencils.

ANSWER 6: C

---

Problem 7:

We are told

\[
3000 \text{ lire}=\$1.60.
\]

We want the number of lire for \(\$1.00\).

Set up the rate:

\[
\frac{3000}{1.60}
\]

Since

\[
1.60=\frac{16}{10},
\]

we get

\[
3000\div 1.60=3000\div \frac{16}{10}=3000\cdot \frac{10}{16}.
\]

\[
=30000\div 16=1875.
\]

So the traveler receives 1875 lire for \(\$1.00\).

ANSWER 7: D

---

Problem 8:

We need four-digit numbers greater than 1000 using the digits of 2012.

The digits are

\[
2,\ 0,\ 1,\ 2.
\]

There are 4 digits, but the digit 2 is repeated. So the total number of distinct arrangements is

\[
\frac{4!}{2!}=\frac{24}{2}=12.
\]

But some arrangements start with 0, and those are not four-digit numbers.

If 0 is first, we arrange the remaining digits \(2,1,2\):

\[
\frac{3!}{2!}=3.
\]

So the number of valid four-digit numbers is

\[
12-3=9.
\]

All of these are greater than 1000, so the answer is 9.

ANSWER 8: D

---

Problem 9:

There are 108 students, and each eats an average of 2 cookies.

So the total number of cookies needed is

\[
108\cdot 2=216.
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough full pans to make at least 216 cookies:

\[
216\div 15=14.4.
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is

\[
15\cdot 2=30.
\]

Eggs are bought by the half-dozen, and one half-dozen is 6 eggs.

\[
30\div 6=5.
\]

So Walter should buy 5 half-dozens.

Choice E, 15, is the number of pans, not the number of half-dozens of eggs.

ANSWER 9: C

---

Problem 10:

We need to simplify

\[
100\times 19.98\times 1.998\times 1000.
\]

Group the factors conveniently:

\[
(100\times 19.98)(1.998\times 1000).
\]

Now compute each group:

\[
100\times 19.98=1998
\]

and

\[
1.998\times 1000=1998.
\]

So the expression becomes

\[
1998\times 1998=(1998)^2.
\]

ANSWER 10: D

---

Problem 11:

You have \(\$10\), but you plan to give the driver a \(\$2\) tip.

So the actual taxi fare can be

\[
10-2=8
\]

dollars.

The first \(\frac12\) mile costs \(\$2.40\).

Money remaining after the first half-mile:

\[
8.00-2.40=5.60.
\]

Additional distance costs \(\$0.20\) per \(0.1\) mile.

The number of additional \(0.1\)-mile segments is

\[
5.60\div 0.20=28.
\]

So the additional distance is

\[
28\cdot 0.1=2.8
\]

miles.

Total distance:

\[
0.5+2.8=3.3
\]

miles.

ANSWER 11: C

---

Problem 12:

The pizza has diameter 12 inches, so its radius is

\[
6
\]

inches.

Six pepperoni circles fit exactly across the diameter of the pizza. Therefore each pepperoni circle has diameter

\[
12\div 6=2
\]

inches, so each pepperoni has radius

\[
1
\]

inch.

Area of the pizza:

\[
\pi\cdot 6^2=36\p

**Problem 1:**
The garbled text `√ 16√ 81 √ 81√ 16` is a result of extracting the mathematical fraction $\frac{\sqrt{16}}{\sqrt{81}}$ (or $\sqrt{\frac{16}{81}}$) from a document. 

We are asked to evaluate this expression:
$\frac{\sqrt{16}}{\sqrt{81}}$

First, we find the square roots of the perfect squares in the numerator and the denominator:
$\sqrt{16} = 4$
$\sqrt{81} = 9$

Substituting these values back into the fraction, we get:
$\frac{4}{9}$

Looking at the choices, this matches option A.

ANSWER 1: A

**Problem 2:**
Let $C$ be the total capacity of the tank in gallons. 
We are told that the tank is initially one-third full, which is $\frac{1}{3}C$. 
When 4 gallons are added, the tank becomes one-half full, which is $\frac{1}{2}C$. 

