🔬 Model Lab

I'll work through each problem systematically.

**Problem 1:**

**What is being asked:** How many socks must be removed to guarantee 5 socks of the same color?

**Setup:** There are 3 colors (red, white, blue). In the worst case scenario, the Martian pulls out socks as evenly distributed among colors as possible before getting 5 of one color.

**Working:** 
- Worst case: 4 red, 4 white, 4 blue = 12 socks (still no color with 5)
- The 13th sock must give us a 5th sock of some color

**Answer 1: D**

---

**Problem 2:**

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

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

**Working:**
- C(4,3) = 4 ways
- The 4 ways correspond to leaving out each person: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred}

**Answer 2: B**

---

**Problem 3:**

**What is being asked:** Which pair does NOT have a product equal to 36?

**Working:** Check each pair:
- A. {−4, −9}: (−4)(−9) = 36 ✓
- B. {−3, −12}: (−3)(−12) = 36 ✓
- C. {1/2, −72}: (1/2)(−72) = −36 ✗
- D. {1, 36}: (1)(36) = 36 ✓
- E. {3/2, 24}: (3/2)(24) = 36 ✓

**Answer 3: C**

---

**Problem 4:**

**What is being asked:** How high is the stool?

**Setup:** Convert all measurements to centimeters:
- Ceiling height: 240 cm
- Light bulb location: 240 − 10 = 230 cm from floor
- Alice's height: 150 cm
- Alice's reach above her head: 46 cm
- Alice's reach without stool: 150 + 46 = 196 cm

**Working:**
- With stool, Alice can just reach the bulb at 230 cm
- Stool height + 196 cm = 230 cm
- Stool height = 230 − 196 = 34 cm

**Answer 4: B**

---

**Problem 5:**

**What is being asked:** How many fish total are in the lake?

**Setup:** Use proportions. In the sample: 30 trout out of 180 fish. In the lake: 250 trout out of total fish.

**Working:**
- 30/180 = 250/x
- 1/6 = 250/x
- x = 250 × 6 = 1500 fish

**Answer 5: B**

---

**Problem 6:**

**What is being asked:** How much could Karl have saved with the 20% off sale?

**Setup:** 
- Original cost: 5 folders × $2.50 = $12.50
- Sale discount: 20% off

**Working:**
- 20% of $12.50 = 0.20 × $12.50 = $2.50

**Answer 6: C**

---

**Problem 7:**

**What is being asked:** How many minutes to paint 300 dimples at 2 seconds each?

**Working:**
- Total time: 300 × 2 = 600 seconds
- Convert to minutes: 600 ÷ 60 = 10 minutes

**Answer 7: D**

---

**Problem 8:**

**What is being asked:** How many combinations of $5 bills and $2 bills make $17?

**Setup:** Let x = number of $5 bills, y = number of $2 bills
- 5x + 2y = 17

**Working:**
- x = 0: 2y = 17 → y = 8.5 (not integer)
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 (not integer)
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x ≥ 4: 5x ≥ 20 > 17 (impossible)

Valid combinations: (1,$5 and 6,$2) and (3,$5 and 1,$2) = 2 combinations

**Answer 8: A**

---

**Problem 9:**

**What is being asked:** When did the contest end, starting at noon and lasting 1000 minutes?

**Working:**
- 1000 minutes ÷ 60 = 16 hours and 40 minutes
- Noon + 16 hours 40 minutes = 4:40 p.m. the same day
- But we need to continue: 4:40 p.m. + additional time consideration
- Actually: 1000 ÷ 60 = 16 remainder 40, so 16 hours 40 minutes
- Noon + 16 hours 40 minutes = 4:40 a.m. the next day

**Answer 9: D**

---

**Problem 10:**

**What is being asked:** Calculate (8 × 4 + 2) − (8 + 4 × 2)

**Working:**
- First parentheses: 8 × 4 + 2 = 32 + 2 = 34
- Second parentheses: 8 + 4 × 2 = 8 + 8 = 16
- Result: 34 − 16 = 18

**Answer 10: D**

---

**Problem 11:**

**What is being asked:** Convert 11 hours 5 minutes to minutes

**Working:**
- 11 hours = 11 × 60 = 660 minutes
- Total: 660 + 5 = 665 minutes

**Answer 11: C**

---

**Problem 12:**

**What is being asked:** In how many ways can the ghost enter by one window and leave by a different window?

**Working:**
- Choices for entry: 6 windows
- Choices for exit: 5 windows (must be different from entry)
- Total ways: 6 × 5 = 30

**Answer 12: D**

Problem 1:

To be certain of getting 5 socks of the same color, we use the pigeonhole principle.

There are 3 colors: red, white, and blue.  
To avoid getting 5 of any one color for as long as possible, Martian could take at most 4 of each color:

- 4 red
- 4 white
- 4 blue

That gives a total of \(4+4+4=12\) socks without yet having 5 of the same color.

The very next sock, the 13th, must make one color appear at least 5 times.

So the Martian must remove 13 socks.

ANSWER 1: D

Problem 2:

There are 4 players: Lance, Sally, Joy, and Fred.

We need to choose 3 starters from these 4 players. Since order does not matter, we count combinations:

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

Indeed, the 3-starter groups are:

- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred

So there are 4 ways.

ANSWER 2: B

Problem 3:

We check each product.

- A: \((-4)(-9)=36\)
- B: \((-3)(-12)=36\)
- C: \(\frac12 \cdot (-72)=-36\), not 36
- D: \(1\cdot 36=36\)
- E: \(\frac32 \cdot 24=36\)

Only choice C does **not** have product 36.

ANSWER 3: C

Problem 4:

First find how high the bulb is above the floor.

The ceiling is 2.4 meters above the floor, which is \(240\) centimeters.  
The bulb is 10 centimeters below the ceiling, so it is:

\[
240 - 10 = 230 \text{ cm}
\]

Now Alice’s height is 1.5 meters = 150 cm.  
She can reach 46 cm above her head, so her reach without the stool is:

\[
150 + 46 = 196 \text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230 - 196 = 34 \text{ cm}
\]

So the stool is 34 cm high.

ANSWER 4: B

Problem 5:

The sample has 30 trout out of 180 fish, so the fraction of trout is:

\[
\frac{30}{180}=\frac{1}{6}
\]

If 250 trout is \(\frac16\) of the total lake population, then the total number of fish is:

\[
250 \div \frac16 = 250 \cdot 6 = 1500
\]

So there are 1500 fish in the lake.

ANSWER 5: B

Problem 6:

Karl bought 5 folders at \$2.50 each, so the total cost was:

\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:

\[
0.20 \times 12.50 = 2.50
\]

So he could have saved \$2.50.

ANSWER 6: C

Problem 7:

He paints 300 dimples, and each dimple takes 2 seconds.

Total time in seconds:

\[
300 \times 2 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10
\]

So he needs 10 minutes.

ANSWER 7: D

Problem 8:

We want combinations of 5-dollar bills and 2-dollar bills that total \$17.

Let \(x\) be the number of \$5 bills and \(y\) the number of \$2 bills. Then:

\[
5x + 2y = 17
\]

Try values of \(x\):

- \(x=1\): \(17-5=12\), so \(y=6\)
- \(x=2\): \(17-10=7\), not even, so impossible
- \(x=3\): \(17-15=2\), so \(y=1\)

These are the only nonnegative integer solutions.

So there are 2 combinations.

ANSWER 8: A

Problem 9:

The contest ended 1000 minutes after noon.

Convert 1000 minutes to hours and minutes:

\[
1000 \div 60 = 16 \text{ hours with } 40 \text{ minutes left}
\]

So:

- Noon + 12 hours = midnight
- plus 4 more hours = 4:00 a.m.
- plus 40 minutes = 4:40 a.m.

So the contest ended at 4:40 a.m.

ANSWER 9: D

Problem 10:

Compute each part using order of operations:

\[
(8\times 4 + 2) - (8 + 4\times 2)
\]

First parentheses:

\[
8\times 4 + 2 = 32 + 2 = 34
\]

Second parentheses:

\[
8 + 4\times 2 = 8 + 8 = 16
\]

Now subtract:

\[
34 - 16 = 18
\]

So the value is 18.

ANSWER 10: D

Problem 11:

We convert 11 hours and 5 minutes into minutes.

\[
11 \times 60 = 660
\]

Then add 5 minutes:

\[
660 + 5 = 665
\]

So the match lasted 665 minutes.

ANSWER 11: C

Problem 12:

Georgie must enter by one window and leave by a different window.

There are 6 choices for the entering window.  
After choosing one, there are 5 remaining choices for the exit window.

So the total number of ways is:

\[
6 \times 5 = 30
\]

Therefore, there are 30 ways.

ANSWER 12: D

Problem 1:  
We need the minimum number of socks to remove to guarantee there are 5 socks of the same color, given 3 colors (red, white, blue) and at least 5 of each color.

Worst case: you could first pull 4 socks of each color without reaching 5 of any color.  
That would be:
- 4 red + 4 white + 4 blue = 12 socks pulled

After 12 socks, you have at most 4 of each color. The next sock (the 13th) must be one of the 3 colors, and by the pigeonhole principle it makes one color reach 5.

So the Martian must remove **13** socks.

Tempting wrong choices:
- 12 isn’t enough because it’s possible to have only 4 of each color.
- Any number less than 13 can’t force a fifth sock of some color.

ANSWER 1: D


Problem 2:  
We have 4 students: Lance, Sally, Joy, Fred. We must choose 3 starters.

Number of ways to choose 3 out of 4 is:
\[
\binom{4}{3} = 4
\]

(You can also think: leave out one student; there are 4 choices for who to leave out.)

ANSWER 2: B


Problem 3:  
We check which pair does **NOT** multiply to 36.

Choices:
A. \(\{-4,-9\}\): \((-4)(-9)=36\) ✅  
B. \(\{-3,-12\}\): \((-3)(-12)=36\) ✅  
C. \(\left\{\frac{1}{2},-72\right\}\): \(\frac{1}{2}\cdot(-72)=-36\) ❌ not 36  
D. \(\{1,36\}\): \(1\cdot 36=36\) ✅  
E. \(\left\{\frac{3}{2},24\right\}\): \(\frac{3}{2}\cdot 24=3\cdot 12=36\) ✅

So the one that does not have product 36 is choice C.

ANSWER 3: C


Problem 4:  
Let’s convert everything to the same reference (floor).

- Ceiling is 2.4 m above the floor = 240 cm  
- The bulb is 10 cm below the ceiling: height of bulb above floor:
\[
240 - 10 = 230 \text{ cm}
\]
- Alice is 1.5 m tall = 150 cm  
- She can reach 46 cm above the top of her head, so from her top to the bulb:
\[
150 + 46 = 196 \text{ cm (if she stands on the floor)}
\]

She is standing on a stool, so her total height is (stool height + 150). She can reach:
\[
(\text{stool height} + 150) + 46
\]
Set equal to bulb height 230:
\[
\text{stool height} + 196 = 230
\]
\[
\text{stool height} = 34
\]

ANSWER 4: B


Problem 5:  
We know:
- Lake has 250 trout.
- Total fish in lake = \(T\)
- Sample has 180 fish, and 30 are trout.
So sample trout-to-total ratio equals lake trout-to-total ratio:
\[
\frac{30}{180} = \frac{250}{T}
\]
Simplify \(\frac{30}{180}=\frac{1}{6}\):
\[
\frac{1}{6} = \frac{250}{T}
\]
\[
T = 250 \cdot 6 = 1500
\]

ANSWER 5: B


Problem 6:  
Cost per folder originally: $2.50  
He bought 5 folders:
\[
5 \cdot 2.50 = 12.50
\]
A 20%-off sale means paying 80% of the original price:
\[
0.80 \cdot 12.50 = 10.00
\]
Savings:
\[
12.50 - 10.00 = 2.50
\]

ANSWER 6: C


Problem 7:  
Paint 300 dimples, 2 seconds per dimple.
Total time in seconds:
\[
300 \cdot 2 = 600 \text{ seconds}
\]
Convert to minutes (60 seconds per minute):
\[
600 \div 60 = 10 \text{ minutes}
\]

ANSWER 7: D


Problem 8:  
We need numbers of $5 bills and $2 bills that total $17 (order doesn’t matter).

Let \(x\) = number of $5 bills, \(y\) = number of $2 bills.
\[
5x + 2y = 17
\]
Try \(x=0\) to \(x=3\) (since \(4\cdot 5=20>17\)):

- \(x=0\): \(2y=17\) not possible  
- \(x=1\): \(5 + 2y=17 \Rightarrow 2y=12 \Rightarrow y=6\)  
- \(x=2\): \(10 + 2y=17 \Rightarrow 2y=7\) not possible  
- \(x=3\): \(15 + 2y=17 \Rightarrow 2y=2 \Rightarrow y=1\)

So there are 2 combinations: \((x,y)=(1,6)\) and \((3,1)\).

ANSWER 8: A


Problem 9:  
The contest starts at noon. It ends 1000 minutes later.

Convert 1000 minutes to hours:
\[
1000 \div 60 = 16 \text{ hours remainder } 40 \text{ minutes}
\]
Noon + 16 hours = 4:00 a.m. (next day)  
Add 40 minutes:
\[
4:00 \text{ a.m.} + 40 \text{ min} = 4:40 \text{ a.m.}
\]

ANSWER 9: D


Problem 10:  
Compute:
\[
(8 \times 4 + 2) - (8 + 4 \times 2)
\]
First parentheses:
\[
8\times 4 = 32,\quad 32+2=34
\]
Second parentheses:
\[
4\times 2 = 8,\quad 8+8=16
\]
Now subtract:
\[
34 - 16 = 18
\]

ANSWER 10: D


Problem 11:  
11 hours and 5 minutes total.
Convert hours to minutes:
\[
11 \cdot 60 = 660
\]
Add 5 minutes:
\[
660 + 5 = 665
\]

ANSWER 11: C


Problem 12:  
There are 6 windows. Georgie enters through one window and must leave through a different window.

Choose entering window: 6 choices.  
Choose leaving window: 5 choices (cannot be the same).

Total:
\[
6 \cdot 5 = 30
\]

ANSWER 12: D

Problem 1:
To be certain of having 5 socks of the same color, we consider the "worst-case scenario" (Pigeonhole Principle). If we have 3 colors (red, white, blue), we could pick 4 red, 4 white, and 4 blue socks without having 5 of any one color. This totals $4 + 4 + 4 = 12$ socks. The very next sock (the 13th) must be either red, white, or blue, which will complete a set of 5 for that color.
ANSWER 1: D

Problem 2:
We need to choose 3 starters from a group of 4 people (Lance, Sally, Joy, Fred). This is a combination problem denoted as $\binom{4}{3}$. Choosing 3 people to be on the team is the same as choosing 1 person to be left out. There are 4 ways to choose who is left out (Lance, Sally, Joy, or Fred).
ANSWER 2: B

Problem 3:
We check the products:
A: $(-4) \times (-9) = 36$
B: $(-3) \times (-12) = 36$
C: $\frac{1}{2} \times (-72) = -36$
D: $1 \times 36 = 36$
E: $\frac{3}{2} \times 24 = 3 \times 12 = 36$
Choice C results in $-36$, not $36$.
ANSWER 3: C

Problem 4:
Ceiling height = 2.4 m = 240 cm. Light bulb height = 240 - 10 = 230 cm from the floor.
Alice's reach = 150 cm (height) + 46 cm (reach) = 196 cm.
Stool height = (Bulb height) - (Alice's reach) = 230 - 196 = 34 cm.
ANSWER 4: B

Problem 5:
Let $T$ be the total number of fish in the lake. The ratio of trout to total fish is $\frac{30}{180} = \frac{1}{6}$.
Set up the proportion: $\frac{250}{T} = \frac{1}{6}$.
$T = 250 \times 6 = 1500$.
ANSWER 5: B

Problem 6:
Total cost = 5 folders × $2.50 = $12.50.
Savings = 20% of $12.50 = 0.20 \times 12.50 = $2.50.
Alternatively, 20% off one folder is $0.20 \times 2.50 = $0.50. For 5 folders, $0.50 \times 5 = $2.50.
ANSWER 6: C

Problem 7:
Total time in seconds = 300 dimples × 2 seconds/dimple = 600 seconds.
Total time in minutes = 600 seconds / 60 seconds/minute = 10 minutes.
ANSWER 7: D

Problem 8:
Let $x$ be the number of $5 bills and $y$ be the number of $2 bills. $5x + 2y = 17$.
If $x=1$, $5+2y=17 \Rightarrow 2y=12 \Rightarrow y=6$. (1, 6)
If $x=3$, $15+2y=17 \Rightarrow 2y=2 \Rightarrow y=1$. (3, 1)
If $x=0$ or $x=2$, $2y$ would be 17 or 7, which are not divisible by 2.
There are 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours and 40 minutes (since $1000 \div 60 = 16$ remainder 40).
Noon + 12 hours = midnight.
Midnight + 4 hours and 40 minutes = 4:40 a.m.
ANSWER 9: D

Problem 10:
$(8 \times 4 + 2) - (8 + 4 \times 2) = (32 + 2) - (8 + 8) = 34 - 16 = 18$.
ANSWER 10: D

Problem 11:
11 hours = $11 \times 60 = 660$ minutes.
Total = 660 + 5 = 665 minutes.
ANSWER 11: C

Problem 12:
There are 6 choices for the entry window. Since the exit window must be different, there are 5 choices for the exit window.
Total ways = $6 \times 5 = 30$.
ANSWER 12: D

Problem 1:
This is a pigeonhole principle problem. To guarantee at least 5 socks of one color, first consider the worst-case scenario of taking as many socks as possible without having 5 of any single color. That means taking exactly 4 socks of each of the 3 colors, for a total of \(4 \times 3 = 12\) socks. The next sock (the 13th) must be a 5th sock of one of the colors. Thus 13 socks are required. (Fewer than 13 leaves open the possibility of only 4 of each color.)

ANSWER 1: D

Problem 2:
The number of ways to choose 3 starters out of the 4 available players is the combination \(\binom{4}{3} = 4\). Each choice simply leaves out one of the four players.

ANSWER 2: B

Problem 3:
Compute the product for each pair:
- A: \((-4) \times (-9) = 36\)
- B: \((-3) \times (-12) = 36\)
- C: \(\frac12 \times (-72) = -36\) (not 36)
- D: \(1 \times 36 = 36\)
- E: \(\frac32 \times 24 = 36\)

Only pair C fails to give a product of 36.

ANSWER 3: C

Problem 4:
Convert all lengths to centimeters. The ceiling is 240 cm above the floor, so the bulb is 230 cm above the floor. Alice's reach is \(150 + 46 = 196\) cm. The stool must therefore supply the remaining \(230 - 196 = 34\) cm.

ANSWER 4: B

Problem 5:
The sample shows that trout comprise \(\frac{30}{180} = \frac16\) of all fish. With 250 trout in the lake, the total number of fish is \(250 \div \frac16 = 1500\).

ANSWER 5: B

Problem 6:
Five folders cost \(5 \times 2.50 = 12.50\). A 20 % discount saves \(0.20 \times 12.50 = 2.50\).

ANSWER 6: C

Problem 7:
Painting 300 dimples at 2 seconds each requires \(300 \times 2 = 600\) seconds, or \(600 \div 60 = 10\) minutes.

ANSWER 7: D

Problem 8:
Solve \(5x + 2y = 17\) for non-negative integers \(x, y\):
- \(x = 1\), \(y = 6\) works;
- \(x = 3\), \(y = 1\) works.
No other non-negative integer solutions exist, so there are exactly two combinations.

ANSWER 8: A

Problem 9:
Divide 1000 minutes by 60: \(1000 = 16\) hours + 40 minutes. Adding 16 hours to noon yields 4:00 a.m.; adding 40 minutes yields 4:40 a.m.

ANSWER 9: D

Problem 10:
Left side: \(8 \times 4 + 2 = 34\). Right side: \(8 + 4 \times 2 = 16\). Difference: \(34 - 16 = 18\).

ANSWER 10: D

Problem 11:
Convert 11 hours 5 minutes to minutes: \(11 \times 60 + 5 = 665\).

ANSWER 11: C

Problem 12:
There are 6 choices for the entry window and then 5 remaining choices for the exit window, giving \(6 \times 5 = 30\) ordered pairs.

ANSWER 12: D

Problem 1:
The problem is asking us to determine the minimum number of socks the Martian must remove from the drawer to guarantee that there are 5 socks of the same color. 
To ensure that there are 5 socks of the same color, we must consider the worst-case scenario where the Martian draws as many socks as possible without getting 5 of the same color. 
The Martian can draw at most 4 socks of each color without having 5 socks of the same color. 
So, the Martian can draw 4 red, 4 white, and 4 blue socks, for a total of 4 + 4 + 4 = 12 socks. 
At this point, the Martian still doesn't have 5 socks of the same color. 
However, drawing one more sock will guarantee that there are 5 socks of the same color. 
Therefore, the Martian must draw at least 12 + 1 = 13 socks.
Let's examine the choices: 
A. 6 is too low because the Martian could draw 3 red, 2 white, and 1 blue, or other combinations that don't guarantee 5 socks of the same color. 
B. 9 is still too low because the Martian could draw 4 red, 3 white, and 2 blue, or other similar combinations. 
C. 12 is the maximum number of socks the Martian can draw without having 5 socks of the same color. 
D. 13 is the correct answer because it is one more than the maximum number of socks that can be drawn without having 5 socks of the same color. 
E. 15 is more than necessary.
ANSWER 1: D

Problem 2:
The problem is asking us to find the number of ways to choose 3 starters from a team of 4 players: Lance, Sally, Joy, and Fred. 
This is a combination problem, because the order in which the starters are chosen does not matter. 
The number of ways to choose 3 starters from 4 players can be calculated using the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of players and k is the number of starters. 
Here, n = 4 and k = 3, so C(4, 3) = 4! / (3!(4-3)!) = 4.
Let's list the possible combinations to verify: 
- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred
There are indeed 4 combinations.
Let's examine the choices: 
A. 2 is too low. 
B. 4 matches our calculation. 
C. 6 is too high. 
D. 8 is too high. 
E. 10 is too high.
ANSWER 2: B

Problem 3:
The problem is asking us to identify the pair of numbers that does NOT have a product equal to 36. 
Let's check each pair: 
A. {−4, −9}: (−4) × (−9) = 36
B. {−3, −12}: (−3) × (−12) = 36
C. { 1 2 , −72}: (1/2) × (−72) = −36
D. {1, 36}: 1 × 36 = 36
E. { 3 2 , 24}: (3/2) × 24 = 36
The pair that does NOT have a product equal to 36 is C.
Let's examine the choices: 
A. {−4, −9} has a product of 36. 
B. {−3, −12} has a product of 36. 
C. { 1 2 , −72} has a product of −36, not 36. 
D. {1, 36} has a product of 36. 
E. { 3 2 , 24} has a product of 36.
ANSWER 3: C

Problem 4:
The problem is asking us to find the height of the stool that Alice needs to stand on to reach the light bulb. 
First, we need to convert the height of the ceiling from meters to centimeters: 2.4 meters = 240 centimeters. 
The light bulb is 10 centimeters below the ceiling, so its height above the floor is 240 - 10 = 230 centimeters. 
Alice is 1.5 meters = 150 centimeters tall and can reach 46 centimeters above her head, so her total reach is 150 + 46 = 196 centimeters. 
To find the height of the stool, we subtract Alice's total reach from the height of the light bulb: 230 - 196 = 34 centimeters.
Let's examine the choices: 
A. 32 is too low because 230 - (196 + 32) = 2, which is not enough to reach the light bulb. 
B. 34 is the correct height. 
C. 36 is too high. 
D. 38 is too high. 
E. 40 is too high.
ANSWER 4: B

Problem 5:
The problem is asking us to find the total number of fish in the lake based on the ratio of trout in the sample to the total number of fish in the sample. 
The sample contains 30 trout out of 180 fish, so the ratio of trout to total fish in the sample is 30/180. 
We are given that this ratio is the same as the ratio of trout to total fish in the lake. 
Let's denote the total number of fish in the lake as x. 
The ratio of trout to total fish in the lake is 250/x. 
Setting the two ratios equal, we get 30/180 = 250/x. 
Cross-multiplying, we have 30x = 250 × 180. 
Simplifying, 30x = 45000. 
Dividing both sides by 30, x = 1500.
Let's examine the choices: 
A. 1250 is too low. 
B. 1500 matches our calculation. 
C. 1750 is too high. 
D. 1800 is too high. 
E. 2000 is too high.
ANSWER 5: B

Problem 6:
The problem is asking us to find out how much Karl could have saved by waiting a day to buy the folders at a 20%-off sale. 
The original price of 5 folders is 5 × $2.50 = $12.50. 
At a 20%-off sale, the discount is 20% of $12.50, which is 0.20 × $12.50 = $2.50. 
So, Karl could have saved $2.50.
Let's examine the choices: 
A. $1.00 is too low. 
B. $2.00 is too low. 
C. $2.50 matches our calculation. 
D. $2.75 is too high. 
E. $5.00 is too high.
ANSWER 6: C

Problem 7:
The problem is asking us to find the time it will take Casey to paint 300 dimples on a golf ball. 
Casey takes 2 seconds to paint one dimple, so to paint 300 dimples, he will take 300 × 2 = 600 seconds. 
To convert seconds to minutes, we divide by 60: 600 / 60 = 10 minutes.
Let's examine the choices: 
A. 4 minutes is too low. 
B. 6 minutes is too low. 
C. 8 minutes is too low. 
D. 10 minutes matches our calculation. 
E. 12 minutes is too high.
ANSWER 7: D

Problem 8:
The problem is asking us to find the number of different combinations of $5 bills and $2 bills that can be used to make a total of $17. 
Let's list the possible combinations: 
1. 3 $5 bills = $15, and $2 bill = $2, total = $17 (1 combination: 3 $5 bills and 1 $2 bill)
2. 1 $5 bill = $5, and 6 $2 bills = $12, total = $17 (1 combination: 1 $5 bill and 6 $2 bills)
We should also check if there are any other combinations: 
- Using 0 $5 bills, we need 8.5 $2 bills, which is not possible since we can't have a fraction of a bill.
- Using 2 $5 bills = $10, we need $7 more, which can be made with 3.5 $2 bills, again not possible.
- Using 4 $5 bills = $20, which is more than $17.

Problem 1:
We want the minimum number of socks to guarantee 5 of the same color. With 3 colors (red, white, blue), the worst case is picking 4 of each color: 4+4+4 = 12 socks. The 13th sock must create a group of 5 of one color. So 13 socks are needed.
ANSWER 1: D

Problem 2:
We choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways is the combination C(4,3) = 4.
ANSWER 2: B

Problem 3:
Compute each product:
A: (−4)×(−9) = 36
B: (−3)×(−12) = 36
C: (1/2)×(−72) = −36
D: 1×36 = 36
E: (3/2)×24 = 36
Only C does not equal 36.
ANSWER 3: C

Problem 4:
Ceiling = 2.4 m = 240 cm. Light bulb is 10 cm below ceiling, so its height = 230 cm.
Alice's height = 1.5 m = 150 cm. Reach above head = 46 cm, so reach without stool = 196 cm.
Stool height needed = 230 − 196 = 34 cm.
ANSWER 4: B

Problem 5:
Sample ratio: 30 trout / 180 fish = 1/6. Lake has 250 trout, which is 1/6 of total fish. Total fish = 250 × 6 = 1500.
ANSWER 5: B

Problem 6:
Cost without sale: 5 × $2.50 = $12.50.
With 20% off, each folder costs $2.50 × 0.8 = $2.00. Total = 5 × $2.00 = $10.00.
Savings = $12.50 − $10.00 = $2.50.
ANSWER 6: C

Problem 7:
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.
ANSWER 7: D

Problem 8:
Find non-negative integers x ($5 bills) and y ($2 bills) such that 5x + 2y = 17.
x=1 → 5 + 2y = 17 → y=6.
x=3 → 15 + 2y = 17 → y=1.
No other x works. So 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours 40 minutes (16×60 = 960, remainder 40).
Noon + 16 hours = 4:00 a.m. next day. Add 40 minutes → 4:40 a.m.
ANSWER 9: D

Problem 10:
(8 × 4 + 2) = 32 + 2 = 34.
(8 + 4 × 2) = 8 + 8 = 16.
34 − 16 = 18.
ANSWER 10: D

Problem 11:
11 hours × 60 = 660 minutes. Plus 5 minutes = 665 minutes.
ANSWER 11: C

Problem 12:
6 choices for entry window, then 5 choices for a different exit window. Total = 6 × 5 = 30.
ANSWER 12: D

Problem 1:
We need to find the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). Using the Pigeonhole Principle, we consider the worst-case scenario: the Martian pulls out as many socks as possible *without* getting 5 of any one color. That means getting exactly 4 of each color:
- 4 red + 4 white + 4 blue = 12 socks
After 12 socks, the Martian could have 4 of each color and still not have 5 matching. The very next sock (the 13th) must be red, white, or blue, pushing the count of that color to 5. So 13 socks guarantees 5 of the same color.
(The "five-legged" detail is flavor text explaining why the Martian needs 5 socks.)

ANSWER 1: D

Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). This is a combination problem since order doesn't matter:
C(4, 3) = 4! / (3! × 1!) = 4
We can also list them: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — exactly 4 ways.

ANSWER 2: B

Problem 3:
We check each product:
- A: (−4) × (−9) = 36 ✓
- B: (−3) × (−12) = 36 ✓
- C: (1/2) × (−72) = −36 ✗ (This equals −36, NOT 36)
- D: 1 × 36 = 36 ✓
- E: (3/2) × 24 = 36 ✓
Choice C gives −36, which does not equal 36.

ANSWER 3: C

Problem 4:
First, convert everything to centimeters:
- Ceiling height: 2.4 m = 240 cm
- Light bulb position: 240 − 10 = 230 cm above the floor
- Alice's height: 1.5 m = 150 cm
- Alice's reach above her head: 46 cm
- Alice's maximum reach (without stool): 150 + 46 = 196 cm above the floor

Let h = height of the stool. Standing on the stool, her reach is h + 196 cm. She can "just reach" the bulb:
h + 196 = 230
h = 230 − 196 = 34 cm

ANSWER 4: B

Problem 5:
The ratio of trout to total fish in the sample should equal the ratio in the lake:
- Sample ratio: 30 trout / 180 fish = 1/6
- Lake ratio: 250 trout / Total fish = 1/6

Setting up the proportion: 250 / Total = 1/6
Total = 250 × 6 = 1500 fish

ANSWER 5: B

Problem 6:
Karl bought 5 folders at $2.50 each:
Total cost = 5 × $2.50 = $12.50
A 20%-off sale means he would save 20% of the total:
Savings = 0.20 × $12.50 = $2.50

ANSWER 6: C

Problem 7:
Total time in seconds: 300 dimples × 2 seconds/dimple = 600 seconds
Convert to minutes: 600 seconds ÷ 60 seconds/minute = 10 minutes

ANSWER 7: D

Problem 8:
We need non-negative integer solutions to 5x + 2y = 17, where x = number of $5 bills and y = number of $2 bills.
- x = 0: 2y = 17 → y = 8.5 (not an integer) ✗
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 ✗
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x = 4: 20 > 17 (too large) ✗

Only 2 valid combinations: (1 five + 6 twos) and (3 fives + 1 two).

ANSWER 8: A

Problem 9:
Convert 1000 minutes to hours and minutes:
1000 ÷ 60 = 16 remainder 40, so 1000 minutes = 16 hours 40 minutes.
Starting at noon (12:00 PM):
- 12 hours later = midnight (12:00 AM)
- 4 more hours = 4:00 AM
- 40 more minutes = 4:40 AM

ANSWER 9: D

Problem 10:
Evaluate each part using order of operations (multiplication before addition):
- First part: (8 × 4 + 2) = (32 + 2) = 34
- Second part: (8 + 4 × 2) = (8 + 8) = 16
- Difference: 34 − 16 = 18

ANSWER 10: D

Problem 11:
Convert 11 hours and 5 minutes to minutes:
11 hours × 60 minutes/hour = 660 minutes
660 + 5 = 665 minutes

ANSWER 11: C

Problem 12:
Georgie enters through one window and leaves through a different one.
- Choices for entry window: 6
- Choices for exit window (must be different): 5
Total ways = 6 × 5 = 30

ANSWER 12: D

Problem 1:
The Martian needs to be **certain** of having 5 socks of the same color. To guarantee this, we must consider the worst-case scenario: the Martian picks as many socks as possible **without** yet having 5 of any single color.  
With 3 colors (red, white, blue), the worst case is picking 4 socks of each color:  
\(4 + 4 + 4 = 12\) socks.  
At this point the Martian has 4 of every color but still does not have 5 of any one color. The very next sock—the 13th—must be red, white, or blue, and will necessarily create a set of 5 of that color.  
Thus 13 socks are required.  
(Choice A is too small; with 6 socks one could have 2,2,2. Choice B allows 3,3,3. Choice C allows 4,4,4. Choice E is more than the minimum guaranteed number.)

ANSWER 1: D

---

Problem 2:
We need to choose a group of 3 starters from the 4 students: Lance, Sally, Joy, and Fred. Since the three starters form a team and order does not matter, this is a combination problem.  
\[
\binom{4}{3} = \frac{4!}{3!\,1!} = 4
\]  
Equivalently, we can think of choosing which **one** of the four students will **not** start; there are 4 such choices.  
(Choice A is too small. Choice C is \(\binom{4}{2}\). Choices D and E do not arise from counting 3-person teams from 4 people.)

ANSWER 2: B

---

Problem 3:
We check the product of each pair to see which is not 36.

- **A.** \((-4)(-9) = 36\)  
- **B.** \((-3)(-12) = 36\)  
- **C.** \(\left(\frac12\right)(-72) = -36\)  
- **D.** \((1)(36) = 36\)  
- **E.** \(\left(\frac32\right)(24) = 36\)

Only pair **C** gives \(-36\), not \(36\).

ANSWER 3: C

---

Problem 4:
First convert all measurements to centimeters.  
- Ceiling height: \(2.4\text{ m} = 240\text{ cm}\).  
- Light bulb is \(10\text{ cm}\) below the ceiling, so its height above the floor is \(240 - 10 = 230\text{ cm}\).  

Alice’s reach:  
- Height: \(1.5\text{ m} = 150\text{ cm}\).  
- She can reach \(46\text{ cm}\) above her head, so her standing reach is \(150 + 46 = 196\text{ cm}\).  

The stool must supply the remaining height:  
\[
230 - 196 = 34\text{ cm}.
\]

ANSWER 4: B

---

Problem 5:
Let \(T\) be the total number of fish in the lake. The problem states that the ratio of trout to total fish is the same in the sample and in the lake.  
\[
\frac{250}{T} = \frac{30}{180}
\]  
Simplify the right side: \(\frac{30}{180} = \frac16\).  
\[
\frac{250}{T} = \frac16 \quad\Longrightarrow\quad T = 250 \times 6 = 1500.
\]

ANSWER 5: B

---

Problem 6:
Karl bought 5 folders at \(\$2.50\) each.  
Total cost: \(5 \times \$2.50 = \$12.50\).  
A 20% discount would have saved him 20% of this total:  
\[
0.20 \times \$12.50 = \$2.50.
\]

ANSWER 6: C

---

Problem 7:
Casey paints 300 dimples at 2 seconds per dimple.  
Total time in seconds: \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(\frac{600}{60} = 10\) minutes.

ANSWER 7: D

---

Problem 8:
Let \(f\) be the number of \(\$5\) bills and \(t\) be the number of \(\$2\) bills. We need non-negative integer solutions to  
\[
5f + 2t = 17.
\]  
Test values of \(f\):  
- \(f=0\): \(2t=17\) → no integer solution.  
- \(f=1\): \(5 + 2t = 17 \Rightarrow t=6\).  **(Valid)**
- \(f=2\): \(10 + 2t = 17 \Rightarrow t=3.5\).  (Invalid)
- \(f=3\): \(15 + 2t = 17 \Rightarrow t=1\).  **(Valid)**
- \(f=4\): \(20 > 17\) → stop.

There are exactly **2** combinations: \((1\text{ five}, 6\text{ twos})\) and \((3\text{ fives}, 1\text{ two})\).

ANSWER 8: A

---

Problem 9:
The contest starts at noon (12:00 p.m.). We add 1000 minutes.  
Divide 1000 by 60:  
\[
1000 = 16 \times 60 + 40,
\]  
so 1000 minutes = 16 hours and 40 minutes.  

Noon + 12 hours = midnight.  
Midnight + 4 hours 40 minutes = 4:40 a.m.

ANSWER 9: D

---

Problem 10:
Evaluate using the correct order of operations (multiplication before addition).

First parentheses:  
\[
8 \times 4 + 2 = 32 + 2 = 34.
\]

Second parentheses:  
\[
8 + 4 \times 2 = 8 + 8 = 16.
\]

Subtract:  
\[
34 - 16 = 18.
\]

ANSWER 10: D

---

Problem 11:
Convert 11 hours to minutes and add the remaining 5 minutes.  
\[
11 \times 60 = 660\text{ minutes}.
\]  
\[
660 + 5 = 665\text{ minutes}.
\]

ANSWER 11: C

---

Problem 12:
Georgie enters through one of 6 windows and must leave through a **different** window.  
- 6 choices for the entry window.  
- For each entry choice, 5 remaining choices for the exit window.  

Total ways:  
\[
6 \times 5 = 30.
\]  
(Choice E, 36, would count entering and leaving through the same window, which is not allowed.)

ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). The worst-case scenario is to draw as many socks as possible without having 5 of any one color. That would be 4 red, 4 white, and 4 blue, for a total of 12 socks. The next sock drawn (the 13th) must be one of the three colors, giving that color 5 socks. Therefore, 13 socks are required. The five-legged Martian is irrelevant to the count.  
Answer: D

Problem 2:
We have 4 people (Lance, Sally, Joy, Fred) and need to choose 3 starters. Since the starters are not assigned to specific positions, order does not matter. The number of ways is the combination \( \binom{4}{3} = 4 \).  
Answer: B

Problem 3:
Compute each product:  
A. \((-4) \times (-9) = 36\)  
B. \((-3) \times (-12) = 36\)  
C. \(\frac{1}{2} \times (-72) = -36\)  
D. \(1 \times 36 = 36\)  
E. \(\frac{3}{2} \times 24 = 36\)  
Only pair C gives a product not equal to 36.  
Answer: C

Problem 4:
Convert all measurements to centimeters:  
Ceiling height = 2.4 m = 240 cm.  
Light bulb is 10 cm below ceiling → height = 240 – 10 = 230 cm.  
Alice’s height = 1.5 m = 150 cm.  
Her reach above her head = 46 cm, so her maximum reach without a stool = 150 + 46 = 196 cm.  
Let \(h\) be the stool height. Standing on the stool raises her reach to \(196 + h\). This must equal 230 cm:  
\(196 + h = 230\) → \(h = 34\) cm.  
Answer: B

Problem 5:
In the sample, 30 out of 180 fish are trout, so the ratio trout : total = \(30/180 = 1/6\).  
Assuming the same ratio in the lake, if the lake has 250 trout, then the total number of fish is \(250 \times 6 = 1500\).  
Answer: B

Problem 6:
Original cost for 5 folders: \(5 \times 2.50 = \$12.50\).  
A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
Thus, by waiting a day, Karl could have saved $2.50.  
Answer: C

Problem 7:
Time to paint one dimple = 2 seconds.  
Total time for 300 dimples = \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(600 \div 60 = 10\) minutes.  
Answer: D

Problem 8:
We need nonnegative integers \(a\) (number of $5 bills) and \(b\) (number of $2 bills) such that \(5a + 2b = 17\).  
Solve by testing values of \(b\):  
\(b = 1\) → \(5a = 15\) → \(a = 3\) → (3,1) works.  
\(b = 6\) → \(5a = 5\) → \(a = 1\) → (1,6) works.  
No other values give integer \(a\). Thus there are 2 combinations.  
Answer: A

Problem 9:
1000 minutes = \(1000 \div 60 = 16\) hours and \(40\) minutes (since \(60 \times 16 = 960\), remainder 40).  
Starting at noon (12:00 PM), add 12 hours → midnight (12:00 AM). Add the remaining 4 hours → 4:00 AM. Then add 40 minutes → 4:40 AM.  
Answer: D

Problem 10:
Evaluate using order of operations:  
First parentheses: \(8 \times 4 + 2 = 32 + 2 = 34\).  
Second parentheses: \(8 + 4 \times 2 = 8 + 8 = 16\).  
Subtract: \(34 - 16 = 18\).  
Answer: D

Problem 11:
11 hours = \(11 \times 60 = 660\) minutes.  
Add 5 minutes → \(660 + 5 = 665\) minutes.  
Answer: C

Problem 12:
Georgie must choose an entry window (6 choices) and a different exit window (5 choices). Since entering and leaving are distinct actions, the total number of ways is \(6 \times 5 = 30\).  
Answer: D

ANSWER 1: D
ANSWER 2: B
ANSWER 3: C
ANSWER 4: B
ANSWER 5: B
ANSWER 6: C
ANSWER 7: D
ANSWER 8: A
ANSWER 9: D
ANSWER 10: D
ANSWER 11: C
ANSWER 12: D

Problem 1:  
The Martian wants to be certain of having 5 socks of the same color. There are three colors: red, white, and blue. In the worst-case scenario, the Martian could pick 4 socks of each color before getting a fifth of any one color. That would be \(4 \times 3 = 12\) socks, and still no color has 5 socks. The next sock drawn (the 13th) must complete a set of 5 of some color, because there are only three colors. Therefore, the Martian must remove **13** socks to be certain of having 5 socks of the same color.  
ANSWER 1: D

Problem 2:  
Ms. Hamilton's class has 4 available players: Lance, Sally, Joy, and Fred. The number of ways to choose 3 starters from 4 players is given by the combination \(\binom{4}{3} = 4\). Thus, there are **4** possible sets of three starters.  
ANSWER 2: B

Problem 3:  
We check each pair's product:  
- A: \((-4) \times (-9) = 36\)  
- B: \((-3) \times (-12) = 36\)  
- C: \(\frac{1}{2} \times (-72) = -36\) (not 36)  
- D: \(1 \times 36 = 36\)  
- E: \(\frac{3}{2} \times 24 = 36\)  
Only pair **C** does not have a product of 36.  
ANSWER 3: C

Problem 4:  
The ceiling is 2.4 m above the floor, and the bulb is 10 cm (0.1 m) below the ceiling, so the bulb's height is \(2.4 - 0.1 = 2.3\) m. Alice's height is 1.5 m, and she can reach 46 cm (0.46 m) above her head, so her reach height from the floor is \(1.5 + 0.46 = 1.96\) m. When standing on a stool, her hands are at the stool height plus her reach height. Setting this equal to the bulb height:  
\[
\text{stool height} + 1.96 = 2.3 \Rightarrow \text{stool height} = 0.34 \text{ m} = 34 \text{ cm}.
\]  
ANSWER 4: B

Problem 5:  
In the sample, the ratio of trout to total fish is \(\frac{30}{180} = \frac{1}{6}\). Assuming the same ratio in the lake, let \(T\) be the total number of fish in the lake. Then \(\frac{250}{T} = \frac{1}{6}\), so \(T = 250 \times 6 = 1500\).  
ANSWER 5: B

Problem 6:  
Karl bought 5 folders at $2.50 each, so the original cost is \(5 \times 2.50 = \$12.50\). With a 20% discount, the cost becomes \(12.50 \times 0.80 = \$10.00\). The savings is \(12.50 - 10.00 = \$2.50\).  
ANSWER 6: C

Problem 7:  
Painting 300 dimples at 2 seconds each takes \(300 \times 2 = 600\) seconds. Converting to minutes: \(600 \div 60 = 10\) minutes.  
ANSWER 7: D

Problem 8:  
We need nonnegative integers \(x\) (number of $5 bills) and \(y\) (number of $2 bills) such that \(5x + 2y = 17\). Solving:  
- \(y = 1 \Rightarrow 5x = 15 \Rightarrow x = 3\)  
- \(y = 6 \Rightarrow 5x = 5 \Rightarrow x = 1\)  
- \(y = 11 \Rightarrow 5x = -5\) (invalid)  
Thus, there are **2** valid combinations: (3 fives, 1 two) and (1 five, 6 twos).  
ANSWER 8: A

Problem 9:  
1000 minutes equals \(16\) hours and \(40\) minutes (since \(16 \times 60 = 960\), remainder \(40\)). Starting at noon, adding \(16\) hours gives 4:00 AM the next day, and adding \(40\) minutes gives 4:40 AM.  
ANSWER 9: D

Problem 10:  
Compute inside parentheses:  
\(8 \times 4 + 2 = 32 + 2 = 34\)  
\(8 + 4 \times 2 = 8 + 8 = 16\)  
Then \(34 - 16 = 18\).  
ANSWER 10: D

Problem 11:  
11 hours is \(11 \times 60 = 660\) minutes. Adding 5 minutes gives \(660 + 5 = 665\) minutes.  
ANSWER 11: C

Problem 12:  
Georgie can enter through any of the 6 windows. For exiting, he must choose a different window, so there are 5 choices. The total number of ways is \(6 \times 5 = 30\).  
ANSWER 12: D

Problem 1:  
The Martian needs to ensure there are 5 socks of the same color. The worst-case scenario is pulling the maximum number of socks without 5 of any color. Since there are 3 colors, this would be 4 socks of each color (4 × 3 = 12 socks). The next sock (the 13th) must match one color, giving 5 socks of that color.  
**ANSWER 1: D**  

Problem 2:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is C(4,3) = 4.  
**ANSWER 2: B** (Note: C(4,3)=4, but rechecking: the team is chosen from 4 specific people, so combinations are 4. Answer B=4)  

Problem 3:  
Check each pair:  
A. (−4)(−9) = 36  
B. (−3)(−12) = 36  
C. (1/2)(−72) = -36 (≠36)  
D. (1)(36) = 36  
E. (3/2)(24) = 36  
The pair that does NOT equal 36 is C.  
**ANSWER 3: E** (Re-evaluated: The question asks which does NOT equal 36. Option E: (3/2)*24=36, so it is valid. Option C: (1/2)*(-72)=-36, so C is the correct choice for NOT equal. Correction: **ANSWER 3: C** [Original had a typo, but computation confirms C is the answer])  

*Correction: For Problem 3, Option C: (1/2, -72) likely means (1/2) × (-72) = -36 ≠ 36. Option E: (3/2) × 24 = 36. Thus, C is the answer.*  
**ANSWER 3: C**  

Problem 4:  
Ceiling height: 2.4 m = 240 cm. Light bulb location: 240 - 10 = 230 cm from floor. Alice’s reach: 150 cm + 46 cm = 196 cm. Stool height = 230 - 196 = 34 cm.  
**ANSWER 4: B**  

Problem 5:  
Ratio of trout in sample: 30/180 = 1/6. Total trout: 250. Total fish = 250 × 6 = 1500.  
**ANSWER 5: B**  

Problem 6:  
Original cost: 5 × $2.50 = $12.50. 20% off: $12.50 × 0.20 = $2.50 saved.  
**ANSWER 6: C** (Note: The question asks how much Karl could have saved, which is $2.50, not $2.00. Original had a mistake; correction: 20% of $12.50 is $2.50, so answer C)  

Problem 7:  
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.  
**ANSWER 7: D**  

Problem 8:  
Let $5 bills = x, $2 bills = y. 5x + 2y = 17. Possible solutions:  
- x=3, y=1 (15+2=17)  
- x=1, y=6 (5+12=17)  
Only 2 combinations.  
**ANSWER 8: A**  

Problem 9:  
1000 minutes = 16 hours 40 minutes. Starting at noon: noon + 16h40m = 4:40 a.m. next day.  
**ANSWER 9: D**  

Problem 10:  
(8×4 + 2) = 34, (8 + 4×2) = 16. 34 - 16 = 18.  
**ANSWER 10: D**  

Problem 11:  
11 hours = 660 minutes. 660 + 5 = 665 minutes.  
**ANSWER 11: C**  

Problem 12:  
Georgie can enter through 6 windows and exit through 5 others. Total ways: 6 × 5 = 30.  
**ANSWER 12: D**  

---  
**Final Answers**  
ANSWER 1: D  
ANSWER 2: B  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: B  
ANSWER 6: C  
ANSWER 7: D  
ANSWER 8: A  
ANSWER 9: D  
ANSWER 10: D  
ANSWER 11: C  
ANSWER 12: D

### Problem 1:
This is a classic pigeonhole principle problem. We need to find the minimum number of socks the Martian must pull to guarantee 5 socks of the same color, with 3 total colors. The worst-case scenario is that the Martian pulls as many socks as possible without getting 5 of any color: that is 4 socks of each color, for a total of 3*4=12 socks. The next sock pulled (the 13th) has to be one of the three colors, pushing that color to 5 socks. Tempting wrong answers: C (12 is only the number of socks that avoids 5 of the same color, not enough to guarantee the requirement), E (15 is the total if you took 5 of each color, not the minimum to guarantee).
ANSWER 1: D

---

### Problem 2:
We need to count the number of ways to choose 3 starters from 4 people, where order on the team does not matter. This is a combination problem: C(n,k) = n!/(k!(n-k)!). For n=4, k=3: C(4,3) = 4. We can also list the teams by who is left out: leaving out Lance, Sally, Joy, or Fred gives exactly 4 unique teams. Tempting wrong answer: C (6 is the count if order of selection mattered, which it does not for a team of starters).
ANSWER 2: B

---

### Problem 3:
We calculate the product of each pair to find which does not equal 36:
- A: (-4)*(-9)=36 ✔️
- B: (-3)*(-12)=36 ✔️
- C: (1/2)*(-72) = -36 ❌
- D: 1*36=36 ✔️
- E: (3/2)*24=36 ✔️
Only pair C has a product of -36, not 36.
ANSWER 3: C

---

### Problem 4:
First convert all units to centimeters to avoid conversion errors:
- Ceiling height: 2.4m = 240cm. Light bulb is 10cm below ceiling, so bulb height from floor = 240-10=230cm.
- Alice's height: 1.5m=150cm, her reach above her head is 46cm, so total reach without stool = 150+46=196cm.
- Stool height = 230-196=34cm.
Tempting wrong answers come from unit conversion mistakes, e.g., mixing up meters and centimeters.
ANSWER 4: B

---

### Problem 5:
This is a proportion problem. The ratio of trout to total fish is the same in the sample and the lake. In the sample, 30 trout out of 180 total fish = 1/6. Let T be total fish in the lake: 250/T = 1/6 → T=250*6=1500.
ANSWER 5: B

---

### Problem 6:
Karl bought 5 folders at $2.50 each, total original cost = 5*2.50=$12.50. He saves 20% of the total cost by waiting for the sale: 0.2*12.50=$2.50. This can also be calculated per folder: 20% of $2.50 is $0.50 saved per folder, 5*0.50=$2.50 total saved.
ANSWER 6: C

---

### Problem 7:
300 dimples * 2 seconds per dimple = 600 total seconds. Convert to minutes: 60 seconds per minute, so 600/60=10 minutes.
ANSWER 7: D

---

### Problem 8:
We find all non-negative integer pairs (x,y) where x=number of $5 bills, y=number of $2 bills, and 5x+2y=17. Only odd values of x work (since $5 is odd, $2 is even, we need an odd sum to get 17):
- x=1: 5+2y=17 → y=6 (valid)
- x=3: 15+2y=17 → y=1 (valid)
Even x values (0,2) leave an odd remainder that cannot be formed with $2 bills, so only 2 valid combinations.
ANSWER 8: A

---

### Problem 9:
Convert 1000 minutes to hours and minutes: 60 minutes per hour, 1000/60=16 hours and 40 minutes. Adding this to noon (12pm): 12pm +16h40m = 4:40am the next day.
ANSWER 9: D

---

### Problem 10:
Use order of operations (PEMDAS) to calculate:
First parenthesis: 8*4 +2 = 32+2=34
Second parenthesis: 8 +4*2=8+8=16
Subtract: 34-16=18. The tempting wrong answer C (10) comes from incorrectly adding before multiplying in the second parenthesis.
ANSWER 10: D

---

### Problem 11:
Convert 11 hours 5 minutes to total minutes: 11*60=660 minutes, plus 5 extra minutes = 665 minutes.
ANSWER 11: C

---

### Problem 12:
Georgie has 6 choices for an entry window, and 5 remaining choices for an exit window (must use a different window). Total ways = 6*5=30. Tempting wrong answer E (36) counts entering and leaving through the same window, which is not allowed; B (15) incorrectly treats entry/exit as unordered, which they are not.
ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color (red, white, blue). This is a pigeonhole principle problem: the worst-case scenario is pulling the maximum number of socks without getting 5 of any color. For each of the 3 colors, we can pull 4 socks (since 4 < 5), so 4×3=12 socks. The next sock pulled will necessarily be the 5th of one color. So total socks needed: 12 + 1 = 13.
ANSWER 1: D
Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). Since the problem does not mention distinct positions for starters, order does not matter, so we use combinations. The number of ways to choose 3 people from 4 is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n=4 \) and \( k=3 \):
\( \binom{4}{3} = \frac{4!}{3!1!} = 4 \). This is equivalent to counting how many people we leave out (4 choices, one for each person not selected).
ANSWER 2: B
Problem 3:
We check the product of each pair:
- A: \( (-4) \times (-9) = 36 \)
- B: \( (-3) \times (-12) = 36 \)
- C: \( \frac{1}{2} \times (-72) = -36 \neq 36 \)
- D: \( 1 \times 36 = 36 \)
- E: \( \frac{3}{2} \times 24 = 36 \)
Only pair C does not have a product of 36.
ANSWER 3: C
Problem 4:
Convert all units to centimeters:
- Ceiling height: \( 2.4 \, \text{m} = 240 \, \text{cm} \)
- Light bulb height from floor: \( 240 - 10 = 230 \, \text{cm} \)
- Alice's reach from floor (without stool): \( 1.5 \, \text{m} + 0.46 \, \text{m} = 150 + 46 = 196 \, \text{cm} \)
Let \( h \) be the stool height. The total reach with the stool equals the light bulb height:
\( h + 196 = 230 \)
\( h = 230 - 196 = 34 \, \text{cm} \)
ANSWER 4: B
Problem 5:
The ratio of trout to total fish in the sample equals the ratio in the lake. The sample has \( \frac{30}{180} = \frac{1}{6} \) trout. Let \( T \) be the total number of fish in the lake. We know there are 250 trout, so:
\( \frac{250}{T} = \frac{1}{6} \)
\( T = 250 \times 6 = 1500 \)
ANSWER 5: B
Problem 6:
Karl bought 5 folders at $2.50 each. The 20% discount applies to each folder, so the savings per folder is \( 2.50 \times 0.20 = 0.50 \). Total savings for 5 folders: \( 5 \times 0.50 = 2.50 \).
ANSWER 6: C
Problem 7:
Total time to paint 300 dimples: \( 300 \times 2 = 600 \) seconds. Convert to minutes: \( 600 \div 60 = 10 \) minutes.
ANSWER 7: D
Problem 8:
Let \( x \) = number of $5 bills, \( y \) = number of $2 bills. We need \( 5x + 2y = 17 \), where \( x,y \geq 0 \) integers. Test non-negative integer values of \( x \):
- \( x=0 \): \( 2y=17 \) → \( y=8.5 \) (not integer, invalid)
- \( x=1 \): \( 5 + 2y=17 \) → \( 2y=12 \) → \( y=6 \) (valid: 1 $5 + 6 $2 = $17)
- \( x=2 \): \( 10 + 2y=17 \) → \( 2y=7 \) → \( y=3.5 \) (invalid)
- \( x=3 \): \( 15 + 2y=17 \) → \( 2y=2 \) → \( y=1 \) (valid: 3 $5 + 1 $2 = $17)
- \( x=4 \): \( 20 > 17 \) (stop, no more valid \( x \))
Only 2 valid combinations.
ANSWER 8: A
Problem 9:
Convert 1000 minutes to hours and minutes: \( 1000 \div 60 = 16 \) hours with a remainder of 40 minutes (since \( 16 \times 60 = 960 \), \( 1000 - 960 = 40 \)). The contest starts at noon, so adding 16 hours 40 minutes: noon + 12 hours = midnight, plus 4 hours 40 minutes = 4:40 a.m.
ANSWER 9: D
Problem 10:
Use order of operations (multiplication before addition):
First parentheses: \( 8 \times 4 + 2 = 32 + 2 = 34 \)
Second parentheses: \( 8 + 4 \times 2 = 8 + 8 = 16 \)
Subtract: \( 34 - 16 = 18 \)
ANSWER 10: D
Problem 11:
Convert 11 hours 5 minutes to total minutes: \( 11 \times 60 + 5 = 660 + 5 = 665 \)
ANSWER 11: C
Problem 12:
Georgie chooses an entrance window first (6 options), then a different exit window (5 remaining options, since exit cannot equal entrance). Since entrance and exit are distinct actions, order matters, so total ways: \( 6 \times 5 = 30 \).
ANSWER 12: D

I'll work through each problem systematically.

**Problem 1:**

**What is being asked:** How many socks must be removed to guarantee 5 socks of the same color?

**Setup:** There are 3 colors (red, white, blue). In the worst case scenario, the Martian pulls out socks as evenly distributed among colors as possible before getting 5 of one color.

**Working:** 
- Worst case: 4 red, 4 white, 4 blue = 12 socks (still no color with 5)
- The 13th sock must give us a 5th sock of some color

**Answer 1: D**

---

**Problem 2:**

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

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

**Working:**
- C(4,3) = 4 ways
- The 4 ways correspond to leaving out each person: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred}

**Answer 2: B**

---

**Problem 3:**

**What is being asked:** Which pair does NOT have a product equal to 36?

**Working:** Check each pair:
- A. {−4, −9}: (−4)(−9) = 36 ✓
- B. {−3, −12}: (−3)(−12) = 36 ✓
- C. {1/2, −72}: (1/2)(−72) = −36 ✗
- D. {1, 36}: (1)(36) = 36 ✓
- E. {3/2, 24}: (3/2)(24) = 36 ✓

**Answer 3: C**

---

**Problem 4:**

**What is being asked:** How high is the stool?

**Setup:** Convert all measurements to centimeters:
- Ceiling height: 240 cm
- Light bulb location: 240 − 10 = 230 cm from floor
- Alice's height: 150 cm
- Alice's reach above her head: 46 cm
- Alice's reach without stool: 150 + 46 = 196 cm

**Working:**
- With stool, Alice can just reach the bulb at 230 cm
- Stool height + 196 cm = 230 cm
- Stool height = 230 − 196 = 34 cm

**Answer 4: B**

---

**Problem 5:**

**What is being asked:** How many fish total are in the lake?

**Setup:** Use proportions. In the sample: 30 trout out of 180 fish. In the lake: 250 trout out of total fish.

**Working:**
- 30/180 = 250/x
- 1/6 = 250/x
- x = 250 × 6 = 1500 fish

**Answer 5: B**

---

**Problem 6:**

**What is being asked:** How much could Karl have saved with the 20% off sale?

**Setup:** 
- Original cost: 5 folders × $2.50 = $12.50
- Sale discount: 20% off

**Working:**
- 20% of $12.50 = 0.20 × $12.50 = $2.50

**Answer 6: C**

---

**Problem 7:**

**What is being asked:** How many minutes to paint 300 dimples at 2 seconds each?

**Working:**
- Total time: 300 × 2 = 600 seconds
- Convert to minutes: 600 ÷ 60 = 10 minutes

**Answer 7: D**

---

**Problem 8:**

**What is being asked:** How many combinations of $5 bills and $2 bills make $17?

**Setup:** Let x = number of $5 bills, y = number of $2 bills
- 5x + 2y = 17

**Working:**
- x = 0: 2y = 17 → y = 8.5 (not integer)
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 (not integer)
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x ≥ 4: 5x ≥ 20 > 17 (impossible)

Valid combinations: (1,$5 and 6,$2) and (3,$5 and 1,$2) = 2 combinations

**Answer 8: A**

---

**Problem 9:**

**What is being asked:** When did the contest end, starting at noon and lasting 1000 minutes?

**Working:**
- 1000 minutes ÷ 60 = 16 hours and 40 minutes
- Noon + 16 hours 40 minutes = 4:40 p.m. the same day
- But we need to continue: 4:40 p.m. + additional time consideration
- Actually: 1000 ÷ 60 = 16 remainder 40, so 16 hours 40 minutes
- Noon + 16 hours 40 minutes = 4:40 a.m. the next day

**Answer 9: D**

---

**Problem 10:**

**What is being asked:** Calculate (8 × 4 + 2) − (8 + 4 × 2)

**Working:**
- First parentheses: 8 × 4 + 2 = 32 + 2 = 34
- Second parentheses: 8 + 4 × 2 = 8 + 8 = 16
- Result: 34 − 16 = 18

**Answer 10: D**

---

**Problem 11:**

**What is being asked:** Convert 11 hours 5 minutes to minutes

**Working:**
- 11 hours = 11 × 60 = 660 minutes
- Total: 660 + 5 = 665 minutes

**Answer 11: C**

---

**Problem 12:**

**What is being asked:** In how many ways can the ghost enter by one window and leave by a different window?

**Working:**
- Choices for entry: 6 windows
- Choices for exit: 5 windows (must be different from entry)
- Total ways: 6 × 5 = 30

**Answer 12: D**

Problem 1:

To be certain of getting 5 socks of the same color, we use the pigeonhole principle.

There are 3 colors: red, white, and blue.  
To avoid getting 5 of any one color for as long as possible, Martian could take at most 4 of each color:

- 4 red
- 4 white
- 4 blue

That gives a total of \(4+4+4=12\) socks without yet having 5 of the same color.

The very next sock, the 13th, must make one color appear at least 5 times.

So the Martian must remove 13 socks.

ANSWER 1: D

Problem 2:

There are 4 players: Lance, Sally, Joy, and Fred.

We need to choose 3 starters from these 4 players. Since order does not matter, we count combinations:

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

Indeed, the 3-starter groups are:

- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred

So there are 4 ways.

ANSWER 2: B

Problem 3:

We check each product.

- A: \((-4)(-9)=36\)
- B: \((-3)(-12)=36\)
- C: \(\frac12 \cdot (-72)=-36\), not 36
- D: \(1\cdot 36=36\)
- E: \(\frac32 \cdot 24=36\)

Only choice C does **not** have product 36.

ANSWER 3: C

Problem 4:

First find how high the bulb is above the floor.

The ceiling is 2.4 meters above the floor, which is \(240\) centimeters.  
The bulb is 10 centimeters below the ceiling, so it is:

\[
240 - 10 = 230 \text{ cm}
\]

Now Alice’s height is 1.5 meters = 150 cm.  
She can reach 46 cm above her head, so her reach without the stool is:

\[
150 + 46 = 196 \text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230 - 196 = 34 \text{ cm}
\]

So the stool is 34 cm high.

ANSWER 4: B

Problem 5:

The sample has 30 trout out of 180 fish, so the fraction of trout is:

\[
\frac{30}{180}=\frac{1}{6}
\]

If 250 trout is \(\frac16\) of the total lake population, then the total number of fish is:

\[
250 \div \frac16 = 250 \cdot 6 = 1500
\]

So there are 1500 fish in the lake.

ANSWER 5: B

Problem 6:

Karl bought 5 folders at \$2.50 each, so the total cost was:

\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:

\[
0.20 \times 12.50 = 2.50
\]

So he could have saved \$2.50.

ANSWER 6: C

Problem 7:

He paints 300 dimples, and each dimple takes 2 seconds.

Total time in seconds:

\[
300 \times 2 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10
\]

So he needs 10 minutes.

ANSWER 7: D

Problem 8:

We want combinations of 5-dollar bills and 2-dollar bills that total \$17.

Let \(x\) be the number of \$5 bills and \(y\) the number of \$2 bills. Then:

\[
5x + 2y = 17
\]

Try values of \(x\):

- \(x=1\): \(17-5=12\), so \(y=6\)
- \(x=2\): \(17-10=7\), not even, so impossible
- \(x=3\): \(17-15=2\), so \(y=1\)

These are the only nonnegative integer solutions.

So there are 2 combinations.

ANSWER 8: A

Problem 9:

The contest ended 1000 minutes after noon.

Convert 1000 minutes to hours and minutes:

\[
1000 \div 60 = 16 \text{ hours with } 40 \text{ minutes left}
\]

So:

- Noon + 12 hours = midnight
- plus 4 more hours = 4:00 a.m.
- plus 40 minutes = 4:40 a.m.

So the contest ended at 4:40 a.m.

ANSWER 9: D

Problem 10:

Compute each part using order of operations:

\[
(8\times 4 + 2) - (8 + 4\times 2)
\]

First parentheses:

\[
8\times 4 + 2 = 32 + 2 = 34
\]

Second parentheses:

\[
8 + 4\times 2 = 8 + 8 = 16
\]

Now subtract:

\[
34 - 16 = 18
\]

So the value is 18.

ANSWER 10: D

Problem 11:

We convert 11 hours and 5 minutes into minutes.

\[
11 \times 60 = 660
\]

Then add 5 minutes:

\[
660 + 5 = 665
\]

So the match lasted 665 minutes.

ANSWER 11: C

Problem 12:

Georgie must enter by one window and leave by a different window.

There are 6 choices for the entering window.  
After choosing one, there are 5 remaining choices for the exit window.

So the total number of ways is:

\[
6 \times 5 = 30
\]

Therefore, there are 30 ways.

ANSWER 12: D

Problem 1:  
We need the minimum number of socks to remove to guarantee there are 5 socks of the same color, given 3 colors (red, white, blue) and at least 5 of each color.

Worst case: you could first pull 4 socks of each color without reaching 5 of any color.  
That would be:
- 4 red + 4 white + 4 blue = 12 socks pulled

After 12 socks, you have at most 4 of each color. The next sock (the 13th) must be one of the 3 colors, and by the pigeonhole principle it makes one color reach 5.

So the Martian must remove **13** socks.

Tempting wrong choices:
- 12 isn’t enough because it’s possible to have only 4 of each color.
- Any number less than 13 can’t force a fifth sock of some color.

ANSWER 1: D


Problem 2:  
We have 4 students: Lance, Sally, Joy, Fred. We must choose 3 starters.

Number of ways to choose 3 out of 4 is:
\[
\binom{4}{3} = 4
\]

(You can also think: leave out one student; there are 4 choices for who to leave out.)

ANSWER 2: B


Problem 3:  
We check which pair does **NOT** multiply to 36.

Choices:
A. \(\{-4,-9\}\): \((-4)(-9)=36\) ✅  
B. \(\{-3,-12\}\): \((-3)(-12)=36\) ✅  
C. \(\left\{\frac{1}{2},-72\right\}\): \(\frac{1}{2}\cdot(-72)=-36\) ❌ not 36  
D. \(\{1,36\}\): \(1\cdot 36=36\) ✅  
E. \(\left\{\frac{3}{2},24\right\}\): \(\frac{3}{2}\cdot 24=3\cdot 12=36\) ✅

So the one that does not have product 36 is choice C.

ANSWER 3: C


Problem 4:  
Let’s convert everything to the same reference (floor).

- Ceiling is 2.4 m above the floor = 240 cm  
- The bulb is 10 cm below the ceiling: height of bulb above floor:
\[
240 - 10 = 230 \text{ cm}
\]
- Alice is 1.5 m tall = 150 cm  
- She can reach 46 cm above the top of her head, so from her top to the bulb:
\[
150 + 46 = 196 \text{ cm (if she stands on the floor)}
\]

She is standing on a stool, so her total height is (stool height + 150). She can reach:
\[
(\text{stool height} + 150) + 46
\]
Set equal to bulb height 230:
\[
\text{stool height} + 196 = 230
\]
\[
\text{stool height} = 34
\]

ANSWER 4: B


Problem 5:  
We know:
- Lake has 250 trout.
- Total fish in lake = \(T\)
- Sample has 180 fish, and 30 are trout.
So sample trout-to-total ratio equals lake trout-to-total ratio:
\[
\frac{30}{180} = \frac{250}{T}
\]
Simplify \(\frac{30}{180}=\frac{1}{6}\):
\[
\frac{1}{6} = \frac{250}{T}
\]
\[
T = 250 \cdot 6 = 1500
\]

ANSWER 5: B


Problem 6:  
Cost per folder originally: $2.50  
He bought 5 folders:
\[
5 \cdot 2.50 = 12.50
\]
A 20%-off sale means paying 80% of the original price:
\[
0.80 \cdot 12.50 = 10.00
\]
Savings:
\[
12.50 - 10.00 = 2.50
\]

ANSWER 6: C


Problem 7:  
Paint 300 dimples, 2 seconds per dimple.
Total time in seconds:
\[
300 \cdot 2 = 600 \text{ seconds}
\]
Convert to minutes (60 seconds per minute):
\[
600 \div 60 = 10 \text{ minutes}
\]

ANSWER 7: D


Problem 8:  
We need numbers of $5 bills and $2 bills that total $17 (order doesn’t matter).

Let \(x\) = number of $5 bills, \(y\) = number of $2 bills.
\[
5x + 2y = 17
\]
Try \(x=0\) to \(x=3\) (since \(4\cdot 5=20>17\)):

- \(x=0\): \(2y=17\) not possible  
- \(x=1\): \(5 + 2y=17 \Rightarrow 2y=12 \Rightarrow y=6\)  
- \(x=2\): \(10 + 2y=17 \Rightarrow 2y=7\) not possible  
- \(x=3\): \(15 + 2y=17 \Rightarrow 2y=2 \Rightarrow y=1\)

So there are 2 combinations: \((x,y)=(1,6)\) and \((3,1)\).

ANSWER 8: A


Problem 9:  
The contest starts at noon. It ends 1000 minutes later.

Convert 1000 minutes to hours:
\[
1000 \div 60 = 16 \text{ hours remainder } 40 \text{ minutes}
\]
Noon + 16 hours = 4:00 a.m. (next day)  
Add 40 minutes:
\[
4:00 \text{ a.m.} + 40 \text{ min} = 4:40 \text{ a.m.}
\]

ANSWER 9: D


Problem 10:  
Compute:
\[
(8 \times 4 + 2) - (8 + 4 \times 2)
\]
First parentheses:
\[
8\times 4 = 32,\quad 32+2=34
\]
Second parentheses:
\[
4\times 2 = 8,\quad 8+8=16
\]
Now subtract:
\[
34 - 16 = 18
\]

ANSWER 10: D


Problem 11:  
11 hours and 5 minutes total.
Convert hours to minutes:
\[
11 \cdot 60 = 660
\]
Add 5 minutes:
\[
660 + 5 = 665
\]

ANSWER 11: C


Problem 12:  
There are 6 windows. Georgie enters through one window and must leave through a different window.

Choose entering window: 6 choices.  
Choose leaving window: 5 choices (cannot be the same).

Total:
\[
6 \cdot 5 = 30
\]

ANSWER 12: D

Problem 1:
To be certain of having 5 socks of the same color, we consider the "worst-case scenario" (Pigeonhole Principle). If we have 3 colors (red, white, blue), we could pick 4 red, 4 white, and 4 blue socks without having 5 of any one color. This totals $4 + 4 + 4 = 12$ socks. The very next sock (the 13th) must be either red, white, or blue, which will complete a set of 5 for that color.
ANSWER 1: D

Problem 2:
We need to choose 3 starters from a group of 4 people (Lance, Sally, Joy, Fred). This is a combination problem denoted as $\binom{4}{3}$. Choosing 3 people to be on the team is the same as choosing 1 person to be left out. There are 4 ways to choose who is left out (Lance, Sally, Joy, or Fred).
ANSWER 2: B

Problem 3:
We check the products:
A: $(-4) \times (-9) = 36$
B: $(-3) \times (-12) = 36$
C: $\frac{1}{2} \times (-72) = -36$
D: $1 \times 36 = 36$
E: $\frac{3}{2} \times 24 = 3 \times 12 = 36$
Choice C results in $-36$, not $36$.
ANSWER 3: C

Problem 4:
Ceiling height = 2.4 m = 240 cm. Light bulb height = 240 - 10 = 230 cm from the floor.
Alice's reach = 150 cm (height) + 46 cm (reach) = 196 cm.
Stool height = (Bulb height) - (Alice's reach) = 230 - 196 = 34 cm.
ANSWER 4: B

Problem 5:
Let $T$ be the total number of fish in the lake. The ratio of trout to total fish is $\frac{30}{180} = \frac{1}{6}$.
Set up the proportion: $\frac{250}{T} = \frac{1}{6}$.
$T = 250 \times 6 = 1500$.
ANSWER 5: B

Problem 6:
Total cost = 5 folders × $2.50 = $12.50.
Savings = 20% of $12.50 = 0.20 \times 12.50 = $2.50.
Alternatively, 20% off one folder is $0.20 \times 2.50 = $0.50. For 5 folders, $0.50 \times 5 = $2.50.
ANSWER 6: C

Problem 7:
Total time in seconds = 300 dimples × 2 seconds/dimple = 600 seconds.
Total time in minutes = 600 seconds / 60 seconds/minute = 10 minutes.
ANSWER 7: D

Problem 8:
Let $x$ be the number of $5 bills and $y$ be the number of $2 bills. $5x + 2y = 17$.
If $x=1$, $5+2y=17 \Rightarrow 2y=12 \Rightarrow y=6$. (1, 6)
If $x=3$, $15+2y=17 \Rightarrow 2y=2 \Rightarrow y=1$. (3, 1)
If $x=0$ or $x=2$, $2y$ would be 17 or 7, which are not divisible by 2.
There are 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours and 40 minutes (since $1000 \div 60 = 16$ remainder 40).
Noon + 12 hours = midnight.
Midnight + 4 hours and 40 minutes = 4:40 a.m.
ANSWER 9: D

Problem 10:
$(8 \times 4 + 2) - (8 + 4 \times 2) = (32 + 2) - (8 + 8) = 34 - 16 = 18$.
ANSWER 10: D

Problem 11:
11 hours = $11 \times 60 = 660$ minutes.
Total = 660 + 5 = 665 minutes.
ANSWER 11: C

Problem 12:
There are 6 choices for the entry window. Since the exit window must be different, there are 5 choices for the exit window.
Total ways = $6 \times 5 = 30$.
ANSWER 12: D

Problem 1:
This is a pigeonhole principle problem. To guarantee at least 5 socks of one color, first consider the worst-case scenario of taking as many socks as possible without having 5 of any single color. That means taking exactly 4 socks of each of the 3 colors, for a total of \(4 \times 3 = 12\) socks. The next sock (the 13th) must be a 5th sock of one of the colors. Thus 13 socks are required. (Fewer than 13 leaves open the possibility of only 4 of each color.)

ANSWER 1: D

Problem 2:
The number of ways to choose 3 starters out of the 4 available players is the combination \(\binom{4}{3} = 4\). Each choice simply leaves out one of the four players.

ANSWER 2: B

Problem 3:
Compute the product for each pair:
- A: \((-4) \times (-9) = 36\)
- B: \((-3) \times (-12) = 36\)
- C: \(\frac12 \times (-72) = -36\) (not 36)
- D: \(1 \times 36 = 36\)
- E: \(\frac32 \times 24 = 36\)

Only pair C fails to give a product of 36.

ANSWER 3: C

Problem 4:
Convert all lengths to centimeters. The ceiling is 240 cm above the floor, so the bulb is 230 cm above the floor. Alice's reach is \(150 + 46 = 196\) cm. The stool must therefore supply the remaining \(230 - 196 = 34\) cm.

ANSWER 4: B

Problem 5:
The sample shows that trout comprise \(\frac{30}{180} = \frac16\) of all fish. With 250 trout in the lake, the total number of fish is \(250 \div \frac16 = 1500\).

ANSWER 5: B

Problem 6:
Five folders cost \(5 \times 2.50 = 12.50\). A 20 % discount saves \(0.20 \times 12.50 = 2.50\).

ANSWER 6: C

Problem 7:
Painting 300 dimples at 2 seconds each requires \(300 \times 2 = 600\) seconds, or \(600 \div 60 = 10\) minutes.

ANSWER 7: D

Problem 8:
Solve \(5x + 2y = 17\) for non-negative integers \(x, y\):
- \(x = 1\), \(y = 6\) works;
- \(x = 3\), \(y = 1\) works.
No other non-negative integer solutions exist, so there are exactly two combinations.

ANSWER 8: A

Problem 9:
Divide 1000 minutes by 60: \(1000 = 16\) hours + 40 minutes. Adding 16 hours to noon yields 4:00 a.m.; adding 40 minutes yields 4:40 a.m.

ANSWER 9: D

Problem 10:
Left side: \(8 \times 4 + 2 = 34\). Right side: \(8 + 4 \times 2 = 16\). Difference: \(34 - 16 = 18\).

ANSWER 10: D

Problem 11:
Convert 11 hours 5 minutes to minutes: \(11 \times 60 + 5 = 665\).

ANSWER 11: C

Problem 12:
There are 6 choices for the entry window and then 5 remaining choices for the exit window, giving \(6 \times 5 = 30\) ordered pairs.

ANSWER 12: D

Problem 1:
The problem is asking us to determine the minimum number of socks the Martian must remove from the drawer to guarantee that there are 5 socks of the same color. 
To ensure that there are 5 socks of the same color, we must consider the worst-case scenario where the Martian draws as many socks as possible without getting 5 of the same color. 
The Martian can draw at most 4 socks of each color without having 5 socks of the same color. 
So, the Martian can draw 4 red, 4 white, and 4 blue socks, for a total of 4 + 4 + 4 = 12 socks. 
At this point, the Martian still doesn't have 5 socks of the same color. 
However, drawing one more sock will guarantee that there are 5 socks of the same color. 
Therefore, the Martian must draw at least 12 + 1 = 13 socks.
Let's examine the choices: 
A. 6 is too low because the Martian could draw 3 red, 2 white, and 1 blue, or other combinations that don't guarantee 5 socks of the same color. 
B. 9 is still too low because the Martian could draw 4 red, 3 white, and 2 blue, or other similar combinations. 
C. 12 is the maximum number of socks the Martian can draw without having 5 socks of the same color. 
D. 13 is the correct answer because it is one more than the maximum number of socks that can be drawn without having 5 socks of the same color. 
E. 15 is more than necessary.
ANSWER 1: D

Problem 2:
The problem is asking us to find the number of ways to choose 3 starters from a team of 4 players: Lance, Sally, Joy, and Fred. 
This is a combination problem, because the order in which the starters are chosen does not matter. 
The number of ways to choose 3 starters from 4 players can be calculated using the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of players and k is the number of starters. 
Here, n = 4 and k = 3, so C(4, 3) = 4! / (3!(4-3)!) = 4.
Let's list the possible combinations to verify: 
- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred
There are indeed 4 combinations.
Let's examine the choices: 
A. 2 is too low. 
B. 4 matches our calculation. 
C. 6 is too high. 
D. 8 is too high. 
E. 10 is too high.
ANSWER 2: B

Problem 3:
The problem is asking us to identify the pair of numbers that does NOT have a product equal to 36. 
Let's check each pair: 
A. {−4, −9}: (−4) × (−9) = 36
B. {−3, −12}: (−3) × (−12) = 36
C. { 1 2 , −72}: (1/2) × (−72) = −36
D. {1, 36}: 1 × 36 = 36
E. { 3 2 , 24}: (3/2) × 24 = 36
The pair that does NOT have a product equal to 36 is C.
Let's examine the choices: 
A. {−4, −9} has a product of 36. 
B. {−3, −12} has a product of 36. 
C. { 1 2 , −72} has a product of −36, not 36. 
D. {1, 36} has a product of 36. 
E. { 3 2 , 24} has a product of 36.
ANSWER 3: C

Problem 4:
The problem is asking us to find the height of the stool that Alice needs to stand on to reach the light bulb. 
First, we need to convert the height of the ceiling from meters to centimeters: 2.4 meters = 240 centimeters. 
The light bulb is 10 centimeters below the ceiling, so its height above the floor is 240 - 10 = 230 centimeters. 
Alice is 1.5 meters = 150 centimeters tall and can reach 46 centimeters above her head, so her total reach is 150 + 46 = 196 centimeters. 
To find the height of the stool, we subtract Alice's total reach from the height of the light bulb: 230 - 196 = 34 centimeters.
Let's examine the choices: 
A. 32 is too low because 230 - (196 + 32) = 2, which is not enough to reach the light bulb. 
B. 34 is the correct height. 
C. 36 is too high. 
D. 38 is too high. 
E. 40 is too high.
ANSWER 4: B

Problem 5:
The problem is asking us to find the total number of fish in the lake based on the ratio of trout in the sample to the total number of fish in the sample. 
The sample contains 30 trout out of 180 fish, so the ratio of trout to total fish in the sample is 30/180. 
We are given that this ratio is the same as the ratio of trout to total fish in the lake. 
Let's denote the total number of fish in the lake as x. 
The ratio of trout to total fish in the lake is 250/x. 
Setting the two ratios equal, we get 30/180 = 250/x. 
Cross-multiplying, we have 30x = 250 × 180. 
Simplifying, 30x = 45000. 
Dividing both sides by 30, x = 1500.
Let's examine the choices: 
A. 1250 is too low. 
B. 1500 matches our calculation. 
C. 1750 is too high. 
D. 1800 is too high. 
E. 2000 is too high.
ANSWER 5: B

Problem 6:
The problem is asking us to find out how much Karl could have saved by waiting a day to buy the folders at a 20%-off sale. 
The original price of 5 folders is 5 × $2.50 = $12.50. 
At a 20%-off sale, the discount is 20% of $12.50, which is 0.20 × $12.50 = $2.50. 
So, Karl could have saved $2.50.
Let's examine the choices: 
A. $1.00 is too low. 
B. $2.00 is too low. 
C. $2.50 matches our calculation. 
D. $2.75 is too high. 
E. $5.00 is too high.
ANSWER 6: C

Problem 7:
The problem is asking us to find the time it will take Casey to paint 300 dimples on a golf ball. 
Casey takes 2 seconds to paint one dimple, so to paint 300 dimples, he will take 300 × 2 = 600 seconds. 
To convert seconds to minutes, we divide by 60: 600 / 60 = 10 minutes.
Let's examine the choices: 
A. 4 minutes is too low. 
B. 6 minutes is too low. 
C. 8 minutes is too low. 
D. 10 minutes matches our calculation. 
E. 12 minutes is too high.
ANSWER 7: D

Problem 8:
The problem is asking us to find the number of different combinations of $5 bills and $2 bills that can be used to make a total of $17. 
Let's list the possible combinations: 
1. 3 $5 bills = $15, and $2 bill = $2, total = $17 (1 combination: 3 $5 bills and 1 $2 bill)
2. 1 $5 bill = $5, and 6 $2 bills = $12, total = $17 (1 combination: 1 $5 bill and 6 $2 bills)
We should also check if there are any other combinations: 
- Using 0 $5 bills, we need 8.5 $2 bills, which is not possible since we can't have a fraction of a bill.
- Using 2 $5 bills = $10, we need $7 more, which can be made with 3.5 $2 bills, again not possible.
- Using 4 $5 bills = $20, which is more than $17.

Problem 1:
We want the minimum number of socks to guarantee 5 of the same color. With 3 colors (red, white, blue), the worst case is picking 4 of each color: 4+4+4 = 12 socks. The 13th sock must create a group of 5 of one color. So 13 socks are needed.
ANSWER 1: D

Problem 2:
We choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways is the combination C(4,3) = 4.
ANSWER 2: B

Problem 3:
Compute each product:
A: (−4)×(−9) = 36
B: (−3)×(−12) = 36
C: (1/2)×(−72) = −36
D: 1×36 = 36
E: (3/2)×24 = 36
Only C does not equal 36.
ANSWER 3: C

Problem 4:
Ceiling = 2.4 m = 240 cm. Light bulb is 10 cm below ceiling, so its height = 230 cm.
Alice's height = 1.5 m = 150 cm. Reach above head = 46 cm, so reach without stool = 196 cm.
Stool height needed = 230 − 196 = 34 cm.
ANSWER 4: B

Problem 5:
Sample ratio: 30 trout / 180 fish = 1/6. Lake has 250 trout, which is 1/6 of total fish. Total fish = 250 × 6 = 1500.
ANSWER 5: B

Problem 6:
Cost without sale: 5 × $2.50 = $12.50.
With 20% off, each folder costs $2.50 × 0.8 = $2.00. Total = 5 × $2.00 = $10.00.
Savings = $12.50 − $10.00 = $2.50.
ANSWER 6: C

Problem 7:
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.
ANSWER 7: D

Problem 8:
Find non-negative integers x ($5 bills) and y ($2 bills) such that 5x + 2y = 17.
x=1 → 5 + 2y = 17 → y=6.
x=3 → 15 + 2y = 17 → y=1.
No other x works. So 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours 40 minutes (16×60 = 960, remainder 40).
Noon + 16 hours = 4:00 a.m. next day. Add 40 minutes → 4:40 a.m.
ANSWER 9: D

Problem 10:
(8 × 4 + 2) = 32 + 2 = 34.
(8 + 4 × 2) = 8 + 8 = 16.
34 − 16 = 18.
ANSWER 10: D

Problem 11:
11 hours × 60 = 660 minutes. Plus 5 minutes = 665 minutes.
ANSWER 11: C

Problem 12:
6 choices for entry window, then 5 choices for a different exit window. Total = 6 × 5 = 30.
ANSWER 12: D

Problem 1:
We need to find the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). Using the Pigeonhole Principle, we consider the worst-case scenario: the Martian pulls out as many socks as possible *without* getting 5 of any one color. That means getting exactly 4 of each color:
- 4 red + 4 white + 4 blue = 12 socks
After 12 socks, the Martian could have 4 of each color and still not have 5 matching. The very next sock (the 13th) must be red, white, or blue, pushing the count of that color to 5. So 13 socks guarantees 5 of the same color.
(The "five-legged" detail is flavor text explaining why the Martian needs 5 socks.)

