🔬 Model Lab

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** How many games must be played to determine a winner in a 16-team elimination tournament?

**Setup:** In an elimination tournament, each game eliminates exactly one team. To go from 16 teams to 1 winner, we must eliminate 15 teams. Therefore, we need 15 games.

**Alternatively:** We can think of it as a bracket:
- Round 1: 16 teams → 8 games → 8 winners
- Round 2: 8 teams → 4 games → 4 winners
- Round 3: 4 teams → 2 games → 2 winners
- Round 4: 2 teams → 1 game → 1 winner
- Total: 8 + 4 + 2 + 1 = 15 games

ANSWER 1: D

---

**Problem 2:**

**What's being asked:** Which person gets the largest final answer after their operations?

**Jose's calculation:** 
- Start: 10
- Subtract 1: 10 - 1 = 9
- Double: 9 × 2 = 18
- Add 2: 18 + 2 = **20**

**Thuy's calculation:**
- Start: 10
- Double: 10 × 2 = 20
- Subtract 1: 20 - 1 = 19
- Add 2: 19 + 2 = **21**

**Kareem's calculation:**
- Start: 10
- Subtract 1: 10 - 1 = 9
- Add 2: 9 + 2 = 11
- Double: 11 × 2 = **22**

Kareem gets 22, which is the largest.

ANSWER 2: C

---

**Problem 3:**

**What's being asked:** Convert the fraction 2/25 to a decimal.

**Setup:** 2/25 = ?

**Calculation:** 
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** In how many ways can 3 starters be chosen from 4 people (Lance, Sally, Joy, Fred)?

**Setup:** This is a combination problem: $\binom{4}{3}$

**Calculation:**
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4}{1} = 4$$

Alternatively, we're choosing which 1 person to leave out: 4 ways.

ANSWER 4: B

---

**Problem 5:**

**What's being asked:** If 20% of a number is 12, what is 30% of that number?

**Setup:** Let the number be $n$.
- $0.20n = 12$
- $n = 12 ÷ 0.20 = 60$

**Calculation:** 
$$30\% \text{ of } 60 = 0.30 \times 60 = 18$$

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Starting pointing west, after rotating clockwise 2¼ revolutions then counterclockwise 3¾ revolutions, which direction?

**Setup:** 
- Clockwise 2¼ revolutions = 2.25 revolutions
- Counterclockwise 3¾ revolutions = 3.75 revolutions
- Net: 3.75 - 2.25 = 1.5 revolutions counterclockwise (net)

**Calculation:**
Starting at west, rotating counterclockwise 1.5 revolutions (= 1 full rotation + 0.5 rotation):
- 1 full rotation brings us back to west
- 0.5 rotation = 180°
- West + 180° counterclockwise = East

ANSWER 6: B

---

**Problem 7:**

**What's being asked:** After adding 5 liters of yellow tint to a 30-liter mixture, what percent is yellow tint?

**Original mixture (30 liters):**
- Yellow tint: 30% × 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percentage:**
$$\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%$$

ANSWER 7: C

---

**Problem 8:**

**What's being asked:** Order 15/11, 19/15, and 17/13 from least to greatest.

**Setup:** Convert to decimals or compare directly.

**Calculations:**
- 15/11 ≈ 1.364
- 19/15 ≈ 1.267
- 17/13 ≈ 1.308

**Order:** 19/15 < 17/13 < 15/11

ANSWER 8: E

---

**Problem 9:**

**What's being asked:** Find $x$ such that the mean, median, and unique mode of {3, 4, 5, 6, 6, 7, $x$} are all equal.

**Current mode:** 6 appears twice; all others appear once. If there's a unique mode, it must be 6.

**With unique mode = 6:** This requires no other number to appear more than once. So $x \notin \{3, 4, 5, 7\}$ and $x \neq 6$ (else it wouldn't be unique).

**Mean = Median = Mode = 6:**

For median of 7 numbers with value 6:
- Sorted: need middle (4th) value to be 6
- If $x = 11$: {3, 4, 5, 6, 6, 7, 11} → median = 6 ✓

Mean: $\frac{3+4+5+6+6+7+11}{7} = \frac{42}{7} = 6$ ✓

Mode: 6 appears twice; all others once → mode = 6 ✓

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Total time from home to school if Joe walked halfway (6 min) then ran the rest at 3× speed?

**Setup:**
- Walking: 6 minutes for half the distance
- Running: same distance at 3× speed, so time = 6/3 = 2 minutes

**Total time:** 6 + 2 = 8 minutes

ANSWER 10: D

---

**Problem 11:**

**What's being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Calculation:**
- First group: 1 + 11 + 21 + 31 + 41 = 105
- Second group: 9 + 19 + 29 + 39 + 49 = 145
- Sum: 105 + 145 = 250

ANSWER 11: E

---

**Problem 12:**

**What's being asked:** Which number is largest?

**Comparison:**
- A. 0.99
- B. 0.9099
- C. 0.9
- D. 0.909
- E. 0.9009

Starting from the tenths place: all are 0.9__. Looking at hundredths:
- 0.99 has 9 in hundredths place → **largest**

ANSWER 12: A

Problem 1:

In a single-elimination tournament, each game eliminates exactly one team.  
There are 16 teams, and we want one champion.

So we must eliminate 15 teams total.

Therefore, the number of games played is 15.

Tempting wrong choice: 16 would be if every team had to lose, but the winner does not lose.

ANSWER 1: D

Problem 2:

Compute each person’s result carefully.

- Jose: start with 10  
  subtract 1 → 9  
  double → 18  
  add 2 → 20

- Thuy: start with 10  
  double → 20  
  subtract 1 → 19  
  add 2 → 21

- Kareem: start with 10  
  subtract 1 → 9  
  add 2 → 11  
  double → 22

Compare:
- Jose = 20
- Thuy = 21
- Kareem = 22

Kareem gets the largest final answer.

ANSWER 2: C

Problem 3:

\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

So the correct decimal is 0.08.

ANSWER 3: B

Problem 4:

We are choosing 3 starters from 4 players: Lance, Sally, Joy, and Fred.

This is a combination problem:
\[
\binom{4}{3} = 4
\]

You can also see it by noticing that choosing 3 starters is the same as choosing which 1 player sits out, and there are 4 choices.

ANSWER 4: B

Problem 5:

Let the number be \(n\).

20% of \(n\) is 12, so:
\[
0.2n = 12
\]
\[
n = 60
\]

Now find 30% of 60:
\[
0.3 \times 60 = 18
\]

ANSWER 5: B

Problem 6:

The spinner starts pointing west.

First move: clockwise \(2\frac{1}{4}\) revolutions.  
A full revolution brings it back to west, so only the extra \(\frac{1}{4}\) revolution matters.

From west, a quarter-turn clockwise points it to north.

Second move: counterclockwise \(3\frac{3}{4}\) revolutions.  
Again, the 3 full revolutions do nothing, so this is just \(\frac{3}{4}\) of a revolution counterclockwise, which is the same as \(\frac{1}{4}\) of a revolution clockwise.

From north, a quarter-turn clockwise points east.

So the spinner ends pointing east.

ANSWER 6: B

Problem 7:

Start with 30 liters total.

Yellow tint is 30% of 30 liters:
\[
0.30 \times 30 = 9 \text{ liters}
\]

Then 5 liters of yellow tint are added, so yellow becomes:
\[
9 + 5 = 14 \text{ liters}
\]

New total mixture:
\[
30 + 5 = 35 \text{ liters}
\]

Percent yellow in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.4 = 40\%
\]

ANSWER 7: C

Problem 8:

The fractions are:
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]

Approximate each one:

\[
\frac{15}{11} \approx 1.364
\]
\[
\frac{17}{13} \approx 1.308
\]
\[
\frac{19}{15} \approx 1.267
\]

So from least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

That matches choice E.

ANSWER 8: E

Problem 9:

The numbers are:
\[
3,4,5,6,6,7,x
\]

We are told the mean, median, and unique mode are all equal.

Since 6 appears twice and every other listed number appears once, the unique mode is 6.

So the median and mean must also be 6.

The median of 7 numbers is the 4th number when ordered.  
The sorted list is:
\[
3,4,5,6,6,7,x
\]
To make the median 6, \(x\) must be 6 or larger, but we also need the mean to be 6.

Sum of the known numbers:
\[
3+4+5+6+6+7 = 31
\]

For the mean to be 6, total sum must be:
\[
7 \times 6 = 42
\]

So:
\[
31 + x = 42
\]
\[
x = 11
\]

Check:
- mode = 6
- median = 6
- mean = 42/7 = 6

Works perfectly.

ANSWER 9: D

Problem 10:

Joe walked half the distance in 6 minutes.  
He then ran the other half at 3 times his walking speed.

If he runs 3 times as fast, then the time for the same distance is 1/3 as long.

So the running time is:
\[
6 \div 3 = 2 \text{ minutes}
\]

Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

Problem 11:

Add the first group:
\[
1+11+21+31+41 = 105
\]

Add the second group:
\[
9+19+29+39+49 = 145
\]

Now add the totals:
\[
105 + 145 = 250
\]

ANSWER 11: E

Problem 12:

Compare the decimals:

- \(0.99\)
- \(0.9099\)
- \(0.9 = 0.9000\)
- \(0.909\)
- \(0.9009\)

The largest is clearly \(0.99\), since it is closest to 1.

ANSWER 12: A

Problem 1:  
We have a three-person-team tournament where **the losing team of each game is eliminated**. With **16 teams**, how many games are played to determine the winner?

To find the number of games in a single-elimination tournament:  
- Each game eliminates exactly **one** team (the loser).  
- To go from **16 teams to 1 winning team**, we must eliminate **15 teams**.  
- Therefore, the number of games is **15**.

So the correct choice is **D**.

ANSWER 1: D


Problem 2:  
Jose, Thuy, and Kareem all start at 10. We compute each final value.

**Jose:**  
- Subtract 1: \(10 - 1 = 9\)  
- Double: \(2 \cdot 9 = 18\)  
- Add 2: \(18 + 2 = 20\)  
Final: **20**

**Thuy:**  
- Double 10: \(2 \cdot 10 = 20\)  
- Subtract 1: \(20 - 1 = 19\)  
- Add 2: \(19 + 2 = 21\)  
Final: **21**

**Kareem:**  
- Subtract 1: \(10 - 1 = 9\)  
- Add 2: \(9 + 2 = 11\)  
- Double: \(2 \cdot 11 = 22\)  
Final: **22**

Largest is **22**, which is Kareem.

ANSWER 2: C


Problem 3:  
We need to evaluate:  
\[
\frac{2}{25}
\]
Convert to decimals:  
\[
\frac{2}{25} = 2 \div 25
\]
Since \(25 \cdot 0.08 = 2\):  
\[
2 \div 25 = 0.08
\]
So the correct choice is **B** (.08).

ANSWER 3: B


Problem 4:  
We have 4 people chosen: Lance, Sally, Joy, Fred. We need to choose **3 starters**.

Number of ways:
\[
\binom{4}{3} = 4
\]
List them conceptually: leaving out each one individually gives 4 choices.

Correct choice: **B**.

ANSWER 4: B


Problem 5:  
“If 20% of a number is 12, what is 30% of the same number?”

Let the number be \(N\).  
\[
0.20N = 12
\]
\[
N = \frac{12}{0.20} = 60
\]
Now compute \(30\%\) of 60:
\[
0.30 \cdot 60 = 18
\]
Correct choice: **B**.

ANSWER 5: B


Problem 6:  
Spinner starts pointing **west**.

Chenille moves it:
- **Clockwise** \(2\frac{1}{4}\) revolutions  
- then **Counterclockwise** \(3\frac{3}{4}\) revolutions

Net rotation (clockwise minus counterclockwise):
\[
2\frac{1}{4} - 3\frac{3}{4}
\]
Convert to improper fractions:
\[
2\frac{1}{4} = \frac{9}{4}, \quad 3\frac{3}{4} = \frac{15}{4}
\]
\[
\frac{9}{4} - \frac{15}{4} = -\frac{6}{4} = -\frac{3}{2}
\]
The negative means **net counterclockwise** by \(\frac{3}{2}\) revolutions.

\(\frac{3}{2} = 1 + \frac{1}{2}\), so this is:
- 1 full turn (brings it back where it started)
- plus \(\frac{1}{2}\) turn counterclockwise

Half a revolution from **west** points **east** (opposite direction).

Correct answer: **B (east)**.

ANSWER 6: B


Problem 7:  
Original mixture: 30 L with  
- 25% red (doesn’t matter directly)
- 30% yellow  
- 45% water

Yellow amount initially:
\[
0.30 \cdot 30 = 9 \text{ liters}
\]
Add 5 liters of yellow tint, so new yellow amount:
\[
9 + 5 = 14
\]
New total volume:
\[
30 + 5 = 35
\]
Percent yellow in new mixture:
\[
\frac{14}{35} = 0.4 = 40\%
\]
Correct choice: **C**.

ANSWER 7: C


Problem 8:  
Order fractions least to greatest:
\[
\frac{15}{11},\ \frac{19}{15},\ \frac{17}{13}
\]
A good way: compare values by cross-multiplication or decimal approximations.

Compute approximate decimals:
\[
\frac{15}{11} \approx 1.3636
\]
\[
\frac{19}{15} \approx 1.2667
\]
\[
\frac{17}{13} \approx 1.3077
\]
Least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]
Check choice: that matches **E**:
- \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11}\)

ANSWER 8: E


Problem 9:  
Numbers: \(3,4,5,6,6,7,x\) (7 positive integers).  
Mean = median = unique mode.

**Step 1: Mode**  
The only number that repeats is **6**, occurring twice. So the **unique mode = 6**.

Thus median must be 6 as well.

**Step 2: Median = 6**  
For 7 numbers sorted, the median is the **4th** number.

Sorted list must have the 4th term equal to 6. We already have two 6’s and surrounding numbers 3,4,5,7.

Currently without \(x\), the ordered portion is:
\[
3,4,5,6,6,7,\ ?
\]
The 4th number here is 6 already. To keep the 4th number as 6, \(x\) must not be less than 6 (otherwise it would shift positions so 4th might change).

If \(x = 6\), then 6 occurs 3 times, still mode 6 (unique mode). Then list:
\[
3,4,5,6,6,6,7
\]
Median (4th) is 6—works.

If \(x=7\), list:
\[
3,4,5,6,6,7,7
\]
Mode would be 6? Actually 7 repeats too, so mode not unique.

If \(x>7\), list:
\[
3,4,5,6,6,7,x
\]
Median still 6, but mode would still be 6 (unique), since only 6 repeats. That would also work—wait, but then 6 is still unique mode, yes. However we also need **mean = 6**.

So use mean condition.

Sum of known numbers:
\[
3+4+5+6+6+7 = 31
\]
Total sum including \(x\): \(31 + x\)

Mean:
\[
\frac{31+x}{7} = 6
\]
\[
31 + x = 42
\]
\[
x = 11
\]

Check uniqueness of mode: 6 occurs twice, no other repeats (since x=11 is different). Unique mode is 6—consistent.

ANSWER 9: D


Problem 10:  
Joe walks half the distance, then runs the rest.  
He runs **3 times as fast** as he walks.

Given: time to walk half way = **6 minutes**.  
Let walking speed be \(v\). Then walking time for half distance \(=\) 6 minutes.

Let half the distance be \(d\). Then:
\[
d = v \cdot 6
\]
Running speed is \(3v\).  
Remaining distance is also \(d\), so running time:
\[
\text{time} = \frac{d}{3v} = \frac{v\cdot 6}{3v} = 2
\]
Total time:
\[
6 + 2 = 8
\]
Correct choice: **D**.

ANSWER 10: D


Problem 11:  
Compute:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]

Group each set:

First set:
\[
1+11+21+31+41
\]
These are all of the form \(10k+1\):  
\(1, 11, 21, 31, 41\) correspond to \(k=0,1,2,3,4\).

Sum:
\[
(1+11+21+31+41) = (1+11+21+31+41)
\]
Compute directly:
\[
(1+11)=12,\ (21+31)=52,\ 12+52=64,\ 64+41=105
\]

Second set:
\[
9+19+29+39+49
\]
Direct sum:
\[
(9+19)=28,\ (29+39)=68,\ 28+68=96,\ 96+49=145
\]

Total:
\[
105 + 145 = 250
\]

Correct choice: **E**.

ANSWER 11: E


Problem 12:  
Which number is largest?

Options:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare decimals:  
- .99 is the largest among all because it is greater than .9 (and greater than any number starting 0.90... but with fewer 9s afterward).

So .99 > .9099 > .909 > .9009 > .9.

Correct choice: **A**.

ANSWER 12: A

Problem 1:
In a single-elimination tournament, every game results in exactly one team being eliminated. To determine a single winner from 16 teams, 15 teams must be eliminated. Since each game eliminates one team, it takes exactly 15 games.
ANSWER 1: D

Problem 2:
- Jose: (10 - 1) = 9; 9 * 2 = 18; 18 + 2 = 20.
- Thuy: (10 * 2) = 20; 20 - 1 = 19; 19 + 2 = 21.
- Kareem: (10 - 1) = 9; 9 + 2 = 11; 11 * 2 = 22.
Kareem has the largest answer (22).
ANSWER 2: C

Problem 3:
To convert 2/25 to a decimal, multiply the numerator and denominator by 4 to get a denominator of 100: (2 * 4) / (25 * 4) = 8 / 100 = 0.08.
ANSWER 3: B

Problem 4:
We need to choose 3 starters from a group of 4 (Lance, Sally, Joy, Fred). This is a combination problem denoted as 4C3. Choosing 3 people to be on the team is the same as choosing 1 person to be left off the team. There are 4 ways to choose who is left off.
ANSWER 4: B

Problem 5:
Let $x$ be the number. $0.20x = 12$. Dividing by 0.20, $x = 12 / 0.2 = 60$. We want 30% of $x$: $0.30 * 60 = 18$.
ANSWER 5: B

Problem 6:
Start at 0 revolutions. Clockwise is positive, counterclockwise is negative.
$2 \frac{1}{4} - 3 \frac{3}{4} = 2.25 - 3.75 = -1.5$ revolutions.
-1.5 revolutions is the same as -0.5 revolutions (since -1 is a full circle). -0.5 revolutions (counterclockwise) from West points to East.
ANSWER 6: B

Problem 7:
Original yellow tint: 30% of 30 liters = 9 liters.
New mixture: 30 + 5 = 35 liters.
New yellow tint: 9 + 5 = 14 liters.
Percent: (14 / 35) * 100 = (2 / 5) * 100 = 40%.
ANSWER 7: C

Problem 8:
Convert to decimals or compare differences from 1:
15/11 = 1 + 4/11 ≈ 1.3636
19/15 = 1 + 4/15 ≈ 1.2666
17/13 = 1 + 4/13 ≈ 1.3076
Comparing the fractions 4/11, 4/13, and 4/15: the larger the denominator, the smaller the fraction. Thus, 4/15 < 4/13 < 4/11.
Order: 19/15 < 17/13 < 15/11.
ANSWER 8: E

Problem 9:
The set is {3, 4, 5, 6, 6, 7, x}. The mode is 6 (it appears twice). For the mean and median to be 6:
Median: With 7 numbers, the 4th number must be 6. Since the sorted list is 3, 4, 5, 6, 6, 7, x, if $x > 6$, the 4th number is 6.
Mean: (3+4+5+6+6+7+x) / 7 = 6 => 31 + x = 42 => x = 11.
Checking: {3, 4, 5, 6, 6, 7, 11}. Median is 6. Mean is 42/7 = 6. Mode is 6.
ANSWER 9: D

Problem 10:
Walking time = 6 minutes. Running speed = 3 * walking speed.
Time taken = Distance / Speed. Since speed is 3 times faster, time is 1/3 of the walking time.
Running time = 6 / 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
ANSWER 10: D

Problem 11:
Sum 1: (1+11+21+31+41) = 105.
Sum 2: (9+19+29+39+49) = 145.
105 + 145 = 250.
ANSWER 11: E

Problem 12:
Compare digit by digit:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
0.99 is the largest.
ANSWER 12: A

Problem 1:
This is a single-elimination tournament with 16 teams. Exactly one team is eliminated per game, and 15 teams must be eliminated to leave a single winner. Thus exactly 15 games are required.

ANSWER 1: D

Problem 2:
Jose evaluates \((10-1)\times2+2=20\).
Thuy evaluates \((10\times2)-1+2=21\).
Kareem evaluates \((10-1+2)\times2=22\).
Kareem obtains the strictly largest value.

ANSWER 2: C

Problem 3:
Division yields the decimal expansion \(2\div25=0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters out of 4 distinct players is the binomial coefficient \(\binom{4}{3}=4\).

ANSWER 4: B

Problem 5:
If 20% of a number equals 12, the number itself is \(12/0.2=60\). Then 30% of 60 is \(0.3\times60=18\).

ANSWER 5: B

Problem 6:
A net rotation of \(+2.25-3.75=-1.5\) revolutions is equivalent to 1.5 revolutions counterclockwise. From an initial westward orientation, 1.5 revolutions counterclockwise lands exactly on east.

ANSWER 6: B

Problem 7:
The original mixture contains \(0.30\times30=9\) liters of yellow tint. Adding 5 liters raises the yellow volume to 14 liters and the total volume to 35 liters, so the new percentage is \(14/35=40\%\).

ANSWER 7: C

Problem 8:
Decimal approximations are \(15/11\approx1.3636\), \(19/15\approx1.2667\), and \(17/13\approx1.3077\). Ordering these values produces \(19/15<17/13<15/11\).

ANSWER 8: E

Problem 9:
The unique mode must be 6, forcing the mean also to equal 6. The sum of the seven numbers is therefore 42. The six given numbers sum to 31, so \(x=11\). With this value the ordered list is 3,4,5,6,6,7,11 whose median is likewise 6 and whose unique mode remains 6.

ANSWER 9: D

Problem 10:
Let the total distance be \(2d\). Walking time for distance \(d\) is 6 min, so running time for the remaining distance \(d\) at triple speed is \(6/3=2\) min. The total time is therefore 8 min.

ANSWER 10: D

Problem 11:
Each pair sums to a multiple of 10: \(1+9=10\), \(11+19=30\), \(21+29=50\), \(31+39=70\), \(41+49=90\). Their total is 250. Equivalently each five-term arithmetic series sums to 105 and 145 respectively, again yielding 250.

ANSWER 11: E

Problem 12:
Direct comparison of the five decimals shows that 0.99 is strictly larger than each of the others.

ANSWER 12: A

Problem 1:
The problem asks us to determine the number of games that will be played to determine the winner of a basketball tournament with 16 teams, where the losing team of each game is eliminated.

## Step 1
To determine the number of games needed to find the winner, we first note that in a single-elimination tournament, each game eliminates one team.

## Step 2
Since there can only be one winner, all other teams must be eliminated. Therefore, the number of games required is equal to the number of teams minus one.

## Step 3
We have 16 teams competing in the tournament. So, the number of games needed to determine the winner is 16 - 1 = 15.

## Step 4
Let's examine the given choices to rule out the incorrect options. A. 4 is too low because it doesn't account for all the teams. B. 7 is also too low for the same reason. C. 8 is still too low. D. 15 matches our calculation. E. 16 is one more than our calculation.

The final answer is supported by our step-by-step reasoning.
ANSWER 1: D

Problem 2:
The problem asks us to determine who among Jose, Thuy, and Kareem gets the largest final answer after performing a series of operations starting with the number 10.

## Step 1
Let's calculate Jose's final answer: Jose subtracts 1 from 10 to get 9, doubles it to get 18, and then adds 2 to get 20.

## Step 2
Now, let's calculate Thuy's final answer: Thuy doubles 10 to get 20, subtracts 1 to get 19, and then adds 2 to get 21.

## Step 3
Next, let's calculate Kareem's final answer: Kareem subtracts 1 from 10 to get 9, adds 2 to get 11, and then doubles it to get 22.

## Step 4
Comparing the final answers: Jose got 20, Thuy got 21, and Kareem got 22. The largest final answer is 22, which belongs to Kareem.

## Step 5
Let's examine the given choices. A. Jose is incorrect because his answer is 20. B. Thuy is incorrect because her answer is 21. C. Kareem is correct because his answer is 22. D. Jose and Thuy is incorrect because Kareem's answer is larger. E. Thuy and Kareem is incorrect because Jose's answer is not the largest.

The final answer is supported by our step-by-step reasoning.
ANSWER 2: C

Problem 3:
The problem asks us to convert the fraction 2/25 into a decimal.

## Step 1
To convert 2/25 into a decimal, we divide the numerator by the denominator: 2 ÷ 25 = 0.08.

## Step 2
Let's examine the given choices to identify the correct decimal representation. A. .008 is too small. B. .08 matches our calculation. C. .8 is too large. D. 1.25 is much too large. E. 12.5 is also much too large.

The final answer is supported by our step-by-step reasoning.
ANSWER 3: B

Problem 4:
The problem asks us to determine the number of ways to choose 3 starters from a team of 4 members: Lance, Sally, Joy, and Fred.

## Step 1
This is a combination problem, because the order in which the starters are chosen does not matter.

## Step 2
The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial, the product of all positive integers up to that number.

## Step 3
Here, n = 4 (total team members) and k = 3 (number of starters to be chosen). So, C(4, 3) = 4! / [3!(4 - 3)!] = 4! / (3! * 1!) = (4 * 3 * 2 * 1) / ((3 * 2 * 1) * 1) = 4.

## Step 4
Let's examine the given choices. A. 2 is too low. B. 4 matches our calculation. C. 6 is too high. D. 8 is also too high. E. 10 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 4: B

Problem 5:
The problem asks us to find 30% of a number, given that 20% of the same number is 12.

## Step 1
First, we need to find the number. We know that 20% of the number is 12. So, 0.20 * number = 12.

## Step 2
To find the number, we divide both sides of the equation by 0.20: number = 12 / 0.20 = 60.

## Step 3
Now, we need to calculate 30% of 60: 0.30 * 60 = 18.

## Step 4
Let's examine the given choices. A. 15 is too low. B. 18 matches our calculation. C. 20 is too high. D. 24 is also too high. E. 30 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 5: B

Problem 6:
The problem asks us to determine the direction the spinner points after Chenille moves it clockwise 2 1/4 revolutions and then counterclockwise 3 3/4 revolutions, starting from pointing west.

## Step 1
First, let's understand the effect of the movements. A full revolution is 360 degrees or 4 quarters. So, 2 1/4 revolutions clockwise is equivalent to 9 quarters clockwise.

## Step 2
9 quarters clockwise is the same as 1 quarter clockwise (since 8 quarters make 2 full revolutions, which brings the spinner back to the original direction). So, after the first move, the spinner points north.

## Step 3
Then, Chenille moves it counterclockwise 3 3/4 revolutions, which is equivalent to 15 quarters counterclockwise.

## Step 4
15 quarters counterclockwise is the same as 3 quarters counterclockwise (since 12 quarters make 3 full revolutions). So, moving 3 quarters counterclockwise from north results in pointing west.

## Step 5
Let's examine the given choices. A. north is incorrect because after the second move, it's not north. B. east is incorrect. C. south is incorrect. D. west matches our calculation. E. northwest is incorrect.

The final answer is supported by our step-by-step reasoning.
ANSWER 6: D

Problem 7:
The problem asks us to find the percentage of yellow tint in a new mixture after adding 5 liters of yellow tint to an original 30-liter mixture that is 25% red tint, 30% yellow tint, and 45% water.

## Step 1
First, we calculate the amount of yellow tint in the original mixture: 30% of 30 liters = 0.30 * 30 = 9 liters.

## Step 2
After adding 5 liters of yellow tint, the total amount of yellow tint becomes 9 + 5 = 14 liters.

## Step 3
The total volume of the new mixture is 30 + 5 = 35 liters.

## Step 4
To find the percentage of yellow tint in the new mixture, we divide the amount of yellow tint by the total volume and multiply by 100: (14 / 35) * 100 = 40%

Problem 1:
In a single-elimination tournament, each game eliminates exactly one team. To eliminate all but the champion, the number of games needed is one fewer than the number of teams. With 16 teams, the number of games is \(16 - 1 = 15\).

ANSWER 1: D

Problem 2:
Jose: \(10 - 1 = 9\), \(9 \times 2 = 18\), \(18 + 2 = 20\).
Thuy: \(10 \times 2 = 20\), \(20 - 1 = 19\), \(19 + 2 = 21\).
Kareem: \(10 - 1 = 9\), \(9 + 2 = 11\), \(11 \times 2 = 22\).
Kareem’s final answer of 22 is the largest.

ANSWER 2: C

Problem 3:
Divide 2 by 25: \(2 \div 25 = 0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters from 4 people is the combination \(\binom{4}{3} = 4\).

ANSWER 4: B

Problem 5:
Let the number be \(n\). \(20\%\) of \(n\) is \(0.2n = 12\), so \(n = 60\). Then \(30\%\) of \(60 = 0.3 \times 60 = 18\).

ANSWER 5: B

Problem 6:
Initial direction: west.
Clockwise \(2\frac{1}{4}\) revolutions: \(2\frac{1}{4} = 2.25\) revolutions clockwise.
Counterclockwise \(3\frac{3}{4}\) revolutions: \(3\frac{3}{4} = 3.75\) revolutions counterclockwise.
Net revolution = \(2.25 - 3.75 = -1.5\) revolutions, i.e., \(1.5\) revolutions counterclockwise.
A full revolution (1.0) returns to west. The remaining 0.5 revolution counterclockwise from west is a half-turn, which points to east.

ANSWER 6: B

Problem 7:
Original mixture: 30 liters, yellow tint = \(30\% \times 30 = 9\) liters.
Add 5 liters of yellow tint: new yellow = \(9 + 5 = 14\) liters; new total volume = \(30 + 5 = 35\) liters.
Percent yellow = \(\frac{14}{35} \times 100\% = 40\%\).

ANSWER 7: C

Problem 8:
Convert each fraction to a decimal for comparison:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
Order from least to greatest: \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

ANSWER 8: E

Problem 9:
The data set: 3, 4, 5, 6, 6, 7, \(x\). The mode is the most frequent number. With two 6’s, 6 is the mode unless \(x\) equals another number that appears twice, which would create a second mode, violating the “unique mode” condition. Therefore, \(x\) must not be 3, 4, 5, or 7. It could be 6, but if \(x=6\), the median is 6, while the mean is \((3+4+5+6+6+6+7)/7 = 37/7 \approx 5.29\), so mean ≠ median.  
For the mean to equal the median (which is 6, the middle of seven sorted numbers if \(x > 6\) or \(x=6\) only if placed correctly), we set the mean to 6. Sum without \(x\) = \(3+4+5+6+6+7 = 31\). Total needed for mean 6 = \(7 \times 6 = 42\). Thus \(x = 42 - 31 = 11\).  
Sorted: 3, 4, 5, 6, 6, 7, 11. Median = 6, mode = 6, mean = 6. Unique mode holds.

ANSWER 9: D

Problem 10:
Walking half the distance took 6 minutes, so walking speed = \(\frac{\text{half distance}}{6}\).
Running speed is 3 times walking speed. Time to run the second half = \(\frac{\text{half distance}}{3 \times (\text{half distance}/6)} = \frac{6}{3} = 2\) minutes.
Total time = \(6 + 2 = 8\) minutes.

ANSWER 10: D

Problem 11:
Pair the terms: \(1+49 = 50\), \(11+39 = 50\), \(21+29 = 50\), \(31+19 = 50\), \(41+9 = 50\). Five pairs of 50 give \(5 \times 50 = 250\).

ANSWER 11: E

Problem 12:
Compare by extending decimals to the same number of places:
\(0.99   = 0.9900\)
\(0.9099 = 0.9099\)
\(0.9    = 0.9000\)
\(0.909  = 0.9090\)
\(0.9009 = 0.9009\)
Clearly \(0.99 > 0.9099 > 0.909 > 0.9009 > 0.9\). The largest is \(0.99\).

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key Insight:** In a single-elimination tournament, every game eliminates exactly one team. To crown a single winner from 16 teams, we must eliminate 16 − 1 = 15 teams. Therefore, exactly 15 games are needed.

**Verification by rounds:**
- Round 1: 16 teams → 8 games → 8 teams left
- Round 2: 8 teams → 4 games → 4 teams left
- Round 3: 4 teams → 2 games → 2 teams left
- Round 4 (Final): 2 teams → 1 game → 1 winner

Total = 8 + 4 + 2 + 1 = 15 games.

Choices A (4), B (7), and C (8) are too small. Choice E (16) would mean every team is eliminated, leaving no winner.

ANSWER 1: D

---

Problem 2:
We track each person's calculation step by step, all starting from 10.

**Jose:** subtract 1 → 9; double → 18; add 2 → **20**
**Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
**Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

Comparing: 20 < 21 < 22. Kareem gets the largest final answer.

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

Choice A (.008) would be 8/1000. Choice C (.8) would be 8/10. Choices D and E are greater than 1, but 2/25 is clearly less than 1.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination problem — the order of selection doesn't matter:
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4$$

Equivalently, choosing 3 people to start is the same as choosing 1 person to sit out. There are 4 people, so 4 ways to pick who sits out.

The 4 possible starting lineups are: {L,S,J}, {L,S,F}, {L,J,F}, {S,J,F}.

ANSWER 4: B

---

Problem 5:
We're told 20% of a number is 12. Let the number be $x$.

$$0.20x = 12 \implies x = \frac{12}{0.20} = 60$$

Now find 30% of 60:
$$0.30 \times 60 = 18$$

**Shortcut:** 30% is $\frac{30}{20} = 1.5$ times 20%, so the answer is $1.5 \times 12 = 18$.

Choice A (15) would be 25%. Choice D (24) would be 40%. Choice E (30) would be 50%.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing **West**.

**Move 1:** Clockwise $2\frac{1}{4}$ revolutions.
- The 2 full revolutions bring it back to West.
- The extra $\frac{1}{4}$ revolution clockwise from West: West → **North**.

**Move 2:** Counterclockwise $3\frac{3}{4}$ revolutions.
- The 3 full revolutions bring it back to North.
- The extra $\frac{3}{4}$ revolution counterclockwise from North:
  - $\frac{1}{4}$ CCW from North → West
  - $\frac{2}{4}$ CCW from North → South
  - $\frac{3}{4}$ CCW from North → **East**

The spinner points East.

ANSWER 6: B

---

Problem 7:
Original mixture: 30 liters total.
- Red tint: 25% of 30 = 7.5 liters
- Yellow tint: 30% of 30 = 9 liters
- Water: 45% of 30 = 13.5 liters

Add 5 liters of yellow tint:
- New yellow tint = 9 + 5 = 14 liters
- New total mixture = 30 + 5 = 35 liters

Percent yellow tint in new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

Choice B (35%) is a trap for those who just add 5% to 30%. Choice D (45%) is the original water percentage.

ANSWER 7: C

---

Problem 8:
We compare $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$.

Rewrite each as a mixed number:
$$\frac{15}{11} = 1 + \frac{4}{11}, \quad \frac{19}{15} = 1 + \frac{4}{15}, \quad \frac{17}{13} = 1 + \frac{4}{13}$$

Since all three have the same numerator (4) in their fractional parts, we compare denominators. For fractions with equal numerators, a **larger denominator** means a **smaller fraction**:
$$15 > 13 > 11 \implies \frac{4}{15} < \frac{4}{13} < \frac{4}{11}$$

Therefore:
$$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$$

ANSWER 8: E

---

Problem 9:
The data set is: 3, 4, 5, 6, 6, 7, and $x$. The mean, median, and unique mode are all equal.

**Mode:** Since 6 already appears twice and all others appear once, the unique mode must be 6 (if $x$ were 3, 4, 5, or 7, we'd have two modes, violating "unique mode"). So mode = 6.

**Mean = 6:**
$$\frac{3 + 4 + 5 + 6 + 6 + 7 + x}{7} = 6$$
$$\frac{31 + x}{7} = 6 \implies 31 + x = 42 \implies x = 11$$

**Check Median:** With $x = 11$, the sorted list is: 3, 4, 5, **6**, 6, 7, 11. The median (4th value) is 6. ✓

All three measures equal 6. Choice B (6) would make 6 appear three times, but the mean would be $(31+6)/7 = 37/7 \neq 6$.

ANSWER 9: D

---

Problem 10:
Joe walked half the distance in 6 minutes. He then ran the remaining half at 3 times his walking speed.

Let walking speed = $v$. The half-distance = $6v$.

Running speed = $3v$. Time to run the second half:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{6v}{3v} = 2 \text{ minutes}$$

Total time = 6 + 2 = **8 minutes**.

Choice A (7) would result from incorrectly subtracting. Choices B, C, E involve unnecessary decimal complications.

ANSWER 10: D

---

Problem 11:
Compute $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$.

Pair the terms from the first group with terms from the second group (first with last, etc.):
- $1 + 49 = 50$
- $11 + 39 = 50$
- $21 + 29 = 50$
- $31 + 19 = 50$
- $41 + 9 = 50$

Total = $50 \times 5 = 250$.

Choice C (200) is a common error from miscounting pairs. Choice D (249) comes from arithmetic mistakes.

ANSWER 11: E

---

Problem 12:
Compare the five decimals by writing them to four decimal places:

- A: 0.9900
- B: 0.9099
- C: 0.9000
- D: 0.9090
- E: 0.9009

Comparing digit by digit from the tenths place: all have 9 in the tenths place. In the hundredths place, A has **9** while all others have **0**. So A is immediately the largest.

$0.99 > 0.9099 > 0.909 > 0.9009 > 0.9$

ANSWER 12: A

 **Problem 1:**
We need to find the number of games played in a single-elimination tournament with 16 teams. In single elimination, every game eliminates exactly one team. To determine a winner, 15 of the 16 teams must be eliminated. Therefore, exactly 15 games are required.  
(Another way: with 16 teams there are 8 games in the first round, 4 in the second, 2 in the third, and 1 final game: \(8+4+2+1=15\).)

ANSWER 1: D

---

**Problem 2:**
We compute each person’s final answer step by step, starting from 10.

- **Jose:**  
  \(10 - 1 = 9\)  
  \(9 \times 2 = 18\)  
  \(18 + 2 = 20\)

- **Thuy:**  
  \(10 \times 2 = 20\)  
  \(20 - 1 = 19\)  
  \(19 + 2 = 21\)

- **Kareem:**  
  \(10 - 1 = 9\)  
  \(9 + 2 = 11\)  
  \(11 \times 2 = 22\)

Comparing the results: \(20 < 21 < 22\). Kareem’s answer is the largest.

ANSWER 2: C

---

**Problem 3:**
We convert the fraction \(\frac{2}{25}\) to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100:
\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

ANSWER 3: B

---

**Problem 4:**
We must choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways to choose 3 from 4 is the combination \(\binom{4}{3}\):
\[
\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4
\]
Equivalently, choosing 3 starters is the same as choosing 1 person to sit out, and there are 4 choices for who sits out.

ANSWER 4: B

---

**Problem 5:**
Let the number be \(N\). We are given \(0.20N = 12\). Solving for \(N\):
\[
N = \frac{12}{0.20} = 60
\]
Now find 30% of 60:
\[
0.30 \times 60 = 18
\]
(Shortcut: 30% is \(1.5\) times 20%, so \(1.5 \times 12 = 18\).)

ANSWER 5: B

---

**Problem 6:**
The spinner starts pointing west.  
- Clockwise \(2\frac14\) revolutions: 2 full revolutions bring it back to west, then \(\frac14\) turn clockwise from west points it **north**.  
- Counterclockwise \(3\frac34\) revolutions: 3 full revolutions bring it back to north, then \(\frac34\) turn counterclockwise from north goes through west, south, and ends at **east**.

Net movement: \(3\frac34 - 2\frac14 = 1\frac12\) revolutions counterclockwise. One full revolution returns to west; an additional \(\frac12\) revolution counterclockwise from west points to **east**.

ANSWER 6: B

---

**Problem 7:**
Original mixture: 30 liters.  
- Yellow tint originally: \(30\% \times 30 = 9\) liters.  
After adding 5 liters of yellow tint:  
- New yellow tint: \(9 + 5 = 14\) liters.  
- New total mixture: \(30 + 5 = 35\) liters.  

Percent of yellow tint in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%
\]

ANSWER 7: C

---

**Problem 8:**
Compare \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\). Each is greater than 1. Write each as \(1\) plus a proper fraction:
\[
\frac{15}{11} = 1 + \frac{4}{11}, \qquad
\frac{19}{15} = 1 + \frac{4}{15}, \qquad
\frac{17}{13} = 1 + \frac{4}{13}
\]
Since the numerators are the same (4), the fraction with the largest denominator is smallest:
\[
\frac{4}{15} < \frac{4}{13} < \frac{4}{11}
\]
Adding 1 preserves the order:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

ANSWER 8: E

---

**Problem 9:**
The data set is \(3, 4, 5, 6, 6, 7, x\).  
- The current mode is 6 (appears twice). For 6 to remain the **unique** mode, \(x\) must not equal 3, 4, 5, or 7 (which would create a second mode).  
- We are told mean = median = mode = 6.  

Set the mean equal to 6:
\[
\frac{3+4+5+6+6+7+x}{7} = 6 \implies \frac{31+x}{7} = 6 \implies 31+x = 42 \implies x = 11
\]

Check: With \(x=11\), the sorted list is \(3, 4, 5, 6, 6, 7, 11\).  
- Median (4th term) = 6.  
- Mode = 6 (unique).  
- Mean = 6.  

All conditions are satisfied.

ANSWER 9: D

---

**Problem 10:**
Joe walks half the distance in 6 minutes. He runs the second half at 3 times his walking speed. Since distance is the same and speed is tripled, the time required is divided by 3:
\[
\text{Running time} = \frac{6}{3} = 2 \text{ minutes}
\]
Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

---

**Problem 11:**
Add the two groups:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]
Pair the terms to make convenient sums:
\[
(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 5 \times 50 = 250
\]

ANSWER 11: E

---

**Problem 12:**
Compare the decimals by writing them with the same number of places:
- A. \(0.9900\)
- B. \(0.9099\)
- C. \(0.9000\)
- D. \(0.9090\)
- E. \(0.9009\)

The tenths digit is 9 for all. Looking at the hundredths digit: A has 9, while B, D have 0 and C, E have 0. Thus \(0.9900\) is clearly the largest.

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key insight:** Each game eliminates exactly one team. To determine a winner from 16 teams, we must eliminate 15 teams (everyone except the champion).

Therefore, exactly 15 games must be played.

*Verification by rounds:* Round 1: 8 games (16→8 teams), Round 2: 4 games (8→4), Round 3: 2 games (4→2), Round 4: 1 game (2→1). Total = 8+4+2+1 = 15.

The tempting wrong answer E (16) counts teams instead of eliminations.

ANSWER 1: D

---

Problem 2:
We need to compute each person's final answer starting from 10.

**Jose:** Start with 10 → subtract 1: 10−1 = 9 → double: 9×2 = 18 → add 2: 18+2 = **20**

**Thuy:** Start with 10 → double: 10×2 = 20 → subtract 1: 20−1 = 19 → add 2: 19+2 = **21**

**Kareem:** Start with 10 → subtract 1: 10−1 = 9 → add 2: 9+2 = 11 → double: 11×2 = **22**

Kareem gets the largest final answer (22).

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

The tempting wrong answer A (0.008) would be 8/1000, and C (0.8) would be 8/10.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination: C(4,3) = 4!/(3! × 1!) = 4

Listing them out: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — that's 4 ways.

The tempting wrong answer C (6) would be C(4,2), choosing who's *not* a starter, but that also equals 4... Actually C(4,2)=6 is wrong here; we're choosing 3 from 4.

ANSWER 4: B

---

Problem 5:
We know 20% of a number is 12. We need to find 30% of the same number.

**Step 1:** Find the number.
$$0.20 \times n = 12 \implies n = \frac{12}{0.20} = 60$$

**Step 2:** Find 30% of 60.
$$0.30 \times 60 = 18$$

*Shortcut:* Since 30% is 1.5 times 20%, the answer is 1.5 × 12 = 18.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing west. We need to find the net rotation.

**Clockwise:** 2¼ revolutions
**Counterclockwise:** 3¾ revolutions

**Net movement:** 2¼ − 3¾ = −1½ revolutions (negative means counterclockwise)

So the spinner moves 1½ revolutions counterclockwise from west.

- 1 full revolution counterclockwise returns to **west**
- ½ revolution counterclockwise from west goes to **east**

(Counterclockwise from west: ¼ turn → south, ½ turn → east)

ANSWER 6: B

---

Problem 7:
We need to find the percent of yellow tint after adding 5 liters of yellow tint.

**Original mixture (30 liters):**
- Yellow tint: 30% of 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percent yellow tint:** (14/35) × 100 = 40%

The tempting wrong answer B (35) might come from simply adding 5% to 30%.

ANSWER 7: C

---

Problem 8:
We need to order 15/11, 19/15, and 17/13 from least to greatest.

Computing each as a decimal:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3077...

From least to greatest: **19/15 < 17/13 < 15/11**

*Alternative method:* Note each is 1 + (numerator − denominator)/denominator:
- 15/11 = 1 + 4/11
- 19/15 = 1 + 4/15
- 17/13 = 1 + 4/13

Since 4/15 < 4/13 < 4/11, the order is confirmed.

ANSWER 8: E

---

Problem 9:
The positive integers 3, 4, 5, 6, 6, 7, and x have equal mean, median, and unique mode.

**Mode:** Since 6 already appears twice and no other number does, the unique mode must be 6. So x ≠ 3, 4, 5, or 7 (which would create a second mode).

**Median:** With 7 numbers, the median is the 4th value when sorted. For the median to be 6, we need x ≥ 6 (so that 6 is the 4th value).

- If x = 6: sorted = 3, 4, 5, 6, 6, 6, 7 → median = 6 ✓
- If x > 7: sorted = 3, 4, 5, 6, 6, 7, x → median = 6 ✓

**Mean:** (3 + 4 + 5 + 6 + 6 + 7 + x)/7 = (31 + x)/7

Setting mean = 6: (31 + x)/7 = 6 → 31 + x = 42 → **x = 11**

**Verification:** Numbers: 3, 4, 5, 6, 6, 7, 11. Mean = 42/7 = 6, Median = 6, Mode = 6 ✓

ANSWER 9: D

---

Problem 10:
Joe walks halfway, then runs the rest. We need total time.

**Walking:** Half the distance takes 6 minutes.

**Running:** He runs 3 times as fast as he walks, so he covers the same distance in 1/3 the time.
- Time to run the second half = 6 ÷ 3 = 2 minutes

**Total time:** 6 + 2 = **8 minutes**

The tempting wrong answers come from setting up the problem incorrectly or mixing up distance and time relationships.

ANSWER 10: D

---

Problem 11:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

**First group:** 1 + 11 + 21 + 31 + 41 = 105

**Second group:** 9 + 19 + 29 + 39 + 49 = 145

**Total:** 105 + 145 = **250**

*Elegant method:* Pair terms: (1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250

ANSWER 11: E

---

Problem 12:
We need to find the largest number among the options. Writing each with the same number of decimal places:

- A. 0.99 = **0.9900**
- B. 0.9099 = 0.9099
- C. 0.9 = 0.9000
- D. 0.909 = 0.9090
- E. 0.9009 = 0.9009

Comparing digit by digit after the decimal point:
- First digit: all are 9
- Second digit: A has **9**, all others have 0

Since A has a 9 in the second decimal place while all others have 0, **0.99 is the largest**.

ANSWER 12: A

**Problem 1:**  
The tournament is single‑elimination. Starting with 16 teams, every game eliminates exactly one team. To reduce from 16 teams to a single champion we must eliminate 15 teams, and each elimination corresponds to one game.  

Number of games = 16 − 1 = **15**.  

ANSWER 1: D  

---

**Problem 2:**  
Start with 10 for each person.

- **Jose:**  
  Subtract 1 → 9; double → 18; add 2 → 20.  

- **Thuy:**  
  Double → 20; subtract 1 → 19; add 2 → 21.  

- **Kareem:**  
  Subtract 1 → 9; add 2 → 11; double → 22.  

The largest final value is 22, obtained by Kareem.  

ANSWER 2: C  

---

**Problem 3:**  

\[
\frac{2}{25}=0.08
\]

(2 ÷ 25 = 0.08).  

ANSWER 3: B  

---

**Problem 4:**  
We need to choose 3 starters from the 4 available players:  

\[
\binom{4}{3}=4
\]

ways.  

ANSWER 4: B  

---

**Problem 5:**  
Let the number be \(N\).  
\(0.20N = 12 \Rightarrow N = \frac{12}{0.20}=60\).  
\(30\%\) of \(N\) is \(0.30 \times 60 = 18\).  

ANSWER 5: B  

---

**Problem 6:**  
Interpret the fractions as mixed numbers:  

- Clockwise \(2\frac14 = 2.25\) revolutions.  
- Counter‑clockwise \(3\frac34 = 3.75\) revolutions.  

Net rotation (counter‑clockwise positive) = \(3.75 - 2.25 = 1.5\) revolutions.  

A full revolution returns to the same direction, so we only care about the extra \(0.5\) revolution.  
\(0.5\) revolution = 180° counter‑clockwise.  

Starting from west, turning 180° counter‑clockwise points to **east**.  

ANSWER 6: B  

---

**Problem 7:**  
Original yellow = \(30\%\) of 30 L = \(0.30 \times 30 = 9\) L.  
Add 5 L yellow → new yellow = \(9+5 = 14\) L.  
New total volume = \(30+5 = 35\) L.  

Percent yellow = \(\frac{14}{35}\times100 = 40\%\).  

ANSWER 7: C  

---

**Problem 8:**  
Compute the three fractions:

\[
\frac{15}{11}=1.3636\ldots,\qquad
\frac{19}{15}=1.2666\ldots,\qquad
\frac{17}{13}=1.3077\ldots
\]

Ordered from least to greatest:  

\(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

This matches option **E**.  

ANSWER 8: E  

---

**Problem 9:**  
Current list: \(3,4,5,6,6,7\). The unique mode is 6 (appears twice).  
For the mode to remain unique and equal to the mean and median, \(x\) must not create a tie.  
Let the mean = median = mode = \(m\).  

Sum of numbers = \(31 + x\).  
Mean = \(\frac{31 + x}{7}=m\).  
Median of the seven numbers is the 4th after sorting; with \(x>7\) the median stays 6, so \(m=6\).  

Thus \(\frac{31 + x}{7}=6 \Rightarrow 31 + x = 42 \Rightarrow x = 11\).

Check: sorted list becomes \(3,4,5,6,6,7,11\) → median = 6, mode = 6, mean = 42/7 = 6. All equal.  

ANSWER 9: D  

---

**Problem 10:**  
Let the walking speed be \(v\) and the half‑distance be \(d/2\).  

Walking time for half‑distance: \(\frac{d/2}{v}=6\) min ⇒ \(\frac{d}{v}=12\) min.  

Running speed = \(3v\).  
Running time for the second half: \(\frac{d/2}{3v}= \frac{d}{6v}= \frac{1}{6}\times\frac{d}{v}= \frac{12}{6}=2\) min.  

Total time = \(6 + 2 = 8\) minutes.  

ANSWER 10: D  

---

**Problem 11:**  

\[
\begin{aligned}
1+11+21+31+41 &= 105,\\
9+19+29+39+49 &= 145,\\
\text{Total} &= 105+145 = 250.
\end{aligned}
\]

ANSWER 11: E  

---

**Problem 12:**  
Write the numbers with four decimal places:

- .99 = 0.9900  
- .9099 = 0.9099  
- .9 = 0.9000  
- .909 = 0.9090  
- .9009 = 0.9009  

The largest is 0.9900, i.e., **.99**.  

ANSWER 12: A  

Problem 1:  
The tournament starts with 16 teams. Each game eliminates one team, and the losing team is eliminated. To determine the winner, all teams except one must be eliminated. Thus, 15 teams must be eliminated, requiring 15 games.  
- **Why not other choices?** A (4) is too few; B (7), C (8) are also insufficient; E (16) would imply one game per team without elimination, which is incorrect.  
ANSWER 1: D  

Problem 2:  
- **Jose**: Starts with 10 → subtracts 1 (9) → doubles (18) → adds 2 → **20**.  
- **Thuy**: Starts with 10 → doubles (20) → subtracts 1 (19) → adds 2 → **21**.  
- **Kareem**: Starts with 10 → subtracts 1 (9) → adds 2 (11) → doubles → **22**.  
Kareem has the largest final answer.  
- **Why not other choices?** A (Jose) and B (Thuy) are smaller; D and E pair incorrect winners.  
ANSWER 2: C  

Problem 3:  
\( \frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08 \).  
- **Why not other choices?** A (.008) is \( \frac{2}{250} \); C (.8) is \( \frac{4}{5} \); D and E are larger decimals.  
ANSWER 3: B  

Problem 4:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is \( \binom{4}{3} = 4 \).  
- **Why not other choices?** A (2) is too few; C (6) is \( \binom{4}{2} \), which is incorrect; D (8) and E (10) are overcounts.  
ANSWER 4: B  

Problem 5:  
Let the number be \( x \). If 20% of \( x \) is 12, then \( 0.2x = 12 \), so \( x = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).  
- **Why not other choices?** A (15) is 25% of 60; C (20) is \( \frac{1}{3} \) of 60; D (24) is 40%; E (30) is 50%.  
ANSWER 5: B  

Problem 6:  
- A full revolution is 360°. Clockwise \( 2 \frac{1}{4} \) revolutions = \( 2.25 \times 360 = 810° \).  
- Counterclockwise \( 3 \frac{3}{4} \) revolutions = \( -3.75 \times 360 = -1350° \).  
- Net movement: \( 810° - 1350° = -540° \).  
- \( -540° \mod 360° = 180° \) (since \( -540 + 720 = 180° \)), which points west → south? Wait, initial direction is west.  
- West (270°) + 810° = 1080° → 1080° mod 360° = 0° (north). Then -1350° → 0° - 1350° = -1350° mod 360° = 90° (east)? Let's reframe:  
- Start at west (270°). Clockwise 2.25 rev: 270° + 810° = 1080° mod 360° = 0° (north). Then counterclockwise 3.75 rev: 0° - 1350° = -1350° mod 360° = 90° (east).  
- **Why not other choices?** A (north) is initial after clockwise; B (east) is final; C (south) is 180°; D (west) is start; E (northwest) is not a cardinal direction.  
ANSWER 6: B  

Problem 7:  
Original mixture: 30 liters, 30% yellow tint → 9 liters yellow. Add 5 liters yellow → total yellow = 14 liters. New mixture = 35 liters. Percent yellow = \( \frac{14}{35} \times 100 = 40\% \).  
- **Why not other choices?** A (25) is original yellow%; B (35) is before addition; D (45) and E (50) are overestimates.  
ANSWER 7: E (Note: Correction based on computation: 14/35=0.4 → 40%, so answer is E? Wait, 14/35=0.4 → 40%, which is E. But earlier I thought B, but B is 35, which is wrong. Original yellow is 9, add 5 → 14. Total mixture 30+5=35. 14/35=0.4 → 40%. So E. But in choices E is 50? Wait, no: choices are A.25 B.35 C.40 D.45 E.50. So 40% is C. I misread. Correction: 14/35=0.4 → 40%, which is C.  
**Correction**: 14/35=0.4 → 40%, so answer is C.  
ANSWER 7: C  

Problem 8:  
Compare \( \frac{15}{11} \approx 1.3636 \), \( \frac{19}{15} \approx 1.2667 \), \( \frac{17}{13} \approx 1.3077 \). Order: \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11} \).  
- **Why not other choices?** A, B, C, D have incorrect orders.  
ANSWER 8: E  

Problem 9:  
The set is {3, 4, 5, 6, 6, 7, x}. Mean = median = mode = 6.  
- Mode is 6 (unique), so 6 must appear most.  
- Median is 6 (4th value in ordered list: 3,4,5,6,6,7,x → if x≥6, ordered is ...,6,6,x or ...,6,x,6? Must have 4th value 6.  
- If x=6, set is {3,4,5,6,6,6,7}: median=6, mode=6, mean=(3+4+5+6+6+6+7)/7=37/7≈5.28 ≠6.  
- If x=11: set {3,4,5,6,6,7,11}: mean=(3+4+5+6+6+7+11)/7=42/7=6, median=6, mode=6.  
- **Why not other choices?** A (5): mean=36/7≈5.14; B (6): mean≈5.28; C (7): mean=38/7≈5.43; D (11) works.  
ANSWER 9: D  

Problem 10:  
Joe walks half the distance in 6 minutes. Let walking speed be \( v \), so distance = \( 6v \). He runs at \( 3v \), so time to run half distance = \( \frac{6v}{3v} = 2 \) minutes. Total time = 6 + 2 = 8 minutes.  
- **Why not other choices?** A (7), B (7.3), C (7.7), E (8.3) are not exact.  
ANSWER 10: D  

Problem 11:  
First group: 1+11+21+31+41 = 105. Second group: 9+19+29+39+49 = 145. Total = 105 + 145 = 250.  
- **Why not other choices?** A (150), B (199), C (200), D (249) are incorrect sums.  
ANSWER 11: E  

Problem 12:  
Compare decimals: A (.99), B (.9099), C (.9), D (.909), E (.9009). Largest is .99.  
- **Why not other choices?** B, C, D, E are all smaller than .99.  
ANSWER 12: A

### Problem 1:
We need to find the total number of games to determine a winner in a 16-team single-elimination tournament, where every game eliminates exactly one losing team. To get 1 winner, we must eliminate 16 - 1 = 15 teams. Since each game eliminates 1 team, we need 15 total games. Wrong choices: A is the number of bracket rounds, not games; C is half the number of teams; E is the total number of teams, not games.
ANSWER 1: D

---

### Problem 2:
All three start with 10, we calculate each person's final result step-by-step:
- Jose: (10 - 1) * 2 + 2 = 9*2 +2 = 20
- Thuy: (10*2) -1 +2 = 20-1+2=21
- Kareem: (10-1 +2)*2 = 11*2=22
Kareem has the largest final answer. Wrong choices stem from misordering the operations for each person.
ANSWER 2: C

---

### Problem3:
Convert 2/25 to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100: (2*4)/(25*4) = 8/100 = 0.08. Wrong choices: A is 2/250, C is 20/25.
ANSWER 3: B

---

### Problem4:
We choose 3 starters from 4 people, which is a combination problem (order of starters does not matter here). The number of ways to choose 3 from 4 is equivalent to choosing 1 person to leave off the starting lineup: there are 4 people, so 4 unique lineups, each omitting one person. Wrong choice C (6) is the number of permutations, not combinations.
ANSWER 4: B

---

### Problem5:
If 20% of a number is 12, first find the full number: 0.2*N=12 → N=12/0.2=60. 30% of 60 is 0.3*60=18. Wrong choices: A is 25% of 60, D is 40% of 60.
ANSWER 5: B

---

### Problem6:
The spinner starts pointing west. Net movement: 2.25 clockwise revolutions minus 3.75 clockwise (equivalent to 3.75 counterclockwise) = -1.5 total clockwise revolutions, or 1.5 counterclockwise revolutions. Full revolutions cancel out, so we only have a net 0.5 (180 degree) counterclockwise turn from west. A 180-degree turn from west points to east.
ANSWER 6: B

---

### Problem7:
Original 30L mixture has 30% yellow tint: 0.3*30=9L of yellow. Add 5L of yellow, so new yellow is 14L, new total mixture is 35L. The new percentage of yellow is (14/35)*100=40%. Wrong choice: people often forget to add 5L to the total mixture volume, leading to an incorrect higher percentage.
ANSWER 7: C

---

### Problem8:
Rewrite each fraction to compare easily: 15/11 = 1 + 4/11, 19/15=1+4/15, 17/13=1+4/13. For equal numerators, the larger the denominator, the smaller the fraction. So 4/15 < 4/13 < 4/11, so the full order from least to greatest is 19/15 < 17/13 <15/11.
ANSWER 8: E

---

### Problem9:
The dataset is [3,4,5,6,6,7,x]. The unique mode is currently 6 (only value appearing twice, so it must stay the only unique mode). For the mean, median, and mode to all be 6: the sum of the dataset must be 7*6=42. The sum of the existing numbers is 3+4+5+6+6+7=31, so x=42-31=11. Sorted dataset [3,4,5,6,6,7,11] has median 6, unique mode 6, mean 6, which fits. Wrong choice B (x=6) gives a mean of ~5.28, not 6.
ANSWER 9: D

---

### Problem10:
Joe walks half the distance in 6 minutes. His running speed is 3x his walking speed, so time to cover the same distance is 1/3 of walking time. Time to run the second half: 6/3=2 minutes. Total time: 6+2=8 minutes.
ANSWER 10: D

---

### Problem11:
Pair terms to simplify addition: (1+49)+(11+39)+(21+29)+(31+19)+(41+9) = 50+50+50+50+50=250.
ANSWER 11: E

---

### Problem12:
Line up all decimals to compare: A=0.9900, B=0.9099, C=0.9000, D=0.9090, E=0.9009. The largest value is 0.99, option A.
ANSWER 12: A

Problem 1:  
In a single-elimination tournament, each game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams (all except 1). Thus, 15 games are played.  
ANSWER 1: D  

Problem 2:  
Calculate each person's result starting from 10:  
- Jose: \(10 -1 =9\), double → \(9*2=18\), add 2 → \(18+2=20\)  
- Thuy: Double 10 → \(10*2=20\), subtract 1 → \(20-1=19\), add 2 → \(19+2=21\)  
- Kareem: \(10-1=9\), add 2 → \(9+2=11\), double → \(11*2=22\)  
Kareem has the largest result.  
ANSWER 2: C  

Problem 3:  
Convert \(\frac{2}{25}\) to decimal: \(2\div25=0.08\).  
ANSWER 3: B  

Problem 4:  
Choosing 3 starters from 4 people is a combination (order does not matter). The number of ways is \(\binom{4}{3}=4\) (or list: {Lance,Sally,Joy}, {Lance,Sally,Fred}, {Lance,Joy,Fred}, {Sally,Joy,Fred}).  
ANSWER 4: B  

Problem 5:  
Let the number be \(x\). \(20\%x=12\) → \(0.2x=12\) → \(x=60\). \(30\%\ of\ 60=0.3*60=18\).  
ANSWER 5: B  

Problem 6:  
Model directions with standard angles: 0°=east, 90°=north, 180°=west, 270°=south. Initial direction is west (180°).  
- Clockwise \(2\frac{1}{4}\) revolutions: \(2.25*(-360°)=-810°\), new angle: \(180-810=-630°\)  
- Counterclockwise \(3\frac{3}{4}\) revolutions: \(3.75*(360°)=+1350°\), total angle: \(-630+1350=720°\)  
- \(720°\mod360°=0°\), which is east.  
ANSWER 6: B  

Problem 7:  
Original yellow tint: \(30\%\ of\ 30=0.3*30=9\) liters. Add 5 liters, total yellow = \(9+5=14\) liters. New total mixture: \(30+5=35\) liters. Percent yellow: \(\frac{14}{35}*100\%=40\%\).  
ANSWER 7: C  

Problem 8:  
Compare fractions via cross-multiplication or decimals:  
- \(\frac{19}{15}\approx1.267\), \(\frac{17}{13}\approx1.308\), \(\frac{15}{11}\approx1.364\)  
Order from least to greatest: \(\frac{19}{15}<\frac{17}{13}<\frac{15}{11}\).  
ANSWER 8: E  

Problem 9:  
The unique mode is 6 (appears twice, others once). Thus mean and median equal 6.  
- Median of 7 numbers is the 4th term when sorted, so 4th term =6 → \(x\geq6\).  
- Mean = \(\frac{3+4+5+6+6+7+x}{7}=6\) → \(31+x=42\) → \(x=11\).  
Check: Sorted list [3,4,5,6,6,7,11], median=6, mode=6 (unique), mean=6.  
ANSWER 9: D  

Problem 10:  
Walking half the distance takes 6 minutes, so full walking time would be 12 minutes. Running speed is 3x walking speed, so time for the second half is \(\frac{6}{3}=2\) minutes. Total time: \(6+2=8\) minutes.  
ANSWER 10: D  

Problem 11:  
Sum first group: \(1+11+21+31+41=105\) (arithmetic series: \(\frac{5}{2}(1+41)=105\)).  
Sum second group: \(9+19+29+39+49=145\) (\(\frac{5}{2}(9+49)=145\)).  
Total: \(105+145=250\).  
ANSWER 11: E  

Problem 12:  
Compare decimals digit by digit:  
- A=0.99, B=0.9099, C=0.9, D=0.909, E=0.9009  
A has the highest second decimal digit (9 vs 0 for others), so it is the largest.  
ANSWER 12: A

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** How many games must be played to determine a winner in a 16-team elimination tournament?

**Setup:** In an elimination tournament, each game eliminates exactly one team. To go from 16 teams to 1 winner, we must eliminate 15 teams. Therefore, we need 15 games.

**Alternatively:** We can think of it as a bracket:
- Round 1: 16 teams → 8 games → 8 winners
- Round 2: 8 teams → 4 games → 4 winners
- Round 3: 4 teams → 2 games → 2 winners
- Round 4: 2 teams → 1 game → 1 winner
- Total: 8 + 4 + 2 + 1 = 15 games

ANSWER 1: D

---

**Problem 2:**

**What's being asked:** Which person gets the largest final answer after their operations?

**Jose's calculation:** 
- Start: 10
- Subtract 1: 10 - 1 = 9
- Double: 9 × 2 = 18
- Add 2: 18 + 2 = **20**

**Thuy's calculation:**
- Start: 10
- Double: 10 × 2 = 20
- Subtract 1: 20 - 1 = 19
- Add 2: 19 + 2 = **21**

**Kareem's calculation:**
- Start: 10
- Subtract 1: 10 - 1 = 9
- Add 2: 9 + 2 = 11
- Double: 11 × 2 = **22**

Kareem gets 22, which is the largest.

ANSWER 2: C

---

**Problem 3:**

**What's being asked:** Convert the fraction 2/25 to a decimal.

**Setup:** 2/25 = ?

**Calculation:** 
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** In how many ways can 3 starters be chosen from 4 people (Lance, Sally, Joy, Fred)?

**Setup:** This is a combination problem: $\binom{4}{3}$

**Calculation:**
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4}{1} = 4$$

Alternatively, we're choosing which 1 person to leave out: 4 ways.

ANSWER 4: B

---

**Problem 5:**

**What's being asked:** If 20% of a number is 12, what is 30% of that number?

**Setup:** Let the number be $n$.
- $0.20n = 12$
- $n = 12 ÷ 0.20 = 60$

**Calculation:** 
$$30\% \text{ of } 60 = 0.30 \times 60 = 18$$

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Starting pointing west, after rotating clockwise 2¼ revolutions then counterclockwise 3¾ revolutions, which direction?

**Setup:** 
- Clockwise 2¼ revolutions = 2.25 revolutions
- Counterclockwise 3¾ revolutions = 3.75 revolutions
- Net: 3.75 - 2.25 = 1.5 revolutions counterclockwise (net)

**Calculation:**
Starting at west, rotating counterclockwise 1.5 revolutions (= 1 full rotation + 0.5 rotation):
- 1 full rotation brings us back to west
- 0.5 rotation = 180°
- West + 180° counterclockwise = East

ANSWER 6: B

---

**Problem 7:**

**What's being asked:** After adding 5 liters of yellow tint to a 30-liter mixture, what percent is yellow tint?

**Original mixture (30 liters):**
- Yellow tint: 30% × 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percentage:**
$$\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%$$

ANSWER 7: C

---

**Problem 8:**

**What's being asked:** Order 15/11, 19/15, and 17/13 from least to greatest.

**Setup:** Convert to decimals or compare directly.

**Calculations:**
- 15/11 ≈ 1.364
- 19/15 ≈ 1.267
- 17/13 ≈ 1.308

**Order:** 19/15 < 17/13 < 15/11

ANSWER 8: E

---

**Problem 9:**

**What's being asked:** Find $x$ such that the mean, median, and unique mode of {3, 4, 5, 6, 6, 7, $x$} are all equal.

**Current mode:** 6 appears twice; all others appear once. If there's a unique mode, it must be 6.

**With unique mode = 6:** This requires no other number to appear more than once. So $x \notin \{3, 4, 5, 7\}$ and $x \neq 6$ (else it wouldn't be unique).

**Mean = Median = Mode = 6:**

For median of 7 numbers with value 6:
- Sorted: need middle (4th) value to be 6
- If $x = 11$: {3, 4, 5, 6, 6, 7, 11} → median = 6 ✓

Mean: $\frac{3+4+5+6+6+7+11}{7} = \frac{42}{7} = 6$ ✓

Mode: 6 appears twice; all others once → mode = 6 ✓

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Total time from home to school if Joe walked halfway (6 min) then ran the rest at 3× speed?

**Setup:**
- Walking: 6 minutes for half the distance
- Running: same distance at 3× speed, so time = 6/3 = 2 minutes

**Total time:** 6 + 2 = 8 minutes

ANSWER 10: D

---

**Problem 11:**

**What's being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Calculation:**
- First group: 1 + 11 + 21 + 31 + 41 = 105
- Second group: 9 + 19 + 29 + 39 + 49 = 145
- Sum: 105 + 145 = 250

ANSWER 11: E

---

**Problem 12:**

**What's being asked:** Which number is largest?

**Comparison:**
- A. 0.99
- B. 0.9099
- C. 0.9
- D. 0.909
- E. 0.9009

Starting from the tenths place: all are 0.9__. Looking at hundredths:
- 0.99 has 9 in hundredths place → **largest**

ANSWER 12: A

Problem 1:

In a single-elimination tournament, each game eliminates exactly one team.  
There are 16 teams, and we want one champion.

So we must eliminate 15 teams total.

Therefore, the number of games played is 15.

Tempting wrong choice: 16 would be if every team had to lose, but the winner does not lose.

ANSWER 1: D

Problem 2:

Compute each person’s result carefully.

- Jose: start with 10  
  subtract 1 → 9  
  double → 18  
  add 2 → 20

- Thuy: start with 10  
  double → 20  
  subtract 1 → 19  
  add 2 → 21

- Kareem: start with 10  
  subtract 1 → 9  
  add 2 → 11  
  double → 22

Compare:
- Jose = 20
- Thuy = 21
- Kareem = 22

Kareem gets the largest final answer.

ANSWER 2: C

Problem 3:

\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

So the correct decimal is 0.08.

ANSWER 3: B

Problem 4:

We are choosing 3 starters from 4 players: Lance, Sally, Joy, and Fred.

This is a combination problem:
\[
\binom{4}{3} = 4
\]

You can also see it by noticing that choosing 3 starters is the same as choosing which 1 player sits out, and there are 4 choices.

ANSWER 4: B

Problem 5:

Let the number be \(n\).

20% of \(n\) is 12, so:
\[
0.2n = 12
\]
\[
n = 60
\]

Now find 30% of 60:
\[
0.3 \times 60 = 18
\]

ANSWER 5: B

Problem 6:

The spinner starts pointing west.

First move: clockwise \(2\frac{1}{4}\) revolutions.  
A full revolution brings it back to west, so only the extra \(\frac{1}{4}\) revolution matters.

From west, a quarter-turn clockwise points it to north.

Second move: counterclockwise \(3\frac{3}{4}\) revolutions.  
Again, the 3 full revolutions do nothing, so this is just \(\frac{3}{4}\) of a revolution counterclockwise, which is the same as \(\frac{1}{4}\) of a revolution clockwise.

From north, a quarter-turn clockwise points east.

So the spinner ends pointing east.

ANSWER 6: B

Problem 7:

Start with 30 liters total.

Yellow tint is 30% of 30 liters:
\[
0.30 \times 30 = 9 \text{ liters}
\]

Then 5 liters of yellow tint are added, so yellow becomes:
\[
9 + 5 = 14 \text{ liters}
\]

New total mixture:
\[
30 + 5 = 35 \text{ liters}
\]

Percent yellow in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.4 = 40\%
\]

ANSWER 7: C

Problem 8:

The fractions are:
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]

Approximate each one:

\[
\frac{15}{11} \approx 1.364
\]
\[
\frac{17}{13} \approx 1.308
\]
\[
\frac{19}{15} \approx 1.267
\]

So from least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

That matches choice E.

ANSWER 8: E

Problem 9:

The numbers are:
\[
3,4,5,6,6,7,x
\]

We are told the mean, median, and unique mode are all equal.

Since 6 appears twice and every other listed number appears once, the unique mode is 6.

So the median and mean must also be 6.

The median of 7 numbers is the 4th number when ordered.  
The sorted list is:
\[
3,4,5,6,6,7,x
\]
To make the median 6, \(x\) must be 6 or larger, but we also need the mean to be 6.

Sum of the known numbers:
\[
3+4+5+6+6+7 = 31
\]

For the mean to be 6, total sum must be:
\[
7 \times 6 = 42
\]

So:
\[
31 + x = 42
\]
\[
x = 11
\]

Check:
- mode = 6
- median = 6
- mean = 42/7 = 6

Works perfectly.

ANSWER 9: D

Problem 10:

Joe walked half the distance in 6 minutes.  
He then ran the other half at 3 times his walking speed.

If he runs 3 times as fast, then the time for the same distance is 1/3 as long.

So the running time is:
\[
6 \div 3 = 2 \text{ minutes}
\]

Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

Problem 11:

Add the first group:
\[
1+11+21+31+41 = 105
\]

Add the second group:
\[
9+19+29+39+49 = 145
\]

Now add the totals:
\[
105 + 145 = 250
\]

ANSWER 11: E

Problem 12:

Compare the decimals:

- \(0.99\)
- \(0.9099\)
- \(0.9 = 0.9000\)
- \(0.909\)
- \(0.9009\)

The largest is clearly \(0.99\), since it is closest to 1.

ANSWER 12: A

Problem 1:  
We have a three-person-team tournament where **the losing team of each game is eliminated**. With **16 teams**, how many games are played to determine the winner?

To find the number of games in a single-elimination tournament:  
- Each game eliminates exactly **one** team (the loser).  
- To go from **16 teams to 1 winning team**, we must eliminate **15 teams**.  
- Therefore, the number of games is **15**.

So the correct choice is **D**.

ANSWER 1: D


Problem 2:  
Jose, Thuy, and Kareem all start at 10. We compute each final value.

**Jose:**  
- Subtract 1: \(10 - 1 = 9\)  
- Double: \(2 \cdot 9 = 18\)  
- Add 2: \(18 + 2 = 20\)  
Final: **20**

**Thuy:**  
- Double 10: \(2 \cdot 10 = 20\)  
- Subtract 1: \(20 - 1 = 19\)  
- Add 2: \(19 + 2 = 21\)  
Final: **21**

**Kareem:**  
- Subtract 1: \(10 - 1 = 9\)  
- Add 2: \(9 + 2 = 11\)  
- Double: \(2 \cdot 11 = 22\)  
Final: **22**

Largest is **22**, which is Kareem.

ANSWER 2: C


Problem 3:  
We need to evaluate:  
\[
\frac{2}{25}
\]
Convert to decimals:  
\[
\frac{2}{25} = 2 \div 25
\]
Since \(25 \cdot 0.08 = 2\):  
\[
2 \div 25 = 0.08
\]
So the correct choice is **B** (.08).

ANSWER 3: B


Problem 4:  
We have 4 people chosen: Lance, Sally, Joy, Fred. We need to choose **3 starters**.

Number of ways:
\[
\binom{4}{3} = 4
\]
List them conceptually: leaving out each one individually gives 4 choices.

Correct choice: **B**.

ANSWER 4: B


Problem 5:  
“If 20% of a number is 12, what is 30% of the same number?”

Let the number be \(N\).  
\[
0.20N = 12
\]
\[
N = \frac{12}{0.20} = 60
\]
Now compute \(30\%\) of 60:
\[
0.30 \cdot 60 = 18
\]
Correct choice: **B**.

ANSWER 5: B


Problem 6:  
Spinner starts pointing **west**.

Chenille moves it:
- **Clockwise** \(2\frac{1}{4}\) revolutions  
- then **Counterclockwise** \(3\frac{3}{4}\) revolutions

Net rotation (clockwise minus counterclockwise):
\[
2\frac{1}{4} - 3\frac{3}{4}
\]
Convert to improper fractions:
\[
2\frac{1}{4} = \frac{9}{4}, \quad 3\frac{3}{4} = \frac{15}{4}
\]
\[
\frac{9}{4} - \frac{15}{4} = -\frac{6}{4} = -\frac{3}{2}
\]
The negative means **net counterclockwise** by \(\frac{3}{2}\) revolutions.

\(\frac{3}{2} = 1 + \frac{1}{2}\), so this is:
- 1 full turn (brings it back where it started)
- plus \(\frac{1}{2}\) turn counterclockwise

Half a revolution from **west** points **east** (opposite direction).

Correct answer: **B (east)**.

ANSWER 6: B


Problem 7:  
Original mixture: 30 L with  
- 25% red (doesn’t matter directly)
- 30% yellow  
- 45% water

Yellow amount initially:
\[
0.30 \cdot 30 = 9 \text{ liters}
\]
Add 5 liters of yellow tint, so new yellow amount:
\[
9 + 5 = 14
\]
New total volume:
\[
30 + 5 = 35
\]
Percent yellow in new mixture:
\[
\frac{14}{35} = 0.4 = 40\%
\]
Correct choice: **C**.

ANSWER 7: C


Problem 8:  
Order fractions least to greatest:
\[
\frac{15}{11},\ \frac{19}{15},\ \frac{17}{13}
\]
A good way: compare values by cross-multiplication or decimal approximations.

Compute approximate decimals:
\[
\frac{15}{11} \approx 1.3636
\]
\[
\frac{19}{15} \approx 1.2667
\]
\[
\frac{17}{13} \approx 1.3077
\]
Least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]
Check choice: that matches **E**:
- \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11}\)

ANSWER 8: E


Problem 9:  
Numbers: \(3,4,5,6,6,7,x\) (7 positive integers).  
Mean = median = unique mode.

**Step 1: Mode**  
The only number that repeats is **6**, occurring twice. So the **unique mode = 6**.

Thus median must be 6 as well.

**Step 2: Median = 6**  
For 7 numbers sorted, the median is the **4th** number.

Sorted list must have the 4th term equal to 6. We already have two 6’s and surrounding numbers 3,4,5,7.

Currently without \(x\), the ordered portion is:
\[
3,4,5,6,6,7,\ ?
\]
The 4th number here is 6 already. To keep the 4th number as 6, \(x\) must not be less than 6 (otherwise it would shift positions so 4th might change).

If \(x = 6\), then 6 occurs 3 times, still mode 6 (unique mode). Then list:
\[
3,4,5,6,6,6,7
\]
Median (4th) is 6—works.

If \(x=7\), list:
\[
3,4,5,6,6,7,7
\]
Mode would be 6? Actually 7 repeats too, so mode not unique.

If \(x>7\), list:
\[
3,4,5,6,6,7,x
\]
Median still 6, but mode would still be 6 (unique), since only 6 repeats. That would also work—wait, but then 6 is still unique mode, yes. However we also need **mean = 6**.

So use mean condition.

Sum of known numbers:
\[
3+4+5+6+6+7 = 31
\]
Total sum including \(x\): \(31 + x\)

Mean:
\[
\frac{31+x}{7} = 6
\]
\[
31 + x = 42
\]
\[
x = 11
\]

Check uniqueness of mode: 6 occurs twice, no other repeats (since x=11 is different). Unique mode is 6—consistent.

ANSWER 9: D


Problem 10:  
Joe walks half the distance, then runs the rest.  
He runs **3 times as fast** as he walks.

Given: time to walk half way = **6 minutes**.  
Let walking speed be \(v\). Then walking time for half distance \(=\) 6 minutes.

Let half the distance be \(d\). Then:
\[
d = v \cdot 6
\]
Running speed is \(3v\).  
Remaining distance is also \(d\), so running time:
\[
\text{time} = \frac{d}{3v} = \frac{v\cdot 6}{3v} = 2
\]
Total time:
\[
6 + 2 = 8
\]
Correct choice: **D**.

ANSWER 10: D


Problem 11:  
Compute:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]

Group each set:

First set:
\[
1+11+21+31+41
\]
These are all of the form \(10k+1\):  
\(1, 11, 21, 31, 41\) correspond to \(k=0,1,2,3,4\).

Sum:
\[
(1+11+21+31+41) = (1+11+21+31+41)
\]
Compute directly:
\[
(1+11)=12,\ (21+31)=52,\ 12+52=64,\ 64+41=105
\]

Second set:
\[
9+19+29+39+49
\]
Direct sum:
\[
(9+19)=28,\ (29+39)=68,\ 28+68=96,\ 96+49=145
\]

Total:
\[
105 + 145 = 250
\]

Correct choice: **E**.

ANSWER 11: E


Problem 12:  
Which number is largest?

Options:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare decimals:  
- .99 is the largest among all because it is greater than .9 (and greater than any number starting 0.90... but with fewer 9s afterward).

So .99 > .9099 > .909 > .9009 > .9.

Correct choice: **A**.

ANSWER 12: A

Problem 1:
In a single-elimination tournament, every game results in exactly one team being eliminated. To determine a single winner from 16 teams, 15 teams must be eliminated. Since each game eliminates one team, it takes exactly 15 games.
ANSWER 1: D

Problem 2:
- Jose: (10 - 1) = 9; 9 * 2 = 18; 18 + 2 = 20.
- Thuy: (10 * 2) = 20; 20 - 1 = 19; 19 + 2 = 21.
- Kareem: (10 - 1) = 9; 9 + 2 = 11; 11 * 2 = 22.
Kareem has the largest answer (22).
ANSWER 2: C

Problem 3:
To convert 2/25 to a decimal, multiply the numerator and denominator by 4 to get a denominator of 100: (2 * 4) / (25 * 4) = 8 / 100 = 0.08.
ANSWER 3: B

Problem 4:
We need to choose 3 starters from a group of 4 (Lance, Sally, Joy, Fred). This is a combination problem denoted as 4C3. Choosing 3 people to be on the team is the same as choosing 1 person to be left off the team. There are 4 ways to choose who is left off.
ANSWER 4: B

Problem 5:
Let $x$ be the number. $0.20x = 12$. Dividing by 0.20, $x = 12 / 0.2 = 60$. We want 30% of $x$: $0.30 * 60 = 18$.
ANSWER 5: B

Problem 6:
Start at 0 revolutions. Clockwise is positive, counterclockwise is negative.
$2 \frac{1}{4} - 3 \frac{3}{4} = 2.25 - 3.75 = -1.5$ revolutions.
-1.5 revolutions is the same as -0.5 revolutions (since -1 is a full circle). -0.5 revolutions (counterclockwise) from West points to East.
ANSWER 6: B

Problem 7:
Original yellow tint: 30% of 30 liters = 9 liters.
New mixture: 30 + 5 = 35 liters.
New yellow tint: 9 + 5 = 14 liters.
Percent: (14 / 35) * 100 = (2 / 5) * 100 = 40%.
ANSWER 7: C

Problem 8:
Convert to decimals or compare differences from 1:
15/11 = 1 + 4/11 ≈ 1.3636
19/15 = 1 + 4/15 ≈ 1.2666
17/13 = 1 + 4/13 ≈ 1.3076
Comparing the fractions 4/11, 4/13, and 4/15: the larger the denominator, the smaller the fraction. Thus, 4/15 < 4/13 < 4/11.
Order: 19/15 < 17/13 < 15/11.
ANSWER 8: E

Problem 9:
The set is {3, 4, 5, 6, 6, 7, x}. The mode is 6 (it appears twice). For the mean and median to be 6:
Median: With 7 numbers, the 4th number must be 6. Since the sorted list is 3, 4, 5, 6, 6, 7, x, if $x > 6$, the 4th number is 6.
Mean: (3+4+5+6+6+7+x) / 7 = 6 => 31 + x = 42 => x = 11.
Checking: {3, 4, 5, 6, 6, 7, 11}. Median is 6. Mean is 42/7 = 6. Mode is 6.
ANSWER 9: D

Problem 10:
Walking time = 6 minutes. Running speed = 3 * walking speed.
Time taken = Distance / Speed. Since speed is 3 times faster, time is 1/3 of the walking time.
Running time = 6 / 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
ANSWER 10: D

Problem 11:
Sum 1: (1+11+21+31+41) = 105.
Sum 2: (9+19+29+39+49) = 145.
105 + 145 = 250.
ANSWER 11: E

Problem 12:
Compare digit by digit:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
0.99 is the largest.
ANSWER 12: A

Problem 1:
This is a single-elimination tournament with 16 teams. Exactly one team is eliminated per game, and 15 teams must be eliminated to leave a single winner. Thus exactly 15 games are required.

ANSWER 1: D

Problem 2:
Jose evaluates \((10-1)\times2+2=20\).
Thuy evaluates \((10\times2)-1+2=21\).
Kareem evaluates \((10-1+2)\times2=22\).
Kareem obtains the strictly largest value.

ANSWER 2: C

Problem 3:
Division yields the decimal expansion \(2\div25=0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters out of 4 distinct players is the binomial coefficient \(\binom{4}{3}=4\).

ANSWER 4: B

Problem 5:
If 20% of a number equals 12, the number itself is \(12/0.2=60\). Then 30% of 60 is \(0.3\times60=18\).

ANSWER 5: B

Problem 6:
A net rotation of \(+2.25-3.75=-1.5\) revolutions is equivalent to 1.5 revolutions counterclockwise. From an initial westward orientation, 1.5 revolutions counterclockwise lands exactly on east.

ANSWER 6: B

Problem 7:
The original mixture contains \(0.30\times30=9\) liters of yellow tint. Adding 5 liters raises the yellow volume to 14 liters and the total volume to 35 liters, so the new percentage is \(14/35=40\%\).

ANSWER 7: C

Problem 8:
Decimal approximations are \(15/11\approx1.3636\), \(19/15\approx1.2667\), and \(17/13\approx1.3077\). Ordering these values produces \(19/15<17/13<15/11\).

ANSWER 8: E

Problem 9:
The unique mode must be 6, forcing the mean also to equal 6. The sum of the seven numbers is therefore 42. The six given numbers sum to 31, so \(x=11\). With this value the ordered list is 3,4,5,6,6,7,11 whose median is likewise 6 and whose unique mode remains 6.

ANSWER 9: D

Problem 10:
Let the total distance be \(2d\). Walking time for distance \(d\) is 6 min, so running time for the remaining distance \(d\) at triple speed is \(6/3=2\) min. The total time is therefore 8 min.

ANSWER 10: D

Problem 11:
Each pair sums to a multiple of 10: \(1+9=10\), \(11+19=30\), \(21+29=50\), \(31+39=70\), \(41+49=90\). Their total is 250. Equivalently each five-term arithmetic series sums to 105 and 145 respectively, again yielding 250.

ANSWER 11: E

Problem 12:
Direct comparison of the five decimals shows that 0.99 is strictly larger than each of the others.

ANSWER 12: A

Problem 1:
The problem asks us to determine the number of games that will be played to determine the winner of a basketball tournament with 16 teams, where the losing team of each game is eliminated.

## Step 1
To determine the number of games needed to find the winner, we first note that in a single-elimination tournament, each game eliminates one team.

## Step 2
Since there can only be one winner, all other teams must be eliminated. Therefore, the number of games required is equal to the number of teams minus one.

## Step 3
We have 16 teams competing in the tournament. So, the number of games needed to determine the winner is 16 - 1 = 15.

## Step 4
Let's examine the given choices to rule out the incorrect options. A. 4 is too low because it doesn't account for all the teams. B. 7 is also too low for the same reason. C. 8 is still too low. D. 15 matches our calculation. E. 16 is one more than our calculation.

The final answer is supported by our step-by-step reasoning.
ANSWER 1: D

Problem 2:
The problem asks us to determine who among Jose, Thuy, and Kareem gets the largest final answer after performing a series of operations starting with the number 10.

## Step 1
Let's calculate Jose's final answer: Jose subtracts 1 from 10 to get 9, doubles it to get 18, and then adds 2 to get 20.

## Step 2
Now, let's calculate Thuy's final answer: Thuy doubles 10 to get 20, subtracts 1 to get 19, and then adds 2 to get 21.

## Step 3
Next, let's calculate Kareem's final answer: Kareem subtracts 1 from 10 to get 9, adds 2 to get 11, and then doubles it to get 22.

## Step 4
Comparing the final answers: Jose got 20, Thuy got 21, and Kareem got 22. The largest final answer is 22, which belongs to Kareem.

## Step 5
Let's examine the given choices. A. Jose is incorrect because his answer is 20. B. Thuy is incorrect because her answer is 21. C. Kareem is correct because his answer is 22. D. Jose and Thuy is incorrect because Kareem's answer is larger. E. Thuy and Kareem is incorrect because Jose's answer is not the largest.

The final answer is supported by our step-by-step reasoning.
ANSWER 2: C

Problem 3:
The problem asks us to convert the fraction 2/25 into a decimal.

## Step 1
To convert 2/25 into a decimal, we divide the numerator by the denominator: 2 ÷ 25 = 0.08.

## Step 2
Let's examine the given choices to identify the correct decimal representation. A. .008 is too small. B. .08 matches our calculation. C. .8 is too large. D. 1.25 is much too large. E. 12.5 is also much too large.

The final answer is supported by our step-by-step reasoning.
ANSWER 3: B

Problem 4:
The problem asks us to determine the number of ways to choose 3 starters from a team of 4 members: Lance, Sally, Joy, and Fred.

## Step 1
This is a combination problem, because the order in which the starters are chosen does not matter.

## Step 2
The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial, the product of all positive integers up to that number.

## Step 3
Here, n = 4 (total team members) and k = 3 (number of starters to be chosen). So, C(4, 3) = 4! / [3!(4 - 3)!] = 4! / (3! * 1!) = (4 * 3 * 2 * 1) / ((3 * 2 * 1) * 1) = 4.

## Step 4
Let's examine the given choices. A. 2 is too low. B. 4 matches our calculation. C. 6 is too high. D. 8 is also too high. E. 10 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 4: B

Problem 5:
The problem asks us to find 30% of a number, given that 20% of the same number is 12.

## Step 1
First, we need to find the number. We know that 20% of the number is 12. So, 0.20 * number = 12.

## Step 2
To find the number, we divide both sides of the equation by 0.20: number = 12 / 0.20 = 60.

## Step 3
Now, we need to calculate 30% of 60: 0.30 * 60 = 18.

## Step 4
Let's examine the given choices. A. 15 is too low. B. 18 matches our calculation. C. 20 is too high. D. 24 is also too high. E. 30 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 5: B

Problem 6:
The problem asks us to determine the direction the spinner points after Chenille moves it clockwise 2 1/4 revolutions and then counterclockwise 3 3/4 revolutions, starting from pointing west.

## Step 1
First, let's understand the effect of the movements. A full revolution is 360 degrees or 4 quarters. So, 2 1/4 revolutions clockwise is equivalent to 9 quarters clockwise.

## Step 2
9 quarters clockwise is the same as 1 quarter clockwise (since 8 quarters make 2 full revolutions, which brings the spinner back to the original direction). So, after the first move, the spinner points north.

## Step 3
Then, Chenille moves it counterclockwise 3 3/4 revolutions, which is equivalent to 15 quarters counterclockwise.

## Step 4
15 quarters counterclockwise is the same as 3 quarters counterclockwise (since 12 quarters make 3 full revolutions). So, moving 3 quarters counterclockwise from north results in pointing west.

## Step 5
Let's examine the given choices. A. north is incorrect because after the second move, it's not north. B. east is incorrect. C. south is incorrect. D. west matches our calculation. E. northwest is incorrect.

The final answer is supported by our step-by-step reasoning.
ANSWER 6: D

Problem 7:
The problem asks us to find the percentage of yellow tint in a new mixture after adding 5 liters of yellow tint to an original 30-liter mixture that is 25% red tint, 30% yellow tint, and 45% water.

## Step 1
First, we calculate the amount of yellow tint in the original mixture: 30% of 30 liters = 0.30 * 30 = 9 liters.

## Step 2
After adding 5 liters of yellow tint, the total amount of yellow tint becomes 9 + 5 = 14 liters.

## Step 3
The total volume of the new mixture is 30 + 5 = 35 liters.

## Step 4
To find the percentage of yellow tint in the new mixture, we divide the amount of yellow tint by the total volume and multiply by 100: (14 / 35) * 100 = 40%

Problem 1:
In a single-elimination tournament, each game eliminates exactly one team. To eliminate all but the champion, the number of games needed is one fewer than the number of teams. With 16 teams, the number of games is \(16 - 1 = 15\).

ANSWER 1: D

Problem 2:
Jose: \(10 - 1 = 9\), \(9 \times 2 = 18\), \(18 + 2 = 20\).
Thuy: \(10 \times 2 = 20\), \(20 - 1 = 19\), \(19 + 2 = 21\).
Kareem: \(10 - 1 = 9\), \(9 + 2 = 11\), \(11 \times 2 = 22\).
Kareem’s final answer of 22 is the largest.

ANSWER 2: C

Problem 3:
Divide 2 by 25: \(2 \div 25 = 0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters from 4 people is the combination \(\binom{4}{3} = 4\).

ANSWER 4: B

Problem 5:
Let the number be \(n\). \(20\%\) of \(n\) is \(0.2n = 12\), so \(n = 60\). Then \(30\%\) of \(60 = 0.3 \times 60 = 18\).

ANSWER 5: B

Problem 6:
Initial direction: west.
Clockwise \(2\frac{1}{4}\) revolutions: \(2\frac{1}{4} = 2.25\) revolutions clockwise.
Counterclockwise \(3\frac{3}{4}\) revolutions: \(3\frac{3}{4} = 3.75\) revolutions counterclockwise.
Net revolution = \(2.25 - 3.75 = -1.5\) revolutions, i.e., \(1.5\) revolutions counterclockwise.
A full revolution (1.0) returns to west. The remaining 0.5 revolution counterclockwise from west is a half-turn, which points to east.

ANSWER 6: B

Problem 7:
Original mixture: 30 liters, yellow tint = \(30\% \times 30 = 9\) liters.
Add 5 liters of yellow tint: new yellow = \(9 + 5 = 14\) liters; new total volume = \(30 + 5 = 35\) liters.
Percent yellow = \(\frac{14}{35} \times 100\% = 40\%\).

ANSWER 7: C

Problem 8:
Convert each fraction to a decimal for comparison:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
Order from least to greatest: \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

ANSWER 8: E

Problem 9:
The data set: 3, 4, 5, 6, 6, 7, \(x\). The mode is the most frequent number. With two 6’s, 6 is the mode unless \(x\) equals another number that appears twice, which would create a second mode, violating the “unique mode” condition. Therefore, \(x\) must not be 3, 4, 5, or 7. It could be 6, but if \(x=6\), the median is 6, while the mean is \((3+4+5+6+6+6+7)/7 = 37/7 \approx 5.29\), so mean ≠ median.  
For the mean to equal the median (which is 6, the middle of seven sorted numbers if \(x > 6\) or \(x=6\) only if placed correctly), we set the mean to 6. Sum without \(x\) = \(3+4+5+6+6+7 = 31\). Total needed for mean 6 = \(7 \times 6 = 42\). Thus \(x = 42 - 31 = 11\).  
Sorted: 3, 4, 5, 6, 6, 7, 11. Median = 6, mode = 6, mean = 6. Unique mode holds.

ANSWER 9: D

Problem 10:
Walking half the distance took 6 minutes, so walking speed = \(\frac{\text{half distance}}{6}\).
Running speed is 3 times walking speed. Time to run the second half = \(\frac{\text{half distance}}{3 \times (\text{half distance}/6)} = \frac{6}{3} = 2\) minutes.
Total time = \(6 + 2 = 8\) minutes.

ANSWER 10: D

Problem 11:
Pair the terms: \(1+49 = 50\), \(11+39 = 50\), \(21+29 = 50\), \(31+19 = 50\), \(41+9 = 50\). Five pairs of 50 give \(5 \times 50 = 250\).

ANSWER 11: E

Problem 12:
Compare by extending decimals to the same number of places:
\(0.99   = 0.9900\)
\(0.9099 = 0.9099\)
\(0.9    = 0.9000\)
\(0.909  = 0.9090\)
\(0.9009 = 0.9009\)
Clearly \(0.99 > 0.9099 > 0.909 > 0.9009 > 0.9\). The largest is \(0.99\).

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key Insight:** In a single-elimination tournament, every game eliminates exactly one team. To crown a single winner from 16 teams, we must eliminate 16 − 1 = 15 teams. Therefore, exactly 15 games are needed.

**Verification by rounds:**
- Round 1: 16 teams → 8 games → 8 teams left
- Round 2: 8 teams → 4 games → 4 teams left
- Round 3: 4 teams → 2 games → 2 teams left
- Round 4 (Final): 2 teams → 1 game → 1 winner

Total = 8 + 4 + 2 + 1 = 15 games.

Choices A (4), B (7), and C (8) are too small. Choice E (16) would mean every team is eliminated, leaving no winner.

ANSWER 1: D

---

Problem 2:
We track each person's calculation step by step, all starting from 10.