We can set up the following equation:
$\frac{1}{3}C + 4 = \frac{1}{2}C$

To solve for $C$, we subtract $\frac{1}{3}C$ from both sides:
$4 = \frac{1}{2}C - \frac{1}{3}C$

Find a common denominator to subtract the fractions (which is 6):
$4 = \frac{3}{6}C - \frac{2}{6}C$
$4 = \frac{1}{6}C$

Multiply both sides by 6:
$C = 24$

The capacity of the tank is 24 gallons.

ANSWER 2: D

**Problem 3:**
Let $N$ be the total number of ninth graders and $S$ be the total number of sixth graders.
The number of ninth graders who have a buddy is $\frac{1}{3}N$.
The number of sixth graders who have a buddy is $\frac{2}{5}S$.

Since each paired ninth grader is matched with exactly one sixth grader, the number of paired ninth graders must equal the number of paired sixth graders:
$\frac{1}{3}N = \frac{2}{5}S$

We can express $N$ in terms of $S$ by multiplying both sides by 3:
$N = \frac{6}{5}S$

The total number of students in the program is:
$N + S = \frac{6}{5}S + S = \frac{11}{5}S$

The total number of students who have a buddy is the sum of the paired ninth graders and paired sixth graders:
$\frac{2}{5}S + \frac{2}{5}S = \frac{4}{5}S$

The fraction of the total number of students who have a buddy is:
$\frac{\text{Total paired}}{\text{Total students}} = \frac{\frac{4}{5}S}{\frac{11}{5}S} = \frac{4}{11}$

ANSWER 3: B

**Problem 4:**
We are looking for the number of 4-digit integers (from 1000 to 9999) where all four digits are distinct.
We can determine this by finding the number of choices for each digit place (thousands, hundreds, tens, units):
1. **Thousands digit:** It cannot be 0 (otherwise it wouldn't be a 4-digit number). So, it can be any digit from 1 to 9. That gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, but it must be different from the thousands digit. That gives us **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, but it must be different from the first two digits. That gives us **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, but it must be different from the first three digits. That gives us **7 choices**.

To find the total number of such integers, we multiply the number of choices together:
$9 \times 9 \times 8 \times 7 = 81 \times 56 = 4536$

ANSWER 4: B

**Problem 5:**
Let's find the total difference between Jenny's points and Blake's points across all four tests.
Compared to Blake, Jenny's score differences were:
Test 1: $+10$
Test 2: $-10$
Test 3: $+20$
Test 4: $+20$

Adding these differences together gives the total amount by which Jenny's overall score exceeded Blake's:
$10 - 10 + 20 + 20 = 40$ points

Since the average is the total score divided by the number of tests (4), Jenny's average exceeds Blake's average by:
$\frac{40}{4} = 10$ points

ANSWER 5: A

**Problem 6:**
Let $p$ be the price of a single pencil in cents. Since both Jamar and Sharona bought the same pencils, $p$ must be a common divisor of the amounts they paid (in cents).
Jamar paid $143$ cents, and Sharona paid $187$ cents.

Let's find the prime factorizations of 143 and 187:
$143 = 11 \times 13$
$187 = 11 \times 17$

The common divisors of 143 and 187 are 1 and 11. Since the problem states that the pencils cost *more than a penny each*, the price of one pencil must be $p = 11$ cents.

Now we can find out how many pencils each bought:
Jamar bought: $\frac{143}{11} = 13$ pencils
Sharona bought: $\frac{187}{11} = 17$ pencils

Sharona bought $17 - 13 = 4$ more pencils than Jamar.

ANSWER 6: C

**Problem 7:**
We are given the exchange rate:
$\$1.60 = 3000 \text{ lire}$

To find out how many lire the traveler will receive for $\$1.00$, we divide the amount of lire by $1.60$:
$\text{Lire for } \$1.00 = \frac{3000}{1.60}$

To make the division easier, we can multiply the numerator and the denominator by 10:
$\frac{30000}{16}$

Now, we simplify the fraction:
$\frac{30000}{16} = \frac{15000}{8} = \frac{7500}{4} = \frac{3750}{2} = 1875$

The traveler will receive 1875 lire for $\$1.00$.