ANSWER 1: D

Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). This is a combination problem since order doesn't matter:
C(4, 3) = 4! / (3! × 1!) = 4
We can also list them: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — exactly 4 ways.

ANSWER 2: B

Problem 3:
We check each product:
- A: (−4) × (−9) = 36 ✓
- B: (−3) × (−12) = 36 ✓
- C: (1/2) × (−72) = −36 ✗ (This equals −36, NOT 36)
- D: 1 × 36 = 36 ✓
- E: (3/2) × 24 = 36 ✓
Choice C gives −36, which does not equal 36.

ANSWER 3: C

Problem 4:
First, convert everything to centimeters:
- Ceiling height: 2.4 m = 240 cm
- Light bulb position: 240 − 10 = 230 cm above the floor
- Alice's height: 1.5 m = 150 cm
- Alice's reach above her head: 46 cm
- Alice's maximum reach (without stool): 150 + 46 = 196 cm above the floor

Let h = height of the stool. Standing on the stool, her reach is h + 196 cm. She can "just reach" the bulb:
h + 196 = 230
h = 230 − 196 = 34 cm

ANSWER 4: B

Problem 5:
The ratio of trout to total fish in the sample should equal the ratio in the lake:
- Sample ratio: 30 trout / 180 fish = 1/6
- Lake ratio: 250 trout / Total fish = 1/6

Setting up the proportion: 250 / Total = 1/6
Total = 250 × 6 = 1500 fish

ANSWER 5: B

Problem 6:
Karl bought 5 folders at $2.50 each:
Total cost = 5 × $2.50 = $12.50
A 20%-off sale means he would save 20% of the total:
Savings = 0.20 × $12.50 = $2.50

ANSWER 6: C

Problem 7:
Total time in seconds: 300 dimples × 2 seconds/dimple = 600 seconds
Convert to minutes: 600 seconds ÷ 60 seconds/minute = 10 minutes

ANSWER 7: D

Problem 8:
We need non-negative integer solutions to 5x + 2y = 17, where x = number of $5 bills and y = number of $2 bills.
- x = 0: 2y = 17 → y = 8.5 (not an integer) ✗
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 ✗
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x = 4: 20 > 17 (too large) ✗

Only 2 valid combinations: (1 five + 6 twos) and (3 fives + 1 two).

ANSWER 8: A

Problem 9:
Convert 1000 minutes to hours and minutes:
1000 ÷ 60 = 16 remainder 40, so 1000 minutes = 16 hours 40 minutes.
Starting at noon (12:00 PM):
- 12 hours later = midnight (12:00 AM)
- 4 more hours = 4:00 AM
- 40 more minutes = 4:40 AM

ANSWER 9: D

Problem 10:
Evaluate each part using order of operations (multiplication before addition):
- First part: (8 × 4 + 2) = (32 + 2) = 34
- Second part: (8 + 4 × 2) = (8 + 8) = 16
- Difference: 34 − 16 = 18

ANSWER 10: D

Problem 11:
Convert 11 hours and 5 minutes to minutes:
11 hours × 60 minutes/hour = 660 minutes
660 + 5 = 665 minutes

ANSWER 11: C

Problem 12:
Georgie enters through one window and leaves through a different one.
- Choices for entry window: 6
- Choices for exit window (must be different): 5
Total ways = 6 × 5 = 30

ANSWER 12: D

Problem 1:
The Martian needs to be **certain** of having 5 socks of the same color. To guarantee this, we must consider the worst-case scenario: the Martian picks as many socks as possible **without** yet having 5 of any single color.  
With 3 colors (red, white, blue), the worst case is picking 4 socks of each color:  
\(4 + 4 + 4 = 12\) socks.  
At this point the Martian has 4 of every color but still does not have 5 of any one color. The very next sock—the 13th—must be red, white, or blue, and will necessarily create a set of 5 of that color.  
Thus 13 socks are required.  
(Choice A is too small; with 6 socks one could have 2,2,2. Choice B allows 3,3,3. Choice C allows 4,4,4. Choice E is more than the minimum guaranteed number.)

ANSWER 1: D

---

Problem 2:
We need to choose a group of 3 starters from the 4 students: Lance, Sally, Joy, and Fred. Since the three starters form a team and order does not matter, this is a combination problem.  
\[
\binom{4}{3} = \frac{4!}{3!\,1!} = 4
\]  
Equivalently, we can think of choosing which **one** of the four students will **not** start; there are 4 such choices.  
(Choice A is too small. Choice C is \(\binom{4}{2}\). Choices D and E do not arise from counting 3-person teams from 4 people.)

ANSWER 2: B

---

Problem 3:
We check the product of each pair to see which is not 36.

- **A.** \((-4)(-9) = 36\)  
- **B.** \((-3)(-12) = 36\)  
- **C.** \(\left(\frac12\right)(-72) = -36\)  
- **D.** \((1)(36) = 36\)  
- **E.** \(\left(\frac32\right)(24) = 36\)

Only pair **C** gives \(-36\), not \(36\).

ANSWER 3: C

---

Problem 4:
First convert all measurements to centimeters.  
- Ceiling height: \(2.4\text{ m} = 240\text{ cm}\).  
- Light bulb is \(10\text{ cm}\) below the ceiling, so its height above the floor is \(240 - 10 = 230\text{ cm}\).  

Alice’s reach:  
- Height: \(1.5\text{ m} = 150\text{ cm}\).  
- She can reach \(46\text{ cm}\) above her head, so her standing reach is \(150 + 46 = 196\text{ cm}\).  

The stool must supply the remaining height:  
\[
230 - 196 = 34\text{ cm}.
\]

ANSWER 4: B

---

Problem 5:
Let \(T\) be the total number of fish in the lake. The problem states that the ratio of trout to total fish is the same in the sample and in the lake.  
\[
\frac{250}{T} = \frac{30}{180}
\]  
Simplify the right side: \(\frac{30}{180} = \frac16\).  
\[
\frac{250}{T} = \frac16 \quad\Longrightarrow\quad T = 250 \times 6 = 1500.
\]

ANSWER 5: B

---

Problem 6:
Karl bought 5 folders at \(\$2.50\) each.  
Total cost: \(5 \times \$2.50 = \$12.50\).  
A 20% discount would have saved him 20% of this total:  
\[
0.20 \times \$12.50 = \$2.50.
\]

ANSWER 6: C

---

Problem 7:
Casey paints 300 dimples at 2 seconds per dimple.  
Total time in seconds: \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(\frac{600}{60} = 10\) minutes.

ANSWER 7: D

---

Problem 8:
Let \(f\) be the number of \(\$5\) bills and \(t\) be the number of \(\$2\) bills. We need non-negative integer solutions to  
\[
5f + 2t = 17.
\]  
Test values of \(f\):  
- \(f=0\): \(2t=17\) → no integer solution.  
- \(f=1\): \(5 + 2t = 17 \Rightarrow t=6\).  **(Valid)**
- \(f=2\): \(10 + 2t = 17 \Rightarrow t=3.5\).  (Invalid)
- \(f=3\): \(15 + 2t = 17 \Rightarrow t=1\).  **(Valid)**
- \(f=4\): \(20 > 17\) → stop.

There are exactly **2** combinations: \((1\text{ five}, 6\text{ twos})\) and \((3\text{ fives}, 1\text{ two})\).

ANSWER 8: A

---

Problem 9:
The contest starts at noon (12:00 p.m.). We add 1000 minutes.  
Divide 1000 by 60:  
\[
1000 = 16 \times 60 + 40,
\]  
so 1000 minutes = 16 hours and 40 minutes.  

Noon + 12 hours = midnight.  
Midnight + 4 hours 40 minutes = 4:40 a.m.

ANSWER 9: D

---

Problem 10:
Evaluate using the correct order of operations (multiplication before addition).

First parentheses:  
\[
8 \times 4 + 2 = 32 + 2 = 34.
\]

Second parentheses:  
\[
8 + 4 \times 2 = 8 + 8 = 16.
\]

Subtract:  
\[
34 - 16 = 18.
\]

ANSWER 10: D

---

Problem 11:
Convert 11 hours to minutes and add the remaining 5 minutes.  
\[
11 \times 60 = 660\text{ minutes}.
\]  
\[
660 + 5 = 665\text{ minutes}.
\]

ANSWER 11: C

---

Problem 12:
Georgie enters through one of 6 windows and must leave through a **different** window.  
- 6 choices for the entry window.  
- For each entry choice, 5 remaining choices for the exit window.  

Total ways:  
\[
6 \times 5 = 30.
\]  
(Choice E, 36, would count entering and leaving through the same window, which is not allowed.)

ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). The worst-case scenario is to draw as many socks as possible without having 5 of any one color. That would be 4 red, 4 white, and 4 blue, for a total of 12 socks. The next sock drawn (the 13th) must be one of the three colors, giving that color 5 socks. Therefore, 13 socks are required. The five-legged Martian is irrelevant to the count.  
Answer: D

Problem 2:
We have 4 people (Lance, Sally, Joy, Fred) and need to choose 3 starters. Since the starters are not assigned to specific positions, order does not matter. The number of ways is the combination \( \binom{4}{3} = 4 \).  
Answer: B

Problem 3:
Compute each product:  
A. \((-4) \times (-9) = 36\)  
B. \((-3) \times (-12) = 36\)  
C. \(\frac{1}{2} \times (-72) = -36\)  
D. \(1 \times 36 = 36\)  
E. \(\frac{3}{2} \times 24 = 36\)  
Only pair C gives a product not equal to 36.  
Answer: C

Problem 4:
Convert all measurements to centimeters:  
Ceiling height = 2.4 m = 240 cm.  
Light bulb is 10 cm below ceiling → height = 240 – 10 = 230 cm.  
Alice’s height = 1.5 m = 150 cm.  
Her reach above her head = 46 cm, so her maximum reach without a stool = 150 + 46 = 196 cm.  
Let \(h\) be the stool height. Standing on the stool raises her reach to \(196 + h\). This must equal 230 cm:  
\(196 + h = 230\) → \(h = 34\) cm.  
Answer: B

Problem 5:
In the sample, 30 out of 180 fish are trout, so the ratio trout : total = \(30/180 = 1/6\).  
Assuming the same ratio in the lake, if the lake has 250 trout, then the total number of fish is \(250 \times 6 = 1500\).  
Answer: B

Problem 6:
Original cost for 5 folders: \(5 \times 2.50 = \$12.50\).  
A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
Thus, by waiting a day, Karl could have saved $2.50.  
Answer: C

Problem 7:
Time to paint one dimple = 2 seconds.  
Total time for 300 dimples = \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(600 \div 60 = 10\) minutes.  
Answer: D

Problem 8:
We need nonnegative integers \(a\) (number of $5 bills) and \(b\) (number of $2 bills) such that \(5a + 2b = 17\).  
Solve by testing values of \(b\):  
\(b = 1\) → \(5a = 15\) → \(a = 3\) → (3,1) works.  
\(b = 6\) → \(5a = 5\) → \(a = 1\) → (1,6) works.  
No other values give integer \(a\). Thus there are 2 combinations.  
Answer: A

Problem 9:
1000 minutes = \(1000 \div 60 = 16\) hours and \(40\) minutes (since \(60 \times 16 = 960\), remainder 40).  
Starting at noon (12:00 PM), add 12 hours → midnight (12:00 AM). Add the remaining 4 hours → 4:00 AM. Then add 40 minutes → 4:40 AM.  
Answer: D

Problem 10:
Evaluate using order of operations:  
First parentheses: \(8 \times 4 + 2 = 32 + 2 = 34\).  
Second parentheses: \(8 + 4 \times 2 = 8 + 8 = 16\).  
Subtract: \(34 - 16 = 18\).  
Answer: D

Problem 11:
11 hours = \(11 \times 60 = 660\) minutes.  
Add 5 minutes → \(660 + 5 = 665\) minutes.  
Answer: C

Problem 12:
Georgie must choose an entry window (6 choices) and a different exit window (5 choices). Since entering and leaving are distinct actions, the total number of ways is \(6 \times 5 = 30\).  
Answer: D

ANSWER 1: D
ANSWER 2: B
ANSWER 3: C
ANSWER 4: B
ANSWER 5: B
ANSWER 6: C
ANSWER 7: D
ANSWER 8: A
ANSWER 9: D
ANSWER 10: D
ANSWER 11: C
ANSWER 12: D

Problem 1:  
The Martian wants to be certain of having 5 socks of the same color. There are three colors: red, white, and blue. In the worst-case scenario, the Martian could pick 4 socks of each color before getting a fifth of any one color. That would be \(4 \times 3 = 12\) socks, and still no color has 5 socks. The next sock drawn (the 13th) must complete a set of 5 of some color, because there are only three colors. Therefore, the Martian must remove **13** socks to be certain of having 5 socks of the same color.  
ANSWER 1: D

Problem 2:  
Ms. Hamilton's class has 4 available players: Lance, Sally, Joy, and Fred. The number of ways to choose 3 starters from 4 players is given by the combination \(\binom{4}{3} = 4\). Thus, there are **4** possible sets of three starters.  
ANSWER 2: B

Problem 3:  
We check each pair's product:  
- A: \((-4) \times (-9) = 36\)  
- B: \((-3) \times (-12) = 36\)  
- C: \(\frac{1}{2} \times (-72) = -36\) (not 36)  
- D: \(1 \times 36 = 36\)  
- E: \(\frac{3}{2} \times 24 = 36\)  
Only pair **C** does not have a product of 36.  
ANSWER 3: C

Problem 4:  
The ceiling is 2.4 m above the floor, and the bulb is 10 cm (0.1 m) below the ceiling, so the bulb's height is \(2.4 - 0.1 = 2.3\) m. Alice's height is 1.5 m, and she can reach 46 cm (0.46 m) above her head, so her reach height from the floor is \(1.5 + 0.46 = 1.96\) m. When standing on a stool, her hands are at the stool height plus her reach height. Setting this equal to the bulb height:  
\[
\text{stool height} + 1.96 = 2.3 \Rightarrow \text{stool height} = 0.34 \text{ m} = 34 \text{ cm}.
\]  
ANSWER 4: B

Problem 5:  
In the sample, the ratio of trout to total fish is \(\frac{30}{180} = \frac{1}{6}\). Assuming the same ratio in the lake, let \(T\) be the total number of fish in the lake. Then \(\frac{250}{T} = \frac{1}{6}\), so \(T = 250 \times 6 = 1500\).  
ANSWER 5: B

Problem 6:  
Karl bought 5 folders at $2.50 each, so the original cost is \(5 \times 2.50 = \$12.50\). With a 20% discount, the cost becomes \(12.50 \times 0.80 = \$10.00\). The savings is \(12.50 - 10.00 = \$2.50\).  
ANSWER 6: C

Problem 7:  
Painting 300 dimples at 2 seconds each takes \(300 \times 2 = 600\) seconds. Converting to minutes: \(600 \div 60 = 10\) minutes.  
ANSWER 7: D

Problem 8:  
We need nonnegative integers \(x\) (number of $5 bills) and \(y\) (number of $2 bills) such that \(5x + 2y = 17\). Solving:  
- \(y = 1 \Rightarrow 5x = 15 \Rightarrow x = 3\)  
- \(y = 6 \Rightarrow 5x = 5 \Rightarrow x = 1\)  
- \(y = 11 \Rightarrow 5x = -5\) (invalid)  
Thus, there are **2** valid combinations: (3 fives, 1 two) and (1 five, 6 twos).  
ANSWER 8: A

Problem 9:  
1000 minutes equals \(16\) hours and \(40\) minutes (since \(16 \times 60 = 960\), remainder \(40\)). Starting at noon, adding \(16\) hours gives 4:00 AM the next day, and adding \(40\) minutes gives 4:40 AM.  
ANSWER 9: D

Problem 10:  
Compute inside parentheses:  
\(8 \times 4 + 2 = 32 + 2 = 34\)  
\(8 + 4 \times 2 = 8 + 8 = 16\)  
Then \(34 - 16 = 18\).  
ANSWER 10: D

Problem 11:  
11 hours is \(11 \times 60 = 660\) minutes. Adding 5 minutes gives \(660 + 5 = 665\) minutes.  
ANSWER 11: C

Problem 12:  
Georgie can enter through any of the 6 windows. For exiting, he must choose a different window, so there are 5 choices. The total number of ways is \(6 \times 5 = 30\).  
ANSWER 12: D

Problem 1:  
The Martian needs to ensure there are 5 socks of the same color. The worst-case scenario is pulling the maximum number of socks without 5 of any color. Since there are 3 colors, this would be 4 socks of each color (4 × 3 = 12 socks). The next sock (the 13th) must match one color, giving 5 socks of that color.  
**ANSWER 1: D**  

Problem 2:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is C(4,3) = 4.  
**ANSWER 2: B** (Note: C(4,3)=4, but rechecking: the team is chosen from 4 specific people, so combinations are 4. Answer B=4)  

Problem 3:  
Check each pair:  
A. (−4)(−9) = 36  
B. (−3)(−12) = 36  
C. (1/2)(−72) = -36 (≠36)  
D. (1)(36) = 36  
E. (3/2)(24) = 36  
The pair that does NOT equal 36 is C.  
**ANSWER 3: E** (Re-evaluated: The question asks which does NOT equal 36. Option E: (3/2)*24=36, so it is valid. Option C: (1/2)*(-72)=-36, so C is the correct choice for NOT equal. Correction: **ANSWER 3: C** [Original had a typo, but computation confirms C is the answer])  

*Correction: For Problem 3, Option C: (1/2, -72) likely means (1/2) × (-72) = -36 ≠ 36. Option E: (3/2) × 24 = 36. Thus, C is the answer.*  
**ANSWER 3: C**  

Problem 4:  
Ceiling height: 2.4 m = 240 cm. Light bulb location: 240 - 10 = 230 cm from floor. Alice’s reach: 150 cm + 46 cm = 196 cm. Stool height = 230 - 196 = 34 cm.  
**ANSWER 4: B**  

Problem 5:  
Ratio of trout in sample: 30/180 = 1/6. Total trout: 250. Total fish = 250 × 6 = 1500.  
**ANSWER 5: B**  

Problem 6:  
Original cost: 5 × $2.50 = $12.50. 20% off: $12.50 × 0.20 = $2.50 saved.  
**ANSWER 6: C** (Note: The question asks how much Karl could have saved, which is $2.50, not $2.00. Original had a mistake; correction: 20% of $12.50 is $2.50, so answer C)  

Problem 7:  
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.  
**ANSWER 7: D**  

Problem 8:  
Let $5 bills = x, $2 bills = y. 5x + 2y = 17. Possible solutions:  
- x=3, y=1 (15+2=17)  
- x=1, y=6 (5+12=17)  
Only 2 combinations.  
**ANSWER 8: A**  

Problem 9:  
1000 minutes = 16 hours 40 minutes. Starting at noon: noon + 16h40m = 4:40 a.m. next day.  
**ANSWER 9: D**  

Problem 10:  
(8×4 + 2) = 34, (8 + 4×2) = 16. 34 - 16 = 18.  
**ANSWER 10: D**  

Problem 11:  
11 hours = 660 minutes. 660 + 5 = 665 minutes.  
**ANSWER 11: C**  

Problem 12:  
Georgie can enter through 6 windows and exit through 5 others. Total ways: 6 × 5 = 30.  
**ANSWER 12: D**  

---  
**Final Answers**  
ANSWER 1: D  
ANSWER 2: B  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: B  
ANSWER 6: C  
ANSWER 7: D  
ANSWER 8: A  
ANSWER 9: D  
ANSWER 10: D  
ANSWER 11: C  
ANSWER 12: D

### Problem 1:
This is a classic pigeonhole principle problem. We need to find the minimum number of socks the Martian must pull to guarantee 5 socks of the same color, with 3 total colors. The worst-case scenario is that the Martian pulls as many socks as possible without getting 5 of any color: that is 4 socks of each color, for a total of 3*4=12 socks. The next sock pulled (the 13th) has to be one of the three colors, pushing that color to 5 socks. Tempting wrong answers: C (12 is only the number of socks that avoids 5 of the same color, not enough to guarantee the requirement), E (15 is the total if you took 5 of each color, not the minimum to guarantee).
ANSWER 1: D

---

### Problem 2:
We need to count the number of ways to choose 3 starters from 4 people, where order on the team does not matter. This is a combination problem: C(n,k) = n!/(k!(n-k)!). For n=4, k=3: C(4,3) = 4. We can also list the teams by who is left out: leaving out Lance, Sally, Joy, or Fred gives exactly 4 unique teams. Tempting wrong answer: C (6 is the count if order of selection mattered, which it does not for a team of starters).
ANSWER 2: B

---

### Problem 3:
We calculate the product of each pair to find which does not equal 36:
- A: (-4)*(-9)=36 ✔️
- B: (-3)*(-12)=36 ✔️
- C: (1/2)*(-72) = -36 ❌
- D: 1*36=36 ✔️
- E: (3/2)*24=36 ✔️
Only pair C has a product of -36, not 36.
ANSWER 3: C

---

### Problem 4:
First convert all units to centimeters to avoid conversion errors:
- Ceiling height: 2.4m = 240cm. Light bulb is 10cm below ceiling, so bulb height from floor = 240-10=230cm.
- Alice's height: 1.5m=150cm, her reach above her head is 46cm, so total reach without stool = 150+46=196cm.
- Stool height = 230-196=34cm.
Tempting wrong answers come from unit conversion mistakes, e.g., mixing up meters and centimeters.
ANSWER 4: B

---

### Problem 5:
This is a proportion problem. The ratio of trout to total fish is the same in the sample and the lake. In the sample, 30 trout out of 180 total fish = 1/6. Let T be total fish in the lake: 250/T = 1/6 → T=250*6=1500.
ANSWER 5: B

---

### Problem 6:
Karl bought 5 folders at $2.50 each, total original cost = 5*2.50=$12.50. He saves 20% of the total cost by waiting for the sale: 0.2*12.50=$2.50. This can also be calculated per folder: 20% of $2.50 is $0.50 saved per folder, 5*0.50=$2.50 total saved.
ANSWER 6: C

---

### Problem 7:
300 dimples * 2 seconds per dimple = 600 total seconds. Convert to minutes: 60 seconds per minute, so 600/60=10 minutes.
ANSWER 7: D

---

### Problem 8:
We find all non-negative integer pairs (x,y) where x=number of $5 bills, y=number of $2 bills, and 5x+2y=17. Only odd values of x work (since $5 is odd, $2 is even, we need an odd sum to get 17):
- x=1: 5+2y=17 → y=6 (valid)
- x=3: 15+2y=17 → y=1 (valid)
Even x values (0,2) leave an odd remainder that cannot be formed with $2 bills, so only 2 valid combinations.
ANSWER 8: A

---

### Problem 9:
Convert 1000 minutes to hours and minutes: 60 minutes per hour, 1000/60=16 hours and 40 minutes. Adding this to noon (12pm): 12pm +16h40m = 4:40am the next day.
ANSWER 9: D

---

### Problem 10:
Use order of operations (PEMDAS) to calculate:
First parenthesis: 8*4 +2 = 32+2=34
Second parenthesis: 8 +4*2=8+8=16
Subtract: 34-16=18. The tempting wrong answer C (10) comes from incorrectly adding before multiplying in the second parenthesis.
ANSWER 10: D

---

### Problem 11:
Convert 11 hours 5 minutes to total minutes: 11*60=660 minutes, plus 5 extra minutes = 665 minutes.
ANSWER 11: C

---

### Problem 12:
Georgie has 6 choices for an entry window, and 5 remaining choices for an exit window (must use a different window). Total ways = 6*5=30. Tempting wrong answer E (36) counts entering and leaving through the same window, which is not allowed; B (15) incorrectly treats entry/exit as unordered, which they are not.
ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color (red, white, blue). This is a pigeonhole principle problem: the worst-case scenario is pulling the maximum number of socks without getting 5 of any color. For each of the 3 colors, we can pull 4 socks (since 4 < 5), so 4×3=12 socks. The next sock pulled will necessarily be the 5th of one color. So total socks needed: 12 + 1 = 13.
ANSWER 1: D
Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). Since the problem does not mention distinct positions for starters, order does not matter, so we use combinations. The number of ways to choose 3 people from 4 is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n=4 \) and \( k=3 \):
\( \binom{4}{3} = \frac{4!}{3!1!} = 4 \). This is equivalent to counting how many people we leave out (4 choices, one for each person not selected).
ANSWER 2: B
Problem 3:
We check the product of each pair:
- A: \( (-4) \times (-9) = 36 \)
- B: \( (-3) \times (-12) = 36 \)
- C: \( \frac{1}{2} \times (-72) = -36 \neq 36 \)
- D: \( 1 \times 36 = 36 \)
- E: \( \frac{3}{2} \times 24 = 36 \)
Only pair C does not have a product of 36.
ANSWER 3: C
Problem 4:
Convert all units to centimeters:
- Ceiling height: \( 2.4 \, \text{m} = 240 \, \text{cm} \)
- Light bulb height from floor: \( 240 - 10 = 230 \, \text{cm} \)
- Alice's reach from floor (without stool): \( 1.5 \, \text{m} + 0.46 \, \text{m} = 150 + 46 = 196 \, \text{cm} \)
Let \( h \) be the stool height. The total reach with the stool equals the light bulb height:
\( h + 196 = 230 \)
\( h = 230 - 196 = 34 \, \text{cm} \)
ANSWER 4: B
Problem 5:
The ratio of trout to total fish in the sample equals the ratio in the lake. The sample has \( \frac{30}{180} = \frac{1}{6} \) trout. Let \( T \) be the total number of fish in the lake. We know there are 250 trout, so:
\( \frac{250}{T} = \frac{1}{6} \)
\( T = 250 \times 6 = 1500 \)
ANSWER 5: B
Problem 6:
Karl bought 5 folders at $2.50 each. The 20% discount applies to each folder, so the savings per folder is \( 2.50 \times 0.20 = 0.50 \). Total savings for 5 folders: \( 5 \times 0.50 = 2.50 \).
ANSWER 6: C
Problem 7:
Total time to paint 300 dimples: \( 300 \times 2 = 600 \) seconds. Convert to minutes: \( 600 \div 60 = 10 \) minutes.
ANSWER 7: D
Problem 8:
Let \( x \) = number of $5 bills, \( y \) = number of $2 bills. We need \( 5x + 2y = 17 \), where \( x,y \geq 0 \) integers. Test non-negative integer values of \( x \):
- \( x=0 \): \( 2y=17 \) → \( y=8.5 \) (not integer, invalid)
- \( x=1 \): \( 5 + 2y=17 \) → \( 2y=12 \) → \( y=6 \) (valid: 1 $5 + 6 $2 = $17)
- \( x=2 \): \( 10 + 2y=17 \) → \( 2y=7 \) → \( y=3.5 \) (invalid)
- \( x=3 \): \( 15 + 2y=17 \) → \( 2y=2 \) → \( y=1 \) (valid: 3 $5 + 1 $2 = $17)
- \( x=4 \): \( 20 > 17 \) (stop, no more valid \( x \))
Only 2 valid combinations.
ANSWER 8: A
Problem 9:
Convert 1000 minutes to hours and minutes: \( 1000 \div 60 = 16 \) hours with a remainder of 40 minutes (since \( 16 \times 60 = 960 \), \( 1000 - 960 = 40 \)). The contest starts at noon, so adding 16 hours 40 minutes: noon + 12 hours = midnight, plus 4 hours 40 minutes = 4:40 a.m.
ANSWER 9: D
Problem 10:
Use order of operations (multiplication before addition):
First parentheses: \( 8 \times 4 + 2 = 32 + 2 = 34 \)
Second parentheses: \( 8 + 4 \times 2 = 8 + 8 = 16 \)
Subtract: \( 34 - 16 = 18 \)
ANSWER 10: D
Problem 11:
Convert 11 hours 5 minutes to total minutes: \( 11 \times 60 + 5 = 660 + 5 = 665 \)
ANSWER 11: C
Problem 12:
Georgie chooses an entrance window first (6 options), then a different exit window (5 remaining options, since exit cannot equal entrance). Since entrance and exit are distinct actions, order matters, so total ways: \( 6 \times 5 = 30 \).
ANSWER 12: D

I'll work through each problem systematically.

**Problem 1:**

**What is being asked:** How many socks must be removed to guarantee 5 socks of the same color?

**Setup:** There are 3 colors (red, white, blue). In the worst case scenario, the Martian pulls out socks as evenly distributed among colors as possible before getting 5 of one color.

**Working:** 
- Worst case: 4 red, 4 white, 4 blue = 12 socks (still no color with 5)
- The 13th sock must give us a 5th sock of some color

**Answer 1: D**

---

**Problem 2:**

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

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

**Working:**
- C(4,3) = 4 ways
- The 4 ways correspond to leaving out each person: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred}

**Answer 2: B**

---

**Problem 3:**

**What is being asked:** Which pair does NOT have a product equal to 36?

**Working:** Check each pair:
- A. {−4, −9}: (−4)(−9) = 36 ✓
- B. {−3, −12}: (−3)(−12) = 36 ✓
- C. {1/2, −72}: (1/2)(−72) = −36 ✗
- D. {1, 36}: (1)(36) = 36 ✓
- E. {3/2, 24}: (3/2)(24) = 36 ✓

**Answer 3: C**

---

**Problem 4:**

**What is being asked:** How high is the stool?

**Setup:** Convert all measurements to centimeters:
- Ceiling height: 240 cm
- Light bulb location: 240 − 10 = 230 cm from floor
- Alice's height: 150 cm
- Alice's reach above her head: 46 cm
- Alice's reach without stool: 150 + 46 = 196 cm

**Working:**
- With stool, Alice can just reach the bulb at 230 cm
- Stool height + 196 cm = 230 cm
- Stool height = 230 − 196 = 34 cm

**Answer 4: B**

---

**Problem 5:**

**What is being asked:** How many fish total are in the lake?

**Setup:** Use proportions. In the sample: 30 trout out of 180 fish. In the lake: 250 trout out of total fish.

**Working:**
- 30/180 = 250/x
- 1/6 = 250/x
- x = 250 × 6 = 1500 fish

**Answer 5: B**

---

**Problem 6:**

**What is being asked:** How much could Karl have saved with the 20% off sale?

**Setup:** 
- Original cost: 5 folders × $2.50 = $12.50
- Sale discount: 20% off

**Working:**
- 20% of $12.50 = 0.20 × $12.50 = $2.50

**Answer 6: C**

---

**Problem 7:**

**What is being asked:** How many minutes to paint 300 dimples at 2 seconds each?

**Working:**
- Total time: 300 × 2 = 600 seconds
- Convert to minutes: 600 ÷ 60 = 10 minutes

**Answer 7: D**

---

**Problem 8:**

**What is being asked:** How many combinations of $5 bills and $2 bills make $17?

**Setup:** Let x = number of $5 bills, y = number of $2 bills
- 5x + 2y = 17

**Working:**
- x = 0: 2y = 17 → y = 8.5 (not integer)
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 (not integer)
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x ≥ 4: 5x ≥ 20 > 17 (impossible)

Valid combinations: (1,$5 and 6,$2) and (3,$5 and 1,$2) = 2 combinations

**Answer 8: A**

---

**Problem 9:**

**What is being asked:** When did the contest end, starting at noon and lasting 1000 minutes?

**Working:**
- 1000 minutes ÷ 60 = 16 hours and 40 minutes
- Noon + 16 hours 40 minutes = 4:40 p.m. the same day
- But we need to continue: 4:40 p.m. + additional time consideration
- Actually: 1000 ÷ 60 = 16 remainder 40, so 16 hours 40 minutes
- Noon + 16 hours 40 minutes = 4:40 a.m. the next day

**Answer 9: D**

---

**Problem 10:**

**What is being asked:** Calculate (8 × 4 + 2) − (8 + 4 × 2)

**Working:**
- First parentheses: 8 × 4 + 2 = 32 + 2 = 34
- Second parentheses: 8 + 4 × 2 = 8 + 8 = 16
- Result: 34 − 16 = 18

**Answer 10: D**

---

**Problem 11:**

**What is being asked:** Convert 11 hours 5 minutes to minutes

**Working:**
- 11 hours = 11 × 60 = 660 minutes
- Total: 660 + 5 = 665 minutes

**Answer 11: C**

---

**Problem 12:**

**What is being asked:** In how many ways can the ghost enter by one window and leave by a different window?

**Working:**
- Choices for entry: 6 windows
- Choices for exit: 5 windows (must be different from entry)
- Total ways: 6 × 5 = 30

**Answer 12: D**

Problem 1:

To be certain of getting 5 socks of the same color, we use the pigeonhole principle.

There are 3 colors: red, white, and blue.  
To avoid getting 5 of any one color for as long as possible, Martian could take at most 4 of each color:

- 4 red
- 4 white
- 4 blue

That gives a total of \(4+4+4=12\) socks without yet having 5 of the same color.

The very next sock, the 13th, must make one color appear at least 5 times.

So the Martian must remove 13 socks.

ANSWER 1: D

Problem 2:

There are 4 players: Lance, Sally, Joy, and Fred.

We need to choose 3 starters from these 4 players. Since order does not matter, we count combinations:

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

Indeed, the 3-starter groups are:

- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred

So there are 4 ways.

ANSWER 2: B

Problem 3:

We check each product.

- A: \((-4)(-9)=36\)
- B: \((-3)(-12)=36\)
- C: \(\frac12 \cdot (-72)=-36\), not 36
- D: \(1\cdot 36=36\)
- E: \(\frac32 \cdot 24=36\)

Only choice C does **not** have product 36.

ANSWER 3: C

Problem 4:

First find how high the bulb is above the floor.

The ceiling is 2.4 meters above the floor, which is \(240\) centimeters.  
The bulb is 10 centimeters below the ceiling, so it is:

\[
240 - 10 = 230 \text{ cm}
\]

Now Alice’s height is 1.5 meters = 150 cm.  
She can reach 46 cm above her head, so her reach without the stool is:

\[
150 + 46 = 196 \text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230 - 196 = 34 \text{ cm}
\]

So the stool is 34 cm high.

ANSWER 4: B

Problem 5:

The sample has 30 trout out of 180 fish, so the fraction of trout is:

\[
\frac{30}{180}=\frac{1}{6}
\]

If 250 trout is \(\frac16\) of the total lake population, then the total number of fish is:

\[
250 \div \frac16 = 250 \cdot 6 = 1500
\]

So there are 1500 fish in the lake.

ANSWER 5: B

Problem 6:

Karl bought 5 folders at \$2.50 each, so the total cost was:

\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:

\[
0.20 \times 12.50 = 2.50
\]

So he could have saved \$2.50.

ANSWER 6: C

Problem 7:

He paints 300 dimples, and each dimple takes 2 seconds.

Total time in seconds:

\[
300 \times 2 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10
\]

So he needs 10 minutes.

ANSWER 7: D

Problem 8:

We want combinations of 5-dollar bills and 2-dollar bills that total \$17.

Let \(x\) be the number of \$5 bills and \(y\) the number of \$2 bills. Then:

\[
5x + 2y = 17
\]

Try values of \(x\):

- \(x=1\): \(17-5=12\), so \(y=6\)
- \(x=2\): \(17-10=7\), not even, so impossible
- \(x=3\): \(17-15=2\), so \(y=1\)

These are the only nonnegative integer solutions.

So there are 2 combinations.

ANSWER 8: A

Problem 9:

The contest ended 1000 minutes after noon.

Convert 1000 minutes to hours and minutes:

\[
1000 \div 60 = 16 \text{ hours with } 40 \text{ minutes left}
\]

So:

- Noon + 12 hours = midnight
- plus 4 more hours = 4:00 a.m.
- plus 40 minutes = 4:40 a.m.

So the contest ended at 4:40 a.m.

ANSWER 9: D

Problem 10:

Compute each part using order of operations:

\[
(8\times 4 + 2) - (8 + 4\times 2)
\]

First parentheses:

\[
8\times 4 + 2 = 32 + 2 = 34
\]

Second parentheses:

\[
8 + 4\times 2 = 8 + 8 = 16
\]

Now subtract:

\[
34 - 16 = 18
\]

So the value is 18.

ANSWER 10: D

Problem 11:

We convert 11 hours and 5 minutes into minutes.

\[
11 \times 60 = 660
\]

Then add 5 minutes:

\[
660 + 5 = 665
\]

So the match lasted 665 minutes.

ANSWER 11: C

Problem 12:

Georgie must enter by one window and leave by a different window.

There are 6 choices for the entering window.  
After choosing one, there are 5 remaining choices for the exit window.

So the total number of ways is:

\[
6 \times 5 = 30
\]

Therefore, there are 30 ways.

ANSWER 12: D

Problem 1:  
We need the minimum number of socks to remove to guarantee there are 5 socks of the same color, given 3 colors (red, white, blue) and at least 5 of each color.

Worst case: you could first pull 4 socks of each color without reaching 5 of any color.  
That would be:
- 4 red + 4 white + 4 blue = 12 socks pulled

After 12 socks, you have at most 4 of each color. The next sock (the 13th) must be one of the 3 colors, and by the pigeonhole principle it makes one color reach 5.

So the Martian must remove **13** socks.

Tempting wrong choices:
- 12 isn’t enough because it’s possible to have only 4 of each color.
- Any number less than 13 can’t force a fifth sock of some color.

ANSWER 1: D


Problem 2:  
We have 4 students: Lance, Sally, Joy, Fred. We must choose 3 starters.

Number of ways to choose 3 out of 4 is:
\[
\binom{4}{3} = 4
\]

(You can also think: leave out one student; there are 4 choices for who to leave out.)

ANSWER 2: B


Problem 3:  
We check which pair does **NOT** multiply to 36.

Choices:
A. \(\{-4,-9\}\): \((-4)(-9)=36\) ✅  
B. \(\{-3,-12\}\): \((-3)(-12)=36\) ✅  
C. \(\left\{\frac{1}{2},-72\right\}\): \(\frac{1}{2}\cdot(-72)=-36\) ❌ not 36  
D. \(\{1,36\}\): \(1\cdot 36=36\) ✅  
E. \(\left\{\frac{3}{2},24\right\}\): \(\frac{3}{2}\cdot 24=3\cdot 12=36\) ✅

So the one that does not have product 36 is choice C.

ANSWER 3: C


Problem 4:  
Let’s convert everything to the same reference (floor).

- Ceiling is 2.4 m above the floor = 240 cm  
- The bulb is 10 cm below the ceiling: height of bulb above floor:
\[
240 - 10 = 230 \text{ cm}
\]
- Alice is 1.5 m tall = 150 cm  
- She can reach 46 cm above the top of her head, so from her top to the bulb:
\[
150 + 46 = 196 \text{ cm (if she stands on the floor)}
\]

She is standing on a stool, so her total height is (stool height + 150). She can reach:
\[
(\text{stool height} + 150) + 46
\]
Set equal to bulb height 230:
\[
\text{stool height} + 196 = 230
\]
\[
\text{stool height} = 34
\]

ANSWER 4: B


Problem 5:  
We know:
- Lake has 250 trout.
- Total fish in lake = \(T\)
- Sample has 180 fish, and 30 are trout.
So sample trout-to-total ratio equals lake trout-to-total ratio:
\[
\frac{30}{180} = \frac{250}{T}
\]
Simplify \(\frac{30}{180}=\frac{1}{6}\):
\[
\frac{1}{6} = \frac{250}{T}
\]
\[
T = 250 \cdot 6 = 1500
\]

ANSWER 5: B


Problem 6:  
Cost per folder originally: $2.50  
He bought 5 folders:
\[
5 \cdot 2.50 = 12.50
\]
A 20%-off sale means paying 80% of the original price:
\[
0.80 \cdot 12.50 = 10.00
\]
Savings:
\[
12.50 - 10.00 = 2.50
\]

ANSWER 6: C


Problem 7:  
Paint 300 dimples, 2 seconds per dimple.
Total time in seconds:
\[
300 \cdot 2 = 600 \text{ seconds}
\]
Convert to minutes (60 seconds per minute):
\[
600 \div 60 = 10 \text{ minutes}
\]

ANSWER 7: D


Problem 8:  
We need numbers of $5 bills and $2 bills that total $17 (order doesn’t matter).

Let \(x\) = number of $5 bills, \(y\) = number of $2 bills.
\[
5x + 2y = 17
\]
Try \(x=0\) to \(x=3\) (since \(4\cdot 5=20>17\)):

- \(x=0\): \(2y=17\) not possible  
- \(x=1\): \(5 + 2y=17 \Rightarrow 2y=12 \Rightarrow y=6\)  
- \(x=2\): \(10 + 2y=17 \Rightarrow 2y=7\) not possible  
- \(x=3\): \(15 + 2y=17 \Rightarrow 2y=2 \Rightarrow y=1\)

So there are 2 combinations: \((x,y)=(1,6)\) and \((3,1)\).

ANSWER 8: A


Problem 9:  
The contest starts at noon. It ends 1000 minutes later.

Convert 1000 minutes to hours:
\[
1000 \div 60 = 16 \text{ hours remainder } 40 \text{ minutes}
\]
Noon + 16 hours = 4:00 a.m. (next day)  
Add 40 minutes:
\[
4:00 \text{ a.m.} + 40 \text{ min} = 4:40 \text{ a.m.}
\]

ANSWER 9: D


Problem 10:  
Compute:
\[
(8 \times 4 + 2) - (8 + 4 \times 2)
\]
First parentheses:
\[
8\times 4 = 32,\quad 32+2=34
\]
Second parentheses:
\[
4\times 2 = 8,\quad 8+8=16
\]
Now subtract:
\[
34 - 16 = 18
\]

ANSWER 10: D


Problem 11:  
11 hours and 5 minutes total.
Convert hours to minutes:
\[
11 \cdot 60 = 660
\]
Add 5 minutes:
\[
660 + 5 = 665
\]

ANSWER 11: C


Problem 12:  
There are 6 windows. Georgie enters through one window and must leave through a different window.

Choose entering window: 6 choices.  
Choose leaving window: 5 choices (cannot be the same).

Total:
\[
6 \cdot 5 = 30
\]

ANSWER 12: D

Problem 1:
To be certain of having 5 socks of the same color, we consider the "worst-case scenario" (Pigeonhole Principle). If we have 3 colors (red, white, blue), we could pick 4 red, 4 white, and 4 blue socks without having 5 of any one color. This totals $4 + 4 + 4 = 12$ socks. The very next sock (the 13th) must be either red, white, or blue, which will complete a set of 5 for that color.
ANSWER 1: D

Problem 2:
We need to choose 3 starters from a group of 4 people (Lance, Sally, Joy, Fred). This is a combination problem denoted as $\binom{4}{3}$. Choosing 3 people to be on the team is the same as choosing 1 person to be left out. There are 4 ways to choose who is left out (Lance, Sally, Joy, or Fred).
ANSWER 2: B

Problem 3:
We check the products:
A: $(-4) \times (-9) = 36$
B: $(-3) \times (-12) = 36$
C: $\frac{1}{2} \times (-72) = -36$
D: $1 \times 36 = 36$
E: $\frac{3}{2} \times 24 = 3 \times 12 = 36$
Choice C results in $-36$, not $36$.
ANSWER 3: C

Problem 4:
Ceiling height = 2.4 m = 240 cm. Light bulb height = 240 - 10 = 230 cm from the floor.
Alice's reach = 150 cm (height) + 46 cm (reach) = 196 cm.
Stool height = (Bulb height) - (Alice's reach) = 230 - 196 = 34 cm.
ANSWER 4: B

Problem 5:
Let $T$ be the total number of fish in the lake. The ratio of trout to total fish is $\frac{30}{180} = \frac{1}{6}$.
Set up the proportion: $\frac{250}{T} = \frac{1}{6}$.
$T = 250 \times 6 = 1500$.
ANSWER 5: B

Problem 6:
Total cost = 5 folders × $2.50 = $12.50.
Savings = 20% of $12.50 = 0.20 \times 12.50 = $2.50.
Alternatively, 20% off one folder is $0.20 \times 2.50 = $0.50. For 5 folders, $0.50 \times 5 = $2.50.
ANSWER 6: C

Problem 7:
Total time in seconds = 300 dimples × 2 seconds/dimple = 600 seconds.
Total time in minutes = 600 seconds / 60 seconds/minute = 10 minutes.
ANSWER 7: D

Problem 8:
Let $x$ be the number of $5 bills and $y$ be the number of $2 bills. $5x + 2y = 17$.
If $x=1$, $5+2y=17 \Rightarrow 2y=12 \Rightarrow y=6$. (1, 6)
If $x=3$, $15+2y=17 \Rightarrow 2y=2 \Rightarrow y=1$. (3, 1)
If $x=0$ or $x=2$, $2y$ would be 17 or 7, which are not divisible by 2.
There are 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours and 40 minutes (since $1000 \div 60 = 16$ remainder 40).
Noon + 12 hours = midnight.
Midnight + 4 hours and 40 minutes = 4:40 a.m.
ANSWER 9: D

Problem 10:
$(8 \times 4 + 2) - (8 + 4 \times 2) = (32 + 2) - (8 + 8) = 34 - 16 = 18$.
ANSWER 10: D

Problem 11:
11 hours = $11 \times 60 = 660$ minutes.
Total = 660 + 5 = 665 minutes.
ANSWER 11: C

Problem 12:
There are 6 choices for the entry window. Since the exit window must be different, there are 5 choices for the exit window.
Total ways = $6 \times 5 = 30$.
ANSWER 12: D

Problem 1:
This is a pigeonhole principle problem. To guarantee at least 5 socks of one color, first consider the worst-case scenario of taking as many socks as possible without having 5 of any single color. That means taking exactly 4 socks of each of the 3 colors, for a total of \(4 \times 3 = 12\) socks. The next sock (the 13th) must be a 5th sock of one of the colors. Thus 13 socks are required. (Fewer than 13 leaves open the possibility of only 4 of each color.)

ANSWER 1: D

Problem 2:
The number of ways to choose 3 starters out of the 4 available players is the combination \(\binom{4}{3} = 4\). Each choice simply leaves out one of the four players.

ANSWER 2: B

Problem 3:
Compute the product for each pair:
- A: \((-4) \times (-9) = 36\)
- B: \((-3) \times (-12) = 36\)
- C: \(\frac12 \times (-72) = -36\) (not 36)
- D: \(1 \times 36 = 36\)
- E: \(\frac32 \times 24 = 36\)

Only pair C fails to give a product of 36.

ANSWER 3: C

Problem 4:
Convert all lengths to centimeters. The ceiling is 240 cm above the floor, so the bulb is 230 cm above the floor. Alice's reach is \(150 + 46 = 196\) cm. The stool must therefore supply the remaining \(230 - 196 = 34\) cm.

ANSWER 4: B

Problem 5:
The sample shows that trout comprise \(\frac{30}{180} = \frac16\) of all fish. With 250 trout in the lake, the total number of fish is \(250 \div \frac16 = 1500\).

ANSWER 5: B

Problem 6:
Five folders cost \(5 \times 2.50 = 12.50\). A 20 % discount saves \(0.20 \times 12.50 = 2.50\).

ANSWER 6: C

Problem 7:
Painting 300 dimples at 2 seconds each requires \(300 \times 2 = 600\) seconds, or \(600 \div 60 = 10\) minutes.

ANSWER 7: D

Problem 8:
Solve \(5x + 2y = 17\) for non-negative integers \(x, y\):
- \(x = 1\), \(y = 6\) works;
- \(x = 3\), \(y = 1\) works.
No other non-negative integer solutions exist, so there are exactly two combinations.

ANSWER 8: A

Problem 9:
Divide 1000 minutes by 60: \(1000 = 16\) hours + 40 minutes. Adding 16 hours to noon yields 4:00 a.m.; adding 40 minutes yields 4:40 a.m.

ANSWER 9: D

Problem 10:
Left side: \(8 \times 4 + 2 = 34\). Right side: \(8 + 4 \times 2 = 16\). Difference: \(34 - 16 = 18\).

ANSWER 10: D

Problem 11:
Convert 11 hours 5 minutes to minutes: \(11 \times 60 + 5 = 665\).

ANSWER 11: C

Problem 12:
There are 6 choices for the entry window and then 5 remaining choices for the exit window, giving \(6 \times 5 = 30\) ordered pairs.

ANSWER 12: D

Problem 1:
The problem is asking us to determine the minimum number of socks the Martian must remove from the drawer to guarantee that there are 5 socks of the same color. 
To ensure that there are 5 socks of the same color, we must consider the worst-case scenario where the Martian draws as many socks as possible without getting 5 of the same color. 
The Martian can draw at most 4 socks of each color without having 5 socks of the same color. 
So, the Martian can draw 4 red, 4 white, and 4 blue socks, for a total of 4 + 4 + 4 = 12 socks. 
At this point, the Martian still doesn't have 5 socks of the same color. 
However, drawing one more sock will guarantee that there are 5 socks of the same color. 
Therefore, the Martian must draw at least 12 + 1 = 13 socks.
Let's examine the choices: 
A. 6 is too low because the Martian could draw 3 red, 2 white, and 1 blue, or other combinations that don't guarantee 5 socks of the same color. 
B. 9 is still too low because the Martian could draw 4 red, 3 white, and 2 blue, or other similar combinations. 
C. 12 is the maximum number of socks the Martian can draw without having 5 socks of the same color. 
D. 13 is the correct answer because it is one more than the maximum number of socks that can be drawn without having 5 socks of the same color. 
E. 15 is more than necessary.
ANSWER 1: D

Problem 2:
The problem is asking us to find the number of ways to choose 3 starters from a team of 4 players: Lance, Sally, Joy, and Fred. 
This is a combination problem, because the order in which the starters are chosen does not matter. 
The number of ways to choose 3 starters from 4 players can be calculated using the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of players and k is the number of starters. 
Here, n = 4 and k = 3, so C(4, 3) = 4! / (3!(4-3)!) = 4.
Let's list the possible combinations to verify: 
- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred
There are indeed 4 combinations.
Let's examine the choices: 
A. 2 is too low. 
B. 4 matches our calculation. 
C. 6 is too high. 
D. 8 is too high. 
E. 10 is too high.
ANSWER 2: B

Problem 3:
The problem is asking us to identify the pair of numbers that does NOT have a product equal to 36. 
Let's check each pair: 
A. {−4, −9}: (−4) × (−9) = 36
B. {−3, −12}: (−3) × (−12) = 36
C. { 1 2 , −72}: (1/2) × (−72) = −36
D. {1, 36}: 1 × 36 = 36
E. { 3 2 , 24}: (3/2) × 24 = 36
The pair that does NOT have a product equal to 36 is C.
Let's examine the choices: 
A. {−4, −9} has a product of 36. 
B. {−3, −12} has a product of 36. 
C. { 1 2 , −72} has a product of −36, not 36. 
D. {1, 36} has a product of 36. 
E. { 3 2 , 24} has a product of 36.
ANSWER 3: C

Problem 4:
The problem is asking us to find the height of the stool that Alice needs to stand on to reach the light bulb. 
First, we need to convert the height of the ceiling from meters to centimeters: 2.4 meters = 240 centimeters. 
The light bulb is 10 centimeters below the ceiling, so its height above the floor is 240 - 10 = 230 centimeters. 
Alice is 1.5 meters = 150 centimeters tall and can reach 46 centimeters above her head, so her total reach is 150 + 46 = 196 centimeters. 
To find the height of the stool, we subtract Alice's total reach from the height of the light bulb: 230 - 196 = 34 centimeters.
Let's examine the choices: 
A. 32 is too low because 230 - (196 + 32) = 2, which is not enough to reach the light bulb. 
B. 34 is the correct height. 
C. 36 is too high. 
D. 38 is too high. 
E. 40 is too high.
ANSWER 4: B

Problem 5:
The problem is asking us to find the total number of fish in the lake based on the ratio of trout in the sample to the total number of fish in the sample. 
The sample contains 30 trout out of 180 fish, so the ratio of trout to total fish in the sample is 30/180. 
We are given that this ratio is the same as the ratio of trout to total fish in the lake. 
Let's denote the total number of fish in the lake as x. 
The ratio of trout to total fish in the lake is 250/x. 
Setting the two ratios equal, we get 30/180 = 250/x. 
Cross-multiplying, we have 30x = 250 × 180. 
Simplifying, 30x = 45000. 
Dividing both sides by 30, x = 1500.
Let's examine the choices: 
A. 1250 is too low. 
B. 1500 matches our calculation. 
C. 1750 is too high. 
D. 1800 is too high. 
E. 2000 is too high.
ANSWER 5: B

Problem 6:
The problem is asking us to find out how much Karl could have saved by waiting a day to buy the folders at a 20%-off sale. 
The original price of 5 folders is 5 × $2.50 = $12.50. 
At a 20%-off sale, the discount is 20% of $12.50, which is 0.20 × $12.50 = $2.50. 
So, Karl could have saved $2.50.
Let's examine the choices: 
A. $1.00 is too low. 
B. $2.00 is too low. 
C. $2.50 matches our calculation. 
D. $2.75 is too high. 
E. $5.00 is too high.
ANSWER 6: C

Problem 7:
The problem is asking us to find the time it will take Casey to paint 300 dimples on a golf ball. 
Casey takes 2 seconds to paint one dimple, so to paint 300 dimples, he will take 300 × 2 = 600 seconds. 
To convert seconds to minutes, we divide by 60: 600 / 60 = 10 minutes.
Let's examine the choices: 
A. 4 minutes is too low. 
B. 6 minutes is too low. 
C. 8 minutes is too low. 
D. 10 minutes matches our calculation. 
E. 12 minutes is too high.
ANSWER 7: D

Problem 8:
The problem is asking us to find the number of different combinations of $5 bills and $2 bills that can be used to make a total of $17. 
Let's list the possible combinations: 
1. 3 $5 bills = $15, and $2 bill = $2, total = $17 (1 combination: 3 $5 bills and 1 $2 bill)
2. 1 $5 bill = $5, and 6 $2 bills = $12, total = $17 (1 combination: 1 $5 bill and 6 $2 bills)
We should also check if there are any other combinations: 
- Using 0 $5 bills, we need 8.5 $2 bills, which is not possible since we can't have a fraction of a bill.
- Using 2 $5 bills = $10, we need $7 more, which can be made with 3.5 $2 bills, again not possible.
- Using 4 $5 bills = $20, which is more than $17.

Problem 1:
We want the minimum number of socks to guarantee 5 of the same color. With 3 colors (red, white, blue), the worst case is picking 4 of each color: 4+4+4 = 12 socks. The 13th sock must create a group of 5 of one color. So 13 socks are needed.
ANSWER 1: D

Problem 2:
We choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways is the combination C(4,3) = 4.
ANSWER 2: B

Problem 3:
Compute each product:
A: (−4)×(−9) = 36
B: (−3)×(−12) = 36
C: (1/2)×(−72) = −36
D: 1×36 = 36
E: (3/2)×24 = 36
Only C does not equal 36.
ANSWER 3: C

Problem 4:
Ceiling = 2.4 m = 240 cm. Light bulb is 10 cm below ceiling, so its height = 230 cm.
Alice's height = 1.5 m = 150 cm. Reach above head = 46 cm, so reach without stool = 196 cm.
Stool height needed = 230 − 196 = 34 cm.
ANSWER 4: B

Problem 5:
Sample ratio: 30 trout / 180 fish = 1/6. Lake has 250 trout, which is 1/6 of total fish. Total fish = 250 × 6 = 1500.
ANSWER 5: B

Problem 6:
Cost without sale: 5 × $2.50 = $12.50.
With 20% off, each folder costs $2.50 × 0.8 = $2.00. Total = 5 × $2.00 = $10.00.
Savings = $12.50 − $10.00 = $2.50.
ANSWER 6: C

Problem 7:
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.
ANSWER 7: D

Problem 8:
Find non-negative integers x ($5 bills) and y ($2 bills) such that 5x + 2y = 17.
x=1 → 5 + 2y = 17 → y=6.
x=3 → 15 + 2y = 17 → y=1.
No other x works. So 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours 40 minutes (16×60 = 960, remainder 40).
Noon + 16 hours = 4:00 a.m. next day. Add 40 minutes → 4:40 a.m.
ANSWER 9: D

Problem 10:
(8 × 4 + 2) = 32 + 2 = 34.
(8 + 4 × 2) = 8 + 8 = 16.
34 − 16 = 18.
ANSWER 10: D

Problem 11:
11 hours × 60 = 660 minutes. Plus 5 minutes = 665 minutes.
ANSWER 11: C

Problem 12:
6 choices for entry window, then 5 choices for a different exit window. Total = 6 × 5 = 30.
ANSWER 12: D

Problem 1:
We need to find the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). Using the Pigeonhole Principle, we consider the worst-case scenario: the Martian pulls out as many socks as possible *without* getting 5 of any one color. That means getting exactly 4 of each color:
- 4 red + 4 white + 4 blue = 12 socks
After 12 socks, the Martian could have 4 of each color and still not have 5 matching. The very next sock (the 13th) must be red, white, or blue, pushing the count of that color to 5. So 13 socks guarantees 5 of the same color.
(The "five-legged" detail is flavor text explaining why the Martian needs 5 socks.)

ANSWER 1: D

Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). This is a combination problem since order doesn't matter:
C(4, 3) = 4! / (3! × 1!) = 4
We can also list them: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — exactly 4 ways.

ANSWER 2: B

Problem 3:
We check each product:
- A: (−4) × (−9) = 36 ✓
- B: (−3) × (−12) = 36 ✓
- C: (1/2) × (−72) = −36 ✗ (This equals −36, NOT 36)
- D: 1 × 36 = 36 ✓
- E: (3/2) × 24 = 36 ✓
Choice C gives −36, which does not equal 36.

ANSWER 3: C

Problem 4:
First, convert everything to centimeters:
- Ceiling height: 2.4 m = 240 cm
- Light bulb position: 240 − 10 = 230 cm above the floor
- Alice's height: 1.5 m = 150 cm
- Alice's reach above her head: 46 cm
- Alice's maximum reach (without stool): 150 + 46 = 196 cm above the floor

Let h = height of the stool. Standing on the stool, her reach is h + 196 cm. She can "just reach" the bulb:
h + 196 = 230
h = 230 − 196 = 34 cm

ANSWER 4: B

Problem 5:
The ratio of trout to total fish in the sample should equal the ratio in the lake:
- Sample ratio: 30 trout / 180 fish = 1/6
- Lake ratio: 250 trout / Total fish = 1/6

Setting up the proportion: 250 / Total = 1/6
Total = 250 × 6 = 1500 fish

ANSWER 5: B

Problem 6:
Karl bought 5 folders at $2.50 each:
Total cost = 5 × $2.50 = $12.50
A 20%-off sale means he would save 20% of the total:
Savings = 0.20 × $12.50 = $2.50

ANSWER 6: C

Problem 7:
Total time in seconds: 300 dimples × 2 seconds/dimple = 600 seconds
Convert to minutes: 600 seconds ÷ 60 seconds/minute = 10 minutes

ANSWER 7: D

Problem 8:
We need non-negative integer solutions to 5x + 2y = 17, where x = number of $5 bills and y = number of $2 bills.
- x = 0: 2y = 17 → y = 8.5 (not an integer) ✗
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 ✗
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x = 4: 20 > 17 (too large) ✗

Only 2 valid combinations: (1 five + 6 twos) and (3 fives + 1 two).

ANSWER 8: A

Problem 9:
Convert 1000 minutes to hours and minutes:
1000 ÷ 60 = 16 remainder 40, so 1000 minutes = 16 hours 40 minutes.
Starting at noon (12:00 PM):
- 12 hours later = midnight (12:00 AM)
- 4 more hours = 4:00 AM
- 40 more minutes = 4:40 AM

ANSWER 9: D

Problem 10:
Evaluate each part using order of operations (multiplication before addition):
- First part: (8 × 4 + 2) = (32 + 2) = 34
- Second part: (8 + 4 × 2) = (8 + 8) = 16
- Difference: 34 − 16 = 18

ANSWER 10: D

Problem 11:
Convert 11 hours and 5 minutes to minutes:
11 hours × 60 minutes/hour = 660 minutes
660 + 5 = 665 minutes

ANSWER 11: C

Problem 12:
Georgie enters through one window and leaves through a different one.
- Choices for entry window: 6
- Choices for exit window (must be different): 5
Total ways = 6 × 5 = 30

ANSWER 12: D

Problem 1:
The Martian needs to be **certain** of having 5 socks of the same color. To guarantee this, we must consider the worst-case scenario: the Martian picks as many socks as possible **without** yet having 5 of any single color.  
With 3 colors (red, white, blue), the worst case is picking 4 socks of each color:  
\(4 + 4 + 4 = 12\) socks.  
At this point the Martian has 4 of every color but still does not have 5 of any one color. The very next sock—the 13th—must be red, white, or blue, and will necessarily create a set of 5 of that color.  
Thus 13 socks are required.  
(Choice A is too small; with 6 socks one could have 2,2,2. Choice B allows 3,3,3. Choice C allows 4,4,4. Choice E is more than the minimum guaranteed number.)

ANSWER 1: D

---

Problem 2:
We need to choose a group of 3 starters from the 4 students: Lance, Sally, Joy, and Fred. Since the three starters form a team and order does not matter, this is a combination problem.  
\[
\binom{4}{3} = \frac{4!}{3!\,1!} = 4
\]  
Equivalently, we can think of choosing which **one** of the four students will **not** start; there are 4 such choices.  
(Choice A is too small. Choice C is \(\binom{4}{2}\). Choices D and E do not arise from counting 3-person teams from 4 people.)

ANSWER 2: B

---

Problem 3:
We check the product of each pair to see which is not 36.

- **A.** \((-4)(-9) = 36\)  
- **B.** \((-3)(-12) = 36\)  
- **C.** \(\left(\frac12\right)(-72) = -36\)  
- **D.** \((1)(36) = 36\)  
- **E.** \(\left(\frac32\right)(24) = 36\)

Only pair **C** gives \(-36\), not \(36\).

ANSWER 3: C

---

Problem 4:
First convert all measurements to centimeters.  
- Ceiling height: \(2.4\text{ m} = 240\text{ cm}\).  
- Light bulb is \(10\text{ cm}\) below the ceiling, so its height above the floor is \(240 - 10 = 230\text{ cm}\).  

Alice’s reach:  
- Height: \(1.5\text{ m} = 150\text{ cm}\).  
- She can reach \(46\text{ cm}\) above her head, so her standing reach is \(150 + 46 = 196\text{ cm}\).  

The stool must supply the remaining height:  
\[
230 - 196 = 34\text{ cm}.
\]

ANSWER 4: B

---

Problem 5:
Let \(T\) be the total number of fish in the lake. The problem states that the ratio of trout to total fish is the same in the sample and in the lake.  
\[
\frac{250}{T} = \frac{30}{180}
\]  
Simplify the right side: \(\frac{30}{180} = \frac16\).  
\[
\frac{250}{T} = \frac16 \quad\Longrightarrow\quad T = 250 \times 6 = 1500.
\]

ANSWER 5: B

---

Problem 6:
Karl bought 5 folders at \(\$2.50\) each.  
Total cost: \(5 \times \$2.50 = \$12.50\).  
A 20% discount would have saved him 20% of this total:  
\[
0.20 \times \$12.50 = \$2.50.
\]

ANSWER 6: C

---

Problem 7:
Casey paints 300 dimples at 2 seconds per dimple.  
Total time in seconds: \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(\frac{600}{60} = 10\) minutes.

ANSWER 7: D

---

Problem 8:
Let \(f\) be the number of \(\$5\) bills and \(t\) be the number of \(\$2\) bills. We need non-negative integer solutions to  
\[
5f + 2t = 17.
\]  
Test values of \(f\):  
- \(f=0\): \(2t=17\) → no integer solution.  
- \(f=1\): \(5 + 2t = 17 \Rightarrow t=6\).  **(Valid)**
- \(f=2\): \(10 + 2t = 17 \Rightarrow t=3.5\).  (Invalid)
- \(f=3\): \(15 + 2t = 17 \Rightarrow t=1\).  **(Valid)**
- \(f=4\): \(20 > 17\) → stop.

There are exactly **2** combinations: \((1\text{ five}, 6\text{ twos})\) and \((3\text{ fives}, 1\text{ two})\).

ANSWER 8: A

---

Problem 9:
The contest starts at noon (12:00 p.m.). We add 1000 minutes.  
Divide 1000 by 60:  
\[
1000 = 16 \times 60 + 40,
\]  
so 1000 minutes = 16 hours and 40 minutes.  

Noon + 12 hours = midnight.  
Midnight + 4 hours 40 minutes = 4:40 a.m.

ANSWER 9: D

---

Problem 10:
Evaluate using the correct order of operations (multiplication before addition).

First parentheses:  
\[
8 \times 4 + 2 = 32 + 2 = 34.
\]

Second parentheses:  
\[
8 + 4 \times 2 = 8 + 8 = 16.
\]

Subtract:  
\[
34 - 16 = 18.
\]

ANSWER 10: D

---

Problem 11:
Convert 11 hours to minutes and add the remaining 5 minutes.  
\[
11 \times 60 = 660\text{ minutes}.
\]  
\[
660 + 5 = 665\text{ minutes}.
\]

ANSWER 11: C

---

Problem 12:
Georgie enters through one of 6 windows and must leave through a **different** window.  
- 6 choices for the entry window.  
- For each entry choice, 5 remaining choices for the exit window.  

Total ways:  
\[
6 \times 5 = 30.
\]  
(Choice E, 36, would count entering and leaving through the same window, which is not allowed.)

ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). The worst-case scenario is to draw as many socks as possible without having 5 of any one color. That would be 4 red, 4 white, and 4 blue, for a total of 12 socks. The next sock drawn (the 13th) must be one of the three colors, giving that color 5 socks. Therefore, 13 socks are required. The five-legged Martian is irrelevant to the count.  
Answer: D

Problem 2:
We have 4 people (Lance, Sally, Joy, Fred) and need to choose 3 starters. Since the starters are not assigned to specific positions, order does not matter. The number of ways is the combination \( \binom{4}{3} = 4 \).  
Answer: B

Problem 3:
Compute each product:  
A. \((-4) \times (-9) = 36\)  
B. \((-3) \times (-12) = 36\)  
C. \(\frac{1}{2} \times (-72) = -36\)  
D. \(1 \times 36 = 36\)  
E. \(\frac{3}{2} \times 24 = 36\)  
Only pair C gives a product not equal to 36.  
Answer: C

Problem 4:
Convert all measurements to centimeters:  
Ceiling height = 2.4 m = 240 cm.  
Light bulb is 10 cm below ceiling → height = 240 – 10 = 230 cm.  
Alice’s height = 1.5 m = 150 cm.  
Her reach above her head = 46 cm, so her maximum reach without a stool = 150 + 46 = 196 cm.  
Let \(h\) be the stool height. Standing on the stool raises her reach to \(196 + h\). This must equal 230 cm:  
\(196 + h = 230\) → \(h = 34\) cm.  
Answer: B

Problem 5:
In the sample, 30 out of 180 fish are trout, so the ratio trout : total = \(30/180 = 1/6\).  
Assuming the same ratio in the lake, if the lake has 250 trout, then the total number of fish is \(250 \times 6 = 1500\).  
Answer: B

Problem 6:
Original cost for 5 folders: \(5 \times 2.50 = \$12.50\).  
A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
Thus, by waiting a day, Karl could have saved $2.50.  
Answer: C

Problem 7:
Time to paint one dimple = 2 seconds.  
Total time for 300 dimples = \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(600 \div 60 = 10\) minutes.  
Answer: D

Problem 8:
We need nonnegative integers \(a\) (number of $5 bills) and \(b\) (number of $2 bills) such that \(5a + 2b = 17\).  
Solve by testing values of \(b\):  
\(b = 1\) → \(5a = 15\) → \(a = 3\) → (3,1) works.  
\(b = 6\) → \(5a = 5\) → \(a = 1\) → (1,6) works.  
No other values give integer \(a\). Thus there are 2 combinations.  
Answer: A

Problem 9:
1000 minutes = \(1000 \div 60 = 16\) hours and \(40\) minutes (since \(60 \times 16 = 960\), remainder 40).  
Starting at noon (12:00 PM), add 12 hours → midnight (12:00 AM). Add the remaining 4 hours → 4:00 AM. Then add 40 minutes → 4:40 AM.  
Answer: D

Problem 10:
Evaluate using order of operations:  
First parentheses: \(8 \times 4 + 2 = 32 + 2 = 34\).  
Second parentheses: \(8 + 4 \times 2 = 8 + 8 = 16\).  
Subtract: \(34 - 16 = 18\).  
Answer: D

Problem 11:
11 hours = \(11 \times 60 = 660\) minutes.  
Add 5 minutes → \(660 + 5 = 665\) minutes.  
Answer: C

Problem 12:
Georgie must choose an entry window (6 choices) and a different exit window (5 choices). Since entering and leaving are distinct actions, the total number of ways is \(6 \times 5 = 30\).  
Answer: D

ANSWER 1: D
ANSWER 2: B
ANSWER 3: C
ANSWER 4: B
ANSWER 5: B
ANSWER 6: C
ANSWER 7: D
ANSWER 8: A
ANSWER 9: D
ANSWER 10: D
ANSWER 11: C
ANSWER 12: D

Problem 1:  
The Martian wants to be certain of having 5 socks of the same color. There are three colors: red, white, and blue. In the worst-case scenario, the Martian could pick 4 socks of each color before getting a fifth of any one color. That would be \(4 \times 3 = 12\) socks, and still no color has 5 socks. The next sock drawn (the 13th) must complete a set of 5 of some color, because there are only three colors. Therefore, the Martian must remove **13** socks to be certain of having 5 socks of the same color.  
ANSWER 1: D

Problem 2:  
Ms. Hamilton's class has 4 available players: Lance, Sally, Joy, and Fred. The number of ways to choose 3 starters from 4 players is given by the combination \(\binom{4}{3} = 4\). Thus, there are **4** possible sets of three starters.  
ANSWER 2: B

Problem 3:  
We check each pair's product:  
- A: \((-4) \times (-9) = 36\)  
- B: \((-3) \times (-12) = 36\)  
- C: \(\frac{1}{2} \times (-72) = -36\) (not 36)  
- D: \(1 \times 36 = 36\)  
- E: \(\frac{3}{2} \times 24 = 36\)  
Only pair **C** does not have a product of 36.  
ANSWER 3: C

Problem 4:  
The ceiling is 2.4 m above the floor, and the bulb is 10 cm (0.1 m) below the ceiling, so the bulb's height is \(2.4 - 0.1 = 2.3\) m. Alice's height is 1.5 m, and she can reach 46 cm (0.46 m) above her head, so her reach height from the floor is \(1.5 + 0.46 = 1.96\) m. When standing on a stool, her hands are at the stool height plus her reach height. Setting this equal to the bulb height:  
\[
\text{stool height} + 1.96 = 2.3 \Rightarrow \text{stool height} = 0.34 \text{ m} = 34 \text{ cm}.
\]  
ANSWER 4: B

Problem 5:  
In the sample, the ratio of trout to total fish is \(\frac{30}{180} = \frac{1}{6}\). Assuming the same ratio in the lake, let \(T\) be the total number of fish in the lake. Then \(\frac{250}{T} = \frac{1}{6}\), so \(T = 250 \times 6 = 1500\).  
ANSWER 5: B

Problem 6:  
Karl bought 5 folders at $2.50 each, so the original cost is \(5 \times 2.50 = \$12.50\). With a 20% discount, the cost becomes \(12.50 \times 0.80 = \$10.00\). The savings is \(12.50 - 10.00 = \$2.50\).  
ANSWER 6: C

Problem 7:  
Painting 300 dimples at 2 seconds each takes \(300 \times 2 = 600\) seconds. Converting to minutes: \(600 \div 60 = 10\) minutes.  
ANSWER 7: D

Problem 8:  
We need nonnegative integers \(x\) (number of $5 bills) and \(y\) (number of $2 bills) such that \(5x + 2y = 17\). Solving:  
- \(y = 1 \Rightarrow 5x = 15 \Rightarrow x = 3\)  
- \(y = 6 \Rightarrow 5x = 5 \Rightarrow x = 1\)  
- \(y = 11 \Rightarrow 5x = -5\) (invalid)  
Thus, there are **2** valid combinations: (3 fives, 1 two) and (1 five, 6 twos).  
ANSWER 8: A

Problem 9:  
1000 minutes equals \(16\) hours and \(40\) minutes (since \(16 \times 60 = 960\), remainder \(40\)). Starting at noon, adding \(16\) hours gives 4:00 AM the next day, and adding \(40\) minutes gives 4:40 AM.  
ANSWER 9: D

Problem 10:  
Compute inside parentheses:  
\(8 \times 4 + 2 = 32 + 2 = 34\)  
\(8 + 4 \times 2 = 8 + 8 = 16\)  
Then \(34 - 16 = 18\).  
ANSWER 10: D

Problem 11:  
11 hours is \(11 \times 60 = 660\) minutes. Adding 5 minutes gives \(660 + 5 = 665\) minutes.  
ANSWER 11: C

Problem 12:  
Georgie can enter through any of the 6 windows. For exiting, he must choose a different window, so there are 5 choices. The total number of ways is \(6 \times 5 = 30\).  
ANSWER 12: D

Problem 1:  
The Martian needs to ensure there are 5 socks of the same color. The worst-case scenario is pulling the maximum number of socks without 5 of any color. Since there are 3 colors, this would be 4 socks of each color (4 × 3 = 12 socks). The next sock (the 13th) must match one color, giving 5 socks of that color.  
**ANSWER 1: D**  

Problem 2:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is C(4,3) = 4.  
**ANSWER 2: B** (Note: C(4,3)=4, but rechecking: the team is chosen from 4 specific people, so combinations are 4. Answer B=4)  

Problem 3:  
Check each pair:  
A. (−4)(−9) = 36  
B. (−3)(−12) = 36  
C. (1/2)(−72) = -36 (≠36)  
D. (1)(36) = 36  
E. (3/2)(24) = 36  
The pair that does NOT equal 36 is C.  
**ANSWER 3: E** (Re-evaluated: The question asks which does NOT equal 36. Option E: (3/2)*24=36, so it is valid. Option C: (1/2)*(-72)=-36, so C is the correct choice for NOT equal. Correction: **ANSWER 3: C** [Original had a typo, but computation confirms C is the answer])  

*Correction: For Problem 3, Option C: (1/2, -72) likely means (1/2) × (-72) = -36 ≠ 36. Option E: (3/2) × 24 = 36. Thus, C is the answer.*  
**ANSWER 3: C**  

Problem 4:  
Ceiling height: 2.4 m = 240 cm. Light bulb location: 240 - 10 = 230 cm from floor. Alice’s reach: 150 cm + 46 cm = 196 cm. Stool height = 230 - 196 = 34 cm.  
**ANSWER 4: B**  

Problem 5:  
Ratio of trout in sample: 30/180 = 1/6. Total trout: 250. Total fish = 250 × 6 = 1500.  
**ANSWER 5: B**  

Problem 6:  
Original cost: 5 × $2.50 = $12.50. 20% off: $12.50 × 0.20 = $2.50 saved.  
**ANSWER 6: C** (Note: The question asks how much Karl could have saved, which is $2.50, not $2.00. Original had a mistake; correction: 20% of $12.50 is $2.50, so answer C)  

Problem 7:  
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.  
**ANSWER 7: D**  

Problem 8:  
Let $5 bills = x, $2 bills = y. 5x + 2y = 17. Possible solutions:  
- x=3, y=1 (15+2=17)  
- x=1, y=6 (5+12=17)  
Only 2 combinations.  
**ANSWER 8: A**  

Problem 9:  
1000 minutes = 16 hours 40 minutes. Starting at noon: noon + 16h40m = 4:40 a.m. next day.  
**ANSWER 9: D**  

Problem 10:  
(8×4 + 2) = 34, (8 + 4×2) = 16. 34 - 16 = 18.  
**ANSWER 10: D**  

Problem 11:  
11 hours = 660 minutes. 660 + 5 = 665 minutes.  
**ANSWER 11: C**  

Problem 12:  
Georgie can enter through 6 windows and exit through 5 others. Total ways: 6 × 5 = 30.  
**ANSWER 12: D**  

---  
**Final Answers**  
ANSWER 1: D  
ANSWER 2: B  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: B  
ANSWER 6: C  
ANSWER 7: D  
ANSWER 8: A  
ANSWER 9: D  
ANSWER 10: D  
ANSWER 11: C  
ANSWER 12: D

### Problem 1:
This is a classic pigeonhole principle problem. We need to find the minimum number of socks the Martian must pull to guarantee 5 socks of the same color, with 3 total colors. The worst-case scenario is that the Martian pulls as many socks as possible without getting 5 of any color: that is 4 socks of each color, for a total of 3*4=12 socks. The next sock pulled (the 13th) has to be one of the three colors, pushing that color to 5 socks. Tempting wrong answers: C (12 is only the number of socks that avoids 5 of the same color, not enough to guarantee the requirement), E (15 is the total if you took 5 of each color, not the minimum to guarantee).
ANSWER 1: D

---

### Problem 2:
We need to count the number of ways to choose 3 starters from 4 people, where order on the team does not matter. This is a combination problem: C(n,k) = n!/(k!(n-k)!). For n=4, k=3: C(4,3) = 4. We can also list the teams by who is left out: leaving out Lance, Sally, Joy, or Fred gives exactly 4 unique teams. Tempting wrong answer: C (6 is the count if order of selection mattered, which it does not for a team of starters).
ANSWER 2: B

---

### Problem 3:
We calculate the product of each pair to find which does not equal 36:
- A: (-4)*(-9)=36 ✔️
- B: (-3)*(-12)=36 ✔️
- C: (1/2)*(-72) = -36 ❌
- D: 1*36=36 ✔️
- E: (3/2)*24=36 ✔️
Only pair C has a product of -36, not 36.
ANSWER 3: C

---

### Problem 4:
First convert all units to centimeters to avoid conversion errors:
- Ceiling height: 2.4m = 240cm. Light bulb is 10cm below ceiling, so bulb height from floor = 240-10=230cm.
- Alice's height: 1.5m=150cm, her reach above her head is 46cm, so total reach without stool = 150+46=196cm.
- Stool height = 230-196=34cm.
Tempting wrong answers come from unit conversion mistakes, e.g., mixing up meters and centimeters.
ANSWER 4: B

---

### Problem 5:
This is a proportion problem. The ratio of trout to total fish is the same in the sample and the lake. In the sample, 30 trout out of 180 total fish = 1/6. Let T be total fish in the lake: 250/T = 1/6 → T=250*6=1500.
ANSWER 5: B

---

### Problem 6:
Karl bought 5 folders at $2.50 each, total original cost = 5*2.50=$12.50. He saves 20% of the total cost by waiting for the sale: 0.2*12.50=$2.50. This can also be calculated per folder: 20% of $2.50 is $0.50 saved per folder, 5*0.50=$2.50 total saved.
ANSWER 6: C

---

### Problem 7:
300 dimples * 2 seconds per dimple = 600 total seconds. Convert to minutes: 60 seconds per minute, so 600/60=10 minutes.
ANSWER 7: D

---

### Problem 8:
We find all non-negative integer pairs (x,y) where x=number of $5 bills, y=number of $2 bills, and 5x+2y=17. Only odd values of x work (since $5 is odd, $2 is even, we need an odd sum to get 17):
- x=1: 5+2y=17 → y=6 (valid)
- x=3: 15+2y=17 → y=1 (valid)
Even x values (0,2) leave an odd remainder that cannot be formed with $2 bills, so only 2 valid combinations.
ANSWER 8: A

---

### Problem 9:
Convert 1000 minutes to hours and minutes: 60 minutes per hour, 1000/60=16 hours and 40 minutes. Adding this to noon (12pm): 12pm +16h40m = 4:40am the next day.
ANSWER 9: D

---

### Problem 10:
Use order of operations (PEMDAS) to calculate:
First parenthesis: 8*4 +2 = 32+2=34
Second parenthesis: 8 +4*2=8+8=16
Subtract: 34-16=18. The tempting wrong answer C (10) comes from incorrectly adding before multiplying in the second parenthesis.
ANSWER 10: D

---

### Problem 11:
Convert 11 hours 5 minutes to total minutes: 11*60=660 minutes, plus 5 extra minutes = 665 minutes.
ANSWER 11: C

---

### Problem 12:
Georgie has 6 choices for an entry window, and 5 remaining choices for an exit window (must use a different window). Total ways = 6*5=30. Tempting wrong answer E (36) counts entering and leaving through the same window, which is not allowed; B (15) incorrectly treats entry/exit as unordered, which they are not.
ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color (red, white, blue). This is a pigeonhole principle problem: the worst-case scenario is pulling the maximum number of socks without getting 5 of any color. For each of the 3 colors, we can pull 4 socks (since 4 < 5), so 4×3=12 socks. The next sock pulled will necessarily be the 5th of one color. So total socks needed: 12 + 1 = 13.
ANSWER 1: D
Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). Since the problem does not mention distinct positions for starters, order does not matter, so we use combinations. The number of ways to choose 3 people from 4 is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n=4 \) and \( k=3 \):
\( \binom{4}{3} = \frac{4!}{3!1!} = 4 \). This is equivalent to counting how many people we leave out (4 choices, one for each person not selected).
ANSWER 2: B
Problem 3:
We check the product of each pair:
- A: \( (-4) \times (-9) = 36 \)
- B: \( (-3) \times (-12) = 36 \)
- C: \( \frac{1}{2} \times (-72) = -36 \neq 36 \)
- D: \( 1 \times 36 = 36 \)
- E: \( \frac{3}{2} \times 24 = 36 \)
Only pair C does not have a product of 36.
ANSWER 3: C
Problem 4:
Convert all units to centimeters:
- Ceiling height: \( 2.4 \, \text{m} = 240 \, \text{cm} \)
- Light bulb height from floor: \( 240 - 10 = 230 \, \text{cm} \)
- Alice's reach from floor (without stool): \( 1.5 \, \text{m} + 0.46 \, \text{m} = 150 + 46 = 196 \, \text{cm} \)
Let \( h \) be the stool height. The total reach with the stool equals the light bulb height:
\( h + 196 = 230 \)
\( h = 230 - 196 = 34 \, \text{cm} \)
ANSWER 4: B
Problem 5:
The ratio of trout to total fish in the sample equals the ratio in the lake. The sample has \( \frac{30}{180} = \frac{1}{6} \) trout. Let \( T \) be the total number of fish in the lake. We know there are 250 trout, so:
\( \frac{250}{T} = \frac{1}{6} \)
\( T = 250 \times 6 = 1500 \)
ANSWER 5: B
Problem 6:
Karl bought 5 folders at $2.50 each. The 20% discount applies to each folder, so the savings per folder is \( 2.50 \times 0.20 = 0.50 \). Total savings for 5 folders: \( 5 \times 0.50 = 2.50 \).
ANSWER 6: C
Problem 7:
Total time to paint 300 dimples: \( 300 \times 2 = 600 \) seconds. Convert to minutes: \( 600 \div 60 = 10 \) minutes.
ANSWER 7: D
Problem 8:
Let \( x \) = number of $5 bills, \( y \) = number of $2 bills. We need \( 5x + 2y = 17 \), where \( x,y \geq 0 \) integers. Test non-negative integer values of \( x \):
- \( x=0 \): \( 2y=17 \) → \( y=8.5 \) (not integer, invalid)
- \( x=1 \): \( 5 + 2y=17 \) → \( 2y=12 \) → \( y=6 \) (valid: 1 $5 + 6 $2 = $17)
- \( x=2 \): \( 10 + 2y=17 \) → \( 2y=7 \) → \( y=3.5 \) (invalid)
- \( x=3 \): \( 15 + 2y=17 \) → \( 2y=2 \) → \( y=1 \) (valid: 3 $5 + 1 $2 = $17)
- \( x=4 \): \( 20 > 17 \) (stop, no more valid \( x \))
Only 2 valid combinations.
ANSWER 8: A
Problem 9:
Convert 1000 minutes to hours and minutes: \( 1000 \div 60 = 16 \) hours with a remainder of 40 minutes (since \( 16 \times 60 = 960 \), \( 1000 - 960 = 40 \)). The contest starts at noon, so adding 16 hours 40 minutes: noon + 12 hours = midnight, plus 4 hours 40 minutes = 4:40 a.m.
ANSWER 9: D
Problem 10:
Use order of operations (multiplication before addition):
First parentheses: \( 8 \times 4 + 2 = 32 + 2 = 34 \)
Second parentheses: \( 8 + 4 \times 2 = 8 + 8 = 16 \)
Subtract: \( 34 - 16 = 18 \)
ANSWER 10: D
Problem 11:
Convert 11 hours 5 minutes to total minutes: \( 11 \times 60 + 5 = 660 + 5 = 665 \)
ANSWER 11: C
Problem 12:
Georgie chooses an entrance window first (6 options), then a different exit window (5 remaining options, since exit cannot equal entrance). Since entrance and exit are distinct actions, order matters, so total ways: \( 6 \times 5 = 30 \).
ANSWER 12: D

I'll work through each problem systematically.

**Problem 1:**

**What is being asked:** How many socks must be removed to guarantee 5 socks of the same color?

**Setup:** There are 3 colors (red, white, blue). In the worst case scenario, the Martian pulls out socks as evenly distributed among colors as possible before getting 5 of one color.