**Jose:** subtract 1 → 9; double → 18; add 2 → **20**
**Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
**Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

Comparing: 20 < 21 < 22. Kareem gets the largest final answer.

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

Choice A (.008) would be 8/1000. Choice C (.8) would be 8/10. Choices D and E are greater than 1, but 2/25 is clearly less than 1.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination problem — the order of selection doesn't matter:
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4$$

Equivalently, choosing 3 people to start is the same as choosing 1 person to sit out. There are 4 people, so 4 ways to pick who sits out.

The 4 possible starting lineups are: {L,S,J}, {L,S,F}, {L,J,F}, {S,J,F}.

ANSWER 4: B

---

Problem 5:
We're told 20% of a number is 12. Let the number be $x$.

$$0.20x = 12 \implies x = \frac{12}{0.20} = 60$$

Now find 30% of 60:
$$0.30 \times 60 = 18$$

**Shortcut:** 30% is $\frac{30}{20} = 1.5$ times 20%, so the answer is $1.5 \times 12 = 18$.

Choice A (15) would be 25%. Choice D (24) would be 40%. Choice E (30) would be 50%.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing **West**.

**Move 1:** Clockwise $2\frac{1}{4}$ revolutions.
- The 2 full revolutions bring it back to West.
- The extra $\frac{1}{4}$ revolution clockwise from West: West → **North**.

**Move 2:** Counterclockwise $3\frac{3}{4}$ revolutions.
- The 3 full revolutions bring it back to North.
- The extra $\frac{3}{4}$ revolution counterclockwise from North:
  - $\frac{1}{4}$ CCW from North → West
  - $\frac{2}{4}$ CCW from North → South
  - $\frac{3}{4}$ CCW from North → **East**

The spinner points East.

ANSWER 6: B

---

Problem 7:
Original mixture: 30 liters total.
- Red tint: 25% of 30 = 7.5 liters
- Yellow tint: 30% of 30 = 9 liters
- Water: 45% of 30 = 13.5 liters

Add 5 liters of yellow tint:
- New yellow tint = 9 + 5 = 14 liters
- New total mixture = 30 + 5 = 35 liters

Percent yellow tint in new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

Choice B (35%) is a trap for those who just add 5% to 30%. Choice D (45%) is the original water percentage.

ANSWER 7: C

---

Problem 8:
We compare $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$.

Rewrite each as a mixed number:
$$\frac{15}{11} = 1 + \frac{4}{11}, \quad \frac{19}{15} = 1 + \frac{4}{15}, \quad \frac{17}{13} = 1 + \frac{4}{13}$$

Since all three have the same numerator (4) in their fractional parts, we compare denominators. For fractions with equal numerators, a **larger denominator** means a **smaller fraction**:
$$15 > 13 > 11 \implies \frac{4}{15} < \frac{4}{13} < \frac{4}{11}$$

Therefore:
$$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$$

ANSWER 8: E

---

Problem 9:
The data set is: 3, 4, 5, 6, 6, 7, and $x$. The mean, median, and unique mode are all equal.

**Mode:** Since 6 already appears twice and all others appear once, the unique mode must be 6 (if $x$ were 3, 4, 5, or 7, we'd have two modes, violating "unique mode"). So mode = 6.

**Mean = 6:**
$$\frac{3 + 4 + 5 + 6 + 6 + 7 + x}{7} = 6$$
$$\frac{31 + x}{7} = 6 \implies 31 + x = 42 \implies x = 11$$

**Check Median:** With $x = 11$, the sorted list is: 3, 4, 5, **6**, 6, 7, 11. The median (4th value) is 6. ✓

All three measures equal 6. Choice B (6) would make 6 appear three times, but the mean would be $(31+6)/7 = 37/7 \neq 6$.

ANSWER 9: D

---

Problem 10:
Joe walked half the distance in 6 minutes. He then ran the remaining half at 3 times his walking speed.

Let walking speed = $v$. The half-distance = $6v$.

Running speed = $3v$. Time to run the second half:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{6v}{3v} = 2 \text{ minutes}$$

Total time = 6 + 2 = **8 minutes**.

Choice A (7) would result from incorrectly subtracting. Choices B, C, E involve unnecessary decimal complications.

ANSWER 10: D

---

Problem 11:
Compute $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$.

Pair the terms from the first group with terms from the second group (first with last, etc.):
- $1 + 49 = 50$
- $11 + 39 = 50$
- $21 + 29 = 50$
- $31 + 19 = 50$
- $41 + 9 = 50$

Total = $50 \times 5 = 250$.

Choice C (200) is a common error from miscounting pairs. Choice D (249) comes from arithmetic mistakes.

ANSWER 11: E

---

Problem 12:
Compare the five decimals by writing them to four decimal places:

- A: 0.9900
- B: 0.9099
- C: 0.9000
- D: 0.9090
- E: 0.9009

Comparing digit by digit from the tenths place: all have 9 in the tenths place. In the hundredths place, A has **9** while all others have **0**. So A is immediately the largest.

$0.99 > 0.9099 > 0.909 > 0.9009 > 0.9$

ANSWER 12: A

 **Problem 1:**
We need to find the number of games played in a single-elimination tournament with 16 teams. In single elimination, every game eliminates exactly one team. To determine a winner, 15 of the 16 teams must be eliminated. Therefore, exactly 15 games are required.  
(Another way: with 16 teams there are 8 games in the first round, 4 in the second, 2 in the third, and 1 final game: \(8+4+2+1=15\).)

ANSWER 1: D

---

**Problem 2:**
We compute each person’s final answer step by step, starting from 10.

- **Jose:**  
  \(10 - 1 = 9\)  
  \(9 \times 2 = 18\)  
  \(18 + 2 = 20\)

- **Thuy:**  
  \(10 \times 2 = 20\)  
  \(20 - 1 = 19\)  
  \(19 + 2 = 21\)

- **Kareem:**  
  \(10 - 1 = 9\)  
  \(9 + 2 = 11\)  
  \(11 \times 2 = 22\)

Comparing the results: \(20 < 21 < 22\). Kareem’s answer is the largest.

ANSWER 2: C

---

**Problem 3:**
We convert the fraction \(\frac{2}{25}\) to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100:
\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

ANSWER 3: B

---

**Problem 4:**
We must choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways to choose 3 from 4 is the combination \(\binom{4}{3}\):
\[
\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4
\]
Equivalently, choosing 3 starters is the same as choosing 1 person to sit out, and there are 4 choices for who sits out.

ANSWER 4: B

---

**Problem 5:**
Let the number be \(N\). We are given \(0.20N = 12\). Solving for \(N\):
\[
N = \frac{12}{0.20} = 60
\]
Now find 30% of 60:
\[
0.30 \times 60 = 18
\]
(Shortcut: 30% is \(1.5\) times 20%, so \(1.5 \times 12 = 18\).)

ANSWER 5: B

---

**Problem 6:**
The spinner starts pointing west.  
- Clockwise \(2\frac14\) revolutions: 2 full revolutions bring it back to west, then \(\frac14\) turn clockwise from west points it **north**.  
- Counterclockwise \(3\frac34\) revolutions: 3 full revolutions bring it back to north, then \(\frac34\) turn counterclockwise from north goes through west, south, and ends at **east**.

Net movement: \(3\frac34 - 2\frac14 = 1\frac12\) revolutions counterclockwise. One full revolution returns to west; an additional \(\frac12\) revolution counterclockwise from west points to **east**.

ANSWER 6: B

---

**Problem 7:**
Original mixture: 30 liters.  
- Yellow tint originally: \(30\% \times 30 = 9\) liters.  
After adding 5 liters of yellow tint:  
- New yellow tint: \(9 + 5 = 14\) liters.  
- New total mixture: \(30 + 5 = 35\) liters.  

Percent of yellow tint in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%
\]

ANSWER 7: C

---

**Problem 8:**
Compare \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\). Each is greater than 1. Write each as \(1\) plus a proper fraction:
\[
\frac{15}{11} = 1 + \frac{4}{11}, \qquad
\frac{19}{15} = 1 + \frac{4}{15}, \qquad
\frac{17}{13} = 1 + \frac{4}{13}
\]
Since the numerators are the same (4), the fraction with the largest denominator is smallest:
\[
\frac{4}{15} < \frac{4}{13} < \frac{4}{11}
\]
Adding 1 preserves the order:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

ANSWER 8: E

---

**Problem 9:**
The data set is \(3, 4, 5, 6, 6, 7, x\).  
- The current mode is 6 (appears twice). For 6 to remain the **unique** mode, \(x\) must not equal 3, 4, 5, or 7 (which would create a second mode).  
- We are told mean = median = mode = 6.  

Set the mean equal to 6:
\[
\frac{3+4+5+6+6+7+x}{7} = 6 \implies \frac{31+x}{7} = 6 \implies 31+x = 42 \implies x = 11
\]

Check: With \(x=11\), the sorted list is \(3, 4, 5, 6, 6, 7, 11\).  
- Median (4th term) = 6.  
- Mode = 6 (unique).  
- Mean = 6.  

All conditions are satisfied.

ANSWER 9: D

---

**Problem 10:**
Joe walks half the distance in 6 minutes. He runs the second half at 3 times his walking speed. Since distance is the same and speed is tripled, the time required is divided by 3:
\[
\text{Running time} = \frac{6}{3} = 2 \text{ minutes}
\]
Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

---

**Problem 11:**
Add the two groups:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]
Pair the terms to make convenient sums:
\[
(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 5 \times 50 = 250
\]

ANSWER 11: E

---

**Problem 12:**
Compare the decimals by writing them with the same number of places:
- A. \(0.9900\)
- B. \(0.9099\)
- C. \(0.9000\)
- D. \(0.9090\)
- E. \(0.9009\)

The tenths digit is 9 for all. Looking at the hundredths digit: A has 9, while B, D have 0 and C, E have 0. Thus \(0.9900\) is clearly the largest.

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key insight:** Each game eliminates exactly one team. To determine a winner from 16 teams, we must eliminate 15 teams (everyone except the champion).

Therefore, exactly 15 games must be played.

*Verification by rounds:* Round 1: 8 games (16→8 teams), Round 2: 4 games (8→4), Round 3: 2 games (4→2), Round 4: 1 game (2→1). Total = 8+4+2+1 = 15.

The tempting wrong answer E (16) counts teams instead of eliminations.

ANSWER 1: D

---

Problem 2:
We need to compute each person's final answer starting from 10.

**Jose:** Start with 10 → subtract 1: 10−1 = 9 → double: 9×2 = 18 → add 2: 18+2 = **20**

**Thuy:** Start with 10 → double: 10×2 = 20 → subtract 1: 20−1 = 19 → add 2: 19+2 = **21**

**Kareem:** Start with 10 → subtract 1: 10−1 = 9 → add 2: 9+2 = 11 → double: 11×2 = **22**

Kareem gets the largest final answer (22).

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

The tempting wrong answer A (0.008) would be 8/1000, and C (0.8) would be 8/10.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination: C(4,3) = 4!/(3! × 1!) = 4

Listing them out: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — that's 4 ways.

The tempting wrong answer C (6) would be C(4,2), choosing who's *not* a starter, but that also equals 4... Actually C(4,2)=6 is wrong here; we're choosing 3 from 4.

ANSWER 4: B

---

Problem 5:
We know 20% of a number is 12. We need to find 30% of the same number.

**Step 1:** Find the number.
$$0.20 \times n = 12 \implies n = \frac{12}{0.20} = 60$$

**Step 2:** Find 30% of 60.
$$0.30 \times 60 = 18$$

*Shortcut:* Since 30% is 1.5 times 20%, the answer is 1.5 × 12 = 18.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing west. We need to find the net rotation.

**Clockwise:** 2¼ revolutions
**Counterclockwise:** 3¾ revolutions

**Net movement:** 2¼ − 3¾ = −1½ revolutions (negative means counterclockwise)

So the spinner moves 1½ revolutions counterclockwise from west.

- 1 full revolution counterclockwise returns to **west**
- ½ revolution counterclockwise from west goes to **east**

(Counterclockwise from west: ¼ turn → south, ½ turn → east)

ANSWER 6: B

---

Problem 7:
We need to find the percent of yellow tint after adding 5 liters of yellow tint.

**Original mixture (30 liters):**
- Yellow tint: 30% of 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percent yellow tint:** (14/35) × 100 = 40%

The tempting wrong answer B (35) might come from simply adding 5% to 30%.

ANSWER 7: C

---

Problem 8:
We need to order 15/11, 19/15, and 17/13 from least to greatest.

Computing each as a decimal:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3077...

From least to greatest: **19/15 < 17/13 < 15/11**

*Alternative method:* Note each is 1 + (numerator − denominator)/denominator:
- 15/11 = 1 + 4/11
- 19/15 = 1 + 4/15
- 17/13 = 1 + 4/13

Since 4/15 < 4/13 < 4/11, the order is confirmed.

ANSWER 8: E

---

Problem 9:
The positive integers 3, 4, 5, 6, 6, 7, and x have equal mean, median, and unique mode.

**Mode:** Since 6 already appears twice and no other number does, the unique mode must be 6. So x ≠ 3, 4, 5, or 7 (which would create a second mode).

**Median:** With 7 numbers, the median is the 4th value when sorted. For the median to be 6, we need x ≥ 6 (so that 6 is the 4th value).

- If x = 6: sorted = 3, 4, 5, 6, 6, 6, 7 → median = 6 ✓
- If x > 7: sorted = 3, 4, 5, 6, 6, 7, x → median = 6 ✓

**Mean:** (3 + 4 + 5 + 6 + 6 + 7 + x)/7 = (31 + x)/7

Setting mean = 6: (31 + x)/7 = 6 → 31 + x = 42 → **x = 11**

**Verification:** Numbers: 3, 4, 5, 6, 6, 7, 11. Mean = 42/7 = 6, Median = 6, Mode = 6 ✓

ANSWER 9: D

---

Problem 10:
Joe walks halfway, then runs the rest. We need total time.

**Walking:** Half the distance takes 6 minutes.

**Running:** He runs 3 times as fast as he walks, so he covers the same distance in 1/3 the time.
- Time to run the second half = 6 ÷ 3 = 2 minutes

**Total time:** 6 + 2 = **8 minutes**

The tempting wrong answers come from setting up the problem incorrectly or mixing up distance and time relationships.

ANSWER 10: D

---

Problem 11:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

**First group:** 1 + 11 + 21 + 31 + 41 = 105

**Second group:** 9 + 19 + 29 + 39 + 49 = 145

**Total:** 105 + 145 = **250**

*Elegant method:* Pair terms: (1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250

ANSWER 11: E

---

Problem 12:
We need to find the largest number among the options. Writing each with the same number of decimal places:

- A. 0.99 = **0.9900**
- B. 0.9099 = 0.9099
- C. 0.9 = 0.9000
- D. 0.909 = 0.9090
- E. 0.9009 = 0.9009

Comparing digit by digit after the decimal point:
- First digit: all are 9
- Second digit: A has **9**, all others have 0

Since A has a 9 in the second decimal place while all others have 0, **0.99 is the largest**.

ANSWER 12: A

**Problem 1:**  
The tournament is single‑elimination. Starting with 16 teams, every game eliminates exactly one team. To reduce from 16 teams to a single champion we must eliminate 15 teams, and each elimination corresponds to one game.  

Number of games = 16 − 1 = **15**.  

ANSWER 1: D  

---

**Problem 2:**  
Start with 10 for each person.

- **Jose:**  
  Subtract 1 → 9; double → 18; add 2 → 20.  

- **Thuy:**  
  Double → 20; subtract 1 → 19; add 2 → 21.  

- **Kareem:**  
  Subtract 1 → 9; add 2 → 11; double → 22.  

The largest final value is 22, obtained by Kareem.  

ANSWER 2: C  

---

**Problem 3:**  

\[
\frac{2}{25}=0.08
\]

(2 ÷ 25 = 0.08).  

ANSWER 3: B  

---

**Problem 4:**  
We need to choose 3 starters from the 4 available players:  

\[
\binom{4}{3}=4
\]

ways.  

ANSWER 4: B  

---

**Problem 5:**  
Let the number be \(N\).  
\(0.20N = 12 \Rightarrow N = \frac{12}{0.20}=60\).  
\(30\%\) of \(N\) is \(0.30 \times 60 = 18\).  

ANSWER 5: B  

---

**Problem 6:**  
Interpret the fractions as mixed numbers:  

- Clockwise \(2\frac14 = 2.25\) revolutions.  
- Counter‑clockwise \(3\frac34 = 3.75\) revolutions.  

Net rotation (counter‑clockwise positive) = \(3.75 - 2.25 = 1.5\) revolutions.  

A full revolution returns to the same direction, so we only care about the extra \(0.5\) revolution.  
\(0.5\) revolution = 180° counter‑clockwise.  

Starting from west, turning 180° counter‑clockwise points to **east**.  

ANSWER 6: B  

---

**Problem 7:**  
Original yellow = \(30\%\) of 30 L = \(0.30 \times 30 = 9\) L.  
Add 5 L yellow → new yellow = \(9+5 = 14\) L.  
New total volume = \(30+5 = 35\) L.  

Percent yellow = \(\frac{14}{35}\times100 = 40\%\).  

ANSWER 7: C  

---

**Problem 8:**  
Compute the three fractions:

\[
\frac{15}{11}=1.3636\ldots,\qquad
\frac{19}{15}=1.2666\ldots,\qquad
\frac{17}{13}=1.3077\ldots
\]

Ordered from least to greatest:  

\(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

This matches option **E**.  

ANSWER 8: E  

---

**Problem 9:**  
Current list: \(3,4,5,6,6,7\). The unique mode is 6 (appears twice).  
For the mode to remain unique and equal to the mean and median, \(x\) must not create a tie.  
Let the mean = median = mode = \(m\).  

Sum of numbers = \(31 + x\).  
Mean = \(\frac{31 + x}{7}=m\).  
Median of the seven numbers is the 4th after sorting; with \(x>7\) the median stays 6, so \(m=6\).  

Thus \(\frac{31 + x}{7}=6 \Rightarrow 31 + x = 42 \Rightarrow x = 11\).

Check: sorted list becomes \(3,4,5,6,6,7,11\) → median = 6, mode = 6, mean = 42/7 = 6. All equal.  

ANSWER 9: D  

---

**Problem 10:**  
Let the walking speed be \(v\) and the half‑distance be \(d/2\).  

Walking time for half‑distance: \(\frac{d/2}{v}=6\) min ⇒ \(\frac{d}{v}=12\) min.  

Running speed = \(3v\).  
Running time for the second half: \(\frac{d/2}{3v}= \frac{d}{6v}= \frac{1}{6}\times\frac{d}{v}= \frac{12}{6}=2\) min.  

Total time = \(6 + 2 = 8\) minutes.  

ANSWER 10: D  

---

**Problem 11:**  

\[
\begin{aligned}
1+11+21+31+41 &= 105,\\
9+19+29+39+49 &= 145,\\
\text{Total} &= 105+145 = 250.
\end{aligned}
\]

ANSWER 11: E  

---

**Problem 12:**  
Write the numbers with four decimal places:

- .99 = 0.9900  
- .9099 = 0.9099  
- .9 = 0.9000  
- .909 = 0.9090  
- .9009 = 0.9009  

The largest is 0.9900, i.e., **.99**.  

ANSWER 12: A  

Problem 1:  
The tournament starts with 16 teams. Each game eliminates one team, and the losing team is eliminated. To determine the winner, all teams except one must be eliminated. Thus, 15 teams must be eliminated, requiring 15 games.  
- **Why not other choices?** A (4) is too few; B (7), C (8) are also insufficient; E (16) would imply one game per team without elimination, which is incorrect.  
ANSWER 1: D  

Problem 2:  
- **Jose**: Starts with 10 → subtracts 1 (9) → doubles (18) → adds 2 → **20**.  
- **Thuy**: Starts with 10 → doubles (20) → subtracts 1 (19) → adds 2 → **21**.  
- **Kareem**: Starts with 10 → subtracts 1 (9) → adds 2 (11) → doubles → **22**.  
Kareem has the largest final answer.  
- **Why not other choices?** A (Jose) and B (Thuy) are smaller; D and E pair incorrect winners.  
ANSWER 2: C  

Problem 3:  
\( \frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08 \).  
- **Why not other choices?** A (.008) is \( \frac{2}{250} \); C (.8) is \( \frac{4}{5} \); D and E are larger decimals.  
ANSWER 3: B  

Problem 4:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is \( \binom{4}{3} = 4 \).  
- **Why not other choices?** A (2) is too few; C (6) is \( \binom{4}{2} \), which is incorrect; D (8) and E (10) are overcounts.  
ANSWER 4: B  

Problem 5:  
Let the number be \( x \). If 20% of \( x \) is 12, then \( 0.2x = 12 \), so \( x = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).  
- **Why not other choices?** A (15) is 25% of 60; C (20) is \( \frac{1}{3} \) of 60; D (24) is 40%; E (30) is 50%.  
ANSWER 5: B  

Problem 6:  
- A full revolution is 360°. Clockwise \( 2 \frac{1}{4} \) revolutions = \( 2.25 \times 360 = 810° \).  
- Counterclockwise \( 3 \frac{3}{4} \) revolutions = \( -3.75 \times 360 = -1350° \).  
- Net movement: \( 810° - 1350° = -540° \).  
- \( -540° \mod 360° = 180° \) (since \( -540 + 720 = 180° \)), which points west → south? Wait, initial direction is west.  
- West (270°) + 810° = 1080° → 1080° mod 360° = 0° (north). Then -1350° → 0° - 1350° = -1350° mod 360° = 90° (east)? Let's reframe:  
- Start at west (270°). Clockwise 2.25 rev: 270° + 810° = 1080° mod 360° = 0° (north). Then counterclockwise 3.75 rev: 0° - 1350° = -1350° mod 360° = 90° (east).  
- **Why not other choices?** A (north) is initial after clockwise; B (east) is final; C (south) is 180°; D (west) is start; E (northwest) is not a cardinal direction.  
ANSWER 6: B  

Problem 7:  
Original mixture: 30 liters, 30% yellow tint → 9 liters yellow. Add 5 liters yellow → total yellow = 14 liters. New mixture = 35 liters. Percent yellow = \( \frac{14}{35} \times 100 = 40\% \).  
- **Why not other choices?** A (25) is original yellow%; B (35) is before addition; D (45) and E (50) are overestimates.  
ANSWER 7: E (Note: Correction based on computation: 14/35=0.4 → 40%, so answer is E? Wait, 14/35=0.4 → 40%, which is E. But earlier I thought B, but B is 35, which is wrong. Original yellow is 9, add 5 → 14. Total mixture 30+5=35. 14/35=0.4 → 40%. So E. But in choices E is 50? Wait, no: choices are A.25 B.35 C.40 D.45 E.50. So 40% is C. I misread. Correction: 14/35=0.4 → 40%, which is C.  
**Correction**: 14/35=0.4 → 40%, so answer is C.  
ANSWER 7: C  

Problem 8:  
Compare \( \frac{15}{11} \approx 1.3636 \), \( \frac{19}{15} \approx 1.2667 \), \( \frac{17}{13} \approx 1.3077 \). Order: \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11} \).  
- **Why not other choices?** A, B, C, D have incorrect orders.  
ANSWER 8: E  

Problem 9:  
The set is {3, 4, 5, 6, 6, 7, x}. Mean = median = mode = 6.  
- Mode is 6 (unique), so 6 must appear most.  
- Median is 6 (4th value in ordered list: 3,4,5,6,6,7,x → if x≥6, ordered is ...,6,6,x or ...,6,x,6? Must have 4th value 6.  
- If x=6, set is {3,4,5,6,6,6,7}: median=6, mode=6, mean=(3+4+5+6+6+6+7)/7=37/7≈5.28 ≠6.  
- If x=11: set {3,4,5,6,6,7,11}: mean=(3+4+5+6+6+7+11)/7=42/7=6, median=6, mode=6.  
- **Why not other choices?** A (5): mean=36/7≈5.14; B (6): mean≈5.28; C (7): mean=38/7≈5.43; D (11) works.  
ANSWER 9: D  

Problem 10:  
Joe walks half the distance in 6 minutes. Let walking speed be \( v \), so distance = \( 6v \). He runs at \( 3v \), so time to run half distance = \( \frac{6v}{3v} = 2 \) minutes. Total time = 6 + 2 = 8 minutes.  
- **Why not other choices?** A (7), B (7.3), C (7.7), E (8.3) are not exact.  
ANSWER 10: D  

Problem 11:  
First group: 1+11+21+31+41 = 105. Second group: 9+19+29+39+49 = 145. Total = 105 + 145 = 250.  
- **Why not other choices?** A (150), B (199), C (200), D (249) are incorrect sums.  
ANSWER 11: E  

Problem 12:  
Compare decimals: A (.99), B (.9099), C (.9), D (.909), E (.9009). Largest is .99.  
- **Why not other choices?** B, C, D, E are all smaller than .99.  
ANSWER 12: A

### Problem 1:
We need to find the total number of games to determine a winner in a 16-team single-elimination tournament, where every game eliminates exactly one losing team. To get 1 winner, we must eliminate 16 - 1 = 15 teams. Since each game eliminates 1 team, we need 15 total games. Wrong choices: A is the number of bracket rounds, not games; C is half the number of teams; E is the total number of teams, not games.
ANSWER 1: D

---

### Problem 2:
All three start with 10, we calculate each person's final result step-by-step:
- Jose: (10 - 1) * 2 + 2 = 9*2 +2 = 20
- Thuy: (10*2) -1 +2 = 20-1+2=21
- Kareem: (10-1 +2)*2 = 11*2=22
Kareem has the largest final answer. Wrong choices stem from misordering the operations for each person.
ANSWER 2: C

---

### Problem3:
Convert 2/25 to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100: (2*4)/(25*4) = 8/100 = 0.08. Wrong choices: A is 2/250, C is 20/25.
ANSWER 3: B

---

### Problem4:
We choose 3 starters from 4 people, which is a combination problem (order of starters does not matter here). The number of ways to choose 3 from 4 is equivalent to choosing 1 person to leave off the starting lineup: there are 4 people, so 4 unique lineups, each omitting one person. Wrong choice C (6) is the number of permutations, not combinations.
ANSWER 4: B

---

### Problem5:
If 20% of a number is 12, first find the full number: 0.2*N=12 → N=12/0.2=60. 30% of 60 is 0.3*60=18. Wrong choices: A is 25% of 60, D is 40% of 60.
ANSWER 5: B

---

### Problem6:
The spinner starts pointing west. Net movement: 2.25 clockwise revolutions minus 3.75 clockwise (equivalent to 3.75 counterclockwise) = -1.5 total clockwise revolutions, or 1.5 counterclockwise revolutions. Full revolutions cancel out, so we only have a net 0.5 (180 degree) counterclockwise turn from west. A 180-degree turn from west points to east.
ANSWER 6: B

---

### Problem7:
Original 30L mixture has 30% yellow tint: 0.3*30=9L of yellow. Add 5L of yellow, so new yellow is 14L, new total mixture is 35L. The new percentage of yellow is (14/35)*100=40%. Wrong choice: people often forget to add 5L to the total mixture volume, leading to an incorrect higher percentage.
ANSWER 7: C

---

### Problem8:
Rewrite each fraction to compare easily: 15/11 = 1 + 4/11, 19/15=1+4/15, 17/13=1+4/13. For equal numerators, the larger the denominator, the smaller the fraction. So 4/15 < 4/13 < 4/11, so the full order from least to greatest is 19/15 < 17/13 <15/11.
ANSWER 8: E

---

### Problem9:
The dataset is [3,4,5,6,6,7,x]. The unique mode is currently 6 (only value appearing twice, so it must stay the only unique mode). For the mean, median, and mode to all be 6: the sum of the dataset must be 7*6=42. The sum of the existing numbers is 3+4+5+6+6+7=31, so x=42-31=11. Sorted dataset [3,4,5,6,6,7,11] has median 6, unique mode 6, mean 6, which fits. Wrong choice B (x=6) gives a mean of ~5.28, not 6.
ANSWER 9: D

---

### Problem10:
Joe walks half the distance in 6 minutes. His running speed is 3x his walking speed, so time to cover the same distance is 1/3 of walking time. Time to run the second half: 6/3=2 minutes. Total time: 6+2=8 minutes.
ANSWER 10: D

---

### Problem11:
Pair terms to simplify addition: (1+49)+(11+39)+(21+29)+(31+19)+(41+9) = 50+50+50+50+50=250.
ANSWER 11: E

---

### Problem12:
Line up all decimals to compare: A=0.9900, B=0.9099, C=0.9000, D=0.9090, E=0.9009. The largest value is 0.99, option A.
ANSWER 12: A

Problem 1:  
In a single-elimination tournament, each game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams (all except 1). Thus, 15 games are played.  
ANSWER 1: D  

Problem 2:  
Calculate each person's result starting from 10:  
- Jose: \(10 -1 =9\), double → \(9*2=18\), add 2 → \(18+2=20\)  
- Thuy: Double 10 → \(10*2=20\), subtract 1 → \(20-1=19\), add 2 → \(19+2=21\)  
- Kareem: \(10-1=9\), add 2 → \(9+2=11\), double → \(11*2=22\)  
Kareem has the largest result.  
ANSWER 2: C  

Problem 3:  
Convert \(\frac{2}{25}\) to decimal: \(2\div25=0.08\).  
ANSWER 3: B  

Problem 4:  
Choosing 3 starters from 4 people is a combination (order does not matter). The number of ways is \(\binom{4}{3}=4\) (or list: {Lance,Sally,Joy}, {Lance,Sally,Fred}, {Lance,Joy,Fred}, {Sally,Joy,Fred}).  
ANSWER 4: B  

Problem 5:  
Let the number be \(x\). \(20\%x=12\) → \(0.2x=12\) → \(x=60\). \(30\%\ of\ 60=0.3*60=18\).  
ANSWER 5: B  

Problem 6:  
Model directions with standard angles: 0°=east, 90°=north, 180°=west, 270°=south. Initial direction is west (180°).  
- Clockwise \(2\frac{1}{4}\) revolutions: \(2.25*(-360°)=-810°\), new angle: \(180-810=-630°\)  
- Counterclockwise \(3\frac{3}{4}\) revolutions: \(3.75*(360°)=+1350°\), total angle: \(-630+1350=720°\)  
- \(720°\mod360°=0°\), which is east.  
ANSWER 6: B  

Problem 7:  
Original yellow tint: \(30\%\ of\ 30=0.3*30=9\) liters. Add 5 liters, total yellow = \(9+5=14\) liters. New total mixture: \(30+5=35\) liters. Percent yellow: \(\frac{14}{35}*100\%=40\%\).  
ANSWER 7: C  

Problem 8:  
Compare fractions via cross-multiplication or decimals:  
- \(\frac{19}{15}\approx1.267\), \(\frac{17}{13}\approx1.308\), \(\frac{15}{11}\approx1.364\)  
Order from least to greatest: \(\frac{19}{15}<\frac{17}{13}<\frac{15}{11}\).  
ANSWER 8: E  

Problem 9:  
The unique mode is 6 (appears twice, others once). Thus mean and median equal 6.  
- Median of 7 numbers is the 4th term when sorted, so 4th term =6 → \(x\geq6\).  
- Mean = \(\frac{3+4+5+6+6+7+x}{7}=6\) → \(31+x=42\) → \(x=11\).  
Check: Sorted list [3,4,5,6,6,7,11], median=6, mode=6 (unique), mean=6.  
ANSWER 9: D  

Problem 10:  
Walking half the distance takes 6 minutes, so full walking time would be 12 minutes. Running speed is 3x walking speed, so time for the second half is \(\frac{6}{3}=2\) minutes. Total time: \(6+2=8\) minutes.  
ANSWER 10: D  

Problem 11:  
Sum first group: \(1+11+21+31+41=105\) (arithmetic series: \(\frac{5}{2}(1+41)=105\)).  
Sum second group: \(9+19+29+39+49=145\) (\(\frac{5}{2}(9+49)=145\)).  
Total: \(105+145=250\).  
ANSWER 11: E  

Problem 12:  
Compare decimals digit by digit:  
- A=0.99, B=0.9099, C=0.9, D=0.909, E=0.9009  
A has the highest second decimal digit (9 vs 0 for others), so it is the largest.  
ANSWER 12: A

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** How many games must be played to determine a winner in a 16-team elimination tournament?

**Setup:** In an elimination tournament, each game eliminates exactly one team. To go from 16 teams to 1 winner, we must eliminate 15 teams. Therefore, we need 15 games.

**Alternatively:** We can think of it as a bracket:
- Round 1: 16 teams → 8 games → 8 winners
- Round 2: 8 teams → 4 games → 4 winners
- Round 3: 4 teams → 2 games → 2 winners
- Round 4: 2 teams → 1 game → 1 winner
- Total: 8 + 4 + 2 + 1 = 15 games

ANSWER 1: D

---

**Problem 2:**

**What's being asked:** Which person gets the largest final answer after their operations?

**Jose's calculation:** 
- Start: 10
- Subtract 1: 10 - 1 = 9
- Double: 9 × 2 = 18
- Add 2: 18 + 2 = **20**

**Thuy's calculation:**
- Start: 10
- Double: 10 × 2 = 20
- Subtract 1: 20 - 1 = 19
- Add 2: 19 + 2 = **21**

**Kareem's calculation:**
- Start: 10
- Subtract 1: 10 - 1 = 9
- Add 2: 9 + 2 = 11
- Double: 11 × 2 = **22**

Kareem gets 22, which is the largest.

ANSWER 2: C

---

**Problem 3:**

**What's being asked:** Convert the fraction 2/25 to a decimal.

**Setup:** 2/25 = ?

**Calculation:** 
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** In how many ways can 3 starters be chosen from 4 people (Lance, Sally, Joy, Fred)?

**Setup:** This is a combination problem: $\binom{4}{3}$

**Calculation:**
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4}{1} = 4$$

Alternatively, we're choosing which 1 person to leave out: 4 ways.

ANSWER 4: B

---

**Problem 5:**

**What's being asked:** If 20% of a number is 12, what is 30% of that number?

**Setup:** Let the number be $n$.
- $0.20n = 12$
- $n = 12 ÷ 0.20 = 60$

**Calculation:** 
$$30\% \text{ of } 60 = 0.30 \times 60 = 18$$

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Starting pointing west, after rotating clockwise 2¼ revolutions then counterclockwise 3¾ revolutions, which direction?

**Setup:** 
- Clockwise 2¼ revolutions = 2.25 revolutions
- Counterclockwise 3¾ revolutions = 3.75 revolutions
- Net: 3.75 - 2.25 = 1.5 revolutions counterclockwise (net)

**Calculation:**
Starting at west, rotating counterclockwise 1.5 revolutions (= 1 full rotation + 0.5 rotation):
- 1 full rotation brings us back to west
- 0.5 rotation = 180°
- West + 180° counterclockwise = East

ANSWER 6: B

---

**Problem 7:**

**What's being asked:** After adding 5 liters of yellow tint to a 30-liter mixture, what percent is yellow tint?

**Original mixture (30 liters):**
- Yellow tint: 30% × 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percentage:**
$$\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%$$

ANSWER 7: C

---

**Problem 8:**

**What's being asked:** Order 15/11, 19/15, and 17/13 from least to greatest.

**Setup:** Convert to decimals or compare directly.

**Calculations:**
- 15/11 ≈ 1.364
- 19/15 ≈ 1.267
- 17/13 ≈ 1.308

**Order:** 19/15 < 17/13 < 15/11

ANSWER 8: E

---

**Problem 9:**

**What's being asked:** Find $x$ such that the mean, median, and unique mode of {3, 4, 5, 6, 6, 7, $x$} are all equal.

**Current mode:** 6 appears twice; all others appear once. If there's a unique mode, it must be 6.

**With unique mode = 6:** This requires no other number to appear more than once. So $x \notin \{3, 4, 5, 7\}$ and $x \neq 6$ (else it wouldn't be unique).

**Mean = Median = Mode = 6:**

For median of 7 numbers with value 6:
- Sorted: need middle (4th) value to be 6
- If $x = 11$: {3, 4, 5, 6, 6, 7, 11} → median = 6 ✓

Mean: $\frac{3+4+5+6+6+7+11}{7} = \frac{42}{7} = 6$ ✓

Mode: 6 appears twice; all others once → mode = 6 ✓

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Total time from home to school if Joe walked halfway (6 min) then ran the rest at 3× speed?

**Setup:**
- Walking: 6 minutes for half the distance
- Running: same distance at 3× speed, so time = 6/3 = 2 minutes

**Total time:** 6 + 2 = 8 minutes

ANSWER 10: D

---

**Problem 11:**

**What's being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Calculation:**
- First group: 1 + 11 + 21 + 31 + 41 = 105
- Second group: 9 + 19 + 29 + 39 + 49 = 145
- Sum: 105 + 145 = 250

ANSWER 11: E

---

**Problem 12:**

**What's being asked:** Which number is largest?

**Comparison:**
- A. 0.99
- B. 0.9099
- C. 0.9
- D. 0.909
- E. 0.9009

Starting from the tenths place: all are 0.9__. Looking at hundredths:
- 0.99 has 9 in hundredths place → **largest**

ANSWER 12: A

Problem 1:

In a single-elimination tournament, each game eliminates exactly one team.  
There are 16 teams, and we want one champion.

So we must eliminate 15 teams total.

Therefore, the number of games played is 15.

Tempting wrong choice: 16 would be if every team had to lose, but the winner does not lose.

ANSWER 1: D

Problem 2:

Compute each person’s result carefully.

- Jose: start with 10  
  subtract 1 → 9  
  double → 18  
  add 2 → 20

- Thuy: start with 10  
  double → 20  
  subtract 1 → 19  
  add 2 → 21

- Kareem: start with 10  
  subtract 1 → 9  
  add 2 → 11  
  double → 22

Compare:
- Jose = 20
- Thuy = 21
- Kareem = 22

Kareem gets the largest final answer.

ANSWER 2: C

Problem 3:

\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

So the correct decimal is 0.08.

ANSWER 3: B

Problem 4:

We are choosing 3 starters from 4 players: Lance, Sally, Joy, and Fred.

This is a combination problem:
\[
\binom{4}{3} = 4
\]

You can also see it by noticing that choosing 3 starters is the same as choosing which 1 player sits out, and there are 4 choices.

ANSWER 4: B

Problem 5:

Let the number be \(n\).

20% of \(n\) is 12, so:
\[
0.2n = 12
\]
\[
n = 60
\]

Now find 30% of 60:
\[
0.3 \times 60 = 18
\]

ANSWER 5: B

Problem 6:

The spinner starts pointing west.

First move: clockwise \(2\frac{1}{4}\) revolutions.  
A full revolution brings it back to west, so only the extra \(\frac{1}{4}\) revolution matters.

From west, a quarter-turn clockwise points it to north.

Second move: counterclockwise \(3\frac{3}{4}\) revolutions.  
Again, the 3 full revolutions do nothing, so this is just \(\frac{3}{4}\) of a revolution counterclockwise, which is the same as \(\frac{1}{4}\) of a revolution clockwise.

From north, a quarter-turn clockwise points east.

So the spinner ends pointing east.

ANSWER 6: B

Problem 7:

Start with 30 liters total.

Yellow tint is 30% of 30 liters:
\[
0.30 \times 30 = 9 \text{ liters}
\]

Then 5 liters of yellow tint are added, so yellow becomes:
\[
9 + 5 = 14 \text{ liters}
\]

New total mixture:
\[
30 + 5 = 35 \text{ liters}
\]

Percent yellow in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.4 = 40\%
\]

ANSWER 7: C

Problem 8:

The fractions are:
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]

Approximate each one:

\[
\frac{15}{11} \approx 1.364
\]
\[
\frac{17}{13} \approx 1.308
\]
\[
\frac{19}{15} \approx 1.267
\]

So from least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

That matches choice E.

ANSWER 8: E

Problem 9:

The numbers are:
\[
3,4,5,6,6,7,x
\]

We are told the mean, median, and unique mode are all equal.

Since 6 appears twice and every other listed number appears once, the unique mode is 6.

So the median and mean must also be 6.

The median of 7 numbers is the 4th number when ordered.  
The sorted list is:
\[
3,4,5,6,6,7,x
\]
To make the median 6, \(x\) must be 6 or larger, but we also need the mean to be 6.

Sum of the known numbers:
\[
3+4+5+6+6+7 = 31
\]

For the mean to be 6, total sum must be:
\[
7 \times 6 = 42
\]

So:
\[
31 + x = 42
\]
\[
x = 11
\]

Check:
- mode = 6
- median = 6
- mean = 42/7 = 6

Works perfectly.

ANSWER 9: D

Problem 10:

Joe walked half the distance in 6 minutes.  
He then ran the other half at 3 times his walking speed.

If he runs 3 times as fast, then the time for the same distance is 1/3 as long.

So the running time is:
\[
6 \div 3 = 2 \text{ minutes}
\]

Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

Problem 11:

Add the first group:
\[
1+11+21+31+41 = 105
\]

Add the second group:
\[
9+19+29+39+49 = 145
\]

Now add the totals:
\[
105 + 145 = 250
\]

ANSWER 11: E

Problem 12:

Compare the decimals:

- \(0.99\)
- \(0.9099\)
- \(0.9 = 0.9000\)
- \(0.909\)
- \(0.9009\)

The largest is clearly \(0.99\), since it is closest to 1.

ANSWER 12: A

Problem 1:  
We have a three-person-team tournament where **the losing team of each game is eliminated**. With **16 teams**, how many games are played to determine the winner?

To find the number of games in a single-elimination tournament:  
- Each game eliminates exactly **one** team (the loser).  
- To go from **16 teams to 1 winning team**, we must eliminate **15 teams**.  
- Therefore, the number of games is **15**.

So the correct choice is **D**.

ANSWER 1: D


Problem 2:  
Jose, Thuy, and Kareem all start at 10. We compute each final value.

**Jose:**  
- Subtract 1: \(10 - 1 = 9\)  
- Double: \(2 \cdot 9 = 18\)  
- Add 2: \(18 + 2 = 20\)  
Final: **20**

**Thuy:**  
- Double 10: \(2 \cdot 10 = 20\)  
- Subtract 1: \(20 - 1 = 19\)  
- Add 2: \(19 + 2 = 21\)  
Final: **21**

**Kareem:**  
- Subtract 1: \(10 - 1 = 9\)  
- Add 2: \(9 + 2 = 11\)  
- Double: \(2 \cdot 11 = 22\)  
Final: **22**

Largest is **22**, which is Kareem.

ANSWER 2: C


Problem 3:  
We need to evaluate:  
\[
\frac{2}{25}
\]
Convert to decimals:  
\[
\frac{2}{25} = 2 \div 25
\]
Since \(25 \cdot 0.08 = 2\):  
\[
2 \div 25 = 0.08
\]
So the correct choice is **B** (.08).

ANSWER 3: B


Problem 4:  
We have 4 people chosen: Lance, Sally, Joy, Fred. We need to choose **3 starters**.

Number of ways:
\[
\binom{4}{3} = 4
\]
List them conceptually: leaving out each one individually gives 4 choices.

Correct choice: **B**.

ANSWER 4: B


Problem 5:  
“If 20% of a number is 12, what is 30% of the same number?”

Let the number be \(N\).  
\[
0.20N = 12
\]
\[
N = \frac{12}{0.20} = 60
\]
Now compute \(30\%\) of 60:
\[
0.30 \cdot 60 = 18
\]
Correct choice: **B**.

ANSWER 5: B


Problem 6:  
Spinner starts pointing **west**.

Chenille moves it:
- **Clockwise** \(2\frac{1}{4}\) revolutions  
- then **Counterclockwise** \(3\frac{3}{4}\) revolutions

Net rotation (clockwise minus counterclockwise):
\[
2\frac{1}{4} - 3\frac{3}{4}
\]
Convert to improper fractions:
\[
2\frac{1}{4} = \frac{9}{4}, \quad 3\frac{3}{4} = \frac{15}{4}
\]
\[
\frac{9}{4} - \frac{15}{4} = -\frac{6}{4} = -\frac{3}{2}
\]
The negative means **net counterclockwise** by \(\frac{3}{2}\) revolutions.

\(\frac{3}{2} = 1 + \frac{1}{2}\), so this is:
- 1 full turn (brings it back where it started)
- plus \(\frac{1}{2}\) turn counterclockwise

Half a revolution from **west** points **east** (opposite direction).

Correct answer: **B (east)**.

ANSWER 6: B


Problem 7:  
Original mixture: 30 L with  
- 25% red (doesn’t matter directly)
- 30% yellow  
- 45% water

Yellow amount initially:
\[
0.30 \cdot 30 = 9 \text{ liters}
\]
Add 5 liters of yellow tint, so new yellow amount:
\[
9 + 5 = 14
\]
New total volume:
\[
30 + 5 = 35
\]
Percent yellow in new mixture:
\[
\frac{14}{35} = 0.4 = 40\%
\]
Correct choice: **C**.

ANSWER 7: C


Problem 8:  
Order fractions least to greatest:
\[
\frac{15}{11},\ \frac{19}{15},\ \frac{17}{13}
\]
A good way: compare values by cross-multiplication or decimal approximations.

Compute approximate decimals:
\[
\frac{15}{11} \approx 1.3636
\]
\[
\frac{19}{15} \approx 1.2667
\]
\[
\frac{17}{13} \approx 1.3077
\]
Least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]
Check choice: that matches **E**:
- \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11}\)

ANSWER 8: E


Problem 9:  
Numbers: \(3,4,5,6,6,7,x\) (7 positive integers).  
Mean = median = unique mode.

**Step 1: Mode**  
The only number that repeats is **6**, occurring twice. So the **unique mode = 6**.

Thus median must be 6 as well.

**Step 2: Median = 6**  
For 7 numbers sorted, the median is the **4th** number.

Sorted list must have the 4th term equal to 6. We already have two 6’s and surrounding numbers 3,4,5,7.

Currently without \(x\), the ordered portion is:
\[
3,4,5,6,6,7,\ ?
\]
The 4th number here is 6 already. To keep the 4th number as 6, \(x\) must not be less than 6 (otherwise it would shift positions so 4th might change).

If \(x = 6\), then 6 occurs 3 times, still mode 6 (unique mode). Then list:
\[
3,4,5,6,6,6,7
\]
Median (4th) is 6—works.

If \(x=7\), list:
\[
3,4,5,6,6,7,7
\]
Mode would be 6? Actually 7 repeats too, so mode not unique.

If \(x>7\), list:
\[
3,4,5,6,6,7,x
\]
Median still 6, but mode would still be 6 (unique), since only 6 repeats. That would also work—wait, but then 6 is still unique mode, yes. However we also need **mean = 6**.

So use mean condition.

Sum of known numbers:
\[
3+4+5+6+6+7 = 31
\]
Total sum including \(x\): \(31 + x\)

Mean:
\[
\frac{31+x}{7} = 6
\]
\[
31 + x = 42
\]
\[
x = 11
\]

Check uniqueness of mode: 6 occurs twice, no other repeats (since x=11 is different). Unique mode is 6—consistent.

ANSWER 9: D


Problem 10:  
Joe walks half the distance, then runs the rest.  
He runs **3 times as fast** as he walks.

Given: time to walk half way = **6 minutes**.  
Let walking speed be \(v\). Then walking time for half distance \(=\) 6 minutes.

Let half the distance be \(d\). Then:
\[
d = v \cdot 6
\]
Running speed is \(3v\).  
Remaining distance is also \(d\), so running time:
\[
\text{time} = \frac{d}{3v} = \frac{v\cdot 6}{3v} = 2
\]
Total time:
\[
6 + 2 = 8
\]
Correct choice: **D**.

ANSWER 10: D


Problem 11:  
Compute:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]

Group each set:

First set:
\[
1+11+21+31+41
\]
These are all of the form \(10k+1\):  
\(1, 11, 21, 31, 41\) correspond to \(k=0,1,2,3,4\).

Sum:
\[
(1+11+21+31+41) = (1+11+21+31+41)
\]
Compute directly:
\[
(1+11)=12,\ (21+31)=52,\ 12+52=64,\ 64+41=105
\]

Second set:
\[
9+19+29+39+49
\]
Direct sum:
\[
(9+19)=28,\ (29+39)=68,\ 28+68=96,\ 96+49=145
\]

Total:
\[
105 + 145 = 250
\]

Correct choice: **E**.

ANSWER 11: E


Problem 12:  
Which number is largest?

Options:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare decimals:  
- .99 is the largest among all because it is greater than .9 (and greater than any number starting 0.90... but with fewer 9s afterward).

So .99 > .9099 > .909 > .9009 > .9.

Correct choice: **A**.

ANSWER 12: A

Problem 1:
In a single-elimination tournament, every game results in exactly one team being eliminated. To determine a single winner from 16 teams, 15 teams must be eliminated. Since each game eliminates one team, it takes exactly 15 games.
ANSWER 1: D

Problem 2:
- Jose: (10 - 1) = 9; 9 * 2 = 18; 18 + 2 = 20.
- Thuy: (10 * 2) = 20; 20 - 1 = 19; 19 + 2 = 21.
- Kareem: (10 - 1) = 9; 9 + 2 = 11; 11 * 2 = 22.
Kareem has the largest answer (22).
ANSWER 2: C

Problem 3:
To convert 2/25 to a decimal, multiply the numerator and denominator by 4 to get a denominator of 100: (2 * 4) / (25 * 4) = 8 / 100 = 0.08.
ANSWER 3: B

Problem 4:
We need to choose 3 starters from a group of 4 (Lance, Sally, Joy, Fred). This is a combination problem denoted as 4C3. Choosing 3 people to be on the team is the same as choosing 1 person to be left off the team. There are 4 ways to choose who is left off.
ANSWER 4: B

Problem 5:
Let $x$ be the number. $0.20x = 12$. Dividing by 0.20, $x = 12 / 0.2 = 60$. We want 30% of $x$: $0.30 * 60 = 18$.
ANSWER 5: B

Problem 6:
Start at 0 revolutions. Clockwise is positive, counterclockwise is negative.
$2 \frac{1}{4} - 3 \frac{3}{4} = 2.25 - 3.75 = -1.5$ revolutions.
-1.5 revolutions is the same as -0.5 revolutions (since -1 is a full circle). -0.5 revolutions (counterclockwise) from West points to East.
ANSWER 6: B

Problem 7:
Original yellow tint: 30% of 30 liters = 9 liters.
New mixture: 30 + 5 = 35 liters.
New yellow tint: 9 + 5 = 14 liters.
Percent: (14 / 35) * 100 = (2 / 5) * 100 = 40%.
ANSWER 7: C

Problem 8:
Convert to decimals or compare differences from 1:
15/11 = 1 + 4/11 ≈ 1.3636
19/15 = 1 + 4/15 ≈ 1.2666
17/13 = 1 + 4/13 ≈ 1.3076
Comparing the fractions 4/11, 4/13, and 4/15: the larger the denominator, the smaller the fraction. Thus, 4/15 < 4/13 < 4/11.
Order: 19/15 < 17/13 < 15/11.
ANSWER 8: E

Problem 9:
The set is {3, 4, 5, 6, 6, 7, x}. The mode is 6 (it appears twice). For the mean and median to be 6:
Median: With 7 numbers, the 4th number must be 6. Since the sorted list is 3, 4, 5, 6, 6, 7, x, if $x > 6$, the 4th number is 6.
Mean: (3+4+5+6+6+7+x) / 7 = 6 => 31 + x = 42 => x = 11.
Checking: {3, 4, 5, 6, 6, 7, 11}. Median is 6. Mean is 42/7 = 6. Mode is 6.
ANSWER 9: D

Problem 10:
Walking time = 6 minutes. Running speed = 3 * walking speed.
Time taken = Distance / Speed. Since speed is 3 times faster, time is 1/3 of the walking time.
Running time = 6 / 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
ANSWER 10: D

Problem 11:
Sum 1: (1+11+21+31+41) = 105.
Sum 2: (9+19+29+39+49) = 145.
105 + 145 = 250.
ANSWER 11: E

Problem 12:
Compare digit by digit:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
0.99 is the largest.
ANSWER 12: A

Problem 1:
This is a single-elimination tournament with 16 teams. Exactly one team is eliminated per game, and 15 teams must be eliminated to leave a single winner. Thus exactly 15 games are required.

ANSWER 1: D

Problem 2:
Jose evaluates \((10-1)\times2+2=20\).
Thuy evaluates \((10\times2)-1+2=21\).
Kareem evaluates \((10-1+2)\times2=22\).
Kareem obtains the strictly largest value.

ANSWER 2: C

Problem 3:
Division yields the decimal expansion \(2\div25=0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters out of 4 distinct players is the binomial coefficient \(\binom{4}{3}=4\).

ANSWER 4: B

Problem 5:
If 20% of a number equals 12, the number itself is \(12/0.2=60\). Then 30% of 60 is \(0.3\times60=18\).

ANSWER 5: B

Problem 6:
A net rotation of \(+2.25-3.75=-1.5\) revolutions is equivalent to 1.5 revolutions counterclockwise. From an initial westward orientation, 1.5 revolutions counterclockwise lands exactly on east.

ANSWER 6: B

Problem 7:
The original mixture contains \(0.30\times30=9\) liters of yellow tint. Adding 5 liters raises the yellow volume to 14 liters and the total volume to 35 liters, so the new percentage is \(14/35=40\%\).

ANSWER 7: C

Problem 8:
Decimal approximations are \(15/11\approx1.3636\), \(19/15\approx1.2667\), and \(17/13\approx1.3077\). Ordering these values produces \(19/15<17/13<15/11\).

ANSWER 8: E

Problem 9:
The unique mode must be 6, forcing the mean also to equal 6. The sum of the seven numbers is therefore 42. The six given numbers sum to 31, so \(x=11\). With this value the ordered list is 3,4,5,6,6,7,11 whose median is likewise 6 and whose unique mode remains 6.

ANSWER 9: D

Problem 10:
Let the total distance be \(2d\). Walking time for distance \(d\) is 6 min, so running time for the remaining distance \(d\) at triple speed is \(6/3=2\) min. The total time is therefore 8 min.

ANSWER 10: D

Problem 11:
Each pair sums to a multiple of 10: \(1+9=10\), \(11+19=30\), \(21+29=50\), \(31+39=70\), \(41+49=90\). Their total is 250. Equivalently each five-term arithmetic series sums to 105 and 145 respectively, again yielding 250.

ANSWER 11: E

Problem 12:
Direct comparison of the five decimals shows that 0.99 is strictly larger than each of the others.

ANSWER 12: A

Problem 1:
The problem asks us to determine the number of games that will be played to determine the winner of a basketball tournament with 16 teams, where the losing team of each game is eliminated.

## Step 1
To determine the number of games needed to find the winner, we first note that in a single-elimination tournament, each game eliminates one team.

## Step 2
Since there can only be one winner, all other teams must be eliminated. Therefore, the number of games required is equal to the number of teams minus one.

## Step 3
We have 16 teams competing in the tournament. So, the number of games needed to determine the winner is 16 - 1 = 15.

## Step 4
Let's examine the given choices to rule out the incorrect options. A. 4 is too low because it doesn't account for all the teams. B. 7 is also too low for the same reason. C. 8 is still too low. D. 15 matches our calculation. E. 16 is one more than our calculation.

The final answer is supported by our step-by-step reasoning.
ANSWER 1: D

Problem 2:
The problem asks us to determine who among Jose, Thuy, and Kareem gets the largest final answer after performing a series of operations starting with the number 10.

## Step 1
Let's calculate Jose's final answer: Jose subtracts 1 from 10 to get 9, doubles it to get 18, and then adds 2 to get 20.

## Step 2
Now, let's calculate Thuy's final answer: Thuy doubles 10 to get 20, subtracts 1 to get 19, and then adds 2 to get 21.

## Step 3
Next, let's calculate Kareem's final answer: Kareem subtracts 1 from 10 to get 9, adds 2 to get 11, and then doubles it to get 22.

## Step 4
Comparing the final answers: Jose got 20, Thuy got 21, and Kareem got 22. The largest final answer is 22, which belongs to Kareem.

## Step 5
Let's examine the given choices. A. Jose is incorrect because his answer is 20. B. Thuy is incorrect because her answer is 21. C. Kareem is correct because his answer is 22. D. Jose and Thuy is incorrect because Kareem's answer is larger. E. Thuy and Kareem is incorrect because Jose's answer is not the largest.

The final answer is supported by our step-by-step reasoning.
ANSWER 2: C

Problem 3:
The problem asks us to convert the fraction 2/25 into a decimal.

## Step 1
To convert 2/25 into a decimal, we divide the numerator by the denominator: 2 ÷ 25 = 0.08.

## Step 2
Let's examine the given choices to identify the correct decimal representation. A. .008 is too small. B. .08 matches our calculation. C. .8 is too large. D. 1.25 is much too large. E. 12.5 is also much too large.

The final answer is supported by our step-by-step reasoning.
ANSWER 3: B

Problem 4:
The problem asks us to determine the number of ways to choose 3 starters from a team of 4 members: Lance, Sally, Joy, and Fred.

## Step 1
This is a combination problem, because the order in which the starters are chosen does not matter.

## Step 2
The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial, the product of all positive integers up to that number.

## Step 3
Here, n = 4 (total team members) and k = 3 (number of starters to be chosen). So, C(4, 3) = 4! / [3!(4 - 3)!] = 4! / (3! * 1!) = (4 * 3 * 2 * 1) / ((3 * 2 * 1) * 1) = 4.

## Step 4
Let's examine the given choices. A. 2 is too low. B. 4 matches our calculation. C. 6 is too high. D. 8 is also too high. E. 10 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 4: B

Problem 5:
The problem asks us to find 30% of a number, given that 20% of the same number is 12.

## Step 1
First, we need to find the number. We know that 20% of the number is 12. So, 0.20 * number = 12.

## Step 2
To find the number, we divide both sides of the equation by 0.20: number = 12 / 0.20 = 60.

## Step 3
Now, we need to calculate 30% of 60: 0.30 * 60 = 18.

## Step 4
Let's examine the given choices. A. 15 is too low. B. 18 matches our calculation. C. 20 is too high. D. 24 is also too high. E. 30 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 5: B

Problem 6:
The problem asks us to determine the direction the spinner points after Chenille moves it clockwise 2 1/4 revolutions and then counterclockwise 3 3/4 revolutions, starting from pointing west.

## Step 1
First, let's understand the effect of the movements. A full revolution is 360 degrees or 4 quarters. So, 2 1/4 revolutions clockwise is equivalent to 9 quarters clockwise.

## Step 2
9 quarters clockwise is the same as 1 quarter clockwise (since 8 quarters make 2 full revolutions, which brings the spinner back to the original direction). So, after the first move, the spinner points north.

## Step 3
Then, Chenille moves it counterclockwise 3 3/4 revolutions, which is equivalent to 15 quarters counterclockwise.

## Step 4
15 quarters counterclockwise is the same as 3 quarters counterclockwise (since 12 quarters make 3 full revolutions). So, moving 3 quarters counterclockwise from north results in pointing west.

## Step 5
Let's examine the given choices. A. north is incorrect because after the second move, it's not north. B. east is incorrect. C. south is incorrect. D. west matches our calculation. E. northwest is incorrect.

The final answer is supported by our step-by-step reasoning.
ANSWER 6: D

Problem 7:
The problem asks us to find the percentage of yellow tint in a new mixture after adding 5 liters of yellow tint to an original 30-liter mixture that is 25% red tint, 30% yellow tint, and 45% water.

## Step 1
First, we calculate the amount of yellow tint in the original mixture: 30% of 30 liters = 0.30 * 30 = 9 liters.

## Step 2
After adding 5 liters of yellow tint, the total amount of yellow tint becomes 9 + 5 = 14 liters.

## Step 3
The total volume of the new mixture is 30 + 5 = 35 liters.

## Step 4
To find the percentage of yellow tint in the new mixture, we divide the amount of yellow tint by the total volume and multiply by 100: (14 / 35) * 100 = 40%

Problem 1:
In a single-elimination tournament, each game eliminates exactly one team. To eliminate all but the champion, the number of games needed is one fewer than the number of teams. With 16 teams, the number of games is \(16 - 1 = 15\).

ANSWER 1: D

Problem 2:
Jose: \(10 - 1 = 9\), \(9 \times 2 = 18\), \(18 + 2 = 20\).
Thuy: \(10 \times 2 = 20\), \(20 - 1 = 19\), \(19 + 2 = 21\).
Kareem: \(10 - 1 = 9\), \(9 + 2 = 11\), \(11 \times 2 = 22\).
Kareem’s final answer of 22 is the largest.

ANSWER 2: C

Problem 3:
Divide 2 by 25: \(2 \div 25 = 0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters from 4 people is the combination \(\binom{4}{3} = 4\).

ANSWER 4: B

Problem 5:
Let the number be \(n\). \(20\%\) of \(n\) is \(0.2n = 12\), so \(n = 60\). Then \(30\%\) of \(60 = 0.3 \times 60 = 18\).

ANSWER 5: B

Problem 6:
Initial direction: west.
Clockwise \(2\frac{1}{4}\) revolutions: \(2\frac{1}{4} = 2.25\) revolutions clockwise.
Counterclockwise \(3\frac{3}{4}\) revolutions: \(3\frac{3}{4} = 3.75\) revolutions counterclockwise.
Net revolution = \(2.25 - 3.75 = -1.5\) revolutions, i.e., \(1.5\) revolutions counterclockwise.
A full revolution (1.0) returns to west. The remaining 0.5 revolution counterclockwise from west is a half-turn, which points to east.

ANSWER 6: B

Problem 7:
Original mixture: 30 liters, yellow tint = \(30\% \times 30 = 9\) liters.
Add 5 liters of yellow tint: new yellow = \(9 + 5 = 14\) liters; new total volume = \(30 + 5 = 35\) liters.
Percent yellow = \(\frac{14}{35} \times 100\% = 40\%\).

ANSWER 7: C

Problem 8:
Convert each fraction to a decimal for comparison:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
Order from least to greatest: \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

ANSWER 8: E

Problem 9:
The data set: 3, 4, 5, 6, 6, 7, \(x\). The mode is the most frequent number. With two 6’s, 6 is the mode unless \(x\) equals another number that appears twice, which would create a second mode, violating the “unique mode” condition. Therefore, \(x\) must not be 3, 4, 5, or 7. It could be 6, but if \(x=6\), the median is 6, while the mean is \((3+4+5+6+6+6+7)/7 = 37/7 \approx 5.29\), so mean ≠ median.  
For the mean to equal the median (which is 6, the middle of seven sorted numbers if \(x > 6\) or \(x=6\) only if placed correctly), we set the mean to 6. Sum without \(x\) = \(3+4+5+6+6+7 = 31\). Total needed for mean 6 = \(7 \times 6 = 42\). Thus \(x = 42 - 31 = 11\).  
Sorted: 3, 4, 5, 6, 6, 7, 11. Median = 6, mode = 6, mean = 6. Unique mode holds.

ANSWER 9: D

Problem 10:
Walking half the distance took 6 minutes, so walking speed = \(\frac{\text{half distance}}{6}\).
Running speed is 3 times walking speed. Time to run the second half = \(\frac{\text{half distance}}{3 \times (\text{half distance}/6)} = \frac{6}{3} = 2\) minutes.
Total time = \(6 + 2 = 8\) minutes.

ANSWER 10: D

Problem 11:
Pair the terms: \(1+49 = 50\), \(11+39 = 50\), \(21+29 = 50\), \(31+19 = 50\), \(41+9 = 50\). Five pairs of 50 give \(5 \times 50 = 250\).

ANSWER 11: E

Problem 12:
Compare by extending decimals to the same number of places:
\(0.99   = 0.9900\)
\(0.9099 = 0.9099\)
\(0.9    = 0.9000\)
\(0.909  = 0.9090\)
\(0.9009 = 0.9009\)
Clearly \(0.99 > 0.9099 > 0.909 > 0.9009 > 0.9\). The largest is \(0.99\).

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key Insight:** In a single-elimination tournament, every game eliminates exactly one team. To crown a single winner from 16 teams, we must eliminate 16 − 1 = 15 teams. Therefore, exactly 15 games are needed.

**Verification by rounds:**
- Round 1: 16 teams → 8 games → 8 teams left
- Round 2: 8 teams → 4 games → 4 teams left
- Round 3: 4 teams → 2 games → 2 teams left
- Round 4 (Final): 2 teams → 1 game → 1 winner

Total = 8 + 4 + 2 + 1 = 15 games.

Choices A (4), B (7), and C (8) are too small. Choice E (16) would mean every team is eliminated, leaving no winner.

ANSWER 1: D

---

Problem 2:
We track each person's calculation step by step, all starting from 10.

**Jose:** subtract 1 → 9; double → 18; add 2 → **20**
**Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
**Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

Comparing: 20 < 21 < 22. Kareem gets the largest final answer.

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

Choice A (.008) would be 8/1000. Choice C (.8) would be 8/10. Choices D and E are greater than 1, but 2/25 is clearly less than 1.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination problem — the order of selection doesn't matter:
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4$$

Equivalently, choosing 3 people to start is the same as choosing 1 person to sit out. There are 4 people, so 4 ways to pick who sits out.

The 4 possible starting lineups are: {L,S,J}, {L,S,F}, {L,J,F}, {S,J,F}.

ANSWER 4: B

---

Problem 5:
We're told 20% of a number is 12. Let the number be $x$.

$$0.20x = 12 \implies x = \frac{12}{0.20} = 60$$

Now find 30% of 60:
$$0.30 \times 60 = 18$$

**Shortcut:** 30% is $\frac{30}{20} = 1.5$ times 20%, so the answer is $1.5 \times 12 = 18$.

Choice A (15) would be 25%. Choice D (24) would be 40%. Choice E (30) would be 50%.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing **West**.

**Move 1:** Clockwise $2\frac{1}{4}$ revolutions.
- The 2 full revolutions bring it back to West.
- The extra $\frac{1}{4}$ revolution clockwise from West: West → **North**.

**Move 2:** Counterclockwise $3\frac{3}{4}$ revolutions.
- The 3 full revolutions bring it back to North.
- The extra $\frac{3}{4}$ revolution counterclockwise from North:
  - $\frac{1}{4}$ CCW from North → West
  - $\frac{2}{4}$ CCW from North → South
  - $\frac{3}{4}$ CCW from North → **East**

The spinner points East.

ANSWER 6: B

---

Problem 7:
Original mixture: 30 liters total.
- Red tint: 25% of 30 = 7.5 liters
- Yellow tint: 30% of 30 = 9 liters
- Water: 45% of 30 = 13.5 liters

Add 5 liters of yellow tint:
- New yellow tint = 9 + 5 = 14 liters
- New total mixture = 30 + 5 = 35 liters

Percent yellow tint in new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

Choice B (35%) is a trap for those who just add 5% to 30%. Choice D (45%) is the original water percentage.

ANSWER 7: C

---

Problem 8:
We compare $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$.

Rewrite each as a mixed number:
$$\frac{15}{11} = 1 + \frac{4}{11}, \quad \frac{19}{15} = 1 + \frac{4}{15}, \quad \frac{17}{13} = 1 + \frac{4}{13}$$

Since all three have the same numerator (4) in their fractional parts, we compare denominators. For fractions with equal numerators, a **larger denominator** means a **smaller fraction**:
$$15 > 13 > 11 \implies \frac{4}{15} < \frac{4}{13} < \frac{4}{11}$$

Therefore:
$$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$$

ANSWER 8: E

---

Problem 9:
The data set is: 3, 4, 5, 6, 6, 7, and $x$. The mean, median, and unique mode are all equal.

**Mode:** Since 6 already appears twice and all others appear once, the unique mode must be 6 (if $x$ were 3, 4, 5, or 7, we'd have two modes, violating "unique mode"). So mode = 6.

**Mean = 6:**
$$\frac{3 + 4 + 5 + 6 + 6 + 7 + x}{7} = 6$$
$$\frac{31 + x}{7} = 6 \implies 31 + x = 42 \implies x = 11$$

**Check Median:** With $x = 11$, the sorted list is: 3, 4, 5, **6**, 6, 7, 11. The median (4th value) is 6. ✓

All three measures equal 6. Choice B (6) would make 6 appear three times, but the mean would be $(31+6)/7 = 37/7 \neq 6$.

ANSWER 9: D

---

Problem 10:
Joe walked half the distance in 6 minutes. He then ran the remaining half at 3 times his walking speed.

Let walking speed = $v$. The half-distance = $6v$.

Running speed = $3v$. Time to run the second half:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{6v}{3v} = 2 \text{ minutes}$$

Total time = 6 + 2 = **8 minutes**.

Choice A (7) would result from incorrectly subtracting. Choices B, C, E involve unnecessary decimal complications.

ANSWER 10: D

---

Problem 11:
Compute $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$.

Pair the terms from the first group with terms from the second group (first with last, etc.):
- $1 + 49 = 50$
- $11 + 39 = 50$
- $21 + 29 = 50$
- $31 + 19 = 50$
- $41 + 9 = 50$

Total = $50 \times 5 = 250$.

Choice C (200) is a common error from miscounting pairs. Choice D (249) comes from arithmetic mistakes.

ANSWER 11: E

---

Problem 12:
Compare the five decimals by writing them to four decimal places:

- A: 0.9900
- B: 0.9099
- C: 0.9000
- D: 0.9090
- E: 0.9009

Comparing digit by digit from the tenths place: all have 9 in the tenths place. In the hundredths place, A has **9** while all others have **0**. So A is immediately the largest.

$0.99 > 0.9099 > 0.909 > 0.9009 > 0.9$

ANSWER 12: A

 **Problem 1:**
We need to find the number of games played in a single-elimination tournament with 16 teams. In single elimination, every game eliminates exactly one team. To determine a winner, 15 of the 16 teams must be eliminated. Therefore, exactly 15 games are required.  
(Another way: with 16 teams there are 8 games in the first round, 4 in the second, 2 in the third, and 1 final game: \(8+4+2+1=15\).)

ANSWER 1: D

---

**Problem 2:**
We compute each person’s final answer step by step, starting from 10.

- **Jose:**  
  \(10 - 1 = 9\)  
  \(9 \times 2 = 18\)  
  \(18 + 2 = 20\)

- **Thuy:**  
  \(10 \times 2 = 20\)  
  \(20 - 1 = 19\)  
  \(19 + 2 = 21\)

- **Kareem:**  
  \(10 - 1 = 9\)  
  \(9 + 2 = 11\)  
  \(11 \times 2 = 22\)

Comparing the results: \(20 < 21 < 22\). Kareem’s answer is the largest.

ANSWER 2: C

---

**Problem 3:**
We convert the fraction \(\frac{2}{25}\) to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100:
\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

ANSWER 3: B

---

**Problem 4:**
We must choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways to choose 3 from 4 is the combination \(\binom{4}{3}\):
\[
\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4
\]
Equivalently, choosing 3 starters is the same as choosing 1 person to sit out, and there are 4 choices for who sits out.

ANSWER 4: B

---

**Problem 5:**
Let the number be \(N\). We are given \(0.20N = 12\). Solving for \(N\):
\[
N = \frac{12}{0.20} = 60
\]
Now find 30% of 60:
\[
0.30 \times 60 = 18
\]
(Shortcut: 30% is \(1.5\) times 20%, so \(1.5 \times 12 = 18\).)

ANSWER 5: B

---

**Problem 6:**
The spinner starts pointing west.  
- Clockwise \(2\frac14\) revolutions: 2 full revolutions bring it back to west, then \(\frac14\) turn clockwise from west points it **north**.  
- Counterclockwise \(3\frac34\) revolutions: 3 full revolutions bring it back to north, then \(\frac34\) turn counterclockwise from north goes through west, south, and ends at **east**.

Net movement: \(3\frac34 - 2\frac14 = 1\frac12\) revolutions counterclockwise. One full revolution returns to west; an additional \(\frac12\) revolution counterclockwise from west points to **east**.

ANSWER 6: B

---

**Problem 7:**
Original mixture: 30 liters.  
- Yellow tint originally: \(30\% \times 30 = 9\) liters.  
After adding 5 liters of yellow tint:  
- New yellow tint: \(9 + 5 = 14\) liters.  
- New total mixture: \(30 + 5 = 35\) liters.  

Percent of yellow tint in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%
\]

ANSWER 7: C

---

**Problem 8:**
Compare \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\). Each is greater than 1. Write each as \(1\) plus a proper fraction:
\[
\frac{15}{11} = 1 + \frac{4}{11}, \qquad
\frac{19}{15} = 1 + \frac{4}{15}, \qquad
\frac{17}{13} = 1 + \frac{4}{13}
\]
Since the numerators are the same (4), the fraction with the largest denominator is smallest:
\[
\frac{4}{15} < \frac{4}{13} < \frac{4}{11}
\]
Adding 1 preserves the order:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

ANSWER 8: E

---

**Problem 9:**
The data set is \(3, 4, 5, 6, 6, 7, x\).  
- The current mode is 6 (appears twice). For 6 to remain the **unique** mode, \(x\) must not equal 3, 4, 5, or 7 (which would create a second mode).  
- We are told mean = median = mode = 6.  

Set the mean equal to 6:
\[
\frac{3+4+5+6+6+7+x}{7} = 6 \implies \frac{31+x}{7} = 6 \implies 31+x = 42 \implies x = 11
\]

Check: With \(x=11\), the sorted list is \(3, 4, 5, 6, 6, 7, 11\).  
- Median (4th term) = 6.  
- Mode = 6 (unique).  
- Mean = 6.  

All conditions are satisfied.

ANSWER 9: D

---

**Problem 10:**
Joe walks half the distance in 6 minutes. He runs the second half at 3 times his walking speed. Since distance is the same and speed is tripled, the time required is divided by 3:
\[
\text{Running time} = \frac{6}{3} = 2 \text{ minutes}
\]
Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

---

**Problem 11:**
Add the two groups:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]
Pair the terms to make convenient sums:
\[
(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 5 \times 50 = 250
\]

ANSWER 11: E

---

**Problem 12:**
Compare the decimals by writing them with the same number of places:
- A. \(0.9900\)
- B. \(0.9099\)
- C. \(0.9000\)
- D. \(0.9090\)
- E. \(0.9009\)

The tenths digit is 9 for all. Looking at the hundredths digit: A has 9, while B, D have 0 and C, E have 0. Thus \(0.9900\) is clearly the largest.

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key insight:** Each game eliminates exactly one team. To determine a winner from 16 teams, we must eliminate 15 teams (everyone except the champion).

Therefore, exactly 15 games must be played.

*Verification by rounds:* Round 1: 8 games (16→8 teams), Round 2: 4 games (8→4), Round 3: 2 games (4→2), Round 4: 1 game (2→1). Total = 8+4+2+1 = 15.

The tempting wrong answer E (16) counts teams instead of eliminations.

ANSWER 1: D

---

Problem 2:
We need to compute each person's final answer starting from 10.

**Jose:** Start with 10 → subtract 1: 10−1 = 9 → double: 9×2 = 18 → add 2: 18+2 = **20**

**Thuy:** Start with 10 → double: 10×2 = 20 → subtract 1: 20−1 = 19 → add 2: 19+2 = **21**

**Kareem:** Start with 10 → subtract 1: 10−1 = 9 → add 2: 9+2 = 11 → double: 11×2 = **22**

Kareem gets the largest final answer (22).

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

The tempting wrong answer A (0.008) would be 8/1000, and C (0.8) would be 8/10.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination: C(4,3) = 4!/(3! × 1!) = 4

Listing them out: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — that's 4 ways.

The tempting wrong answer C (6) would be C(4,2), choosing who's *not* a starter, but that also equals 4... Actually C(4,2)=6 is wrong here; we're choosing 3 from 4.

ANSWER 4: B

---

Problem 5:
We know 20% of a number is 12. We need to find 30% of the same number.

**Step 1:** Find the number.
$$0.20 \times n = 12 \implies n = \frac{12}{0.20} = 60$$

**Step 2:** Find 30% of 60.
$$0.30 \times 60 = 18$$

*Shortcut:* Since 30% is 1.5 times 20%, the answer is 1.5 × 12 = 18.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing west. We need to find the net rotation.

**Clockwise:** 2¼ revolutions
**Counterclockwise:** 3¾ revolutions

**Net movement:** 2¼ − 3¾ = −1½ revolutions (negative means counterclockwise)

So the spinner moves 1½ revolutions counterclockwise from west.

- 1 full revolution counterclockwise returns to **west**
- ½ revolution counterclockwise from west goes to **east**

(Counterclockwise from west: ¼ turn → south, ½ turn → east)

ANSWER 6: B

---

Problem 7:
We need to find the percent of yellow tint after adding 5 liters of yellow tint.

**Original mixture (30 liters):**
- Yellow tint: 30% of 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percent yellow tint:** (14/35) × 100 = 40%

The tempting wrong answer B (35) might come from simply adding 5% to 30%.

ANSWER 7: C

---

Problem 8:
We need to order 15/11, 19/15, and 17/13 from least to greatest.

Computing each as a decimal:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3077...

From least to greatest: **19/15 < 17/13 < 15/11**

*Alternative method:* Note each is 1 + (numerator − denominator)/denominator:
- 15/11 = 1 + 4/11
- 19/15 = 1 + 4/15
- 17/13 = 1 + 4/13

Since 4/15 < 4/13 < 4/11, the order is confirmed.

ANSWER 8: E

---

Problem 9:
The positive integers 3, 4, 5, 6, 6, 7, and x have equal mean, median, and unique mode.

**Mode:** Since 6 already appears twice and no other number does, the unique mode must be 6. So x ≠ 3, 4, 5, or 7 (which would create a second mode).

**Median:** With 7 numbers, the median is the 4th value when sorted. For the median to be 6, we need x ≥ 6 (so that 6 is the 4th value).

- If x = 6: sorted = 3, 4, 5, 6, 6, 6, 7 → median = 6 ✓
- If x > 7: sorted = 3, 4, 5, 6, 6, 7, x → median = 6 ✓

**Mean:** (3 + 4 + 5 + 6 + 6 + 7 + x)/7 = (31 + x)/7

Setting mean = 6: (31 + x)/7 = 6 → 31 + x = 42 → **x = 11**

**Verification:** Numbers: 3, 4, 5, 6, 6, 7, 11. Mean = 42/7 = 6, Median = 6, Mode = 6 ✓

ANSWER 9: D

---

Problem 10:
Joe walks halfway, then runs the rest. We need total time.

**Walking:** Half the distance takes 6 minutes.

**Running:** He runs 3 times as fast as he walks, so he covers the same distance in 1/3 the time.
- Time to run the second half = 6 ÷ 3 = 2 minutes

**Total time:** 6 + 2 = **8 minutes**

The tempting wrong answers come from setting up the problem incorrectly or mixing up distance and time relationships.

ANSWER 10: D

---

Problem 11:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

**First group:** 1 + 11 + 21 + 31 + 41 = 105

**Second group:** 9 + 19 + 29 + 39 + 49 = 145

**Total:** 105 + 145 = **250**

*Elegant method:* Pair terms: (1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250

ANSWER 11: E

---

Problem 12:
We need to find the largest number among the options. Writing each with the same number of decimal places:

- A. 0.99 = **0.9900**
- B. 0.9099 = 0.9099
- C. 0.9 = 0.9000
- D. 0.909 = 0.9090
- E. 0.9009 = 0.9009

Comparing digit by digit after the decimal point:
- First digit: all are 9
- Second digit: A has **9**, all others have 0

Since A has a 9 in the second decimal place while all others have 0, **0.99 is the largest**.

ANSWER 12: A

**Problem 1:**  
The tournament is single‑elimination. Starting with 16 teams, every game eliminates exactly one team. To reduce from 16 teams to a single champion we must eliminate 15 teams, and each elimination corresponds to one game.  

Number of games = 16 − 1 = **15**.  

ANSWER 1: D  

---

**Problem 2:**  
Start with 10 for each person.

- **Jose:**  
  Subtract 1 → 9; double → 18; add 2 → 20.  

- **Thuy:**  
  Double → 20; subtract 1 → 19; add 2 → 21.  

- **Kareem:**  
  Subtract 1 → 9; add 2 → 11; double → 22.  

The largest final value is 22, obtained by Kareem.  

ANSWER 2: C  

---

**Problem 3:**  

\[
\frac{2}{25}=0.08
\]

(2 ÷ 25 = 0.08).  

ANSWER 3: B  

---

**Problem 4:**  
We need to choose 3 starters from the 4 available players:  

\[
\binom{4}{3}=4
\]

ways.  

ANSWER 4: B  

---

**Problem 5:**  
Let the number be \(N\).  
\(0.20N = 12 \Rightarrow N = \frac{12}{0.20}=60\).  
\(30\%\) of \(N\) is \(0.30 \times 60 = 18\).  

ANSWER 5: B  

---

**Problem 6:**  
Interpret the fractions as mixed numbers:  

- Clockwise \(2\frac14 = 2.25\) revolutions.  
- Counter‑clockwise \(3\frac34 = 3.75\) revolutions.  

Net rotation (counter‑clockwise positive) = \(3.75 - 2.25 = 1.5\) revolutions.  

A full revolution returns to the same direction, so we only care about the extra \(0.5\) revolution.  
\(0.5\) revolution = 180° counter‑clockwise.  

Starting from west, turning 180° counter‑clockwise points to **east**.  

ANSWER 6: B  

---

**Problem 7:**  
Original yellow = \(30\%\) of 30 L = \(0.30 \times 30 = 9\) L.  
Add 5 L yellow → new yellow = \(9+5 = 14\) L.  
New total volume = \(30+5 = 35\) L.  

Percent yellow = \(\frac{14}{35}\times100 = 40\%\).  

ANSWER 7: C  

---

**Problem 8:**  
Compute the three fractions:

\[
\frac{15}{11}=1.3636\ldots,\qquad
\frac{19}{15}=1.2666\ldots,\qquad
\frac{17}{13}=1.3077\ldots
\]

Ordered from least to greatest:  

\(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

This matches option **E**.  

ANSWER 8: E  

---

**Problem 9:**  
Current list: \(3,4,5,6,6,7\). The unique mode is 6 (appears twice).  
For the mode to remain unique and equal to the mean and median, \(x\) must not create a tie.  
Let the mean = median = mode = \(m\).  

Sum of numbers = \(31 + x\).  
Mean = \(\frac{31 + x}{7}=m\).  
Median of the seven numbers is the 4th after sorting; with \(x>7\) the median stays 6, so \(m=6\).  

Thus \(\frac{31 + x}{7}=6 \Rightarrow 31 + x = 42 \Rightarrow x = 11\).

Check: sorted list becomes \(3,4,5,6,6,7,11\) → median = 6, mode = 6, mean = 42/7 = 6. All equal.  

ANSWER 9: D  

---

**Problem 10:**  
Let the walking speed be \(v\) and the half‑distance be \(d/2\).  

Walking time for half‑distance: \(\frac{d/2}{v}=6\) min ⇒ \(\frac{d}{v}=12\) min.  

Running speed = \(3v\).  
Running time for the second half: \(\frac{d/2}{3v}= \frac{d}{6v}= \frac{1}{6}\times\frac{d}{v}= \frac{12}{6}=2\) min.  

Total time = \(6 + 2 = 8\) minutes.  

ANSWER 10: D  

---

**Problem 11:**  

\[
\begin{aligned}
1+11+21+31+41 &= 105,\\
9+19+29+39+49 &= 145,\\
\text{Total} &= 105+145 = 250.
\end{aligned}
\]

ANSWER 11: E  

---

**Problem 12:**  
Write the numbers with four decimal places:

- .99 = 0.9900  
- .9099 = 0.9099  
- .9 = 0.9000  
- .909 = 0.9090  
- .9009 = 0.9009  

The largest is 0.9900, i.e., **.99**.  

ANSWER 12: A  

Problem 1:  
The tournament starts with 16 teams. Each game eliminates one team, and the losing team is eliminated. To determine the winner, all teams except one must be eliminated. Thus, 15 teams must be eliminated, requiring 15 games.  
- **Why not other choices?** A (4) is too few; B (7), C (8) are also insufficient; E (16) would imply one game per team without elimination, which is incorrect.  
ANSWER 1: D  

Problem 2:  
- **Jose**: Starts with 10 → subtracts 1 (9) → doubles (18) → adds 2 → **20**.  
- **Thuy**: Starts with 10 → doubles (20) → subtracts 1 (19) → adds 2 → **21**.  
- **Kareem**: Starts with 10 → subtracts 1 (9) → adds 2 (11) → doubles → **22**.  
Kareem has the largest final answer.  
- **Why not other choices?** A (Jose) and B (Thuy) are smaller; D and E pair incorrect winners.  
ANSWER 2: C  

Problem 3:  
\( \frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08 \).  
- **Why not other choices?** A (.008) is \( \frac{2}{250} \); C (.8) is \( \frac{4}{5} \); D and E are larger decimals.  
ANSWER 3: B  

Problem 4:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is \( \binom{4}{3} = 4 \).  
- **Why not other choices?** A (2) is too few; C (6) is \( \binom{4}{2} \), which is incorrect; D (8) and E (10) are overcounts.  
ANSWER 4: B  

Problem 5:  
Let the number be \( x \). If 20% of \( x \) is 12, then \( 0.2x = 12 \), so \( x = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).  
- **Why not other choices?** A (15) is 25% of 60; C (20) is \( \frac{1}{3} \) of 60; D (24) is 40%; E (30) is 50%.  
ANSWER 5: B  

Problem 6:  
- A full revolution is 360°. Clockwise \( 2 \frac{1}{4} \) revolutions = \( 2.25 \times 360 = 810° \).  
- Counterclockwise \( 3 \frac{3}{4} \) revolutions = \( -3.75 \times 360 = -1350° \).  
- Net movement: \( 810° - 1350° = -540° \).  
- \( -540° \mod 360° = 180° \) (since \( -540 + 720 = 180° \)), which points west → south? Wait, initial direction is west.  
- West (270°) + 810° = 1080° → 1080° mod 360° = 0° (north). Then -1350° → 0° - 1350° = -1350° mod 360° = 90° (east)? Let's reframe:  
- Start at west (270°). Clockwise 2.25 rev: 270° + 810° = 1080° mod 360° = 0° (north). Then counterclockwise 3.75 rev: 0° - 1350° = -1350° mod 360° = 90° (east).  
- **Why not other choices?** A (north) is initial after clockwise; B (east) is final; C (south) is 180°; D (west) is start; E (northwest) is not a cardinal direction.  
ANSWER 6: B  

Problem 7:  
Original mixture: 30 liters, 30% yellow tint → 9 liters yellow. Add 5 liters yellow → total yellow = 14 liters. New mixture = 35 liters. Percent yellow = \( \frac{14}{35} \times 100 = 40\% \).  
- **Why not other choices?** A (25) is original yellow%; B (35) is before addition; D (45) and E (50) are overestimates.  
ANSWER 7: E (Note: Correction based on computation: 14/35=0.4 → 40%, so answer is E? Wait, 14/35=0.4 → 40%, which is E. But earlier I thought B, but B is 35, which is wrong. Original yellow is 9, add 5 → 14. Total mixture 30+5=35. 14/35=0.4 → 40%. So E. But in choices E is 50? Wait, no: choices are A.25 B.35 C.40 D.45 E.50. So 40% is C. I misread. Correction: 14/35=0.4 → 40%, which is C.  
**Correction**: 14/35=0.4 → 40%, so answer is C.  
ANSWER 7: C  

Problem 8:  
Compare \( \frac{15}{11} \approx 1.3636 \), \( \frac{19}{15} \approx 1.2667 \), \( \frac{17}{13} \approx 1.3077 \). Order: \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11} \).  
- **Why not other choices?** A, B, C, D have incorrect orders.  
ANSWER 8: E  

Problem 9:  
The set is {3, 4, 5, 6, 6, 7, x}. Mean = median = mode = 6.  
- Mode is 6 (unique), so 6 must appear most.  
- Median is 6 (4th value in ordered list: 3,4,5,6,6,7,x → if x≥6, ordered is ...,6,6,x or ...,6,x,6? Must have 4th value 6.  
- If x=6, set is {3,4,5,6,6,6,7}: median=6, mode=6, mean=(3+4+5+6+6+6+7)/7=37/7≈5.28 ≠6.  
- If x=11: set {3,4,5,6,6,7,11}: mean=(3+4+5+6+6+7+11)/7=42/7=6, median=6, mode=6.  
- **Why not other choices?** A (5): mean=36/7≈5.14; B (6): mean≈5.28; C (7): mean=38/7≈5.43; D (11) works.  
ANSWER 9: D  

Problem 10:  
Joe walks half the distance in 6 minutes. Let walking speed be \( v \), so distance = \( 6v \). He runs at \( 3v \), so time to run half distance = \( \frac{6v}{3v} = 2 \) minutes. Total time = 6 + 2 = 8 minutes.  
- **Why not other choices?** A (7), B (7.3), C (7.7), E (8.3) are not exact.  
ANSWER 10: D  

Problem 11:  
First group: 1+11+21+31+41 = 105. Second group: 9+19+29+39+49 = 145. Total = 105 + 145 = 250.  
- **Why not other choices?** A (150), B (199), C (200), D (249) are incorrect sums.  
ANSWER 11: E  

Problem 12:  
Compare decimals: A (.99), B (.9099), C (.9), D (.909), E (.9009). Largest is .99.  
- **Why not other choices?** B, C, D, E are all smaller than .99.  
ANSWER 12: A

### Problem 1:
We need to find the total number of games to determine a winner in a 16-team single-elimination tournament, where every game eliminates exactly one losing team. To get 1 winner, we must eliminate 16 - 1 = 15 teams. Since each game eliminates 1 team, we need 15 total games. Wrong choices: A is the number of bracket rounds, not games; C is half the number of teams; E is the total number of teams, not games.
ANSWER 1: D

---

### Problem 2:
All three start with 10, we calculate each person's final result step-by-step:
- Jose: (10 - 1) * 2 + 2 = 9*2 +2 = 20
- Thuy: (10*2) -1 +2 = 20-1+2=21
- Kareem: (10-1 +2)*2 = 11*2=22
Kareem has the largest final answer. Wrong choices stem from misordering the operations for each person.
ANSWER 2: C

---

### Problem3:
Convert 2/25 to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100: (2*4)/(25*4) = 8/100 = 0.08. Wrong choices: A is 2/250, C is 20/25.
ANSWER 3: B

---

### Problem4:
We choose 3 starters from 4 people, which is a combination problem (order of starters does not matter here). The number of ways to choose 3 from 4 is equivalent to choosing 1 person to leave off the starting lineup: there are 4 people, so 4 unique lineups, each omitting one person. Wrong choice C (6) is the number of permutations, not combinations.
ANSWER 4: B

---

### Problem5:
If 20% of a number is 12, first find the full number: 0.2*N=12 → N=12/0.2=60. 30% of 60 is 0.3*60=18. Wrong choices: A is 25% of 60, D is 40% of 60.
ANSWER 5: B

---

### Problem6:
The spinner starts pointing west. Net movement: 2.25 clockwise revolutions minus 3.75 clockwise (equivalent to 3.75 counterclockwise) = -1.5 total clockwise revolutions, or 1.5 counterclockwise revolutions. Full revolutions cancel out, so we only have a net 0.5 (180 degree) counterclockwise turn from west. A 180-degree turn from west points to east.
ANSWER 6: B

---

### Problem7:
Original 30L mixture has 30% yellow tint: 0.3*30=9L of yellow. Add 5L of yellow, so new yellow is 14L, new total mixture is 35L. The new percentage of yellow is (14/35)*100=40%. Wrong choice: people often forget to add 5L to the total mixture volume, leading to an incorrect higher percentage.
ANSWER 7: C

---

### Problem8:
Rewrite each fraction to compare easily: 15/11 = 1 + 4/11, 19/15=1+4/15, 17/13=1+4/13. For equal numerators, the larger the denominator, the smaller the fraction. So 4/15 < 4/13 < 4/11, so the full order from least to greatest is 19/15 < 17/13 <15/11.
ANSWER 8: E

---

### Problem9:
The dataset is [3,4,5,6,6,7,x]. The unique mode is currently 6 (only value appearing twice, so it must stay the only unique mode). For the mean, median, and mode to all be 6: the sum of the dataset must be 7*6=42. The sum of the existing numbers is 3+4+5+6+6+7=31, so x=42-31=11. Sorted dataset [3,4,5,6,6,7,11] has median 6, unique mode 6, mean 6, which fits. Wrong choice B (x=6) gives a mean of ~5.28, not 6.
ANSWER 9: D

---

### Problem10:
Joe walks half the distance in 6 minutes. His running speed is 3x his walking speed, so time to cover the same distance is 1/3 of walking time. Time to run the second half: 6/3=2 minutes. Total time: 6+2=8 minutes.
ANSWER 10: D

---

### Problem11:
Pair terms to simplify addition: (1+49)+(11+39)+(21+29)+(31+19)+(41+9) = 50+50+50+50+50=250.
ANSWER 11: E

---

### Problem12:
Line up all decimals to compare: A=0.9900, B=0.9099, C=0.9000, D=0.9090, E=0.9009. The largest value is 0.99, option A.
ANSWER 12: A

Problem 1:  
In a single-elimination tournament, each game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams (all except 1). Thus, 15 games are played.  
ANSWER 1: D  

Problem 2:  
Calculate each person's result starting from 10:  
- Jose: \(10 -1 =9\), double → \(9*2=18\), add 2 → \(18+2=20\)  
- Thuy: Double 10 → \(10*2=20\), subtract 1 → \(20-1=19\), add 2 → \(19+2=21\)  
- Kareem: \(10-1=9\), add 2 → \(9+2=11\), double → \(11*2=22\)  
Kareem has the largest result.  
ANSWER 2: C  

Problem 3:  
Convert \(\frac{2}{25}\) to decimal: \(2\div25=0.08\).  
ANSWER 3: B  

Problem 4:  
Choosing 3 starters from 4 people is a combination (order does not matter). The number of ways is \(\binom{4}{3}=4\) (or list: {Lance,Sally,Joy}, {Lance,Sally,Fred}, {Lance,Joy,Fred}, {Sally,Joy,Fred}).  
ANSWER 4: B  

Problem 5:  
Let the number be \(x\). \(20\%x=12\) → \(0.2x=12\) → \(x=60\). \(30\%\ of\ 60=0.3*60=18\).  
ANSWER 5: B  

Problem 6:  
Model directions with standard angles: 0°=east, 90°=north, 180°=west, 270°=south. Initial direction is west (180°).  
- Clockwise \(2\frac{1}{4}\) revolutions: \(2.25*(-360°)=-810°\), new angle: \(180-810=-630°\)  
- Counterclockwise \(3\frac{3}{4}\) revolutions: \(3.75*(360°)=+1350°\), total angle: \(-630+1350=720°\)  
- \(720°\mod360°=0°\), which is east.  
ANSWER 6: B  

Problem 7:  
Original yellow tint: \(30\%\ of\ 30=0.3*30=9\) liters. Add 5 liters, total yellow = \(9+5=14\) liters. New total mixture: \(30+5=35\) liters. Percent yellow: \(\frac{14}{35}*100\%=40\%\).  
ANSWER 7: C  

Problem 8:  
Compare fractions via cross-multiplication or decimals:  
- \(\frac{19}{15}\approx1.267\), \(\frac{17}{13}\approx1.308\), \(\frac{15}{11}\approx1.364\)  
Order from least to greatest: \(\frac{19}{15}<\frac{17}{13}<\frac{15}{11}\).  
ANSWER 8: E  

Problem 9:  
The unique mode is 6 (appears twice, others once). Thus mean and median equal 6.  
- Median of 7 numbers is the 4th term when sorted, so 4th term =6 → \(x\geq6\).  
- Mean = \(\frac{3+4+5+6+6+7+x}{7}=6\) → \(31+x=42\) → \(x=11\).  
Check: Sorted list [3,4,5,6,6,7,11], median=6, mode=6 (unique), mean=6.  
ANSWER 9: D  

Problem 10:  
Walking half the distance takes 6 minutes, so full walking time would be 12 minutes. Running speed is 3x walking speed, so time for the second half is \(\frac{6}{3}=2\) minutes. Total time: \(6+2=8\) minutes.  
ANSWER 10: D  

Problem 11:  
Sum first group: \(1+11+21+31+41=105\) (arithmetic series: \(\frac{5}{2}(1+41)=105\)).  
Sum second group: \(9+19+29+39+49=145\) (\(\frac{5}{2}(9+49)=145\)).  
Total: \(105+145=250\).  
ANSWER 11: E  

Problem 12:  
Compare decimals digit by digit:  
- A=0.99, B=0.9099, C=0.9, D=0.909, E=0.9009  
A has the highest second decimal digit (9 vs 0 for others), so it is the largest.  
ANSWER 12: A

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** How many games must be played to determine a winner in a 16-team elimination tournament?

**Setup:** In an elimination tournament, each game eliminates exactly one team. To go from 16 teams to 1 winner, we must eliminate 15 teams. Therefore, we need 15 games.