ANSWER 7: D

**Problem 8:**
We need to form 4-digit numbers greater than 1000 using the digits 2, 0, 1, and 2.
For the number to be a valid 4-digit number (and greater than 1000), the first digit (thousands place) cannot be 0. Thus, the first digit must be either 1 or 2.

**Case 1: The first digit is 1.**
The remaining digits to arrange are 0, 2, and 2. 
The number of unique arrangements of these three digits is:
$\frac{3!}{2!} = 3$
These numbers are: 1022, 1202, and 1220.

**Case 2: The first digit is 2.**
The remaining digits to arrange are 0, 1, and 2.
The number of unique arrangements of these three distinct digits is:
$3! = 6$
These numbers are: 2012, 2021, 2102, 2120, 2201, and 2210.

Adding the possibilities from both cases gives a total of $3 + 6 = 9$ numbers.

ANSWER 8: D

**Problem 9:**
First, we calculate the total number of cookies needed for the students:
$108 \text{ students} \times 2 \text{ cookies/student} = 216 \text{ cookies}$

Next, we determine how many pans of cookies are required. Each pan makes 15 cookies, and only full recipes (full pans) are made:
$\frac{216}{15} = 14.4$
Since Walter can only make full pans, he must round up and make 15 pans.

Now, we calculate the total number of eggs needed. The recipe calls fo

Problem 1:
What is the value of this expression? √ 16√ 81 √ 81√ 16
To solve this problem, we need to simplify the given expression step by step.

First, let's simplify the square roots: √16 = 4 and √81 = 9.

Now, substitute these values into the expression: 
√ 16√ 81 √ 81√ 16 = 4 × 9 9 × 4

Next, simplify the expression: 
4 × 9 9 × 4 = 36 36 = 1

Let's examine the answer choices to rule out the incorrect ones:
A. 4/9 = 0.44 (not equal to 1)
B. 2/3 = 0.67 (not equal to 1)
C. 1 = 1 (matches our result)
D. 3/2 = 1.5 (not equal to 1)
E. 9/4 = 2.25 (not equal to 1)

The only correct answer is C.

ANSWER 1: C

Problem 2:
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
Let's denote the capacity of the tank as x gallons.

The tank is initially one-third full, so it contains x/3 gallons of liquid.

After adding 4 gallons, the tank becomes one-half full, so it contains x/2 gallons of liquid.

We can set up an equation based on this information: 
x/3 + 4 = x/2

To solve for x, first multiply the entire equation by 6 to eliminate the fractions: 
2x + 24 = 3x

Now, subtract 2x from both sides: 
24 = x

Let's examine the answer choices to rule out the incorrect ones:
A. 8: 8/3 + 4 = 6.67, and 8/2 = 4 (not equal)
B. 12: 12/3 + 4 = 8, and 12/2 = 6 (not equal)
C. 20: 20/3 + 4 = 10.67, and 20/2 = 10 (not equal)
D. 24: 24/3 + 4 = 12, and 24/2 = 12 (matches our result)
E. 48: 48/3 + 4 = 20, and 48/2 = 24 (not equal)

The correct answer is D.

ANSWER 2: D

Problem 3:
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 1/3 of all the ninth graders are paired with 2/5 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Let's denote the number of sixth graders as S and the number of ninth graders as N.

We are given that 1/3 of N is paired with 2/5 of S. Since the number of pairs is the same, we can set up the following equation: 
(1/3)N = (2/5)S

To simplify, multiply both sides by 15 to eliminate the fractions: 
5N = 6S

The total number of students is S + N. We need to find the fraction of students who have a buddy, which is (2/5)S + (1/3)N.

Since (1/3)N = (2/5)S, we can simplify this to 2 × (2/5)S or 2 × (1/3)N.

Let's express N in terms of S: 
5N = 6S => N = (6/5)S

The total number of students is S + (6/5)S = (11/5)S.

The number of students with a buddy is 2 × (2/5)S = (4/5)S.

Now, we can find the fraction of students who have a buddy: 
((4/5)S) / ((11/5)S) = 4/11

Let's examine the answer choices to rule out the incorrect ones:
A. 2/15 = 0.133 (not equal to 4/11)
B. 4/11 = 0.364 (matches our result)
C. 11/30 = 0.367 (not equal to 4/11)
D. 3/8 = 0.375 (not equal to 4/11)
E. 11/15 = 0.733 (not equal to 4/11)

The correct answer is B.