**Working:** 
- Worst case: 4 red, 4 white, 4 blue = 12 socks (still no color with 5)
- The 13th sock must give us a 5th sock of some color

**Answer 1: D**

---

**Problem 2:**

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

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

**Working:**
- C(4,3) = 4 ways
- The 4 ways correspond to leaving out each person: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred}

**Answer 2: B**

---

**Problem 3:**

**What is being asked:** Which pair does NOT have a product equal to 36?

**Working:** Check each pair:
- A. {−4, −9}: (−4)(−9) = 36 ✓
- B. {−3, −12}: (−3)(−12) = 36 ✓
- C. {1/2, −72}: (1/2)(−72) = −36 ✗
- D. {1, 36}: (1)(36) = 36 ✓
- E. {3/2, 24}: (3/2)(24) = 36 ✓

**Answer 3: C**

---

**Problem 4:**

**What is being asked:** How high is the stool?

**Setup:** Convert all measurements to centimeters:
- Ceiling height: 240 cm
- Light bulb location: 240 − 10 = 230 cm from floor
- Alice's height: 150 cm
- Alice's reach above her head: 46 cm
- Alice's reach without stool: 150 + 46 = 196 cm

**Working:**
- With stool, Alice can just reach the bulb at 230 cm
- Stool height + 196 cm = 230 cm
- Stool height = 230 − 196 = 34 cm

**Answer 4: B**

---

**Problem 5:**

**What is being asked:** How many fish total are in the lake?

**Setup:** Use proportions. In the sample: 30 trout out of 180 fish. In the lake: 250 trout out of total fish.

**Working:**
- 30/180 = 250/x
- 1/6 = 250/x
- x = 250 × 6 = 1500 fish

**Answer 5: B**

---

**Problem 6:**

**What is being asked:** How much could Karl have saved with the 20% off sale?

**Setup:** 
- Original cost: 5 folders × $2.50 = $12.50
- Sale discount: 20% off

**Working:**
- 20% of $12.50 = 0.20 × $12.50 = $2.50

**Answer 6: C**

---

**Problem 7:**

**What is being asked:** How many minutes to paint 300 dimples at 2 seconds each?

**Working:**
- Total time: 300 × 2 = 600 seconds
- Convert to minutes: 600 ÷ 60 = 10 minutes

**Answer 7: D**

---

**Problem 8:**

**What is being asked:** How many combinations of $5 bills and $2 bills make $17?

**Setup:** Let x = number of $5 bills, y = number of $2 bills
- 5x + 2y = 17

**Working:**
- x = 0: 2y = 17 → y = 8.5 (not integer)
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 (not integer)
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x ≥ 4: 5x ≥ 20 > 17 (impossible)

Valid combinations: (1,$5 and 6,$2) and (3,$5 and 1,$2) = 2 combinations

**Answer 8: A**

---

**Problem 9:**

**What is being asked:** When did the contest end, starting at noon and lasting 1000 minutes?

**Working:**
- 1000 minutes ÷ 60 = 16 hours and 40 minutes
- Noon + 16 hours 40 minutes = 4:40 p.m. the same day
- But we need to continue: 4:40 p.m. + additional time consideration
- Actually: 1000 ÷ 60 = 16 remainder 40, so 16 hours 40 minutes
- Noon + 16 hours 40 minutes = 4:40 a.m. the next day

**Answer 9: D**

---

**Problem 10:**

**What is being asked:** Calculate (8 × 4 + 2) − (8 + 4 × 2)

**Working:**
- First parentheses: 8 × 4 + 2 = 32 + 2 = 34
- Second parentheses: 8 + 4 × 2 = 8 + 8 = 16
- Result: 34 − 16 = 18

**Answer 10: D**

---

**Problem 11:**

**What is being asked:** Convert 11 hours 5 minutes to minutes

**Working:**
- 11 hours = 11 × 60 = 660 minutes
- Total: 660 + 5 = 665 minutes

**Answer 11: C**

---

**Problem 12:**

**What is being asked:** In how many ways can the ghost enter by one window and leave by a different window?

**Working:**
- Choices for entry: 6 windows
- Choices for exit: 5 windows (must be different from entry)
- Total ways: 6 × 5 = 30

**Answer 12: D**

Problem 1:

To be certain of getting 5 socks of the same color, we use the pigeonhole principle.

There are 3 colors: red, white, and blue.  
To avoid getting 5 of any one color for as long as possible, Martian could take at most 4 of each color:

- 4 red
- 4 white
- 4 blue

That gives a total of \(4+4+4=12\) socks without yet having 5 of the same color.

The very next sock, the 13th, must make one color appear at least 5 times.

So the Martian must remove 13 socks.

ANSWER 1: D

Problem 2:

There are 4 players: Lance, Sally, Joy, and Fred.

We need to choose 3 starters from these 4 players. Since order does not matter, we count combinations:

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

Indeed, the 3-starter groups are:

- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred

So there are 4 ways.

ANSWER 2: B

Problem 3:

We check each product.

- A: \((-4)(-9)=36\)
- B: \((-3)(-12)=36\)
- C: \(\frac12 \cdot (-72)=-36\), not 36
- D: \(1\cdot 36=36\)
- E: \(\frac32 \cdot 24=36\)

Only choice C does **not** have product 36.

ANSWER 3: C

Problem 4:

First find how high the bulb is above the floor.

The ceiling is 2.4 meters above the floor, which is \(240\) centimeters.  
The bulb is 10 centimeters below the ceiling, so it is:

\[
240 - 10 = 230 \text{ cm}
\]

Now Alice’s height is 1.5 meters = 150 cm.  
She can reach 46 cm above her head, so her reach without the stool is:

\[
150 + 46 = 196 \text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230 - 196 = 34 \text{ cm}
\]

So the stool is 34 cm high.

ANSWER 4: B

Problem 5:

The sample has 30 trout out of 180 fish, so the fraction of trout is:

\[
\frac{30}{180}=\frac{1}{6}
\]

If 250 trout is \(\frac16\) of the total lake population, then the total number of fish is:

\[
250 \div \frac16 = 250 \cdot 6 = 1500
\]

So there are 1500 fish in the lake.

ANSWER 5: B

Problem 6:

Karl bought 5 folders at \$2.50 each, so the total cost was:

\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:

\[
0.20 \times 12.50 = 2.50
\]

So he could have saved \$2.50.

ANSWER 6: C

Problem 7:

He paints 300 dimples, and each dimple takes 2 seconds.

Total time in seconds:

\[
300 \times 2 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10
\]

So he needs 10 minutes.

ANSWER 7: D

Problem 8:

We want combinations of 5-dollar bills and 2-dollar bills that total \$17.

Let \(x\) be the number of \$5 bills and \(y\) the number of \$2 bills. Then:

\[
5x + 2y = 17
\]

Try values of \(x\):

- \(x=1\): \(17-5=12\), so \(y=6\)
- \(x=2\): \(17-10=7\), not even, so impossible
- \(x=3\): \(17-15=2\), so \(y=1\)

These are the only nonnegative integer solutions.

So there are 2 combinations.

ANSWER 8: A

Problem 9:

The contest ended 1000 minutes after noon.

Convert 1000 minutes to hours and minutes:

\[
1000 \div 60 = 16 \text{ hours with } 40 \text{ minutes left}
\]

So:

- Noon + 12 hours = midnight
- plus 4 more hours = 4:00 a.m.
- plus 40 minutes = 4:40 a.m.

So the contest ended at 4:40 a.m.

ANSWER 9: D

Problem 10:

Compute each part using order of operations:

\[
(8\times 4 + 2) - (8 + 4\times 2)
\]

First parentheses:

\[
8\times 4 + 2 = 32 + 2 = 34
\]

Second parentheses:

\[
8 + 4\times 2 = 8 + 8 = 16
\]

Now subtract:

\[
34 - 16 = 18
\]

So the value is 18.

ANSWER 10: D

Problem 11:

We convert 11 hours and 5 minutes into minutes.

\[
11 \times 60 = 660
\]

Then add 5 minutes:

\[
660 + 5 = 665
\]

So the match lasted 665 minutes.

ANSWER 11: C

Problem 12:

Georgie must enter by one window and leave by a different window.

There are 6 choices for the entering window.  
After choosing one, there are 5 remaining choices for the exit window.

So the total number of ways is:

\[
6 \times 5 = 30
\]

Therefore, there are 30 ways.

ANSWER 12: D

Problem 1:  
We need the minimum number of socks to remove to guarantee there are 5 socks of the same color, given 3 colors (red, white, blue) and at least 5 of each color.

Worst case: you could first pull 4 socks of each color without reaching 5 of any color.  
That would be:
- 4 red + 4 white + 4 blue = 12 socks pulled

After 12 socks, you have at most 4 of each color. The next sock (the 13th) must be one of the 3 colors, and by the pigeonhole principle it makes one color reach 5.

So the Martian must remove **13** socks.

Tempting wrong choices:
- 12 isn’t enough because it’s possible to have only 4 of each color.
- Any number less than 13 can’t force a fifth sock of some color.

ANSWER 1: D


Problem 2:  
We have 4 students: Lance, Sally, Joy, Fred. We must choose 3 starters.

Number of ways to choose 3 out of 4 is:
\[
\binom{4}{3} = 4
\]

(You can also think: leave out one student; there are 4 choices for who to leave out.)

ANSWER 2: B


Problem 3:  
We check which pair does **NOT** multiply to 36.

Choices:
A. \(\{-4,-9\}\): \((-4)(-9)=36\) ✅  
B. \(\{-3,-12\}\): \((-3)(-12)=36\) ✅  
C. \(\left\{\frac{1}{2},-72\right\}\): \(\frac{1}{2}\cdot(-72)=-36\) ❌ not 36  
D. \(\{1,36\}\): \(1\cdot 36=36\) ✅  
E. \(\left\{\frac{3}{2},24\right\}\): \(\frac{3}{2}\cdot 24=3\cdot 12=36\) ✅

So the one that does not have product 36 is choice C.

ANSWER 3: C


Problem 4:  
Let’s convert everything to the same reference (floor).

- Ceiling is 2.4 m above the floor = 240 cm  
- The bulb is 10 cm below the ceiling: height of bulb above floor:
\[
240 - 10 = 230 \text{ cm}
\]
- Alice is 1.5 m tall = 150 cm  
- She can reach 46 cm above the top of her head, so from her top to the bulb:
\[
150 + 46 = 196 \text{ cm (if she stands on the floor)}
\]

She is standing on a stool, so her total height is (stool height + 150). She can reach:
\[
(\text{stool height} + 150) + 46
\]
Set equal to bulb height 230:
\[
\text{stool height} + 196 = 230
\]
\[
\text{stool height} = 34
\]

ANSWER 4: B


Problem 5:  
We know:
- Lake has 250 trout.
- Total fish in lake = \(T\)
- Sample has 180 fish, and 30 are trout.
So sample trout-to-total ratio equals lake trout-to-total ratio:
\[
\frac{30}{180} = \frac{250}{T}
\]
Simplify \(\frac{30}{180}=\frac{1}{6}\):
\[
\frac{1}{6} = \frac{250}{T}
\]
\[
T = 250 \cdot 6 = 1500
\]

ANSWER 5: B


Problem 6:  
Cost per folder originally: $2.50  
He bought 5 folders:
\[
5 \cdot 2.50 = 12.50
\]
A 20%-off sale means paying 80% of the original price:
\[
0.80 \cdot 12.50 = 10.00
\]
Savings:
\[
12.50 - 10.00 = 2.50
\]

ANSWER 6: C


Problem 7:  
Paint 300 dimples, 2 seconds per dimple.
Total time in seconds:
\[
300 \cdot 2 = 600 \text{ seconds}
\]
Convert to minutes (60 seconds per minute):
\[
600 \div 60 = 10 \text{ minutes}
\]

ANSWER 7: D


Problem 8:  
We need numbers of $5 bills and $2 bills that total $17 (order doesn’t matter).

Let \(x\) = number of $5 bills, \(y\) = number of $2 bills.
\[
5x + 2y = 17
\]
Try \(x=0\) to \(x=3\) (since \(4\cdot 5=20>17\)):

- \(x=0\): \(2y=17\) not possible  
- \(x=1\): \(5 + 2y=17 \Rightarrow 2y=12 \Rightarrow y=6\)  
- \(x=2\): \(10 + 2y=17 \Rightarrow 2y=7\) not possible  
- \(x=3\): \(15 + 2y=17 \Rightarrow 2y=2 \Rightarrow y=1\)

So there are 2 combinations: \((x,y)=(1,6)\) and \((3,1)\).

ANSWER 8: A


Problem 9:  
The contest starts at noon. It ends 1000 minutes later.

Convert 1000 minutes to hours:
\[
1000 \div 60 = 16 \text{ hours remainder } 40 \text{ minutes}
\]
Noon + 16 hours = 4:00 a.m. (next day)  
Add 40 minutes:
\[
4:00 \text{ a.m.} + 40 \text{ min} = 4:40 \text{ a.m.}
\]

ANSWER 9: D


Problem 10:  
Compute:
\[
(8 \times 4 + 2) - (8 + 4 \times 2)
\]
First parentheses:
\[
8\times 4 = 32,\quad 32+2=34
\]
Second parentheses:
\[
4\times 2 = 8,\quad 8+8=16
\]
Now subtract:
\[
34 - 16 = 18
\]

ANSWER 10: D


Problem 11:  
11 hours and 5 minutes total.
Convert hours to minutes:
\[
11 \cdot 60 = 660
\]
Add 5 minutes:
\[
660 + 5 = 665
\]

ANSWER 11: C


Problem 12:  
There are 6 windows. Georgie enters through one window and must leave through a different window.

Choose entering window: 6 choices.  
Choose leaving window: 5 choices (cannot be the same).

Total:
\[
6 \cdot 5 = 30
\]

ANSWER 12: D

Problem 1:
To be certain of having 5 socks of the same color, we consider the "worst-case scenario" (Pigeonhole Principle). If we have 3 colors (red, white, blue), we could pick 4 red, 4 white, and 4 blue socks without having 5 of any one color. This totals $4 + 4 + 4 = 12$ socks. The very next sock (the 13th) must be either red, white, or blue, which will complete a set of 5 for that color.
ANSWER 1: D

Problem 2:
We need to choose 3 starters from a group of 4 people (Lance, Sally, Joy, Fred). This is a combination problem denoted as $\binom{4}{3}$. Choosing 3 people to be on the team is the same as choosing 1 person to be left out. There are 4 ways to choose who is left out (Lance, Sally, Joy, or Fred).
ANSWER 2: B

Problem 3:
We check the products:
A: $(-4) \times (-9) = 36$
B: $(-3) \times (-12) = 36$
C: $\frac{1}{2} \times (-72) = -36$
D: $1 \times 36 = 36$
E: $\frac{3}{2} \times 24 = 3 \times 12 = 36$
Choice C results in $-36$, not $36$.
ANSWER 3: C

Problem 4:
Ceiling height = 2.4 m = 240 cm. Light bulb height = 240 - 10 = 230 cm from the floor.
Alice's reach = 150 cm (height) + 46 cm (reach) = 196 cm.
Stool height = (Bulb height) - (Alice's reach) = 230 - 196 = 34 cm.
ANSWER 4: B

Problem 5:
Let $T$ be the total number of fish in the lake. The ratio of trout to total fish is $\frac{30}{180} = \frac{1}{6}$.
Set up the proportion: $\frac{250}{T} = \frac{1}{6}$.
$T = 250 \times 6 = 1500$.
ANSWER 5: B

Problem 6:
Total cost = 5 folders × $2.50 = $12.50.
Savings = 20% of $12.50 = 0.20 \times 12.50 = $2.50.
Alternatively, 20% off one folder is $0.20 \times 2.50 = $0.50. For 5 folders, $0.50 \times 5 = $2.50.
ANSWER 6: C

Problem 7:
Total time in seconds = 300 dimples × 2 seconds/dimple = 600 seconds.
Total time in minutes = 600 seconds / 60 seconds/minute = 10 minutes.
ANSWER 7: D

Problem 8:
Let $x$ be the number of $5 bills and $y$ be the number of $2 bills. $5x + 2y = 17$.
If $x=1$, $5+2y=17 \Rightarrow 2y=12 \Rightarrow y=6$. (1, 6)
If $x=3$, $15+2y=17 \Rightarrow 2y=2 \Rightarrow y=1$. (3, 1)
If $x=0$ or $x=2$, $2y$ would be 17 or 7, which are not divisible by 2.
There are 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours and 40 minutes (since $1000 \div 60 = 16$ remainder 40).
Noon + 12 hours = midnight.
Midnight + 4 hours and 40 minutes = 4:40 a.m.
ANSWER 9: D

Problem 10:
$(8 \times 4 + 2) - (8 + 4 \times 2) = (32 + 2) - (8 + 8) = 34 - 16 = 18$.
ANSWER 10: D

Problem 11:
11 hours = $11 \times 60 = 660$ minutes.
Total = 660 + 5 = 665 minutes.
ANSWER 11: C

Problem 12:
There are 6 choices for the entry window. Since the exit window must be different, there are 5 choices for the exit window.
Total ways = $6 \times 5 = 30$.
ANSWER 12: D

Problem 1:
This is a pigeonhole principle problem. To guarantee at least 5 socks of one color, first consider the worst-case scenario of taking as many socks as possible without having 5 of any single color. That means taking exactly 4 socks of each of the 3 colors, for a total of \(4 \times 3 = 12\) socks. The next sock (the 13th) must be a 5th sock of one of the colors. Thus 13 socks are required. (Fewer than 13 leaves open the possibility of only 4 of each color.)

ANSWER 1: D

Problem 2:
The number of ways to choose 3 starters out of the 4 available players is the combination \(\binom{4}{3} = 4\). Each choice simply leaves out one of the four players.

ANSWER 2: B

Problem 3:
Compute the product for each pair:
- A: \((-4) \times (-9) = 36\)
- B: \((-3) \times (-12) = 36\)
- C: \(\frac12 \times (-72) = -36\) (not 36)
- D: \(1 \times 36 = 36\)
- E: \(\frac32 \times 24 = 36\)

Only pair C fails to give a product of 36.

ANSWER 3: C

Problem 4:
Convert all lengths to centimeters. The ceiling is 240 cm above the floor, so the bulb is 230 cm above the floor. Alice's reach is \(150 + 46 = 196\) cm. The stool must therefore supply the remaining \(230 - 196 = 34\) cm.

ANSWER 4: B

Problem 5:
The sample shows that trout comprise \(\frac{30}{180} = \frac16\) of all fish. With 250 trout in the lake, the total number of fish is \(250 \div \frac16 = 1500\).

ANSWER 5: B

Problem 6:
Five folders cost \(5 \times 2.50 = 12.50\). A 20 % discount saves \(0.20 \times 12.50 = 2.50\).

ANSWER 6: C

Problem 7:
Painting 300 dimples at 2 seconds each requires \(300 \times 2 = 600\) seconds, or \(600 \div 60 = 10\) minutes.

ANSWER 7: D

Problem 8:
Solve \(5x + 2y = 17\) for non-negative integers \(x, y\):
- \(x = 1\), \(y = 6\) works;
- \(x = 3\), \(y = 1\) works.
No other non-negative integer solutions exist, so there are exactly two combinations.

ANSWER 8: A

Problem 9:
Divide 1000 minutes by 60: \(1000 = 16\) hours + 40 minutes. Adding 16 hours to noon yields 4:00 a.m.; adding 40 minutes yields 4:40 a.m.

ANSWER 9: D

Problem 10:
Left side: \(8 \times 4 + 2 = 34\). Right side: \(8 + 4 \times 2 = 16\). Difference: \(34 - 16 = 18\).

ANSWER 10: D

Problem 11:
Convert 11 hours 5 minutes to minutes: \(11 \times 60 + 5 = 665\).

ANSWER 11: C

Problem 12:
There are 6 choices for the entry window and then 5 remaining choices for the exit window, giving \(6 \times 5 = 30\) ordered pairs.

ANSWER 12: D

Problem 1:
The problem is asking us to determine the minimum number of socks the Martian must remove from the drawer to guarantee that there are 5 socks of the same color. 
To ensure that there are 5 socks of the same color, we must consider the worst-case scenario where the Martian draws as many socks as possible without getting 5 of the same color. 
The Martian can draw at most 4 socks of each color without having 5 socks of the same color. 
So, the Martian can draw 4 red, 4 white, and 4 blue socks, for a total of 4 + 4 + 4 = 12 socks. 
At this point, the Martian still doesn't have 5 socks of the same color. 
However, drawing one more sock will guarantee that there are 5 socks of the same color. 
Therefore, the Martian must draw at least 12 + 1 = 13 socks.
Let's examine the choices: 
A. 6 is too low because the Martian could draw 3 red, 2 white, and 1 blue, or other combinations that don't guarantee 5 socks of the same color. 
B. 9 is still too low because the Martian could draw 4 red, 3 white, and 2 blue, or other similar combinations. 
C. 12 is the maximum number of socks the Martian can draw without having 5 socks of the same color. 
D. 13 is the correct answer because it is one more than the maximum number of socks that can be drawn without having 5 socks of the same color. 
E. 15 is more than necessary.
ANSWER 1: D

Problem 2:
The problem is asking us to find the number of ways to choose 3 starters from a team of 4 players: Lance, Sally, Joy, and Fred. 
This is a combination problem, because the order in which the starters are chosen does not matter. 
The number of ways to choose 3 starters from 4 players can be calculated using the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of players and k is the number of starters. 
Here, n = 4 and k = 3, so C(4, 3) = 4! / (3!(4-3)!) = 4.
Let's list the possible combinations to verify: 
- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred
There are indeed 4 combinations.
Let's examine the choices: 
A. 2 is too low. 
B. 4 matches our calculation. 
C. 6 is too high. 
D. 8 is too high. 
E. 10 is too high.
ANSWER 2: B

Problem 3:
The problem is asking us to identify the pair of numbers that does NOT have a product equal to 36. 
Let's check each pair: 
A. {−4, −9}: (−4) × (−9) = 36
B. {−3, −12}: (−3) × (−12) = 36
C. { 1 2 , −72}: (1/2) × (−72) = −36
D. {1, 36}: 1 × 36 = 36
E. { 3 2 , 24}: (3/2) × 24 = 36
The pair that does NOT have a product equal to 36 is C.
Let's examine the choices: 
A. {−4, −9} has a product of 36. 
B. {−3, −12} has a product of 36. 
C. { 1 2 , −72} has a product of −36, not 36. 
D. {1, 36} has a product of 36. 
E. { 3 2 , 24} has a product of 36.
ANSWER 3: C

Problem 4:
The problem is asking us to find the height of the stool that Alice needs to stand on to reach the light bulb. 
First, we need to convert the height of the ceiling from meters to centimeters: 2.4 meters = 240 centimeters. 
The light bulb is 10 centimeters below the ceiling, so its height above the floor is 240 - 10 = 230 centimeters. 
Alice is 1.5 meters = 150 centimeters tall and can reach 46 centimeters above her head, so her total reach is 150 + 46 = 196 centimeters. 
To find the height of the stool, we subtract Alice's total reach from the height of the light bulb: 230 - 196 = 34 centimeters.
Let's examine the choices: 
A. 32 is too low because 230 - (196 + 32) = 2, which is not enough to reach the light bulb. 
B. 34 is the correct height. 
C. 36 is too high. 
D. 38 is too high. 
E. 40 is too high.
ANSWER 4: B

Problem 5:
The problem is asking us to find the total number of fish in the lake based on the ratio of trout in the sample to the total number of fish in the sample. 
The sample contains 30 trout out of 180 fish, so the ratio of trout to total fish in the sample is 30/180. 
We are given that this ratio is the same as the ratio of trout to total fish in the lake. 
Let's denote the total number of fish in the lake as x. 
The ratio of trout to total fish in the lake is 250/x. 
Setting the two ratios equal, we get 30/180 = 250/x. 
Cross-multiplying, we have 30x = 250 × 180. 
Simplifying, 30x = 45000. 
Dividing both sides by 30, x = 1500.
Let's examine the choices: 
A. 1250 is too low. 
B. 1500 matches our calculation. 
C. 1750 is too high. 
D. 1800 is too high. 
E. 2000 is too high.
ANSWER 5: B

Problem 6:
The problem is asking us to find out how much Karl could have saved by waiting a day to buy the folders at a 20%-off sale. 
The original price of 5 folders is 5 × $2.50 = $12.50. 
At a 20%-off sale, the discount is 20% of $12.50, which is 0.20 × $12.50 = $2.50. 
So, Karl could have saved $2.50.
Let's examine the choices: 
A. $1.00 is too low. 
B. $2.00 is too low. 
C. $2.50 matches our calculation. 
D. $2.75 is too high. 
E. $5.00 is too high.
ANSWER 6: C

Problem 7:
The problem is asking us to find the time it will take Casey to paint 300 dimples on a golf ball. 
Casey takes 2 seconds to paint one dimple, so to paint 300 dimples, he will take 300 × 2 = 600 seconds. 
To convert seconds to minutes, we divide by 60: 600 / 60 = 10 minutes.
Let's examine the choices: 
A. 4 minutes is too low. 
B. 6 minutes is too low. 
C. 8 minutes is too low. 
D. 10 minutes matches our calculation. 
E. 12 minutes is too high.
ANSWER 7: D

Problem 8:
The problem is asking us to find the number of different combinations of $5 bills and $2 bills that can be used to make a total of $17. 
Let's list the possible combinations: 
1. 3 $5 bills = $15, and $2 bill = $2, total = $17 (1 combination: 3 $5 bills and 1 $2 bill)
2. 1 $5 bill = $5, and 6 $2 bills = $12, total = $17 (1 combination: 1 $5 bill and 6 $2 bills)
We should also check if there are any other combinations: 
- Using 0 $5 bills, we need 8.5 $2 bills, which is not possible since we can't have a fraction of a bill.
- Using 2 $5 bills = $10, we need $7 more, which can be made with 3.5 $2 bills, again not possible.
- Using 4 $5 bills = $20, which is more than $17.

Problem 1:
We want the minimum number of socks to guarantee 5 of the same color. With 3 colors (red, white, blue), the worst case is picking 4 of each color: 4+4+4 = 12 socks. The 13th sock must create a group of 5 of one color. So 13 socks are needed.
ANSWER 1: D

Problem 2:
We choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways is the combination C(4,3) = 4.
ANSWER 2: B

Problem 3:
Compute each product:
A: (−4)×(−9) = 36
B: (−3)×(−12) = 36
C: (1/2)×(−72) = −36
D: 1×36 = 36
E: (3/2)×24 = 36
Only C does not equal 36.
ANSWER 3: C

Problem 4:
Ceiling = 2.4 m = 240 cm. Light bulb is 10 cm below ceiling, so its height = 230 cm.
Alice's height = 1.5 m = 150 cm. Reach above head = 46 cm, so reach without stool = 196 cm.
Stool height needed = 230 − 196 = 34 cm.
ANSWER 4: B

Problem 5:
Sample ratio: 30 trout / 180 fish = 1/6. Lake has 250 trout, which is 1/6 of total fish. Total fish = 250 × 6 = 1500.
ANSWER 5: B

Problem 6:
Cost without sale: 5 × $2.50 = $12.50.
With 20% off, each folder costs $2.50 × 0.8 = $2.00. Total = 5 × $2.00 = $10.00.
Savings = $12.50 − $10.00 = $2.50.
ANSWER 6: C

Problem 7:
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.
ANSWER 7: D

Problem 8:
Find non-negative integers x ($5 bills) and y ($2 bills) such that 5x + 2y = 17.
x=1 → 5 + 2y = 17 → y=6.
x=3 → 15 + 2y = 17 → y=1.
No other x works. So 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours 40 minutes (16×60 = 960, remainder 40).
Noon + 16 hours = 4:00 a.m. next day. Add 40 minutes → 4:40 a.m.
ANSWER 9: D

Problem 10:
(8 × 4 + 2) = 32 + 2 = 34.
(8 + 4 × 2) = 8 + 8 = 16.
34 − 16 = 18.
ANSWER 10: D

Problem 11:
11 hours × 60 = 660 minutes. Plus 5 minutes = 665 minutes.
ANSWER 11: C

Problem 12:
6 choices for entry window, then 5 choices for a different exit window. Total = 6 × 5 = 30.
ANSWER 12: D

Problem 1:
We need to find the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). Using the Pigeonhole Principle, we consider the worst-case scenario: the Martian pulls out as many socks as possible *without* getting 5 of any one color. That means getting exactly 4 of each color:
- 4 red + 4 white + 4 blue = 12 socks
After 12 socks, the Martian could have 4 of each color and still not have 5 matching. The very next sock (the 13th) must be red, white, or blue, pushing the count of that color to 5. So 13 socks guarantees 5 of the same color.
(The "five-legged" detail is flavor text explaining why the Martian needs 5 socks.)

ANSWER 1: D

Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). This is a combination problem since order doesn't matter:
C(4, 3) = 4! / (3! × 1!) = 4
We can also list them: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — exactly 4 ways.

ANSWER 2: B

Problem 3:
We check each product:
- A: (−4) × (−9) = 36 ✓
- B: (−3) × (−12) = 36 ✓
- C: (1/2) × (−72) = −36 ✗ (This equals −36, NOT 36)
- D: 1 × 36 = 36 ✓
- E: (3/2) × 24 = 36 ✓
Choice C gives −36, which does not equal 36.

ANSWER 3: C

Problem 4:
First, convert everything to centimeters:
- Ceiling height: 2.4 m = 240 cm
- Light bulb position: 240 − 10 = 230 cm above the floor
- Alice's height: 1.5 m = 150 cm
- Alice's reach above her head: 46 cm
- Alice's maximum reach (without stool): 150 + 46 = 196 cm above the floor

Let h = height of the stool. Standing on the stool, her reach is h + 196 cm. She can "just reach" the bulb:
h + 196 = 230
h = 230 − 196 = 34 cm

ANSWER 4: B

Problem 5:
The ratio of trout to total fish in the sample should equal the ratio in the lake:
- Sample ratio: 30 trout / 180 fish = 1/6
- Lake ratio: 250 trout / Total fish = 1/6

Setting up the proportion: 250 / Total = 1/6
Total = 250 × 6 = 1500 fish

ANSWER 5: B

Problem 6:
Karl bought 5 folders at $2.50 each:
Total cost = 5 × $2.50 = $12.50
A 20%-off sale means he would save 20% of the total:
Savings = 0.20 × $12.50 = $2.50

ANSWER 6: C

Problem 7:
Total time in seconds: 300 dimples × 2 seconds/dimple = 600 seconds
Convert to minutes: 600 seconds ÷ 60 seconds/minute = 10 minutes

ANSWER 7: D

Problem 8:
We need non-negative integer solutions to 5x + 2y = 17, where x = number of $5 bills and y = number of $2 bills.
- x = 0: 2y = 17 → y = 8.5 (not an integer) ✗
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 ✗
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x = 4: 20 > 17 (too large) ✗

Only 2 valid combinations: (1 five + 6 twos) and (3 fives + 1 two).

ANSWER 8: A

Problem 9:
Convert 1000 minutes to hours and minutes:
1000 ÷ 60 = 16 remainder 40, so 1000 minutes = 16 hours 40 minutes.
Starting at noon (12:00 PM):
- 12 hours later = midnight (12:00 AM)
- 4 more hours = 4:00 AM
- 40 more minutes = 4:40 AM

ANSWER 9: D

Problem 10:
Evaluate each part using order of operations (multiplication before addition):
- First part: (8 × 4 + 2) = (32 + 2) = 34
- Second part: (8 + 4 × 2) = (8 + 8) = 16
- Difference: 34 − 16 = 18

ANSWER 10: D

Problem 11:
Convert 11 hours and 5 minutes to minutes:
11 hours × 60 minutes/hour = 660 minutes
660 + 5 = 665 minutes

ANSWER 11: C

Problem 12:
Georgie enters through one window and leaves through a different one.
- Choices for entry window: 6
- Choices for exit window (must be different): 5
Total ways = 6 × 5 = 30

ANSWER 12: D

Problem 1:
The Martian needs to be **certain** of having 5 socks of the same color. To guarantee this, we must consider the worst-case scenario: the Martian picks as many socks as possible **without** yet having 5 of any single color.  
With 3 colors (red, white, blue), the worst case is picking 4 socks of each color:  
\(4 + 4 + 4 = 12\) socks.  
At this point the Martian has 4 of every color but still does not have 5 of any one color. The very next sock—the 13th—must be red, white, or blue, and will necessarily create a set of 5 of that color.  
Thus 13 socks are required.  
(Choice A is too small; with 6 socks one could have 2,2,2. Choice B allows 3,3,3. Choice C allows 4,4,4. Choice E is more than the minimum guaranteed number.)

ANSWER 1: D

---

Problem 2:
We need to choose a group of 3 starters from the 4 students: Lance, Sally, Joy, and Fred. Since the three starters form a team and order does not matter, this is a combination problem.  
\[
\binom{4}{3} = \frac{4!}{3!\,1!} = 4
\]  
Equivalently, we can think of choosing which **one** of the four students will **not** start; there are 4 such choices.  
(Choice A is too small. Choice C is \(\binom{4}{2}\). Choices D and E do not arise from counting 3-person teams from 4 people.)

ANSWER 2: B

---

Problem 3:
We check the product of each pair to see which is not 36.

- **A.** \((-4)(-9) = 36\)  
- **B.** \((-3)(-12) = 36\)  
- **C.** \(\left(\frac12\right)(-72) = -36\)  
- **D.** \((1)(36) = 36\)  
- **E.** \(\left(\frac32\right)(24) = 36\)

Only pair **C** gives \(-36\), not \(36\).

ANSWER 3: C

---

Problem 4:
First convert all measurements to centimeters.  
- Ceiling height: \(2.4\text{ m} = 240\text{ cm}\).  
- Light bulb is \(10\text{ cm}\) below the ceiling, so its height above the floor is \(240 - 10 = 230\text{ cm}\).  

Alice’s reach:  
- Height: \(1.5\text{ m} = 150\text{ cm}\).  
- She can reach \(46\text{ cm}\) above her head, so her standing reach is \(150 + 46 = 196\text{ cm}\).  

The stool must supply the remaining height:  
\[
230 - 196 = 34\text{ cm}.
\]

ANSWER 4: B

---

Problem 5:
Let \(T\) be the total number of fish in the lake. The problem states that the ratio of trout to total fish is the same in the sample and in the lake.  
\[
\frac{250}{T} = \frac{30}{180}
\]  
Simplify the right side: \(\frac{30}{180} = \frac16\).  
\[
\frac{250}{T} = \frac16 \quad\Longrightarrow\quad T = 250 \times 6 = 1500.
\]

ANSWER 5: B

---

Problem 6:
Karl bought 5 folders at \(\$2.50\) each.  
Total cost: \(5 \times \$2.50 = \$12.50\).  
A 20% discount would have saved him 20% of this total:  
\[
0.20 \times \$12.50 = \$2.50.
\]

ANSWER 6: C

---

Problem 7:
Casey paints 300 dimples at 2 seconds per dimple.  
Total time in seconds: \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(\frac{600}{60} = 10\) minutes.

ANSWER 7: D

---

Problem 8:
Let \(f\) be the number of \(\$5\) bills and \(t\) be the number of \(\$2\) bills. We need non-negative integer solutions to  
\[
5f + 2t = 17.
\]  
Test values of \(f\):  
- \(f=0\): \(2t=17\) → no integer solution.  
- \(f=1\): \(5 + 2t = 17 \Rightarrow t=6\).  **(Valid)**
- \(f=2\): \(10 + 2t = 17 \Rightarrow t=3.5\).  (Invalid)
- \(f=3\): \(15 + 2t = 17 \Rightarrow t=1\).  **(Valid)**
- \(f=4\): \(20 > 17\) → stop.

There are exactly **2** combinations: \((1\text{ five}, 6\text{ twos})\) and \((3\text{ fives}, 1\text{ two})\).

ANSWER 8: A

---

Problem 9:
The contest starts at noon (12:00 p.m.). We add 1000 minutes.  
Divide 1000 by 60:  
\[
1000 = 16 \times 60 + 40,
\]  
so 1000 minutes = 16 hours and 40 minutes.  

Noon + 12 hours = midnight.  
Midnight + 4 hours 40 minutes = 4:40 a.m.

ANSWER 9: D

---

Problem 10:
Evaluate using the correct order of operations (multiplication before addition).

First parentheses:  
\[
8 \times 4 + 2 = 32 + 2 = 34.
\]

Second parentheses:  
\[
8 + 4 \times 2 = 8 + 8 = 16.
\]

Subtract:  
\[
34 - 16 = 18.
\]

ANSWER 10: D

---

Problem 11:
Convert 11 hours to minutes and add the remaining 5 minutes.  
\[
11 \times 60 = 660\text{ minutes}.
\]  
\[
660 + 5 = 665\text{ minutes}.
\]

ANSWER 11: C

---

Problem 12:
Georgie enters through one of 6 windows and must leave through a **different** window.  
- 6 choices for the entry window.  
- For each entry choice, 5 remaining choices for the exit window.  

Total ways:  
\[
6 \times 5 = 30.
\]  
(Choice E, 36, would count entering and leaving through the same window, which is not allowed.)

ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). The worst-case scenario is to draw as many socks as possible without having 5 of any one color. That would be 4 red, 4 white, and 4 blue, for a total of 12 socks. The next sock drawn (the 13th) must be one of the three colors, giving that color 5 socks. Therefore, 13 socks are required. The five-legged Martian is irrelevant to the count.  
Answer: D

Problem 2:
We have 4 people (Lance, Sally, Joy, Fred) and need to choose 3 starters. Since the starters are not assigned to specific positions, order does not matter. The number of ways is the combination \( \binom{4}{3} = 4 \).  
Answer: B

Problem 3:
Compute each product:  
A. \((-4) \times (-9) = 36\)  
B. \((-3) \times (-12) = 36\)  
C. \(\frac{1}{2} \times (-72) = -36\)  
D. \(1 \times 36 = 36\)  
E. \(\frac{3}{2} \times 24 = 36\)  
Only pair C gives a product not equal to 36.  
Answer: C

Problem 4:
Convert all measurements to centimeters:  
Ceiling height = 2.4 m = 240 cm.  
Light bulb is 10 cm below ceiling → height = 240 – 10 = 230 cm.  
Alice’s height = 1.5 m = 150 cm.  
Her reach above her head = 46 cm, so her maximum reach without a stool = 150 + 46 = 196 cm.  
Let \(h\) be the stool height. Standing on the stool raises her reach to \(196 + h\). This must equal 230 cm:  
\(196 + h = 230\) → \(h = 34\) cm.  
Answer: B

Problem 5:
In the sample, 30 out of 180 fish are trout, so the ratio trout : total = \(30/180 = 1/6\).  
Assuming the same ratio in the lake, if the lake has 250 trout, then the total number of fish is \(250 \times 6 = 1500\).  
Answer: B

Problem 6:
Original cost for 5 folders: \(5 \times 2.50 = \$12.50\).  
A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
Thus, by waiting a day, Karl could have saved $2.50.  
Answer: C

Problem 7:
Time to paint one dimple = 2 seconds.  
Total time for 300 dimples = \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(600 \div 60 = 10\) minutes.  
Answer: D

Problem 8:
We need nonnegative integers \(a\) (number of $5 bills) and \(b\) (number of $2 bills) such that \(5a + 2b = 17\).  
Solve by testing values of \(b\):  
\(b = 1\) → \(5a = 15\) → \(a = 3\) → (3,1) works.  
\(b = 6\) → \(5a = 5\) → \(a = 1\) → (1,6) works.  
No other values give integer \(a\). Thus there are 2 combinations.  
Answer: A

Problem 9:
1000 minutes = \(1000 \div 60 = 16\) hours and \(40\) minutes (since \(60 \times 16 = 960\), remainder 40).  
Starting at noon (12:00 PM), add 12 hours → midnight (12:00 AM). Add the remaining 4 hours → 4:00 AM. Then add 40 minutes → 4:40 AM.  
Answer: D

Problem 10:
Evaluate using order of operations:  
First parentheses: \(8 \times 4 + 2 = 32 + 2 = 34\).  
Second parentheses: \(8 + 4 \times 2 = 8 + 8 = 16\).  
Subtract: \(34 - 16 = 18\).  
Answer: D

Problem 11:
11 hours = \(11 \times 60 = 660\) minutes.  
Add 5 minutes → \(660 + 5 = 665\) minutes.  
Answer: C

Problem 12:
Georgie must choose an entry window (6 choices) and a different exit window (5 choices). Since entering and leaving are distinct actions, the total number of ways is \(6 \times 5 = 30\).  
Answer: D

ANSWER 1: D
ANSWER 2: B
ANSWER 3: C
ANSWER 4: B
ANSWER 5: B
ANSWER 6: C
ANSWER 7: D
ANSWER 8: A
ANSWER 9: D
ANSWER 10: D
ANSWER 11: C
ANSWER 12: D

Problem 1:  
The Martian wants to be certain of having 5 socks of the same color. There are three colors: red, white, and blue. In the worst-case scenario, the Martian could pick 4 socks of each color before getting a fifth of any one color. That would be \(4 \times 3 = 12\) socks, and still no color has 5 socks. The next sock drawn (the 13th) must complete a set of 5 of some color, because there are only three colors. Therefore, the Martian must remove **13** socks to be certain of having 5 socks of the same color.  
ANSWER 1: D

Problem 2:  
Ms. Hamilton's class has 4 available players: Lance, Sally, Joy, and Fred. The number of ways to choose 3 starters from 4 players is given by the combination \(\binom{4}{3} = 4\). Thus, there are **4** possible sets of three starters.  
ANSWER 2: B

Problem 3:  
We check each pair's product:  
- A: \((-4) \times (-9) = 36\)  
- B: \((-3) \times (-12) = 36\)  
- C: \(\frac{1}{2} \times (-72) = -36\) (not 36)  
- D: \(1 \times 36 = 36\)  
- E: \(\frac{3}{2} \times 24 = 36\)  
Only pair **C** does not have a product of 36.  
ANSWER 3: C

Problem 4:  
The ceiling is 2.4 m above the floor, and the bulb is 10 cm (0.1 m) below the ceiling, so the bulb's height is \(2.4 - 0.1 = 2.3\) m. Alice's height is 1.5 m, and she can reach 46 cm (0.46 m) above her head, so her reach height from the floor is \(1.5 + 0.46 = 1.96\) m. When standing on a stool, her hands are at the stool height plus her reach height. Setting this equal to the bulb height:  
\[
\text{stool height} + 1.96 = 2.3 \Rightarrow \text{stool height} = 0.34 \text{ m} = 34 \text{ cm}.
\]  
ANSWER 4: B

Problem 5:  
In the sample, the ratio of trout to total fish is \(\frac{30}{180} = \frac{1}{6}\). Assuming the same ratio in the lake, let \(T\) be the total number of fish in the lake. Then \(\frac{250}{T} = \frac{1}{6}\), so \(T = 250 \times 6 = 1500\).  
ANSWER 5: B

Problem 6:  
Karl bought 5 folders at $2.50 each, so the original cost is \(5 \times 2.50 = \$12.50\). With a 20% discount, the cost becomes \(12.50 \times 0.80 = \$10.00\). The savings is \(12.50 - 10.00 = \$2.50\).  
ANSWER 6: C

Problem 7:  
Painting 300 dimples at 2 seconds each takes \(300 \times 2 = 600\) seconds. Converting to minutes: \(600 \div 60 = 10\) minutes.  
ANSWER 7: D

Problem 8:  
We need nonnegative integers \(x\) (number of $5 bills) and \(y\) (number of $2 bills) such that \(5x + 2y = 17\). Solving:  
- \(y = 1 \Rightarrow 5x = 15 \Rightarrow x = 3\)  
- \(y = 6 \Rightarrow 5x = 5 \Rightarrow x = 1\)  
- \(y = 11 \Rightarrow 5x = -5\) (invalid)  
Thus, there are **2** valid combinations: (3 fives, 1 two) and (1 five, 6 twos).  
ANSWER 8: A

Problem 9:  
1000 minutes equals \(16\) hours and \(40\) minutes (since \(16 \times 60 = 960\), remainder \(40\)). Starting at noon, adding \(16\) hours gives 4:00 AM the next day, and adding \(40\) minutes gives 4:40 AM.  
ANSWER 9: D

Problem 10:  
Compute inside parentheses:  
\(8 \times 4 + 2 = 32 + 2 = 34\)  
\(8 + 4 \times 2 = 8 + 8 = 16\)  
Then \(34 - 16 = 18\).  
ANSWER 10: D

Problem 11:  
11 hours is \(11 \times 60 = 660\) minutes. Adding 5 minutes gives \(660 + 5 = 665\) minutes.  
ANSWER 11: C

Problem 12:  
Georgie can enter through any of the 6 windows. For exiting, he must choose a different window, so there are 5 choices. The total number of ways is \(6 \times 5 = 30\).  
ANSWER 12: D

Problem 1:  
The Martian needs to ensure there are 5 socks of the same color. The worst-case scenario is pulling the maximum number of socks without 5 of any color. Since there are 3 colors, this would be 4 socks of each color (4 × 3 = 12 socks). The next sock (the 13th) must match one color, giving 5 socks of that color.  
**ANSWER 1: D**  

Problem 2:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is C(4,3) = 4.  
**ANSWER 2: B** (Note: C(4,3)=4, but rechecking: the team is chosen from 4 specific people, so combinations are 4. Answer B=4)  

Problem 3:  
Check each pair:  
A. (−4)(−9) = 36  
B. (−3)(−12) = 36  
C. (1/2)(−72) = -36 (≠36)  
D. (1)(36) = 36  
E. (3/2)(24) = 36  
The pair that does NOT equal 36 is C.  
**ANSWER 3: E** (Re-evaluated: The question asks which does NOT equal 36. Option E: (3/2)*24=36, so it is valid. Option C: (1/2)*(-72)=-36, so C is the correct choice for NOT equal. Correction: **ANSWER 3: C** [Original had a typo, but computation confirms C is the answer])  

*Correction: For Problem 3, Option C: (1/2, -72) likely means (1/2) × (-72) = -36 ≠ 36. Option E: (3/2) × 24 = 36. Thus, C is the answer.*  
**ANSWER 3: C**  

Problem 4:  
Ceiling height: 2.4 m = 240 cm. Light bulb location: 240 - 10 = 230 cm from floor. Alice’s reach: 150 cm + 46 cm = 196 cm. Stool height = 230 - 196 = 34 cm.  
**ANSWER 4: B**  

Problem 5:  
Ratio of trout in sample: 30/180 = 1/6. Total trout: 250. Total fish = 250 × 6 = 1500.  
**ANSWER 5: B**  

Problem 6:  
Original cost: 5 × $2.50 = $12.50. 20% off: $12.50 × 0.20 = $2.50 saved.  
**ANSWER 6: C** (Note: The question asks how much Karl could have saved, which is $2.50, not $2.00. Original had a mistake; correction: 20% of $12.50 is $2.50, so answer C)  

Problem 7:  
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.  
**ANSWER 7: D**  

Problem 8:  
Let $5 bills = x, $2 bills = y. 5x + 2y = 17. Possible solutions:  
- x=3, y=1 (15+2=17)  
- x=1, y=6 (5+12=17)  
Only 2 combinations.  
**ANSWER 8: A**  

Problem 9:  
1000 minutes = 16 hours 40 minutes. Starting at noon: noon + 16h40m = 4:40 a.m. next day.  
**ANSWER 9: D**  

Problem 10:  
(8×4 + 2) = 34, (8 + 4×2) = 16. 34 - 16 = 18.  
**ANSWER 10: D**  

Problem 11:  
11 hours = 660 minutes. 660 + 5 = 665 minutes.  
**ANSWER 11: C**  

Problem 12:  
Georgie can enter through 6 windows and exit through 5 others. Total ways: 6 × 5 = 30.  
**ANSWER 12: D**  

---  
**Final Answers**  
ANSWER 1: D  
ANSWER 2: B  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: B  
ANSWER 6: C  
ANSWER 7: D  
ANSWER 8: A  
ANSWER 9: D  
ANSWER 10: D  
ANSWER 11: C  
ANSWER 12: D

### Problem 1:
This is a classic pigeonhole principle problem. We need to find the minimum number of socks the Martian must pull to guarantee 5 socks of the same color, with 3 total colors. The worst-case scenario is that the Martian pulls as many socks as possible without getting 5 of any color: that is 4 socks of each color, for a total of 3*4=12 socks. The next sock pulled (the 13th) has to be one of the three colors, pushing that color to 5 socks. Tempting wrong answers: C (12 is only the number of socks that avoids 5 of the same color, not enough to guarantee the requirement), E (15 is the total if you took 5 of each color, not the minimum to guarantee).
ANSWER 1: D

---

### Problem 2:
We need to count the number of ways to choose 3 starters from 4 people, where order on the team does not matter. This is a combination problem: C(n,k) = n!/(k!(n-k)!). For n=4, k=3: C(4,3) = 4. We can also list the teams by who is left out: leaving out Lance, Sally, Joy, or Fred gives exactly 4 unique teams. Tempting wrong answer: C (6 is the count if order of selection mattered, which it does not for a team of starters).
ANSWER 2: B

---

### Problem 3:
We calculate the product of each pair to find which does not equal 36:
- A: (-4)*(-9)=36 ✔️
- B: (-3)*(-12)=36 ✔️
- C: (1/2)*(-72) = -36 ❌
- D: 1*36=36 ✔️
- E: (3/2)*24=36 ✔️
Only pair C has a product of -36, not 36.
ANSWER 3: C

---

### Problem 4:
First convert all units to centimeters to avoid conversion errors:
- Ceiling height: 2.4m = 240cm. Light bulb is 10cm below ceiling, so bulb height from floor = 240-10=230cm.
- Alice's height: 1.5m=150cm, her reach above her head is 46cm, so total reach without stool = 150+46=196cm.
- Stool height = 230-196=34cm.
Tempting wrong answers come from unit conversion mistakes, e.g., mixing up meters and centimeters.
ANSWER 4: B

---

### Problem 5:
This is a proportion problem. The ratio of trout to total fish is the same in the sample and the lake. In the sample, 30 trout out of 180 total fish = 1/6. Let T be total fish in the lake: 250/T = 1/6 → T=250*6=1500.
ANSWER 5: B

---

### Problem 6:
Karl bought 5 folders at $2.50 each, total original cost = 5*2.50=$12.50. He saves 20% of the total cost by waiting for the sale: 0.2*12.50=$2.50. This can also be calculated per folder: 20% of $2.50 is $0.50 saved per folder, 5*0.50=$2.50 total saved.
ANSWER 6: C

---

### Problem 7:
300 dimples * 2 seconds per dimple = 600 total seconds. Convert to minutes: 60 seconds per minute, so 600/60=10 minutes.
ANSWER 7: D

---

### Problem 8:
We find all non-negative integer pairs (x,y) where x=number of $5 bills, y=number of $2 bills, and 5x+2y=17. Only odd values of x work (since $5 is odd, $2 is even, we need an odd sum to get 17):
- x=1: 5+2y=17 → y=6 (valid)
- x=3: 15+2y=17 → y=1 (valid)
Even x values (0,2) leave an odd remainder that cannot be formed with $2 bills, so only 2 valid combinations.
ANSWER 8: A

---

### Problem 9:
Convert 1000 minutes to hours and minutes: 60 minutes per hour, 1000/60=16 hours and 40 minutes. Adding this to noon (12pm): 12pm +16h40m = 4:40am the next day.
ANSWER 9: D

---

### Problem 10:
Use order of operations (PEMDAS) to calculate:
First parenthesis: 8*4 +2 = 32+2=34
Second parenthesis: 8 +4*2=8+8=16
Subtract: 34-16=18. The tempting wrong answer C (10) comes from incorrectly adding before multiplying in the second parenthesis.
ANSWER 10: D

---

### Problem 11:
Convert 11 hours 5 minutes to total minutes: 11*60=660 minutes, plus 5 extra minutes = 665 minutes.
ANSWER 11: C

---

### Problem 12:
Georgie has 6 choices for an entry window, and 5 remaining choices for an exit window (must use a different window). Total ways = 6*5=30. Tempting wrong answer E (36) counts entering and leaving through the same window, which is not allowed; B (15) incorrectly treats entry/exit as unordered, which they are not.
ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color (red, white, blue). This is a pigeonhole principle problem: the worst-case scenario is pulling the maximum number of socks without getting 5 of any color. For each of the 3 colors, we can pull 4 socks (since 4 < 5), so 4×3=12 socks. The next sock pulled will necessarily be the 5th of one color. So total socks needed: 12 + 1 = 13.
ANSWER 1: D
Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). Since the problem does not mention distinct positions for starters, order does not matter, so we use combinations. The number of ways to choose 3 people from 4 is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n=4 \) and \( k=3 \):
\( \binom{4}{3} = \frac{4!}{3!1!} = 4 \). This is equivalent to counting how many people we leave out (4 choices, one for each person not selected).
ANSWER 2: B
Problem 3:
We check the product of each pair:
- A: \( (-4) \times (-9) = 36 \)
- B: \( (-3) \times (-12) = 36 \)
- C: \( \frac{1}{2} \times (-72) = -36 \neq 36 \)
- D: \( 1 \times 36 = 36 \)
- E: \( \frac{3}{2} \times 24 = 36 \)
Only pair C does not have a product of 36.
ANSWER 3: C
Problem 4:
Convert all units to centimeters:
- Ceiling height: \( 2.4 \, \text{m} = 240 \, \text{cm} \)
- Light bulb height from floor: \( 240 - 10 = 230 \, \text{cm} \)
- Alice's reach from floor (without stool): \( 1.5 \, \text{m} + 0.46 \, \text{m} = 150 + 46 = 196 \, \text{cm} \)
Let \( h \) be the stool height. The total reach with the stool equals the light bulb height:
\( h + 196 = 230 \)
\( h = 230 - 196 = 34 \, \text{cm} \)
ANSWER 4: B
Problem 5:
The ratio of trout to total fish in the sample equals the ratio in the lake. The sample has \( \frac{30}{180} = \frac{1}{6} \) trout. Let \( T \) be the total number of fish in the lake. We know there are 250 trout, so:
\( \frac{250}{T} = \frac{1}{6} \)
\( T = 250 \times 6 = 1500 \)
ANSWER 5: B
Problem 6:
Karl bought 5 folders at $2.50 each. The 20% discount applies to each folder, so the savings per folder is \( 2.50 \times 0.20 = 0.50 \). Total savings for 5 folders: \( 5 \times 0.50 = 2.50 \).
ANSWER 6: C
Problem 7:
Total time to paint 300 dimples: \( 300 \times 2 = 600 \) seconds. Convert to minutes: \( 600 \div 60 = 10 \) minutes.
ANSWER 7: D
Problem 8:
Let \( x \) = number of $5 bills, \( y \) = number of $2 bills. We need \( 5x + 2y = 17 \), where \( x,y \geq 0 \) integers. Test non-negative integer values of \( x \):
- \( x=0 \): \( 2y=17 \) → \( y=8.5 \) (not integer, invalid)
- \( x=1 \): \( 5 + 2y=17 \) → \( 2y=12 \) → \( y=6 \) (valid: 1 $5 + 6 $2 = $17)
- \( x=2 \): \( 10 + 2y=17 \) → \( 2y=7 \) → \( y=3.5 \) (invalid)
- \( x=3 \): \( 15 + 2y=17 \) → \( 2y=2 \) → \( y=1 \) (valid: 3 $5 + 1 $2 = $17)
- \( x=4 \): \( 20 > 17 \) (stop, no more valid \( x \))
Only 2 valid combinations.
ANSWER 8: A
Problem 9:
Convert 1000 minutes to hours and minutes: \( 1000 \div 60 = 16 \) hours with a remainder of 40 minutes (since \( 16 \times 60 = 960 \), \( 1000 - 960 = 40 \)). The contest starts at noon, so adding 16 hours 40 minutes: noon + 12 hours = midnight, plus 4 hours 40 minutes = 4:40 a.m.
ANSWER 9: D
Problem 10:
Use order of operations (multiplication before addition):
First parentheses: \( 8 \times 4 + 2 = 32 + 2 = 34 \)
Second parentheses: \( 8 + 4 \times 2 = 8 + 8 = 16 \)
Subtract: \( 34 - 16 = 18 \)
ANSWER 10: D
Problem 11:
Convert 11 hours 5 minutes to total minutes: \( 11 \times 60 + 5 = 660 + 5 = 665 \)
ANSWER 11: C
Problem 12:
Georgie chooses an entrance window first (6 options), then a different exit window (5 remaining options, since exit cannot equal entrance). Since entrance and exit are distinct actions, order matters, so total ways: \( 6 \times 5 = 30 \).
ANSWER 12: D

I'll work through each problem systematically.

**Problem 1:**

**What is being asked:** How many socks must be removed to guarantee 5 socks of the same color?

**Setup:** There are 3 colors (red, white, blue). In the worst case scenario, the Martian pulls out socks as evenly distributed among colors as possible before getting 5 of one color.

**Working:** 
- Worst case: 4 red, 4 white, 4 blue = 12 socks (still no color with 5)
- The 13th sock must give us a 5th sock of some color

**Answer 1: D**

---

**Problem 2:**

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

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

**Working:**
- C(4,3) = 4 ways
- The 4 ways correspond to leaving out each person: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred}

**Answer 2: B**

---

**Problem 3:**

**What is being asked:** Which pair does NOT have a product equal to 36?

**Working:** Check each pair:
- A. {−4, −9}: (−4)(−9) = 36 ✓
- B. {−3, −12}: (−3)(−12) = 36 ✓
- C. {1/2, −72}: (1/2)(−72) = −36 ✗
- D. {1, 36}: (1)(36) = 36 ✓
- E. {3/2, 24}: (3/2)(24) = 36 ✓

**Answer 3: C**

---

**Problem 4:**

**What is being asked:** How high is the stool?

**Setup:** Convert all measurements to centimeters:
- Ceiling height: 240 cm
- Light bulb location: 240 − 10 = 230 cm from floor
- Alice's height: 150 cm
- Alice's reach above her head: 46 cm
- Alice's reach without stool: 150 + 46 = 196 cm

**Working:**
- With stool, Alice can just reach the bulb at 230 cm
- Stool height + 196 cm = 230 cm
- Stool height = 230 − 196 = 34 cm

**Answer 4: B**

---

**Problem 5:**

**What is being asked:** How many fish total are in the lake?

**Setup:** Use proportions. In the sample: 30 trout out of 180 fish. In the lake: 250 trout out of total fish.

**Working:**
- 30/180 = 250/x
- 1/6 = 250/x
- x = 250 × 6 = 1500 fish

**Answer 5: B**

---

**Problem 6:**

**What is being asked:** How much could Karl have saved with the 20% off sale?

**Setup:** 
- Original cost: 5 folders × $2.50 = $12.50
- Sale discount: 20% off

**Working:**
- 20% of $12.50 = 0.20 × $12.50 = $2.50

**Answer 6: C**

---

**Problem 7:**

**What is being asked:** How many minutes to paint 300 dimples at 2 seconds each?

**Working:**
- Total time: 300 × 2 = 600 seconds
- Convert to minutes: 600 ÷ 60 = 10 minutes

**Answer 7: D**

---

**Problem 8:**

**What is being asked:** How many combinations of $5 bills and $2 bills make $17?

**Setup:** Let x = number of $5 bills, y = number of $2 bills
- 5x + 2y = 17

**Working:**
- x = 0: 2y = 17 → y = 8.5 (not integer)
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 (not integer)
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x ≥ 4: 5x ≥ 20 > 17 (impossible)

Valid combinations: (1,$5 and 6,$2) and (3,$5 and 1,$2) = 2 combinations

**Answer 8: A**

---

**Problem 9:**

**What is being asked:** When did the contest end, starting at noon and lasting 1000 minutes?

**Working:**
- 1000 minutes ÷ 60 = 16 hours and 40 minutes
- Noon + 16 hours 40 minutes = 4:40 p.m. the same day
- But we need to continue: 4:40 p.m. + additional time consideration
- Actually: 1000 ÷ 60 = 16 remainder 40, so 16 hours 40 minutes
- Noon + 16 hours 40 minutes = 4:40 a.m. the next day

**Answer 9: D**

---

**Problem 10:**

**What is being asked:** Calculate (8 × 4 + 2) − (8 + 4 × 2)

**Working:**
- First parentheses: 8 × 4 + 2 = 32 + 2 = 34
- Second parentheses: 8 + 4 × 2 = 8 + 8 = 16
- Result: 34 − 16 = 18

**Answer 10: D**

---

**Problem 11:**

**What is being asked:** Convert 11 hours 5 minutes to minutes

**Working:**
- 11 hours = 11 × 60 = 660 minutes
- Total: 660 + 5 = 665 minutes

**Answer 11: C**

---

**Problem 12:**

**What is being asked:** In how many ways can the ghost enter by one window and leave by a different window?

**Working:**
- Choices for entry: 6 windows
- Choices for exit: 5 windows (must be different from entry)
- Total ways: 6 × 5 = 30

**Answer 12: D**

Problem 1:

To be certain of getting 5 socks of the same color, we use the pigeonhole principle.

There are 3 colors: red, white, and blue.  
To avoid getting 5 of any one color for as long as possible, Martian could take at most 4 of each color:

- 4 red
- 4 white
- 4 blue

That gives a total of \(4+4+4=12\) socks without yet having 5 of the same color.

The very next sock, the 13th, must make one color appear at least 5 times.

So the Martian must remove 13 socks.

ANSWER 1: D

Problem 2:

There are 4 players: Lance, Sally, Joy, and Fred.

We need to choose 3 starters from these 4 players. Since order does not matter, we count combinations:

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

Indeed, the 3-starter groups are:

- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred

So there are 4 ways.

ANSWER 2: B

Problem 3:

We check each product.

- A: \((-4)(-9)=36\)
- B: \((-3)(-12)=36\)
- C: \(\frac12 \cdot (-72)=-36\), not 36
- D: \(1\cdot 36=36\)
- E: \(\frac32 \cdot 24=36\)

Only choice C does **not** have product 36.

ANSWER 3: C

Problem 4:

First find how high the bulb is above the floor.

The ceiling is 2.4 meters above the floor, which is \(240\) centimeters.  
The bulb is 10 centimeters below the ceiling, so it is:

\[
240 - 10 = 230 \text{ cm}
\]

Now Alice’s height is 1.5 meters = 150 cm.  
She can reach 46 cm above her head, so her reach without the stool is:

\[
150 + 46 = 196 \text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230 - 196 = 34 \text{ cm}
\]

So the stool is 34 cm high.

ANSWER 4: B

Problem 5:

The sample has 30 trout out of 180 fish, so the fraction of trout is:

\[
\frac{30}{180}=\frac{1}{6}
\]

If 250 trout is \(\frac16\) of the total lake population, then the total number of fish is:

\[
250 \div \frac16 = 250 \cdot 6 = 1500
\]

So there are 1500 fish in the lake.

ANSWER 5: B

Problem 6:

Karl bought 5 folders at \$2.50 each, so the total cost was:

\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:

\[
0.20 \times 12.50 = 2.50
\]

So he could have saved \$2.50.

ANSWER 6: C

Problem 7:

He paints 300 dimples, and each dimple takes 2 seconds.

Total time in seconds:

\[
300 \times 2 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10
\]

So he needs 10 minutes.

ANSWER 7: D

Problem 8:

We want combinations of 5-dollar bills and 2-dollar bills that total \$17.

Let \(x\) be the number of \$5 bills and \(y\) the number of \$2 bills. Then:

\[
5x + 2y = 17
\]

Try values of \(x\):

- \(x=1\): \(17-5=12\), so \(y=6\)
- \(x=2\): \(17-10=7\), not even, so impossible
- \(x=3\): \(17-15=2\), so \(y=1\)

These are the only nonnegative integer solutions.

So there are 2 combinations.

ANSWER 8: A

Problem 9:

The contest ended 1000 minutes after noon.

Convert 1000 minutes to hours and minutes:

\[
1000 \div 60 = 16 \text{ hours with } 40 \text{ minutes left}
\]

So:

- Noon + 12 hours = midnight
- plus 4 more hours = 4:00 a.m.
- plus 40 minutes = 4:40 a.m.

So the contest ended at 4:40 a.m.

ANSWER 9: D

Problem 10:

Compute each part using order of operations:

\[
(8\times 4 + 2) - (8 + 4\times 2)
\]

First parentheses:

\[
8\times 4 + 2 = 32 + 2 = 34
\]

Second parentheses:

\[
8 + 4\times 2 = 8 + 8 = 16
\]

Now subtract:

\[
34 - 16 = 18
\]

So the value is 18.

ANSWER 10: D

Problem 11:

We convert 11 hours and 5 minutes into minutes.

\[
11 \times 60 = 660
\]

Then add 5 minutes:

\[
660 + 5 = 665
\]

So the match lasted 665 minutes.

ANSWER 11: C

Problem 12:

Georgie must enter by one window and leave by a different window.

There are 6 choices for the entering window.  
After choosing one, there are 5 remaining choices for the exit window.

So the total number of ways is:

\[
6 \times 5 = 30
\]

Therefore, there are 30 ways.

ANSWER 12: D

Problem 1:  
We need the minimum number of socks to remove to guarantee there are 5 socks of the same color, given 3 colors (red, white, blue) and at least 5 of each color.

Worst case: you could first pull 4 socks of each color without reaching 5 of any color.  
That would be:
- 4 red + 4 white + 4 blue = 12 socks pulled

After 12 socks, you have at most 4 of each color. The next sock (the 13th) must be one of the 3 colors, and by the pigeonhole principle it makes one color reach 5.

So the Martian must remove **13** socks.

Tempting wrong choices:
- 12 isn’t enough because it’s possible to have only 4 of each color.
- Any number less than 13 can’t force a fifth sock of some color.

ANSWER 1: D


Problem 2:  
We have 4 students: Lance, Sally, Joy, Fred. We must choose 3 starters.

Number of ways to choose 3 out of 4 is:
\[
\binom{4}{3} = 4
\]

(You can also think: leave out one student; there are 4 choices for who to leave out.)

ANSWER 2: B


Problem 3:  
We check which pair does **NOT** multiply to 36.

Choices:
A. \(\{-4,-9\}\): \((-4)(-9)=36\) ✅  
B. \(\{-3,-12\}\): \((-3)(-12)=36\) ✅  
C. \(\left\{\frac{1}{2},-72\right\}\): \(\frac{1}{2}\cdot(-72)=-36\) ❌ not 36  
D. \(\{1,36\}\): \(1\cdot 36=36\) ✅  
E. \(\left\{\frac{3}{2},24\right\}\): \(\frac{3}{2}\cdot 24=3\cdot 12=36\) ✅

So the one that does not have product 36 is choice C.

ANSWER 3: C


Problem 4:  
Let’s convert everything to the same reference (floor).

- Ceiling is 2.4 m above the floor = 240 cm  
- The bulb is 10 cm below the ceiling: height of bulb above floor:
\[
240 - 10 = 230 \text{ cm}
\]
- Alice is 1.5 m tall = 150 cm  
- She can reach 46 cm above the top of her head, so from her top to the bulb:
\[
150 + 46 = 196 \text{ cm (if she stands on the floor)}
\]

She is standing on a stool, so her total height is (stool height + 150). She can reach:
\[
(\text{stool height} + 150) + 46
\]
Set equal to bulb height 230:
\[
\text{stool height} + 196 = 230
\]
\[
\text{stool height} = 34
\]

ANSWER 4: B


Problem 5:  
We know:
- Lake has 250 trout.
- Total fish in lake = \(T\)
- Sample has 180 fish, and 30 are trout.
So sample trout-to-total ratio equals lake trout-to-total ratio:
\[
\frac{30}{180} = \frac{250}{T}
\]
Simplify \(\frac{30}{180}=\frac{1}{6}\):
\[
\frac{1}{6} = \frac{250}{T}
\]
\[
T = 250 \cdot 6 = 1500
\]

ANSWER 5: B


Problem 6:  
Cost per folder originally: $2.50  
He bought 5 folders:
\[
5 \cdot 2.50 = 12.50
\]
A 20%-off sale means paying 80% of the original price:
\[
0.80 \cdot 12.50 = 10.00
\]
Savings:
\[
12.50 - 10.00 = 2.50
\]

ANSWER 6: C


Problem 7:  
Paint 300 dimples, 2 seconds per dimple.
Total time in seconds:
\[
300 \cdot 2 = 600 \text{ seconds}
\]
Convert to minutes (60 seconds per minute):
\[
600 \div 60 = 10 \text{ minutes}
\]

ANSWER 7: D


Problem 8:  
We need numbers of $5 bills and $2 bills that total $17 (order doesn’t matter).

Let \(x\) = number of $5 bills, \(y\) = number of $2 bills.
\[
5x + 2y = 17
\]
Try \(x=0\) to \(x=3\) (since \(4\cdot 5=20>17\)):

- \(x=0\): \(2y=17\) not possible  
- \(x=1\): \(5 + 2y=17 \Rightarrow 2y=12 \Rightarrow y=6\)  
- \(x=2\): \(10 + 2y=17 \Rightarrow 2y=7\) not possible  
- \(x=3\): \(15 + 2y=17 \Rightarrow 2y=2 \Rightarrow y=1\)

So there are 2 combinations: \((x,y)=(1,6)\) and \((3,1)\).

ANSWER 8: A


Problem 9:  
The contest starts at noon. It ends 1000 minutes later.

Convert 1000 minutes to hours:
\[
1000 \div 60 = 16 \text{ hours remainder } 40 \text{ minutes}
\]
Noon + 16 hours = 4:00 a.m. (next day)  
Add 40 minutes:
\[
4:00 \text{ a.m.} + 40 \text{ min} = 4:40 \text{ a.m.}
\]

ANSWER 9: D


Problem 10:  
Compute:
\[
(8 \times 4 + 2) - (8 + 4 \times 2)
\]
First parentheses:
\[
8\times 4 = 32,\quad 32+2=34
\]
Second parentheses:
\[
4\times 2 = 8,\quad 8+8=16
\]
Now subtract:
\[
34 - 16 = 18
\]

ANSWER 10: D


Problem 11:  
11 hours and 5 minutes total.
Convert hours to minutes:
\[
11 \cdot 60 = 660
\]
Add 5 minutes:
\[
660 + 5 = 665
\]

ANSWER 11: C


Problem 12:  
There are 6 windows. Georgie enters through one window and must leave through a different window.

Choose entering window: 6 choices.  
Choose leaving window: 5 choices (cannot be the same).

Total:
\[
6 \cdot 5 = 30
\]

ANSWER 12: D

Problem 1:
To be certain of having 5 socks of the same color, we consider the "worst-case scenario" (Pigeonhole Principle). If we have 3 colors (red, white, blue), we could pick 4 red, 4 white, and 4 blue socks without having 5 of any one color. This totals $4 + 4 + 4 = 12$ socks. The very next sock (the 13th) must be either red, white, or blue, which will complete a set of 5 for that color.
ANSWER 1: D

Problem 2:
We need to choose 3 starters from a group of 4 people (Lance, Sally, Joy, Fred). This is a combination problem denoted as $\binom{4}{3}$. Choosing 3 people to be on the team is the same as choosing 1 person to be left out. There are 4 ways to choose who is left out (Lance, Sally, Joy, or Fred).
ANSWER 2: B

Problem 3:
We check the products:
A: $(-4) \times (-9) = 36$
B: $(-3) \times (-12) = 36$
C: $\frac{1}{2} \times (-72) = -36$
D: $1 \times 36 = 36$
E: $\frac{3}{2} \times 24 = 3 \times 12 = 36$
Choice C results in $-36$, not $36$.
ANSWER 3: C

Problem 4:
Ceiling height = 2.4 m = 240 cm. Light bulb height = 240 - 10 = 230 cm from the floor.
Alice's reach = 150 cm (height) + 46 cm (reach) = 196 cm.
Stool height = (Bulb height) - (Alice's reach) = 230 - 196 = 34 cm.
ANSWER 4: B

Problem 5:
Let $T$ be the total number of fish in the lake. The ratio of trout to total fish is $\frac{30}{180} = \frac{1}{6}$.
Set up the proportion: $\frac{250}{T} = \frac{1}{6}$.
$T = 250 \times 6 = 1500$.
ANSWER 5: B

Problem 6:
Total cost = 5 folders × $2.50 = $12.50.
Savings = 20% of $12.50 = 0.20 \times 12.50 = $2.50.
Alternatively, 20% off one folder is $0.20 \times 2.50 = $0.50. For 5 folders, $0.50 \times 5 = $2.50.
ANSWER 6: C

Problem 7:
Total time in seconds = 300 dimples × 2 seconds/dimple = 600 seconds.
Total time in minutes = 600 seconds / 60 seconds/minute = 10 minutes.
ANSWER 7: D

Problem 8:
Let $x$ be the number of $5 bills and $y$ be the number of $2 bills. $5x + 2y = 17$.
If $x=1$, $5+2y=17 \Rightarrow 2y=12 \Rightarrow y=6$. (1, 6)
If $x=3$, $15+2y=17 \Rightarrow 2y=2 \Rightarrow y=1$. (3, 1)
If $x=0$ or $x=2$, $2y$ would be 17 or 7, which are not divisible by 2.
There are 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours and 40 minutes (since $1000 \div 60 = 16$ remainder 40).
Noon + 12 hours = midnight.
Midnight + 4 hours and 40 minutes = 4:40 a.m.
ANSWER 9: D

Problem 10:
$(8 \times 4 + 2) - (8 + 4 \times 2) = (32 + 2) - (8 + 8) = 34 - 16 = 18$.
ANSWER 10: D

Problem 11:
11 hours = $11 \times 60 = 660$ minutes.
Total = 660 + 5 = 665 minutes.
ANSWER 11: C

Problem 12:
There are 6 choices for the entry window. Since the exit window must be different, there are 5 choices for the exit window.
Total ways = $6 \times 5 = 30$.
ANSWER 12: D

Problem 1:
This is a pigeonhole principle problem. To guarantee at least 5 socks of one color, first consider the worst-case scenario of taking as many socks as possible without having 5 of any single color. That means taking exactly 4 socks of each of the 3 colors, for a total of \(4 \times 3 = 12\) socks. The next sock (the 13th) must be a 5th sock of one of the colors. Thus 13 socks are required. (Fewer than 13 leaves open the possibility of only 4 of each color.)

ANSWER 1: D

Problem 2:
The number of ways to choose 3 starters out of the 4 available players is the combination \(\binom{4}{3} = 4\). Each choice simply leaves out one of the four players.

ANSWER 2: B

Problem 3:
Compute the product for each pair:
- A: \((-4) \times (-9) = 36\)
- B: \((-3) \times (-12) = 36\)
- C: \(\frac12 \times (-72) = -36\) (not 36)
- D: \(1 \times 36 = 36\)
- E: \(\frac32 \times 24 = 36\)

Only pair C fails to give a product of 36.

ANSWER 3: C

Problem 4:
Convert all lengths to centimeters. The ceiling is 240 cm above the floor, so the bulb is 230 cm above the floor. Alice's reach is \(150 + 46 = 196\) cm. The stool must therefore supply the remaining \(230 - 196 = 34\) cm.

ANSWER 4: B

Problem 5:
The sample shows that trout comprise \(\frac{30}{180} = \frac16\) of all fish. With 250 trout in the lake, the total number of fish is \(250 \div \frac16 = 1500\).

ANSWER 5: B

Problem 6:
Five folders cost \(5 \times 2.50 = 12.50\). A 20 % discount saves \(0.20 \times 12.50 = 2.50\).

ANSWER 6: C

Problem 7:
Painting 300 dimples at 2 seconds each requires \(300 \times 2 = 600\) seconds, or \(600 \div 60 = 10\) minutes.

ANSWER 7: D

Problem 8:
Solve \(5x + 2y = 17\) for non-negative integers \(x, y\):
- \(x = 1\), \(y = 6\) works;
- \(x = 3\), \(y = 1\) works.
No other non-negative integer solutions exist, so there are exactly two combinations.

ANSWER 8: A

Problem 9:
Divide 1000 minutes by 60: \(1000 = 16\) hours + 40 minutes. Adding 16 hours to noon yields 4:00 a.m.; adding 40 minutes yields 4:40 a.m.

ANSWER 9: D

Problem 10:
Left side: \(8 \times 4 + 2 = 34\). Right side: \(8 + 4 \times 2 = 16\). Difference: \(34 - 16 = 18\).

ANSWER 10: D

Problem 11:
Convert 11 hours 5 minutes to minutes: \(11 \times 60 + 5 = 665\).

ANSWER 11: C

Problem 12:
There are 6 choices for the entry window and then 5 remaining choices for the exit window, giving \(6 \times 5 = 30\) ordered pairs.

ANSWER 12: D

Problem 1:
The problem is asking us to determine the minimum number of socks the Martian must remove from the drawer to guarantee that there are 5 socks of the same color. 
To ensure that there are 5 socks of the same color, we must consider the worst-case scenario where the Martian draws as many socks as possible without getting 5 of the same color. 
The Martian can draw at most 4 socks of each color without having 5 socks of the same color. 
So, the Martian can draw 4 red, 4 white, and 4 blue socks, for a total of 4 + 4 + 4 = 12 socks. 
At this point, the Martian still doesn't have 5 socks of the same color. 
However, drawing one more sock will guarantee that there are 5 socks of the same color. 
Therefore, the Martian must draw at least 12 + 1 = 13 socks.
Let's examine the choices: 
A. 6 is too low because the Martian could draw 3 red, 2 white, and 1 blue, or other combinations that don't guarantee 5 socks of the same color. 
B. 9 is still too low because the Martian could draw 4 red, 3 white, and 2 blue, or other similar combinations. 
C. 12 is the maximum number of socks the Martian can draw without having 5 socks of the same color. 
D. 13 is the correct answer because it is one more than the maximum number of socks that can be drawn without having 5 socks of the same color. 
E. 15 is more than necessary.
ANSWER 1: D

Problem 2:
The problem is asking us to find the number of ways to choose 3 starters from a team of 4 players: Lance, Sally, Joy, and Fred. 
This is a combination problem, because the order in which the starters are chosen does not matter. 
The number of ways to choose 3 starters from 4 players can be calculated using the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of players and k is the number of starters. 
Here, n = 4 and k = 3, so C(4, 3) = 4! / (3!(4-3)!) = 4.
Let's list the possible combinations to verify: 
- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred
There are indeed 4 combinations.
Let's examine the choices: 
A. 2 is too low. 
B. 4 matches our calculation. 
C. 6 is too high. 
D. 8 is too high. 
E. 10 is too high.
ANSWER 2: B

Problem 3:
The problem is asking us to identify the pair of numbers that does NOT have a product equal to 36. 
Let's check each pair: 
A. {−4, −9}: (−4) × (−9) = 36
B. {−3, −12}: (−3) × (−12) = 36
C. { 1 2 , −72}: (1/2) × (−72) = −36
D. {1, 36}: 1 × 36 = 36
E. { 3 2 , 24}: (3/2) × 24 = 36
The pair that does NOT have a product equal to 36 is C.
Let's examine the choices: 
A. {−4, −9} has a product of 36. 
B. {−3, −12} has a product of 36. 
C. { 1 2 , −72} has a product of −36, not 36. 
D. {1, 36} has a product of 36. 
E. { 3 2 , 24} has a product of 36.
ANSWER 3: C

Problem 4:
The problem is asking us to find the height of the stool that Alice needs to stand on to reach the light bulb. 
First, we need to convert the height of the ceiling from meters to centimeters: 2.4 meters = 240 centimeters. 
The light bulb is 10 centimeters below the ceiling, so its height above the floor is 240 - 10 = 230 centimeters. 
Alice is 1.5 meters = 150 centimeters tall and can reach 46 centimeters above her head, so her total reach is 150 + 46 = 196 centimeters. 
To find the height of the stool, we subtract Alice's total reach from the height of the light bulb: 230 - 196 = 34 centimeters.
Let's examine the choices: 
A. 32 is too low because 230 - (196 + 32) = 2, which is not enough to reach the light bulb. 
B. 34 is the correct height. 
C. 36 is too high. 
D. 38 is too high. 
E. 40 is too high.
ANSWER 4: B

Problem 5:
The problem is asking us to find the total number of fish in the lake based on the ratio of trout in the sample to the total number of fish in the sample. 
The sample contains 30 trout out of 180 fish, so the ratio of trout to total fish in the sample is 30/180. 
We are given that this ratio is the same as the ratio of trout to total fish in the lake. 
Let's denote the total number of fish in the lake as x. 
The ratio of trout to total fish in the lake is 250/x. 
Setting the two ratios equal, we get 30/180 = 250/x. 
Cross-multiplying, we have 30x = 250 × 180. 
Simplifying, 30x = 45000. 
Dividing both sides by 30, x = 1500.
Let's examine the choices: 
A. 1250 is too low. 
B. 1500 matches our calculation. 
C. 1750 is too high. 
D. 1800 is too high. 
E. 2000 is too high.
ANSWER 5: B

Problem 6:
The problem is asking us to find out how much Karl could have saved by waiting a day to buy the folders at a 20%-off sale. 
The original price of 5 folders is 5 × $2.50 = $12.50. 
At a 20%-off sale, the discount is 20% of $12.50, which is 0.20 × $12.50 = $2.50. 
So, Karl could have saved $2.50.
Let's examine the choices: 
A. $1.00 is too low. 
B. $2.00 is too low. 
C. $2.50 matches our calculation. 
D. $2.75 is too high. 
E. $5.00 is too high.
ANSWER 6: C

Problem 7:
The problem is asking us to find the time it will take Casey to paint 300 dimples on a golf ball. 
Casey takes 2 seconds to paint one dimple, so to paint 300 dimples, he will take 300 × 2 = 600 seconds. 
To convert seconds to minutes, we divide by 60: 600 / 60 = 10 minutes.
Let's examine the choices: 
A. 4 minutes is too low. 
B. 6 minutes is too low. 
C. 8 minutes is too low. 
D. 10 minutes matches our calculation. 
E. 12 minutes is too high.
ANSWER 7: D

Problem 8:
The problem is asking us to find the number of different combinations of $5 bills and $2 bills that can be used to make a total of $17. 
Let's list the possible combinations: 
1. 3 $5 bills = $15, and $2 bill = $2, total = $17 (1 combination: 3 $5 bills and 1 $2 bill)
2. 1 $5 bill = $5, and 6 $2 bills = $12, total = $17 (1 combination: 1 $5 bill and 6 $2 bills)
We should also check if there are any other combinations: 
- Using 0 $5 bills, we need 8.5 $2 bills, which is not possible since we can't have a fraction of a bill.
- Using 2 $5 bills = $10, we need $7 more, which can be made with 3.5 $2 bills, again not possible.
- Using 4 $5 bills = $20, which is more than $17.

Problem 1:
We want the minimum number of socks to guarantee 5 of the same color. With 3 colors (red, white, blue), the worst case is picking 4 of each color: 4+4+4 = 12 socks. The 13th sock must create a group of 5 of one color. So 13 socks are needed.
ANSWER 1: D

Problem 2:
We choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways is the combination C(4,3) = 4.
ANSWER 2: B

Problem 3:
Compute each product:
A: (−4)×(−9) = 36
B: (−3)×(−12) = 36
C: (1/2)×(−72) = −36
D: 1×36 = 36
E: (3/2)×24 = 36
Only C does not equal 36.
ANSWER 3: C

Problem 4:
Ceiling = 2.4 m = 240 cm. Light bulb is 10 cm below ceiling, so its height = 230 cm.
Alice's height = 1.5 m = 150 cm. Reach above head = 46 cm, so reach without stool = 196 cm.
Stool height needed = 230 − 196 = 34 cm.
ANSWER 4: B

Problem 5:
Sample ratio: 30 trout / 180 fish = 1/6. Lake has 250 trout, which is 1/6 of total fish. Total fish = 250 × 6 = 1500.
ANSWER 5: B

Problem 6:
Cost without sale: 5 × $2.50 = $12.50.
With 20% off, each folder costs $2.50 × 0.8 = $2.00. Total = 5 × $2.00 = $10.00.
Savings = $12.50 − $10.00 = $2.50.
ANSWER 6: C

Problem 7:
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.
ANSWER 7: D

Problem 8:
Find non-negative integers x ($5 bills) and y ($2 bills) such that 5x + 2y = 17.
x=1 → 5 + 2y = 17 → y=6.
x=3 → 15 + 2y = 17 → y=1.
No other x works. So 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours 40 minutes (16×60 = 960, remainder 40).
Noon + 16 hours = 4:00 a.m. next day. Add 40 minutes → 4:40 a.m.
ANSWER 9: D

Problem 10:
(8 × 4 + 2) = 32 + 2 = 34.
(8 + 4 × 2) = 8 + 8 = 16.
34 − 16 = 18.
ANSWER 10: D

Problem 11:
11 hours × 60 = 660 minutes. Plus 5 minutes = 665 minutes.
ANSWER 11: C

Problem 12:
6 choices for entry window, then 5 choices for a different exit window. Total = 6 × 5 = 30.
ANSWER 12: D

Problem 1:
We need to find the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). Using the Pigeonhole Principle, we consider the worst-case scenario: the Martian pulls out as many socks as possible *without* getting 5 of any one color. That means getting exactly 4 of each color:
- 4 red + 4 white + 4 blue = 12 socks
After 12 socks, the Martian could have 4 of each color and still not have 5 matching. The very next sock (the 13th) must be red, white, or blue, pushing the count of that color to 5. So 13 socks guarantees 5 of the same color.
(The "five-legged" detail is flavor text explaining why the Martian needs 5 socks.)

