🔬 Model Lab

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

**Problem 1:**

What is being asked: By how many centimeters does the water level rise when a 1000 cm³ rock is submerged?

Setup: The base area is 100 cm × 40 cm = 4000 cm². When the rock is submerged, it displaces 1000 cm³ of water.

Water level rise = Volume displaced ÷ Base area = 1000 ÷ 4000 = 0.25 cm

ANSWER 1: A

---

**Problem 2:**

What is being asked: If cards are distributed evenly, how many cards per player when the player count increases?

Setup: Initially 4 players (Annika + 3 friends) each get 15 cards.
Total cards = 4 × 15 = 60 cards

With 2 more friends joining: 4 + 2 = 6 players
Cards per player = 60 ÷ 6 = 10 cards

ANSWER 2: C

---

**Problem 3:**

What is being asked: Evaluate the operation 3 1 4 2 using the rule a c b d = a·d − b·c

Setup: With a = 3, c = 1, b = 4, d = 2:
3 1 4 2 = 3·2 − 4·1 = 6 − 4 = 2

ANSWER 3: E

---

**Problem 4:**

What is being asked: What is the lowest score on test 3 that allows Isabella to achieve a 95 average on four tests?

Setup: Goal average is 95, so total needed = 95 × 4 = 380 points
First two tests: 97 + 91 = 188 points
Points needed from tests 3 and 4: 380 − 188 = 192 points

Let x = score on test 3. She can still reach her goal if she can score 100 on test 4:
x + 100 ≥ 192
x ≥ 92

The lowest possible score on test 3 is 92.

ANSWER 4: B

---

**Problem 5:**

What is being asked: How many apples did Bridget buy originally?

Setup: Let x = original number of apples
- She gave half to Ann: x/2 apples given away
- Remaining: x/2
- She gave 3 to Cassie: x/2 − 3
- She kept 4 for herself: x/2 − 3 = 4

Solving: x/2 = 7, so x = 14

ANSWER 5: E

---

**Problem 6:**

What is being asked: How many marbles are in the jar?

Setup: Let R, G, B = numbers of red, green, blue marbles; T = total
- "All but 6 are red": G + B = 6
- "All but 8 are green": R + B = 8
- "All but 4 are blue": R + G = 4

Adding all three equations: 2(R + G + B) = 6 + 8 + 4 = 18
Therefore: R + G + B = 9

ANSWER 6: C

---

**Problem 7:**

What is being asked: What is the sum of the two smallest prime factors of 250?

Setup: Factor 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5³

Prime factors: 2 and 5
Sum = 2 + 5 = 7

ANSWER 7: C

---

**Problem 8:**

What is being asked: How many different sums are possible?

Setup: Bag A: {1, 3, 5}, Bag B: {2, 4, 6}
All possible sums:
- 1 + 2 = 3
- 1 + 4 = 5
- 1 + 6 = 7
- 3 + 2 = 5 (duplicate)
- 3 + 4 = 7 (duplicate)
- 3 + 6 = 9
- 5 + 2 = 7 (duplicate)
- 5 + 4 = 9 (duplicate)
- 5 + 6 = 11

Distinct sums: {3, 5, 7, 9, 11} = 5 different values

ANSWER 8: B

---

**Problem 9:**

What is being asked: How many seating arrangements with boys at ends and girls in middle?

Setup: 2 boys at the ends can be arranged in 2! = 2 ways
3 girls in the middle can be arranged in 3! = 6 ways
Total arrangements = 2 × 6 = 12

ANSWER 9: E

---

**Problem 10:**

What is being asked: Evaluate 16 + 8 ÷ 4 − 2

Setup: Using order of operations (division before addition/subtraction):
16 + (8 ÷ 4) − 2 = 16 + 2 − 2 = 16

ANSWER 10: D

---

**Problem 11:**

What is being asked: Evaluate √(16 · √(8 · √4))

Setup: Work from the inside out:
- √4 = 2
- 8 · √4 = 8 · 2 = 16
- √(8 · √4) = √16 = 4
- 16 · √(8 · √4) = 16 · 4 = 64
- √(16 · √(8 · √4)) = √64 = 8

ANSWER 11: C

---

**Problem 12:**

What is being asked: Which decimal is largest?

Setup: Compare decimals:
- A: 0.99
- B: 0.9099
- C: 0.9
- D: 0.909
- E: 0.9009

Comparing: 0.99 > 0.9099 > 0.909 > 0.9009 > 0.9

The largest is 0.99

ANSWER 12: A

Problem 1:

The aquarium has a rectangular base of \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is

\[
100 \cdot 40 = 4000 \text{ cm}^2.
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

If the water rises by \(h\) centimeters, then

\[
4000h = 1000.
\]

So

\[
h = \frac{1000}{4000} = \frac14 = 0.25.
\]

So the water level rises \(0.25\) cm.

ANSWER 1: A

---

Problem 2:

Annika and 3 friends means there are

\[
4 \text{ players}.
\]

If each player gets 15 cards, then the total number of cards is

\[
4 \cdot 15 = 60.
\]

Now 2 more friends join, so there are

\[
6 \text{ players}.
\]

The same 60 cards are shared evenly, so each gets

\[
60 \div 6 = 10.
\]

ANSWER 2: C

---

Problem 3:

We are told that if the four numbers are \(a, b, c, d\), then

\[
a \ b \ c \ d = a\cdot d - b\cdot c.
\]

So for \(3,1,4,2\),

\[
3 \ 1 \ 4 \ 2 = 3\cdot 2 - 1\cdot 4 = 6 - 4 = 2.
\]

ANSWER 3: E

---

Problem 4:

She wants an average of 95 on 4 tests, so the total of the 4 scores must be

\[
4 \cdot 95 = 380.
\]

Her first two scores are 97 and 91, which add to

\[
97+91=188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To make the total 380, the fourth score would have to be

\[
380 - (188+x) = 192 - x.
\]

For this to be possible, the fourth score must be at most 100:

\[
192 - x \le 100.
\]

So

\[
x \ge 92.
\]

The lowest possible third score is 92.

ANSWER 4: B

---

Problem 5:

Let the number of apples Bridget bought be \(x\).

She gave half to Ann, so she kept the other half:

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

Then she gave Cassie 3 apples and kept 4 apples, so before giving Cassie the apples she had

\[
3+4=7.
\]

Thus,

\[
\frac{x}{2}=7 \quad \Rightarrow \quad x=14.
\]

ANSWER 5: E

---

Problem 6:

Let the total number of marbles be \(n\).

- “All but 6 are red” means the number of red marbles is \(n-6\).
- “All but 8 are green” means the number of green marbles is \(n-8\).
- “All but 4 are blue” means the number of blue marbles is \(n-4\).

Since every marble is one of these colors, their counts add to \(n\):

\[
(n-6)+(n-8)+(n-4)=n.
\]

Simplify:

\[
3n-18=n
\]

\[
2n=18
\]

\[
n=9.
\]

ANSWER 6: C

---

Problem 7:

We need the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3.
\]

So the prime factors are 2 and 5. The two smallest prime factors are 2 and 5, and their sum is

\[
2+5=7.
\]

ANSWER 7: C

---

Problem 8:

Bag A has \(1,3,5\); Bag B has \(2,4,6\).

Possible sums:

- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums are

\[
3,5,7,9,11
\]

which is 5 different values.

ANSWER 8: B

---

Problem 9:

There are 2 boys and 3 girls, and the picture must have:

- a boy at each end,
- the 3 girls in the middle.

First, arrange the 2 boys at the ends. They can be switched in

\[
2! = 2
\]

ways.

Then arrange the 3 girls in the middle in

\[
3! = 6
\]

ways.

So the total number of arrangements is

\[
2 \cdot 6 = 12.
\]

ANSWER 9: E

---

Problem 10:

The expression is

\[
16 + 8\cdot 4 - 2.
\]

Use order of operations:

\[
8\cdot 4=32.
\]

Then

\[
16+32-2=46.
\]

But 46 is not among the choices, so the intended expression must be read as

\[
16 + 8 \div 4 - 2.
\]

Then

\[
8\div 4 = 2,
\]

so

\[
16+2-2=16.
\]

ANSWER 10: D

---

Problem 11:

We evaluate from the inside out:

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

First,

\[
\sqrt4=2.
\]

So the expression becomes

\[
\sqrt{16\cdot \sqrt{8\cdot 2}}
=
\sqrt{16\cdot \sqrt{16}}.
\]

Since \(\sqrt{16}=4\),

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C

---

Problem 12:

Compare the decimals:

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

The largest is the one closest to 1. Since \(0.99\) is greater than any number beginning with \(0.90\), it is the largest.

ANSWER 12: A

Problem 1:  
We need the rise in water level when a rock is placed in the aquarium and submerged.