ANSWER 3: B

Problem 4:
How many integers between 1000 and 9999 have four distinct digits?
To solve this problem, let's consider the constraints for each digit.

The first digit cannot be 0, so there are 9 possibilities (1-9).

The second digit can be any digit except the one chosen for the first digit, so there are 9 possibilities.

The third digit can be any digit except the two chosen for the first and second digits, so there are 8 possibilities.

The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 possibilities.

Now, we can calculate the total number of integers with four distinct digits: 
9 × 9 × 8 × 7 = 4536

Let's examine the answer choices to rule out the incorrect ones:
A. 3024 (not equal to 4536)
B. 4536 (matches our result)
C. 5040 (not equal to 4536)
D. 6480 (not equal to 4536)
E. 6561 (not equal to 4536)

The correct answer is B.

ANSWER 4: B

Problem 5:
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Compared with Blake, Jenny scored 10 points higher on the first test, 10 points lower on the second, and 20 points higher on each of the third and fourth.
Let's denote Blake's scores on the four tests as B1, B2, B3, and B4.

Blake's average score is 78, so (B1 + B2 + B3 + B4)/4 = 78.

Jenny's scores are B1 + 10, B2 - 10, B3 + 20, and B4 + 20.

Jenny's total score is (B1 + 10) + (B2 - 10) + (B3 + 20) + (B4 + 20) = B1 + B2 + B3 + B4 + 40.

Since B1 + B2 + B3 + B4 = 78 × 4 = 312, Jenny's total score is 312 + 40 = 352.

Jenny's average score is 352/4 = 88.

The difference between Jenny's average and Blake's average is 88 - 78 = 10.

Let's examine the answer choices to rule out the incorrect ones:
A. 10 (matches our result)
B. 15 (not equal to 10)
C. 20 (not equal to 10)
D. 25 (not equal to 10)
E. 40 (not equal to 10)

The correct answer is A.

ANSWER 5: A

Problem 6:
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
Let's denote the cost of one pencil as x cents.

Since the cost is more than a penny, x > 1.

Jamar paid 143 cents, so the number of pencils he bought is 143/x.

Sharona paid 187 cents, so the number of pencils she bought is 187/x.

The difference in the number of pencils bought is (187/x) - (143/x) = (187 - 143)/x = 44/x.

Since x is a factor of both 143 and 187, we need to find the common factors.

The prime factorization of 143 is 11 × 13, and the prime factorization of 187 is 11 × 17.

The common factor is 11, so x = 11.

Now, we can calculate the difference: 
(187/11) - (143/11) = 17 - 13 = 4

Let's examine the answer choices to rule out the incorrect ones:
A. 2 (not equal to 4)
B. 3 (not equal to 4)
C. 4 (matches our result)
D. 5 (not equal to 4)
E. 6 (not equal to 4)

The correct answer is C.

ANSWER 6: C

Problem 7:
An American traveling in Italy wishes to exchange American dollars for Italian lire. If 3000 lire = $1.60, how much lire will 

Problem 1:

We are evaluating the expression

\[
\frac{\sqrt{16\sqrt{81}}}{\sqrt{81\sqrt{16}}}.
\]

First simplify the square roots inside:

\[
\sqrt{81}=9,\qquad \sqrt{16}=4.
\]

So the numerator is

\[
\sqrt{16\sqrt{81}}=\sqrt{16\cdot 9}=\sqrt{144}=12.
\]

The denominator is

\[
\sqrt{81\sqrt{16}}=\sqrt{81\cdot 4}=\sqrt{324}=18.
\]

Thus the value is

\[
\frac{12}{18}=\frac{2}{3}.
\]

Choice C, \(1\), would come from incorrectly canceling the two expressions as if they were the same, but the numbers inside the square roots are different.

ANSWER 1: B

---

Problem 2:

Let the capacity of the tank be \(x\) gallons.

The tank starts one-third full:

\[
\frac{x}{3}
\]

After adding 4 gallons, it is one-half full:

\[
\frac{x}{2}
\]

So the added amount is the difference:

\[
\frac{x}{2}-\frac{x}{3}=4
\]

Find a common denominator:

\[
\frac{3x}{6}-\frac{2x}{6}=4
\]

\[
\frac{x}{6}=4
\]

\[
x=24
\]

So the tank holds 24 gallons.