ANSWER 1: D

Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). This is a combination problem since order doesn't matter:
C(4, 3) = 4! / (3! × 1!) = 4
We can also list them: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — exactly 4 ways.

ANSWER 2: B

Problem 3:
We check each product:
- A: (−4) × (−9) = 36 ✓
- B: (−3) × (−12) = 36 ✓
- C: (1/2) × (−72) = −36 ✗ (This equals −36, NOT 36)
- D: 1 × 36 = 36 ✓
- E: (3/2) × 24 = 36 ✓
Choice C gives −36, which does not equal 36.

ANSWER 3: C

Problem 4:
First, convert everything to centimeters:
- Ceiling height: 2.4 m = 240 cm
- Light bulb position: 240 − 10 = 230 cm above the floor
- Alice's height: 1.5 m = 150 cm
- Alice's reach above her head: 46 cm
- Alice's maximum reach (without stool): 150 + 46 = 196 cm above the floor

Let h = height of the stool. Standing on the stool, her reach is h + 196 cm. She can "just reach" the bulb:
h + 196 = 230
h = 230 − 196 = 34 cm

ANSWER 4: B

Problem 5:
The ratio of trout to total fish in the sample should equal the ratio in the lake:
- Sample ratio: 30 trout / 180 fish = 1/6
- Lake ratio: 250 trout / Total fish = 1/6

Setting up the proportion: 250 / Total = 1/6
Total = 250 × 6 = 1500 fish

ANSWER 5: B

Problem 6:
Karl bought 5 folders at $2.50 each:
Total cost = 5 × $2.50 = $12.50
A 20%-off sale means he would save 20% of the total:
Savings = 0.20 × $12.50 = $2.50

ANSWER 6: C

Problem 7:
Total time in seconds: 300 dimples × 2 seconds/dimple = 600 seconds
Convert to minutes: 600 seconds ÷ 60 seconds/minute = 10 minutes

ANSWER 7: D

Problem 8:
We need non-negative integer solutions to 5x + 2y = 17, where x = number of $5 bills and y = number of $2 bills.
- x = 0: 2y = 17 → y = 8.5 (not an integer) ✗
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 ✗
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x = 4: 20 > 17 (too large) ✗

Only 2 valid combinations: (1 five + 6 twos) and (3 fives + 1 two).

ANSWER 8: A

Problem 9:
Convert 1000 minutes to hours and minutes:
1000 ÷ 60 = 16 remainder 40, so 1000 minutes = 16 hours 40 minutes.
Starting at noon (12:00 PM):
- 12 hours later = midnight (12:00 AM)
- 4 more hours = 4:00 AM
- 40 more minutes = 4:40 AM

ANSWER 9: D

Problem 10:
Evaluate each part using order of operations (multiplication before addition):
- First part: (8 × 4 + 2) = (32 + 2) = 34
- Second part: (8 + 4 × 2) = (8 + 8) = 16
- Difference: 34 − 16 = 18

ANSWER 10: D

Problem 11:
Convert 11 hours and 5 minutes to minutes:
11 hours × 60 minutes/hour = 660 minutes
660 + 5 = 665 minutes

ANSWER 11: C

Problem 12:
Georgie enters through one window and leaves through a different one.
- Choices for entry window: 6
- Choices for exit window (must be different): 5
Total ways = 6 × 5 = 30

ANSWER 12: D

Problem 1:
The Martian needs to be **certain** of having 5 socks of the same color. To guarantee this, we must consider the worst-case scenario: the Martian picks as many socks as possible **without** yet having 5 of any single color.  
With 3 colors (red, white, blue), the worst case is picking 4 socks of each color:  
\(4 + 4 + 4 = 12\) socks.  
At this point the Martian has 4 of every color but still does not have 5 of any one color. The very next sock—the 13th—must be red, white, or blue, and will necessarily create a set of 5 of that color.  
Thus 13 socks are required.  
(Choice A is too small; with 6 socks one could have 2,2,2. Choice B allows 3,3,3. Choice C allows 4,4,4. Choice E is more than the minimum guaranteed number.)

ANSWER 1: D

---

Problem 2:
We need to choose a group of 3 starters from the 4 students: Lance, Sally, Joy, and Fred. Since the three starters form a team and order does not matter, this is a combination problem.  
\[
\binom{4}{3} = \frac{4!}{3!\,1!} = 4
\]  
Equivalently, we can think of choosing which **one** of the four students will **not** start; there are 4 such choices.  
(Choice A is too small. Choice C is \(\binom{4}{2}\). Choices D and E do not arise from counting 3-person teams from 4 people.)

ANSWER 2: B

---

Problem 3:
We check the product of each pair to see which is not 36.

- **A.** \((-4)(-9) = 36\)  
- **B.** \((-3)(-12) = 36\)  
- **C.** \(\left(\frac12\right)(-72) = -36\)  
- **D.** \((1)(36) = 36\)  
- **E.** \(\left(\frac32\right)(24) = 36\)

Only pair **C** gives \(-36\), not \(36\).

ANSWER 3: C

---

Problem 4:
First convert all measurements to centimeters.  
- Ceiling height: \(2.4\text{ m} = 240\text{ cm}\).  
- Light bulb is \(10\text{ cm}\) below the ceiling, so its height above the floor is \(240 - 10 = 230\text{ cm}\).  

Alice’s reach:  
- Height: \(1.5\text{ m} = 150\text{ cm}\).  
- She can reach \(46\text{ cm}\) above her head, so her standing reach is \(150 + 46 = 196\text{ cm}\).  

The stool must supply the remaining height:  
\[
230 - 196 = 34\text{ cm}.
\]

ANSWER 4: B

---

Problem 5:
Let \(T\) be the total number of fish in the lake. The problem states that the ratio of trout to total fish is the same in the sample and in the lake.  
\[
\frac{250}{T} = \frac{30}{180}
\]  
Simplify the right side: \(\frac{30}{180} = \frac16\).  
\[
\frac{250}{T} = \frac16 \quad\Longrightarrow\quad T = 250 \times 6 = 1500.
\]

ANSWER 5: B

---

Problem 6:
Karl bought 5 folders at \(\$2.50\) each.  
Total cost: \(5 \times \$2.50 = \$12.50\).  
A 20% discount would have saved him 20% of this total:  
\[
0.20 \times \$12.50 = \$2.50.
\]

ANSWER 6: C

---

Problem 7:
Casey paints 300 dimples at 2 seconds per dimple.  
Total time in seconds: \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(\frac{600}{60} = 10\) minutes.

ANSWER 7: D

---

Problem 8:
Let \(f\) be the number of \(\$5\) bills and \(t\) be the number of \(\$2\) bills. We need non-negative integer solutions to  
\[
5f + 2t = 17.
\]  
Test values of \(f\):  
- \(f=0\): \(2t=17\) → no integer solution.  
- \(f=1\): \(5 + 2t = 17 \Rightarrow t=6\).  **(Valid)**
- \(f=2\): \(10 + 2t = 17 \Rightarrow t=3.5\).  (Invalid)
- \(f=3\): \(15 + 2t = 17 \Rightarrow t=1\).  **(Valid)**
- \(f=4\): \(20 > 17\) → stop.

There are exactly **2** combinations: \((1\text{ five}, 6\text{ twos})\) and \((3\text{ fives}, 1\text{ two})\).

ANSWER 8: A

---

Problem 9:
The contest starts at noon (12:00 p.m.). We add 1000 minutes.  
Divide 1000 by 60:  
\[
1000 = 16 \times 60 + 40,
\]  
so 1000 minutes = 16 hours and 40 minutes.  

Noon + 12 hours = midnight.  
Midnight + 4 hours 40 minutes = 4:40 a.m.

ANSWER 9: D

---

Problem 10:
Evaluate using the correct order of operations (multiplication before addition).

First parentheses:  
\[
8 \times 4 + 2 = 32 + 2 = 34.
\]

Second parentheses:  
\[
8 + 4 \times 2 = 8 + 8 = 16.
\]

Subtract:  
\[
34 - 16 = 18.
\]

ANSWER 10: D

---

Problem 11:
Convert 11 hours to minutes and add the remaining 5 minutes.  
\[
11 \times 60 = 660\text{ minutes}.
\]  
\[
660 + 5 = 665\text{ minutes}.
\]

ANSWER 11: C

---

Problem 12:
Georgie enters through one of 6 windows and must leave through a **different** window.  
- 6 choices for the entry window.  
- For each entry choice, 5 remaining choices for the exit window.  

Total ways:  
\[
6 \times 5 = 30.
\]  
(Choice E, 36, would count entering and leaving through the same window, which is not allowed.)

ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). The worst-case scenario is to draw as many socks as possible without having 5 of any one color. That would be 4 red, 4 white, and 4 blue, for a total of 12 socks. The next sock drawn (the 13th) must be one of the three colors, giving that color 5 socks. Therefore, 13 socks are required. The five-legged Martian is irrelevant to the count.  
Answer: D

Problem 2:
We have 4 people (Lance, Sally, Joy, Fred) and need to choose 3 starters. Since the starters are not assigned to specific positions, order does not matter. The number of ways is the combination \( \binom{4}{3} = 4 \).  
Answer: B

Problem 3:
Compute each product:  
A. \((-4) \times (-9) = 36\)  
B. \((-3) \times (-12) = 36\)  
C. \(\frac{1}{2} \times (-72) = -36\)  
D. \(1 \times 36 = 36\)  
E. \(\frac{3}{2} \times 24 = 36\)  
Only pair C gives a product not equal to 36.  
Answer: C

Problem 4:
Convert all measurements to centimeters:  
Ceiling height = 2.4 m = 240 cm.  
Light bulb is 10 cm below ceiling → height = 240 – 10 = 230 cm.  
Alice’s height = 1.5 m = 150 cm.  
Her reach above her head = 46 cm, so her maximum reach without a stool = 150 + 46 = 196 cm.  
Let \(h\) be the stool height. Standing on the stool raises her reach to \(196 + h\). This must equal 230 cm:  
\(196 + h = 230\) → \(h = 34\) cm.  
Answer: B

Problem 5:
In the sample, 30 out of 180 fish are trout, so the ratio trout : total = \(30/180 = 1/6\).  
Assuming the same ratio in the lake, if the lake has 250 trout, then the total number of fish is \(250 \times 6 = 1500\).  
Answer: B

Problem 6:
Original cost for 5 folders: \(5 \times 2.50 = \$12.50\).  
A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
Thus, by waiting a day, Karl could have saved $2.50.  
Answer: C

Problem 7:
Time to paint one dimple = 2 seconds.  
Total time for 300 dimples = \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(600 \div 60 = 10\) minutes.  
Answer: D

Problem 8:
We need nonnegative integers \(a\) (number of $5 bills) and \(b\) (number of $2 bills) such that \(5a + 2b = 17\).  
Solve by testing values of \(b\):  
\(b = 1\) → \(5a = 15\) → \(a = 3\) → (3,1) works.  
\(b = 6\) → \(5a = 5\) → \(a = 1\) → (1,6) works.  
No other values give integer \(a\). Thus there are 2 combinations.  
Answer: A

Problem 9:
1000 minutes = \(1000 \div 60 = 16\) hours and \(40\) minutes (since \(60 \times 16 = 960\), remainder 40).  
Starting at noon (12:00 PM), add 12 hours → midnight (12:00 AM). Add the remaining 4 hours → 4:00 AM. Then add 40 minutes → 4:40 AM.  
Answer: D

Problem 10:
Evaluate using order of operations:  
First parentheses: \(8 \times 4 + 2 = 32 + 2 = 34\).  
Second parentheses: \(8 + 4 \times 2 = 8 + 8 = 16\).  
Subtract: \(34 - 16 = 18\).  
Answer: D

Problem 11:
11 hours = \(11 \times 60 = 660\) minutes.  
Add 5 minutes → \(660 + 5 = 665\) minutes.  
Answer: C

Problem 12:
Georgie must choose an entry window (6 choices) and a different exit window (5 choices). Since entering and leaving are distinct actions, the total number of ways is \(6 \times 5 = 30\).  
Answer: D

ANSWER 1: D
ANSWER 2: B
ANSWER 3: C
ANSWER 4: B
ANSWER 5: B
ANSWER 6: C
ANSWER 7: D
ANSWER 8: A
ANSWER 9: D
ANSWER 10: D
ANSWER 11: C
ANSWER 12: D

Problem 1:  
The Martian wants to be certain of having 5 socks of the same color. There are three colors: red, white, and blue. In the worst-case scenario, the Martian could pick 4 socks of each color before getting a fifth of any one color. That would be \(4 \times 3 = 12\) socks, and still no color has 5 socks. The next sock drawn (the 13th) must complete a set of 5 of some color, because there are only three colors. Therefore, the Martian must remove **13** socks to be certain of having 5 socks of the same color.  
ANSWER 1: D

Problem 2:  
Ms. Hamilton's class has 4 available players: Lance, Sally, Joy, and Fred. The number of ways to choose 3 starters from 4 players is given by the combination \(\binom{4}{3} = 4\). Thus, there are **4** possible sets of three starters.  
ANSWER 2: B

Problem 3:  
We check each pair's product:  
- A: \((-4) \times (-9) = 36\)  
- B: \((-3) \times (-12) = 36\)  
- C: \(\frac{1}{2} \times (-72) = -36\) (not 36)  
- D: \(1 \times 36 = 36\)  
- E: \(\frac{3}{2} \times 24 = 36\)  
Only pair **C** does not have a product of 36.  
ANSWER 3: C

Problem 4:  
The ceiling is 2.4 m above the floor, and the bulb is 10 cm (0.1 m) below the ceiling, so the bulb's height is \(2.4 - 0.1 = 2.3\) m. Alice's height is 1.5 m, and she can reach 46 cm (0.46 m) above her head, so her reach height from the floor is \(1.5 + 0.46 = 1.96\) m. When standing on a stool, her hands are at the stool height plus her reach height. Setting this equal to the bulb height:  
\[
\text{stool height} + 1.96 = 2.3 \Rightarrow \text{stool height} = 0.34 \text{ m} = 34 \text{ cm}.
\]  
ANSWER 4: B

Problem 5:  
In the sample, the ratio of trout to total fish is \(\frac{30}{180} = \frac{1}{6}\). Assuming the same ratio in the lake, let \(T\) be the total number of fish in the lake. Then \(\frac{250}{T} = \frac{1}{6}\), so \(T = 250 \times 6 = 1500\).  
ANSWER 5: B

Problem 6:  
Karl bought 5 folders at $2.50 each, so the original cost is \(5 \times 2.50 = \$12.50\). With a 20% discount, the cost becomes \(12.50 \times 0.80 = \$10.00\). The savings is \(12.50 - 10.00 = \$2.50\).  
ANSWER 6: C

Problem 7:  
Painting 300 dimples at 2 seconds each takes \(300 \times 2 = 600\) seconds. Converting to minutes: \(600 \div 60 = 10\) minutes.  
ANSWER 7: D

Problem 8:  
We need nonnegative integers \(x\) (number of $5 bills) and \(y\) (number of $2 bills) such that \(5x + 2y = 17\). Solving:  
- \(y = 1 \Rightarrow 5x = 15 \Rightarrow x = 3\)  
- \(y = 6 \Rightarrow 5x = 5 \Rightarrow x = 1\)  
- \(y = 11 \Rightarrow 5x = -5\) (invalid)  
Thus, there are **2** valid combinations: (3 fives, 1 two) and (1 five, 6 twos).  
ANSWER 8: A

Problem 9:  
1000 minutes equals \(16\) hours and \(40\) minutes (since \(16 \times 60 = 960\), remainder \(40\)). Starting at noon, adding \(16\) hours gives 4:00 AM the next day, and adding \(40\) minutes gives 4:40 AM.  
ANSWER 9: D

Problem 10:  
Compute inside parentheses:  
\(8 \times 4 + 2 = 32 + 2 = 34\)  
\(8 + 4 \times 2 = 8 + 8 = 16\)  
Then \(34 - 16 = 18\).  
ANSWER 10: D

Problem 11:  
11 hours is \(11 \times 60 = 660\) minutes. Adding 5 minutes gives \(660 + 5 = 665\) minutes.  
ANSWER 11: C

Problem 12:  
Georgie can enter through any of the 6 windows. For exiting, he must choose a different window, so there are 5 choices. The total number of ways is \(6 \times 5 = 30\).  
ANSWER 12: D

Problem 1:  
The Martian needs to ensure there are 5 socks of the same color. The worst-case scenario is pulling the maximum number of socks without 5 of any color. Since there are 3 colors, this would be 4 socks of each color (4 × 3 = 12 socks). The next sock (the 13th) must match one color, giving 5 socks of that color.  
**ANSWER 1: D**  

Problem 2:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is C(4,3) = 4.  
**ANSWER 2: B** (Note: C(4,3)=4, but rechecking: the team is chosen from 4 specific people, so combinations are 4. Answer B=4)  

Problem 3:  
Check each pair:  
A. (−4)(−9) = 36  
B. (−3)(−12) = 36  
C. (1/2)(−72) = -36 (≠36)  
D. (1)(36) = 36  
E. (3/2)(24) = 36  
The pair that does NOT equal 36 is C.  
**ANSWER 3: E** (Re-evaluated: The question asks which does NOT equal 36. Option E: (3/2)*24=36, so it is valid. Option C: (1/2)*(-72)=-36, so C is the correct choice for NOT equal. Correction: **ANSWER 3: C** [Original had a typo, but computation confirms C is the answer])  

*Correction: For Problem 3, Option C: (1/2, -72) likely means (1/2) × (-72) = -36 ≠ 36. Option E: (3/2) × 24 = 36. Thus, C is the answer.*  
**ANSWER 3: C**  

Problem 4:  
Ceiling height: 2.4 m = 240 cm. Light bulb location: 240 - 10 = 230 cm from floor. Alice’s reach: 150 cm + 46 cm = 196 cm. Stool height = 230 - 196 = 34 cm.  
**ANSWER 4: B**  

Problem 5:  
Ratio of trout in sample: 30/180 = 1/6. Total trout: 250. Total fish = 250 × 6 = 1500.  
**ANSWER 5: B**  

Problem 6:  
Original cost: 5 × $2.50 = $12.50. 20% off: $12.50 × 0.20 = $2.50 saved.  
**ANSWER 6: C** (Note: The question asks how much Karl could have saved, which is $2.50, not $2.00. Original had a mistake; correction: 20% of $12.50 is $2.50, so answer C)  

Problem 7:  
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.  
**ANSWER 7: D**  

Problem 8:  
Let $5 bills = x, $2 bills = y. 5x + 2y = 17. Possible solutions:  
- x=3, y=1 (15+2=17)  
- x=1, y=6 (5+12=17)  
Only 2 combinations.  
**ANSWER 8: A**  

Problem 9:  
1000 minutes = 16 hours 40 minutes. Starting at noon: noon + 16h40m = 4:40 a.m. next day.  
**ANSWER 9: D**  

Problem 10:  
(8×4 + 2) = 34, (8 + 4×2) = 16. 34 - 16 = 18.  
**ANSWER 10: D**  

Problem 11:  
11 hours = 660 minutes. 660 + 5 = 665 minutes.  
**ANSWER 11: C**  

Problem 12:  
Georgie can enter through 6 windows and exit through 5 others. Total ways: 6 × 5 = 30.  
**ANSWER 12: D**  

---  
**Final Answers**  
ANSWER 1: D  
ANSWER 2: B  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: B  
ANSWER 6: C  
ANSWER 7: D  
ANSWER 8: A  
ANSWER 9: D  
ANSWER 10: D  
ANSWER 11: C  
ANSWER 12: D

### Problem 1:
This is a classic pigeonhole principle problem. We need to find the minimum number of socks the Martian must pull to guarantee 5 socks of the same color, with 3 total colors. The worst-case scenario is that the Martian pulls as many socks as possible without getting 5 of any color: that is 4 socks of each color, for a total of 3*4=12 socks. The next sock pulled (the 13th) has to be one of the three colors, pushing that color to 5 socks. Tempting wrong answers: C (12 is only the number of socks that avoids 5 of the same color, not enough to guarantee the requirement), E (15 is the total if you took 5 of each color, not the minimum to guarantee).
ANSWER 1: D

---

### Problem 2:
We need to count the number of ways to choose 3 starters from 4 people, where order on the team does not matter. This is a combination problem: C(n,k) = n!/(k!(n-k)!). For n=4, k=3: C(4,3) = 4. We can also list the teams by who is left out: leaving out Lance, Sally, Joy, or Fred gives exactly 4 unique teams. Tempting wrong answer: C (6 is the count if order of selection mattered, which it does not for a team of starters).
ANSWER 2: B

---

### Problem 3:
We calculate the product of each pair to find which does not equal 36:
- A: (-4)*(-9)=36 ✔️
- B: (-3)*(-12)=36 ✔️
- C: (1/2)*(-72) = -36 ❌
- D: 1*36=36 ✔️
- E: (3/2)*24=36 ✔️
Only pair C has a product of -36, not 36.
ANSWER 3: C

---

### Problem 4:
First convert all units to centimeters to avoid conversion errors:
- Ceiling height: 2.4m = 240cm. Light bulb is 10cm below ceiling, so bulb height from floor = 240-10=230cm.
- Alice's height: 1.5m=150cm, her reach above her head is 46cm, so total reach without stool = 150+46=196cm.
- Stool height = 230-196=34cm.
Tempting wrong answers come from unit conversion mistakes, e.g., mixing up meters and centimeters.
ANSWER 4: B

---

### Problem 5:
This is a proportion problem. The ratio of trout to total fish is the same in the sample and the lake. In the sample, 30 trout out of 180 total fish = 1/6. Let T be total fish in the lake: 250/T = 1/6 → T=250*6=1500.
ANSWER 5: B

---

### Problem 6:
Karl bought 5 folders at $2.50 each, total original cost = 5*2.50=$12.50. He saves 20% of the total cost by waiting for the sale: 0.2*12.50=$2.50. This can also be calculated per folder: 20% of $2.50 is $0.50 saved per folder, 5*0.50=$2.50 total saved.
ANSWER 6: C

---

### Problem 7:
300 dimples * 2 seconds per dimple = 600 total seconds. Convert to minutes: 60 seconds per minute, so 600/60=10 minutes.
ANSWER 7: D

---

### Problem 8:
We find all non-negative integer pairs (x,y) where x=number of $5 bills, y=number of $2 bills, and 5x+2y=17. Only odd values of x work (since $5 is odd, $2 is even, we need an odd sum to get 17):
- x=1: 5+2y=17 → y=6 (valid)
- x=3: 15+2y=17 → y=1 (valid)
Even x values (0,2) leave an odd remainder that cannot be formed with $2 bills, so only 2 valid combinations.
ANSWER 8: A

---

### Problem 9:
Convert 1000 minutes to hours and minutes: 60 minutes per hour, 1000/60=16 hours and 40 minutes. Adding this to noon (12pm): 12pm +16h40m = 4:40am the next day.
ANSWER 9: D

---

### Problem 10:
Use order of operations (PEMDAS) to calculate:
First parenthesis: 8*4 +2 = 32+2=34
Second parenthesis: 8 +4*2=8+8=16
Subtract: 34-16=18. The tempting wrong answer C (10) comes from incorrectly adding before multiplying in the second parenthesis.
ANSWER 10: D

---

### Problem 11:
Convert 11 hours 5 minutes to total minutes: 11*60=660 minutes, plus 5 extra minutes = 665 minutes.
ANSWER 11: C

---

### Problem 12:
Georgie has 6 choices for an entry window, and 5 remaining choices for an exit window (must use a different window). Total ways = 6*5=30. Tempting wrong answer E (36) counts entering and leaving through the same window, which is not allowed; B (15) incorrectly treats entry/exit as unordered, which they are not.
ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color (red, white, blue). This is a pigeonhole principle problem: the worst-case scenario is pulling the maximum number of socks without getting 5 of any color. For each of the 3 colors, we can pull 4 socks (since 4 < 5), so 4×3=12 socks. The next sock pulled will necessarily be the 5th of one color. So total socks needed: 12 + 1 = 13.
ANSWER 1: D
Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). Since the problem does not mention distinct positions for starters, order does not matter, so we use combinations. The number of ways to choose 3 people from 4 is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n=4 \) and \( k=3 \):
\( \binom{4}{3} = \frac{4!}{3!1!} = 4 \). This is equivalent to counting how many people we leave out (4 choices, one for each person not selected).
ANSWER 2: B
Problem 3:
We check the product of each pair:
- A: \( (-4) \times (-9) = 36 \)
- B: \( (-3) \times (-12) = 36 \)
- C: \( \frac{1}{2} \times (-72) = -36 \neq 36 \)
- D: \( 1 \times 36 = 36 \)
- E: \( \frac{3}{2} \times 24 = 36 \)
Only pair C does not have a product of 36.
ANSWER 3: C
Problem 4:
Convert all units to centimeters:
- Ceiling height: \( 2.4 \, \text{m} = 240 \, \text{cm} \)
- Light bulb height from floor: \( 240 - 10 = 230 \, \text{cm} \)
- Alice's reach from floor (without stool): \( 1.5 \, \text{m} + 0.46 \, \text{m} = 150 + 46 = 196 \, \text{cm} \)
Let \( h \) be the stool height. The total reach with the stool equals the light bulb height:
\( h + 196 = 230 \)
\( h = 230 - 196 = 34 \, \text{cm} \)
ANSWER 4: B
Problem 5:
The ratio of trout to total fish in the sample equals the ratio in the lake. The sample has \( \frac{30}{180} = \frac{1}{6} \) trout. Let \( T \) be the total number of fish in the lake. We know there are 250 trout, so:
\( \frac{250}{T} = \frac{1}{6} \)
\( T = 250 \times 6 = 1500 \)
ANSWER 5: B
Problem 6:
Karl bought 5 folders at $2.50 each. The 20% discount applies to each folder, so the savings per folder is \( 2.50 \times 0.20 = 0.50 \). Total savings for 5 folders: \( 5 \times 0.50 = 2.50 \).
ANSWER 6: C
Problem 7:
Total time to paint 300 dimples: \( 300 \times 2 = 600 \) seconds. Convert to minutes: \( 600 \div 60 = 10 \) minutes.
ANSWER 7: D
Problem 8:
Let \( x \) = number of $5 bills, \( y \) = number of $2 bills. We need \( 5x + 2y = 17 \), where \( x,y \geq 0 \) integers. Test non-negative integer values of \( x \):
- \( x=0 \): \( 2y=17 \) → \( y=8.5 \) (not integer, invalid)
- \( x=1 \): \( 5 + 2y=17 \) → \( 2y=12 \) → \( y=6 \) (valid: 1 $5 + 6 $2 = $17)
- \( x=2 \): \( 10 + 2y=17 \) → \( 2y=7 \) → \( y=3.5 \) (invalid)
- \( x=3 \): \( 15 + 2y=17 \) → \( 2y=2 \) → \( y=1 \) (valid: 3 $5 + 1 $2 = $17)
- \( x=4 \): \( 20 > 17 \) (stop, no more valid \( x \))
Only 2 valid combinations.
ANSWER 8: A
Problem 9:
Convert 1000 minutes to hours and minutes: \( 1000 \div 60 = 16 \) hours with a remainder of 40 minutes (since \( 16 \times 60 = 960 \), \( 1000 - 960 = 40 \)). The contest starts at noon, so adding 16 hours 40 minutes: noon + 12 hours = midnight, plus 4 hours 40 minutes = 4:40 a.m.
ANSWER 9: D
Problem 10:
Use order of operations (multiplication before addition):
First parentheses: \( 8 \times 4 + 2 = 32 + 2 = 34 \)
Second parentheses: \( 8 + 4 \times 2 = 8 + 8 = 16 \)
Subtract: \( 34 - 16 = 18 \)
ANSWER 10: D
Problem 11:
Convert 11 hours 5 minutes to total minutes: \( 11 \times 60 + 5 = 660 + 5 = 665 \)
ANSWER 11: C
Problem 12:
Georgie chooses an entrance window first (6 options), then a different exit window (5 remaining options, since exit cannot equal entrance). Since entrance and exit are distinct actions, order matters, so total ways: \( 6 \times 5 = 30 \).
ANSWER 12: D

I'll work through each problem systematically.

**Problem 1:**

**What is being asked:** How many socks must be removed to guarantee 5 socks of the same color?

**Setup:** There are 3 colors (red, white, blue). In the worst case scenario, the Martian pulls out socks as evenly distributed among colors as possible before getting 5 of one color.

**Working:** 
- Worst case: 4 red, 4 white, 4 blue = 12 socks (still no color with 5)
- The 13th sock must give us a 5th sock of some color

**Answer 1: D**

---

**Problem 2:**

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

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

**Working:**
- C(4,3) = 4 ways
- The 4 ways correspond to leaving out each person: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred}

**Answer 2: B**

---

**Problem 3:**

**What is being asked:** Which pair does NOT have a product equal to 36?

**Working:** Check each pair:
- A. {−4, −9}: (−4)(−9) = 36 ✓
- B. {−3, −12}: (−3)(−12) = 36 ✓
- C. {1/2, −72}: (1/2)(−72) = −36 ✗
- D. {1, 36}: (1)(36) = 36 ✓
- E. {3/2, 24}: (3/2)(24) = 36 ✓

**Answer 3: C**

---

**Problem 4:**

**What is being asked:** How high is the stool?

**Setup:** Convert all measurements to centimeters:
- Ceiling height: 240 cm
- Light bulb location: 240 − 10 = 230 cm from floor
- Alice's height: 150 cm
- Alice's reach above her head: 46 cm
- Alice's reach without stool: 150 + 46 = 196 cm

**Working:**
- With stool, Alice can just reach the bulb at 230 cm
- Stool height + 196 cm = 230 cm
- Stool height = 230 − 196 = 34 cm

**Answer 4: B**

---

**Problem 5:**

**What is being asked:** How many fish total are in the lake?

**Setup:** Use proportions. In the sample: 30 trout out of 180 fish. In the lake: 250 trout out of total fish.

**Working:**
- 30/180 = 250/x
- 1/6 = 250/x
- x = 250 × 6 = 1500 fish

**Answer 5: B**

---

**Problem 6:**

**What is being asked:** How much could Karl have saved with the 20% off sale?

**Setup:** 
- Original cost: 5 folders × $2.50 = $12.50
- Sale discount: 20% off

**Working:**
- 20% of $12.50 = 0.20 × $12.50 = $2.50

**Answer 6: C**

---

**Problem 7:**

**What is being asked:** How many minutes to paint 300 dimples at 2 seconds each?

**Working:**
- Total time: 300 × 2 = 600 seconds
- Convert to minutes: 600 ÷ 60 = 10 minutes

**Answer 7: D**

---

**Problem 8:**

**What is being asked:** How many combinations of $5 bills and $2 bills make $17?

**Setup:** Let x = number of $5 bills, y = number of $2 bills
- 5x + 2y = 17

**Working:**
- x = 0: 2y = 17 → y = 8.5 (not integer)
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 (not integer)
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x ≥ 4: 5x ≥ 20 > 17 (impossible)

Valid combinations: (1,$5 and 6,$2) and (3,$5 and 1,$2) = 2 combinations

**Answer 8: A**

---

**Problem 9:**

**What is being asked:** When did the contest end, starting at noon and lasting 1000 minutes?

**Working:**
- 1000 minutes ÷ 60 = 16 hours and 40 minutes
- Noon + 16 hours 40 minutes = 4:40 p.m. the same day
- But we need to continue: 4:40 p.m. + additional time consideration
- Actually: 1000 ÷ 60 = 16 remainder 40, so 16 hours 40 minutes
- Noon + 16 hours 40 minutes = 4:40 a.m. the next day

**Answer 9: D**

---

**Problem 10:**

**What is being asked:** Calculate (8 × 4 + 2) − (8 + 4 × 2)

**Working:**
- First parentheses: 8 × 4 + 2 = 32 + 2 = 34
- Second parentheses: 8 + 4 × 2 = 8 + 8 = 16
- Result: 34 − 16 = 18

**Answer 10: D**

---

**Problem 11:**

**What is being asked:** Convert 11 hours 5 minutes to minutes

**Working:**
- 11 hours = 11 × 60 = 660 minutes
- Total: 660 + 5 = 665 minutes

**Answer 11: C**

---

**Problem 12:**

**What is being asked:** In how many ways can the ghost enter by one window and leave by a different window?

**Working:**
- Choices for entry: 6 windows
- Choices for exit: 5 windows (must be different from entry)
- Total ways: 6 × 5 = 30

**Answer 12: D**

Problem 1:

To be certain of getting 5 socks of the same color, we use the pigeonhole principle.

There are 3 colors: red, white, and blue.  
To avoid getting 5 of any one color for as long as possible, Martian could take at most 4 of each color:

- 4 red
- 4 white
- 4 blue

That gives a total of \(4+4+4=12\) socks without yet having 5 of the same color.

The very next sock, the 13th, must make one color appear at least 5 times.

So the Martian must remove 13 socks.

ANSWER 1: D

Problem 2:

There are 4 players: Lance, Sally, Joy, and Fred.

We need to choose 3 starters from these 4 players. Since order does not matter, we count combinations:

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

Indeed, the 3-starter groups are:

- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred

So there are 4 ways.

ANSWER 2: B

Problem 3:

We check each product.

- A: \((-4)(-9)=36\)
- B: \((-3)(-12)=36\)
- C: \(\frac12 \cdot (-72)=-36\), not 36
- D: \(1\cdot 36=36\)
- E: \(\frac32 \cdot 24=36\)

Only choice C does **not** have product 36.

ANSWER 3: C

Problem 4:

First find how high the bulb is above the floor.

The ceiling is 2.4 meters above the floor, which is \(240\) centimeters.  
The bulb is 10 centimeters below the ceiling, so it is:

\[
240 - 10 = 230 \text{ cm}
\]

Now Alice’s height is 1.5 meters = 150 cm.  
She can reach 46 cm above her head, so her reach without the stool is:

\[
150 + 46 = 196 \text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230 - 196 = 34 \text{ cm}
\]

So the stool is 34 cm high.

ANSWER 4: B

Problem 5:

The sample has 30 trout out of 180 fish, so the fraction of trout is:

\[
\frac{30}{180}=\frac{1}{6}
\]

If 250 trout is \(\frac16\) of the total lake population, then the total number of fish is:

\[
250 \div \frac16 = 250 \cdot 6 = 1500
\]

So there are 1500 fish in the lake.

ANSWER 5: B

Problem 6:

Karl bought 5 folders at \$2.50 each, so the total cost was:

\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:

\[
0.20 \times 12.50 = 2.50
\]

So he could have saved \$2.50.

ANSWER 6: C

Problem 7:

He paints 300 dimples, and each dimple takes 2 seconds.

Total time in seconds:

\[
300 \times 2 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10
\]

So he needs 10 minutes.

ANSWER 7: D

Problem 8:

We want combinations of 5-dollar bills and 2-dollar bills that total \$17.

Let \(x\) be the number of \$5 bills and \(y\) the number of \$2 bills. Then:

\[
5x + 2y = 17
\]

Try values of \(x\):

- \(x=1\): \(17-5=12\), so \(y=6\)
- \(x=2\): \(17-10=7\), not even, so impossible
- \(x=3\): \(17-15=2\), so \(y=1\)

These are the only nonnegative integer solutions.

So there are 2 combinations.

ANSWER 8: A

Problem 9:

The contest ended 1000 minutes after noon.

Convert 1000 minutes to hours and minutes:

\[
1000 \div 60 = 16 \text{ hours with } 40 \text{ minutes left}
\]

So:

- Noon + 12 hours = midnight
- plus 4 more hours = 4:00 a.m.
- plus 40 minutes = 4:40 a.m.

So the contest ended at 4:40 a.m.

ANSWER 9: D

Problem 10:

Compute each part using order of operations:

\[
(8\times 4 + 2) - (8 + 4\times 2)
\]

First parentheses:

\[
8\times 4 + 2 = 32 + 2 = 34
\]

Second parentheses:

\[
8 + 4\times 2 = 8 + 8 = 16
\]

Now subtract:

\[
34 - 16 = 18
\]

So the value is 18.

ANSWER 10: D

Problem 11:

We convert 11 hours and 5 minutes into minutes.

\[
11 \times 60 = 660
\]

Then add 5 minutes:

\[
660 + 5 = 665
\]

So the match lasted 665 minutes.

ANSWER 11: C

Problem 12:

Georgie must enter by one window and leave by a different window.

There are 6 choices for the entering window.  
After choosing one, there are 5 remaining choices for the exit window.

So the total number of ways is:

\[
6 \times 5 = 30
\]

Therefore, there are 30 ways.

ANSWER 12: D

Problem 1:  
We need the minimum number of socks to remove to guarantee there are 5 socks of the same color, given 3 colors (red, white, blue) and at least 5 of each color.

Worst case: you could first pull 4 socks of each color without reaching 5 of any color.  
That would be:
- 4 red + 4 white + 4 blue = 12 socks pulled

After 12 socks, you have at most 4 of each color. The next sock (the 13th) must be one of the 3 colors, and by the pigeonhole principle it makes one color reach 5.

So the Martian must remove **13** socks.

Tempting wrong choices:
- 12 isn’t enough because it’s possible to have only 4 of each color.
- Any number less than 13 can’t force a fifth sock of some color.

ANSWER 1: D


Problem 2:  
We have 4 students: Lance, Sally, Joy, Fred. We must choose 3 starters.

Number of ways to choose 3 out of 4 is:
\[
\binom{4}{3} = 4
\]

(You can also think: leave out one student; there are 4 choices for who to leave out.)

ANSWER 2: B


Problem 3:  
We check which pair does **NOT** multiply to 36.

Choices:
A. \(\{-4,-9\}\): \((-4)(-9)=36\) ✅  
B. \(\{-3,-12\}\): \((-3)(-12)=36\) ✅  
C. \(\left\{\frac{1}{2},-72\right\}\): \(\frac{1}{2}\cdot(-72)=-36\) ❌ not 36  
D. \(\{1,36\}\): \(1\cdot 36=36\) ✅  
E. \(\left\{\frac{3}{2},24\right\}\): \(\frac{3}{2}\cdot 24=3\cdot 12=36\) ✅

So the one that does not have product 36 is choice C.

ANSWER 3: C


Problem 4:  
Let’s convert everything to the same reference (floor).

- Ceiling is 2.4 m above the floor = 240 cm  
- The bulb is 10 cm below the ceiling: height of bulb above floor:
\[
240 - 10 = 230 \text{ cm}
\]
- Alice is 1.5 m tall = 150 cm  
- She can reach 46 cm above the top of her head, so from her top to the bulb:
\[
150 + 46 = 196 \text{ cm (if she stands on the floor)}
\]

She is standing on a stool, so her total height is (stool height + 150). She can reach:
\[
(\text{stool height} + 150) + 46
\]
Set equal to bulb height 230:
\[
\text{stool height} + 196 = 230
\]
\[
\text{stool height} = 34
\]

ANSWER 4: B


Problem 5:  
We know:
- Lake has 250 trout.
- Total fish in lake = \(T\)
- Sample has 180 fish, and 30 are trout.
So sample trout-to-total ratio equals lake trout-to-total ratio:
\[
\frac{30}{180} = \frac{250}{T}
\]
Simplify \(\frac{30}{180}=\frac{1}{6}\):
\[
\frac{1}{6} = \frac{250}{T}
\]
\[
T = 250 \cdot 6 = 1500
\]

ANSWER 5: B


Problem 6:  
Cost per folder originally: $2.50  
He bought 5 folders:
\[
5 \cdot 2.50 = 12.50
\]
A 20%-off sale means paying 80% of the original price:
\[
0.80 \cdot 12.50 = 10.00
\]
Savings:
\[
12.50 - 10.00 = 2.50
\]

ANSWER 6: C


Problem 7:  
Paint 300 dimples, 2 seconds per dimple.
Total time in seconds:
\[
300 \cdot 2 = 600 \text{ seconds}
\]
Convert to minutes (60 seconds per minute):
\[
600 \div 60 = 10 \text{ minutes}
\]

ANSWER 7: D


Problem 8:  
We need numbers of $5 bills and $2 bills that total $17 (order doesn’t matter).

Let \(x\) = number of $5 bills, \(y\) = number of $2 bills.
\[
5x + 2y = 17
\]
Try \(x=0\) to \(x=3\) (since \(4\cdot 5=20>17\)):

- \(x=0\): \(2y=17\) not possible  
- \(x=1\): \(5 + 2y=17 \Rightarrow 2y=12 \Rightarrow y=6\)  
- \(x=2\): \(10 + 2y=17 \Rightarrow 2y=7\) not possible  
- \(x=3\): \(15 + 2y=17 \Rightarrow 2y=2 \Rightarrow y=1\)

So there are 2 combinations: \((x,y)=(1,6)\) and \((3,1)\).

ANSWER 8: A


Problem 9:  
The contest starts at noon. It ends 1000 minutes later.

Convert 1000 minutes to hours:
\[
1000 \div 60 = 16 \text{ hours remainder } 40 \text{ minutes}
\]
Noon + 16 hours = 4:00 a.m. (next day)  
Add 40 minutes:
\[
4:00 \text{ a.m.} + 40 \text{ min} = 4:40 \text{ a.m.}
\]

ANSWER 9: D


Problem 10:  
Compute:
\[
(8 \times 4 + 2) - (8 + 4 \times 2)
\]
First parentheses:
\[
8\times 4 = 32,\quad 32+2=34
\]
Second parentheses:
\[
4\times 2 = 8,\quad 8+8=16
\]
Now subtract:
\[
34 - 16 = 18
\]

ANSWER 10: D


Problem 11:  
11 hours and 5 minutes total.
Convert hours to minutes:
\[
11 \cdot 60 = 660
\]
Add 5 minutes:
\[
660 + 5 = 665
\]

ANSWER 11: C


Problem 12:  
There are 6 windows. Georgie enters through one window and must leave through a different window.

Choose entering window: 6 choices.  
Choose leaving window: 5 choices (cannot be the same).

Total:
\[
6 \cdot 5 = 30
\]

ANSWER 12: D

Problem 1:
To be certain of having 5 socks of the same color, we consider the "worst-case scenario" (Pigeonhole Principle). If we have 3 colors (red, white, blue), we could pick 4 red, 4 white, and 4 blue socks without having 5 of any one color. This totals $4 + 4 + 4 = 12$ socks. The very next sock (the 13th) must be either red, white, or blue, which will complete a set of 5 for that color.
ANSWER 1: D

Problem 2:
We need to choose 3 starters from a group of 4 people (Lance, Sally, Joy, Fred). This is a combination problem denoted as $\binom{4}{3}$. Choosing 3 people to be on the team is the same as choosing 1 person to be left out. There are 4 ways to choose who is left out (Lance, Sally, Joy, or Fred).
ANSWER 2: B

Problem 3:
We check the products:
A: $(-4) \times (-9) = 36$
B: $(-3) \times (-12) = 36$
C: $\frac{1}{2} \times (-72) = -36$
D: $1 \times 36 = 36$
E: $\frac{3}{2} \times 24 = 3 \times 12 = 36$
Choice C results in $-36$, not $36$.
ANSWER 3: C

Problem 4:
Ceiling height = 2.4 m = 240 cm. Light bulb height = 240 - 10 = 230 cm from the floor.
Alice's reach = 150 cm (height) + 46 cm (reach) = 196 cm.
Stool height = (Bulb height) - (Alice's reach) = 230 - 196 = 34 cm.
ANSWER 4: B

Problem 5:
Let $T$ be the total number of fish in the lake. The ratio of trout to total fish is $\frac{30}{180} = \frac{1}{6}$.
Set up the proportion: $\frac{250}{T} = \frac{1}{6}$.
$T = 250 \times 6 = 1500$.
ANSWER 5: B

Problem 6:
Total cost = 5 folders × $2.50 = $12.50.
Savings = 20% of $12.50 = 0.20 \times 12.50 = $2.50.
Alternatively, 20% off one folder is $0.20 \times 2.50 = $0.50. For 5 folders, $0.50 \times 5 = $2.50.
ANSWER 6: C

Problem 7:
Total time in seconds = 300 dimples × 2 seconds/dimple = 600 seconds.
Total time in minutes = 600 seconds / 60 seconds/minute = 10 minutes.
ANSWER 7: D

Problem 8:
Let $x$ be the number of $5 bills and $y$ be the number of $2 bills. $5x + 2y = 17$.
If $x=1$, $5+2y=17 \Rightarrow 2y=12 \Rightarrow y=6$. (1, 6)
If $x=3$, $15+2y=17 \Rightarrow 2y=2 \Rightarrow y=1$. (3, 1)
If $x=0$ or $x=2$, $2y$ would be 17 or 7, which are not divisible by 2.
There are 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours and 40 minutes (since $1000 \div 60 = 16$ remainder 40).
Noon + 12 hours = midnight.
Midnight + 4 hours and 40 minutes = 4:40 a.m.
ANSWER 9: D

Problem 10:
$(8 \times 4 + 2) - (8 + 4 \times 2) = (32 + 2) - (8 + 8) = 34 - 16 = 18$.
ANSWER 10: D

Problem 11:
11 hours = $11 \times 60 = 660$ minutes.
Total = 660 + 5 = 665 minutes.
ANSWER 11: C

Problem 12:
There are 6 choices for the entry window. Since the exit window must be different, there are 5 choices for the exit window.
Total ways = $6 \times 5 = 30$.
ANSWER 12: D

Problem 1:
This is a pigeonhole principle problem. To guarantee at least 5 socks of one color, first consider the worst-case scenario of taking as many socks as possible without having 5 of any single color. That means taking exactly 4 socks of each of the 3 colors, for a total of \(4 \times 3 = 12\) socks. The next sock (the 13th) must be a 5th sock of one of the colors. Thus 13 socks are required. (Fewer than 13 leaves open the possibility of only 4 of each color.)

ANSWER 1: D

Problem 2:
The number of ways to choose 3 starters out of the 4 available players is the combination \(\binom{4}{3} = 4\). Each choice simply leaves out one of the four players.

ANSWER 2: B

Problem 3:
Compute the product for each pair:
- A: \((-4) \times (-9) = 36\)
- B: \((-3) \times (-12) = 36\)
- C: \(\frac12 \times (-72) = -36\) (not 36)
- D: \(1 \times 36 = 36\)
- E: \(\frac32 \times 24 = 36\)

Only pair C fails to give a product of 36.

ANSWER 3: C

Problem 4:
Convert all lengths to centimeters. The ceiling is 240 cm above the floor, so the bulb is 230 cm above the floor. Alice's reach is \(150 + 46 = 196\) cm. The stool must therefore supply the remaining \(230 - 196 = 34\) cm.

ANSWER 4: B

Problem 5:
The sample shows that trout comprise \(\frac{30}{180} = \frac16\) of all fish. With 250 trout in the lake, the total number of fish is \(250 \div \frac16 = 1500\).

ANSWER 5: B

Problem 6:
Five folders cost \(5 \times 2.50 = 12.50\). A 20 % discount saves \(0.20 \times 12.50 = 2.50\).

ANSWER 6: C

Problem 7:
Painting 300 dimples at 2 seconds each requires \(300 \times 2 = 600\) seconds, or \(600 \div 60 = 10\) minutes.

ANSWER 7: D

Problem 8:
Solve \(5x + 2y = 17\) for non-negative integers \(x, y\):
- \(x = 1\), \(y = 6\) works;
- \(x = 3\), \(y = 1\) works.
No other non-negative integer solutions exist, so there are exactly two combinations.

ANSWER 8: A

Problem 9:
Divide 1000 minutes by 60: \(1000 = 16\) hours + 40 minutes. Adding 16 hours to noon yields 4:00 a.m.; adding 40 minutes yields 4:40 a.m.

ANSWER 9: D

Problem 10:
Left side: \(8 \times 4 + 2 = 34\). Right side: \(8 + 4 \times 2 = 16\). Difference: \(34 - 16 = 18\).

ANSWER 10: D

Problem 11:
Convert 11 hours 5 minutes to minutes: \(11 \times 60 + 5 = 665\).

ANSWER 11: C

Problem 12:
There are 6 choices for the entry window and then 5 remaining choices for the exit window, giving \(6 \times 5 = 30\) ordered pairs.

ANSWER 12: D

Problem 1:
The problem is asking us to determine the minimum number of socks the Martian must remove from the drawer to guarantee that there are 5 socks of the same color. 
To ensure that there are 5 socks of the same color, we must consider the worst-case scenario where the Martian draws as many socks as possible without getting 5 of the same color. 
The Martian can draw at most 4 socks of each color without having 5 socks of the same color. 
So, the Martian can draw 4 red, 4 white, and 4 blue socks, for a total of 4 + 4 + 4 = 12 socks. 
At this point, the Martian still doesn't have 5 socks of the same color. 
However, drawing one more sock will guarantee that there are 5 socks of the same color. 
Therefore, the Martian must draw at least 12 + 1 = 13 socks.
Let's examine the choices: 
A. 6 is too low because the Martian could draw 3 red, 2 white, and 1 blue, or other combinations that don't guarantee 5 socks of the same color. 
B. 9 is still too low because the Martian could draw 4 red, 3 white, and 2 blue, or other similar combinations. 
C. 12 is the maximum number of socks the Martian can draw without having 5 socks of the same color. 
D. 13 is the correct answer because it is one more than the maximum number of socks that can be drawn without having 5 socks of the same color. 
E. 15 is more than necessary.
ANSWER 1: D

Problem 2:
The problem is asking us to find the number of ways to choose 3 starters from a team of 4 players: Lance, Sally, Joy, and Fred. 
This is a combination problem, because the order in which the starters are chosen does not matter. 
The number of ways to choose 3 starters from 4 players can be calculated using the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of players and k is the number of starters. 
Here, n = 4 and k = 3, so C(4, 3) = 4! / (3!(4-3)!) = 4.
Let's list the possible combinations to verify: 
- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred
There are indeed 4 combinations.
Let's examine the choices: 
A. 2 is too low. 
B. 4 matches our calculation. 
C. 6 is too high. 
D. 8 is too high. 
E. 10 is too high.
ANSWER 2: B

Problem 3:
The problem is asking us to identify the pair of numbers that does NOT have a product equal to 36. 
Let's check each pair: 
A. {−4, −9}: (−4) × (−9) = 36
B. {−3, −12}: (−3) × (−12) = 36
C. { 1 2 , −72}: (1/2) × (−72) = −36
D. {1, 36}: 1 × 36 = 36
E. { 3 2 , 24}: (3/2) × 24 = 36
The pair that does NOT have a product equal to 36 is C.
Let's examine the choices: 
A. {−4, −9} has a product of 36. 
B. {−3, −12} has a product of 36. 
C. { 1 2 , −72} has a product of −36, not 36. 
D. {1, 36} has a product of 36. 
E. { 3 2 , 24} has a product of 36.
ANSWER 3: C

Problem 4:
The problem is asking us to find the height of the stool that Alice needs to stand on to reach the light bulb. 
First, we need to convert the height of the ceiling from meters to centimeters: 2.4 meters = 240 centimeters. 
The light bulb is 10 centimeters below the ceiling, so its height above the floor is 240 - 10 = 230 centimeters. 
Alice is 1.5 meters = 150 centimeters tall and can reach 46 centimeters above her head, so her total reach is 150 + 46 = 196 centimeters. 
To find the height of the stool, we subtract Alice's total reach from the height of the light bulb: 230 - 196 = 34 centimeters.
Let's examine the choices: 
A. 32 is too low because 230 - (196 + 32) = 2, which is not enough to reach the light bulb. 
B. 34 is the correct height. 
C. 36 is too high. 
D. 38 is too high. 
E. 40 is too high.
ANSWER 4: B

Problem 5:
The problem is asking us to find the total number of fish in the lake based on the ratio of trout in the sample to the total number of fish in the sample. 
The sample contains 30 trout out of 180 fish, so the ratio of trout to total fish in the sample is 30/180. 
We are given that this ratio is the same as the ratio of trout to total fish in the lake. 
Let's denote the total number of fish in the lake as x. 
The ratio of trout to total fish in the lake is 250/x. 
Setting the two ratios equal, we get 30/180 = 250/x. 
Cross-multiplying, we have 30x = 250 × 180. 
Simplifying, 30x = 45000. 
Dividing both sides by 30, x = 1500.
Let's examine the choices: 
A. 1250 is too low. 
B. 1500 matches our calculation. 
C. 1750 is too high. 
D. 1800 is too high. 
E. 2000 is too high.
ANSWER 5: B

Problem 6:
The problem is asking us to find out how much Karl could have saved by waiting a day to buy the folders at a 20%-off sale. 
The original price of 5 folders is 5 × $2.50 = $12.50. 
At a 20%-off sale, the discount is 20% of $12.50, which is 0.20 × $12.50 = $2.50. 
So, Karl could have saved $2.50.
Let's examine the choices: 
A. $1.00 is too low. 
B. $2.00 is too low. 
C. $2.50 matches our calculation. 
D. $2.75 is too high. 
E. $5.00 is too high.
ANSWER 6: C

Problem 7:
The problem is asking us to find the time it will take Casey to paint 300 dimples on a golf ball. 
Casey takes 2 seconds to paint one dimple, so to paint 300 dimples, he will take 300 × 2 = 600 seconds. 
To convert seconds to minutes, we divide by 60: 600 / 60 = 10 minutes.
Let's examine the choices: 
A. 4 minutes is too low. 
B. 6 minutes is too low. 
C. 8 minutes is too low. 
D. 10 minutes matches our calculation. 
E. 12 minutes is too high.
ANSWER 7: D

Problem 8:
The problem is asking us to find the number of different combinations of $5 bills and $2 bills that can be used to make a total of $17. 
Let's list the possible combinations: 
1. 3 $5 bills = $15, and $2 bill = $2, total = $17 (1 combination: 3 $5 bills and 1 $2 bill)
2. 1 $5 bill = $5, and 6 $2 bills = $12, total = $17 (1 combination: 1 $5 bill and 6 $2 bills)
We should also check if there are any other combinations: 
- Using 0 $5 bills, we need 8.5 $2 bills, which is not possible since we can't have a fraction of a bill.
- Using 2 $5 bills = $10, we need $7 more, which can be made with 3.5 $2 bills, again not possible.
- Using 4 $5 bills = $20, which is more than $17.

Problem 1:
We want the minimum number of socks to guarantee 5 of the same color. With 3 colors (red, white, blue), the worst case is picking 4 of each color: 4+4+4 = 12 socks. The 13th sock must create a group of 5 of one color. So 13 socks are needed.
ANSWER 1: D

Problem 2:
We choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways is the combination C(4,3) = 4.
ANSWER 2: B

Problem 3:
Compute each product:
A: (−4)×(−9) = 36
B: (−3)×(−12) = 36
C: (1/2)×(−72) = −36
D: 1×36 = 36
E: (3/2)×24 = 36
Only C does not equal 36.
ANSWER 3: C

Problem 4:
Ceiling = 2.4 m = 240 cm. Light bulb is 10 cm below ceiling, so its height = 230 cm.
Alice's height = 1.5 m = 150 cm. Reach above head = 46 cm, so reach without stool = 196 cm.
Stool height needed = 230 − 196 = 34 cm.
ANSWER 4: B

Problem 5:
Sample ratio: 30 trout / 180 fish = 1/6. Lake has 250 trout, which is 1/6 of total fish. Total fish = 250 × 6 = 1500.
ANSWER 5: B

Problem 6:
Cost without sale: 5 × $2.50 = $12.50.
With 20% off, each folder costs $2.50 × 0.8 = $2.00. Total = 5 × $2.00 = $10.00.
Savings = $12.50 − $10.00 = $2.50.
ANSWER 6: C

Problem 7:
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.
ANSWER 7: D

Problem 8:
Find non-negative integers x ($5 bills) and y ($2 bills) such that 5x + 2y = 17.
x=1 → 5 + 2y = 17 → y=6.
x=3 → 15 + 2y = 17 → y=1.
No other x works. So 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours 40 minutes (16×60 = 960, remainder 40).
Noon + 16 hours = 4:00 a.m. next day. Add 40 minutes → 4:40 a.m.
ANSWER 9: D

Problem 10:
(8 × 4 + 2) = 32 + 2 = 34.
(8 + 4 × 2) = 8 + 8 = 16.
34 − 16 = 18.
ANSWER 10: D

Problem 11:
11 hours × 60 = 660 minutes. Plus 5 minutes = 665 minutes.
ANSWER 11: C

Problem 12:
6 choices for entry window, then 5 choices for a different exit window. Total = 6 × 5 = 30.
ANSWER 12: D

Problem 1:
We need to find the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). Using the Pigeonhole Principle, we consider the worst-case scenario: the Martian pulls out as many socks as possible *without* getting 5 of any one color. That means getting exactly 4 of each color:
- 4 red + 4 white + 4 blue = 12 socks
After 12 socks, the Martian could have 4 of each color and still not have 5 matching. The very next sock (the 13th) must be red, white, or blue, pushing the count of that color to 5. So 13 socks guarantees 5 of the same color.
(The "five-legged" detail is flavor text explaining why the Martian needs 5 socks.)

ANSWER 1: D

Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). This is a combination problem since order doesn't matter:
C(4, 3) = 4! / (3! × 1!) = 4
We can also list them: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — exactly 4 ways.

ANSWER 2: B

Problem 3:
We check each product:
- A: (−4) × (−9) = 36 ✓
- B: (−3) × (−12) = 36 ✓
- C: (1/2) × (−72) = −36 ✗ (This equals −36, NOT 36)
- D: 1 × 36 = 36 ✓
- E: (3/2) × 24 = 36 ✓
Choice C gives −36, which does not equal 36.

ANSWER 3: C

Problem 4:
First, convert everything to centimeters:
- Ceiling height: 2.4 m = 240 cm
- Light bulb position: 240 − 10 = 230 cm above the floor
- Alice's height: 1.5 m = 150 cm
- Alice's reach above her head: 46 cm
- Alice's maximum reach (without stool): 150 + 46 = 196 cm above the floor

Let h = height of the stool. Standing on the stool, her reach is h + 196 cm. She can "just reach" the bulb:
h + 196 = 230
h = 230 − 196 = 34 cm

ANSWER 4: B

Problem 5:
The ratio of trout to total fish in the sample should equal the ratio in the lake:
- Sample ratio: 30 trout / 180 fish = 1/6
- Lake ratio: 250 trout / Total fish = 1/6

Setting up the proportion: 250 / Total = 1/6
Total = 250 × 6 = 1500 fish

ANSWER 5: B

Problem 6:
Karl bought 5 folders at $2.50 each:
Total cost = 5 × $2.50 = $12.50
A 20%-off sale means he would save 20% of the total:
Savings = 0.20 × $12.50 = $2.50

ANSWER 6: C

Problem 7:
Total time in seconds: 300 dimples × 2 seconds/dimple = 600 seconds
Convert to minutes: 600 seconds ÷ 60 seconds/minute = 10 minutes

ANSWER 7: D

Problem 8:
We need non-negative integer solutions to 5x + 2y = 17, where x = number of $5 bills and y = number of $2 bills.
- x = 0: 2y = 17 → y = 8.5 (not an integer) ✗
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 ✗
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x = 4: 20 > 17 (too large) ✗

Only 2 valid combinations: (1 five + 6 twos) and (3 fives + 1 two).

ANSWER 8: A

Problem 9:
Convert 1000 minutes to hours and minutes:
1000 ÷ 60 = 16 remainder 40, so 1000 minutes = 16 hours 40 minutes.
Starting at noon (12:00 PM):
- 12 hours later = midnight (12:00 AM)
- 4 more hours = 4:00 AM
- 40 more minutes = 4:40 AM

ANSWER 9: D

Problem 10:
Evaluate each part using order of operations (multiplication before addition):
- First part: (8 × 4 + 2) = (32 + 2) = 34
- Second part: (8 + 4 × 2) = (8 + 8) = 16
- Difference: 34 − 16 = 18

ANSWER 10: D

Problem 11:
Convert 11 hours and 5 minutes to minutes:
11 hours × 60 minutes/hour = 660 minutes
660 + 5 = 665 minutes

ANSWER 11: C

Problem 12:
Georgie enters through one window and leaves through a different one.
- Choices for entry window: 6
- Choices for exit window (must be different): 5
Total ways = 6 × 5 = 30

ANSWER 12: D

Problem 1:
The Martian needs to be **certain** of having 5 socks of the same color. To guarantee this, we must consider the worst-case scenario: the Martian picks as many socks as possible **without** yet having 5 of any single color.  
With 3 colors (red, white, blue), the worst case is picking 4 socks of each color:  
\(4 + 4 + 4 = 12\) socks.  
At this point the Martian has 4 of every color but still does not have 5 of any one color. The very next sock—the 13th—must be red, white, or blue, and will necessarily create a set of 5 of that color.  
Thus 13 socks are required.  
(Choice A is too small; with 6 socks one could have 2,2,2. Choice B allows 3,3,3. Choice C allows 4,4,4. Choice E is more than the minimum guaranteed number.)

ANSWER 1: D

---

Problem 2:
We need to choose a group of 3 starters from the 4 students: Lance, Sally, Joy, and Fred. Since the three starters form a team and order does not matter, this is a combination problem.  
\[
\binom{4}{3} = \frac{4!}{3!\,1!} = 4
\]  
Equivalently, we can think of choosing which **one** of the four students will **not** start; there are 4 such choices.  
(Choice A is too small. Choice C is \(\binom{4}{2}\). Choices D and E do not arise from counting 3-person teams from 4 people.)

ANSWER 2: B

---

Problem 3:
We check the product of each pair to see which is not 36.

- **A.** \((-4)(-9) = 36\)  
- **B.** \((-3)(-12) = 36\)  
- **C.** \(\left(\frac12\right)(-72) = -36\)  
- **D.** \((1)(36) = 36\)  
- **E.** \(\left(\frac32\right)(24) = 36\)

Only pair **C** gives \(-36\), not \(36\).

ANSWER 3: C

---

Problem 4:
First convert all measurements to centimeters.  
- Ceiling height: \(2.4\text{ m} = 240\text{ cm}\).  
- Light bulb is \(10\text{ cm}\) below the ceiling, so its height above the floor is \(240 - 10 = 230\text{ cm}\).  

Alice’s reach:  
- Height: \(1.5\text{ m} = 150\text{ cm}\).  
- She can reach \(46\text{ cm}\) above her head, so her standing reach is \(150 + 46 = 196\text{ cm}\).  

The stool must supply the remaining height:  
\[
230 - 196 = 34\text{ cm}.
\]

ANSWER 4: B

---

Problem 5:
Let \(T\) be the total number of fish in the lake. The problem states that the ratio of trout to total fish is the same in the sample and in the lake.  
\[
\frac{250}{T} = \frac{30}{180}
\]  
Simplify the right side: \(\frac{30}{180} = \frac16\).  
\[
\frac{250}{T} = \frac16 \quad\Longrightarrow\quad T = 250 \times 6 = 1500.
\]

ANSWER 5: B

---

Problem 6:
Karl bought 5 folders at \(\$2.50\) each.  
Total cost: \(5 \times \$2.50 = \$12.50\).  
A 20% discount would have saved him 20% of this total:  
\[
0.20 \times \$12.50 = \$2.50.
\]

ANSWER 6: C

---

Problem 7:
Casey paints 300 dimples at 2 seconds per dimple.  
Total time in seconds: \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(\frac{600}{60} = 10\) minutes.

ANSWER 7: D

---

Problem 8:
Let \(f\) be the number of \(\$5\) bills and \(t\) be the number of \(\$2\) bills. We need non-negative integer solutions to  
\[
5f + 2t = 17.
\]  
Test values of \(f\):  
- \(f=0\): \(2t=17\) → no integer solution.  
- \(f=1\): \(5 + 2t = 17 \Rightarrow t=6\).  **(Valid)**
- \(f=2\): \(10 + 2t = 17 \Rightarrow t=3.5\).  (Invalid)
- \(f=3\): \(15 + 2t = 17 \Rightarrow t=1\).  **(Valid)**
- \(f=4\): \(20 > 17\) → stop.

There are exactly **2** combinations: \((1\text{ five}, 6\text{ twos})\) and \((3\text{ fives}, 1\text{ two})\).

ANSWER 8: A

---

Problem 9:
The contest starts at noon (12:00 p.m.). We add 1000 minutes.  
Divide 1000 by 60:  
\[
1000 = 16 \times 60 + 40,
\]  
so 1000 minutes = 16 hours and 40 minutes.  

Noon + 12 hours = midnight.  
Midnight + 4 hours 40 minutes = 4:40 a.m.

ANSWER 9: D

---

Problem 10:
Evaluate using the correct order of operations (multiplication before addition).

First parentheses:  
\[
8 \times 4 + 2 = 32 + 2 = 34.
\]

Second parentheses:  
\[
8 + 4 \times 2 = 8 + 8 = 16.
\]

Subtract:  
\[
34 - 16 = 18.
\]

ANSWER 10: D

---

Problem 11:
Convert 11 hours to minutes and add the remaining 5 minutes.  
\[
11 \times 60 = 660\text{ minutes}.
\]  
\[
660 + 5 = 665\text{ minutes}.
\]

ANSWER 11: C

---

Problem 12:
Georgie enters through one of 6 windows and must leave through a **different** window.  
- 6 choices for the entry window.  
- For each entry choice, 5 remaining choices for the exit window.  

Total ways:  
\[
6 \times 5 = 30.
\]  
(Choice E, 36, would count entering and leaving through the same window, which is not allowed.)

ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). The worst-case scenario is to draw as many socks as possible without having 5 of any one color. That would be 4 red, 4 white, and 4 blue, for a total of 12 socks. The next sock drawn (the 13th) must be one of the three colors, giving that color 5 socks. Therefore, 13 socks are required. The five-legged Martian is irrelevant to the count.  
Answer: D

Problem 2:
We have 4 people (Lance, Sally, Joy, Fred) and need to choose 3 starters. Since the starters are not assigned to specific positions, order does not matter. The number of ways is the combination \( \binom{4}{3} = 4 \).  
Answer: B

Problem 3:
Compute each product:  
A. \((-4) \times (-9) = 36\)  
B. \((-3) \times (-12) = 36\)  
C. \(\frac{1}{2} \times (-72) = -36\)  
D. \(1 \times 36 = 36\)  
E. \(\frac{3}{2} \times 24 = 36\)  
Only pair C gives a product not equal to 36.  
Answer: C

Problem 4:
Convert all measurements to centimeters:  
Ceiling height = 2.4 m = 240 cm.  
Light bulb is 10 cm below ceiling → height = 240 – 10 = 230 cm.  
Alice’s height = 1.5 m = 150 cm.  
Her reach above her head = 46 cm, so her maximum reach without a stool = 150 + 46 = 196 cm.  
Let \(h\) be the stool height. Standing on the stool raises her reach to \(196 + h\). This must equal 230 cm:  
\(196 + h = 230\) → \(h = 34\) cm.  
Answer: B

Problem 5:
In the sample, 30 out of 180 fish are trout, so the ratio trout : total = \(30/180 = 1/6\).  
Assuming the same ratio in the lake, if the lake has 250 trout, then the total number of fish is \(250 \times 6 = 1500\).  
Answer: B

Problem 6:
Original cost for 5 folders: \(5 \times 2.50 = \$12.50\).  
A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
Thus, by waiting a day, Karl could have saved $2.50.  
Answer: C

Problem 7:
Time to paint one dimple = 2 seconds.  
Total time for 300 dimples = \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(600 \div 60 = 10\) minutes.  
Answer: D

Problem 8:
We need nonnegative integers \(a\) (number of $5 bills) and \(b\) (number of $2 bills) such that \(5a + 2b = 17\).  
Solve by testing values of \(b\):  
\(b = 1\) → \(5a = 15\) → \(a = 3\) → (3,1) works.  
\(b = 6\) → \(5a = 5\) → \(a = 1\) → (1,6) works.  
No other values give integer \(a\). Thus there are 2 combinations.  
Answer: A

Problem 9:
1000 minutes = \(1000 \div 60 = 16\) hours and \(40\) minutes (since \(60 \times 16 = 960\), remainder 40).  
Starting at noon (12:00 PM), add 12 hours → midnight (12:00 AM). Add the remaining 4 hours → 4:00 AM. Then add 40 minutes → 4:40 AM.  
Answer: D

Problem 10:
Evaluate using order of operations:  
First parentheses: \(8 \times 4 + 2 = 32 + 2 = 34\).  
Second parentheses: \(8 + 4 \times 2 = 8 + 8 = 16\).  
Subtract: \(34 - 16 = 18\).  
Answer: D

Problem 11:
11 hours = \(11 \times 60 = 660\) minutes.  
Add 5 minutes → \(660 + 5 = 665\) minutes.  
Answer: C

Problem 12:
Georgie must choose an entry window (6 choices) and a different exit window (5 choices). Since entering and leaving are distinct actions, the total number of ways is \(6 \times 5 = 30\).  
Answer: D

ANSWER 1: D
ANSWER 2: B
ANSWER 3: C
ANSWER 4: B
ANSWER 5: B
ANSWER 6: C
ANSWER 7: D
ANSWER 8: A
ANSWER 9: D
ANSWER 10: D
ANSWER 11: C
ANSWER 12: D

Problem 1:  
The Martian wants to be certain of having 5 socks of the same color. There are three colors: red, white, and blue. In the worst-case scenario, the Martian could pick 4 socks of each color before getting a fifth of any one color. That would be \(4 \times 3 = 12\) socks, and still no color has 5 socks. The next sock drawn (the 13th) must complete a set of 5 of some color, because there are only three colors. Therefore, the Martian must remove **13** socks to be certain of having 5 socks of the same color.  
ANSWER 1: D

Problem 2:  
Ms. Hamilton's class has 4 available players: Lance, Sally, Joy, and Fred. The number of ways to choose 3 starters from 4 players is given by the combination \(\binom{4}{3} = 4\). Thus, there are **4** possible sets of three starters.  
ANSWER 2: B

Problem 3:  
We check each pair's product:  
- A: \((-4) \times (-9) = 36\)  
- B: \((-3) \times (-12) = 36\)  
- C: \(\frac{1}{2} \times (-72) = -36\) (not 36)  
- D: \(1 \times 36 = 36\)  
- E: \(\frac{3}{2} \times 24 = 36\)  
Only pair **C** does not have a product of 36.  
ANSWER 3: C

Problem 4:  
The ceiling is 2.4 m above the floor, and the bulb is 10 cm (0.1 m) below the ceiling, so the bulb's height is \(2.4 - 0.1 = 2.3\) m. Alice's height is 1.5 m, and she can reach 46 cm (0.46 m) above her head, so her reach height from the floor is \(1.5 + 0.46 = 1.96\) m. When standing on a stool, her hands are at the stool height plus her reach height. Setting this equal to the bulb height:  
\[
\text{stool height} + 1.96 = 2.3 \Rightarrow \text{stool height} = 0.34 \text{ m} = 34 \text{ cm}.
\]  
ANSWER 4: B

Problem 5:  
In the sample, the ratio of trout to total fish is \(\frac{30}{180} = \frac{1}{6}\). Assuming the same ratio in the lake, let \(T\) be the total number of fish in the lake. Then \(\frac{250}{T} = \frac{1}{6}\), so \(T = 250 \times 6 = 1500\).  
ANSWER 5: B

Problem 6:  
Karl bought 5 folders at $2.50 each, so the original cost is \(5 \times 2.50 = \$12.50\). With a 20% discount, the cost becomes \(12.50 \times 0.80 = \$10.00\). The savings is \(12.50 - 10.00 = \$2.50\).  
ANSWER 6: C

Problem 7:  
Painting 300 dimples at 2 seconds each takes \(300 \times 2 = 600\) seconds. Converting to minutes: \(600 \div 60 = 10\) minutes.  
ANSWER 7: D

Problem 8:  
We need nonnegative integers \(x\) (number of $5 bills) and \(y\) (number of $2 bills) such that \(5x + 2y = 17\). Solving:  
- \(y = 1 \Rightarrow 5x = 15 \Rightarrow x = 3\)  
- \(y = 6 \Rightarrow 5x = 5 \Rightarrow x = 1\)  
- \(y = 11 \Rightarrow 5x = -5\) (invalid)  
Thus, there are **2** valid combinations: (3 fives, 1 two) and (1 five, 6 twos).  
ANSWER 8: A

Problem 9:  
1000 minutes equals \(16\) hours and \(40\) minutes (since \(16 \times 60 = 960\), remainder \(40\)). Starting at noon, adding \(16\) hours gives 4:00 AM the next day, and adding \(40\) minutes gives 4:40 AM.  
ANSWER 9: D

Problem 10:  
Compute inside parentheses:  
\(8 \times 4 + 2 = 32 + 2 = 34\)  
\(8 + 4 \times 2 = 8 + 8 = 16\)  
Then \(34 - 16 = 18\).  
ANSWER 10: D

Problem 11:  
11 hours is \(11 \times 60 = 660\) minutes. Adding 5 minutes gives \(660 + 5 = 665\) minutes.  
ANSWER 11: C

Problem 12:  
Georgie can enter through any of the 6 windows. For exiting, he must choose a different window, so there are 5 choices. The total number of ways is \(6 \times 5 = 30\).  
ANSWER 12: D

Problem 1:  
The Martian needs to ensure there are 5 socks of the same color. The worst-case scenario is pulling the maximum number of socks without 5 of any color. Since there are 3 colors, this would be 4 socks of each color (4 × 3 = 12 socks). The next sock (the 13th) must match one color, giving 5 socks of that color.  
**ANSWER 1: D**  

Problem 2:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is C(4,3) = 4.  
**ANSWER 2: B** (Note: C(4,3)=4, but rechecking: the team is chosen from 4 specific people, so combinations are 4. Answer B=4)  

Problem 3:  
Check each pair:  
A. (−4)(−9) = 36  
B. (−3)(−12) = 36  
C. (1/2)(−72) = -36 (≠36)  
D. (1)(36) = 36  
E. (3/2)(24) = 36  
The pair that does NOT equal 36 is C.  
**ANSWER 3: E** (Re-evaluated: The question asks which does NOT equal 36. Option E: (3/2)*24=36, so it is valid. Option C: (1/2)*(-72)=-36, so C is the correct choice for NOT equal. Correction: **ANSWER 3: C** [Original had a typo, but computation confirms C is the answer])  

*Correction: For Problem 3, Option C: (1/2, -72) likely means (1/2) × (-72) = -36 ≠ 36. Option E: (3/2) × 24 = 36. Thus, C is the answer.*  
**ANSWER 3: C**  

Problem 4:  
Ceiling height: 2.4 m = 240 cm. Light bulb location: 240 - 10 = 230 cm from floor. Alice’s reach: 150 cm + 46 cm = 196 cm. Stool height = 230 - 196 = 34 cm.  
**ANSWER 4: B**  

Problem 5:  
Ratio of trout in sample: 30/180 = 1/6. Total trout: 250. Total fish = 250 × 6 = 1500.  
**ANSWER 5: B**  

Problem 6:  
Original cost: 5 × $2.50 = $12.50. 20% off: $12.50 × 0.20 = $2.50 saved.  
**ANSWER 6: C** (Note: The question asks how much Karl could have saved, which is $2.50, not $2.00. Original had a mistake; correction: 20% of $12.50 is $2.50, so answer C)  

Problem 7:  
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.  
**ANSWER 7: D**  

Problem 8:  
Let $5 bills = x, $2 bills = y. 5x + 2y = 17. Possible solutions:  
- x=3, y=1 (15+2=17)  
- x=1, y=6 (5+12=17)  
Only 2 combinations.  
**ANSWER 8: A**  

Problem 9:  
1000 minutes = 16 hours 40 minutes. Starting at noon: noon + 16h40m = 4:40 a.m. next day.  
**ANSWER 9: D**  

Problem 10:  
(8×4 + 2) = 34, (8 + 4×2) = 16. 34 - 16 = 18.  
**ANSWER 10: D**  

Problem 11:  
11 hours = 660 minutes. 660 + 5 = 665 minutes.  
**ANSWER 11: C**  

Problem 12:  
Georgie can enter through 6 windows and exit through 5 others. Total ways: 6 × 5 = 30.  
**ANSWER 12: D**  

---  
**Final Answers**  
ANSWER 1: D  
ANSWER 2: B  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: B  
ANSWER 6: C  
ANSWER 7: D  
ANSWER 8: A  
ANSWER 9: D  
ANSWER 10: D  
ANSWER 11: C  
ANSWER 12: D

### Problem 1:
This is a classic pigeonhole principle problem. We need to find the minimum number of socks the Martian must pull to guarantee 5 socks of the same color, with 3 total colors. The worst-case scenario is that the Martian pulls as many socks as possible without getting 5 of any color: that is 4 socks of each color, for a total of 3*4=12 socks. The next sock pulled (the 13th) has to be one of the three colors, pushing that color to 5 socks. Tempting wrong answers: C (12 is only the number of socks that avoids 5 of the same color, not enough to guarantee the requirement), E (15 is the total if you took 5 of each color, not the minimum to guarantee).
ANSWER 1: D

---

### Problem 2:
We need to count the number of ways to choose 3 starters from 4 people, where order on the team does not matter. This is a combination problem: C(n,k) = n!/(k!(n-k)!). For n=4, k=3: C(4,3) = 4. We can also list the teams by who is left out: leaving out Lance, Sally, Joy, or Fred gives exactly 4 unique teams. Tempting wrong answer: C (6 is the count if order of selection mattered, which it does not for a team of starters).
ANSWER 2: B

---

### Problem 3:
We calculate the product of each pair to find which does not equal 36:
- A: (-4)*(-9)=36 ✔️
- B: (-3)*(-12)=36 ✔️
- C: (1/2)*(-72) = -36 ❌
- D: 1*36=36 ✔️
- E: (3/2)*24=36 ✔️
Only pair C has a product of -36, not 36.
ANSWER 3: C

---

### Problem 4:
First convert all units to centimeters to avoid conversion errors:
- Ceiling height: 2.4m = 240cm. Light bulb is 10cm below ceiling, so bulb height from floor = 240-10=230cm.
- Alice's height: 1.5m=150cm, her reach above her head is 46cm, so total reach without stool = 150+46=196cm.
- Stool height = 230-196=34cm.
Tempting wrong answers come from unit conversion mistakes, e.g., mixing up meters and centimeters.
ANSWER 4: B

---

### Problem 5:
This is a proportion problem. The ratio of trout to total fish is the same in the sample and the lake. In the sample, 30 trout out of 180 total fish = 1/6. Let T be total fish in the lake: 250/T = 1/6 → T=250*6=1500.
ANSWER 5: B

---

### Problem 6:
Karl bought 5 folders at $2.50 each, total original cost = 5*2.50=$12.50. He saves 20% of the total cost by waiting for the sale: 0.2*12.50=$2.50. This can also be calculated per folder: 20% of $2.50 is $0.50 saved per folder, 5*0.50=$2.50 total saved.
ANSWER 6: C

---

### Problem 7:
300 dimples * 2 seconds per dimple = 600 total seconds. Convert to minutes: 60 seconds per minute, so 600/60=10 minutes.
ANSWER 7: D

---

### Problem 8:
We find all non-negative integer pairs (x,y) where x=number of $5 bills, y=number of $2 bills, and 5x+2y=17. Only odd values of x work (since $5 is odd, $2 is even, we need an odd sum to get 17):
- x=1: 5+2y=17 → y=6 (valid)
- x=3: 15+2y=17 → y=1 (valid)
Even x values (0,2) leave an odd remainder that cannot be formed with $2 bills, so only 2 valid combinations.
ANSWER 8: A

---

### Problem 9:
Convert 1000 minutes to hours and minutes: 60 minutes per hour, 1000/60=16 hours and 40 minutes. Adding this to noon (12pm): 12pm +16h40m = 4:40am the next day.
ANSWER 9: D

---

### Problem 10:
Use order of operations (PEMDAS) to calculate:
First parenthesis: 8*4 +2 = 32+2=34
Second parenthesis: 8 +4*2=8+8=16
Subtract: 34-16=18. The tempting wrong answer C (10) comes from incorrectly adding before multiplying in the second parenthesis.
ANSWER 10: D

---

### Problem 11:
Convert 11 hours 5 minutes to total minutes: 11*60=660 minutes, plus 5 extra minutes = 665 minutes.
ANSWER 11: C

---

### Problem 12:
Georgie has 6 choices for an entry window, and 5 remaining choices for an exit window (must use a different window). Total ways = 6*5=30. Tempting wrong answer E (36) counts entering and leaving through the same window, which is not allowed; B (15) incorrectly treats entry/exit as unordered, which they are not.
ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color (red, white, blue). This is a pigeonhole principle problem: the worst-case scenario is pulling the maximum number of socks without getting 5 of any color. For each of the 3 colors, we can pull 4 socks (since 4 < 5), so 4×3=12 socks. The next sock pulled will necessarily be the 5th of one color. So total socks needed: 12 + 1 = 13.
ANSWER 1: D
Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). Since the problem does not mention distinct positions for starters, order does not matter, so we use combinations. The number of ways to choose 3 people from 4 is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n=4 \) and \( k=3 \):
\( \binom{4}{3} = \frac{4!}{3!1!} = 4 \). This is equivalent to counting how many people we leave out (4 choices, one for each person not selected).
ANSWER 2: B
Problem 3:
We check the product of each pair:
- A: \( (-4) \times (-9) = 36 \)
- B: \( (-3) \times (-12) = 36 \)
- C: \( \frac{1}{2} \times (-72) = -36 \neq 36 \)
- D: \( 1 \times 36 = 36 \)
- E: \( \frac{3}{2} \times 24 = 36 \)
Only pair C does not have a product of 36.
ANSWER 3: C
Problem 4:
Convert all units to centimeters:
- Ceiling height: \( 2.4 \, \text{m} = 240 \, \text{cm} \)
- Light bulb height from floor: \( 240 - 10 = 230 \, \text{cm} \)
- Alice's reach from floor (without stool): \( 1.5 \, \text{m} + 0.46 \, \text{m} = 150 + 46 = 196 \, \text{cm} \)
Let \( h \) be the stool height. The total reach with the stool equals the light bulb height:
\( h + 196 = 230 \)
\( h = 230 - 196 = 34 \, \text{cm} \)
ANSWER 4: B
Problem 5:
The ratio of trout to total fish in the sample equals the ratio in the lake. The sample has \( \frac{30}{180} = \frac{1}{6} \) trout. Let \( T \) be the total number of fish in the lake. We know there are 250 trout, so:
\( \frac{250}{T} = \frac{1}{6} \)
\( T = 250 \times 6 = 1500 \)
ANSWER 5: B
Problem 6:
Karl bought 5 folders at $2.50 each. The 20% discount applies to each folder, so the savings per folder is \( 2.50 \times 0.20 = 0.50 \). Total savings for 5 folders: \( 5 \times 0.50 = 2.50 \).
ANSWER 6: C
Problem 7:
Total time to paint 300 dimples: \( 300 \times 2 = 600 \) seconds. Convert to minutes: \( 600 \div 60 = 10 \) minutes.
ANSWER 7: D
Problem 8:
Let \( x \) = number of $5 bills, \( y \) = number of $2 bills. We need \( 5x + 2y = 17 \), where \( x,y \geq 0 \) integers. Test non-negative integer values of \( x \):
- \( x=0 \): \( 2y=17 \) → \( y=8.5 \) (not integer, invalid)
- \( x=1 \): \( 5 + 2y=17 \) → \( 2y=12 \) → \( y=6 \) (valid: 1 $5 + 6 $2 = $17)
- \( x=2 \): \( 10 + 2y=17 \) → \( 2y=7 \) → \( y=3.5 \) (invalid)
- \( x=3 \): \( 15 + 2y=17 \) → \( 2y=2 \) → \( y=1 \) (valid: 3 $5 + 1 $2 = $17)
- \( x=4 \): \( 20 > 17 \) (stop, no more valid \( x \))
Only 2 valid combinations.
ANSWER 8: A
Problem 9:
Convert 1000 minutes to hours and minutes: \( 1000 \div 60 = 16 \) hours with a remainder of 40 minutes (since \( 16 \times 60 = 960 \), \( 1000 - 960 = 40 \)). The contest starts at noon, so adding 16 hours 40 minutes: noon + 12 hours = midnight, plus 4 hours 40 minutes = 4:40 a.m.
ANSWER 9: D
Problem 10:
Use order of operations (multiplication before addition):
First parentheses: \( 8 \times 4 + 2 = 32 + 2 = 34 \)
Second parentheses: \( 8 + 4 \times 2 = 8 + 8 = 16 \)
Subtract: \( 34 - 16 = 18 \)
ANSWER 10: D
Problem 11:
Convert 11 hours 5 minutes to total minutes: \( 11 \times 60 + 5 = 660 + 5 = 665 \)
ANSWER 11: C
Problem 12:
Georgie chooses an entrance window first (6 options), then a different exit window (5 remaining options, since exit cannot equal entrance). Since entrance and exit are distinct actions, order matters, so total ways: \( 6 \times 5 = 30 \).
ANSWER 12: D

I'll work through each problem systematically.

**Problem 1:**

**What is being asked:** How many socks must be removed to guarantee 5 socks of the same color?

**Setup:** There are 3 colors (red, white, blue). In the worst case scenario, the Martian pulls out socks as evenly distributed among colors as possible before getting 5 of one color.