**Alternatively:** We can think of it as a bracket:
- Round 1: 16 teams → 8 games → 8 winners
- Round 2: 8 teams → 4 games → 4 winners
- Round 3: 4 teams → 2 games → 2 winners
- Round 4: 2 teams → 1 game → 1 winner
- Total: 8 + 4 + 2 + 1 = 15 games

ANSWER 1: D

---

**Problem 2:**

**What's being asked:** Which person gets the largest final answer after their operations?

**Jose's calculation:** 
- Start: 10
- Subtract 1: 10 - 1 = 9
- Double: 9 × 2 = 18
- Add 2: 18 + 2 = **20**

**Thuy's calculation:**
- Start: 10
- Double: 10 × 2 = 20
- Subtract 1: 20 - 1 = 19
- Add 2: 19 + 2 = **21**

**Kareem's calculation:**
- Start: 10
- Subtract 1: 10 - 1 = 9
- Add 2: 9 + 2 = 11
- Double: 11 × 2 = **22**

Kareem gets 22, which is the largest.

ANSWER 2: C

---

**Problem 3:**

**What's being asked:** Convert the fraction 2/25 to a decimal.

**Setup:** 2/25 = ?

**Calculation:** 
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** In how many ways can 3 starters be chosen from 4 people (Lance, Sally, Joy, Fred)?

**Setup:** This is a combination problem: $\binom{4}{3}$

**Calculation:**
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4}{1} = 4$$

Alternatively, we're choosing which 1 person to leave out: 4 ways.

ANSWER 4: B

---

**Problem 5:**

**What's being asked:** If 20% of a number is 12, what is 30% of that number?

**Setup:** Let the number be $n$.
- $0.20n = 12$
- $n = 12 ÷ 0.20 = 60$

**Calculation:** 
$$30\% \text{ of } 60 = 0.30 \times 60 = 18$$

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Starting pointing west, after rotating clockwise 2¼ revolutions then counterclockwise 3¾ revolutions, which direction?

**Setup:** 
- Clockwise 2¼ revolutions = 2.25 revolutions
- Counterclockwise 3¾ revolutions = 3.75 revolutions
- Net: 3.75 - 2.25 = 1.5 revolutions counterclockwise (net)

**Calculation:**
Starting at west, rotating counterclockwise 1.5 revolutions (= 1 full rotation + 0.5 rotation):
- 1 full rotation brings us back to west
- 0.5 rotation = 180°
- West + 180° counterclockwise = East

ANSWER 6: B

---

**Problem 7:**

**What's being asked:** After adding 5 liters of yellow tint to a 30-liter mixture, what percent is yellow tint?

**Original mixture (30 liters):**
- Yellow tint: 30% × 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percentage:**
$$\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%$$

ANSWER 7: C

---

**Problem 8:**

**What's being asked:** Order 15/11, 19/15, and 17/13 from least to greatest.

**Setup:** Convert to decimals or compare directly.

**Calculations:**
- 15/11 ≈ 1.364
- 19/15 ≈ 1.267
- 17/13 ≈ 1.308

**Order:** 19/15 < 17/13 < 15/11

ANSWER 8: E

---

**Problem 9:**

**What's being asked:** Find $x$ such that the mean, median, and unique mode of {3, 4, 5, 6, 6, 7, $x$} are all equal.

**Current mode:** 6 appears twice; all others appear once. If there's a unique mode, it must be 6.

**With unique mode = 6:** This requires no other number to appear more than once. So $x \notin \{3, 4, 5, 7\}$ and $x \neq 6$ (else it wouldn't be unique).

**Mean = Median = Mode = 6:**

For median of 7 numbers with value 6:
- Sorted: need middle (4th) value to be 6
- If $x = 11$: {3, 4, 5, 6, 6, 7, 11} → median = 6 ✓

Mean: $\frac{3+4+5+6+6+7+11}{7} = \frac{42}{7} = 6$ ✓

Mode: 6 appears twice; all others once → mode = 6 ✓

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Total time from home to school if Joe walked halfway (6 min) then ran the rest at 3× speed?

**Setup:**
- Walking: 6 minutes for half the distance
- Running: same distance at 3× speed, so time = 6/3 = 2 minutes

**Total time:** 6 + 2 = 8 minutes

ANSWER 10: D

---

**Problem 11:**

**What's being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Calculation:**
- First group: 1 + 11 + 21 + 31 + 41 = 105
- Second group: 9 + 19 + 29 + 39 + 49 = 145
- Sum: 105 + 145 = 250

ANSWER 11: E

---

**Problem 12:**

**What's being asked:** Which number is largest?

**Comparison:**
- A. 0.99
- B. 0.9099
- C. 0.9
- D. 0.909
- E. 0.9009

Starting from the tenths place: all are 0.9__. Looking at hundredths:
- 0.99 has 9 in hundredths place → **largest**

ANSWER 12: A

Problem 1:

In a single-elimination tournament, each game eliminates exactly one team.  
There are 16 teams, and we want one champion.

So we must eliminate 15 teams total.

Therefore, the number of games played is 15.

Tempting wrong choice: 16 would be if every team had to lose, but the winner does not lose.

ANSWER 1: D

Problem 2:

Compute each person’s result carefully.

- Jose: start with 10  
  subtract 1 → 9  
  double → 18  
  add 2 → 20

- Thuy: start with 10  
  double → 20  
  subtract 1 → 19  
  add 2 → 21

- Kareem: start with 10  
  subtract 1 → 9  
  add 2 → 11  
  double → 22

Compare:
- Jose = 20
- Thuy = 21
- Kareem = 22

Kareem gets the largest final answer.

ANSWER 2: C

Problem 3:

\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

So the correct decimal is 0.08.

ANSWER 3: B

Problem 4:

We are choosing 3 starters from 4 players: Lance, Sally, Joy, and Fred.

This is a combination problem:
\[
\binom{4}{3} = 4
\]

You can also see it by noticing that choosing 3 starters is the same as choosing which 1 player sits out, and there are 4 choices.

ANSWER 4: B

Problem 5:

Let the number be \(n\).

20% of \(n\) is 12, so:
\[
0.2n = 12
\]
\[
n = 60
\]

Now find 30% of 60:
\[
0.3 \times 60 = 18
\]

ANSWER 5: B

Problem 6:

The spinner starts pointing west.

First move: clockwise \(2\frac{1}{4}\) revolutions.  
A full revolution brings it back to west, so only the extra \(\frac{1}{4}\) revolution matters.

From west, a quarter-turn clockwise points it to north.

Second move: counterclockwise \(3\frac{3}{4}\) revolutions.  
Again, the 3 full revolutions do nothing, so this is just \(\frac{3}{4}\) of a revolution counterclockwise, which is the same as \(\frac{1}{4}\) of a revolution clockwise.

From north, a quarter-turn clockwise points east.

So the spinner ends pointing east.

ANSWER 6: B

Problem 7:

Start with 30 liters total.

Yellow tint is 30% of 30 liters:
\[
0.30 \times 30 = 9 \text{ liters}
\]

Then 5 liters of yellow tint are added, so yellow becomes:
\[
9 + 5 = 14 \text{ liters}
\]

New total mixture:
\[
30 + 5 = 35 \text{ liters}
\]

Percent yellow in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.4 = 40\%
\]

ANSWER 7: C

Problem 8:

The fractions are:
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]

Approximate each one:

\[
\frac{15}{11} \approx 1.364
\]
\[
\frac{17}{13} \approx 1.308
\]
\[
\frac{19}{15} \approx 1.267
\]

So from least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

That matches choice E.

ANSWER 8: E

Problem 9:

The numbers are:
\[
3,4,5,6,6,7,x
\]

We are told the mean, median, and unique mode are all equal.

Since 6 appears twice and every other listed number appears once, the unique mode is 6.

So the median and mean must also be 6.

The median of 7 numbers is the 4th number when ordered.  
The sorted list is:
\[
3,4,5,6,6,7,x
\]
To make the median 6, \(x\) must be 6 or larger, but we also need the mean to be 6.

Sum of the known numbers:
\[
3+4+5+6+6+7 = 31
\]

For the mean to be 6, total sum must be:
\[
7 \times 6 = 42
\]

So:
\[
31 + x = 42
\]
\[
x = 11
\]

Check:
- mode = 6
- median = 6
- mean = 42/7 = 6

Works perfectly.

ANSWER 9: D

Problem 10:

Joe walked half the distance in 6 minutes.  
He then ran the other half at 3 times his walking speed.

If he runs 3 times as fast, then the time for the same distance is 1/3 as long.

So the running time is:
\[
6 \div 3 = 2 \text{ minutes}
\]

Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

Problem 11:

Add the first group:
\[
1+11+21+31+41 = 105
\]

Add the second group:
\[
9+19+29+39+49 = 145
\]

Now add the totals:
\[
105 + 145 = 250
\]

ANSWER 11: E

Problem 12:

Compare the decimals:

- \(0.99\)
- \(0.9099\)
- \(0.9 = 0.9000\)
- \(0.909\)
- \(0.9009\)

The largest is clearly \(0.99\), since it is closest to 1.

ANSWER 12: A

Problem 1:  
We have a three-person-team tournament where **the losing team of each game is eliminated**. With **16 teams**, how many games are played to determine the winner?

To find the number of games in a single-elimination tournament:  
- Each game eliminates exactly **one** team (the loser).  
- To go from **16 teams to 1 winning team**, we must eliminate **15 teams**.  
- Therefore, the number of games is **15**.

So the correct choice is **D**.

ANSWER 1: D


Problem 2:  
Jose, Thuy, and Kareem all start at 10. We compute each final value.

**Jose:**  
- Subtract 1: \(10 - 1 = 9\)  
- Double: \(2 \cdot 9 = 18\)  
- Add 2: \(18 + 2 = 20\)  
Final: **20**

**Thuy:**  
- Double 10: \(2 \cdot 10 = 20\)  
- Subtract 1: \(20 - 1 = 19\)  
- Add 2: \(19 + 2 = 21\)  
Final: **21**

**Kareem:**  
- Subtract 1: \(10 - 1 = 9\)  
- Add 2: \(9 + 2 = 11\)  
- Double: \(2 \cdot 11 = 22\)  
Final: **22**

Largest is **22**, which is Kareem.

ANSWER 2: C


Problem 3:  
We need to evaluate:  
\[
\frac{2}{25}
\]
Convert to decimals:  
\[
\frac{2}{25} = 2 \div 25
\]
Since \(25 \cdot 0.08 = 2\):  
\[
2 \div 25 = 0.08
\]
So the correct choice is **B** (.08).

ANSWER 3: B


Problem 4:  
We have 4 people chosen: Lance, Sally, Joy, Fred. We need to choose **3 starters**.

Number of ways:
\[
\binom{4}{3} = 4
\]
List them conceptually: leaving out each one individually gives 4 choices.

Correct choice: **B**.

ANSWER 4: B


Problem 5:  
“If 20% of a number is 12, what is 30% of the same number?”

Let the number be \(N\).  
\[
0.20N = 12
\]
\[
N = \frac{12}{0.20} = 60
\]
Now compute \(30\%\) of 60:
\[
0.30 \cdot 60 = 18
\]
Correct choice: **B**.

ANSWER 5: B


Problem 6:  
Spinner starts pointing **west**.

Chenille moves it:
- **Clockwise** \(2\frac{1}{4}\) revolutions  
- then **Counterclockwise** \(3\frac{3}{4}\) revolutions

Net rotation (clockwise minus counterclockwise):
\[
2\frac{1}{4} - 3\frac{3}{4}
\]
Convert to improper fractions:
\[
2\frac{1}{4} = \frac{9}{4}, \quad 3\frac{3}{4} = \frac{15}{4}
\]
\[
\frac{9}{4} - \frac{15}{4} = -\frac{6}{4} = -\frac{3}{2}
\]
The negative means **net counterclockwise** by \(\frac{3}{2}\) revolutions.

\(\frac{3}{2} = 1 + \frac{1}{2}\), so this is:
- 1 full turn (brings it back where it started)
- plus \(\frac{1}{2}\) turn counterclockwise

Half a revolution from **west** points **east** (opposite direction).

Correct answer: **B (east)**.

ANSWER 6: B


Problem 7:  
Original mixture: 30 L with  
- 25% red (doesn’t matter directly)
- 30% yellow  
- 45% water

Yellow amount initially:
\[
0.30 \cdot 30 = 9 \text{ liters}
\]
Add 5 liters of yellow tint, so new yellow amount:
\[
9 + 5 = 14
\]
New total volume:
\[
30 + 5 = 35
\]
Percent yellow in new mixture:
\[
\frac{14}{35} = 0.4 = 40\%
\]
Correct choice: **C**.

ANSWER 7: C


Problem 8:  
Order fractions least to greatest:
\[
\frac{15}{11},\ \frac{19}{15},\ \frac{17}{13}
\]
A good way: compare values by cross-multiplication or decimal approximations.

Compute approximate decimals:
\[
\frac{15}{11} \approx 1.3636
\]
\[
\frac{19}{15} \approx 1.2667
\]
\[
\frac{17}{13} \approx 1.3077
\]
Least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]
Check choice: that matches **E**:
- \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11}\)

ANSWER 8: E


Problem 9:  
Numbers: \(3,4,5,6,6,7,x\) (7 positive integers).  
Mean = median = unique mode.

**Step 1: Mode**  
The only number that repeats is **6**, occurring twice. So the **unique mode = 6**.

Thus median must be 6 as well.

**Step 2: Median = 6**  
For 7 numbers sorted, the median is the **4th** number.

Sorted list must have the 4th term equal to 6. We already have two 6’s and surrounding numbers 3,4,5,7.

Currently without \(x\), the ordered portion is:
\[
3,4,5,6,6,7,\ ?
\]
The 4th number here is 6 already. To keep the 4th number as 6, \(x\) must not be less than 6 (otherwise it would shift positions so 4th might change).

If \(x = 6\), then 6 occurs 3 times, still mode 6 (unique mode). Then list:
\[
3,4,5,6,6,6,7
\]
Median (4th) is 6—works.

If \(x=7\), list:
\[
3,4,5,6,6,7,7
\]
Mode would be 6? Actually 7 repeats too, so mode not unique.

If \(x>7\), list:
\[
3,4,5,6,6,7,x
\]
Median still 6, but mode would still be 6 (unique), since only 6 repeats. That would also work—wait, but then 6 is still unique mode, yes. However we also need **mean = 6**.

So use mean condition.

Sum of known numbers:
\[
3+4+5+6+6+7 = 31
\]
Total sum including \(x\): \(31 + x\)

Mean:
\[
\frac{31+x}{7} = 6
\]
\[
31 + x = 42
\]
\[
x = 11
\]

Check uniqueness of mode: 6 occurs twice, no other repeats (since x=11 is different). Unique mode is 6—consistent.

ANSWER 9: D


Problem 10:  
Joe walks half the distance, then runs the rest.  
He runs **3 times as fast** as he walks.

Given: time to walk half way = **6 minutes**.  
Let walking speed be \(v\). Then walking time for half distance \(=\) 6 minutes.

Let half the distance be \(d\). Then:
\[
d = v \cdot 6
\]
Running speed is \(3v\).  
Remaining distance is also \(d\), so running time:
\[
\text{time} = \frac{d}{3v} = \frac{v\cdot 6}{3v} = 2
\]
Total time:
\[
6 + 2 = 8
\]
Correct choice: **D**.

ANSWER 10: D


Problem 11:  
Compute:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]

Group each set:

First set:
\[
1+11+21+31+41
\]
These are all of the form \(10k+1\):  
\(1, 11, 21, 31, 41\) correspond to \(k=0,1,2,3,4\).

Sum:
\[
(1+11+21+31+41) = (1+11+21+31+41)
\]
Compute directly:
\[
(1+11)=12,\ (21+31)=52,\ 12+52=64,\ 64+41=105
\]

Second set:
\[
9+19+29+39+49
\]
Direct sum:
\[
(9+19)=28,\ (29+39)=68,\ 28+68=96,\ 96+49=145
\]

Total:
\[
105 + 145 = 250
\]

Correct choice: **E**.

ANSWER 11: E


Problem 12:  
Which number is largest?

Options:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare decimals:  
- .99 is the largest among all because it is greater than .9 (and greater than any number starting 0.90... but with fewer 9s afterward).

So .99 > .9099 > .909 > .9009 > .9.

Correct choice: **A**.

ANSWER 12: A

Problem 1:
In a single-elimination tournament, every game results in exactly one team being eliminated. To determine a single winner from 16 teams, 15 teams must be eliminated. Since each game eliminates one team, it takes exactly 15 games.
ANSWER 1: D

Problem 2:
- Jose: (10 - 1) = 9; 9 * 2 = 18; 18 + 2 = 20.
- Thuy: (10 * 2) = 20; 20 - 1 = 19; 19 + 2 = 21.
- Kareem: (10 - 1) = 9; 9 + 2 = 11; 11 * 2 = 22.
Kareem has the largest answer (22).
ANSWER 2: C

Problem 3:
To convert 2/25 to a decimal, multiply the numerator and denominator by 4 to get a denominator of 100: (2 * 4) / (25 * 4) = 8 / 100 = 0.08.
ANSWER 3: B

Problem 4:
We need to choose 3 starters from a group of 4 (Lance, Sally, Joy, Fred). This is a combination problem denoted as 4C3. Choosing 3 people to be on the team is the same as choosing 1 person to be left off the team. There are 4 ways to choose who is left off.
ANSWER 4: B

Problem 5:
Let $x$ be the number. $0.20x = 12$. Dividing by 0.20, $x = 12 / 0.2 = 60$. We want 30% of $x$: $0.30 * 60 = 18$.
ANSWER 5: B

Problem 6:
Start at 0 revolutions. Clockwise is positive, counterclockwise is negative.
$2 \frac{1}{4} - 3 \frac{3}{4} = 2.25 - 3.75 = -1.5$ revolutions.
-1.5 revolutions is the same as -0.5 revolutions (since -1 is a full circle). -0.5 revolutions (counterclockwise) from West points to East.
ANSWER 6: B

Problem 7:
Original yellow tint: 30% of 30 liters = 9 liters.
New mixture: 30 + 5 = 35 liters.
New yellow tint: 9 + 5 = 14 liters.
Percent: (14 / 35) * 100 = (2 / 5) * 100 = 40%.
ANSWER 7: C

Problem 8:
Convert to decimals or compare differences from 1:
15/11 = 1 + 4/11 ≈ 1.3636
19/15 = 1 + 4/15 ≈ 1.2666
17/13 = 1 + 4/13 ≈ 1.3076
Comparing the fractions 4/11, 4/13, and 4/15: the larger the denominator, the smaller the fraction. Thus, 4/15 < 4/13 < 4/11.
Order: 19/15 < 17/13 < 15/11.
ANSWER 8: E

Problem 9:
The set is {3, 4, 5, 6, 6, 7, x}. The mode is 6 (it appears twice). For the mean and median to be 6:
Median: With 7 numbers, the 4th number must be 6. Since the sorted list is 3, 4, 5, 6, 6, 7, x, if $x > 6$, the 4th number is 6.
Mean: (3+4+5+6+6+7+x) / 7 = 6 => 31 + x = 42 => x = 11.
Checking: {3, 4, 5, 6, 6, 7, 11}. Median is 6. Mean is 42/7 = 6. Mode is 6.
ANSWER 9: D

Problem 10:
Walking time = 6 minutes. Running speed = 3 * walking speed.
Time taken = Distance / Speed. Since speed is 3 times faster, time is 1/3 of the walking time.
Running time = 6 / 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
ANSWER 10: D

Problem 11:
Sum 1: (1+11+21+31+41) = 105.
Sum 2: (9+19+29+39+49) = 145.
105 + 145 = 250.
ANSWER 11: E

Problem 12:
Compare digit by digit:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
0.99 is the largest.
ANSWER 12: A

Problem 1:
This is a single-elimination tournament with 16 teams. Exactly one team is eliminated per game, and 15 teams must be eliminated to leave a single winner. Thus exactly 15 games are required.

ANSWER 1: D

Problem 2:
Jose evaluates \((10-1)\times2+2=20\).
Thuy evaluates \((10\times2)-1+2=21\).
Kareem evaluates \((10-1+2)\times2=22\).
Kareem obtains the strictly largest value.

ANSWER 2: C

Problem 3:
Division yields the decimal expansion \(2\div25=0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters out of 4 distinct players is the binomial coefficient \(\binom{4}{3}=4\).

ANSWER 4: B

Problem 5:
If 20% of a number equals 12, the number itself is \(12/0.2=60\). Then 30% of 60 is \(0.3\times60=18\).

ANSWER 5: B

Problem 6:
A net rotation of \(+2.25-3.75=-1.5\) revolutions is equivalent to 1.5 revolutions counterclockwise. From an initial westward orientation, 1.5 revolutions counterclockwise lands exactly on east.

ANSWER 6: B

Problem 7:
The original mixture contains \(0.30\times30=9\) liters of yellow tint. Adding 5 liters raises the yellow volume to 14 liters and the total volume to 35 liters, so the new percentage is \(14/35=40\%\).

ANSWER 7: C

Problem 8:
Decimal approximations are \(15/11\approx1.3636\), \(19/15\approx1.2667\), and \(17/13\approx1.3077\). Ordering these values produces \(19/15<17/13<15/11\).

ANSWER 8: E

Problem 9:
The unique mode must be 6, forcing the mean also to equal 6. The sum of the seven numbers is therefore 42. The six given numbers sum to 31, so \(x=11\). With this value the ordered list is 3,4,5,6,6,7,11 whose median is likewise 6 and whose unique mode remains 6.

ANSWER 9: D

Problem 10:
Let the total distance be \(2d\). Walking time for distance \(d\) is 6 min, so running time for the remaining distance \(d\) at triple speed is \(6/3=2\) min. The total time is therefore 8 min.

ANSWER 10: D

Problem 11:
Each pair sums to a multiple of 10: \(1+9=10\), \(11+19=30\), \(21+29=50\), \(31+39=70\), \(41+49=90\). Their total is 250. Equivalently each five-term arithmetic series sums to 105 and 145 respectively, again yielding 250.

ANSWER 11: E

Problem 12:
Direct comparison of the five decimals shows that 0.99 is strictly larger than each of the others.

ANSWER 12: A

Problem 1:
The problem asks us to determine the number of games that will be played to determine the winner of a basketball tournament with 16 teams, where the losing team of each game is eliminated.

## Step 1
To determine the number of games needed to find the winner, we first note that in a single-elimination tournament, each game eliminates one team.

## Step 2
Since there can only be one winner, all other teams must be eliminated. Therefore, the number of games required is equal to the number of teams minus one.

## Step 3
We have 16 teams competing in the tournament. So, the number of games needed to determine the winner is 16 - 1 = 15.

## Step 4
Let's examine the given choices to rule out the incorrect options. A. 4 is too low because it doesn't account for all the teams. B. 7 is also too low for the same reason. C. 8 is still too low. D. 15 matches our calculation. E. 16 is one more than our calculation.

The final answer is supported by our step-by-step reasoning.
ANSWER 1: D

Problem 2:
The problem asks us to determine who among Jose, Thuy, and Kareem gets the largest final answer after performing a series of operations starting with the number 10.

## Step 1
Let's calculate Jose's final answer: Jose subtracts 1 from 10 to get 9, doubles it to get 18, and then adds 2 to get 20.

## Step 2
Now, let's calculate Thuy's final answer: Thuy doubles 10 to get 20, subtracts 1 to get 19, and then adds 2 to get 21.

## Step 3
Next, let's calculate Kareem's final answer: Kareem subtracts 1 from 10 to get 9, adds 2 to get 11, and then doubles it to get 22.

## Step 4
Comparing the final answers: Jose got 20, Thuy got 21, and Kareem got 22. The largest final answer is 22, which belongs to Kareem.

## Step 5
Let's examine the given choices. A. Jose is incorrect because his answer is 20. B. Thuy is incorrect because her answer is 21. C. Kareem is correct because his answer is 22. D. Jose and Thuy is incorrect because Kareem's answer is larger. E. Thuy and Kareem is incorrect because Jose's answer is not the largest.

The final answer is supported by our step-by-step reasoning.
ANSWER 2: C

Problem 3:
The problem asks us to convert the fraction 2/25 into a decimal.

## Step 1
To convert 2/25 into a decimal, we divide the numerator by the denominator: 2 ÷ 25 = 0.08.

## Step 2
Let's examine the given choices to identify the correct decimal representation. A. .008 is too small. B. .08 matches our calculation. C. .8 is too large. D. 1.25 is much too large. E. 12.5 is also much too large.

The final answer is supported by our step-by-step reasoning.
ANSWER 3: B

Problem 4:
The problem asks us to determine the number of ways to choose 3 starters from a team of 4 members: Lance, Sally, Joy, and Fred.

## Step 1
This is a combination problem, because the order in which the starters are chosen does not matter.

## Step 2
The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial, the product of all positive integers up to that number.

## Step 3
Here, n = 4 (total team members) and k = 3 (number of starters to be chosen). So, C(4, 3) = 4! / [3!(4 - 3)!] = 4! / (3! * 1!) = (4 * 3 * 2 * 1) / ((3 * 2 * 1) * 1) = 4.

## Step 4
Let's examine the given choices. A. 2 is too low. B. 4 matches our calculation. C. 6 is too high. D. 8 is also too high. E. 10 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 4: B

Problem 5:
The problem asks us to find 30% of a number, given that 20% of the same number is 12.

## Step 1
First, we need to find the number. We know that 20% of the number is 12. So, 0.20 * number = 12.

## Step 2
To find the number, we divide both sides of the equation by 0.20: number = 12 / 0.20 = 60.

## Step 3
Now, we need to calculate 30% of 60: 0.30 * 60 = 18.

## Step 4
Let's examine the given choices. A. 15 is too low. B. 18 matches our calculation. C. 20 is too high. D. 24 is also too high. E. 30 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 5: B

Problem 6:
The problem asks us to determine the direction the spinner points after Chenille moves it clockwise 2 1/4 revolutions and then counterclockwise 3 3/4 revolutions, starting from pointing west.

## Step 1
First, let's understand the effect of the movements. A full revolution is 360 degrees or 4 quarters. So, 2 1/4 revolutions clockwise is equivalent to 9 quarters clockwise.

## Step 2
9 quarters clockwise is the same as 1 quarter clockwise (since 8 quarters make 2 full revolutions, which brings the spinner back to the original direction). So, after the first move, the spinner points north.

## Step 3
Then, Chenille moves it counterclockwise 3 3/4 revolutions, which is equivalent to 15 quarters counterclockwise.

## Step 4
15 quarters counterclockwise is the same as 3 quarters counterclockwise (since 12 quarters make 3 full revolutions). So, moving 3 quarters counterclockwise from north results in pointing west.

## Step 5
Let's examine the given choices. A. north is incorrect because after the second move, it's not north. B. east is incorrect. C. south is incorrect. D. west matches our calculation. E. northwest is incorrect.

The final answer is supported by our step-by-step reasoning.
ANSWER 6: D

Problem 7:
The problem asks us to find the percentage of yellow tint in a new mixture after adding 5 liters of yellow tint to an original 30-liter mixture that is 25% red tint, 30% yellow tint, and 45% water.

## Step 1
First, we calculate the amount of yellow tint in the original mixture: 30% of 30 liters = 0.30 * 30 = 9 liters.

## Step 2
After adding 5 liters of yellow tint, the total amount of yellow tint becomes 9 + 5 = 14 liters.

## Step 3
The total volume of the new mixture is 30 + 5 = 35 liters.

## Step 4
To find the percentage of yellow tint in the new mixture, we divide the amount of yellow tint by the total volume and multiply by 100: (14 / 35) * 100 = 40%

Problem 1:
In a single-elimination tournament, each game eliminates exactly one team. To eliminate all but the champion, the number of games needed is one fewer than the number of teams. With 16 teams, the number of games is \(16 - 1 = 15\).

ANSWER 1: D

Problem 2:
Jose: \(10 - 1 = 9\), \(9 \times 2 = 18\), \(18 + 2 = 20\).
Thuy: \(10 \times 2 = 20\), \(20 - 1 = 19\), \(19 + 2 = 21\).
Kareem: \(10 - 1 = 9\), \(9 + 2 = 11\), \(11 \times 2 = 22\).
Kareem’s final answer of 22 is the largest.

ANSWER 2: C

Problem 3:
Divide 2 by 25: \(2 \div 25 = 0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters from 4 people is the combination \(\binom{4}{3} = 4\).

ANSWER 4: B

Problem 5:
Let the number be \(n\). \(20\%\) of \(n\) is \(0.2n = 12\), so \(n = 60\). Then \(30\%\) of \(60 = 0.3 \times 60 = 18\).

ANSWER 5: B

Problem 6:
Initial direction: west.
Clockwise \(2\frac{1}{4}\) revolutions: \(2\frac{1}{4} = 2.25\) revolutions clockwise.
Counterclockwise \(3\frac{3}{4}\) revolutions: \(3\frac{3}{4} = 3.75\) revolutions counterclockwise.
Net revolution = \(2.25 - 3.75 = -1.5\) revolutions, i.e., \(1.5\) revolutions counterclockwise.
A full revolution (1.0) returns to west. The remaining 0.5 revolution counterclockwise from west is a half-turn, which points to east.

ANSWER 6: B

Problem 7:
Original mixture: 30 liters, yellow tint = \(30\% \times 30 = 9\) liters.
Add 5 liters of yellow tint: new yellow = \(9 + 5 = 14\) liters; new total volume = \(30 + 5 = 35\) liters.
Percent yellow = \(\frac{14}{35} \times 100\% = 40\%\).

ANSWER 7: C

Problem 8:
Convert each fraction to a decimal for comparison:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
Order from least to greatest: \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

ANSWER 8: E

Problem 9:
The data set: 3, 4, 5, 6, 6, 7, \(x\). The mode is the most frequent number. With two 6’s, 6 is the mode unless \(x\) equals another number that appears twice, which would create a second mode, violating the “unique mode” condition. Therefore, \(x\) must not be 3, 4, 5, or 7. It could be 6, but if \(x=6\), the median is 6, while the mean is \((3+4+5+6+6+6+7)/7 = 37/7 \approx 5.29\), so mean ≠ median.  
For the mean to equal the median (which is 6, the middle of seven sorted numbers if \(x > 6\) or \(x=6\) only if placed correctly), we set the mean to 6. Sum without \(x\) = \(3+4+5+6+6+7 = 31\). Total needed for mean 6 = \(7 \times 6 = 42\). Thus \(x = 42 - 31 = 11\).  
Sorted: 3, 4, 5, 6, 6, 7, 11. Median = 6, mode = 6, mean = 6. Unique mode holds.

ANSWER 9: D

Problem 10:
Walking half the distance took 6 minutes, so walking speed = \(\frac{\text{half distance}}{6}\).
Running speed is 3 times walking speed. Time to run the second half = \(\frac{\text{half distance}}{3 \times (\text{half distance}/6)} = \frac{6}{3} = 2\) minutes.
Total time = \(6 + 2 = 8\) minutes.

ANSWER 10: D

Problem 11:
Pair the terms: \(1+49 = 50\), \(11+39 = 50\), \(21+29 = 50\), \(31+19 = 50\), \(41+9 = 50\). Five pairs of 50 give \(5 \times 50 = 250\).

ANSWER 11: E

Problem 12:
Compare by extending decimals to the same number of places:
\(0.99   = 0.9900\)
\(0.9099 = 0.9099\)
\(0.9    = 0.9000\)
\(0.909  = 0.9090\)
\(0.9009 = 0.9009\)
Clearly \(0.99 > 0.9099 > 0.909 > 0.9009 > 0.9\). The largest is \(0.99\).

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key Insight:** In a single-elimination tournament, every game eliminates exactly one team. To crown a single winner from 16 teams, we must eliminate 16 − 1 = 15 teams. Therefore, exactly 15 games are needed.

**Verification by rounds:**
- Round 1: 16 teams → 8 games → 8 teams left
- Round 2: 8 teams → 4 games → 4 teams left
- Round 3: 4 teams → 2 games → 2 teams left
- Round 4 (Final): 2 teams → 1 game → 1 winner

Total = 8 + 4 + 2 + 1 = 15 games.

Choices A (4), B (7), and C (8) are too small. Choice E (16) would mean every team is eliminated, leaving no winner.

ANSWER 1: D

---

Problem 2:
We track each person's calculation step by step, all starting from 10.

**Jose:** subtract 1 → 9; double → 18; add 2 → **20**
**Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
**Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

Comparing: 20 < 21 < 22. Kareem gets the largest final answer.

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

Choice A (.008) would be 8/1000. Choice C (.8) would be 8/10. Choices D and E are greater than 1, but 2/25 is clearly less than 1.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination problem — the order of selection doesn't matter:
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4$$

Equivalently, choosing 3 people to start is the same as choosing 1 person to sit out. There are 4 people, so 4 ways to pick who sits out.

The 4 possible starting lineups are: {L,S,J}, {L,S,F}, {L,J,F}, {S,J,F}.

ANSWER 4: B

---

Problem 5:
We're told 20% of a number is 12. Let the number be $x$.

$$0.20x = 12 \implies x = \frac{12}{0.20} = 60$$

Now find 30% of 60:
$$0.30 \times 60 = 18$$

**Shortcut:** 30% is $\frac{30}{20} = 1.5$ times 20%, so the answer is $1.5 \times 12 = 18$.

Choice A (15) would be 25%. Choice D (24) would be 40%. Choice E (30) would be 50%.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing **West**.

**Move 1:** Clockwise $2\frac{1}{4}$ revolutions.
- The 2 full revolutions bring it back to West.
- The extra $\frac{1}{4}$ revolution clockwise from West: West → **North**.

**Move 2:** Counterclockwise $3\frac{3}{4}$ revolutions.
- The 3 full revolutions bring it back to North.
- The extra $\frac{3}{4}$ revolution counterclockwise from North:
  - $\frac{1}{4}$ CCW from North → West
  - $\frac{2}{4}$ CCW from North → South
  - $\frac{3}{4}$ CCW from North → **East**

The spinner points East.

ANSWER 6: B

---

Problem 7:
Original mixture: 30 liters total.
- Red tint: 25% of 30 = 7.5 liters
- Yellow tint: 30% of 30 = 9 liters
- Water: 45% of 30 = 13.5 liters

Add 5 liters of yellow tint:
- New yellow tint = 9 + 5 = 14 liters
- New total mixture = 30 + 5 = 35 liters

Percent yellow tint in new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

Choice B (35%) is a trap for those who just add 5% to 30%. Choice D (45%) is the original water percentage.

ANSWER 7: C

---

Problem 8:
We compare $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$.

Rewrite each as a mixed number:
$$\frac{15}{11} = 1 + \frac{4}{11}, \quad \frac{19}{15} = 1 + \frac{4}{15}, \quad \frac{17}{13} = 1 + \frac{4}{13}$$

Since all three have the same numerator (4) in their fractional parts, we compare denominators. For fractions with equal numerators, a **larger denominator** means a **smaller fraction**:
$$15 > 13 > 11 \implies \frac{4}{15} < \frac{4}{13} < \frac{4}{11}$$

Therefore:
$$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$$

ANSWER 8: E

---

Problem 9:
The data set is: 3, 4, 5, 6, 6, 7, and $x$. The mean, median, and unique mode are all equal.

**Mode:** Since 6 already appears twice and all others appear once, the unique mode must be 6 (if $x$ were 3, 4, 5, or 7, we'd have two modes, violating "unique mode"). So mode = 6.

**Mean = 6:**
$$\frac{3 + 4 + 5 + 6 + 6 + 7 + x}{7} = 6$$
$$\frac{31 + x}{7} = 6 \implies 31 + x = 42 \implies x = 11$$

**Check Median:** With $x = 11$, the sorted list is: 3, 4, 5, **6**, 6, 7, 11. The median (4th value) is 6. ✓

All three measures equal 6. Choice B (6) would make 6 appear three times, but the mean would be $(31+6)/7 = 37/7 \neq 6$.

ANSWER 9: D

---

Problem 10:
Joe walked half the distance in 6 minutes. He then ran the remaining half at 3 times his walking speed.

Let walking speed = $v$. The half-distance = $6v$.

Running speed = $3v$. Time to run the second half:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{6v}{3v} = 2 \text{ minutes}$$

Total time = 6 + 2 = **8 minutes**.

Choice A (7) would result from incorrectly subtracting. Choices B, C, E involve unnecessary decimal complications.

ANSWER 10: D

---

Problem 11:
Compute $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$.

Pair the terms from the first group with terms from the second group (first with last, etc.):
- $1 + 49 = 50$
- $11 + 39 = 50$
- $21 + 29 = 50$
- $31 + 19 = 50$
- $41 + 9 = 50$

Total = $50 \times 5 = 250$.

Choice C (200) is a common error from miscounting pairs. Choice D (249) comes from arithmetic mistakes.

ANSWER 11: E

---

Problem 12:
Compare the five decimals by writing them to four decimal places:

- A: 0.9900
- B: 0.9099
- C: 0.9000
- D: 0.9090
- E: 0.9009

Comparing digit by digit from the tenths place: all have 9 in the tenths place. In the hundredths place, A has **9** while all others have **0**. So A is immediately the largest.

$0.99 > 0.9099 > 0.909 > 0.9009 > 0.9$

ANSWER 12: A

 **Problem 1:**
We need to find the number of games played in a single-elimination tournament with 16 teams. In single elimination, every game eliminates exactly one team. To determine a winner, 15 of the 16 teams must be eliminated. Therefore, exactly 15 games are required.  
(Another way: with 16 teams there are 8 games in the first round, 4 in the second, 2 in the third, and 1 final game: \(8+4+2+1=15\).)

ANSWER 1: D

---

**Problem 2:**
We compute each person’s final answer step by step, starting from 10.

- **Jose:**  
  \(10 - 1 = 9\)  
  \(9 \times 2 = 18\)  
  \(18 + 2 = 20\)

- **Thuy:**  
  \(10 \times 2 = 20\)  
  \(20 - 1 = 19\)  
  \(19 + 2 = 21\)

- **Kareem:**  
  \(10 - 1 = 9\)  
  \(9 + 2 = 11\)  
  \(11 \times 2 = 22\)

Comparing the results: \(20 < 21 < 22\). Kareem’s answer is the largest.

ANSWER 2: C

---

**Problem 3:**
We convert the fraction \(\frac{2}{25}\) to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100:
\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

ANSWER 3: B

---

**Problem 4:**
We must choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways to choose 3 from 4 is the combination \(\binom{4}{3}\):
\[
\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4
\]
Equivalently, choosing 3 starters is the same as choosing 1 person to sit out, and there are 4 choices for who sits out.

ANSWER 4: B

---

**Problem 5:**
Let the number be \(N\). We are given \(0.20N = 12\). Solving for \(N\):
\[
N = \frac{12}{0.20} = 60
\]
Now find 30% of 60:
\[
0.30 \times 60 = 18
\]
(Shortcut: 30% is \(1.5\) times 20%, so \(1.5 \times 12 = 18\).)

ANSWER 5: B

---

**Problem 6:**
The spinner starts pointing west.  
- Clockwise \(2\frac14\) revolutions: 2 full revolutions bring it back to west, then \(\frac14\) turn clockwise from west points it **north**.  
- Counterclockwise \(3\frac34\) revolutions: 3 full revolutions bring it back to north, then \(\frac34\) turn counterclockwise from north goes through west, south, and ends at **east**.

Net movement: \(3\frac34 - 2\frac14 = 1\frac12\) revolutions counterclockwise. One full revolution returns to west; an additional \(\frac12\) revolution counterclockwise from west points to **east**.

ANSWER 6: B

---

**Problem 7:**
Original mixture: 30 liters.  
- Yellow tint originally: \(30\% \times 30 = 9\) liters.  
After adding 5 liters of yellow tint:  
- New yellow tint: \(9 + 5 = 14\) liters.  
- New total mixture: \(30 + 5 = 35\) liters.  

Percent of yellow tint in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%
\]

ANSWER 7: C

---

**Problem 8:**
Compare \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\). Each is greater than 1. Write each as \(1\) plus a proper fraction:
\[
\frac{15}{11} = 1 + \frac{4}{11}, \qquad
\frac{19}{15} = 1 + \frac{4}{15}, \qquad
\frac{17}{13} = 1 + \frac{4}{13}
\]
Since the numerators are the same (4), the fraction with the largest denominator is smallest:
\[
\frac{4}{15} < \frac{4}{13} < \frac{4}{11}
\]
Adding 1 preserves the order:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

ANSWER 8: E

---

**Problem 9:**
The data set is \(3, 4, 5, 6, 6, 7, x\).  
- The current mode is 6 (appears twice). For 6 to remain the **unique** mode, \(x\) must not equal 3, 4, 5, or 7 (which would create a second mode).  
- We are told mean = median = mode = 6.  

Set the mean equal to 6:
\[
\frac{3+4+5+6+6+7+x}{7} = 6 \implies \frac{31+x}{7} = 6 \implies 31+x = 42 \implies x = 11
\]

Check: With \(x=11\), the sorted list is \(3, 4, 5, 6, 6, 7, 11\).  
- Median (4th term) = 6.  
- Mode = 6 (unique).  
- Mean = 6.  

All conditions are satisfied.

ANSWER 9: D

---

**Problem 10:**
Joe walks half the distance in 6 minutes. He runs the second half at 3 times his walking speed. Since distance is the same and speed is tripled, the time required is divided by 3:
\[
\text{Running time} = \frac{6}{3} = 2 \text{ minutes}
\]
Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

---

**Problem 11:**
Add the two groups:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]
Pair the terms to make convenient sums:
\[
(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 5 \times 50 = 250
\]

ANSWER 11: E

---

**Problem 12:**
Compare the decimals by writing them with the same number of places:
- A. \(0.9900\)
- B. \(0.9099\)
- C. \(0.9000\)
- D. \(0.9090\)
- E. \(0.9009\)

The tenths digit is 9 for all. Looking at the hundredths digit: A has 9, while B, D have 0 and C, E have 0. Thus \(0.9900\) is clearly the largest.

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key insight:** Each game eliminates exactly one team. To determine a winner from 16 teams, we must eliminate 15 teams (everyone except the champion).

Therefore, exactly 15 games must be played.

*Verification by rounds:* Round 1: 8 games (16→8 teams), Round 2: 4 games (8→4), Round 3: 2 games (4→2), Round 4: 1 game (2→1). Total = 8+4+2+1 = 15.

The tempting wrong answer E (16) counts teams instead of eliminations.

ANSWER 1: D

---

Problem 2:
We need to compute each person's final answer starting from 10.

**Jose:** Start with 10 → subtract 1: 10−1 = 9 → double: 9×2 = 18 → add 2: 18+2 = **20**

**Thuy:** Start with 10 → double: 10×2 = 20 → subtract 1: 20−1 = 19 → add 2: 19+2 = **21**

**Kareem:** Start with 10 → subtract 1: 10−1 = 9 → add 2: 9+2 = 11 → double: 11×2 = **22**

Kareem gets the largest final answer (22).

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

The tempting wrong answer A (0.008) would be 8/1000, and C (0.8) would be 8/10.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination: C(4,3) = 4!/(3! × 1!) = 4

Listing them out: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — that's 4 ways.

The tempting wrong answer C (6) would be C(4,2), choosing who's *not* a starter, but that also equals 4... Actually C(4,2)=6 is wrong here; we're choosing 3 from 4.

ANSWER 4: B

---

Problem 5:
We know 20% of a number is 12. We need to find 30% of the same number.

**Step 1:** Find the number.
$$0.20 \times n = 12 \implies n = \frac{12}{0.20} = 60$$

**Step 2:** Find 30% of 60.
$$0.30 \times 60 = 18$$

*Shortcut:* Since 30% is 1.5 times 20%, the answer is 1.5 × 12 = 18.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing west. We need to find the net rotation.

**Clockwise:** 2¼ revolutions
**Counterclockwise:** 3¾ revolutions

**Net movement:** 2¼ − 3¾ = −1½ revolutions (negative means counterclockwise)

So the spinner moves 1½ revolutions counterclockwise from west.

- 1 full revolution counterclockwise returns to **west**
- ½ revolution counterclockwise from west goes to **east**

(Counterclockwise from west: ¼ turn → south, ½ turn → east)

ANSWER 6: B

---

Problem 7:
We need to find the percent of yellow tint after adding 5 liters of yellow tint.

**Original mixture (30 liters):**
- Yellow tint: 30% of 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percent yellow tint:** (14/35) × 100 = 40%

The tempting wrong answer B (35) might come from simply adding 5% to 30%.

ANSWER 7: C

---

Problem 8:
We need to order 15/11, 19/15, and 17/13 from least to greatest.

Computing each as a decimal:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3077...

From least to greatest: **19/15 < 17/13 < 15/11**

*Alternative method:* Note each is 1 + (numerator − denominator)/denominator:
- 15/11 = 1 + 4/11
- 19/15 = 1 + 4/15
- 17/13 = 1 + 4/13

Since 4/15 < 4/13 < 4/11, the order is confirmed.

ANSWER 8: E

---

Problem 9:
The positive integers 3, 4, 5, 6, 6, 7, and x have equal mean, median, and unique mode.

**Mode:** Since 6 already appears twice and no other number does, the unique mode must be 6. So x ≠ 3, 4, 5, or 7 (which would create a second mode).

**Median:** With 7 numbers, the median is the 4th value when sorted. For the median to be 6, we need x ≥ 6 (so that 6 is the 4th value).

- If x = 6: sorted = 3, 4, 5, 6, 6, 6, 7 → median = 6 ✓
- If x > 7: sorted = 3, 4, 5, 6, 6, 7, x → median = 6 ✓

**Mean:** (3 + 4 + 5 + 6 + 6 + 7 + x)/7 = (31 + x)/7

Setting mean = 6: (31 + x)/7 = 6 → 31 + x = 42 → **x = 11**

**Verification:** Numbers: 3, 4, 5, 6, 6, 7, 11. Mean = 42/7 = 6, Median = 6, Mode = 6 ✓

ANSWER 9: D

---

Problem 10:
Joe walks halfway, then runs the rest. We need total time.

**Walking:** Half the distance takes 6 minutes.

**Running:** He runs 3 times as fast as he walks, so he covers the same distance in 1/3 the time.
- Time to run the second half = 6 ÷ 3 = 2 minutes

**Total time:** 6 + 2 = **8 minutes**

The tempting wrong answers come from setting up the problem incorrectly or mixing up distance and time relationships.

ANSWER 10: D

---

Problem 11:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

**First group:** 1 + 11 + 21 + 31 + 41 = 105

**Second group:** 9 + 19 + 29 + 39 + 49 = 145

**Total:** 105 + 145 = **250**

*Elegant method:* Pair terms: (1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250

ANSWER 11: E

---

Problem 12:
We need to find the largest number among the options. Writing each with the same number of decimal places:

- A. 0.99 = **0.9900**
- B. 0.9099 = 0.9099
- C. 0.9 = 0.9000
- D. 0.909 = 0.9090
- E. 0.9009 = 0.9009

Comparing digit by digit after the decimal point:
- First digit: all are 9
- Second digit: A has **9**, all others have 0

Since A has a 9 in the second decimal place while all others have 0, **0.99 is the largest**.

ANSWER 12: A

**Problem 1:**  
The tournament is single‑elimination. Starting with 16 teams, every game eliminates exactly one team. To reduce from 16 teams to a single champion we must eliminate 15 teams, and each elimination corresponds to one game.  

Number of games = 16 − 1 = **15**.  

ANSWER 1: D  

---

**Problem 2:**  
Start with 10 for each person.

- **Jose:**  
  Subtract 1 → 9; double → 18; add 2 → 20.  

- **Thuy:**  
  Double → 20; subtract 1 → 19; add 2 → 21.  

- **Kareem:**  
  Subtract 1 → 9; add 2 → 11; double → 22.  

The largest final value is 22, obtained by Kareem.  

ANSWER 2: C  

---

**Problem 3:**  

\[
\frac{2}{25}=0.08
\]

(2 ÷ 25 = 0.08).  

ANSWER 3: B  

---

**Problem 4:**  
We need to choose 3 starters from the 4 available players:  

\[
\binom{4}{3}=4
\]

ways.  

ANSWER 4: B  

---

**Problem 5:**  
Let the number be \(N\).  
\(0.20N = 12 \Rightarrow N = \frac{12}{0.20}=60\).  
\(30\%\) of \(N\) is \(0.30 \times 60 = 18\).  

ANSWER 5: B  

---

**Problem 6:**  
Interpret the fractions as mixed numbers:  

- Clockwise \(2\frac14 = 2.25\) revolutions.  
- Counter‑clockwise \(3\frac34 = 3.75\) revolutions.  

Net rotation (counter‑clockwise positive) = \(3.75 - 2.25 = 1.5\) revolutions.  

A full revolution returns to the same direction, so we only care about the extra \(0.5\) revolution.  
\(0.5\) revolution = 180° counter‑clockwise.  

Starting from west, turning 180° counter‑clockwise points to **east**.  

ANSWER 6: B  

---

**Problem 7:**  
Original yellow = \(30\%\) of 30 L = \(0.30 \times 30 = 9\) L.  
Add 5 L yellow → new yellow = \(9+5 = 14\) L.  
New total volume = \(30+5 = 35\) L.  

Percent yellow = \(\frac{14}{35}\times100 = 40\%\).  

ANSWER 7: C  

---

**Problem 8:**  
Compute the three fractions:

\[
\frac{15}{11}=1.3636\ldots,\qquad
\frac{19}{15}=1.2666\ldots,\qquad
\frac{17}{13}=1.3077\ldots
\]

Ordered from least to greatest:  

\(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

This matches option **E**.  

ANSWER 8: E  

---

**Problem 9:**  
Current list: \(3,4,5,6,6,7\). The unique mode is 6 (appears twice).  
For the mode to remain unique and equal to the mean and median, \(x\) must not create a tie.  
Let the mean = median = mode = \(m\).  

Sum of numbers = \(31 + x\).  
Mean = \(\frac{31 + x}{7}=m\).  
Median of the seven numbers is the 4th after sorting; with \(x>7\) the median stays 6, so \(m=6\).  

Thus \(\frac{31 + x}{7}=6 \Rightarrow 31 + x = 42 \Rightarrow x = 11\).

Check: sorted list becomes \(3,4,5,6,6,7,11\) → median = 6, mode = 6, mean = 42/7 = 6. All equal.  

ANSWER 9: D  

---

**Problem 10:**  
Let the walking speed be \(v\) and the half‑distance be \(d/2\).  

Walking time for half‑distance: \(\frac{d/2}{v}=6\) min ⇒ \(\frac{d}{v}=12\) min.  

Running speed = \(3v\).  
Running time for the second half: \(\frac{d/2}{3v}= \frac{d}{6v}= \frac{1}{6}\times\frac{d}{v}= \frac{12}{6}=2\) min.  

Total time = \(6 + 2 = 8\) minutes.  

ANSWER 10: D  

---

**Problem 11:**  

\[
\begin{aligned}
1+11+21+31+41 &= 105,\\
9+19+29+39+49 &= 145,\\
\text{Total} &= 105+145 = 250.
\end{aligned}
\]

ANSWER 11: E  

---

**Problem 12:**  
Write the numbers with four decimal places:

- .99 = 0.9900  
- .9099 = 0.9099  
- .9 = 0.9000  
- .909 = 0.9090  
- .9009 = 0.9009  

The largest is 0.9900, i.e., **.99**.  

ANSWER 12: A  

Problem 1:  
The tournament starts with 16 teams. Each game eliminates one team, and the losing team is eliminated. To determine the winner, all teams except one must be eliminated. Thus, 15 teams must be eliminated, requiring 15 games.  
- **Why not other choices?** A (4) is too few; B (7), C (8) are also insufficient; E (16) would imply one game per team without elimination, which is incorrect.  
ANSWER 1: D  

Problem 2:  
- **Jose**: Starts with 10 → subtracts 1 (9) → doubles (18) → adds 2 → **20**.  
- **Thuy**: Starts with 10 → doubles (20) → subtracts 1 (19) → adds 2 → **21**.  
- **Kareem**: Starts with 10 → subtracts 1 (9) → adds 2 (11) → doubles → **22**.  
Kareem has the largest final answer.  
- **Why not other choices?** A (Jose) and B (Thuy) are smaller; D and E pair incorrect winners.  
ANSWER 2: C  

Problem 3:  
\( \frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08 \).  
- **Why not other choices?** A (.008) is \( \frac{2}{250} \); C (.8) is \( \frac{4}{5} \); D and E are larger decimals.  
ANSWER 3: B  

Problem 4:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is \( \binom{4}{3} = 4 \).  
- **Why not other choices?** A (2) is too few; C (6) is \( \binom{4}{2} \), which is incorrect; D (8) and E (10) are overcounts.  
ANSWER 4: B  

Problem 5:  
Let the number be \( x \). If 20% of \( x \) is 12, then \( 0.2x = 12 \), so \( x = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).  
- **Why not other choices?** A (15) is 25% of 60; C (20) is \( \frac{1}{3} \) of 60; D (24) is 40%; E (30) is 50%.  
ANSWER 5: B  

Problem 6:  
- A full revolution is 360°. Clockwise \( 2 \frac{1}{4} \) revolutions = \( 2.25 \times 360 = 810° \).  
- Counterclockwise \( 3 \frac{3}{4} \) revolutions = \( -3.75 \times 360 = -1350° \).  
- Net movement: \( 810° - 1350° = -540° \).  
- \( -540° \mod 360° = 180° \) (since \( -540 + 720 = 180° \)), which points west → south? Wait, initial direction is west.  
- West (270°) + 810° = 1080° → 1080° mod 360° = 0° (north). Then -1350° → 0° - 1350° = -1350° mod 360° = 90° (east)? Let's reframe:  
- Start at west (270°). Clockwise 2.25 rev: 270° + 810° = 1080° mod 360° = 0° (north). Then counterclockwise 3.75 rev: 0° - 1350° = -1350° mod 360° = 90° (east).  
- **Why not other choices?** A (north) is initial after clockwise; B (east) is final; C (south) is 180°; D (west) is start; E (northwest) is not a cardinal direction.  
ANSWER 6: B  

Problem 7:  
Original mixture: 30 liters, 30% yellow tint → 9 liters yellow. Add 5 liters yellow → total yellow = 14 liters. New mixture = 35 liters. Percent yellow = \( \frac{14}{35} \times 100 = 40\% \).  
- **Why not other choices?** A (25) is original yellow%; B (35) is before addition; D (45) and E (50) are overestimates.  
ANSWER 7: E (Note: Correction based on computation: 14/35=0.4 → 40%, so answer is E? Wait, 14/35=0.4 → 40%, which is E. But earlier I thought B, but B is 35, which is wrong. Original yellow is 9, add 5 → 14. Total mixture 30+5=35. 14/35=0.4 → 40%. So E. But in choices E is 50? Wait, no: choices are A.25 B.35 C.40 D.45 E.50. So 40% is C. I misread. Correction: 14/35=0.4 → 40%, which is C.  
**Correction**: 14/35=0.4 → 40%, so answer is C.  
ANSWER 7: C  

Problem 8:  
Compare \( \frac{15}{11} \approx 1.3636 \), \( \frac{19}{15} \approx 1.2667 \), \( \frac{17}{13} \approx 1.3077 \). Order: \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11} \).  
- **Why not other choices?** A, B, C, D have incorrect orders.  
ANSWER 8: E  

Problem 9:  
The set is {3, 4, 5, 6, 6, 7, x}. Mean = median = mode = 6.  
- Mode is 6 (unique), so 6 must appear most.  
- Median is 6 (4th value in ordered list: 3,4,5,6,6,7,x → if x≥6, ordered is ...,6,6,x or ...,6,x,6? Must have 4th value 6.  
- If x=6, set is {3,4,5,6,6,6,7}: median=6, mode=6, mean=(3+4+5+6+6+6+7)/7=37/7≈5.28 ≠6.  
- If x=11: set {3,4,5,6,6,7,11}: mean=(3+4+5+6+6+7+11)/7=42/7=6, median=6, mode=6.  
- **Why not other choices?** A (5): mean=36/7≈5.14; B (6): mean≈5.28; C (7): mean=38/7≈5.43; D (11) works.  
ANSWER 9: D  

Problem 10:  
Joe walks half the distance in 6 minutes. Let walking speed be \( v \), so distance = \( 6v \). He runs at \( 3v \), so time to run half distance = \( \frac{6v}{3v} = 2 \) minutes. Total time = 6 + 2 = 8 minutes.  
- **Why not other choices?** A (7), B (7.3), C (7.7), E (8.3) are not exact.  
ANSWER 10: D  

Problem 11:  
First group: 1+11+21+31+41 = 105. Second group: 9+19+29+39+49 = 145. Total = 105 + 145 = 250.  
- **Why not other choices?** A (150), B (199), C (200), D (249) are incorrect sums.  
ANSWER 11: E  

Problem 12:  
Compare decimals: A (.99), B (.9099), C (.9), D (.909), E (.9009). Largest is .99.  
- **Why not other choices?** B, C, D, E are all smaller than .99.  
ANSWER 12: A

### Problem 1:
We need to find the total number of games to determine a winner in a 16-team single-elimination tournament, where every game eliminates exactly one losing team. To get 1 winner, we must eliminate 16 - 1 = 15 teams. Since each game eliminates 1 team, we need 15 total games. Wrong choices: A is the number of bracket rounds, not games; C is half the number of teams; E is the total number of teams, not games.
ANSWER 1: D

---

### Problem 2:
All three start with 10, we calculate each person's final result step-by-step:
- Jose: (10 - 1) * 2 + 2 = 9*2 +2 = 20
- Thuy: (10*2) -1 +2 = 20-1+2=21
- Kareem: (10-1 +2)*2 = 11*2=22
Kareem has the largest final answer. Wrong choices stem from misordering the operations for each person.
ANSWER 2: C

---

### Problem3:
Convert 2/25 to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100: (2*4)/(25*4) = 8/100 = 0.08. Wrong choices: A is 2/250, C is 20/25.
ANSWER 3: B

---

### Problem4:
We choose 3 starters from 4 people, which is a combination problem (order of starters does not matter here). The number of ways to choose 3 from 4 is equivalent to choosing 1 person to leave off the starting lineup: there are 4 people, so 4 unique lineups, each omitting one person. Wrong choice C (6) is the number of permutations, not combinations.
ANSWER 4: B

---

### Problem5:
If 20% of a number is 12, first find the full number: 0.2*N=12 → N=12/0.2=60. 30% of 60 is 0.3*60=18. Wrong choices: A is 25% of 60, D is 40% of 60.
ANSWER 5: B

---

### Problem6:
The spinner starts pointing west. Net movement: 2.25 clockwise revolutions minus 3.75 clockwise (equivalent to 3.75 counterclockwise) = -1.5 total clockwise revolutions, or 1.5 counterclockwise revolutions. Full revolutions cancel out, so we only have a net 0.5 (180 degree) counterclockwise turn from west. A 180-degree turn from west points to east.
ANSWER 6: B

---

### Problem7:
Original 30L mixture has 30% yellow tint: 0.3*30=9L of yellow. Add 5L of yellow, so new yellow is 14L, new total mixture is 35L. The new percentage of yellow is (14/35)*100=40%. Wrong choice: people often forget to add 5L to the total mixture volume, leading to an incorrect higher percentage.
ANSWER 7: C

---

### Problem8:
Rewrite each fraction to compare easily: 15/11 = 1 + 4/11, 19/15=1+4/15, 17/13=1+4/13. For equal numerators, the larger the denominator, the smaller the fraction. So 4/15 < 4/13 < 4/11, so the full order from least to greatest is 19/15 < 17/13 <15/11.
ANSWER 8: E

---

### Problem9:
The dataset is [3,4,5,6,6,7,x]. The unique mode is currently 6 (only value appearing twice, so it must stay the only unique mode). For the mean, median, and mode to all be 6: the sum of the dataset must be 7*6=42. The sum of the existing numbers is 3+4+5+6+6+7=31, so x=42-31=11. Sorted dataset [3,4,5,6,6,7,11] has median 6, unique mode 6, mean 6, which fits. Wrong choice B (x=6) gives a mean of ~5.28, not 6.
ANSWER 9: D

---

### Problem10:
Joe walks half the distance in 6 minutes. His running speed is 3x his walking speed, so time to cover the same distance is 1/3 of walking time. Time to run the second half: 6/3=2 minutes. Total time: 6+2=8 minutes.
ANSWER 10: D

---

### Problem11:
Pair terms to simplify addition: (1+49)+(11+39)+(21+29)+(31+19)+(41+9) = 50+50+50+50+50=250.
ANSWER 11: E

---

### Problem12:
Line up all decimals to compare: A=0.9900, B=0.9099, C=0.9000, D=0.9090, E=0.9009. The largest value is 0.99, option A.
ANSWER 12: A

Problem 1:  
In a single-elimination tournament, each game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams (all except 1). Thus, 15 games are played.  
ANSWER 1: D  

Problem 2:  
Calculate each person's result starting from 10:  
- Jose: \(10 -1 =9\), double → \(9*2=18\), add 2 → \(18+2=20\)  
- Thuy: Double 10 → \(10*2=20\), subtract 1 → \(20-1=19\), add 2 → \(19+2=21\)  
- Kareem: \(10-1=9\), add 2 → \(9+2=11\), double → \(11*2=22\)  
Kareem has the largest result.  
ANSWER 2: C  

Problem 3:  
Convert \(\frac{2}{25}\) to decimal: \(2\div25=0.08\).  
ANSWER 3: B  

Problem 4:  
Choosing 3 starters from 4 people is a combination (order does not matter). The number of ways is \(\binom{4}{3}=4\) (or list: {Lance,Sally,Joy}, {Lance,Sally,Fred}, {Lance,Joy,Fred}, {Sally,Joy,Fred}).  
ANSWER 4: B  

Problem 5:  
Let the number be \(x\). \(20\%x=12\) → \(0.2x=12\) → \(x=60\). \(30\%\ of\ 60=0.3*60=18\).  
ANSWER 5: B  

Problem 6:  
Model directions with standard angles: 0°=east, 90°=north, 180°=west, 270°=south. Initial direction is west (180°).  
- Clockwise \(2\frac{1}{4}\) revolutions: \(2.25*(-360°)=-810°\), new angle: \(180-810=-630°\)  
- Counterclockwise \(3\frac{3}{4}\) revolutions: \(3.75*(360°)=+1350°\), total angle: \(-630+1350=720°\)  
- \(720°\mod360°=0°\), which is east.  
ANSWER 6: B  

Problem 7:  
Original yellow tint: \(30\%\ of\ 30=0.3*30=9\) liters. Add 5 liters, total yellow = \(9+5=14\) liters. New total mixture: \(30+5=35\) liters. Percent yellow: \(\frac{14}{35}*100\%=40\%\).  
ANSWER 7: C  

Problem 8:  
Compare fractions via cross-multiplication or decimals:  
- \(\frac{19}{15}\approx1.267\), \(\frac{17}{13}\approx1.308\), \(\frac{15}{11}\approx1.364\)  
Order from least to greatest: \(\frac{19}{15}<\frac{17}{13}<\frac{15}{11}\).  
ANSWER 8: E  

Problem 9:  
The unique mode is 6 (appears twice, others once). Thus mean and median equal 6.  
- Median of 7 numbers is the 4th term when sorted, so 4th term =6 → \(x\geq6\).  
- Mean = \(\frac{3+4+5+6+6+7+x}{7}=6\) → \(31+x=42\) → \(x=11\).  
Check: Sorted list [3,4,5,6,6,7,11], median=6, mode=6 (unique), mean=6.  
ANSWER 9: D  

Problem 10:  
Walking half the distance takes 6 minutes, so full walking time would be 12 minutes. Running speed is 3x walking speed, so time for the second half is \(\frac{6}{3}=2\) minutes. Total time: \(6+2=8\) minutes.  
ANSWER 10: D  

Problem 11:  
Sum first group: \(1+11+21+31+41=105\) (arithmetic series: \(\frac{5}{2}(1+41)=105\)).  
Sum second group: \(9+19+29+39+49=145\) (\(\frac{5}{2}(9+49)=145\)).  
Total: \(105+145=250\).  
ANSWER 11: E  

Problem 12:  
Compare decimals digit by digit:  
- A=0.99, B=0.9099, C=0.9, D=0.909, E=0.9009  
A has the highest second decimal digit (9 vs 0 for others), so it is the largest.  
ANSWER 12: A

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** How many games must be played to determine a winner in a 16-team elimination tournament?

**Setup:** In an elimination tournament, each game eliminates exactly one team. To go from 16 teams to 1 winner, we must eliminate 15 teams. Therefore, we need 15 games.

**Alternatively:** We can think of it as a bracket:
- Round 1: 16 teams → 8 games → 8 winners
- Round 2: 8 teams → 4 games → 4 winners
- Round 3: 4 teams → 2 games → 2 winners
- Round 4: 2 teams → 1 game → 1 winner
- Total: 8 + 4 + 2 + 1 = 15 games

ANSWER 1: D

---

**Problem 2:**

**What's being asked:** Which person gets the largest final answer after their operations?

**Jose's calculation:** 
- Start: 10
- Subtract 1: 10 - 1 = 9
- Double: 9 × 2 = 18
- Add 2: 18 + 2 = **20**

**Thuy's calculation:**
- Start: 10
- Double: 10 × 2 = 20
- Subtract 1: 20 - 1 = 19
- Add 2: 19 + 2 = **21**

**Kareem's calculation:**
- Start: 10
- Subtract 1: 10 - 1 = 9
- Add 2: 9 + 2 = 11
- Double: 11 × 2 = **22**

Kareem gets 22, which is the largest.

ANSWER 2: C

---

**Problem 3:**

**What's being asked:** Convert the fraction 2/25 to a decimal.

**Setup:** 2/25 = ?

**Calculation:** 
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** In how many ways can 3 starters be chosen from 4 people (Lance, Sally, Joy, Fred)?

**Setup:** This is a combination problem: $\binom{4}{3}$

**Calculation:**
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4}{1} = 4$$

Alternatively, we're choosing which 1 person to leave out: 4 ways.

ANSWER 4: B

---

**Problem 5:**

**What's being asked:** If 20% of a number is 12, what is 30% of that number?

**Setup:** Let the number be $n$.
- $0.20n = 12$
- $n = 12 ÷ 0.20 = 60$

**Calculation:** 
$$30\% \text{ of } 60 = 0.30 \times 60 = 18$$

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Starting pointing west, after rotating clockwise 2¼ revolutions then counterclockwise 3¾ revolutions, which direction?

**Setup:** 
- Clockwise 2¼ revolutions = 2.25 revolutions
- Counterclockwise 3¾ revolutions = 3.75 revolutions
- Net: 3.75 - 2.25 = 1.5 revolutions counterclockwise (net)

**Calculation:**
Starting at west, rotating counterclockwise 1.5 revolutions (= 1 full rotation + 0.5 rotation):
- 1 full rotation brings us back to west
- 0.5 rotation = 180°
- West + 180° counterclockwise = East

ANSWER 6: B

---

**Problem 7:**

**What's being asked:** After adding 5 liters of yellow tint to a 30-liter mixture, what percent is yellow tint?

**Original mixture (30 liters):**
- Yellow tint: 30% × 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percentage:**
$$\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%$$

ANSWER 7: C

---

**Problem 8:**

**What's being asked:** Order 15/11, 19/15, and 17/13 from least to greatest.

**Setup:** Convert to decimals or compare directly.

**Calculations:**
- 15/11 ≈ 1.364
- 19/15 ≈ 1.267
- 17/13 ≈ 1.308

**Order:** 19/15 < 17/13 < 15/11

ANSWER 8: E

---

**Problem 9:**

**What's being asked:** Find $x$ such that the mean, median, and unique mode of {3, 4, 5, 6, 6, 7, $x$} are all equal.

**Current mode:** 6 appears twice; all others appear once. If there's a unique mode, it must be 6.

**With unique mode = 6:** This requires no other number to appear more than once. So $x \notin \{3, 4, 5, 7\}$ and $x \neq 6$ (else it wouldn't be unique).

**Mean = Median = Mode = 6:**

For median of 7 numbers with value 6:
- Sorted: need middle (4th) value to be 6
- If $x = 11$: {3, 4, 5, 6, 6, 7, 11} → median = 6 ✓

Mean: $\frac{3+4+5+6+6+7+11}{7} = \frac{42}{7} = 6$ ✓

Mode: 6 appears twice; all others once → mode = 6 ✓

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Total time from home to school if Joe walked halfway (6 min) then ran the rest at 3× speed?

**Setup:**
- Walking: 6 minutes for half the distance
- Running: same distance at 3× speed, so time = 6/3 = 2 minutes

**Total time:** 6 + 2 = 8 minutes

ANSWER 10: D

---

**Problem 11:**

**What's being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Calculation:**
- First group: 1 + 11 + 21 + 31 + 41 = 105
- Second group: 9 + 19 + 29 + 39 + 49 = 145
- Sum: 105 + 145 = 250

ANSWER 11: E

---

**Problem 12:**

**What's being asked:** Which number is largest?

**Comparison:**
- A. 0.99
- B. 0.9099
- C. 0.9
- D. 0.909
- E. 0.9009

Starting from the tenths place: all are 0.9__. Looking at hundredths:
- 0.99 has 9 in hundredths place → **largest**

ANSWER 12: A

Problem 1:

In a single-elimination tournament, each game eliminates exactly one team.  
There are 16 teams, and we want one champion.

So we must eliminate 15 teams total.

Therefore, the number of games played is 15.

Tempting wrong choice: 16 would be if every team had to lose, but the winner does not lose.

ANSWER 1: D

Problem 2:

Compute each person’s result carefully.

- Jose: start with 10  
  subtract 1 → 9  
  double → 18  
  add 2 → 20

- Thuy: start with 10  
  double → 20  
  subtract 1 → 19  
  add 2 → 21

- Kareem: start with 10  
  subtract 1 → 9  
  add 2 → 11  
  double → 22

Compare:
- Jose = 20
- Thuy = 21
- Kareem = 22

Kareem gets the largest final answer.

ANSWER 2: C

Problem 3:

\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

So the correct decimal is 0.08.

ANSWER 3: B

Problem 4:

We are choosing 3 starters from 4 players: Lance, Sally, Joy, and Fred.

This is a combination problem:
\[
\binom{4}{3} = 4
\]

You can also see it by noticing that choosing 3 starters is the same as choosing which 1 player sits out, and there are 4 choices.

ANSWER 4: B

Problem 5:

Let the number be \(n\).

20% of \(n\) is 12, so:
\[
0.2n = 12
\]
\[
n = 60
\]

Now find 30% of 60:
\[
0.3 \times 60 = 18
\]

ANSWER 5: B

Problem 6:

The spinner starts pointing west.

First move: clockwise \(2\frac{1}{4}\) revolutions.  
A full revolution brings it back to west, so only the extra \(\frac{1}{4}\) revolution matters.

From west, a quarter-turn clockwise points it to north.

Second move: counterclockwise \(3\frac{3}{4}\) revolutions.  
Again, the 3 full revolutions do nothing, so this is just \(\frac{3}{4}\) of a revolution counterclockwise, which is the same as \(\frac{1}{4}\) of a revolution clockwise.

From north, a quarter-turn clockwise points east.

So the spinner ends pointing east.

ANSWER 6: B

Problem 7:

Start with 30 liters total.

Yellow tint is 30% of 30 liters:
\[
0.30 \times 30 = 9 \text{ liters}
\]

Then 5 liters of yellow tint are added, so yellow becomes:
\[
9 + 5 = 14 \text{ liters}
\]

New total mixture:
\[
30 + 5 = 35 \text{ liters}
\]

Percent yellow in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.4 = 40\%
\]

ANSWER 7: C

Problem 8:

The fractions are:
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]

Approximate each one:

\[
\frac{15}{11} \approx 1.364
\]
\[
\frac{17}{13} \approx 1.308
\]
\[
\frac{19}{15} \approx 1.267
\]

So from least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

That matches choice E.

ANSWER 8: E

Problem 9:

The numbers are:
\[
3,4,5,6,6,7,x
\]

We are told the mean, median, and unique mode are all equal.

Since 6 appears twice and every other listed number appears once, the unique mode is 6.

So the median and mean must also be 6.

The median of 7 numbers is the 4th number when ordered.  
The sorted list is:
\[
3,4,5,6,6,7,x
\]
To make the median 6, \(x\) must be 6 or larger, but we also need the mean to be 6.

Sum of the known numbers:
\[
3+4+5+6+6+7 = 31
\]

For the mean to be 6, total sum must be:
\[
7 \times 6 = 42
\]

So:
\[
31 + x = 42
\]
\[
x = 11
\]

Check:
- mode = 6
- median = 6
- mean = 42/7 = 6

Works perfectly.

ANSWER 9: D

Problem 10:

Joe walked half the distance in 6 minutes.  
He then ran the other half at 3 times his walking speed.

If he runs 3 times as fast, then the time for the same distance is 1/3 as long.

So the running time is:
\[
6 \div 3 = 2 \text{ minutes}
\]

Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

Problem 11:

Add the first group:
\[
1+11+21+31+41 = 105
\]

Add the second group:
\[
9+19+29+39+49 = 145
\]

Now add the totals:
\[
105 + 145 = 250
\]

ANSWER 11: E

Problem 12:

Compare the decimals:

- \(0.99\)
- \(0.9099\)
- \(0.9 = 0.9000\)
- \(0.909\)
- \(0.9009\)

The largest is clearly \(0.99\), since it is closest to 1.

ANSWER 12: A

Problem 1:  
We have a three-person-team tournament where **the losing team of each game is eliminated**. With **16 teams**, how many games are played to determine the winner?

To find the number of games in a single-elimination tournament:  
- Each game eliminates exactly **one** team (the loser).  
- To go from **16 teams to 1 winning team**, we must eliminate **15 teams**.  
- Therefore, the number of games is **15**.

So the correct choice is **D**.

ANSWER 1: D


Problem 2:  
Jose, Thuy, and Kareem all start at 10. We compute each final value.

**Jose:**  
- Subtract 1: \(10 - 1 = 9\)  
- Double: \(2 \cdot 9 = 18\)  
- Add 2: \(18 + 2 = 20\)  
Final: **20**

**Thuy:**  
- Double 10: \(2 \cdot 10 = 20\)  
- Subtract 1: \(20 - 1 = 19\)  
- Add 2: \(19 + 2 = 21\)  
Final: **21**

**Kareem:**  
- Subtract 1: \(10 - 1 = 9\)  
- Add 2: \(9 + 2 = 11\)  
- Double: \(2 \cdot 11 = 22\)  
Final: **22**

Largest is **22**, which is Kareem.

ANSWER 2: C


Problem 3:  
We need to evaluate:  
\[
\frac{2}{25}
\]
Convert to decimals:  
\[
\frac{2}{25} = 2 \div 25
\]
Since \(25 \cdot 0.08 = 2\):  
\[
2 \div 25 = 0.08
\]
So the correct choice is **B** (.08).

ANSWER 3: B


Problem 4:  
We have 4 people chosen: Lance, Sally, Joy, Fred. We need to choose **3 starters**.

Number of ways:
\[
\binom{4}{3} = 4
\]
List them conceptually: leaving out each one individually gives 4 choices.

Correct choice: **B**.

ANSWER 4: B


Problem 5:  
“If 20% of a number is 12, what is 30% of the same number?”

Let the number be \(N\).  
\[
0.20N = 12
\]
\[
N = \frac{12}{0.20} = 60
\]
Now compute \(30\%\) of 60:
\[
0.30 \cdot 60 = 18
\]
Correct choice: **B**.

ANSWER 5: B


Problem 6:  
Spinner starts pointing **west**.

Chenille moves it:
- **Clockwise** \(2\frac{1}{4}\) revolutions  
- then **Counterclockwise** \(3\frac{3}{4}\) revolutions

Net rotation (clockwise minus counterclockwise):
\[
2\frac{1}{4} - 3\frac{3}{4}
\]
Convert to improper fractions:
\[
2\frac{1}{4} = \frac{9}{4}, \quad 3\frac{3}{4} = \frac{15}{4}
\]
\[
\frac{9}{4} - \frac{15}{4} = -\frac{6}{4} = -\frac{3}{2}
\]
The negative means **net counterclockwise** by \(\frac{3}{2}\) revolutions.

\(\frac{3}{2} = 1 + \frac{1}{2}\), so this is:
- 1 full turn (brings it back where it started)
- plus \(\frac{1}{2}\) turn counterclockwise

Half a revolution from **west** points **east** (opposite direction).

Correct answer: **B (east)**.

ANSWER 6: B


Problem 7:  
Original mixture: 30 L with  
- 25% red (doesn’t matter directly)
- 30% yellow  
- 45% water

Yellow amount initially:
\[
0.30 \cdot 30 = 9 \text{ liters}
\]
Add 5 liters of yellow tint, so new yellow amount:
\[
9 + 5 = 14
\]
New total volume:
\[
30 + 5 = 35
\]
Percent yellow in new mixture:
\[
\frac{14}{35} = 0.4 = 40\%
\]
Correct choice: **C**.

ANSWER 7: C


Problem 8:  
Order fractions least to greatest:
\[
\frac{15}{11},\ \frac{19}{15},\ \frac{17}{13}
\]
A good way: compare values by cross-multiplication or decimal approximations.

Compute approximate decimals:
\[
\frac{15}{11} \approx 1.3636
\]
\[
\frac{19}{15} \approx 1.2667
\]
\[
\frac{17}{13} \approx 1.3077
\]
Least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]
Check choice: that matches **E**:
- \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11}\)

ANSWER 8: E


Problem 9:  
Numbers: \(3,4,5,6,6,7,x\) (7 positive integers).  
Mean = median = unique mode.

**Step 1: Mode**  
The only number that repeats is **6**, occurring twice. So the **unique mode = 6**.

Thus median must be 6 as well.

**Step 2: Median = 6**  
For 7 numbers sorted, the median is the **4th** number.

Sorted list must have the 4th term equal to 6. We already have two 6’s and surrounding numbers 3,4,5,7.

Currently without \(x\), the ordered portion is:
\[
3,4,5,6,6,7,\ ?
\]
The 4th number here is 6 already. To keep the 4th number as 6, \(x\) must not be less than 6 (otherwise it would shift positions so 4th might change).

If \(x = 6\), then 6 occurs 3 times, still mode 6 (unique mode). Then list:
\[
3,4,5,6,6,6,7
\]
Median (4th) is 6—works.

If \(x=7\), list:
\[
3,4,5,6,6,7,7
\]
Mode would be 6? Actually 7 repeats too, so mode not unique.

If \(x>7\), list:
\[
3,4,5,6,6,7,x
\]
Median still 6, but mode would still be 6 (unique), since only 6 repeats. That would also work—wait, but then 6 is still unique mode, yes. However we also need **mean = 6**.

So use mean condition.

Sum of known numbers:
\[
3+4+5+6+6+7 = 31
\]
Total sum including \(x\): \(31 + x\)

Mean:
\[
\frac{31+x}{7} = 6
\]
\[
31 + x = 42
\]
\[
x = 11
\]

Check uniqueness of mode: 6 occurs twice, no other repeats (since x=11 is different). Unique mode is 6—consistent.

ANSWER 9: D


Problem 10:  
Joe walks half the distance, then runs the rest.  
He runs **3 times as fast** as he walks.

Given: time to walk half way = **6 minutes**.  
Let walking speed be \(v\). Then walking time for half distance \(=\) 6 minutes.

Let half the distance be \(d\). Then:
\[
d = v \cdot 6
\]
Running speed is \(3v\).  
Remaining distance is also \(d\), so running time:
\[
\text{time} = \frac{d}{3v} = \frac{v\cdot 6}{3v} = 2
\]
Total time:
\[
6 + 2 = 8
\]
Correct choice: **D**.

ANSWER 10: D


Problem 11:  
Compute:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]

Group each set:

First set:
\[
1+11+21+31+41
\]
These are all of the form \(10k+1\):  
\(1, 11, 21, 31, 41\) correspond to \(k=0,1,2,3,4\).

Sum:
\[
(1+11+21+31+41) = (1+11+21+31+41)
\]
Compute directly:
\[
(1+11)=12,\ (21+31)=52,\ 12+52=64,\ 64+41=105
\]

Second set:
\[
9+19+29+39+49
\]
Direct sum:
\[
(9+19)=28,\ (29+39)=68,\ 28+68=96,\ 96+49=145
\]

Total:
\[
105 + 145 = 250
\]

Correct choice: **E**.

ANSWER 11: E


Problem 12:  
Which number is largest?

Options:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare decimals:  
- .99 is the largest among all because it is greater than .9 (and greater than any number starting 0.90... but with fewer 9s afterward).

So .99 > .9099 > .909 > .9009 > .9.

Correct choice: **A**.

ANSWER 12: A

Problem 1:
In a single-elimination tournament, every game results in exactly one team being eliminated. To determine a single winner from 16 teams, 15 teams must be eliminated. Since each game eliminates one team, it takes exactly 15 games.
ANSWER 1: D

Problem 2:
- Jose: (10 - 1) = 9; 9 * 2 = 18; 18 + 2 = 20.
- Thuy: (10 * 2) = 20; 20 - 1 = 19; 19 + 2 = 21.
- Kareem: (10 - 1) = 9; 9 + 2 = 11; 11 * 2 = 22.
Kareem has the largest answer (22).
ANSWER 2: C

Problem 3:
To convert 2/25 to a decimal, multiply the numerator and denominator by 4 to get a denominator of 100: (2 * 4) / (25 * 4) = 8 / 100 = 0.08.
ANSWER 3: B

Problem 4:
We need to choose 3 starters from a group of 4 (Lance, Sally, Joy, Fred). This is a combination problem denoted as 4C3. Choosing 3 people to be on the team is the same as choosing 1 person to be left off the team. There are 4 ways to choose who is left off.
ANSWER 4: B

Problem 5:
Let $x$ be the number. $0.20x = 12$. Dividing by 0.20, $x = 12 / 0.2 = 60$. We want 30% of $x$: $0.30 * 60 = 18$.
ANSWER 5: B

Problem 6:
Start at 0 revolutions. Clockwise is positive, counterclockwise is negative.
$2 \frac{1}{4} - 3 \frac{3}{4} = 2.25 - 3.75 = -1.5$ revolutions.
-1.5 revolutions is the same as -0.5 revolutions (since -1 is a full circle). -0.5 revolutions (counterclockwise) from West points to East.
ANSWER 6: B

Problem 7:
Original yellow tint: 30% of 30 liters = 9 liters.
New mixture: 30 + 5 = 35 liters.
New yellow tint: 9 + 5 = 14 liters.
Percent: (14 / 35) * 100 = (2 / 5) * 100 = 40%.
ANSWER 7: C

Problem 8:
Convert to decimals or compare differences from 1:
15/11 = 1 + 4/11 ≈ 1.3636
19/15 = 1 + 4/15 ≈ 1.2666
17/13 = 1 + 4/13 ≈ 1.3076
Comparing the fractions 4/11, 4/13, and 4/15: the larger the denominator, the smaller the fraction. Thus, 4/15 < 4/13 < 4/11.
Order: 19/15 < 17/13 < 15/11.
ANSWER 8: E

Problem 9:
The set is {3, 4, 5, 6, 6, 7, x}. The mode is 6 (it appears twice). For the mean and median to be 6:
Median: With 7 numbers, the 4th number must be 6. Since the sorted list is 3, 4, 5, 6, 6, 7, x, if $x > 6$, the 4th number is 6.
Mean: (3+4+5+6+6+7+x) / 7 = 6 => 31 + x = 42 => x = 11.
Checking: {3, 4, 5, 6, 6, 7, 11}. Median is 6. Mean is 42/7 = 6. Mode is 6.
ANSWER 9: D

Problem 10:
Walking time = 6 minutes. Running speed = 3 * walking speed.
Time taken = Distance / Speed. Since speed is 3 times faster, time is 1/3 of the walking time.
Running time = 6 / 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
ANSWER 10: D

Problem 11:
Sum 1: (1+11+21+31+41) = 105.
Sum 2: (9+19+29+39+49) = 145.
105 + 145 = 250.
ANSWER 11: E

Problem 12:
Compare digit by digit:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
0.99 is the largest.
ANSWER 12: A

Problem 1:
This is a single-elimination tournament with 16 teams. Exactly one team is eliminated per game, and 15 teams must be eliminated to leave a single winner. Thus exactly 15 games are required.

ANSWER 1: D

Problem 2:
Jose evaluates \((10-1)\times2+2=20\).
Thuy evaluates \((10\times2)-1+2=21\).
Kareem evaluates \((10-1+2)\times2=22\).
Kareem obtains the strictly largest value.

ANSWER 2: C

Problem 3:
Division yields the decimal expansion \(2\div25=0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters out of 4 distinct players is the binomial coefficient \(\binom{4}{3}=4\).

ANSWER 4: B

Problem 5:
If 20% of a number equals 12, the number itself is \(12/0.2=60\). Then 30% of 60 is \(0.3\times60=18\).

ANSWER 5: B

Problem 6:
A net rotation of \(+2.25-3.75=-1.5\) revolutions is equivalent to 1.5 revolutions counterclockwise. From an initial westward orientation, 1.5 revolutions counterclockwise lands exactly on east.

ANSWER 6: B

Problem 7:
The original mixture contains \(0.30\times30=9\) liters of yellow tint. Adding 5 liters raises the yellow volume to 14 liters and the total volume to 35 liters, so the new percentage is \(14/35=40\%\).

ANSWER 7: C

Problem 8:
Decimal approximations are \(15/11\approx1.3636\), \(19/15\approx1.2667\), and \(17/13\approx1.3077\). Ordering these values produces \(19/15<17/13<15/11\).

ANSWER 8: E

Problem 9:
The unique mode must be 6, forcing the mean also to equal 6. The sum of the seven numbers is therefore 42. The six given numbers sum to 31, so \(x=11\). With this value the ordered list is 3,4,5,6,6,7,11 whose median is likewise 6 and whose unique mode remains 6.

ANSWER 9: D

Problem 10:
Let the total distance be \(2d\). Walking time for distance \(d\) is 6 min, so running time for the remaining distance \(d\) at triple speed is \(6/3=2\) min. The total time is therefore 8 min.

ANSWER 10: D

Problem 11:
Each pair sums to a multiple of 10: \(1+9=10\), \(11+19=30\), \(21+29=50\), \(31+39=70\), \(41+49=90\). Their total is 250. Equivalently each five-term arithmetic series sums to 105 and 145 respectively, again yielding 250.

ANSWER 11: E

Problem 12:
Direct comparison of the five decimals shows that 0.99 is strictly larger than each of the others.

ANSWER 12: A

Problem 1:
The problem asks us to determine the number of games that will be played to determine the winner of a basketball tournament with 16 teams, where the losing team of each game is eliminated.

## Step 1
To determine the number of games needed to find the winner, we first note that in a single-elimination tournament, each game eliminates one team.

## Step 2
Since there can only be one winner, all other teams must be eliminated. Therefore, the number of games required is equal to the number of teams minus one.

## Step 3
We have 16 teams competing in the tournament. So, the number of games needed to determine the winner is 16 - 1 = 15.

## Step 4
Let's examine the given choices to rule out the incorrect options. A. 4 is too low because it doesn't account for all the teams. B. 7 is also too low for the same reason. C. 8 is still too low. D. 15 matches our calculation. E. 16 is one more than our calculation.

The final answer is supported by our step-by-step reasoning.
ANSWER 1: D

Problem 2:
The problem asks us to determine who among Jose, Thuy, and Kareem gets the largest final answer after performing a series of operations starting with the number 10.

## Step 1
Let's calculate Jose's final answer: Jose subtracts 1 from 10 to get 9, doubles it to get 18, and then adds 2 to get 20.

## Step 2
Now, let's calculate Thuy's final answer: Thuy doubles 10 to get 20, subtracts 1 to get 19, and then adds 2 to get 21.

## Step 3
Next, let's calculate Kareem's final answer: Kareem subtracts 1 from 10 to get 9, adds 2 to get 11, and then doubles it to get 22.

## Step 4
Comparing the final answers: Jose got 20, Thuy got 21, and Kareem got 22. The largest final answer is 22, which belongs to Kareem.

## Step 5
Let's examine the given choices. A. Jose is incorrect because his answer is 20. B. Thuy is incorrect because her answer is 21. C. Kareem is correct because his answer is 22. D. Jose and Thuy is incorrect because Kareem's answer is larger. E. Thuy and Kareem is incorrect because Jose's answer is not the largest.

The final answer is supported by our step-by-step reasoning.
ANSWER 2: C

Problem 3:
The problem asks us to convert the fraction 2/25 into a decimal.

## Step 1
To convert 2/25 into a decimal, we divide the numerator by the denominator: 2 ÷ 25 = 0.08.

## Step 2
Let's examine the given choices to identify the correct decimal representation. A. .008 is too small. B. .08 matches our calculation. C. .8 is too large. D. 1.25 is much too large. E. 12.5 is also much too large.

The final answer is supported by our step-by-step reasoning.
ANSWER 3: B

Problem 4:
The problem asks us to determine the number of ways to choose 3 starters from a team of 4 members: Lance, Sally, Joy, and Fred.

## Step 1
This is a combination problem, because the order in which the starters are chosen does not matter.

## Step 2
The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial, the product of all positive integers up to that number.

## Step 3
Here, n = 4 (total team members) and k = 3 (number of starters to be chosen). So, C(4, 3) = 4! / [3!(4 - 3)!] = 4! / (3! * 1!) = (4 * 3 * 2 * 1) / ((3 * 2 * 1) * 1) = 4.

## Step 4
Let's examine the given choices. A. 2 is too low. B. 4 matches our calculation. C. 6 is too high. D. 8 is also too high. E. 10 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 4: B

Problem 5:
The problem asks us to find 30% of a number, given that 20% of the same number is 12.

## Step 1
First, we need to find the number. We know that 20% of the number is 12. So, 0.20 * number = 12.

## Step 2
To find the number, we divide both sides of the equation by 0.20: number = 12 / 0.20 = 60.

## Step 3
Now, we need to calculate 30% of 60: 0.30 * 60 = 18.

## Step 4
Let's examine the given choices. A. 15 is too low. B. 18 matches our calculation. C. 20 is too high. D. 24 is also too high. E. 30 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 5: B

Problem 6:
The problem asks us to determine the direction the spinner points after Chenille moves it clockwise 2 1/4 revolutions and then counterclockwise 3 3/4 revolutions, starting from pointing west.

## Step 1
First, let's understand the effect of the movements. A full revolution is 360 degrees or 4 quarters. So, 2 1/4 revolutions clockwise is equivalent to 9 quarters clockwise.

## Step 2
9 quarters clockwise is the same as 1 quarter clockwise (since 8 quarters make 2 full revolutions, which brings the spinner back to the original direction). So, after the first move, the spinner points north.

## Step 3
Then, Chenille moves it counterclockwise 3 3/4 revolutions, which is equivalent to 15 quarters counterclockwise.

## Step 4
15 quarters counterclockwise is the same as 3 quarters counterclockwise (since 12 quarters make 3 full revolutions). So, moving 3 quarters counterclockwise from north results in pointing west.

## Step 5
Let's examine the given choices. A. north is incorrect because after the second move, it's not north. B. east is incorrect. C. south is incorrect. D. west matches our calculation. E. northwest is incorrect.

The final answer is supported by our step-by-step reasoning.
ANSWER 6: D

Problem 7:
The problem asks us to find the percentage of yellow tint in a new mixture after adding 5 liters of yellow tint to an original 30-liter mixture that is 25% red tint, 30% yellow tint, and 45% water.

## Step 1
First, we calculate the amount of yellow tint in the original mixture: 30% of 30 liters = 0.30 * 30 = 9 liters.

## Step 2
After adding 5 liters of yellow tint, the total amount of yellow tint becomes 9 + 5 = 14 liters.

## Step 3
The total volume of the new mixture is 30 + 5 = 35 liters.

## Step 4
To find the percentage of yellow tint in the new mixture, we divide the amount of yellow tint by the total volume and multiply by 100: (14 / 35) * 100 = 40%

Problem 1:
In a single-elimination tournament, each game eliminates exactly one team. To eliminate all but the champion, the number of games needed is one fewer than the number of teams. With 16 teams, the number of games is \(16 - 1 = 15\).

ANSWER 1: D

Problem 2:
Jose: \(10 - 1 = 9\), \(9 \times 2 = 18\), \(18 + 2 = 20\).
Thuy: \(10 \times 2 = 20\), \(20 - 1 = 19\), \(19 + 2 = 21\).
Kareem: \(10 - 1 = 9\), \(9 + 2 = 11\), \(11 \times 2 = 22\).
Kareem’s final answer of 22 is the largest.

ANSWER 2: C

Problem 3:
Divide 2 by 25: \(2 \div 25 = 0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters from 4 people is the combination \(\binom{4}{3} = 4\).

ANSWER 4: B

Problem 5:
Let the number be \(n\). \(20\%\) of \(n\) is \(0.2n = 12\), so \(n = 60\). Then \(30\%\) of \(60 = 0.3 \times 60 = 18\).

ANSWER 5: B

Problem 6:
Initial direction: west.
Clockwise \(2\frac{1}{4}\) revolutions: \(2\frac{1}{4} = 2.25\) revolutions clockwise.
Counterclockwise \(3\frac{3}{4}\) revolutions: \(3\frac{3}{4} = 3.75\) revolutions counterclockwise.
Net revolution = \(2.25 - 3.75 = -1.5\) revolutions, i.e., \(1.5\) revolutions counterclockwise.
A full revolution (1.0) returns to west. The remaining 0.5 revolution counterclockwise from west is a half-turn, which points to east.

ANSWER 6: B

Problem 7:
Original mixture: 30 liters, yellow tint = \(30\% \times 30 = 9\) liters.
Add 5 liters of yellow tint: new yellow = \(9 + 5 = 14\) liters; new total volume = \(30 + 5 = 35\) liters.
Percent yellow = \(\frac{14}{35} \times 100\% = 40\%\).

ANSWER 7: C

Problem 8:
Convert each fraction to a decimal for comparison:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
Order from least to greatest: \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

ANSWER 8: E

Problem 9:
The data set: 3, 4, 5, 6, 6, 7, \(x\). The mode is the most frequent number. With two 6’s, 6 is the mode unless \(x\) equals another number that appears twice, which would create a second mode, violating the “unique mode” condition. Therefore, \(x\) must not be 3, 4, 5, or 7. It could be 6, but if \(x=6\), the median is 6, while the mean is \((3+4+5+6+6+6+7)/7 = 37/7 \approx 5.29\), so mean ≠ median.  
For the mean to equal the median (which is 6, the middle of seven sorted numbers if \(x > 6\) or \(x=6\) only if placed correctly), we set the mean to 6. Sum without \(x\) = \(3+4+5+6+6+7 = 31\). Total needed for mean 6 = \(7 \times 6 = 42\). Thus \(x = 42 - 31 = 11\).  
Sorted: 3, 4, 5, 6, 6, 7, 11. Median = 6, mode = 6, mean = 6. Unique mode holds.

ANSWER 9: D

Problem 10:
Walking half the distance took 6 minutes, so walking speed = \(\frac{\text{half distance}}{6}\).
Running speed is 3 times walking speed. Time to run the second half = \(\frac{\text{half distance}}{3 \times (\text{half distance}/6)} = \frac{6}{3} = 2\) minutes.
Total time = \(6 + 2 = 8\) minutes.

ANSWER 10: D

Problem 11:
Pair the terms: \(1+49 = 50\), \(11+39 = 50\), \(21+29 = 50\), \(31+19 = 50\), \(41+9 = 50\). Five pairs of 50 give \(5 \times 50 = 250\).

ANSWER 11: E

Problem 12:
Compare by extending decimals to the same number of places:
\(0.99   = 0.9900\)
\(0.9099 = 0.9099\)
\(0.9    = 0.9000\)
\(0.909  = 0.9090\)
\(0.9009 = 0.9009\)
Clearly \(0.99 > 0.9099 > 0.909 > 0.9009 > 0.9\). The largest is \(0.99\).

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key Insight:** In a single-elimination tournament, every game eliminates exactly one team. To crown a single winner from 16 teams, we must eliminate 16 − 1 = 15 teams. Therefore, exactly 15 games are needed.

**Verification by rounds:**
- Round 1: 16 teams → 8 games → 8 teams left
- Round 2: 8 teams → 4 games → 4 teams left
- Round 3: 4 teams → 2 games → 2 teams left
- Round 4 (Final): 2 teams → 1 game → 1 winner

Total = 8 + 4 + 2 + 1 = 15 games.

Choices A (4), B (7), and C (8) are too small. Choice E (16) would mean every team is eliminated, leaving no winner.

ANSWER 1: D

---

Problem 2:
We track each person's calculation step by step, all starting from 10.

**Jose:** subtract 1 → 9; double → 18; add 2 → **20**
**Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
**Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

Comparing: 20 < 21 < 22. Kareem gets the largest final answer.

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

Choice A (.008) would be 8/1000. Choice C (.8) would be 8/10. Choices D and E are greater than 1, but 2/25 is clearly less than 1.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination problem — the order of selection doesn't matter:
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4$$

Equivalently, choosing 3 people to start is the same as choosing 1 person to sit out. There are 4 people, so 4 ways to pick who sits out.