- Base dimensions: \(100\text{ cm} \times 40\text{ cm}\), so base area  
  \[
  A = 100\cdot 40 = 4000\text{ cm}^2.
  \]
- The rock is completely submerged, so the water volume displaced equals the rock’s volume:  
  \[
  V = 1000\text{ cm}^3.
  \]
- Water level rise \(\Delta h\) satisfies \(A\Delta h = V\):  
  \[
  4000\cdot \Delta h = 1000 \implies \Delta h = \frac{1000}{4000}=0.25\text{ cm}.
  \]

So the rise is \(0.25\) cm.  
Tempting wrong idea: using the *remaining* depth \(37\) cm or checking if it overflows—none of that matters; only displaced volume matters.

ANSWER 1: A  

---

Problem 2:  
All cards are dealt evenly. Let the total number of cards be \(T\).

Initially, Annika + 3 friends = 4 players, and each gets 15 cards:  
\[
T = 4\cdot 15 = 60.
\]
Next game: 2 more friends join, so total players:
\[
4+2=6.
\]
Each player gets:
\[
\frac{60}{6}=10.
\]

ANSWER 2: C  

---

Problem 3:  
Given the rule \(a\,b\,d\,c = a\cdot d - b\cdot c\).  
The expression “\(3\ 1\ 4\ 2\)” corresponds to:
- \(a=3\)
- \(b=1\)
- \(d=4\)
- \(c=2\)

Compute:
\[
a\cdot d - b\cdot c = 3\cdot 4 - 1\cdot 2 = 12-2=10.
\]

But notice the choices are only \(-2,-1,0,1,2\), so we must interpret the placeholder correctly:  
The problem states: **“If \(a b c d = a\cdot d - b\cdot c\)”**.

So “\(3\ 1\ 4\ 2\)” means:
- \(a=3,\ b=1,\ c=4,\ d=2\)

Then:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 1\cdot 4 = 6-4=2.
\]

ANSWER 3: E  

---

Problem 4:  
Goal average is 95 over 4 tests, so total points needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
Let the third test score be \(x\), and the fourth be \(y\). Then:
\[
188 + x + y = 380 \implies x+y=192.
\]
We want the **lowest possible** \(x\) such that she can still reach the goal. That means make \(y\) as large as possible (since grades are at most 100).
Maximum \(y=100\). Then:
\[
x + 100 = 192 \implies x = 92.
\]

Check: then the four scores are \(97,91,92,100\) which sum to \(380\), average \(95\). Works.

ANSWER 4: B  

---

Problem 5:  
Let the total number of apples be \(N\).

- She gave half to Ann, so Ann got \(N/2\), leaving \(N/2\).
- Then she gave Cassie 3 apples, and kept 4 apples. That means after giving Cassie 3, her remaining is 4:
  \[
  \frac{N}{2} - 3 = 4 \implies \frac{N}{2}=7 \implies N=14.
  \]

ANSWER 5: E  

---

Problem 6:  
“All but 6 are red” means: number of red marbles is 6.  
Similarly:
- Red = 6 (since all but 6 are red → 6 red)
- Green: all but 8 are green → 8 green
- Blue: all but 4 are blue → 4 blue

Total marbles:
\[
6+8+4=18.
\]
But none of the choices are 18, so interpret carefully:  
“All but 6 are red” means the number of red marbles is **(total − 6)**, not 6.

Let total be \(T\). Then:
- Red marbles = \(T-6\)
- Green marbles = \(T-8\)
- Blue marbles = \(T-4\)

Sum:
\[
(T-6)+(T-8)+(T-4)=T.
\]
\[
3T - (6+8+4)=T
\]
\[
3T-18=T \implies 2T=18 \implies T=9.
\]

ANSWER 6: C  

---

Problem 7:  
We want the sum of the **two smallest prime factors** of 250.

Factor \(250\):
\[
250 = 25\cdot 10 = 5^2 \cdot 2\cdot 5 = 2\cdot 5^3.
\]
Prime factors are \(2\) and \(5\). The two smallest are \(2\) and \(5\).

Sum:
\[
2+5=7.
\]

ANSWER 7: D  

---

Problem 8:  
Bag A: \(\{1,3,5\}\)  
Bag B: \(\{2,4,6\}\)

Possible sums:
- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums: \(\{3,5,7,9,11\}\) → that’s 5 values.

ANSWER 8: B  

---

Problem 9:  
We need arrangements of 5 people in a row with:
- a boy at each end (positions 1 and 5)
- the three girls in the middle (positions 2,3,4)

There are 2 boys: they can be placed at the ends in
\[
2!\ = 2
\]
ways (left boy/right boy).

The 3 girls can be arranged in the three middle spots in
\[
3! = 6
\]
ways.

Total arrangements:
\[
2\cdot 6 = 12.
\]

ANSWER 9: E  

---

Problem 10:  
Expression: \(16 + 8^4 - 2\).

Compute in order:
- \(8^4 = 4096\)

So:
\[
16 + 4096 - 2 = 4110.
\]
No answer choice matches, so we interpret the formatting as \(16 + 8 \sqrt{4} - 2\) or \(16 + 8^4 - 2\)?  
Given the AMC/JH style, likely it is:
\[
16 + 8\sqrt{4} - 2.
\]
Then:
\[
\sqrt{4}=2 \implies 8\sqrt{4}=8\cdot 2=16.
\]
So:
\[
16 + 16 - 2 = 30,
\]
still not in choices.

Another common formatting is:
\[
16 + 8^{4} - 2 \quad \text{(still not)}
\]

The only choiceable consistent possibility is:
\[
16 + \sqrt{8}\,^4 - 2 \text{ (nonsense)}
\]

Let’s use the most plausible literal from the text: **“16 + 8 4 − 2”** often means \(16 + 8^4 - 2\) if the caret was lost. But then no choice.

Given typical AMC problems, the intended is usually:
\[
16 + 8 - 4 - 2 = 18
\]
also not.

Because the answer choices are only \(4,8,12,16,20\), the intended expression must simplify to one of those. The most natural is:
\[
16 + 8/4 - 2 = 16 + 2 - 2 = 16.
\]
This matches a common missing symbol: “8 4” often means \(8/4\).

So:
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D  

---

Problem 11:  
Evaluate:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}.
\]

Work from inside out:

1) \(\sqrt{4}=2\)

2) Inside next root:
\[
8\cdot \sqrt{4} = 8\cdot 2 = 16
\]
So:
\[
\sqrt{8\cdot \sqrt{4}} = \sqrt{16}=4.
\]

3) Now plug into the outer expression:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}=\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C  

---

Problem 12:  
Largest number among:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare by decimal places:

- .99 is 0.99
- .9099 is 0.9099
- .9 is 0.9000
- .909 is 0.9090
- .9009 is 0.9009

Clearly \(0.99\) is greater than all the numbers less than \(0.91\).

ANSWER 12: A

Problem 1:
The volume of the water displaced by the rock is equal to the volume of the rock itself, which is $1000 \text{ cm}^3$. The aquarium has a rectangular base with area $A = 100 \text{ cm} \times 40 \text{ cm} = 4000 \text{ cm}^2$. When the rock is submerged, the water level rises by height $h$ such that the volume of the displaced water is $A \times h$. 
Setting this equal to the rock's volume: $4000 \times h = 1000$.
Solving for $h$: $h = \frac{1000}{4000} = \frac{1}{4} = 0.25 \text{ cm}$.
ANSWER 1: A

Problem 2:
Annika and 3 friends make 4 players total. Each received 15 cards, so the total number of cards is $4 \times 15 = 60$. If 2 more friends join, the total number of players becomes $4 + 2 = 6$. The cards are distributed evenly among the 6 players, so each player gets $60 \div 6 = 10$ cards.
ANSWER 2: C

Problem 3:
The problem defines the operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$.
For the expression $\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix}$, we have $a=3, b=1, c=4, d=2$.
Calculating: $(3 \times 2) - (1 \times 4) = 6 - 4 = 2$.
ANSWER 3: E

Problem 4:
Isabella needs an average of 95 over 4 tests, meaning the total sum of her scores must be $95 \times 4 = 380$. Her first two scores are 97 and 91, totaling $97 + 91 = 188$. Let $x$ be the score on the third test and $y$ be the score on the fourth. We have $188 + x + y = 380$, so $x + y = 192$. To make $x$ as small as possible, we must make $y$ as large as possible. Since the maximum score on a test is 100, we set $y = 100$. Then $x + 100 = 192$, so $x = 92$.
ANSWER 4: B

Problem 5:
Let $x$ be the total number of apples Bridget bought. She gave half away, leaving $\frac{x}{2}$. Then she gave 3 away, leaving $\frac{x}{2} - 3$. She kept 4, so $\frac{x}{2} - 3 = 4$. Adding 3 to both sides, $\frac{x}{2} = 7$. Multiplying by 2, $x = 14$.
ANSWER 5: E

Problem 6:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R + G + B$.
"All but 6 are red" means $G + B = 6$.
"All but 8 are green" means $R + B = 8$.
"All but 4 are blue" means $R + G = 4$.
Adding these three equations: $2(R + G + B) = 6 + 8 + 4 = 18$.
So $R + G + B = 18 \div 2 = 9$.
ANSWER 6: C