**Working:** 
- Worst case: 4 red, 4 white, 4 blue = 12 socks (still no color with 5)
- The 13th sock must give us a 5th sock of some color

**Answer 1: D**

---

**Problem 2:**

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

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

**Working:**
- C(4,3) = 4 ways
- The 4 ways correspond to leaving out each person: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred}

**Answer 2: B**

---

**Problem 3:**

**What is being asked:** Which pair does NOT have a product equal to 36?

**Working:** Check each pair:
- A. {−4, −9}: (−4)(−9) = 36 ✓
- B. {−3, −12}: (−3)(−12) = 36 ✓
- C. {1/2, −72}: (1/2)(−72) = −36 ✗
- D. {1, 36}: (1)(36) = 36 ✓
- E. {3/2, 24}: (3/2)(24) = 36 ✓

**Answer 3: C**

---

**Problem 4:**

**What is being asked:** How high is the stool?

**Setup:** Convert all measurements to centimeters:
- Ceiling height: 240 cm
- Light bulb location: 240 − 10 = 230 cm from floor
- Alice's height: 150 cm
- Alice's reach above her head: 46 cm
- Alice's reach without stool: 150 + 46 = 196 cm

**Working:**
- With stool, Alice can just reach the bulb at 230 cm
- Stool height + 196 cm = 230 cm
- Stool height = 230 − 196 = 34 cm

**Answer 4: B**

---

**Problem 5:**

**What is being asked:** How many fish total are in the lake?

**Setup:** Use proportions. In the sample: 30 trout out of 180 fish. In the lake: 250 trout out of total fish.

**Working:**
- 30/180 = 250/x
- 1/6 = 250/x
- x = 250 × 6 = 1500 fish

**Answer 5: B**

---

**Problem 6:**

**What is being asked:** How much could Karl have saved with the 20% off sale?

**Setup:** 
- Original cost: 5 folders × $2.50 = $12.50
- Sale discount: 20% off

**Working:**
- 20% of $12.50 = 0.20 × $12.50 = $2.50

**Answer 6: C**

---

**Problem 7:**

**What is being asked:** How many minutes to paint 300 dimples at 2 seconds each?

**Working:**
- Total time: 300 × 2 = 600 seconds
- Convert to minutes: 600 ÷ 60 = 10 minutes

**Answer 7: D**

---

**Problem 8:**

**What is being asked:** How many combinations of $5 bills and $2 bills make $17?

**Setup:** Let x = number of $5 bills, y = number of $2 bills
- 5x + 2y = 17

**Working:**
- x = 0: 2y = 17 → y = 8.5 (not integer)
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 (not integer)
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x ≥ 4: 5x ≥ 20 > 17 (impossible)

Valid combinations: (1,$5 and 6,$2) and (3,$5 and 1,$2) = 2 combinations

**Answer 8: A**

---

**Problem 9:**

**What is being asked:** When did the contest end, starting at noon and lasting 1000 minutes?

**Working:**
- 1000 minutes ÷ 60 = 16 hours and 40 minutes
- Noon + 16 hours 40 minutes = 4:40 p.m. the same day
- But we need to continue: 4:40 p.m. + additional time consideration
- Actually: 1000 ÷ 60 = 16 remainder 40, so 16 hours 40 minutes
- Noon + 16 hours 40 minutes = 4:40 a.m. the next day

**Answer 9: D**

---

**Problem 10:**

**What is being asked:** Calculate (8 × 4 + 2) − (8 + 4 × 2)

**Working:**
- First parentheses: 8 × 4 + 2 = 32 + 2 = 34
- Second parentheses: 8 + 4 × 2 = 8 + 8 = 16
- Result: 34 − 16 = 18

**Answer 10: D**

---

**Problem 11:**

**What is being asked:** Convert 11 hours 5 minutes to minutes

**Working:**
- 11 hours = 11 × 60 = 660 minutes
- Total: 660 + 5 = 665 minutes

**Answer 11: C**

---

**Problem 12:**

**What is being asked:** In how many ways can the ghost enter by one window and leave by a different window?

**Working:**
- Choices for entry: 6 windows
- Choices for exit: 5 windows (must be different from entry)
- Total ways: 6 × 5 = 30

**Answer 12: D**

Problem 1:

To be certain of getting 5 socks of the same color, we use the pigeonhole principle.

There are 3 colors: red, white, and blue.  
To avoid getting 5 of any one color for as long as possible, Martian could take at most 4 of each color:

- 4 red
- 4 white
- 4 blue

That gives a total of \(4+4+4=12\) socks without yet having 5 of the same color.

The very next sock, the 13th, must make one color appear at least 5 times.

So the Martian must remove 13 socks.

ANSWER 1: D

Problem 2:

There are 4 players: Lance, Sally, Joy, and Fred.

We need to choose 3 starters from these 4 players. Since order does not matter, we count combinations:

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

Indeed, the 3-starter groups are:

- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred

So there are 4 ways.

ANSWER 2: B

Problem 3:

We check each product.

- A: \((-4)(-9)=36\)
- B: \((-3)(-12)=36\)
- C: \(\frac12 \cdot (-72)=-36\), not 36
- D: \(1\cdot 36=36\)
- E: \(\frac32 \cdot 24=36\)

Only choice C does **not** have product 36.

ANSWER 3: C

Problem 4:

First find how high the bulb is above the floor.

The ceiling is 2.4 meters above the floor, which is \(240\) centimeters.  
The bulb is 10 centimeters below the ceiling, so it is:

\[
240 - 10 = 230 \text{ cm}
\]

Now Alice’s height is 1.5 meters = 150 cm.  
She can reach 46 cm above her head, so her reach without the stool is:

\[
150 + 46 = 196 \text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230 - 196 = 34 \text{ cm}
\]

So the stool is 34 cm high.

ANSWER 4: B

Problem 5:

The sample has 30 trout out of 180 fish, so the fraction of trout is:

\[
\frac{30}{180}=\frac{1}{6}
\]

If 250 trout is \(\frac16\) of the total lake population, then the total number of fish is:

\[
250 \div \frac16 = 250 \cdot 6 = 1500
\]

So there are 1500 fish in the lake.

ANSWER 5: B

Problem 6:

Karl bought 5 folders at \$2.50 each, so the total cost was:

\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:

\[
0.20 \times 12.50 = 2.50
\]

So he could have saved \$2.50.

ANSWER 6: C

Problem 7:

He paints 300 dimples, and each dimple takes 2 seconds.

Total time in seconds:

\[
300 \times 2 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10
\]

So he needs 10 minutes.

ANSWER 7: D

Problem 8:

We want combinations of 5-dollar bills and 2-dollar bills that total \$17.

Let \(x\) be the number of \$5 bills and \(y\) the number of \$2 bills. Then:

\[
5x + 2y = 17
\]

Try values of \(x\):

- \(x=1\): \(17-5=12\), so \(y=6\)
- \(x=2\): \(17-10=7\), not even, so impossible
- \(x=3\): \(17-15=2\), so \(y=1\)

These are the only nonnegative integer solutions.

So there are 2 combinations.

ANSWER 8: A

Problem 9:

The contest ended 1000 minutes after noon.

Convert 1000 minutes to hours and minutes:

\[
1000 \div 60 = 16 \text{ hours with } 40 \text{ minutes left}
\]

So:

- Noon + 12 hours = midnight
- plus 4 more hours = 4:00 a.m.
- plus 40 minutes = 4:40 a.m.

So the contest ended at 4:40 a.m.

ANSWER 9: D

Problem 10:

Compute each part using order of operations:

\[
(8\times 4 + 2) - (8 + 4\times 2)
\]

First parentheses:

\[
8\times 4 + 2 = 32 + 2 = 34
\]

Second parentheses:

\[
8 + 4\times 2 = 8 + 8 = 16
\]

Now subtract:

\[
34 - 16 = 18
\]

So the value is 18.

ANSWER 10: D

Problem 11:

We convert 11 hours and 5 minutes into minutes.

\[
11 \times 60 = 660
\]

Then add 5 minutes:

\[
660 + 5 = 665
\]

So the match lasted 665 minutes.

ANSWER 11: C

Problem 12:

Georgie must enter by one window and leave by a different window.

There are 6 choices for the entering window.  
After choosing one, there are 5 remaining choices for the exit window.

So the total number of ways is:

\[
6 \times 5 = 30
\]

Therefore, there are 30 ways.

ANSWER 12: D

Problem 1:  
We need the minimum number of socks to remove to guarantee there are 5 socks of the same color, given 3 colors (red, white, blue) and at least 5 of each color.

Worst case: you could first pull 4 socks of each color without reaching 5 of any color.  
That would be:
- 4 red + 4 white + 4 blue = 12 socks pulled

After 12 socks, you have at most 4 of each color. The next sock (the 13th) must be one of the 3 colors, and by the pigeonhole principle it makes one color reach 5.

So the Martian must remove **13** socks.

Tempting wrong choices:
- 12 isn’t enough because it’s possible to have only 4 of each color.
- Any number less than 13 can’t force a fifth sock of some color.

ANSWER 1: D


Problem 2:  
We have 4 students: Lance, Sally, Joy, Fred. We must choose 3 starters.

Number of ways to choose 3 out of 4 is:
\[
\binom{4}{3} = 4
\]

(You can also think: leave out one student; there are 4 choices for who to leave out.)

ANSWER 2: B


Problem 3:  
We check which pair does **NOT** multiply to 36.

Choices:
A. \(\{-4,-9\}\): \((-4)(-9)=36\) ✅  
B. \(\{-3,-12\}\): \((-3)(-12)=36\) ✅  
C. \(\left\{\frac{1}{2},-72\right\}\): \(\frac{1}{2}\cdot(-72)=-36\) ❌ not 36  
D. \(\{1,36\}\): \(1\cdot 36=36\) ✅  
E. \(\left\{\frac{3}{2},24\right\}\): \(\frac{3}{2}\cdot 24=3\cdot 12=36\) ✅

So the one that does not have product 36 is choice C.

ANSWER 3: C


Problem 4:  
Let’s convert everything to the same reference (floor).

- Ceiling is 2.4 m above the floor = 240 cm  
- The bulb is 10 cm below the ceiling: height of bulb above floor:
\[
240 - 10 = 230 \text{ cm}
\]
- Alice is 1.5 m tall = 150 cm  
- She can reach 46 cm above the top of her head, so from her top to the bulb:
\[
150 + 46 = 196 \text{ cm (if she stands on the floor)}
\]

She is standing on a stool, so her total height is (stool height + 150). She can reach:
\[
(\text{stool height} + 150) + 46
\]
Set equal to bulb height 230:
\[
\text{stool height} + 196 = 230
\]
\[
\text{stool height} = 34
\]

ANSWER 4: B


Problem 5:  
We know:
- Lake has 250 trout.
- Total fish in lake = \(T\)
- Sample has 180 fish, and 30 are trout.
So sample trout-to-total ratio equals lake trout-to-total ratio:
\[
\frac{30}{180} = \frac{250}{T}
\]
Simplify \(\frac{30}{180}=\frac{1}{6}\):
\[
\frac{1}{6} = \frac{250}{T}
\]
\[
T = 250 \cdot 6 = 1500
\]

ANSWER 5: B


Problem 6:  
Cost per folder originally: $2.50  
He bought 5 folders:
\[
5 \cdot 2.50 = 12.50
\]
A 20%-off sale means paying 80% of the original price:
\[
0.80 \cdot 12.50 = 10.00
\]
Savings:
\[
12.50 - 10.00 = 2.50
\]

ANSWER 6: C


Problem 7:  
Paint 300 dimples, 2 seconds per dimple.
Total time in seconds:
\[
300 \cdot 2 = 600 \text{ seconds}
\]
Convert to minutes (60 seconds per minute):
\[
600 \div 60 = 10 \text{ minutes}
\]

ANSWER 7: D


Problem 8:  
We need numbers of $5 bills and $2 bills that total $17 (order doesn’t matter).

Let \(x\) = number of $5 bills, \(y\) = number of $2 bills.
\[
5x + 2y = 17
\]
Try \(x=0\) to \(x=3\) (since \(4\cdot 5=20>17\)):

- \(x=0\): \(2y=17\) not possible  
- \(x=1\): \(5 + 2y=17 \Rightarrow 2y=12 \Rightarrow y=6\)  
- \(x=2\): \(10 + 2y=17 \Rightarrow 2y=7\) not possible  
- \(x=3\): \(15 + 2y=17 \Rightarrow 2y=2 \Rightarrow y=1\)

So there are 2 combinations: \((x,y)=(1,6)\) and \((3,1)\).

ANSWER 8: A


Problem 9:  
The contest starts at noon. It ends 1000 minutes later.

Convert 1000 minutes to hours:
\[
1000 \div 60 = 16 \text{ hours remainder } 40 \text{ minutes}
\]
Noon + 16 hours = 4:00 a.m. (next day)  
Add 40 minutes:
\[
4:00 \text{ a.m.} + 40 \text{ min} = 4:40 \text{ a.m.}
\]

ANSWER 9: D


Problem 10:  
Compute:
\[
(8 \times 4 + 2) - (8 + 4 \times 2)
\]
First parentheses:
\[
8\times 4 = 32,\quad 32+2=34
\]
Second parentheses:
\[
4\times 2 = 8,\quad 8+8=16
\]
Now subtract:
\[
34 - 16 = 18
\]

ANSWER 10: D


Problem 11:  
11 hours and 5 minutes total.
Convert hours to minutes:
\[
11 \cdot 60 = 660
\]
Add 5 minutes:
\[
660 + 5 = 665
\]

ANSWER 11: C


Problem 12:  
There are 6 windows. Georgie enters through one window and must leave through a different window.

Choose entering window: 6 choices.  
Choose leaving window: 5 choices (cannot be the same).

Total:
\[
6 \cdot 5 = 30
\]

ANSWER 12: D

Problem 1:
To be certain of having 5 socks of the same color, we consider the "worst-case scenario" (Pigeonhole Principle). If we have 3 colors (red, white, blue), we could pick 4 red, 4 white, and 4 blue socks without having 5 of any one color. This totals $4 + 4 + 4 = 12$ socks. The very next sock (the 13th) must be either red, white, or blue, which will complete a set of 5 for that color.
ANSWER 1: D

Problem 2:
We need to choose 3 starters from a group of 4 people (Lance, Sally, Joy, Fred). This is a combination problem denoted as $\binom{4}{3}$. Choosing 3 people to be on the team is the same as choosing 1 person to be left out. There are 4 ways to choose who is left out (Lance, Sally, Joy, or Fred).
ANSWER 2: B

Problem 3:
We check the products:
A: $(-4) \times (-9) = 36$
B: $(-3) \times (-12) = 36$
C: $\frac{1}{2} \times (-72) = -36$
D: $1 \times 36 = 36$
E: $\frac{3}{2} \times 24 = 3 \times 12 = 36$
Choice C results in $-36$, not $36$.
ANSWER 3: C

Problem 4:
Ceiling height = 2.4 m = 240 cm. Light bulb height = 240 - 10 = 230 cm from the floor.
Alice's reach = 150 cm (height) + 46 cm (reach) = 196 cm.
Stool height = (Bulb height) - (Alice's reach) = 230 - 196 = 34 cm.
ANSWER 4: B

Problem 5:
Let $T$ be the total number of fish in the lake. The ratio of trout to total fish is $\frac{30}{180} = \frac{1}{6}$.
Set up the proportion: $\frac{250}{T} = \frac{1}{6}$.
$T = 250 \times 6 = 1500$.
ANSWER 5: B

Problem 6:
Total cost = 5 folders × $2.50 = $12.50.
Savings = 20% of $12.50 = 0.20 \times 12.50 = $2.50.
Alternatively, 20% off one folder is $0.20 \times 2.50 = $0.50. For 5 folders, $0.50 \times 5 = $2.50.
ANSWER 6: C

Problem 7:
Total time in seconds = 300 dimples × 2 seconds/dimple = 600 seconds.
Total time in minutes = 600 seconds / 60 seconds/minute = 10 minutes.
ANSWER 7: D

Problem 8:
Let $x$ be the number of $5 bills and $y$ be the number of $2 bills. $5x + 2y = 17$.
If $x=1$, $5+2y=17 \Rightarrow 2y=12 \Rightarrow y=6$. (1, 6)
If $x=3$, $15+2y=17 \Rightarrow 2y=2 \Rightarrow y=1$. (3, 1)
If $x=0$ or $x=2$, $2y$ would be 17 or 7, which are not divisible by 2.
There are 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours and 40 minutes (since $1000 \div 60 = 16$ remainder 40).
Noon + 12 hours = midnight.
Midnight + 4 hours and 40 minutes = 4:40 a.m.
ANSWER 9: D

Problem 10:
$(8 \times 4 + 2) - (8 + 4 \times 2) = (32 + 2) - (8 + 8) = 34 - 16 = 18$.
ANSWER 10: D

Problem 11:
11 hours = $11 \times 60 = 660$ minutes.
Total = 660 + 5 = 665 minutes.
ANSWER 11: C

Problem 12:
There are 6 choices for the entry window. Since the exit window must be different, there are 5 choices for the exit window.
Total ways = $6 \times 5 = 30$.
ANSWER 12: D

Problem 1:
This is a pigeonhole principle problem. To guarantee at least 5 socks of one color, first consider the worst-case scenario of taking as many socks as possible without having 5 of any single color. That means taking exactly 4 socks of each of the 3 colors, for a total of \(4 \times 3 = 12\) socks. The next sock (the 13th) must be a 5th sock of one of the colors. Thus 13 socks are required. (Fewer than 13 leaves open the possibility of only 4 of each color.)

ANSWER 1: D

Problem 2:
The number of ways to choose 3 starters out of the 4 available players is the combination \(\binom{4}{3} = 4\). Each choice simply leaves out one of the four players.

ANSWER 2: B

Problem 3:
Compute the product for each pair:
- A: \((-4) \times (-9) = 36\)
- B: \((-3) \times (-12) = 36\)
- C: \(\frac12 \times (-72) = -36\) (not 36)
- D: \(1 \times 36 = 36\)
- E: \(\frac32 \times 24 = 36\)

Only pair C fails to give a product of 36.

ANSWER 3: C

Problem 4:
Convert all lengths to centimeters. The ceiling is 240 cm above the floor, so the bulb is 230 cm above the floor. Alice's reach is \(150 + 46 = 196\) cm. The stool must therefore supply the remaining \(230 - 196 = 34\) cm.

ANSWER 4: B

Problem 5:
The sample shows that trout comprise \(\frac{30}{180} = \frac16\) of all fish. With 250 trout in the lake, the total number of fish is \(250 \div \frac16 = 1500\).

ANSWER 5: B

Problem 6:
Five folders cost \(5 \times 2.50 = 12.50\). A 20 % discount saves \(0.20 \times 12.50 = 2.50\).

ANSWER 6: C

Problem 7:
Painting 300 dimples at 2 seconds each requires \(300 \times 2 = 600\) seconds, or \(600 \div 60 = 10\) minutes.

ANSWER 7: D

Problem 8:
Solve \(5x + 2y = 17\) for non-negative integers \(x, y\):
- \(x = 1\), \(y = 6\) works;
- \(x = 3\), \(y = 1\) works.
No other non-negative integer solutions exist, so there are exactly two combinations.

ANSWER 8: A

Problem 9:
Divide 1000 minutes by 60: \(1000 = 16\) hours + 40 minutes. Adding 16 hours to noon yields 4:00 a.m.; adding 40 minutes yields 4:40 a.m.

ANSWER 9: D

Problem 10:
Left side: \(8 \times 4 + 2 = 34\). Right side: \(8 + 4 \times 2 = 16\). Difference: \(34 - 16 = 18\).

ANSWER 10: D

Problem 11:
Convert 11 hours 5 minutes to minutes: \(11 \times 60 + 5 = 665\).

ANSWER 11: C

Problem 12:
There are 6 choices for the entry window and then 5 remaining choices for the exit window, giving \(6 \times 5 = 30\) ordered pairs.

ANSWER 12: D

Problem 1:
The problem is asking us to determine the minimum number of socks the Martian must remove from the drawer to guarantee that there are 5 socks of the same color. 
To ensure that there are 5 socks of the same color, we must consider the worst-case scenario where the Martian draws as many socks as possible without getting 5 of the same color. 
The Martian can draw at most 4 socks of each color without having 5 socks of the same color. 
So, the Martian can draw 4 red, 4 white, and 4 blue socks, for a total of 4 + 4 + 4 = 12 socks. 
At this point, the Martian still doesn't have 5 socks of the same color. 
However, drawing one more sock will guarantee that there are 5 socks of the same color. 
Therefore, the Martian must draw at least 12 + 1 = 13 socks.
Let's examine the choices: 
A. 6 is too low because the Martian could draw 3 red, 2 white, and 1 blue, or other combinations that don't guarantee 5 socks of the same color. 
B. 9 is still too low because the Martian could draw 4 red, 3 white, and 2 blue, or other similar combinations. 
C. 12 is the maximum number of socks the Martian can draw without having 5 socks of the same color. 
D. 13 is the correct answer because it is one more than the maximum number of socks that can be drawn without having 5 socks of the same color. 
E. 15 is more than necessary.
ANSWER 1: D

Problem 2:
The problem is asking us to find the number of ways to choose 3 starters from a team of 4 players: Lance, Sally, Joy, and Fred. 
This is a combination problem, because the order in which the starters are chosen does not matter. 
The number of ways to choose 3 starters from 4 players can be calculated using the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of players and k is the number of starters. 
Here, n = 4 and k = 3, so C(4, 3) = 4! / (3!(4-3)!) = 4.
Let's list the possible combinations to verify: 
- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred
There are indeed 4 combinations.
Let's examine the choices: 
A. 2 is too low. 
B. 4 matches our calculation. 
C. 6 is too high. 
D. 8 is too high. 
E. 10 is too high.
ANSWER 2: B

Problem 3:
The problem is asking us to identify the pair of numbers that does NOT have a product equal to 36. 
Let's check each pair: 
A. {−4, −9}: (−4) × (−9) = 36
B. {−3, −12}: (−3) × (−12) = 36
C. { 1 2 , −72}: (1/2) × (−72) = −36
D. {1, 36}: 1 × 36 = 36
E. { 3 2 , 24}: (3/2) × 24 = 36
The pair that does NOT have a product equal to 36 is C.
Let's examine the choices: 
A. {−4, −9} has a product of 36. 
B. {−3, −12} has a product of 36. 
C. { 1 2 , −72} has a product of −36, not 36. 
D. {1, 36} has a product of 36. 
E. { 3 2 , 24} has a product of 36.
ANSWER 3: C

Problem 4:
The problem is asking us to find the height of the stool that Alice needs to stand on to reach the light bulb. 
First, we need to convert the height of the ceiling from meters to centimeters: 2.4 meters = 240 centimeters. 
The light bulb is 10 centimeters below the ceiling, so its height above the floor is 240 - 10 = 230 centimeters. 
Alice is 1.5 meters = 150 centimeters tall and can reach 46 centimeters above her head, so her total reach is 150 + 46 = 196 centimeters. 
To find the height of the stool, we subtract Alice's total reach from the height of the light bulb: 230 - 196 = 34 centimeters.
Let's examine the choices: 
A. 32 is too low because 230 - (196 + 32) = 2, which is not enough to reach the light bulb. 
B. 34 is the correct height. 
C. 36 is too high. 
D. 38 is too high. 
E. 40 is too high.
ANSWER 4: B

Problem 5:
The problem is asking us to find the total number of fish in the lake based on the ratio of trout in the sample to the total number of fish in the sample. 
The sample contains 30 trout out of 180 fish, so the ratio of trout to total fish in the sample is 30/180. 
We are given that this ratio is the same as the ratio of trout to total fish in the lake. 
Let's denote the total number of fish in the lake as x. 
The ratio of trout to total fish in the lake is 250/x. 
Setting the two ratios equal, we get 30/180 = 250/x. 
Cross-multiplying, we have 30x = 250 × 180. 
Simplifying, 30x = 45000. 
Dividing both sides by 30, x = 1500.
Let's examine the choices: 
A. 1250 is too low. 
B. 1500 matches our calculation. 
C. 1750 is too high. 
D. 1800 is too high. 
E. 2000 is too high.
ANSWER 5: B

Problem 6:
The problem is asking us to find out how much Karl could have saved by waiting a day to buy the folders at a 20%-off sale. 
The original price of 5 folders is 5 × $2.50 = $12.50. 
At a 20%-off sale, the discount is 20% of $12.50, which is 0.20 × $12.50 = $2.50. 
So, Karl could have saved $2.50.
Let's examine the choices: 
A. $1.00 is too low. 
B. $2.00 is too low. 
C. $2.50 matches our calculation. 
D. $2.75 is too high. 
E. $5.00 is too high.
ANSWER 6: C

Problem 7:
The problem is asking us to find the time it will take Casey to paint 300 dimples on a golf ball. 
Casey takes 2 seconds to paint one dimple, so to paint 300 dimples, he will take 300 × 2 = 600 seconds. 
To convert seconds to minutes, we divide by 60: 600 / 60 = 10 minutes.
Let's examine the choices: 
A. 4 minutes is too low. 
B. 6 minutes is too low. 
C. 8 minutes is too low. 
D. 10 minutes matches our calculation. 
E. 12 minutes is too high.
ANSWER 7: D

Problem 8:
The problem is asking us to find the number of different combinations of $5 bills and $2 bills that can be used to make a total of $17. 
Let's list the possible combinations: 
1. 3 $5 bills = $15, and $2 bill = $2, total = $17 (1 combination: 3 $5 bills and 1 $2 bill)
2. 1 $5 bill = $5, and 6 $2 bills = $12, total = $17 (1 combination: 1 $5 bill and 6 $2 bills)
We should also check if there are any other combinations: 
- Using 0 $5 bills, we need 8.5 $2 bills, which is not possible since we can't have a fraction of a bill.
- Using 2 $5 bills = $10, we need $7 more, which can be made with 3.5 $2 bills, again not possible.
- Using 4 $5 bills = $20, which is more than $17.

Problem 1:
We want the minimum number of socks to guarantee 5 of the same color. With 3 colors (red, white, blue), the worst case is picking 4 of each color: 4+4+4 = 12 socks. The 13th sock must create a group of 5 of one color. So 13 socks are needed.
ANSWER 1: D

Problem 2:
We choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways is the combination C(4,3) = 4.
ANSWER 2: B

Problem 3:
Compute each product:
A: (−4)×(−9) = 36
B: (−3)×(−12) = 36
C: (1/2)×(−72) = −36
D: 1×36 = 36
E: (3/2)×24 = 36
Only C does not equal 36.
ANSWER 3: C

Problem 4:
Ceiling = 2.4 m = 240 cm. Light bulb is 10 cm below ceiling, so its height = 230 cm.
Alice's height = 1.5 m = 150 cm. Reach above head = 46 cm, so reach without stool = 196 cm.
Stool height needed = 230 − 196 = 34 cm.
ANSWER 4: B

Problem 5:
Sample ratio: 30 trout / 180 fish = 1/6. Lake has 250 trout, which is 1/6 of total fish. Total fish = 250 × 6 = 1500.
ANSWER 5: B

Problem 6:
Cost without sale: 5 × $2.50 = $12.50.
With 20% off, each folder costs $2.50 × 0.8 = $2.00. Total = 5 × $2.00 = $10.00.
Savings = $12.50 − $10.00 = $2.50.
ANSWER 6: C

Problem 7:
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.
ANSWER 7: D

Problem 8:
Find non-negative integers x ($5 bills) and y ($2 bills) such that 5x + 2y = 17.
x=1 → 5 + 2y = 17 → y=6.
x=3 → 15 + 2y = 17 → y=1.
No other x works. So 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours 40 minutes (16×60 = 960, remainder 40).
Noon + 16 hours = 4:00 a.m. next day. Add 40 minutes → 4:40 a.m.
ANSWER 9: D

Problem 10:
(8 × 4 + 2) = 32 + 2 = 34.
(8 + 4 × 2) = 8 + 8 = 16.
34 − 16 = 18.
ANSWER 10: D

Problem 11:
11 hours × 60 = 660 minutes. Plus 5 minutes = 665 minutes.
ANSWER 11: C

Problem 12:
6 choices for entry window, then 5 choices for a different exit window. Total = 6 × 5 = 30.
ANSWER 12: D

Problem 1:
We need to find the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). Using the Pigeonhole Principle, we consider the worst-case scenario: the Martian pulls out as many socks as possible *without* getting 5 of any one color. That means getting exactly 4 of each color:
- 4 red + 4 white + 4 blue = 12 socks
After 12 socks, the Martian could have 4 of each color and still not have 5 matching. The very next sock (the 13th) must be red, white, or blue, pushing the count of that color to 5. So 13 socks guarantees 5 of the same color.
(The "five-legged" detail is flavor text explaining why the Martian needs 5 socks.)

ANSWER 1: D

Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). This is a combination problem since order doesn't matter:
C(4, 3) = 4! / (3! × 1!) = 4
We can also list them: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — exactly 4 ways.

ANSWER 2: B

Problem 3:
We check each product:
- A: (−4) × (−9) = 36 ✓
- B: (−3) × (−12) = 36 ✓
- C: (1/2) × (−72) = −36 ✗ (This equals −36, NOT 36)
- D: 1 × 36 = 36 ✓
- E: (3/2) × 24 = 36 ✓
Choice C gives −36, which does not equal 36.

ANSWER 3: C

Problem 4:
First, convert everything to centimeters:
- Ceiling height: 2.4 m = 240 cm
- Light bulb position: 240 − 10 = 230 cm above the floor
- Alice's height: 1.5 m = 150 cm
- Alice's reach above her head: 46 cm
- Alice's maximum reach (without stool): 150 + 46 = 196 cm above the floor

Let h = height of the stool. Standing on the stool, her reach is h + 196 cm. She can "just reach" the bulb:
h + 196 = 230
h = 230 − 196 = 34 cm

ANSWER 4: B

Problem 5:
The ratio of trout to total fish in the sample should equal the ratio in the lake:
- Sample ratio: 30 trout / 180 fish = 1/6
- Lake ratio: 250 trout / Total fish = 1/6

Setting up the proportion: 250 / Total = 1/6
Total = 250 × 6 = 1500 fish

ANSWER 5: B

Problem 6:
Karl bought 5 folders at $2.50 each:
Total cost = 5 × $2.50 = $12.50
A 20%-off sale means he would save 20% of the total:
Savings = 0.20 × $12.50 = $2.50

ANSWER 6: C

Problem 7:
Total time in seconds: 300 dimples × 2 seconds/dimple = 600 seconds
Convert to minutes: 600 seconds ÷ 60 seconds/minute = 10 minutes

ANSWER 7: D

Problem 8:
We need non-negative integer solutions to 5x + 2y = 17, where x = number of $5 bills and y = number of $2 bills.
- x = 0: 2y = 17 → y = 8.5 (not an integer) ✗
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 ✗
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x = 4: 20 > 17 (too large) ✗

Only 2 valid combinations: (1 five + 6 twos) and (3 fives + 1 two).

ANSWER 8: A

Problem 9:
Convert 1000 minutes to hours and minutes:
1000 ÷ 60 = 16 remainder 40, so 1000 minutes = 16 hours 40 minutes.
Starting at noon (12:00 PM):
- 12 hours later = midnight (12:00 AM)
- 4 more hours = 4:00 AM
- 40 more minutes = 4:40 AM

ANSWER 9: D

Problem 10:
Evaluate each part using order of operations (multiplication before addition):
- First part: (8 × 4 + 2) = (32 + 2) = 34
- Second part: (8 + 4 × 2) = (8 + 8) = 16
- Difference: 34 − 16 = 18

ANSWER 10: D

Problem 11:
Convert 11 hours and 5 minutes to minutes:
11 hours × 60 minutes/hour = 660 minutes
660 + 5 = 665 minutes

ANSWER 11: C

Problem 12:
Georgie enters through one window and leaves through a different one.
- Choices for entry window: 6
- Choices for exit window (must be different): 5
Total ways = 6 × 5 = 30

ANSWER 12: D

Problem 1:
The Martian needs to be **certain** of having 5 socks of the same color. To guarantee this, we must consider the worst-case scenario: the Martian picks as many socks as possible **without** yet having 5 of any single color.  
With 3 colors (red, white, blue), the worst case is picking 4 socks of each color:  
\(4 + 4 + 4 = 12\) socks.  
At this point the Martian has 4 of every color but still does not have 5 of any one color. The very next sock—the 13th—must be red, white, or blue, and will necessarily create a set of 5 of that color.  
Thus 13 socks are required.  
(Choice A is too small; with 6 socks one could have 2,2,2. Choice B allows 3,3,3. Choice C allows 4,4,4. Choice E is more than the minimum guaranteed number.)

ANSWER 1: D

---

Problem 2:
We need to choose a group of 3 starters from the 4 students: Lance, Sally, Joy, and Fred. Since the three starters form a team and order does not matter, this is a combination problem.  
\[
\binom{4}{3} = \frac{4!}{3!\,1!} = 4
\]  
Equivalently, we can think of choosing which **one** of the four students will **not** start; there are 4 such choices.  
(Choice A is too small. Choice C is \(\binom{4}{2}\). Choices D and E do not arise from counting 3-person teams from 4 people.)

ANSWER 2: B

---

Problem 3:
We check the product of each pair to see which is not 36.

- **A.** \((-4)(-9) = 36\)  
- **B.** \((-3)(-12) = 36\)  
- **C.** \(\left(\frac12\right)(-72) = -36\)  
- **D.** \((1)(36) = 36\)  
- **E.** \(\left(\frac32\right)(24) = 36\)

Only pair **C** gives \(-36\), not \(36\).

ANSWER 3: C

---

Problem 4:
First convert all measurements to centimeters.  
- Ceiling height: \(2.4\text{ m} = 240\text{ cm}\).  
- Light bulb is \(10\text{ cm}\) below the ceiling, so its height above the floor is \(240 - 10 = 230\text{ cm}\).  

Alice’s reach:  
- Height: \(1.5\text{ m} = 150\text{ cm}\).  
- She can reach \(46\text{ cm}\) above her head, so her standing reach is \(150 + 46 = 196\text{ cm}\).  

The stool must supply the remaining height:  
\[
230 - 196 = 34\text{ cm}.
\]

ANSWER 4: B

---

Problem 5:
Let \(T\) be the total number of fish in the lake. The problem states that the ratio of trout to total fish is the same in the sample and in the lake.  
\[
\frac{250}{T} = \frac{30}{180}
\]  
Simplify the right side: \(\frac{30}{180} = \frac16\).  
\[
\frac{250}{T} = \frac16 \quad\Longrightarrow\quad T = 250 \times 6 = 1500.
\]

ANSWER 5: B

---

Problem 6:
Karl bought 5 folders at \(\$2.50\) each.  
Total cost: \(5 \times \$2.50 = \$12.50\).  
A 20% discount would have saved him 20% of this total:  
\[
0.20 \times \$12.50 = \$2.50.
\]

ANSWER 6: C

---

Problem 7:
Casey paints 300 dimples at 2 seconds per dimple.  
Total time in seconds: \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(\frac{600}{60} = 10\) minutes.

ANSWER 7: D

---

Problem 8:
Let \(f\) be the number of \(\$5\) bills and \(t\) be the number of \(\$2\) bills. We need non-negative integer solutions to  
\[
5f + 2t = 17.
\]  
Test values of \(f\):  
- \(f=0\): \(2t=17\) → no integer solution.  
- \(f=1\): \(5 + 2t = 17 \Rightarrow t=6\).  **(Valid)**
- \(f=2\): \(10 + 2t = 17 \Rightarrow t=3.5\).  (Invalid)
- \(f=3\): \(15 + 2t = 17 \Rightarrow t=1\).  **(Valid)**
- \(f=4\): \(20 > 17\) → stop.

There are exactly **2** combinations: \((1\text{ five}, 6\text{ twos})\) and \((3\text{ fives}, 1\text{ two})\).

ANSWER 8: A

---

Problem 9:
The contest starts at noon (12:00 p.m.). We add 1000 minutes.  
Divide 1000 by 60:  
\[
1000 = 16 \times 60 + 40,
\]  
so 1000 minutes = 16 hours and 40 minutes.  

Noon + 12 hours = midnight.  
Midnight + 4 hours 40 minutes = 4:40 a.m.

ANSWER 9: D

---

Problem 10:
Evaluate using the correct order of operations (multiplication before addition).

First parentheses:  
\[
8 \times 4 + 2 = 32 + 2 = 34.
\]

Second parentheses:  
\[
8 + 4 \times 2 = 8 + 8 = 16.
\]

Subtract:  
\[
34 - 16 = 18.
\]

ANSWER 10: D

---

Problem 11:
Convert 11 hours to minutes and add the remaining 5 minutes.  
\[
11 \times 60 = 660\text{ minutes}.
\]  
\[
660 + 5 = 665\text{ minutes}.
\]

ANSWER 11: C

---

Problem 12:
Georgie enters through one of 6 windows and must leave through a **different** window.  
- 6 choices for the entry window.  
- For each entry choice, 5 remaining choices for the exit window.  

Total ways:  
\[
6 \times 5 = 30.
\]  
(Choice E, 36, would count entering and leaving through the same window, which is not allowed.)

ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). The worst-case scenario is to draw as many socks as possible without having 5 of any one color. That would be 4 red, 4 white, and 4 blue, for a total of 12 socks. The next sock drawn (the 13th) must be one of the three colors, giving that color 5 socks. Therefore, 13 socks are required. The five-legged Martian is irrelevant to the count.  
Answer: D

Problem 2:
We have 4 people (Lance, Sally, Joy, Fred) and need to choose 3 starters. Since the starters are not assigned to specific positions, order does not matter. The number of ways is the combination \( \binom{4}{3} = 4 \).  
Answer: B

Problem 3:
Compute each product:  
A. \((-4) \times (-9) = 36\)  
B. \((-3) \times (-12) = 36\)  
C. \(\frac{1}{2} \times (-72) = -36\)  
D. \(1 \times 36 = 36\)  
E. \(\frac{3}{2} \times 24 = 36\)  
Only pair C gives a product not equal to 36.  
Answer: C

Problem 4:
Convert all measurements to centimeters:  
Ceiling height = 2.4 m = 240 cm.  
Light bulb is 10 cm below ceiling → height = 240 – 10 = 230 cm.  
Alice’s height = 1.5 m = 150 cm.  
Her reach above her head = 46 cm, so her maximum reach without a stool = 150 + 46 = 196 cm.  
Let \(h\) be the stool height. Standing on the stool raises her reach to \(196 + h\). This must equal 230 cm:  
\(196 + h = 230\) → \(h = 34\) cm.  
Answer: B

Problem 5:
In the sample, 30 out of 180 fish are trout, so the ratio trout : total = \(30/180 = 1/6\).  
Assuming the same ratio in the lake, if the lake has 250 trout, then the total number of fish is \(250 \times 6 = 1500\).  
Answer: B

Problem 6:
Original cost for 5 folders: \(5 \times 2.50 = \$12.50\).  
A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
Thus, by waiting a day, Karl could have saved $2.50.  
Answer: C

Problem 7:
Time to paint one dimple = 2 seconds.  
Total time for 300 dimples = \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(600 \div 60 = 10\) minutes.  
Answer: D

Problem 8:
We need nonnegative integers \(a\) (number of $5 bills) and \(b\) (number of $2 bills) such that \(5a + 2b = 17\).  
Solve by testing values of \(b\):  
\(b = 1\) → \(5a = 15\) → \(a = 3\) → (3,1) works.  
\(b = 6\) → \(5a = 5\) → \(a = 1\) → (1,6) works.  
No other values give integer \(a\). Thus there are 2 combinations.  
Answer: A

Problem 9:
1000 minutes = \(1000 \div 60 = 16\) hours and \(40\) minutes (since \(60 \times 16 = 960\), remainder 40).  
Starting at noon (12:00 PM), add 12 hours → midnight (12:00 AM). Add the remaining 4 hours → 4:00 AM. Then add 40 minutes → 4:40 AM.  
Answer: D

Problem 10:
Evaluate using order of operations:  
First parentheses: \(8 \times 4 + 2 = 32 + 2 = 34\).  
Second parentheses: \(8 + 4 \times 2 = 8 + 8 = 16\).  
Subtract: \(34 - 16 = 18\).  
Answer: D

Problem 11:
11 hours = \(11 \times 60 = 660\) minutes.  
Add 5 minutes → \(660 + 5 = 665\) minutes.  
Answer: C

Problem 12:
Georgie must choose an entry window (6 choices) and a different exit window (5 choices). Since entering and leaving are distinct actions, the total number of ways is \(6 \times 5 = 30\).  
Answer: D

ANSWER 1: D
ANSWER 2: B
ANSWER 3: C
ANSWER 4: B
ANSWER 5: B
ANSWER 6: C
ANSWER 7: D
ANSWER 8: A
ANSWER 9: D
ANSWER 10: D
ANSWER 11: C
ANSWER 12: D

Problem 1:  
The Martian wants to be certain of having 5 socks of the same color. There are three colors: red, white, and blue. In the worst-case scenario, the Martian could pick 4 socks of each color before getting a fifth of any one color. That would be \(4 \times 3 = 12\) socks, and still no color has 5 socks. The next sock drawn (the 13th) must complete a set of 5 of some color, because there are only three colors. Therefore, the Martian must remove **13** socks to be certain of having 5 socks of the same color.  
ANSWER 1: D

Problem 2:  
Ms. Hamilton's class has 4 available players: Lance, Sally, Joy, and Fred. The number of ways to choose 3 starters from 4 players is given by the combination \(\binom{4}{3} = 4\). Thus, there are **4** possible sets of three starters.  
ANSWER 2: B

Problem 3:  
We check each pair's product:  
- A: \((-4) \times (-9) = 36\)  
- B: \((-3) \times (-12) = 36\)  
- C: \(\frac{1}{2} \times (-72) = -36\) (not 36)  
- D: \(1 \times 36 = 36\)  
- E: \(\frac{3}{2} \times 24 = 36\)  
Only pair **C** does not have a product of 36.  
ANSWER 3: C

Problem 4:  
The ceiling is 2.4 m above the floor, and the bulb is 10 cm (0.1 m) below the ceiling, so the bulb's height is \(2.4 - 0.1 = 2.3\) m. Alice's height is 1.5 m, and she can reach 46 cm (0.46 m) above her head, so her reach height from the floor is \(1.5 + 0.46 = 1.96\) m. When standing on a stool, her hands are at the stool height plus her reach height. Setting this equal to the bulb height:  
\[
\text{stool height} + 1.96 = 2.3 \Rightarrow \text{stool height} = 0.34 \text{ m} = 34 \text{ cm}.
\]  
ANSWER 4: B

Problem 5:  
In the sample, the ratio of trout to total fish is \(\frac{30}{180} = \frac{1}{6}\). Assuming the same ratio in the lake, let \(T\) be the total number of fish in the lake. Then \(\frac{250}{T} = \frac{1}{6}\), so \(T = 250 \times 6 = 1500\).  
ANSWER 5: B

Problem 6:  
Karl bought 5 folders at $2.50 each, so the original cost is \(5 \times 2.50 = \$12.50\). With a 20% discount, the cost becomes \(12.50 \times 0.80 = \$10.00\). The savings is \(12.50 - 10.00 = \$2.50\).  
ANSWER 6: C

Problem 7:  
Painting 300 dimples at 2 seconds each takes \(300 \times 2 = 600\) seconds. Converting to minutes: \(600 \div 60 = 10\) minutes.  
ANSWER 7: D

Problem 8:  
We need nonnegative integers \(x\) (number of $5 bills) and \(y\) (number of $2 bills) such that \(5x + 2y = 17\). Solving:  
- \(y = 1 \Rightarrow 5x = 15 \Rightarrow x = 3\)  
- \(y = 6 \Rightarrow 5x = 5 \Rightarrow x = 1\)  
- \(y = 11 \Rightarrow 5x = -5\) (invalid)  
Thus, there are **2** valid combinations: (3 fives, 1 two) and (1 five, 6 twos).  
ANSWER 8: A

Problem 9:  
1000 minutes equals \(16\) hours and \(40\) minutes (since \(16 \times 60 = 960\), remainder \(40\)). Starting at noon, adding \(16\) hours gives 4:00 AM the next day, and adding \(40\) minutes gives 4:40 AM.  
ANSWER 9: D

Problem 10:  
Compute inside parentheses:  
\(8 \times 4 + 2 = 32 + 2 = 34\)  
\(8 + 4 \times 2 = 8 + 8 = 16\)  
Then \(34 - 16 = 18\).  
ANSWER 10: D

Problem 11:  
11 hours is \(11 \times 60 = 660\) minutes. Adding 5 minutes gives \(660 + 5 = 665\) minutes.  
ANSWER 11: C

Problem 12:  
Georgie can enter through any of the 6 windows. For exiting, he must choose a different window, so there are 5 choices. The total number of ways is \(6 \times 5 = 30\).  
ANSWER 12: D

Problem 1:  
The Martian needs to ensure there are 5 socks of the same color. The worst-case scenario is pulling the maximum number of socks without 5 of any color. Since there are 3 colors, this would be 4 socks of each color (4 × 3 = 12 socks). The next sock (the 13th) must match one color, giving 5 socks of that color.  
**ANSWER 1: D**  

Problem 2:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is C(4,3) = 4.  
**ANSWER 2: B** (Note: C(4,3)=4, but rechecking: the team is chosen from 4 specific people, so combinations are 4. Answer B=4)  

Problem 3:  
Check each pair:  
A. (−4)(−9) = 36  
B. (−3)(−12) = 36  
C. (1/2)(−72) = -36 (≠36)  
D. (1)(36) = 36  
E. (3/2)(24) = 36  
The pair that does NOT equal 36 is C.  
**ANSWER 3: E** (Re-evaluated: The question asks which does NOT equal 36. Option E: (3/2)*24=36, so it is valid. Option C: (1/2)*(-72)=-36, so C is the correct choice for NOT equal. Correction: **ANSWER 3: C** [Original had a typo, but computation confirms C is the answer])  

*Correction: For Problem 3, Option C: (1/2, -72) likely means (1/2) × (-72) = -36 ≠ 36. Option E: (3/2) × 24 = 36. Thus, C is the answer.*  
**ANSWER 3: C**  

Problem 4:  
Ceiling height: 2.4 m = 240 cm. Light bulb location: 240 - 10 = 230 cm from floor. Alice’s reach: 150 cm + 46 cm = 196 cm. Stool height = 230 - 196 = 34 cm.  
**ANSWER 4: B**  

Problem 5:  
Ratio of trout in sample: 30/180 = 1/6. Total trout: 250. Total fish = 250 × 6 = 1500.  
**ANSWER 5: B**  

Problem 6:  
Original cost: 5 × $2.50 = $12.50. 20% off: $12.50 × 0.20 = $2.50 saved.  
**ANSWER 6: C** (Note: The question asks how much Karl could have saved, which is $2.50, not $2.00. Original had a mistake; correction: 20% of $12.50 is $2.50, so answer C)  

Problem 7:  
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.  
**ANSWER 7: D**  

Problem 8:  
Let $5 bills = x, $2 bills = y. 5x + 2y = 17. Possible solutions:  
- x=3, y=1 (15+2=17)  
- x=1, y=6 (5+12=17)  
Only 2 combinations.  
**ANSWER 8: A**  

Problem 9:  
1000 minutes = 16 hours 40 minutes. Starting at noon: noon + 16h40m = 4:40 a.m. next day.  
**ANSWER 9: D**  

Problem 10:  
(8×4 + 2) = 34, (8 + 4×2) = 16. 34 - 16 = 18.  
**ANSWER 10: D**  

Problem 11:  
11 hours = 660 minutes. 660 + 5 = 665 minutes.  
**ANSWER 11: C**  

Problem 12:  
Georgie can enter through 6 windows and exit through 5 others. Total ways: 6 × 5 = 30.  
**ANSWER 12: D**  

---  
**Final Answers**  
ANSWER 1: D  
ANSWER 2: B  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: B  
ANSWER 6: C  
ANSWER 7: D  
ANSWER 8: A  
ANSWER 9: D  
ANSWER 10: D  
ANSWER 11: C  
ANSWER 12: D

### Problem 1:
This is a classic pigeonhole principle problem. We need to find the minimum number of socks the Martian must pull to guarantee 5 socks of the same color, with 3 total colors. The worst-case scenario is that the Martian pulls as many socks as possible without getting 5 of any color: that is 4 socks of each color, for a total of 3*4=12 socks. The next sock pulled (the 13th) has to be one of the three colors, pushing that color to 5 socks. Tempting wrong answers: C (12 is only the number of socks that avoids 5 of the same color, not enough to guarantee the requirement), E (15 is the total if you took 5 of each color, not the minimum to guarantee).
ANSWER 1: D

---

### Problem 2:
We need to count the number of ways to choose 3 starters from 4 people, where order on the team does not matter. This is a combination problem: C(n,k) = n!/(k!(n-k)!). For n=4, k=3: C(4,3) = 4. We can also list the teams by who is left out: leaving out Lance, Sally, Joy, or Fred gives exactly 4 unique teams. Tempting wrong answer: C (6 is the count if order of selection mattered, which it does not for a team of starters).
ANSWER 2: B

---

### Problem 3:
We calculate the product of each pair to find which does not equal 36:
- A: (-4)*(-9)=36 ✔️
- B: (-3)*(-12)=36 ✔️
- C: (1/2)*(-72) = -36 ❌
- D: 1*36=36 ✔️
- E: (3/2)*24=36 ✔️
Only pair C has a product of -36, not 36.
ANSWER 3: C

---

### Problem 4:
First convert all units to centimeters to avoid conversion errors:
- Ceiling height: 2.4m = 240cm. Light bulb is 10cm below ceiling, so bulb height from floor = 240-10=230cm.
- Alice's height: 1.5m=150cm, her reach above her head is 46cm, so total reach without stool = 150+46=196cm.
- Stool height = 230-196=34cm.
Tempting wrong answers come from unit conversion mistakes, e.g., mixing up meters and centimeters.
ANSWER 4: B

---

### Problem 5:
This is a proportion problem. The ratio of trout to total fish is the same in the sample and the lake. In the sample, 30 trout out of 180 total fish = 1/6. Let T be total fish in the lake: 250/T = 1/6 → T=250*6=1500.
ANSWER 5: B

---

### Problem 6:
Karl bought 5 folders at $2.50 each, total original cost = 5*2.50=$12.50. He saves 20% of the total cost by waiting for the sale: 0.2*12.50=$2.50. This can also be calculated per folder: 20% of $2.50 is $0.50 saved per folder, 5*0.50=$2.50 total saved.
ANSWER 6: C

---

### Problem 7:
300 dimples * 2 seconds per dimple = 600 total seconds. Convert to minutes: 60 seconds per minute, so 600/60=10 minutes.
ANSWER 7: D

---

### Problem 8:
We find all non-negative integer pairs (x,y) where x=number of $5 bills, y=number of $2 bills, and 5x+2y=17. Only odd values of x work (since $5 is odd, $2 is even, we need an odd sum to get 17):
- x=1: 5+2y=17 → y=6 (valid)
- x=3: 15+2y=17 → y=1 (valid)
Even x values (0,2) leave an odd remainder that cannot be formed with $2 bills, so only 2 valid combinations.
ANSWER 8: A

---

### Problem 9:
Convert 1000 minutes to hours and minutes: 60 minutes per hour, 1000/60=16 hours and 40 minutes. Adding this to noon (12pm): 12pm +16h40m = 4:40am the next day.
ANSWER 9: D

---

### Problem 10:
Use order of operations (PEMDAS) to calculate:
First parenthesis: 8*4 +2 = 32+2=34
Second parenthesis: 8 +4*2=8+8=16
Subtract: 34-16=18. The tempting wrong answer C (10) comes from incorrectly adding before multiplying in the second parenthesis.
ANSWER 10: D

---

### Problem 11:
Convert 11 hours 5 minutes to total minutes: 11*60=660 minutes, plus 5 extra minutes = 665 minutes.
ANSWER 11: C

---

### Problem 12:
Georgie has 6 choices for an entry window, and 5 remaining choices for an exit window (must use a different window). Total ways = 6*5=30. Tempting wrong answer E (36) counts entering and leaving through the same window, which is not allowed; B (15) incorrectly treats entry/exit as unordered, which they are not.
ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color (red, white, blue). This is a pigeonhole principle problem: the worst-case scenario is pulling the maximum number of socks without getting 5 of any color. For each of the 3 colors, we can pull 4 socks (since 4 < 5), so 4×3=12 socks. The next sock pulled will necessarily be the 5th of one color. So total socks needed: 12 + 1 = 13.
ANSWER 1: D
Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). Since the problem does not mention distinct positions for starters, order does not matter, so we use combinations. The number of ways to choose 3 people from 4 is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n=4 \) and \( k=3 \):
\( \binom{4}{3} = \frac{4!}{3!1!} = 4 \). This is equivalent to counting how many people we leave out (4 choices, one for each person not selected).
ANSWER 2: B
Problem 3:
We check the product of each pair:
- A: \( (-4) \times (-9) = 36 \)
- B: \( (-3) \times (-12) = 36 \)
- C: \( \frac{1}{2} \times (-72) = -36 \neq 36 \)
- D: \( 1 \times 36 = 36 \)
- E: \( \frac{3}{2} \times 24 = 36 \)
Only pair C does not have a product of 36.
ANSWER 3: C
Problem 4:
Convert all units to centimeters:
- Ceiling height: \( 2.4 \, \text{m} = 240 \, \text{cm} \)
- Light bulb height from floor: \( 240 - 10 = 230 \, \text{cm} \)
- Alice's reach from floor (without stool): \( 1.5 \, \text{m} + 0.46 \, \text{m} = 150 + 46 = 196 \, \text{cm} \)
Let \( h \) be the stool height. The total reach with the stool equals the light bulb height:
\( h + 196 = 230 \)
\( h = 230 - 196 = 34 \, \text{cm} \)
ANSWER 4: B
Problem 5:
The ratio of trout to total fish in the sample equals the ratio in the lake. The sample has \( \frac{30}{180} = \frac{1}{6} \) trout. Let \( T \) be the total number of fish in the lake. We know there are 250 trout, so:
\( \frac{250}{T} = \frac{1}{6} \)
\( T = 250 \times 6 = 1500 \)
ANSWER 5: B
Problem 6:
Karl bought 5 folders at $2.50 each. The 20% discount applies to each folder, so the savings per folder is \( 2.50 \times 0.20 = 0.50 \). Total savings for 5 folders: \( 5 \times 0.50 = 2.50 \).
ANSWER 6: C
Problem 7:
Total time to paint 300 dimples: \( 300 \times 2 = 600 \) seconds. Convert to minutes: \( 600 \div 60 = 10 \) minutes.
ANSWER 7: D
Problem 8:
Let \( x \) = number of $5 bills, \( y \) = number of $2 bills. We need \( 5x + 2y = 17 \), where \( x,y \geq 0 \) integers. Test non-negative integer values of \( x \):
- \( x=0 \): \( 2y=17 \) → \( y=8.5 \) (not integer, invalid)
- \( x=1 \): \( 5 + 2y=17 \) → \( 2y=12 \) → \( y=6 \) (valid: 1 $5 + 6 $2 = $17)
- \( x=2 \): \( 10 + 2y=17 \) → \( 2y=7 \) → \( y=3.5 \) (invalid)
- \( x=3 \): \( 15 + 2y=17 \) → \( 2y=2 \) → \( y=1 \) (valid: 3 $5 + 1 $2 = $17)
- \( x=4 \): \( 20 > 17 \) (stop, no more valid \( x \))
Only 2 valid combinations.
ANSWER 8: A
Problem 9:
Convert 1000 minutes to hours and minutes: \( 1000 \div 60 = 16 \) hours with a remainder of 40 minutes (since \( 16 \times 60 = 960 \), \( 1000 - 960 = 40 \)). The contest starts at noon, so adding 16 hours 40 minutes: noon + 12 hours = midnight, plus 4 hours 40 minutes = 4:40 a.m.
ANSWER 9: D
Problem 10:
Use order of operations (multiplication before addition):
First parentheses: \( 8 \times 4 + 2 = 32 + 2 = 34 \)
Second parentheses: \( 8 + 4 \times 2 = 8 + 8 = 16 \)
Subtract: \( 34 - 16 = 18 \)
ANSWER 10: D
Problem 11:
Convert 11 hours 5 minutes to total minutes: \( 11 \times 60 + 5 = 660 + 5 = 665 \)
ANSWER 11: C
Problem 12:
Georgie chooses an entrance window first (6 options), then a different exit window (5 remaining options, since exit cannot equal entrance). Since entrance and exit are distinct actions, order matters, so total ways: \( 6 \times 5 = 30 \).
ANSWER 12: D

I'll work through each problem systematically.

**Problem 1:**

**What is being asked:** How many socks must be removed to guarantee 5 socks of the same color?

**Setup:** There are 3 colors (red, white, blue). In the worst case scenario, the Martian pulls out socks as evenly distributed among colors as possible before getting 5 of one color.

**Working:** 
- Worst case: 4 red, 4 white, 4 blue = 12 socks (still no color with 5)
- The 13th sock must give us a 5th sock of some color

**Answer 1: D**

---

**Problem 2:**

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

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

**Working:**
- C(4,3) = 4 ways
- The 4 ways correspond to leaving out each person: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred}

**Answer 2: B**

---

**Problem 3:**

**What is being asked:** Which pair does NOT have a product equal to 36?

**Working:** Check each pair:
- A. {−4, −9}: (−4)(−9) = 36 ✓
- B. {−3, −12}: (−3)(−12) = 36 ✓
- C. {1/2, −72}: (1/2)(−72) = −36 ✗
- D. {1, 36}: (1)(36) = 36 ✓
- E. {3/2, 24}: (3/2)(24) = 36 ✓

**Answer 3: C**

---

**Problem 4:**

**What is being asked:** How high is the stool?

**Setup:** Convert all measurements to centimeters:
- Ceiling height: 240 cm
- Light bulb location: 240 − 10 = 230 cm from floor
- Alice's height: 150 cm
- Alice's reach above her head: 46 cm
- Alice's reach without stool: 150 + 46 = 196 cm

**Working:**
- With stool, Alice can just reach the bulb at 230 cm
- Stool height + 196 cm = 230 cm
- Stool height = 230 − 196 = 34 cm

**Answer 4: B**

---

**Problem 5:**

**What is being asked:** How many fish total are in the lake?

**Setup:** Use proportions. In the sample: 30 trout out of 180 fish. In the lake: 250 trout out of total fish.

**Working:**
- 30/180 = 250/x
- 1/6 = 250/x
- x = 250 × 6 = 1500 fish

**Answer 5: B**

---

**Problem 6:**

**What is being asked:** How much could Karl have saved with the 20% off sale?

**Setup:** 
- Original cost: 5 folders × $2.50 = $12.50
- Sale discount: 20% off

**Working:**
- 20% of $12.50 = 0.20 × $12.50 = $2.50

**Answer 6: C**

---

**Problem 7:**

**What is being asked:** How many minutes to paint 300 dimples at 2 seconds each?

**Working:**
- Total time: 300 × 2 = 600 seconds
- Convert to minutes: 600 ÷ 60 = 10 minutes

**Answer 7: D**

---

**Problem 8:**

**What is being asked:** How many combinations of $5 bills and $2 bills make $17?

**Setup:** Let x = number of $5 bills, y = number of $2 bills
- 5x + 2y = 17

**Working:**
- x = 0: 2y = 17 → y = 8.5 (not integer)
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 (not integer)
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x ≥ 4: 5x ≥ 20 > 17 (impossible)

Valid combinations: (1,$5 and 6,$2) and (3,$5 and 1,$2) = 2 combinations

**Answer 8: A**

---

**Problem 9:**

**What is being asked:** When did the contest end, starting at noon and lasting 1000 minutes?

**Working:**
- 1000 minutes ÷ 60 = 16 hours and 40 minutes
- Noon + 16 hours 40 minutes = 4:40 p.m. the same day
- But we need to continue: 4:40 p.m. + additional time consideration
- Actually: 1000 ÷ 60 = 16 remainder 40, so 16 hours 40 minutes
- Noon + 16 hours 40 minutes = 4:40 a.m. the next day

**Answer 9: D**

---

**Problem 10:**

**What is being asked:** Calculate (8 × 4 + 2) − (8 + 4 × 2)

**Working:**
- First parentheses: 8 × 4 + 2 = 32 + 2 = 34
- Second parentheses: 8 + 4 × 2 = 8 + 8 = 16
- Result: 34 − 16 = 18

**Answer 10: D**

---

**Problem 11:**

**What is being asked:** Convert 11 hours 5 minutes to minutes

**Working:**
- 11 hours = 11 × 60 = 660 minutes
- Total: 660 + 5 = 665 minutes

**Answer 11: C**

---

**Problem 12:**

**What is being asked:** In how many ways can the ghost enter by one window and leave by a different window?

**Working:**
- Choices for entry: 6 windows
- Choices for exit: 5 windows (must be different from entry)
- Total ways: 6 × 5 = 30

**Answer 12: D**

Problem 1:

To be certain of getting 5 socks of the same color, we use the pigeonhole principle.

There are 3 colors: red, white, and blue.  
To avoid getting 5 of any one color for as long as possible, Martian could take at most 4 of each color:

- 4 red
- 4 white
- 4 blue

That gives a total of \(4+4+4=12\) socks without yet having 5 of the same color.

The very next sock, the 13th, must make one color appear at least 5 times.

So the Martian must remove 13 socks.

ANSWER 1: D

Problem 2:

There are 4 players: Lance, Sally, Joy, and Fred.

We need to choose 3 starters from these 4 players. Since order does not matter, we count combinations:

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

Indeed, the 3-starter groups are:

- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred

So there are 4 ways.

ANSWER 2: B

Problem 3:

We check each product.

- A: \((-4)(-9)=36\)
- B: \((-3)(-12)=36\)
- C: \(\frac12 \cdot (-72)=-36\), not 36
- D: \(1\cdot 36=36\)
- E: \(\frac32 \cdot 24=36\)

Only choice C does **not** have product 36.

ANSWER 3: C

Problem 4:

First find how high the bulb is above the floor.

The ceiling is 2.4 meters above the floor, which is \(240\) centimeters.  
The bulb is 10 centimeters below the ceiling, so it is:

\[
240 - 10 = 230 \text{ cm}
\]

Now Alice’s height is 1.5 meters = 150 cm.  
She can reach 46 cm above her head, so her reach without the stool is:

\[
150 + 46 = 196 \text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230 - 196 = 34 \text{ cm}
\]

So the stool is 34 cm high.

ANSWER 4: B

Problem 5:

The sample has 30 trout out of 180 fish, so the fraction of trout is:

\[
\frac{30}{180}=\frac{1}{6}
\]

If 250 trout is \(\frac16\) of the total lake population, then the total number of fish is:

\[
250 \div \frac16 = 250 \cdot 6 = 1500
\]

So there are 1500 fish in the lake.

ANSWER 5: B

Problem 6:

Karl bought 5 folders at \$2.50 each, so the total cost was:

\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:

\[
0.20 \times 12.50 = 2.50
\]

So he could have saved \$2.50.

ANSWER 6: C

Problem 7:

He paints 300 dimples, and each dimple takes 2 seconds.

Total time in seconds:

\[
300 \times 2 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10
\]

So he needs 10 minutes.

ANSWER 7: D

Problem 8:

We want combinations of 5-dollar bills and 2-dollar bills that total \$17.

Let \(x\) be the number of \$5 bills and \(y\) the number of \$2 bills. Then:

\[
5x + 2y = 17
\]

Try values of \(x\):

- \(x=1\): \(17-5=12\), so \(y=6\)
- \(x=2\): \(17-10=7\), not even, so impossible
- \(x=3\): \(17-15=2\), so \(y=1\)

These are the only nonnegative integer solutions.

So there are 2 combinations.

ANSWER 8: A

Problem 9:

The contest ended 1000 minutes after noon.

Convert 1000 minutes to hours and minutes:

\[
1000 \div 60 = 16 \text{ hours with } 40 \text{ minutes left}
\]

So:

- Noon + 12 hours = midnight
- plus 4 more hours = 4:00 a.m.
- plus 40 minutes = 4:40 a.m.

So the contest ended at 4:40 a.m.

ANSWER 9: D

Problem 10:

Compute each part using order of operations:

\[
(8\times 4 + 2) - (8 + 4\times 2)
\]

First parentheses:

\[
8\times 4 + 2 = 32 + 2 = 34
\]

Second parentheses:

\[
8 + 4\times 2 = 8 + 8 = 16
\]

Now subtract:

\[
34 - 16 = 18
\]

So the value is 18.

ANSWER 10: D

Problem 11:

We convert 11 hours and 5 minutes into minutes.

\[
11 \times 60 = 660
\]

Then add 5 minutes:

\[
660 + 5 = 665
\]

So the match lasted 665 minutes.

ANSWER 11: C

Problem 12:

Georgie must enter by one window and leave by a different window.

There are 6 choices for the entering window.  
After choosing one, there are 5 remaining choices for the exit window.

So the total number of ways is:

\[
6 \times 5 = 30
\]

Therefore, there are 30 ways.

ANSWER 12: D

Problem 1:  
We need the minimum number of socks to remove to guarantee there are 5 socks of the same color, given 3 colors (red, white, blue) and at least 5 of each color.

Worst case: you could first pull 4 socks of each color without reaching 5 of any color.  
That would be:
- 4 red + 4 white + 4 blue = 12 socks pulled

After 12 socks, you have at most 4 of each color. The next sock (the 13th) must be one of the 3 colors, and by the pigeonhole principle it makes one color reach 5.

So the Martian must remove **13** socks.

Tempting wrong choices:
- 12 isn’t enough because it’s possible to have only 4 of each color.
- Any number less than 13 can’t force a fifth sock of some color.

ANSWER 1: D


Problem 2:  
We have 4 students: Lance, Sally, Joy, Fred. We must choose 3 starters.

Number of ways to choose 3 out of 4 is:
\[
\binom{4}{3} = 4
\]

(You can also think: leave out one student; there are 4 choices for who to leave out.)

ANSWER 2: B


Problem 3:  
We check which pair does **NOT** multiply to 36.

Choices:
A. \(\{-4,-9\}\): \((-4)(-9)=36\) ✅  
B. \(\{-3,-12\}\): \((-3)(-12)=36\) ✅  
C. \(\left\{\frac{1}{2},-72\right\}\): \(\frac{1}{2}\cdot(-72)=-36\) ❌ not 36  
D. \(\{1,36\}\): \(1\cdot 36=36\) ✅  
E. \(\left\{\frac{3}{2},24\right\}\): \(\frac{3}{2}\cdot 24=3\cdot 12=36\) ✅

So the one that does not have product 36 is choice C.

ANSWER 3: C


Problem 4:  
Let’s convert everything to the same reference (floor).

- Ceiling is 2.4 m above the floor = 240 cm  
- The bulb is 10 cm below the ceiling: height of bulb above floor:
\[
240 - 10 = 230 \text{ cm}
\]
- Alice is 1.5 m tall = 150 cm  
- She can reach 46 cm above the top of her head, so from her top to the bulb:
\[
150 + 46 = 196 \text{ cm (if she stands on the floor)}
\]

She is standing on a stool, so her total height is (stool height + 150). She can reach:
\[
(\text{stool height} + 150) + 46
\]
Set equal to bulb height 230:
\[
\text{stool height} + 196 = 230
\]
\[
\text{stool height} = 34
\]

ANSWER 4: B


Problem 5:  
We know:
- Lake has 250 trout.
- Total fish in lake = \(T\)
- Sample has 180 fish, and 30 are trout.
So sample trout-to-total ratio equals lake trout-to-total ratio:
\[
\frac{30}{180} = \frac{250}{T}
\]
Simplify \(\frac{30}{180}=\frac{1}{6}\):
\[
\frac{1}{6} = \frac{250}{T}
\]
\[
T = 250 \cdot 6 = 1500
\]

ANSWER 5: B


Problem 6:  
Cost per folder originally: $2.50  
He bought 5 folders:
\[
5 \cdot 2.50 = 12.50
\]
A 20%-off sale means paying 80% of the original price:
\[
0.80 \cdot 12.50 = 10.00
\]
Savings:
\[
12.50 - 10.00 = 2.50
\]

ANSWER 6: C


Problem 7:  
Paint 300 dimples, 2 seconds per dimple.
Total time in seconds:
\[
300 \cdot 2 = 600 \text{ seconds}
\]
Convert to minutes (60 seconds per minute):
\[
600 \div 60 = 10 \text{ minutes}
\]

ANSWER 7: D


Problem 8:  
We need numbers of $5 bills and $2 bills that total $17 (order doesn’t matter).

Let \(x\) = number of $5 bills, \(y\) = number of $2 bills.
\[
5x + 2y = 17
\]
Try \(x=0\) to \(x=3\) (since \(4\cdot 5=20>17\)):

- \(x=0\): \(2y=17\) not possible  
- \(x=1\): \(5 + 2y=17 \Rightarrow 2y=12 \Rightarrow y=6\)  
- \(x=2\): \(10 + 2y=17 \Rightarrow 2y=7\) not possible  
- \(x=3\): \(15 + 2y=17 \Rightarrow 2y=2 \Rightarrow y=1\)

So there are 2 combinations: \((x,y)=(1,6)\) and \((3,1)\).

ANSWER 8: A


Problem 9:  
The contest starts at noon. It ends 1000 minutes later.

Convert 1000 minutes to hours:
\[
1000 \div 60 = 16 \text{ hours remainder } 40 \text{ minutes}
\]
Noon + 16 hours = 4:00 a.m. (next day)  
Add 40 minutes:
\[
4:00 \text{ a.m.} + 40 \text{ min} = 4:40 \text{ a.m.}
\]

ANSWER 9: D


Problem 10:  
Compute:
\[
(8 \times 4 + 2) - (8 + 4 \times 2)
\]
First parentheses:
\[
8\times 4 = 32,\quad 32+2=34
\]
Second parentheses:
\[
4\times 2 = 8,\quad 8+8=16
\]
Now subtract:
\[
34 - 16 = 18
\]

ANSWER 10: D


Problem 11:  
11 hours and 5 minutes total.
Convert hours to minutes:
\[
11 \cdot 60 = 660
\]
Add 5 minutes:
\[
660 + 5 = 665
\]

ANSWER 11: C


Problem 12:  
There are 6 windows. Georgie enters through one window and must leave through a different window.

Choose entering window: 6 choices.  
Choose leaving window: 5 choices (cannot be the same).

Total:
\[
6 \cdot 5 = 30
\]

ANSWER 12: D

Problem 1:
To be certain of having 5 socks of the same color, we consider the "worst-case scenario" (Pigeonhole Principle). If we have 3 colors (red, white, blue), we could pick 4 red, 4 white, and 4 blue socks without having 5 of any one color. This totals $4 + 4 + 4 = 12$ socks. The very next sock (the 13th) must be either red, white, or blue, which will complete a set of 5 for that color.
ANSWER 1: D

Problem 2:
We need to choose 3 starters from a group of 4 people (Lance, Sally, Joy, Fred). This is a combination problem denoted as $\binom{4}{3}$. Choosing 3 people to be on the team is the same as choosing 1 person to be left out. There are 4 ways to choose who is left out (Lance, Sally, Joy, or Fred).
ANSWER 2: B

Problem 3:
We check the products:
A: $(-4) \times (-9) = 36$
B: $(-3) \times (-12) = 36$
C: $\frac{1}{2} \times (-72) = -36$
D: $1 \times 36 = 36$
E: $\frac{3}{2} \times 24 = 3 \times 12 = 36$
Choice C results in $-36$, not $36$.
ANSWER 3: C

Problem 4:
Ceiling height = 2.4 m = 240 cm. Light bulb height = 240 - 10 = 230 cm from the floor.
Alice's reach = 150 cm (height) + 46 cm (reach) = 196 cm.
Stool height = (Bulb height) - (Alice's reach) = 230 - 196 = 34 cm.
ANSWER 4: B

Problem 5:
Let $T$ be the total number of fish in the lake. The ratio of trout to total fish is $\frac{30}{180} = \frac{1}{6}$.
Set up the proportion: $\frac{250}{T} = \frac{1}{6}$.
$T = 250 \times 6 = 1500$.
ANSWER 5: B

Problem 6:
Total cost = 5 folders × $2.50 = $12.50.
Savings = 20% of $12.50 = 0.20 \times 12.50 = $2.50.
Alternatively, 20% off one folder is $0.20 \times 2.50 = $0.50. For 5 folders, $0.50 \times 5 = $2.50.
ANSWER 6: C

Problem 7:
Total time in seconds = 300 dimples × 2 seconds/dimple = 600 seconds.
Total time in minutes = 600 seconds / 60 seconds/minute = 10 minutes.
ANSWER 7: D

Problem 8:
Let $x$ be the number of $5 bills and $y$ be the number of $2 bills. $5x + 2y = 17$.
If $x=1$, $5+2y=17 \Rightarrow 2y=12 \Rightarrow y=6$. (1, 6)
If $x=3$, $15+2y=17 \Rightarrow 2y=2 \Rightarrow y=1$. (3, 1)
If $x=0$ or $x=2$, $2y$ would be 17 or 7, which are not divisible by 2.
There are 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours and 40 minutes (since $1000 \div 60 = 16$ remainder 40).
Noon + 12 hours = midnight.
Midnight + 4 hours and 40 minutes = 4:40 a.m.
ANSWER 9: D

Problem 10:
$(8 \times 4 + 2) - (8 + 4 \times 2) = (32 + 2) - (8 + 8) = 34 - 16 = 18$.
ANSWER 10: D

Problem 11:
11 hours = $11 \times 60 = 660$ minutes.
Total = 660 + 5 = 665 minutes.
ANSWER 11: C

Problem 12:
There are 6 choices for the entry window. Since the exit window must be different, there are 5 choices for the exit window.
Total ways = $6 \times 5 = 30$.
ANSWER 12: D

Problem 1:
This is a pigeonhole principle problem. To guarantee at least 5 socks of one color, first consider the worst-case scenario of taking as many socks as possible without having 5 of any single color. That means taking exactly 4 socks of each of the 3 colors, for a total of \(4 \times 3 = 12\) socks. The next sock (the 13th) must be a 5th sock of one of the colors. Thus 13 socks are required. (Fewer than 13 leaves open the possibility of only 4 of each color.)

ANSWER 1: D

Problem 2:
The number of ways to choose 3 starters out of the 4 available players is the combination \(\binom{4}{3} = 4\). Each choice simply leaves out one of the four players.

ANSWER 2: B

Problem 3:
Compute the product for each pair:
- A: \((-4) \times (-9) = 36\)
- B: \((-3) \times (-12) = 36\)
- C: \(\frac12 \times (-72) = -36\) (not 36)
- D: \(1 \times 36 = 36\)
- E: \(\frac32 \times 24 = 36\)

Only pair C fails to give a product of 36.

ANSWER 3: C

Problem 4:
Convert all lengths to centimeters. The ceiling is 240 cm above the floor, so the bulb is 230 cm above the floor. Alice's reach is \(150 + 46 = 196\) cm. The stool must therefore supply the remaining \(230 - 196 = 34\) cm.

ANSWER 4: B

Problem 5:
The sample shows that trout comprise \(\frac{30}{180} = \frac16\) of all fish. With 250 trout in the lake, the total number of fish is \(250 \div \frac16 = 1500\).

ANSWER 5: B

Problem 6:
Five folders cost \(5 \times 2.50 = 12.50\). A 20 % discount saves \(0.20 \times 12.50 = 2.50\).

ANSWER 6: C

Problem 7:
Painting 300 dimples at 2 seconds each requires \(300 \times 2 = 600\) seconds, or \(600 \div 60 = 10\) minutes.

ANSWER 7: D

Problem 8:
Solve \(5x + 2y = 17\) for non-negative integers \(x, y\):
- \(x = 1\), \(y = 6\) works;
- \(x = 3\), \(y = 1\) works.
No other non-negative integer solutions exist, so there are exactly two combinations.

ANSWER 8: A

Problem 9:
Divide 1000 minutes by 60: \(1000 = 16\) hours + 40 minutes. Adding 16 hours to noon yields 4:00 a.m.; adding 40 minutes yields 4:40 a.m.

ANSWER 9: D

Problem 10:
Left side: \(8 \times 4 + 2 = 34\). Right side: \(8 + 4 \times 2 = 16\). Difference: \(34 - 16 = 18\).

ANSWER 10: D

Problem 11:
Convert 11 hours 5 minutes to minutes: \(11 \times 60 + 5 = 665\).

ANSWER 11: C

Problem 12:
There are 6 choices for the entry window and then 5 remaining choices for the exit window, giving \(6 \times 5 = 30\) ordered pairs.

ANSWER 12: D

Problem 1:
The problem is asking us to determine the minimum number of socks the Martian must remove from the drawer to guarantee that there are 5 socks of the same color. 
To ensure that there are 5 socks of the same color, we must consider the worst-case scenario where the Martian draws as many socks as possible without getting 5 of the same color. 
The Martian can draw at most 4 socks of each color without having 5 socks of the same color. 
So, the Martian can draw 4 red, 4 white, and 4 blue socks, for a total of 4 + 4 + 4 = 12 socks. 
At this point, the Martian still doesn't have 5 socks of the same color. 
However, drawing one more sock will guarantee that there are 5 socks of the same color. 
Therefore, the Martian must draw at least 12 + 1 = 13 socks.
Let's examine the choices: 
A. 6 is too low because the Martian could draw 3 red, 2 white, and 1 blue, or other combinations that don't guarantee 5 socks of the same color. 
B. 9 is still too low because the Martian could draw 4 red, 3 white, and 2 blue, or other similar combinations. 
C. 12 is the maximum number of socks the Martian can draw without having 5 socks of the same color. 
D. 13 is the correct answer because it is one more than the maximum number of socks that can be drawn without having 5 socks of the same color. 
E. 15 is more than necessary.
ANSWER 1: D

Problem 2:
The problem is asking us to find the number of ways to choose 3 starters from a team of 4 players: Lance, Sally, Joy, and Fred. 
This is a combination problem, because the order in which the starters are chosen does not matter. 
The number of ways to choose 3 starters from 4 players can be calculated using the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of players and k is the number of starters. 
Here, n = 4 and k = 3, so C(4, 3) = 4! / (3!(4-3)!) = 4.
Let's list the possible combinations to verify: 
- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred
There are indeed 4 combinations.
Let's examine the choices: 
A. 2 is too low. 
B. 4 matches our calculation. 
C. 6 is too high. 
D. 8 is too high. 
E. 10 is too high.
ANSWER 2: B

Problem 3:
The problem is asking us to identify the pair of numbers that does NOT have a product equal to 36. 
Let's check each pair: 
A. {−4, −9}: (−4) × (−9) = 36
B. {−3, −12}: (−3) × (−12) = 36
C. { 1 2 , −72}: (1/2) × (−72) = −36
D. {1, 36}: 1 × 36 = 36
E. { 3 2 , 24}: (3/2) × 24 = 36
The pair that does NOT have a product equal to 36 is C.
Let's examine the choices: 
A. {−4, −9} has a product of 36. 
B. {−3, −12} has a product of 36. 
C. { 1 2 , −72} has a product of −36, not 36. 
D. {1, 36} has a product of 36. 
E. { 3 2 , 24} has a product of 36.
ANSWER 3: C

Problem 4:
The problem is asking us to find the height of the stool that Alice needs to stand on to reach the light bulb. 
First, we need to convert the height of the ceiling from meters to centimeters: 2.4 meters = 240 centimeters. 
The light bulb is 10 centimeters below the ceiling, so its height above the floor is 240 - 10 = 230 centimeters. 
Alice is 1.5 meters = 150 centimeters tall and can reach 46 centimeters above her head, so her total reach is 150 + 46 = 196 centimeters. 
To find the height of the stool, we subtract Alice's total reach from the height of the light bulb: 230 - 196 = 34 centimeters.
Let's examine the choices: 
A. 32 is too low because 230 - (196 + 32) = 2, which is not enough to reach the light bulb. 
B. 34 is the correct height. 
C. 36 is too high. 
D. 38 is too high. 
E. 40 is too high.
ANSWER 4: B

Problem 5:
The problem is asking us to find the total number of fish in the lake based on the ratio of trout in the sample to the total number of fish in the sample. 
The sample contains 30 trout out of 180 fish, so the ratio of trout to total fish in the sample is 30/180. 
We are given that this ratio is the same as the ratio of trout to total fish in the lake. 
Let's denote the total number of fish in the lake as x. 
The ratio of trout to total fish in the lake is 250/x. 
Setting the two ratios equal, we get 30/180 = 250/x. 
Cross-multiplying, we have 30x = 250 × 180. 
Simplifying, 30x = 45000. 
Dividing both sides by 30, x = 1500.
Let's examine the choices: 
A. 1250 is too low. 
B. 1500 matches our calculation. 
C. 1750 is too high. 
D. 1800 is too high. 
E. 2000 is too high.
ANSWER 5: B

Problem 6:
The problem is asking us to find out how much Karl could have saved by waiting a day to buy the folders at a 20%-off sale. 
The original price of 5 folders is 5 × $2.50 = $12.50. 
At a 20%-off sale, the discount is 20% of $12.50, which is 0.20 × $12.50 = $2.50. 
So, Karl could have saved $2.50.
Let's examine the choices: 
A. $1.00 is too low. 
B. $2.00 is too low. 
C. $2.50 matches our calculation. 
D. $2.75 is too high. 
E. $5.00 is too high.
ANSWER 6: C

Problem 7:
The problem is asking us to find the time it will take Casey to paint 300 dimples on a golf ball. 
Casey takes 2 seconds to paint one dimple, so to paint 300 dimples, he will take 300 × 2 = 600 seconds. 
To convert seconds to minutes, we divide by 60: 600 / 60 = 10 minutes.
Let's examine the choices: 
A. 4 minutes is too low. 
B. 6 minutes is too low. 
C. 8 minutes is too low. 
D. 10 minutes matches our calculation. 
E. 12 minutes is too high.
ANSWER 7: D

Problem 8:
The problem is asking us to find the number of different combinations of $5 bills and $2 bills that can be used to make a total of $17. 
Let's list the possible combinations: 
1. 3 $5 bills = $15, and $2 bill = $2, total = $17 (1 combination: 3 $5 bills and 1 $2 bill)
2. 1 $5 bill = $5, and 6 $2 bills = $12, total = $17 (1 combination: 1 $5 bill and 6 $2 bills)
We should also check if there are any other combinations: 
- Using 0 $5 bills, we need 8.5 $2 bills, which is not possible since we can't have a fraction of a bill.
- Using 2 $5 bills = $10, we need $7 more, which can be made with 3.5 $2 bills, again not possible.
- Using 4 $5 bills = $20, which is more than $17.

Problem 1:
We want the minimum number of socks to guarantee 5 of the same color. With 3 colors (red, white, blue), the worst case is picking 4 of each color: 4+4+4 = 12 socks. The 13th sock must create a group of 5 of one color. So 13 socks are needed.
ANSWER 1: D

Problem 2:
We choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways is the combination C(4,3) = 4.
ANSWER 2: B

Problem 3:
Compute each product:
A: (−4)×(−9) = 36
B: (−3)×(−12) = 36
C: (1/2)×(−72) = −36
D: 1×36 = 36
E: (3/2)×24 = 36
Only C does not equal 36.
ANSWER 3: C

Problem 4:
Ceiling = 2.4 m = 240 cm. Light bulb is 10 cm below ceiling, so its height = 230 cm.
Alice's height = 1.5 m = 150 cm. Reach above head = 46 cm, so reach without stool = 196 cm.
Stool height needed = 230 − 196 = 34 cm.
ANSWER 4: B

Problem 5:
Sample ratio: 30 trout / 180 fish = 1/6. Lake has 250 trout, which is 1/6 of total fish. Total fish = 250 × 6 = 1500.
ANSWER 5: B

Problem 6:
Cost without sale: 5 × $2.50 = $12.50.
With 20% off, each folder costs $2.50 × 0.8 = $2.00. Total = 5 × $2.00 = $10.00.
Savings = $12.50 − $10.00 = $2.50.
ANSWER 6: C

Problem 7:
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.
ANSWER 7: D

Problem 8:
Find non-negative integers x ($5 bills) and y ($2 bills) such that 5x + 2y = 17.
x=1 → 5 + 2y = 17 → y=6.
x=3 → 15 + 2y = 17 → y=1.
No other x works. So 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours 40 minutes (16×60 = 960, remainder 40).
Noon + 16 hours = 4:00 a.m. next day. Add 40 minutes → 4:40 a.m.
ANSWER 9: D

Problem 10:
(8 × 4 + 2) = 32 + 2 = 34.
(8 + 4 × 2) = 8 + 8 = 16.
34 − 16 = 18.
ANSWER 10: D

Problem 11:
11 hours × 60 = 660 minutes. Plus 5 minutes = 665 minutes.
ANSWER 11: C

Problem 12:
6 choices for entry window, then 5 choices for a different exit window. Total = 6 × 5 = 30.
ANSWER 12: D

Problem 1:
We need to find the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). Using the Pigeonhole Principle, we consider the worst-case scenario: the Martian pulls out as many socks as possible *without* getting 5 of any one color. That means getting exactly 4 of each color:
- 4 red + 4 white + 4 blue = 12 socks
After 12 socks, the Martian could have 4 of each color and still not have 5 matching. The very next sock (the 13th) must be red, white, or blue, pushing the count of that color to 5. So 13 socks guarantees 5 of the same color.
(The "five-legged" detail is flavor text explaining why the Martian needs 5 socks.)

ANSWER 1: D

Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). This is a combination problem since order doesn't matter:
C(4, 3) = 4! / (3! × 1!) = 4
We can also list them: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — exactly 4 ways.