The 4 possible starting lineups are: {L,S,J}, {L,S,F}, {L,J,F}, {S,J,F}.

ANSWER 4: B

---

Problem 5:
We're told 20% of a number is 12. Let the number be $x$.

$$0.20x = 12 \implies x = \frac{12}{0.20} = 60$$

Now find 30% of 60:
$$0.30 \times 60 = 18$$

**Shortcut:** 30% is $\frac{30}{20} = 1.5$ times 20%, so the answer is $1.5 \times 12 = 18$.

Choice A (15) would be 25%. Choice D (24) would be 40%. Choice E (30) would be 50%.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing **West**.

**Move 1:** Clockwise $2\frac{1}{4}$ revolutions.
- The 2 full revolutions bring it back to West.
- The extra $\frac{1}{4}$ revolution clockwise from West: West → **North**.

**Move 2:** Counterclockwise $3\frac{3}{4}$ revolutions.
- The 3 full revolutions bring it back to North.
- The extra $\frac{3}{4}$ revolution counterclockwise from North:
  - $\frac{1}{4}$ CCW from North → West
  - $\frac{2}{4}$ CCW from North → South
  - $\frac{3}{4}$ CCW from North → **East**

The spinner points East.

ANSWER 6: B

---

Problem 7:
Original mixture: 30 liters total.
- Red tint: 25% of 30 = 7.5 liters
- Yellow tint: 30% of 30 = 9 liters
- Water: 45% of 30 = 13.5 liters

Add 5 liters of yellow tint:
- New yellow tint = 9 + 5 = 14 liters
- New total mixture = 30 + 5 = 35 liters

Percent yellow tint in new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

Choice B (35%) is a trap for those who just add 5% to 30%. Choice D (45%) is the original water percentage.

ANSWER 7: C

---

Problem 8:
We compare $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$.

Rewrite each as a mixed number:
$$\frac{15}{11} = 1 + \frac{4}{11}, \quad \frac{19}{15} = 1 + \frac{4}{15}, \quad \frac{17}{13} = 1 + \frac{4}{13}$$

Since all three have the same numerator (4) in their fractional parts, we compare denominators. For fractions with equal numerators, a **larger denominator** means a **smaller fraction**:
$$15 > 13 > 11 \implies \frac{4}{15} < \frac{4}{13} < \frac{4}{11}$$

Therefore:
$$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$$

ANSWER 8: E

---

Problem 9:
The data set is: 3, 4, 5, 6, 6, 7, and $x$. The mean, median, and unique mode are all equal.

**Mode:** Since 6 already appears twice and all others appear once, the unique mode must be 6 (if $x$ were 3, 4, 5, or 7, we'd have two modes, violating "unique mode"). So mode = 6.

**Mean = 6:**
$$\frac{3 + 4 + 5 + 6 + 6 + 7 + x}{7} = 6$$
$$\frac{31 + x}{7} = 6 \implies 31 + x = 42 \implies x = 11$$

**Check Median:** With $x = 11$, the sorted list is: 3, 4, 5, **6**, 6, 7, 11. The median (4th value) is 6. ✓

All three measures equal 6. Choice B (6) would make 6 appear three times, but the mean would be $(31+6)/7 = 37/7 \neq 6$.

ANSWER 9: D

---

Problem 10:
Joe walked half the distance in 6 minutes. He then ran the remaining half at 3 times his walking speed.

Let walking speed = $v$. The half-distance = $6v$.

Running speed = $3v$. Time to run the second half:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{6v}{3v} = 2 \text{ minutes}$$

Total time = 6 + 2 = **8 minutes**.

Choice A (7) would result from incorrectly subtracting. Choices B, C, E involve unnecessary decimal complications.

ANSWER 10: D

---

Problem 11:
Compute $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$.

Pair the terms from the first group with terms from the second group (first with last, etc.):
- $1 + 49 = 50$
- $11 + 39 = 50$
- $21 + 29 = 50$
- $31 + 19 = 50$
- $41 + 9 = 50$

Total = $50 \times 5 = 250$.

Choice C (200) is a common error from miscounting pairs. Choice D (249) comes from arithmetic mistakes.

ANSWER 11: E

---

Problem 12:
Compare the five decimals by writing them to four decimal places:

- A: 0.9900
- B: 0.9099
- C: 0.9000
- D: 0.9090
- E: 0.9009

Comparing digit by digit from the tenths place: all have 9 in the tenths place. In the hundredths place, A has **9** while all others have **0**. So A is immediately the largest.

$0.99 > 0.9099 > 0.909 > 0.9009 > 0.9$

ANSWER 12: A

 **Problem 1:**
We need to find the number of games played in a single-elimination tournament with 16 teams. In single elimination, every game eliminates exactly one team. To determine a winner, 15 of the 16 teams must be eliminated. Therefore, exactly 15 games are required.  
(Another way: with 16 teams there are 8 games in the first round, 4 in the second, 2 in the third, and 1 final game: \(8+4+2+1=15\).)

ANSWER 1: D

---

**Problem 2:**
We compute each person’s final answer step by step, starting from 10.

- **Jose:**  
  \(10 - 1 = 9\)  
  \(9 \times 2 = 18\)  
  \(18 + 2 = 20\)

- **Thuy:**  
  \(10 \times 2 = 20\)  
  \(20 - 1 = 19\)  
  \(19 + 2 = 21\)

- **Kareem:**  
  \(10 - 1 = 9\)  
  \(9 + 2 = 11\)  
  \(11 \times 2 = 22\)

Comparing the results: \(20 < 21 < 22\). Kareem’s answer is the largest.

ANSWER 2: C

---

**Problem 3:**
We convert the fraction \(\frac{2}{25}\) to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100:
\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

ANSWER 3: B

---

**Problem 4:**
We must choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways to choose 3 from 4 is the combination \(\binom{4}{3}\):
\[
\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4
\]
Equivalently, choosing 3 starters is the same as choosing 1 person to sit out, and there are 4 choices for who sits out.

ANSWER 4: B

---

**Problem 5:**
Let the number be \(N\). We are given \(0.20N = 12\). Solving for \(N\):
\[
N = \frac{12}{0.20} = 60
\]
Now find 30% of 60:
\[
0.30 \times 60 = 18
\]
(Shortcut: 30% is \(1.5\) times 20%, so \(1.5 \times 12 = 18\).)

ANSWER 5: B

---

**Problem 6:**
The spinner starts pointing west.  
- Clockwise \(2\frac14\) revolutions: 2 full revolutions bring it back to west, then \(\frac14\) turn clockwise from west points it **north**.  
- Counterclockwise \(3\frac34\) revolutions: 3 full revolutions bring it back to north, then \(\frac34\) turn counterclockwise from north goes through west, south, and ends at **east**.

Net movement: \(3\frac34 - 2\frac14 = 1\frac12\) revolutions counterclockwise. One full revolution returns to west; an additional \(\frac12\) revolution counterclockwise from west points to **east**.

ANSWER 6: B

---

**Problem 7:**
Original mixture: 30 liters.  
- Yellow tint originally: \(30\% \times 30 = 9\) liters.  
After adding 5 liters of yellow tint:  
- New yellow tint: \(9 + 5 = 14\) liters.  
- New total mixture: \(30 + 5 = 35\) liters.  

Percent of yellow tint in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%
\]

ANSWER 7: C

---

**Problem 8:**
Compare \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\). Each is greater than 1. Write each as \(1\) plus a proper fraction:
\[
\frac{15}{11} = 1 + \frac{4}{11}, \qquad
\frac{19}{15} = 1 + \frac{4}{15}, \qquad
\frac{17}{13} = 1 + \frac{4}{13}
\]
Since the numerators are the same (4), the fraction with the largest denominator is smallest:
\[
\frac{4}{15} < \frac{4}{13} < \frac{4}{11}
\]
Adding 1 preserves the order:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

ANSWER 8: E

---

**Problem 9:**
The data set is \(3, 4, 5, 6, 6, 7, x\).  
- The current mode is 6 (appears twice). For 6 to remain the **unique** mode, \(x\) must not equal 3, 4, 5, or 7 (which would create a second mode).  
- We are told mean = median = mode = 6.  

Set the mean equal to 6:
\[
\frac{3+4+5+6+6+7+x}{7} = 6 \implies \frac{31+x}{7} = 6 \implies 31+x = 42 \implies x = 11
\]

Check: With \(x=11\), the sorted list is \(3, 4, 5, 6, 6, 7, 11\).  
- Median (4th term) = 6.  
- Mode = 6 (unique).  
- Mean = 6.  

All conditions are satisfied.

ANSWER 9: D

---

**Problem 10:**
Joe walks half the distance in 6 minutes. He runs the second half at 3 times his walking speed. Since distance is the same and speed is tripled, the time required is divided by 3:
\[
\text{Running time} = \frac{6}{3} = 2 \text{ minutes}
\]
Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

---

**Problem 11:**
Add the two groups:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]
Pair the terms to make convenient sums:
\[
(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 5 \times 50 = 250
\]

ANSWER 11: E

---

**Problem 12:**
Compare the decimals by writing them with the same number of places:
- A. \(0.9900\)
- B. \(0.9099\)
- C. \(0.9000\)
- D. \(0.9090\)
- E. \(0.9009\)

The tenths digit is 9 for all. Looking at the hundredths digit: A has 9, while B, D have 0 and C, E have 0. Thus \(0.9900\) is clearly the largest.

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key insight:** Each game eliminates exactly one team. To determine a winner from 16 teams, we must eliminate 15 teams (everyone except the champion).

Therefore, exactly 15 games must be played.

*Verification by rounds:* Round 1: 8 games (16→8 teams), Round 2: 4 games (8→4), Round 3: 2 games (4→2), Round 4: 1 game (2→1). Total = 8+4+2+1 = 15.

The tempting wrong answer E (16) counts teams instead of eliminations.

ANSWER 1: D

---

Problem 2:
We need to compute each person's final answer starting from 10.

**Jose:** Start with 10 → subtract 1: 10−1 = 9 → double: 9×2 = 18 → add 2: 18+2 = **20**

**Thuy:** Start with 10 → double: 10×2 = 20 → subtract 1: 20−1 = 19 → add 2: 19+2 = **21**

**Kareem:** Start with 10 → subtract 1: 10−1 = 9 → add 2: 9+2 = 11 → double: 11×2 = **22**

Kareem gets the largest final answer (22).

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

The tempting wrong answer A (0.008) would be 8/1000, and C (0.8) would be 8/10.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination: C(4,3) = 4!/(3! × 1!) = 4

Listing them out: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — that's 4 ways.

The tempting wrong answer C (6) would be C(4,2), choosing who's *not* a starter, but that also equals 4... Actually C(4,2)=6 is wrong here; we're choosing 3 from 4.

ANSWER 4: B

---

Problem 5:
We know 20% of a number is 12. We need to find 30% of the same number.

**Step 1:** Find the number.
$$0.20 \times n = 12 \implies n = \frac{12}{0.20} = 60$$

**Step 2:** Find 30% of 60.
$$0.30 \times 60 = 18$$

*Shortcut:* Since 30% is 1.5 times 20%, the answer is 1.5 × 12 = 18.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing west. We need to find the net rotation.

**Clockwise:** 2¼ revolutions
**Counterclockwise:** 3¾ revolutions

**Net movement:** 2¼ − 3¾ = −1½ revolutions (negative means counterclockwise)

So the spinner moves 1½ revolutions counterclockwise from west.

- 1 full revolution counterclockwise returns to **west**
- ½ revolution counterclockwise from west goes to **east**

(Counterclockwise from west: ¼ turn → south, ½ turn → east)

ANSWER 6: B

---

Problem 7:
We need to find the percent of yellow tint after adding 5 liters of yellow tint.

**Original mixture (30 liters):**
- Yellow tint: 30% of 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percent yellow tint:** (14/35) × 100 = 40%

The tempting wrong answer B (35) might come from simply adding 5% to 30%.

ANSWER 7: C

---

Problem 8:
We need to order 15/11, 19/15, and 17/13 from least to greatest.

Computing each as a decimal:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3077...

From least to greatest: **19/15 < 17/13 < 15/11**

*Alternative method:* Note each is 1 + (numerator − denominator)/denominator:
- 15/11 = 1 + 4/11
- 19/15 = 1 + 4/15
- 17/13 = 1 + 4/13

Since 4/15 < 4/13 < 4/11, the order is confirmed.

ANSWER 8: E

---

Problem 9:
The positive integers 3, 4, 5, 6, 6, 7, and x have equal mean, median, and unique mode.

**Mode:** Since 6 already appears twice and no other number does, the unique mode must be 6. So x ≠ 3, 4, 5, or 7 (which would create a second mode).

**Median:** With 7 numbers, the median is the 4th value when sorted. For the median to be 6, we need x ≥ 6 (so that 6 is the 4th value).

- If x = 6: sorted = 3, 4, 5, 6, 6, 6, 7 → median = 6 ✓
- If x > 7: sorted = 3, 4, 5, 6, 6, 7, x → median = 6 ✓

**Mean:** (3 + 4 + 5 + 6 + 6 + 7 + x)/7 = (31 + x)/7

Setting mean = 6: (31 + x)/7 = 6 → 31 + x = 42 → **x = 11**

**Verification:** Numbers: 3, 4, 5, 6, 6, 7, 11. Mean = 42/7 = 6, Median = 6, Mode = 6 ✓

ANSWER 9: D

---

Problem 10:
Joe walks halfway, then runs the rest. We need total time.

**Walking:** Half the distance takes 6 minutes.

**Running:** He runs 3 times as fast as he walks, so he covers the same distance in 1/3 the time.
- Time to run the second half = 6 ÷ 3 = 2 minutes

**Total time:** 6 + 2 = **8 minutes**

The tempting wrong answers come from setting up the problem incorrectly or mixing up distance and time relationships.

ANSWER 10: D

---

Problem 11:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

**First group:** 1 + 11 + 21 + 31 + 41 = 105

**Second group:** 9 + 19 + 29 + 39 + 49 = 145

**Total:** 105 + 145 = **250**

*Elegant method:* Pair terms: (1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250

ANSWER 11: E

---

Problem 12:
We need to find the largest number among the options. Writing each with the same number of decimal places:

- A. 0.99 = **0.9900**
- B. 0.9099 = 0.9099
- C. 0.9 = 0.9000
- D. 0.909 = 0.9090
- E. 0.9009 = 0.9009

Comparing digit by digit after the decimal point:
- First digit: all are 9
- Second digit: A has **9**, all others have 0

Since A has a 9 in the second decimal place while all others have 0, **0.99 is the largest**.

ANSWER 12: A

**Problem 1:**  
The tournament is single‑elimination. Starting with 16 teams, every game eliminates exactly one team. To reduce from 16 teams to a single champion we must eliminate 15 teams, and each elimination corresponds to one game.  

Number of games = 16 − 1 = **15**.  

ANSWER 1: D  

---

**Problem 2:**  
Start with 10 for each person.

- **Jose:**  
  Subtract 1 → 9; double → 18; add 2 → 20.  

- **Thuy:**  
  Double → 20; subtract 1 → 19; add 2 → 21.  

- **Kareem:**  
  Subtract 1 → 9; add 2 → 11; double → 22.  

The largest final value is 22, obtained by Kareem.  

ANSWER 2: C  

---

**Problem 3:**  

\[
\frac{2}{25}=0.08
\]

(2 ÷ 25 = 0.08).  

ANSWER 3: B  

---

**Problem 4:**  
We need to choose 3 starters from the 4 available players:  

\[
\binom{4}{3}=4
\]

ways.  

ANSWER 4: B  

---

**Problem 5:**  
Let the number be \(N\).  
\(0.20N = 12 \Rightarrow N = \frac{12}{0.20}=60\).  
\(30\%\) of \(N\) is \(0.30 \times 60 = 18\).  

ANSWER 5: B  

---

**Problem 6:**  
Interpret the fractions as mixed numbers:  

- Clockwise \(2\frac14 = 2.25\) revolutions.  
- Counter‑clockwise \(3\frac34 = 3.75\) revolutions.  

Net rotation (counter‑clockwise positive) = \(3.75 - 2.25 = 1.5\) revolutions.  

A full revolution returns to the same direction, so we only care about the extra \(0.5\) revolution.  
\(0.5\) revolution = 180° counter‑clockwise.  

Starting from west, turning 180° counter‑clockwise points to **east**.  

ANSWER 6: B  

---

**Problem 7:**  
Original yellow = \(30\%\) of 30 L = \(0.30 \times 30 = 9\) L.  
Add 5 L yellow → new yellow = \(9+5 = 14\) L.  
New total volume = \(30+5 = 35\) L.  

Percent yellow = \(\frac{14}{35}\times100 = 40\%\).  

ANSWER 7: C  

---

**Problem 8:**  
Compute the three fractions:

\[
\frac{15}{11}=1.3636\ldots,\qquad
\frac{19}{15}=1.2666\ldots,\qquad
\frac{17}{13}=1.3077\ldots
\]

Ordered from least to greatest:  

\(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

This matches option **E**.  

ANSWER 8: E  

---

**Problem 9:**  
Current list: \(3,4,5,6,6,7\). The unique mode is 6 (appears twice).  
For the mode to remain unique and equal to the mean and median, \(x\) must not create a tie.  
Let the mean = median = mode = \(m\).  

Sum of numbers = \(31 + x\).  
Mean = \(\frac{31 + x}{7}=m\).  
Median of the seven numbers is the 4th after sorting; with \(x>7\) the median stays 6, so \(m=6\).  

Thus \(\frac{31 + x}{7}=6 \Rightarrow 31 + x = 42 \Rightarrow x = 11\).

Check: sorted list becomes \(3,4,5,6,6,7,11\) → median = 6, mode = 6, mean = 42/7 = 6. All equal.  

ANSWER 9: D  

---

**Problem 10:**  
Let the walking speed be \(v\) and the half‑distance be \(d/2\).  

Walking time for half‑distance: \(\frac{d/2}{v}=6\) min ⇒ \(\frac{d}{v}=12\) min.  

Running speed = \(3v\).  
Running time for the second half: \(\frac{d/2}{3v}= \frac{d}{6v}= \frac{1}{6}\times\frac{d}{v}= \frac{12}{6}=2\) min.  

Total time = \(6 + 2 = 8\) minutes.  

ANSWER 10: D  

---

**Problem 11:**  

\[
\begin{aligned}
1+11+21+31+41 &= 105,\\
9+19+29+39+49 &= 145,\\
\text{Total} &= 105+145 = 250.
\end{aligned}
\]

ANSWER 11: E  

---

**Problem 12:**  
Write the numbers with four decimal places:

- .99 = 0.9900  
- .9099 = 0.9099  
- .9 = 0.9000  
- .909 = 0.9090  
- .9009 = 0.9009  

The largest is 0.9900, i.e., **.99**.  

ANSWER 12: A  

Problem 1:  
The tournament starts with 16 teams. Each game eliminates one team, and the losing team is eliminated. To determine the winner, all teams except one must be eliminated. Thus, 15 teams must be eliminated, requiring 15 games.  
- **Why not other choices?** A (4) is too few; B (7), C (8) are also insufficient; E (16) would imply one game per team without elimination, which is incorrect.  
ANSWER 1: D  

Problem 2:  
- **Jose**: Starts with 10 → subtracts 1 (9) → doubles (18) → adds 2 → **20**.  
- **Thuy**: Starts with 10 → doubles (20) → subtracts 1 (19) → adds 2 → **21**.  
- **Kareem**: Starts with 10 → subtracts 1 (9) → adds 2 (11) → doubles → **22**.  
Kareem has the largest final answer.  
- **Why not other choices?** A (Jose) and B (Thuy) are smaller; D and E pair incorrect winners.  
ANSWER 2: C  

Problem 3:  
\( \frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08 \).  
- **Why not other choices?** A (.008) is \( \frac{2}{250} \); C (.8) is \( \frac{4}{5} \); D and E are larger decimals.  
ANSWER 3: B  

Problem 4:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is \( \binom{4}{3} = 4 \).  
- **Why not other choices?** A (2) is too few; C (6) is \( \binom{4}{2} \), which is incorrect; D (8) and E (10) are overcounts.  
ANSWER 4: B  

Problem 5:  
Let the number be \( x \). If 20% of \( x \) is 12, then \( 0.2x = 12 \), so \( x = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).  
- **Why not other choices?** A (15) is 25% of 60; C (20) is \( \frac{1}{3} \) of 60; D (24) is 40%; E (30) is 50%.  
ANSWER 5: B  

Problem 6:  
- A full revolution is 360°. Clockwise \( 2 \frac{1}{4} \) revolutions = \( 2.25 \times 360 = 810° \).  
- Counterclockwise \( 3 \frac{3}{4} \) revolutions = \( -3.75 \times 360 = -1350° \).  
- Net movement: \( 810° - 1350° = -540° \).  
- \( -540° \mod 360° = 180° \) (since \( -540 + 720 = 180° \)), which points west → south? Wait, initial direction is west.  
- West (270°) + 810° = 1080° → 1080° mod 360° = 0° (north). Then -1350° → 0° - 1350° = -1350° mod 360° = 90° (east)? Let's reframe:  
- Start at west (270°). Clockwise 2.25 rev: 270° + 810° = 1080° mod 360° = 0° (north). Then counterclockwise 3.75 rev: 0° - 1350° = -1350° mod 360° = 90° (east).  
- **Why not other choices?** A (north) is initial after clockwise; B (east) is final; C (south) is 180°; D (west) is start; E (northwest) is not a cardinal direction.  
ANSWER 6: B  

Problem 7:  
Original mixture: 30 liters, 30% yellow tint → 9 liters yellow. Add 5 liters yellow → total yellow = 14 liters. New mixture = 35 liters. Percent yellow = \( \frac{14}{35} \times 100 = 40\% \).  
- **Why not other choices?** A (25) is original yellow%; B (35) is before addition; D (45) and E (50) are overestimates.  
ANSWER 7: E (Note: Correction based on computation: 14/35=0.4 → 40%, so answer is E? Wait, 14/35=0.4 → 40%, which is E. But earlier I thought B, but B is 35, which is wrong. Original yellow is 9, add 5 → 14. Total mixture 30+5=35. 14/35=0.4 → 40%. So E. But in choices E is 50? Wait, no: choices are A.25 B.35 C.40 D.45 E.50. So 40% is C. I misread. Correction: 14/35=0.4 → 40%, which is C.  
**Correction**: 14/35=0.4 → 40%, so answer is C.  
ANSWER 7: C  

Problem 8:  
Compare \( \frac{15}{11} \approx 1.3636 \), \( \frac{19}{15} \approx 1.2667 \), \( \frac{17}{13} \approx 1.3077 \). Order: \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11} \).  
- **Why not other choices?** A, B, C, D have incorrect orders.  
ANSWER 8: E  

Problem 9:  
The set is {3, 4, 5, 6, 6, 7, x}. Mean = median = mode = 6.  
- Mode is 6 (unique), so 6 must appear most.  
- Median is 6 (4th value in ordered list: 3,4,5,6,6,7,x → if x≥6, ordered is ...,6,6,x or ...,6,x,6? Must have 4th value 6.  
- If x=6, set is {3,4,5,6,6,6,7}: median=6, mode=6, mean=(3+4+5+6+6+6+7)/7=37/7≈5.28 ≠6.  
- If x=11: set {3,4,5,6,6,7,11}: mean=(3+4+5+6+6+7+11)/7=42/7=6, median=6, mode=6.  
- **Why not other choices?** A (5): mean=36/7≈5.14; B (6): mean≈5.28; C (7): mean=38/7≈5.43; D (11) works.  
ANSWER 9: D  

Problem 10:  
Joe walks half the distance in 6 minutes. Let walking speed be \( v \), so distance = \( 6v \). He runs at \( 3v \), so time to run half distance = \( \frac{6v}{3v} = 2 \) minutes. Total time = 6 + 2 = 8 minutes.  
- **Why not other choices?** A (7), B (7.3), C (7.7), E (8.3) are not exact.  
ANSWER 10: D  

Problem 11:  
First group: 1+11+21+31+41 = 105. Second group: 9+19+29+39+49 = 145. Total = 105 + 145 = 250.  
- **Why not other choices?** A (150), B (199), C (200), D (249) are incorrect sums.  
ANSWER 11: E  

Problem 12:  
Compare decimals: A (.99), B (.9099), C (.9), D (.909), E (.9009). Largest is .99.  
- **Why not other choices?** B, C, D, E are all smaller than .99.  
ANSWER 12: A

### Problem 1:
We need to find the total number of games to determine a winner in a 16-team single-elimination tournament, where every game eliminates exactly one losing team. To get 1 winner, we must eliminate 16 - 1 = 15 teams. Since each game eliminates 1 team, we need 15 total games. Wrong choices: A is the number of bracket rounds, not games; C is half the number of teams; E is the total number of teams, not games.
ANSWER 1: D

---

### Problem 2:
All three start with 10, we calculate each person's final result step-by-step:
- Jose: (10 - 1) * 2 + 2 = 9*2 +2 = 20
- Thuy: (10*2) -1 +2 = 20-1+2=21
- Kareem: (10-1 +2)*2 = 11*2=22
Kareem has the largest final answer. Wrong choices stem from misordering the operations for each person.
ANSWER 2: C

---

### Problem3:
Convert 2/25 to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100: (2*4)/(25*4) = 8/100 = 0.08. Wrong choices: A is 2/250, C is 20/25.
ANSWER 3: B

---

### Problem4:
We choose 3 starters from 4 people, which is a combination problem (order of starters does not matter here). The number of ways to choose 3 from 4 is equivalent to choosing 1 person to leave off the starting lineup: there are 4 people, so 4 unique lineups, each omitting one person. Wrong choice C (6) is the number of permutations, not combinations.
ANSWER 4: B

---

### Problem5:
If 20% of a number is 12, first find the full number: 0.2*N=12 → N=12/0.2=60. 30% of 60 is 0.3*60=18. Wrong choices: A is 25% of 60, D is 40% of 60.
ANSWER 5: B

---

### Problem6:
The spinner starts pointing west. Net movement: 2.25 clockwise revolutions minus 3.75 clockwise (equivalent to 3.75 counterclockwise) = -1.5 total clockwise revolutions, or 1.5 counterclockwise revolutions. Full revolutions cancel out, so we only have a net 0.5 (180 degree) counterclockwise turn from west. A 180-degree turn from west points to east.
ANSWER 6: B

---

### Problem7:
Original 30L mixture has 30% yellow tint: 0.3*30=9L of yellow. Add 5L of yellow, so new yellow is 14L, new total mixture is 35L. The new percentage of yellow is (14/35)*100=40%. Wrong choice: people often forget to add 5L to the total mixture volume, leading to an incorrect higher percentage.
ANSWER 7: C

---

### Problem8:
Rewrite each fraction to compare easily: 15/11 = 1 + 4/11, 19/15=1+4/15, 17/13=1+4/13. For equal numerators, the larger the denominator, the smaller the fraction. So 4/15 < 4/13 < 4/11, so the full order from least to greatest is 19/15 < 17/13 <15/11.
ANSWER 8: E

---

### Problem9:
The dataset is [3,4,5,6,6,7,x]. The unique mode is currently 6 (only value appearing twice, so it must stay the only unique mode). For the mean, median, and mode to all be 6: the sum of the dataset must be 7*6=42. The sum of the existing numbers is 3+4+5+6+6+7=31, so x=42-31=11. Sorted dataset [3,4,5,6,6,7,11] has median 6, unique mode 6, mean 6, which fits. Wrong choice B (x=6) gives a mean of ~5.28, not 6.
ANSWER 9: D

---

### Problem10:
Joe walks half the distance in 6 minutes. His running speed is 3x his walking speed, so time to cover the same distance is 1/3 of walking time. Time to run the second half: 6/3=2 minutes. Total time: 6+2=8 minutes.
ANSWER 10: D

---

### Problem11:
Pair terms to simplify addition: (1+49)+(11+39)+(21+29)+(31+19)+(41+9) = 50+50+50+50+50=250.
ANSWER 11: E

---

### Problem12:
Line up all decimals to compare: A=0.9900, B=0.9099, C=0.9000, D=0.9090, E=0.9009. The largest value is 0.99, option A.
ANSWER 12: A

Problem 1:  
In a single-elimination tournament, each game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams (all except 1). Thus, 15 games are played.  
ANSWER 1: D  

Problem 2:  
Calculate each person's result starting from 10:  
- Jose: \(10 -1 =9\), double → \(9*2=18\), add 2 → \(18+2=20\)  
- Thuy: Double 10 → \(10*2=20\), subtract 1 → \(20-1=19\), add 2 → \(19+2=21\)  
- Kareem: \(10-1=9\), add 2 → \(9+2=11\), double → \(11*2=22\)  
Kareem has the largest result.  
ANSWER 2: C  

Problem 3:  
Convert \(\frac{2}{25}\) to decimal: \(2\div25=0.08\).  
ANSWER 3: B  

Problem 4:  
Choosing 3 starters from 4 people is a combination (order does not matter). The number of ways is \(\binom{4}{3}=4\) (or list: {Lance,Sally,Joy}, {Lance,Sally,Fred}, {Lance,Joy,Fred}, {Sally,Joy,Fred}).  
ANSWER 4: B  

Problem 5:  
Let the number be \(x\). \(20\%x=12\) → \(0.2x=12\) → \(x=60\). \(30\%\ of\ 60=0.3*60=18\).  
ANSWER 5: B  

Problem 6:  
Model directions with standard angles: 0°=east, 90°=north, 180°=west, 270°=south. Initial direction is west (180°).  
- Clockwise \(2\frac{1}{4}\) revolutions: \(2.25*(-360°)=-810°\), new angle: \(180-810=-630°\)  
- Counterclockwise \(3\frac{3}{4}\) revolutions: \(3.75*(360°)=+1350°\), total angle: \(-630+1350=720°\)  
- \(720°\mod360°=0°\), which is east.  
ANSWER 6: B  

Problem 7:  
Original yellow tint: \(30\%\ of\ 30=0.3*30=9\) liters. Add 5 liters, total yellow = \(9+5=14\) liters. New total mixture: \(30+5=35\) liters. Percent yellow: \(\frac{14}{35}*100\%=40\%\).  
ANSWER 7: C  

Problem 8:  
Compare fractions via cross-multiplication or decimals:  
- \(\frac{19}{15}\approx1.267\), \(\frac{17}{13}\approx1.308\), \(\frac{15}{11}\approx1.364\)  
Order from least to greatest: \(\frac{19}{15}<\frac{17}{13}<\frac{15}{11}\).  
ANSWER 8: E  

Problem 9:  
The unique mode is 6 (appears twice, others once). Thus mean and median equal 6.  
- Median of 7 numbers is the 4th term when sorted, so 4th term =6 → \(x\geq6\).  
- Mean = \(\frac{3+4+5+6+6+7+x}{7}=6\) → \(31+x=42\) → \(x=11\).  
Check: Sorted list [3,4,5,6,6,7,11], median=6, mode=6 (unique), mean=6.  
ANSWER 9: D  

Problem 10:  
Walking half the distance takes 6 minutes, so full walking time would be 12 minutes. Running speed is 3x walking speed, so time for the second half is \(\frac{6}{3}=2\) minutes. Total time: \(6+2=8\) minutes.  
ANSWER 10: D  

Problem 11:  
Sum first group: \(1+11+21+31+41=105\) (arithmetic series: \(\frac{5}{2}(1+41)=105\)).  
Sum second group: \(9+19+29+39+49=145\) (\(\frac{5}{2}(9+49)=145\)).  
Total: \(105+145=250\).  
ANSWER 11: E  

Problem 12:  
Compare decimals digit by digit:  
- A=0.99, B=0.9099, C=0.9, D=0.909, E=0.9009  
A has the highest second decimal digit (9 vs 0 for others), so it is the largest.  
ANSWER 12: A

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** How many games must be played to determine a winner in a 16-team elimination tournament?

**Setup:** In an elimination tournament, each game eliminates exactly one team. To go from 16 teams to 1 winner, we must eliminate 15 teams. Therefore, we need 15 games.

**Alternatively:** We can think of it as a bracket:
- Round 1: 16 teams → 8 games → 8 winners
- Round 2: 8 teams → 4 games → 4 winners
- Round 3: 4 teams → 2 games → 2 winners
- Round 4: 2 teams → 1 game → 1 winner
- Total: 8 + 4 + 2 + 1 = 15 games

ANSWER 1: D

---

**Problem 2:**

**What's being asked:** Which person gets the largest final answer after their operations?

**Jose's calculation:** 
- Start: 10
- Subtract 1: 10 - 1 = 9
- Double: 9 × 2 = 18
- Add 2: 18 + 2 = **20**

**Thuy's calculation:**
- Start: 10
- Double: 10 × 2 = 20
- Subtract 1: 20 - 1 = 19
- Add 2: 19 + 2 = **21**

**Kareem's calculation:**
- Start: 10
- Subtract 1: 10 - 1 = 9
- Add 2: 9 + 2 = 11
- Double: 11 × 2 = **22**

Kareem gets 22, which is the largest.

ANSWER 2: C

---

**Problem 3:**

**What's being asked:** Convert the fraction 2/25 to a decimal.

**Setup:** 2/25 = ?

**Calculation:** 
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** In how many ways can 3 starters be chosen from 4 people (Lance, Sally, Joy, Fred)?

**Setup:** This is a combination problem: $\binom{4}{3}$

**Calculation:**
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4}{1} = 4$$

Alternatively, we're choosing which 1 person to leave out: 4 ways.

ANSWER 4: B

---

**Problem 5:**

**What's being asked:** If 20% of a number is 12, what is 30% of that number?

**Setup:** Let the number be $n$.
- $0.20n = 12$
- $n = 12 ÷ 0.20 = 60$

**Calculation:** 
$$30\% \text{ of } 60 = 0.30 \times 60 = 18$$

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Starting pointing west, after rotating clockwise 2¼ revolutions then counterclockwise 3¾ revolutions, which direction?

**Setup:** 
- Clockwise 2¼ revolutions = 2.25 revolutions
- Counterclockwise 3¾ revolutions = 3.75 revolutions
- Net: 3.75 - 2.25 = 1.5 revolutions counterclockwise (net)

**Calculation:**
Starting at west, rotating counterclockwise 1.5 revolutions (= 1 full rotation + 0.5 rotation):
- 1 full rotation brings us back to west
- 0.5 rotation = 180°
- West + 180° counterclockwise = East

ANSWER 6: B

---

**Problem 7:**

**What's being asked:** After adding 5 liters of yellow tint to a 30-liter mixture, what percent is yellow tint?

**Original mixture (30 liters):**
- Yellow tint: 30% × 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percentage:**
$$\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%$$

ANSWER 7: C

---

**Problem 8:**

**What's being asked:** Order 15/11, 19/15, and 17/13 from least to greatest.

**Setup:** Convert to decimals or compare directly.

**Calculations:**
- 15/11 ≈ 1.364
- 19/15 ≈ 1.267
- 17/13 ≈ 1.308

**Order:** 19/15 < 17/13 < 15/11

ANSWER 8: E

---

**Problem 9:**

**What's being asked:** Find $x$ such that the mean, median, and unique mode of {3, 4, 5, 6, 6, 7, $x$} are all equal.

**Current mode:** 6 appears twice; all others appear once. If there's a unique mode, it must be 6.

**With unique mode = 6:** This requires no other number to appear more than once. So $x \notin \{3, 4, 5, 7\}$ and $x \neq 6$ (else it wouldn't be unique).

**Mean = Median = Mode = 6:**

For median of 7 numbers with value 6:
- Sorted: need middle (4th) value to be 6
- If $x = 11$: {3, 4, 5, 6, 6, 7, 11} → median = 6 ✓

Mean: $\frac{3+4+5+6+6+7+11}{7} = \frac{42}{7} = 6$ ✓

Mode: 6 appears twice; all others once → mode = 6 ✓

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Total time from home to school if Joe walked halfway (6 min) then ran the rest at 3× speed?

**Setup:**
- Walking: 6 minutes for half the distance
- Running: same distance at 3× speed, so time = 6/3 = 2 minutes

**Total time:** 6 + 2 = 8 minutes

ANSWER 10: D

---

**Problem 11:**

**What's being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Calculation:**
- First group: 1 + 11 + 21 + 31 + 41 = 105
- Second group: 9 + 19 + 29 + 39 + 49 = 145
- Sum: 105 + 145 = 250

ANSWER 11: E

---

**Problem 12:**

**What's being asked:** Which number is largest?

**Comparison:**
- A. 0.99
- B. 0.9099
- C. 0.9
- D. 0.909
- E. 0.9009

Starting from the tenths place: all are 0.9__. Looking at hundredths:
- 0.99 has 9 in hundredths place → **largest**

ANSWER 12: A

Problem 1:

In a single-elimination tournament, each game eliminates exactly one team.  
There are 16 teams, and we want one champion.

So we must eliminate 15 teams total.

Therefore, the number of games played is 15.

Tempting wrong choice: 16 would be if every team had to lose, but the winner does not lose.

ANSWER 1: D

Problem 2:

Compute each person’s result carefully.

- Jose: start with 10  
  subtract 1 → 9  
  double → 18  
  add 2 → 20

- Thuy: start with 10  
  double → 20  
  subtract 1 → 19  
  add 2 → 21

- Kareem: start with 10  
  subtract 1 → 9  
  add 2 → 11  
  double → 22

Compare:
- Jose = 20
- Thuy = 21
- Kareem = 22

Kareem gets the largest final answer.

ANSWER 2: C

Problem 3:

\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

So the correct decimal is 0.08.

ANSWER 3: B

Problem 4:

We are choosing 3 starters from 4 players: Lance, Sally, Joy, and Fred.

This is a combination problem:
\[
\binom{4}{3} = 4
\]

You can also see it by noticing that choosing 3 starters is the same as choosing which 1 player sits out, and there are 4 choices.

ANSWER 4: B

Problem 5:

Let the number be \(n\).

20% of \(n\) is 12, so:
\[
0.2n = 12
\]
\[
n = 60
\]

Now find 30% of 60:
\[
0.3 \times 60 = 18
\]

ANSWER 5: B

Problem 6:

The spinner starts pointing west.

First move: clockwise \(2\frac{1}{4}\) revolutions.  
A full revolution brings it back to west, so only the extra \(\frac{1}{4}\) revolution matters.

From west, a quarter-turn clockwise points it to north.

Second move: counterclockwise \(3\frac{3}{4}\) revolutions.  
Again, the 3 full revolutions do nothing, so this is just \(\frac{3}{4}\) of a revolution counterclockwise, which is the same as \(\frac{1}{4}\) of a revolution clockwise.

From north, a quarter-turn clockwise points east.

So the spinner ends pointing east.

ANSWER 6: B

Problem 7:

Start with 30 liters total.

Yellow tint is 30% of 30 liters:
\[
0.30 \times 30 = 9 \text{ liters}
\]

Then 5 liters of yellow tint are added, so yellow becomes:
\[
9 + 5 = 14 \text{ liters}
\]

New total mixture:
\[
30 + 5 = 35 \text{ liters}
\]

Percent yellow in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.4 = 40\%
\]

ANSWER 7: C

Problem 8:

The fractions are:
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]

Approximate each one:

\[
\frac{15}{11} \approx 1.364
\]
\[
\frac{17}{13} \approx 1.308
\]
\[
\frac{19}{15} \approx 1.267
\]

So from least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

That matches choice E.

ANSWER 8: E

Problem 9:

The numbers are:
\[
3,4,5,6,6,7,x
\]

We are told the mean, median, and unique mode are all equal.

Since 6 appears twice and every other listed number appears once, the unique mode is 6.

So the median and mean must also be 6.

The median of 7 numbers is the 4th number when ordered.  
The sorted list is:
\[
3,4,5,6,6,7,x
\]
To make the median 6, \(x\) must be 6 or larger, but we also need the mean to be 6.

Sum of the known numbers:
\[
3+4+5+6+6+7 = 31
\]

For the mean to be 6, total sum must be:
\[
7 \times 6 = 42
\]

So:
\[
31 + x = 42
\]
\[
x = 11
\]

Check:
- mode = 6
- median = 6
- mean = 42/7 = 6

Works perfectly.

ANSWER 9: D

Problem 10:

Joe walked half the distance in 6 minutes.  
He then ran the other half at 3 times his walking speed.

If he runs 3 times as fast, then the time for the same distance is 1/3 as long.

So the running time is:
\[
6 \div 3 = 2 \text{ minutes}
\]

Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

Problem 11:

Add the first group:
\[
1+11+21+31+41 = 105
\]

Add the second group:
\[
9+19+29+39+49 = 145
\]

Now add the totals:
\[
105 + 145 = 250
\]

ANSWER 11: E

Problem 12:

Compare the decimals:

- \(0.99\)
- \(0.9099\)
- \(0.9 = 0.9000\)
- \(0.909\)
- \(0.9009\)

The largest is clearly \(0.99\), since it is closest to 1.

ANSWER 12: A

Problem 1:  
We have a three-person-team tournament where **the losing team of each game is eliminated**. With **16 teams**, how many games are played to determine the winner?

To find the number of games in a single-elimination tournament:  
- Each game eliminates exactly **one** team (the loser).  
- To go from **16 teams to 1 winning team**, we must eliminate **15 teams**.  
- Therefore, the number of games is **15**.

So the correct choice is **D**.

ANSWER 1: D


Problem 2:  
Jose, Thuy, and Kareem all start at 10. We compute each final value.

**Jose:**  
- Subtract 1: \(10 - 1 = 9\)  
- Double: \(2 \cdot 9 = 18\)  
- Add 2: \(18 + 2 = 20\)  
Final: **20**

**Thuy:**  
- Double 10: \(2 \cdot 10 = 20\)  
- Subtract 1: \(20 - 1 = 19\)  
- Add 2: \(19 + 2 = 21\)  
Final: **21**

**Kareem:**  
- Subtract 1: \(10 - 1 = 9\)  
- Add 2: \(9 + 2 = 11\)  
- Double: \(2 \cdot 11 = 22\)  
Final: **22**

Largest is **22**, which is Kareem.

ANSWER 2: C


Problem 3:  
We need to evaluate:  
\[
\frac{2}{25}
\]
Convert to decimals:  
\[
\frac{2}{25} = 2 \div 25
\]
Since \(25 \cdot 0.08 = 2\):  
\[
2 \div 25 = 0.08
\]
So the correct choice is **B** (.08).

ANSWER 3: B


Problem 4:  
We have 4 people chosen: Lance, Sally, Joy, Fred. We need to choose **3 starters**.

Number of ways:
\[
\binom{4}{3} = 4
\]
List them conceptually: leaving out each one individually gives 4 choices.

Correct choice: **B**.

ANSWER 4: B


Problem 5:  
“If 20% of a number is 12, what is 30% of the same number?”

Let the number be \(N\).  
\[
0.20N = 12
\]
\[
N = \frac{12}{0.20} = 60
\]
Now compute \(30\%\) of 60:
\[
0.30 \cdot 60 = 18
\]
Correct choice: **B**.

ANSWER 5: B


Problem 6:  
Spinner starts pointing **west**.

Chenille moves it:
- **Clockwise** \(2\frac{1}{4}\) revolutions  
- then **Counterclockwise** \(3\frac{3}{4}\) revolutions

Net rotation (clockwise minus counterclockwise):
\[
2\frac{1}{4} - 3\frac{3}{4}
\]
Convert to improper fractions:
\[
2\frac{1}{4} = \frac{9}{4}, \quad 3\frac{3}{4} = \frac{15}{4}
\]
\[
\frac{9}{4} - \frac{15}{4} = -\frac{6}{4} = -\frac{3}{2}
\]
The negative means **net counterclockwise** by \(\frac{3}{2}\) revolutions.

\(\frac{3}{2} = 1 + \frac{1}{2}\), so this is:
- 1 full turn (brings it back where it started)
- plus \(\frac{1}{2}\) turn counterclockwise

Half a revolution from **west** points **east** (opposite direction).

Correct answer: **B (east)**.

ANSWER 6: B


Problem 7:  
Original mixture: 30 L with  
- 25% red (doesn’t matter directly)
- 30% yellow  
- 45% water

Yellow amount initially:
\[
0.30 \cdot 30 = 9 \text{ liters}
\]
Add 5 liters of yellow tint, so new yellow amount:
\[
9 + 5 = 14
\]
New total volume:
\[
30 + 5 = 35
\]
Percent yellow in new mixture:
\[
\frac{14}{35} = 0.4 = 40\%
\]
Correct choice: **C**.

ANSWER 7: C


Problem 8:  
Order fractions least to greatest:
\[
\frac{15}{11},\ \frac{19}{15},\ \frac{17}{13}
\]
A good way: compare values by cross-multiplication or decimal approximations.

Compute approximate decimals:
\[
\frac{15}{11} \approx 1.3636
\]
\[
\frac{19}{15} \approx 1.2667
\]
\[
\frac{17}{13} \approx 1.3077
\]
Least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]
Check choice: that matches **E**:
- \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11}\)

ANSWER 8: E


Problem 9:  
Numbers: \(3,4,5,6,6,7,x\) (7 positive integers).  
Mean = median = unique mode.

**Step 1: Mode**  
The only number that repeats is **6**, occurring twice. So the **unique mode = 6**.

Thus median must be 6 as well.

**Step 2: Median = 6**  
For 7 numbers sorted, the median is the **4th** number.

Sorted list must have the 4th term equal to 6. We already have two 6’s and surrounding numbers 3,4,5,7.

Currently without \(x\), the ordered portion is:
\[
3,4,5,6,6,7,\ ?
\]
The 4th number here is 6 already. To keep the 4th number as 6, \(x\) must not be less than 6 (otherwise it would shift positions so 4th might change).

If \(x = 6\), then 6 occurs 3 times, still mode 6 (unique mode). Then list:
\[
3,4,5,6,6,6,7
\]
Median (4th) is 6—works.

If \(x=7\), list:
\[
3,4,5,6,6,7,7
\]
Mode would be 6? Actually 7 repeats too, so mode not unique.

If \(x>7\), list:
\[
3,4,5,6,6,7,x
\]
Median still 6, but mode would still be 6 (unique), since only 6 repeats. That would also work—wait, but then 6 is still unique mode, yes. However we also need **mean = 6**.

So use mean condition.

Sum of known numbers:
\[
3+4+5+6+6+7 = 31
\]
Total sum including \(x\): \(31 + x\)

Mean:
\[
\frac{31+x}{7} = 6
\]
\[
31 + x = 42
\]
\[
x = 11
\]

Check uniqueness of mode: 6 occurs twice, no other repeats (since x=11 is different). Unique mode is 6—consistent.

ANSWER 9: D


Problem 10:  
Joe walks half the distance, then runs the rest.  
He runs **3 times as fast** as he walks.

Given: time to walk half way = **6 minutes**.  
Let walking speed be \(v\). Then walking time for half distance \(=\) 6 minutes.

Let half the distance be \(d\). Then:
\[
d = v \cdot 6
\]
Running speed is \(3v\).  
Remaining distance is also \(d\), so running time:
\[
\text{time} = \frac{d}{3v} = \frac{v\cdot 6}{3v} = 2
\]
Total time:
\[
6 + 2 = 8
\]
Correct choice: **D**.

ANSWER 10: D


Problem 11:  
Compute:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]

Group each set:

First set:
\[
1+11+21+31+41
\]
These are all of the form \(10k+1\):  
\(1, 11, 21, 31, 41\) correspond to \(k=0,1,2,3,4\).

Sum:
\[
(1+11+21+31+41) = (1+11+21+31+41)
\]
Compute directly:
\[
(1+11)=12,\ (21+31)=52,\ 12+52=64,\ 64+41=105
\]

Second set:
\[
9+19+29+39+49
\]
Direct sum:
\[
(9+19)=28,\ (29+39)=68,\ 28+68=96,\ 96+49=145
\]

Total:
\[
105 + 145 = 250
\]

Correct choice: **E**.

ANSWER 11: E


Problem 12:  
Which number is largest?

Options:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare decimals:  
- .99 is the largest among all because it is greater than .9 (and greater than any number starting 0.90... but with fewer 9s afterward).

So .99 > .9099 > .909 > .9009 > .9.

Correct choice: **A**.

ANSWER 12: A

Problem 1:
In a single-elimination tournament, every game results in exactly one team being eliminated. To determine a single winner from 16 teams, 15 teams must be eliminated. Since each game eliminates one team, it takes exactly 15 games.
ANSWER 1: D

Problem 2:
- Jose: (10 - 1) = 9; 9 * 2 = 18; 18 + 2 = 20.
- Thuy: (10 * 2) = 20; 20 - 1 = 19; 19 + 2 = 21.
- Kareem: (10 - 1) = 9; 9 + 2 = 11; 11 * 2 = 22.
Kareem has the largest answer (22).
ANSWER 2: C

Problem 3:
To convert 2/25 to a decimal, multiply the numerator and denominator by 4 to get a denominator of 100: (2 * 4) / (25 * 4) = 8 / 100 = 0.08.
ANSWER 3: B

Problem 4:
We need to choose 3 starters from a group of 4 (Lance, Sally, Joy, Fred). This is a combination problem denoted as 4C3. Choosing 3 people to be on the team is the same as choosing 1 person to be left off the team. There are 4 ways to choose who is left off.
ANSWER 4: B

Problem 5:
Let $x$ be the number. $0.20x = 12$. Dividing by 0.20, $x = 12 / 0.2 = 60$. We want 30% of $x$: $0.30 * 60 = 18$.
ANSWER 5: B

Problem 6:
Start at 0 revolutions. Clockwise is positive, counterclockwise is negative.
$2 \frac{1}{4} - 3 \frac{3}{4} = 2.25 - 3.75 = -1.5$ revolutions.
-1.5 revolutions is the same as -0.5 revolutions (since -1 is a full circle). -0.5 revolutions (counterclockwise) from West points to East.
ANSWER 6: B

Problem 7:
Original yellow tint: 30% of 30 liters = 9 liters.
New mixture: 30 + 5 = 35 liters.
New yellow tint: 9 + 5 = 14 liters.
Percent: (14 / 35) * 100 = (2 / 5) * 100 = 40%.
ANSWER 7: C

Problem 8:
Convert to decimals or compare differences from 1:
15/11 = 1 + 4/11 ≈ 1.3636
19/15 = 1 + 4/15 ≈ 1.2666
17/13 = 1 + 4/13 ≈ 1.3076
Comparing the fractions 4/11, 4/13, and 4/15: the larger the denominator, the smaller the fraction. Thus, 4/15 < 4/13 < 4/11.
Order: 19/15 < 17/13 < 15/11.
ANSWER 8: E

Problem 9:
The set is {3, 4, 5, 6, 6, 7, x}. The mode is 6 (it appears twice). For the mean and median to be 6:
Median: With 7 numbers, the 4th number must be 6. Since the sorted list is 3, 4, 5, 6, 6, 7, x, if $x > 6$, the 4th number is 6.
Mean: (3+4+5+6+6+7+x) / 7 = 6 => 31 + x = 42 => x = 11.
Checking: {3, 4, 5, 6, 6, 7, 11}. Median is 6. Mean is 42/7 = 6. Mode is 6.
ANSWER 9: D

Problem 10:
Walking time = 6 minutes. Running speed = 3 * walking speed.
Time taken = Distance / Speed. Since speed is 3 times faster, time is 1/3 of the walking time.
Running time = 6 / 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
ANSWER 10: D

Problem 11:
Sum 1: (1+11+21+31+41) = 105.
Sum 2: (9+19+29+39+49) = 145.
105 + 145 = 250.
ANSWER 11: E

Problem 12:
Compare digit by digit:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
0.99 is the largest.
ANSWER 12: A

Problem 1:
This is a single-elimination tournament with 16 teams. Exactly one team is eliminated per game, and 15 teams must be eliminated to leave a single winner. Thus exactly 15 games are required.

ANSWER 1: D

Problem 2:
Jose evaluates \((10-1)\times2+2=20\).
Thuy evaluates \((10\times2)-1+2=21\).
Kareem evaluates \((10-1+2)\times2=22\).
Kareem obtains the strictly largest value.

ANSWER 2: C

Problem 3:
Division yields the decimal expansion \(2\div25=0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters out of 4 distinct players is the binomial coefficient \(\binom{4}{3}=4\).

ANSWER 4: B

Problem 5:
If 20% of a number equals 12, the number itself is \(12/0.2=60\). Then 30% of 60 is \(0.3\times60=18\).

ANSWER 5: B

Problem 6:
A net rotation of \(+2.25-3.75=-1.5\) revolutions is equivalent to 1.5 revolutions counterclockwise. From an initial westward orientation, 1.5 revolutions counterclockwise lands exactly on east.

ANSWER 6: B

Problem 7:
The original mixture contains \(0.30\times30=9\) liters of yellow tint. Adding 5 liters raises the yellow volume to 14 liters and the total volume to 35 liters, so the new percentage is \(14/35=40\%\).

ANSWER 7: C

Problem 8:
Decimal approximations are \(15/11\approx1.3636\), \(19/15\approx1.2667\), and \(17/13\approx1.3077\). Ordering these values produces \(19/15<17/13<15/11\).

ANSWER 8: E

Problem 9:
The unique mode must be 6, forcing the mean also to equal 6. The sum of the seven numbers is therefore 42. The six given numbers sum to 31, so \(x=11\). With this value the ordered list is 3,4,5,6,6,7,11 whose median is likewise 6 and whose unique mode remains 6.

ANSWER 9: D

Problem 10:
Let the total distance be \(2d\). Walking time for distance \(d\) is 6 min, so running time for the remaining distance \(d\) at triple speed is \(6/3=2\) min. The total time is therefore 8 min.

ANSWER 10: D

Problem 11:
Each pair sums to a multiple of 10: \(1+9=10\), \(11+19=30\), \(21+29=50\), \(31+39=70\), \(41+49=90\). Their total is 250. Equivalently each five-term arithmetic series sums to 105 and 145 respectively, again yielding 250.

ANSWER 11: E

Problem 12:
Direct comparison of the five decimals shows that 0.99 is strictly larger than each of the others.

ANSWER 12: A

Problem 1:
The problem asks us to determine the number of games that will be played to determine the winner of a basketball tournament with 16 teams, where the losing team of each game is eliminated.

## Step 1
To determine the number of games needed to find the winner, we first note that in a single-elimination tournament, each game eliminates one team.

## Step 2
Since there can only be one winner, all other teams must be eliminated. Therefore, the number of games required is equal to the number of teams minus one.

## Step 3
We have 16 teams competing in the tournament. So, the number of games needed to determine the winner is 16 - 1 = 15.

## Step 4
Let's examine the given choices to rule out the incorrect options. A. 4 is too low because it doesn't account for all the teams. B. 7 is also too low for the same reason. C. 8 is still too low. D. 15 matches our calculation. E. 16 is one more than our calculation.

The final answer is supported by our step-by-step reasoning.
ANSWER 1: D

Problem 2:
The problem asks us to determine who among Jose, Thuy, and Kareem gets the largest final answer after performing a series of operations starting with the number 10.

## Step 1
Let's calculate Jose's final answer: Jose subtracts 1 from 10 to get 9, doubles it to get 18, and then adds 2 to get 20.

## Step 2
Now, let's calculate Thuy's final answer: Thuy doubles 10 to get 20, subtracts 1 to get 19, and then adds 2 to get 21.

## Step 3
Next, let's calculate Kareem's final answer: Kareem subtracts 1 from 10 to get 9, adds 2 to get 11, and then doubles it to get 22.

## Step 4
Comparing the final answers: Jose got 20, Thuy got 21, and Kareem got 22. The largest final answer is 22, which belongs to Kareem.

## Step 5
Let's examine the given choices. A. Jose is incorrect because his answer is 20. B. Thuy is incorrect because her answer is 21. C. Kareem is correct because his answer is 22. D. Jose and Thuy is incorrect because Kareem's answer is larger. E. Thuy and Kareem is incorrect because Jose's answer is not the largest.

The final answer is supported by our step-by-step reasoning.
ANSWER 2: C

Problem 3:
The problem asks us to convert the fraction 2/25 into a decimal.

## Step 1
To convert 2/25 into a decimal, we divide the numerator by the denominator: 2 ÷ 25 = 0.08.

## Step 2
Let's examine the given choices to identify the correct decimal representation. A. .008 is too small. B. .08 matches our calculation. C. .8 is too large. D. 1.25 is much too large. E. 12.5 is also much too large.

The final answer is supported by our step-by-step reasoning.
ANSWER 3: B

Problem 4:
The problem asks us to determine the number of ways to choose 3 starters from a team of 4 members: Lance, Sally, Joy, and Fred.

## Step 1
This is a combination problem, because the order in which the starters are chosen does not matter.

## Step 2
The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial, the product of all positive integers up to that number.

## Step 3
Here, n = 4 (total team members) and k = 3 (number of starters to be chosen). So, C(4, 3) = 4! / [3!(4 - 3)!] = 4! / (3! * 1!) = (4 * 3 * 2 * 1) / ((3 * 2 * 1) * 1) = 4.

## Step 4
Let's examine the given choices. A. 2 is too low. B. 4 matches our calculation. C. 6 is too high. D. 8 is also too high. E. 10 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 4: B

Problem 5:
The problem asks us to find 30% of a number, given that 20% of the same number is 12.

## Step 1
First, we need to find the number. We know that 20% of the number is 12. So, 0.20 * number = 12.

## Step 2
To find the number, we divide both sides of the equation by 0.20: number = 12 / 0.20 = 60.

## Step 3
Now, we need to calculate 30% of 60: 0.30 * 60 = 18.

## Step 4
Let's examine the given choices. A. 15 is too low. B. 18 matches our calculation. C. 20 is too high. D. 24 is also too high. E. 30 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 5: B

Problem 6:
The problem asks us to determine the direction the spinner points after Chenille moves it clockwise 2 1/4 revolutions and then counterclockwise 3 3/4 revolutions, starting from pointing west.

## Step 1
First, let's understand the effect of the movements. A full revolution is 360 degrees or 4 quarters. So, 2 1/4 revolutions clockwise is equivalent to 9 quarters clockwise.

## Step 2
9 quarters clockwise is the same as 1 quarter clockwise (since 8 quarters make 2 full revolutions, which brings the spinner back to the original direction). So, after the first move, the spinner points north.

## Step 3
Then, Chenille moves it counterclockwise 3 3/4 revolutions, which is equivalent to 15 quarters counterclockwise.

## Step 4
15 quarters counterclockwise is the same as 3 quarters counterclockwise (since 12 quarters make 3 full revolutions). So, moving 3 quarters counterclockwise from north results in pointing west.

## Step 5
Let's examine the given choices. A. north is incorrect because after the second move, it's not north. B. east is incorrect. C. south is incorrect. D. west matches our calculation. E. northwest is incorrect.

The final answer is supported by our step-by-step reasoning.
ANSWER 6: D

Problem 7:
The problem asks us to find the percentage of yellow tint in a new mixture after adding 5 liters of yellow tint to an original 30-liter mixture that is 25% red tint, 30% yellow tint, and 45% water.

## Step 1
First, we calculate the amount of yellow tint in the original mixture: 30% of 30 liters = 0.30 * 30 = 9 liters.

## Step 2
After adding 5 liters of yellow tint, the total amount of yellow tint becomes 9 + 5 = 14 liters.

## Step 3
The total volume of the new mixture is 30 + 5 = 35 liters.

## Step 4
To find the percentage of yellow tint in the new mixture, we divide the amount of yellow tint by the total volume and multiply by 100: (14 / 35) * 100 = 40%

Problem 1:
In a single-elimination tournament, each game eliminates exactly one team. To eliminate all but the champion, the number of games needed is one fewer than the number of teams. With 16 teams, the number of games is \(16 - 1 = 15\).

ANSWER 1: D

Problem 2:
Jose: \(10 - 1 = 9\), \(9 \times 2 = 18\), \(18 + 2 = 20\).
Thuy: \(10 \times 2 = 20\), \(20 - 1 = 19\), \(19 + 2 = 21\).
Kareem: \(10 - 1 = 9\), \(9 + 2 = 11\), \(11 \times 2 = 22\).
Kareem’s final answer of 22 is the largest.

ANSWER 2: C

Problem 3:
Divide 2 by 25: \(2 \div 25 = 0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters from 4 people is the combination \(\binom{4}{3} = 4\).

ANSWER 4: B

Problem 5:
Let the number be \(n\). \(20\%\) of \(n\) is \(0.2n = 12\), so \(n = 60\). Then \(30\%\) of \(60 = 0.3 \times 60 = 18\).

ANSWER 5: B

Problem 6:
Initial direction: west.
Clockwise \(2\frac{1}{4}\) revolutions: \(2\frac{1}{4} = 2.25\) revolutions clockwise.
Counterclockwise \(3\frac{3}{4}\) revolutions: \(3\frac{3}{4} = 3.75\) revolutions counterclockwise.
Net revolution = \(2.25 - 3.75 = -1.5\) revolutions, i.e., \(1.5\) revolutions counterclockwise.
A full revolution (1.0) returns to west. The remaining 0.5 revolution counterclockwise from west is a half-turn, which points to east.

ANSWER 6: B

Problem 7:
Original mixture: 30 liters, yellow tint = \(30\% \times 30 = 9\) liters.
Add 5 liters of yellow tint: new yellow = \(9 + 5 = 14\) liters; new total volume = \(30 + 5 = 35\) liters.
Percent yellow = \(\frac{14}{35} \times 100\% = 40\%\).

ANSWER 7: C

Problem 8:
Convert each fraction to a decimal for comparison:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
Order from least to greatest: \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

ANSWER 8: E

Problem 9:
The data set: 3, 4, 5, 6, 6, 7, \(x\). The mode is the most frequent number. With two 6’s, 6 is the mode unless \(x\) equals another number that appears twice, which would create a second mode, violating the “unique mode” condition. Therefore, \(x\) must not be 3, 4, 5, or 7. It could be 6, but if \(x=6\), the median is 6, while the mean is \((3+4+5+6+6+6+7)/7 = 37/7 \approx 5.29\), so mean ≠ median.  
For the mean to equal the median (which is 6, the middle of seven sorted numbers if \(x > 6\) or \(x=6\) only if placed correctly), we set the mean to 6. Sum without \(x\) = \(3+4+5+6+6+7 = 31\). Total needed for mean 6 = \(7 \times 6 = 42\). Thus \(x = 42 - 31 = 11\).  
Sorted: 3, 4, 5, 6, 6, 7, 11. Median = 6, mode = 6, mean = 6. Unique mode holds.

ANSWER 9: D

Problem 10:
Walking half the distance took 6 minutes, so walking speed = \(\frac{\text{half distance}}{6}\).
Running speed is 3 times walking speed. Time to run the second half = \(\frac{\text{half distance}}{3 \times (\text{half distance}/6)} = \frac{6}{3} = 2\) minutes.
Total time = \(6 + 2 = 8\) minutes.

ANSWER 10: D

Problem 11:
Pair the terms: \(1+49 = 50\), \(11+39 = 50\), \(21+29 = 50\), \(31+19 = 50\), \(41+9 = 50\). Five pairs of 50 give \(5 \times 50 = 250\).

ANSWER 11: E

Problem 12:
Compare by extending decimals to the same number of places:
\(0.99   = 0.9900\)
\(0.9099 = 0.9099\)
\(0.9    = 0.9000\)
\(0.909  = 0.9090\)
\(0.9009 = 0.9009\)
Clearly \(0.99 > 0.9099 > 0.909 > 0.9009 > 0.9\). The largest is \(0.99\).

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key Insight:** In a single-elimination tournament, every game eliminates exactly one team. To crown a single winner from 16 teams, we must eliminate 16 − 1 = 15 teams. Therefore, exactly 15 games are needed.

**Verification by rounds:**
- Round 1: 16 teams → 8 games → 8 teams left
- Round 2: 8 teams → 4 games → 4 teams left
- Round 3: 4 teams → 2 games → 2 teams left
- Round 4 (Final): 2 teams → 1 game → 1 winner

Total = 8 + 4 + 2 + 1 = 15 games.

Choices A (4), B (7), and C (8) are too small. Choice E (16) would mean every team is eliminated, leaving no winner.

ANSWER 1: D

---

Problem 2:
We track each person's calculation step by step, all starting from 10.

**Jose:** subtract 1 → 9; double → 18; add 2 → **20**
**Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
**Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

Comparing: 20 < 21 < 22. Kareem gets the largest final answer.

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

Choice A (.008) would be 8/1000. Choice C (.8) would be 8/10. Choices D and E are greater than 1, but 2/25 is clearly less than 1.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination problem — the order of selection doesn't matter:
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4$$

Equivalently, choosing 3 people to start is the same as choosing 1 person to sit out. There are 4 people, so 4 ways to pick who sits out.

The 4 possible starting lineups are: {L,S,J}, {L,S,F}, {L,J,F}, {S,J,F}.

ANSWER 4: B

---

Problem 5:
We're told 20% of a number is 12. Let the number be $x$.

$$0.20x = 12 \implies x = \frac{12}{0.20} = 60$$

Now find 30% of 60:
$$0.30 \times 60 = 18$$

**Shortcut:** 30% is $\frac{30}{20} = 1.5$ times 20%, so the answer is $1.5 \times 12 = 18$.

Choice A (15) would be 25%. Choice D (24) would be 40%. Choice E (30) would be 50%.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing **West**.

**Move 1:** Clockwise $2\frac{1}{4}$ revolutions.
- The 2 full revolutions bring it back to West.
- The extra $\frac{1}{4}$ revolution clockwise from West: West → **North**.

**Move 2:** Counterclockwise $3\frac{3}{4}$ revolutions.
- The 3 full revolutions bring it back to North.
- The extra $\frac{3}{4}$ revolution counterclockwise from North:
  - $\frac{1}{4}$ CCW from North → West
  - $\frac{2}{4}$ CCW from North → South
  - $\frac{3}{4}$ CCW from North → **East**

The spinner points East.

ANSWER 6: B

---

Problem 7:
Original mixture: 30 liters total.
- Red tint: 25% of 30 = 7.5 liters
- Yellow tint: 30% of 30 = 9 liters
- Water: 45% of 30 = 13.5 liters

Add 5 liters of yellow tint:
- New yellow tint = 9 + 5 = 14 liters
- New total mixture = 30 + 5 = 35 liters

Percent yellow tint in new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

Choice B (35%) is a trap for those who just add 5% to 30%. Choice D (45%) is the original water percentage.

ANSWER 7: C

---

Problem 8:
We compare $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$.

Rewrite each as a mixed number:
$$\frac{15}{11} = 1 + \frac{4}{11}, \quad \frac{19}{15} = 1 + \frac{4}{15}, \quad \frac{17}{13} = 1 + \frac{4}{13}$$

Since all three have the same numerator (4) in their fractional parts, we compare denominators. For fractions with equal numerators, a **larger denominator** means a **smaller fraction**:
$$15 > 13 > 11 \implies \frac{4}{15} < \frac{4}{13} < \frac{4}{11}$$

Therefore:
$$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$$

ANSWER 8: E

---

Problem 9:
The data set is: 3, 4, 5, 6, 6, 7, and $x$. The mean, median, and unique mode are all equal.

**Mode:** Since 6 already appears twice and all others appear once, the unique mode must be 6 (if $x$ were 3, 4, 5, or 7, we'd have two modes, violating "unique mode"). So mode = 6.

**Mean = 6:**
$$\frac{3 + 4 + 5 + 6 + 6 + 7 + x}{7} = 6$$
$$\frac{31 + x}{7} = 6 \implies 31 + x = 42 \implies x = 11$$

**Check Median:** With $x = 11$, the sorted list is: 3, 4, 5, **6**, 6, 7, 11. The median (4th value) is 6. ✓

All three measures equal 6. Choice B (6) would make 6 appear three times, but the mean would be $(31+6)/7 = 37/7 \neq 6$.

ANSWER 9: D

---

Problem 10:
Joe walked half the distance in 6 minutes. He then ran the remaining half at 3 times his walking speed.

Let walking speed = $v$. The half-distance = $6v$.

Running speed = $3v$. Time to run the second half:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{6v}{3v} = 2 \text{ minutes}$$

Total time = 6 + 2 = **8 minutes**.

Choice A (7) would result from incorrectly subtracting. Choices B, C, E involve unnecessary decimal complications.

ANSWER 10: D

---

Problem 11:
Compute $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$.

Pair the terms from the first group with terms from the second group (first with last, etc.):
- $1 + 49 = 50$
- $11 + 39 = 50$
- $21 + 29 = 50$
- $31 + 19 = 50$
- $41 + 9 = 50$

Total = $50 \times 5 = 250$.

Choice C (200) is a common error from miscounting pairs. Choice D (249) comes from arithmetic mistakes.

ANSWER 11: E

---

Problem 12:
Compare the five decimals by writing them to four decimal places:

- A: 0.9900
- B: 0.9099
- C: 0.9000
- D: 0.9090
- E: 0.9009

Comparing digit by digit from the tenths place: all have 9 in the tenths place. In the hundredths place, A has **9** while all others have **0**. So A is immediately the largest.

$0.99 > 0.9099 > 0.909 > 0.9009 > 0.9$

ANSWER 12: A

 **Problem 1:**
We need to find the number of games played in a single-elimination tournament with 16 teams. In single elimination, every game eliminates exactly one team. To determine a winner, 15 of the 16 teams must be eliminated. Therefore, exactly 15 games are required.  
(Another way: with 16 teams there are 8 games in the first round, 4 in the second, 2 in the third, and 1 final game: \(8+4+2+1=15\).)

ANSWER 1: D

---

**Problem 2:**
We compute each person’s final answer step by step, starting from 10.

- **Jose:**  
  \(10 - 1 = 9\)  
  \(9 \times 2 = 18\)  
  \(18 + 2 = 20\)

- **Thuy:**  
  \(10 \times 2 = 20\)  
  \(20 - 1 = 19\)  
  \(19 + 2 = 21\)

- **Kareem:**  
  \(10 - 1 = 9\)  
  \(9 + 2 = 11\)  
  \(11 \times 2 = 22\)

Comparing the results: \(20 < 21 < 22\). Kareem’s answer is the largest.

ANSWER 2: C

---

**Problem 3:**
We convert the fraction \(\frac{2}{25}\) to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100:
\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

ANSWER 3: B

---

**Problem 4:**
We must choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways to choose 3 from 4 is the combination \(\binom{4}{3}\):
\[
\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4
\]
Equivalently, choosing 3 starters is the same as choosing 1 person to sit out, and there are 4 choices for who sits out.

ANSWER 4: B

---

**Problem 5:**
Let the number be \(N\). We are given \(0.20N = 12\). Solving for \(N\):
\[
N = \frac{12}{0.20} = 60
\]
Now find 30% of 60:
\[
0.30 \times 60 = 18
\]
(Shortcut: 30% is \(1.5\) times 20%, so \(1.5 \times 12 = 18\).)

ANSWER 5: B

---

**Problem 6:**
The spinner starts pointing west.  
- Clockwise \(2\frac14\) revolutions: 2 full revolutions bring it back to west, then \(\frac14\) turn clockwise from west points it **north**.  
- Counterclockwise \(3\frac34\) revolutions: 3 full revolutions bring it back to north, then \(\frac34\) turn counterclockwise from north goes through west, south, and ends at **east**.

Net movement: \(3\frac34 - 2\frac14 = 1\frac12\) revolutions counterclockwise. One full revolution returns to west; an additional \(\frac12\) revolution counterclockwise from west points to **east**.

ANSWER 6: B

---

**Problem 7:**
Original mixture: 30 liters.  
- Yellow tint originally: \(30\% \times 30 = 9\) liters.  
After adding 5 liters of yellow tint:  
- New yellow tint: \(9 + 5 = 14\) liters.  
- New total mixture: \(30 + 5 = 35\) liters.  

Percent of yellow tint in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%
\]

ANSWER 7: C

---

**Problem 8:**
Compare \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\). Each is greater than 1. Write each as \(1\) plus a proper fraction:
\[
\frac{15}{11} = 1 + \frac{4}{11}, \qquad
\frac{19}{15} = 1 + \frac{4}{15}, \qquad
\frac{17}{13} = 1 + \frac{4}{13}
\]
Since the numerators are the same (4), the fraction with the largest denominator is smallest:
\[
\frac{4}{15} < \frac{4}{13} < \frac{4}{11}
\]
Adding 1 preserves the order:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

ANSWER 8: E

---

**Problem 9:**
The data set is \(3, 4, 5, 6, 6, 7, x\).  
- The current mode is 6 (appears twice). For 6 to remain the **unique** mode, \(x\) must not equal 3, 4, 5, or 7 (which would create a second mode).  
- We are told mean = median = mode = 6.  

Set the mean equal to 6:
\[
\frac{3+4+5+6+6+7+x}{7} = 6 \implies \frac{31+x}{7} = 6 \implies 31+x = 42 \implies x = 11
\]

Check: With \(x=11\), the sorted list is \(3, 4, 5, 6, 6, 7, 11\).  
- Median (4th term) = 6.  
- Mode = 6 (unique).  
- Mean = 6.  

All conditions are satisfied.

ANSWER 9: D

---

**Problem 10:**
Joe walks half the distance in 6 minutes. He runs the second half at 3 times his walking speed. Since distance is the same and speed is tripled, the time required is divided by 3:
\[
\text{Running time} = \frac{6}{3} = 2 \text{ minutes}
\]
Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

---

**Problem 11:**
Add the two groups:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]
Pair the terms to make convenient sums:
\[
(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 5 \times 50 = 250
\]

ANSWER 11: E

---

**Problem 12:**
Compare the decimals by writing them with the same number of places:
- A. \(0.9900\)
- B. \(0.9099\)
- C. \(0.9000\)
- D. \(0.9090\)
- E. \(0.9009\)

The tenths digit is 9 for all. Looking at the hundredths digit: A has 9, while B, D have 0 and C, E have 0. Thus \(0.9900\) is clearly the largest.

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key insight:** Each game eliminates exactly one team. To determine a winner from 16 teams, we must eliminate 15 teams (everyone except the champion).

Therefore, exactly 15 games must be played.

*Verification by rounds:* Round 1: 8 games (16→8 teams), Round 2: 4 games (8→4), Round 3: 2 games (4→2), Round 4: 1 game (2→1). Total = 8+4+2+1 = 15.

The tempting wrong answer E (16) counts teams instead of eliminations.

ANSWER 1: D

---

Problem 2:
We need to compute each person's final answer starting from 10.

**Jose:** Start with 10 → subtract 1: 10−1 = 9 → double: 9×2 = 18 → add 2: 18+2 = **20**

**Thuy:** Start with 10 → double: 10×2 = 20 → subtract 1: 20−1 = 19 → add 2: 19+2 = **21**

**Kareem:** Start with 10 → subtract 1: 10−1 = 9 → add 2: 9+2 = 11 → double: 11×2 = **22**

Kareem gets the largest final answer (22).

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

The tempting wrong answer A (0.008) would be 8/1000, and C (0.8) would be 8/10.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination: C(4,3) = 4!/(3! × 1!) = 4

Listing them out: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — that's 4 ways.

The tempting wrong answer C (6) would be C(4,2), choosing who's *not* a starter, but that also equals 4... Actually C(4,2)=6 is wrong here; we're choosing 3 from 4.

ANSWER 4: B

---

Problem 5:
We know 20% of a number is 12. We need to find 30% of the same number.

**Step 1:** Find the number.
$$0.20 \times n = 12 \implies n = \frac{12}{0.20} = 60$$

**Step 2:** Find 30% of 60.
$$0.30 \times 60 = 18$$

*Shortcut:* Since 30% is 1.5 times 20%, the answer is 1.5 × 12 = 18.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing west. We need to find the net rotation.

**Clockwise:** 2¼ revolutions
**Counterclockwise:** 3¾ revolutions

**Net movement:** 2¼ − 3¾ = −1½ revolutions (negative means counterclockwise)

So the spinner moves 1½ revolutions counterclockwise from west.

- 1 full revolution counterclockwise returns to **west**
- ½ revolution counterclockwise from west goes to **east**

(Counterclockwise from west: ¼ turn → south, ½ turn → east)

ANSWER 6: B

---

Problem 7:
We need to find the percent of yellow tint after adding 5 liters of yellow tint.

**Original mixture (30 liters):**
- Yellow tint: 30% of 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percent yellow tint:** (14/35) × 100 = 40%

The tempting wrong answer B (35) might come from simply adding 5% to 30%.

ANSWER 7: C

---

Problem 8:
We need to order 15/11, 19/15, and 17/13 from least to greatest.

Computing each as a decimal:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3077...

From least to greatest: **19/15 < 17/13 < 15/11**

*Alternative method:* Note each is 1 + (numerator − denominator)/denominator:
- 15/11 = 1 + 4/11
- 19/15 = 1 + 4/15
- 17/13 = 1 + 4/13

Since 4/15 < 4/13 < 4/11, the order is confirmed.

ANSWER 8: E

---

Problem 9:
The positive integers 3, 4, 5, 6, 6, 7, and x have equal mean, median, and unique mode.

**Mode:** Since 6 already appears twice and no other number does, the unique mode must be 6. So x ≠ 3, 4, 5, or 7 (which would create a second mode).

**Median:** With 7 numbers, the median is the 4th value when sorted. For the median to be 6, we need x ≥ 6 (so that 6 is the 4th value).

- If x = 6: sorted = 3, 4, 5, 6, 6, 6, 7 → median = 6 ✓
- If x > 7: sorted = 3, 4, 5, 6, 6, 7, x → median = 6 ✓

**Mean:** (3 + 4 + 5 + 6 + 6 + 7 + x)/7 = (31 + x)/7

Setting mean = 6: (31 + x)/7 = 6 → 31 + x = 42 → **x = 11**

**Verification:** Numbers: 3, 4, 5, 6, 6, 7, 11. Mean = 42/7 = 6, Median = 6, Mode = 6 ✓

ANSWER 9: D

---

Problem 10:
Joe walks halfway, then runs the rest. We need total time.

**Walking:** Half the distance takes 6 minutes.

**Running:** He runs 3 times as fast as he walks, so he covers the same distance in 1/3 the time.
- Time to run the second half = 6 ÷ 3 = 2 minutes

**Total time:** 6 + 2 = **8 minutes**

The tempting wrong answers come from setting up the problem incorrectly or mixing up distance and time relationships.

ANSWER 10: D

---

Problem 11:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

**First group:** 1 + 11 + 21 + 31 + 41 = 105

**Second group:** 9 + 19 + 29 + 39 + 49 = 145

**Total:** 105 + 145 = **250**

*Elegant method:* Pair terms: (1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250

ANSWER 11: E

---

Problem 12:
We need to find the largest number among the options. Writing each with the same number of decimal places:

- A. 0.99 = **0.9900**
- B. 0.9099 = 0.9099
- C. 0.9 = 0.9000
- D. 0.909 = 0.9090
- E. 0.9009 = 0.9009

Comparing digit by digit after the decimal point:
- First digit: all are 9
- Second digit: A has **9**, all others have 0

Since A has a 9 in the second decimal place while all others have 0, **0.99 is the largest**.

ANSWER 12: A

**Problem 1:**  
The tournament is single‑elimination. Starting with 16 teams, every game eliminates exactly one team. To reduce from 16 teams to a single champion we must eliminate 15 teams, and each elimination corresponds to one game.  

Number of games = 16 − 1 = **15**.  

ANSWER 1: D  

---

**Problem 2:**  
Start with 10 for each person.

- **Jose:**  
  Subtract 1 → 9; double → 18; add 2 → 20.  

- **Thuy:**  
  Double → 20; subtract 1 → 19; add 2 → 21.  

- **Kareem:**  
  Subtract 1 → 9; add 2 → 11; double → 22.  

The largest final value is 22, obtained by Kareem.  

ANSWER 2: C  

---

**Problem 3:**  

\[
\frac{2}{25}=0.08
\]

(2 ÷ 25 = 0.08).  

ANSWER 3: B  

---

**Problem 4:**  
We need to choose 3 starters from the 4 available players:  

\[
\binom{4}{3}=4
\]

ways.  

ANSWER 4: B  

---

**Problem 5:**  
Let the number be \(N\).  
\(0.20N = 12 \Rightarrow N = \frac{12}{0.20}=60\).  
\(30\%\) of \(N\) is \(0.30 \times 60 = 18\).  

ANSWER 5: B  

---

**Problem 6:**  
Interpret the fractions as mixed numbers:  

- Clockwise \(2\frac14 = 2.25\) revolutions.  
- Counter‑clockwise \(3\frac34 = 3.75\) revolutions.  

Net rotation (counter‑clockwise positive) = \(3.75 - 2.25 = 1.5\) revolutions.  

A full revolution returns to the same direction, so we only care about the extra \(0.5\) revolution.  
\(0.5\) revolution = 180° counter‑clockwise.  

Starting from west, turning 180° counter‑clockwise points to **east**.  

ANSWER 6: B  

---

**Problem 7:**  
Original yellow = \(30\%\) of 30 L = \(0.30 \times 30 = 9\) L.  
Add 5 L yellow → new yellow = \(9+5 = 14\) L.  
New total volume = \(30+5 = 35\) L.  

Percent yellow = \(\frac{14}{35}\times100 = 40\%\).  

ANSWER 7: C  

---

**Problem 8:**  
Compute the three fractions:

\[
\frac{15}{11}=1.3636\ldots,\qquad
\frac{19}{15}=1.2666\ldots,\qquad
\frac{17}{13}=1.3077\ldots
\]

Ordered from least to greatest:  

\(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

This matches option **E**.  

ANSWER 8: E  

---

**Problem 9:**  
Current list: \(3,4,5,6,6,7\). The unique mode is 6 (appears twice).  
For the mode to remain unique and equal to the mean and median, \(x\) must not create a tie.  
Let the mean = median = mode = \(m\).  

Sum of numbers = \(31 + x\).  
Mean = \(\frac{31 + x}{7}=m\).  
Median of the seven numbers is the 4th after sorting; with \(x>7\) the median stays 6, so \(m=6\).  

Thus \(\frac{31 + x}{7}=6 \Rightarrow 31 + x = 42 \Rightarrow x = 11\).

Check: sorted list becomes \(3,4,5,6,6,7,11\) → median = 6, mode = 6, mean = 42/7 = 6. All equal.  

ANSWER 9: D  

---

**Problem 10:**  
Let the walking speed be \(v\) and the half‑distance be \(d/2\).  

Walking time for half‑distance: \(\frac{d/2}{v}=6\) min ⇒ \(\frac{d}{v}=12\) min.  

Running speed = \(3v\).  
Running time for the second half: \(\frac{d/2}{3v}= \frac{d}{6v}= \frac{1}{6}\times\frac{d}{v}= \frac{12}{6}=2\) min.  

Total time = \(6 + 2 = 8\) minutes.  

ANSWER 10: D  

---

**Problem 11:**  

\[
\begin{aligned}
1+11+21+31+41 &= 105,\\
9+19+29+39+49 &= 145,\\
\text{Total} &= 105+145 = 250.
\end{aligned}
\]

ANSWER 11: E  

---

**Problem 12:**  
Write the numbers with four decimal places:

- .99 = 0.9900  
- .9099 = 0.9099  
- .9 = 0.9000  
- .909 = 0.9090  
- .9009 = 0.9009  

The largest is 0.9900, i.e., **.99**.  

ANSWER 12: A  

Problem 1:  
The tournament starts with 16 teams. Each game eliminates one team, and the losing team is eliminated. To determine the winner, all teams except one must be eliminated. Thus, 15 teams must be eliminated, requiring 15 games.  
- **Why not other choices?** A (4) is too few; B (7), C (8) are also insufficient; E (16) would imply one game per team without elimination, which is incorrect.  
ANSWER 1: D  

Problem 2:  
- **Jose**: Starts with 10 → subtracts 1 (9) → doubles (18) → adds 2 → **20**.  
- **Thuy**: Starts with 10 → doubles (20) → subtracts 1 (19) → adds 2 → **21**.  
- **Kareem**: Starts with 10 → subtracts 1 (9) → adds 2 (11) → doubles → **22**.  
Kareem has the largest final answer.  
- **Why not other choices?** A (Jose) and B (Thuy) are smaller; D and E pair incorrect winners.  
ANSWER 2: C  

Problem 3:  
\( \frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08 \).  
- **Why not other choices?** A (.008) is \( \frac{2}{250} \); C (.8) is \( \frac{4}{5} \); D and E are larger decimals.  
ANSWER 3: B  

Problem 4:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is \( \binom{4}{3} = 4 \).  
- **Why not other choices?** A (2) is too few; C (6) is \( \binom{4}{2} \), which is incorrect; D (8) and E (10) are overcounts.  
ANSWER 4: B  

Problem 5:  
Let the number be \( x \). If 20% of \( x \) is 12, then \( 0.2x = 12 \), so \( x = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).  
- **Why not other choices?** A (15) is 25% of 60; C (20) is \( \frac{1}{3} \) of 60; D (24) is 40%; E (30) is 50%.  
ANSWER 5: B  

Problem 6:  
- A full revolution is 360°. Clockwise \( 2 \frac{1}{4} \) revolutions = \( 2.25 \times 360 = 810° \).  
- Counterclockwise \( 3 \frac{3}{4} \) revolutions = \( -3.75 \times 360 = -1350° \).  
- Net movement: \( 810° - 1350° = -540° \).  
- \( -540° \mod 360° = 180° \) (since \( -540 + 720 = 180° \)), which points west → south? Wait, initial direction is west.  
- West (270°) + 810° = 1080° → 1080° mod 360° = 0° (north). Then -1350° → 0° - 1350° = -1350° mod 360° = 90° (east)? Let's reframe:  
- Start at west (270°). Clockwise 2.25 rev: 270° + 810° = 1080° mod 360° = 0° (north). Then counterclockwise 3.75 rev: 0° - 1350° = -1350° mod 360° = 90° (east).  
- **Why not other choices?** A (north) is initial after clockwise; B (east) is final; C (south) is 180°; D (west) is start; E (northwest) is not a cardinal direction.  
ANSWER 6: B  

Problem 7:  
Original mixture: 30 liters, 30% yellow tint → 9 liters yellow. Add 5 liters yellow → total yellow = 14 liters. New mixture = 35 liters. Percent yellow = \( \frac{14}{35} \times 100 = 40\% \).  
- **Why not other choices?** A (25) is original yellow%; B (35) is before addition; D (45) and E (50) are overestimates.  
ANSWER 7: E (Note: Correction based on computation: 14/35=0.4 → 40%, so answer is E? Wait, 14/35=0.4 → 40%, which is E. But earlier I thought B, but B is 35, which is wrong. Original yellow is 9, add 5 → 14. Total mixture 30+5=35. 14/35=0.4 → 40%. So E. But in choices E is 50? Wait, no: choices are A.25 B.35 C.40 D.45 E.50. So 40% is C. I misread. Correction: 14/35=0.4 → 40%, which is C.  
**Correction**: 14/35=0.4 → 40%, so answer is C.  
ANSWER 7: C  

Problem 8:  
Compare \( \frac{15}{11} \approx 1.3636 \), \( \frac{19}{15} \approx 1.2667 \), \( \frac{17}{13} \approx 1.3077 \). Order: \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11} \).  
- **Why not other choices?** A, B, C, D have incorrect orders.  
ANSWER 8: E  

Problem 9:  
The set is {3, 4, 5, 6, 6, 7, x}. Mean = median = mode = 6.  
- Mode is 6 (unique), so 6 must appear most.  
- Median is 6 (4th value in ordered list: 3,4,5,6,6,7,x → if x≥6, ordered is ...,6,6,x or ...,6,x,6? Must have 4th value 6.  
- If x=6, set is {3,4,5,6,6,6,7}: median=6, mode=6, mean=(3+4+5+6+6+6+7)/7=37/7≈5.28 ≠6.  
- If x=11: set {3,4,5,6,6,7,11}: mean=(3+4+5+6+6+7+11)/7=42/7=6, median=6, mode=6.  
- **Why not other choices?** A (5): mean=36/7≈5.14; B (6): mean≈5.28; C (7): mean=38/7≈5.43; D (11) works.  
ANSWER 9: D  

Problem 10:  
Joe walks half the distance in 6 minutes. Let walking speed be \( v \), so distance = \( 6v \). He runs at \( 3v \), so time to run half distance = \( \frac{6v}{3v} = 2 \) minutes. Total time = 6 + 2 = 8 minutes.  
- **Why not other choices?** A (7), B (7.3), C (7.7), E (8.3) are not exact.  
ANSWER 10: D  

Problem 11:  
First group: 1+11+21+31+41 = 105. Second group: 9+19+29+39+49 = 145. Total = 105 + 145 = 250.  
- **Why not other choices?** A (150), B (199), C (200), D (249) are incorrect sums.  
ANSWER 11: E  

Problem 12:  
Compare decimals: A (.99), B (.9099), C (.9), D (.909), E (.9009). Largest is .99.  
- **Why not other choices?** B, C, D, E are all smaller than .99.  
ANSWER 12: A

### Problem 1:
We need to find the total number of games to determine a winner in a 16-team single-elimination tournament, where every game eliminates exactly one losing team. To get 1 winner, we must eliminate 16 - 1 = 15 teams. Since each game eliminates 1 team, we need 15 total games. Wrong choices: A is the number of bracket rounds, not games; C is half the number of teams; E is the total number of teams, not games.
ANSWER 1: D

---

### Problem 2:
All three start with 10, we calculate each person's final result step-by-step:
- Jose: (10 - 1) * 2 + 2 = 9*2 +2 = 20
- Thuy: (10*2) -1 +2 = 20-1+2=21
- Kareem: (10-1 +2)*2 = 11*2=22
Kareem has the largest final answer. Wrong choices stem from misordering the operations for each person.
ANSWER 2: C

---

### Problem3:
Convert 2/25 to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100: (2*4)/(25*4) = 8/100 = 0.08. Wrong choices: A is 2/250, C is 20/25.
ANSWER 3: B

---

### Problem4:
We choose 3 starters from 4 people, which is a combination problem (order of starters does not matter here). The number of ways to choose 3 from 4 is equivalent to choosing 1 person to leave off the starting lineup: there are 4 people, so 4 unique lineups, each omitting one person. Wrong choice C (6) is the number of permutations, not combinations.
ANSWER 4: B

---

### Problem5:
If 20% of a number is 12, first find the full number: 0.2*N=12 → N=12/0.2=60. 30% of 60 is 0.3*60=18. Wrong choices: A is 25% of 60, D is 40% of 60.
ANSWER 5: B

---

### Problem6:
The spinner starts pointing west. Net movement: 2.25 clockwise revolutions minus 3.75 clockwise (equivalent to 3.75 counterclockwise) = -1.5 total clockwise revolutions, or 1.5 counterclockwise revolutions. Full revolutions cancel out, so we only have a net 0.5 (180 degree) counterclockwise turn from west. A 180-degree turn from west points to east.
ANSWER 6: B

---

### Problem7:
Original 30L mixture has 30% yellow tint: 0.3*30=9L of yellow. Add 5L of yellow, so new yellow is 14L, new total mixture is 35L. The new percentage of yellow is (14/35)*100=40%. Wrong choice: people often forget to add 5L to the total mixture volume, leading to an incorrect higher percentage.
ANSWER 7: C

---

### Problem8:
Rewrite each fraction to compare easily: 15/11 = 1 + 4/11, 19/15=1+4/15, 17/13=1+4/13. For equal numerators, the larger the denominator, the smaller the fraction. So 4/15 < 4/13 < 4/11, so the full order from least to greatest is 19/15 < 17/13 <15/11.
ANSWER 8: E

---

### Problem9:
The dataset is [3,4,5,6,6,7,x]. The unique mode is currently 6 (only value appearing twice, so it must stay the only unique mode). For the mean, median, and mode to all be 6: the sum of the dataset must be 7*6=42. The sum of the existing numbers is 3+4+5+6+6+7=31, so x=42-31=11. Sorted dataset [3,4,5,6,6,7,11] has median 6, unique mode 6, mean 6, which fits. Wrong choice B (x=6) gives a mean of ~5.28, not 6.
ANSWER 9: D

---

### Problem10:
Joe walks half the distance in 6 minutes. His running speed is 3x his walking speed, so time to cover the same distance is 1/3 of walking time. Time to run the second half: 6/3=2 minutes. Total time: 6+2=8 minutes.
ANSWER 10: D

---

### Problem11:
Pair terms to simplify addition: (1+49)+(11+39)+(21+29)+(31+19)+(41+9) = 50+50+50+50+50=250.
ANSWER 11: E

---

### Problem12:
Line up all decimals to compare: A=0.9900, B=0.9099, C=0.9000, D=0.9090, E=0.9009. The largest value is 0.99, option A.
ANSWER 12: A

Problem 1:  
In a single-elimination tournament, each game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams (all except 1). Thus, 15 games are played.  
ANSWER 1: D  

Problem 2:  
Calculate each person's result starting from 10:  
- Jose: \(10 -1 =9\), double → \(9*2=18\), add 2 → \(18+2=20\)  
- Thuy: Double 10 → \(10*2=20\), subtract 1 → \(20-1=19\), add 2 → \(19+2=21\)  
- Kareem: \(10-1=9\), add 2 → \(9+2=11\), double → \(11*2=22\)  
Kareem has the largest result.  
ANSWER 2: C  

Problem 3:  
Convert \(\frac{2}{25}\) to decimal: \(2\div25=0.08\).  
ANSWER 3: B  

Problem 4:  
Choosing 3 starters from 4 people is a combination (order does not matter). The number of ways is \(\binom{4}{3}=4\) (or list: {Lance,Sally,Joy}, {Lance,Sally,Fred}, {Lance,Joy,Fred}, {Sally,Joy,Fred}).  
ANSWER 4: B  

Problem 5:  
Let the number be \(x\). \(20\%x=12\) → \(0.2x=12\) → \(x=60\). \(30\%\ of\ 60=0.3*60=18\).  
ANSWER 5: B  

Problem 6:  
Model directions with standard angles: 0°=east, 90°=north, 180°=west, 270°=south. Initial direction is west (180°).  
- Clockwise \(2\frac{1}{4}\) revolutions: \(2.25*(-360°)=-810°\), new angle: \(180-810=-630°\)  
- Counterclockwise \(3\frac{3}{4}\) revolutions: \(3.75*(360°)=+1350°\), total angle: \(-630+1350=720°\)  
- \(720°\mod360°=0°\), which is east.  
ANSWER 6: B  

Problem 7:  
Original yellow tint: \(30\%\ of\ 30=0.3*30=9\) liters. Add 5 liters, total yellow = \(9+5=14\) liters. New total mixture: \(30+5=35\) liters. Percent yellow: \(\frac{14}{35}*100\%=40\%\).  
ANSWER 7: C  

Problem 8:  
Compare fractions via cross-multiplication or decimals:  
- \(\frac{19}{15}\approx1.267\), \(\frac{17}{13}\approx1.308\), \(\frac{15}{11}\approx1.364\)  
Order from least to greatest: \(\frac{19}{15}<\frac{17}{13}<\frac{15}{11}\).  
ANSWER 8: E  

Problem 9:  
The unique mode is 6 (appears twice, others once). Thus mean and median equal 6.  
- Median of 7 numbers is the 4th term when sorted, so 4th term =6 → \(x\geq6\).  
- Mean = \(\frac{3+4+5+6+6+7+x}{7}=6\) → \(31+x=42\) → \(x=11\).  
Check: Sorted list [3,4,5,6,6,7,11], median=6, mode=6 (unique), mean=6.  
ANSWER 9: D  

Problem 10:  
Walking half the distance takes 6 minutes, so full walking time would be 12 minutes. Running speed is 3x walking speed, so time for the second half is \(\frac{6}{3}=2\) minutes. Total time: \(6+2=8\) minutes.  
ANSWER 10: D  

Problem 11:  
Sum first group: \(1+11+21+31+41=105\) (arithmetic series: \(\frac{5}{2}(1+41)=105\)).  
Sum second group: \(9+19+29+39+49=145\) (\(\frac{5}{2}(9+49)=145\)).  
Total: \(105+145=250\).  
ANSWER 11: E  

Problem 12:  
Compare decimals digit by digit:  
- A=0.99, B=0.9099, C=0.9, D=0.909, E=0.9009  
A has the highest second decimal digit (9 vs 0 for others), so it is the largest.  
ANSWER 12: A

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** How many games must be played to determine a winner in a 16-team elimination tournament?

**Setup:** In an elimination tournament, each game eliminates exactly one team. To go from 16 teams to 1 winner, we must eliminate 15 teams. Therefore, we need 15 games.

**Alternatively:** We can think of it as a bracket:
- Round 1: 16 teams → 8 games → 8 winners
- Round 2: 8 teams → 4 games → 4 winners
- Round 3: 4 teams → 2 games → 2 winners
- Round 4: 2 teams → 1 game → 1 winner
- Total: 8 + 4 + 2 + 1 = 15 games

ANSWER 1: D

---

**Problem 2:**

**What's being asked:** Which person gets the largest final answer after their operations?

**Jose's calculation:** 
- Start: 10
- Subtract 1: 10 - 1 = 9
- Double: 9 × 2 = 18
- Add 2: 18 + 2 = **20**

**Thuy's calculation:**
- Start: 10
- Double: 10 × 2 = 20
- Subtract 1: 20 - 1 = 19
- Add 2: 19 + 2 = **21**

**Kareem's calculation:**
- Start: 10
- Subtract 1: 10 - 1 = 9
- Add 2: 9 + 2 = 11
- Double: 11 × 2 = **22**

Kareem gets 22, which is the largest.