Problem 7:
The prime factorization of 250 is $250 = 2 \times 125 = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two prime factors is $2 + 5 = 7$.
ANSWER 7: C

Problem 8:
Bag A has $\{1, 3, 5\}$ and Bag B has $\{2, 4, 6\}$.
Possible sums:
$1+2=3, 1+4=5, 1+6=7$
$3+2=5, 3+4=7, 3+6=9$
$5+2=7, 5+4=9, 5+6=11$
The set of distinct sums is $\{3, 5, 7, 9, 11\}$. There are 5 distinct values.
ANSWER 8: B

Problem 9:
There are 2 boys (B1, B2) and 3 girls (G1, G2, G3). A boy must be at each end. There are 2 choices for the left end and 1 choice for the right end ($2 \times 1 = 2$ ways to place boys). The 3 girls can be arranged in the middle in $3! = 3 \times 2 \times 1 = 6$ ways. Total arrangements = $2 \times 6 = 12$.
ANSWER 9: E

Problem 10:
Following the order of operations (PEMDAS/BODMAS):
$16 + (8 \div 4) - 2 = 16 + 2 - 2 = 16$.
ANSWER 10: D

Problem 11:
Evaluate from the inside out:
$\sqrt{4} = 2$
$\sqrt{8 \times 2} = \sqrt{16} = 4$
$\sqrt{16 \times 4} = \sqrt{64} = 8$
ANSWER 11: C

Problem 12:
To compare, pad with zeros to match the number of decimal places:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
Comparing the values, 0.9900 is the largest.
ANSWER 12: A

Problem 1:
The base of the aquarium measures 100 cm by 40 cm, so its area is \(100 \times 40 = 4000\) cm². The rock has volume 1000 cm³ and is fully submerged, displacing exactly 1000 cm³ of water. The resulting rise in water level is therefore the displaced volume divided by the base area: \(1000 / 4000 = 0.25\) cm. (The original water depth and tank height are irrelevant once the surface area is known.)

ANSWER 1: A

Problem 2:
Annika plus 3 friends gives 4 players, each receiving 15 cards, for a total of \(4 \times 15 = 60\) cards. When 2 more players join, there are 6 players sharing the same 60 cards, so each receives \(60 / 6 = 10\) cards.

ANSWER 2: C

Problem 3:
The given operation evaluates \(a\, c\, b\, d\) as \(a \cdot d - b \cdot c\). Substituting the numbers yields \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).

ANSWER 3: E

Problem 4:
An average of 95 on four 100-point tests requires a total of \(4 \times 95 = 380\) points. The first two scores sum to \(97 + 91 = 188\), leaving 192 points still needed from the last two tests. The fourth test can score at most 100, so the third test must be at least \(192 - 100 = 92\). Any lower score on the third test would make a total of 380 impossible.

ANSWER 4: B

Problem 5:
Let \(t\) be the original number of apples. After giving half to Ann, \(t/2\) remain. After giving 3 more to Cassie, \(t/2 - 3\) remain, and this equals the 4 apples kept. Solving \(t/2 - 3 = 4\) gives \(t/2 = 7\), so \(t = 14\).

ANSWER 5: E

Problem 6:
Let \(m\) be the total number of marbles. Then there are \(m-6\) red, \(m-8\) green, and \(m-4\) blue. Their sum must equal \(m\):
\[
(m-6) + (m-8) + (m-4) = m \implies 3m - 18 = m \implies 2m = 18 \implies m = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies the three “all-but” statements.

ANSWER 6: C

Problem 7:
Factorize \(250 = 2 \cdot 5^3\). The two smallest distinct prime factors are 2 and 5; their sum is 7.

ANSWER 7: C

Problem 8:
All nine possible sums are:
\[
1+2=3,\ 1+4=5,\ 1+6=7,\ 3+2=5,\ 3+4=7,\ 3+6=9,\ 5+2=7,\ 5+4=9,\ 5+6=11.
\]
The distinct values are 3, 5, 7, 9, 11—exactly five.

ANSWER 8: B

Problem 9:
The seating pattern must be boy–girl–girl–girl–boy. The two distinct boys may be placed at the ends in \(2!\) = 2 ways. The three distinct girls may be arranged in the middle three seats in \(3!\) = 6 ways. The total number of arrangements is therefore \(2 \times 6 = 12\).

ANSWER 9: E

Problem 10:
Interpreting the expression with standard order of operations and treating juxtaposition of 8 and 4 as division yields
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D

Problem 11:
Evaluate from the inside outward:
\[
\sqrt{4} = 2, \quad 8 \cdot 2 = 16, \quad \sqrt{16} = 4, \quad 16 \cdot 4 = 64, \quad \sqrt{64} = 8.
\]

ANSWER 11: C

Problem 12:
Compare the decimals directly:
\[
0.99 > 0.9 > 0.9099 > 0.909 > 0.9009.
\]
Thus 0.99 is the largest.

ANSWER 12: A

Problem 1:
The problem asks us to determine by how many centimeters the water level rises when a rock with a volume of 1000 cm³ is submerged in an aquarium filled with water to a depth of 37 cm. The aquarium has a rectangular base measuring 100 cm by 40 cm and a height of 50 cm.

First, we need to calculate the initial volume of water in the aquarium. The volume of water is given by the formula length × width × height. So, the initial volume of water = 100 cm × 40 cm × 37 cm = 148,000 cm³.

When the rock is submerged, it displaces a volume of water equal to its own volume, which is 1000 cm³. This means the water level will rise to accommodate the volume of the rock.

The rise in water level can be calculated by dividing the volume of the rock by the base area of the aquarium. The base area = 100 cm × 40 cm = 4000 cm².

Rise in water level = Volume of rock / Base area = 1000 cm³ / 4000 cm² = 0.25 cm.

Let's examine the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

The calculated rise in water level matches option A.

ANSWER 1: A

Problem 2:
The problem states that in the card game Buffalo Shuffle-o, all cards are distributed evenly among players. Initially, Annika and 3 friends play, and each is dealt 15 cards. We need to find out how many cards each player will be dealt if 2 more friends join the next game.

First, we determine the total number of cards. With 4 players initially, each getting 15 cards, the total number of cards = 4 × 15 = 60.

When 2 more friends join, the total number of players becomes 4 + 2 = 6.

To find out how many cards each player will get, we divide the total number of cards by the new number of players: 60 / 6 = 10.

Let's examine the choices:
A. 8
B. 9
C. 10
D. 11
E. 12

The calculated number of cards per player matches option C.

ANSWER 2: C

Problem 3:
The problem defines a operation as a c b d = a·d − b·c and asks us to find the value of 3 1 4 2.

Using the given definition, we substitute a = 3, b = 1, c = 4, and d = 2 into the equation: 3 1 4 2 = 3·2 − 1·4 = 6 - 4 = 2.

Let's examine the choices:
A. −2
B. −1
C. 0
D. 1
E. 2

The calculated value matches option E.

ANSWER 3: E

Problem 4:
Isabella needs to achieve an average grade of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score she could have made on the third test to still reach her goal.

The total score required for an average of 95 on 4 tests = 95 × 4 = 380.

The sum of her first two test scores = 97 + 91 = 188.

So, the total score needed on the last two tests = 380 - 188 = 192.

To minimize the score on the third test, we maximize the score on the fourth test. The maximum score on the fourth test is 100.

Thus, the minimum score on the third test = 192 - 100 = 92.

Let's examine the choices:
A. 90
B. 92
C. 95
D. 96
E. 97

The calculated minimum score matches option B.

ANSWER 4: B

Problem 5:
Bridget gave half of her apples to Ann, then gave 3 to Cassie, and kept 4 for herself. We need to find out how many apples she originally bought.

Let's denote the original number of apples as x.

After giving half to Ann, Bridget had x/2 apples left.

Then, she gave 3 apples to Cassie and kept 4, so x/2 = 3 + 4 = 7.

Thus, x = 7 × 2 = 14.

Let's examine the choices:
A. 3
B. 4
C. 7
D. 11
E. 14

The calculated original number of apples matches option E.

ANSWER 5: E

Problem 6:
The problem states that in a jar, all but 6 marbles are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T, the number of red marbles as R, the number of green marbles as G, and the number of blue marbles as B.

From the given information:
1. T - R = 6 (all but 6 are red)
2. T - G = 8 (all but 8 are green)
3. T - B = 4 (all but 4 are blue)

Adding these equations gives: 3T - (R + G + B) = 18.

Since R + G + B = T (the total number of marbles is the sum of the marbles of each color), we have:
3T - T = 18, which simplifies to 2T = 18.

Thus, T = 18 / 2 = 9.

Let's examine the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

The calculated total number of marbles matches option C.

ANSWER 6: C

Problem 7:
We need to find the sum of the two smallest prime factors of 250.

First, let's factorize 250 into its prime factors: 250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5.

The sum of these two smallest prime factors = 2 + 5 = 7.

Let's examine the choices:
A. 2
B. 5
C. 7
D. 10
E. 12

The calculated sum matches option C.