ANSWER 2: B

Problem 3:
We check each product:
- A: (−4) × (−9) = 36 ✓
- B: (−3) × (−12) = 36 ✓
- C: (1/2) × (−72) = −36 ✗ (This equals −36, NOT 36)
- D: 1 × 36 = 36 ✓
- E: (3/2) × 24 = 36 ✓
Choice C gives −36, which does not equal 36.

ANSWER 3: C

Problem 4:
First, convert everything to centimeters:
- Ceiling height: 2.4 m = 240 cm
- Light bulb position: 240 − 10 = 230 cm above the floor
- Alice's height: 1.5 m = 150 cm
- Alice's reach above her head: 46 cm
- Alice's maximum reach (without stool): 150 + 46 = 196 cm above the floor

Let h = height of the stool. Standing on the stool, her reach is h + 196 cm. She can "just reach" the bulb:
h + 196 = 230
h = 230 − 196 = 34 cm

ANSWER 4: B

Problem 5:
The ratio of trout to total fish in the sample should equal the ratio in the lake:
- Sample ratio: 30 trout / 180 fish = 1/6
- Lake ratio: 250 trout / Total fish = 1/6

Setting up the proportion: 250 / Total = 1/6
Total = 250 × 6 = 1500 fish

ANSWER 5: B

Problem 6:
Karl bought 5 folders at $2.50 each:
Total cost = 5 × $2.50 = $12.50
A 20%-off sale means he would save 20% of the total:
Savings = 0.20 × $12.50 = $2.50

ANSWER 6: C

Problem 7:
Total time in seconds: 300 dimples × 2 seconds/dimple = 600 seconds
Convert to minutes: 600 seconds ÷ 60 seconds/minute = 10 minutes

ANSWER 7: D

Problem 8:
We need non-negative integer solutions to 5x + 2y = 17, where x = number of $5 bills and y = number of $2 bills.
- x = 0: 2y = 17 → y = 8.5 (not an integer) ✗
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 ✗
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x = 4: 20 > 17 (too large) ✗

Only 2 valid combinations: (1 five + 6 twos) and (3 fives + 1 two).

ANSWER 8: A

Problem 9:
Convert 1000 minutes to hours and minutes:
1000 ÷ 60 = 16 remainder 40, so 1000 minutes = 16 hours 40 minutes.
Starting at noon (12:00 PM):
- 12 hours later = midnight (12:00 AM)
- 4 more hours = 4:00 AM
- 40 more minutes = 4:40 AM

ANSWER 9: D

Problem 10:
Evaluate each part using order of operations (multiplication before addition):
- First part: (8 × 4 + 2) = (32 + 2) = 34
- Second part: (8 + 4 × 2) = (8 + 8) = 16
- Difference: 34 − 16 = 18

ANSWER 10: D

Problem 11:
Convert 11 hours and 5 minutes to minutes:
11 hours × 60 minutes/hour = 660 minutes
660 + 5 = 665 minutes

ANSWER 11: C

Problem 12:
Georgie enters through one window and leaves through a different one.
- Choices for entry window: 6
- Choices for exit window (must be different): 5
Total ways = 6 × 5 = 30

ANSWER 12: D

Problem 1:
The Martian needs to be **certain** of having 5 socks of the same color. To guarantee this, we must consider the worst-case scenario: the Martian picks as many socks as possible **without** yet having 5 of any single color.  
With 3 colors (red, white, blue), the worst case is picking 4 socks of each color:  
\(4 + 4 + 4 = 12\) socks.  
At this point the Martian has 4 of every color but still does not have 5 of any one color. The very next sock—the 13th—must be red, white, or blue, and will necessarily create a set of 5 of that color.  
Thus 13 socks are required.  
(Choice A is too small; with 6 socks one could have 2,2,2. Choice B allows 3,3,3. Choice C allows 4,4,4. Choice E is more than the minimum guaranteed number.)

ANSWER 1: D

---

Problem 2:
We need to choose a group of 3 starters from the 4 students: Lance, Sally, Joy, and Fred. Since the three starters form a team and order does not matter, this is a combination problem.  
\[
\binom{4}{3} = \frac{4!}{3!\,1!} = 4
\]  
Equivalently, we can think of choosing which **one** of the four students will **not** start; there are 4 such choices.  
(Choice A is too small. Choice C is \(\binom{4}{2}\). Choices D and E do not arise from counting 3-person teams from 4 people.)

ANSWER 2: B

---

Problem 3:
We check the product of each pair to see which is not 36.

- **A.** \((-4)(-9) = 36\)  
- **B.** \((-3)(-12) = 36\)  
- **C.** \(\left(\frac12\right)(-72) = -36\)  
- **D.** \((1)(36) = 36\)  
- **E.** \(\left(\frac32\right)(24) = 36\)

Only pair **C** gives \(-36\), not \(36\).

ANSWER 3: C

---

Problem 4:
First convert all measurements to centimeters.  
- Ceiling height: \(2.4\text{ m} = 240\text{ cm}\).  
- Light bulb is \(10\text{ cm}\) below the ceiling, so its height above the floor is \(240 - 10 = 230\text{ cm}\).  

Alice’s reach:  
- Height: \(1.5\text{ m} = 150\text{ cm}\).  
- She can reach \(46\text{ cm}\) above her head, so her standing reach is \(150 + 46 = 196\text{ cm}\).  

The stool must supply the remaining height:  
\[
230 - 196 = 34\text{ cm}.
\]

ANSWER 4: B

---

Problem 5:
Let \(T\) be the total number of fish in the lake. The problem states that the ratio of trout to total fish is the same in the sample and in the lake.  
\[
\frac{250}{T} = \frac{30}{180}
\]  
Simplify the right side: \(\frac{30}{180} = \frac16\).  
\[
\frac{250}{T} = \frac16 \quad\Longrightarrow\quad T = 250 \times 6 = 1500.
\]

ANSWER 5: B

---

Problem 6:
Karl bought 5 folders at \(\$2.50\) each.  
Total cost: \(5 \times \$2.50 = \$12.50\).  
A 20% discount would have saved him 20% of this total:  
\[
0.20 \times \$12.50 = \$2.50.
\]

ANSWER 6: C

---

Problem 7:
Casey paints 300 dimples at 2 seconds per dimple.  
Total time in seconds: \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(\frac{600}{60} = 10\) minutes.

ANSWER 7: D

---

Problem 8:
Let \(f\) be the number of \(\$5\) bills and \(t\) be the number of \(\$2\) bills. We need non-negative integer solutions to  
\[
5f + 2t = 17.
\]  
Test values of \(f\):  
- \(f=0\): \(2t=17\) → no integer solution.  
- \(f=1\): \(5 + 2t = 17 \Rightarrow t=6\).  **(Valid)**
- \(f=2\): \(10 + 2t = 17 \Rightarrow t=3.5\).  (Invalid)
- \(f=3\): \(15 + 2t = 17 \Rightarrow t=1\).  **(Valid)**
- \(f=4\): \(20 > 17\) → stop.

There are exactly **2** combinations: \((1\text{ five}, 6\text{ twos})\) and \((3\text{ fives}, 1\text{ two})\).

ANSWER 8: A

---

Problem 9:
The contest starts at noon (12:00 p.m.). We add 1000 minutes.  
Divide 1000 by 60:  
\[
1000 = 16 \times 60 + 40,
\]  
so 1000 minutes = 16 hours and 40 minutes.  

Noon + 12 hours = midnight.  
Midnight + 4 hours 40 minutes = 4:40 a.m.

ANSWER 9: D

---

Problem 10:
Evaluate using the correct order of operations (multiplication before addition).

First parentheses:  
\[
8 \times 4 + 2 = 32 + 2 = 34.
\]

Second parentheses:  
\[
8 + 4 \times 2 = 8 + 8 = 16.
\]

Subtract:  
\[
34 - 16 = 18.
\]

ANSWER 10: D

---

Problem 11:
Convert 11 hours to minutes and add the remaining 5 minutes.  
\[
11 \times 60 = 660\text{ minutes}.
\]  
\[
660 + 5 = 665\text{ minutes}.
\]

ANSWER 11: C

---

Problem 12:
Georgie enters through one of 6 windows and must leave through a **different** window.  
- 6 choices for the entry window.  
- For each entry choice, 5 remaining choices for the exit window.  

Total ways:  
\[
6 \times 5 = 30.
\]  
(Choice E, 36, would count entering and leaving through the same window, which is not allowed.)

ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). The worst-case scenario is to draw as many socks as possible without having 5 of any one color. That would be 4 red, 4 white, and 4 blue, for a total of 12 socks. The next sock drawn (the 13th) must be one of the three colors, giving that color 5 socks. Therefore, 13 socks are required. The five-legged Martian is irrelevant to the count.  
Answer: D

Problem 2:
We have 4 people (Lance, Sally, Joy, Fred) and need to choose 3 starters. Since the starters are not assigned to specific positions, order does not matter. The number of ways is the combination \( \binom{4}{3} = 4 \).  
Answer: B

Problem 3:
Compute each product:  
A. \((-4) \times (-9) = 36\)  
B. \((-3) \times (-12) = 36\)  
C. \(\frac{1}{2} \times (-72) = -36\)  
D. \(1 \times 36 = 36\)  
E. \(\frac{3}{2} \times 24 = 36\)  
Only pair C gives a product not equal to 36.  
Answer: C

Problem 4:
Convert all measurements to centimeters:  
Ceiling height = 2.4 m = 240 cm.  
Light bulb is 10 cm below ceiling → height = 240 – 10 = 230 cm.  
Alice’s height = 1.5 m = 150 cm.  
Her reach above her head = 46 cm, so her maximum reach without a stool = 150 + 46 = 196 cm.  
Let \(h\) be the stool height. Standing on the stool raises her reach to \(196 + h\). This must equal 230 cm:  
\(196 + h = 230\) → \(h = 34\) cm.  
Answer: B

Problem 5:
In the sample, 30 out of 180 fish are trout, so the ratio trout : total = \(30/180 = 1/6\).  
Assuming the same ratio in the lake, if the lake has 250 trout, then the total number of fish is \(250 \times 6 = 1500\).  
Answer: B

Problem 6:
Original cost for 5 folders: \(5 \times 2.50 = \$12.50\).  
A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
Thus, by waiting a day, Karl could have saved $2.50.  
Answer: C

Problem 7:
Time to paint one dimple = 2 seconds.  
Total time for 300 dimples = \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(600 \div 60 = 10\) minutes.  
Answer: D

Problem 8:
We need nonnegative integers \(a\) (number of $5 bills) and \(b\) (number of $2 bills) such that \(5a + 2b = 17\).  
Solve by testing values of \(b\):  
\(b = 1\) → \(5a = 15\) → \(a = 3\) → (3,1) works.  
\(b = 6\) → \(5a = 5\) → \(a = 1\) → (1,6) works.  
No other values give integer \(a\). Thus there are 2 combinations.  
Answer: A

Problem 9:
1000 minutes = \(1000 \div 60 = 16\) hours and \(40\) minutes (since \(60 \times 16 = 960\), remainder 40).  
Starting at noon (12:00 PM), add 12 hours → midnight (12:00 AM). Add the remaining 4 hours → 4:00 AM. Then add 40 minutes → 4:40 AM.  
Answer: D

Problem 10:
Evaluate using order of operations:  
First parentheses: \(8 \times 4 + 2 = 32 + 2 = 34\).  
Second parentheses: \(8 + 4 \times 2 = 8 + 8 = 16\).  
Subtract: \(34 - 16 = 18\).  
Answer: D

Problem 11:
11 hours = \(11 \times 60 = 660\) minutes.  
Add 5 minutes → \(660 + 5 = 665\) minutes.  
Answer: C

Problem 12:
Georgie must choose an entry window (6 choices) and a different exit window (5 choices). Since entering and leaving are distinct actions, the total number of ways is \(6 \times 5 = 30\).  
Answer: D

ANSWER 1: D
ANSWER 2: B
ANSWER 3: C
ANSWER 4: B
ANSWER 5: B
ANSWER 6: C
ANSWER 7: D
ANSWER 8: A
ANSWER 9: D
ANSWER 10: D
ANSWER 11: C
ANSWER 12: D

Problem 1:  
The Martian wants to be certain of having 5 socks of the same color. There are three colors: red, white, and blue. In the worst-case scenario, the Martian could pick 4 socks of each color before getting a fifth of any one color. That would be \(4 \times 3 = 12\) socks, and still no color has 5 socks. The next sock drawn (the 13th) must complete a set of 5 of some color, because there are only three colors. Therefore, the Martian must remove **13** socks to be certain of having 5 socks of the same color.  
ANSWER 1: D

Problem 2:  
Ms. Hamilton's class has 4 available players: Lance, Sally, Joy, and Fred. The number of ways to choose 3 starters from 4 players is given by the combination \(\binom{4}{3} = 4\). Thus, there are **4** possible sets of three starters.  
ANSWER 2: B

Problem 3:  
We check each pair's product:  
- A: \((-4) \times (-9) = 36\)  
- B: \((-3) \times (-12) = 36\)  
- C: \(\frac{1}{2} \times (-72) = -36\) (not 36)  
- D: \(1 \times 36 = 36\)  
- E: \(\frac{3}{2} \times 24 = 36\)  
Only pair **C** does not have a product of 36.  
ANSWER 3: C

Problem 4:  
The ceiling is 2.4 m above the floor, and the bulb is 10 cm (0.1 m) below the ceiling, so the bulb's height is \(2.4 - 0.1 = 2.3\) m. Alice's height is 1.5 m, and she can reach 46 cm (0.46 m) above her head, so her reach height from the floor is \(1.5 + 0.46 = 1.96\) m. When standing on a stool, her hands are at the stool height plus her reach height. Setting this equal to the bulb height:  
\[
\text{stool height} + 1.96 = 2.3 \Rightarrow \text{stool height} = 0.34 \text{ m} = 34 \text{ cm}.
\]  
ANSWER 4: B

Problem 5:  
In the sample, the ratio of trout to total fish is \(\frac{30}{180} = \frac{1}{6}\). Assuming the same ratio in the lake, let \(T\) be the total number of fish in the lake. Then \(\frac{250}{T} = \frac{1}{6}\), so \(T = 250 \times 6 = 1500\).  
ANSWER 5: B

Problem 6:  
Karl bought 5 folders at $2.50 each, so the original cost is \(5 \times 2.50 = \$12.50\). With a 20% discount, the cost becomes \(12.50 \times 0.80 = \$10.00\). The savings is \(12.50 - 10.00 = \$2.50\).  
ANSWER 6: C

Problem 7:  
Painting 300 dimples at 2 seconds each takes \(300 \times 2 = 600\) seconds. Converting to minutes: \(600 \div 60 = 10\) minutes.  
ANSWER 7: D

Problem 8:  
We need nonnegative integers \(x\) (number of $5 bills) and \(y\) (number of $2 bills) such that \(5x + 2y = 17\). Solving:  
- \(y = 1 \Rightarrow 5x = 15 \Rightarrow x = 3\)  
- \(y = 6 \Rightarrow 5x = 5 \Rightarrow x = 1\)  
- \(y = 11 \Rightarrow 5x = -5\) (invalid)  
Thus, there are **2** valid combinations: (3 fives, 1 two) and (1 five, 6 twos).  
ANSWER 8: A

Problem 9:  
1000 minutes equals \(16\) hours and \(40\) minutes (since \(16 \times 60 = 960\), remainder \(40\)). Starting at noon, adding \(16\) hours gives 4:00 AM the next day, and adding \(40\) minutes gives 4:40 AM.  
ANSWER 9: D

Problem 10:  
Compute inside parentheses:  
\(8 \times 4 + 2 = 32 + 2 = 34\)  
\(8 + 4 \times 2 = 8 + 8 = 16\)  
Then \(34 - 16 = 18\).  
ANSWER 10: D

Problem 11:  
11 hours is \(11 \times 60 = 660\) minutes. Adding 5 minutes gives \(660 + 5 = 665\) minutes.  
ANSWER 11: C

Problem 12:  
Georgie can enter through any of the 6 windows. For exiting, he must choose a different window, so there are 5 choices. The total number of ways is \(6 \times 5 = 30\).  
ANSWER 12: D

Problem 1:  
The Martian needs to ensure there are 5 socks of the same color. The worst-case scenario is pulling the maximum number of socks without 5 of any color. Since there are 3 colors, this would be 4 socks of each color (4 × 3 = 12 socks). The next sock (the 13th) must match one color, giving 5 socks of that color.  
**ANSWER 1: D**  

Problem 2:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is C(4,3) = 4.  
**ANSWER 2: B** (Note: C(4,3)=4, but rechecking: the team is chosen from 4 specific people, so combinations are 4. Answer B=4)  

Problem 3:  
Check each pair:  
A. (−4)(−9) = 36  
B. (−3)(−12) = 36  
C. (1/2)(−72) = -36 (≠36)  
D. (1)(36) = 36  
E. (3/2)(24) = 36  
The pair that does NOT equal 36 is C.  
**ANSWER 3: E** (Re-evaluated: The question asks which does NOT equal 36. Option E: (3/2)*24=36, so it is valid. Option C: (1/2)*(-72)=-36, so C is the correct choice for NOT equal. Correction: **ANSWER 3: C** [Original had a typo, but computation confirms C is the answer])  

*Correction: For Problem 3, Option C: (1/2, -72) likely means (1/2) × (-72) = -36 ≠ 36. Option E: (3/2) × 24 = 36. Thus, C is the answer.*  
**ANSWER 3: C**  

Problem 4:  
Ceiling height: 2.4 m = 240 cm. Light bulb location: 240 - 10 = 230 cm from floor. Alice’s reach: 150 cm + 46 cm = 196 cm. Stool height = 230 - 196 = 34 cm.  
**ANSWER 4: B**  

Problem 5:  
Ratio of trout in sample: 30/180 = 1/6. Total trout: 250. Total fish = 250 × 6 = 1500.  
**ANSWER 5: B**  

Problem 6:  
Original cost: 5 × $2.50 = $12.50. 20% off: $12.50 × 0.20 = $2.50 saved.  
**ANSWER 6: C** (Note: The question asks how much Karl could have saved, which is $2.50, not $2.00. Original had a mistake; correction: 20% of $12.50 is $2.50, so answer C)  

Problem 7:  
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.  
**ANSWER 7: D**  

Problem 8:  
Let $5 bills = x, $2 bills = y. 5x + 2y = 17. Possible solutions:  
- x=3, y=1 (15+2=17)  
- x=1, y=6 (5+12=17)  
Only 2 combinations.  
**ANSWER 8: A**  

Problem 9:  
1000 minutes = 16 hours 40 minutes. Starting at noon: noon + 16h40m = 4:40 a.m. next day.  
**ANSWER 9: D**  

Problem 10:  
(8×4 + 2) = 34, (8 + 4×2) = 16. 34 - 16 = 18.  
**ANSWER 10: D**  

Problem 11:  
11 hours = 660 minutes. 660 + 5 = 665 minutes.  
**ANSWER 11: C**  

Problem 12:  
Georgie can enter through 6 windows and exit through 5 others. Total ways: 6 × 5 = 30.  
**ANSWER 12: D**  

---  
**Final Answers**  
ANSWER 1: D  
ANSWER 2: B  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: B  
ANSWER 6: C  
ANSWER 7: D  
ANSWER 8: A  
ANSWER 9: D  
ANSWER 10: D  
ANSWER 11: C  
ANSWER 12: D

### Problem 1:
This is a classic pigeonhole principle problem. We need to find the minimum number of socks the Martian must pull to guarantee 5 socks of the same color, with 3 total colors. The worst-case scenario is that the Martian pulls as many socks as possible without getting 5 of any color: that is 4 socks of each color, for a total of 3*4=12 socks. The next sock pulled (the 13th) has to be one of the three colors, pushing that color to 5 socks. Tempting wrong answers: C (12 is only the number of socks that avoids 5 of the same color, not enough to guarantee the requirement), E (15 is the total if you took 5 of each color, not the minimum to guarantee).
ANSWER 1: D

---

### Problem 2:
We need to count the number of ways to choose 3 starters from 4 people, where order on the team does not matter. This is a combination problem: C(n,k) = n!/(k!(n-k)!). For n=4, k=3: C(4,3) = 4. We can also list the teams by who is left out: leaving out Lance, Sally, Joy, or Fred gives exactly 4 unique teams. Tempting wrong answer: C (6 is the count if order of selection mattered, which it does not for a team of starters).
ANSWER 2: B

---

### Problem 3:
We calculate the product of each pair to find which does not equal 36:
- A: (-4)*(-9)=36 ✔️
- B: (-3)*(-12)=36 ✔️
- C: (1/2)*(-72) = -36 ❌
- D: 1*36=36 ✔️
- E: (3/2)*24=36 ✔️
Only pair C has a product of -36, not 36.
ANSWER 3: C

---

### Problem 4:
First convert all units to centimeters to avoid conversion errors:
- Ceiling height: 2.4m = 240cm. Light bulb is 10cm below ceiling, so bulb height from floor = 240-10=230cm.
- Alice's height: 1.5m=150cm, her reach above her head is 46cm, so total reach without stool = 150+46=196cm.
- Stool height = 230-196=34cm.
Tempting wrong answers come from unit conversion mistakes, e.g., mixing up meters and centimeters.
ANSWER 4: B

---

### Problem 5:
This is a proportion problem. The ratio of trout to total fish is the same in the sample and the lake. In the sample, 30 trout out of 180 total fish = 1/6. Let T be total fish in the lake: 250/T = 1/6 → T=250*6=1500.
ANSWER 5: B

---

### Problem 6:
Karl bought 5 folders at $2.50 each, total original cost = 5*2.50=$12.50. He saves 20% of the total cost by waiting for the sale: 0.2*12.50=$2.50. This can also be calculated per folder: 20% of $2.50 is $0.50 saved per folder, 5*0.50=$2.50 total saved.
ANSWER 6: C

---

### Problem 7:
300 dimples * 2 seconds per dimple = 600 total seconds. Convert to minutes: 60 seconds per minute, so 600/60=10 minutes.
ANSWER 7: D

---

### Problem 8:
We find all non-negative integer pairs (x,y) where x=number of $5 bills, y=number of $2 bills, and 5x+2y=17. Only odd values of x work (since $5 is odd, $2 is even, we need an odd sum to get 17):
- x=1: 5+2y=17 → y=6 (valid)
- x=3: 15+2y=17 → y=1 (valid)
Even x values (0,2) leave an odd remainder that cannot be formed with $2 bills, so only 2 valid combinations.
ANSWER 8: A

---

### Problem 9:
Convert 1000 minutes to hours and minutes: 60 minutes per hour, 1000/60=16 hours and 40 minutes. Adding this to noon (12pm): 12pm +16h40m = 4:40am the next day.
ANSWER 9: D

---

### Problem 10:
Use order of operations (PEMDAS) to calculate:
First parenthesis: 8*4 +2 = 32+2=34
Second parenthesis: 8 +4*2=8+8=16
Subtract: 34-16=18. The tempting wrong answer C (10) comes from incorrectly adding before multiplying in the second parenthesis.
ANSWER 10: D

---

### Problem 11:
Convert 11 hours 5 minutes to total minutes: 11*60=660 minutes, plus 5 extra minutes = 665 minutes.
ANSWER 11: C

---

### Problem 12:
Georgie has 6 choices for an entry window, and 5 remaining choices for an exit window (must use a different window). Total ways = 6*5=30. Tempting wrong answer E (36) counts entering and leaving through the same window, which is not allowed; B (15) incorrectly treats entry/exit as unordered, which they are not.
ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color (red, white, blue). This is a pigeonhole principle problem: the worst-case scenario is pulling the maximum number of socks without getting 5 of any color. For each of the 3 colors, we can pull 4 socks (since 4 < 5), so 4×3=12 socks. The next sock pulled will necessarily be the 5th of one color. So total socks needed: 12 + 1 = 13.
ANSWER 1: D
Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). Since the problem does not mention distinct positions for starters, order does not matter, so we use combinations. The number of ways to choose 3 people from 4 is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n=4 \) and \( k=3 \):
\( \binom{4}{3} = \frac{4!}{3!1!} = 4 \). This is equivalent to counting how many people we leave out (4 choices, one for each person not selected).
ANSWER 2: B
Problem 3:
We check the product of each pair:
- A: \( (-4) \times (-9) = 36 \)
- B: \( (-3) \times (-12) = 36 \)
- C: \( \frac{1}{2} \times (-72) = -36 \neq 36 \)
- D: \( 1 \times 36 = 36 \)
- E: \( \frac{3}{2} \times 24 = 36 \)
Only pair C does not have a product of 36.
ANSWER 3: C
Problem 4:
Convert all units to centimeters:
- Ceiling height: \( 2.4 \, \text{m} = 240 \, \text{cm} \)
- Light bulb height from floor: \( 240 - 10 = 230 \, \text{cm} \)
- Alice's reach from floor (without stool): \( 1.5 \, \text{m} + 0.46 \, \text{m} = 150 + 46 = 196 \, \text{cm} \)
Let \( h \) be the stool height. The total reach with the stool equals the light bulb height:
\( h + 196 = 230 \)
\( h = 230 - 196 = 34 \, \text{cm} \)
ANSWER 4: B
Problem 5:
The ratio of trout to total fish in the sample equals the ratio in the lake. The sample has \( \frac{30}{180} = \frac{1}{6} \) trout. Let \( T \) be the total number of fish in the lake. We know there are 250 trout, so:
\( \frac{250}{T} = \frac{1}{6} \)
\( T = 250 \times 6 = 1500 \)
ANSWER 5: B
Problem 6:
Karl bought 5 folders at $2.50 each. The 20% discount applies to each folder, so the savings per folder is \( 2.50 \times 0.20 = 0.50 \). Total savings for 5 folders: \( 5 \times 0.50 = 2.50 \).
ANSWER 6: C
Problem 7:
Total time to paint 300 dimples: \( 300 \times 2 = 600 \) seconds. Convert to minutes: \( 600 \div 60 = 10 \) minutes.
ANSWER 7: D
Problem 8:
Let \( x \) = number of $5 bills, \( y \) = number of $2 bills. We need \( 5x + 2y = 17 \), where \( x,y \geq 0 \) integers. Test non-negative integer values of \( x \):
- \( x=0 \): \( 2y=17 \) → \( y=8.5 \) (not integer, invalid)
- \( x=1 \): \( 5 + 2y=17 \) → \( 2y=12 \) → \( y=6 \) (valid: 1 $5 + 6 $2 = $17)
- \( x=2 \): \( 10 + 2y=17 \) → \( 2y=7 \) → \( y=3.5 \) (invalid)
- \( x=3 \): \( 15 + 2y=17 \) → \( 2y=2 \) → \( y=1 \) (valid: 3 $5 + 1 $2 = $17)
- \( x=4 \): \( 20 > 17 \) (stop, no more valid \( x \))
Only 2 valid combinations.
ANSWER 8: A
Problem 9:
Convert 1000 minutes to hours and minutes: \( 1000 \div 60 = 16 \) hours with a remainder of 40 minutes (since \( 16 \times 60 = 960 \), \( 1000 - 960 = 40 \)). The contest starts at noon, so adding 16 hours 40 minutes: noon + 12 hours = midnight, plus 4 hours 40 minutes = 4:40 a.m.
ANSWER 9: D
Problem 10:
Use order of operations (multiplication before addition):
First parentheses: \( 8 \times 4 + 2 = 32 + 2 = 34 \)
Second parentheses: \( 8 + 4 \times 2 = 8 + 8 = 16 \)
Subtract: \( 34 - 16 = 18 \)
ANSWER 10: D
Problem 11:
Convert 11 hours 5 minutes to total minutes: \( 11 \times 60 + 5 = 660 + 5 = 665 \)
ANSWER 11: C
Problem 12:
Georgie chooses an entrance window first (6 options), then a different exit window (5 remaining options, since exit cannot equal entrance). Since entrance and exit are distinct actions, order matters, so total ways: \( 6 \times 5 = 30 \).
ANSWER 12: D

I'll work through each problem systematically.

**Problem 1:**

**What is being asked:** How many socks must be removed to guarantee 5 socks of the same color?

**Setup:** There are 3 colors (red, white, blue). In the worst case scenario, the Martian pulls out socks as evenly distributed among colors as possible before getting 5 of one color.

**Working:** 
- Worst case: 4 red, 4 white, 4 blue = 12 socks (still no color with 5)
- The 13th sock must give us a 5th sock of some color

**Answer 1: D**

---

**Problem 2:**

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

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

**Working:**
- C(4,3) = 4 ways
- The 4 ways correspond to leaving out each person: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred}

**Answer 2: B**

---

**Problem 3:**

**What is being asked:** Which pair does NOT have a product equal to 36?

**Working:** Check each pair:
- A. {−4, −9}: (−4)(−9) = 36 ✓
- B. {−3, −12}: (−3)(−12) = 36 ✓
- C. {1/2, −72}: (1/2)(−72) = −36 ✗
- D. {1, 36}: (1)(36) = 36 ✓
- E. {3/2, 24}: (3/2)(24) = 36 ✓

**Answer 3: C**

---

**Problem 4:**

**What is being asked:** How high is the stool?

**Setup:** Convert all measurements to centimeters:
- Ceiling height: 240 cm
- Light bulb location: 240 − 10 = 230 cm from floor
- Alice's height: 150 cm
- Alice's reach above her head: 46 cm
- Alice's reach without stool: 150 + 46 = 196 cm

**Working:**
- With stool, Alice can just reach the bulb at 230 cm
- Stool height + 196 cm = 230 cm
- Stool height = 230 − 196 = 34 cm

**Answer 4: B**

---

**Problem 5:**

**What is being asked:** How many fish total are in the lake?

**Setup:** Use proportions. In the sample: 30 trout out of 180 fish. In the lake: 250 trout out of total fish.

**Working:**
- 30/180 = 250/x
- 1/6 = 250/x
- x = 250 × 6 = 1500 fish

**Answer 5: B**

---

**Problem 6:**

**What is being asked:** How much could Karl have saved with the 20% off sale?

**Setup:** 
- Original cost: 5 folders × $2.50 = $12.50
- Sale discount: 20% off

**Working:**
- 20% of $12.50 = 0.20 × $12.50 = $2.50

**Answer 6: C**

---

**Problem 7:**

**What is being asked:** How many minutes to paint 300 dimples at 2 seconds each?

**Working:**
- Total time: 300 × 2 = 600 seconds
- Convert to minutes: 600 ÷ 60 = 10 minutes

**Answer 7: D**

---

**Problem 8:**

**What is being asked:** How many combinations of $5 bills and $2 bills make $17?

**Setup:** Let x = number of $5 bills, y = number of $2 bills
- 5x + 2y = 17

**Working:**
- x = 0: 2y = 17 → y = 8.5 (not integer)
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 (not integer)
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x ≥ 4: 5x ≥ 20 > 17 (impossible)

Valid combinations: (1,$5 and 6,$2) and (3,$5 and 1,$2) = 2 combinations

**Answer 8: A**

---

**Problem 9:**

**What is being asked:** When did the contest end, starting at noon and lasting 1000 minutes?

**Working:**
- 1000 minutes ÷ 60 = 16 hours and 40 minutes
- Noon + 16 hours 40 minutes = 4:40 p.m. the same day
- But we need to continue: 4:40 p.m. + additional time consideration
- Actually: 1000 ÷ 60 = 16 remainder 40, so 16 hours 40 minutes
- Noon + 16 hours 40 minutes = 4:40 a.m. the next day

**Answer 9: D**

---

**Problem 10:**

**What is being asked:** Calculate (8 × 4 + 2) − (8 + 4 × 2)

**Working:**
- First parentheses: 8 × 4 + 2 = 32 + 2 = 34
- Second parentheses: 8 + 4 × 2 = 8 + 8 = 16
- Result: 34 − 16 = 18

**Answer 10: D**

---

**Problem 11:**

**What is being asked:** Convert 11 hours 5 minutes to minutes

**Working:**
- 11 hours = 11 × 60 = 660 minutes
- Total: 660 + 5 = 665 minutes

**Answer 11: C**

---

**Problem 12:**

**What is being asked:** In how many ways can the ghost enter by one window and leave by a different window?

**Working:**
- Choices for entry: 6 windows
- Choices for exit: 5 windows (must be different from entry)
- Total ways: 6 × 5 = 30

**Answer 12: D**

Problem 1:

To be certain of getting 5 socks of the same color, we use the pigeonhole principle.

There are 3 colors: red, white, and blue.  
To avoid getting 5 of any one color for as long as possible, Martian could take at most 4 of each color:

- 4 red
- 4 white
- 4 blue

That gives a total of \(4+4+4=12\) socks without yet having 5 of the same color.

The very next sock, the 13th, must make one color appear at least 5 times.

So the Martian must remove 13 socks.

ANSWER 1: D

Problem 2:

There are 4 players: Lance, Sally, Joy, and Fred.

We need to choose 3 starters from these 4 players. Since order does not matter, we count combinations:

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

Indeed, the 3-starter groups are:

- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred

So there are 4 ways.

ANSWER 2: B

Problem 3:

We check each product.

- A: \((-4)(-9)=36\)
- B: \((-3)(-12)=36\)
- C: \(\frac12 \cdot (-72)=-36\), not 36
- D: \(1\cdot 36=36\)
- E: \(\frac32 \cdot 24=36\)

Only choice C does **not** have product 36.

ANSWER 3: C

Problem 4:

First find how high the bulb is above the floor.

The ceiling is 2.4 meters above the floor, which is \(240\) centimeters.  
The bulb is 10 centimeters below the ceiling, so it is:

\[
240 - 10 = 230 \text{ cm}
\]

Now Alice’s height is 1.5 meters = 150 cm.  
She can reach 46 cm above her head, so her reach without the stool is:

\[
150 + 46 = 196 \text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230 - 196 = 34 \text{ cm}
\]

So the stool is 34 cm high.

ANSWER 4: B

Problem 5:

The sample has 30 trout out of 180 fish, so the fraction of trout is:

\[
\frac{30}{180}=\frac{1}{6}
\]

If 250 trout is \(\frac16\) of the total lake population, then the total number of fish is:

\[
250 \div \frac16 = 250 \cdot 6 = 1500
\]

So there are 1500 fish in the lake.

ANSWER 5: B

Problem 6:

Karl bought 5 folders at \$2.50 each, so the total cost was:

\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:

\[
0.20 \times 12.50 = 2.50
\]

So he could have saved \$2.50.

ANSWER 6: C

Problem 7:

He paints 300 dimples, and each dimple takes 2 seconds.

Total time in seconds:

\[
300 \times 2 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10
\]

So he needs 10 minutes.

ANSWER 7: D

Problem 8:

We want combinations of 5-dollar bills and 2-dollar bills that total \$17.

Let \(x\) be the number of \$5 bills and \(y\) the number of \$2 bills. Then:

\[
5x + 2y = 17
\]

Try values of \(x\):

- \(x=1\): \(17-5=12\), so \(y=6\)
- \(x=2\): \(17-10=7\), not even, so impossible
- \(x=3\): \(17-15=2\), so \(y=1\)

These are the only nonnegative integer solutions.

So there are 2 combinations.

ANSWER 8: A

Problem 9:

The contest ended 1000 minutes after noon.

Convert 1000 minutes to hours and minutes:

\[
1000 \div 60 = 16 \text{ hours with } 40 \text{ minutes left}
\]

So:

- Noon + 12 hours = midnight
- plus 4 more hours = 4:00 a.m.
- plus 40 minutes = 4:40 a.m.

So the contest ended at 4:40 a.m.

ANSWER 9: D

Problem 10:

Compute each part using order of operations:

\[
(8\times 4 + 2) - (8 + 4\times 2)
\]

First parentheses:

\[
8\times 4 + 2 = 32 + 2 = 34
\]

Second parentheses:

\[
8 + 4\times 2 = 8 + 8 = 16
\]

Now subtract:

\[
34 - 16 = 18
\]

So the value is 18.

ANSWER 10: D

Problem 11:

We convert 11 hours and 5 minutes into minutes.

\[
11 \times 60 = 660
\]

Then add 5 minutes:

\[
660 + 5 = 665
\]

So the match lasted 665 minutes.

ANSWER 11: C

Problem 12:

Georgie must enter by one window and leave by a different window.

There are 6 choices for the entering window.  
After choosing one, there are 5 remaining choices for the exit window.

So the total number of ways is:

\[
6 \times 5 = 30
\]

Therefore, there are 30 ways.

ANSWER 12: D

Problem 1:  
We need the minimum number of socks to remove to guarantee there are 5 socks of the same color, given 3 colors (red, white, blue) and at least 5 of each color.

Worst case: you could first pull 4 socks of each color without reaching 5 of any color.  
That would be:
- 4 red + 4 white + 4 blue = 12 socks pulled

After 12 socks, you have at most 4 of each color. The next sock (the 13th) must be one of the 3 colors, and by the pigeonhole principle it makes one color reach 5.

So the Martian must remove **13** socks.

Tempting wrong choices:
- 12 isn’t enough because it’s possible to have only 4 of each color.
- Any number less than 13 can’t force a fifth sock of some color.

ANSWER 1: D


Problem 2:  
We have 4 students: Lance, Sally, Joy, Fred. We must choose 3 starters.

Number of ways to choose 3 out of 4 is:
\[
\binom{4}{3} = 4
\]

(You can also think: leave out one student; there are 4 choices for who to leave out.)

ANSWER 2: B


Problem 3:  
We check which pair does **NOT** multiply to 36.

Choices:
A. \(\{-4,-9\}\): \((-4)(-9)=36\) ✅  
B. \(\{-3,-12\}\): \((-3)(-12)=36\) ✅  
C. \(\left\{\frac{1}{2},-72\right\}\): \(\frac{1}{2}\cdot(-72)=-36\) ❌ not 36  
D. \(\{1,36\}\): \(1\cdot 36=36\) ✅  
E. \(\left\{\frac{3}{2},24\right\}\): \(\frac{3}{2}\cdot 24=3\cdot 12=36\) ✅

So the one that does not have product 36 is choice C.

ANSWER 3: C


Problem 4:  
Let’s convert everything to the same reference (floor).

- Ceiling is 2.4 m above the floor = 240 cm  
- The bulb is 10 cm below the ceiling: height of bulb above floor:
\[
240 - 10 = 230 \text{ cm}
\]
- Alice is 1.5 m tall = 150 cm  
- She can reach 46 cm above the top of her head, so from her top to the bulb:
\[
150 + 46 = 196 \text{ cm (if she stands on the floor)}
\]

She is standing on a stool, so her total height is (stool height + 150). She can reach:
\[
(\text{stool height} + 150) + 46
\]
Set equal to bulb height 230:
\[
\text{stool height} + 196 = 230
\]
\[
\text{stool height} = 34
\]

ANSWER 4: B


Problem 5:  
We know:
- Lake has 250 trout.
- Total fish in lake = \(T\)
- Sample has 180 fish, and 30 are trout.
So sample trout-to-total ratio equals lake trout-to-total ratio:
\[
\frac{30}{180} = \frac{250}{T}
\]
Simplify \(\frac{30}{180}=\frac{1}{6}\):
\[
\frac{1}{6} = \frac{250}{T}
\]
\[
T = 250 \cdot 6 = 1500
\]

ANSWER 5: B


Problem 6:  
Cost per folder originally: $2.50  
He bought 5 folders:
\[
5 \cdot 2.50 = 12.50
\]
A 20%-off sale means paying 80% of the original price:
\[
0.80 \cdot 12.50 = 10.00
\]
Savings:
\[
12.50 - 10.00 = 2.50
\]

ANSWER 6: C


Problem 7:  
Paint 300 dimples, 2 seconds per dimple.
Total time in seconds:
\[
300 \cdot 2 = 600 \text{ seconds}
\]
Convert to minutes (60 seconds per minute):
\[
600 \div 60 = 10 \text{ minutes}
\]

ANSWER 7: D


Problem 8:  
We need numbers of $5 bills and $2 bills that total $17 (order doesn’t matter).

Let \(x\) = number of $5 bills, \(y\) = number of $2 bills.
\[
5x + 2y = 17
\]
Try \(x=0\) to \(x=3\) (since \(4\cdot 5=20>17\)):

- \(x=0\): \(2y=17\) not possible  
- \(x=1\): \(5 + 2y=17 \Rightarrow 2y=12 \Rightarrow y=6\)  
- \(x=2\): \(10 + 2y=17 \Rightarrow 2y=7\) not possible  
- \(x=3\): \(15 + 2y=17 \Rightarrow 2y=2 \Rightarrow y=1\)

So there are 2 combinations: \((x,y)=(1,6)\) and \((3,1)\).

ANSWER 8: A


Problem 9:  
The contest starts at noon. It ends 1000 minutes later.

Convert 1000 minutes to hours:
\[
1000 \div 60 = 16 \text{ hours remainder } 40 \text{ minutes}
\]
Noon + 16 hours = 4:00 a.m. (next day)  
Add 40 minutes:
\[
4:00 \text{ a.m.} + 40 \text{ min} = 4:40 \text{ a.m.}
\]

ANSWER 9: D


Problem 10:  
Compute:
\[
(8 \times 4 + 2) - (8 + 4 \times 2)
\]
First parentheses:
\[
8\times 4 = 32,\quad 32+2=34
\]
Second parentheses:
\[
4\times 2 = 8,\quad 8+8=16
\]
Now subtract:
\[
34 - 16 = 18
\]

ANSWER 10: D


Problem 11:  
11 hours and 5 minutes total.
Convert hours to minutes:
\[
11 \cdot 60 = 660
\]
Add 5 minutes:
\[
660 + 5 = 665
\]

ANSWER 11: C


Problem 12:  
There are 6 windows. Georgie enters through one window and must leave through a different window.

Choose entering window: 6 choices.  
Choose leaving window: 5 choices (cannot be the same).

Total:
\[
6 \cdot 5 = 30
\]

ANSWER 12: D

Problem 1:
To be certain of having 5 socks of the same color, we consider the "worst-case scenario" (Pigeonhole Principle). If we have 3 colors (red, white, blue), we could pick 4 red, 4 white, and 4 blue socks without having 5 of any one color. This totals $4 + 4 + 4 = 12$ socks. The very next sock (the 13th) must be either red, white, or blue, which will complete a set of 5 for that color.
ANSWER 1: D

Problem 2:
We need to choose 3 starters from a group of 4 people (Lance, Sally, Joy, Fred). This is a combination problem denoted as $\binom{4}{3}$. Choosing 3 people to be on the team is the same as choosing 1 person to be left out. There are 4 ways to choose who is left out (Lance, Sally, Joy, or Fred).
ANSWER 2: B

Problem 3:
We check the products:
A: $(-4) \times (-9) = 36$
B: $(-3) \times (-12) = 36$
C: $\frac{1}{2} \times (-72) = -36$
D: $1 \times 36 = 36$
E: $\frac{3}{2} \times 24 = 3 \times 12 = 36$
Choice C results in $-36$, not $36$.
ANSWER 3: C

Problem 4:
Ceiling height = 2.4 m = 240 cm. Light bulb height = 240 - 10 = 230 cm from the floor.
Alice's reach = 150 cm (height) + 46 cm (reach) = 196 cm.
Stool height = (Bulb height) - (Alice's reach) = 230 - 196 = 34 cm.
ANSWER 4: B

Problem 5:
Let $T$ be the total number of fish in the lake. The ratio of trout to total fish is $\frac{30}{180} = \frac{1}{6}$.
Set up the proportion: $\frac{250}{T} = \frac{1}{6}$.
$T = 250 \times 6 = 1500$.
ANSWER 5: B

Problem 6:
Total cost = 5 folders × $2.50 = $12.50.
Savings = 20% of $12.50 = 0.20 \times 12.50 = $2.50.
Alternatively, 20% off one folder is $0.20 \times 2.50 = $0.50. For 5 folders, $0.50 \times 5 = $2.50.
ANSWER 6: C

Problem 7:
Total time in seconds = 300 dimples × 2 seconds/dimple = 600 seconds.
Total time in minutes = 600 seconds / 60 seconds/minute = 10 minutes.
ANSWER 7: D

Problem 8:
Let $x$ be the number of $5 bills and $y$ be the number of $2 bills. $5x + 2y = 17$.
If $x=1$, $5+2y=17 \Rightarrow 2y=12 \Rightarrow y=6$. (1, 6)
If $x=3$, $15+2y=17 \Rightarrow 2y=2 \Rightarrow y=1$. (3, 1)
If $x=0$ or $x=2$, $2y$ would be 17 or 7, which are not divisible by 2.
There are 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours and 40 minutes (since $1000 \div 60 = 16$ remainder 40).
Noon + 12 hours = midnight.
Midnight + 4 hours and 40 minutes = 4:40 a.m.
ANSWER 9: D

Problem 10:
$(8 \times 4 + 2) - (8 + 4 \times 2) = (32 + 2) - (8 + 8) = 34 - 16 = 18$.
ANSWER 10: D

Problem 11:
11 hours = $11 \times 60 = 660$ minutes.
Total = 660 + 5 = 665 minutes.
ANSWER 11: C

Problem 12:
There are 6 choices for the entry window. Since the exit window must be different, there are 5 choices for the exit window.
Total ways = $6 \times 5 = 30$.
ANSWER 12: D

Problem 1:
This is a pigeonhole principle problem. To guarantee at least 5 socks of one color, first consider the worst-case scenario of taking as many socks as possible without having 5 of any single color. That means taking exactly 4 socks of each of the 3 colors, for a total of \(4 \times 3 = 12\) socks. The next sock (the 13th) must be a 5th sock of one of the colors. Thus 13 socks are required. (Fewer than 13 leaves open the possibility of only 4 of each color.)

ANSWER 1: D

Problem 2:
The number of ways to choose 3 starters out of the 4 available players is the combination \(\binom{4}{3} = 4\). Each choice simply leaves out one of the four players.

ANSWER 2: B

Problem 3:
Compute the product for each pair:
- A: \((-4) \times (-9) = 36\)
- B: \((-3) \times (-12) = 36\)
- C: \(\frac12 \times (-72) = -36\) (not 36)
- D: \(1 \times 36 = 36\)
- E: \(\frac32 \times 24 = 36\)

Only pair C fails to give a product of 36.

ANSWER 3: C

Problem 4:
Convert all lengths to centimeters. The ceiling is 240 cm above the floor, so the bulb is 230 cm above the floor. Alice's reach is \(150 + 46 = 196\) cm. The stool must therefore supply the remaining \(230 - 196 = 34\) cm.

ANSWER 4: B

Problem 5:
The sample shows that trout comprise \(\frac{30}{180} = \frac16\) of all fish. With 250 trout in the lake, the total number of fish is \(250 \div \frac16 = 1500\).

ANSWER 5: B

Problem 6:
Five folders cost \(5 \times 2.50 = 12.50\). A 20 % discount saves \(0.20 \times 12.50 = 2.50\).

ANSWER 6: C

Problem 7:
Painting 300 dimples at 2 seconds each requires \(300 \times 2 = 600\) seconds, or \(600 \div 60 = 10\) minutes.

ANSWER 7: D

Problem 8:
Solve \(5x + 2y = 17\) for non-negative integers \(x, y\):
- \(x = 1\), \(y = 6\) works;
- \(x = 3\), \(y = 1\) works.
No other non-negative integer solutions exist, so there are exactly two combinations.

ANSWER 8: A

Problem 9:
Divide 1000 minutes by 60: \(1000 = 16\) hours + 40 minutes. Adding 16 hours to noon yields 4:00 a.m.; adding 40 minutes yields 4:40 a.m.

ANSWER 9: D

Problem 10:
Left side: \(8 \times 4 + 2 = 34\). Right side: \(8 + 4 \times 2 = 16\). Difference: \(34 - 16 = 18\).

ANSWER 10: D

Problem 11:
Convert 11 hours 5 minutes to minutes: \(11 \times 60 + 5 = 665\).

ANSWER 11: C

Problem 12:
There are 6 choices for the entry window and then 5 remaining choices for the exit window, giving \(6 \times 5 = 30\) ordered pairs.

ANSWER 12: D

Problem 1:
The problem is asking us to determine the minimum number of socks the Martian must remove from the drawer to guarantee that there are 5 socks of the same color. 
To ensure that there are 5 socks of the same color, we must consider the worst-case scenario where the Martian draws as many socks as possible without getting 5 of the same color. 
The Martian can draw at most 4 socks of each color without having 5 socks of the same color. 
So, the Martian can draw 4 red, 4 white, and 4 blue socks, for a total of 4 + 4 + 4 = 12 socks. 
At this point, the Martian still doesn't have 5 socks of the same color. 
However, drawing one more sock will guarantee that there are 5 socks of the same color. 
Therefore, the Martian must draw at least 12 + 1 = 13 socks.
Let's examine the choices: 
A. 6 is too low because the Martian could draw 3 red, 2 white, and 1 blue, or other combinations that don't guarantee 5 socks of the same color. 
B. 9 is still too low because the Martian could draw 4 red, 3 white, and 2 blue, or other similar combinations. 
C. 12 is the maximum number of socks the Martian can draw without having 5 socks of the same color. 
D. 13 is the correct answer because it is one more than the maximum number of socks that can be drawn without having 5 socks of the same color. 
E. 15 is more than necessary.
ANSWER 1: D

Problem 2:
The problem is asking us to find the number of ways to choose 3 starters from a team of 4 players: Lance, Sally, Joy, and Fred. 
This is a combination problem, because the order in which the starters are chosen does not matter. 
The number of ways to choose 3 starters from 4 players can be calculated using the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of players and k is the number of starters. 
Here, n = 4 and k = 3, so C(4, 3) = 4! / (3!(4-3)!) = 4.
Let's list the possible combinations to verify: 
- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred
There are indeed 4 combinations.
Let's examine the choices: 
A. 2 is too low. 
B. 4 matches our calculation. 
C. 6 is too high. 
D. 8 is too high. 
E. 10 is too high.
ANSWER 2: B

Problem 3:
The problem is asking us to identify the pair of numbers that does NOT have a product equal to 36. 
Let's check each pair: 
A. {−4, −9}: (−4) × (−9) = 36
B. {−3, −12}: (−3) × (−12) = 36
C. { 1 2 , −72}: (1/2) × (−72) = −36
D. {1, 36}: 1 × 36 = 36
E. { 3 2 , 24}: (3/2) × 24 = 36
The pair that does NOT have a product equal to 36 is C.
Let's examine the choices: 
A. {−4, −9} has a product of 36. 
B. {−3, −12} has a product of 36. 
C. { 1 2 , −72} has a product of −36, not 36. 
D. {1, 36} has a product of 36. 
E. { 3 2 , 24} has a product of 36.
ANSWER 3: C

Problem 4:
The problem is asking us to find the height of the stool that Alice needs to stand on to reach the light bulb. 
First, we need to convert the height of the ceiling from meters to centimeters: 2.4 meters = 240 centimeters. 
The light bulb is 10 centimeters below the ceiling, so its height above the floor is 240 - 10 = 230 centimeters. 
Alice is 1.5 meters = 150 centimeters tall and can reach 46 centimeters above her head, so her total reach is 150 + 46 = 196 centimeters. 
To find the height of the stool, we subtract Alice's total reach from the height of the light bulb: 230 - 196 = 34 centimeters.
Let's examine the choices: 
A. 32 is too low because 230 - (196 + 32) = 2, which is not enough to reach the light bulb. 
B. 34 is the correct height. 
C. 36 is too high. 
D. 38 is too high. 
E. 40 is too high.
ANSWER 4: B

Problem 5:
The problem is asking us to find the total number of fish in the lake based on the ratio of trout in the sample to the total number of fish in the sample. 
The sample contains 30 trout out of 180 fish, so the ratio of trout to total fish in the sample is 30/180. 
We are given that this ratio is the same as the ratio of trout to total fish in the lake. 
Let's denote the total number of fish in the lake as x. 
The ratio of trout to total fish in the lake is 250/x. 
Setting the two ratios equal, we get 30/180 = 250/x. 
Cross-multiplying, we have 30x = 250 × 180. 
Simplifying, 30x = 45000. 
Dividing both sides by 30, x = 1500.
Let's examine the choices: 
A. 1250 is too low. 
B. 1500 matches our calculation. 
C. 1750 is too high. 
D. 1800 is too high. 
E. 2000 is too high.
ANSWER 5: B

Problem 6:
The problem is asking us to find out how much Karl could have saved by waiting a day to buy the folders at a 20%-off sale. 
The original price of 5 folders is 5 × $2.50 = $12.50. 
At a 20%-off sale, the discount is 20% of $12.50, which is 0.20 × $12.50 = $2.50. 
So, Karl could have saved $2.50.
Let's examine the choices: 
A. $1.00 is too low. 
B. $2.00 is too low. 
C. $2.50 matches our calculation. 
D. $2.75 is too high. 
E. $5.00 is too high.
ANSWER 6: C

Problem 7:
The problem is asking us to find the time it will take Casey to paint 300 dimples on a golf ball. 
Casey takes 2 seconds to paint one dimple, so to paint 300 dimples, he will take 300 × 2 = 600 seconds. 
To convert seconds to minutes, we divide by 60: 600 / 60 = 10 minutes.
Let's examine the choices: 
A. 4 minutes is too low. 
B. 6 minutes is too low. 
C. 8 minutes is too low. 
D. 10 minutes matches our calculation. 
E. 12 minutes is too high.
ANSWER 7: D

Problem 8:
The problem is asking us to find the number of different combinations of $5 bills and $2 bills that can be used to make a total of $17. 
Let's list the possible combinations: 
1. 3 $5 bills = $15, and $2 bill = $2, total = $17 (1 combination: 3 $5 bills and 1 $2 bill)
2. 1 $5 bill = $5, and 6 $2 bills = $12, total = $17 (1 combination: 1 $5 bill and 6 $2 bills)
We should also check if there are any other combinations: 
- Using 0 $5 bills, we need 8.5 $2 bills, which is not possible since we can't have a fraction of a bill.
- Using 2 $5 bills = $10, we need $7 more, which can be made with 3.5 $2 bills, again not possible.
- Using 4 $5 bills = $20, which is more than $17.

Problem 1:
We want the minimum number of socks to guarantee 5 of the same color. With 3 colors (red, white, blue), the worst case is picking 4 of each color: 4+4+4 = 12 socks. The 13th sock must create a group of 5 of one color. So 13 socks are needed.
ANSWER 1: D

Problem 2:
We choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways is the combination C(4,3) = 4.
ANSWER 2: B

Problem 3:
Compute each product:
A: (−4)×(−9) = 36
B: (−3)×(−12) = 36
C: (1/2)×(−72) = −36
D: 1×36 = 36
E: (3/2)×24 = 36
Only C does not equal 36.
ANSWER 3: C

Problem 4:
Ceiling = 2.4 m = 240 cm. Light bulb is 10 cm below ceiling, so its height = 230 cm.
Alice's height = 1.5 m = 150 cm. Reach above head = 46 cm, so reach without stool = 196 cm.
Stool height needed = 230 − 196 = 34 cm.
ANSWER 4: B

Problem 5:
Sample ratio: 30 trout / 180 fish = 1/6. Lake has 250 trout, which is 1/6 of total fish. Total fish = 250 × 6 = 1500.
ANSWER 5: B

Problem 6:
Cost without sale: 5 × $2.50 = $12.50.
With 20% off, each folder costs $2.50 × 0.8 = $2.00. Total = 5 × $2.00 = $10.00.
Savings = $12.50 − $10.00 = $2.50.
ANSWER 6: C

Problem 7:
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.
ANSWER 7: D

Problem 8:
Find non-negative integers x ($5 bills) and y ($2 bills) such that 5x + 2y = 17.
x=1 → 5 + 2y = 17 → y=6.
x=3 → 15 + 2y = 17 → y=1.
No other x works. So 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours 40 minutes (16×60 = 960, remainder 40).
Noon + 16 hours = 4:00 a.m. next day. Add 40 minutes → 4:40 a.m.
ANSWER 9: D

Problem 10:
(8 × 4 + 2) = 32 + 2 = 34.
(8 + 4 × 2) = 8 + 8 = 16.
34 − 16 = 18.
ANSWER 10: D

Problem 11:
11 hours × 60 = 660 minutes. Plus 5 minutes = 665 minutes.
ANSWER 11: C

Problem 12:
6 choices for entry window, then 5 choices for a different exit window. Total = 6 × 5 = 30.
ANSWER 12: D

Problem 1:
We need to find the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). Using the Pigeonhole Principle, we consider the worst-case scenario: the Martian pulls out as many socks as possible *without* getting 5 of any one color. That means getting exactly 4 of each color:
- 4 red + 4 white + 4 blue = 12 socks
After 12 socks, the Martian could have 4 of each color and still not have 5 matching. The very next sock (the 13th) must be red, white, or blue, pushing the count of that color to 5. So 13 socks guarantees 5 of the same color.
(The "five-legged" detail is flavor text explaining why the Martian needs 5 socks.)

ANSWER 1: D

Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). This is a combination problem since order doesn't matter:
C(4, 3) = 4! / (3! × 1!) = 4
We can also list them: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — exactly 4 ways.

ANSWER 2: B

Problem 3:
We check each product:
- A: (−4) × (−9) = 36 ✓
- B: (−3) × (−12) = 36 ✓
- C: (1/2) × (−72) = −36 ✗ (This equals −36, NOT 36)
- D: 1 × 36 = 36 ✓
- E: (3/2) × 24 = 36 ✓
Choice C gives −36, which does not equal 36.

ANSWER 3: C

Problem 4:
First, convert everything to centimeters:
- Ceiling height: 2.4 m = 240 cm
- Light bulb position: 240 − 10 = 230 cm above the floor
- Alice's height: 1.5 m = 150 cm
- Alice's reach above her head: 46 cm
- Alice's maximum reach (without stool): 150 + 46 = 196 cm above the floor

Let h = height of the stool. Standing on the stool, her reach is h + 196 cm. She can "just reach" the bulb:
h + 196 = 230
h = 230 − 196 = 34 cm

ANSWER 4: B

Problem 5:
The ratio of trout to total fish in the sample should equal the ratio in the lake:
- Sample ratio: 30 trout / 180 fish = 1/6
- Lake ratio: 250 trout / Total fish = 1/6

Setting up the proportion: 250 / Total = 1/6
Total = 250 × 6 = 1500 fish

ANSWER 5: B

Problem 6:
Karl bought 5 folders at $2.50 each:
Total cost = 5 × $2.50 = $12.50
A 20%-off sale means he would save 20% of the total:
Savings = 0.20 × $12.50 = $2.50

ANSWER 6: C

Problem 7:
Total time in seconds: 300 dimples × 2 seconds/dimple = 600 seconds
Convert to minutes: 600 seconds ÷ 60 seconds/minute = 10 minutes

ANSWER 7: D

Problem 8:
We need non-negative integer solutions to 5x + 2y = 17, where x = number of $5 bills and y = number of $2 bills.
- x = 0: 2y = 17 → y = 8.5 (not an integer) ✗
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 ✗
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x = 4: 20 > 17 (too large) ✗

Only 2 valid combinations: (1 five + 6 twos) and (3 fives + 1 two).

ANSWER 8: A

Problem 9:
Convert 1000 minutes to hours and minutes:
1000 ÷ 60 = 16 remainder 40, so 1000 minutes = 16 hours 40 minutes.
Starting at noon (12:00 PM):
- 12 hours later = midnight (12:00 AM)
- 4 more hours = 4:00 AM
- 40 more minutes = 4:40 AM

ANSWER 9: D

Problem 10:
Evaluate each part using order of operations (multiplication before addition):
- First part: (8 × 4 + 2) = (32 + 2) = 34
- Second part: (8 + 4 × 2) = (8 + 8) = 16
- Difference: 34 − 16 = 18

ANSWER 10: D

Problem 11:
Convert 11 hours and 5 minutes to minutes:
11 hours × 60 minutes/hour = 660 minutes
660 + 5 = 665 minutes

ANSWER 11: C

Problem 12:
Georgie enters through one window and leaves through a different one.
- Choices for entry window: 6
- Choices for exit window (must be different): 5
Total ways = 6 × 5 = 30

ANSWER 12: D

Problem 1:
The Martian needs to be **certain** of having 5 socks of the same color. To guarantee this, we must consider the worst-case scenario: the Martian picks as many socks as possible **without** yet having 5 of any single color.  
With 3 colors (red, white, blue), the worst case is picking 4 socks of each color:  
\(4 + 4 + 4 = 12\) socks.  
At this point the Martian has 4 of every color but still does not have 5 of any one color. The very next sock—the 13th—must be red, white, or blue, and will necessarily create a set of 5 of that color.  
Thus 13 socks are required.  
(Choice A is too small; with 6 socks one could have 2,2,2. Choice B allows 3,3,3. Choice C allows 4,4,4. Choice E is more than the minimum guaranteed number.)

ANSWER 1: D

---

Problem 2:
We need to choose a group of 3 starters from the 4 students: Lance, Sally, Joy, and Fred. Since the three starters form a team and order does not matter, this is a combination problem.  
\[
\binom{4}{3} = \frac{4!}{3!\,1!} = 4
\]  
Equivalently, we can think of choosing which **one** of the four students will **not** start; there are 4 such choices.  
(Choice A is too small. Choice C is \(\binom{4}{2}\). Choices D and E do not arise from counting 3-person teams from 4 people.)

ANSWER 2: B

---

Problem 3:
We check the product of each pair to see which is not 36.

- **A.** \((-4)(-9) = 36\)  
- **B.** \((-3)(-12) = 36\)  
- **C.** \(\left(\frac12\right)(-72) = -36\)  
- **D.** \((1)(36) = 36\)  
- **E.** \(\left(\frac32\right)(24) = 36\)

Only pair **C** gives \(-36\), not \(36\).

ANSWER 3: C

---

Problem 4:
First convert all measurements to centimeters.  
- Ceiling height: \(2.4\text{ m} = 240\text{ cm}\).  
- Light bulb is \(10\text{ cm}\) below the ceiling, so its height above the floor is \(240 - 10 = 230\text{ cm}\).  

Alice’s reach:  
- Height: \(1.5\text{ m} = 150\text{ cm}\).  
- She can reach \(46\text{ cm}\) above her head, so her standing reach is \(150 + 46 = 196\text{ cm}\).  

The stool must supply the remaining height:  
\[
230 - 196 = 34\text{ cm}.
\]

ANSWER 4: B

---

Problem 5:
Let \(T\) be the total number of fish in the lake. The problem states that the ratio of trout to total fish is the same in the sample and in the lake.  
\[
\frac{250}{T} = \frac{30}{180}
\]  
Simplify the right side: \(\frac{30}{180} = \frac16\).  
\[
\frac{250}{T} = \frac16 \quad\Longrightarrow\quad T = 250 \times 6 = 1500.
\]

ANSWER 5: B

---

Problem 6:
Karl bought 5 folders at \(\$2.50\) each.  
Total cost: \(5 \times \$2.50 = \$12.50\).  
A 20% discount would have saved him 20% of this total:  
\[
0.20 \times \$12.50 = \$2.50.
\]

ANSWER 6: C

---

Problem 7:
Casey paints 300 dimples at 2 seconds per dimple.  
Total time in seconds: \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(\frac{600}{60} = 10\) minutes.

ANSWER 7: D

---

Problem 8:
Let \(f\) be the number of \(\$5\) bills and \(t\) be the number of \(\$2\) bills. We need non-negative integer solutions to  
\[
5f + 2t = 17.
\]  
Test values of \(f\):  
- \(f=0\): \(2t=17\) → no integer solution.  
- \(f=1\): \(5 + 2t = 17 \Rightarrow t=6\).  **(Valid)**
- \(f=2\): \(10 + 2t = 17 \Rightarrow t=3.5\).  (Invalid)
- \(f=3\): \(15 + 2t = 17 \Rightarrow t=1\).  **(Valid)**
- \(f=4\): \(20 > 17\) → stop.

There are exactly **2** combinations: \((1\text{ five}, 6\text{ twos})\) and \((3\text{ fives}, 1\text{ two})\).

ANSWER 8: A

---

Problem 9:
The contest starts at noon (12:00 p.m.). We add 1000 minutes.  
Divide 1000 by 60:  
\[
1000 = 16 \times 60 + 40,
\]  
so 1000 minutes = 16 hours and 40 minutes.  

Noon + 12 hours = midnight.  
Midnight + 4 hours 40 minutes = 4:40 a.m.

ANSWER 9: D

---

Problem 10:
Evaluate using the correct order of operations (multiplication before addition).

First parentheses:  
\[
8 \times 4 + 2 = 32 + 2 = 34.
\]

Second parentheses:  
\[
8 + 4 \times 2 = 8 + 8 = 16.
\]

Subtract:  
\[
34 - 16 = 18.
\]

ANSWER 10: D

---

Problem 11:
Convert 11 hours to minutes and add the remaining 5 minutes.  
\[
11 \times 60 = 660\text{ minutes}.
\]  
\[
660 + 5 = 665\text{ minutes}.
\]

ANSWER 11: C

---

Problem 12:
Georgie enters through one of 6 windows and must leave through a **different** window.  
- 6 choices for the entry window.  
- For each entry choice, 5 remaining choices for the exit window.  

Total ways:  
\[
6 \times 5 = 30.
\]  
(Choice E, 36, would count entering and leaving through the same window, which is not allowed.)

ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). The worst-case scenario is to draw as many socks as possible without having 5 of any one color. That would be 4 red, 4 white, and 4 blue, for a total of 12 socks. The next sock drawn (the 13th) must be one of the three colors, giving that color 5 socks. Therefore, 13 socks are required. The five-legged Martian is irrelevant to the count.  
Answer: D

Problem 2:
We have 4 people (Lance, Sally, Joy, Fred) and need to choose 3 starters. Since the starters are not assigned to specific positions, order does not matter. The number of ways is the combination \( \binom{4}{3} = 4 \).  
Answer: B

Problem 3:
Compute each product:  
A. \((-4) \times (-9) = 36\)  
B. \((-3) \times (-12) = 36\)  
C. \(\frac{1}{2} \times (-72) = -36\)  
D. \(1 \times 36 = 36\)  
E. \(\frac{3}{2} \times 24 = 36\)  
Only pair C gives a product not equal to 36.  
Answer: C

Problem 4:
Convert all measurements to centimeters:  
Ceiling height = 2.4 m = 240 cm.  
Light bulb is 10 cm below ceiling → height = 240 – 10 = 230 cm.  
Alice’s height = 1.5 m = 150 cm.  
Her reach above her head = 46 cm, so her maximum reach without a stool = 150 + 46 = 196 cm.  
Let \(h\) be the stool height. Standing on the stool raises her reach to \(196 + h\). This must equal 230 cm:  
\(196 + h = 230\) → \(h = 34\) cm.  
Answer: B

Problem 5:
In the sample, 30 out of 180 fish are trout, so the ratio trout : total = \(30/180 = 1/6\).  
Assuming the same ratio in the lake, if the lake has 250 trout, then the total number of fish is \(250 \times 6 = 1500\).  
Answer: B

Problem 6:
Original cost for 5 folders: \(5 \times 2.50 = \$12.50\).  
A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
Thus, by waiting a day, Karl could have saved $2.50.  
Answer: C

Problem 7:
Time to paint one dimple = 2 seconds.  
Total time for 300 dimples = \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(600 \div 60 = 10\) minutes.  
Answer: D

Problem 8:
We need nonnegative integers \(a\) (number of $5 bills) and \(b\) (number of $2 bills) such that \(5a + 2b = 17\).  
Solve by testing values of \(b\):  
\(b = 1\) → \(5a = 15\) → \(a = 3\) → (3,1) works.  
\(b = 6\) → \(5a = 5\) → \(a = 1\) → (1,6) works.  
No other values give integer \(a\). Thus there are 2 combinations.  
Answer: A

Problem 9:
1000 minutes = \(1000 \div 60 = 16\) hours and \(40\) minutes (since \(60 \times 16 = 960\), remainder 40).  
Starting at noon (12:00 PM), add 12 hours → midnight (12:00 AM). Add the remaining 4 hours → 4:00 AM. Then add 40 minutes → 4:40 AM.  
Answer: D

Problem 10:
Evaluate using order of operations:  
First parentheses: \(8 \times 4 + 2 = 32 + 2 = 34\).  
Second parentheses: \(8 + 4 \times 2 = 8 + 8 = 16\).  
Subtract: \(34 - 16 = 18\).  
Answer: D

Problem 11:
11 hours = \(11 \times 60 = 660\) minutes.  
Add 5 minutes → \(660 + 5 = 665\) minutes.  
Answer: C

Problem 12:
Georgie must choose an entry window (6 choices) and a different exit window (5 choices). Since entering and leaving are distinct actions, the total number of ways is \(6 \times 5 = 30\).  
Answer: D

ANSWER 1: D
ANSWER 2: B
ANSWER 3: C
ANSWER 4: B
ANSWER 5: B
ANSWER 6: C
ANSWER 7: D
ANSWER 8: A
ANSWER 9: D
ANSWER 10: D
ANSWER 11: C
ANSWER 12: D

Problem 1:  
The Martian wants to be certain of having 5 socks of the same color. There are three colors: red, white, and blue. In the worst-case scenario, the Martian could pick 4 socks of each color before getting a fifth of any one color. That would be \(4 \times 3 = 12\) socks, and still no color has 5 socks. The next sock drawn (the 13th) must complete a set of 5 of some color, because there are only three colors. Therefore, the Martian must remove **13** socks to be certain of having 5 socks of the same color.  
ANSWER 1: D

Problem 2:  
Ms. Hamilton's class has 4 available players: Lance, Sally, Joy, and Fred. The number of ways to choose 3 starters from 4 players is given by the combination \(\binom{4}{3} = 4\). Thus, there are **4** possible sets of three starters.  
ANSWER 2: B

Problem 3:  
We check each pair's product:  
- A: \((-4) \times (-9) = 36\)  
- B: \((-3) \times (-12) = 36\)  
- C: \(\frac{1}{2} \times (-72) = -36\) (not 36)  
- D: \(1 \times 36 = 36\)  
- E: \(\frac{3}{2} \times 24 = 36\)  
Only pair **C** does not have a product of 36.  
ANSWER 3: C

Problem 4:  
The ceiling is 2.4 m above the floor, and the bulb is 10 cm (0.1 m) below the ceiling, so the bulb's height is \(2.4 - 0.1 = 2.3\) m. Alice's height is 1.5 m, and she can reach 46 cm (0.46 m) above her head, so her reach height from the floor is \(1.5 + 0.46 = 1.96\) m. When standing on a stool, her hands are at the stool height plus her reach height. Setting this equal to the bulb height:  
\[
\text{stool height} + 1.96 = 2.3 \Rightarrow \text{stool height} = 0.34 \text{ m} = 34 \text{ cm}.
\]  
ANSWER 4: B

Problem 5:  
In the sample, the ratio of trout to total fish is \(\frac{30}{180} = \frac{1}{6}\). Assuming the same ratio in the lake, let \(T\) be the total number of fish in the lake. Then \(\frac{250}{T} = \frac{1}{6}\), so \(T = 250 \times 6 = 1500\).  
ANSWER 5: B

Problem 6:  
Karl bought 5 folders at $2.50 each, so the original cost is \(5 \times 2.50 = \$12.50\). With a 20% discount, the cost becomes \(12.50 \times 0.80 = \$10.00\). The savings is \(12.50 - 10.00 = \$2.50\).  
ANSWER 6: C

Problem 7:  
Painting 300 dimples at 2 seconds each takes \(300 \times 2 = 600\) seconds. Converting to minutes: \(600 \div 60 = 10\) minutes.  
ANSWER 7: D

Problem 8:  
We need nonnegative integers \(x\) (number of $5 bills) and \(y\) (number of $2 bills) such that \(5x + 2y = 17\). Solving:  
- \(y = 1 \Rightarrow 5x = 15 \Rightarrow x = 3\)  
- \(y = 6 \Rightarrow 5x = 5 \Rightarrow x = 1\)  
- \(y = 11 \Rightarrow 5x = -5\) (invalid)  
Thus, there are **2** valid combinations: (3 fives, 1 two) and (1 five, 6 twos).  
ANSWER 8: A

Problem 9:  
1000 minutes equals \(16\) hours and \(40\) minutes (since \(16 \times 60 = 960\), remainder \(40\)). Starting at noon, adding \(16\) hours gives 4:00 AM the next day, and adding \(40\) minutes gives 4:40 AM.  
ANSWER 9: D

Problem 10:  
Compute inside parentheses:  
\(8 \times 4 + 2 = 32 + 2 = 34\)  
\(8 + 4 \times 2 = 8 + 8 = 16\)  
Then \(34 - 16 = 18\).  
ANSWER 10: D

Problem 11:  
11 hours is \(11 \times 60 = 660\) minutes. Adding 5 minutes gives \(660 + 5 = 665\) minutes.  
ANSWER 11: C

Problem 12:  
Georgie can enter through any of the 6 windows. For exiting, he must choose a different window, so there are 5 choices. The total number of ways is \(6 \times 5 = 30\).  
ANSWER 12: D

Problem 1:  
The Martian needs to ensure there are 5 socks of the same color. The worst-case scenario is pulling the maximum number of socks without 5 of any color. Since there are 3 colors, this would be 4 socks of each color (4 × 3 = 12 socks). The next sock (the 13th) must match one color, giving 5 socks of that color.  
**ANSWER 1: D**  

Problem 2:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is C(4,3) = 4.  
**ANSWER 2: B** (Note: C(4,3)=4, but rechecking: the team is chosen from 4 specific people, so combinations are 4. Answer B=4)  

Problem 3:  
Check each pair:  
A. (−4)(−9) = 36  
B. (−3)(−12) = 36  
C. (1/2)(−72) = -36 (≠36)  
D. (1)(36) = 36  
E. (3/2)(24) = 36  
The pair that does NOT equal 36 is C.  
**ANSWER 3: E** (Re-evaluated: The question asks which does NOT equal 36. Option E: (3/2)*24=36, so it is valid. Option C: (1/2)*(-72)=-36, so C is the correct choice for NOT equal. Correction: **ANSWER 3: C** [Original had a typo, but computation confirms C is the answer])  

*Correction: For Problem 3, Option C: (1/2, -72) likely means (1/2) × (-72) = -36 ≠ 36. Option E: (3/2) × 24 = 36. Thus, C is the answer.*  
**ANSWER 3: C**  

Problem 4:  
Ceiling height: 2.4 m = 240 cm. Light bulb location: 240 - 10 = 230 cm from floor. Alice’s reach: 150 cm + 46 cm = 196 cm. Stool height = 230 - 196 = 34 cm.  
**ANSWER 4: B**  

Problem 5:  
Ratio of trout in sample: 30/180 = 1/6. Total trout: 250. Total fish = 250 × 6 = 1500.  
**ANSWER 5: B**  

Problem 6:  
Original cost: 5 × $2.50 = $12.50. 20% off: $12.50 × 0.20 = $2.50 saved.  
**ANSWER 6: C** (Note: The question asks how much Karl could have saved, which is $2.50, not $2.00. Original had a mistake; correction: 20% of $12.50 is $2.50, so answer C)  

Problem 7:  
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.  
**ANSWER 7: D**  

Problem 8:  
Let $5 bills = x, $2 bills = y. 5x + 2y = 17. Possible solutions:  
- x=3, y=1 (15+2=17)  
- x=1, y=6 (5+12=17)  
Only 2 combinations.  
**ANSWER 8: A**  

Problem 9:  
1000 minutes = 16 hours 40 minutes. Starting at noon: noon + 16h40m = 4:40 a.m. next day.  
**ANSWER 9: D**  

Problem 10:  
(8×4 + 2) = 34, (8 + 4×2) = 16. 34 - 16 = 18.  
**ANSWER 10: D**  

Problem 11:  
11 hours = 660 minutes. 660 + 5 = 665 minutes.  
**ANSWER 11: C**  

Problem 12:  
Georgie can enter through 6 windows and exit through 5 others. Total ways: 6 × 5 = 30.  
**ANSWER 12: D**  

---  
**Final Answers**  
ANSWER 1: D  
ANSWER 2: B  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: B  
ANSWER 6: C  
ANSWER 7: D  
ANSWER 8: A  
ANSWER 9: D  
ANSWER 10: D  
ANSWER 11: C  
ANSWER 12: D

### Problem 1:
This is a classic pigeonhole principle problem. We need to find the minimum number of socks the Martian must pull to guarantee 5 socks of the same color, with 3 total colors. The worst-case scenario is that the Martian pulls as many socks as possible without getting 5 of any color: that is 4 socks of each color, for a total of 3*4=12 socks. The next sock pulled (the 13th) has to be one of the three colors, pushing that color to 5 socks. Tempting wrong answers: C (12 is only the number of socks that avoids 5 of the same color, not enough to guarantee the requirement), E (15 is the total if you took 5 of each color, not the minimum to guarantee).
ANSWER 1: D

---

### Problem 2:
We need to count the number of ways to choose 3 starters from 4 people, where order on the team does not matter. This is a combination problem: C(n,k) = n!/(k!(n-k)!). For n=4, k=3: C(4,3) = 4. We can also list the teams by who is left out: leaving out Lance, Sally, Joy, or Fred gives exactly 4 unique teams. Tempting wrong answer: C (6 is the count if order of selection mattered, which it does not for a team of starters).
ANSWER 2: B

---

### Problem 3:
We calculate the product of each pair to find which does not equal 36:
- A: (-4)*(-9)=36 ✔️
- B: (-3)*(-12)=36 ✔️
- C: (1/2)*(-72) = -36 ❌
- D: 1*36=36 ✔️
- E: (3/2)*24=36 ✔️
Only pair C has a product of -36, not 36.
ANSWER 3: C

---

### Problem 4:
First convert all units to centimeters to avoid conversion errors:
- Ceiling height: 2.4m = 240cm. Light bulb is 10cm below ceiling, so bulb height from floor = 240-10=230cm.
- Alice's height: 1.5m=150cm, her reach above her head is 46cm, so total reach without stool = 150+46=196cm.
- Stool height = 230-196=34cm.
Tempting wrong answers come from unit conversion mistakes, e.g., mixing up meters and centimeters.
ANSWER 4: B

---

### Problem 5:
This is a proportion problem. The ratio of trout to total fish is the same in the sample and the lake. In the sample, 30 trout out of 180 total fish = 1/6. Let T be total fish in the lake: 250/T = 1/6 → T=250*6=1500.
ANSWER 5: B

---

### Problem 6:
Karl bought 5 folders at $2.50 each, total original cost = 5*2.50=$12.50. He saves 20% of the total cost by waiting for the sale: 0.2*12.50=$2.50. This can also be calculated per folder: 20% of $2.50 is $0.50 saved per folder, 5*0.50=$2.50 total saved.
ANSWER 6: C

---

### Problem 7:
300 dimples * 2 seconds per dimple = 600 total seconds. Convert to minutes: 60 seconds per minute, so 600/60=10 minutes.
ANSWER 7: D

---

### Problem 8:
We find all non-negative integer pairs (x,y) where x=number of $5 bills, y=number of $2 bills, and 5x+2y=17. Only odd values of x work (since $5 is odd, $2 is even, we need an odd sum to get 17):
- x=1: 5+2y=17 → y=6 (valid)
- x=3: 15+2y=17 → y=1 (valid)
Even x values (0,2) leave an odd remainder that cannot be formed with $2 bills, so only 2 valid combinations.
ANSWER 8: A

---

### Problem 9:
Convert 1000 minutes to hours and minutes: 60 minutes per hour, 1000/60=16 hours and 40 minutes. Adding this to noon (12pm): 12pm +16h40m = 4:40am the next day.
ANSWER 9: D

---

### Problem 10:
Use order of operations (PEMDAS) to calculate:
First parenthesis: 8*4 +2 = 32+2=34
Second parenthesis: 8 +4*2=8+8=16
Subtract: 34-16=18. The tempting wrong answer C (10) comes from incorrectly adding before multiplying in the second parenthesis.
ANSWER 10: D

---

### Problem 11:
Convert 11 hours 5 minutes to total minutes: 11*60=660 minutes, plus 5 extra minutes = 665 minutes.
ANSWER 11: C

---

### Problem 12:
Georgie has 6 choices for an entry window, and 5 remaining choices for an exit window (must use a different window). Total ways = 6*5=30. Tempting wrong answer E (36) counts entering and leaving through the same window, which is not allowed; B (15) incorrectly treats entry/exit as unordered, which they are not.
ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color (red, white, blue). This is a pigeonhole principle problem: the worst-case scenario is pulling the maximum number of socks without getting 5 of any color. For each of the 3 colors, we can pull 4 socks (since 4 < 5), so 4×3=12 socks. The next sock pulled will necessarily be the 5th of one color. So total socks needed: 12 + 1 = 13.
ANSWER 1: D
Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). Since the problem does not mention distinct positions for starters, order does not matter, so we use combinations. The number of ways to choose 3 people from 4 is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n=4 \) and \( k=3 \):
\( \binom{4}{3} = \frac{4!}{3!1!} = 4 \). This is equivalent to counting how many people we leave out (4 choices, one for each person not selected).
ANSWER 2: B
Problem 3:
We check the product of each pair:
- A: \( (-4) \times (-9) = 36 \)
- B: \( (-3) \times (-12) = 36 \)
- C: \( \frac{1}{2} \times (-72) = -36 \neq 36 \)
- D: \( 1 \times 36 = 36 \)
- E: \( \frac{3}{2} \times 24 = 36 \)
Only pair C does not have a product of 36.
ANSWER 3: C
Problem 4:
Convert all units to centimeters:
- Ceiling height: \( 2.4 \, \text{m} = 240 \, \text{cm} \)
- Light bulb height from floor: \( 240 - 10 = 230 \, \text{cm} \)
- Alice's reach from floor (without stool): \( 1.5 \, \text{m} + 0.46 \, \text{m} = 150 + 46 = 196 \, \text{cm} \)
Let \( h \) be the stool height. The total reach with the stool equals the light bulb height:
\( h + 196 = 230 \)
\( h = 230 - 196 = 34 \, \text{cm} \)
ANSWER 4: B
Problem 5:
The ratio of trout to total fish in the sample equals the ratio in the lake. The sample has \( \frac{30}{180} = \frac{1}{6} \) trout. Let \( T \) be the total number of fish in the lake. We know there are 250 trout, so:
\( \frac{250}{T} = \frac{1}{6} \)
\( T = 250 \times 6 = 1500 \)
ANSWER 5: B
Problem 6:
Karl bought 5 folders at $2.50 each. The 20% discount applies to each folder, so the savings per folder is \( 2.50 \times 0.20 = 0.50 \). Total savings for 5 folders: \( 5 \times 0.50 = 2.50 \).
ANSWER 6: C
Problem 7:
Total time to paint 300 dimples: \( 300 \times 2 = 600 \) seconds. Convert to minutes: \( 600 \div 60 = 10 \) minutes.
ANSWER 7: D
Problem 8:
Let \( x \) = number of $5 bills, \( y \) = number of $2 bills. We need \( 5x + 2y = 17 \), where \( x,y \geq 0 \) integers. Test non-negative integer values of \( x \):
- \( x=0 \): \( 2y=17 \) → \( y=8.5 \) (not integer, invalid)
- \( x=1 \): \( 5 + 2y=17 \) → \( 2y=12 \) → \( y=6 \) (valid: 1 $5 + 6 $2 = $17)
- \( x=2 \): \( 10 + 2y=17 \) → \( 2y=7 \) → \( y=3.5 \) (invalid)
- \( x=3 \): \( 15 + 2y=17 \) → \( 2y=2 \) → \( y=1 \) (valid: 3 $5 + 1 $2 = $17)
- \( x=4 \): \( 20 > 17 \) (stop, no more valid \( x \))
Only 2 valid combinations.
ANSWER 8: A
Problem 9:
Convert 1000 minutes to hours and minutes: \( 1000 \div 60 = 16 \) hours with a remainder of 40 minutes (since \( 16 \times 60 = 960 \), \( 1000 - 960 = 40 \)). The contest starts at noon, so adding 16 hours 40 minutes: noon + 12 hours = midnight, plus 4 hours 40 minutes = 4:40 a.m.
ANSWER 9: D
Problem 10:
Use order of operations (multiplication before addition):
First parentheses: \( 8 \times 4 + 2 = 32 + 2 = 34 \)
Second parentheses: \( 8 + 4 \times 2 = 8 + 8 = 16 \)
Subtract: \( 34 - 16 = 18 \)
ANSWER 10: D
Problem 11:
Convert 11 hours 5 minutes to total minutes: \( 11 \times 60 + 5 = 660 + 5 = 665 \)
ANSWER 11: C
Problem 12:
Georgie chooses an entrance window first (6 options), then a different exit window (5 remaining options, since exit cannot equal entrance). Since entrance and exit are distinct actions, order matters, so total ways: \( 6 \times 5 = 30 \).
ANSWER 12: D

I'll work through each problem systematically.

**Problem 1:**

**What is being asked:** How many socks must be removed to guarantee 5 socks of the same color?

**Setup:** There are 3 colors (red, white, blue). In the worst case scenario, the Martian pulls out socks as evenly distributed among colors as possible before getting 5 of one color.