ANSWER 2: D

---

Problem 3:

Let the number of paired ninth graders be \(k\). Since \(\frac13\) of all ninth graders are paired, the total number of ninth graders is

\[
3k.
\]

The same \(k\) ninth graders are paired with sixth graders, and this represents \(\frac25\) of all sixth graders. So

\[
\frac25(\text{sixth graders})=k.
\]

Thus the total number of sixth graders is

\[
\frac52k.
\]

The number of students who have a buddy is both members of each pair:

\[
2k.
\]

The total number of sixth and ninth graders is

\[
3k+\frac52k=\frac62k+\frac52k=\frac{11}{2}k.
\]

So the fraction who have a buddy is

\[
\frac{2k}{\frac{11}{2}k}=\frac{2k\cdot 2}{11k}=\frac{4}{11}.
\]

A tempting mistake is to add \(\frac13+\frac25=\frac{11}{15}\), but those fractions are of two different groups, not of the total group.

ANSWER 3: B

---

Problem 4:

We need four-digit integers from 1000 to 9999 with four distinct digits.

For the thousands digit, we can choose any digit from 1 to 9:

\[
9 \text{ choices}
\]

For the hundreds digit, we can choose any digit except the thousands digit. This includes 0, so there are

\[
9 \text{ choices}
\]

For the tens digit, we must avoid the first two digits:

\[
8 \text{ choices}
\]

For the ones digit, we must avoid the first three digits:

\[
7 \text{ choices}
\]

Therefore the total number is

\[
9\cdot 9\cdot 8\cdot 7=4536.
\]

Choice D, 6480, would come from allowing too many choices after the first digit.

ANSWER 4: B

---

Problem 5:

Blake took four tests. Jenny’s scores compared with Blake’s were:

- first test: \(+10\)
- second test: \(-10\)
- third test: \(+20\)
- fourth test: \(+20\)

So Jenny’s total score compared with Blake’s total score is

\[
10-10+20+20=40
\]

points higher.

Since there are 4 tests, Jenny’s average is higher by

\[
\frac{40}{4}=10.
\]

So Jenny’s average exceeds Blake’s by 10 points.

ANSWER 5: A

---

Problem 6:

Jamar paid \(\$1.43=143\) cents.

Sharona paid \(\$1.87=187\) cents.

The pencils cost more than 1 cent each, and both totals must be multiples of the pencil price.

So the pencil price must divide both 143 and 187.

Factor:

\[
143=11\cdot 13
\]

\[
187=11\cdot 17
\]

The common factor greater than 1 is 11, so each pencil costs 11 cents.

Jamar bought

\[
143\div 11=13
\]

pencils.

Sharona bought

\[
187\div 11=17
\]

pencils.

So Sharona bought

\[
17-13=4
\]

more pencils.

ANSWER 6: C

---

Problem 7:

We are told

\[
3000 \text{ lire}=\$1.60.
\]

We want the number of lire for \(\$1.00\).

Set up the rate:

\[
\frac{3000}{1.60}
\]

Since

\[
1.60=\frac{16}{10},
\]

we get

\[
3000\div 1.60=3000\div \frac{16}{10}=3000\cdot \frac{10}{16}.
\]

\[
=30000\div 16=1875.
\]

So the traveler receives 1875 lire for \(\$1.00\).

ANSWER 7: D

---

Problem 8:

We need four-digit numbers greater than 1000 using the digits of 2012.

The digits are

\[
2,\ 0,\ 1,\ 2.
\]

There are 4 digits, but the digit 2 is repeated. So the total number of distinct arrangements is

\[
\frac{4!}{2!}=\frac{24}{2}=12.
\]

But some arrangements start with 0, and those are not four-digit numbers.

If 0 is first, we arrange the remaining digits \(2,1,2\):

\[
\frac{3!}{2!}=3.
\]

So the number of valid four-digit numbers is

\[
12-3=9.
\]

All of these are greater than 1000, so the answer is 9.