ANSWER 7: C

Problem 8:
Bag A contains chips labeled 1, 3, and 5. Bag B contains chips labeled 2, 4, and 6. We need to find how many different values are possible for the sum of the numbers on the chips drawn from each bag.

Let's list the possible sums:
- (1,2) = 3
- (1,4) = 5
- (1,6) = 7
- (3,2) = 5
- (3,4) = 7
- (3,6) = 9
- (5,2) = 7
- (5,4) = 9
- (5,6) = 11

The possible sums are 3, 5, 7, 9, and 11. There are 5 different values.

Let's examine the choices:
A. 4
B. 5
C. 6
D. 7
E. 9

The number of different sums matches option B.

ANSWER 8: B

Problem 9:
The chess team consists of 2 boys and 3 girls. They need to be arranged with a boy at each end and the girls in the middle. We need to find out how many such arrangements are possible.

The boys can be arranged at the ends in 2! = 2 ways (either boy can be on the left or the right).

The 3 girls can be arranged in the 3 middle positions in 3! = 6 ways.

The total number of arrangements = (number of ways to arrange boys) × (number of ways to arrange girls) = 2 × 6 = 12.

Let's examine the choices:
A. 2
B. 4
C. 5
D. 6
E. 12

The calculated total number of arrangements matches option E.

ANSWER 9: E

Problem 10:
The problem asks us to evaluate the expression 16 + 8 ÷ 4 − 2.

Following the order of operations (PEMDAS/BODMAS: Parentheses/Brackets, Exponents/Orders, Multiplication and Division from left to right, Addition and Subtraction from left to right):
1. D

Problem 1:
The water rise is determined by the volume of the rock divided by the base area of the aquarium.  
Base area = \(100 \times 40 = 4000\text{ cm}^2\).  
Volume of rock = \(1000\text{ cm}^3\).  
Rise = \(\frac{1000}{4000} = 0.25\text{ cm}\).  
ANSWER 1: A

Problem 2:
Initially, Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 2: C

Problem 3:
The notation \(\begin{array}{cc} a & c \\ b & d \end{array} = a \cdot d - b \cdot c\).  
For \(\begin{array}{cc} 3 & 1 \\ 4 & 2 \end{array}\): \(a=3, c=1, b=4, d=2\).  
Value = \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).  
ANSWER 3: E

Problem 4:
Total points needed for an average of 95 over 4 tests = \(4 \times 95 = 380\).  
First two tests sum = \(97 + 91 = 188\).  
Remaining points needed = \(380 - 188 = 192\).  
To minimize the third test score, maximize the fourth (100). Then third test \(\ge 192 - 100 = 92\).  
Lowest possible third test score is 92.  
ANSWER 4: B

Problem 5:
Let \(x\) be the number of apples Bridget bought.  
She gave half to Ann, so she kept \(\frac{x}{2}\).  
Then she gave 3 to Cassie and kept 4, so \(\frac{x}{2} = 3 + 4 = 7\).  
Thus \(x = 14\).  
ANSWER 5: E

Problem 6:
Let \(R, G, B\) be the number of red, green, blue marbles.  
"All but 6 are red" \(\Rightarrow G + B = 6\).  
"All but 8 are green" \(\Rightarrow R + B = 8\).  
"All but 4 are blue" \(\Rightarrow R + G = 4\).  
Adding: \((G+B)+(R+B)+(R+G) = 6+8+4 = 18\) \(\Rightarrow 2(R+G+B) = 18\) \(\Rightarrow R+G+B = 9\).  
Total marbles = 9.  
ANSWER 6: C

Problem 7:
Prime factorization of 250: \(250 = 2 \times 5^3\).  
Prime factors are 2 and 5. The two smallest are 2 and 5.  
Sum = \(2 + 5 = 7\).  
ANSWER 7: C

Problem 8:
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}.  
Possible sums:  
1+2=3, 1+4=5, 1+6=7;  
3+2=5, 3+4=7, 3+6=9;  
5+2=7, 5+4=9, 5+6=11.  
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.  
ANSWER 8: B

Problem 9:
Boys must sit at the two ends.  
Arrange boys: 2 choices for left end, 1 for right end → \(2! = 2\) ways.  
Arrange 3 girls in the middle 3 seats: \(3! = 6\) ways.  
Total arrangements = \(2 \times 6 = 12\).  
ANSWER 9: E

Problem 10:
Interpret "8 4" as \(8 \div 4\).  
\(16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16\).  
ANSWER 10: D

Problem 11:
\(\sqrt{4} = 2\).  
\(8 \cdot 2 = 16\), \(\sqrt{16} = 4\).  
\(16 \cdot 4 = 64\), \(\sqrt{64} = 8\).  
ANSWER 11: C

Problem 12:
Compare decimals:  
0.99, 0.9099, 0.9, 0.909, 0.9009.  
0.99 is the largest.  
ANSWER 12: A

# Problem 1:

**What is being asked:** A rock of volume 1000 cm³ is submerged in an aquarium with a 100 cm × 40 cm base. How much does the water level rise?

**Setup:** When the rock is submerged, it displaces a volume of water equal to its own volume. This displaced water spreads across the base of the aquarium, causing the level to rise.

$$\text{Volume displaced} = \text{Base Area} \times \text{Rise in water level}$$

**Computation:**
- Base area = $100 \times 40 = 4000 \text{ cm}^2$
- $1000 = 4000 \times h$
- $h = \frac{1000}{4000} = 0.25 \text{ cm}$

**Check for overflow:** The current depth is 37 cm, and 37 + 0.25 = 37.25 cm, which is well below the 50 cm height. No overflow occurs.

**Ruling out other choices:** Choices B–E would require rock volumes of 2000, 4000, 5000, and 10000 cm³ respectively — all larger than the given 1000 cm³.

ANSWER 1: A

---

# Problem 2:

**What is being asked:** With 4 players, each gets 15 cards. With 6 players, how many cards does each get?

**Setup:** The total number of cards in the deck stays the same.

**Computation:**
- Number of players in first game: Annika + 3 friends = 4 players
- Total cards = $4 \times 15 = 60$ cards
- Number of players in second game: $4 + 2 = 6$ players
- Cards per player = $\frac{60}{6} = 10$

**Ruling out other choices:** Some might forget to count Annika (getting 3 players → 45 cards → 45/5 = 9, choice B) or miscount the new total as 5 players (60/5 = 12, choice E).

ANSWER 2: C

---

# Problem 3:

**What is being asked:** Evaluate the custom operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$ for $a=3, b=1, c=4, d=2$.

**Setup:** Substitute directly into the formula.

**Computation:**
$$\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2$$

**Ruling out other choices:** A common mistake is computing $b \cdot c - a \cdot d = 4 - 6 = -2$ (choice A) or mixing up which values go where.

ANSWER 3: E

---

# Problem 4:

**What is being asked:** Isabella needs an average of 95 on four tests. Her first two scores are 97 and 91. What is the lowest possible third test score so she can still reach her goal?

**Setup:**
- Total points needed: $4 \times 95 = 380$
- Sum of first two tests: $97 + 91 = 188$
- Sum needed from tests 3 and 4: $380 - 188 = 192$

**Computation:**
To minimize the third test score, maximize the fourth test score. The maximum possible score on any test is 100.
$$\text{Minimum test 3} = 192 - 100 = 92$$

**Ruling out other choices:** Choice A (90) would require test 4 to be $192 - 90 = 102$, which is impossible on a 100-point test. Choices C, D, E are higher than necessary.

ANSWER 4: B

---

# Problem 5:

**What is being asked:** Bridget bought apples, gave half to Ann, gave 3 to Cassie, and kept 4. How many did she buy?

**Setup:** Let $x$ = total apples bought. Work through the sequence of events.

**Computation:**
- After giving half to Ann: $\frac{x}{2}$ apples remain
- After giving 3 to Cassie: $\frac{x}{2} - 3$ apples remain
- She keeps 4: $\frac{x}{2} - 3 = 4$
- $\frac{x}{2} = 7$
- $x = 14$

**Verification:** 14 apples → give 7 to Ann → 7 left → give 3 to Cassie → 4 left. ✓

**Ruling out other choices:** Choice C (7) is a trap for those who forget to double the 7 remaining apples. Choice D (11) comes from incorrectly adding $4 + 3 + 4$.

ANSWER 5: E

---

# Problem 6:

**What is being asked:** Find the total number of marbles given conditions about each color.

**Setup:** Let $T$ = total marbles, $R$ = red, $G$ = green, $B$ = blue.
- "All but 6 are red" → $G + B = 6$ → $R = T - 6$
- "All but 8 are green" → $R + B = 8$ → $G = T - 8$
- "All but 4 are blue" → $R + G = 4$ → $B = T - 4$

**Computation:**
$$T = R + G + B = (T - 6) + (T - 8) + (T - 4) = 3T - 18$$
$$2T = 18 \implies T = 9$$

**Verification:** $R = 3, G = 1, B = 5$. Check: $G+B=6$ ✓, $R+B=8$ ✓, $R+G=4$ ✓.

ANSWER 6: C

---

# Problem 7:

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Setup:** Find the prime factorization of 250.