ANSWER 2: C

---

**Problem 3:**

**What's being asked:** Convert the fraction 2/25 to a decimal.

**Setup:** 2/25 = ?

**Calculation:** 
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** In how many ways can 3 starters be chosen from 4 people (Lance, Sally, Joy, Fred)?

**Setup:** This is a combination problem: $\binom{4}{3}$

**Calculation:**
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4}{1} = 4$$

Alternatively, we're choosing which 1 person to leave out: 4 ways.

ANSWER 4: B

---

**Problem 5:**

**What's being asked:** If 20% of a number is 12, what is 30% of that number?

**Setup:** Let the number be $n$.
- $0.20n = 12$
- $n = 12 ÷ 0.20 = 60$

**Calculation:** 
$$30\% \text{ of } 60 = 0.30 \times 60 = 18$$

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Starting pointing west, after rotating clockwise 2¼ revolutions then counterclockwise 3¾ revolutions, which direction?

**Setup:** 
- Clockwise 2¼ revolutions = 2.25 revolutions
- Counterclockwise 3¾ revolutions = 3.75 revolutions
- Net: 3.75 - 2.25 = 1.5 revolutions counterclockwise (net)

**Calculation:**
Starting at west, rotating counterclockwise 1.5 revolutions (= 1 full rotation + 0.5 rotation):
- 1 full rotation brings us back to west
- 0.5 rotation = 180°
- West + 180° counterclockwise = East

ANSWER 6: B

---

**Problem 7:**

**What's being asked:** After adding 5 liters of yellow tint to a 30-liter mixture, what percent is yellow tint?

**Original mixture (30 liters):**
- Yellow tint: 30% × 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percentage:**
$$\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%$$

ANSWER 7: C

---

**Problem 8:**

**What's being asked:** Order 15/11, 19/15, and 17/13 from least to greatest.

**Setup:** Convert to decimals or compare directly.

**Calculations:**
- 15/11 ≈ 1.364
- 19/15 ≈ 1.267
- 17/13 ≈ 1.308

**Order:** 19/15 < 17/13 < 15/11

ANSWER 8: E

---

**Problem 9:**

**What's being asked:** Find $x$ such that the mean, median, and unique mode of {3, 4, 5, 6, 6, 7, $x$} are all equal.

**Current mode:** 6 appears twice; all others appear once. If there's a unique mode, it must be 6.

**With unique mode = 6:** This requires no other number to appear more than once. So $x \notin \{3, 4, 5, 7\}$ and $x \neq 6$ (else it wouldn't be unique).

**Mean = Median = Mode = 6:**

For median of 7 numbers with value 6:
- Sorted: need middle (4th) value to be 6
- If $x = 11$: {3, 4, 5, 6, 6, 7, 11} → median = 6 ✓

Mean: $\frac{3+4+5+6+6+7+11}{7} = \frac{42}{7} = 6$ ✓

Mode: 6 appears twice; all others once → mode = 6 ✓

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Total time from home to school if Joe walked halfway (6 min) then ran the rest at 3× speed?

**Setup:**
- Walking: 6 minutes for half the distance
- Running: same distance at 3× speed, so time = 6/3 = 2 minutes

**Total time:** 6 + 2 = 8 minutes

ANSWER 10: D

---

**Problem 11:**

**What's being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Calculation:**
- First group: 1 + 11 + 21 + 31 + 41 = 105
- Second group: 9 + 19 + 29 + 39 + 49 = 145
- Sum: 105 + 145 = 250

ANSWER 11: E

---

**Problem 12:**

**What's being asked:** Which number is largest?

**Comparison:**
- A. 0.99
- B. 0.9099
- C. 0.9
- D. 0.909
- E. 0.9009

Starting from the tenths place: all are 0.9__. Looking at hundredths:
- 0.99 has 9 in hundredths place → **largest**

ANSWER 12: A

Problem 1:

In a single-elimination tournament, each game eliminates exactly one team.  
There are 16 teams, and we want one champion.

So we must eliminate 15 teams total.

Therefore, the number of games played is 15.

Tempting wrong choice: 16 would be if every team had to lose, but the winner does not lose.

ANSWER 1: D

Problem 2:

Compute each person’s result carefully.

- Jose: start with 10  
  subtract 1 → 9  
  double → 18  
  add 2 → 20

- Thuy: start with 10  
  double → 20  
  subtract 1 → 19  
  add 2 → 21

- Kareem: start with 10  
  subtract 1 → 9  
  add 2 → 11  
  double → 22

Compare:
- Jose = 20
- Thuy = 21
- Kareem = 22

Kareem gets the largest final answer.

ANSWER 2: C

Problem 3:

\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

So the correct decimal is 0.08.

ANSWER 3: B

Problem 4:

We are choosing 3 starters from 4 players: Lance, Sally, Joy, and Fred.

This is a combination problem:
\[
\binom{4}{3} = 4
\]

You can also see it by noticing that choosing 3 starters is the same as choosing which 1 player sits out, and there are 4 choices.

ANSWER 4: B

Problem 5:

Let the number be \(n\).

20% of \(n\) is 12, so:
\[
0.2n = 12
\]
\[
n = 60
\]

Now find 30% of 60:
\[
0.3 \times 60 = 18
\]

ANSWER 5: B

Problem 6:

The spinner starts pointing west.

First move: clockwise \(2\frac{1}{4}\) revolutions.  
A full revolution brings it back to west, so only the extra \(\frac{1}{4}\) revolution matters.

From west, a quarter-turn clockwise points it to north.

Second move: counterclockwise \(3\frac{3}{4}\) revolutions.  
Again, the 3 full revolutions do nothing, so this is just \(\frac{3}{4}\) of a revolution counterclockwise, which is the same as \(\frac{1}{4}\) of a revolution clockwise.

From north, a quarter-turn clockwise points east.

So the spinner ends pointing east.

ANSWER 6: B

Problem 7:

Start with 30 liters total.

Yellow tint is 30% of 30 liters:
\[
0.30 \times 30 = 9 \text{ liters}
\]

Then 5 liters of yellow tint are added, so yellow becomes:
\[
9 + 5 = 14 \text{ liters}
\]

New total mixture:
\[
30 + 5 = 35 \text{ liters}
\]

Percent yellow in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.4 = 40\%
\]

ANSWER 7: C

Problem 8:

The fractions are:
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]

Approximate each one:

\[
\frac{15}{11} \approx 1.364
\]
\[
\frac{17}{13} \approx 1.308
\]
\[
\frac{19}{15} \approx 1.267
\]

So from least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

That matches choice E.

ANSWER 8: E

Problem 9:

The numbers are:
\[
3,4,5,6,6,7,x
\]

We are told the mean, median, and unique mode are all equal.

Since 6 appears twice and every other listed number appears once, the unique mode is 6.

So the median and mean must also be 6.

The median of 7 numbers is the 4th number when ordered.  
The sorted list is:
\[
3,4,5,6,6,7,x
\]
To make the median 6, \(x\) must be 6 or larger, but we also need the mean to be 6.

Sum of the known numbers:
\[
3+4+5+6+6+7 = 31
\]

For the mean to be 6, total sum must be:
\[
7 \times 6 = 42
\]

So:
\[
31 + x = 42
\]
\[
x = 11
\]

Check:
- mode = 6
- median = 6
- mean = 42/7 = 6

Works perfectly.

ANSWER 9: D

Problem 10:

Joe walked half the distance in 6 minutes.  
He then ran the other half at 3 times his walking speed.

If he runs 3 times as fast, then the time for the same distance is 1/3 as long.

So the running time is:
\[
6 \div 3 = 2 \text{ minutes}
\]

Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

Problem 11:

Add the first group:
\[
1+11+21+31+41 = 105
\]

Add the second group:
\[
9+19+29+39+49 = 145
\]

Now add the totals:
\[
105 + 145 = 250
\]

ANSWER 11: E

Problem 12:

Compare the decimals:

- \(0.99\)
- \(0.9099\)
- \(0.9 = 0.9000\)
- \(0.909\)
- \(0.9009\)

The largest is clearly \(0.99\), since it is closest to 1.

ANSWER 12: A

Problem 1:  
We have a three-person-team tournament where **the losing team of each game is eliminated**. With **16 teams**, how many games are played to determine the winner?

To find the number of games in a single-elimination tournament:  
- Each game eliminates exactly **one** team (the loser).  
- To go from **16 teams to 1 winning team**, we must eliminate **15 teams**.  
- Therefore, the number of games is **15**.

So the correct choice is **D**.

ANSWER 1: D


Problem 2:  
Jose, Thuy, and Kareem all start at 10. We compute each final value.

**Jose:**  
- Subtract 1: \(10 - 1 = 9\)  
- Double: \(2 \cdot 9 = 18\)  
- Add 2: \(18 + 2 = 20\)  
Final: **20**

**Thuy:**  
- Double 10: \(2 \cdot 10 = 20\)  
- Subtract 1: \(20 - 1 = 19\)  
- Add 2: \(19 + 2 = 21\)  
Final: **21**

**Kareem:**  
- Subtract 1: \(10 - 1 = 9\)  
- Add 2: \(9 + 2 = 11\)  
- Double: \(2 \cdot 11 = 22\)  
Final: **22**

Largest is **22**, which is Kareem.

ANSWER 2: C


Problem 3:  
We need to evaluate:  
\[
\frac{2}{25}
\]
Convert to decimals:  
\[
\frac{2}{25} = 2 \div 25
\]
Since \(25 \cdot 0.08 = 2\):  
\[
2 \div 25 = 0.08
\]
So the correct choice is **B** (.08).

ANSWER 3: B


Problem 4:  
We have 4 people chosen: Lance, Sally, Joy, Fred. We need to choose **3 starters**.

Number of ways:
\[
\binom{4}{3} = 4
\]
List them conceptually: leaving out each one individually gives 4 choices.

Correct choice: **B**.

ANSWER 4: B


Problem 5:  
“If 20% of a number is 12, what is 30% of the same number?”

Let the number be \(N\).  
\[
0.20N = 12
\]
\[
N = \frac{12}{0.20} = 60
\]
Now compute \(30\%\) of 60:
\[
0.30 \cdot 60 = 18
\]
Correct choice: **B**.

ANSWER 5: B


Problem 6:  
Spinner starts pointing **west**.

Chenille moves it:
- **Clockwise** \(2\frac{1}{4}\) revolutions  
- then **Counterclockwise** \(3\frac{3}{4}\) revolutions

Net rotation (clockwise minus counterclockwise):
\[
2\frac{1}{4} - 3\frac{3}{4}
\]
Convert to improper fractions:
\[
2\frac{1}{4} = \frac{9}{4}, \quad 3\frac{3}{4} = \frac{15}{4}
\]
\[
\frac{9}{4} - \frac{15}{4} = -\frac{6}{4} = -\frac{3}{2}
\]
The negative means **net counterclockwise** by \(\frac{3}{2}\) revolutions.

\(\frac{3}{2} = 1 + \frac{1}{2}\), so this is:
- 1 full turn (brings it back where it started)
- plus \(\frac{1}{2}\) turn counterclockwise

Half a revolution from **west** points **east** (opposite direction).

Correct answer: **B (east)**.

ANSWER 6: B


Problem 7:  
Original mixture: 30 L with  
- 25% red (doesn’t matter directly)
- 30% yellow  
- 45% water

Yellow amount initially:
\[
0.30 \cdot 30 = 9 \text{ liters}
\]
Add 5 liters of yellow tint, so new yellow amount:
\[
9 + 5 = 14
\]
New total volume:
\[
30 + 5 = 35
\]
Percent yellow in new mixture:
\[
\frac{14}{35} = 0.4 = 40\%
\]
Correct choice: **C**.

ANSWER 7: C


Problem 8:  
Order fractions least to greatest:
\[
\frac{15}{11},\ \frac{19}{15},\ \frac{17}{13}
\]
A good way: compare values by cross-multiplication or decimal approximations.

Compute approximate decimals:
\[
\frac{15}{11} \approx 1.3636
\]
\[
\frac{19}{15} \approx 1.2667
\]
\[
\frac{17}{13} \approx 1.3077
\]
Least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]
Check choice: that matches **E**:
- \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11}\)

ANSWER 8: E


Problem 9:  
Numbers: \(3,4,5,6,6,7,x\) (7 positive integers).  
Mean = median = unique mode.

**Step 1: Mode**  
The only number that repeats is **6**, occurring twice. So the **unique mode = 6**.

Thus median must be 6 as well.

**Step 2: Median = 6**  
For 7 numbers sorted, the median is the **4th** number.

Sorted list must have the 4th term equal to 6. We already have two 6’s and surrounding numbers 3,4,5,7.

Currently without \(x\), the ordered portion is:
\[
3,4,5,6,6,7,\ ?
\]
The 4th number here is 6 already. To keep the 4th number as 6, \(x\) must not be less than 6 (otherwise it would shift positions so 4th might change).

If \(x = 6\), then 6 occurs 3 times, still mode 6 (unique mode). Then list:
\[
3,4,5,6,6,6,7
\]
Median (4th) is 6—works.

If \(x=7\), list:
\[
3,4,5,6,6,7,7
\]
Mode would be 6? Actually 7 repeats too, so mode not unique.

If \(x>7\), list:
\[
3,4,5,6,6,7,x
\]
Median still 6, but mode would still be 6 (unique), since only 6 repeats. That would also work—wait, but then 6 is still unique mode, yes. However we also need **mean = 6**.

So use mean condition.

Sum of known numbers:
\[
3+4+5+6+6+7 = 31
\]
Total sum including \(x\): \(31 + x\)

Mean:
\[
\frac{31+x}{7} = 6
\]
\[
31 + x = 42
\]
\[
x = 11
\]

Check uniqueness of mode: 6 occurs twice, no other repeats (since x=11 is different). Unique mode is 6—consistent.

ANSWER 9: D


Problem 10:  
Joe walks half the distance, then runs the rest.  
He runs **3 times as fast** as he walks.

Given: time to walk half way = **6 minutes**.  
Let walking speed be \(v\). Then walking time for half distance \(=\) 6 minutes.

Let half the distance be \(d\). Then:
\[
d = v \cdot 6
\]
Running speed is \(3v\).  
Remaining distance is also \(d\), so running time:
\[
\text{time} = \frac{d}{3v} = \frac{v\cdot 6}{3v} = 2
\]
Total time:
\[
6 + 2 = 8
\]
Correct choice: **D**.

ANSWER 10: D


Problem 11:  
Compute:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]

Group each set:

First set:
\[
1+11+21+31+41
\]
These are all of the form \(10k+1\):  
\(1, 11, 21, 31, 41\) correspond to \(k=0,1,2,3,4\).

Sum:
\[
(1+11+21+31+41) = (1+11+21+31+41)
\]
Compute directly:
\[
(1+11)=12,\ (21+31)=52,\ 12+52=64,\ 64+41=105
\]

Second set:
\[
9+19+29+39+49
\]
Direct sum:
\[
(9+19)=28,\ (29+39)=68,\ 28+68=96,\ 96+49=145
\]

Total:
\[
105 + 145 = 250
\]

Correct choice: **E**.

ANSWER 11: E


Problem 12:  
Which number is largest?

Options:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare decimals:  
- .99 is the largest among all because it is greater than .9 (and greater than any number starting 0.90... but with fewer 9s afterward).

So .99 > .9099 > .909 > .9009 > .9.

Correct choice: **A**.

ANSWER 12: A

Problem 1:
In a single-elimination tournament, every game results in exactly one team being eliminated. To determine a single winner from 16 teams, 15 teams must be eliminated. Since each game eliminates one team, it takes exactly 15 games.
ANSWER 1: D

Problem 2:
- Jose: (10 - 1) = 9; 9 * 2 = 18; 18 + 2 = 20.
- Thuy: (10 * 2) = 20; 20 - 1 = 19; 19 + 2 = 21.
- Kareem: (10 - 1) = 9; 9 + 2 = 11; 11 * 2 = 22.
Kareem has the largest answer (22).
ANSWER 2: C

Problem 3:
To convert 2/25 to a decimal, multiply the numerator and denominator by 4 to get a denominator of 100: (2 * 4) / (25 * 4) = 8 / 100 = 0.08.
ANSWER 3: B

Problem 4:
We need to choose 3 starters from a group of 4 (Lance, Sally, Joy, Fred). This is a combination problem denoted as 4C3. Choosing 3 people to be on the team is the same as choosing 1 person to be left off the team. There are 4 ways to choose who is left off.
ANSWER 4: B

Problem 5:
Let $x$ be the number. $0.20x = 12$. Dividing by 0.20, $x = 12 / 0.2 = 60$. We want 30% of $x$: $0.30 * 60 = 18$.
ANSWER 5: B

Problem 6:
Start at 0 revolutions. Clockwise is positive, counterclockwise is negative.
$2 \frac{1}{4} - 3 \frac{3}{4} = 2.25 - 3.75 = -1.5$ revolutions.
-1.5 revolutions is the same as -0.5 revolutions (since -1 is a full circle). -0.5 revolutions (counterclockwise) from West points to East.
ANSWER 6: B

Problem 7:
Original yellow tint: 30% of 30 liters = 9 liters.
New mixture: 30 + 5 = 35 liters.
New yellow tint: 9 + 5 = 14 liters.
Percent: (14 / 35) * 100 = (2 / 5) * 100 = 40%.
ANSWER 7: C

Problem 8:
Convert to decimals or compare differences from 1:
15/11 = 1 + 4/11 ≈ 1.3636
19/15 = 1 + 4/15 ≈ 1.2666
17/13 = 1 + 4/13 ≈ 1.3076
Comparing the fractions 4/11, 4/13, and 4/15: the larger the denominator, the smaller the fraction. Thus, 4/15 < 4/13 < 4/11.
Order: 19/15 < 17/13 < 15/11.
ANSWER 8: E

Problem 9:
The set is {3, 4, 5, 6, 6, 7, x}. The mode is 6 (it appears twice). For the mean and median to be 6:
Median: With 7 numbers, the 4th number must be 6. Since the sorted list is 3, 4, 5, 6, 6, 7, x, if $x > 6$, the 4th number is 6.
Mean: (3+4+5+6+6+7+x) / 7 = 6 => 31 + x = 42 => x = 11.
Checking: {3, 4, 5, 6, 6, 7, 11}. Median is 6. Mean is 42/7 = 6. Mode is 6.
ANSWER 9: D

Problem 10:
Walking time = 6 minutes. Running speed = 3 * walking speed.
Time taken = Distance / Speed. Since speed is 3 times faster, time is 1/3 of the walking time.
Running time = 6 / 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
ANSWER 10: D

Problem 11:
Sum 1: (1+11+21+31+41) = 105.
Sum 2: (9+19+29+39+49) = 145.
105 + 145 = 250.
ANSWER 11: E

Problem 12:
Compare digit by digit:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
0.99 is the largest.
ANSWER 12: A

Problem 1:
This is a single-elimination tournament with 16 teams. Exactly one team is eliminated per game, and 15 teams must be eliminated to leave a single winner. Thus exactly 15 games are required.

ANSWER 1: D

Problem 2:
Jose evaluates \((10-1)\times2+2=20\).
Thuy evaluates \((10\times2)-1+2=21\).
Kareem evaluates \((10-1+2)\times2=22\).
Kareem obtains the strictly largest value.

ANSWER 2: C

Problem 3:
Division yields the decimal expansion \(2\div25=0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters out of 4 distinct players is the binomial coefficient \(\binom{4}{3}=4\).

ANSWER 4: B

Problem 5:
If 20% of a number equals 12, the number itself is \(12/0.2=60\). Then 30% of 60 is \(0.3\times60=18\).

ANSWER 5: B

Problem 6:
A net rotation of \(+2.25-3.75=-1.5\) revolutions is equivalent to 1.5 revolutions counterclockwise. From an initial westward orientation, 1.5 revolutions counterclockwise lands exactly on east.

ANSWER 6: B

Problem 7:
The original mixture contains \(0.30\times30=9\) liters of yellow tint. Adding 5 liters raises the yellow volume to 14 liters and the total volume to 35 liters, so the new percentage is \(14/35=40\%\).

ANSWER 7: C

Problem 8:
Decimal approximations are \(15/11\approx1.3636\), \(19/15\approx1.2667\), and \(17/13\approx1.3077\). Ordering these values produces \(19/15<17/13<15/11\).

ANSWER 8: E

Problem 9:
The unique mode must be 6, forcing the mean also to equal 6. The sum of the seven numbers is therefore 42. The six given numbers sum to 31, so \(x=11\). With this value the ordered list is 3,4,5,6,6,7,11 whose median is likewise 6 and whose unique mode remains 6.

ANSWER 9: D

Problem 10:
Let the total distance be \(2d\). Walking time for distance \(d\) is 6 min, so running time for the remaining distance \(d\) at triple speed is \(6/3=2\) min. The total time is therefore 8 min.

ANSWER 10: D

Problem 11:
Each pair sums to a multiple of 10: \(1+9=10\), \(11+19=30\), \(21+29=50\), \(31+39=70\), \(41+49=90\). Their total is 250. Equivalently each five-term arithmetic series sums to 105 and 145 respectively, again yielding 250.

ANSWER 11: E

Problem 12:
Direct comparison of the five decimals shows that 0.99 is strictly larger than each of the others.

ANSWER 12: A

Problem 1:
The problem asks us to determine the number of games that will be played to determine the winner of a basketball tournament with 16 teams, where the losing team of each game is eliminated.

## Step 1
To determine the number of games needed to find the winner, we first note that in a single-elimination tournament, each game eliminates one team.

## Step 2
Since there can only be one winner, all other teams must be eliminated. Therefore, the number of games required is equal to the number of teams minus one.

## Step 3
We have 16 teams competing in the tournament. So, the number of games needed to determine the winner is 16 - 1 = 15.

## Step 4
Let's examine the given choices to rule out the incorrect options. A. 4 is too low because it doesn't account for all the teams. B. 7 is also too low for the same reason. C. 8 is still too low. D. 15 matches our calculation. E. 16 is one more than our calculation.

The final answer is supported by our step-by-step reasoning.
ANSWER 1: D

Problem 2:
The problem asks us to determine who among Jose, Thuy, and Kareem gets the largest final answer after performing a series of operations starting with the number 10.

## Step 1
Let's calculate Jose's final answer: Jose subtracts 1 from 10 to get 9, doubles it to get 18, and then adds 2 to get 20.

## Step 2
Now, let's calculate Thuy's final answer: Thuy doubles 10 to get 20, subtracts 1 to get 19, and then adds 2 to get 21.

## Step 3
Next, let's calculate Kareem's final answer: Kareem subtracts 1 from 10 to get 9, adds 2 to get 11, and then doubles it to get 22.

## Step 4
Comparing the final answers: Jose got 20, Thuy got 21, and Kareem got 22. The largest final answer is 22, which belongs to Kareem.

## Step 5
Let's examine the given choices. A. Jose is incorrect because his answer is 20. B. Thuy is incorrect because her answer is 21. C. Kareem is correct because his answer is 22. D. Jose and Thuy is incorrect because Kareem's answer is larger. E. Thuy and Kareem is incorrect because Jose's answer is not the largest.

The final answer is supported by our step-by-step reasoning.
ANSWER 2: C

Problem 3:
The problem asks us to convert the fraction 2/25 into a decimal.

## Step 1
To convert 2/25 into a decimal, we divide the numerator by the denominator: 2 ÷ 25 = 0.08.

## Step 2
Let's examine the given choices to identify the correct decimal representation. A. .008 is too small. B. .08 matches our calculation. C. .8 is too large. D. 1.25 is much too large. E. 12.5 is also much too large.

The final answer is supported by our step-by-step reasoning.
ANSWER 3: B

Problem 4:
The problem asks us to determine the number of ways to choose 3 starters from a team of 4 members: Lance, Sally, Joy, and Fred.

## Step 1
This is a combination problem, because the order in which the starters are chosen does not matter.

## Step 2
The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial, the product of all positive integers up to that number.

## Step 3
Here, n = 4 (total team members) and k = 3 (number of starters to be chosen). So, C(4, 3) = 4! / [3!(4 - 3)!] = 4! / (3! * 1!) = (4 * 3 * 2 * 1) / ((3 * 2 * 1) * 1) = 4.

## Step 4
Let's examine the given choices. A. 2 is too low. B. 4 matches our calculation. C. 6 is too high. D. 8 is also too high. E. 10 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 4: B

Problem 5:
The problem asks us to find 30% of a number, given that 20% of the same number is 12.

## Step 1
First, we need to find the number. We know that 20% of the number is 12. So, 0.20 * number = 12.

## Step 2
To find the number, we divide both sides of the equation by 0.20: number = 12 / 0.20 = 60.

## Step 3
Now, we need to calculate 30% of 60: 0.30 * 60 = 18.

## Step 4
Let's examine the given choices. A. 15 is too low. B. 18 matches our calculation. C. 20 is too high. D. 24 is also too high. E. 30 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 5: B

Problem 6:
The problem asks us to determine the direction the spinner points after Chenille moves it clockwise 2 1/4 revolutions and then counterclockwise 3 3/4 revolutions, starting from pointing west.

## Step 1
First, let's understand the effect of the movements. A full revolution is 360 degrees or 4 quarters. So, 2 1/4 revolutions clockwise is equivalent to 9 quarters clockwise.

## Step 2
9 quarters clockwise is the same as 1 quarter clockwise (since 8 quarters make 2 full revolutions, which brings the spinner back to the original direction). So, after the first move, the spinner points north.

## Step 3
Then, Chenille moves it counterclockwise 3 3/4 revolutions, which is equivalent to 15 quarters counterclockwise.

## Step 4
15 quarters counterclockwise is the same as 3 quarters counterclockwise (since 12 quarters make 3 full revolutions). So, moving 3 quarters counterclockwise from north results in pointing west.

## Step 5
Let's examine the given choices. A. north is incorrect because after the second move, it's not north. B. east is incorrect. C. south is incorrect. D. west matches our calculation. E. northwest is incorrect.

The final answer is supported by our step-by-step reasoning.
ANSWER 6: D

Problem 7:
The problem asks us to find the percentage of yellow tint in a new mixture after adding 5 liters of yellow tint to an original 30-liter mixture that is 25% red tint, 30% yellow tint, and 45% water.

## Step 1
First, we calculate the amount of yellow tint in the original mixture: 30% of 30 liters = 0.30 * 30 = 9 liters.

## Step 2
After adding 5 liters of yellow tint, the total amount of yellow tint becomes 9 + 5 = 14 liters.

## Step 3
The total volume of the new mixture is 30 + 5 = 35 liters.

## Step 4
To find the percentage of yellow tint in the new mixture, we divide the amount of yellow tint by the total volume and multiply by 100: (14 / 35) * 100 = 40%

Problem 1:
In a single-elimination tournament, each game eliminates exactly one team. To eliminate all but the champion, the number of games needed is one fewer than the number of teams. With 16 teams, the number of games is \(16 - 1 = 15\).

ANSWER 1: D

Problem 2:
Jose: \(10 - 1 = 9\), \(9 \times 2 = 18\), \(18 + 2 = 20\).
Thuy: \(10 \times 2 = 20\), \(20 - 1 = 19\), \(19 + 2 = 21\).
Kareem: \(10 - 1 = 9\), \(9 + 2 = 11\), \(11 \times 2 = 22\).
Kareem’s final answer of 22 is the largest.

ANSWER 2: C

Problem 3:
Divide 2 by 25: \(2 \div 25 = 0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters from 4 people is the combination \(\binom{4}{3} = 4\).

ANSWER 4: B

Problem 5:
Let the number be \(n\). \(20\%\) of \(n\) is \(0.2n = 12\), so \(n = 60\). Then \(30\%\) of \(60 = 0.3 \times 60 = 18\).

ANSWER 5: B

Problem 6:
Initial direction: west.
Clockwise \(2\frac{1}{4}\) revolutions: \(2\frac{1}{4} = 2.25\) revolutions clockwise.
Counterclockwise \(3\frac{3}{4}\) revolutions: \(3\frac{3}{4} = 3.75\) revolutions counterclockwise.
Net revolution = \(2.25 - 3.75 = -1.5\) revolutions, i.e., \(1.5\) revolutions counterclockwise.
A full revolution (1.0) returns to west. The remaining 0.5 revolution counterclockwise from west is a half-turn, which points to east.

ANSWER 6: B

Problem 7:
Original mixture: 30 liters, yellow tint = \(30\% \times 30 = 9\) liters.
Add 5 liters of yellow tint: new yellow = \(9 + 5 = 14\) liters; new total volume = \(30 + 5 = 35\) liters.
Percent yellow = \(\frac{14}{35} \times 100\% = 40\%\).

ANSWER 7: C

Problem 8:
Convert each fraction to a decimal for comparison:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
Order from least to greatest: \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

ANSWER 8: E

Problem 9:
The data set: 3, 4, 5, 6, 6, 7, \(x\). The mode is the most frequent number. With two 6’s, 6 is the mode unless \(x\) equals another number that appears twice, which would create a second mode, violating the “unique mode” condition. Therefore, \(x\) must not be 3, 4, 5, or 7. It could be 6, but if \(x=6\), the median is 6, while the mean is \((3+4+5+6+6+6+7)/7 = 37/7 \approx 5.29\), so mean ≠ median.  
For the mean to equal the median (which is 6, the middle of seven sorted numbers if \(x > 6\) or \(x=6\) only if placed correctly), we set the mean to 6. Sum without \(x\) = \(3+4+5+6+6+7 = 31\). Total needed for mean 6 = \(7 \times 6 = 42\). Thus \(x = 42 - 31 = 11\).  
Sorted: 3, 4, 5, 6, 6, 7, 11. Median = 6, mode = 6, mean = 6. Unique mode holds.

ANSWER 9: D

Problem 10:
Walking half the distance took 6 minutes, so walking speed = \(\frac{\text{half distance}}{6}\).
Running speed is 3 times walking speed. Time to run the second half = \(\frac{\text{half distance}}{3 \times (\text{half distance}/6)} = \frac{6}{3} = 2\) minutes.
Total time = \(6 + 2 = 8\) minutes.

ANSWER 10: D

Problem 11:
Pair the terms: \(1+49 = 50\), \(11+39 = 50\), \(21+29 = 50\), \(31+19 = 50\), \(41+9 = 50\). Five pairs of 50 give \(5 \times 50 = 250\).

ANSWER 11: E

Problem 12:
Compare by extending decimals to the same number of places:
\(0.99   = 0.9900\)
\(0.9099 = 0.9099\)
\(0.9    = 0.9000\)
\(0.909  = 0.9090\)
\(0.9009 = 0.9009\)
Clearly \(0.99 > 0.9099 > 0.909 > 0.9009 > 0.9\). The largest is \(0.99\).

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key Insight:** In a single-elimination tournament, every game eliminates exactly one team. To crown a single winner from 16 teams, we must eliminate 16 − 1 = 15 teams. Therefore, exactly 15 games are needed.

**Verification by rounds:**
- Round 1: 16 teams → 8 games → 8 teams left
- Round 2: 8 teams → 4 games → 4 teams left
- Round 3: 4 teams → 2 games → 2 teams left
- Round 4 (Final): 2 teams → 1 game → 1 winner

Total = 8 + 4 + 2 + 1 = 15 games.

Choices A (4), B (7), and C (8) are too small. Choice E (16) would mean every team is eliminated, leaving no winner.

ANSWER 1: D

---

Problem 2:
We track each person's calculation step by step, all starting from 10.

**Jose:** subtract 1 → 9; double → 18; add 2 → **20**
**Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
**Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

Comparing: 20 < 21 < 22. Kareem gets the largest final answer.

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

Choice A (.008) would be 8/1000. Choice C (.8) would be 8/10. Choices D and E are greater than 1, but 2/25 is clearly less than 1.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination problem — the order of selection doesn't matter:
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4$$

Equivalently, choosing 3 people to start is the same as choosing 1 person to sit out. There are 4 people, so 4 ways to pick who sits out.

The 4 possible starting lineups are: {L,S,J}, {L,S,F}, {L,J,F}, {S,J,F}.

ANSWER 4: B

---

Problem 5:
We're told 20% of a number is 12. Let the number be $x$.

$$0.20x = 12 \implies x = \frac{12}{0.20} = 60$$

Now find 30% of 60:
$$0.30 \times 60 = 18$$

**Shortcut:** 30% is $\frac{30}{20} = 1.5$ times 20%, so the answer is $1.5 \times 12 = 18$.

Choice A (15) would be 25%. Choice D (24) would be 40%. Choice E (30) would be 50%.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing **West**.

**Move 1:** Clockwise $2\frac{1}{4}$ revolutions.
- The 2 full revolutions bring it back to West.
- The extra $\frac{1}{4}$ revolution clockwise from West: West → **North**.

**Move 2:** Counterclockwise $3\frac{3}{4}$ revolutions.
- The 3 full revolutions bring it back to North.
- The extra $\frac{3}{4}$ revolution counterclockwise from North:
  - $\frac{1}{4}$ CCW from North → West
  - $\frac{2}{4}$ CCW from North → South
  - $\frac{3}{4}$ CCW from North → **East**

The spinner points East.

ANSWER 6: B

---

Problem 7:
Original mixture: 30 liters total.
- Red tint: 25% of 30 = 7.5 liters
- Yellow tint: 30% of 30 = 9 liters
- Water: 45% of 30 = 13.5 liters

Add 5 liters of yellow tint:
- New yellow tint = 9 + 5 = 14 liters
- New total mixture = 30 + 5 = 35 liters

Percent yellow tint in new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

Choice B (35%) is a trap for those who just add 5% to 30%. Choice D (45%) is the original water percentage.

ANSWER 7: C

---

Problem 8:
We compare $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$.

Rewrite each as a mixed number:
$$\frac{15}{11} = 1 + \frac{4}{11}, \quad \frac{19}{15} = 1 + \frac{4}{15}, \quad \frac{17}{13} = 1 + \frac{4}{13}$$

Since all three have the same numerator (4) in their fractional parts, we compare denominators. For fractions with equal numerators, a **larger denominator** means a **smaller fraction**:
$$15 > 13 > 11 \implies \frac{4}{15} < \frac{4}{13} < \frac{4}{11}$$

Therefore:
$$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$$

ANSWER 8: E

---

Problem 9:
The data set is: 3, 4, 5, 6, 6, 7, and $x$. The mean, median, and unique mode are all equal.

**Mode:** Since 6 already appears twice and all others appear once, the unique mode must be 6 (if $x$ were 3, 4, 5, or 7, we'd have two modes, violating "unique mode"). So mode = 6.

**Mean = 6:**
$$\frac{3 + 4 + 5 + 6 + 6 + 7 + x}{7} = 6$$
$$\frac{31 + x}{7} = 6 \implies 31 + x = 42 \implies x = 11$$

**Check Median:** With $x = 11$, the sorted list is: 3, 4, 5, **6**, 6, 7, 11. The median (4th value) is 6. ✓

All three measures equal 6. Choice B (6) would make 6 appear three times, but the mean would be $(31+6)/7 = 37/7 \neq 6$.

ANSWER 9: D

---

Problem 10:
Joe walked half the distance in 6 minutes. He then ran the remaining half at 3 times his walking speed.

Let walking speed = $v$. The half-distance = $6v$.

Running speed = $3v$. Time to run the second half:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{6v}{3v} = 2 \text{ minutes}$$

Total time = 6 + 2 = **8 minutes**.

Choice A (7) would result from incorrectly subtracting. Choices B, C, E involve unnecessary decimal complications.

ANSWER 10: D

---

Problem 11:
Compute $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$.

Pair the terms from the first group with terms from the second group (first with last, etc.):
- $1 + 49 = 50$
- $11 + 39 = 50$
- $21 + 29 = 50$
- $31 + 19 = 50$
- $41 + 9 = 50$

Total = $50 \times 5 = 250$.

Choice C (200) is a common error from miscounting pairs. Choice D (249) comes from arithmetic mistakes.

ANSWER 11: E

---

Problem 12:
Compare the five decimals by writing them to four decimal places:

- A: 0.9900
- B: 0.9099
- C: 0.9000
- D: 0.9090
- E: 0.9009

Comparing digit by digit from the tenths place: all have 9 in the tenths place. In the hundredths place, A has **9** while all others have **0**. So A is immediately the largest.

$0.99 > 0.9099 > 0.909 > 0.9009 > 0.9$

ANSWER 12: A

 **Problem 1:**
We need to find the number of games played in a single-elimination tournament with 16 teams. In single elimination, every game eliminates exactly one team. To determine a winner, 15 of the 16 teams must be eliminated. Therefore, exactly 15 games are required.  
(Another way: with 16 teams there are 8 games in the first round, 4 in the second, 2 in the third, and 1 final game: \(8+4+2+1=15\).)

ANSWER 1: D

---

**Problem 2:**
We compute each person’s final answer step by step, starting from 10.

- **Jose:**  
  \(10 - 1 = 9\)  
  \(9 \times 2 = 18\)  
  \(18 + 2 = 20\)

- **Thuy:**  
  \(10 \times 2 = 20\)  
  \(20 - 1 = 19\)  
  \(19 + 2 = 21\)

- **Kareem:**  
  \(10 - 1 = 9\)  
  \(9 + 2 = 11\)  
  \(11 \times 2 = 22\)

Comparing the results: \(20 < 21 < 22\). Kareem’s answer is the largest.

ANSWER 2: C

---

**Problem 3:**
We convert the fraction \(\frac{2}{25}\) to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100:
\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

ANSWER 3: B

---

**Problem 4:**
We must choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways to choose 3 from 4 is the combination \(\binom{4}{3}\):
\[
\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4
\]
Equivalently, choosing 3 starters is the same as choosing 1 person to sit out, and there are 4 choices for who sits out.

ANSWER 4: B

---

**Problem 5:**
Let the number be \(N\). We are given \(0.20N = 12\). Solving for \(N\):
\[
N = \frac{12}{0.20} = 60
\]
Now find 30% of 60:
\[
0.30 \times 60 = 18
\]
(Shortcut: 30% is \(1.5\) times 20%, so \(1.5 \times 12 = 18\).)

ANSWER 5: B

---

**Problem 6:**
The spinner starts pointing west.  
- Clockwise \(2\frac14\) revolutions: 2 full revolutions bring it back to west, then \(\frac14\) turn clockwise from west points it **north**.  
- Counterclockwise \(3\frac34\) revolutions: 3 full revolutions bring it back to north, then \(\frac34\) turn counterclockwise from north goes through west, south, and ends at **east**.

Net movement: \(3\frac34 - 2\frac14 = 1\frac12\) revolutions counterclockwise. One full revolution returns to west; an additional \(\frac12\) revolution counterclockwise from west points to **east**.

ANSWER 6: B

---

**Problem 7:**
Original mixture: 30 liters.  
- Yellow tint originally: \(30\% \times 30 = 9\) liters.  
After adding 5 liters of yellow tint:  
- New yellow tint: \(9 + 5 = 14\) liters.  
- New total mixture: \(30 + 5 = 35\) liters.  

Percent of yellow tint in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%
\]

ANSWER 7: C

---

**Problem 8:**
Compare \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\). Each is greater than 1. Write each as \(1\) plus a proper fraction:
\[
\frac{15}{11} = 1 + \frac{4}{11}, \qquad
\frac{19}{15} = 1 + \frac{4}{15}, \qquad
\frac{17}{13} = 1 + \frac{4}{13}
\]
Since the numerators are the same (4), the fraction with the largest denominator is smallest:
\[
\frac{4}{15} < \frac{4}{13} < \frac{4}{11}
\]
Adding 1 preserves the order:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

ANSWER 8: E

---

**Problem 9:**
The data set is \(3, 4, 5, 6, 6, 7, x\).  
- The current mode is 6 (appears twice). For 6 to remain the **unique** mode, \(x\) must not equal 3, 4, 5, or 7 (which would create a second mode).  
- We are told mean = median = mode = 6.  

Set the mean equal to 6:
\[
\frac{3+4+5+6+6+7+x}{7} = 6 \implies \frac{31+x}{7} = 6 \implies 31+x = 42 \implies x = 11
\]

Check: With \(x=11\), the sorted list is \(3, 4, 5, 6, 6, 7, 11\).  
- Median (4th term) = 6.  
- Mode = 6 (unique).  
- Mean = 6.  

All conditions are satisfied.

ANSWER 9: D

---

**Problem 10:**
Joe walks half the distance in 6 minutes. He runs the second half at 3 times his walking speed. Since distance is the same and speed is tripled, the time required is divided by 3:
\[
\text{Running time} = \frac{6}{3} = 2 \text{ minutes}
\]
Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

---

**Problem 11:**
Add the two groups:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]
Pair the terms to make convenient sums:
\[
(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 5 \times 50 = 250
\]

ANSWER 11: E

---

**Problem 12:**
Compare the decimals by writing them with the same number of places:
- A. \(0.9900\)
- B. \(0.9099\)
- C. \(0.9000\)
- D. \(0.9090\)
- E. \(0.9009\)

The tenths digit is 9 for all. Looking at the hundredths digit: A has 9, while B, D have 0 and C, E have 0. Thus \(0.9900\) is clearly the largest.

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key insight:** Each game eliminates exactly one team. To determine a winner from 16 teams, we must eliminate 15 teams (everyone except the champion).

Therefore, exactly 15 games must be played.

*Verification by rounds:* Round 1: 8 games (16→8 teams), Round 2: 4 games (8→4), Round 3: 2 games (4→2), Round 4: 1 game (2→1). Total = 8+4+2+1 = 15.

The tempting wrong answer E (16) counts teams instead of eliminations.

ANSWER 1: D

---

Problem 2:
We need to compute each person's final answer starting from 10.

**Jose:** Start with 10 → subtract 1: 10−1 = 9 → double: 9×2 = 18 → add 2: 18+2 = **20**

**Thuy:** Start with 10 → double: 10×2 = 20 → subtract 1: 20−1 = 19 → add 2: 19+2 = **21**

**Kareem:** Start with 10 → subtract 1: 10−1 = 9 → add 2: 9+2 = 11 → double: 11×2 = **22**

Kareem gets the largest final answer (22).

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

The tempting wrong answer A (0.008) would be 8/1000, and C (0.8) would be 8/10.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination: C(4,3) = 4!/(3! × 1!) = 4

Listing them out: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — that's 4 ways.

The tempting wrong answer C (6) would be C(4,2), choosing who's *not* a starter, but that also equals 4... Actually C(4,2)=6 is wrong here; we're choosing 3 from 4.

ANSWER 4: B

---

Problem 5:
We know 20% of a number is 12. We need to find 30% of the same number.

**Step 1:** Find the number.
$$0.20 \times n = 12 \implies n = \frac{12}{0.20} = 60$$

**Step 2:** Find 30% of 60.
$$0.30 \times 60 = 18$$

*Shortcut:* Since 30% is 1.5 times 20%, the answer is 1.5 × 12 = 18.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing west. We need to find the net rotation.

**Clockwise:** 2¼ revolutions
**Counterclockwise:** 3¾ revolutions

**Net movement:** 2¼ − 3¾ = −1½ revolutions (negative means counterclockwise)

So the spinner moves 1½ revolutions counterclockwise from west.

- 1 full revolution counterclockwise returns to **west**
- ½ revolution counterclockwise from west goes to **east**

(Counterclockwise from west: ¼ turn → south, ½ turn → east)

ANSWER 6: B

---

Problem 7:
We need to find the percent of yellow tint after adding 5 liters of yellow tint.

**Original mixture (30 liters):**
- Yellow tint: 30% of 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percent yellow tint:** (14/35) × 100 = 40%

The tempting wrong answer B (35) might come from simply adding 5% to 30%.

ANSWER 7: C

---

Problem 8:
We need to order 15/11, 19/15, and 17/13 from least to greatest.

Computing each as a decimal:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3077...

From least to greatest: **19/15 < 17/13 < 15/11**

*Alternative method:* Note each is 1 + (numerator − denominator)/denominator:
- 15/11 = 1 + 4/11
- 19/15 = 1 + 4/15
- 17/13 = 1 + 4/13

Since 4/15 < 4/13 < 4/11, the order is confirmed.

ANSWER 8: E

---

Problem 9:
The positive integers 3, 4, 5, 6, 6, 7, and x have equal mean, median, and unique mode.

**Mode:** Since 6 already appears twice and no other number does, the unique mode must be 6. So x ≠ 3, 4, 5, or 7 (which would create a second mode).

**Median:** With 7 numbers, the median is the 4th value when sorted. For the median to be 6, we need x ≥ 6 (so that 6 is the 4th value).

- If x = 6: sorted = 3, 4, 5, 6, 6, 6, 7 → median = 6 ✓
- If x > 7: sorted = 3, 4, 5, 6, 6, 7, x → median = 6 ✓

**Mean:** (3 + 4 + 5 + 6 + 6 + 7 + x)/7 = (31 + x)/7

Setting mean = 6: (31 + x)/7 = 6 → 31 + x = 42 → **x = 11**

**Verification:** Numbers: 3, 4, 5, 6, 6, 7, 11. Mean = 42/7 = 6, Median = 6, Mode = 6 ✓

ANSWER 9: D

---

Problem 10:
Joe walks halfway, then runs the rest. We need total time.

**Walking:** Half the distance takes 6 minutes.

**Running:** He runs 3 times as fast as he walks, so he covers the same distance in 1/3 the time.
- Time to run the second half = 6 ÷ 3 = 2 minutes

**Total time:** 6 + 2 = **8 minutes**

The tempting wrong answers come from setting up the problem incorrectly or mixing up distance and time relationships.

ANSWER 10: D

---

Problem 11:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

**First group:** 1 + 11 + 21 + 31 + 41 = 105

**Second group:** 9 + 19 + 29 + 39 + 49 = 145

**Total:** 105 + 145 = **250**

*Elegant method:* Pair terms: (1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250

ANSWER 11: E

---

Problem 12:
We need to find the largest number among the options. Writing each with the same number of decimal places:

- A. 0.99 = **0.9900**
- B. 0.9099 = 0.9099
- C. 0.9 = 0.9000
- D. 0.909 = 0.9090
- E. 0.9009 = 0.9009

Comparing digit by digit after the decimal point:
- First digit: all are 9
- Second digit: A has **9**, all others have 0

Since A has a 9 in the second decimal place while all others have 0, **0.99 is the largest**.

ANSWER 12: A

**Problem 1:**  
The tournament is single‑elimination. Starting with 16 teams, every game eliminates exactly one team. To reduce from 16 teams to a single champion we must eliminate 15 teams, and each elimination corresponds to one game.  

Number of games = 16 − 1 = **15**.  

ANSWER 1: D  

---

**Problem 2:**  
Start with 10 for each person.

- **Jose:**  
  Subtract 1 → 9; double → 18; add 2 → 20.  

- **Thuy:**  
  Double → 20; subtract 1 → 19; add 2 → 21.  

- **Kareem:**  
  Subtract 1 → 9; add 2 → 11; double → 22.  

The largest final value is 22, obtained by Kareem.  

ANSWER 2: C  

---

**Problem 3:**  

\[
\frac{2}{25}=0.08
\]

(2 ÷ 25 = 0.08).  

ANSWER 3: B  

---

**Problem 4:**  
We need to choose 3 starters from the 4 available players:  

\[
\binom{4}{3}=4
\]

ways.  

ANSWER 4: B  

---

**Problem 5:**  
Let the number be \(N\).  
\(0.20N = 12 \Rightarrow N = \frac{12}{0.20}=60\).  
\(30\%\) of \(N\) is \(0.30 \times 60 = 18\).  

ANSWER 5: B  

---

**Problem 6:**  
Interpret the fractions as mixed numbers:  

- Clockwise \(2\frac14 = 2.25\) revolutions.  
- Counter‑clockwise \(3\frac34 = 3.75\) revolutions.  

Net rotation (counter‑clockwise positive) = \(3.75 - 2.25 = 1.5\) revolutions.  

A full revolution returns to the same direction, so we only care about the extra \(0.5\) revolution.  
\(0.5\) revolution = 180° counter‑clockwise.  

Starting from west, turning 180° counter‑clockwise points to **east**.  

ANSWER 6: B  

---

**Problem 7:**  
Original yellow = \(30\%\) of 30 L = \(0.30 \times 30 = 9\) L.  
Add 5 L yellow → new yellow = \(9+5 = 14\) L.  
New total volume = \(30+5 = 35\) L.  

Percent yellow = \(\frac{14}{35}\times100 = 40\%\).  

ANSWER 7: C  

---

**Problem 8:**  
Compute the three fractions:

\[
\frac{15}{11}=1.3636\ldots,\qquad
\frac{19}{15}=1.2666\ldots,\qquad
\frac{17}{13}=1.3077\ldots
\]

Ordered from least to greatest:  

\(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

This matches option **E**.  

ANSWER 8: E  

---

**Problem 9:**  
Current list: \(3,4,5,6,6,7\). The unique mode is 6 (appears twice).  
For the mode to remain unique and equal to the mean and median, \(x\) must not create a tie.  
Let the mean = median = mode = \(m\).  

Sum of numbers = \(31 + x\).  
Mean = \(\frac{31 + x}{7}=m\).  
Median of the seven numbers is the 4th after sorting; with \(x>7\) the median stays 6, so \(m=6\).  

Thus \(\frac{31 + x}{7}=6 \Rightarrow 31 + x = 42 \Rightarrow x = 11\).

Check: sorted list becomes \(3,4,5,6,6,7,11\) → median = 6, mode = 6, mean = 42/7 = 6. All equal.  

ANSWER 9: D  

---

**Problem 10:**  
Let the walking speed be \(v\) and the half‑distance be \(d/2\).  

Walking time for half‑distance: \(\frac{d/2}{v}=6\) min ⇒ \(\frac{d}{v}=12\) min.  

Running speed = \(3v\).  
Running time for the second half: \(\frac{d/2}{3v}= \frac{d}{6v}= \frac{1}{6}\times\frac{d}{v}= \frac{12}{6}=2\) min.  

Total time = \(6 + 2 = 8\) minutes.  

ANSWER 10: D  

---

**Problem 11:**  

\[
\begin{aligned}
1+11+21+31+41 &= 105,\\
9+19+29+39+49 &= 145,\\
\text{Total} &= 105+145 = 250.
\end{aligned}
\]

ANSWER 11: E  

---

**Problem 12:**  
Write the numbers with four decimal places:

- .99 = 0.9900  
- .9099 = 0.9099  
- .9 = 0.9000  
- .909 = 0.9090  
- .9009 = 0.9009  

The largest is 0.9900, i.e., **.99**.  

ANSWER 12: A  

Problem 1:  
The tournament starts with 16 teams. Each game eliminates one team, and the losing team is eliminated. To determine the winner, all teams except one must be eliminated. Thus, 15 teams must be eliminated, requiring 15 games.  
- **Why not other choices?** A (4) is too few; B (7), C (8) are also insufficient; E (16) would imply one game per team without elimination, which is incorrect.  
ANSWER 1: D  

Problem 2:  
- **Jose**: Starts with 10 → subtracts 1 (9) → doubles (18) → adds 2 → **20**.  
- **Thuy**: Starts with 10 → doubles (20) → subtracts 1 (19) → adds 2 → **21**.  
- **Kareem**: Starts with 10 → subtracts 1 (9) → adds 2 (11) → doubles → **22**.  
Kareem has the largest final answer.  
- **Why not other choices?** A (Jose) and B (Thuy) are smaller; D and E pair incorrect winners.  
ANSWER 2: C  

Problem 3:  
\( \frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08 \).  
- **Why not other choices?** A (.008) is \( \frac{2}{250} \); C (.8) is \( \frac{4}{5} \); D and E are larger decimals.  
ANSWER 3: B  

Problem 4:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is \( \binom{4}{3} = 4 \).  
- **Why not other choices?** A (2) is too few; C (6) is \( \binom{4}{2} \), which is incorrect; D (8) and E (10) are overcounts.  
ANSWER 4: B  

Problem 5:  
Let the number be \( x \). If 20% of \( x \) is 12, then \( 0.2x = 12 \), so \( x = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).  
- **Why not other choices?** A (15) is 25% of 60; C (20) is \( \frac{1}{3} \) of 60; D (24) is 40%; E (30) is 50%.  
ANSWER 5: B  

Problem 6:  
- A full revolution is 360°. Clockwise \( 2 \frac{1}{4} \) revolutions = \( 2.25 \times 360 = 810° \).  
- Counterclockwise \( 3 \frac{3}{4} \) revolutions = \( -3.75 \times 360 = -1350° \).  
- Net movement: \( 810° - 1350° = -540° \).  
- \( -540° \mod 360° = 180° \) (since \( -540 + 720 = 180° \)), which points west → south? Wait, initial direction is west.  
- West (270°) + 810° = 1080° → 1080° mod 360° = 0° (north). Then -1350° → 0° - 1350° = -1350° mod 360° = 90° (east)? Let's reframe:  
- Start at west (270°). Clockwise 2.25 rev: 270° + 810° = 1080° mod 360° = 0° (north). Then counterclockwise 3.75 rev: 0° - 1350° = -1350° mod 360° = 90° (east).  
- **Why not other choices?** A (north) is initial after clockwise; B (east) is final; C (south) is 180°; D (west) is start; E (northwest) is not a cardinal direction.  
ANSWER 6: B  

Problem 7:  
Original mixture: 30 liters, 30% yellow tint → 9 liters yellow. Add 5 liters yellow → total yellow = 14 liters. New mixture = 35 liters. Percent yellow = \( \frac{14}{35} \times 100 = 40\% \).  
- **Why not other choices?** A (25) is original yellow%; B (35) is before addition; D (45) and E (50) are overestimates.  
ANSWER 7: E (Note: Correction based on computation: 14/35=0.4 → 40%, so answer is E? Wait, 14/35=0.4 → 40%, which is E. But earlier I thought B, but B is 35, which is wrong. Original yellow is 9, add 5 → 14. Total mixture 30+5=35. 14/35=0.4 → 40%. So E. But in choices E is 50? Wait, no: choices are A.25 B.35 C.40 D.45 E.50. So 40% is C. I misread. Correction: 14/35=0.4 → 40%, which is C.  
**Correction**: 14/35=0.4 → 40%, so answer is C.  
ANSWER 7: C  

Problem 8:  
Compare \( \frac{15}{11} \approx 1.3636 \), \( \frac{19}{15} \approx 1.2667 \), \( \frac{17}{13} \approx 1.3077 \). Order: \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11} \).  
- **Why not other choices?** A, B, C, D have incorrect orders.  
ANSWER 8: E  

Problem 9:  
The set is {3, 4, 5, 6, 6, 7, x}. Mean = median = mode = 6.  
- Mode is 6 (unique), so 6 must appear most.  
- Median is 6 (4th value in ordered list: 3,4,5,6,6,7,x → if x≥6, ordered is ...,6,6,x or ...,6,x,6? Must have 4th value 6.  
- If x=6, set is {3,4,5,6,6,6,7}: median=6, mode=6, mean=(3+4+5+6+6+6+7)/7=37/7≈5.28 ≠6.  
- If x=11: set {3,4,5,6,6,7,11}: mean=(3+4+5+6+6+7+11)/7=42/7=6, median=6, mode=6.  
- **Why not other choices?** A (5): mean=36/7≈5.14; B (6): mean≈5.28; C (7): mean=38/7≈5.43; D (11) works.  
ANSWER 9: D  

Problem 10:  
Joe walks half the distance in 6 minutes. Let walking speed be \( v \), so distance = \( 6v \). He runs at \( 3v \), so time to run half distance = \( \frac{6v}{3v} = 2 \) minutes. Total time = 6 + 2 = 8 minutes.  
- **Why not other choices?** A (7), B (7.3), C (7.7), E (8.3) are not exact.  
ANSWER 10: D  

Problem 11:  
First group: 1+11+21+31+41 = 105. Second group: 9+19+29+39+49 = 145. Total = 105 + 145 = 250.  
- **Why not other choices?** A (150), B (199), C (200), D (249) are incorrect sums.  
ANSWER 11: E  

Problem 12:  
Compare decimals: A (.99), B (.9099), C (.9), D (.909), E (.9009). Largest is .99.  
- **Why not other choices?** B, C, D, E are all smaller than .99.  
ANSWER 12: A

### Problem 1:
We need to find the total number of games to determine a winner in a 16-team single-elimination tournament, where every game eliminates exactly one losing team. To get 1 winner, we must eliminate 16 - 1 = 15 teams. Since each game eliminates 1 team, we need 15 total games. Wrong choices: A is the number of bracket rounds, not games; C is half the number of teams; E is the total number of teams, not games.
ANSWER 1: D

---

### Problem 2:
All three start with 10, we calculate each person's final result step-by-step:
- Jose: (10 - 1) * 2 + 2 = 9*2 +2 = 20
- Thuy: (10*2) -1 +2 = 20-1+2=21
- Kareem: (10-1 +2)*2 = 11*2=22
Kareem has the largest final answer. Wrong choices stem from misordering the operations for each person.
ANSWER 2: C

---

### Problem3:
Convert 2/25 to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100: (2*4)/(25*4) = 8/100 = 0.08. Wrong choices: A is 2/250, C is 20/25.
ANSWER 3: B

---

### Problem4:
We choose 3 starters from 4 people, which is a combination problem (order of starters does not matter here). The number of ways to choose 3 from 4 is equivalent to choosing 1 person to leave off the starting lineup: there are 4 people, so 4 unique lineups, each omitting one person. Wrong choice C (6) is the number of permutations, not combinations.
ANSWER 4: B

---

### Problem5:
If 20% of a number is 12, first find the full number: 0.2*N=12 → N=12/0.2=60. 30% of 60 is 0.3*60=18. Wrong choices: A is 25% of 60, D is 40% of 60.
ANSWER 5: B

---

### Problem6:
The spinner starts pointing west. Net movement: 2.25 clockwise revolutions minus 3.75 clockwise (equivalent to 3.75 counterclockwise) = -1.5 total clockwise revolutions, or 1.5 counterclockwise revolutions. Full revolutions cancel out, so we only have a net 0.5 (180 degree) counterclockwise turn from west. A 180-degree turn from west points to east.
ANSWER 6: B

---

### Problem7:
Original 30L mixture has 30% yellow tint: 0.3*30=9L of yellow. Add 5L of yellow, so new yellow is 14L, new total mixture is 35L. The new percentage of yellow is (14/35)*100=40%. Wrong choice: people often forget to add 5L to the total mixture volume, leading to an incorrect higher percentage.
ANSWER 7: C

---

### Problem8:
Rewrite each fraction to compare easily: 15/11 = 1 + 4/11, 19/15=1+4/15, 17/13=1+4/13. For equal numerators, the larger the denominator, the smaller the fraction. So 4/15 < 4/13 < 4/11, so the full order from least to greatest is 19/15 < 17/13 <15/11.
ANSWER 8: E

---

### Problem9:
The dataset is [3,4,5,6,6,7,x]. The unique mode is currently 6 (only value appearing twice, so it must stay the only unique mode). For the mean, median, and mode to all be 6: the sum of the dataset must be 7*6=42. The sum of the existing numbers is 3+4+5+6+6+7=31, so x=42-31=11. Sorted dataset [3,4,5,6,6,7,11] has median 6, unique mode 6, mean 6, which fits. Wrong choice B (x=6) gives a mean of ~5.28, not 6.
ANSWER 9: D

---

### Problem10:
Joe walks half the distance in 6 minutes. His running speed is 3x his walking speed, so time to cover the same distance is 1/3 of walking time. Time to run the second half: 6/3=2 minutes. Total time: 6+2=8 minutes.
ANSWER 10: D

---

### Problem11:
Pair terms to simplify addition: (1+49)+(11+39)+(21+29)+(31+19)+(41+9) = 50+50+50+50+50=250.
ANSWER 11: E

---

### Problem12:
Line up all decimals to compare: A=0.9900, B=0.9099, C=0.9000, D=0.9090, E=0.9009. The largest value is 0.99, option A.
ANSWER 12: A

Problem 1:  
In a single-elimination tournament, each game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams (all except 1). Thus, 15 games are played.  
ANSWER 1: D  

Problem 2:  
Calculate each person's result starting from 10:  
- Jose: \(10 -1 =9\), double → \(9*2=18\), add 2 → \(18+2=20\)  
- Thuy: Double 10 → \(10*2=20\), subtract 1 → \(20-1=19\), add 2 → \(19+2=21\)  
- Kareem: \(10-1=9\), add 2 → \(9+2=11\), double → \(11*2=22\)  
Kareem has the largest result.  
ANSWER 2: C  

Problem 3:  
Convert \(\frac{2}{25}\) to decimal: \(2\div25=0.08\).  
ANSWER 3: B  

Problem 4:  
Choosing 3 starters from 4 people is a combination (order does not matter). The number of ways is \(\binom{4}{3}=4\) (or list: {Lance,Sally,Joy}, {Lance,Sally,Fred}, {Lance,Joy,Fred}, {Sally,Joy,Fred}).  
ANSWER 4: B  

Problem 5:  
Let the number be \(x\). \(20\%x=12\) → \(0.2x=12\) → \(x=60\). \(30\%\ of\ 60=0.3*60=18\).  
ANSWER 5: B  

Problem 6:  
Model directions with standard angles: 0°=east, 90°=north, 180°=west, 270°=south. Initial direction is west (180°).  
- Clockwise \(2\frac{1}{4}\) revolutions: \(2.25*(-360°)=-810°\), new angle: \(180-810=-630°\)  
- Counterclockwise \(3\frac{3}{4}\) revolutions: \(3.75*(360°)=+1350°\), total angle: \(-630+1350=720°\)  
- \(720°\mod360°=0°\), which is east.  
ANSWER 6: B  

Problem 7:  
Original yellow tint: \(30\%\ of\ 30=0.3*30=9\) liters. Add 5 liters, total yellow = \(9+5=14\) liters. New total mixture: \(30+5=35\) liters. Percent yellow: \(\frac{14}{35}*100\%=40\%\).  
ANSWER 7: C  

Problem 8:  
Compare fractions via cross-multiplication or decimals:  
- \(\frac{19}{15}\approx1.267\), \(\frac{17}{13}\approx1.308\), \(\frac{15}{11}\approx1.364\)  
Order from least to greatest: \(\frac{19}{15}<\frac{17}{13}<\frac{15}{11}\).  
ANSWER 8: E  

Problem 9:  
The unique mode is 6 (appears twice, others once). Thus mean and median equal 6.  
- Median of 7 numbers is the 4th term when sorted, so 4th term =6 → \(x\geq6\).  
- Mean = \(\frac{3+4+5+6+6+7+x}{7}=6\) → \(31+x=42\) → \(x=11\).  
Check: Sorted list [3,4,5,6,6,7,11], median=6, mode=6 (unique), mean=6.  
ANSWER 9: D  

Problem 10:  
Walking half the distance takes 6 minutes, so full walking time would be 12 minutes. Running speed is 3x walking speed, so time for the second half is \(\frac{6}{3}=2\) minutes. Total time: \(6+2=8\) minutes.  
ANSWER 10: D  

Problem 11:  
Sum first group: \(1+11+21+31+41=105\) (arithmetic series: \(\frac{5}{2}(1+41)=105\)).  
Sum second group: \(9+19+29+39+49=145\) (\(\frac{5}{2}(9+49)=145\)).  
Total: \(105+145=250\).  
ANSWER 11: E  

Problem 12:  
Compare decimals digit by digit:  
- A=0.99, B=0.9099, C=0.9, D=0.909, E=0.9009  
A has the highest second decimal digit (9 vs 0 for others), so it is the largest.  
ANSWER 12: A

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** How many games must be played to determine a winner in a 16-team elimination tournament?

**Setup:** In an elimination tournament, each game eliminates exactly one team. To go from 16 teams to 1 winner, we must eliminate 15 teams. Therefore, we need 15 games.

**Alternatively:** We can think of it as a bracket:
- Round 1: 16 teams → 8 games → 8 winners
- Round 2: 8 teams → 4 games → 4 winners
- Round 3: 4 teams → 2 games → 2 winners
- Round 4: 2 teams → 1 game → 1 winner
- Total: 8 + 4 + 2 + 1 = 15 games

ANSWER 1: D

---

**Problem 2:**

**What's being asked:** Which person gets the largest final answer after their operations?

**Jose's calculation:** 
- Start: 10
- Subtract 1: 10 - 1 = 9
- Double: 9 × 2 = 18
- Add 2: 18 + 2 = **20**

**Thuy's calculation:**
- Start: 10
- Double: 10 × 2 = 20
- Subtract 1: 20 - 1 = 19
- Add 2: 19 + 2 = **21**

**Kareem's calculation:**
- Start: 10
- Subtract 1: 10 - 1 = 9
- Add 2: 9 + 2 = 11
- Double: 11 × 2 = **22**

Kareem gets 22, which is the largest.

ANSWER 2: C

---

**Problem 3:**

**What's being asked:** Convert the fraction 2/25 to a decimal.

**Setup:** 2/25 = ?

**Calculation:** 
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** In how many ways can 3 starters be chosen from 4 people (Lance, Sally, Joy, Fred)?

**Setup:** This is a combination problem: $\binom{4}{3}$

**Calculation:**
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4}{1} = 4$$

Alternatively, we're choosing which 1 person to leave out: 4 ways.

ANSWER 4: B

---

**Problem 5:**

**What's being asked:** If 20% of a number is 12, what is 30% of that number?

**Setup:** Let the number be $n$.
- $0.20n = 12$
- $n = 12 ÷ 0.20 = 60$

**Calculation:** 
$$30\% \text{ of } 60 = 0.30 \times 60 = 18$$

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Starting pointing west, after rotating clockwise 2¼ revolutions then counterclockwise 3¾ revolutions, which direction?

**Setup:** 
- Clockwise 2¼ revolutions = 2.25 revolutions
- Counterclockwise 3¾ revolutions = 3.75 revolutions
- Net: 3.75 - 2.25 = 1.5 revolutions counterclockwise (net)

**Calculation:**
Starting at west, rotating counterclockwise 1.5 revolutions (= 1 full rotation + 0.5 rotation):
- 1 full rotation brings us back to west
- 0.5 rotation = 180°
- West + 180° counterclockwise = East

ANSWER 6: B

---

**Problem 7:**

**What's being asked:** After adding 5 liters of yellow tint to a 30-liter mixture, what percent is yellow tint?

**Original mixture (30 liters):**
- Yellow tint: 30% × 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percentage:**
$$\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%$$

ANSWER 7: C

---

**Problem 8:**

**What's being asked:** Order 15/11, 19/15, and 17/13 from least to greatest.

**Setup:** Convert to decimals or compare directly.

**Calculations:**
- 15/11 ≈ 1.364
- 19/15 ≈ 1.267
- 17/13 ≈ 1.308

**Order:** 19/15 < 17/13 < 15/11

ANSWER 8: E

---

**Problem 9:**

**What's being asked:** Find $x$ such that the mean, median, and unique mode of {3, 4, 5, 6, 6, 7, $x$} are all equal.

**Current mode:** 6 appears twice; all others appear once. If there's a unique mode, it must be 6.

**With unique mode = 6:** This requires no other number to appear more than once. So $x \notin \{3, 4, 5, 7\}$ and $x \neq 6$ (else it wouldn't be unique).

**Mean = Median = Mode = 6:**

For median of 7 numbers with value 6:
- Sorted: need middle (4th) value to be 6
- If $x = 11$: {3, 4, 5, 6, 6, 7, 11} → median = 6 ✓

Mean: $\frac{3+4+5+6+6+7+11}{7} = \frac{42}{7} = 6$ ✓

Mode: 6 appears twice; all others once → mode = 6 ✓

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Total time from home to school if Joe walked halfway (6 min) then ran the rest at 3× speed?

**Setup:**
- Walking: 6 minutes for half the distance
- Running: same distance at 3× speed, so time = 6/3 = 2 minutes

**Total time:** 6 + 2 = 8 minutes

ANSWER 10: D

---

**Problem 11:**

**What's being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Calculation:**
- First group: 1 + 11 + 21 + 31 + 41 = 105
- Second group: 9 + 19 + 29 + 39 + 49 = 145
- Sum: 105 + 145 = 250

ANSWER 11: E

---

**Problem 12:**

**What's being asked:** Which number is largest?

**Comparison:**
- A. 0.99
- B. 0.9099
- C. 0.9
- D. 0.909
- E. 0.9009

Starting from the tenths place: all are 0.9__. Looking at hundredths:
- 0.99 has 9 in hundredths place → **largest**

ANSWER 12: A

Problem 1:

In a single-elimination tournament, each game eliminates exactly one team.  
There are 16 teams, and we want one champion.

So we must eliminate 15 teams total.

Therefore, the number of games played is 15.

Tempting wrong choice: 16 would be if every team had to lose, but the winner does not lose.

ANSWER 1: D

Problem 2:

Compute each person’s result carefully.

- Jose: start with 10  
  subtract 1 → 9  
  double → 18  
  add 2 → 20

- Thuy: start with 10  
  double → 20  
  subtract 1 → 19  
  add 2 → 21

- Kareem: start with 10  
  subtract 1 → 9  
  add 2 → 11  
  double → 22

Compare:
- Jose = 20
- Thuy = 21
- Kareem = 22

Kareem gets the largest final answer.

ANSWER 2: C

Problem 3:

\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

So the correct decimal is 0.08.

ANSWER 3: B

Problem 4:

We are choosing 3 starters from 4 players: Lance, Sally, Joy, and Fred.

This is a combination problem:
\[
\binom{4}{3} = 4
\]

You can also see it by noticing that choosing 3 starters is the same as choosing which 1 player sits out, and there are 4 choices.

ANSWER 4: B

Problem 5:

Let the number be \(n\).

20% of \(n\) is 12, so:
\[
0.2n = 12
\]
\[
n = 60
\]

Now find 30% of 60:
\[
0.3 \times 60 = 18
\]

ANSWER 5: B

Problem 6:

The spinner starts pointing west.

First move: clockwise \(2\frac{1}{4}\) revolutions.  
A full revolution brings it back to west, so only the extra \(\frac{1}{4}\) revolution matters.

From west, a quarter-turn clockwise points it to north.

Second move: counterclockwise \(3\frac{3}{4}\) revolutions.  
Again, the 3 full revolutions do nothing, so this is just \(\frac{3}{4}\) of a revolution counterclockwise, which is the same as \(\frac{1}{4}\) of a revolution clockwise.

From north, a quarter-turn clockwise points east.

So the spinner ends pointing east.

ANSWER 6: B

Problem 7:

Start with 30 liters total.

Yellow tint is 30% of 30 liters:
\[
0.30 \times 30 = 9 \text{ liters}
\]

Then 5 liters of yellow tint are added, so yellow becomes:
\[
9 + 5 = 14 \text{ liters}
\]

New total mixture:
\[
30 + 5 = 35 \text{ liters}
\]

Percent yellow in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.4 = 40\%
\]

ANSWER 7: C

Problem 8:

The fractions are:
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]

Approximate each one:

\[
\frac{15}{11} \approx 1.364
\]
\[
\frac{17}{13} \approx 1.308
\]
\[
\frac{19}{15} \approx 1.267
\]

So from least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

That matches choice E.

ANSWER 8: E

Problem 9:

The numbers are:
\[
3,4,5,6,6,7,x
\]

We are told the mean, median, and unique mode are all equal.

Since 6 appears twice and every other listed number appears once, the unique mode is 6.

So the median and mean must also be 6.

The median of 7 numbers is the 4th number when ordered.  
The sorted list is:
\[
3,4,5,6,6,7,x
\]
To make the median 6, \(x\) must be 6 or larger, but we also need the mean to be 6.

Sum of the known numbers:
\[
3+4+5+6+6+7 = 31
\]

For the mean to be 6, total sum must be:
\[
7 \times 6 = 42
\]

So:
\[
31 + x = 42
\]
\[
x = 11
\]

Check:
- mode = 6
- median = 6
- mean = 42/7 = 6

Works perfectly.

ANSWER 9: D

Problem 10:

Joe walked half the distance in 6 minutes.  
He then ran the other half at 3 times his walking speed.

If he runs 3 times as fast, then the time for the same distance is 1/3 as long.

So the running time is:
\[
6 \div 3 = 2 \text{ minutes}
\]

Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

Problem 11:

Add the first group:
\[
1+11+21+31+41 = 105
\]

Add the second group:
\[
9+19+29+39+49 = 145
\]

Now add the totals:
\[
105 + 145 = 250
\]

ANSWER 11: E

Problem 12:

Compare the decimals:

- \(0.99\)
- \(0.9099\)
- \(0.9 = 0.9000\)
- \(0.909\)
- \(0.9009\)

The largest is clearly \(0.99\), since it is closest to 1.

ANSWER 12: A

Problem 1:  
We have a three-person-team tournament where **the losing team of each game is eliminated**. With **16 teams**, how many games are played to determine the winner?

To find the number of games in a single-elimination tournament:  
- Each game eliminates exactly **one** team (the loser).  
- To go from **16 teams to 1 winning team**, we must eliminate **15 teams**.  
- Therefore, the number of games is **15**.

So the correct choice is **D**.

ANSWER 1: D


Problem 2:  
Jose, Thuy, and Kareem all start at 10. We compute each final value.

**Jose:**  
- Subtract 1: \(10 - 1 = 9\)  
- Double: \(2 \cdot 9 = 18\)  
- Add 2: \(18 + 2 = 20\)  
Final: **20**

**Thuy:**  
- Double 10: \(2 \cdot 10 = 20\)  
- Subtract 1: \(20 - 1 = 19\)  
- Add 2: \(19 + 2 = 21\)  
Final: **21**

**Kareem:**  
- Subtract 1: \(10 - 1 = 9\)  
- Add 2: \(9 + 2 = 11\)  
- Double: \(2 \cdot 11 = 22\)  
Final: **22**

Largest is **22**, which is Kareem.

ANSWER 2: C


Problem 3:  
We need to evaluate:  
\[
\frac{2}{25}
\]
Convert to decimals:  
\[
\frac{2}{25} = 2 \div 25
\]
Since \(25 \cdot 0.08 = 2\):  
\[
2 \div 25 = 0.08
\]
So the correct choice is **B** (.08).

ANSWER 3: B


Problem 4:  
We have 4 people chosen: Lance, Sally, Joy, Fred. We need to choose **3 starters**.

Number of ways:
\[
\binom{4}{3} = 4
\]
List them conceptually: leaving out each one individually gives 4 choices.

Correct choice: **B**.

ANSWER 4: B


Problem 5:  
“If 20% of a number is 12, what is 30% of the same number?”

Let the number be \(N\).  
\[
0.20N = 12
\]
\[
N = \frac{12}{0.20} = 60
\]
Now compute \(30\%\) of 60:
\[
0.30 \cdot 60 = 18
\]
Correct choice: **B**.

ANSWER 5: B


Problem 6:  
Spinner starts pointing **west**.

Chenille moves it:
- **Clockwise** \(2\frac{1}{4}\) revolutions  
- then **Counterclockwise** \(3\frac{3}{4}\) revolutions

Net rotation (clockwise minus counterclockwise):
\[
2\frac{1}{4} - 3\frac{3}{4}
\]
Convert to improper fractions:
\[
2\frac{1}{4} = \frac{9}{4}, \quad 3\frac{3}{4} = \frac{15}{4}
\]
\[
\frac{9}{4} - \frac{15}{4} = -\frac{6}{4} = -\frac{3}{2}
\]
The negative means **net counterclockwise** by \(\frac{3}{2}\) revolutions.

\(\frac{3}{2} = 1 + \frac{1}{2}\), so this is:
- 1 full turn (brings it back where it started)
- plus \(\frac{1}{2}\) turn counterclockwise

Half a revolution from **west** points **east** (opposite direction).

Correct answer: **B (east)**.

ANSWER 6: B


Problem 7:  
Original mixture: 30 L with  
- 25% red (doesn’t matter directly)
- 30% yellow  
- 45% water

Yellow amount initially:
\[
0.30 \cdot 30 = 9 \text{ liters}
\]
Add 5 liters of yellow tint, so new yellow amount:
\[
9 + 5 = 14
\]
New total volume:
\[
30 + 5 = 35
\]
Percent yellow in new mixture:
\[
\frac{14}{35} = 0.4 = 40\%
\]
Correct choice: **C**.

ANSWER 7: C


Problem 8:  
Order fractions least to greatest:
\[
\frac{15}{11},\ \frac{19}{15},\ \frac{17}{13}
\]
A good way: compare values by cross-multiplication or decimal approximations.

Compute approximate decimals:
\[
\frac{15}{11} \approx 1.3636
\]
\[
\frac{19}{15} \approx 1.2667
\]
\[
\frac{17}{13} \approx 1.3077
\]
Least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]
Check choice: that matches **E**:
- \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11}\)

ANSWER 8: E


Problem 9:  
Numbers: \(3,4,5,6,6,7,x\) (7 positive integers).  
Mean = median = unique mode.

**Step 1: Mode**  
The only number that repeats is **6**, occurring twice. So the **unique mode = 6**.

Thus median must be 6 as well.

**Step 2: Median = 6**  
For 7 numbers sorted, the median is the **4th** number.

Sorted list must have the 4th term equal to 6. We already have two 6’s and surrounding numbers 3,4,5,7.

Currently without \(x\), the ordered portion is:
\[
3,4,5,6,6,7,\ ?
\]
The 4th number here is 6 already. To keep the 4th number as 6, \(x\) must not be less than 6 (otherwise it would shift positions so 4th might change).

If \(x = 6\), then 6 occurs 3 times, still mode 6 (unique mode). Then list:
\[
3,4,5,6,6,6,7
\]
Median (4th) is 6—works.

If \(x=7\), list:
\[
3,4,5,6,6,7,7
\]
Mode would be 6? Actually 7 repeats too, so mode not unique.

If \(x>7\), list:
\[
3,4,5,6,6,7,x
\]
Median still 6, but mode would still be 6 (unique), since only 6 repeats. That would also work—wait, but then 6 is still unique mode, yes. However we also need **mean = 6**.

So use mean condition.

Sum of known numbers:
\[
3+4+5+6+6+7 = 31
\]
Total sum including \(x\): \(31 + x\)

Mean:
\[
\frac{31+x}{7} = 6
\]
\[
31 + x = 42
\]
\[
x = 11
\]

Check uniqueness of mode: 6 occurs twice, no other repeats (since x=11 is different). Unique mode is 6—consistent.

ANSWER 9: D


Problem 10:  
Joe walks half the distance, then runs the rest.  
He runs **3 times as fast** as he walks.

Given: time to walk half way = **6 minutes**.  
Let walking speed be \(v\). Then walking time for half distance \(=\) 6 minutes.

Let half the distance be \(d\). Then:
\[
d = v \cdot 6
\]
Running speed is \(3v\).  
Remaining distance is also \(d\), so running time:
\[
\text{time} = \frac{d}{3v} = \frac{v\cdot 6}{3v} = 2
\]
Total time:
\[
6 + 2 = 8
\]
Correct choice: **D**.

ANSWER 10: D


Problem 11:  
Compute:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]

Group each set:

First set:
\[
1+11+21+31+41
\]
These are all of the form \(10k+1\):  
\(1, 11, 21, 31, 41\) correspond to \(k=0,1,2,3,4\).

Sum:
\[
(1+11+21+31+41) = (1+11+21+31+41)
\]
Compute directly:
\[
(1+11)=12,\ (21+31)=52,\ 12+52=64,\ 64+41=105
\]

Second set:
\[
9+19+29+39+49
\]
Direct sum:
\[
(9+19)=28,\ (29+39)=68,\ 28+68=96,\ 96+49=145
\]

Total:
\[
105 + 145 = 250
\]

Correct choice: **E**.

ANSWER 11: E


Problem 12:  
Which number is largest?

Options:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare decimals:  
- .99 is the largest among all because it is greater than .9 (and greater than any number starting 0.90... but with fewer 9s afterward).

So .99 > .9099 > .909 > .9009 > .9.

Correct choice: **A**.

ANSWER 12: A

Problem 1:
In a single-elimination tournament, every game results in exactly one team being eliminated. To determine a single winner from 16 teams, 15 teams must be eliminated. Since each game eliminates one team, it takes exactly 15 games.
ANSWER 1: D

Problem 2:
- Jose: (10 - 1) = 9; 9 * 2 = 18; 18 + 2 = 20.
- Thuy: (10 * 2) = 20; 20 - 1 = 19; 19 + 2 = 21.
- Kareem: (10 - 1) = 9; 9 + 2 = 11; 11 * 2 = 22.
Kareem has the largest answer (22).
ANSWER 2: C

Problem 3:
To convert 2/25 to a decimal, multiply the numerator and denominator by 4 to get a denominator of 100: (2 * 4) / (25 * 4) = 8 / 100 = 0.08.
ANSWER 3: B

Problem 4:
We need to choose 3 starters from a group of 4 (Lance, Sally, Joy, Fred). This is a combination problem denoted as 4C3. Choosing 3 people to be on the team is the same as choosing 1 person to be left off the team. There are 4 ways to choose who is left off.
ANSWER 4: B

Problem 5:
Let $x$ be the number. $0.20x = 12$. Dividing by 0.20, $x = 12 / 0.2 = 60$. We want 30% of $x$: $0.30 * 60 = 18$.
ANSWER 5: B

Problem 6:
Start at 0 revolutions. Clockwise is positive, counterclockwise is negative.
$2 \frac{1}{4} - 3 \frac{3}{4} = 2.25 - 3.75 = -1.5$ revolutions.
-1.5 revolutions is the same as -0.5 revolutions (since -1 is a full circle). -0.5 revolutions (counterclockwise) from West points to East.
ANSWER 6: B

Problem 7:
Original yellow tint: 30% of 30 liters = 9 liters.
New mixture: 30 + 5 = 35 liters.
New yellow tint: 9 + 5 = 14 liters.
Percent: (14 / 35) * 100 = (2 / 5) * 100 = 40%.
ANSWER 7: C

Problem 8:
Convert to decimals or compare differences from 1:
15/11 = 1 + 4/11 ≈ 1.3636
19/15 = 1 + 4/15 ≈ 1.2666
17/13 = 1 + 4/13 ≈ 1.3076
Comparing the fractions 4/11, 4/13, and 4/15: the larger the denominator, the smaller the fraction. Thus, 4/15 < 4/13 < 4/11.
Order: 19/15 < 17/13 < 15/11.
ANSWER 8: E

Problem 9:
The set is {3, 4, 5, 6, 6, 7, x}. The mode is 6 (it appears twice). For the mean and median to be 6:
Median: With 7 numbers, the 4th number must be 6. Since the sorted list is 3, 4, 5, 6, 6, 7, x, if $x > 6$, the 4th number is 6.
Mean: (3+4+5+6+6+7+x) / 7 = 6 => 31 + x = 42 => x = 11.
Checking: {3, 4, 5, 6, 6, 7, 11}. Median is 6. Mean is 42/7 = 6. Mode is 6.
ANSWER 9: D

Problem 10:
Walking time = 6 minutes. Running speed = 3 * walking speed.
Time taken = Distance / Speed. Since speed is 3 times faster, time is 1/3 of the walking time.
Running time = 6 / 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
ANSWER 10: D

Problem 11:
Sum 1: (1+11+21+31+41) = 105.
Sum 2: (9+19+29+39+49) = 145.
105 + 145 = 250.
ANSWER 11: E

Problem 12:
Compare digit by digit:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
0.99 is the largest.
ANSWER 12: A

Problem 1:
This is a single-elimination tournament with 16 teams. Exactly one team is eliminated per game, and 15 teams must be eliminated to leave a single winner. Thus exactly 15 games are required.

ANSWER 1: D

Problem 2:
Jose evaluates \((10-1)\times2+2=20\).
Thuy evaluates \((10\times2)-1+2=21\).
Kareem evaluates \((10-1+2)\times2=22\).
Kareem obtains the strictly largest value.

ANSWER 2: C

Problem 3:
Division yields the decimal expansion \(2\div25=0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters out of 4 distinct players is the binomial coefficient \(\binom{4}{3}=4\).

ANSWER 4: B

Problem 5:
If 20% of a number equals 12, the number itself is \(12/0.2=60\). Then 30% of 60 is \(0.3\times60=18\).

ANSWER 5: B

Problem 6:
A net rotation of \(+2.25-3.75=-1.5\) revolutions is equivalent to 1.5 revolutions counterclockwise. From an initial westward orientation, 1.5 revolutions counterclockwise lands exactly on east.

ANSWER 6: B

Problem 7:
The original mixture contains \(0.30\times30=9\) liters of yellow tint. Adding 5 liters raises the yellow volume to 14 liters and the total volume to 35 liters, so the new percentage is \(14/35=40\%\).

ANSWER 7: C

Problem 8:
Decimal approximations are \(15/11\approx1.3636\), \(19/15\approx1.2667\), and \(17/13\approx1.3077\). Ordering these values produces \(19/15<17/13<15/11\).

ANSWER 8: E

Problem 9:
The unique mode must be 6, forcing the mean also to equal 6. The sum of the seven numbers is therefore 42. The six given numbers sum to 31, so \(x=11\). With this value the ordered list is 3,4,5,6,6,7,11 whose median is likewise 6 and whose unique mode remains 6.

ANSWER 9: D

Problem 10:
Let the total distance be \(2d\). Walking time for distance \(d\) is 6 min, so running time for the remaining distance \(d\) at triple speed is \(6/3=2\) min. The total time is therefore 8 min.

ANSWER 10: D

Problem 11:
Each pair sums to a multiple of 10: \(1+9=10\), \(11+19=30\), \(21+29=50\), \(31+39=70\), \(41+49=90\). Their total is 250. Equivalently each five-term arithmetic series sums to 105 and 145 respectively, again yielding 250.

ANSWER 11: E

Problem 12:
Direct comparison of the five decimals shows that 0.99 is strictly larger than each of the others.

ANSWER 12: A

Problem 1:
The problem asks us to determine the number of games that will be played to determine the winner of a basketball tournament with 16 teams, where the losing team of each game is eliminated.

## Step 1
To determine the number of games needed to find the winner, we first note that in a single-elimination tournament, each game eliminates one team.

## Step 2
Since there can only be one winner, all other teams must be eliminated. Therefore, the number of games required is equal to the number of teams minus one.

## Step 3
We have 16 teams competing in the tournament. So, the number of games needed to determine the winner is 16 - 1 = 15.

## Step 4
Let's examine the given choices to rule out the incorrect options. A. 4 is too low because it doesn't account for all the teams. B. 7 is also too low for the same reason. C. 8 is still too low. D. 15 matches our calculation. E. 16 is one more than our calculation.

The final answer is supported by our step-by-step reasoning.
ANSWER 1: D

Problem 2:
The problem asks us to determine who among Jose, Thuy, and Kareem gets the largest final answer after performing a series of operations starting with the number 10.

## Step 1
Let's calculate Jose's final answer: Jose subtracts 1 from 10 to get 9, doubles it to get 18, and then adds 2 to get 20.

## Step 2
Now, let's calculate Thuy's final answer: Thuy doubles 10 to get 20, subtracts 1 to get 19, and then adds 2 to get 21.

## Step 3
Next, let's calculate Kareem's final answer: Kareem subtracts 1 from 10 to get 9, adds 2 to get 11, and then doubles it to get 22.

## Step 4
Comparing the final answers: Jose got 20, Thuy got 21, and Kareem got 22. The largest final answer is 22, which belongs to Kareem.

## Step 5
Let's examine the given choices. A. Jose is incorrect because his answer is 20. B. Thuy is incorrect because her answer is 21. C. Kareem is correct because his answer is 22. D. Jose and Thuy is incorrect because Kareem's answer is larger. E. Thuy and Kareem is incorrect because Jose's answer is not the largest.

The final answer is supported by our step-by-step reasoning.
ANSWER 2: C

Problem 3:
The problem asks us to convert the fraction 2/25 into a decimal.

## Step 1
To convert 2/25 into a decimal, we divide the numerator by the denominator: 2 ÷ 25 = 0.08.

## Step 2
Let's examine the given choices to identify the correct decimal representation. A. .008 is too small. B. .08 matches our calculation. C. .8 is too large. D. 1.25 is much too large. E. 12.5 is also much too large.

The final answer is supported by our step-by-step reasoning.
ANSWER 3: B

Problem 4:
The problem asks us to determine the number of ways to choose 3 starters from a team of 4 members: Lance, Sally, Joy, and Fred.

## Step 1
This is a combination problem, because the order in which the starters are chosen does not matter.

## Step 2
The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial, the product of all positive integers up to that number.

## Step 3
Here, n = 4 (total team members) and k = 3 (number of starters to be chosen). So, C(4, 3) = 4! / [3!(4 - 3)!] = 4! / (3! * 1!) = (4 * 3 * 2 * 1) / ((3 * 2 * 1) * 1) = 4.

## Step 4
Let's examine the given choices. A. 2 is too low. B. 4 matches our calculation. C. 6 is too high. D. 8 is also too high. E. 10 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 4: B

Problem 5:
The problem asks us to find 30% of a number, given that 20% of the same number is 12.

## Step 1
First, we need to find the number. We know that 20% of the number is 12. So, 0.20 * number = 12.

## Step 2
To find the number, we divide both sides of the equation by 0.20: number = 12 / 0.20 = 60.

## Step 3
Now, we need to calculate 30% of 60: 0.30 * 60 = 18.

## Step 4
Let's examine the given choices. A. 15 is too low. B. 18 matches our calculation. C. 20 is too high. D. 24 is also too high. E. 30 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 5: B

Problem 6:
The problem asks us to determine the direction the spinner points after Chenille moves it clockwise 2 1/4 revolutions and then counterclockwise 3 3/4 revolutions, starting from pointing west.

## Step 1
First, let's understand the effect of the movements. A full revolution is 360 degrees or 4 quarters. So, 2 1/4 revolutions clockwise is equivalent to 9 quarters clockwise.

## Step 2
9 quarters clockwise is the same as 1 quarter clockwise (since 8 quarters make 2 full revolutions, which brings the spinner back to the original direction). So, after the first move, the spinner points north.

## Step 3
Then, Chenille moves it counterclockwise 3 3/4 revolutions, which is equivalent to 15 quarters counterclockwise.

## Step 4
15 quarters counterclockwise is the same as 3 quarters counterclockwise (since 12 quarters make 3 full revolutions). So, moving 3 quarters counterclockwise from north results in pointing west.

## Step 5
Let's examine the given choices. A. north is incorrect because after the second move, it's not north. B. east is incorrect. C. south is incorrect. D. west matches our calculation. E. northwest is incorrect.

The final answer is supported by our step-by-step reasoning.
ANSWER 6: D

Problem 7:
The problem asks us to find the percentage of yellow tint in a new mixture after adding 5 liters of yellow tint to an original 30-liter mixture that is 25% red tint, 30% yellow tint, and 45% water.

## Step 1
First, we calculate the amount of yellow tint in the original mixture: 30% of 30 liters = 0.30 * 30 = 9 liters.

## Step 2
After adding 5 liters of yellow tint, the total amount of yellow tint becomes 9 + 5 = 14 liters.

## Step 3
The total volume of the new mixture is 30 + 5 = 35 liters.

## Step 4
To find the percentage of yellow tint in the new mixture, we divide the amount of yellow tint by the total volume and multiply by 100: (14 / 35) * 100 = 40%

Problem 1:
In a single-elimination tournament, each game eliminates exactly one team. To eliminate all but the champion, the number of games needed is one fewer than the number of teams. With 16 teams, the number of games is \(16 - 1 = 15\).

ANSWER 1: D

Problem 2:
Jose: \(10 - 1 = 9\), \(9 \times 2 = 18\), \(18 + 2 = 20\).
Thuy: \(10 \times 2 = 20\), \(20 - 1 = 19\), \(19 + 2 = 21\).
Kareem: \(10 - 1 = 9\), \(9 + 2 = 11\), \(11 \times 2 = 22\).
Kareem’s final answer of 22 is the largest.

ANSWER 2: C

Problem 3:
Divide 2 by 25: \(2 \div 25 = 0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters from 4 people is the combination \(\binom{4}{3} = 4\).

ANSWER 4: B

Problem 5:
Let the number be \(n\). \(20\%\) of \(n\) is \(0.2n = 12\), so \(n = 60\). Then \(30\%\) of \(60 = 0.3 \times 60 = 18\).

ANSWER 5: B

Problem 6:
Initial direction: west.
Clockwise \(2\frac{1}{4}\) revolutions: \(2\frac{1}{4} = 2.25\) revolutions clockwise.
Counterclockwise \(3\frac{3}{4}\) revolutions: \(3\frac{3}{4} = 3.75\) revolutions counterclockwise.
Net revolution = \(2.25 - 3.75 = -1.5\) revolutions, i.e., \(1.5\) revolutions counterclockwise.
A full revolution (1.0) returns to west. The remaining 0.5 revolution counterclockwise from west is a half-turn, which points to east.

ANSWER 6: B

Problem 7:
Original mixture: 30 liters, yellow tint = \(30\% \times 30 = 9\) liters.
Add 5 liters of yellow tint: new yellow = \(9 + 5 = 14\) liters; new total volume = \(30 + 5 = 35\) liters.
Percent yellow = \(\frac{14}{35} \times 100\% = 40\%\).

ANSWER 7: C

Problem 8:
Convert each fraction to a decimal for comparison:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
Order from least to greatest: \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

ANSWER 8: E

Problem 9:
The data set: 3, 4, 5, 6, 6, 7, \(x\). The mode is the most frequent number. With two 6’s, 6 is the mode unless \(x\) equals another number that appears twice, which would create a second mode, violating the “unique mode” condition. Therefore, \(x\) must not be 3, 4, 5, or 7. It could be 6, but if \(x=6\), the median is 6, while the mean is \((3+4+5+6+6+6+7)/7 = 37/7 \approx 5.29\), so mean ≠ median.  
For the mean to equal the median (which is 6, the middle of seven sorted numbers if \(x > 6\) or \(x=6\) only if placed correctly), we set the mean to 6. Sum without \(x\) = \(3+4+5+6+6+7 = 31\). Total needed for mean 6 = \(7 \times 6 = 42\). Thus \(x = 42 - 31 = 11\).  
Sorted: 3, 4, 5, 6, 6, 7, 11. Median = 6, mode = 6, mean = 6. Unique mode holds.

ANSWER 9: D

Problem 10:
Walking half the distance took 6 minutes, so walking speed = \(\frac{\text{half distance}}{6}\).
Running speed is 3 times walking speed. Time to run the second half = \(\frac{\text{half distance}}{3 \times (\text{half distance}/6)} = \frac{6}{3} = 2\) minutes.
Total time = \(6 + 2 = 8\) minutes.

ANSWER 10: D

Problem 11:
Pair the terms: \(1+49 = 50\), \(11+39 = 50\), \(21+29 = 50\), \(31+19 = 50\), \(41+9 = 50\). Five pairs of 50 give \(5 \times 50 = 250\).

ANSWER 11: E

Problem 12:
Compare by extending decimals to the same number of places:
\(0.99   = 0.9900\)
\(0.9099 = 0.9099\)
\(0.9    = 0.9000\)
\(0.909  = 0.9090\)
\(0.9009 = 0.9009\)
Clearly \(0.99 > 0.9099 > 0.909 > 0.9009 > 0.9\). The largest is \(0.99\).

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key Insight:** In a single-elimination tournament, every game eliminates exactly one team. To crown a single winner from 16 teams, we must eliminate 16 − 1 = 15 teams. Therefore, exactly 15 games are needed.

**Verification by rounds:**
- Round 1: 16 teams → 8 games → 8 teams left
- Round 2: 8 teams → 4 games → 4 teams left
- Round 3: 4 teams → 2 games → 2 teams left
- Round 4 (Final): 2 teams → 1 game → 1 winner

Total = 8 + 4 + 2 + 1 = 15 games.

Choices A (4), B (7), and C (8) are too small. Choice E (16) would mean every team is eliminated, leaving no winner.

ANSWER 1: D

---

Problem 2:
We track each person's calculation step by step, all starting from 10.

**Jose:** subtract 1 → 9; double → 18; add 2 → **20**
**Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
**Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

Comparing: 20 < 21 < 22. Kareem gets the largest final answer.

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

Choice A (.008) would be 8/1000. Choice C (.8) would be 8/10. Choices D and E are greater than 1, but 2/25 is clearly less than 1.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination problem — the order of selection doesn't matter:
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4$$

Equivalently, choosing 3 people to start is the same as choosing 1 person to sit out. There are 4 people, so 4 ways to pick who sits out.

The 4 possible starting lineups are: {L,S,J}, {L,S,F}, {L,J,F}, {S,J,F}.