ANSWER 8: D

---

Problem 9:

There are 108 students, and each eats an average of 2 cookies.

So the total number of cookies needed is

\[
108\cdot 2=216.
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough full pans to make at least 216 cookies:

\[
216\div 15=14.4.
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is

\[
15\cdot 2=30.
\]

Eggs are bought by the half-dozen, and one half-dozen is 6 eggs.

\[
30\div 6=5.
\]

So Walter should buy 5 half-dozens.

Choice E, 15, is the number of pans, not the number of half-dozens of eggs.

ANSWER 9: C

---

Problem 10:

We need to simplify

\[
100\times 19.98\times 1.998\times 1000.
\]

Group the factors conveniently:

\[
(100\times 19.98)(1.998\times 1000).
\]

Now compute each group:

\[
100\times 19.98=1998
\]

and

\[
1.998\times 1000=1998.
\]

So the expression becomes

\[
1998\times 1998=(1998)^2.
\]

ANSWER 10: D

---

Problem 11:

You have \(\$10\), but you plan to give the driver a \(\$2\) tip.

So the actual taxi fare can be

\[
10-2=8
\]

dollars.

The first \(\frac12\) mile costs \(\$2.40\).

Money remaining after the first half-mile:

\[
8.00-2.40=5.60.
\]

Additional distance costs \(\$0.20\) per \(0.1\) mile.

The number of additional \(0.1\)-mile segments is

\[
5.60\div 0.20=28.
\]

So the additional distance is

\[
28\cdot 0.1=2.8
\]

miles.

Total distance:

\[
0.5+2.8=3.3
\]

miles.

ANSWER 11: C

---

Problem 12:

The pizza has diameter 12 inches, so its radius is

\[
6
\]

inches.

Six pepperoni circles fit exactly across the diameter of the pizza. Therefore each pepperoni circle has diameter

\[
12\div 6=2
\]

inches, so each pepperoni has radius

\[
1
\]

inch.

Area of the pizza:

\[
\pi\cdot 6^2=36\p

**Problem 1:**
The garbled text `√ 16√ 81 √ 81√ 16` is a result of extracting the mathematical fraction $\frac{\sqrt{16}}{\sqrt{81}}$ (or $\sqrt{\frac{16}{81}}$) from a document. 

We are asked to evaluate this expression:
$\frac{\sqrt{16}}{\sqrt{81}}$

First, we find the square roots of the perfect squares in the numerator and the denominator:
$\sqrt{16} = 4$
$\sqrt{81} = 9$

Substituting these values back into the fraction, we get:
$\frac{4}{9}$

Looking at the choices, this matches option A.

ANSWER 1: A

**Problem 2:**
Let $C$ be the total capacity of the tank in gallons. 
We are told that the tank is initially one-third full, which is $\frac{1}{3}C$. 
When 4 gallons are added, the tank becomes one-half full, which is $\frac{1}{2}C$. 

We can set up the following equation:
$\frac{1}{3}C + 4 = \frac{1}{2}C$

To solve for $C$, we subtract $\frac{1}{3}C$ from both sides:
$4 = \frac{1}{2}C - \frac{1}{3}C$

Find a common denominator to subtract the fractions (which is 6):
$4 = \frac{3}{6}C - \frac{2}{6}C$
$4 = \frac{1}{6}C$

Multiply both sides by 6:
$C = 24$

The capacity of the tank is 24 gallons.

ANSWER 2: D

**Problem 3:**
Let $N$ be the total number of ninth graders and $S$ be the total number of sixth graders.
The number of ninth graders who have a buddy is $\frac{1}{3}N$.
The number of sixth graders who have a buddy is $\frac{2}{5}S$.