**Computation:**
$$250 = 2 \times 125 = 2 \times 5 \times 25 = 2 \times 5 \times 5 \times 5 = 2 \times 5^3$$

The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.
$$\text{Sum} = 2 + 5 = 7$$

**Ruling out other choices:** Choice A (2) is just one factor. Choice B (5) is just one factor. Choice D (10) would be $2 + 5 + 3$ or $5 + 5$, neither of which applies.

ANSWER 7: C

---

# Problem 8:

**What is being asked:** How many distinct sums can be formed by picking one chip from Bag A {1, 3, 5} and one from Bag B {2, 4, 6}?

**Setup:** List all 9 possible sums and count the unique values.

**Computation:**
| | 2 | 4 | 6 |
|---|---|---|---|
| **1** | 3 | 5 | 7 |
| **3** | 5 | 7 | 9 |
| **5** | 7 | 9 | 11 |

Unique sums: {3, 5, 7, 9, 11} → **5 different values**

**Ruling out other choices:** Choice E (9) is the trap of counting all 9 pairs without removing duplicates. Choice D (7) might come from miscounting.

ANSWER 8: B

---

# Problem 9:

**What is being asked:** How many ways can 2 boys and 3 girls sit in a row with boys at the ends and girls in the middle?

**Setup:** The arrangement must be: **B G G G B**

**Computation:**
- Arrange the 2 boys at the 2 end positions: $2! = 2$ ways
- Arrange the 3 girls in the 3 middle positions: $3! = 6$ ways
- Total arrangements: $2 \times 6 = 12$

**Ruling out other choices:** Choice A (2) only counts the boys' arrangements. Choice D (6) only counts the girls' arrangements. Both must be multiplied together.

ANSWER 9: E

---

# Problem 10:

**What is being asked:** Evaluate $16 + 8 \div 4 - 2$ using order of operations.

**Setup:** Apply PEMDAS — division before addition and subtraction.

**Computation:**
$$16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16$$

**Ruling out other choices:**
- Left-to-right without order of operations: $(16+8) \div 4 - 2 = 6 - 2 = 4$ (choice A)
- $(16+8) \div (4-2) = 24 \div 2 = 12$ (choice C)
- $16 + 8 \div (4-2) = 16 + 4 = 20$ (choice E

 Problem 1:
We need to find how much the water level rises when a rock of volume 1000 cm³ is submerged in a rectangular aquarium.

First, find the area of the aquarium’s base:
\[
100\text{ cm} \times 40\text{ cm} = 4000\text{ cm}^2.
\]

When the rock is completely submerged, it displaces 1000 cm³ of water. This displaced volume spreads out over the base, causing the water level to rise. The rise in height \(h\) satisfies:
\[
\text{base area} \times h = \text{volume of rock}.
\]
So,
\[
h = \frac{1000}{4000} = 0.25\text{ cm}.
\]

We should check that the aquarium does not overflow. The current depth is 37 cm and the height is 50 cm, so there is 13 cm of free space—plenty of room for a 0.25 cm rise.

Tempting wrong choices come from using the wrong area (e.g., using the side wall 100×50 or 40×50) or from dividing by the height 50. Only the base area matters for the level rise.

ANSWER 1: A

---

Problem 2:
We need to find how many cards each player gets when two extra friends join the game.

First, determine the total number of cards. Annika and 3 friends make 4 players, and each gets 15 cards:
\[
4 \times 15 = 60\text{ cards total}.
\]

When 2 more friends join, the number of players becomes:
\[
4 + 2 = 6.
\]

Distribute the 60 cards evenly among 6 players:
\[
60 \div 6 = 10.
\]

A tempting mistake is to subtract 2 from 15, giving 13, which is not an option, or to guess without computing the total deck size. The key is that the total number of cards stays the same.

ANSWER 2: C

---

Problem 3:
We are given the rule \(a\ c\ b\ d = a\cdot d - b\cdot c\). We need to evaluate \(3\ 1\ 4\ 2\).

Matching the pattern:
- \(a = 3\)
- \(c = 1\)
- \(b = 4\)
- \(d = 2\)

Apply the formula:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

A tempting error is to swap the pairs and compute \(3\cdot 1 - 4\cdot 2 = 3 - 8 = -5\), which is not among the choices, or to read the order incorrectly. Careful matching to \(a, c, b, d\) is essential.

ANSWER 3: E

---

Problem 4:
Isabella wants an average of 95 on four 100-point tests. Her first two scores are 97 and 91. We need the lowest possible score on the third test that still allows her to reach her goal.

Total points needed for a 95 average:
\[
4 \times 95 = 380.
\]

Points earned so far:
\[
97 + 91 = 188.
\]

Points still needed from the last two tests:
\[
380 - 188 = 192.
\]

Let the third test score be \(x\) and the fourth be \(y\), with \(x + y = 192\) and each score at most 100. To minimize \(x\), we maximize \(y\). The highest possible score is 100, so:
\[
x = 192 - 100 = 92.
\]

If she scores 92 on the third test, she can score 100 on the fourth and exactly reach her goal. A tempting wrong choice is 95, which would be the answer if she needed the two remaining tests to average 95, but she can shift points to the fourth test.

ANSWER 4: B

---

Problem 5:
We need to find how many apples Bridget originally bought.

Let the original number be \(x\).
- She gave half to Ann, leaving her with \(\frac{x}{2}\).
- She gave 3 to Cassie and kept 4:
\[
\frac{x}{2} - 3 = 4.
\]

Solve:
\[
\frac{x}{2} = 7 \quad\Rightarrow\quad x = 14.
\]

Check: Start with 14. Half is 7, given to Ann. She has 7 left. Gives 3 to Cassie, keeps 4. This matches.

A tempting error is to add \(3 + 4 = 7\) and stop, forgetting that 7 is only what remained after giving away half, so the original amount must be doubled.

ANSWER 5: E

---

Problem 6:
We need the total number of marbles given three conditions about how many are *not* each color.

Let \(R, G, B\) be the numbers of red, green, and blue marbles.
- “All but 6 are red” means non-red marbles equal 6: \(G + B = 6\).
- “All but 8 are green” means non-green marbles equal 8: \(R + B = 8\).
- “All but 4 are blue” means non-blue marbles equal 4: \(R + G = 4\).

Add all three equations:
\[
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
\]
\[
2R + 2G + 2B = 18
\]
\[
R + G + B = 9.
\]

So there are 9 marbles total. (Checking: \(R=3, G=1, B=5\), which satisfies all three statements.)

A tempting mistake is to add \(6+8+4=18\) and choose that, but that sum counts each marble twice.

ANSWER 6: C

---

Problem 7:
We need the sum of the two smallest prime factors of 250.

Factor 250:
\[
250 = 2 \times 125 = 2 \times 5^3.
\]

The prime factors are 2 and 5. The two smallest (and only) distinct prime factors are 2 and 5. Their sum is:
\[
2 + 5 = 7.
\]

A tempting wrong choice is 10, which comes from adding 5 and 5 (the two smallest factors with multiplicity), but the problem asks for the two smallest prime factors, i.e., distinct primes.

ANSWER 7: C

---

Problem 8:
Bag A has \(\{1, 3, 5\}\) and Bag B has \(\{2, 4, 6\}\). We draw one from each and want the number of distinct possible sums.

Compute all sums:
- \(1+2=3\), \(1+4=5\), \(1+6=7\)
- \(3+2=5\), \(3+4=7\), \(3+6=9\)
- \(5+2=7\), \(5+4=9\), \(5+6=11\)

The distinct values are:
\[
3, 5, 7, 9, 11.
\]

That is 5 different values. A tempting error is to count all 9 ordered pairs without removing duplicates.

ANSWER 8: B

---

Problem 9:
We have 2 boys and 3 girls to arrange in a row with a boy at each end and the three girls in the middle.

Positions: B _ _ _ G G G _ _ _ B (but specifically 5 seats: 1,2,3,4,5 with boys at 1 and 5, girls at 2,3,4).

- Choose the boy for the left end and the boy for the right end: \(2! = 2\) ways.
- Arrange the 3 girls in the middle three seats: \(3! = 6\) ways.

Total arrangements:
\[
2 \times 6 = 12.
\]

A tempting mistake is to ignore the constraint and compute \(5! = 120\), or to forget that the boys can swap places.

ANSWER 9: E

---

Problem 10:
Evaluate \(16 + 8 \div 4 - 2\).

Using order of operations (PEMDAS/BODMAS), division comes before addition and subtraction:
\[
8 \div 4 = 2.
\]

Now substitute back:
\[
16 + 2 - 2 = 16.
\]

A tempting error is to work left-to-right doing addition first: \((16+8)\div 4 - 2 = 24\div 4 - 2 = 6 - 2 = 4\), which is choice A. Another is to do \(16 + 2 = 18\), then \(18 - 2 = 16\), which luckily gives the same corr

Problem 1:
We need the rise in water level when a rock of volume 1000 cm³ is placed in an aquarium with base 100 cm by 40 cm. The rock displaces water equal to its volume, causing the water level to rise by Δh such that base area × Δh = rock volume. Base area = 100 cm × 40 cm = 4000 cm². So Δh = 1000 cm³ / 4000 cm² = 0.25 cm. The initial depth is 37 cm, and after rise it becomes 37.25 cm, which is still below the height of 50 cm, so the rock is fully submerged. The rise is 0.25 cm.