**Working:** 
- Worst case: 4 red, 4 white, 4 blue = 12 socks (still no color with 5)
- The 13th sock must give us a 5th sock of some color

**Answer 1: D**

---

**Problem 2:**

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

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

**Working:**
- C(4,3) = 4 ways
- The 4 ways correspond to leaving out each person: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred}

**Answer 2: B**

---

**Problem 3:**

**What is being asked:** Which pair does NOT have a product equal to 36?

**Working:** Check each pair:
- A. {−4, −9}: (−4)(−9) = 36 ✓
- B. {−3, −12}: (−3)(−12) = 36 ✓
- C. {1/2, −72}: (1/2)(−72) = −36 ✗
- D. {1, 36}: (1)(36) = 36 ✓
- E. {3/2, 24}: (3/2)(24) = 36 ✓

**Answer 3: C**

---

**Problem 4:**

**What is being asked:** How high is the stool?

**Setup:** Convert all measurements to centimeters:
- Ceiling height: 240 cm
- Light bulb location: 240 − 10 = 230 cm from floor
- Alice's height: 150 cm
- Alice's reach above her head: 46 cm
- Alice's reach without stool: 150 + 46 = 196 cm

**Working:**
- With stool, Alice can just reach the bulb at 230 cm
- Stool height + 196 cm = 230 cm
- Stool height = 230 − 196 = 34 cm

**Answer 4: B**

---

**Problem 5:**

**What is being asked:** How many fish total are in the lake?

**Setup:** Use proportions. In the sample: 30 trout out of 180 fish. In the lake: 250 trout out of total fish.

**Working:**
- 30/180 = 250/x
- 1/6 = 250/x
- x = 250 × 6 = 1500 fish

**Answer 5: B**

---

**Problem 6:**

**What is being asked:** How much could Karl have saved with the 20% off sale?

**Setup:** 
- Original cost: 5 folders × $2.50 = $12.50
- Sale discount: 20% off

**Working:**
- 20% of $12.50 = 0.20 × $12.50 = $2.50

**Answer 6: C**

---

**Problem 7:**

**What is being asked:** How many minutes to paint 300 dimples at 2 seconds each?

**Working:**
- Total time: 300 × 2 = 600 seconds
- Convert to minutes: 600 ÷ 60 = 10 minutes

**Answer 7: D**

---

**Problem 8:**

**What is being asked:** How many combinations of $5 bills and $2 bills make $17?

**Setup:** Let x = number of $5 bills, y = number of $2 bills
- 5x + 2y = 17

**Working:**
- x = 0: 2y = 17 → y = 8.5 (not integer)
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 (not integer)
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x ≥ 4: 5x ≥ 20 > 17 (impossible)

Valid combinations: (1,$5 and 6,$2) and (3,$5 and 1,$2) = 2 combinations

**Answer 8: A**

---

**Problem 9:**

**What is being asked:** When did the contest end, starting at noon and lasting 1000 minutes?

**Working:**
- 1000 minutes ÷ 60 = 16 hours and 40 minutes
- Noon + 16 hours 40 minutes = 4:40 p.m. the same day
- But we need to continue: 4:40 p.m. + additional time consideration
- Actually: 1000 ÷ 60 = 16 remainder 40, so 16 hours 40 minutes
- Noon + 16 hours 40 minutes = 4:40 a.m. the next day

**Answer 9: D**

---

**Problem 10:**

**What is being asked:** Calculate (8 × 4 + 2) − (8 + 4 × 2)

**Working:**
- First parentheses: 8 × 4 + 2 = 32 + 2 = 34
- Second parentheses: 8 + 4 × 2 = 8 + 8 = 16
- Result: 34 − 16 = 18

**Answer 10: D**

---

**Problem 11:**

**What is being asked:** Convert 11 hours 5 minutes to minutes

**Working:**
- 11 hours = 11 × 60 = 660 minutes
- Total: 660 + 5 = 665 minutes

**Answer 11: C**

---

**Problem 12:**

**What is being asked:** In how many ways can the ghost enter by one window and leave by a different window?

**Working:**
- Choices for entry: 6 windows
- Choices for exit: 5 windows (must be different from entry)
- Total ways: 6 × 5 = 30

**Answer 12: D**

Problem 1:

To be certain of getting 5 socks of the same color, we use the pigeonhole principle.

There are 3 colors: red, white, and blue.  
To avoid getting 5 of any one color for as long as possible, Martian could take at most 4 of each color:

- 4 red
- 4 white
- 4 blue

That gives a total of \(4+4+4=12\) socks without yet having 5 of the same color.

The very next sock, the 13th, must make one color appear at least 5 times.

So the Martian must remove 13 socks.

ANSWER 1: D

Problem 2:

There are 4 players: Lance, Sally, Joy, and Fred.

We need to choose 3 starters from these 4 players. Since order does not matter, we count combinations:

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

Indeed, the 3-starter groups are:

- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred

So there are 4 ways.

ANSWER 2: B

Problem 3:

We check each product.

- A: \((-4)(-9)=36\)
- B: \((-3)(-12)=36\)
- C: \(\frac12 \cdot (-72)=-36\), not 36
- D: \(1\cdot 36=36\)
- E: \(\frac32 \cdot 24=36\)

Only choice C does **not** have product 36.

ANSWER 3: C

Problem 4:

First find how high the bulb is above the floor.

The ceiling is 2.4 meters above the floor, which is \(240\) centimeters.  
The bulb is 10 centimeters below the ceiling, so it is:

\[
240 - 10 = 230 \text{ cm}
\]

Now Alice’s height is 1.5 meters = 150 cm.  
She can reach 46 cm above her head, so her reach without the stool is:

\[
150 + 46 = 196 \text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230 - 196 = 34 \text{ cm}
\]

So the stool is 34 cm high.

ANSWER 4: B

Problem 5:

The sample has 30 trout out of 180 fish, so the fraction of trout is:

\[
\frac{30}{180}=\frac{1}{6}
\]

If 250 trout is \(\frac16\) of the total lake population, then the total number of fish is:

\[
250 \div \frac16 = 250 \cdot 6 = 1500
\]

So there are 1500 fish in the lake.

ANSWER 5: B

Problem 6:

Karl bought 5 folders at \$2.50 each, so the total cost was:

\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:

\[
0.20 \times 12.50 = 2.50
\]

So he could have saved \$2.50.

ANSWER 6: C

Problem 7:

He paints 300 dimples, and each dimple takes 2 seconds.

Total time in seconds:

\[
300 \times 2 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10
\]

So he needs 10 minutes.

ANSWER 7: D

Problem 8:

We want combinations of 5-dollar bills and 2-dollar bills that total \$17.

Let \(x\) be the number of \$5 bills and \(y\) the number of \$2 bills. Then:

\[
5x + 2y = 17
\]

Try values of \(x\):

- \(x=1\): \(17-5=12\), so \(y=6\)
- \(x=2\): \(17-10=7\), not even, so impossible
- \(x=3\): \(17-15=2\), so \(y=1\)

These are the only nonnegative integer solutions.

So there are 2 combinations.

ANSWER 8: A

Problem 9:

The contest ended 1000 minutes after noon.

Convert 1000 minutes to hours and minutes:

\[
1000 \div 60 = 16 \text{ hours with } 40 \text{ minutes left}
\]

So:

- Noon + 12 hours = midnight
- plus 4 more hours = 4:00 a.m.
- plus 40 minutes = 4:40 a.m.

So the contest ended at 4:40 a.m.

ANSWER 9: D

Problem 10:

Compute each part using order of operations:

\[
(8\times 4 + 2) - (8 + 4\times 2)
\]

First parentheses:

\[
8\times 4 + 2 = 32 + 2 = 34
\]

Second parentheses:

\[
8 + 4\times 2 = 8 + 8 = 16
\]

Now subtract:

\[
34 - 16 = 18
\]

So the value is 18.

ANSWER 10: D

Problem 11:

We convert 11 hours and 5 minutes into minutes.

\[
11 \times 60 = 660
\]

Then add 5 minutes:

\[
660 + 5 = 665
\]

So the match lasted 665 minutes.

ANSWER 11: C

Problem 12:

Georgie must enter by one window and leave by a different window.

There are 6 choices for the entering window.  
After choosing one, there are 5 remaining choices for the exit window.

So the total number of ways is:

\[
6 \times 5 = 30
\]

Therefore, there are 30 ways.

ANSWER 12: D

Problem 1:  
We need the minimum number of socks to remove to guarantee there are 5 socks of the same color, given 3 colors (red, white, blue) and at least 5 of each color.

Worst case: you could first pull 4 socks of each color without reaching 5 of any color.  
That would be:
- 4 red + 4 white + 4 blue = 12 socks pulled

After 12 socks, you have at most 4 of each color. The next sock (the 13th) must be one of the 3 colors, and by the pigeonhole principle it makes one color reach 5.

So the Martian must remove **13** socks.

Tempting wrong choices:
- 12 isn’t enough because it’s possible to have only 4 of each color.
- Any number less than 13 can’t force a fifth sock of some color.

ANSWER 1: D


Problem 2:  
We have 4 students: Lance, Sally, Joy, Fred. We must choose 3 starters.

Number of ways to choose 3 out of 4 is:
\[
\binom{4}{3} = 4
\]

(You can also think: leave out one student; there are 4 choices for who to leave out.)

ANSWER 2: B


Problem 3:  
We check which pair does **NOT** multiply to 36.

Choices:
A. \(\{-4,-9\}\): \((-4)(-9)=36\) ✅  
B. \(\{-3,-12\}\): \((-3)(-12)=36\) ✅  
C. \(\left\{\frac{1}{2},-72\right\}\): \(\frac{1}{2}\cdot(-72)=-36\) ❌ not 36  
D. \(\{1,36\}\): \(1\cdot 36=36\) ✅  
E. \(\left\{\frac{3}{2},24\right\}\): \(\frac{3}{2}\cdot 24=3\cdot 12=36\) ✅

So the one that does not have product 36 is choice C.

ANSWER 3: C


Problem 4:  
Let’s convert everything to the same reference (floor).

- Ceiling is 2.4 m above the floor = 240 cm  
- The bulb is 10 cm below the ceiling: height of bulb above floor:
\[
240 - 10 = 230 \text{ cm}
\]
- Alice is 1.5 m tall = 150 cm  
- She can reach 46 cm above the top of her head, so from her top to the bulb:
\[
150 + 46 = 196 \text{ cm (if she stands on the floor)}
\]

She is standing on a stool, so her total height is (stool height + 150). She can reach:
\[
(\text{stool height} + 150) + 46
\]
Set equal to bulb height 230:
\[
\text{stool height} + 196 = 230
\]
\[
\text{stool height} = 34
\]

ANSWER 4: B


Problem 5:  
We know:
- Lake has 250 trout.
- Total fish in lake = \(T\)
- Sample has 180 fish, and 30 are trout.
So sample trout-to-total ratio equals lake trout-to-total ratio:
\[
\frac{30}{180} = \frac{250}{T}
\]
Simplify \(\frac{30}{180}=\frac{1}{6}\):
\[
\frac{1}{6} = \frac{250}{T}
\]
\[
T = 250 \cdot 6 = 1500
\]

ANSWER 5: B


Problem 6:  
Cost per folder originally: $2.50  
He bought 5 folders:
\[
5 \cdot 2.50 = 12.50
\]
A 20%-off sale means paying 80% of the original price:
\[
0.80 \cdot 12.50 = 10.00
\]
Savings:
\[
12.50 - 10.00 = 2.50
\]

ANSWER 6: C


Problem 7:  
Paint 300 dimples, 2 seconds per dimple.
Total time in seconds:
\[
300 \cdot 2 = 600 \text{ seconds}
\]
Convert to minutes (60 seconds per minute):
\[
600 \div 60 = 10 \text{ minutes}
\]

ANSWER 7: D


Problem 8:  
We need numbers of $5 bills and $2 bills that total $17 (order doesn’t matter).

Let \(x\) = number of $5 bills, \(y\) = number of $2 bills.
\[
5x + 2y = 17
\]
Try \(x=0\) to \(x=3\) (since \(4\cdot 5=20>17\)):

- \(x=0\): \(2y=17\) not possible  
- \(x=1\): \(5 + 2y=17 \Rightarrow 2y=12 \Rightarrow y=6\)  
- \(x=2\): \(10 + 2y=17 \Rightarrow 2y=7\) not possible  
- \(x=3\): \(15 + 2y=17 \Rightarrow 2y=2 \Rightarrow y=1\)

So there are 2 combinations: \((x,y)=(1,6)\) and \((3,1)\).

ANSWER 8: A


Problem 9:  
The contest starts at noon. It ends 1000 minutes later.

Convert 1000 minutes to hours:
\[
1000 \div 60 = 16 \text{ hours remainder } 40 \text{ minutes}
\]
Noon + 16 hours = 4:00 a.m. (next day)  
Add 40 minutes:
\[
4:00 \text{ a.m.} + 40 \text{ min} = 4:40 \text{ a.m.}
\]

ANSWER 9: D


Problem 10:  
Compute:
\[
(8 \times 4 + 2) - (8 + 4 \times 2)
\]
First parentheses:
\[
8\times 4 = 32,\quad 32+2=34
\]
Second parentheses:
\[
4\times 2 = 8,\quad 8+8=16
\]
Now subtract:
\[
34 - 16 = 18
\]

ANSWER 10: D


Problem 11:  
11 hours and 5 minutes total.
Convert hours to minutes:
\[
11 \cdot 60 = 660
\]
Add 5 minutes:
\[
660 + 5 = 665
\]

ANSWER 11: C


Problem 12:  
There are 6 windows. Georgie enters through one window and must leave through a different window.

Choose entering window: 6 choices.  
Choose leaving window: 5 choices (cannot be the same).

Total:
\[
6 \cdot 5 = 30
\]

ANSWER 12: D

Problem 1:
To be certain of having 5 socks of the same color, we consider the "worst-case scenario" (Pigeonhole Principle). If we have 3 colors (red, white, blue), we could pick 4 red, 4 white, and 4 blue socks without having 5 of any one color. This totals $4 + 4 + 4 = 12$ socks. The very next sock (the 13th) must be either red, white, or blue, which will complete a set of 5 for that color.
ANSWER 1: D

Problem 2:
We need to choose 3 starters from a group of 4 people (Lance, Sally, Joy, Fred). This is a combination problem denoted as $\binom{4}{3}$. Choosing 3 people to be on the team is the same as choosing 1 person to be left out. There are 4 ways to choose who is left out (Lance, Sally, Joy, or Fred).
ANSWER 2: B

Problem 3:
We check the products:
A: $(-4) \times (-9) = 36$
B: $(-3) \times (-12) = 36$
C: $\frac{1}{2} \times (-72) = -36$
D: $1 \times 36 = 36$
E: $\frac{3}{2} \times 24 = 3 \times 12 = 36$
Choice C results in $-36$, not $36$.
ANSWER 3: C

Problem 4:
Ceiling height = 2.4 m = 240 cm. Light bulb height = 240 - 10 = 230 cm from the floor.
Alice's reach = 150 cm (height) + 46 cm (reach) = 196 cm.
Stool height = (Bulb height) - (Alice's reach) = 230 - 196 = 34 cm.
ANSWER 4: B

Problem 5:
Let $T$ be the total number of fish in the lake. The ratio of trout to total fish is $\frac{30}{180} = \frac{1}{6}$.
Set up the proportion: $\frac{250}{T} = \frac{1}{6}$.
$T = 250 \times 6 = 1500$.
ANSWER 5: B

Problem 6:
Total cost = 5 folders × $2.50 = $12.50.
Savings = 20% of $12.50 = 0.20 \times 12.50 = $2.50.
Alternatively, 20% off one folder is $0.20 \times 2.50 = $0.50. For 5 folders, $0.50 \times 5 = $2.50.
ANSWER 6: C

Problem 7:
Total time in seconds = 300 dimples × 2 seconds/dimple = 600 seconds.
Total time in minutes = 600 seconds / 60 seconds/minute = 10 minutes.
ANSWER 7: D

Problem 8:
Let $x$ be the number of $5 bills and $y$ be the number of $2 bills. $5x + 2y = 17$.
If $x=1$, $5+2y=17 \Rightarrow 2y=12 \Rightarrow y=6$. (1, 6)
If $x=3$, $15+2y=17 \Rightarrow 2y=2 \Rightarrow y=1$. (3, 1)
If $x=0$ or $x=2$, $2y$ would be 17 or 7, which are not divisible by 2.
There are 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours and 40 minutes (since $1000 \div 60 = 16$ remainder 40).
Noon + 12 hours = midnight.
Midnight + 4 hours and 40 minutes = 4:40 a.m.
ANSWER 9: D

Problem 10:
$(8 \times 4 + 2) - (8 + 4 \times 2) = (32 + 2) - (8 + 8) = 34 - 16 = 18$.
ANSWER 10: D

Problem 11:
11 hours = $11 \times 60 = 660$ minutes.
Total = 660 + 5 = 665 minutes.
ANSWER 11: C

Problem 12:
There are 6 choices for the entry window. Since the exit window must be different, there are 5 choices for the exit window.
Total ways = $6 \times 5 = 30$.
ANSWER 12: D

Problem 1:
This is a pigeonhole principle problem. To guarantee at least 5 socks of one color, first consider the worst-case scenario of taking as many socks as possible without having 5 of any single color. That means taking exactly 4 socks of each of the 3 colors, for a total of \(4 \times 3 = 12\) socks. The next sock (the 13th) must be a 5th sock of one of the colors. Thus 13 socks are required. (Fewer than 13 leaves open the possibility of only 4 of each color.)

ANSWER 1: D

Problem 2:
The number of ways to choose 3 starters out of the 4 available players is the combination \(\binom{4}{3} = 4\). Each choice simply leaves out one of the four players.

ANSWER 2: B

Problem 3:
Compute the product for each pair:
- A: \((-4) \times (-9) = 36\)
- B: \((-3) \times (-12) = 36\)
- C: \(\frac12 \times (-72) = -36\) (not 36)
- D: \(1 \times 36 = 36\)
- E: \(\frac32 \times 24 = 36\)

Only pair C fails to give a product of 36.

ANSWER 3: C

Problem 4:
Convert all lengths to centimeters. The ceiling is 240 cm above the floor, so the bulb is 230 cm above the floor. Alice's reach is \(150 + 46 = 196\) cm. The stool must therefore supply the remaining \(230 - 196 = 34\) cm.

ANSWER 4: B

Problem 5:
The sample shows that trout comprise \(\frac{30}{180} = \frac16\) of all fish. With 250 trout in the lake, the total number of fish is \(250 \div \frac16 = 1500\).

ANSWER 5: B

Problem 6:
Five folders cost \(5 \times 2.50 = 12.50\). A 20 % discount saves \(0.20 \times 12.50 = 2.50\).

ANSWER 6: C

Problem 7:
Painting 300 dimples at 2 seconds each requires \(300 \times 2 = 600\) seconds, or \(600 \div 60 = 10\) minutes.

ANSWER 7: D

Problem 8:
Solve \(5x + 2y = 17\) for non-negative integers \(x, y\):
- \(x = 1\), \(y = 6\) works;
- \(x = 3\), \(y = 1\) works.
No other non-negative integer solutions exist, so there are exactly two combinations.

ANSWER 8: A

Problem 9:
Divide 1000 minutes by 60: \(1000 = 16\) hours + 40 minutes. Adding 16 hours to noon yields 4:00 a.m.; adding 40 minutes yields 4:40 a.m.

ANSWER 9: D

Problem 10:
Left side: \(8 \times 4 + 2 = 34\). Right side: \(8 + 4 \times 2 = 16\). Difference: \(34 - 16 = 18\).

ANSWER 10: D

Problem 11:
Convert 11 hours 5 minutes to minutes: \(11 \times 60 + 5 = 665\).

ANSWER 11: C

Problem 12:
There are 6 choices for the entry window and then 5 remaining choices for the exit window, giving \(6 \times 5 = 30\) ordered pairs.

ANSWER 12: D

Problem 1:
The problem is asking us to determine the minimum number of socks the Martian must remove from the drawer to guarantee that there are 5 socks of the same color. 
To ensure that there are 5 socks of the same color, we must consider the worst-case scenario where the Martian draws as many socks as possible without getting 5 of the same color. 
The Martian can draw at most 4 socks of each color without having 5 socks of the same color. 
So, the Martian can draw 4 red, 4 white, and 4 blue socks, for a total of 4 + 4 + 4 = 12 socks. 
At this point, the Martian still doesn't have 5 socks of the same color. 
However, drawing one more sock will guarantee that there are 5 socks of the same color. 
Therefore, the Martian must draw at least 12 + 1 = 13 socks.
Let's examine the choices: 
A. 6 is too low because the Martian could draw 3 red, 2 white, and 1 blue, or other combinations that don't guarantee 5 socks of the same color. 
B. 9 is still too low because the Martian could draw 4 red, 3 white, and 2 blue, or other similar combinations. 
C. 12 is the maximum number of socks the Martian can draw without having 5 socks of the same color. 
D. 13 is the correct answer because it is one more than the maximum number of socks that can be drawn without having 5 socks of the same color. 
E. 15 is more than necessary.
ANSWER 1: D

Problem 2:
The problem is asking us to find the number of ways to choose 3 starters from a team of 4 players: Lance, Sally, Joy, and Fred. 
This is a combination problem, because the order in which the starters are chosen does not matter. 
The number of ways to choose 3 starters from 4 players can be calculated using the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of players and k is the number of starters. 
Here, n = 4 and k = 3, so C(4, 3) = 4! / (3!(4-3)!) = 4.
Let's list the possible combinations to verify: 
- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred
There are indeed 4 combinations.
Let's examine the choices: 
A. 2 is too low. 
B. 4 matches our calculation. 
C. 6 is too high. 
D. 8 is too high. 
E. 10 is too high.
ANSWER 2: B

Problem 3:
The problem is asking us to identify the pair of numbers that does NOT have a product equal to 36. 
Let's check each pair: 
A. {−4, −9}: (−4) × (−9) = 36
B. {−3, −12}: (−3) × (−12) = 36
C. { 1 2 , −72}: (1/2) × (−72) = −36
D. {1, 36}: 1 × 36 = 36
E. { 3 2 , 24}: (3/2) × 24 = 36
The pair that does NOT have a product equal to 36 is C.
Let's examine the choices: 
A. {−4, −9} has a product of 36. 
B. {−3, −12} has a product of 36. 
C. { 1 2 , −72} has a product of −36, not 36. 
D. {1, 36} has a product of 36. 
E. { 3 2 , 24} has a product of 36.
ANSWER 3: C

Problem 4:
The problem is asking us to find the height of the stool that Alice needs to stand on to reach the light bulb. 
First, we need to convert the height of the ceiling from meters to centimeters: 2.4 meters = 240 centimeters. 
The light bulb is 10 centimeters below the ceiling, so its height above the floor is 240 - 10 = 230 centimeters. 
Alice is 1.5 meters = 150 centimeters tall and can reach 46 centimeters above her head, so her total reach is 150 + 46 = 196 centimeters. 
To find the height of the stool, we subtract Alice's total reach from the height of the light bulb: 230 - 196 = 34 centimeters.
Let's examine the choices: 
A. 32 is too low because 230 - (196 + 32) = 2, which is not enough to reach the light bulb. 
B. 34 is the correct height. 
C. 36 is too high. 
D. 38 is too high. 
E. 40 is too high.
ANSWER 4: B

Problem 5:
The problem is asking us to find the total number of fish in the lake based on the ratio of trout in the sample to the total number of fish in the sample. 
The sample contains 30 trout out of 180 fish, so the ratio of trout to total fish in the sample is 30/180. 
We are given that this ratio is the same as the ratio of trout to total fish in the lake. 
Let's denote the total number of fish in the lake as x. 
The ratio of trout to total fish in the lake is 250/x. 
Setting the two ratios equal, we get 30/180 = 250/x. 
Cross-multiplying, we have 30x = 250 × 180. 
Simplifying, 30x = 45000. 
Dividing both sides by 30, x = 1500.
Let's examine the choices: 
A. 1250 is too low. 
B. 1500 matches our calculation. 
C. 1750 is too high. 
D. 1800 is too high. 
E. 2000 is too high.
ANSWER 5: B

Problem 6:
The problem is asking us to find out how much Karl could have saved by waiting a day to buy the folders at a 20%-off sale. 
The original price of 5 folders is 5 × $2.50 = $12.50. 
At a 20%-off sale, the discount is 20% of $12.50, which is 0.20 × $12.50 = $2.50. 
So, Karl could have saved $2.50.
Let's examine the choices: 
A. $1.00 is too low. 
B. $2.00 is too low. 
C. $2.50 matches our calculation. 
D. $2.75 is too high. 
E. $5.00 is too high.
ANSWER 6: C

Problem 7:
The problem is asking us to find the time it will take Casey to paint 300 dimples on a golf ball. 
Casey takes 2 seconds to paint one dimple, so to paint 300 dimples, he will take 300 × 2 = 600 seconds. 
To convert seconds to minutes, we divide by 60: 600 / 60 = 10 minutes.
Let's examine the choices: 
A. 4 minutes is too low. 
B. 6 minutes is too low. 
C. 8 minutes is too low. 
D. 10 minutes matches our calculation. 
E. 12 minutes is too high.
ANSWER 7: D

Problem 8:
The problem is asking us to find the number of different combinations of $5 bills and $2 bills that can be used to make a total of $17. 
Let's list the possible combinations: 
1. 3 $5 bills = $15, and $2 bill = $2, total = $17 (1 combination: 3 $5 bills and 1 $2 bill)
2. 1 $5 bill = $5, and 6 $2 bills = $12, total = $17 (1 combination: 1 $5 bill and 6 $2 bills)
We should also check if there are any other combinations: 
- Using 0 $5 bills, we need 8.5 $2 bills, which is not possible since we can't have a fraction of a bill.
- Using 2 $5 bills = $10, we need $7 more, which can be made with 3.5 $2 bills, again not possible.
- Using 4 $5 bills = $20, which is more than $17.

Problem 1:
We want the minimum number of socks to guarantee 5 of the same color. With 3 colors (red, white, blue), the worst case is picking 4 of each color: 4+4+4 = 12 socks. The 13th sock must create a group of 5 of one color. So 13 socks are needed.
ANSWER 1: D

Problem 2:
We choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways is the combination C(4,3) = 4.
ANSWER 2: B

Problem 3:
Compute each product:
A: (−4)×(−9) = 36
B: (−3)×(−12) = 36
C: (1/2)×(−72) = −36
D: 1×36 = 36
E: (3/2)×24 = 36
Only C does not equal 36.
ANSWER 3: C

Problem 4:
Ceiling = 2.4 m = 240 cm. Light bulb is 10 cm below ceiling, so its height = 230 cm.
Alice's height = 1.5 m = 150 cm. Reach above head = 46 cm, so reach without stool = 196 cm.
Stool height needed = 230 − 196 = 34 cm.
ANSWER 4: B

Problem 5:
Sample ratio: 30 trout / 180 fish = 1/6. Lake has 250 trout, which is 1/6 of total fish. Total fish = 250 × 6 = 1500.
ANSWER 5: B

Problem 6:
Cost without sale: 5 × $2.50 = $12.50.
With 20% off, each folder costs $2.50 × 0.8 = $2.00. Total = 5 × $2.00 = $10.00.
Savings = $12.50 − $10.00 = $2.50.
ANSWER 6: C

Problem 7:
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.
ANSWER 7: D

Problem 8:
Find non-negative integers x ($5 bills) and y ($2 bills) such that 5x + 2y = 17.
x=1 → 5 + 2y = 17 → y=6.
x=3 → 15 + 2y = 17 → y=1.
No other x works. So 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours 40 minutes (16×60 = 960, remainder 40).
Noon + 16 hours = 4:00 a.m. next day. Add 40 minutes → 4:40 a.m.
ANSWER 9: D

Problem 10:
(8 × 4 + 2) = 32 + 2 = 34.
(8 + 4 × 2) = 8 + 8 = 16.
34 − 16 = 18.
ANSWER 10: D

Problem 11:
11 hours × 60 = 660 minutes. Plus 5 minutes = 665 minutes.
ANSWER 11: C

Problem 12:
6 choices for entry window, then 5 choices for a different exit window. Total = 6 × 5 = 30.
ANSWER 12: D

Problem 1:
We need to find the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). Using the Pigeonhole Principle, we consider the worst-case scenario: the Martian pulls out as many socks as possible *without* getting 5 of any one color. That means getting exactly 4 of each color:
- 4 red + 4 white + 4 blue = 12 socks
After 12 socks, the Martian could have 4 of each color and still not have 5 matching. The very next sock (the 13th) must be red, white, or blue, pushing the count of that color to 5. So 13 socks guarantees 5 of the same color.
(The "five-legged" detail is flavor text explaining why the Martian needs 5 socks.)

ANSWER 1: D

Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). This is a combination problem since order doesn't matter:
C(4, 3) = 4! / (3! × 1!) = 4
We can also list them: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — exactly 4 ways.

ANSWER 2: B

Problem 3:
We check each product:
- A: (−4) × (−9) = 36 ✓
- B: (−3) × (−12) = 36 ✓
- C: (1/2) × (−72) = −36 ✗ (This equals −36, NOT 36)
- D: 1 × 36 = 36 ✓
- E: (3/2) × 24 = 36 ✓
Choice C gives −36, which does not equal 36.

ANSWER 3: C

Problem 4:
First, convert everything to centimeters:
- Ceiling height: 2.4 m = 240 cm
- Light bulb position: 240 − 10 = 230 cm above the floor
- Alice's height: 1.5 m = 150 cm
- Alice's reach above her head: 46 cm
- Alice's maximum reach (without stool): 150 + 46 = 196 cm above the floor

Let h = height of the stool. Standing on the stool, her reach is h + 196 cm. She can "just reach" the bulb:
h + 196 = 230
h = 230 − 196 = 34 cm

ANSWER 4: B

Problem 5:
The ratio of trout to total fish in the sample should equal the ratio in the lake:
- Sample ratio: 30 trout / 180 fish = 1/6
- Lake ratio: 250 trout / Total fish = 1/6

Setting up the proportion: 250 / Total = 1/6
Total = 250 × 6 = 1500 fish

ANSWER 5: B

Problem 6:
Karl bought 5 folders at $2.50 each:
Total cost = 5 × $2.50 = $12.50
A 20%-off sale means he would save 20% of the total:
Savings = 0.20 × $12.50 = $2.50

ANSWER 6: C

Problem 7:
Total time in seconds: 300 dimples × 2 seconds/dimple = 600 seconds
Convert to minutes: 600 seconds ÷ 60 seconds/minute = 10 minutes

ANSWER 7: D

Problem 8:
We need non-negative integer solutions to 5x + 2y = 17, where x = number of $5 bills and y = number of $2 bills.
- x = 0: 2y = 17 → y = 8.5 (not an integer) ✗
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 ✗
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x = 4: 20 > 17 (too large) ✗

Only 2 valid combinations: (1 five + 6 twos) and (3 fives + 1 two).

ANSWER 8: A

Problem 9:
Convert 1000 minutes to hours and minutes:
1000 ÷ 60 = 16 remainder 40, so 1000 minutes = 16 hours 40 minutes.
Starting at noon (12:00 PM):
- 12 hours later = midnight (12:00 AM)
- 4 more hours = 4:00 AM
- 40 more minutes = 4:40 AM

ANSWER 9: D

Problem 10:
Evaluate each part using order of operations (multiplication before addition):
- First part: (8 × 4 + 2) = (32 + 2) = 34
- Second part: (8 + 4 × 2) = (8 + 8) = 16
- Difference: 34 − 16 = 18

ANSWER 10: D

Problem 11:
Convert 11 hours and 5 minutes to minutes:
11 hours × 60 minutes/hour = 660 minutes
660 + 5 = 665 minutes

ANSWER 11: C

Problem 12:
Georgie enters through one window and leaves through a different one.
- Choices for entry window: 6
- Choices for exit window (must be different): 5
Total ways = 6 × 5 = 30

ANSWER 12: D

Problem 1:
The Martian needs to be **certain** of having 5 socks of the same color. To guarantee this, we must consider the worst-case scenario: the Martian picks as many socks as possible **without** yet having 5 of any single color.  
With 3 colors (red, white, blue), the worst case is picking 4 socks of each color:  
\(4 + 4 + 4 = 12\) socks.  
At this point the Martian has 4 of every color but still does not have 5 of any one color. The very next sock—the 13th—must be red, white, or blue, and will necessarily create a set of 5 of that color.  
Thus 13 socks are required.  
(Choice A is too small; with 6 socks one could have 2,2,2. Choice B allows 3,3,3. Choice C allows 4,4,4. Choice E is more than the minimum guaranteed number.)

ANSWER 1: D

---

Problem 2:
We need to choose a group of 3 starters from the 4 students: Lance, Sally, Joy, and Fred. Since the three starters form a team and order does not matter, this is a combination problem.  
\[
\binom{4}{3} = \frac{4!}{3!\,1!} = 4
\]  
Equivalently, we can think of choosing which **one** of the four students will **not** start; there are 4 such choices.  
(Choice A is too small. Choice C is \(\binom{4}{2}\). Choices D and E do not arise from counting 3-person teams from 4 people.)

ANSWER 2: B

---

Problem 3:
We check the product of each pair to see which is not 36.

- **A.** \((-4)(-9) = 36\)  
- **B.** \((-3)(-12) = 36\)  
- **C.** \(\left(\frac12\right)(-72) = -36\)  
- **D.** \((1)(36) = 36\)  
- **E.** \(\left(\frac32\right)(24) = 36\)

Only pair **C** gives \(-36\), not \(36\).

ANSWER 3: C

---

Problem 4:
First convert all measurements to centimeters.  
- Ceiling height: \(2.4\text{ m} = 240\text{ cm}\).  
- Light bulb is \(10\text{ cm}\) below the ceiling, so its height above the floor is \(240 - 10 = 230\text{ cm}\).  

Alice’s reach:  
- Height: \(1.5\text{ m} = 150\text{ cm}\).  
- She can reach \(46\text{ cm}\) above her head, so her standing reach is \(150 + 46 = 196\text{ cm}\).  

The stool must supply the remaining height:  
\[
230 - 196 = 34\text{ cm}.
\]

ANSWER 4: B

---

Problem 5:
Let \(T\) be the total number of fish in the lake. The problem states that the ratio of trout to total fish is the same in the sample and in the lake.  
\[
\frac{250}{T} = \frac{30}{180}
\]  
Simplify the right side: \(\frac{30}{180} = \frac16\).  
\[
\frac{250}{T} = \frac16 \quad\Longrightarrow\quad T = 250 \times 6 = 1500.
\]

ANSWER 5: B

---

Problem 6:
Karl bought 5 folders at \(\$2.50\) each.  
Total cost: \(5 \times \$2.50 = \$12.50\).  
A 20% discount would have saved him 20% of this total:  
\[
0.20 \times \$12.50 = \$2.50.
\]

ANSWER 6: C

---

Problem 7:
Casey paints 300 dimples at 2 seconds per dimple.  
Total time in seconds: \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(\frac{600}{60} = 10\) minutes.

ANSWER 7: D

---

Problem 8:
Let \(f\) be the number of \(\$5\) bills and \(t\) be the number of \(\$2\) bills. We need non-negative integer solutions to  
\[
5f + 2t = 17.
\]  
Test values of \(f\):  
- \(f=0\): \(2t=17\) → no integer solution.  
- \(f=1\): \(5 + 2t = 17 \Rightarrow t=6\).  **(Valid)**
- \(f=2\): \(10 + 2t = 17 \Rightarrow t=3.5\).  (Invalid)
- \(f=3\): \(15 + 2t = 17 \Rightarrow t=1\).  **(Valid)**
- \(f=4\): \(20 > 17\) → stop.

There are exactly **2** combinations: \((1\text{ five}, 6\text{ twos})\) and \((3\text{ fives}, 1\text{ two})\).

ANSWER 8: A

---

Problem 9:
The contest starts at noon (12:00 p.m.). We add 1000 minutes.  
Divide 1000 by 60:  
\[
1000 = 16 \times 60 + 40,
\]  
so 1000 minutes = 16 hours and 40 minutes.  

Noon + 12 hours = midnight.  
Midnight + 4 hours 40 minutes = 4:40 a.m.

ANSWER 9: D

---

Problem 10:
Evaluate using the correct order of operations (multiplication before addition).

First parentheses:  
\[
8 \times 4 + 2 = 32 + 2 = 34.
\]

Second parentheses:  
\[
8 + 4 \times 2 = 8 + 8 = 16.
\]

Subtract:  
\[
34 - 16 = 18.
\]

ANSWER 10: D

---

Problem 11:
Convert 11 hours to minutes and add the remaining 5 minutes.  
\[
11 \times 60 = 660\text{ minutes}.
\]  
\[
660 + 5 = 665\text{ minutes}.
\]

ANSWER 11: C

---

Problem 12:
Georgie enters through one of 6 windows and must leave through a **different** window.  
- 6 choices for the entry window.  
- For each entry choice, 5 remaining choices for the exit window.  

Total ways:  
\[
6 \times 5 = 30.
\]  
(Choice E, 36, would count entering and leaving through the same window, which is not allowed.)

ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). The worst-case scenario is to draw as many socks as possible without having 5 of any one color. That would be 4 red, 4 white, and 4 blue, for a total of 12 socks. The next sock drawn (the 13th) must be one of the three colors, giving that color 5 socks. Therefore, 13 socks are required. The five-legged Martian is irrelevant to the count.  
Answer: D

Problem 2:
We have 4 people (Lance, Sally, Joy, Fred) and need to choose 3 starters. Since the starters are not assigned to specific positions, order does not matter. The number of ways is the combination \( \binom{4}{3} = 4 \).  
Answer: B

Problem 3:
Compute each product:  
A. \((-4) \times (-9) = 36\)  
B. \((-3) \times (-12) = 36\)  
C. \(\frac{1}{2} \times (-72) = -36\)  
D. \(1 \times 36 = 36\)  
E. \(\frac{3}{2} \times 24 = 36\)  
Only pair C gives a product not equal to 36.  
Answer: C

Problem 4:
Convert all measurements to centimeters:  
Ceiling height = 2.4 m = 240 cm.  
Light bulb is 10 cm below ceiling → height = 240 – 10 = 230 cm.  
Alice’s height = 1.5 m = 150 cm.  
Her reach above her head = 46 cm, so her maximum reach without a stool = 150 + 46 = 196 cm.  
Let \(h\) be the stool height. Standing on the stool raises her reach to \(196 + h\). This must equal 230 cm:  
\(196 + h = 230\) → \(h = 34\) cm.  
Answer: B

Problem 5:
In the sample, 30 out of 180 fish are trout, so the ratio trout : total = \(30/180 = 1/6\).  
Assuming the same ratio in the lake, if the lake has 250 trout, then the total number of fish is \(250 \times 6 = 1500\).  
Answer: B

Problem 6:
Original cost for 5 folders: \(5 \times 2.50 = \$12.50\).  
A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
Thus, by waiting a day, Karl could have saved $2.50.  
Answer: C

Problem 7:
Time to paint one dimple = 2 seconds.  
Total time for 300 dimples = \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(600 \div 60 = 10\) minutes.  
Answer: D

Problem 8:
We need nonnegative integers \(a\) (number of $5 bills) and \(b\) (number of $2 bills) such that \(5a + 2b = 17\).  
Solve by testing values of \(b\):  
\(b = 1\) → \(5a = 15\) → \(a = 3\) → (3,1) works.  
\(b = 6\) → \(5a = 5\) → \(a = 1\) → (1,6) works.  
No other values give integer \(a\). Thus there are 2 combinations.  
Answer: A

Problem 9:
1000 minutes = \(1000 \div 60 = 16\) hours and \(40\) minutes (since \(60 \times 16 = 960\), remainder 40).  
Starting at noon (12:00 PM), add 12 hours → midnight (12:00 AM). Add the remaining 4 hours → 4:00 AM. Then add 40 minutes → 4:40 AM.  
Answer: D

Problem 10:
Evaluate using order of operations:  
First parentheses: \(8 \times 4 + 2 = 32 + 2 = 34\).  
Second parentheses: \(8 + 4 \times 2 = 8 + 8 = 16\).  
Subtract: \(34 - 16 = 18\).  
Answer: D

Problem 11:
11 hours = \(11 \times 60 = 660\) minutes.  
Add 5 minutes → \(660 + 5 = 665\) minutes.  
Answer: C

Problem 12:
Georgie must choose an entry window (6 choices) and a different exit window (5 choices). Since entering and leaving are distinct actions, the total number of ways is \(6 \times 5 = 30\).  
Answer: D

ANSWER 1: D
ANSWER 2: B
ANSWER 3: C
ANSWER 4: B
ANSWER 5: B
ANSWER 6: C
ANSWER 7: D
ANSWER 8: A
ANSWER 9: D
ANSWER 10: D
ANSWER 11: C
ANSWER 12: D

Problem 1:  
The Martian wants to be certain of having 5 socks of the same color. There are three colors: red, white, and blue. In the worst-case scenario, the Martian could pick 4 socks of each color before getting a fifth of any one color. That would be \(4 \times 3 = 12\) socks, and still no color has 5 socks. The next sock drawn (the 13th) must complete a set of 5 of some color, because there are only three colors. Therefore, the Martian must remove **13** socks to be certain of having 5 socks of the same color.  
ANSWER 1: D

Problem 2:  
Ms. Hamilton's class has 4 available players: Lance, Sally, Joy, and Fred. The number of ways to choose 3 starters from 4 players is given by the combination \(\binom{4}{3} = 4\). Thus, there are **4** possible sets of three starters.  
ANSWER 2: B

Problem 3:  
We check each pair's product:  
- A: \((-4) \times (-9) = 36\)  
- B: \((-3) \times (-12) = 36\)  
- C: \(\frac{1}{2} \times (-72) = -36\) (not 36)  
- D: \(1 \times 36 = 36\)  
- E: \(\frac{3}{2} \times 24 = 36\)  
Only pair **C** does not have a product of 36.  
ANSWER 3: C

Problem 4:  
The ceiling is 2.4 m above the floor, and the bulb is 10 cm (0.1 m) below the ceiling, so the bulb's height is \(2.4 - 0.1 = 2.3\) m. Alice's height is 1.5 m, and she can reach 46 cm (0.46 m) above her head, so her reach height from the floor is \(1.5 + 0.46 = 1.96\) m. When standing on a stool, her hands are at the stool height plus her reach height. Setting this equal to the bulb height:  
\[
\text{stool height} + 1.96 = 2.3 \Rightarrow \text{stool height} = 0.34 \text{ m} = 34 \text{ cm}.
\]  
ANSWER 4: B

Problem 5:  
In the sample, the ratio of trout to total fish is \(\frac{30}{180} = \frac{1}{6}\). Assuming the same ratio in the lake, let \(T\) be the total number of fish in the lake. Then \(\frac{250}{T} = \frac{1}{6}\), so \(T = 250 \times 6 = 1500\).  
ANSWER 5: B

Problem 6:  
Karl bought 5 folders at $2.50 each, so the original cost is \(5 \times 2.50 = \$12.50\). With a 20% discount, the cost becomes \(12.50 \times 0.80 = \$10.00\). The savings is \(12.50 - 10.00 = \$2.50\).  
ANSWER 6: C

Problem 7:  
Painting 300 dimples at 2 seconds each takes \(300 \times 2 = 600\) seconds. Converting to minutes: \(600 \div 60 = 10\) minutes.  
ANSWER 7: D

Problem 8:  
We need nonnegative integers \(x\) (number of $5 bills) and \(y\) (number of $2 bills) such that \(5x + 2y = 17\). Solving:  
- \(y = 1 \Rightarrow 5x = 15 \Rightarrow x = 3\)  
- \(y = 6 \Rightarrow 5x = 5 \Rightarrow x = 1\)  
- \(y = 11 \Rightarrow 5x = -5\) (invalid)  
Thus, there are **2** valid combinations: (3 fives, 1 two) and (1 five, 6 twos).  
ANSWER 8: A

Problem 9:  
1000 minutes equals \(16\) hours and \(40\) minutes (since \(16 \times 60 = 960\), remainder \(40\)). Starting at noon, adding \(16\) hours gives 4:00 AM the next day, and adding \(40\) minutes gives 4:40 AM.  
ANSWER 9: D

Problem 10:  
Compute inside parentheses:  
\(8 \times 4 + 2 = 32 + 2 = 34\)  
\(8 + 4 \times 2 = 8 + 8 = 16\)  
Then \(34 - 16 = 18\).  
ANSWER 10: D

Problem 11:  
11 hours is \(11 \times 60 = 660\) minutes. Adding 5 minutes gives \(660 + 5 = 665\) minutes.  
ANSWER 11: C

Problem 12:  
Georgie can enter through any of the 6 windows. For exiting, he must choose a different window, so there are 5 choices. The total number of ways is \(6 \times 5 = 30\).  
ANSWER 12: D

Problem 1:  
The Martian needs to ensure there are 5 socks of the same color. The worst-case scenario is pulling the maximum number of socks without 5 of any color. Since there are 3 colors, this would be 4 socks of each color (4 × 3 = 12 socks). The next sock (the 13th) must match one color, giving 5 socks of that color.  
**ANSWER 1: D**  

Problem 2:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is C(4,3) = 4.  
**ANSWER 2: B** (Note: C(4,3)=4, but rechecking: the team is chosen from 4 specific people, so combinations are 4. Answer B=4)  

Problem 3:  
Check each pair:  
A. (−4)(−9) = 36  
B. (−3)(−12) = 36  
C. (1/2)(−72) = -36 (≠36)  
D. (1)(36) = 36  
E. (3/2)(24) = 36  
The pair that does NOT equal 36 is C.  
**ANSWER 3: E** (Re-evaluated: The question asks which does NOT equal 36. Option E: (3/2)*24=36, so it is valid. Option C: (1/2)*(-72)=-36, so C is the correct choice for NOT equal. Correction: **ANSWER 3: C** [Original had a typo, but computation confirms C is the answer])  

*Correction: For Problem 3, Option C: (1/2, -72) likely means (1/2) × (-72) = -36 ≠ 36. Option E: (3/2) × 24 = 36. Thus, C is the answer.*  
**ANSWER 3: C**  

Problem 4:  
Ceiling height: 2.4 m = 240 cm. Light bulb location: 240 - 10 = 230 cm from floor. Alice’s reach: 150 cm + 46 cm = 196 cm. Stool height = 230 - 196 = 34 cm.  
**ANSWER 4: B**  

Problem 5:  
Ratio of trout in sample: 30/180 = 1/6. Total trout: 250. Total fish = 250 × 6 = 1500.  
**ANSWER 5: B**  

Problem 6:  
Original cost: 5 × $2.50 = $12.50. 20% off: $12.50 × 0.20 = $2.50 saved.  
**ANSWER 6: C** (Note: The question asks how much Karl could have saved, which is $2.50, not $2.00. Original had a mistake; correction: 20% of $12.50 is $2.50, so answer C)  

Problem 7:  
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.  
**ANSWER 7: D**  

Problem 8:  
Let $5 bills = x, $2 bills = y. 5x + 2y = 17. Possible solutions:  
- x=3, y=1 (15+2=17)  
- x=1, y=6 (5+12=17)  
Only 2 combinations.  
**ANSWER 8: A**  

Problem 9:  
1000 minutes = 16 hours 40 minutes. Starting at noon: noon + 16h40m = 4:40 a.m. next day.  
**ANSWER 9: D**  

Problem 10:  
(8×4 + 2) = 34, (8 + 4×2) = 16. 34 - 16 = 18.  
**ANSWER 10: D**  

Problem 11:  
11 hours = 660 minutes. 660 + 5 = 665 minutes.  
**ANSWER 11: C**  

Problem 12:  
Georgie can enter through 6 windows and exit through 5 others. Total ways: 6 × 5 = 30.  
**ANSWER 12: D**  

---  
**Final Answers**  
ANSWER 1: D  
ANSWER 2: B  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: B  
ANSWER 6: C  
ANSWER 7: D  
ANSWER 8: A  
ANSWER 9: D  
ANSWER 10: D  
ANSWER 11: C  
ANSWER 12: D

### Problem 1:
This is a classic pigeonhole principle problem. We need to find the minimum number of socks the Martian must pull to guarantee 5 socks of the same color, with 3 total colors. The worst-case scenario is that the Martian pulls as many socks as possible without getting 5 of any color: that is 4 socks of each color, for a total of 3*4=12 socks. The next sock pulled (the 13th) has to be one of the three colors, pushing that color to 5 socks. Tempting wrong answers: C (12 is only the number of socks that avoids 5 of the same color, not enough to guarantee the requirement), E (15 is the total if you took 5 of each color, not the minimum to guarantee).
ANSWER 1: D

---

### Problem 2:
We need to count the number of ways to choose 3 starters from 4 people, where order on the team does not matter. This is a combination problem: C(n,k) = n!/(k!(n-k)!). For n=4, k=3: C(4,3) = 4. We can also list the teams by who is left out: leaving out Lance, Sally, Joy, or Fred gives exactly 4 unique teams. Tempting wrong answer: C (6 is the count if order of selection mattered, which it does not for a team of starters).
ANSWER 2: B

---

### Problem 3:
We calculate the product of each pair to find which does not equal 36:
- A: (-4)*(-9)=36 ✔️
- B: (-3)*(-12)=36 ✔️
- C: (1/2)*(-72) = -36 ❌
- D: 1*36=36 ✔️
- E: (3/2)*24=36 ✔️
Only pair C has a product of -36, not 36.
ANSWER 3: C

---

### Problem 4:
First convert all units to centimeters to avoid conversion errors:
- Ceiling height: 2.4m = 240cm. Light bulb is 10cm below ceiling, so bulb height from floor = 240-10=230cm.
- Alice's height: 1.5m=150cm, her reach above her head is 46cm, so total reach without stool = 150+46=196cm.
- Stool height = 230-196=34cm.
Tempting wrong answers come from unit conversion mistakes, e.g., mixing up meters and centimeters.
ANSWER 4: B

---

### Problem 5:
This is a proportion problem. The ratio of trout to total fish is the same in the sample and the lake. In the sample, 30 trout out of 180 total fish = 1/6. Let T be total fish in the lake: 250/T = 1/6 → T=250*6=1500.
ANSWER 5: B

---

### Problem 6:
Karl bought 5 folders at $2.50 each, total original cost = 5*2.50=$12.50. He saves 20% of the total cost by waiting for the sale: 0.2*12.50=$2.50. This can also be calculated per folder: 20% of $2.50 is $0.50 saved per folder, 5*0.50=$2.50 total saved.
ANSWER 6: C

---

### Problem 7:
300 dimples * 2 seconds per dimple = 600 total seconds. Convert to minutes: 60 seconds per minute, so 600/60=10 minutes.
ANSWER 7: D

---

### Problem 8:
We find all non-negative integer pairs (x,y) where x=number of $5 bills, y=number of $2 bills, and 5x+2y=17. Only odd values of x work (since $5 is odd, $2 is even, we need an odd sum to get 17):
- x=1: 5+2y=17 → y=6 (valid)
- x=3: 15+2y=17 → y=1 (valid)
Even x values (0,2) leave an odd remainder that cannot be formed with $2 bills, so only 2 valid combinations.
ANSWER 8: A

---

### Problem 9:
Convert 1000 minutes to hours and minutes: 60 minutes per hour, 1000/60=16 hours and 40 minutes. Adding this to noon (12pm): 12pm +16h40m = 4:40am the next day.
ANSWER 9: D

---

### Problem 10:
Use order of operations (PEMDAS) to calculate:
First parenthesis: 8*4 +2 = 32+2=34
Second parenthesis: 8 +4*2=8+8=16
Subtract: 34-16=18. The tempting wrong answer C (10) comes from incorrectly adding before multiplying in the second parenthesis.
ANSWER 10: D

---

### Problem 11:
Convert 11 hours 5 minutes to total minutes: 11*60=660 minutes, plus 5 extra minutes = 665 minutes.
ANSWER 11: C

---

### Problem 12:
Georgie has 6 choices for an entry window, and 5 remaining choices for an exit window (must use a different window). Total ways = 6*5=30. Tempting wrong answer E (36) counts entering and leaving through the same window, which is not allowed; B (15) incorrectly treats entry/exit as unordered, which they are not.
ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color (red, white, blue). This is a pigeonhole principle problem: the worst-case scenario is pulling the maximum number of socks without getting 5 of any color. For each of the 3 colors, we can pull 4 socks (since 4 < 5), so 4×3=12 socks. The next sock pulled will necessarily be the 5th of one color. So total socks needed: 12 + 1 = 13.
ANSWER 1: D
Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). Since the problem does not mention distinct positions for starters, order does not matter, so we use combinations. The number of ways to choose 3 people from 4 is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n=4 \) and \( k=3 \):
\( \binom{4}{3} = \frac{4!}{3!1!} = 4 \). This is equivalent to counting how many people we leave out (4 choices, one for each person not selected).
ANSWER 2: B
Problem 3:
We check the product of each pair:
- A: \( (-4) \times (-9) = 36 \)
- B: \( (-3) \times (-12) = 36 \)
- C: \( \frac{1}{2} \times (-72) = -36 \neq 36 \)
- D: \( 1 \times 36 = 36 \)
- E: \( \frac{3}{2} \times 24 = 36 \)
Only pair C does not have a product of 36.
ANSWER 3: C
Problem 4:
Convert all units to centimeters:
- Ceiling height: \( 2.4 \, \text{m} = 240 \, \text{cm} \)
- Light bulb height from floor: \( 240 - 10 = 230 \, \text{cm} \)
- Alice's reach from floor (without stool): \( 1.5 \, \text{m} + 0.46 \, \text{m} = 150 + 46 = 196 \, \text{cm} \)
Let \( h \) be the stool height. The total reach with the stool equals the light bulb height:
\( h + 196 = 230 \)
\( h = 230 - 196 = 34 \, \text{cm} \)
ANSWER 4: B
Problem 5:
The ratio of trout to total fish in the sample equals the ratio in the lake. The sample has \( \frac{30}{180} = \frac{1}{6} \) trout. Let \( T \) be the total number of fish in the lake. We know there are 250 trout, so:
\( \frac{250}{T} = \frac{1}{6} \)
\( T = 250 \times 6 = 1500 \)
ANSWER 5: B
Problem 6:
Karl bought 5 folders at $2.50 each. The 20% discount applies to each folder, so the savings per folder is \( 2.50 \times 0.20 = 0.50 \). Total savings for 5 folders: \( 5 \times 0.50 = 2.50 \).
ANSWER 6: C
Problem 7:
Total time to paint 300 dimples: \( 300 \times 2 = 600 \) seconds. Convert to minutes: \( 600 \div 60 = 10 \) minutes.
ANSWER 7: D
Problem 8:
Let \( x \) = number of $5 bills, \( y \) = number of $2 bills. We need \( 5x + 2y = 17 \), where \( x,y \geq 0 \) integers. Test non-negative integer values of \( x \):
- \( x=0 \): \( 2y=17 \) → \( y=8.5 \) (not integer, invalid)
- \( x=1 \): \( 5 + 2y=17 \) → \( 2y=12 \) → \( y=6 \) (valid: 1 $5 + 6 $2 = $17)
- \( x=2 \): \( 10 + 2y=17 \) → \( 2y=7 \) → \( y=3.5 \) (invalid)
- \( x=3 \): \( 15 + 2y=17 \) → \( 2y=2 \) → \( y=1 \) (valid: 3 $5 + 1 $2 = $17)
- \( x=4 \): \( 20 > 17 \) (stop, no more valid \( x \))
Only 2 valid combinations.
ANSWER 8: A
Problem 9:
Convert 1000 minutes to hours and minutes: \( 1000 \div 60 = 16 \) hours with a remainder of 40 minutes (since \( 16 \times 60 = 960 \), \( 1000 - 960 = 40 \)). The contest starts at noon, so adding 16 hours 40 minutes: noon + 12 hours = midnight, plus 4 hours 40 minutes = 4:40 a.m.
ANSWER 9: D
Problem 10:
Use order of operations (multiplication before addition):
First parentheses: \( 8 \times 4 + 2 = 32 + 2 = 34 \)
Second parentheses: \( 8 + 4 \times 2 = 8 + 8 = 16 \)
Subtract: \( 34 - 16 = 18 \)
ANSWER 10: D
Problem 11:
Convert 11 hours 5 minutes to total minutes: \( 11 \times 60 + 5 = 660 + 5 = 665 \)
ANSWER 11: C
Problem 12:
Georgie chooses an entrance window first (6 options), then a different exit window (5 remaining options, since exit cannot equal entrance). Since entrance and exit are distinct actions, order matters, so total ways: \( 6 \times 5 = 30 \).
ANSWER 12: D

I'll work through each problem systematically.

**Problem 1:**

**What is being asked:** How many socks must be removed to guarantee 5 socks of the same color?

**Setup:** There are 3 colors (red, white, blue). In the worst case scenario, the Martian pulls out socks as evenly distributed among colors as possible before getting 5 of one color.

**Working:** 
- Worst case: 4 red, 4 white, 4 blue = 12 socks (still no color with 5)
- The 13th sock must give us a 5th sock of some color

**Answer 1: D**

---

**Problem 2:**

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

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

**Working:**
- C(4,3) = 4 ways
- The 4 ways correspond to leaving out each person: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred}

**Answer 2: B**

---

**Problem 3:**

**What is being asked:** Which pair does NOT have a product equal to 36?

**Working:** Check each pair:
- A. {−4, −9}: (−4)(−9) = 36 ✓
- B. {−3, −12}: (−3)(−12) = 36 ✓
- C. {1/2, −72}: (1/2)(−72) = −36 ✗
- D. {1, 36}: (1)(36) = 36 ✓
- E. {3/2, 24}: (3/2)(24) = 36 ✓

**Answer 3: C**

---

**Problem 4:**

**What is being asked:** How high is the stool?

**Setup:** Convert all measurements to centimeters:
- Ceiling height: 240 cm
- Light bulb location: 240 − 10 = 230 cm from floor
- Alice's height: 150 cm
- Alice's reach above her head: 46 cm
- Alice's reach without stool: 150 + 46 = 196 cm

**Working:**
- With stool, Alice can just reach the bulb at 230 cm
- Stool height + 196 cm = 230 cm
- Stool height = 230 − 196 = 34 cm

**Answer 4: B**

---

**Problem 5:**

**What is being asked:** How many fish total are in the lake?

**Setup:** Use proportions. In the sample: 30 trout out of 180 fish. In the lake: 250 trout out of total fish.

**Working:**
- 30/180 = 250/x
- 1/6 = 250/x
- x = 250 × 6 = 1500 fish

**Answer 5: B**

---

**Problem 6:**

**What is being asked:** How much could Karl have saved with the 20% off sale?

**Setup:** 
- Original cost: 5 folders × $2.50 = $12.50
- Sale discount: 20% off

**Working:**
- 20% of $12.50 = 0.20 × $12.50 = $2.50

**Answer 6: C**

---

**Problem 7:**

**What is being asked:** How many minutes to paint 300 dimples at 2 seconds each?

**Working:**
- Total time: 300 × 2 = 600 seconds
- Convert to minutes: 600 ÷ 60 = 10 minutes

**Answer 7: D**

---

**Problem 8:**

**What is being asked:** How many combinations of $5 bills and $2 bills make $17?

**Setup:** Let x = number of $5 bills, y = number of $2 bills
- 5x + 2y = 17

**Working:**
- x = 0: 2y = 17 → y = 8.5 (not integer)
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 (not integer)
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x ≥ 4: 5x ≥ 20 > 17 (impossible)

Valid combinations: (1,$5 and 6,$2) and (3,$5 and 1,$2) = 2 combinations

**Answer 8: A**

---

**Problem 9:**

**What is being asked:** When did the contest end, starting at noon and lasting 1000 minutes?

**Working:**
- 1000 minutes ÷ 60 = 16 hours and 40 minutes
- Noon + 16 hours 40 minutes = 4:40 p.m. the same day
- But we need to continue: 4:40 p.m. + additional time consideration
- Actually: 1000 ÷ 60 = 16 remainder 40, so 16 hours 40 minutes
- Noon + 16 hours 40 minutes = 4:40 a.m. the next day

**Answer 9: D**

---

**Problem 10:**

**What is being asked:** Calculate (8 × 4 + 2) − (8 + 4 × 2)

**Working:**
- First parentheses: 8 × 4 + 2 = 32 + 2 = 34
- Second parentheses: 8 + 4 × 2 = 8 + 8 = 16
- Result: 34 − 16 = 18

**Answer 10: D**

---

**Problem 11:**

**What is being asked:** Convert 11 hours 5 minutes to minutes

**Working:**
- 11 hours = 11 × 60 = 660 minutes
- Total: 660 + 5 = 665 minutes

**Answer 11: C**

---

**Problem 12:**

**What is being asked:** In how many ways can the ghost enter by one window and leave by a different window?

**Working:**
- Choices for entry: 6 windows
- Choices for exit: 5 windows (must be different from entry)
- Total ways: 6 × 5 = 30

**Answer 12: D**

Problem 1:

To be certain of getting 5 socks of the same color, we use the pigeonhole principle.

There are 3 colors: red, white, and blue.  
To avoid getting 5 of any one color for as long as possible, Martian could take at most 4 of each color:

- 4 red
- 4 white
- 4 blue

That gives a total of \(4+4+4=12\) socks without yet having 5 of the same color.

The very next sock, the 13th, must make one color appear at least 5 times.

So the Martian must remove 13 socks.

ANSWER 1: D

Problem 2:

There are 4 players: Lance, Sally, Joy, and Fred.

We need to choose 3 starters from these 4 players. Since order does not matter, we count combinations:

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

Indeed, the 3-starter groups are:

- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred

So there are 4 ways.

ANSWER 2: B

Problem 3:

We check each product.

- A: \((-4)(-9)=36\)
- B: \((-3)(-12)=36\)
- C: \(\frac12 \cdot (-72)=-36\), not 36
- D: \(1\cdot 36=36\)
- E: \(\frac32 \cdot 24=36\)

Only choice C does **not** have product 36.

ANSWER 3: C

Problem 4:

First find how high the bulb is above the floor.

The ceiling is 2.4 meters above the floor, which is \(240\) centimeters.  
The bulb is 10 centimeters below the ceiling, so it is:

\[
240 - 10 = 230 \text{ cm}
\]

Now Alice’s height is 1.5 meters = 150 cm.  
She can reach 46 cm above her head, so her reach without the stool is:

\[
150 + 46 = 196 \text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230 - 196 = 34 \text{ cm}
\]

So the stool is 34 cm high.

ANSWER 4: B

Problem 5:

The sample has 30 trout out of 180 fish, so the fraction of trout is:

\[
\frac{30}{180}=\frac{1}{6}
\]

If 250 trout is \(\frac16\) of the total lake population, then the total number of fish is:

\[
250 \div \frac16 = 250 \cdot 6 = 1500
\]

So there are 1500 fish in the lake.

ANSWER 5: B

Problem 6:

Karl bought 5 folders at \$2.50 each, so the total cost was:

\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:

\[
0.20 \times 12.50 = 2.50
\]

So he could have saved \$2.50.

ANSWER 6: C

Problem 7:

He paints 300 dimples, and each dimple takes 2 seconds.

Total time in seconds:

\[
300 \times 2 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10
\]

So he needs 10 minutes.

ANSWER 7: D

Problem 8:

We want combinations of 5-dollar bills and 2-dollar bills that total \$17.

Let \(x\) be the number of \$5 bills and \(y\) the number of \$2 bills. Then:

\[
5x + 2y = 17
\]

Try values of \(x\):

- \(x=1\): \(17-5=12\), so \(y=6\)
- \(x=2\): \(17-10=7\), not even, so impossible
- \(x=3\): \(17-15=2\), so \(y=1\)

These are the only nonnegative integer solutions.

So there are 2 combinations.

ANSWER 8: A

Problem 9:

The contest ended 1000 minutes after noon.

Convert 1000 minutes to hours and minutes:

\[
1000 \div 60 = 16 \text{ hours with } 40 \text{ minutes left}
\]

So:

- Noon + 12 hours = midnight
- plus 4 more hours = 4:00 a.m.
- plus 40 minutes = 4:40 a.m.

So the contest ended at 4:40 a.m.

ANSWER 9: D

Problem 10:

Compute each part using order of operations:

\[
(8\times 4 + 2) - (8 + 4\times 2)
\]

First parentheses:

\[
8\times 4 + 2 = 32 + 2 = 34
\]

Second parentheses:

\[
8 + 4\times 2 = 8 + 8 = 16
\]

Now subtract:

\[
34 - 16 = 18
\]

So the value is 18.

ANSWER 10: D

Problem 11:

We convert 11 hours and 5 minutes into minutes.

\[
11 \times 60 = 660
\]

Then add 5 minutes:

\[
660 + 5 = 665
\]

So the match lasted 665 minutes.

ANSWER 11: C

Problem 12:

Georgie must enter by one window and leave by a different window.

There are 6 choices for the entering window.  
After choosing one, there are 5 remaining choices for the exit window.

So the total number of ways is:

\[
6 \times 5 = 30
\]

Therefore, there are 30 ways.

ANSWER 12: D

Problem 1:  
We need the minimum number of socks to remove to guarantee there are 5 socks of the same color, given 3 colors (red, white, blue) and at least 5 of each color.

Worst case: you could first pull 4 socks of each color without reaching 5 of any color.  
That would be:
- 4 red + 4 white + 4 blue = 12 socks pulled

After 12 socks, you have at most 4 of each color. The next sock (the 13th) must be one of the 3 colors, and by the pigeonhole principle it makes one color reach 5.

So the Martian must remove **13** socks.

Tempting wrong choices:
- 12 isn’t enough because it’s possible to have only 4 of each color.
- Any number less than 13 can’t force a fifth sock of some color.

ANSWER 1: D


Problem 2:  
We have 4 students: Lance, Sally, Joy, Fred. We must choose 3 starters.

Number of ways to choose 3 out of 4 is:
\[
\binom{4}{3} = 4
\]

(You can also think: leave out one student; there are 4 choices for who to leave out.)

ANSWER 2: B


Problem 3:  
We check which pair does **NOT** multiply to 36.

Choices:
A. \(\{-4,-9\}\): \((-4)(-9)=36\) ✅  
B. \(\{-3,-12\}\): \((-3)(-12)=36\) ✅  
C. \(\left\{\frac{1}{2},-72\right\}\): \(\frac{1}{2}\cdot(-72)=-36\) ❌ not 36  
D. \(\{1,36\}\): \(1\cdot 36=36\) ✅  
E. \(\left\{\frac{3}{2},24\right\}\): \(\frac{3}{2}\cdot 24=3\cdot 12=36\) ✅

So the one that does not have product 36 is choice C.

ANSWER 3: C


Problem 4:  
Let’s convert everything to the same reference (floor).

- Ceiling is 2.4 m above the floor = 240 cm  
- The bulb is 10 cm below the ceiling: height of bulb above floor:
\[
240 - 10 = 230 \text{ cm}
\]
- Alice is 1.5 m tall = 150 cm  
- She can reach 46 cm above the top of her head, so from her top to the bulb:
\[
150 + 46 = 196 \text{ cm (if she stands on the floor)}
\]

She is standing on a stool, so her total height is (stool height + 150). She can reach:
\[
(\text{stool height} + 150) + 46
\]
Set equal to bulb height 230:
\[
\text{stool height} + 196 = 230
\]
\[
\text{stool height} = 34
\]

ANSWER 4: B


Problem 5:  
We know:
- Lake has 250 trout.
- Total fish in lake = \(T\)
- Sample has 180 fish, and 30 are trout.
So sample trout-to-total ratio equals lake trout-to-total ratio:
\[
\frac{30}{180} = \frac{250}{T}
\]
Simplify \(\frac{30}{180}=\frac{1}{6}\):
\[
\frac{1}{6} = \frac{250}{T}
\]
\[
T = 250 \cdot 6 = 1500
\]

ANSWER 5: B


Problem 6:  
Cost per folder originally: $2.50  
He bought 5 folders:
\[
5 \cdot 2.50 = 12.50
\]
A 20%-off sale means paying 80% of the original price:
\[
0.80 \cdot 12.50 = 10.00
\]
Savings:
\[
12.50 - 10.00 = 2.50
\]

ANSWER 6: C


Problem 7:  
Paint 300 dimples, 2 seconds per dimple.
Total time in seconds:
\[
300 \cdot 2 = 600 \text{ seconds}
\]
Convert to minutes (60 seconds per minute):
\[
600 \div 60 = 10 \text{ minutes}
\]

ANSWER 7: D


Problem 8:  
We need numbers of $5 bills and $2 bills that total $17 (order doesn’t matter).

Let \(x\) = number of $5 bills, \(y\) = number of $2 bills.
\[
5x + 2y = 17
\]
Try \(x=0\) to \(x=3\) (since \(4\cdot 5=20>17\)):

- \(x=0\): \(2y=17\) not possible  
- \(x=1\): \(5 + 2y=17 \Rightarrow 2y=12 \Rightarrow y=6\)  
- \(x=2\): \(10 + 2y=17 \Rightarrow 2y=7\) not possible  
- \(x=3\): \(15 + 2y=17 \Rightarrow 2y=2 \Rightarrow y=1\)

So there are 2 combinations: \((x,y)=(1,6)\) and \((3,1)\).

ANSWER 8: A


Problem 9:  
The contest starts at noon. It ends 1000 minutes later.

Convert 1000 minutes to hours:
\[
1000 \div 60 = 16 \text{ hours remainder } 40 \text{ minutes}
\]
Noon + 16 hours = 4:00 a.m. (next day)  
Add 40 minutes:
\[
4:00 \text{ a.m.} + 40 \text{ min} = 4:40 \text{ a.m.}
\]

ANSWER 9: D


Problem 10:  
Compute:
\[
(8 \times 4 + 2) - (8 + 4 \times 2)
\]
First parentheses:
\[
8\times 4 = 32,\quad 32+2=34
\]
Second parentheses:
\[
4\times 2 = 8,\quad 8+8=16
\]
Now subtract:
\[
34 - 16 = 18
\]

ANSWER 10: D


Problem 11:  
11 hours and 5 minutes total.
Convert hours to minutes:
\[
11 \cdot 60 = 660
\]
Add 5 minutes:
\[
660 + 5 = 665
\]

ANSWER 11: C


Problem 12:  
There are 6 windows. Georgie enters through one window and must leave through a different window.

Choose entering window: 6 choices.  
Choose leaving window: 5 choices (cannot be the same).

Total:
\[
6 \cdot 5 = 30
\]

ANSWER 12: D

Problem 1:
To be certain of having 5 socks of the same color, we consider the "worst-case scenario" (Pigeonhole Principle). If we have 3 colors (red, white, blue), we could pick 4 red, 4 white, and 4 blue socks without having 5 of any one color. This totals $4 + 4 + 4 = 12$ socks. The very next sock (the 13th) must be either red, white, or blue, which will complete a set of 5 for that color.
ANSWER 1: D

Problem 2:
We need to choose 3 starters from a group of 4 people (Lance, Sally, Joy, Fred). This is a combination problem denoted as $\binom{4}{3}$. Choosing 3 people to be on the team is the same as choosing 1 person to be left out. There are 4 ways to choose who is left out (Lance, Sally, Joy, or Fred).
ANSWER 2: B

Problem 3:
We check the products:
A: $(-4) \times (-9) = 36$
B: $(-3) \times (-12) = 36$
C: $\frac{1}{2} \times (-72) = -36$
D: $1 \times 36 = 36$
E: $\frac{3}{2} \times 24 = 3 \times 12 = 36$
Choice C results in $-36$, not $36$.
ANSWER 3: C

Problem 4:
Ceiling height = 2.4 m = 240 cm. Light bulb height = 240 - 10 = 230 cm from the floor.
Alice's reach = 150 cm (height) + 46 cm (reach) = 196 cm.
Stool height = (Bulb height) - (Alice's reach) = 230 - 196 = 34 cm.
ANSWER 4: B

Problem 5:
Let $T$ be the total number of fish in the lake. The ratio of trout to total fish is $\frac{30}{180} = \frac{1}{6}$.
Set up the proportion: $\frac{250}{T} = \frac{1}{6}$.
$T = 250 \times 6 = 1500$.
ANSWER 5: B

Problem 6:
Total cost = 5 folders × $2.50 = $12.50.
Savings = 20% of $12.50 = 0.20 \times 12.50 = $2.50.
Alternatively, 20% off one folder is $0.20 \times 2.50 = $0.50. For 5 folders, $0.50 \times 5 = $2.50.
ANSWER 6: C

Problem 7:
Total time in seconds = 300 dimples × 2 seconds/dimple = 600 seconds.
Total time in minutes = 600 seconds / 60 seconds/minute = 10 minutes.
ANSWER 7: D

Problem 8:
Let $x$ be the number of $5 bills and $y$ be the number of $2 bills. $5x + 2y = 17$.
If $x=1$, $5+2y=17 \Rightarrow 2y=12 \Rightarrow y=6$. (1, 6)
If $x=3$, $15+2y=17 \Rightarrow 2y=2 \Rightarrow y=1$. (3, 1)
If $x=0$ or $x=2$, $2y$ would be 17 or 7, which are not divisible by 2.
There are 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours and 40 minutes (since $1000 \div 60 = 16$ remainder 40).
Noon + 12 hours = midnight.
Midnight + 4 hours and 40 minutes = 4:40 a.m.
ANSWER 9: D

Problem 10:
$(8 \times 4 + 2) - (8 + 4 \times 2) = (32 + 2) - (8 + 8) = 34 - 16 = 18$.
ANSWER 10: D

Problem 11:
11 hours = $11 \times 60 = 660$ minutes.
Total = 660 + 5 = 665 minutes.
ANSWER 11: C

Problem 12:
There are 6 choices for the entry window. Since the exit window must be different, there are 5 choices for the exit window.
Total ways = $6 \times 5 = 30$.
ANSWER 12: D

Problem 1:
This is a pigeonhole principle problem. To guarantee at least 5 socks of one color, first consider the worst-case scenario of taking as many socks as possible without having 5 of any single color. That means taking exactly 4 socks of each of the 3 colors, for a total of \(4 \times 3 = 12\) socks. The next sock (the 13th) must be a 5th sock of one of the colors. Thus 13 socks are required. (Fewer than 13 leaves open the possibility of only 4 of each color.)

ANSWER 1: D

Problem 2:
The number of ways to choose 3 starters out of the 4 available players is the combination \(\binom{4}{3} = 4\). Each choice simply leaves out one of the four players.

ANSWER 2: B

Problem 3:
Compute the product for each pair:
- A: \((-4) \times (-9) = 36\)
- B: \((-3) \times (-12) = 36\)
- C: \(\frac12 \times (-72) = -36\) (not 36)
- D: \(1 \times 36 = 36\)
- E: \(\frac32 \times 24 = 36\)

Only pair C fails to give a product of 36.

ANSWER 3: C

Problem 4:
Convert all lengths to centimeters. The ceiling is 240 cm above the floor, so the bulb is 230 cm above the floor. Alice's reach is \(150 + 46 = 196\) cm. The stool must therefore supply the remaining \(230 - 196 = 34\) cm.

ANSWER 4: B

Problem 5:
The sample shows that trout comprise \(\frac{30}{180} = \frac16\) of all fish. With 250 trout in the lake, the total number of fish is \(250 \div \frac16 = 1500\).

ANSWER 5: B

Problem 6:
Five folders cost \(5 \times 2.50 = 12.50\). A 20 % discount saves \(0.20 \times 12.50 = 2.50\).

ANSWER 6: C

Problem 7:
Painting 300 dimples at 2 seconds each requires \(300 \times 2 = 600\) seconds, or \(600 \div 60 = 10\) minutes.

ANSWER 7: D

Problem 8:
Solve \(5x + 2y = 17\) for non-negative integers \(x, y\):
- \(x = 1\), \(y = 6\) works;
- \(x = 3\), \(y = 1\) works.
No other non-negative integer solutions exist, so there are exactly two combinations.

ANSWER 8: A

Problem 9:
Divide 1000 minutes by 60: \(1000 = 16\) hours + 40 minutes. Adding 16 hours to noon yields 4:00 a.m.; adding 40 minutes yields 4:40 a.m.

ANSWER 9: D

Problem 10:
Left side: \(8 \times 4 + 2 = 34\). Right side: \(8 + 4 \times 2 = 16\). Difference: \(34 - 16 = 18\).

ANSWER 10: D

Problem 11:
Convert 11 hours 5 minutes to minutes: \(11 \times 60 + 5 = 665\).

ANSWER 11: C

Problem 12:
There are 6 choices for the entry window and then 5 remaining choices for the exit window, giving \(6 \times 5 = 30\) ordered pairs.

ANSWER 12: D

Problem 1:
The problem is asking us to determine the minimum number of socks the Martian must remove from the drawer to guarantee that there are 5 socks of the same color. 
To ensure that there are 5 socks of the same color, we must consider the worst-case scenario where the Martian draws as many socks as possible without getting 5 of the same color. 
The Martian can draw at most 4 socks of each color without having 5 socks of the same color. 
So, the Martian can draw 4 red, 4 white, and 4 blue socks, for a total of 4 + 4 + 4 = 12 socks. 
At this point, the Martian still doesn't have 5 socks of the same color. 
However, drawing one more sock will guarantee that there are 5 socks of the same color. 
Therefore, the Martian must draw at least 12 + 1 = 13 socks.
Let's examine the choices: 
A. 6 is too low because the Martian could draw 3 red, 2 white, and 1 blue, or other combinations that don't guarantee 5 socks of the same color. 
B. 9 is still too low because the Martian could draw 4 red, 3 white, and 2 blue, or other similar combinations. 
C. 12 is the maximum number of socks the Martian can draw without having 5 socks of the same color. 
D. 13 is the correct answer because it is one more than the maximum number of socks that can be drawn without having 5 socks of the same color. 
E. 15 is more than necessary.
ANSWER 1: D

Problem 2:
The problem is asking us to find the number of ways to choose 3 starters from a team of 4 players: Lance, Sally, Joy, and Fred. 
This is a combination problem, because the order in which the starters are chosen does not matter. 
The number of ways to choose 3 starters from 4 players can be calculated using the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of players and k is the number of starters. 
Here, n = 4 and k = 3, so C(4, 3) = 4! / (3!(4-3)!) = 4.
Let's list the possible combinations to verify: 
- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred
There are indeed 4 combinations.
Let's examine the choices: 
A. 2 is too low. 
B. 4 matches our calculation. 
C. 6 is too high. 
D. 8 is too high. 
E. 10 is too high.
ANSWER 2: B

Problem 3:
The problem is asking us to identify the pair of numbers that does NOT have a product equal to 36. 
Let's check each pair: 
A. {−4, −9}: (−4) × (−9) = 36
B. {−3, −12}: (−3) × (−12) = 36
C. { 1 2 , −72}: (1/2) × (−72) = −36
D. {1, 36}: 1 × 36 = 36
E. { 3 2 , 24}: (3/2) × 24 = 36
The pair that does NOT have a product equal to 36 is C.
Let's examine the choices: 
A. {−4, −9} has a product of 36. 
B. {−3, −12} has a product of 36. 
C. { 1 2 , −72} has a product of −36, not 36. 
D. {1, 36} has a product of 36. 
E. { 3 2 , 24} has a product of 36.
ANSWER 3: C

Problem 4:
The problem is asking us to find the height of the stool that Alice needs to stand on to reach the light bulb. 
First, we need to convert the height of the ceiling from meters to centimeters: 2.4 meters = 240 centimeters. 
The light bulb is 10 centimeters below the ceiling, so its height above the floor is 240 - 10 = 230 centimeters. 
Alice is 1.5 meters = 150 centimeters tall and can reach 46 centimeters above her head, so her total reach is 150 + 46 = 196 centimeters. 
To find the height of the stool, we subtract Alice's total reach from the height of the light bulb: 230 - 196 = 34 centimeters.
Let's examine the choices: 
A. 32 is too low because 230 - (196 + 32) = 2, which is not enough to reach the light bulb. 
B. 34 is the correct height. 
C. 36 is too high. 
D. 38 is too high. 
E. 40 is too high.
ANSWER 4: B

Problem 5:
The problem is asking us to find the total number of fish in the lake based on the ratio of trout in the sample to the total number of fish in the sample. 
The sample contains 30 trout out of 180 fish, so the ratio of trout to total fish in the sample is 30/180. 
We are given that this ratio is the same as the ratio of trout to total fish in the lake. 
Let's denote the total number of fish in the lake as x. 
The ratio of trout to total fish in the lake is 250/x. 
Setting the two ratios equal, we get 30/180 = 250/x. 
Cross-multiplying, we have 30x = 250 × 180. 
Simplifying, 30x = 45000. 
Dividing both sides by 30, x = 1500.
Let's examine the choices: 
A. 1250 is too low. 
B. 1500 matches our calculation. 
C. 1750 is too high. 
D. 1800 is too high. 
E. 2000 is too high.
ANSWER 5: B

Problem 6:
The problem is asking us to find out how much Karl could have saved by waiting a day to buy the folders at a 20%-off sale. 
The original price of 5 folders is 5 × $2.50 = $12.50. 
At a 20%-off sale, the discount is 20% of $12.50, which is 0.20 × $12.50 = $2.50. 
So, Karl could have saved $2.50.
Let's examine the choices: 
A. $1.00 is too low. 
B. $2.00 is too low. 
C. $2.50 matches our calculation. 
D. $2.75 is too high. 
E. $5.00 is too high.
ANSWER 6: C

Problem 7:
The problem is asking us to find the time it will take Casey to paint 300 dimples on a golf ball. 
Casey takes 2 seconds to paint one dimple, so to paint 300 dimples, he will take 300 × 2 = 600 seconds. 
To convert seconds to minutes, we divide by 60: 600 / 60 = 10 minutes.
Let's examine the choices: 
A. 4 minutes is too low. 
B. 6 minutes is too low. 
C. 8 minutes is too low. 
D. 10 minutes matches our calculation. 
E. 12 minutes is too high.
ANSWER 7: D

Problem 8:
The problem is asking us to find the number of different combinations of $5 bills and $2 bills that can be used to make a total of $17. 
Let's list the possible combinations: 
1. 3 $5 bills = $15, and $2 bill = $2, total = $17 (1 combination: 3 $5 bills and 1 $2 bill)
2. 1 $5 bill = $5, and 6 $2 bills = $12, total = $17 (1 combination: 1 $5 bill and 6 $2 bills)
We should also check if there are any other combinations: 
- Using 0 $5 bills, we need 8.5 $2 bills, which is not possible since we can't have a fraction of a bill.
- Using 2 $5 bills = $10, we need $7 more, which can be made with 3.5 $2 bills, again not possible.
- Using 4 $5 bills = $20, which is more than $17.

Problem 1:
We want the minimum number of socks to guarantee 5 of the same color. With 3 colors (red, white, blue), the worst case is picking 4 of each color: 4+4+4 = 12 socks. The 13th sock must create a group of 5 of one color. So 13 socks are needed.
ANSWER 1: D

Problem 2:
We choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways is the combination C(4,3) = 4.
ANSWER 2: B

Problem 3:
Compute each product:
A: (−4)×(−9) = 36
B: (−3)×(−12) = 36
C: (1/2)×(−72) = −36
D: 1×36 = 36
E: (3/2)×24 = 36
Only C does not equal 36.
ANSWER 3: C

Problem 4:
Ceiling = 2.4 m = 240 cm. Light bulb is 10 cm below ceiling, so its height = 230 cm.
Alice's height = 1.5 m = 150 cm. Reach above head = 46 cm, so reach without stool = 196 cm.
Stool height needed = 230 − 196 = 34 cm.
ANSWER 4: B

Problem 5:
Sample ratio: 30 trout / 180 fish = 1/6. Lake has 250 trout, which is 1/6 of total fish. Total fish = 250 × 6 = 1500.
ANSWER 5: B

Problem 6:
Cost without sale: 5 × $2.50 = $12.50.
With 20% off, each folder costs $2.50 × 0.8 = $2.00. Total = 5 × $2.00 = $10.00.
Savings = $12.50 − $10.00 = $2.50.
ANSWER 6: C

Problem 7:
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.
ANSWER 7: D

Problem 8:
Find non-negative integers x ($5 bills) and y ($2 bills) such that 5x + 2y = 17.
x=1 → 5 + 2y = 17 → y=6.
x=3 → 15 + 2y = 17 → y=1.
No other x works. So 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours 40 minutes (16×60 = 960, remainder 40).
Noon + 16 hours = 4:00 a.m. next day. Add 40 minutes → 4:40 a.m.
ANSWER 9: D

Problem 10:
(8 × 4 + 2) = 32 + 2 = 34.
(8 + 4 × 2) = 8 + 8 = 16.
34 − 16 = 18.
ANSWER 10: D

Problem 11:
11 hours × 60 = 660 minutes. Plus 5 minutes = 665 minutes.
ANSWER 11: C

Problem 12:
6 choices for entry window, then 5 choices for a different exit window. Total = 6 × 5 = 30.
ANSWER 12: D

Problem 1:
We need to find the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). Using the Pigeonhole Principle, we consider the worst-case scenario: the Martian pulls out as many socks as possible *without* getting 5 of any one color. That means getting exactly 4 of each color:
- 4 red + 4 white + 4 blue = 12 socks
After 12 socks, the Martian could have 4 of each color and still not have 5 matching. The very next sock (the 13th) must be red, white, or blue, pushing the count of that color to 5. So 13 socks guarantees 5 of the same color.
(The "five-legged" detail is flavor text explaining why the Martian needs 5 socks.)

ANSWER 1: D

Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). This is a combination problem since order doesn't matter:
C(4, 3) = 4! / (3! × 1!) = 4
We can also list them: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — exactly 4 ways.

ANSWER 2: B

Problem 3:
We check each product:
- A: (−4) × (−9) = 36 ✓
- B: (−3) × (−12) = 36 ✓
- C: (1/2) × (−72) = −36 ✗ (This equals −36, NOT 36)
- D: 1 × 36 = 36 ✓
- E: (3/2) × 24 = 36 ✓
Choice C gives −36, which does not equal 36.