ANSWER 4: B

---

Problem 5:
We're told 20% of a number is 12. Let the number be $x$.

$$0.20x = 12 \implies x = \frac{12}{0.20} = 60$$

Now find 30% of 60:
$$0.30 \times 60 = 18$$

**Shortcut:** 30% is $\frac{30}{20} = 1.5$ times 20%, so the answer is $1.5 \times 12 = 18$.

Choice A (15) would be 25%. Choice D (24) would be 40%. Choice E (30) would be 50%.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing **West**.

**Move 1:** Clockwise $2\frac{1}{4}$ revolutions.
- The 2 full revolutions bring it back to West.
- The extra $\frac{1}{4}$ revolution clockwise from West: West → **North**.

**Move 2:** Counterclockwise $3\frac{3}{4}$ revolutions.
- The 3 full revolutions bring it back to North.
- The extra $\frac{3}{4}$ revolution counterclockwise from North:
  - $\frac{1}{4}$ CCW from North → West
  - $\frac{2}{4}$ CCW from North → South
  - $\frac{3}{4}$ CCW from North → **East**

The spinner points East.

ANSWER 6: B

---

Problem 7:
Original mixture: 30 liters total.
- Red tint: 25% of 30 = 7.5 liters
- Yellow tint: 30% of 30 = 9 liters
- Water: 45% of 30 = 13.5 liters

Add 5 liters of yellow tint:
- New yellow tint = 9 + 5 = 14 liters
- New total mixture = 30 + 5 = 35 liters

Percent yellow tint in new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

Choice B (35%) is a trap for those who just add 5% to 30%. Choice D (45%) is the original water percentage.

ANSWER 7: C

---

Problem 8:
We compare $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$.

Rewrite each as a mixed number:
$$\frac{15}{11} = 1 + \frac{4}{11}, \quad \frac{19}{15} = 1 + \frac{4}{15}, \quad \frac{17}{13} = 1 + \frac{4}{13}$$

Since all three have the same numerator (4) in their fractional parts, we compare denominators. For fractions with equal numerators, a **larger denominator** means a **smaller fraction**:
$$15 > 13 > 11 \implies \frac{4}{15} < \frac{4}{13} < \frac{4}{11}$$

Therefore:
$$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$$

ANSWER 8: E

---

Problem 9:
The data set is: 3, 4, 5, 6, 6, 7, and $x$. The mean, median, and unique mode are all equal.

**Mode:** Since 6 already appears twice and all others appear once, the unique mode must be 6 (if $x$ were 3, 4, 5, or 7, we'd have two modes, violating "unique mode"). So mode = 6.

**Mean = 6:**
$$\frac{3 + 4 + 5 + 6 + 6 + 7 + x}{7} = 6$$
$$\frac{31 + x}{7} = 6 \implies 31 + x = 42 \implies x = 11$$

**Check Median:** With $x = 11$, the sorted list is: 3, 4, 5, **6**, 6, 7, 11. The median (4th value) is 6. ✓

All three measures equal 6. Choice B (6) would make 6 appear three times, but the mean would be $(31+6)/7 = 37/7 \neq 6$.

ANSWER 9: D

---

Problem 10:
Joe walked half the distance in 6 minutes. He then ran the remaining half at 3 times his walking speed.

Let walking speed = $v$. The half-distance = $6v$.

Running speed = $3v$. Time to run the second half:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{6v}{3v} = 2 \text{ minutes}$$

Total time = 6 + 2 = **8 minutes**.

Choice A (7) would result from incorrectly subtracting. Choices B, C, E involve unnecessary decimal complications.

ANSWER 10: D

---

Problem 11:
Compute $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$.

Pair the terms from the first group with terms from the second group (first with last, etc.):
- $1 + 49 = 50$
- $11 + 39 = 50$
- $21 + 29 = 50$
- $31 + 19 = 50$
- $41 + 9 = 50$

Total = $50 \times 5 = 250$.

Choice C (200) is a common error from miscounting pairs. Choice D (249) comes from arithmetic mistakes.

ANSWER 11: E

---

Problem 12:
Compare the five decimals by writing them to four decimal places:

- A: 0.9900
- B: 0.9099
- C: 0.9000
- D: 0.9090
- E: 0.9009

Comparing digit by digit from the tenths place: all have 9 in the tenths place. In the hundredths place, A has **9** while all others have **0**. So A is immediately the largest.

$0.99 > 0.9099 > 0.909 > 0.9009 > 0.9$

ANSWER 12: A

 **Problem 1:**
We need to find the number of games played in a single-elimination tournament with 16 teams. In single elimination, every game eliminates exactly one team. To determine a winner, 15 of the 16 teams must be eliminated. Therefore, exactly 15 games are required.  
(Another way: with 16 teams there are 8 games in the first round, 4 in the second, 2 in the third, and 1 final game: \(8+4+2+1=15\).)

ANSWER 1: D

---

**Problem 2:**
We compute each person’s final answer step by step, starting from 10.

- **Jose:**  
  \(10 - 1 = 9\)  
  \(9 \times 2 = 18\)  
  \(18 + 2 = 20\)

- **Thuy:**  
  \(10 \times 2 = 20\)  
  \(20 - 1 = 19\)  
  \(19 + 2 = 21\)

- **Kareem:**  
  \(10 - 1 = 9\)  
  \(9 + 2 = 11\)  
  \(11 \times 2 = 22\)

Comparing the results: \(20 < 21 < 22\). Kareem’s answer is the largest.

ANSWER 2: C

---

**Problem 3:**
We convert the fraction \(\frac{2}{25}\) to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100:
\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

ANSWER 3: B

---

**Problem 4:**
We must choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways to choose 3 from 4 is the combination \(\binom{4}{3}\):
\[
\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4
\]
Equivalently, choosing 3 starters is the same as choosing 1 person to sit out, and there are 4 choices for who sits out.

ANSWER 4: B

---

**Problem 5:**
Let the number be \(N\). We are given \(0.20N = 12\). Solving for \(N\):
\[
N = \frac{12}{0.20} = 60
\]
Now find 30% of 60:
\[
0.30 \times 60 = 18
\]
(Shortcut: 30% is \(1.5\) times 20%, so \(1.5 \times 12 = 18\).)

ANSWER 5: B

---

**Problem 6:**
The spinner starts pointing west.  
- Clockwise \(2\frac14\) revolutions: 2 full revolutions bring it back to west, then \(\frac14\) turn clockwise from west points it **north**.  
- Counterclockwise \(3\frac34\) revolutions: 3 full revolutions bring it back to north, then \(\frac34\) turn counterclockwise from north goes through west, south, and ends at **east**.

Net movement: \(3\frac34 - 2\frac14 = 1\frac12\) revolutions counterclockwise. One full revolution returns to west; an additional \(\frac12\) revolution counterclockwise from west points to **east**.

ANSWER 6: B

---

**Problem 7:**
Original mixture: 30 liters.  
- Yellow tint originally: \(30\% \times 30 = 9\) liters.  
After adding 5 liters of yellow tint:  
- New yellow tint: \(9 + 5 = 14\) liters.  
- New total mixture: \(30 + 5 = 35\) liters.  

Percent of yellow tint in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%
\]

ANSWER 7: C

---

**Problem 8:**
Compare \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\). Each is greater than 1. Write each as \(1\) plus a proper fraction:
\[
\frac{15}{11} = 1 + \frac{4}{11}, \qquad
\frac{19}{15} = 1 + \frac{4}{15}, \qquad
\frac{17}{13} = 1 + \frac{4}{13}
\]
Since the numerators are the same (4), the fraction with the largest denominator is smallest:
\[
\frac{4}{15} < \frac{4}{13} < \frac{4}{11}
\]
Adding 1 preserves the order:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

ANSWER 8: E

---

**Problem 9:**
The data set is \(3, 4, 5, 6, 6, 7, x\).  
- The current mode is 6 (appears twice). For 6 to remain the **unique** mode, \(x\) must not equal 3, 4, 5, or 7 (which would create a second mode).  
- We are told mean = median = mode = 6.  

Set the mean equal to 6:
\[
\frac{3+4+5+6+6+7+x}{7} = 6 \implies \frac{31+x}{7} = 6 \implies 31+x = 42 \implies x = 11
\]

Check: With \(x=11\), the sorted list is \(3, 4, 5, 6, 6, 7, 11\).  
- Median (4th term) = 6.  
- Mode = 6 (unique).  
- Mean = 6.  

All conditions are satisfied.

ANSWER 9: D

---

**Problem 10:**
Joe walks half the distance in 6 minutes. He runs the second half at 3 times his walking speed. Since distance is the same and speed is tripled, the time required is divided by 3:
\[
\text{Running time} = \frac{6}{3} = 2 \text{ minutes}
\]
Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

---

**Problem 11:**
Add the two groups:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]
Pair the terms to make convenient sums:
\[
(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 5 \times 50 = 250
\]

ANSWER 11: E

---

**Problem 12:**
Compare the decimals by writing them with the same number of places:
- A. \(0.9900\)
- B. \(0.9099\)
- C. \(0.9000\)
- D. \(0.9090\)
- E. \(0.9009\)

The tenths digit is 9 for all. Looking at the hundredths digit: A has 9, while B, D have 0 and C, E have 0. Thus \(0.9900\) is clearly the largest.

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key insight:** Each game eliminates exactly one team. To determine a winner from 16 teams, we must eliminate 15 teams (everyone except the champion).

Therefore, exactly 15 games must be played.

*Verification by rounds:* Round 1: 8 games (16→8 teams), Round 2: 4 games (8→4), Round 3: 2 games (4→2), Round 4: 1 game (2→1). Total = 8+4+2+1 = 15.

The tempting wrong answer E (16) counts teams instead of eliminations.

ANSWER 1: D

---

Problem 2:
We need to compute each person's final answer starting from 10.

**Jose:** Start with 10 → subtract 1: 10−1 = 9 → double: 9×2 = 18 → add 2: 18+2 = **20**

**Thuy:** Start with 10 → double: 10×2 = 20 → subtract 1: 20−1 = 19 → add 2: 19+2 = **21**

**Kareem:** Start with 10 → subtract 1: 10−1 = 9 → add 2: 9+2 = 11 → double: 11×2 = **22**

Kareem gets the largest final answer (22).

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

The tempting wrong answer A (0.008) would be 8/1000, and C (0.8) would be 8/10.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination: C(4,3) = 4!/(3! × 1!) = 4

Listing them out: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — that's 4 ways.

The tempting wrong answer C (6) would be C(4,2), choosing who's *not* a starter, but that also equals 4... Actually C(4,2)=6 is wrong here; we're choosing 3 from 4.

ANSWER 4: B

---

Problem 5:
We know 20% of a number is 12. We need to find 30% of the same number.

**Step 1:** Find the number.
$$0.20 \times n = 12 \implies n = \frac{12}{0.20} = 60$$

**Step 2:** Find 30% of 60.
$$0.30 \times 60 = 18$$

*Shortcut:* Since 30% is 1.5 times 20%, the answer is 1.5 × 12 = 18.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing west. We need to find the net rotation.

**Clockwise:** 2¼ revolutions
**Counterclockwise:** 3¾ revolutions

**Net movement:** 2¼ − 3¾ = −1½ revolutions (negative means counterclockwise)

So the spinner moves 1½ revolutions counterclockwise from west.

- 1 full revolution counterclockwise returns to **west**
- ½ revolution counterclockwise from west goes to **east**

(Counterclockwise from west: ¼ turn → south, ½ turn → east)

ANSWER 6: B

---

Problem 7:
We need to find the percent of yellow tint after adding 5 liters of yellow tint.

**Original mixture (30 liters):**
- Yellow tint: 30% of 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percent yellow tint:** (14/35) × 100 = 40%

The tempting wrong answer B (35) might come from simply adding 5% to 30%.

ANSWER 7: C

---

Problem 8:
We need to order 15/11, 19/15, and 17/13 from least to greatest.

Computing each as a decimal:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3077...

From least to greatest: **19/15 < 17/13 < 15/11**

*Alternative method:* Note each is 1 + (numerator − denominator)/denominator:
- 15/11 = 1 + 4/11
- 19/15 = 1 + 4/15
- 17/13 = 1 + 4/13

Since 4/15 < 4/13 < 4/11, the order is confirmed.

ANSWER 8: E

---

Problem 9:
The positive integers 3, 4, 5, 6, 6, 7, and x have equal mean, median, and unique mode.

**Mode:** Since 6 already appears twice and no other number does, the unique mode must be 6. So x ≠ 3, 4, 5, or 7 (which would create a second mode).

**Median:** With 7 numbers, the median is the 4th value when sorted. For the median to be 6, we need x ≥ 6 (so that 6 is the 4th value).

- If x = 6: sorted = 3, 4, 5, 6, 6, 6, 7 → median = 6 ✓
- If x > 7: sorted = 3, 4, 5, 6, 6, 7, x → median = 6 ✓

**Mean:** (3 + 4 + 5 + 6 + 6 + 7 + x)/7 = (31 + x)/7

Setting mean = 6: (31 + x)/7 = 6 → 31 + x = 42 → **x = 11**

**Verification:** Numbers: 3, 4, 5, 6, 6, 7, 11. Mean = 42/7 = 6, Median = 6, Mode = 6 ✓

ANSWER 9: D

---

Problem 10:
Joe walks halfway, then runs the rest. We need total time.

**Walking:** Half the distance takes 6 minutes.

**Running:** He runs 3 times as fast as he walks, so he covers the same distance in 1/3 the time.
- Time to run the second half = 6 ÷ 3 = 2 minutes

**Total time:** 6 + 2 = **8 minutes**

The tempting wrong answers come from setting up the problem incorrectly or mixing up distance and time relationships.

ANSWER 10: D

---

Problem 11:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

**First group:** 1 + 11 + 21 + 31 + 41 = 105

**Second group:** 9 + 19 + 29 + 39 + 49 = 145

**Total:** 105 + 145 = **250**

*Elegant method:* Pair terms: (1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250

ANSWER 11: E

---

Problem 12:
We need to find the largest number among the options. Writing each with the same number of decimal places:

- A. 0.99 = **0.9900**
- B. 0.9099 = 0.9099
- C. 0.9 = 0.9000
- D. 0.909 = 0.9090
- E. 0.9009 = 0.9009

Comparing digit by digit after the decimal point:
- First digit: all are 9
- Second digit: A has **9**, all others have 0

Since A has a 9 in the second decimal place while all others have 0, **0.99 is the largest**.

ANSWER 12: A

**Problem 1:**  
The tournament is single‑elimination. Starting with 16 teams, every game eliminates exactly one team. To reduce from 16 teams to a single champion we must eliminate 15 teams, and each elimination corresponds to one game.  

Number of games = 16 − 1 = **15**.  

ANSWER 1: D  

---

**Problem 2:**  
Start with 10 for each person.

- **Jose:**  
  Subtract 1 → 9; double → 18; add 2 → 20.  

- **Thuy:**  
  Double → 20; subtract 1 → 19; add 2 → 21.  

- **Kareem:**  
  Subtract 1 → 9; add 2 → 11; double → 22.  

The largest final value is 22, obtained by Kareem.  

ANSWER 2: C  

---

**Problem 3:**  

\[
\frac{2}{25}=0.08
\]

(2 ÷ 25 = 0.08).  

ANSWER 3: B  

---

**Problem 4:**  
We need to choose 3 starters from the 4 available players:  

\[
\binom{4}{3}=4
\]

ways.  

ANSWER 4: B  

---

**Problem 5:**  
Let the number be \(N\).  
\(0.20N = 12 \Rightarrow N = \frac{12}{0.20}=60\).  
\(30\%\) of \(N\) is \(0.30 \times 60 = 18\).  

ANSWER 5: B  

---

**Problem 6:**  
Interpret the fractions as mixed numbers:  

- Clockwise \(2\frac14 = 2.25\) revolutions.  
- Counter‑clockwise \(3\frac34 = 3.75\) revolutions.  

Net rotation (counter‑clockwise positive) = \(3.75 - 2.25 = 1.5\) revolutions.  

A full revolution returns to the same direction, so we only care about the extra \(0.5\) revolution.  
\(0.5\) revolution = 180° counter‑clockwise.  

Starting from west, turning 180° counter‑clockwise points to **east**.  

ANSWER 6: B  

---

**Problem 7:**  
Original yellow = \(30\%\) of 30 L = \(0.30 \times 30 = 9\) L.  
Add 5 L yellow → new yellow = \(9+5 = 14\) L.  
New total volume = \(30+5 = 35\) L.  

Percent yellow = \(\frac{14}{35}\times100 = 40\%\).  

ANSWER 7: C  

---

**Problem 8:**  
Compute the three fractions:

\[
\frac{15}{11}=1.3636\ldots,\qquad
\frac{19}{15}=1.2666\ldots,\qquad
\frac{17}{13}=1.3077\ldots
\]

Ordered from least to greatest:  

\(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

This matches option **E**.  

ANSWER 8: E  

---

**Problem 9:**  
Current list: \(3,4,5,6,6,7\). The unique mode is 6 (appears twice).  
For the mode to remain unique and equal to the mean and median, \(x\) must not create a tie.  
Let the mean = median = mode = \(m\).  

Sum of numbers = \(31 + x\).  
Mean = \(\frac{31 + x}{7}=m\).  
Median of the seven numbers is the 4th after sorting; with \(x>7\) the median stays 6, so \(m=6\).  

Thus \(\frac{31 + x}{7}=6 \Rightarrow 31 + x = 42 \Rightarrow x = 11\).

Check: sorted list becomes \(3,4,5,6,6,7,11\) → median = 6, mode = 6, mean = 42/7 = 6. All equal.  

ANSWER 9: D  

---

**Problem 10:**  
Let the walking speed be \(v\) and the half‑distance be \(d/2\).  

Walking time for half‑distance: \(\frac{d/2}{v}=6\) min ⇒ \(\frac{d}{v}=12\) min.  

Running speed = \(3v\).  
Running time for the second half: \(\frac{d/2}{3v}= \frac{d}{6v}= \frac{1}{6}\times\frac{d}{v}= \frac{12}{6}=2\) min.  

Total time = \(6 + 2 = 8\) minutes.  

ANSWER 10: D  

---

**Problem 11:**  

\[
\begin{aligned}
1+11+21+31+41 &= 105,\\
9+19+29+39+49 &= 145,\\
\text{Total} &= 105+145 = 250.
\end{aligned}
\]

ANSWER 11: E  

---

**Problem 12:**  
Write the numbers with four decimal places:

- .99 = 0.9900  
- .9099 = 0.9099  
- .9 = 0.9000  
- .909 = 0.9090  
- .9009 = 0.9009  

The largest is 0.9900, i.e., **.99**.  

ANSWER 12: A  

Problem 1:  
The tournament starts with 16 teams. Each game eliminates one team, and the losing team is eliminated. To determine the winner, all teams except one must be eliminated. Thus, 15 teams must be eliminated, requiring 15 games.  
- **Why not other choices?** A (4) is too few; B (7), C (8) are also insufficient; E (16) would imply one game per team without elimination, which is incorrect.  
ANSWER 1: D  

Problem 2:  
- **Jose**: Starts with 10 → subtracts 1 (9) → doubles (18) → adds 2 → **20**.  
- **Thuy**: Starts with 10 → doubles (20) → subtracts 1 (19) → adds 2 → **21**.  
- **Kareem**: Starts with 10 → subtracts 1 (9) → adds 2 (11) → doubles → **22**.  
Kareem has the largest final answer.  
- **Why not other choices?** A (Jose) and B (Thuy) are smaller; D and E pair incorrect winners.  
ANSWER 2: C  

Problem 3:  
\( \frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08 \).  
- **Why not other choices?** A (.008) is \( \frac{2}{250} \); C (.8) is \( \frac{4}{5} \); D and E are larger decimals.  
ANSWER 3: B  

Problem 4:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is \( \binom{4}{3} = 4 \).  
- **Why not other choices?** A (2) is too few; C (6) is \( \binom{4}{2} \), which is incorrect; D (8) and E (10) are overcounts.  
ANSWER 4: B  

Problem 5:  
Let the number be \( x \). If 20% of \( x \) is 12, then \( 0.2x = 12 \), so \( x = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).  
- **Why not other choices?** A (15) is 25% of 60; C (20) is \( \frac{1}{3} \) of 60; D (24) is 40%; E (30) is 50%.  
ANSWER 5: B  

Problem 6:  
- A full revolution is 360°. Clockwise \( 2 \frac{1}{4} \) revolutions = \( 2.25 \times 360 = 810° \).  
- Counterclockwise \( 3 \frac{3}{4} \) revolutions = \( -3.75 \times 360 = -1350° \).  
- Net movement: \( 810° - 1350° = -540° \).  
- \( -540° \mod 360° = 180° \) (since \( -540 + 720 = 180° \)), which points west → south? Wait, initial direction is west.  
- West (270°) + 810° = 1080° → 1080° mod 360° = 0° (north). Then -1350° → 0° - 1350° = -1350° mod 360° = 90° (east)? Let's reframe:  
- Start at west (270°). Clockwise 2.25 rev: 270° + 810° = 1080° mod 360° = 0° (north). Then counterclockwise 3.75 rev: 0° - 1350° = -1350° mod 360° = 90° (east).  
- **Why not other choices?** A (north) is initial after clockwise; B (east) is final; C (south) is 180°; D (west) is start; E (northwest) is not a cardinal direction.  
ANSWER 6: B  

Problem 7:  
Original mixture: 30 liters, 30% yellow tint → 9 liters yellow. Add 5 liters yellow → total yellow = 14 liters. New mixture = 35 liters. Percent yellow = \( \frac{14}{35} \times 100 = 40\% \).  
- **Why not other choices?** A (25) is original yellow%; B (35) is before addition; D (45) and E (50) are overestimates.  
ANSWER 7: E (Note: Correction based on computation: 14/35=0.4 → 40%, so answer is E? Wait, 14/35=0.4 → 40%, which is E. But earlier I thought B, but B is 35, which is wrong. Original yellow is 9, add 5 → 14. Total mixture 30+5=35. 14/35=0.4 → 40%. So E. But in choices E is 50? Wait, no: choices are A.25 B.35 C.40 D.45 E.50. So 40% is C. I misread. Correction: 14/35=0.4 → 40%, which is C.  
**Correction**: 14/35=0.4 → 40%, so answer is C.  
ANSWER 7: C  

Problem 8:  
Compare \( \frac{15}{11} \approx 1.3636 \), \( \frac{19}{15} \approx 1.2667 \), \( \frac{17}{13} \approx 1.3077 \). Order: \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11} \).  
- **Why not other choices?** A, B, C, D have incorrect orders.  
ANSWER 8: E  

Problem 9:  
The set is {3, 4, 5, 6, 6, 7, x}. Mean = median = mode = 6.  
- Mode is 6 (unique), so 6 must appear most.  
- Median is 6 (4th value in ordered list: 3,4,5,6,6,7,x → if x≥6, ordered is ...,6,6,x or ...,6,x,6? Must have 4th value 6.  
- If x=6, set is {3,4,5,6,6,6,7}: median=6, mode=6, mean=(3+4+5+6+6+6+7)/7=37/7≈5.28 ≠6.  
- If x=11: set {3,4,5,6,6,7,11}: mean=(3+4+5+6+6+7+11)/7=42/7=6, median=6, mode=6.  
- **Why not other choices?** A (5): mean=36/7≈5.14; B (6): mean≈5.28; C (7): mean=38/7≈5.43; D (11) works.  
ANSWER 9: D  

Problem 10:  
Joe walks half the distance in 6 minutes. Let walking speed be \( v \), so distance = \( 6v \). He runs at \( 3v \), so time to run half distance = \( \frac{6v}{3v} = 2 \) minutes. Total time = 6 + 2 = 8 minutes.  
- **Why not other choices?** A (7), B (7.3), C (7.7), E (8.3) are not exact.  
ANSWER 10: D  

Problem 11:  
First group: 1+11+21+31+41 = 105. Second group: 9+19+29+39+49 = 145. Total = 105 + 145 = 250.  
- **Why not other choices?** A (150), B (199), C (200), D (249) are incorrect sums.  
ANSWER 11: E  

Problem 12:  
Compare decimals: A (.99), B (.9099), C (.9), D (.909), E (.9009). Largest is .99.  
- **Why not other choices?** B, C, D, E are all smaller than .99.  
ANSWER 12: A

### Problem 1:
We need to find the total number of games to determine a winner in a 16-team single-elimination tournament, where every game eliminates exactly one losing team. To get 1 winner, we must eliminate 16 - 1 = 15 teams. Since each game eliminates 1 team, we need 15 total games. Wrong choices: A is the number of bracket rounds, not games; C is half the number of teams; E is the total number of teams, not games.
ANSWER 1: D

---

### Problem 2:
All three start with 10, we calculate each person's final result step-by-step:
- Jose: (10 - 1) * 2 + 2 = 9*2 +2 = 20
- Thuy: (10*2) -1 +2 = 20-1+2=21
- Kareem: (10-1 +2)*2 = 11*2=22
Kareem has the largest final answer. Wrong choices stem from misordering the operations for each person.
ANSWER 2: C

---

### Problem3:
Convert 2/25 to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100: (2*4)/(25*4) = 8/100 = 0.08. Wrong choices: A is 2/250, C is 20/25.
ANSWER 3: B

---

### Problem4:
We choose 3 starters from 4 people, which is a combination problem (order of starters does not matter here). The number of ways to choose 3 from 4 is equivalent to choosing 1 person to leave off the starting lineup: there are 4 people, so 4 unique lineups, each omitting one person. Wrong choice C (6) is the number of permutations, not combinations.
ANSWER 4: B

---

### Problem5:
If 20% of a number is 12, first find the full number: 0.2*N=12 → N=12/0.2=60. 30% of 60 is 0.3*60=18. Wrong choices: A is 25% of 60, D is 40% of 60.
ANSWER 5: B

---

### Problem6:
The spinner starts pointing west. Net movement: 2.25 clockwise revolutions minus 3.75 clockwise (equivalent to 3.75 counterclockwise) = -1.5 total clockwise revolutions, or 1.5 counterclockwise revolutions. Full revolutions cancel out, so we only have a net 0.5 (180 degree) counterclockwise turn from west. A 180-degree turn from west points to east.
ANSWER 6: B

---

### Problem7:
Original 30L mixture has 30% yellow tint: 0.3*30=9L of yellow. Add 5L of yellow, so new yellow is 14L, new total mixture is 35L. The new percentage of yellow is (14/35)*100=40%. Wrong choice: people often forget to add 5L to the total mixture volume, leading to an incorrect higher percentage.
ANSWER 7: C

---

### Problem8:
Rewrite each fraction to compare easily: 15/11 = 1 + 4/11, 19/15=1+4/15, 17/13=1+4/13. For equal numerators, the larger the denominator, the smaller the fraction. So 4/15 < 4/13 < 4/11, so the full order from least to greatest is 19/15 < 17/13 <15/11.
ANSWER 8: E

---

### Problem9:
The dataset is [3,4,5,6,6,7,x]. The unique mode is currently 6 (only value appearing twice, so it must stay the only unique mode). For the mean, median, and mode to all be 6: the sum of the dataset must be 7*6=42. The sum of the existing numbers is 3+4+5+6+6+7=31, so x=42-31=11. Sorted dataset [3,4,5,6,6,7,11] has median 6, unique mode 6, mean 6, which fits. Wrong choice B (x=6) gives a mean of ~5.28, not 6.
ANSWER 9: D

---

### Problem10:
Joe walks half the distance in 6 minutes. His running speed is 3x his walking speed, so time to cover the same distance is 1/3 of walking time. Time to run the second half: 6/3=2 minutes. Total time: 6+2=8 minutes.
ANSWER 10: D

---

### Problem11:
Pair terms to simplify addition: (1+49)+(11+39)+(21+29)+(31+19)+(41+9) = 50+50+50+50+50=250.
ANSWER 11: E

---

### Problem12:
Line up all decimals to compare: A=0.9900, B=0.9099, C=0.9000, D=0.9090, E=0.9009. The largest value is 0.99, option A.
ANSWER 12: A

Problem 1:  
In a single-elimination tournament, each game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams (all except 1). Thus, 15 games are played.  
ANSWER 1: D  

Problem 2:  
Calculate each person's result starting from 10:  
- Jose: \(10 -1 =9\), double → \(9*2=18\), add 2 → \(18+2=20\)  
- Thuy: Double 10 → \(10*2=20\), subtract 1 → \(20-1=19\), add 2 → \(19+2=21\)  
- Kareem: \(10-1=9\), add 2 → \(9+2=11\), double → \(11*2=22\)  
Kareem has the largest result.  
ANSWER 2: C  

Problem 3:  
Convert \(\frac{2}{25}\) to decimal: \(2\div25=0.08\).  
ANSWER 3: B  

Problem 4:  
Choosing 3 starters from 4 people is a combination (order does not matter). The number of ways is \(\binom{4}{3}=4\) (or list: {Lance,Sally,Joy}, {Lance,Sally,Fred}, {Lance,Joy,Fred}, {Sally,Joy,Fred}).  
ANSWER 4: B  

Problem 5:  
Let the number be \(x\). \(20\%x=12\) → \(0.2x=12\) → \(x=60\). \(30\%\ of\ 60=0.3*60=18\).  
ANSWER 5: B  

Problem 6:  
Model directions with standard angles: 0°=east, 90°=north, 180°=west, 270°=south. Initial direction is west (180°).  
- Clockwise \(2\frac{1}{4}\) revolutions: \(2.25*(-360°)=-810°\), new angle: \(180-810=-630°\)  
- Counterclockwise \(3\frac{3}{4}\) revolutions: \(3.75*(360°)=+1350°\), total angle: \(-630+1350=720°\)  
- \(720°\mod360°=0°\), which is east.  
ANSWER 6: B  

Problem 7:  
Original yellow tint: \(30\%\ of\ 30=0.3*30=9\) liters. Add 5 liters, total yellow = \(9+5=14\) liters. New total mixture: \(30+5=35\) liters. Percent yellow: \(\frac{14}{35}*100\%=40\%\).  
ANSWER 7: C  

Problem 8:  
Compare fractions via cross-multiplication or decimals:  
- \(\frac{19}{15}\approx1.267\), \(\frac{17}{13}\approx1.308\), \(\frac{15}{11}\approx1.364\)  
Order from least to greatest: \(\frac{19}{15}<\frac{17}{13}<\frac{15}{11}\).  
ANSWER 8: E  

Problem 9:  
The unique mode is 6 (appears twice, others once). Thus mean and median equal 6.  
- Median of 7 numbers is the 4th term when sorted, so 4th term =6 → \(x\geq6\).  
- Mean = \(\frac{3+4+5+6+6+7+x}{7}=6\) → \(31+x=42\) → \(x=11\).  
Check: Sorted list [3,4,5,6,6,7,11], median=6, mode=6 (unique), mean=6.  
ANSWER 9: D  

Problem 10:  
Walking half the distance takes 6 minutes, so full walking time would be 12 minutes. Running speed is 3x walking speed, so time for the second half is \(\frac{6}{3}=2\) minutes. Total time: \(6+2=8\) minutes.  
ANSWER 10: D  

Problem 11:  
Sum first group: \(1+11+21+31+41=105\) (arithmetic series: \(\frac{5}{2}(1+41)=105\)).  
Sum second group: \(9+19+29+39+49=145\) (\(\frac{5}{2}(9+49)=145\)).  
Total: \(105+145=250\).  
ANSWER 11: E  

Problem 12:  
Compare decimals digit by digit:  
- A=0.99, B=0.9099, C=0.9, D=0.909, E=0.9009  
A has the highest second decimal digit (9 vs 0 for others), so it is the largest.  
ANSWER 12: A

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** How many games must be played to determine a winner in a 16-team elimination tournament?

**Setup:** In an elimination tournament, each game eliminates exactly one team. To go from 16 teams to 1 winner, we must eliminate 15 teams. Therefore, we need 15 games.

**Alternatively:** We can think of it as a bracket:
- Round 1: 16 teams → 8 games → 8 winners
- Round 2: 8 teams → 4 games → 4 winners
- Round 3: 4 teams → 2 games → 2 winners
- Round 4: 2 teams → 1 game → 1 winner
- Total: 8 + 4 + 2 + 1 = 15 games

ANSWER 1: D

---

**Problem 2:**

**What's being asked:** Which person gets the largest final answer after their operations?

**Jose's calculation:** 
- Start: 10
- Subtract 1: 10 - 1 = 9
- Double: 9 × 2 = 18
- Add 2: 18 + 2 = **20**

**Thuy's calculation:**
- Start: 10
- Double: 10 × 2 = 20
- Subtract 1: 20 - 1 = 19
- Add 2: 19 + 2 = **21**

**Kareem's calculation:**
- Start: 10
- Subtract 1: 10 - 1 = 9
- Add 2: 9 + 2 = 11
- Double: 11 × 2 = **22**

Kareem gets 22, which is the largest.

ANSWER 2: C

---

**Problem 3:**

**What's being asked:** Convert the fraction 2/25 to a decimal.

**Setup:** 2/25 = ?

**Calculation:** 
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** In how many ways can 3 starters be chosen from 4 people (Lance, Sally, Joy, Fred)?

**Setup:** This is a combination problem: $\binom{4}{3}$

**Calculation:**
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4}{1} = 4$$

Alternatively, we're choosing which 1 person to leave out: 4 ways.

ANSWER 4: B

---

**Problem 5:**

**What's being asked:** If 20% of a number is 12, what is 30% of that number?

**Setup:** Let the number be $n$.
- $0.20n = 12$
- $n = 12 ÷ 0.20 = 60$

**Calculation:** 
$$30\% \text{ of } 60 = 0.30 \times 60 = 18$$

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Starting pointing west, after rotating clockwise 2¼ revolutions then counterclockwise 3¾ revolutions, which direction?

**Setup:** 
- Clockwise 2¼ revolutions = 2.25 revolutions
- Counterclockwise 3¾ revolutions = 3.75 revolutions
- Net: 3.75 - 2.25 = 1.5 revolutions counterclockwise (net)

**Calculation:**
Starting at west, rotating counterclockwise 1.5 revolutions (= 1 full rotation + 0.5 rotation):
- 1 full rotation brings us back to west
- 0.5 rotation = 180°
- West + 180° counterclockwise = East

ANSWER 6: B

---

**Problem 7:**

**What's being asked:** After adding 5 liters of yellow tint to a 30-liter mixture, what percent is yellow tint?

**Original mixture (30 liters):**
- Yellow tint: 30% × 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percentage:**
$$\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%$$

ANSWER 7: C

---

**Problem 8:**

**What's being asked:** Order 15/11, 19/15, and 17/13 from least to greatest.

**Setup:** Convert to decimals or compare directly.

**Calculations:**
- 15/11 ≈ 1.364
- 19/15 ≈ 1.267
- 17/13 ≈ 1.308

**Order:** 19/15 < 17/13 < 15/11

ANSWER 8: E

---

**Problem 9:**

**What's being asked:** Find $x$ such that the mean, median, and unique mode of {3, 4, 5, 6, 6, 7, $x$} are all equal.

**Current mode:** 6 appears twice; all others appear once. If there's a unique mode, it must be 6.

**With unique mode = 6:** This requires no other number to appear more than once. So $x \notin \{3, 4, 5, 7\}$ and $x \neq 6$ (else it wouldn't be unique).

**Mean = Median = Mode = 6:**

For median of 7 numbers with value 6:
- Sorted: need middle (4th) value to be 6
- If $x = 11$: {3, 4, 5, 6, 6, 7, 11} → median = 6 ✓

Mean: $\frac{3+4+5+6+6+7+11}{7} = \frac{42}{7} = 6$ ✓

Mode: 6 appears twice; all others once → mode = 6 ✓

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Total time from home to school if Joe walked halfway (6 min) then ran the rest at 3× speed?

**Setup:**
- Walking: 6 minutes for half the distance
- Running: same distance at 3× speed, so time = 6/3 = 2 minutes

**Total time:** 6 + 2 = 8 minutes

ANSWER 10: D

---

**Problem 11:**

**What's being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Calculation:**
- First group: 1 + 11 + 21 + 31 + 41 = 105
- Second group: 9 + 19 + 29 + 39 + 49 = 145
- Sum: 105 + 145 = 250

ANSWER 11: E

---

**Problem 12:**

**What's being asked:** Which number is largest?

**Comparison:**
- A. 0.99
- B. 0.9099
- C. 0.9
- D. 0.909
- E. 0.9009

Starting from the tenths place: all are 0.9__. Looking at hundredths:
- 0.99 has 9 in hundredths place → **largest**

ANSWER 12: A

Problem 1:

In a single-elimination tournament, each game eliminates exactly one team.  
There are 16 teams, and we want one champion.

So we must eliminate 15 teams total.

Therefore, the number of games played is 15.

Tempting wrong choice: 16 would be if every team had to lose, but the winner does not lose.

ANSWER 1: D

Problem 2:

Compute each person’s result carefully.

- Jose: start with 10  
  subtract 1 → 9  
  double → 18  
  add 2 → 20

- Thuy: start with 10  
  double → 20  
  subtract 1 → 19  
  add 2 → 21

- Kareem: start with 10  
  subtract 1 → 9  
  add 2 → 11  
  double → 22

Compare:
- Jose = 20
- Thuy = 21
- Kareem = 22

Kareem gets the largest final answer.

ANSWER 2: C

Problem 3:

\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

So the correct decimal is 0.08.

ANSWER 3: B

Problem 4:

We are choosing 3 starters from 4 players: Lance, Sally, Joy, and Fred.

This is a combination problem:
\[
\binom{4}{3} = 4
\]

You can also see it by noticing that choosing 3 starters is the same as choosing which 1 player sits out, and there are 4 choices.

ANSWER 4: B

Problem 5:

Let the number be \(n\).

20% of \(n\) is 12, so:
\[
0.2n = 12
\]
\[
n = 60
\]

Now find 30% of 60:
\[
0.3 \times 60 = 18
\]

ANSWER 5: B

Problem 6:

The spinner starts pointing west.

First move: clockwise \(2\frac{1}{4}\) revolutions.  
A full revolution brings it back to west, so only the extra \(\frac{1}{4}\) revolution matters.

From west, a quarter-turn clockwise points it to north.

Second move: counterclockwise \(3\frac{3}{4}\) revolutions.  
Again, the 3 full revolutions do nothing, so this is just \(\frac{3}{4}\) of a revolution counterclockwise, which is the same as \(\frac{1}{4}\) of a revolution clockwise.

From north, a quarter-turn clockwise points east.

So the spinner ends pointing east.

ANSWER 6: B

Problem 7:

Start with 30 liters total.

Yellow tint is 30% of 30 liters:
\[
0.30 \times 30 = 9 \text{ liters}
\]

Then 5 liters of yellow tint are added, so yellow becomes:
\[
9 + 5 = 14 \text{ liters}
\]

New total mixture:
\[
30 + 5 = 35 \text{ liters}
\]

Percent yellow in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.4 = 40\%
\]

ANSWER 7: C

Problem 8:

The fractions are:
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]

Approximate each one:

\[
\frac{15}{11} \approx 1.364
\]
\[
\frac{17}{13} \approx 1.308
\]
\[
\frac{19}{15} \approx 1.267
\]

So from least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

That matches choice E.

ANSWER 8: E

Problem 9:

The numbers are:
\[
3,4,5,6,6,7,x
\]

We are told the mean, median, and unique mode are all equal.

Since 6 appears twice and every other listed number appears once, the unique mode is 6.

So the median and mean must also be 6.

The median of 7 numbers is the 4th number when ordered.  
The sorted list is:
\[
3,4,5,6,6,7,x
\]
To make the median 6, \(x\) must be 6 or larger, but we also need the mean to be 6.

Sum of the known numbers:
\[
3+4+5+6+6+7 = 31
\]

For the mean to be 6, total sum must be:
\[
7 \times 6 = 42
\]

So:
\[
31 + x = 42
\]
\[
x = 11
\]

Check:
- mode = 6
- median = 6
- mean = 42/7 = 6

Works perfectly.

ANSWER 9: D

Problem 10:

Joe walked half the distance in 6 minutes.  
He then ran the other half at 3 times his walking speed.

If he runs 3 times as fast, then the time for the same distance is 1/3 as long.

So the running time is:
\[
6 \div 3 = 2 \text{ minutes}
\]

Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

Problem 11:

Add the first group:
\[
1+11+21+31+41 = 105
\]

Add the second group:
\[
9+19+29+39+49 = 145
\]

Now add the totals:
\[
105 + 145 = 250
\]

ANSWER 11: E

Problem 12:

Compare the decimals:

- \(0.99\)
- \(0.9099\)
- \(0.9 = 0.9000\)
- \(0.909\)
- \(0.9009\)

The largest is clearly \(0.99\), since it is closest to 1.

ANSWER 12: A

Problem 1:  
We have a three-person-team tournament where **the losing team of each game is eliminated**. With **16 teams**, how many games are played to determine the winner?

To find the number of games in a single-elimination tournament:  
- Each game eliminates exactly **one** team (the loser).  
- To go from **16 teams to 1 winning team**, we must eliminate **15 teams**.  
- Therefore, the number of games is **15**.

So the correct choice is **D**.

ANSWER 1: D


Problem 2:  
Jose, Thuy, and Kareem all start at 10. We compute each final value.

**Jose:**  
- Subtract 1: \(10 - 1 = 9\)  
- Double: \(2 \cdot 9 = 18\)  
- Add 2: \(18 + 2 = 20\)  
Final: **20**

**Thuy:**  
- Double 10: \(2 \cdot 10 = 20\)  
- Subtract 1: \(20 - 1 = 19\)  
- Add 2: \(19 + 2 = 21\)  
Final: **21**

**Kareem:**  
- Subtract 1: \(10 - 1 = 9\)  
- Add 2: \(9 + 2 = 11\)  
- Double: \(2 \cdot 11 = 22\)  
Final: **22**

Largest is **22**, which is Kareem.

ANSWER 2: C


Problem 3:  
We need to evaluate:  
\[
\frac{2}{25}
\]
Convert to decimals:  
\[
\frac{2}{25} = 2 \div 25
\]
Since \(25 \cdot 0.08 = 2\):  
\[
2 \div 25 = 0.08
\]
So the correct choice is **B** (.08).

ANSWER 3: B


Problem 4:  
We have 4 people chosen: Lance, Sally, Joy, Fred. We need to choose **3 starters**.

Number of ways:
\[
\binom{4}{3} = 4
\]
List them conceptually: leaving out each one individually gives 4 choices.

Correct choice: **B**.

ANSWER 4: B


Problem 5:  
“If 20% of a number is 12, what is 30% of the same number?”

Let the number be \(N\).  
\[
0.20N = 12
\]
\[
N = \frac{12}{0.20} = 60
\]
Now compute \(30\%\) of 60:
\[
0.30 \cdot 60 = 18
\]
Correct choice: **B**.

ANSWER 5: B


Problem 6:  
Spinner starts pointing **west**.

Chenille moves it:
- **Clockwise** \(2\frac{1}{4}\) revolutions  
- then **Counterclockwise** \(3\frac{3}{4}\) revolutions

Net rotation (clockwise minus counterclockwise):
\[
2\frac{1}{4} - 3\frac{3}{4}
\]
Convert to improper fractions:
\[
2\frac{1}{4} = \frac{9}{4}, \quad 3\frac{3}{4} = \frac{15}{4}
\]
\[
\frac{9}{4} - \frac{15}{4} = -\frac{6}{4} = -\frac{3}{2}
\]
The negative means **net counterclockwise** by \(\frac{3}{2}\) revolutions.

\(\frac{3}{2} = 1 + \frac{1}{2}\), so this is:
- 1 full turn (brings it back where it started)
- plus \(\frac{1}{2}\) turn counterclockwise

Half a revolution from **west** points **east** (opposite direction).

Correct answer: **B (east)**.

ANSWER 6: B


Problem 7:  
Original mixture: 30 L with  
- 25% red (doesn’t matter directly)
- 30% yellow  
- 45% water

Yellow amount initially:
\[
0.30 \cdot 30 = 9 \text{ liters}
\]
Add 5 liters of yellow tint, so new yellow amount:
\[
9 + 5 = 14
\]
New total volume:
\[
30 + 5 = 35
\]
Percent yellow in new mixture:
\[
\frac{14}{35} = 0.4 = 40\%
\]
Correct choice: **C**.

ANSWER 7: C


Problem 8:  
Order fractions least to greatest:
\[
\frac{15}{11},\ \frac{19}{15},\ \frac{17}{13}
\]
A good way: compare values by cross-multiplication or decimal approximations.

Compute approximate decimals:
\[
\frac{15}{11} \approx 1.3636
\]
\[
\frac{19}{15} \approx 1.2667
\]
\[
\frac{17}{13} \approx 1.3077
\]
Least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]
Check choice: that matches **E**:
- \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11}\)

ANSWER 8: E


Problem 9:  
Numbers: \(3,4,5,6,6,7,x\) (7 positive integers).  
Mean = median = unique mode.

**Step 1: Mode**  
The only number that repeats is **6**, occurring twice. So the **unique mode = 6**.

Thus median must be 6 as well.

**Step 2: Median = 6**  
For 7 numbers sorted, the median is the **4th** number.

Sorted list must have the 4th term equal to 6. We already have two 6’s and surrounding numbers 3,4,5,7.

Currently without \(x\), the ordered portion is:
\[
3,4,5,6,6,7,\ ?
\]
The 4th number here is 6 already. To keep the 4th number as 6, \(x\) must not be less than 6 (otherwise it would shift positions so 4th might change).

If \(x = 6\), then 6 occurs 3 times, still mode 6 (unique mode). Then list:
\[
3,4,5,6,6,6,7
\]
Median (4th) is 6—works.

If \(x=7\), list:
\[
3,4,5,6,6,7,7
\]
Mode would be 6? Actually 7 repeats too, so mode not unique.

If \(x>7\), list:
\[
3,4,5,6,6,7,x
\]
Median still 6, but mode would still be 6 (unique), since only 6 repeats. That would also work—wait, but then 6 is still unique mode, yes. However we also need **mean = 6**.

So use mean condition.

Sum of known numbers:
\[
3+4+5+6+6+7 = 31
\]
Total sum including \(x\): \(31 + x\)

Mean:
\[
\frac{31+x}{7} = 6
\]
\[
31 + x = 42
\]
\[
x = 11
\]

Check uniqueness of mode: 6 occurs twice, no other repeats (since x=11 is different). Unique mode is 6—consistent.

ANSWER 9: D


Problem 10:  
Joe walks half the distance, then runs the rest.  
He runs **3 times as fast** as he walks.

Given: time to walk half way = **6 minutes**.  
Let walking speed be \(v\). Then walking time for half distance \(=\) 6 minutes.

Let half the distance be \(d\). Then:
\[
d = v \cdot 6
\]
Running speed is \(3v\).  
Remaining distance is also \(d\), so running time:
\[
\text{time} = \frac{d}{3v} = \frac{v\cdot 6}{3v} = 2
\]
Total time:
\[
6 + 2 = 8
\]
Correct choice: **D**.

ANSWER 10: D


Problem 11:  
Compute:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]

Group each set:

First set:
\[
1+11+21+31+41
\]
These are all of the form \(10k+1\):  
\(1, 11, 21, 31, 41\) correspond to \(k=0,1,2,3,4\).

Sum:
\[
(1+11+21+31+41) = (1+11+21+31+41)
\]
Compute directly:
\[
(1+11)=12,\ (21+31)=52,\ 12+52=64,\ 64+41=105
\]

Second set:
\[
9+19+29+39+49
\]
Direct sum:
\[
(9+19)=28,\ (29+39)=68,\ 28+68=96,\ 96+49=145
\]

Total:
\[
105 + 145 = 250
\]

Correct choice: **E**.

ANSWER 11: E


Problem 12:  
Which number is largest?

Options:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare decimals:  
- .99 is the largest among all because it is greater than .9 (and greater than any number starting 0.90... but with fewer 9s afterward).

So .99 > .9099 > .909 > .9009 > .9.

Correct choice: **A**.

ANSWER 12: A

Problem 1:
In a single-elimination tournament, every game results in exactly one team being eliminated. To determine a single winner from 16 teams, 15 teams must be eliminated. Since each game eliminates one team, it takes exactly 15 games.
ANSWER 1: D

Problem 2:
- Jose: (10 - 1) = 9; 9 * 2 = 18; 18 + 2 = 20.
- Thuy: (10 * 2) = 20; 20 - 1 = 19; 19 + 2 = 21.
- Kareem: (10 - 1) = 9; 9 + 2 = 11; 11 * 2 = 22.
Kareem has the largest answer (22).
ANSWER 2: C

Problem 3:
To convert 2/25 to a decimal, multiply the numerator and denominator by 4 to get a denominator of 100: (2 * 4) / (25 * 4) = 8 / 100 = 0.08.
ANSWER 3: B

Problem 4:
We need to choose 3 starters from a group of 4 (Lance, Sally, Joy, Fred). This is a combination problem denoted as 4C3. Choosing 3 people to be on the team is the same as choosing 1 person to be left off the team. There are 4 ways to choose who is left off.
ANSWER 4: B

Problem 5:
Let $x$ be the number. $0.20x = 12$. Dividing by 0.20, $x = 12 / 0.2 = 60$. We want 30% of $x$: $0.30 * 60 = 18$.
ANSWER 5: B

Problem 6:
Start at 0 revolutions. Clockwise is positive, counterclockwise is negative.
$2 \frac{1}{4} - 3 \frac{3}{4} = 2.25 - 3.75 = -1.5$ revolutions.
-1.5 revolutions is the same as -0.5 revolutions (since -1 is a full circle). -0.5 revolutions (counterclockwise) from West points to East.
ANSWER 6: B

Problem 7:
Original yellow tint: 30% of 30 liters = 9 liters.
New mixture: 30 + 5 = 35 liters.
New yellow tint: 9 + 5 = 14 liters.
Percent: (14 / 35) * 100 = (2 / 5) * 100 = 40%.
ANSWER 7: C

Problem 8:
Convert to decimals or compare differences from 1:
15/11 = 1 + 4/11 ≈ 1.3636
19/15 = 1 + 4/15 ≈ 1.2666
17/13 = 1 + 4/13 ≈ 1.3076
Comparing the fractions 4/11, 4/13, and 4/15: the larger the denominator, the smaller the fraction. Thus, 4/15 < 4/13 < 4/11.
Order: 19/15 < 17/13 < 15/11.
ANSWER 8: E

Problem 9:
The set is {3, 4, 5, 6, 6, 7, x}. The mode is 6 (it appears twice). For the mean and median to be 6:
Median: With 7 numbers, the 4th number must be 6. Since the sorted list is 3, 4, 5, 6, 6, 7, x, if $x > 6$, the 4th number is 6.
Mean: (3+4+5+6+6+7+x) / 7 = 6 => 31 + x = 42 => x = 11.
Checking: {3, 4, 5, 6, 6, 7, 11}. Median is 6. Mean is 42/7 = 6. Mode is 6.
ANSWER 9: D

Problem 10:
Walking time = 6 minutes. Running speed = 3 * walking speed.
Time taken = Distance / Speed. Since speed is 3 times faster, time is 1/3 of the walking time.
Running time = 6 / 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
ANSWER 10: D

Problem 11:
Sum 1: (1+11+21+31+41) = 105.
Sum 2: (9+19+29+39+49) = 145.
105 + 145 = 250.
ANSWER 11: E

Problem 12:
Compare digit by digit:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
0.99 is the largest.
ANSWER 12: A

Problem 1:
This is a single-elimination tournament with 16 teams. Exactly one team is eliminated per game, and 15 teams must be eliminated to leave a single winner. Thus exactly 15 games are required.

ANSWER 1: D

Problem 2:
Jose evaluates \((10-1)\times2+2=20\).
Thuy evaluates \((10\times2)-1+2=21\).
Kareem evaluates \((10-1+2)\times2=22\).
Kareem obtains the strictly largest value.

ANSWER 2: C

Problem 3:
Division yields the decimal expansion \(2\div25=0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters out of 4 distinct players is the binomial coefficient \(\binom{4}{3}=4\).

ANSWER 4: B

Problem 5:
If 20% of a number equals 12, the number itself is \(12/0.2=60\). Then 30% of 60 is \(0.3\times60=18\).

ANSWER 5: B

Problem 6:
A net rotation of \(+2.25-3.75=-1.5\) revolutions is equivalent to 1.5 revolutions counterclockwise. From an initial westward orientation, 1.5 revolutions counterclockwise lands exactly on east.

ANSWER 6: B

Problem 7:
The original mixture contains \(0.30\times30=9\) liters of yellow tint. Adding 5 liters raises the yellow volume to 14 liters and the total volume to 35 liters, so the new percentage is \(14/35=40\%\).

ANSWER 7: C

Problem 8:
Decimal approximations are \(15/11\approx1.3636\), \(19/15\approx1.2667\), and \(17/13\approx1.3077\). Ordering these values produces \(19/15<17/13<15/11\).

ANSWER 8: E

Problem 9:
The unique mode must be 6, forcing the mean also to equal 6. The sum of the seven numbers is therefore 42. The six given numbers sum to 31, so \(x=11\). With this value the ordered list is 3,4,5,6,6,7,11 whose median is likewise 6 and whose unique mode remains 6.

ANSWER 9: D

Problem 10:
Let the total distance be \(2d\). Walking time for distance \(d\) is 6 min, so running time for the remaining distance \(d\) at triple speed is \(6/3=2\) min. The total time is therefore 8 min.

ANSWER 10: D

Problem 11:
Each pair sums to a multiple of 10: \(1+9=10\), \(11+19=30\), \(21+29=50\), \(31+39=70\), \(41+49=90\). Their total is 250. Equivalently each five-term arithmetic series sums to 105 and 145 respectively, again yielding 250.

ANSWER 11: E

Problem 12:
Direct comparison of the five decimals shows that 0.99 is strictly larger than each of the others.

ANSWER 12: A

Problem 1:
The problem asks us to determine the number of games that will be played to determine the winner of a basketball tournament with 16 teams, where the losing team of each game is eliminated.

## Step 1
To determine the number of games needed to find the winner, we first note that in a single-elimination tournament, each game eliminates one team.

## Step 2
Since there can only be one winner, all other teams must be eliminated. Therefore, the number of games required is equal to the number of teams minus one.

## Step 3
We have 16 teams competing in the tournament. So, the number of games needed to determine the winner is 16 - 1 = 15.

## Step 4
Let's examine the given choices to rule out the incorrect options. A. 4 is too low because it doesn't account for all the teams. B. 7 is also too low for the same reason. C. 8 is still too low. D. 15 matches our calculation. E. 16 is one more than our calculation.

The final answer is supported by our step-by-step reasoning.
ANSWER 1: D

Problem 2:
The problem asks us to determine who among Jose, Thuy, and Kareem gets the largest final answer after performing a series of operations starting with the number 10.

## Step 1
Let's calculate Jose's final answer: Jose subtracts 1 from 10 to get 9, doubles it to get 18, and then adds 2 to get 20.

## Step 2
Now, let's calculate Thuy's final answer: Thuy doubles 10 to get 20, subtracts 1 to get 19, and then adds 2 to get 21.

## Step 3
Next, let's calculate Kareem's final answer: Kareem subtracts 1 from 10 to get 9, adds 2 to get 11, and then doubles it to get 22.

## Step 4
Comparing the final answers: Jose got 20, Thuy got 21, and Kareem got 22. The largest final answer is 22, which belongs to Kareem.

## Step 5
Let's examine the given choices. A. Jose is incorrect because his answer is 20. B. Thuy is incorrect because her answer is 21. C. Kareem is correct because his answer is 22. D. Jose and Thuy is incorrect because Kareem's answer is larger. E. Thuy and Kareem is incorrect because Jose's answer is not the largest.

The final answer is supported by our step-by-step reasoning.
ANSWER 2: C

Problem 3:
The problem asks us to convert the fraction 2/25 into a decimal.

## Step 1
To convert 2/25 into a decimal, we divide the numerator by the denominator: 2 ÷ 25 = 0.08.

## Step 2
Let's examine the given choices to identify the correct decimal representation. A. .008 is too small. B. .08 matches our calculation. C. .8 is too large. D. 1.25 is much too large. E. 12.5 is also much too large.

The final answer is supported by our step-by-step reasoning.
ANSWER 3: B

Problem 4:
The problem asks us to determine the number of ways to choose 3 starters from a team of 4 members: Lance, Sally, Joy, and Fred.

## Step 1
This is a combination problem, because the order in which the starters are chosen does not matter.

## Step 2
The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial, the product of all positive integers up to that number.

## Step 3
Here, n = 4 (total team members) and k = 3 (number of starters to be chosen). So, C(4, 3) = 4! / [3!(4 - 3)!] = 4! / (3! * 1!) = (4 * 3 * 2 * 1) / ((3 * 2 * 1) * 1) = 4.

## Step 4
Let's examine the given choices. A. 2 is too low. B. 4 matches our calculation. C. 6 is too high. D. 8 is also too high. E. 10 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 4: B

Problem 5:
The problem asks us to find 30% of a number, given that 20% of the same number is 12.

## Step 1
First, we need to find the number. We know that 20% of the number is 12. So, 0.20 * number = 12.

## Step 2
To find the number, we divide both sides of the equation by 0.20: number = 12 / 0.20 = 60.

## Step 3
Now, we need to calculate 30% of 60: 0.30 * 60 = 18.

## Step 4
Let's examine the given choices. A. 15 is too low. B. 18 matches our calculation. C. 20 is too high. D. 24 is also too high. E. 30 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 5: B

Problem 6:
The problem asks us to determine the direction the spinner points after Chenille moves it clockwise 2 1/4 revolutions and then counterclockwise 3 3/4 revolutions, starting from pointing west.

## Step 1
First, let's understand the effect of the movements. A full revolution is 360 degrees or 4 quarters. So, 2 1/4 revolutions clockwise is equivalent to 9 quarters clockwise.

## Step 2
9 quarters clockwise is the same as 1 quarter clockwise (since 8 quarters make 2 full revolutions, which brings the spinner back to the original direction). So, after the first move, the spinner points north.

## Step 3
Then, Chenille moves it counterclockwise 3 3/4 revolutions, which is equivalent to 15 quarters counterclockwise.

## Step 4
15 quarters counterclockwise is the same as 3 quarters counterclockwise (since 12 quarters make 3 full revolutions). So, moving 3 quarters counterclockwise from north results in pointing west.

## Step 5
Let's examine the given choices. A. north is incorrect because after the second move, it's not north. B. east is incorrect. C. south is incorrect. D. west matches our calculation. E. northwest is incorrect.

The final answer is supported by our step-by-step reasoning.
ANSWER 6: D

Problem 7:
The problem asks us to find the percentage of yellow tint in a new mixture after adding 5 liters of yellow tint to an original 30-liter mixture that is 25% red tint, 30% yellow tint, and 45% water.

## Step 1
First, we calculate the amount of yellow tint in the original mixture: 30% of 30 liters = 0.30 * 30 = 9 liters.

## Step 2
After adding 5 liters of yellow tint, the total amount of yellow tint becomes 9 + 5 = 14 liters.

## Step 3
The total volume of the new mixture is 30 + 5 = 35 liters.

## Step 4
To find the percentage of yellow tint in the new mixture, we divide the amount of yellow tint by the total volume and multiply by 100: (14 / 35) * 100 = 40%

Problem 1:
In a single-elimination tournament, each game eliminates exactly one team. To eliminate all but the champion, the number of games needed is one fewer than the number of teams. With 16 teams, the number of games is \(16 - 1 = 15\).

ANSWER 1: D

Problem 2:
Jose: \(10 - 1 = 9\), \(9 \times 2 = 18\), \(18 + 2 = 20\).
Thuy: \(10 \times 2 = 20\), \(20 - 1 = 19\), \(19 + 2 = 21\).
Kareem: \(10 - 1 = 9\), \(9 + 2 = 11\), \(11 \times 2 = 22\).
Kareem’s final answer of 22 is the largest.

ANSWER 2: C

Problem 3:
Divide 2 by 25: \(2 \div 25 = 0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters from 4 people is the combination \(\binom{4}{3} = 4\).

ANSWER 4: B

Problem 5:
Let the number be \(n\). \(20\%\) of \(n\) is \(0.2n = 12\), so \(n = 60\). Then \(30\%\) of \(60 = 0.3 \times 60 = 18\).

ANSWER 5: B

Problem 6:
Initial direction: west.
Clockwise \(2\frac{1}{4}\) revolutions: \(2\frac{1}{4} = 2.25\) revolutions clockwise.
Counterclockwise \(3\frac{3}{4}\) revolutions: \(3\frac{3}{4} = 3.75\) revolutions counterclockwise.
Net revolution = \(2.25 - 3.75 = -1.5\) revolutions, i.e., \(1.5\) revolutions counterclockwise.
A full revolution (1.0) returns to west. The remaining 0.5 revolution counterclockwise from west is a half-turn, which points to east.

ANSWER 6: B

Problem 7:
Original mixture: 30 liters, yellow tint = \(30\% \times 30 = 9\) liters.
Add 5 liters of yellow tint: new yellow = \(9 + 5 = 14\) liters; new total volume = \(30 + 5 = 35\) liters.
Percent yellow = \(\frac{14}{35} \times 100\% = 40\%\).

ANSWER 7: C

Problem 8:
Convert each fraction to a decimal for comparison:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
Order from least to greatest: \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

ANSWER 8: E

Problem 9:
The data set: 3, 4, 5, 6, 6, 7, \(x\). The mode is the most frequent number. With two 6’s, 6 is the mode unless \(x\) equals another number that appears twice, which would create a second mode, violating the “unique mode” condition. Therefore, \(x\) must not be 3, 4, 5, or 7. It could be 6, but if \(x=6\), the median is 6, while the mean is \((3+4+5+6+6+6+7)/7 = 37/7 \approx 5.29\), so mean ≠ median.  
For the mean to equal the median (which is 6, the middle of seven sorted numbers if \(x > 6\) or \(x=6\) only if placed correctly), we set the mean to 6. Sum without \(x\) = \(3+4+5+6+6+7 = 31\). Total needed for mean 6 = \(7 \times 6 = 42\). Thus \(x = 42 - 31 = 11\).  
Sorted: 3, 4, 5, 6, 6, 7, 11. Median = 6, mode = 6, mean = 6. Unique mode holds.

ANSWER 9: D

Problem 10:
Walking half the distance took 6 minutes, so walking speed = \(\frac{\text{half distance}}{6}\).
Running speed is 3 times walking speed. Time to run the second half = \(\frac{\text{half distance}}{3 \times (\text{half distance}/6)} = \frac{6}{3} = 2\) minutes.
Total time = \(6 + 2 = 8\) minutes.

ANSWER 10: D

Problem 11:
Pair the terms: \(1+49 = 50\), \(11+39 = 50\), \(21+29 = 50\), \(31+19 = 50\), \(41+9 = 50\). Five pairs of 50 give \(5 \times 50 = 250\).

ANSWER 11: E

Problem 12:
Compare by extending decimals to the same number of places:
\(0.99   = 0.9900\)
\(0.9099 = 0.9099\)
\(0.9    = 0.9000\)
\(0.909  = 0.9090\)
\(0.9009 = 0.9009\)
Clearly \(0.99 > 0.9099 > 0.909 > 0.9009 > 0.9\). The largest is \(0.99\).

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key Insight:** In a single-elimination tournament, every game eliminates exactly one team. To crown a single winner from 16 teams, we must eliminate 16 − 1 = 15 teams. Therefore, exactly 15 games are needed.

**Verification by rounds:**
- Round 1: 16 teams → 8 games → 8 teams left
- Round 2: 8 teams → 4 games → 4 teams left
- Round 3: 4 teams → 2 games → 2 teams left
- Round 4 (Final): 2 teams → 1 game → 1 winner

Total = 8 + 4 + 2 + 1 = 15 games.

Choices A (4), B (7), and C (8) are too small. Choice E (16) would mean every team is eliminated, leaving no winner.

ANSWER 1: D

---

Problem 2:
We track each person's calculation step by step, all starting from 10.

**Jose:** subtract 1 → 9; double → 18; add 2 → **20**
**Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
**Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

Comparing: 20 < 21 < 22. Kareem gets the largest final answer.

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

Choice A (.008) would be 8/1000. Choice C (.8) would be 8/10. Choices D and E are greater than 1, but 2/25 is clearly less than 1.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination problem — the order of selection doesn't matter:
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4$$

Equivalently, choosing 3 people to start is the same as choosing 1 person to sit out. There are 4 people, so 4 ways to pick who sits out.

The 4 possible starting lineups are: {L,S,J}, {L,S,F}, {L,J,F}, {S,J,F}.

ANSWER 4: B

---

Problem 5:
We're told 20% of a number is 12. Let the number be $x$.

$$0.20x = 12 \implies x = \frac{12}{0.20} = 60$$

Now find 30% of 60:
$$0.30 \times 60 = 18$$

**Shortcut:** 30% is $\frac{30}{20} = 1.5$ times 20%, so the answer is $1.5 \times 12 = 18$.

Choice A (15) would be 25%. Choice D (24) would be 40%. Choice E (30) would be 50%.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing **West**.

**Move 1:** Clockwise $2\frac{1}{4}$ revolutions.
- The 2 full revolutions bring it back to West.
- The extra $\frac{1}{4}$ revolution clockwise from West: West → **North**.

**Move 2:** Counterclockwise $3\frac{3}{4}$ revolutions.
- The 3 full revolutions bring it back to North.
- The extra $\frac{3}{4}$ revolution counterclockwise from North:
  - $\frac{1}{4}$ CCW from North → West
  - $\frac{2}{4}$ CCW from North → South
  - $\frac{3}{4}$ CCW from North → **East**

The spinner points East.

ANSWER 6: B

---

Problem 7:
Original mixture: 30 liters total.
- Red tint: 25% of 30 = 7.5 liters
- Yellow tint: 30% of 30 = 9 liters
- Water: 45% of 30 = 13.5 liters

Add 5 liters of yellow tint:
- New yellow tint = 9 + 5 = 14 liters
- New total mixture = 30 + 5 = 35 liters

Percent yellow tint in new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

Choice B (35%) is a trap for those who just add 5% to 30%. Choice D (45%) is the original water percentage.

ANSWER 7: C

---

Problem 8:
We compare $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$.

Rewrite each as a mixed number:
$$\frac{15}{11} = 1 + \frac{4}{11}, \quad \frac{19}{15} = 1 + \frac{4}{15}, \quad \frac{17}{13} = 1 + \frac{4}{13}$$

Since all three have the same numerator (4) in their fractional parts, we compare denominators. For fractions with equal numerators, a **larger denominator** means a **smaller fraction**:
$$15 > 13 > 11 \implies \frac{4}{15} < \frac{4}{13} < \frac{4}{11}$$

Therefore:
$$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$$

ANSWER 8: E

---

Problem 9:
The data set is: 3, 4, 5, 6, 6, 7, and $x$. The mean, median, and unique mode are all equal.

**Mode:** Since 6 already appears twice and all others appear once, the unique mode must be 6 (if $x$ were 3, 4, 5, or 7, we'd have two modes, violating "unique mode"). So mode = 6.

**Mean = 6:**
$$\frac{3 + 4 + 5 + 6 + 6 + 7 + x}{7} = 6$$
$$\frac{31 + x}{7} = 6 \implies 31 + x = 42 \implies x = 11$$

**Check Median:** With $x = 11$, the sorted list is: 3, 4, 5, **6**, 6, 7, 11. The median (4th value) is 6. ✓

All three measures equal 6. Choice B (6) would make 6 appear three times, but the mean would be $(31+6)/7 = 37/7 \neq 6$.

ANSWER 9: D

---

Problem 10:
Joe walked half the distance in 6 minutes. He then ran the remaining half at 3 times his walking speed.

Let walking speed = $v$. The half-distance = $6v$.

Running speed = $3v$. Time to run the second half:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{6v}{3v} = 2 \text{ minutes}$$

Total time = 6 + 2 = **8 minutes**.

Choice A (7) would result from incorrectly subtracting. Choices B, C, E involve unnecessary decimal complications.

ANSWER 10: D

---

Problem 11:
Compute $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$.

Pair the terms from the first group with terms from the second group (first with last, etc.):
- $1 + 49 = 50$
- $11 + 39 = 50$
- $21 + 29 = 50$
- $31 + 19 = 50$
- $41 + 9 = 50$

Total = $50 \times 5 = 250$.

Choice C (200) is a common error from miscounting pairs. Choice D (249) comes from arithmetic mistakes.

ANSWER 11: E

---

Problem 12:
Compare the five decimals by writing them to four decimal places:

- A: 0.9900
- B: 0.9099
- C: 0.9000
- D: 0.9090
- E: 0.9009

Comparing digit by digit from the tenths place: all have 9 in the tenths place. In the hundredths place, A has **9** while all others have **0**. So A is immediately the largest.

$0.99 > 0.9099 > 0.909 > 0.9009 > 0.9$

ANSWER 12: A

 **Problem 1:**
We need to find the number of games played in a single-elimination tournament with 16 teams. In single elimination, every game eliminates exactly one team. To determine a winner, 15 of the 16 teams must be eliminated. Therefore, exactly 15 games are required.  
(Another way: with 16 teams there are 8 games in the first round, 4 in the second, 2 in the third, and 1 final game: \(8+4+2+1=15\).)

ANSWER 1: D

---

**Problem 2:**
We compute each person’s final answer step by step, starting from 10.

- **Jose:**  
  \(10 - 1 = 9\)  
  \(9 \times 2 = 18\)  
  \(18 + 2 = 20\)

- **Thuy:**  
  \(10 \times 2 = 20\)  
  \(20 - 1 = 19\)  
  \(19 + 2 = 21\)

- **Kareem:**  
  \(10 - 1 = 9\)  
  \(9 + 2 = 11\)  
  \(11 \times 2 = 22\)

Comparing the results: \(20 < 21 < 22\). Kareem’s answer is the largest.

ANSWER 2: C

---

**Problem 3:**
We convert the fraction \(\frac{2}{25}\) to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100:
\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

ANSWER 3: B

---

**Problem 4:**
We must choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways to choose 3 from 4 is the combination \(\binom{4}{3}\):
\[
\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4
\]
Equivalently, choosing 3 starters is the same as choosing 1 person to sit out, and there are 4 choices for who sits out.

ANSWER 4: B

---

**Problem 5:**
Let the number be \(N\). We are given \(0.20N = 12\). Solving for \(N\):
\[
N = \frac{12}{0.20} = 60
\]
Now find 30% of 60:
\[
0.30 \times 60 = 18
\]
(Shortcut: 30% is \(1.5\) times 20%, so \(1.5 \times 12 = 18\).)

ANSWER 5: B

---

**Problem 6:**
The spinner starts pointing west.  
- Clockwise \(2\frac14\) revolutions: 2 full revolutions bring it back to west, then \(\frac14\) turn clockwise from west points it **north**.  
- Counterclockwise \(3\frac34\) revolutions: 3 full revolutions bring it back to north, then \(\frac34\) turn counterclockwise from north goes through west, south, and ends at **east**.

Net movement: \(3\frac34 - 2\frac14 = 1\frac12\) revolutions counterclockwise. One full revolution returns to west; an additional \(\frac12\) revolution counterclockwise from west points to **east**.

ANSWER 6: B

---

**Problem 7:**
Original mixture: 30 liters.  
- Yellow tint originally: \(30\% \times 30 = 9\) liters.  
After adding 5 liters of yellow tint:  
- New yellow tint: \(9 + 5 = 14\) liters.  
- New total mixture: \(30 + 5 = 35\) liters.  

Percent of yellow tint in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%
\]

ANSWER 7: C

---

**Problem 8:**
Compare \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\). Each is greater than 1. Write each as \(1\) plus a proper fraction:
\[
\frac{15}{11} = 1 + \frac{4}{11}, \qquad
\frac{19}{15} = 1 + \frac{4}{15}, \qquad
\frac{17}{13} = 1 + \frac{4}{13}
\]
Since the numerators are the same (4), the fraction with the largest denominator is smallest:
\[
\frac{4}{15} < \frac{4}{13} < \frac{4}{11}
\]
Adding 1 preserves the order:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

ANSWER 8: E

---

**Problem 9:**
The data set is \(3, 4, 5, 6, 6, 7, x\).  
- The current mode is 6 (appears twice). For 6 to remain the **unique** mode, \(x\) must not equal 3, 4, 5, or 7 (which would create a second mode).  
- We are told mean = median = mode = 6.  

Set the mean equal to 6:
\[
\frac{3+4+5+6+6+7+x}{7} = 6 \implies \frac{31+x}{7} = 6 \implies 31+x = 42 \implies x = 11
\]

Check: With \(x=11\), the sorted list is \(3, 4, 5, 6, 6, 7, 11\).  
- Median (4th term) = 6.  
- Mode = 6 (unique).  
- Mean = 6.  

All conditions are satisfied.

ANSWER 9: D

---

**Problem 10:**
Joe walks half the distance in 6 minutes. He runs the second half at 3 times his walking speed. Since distance is the same and speed is tripled, the time required is divided by 3:
\[
\text{Running time} = \frac{6}{3} = 2 \text{ minutes}
\]
Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

---

**Problem 11:**
Add the two groups:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]
Pair the terms to make convenient sums:
\[
(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 5 \times 50 = 250
\]

ANSWER 11: E

---

**Problem 12:**
Compare the decimals by writing them with the same number of places:
- A. \(0.9900\)
- B. \(0.9099\)
- C. \(0.9000\)
- D. \(0.9090\)
- E. \(0.9009\)

The tenths digit is 9 for all. Looking at the hundredths digit: A has 9, while B, D have 0 and C, E have 0. Thus \(0.9900\) is clearly the largest.

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key insight:** Each game eliminates exactly one team. To determine a winner from 16 teams, we must eliminate 15 teams (everyone except the champion).

Therefore, exactly 15 games must be played.

*Verification by rounds:* Round 1: 8 games (16→8 teams), Round 2: 4 games (8→4), Round 3: 2 games (4→2), Round 4: 1 game (2→1). Total = 8+4+2+1 = 15.

The tempting wrong answer E (16) counts teams instead of eliminations.

ANSWER 1: D

---

Problem 2:
We need to compute each person's final answer starting from 10.

**Jose:** Start with 10 → subtract 1: 10−1 = 9 → double: 9×2 = 18 → add 2: 18+2 = **20**

**Thuy:** Start with 10 → double: 10×2 = 20 → subtract 1: 20−1 = 19 → add 2: 19+2 = **21**

**Kareem:** Start with 10 → subtract 1: 10−1 = 9 → add 2: 9+2 = 11 → double: 11×2 = **22**

Kareem gets the largest final answer (22).

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

The tempting wrong answer A (0.008) would be 8/1000, and C (0.8) would be 8/10.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination: C(4,3) = 4!/(3! × 1!) = 4

Listing them out: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — that's 4 ways.

The tempting wrong answer C (6) would be C(4,2), choosing who's *not* a starter, but that also equals 4... Actually C(4,2)=6 is wrong here; we're choosing 3 from 4.

ANSWER 4: B

---

Problem 5:
We know 20% of a number is 12. We need to find 30% of the same number.

**Step 1:** Find the number.
$$0.20 \times n = 12 \implies n = \frac{12}{0.20} = 60$$

**Step 2:** Find 30% of 60.
$$0.30 \times 60 = 18$$

*Shortcut:* Since 30% is 1.5 times 20%, the answer is 1.5 × 12 = 18.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing west. We need to find the net rotation.

**Clockwise:** 2¼ revolutions
**Counterclockwise:** 3¾ revolutions

**Net movement:** 2¼ − 3¾ = −1½ revolutions (negative means counterclockwise)

So the spinner moves 1½ revolutions counterclockwise from west.

- 1 full revolution counterclockwise returns to **west**
- ½ revolution counterclockwise from west goes to **east**

(Counterclockwise from west: ¼ turn → south, ½ turn → east)

ANSWER 6: B

---

Problem 7:
We need to find the percent of yellow tint after adding 5 liters of yellow tint.

**Original mixture (30 liters):**
- Yellow tint: 30% of 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percent yellow tint:** (14/35) × 100 = 40%

The tempting wrong answer B (35) might come from simply adding 5% to 30%.

ANSWER 7: C

---

Problem 8:
We need to order 15/11, 19/15, and 17/13 from least to greatest.

Computing each as a decimal:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3077...

From least to greatest: **19/15 < 17/13 < 15/11**

*Alternative method:* Note each is 1 + (numerator − denominator)/denominator:
- 15/11 = 1 + 4/11
- 19/15 = 1 + 4/15
- 17/13 = 1 + 4/13

Since 4/15 < 4/13 < 4/11, the order is confirmed.

ANSWER 8: E

---

Problem 9:
The positive integers 3, 4, 5, 6, 6, 7, and x have equal mean, median, and unique mode.

**Mode:** Since 6 already appears twice and no other number does, the unique mode must be 6. So x ≠ 3, 4, 5, or 7 (which would create a second mode).

**Median:** With 7 numbers, the median is the 4th value when sorted. For the median to be 6, we need x ≥ 6 (so that 6 is the 4th value).

- If x = 6: sorted = 3, 4, 5, 6, 6, 6, 7 → median = 6 ✓
- If x > 7: sorted = 3, 4, 5, 6, 6, 7, x → median = 6 ✓

**Mean:** (3 + 4 + 5 + 6 + 6 + 7 + x)/7 = (31 + x)/7

Setting mean = 6: (31 + x)/7 = 6 → 31 + x = 42 → **x = 11**

**Verification:** Numbers: 3, 4, 5, 6, 6, 7, 11. Mean = 42/7 = 6, Median = 6, Mode = 6 ✓

ANSWER 9: D

---

Problem 10:
Joe walks halfway, then runs the rest. We need total time.

**Walking:** Half the distance takes 6 minutes.

**Running:** He runs 3 times as fast as he walks, so he covers the same distance in 1/3 the time.
- Time to run the second half = 6 ÷ 3 = 2 minutes

**Total time:** 6 + 2 = **8 minutes**

The tempting wrong answers come from setting up the problem incorrectly or mixing up distance and time relationships.

ANSWER 10: D

---

Problem 11:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

**First group:** 1 + 11 + 21 + 31 + 41 = 105

**Second group:** 9 + 19 + 29 + 39 + 49 = 145

**Total:** 105 + 145 = **250**

*Elegant method:* Pair terms: (1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250

ANSWER 11: E

---

Problem 12:
We need to find the largest number among the options. Writing each with the same number of decimal places:

- A. 0.99 = **0.9900**
- B. 0.9099 = 0.9099
- C. 0.9 = 0.9000
- D. 0.909 = 0.9090
- E. 0.9009 = 0.9009

Comparing digit by digit after the decimal point:
- First digit: all are 9
- Second digit: A has **9**, all others have 0

Since A has a 9 in the second decimal place while all others have 0, **0.99 is the largest**.

ANSWER 12: A

**Problem 1:**  
The tournament is single‑elimination. Starting with 16 teams, every game eliminates exactly one team. To reduce from 16 teams to a single champion we must eliminate 15 teams, and each elimination corresponds to one game.  

Number of games = 16 − 1 = **15**.  

ANSWER 1: D  

---

**Problem 2:**  
Start with 10 for each person.

- **Jose:**  
  Subtract 1 → 9; double → 18; add 2 → 20.  

- **Thuy:**  
  Double → 20; subtract 1 → 19; add 2 → 21.  

- **Kareem:**  
  Subtract 1 → 9; add 2 → 11; double → 22.  

The largest final value is 22, obtained by Kareem.  

ANSWER 2: C  

---

**Problem 3:**  

\[
\frac{2}{25}=0.08
\]

(2 ÷ 25 = 0.08).  

ANSWER 3: B  

---

**Problem 4:**  
We need to choose 3 starters from the 4 available players:  

\[
\binom{4}{3}=4
\]

ways.  

ANSWER 4: B  

---

**Problem 5:**  
Let the number be \(N\).  
\(0.20N = 12 \Rightarrow N = \frac{12}{0.20}=60\).  
\(30\%\) of \(N\) is \(0.30 \times 60 = 18\).  

ANSWER 5: B  

---

**Problem 6:**  
Interpret the fractions as mixed numbers:  

- Clockwise \(2\frac14 = 2.25\) revolutions.  
- Counter‑clockwise \(3\frac34 = 3.75\) revolutions.  

Net rotation (counter‑clockwise positive) = \(3.75 - 2.25 = 1.5\) revolutions.  

A full revolution returns to the same direction, so we only care about the extra \(0.5\) revolution.  
\(0.5\) revolution = 180° counter‑clockwise.  

Starting from west, turning 180° counter‑clockwise points to **east**.  

ANSWER 6: B  

---

**Problem 7:**  
Original yellow = \(30\%\) of 30 L = \(0.30 \times 30 = 9\) L.  
Add 5 L yellow → new yellow = \(9+5 = 14\) L.  
New total volume = \(30+5 = 35\) L.  

Percent yellow = \(\frac{14}{35}\times100 = 40\%\).  

ANSWER 7: C  

---

**Problem 8:**  
Compute the three fractions:

\[
\frac{15}{11}=1.3636\ldots,\qquad
\frac{19}{15}=1.2666\ldots,\qquad
\frac{17}{13}=1.3077\ldots
\]

Ordered from least to greatest:  

\(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

This matches option **E**.  

ANSWER 8: E  

---

**Problem 9:**  
Current list: \(3,4,5,6,6,7\). The unique mode is 6 (appears twice).  
For the mode to remain unique and equal to the mean and median, \(x\) must not create a tie.  
Let the mean = median = mode = \(m\).  

Sum of numbers = \(31 + x\).  
Mean = \(\frac{31 + x}{7}=m\).  
Median of the seven numbers is the 4th after sorting; with \(x>7\) the median stays 6, so \(m=6\).  

Thus \(\frac{31 + x}{7}=6 \Rightarrow 31 + x = 42 \Rightarrow x = 11\).

Check: sorted list becomes \(3,4,5,6,6,7,11\) → median = 6, mode = 6, mean = 42/7 = 6. All equal.  

ANSWER 9: D  

---

**Problem 10:**  
Let the walking speed be \(v\) and the half‑distance be \(d/2\).  

Walking time for half‑distance: \(\frac{d/2}{v}=6\) min ⇒ \(\frac{d}{v}=12\) min.  

Running speed = \(3v\).  
Running time for the second half: \(\frac{d/2}{3v}= \frac{d}{6v}= \frac{1}{6}\times\frac{d}{v}= \frac{12}{6}=2\) min.  

Total time = \(6 + 2 = 8\) minutes.  

ANSWER 10: D  

---

**Problem 11:**  

\[
\begin{aligned}
1+11+21+31+41 &= 105,\\
9+19+29+39+49 &= 145,\\
\text{Total} &= 105+145 = 250.
\end{aligned}
\]

ANSWER 11: E  

---

**Problem 12:**  
Write the numbers with four decimal places:

- .99 = 0.9900  
- .9099 = 0.9099  
- .9 = 0.9000  
- .909 = 0.9090  
- .9009 = 0.9009  

The largest is 0.9900, i.e., **.99**.  

ANSWER 12: A  

Problem 1:  
The tournament starts with 16 teams. Each game eliminates one team, and the losing team is eliminated. To determine the winner, all teams except one must be eliminated. Thus, 15 teams must be eliminated, requiring 15 games.  
- **Why not other choices?** A (4) is too few; B (7), C (8) are also insufficient; E (16) would imply one game per team without elimination, which is incorrect.  
ANSWER 1: D  

Problem 2:  
- **Jose**: Starts with 10 → subtracts 1 (9) → doubles (18) → adds 2 → **20**.  
- **Thuy**: Starts with 10 → doubles (20) → subtracts 1 (19) → adds 2 → **21**.  
- **Kareem**: Starts with 10 → subtracts 1 (9) → adds 2 (11) → doubles → **22**.  
Kareem has the largest final answer.  
- **Why not other choices?** A (Jose) and B (Thuy) are smaller; D and E pair incorrect winners.  
ANSWER 2: C  

Problem 3:  
\( \frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08 \).  
- **Why not other choices?** A (.008) is \( \frac{2}{250} \); C (.8) is \( \frac{4}{5} \); D and E are larger decimals.  
ANSWER 3: B  

Problem 4:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is \( \binom{4}{3} = 4 \).  
- **Why not other choices?** A (2) is too few; C (6) is \( \binom{4}{2} \), which is incorrect; D (8) and E (10) are overcounts.  
ANSWER 4: B  

Problem 5:  
Let the number be \( x \). If 20% of \( x \) is 12, then \( 0.2x = 12 \), so \( x = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).  
- **Why not other choices?** A (15) is 25% of 60; C (20) is \( \frac{1}{3} \) of 60; D (24) is 40%; E (30) is 50%.  
ANSWER 5: B  

Problem 6:  
- A full revolution is 360°. Clockwise \( 2 \frac{1}{4} \) revolutions = \( 2.25 \times 360 = 810° \).  
- Counterclockwise \( 3 \frac{3}{4} \) revolutions = \( -3.75 \times 360 = -1350° \).  
- Net movement: \( 810° - 1350° = -540° \).  
- \( -540° \mod 360° = 180° \) (since \( -540 + 720 = 180° \)), which points west → south? Wait, initial direction is west.  
- West (270°) + 810° = 1080° → 1080° mod 360° = 0° (north). Then -1350° → 0° - 1350° = -1350° mod 360° = 90° (east)? Let's reframe:  
- Start at west (270°). Clockwise 2.25 rev: 270° + 810° = 1080° mod 360° = 0° (north). Then counterclockwise 3.75 rev: 0° - 1350° = -1350° mod 360° = 90° (east).  
- **Why not other choices?** A (north) is initial after clockwise; B (east) is final; C (south) is 180°; D (west) is start; E (northwest) is not a cardinal direction.  
ANSWER 6: B  

Problem 7:  
Original mixture: 30 liters, 30% yellow tint → 9 liters yellow. Add 5 liters yellow → total yellow = 14 liters. New mixture = 35 liters. Percent yellow = \( \frac{14}{35} \times 100 = 40\% \).  
- **Why not other choices?** A (25) is original yellow%; B (35) is before addition; D (45) and E (50) are overestimates.  
ANSWER 7: E (Note: Correction based on computation: 14/35=0.4 → 40%, so answer is E? Wait, 14/35=0.4 → 40%, which is E. But earlier I thought B, but B is 35, which is wrong. Original yellow is 9, add 5 → 14. Total mixture 30+5=35. 14/35=0.4 → 40%. So E. But in choices E is 50? Wait, no: choices are A.25 B.35 C.40 D.45 E.50. So 40% is C. I misread. Correction: 14/35=0.4 → 40%, which is C.  
**Correction**: 14/35=0.4 → 40%, so answer is C.  
ANSWER 7: C  

Problem 8:  
Compare \( \frac{15}{11} \approx 1.3636 \), \( \frac{19}{15} \approx 1.2667 \), \( \frac{17}{13} \approx 1.3077 \). Order: \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11} \).  
- **Why not other choices?** A, B, C, D have incorrect orders.  
ANSWER 8: E  

Problem 9:  
The set is {3, 4, 5, 6, 6, 7, x}. Mean = median = mode = 6.  
- Mode is 6 (unique), so 6 must appear most.  
- Median is 6 (4th value in ordered list: 3,4,5,6,6,7,x → if x≥6, ordered is ...,6,6,x or ...,6,x,6? Must have 4th value 6.  
- If x=6, set is {3,4,5,6,6,6,7}: median=6, mode=6, mean=(3+4+5+6+6+6+7)/7=37/7≈5.28 ≠6.  
- If x=11: set {3,4,5,6,6,7,11}: mean=(3+4+5+6+6+7+11)/7=42/7=6, median=6, mode=6.  
- **Why not other choices?** A (5): mean=36/7≈5.14; B (6): mean≈5.28; C (7): mean=38/7≈5.43; D (11) works.  
ANSWER 9: D  

Problem 10:  
Joe walks half the distance in 6 minutes. Let walking speed be \( v \), so distance = \( 6v \). He runs at \( 3v \), so time to run half distance = \( \frac{6v}{3v} = 2 \) minutes. Total time = 6 + 2 = 8 minutes.  
- **Why not other choices?** A (7), B (7.3), C (7.7), E (8.3) are not exact.  
ANSWER 10: D  

Problem 11:  
First group: 1+11+21+31+41 = 105. Second group: 9+19+29+39+49 = 145. Total = 105 + 145 = 250.  
- **Why not other choices?** A (150), B (199), C (200), D (249) are incorrect sums.  
ANSWER 11: E  

Problem 12:  
Compare decimals: A (.99), B (.9099), C (.9), D (.909), E (.9009). Largest is .99.  
- **Why not other choices?** B, C, D, E are all smaller than .99.  
ANSWER 12: A

### Problem 1:
We need to find the total number of games to determine a winner in a 16-team single-elimination tournament, where every game eliminates exactly one losing team. To get 1 winner, we must eliminate 16 - 1 = 15 teams. Since each game eliminates 1 team, we need 15 total games. Wrong choices: A is the number of bracket rounds, not games; C is half the number of teams; E is the total number of teams, not games.
ANSWER 1: D

---

### Problem 2:
All three start with 10, we calculate each person's final result step-by-step:
- Jose: (10 - 1) * 2 + 2 = 9*2 +2 = 20
- Thuy: (10*2) -1 +2 = 20-1+2=21
- Kareem: (10-1 +2)*2 = 11*2=22
Kareem has the largest final answer. Wrong choices stem from misordering the operations for each person.
ANSWER 2: C

---

### Problem3:
Convert 2/25 to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100: (2*4)/(25*4) = 8/100 = 0.08. Wrong choices: A is 2/250, C is 20/25.
ANSWER 3: B

---

### Problem4:
We choose 3 starters from 4 people, which is a combination problem (order of starters does not matter here). The number of ways to choose 3 from 4 is equivalent to choosing 1 person to leave off the starting lineup: there are 4 people, so 4 unique lineups, each omitting one person. Wrong choice C (6) is the number of permutations, not combinations.
ANSWER 4: B

---

### Problem5:
If 20% of a number is 12, first find the full number: 0.2*N=12 → N=12/0.2=60. 30% of 60 is 0.3*60=18. Wrong choices: A is 25% of 60, D is 40% of 60.
ANSWER 5: B

---

### Problem6:
The spinner starts pointing west. Net movement: 2.25 clockwise revolutions minus 3.75 clockwise (equivalent to 3.75 counterclockwise) = -1.5 total clockwise revolutions, or 1.5 counterclockwise revolutions. Full revolutions cancel out, so we only have a net 0.5 (180 degree) counterclockwise turn from west. A 180-degree turn from west points to east.
ANSWER 6: B

---

### Problem7:
Original 30L mixture has 30% yellow tint: 0.3*30=9L of yellow. Add 5L of yellow, so new yellow is 14L, new total mixture is 35L. The new percentage of yellow is (14/35)*100=40%. Wrong choice: people often forget to add 5L to the total mixture volume, leading to an incorrect higher percentage.
ANSWER 7: C

---

### Problem8:
Rewrite each fraction to compare easily: 15/11 = 1 + 4/11, 19/15=1+4/15, 17/13=1+4/13. For equal numerators, the larger the denominator, the smaller the fraction. So 4/15 < 4/13 < 4/11, so the full order from least to greatest is 19/15 < 17/13 <15/11.
ANSWER 8: E

---

### Problem9:
The dataset is [3,4,5,6,6,7,x]. The unique mode is currently 6 (only value appearing twice, so it must stay the only unique mode). For the mean, median, and mode to all be 6: the sum of the dataset must be 7*6=42. The sum of the existing numbers is 3+4+5+6+6+7=31, so x=42-31=11. Sorted dataset [3,4,5,6,6,7,11] has median 6, unique mode 6, mean 6, which fits. Wrong choice B (x=6) gives a mean of ~5.28, not 6.
ANSWER 9: D

---

### Problem10:
Joe walks half the distance in 6 minutes. His running speed is 3x his walking speed, so time to cover the same distance is 1/3 of walking time. Time to run the second half: 6/3=2 minutes. Total time: 6+2=8 minutes.
ANSWER 10: D

---

### Problem11:
Pair terms to simplify addition: (1+49)+(11+39)+(21+29)+(31+19)+(41+9) = 50+50+50+50+50=250.
ANSWER 11: E

---

### Problem12:
Line up all decimals to compare: A=0.9900, B=0.9099, C=0.9000, D=0.9090, E=0.9009. The largest value is 0.99, option A.
ANSWER 12: A

Problem 1:  
In a single-elimination tournament, each game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams (all except 1). Thus, 15 games are played.  
ANSWER 1: D  

Problem 2:  
Calculate each person's result starting from 10:  
- Jose: \(10 -1 =9\), double → \(9*2=18\), add 2 → \(18+2=20\)  
- Thuy: Double 10 → \(10*2=20\), subtract 1 → \(20-1=19\), add 2 → \(19+2=21\)  
- Kareem: \(10-1=9\), add 2 → \(9+2=11\), double → \(11*2=22\)  
Kareem has the largest result.  
ANSWER 2: C  

Problem 3:  
Convert \(\frac{2}{25}\) to decimal: \(2\div25=0.08\).  
ANSWER 3: B  

Problem 4:  
Choosing 3 starters from 4 people is a combination (order does not matter). The number of ways is \(\binom{4}{3}=4\) (or list: {Lance,Sally,Joy}, {Lance,Sally,Fred}, {Lance,Joy,Fred}, {Sally,Joy,Fred}).  
ANSWER 4: B  

Problem 5:  
Let the number be \(x\). \(20\%x=12\) → \(0.2x=12\) → \(x=60\). \(30\%\ of\ 60=0.3*60=18\).  
ANSWER 5: B  

Problem 6:  
Model directions with standard angles: 0°=east, 90°=north, 180°=west, 270°=south. Initial direction is west (180°).  
- Clockwise \(2\frac{1}{4}\) revolutions: \(2.25*(-360°)=-810°\), new angle: \(180-810=-630°\)  
- Counterclockwise \(3\frac{3}{4}\) revolutions: \(3.75*(360°)=+1350°\), total angle: \(-630+1350=720°\)  
- \(720°\mod360°=0°\), which is east.  
ANSWER 6: B  

Problem 7:  
Original yellow tint: \(30\%\ of\ 30=0.3*30=9\) liters. Add 5 liters, total yellow = \(9+5=14\) liters. New total mixture: \(30+5=35\) liters. Percent yellow: \(\frac{14}{35}*100\%=40\%\).  
ANSWER 7: C  

Problem 8:  
Compare fractions via cross-multiplication or decimals:  
- \(\frac{19}{15}\approx1.267\), \(\frac{17}{13}\approx1.308\), \(\frac{15}{11}\approx1.364\)  
Order from least to greatest: \(\frac{19}{15}<\frac{17}{13}<\frac{15}{11}\).  
ANSWER 8: E  

Problem 9:  
The unique mode is 6 (appears twice, others once). Thus mean and median equal 6.  
- Median of 7 numbers is the 4th term when sorted, so 4th term =6 → \(x\geq6\).  
- Mean = \(\frac{3+4+5+6+6+7+x}{7}=6\) → \(31+x=42\) → \(x=11\).  
Check: Sorted list [3,4,5,6,6,7,11], median=6, mode=6 (unique), mean=6.  
ANSWER 9: D  

Problem 10:  
Walking half the distance takes 6 minutes, so full walking time would be 12 minutes. Running speed is 3x walking speed, so time for the second half is \(\frac{6}{3}=2\) minutes. Total time: \(6+2=8\) minutes.  
ANSWER 10: D  

Problem 11:  
Sum first group: \(1+11+21+31+41=105\) (arithmetic series: \(\frac{5}{2}(1+41)=105\)).  
Sum second group: \(9+19+29+39+49=145\) (\(\frac{5}{2}(9+49)=145\)).  
Total: \(105+145=250\).  
ANSWER 11: E  

Problem 12:  
Compare decimals digit by digit:  
- A=0.99, B=0.9099, C=0.9, D=0.909, E=0.9009  
A has the highest second decimal digit (9 vs 0 for others), so it is the largest.  
ANSWER 12: A

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** How many games must be played to determine a winner in a 16-team elimination tournament?

**Setup:** In an elimination tournament, each game eliminates exactly one team. To go from 16 teams to 1 winner, we must eliminate 15 teams. Therefore, we need 15 games.

**Alternatively:** We can think of it as a bracket:
- Round 1: 16 teams → 8 games → 8 winners
- Round 2: 8 teams → 4 games → 4 winners
- Round 3: 4 teams → 2 games → 2 winners
- Round 4: 2 teams → 1 game → 1 winner
- Total: 8 + 4 + 2 + 1 = 15 games

ANSWER 1: D

---

**Problem 2:**

**What's being asked:** Which person gets the largest final answer after their operations?

**Jose's calculation:** 
- Start: 10
- Subtract 1: 10 - 1 = 9
- Double: 9 × 2 = 18
- Add 2: 18 + 2 = **20**

**Thuy's calculation:**
- Start: 10
- Double: 10 × 2 = 20
- Subtract 1: 20 - 1 = 19
- Add 2: 19 + 2 = **21**

**Kareem's calculation:**
- Start: 10
- Subtract 1: 10 - 1 = 9
- Add 2: 9 + 2 = 11
- Double: 11 × 2 = **22**

Kareem gets 22, which is the largest.