ANSWER 1: A

Problem 2:
When Annika and 3 friends play, there are 4 players, each dealt 15 cards. Total cards = 4 × 15 = 60. In the next game, 2 more friends join, making 6 players. Since the same total number of cards is used, each player gets 60 / 6 = 10 cards.

ANSWER 2: C

Problem 3:
The operation is defined as a c b d = a·d − b·c. For the numbers 3 1 4 2, we have a = 3, c = 1, b = 4, d = 2. Compute a·d − b·c = 3×2 − 4×1 = 6 − 4 = 2.

ANSWER 3: E

Problem 4:
Isabella wants an average of 95 over four tests, so total points needed = 4 × 95 = 380. Her first two scores sum to 97 + 91 = 188. Let the third score be x. After the third test, she can still reach the goal if it is possible to score at most 100 on the fourth test to reach 380. That requires 188 + x + 100 ≥ 380 → x ≥ 380 − 288 = 92. The smallest integer x satisfying this is 92. If x = 91, then even a perfect 100 on the fourth gives 188+91+100 = 379 < 380. So the lowest possible third score is 92.

ANSWER 4: B

Problem 5:
Let the number of apples Bridget bought be N. She gave half to Ann, leaving N/2. Then she gave Cassie 3 apples, leaving N/2 − 3. She keeps 4 apples, so N/2 − 3 = 4 → N/2 = 7 → N = 14.

ANSWER 5: E

Problem 6:
Let T be the total number of marbles. "All but 6 are red" means T − R = 6 → R = T − 6. Similarly, T − G = 8 → G = T − 8, and T − B = 4 → B = T − 4. Since R + G + B = T, we have (T−6)+(T−8)+(T−4) = T → 3T − 18 = T → 2T = 18 → T = 9.

ANSWER 6: C

Problem 7:
Factor 250: 250 = 2 × 5³. Its prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5, and their sum is 7.

ANSWER 7: C

Problem 8:
List all possible sums from drawing one chip from Bag A {1,3,5} and one from Bag B {2,4,6}:
1+2=3, 1+4=5, 1+6=7;
3+2=5, 3+4=7, 3+6=9;
5+2=7, 5+4=9, 5+6=11.
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.

ANSWER 8: B

Problem 9:
We need arrangements of 2 boys and 3 girls in a row with a boy at each end. Choose which boy sits on the left end and which on the right: 2! = 2 ways. The three girls occupy the three middle seats and can be arranged in 3! = 6 ways. Total arrangements = 2 × 6 = 12.

ANSWER 9: E

Problem 10:
The expression is 16 + 8 ÷ 4 − 2 (the missing operator is division, as typical for order-of-operations problems). Following order of operations, division before addition and subtraction: 8 ÷ 4 = 2. Then 16 + 2 = 18, and 18 − 2 = 16.

ANSWER 10: D

Problem 11:
Evaluate step by step: √4 = 2. Then 8 · 2 = 16, so √(8·√4) = √16 = 4. Then 16 · 4 = 64, so √(16·√(8·√4)) = √64 = 8.

ANSWER 11: C

Problem 12:
Compare the decimal numbers:
A. 0.99
B. 0.9099
C. 0.9
D. 0.909
E. 0.9009
Clearly 0.99 is the largest because it is greater than 0.9 and all the others are less than 0.91.

ANSWER 12: A

Problem 1:
The aquarium’s base has area  
\(A = 100\text{ cm} \times 40\text{ cm}=4000\text{ cm}^2\).  
When the rock of volume \(V=1000\text{ cm}^3\) is fully submerged, the water rises by a height \(h\) such that  

\[
A\cdot h = V \quad\Longrightarrow\quad h = \frac{1000}{4000}=0.25\text{ cm}.
\]

The rise cannot be any of the larger listed values (0.5 cm, 1 cm, …) because those would require a volume larger than the rock’s.  

ANSWER 1: A  

---

Problem 2:
With 4 players each receiving 15 cards, the total number of cards is  

\[
4 \times 15 = 60.
\]

If 2 more friends join, there are \(4+2 = 6\) players.  The cards are still all the cards, so each player receives  

\[
\frac{60}{6}=10\text{ cards}.
\]

The other answer choices (8, 9, 11, 12) would not distribute the 60 cards evenly.  

ANSWER 2: C  

---

Problem 3:
The operation is defined by  

\[
a\;c\;b\;d = a\cdot d - b\cdot c.
\]

For the numbers \(3\;1\;4\;2\) we have \(a=3,\;c=1,\;b=4,\;d=2\).  Hence  

\[
3\;1\;4\;2 = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

The other options \(-2,-1,0,1\) would correspond to different combinations that do not satisfy the given formula.  

ANSWER 3: E  

---

Problem 4:
To average 95 over four 100‑point tests, the total points needed are  

\[
4 \times 95 = 380.
\]

After the first two tests Isabella has  

\[
97 + 91 = 188\text{ points}.
\]

Let the third test score be \(x\).  Then the fourth test must supply  

\[
380 - (188 + x) = 192 - x \text{ points}.
\]

The fourth test can give at most 100 points, so we need  

\[
192 - x \le 100 \quad\Longrightarrow\quad x \ge 92.
\]

If she scores \(92\) on the third test, she would need a perfect 100 on the fourth, which is possible.  Any lower score would require more than 100 points on the fourth test, which is impossible.  

Thus the lowest possible third‑test score is **92**.  

ANSWER 4: B  

---

Problem 5:
Let the original number of apples be \(N\).  
She gave half of them to Ann: \(\frac{N}{2}\).  
She then gave Cassie 3 apples and kept 4, leaving  

\[
N - \frac{N}{2} - 3 = 4.
\]

Simplify: \(\frac{N}{2} - 3 = 4 \Rightarrow \frac{N}{2}=7 \Rightarrow N=14\).

If the total had been any of the other answer choices, the steps would not balance (e.g., 11 would leave 2 apples after giving Cassie 3).  

ANSWER 5: E  

---

Problem 6:
Let the total number of marbles be \(T\).  
- “All but 6 are red’’ → red = \(T-6\).  
- “All but 8 are green’’ → green = \(T-8\).  
- “All but 4 are blue’’ → blue = \(T-4\).

These three colors account for all marbles, so  

\[
(T-6)+(T-8)+(T-4)=T \;\Longrightarrow\; 3T-18=T \;\Longrightarrow\; 2T=18 \;\Longrightarrow\; T=9.
\]

The other totals (6, 8, 10, 12) would not satisfy the three “all but…’’ statements simultaneously.  

ANSWER 6: C  

---

Problem 7:
Factor \(250\):  

\[
250 = 2 \times 5^3.
\]

The prime factors are \(2\) and \(5\).  The two smallest are \(2\) and \(5\), whose sum is  

\[
2+5=7.
\]

Choices 2, 5, 10, 12 are either single prime factors or incorrect sums.  

ANSWER 7: C  

---

Problem 8:
Bag A contains \(\{1,3,5\}\); Bag B contains \(\{2,4,6\}\).  
All possible sums \(a+b\) are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The distinct results are \(\{3,5,7,9,11\}\), a total of **5** different sums.  

The other answer choices (4, 6, 7, 9) do not match the count of distinct sums.  

ANSWER 8: B  

---

Problem 9:
Place the two boys at the two ends and the three girls in the middle three seats.

- The two boys can be ordered in \(2!\) ways (left‑right swap).  
- The three girls can be arranged among the three middle seats in \(3! = 6\) ways.

Total arrangements  

\[
2! \times 3! = 2 \times 6 = 12.
\]

Choices 2, 4, 5, 6 are too small; 12 accounts for all possible permutations under the given constraints.  

ANSWER 9: E  

---

Problem 10:
Interpret the expression as  

\[
16 + 8 \div 4 - 2.
\]

Division comes before addition and subtraction, so  

\[
8 \div 4 = 2 \quad\Longrightarrow\quad 16 + 2 - 2 = 16.
\]

The other options (4, 8, 12, 20) would result only if the order of operations were misinterpreted.  

ANSWER 10: D  

---

Problem 11:
Simplify from the innermost radical outward.

\[
\sqrt{4}=2,\qquad 
\sqrt{8\cdot\sqrt{4}} = \sqrt{8\cdot2}= \sqrt{16}=4.
\]

Now  

\[
\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{16\cdot4}= \sqrt{64}=8.
\]

Thus the value is 8, matching choice C.  

ANSWER 11: C  

---

Problem 12:
Compare the decimal values:

- A. \(0.99\)
- B. \(0.9099\)
- C. \(0.9\)
- D. \(0.909\)
- E. \(0.9009\)

Clearly \(0.99\) is greater than \(0.9099\), which is greater than \(0.909\), etc.  
Hence the largest number is \(0.99\).  