Since each paired ninth grader is matched with exactly one sixth grader, the number of paired ninth graders must equal the number of paired sixth graders:
$\frac{1}{3}N = \frac{2}{5}S$

We can express $N$ in terms of $S$ by multiplying both sides by 3:
$N = \frac{6}{5}S$

The total number of students in the program is:
$N + S = \frac{6}{5}S + S = \frac{11}{5}S$

The total number of students who have a buddy is the sum of the paired ninth graders and paired sixth graders:
$\frac{2}{5}S + \frac{2}{5}S = \frac{4}{5}S$

The fraction of the total number of students who have a buddy is:
$\frac{\text{Total paired}}{\text{Total students}} = \frac{\frac{4}{5}S}{\frac{11}{5}S} = \frac{4}{11}$

ANSWER 3: B

**Problem 4:**
We are looking for the number of 4-digit integers (from 1000 to 9999) where all four digits are distinct.
We can determine this by finding the number of choices for each digit place (thousands, hundreds, tens, units):
1. **Thousands digit:** It cannot be 0 (otherwise it wouldn't be a 4-digit number). So, it can be any digit from 1 to 9. That gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, but it must be different from the thousands digit. That gives us **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, but it must be different from the first two digits. That gives us **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, but it must be different from the first three digits. That gives us **7 choices**.

To find the total number of such integers, we multiply the number of choices together:
$9 \times 9 \times 8 \times 7 = 81 \times 56 = 4536$

ANSWER 4: B

**Problem 5:**
Let's find the total difference between Jenny's points and Blake's points across all four tests.
Compared to Blake, Jenny's score differences were:
Test 1: $+10$
Test 2: $-10$
Test 3: $+20$
Test 4: $+20$

Adding these differences together gives the total amount by which Jenny's overall score exceeded Blake's:
$10 - 10 + 20 + 20 = 40$ points

Since the average is the total score divided by the number of tests (4), Jenny's average exceeds Blake's average by:
$\frac{40}{4} = 10$ points

ANSWER 5: A

**Problem 6:**
Let $p$ be the price of a single pencil in cents. Since both Jamar and Sharona bought the same pencils, $p$ must be a common divisor of the amounts they paid (in cents).
Jamar paid $143$ cents, and Sharona paid $187$ cents.

Let's find the prime factorizations of 143 and 187:
$143 = 11 \times 13$
$187 = 11 \times 17$

The common divisors of 143 and 187 are 1 and 11. Since the problem states that the pencils cost *more than a penny each*, the price of one pencil must be $p = 11$ cents.

Now we can find out how many pencils each bought:
Jamar bought: $\frac{143}{11} = 13$ pencils
Sharona bought: $\frac{187}{11} = 17$ pencils

Sharona bought $17 - 13 = 4$ more pencils than Jamar.

ANSWER 6: C

**Problem 7:**
We are given the exchange rate:
$\$1.60 = 3000 \text{ lire}$

To find out how many lire the traveler will receive for $\$1.00$, we divide the amount of lire by $1.60$:
$\text{Lire for } \$1.00 = \frac{3000}{1.60}$

To make the division easier, we can multiply the numerator and the denominator by 10:
$\frac{30000}{16}$

Now, we simplify the fraction:
$\frac{30000}{16} = \frac{15000}{8} = \frac{7500}{4} = \frac{3750}{2} = 1875$

The traveler will receive 1875 lire for $\$1.00$.

ANSWER 7: D

**Problem 8:**
We need to form 4-digit numbers greater than 1000 using the digits 2, 0, 1, and 2.
For the number to be a valid 4-digit number (and greater than 1000), the first digit (thousands place) cannot be 0. Thus, the first digit must be either 1 or 2.

**Case 1: The first digit is 1.**
The remaining digits to arrange are 0, 2, and 2. 
The number of unique arrangements of these three digits is:
$\frac{3!}{2!} = 3$
These numbers are: 1022, 1202, and 1220.

**Case 2: The first digit is 2.**
The remaining digits to arrange are 0, 1, and 2.
The number of unique arrangements of these three distinct digits is:
$3! = 6$
These numbers are: 2012, 2021, 2102, 2120, 2201, and 2210.

Adding the possibilities from both cases gives a total of $3 + 6 = 9$ numbers.

ANSWER 8: D

**Problem 9:**
First, we calculate the total number of cookies needed for the students:
$108 \text{ students} \times 2 \text{ cookies/student} = 216 \text{ cookies}$

Next, we determine how many pans of cookies are required. Each pan makes 15 cookies, and only full recipes (full pans) are made:
$\frac{216}{15} = 14.4$
Since Walter can only make full pans, he must round up and make 15 pans.

Now, we calculate the total number of eggs needed. The recipe calls fo