ANSWER 2: C

---

**Problem 3:**

**What's being asked:** Convert the fraction 2/25 to a decimal.

**Setup:** 2/25 = ?

**Calculation:** 
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** In how many ways can 3 starters be chosen from 4 people (Lance, Sally, Joy, Fred)?

**Setup:** This is a combination problem: $\binom{4}{3}$

**Calculation:**
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4}{1} = 4$$

Alternatively, we're choosing which 1 person to leave out: 4 ways.

ANSWER 4: B

---

**Problem 5:**

**What's being asked:** If 20% of a number is 12, what is 30% of that number?

**Setup:** Let the number be $n$.
- $0.20n = 12$
- $n = 12 ÷ 0.20 = 60$

**Calculation:** 
$$30\% \text{ of } 60 = 0.30 \times 60 = 18$$

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Starting pointing west, after rotating clockwise 2¼ revolutions then counterclockwise 3¾ revolutions, which direction?

**Setup:** 
- Clockwise 2¼ revolutions = 2.25 revolutions
- Counterclockwise 3¾ revolutions = 3.75 revolutions
- Net: 3.75 - 2.25 = 1.5 revolutions counterclockwise (net)

**Calculation:**
Starting at west, rotating counterclockwise 1.5 revolutions (= 1 full rotation + 0.5 rotation):
- 1 full rotation brings us back to west
- 0.5 rotation = 180°
- West + 180° counterclockwise = East

ANSWER 6: B

---

**Problem 7:**

**What's being asked:** After adding 5 liters of yellow tint to a 30-liter mixture, what percent is yellow tint?

**Original mixture (30 liters):**
- Yellow tint: 30% × 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percentage:**
$$\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%$$

ANSWER 7: C

---

**Problem 8:**

**What's being asked:** Order 15/11, 19/15, and 17/13 from least to greatest.

**Setup:** Convert to decimals or compare directly.

**Calculations:**
- 15/11 ≈ 1.364
- 19/15 ≈ 1.267
- 17/13 ≈ 1.308

**Order:** 19/15 < 17/13 < 15/11

ANSWER 8: E

---

**Problem 9:**

**What's being asked:** Find $x$ such that the mean, median, and unique mode of {3, 4, 5, 6, 6, 7, $x$} are all equal.

**Current mode:** 6 appears twice; all others appear once. If there's a unique mode, it must be 6.

**With unique mode = 6:** This requires no other number to appear more than once. So $x \notin \{3, 4, 5, 7\}$ and $x \neq 6$ (else it wouldn't be unique).

**Mean = Median = Mode = 6:**

For median of 7 numbers with value 6:
- Sorted: need middle (4th) value to be 6
- If $x = 11$: {3, 4, 5, 6, 6, 7, 11} → median = 6 ✓

Mean: $\frac{3+4+5+6+6+7+11}{7} = \frac{42}{7} = 6$ ✓

Mode: 6 appears twice; all others once → mode = 6 ✓

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Total time from home to school if Joe walked halfway (6 min) then ran the rest at 3× speed?

**Setup:**
- Walking: 6 minutes for half the distance
- Running: same distance at 3× speed, so time = 6/3 = 2 minutes

**Total time:** 6 + 2 = 8 minutes

ANSWER 10: D

---

**Problem 11:**

**What's being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Calculation:**
- First group: 1 + 11 + 21 + 31 + 41 = 105
- Second group: 9 + 19 + 29 + 39 + 49 = 145
- Sum: 105 + 145 = 250

ANSWER 11: E

---

**Problem 12:**

**What's being asked:** Which number is largest?

**Comparison:**
- A. 0.99
- B. 0.9099
- C. 0.9
- D. 0.909
- E. 0.9009

Starting from the tenths place: all are 0.9__. Looking at hundredths:
- 0.99 has 9 in hundredths place → **largest**

ANSWER 12: A

Problem 1:

In a single-elimination tournament, each game eliminates exactly one team.  
There are 16 teams, and we want one champion.

So we must eliminate 15 teams total.

Therefore, the number of games played is 15.

Tempting wrong choice: 16 would be if every team had to lose, but the winner does not lose.

ANSWER 1: D

Problem 2:

Compute each person’s result carefully.

- Jose: start with 10  
  subtract 1 → 9  
  double → 18  
  add 2 → 20

- Thuy: start with 10  
  double → 20  
  subtract 1 → 19  
  add 2 → 21

- Kareem: start with 10  
  subtract 1 → 9  
  add 2 → 11  
  double → 22

Compare:
- Jose = 20
- Thuy = 21
- Kareem = 22

Kareem gets the largest final answer.

ANSWER 2: C

Problem 3:

\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

So the correct decimal is 0.08.

ANSWER 3: B

Problem 4:

We are choosing 3 starters from 4 players: Lance, Sally, Joy, and Fred.

This is a combination problem:
\[
\binom{4}{3} = 4
\]

You can also see it by noticing that choosing 3 starters is the same as choosing which 1 player sits out, and there are 4 choices.

ANSWER 4: B

Problem 5:

Let the number be \(n\).

20% of \(n\) is 12, so:
\[
0.2n = 12
\]
\[
n = 60
\]

Now find 30% of 60:
\[
0.3 \times 60 = 18
\]

ANSWER 5: B

Problem 6:

The spinner starts pointing west.

First move: clockwise \(2\frac{1}{4}\) revolutions.  
A full revolution brings it back to west, so only the extra \(\frac{1}{4}\) revolution matters.

From west, a quarter-turn clockwise points it to north.

Second move: counterclockwise \(3\frac{3}{4}\) revolutions.  
Again, the 3 full revolutions do nothing, so this is just \(\frac{3}{4}\) of a revolution counterclockwise, which is the same as \(\frac{1}{4}\) of a revolution clockwise.

From north, a quarter-turn clockwise points east.

So the spinner ends pointing east.

ANSWER 6: B

Problem 7:

Start with 30 liters total.

Yellow tint is 30% of 30 liters:
\[
0.30 \times 30 = 9 \text{ liters}
\]

Then 5 liters of yellow tint are added, so yellow becomes:
\[
9 + 5 = 14 \text{ liters}
\]

New total mixture:
\[
30 + 5 = 35 \text{ liters}
\]

Percent yellow in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.4 = 40\%
\]

ANSWER 7: C

Problem 8:

The fractions are:
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]

Approximate each one:

\[
\frac{15}{11} \approx 1.364
\]
\[
\frac{17}{13} \approx 1.308
\]
\[
\frac{19}{15} \approx 1.267
\]

So from least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

That matches choice E.

ANSWER 8: E

Problem 9:

The numbers are:
\[
3,4,5,6,6,7,x
\]

We are told the mean, median, and unique mode are all equal.

Since 6 appears twice and every other listed number appears once, the unique mode is 6.

So the median and mean must also be 6.

The median of 7 numbers is the 4th number when ordered.  
The sorted list is:
\[
3,4,5,6,6,7,x
\]
To make the median 6, \(x\) must be 6 or larger, but we also need the mean to be 6.

Sum of the known numbers:
\[
3+4+5+6+6+7 = 31
\]

For the mean to be 6, total sum must be:
\[
7 \times 6 = 42
\]

So:
\[
31 + x = 42
\]
\[
x = 11
\]

Check:
- mode = 6
- median = 6
- mean = 42/7 = 6

Works perfectly.

ANSWER 9: D

Problem 10:

Joe walked half the distance in 6 minutes.  
He then ran the other half at 3 times his walking speed.

If he runs 3 times as fast, then the time for the same distance is 1/3 as long.

So the running time is:
\[
6 \div 3 = 2 \text{ minutes}
\]

Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

Problem 11:

Add the first group:
\[
1+11+21+31+41 = 105
\]

Add the second group:
\[
9+19+29+39+49 = 145
\]

Now add the totals:
\[
105 + 145 = 250
\]

ANSWER 11: E

Problem 12:

Compare the decimals:

- \(0.99\)
- \(0.9099\)
- \(0.9 = 0.9000\)
- \(0.909\)
- \(0.9009\)

The largest is clearly \(0.99\), since it is closest to 1.

ANSWER 12: A

Problem 1:  
We have a three-person-team tournament where **the losing team of each game is eliminated**. With **16 teams**, how many games are played to determine the winner?

To find the number of games in a single-elimination tournament:  
- Each game eliminates exactly **one** team (the loser).  
- To go from **16 teams to 1 winning team**, we must eliminate **15 teams**.  
- Therefore, the number of games is **15**.

So the correct choice is **D**.

ANSWER 1: D


Problem 2:  
Jose, Thuy, and Kareem all start at 10. We compute each final value.

**Jose:**  
- Subtract 1: \(10 - 1 = 9\)  
- Double: \(2 \cdot 9 = 18\)  
- Add 2: \(18 + 2 = 20\)  
Final: **20**

**Thuy:**  
- Double 10: \(2 \cdot 10 = 20\)  
- Subtract 1: \(20 - 1 = 19\)  
- Add 2: \(19 + 2 = 21\)  
Final: **21**

**Kareem:**  
- Subtract 1: \(10 - 1 = 9\)  
- Add 2: \(9 + 2 = 11\)  
- Double: \(2 \cdot 11 = 22\)  
Final: **22**

Largest is **22**, which is Kareem.

ANSWER 2: C


Problem 3:  
We need to evaluate:  
\[
\frac{2}{25}
\]
Convert to decimals:  
\[
\frac{2}{25} = 2 \div 25
\]
Since \(25 \cdot 0.08 = 2\):  
\[
2 \div 25 = 0.08
\]
So the correct choice is **B** (.08).

ANSWER 3: B


Problem 4:  
We have 4 people chosen: Lance, Sally, Joy, Fred. We need to choose **3 starters**.

Number of ways:
\[
\binom{4}{3} = 4
\]
List them conceptually: leaving out each one individually gives 4 choices.

Correct choice: **B**.

ANSWER 4: B


Problem 5:  
“If 20% of a number is 12, what is 30% of the same number?”

Let the number be \(N\).  
\[
0.20N = 12
\]
\[
N = \frac{12}{0.20} = 60
\]
Now compute \(30\%\) of 60:
\[
0.30 \cdot 60 = 18
\]
Correct choice: **B**.

ANSWER 5: B


Problem 6:  
Spinner starts pointing **west**.

Chenille moves it:
- **Clockwise** \(2\frac{1}{4}\) revolutions  
- then **Counterclockwise** \(3\frac{3}{4}\) revolutions

Net rotation (clockwise minus counterclockwise):
\[
2\frac{1}{4} - 3\frac{3}{4}
\]
Convert to improper fractions:
\[
2\frac{1}{4} = \frac{9}{4}, \quad 3\frac{3}{4} = \frac{15}{4}
\]
\[
\frac{9}{4} - \frac{15}{4} = -\frac{6}{4} = -\frac{3}{2}
\]
The negative means **net counterclockwise** by \(\frac{3}{2}\) revolutions.

\(\frac{3}{2} = 1 + \frac{1}{2}\), so this is:
- 1 full turn (brings it back where it started)
- plus \(\frac{1}{2}\) turn counterclockwise

Half a revolution from **west** points **east** (opposite direction).

Correct answer: **B (east)**.

ANSWER 6: B


Problem 7:  
Original mixture: 30 L with  
- 25% red (doesn’t matter directly)
- 30% yellow  
- 45% water

Yellow amount initially:
\[
0.30 \cdot 30 = 9 \text{ liters}
\]
Add 5 liters of yellow tint, so new yellow amount:
\[
9 + 5 = 14
\]
New total volume:
\[
30 + 5 = 35
\]
Percent yellow in new mixture:
\[
\frac{14}{35} = 0.4 = 40\%
\]
Correct choice: **C**.

ANSWER 7: C


Problem 8:  
Order fractions least to greatest:
\[
\frac{15}{11},\ \frac{19}{15},\ \frac{17}{13}
\]
A good way: compare values by cross-multiplication or decimal approximations.

Compute approximate decimals:
\[
\frac{15}{11} \approx 1.3636
\]
\[
\frac{19}{15} \approx 1.2667
\]
\[
\frac{17}{13} \approx 1.3077
\]
Least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]
Check choice: that matches **E**:
- \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11}\)

ANSWER 8: E


Problem 9:  
Numbers: \(3,4,5,6,6,7,x\) (7 positive integers).  
Mean = median = unique mode.

**Step 1: Mode**  
The only number that repeats is **6**, occurring twice. So the **unique mode = 6**.

Thus median must be 6 as well.

**Step 2: Median = 6**  
For 7 numbers sorted, the median is the **4th** number.

Sorted list must have the 4th term equal to 6. We already have two 6’s and surrounding numbers 3,4,5,7.

Currently without \(x\), the ordered portion is:
\[
3,4,5,6,6,7,\ ?
\]
The 4th number here is 6 already. To keep the 4th number as 6, \(x\) must not be less than 6 (otherwise it would shift positions so 4th might change).

If \(x = 6\), then 6 occurs 3 times, still mode 6 (unique mode). Then list:
\[
3,4,5,6,6,6,7
\]
Median (4th) is 6—works.

If \(x=7\), list:
\[
3,4,5,6,6,7,7
\]
Mode would be 6? Actually 7 repeats too, so mode not unique.

If \(x>7\), list:
\[
3,4,5,6,6,7,x
\]
Median still 6, but mode would still be 6 (unique), since only 6 repeats. That would also work—wait, but then 6 is still unique mode, yes. However we also need **mean = 6**.

So use mean condition.

Sum of known numbers:
\[
3+4+5+6+6+7 = 31
\]
Total sum including \(x\): \(31 + x\)

Mean:
\[
\frac{31+x}{7} = 6
\]
\[
31 + x = 42
\]
\[
x = 11
\]

Check uniqueness of mode: 6 occurs twice, no other repeats (since x=11 is different). Unique mode is 6—consistent.

ANSWER 9: D


Problem 10:  
Joe walks half the distance, then runs the rest.  
He runs **3 times as fast** as he walks.

Given: time to walk half way = **6 minutes**.  
Let walking speed be \(v\). Then walking time for half distance \(=\) 6 minutes.

Let half the distance be \(d\). Then:
\[
d = v \cdot 6
\]
Running speed is \(3v\).  
Remaining distance is also \(d\), so running time:
\[
\text{time} = \frac{d}{3v} = \frac{v\cdot 6}{3v} = 2
\]
Total time:
\[
6 + 2 = 8
\]
Correct choice: **D**.

ANSWER 10: D


Problem 11:  
Compute:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]

Group each set:

First set:
\[
1+11+21+31+41
\]
These are all of the form \(10k+1\):  
\(1, 11, 21, 31, 41\) correspond to \(k=0,1,2,3,4\).

Sum:
\[
(1+11+21+31+41) = (1+11+21+31+41)
\]
Compute directly:
\[
(1+11)=12,\ (21+31)=52,\ 12+52=64,\ 64+41=105
\]

Second set:
\[
9+19+29+39+49
\]
Direct sum:
\[
(9+19)=28,\ (29+39)=68,\ 28+68=96,\ 96+49=145
\]

Total:
\[
105 + 145 = 250
\]

Correct choice: **E**.

ANSWER 11: E


Problem 12:  
Which number is largest?

Options:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare decimals:  
- .99 is the largest among all because it is greater than .9 (and greater than any number starting 0.90... but with fewer 9s afterward).

So .99 > .9099 > .909 > .9009 > .9.

Correct choice: **A**.

ANSWER 12: A

Problem 1:
In a single-elimination tournament, every game results in exactly one team being eliminated. To determine a single winner from 16 teams, 15 teams must be eliminated. Since each game eliminates one team, it takes exactly 15 games.
ANSWER 1: D

Problem 2:
- Jose: (10 - 1) = 9; 9 * 2 = 18; 18 + 2 = 20.
- Thuy: (10 * 2) = 20; 20 - 1 = 19; 19 + 2 = 21.
- Kareem: (10 - 1) = 9; 9 + 2 = 11; 11 * 2 = 22.
Kareem has the largest answer (22).
ANSWER 2: C

Problem 3:
To convert 2/25 to a decimal, multiply the numerator and denominator by 4 to get a denominator of 100: (2 * 4) / (25 * 4) = 8 / 100 = 0.08.
ANSWER 3: B

Problem 4:
We need to choose 3 starters from a group of 4 (Lance, Sally, Joy, Fred). This is a combination problem denoted as 4C3. Choosing 3 people to be on the team is the same as choosing 1 person to be left off the team. There are 4 ways to choose who is left off.
ANSWER 4: B

Problem 5:
Let $x$ be the number. $0.20x = 12$. Dividing by 0.20, $x = 12 / 0.2 = 60$. We want 30% of $x$: $0.30 * 60 = 18$.
ANSWER 5: B

Problem 6:
Start at 0 revolutions. Clockwise is positive, counterclockwise is negative.
$2 \frac{1}{4} - 3 \frac{3}{4} = 2.25 - 3.75 = -1.5$ revolutions.
-1.5 revolutions is the same as -0.5 revolutions (since -1 is a full circle). -0.5 revolutions (counterclockwise) from West points to East.
ANSWER 6: B

Problem 7:
Original yellow tint: 30% of 30 liters = 9 liters.
New mixture: 30 + 5 = 35 liters.
New yellow tint: 9 + 5 = 14 liters.
Percent: (14 / 35) * 100 = (2 / 5) * 100 = 40%.
ANSWER 7: C

Problem 8:
Convert to decimals or compare differences from 1:
15/11 = 1 + 4/11 ≈ 1.3636
19/15 = 1 + 4/15 ≈ 1.2666
17/13 = 1 + 4/13 ≈ 1.3076
Comparing the fractions 4/11, 4/13, and 4/15: the larger the denominator, the smaller the fraction. Thus, 4/15 < 4/13 < 4/11.
Order: 19/15 < 17/13 < 15/11.
ANSWER 8: E

Problem 9:
The set is {3, 4, 5, 6, 6, 7, x}. The mode is 6 (it appears twice). For the mean and median to be 6:
Median: With 7 numbers, the 4th number must be 6. Since the sorted list is 3, 4, 5, 6, 6, 7, x, if $x > 6$, the 4th number is 6.
Mean: (3+4+5+6+6+7+x) / 7 = 6 => 31 + x = 42 => x = 11.
Checking: {3, 4, 5, 6, 6, 7, 11}. Median is 6. Mean is 42/7 = 6. Mode is 6.
ANSWER 9: D

Problem 10:
Walking time = 6 minutes. Running speed = 3 * walking speed.
Time taken = Distance / Speed. Since speed is 3 times faster, time is 1/3 of the walking time.
Running time = 6 / 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
ANSWER 10: D

Problem 11:
Sum 1: (1+11+21+31+41) = 105.
Sum 2: (9+19+29+39+49) = 145.
105 + 145 = 250.
ANSWER 11: E

Problem 12:
Compare digit by digit:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
0.99 is the largest.
ANSWER 12: A

Problem 1:
This is a single-elimination tournament with 16 teams. Exactly one team is eliminated per game, and 15 teams must be eliminated to leave a single winner. Thus exactly 15 games are required.

ANSWER 1: D

Problem 2:
Jose evaluates \((10-1)\times2+2=20\).
Thuy evaluates \((10\times2)-1+2=21\).
Kareem evaluates \((10-1+2)\times2=22\).
Kareem obtains the strictly largest value.

ANSWER 2: C

Problem 3:
Division yields the decimal expansion \(2\div25=0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters out of 4 distinct players is the binomial coefficient \(\binom{4}{3}=4\).

ANSWER 4: B

Problem 5:
If 20% of a number equals 12, the number itself is \(12/0.2=60\). Then 30% of 60 is \(0.3\times60=18\).

ANSWER 5: B

Problem 6:
A net rotation of \(+2.25-3.75=-1.5\) revolutions is equivalent to 1.5 revolutions counterclockwise. From an initial westward orientation, 1.5 revolutions counterclockwise lands exactly on east.

ANSWER 6: B

Problem 7:
The original mixture contains \(0.30\times30=9\) liters of yellow tint. Adding 5 liters raises the yellow volume to 14 liters and the total volume to 35 liters, so the new percentage is \(14/35=40\%\).

ANSWER 7: C

Problem 8:
Decimal approximations are \(15/11\approx1.3636\), \(19/15\approx1.2667\), and \(17/13\approx1.3077\). Ordering these values produces \(19/15<17/13<15/11\).

ANSWER 8: E

Problem 9:
The unique mode must be 6, forcing the mean also to equal 6. The sum of the seven numbers is therefore 42. The six given numbers sum to 31, so \(x=11\). With this value the ordered list is 3,4,5,6,6,7,11 whose median is likewise 6 and whose unique mode remains 6.

ANSWER 9: D

Problem 10:
Let the total distance be \(2d\). Walking time for distance \(d\) is 6 min, so running time for the remaining distance \(d\) at triple speed is \(6/3=2\) min. The total time is therefore 8 min.

ANSWER 10: D

Problem 11:
Each pair sums to a multiple of 10: \(1+9=10\), \(11+19=30\), \(21+29=50\), \(31+39=70\), \(41+49=90\). Their total is 250. Equivalently each five-term arithmetic series sums to 105 and 145 respectively, again yielding 250.

ANSWER 11: E

Problem 12:
Direct comparison of the five decimals shows that 0.99 is strictly larger than each of the others.

ANSWER 12: A

Problem 1:
The problem asks us to determine the number of games that will be played to determine the winner of a basketball tournament with 16 teams, where the losing team of each game is eliminated.

## Step 1
To determine the number of games needed to find the winner, we first note that in a single-elimination tournament, each game eliminates one team.

## Step 2
Since there can only be one winner, all other teams must be eliminated. Therefore, the number of games required is equal to the number of teams minus one.

## Step 3
We have 16 teams competing in the tournament. So, the number of games needed to determine the winner is 16 - 1 = 15.

## Step 4
Let's examine the given choices to rule out the incorrect options. A. 4 is too low because it doesn't account for all the teams. B. 7 is also too low for the same reason. C. 8 is still too low. D. 15 matches our calculation. E. 16 is one more than our calculation.

The final answer is supported by our step-by-step reasoning.
ANSWER 1: D

Problem 2:
The problem asks us to determine who among Jose, Thuy, and Kareem gets the largest final answer after performing a series of operations starting with the number 10.

## Step 1
Let's calculate Jose's final answer: Jose subtracts 1 from 10 to get 9, doubles it to get 18, and then adds 2 to get 20.

## Step 2
Now, let's calculate Thuy's final answer: Thuy doubles 10 to get 20, subtracts 1 to get 19, and then adds 2 to get 21.

## Step 3
Next, let's calculate Kareem's final answer: Kareem subtracts 1 from 10 to get 9, adds 2 to get 11, and then doubles it to get 22.

## Step 4
Comparing the final answers: Jose got 20, Thuy got 21, and Kareem got 22. The largest final answer is 22, which belongs to Kareem.

## Step 5
Let's examine the given choices. A. Jose is incorrect because his answer is 20. B. Thuy is incorrect because her answer is 21. C. Kareem is correct because his answer is 22. D. Jose and Thuy is incorrect because Kareem's answer is larger. E. Thuy and Kareem is incorrect because Jose's answer is not the largest.

The final answer is supported by our step-by-step reasoning.
ANSWER 2: C

Problem 3:
The problem asks us to convert the fraction 2/25 into a decimal.

## Step 1
To convert 2/25 into a decimal, we divide the numerator by the denominator: 2 ÷ 25 = 0.08.

## Step 2
Let's examine the given choices to identify the correct decimal representation. A. .008 is too small. B. .08 matches our calculation. C. .8 is too large. D. 1.25 is much too large. E. 12.5 is also much too large.

The final answer is supported by our step-by-step reasoning.
ANSWER 3: B

Problem 4:
The problem asks us to determine the number of ways to choose 3 starters from a team of 4 members: Lance, Sally, Joy, and Fred.

## Step 1
This is a combination problem, because the order in which the starters are chosen does not matter.

## Step 2
The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial, the product of all positive integers up to that number.

## Step 3
Here, n = 4 (total team members) and k = 3 (number of starters to be chosen). So, C(4, 3) = 4! / [3!(4 - 3)!] = 4! / (3! * 1!) = (4 * 3 * 2 * 1) / ((3 * 2 * 1) * 1) = 4.

## Step 4
Let's examine the given choices. A. 2 is too low. B. 4 matches our calculation. C. 6 is too high. D. 8 is also too high. E. 10 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 4: B

Problem 5:
The problem asks us to find 30% of a number, given that 20% of the same number is 12.

## Step 1
First, we need to find the number. We know that 20% of the number is 12. So, 0.20 * number = 12.

## Step 2
To find the number, we divide both sides of the equation by 0.20: number = 12 / 0.20 = 60.

## Step 3
Now, we need to calculate 30% of 60: 0.30 * 60 = 18.

## Step 4
Let's examine the given choices. A. 15 is too low. B. 18 matches our calculation. C. 20 is too high. D. 24 is also too high. E. 30 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 5: B

Problem 6:
The problem asks us to determine the direction the spinner points after Chenille moves it clockwise 2 1/4 revolutions and then counterclockwise 3 3/4 revolutions, starting from pointing west.

## Step 1
First, let's understand the effect of the movements. A full revolution is 360 degrees or 4 quarters. So, 2 1/4 revolutions clockwise is equivalent to 9 quarters clockwise.

## Step 2
9 quarters clockwise is the same as 1 quarter clockwise (since 8 quarters make 2 full revolutions, which brings the spinner back to the original direction). So, after the first move, the spinner points north.

## Step 3
Then, Chenille moves it counterclockwise 3 3/4 revolutions, which is equivalent to 15 quarters counterclockwise.

## Step 4
15 quarters counterclockwise is the same as 3 quarters counterclockwise (since 12 quarters make 3 full revolutions). So, moving 3 quarters counterclockwise from north results in pointing west.

## Step 5
Let's examine the given choices. A. north is incorrect because after the second move, it's not north. B. east is incorrect. C. south is incorrect. D. west matches our calculation. E. northwest is incorrect.

The final answer is supported by our step-by-step reasoning.
ANSWER 6: D

Problem 7:
The problem asks us to find the percentage of yellow tint in a new mixture after adding 5 liters of yellow tint to an original 30-liter mixture that is 25% red tint, 30% yellow tint, and 45% water.

## Step 1
First, we calculate the amount of yellow tint in the original mixture: 30% of 30 liters = 0.30 * 30 = 9 liters.

## Step 2
After adding 5 liters of yellow tint, the total amount of yellow tint becomes 9 + 5 = 14 liters.

## Step 3
The total volume of the new mixture is 30 + 5 = 35 liters.

## Step 4
To find the percentage of yellow tint in the new mixture, we divide the amount of yellow tint by the total volume and multiply by 100: (14 / 35) * 100 = 40%

Problem 1:
In a single-elimination tournament, each game eliminates exactly one team. To eliminate all but the champion, the number of games needed is one fewer than the number of teams. With 16 teams, the number of games is \(16 - 1 = 15\).

ANSWER 1: D

Problem 2:
Jose: \(10 - 1 = 9\), \(9 \times 2 = 18\), \(18 + 2 = 20\).
Thuy: \(10 \times 2 = 20\), \(20 - 1 = 19\), \(19 + 2 = 21\).
Kareem: \(10 - 1 = 9\), \(9 + 2 = 11\), \(11 \times 2 = 22\).
Kareem’s final answer of 22 is the largest.

ANSWER 2: C

Problem 3:
Divide 2 by 25: \(2 \div 25 = 0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters from 4 people is the combination \(\binom{4}{3} = 4\).

ANSWER 4: B

Problem 5:
Let the number be \(n\). \(20\%\) of \(n\) is \(0.2n = 12\), so \(n = 60\). Then \(30\%\) of \(60 = 0.3 \times 60 = 18\).

ANSWER 5: B

Problem 6:
Initial direction: west.
Clockwise \(2\frac{1}{4}\) revolutions: \(2\frac{1}{4} = 2.25\) revolutions clockwise.
Counterclockwise \(3\frac{3}{4}\) revolutions: \(3\frac{3}{4} = 3.75\) revolutions counterclockwise.
Net revolution = \(2.25 - 3.75 = -1.5\) revolutions, i.e., \(1.5\) revolutions counterclockwise.
A full revolution (1.0) returns to west. The remaining 0.5 revolution counterclockwise from west is a half-turn, which points to east.

ANSWER 6: B

Problem 7:
Original mixture: 30 liters, yellow tint = \(30\% \times 30 = 9\) liters.
Add 5 liters of yellow tint: new yellow = \(9 + 5 = 14\) liters; new total volume = \(30 + 5 = 35\) liters.
Percent yellow = \(\frac{14}{35} \times 100\% = 40\%\).

ANSWER 7: C

Problem 8:
Convert each fraction to a decimal for comparison:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
Order from least to greatest: \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

ANSWER 8: E

Problem 9:
The data set: 3, 4, 5, 6, 6, 7, \(x\). The mode is the most frequent number. With two 6’s, 6 is the mode unless \(x\) equals another number that appears twice, which would create a second mode, violating the “unique mode” condition. Therefore, \(x\) must not be 3, 4, 5, or 7. It could be 6, but if \(x=6\), the median is 6, while the mean is \((3+4+5+6+6+6+7)/7 = 37/7 \approx 5.29\), so mean ≠ median.  
For the mean to equal the median (which is 6, the middle of seven sorted numbers if \(x > 6\) or \(x=6\) only if placed correctly), we set the mean to 6. Sum without \(x\) = \(3+4+5+6+6+7 = 31\). Total needed for mean 6 = \(7 \times 6 = 42\). Thus \(x = 42 - 31 = 11\).  
Sorted: 3, 4, 5, 6, 6, 7, 11. Median = 6, mode = 6, mean = 6. Unique mode holds.

ANSWER 9: D

Problem 10:
Walking half the distance took 6 minutes, so walking speed = \(\frac{\text{half distance}}{6}\).
Running speed is 3 times walking speed. Time to run the second half = \(\frac{\text{half distance}}{3 \times (\text{half distance}/6)} = \frac{6}{3} = 2\) minutes.
Total time = \(6 + 2 = 8\) minutes.

ANSWER 10: D

Problem 11:
Pair the terms: \(1+49 = 50\), \(11+39 = 50\), \(21+29 = 50\), \(31+19 = 50\), \(41+9 = 50\). Five pairs of 50 give \(5 \times 50 = 250\).

ANSWER 11: E

Problem 12:
Compare by extending decimals to the same number of places:
\(0.99   = 0.9900\)
\(0.9099 = 0.9099\)
\(0.9    = 0.9000\)
\(0.909  = 0.9090\)
\(0.9009 = 0.9009\)
Clearly \(0.99 > 0.9099 > 0.909 > 0.9009 > 0.9\). The largest is \(0.99\).

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key Insight:** In a single-elimination tournament, every game eliminates exactly one team. To crown a single winner from 16 teams, we must eliminate 16 − 1 = 15 teams. Therefore, exactly 15 games are needed.

**Verification by rounds:**
- Round 1: 16 teams → 8 games → 8 teams left
- Round 2: 8 teams → 4 games → 4 teams left
- Round 3: 4 teams → 2 games → 2 teams left
- Round 4 (Final): 2 teams → 1 game → 1 winner

Total = 8 + 4 + 2 + 1 = 15 games.

Choices A (4), B (7), and C (8) are too small. Choice E (16) would mean every team is eliminated, leaving no winner.

ANSWER 1: D

---

Problem 2:
We track each person's calculation step by step, all starting from 10.

**Jose:** subtract 1 → 9; double → 18; add 2 → **20**
**Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
**Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

Comparing: 20 < 21 < 22. Kareem gets the largest final answer.

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

Choice A (.008) would be 8/1000. Choice C (.8) would be 8/10. Choices D and E are greater than 1, but 2/25 is clearly less than 1.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination problem — the order of selection doesn't matter:
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4$$

Equivalently, choosing 3 people to start is the same as choosing 1 person to sit out. There are 4 people, so 4 ways to pick who sits out.

The 4 possible starting lineups are: {L,S,J}, {L,S,F}, {L,J,F}, {S,J,F}.

ANSWER 4: B

---

Problem 5:
We're told 20% of a number is 12. Let the number be $x$.

$$0.20x = 12 \implies x = \frac{12}{0.20} = 60$$

Now find 30% of 60:
$$0.30 \times 60 = 18$$

**Shortcut:** 30% is $\frac{30}{20} = 1.5$ times 20%, so the answer is $1.5 \times 12 = 18$.

Choice A (15) would be 25%. Choice D (24) would be 40%. Choice E (30) would be 50%.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing **West**.

**Move 1:** Clockwise $2\frac{1}{4}$ revolutions.
- The 2 full revolutions bring it back to West.
- The extra $\frac{1}{4}$ revolution clockwise from West: West → **North**.

**Move 2:** Counterclockwise $3\frac{3}{4}$ revolutions.
- The 3 full revolutions bring it back to North.
- The extra $\frac{3}{4}$ revolution counterclockwise from North:
  - $\frac{1}{4}$ CCW from North → West
  - $\frac{2}{4}$ CCW from North → South
  - $\frac{3}{4}$ CCW from North → **East**

The spinner points East.

ANSWER 6: B

---

Problem 7:
Original mixture: 30 liters total.
- Red tint: 25% of 30 = 7.5 liters
- Yellow tint: 30% of 30 = 9 liters
- Water: 45% of 30 = 13.5 liters

Add 5 liters of yellow tint:
- New yellow tint = 9 + 5 = 14 liters
- New total mixture = 30 + 5 = 35 liters

Percent yellow tint in new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

Choice B (35%) is a trap for those who just add 5% to 30%. Choice D (45%) is the original water percentage.

ANSWER 7: C

---

Problem 8:
We compare $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$.

Rewrite each as a mixed number:
$$\frac{15}{11} = 1 + \frac{4}{11}, \quad \frac{19}{15} = 1 + \frac{4}{15}, \quad \frac{17}{13} = 1 + \frac{4}{13}$$

Since all three have the same numerator (4) in their fractional parts, we compare denominators. For fractions with equal numerators, a **larger denominator** means a **smaller fraction**:
$$15 > 13 > 11 \implies \frac{4}{15} < \frac{4}{13} < \frac{4}{11}$$

Therefore:
$$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$$

ANSWER 8: E

---

Problem 9:
The data set is: 3, 4, 5, 6, 6, 7, and $x$. The mean, median, and unique mode are all equal.

**Mode:** Since 6 already appears twice and all others appear once, the unique mode must be 6 (if $x$ were 3, 4, 5, or 7, we'd have two modes, violating "unique mode"). So mode = 6.

**Mean = 6:**
$$\frac{3 + 4 + 5 + 6 + 6 + 7 + x}{7} = 6$$
$$\frac{31 + x}{7} = 6 \implies 31 + x = 42 \implies x = 11$$

**Check Median:** With $x = 11$, the sorted list is: 3, 4, 5, **6**, 6, 7, 11. The median (4th value) is 6. ✓

All three measures equal 6. Choice B (6) would make 6 appear three times, but the mean would be $(31+6)/7 = 37/7 \neq 6$.

ANSWER 9: D

---

Problem 10:
Joe walked half the distance in 6 minutes. He then ran the remaining half at 3 times his walking speed.

Let walking speed = $v$. The half-distance = $6v$.

Running speed = $3v$. Time to run the second half:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{6v}{3v} = 2 \text{ minutes}$$

Total time = 6 + 2 = **8 minutes**.

Choice A (7) would result from incorrectly subtracting. Choices B, C, E involve unnecessary decimal complications.

ANSWER 10: D

---

Problem 11:
Compute $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$.

Pair the terms from the first group with terms from the second group (first with last, etc.):
- $1 + 49 = 50$
- $11 + 39 = 50$
- $21 + 29 = 50$
- $31 + 19 = 50$
- $41 + 9 = 50$

Total = $50 \times 5 = 250$.

Choice C (200) is a common error from miscounting pairs. Choice D (249) comes from arithmetic mistakes.

ANSWER 11: E

---

Problem 12:
Compare the five decimals by writing them to four decimal places:

- A: 0.9900
- B: 0.9099
- C: 0.9000
- D: 0.9090
- E: 0.9009

Comparing digit by digit from the tenths place: all have 9 in the tenths place. In the hundredths place, A has **9** while all others have **0**. So A is immediately the largest.

$0.99 > 0.9099 > 0.909 > 0.9009 > 0.9$

ANSWER 12: A

 **Problem 1:**
We need to find the number of games played in a single-elimination tournament with 16 teams. In single elimination, every game eliminates exactly one team. To determine a winner, 15 of the 16 teams must be eliminated. Therefore, exactly 15 games are required.  
(Another way: with 16 teams there are 8 games in the first round, 4 in the second, 2 in the third, and 1 final game: \(8+4+2+1=15\).)

ANSWER 1: D

---

**Problem 2:**
We compute each person’s final answer step by step, starting from 10.

- **Jose:**  
  \(10 - 1 = 9\)  
  \(9 \times 2 = 18\)  
  \(18 + 2 = 20\)

- **Thuy:**  
  \(10 \times 2 = 20\)  
  \(20 - 1 = 19\)  
  \(19 + 2 = 21\)

- **Kareem:**  
  \(10 - 1 = 9\)  
  \(9 + 2 = 11\)  
  \(11 \times 2 = 22\)

Comparing the results: \(20 < 21 < 22\). Kareem’s answer is the largest.

ANSWER 2: C

---

**Problem 3:**
We convert the fraction \(\frac{2}{25}\) to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100:
\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

ANSWER 3: B

---

**Problem 4:**
We must choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways to choose 3 from 4 is the combination \(\binom{4}{3}\):
\[
\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4
\]
Equivalently, choosing 3 starters is the same as choosing 1 person to sit out, and there are 4 choices for who sits out.

ANSWER 4: B

---

**Problem 5:**
Let the number be \(N\). We are given \(0.20N = 12\). Solving for \(N\):
\[
N = \frac{12}{0.20} = 60
\]
Now find 30% of 60:
\[
0.30 \times 60 = 18
\]
(Shortcut: 30% is \(1.5\) times 20%, so \(1.5 \times 12 = 18\).)

ANSWER 5: B

---

**Problem 6:**
The spinner starts pointing west.  
- Clockwise \(2\frac14\) revolutions: 2 full revolutions bring it back to west, then \(\frac14\) turn clockwise from west points it **north**.  
- Counterclockwise \(3\frac34\) revolutions: 3 full revolutions bring it back to north, then \(\frac34\) turn counterclockwise from north goes through west, south, and ends at **east**.

Net movement: \(3\frac34 - 2\frac14 = 1\frac12\) revolutions counterclockwise. One full revolution returns to west; an additional \(\frac12\) revolution counterclockwise from west points to **east**.

ANSWER 6: B

---

**Problem 7:**
Original mixture: 30 liters.  
- Yellow tint originally: \(30\% \times 30 = 9\) liters.  
After adding 5 liters of yellow tint:  
- New yellow tint: \(9 + 5 = 14\) liters.  
- New total mixture: \(30 + 5 = 35\) liters.  

Percent of yellow tint in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%
\]

ANSWER 7: C

---

**Problem 8:**
Compare \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\). Each is greater than 1. Write each as \(1\) plus a proper fraction:
\[
\frac{15}{11} = 1 + \frac{4}{11}, \qquad
\frac{19}{15} = 1 + \frac{4}{15}, \qquad
\frac{17}{13} = 1 + \frac{4}{13}
\]
Since the numerators are the same (4), the fraction with the largest denominator is smallest:
\[
\frac{4}{15} < \frac{4}{13} < \frac{4}{11}
\]
Adding 1 preserves the order:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

ANSWER 8: E

---

**Problem 9:**
The data set is \(3, 4, 5, 6, 6, 7, x\).  
- The current mode is 6 (appears twice). For 6 to remain the **unique** mode, \(x\) must not equal 3, 4, 5, or 7 (which would create a second mode).  
- We are told mean = median = mode = 6.  

Set the mean equal to 6:
\[
\frac{3+4+5+6+6+7+x}{7} = 6 \implies \frac{31+x}{7} = 6 \implies 31+x = 42 \implies x = 11
\]

Check: With \(x=11\), the sorted list is \(3, 4, 5, 6, 6, 7, 11\).  
- Median (4th term) = 6.  
- Mode = 6 (unique).  
- Mean = 6.  

All conditions are satisfied.

ANSWER 9: D

---

**Problem 10:**
Joe walks half the distance in 6 minutes. He runs the second half at 3 times his walking speed. Since distance is the same and speed is tripled, the time required is divided by 3:
\[
\text{Running time} = \frac{6}{3} = 2 \text{ minutes}
\]
Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

---

**Problem 11:**
Add the two groups:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]
Pair the terms to make convenient sums:
\[
(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 5 \times 50 = 250
\]

ANSWER 11: E

---

**Problem 12:**
Compare the decimals by writing them with the same number of places:
- A. \(0.9900\)
- B. \(0.9099\)
- C. \(0.9000\)
- D. \(0.9090\)
- E. \(0.9009\)

The tenths digit is 9 for all. Looking at the hundredths digit: A has 9, while B, D have 0 and C, E have 0. Thus \(0.9900\) is clearly the largest.

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key insight:** Each game eliminates exactly one team. To determine a winner from 16 teams, we must eliminate 15 teams (everyone except the champion).

Therefore, exactly 15 games must be played.

*Verification by rounds:* Round 1: 8 games (16→8 teams), Round 2: 4 games (8→4), Round 3: 2 games (4→2), Round 4: 1 game (2→1). Total = 8+4+2+1 = 15.

The tempting wrong answer E (16) counts teams instead of eliminations.

ANSWER 1: D

---

Problem 2:
We need to compute each person's final answer starting from 10.

**Jose:** Start with 10 → subtract 1: 10−1 = 9 → double: 9×2 = 18 → add 2: 18+2 = **20**

**Thuy:** Start with 10 → double: 10×2 = 20 → subtract 1: 20−1 = 19 → add 2: 19+2 = **21**

**Kareem:** Start with 10 → subtract 1: 10−1 = 9 → add 2: 9+2 = 11 → double: 11×2 = **22**

Kareem gets the largest final answer (22).

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

The tempting wrong answer A (0.008) would be 8/1000, and C (0.8) would be 8/10.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination: C(4,3) = 4!/(3! × 1!) = 4

Listing them out: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — that's 4 ways.

The tempting wrong answer C (6) would be C(4,2), choosing who's *not* a starter, but that also equals 4... Actually C(4,2)=6 is wrong here; we're choosing 3 from 4.

ANSWER 4: B

---

Problem 5:
We know 20% of a number is 12. We need to find 30% of the same number.

**Step 1:** Find the number.
$$0.20 \times n = 12 \implies n = \frac{12}{0.20} = 60$$

**Step 2:** Find 30% of 60.
$$0.30 \times 60 = 18$$

*Shortcut:* Since 30% is 1.5 times 20%, the answer is 1.5 × 12 = 18.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing west. We need to find the net rotation.

**Clockwise:** 2¼ revolutions
**Counterclockwise:** 3¾ revolutions

**Net movement:** 2¼ − 3¾ = −1½ revolutions (negative means counterclockwise)

So the spinner moves 1½ revolutions counterclockwise from west.

- 1 full revolution counterclockwise returns to **west**
- ½ revolution counterclockwise from west goes to **east**

(Counterclockwise from west: ¼ turn → south, ½ turn → east)

ANSWER 6: B

---

Problem 7:
We need to find the percent of yellow tint after adding 5 liters of yellow tint.

**Original mixture (30 liters):**
- Yellow tint: 30% of 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percent yellow tint:** (14/35) × 100 = 40%

The tempting wrong answer B (35) might come from simply adding 5% to 30%.

ANSWER 7: C

---

Problem 8:
We need to order 15/11, 19/15, and 17/13 from least to greatest.

Computing each as a decimal:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3077...

From least to greatest: **19/15 < 17/13 < 15/11**

*Alternative method:* Note each is 1 + (numerator − denominator)/denominator:
- 15/11 = 1 + 4/11
- 19/15 = 1 + 4/15
- 17/13 = 1 + 4/13

Since 4/15 < 4/13 < 4/11, the order is confirmed.

ANSWER 8: E

---

Problem 9:
The positive integers 3, 4, 5, 6, 6, 7, and x have equal mean, median, and unique mode.

**Mode:** Since 6 already appears twice and no other number does, the unique mode must be 6. So x ≠ 3, 4, 5, or 7 (which would create a second mode).

**Median:** With 7 numbers, the median is the 4th value when sorted. For the median to be 6, we need x ≥ 6 (so that 6 is the 4th value).

- If x = 6: sorted = 3, 4, 5, 6, 6, 6, 7 → median = 6 ✓
- If x > 7: sorted = 3, 4, 5, 6, 6, 7, x → median = 6 ✓

**Mean:** (3 + 4 + 5 + 6 + 6 + 7 + x)/7 = (31 + x)/7

Setting mean = 6: (31 + x)/7 = 6 → 31 + x = 42 → **x = 11**

**Verification:** Numbers: 3, 4, 5, 6, 6, 7, 11. Mean = 42/7 = 6, Median = 6, Mode = 6 ✓

ANSWER 9: D

---

Problem 10:
Joe walks halfway, then runs the rest. We need total time.

**Walking:** Half the distance takes 6 minutes.

**Running:** He runs 3 times as fast as he walks, so he covers the same distance in 1/3 the time.
- Time to run the second half = 6 ÷ 3 = 2 minutes

**Total time:** 6 + 2 = **8 minutes**

The tempting wrong answers come from setting up the problem incorrectly or mixing up distance and time relationships.

ANSWER 10: D

---

Problem 11:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

**First group:** 1 + 11 + 21 + 31 + 41 = 105

**Second group:** 9 + 19 + 29 + 39 + 49 = 145

**Total:** 105 + 145 = **250**

*Elegant method:* Pair terms: (1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250

ANSWER 11: E

---

Problem 12:
We need to find the largest number among the options. Writing each with the same number of decimal places:

- A. 0.99 = **0.9900**
- B. 0.9099 = 0.9099
- C. 0.9 = 0.9000
- D. 0.909 = 0.9090
- E. 0.9009 = 0.9009

Comparing digit by digit after the decimal point:
- First digit: all are 9
- Second digit: A has **9**, all others have 0

Since A has a 9 in the second decimal place while all others have 0, **0.99 is the largest**.

ANSWER 12: A

**Problem 1:**  
The tournament is single‑elimination. Starting with 16 teams, every game eliminates exactly one team. To reduce from 16 teams to a single champion we must eliminate 15 teams, and each elimination corresponds to one game.  

Number of games = 16 − 1 = **15**.  

ANSWER 1: D  

---

**Problem 2:**  
Start with 10 for each person.

- **Jose:**  
  Subtract 1 → 9; double → 18; add 2 → 20.  

- **Thuy:**  
  Double → 20; subtract 1 → 19; add 2 → 21.  

- **Kareem:**  
  Subtract 1 → 9; add 2 → 11; double → 22.  

The largest final value is 22, obtained by Kareem.  

ANSWER 2: C  

---

**Problem 3:**  

\[
\frac{2}{25}=0.08
\]

(2 ÷ 25 = 0.08).  

ANSWER 3: B  

---

**Problem 4:**  
We need to choose 3 starters from the 4 available players:  

\[
\binom{4}{3}=4
\]

ways.  

ANSWER 4: B  

---

**Problem 5:**  
Let the number be \(N\).  
\(0.20N = 12 \Rightarrow N = \frac{12}{0.20}=60\).  
\(30\%\) of \(N\) is \(0.30 \times 60 = 18\).  

ANSWER 5: B  

---

**Problem 6:**  
Interpret the fractions as mixed numbers:  

- Clockwise \(2\frac14 = 2.25\) revolutions.  
- Counter‑clockwise \(3\frac34 = 3.75\) revolutions.  

Net rotation (counter‑clockwise positive) = \(3.75 - 2.25 = 1.5\) revolutions.  

A full revolution returns to the same direction, so we only care about the extra \(0.5\) revolution.  
\(0.5\) revolution = 180° counter‑clockwise.  

Starting from west, turning 180° counter‑clockwise points to **east**.  

ANSWER 6: B  

---

**Problem 7:**  
Original yellow = \(30\%\) of 30 L = \(0.30 \times 30 = 9\) L.  
Add 5 L yellow → new yellow = \(9+5 = 14\) L.  
New total volume = \(30+5 = 35\) L.  

Percent yellow = \(\frac{14}{35}\times100 = 40\%\).  

ANSWER 7: C  

---

**Problem 8:**  
Compute the three fractions:

\[
\frac{15}{11}=1.3636\ldots,\qquad
\frac{19}{15}=1.2666\ldots,\qquad
\frac{17}{13}=1.3077\ldots
\]

Ordered from least to greatest:  

\(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

This matches option **E**.  

ANSWER 8: E  

---

**Problem 9:**  
Current list: \(3,4,5,6,6,7\). The unique mode is 6 (appears twice).  
For the mode to remain unique and equal to the mean and median, \(x\) must not create a tie.  
Let the mean = median = mode = \(m\).  

Sum of numbers = \(31 + x\).  
Mean = \(\frac{31 + x}{7}=m\).  
Median of the seven numbers is the 4th after sorting; with \(x>7\) the median stays 6, so \(m=6\).  

Thus \(\frac{31 + x}{7}=6 \Rightarrow 31 + x = 42 \Rightarrow x = 11\).

Check: sorted list becomes \(3,4,5,6,6,7,11\) → median = 6, mode = 6, mean = 42/7 = 6. All equal.  

ANSWER 9: D  

---

**Problem 10:**  
Let the walking speed be \(v\) and the half‑distance be \(d/2\).  

Walking time for half‑distance: \(\frac{d/2}{v}=6\) min ⇒ \(\frac{d}{v}=12\) min.  

Running speed = \(3v\).  
Running time for the second half: \(\frac{d/2}{3v}= \frac{d}{6v}= \frac{1}{6}\times\frac{d}{v}= \frac{12}{6}=2\) min.  

Total time = \(6 + 2 = 8\) minutes.  

ANSWER 10: D  

---

**Problem 11:**  

\[
\begin{aligned}
1+11+21+31+41 &= 105,\\
9+19+29+39+49 &= 145,\\
\text{Total} &= 105+145 = 250.
\end{aligned}
\]

ANSWER 11: E  

---

**Problem 12:**  
Write the numbers with four decimal places:

- .99 = 0.9900  
- .9099 = 0.9099  
- .9 = 0.9000  
- .909 = 0.9090  
- .9009 = 0.9009  

The largest is 0.9900, i.e., **.99**.  

ANSWER 12: A  

Problem 1:  
The tournament starts with 16 teams. Each game eliminates one team, and the losing team is eliminated. To determine the winner, all teams except one must be eliminated. Thus, 15 teams must be eliminated, requiring 15 games.  
- **Why not other choices?** A (4) is too few; B (7), C (8) are also insufficient; E (16) would imply one game per team without elimination, which is incorrect.  
ANSWER 1: D  

Problem 2:  
- **Jose**: Starts with 10 → subtracts 1 (9) → doubles (18) → adds 2 → **20**.  
- **Thuy**: Starts with 10 → doubles (20) → subtracts 1 (19) → adds 2 → **21**.  
- **Kareem**: Starts with 10 → subtracts 1 (9) → adds 2 (11) → doubles → **22**.  
Kareem has the largest final answer.  
- **Why not other choices?** A (Jose) and B (Thuy) are smaller; D and E pair incorrect winners.  
ANSWER 2: C  

Problem 3:  
\( \frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08 \).  
- **Why not other choices?** A (.008) is \( \frac{2}{250} \); C (.8) is \( \frac{4}{5} \); D and E are larger decimals.  
ANSWER 3: B  

Problem 4:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is \( \binom{4}{3} = 4 \).  
- **Why not other choices?** A (2) is too few; C (6) is \( \binom{4}{2} \), which is incorrect; D (8) and E (10) are overcounts.  
ANSWER 4: B  

Problem 5:  
Let the number be \( x \). If 20% of \( x \) is 12, then \( 0.2x = 12 \), so \( x = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).  
- **Why not other choices?** A (15) is 25% of 60; C (20) is \( \frac{1}{3} \) of 60; D (24) is 40%; E (30) is 50%.  
ANSWER 5: B  

Problem 6:  
- A full revolution is 360°. Clockwise \( 2 \frac{1}{4} \) revolutions = \( 2.25 \times 360 = 810° \).  
- Counterclockwise \( 3 \frac{3}{4} \) revolutions = \( -3.75 \times 360 = -1350° \).  
- Net movement: \( 810° - 1350° = -540° \).  
- \( -540° \mod 360° = 180° \) (since \( -540 + 720 = 180° \)), which points west → south? Wait, initial direction is west.  
- West (270°) + 810° = 1080° → 1080° mod 360° = 0° (north). Then -1350° → 0° - 1350° = -1350° mod 360° = 90° (east)? Let's reframe:  
- Start at west (270°). Clockwise 2.25 rev: 270° + 810° = 1080° mod 360° = 0° (north). Then counterclockwise 3.75 rev: 0° - 1350° = -1350° mod 360° = 90° (east).  
- **Why not other choices?** A (north) is initial after clockwise; B (east) is final; C (south) is 180°; D (west) is start; E (northwest) is not a cardinal direction.  
ANSWER 6: B  

Problem 7:  
Original mixture: 30 liters, 30% yellow tint → 9 liters yellow. Add 5 liters yellow → total yellow = 14 liters. New mixture = 35 liters. Percent yellow = \( \frac{14}{35} \times 100 = 40\% \).  
- **Why not other choices?** A (25) is original yellow%; B (35) is before addition; D (45) and E (50) are overestimates.  
ANSWER 7: E (Note: Correction based on computation: 14/35=0.4 → 40%, so answer is E? Wait, 14/35=0.4 → 40%, which is E. But earlier I thought B, but B is 35, which is wrong. Original yellow is 9, add 5 → 14. Total mixture 30+5=35. 14/35=0.4 → 40%. So E. But in choices E is 50? Wait, no: choices are A.25 B.35 C.40 D.45 E.50. So 40% is C. I misread. Correction: 14/35=0.4 → 40%, which is C.  
**Correction**: 14/35=0.4 → 40%, so answer is C.  
ANSWER 7: C  

Problem 8:  
Compare \( \frac{15}{11} \approx 1.3636 \), \( \frac{19}{15} \approx 1.2667 \), \( \frac{17}{13} \approx 1.3077 \). Order: \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11} \).  
- **Why not other choices?** A, B, C, D have incorrect orders.  
ANSWER 8: E  

Problem 9:  
The set is {3, 4, 5, 6, 6, 7, x}. Mean = median = mode = 6.  
- Mode is 6 (unique), so 6 must appear most.  
- Median is 6 (4th value in ordered list: 3,4,5,6,6,7,x → if x≥6, ordered is ...,6,6,x or ...,6,x,6? Must have 4th value 6.  
- If x=6, set is {3,4,5,6,6,6,7}: median=6, mode=6, mean=(3+4+5+6+6+6+7)/7=37/7≈5.28 ≠6.  
- If x=11: set {3,4,5,6,6,7,11}: mean=(3+4+5+6+6+7+11)/7=42/7=6, median=6, mode=6.  
- **Why not other choices?** A (5): mean=36/7≈5.14; B (6): mean≈5.28; C (7): mean=38/7≈5.43; D (11) works.  
ANSWER 9: D  

Problem 10:  
Joe walks half the distance in 6 minutes. Let walking speed be \( v \), so distance = \( 6v \). He runs at \( 3v \), so time to run half distance = \( \frac{6v}{3v} = 2 \) minutes. Total time = 6 + 2 = 8 minutes.  
- **Why not other choices?** A (7), B (7.3), C (7.7), E (8.3) are not exact.  
ANSWER 10: D  

Problem 11:  
First group: 1+11+21+31+41 = 105. Second group: 9+19+29+39+49 = 145. Total = 105 + 145 = 250.  
- **Why not other choices?** A (150), B (199), C (200), D (249) are incorrect sums.  
ANSWER 11: E  

Problem 12:  
Compare decimals: A (.99), B (.9099), C (.9), D (.909), E (.9009). Largest is .99.  
- **Why not other choices?** B, C, D, E are all smaller than .99.  
ANSWER 12: A

### Problem 1:
We need to find the total number of games to determine a winner in a 16-team single-elimination tournament, where every game eliminates exactly one losing team. To get 1 winner, we must eliminate 16 - 1 = 15 teams. Since each game eliminates 1 team, we need 15 total games. Wrong choices: A is the number of bracket rounds, not games; C is half the number of teams; E is the total number of teams, not games.
ANSWER 1: D

---

### Problem 2:
All three start with 10, we calculate each person's final result step-by-step:
- Jose: (10 - 1) * 2 + 2 = 9*2 +2 = 20
- Thuy: (10*2) -1 +2 = 20-1+2=21
- Kareem: (10-1 +2)*2 = 11*2=22
Kareem has the largest final answer. Wrong choices stem from misordering the operations for each person.
ANSWER 2: C

---

### Problem3:
Convert 2/25 to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100: (2*4)/(25*4) = 8/100 = 0.08. Wrong choices: A is 2/250, C is 20/25.
ANSWER 3: B

---

### Problem4:
We choose 3 starters from 4 people, which is a combination problem (order of starters does not matter here). The number of ways to choose 3 from 4 is equivalent to choosing 1 person to leave off the starting lineup: there are 4 people, so 4 unique lineups, each omitting one person. Wrong choice C (6) is the number of permutations, not combinations.
ANSWER 4: B

---

### Problem5:
If 20% of a number is 12, first find the full number: 0.2*N=12 → N=12/0.2=60. 30% of 60 is 0.3*60=18. Wrong choices: A is 25% of 60, D is 40% of 60.
ANSWER 5: B

---

### Problem6:
The spinner starts pointing west. Net movement: 2.25 clockwise revolutions minus 3.75 clockwise (equivalent to 3.75 counterclockwise) = -1.5 total clockwise revolutions, or 1.5 counterclockwise revolutions. Full revolutions cancel out, so we only have a net 0.5 (180 degree) counterclockwise turn from west. A 180-degree turn from west points to east.
ANSWER 6: B

---

### Problem7:
Original 30L mixture has 30% yellow tint: 0.3*30=9L of yellow. Add 5L of yellow, so new yellow is 14L, new total mixture is 35L. The new percentage of yellow is (14/35)*100=40%. Wrong choice: people often forget to add 5L to the total mixture volume, leading to an incorrect higher percentage.
ANSWER 7: C

---

### Problem8:
Rewrite each fraction to compare easily: 15/11 = 1 + 4/11, 19/15=1+4/15, 17/13=1+4/13. For equal numerators, the larger the denominator, the smaller the fraction. So 4/15 < 4/13 < 4/11, so the full order from least to greatest is 19/15 < 17/13 <15/11.
ANSWER 8: E

---

### Problem9:
The dataset is [3,4,5,6,6,7,x]. The unique mode is currently 6 (only value appearing twice, so it must stay the only unique mode). For the mean, median, and mode to all be 6: the sum of the dataset must be 7*6=42. The sum of the existing numbers is 3+4+5+6+6+7=31, so x=42-31=11. Sorted dataset [3,4,5,6,6,7,11] has median 6, unique mode 6, mean 6, which fits. Wrong choice B (x=6) gives a mean of ~5.28, not 6.
ANSWER 9: D

---

### Problem10:
Joe walks half the distance in 6 minutes. His running speed is 3x his walking speed, so time to cover the same distance is 1/3 of walking time. Time to run the second half: 6/3=2 minutes. Total time: 6+2=8 minutes.
ANSWER 10: D

---

### Problem11:
Pair terms to simplify addition: (1+49)+(11+39)+(21+29)+(31+19)+(41+9) = 50+50+50+50+50=250.
ANSWER 11: E

---

### Problem12:
Line up all decimals to compare: A=0.9900, B=0.9099, C=0.9000, D=0.9090, E=0.9009. The largest value is 0.99, option A.
ANSWER 12: A

Problem 1:  
In a single-elimination tournament, each game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams (all except 1). Thus, 15 games are played.  
ANSWER 1: D  

Problem 2:  
Calculate each person's result starting from 10:  
- Jose: \(10 -1 =9\), double → \(9*2=18\), add 2 → \(18+2=20\)  
- Thuy: Double 10 → \(10*2=20\), subtract 1 → \(20-1=19\), add 2 → \(19+2=21\)  
- Kareem: \(10-1=9\), add 2 → \(9+2=11\), double → \(11*2=22\)  
Kareem has the largest result.  
ANSWER 2: C  

Problem 3:  
Convert \(\frac{2}{25}\) to decimal: \(2\div25=0.08\).  
ANSWER 3: B  

Problem 4:  
Choosing 3 starters from 4 people is a combination (order does not matter). The number of ways is \(\binom{4}{3}=4\) (or list: {Lance,Sally,Joy}, {Lance,Sally,Fred}, {Lance,Joy,Fred}, {Sally,Joy,Fred}).  
ANSWER 4: B  

Problem 5:  
Let the number be \(x\). \(20\%x=12\) → \(0.2x=12\) → \(x=60\). \(30\%\ of\ 60=0.3*60=18\).  
ANSWER 5: B  

Problem 6:  
Model directions with standard angles: 0°=east, 90°=north, 180°=west, 270°=south. Initial direction is west (180°).  
- Clockwise \(2\frac{1}{4}\) revolutions: \(2.25*(-360°)=-810°\), new angle: \(180-810=-630°\)  
- Counterclockwise \(3\frac{3}{4}\) revolutions: \(3.75*(360°)=+1350°\), total angle: \(-630+1350=720°\)  
- \(720°\mod360°=0°\), which is east.  
ANSWER 6: B  

Problem 7:  
Original yellow tint: \(30\%\ of\ 30=0.3*30=9\) liters. Add 5 liters, total yellow = \(9+5=14\) liters. New total mixture: \(30+5=35\) liters. Percent yellow: \(\frac{14}{35}*100\%=40\%\).  
ANSWER 7: C  

Problem 8:  
Compare fractions via cross-multiplication or decimals:  
- \(\frac{19}{15}\approx1.267\), \(\frac{17}{13}\approx1.308\), \(\frac{15}{11}\approx1.364\)  
Order from least to greatest: \(\frac{19}{15}<\frac{17}{13}<\frac{15}{11}\).  
ANSWER 8: E  

Problem 9:  
The unique mode is 6 (appears twice, others once). Thus mean and median equal 6.  
- Median of 7 numbers is the 4th term when sorted, so 4th term =6 → \(x\geq6\).  
- Mean = \(\frac{3+4+5+6+6+7+x}{7}=6\) → \(31+x=42\) → \(x=11\).  
Check: Sorted list [3,4,5,6,6,7,11], median=6, mode=6 (unique), mean=6.  
ANSWER 9: D  

Problem 10:  
Walking half the distance takes 6 minutes, so full walking time would be 12 minutes. Running speed is 3x walking speed, so time for the second half is \(\frac{6}{3}=2\) minutes. Total time: \(6+2=8\) minutes.  
ANSWER 10: D  

Problem 11:  
Sum first group: \(1+11+21+31+41=105\) (arithmetic series: \(\frac{5}{2}(1+41)=105\)).  
Sum second group: \(9+19+29+39+49=145\) (\(\frac{5}{2}(9+49)=145\)).  
Total: \(105+145=250\).  
ANSWER 11: E  

Problem 12:  
Compare decimals digit by digit:  
- A=0.99, B=0.9099, C=0.9, D=0.909, E=0.9009  
A has the highest second decimal digit (9 vs 0 for others), so it is the largest.  
ANSWER 12: A

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** How many games must be played to determine a winner in a 16-team elimination tournament?

**Setup:** In an elimination tournament, each game eliminates exactly one team. To go from 16 teams to 1 winner, we must eliminate 15 teams. Therefore, we need 15 games.

**Alternatively:** We can think of it as a bracket:
- Round 1: 16 teams → 8 games → 8 winners
- Round 2: 8 teams → 4 games → 4 winners
- Round 3: 4 teams → 2 games → 2 winners
- Round 4: 2 teams → 1 game → 1 winner
- Total: 8 + 4 + 2 + 1 = 15 games

ANSWER 1: D

---

**Problem 2:**

**What's being asked:** Which person gets the largest final answer after their operations?

**Jose's calculation:** 
- Start: 10
- Subtract 1: 10 - 1 = 9
- Double: 9 × 2 = 18
- Add 2: 18 + 2 = **20**

**Thuy's calculation:**
- Start: 10
- Double: 10 × 2 = 20
- Subtract 1: 20 - 1 = 19
- Add 2: 19 + 2 = **21**

**Kareem's calculation:**
- Start: 10
- Subtract 1: 10 - 1 = 9
- Add 2: 9 + 2 = 11
- Double: 11 × 2 = **22**

Kareem gets 22, which is the largest.

ANSWER 2: C

---

**Problem 3:**

**What's being asked:** Convert the fraction 2/25 to a decimal.

**Setup:** 2/25 = ?

**Calculation:** 
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** In how many ways can 3 starters be chosen from 4 people (Lance, Sally, Joy, Fred)?

**Setup:** This is a combination problem: $\binom{4}{3}$

**Calculation:**
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4}{1} = 4$$

Alternatively, we're choosing which 1 person to leave out: 4 ways.

ANSWER 4: B

---

**Problem 5:**

**What's being asked:** If 20% of a number is 12, what is 30% of that number?

**Setup:** Let the number be $n$.
- $0.20n = 12$
- $n = 12 ÷ 0.20 = 60$

**Calculation:** 
$$30\% \text{ of } 60 = 0.30 \times 60 = 18$$

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Starting pointing west, after rotating clockwise 2¼ revolutions then counterclockwise 3¾ revolutions, which direction?

**Setup:** 
- Clockwise 2¼ revolutions = 2.25 revolutions
- Counterclockwise 3¾ revolutions = 3.75 revolutions
- Net: 3.75 - 2.25 = 1.5 revolutions counterclockwise (net)

**Calculation:**
Starting at west, rotating counterclockwise 1.5 revolutions (= 1 full rotation + 0.5 rotation):
- 1 full rotation brings us back to west
- 0.5 rotation = 180°
- West + 180° counterclockwise = East

ANSWER 6: B

---

**Problem 7:**

**What's being asked:** After adding 5 liters of yellow tint to a 30-liter mixture, what percent is yellow tint?

**Original mixture (30 liters):**
- Yellow tint: 30% × 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percentage:**
$$\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%$$

ANSWER 7: C

---

**Problem 8:**

**What's being asked:** Order 15/11, 19/15, and 17/13 from least to greatest.

**Setup:** Convert to decimals or compare directly.

**Calculations:**
- 15/11 ≈ 1.364
- 19/15 ≈ 1.267
- 17/13 ≈ 1.308

**Order:** 19/15 < 17/13 < 15/11

ANSWER 8: E

---

**Problem 9:**

**What's being asked:** Find $x$ such that the mean, median, and unique mode of {3, 4, 5, 6, 6, 7, $x$} are all equal.

**Current mode:** 6 appears twice; all others appear once. If there's a unique mode, it must be 6.

**With unique mode = 6:** This requires no other number to appear more than once. So $x \notin \{3, 4, 5, 7\}$ and $x \neq 6$ (else it wouldn't be unique).

**Mean = Median = Mode = 6:**

For median of 7 numbers with value 6:
- Sorted: need middle (4th) value to be 6
- If $x = 11$: {3, 4, 5, 6, 6, 7, 11} → median = 6 ✓

Mean: $\frac{3+4+5+6+6+7+11}{7} = \frac{42}{7} = 6$ ✓

Mode: 6 appears twice; all others once → mode = 6 ✓

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Total time from home to school if Joe walked halfway (6 min) then ran the rest at 3× speed?

**Setup:**
- Walking: 6 minutes for half the distance
- Running: same distance at 3× speed, so time = 6/3 = 2 minutes

**Total time:** 6 + 2 = 8 minutes

ANSWER 10: D

---

**Problem 11:**

**What's being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Calculation:**
- First group: 1 + 11 + 21 + 31 + 41 = 105
- Second group: 9 + 19 + 29 + 39 + 49 = 145
- Sum: 105 + 145 = 250

ANSWER 11: E

---

**Problem 12:**

**What's being asked:** Which number is largest?

**Comparison:**
- A. 0.99
- B. 0.9099
- C. 0.9
- D. 0.909
- E. 0.9009

Starting from the tenths place: all are 0.9__. Looking at hundredths:
- 0.99 has 9 in hundredths place → **largest**

ANSWER 12: A

Problem 1:

In a single-elimination tournament, each game eliminates exactly one team.  
There are 16 teams, and we want one champion.

So we must eliminate 15 teams total.

Therefore, the number of games played is 15.

Tempting wrong choice: 16 would be if every team had to lose, but the winner does not lose.

ANSWER 1: D

Problem 2:

Compute each person’s result carefully.

- Jose: start with 10  
  subtract 1 → 9  
  double → 18  
  add 2 → 20

- Thuy: start with 10  
  double → 20  
  subtract 1 → 19  
  add 2 → 21

- Kareem: start with 10  
  subtract 1 → 9  
  add 2 → 11  
  double → 22

Compare:
- Jose = 20
- Thuy = 21
- Kareem = 22

Kareem gets the largest final answer.

ANSWER 2: C

Problem 3:

\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

So the correct decimal is 0.08.

ANSWER 3: B

Problem 4:

We are choosing 3 starters from 4 players: Lance, Sally, Joy, and Fred.

This is a combination problem:
\[
\binom{4}{3} = 4
\]

You can also see it by noticing that choosing 3 starters is the same as choosing which 1 player sits out, and there are 4 choices.

ANSWER 4: B

Problem 5:

Let the number be \(n\).

20% of \(n\) is 12, so:
\[
0.2n = 12
\]
\[
n = 60
\]

Now find 30% of 60:
\[
0.3 \times 60 = 18
\]

ANSWER 5: B

Problem 6:

The spinner starts pointing west.

First move: clockwise \(2\frac{1}{4}\) revolutions.  
A full revolution brings it back to west, so only the extra \(\frac{1}{4}\) revolution matters.

From west, a quarter-turn clockwise points it to north.

Second move: counterclockwise \(3\frac{3}{4}\) revolutions.  
Again, the 3 full revolutions do nothing, so this is just \(\frac{3}{4}\) of a revolution counterclockwise, which is the same as \(\frac{1}{4}\) of a revolution clockwise.

From north, a quarter-turn clockwise points east.

So the spinner ends pointing east.

ANSWER 6: B

Problem 7:

Start with 30 liters total.

Yellow tint is 30% of 30 liters:
\[
0.30 \times 30 = 9 \text{ liters}
\]

Then 5 liters of yellow tint are added, so yellow becomes:
\[
9 + 5 = 14 \text{ liters}
\]

New total mixture:
\[
30 + 5 = 35 \text{ liters}
\]

Percent yellow in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.4 = 40\%
\]

ANSWER 7: C

Problem 8:

The fractions are:
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]

Approximate each one:

\[
\frac{15}{11} \approx 1.364
\]
\[
\frac{17}{13} \approx 1.308
\]
\[
\frac{19}{15} \approx 1.267
\]

So from least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

That matches choice E.

ANSWER 8: E

Problem 9:

The numbers are:
\[
3,4,5,6,6,7,x
\]

We are told the mean, median, and unique mode are all equal.

Since 6 appears twice and every other listed number appears once, the unique mode is 6.

So the median and mean must also be 6.

The median of 7 numbers is the 4th number when ordered.  
The sorted list is:
\[
3,4,5,6,6,7,x
\]
To make the median 6, \(x\) must be 6 or larger, but we also need the mean to be 6.

Sum of the known numbers:
\[
3+4+5+6+6+7 = 31
\]

For the mean to be 6, total sum must be:
\[
7 \times 6 = 42
\]

So:
\[
31 + x = 42
\]
\[
x = 11
\]

Check:
- mode = 6
- median = 6
- mean = 42/7 = 6

Works perfectly.

ANSWER 9: D

Problem 10:

Joe walked half the distance in 6 minutes.  
He then ran the other half at 3 times his walking speed.

If he runs 3 times as fast, then the time for the same distance is 1/3 as long.

So the running time is:
\[
6 \div 3 = 2 \text{ minutes}
\]

Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

Problem 11:

Add the first group:
\[
1+11+21+31+41 = 105
\]

Add the second group:
\[
9+19+29+39+49 = 145
\]

Now add the totals:
\[
105 + 145 = 250
\]

ANSWER 11: E

Problem 12:

Compare the decimals:

- \(0.99\)
- \(0.9099\)
- \(0.9 = 0.9000\)
- \(0.909\)
- \(0.9009\)

The largest is clearly \(0.99\), since it is closest to 1.

ANSWER 12: A

Problem 1:  
We have a three-person-team tournament where **the losing team of each game is eliminated**. With **16 teams**, how many games are played to determine the winner?

To find the number of games in a single-elimination tournament:  
- Each game eliminates exactly **one** team (the loser).  
- To go from **16 teams to 1 winning team**, we must eliminate **15 teams**.  
- Therefore, the number of games is **15**.

So the correct choice is **D**.

ANSWER 1: D


Problem 2:  
Jose, Thuy, and Kareem all start at 10. We compute each final value.

**Jose:**  
- Subtract 1: \(10 - 1 = 9\)  
- Double: \(2 \cdot 9 = 18\)  
- Add 2: \(18 + 2 = 20\)  
Final: **20**

**Thuy:**  
- Double 10: \(2 \cdot 10 = 20\)  
- Subtract 1: \(20 - 1 = 19\)  
- Add 2: \(19 + 2 = 21\)  
Final: **21**

**Kareem:**  
- Subtract 1: \(10 - 1 = 9\)  
- Add 2: \(9 + 2 = 11\)  
- Double: \(2 \cdot 11 = 22\)  
Final: **22**

Largest is **22**, which is Kareem.

ANSWER 2: C


Problem 3:  
We need to evaluate:  
\[
\frac{2}{25}
\]
Convert to decimals:  
\[
\frac{2}{25} = 2 \div 25
\]
Since \(25 \cdot 0.08 = 2\):  
\[
2 \div 25 = 0.08
\]
So the correct choice is **B** (.08).

ANSWER 3: B


Problem 4:  
We have 4 people chosen: Lance, Sally, Joy, Fred. We need to choose **3 starters**.

Number of ways:
\[
\binom{4}{3} = 4
\]
List them conceptually: leaving out each one individually gives 4 choices.

Correct choice: **B**.

ANSWER 4: B


Problem 5:  
“If 20% of a number is 12, what is 30% of the same number?”

Let the number be \(N\).  
\[
0.20N = 12
\]
\[
N = \frac{12}{0.20} = 60
\]
Now compute \(30\%\) of 60:
\[
0.30 \cdot 60 = 18
\]
Correct choice: **B**.

ANSWER 5: B


Problem 6:  
Spinner starts pointing **west**.

Chenille moves it:
- **Clockwise** \(2\frac{1}{4}\) revolutions  
- then **Counterclockwise** \(3\frac{3}{4}\) revolutions

Net rotation (clockwise minus counterclockwise):
\[
2\frac{1}{4} - 3\frac{3}{4}
\]
Convert to improper fractions:
\[
2\frac{1}{4} = \frac{9}{4}, \quad 3\frac{3}{4} = \frac{15}{4}
\]
\[
\frac{9}{4} - \frac{15}{4} = -\frac{6}{4} = -\frac{3}{2}
\]
The negative means **net counterclockwise** by \(\frac{3}{2}\) revolutions.

\(\frac{3}{2} = 1 + \frac{1}{2}\), so this is:
- 1 full turn (brings it back where it started)
- plus \(\frac{1}{2}\) turn counterclockwise

Half a revolution from **west** points **east** (opposite direction).

Correct answer: **B (east)**.

ANSWER 6: B


Problem 7:  
Original mixture: 30 L with  
- 25% red (doesn’t matter directly)
- 30% yellow  
- 45% water

Yellow amount initially:
\[
0.30 \cdot 30 = 9 \text{ liters}
\]
Add 5 liters of yellow tint, so new yellow amount:
\[
9 + 5 = 14
\]
New total volume:
\[
30 + 5 = 35
\]
Percent yellow in new mixture:
\[
\frac{14}{35} = 0.4 = 40\%
\]
Correct choice: **C**.

ANSWER 7: C


Problem 8:  
Order fractions least to greatest:
\[
\frac{15}{11},\ \frac{19}{15},\ \frac{17}{13}
\]
A good way: compare values by cross-multiplication or decimal approximations.

Compute approximate decimals:
\[
\frac{15}{11} \approx 1.3636
\]
\[
\frac{19}{15} \approx 1.2667
\]
\[
\frac{17}{13} \approx 1.3077
\]
Least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]
Check choice: that matches **E**:
- \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11}\)

ANSWER 8: E


Problem 9:  
Numbers: \(3,4,5,6,6,7,x\) (7 positive integers).  
Mean = median = unique mode.

**Step 1: Mode**  
The only number that repeats is **6**, occurring twice. So the **unique mode = 6**.

Thus median must be 6 as well.

**Step 2: Median = 6**  
For 7 numbers sorted, the median is the **4th** number.

Sorted list must have the 4th term equal to 6. We already have two 6’s and surrounding numbers 3,4,5,7.

Currently without \(x\), the ordered portion is:
\[
3,4,5,6,6,7,\ ?
\]
The 4th number here is 6 already. To keep the 4th number as 6, \(x\) must not be less than 6 (otherwise it would shift positions so 4th might change).

If \(x = 6\), then 6 occurs 3 times, still mode 6 (unique mode). Then list:
\[
3,4,5,6,6,6,7
\]
Median (4th) is 6—works.

If \(x=7\), list:
\[
3,4,5,6,6,7,7
\]
Mode would be 6? Actually 7 repeats too, so mode not unique.

If \(x>7\), list:
\[
3,4,5,6,6,7,x
\]
Median still 6, but mode would still be 6 (unique), since only 6 repeats. That would also work—wait, but then 6 is still unique mode, yes. However we also need **mean = 6**.

So use mean condition.

Sum of known numbers:
\[
3+4+5+6+6+7 = 31
\]
Total sum including \(x\): \(31 + x\)

Mean:
\[
\frac{31+x}{7} = 6
\]
\[
31 + x = 42
\]
\[
x = 11
\]

Check uniqueness of mode: 6 occurs twice, no other repeats (since x=11 is different). Unique mode is 6—consistent.

ANSWER 9: D


Problem 10:  
Joe walks half the distance, then runs the rest.  
He runs **3 times as fast** as he walks.

Given: time to walk half way = **6 minutes**.  
Let walking speed be \(v\). Then walking time for half distance \(=\) 6 minutes.

Let half the distance be \(d\). Then:
\[
d = v \cdot 6
\]
Running speed is \(3v\).  
Remaining distance is also \(d\), so running time:
\[
\text{time} = \frac{d}{3v} = \frac{v\cdot 6}{3v} = 2
\]
Total time:
\[
6 + 2 = 8
\]
Correct choice: **D**.

ANSWER 10: D


Problem 11:  
Compute:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]

Group each set:

First set:
\[
1+11+21+31+41
\]
These are all of the form \(10k+1\):  
\(1, 11, 21, 31, 41\) correspond to \(k=0,1,2,3,4\).

Sum:
\[
(1+11+21+31+41) = (1+11+21+31+41)
\]
Compute directly:
\[
(1+11)=12,\ (21+31)=52,\ 12+52=64,\ 64+41=105
\]

Second set:
\[
9+19+29+39+49
\]
Direct sum:
\[
(9+19)=28,\ (29+39)=68,\ 28+68=96,\ 96+49=145
\]

Total:
\[
105 + 145 = 250
\]

Correct choice: **E**.

ANSWER 11: E


Problem 12:  
Which number is largest?

Options:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare decimals:  
- .99 is the largest among all because it is greater than .9 (and greater than any number starting 0.90... but with fewer 9s afterward).

So .99 > .9099 > .909 > .9009 > .9.

Correct choice: **A**.

ANSWER 12: A

Problem 1:
In a single-elimination tournament, every game results in exactly one team being eliminated. To determine a single winner from 16 teams, 15 teams must be eliminated. Since each game eliminates one team, it takes exactly 15 games.
ANSWER 1: D

Problem 2:
- Jose: (10 - 1) = 9; 9 * 2 = 18; 18 + 2 = 20.
- Thuy: (10 * 2) = 20; 20 - 1 = 19; 19 + 2 = 21.
- Kareem: (10 - 1) = 9; 9 + 2 = 11; 11 * 2 = 22.
Kareem has the largest answer (22).
ANSWER 2: C

Problem 3:
To convert 2/25 to a decimal, multiply the numerator and denominator by 4 to get a denominator of 100: (2 * 4) / (25 * 4) = 8 / 100 = 0.08.
ANSWER 3: B

Problem 4:
We need to choose 3 starters from a group of 4 (Lance, Sally, Joy, Fred). This is a combination problem denoted as 4C3. Choosing 3 people to be on the team is the same as choosing 1 person to be left off the team. There are 4 ways to choose who is left off.
ANSWER 4: B

Problem 5:
Let $x$ be the number. $0.20x = 12$. Dividing by 0.20, $x = 12 / 0.2 = 60$. We want 30% of $x$: $0.30 * 60 = 18$.
ANSWER 5: B

Problem 6:
Start at 0 revolutions. Clockwise is positive, counterclockwise is negative.
$2 \frac{1}{4} - 3 \frac{3}{4} = 2.25 - 3.75 = -1.5$ revolutions.
-1.5 revolutions is the same as -0.5 revolutions (since -1 is a full circle). -0.5 revolutions (counterclockwise) from West points to East.
ANSWER 6: B

Problem 7:
Original yellow tint: 30% of 30 liters = 9 liters.
New mixture: 30 + 5 = 35 liters.
New yellow tint: 9 + 5 = 14 liters.
Percent: (14 / 35) * 100 = (2 / 5) * 100 = 40%.
ANSWER 7: C

Problem 8:
Convert to decimals or compare differences from 1:
15/11 = 1 + 4/11 ≈ 1.3636
19/15 = 1 + 4/15 ≈ 1.2666
17/13 = 1 + 4/13 ≈ 1.3076
Comparing the fractions 4/11, 4/13, and 4/15: the larger the denominator, the smaller the fraction. Thus, 4/15 < 4/13 < 4/11.
Order: 19/15 < 17/13 < 15/11.
ANSWER 8: E

Problem 9:
The set is {3, 4, 5, 6, 6, 7, x}. The mode is 6 (it appears twice). For the mean and median to be 6:
Median: With 7 numbers, the 4th number must be 6. Since the sorted list is 3, 4, 5, 6, 6, 7, x, if $x > 6$, the 4th number is 6.
Mean: (3+4+5+6+6+7+x) / 7 = 6 => 31 + x = 42 => x = 11.
Checking: {3, 4, 5, 6, 6, 7, 11}. Median is 6. Mean is 42/7 = 6. Mode is 6.
ANSWER 9: D

Problem 10:
Walking time = 6 minutes. Running speed = 3 * walking speed.
Time taken = Distance / Speed. Since speed is 3 times faster, time is 1/3 of the walking time.
Running time = 6 / 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
ANSWER 10: D

Problem 11:
Sum 1: (1+11+21+31+41) = 105.
Sum 2: (9+19+29+39+49) = 145.
105 + 145 = 250.
ANSWER 11: E

Problem 12:
Compare digit by digit:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
0.99 is the largest.
ANSWER 12: A

Problem 1:
This is a single-elimination tournament with 16 teams. Exactly one team is eliminated per game, and 15 teams must be eliminated to leave a single winner. Thus exactly 15 games are required.

ANSWER 1: D

Problem 2:
Jose evaluates \((10-1)\times2+2=20\).
Thuy evaluates \((10\times2)-1+2=21\).
Kareem evaluates \((10-1+2)\times2=22\).
Kareem obtains the strictly largest value.

ANSWER 2: C

Problem 3:
Division yields the decimal expansion \(2\div25=0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters out of 4 distinct players is the binomial coefficient \(\binom{4}{3}=4\).

ANSWER 4: B

Problem 5:
If 20% of a number equals 12, the number itself is \(12/0.2=60\). Then 30% of 60 is \(0.3\times60=18\).

ANSWER 5: B

Problem 6:
A net rotation of \(+2.25-3.75=-1.5\) revolutions is equivalent to 1.5 revolutions counterclockwise. From an initial westward orientation, 1.5 revolutions counterclockwise lands exactly on east.

ANSWER 6: B

Problem 7:
The original mixture contains \(0.30\times30=9\) liters of yellow tint. Adding 5 liters raises the yellow volume to 14 liters and the total volume to 35 liters, so the new percentage is \(14/35=40\%\).

ANSWER 7: C

Problem 8:
Decimal approximations are \(15/11\approx1.3636\), \(19/15\approx1.2667\), and \(17/13\approx1.3077\). Ordering these values produces \(19/15<17/13<15/11\).

ANSWER 8: E

Problem 9:
The unique mode must be 6, forcing the mean also to equal 6. The sum of the seven numbers is therefore 42. The six given numbers sum to 31, so \(x=11\). With this value the ordered list is 3,4,5,6,6,7,11 whose median is likewise 6 and whose unique mode remains 6.

ANSWER 9: D

Problem 10:
Let the total distance be \(2d\). Walking time for distance \(d\) is 6 min, so running time for the remaining distance \(d\) at triple speed is \(6/3=2\) min. The total time is therefore 8 min.

ANSWER 10: D

Problem 11:
Each pair sums to a multiple of 10: \(1+9=10\), \(11+19=30\), \(21+29=50\), \(31+39=70\), \(41+49=90\). Their total is 250. Equivalently each five-term arithmetic series sums to 105 and 145 respectively, again yielding 250.

ANSWER 11: E

Problem 12:
Direct comparison of the five decimals shows that 0.99 is strictly larger than each of the others.

ANSWER 12: A

Problem 1:
The problem asks us to determine the number of games that will be played to determine the winner of a basketball tournament with 16 teams, where the losing team of each game is eliminated.

## Step 1
To determine the number of games needed to find the winner, we first note that in a single-elimination tournament, each game eliminates one team.

## Step 2
Since there can only be one winner, all other teams must be eliminated. Therefore, the number of games required is equal to the number of teams minus one.

## Step 3
We have 16 teams competing in the tournament. So, the number of games needed to determine the winner is 16 - 1 = 15.

## Step 4
Let's examine the given choices to rule out the incorrect options. A. 4 is too low because it doesn't account for all the teams. B. 7 is also too low for the same reason. C. 8 is still too low. D. 15 matches our calculation. E. 16 is one more than our calculation.

The final answer is supported by our step-by-step reasoning.
ANSWER 1: D

Problem 2:
The problem asks us to determine who among Jose, Thuy, and Kareem gets the largest final answer after performing a series of operations starting with the number 10.

## Step 1
Let's calculate Jose's final answer: Jose subtracts 1 from 10 to get 9, doubles it to get 18, and then adds 2 to get 20.

## Step 2
Now, let's calculate Thuy's final answer: Thuy doubles 10 to get 20, subtracts 1 to get 19, and then adds 2 to get 21.

## Step 3
Next, let's calculate Kareem's final answer: Kareem subtracts 1 from 10 to get 9, adds 2 to get 11, and then doubles it to get 22.

## Step 4
Comparing the final answers: Jose got 20, Thuy got 21, and Kareem got 22. The largest final answer is 22, which belongs to Kareem.

## Step 5
Let's examine the given choices. A. Jose is incorrect because his answer is 20. B. Thuy is incorrect because her answer is 21. C. Kareem is correct because his answer is 22. D. Jose and Thuy is incorrect because Kareem's answer is larger. E. Thuy and Kareem is incorrect because Jose's answer is not the largest.

The final answer is supported by our step-by-step reasoning.
ANSWER 2: C

Problem 3:
The problem asks us to convert the fraction 2/25 into a decimal.

## Step 1
To convert 2/25 into a decimal, we divide the numerator by the denominator: 2 ÷ 25 = 0.08.

## Step 2
Let's examine the given choices to identify the correct decimal representation. A. .008 is too small. B. .08 matches our calculation. C. .8 is too large. D. 1.25 is much too large. E. 12.5 is also much too large.

The final answer is supported by our step-by-step reasoning.
ANSWER 3: B

Problem 4:
The problem asks us to determine the number of ways to choose 3 starters from a team of 4 members: Lance, Sally, Joy, and Fred.

## Step 1
This is a combination problem, because the order in which the starters are chosen does not matter.

## Step 2
The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial, the product of all positive integers up to that number.

## Step 3
Here, n = 4 (total team members) and k = 3 (number of starters to be chosen). So, C(4, 3) = 4! / [3!(4 - 3)!] = 4! / (3! * 1!) = (4 * 3 * 2 * 1) / ((3 * 2 * 1) * 1) = 4.

## Step 4
Let's examine the given choices. A. 2 is too low. B. 4 matches our calculation. C. 6 is too high. D. 8 is also too high. E. 10 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 4: B

Problem 5:
The problem asks us to find 30% of a number, given that 20% of the same number is 12.

## Step 1
First, we need to find the number. We know that 20% of the number is 12. So, 0.20 * number = 12.

## Step 2
To find the number, we divide both sides of the equation by 0.20: number = 12 / 0.20 = 60.

## Step 3
Now, we need to calculate 30% of 60: 0.30 * 60 = 18.

## Step 4
Let's examine the given choices. A. 15 is too low. B. 18 matches our calculation. C. 20 is too high. D. 24 is also too high. E. 30 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 5: B

Problem 6:
The problem asks us to determine the direction the spinner points after Chenille moves it clockwise 2 1/4 revolutions and then counterclockwise 3 3/4 revolutions, starting from pointing west.

## Step 1
First, let's understand the effect of the movements. A full revolution is 360 degrees or 4 quarters. So, 2 1/4 revolutions clockwise is equivalent to 9 quarters clockwise.

## Step 2
9 quarters clockwise is the same as 1 quarter clockwise (since 8 quarters make 2 full revolutions, which brings the spinner back to the original direction). So, after the first move, the spinner points north.

## Step 3
Then, Chenille moves it counterclockwise 3 3/4 revolutions, which is equivalent to 15 quarters counterclockwise.

## Step 4
15 quarters counterclockwise is the same as 3 quarters counterclockwise (since 12 quarters make 3 full revolutions). So, moving 3 quarters counterclockwise from north results in pointing west.

## Step 5
Let's examine the given choices. A. north is incorrect because after the second move, it's not north. B. east is incorrect. C. south is incorrect. D. west matches our calculation. E. northwest is incorrect.

The final answer is supported by our step-by-step reasoning.
ANSWER 6: D

Problem 7:
The problem asks us to find the percentage of yellow tint in a new mixture after adding 5 liters of yellow tint to an original 30-liter mixture that is 25% red tint, 30% yellow tint, and 45% water.

## Step 1
First, we calculate the amount of yellow tint in the original mixture: 30% of 30 liters = 0.30 * 30 = 9 liters.

## Step 2
After adding 5 liters of yellow tint, the total amount of yellow tint becomes 9 + 5 = 14 liters.

## Step 3
The total volume of the new mixture is 30 + 5 = 35 liters.

## Step 4
To find the percentage of yellow tint in the new mixture, we divide the amount of yellow tint by the total volume and multiply by 100: (14 / 35) * 100 = 40%

Problem 1:
In a single-elimination tournament, each game eliminates exactly one team. To eliminate all but the champion, the number of games needed is one fewer than the number of teams. With 16 teams, the number of games is \(16 - 1 = 15\).

ANSWER 1: D

Problem 2:
Jose: \(10 - 1 = 9\), \(9 \times 2 = 18\), \(18 + 2 = 20\).
Thuy: \(10 \times 2 = 20\), \(20 - 1 = 19\), \(19 + 2 = 21\).
Kareem: \(10 - 1 = 9\), \(9 + 2 = 11\), \(11 \times 2 = 22\).
Kareem’s final answer of 22 is the largest.

ANSWER 2: C

Problem 3:
Divide 2 by 25: \(2 \div 25 = 0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters from 4 people is the combination \(\binom{4}{3} = 4\).

ANSWER 4: B

Problem 5:
Let the number be \(n\). \(20\%\) of \(n\) is \(0.2n = 12\), so \(n = 60\). Then \(30\%\) of \(60 = 0.3 \times 60 = 18\).

ANSWER 5: B

Problem 6:
Initial direction: west.
Clockwise \(2\frac{1}{4}\) revolutions: \(2\frac{1}{4} = 2.25\) revolutions clockwise.
Counterclockwise \(3\frac{3}{4}\) revolutions: \(3\frac{3}{4} = 3.75\) revolutions counterclockwise.
Net revolution = \(2.25 - 3.75 = -1.5\) revolutions, i.e., \(1.5\) revolutions counterclockwise.
A full revolution (1.0) returns to west. The remaining 0.5 revolution counterclockwise from west is a half-turn, which points to east.

ANSWER 6: B

Problem 7:
Original mixture: 30 liters, yellow tint = \(30\% \times 30 = 9\) liters.
Add 5 liters of yellow tint: new yellow = \(9 + 5 = 14\) liters; new total volume = \(30 + 5 = 35\) liters.
Percent yellow = \(\frac{14}{35} \times 100\% = 40\%\).

ANSWER 7: C

Problem 8:
Convert each fraction to a decimal for comparison:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
Order from least to greatest: \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

ANSWER 8: E

Problem 9:
The data set: 3, 4, 5, 6, 6, 7, \(x\). The mode is the most frequent number. With two 6’s, 6 is the mode unless \(x\) equals another number that appears twice, which would create a second mode, violating the “unique mode” condition. Therefore, \(x\) must not be 3, 4, 5, or 7. It could be 6, but if \(x=6\), the median is 6, while the mean is \((3+4+5+6+6+6+7)/7 = 37/7 \approx 5.29\), so mean ≠ median.  
For the mean to equal the median (which is 6, the middle of seven sorted numbers if \(x > 6\) or \(x=6\) only if placed correctly), we set the mean to 6. Sum without \(x\) = \(3+4+5+6+6+7 = 31\). Total needed for mean 6 = \(7 \times 6 = 42\). Thus \(x = 42 - 31 = 11\).  
Sorted: 3, 4, 5, 6, 6, 7, 11. Median = 6, mode = 6, mean = 6. Unique mode holds.

ANSWER 9: D

Problem 10:
Walking half the distance took 6 minutes, so walking speed = \(\frac{\text{half distance}}{6}\).
Running speed is 3 times walking speed. Time to run the second half = \(\frac{\text{half distance}}{3 \times (\text{half distance}/6)} = \frac{6}{3} = 2\) minutes.
Total time = \(6 + 2 = 8\) minutes.

ANSWER 10: D

Problem 11:
Pair the terms: \(1+49 = 50\), \(11+39 = 50\), \(21+29 = 50\), \(31+19 = 50\), \(41+9 = 50\). Five pairs of 50 give \(5 \times 50 = 250\).

ANSWER 11: E

Problem 12:
Compare by extending decimals to the same number of places:
\(0.99   = 0.9900\)
\(0.9099 = 0.9099\)
\(0.9    = 0.9000\)
\(0.909  = 0.9090\)
\(0.9009 = 0.9009\)
Clearly \(0.99 > 0.9099 > 0.909 > 0.9009 > 0.9\). The largest is \(0.99\).

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key Insight:** In a single-elimination tournament, every game eliminates exactly one team. To crown a single winner from 16 teams, we must eliminate 16 − 1 = 15 teams. Therefore, exactly 15 games are needed.

**Verification by rounds:**
- Round 1: 16 teams → 8 games → 8 teams left
- Round 2: 8 teams → 4 games → 4 teams left
- Round 3: 4 teams → 2 games → 2 teams left
- Round 4 (Final): 2 teams → 1 game → 1 winner

Total = 8 + 4 + 2 + 1 = 15 games.

Choices A (4), B (7), and C (8) are too small. Choice E (16) would mean every team is eliminated, leaving no winner.

ANSWER 1: D

---

Problem 2:
We track each person's calculation step by step, all starting from 10.

**Jose:** subtract 1 → 9; double → 18; add 2 → **20**
**Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
**Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

Comparing: 20 < 21 < 22. Kareem gets the largest final answer.

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

Choice A (.008) would be 8/1000. Choice C (.8) would be 8/10. Choices D and E are greater than 1, but 2/25 is clearly less than 1.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination problem — the order of selection doesn't matter:
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4$$

Equivalently, choosing 3 people to start is the same as choosing 1 person to sit out. There are 4 people, so 4 ways to pick who sits out.

The 4 possible starting lineups are: {L,S,J}, {L,S,F}, {L,J,F}, {S,J,F}.

ANSWER 4: B

---

Problem 5:
We're told 20% of a number is 12. Let the number be $x$.

$$0.20x = 12 \implies x = \frac{12}{0.20} = 60$$

Now find 30% of 60:
$$0.30 \times 60 = 18$$

**Shortcut:** 30% is $\frac{30}{20} = 1.5$ times 20%, so the answer is $1.5 \times 12 = 18$.

Choice A (15) would be 25%. Choice D (24) would be 40%. Choice E (30) would be 50%.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing **West**.