ANSWER 12: A

Problem 1:  
An aquarium has a rectangular base of 100 cm by 40 cm and height 50 cm, filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is submerged. By how many centimeters does the water level rise?  

The base area is \(100 \times 40 = 4000 \, \text{cm}^2\). The rock’s volume is 1000 cm³, so the rise in water level is \(\frac{1000}{4000} = 0.25 \, \text{cm}\).  

Tempting choices: 0.5 (B) or 1 (C) might be guessed if base area is miscalculated.  

ANSWER 1: A  

Problem 2:  
Annika and 3 friends (4 players) get 15 cards each. Total cards = \(4 \times 15 = 60\). If 2 more join (6 players), each gets \(\frac{60}{6} = 10\).  

Tempting choices: 8 (A) if total cards miscalculated as 48.  

ANSWER 2: C  

Problem 3:  
Operation defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\): \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\), but the order is \(a c b d\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). However, the problem writes "a c b d" which is interpreted as \(a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). But the choices include 2 (E). Re-evaluating: the operation is \(a c b d = a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\).  

Tempting choices: -2 (A) if signs reversed.  

ANSWER 3: A (Note: Correction based on standard interpretation: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\), but the correct answer based on the operation defined is actually \(3 \cdot 2 - 1 \cdot 4 = 2\), but the problem writes "a c b d", which is \(a \cdot d - b \cdot c\), so 2. However, the answer is 2, but the choices include 2 as E. The correct calculation is 2, so E. Correction: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\).)  

**Correction:** The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\), this is \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\).  

ANSWER 3: E  

Problem 4:  
Isabella wants an average of 95 on four 100-point tests. Total points needed: \(4 \times 95 = 380\). First two scores: 97 + 91 = 188. Let third score be \(x\), fourth be \(y\). Then \(188 + x + y = 380\), so \(x + y = 192\). To minimize \(x\), maximize \(y\): \(y = 100\), so \(x = 92\).  

Tempting choices: 90 (A) if \(y\) not maximized.  

ANSWER 4: B  

Problem 5:  
Bridget gave half apples to Ann, then 3 to Cassie, keeping 4. Let total be \(x\). After giving half to Ann: \(\frac{x}{2}\) left. Then gives 3 to Cassie: \(\frac{x}{2} - 3 = 4\). So \(\frac{x}{2} = 7\), \(x = 14\).  

Tempting choices: 7 (C) if half misinterpreted.  

ANSWER 5: E  

Problem 6:  
Let \(r, g, b\) be red, green, blue marbles. Then:  
- All but 6 are red: \(g + b = 6\)  
- All but 8 are green: \(r + b = 8\)  
- All but 4 are blue: \(r + g = 4\)  
Adding: \(2(r + g + b) = 18\), so total \(r + g + b = 9\).  

Tempting choices: 6 (A) if equations misread.  

ANSWER 6: C  

Problem 7:  
Prime factors of 250: \(250 = 2 \times 5^3\). Two smallest: 2 and 5. Sum: \(2 + 5 = 7\).  

Tempting choices: 5 (B) if only 5 considered.  

ANSWER 7: C  

Problem 8:  
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}. Possible sums:  
- 1+2=3, 1+4=5, 1+6=7  
- 3+2=5, 3+4=7, 3+6=9  
- 5+2=7, 5+4=9, 5+6=11  
Unique sums: 3, 5, 7, 9, 11 → 5 values.  

Tempting choices: 6 (C) if duplicates counted.  

ANSWER 8: B  

Problem 9:  
Two boys at ends: 2 ways. Three girls in middle: \(3! = 6\) ways. Total: \(2 \times 6 = 12\).  

Tempting choices: 6 (D) if boys not permuted.  

ANSWER 9: E  

Problem 10:  
Expression: \(16 + 8 \times 4 - 2\). Order: \(8 \times 4 = 32\), then \(16 + 32 - 2 = 46\)? Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 \times 4 - 2\): \(8 \times 4 = 32\), \(16 + 32 = 48\), \(48 - 2 = 46\). But choices max at 20. Re-interpreting: "16 + 8 4 - 2" might mean \(16 + 8 - 4 - 2 = 18\), not in choices. Alternatively, if "8 4" is 8×4, but 46 not in choices. The expression is likely \(16 + 8 - 4 - 2 = 18\), but not in choices. Correction: The expression is "16 + 8 4 − 2", which might be \(16 + (8 \times 4) - 2 = 16 + 32 - 2 = 46\), but not in choices. Re-reading: It’s "16 + 8 4 − 2", which could be \(16 + 8 - 4 - 2 = 18\), but not in choices. The intended expression is likely \(16 + 8 \times 4 - 2 = 46\), but choices max at 20. Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 - 4 - 2 = 18\), but not in choices. Alternatively, if "8 4" is 84, but 16+84-2=98. Not matching. The expression is likely \( \frac{16 + 8}{4} - 2 = \frac{24}{4} - 2 = 6 - 2 = 4 \).  

Tempting choices: 8 (B) if order misapplied.  

ANSWER 10: A  

Problem 11:  
Expression: \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\).  
- \(\sqrt{4} = 2\)  
- \(\sqrt{8 \cdot 2} = \sqrt{16} = 4\)  
- \(\sqrt{16 \cdot 4} = \sqrt{64} = 8\)  

Tempting choices: 4 (A) if innermost not simplified.  

ANSWER 11: C  

Problem 12:  
Compare decimals:  
- A: 0.99  
- B: 0.9099  
- C: 0.9  
- D: 0.909  
- E: 0.9009  
Largest is 0.99.  

Tempting choices: 0.9099 (B) if decimals misaligned.  

ANSWER 12: A  

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

### Problem 1:
What is asked: An aquarium with a 100cm×40cm rectangular base is filled with water to 37cm depth. A 1000 cm³ rock is submerged; how much does the water level rise?
Step-by-step: When a submerged object displaces water, its volume equals the additional volume of water in the aquarium. The rise in water level Δh satisfies: Volume of rock = base area of aquarium × Δh.
Base area = 100×40 = 4000 cm².
Δh = 1000 / 4000 = 0.25 cm.
Tempting wrong answers: 0.5 (if base area is miscalculated as 2000 cm²) or 1 (wrong division).
ANSWER 1: A

---

### Problem 2:
What is asked: Annika plus 3 friends (4 total players) each get 15 cards. If 2 more join, how many cards does each player get?
Step-by-step: First calculate total number of cards: 4 players × 15 cards = 60 total cards. After 2 join, total players = 4+2=6. Cards per player = 60 / 6 = 10.
Tempting wrong answers: 9 (if you incorrectly count only 3 initial players, leading to 45 total cards) or 12 (wrong division).
ANSWER 2: C

---

### Problem 3:
What is asked: The custom notation a c b d = a·d − b·c. What is the value of 3 1 4 2?
Step-by-step: Match the values to the definition: a=3, c=1, b=4, d=2. Calculate: 3·2 - 4·1 = 6 - 4 = 2.
Tempting wrong answers: -2 (swap a and b incorrectly) or 0 (miscalculate the multiplication).
ANSWER 3: E

---

### Problem 4:
What is asked: Isabella needs a 95 average on 4 100-point tests. She scored 97 and 91 on the first two. What is the lowest possible third test score that lets her reach her goal?
Step-by-step: Total points needed for 95 average: 4×95 = 380 points. Sum of first two tests: 97+91=188. To minimize the third test score, maximize the fourth test (max 100 points). Total points left for tests 3 and 4: 380-188=192. Minimum third test = 192 - 100 = 92.
Tempting wrong answers: 96 (the average of the remaining two tests, not the minimum possible third score).
ANSWER 4: B

---

### Problem 5:
What is asked: Bridget gave half her apples to Ann, 3 to Cassie, and kept 4. How many apples did she buy originally?
Step-by-step: Let N = original number of apples. After giving half to Ann, she has N/2 left. Then N/2 - 3 = 4 → N/2=7 → N=14. Verify: 14 → 7 left after Ann, minus 3 for Cassie = 4 kept, which matches.
Tempting wrong answers: 7 (the number of apples Bridget had after giving half to Ann, not the original total).
ANSWER 5: E

---

### Problem 6:
What is asked: A jar has red, green, blue marbles. All but 6 are red, all but 8 are green, all but 4 are blue. How many total marbles?
Step-by-step: Let T=total marbles, R=red, G=green, B=blue. The conditions translate to: G+B=6, R+B=8, R+G=4. Add all equations: 2(R+G+B)=18 → R+G+B=T=9. Verify: 3 red, 1 green, 5 blue sum to 9, matching all conditions.
Tempting wrong answers: 18 (the sum of 6+8+4, not halved to get total marbles).
ANSWER 6: C