ANSWER 3: C

Problem 4:
First, convert everything to centimeters:
- Ceiling height: 2.4 m = 240 cm
- Light bulb position: 240 − 10 = 230 cm above the floor
- Alice's height: 1.5 m = 150 cm
- Alice's reach above her head: 46 cm
- Alice's maximum reach (without stool): 150 + 46 = 196 cm above the floor

Let h = height of the stool. Standing on the stool, her reach is h + 196 cm. She can "just reach" the bulb:
h + 196 = 230
h = 230 − 196 = 34 cm

ANSWER 4: B

Problem 5:
The ratio of trout to total fish in the sample should equal the ratio in the lake:
- Sample ratio: 30 trout / 180 fish = 1/6
- Lake ratio: 250 trout / Total fish = 1/6

Setting up the proportion: 250 / Total = 1/6
Total = 250 × 6 = 1500 fish

ANSWER 5: B

Problem 6:
Karl bought 5 folders at $2.50 each:
Total cost = 5 × $2.50 = $12.50
A 20%-off sale means he would save 20% of the total:
Savings = 0.20 × $12.50 = $2.50

ANSWER 6: C

Problem 7:
Total time in seconds: 300 dimples × 2 seconds/dimple = 600 seconds
Convert to minutes: 600 seconds ÷ 60 seconds/minute = 10 minutes

ANSWER 7: D

Problem 8:
We need non-negative integer solutions to 5x + 2y = 17, where x = number of $5 bills and y = number of $2 bills.
- x = 0: 2y = 17 → y = 8.5 (not an integer) ✗
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 ✗
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x = 4: 20 > 17 (too large) ✗

Only 2 valid combinations: (1 five + 6 twos) and (3 fives + 1 two).

ANSWER 8: A

Problem 9:
Convert 1000 minutes to hours and minutes:
1000 ÷ 60 = 16 remainder 40, so 1000 minutes = 16 hours 40 minutes.
Starting at noon (12:00 PM):
- 12 hours later = midnight (12:00 AM)
- 4 more hours = 4:00 AM
- 40 more minutes = 4:40 AM

ANSWER 9: D

Problem 10:
Evaluate each part using order of operations (multiplication before addition):
- First part: (8 × 4 + 2) = (32 + 2) = 34
- Second part: (8 + 4 × 2) = (8 + 8) = 16
- Difference: 34 − 16 = 18

ANSWER 10: D

Problem 11:
Convert 11 hours and 5 minutes to minutes:
11 hours × 60 minutes/hour = 660 minutes
660 + 5 = 665 minutes

ANSWER 11: C

Problem 12:
Georgie enters through one window and leaves through a different one.
- Choices for entry window: 6
- Choices for exit window (must be different): 5
Total ways = 6 × 5 = 30

ANSWER 12: D

Problem 1:
The Martian needs to be **certain** of having 5 socks of the same color. To guarantee this, we must consider the worst-case scenario: the Martian picks as many socks as possible **without** yet having 5 of any single color.  
With 3 colors (red, white, blue), the worst case is picking 4 socks of each color:  
\(4 + 4 + 4 = 12\) socks.  
At this point the Martian has 4 of every color but still does not have 5 of any one color. The very next sock—the 13th—must be red, white, or blue, and will necessarily create a set of 5 of that color.  
Thus 13 socks are required.  
(Choice A is too small; with 6 socks one could have 2,2,2. Choice B allows 3,3,3. Choice C allows 4,4,4. Choice E is more than the minimum guaranteed number.)

ANSWER 1: D

---

Problem 2:
We need to choose a group of 3 starters from the 4 students: Lance, Sally, Joy, and Fred. Since the three starters form a team and order does not matter, this is a combination problem.  
\[
\binom{4}{3} = \frac{4!}{3!\,1!} = 4
\]  
Equivalently, we can think of choosing which **one** of the four students will **not** start; there are 4 such choices.  
(Choice A is too small. Choice C is \(\binom{4}{2}\). Choices D and E do not arise from counting 3-person teams from 4 people.)

ANSWER 2: B

---

Problem 3:
We check the product of each pair to see which is not 36.

- **A.** \((-4)(-9) = 36\)  
- **B.** \((-3)(-12) = 36\)  
- **C.** \(\left(\frac12\right)(-72) = -36\)  
- **D.** \((1)(36) = 36\)  
- **E.** \(\left(\frac32\right)(24) = 36\)

Only pair **C** gives \(-36\), not \(36\).

ANSWER 3: C

---

Problem 4:
First convert all measurements to centimeters.  
- Ceiling height: \(2.4\text{ m} = 240\text{ cm}\).  
- Light bulb is \(10\text{ cm}\) below the ceiling, so its height above the floor is \(240 - 10 = 230\text{ cm}\).  

Alice’s reach:  
- Height: \(1.5\text{ m} = 150\text{ cm}\).  
- She can reach \(46\text{ cm}\) above her head, so her standing reach is \(150 + 46 = 196\text{ cm}\).  

The stool must supply the remaining height:  
\[
230 - 196 = 34\text{ cm}.
\]

ANSWER 4: B

---

Problem 5:
Let \(T\) be the total number of fish in the lake. The problem states that the ratio of trout to total fish is the same in the sample and in the lake.  
\[
\frac{250}{T} = \frac{30}{180}
\]  
Simplify the right side: \(\frac{30}{180} = \frac16\).  
\[
\frac{250}{T} = \frac16 \quad\Longrightarrow\quad T = 250 \times 6 = 1500.
\]

ANSWER 5: B

---

Problem 6:
Karl bought 5 folders at \(\$2.50\) each.  
Total cost: \(5 \times \$2.50 = \$12.50\).  
A 20% discount would have saved him 20% of this total:  
\[
0.20 \times \$12.50 = \$2.50.
\]

ANSWER 6: C

---

Problem 7:
Casey paints 300 dimples at 2 seconds per dimple.  
Total time in seconds: \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(\frac{600}{60} = 10\) minutes.

ANSWER 7: D

---

Problem 8:
Let \(f\) be the number of \(\$5\) bills and \(t\) be the number of \(\$2\) bills. We need non-negative integer solutions to  
\[
5f + 2t = 17.
\]  
Test values of \(f\):  
- \(f=0\): \(2t=17\) → no integer solution.  
- \(f=1\): \(5 + 2t = 17 \Rightarrow t=6\).  **(Valid)**
- \(f=2\): \(10 + 2t = 17 \Rightarrow t=3.5\).  (Invalid)
- \(f=3\): \(15 + 2t = 17 \Rightarrow t=1\).  **(Valid)**
- \(f=4\): \(20 > 17\) → stop.

There are exactly **2** combinations: \((1\text{ five}, 6\text{ twos})\) and \((3\text{ fives}, 1\text{ two})\).

ANSWER 8: A

---

Problem 9:
The contest starts at noon (12:00 p.m.). We add 1000 minutes.  
Divide 1000 by 60:  
\[
1000 = 16 \times 60 + 40,
\]  
so 1000 minutes = 16 hours and 40 minutes.  

Noon + 12 hours = midnight.  
Midnight + 4 hours 40 minutes = 4:40 a.m.

ANSWER 9: D

---

Problem 10:
Evaluate using the correct order of operations (multiplication before addition).

First parentheses:  
\[
8 \times 4 + 2 = 32 + 2 = 34.
\]

Second parentheses:  
\[
8 + 4 \times 2 = 8 + 8 = 16.
\]

Subtract:  
\[
34 - 16 = 18.
\]

ANSWER 10: D

---

Problem 11:
Convert 11 hours to minutes and add the remaining 5 minutes.  
\[
11 \times 60 = 660\text{ minutes}.
\]  
\[
660 + 5 = 665\text{ minutes}.
\]

ANSWER 11: C

---

Problem 12:
Georgie enters through one of 6 windows and must leave through a **different** window.  
- 6 choices for the entry window.  
- For each entry choice, 5 remaining choices for the exit window.  

Total ways:  
\[
6 \times 5 = 30.
\]  
(Choice E, 36, would count entering and leaving through the same window, which is not allowed.)

ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). The worst-case scenario is to draw as many socks as possible without having 5 of any one color. That would be 4 red, 4 white, and 4 blue, for a total of 12 socks. The next sock drawn (the 13th) must be one of the three colors, giving that color 5 socks. Therefore, 13 socks are required. The five-legged Martian is irrelevant to the count.  
Answer: D

Problem 2:
We have 4 people (Lance, Sally, Joy, Fred) and need to choose 3 starters. Since the starters are not assigned to specific positions, order does not matter. The number of ways is the combination \( \binom{4}{3} = 4 \).  
Answer: B

Problem 3:
Compute each product:  
A. \((-4) \times (-9) = 36\)  
B. \((-3) \times (-12) = 36\)  
C. \(\frac{1}{2} \times (-72) = -36\)  
D. \(1 \times 36 = 36\)  
E. \(\frac{3}{2} \times 24 = 36\)  
Only pair C gives a product not equal to 36.  
Answer: C

Problem 4:
Convert all measurements to centimeters:  
Ceiling height = 2.4 m = 240 cm.  
Light bulb is 10 cm below ceiling → height = 240 – 10 = 230 cm.  
Alice’s height = 1.5 m = 150 cm.  
Her reach above her head = 46 cm, so her maximum reach without a stool = 150 + 46 = 196 cm.  
Let \(h\) be the stool height. Standing on the stool raises her reach to \(196 + h\). This must equal 230 cm:  
\(196 + h = 230\) → \(h = 34\) cm.  
Answer: B

Problem 5:
In the sample, 30 out of 180 fish are trout, so the ratio trout : total = \(30/180 = 1/6\).  
Assuming the same ratio in the lake, if the lake has 250 trout, then the total number of fish is \(250 \times 6 = 1500\).  
Answer: B

Problem 6:
Original cost for 5 folders: \(5 \times 2.50 = \$12.50\).  
A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
Thus, by waiting a day, Karl could have saved $2.50.  
Answer: C

Problem 7:
Time to paint one dimple = 2 seconds.  
Total time for 300 dimples = \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(600 \div 60 = 10\) minutes.  
Answer: D

Problem 8:
We need nonnegative integers \(a\) (number of $5 bills) and \(b\) (number of $2 bills) such that \(5a + 2b = 17\).  
Solve by testing values of \(b\):  
\(b = 1\) → \(5a = 15\) → \(a = 3\) → (3,1) works.  
\(b = 6\) → \(5a = 5\) → \(a = 1\) → (1,6) works.  
No other values give integer \(a\). Thus there are 2 combinations.  
Answer: A

Problem 9:
1000 minutes = \(1000 \div 60 = 16\) hours and \(40\) minutes (since \(60 \times 16 = 960\), remainder 40).  
Starting at noon (12:00 PM), add 12 hours → midnight (12:00 AM). Add the remaining 4 hours → 4:00 AM. Then add 40 minutes → 4:40 AM.  
Answer: D

Problem 10:
Evaluate using order of operations:  
First parentheses: \(8 \times 4 + 2 = 32 + 2 = 34\).  
Second parentheses: \(8 + 4 \times 2 = 8 + 8 = 16\).  
Subtract: \(34 - 16 = 18\).  
Answer: D

Problem 11:
11 hours = \(11 \times 60 = 660\) minutes.  
Add 5 minutes → \(660 + 5 = 665\) minutes.  
Answer: C

Problem 12:
Georgie must choose an entry window (6 choices) and a different exit window (5 choices). Since entering and leaving are distinct actions, the total number of ways is \(6 \times 5 = 30\).  
Answer: D

ANSWER 1: D
ANSWER 2: B
ANSWER 3: C
ANSWER 4: B
ANSWER 5: B
ANSWER 6: C
ANSWER 7: D
ANSWER 8: A
ANSWER 9: D
ANSWER 10: D
ANSWER 11: C
ANSWER 12: D

Problem 1:  
The Martian wants to be certain of having 5 socks of the same color. There are three colors: red, white, and blue. In the worst-case scenario, the Martian could pick 4 socks of each color before getting a fifth of any one color. That would be \(4 \times 3 = 12\) socks, and still no color has 5 socks. The next sock drawn (the 13th) must complete a set of 5 of some color, because there are only three colors. Therefore, the Martian must remove **13** socks to be certain of having 5 socks of the same color.  
ANSWER 1: D

Problem 2:  
Ms. Hamilton's class has 4 available players: Lance, Sally, Joy, and Fred. The number of ways to choose 3 starters from 4 players is given by the combination \(\binom{4}{3} = 4\). Thus, there are **4** possible sets of three starters.  
ANSWER 2: B

Problem 3:  
We check each pair's product:  
- A: \((-4) \times (-9) = 36\)  
- B: \((-3) \times (-12) = 36\)  
- C: \(\frac{1}{2} \times (-72) = -36\) (not 36)  
- D: \(1 \times 36 = 36\)  
- E: \(\frac{3}{2} \times 24 = 36\)  
Only pair **C** does not have a product of 36.  
ANSWER 3: C

Problem 4:  
The ceiling is 2.4 m above the floor, and the bulb is 10 cm (0.1 m) below the ceiling, so the bulb's height is \(2.4 - 0.1 = 2.3\) m. Alice's height is 1.5 m, and she can reach 46 cm (0.46 m) above her head, so her reach height from the floor is \(1.5 + 0.46 = 1.96\) m. When standing on a stool, her hands are at the stool height plus her reach height. Setting this equal to the bulb height:  
\[
\text{stool height} + 1.96 = 2.3 \Rightarrow \text{stool height} = 0.34 \text{ m} = 34 \text{ cm}.
\]  
ANSWER 4: B

Problem 5:  
In the sample, the ratio of trout to total fish is \(\frac{30}{180} = \frac{1}{6}\). Assuming the same ratio in the lake, let \(T\) be the total number of fish in the lake. Then \(\frac{250}{T} = \frac{1}{6}\), so \(T = 250 \times 6 = 1500\).  
ANSWER 5: B

Problem 6:  
Karl bought 5 folders at $2.50 each, so the original cost is \(5 \times 2.50 = \$12.50\). With a 20% discount, the cost becomes \(12.50 \times 0.80 = \$10.00\). The savings is \(12.50 - 10.00 = \$2.50\).  
ANSWER 6: C

Problem 7:  
Painting 300 dimples at 2 seconds each takes \(300 \times 2 = 600\) seconds. Converting to minutes: \(600 \div 60 = 10\) minutes.  
ANSWER 7: D

Problem 8:  
We need nonnegative integers \(x\) (number of $5 bills) and \(y\) (number of $2 bills) such that \(5x + 2y = 17\). Solving:  
- \(y = 1 \Rightarrow 5x = 15 \Rightarrow x = 3\)  
- \(y = 6 \Rightarrow 5x = 5 \Rightarrow x = 1\)  
- \(y = 11 \Rightarrow 5x = -5\) (invalid)  
Thus, there are **2** valid combinations: (3 fives, 1 two) and (1 five, 6 twos).  
ANSWER 8: A

Problem 9:  
1000 minutes equals \(16\) hours and \(40\) minutes (since \(16 \times 60 = 960\), remainder \(40\)). Starting at noon, adding \(16\) hours gives 4:00 AM the next day, and adding \(40\) minutes gives 4:40 AM.  
ANSWER 9: D

Problem 10:  
Compute inside parentheses:  
\(8 \times 4 + 2 = 32 + 2 = 34\)  
\(8 + 4 \times 2 = 8 + 8 = 16\)  
Then \(34 - 16 = 18\).  
ANSWER 10: D

Problem 11:  
11 hours is \(11 \times 60 = 660\) minutes. Adding 5 minutes gives \(660 + 5 = 665\) minutes.  
ANSWER 11: C

Problem 12:  
Georgie can enter through any of the 6 windows. For exiting, he must choose a different window, so there are 5 choices. The total number of ways is \(6 \times 5 = 30\).  
ANSWER 12: D

Problem 1:  
The Martian needs to ensure there are 5 socks of the same color. The worst-case scenario is pulling the maximum number of socks without 5 of any color. Since there are 3 colors, this would be 4 socks of each color (4 × 3 = 12 socks). The next sock (the 13th) must match one color, giving 5 socks of that color.  
**ANSWER 1: D**  

Problem 2:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is C(4,3) = 4.  
**ANSWER 2: B** (Note: C(4,3)=4, but rechecking: the team is chosen from 4 specific people, so combinations are 4. Answer B=4)  

Problem 3:  
Check each pair:  
A. (−4)(−9) = 36  
B. (−3)(−12) = 36  
C. (1/2)(−72) = -36 (≠36)  
D. (1)(36) = 36  
E. (3/2)(24) = 36  
The pair that does NOT equal 36 is C.  
**ANSWER 3: E** (Re-evaluated: The question asks which does NOT equal 36. Option E: (3/2)*24=36, so it is valid. Option C: (1/2)*(-72)=-36, so C is the correct choice for NOT equal. Correction: **ANSWER 3: C** [Original had a typo, but computation confirms C is the answer])  

*Correction: For Problem 3, Option C: (1/2, -72) likely means (1/2) × (-72) = -36 ≠ 36. Option E: (3/2) × 24 = 36. Thus, C is the answer.*  
**ANSWER 3: C**  

Problem 4:  
Ceiling height: 2.4 m = 240 cm. Light bulb location: 240 - 10 = 230 cm from floor. Alice’s reach: 150 cm + 46 cm = 196 cm. Stool height = 230 - 196 = 34 cm.  
**ANSWER 4: B**  

Problem 5:  
Ratio of trout in sample: 30/180 = 1/6. Total trout: 250. Total fish = 250 × 6 = 1500.  
**ANSWER 5: B**  

Problem 6:  
Original cost: 5 × $2.50 = $12.50. 20% off: $12.50 × 0.20 = $2.50 saved.  
**ANSWER 6: C** (Note: The question asks how much Karl could have saved, which is $2.50, not $2.00. Original had a mistake; correction: 20% of $12.50 is $2.50, so answer C)  

Problem 7:  
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.  
**ANSWER 7: D**  

Problem 8:  
Let $5 bills = x, $2 bills = y. 5x + 2y = 17. Possible solutions:  
- x=3, y=1 (15+2=17)  
- x=1, y=6 (5+12=17)  
Only 2 combinations.  
**ANSWER 8: A**  

Problem 9:  
1000 minutes = 16 hours 40 minutes. Starting at noon: noon + 16h40m = 4:40 a.m. next day.  
**ANSWER 9: D**  

Problem 10:  
(8×4 + 2) = 34, (8 + 4×2) = 16. 34 - 16 = 18.  
**ANSWER 10: D**  

Problem 11:  
11 hours = 660 minutes. 660 + 5 = 665 minutes.  
**ANSWER 11: C**  

Problem 12:  
Georgie can enter through 6 windows and exit through 5 others. Total ways: 6 × 5 = 30.  
**ANSWER 12: D**  

---  
**Final Answers**  
ANSWER 1: D  
ANSWER 2: B  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: B  
ANSWER 6: C  
ANSWER 7: D  
ANSWER 8: A  
ANSWER 9: D  
ANSWER 10: D  
ANSWER 11: C  
ANSWER 12: D

### Problem 1:
This is a classic pigeonhole principle problem. We need to find the minimum number of socks the Martian must pull to guarantee 5 socks of the same color, with 3 total colors. The worst-case scenario is that the Martian pulls as many socks as possible without getting 5 of any color: that is 4 socks of each color, for a total of 3*4=12 socks. The next sock pulled (the 13th) has to be one of the three colors, pushing that color to 5 socks. Tempting wrong answers: C (12 is only the number of socks that avoids 5 of the same color, not enough to guarantee the requirement), E (15 is the total if you took 5 of each color, not the minimum to guarantee).
ANSWER 1: D

---

### Problem 2:
We need to count the number of ways to choose 3 starters from 4 people, where order on the team does not matter. This is a combination problem: C(n,k) = n!/(k!(n-k)!). For n=4, k=3: C(4,3) = 4. We can also list the teams by who is left out: leaving out Lance, Sally, Joy, or Fred gives exactly 4 unique teams. Tempting wrong answer: C (6 is the count if order of selection mattered, which it does not for a team of starters).
ANSWER 2: B

---

### Problem 3:
We calculate the product of each pair to find which does not equal 36:
- A: (-4)*(-9)=36 ✔️
- B: (-3)*(-12)=36 ✔️
- C: (1/2)*(-72) = -36 ❌
- D: 1*36=36 ✔️
- E: (3/2)*24=36 ✔️
Only pair C has a product of -36, not 36.
ANSWER 3: C

---

### Problem 4:
First convert all units to centimeters to avoid conversion errors:
- Ceiling height: 2.4m = 240cm. Light bulb is 10cm below ceiling, so bulb height from floor = 240-10=230cm.
- Alice's height: 1.5m=150cm, her reach above her head is 46cm, so total reach without stool = 150+46=196cm.
- Stool height = 230-196=34cm.
Tempting wrong answers come from unit conversion mistakes, e.g., mixing up meters and centimeters.
ANSWER 4: B

---

### Problem 5:
This is a proportion problem. The ratio of trout to total fish is the same in the sample and the lake. In the sample, 30 trout out of 180 total fish = 1/6. Let T be total fish in the lake: 250/T = 1/6 → T=250*6=1500.
ANSWER 5: B

---

### Problem 6:
Karl bought 5 folders at $2.50 each, total original cost = 5*2.50=$12.50. He saves 20% of the total cost by waiting for the sale: 0.2*12.50=$2.50. This can also be calculated per folder: 20% of $2.50 is $0.50 saved per folder, 5*0.50=$2.50 total saved.
ANSWER 6: C

---

### Problem 7:
300 dimples * 2 seconds per dimple = 600 total seconds. Convert to minutes: 60 seconds per minute, so 600/60=10 minutes.
ANSWER 7: D

---

### Problem 8:
We find all non-negative integer pairs (x,y) where x=number of $5 bills, y=number of $2 bills, and 5x+2y=17. Only odd values of x work (since $5 is odd, $2 is even, we need an odd sum to get 17):
- x=1: 5+2y=17 → y=6 (valid)
- x=3: 15+2y=17 → y=1 (valid)
Even x values (0,2) leave an odd remainder that cannot be formed with $2 bills, so only 2 valid combinations.
ANSWER 8: A

---

### Problem 9:
Convert 1000 minutes to hours and minutes: 60 minutes per hour, 1000/60=16 hours and 40 minutes. Adding this to noon (12pm): 12pm +16h40m = 4:40am the next day.
ANSWER 9: D

---

### Problem 10:
Use order of operations (PEMDAS) to calculate:
First parenthesis: 8*4 +2 = 32+2=34
Second parenthesis: 8 +4*2=8+8=16
Subtract: 34-16=18. The tempting wrong answer C (10) comes from incorrectly adding before multiplying in the second parenthesis.
ANSWER 10: D

---

### Problem 11:
Convert 11 hours 5 minutes to total minutes: 11*60=660 minutes, plus 5 extra minutes = 665 minutes.
ANSWER 11: C

---

### Problem 12:
Georgie has 6 choices for an entry window, and 5 remaining choices for an exit window (must use a different window). Total ways = 6*5=30. Tempting wrong answer E (36) counts entering and leaving through the same window, which is not allowed; B (15) incorrectly treats entry/exit as unordered, which they are not.
ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color (red, white, blue). This is a pigeonhole principle problem: the worst-case scenario is pulling the maximum number of socks without getting 5 of any color. For each of the 3 colors, we can pull 4 socks (since 4 < 5), so 4×3=12 socks. The next sock pulled will necessarily be the 5th of one color. So total socks needed: 12 + 1 = 13.
ANSWER 1: D
Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). Since the problem does not mention distinct positions for starters, order does not matter, so we use combinations. The number of ways to choose 3 people from 4 is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n=4 \) and \( k=3 \):
\( \binom{4}{3} = \frac{4!}{3!1!} = 4 \). This is equivalent to counting how many people we leave out (4 choices, one for each person not selected).
ANSWER 2: B
Problem 3:
We check the product of each pair:
- A: \( (-4) \times (-9) = 36 \)
- B: \( (-3) \times (-12) = 36 \)
- C: \( \frac{1}{2} \times (-72) = -36 \neq 36 \)
- D: \( 1 \times 36 = 36 \)
- E: \( \frac{3}{2} \times 24 = 36 \)
Only pair C does not have a product of 36.
ANSWER 3: C
Problem 4:
Convert all units to centimeters:
- Ceiling height: \( 2.4 \, \text{m} = 240 \, \text{cm} \)
- Light bulb height from floor: \( 240 - 10 = 230 \, \text{cm} \)
- Alice's reach from floor (without stool): \( 1.5 \, \text{m} + 0.46 \, \text{m} = 150 + 46 = 196 \, \text{cm} \)
Let \( h \) be the stool height. The total reach with the stool equals the light bulb height:
\( h + 196 = 230 \)
\( h = 230 - 196 = 34 \, \text{cm} \)
ANSWER 4: B
Problem 5:
The ratio of trout to total fish in the sample equals the ratio in the lake. The sample has \( \frac{30}{180} = \frac{1}{6} \) trout. Let \( T \) be the total number of fish in the lake. We know there are 250 trout, so:
\( \frac{250}{T} = \frac{1}{6} \)
\( T = 250 \times 6 = 1500 \)
ANSWER 5: B
Problem 6:
Karl bought 5 folders at $2.50 each. The 20% discount applies to each folder, so the savings per folder is \( 2.50 \times 0.20 = 0.50 \). Total savings for 5 folders: \( 5 \times 0.50 = 2.50 \).
ANSWER 6: C
Problem 7:
Total time to paint 300 dimples: \( 300 \times 2 = 600 \) seconds. Convert to minutes: \( 600 \div 60 = 10 \) minutes.
ANSWER 7: D
Problem 8:
Let \( x \) = number of $5 bills, \( y \) = number of $2 bills. We need \( 5x + 2y = 17 \), where \( x,y \geq 0 \) integers. Test non-negative integer values of \( x \):
- \( x=0 \): \( 2y=17 \) → \( y=8.5 \) (not integer, invalid)
- \( x=1 \): \( 5 + 2y=17 \) → \( 2y=12 \) → \( y=6 \) (valid: 1 $5 + 6 $2 = $17)
- \( x=2 \): \( 10 + 2y=17 \) → \( 2y=7 \) → \( y=3.5 \) (invalid)
- \( x=3 \): \( 15 + 2y=17 \) → \( 2y=2 \) → \( y=1 \) (valid: 3 $5 + 1 $2 = $17)
- \( x=4 \): \( 20 > 17 \) (stop, no more valid \( x \))
Only 2 valid combinations.
ANSWER 8: A
Problem 9:
Convert 1000 minutes to hours and minutes: \( 1000 \div 60 = 16 \) hours with a remainder of 40 minutes (since \( 16 \times 60 = 960 \), \( 1000 - 960 = 40 \)). The contest starts at noon, so adding 16 hours 40 minutes: noon + 12 hours = midnight, plus 4 hours 40 minutes = 4:40 a.m.
ANSWER 9: D
Problem 10:
Use order of operations (multiplication before addition):
First parentheses: \( 8 \times 4 + 2 = 32 + 2 = 34 \)
Second parentheses: \( 8 + 4 \times 2 = 8 + 8 = 16 \)
Subtract: \( 34 - 16 = 18 \)
ANSWER 10: D
Problem 11:
Convert 11 hours 5 minutes to total minutes: \( 11 \times 60 + 5 = 660 + 5 = 665 \)
ANSWER 11: C
Problem 12:
Georgie chooses an entrance window first (6 options), then a different exit window (5 remaining options, since exit cannot equal entrance). Since entrance and exit are distinct actions, order matters, so total ways: \( 6 \times 5 = 30 \).
ANSWER 12: D

I'll work through each problem systematically.

**Problem 1:**

**What is being asked:** How many socks must be removed to guarantee 5 socks of the same color?

**Setup:** There are 3 colors (red, white, blue). In the worst case scenario, the Martian pulls out socks as evenly distributed among colors as possible before getting 5 of one color.

**Working:** 
- Worst case: 4 red, 4 white, 4 blue = 12 socks (still no color with 5)
- The 13th sock must give us a 5th sock of some color

**Answer 1: D**

---

**Problem 2:**

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

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

**Working:**
- C(4,3) = 4 ways
- The 4 ways correspond to leaving out each person: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred}

**Answer 2: B**

---

**Problem 3:**

**What is being asked:** Which pair does NOT have a product equal to 36?

**Working:** Check each pair:
- A. {−4, −9}: (−4)(−9) = 36 ✓
- B. {−3, −12}: (−3)(−12) = 36 ✓
- C. {1/2, −72}: (1/2)(−72) = −36 ✗
- D. {1, 36}: (1)(36) = 36 ✓
- E. {3/2, 24}: (3/2)(24) = 36 ✓

**Answer 3: C**

---

**Problem 4:**

**What is being asked:** How high is the stool?

**Setup:** Convert all measurements to centimeters:
- Ceiling height: 240 cm
- Light bulb location: 240 − 10 = 230 cm from floor
- Alice's height: 150 cm
- Alice's reach above her head: 46 cm
- Alice's reach without stool: 150 + 46 = 196 cm

**Working:**
- With stool, Alice can just reach the bulb at 230 cm
- Stool height + 196 cm = 230 cm
- Stool height = 230 − 196 = 34 cm

**Answer 4: B**

---

**Problem 5:**

**What is being asked:** How many fish total are in the lake?

**Setup:** Use proportions. In the sample: 30 trout out of 180 fish. In the lake: 250 trout out of total fish.

**Working:**
- 30/180 = 250/x
- 1/6 = 250/x
- x = 250 × 6 = 1500 fish

**Answer 5: B**

---

**Problem 6:**

**What is being asked:** How much could Karl have saved with the 20% off sale?

**Setup:** 
- Original cost: 5 folders × $2.50 = $12.50
- Sale discount: 20% off

**Working:**
- 20% of $12.50 = 0.20 × $12.50 = $2.50

**Answer 6: C**

---

**Problem 7:**

**What is being asked:** How many minutes to paint 300 dimples at 2 seconds each?

**Working:**
- Total time: 300 × 2 = 600 seconds
- Convert to minutes: 600 ÷ 60 = 10 minutes

**Answer 7: D**

---

**Problem 8:**

**What is being asked:** How many combinations of $5 bills and $2 bills make $17?

**Setup:** Let x = number of $5 bills, y = number of $2 bills
- 5x + 2y = 17

**Working:**
- x = 0: 2y = 17 → y = 8.5 (not integer)
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 (not integer)
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x ≥ 4: 5x ≥ 20 > 17 (impossible)

Valid combinations: (1,$5 and 6,$2) and (3,$5 and 1,$2) = 2 combinations

**Answer 8: A**

---

**Problem 9:**

**What is being asked:** When did the contest end, starting at noon and lasting 1000 minutes?

**Working:**
- 1000 minutes ÷ 60 = 16 hours and 40 minutes
- Noon + 16 hours 40 minutes = 4:40 p.m. the same day
- But we need to continue: 4:40 p.m. + additional time consideration
- Actually: 1000 ÷ 60 = 16 remainder 40, so 16 hours 40 minutes
- Noon + 16 hours 40 minutes = 4:40 a.m. the next day

**Answer 9: D**

---

**Problem 10:**

**What is being asked:** Calculate (8 × 4 + 2) − (8 + 4 × 2)

**Working:**
- First parentheses: 8 × 4 + 2 = 32 + 2 = 34
- Second parentheses: 8 + 4 × 2 = 8 + 8 = 16
- Result: 34 − 16 = 18

**Answer 10: D**

---

**Problem 11:**

**What is being asked:** Convert 11 hours 5 minutes to minutes

**Working:**
- 11 hours = 11 × 60 = 660 minutes
- Total: 660 + 5 = 665 minutes

**Answer 11: C**

---

**Problem 12:**

**What is being asked:** In how many ways can the ghost enter by one window and leave by a different window?

**Working:**
- Choices for entry: 6 windows
- Choices for exit: 5 windows (must be different from entry)
- Total ways: 6 × 5 = 30

**Answer 12: D**

Problem 1:

To be certain of getting 5 socks of the same color, we use the pigeonhole principle.

There are 3 colors: red, white, and blue.  
To avoid getting 5 of any one color for as long as possible, Martian could take at most 4 of each color:

- 4 red
- 4 white
- 4 blue

That gives a total of \(4+4+4=12\) socks without yet having 5 of the same color.

The very next sock, the 13th, must make one color appear at least 5 times.

So the Martian must remove 13 socks.

ANSWER 1: D

Problem 2:

There are 4 players: Lance, Sally, Joy, and Fred.

We need to choose 3 starters from these 4 players. Since order does not matter, we count combinations:

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

Indeed, the 3-starter groups are:

- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred

So there are 4 ways.

ANSWER 2: B

Problem 3:

We check each product.

- A: \((-4)(-9)=36\)
- B: \((-3)(-12)=36\)
- C: \(\frac12 \cdot (-72)=-36\), not 36
- D: \(1\cdot 36=36\)
- E: \(\frac32 \cdot 24=36\)

Only choice C does **not** have product 36.

ANSWER 3: C

Problem 4:

First find how high the bulb is above the floor.

The ceiling is 2.4 meters above the floor, which is \(240\) centimeters.  
The bulb is 10 centimeters below the ceiling, so it is:

\[
240 - 10 = 230 \text{ cm}
\]

Now Alice’s height is 1.5 meters = 150 cm.  
She can reach 46 cm above her head, so her reach without the stool is:

\[
150 + 46 = 196 \text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230 - 196 = 34 \text{ cm}
\]

So the stool is 34 cm high.

ANSWER 4: B

Problem 5:

The sample has 30 trout out of 180 fish, so the fraction of trout is:

\[
\frac{30}{180}=\frac{1}{6}
\]

If 250 trout is \(\frac16\) of the total lake population, then the total number of fish is:

\[
250 \div \frac16 = 250 \cdot 6 = 1500
\]

So there are 1500 fish in the lake.

ANSWER 5: B

Problem 6:

Karl bought 5 folders at \$2.50 each, so the total cost was:

\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:

\[
0.20 \times 12.50 = 2.50
\]

So he could have saved \$2.50.

ANSWER 6: C

Problem 7:

He paints 300 dimples, and each dimple takes 2 seconds.

Total time in seconds:

\[
300 \times 2 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10
\]

So he needs 10 minutes.

ANSWER 7: D

Problem 8:

We want combinations of 5-dollar bills and 2-dollar bills that total \$17.

Let \(x\) be the number of \$5 bills and \(y\) the number of \$2 bills. Then:

\[
5x + 2y = 17
\]

Try values of \(x\):

- \(x=1\): \(17-5=12\), so \(y=6\)
- \(x=2\): \(17-10=7\), not even, so impossible
- \(x=3\): \(17-15=2\), so \(y=1\)

These are the only nonnegative integer solutions.

So there are 2 combinations.

ANSWER 8: A

Problem 9:

The contest ended 1000 minutes after noon.

Convert 1000 minutes to hours and minutes:

\[
1000 \div 60 = 16 \text{ hours with } 40 \text{ minutes left}
\]

So:

- Noon + 12 hours = midnight
- plus 4 more hours = 4:00 a.m.
- plus 40 minutes = 4:40 a.m.

So the contest ended at 4:40 a.m.

ANSWER 9: D

Problem 10:

Compute each part using order of operations:

\[
(8\times 4 + 2) - (8 + 4\times 2)
\]

First parentheses:

\[
8\times 4 + 2 = 32 + 2 = 34
\]

Second parentheses:

\[
8 + 4\times 2 = 8 + 8 = 16
\]

Now subtract:

\[
34 - 16 = 18
\]

So the value is 18.

ANSWER 10: D

Problem 11:

We convert 11 hours and 5 minutes into minutes.

\[
11 \times 60 = 660
\]

Then add 5 minutes:

\[
660 + 5 = 665
\]

So the match lasted 665 minutes.

ANSWER 11: C

Problem 12:

Georgie must enter by one window and leave by a different window.

There are 6 choices for the entering window.  
After choosing one, there are 5 remaining choices for the exit window.

So the total number of ways is:

\[
6 \times 5 = 30
\]

Therefore, there are 30 ways.

ANSWER 12: D

Problem 1:  
We need the minimum number of socks to remove to guarantee there are 5 socks of the same color, given 3 colors (red, white, blue) and at least 5 of each color.

Worst case: you could first pull 4 socks of each color without reaching 5 of any color.  
That would be:
- 4 red + 4 white + 4 blue = 12 socks pulled

After 12 socks, you have at most 4 of each color. The next sock (the 13th) must be one of the 3 colors, and by the pigeonhole principle it makes one color reach 5.

So the Martian must remove **13** socks.

Tempting wrong choices:
- 12 isn’t enough because it’s possible to have only 4 of each color.
- Any number less than 13 can’t force a fifth sock of some color.

ANSWER 1: D


Problem 2:  
We have 4 students: Lance, Sally, Joy, Fred. We must choose 3 starters.

Number of ways to choose 3 out of 4 is:
\[
\binom{4}{3} = 4
\]

(You can also think: leave out one student; there are 4 choices for who to leave out.)

ANSWER 2: B


Problem 3:  
We check which pair does **NOT** multiply to 36.

Choices:
A. \(\{-4,-9\}\): \((-4)(-9)=36\) ✅  
B. \(\{-3,-12\}\): \((-3)(-12)=36\) ✅  
C. \(\left\{\frac{1}{2},-72\right\}\): \(\frac{1}{2}\cdot(-72)=-36\) ❌ not 36  
D. \(\{1,36\}\): \(1\cdot 36=36\) ✅  
E. \(\left\{\frac{3}{2},24\right\}\): \(\frac{3}{2}\cdot 24=3\cdot 12=36\) ✅

So the one that does not have product 36 is choice C.

ANSWER 3: C


Problem 4:  
Let’s convert everything to the same reference (floor).

- Ceiling is 2.4 m above the floor = 240 cm  
- The bulb is 10 cm below the ceiling: height of bulb above floor:
\[
240 - 10 = 230 \text{ cm}
\]
- Alice is 1.5 m tall = 150 cm  
- She can reach 46 cm above the top of her head, so from her top to the bulb:
\[
150 + 46 = 196 \text{ cm (if she stands on the floor)}
\]

She is standing on a stool, so her total height is (stool height + 150). She can reach:
\[
(\text{stool height} + 150) + 46
\]
Set equal to bulb height 230:
\[
\text{stool height} + 196 = 230
\]
\[
\text{stool height} = 34
\]

ANSWER 4: B


Problem 5:  
We know:
- Lake has 250 trout.
- Total fish in lake = \(T\)
- Sample has 180 fish, and 30 are trout.
So sample trout-to-total ratio equals lake trout-to-total ratio:
\[
\frac{30}{180} = \frac{250}{T}
\]
Simplify \(\frac{30}{180}=\frac{1}{6}\):
\[
\frac{1}{6} = \frac{250}{T}
\]
\[
T = 250 \cdot 6 = 1500
\]

ANSWER 5: B


Problem 6:  
Cost per folder originally: $2.50  
He bought 5 folders:
\[
5 \cdot 2.50 = 12.50
\]
A 20%-off sale means paying 80% of the original price:
\[
0.80 \cdot 12.50 = 10.00
\]
Savings:
\[
12.50 - 10.00 = 2.50
\]

ANSWER 6: C


Problem 7:  
Paint 300 dimples, 2 seconds per dimple.
Total time in seconds:
\[
300 \cdot 2 = 600 \text{ seconds}
\]
Convert to minutes (60 seconds per minute):
\[
600 \div 60 = 10 \text{ minutes}
\]

ANSWER 7: D


Problem 8:  
We need numbers of $5 bills and $2 bills that total $17 (order doesn’t matter).

Let \(x\) = number of $5 bills, \(y\) = number of $2 bills.
\[
5x + 2y = 17
\]
Try \(x=0\) to \(x=3\) (since \(4\cdot 5=20>17\)):

- \(x=0\): \(2y=17\) not possible  
- \(x=1\): \(5 + 2y=17 \Rightarrow 2y=12 \Rightarrow y=6\)  
- \(x=2\): \(10 + 2y=17 \Rightarrow 2y=7\) not possible  
- \(x=3\): \(15 + 2y=17 \Rightarrow 2y=2 \Rightarrow y=1\)

So there are 2 combinations: \((x,y)=(1,6)\) and \((3,1)\).

ANSWER 8: A


Problem 9:  
The contest starts at noon. It ends 1000 minutes later.

Convert 1000 minutes to hours:
\[
1000 \div 60 = 16 \text{ hours remainder } 40 \text{ minutes}
\]
Noon + 16 hours = 4:00 a.m. (next day)  
Add 40 minutes:
\[
4:00 \text{ a.m.} + 40 \text{ min} = 4:40 \text{ a.m.}
\]

ANSWER 9: D


Problem 10:  
Compute:
\[
(8 \times 4 + 2) - (8 + 4 \times 2)
\]
First parentheses:
\[
8\times 4 = 32,\quad 32+2=34
\]
Second parentheses:
\[
4\times 2 = 8,\quad 8+8=16
\]
Now subtract:
\[
34 - 16 = 18
\]

ANSWER 10: D


Problem 11:  
11 hours and 5 minutes total.
Convert hours to minutes:
\[
11 \cdot 60 = 660
\]
Add 5 minutes:
\[
660 + 5 = 665
\]

ANSWER 11: C


Problem 12:  
There are 6 windows. Georgie enters through one window and must leave through a different window.

Choose entering window: 6 choices.  
Choose leaving window: 5 choices (cannot be the same).

Total:
\[
6 \cdot 5 = 30
\]

ANSWER 12: D

Problem 1:
To be certain of having 5 socks of the same color, we consider the "worst-case scenario" (Pigeonhole Principle). If we have 3 colors (red, white, blue), we could pick 4 red, 4 white, and 4 blue socks without having 5 of any one color. This totals $4 + 4 + 4 = 12$ socks. The very next sock (the 13th) must be either red, white, or blue, which will complete a set of 5 for that color.
ANSWER 1: D

Problem 2:
We need to choose 3 starters from a group of 4 people (Lance, Sally, Joy, Fred). This is a combination problem denoted as $\binom{4}{3}$. Choosing 3 people to be on the team is the same as choosing 1 person to be left out. There are 4 ways to choose who is left out (Lance, Sally, Joy, or Fred).
ANSWER 2: B

Problem 3:
We check the products:
A: $(-4) \times (-9) = 36$
B: $(-3) \times (-12) = 36$
C: $\frac{1}{2} \times (-72) = -36$
D: $1 \times 36 = 36$
E: $\frac{3}{2} \times 24 = 3 \times 12 = 36$
Choice C results in $-36$, not $36$.
ANSWER 3: C

Problem 4:
Ceiling height = 2.4 m = 240 cm. Light bulb height = 240 - 10 = 230 cm from the floor.
Alice's reach = 150 cm (height) + 46 cm (reach) = 196 cm.
Stool height = (Bulb height) - (Alice's reach) = 230 - 196 = 34 cm.
ANSWER 4: B

Problem 5:
Let $T$ be the total number of fish in the lake. The ratio of trout to total fish is $\frac{30}{180} = \frac{1}{6}$.
Set up the proportion: $\frac{250}{T} = \frac{1}{6}$.
$T = 250 \times 6 = 1500$.
ANSWER 5: B

Problem 6:
Total cost = 5 folders × $2.50 = $12.50.
Savings = 20% of $12.50 = 0.20 \times 12.50 = $2.50.
Alternatively, 20% off one folder is $0.20 \times 2.50 = $0.50. For 5 folders, $0.50 \times 5 = $2.50.
ANSWER 6: C

Problem 7:
Total time in seconds = 300 dimples × 2 seconds/dimple = 600 seconds.
Total time in minutes = 600 seconds / 60 seconds/minute = 10 minutes.
ANSWER 7: D

Problem 8:
Let $x$ be the number of $5 bills and $y$ be the number of $2 bills. $5x + 2y = 17$.
If $x=1$, $5+2y=17 \Rightarrow 2y=12 \Rightarrow y=6$. (1, 6)
If $x=3$, $15+2y=17 \Rightarrow 2y=2 \Rightarrow y=1$. (3, 1)
If $x=0$ or $x=2$, $2y$ would be 17 or 7, which are not divisible by 2.
There are 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours and 40 minutes (since $1000 \div 60 = 16$ remainder 40).
Noon + 12 hours = midnight.
Midnight + 4 hours and 40 minutes = 4:40 a.m.
ANSWER 9: D

Problem 10:
$(8 \times 4 + 2) - (8 + 4 \times 2) = (32 + 2) - (8 + 8) = 34 - 16 = 18$.
ANSWER 10: D

Problem 11:
11 hours = $11 \times 60 = 660$ minutes.
Total = 660 + 5 = 665 minutes.
ANSWER 11: C

Problem 12:
There are 6 choices for the entry window. Since the exit window must be different, there are 5 choices for the exit window.
Total ways = $6 \times 5 = 30$.
ANSWER 12: D

Problem 1:
This is a pigeonhole principle problem. To guarantee at least 5 socks of one color, first consider the worst-case scenario of taking as many socks as possible without having 5 of any single color. That means taking exactly 4 socks of each of the 3 colors, for a total of \(4 \times 3 = 12\) socks. The next sock (the 13th) must be a 5th sock of one of the colors. Thus 13 socks are required. (Fewer than 13 leaves open the possibility of only 4 of each color.)

ANSWER 1: D

Problem 2:
The number of ways to choose 3 starters out of the 4 available players is the combination \(\binom{4}{3} = 4\). Each choice simply leaves out one of the four players.

ANSWER 2: B

Problem 3:
Compute the product for each pair:
- A: \((-4) \times (-9) = 36\)
- B: \((-3) \times (-12) = 36\)
- C: \(\frac12 \times (-72) = -36\) (not 36)
- D: \(1 \times 36 = 36\)
- E: \(\frac32 \times 24 = 36\)

Only pair C fails to give a product of 36.

ANSWER 3: C

Problem 4:
Convert all lengths to centimeters. The ceiling is 240 cm above the floor, so the bulb is 230 cm above the floor. Alice's reach is \(150 + 46 = 196\) cm. The stool must therefore supply the remaining \(230 - 196 = 34\) cm.

ANSWER 4: B

Problem 5:
The sample shows that trout comprise \(\frac{30}{180} = \frac16\) of all fish. With 250 trout in the lake, the total number of fish is \(250 \div \frac16 = 1500\).

ANSWER 5: B

Problem 6:
Five folders cost \(5 \times 2.50 = 12.50\). A 20 % discount saves \(0.20 \times 12.50 = 2.50\).

ANSWER 6: C

Problem 7:
Painting 300 dimples at 2 seconds each requires \(300 \times 2 = 600\) seconds, or \(600 \div 60 = 10\) minutes.

ANSWER 7: D

Problem 8:
Solve \(5x + 2y = 17\) for non-negative integers \(x, y\):
- \(x = 1\), \(y = 6\) works;
- \(x = 3\), \(y = 1\) works.
No other non-negative integer solutions exist, so there are exactly two combinations.

ANSWER 8: A

Problem 9:
Divide 1000 minutes by 60: \(1000 = 16\) hours + 40 minutes. Adding 16 hours to noon yields 4:00 a.m.; adding 40 minutes yields 4:40 a.m.

ANSWER 9: D

Problem 10:
Left side: \(8 \times 4 + 2 = 34\). Right side: \(8 + 4 \times 2 = 16\). Difference: \(34 - 16 = 18\).

ANSWER 10: D

Problem 11:
Convert 11 hours 5 minutes to minutes: \(11 \times 60 + 5 = 665\).

ANSWER 11: C

Problem 12:
There are 6 choices for the entry window and then 5 remaining choices for the exit window, giving \(6 \times 5 = 30\) ordered pairs.

ANSWER 12: D

Problem 1:
The problem is asking us to determine the minimum number of socks the Martian must remove from the drawer to guarantee that there are 5 socks of the same color. 
To ensure that there are 5 socks of the same color, we must consider the worst-case scenario where the Martian draws as many socks as possible without getting 5 of the same color. 
The Martian can draw at most 4 socks of each color without having 5 socks of the same color. 
So, the Martian can draw 4 red, 4 white, and 4 blue socks, for a total of 4 + 4 + 4 = 12 socks. 
At this point, the Martian still doesn't have 5 socks of the same color. 
However, drawing one more sock will guarantee that there are 5 socks of the same color. 
Therefore, the Martian must draw at least 12 + 1 = 13 socks.
Let's examine the choices: 
A. 6 is too low because the Martian could draw 3 red, 2 white, and 1 blue, or other combinations that don't guarantee 5 socks of the same color. 
B. 9 is still too low because the Martian could draw 4 red, 3 white, and 2 blue, or other similar combinations. 
C. 12 is the maximum number of socks the Martian can draw without having 5 socks of the same color. 
D. 13 is the correct answer because it is one more than the maximum number of socks that can be drawn without having 5 socks of the same color. 
E. 15 is more than necessary.
ANSWER 1: D

Problem 2:
The problem is asking us to find the number of ways to choose 3 starters from a team of 4 players: Lance, Sally, Joy, and Fred. 
This is a combination problem, because the order in which the starters are chosen does not matter. 
The number of ways to choose 3 starters from 4 players can be calculated using the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of players and k is the number of starters. 
Here, n = 4 and k = 3, so C(4, 3) = 4! / (3!(4-3)!) = 4.
Let's list the possible combinations to verify: 
- Lance, Sally, Joy
- Lance, Sally, Fred
- Lance, Joy, Fred
- Sally, Joy, Fred
There are indeed 4 combinations.
Let's examine the choices: 
A. 2 is too low. 
B. 4 matches our calculation. 
C. 6 is too high. 
D. 8 is too high. 
E. 10 is too high.
ANSWER 2: B

Problem 3:
The problem is asking us to identify the pair of numbers that does NOT have a product equal to 36. 
Let's check each pair: 
A. {−4, −9}: (−4) × (−9) = 36
B. {−3, −12}: (−3) × (−12) = 36
C. { 1 2 , −72}: (1/2) × (−72) = −36
D. {1, 36}: 1 × 36 = 36
E. { 3 2 , 24}: (3/2) × 24 = 36
The pair that does NOT have a product equal to 36 is C.
Let's examine the choices: 
A. {−4, −9} has a product of 36. 
B. {−3, −12} has a product of 36. 
C. { 1 2 , −72} has a product of −36, not 36. 
D. {1, 36} has a product of 36. 
E. { 3 2 , 24} has a product of 36.
ANSWER 3: C

Problem 4:
The problem is asking us to find the height of the stool that Alice needs to stand on to reach the light bulb. 
First, we need to convert the height of the ceiling from meters to centimeters: 2.4 meters = 240 centimeters. 
The light bulb is 10 centimeters below the ceiling, so its height above the floor is 240 - 10 = 230 centimeters. 
Alice is 1.5 meters = 150 centimeters tall and can reach 46 centimeters above her head, so her total reach is 150 + 46 = 196 centimeters. 
To find the height of the stool, we subtract Alice's total reach from the height of the light bulb: 230 - 196 = 34 centimeters.
Let's examine the choices: 
A. 32 is too low because 230 - (196 + 32) = 2, which is not enough to reach the light bulb. 
B. 34 is the correct height. 
C. 36 is too high. 
D. 38 is too high. 
E. 40 is too high.
ANSWER 4: B

Problem 5:
The problem is asking us to find the total number of fish in the lake based on the ratio of trout in the sample to the total number of fish in the sample. 
The sample contains 30 trout out of 180 fish, so the ratio of trout to total fish in the sample is 30/180. 
We are given that this ratio is the same as the ratio of trout to total fish in the lake. 
Let's denote the total number of fish in the lake as x. 
The ratio of trout to total fish in the lake is 250/x. 
Setting the two ratios equal, we get 30/180 = 250/x. 
Cross-multiplying, we have 30x = 250 × 180. 
Simplifying, 30x = 45000. 
Dividing both sides by 30, x = 1500.
Let's examine the choices: 
A. 1250 is too low. 
B. 1500 matches our calculation. 
C. 1750 is too high. 
D. 1800 is too high. 
E. 2000 is too high.
ANSWER 5: B

Problem 6:
The problem is asking us to find out how much Karl could have saved by waiting a day to buy the folders at a 20%-off sale. 
The original price of 5 folders is 5 × $2.50 = $12.50. 
At a 20%-off sale, the discount is 20% of $12.50, which is 0.20 × $12.50 = $2.50. 
So, Karl could have saved $2.50.
Let's examine the choices: 
A. $1.00 is too low. 
B. $2.00 is too low. 
C. $2.50 matches our calculation. 
D. $2.75 is too high. 
E. $5.00 is too high.
ANSWER 6: C

Problem 7:
The problem is asking us to find the time it will take Casey to paint 300 dimples on a golf ball. 
Casey takes 2 seconds to paint one dimple, so to paint 300 dimples, he will take 300 × 2 = 600 seconds. 
To convert seconds to minutes, we divide by 60: 600 / 60 = 10 minutes.
Let's examine the choices: 
A. 4 minutes is too low. 
B. 6 minutes is too low. 
C. 8 minutes is too low. 
D. 10 minutes matches our calculation. 
E. 12 minutes is too high.
ANSWER 7: D

Problem 8:
The problem is asking us to find the number of different combinations of $5 bills and $2 bills that can be used to make a total of $17. 
Let's list the possible combinations: 
1. 3 $5 bills = $15, and $2 bill = $2, total = $17 (1 combination: 3 $5 bills and 1 $2 bill)
2. 1 $5 bill = $5, and 6 $2 bills = $12, total = $17 (1 combination: 1 $5 bill and 6 $2 bills)
We should also check if there are any other combinations: 
- Using 0 $5 bills, we need 8.5 $2 bills, which is not possible since we can't have a fraction of a bill.
- Using 2 $5 bills = $10, we need $7 more, which can be made with 3.5 $2 bills, again not possible.
- Using 4 $5 bills = $20, which is more than $17.

Problem 1:
We want the minimum number of socks to guarantee 5 of the same color. With 3 colors (red, white, blue), the worst case is picking 4 of each color: 4+4+4 = 12 socks. The 13th sock must create a group of 5 of one color. So 13 socks are needed.
ANSWER 1: D

Problem 2:
We choose 3 starters from 4 people (Lance, Sally, Joy, Fred). The number of ways is the combination C(4,3) = 4.
ANSWER 2: B

Problem 3:
Compute each product:
A: (−4)×(−9) = 36
B: (−3)×(−12) = 36
C: (1/2)×(−72) = −36
D: 1×36 = 36
E: (3/2)×24 = 36
Only C does not equal 36.
ANSWER 3: C

Problem 4:
Ceiling = 2.4 m = 240 cm. Light bulb is 10 cm below ceiling, so its height = 230 cm.
Alice's height = 1.5 m = 150 cm. Reach above head = 46 cm, so reach without stool = 196 cm.
Stool height needed = 230 − 196 = 34 cm.
ANSWER 4: B

Problem 5:
Sample ratio: 30 trout / 180 fish = 1/6. Lake has 250 trout, which is 1/6 of total fish. Total fish = 250 × 6 = 1500.
ANSWER 5: B

Problem 6:
Cost without sale: 5 × $2.50 = $12.50.
With 20% off, each folder costs $2.50 × 0.8 = $2.00. Total = 5 × $2.00 = $10.00.
Savings = $12.50 − $10.00 = $2.50.
ANSWER 6: C

Problem 7:
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.
ANSWER 7: D

Problem 8:
Find non-negative integers x ($5 bills) and y ($2 bills) such that 5x + 2y = 17.
x=1 → 5 + 2y = 17 → y=6.
x=3 → 15 + 2y = 17 → y=1.
No other x works. So 2 combinations.
ANSWER 8: A

Problem 9:
1000 minutes = 16 hours 40 minutes (16×60 = 960, remainder 40).
Noon + 16 hours = 4:00 a.m. next day. Add 40 minutes → 4:40 a.m.
ANSWER 9: D

Problem 10:
(8 × 4 + 2) = 32 + 2 = 34.
(8 + 4 × 2) = 8 + 8 = 16.
34 − 16 = 18.
ANSWER 10: D

Problem 11:
11 hours × 60 = 660 minutes. Plus 5 minutes = 665 minutes.
ANSWER 11: C

Problem 12:
6 choices for entry window, then 5 choices for a different exit window. Total = 6 × 5 = 30.
ANSWER 12: D

Problem 1:
We need to find the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). Using the Pigeonhole Principle, we consider the worst-case scenario: the Martian pulls out as many socks as possible *without* getting 5 of any one color. That means getting exactly 4 of each color:
- 4 red + 4 white + 4 blue = 12 socks
After 12 socks, the Martian could have 4 of each color and still not have 5 matching. The very next sock (the 13th) must be red, white, or blue, pushing the count of that color to 5. So 13 socks guarantees 5 of the same color.
(The "five-legged" detail is flavor text explaining why the Martian needs 5 socks.)

ANSWER 1: D

Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). This is a combination problem since order doesn't matter:
C(4, 3) = 4! / (3! × 1!) = 4
We can also list them: {Lance, Sally, Joy}, {Lance, Sally, Fred}, {Lance, Joy, Fred}, {Sally, Joy, Fred} — exactly 4 ways.

ANSWER 2: B

Problem 3:
We check each product:
- A: (−4) × (−9) = 36 ✓
- B: (−3) × (−12) = 36 ✓
- C: (1/2) × (−72) = −36 ✗ (This equals −36, NOT 36)
- D: 1 × 36 = 36 ✓
- E: (3/2) × 24 = 36 ✓
Choice C gives −36, which does not equal 36.

ANSWER 3: C

Problem 4:
First, convert everything to centimeters:
- Ceiling height: 2.4 m = 240 cm
- Light bulb position: 240 − 10 = 230 cm above the floor
- Alice's height: 1.5 m = 150 cm
- Alice's reach above her head: 46 cm
- Alice's maximum reach (without stool): 150 + 46 = 196 cm above the floor

Let h = height of the stool. Standing on the stool, her reach is h + 196 cm. She can "just reach" the bulb:
h + 196 = 230
h = 230 − 196 = 34 cm

ANSWER 4: B

Problem 5:
The ratio of trout to total fish in the sample should equal the ratio in the lake:
- Sample ratio: 30 trout / 180 fish = 1/6
- Lake ratio: 250 trout / Total fish = 1/6

Setting up the proportion: 250 / Total = 1/6
Total = 250 × 6 = 1500 fish

ANSWER 5: B

Problem 6:
Karl bought 5 folders at $2.50 each:
Total cost = 5 × $2.50 = $12.50
A 20%-off sale means he would save 20% of the total:
Savings = 0.20 × $12.50 = $2.50

ANSWER 6: C

Problem 7:
Total time in seconds: 300 dimples × 2 seconds/dimple = 600 seconds
Convert to minutes: 600 seconds ÷ 60 seconds/minute = 10 minutes

ANSWER 7: D

Problem 8:
We need non-negative integer solutions to 5x + 2y = 17, where x = number of $5 bills and y = number of $2 bills.
- x = 0: 2y = 17 → y = 8.5 (not an integer) ✗
- x = 1: 5 + 2y = 17 → 2y = 12 → y = 6 ✓
- x = 2: 10 + 2y = 17 → 2y = 7 → y = 3.5 ✗
- x = 3: 15 + 2y = 17 → 2y = 2 → y = 1 ✓
- x = 4: 20 > 17 (too large) ✗

Only 2 valid combinations: (1 five + 6 twos) and (3 fives + 1 two).

ANSWER 8: A

Problem 9:
Convert 1000 minutes to hours and minutes:
1000 ÷ 60 = 16 remainder 40, so 1000 minutes = 16 hours 40 minutes.
Starting at noon (12:00 PM):
- 12 hours later = midnight (12:00 AM)
- 4 more hours = 4:00 AM
- 40 more minutes = 4:40 AM

ANSWER 9: D

Problem 10:
Evaluate each part using order of operations (multiplication before addition):
- First part: (8 × 4 + 2) = (32 + 2) = 34
- Second part: (8 + 4 × 2) = (8 + 8) = 16
- Difference: 34 − 16 = 18

ANSWER 10: D

Problem 11:
Convert 11 hours and 5 minutes to minutes:
11 hours × 60 minutes/hour = 660 minutes
660 + 5 = 665 minutes

ANSWER 11: C

Problem 12:
Georgie enters through one window and leaves through a different one.
- Choices for entry window: 6
- Choices for exit window (must be different): 5
Total ways = 6 × 5 = 30

ANSWER 12: D

Problem 1:
The Martian needs to be **certain** of having 5 socks of the same color. To guarantee this, we must consider the worst-case scenario: the Martian picks as many socks as possible **without** yet having 5 of any single color.  
With 3 colors (red, white, blue), the worst case is picking 4 socks of each color:  
\(4 + 4 + 4 = 12\) socks.  
At this point the Martian has 4 of every color but still does not have 5 of any one color. The very next sock—the 13th—must be red, white, or blue, and will necessarily create a set of 5 of that color.  
Thus 13 socks are required.  
(Choice A is too small; with 6 socks one could have 2,2,2. Choice B allows 3,3,3. Choice C allows 4,4,4. Choice E is more than the minimum guaranteed number.)

ANSWER 1: D

---

Problem 2:
We need to choose a group of 3 starters from the 4 students: Lance, Sally, Joy, and Fred. Since the three starters form a team and order does not matter, this is a combination problem.  
\[
\binom{4}{3} = \frac{4!}{3!\,1!} = 4
\]  
Equivalently, we can think of choosing which **one** of the four students will **not** start; there are 4 such choices.  
(Choice A is too small. Choice C is \(\binom{4}{2}\). Choices D and E do not arise from counting 3-person teams from 4 people.)

ANSWER 2: B

---

Problem 3:
We check the product of each pair to see which is not 36.

- **A.** \((-4)(-9) = 36\)  
- **B.** \((-3)(-12) = 36\)  
- **C.** \(\left(\frac12\right)(-72) = -36\)  
- **D.** \((1)(36) = 36\)  
- **E.** \(\left(\frac32\right)(24) = 36\)

Only pair **C** gives \(-36\), not \(36\).

ANSWER 3: C

---

Problem 4:
First convert all measurements to centimeters.  
- Ceiling height: \(2.4\text{ m} = 240\text{ cm}\).  
- Light bulb is \(10\text{ cm}\) below the ceiling, so its height above the floor is \(240 - 10 = 230\text{ cm}\).  

Alice’s reach:  
- Height: \(1.5\text{ m} = 150\text{ cm}\).  
- She can reach \(46\text{ cm}\) above her head, so her standing reach is \(150 + 46 = 196\text{ cm}\).  

The stool must supply the remaining height:  
\[
230 - 196 = 34\text{ cm}.
\]

ANSWER 4: B

---

Problem 5:
Let \(T\) be the total number of fish in the lake. The problem states that the ratio of trout to total fish is the same in the sample and in the lake.  
\[
\frac{250}{T} = \frac{30}{180}
\]  
Simplify the right side: \(\frac{30}{180} = \frac16\).  
\[
\frac{250}{T} = \frac16 \quad\Longrightarrow\quad T = 250 \times 6 = 1500.
\]

ANSWER 5: B

---

Problem 6:
Karl bought 5 folders at \(\$2.50\) each.  
Total cost: \(5 \times \$2.50 = \$12.50\).  
A 20% discount would have saved him 20% of this total:  
\[
0.20 \times \$12.50 = \$2.50.
\]

ANSWER 6: C

---

Problem 7:
Casey paints 300 dimples at 2 seconds per dimple.  
Total time in seconds: \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(\frac{600}{60} = 10\) minutes.

ANSWER 7: D

---

Problem 8:
Let \(f\) be the number of \(\$5\) bills and \(t\) be the number of \(\$2\) bills. We need non-negative integer solutions to  
\[
5f + 2t = 17.
\]  
Test values of \(f\):  
- \(f=0\): \(2t=17\) → no integer solution.  
- \(f=1\): \(5 + 2t = 17 \Rightarrow t=6\).  **(Valid)**
- \(f=2\): \(10 + 2t = 17 \Rightarrow t=3.5\).  (Invalid)
- \(f=3\): \(15 + 2t = 17 \Rightarrow t=1\).  **(Valid)**
- \(f=4\): \(20 > 17\) → stop.

There are exactly **2** combinations: \((1\text{ five}, 6\text{ twos})\) and \((3\text{ fives}, 1\text{ two})\).

ANSWER 8: A

---

Problem 9:
The contest starts at noon (12:00 p.m.). We add 1000 minutes.  
Divide 1000 by 60:  
\[
1000 = 16 \times 60 + 40,
\]  
so 1000 minutes = 16 hours and 40 minutes.  

Noon + 12 hours = midnight.  
Midnight + 4 hours 40 minutes = 4:40 a.m.

ANSWER 9: D

---

Problem 10:
Evaluate using the correct order of operations (multiplication before addition).

First parentheses:  
\[
8 \times 4 + 2 = 32 + 2 = 34.
\]

Second parentheses:  
\[
8 + 4 \times 2 = 8 + 8 = 16.
\]

Subtract:  
\[
34 - 16 = 18.
\]

ANSWER 10: D

---

Problem 11:
Convert 11 hours to minutes and add the remaining 5 minutes.  
\[
11 \times 60 = 660\text{ minutes}.
\]  
\[
660 + 5 = 665\text{ minutes}.
\]

ANSWER 11: C

---

Problem 12:
Georgie enters through one of 6 windows and must leave through a **different** window.  
- 6 choices for the entry window.  
- For each entry choice, 5 remaining choices for the exit window.  

Total ways:  
\[
6 \times 5 = 30.
\]  
(Choice E, 36, would count entering and leaving through the same window, which is not allowed.)

ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color. There are 3 colors (red, white, blue). The worst-case scenario is to draw as many socks as possible without having 5 of any one color. That would be 4 red, 4 white, and 4 blue, for a total of 12 socks. The next sock drawn (the 13th) must be one of the three colors, giving that color 5 socks. Therefore, 13 socks are required. The five-legged Martian is irrelevant to the count.  
Answer: D

Problem 2:
We have 4 people (Lance, Sally, Joy, Fred) and need to choose 3 starters. Since the starters are not assigned to specific positions, order does not matter. The number of ways is the combination \( \binom{4}{3} = 4 \).  
Answer: B

Problem 3:
Compute each product:  
A. \((-4) \times (-9) = 36\)  
B. \((-3) \times (-12) = 36\)  
C. \(\frac{1}{2} \times (-72) = -36\)  
D. \(1 \times 36 = 36\)  
E. \(\frac{3}{2} \times 24 = 36\)  
Only pair C gives a product not equal to 36.  
Answer: C

Problem 4:
Convert all measurements to centimeters:  
Ceiling height = 2.4 m = 240 cm.  
Light bulb is 10 cm below ceiling → height = 240 – 10 = 230 cm.  
Alice’s height = 1.5 m = 150 cm.  
Her reach above her head = 46 cm, so her maximum reach without a stool = 150 + 46 = 196 cm.  
Let \(h\) be the stool height. Standing on the stool raises her reach to \(196 + h\). This must equal 230 cm:  
\(196 + h = 230\) → \(h = 34\) cm.  
Answer: B

Problem 5:
In the sample, 30 out of 180 fish are trout, so the ratio trout : total = \(30/180 = 1/6\).  
Assuming the same ratio in the lake, if the lake has 250 trout, then the total number of fish is \(250 \times 6 = 1500\).  
Answer: B

Problem 6:
Original cost for 5 folders: \(5 \times 2.50 = \$12.50\).  
A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
Thus, by waiting a day, Karl could have saved $2.50.  
Answer: C

Problem 7:
Time to paint one dimple = 2 seconds.  
Total time for 300 dimples = \(300 \times 2 = 600\) seconds.  
Convert to minutes: \(600 \div 60 = 10\) minutes.  
Answer: D

Problem 8:
We need nonnegative integers \(a\) (number of $5 bills) and \(b\) (number of $2 bills) such that \(5a + 2b = 17\).  
Solve by testing values of \(b\):  
\(b = 1\) → \(5a = 15\) → \(a = 3\) → (3,1) works.  
\(b = 6\) → \(5a = 5\) → \(a = 1\) → (1,6) works.  
No other values give integer \(a\). Thus there are 2 combinations.  
Answer: A

Problem 9:
1000 minutes = \(1000 \div 60 = 16\) hours and \(40\) minutes (since \(60 \times 16 = 960\), remainder 40).  
Starting at noon (12:00 PM), add 12 hours → midnight (12:00 AM). Add the remaining 4 hours → 4:00 AM. Then add 40 minutes → 4:40 AM.  
Answer: D

Problem 10:
Evaluate using order of operations:  
First parentheses: \(8 \times 4 + 2 = 32 + 2 = 34\).  
Second parentheses: \(8 + 4 \times 2 = 8 + 8 = 16\).  
Subtract: \(34 - 16 = 18\).  
Answer: D

Problem 11:
11 hours = \(11 \times 60 = 660\) minutes.  
Add 5 minutes → \(660 + 5 = 665\) minutes.  
Answer: C

Problem 12:
Georgie must choose an entry window (6 choices) and a different exit window (5 choices). Since entering and leaving are distinct actions, the total number of ways is \(6 \times 5 = 30\).  
Answer: D

ANSWER 1: D
ANSWER 2: B
ANSWER 3: C
ANSWER 4: B
ANSWER 5: B
ANSWER 6: C
ANSWER 7: D
ANSWER 8: A
ANSWER 9: D
ANSWER 10: D
ANSWER 11: C
ANSWER 12: D

Problem 1:  
The Martian wants to be certain of having 5 socks of the same color. There are three colors: red, white, and blue. In the worst-case scenario, the Martian could pick 4 socks of each color before getting a fifth of any one color. That would be \(4 \times 3 = 12\) socks, and still no color has 5 socks. The next sock drawn (the 13th) must complete a set of 5 of some color, because there are only three colors. Therefore, the Martian must remove **13** socks to be certain of having 5 socks of the same color.  
ANSWER 1: D

Problem 2:  
Ms. Hamilton's class has 4 available players: Lance, Sally, Joy, and Fred. The number of ways to choose 3 starters from 4 players is given by the combination \(\binom{4}{3} = 4\). Thus, there are **4** possible sets of three starters.  
ANSWER 2: B

Problem 3:  
We check each pair's product:  
- A: \((-4) \times (-9) = 36\)  
- B: \((-3) \times (-12) = 36\)  
- C: \(\frac{1}{2} \times (-72) = -36\) (not 36)  
- D: \(1 \times 36 = 36\)  
- E: \(\frac{3}{2} \times 24 = 36\)  
Only pair **C** does not have a product of 36.  
ANSWER 3: C

Problem 4:  
The ceiling is 2.4 m above the floor, and the bulb is 10 cm (0.1 m) below the ceiling, so the bulb's height is \(2.4 - 0.1 = 2.3\) m. Alice's height is 1.5 m, and she can reach 46 cm (0.46 m) above her head, so her reach height from the floor is \(1.5 + 0.46 = 1.96\) m. When standing on a stool, her hands are at the stool height plus her reach height. Setting this equal to the bulb height:  
\[
\text{stool height} + 1.96 = 2.3 \Rightarrow \text{stool height} = 0.34 \text{ m} = 34 \text{ cm}.
\]  
ANSWER 4: B

Problem 5:  
In the sample, the ratio of trout to total fish is \(\frac{30}{180} = \frac{1}{6}\). Assuming the same ratio in the lake, let \(T\) be the total number of fish in the lake. Then \(\frac{250}{T} = \frac{1}{6}\), so \(T = 250 \times 6 = 1500\).  
ANSWER 5: B

Problem 6:  
Karl bought 5 folders at $2.50 each, so the original cost is \(5 \times 2.50 = \$12.50\). With a 20% discount, the cost becomes \(12.50 \times 0.80 = \$10.00\). The savings is \(12.50 - 10.00 = \$2.50\).  
ANSWER 6: C

Problem 7:  
Painting 300 dimples at 2 seconds each takes \(300 \times 2 = 600\) seconds. Converting to minutes: \(600 \div 60 = 10\) minutes.  
ANSWER 7: D

Problem 8:  
We need nonnegative integers \(x\) (number of $5 bills) and \(y\) (number of $2 bills) such that \(5x + 2y = 17\). Solving:  
- \(y = 1 \Rightarrow 5x = 15 \Rightarrow x = 3\)  
- \(y = 6 \Rightarrow 5x = 5 \Rightarrow x = 1\)  
- \(y = 11 \Rightarrow 5x = -5\) (invalid)  
Thus, there are **2** valid combinations: (3 fives, 1 two) and (1 five, 6 twos).  
ANSWER 8: A

Problem 9:  
1000 minutes equals \(16\) hours and \(40\) minutes (since \(16 \times 60 = 960\), remainder \(40\)). Starting at noon, adding \(16\) hours gives 4:00 AM the next day, and adding \(40\) minutes gives 4:40 AM.  
ANSWER 9: D

Problem 10:  
Compute inside parentheses:  
\(8 \times 4 + 2 = 32 + 2 = 34\)  
\(8 + 4 \times 2 = 8 + 8 = 16\)  
Then \(34 - 16 = 18\).  
ANSWER 10: D

Problem 11:  
11 hours is \(11 \times 60 = 660\) minutes. Adding 5 minutes gives \(660 + 5 = 665\) minutes.  
ANSWER 11: C

Problem 12:  
Georgie can enter through any of the 6 windows. For exiting, he must choose a different window, so there are 5 choices. The total number of ways is \(6 \times 5 = 30\).  
ANSWER 12: D

Problem 1:  
The Martian needs to ensure there are 5 socks of the same color. The worst-case scenario is pulling the maximum number of socks without 5 of any color. Since there are 3 colors, this would be 4 socks of each color (4 × 3 = 12 socks). The next sock (the 13th) must match one color, giving 5 socks of that color.  
**ANSWER 1: D**  

Problem 2:  
There are 4 students (Lance, Sally, Joy, Fred), and we need to choose 3 starters. The number of combinations is C(4,3) = 4.  
**ANSWER 2: B** (Note: C(4,3)=4, but rechecking: the team is chosen from 4 specific people, so combinations are 4. Answer B=4)  

Problem 3:  
Check each pair:  
A. (−4)(−9) = 36  
B. (−3)(−12) = 36  
C. (1/2)(−72) = -36 (≠36)  
D. (1)(36) = 36  
E. (3/2)(24) = 36  
The pair that does NOT equal 36 is C.  
**ANSWER 3: E** (Re-evaluated: The question asks which does NOT equal 36. Option E: (3/2)*24=36, so it is valid. Option C: (1/2)*(-72)=-36, so C is the correct choice for NOT equal. Correction: **ANSWER 3: C** [Original had a typo, but computation confirms C is the answer])  

*Correction: For Problem 3, Option C: (1/2, -72) likely means (1/2) × (-72) = -36 ≠ 36. Option E: (3/2) × 24 = 36. Thus, C is the answer.*  
**ANSWER 3: C**  

Problem 4:  
Ceiling height: 2.4 m = 240 cm. Light bulb location: 240 - 10 = 230 cm from floor. Alice’s reach: 150 cm + 46 cm = 196 cm. Stool height = 230 - 196 = 34 cm.  
**ANSWER 4: B**  

Problem 5:  
Ratio of trout in sample: 30/180 = 1/6. Total trout: 250. Total fish = 250 × 6 = 1500.  
**ANSWER 5: B**  

Problem 6:  
Original cost: 5 × $2.50 = $12.50. 20% off: $12.50 × 0.20 = $2.50 saved.  
**ANSWER 6: C** (Note: The question asks how much Karl could have saved, which is $2.50, not $2.00. Original had a mistake; correction: 20% of $12.50 is $2.50, so answer C)  

Problem 7:  
300 dimples × 2 seconds = 600 seconds. 600 ÷ 60 = 10 minutes.  
**ANSWER 7: D**  

Problem 8:  
Let $5 bills = x, $2 bills = y. 5x + 2y = 17. Possible solutions:  
- x=3, y=1 (15+2=17)  
- x=1, y=6 (5+12=17)  
Only 2 combinations.  
**ANSWER 8: A**  

Problem 9:  
1000 minutes = 16 hours 40 minutes. Starting at noon: noon + 16h40m = 4:40 a.m. next day.  
**ANSWER 9: D**  

Problem 10:  
(8×4 + 2) = 34, (8 + 4×2) = 16. 34 - 16 = 18.  
**ANSWER 10: D**  

Problem 11:  
11 hours = 660 minutes. 660 + 5 = 665 minutes.  
**ANSWER 11: C**  

Problem 12:  
Georgie can enter through 6 windows and exit through 5 others. Total ways: 6 × 5 = 30.  
**ANSWER 12: D**  

---  
**Final Answers**  
ANSWER 1: D  
ANSWER 2: B  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: B  
ANSWER 6: C  
ANSWER 7: D  
ANSWER 8: A  
ANSWER 9: D  
ANSWER 10: D  
ANSWER 11: C  
ANSWER 12: D

### Problem 1:
This is a classic pigeonhole principle problem. We need to find the minimum number of socks the Martian must pull to guarantee 5 socks of the same color, with 3 total colors. The worst-case scenario is that the Martian pulls as many socks as possible without getting 5 of any color: that is 4 socks of each color, for a total of 3*4=12 socks. The next sock pulled (the 13th) has to be one of the three colors, pushing that color to 5 socks. Tempting wrong answers: C (12 is only the number of socks that avoids 5 of the same color, not enough to guarantee the requirement), E (15 is the total if you took 5 of each color, not the minimum to guarantee).
ANSWER 1: D

---

### Problem 2:
We need to count the number of ways to choose 3 starters from 4 people, where order on the team does not matter. This is a combination problem: C(n,k) = n!/(k!(n-k)!). For n=4, k=3: C(4,3) = 4. We can also list the teams by who is left out: leaving out Lance, Sally, Joy, or Fred gives exactly 4 unique teams. Tempting wrong answer: C (6 is the count if order of selection mattered, which it does not for a team of starters).
ANSWER 2: B

---

### Problem 3:
We calculate the product of each pair to find which does not equal 36:
- A: (-4)*(-9)=36 ✔️
- B: (-3)*(-12)=36 ✔️
- C: (1/2)*(-72) = -36 ❌
- D: 1*36=36 ✔️
- E: (3/2)*24=36 ✔️
Only pair C has a product of -36, not 36.
ANSWER 3: C

---

### Problem 4:
First convert all units to centimeters to avoid conversion errors:
- Ceiling height: 2.4m = 240cm. Light bulb is 10cm below ceiling, so bulb height from floor = 240-10=230cm.
- Alice's height: 1.5m=150cm, her reach above her head is 46cm, so total reach without stool = 150+46=196cm.
- Stool height = 230-196=34cm.
Tempting wrong answers come from unit conversion mistakes, e.g., mixing up meters and centimeters.
ANSWER 4: B

---

### Problem 5:
This is a proportion problem. The ratio of trout to total fish is the same in the sample and the lake. In the sample, 30 trout out of 180 total fish = 1/6. Let T be total fish in the lake: 250/T = 1/6 → T=250*6=1500.
ANSWER 5: B

---

### Problem 6:
Karl bought 5 folders at $2.50 each, total original cost = 5*2.50=$12.50. He saves 20% of the total cost by waiting for the sale: 0.2*12.50=$2.50. This can also be calculated per folder: 20% of $2.50 is $0.50 saved per folder, 5*0.50=$2.50 total saved.
ANSWER 6: C

---

### Problem 7:
300 dimples * 2 seconds per dimple = 600 total seconds. Convert to minutes: 60 seconds per minute, so 600/60=10 minutes.
ANSWER 7: D

---

### Problem 8:
We find all non-negative integer pairs (x,y) where x=number of $5 bills, y=number of $2 bills, and 5x+2y=17. Only odd values of x work (since $5 is odd, $2 is even, we need an odd sum to get 17):
- x=1: 5+2y=17 → y=6 (valid)
- x=3: 15+2y=17 → y=1 (valid)
Even x values (0,2) leave an odd remainder that cannot be formed with $2 bills, so only 2 valid combinations.
ANSWER 8: A

---

### Problem 9:
Convert 1000 minutes to hours and minutes: 60 minutes per hour, 1000/60=16 hours and 40 minutes. Adding this to noon (12pm): 12pm +16h40m = 4:40am the next day.
ANSWER 9: D

---

### Problem 10:
Use order of operations (PEMDAS) to calculate:
First parenthesis: 8*4 +2 = 32+2=34
Second parenthesis: 8 +4*2=8+8=16
Subtract: 34-16=18. The tempting wrong answer C (10) comes from incorrectly adding before multiplying in the second parenthesis.
ANSWER 10: D

---

### Problem 11:
Convert 11 hours 5 minutes to total minutes: 11*60=660 minutes, plus 5 extra minutes = 665 minutes.
ANSWER 11: C

---

### Problem 12:
Georgie has 6 choices for an entry window, and 5 remaining choices for an exit window (must use a different window). Total ways = 6*5=30. Tempting wrong answer E (36) counts entering and leaving through the same window, which is not allowed; B (15) incorrectly treats entry/exit as unordered, which they are not.
ANSWER 12: D

Problem 1:
We need the minimum number of socks to guarantee 5 of the same color (red, white, blue). This is a pigeonhole principle problem: the worst-case scenario is pulling the maximum number of socks without getting 5 of any color. For each of the 3 colors, we can pull 4 socks (since 4 < 5), so 4×3=12 socks. The next sock pulled will necessarily be the 5th of one color. So total socks needed: 12 + 1 = 13.
ANSWER 1: D
Problem 2:
We need to choose 3 starters from 4 people (Lance, Sally, Joy, Fred). Since the problem does not mention distinct positions for starters, order does not matter, so we use combinations. The number of ways to choose 3 people from 4 is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n=4 \) and \( k=3 \):
\( \binom{4}{3} = \frac{4!}{3!1!} = 4 \). This is equivalent to counting how many people we leave out (4 choices, one for each person not selected).
ANSWER 2: B
Problem 3:
We check the product of each pair:
- A: \( (-4) \times (-9) = 36 \)
- B: \( (-3) \times (-12) = 36 \)
- C: \( \frac{1}{2} \times (-72) = -36 \neq 36 \)
- D: \( 1 \times 36 = 36 \)
- E: \( \frac{3}{2} \times 24 = 36 \)
Only pair C does not have a product of 36.
ANSWER 3: C
Problem 4:
Convert all units to centimeters:
- Ceiling height: \( 2.4 \, \text{m} = 240 \, \text{cm} \)
- Light bulb height from floor: \( 240 - 10 = 230 \, \text{cm} \)
- Alice's reach from floor (without stool): \( 1.5 \, \text{m} + 0.46 \, \text{m} = 150 + 46 = 196 \, \text{cm} \)
Let \( h \) be the stool height. The total reach with the stool equals the light bulb height:
\( h + 196 = 230 \)
\( h = 230 - 196 = 34 \, \text{cm} \)
ANSWER 4: B
Problem 5:
The ratio of trout to total fish in the sample equals the ratio in the lake. The sample has \( \frac{30}{180} = \frac{1}{6} \) trout. Let \( T \) be the total number of fish in the lake. We know there are 250 trout, so:
\( \frac{250}{T} = \frac{1}{6} \)
\( T = 250 \times 6 = 1500 \)
ANSWER 5: B
Problem 6:
Karl bought 5 folders at $2.50 each. The 20% discount applies to each folder, so the savings per folder is \( 2.50 \times 0.20 = 0.50 \). Total savings for 5 folders: \( 5 \times 0.50 = 2.50 \).
ANSWER 6: C
Problem 7:
Total time to paint 300 dimples: \( 300 \times 2 = 600 \) seconds. Convert to minutes: \( 600 \div 60 = 10 \) minutes.
ANSWER 7: D
Problem 8:
Let \( x \) = number of $5 bills, \( y \) = number of $2 bills. We need \( 5x + 2y = 17 \), where \( x,y \geq 0 \) integers. Test non-negative integer values of \( x \):
- \( x=0 \): \( 2y=17 \) → \( y=8.5 \) (not integer, invalid)
- \( x=1 \): \( 5 + 2y=17 \) → \( 2y=12 \) → \( y=6 \) (valid: 1 $5 + 6 $2 = $17)
- \( x=2 \): \( 10 + 2y=17 \) → \( 2y=7 \) → \( y=3.5 \) (invalid)
- \( x=3 \): \( 15 + 2y=17 \) → \( 2y=2 \) → \( y=1 \) (valid: 3 $5 + 1 $2 = $17)
- \( x=4 \): \( 20 > 17 \) (stop, no more valid \( x \))
Only 2 valid combinations.
ANSWER 8: A
Problem 9:
Convert 1000 minutes to hours and minutes: \( 1000 \div 60 = 16 \) hours with a remainder of 40 minutes (since \( 16 \times 60 = 960 \), \( 1000 - 960 = 40 \)). The contest starts at noon, so adding 16 hours 40 minutes: noon + 12 hours = midnight, plus 4 hours 40 minutes = 4:40 a.m.
ANSWER 9: D
Problem 10:
Use order of operations (multiplication before addition):
First parentheses: \( 8 \times 4 + 2 = 32 + 2 = 34 \)
Second parentheses: \( 8 + 4 \times 2 = 8 + 8 = 16 \)
Subtract: \( 34 - 16 = 18 \)
ANSWER 10: D
Problem 11:
Convert 11 hours 5 minutes to total minutes: \( 11 \times 60 + 5 = 660 + 5 = 665 \)
ANSWER 11: C
Problem 12:
Georgie chooses an entrance window first (6 options), then a different exit window (5 remaining options, since exit cannot equal entrance). Since entrance and exit are distinct actions, order matters, so total ways: \( 6 \times 5 = 30 \).
ANSWER 12: D