**Move 1:** Clockwise $2\frac{1}{4}$ revolutions.
- The 2 full revolutions bring it back to West.
- The extra $\frac{1}{4}$ revolution clockwise from West: West → **North**.

**Move 2:** Counterclockwise $3\frac{3}{4}$ revolutions.
- The 3 full revolutions bring it back to North.
- The extra $\frac{3}{4}$ revolution counterclockwise from North:
  - $\frac{1}{4}$ CCW from North → West
  - $\frac{2}{4}$ CCW from North → South
  - $\frac{3}{4}$ CCW from North → **East**

The spinner points East.

ANSWER 6: B

---

Problem 7:
Original mixture: 30 liters total.
- Red tint: 25% of 30 = 7.5 liters
- Yellow tint: 30% of 30 = 9 liters
- Water: 45% of 30 = 13.5 liters

Add 5 liters of yellow tint:
- New yellow tint = 9 + 5 = 14 liters
- New total mixture = 30 + 5 = 35 liters

Percent yellow tint in new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

Choice B (35%) is a trap for those who just add 5% to 30%. Choice D (45%) is the original water percentage.

ANSWER 7: C

---

Problem 8:
We compare $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$.

Rewrite each as a mixed number:
$$\frac{15}{11} = 1 + \frac{4}{11}, \quad \frac{19}{15} = 1 + \frac{4}{15}, \quad \frac{17}{13} = 1 + \frac{4}{13}$$

Since all three have the same numerator (4) in their fractional parts, we compare denominators. For fractions with equal numerators, a **larger denominator** means a **smaller fraction**:
$$15 > 13 > 11 \implies \frac{4}{15} < \frac{4}{13} < \frac{4}{11}$$

Therefore:
$$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$$

ANSWER 8: E

---

Problem 9:
The data set is: 3, 4, 5, 6, 6, 7, and $x$. The mean, median, and unique mode are all equal.

**Mode:** Since 6 already appears twice and all others appear once, the unique mode must be 6 (if $x$ were 3, 4, 5, or 7, we'd have two modes, violating "unique mode"). So mode = 6.

**Mean = 6:**
$$\frac{3 + 4 + 5 + 6 + 6 + 7 + x}{7} = 6$$
$$\frac{31 + x}{7} = 6 \implies 31 + x = 42 \implies x = 11$$

**Check Median:** With $x = 11$, the sorted list is: 3, 4, 5, **6**, 6, 7, 11. The median (4th value) is 6. ✓

All three measures equal 6. Choice B (6) would make 6 appear three times, but the mean would be $(31+6)/7 = 37/7 \neq 6$.

ANSWER 9: D

---

Problem 10:
Joe walked half the distance in 6 minutes. He then ran the remaining half at 3 times his walking speed.

Let walking speed = $v$. The half-distance = $6v$.

Running speed = $3v$. Time to run the second half:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{6v}{3v} = 2 \text{ minutes}$$

Total time = 6 + 2 = **8 minutes**.

Choice A (7) would result from incorrectly subtracting. Choices B, C, E involve unnecessary decimal complications.

ANSWER 10: D

---

Problem 11:
Compute $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$.

Pair the terms from the first group with terms from the second group (first with last, etc.):
- $1 + 49 = 50$
- $11 + 39 = 50$
- $21 + 29 = 50$
- $31 + 19 = 50$
- $41 + 9 = 50$

Total = $50 \times 5 = 250$.

Choice C (200) is a common error from miscounting pairs. Choice D (249) comes from arithmetic mistakes.

ANSWER 11: E

---

Problem 12:
Compare the five decimals by writing them to four decimal places:

- A: 0.9900
- B: 0.9099
- C: 0.9000
- D: 0.9090
- E: 0.9009

Comparing digit by digit from the tenths place: all have 9 in the tenths place. In the hundredths place, A has **9** while all others have **0**. So A is immediately the largest.

$0.99 > 0.9099 > 0.909 > 0.9009 > 0.9$

ANSWER 12: A

 **Problem 1:**
We need to find the number of games played in a single-elimination tournament with 16 teams. In single elimination, every game eliminates exactly one team. To determine a winner, 15 of the 16 teams must be eliminated. Therefore, exactly 15 games are required.  
(Another way: with 16 teams there are 8 games in the first round, 4 in the second, 2 in the third, and 1 final game: \(8+4+2+1=15\).)

ANSWER 1: D

---

**Problem 2:**
We compute each person’s final answer step by step, starting from 10.

- **Jose:**  
  \(10 - 1 = 9\)  
  \(9 \times 2 = 18\)  
  \(18 + 2 = 20\)

- **Thuy:**  
  \(10 \times 2 = 20\)  
  \(20 - 1 = 19\)  
  \(19 + 2 = 21\)

- **Kareem:**  
  \(10 - 1 = 9\)  
  \(9 + 2 = 11\)  
  \(11 \times 2 = 22\)

Comparing the results: \(20 < 21 < 22\). Kareem’s answer is the largest.

ANSWER 2: C

---

**Problem 3:**
We convert the fraction \(\frac{2}{25}\) to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100:
\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

ANSWER 3: B

---

**Problem 4:**
We must choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways to choose 3 from 4 is the combination \(\binom{4}{3}\):
\[
\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4
\]
Equivalently, choosing 3 starters is the same as choosing 1 person to sit out, and there are 4 choices for who sits out.

ANSWER 4: B

---

**Problem 5:**
Let the number be \(N\). We are given \(0.20N = 12\). Solving for \(N\):
\[
N = \frac{12}{0.20} = 60
\]
Now find 30% of 60:
\[
0.30 \times 60 = 18
\]
(Shortcut: 30% is \(1.5\) times 20%, so \(1.5 \times 12 = 18\).)

ANSWER 5: B

---

**Problem 6:**
The spinner starts pointing west.  
- Clockwise \(2\frac14\) revolutions: 2 full revolutions bring it back to west, then \(\frac14\) turn clockwise from west points it **north**.  
- Counterclockwise \(3\frac34\) revolutions: 3 full revolutions bring it back to north, then \(\frac34\) turn counterclockwise from north goes through west, south, and ends at **east**.

Net movement: \(3\frac34 - 2\frac14 = 1\frac12\) revolutions counterclockwise. One full revolution returns to west; an additional \(\frac12\) revolution counterclockwise from west points to **east**.

ANSWER 6: B

---

**Problem 7:**
Original mixture: 30 liters.  
- Yellow tint originally: \(30\% \times 30 = 9\) liters.  
After adding 5 liters of yellow tint:  
- New yellow tint: \(9 + 5 = 14\) liters.  
- New total mixture: \(30 + 5 = 35\) liters.  

Percent of yellow tint in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%
\]

ANSWER 7: C

---

**Problem 8:**
Compare \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\). Each is greater than 1. Write each as \(1\) plus a proper fraction:
\[
\frac{15}{11} = 1 + \frac{4}{11}, \qquad
\frac{19}{15} = 1 + \frac{4}{15}, \qquad
\frac{17}{13} = 1 + \frac{4}{13}
\]
Since the numerators are the same (4), the fraction with the largest denominator is smallest:
\[
\frac{4}{15} < \frac{4}{13} < \frac{4}{11}
\]
Adding 1 preserves the order:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

ANSWER 8: E

---

**Problem 9:**
The data set is \(3, 4, 5, 6, 6, 7, x\).  
- The current mode is 6 (appears twice). For 6 to remain the **unique** mode, \(x\) must not equal 3, 4, 5, or 7 (which would create a second mode).  
- We are told mean = median = mode = 6.  

Set the mean equal to 6:
\[
\frac{3+4+5+6+6+7+x}{7} = 6 \implies \frac{31+x}{7} = 6 \implies 31+x = 42 \implies x = 11
\]

Check: With \(x=11\), the sorted list is \(3, 4, 5, 6, 6, 7, 11\).  
- Median (4th term) = 6.  
- Mode = 6 (unique).  
- Mean = 6.  

All conditions are satisfied.

ANSWER 9: D

---

**Problem 10:**
Joe walks half the distance in 6 minutes. He runs the second half at 3 times his walking speed. Since distance is the same and speed is tripled, the time required is divided by 3:
\[
\text{Running time} = \frac{6}{3} = 2 \text{ minutes}
\]
Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

---

**Problem 11:**
Add the two groups:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]
Pair the terms to make convenient sums:
\[
(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 5 \times 50 = 250
\]

ANSWER 11: E

---

**Problem 12:**
Compare the decimals by writing them with the same number of places:
- A. \(0.9900\)
- B. \(0.9099\)
- C. \(0.9000\)
- D. \(0.9090\)
- E. \(0.9009\)

The tenths digit is 9 for all. Looking at the hundredths digit: A has 9, while B, D have 0 and C, E have 0. Thus \(0.9900\) is clearly the largest.

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key insight:** Each game eliminates exactly one team. To determine a winner from 16 teams, we must eliminate 15 teams (everyone except the champion).

Therefore, exactly 15 games must be played.

*Verification by rounds:* Round 1: 8 games (16→8 teams), Round 2: 4 games (8→4), Round 3: 2 games (4→2), Round 4: 1 game (2→1). Total = 8+4+2+1 = 15.

The tempting wrong answer E (16) counts teams instead of eliminations.

ANSWER 1: D

---

Problem 2:
We need to compute each person's final answer starting from 10.

**Jose:** Start with 10 → subtract 1: 10−1 = 9 → double: 9×2 = 18 → add 2: 18+2 = **20**

**Thuy:** Start with 10 → double: 10×2 = 20 → subtract 1: 20−1 = 19 → add 2: 19+2 = **21**

**Kareem:** Start with 10 → subtract 1: 10−1 = 9 → add 2: 9+2 = 11 → double: 11×2 = **22**

Kareem gets the largest final answer (22).

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

The tempting wrong answer A (0.008) would be 8/1000, and C (0.8) would be 8/10.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination: C(4,3) = 4!/(3! × 1!) = 4

Listing them out: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — that's 4 ways.

The tempting wrong answer C (6) would be C(4,2), choosing who's *not* a starter, but that also equals 4... Actually C(4,2)=6 is wrong here; we're choosing 3 from 4.

ANSWER 4: B

---

Problem 5:
We know 20% of a number is 12. We need to find 30% of the same number.

**Step 1:** Find the number.
$$0.20 \times n = 12 \implies n = \frac{12}{0.20} = 60$$

**Step 2:** Find 30% of 60.
$$0.30 \times 60 = 18$$

*Shortcut:* Since 30% is 1.5 times 20%, the answer is 1.5 × 12 = 18.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing west. We need to find the net rotation.

**Clockwise:** 2¼ revolutions
**Counterclockwise:** 3¾ revolutions

**Net movement:** 2¼ − 3¾ = −1½ revolutions (negative means counterclockwise)

So the spinner moves 1½ revolutions counterclockwise from west.

- 1 full revolution counterclockwise returns to **west**
- ½ revolution counterclockwise from west goes to **east**

(Counterclockwise from west: ¼ turn → south, ½ turn → east)

ANSWER 6: B

---

Problem 7:
We need to find the percent of yellow tint after adding 5 liters of yellow tint.

**Original mixture (30 liters):**
- Yellow tint: 30% of 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percent yellow tint:** (14/35) × 100 = 40%

The tempting wrong answer B (35) might come from simply adding 5% to 30%.

ANSWER 7: C

---

Problem 8:
We need to order 15/11, 19/15, and 17/13 from least to greatest.

Computing each as a decimal:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3077...

From least to greatest: **19/15 < 17/13 < 15/11**

*Alternative method:* Note each is 1 + (numerator − denominator)/denominator:
- 15/11 = 1 + 4/11
- 19/15 = 1 + 4/15
- 17/13 = 1 + 4/13

Since 4/15 < 4/13 < 4/11, the order is confirmed.

ANSWER 8: E

---

Problem 9:
The positive integers 3, 4, 5, 6, 6, 7, and x have equal mean, median, and unique mode.

**Mode:** Since 6 already appears twice and no other number does, the unique mode must be 6. So x ≠ 3, 4, 5, or 7 (which would create a second mode).

**Median:** With 7 numbers, the median is the 4th value when sorted. For the median to be 6, we need x ≥ 6 (so that 6 is the 4th value).

- If x = 6: sorted = 3, 4, 5, 6, 6, 6, 7 → median = 6 ✓
- If x > 7: sorted = 3, 4, 5, 6, 6, 7, x → median = 6 ✓

**Mean:** (3 + 4 + 5 + 6 + 6 + 7 + x)/7 = (31 + x)/7

Setting mean = 6: (31 + x)/7 = 6 → 31 + x = 42 → **x = 11**

**Verification:** Numbers: 3, 4, 5, 6, 6, 7, 11. Mean = 42/7 = 6, Median = 6, Mode = 6 ✓

ANSWER 9: D

---

Problem 10:
Joe walks halfway, then runs the rest. We need total time.

**Walking:** Half the distance takes 6 minutes.

**Running:** He runs 3 times as fast as he walks, so he covers the same distance in 1/3 the time.
- Time to run the second half = 6 ÷ 3 = 2 minutes

**Total time:** 6 + 2 = **8 minutes**

The tempting wrong answers come from setting up the problem incorrectly or mixing up distance and time relationships.

ANSWER 10: D

---

Problem 11:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

**First group:** 1 + 11 + 21 + 31 + 41 = 105

**Second group:** 9 + 19 + 29 + 39 + 49 = 145

**Total:** 105 + 145 = **250**

*Elegant method:* Pair terms: (1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250

ANSWER 11: E

---

Problem 12:
We need to find the largest number among the options. Writing each with the same number of decimal places:

- A. 0.99 = **0.9900**
- B. 0.9099 = 0.9099
- C. 0.9 = 0.9000
- D. 0.909 = 0.9090
- E. 0.9009 = 0.9009

Comparing digit by digit after the decimal point:
- First digit: all are 9
- Second digit: A has **9**, all others have 0

Since A has a 9 in the second decimal place while all others have 0, **0.99 is the largest**.

ANSWER 12: A

**Problem 1:**  
The tournament is single‑elimination. Starting with 16 teams, every game eliminates exactly one team. To reduce from 16 teams to a single champion we must eliminate 15 teams, and each elimination corresponds to one game.  

Number of games = 16 − 1 = **15**.  

ANSWER 1: D  

---

**Problem 2:**  
Start with 10 for each person.

- **Jose:**  
  Subtract 1 → 9; double → 18; add 2 → 20.  

- **Thuy:**  
  Double → 20; subtract 1 → 19; add 2 → 21.  

- **Kareem:**  
  Subtract 1 → 9; add 2 → 11; double → 22.  

The largest final value is 22, obtained by Kareem.  

ANSWER 2: C  

---

**Problem 3:**  

\[
\frac{2}{25}=0.08
\]

(2 ÷ 25 = 0.08).  

ANSWER 3: B  

---

**Problem 4:**  
We need to choose 3 starters from the 4 available players:  

\[
\binom{4}{3}=4
\]

ways.  

ANSWER 4: B  

---

**Problem 5:**  
Let the number be \(N\).  
\(0.20N = 12 \Rightarrow N = \frac{12}{0.20}=60\).  
\(30\%\) of \(N\) is \(0.30 \times 60 = 18\).  

ANSWER 5: B  

---

**Problem 6:**  
Interpret the fractions as mixed numbers:  

- Clockwise \(2\frac14 = 2.25\) revolutions.  
- Counter‑clockwise \(3\frac34 = 3.75\) revolutions.  

Net rotation (counter‑clockwise positive) = \(3.75 - 2.25 = 1.5\) revolutions.  

A full revolution returns to the same direction, so we only care about the extra \(0.5\) revolution.  
\(0.5\) revolution = 180° counter‑clockwise.  

Starting from west, turning 180° counter‑clockwise points to **east**.  

ANSWER 6: B  

---

**Problem 7:**  
Original yellow = \(30\%\) of 30 L = \(0.30 \times 30 = 9\) L.  
Add 5 L yellow → new yellow = \(9+5 = 14\) L.  
New total volume = \(30+5 = 35\) L.  

Percent yellow = \(\frac{14}{35}\times100 = 40\%\).  

ANSWER 7: C  

---

**Problem 8:**  
Compute the three fractions:

\[
\frac{15}{11}=1.3636\ldots,\qquad
\frac{19}{15}=1.2666\ldots,\qquad
\frac{17}{13}=1.3077\ldots
\]

Ordered from least to greatest:  

\(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

This matches option **E**.  

ANSWER 8: E  

---

**Problem 9:**  
Current list: \(3,4,5,6,6,7\). The unique mode is 6 (appears twice).  
For the mode to remain unique and equal to the mean and median, \(x\) must not create a tie.  
Let the mean = median = mode = \(m\).  

Sum of numbers = \(31 + x\).  
Mean = \(\frac{31 + x}{7}=m\).  
Median of the seven numbers is the 4th after sorting; with \(x>7\) the median stays 6, so \(m=6\).  

Thus \(\frac{31 + x}{7}=6 \Rightarrow 31 + x = 42 \Rightarrow x = 11\).

Check: sorted list becomes \(3,4,5,6,6,7,11\) → median = 6, mode = 6, mean = 42/7 = 6. All equal.  

ANSWER 9: D  

---

**Problem 10:**  
Let the walking speed be \(v\) and the half‑distance be \(d/2\).  

Walking time for half‑distance: \(\frac{d/2}{v}=6\) min ⇒ \(\frac{d}{v}=12\) min.  

Running speed = \(3v\).  
Running time for the second half: \(\frac{d/2}{3v}= \frac{d}{6v}= \frac{1}{6}\times\frac{d}{v}= \frac{12}{6}=2\) min.  

Total time = \(6 + 2 = 8\) minutes.  

ANSWER 10: D  

---

**Problem 11:**  

\[
\begin{aligned}
1+11+21+31+41 &= 105,\\
9+19+29+39+49 &= 145,\\
\text{Total} &= 105+145 = 250.
\end{aligned}
\]

ANSWER 11: E  

---

**Problem 12:**  
Write the numbers with four decimal places:

- .99 = 0.9900  
- .9099 = 0.9099  
- .9 = 0.9000  
- .909 = 0.9090  
- .9009 = 0.9009  

The largest is 0.9900, i.e., **.99**.  

ANSWER 12: A  

Problem 1:  
The tournament starts with 16 teams. Each game eliminates one team, and the losing team is eliminated. To determine the winner, all teams except one must be eliminated. Thus, 15 teams must be eliminated, requiring 15 games.  
- **Why not other choices?** A (4) is too few; B (7), C (8) are also insufficient; E (16) would imply one game per team without elimination, which is incorrect.  
ANSWER 1: D  

Problem 2:  
- **Jose**: Starts with 10 → subtracts 1 (9) → doubles (18) → adds 2 → **20**.  
- **Thuy**: Starts with 10 → doubles (20) → subtracts 1 (19) → adds 2 → **21**.  
- **Kareem**: Starts with 10 → subtracts 1 (9) → adds 2 (11) → doubles → **22**.  
Kareem has the largest final answer.  
- **Why not other choices?** A (Jose) and B (Thuy) are smaller; D and E pair incorrect winners.  
ANSWER 2: C  

Problem 3:  
\( \frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08 \).  
- **Why not other choices?** A (.008) is \( \frac{2}{250} \); C (.8) is \( \frac{4}{5} \); D and E are larger decimals.  
ANSWER 3: B  

Problem 4:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is \( \binom{4}{3} = 4 \).  
- **Why not other choices?** A (2) is too few; C (6) is \( \binom{4}{2} \), which is incorrect; D (8) and E (10) are overcounts.  
ANSWER 4: B  

Problem 5:  
Let the number be \( x \). If 20% of \( x \) is 12, then \( 0.2x = 12 \), so \( x = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).  
- **Why not other choices?** A (15) is 25% of 60; C (20) is \( \frac{1}{3} \) of 60; D (24) is 40%; E (30) is 50%.  
ANSWER 5: B  

Problem 6:  
- A full revolution is 360°. Clockwise \( 2 \frac{1}{4} \) revolutions = \( 2.25 \times 360 = 810° \).  
- Counterclockwise \( 3 \frac{3}{4} \) revolutions = \( -3.75 \times 360 = -1350° \).  
- Net movement: \( 810° - 1350° = -540° \).  
- \( -540° \mod 360° = 180° \) (since \( -540 + 720 = 180° \)), which points west → south? Wait, initial direction is west.  
- West (270°) + 810° = 1080° → 1080° mod 360° = 0° (north). Then -1350° → 0° - 1350° = -1350° mod 360° = 90° (east)? Let's reframe:  
- Start at west (270°). Clockwise 2.25 rev: 270° + 810° = 1080° mod 360° = 0° (north). Then counterclockwise 3.75 rev: 0° - 1350° = -1350° mod 360° = 90° (east).  
- **Why not other choices?** A (north) is initial after clockwise; B (east) is final; C (south) is 180°; D (west) is start; E (northwest) is not a cardinal direction.  
ANSWER 6: B  

Problem 7:  
Original mixture: 30 liters, 30% yellow tint → 9 liters yellow. Add 5 liters yellow → total yellow = 14 liters. New mixture = 35 liters. Percent yellow = \( \frac{14}{35} \times 100 = 40\% \).  
- **Why not other choices?** A (25) is original yellow%; B (35) is before addition; D (45) and E (50) are overestimates.  
ANSWER 7: E (Note: Correction based on computation: 14/35=0.4 → 40%, so answer is E? Wait, 14/35=0.4 → 40%, which is E. But earlier I thought B, but B is 35, which is wrong. Original yellow is 9, add 5 → 14. Total mixture 30+5=35. 14/35=0.4 → 40%. So E. But in choices E is 50? Wait, no: choices are A.25 B.35 C.40 D.45 E.50. So 40% is C. I misread. Correction: 14/35=0.4 → 40%, which is C.  
**Correction**: 14/35=0.4 → 40%, so answer is C.  
ANSWER 7: C  

Problem 8:  
Compare \( \frac{15}{11} \approx 1.3636 \), \( \frac{19}{15} \approx 1.2667 \), \( \frac{17}{13} \approx 1.3077 \). Order: \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11} \).  
- **Why not other choices?** A, B, C, D have incorrect orders.  
ANSWER 8: E  

Problem 9:  
The set is {3, 4, 5, 6, 6, 7, x}. Mean = median = mode = 6.  
- Mode is 6 (unique), so 6 must appear most.  
- Median is 6 (4th value in ordered list: 3,4,5,6,6,7,x → if x≥6, ordered is ...,6,6,x or ...,6,x,6? Must have 4th value 6.  
- If x=6, set is {3,4,5,6,6,6,7}: median=6, mode=6, mean=(3+4+5+6+6+6+7)/7=37/7≈5.28 ≠6.  
- If x=11: set {3,4,5,6,6,7,11}: mean=(3+4+5+6+6+7+11)/7=42/7=6, median=6, mode=6.  
- **Why not other choices?** A (5): mean=36/7≈5.14; B (6): mean≈5.28; C (7): mean=38/7≈5.43; D (11) works.  
ANSWER 9: D  

Problem 10:  
Joe walks half the distance in 6 minutes. Let walking speed be \( v \), so distance = \( 6v \). He runs at \( 3v \), so time to run half distance = \( \frac{6v}{3v} = 2 \) minutes. Total time = 6 + 2 = 8 minutes.  
- **Why not other choices?** A (7), B (7.3), C (7.7), E (8.3) are not exact.  
ANSWER 10: D  

Problem 11:  
First group: 1+11+21+31+41 = 105. Second group: 9+19+29+39+49 = 145. Total = 105 + 145 = 250.  
- **Why not other choices?** A (150), B (199), C (200), D (249) are incorrect sums.  
ANSWER 11: E  

Problem 12:  
Compare decimals: A (.99), B (.9099), C (.9), D (.909), E (.9009). Largest is .99.  
- **Why not other choices?** B, C, D, E are all smaller than .99.  
ANSWER 12: A

### Problem 1:
We need to find the total number of games to determine a winner in a 16-team single-elimination tournament, where every game eliminates exactly one losing team. To get 1 winner, we must eliminate 16 - 1 = 15 teams. Since each game eliminates 1 team, we need 15 total games. Wrong choices: A is the number of bracket rounds, not games; C is half the number of teams; E is the total number of teams, not games.
ANSWER 1: D

---

### Problem 2:
All three start with 10, we calculate each person's final result step-by-step:
- Jose: (10 - 1) * 2 + 2 = 9*2 +2 = 20
- Thuy: (10*2) -1 +2 = 20-1+2=21
- Kareem: (10-1 +2)*2 = 11*2=22
Kareem has the largest final answer. Wrong choices stem from misordering the operations for each person.
ANSWER 2: C

---

### Problem3:
Convert 2/25 to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100: (2*4)/(25*4) = 8/100 = 0.08. Wrong choices: A is 2/250, C is 20/25.
ANSWER 3: B

---

### Problem4:
We choose 3 starters from 4 people, which is a combination problem (order of starters does not matter here). The number of ways to choose 3 from 4 is equivalent to choosing 1 person to leave off the starting lineup: there are 4 people, so 4 unique lineups, each omitting one person. Wrong choice C (6) is the number of permutations, not combinations.
ANSWER 4: B

---

### Problem5:
If 20% of a number is 12, first find the full number: 0.2*N=12 → N=12/0.2=60. 30% of 60 is 0.3*60=18. Wrong choices: A is 25% of 60, D is 40% of 60.
ANSWER 5: B

---

### Problem6:
The spinner starts pointing west. Net movement: 2.25 clockwise revolutions minus 3.75 clockwise (equivalent to 3.75 counterclockwise) = -1.5 total clockwise revolutions, or 1.5 counterclockwise revolutions. Full revolutions cancel out, so we only have a net 0.5 (180 degree) counterclockwise turn from west. A 180-degree turn from west points to east.
ANSWER 6: B

---

### Problem7:
Original 30L mixture has 30% yellow tint: 0.3*30=9L of yellow. Add 5L of yellow, so new yellow is 14L, new total mixture is 35L. The new percentage of yellow is (14/35)*100=40%. Wrong choice: people often forget to add 5L to the total mixture volume, leading to an incorrect higher percentage.
ANSWER 7: C

---

### Problem8:
Rewrite each fraction to compare easily: 15/11 = 1 + 4/11, 19/15=1+4/15, 17/13=1+4/13. For equal numerators, the larger the denominator, the smaller the fraction. So 4/15 < 4/13 < 4/11, so the full order from least to greatest is 19/15 < 17/13 <15/11.
ANSWER 8: E

---

### Problem9:
The dataset is [3,4,5,6,6,7,x]. The unique mode is currently 6 (only value appearing twice, so it must stay the only unique mode). For the mean, median, and mode to all be 6: the sum of the dataset must be 7*6=42. The sum of the existing numbers is 3+4+5+6+6+7=31, so x=42-31=11. Sorted dataset [3,4,5,6,6,7,11] has median 6, unique mode 6, mean 6, which fits. Wrong choice B (x=6) gives a mean of ~5.28, not 6.
ANSWER 9: D

---

### Problem10:
Joe walks half the distance in 6 minutes. His running speed is 3x his walking speed, so time to cover the same distance is 1/3 of walking time. Time to run the second half: 6/3=2 minutes. Total time: 6+2=8 minutes.
ANSWER 10: D

---

### Problem11:
Pair terms to simplify addition: (1+49)+(11+39)+(21+29)+(31+19)+(41+9) = 50+50+50+50+50=250.
ANSWER 11: E

---

### Problem12:
Line up all decimals to compare: A=0.9900, B=0.9099, C=0.9000, D=0.9090, E=0.9009. The largest value is 0.99, option A.
ANSWER 12: A

Problem 1:  
In a single-elimination tournament, each game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams (all except 1). Thus, 15 games are played.  
ANSWER 1: D  

Problem 2:  
Calculate each person's result starting from 10:  
- Jose: \(10 -1 =9\), double → \(9*2=18\), add 2 → \(18+2=20\)  
- Thuy: Double 10 → \(10*2=20\), subtract 1 → \(20-1=19\), add 2 → \(19+2=21\)  
- Kareem: \(10-1=9\), add 2 → \(9+2=11\), double → \(11*2=22\)  
Kareem has the largest result.  
ANSWER 2: C  

Problem 3:  
Convert \(\frac{2}{25}\) to decimal: \(2\div25=0.08\).  
ANSWER 3: B  

Problem 4:  
Choosing 3 starters from 4 people is a combination (order does not matter). The number of ways is \(\binom{4}{3}=4\) (or list: {Lance,Sally,Joy}, {Lance,Sally,Fred}, {Lance,Joy,Fred}, {Sally,Joy,Fred}).  
ANSWER 4: B  

Problem 5:  
Let the number be \(x\). \(20\%x=12\) → \(0.2x=12\) → \(x=60\). \(30\%\ of\ 60=0.3*60=18\).  
ANSWER 5: B  

Problem 6:  
Model directions with standard angles: 0°=east, 90°=north, 180°=west, 270°=south. Initial direction is west (180°).  
- Clockwise \(2\frac{1}{4}\) revolutions: \(2.25*(-360°)=-810°\), new angle: \(180-810=-630°\)  
- Counterclockwise \(3\frac{3}{4}\) revolutions: \(3.75*(360°)=+1350°\), total angle: \(-630+1350=720°\)  
- \(720°\mod360°=0°\), which is east.  
ANSWER 6: B  

Problem 7:  
Original yellow tint: \(30\%\ of\ 30=0.3*30=9\) liters. Add 5 liters, total yellow = \(9+5=14\) liters. New total mixture: \(30+5=35\) liters. Percent yellow: \(\frac{14}{35}*100\%=40\%\).  
ANSWER 7: C  

Problem 8:  
Compare fractions via cross-multiplication or decimals:  
- \(\frac{19}{15}\approx1.267\), \(\frac{17}{13}\approx1.308\), \(\frac{15}{11}\approx1.364\)  
Order from least to greatest: \(\frac{19}{15}<\frac{17}{13}<\frac{15}{11}\).  
ANSWER 8: E  

Problem 9:  
The unique mode is 6 (appears twice, others once). Thus mean and median equal 6.  
- Median of 7 numbers is the 4th term when sorted, so 4th term =6 → \(x\geq6\).  
- Mean = \(\frac{3+4+5+6+6+7+x}{7}=6\) → \(31+x=42\) → \(x=11\).  
Check: Sorted list [3,4,5,6,6,7,11], median=6, mode=6 (unique), mean=6.  
ANSWER 9: D  

Problem 10:  
Walking half the distance takes 6 minutes, so full walking time would be 12 minutes. Running speed is 3x walking speed, so time for the second half is \(\frac{6}{3}=2\) minutes. Total time: \(6+2=8\) minutes.  
ANSWER 10: D  

Problem 11:  
Sum first group: \(1+11+21+31+41=105\) (arithmetic series: \(\frac{5}{2}(1+41)=105\)).  
Sum second group: \(9+19+29+39+49=145\) (\(\frac{5}{2}(9+49)=145\)).  
Total: \(105+145=250\).  
ANSWER 11: E  

Problem 12:  
Compare decimals digit by digit:  
- A=0.99, B=0.9099, C=0.9, D=0.909, E=0.9009  
A has the highest second decimal digit (9 vs 0 for others), so it is the largest.  
ANSWER 12: A

I'll solve each problem step-by-step.

---

**Problem 1:**

**What's being asked:** How many games must be played to determine a winner in a 16-team elimination tournament?

**Setup:** In an elimination tournament, each game eliminates exactly one team. To go from 16 teams to 1 winner, we must eliminate 15 teams. Therefore, we need 15 games.

**Alternatively:** We can think of it as a bracket:
- Round 1: 16 teams → 8 games → 8 winners
- Round 2: 8 teams → 4 games → 4 winners
- Round 3: 4 teams → 2 games → 2 winners
- Round 4: 2 teams → 1 game → 1 winner
- Total: 8 + 4 + 2 + 1 = 15 games

ANSWER 1: D

---

**Problem 2:**

**What's being asked:** Which person gets the largest final answer after their operations?

**Jose's calculation:** 
- Start: 10
- Subtract 1: 10 - 1 = 9
- Double: 9 × 2 = 18
- Add 2: 18 + 2 = **20**

**Thuy's calculation:**
- Start: 10
- Double: 10 × 2 = 20
- Subtract 1: 20 - 1 = 19
- Add 2: 19 + 2 = **21**

**Kareem's calculation:**
- Start: 10
- Subtract 1: 10 - 1 = 9
- Add 2: 9 + 2 = 11
- Double: 11 × 2 = **22**

Kareem gets 22, which is the largest.

ANSWER 2: C

---

**Problem 3:**

**What's being asked:** Convert the fraction 2/25 to a decimal.

**Setup:** 2/25 = ?

**Calculation:** 
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

ANSWER 3: B

---

**Problem 4:**

**What's being asked:** In how many ways can 3 starters be chosen from 4 people (Lance, Sally, Joy, Fred)?

**Setup:** This is a combination problem: $\binom{4}{3}$

**Calculation:**
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4}{1} = 4$$

Alternatively, we're choosing which 1 person to leave out: 4 ways.

ANSWER 4: B

---

**Problem 5:**

**What's being asked:** If 20% of a number is 12, what is 30% of that number?

**Setup:** Let the number be $n$.
- $0.20n = 12$
- $n = 12 ÷ 0.20 = 60$

**Calculation:** 
$$30\% \text{ of } 60 = 0.30 \times 60 = 18$$

ANSWER 5: B

---

**Problem 6:**

**What's being asked:** Starting pointing west, after rotating clockwise 2¼ revolutions then counterclockwise 3¾ revolutions, which direction?

**Setup:** 
- Clockwise 2¼ revolutions = 2.25 revolutions
- Counterclockwise 3¾ revolutions = 3.75 revolutions
- Net: 3.75 - 2.25 = 1.5 revolutions counterclockwise (net)

**Calculation:**
Starting at west, rotating counterclockwise 1.5 revolutions (= 1 full rotation + 0.5 rotation):
- 1 full rotation brings us back to west
- 0.5 rotation = 180°
- West + 180° counterclockwise = East

ANSWER 6: B

---

**Problem 7:**

**What's being asked:** After adding 5 liters of yellow tint to a 30-liter mixture, what percent is yellow tint?

**Original mixture (30 liters):**
- Yellow tint: 30% × 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percentage:**
$$\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%$$

ANSWER 7: C

---

**Problem 8:**

**What's being asked:** Order 15/11, 19/15, and 17/13 from least to greatest.

**Setup:** Convert to decimals or compare directly.

**Calculations:**
- 15/11 ≈ 1.364
- 19/15 ≈ 1.267
- 17/13 ≈ 1.308

**Order:** 19/15 < 17/13 < 15/11

ANSWER 8: E

---

**Problem 9:**

**What's being asked:** Find $x$ such that the mean, median, and unique mode of {3, 4, 5, 6, 6, 7, $x$} are all equal.

**Current mode:** 6 appears twice; all others appear once. If there's a unique mode, it must be 6.

**With unique mode = 6:** This requires no other number to appear more than once. So $x \notin \{3, 4, 5, 7\}$ and $x \neq 6$ (else it wouldn't be unique).

**Mean = Median = Mode = 6:**

For median of 7 numbers with value 6:
- Sorted: need middle (4th) value to be 6
- If $x = 11$: {3, 4, 5, 6, 6, 7, 11} → median = 6 ✓

Mean: $\frac{3+4+5+6+6+7+11}{7} = \frac{42}{7} = 6$ ✓

Mode: 6 appears twice; all others once → mode = 6 ✓

ANSWER 9: D

---

**Problem 10:**

**What's being asked:** Total time from home to school if Joe walked halfway (6 min) then ran the rest at 3× speed?

**Setup:**
- Walking: 6 minutes for half the distance
- Running: same distance at 3× speed, so time = 6/3 = 2 minutes

**Total time:** 6 + 2 = 8 minutes

ANSWER 10: D

---

**Problem 11:**

**What's being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Calculation:**
- First group: 1 + 11 + 21 + 31 + 41 = 105
- Second group: 9 + 19 + 29 + 39 + 49 = 145
- Sum: 105 + 145 = 250

ANSWER 11: E

---

**Problem 12:**

**What's being asked:** Which number is largest?

**Comparison:**
- A. 0.99
- B. 0.9099
- C. 0.9
- D. 0.909
- E. 0.9009

Starting from the tenths place: all are 0.9__. Looking at hundredths:
- 0.99 has 9 in hundredths place → **largest**

ANSWER 12: A

Problem 1:

In a single-elimination tournament, each game eliminates exactly one team.  
There are 16 teams, and we want one champion.

So we must eliminate 15 teams total.

Therefore, the number of games played is 15.

Tempting wrong choice: 16 would be if every team had to lose, but the winner does not lose.

ANSWER 1: D

Problem 2:

Compute each person’s result carefully.

- Jose: start with 10  
  subtract 1 → 9  
  double → 18  
  add 2 → 20

- Thuy: start with 10  
  double → 20  
  subtract 1 → 19  
  add 2 → 21

- Kareem: start with 10  
  subtract 1 → 9  
  add 2 → 11  
  double → 22

Compare:
- Jose = 20
- Thuy = 21
- Kareem = 22

Kareem gets the largest final answer.

ANSWER 2: C

Problem 3:

\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

So the correct decimal is 0.08.

ANSWER 3: B

Problem 4:

We are choosing 3 starters from 4 players: Lance, Sally, Joy, and Fred.

This is a combination problem:
\[
\binom{4}{3} = 4
\]

You can also see it by noticing that choosing 3 starters is the same as choosing which 1 player sits out, and there are 4 choices.

ANSWER 4: B

Problem 5:

Let the number be \(n\).

20% of \(n\) is 12, so:
\[
0.2n = 12
\]
\[
n = 60
\]

Now find 30% of 60:
\[
0.3 \times 60 = 18
\]

ANSWER 5: B

Problem 6:

The spinner starts pointing west.

First move: clockwise \(2\frac{1}{4}\) revolutions.  
A full revolution brings it back to west, so only the extra \(\frac{1}{4}\) revolution matters.

From west, a quarter-turn clockwise points it to north.

Second move: counterclockwise \(3\frac{3}{4}\) revolutions.  
Again, the 3 full revolutions do nothing, so this is just \(\frac{3}{4}\) of a revolution counterclockwise, which is the same as \(\frac{1}{4}\) of a revolution clockwise.

From north, a quarter-turn clockwise points east.

So the spinner ends pointing east.

ANSWER 6: B

Problem 7:

Start with 30 liters total.

Yellow tint is 30% of 30 liters:
\[
0.30 \times 30 = 9 \text{ liters}
\]

Then 5 liters of yellow tint are added, so yellow becomes:
\[
9 + 5 = 14 \text{ liters}
\]

New total mixture:
\[
30 + 5 = 35 \text{ liters}
\]

Percent yellow in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.4 = 40\%
\]

ANSWER 7: C

Problem 8:

The fractions are:
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]

Approximate each one:

\[
\frac{15}{11} \approx 1.364
\]
\[
\frac{17}{13} \approx 1.308
\]
\[
\frac{19}{15} \approx 1.267
\]

So from least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

That matches choice E.

ANSWER 8: E

Problem 9:

The numbers are:
\[
3,4,5,6,6,7,x
\]

We are told the mean, median, and unique mode are all equal.

Since 6 appears twice and every other listed number appears once, the unique mode is 6.

So the median and mean must also be 6.

The median of 7 numbers is the 4th number when ordered.  
The sorted list is:
\[
3,4,5,6,6,7,x
\]
To make the median 6, \(x\) must be 6 or larger, but we also need the mean to be 6.

Sum of the known numbers:
\[
3+4+5+6+6+7 = 31
\]

For the mean to be 6, total sum must be:
\[
7 \times 6 = 42
\]

So:
\[
31 + x = 42
\]
\[
x = 11
\]

Check:
- mode = 6
- median = 6
- mean = 42/7 = 6

Works perfectly.

ANSWER 9: D

Problem 10:

Joe walked half the distance in 6 minutes.  
He then ran the other half at 3 times his walking speed.

If he runs 3 times as fast, then the time for the same distance is 1/3 as long.

So the running time is:
\[
6 \div 3 = 2 \text{ minutes}
\]

Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

Problem 11:

Add the first group:
\[
1+11+21+31+41 = 105
\]

Add the second group:
\[
9+19+29+39+49 = 145
\]

Now add the totals:
\[
105 + 145 = 250
\]

ANSWER 11: E

Problem 12:

Compare the decimals:

- \(0.99\)
- \(0.9099\)
- \(0.9 = 0.9000\)
- \(0.909\)
- \(0.9009\)

The largest is clearly \(0.99\), since it is closest to 1.

ANSWER 12: A

Problem 1:  
We have a three-person-team tournament where **the losing team of each game is eliminated**. With **16 teams**, how many games are played to determine the winner?

To find the number of games in a single-elimination tournament:  
- Each game eliminates exactly **one** team (the loser).  
- To go from **16 teams to 1 winning team**, we must eliminate **15 teams**.  
- Therefore, the number of games is **15**.

So the correct choice is **D**.

ANSWER 1: D


Problem 2:  
Jose, Thuy, and Kareem all start at 10. We compute each final value.

**Jose:**  
- Subtract 1: \(10 - 1 = 9\)  
- Double: \(2 \cdot 9 = 18\)  
- Add 2: \(18 + 2 = 20\)  
Final: **20**

**Thuy:**  
- Double 10: \(2 \cdot 10 = 20\)  
- Subtract 1: \(20 - 1 = 19\)  
- Add 2: \(19 + 2 = 21\)  
Final: **21**

**Kareem:**  
- Subtract 1: \(10 - 1 = 9\)  
- Add 2: \(9 + 2 = 11\)  
- Double: \(2 \cdot 11 = 22\)  
Final: **22**

Largest is **22**, which is Kareem.

ANSWER 2: C


Problem 3:  
We need to evaluate:  
\[
\frac{2}{25}
\]
Convert to decimals:  
\[
\frac{2}{25} = 2 \div 25
\]
Since \(25 \cdot 0.08 = 2\):  
\[
2 \div 25 = 0.08
\]
So the correct choice is **B** (.08).

ANSWER 3: B


Problem 4:  
We have 4 people chosen: Lance, Sally, Joy, Fred. We need to choose **3 starters**.

Number of ways:
\[
\binom{4}{3} = 4
\]
List them conceptually: leaving out each one individually gives 4 choices.

Correct choice: **B**.

ANSWER 4: B


Problem 5:  
“If 20% of a number is 12, what is 30% of the same number?”

Let the number be \(N\).  
\[
0.20N = 12
\]
\[
N = \frac{12}{0.20} = 60
\]
Now compute \(30\%\) of 60:
\[
0.30 \cdot 60 = 18
\]
Correct choice: **B**.

ANSWER 5: B


Problem 6:  
Spinner starts pointing **west**.

Chenille moves it:
- **Clockwise** \(2\frac{1}{4}\) revolutions  
- then **Counterclockwise** \(3\frac{3}{4}\) revolutions

Net rotation (clockwise minus counterclockwise):
\[
2\frac{1}{4} - 3\frac{3}{4}
\]
Convert to improper fractions:
\[
2\frac{1}{4} = \frac{9}{4}, \quad 3\frac{3}{4} = \frac{15}{4}
\]
\[
\frac{9}{4} - \frac{15}{4} = -\frac{6}{4} = -\frac{3}{2}
\]
The negative means **net counterclockwise** by \(\frac{3}{2}\) revolutions.

\(\frac{3}{2} = 1 + \frac{1}{2}\), so this is:
- 1 full turn (brings it back where it started)
- plus \(\frac{1}{2}\) turn counterclockwise

Half a revolution from **west** points **east** (opposite direction).

Correct answer: **B (east)**.

ANSWER 6: B


Problem 7:  
Original mixture: 30 L with  
- 25% red (doesn’t matter directly)
- 30% yellow  
- 45% water

Yellow amount initially:
\[
0.30 \cdot 30 = 9 \text{ liters}
\]
Add 5 liters of yellow tint, so new yellow amount:
\[
9 + 5 = 14
\]
New total volume:
\[
30 + 5 = 35
\]
Percent yellow in new mixture:
\[
\frac{14}{35} = 0.4 = 40\%
\]
Correct choice: **C**.

ANSWER 7: C


Problem 8:  
Order fractions least to greatest:
\[
\frac{15}{11},\ \frac{19}{15},\ \frac{17}{13}
\]
A good way: compare values by cross-multiplication or decimal approximations.

Compute approximate decimals:
\[
\frac{15}{11} \approx 1.3636
\]
\[
\frac{19}{15} \approx 1.2667
\]
\[
\frac{17}{13} \approx 1.3077
\]
Least to greatest:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]
Check choice: that matches **E**:
- \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11}\)

ANSWER 8: E


Problem 9:  
Numbers: \(3,4,5,6,6,7,x\) (7 positive integers).  
Mean = median = unique mode.

**Step 1: Mode**  
The only number that repeats is **6**, occurring twice. So the **unique mode = 6**.

Thus median must be 6 as well.

**Step 2: Median = 6**  
For 7 numbers sorted, the median is the **4th** number.

Sorted list must have the 4th term equal to 6. We already have two 6’s and surrounding numbers 3,4,5,7.

Currently without \(x\), the ordered portion is:
\[
3,4,5,6,6,7,\ ?
\]
The 4th number here is 6 already. To keep the 4th number as 6, \(x\) must not be less than 6 (otherwise it would shift positions so 4th might change).

If \(x = 6\), then 6 occurs 3 times, still mode 6 (unique mode). Then list:
\[
3,4,5,6,6,6,7
\]
Median (4th) is 6—works.

If \(x=7\), list:
\[
3,4,5,6,6,7,7
\]
Mode would be 6? Actually 7 repeats too, so mode not unique.

If \(x>7\), list:
\[
3,4,5,6,6,7,x
\]
Median still 6, but mode would still be 6 (unique), since only 6 repeats. That would also work—wait, but then 6 is still unique mode, yes. However we also need **mean = 6**.

So use mean condition.

Sum of known numbers:
\[
3+4+5+6+6+7 = 31
\]
Total sum including \(x\): \(31 + x\)

Mean:
\[
\frac{31+x}{7} = 6
\]
\[
31 + x = 42
\]
\[
x = 11
\]

Check uniqueness of mode: 6 occurs twice, no other repeats (since x=11 is different). Unique mode is 6—consistent.

ANSWER 9: D


Problem 10:  
Joe walks half the distance, then runs the rest.  
He runs **3 times as fast** as he walks.

Given: time to walk half way = **6 minutes**.  
Let walking speed be \(v\). Then walking time for half distance \(=\) 6 minutes.

Let half the distance be \(d\). Then:
\[
d = v \cdot 6
\]
Running speed is \(3v\).  
Remaining distance is also \(d\), so running time:
\[
\text{time} = \frac{d}{3v} = \frac{v\cdot 6}{3v} = 2
\]
Total time:
\[
6 + 2 = 8
\]
Correct choice: **D**.

ANSWER 10: D


Problem 11:  
Compute:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]

Group each set:

First set:
\[
1+11+21+31+41
\]
These are all of the form \(10k+1\):  
\(1, 11, 21, 31, 41\) correspond to \(k=0,1,2,3,4\).

Sum:
\[
(1+11+21+31+41) = (1+11+21+31+41)
\]
Compute directly:
\[
(1+11)=12,\ (21+31)=52,\ 12+52=64,\ 64+41=105
\]

Second set:
\[
9+19+29+39+49
\]
Direct sum:
\[
(9+19)=28,\ (29+39)=68,\ 28+68=96,\ 96+49=145
\]

Total:
\[
105 + 145 = 250
\]

Correct choice: **E**.

ANSWER 11: E


Problem 12:  
Which number is largest?

Options:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare decimals:  
- .99 is the largest among all because it is greater than .9 (and greater than any number starting 0.90... but with fewer 9s afterward).

So .99 > .9099 > .909 > .9009 > .9.

Correct choice: **A**.

ANSWER 12: A

Problem 1:
In a single-elimination tournament, every game results in exactly one team being eliminated. To determine a single winner from 16 teams, 15 teams must be eliminated. Since each game eliminates one team, it takes exactly 15 games.
ANSWER 1: D

Problem 2:
- Jose: (10 - 1) = 9; 9 * 2 = 18; 18 + 2 = 20.
- Thuy: (10 * 2) = 20; 20 - 1 = 19; 19 + 2 = 21.
- Kareem: (10 - 1) = 9; 9 + 2 = 11; 11 * 2 = 22.
Kareem has the largest answer (22).
ANSWER 2: C

Problem 3:
To convert 2/25 to a decimal, multiply the numerator and denominator by 4 to get a denominator of 100: (2 * 4) / (25 * 4) = 8 / 100 = 0.08.
ANSWER 3: B

Problem 4:
We need to choose 3 starters from a group of 4 (Lance, Sally, Joy, Fred). This is a combination problem denoted as 4C3. Choosing 3 people to be on the team is the same as choosing 1 person to be left off the team. There are 4 ways to choose who is left off.
ANSWER 4: B

Problem 5:
Let $x$ be the number. $0.20x = 12$. Dividing by 0.20, $x = 12 / 0.2 = 60$. We want 30% of $x$: $0.30 * 60 = 18$.
ANSWER 5: B

Problem 6:
Start at 0 revolutions. Clockwise is positive, counterclockwise is negative.
$2 \frac{1}{4} - 3 \frac{3}{4} = 2.25 - 3.75 = -1.5$ revolutions.
-1.5 revolutions is the same as -0.5 revolutions (since -1 is a full circle). -0.5 revolutions (counterclockwise) from West points to East.
ANSWER 6: B

Problem 7:
Original yellow tint: 30% of 30 liters = 9 liters.
New mixture: 30 + 5 = 35 liters.
New yellow tint: 9 + 5 = 14 liters.
Percent: (14 / 35) * 100 = (2 / 5) * 100 = 40%.
ANSWER 7: C

Problem 8:
Convert to decimals or compare differences from 1:
15/11 = 1 + 4/11 ≈ 1.3636
19/15 = 1 + 4/15 ≈ 1.2666
17/13 = 1 + 4/13 ≈ 1.3076
Comparing the fractions 4/11, 4/13, and 4/15: the larger the denominator, the smaller the fraction. Thus, 4/15 < 4/13 < 4/11.
Order: 19/15 < 17/13 < 15/11.
ANSWER 8: E

Problem 9:
The set is {3, 4, 5, 6, 6, 7, x}. The mode is 6 (it appears twice). For the mean and median to be 6:
Median: With 7 numbers, the 4th number must be 6. Since the sorted list is 3, 4, 5, 6, 6, 7, x, if $x > 6$, the 4th number is 6.
Mean: (3+4+5+6+6+7+x) / 7 = 6 => 31 + x = 42 => x = 11.
Checking: {3, 4, 5, 6, 6, 7, 11}. Median is 6. Mean is 42/7 = 6. Mode is 6.
ANSWER 9: D

Problem 10:
Walking time = 6 minutes. Running speed = 3 * walking speed.
Time taken = Distance / Speed. Since speed is 3 times faster, time is 1/3 of the walking time.
Running time = 6 / 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
ANSWER 10: D

Problem 11:
Sum 1: (1+11+21+31+41) = 105.
Sum 2: (9+19+29+39+49) = 145.
105 + 145 = 250.
ANSWER 11: E

Problem 12:
Compare digit by digit:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
0.99 is the largest.
ANSWER 12: A

Problem 1:
This is a single-elimination tournament with 16 teams. Exactly one team is eliminated per game, and 15 teams must be eliminated to leave a single winner. Thus exactly 15 games are required.

ANSWER 1: D

Problem 2:
Jose evaluates \((10-1)\times2+2=20\).
Thuy evaluates \((10\times2)-1+2=21\).
Kareem evaluates \((10-1+2)\times2=22\).
Kareem obtains the strictly largest value.

ANSWER 2: C

Problem 3:
Division yields the decimal expansion \(2\div25=0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters out of 4 distinct players is the binomial coefficient \(\binom{4}{3}=4\).

ANSWER 4: B

Problem 5:
If 20% of a number equals 12, the number itself is \(12/0.2=60\). Then 30% of 60 is \(0.3\times60=18\).

ANSWER 5: B

Problem 6:
A net rotation of \(+2.25-3.75=-1.5\) revolutions is equivalent to 1.5 revolutions counterclockwise. From an initial westward orientation, 1.5 revolutions counterclockwise lands exactly on east.

ANSWER 6: B

Problem 7:
The original mixture contains \(0.30\times30=9\) liters of yellow tint. Adding 5 liters raises the yellow volume to 14 liters and the total volume to 35 liters, so the new percentage is \(14/35=40\%\).

ANSWER 7: C

Problem 8:
Decimal approximations are \(15/11\approx1.3636\), \(19/15\approx1.2667\), and \(17/13\approx1.3077\). Ordering these values produces \(19/15<17/13<15/11\).

ANSWER 8: E

Problem 9:
The unique mode must be 6, forcing the mean also to equal 6. The sum of the seven numbers is therefore 42. The six given numbers sum to 31, so \(x=11\). With this value the ordered list is 3,4,5,6,6,7,11 whose median is likewise 6 and whose unique mode remains 6.

ANSWER 9: D

Problem 10:
Let the total distance be \(2d\). Walking time for distance \(d\) is 6 min, so running time for the remaining distance \(d\) at triple speed is \(6/3=2\) min. The total time is therefore 8 min.

ANSWER 10: D

Problem 11:
Each pair sums to a multiple of 10: \(1+9=10\), \(11+19=30\), \(21+29=50\), \(31+39=70\), \(41+49=90\). Their total is 250. Equivalently each five-term arithmetic series sums to 105 and 145 respectively, again yielding 250.

ANSWER 11: E

Problem 12:
Direct comparison of the five decimals shows that 0.99 is strictly larger than each of the others.

ANSWER 12: A

Problem 1:
The problem asks us to determine the number of games that will be played to determine the winner of a basketball tournament with 16 teams, where the losing team of each game is eliminated.

## Step 1
To determine the number of games needed to find the winner, we first note that in a single-elimination tournament, each game eliminates one team.

## Step 2
Since there can only be one winner, all other teams must be eliminated. Therefore, the number of games required is equal to the number of teams minus one.

## Step 3
We have 16 teams competing in the tournament. So, the number of games needed to determine the winner is 16 - 1 = 15.

## Step 4
Let's examine the given choices to rule out the incorrect options. A. 4 is too low because it doesn't account for all the teams. B. 7 is also too low for the same reason. C. 8 is still too low. D. 15 matches our calculation. E. 16 is one more than our calculation.

The final answer is supported by our step-by-step reasoning.
ANSWER 1: D

Problem 2:
The problem asks us to determine who among Jose, Thuy, and Kareem gets the largest final answer after performing a series of operations starting with the number 10.

## Step 1
Let's calculate Jose's final answer: Jose subtracts 1 from 10 to get 9, doubles it to get 18, and then adds 2 to get 20.

## Step 2
Now, let's calculate Thuy's final answer: Thuy doubles 10 to get 20, subtracts 1 to get 19, and then adds 2 to get 21.

## Step 3
Next, let's calculate Kareem's final answer: Kareem subtracts 1 from 10 to get 9, adds 2 to get 11, and then doubles it to get 22.

## Step 4
Comparing the final answers: Jose got 20, Thuy got 21, and Kareem got 22. The largest final answer is 22, which belongs to Kareem.

## Step 5
Let's examine the given choices. A. Jose is incorrect because his answer is 20. B. Thuy is incorrect because her answer is 21. C. Kareem is correct because his answer is 22. D. Jose and Thuy is incorrect because Kareem's answer is larger. E. Thuy and Kareem is incorrect because Jose's answer is not the largest.

The final answer is supported by our step-by-step reasoning.
ANSWER 2: C

Problem 3:
The problem asks us to convert the fraction 2/25 into a decimal.

## Step 1
To convert 2/25 into a decimal, we divide the numerator by the denominator: 2 ÷ 25 = 0.08.

## Step 2
Let's examine the given choices to identify the correct decimal representation. A. .008 is too small. B. .08 matches our calculation. C. .8 is too large. D. 1.25 is much too large. E. 12.5 is also much too large.

The final answer is supported by our step-by-step reasoning.
ANSWER 3: B

Problem 4:
The problem asks us to determine the number of ways to choose 3 starters from a team of 4 members: Lance, Sally, Joy, and Fred.

## Step 1
This is a combination problem, because the order in which the starters are chosen does not matter.

## Step 2
The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial, the product of all positive integers up to that number.

## Step 3
Here, n = 4 (total team members) and k = 3 (number of starters to be chosen). So, C(4, 3) = 4! / [3!(4 - 3)!] = 4! / (3! * 1!) = (4 * 3 * 2 * 1) / ((3 * 2 * 1) * 1) = 4.

## Step 4
Let's examine the given choices. A. 2 is too low. B. 4 matches our calculation. C. 6 is too high. D. 8 is also too high. E. 10 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 4: B

Problem 5:
The problem asks us to find 30% of a number, given that 20% of the same number is 12.

## Step 1
First, we need to find the number. We know that 20% of the number is 12. So, 0.20 * number = 12.

## Step 2
To find the number, we divide both sides of the equation by 0.20: number = 12 / 0.20 = 60.

## Step 3
Now, we need to calculate 30% of 60: 0.30 * 60 = 18.

## Step 4
Let's examine the given choices. A. 15 is too low. B. 18 matches our calculation. C. 20 is too high. D. 24 is also too high. E. 30 is much too high.

The final answer is supported by our step-by-step reasoning.
ANSWER 5: B

Problem 6:
The problem asks us to determine the direction the spinner points after Chenille moves it clockwise 2 1/4 revolutions and then counterclockwise 3 3/4 revolutions, starting from pointing west.

## Step 1
First, let's understand the effect of the movements. A full revolution is 360 degrees or 4 quarters. So, 2 1/4 revolutions clockwise is equivalent to 9 quarters clockwise.

## Step 2
9 quarters clockwise is the same as 1 quarter clockwise (since 8 quarters make 2 full revolutions, which brings the spinner back to the original direction). So, after the first move, the spinner points north.

## Step 3
Then, Chenille moves it counterclockwise 3 3/4 revolutions, which is equivalent to 15 quarters counterclockwise.

## Step 4
15 quarters counterclockwise is the same as 3 quarters counterclockwise (since 12 quarters make 3 full revolutions). So, moving 3 quarters counterclockwise from north results in pointing west.

## Step 5
Let's examine the given choices. A. north is incorrect because after the second move, it's not north. B. east is incorrect. C. south is incorrect. D. west matches our calculation. E. northwest is incorrect.

The final answer is supported by our step-by-step reasoning.
ANSWER 6: D

Problem 7:
The problem asks us to find the percentage of yellow tint in a new mixture after adding 5 liters of yellow tint to an original 30-liter mixture that is 25% red tint, 30% yellow tint, and 45% water.

## Step 1
First, we calculate the amount of yellow tint in the original mixture: 30% of 30 liters = 0.30 * 30 = 9 liters.

## Step 2
After adding 5 liters of yellow tint, the total amount of yellow tint becomes 9 + 5 = 14 liters.

## Step 3
The total volume of the new mixture is 30 + 5 = 35 liters.

## Step 4
To find the percentage of yellow tint in the new mixture, we divide the amount of yellow tint by the total volume and multiply by 100: (14 / 35) * 100 = 40%

Problem 1:
In a single-elimination tournament, each game eliminates exactly one team. To eliminate all but the champion, the number of games needed is one fewer than the number of teams. With 16 teams, the number of games is \(16 - 1 = 15\).

ANSWER 1: D

Problem 2:
Jose: \(10 - 1 = 9\), \(9 \times 2 = 18\), \(18 + 2 = 20\).
Thuy: \(10 \times 2 = 20\), \(20 - 1 = 19\), \(19 + 2 = 21\).
Kareem: \(10 - 1 = 9\), \(9 + 2 = 11\), \(11 \times 2 = 22\).
Kareem’s final answer of 22 is the largest.

ANSWER 2: C

Problem 3:
Divide 2 by 25: \(2 \div 25 = 0.08\).

ANSWER 3: B

Problem 4:
The number of ways to choose 3 starters from 4 people is the combination \(\binom{4}{3} = 4\).

ANSWER 4: B

Problem 5:
Let the number be \(n\). \(20\%\) of \(n\) is \(0.2n = 12\), so \(n = 60\). Then \(30\%\) of \(60 = 0.3 \times 60 = 18\).

ANSWER 5: B

Problem 6:
Initial direction: west.
Clockwise \(2\frac{1}{4}\) revolutions: \(2\frac{1}{4} = 2.25\) revolutions clockwise.
Counterclockwise \(3\frac{3}{4}\) revolutions: \(3\frac{3}{4} = 3.75\) revolutions counterclockwise.
Net revolution = \(2.25 - 3.75 = -1.5\) revolutions, i.e., \(1.5\) revolutions counterclockwise.
A full revolution (1.0) returns to west. The remaining 0.5 revolution counterclockwise from west is a half-turn, which points to east.

ANSWER 6: B

Problem 7:
Original mixture: 30 liters, yellow tint = \(30\% \times 30 = 9\) liters.
Add 5 liters of yellow tint: new yellow = \(9 + 5 = 14\) liters; new total volume = \(30 + 5 = 35\) liters.
Percent yellow = \(\frac{14}{35} \times 100\% = 40\%\).

ANSWER 7: C

Problem 8:
Convert each fraction to a decimal for comparison:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
Order from least to greatest: \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

ANSWER 8: E

Problem 9:
The data set: 3, 4, 5, 6, 6, 7, \(x\). The mode is the most frequent number. With two 6’s, 6 is the mode unless \(x\) equals another number that appears twice, which would create a second mode, violating the “unique mode” condition. Therefore, \(x\) must not be 3, 4, 5, or 7. It could be 6, but if \(x=6\), the median is 6, while the mean is \((3+4+5+6+6+6+7)/7 = 37/7 \approx 5.29\), so mean ≠ median.  
For the mean to equal the median (which is 6, the middle of seven sorted numbers if \(x > 6\) or \(x=6\) only if placed correctly), we set the mean to 6. Sum without \(x\) = \(3+4+5+6+6+7 = 31\). Total needed for mean 6 = \(7 \times 6 = 42\). Thus \(x = 42 - 31 = 11\).  
Sorted: 3, 4, 5, 6, 6, 7, 11. Median = 6, mode = 6, mean = 6. Unique mode holds.

ANSWER 9: D

Problem 10:
Walking half the distance took 6 minutes, so walking speed = \(\frac{\text{half distance}}{6}\).
Running speed is 3 times walking speed. Time to run the second half = \(\frac{\text{half distance}}{3 \times (\text{half distance}/6)} = \frac{6}{3} = 2\) minutes.
Total time = \(6 + 2 = 8\) minutes.

ANSWER 10: D

Problem 11:
Pair the terms: \(1+49 = 50\), \(11+39 = 50\), \(21+29 = 50\), \(31+19 = 50\), \(41+9 = 50\). Five pairs of 50 give \(5 \times 50 = 250\).

ANSWER 11: E

Problem 12:
Compare by extending decimals to the same number of places:
\(0.99   = 0.9900\)
\(0.9099 = 0.9099\)
\(0.9    = 0.9000\)
\(0.909  = 0.9090\)
\(0.9009 = 0.9009\)
Clearly \(0.99 > 0.9099 > 0.909 > 0.9009 > 0.9\). The largest is \(0.99\).

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key Insight:** In a single-elimination tournament, every game eliminates exactly one team. To crown a single winner from 16 teams, we must eliminate 16 − 1 = 15 teams. Therefore, exactly 15 games are needed.

**Verification by rounds:**
- Round 1: 16 teams → 8 games → 8 teams left
- Round 2: 8 teams → 4 games → 4 teams left
- Round 3: 4 teams → 2 games → 2 teams left
- Round 4 (Final): 2 teams → 1 game → 1 winner

Total = 8 + 4 + 2 + 1 = 15 games.

Choices A (4), B (7), and C (8) are too small. Choice E (16) would mean every team is eliminated, leaving no winner.

ANSWER 1: D

---

Problem 2:
We track each person's calculation step by step, all starting from 10.

**Jose:** subtract 1 → 9; double → 18; add 2 → **20**
**Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
**Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

Comparing: 20 < 21 < 22. Kareem gets the largest final answer.

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

Choice A (.008) would be 8/1000. Choice C (.8) would be 8/10. Choices D and E are greater than 1, but 2/25 is clearly less than 1.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination problem — the order of selection doesn't matter:
$$\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4$$

Equivalently, choosing 3 people to start is the same as choosing 1 person to sit out. There are 4 people, so 4 ways to pick who sits out.

The 4 possible starting lineups are: {L,S,J}, {L,S,F}, {L,J,F}, {S,J,F}.

ANSWER 4: B

---

Problem 5:
We're told 20% of a number is 12. Let the number be $x$.

$$0.20x = 12 \implies x = \frac{12}{0.20} = 60$$

Now find 30% of 60:
$$0.30 \times 60 = 18$$

**Shortcut:** 30% is $\frac{30}{20} = 1.5$ times 20%, so the answer is $1.5 \times 12 = 18$.

Choice A (15) would be 25%. Choice D (24) would be 40%. Choice E (30) would be 50%.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing **West**.

**Move 1:** Clockwise $2\frac{1}{4}$ revolutions.
- The 2 full revolutions bring it back to West.
- The extra $\frac{1}{4}$ revolution clockwise from West: West → **North**.

**Move 2:** Counterclockwise $3\frac{3}{4}$ revolutions.
- The 3 full revolutions bring it back to North.
- The extra $\frac{3}{4}$ revolution counterclockwise from North:
  - $\frac{1}{4}$ CCW from North → West
  - $\frac{2}{4}$ CCW from North → South
  - $\frac{3}{4}$ CCW from North → **East**

The spinner points East.

ANSWER 6: B

---

Problem 7:
Original mixture: 30 liters total.
- Red tint: 25% of 30 = 7.5 liters
- Yellow tint: 30% of 30 = 9 liters
- Water: 45% of 30 = 13.5 liters

Add 5 liters of yellow tint:
- New yellow tint = 9 + 5 = 14 liters
- New total mixture = 30 + 5 = 35 liters

Percent yellow tint in new mixture:
$$\frac{14}{35} = \frac{2}{5} = 40\%$$

Choice B (35%) is a trap for those who just add 5% to 30%. Choice D (45%) is the original water percentage.

ANSWER 7: C

---

Problem 8:
We compare $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$.

Rewrite each as a mixed number:
$$\frac{15}{11} = 1 + \frac{4}{11}, \quad \frac{19}{15} = 1 + \frac{4}{15}, \quad \frac{17}{13} = 1 + \frac{4}{13}$$

Since all three have the same numerator (4) in their fractional parts, we compare denominators. For fractions with equal numerators, a **larger denominator** means a **smaller fraction**:
$$15 > 13 > 11 \implies \frac{4}{15} < \frac{4}{13} < \frac{4}{11}$$

Therefore:
$$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$$

ANSWER 8: E

---

Problem 9:
The data set is: 3, 4, 5, 6, 6, 7, and $x$. The mean, median, and unique mode are all equal.

**Mode:** Since 6 already appears twice and all others appear once, the unique mode must be 6 (if $x$ were 3, 4, 5, or 7, we'd have two modes, violating "unique mode"). So mode = 6.

**Mean = 6:**
$$\frac{3 + 4 + 5 + 6 + 6 + 7 + x}{7} = 6$$
$$\frac{31 + x}{7} = 6 \implies 31 + x = 42 \implies x = 11$$

**Check Median:** With $x = 11$, the sorted list is: 3, 4, 5, **6**, 6, 7, 11. The median (4th value) is 6. ✓

All three measures equal 6. Choice B (6) would make 6 appear three times, but the mean would be $(31+6)/7 = 37/7 \neq 6$.

ANSWER 9: D

---

Problem 10:
Joe walked half the distance in 6 minutes. He then ran the remaining half at 3 times his walking speed.

Let walking speed = $v$. The half-distance = $6v$.

Running speed = $3v$. Time to run the second half:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{6v}{3v} = 2 \text{ minutes}$$

Total time = 6 + 2 = **8 minutes**.

Choice A (7) would result from incorrectly subtracting. Choices B, C, E involve unnecessary decimal complications.

ANSWER 10: D

---

Problem 11:
Compute $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$.

Pair the terms from the first group with terms from the second group (first with last, etc.):
- $1 + 49 = 50$
- $11 + 39 = 50$
- $21 + 29 = 50$
- $31 + 19 = 50$
- $41 + 9 = 50$

Total = $50 \times 5 = 250$.

Choice C (200) is a common error from miscounting pairs. Choice D (249) comes from arithmetic mistakes.

ANSWER 11: E

---

Problem 12:
Compare the five decimals by writing them to four decimal places:

- A: 0.9900
- B: 0.9099
- C: 0.9000
- D: 0.9090
- E: 0.9009

Comparing digit by digit from the tenths place: all have 9 in the tenths place. In the hundredths place, A has **9** while all others have **0**. So A is immediately the largest.

$0.99 > 0.9099 > 0.909 > 0.9009 > 0.9$

ANSWER 12: A

 **Problem 1:**
We need to find the number of games played in a single-elimination tournament with 16 teams. In single elimination, every game eliminates exactly one team. To determine a winner, 15 of the 16 teams must be eliminated. Therefore, exactly 15 games are required.  
(Another way: with 16 teams there are 8 games in the first round, 4 in the second, 2 in the third, and 1 final game: \(8+4+2+1=15\).)

ANSWER 1: D

---

**Problem 2:**
We compute each person’s final answer step by step, starting from 10.

- **Jose:**  
  \(10 - 1 = 9\)  
  \(9 \times 2 = 18\)  
  \(18 + 2 = 20\)

- **Thuy:**  
  \(10 \times 2 = 20\)  
  \(20 - 1 = 19\)  
  \(19 + 2 = 21\)

- **Kareem:**  
  \(10 - 1 = 9\)  
  \(9 + 2 = 11\)  
  \(11 \times 2 = 22\)

Comparing the results: \(20 < 21 < 22\). Kareem’s answer is the largest.

ANSWER 2: C

---

**Problem 3:**
We convert the fraction \(\frac{2}{25}\) to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100:
\[
\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08
\]

ANSWER 3: B

---

**Problem 4:**
We must choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways to choose 3 from 4 is the combination \(\binom{4}{3}\):
\[
\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4
\]
Equivalently, choosing 3 starters is the same as choosing 1 person to sit out, and there are 4 choices for who sits out.

ANSWER 4: B

---

**Problem 5:**
Let the number be \(N\). We are given \(0.20N = 12\). Solving for \(N\):
\[
N = \frac{12}{0.20} = 60
\]
Now find 30% of 60:
\[
0.30 \times 60 = 18
\]
(Shortcut: 30% is \(1.5\) times 20%, so \(1.5 \times 12 = 18\).)

ANSWER 5: B

---

**Problem 6:**
The spinner starts pointing west.  
- Clockwise \(2\frac14\) revolutions: 2 full revolutions bring it back to west, then \(\frac14\) turn clockwise from west points it **north**.  
- Counterclockwise \(3\frac34\) revolutions: 3 full revolutions bring it back to north, then \(\frac34\) turn counterclockwise from north goes through west, south, and ends at **east**.

Net movement: \(3\frac34 - 2\frac14 = 1\frac12\) revolutions counterclockwise. One full revolution returns to west; an additional \(\frac12\) revolution counterclockwise from west points to **east**.

ANSWER 6: B

---

**Problem 7:**
Original mixture: 30 liters.  
- Yellow tint originally: \(30\% \times 30 = 9\) liters.  
After adding 5 liters of yellow tint:  
- New yellow tint: \(9 + 5 = 14\) liters.  
- New total mixture: \(30 + 5 = 35\) liters.  

Percent of yellow tint in the new mixture:
\[
\frac{14}{35} = \frac{2}{5} = 0.40 = 40\%
\]

ANSWER 7: C

---

**Problem 8:**
Compare \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\). Each is greater than 1. Write each as \(1\) plus a proper fraction:
\[
\frac{15}{11} = 1 + \frac{4}{11}, \qquad
\frac{19}{15} = 1 + \frac{4}{15}, \qquad
\frac{17}{13} = 1 + \frac{4}{13}
\]
Since the numerators are the same (4), the fraction with the largest denominator is smallest:
\[
\frac{4}{15} < \frac{4}{13} < \frac{4}{11}
\]
Adding 1 preserves the order:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}
\]

ANSWER 8: E

---

**Problem 9:**
The data set is \(3, 4, 5, 6, 6, 7, x\).  
- The current mode is 6 (appears twice). For 6 to remain the **unique** mode, \(x\) must not equal 3, 4, 5, or 7 (which would create a second mode).  
- We are told mean = median = mode = 6.  

Set the mean equal to 6:
\[
\frac{3+4+5+6+6+7+x}{7} = 6 \implies \frac{31+x}{7} = 6 \implies 31+x = 42 \implies x = 11
\]

Check: With \(x=11\), the sorted list is \(3, 4, 5, 6, 6, 7, 11\).  
- Median (4th term) = 6.  
- Mode = 6 (unique).  
- Mean = 6.  

All conditions are satisfied.

ANSWER 9: D

---

**Problem 10:**
Joe walks half the distance in 6 minutes. He runs the second half at 3 times his walking speed. Since distance is the same and speed is tripled, the time required is divided by 3:
\[
\text{Running time} = \frac{6}{3} = 2 \text{ minutes}
\]
Total time:
\[
6 + 2 = 8 \text{ minutes}
\]

ANSWER 10: D

---

**Problem 11:**
Add the two groups:
\[
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)
\]
Pair the terms to make convenient sums:
\[
(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 5 \times 50 = 250
\]

ANSWER 11: E

---

**Problem 12:**
Compare the decimals by writing them with the same number of places:
- A. \(0.9900\)
- B. \(0.9099\)
- C. \(0.9000\)
- D. \(0.9090\)
- E. \(0.9009\)

The tenths digit is 9 for all. Looking at the hundredths digit: A has 9, while B, D have 0 and C, E have 0. Thus \(0.9900\) is clearly the largest.

ANSWER 12: A

Problem 1:
We need to find how many games are played in a single-elimination tournament with 16 teams.

**Key insight:** Each game eliminates exactly one team. To determine a winner from 16 teams, we must eliminate 15 teams (everyone except the champion).

Therefore, exactly 15 games must be played.

*Verification by rounds:* Round 1: 8 games (16→8 teams), Round 2: 4 games (8→4), Round 3: 2 games (4→2), Round 4: 1 game (2→1). Total = 8+4+2+1 = 15.

The tempting wrong answer E (16) counts teams instead of eliminations.

ANSWER 1: D

---

Problem 2:
We need to compute each person's final answer starting from 10.

**Jose:** Start with 10 → subtract 1: 10−1 = 9 → double: 9×2 = 18 → add 2: 18+2 = **20**

**Thuy:** Start with 10 → double: 10×2 = 20 → subtract 1: 20−1 = 19 → add 2: 19+2 = **21**

**Kareem:** Start with 10 → subtract 1: 10−1 = 9 → add 2: 9+2 = 11 → double: 11×2 = **22**

Kareem gets the largest final answer (22).

ANSWER 2: C

---

Problem 3:
We need to convert 2/25 to a decimal.

Multiply numerator and denominator by 4 to get a denominator of 100:
$$\frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08$$

The tempting wrong answer A (0.008) would be 8/1000, and C (0.8) would be 8/10.

ANSWER 3: B

---

Problem 4:
We need to choose 3 starters from 4 people: Lance, Sally, Joy, and Fred.

This is a combination: C(4,3) = 4!/(3! × 1!) = 4

Listing them out: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — that's 4 ways.

The tempting wrong answer C (6) would be C(4,2), choosing who's *not* a starter, but that also equals 4... Actually C(4,2)=6 is wrong here; we're choosing 3 from 4.

ANSWER 4: B

---

Problem 5:
We know 20% of a number is 12. We need to find 30% of the same number.

**Step 1:** Find the number.
$$0.20 \times n = 12 \implies n = \frac{12}{0.20} = 60$$

**Step 2:** Find 30% of 60.
$$0.30 \times 60 = 18$$

*Shortcut:* Since 30% is 1.5 times 20%, the answer is 1.5 × 12 = 18.

ANSWER 5: B

---

Problem 6:
The spinner starts pointing west. We need to find the net rotation.

**Clockwise:** 2¼ revolutions
**Counterclockwise:** 3¾ revolutions

**Net movement:** 2¼ − 3¾ = −1½ revolutions (negative means counterclockwise)

So the spinner moves 1½ revolutions counterclockwise from west.

- 1 full revolution counterclockwise returns to **west**
- ½ revolution counterclockwise from west goes to **east**

(Counterclockwise from west: ¼ turn → south, ½ turn → east)

ANSWER 6: B

---

Problem 7:
We need to find the percent of yellow tint after adding 5 liters of yellow tint.

**Original mixture (30 liters):**
- Yellow tint: 30% of 30 = 9 liters

**New mixture:**
- Yellow tint: 9 + 5 = 14 liters
- Total volume: 30 + 5 = 35 liters

**Percent yellow tint:** (14/35) × 100 = 40%

The tempting wrong answer B (35) might come from simply adding 5% to 30%.

ANSWER 7: C

---

Problem 8:
We need to order 15/11, 19/15, and 17/13 from least to greatest.

Computing each as a decimal:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3077...

From least to greatest: **19/15 < 17/13 < 15/11**

*Alternative method:* Note each is 1 + (numerator − denominator)/denominator:
- 15/11 = 1 + 4/11
- 19/15 = 1 + 4/15
- 17/13 = 1 + 4/13

Since 4/15 < 4/13 < 4/11, the order is confirmed.

ANSWER 8: E

---

Problem 9:
The positive integers 3, 4, 5, 6, 6, 7, and x have equal mean, median, and unique mode.

**Mode:** Since 6 already appears twice and no other number does, the unique mode must be 6. So x ≠ 3, 4, 5, or 7 (which would create a second mode).

**Median:** With 7 numbers, the median is the 4th value when sorted. For the median to be 6, we need x ≥ 6 (so that 6 is the 4th value).

- If x = 6: sorted = 3, 4, 5, 6, 6, 6, 7 → median = 6 ✓
- If x > 7: sorted = 3, 4, 5, 6, 6, 7, x → median = 6 ✓

**Mean:** (3 + 4 + 5 + 6 + 6 + 7 + x)/7 = (31 + x)/7

Setting mean = 6: (31 + x)/7 = 6 → 31 + x = 42 → **x = 11**

**Verification:** Numbers: 3, 4, 5, 6, 6, 7, 11. Mean = 42/7 = 6, Median = 6, Mode = 6 ✓

ANSWER 9: D

---

Problem 10:
Joe walks halfway, then runs the rest. We need total time.

**Walking:** Half the distance takes 6 minutes.

**Running:** He runs 3 times as fast as he walks, so he covers the same distance in 1/3 the time.
- Time to run the second half = 6 ÷ 3 = 2 minutes

**Total time:** 6 + 2 = **8 minutes**

The tempting wrong answers come from setting up the problem incorrectly or mixing up distance and time relationships.

ANSWER 10: D

---

Problem 11:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

**First group:** 1 + 11 + 21 + 31 + 41 = 105

**Second group:** 9 + 19 + 29 + 39 + 49 = 145

**Total:** 105 + 145 = **250**

*Elegant method:* Pair terms: (1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250

ANSWER 11: E

---

Problem 12:
We need to find the largest number among the options. Writing each with the same number of decimal places:

- A. 0.99 = **0.9900**
- B. 0.9099 = 0.9099
- C. 0.9 = 0.9000
- D. 0.909 = 0.9090
- E. 0.9009 = 0.9009

Comparing digit by digit after the decimal point:
- First digit: all are 9
- Second digit: A has **9**, all others have 0

Since A has a 9 in the second decimal place while all others have 0, **0.99 is the largest**.

ANSWER 12: A

**Problem 1:**  
The tournament is single‑elimination. Starting with 16 teams, every game eliminates exactly one team. To reduce from 16 teams to a single champion we must eliminate 15 teams, and each elimination corresponds to one game.  

Number of games = 16 − 1 = **15**.  

ANSWER 1: D  

---

**Problem 2:**  
Start with 10 for each person.

- **Jose:**  
  Subtract 1 → 9; double → 18; add 2 → 20.  

- **Thuy:**  
  Double → 20; subtract 1 → 19; add 2 → 21.  

- **Kareem:**  
  Subtract 1 → 9; add 2 → 11; double → 22.  

The largest final value is 22, obtained by Kareem.  

ANSWER 2: C  

---

**Problem 3:**  

\[
\frac{2}{25}=0.08
\]

(2 ÷ 25 = 0.08).  

ANSWER 3: B  

---

**Problem 4:**  
We need to choose 3 starters from the 4 available players:  

\[
\binom{4}{3}=4
\]

ways.  

ANSWER 4: B  

---

**Problem 5:**  
Let the number be \(N\).  
\(0.20N = 12 \Rightarrow N = \frac{12}{0.20}=60\).  
\(30\%\) of \(N\) is \(0.30 \times 60 = 18\).  

ANSWER 5: B  

---

**Problem 6:**  
Interpret the fractions as mixed numbers:  

- Clockwise \(2\frac14 = 2.25\) revolutions.  
- Counter‑clockwise \(3\frac34 = 3.75\) revolutions.  

Net rotation (counter‑clockwise positive) = \(3.75 - 2.25 = 1.5\) revolutions.  

A full revolution returns to the same direction, so we only care about the extra \(0.5\) revolution.  
\(0.5\) revolution = 180° counter‑clockwise.  

Starting from west, turning 180° counter‑clockwise points to **east**.  

ANSWER 6: B  

---

**Problem 7:**  
Original yellow = \(30\%\) of 30 L = \(0.30 \times 30 = 9\) L.  
Add 5 L yellow → new yellow = \(9+5 = 14\) L.  
New total volume = \(30+5 = 35\) L.  

Percent yellow = \(\frac{14}{35}\times100 = 40\%\).  

ANSWER 7: C  

---

**Problem 8:**  
Compute the three fractions:

\[
\frac{15}{11}=1.3636\ldots,\qquad
\frac{19}{15}=1.2666\ldots,\qquad
\frac{17}{13}=1.3077\ldots
\]

Ordered from least to greatest:  

\(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).

This matches option **E**.  

ANSWER 8: E  

---

**Problem 9:**  
Current list: \(3,4,5,6,6,7\). The unique mode is 6 (appears twice).  
For the mode to remain unique and equal to the mean and median, \(x\) must not create a tie.  
Let the mean = median = mode = \(m\).  

Sum of numbers = \(31 + x\).  
Mean = \(\frac{31 + x}{7}=m\).  
Median of the seven numbers is the 4th after sorting; with \(x>7\) the median stays 6, so \(m=6\).  

Thus \(\frac{31 + x}{7}=6 \Rightarrow 31 + x = 42 \Rightarrow x = 11\).

Check: sorted list becomes \(3,4,5,6,6,7,11\) → median = 6, mode = 6, mean = 42/7 = 6. All equal.  

ANSWER 9: D  

---

**Problem 10:**  
Let the walking speed be \(v\) and the half‑distance be \(d/2\).  

Walking time for half‑distance: \(\frac{d/2}{v}=6\) min ⇒ \(\frac{d}{v}=12\) min.  

Running speed = \(3v\).  
Running time for the second half: \(\frac{d/2}{3v}= \frac{d}{6v}= \frac{1}{6}\times\frac{d}{v}= \frac{12}{6}=2\) min.  

Total time = \(6 + 2 = 8\) minutes.  

ANSWER 10: D  

---

**Problem 11:**  

\[
\begin{aligned}
1+11+21+31+41 &= 105,\\
9+19+29+39+49 &= 145,\\
\text{Total} &= 105+145 = 250.
\end{aligned}
\]

ANSWER 11: E  

---

**Problem 12:**  
Write the numbers with four decimal places:

- .99 = 0.9900  
- .9099 = 0.9099  
- .9 = 0.9000  
- .909 = 0.9090  
- .9009 = 0.9009  

The largest is 0.9900, i.e., **.99**.  

ANSWER 12: A  

Problem 1:  
The tournament starts with 16 teams. Each game eliminates one team, and the losing team is eliminated. To determine the winner, all teams except one must be eliminated. Thus, 15 teams must be eliminated, requiring 15 games.  
- **Why not other choices?** A (4) is too few; B (7), C (8) are also insufficient; E (16) would imply one game per team without elimination, which is incorrect.  
ANSWER 1: D  

Problem 2:  
- **Jose**: Starts with 10 → subtracts 1 (9) → doubles (18) → adds 2 → **20**.  
- **Thuy**: Starts with 10 → doubles (20) → subtracts 1 (19) → adds 2 → **21**.  
- **Kareem**: Starts with 10 → subtracts 1 (9) → adds 2 (11) → doubles → **22**.  
Kareem has the largest final answer.  
- **Why not other choices?** A (Jose) and B (Thuy) are smaller; D and E pair incorrect winners.  
ANSWER 2: C  

Problem 3:  
\( \frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08 \).  
- **Why not other choices?** A (.008) is \( \frac{2}{250} \); C (.8) is \( \frac{4}{5} \); D and E are larger decimals.  
ANSWER 3: B  

Problem 4:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is \( \binom{4}{3} = 4 \).  
- **Why not other choices?** A (2) is too few; C (6) is \( \binom{4}{2} \), which is incorrect; D (8) and E (10) are overcounts.  
ANSWER 4: B  

Problem 5:  
Let the number be \( x \). If 20% of \( x \) is 12, then \( 0.2x = 12 \), so \( x = 60 \). 30% of 60 is \( 0.3 \times 60 = 18 \).  
- **Why not other choices?** A (15) is 25% of 60; C (20) is \( \frac{1}{3} \) of 60; D (24) is 40%; E (30) is 50%.  
ANSWER 5: B  

Problem 6:  
- A full revolution is 360°. Clockwise \( 2 \frac{1}{4} \) revolutions = \( 2.25 \times 360 = 810° \).  
- Counterclockwise \( 3 \frac{3}{4} \) revolutions = \( -3.75 \times 360 = -1350° \).  
- Net movement: \( 810° - 1350° = -540° \).  
- \( -540° \mod 360° = 180° \) (since \( -540 + 720 = 180° \)), which points west → south? Wait, initial direction is west.  
- West (270°) + 810° = 1080° → 1080° mod 360° = 0° (north). Then -1350° → 0° - 1350° = -1350° mod 360° = 90° (east)? Let's reframe:  
- Start at west (270°). Clockwise 2.25 rev: 270° + 810° = 1080° mod 360° = 0° (north). Then counterclockwise 3.75 rev: 0° - 1350° = -1350° mod 360° = 90° (east).  
- **Why not other choices?** A (north) is initial after clockwise; B (east) is final; C (south) is 180°; D (west) is start; E (northwest) is not a cardinal direction.  
ANSWER 6: B  

Problem 7:  
Original mixture: 30 liters, 30% yellow tint → 9 liters yellow. Add 5 liters yellow → total yellow = 14 liters. New mixture = 35 liters. Percent yellow = \( \frac{14}{35} \times 100 = 40\% \).  
- **Why not other choices?** A (25) is original yellow%; B (35) is before addition; D (45) and E (50) are overestimates.  
ANSWER 7: E (Note: Correction based on computation: 14/35=0.4 → 40%, so answer is E? Wait, 14/35=0.4 → 40%, which is E. But earlier I thought B, but B is 35, which is wrong. Original yellow is 9, add 5 → 14. Total mixture 30+5=35. 14/35=0.4 → 40%. So E. But in choices E is 50? Wait, no: choices are A.25 B.35 C.40 D.45 E.50. So 40% is C. I misread. Correction: 14/35=0.4 → 40%, which is C.  
**Correction**: 14/35=0.4 → 40%, so answer is C.  
ANSWER 7: C  

Problem 8:  
Compare \( \frac{15}{11} \approx 1.3636 \), \( \frac{19}{15} \approx 1.2667 \), \( \frac{17}{13} \approx 1.3077 \). Order: \( \frac{19}{15} < \frac{17}{13} < \frac{15}{11} \).  
- **Why not other choices?** A, B, C, D have incorrect orders.  
ANSWER 8: E  

Problem 9:  
The set is {3, 4, 5, 6, 6, 7, x}. Mean = median = mode = 6.  
- Mode is 6 (unique), so 6 must appear most.  
- Median is 6 (4th value in ordered list: 3,4,5,6,6,7,x → if x≥6, ordered is ...,6,6,x or ...,6,x,6? Must have 4th value 6.  
- If x=6, set is {3,4,5,6,6,6,7}: median=6, mode=6, mean=(3+4+5+6+6+6+7)/7=37/7≈5.28 ≠6.  
- If x=11: set {3,4,5,6,6,7,11}: mean=(3+4+5+6+6+7+11)/7=42/7=6, median=6, mode=6.  
- **Why not other choices?** A (5): mean=36/7≈5.14; B (6): mean≈5.28; C (7): mean=38/7≈5.43; D (11) works.  
ANSWER 9: D  

Problem 10:  
Joe walks half the distance in 6 minutes. Let walking speed be \( v \), so distance = \( 6v \). He runs at \( 3v \), so time to run half distance = \( \frac{6v}{3v} = 2 \) minutes. Total time = 6 + 2 = 8 minutes.  
- **Why not other choices?** A (7), B (7.3), C (7.7), E (8.3) are not exact.  
ANSWER 10: D  

Problem 11:  
First group: 1+11+21+31+41 = 105. Second group: 9+19+29+39+49 = 145. Total = 105 + 145 = 250.  
- **Why not other choices?** A (150), B (199), C (200), D (249) are incorrect sums.  
ANSWER 11: E  

Problem 12:  
Compare decimals: A (.99), B (.9099), C (.9), D (.909), E (.9009). Largest is .99.  
- **Why not other choices?** B, C, D, E are all smaller than .99.  
ANSWER 12: A

### Problem 1:
We need to find the total number of games to determine a winner in a 16-team single-elimination tournament, where every game eliminates exactly one losing team. To get 1 winner, we must eliminate 16 - 1 = 15 teams. Since each game eliminates 1 team, we need 15 total games. Wrong choices: A is the number of bracket rounds, not games; C is half the number of teams; E is the total number of teams, not games.
ANSWER 1: D

---

### Problem 2:
All three start with 10, we calculate each person's final result step-by-step:
- Jose: (10 - 1) * 2 + 2 = 9*2 +2 = 20
- Thuy: (10*2) -1 +2 = 20-1+2=21
- Kareem: (10-1 +2)*2 = 11*2=22
Kareem has the largest final answer. Wrong choices stem from misordering the operations for each person.
ANSWER 2: C

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### Problem3:
Convert 2/25 to a decimal. Multiply numerator and denominator by 4 to get a denominator of 100: (2*4)/(25*4) = 8/100 = 0.08. Wrong choices: A is 2/250, C is 20/25.
ANSWER 3: B

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### Problem4:
We choose 3 starters from 4 people, which is a combination problem (order of starters does not matter here). The number of ways to choose 3 from 4 is equivalent to choosing 1 person to leave off the starting lineup: there are 4 people, so 4 unique lineups, each omitting one person. Wrong choice C (6) is the number of permutations, not combinations.
ANSWER 4: B

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### Problem5:
If 20% of a number is 12, first find the full number: 0.2*N=12 → N=12/0.2=60. 30% of 60 is 0.3*60=18. Wrong choices: A is 25% of 60, D is 40% of 60.
ANSWER 5: B

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### Problem6:
The spinner starts pointing west. Net movement: 2.25 clockwise revolutions minus 3.75 clockwise (equivalent to 3.75 counterclockwise) = -1.5 total clockwise revolutions, or 1.5 counterclockwise revolutions. Full revolutions cancel out, so we only have a net 0.5 (180 degree) counterclockwise turn from west. A 180-degree turn from west points to east.
ANSWER 6: B

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### Problem7:
Original 30L mixture has 30% yellow tint: 0.3*30=9L of yellow. Add 5L of yellow, so new yellow is 14L, new total mixture is 35L. The new percentage of yellow is (14/35)*100=40%. Wrong choice: people often forget to add 5L to the total mixture volume, leading to an incorrect higher percentage.
ANSWER 7: C

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### Problem8:
Rewrite each fraction to compare easily: 15/11 = 1 + 4/11, 19/15=1+4/15, 17/13=1+4/13. For equal numerators, the larger the denominator, the smaller the fraction. So 4/15 < 4/13 < 4/11, so the full order from least to greatest is 19/15 < 17/13 <15/11.
ANSWER 8: E

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### Problem9:
The dataset is [3,4,5,6,6,7,x]. The unique mode is currently 6 (only value appearing twice, so it must stay the only unique mode). For the mean, median, and mode to all be 6: the sum of the dataset must be 7*6=42. The sum of the existing numbers is 3+4+5+6+6+7=31, so x=42-31=11. Sorted dataset [3,4,5,6,6,7,11] has median 6, unique mode 6, mean 6, which fits. Wrong choice B (x=6) gives a mean of ~5.28, not 6.
ANSWER 9: D

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### Problem10:
Joe walks half the distance in 6 minutes. His running speed is 3x his walking speed, so time to cover the same distance is 1/3 of walking time. Time to run the second half: 6/3=2 minutes. Total time: 6+2=8 minutes.
ANSWER 10: D

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### Problem11:
Pair terms to simplify addition: (1+49)+(11+39)+(21+29)+(31+19)+(41+9) = 50+50+50+50+50=250.
ANSWER 11: E

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### Problem12:
Line up all decimals to compare: A=0.9900, B=0.9099, C=0.9000, D=0.9090, E=0.9009. The largest value is 0.99, option A.
ANSWER 12: A

Problem 1:  
In a single-elimination tournament, each game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams (all except 1). Thus, 15 games are played.  
ANSWER 1: D  

Problem 2:  
Calculate each person's result starting from 10:  
- Jose: \(10 -1 =9\), double → \(9*2=18\), add 2 → \(18+2=20\)  
- Thuy: Double 10 → \(10*2=20\), subtract 1 → \(20-1=19\), add 2 → \(19+2=21\)  
- Kareem: \(10-1=9\), add 2 → \(9+2=11\), double → \(11*2=22\)  
Kareem has the largest result.  
ANSWER 2: C  

Problem 3:  
Convert \(\frac{2}{25}\) to decimal: \(2\div25=0.08\).  
ANSWER 3: B  

Problem 4:  
Choosing 3 starters from 4 people is a combination (order does not matter). The number of ways is \(\binom{4}{3}=4\) (or list: {Lance,Sally,Joy}, {Lance,Sally,Fred}, {Lance,Joy,Fred}, {Sally,Joy,Fred}).  
ANSWER 4: B  

Problem 5:  
Let the number be \(x\). \(20\%x=12\) → \(0.2x=12\) → \(x=60\). \(30\%\ of\ 60=0.3*60=18\).  
ANSWER 5: B  

Problem 6:  
Model directions with standard angles: 0°=east, 90°=north, 180°=west, 270°=south. Initial direction is west (180°).  
- Clockwise \(2\frac{1}{4}\) revolutions: \(2.25*(-360°)=-810°\), new angle: \(180-810=-630°\)  
- Counterclockwise \(3\frac{3}{4}\) revolutions: \(3.75*(360°)=+1350°\), total angle: \(-630+1350=720°\)  
- \(720°\mod360°=0°\), which is east.  
ANSWER 6: B  

Problem 7:  
Original yellow tint: \(30\%\ of\ 30=0.3*30=9\) liters. Add 5 liters, total yellow = \(9+5=14\) liters. New total mixture: \(30+5=35\) liters. Percent yellow: \(\frac{14}{35}*100\%=40\%\).  
ANSWER 7: C  

Problem 8:  
Compare fractions via cross-multiplication or decimals:  
- \(\frac{19}{15}\approx1.267\), \(\frac{17}{13}\approx1.308\), \(\frac{15}{11}\approx1.364\)  
Order from least to greatest: \(\frac{19}{15}<\frac{17}{13}<\frac{15}{11}\).  
ANSWER 8: E  

Problem 9:  
The unique mode is 6 (appears twice, others once). Thus mean and median equal 6.  
- Median of 7 numbers is the 4th term when sorted, so 4th term =6 → \(x\geq6\).  
- Mean = \(\frac{3+4+5+6+6+7+x}{7}=6\) → \(31+x=42\) → \(x=11\).  
Check: Sorted list [3,4,5,6,6,7,11], median=6, mode=6 (unique), mean=6.  
ANSWER 9: D  

Problem 10:  
Walking half the distance takes 6 minutes, so full walking time would be 12 minutes. Running speed is 3x walking speed, so time for the second half is \(\frac{6}{3}=2\) minutes. Total time: \(6+2=8\) minutes.  
ANSWER 10: D  

Problem 11:  
Sum first group: \(1+11+21+31+41=105\) (arithmetic series: \(\frac{5}{2}(1+41)=105\)).  
Sum second group: \(9+19+29+39+49=145\) (\(\frac{5}{2}(9+49)=145\)).  
Total: \(105+145=250\).  
ANSWER 11: E  

Problem 12:  
Compare decimals digit by digit:  
- A=0.99, B=0.9099, C=0.9, D=0.909, E=0.9009  
A has the highest second decimal digit (9 vs 0 for others), so it is the largest.  
ANSWER 12: A