---

### Problem 7:
What is asked: What is the sum of the two smallest prime factors of 250?
Step-by-step: Prime factorization of 250: 250=2×5³. The prime factors are 2 and 5. Their sum is 2+5=7.
Tempting wrong answers: 10 (sum of 5+5, or wrong addition of 2+5).
ANSWER 7: C

---

### Problem 8:
What is asked: Bag A has chips 1,3,5; Bag B has 2,4,6. One chip drawn from each bag, how many distinct sum values are possible?
Step-by-step: List all sums: 1+2=3, 1+4=5,1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7,5+4=9,5+6=11. Distinct sums: 3,5,7,9,11 → 5 total.
Tempting wrong answers: 9 (total number of unordered pairs, not distinct sums).
ANSWER 8: B

---

### Problem 9:
What is asked: 2 boys and 3 girls line up for a photo, with a boy at each end and girls in the middle. How many arrangements are possible?
Step-by-step: Arrange the boys first: 2 boys can be ordered at the ends in 2! = 2 ways. The 3 middle girls can be ordered in 3! =6 ways. Total arrangements: 2×6=12.
Tempting wrong answers: 6 (only the number of ways to arrange the girls, ignoring the boys).
ANSWER 9: E

---

### Problem 10:
What is asked: Evaluate (16+8)/(4-2) (formatted as a fraction in the original problem).
Step-by-step: Calculate numerator:16+8=24, denominator:4-2=2. 24/2=12.
Tempting wrong answers:16 (if order of operations is misapplied to calculate 16 + 8/4 -2=16).
ANSWER 10: C

---

### Problem 11:
What is asked: Evaluate √(16 · √(8 · √4)).
Step-by-step: Calculate from the innermost root first: √4=2. Next: 8×2=16, √16=4. Next:16×4=64, √64=8.
Tempting wrong answers: 4√2 (incorrect root calculation).
ANSWER 11: C

---

### Problem 12:
What is asked: Which number is largest: 0.99, 0.9099, 0.9, 0.909, 0.9009.
Step-by-step: Line up all decimals to compare place values: 0.9900 > 0.9099 > 0.9090 > 0.9009 > 0.9000. The largest is 0.99.
Tempting wrong answers: 0.9099 (the second largest value).
ANSWER 12: A

### Problem 1:
We need to find how much the water level rises when a 1000 cm³ rock is submerged in the aquarium.
1.  The volume of displaced water equals the volume of the submerged rock: 1000 cm³.
2.  The base area of the aquarium is \(100 \, \text{cm} \times 40 \, \text{cm} = 4000 \, \text{cm}^2\).
3.  Let \(h\) be the rise in water level. The volume of the displaced water is also equal to base area × rise in height: \(4000h = 1000\).
4.  Solve for \(h\): \(h = \frac{1000}{4000} = 0.25\) cm.
ANSWER 1: A
---
### Problem 2:
We need to find the number of cards per player when 2 more friends join.
1.  Original number of players: Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).
2.  New number of players: 4 + 2 = 6 players. Total cards remain 60.
3.  Cards per player: \(\frac{60}{6} = 10\).
ANSWER 2: C
---
### Problem 3:
The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). Compute \(3 \, 1 \, 4 \, 2\).
1.  Identify \(a=3, c=1, b=4, d=2\).
2.  Substitute into the formula: \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).
ANSWER 3: E
---
### Problem 4:
Find the lowest possible third test score to get an average of 95 on 4 tests.
1.  Total points needed for an average of 95: \(4 \times 95 = 380\).
2.  Sum of first two test scores: \(97 + 91 = 188\).
3.  Let \(x\) = third test score, \(y\) = fourth test score (max 100, since tests are 100-point). We need \(188 + x + y = 380\), so \(x + y = 192\).
4.  To minimize \(x\), maximize \(y = 100\). Then \(x = 192 - 100 = 92\).
ANSWER 4: B
---
### Problem 5:
Find the original number of apples Bridget bought.
1.  Let \(x\) = total apples. After giving half to Ann, she has \(\frac{x}{2}\) left.
2.  After giving 3 to Cassie, she has \(\frac{x}{2} - 3\) left, which equals 4 (the apples she kept).
3.  Solve: \(\frac{x}{2} - 3 = 4 \implies \frac{x}{2} =7 \implies x=14\).
ANSWER 5: E
---
### Problem 6:
Find the total number of marbles.
1.  Let \(R, G, B\) be the number of red, green, blue marbles, total \(T = R+G+B\).
2.  Translate the statements:
    - All but 6 are red: \(G + B = 6\)
    - All but 8 are green: \(R + B = 8\)
    - All but 4 are blue: \(R + G =4\)
3.  Add all three equations: \(2(R+G+B) = 6+8+4=18 \implies 2T=18 \implies T=9\).
ANSWER 6: C
---
### Problem 7:
Find the sum of the two smallest prime factors of 250.
1.  Factorize 250: \(250 = 2 \times 125 = 2 \times 5^3\).
2.  The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.
3.  Sum: \(2 +5=7\).
ANSWER 7: C
---
### Problem 8:
Find the number of distinct sums when drawing one chip from each bag.
1.  List all possible pairs and sums:
    - Bag A (1,3,5) + Bag B (2,4,6):
      \(1+2=3, 1+4=5, 1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7, 5+4=9, 5+6=11\)
2.  Distinct sums: 3, 5, 7, 9, 11. There are 5 unique values.
ANSWER 8: B
---
### Problem 9:
Find the number of arrangements with boys at each end and girls in the middle.
1.  Arrange the 2 boys at the ends: there are \(2! = 2\) ways (order matters: which boy is left/right).
2.  Arrange the 3 girls in the middle 3 seats: \(3! = 6\) ways.
3.  Total arrangements: \(2 \times 6 =12\).
ANSWER 9: E
---
### Problem 10:
Compute \(16 + 8 \, 4 - 2\). The missing operator between 8 and 4 is division (standard order of operations, common middle school problem):
1.  Interpret as \(16 + \frac{8}{4} - 2\).
2.  Compute division first: \(\frac{8}{4}=2\).
3.  Then add/subtract left to right: \(16 + 2 - 2 =16\).
*(Note: If interpreted as \(16 +8 -4 \times 2\), order of operations gives \(4 \times 2=8\), so \(16+8-8=16\), same result.)*
ANSWER 10: D
---
### Problem 11:
Compute \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\) by working from the innermost root outward.
1.  Innermost: \(\sqrt{4} = 2\).
2.  Next layer: \(\sqrt{8 \cdot 2} = \sqrt{16} =4\).
3.  Outermost: \(\sqrt{16 \cdot 4} = \sqrt{64} =8\).
ANSWER 11: C
---
### Problem 12:
Find the largest decimal.
1.  Rewrite all numbers with 4 decimal places for easy comparison:
    - A: \(0.9900\)
    - B: \(0.9099\)
    - C: \(0.9000\)
    - D: \(0.9090\)
    - E: \(0.9009\)
2.  Compare digit by digit: all have 9 in the tenths place. A has 9 in the hundredths place, while all others have 0. Thus A is the largest.
ANSWER 12: A

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

**Problem 1:**

What is being asked: By how many centimeters does the water level rise when a 1000 cm³ rock is submerged?

Setup: The base area is 100 cm × 40 cm = 4000 cm². When the rock is submerged, it displaces 1000 cm³ of water.

Water level rise = Volume displaced ÷ Base area = 1000 ÷ 4000 = 0.25 cm

ANSWER 1: A

---

**Problem 2:**

What is being asked: If cards are distributed evenly, how many cards per player when the player count increases?

Setup: Initially 4 players (Annika + 3 friends) each get 15 cards.
Total cards = 4 × 15 = 60 cards

With 2 more friends joining: 4 + 2 = 6 players
Cards per player = 60 ÷ 6 = 10 cards

ANSWER 2: C

---

**Problem 3:**

What is being asked: Evaluate the operation 3 1 4 2 using the rule a c b d = a·d − b·c

Setup: With a = 3, c = 1, b = 4, d = 2:
3 1 4 2 = 3·2 − 4·1 = 6 − 4 = 2

ANSWER 3: E

---

**Problem 4:**

What is being asked: What is the lowest score on test 3 that allows Isabella to achieve a 95 average on four tests?

Setup: Goal average is 95, so total needed = 95 × 4 = 380 points
First two tests: 97 + 91 = 188 points
Points needed from tests 3 and 4: 380 − 188 = 192 points

Let x = score on test 3. She can still reach her goal if she can score 100 on test 4:
x + 100 ≥ 192
x ≥ 92

The lowest possible score on test 3 is 92.

ANSWER 4: B

---

**Problem 5:**

What is being asked: How many apples did Bridget buy originally?

Setup: Let x = original number of apples
- She gave half to Ann: x/2 apples given away
- Remaining: x/2
- She gave 3 to Cassie: x/2 − 3
- She kept 4 for herself: x/2 − 3 = 4

Solving: x/2 = 7, so x = 14

ANSWER 5: E

---

**Problem 6:**

What is being asked: How many marbles are in the jar?

Setup: Let R, G, B = numbers of red, green, blue marbles; T = total
- "All but 6 are red": G + B = 6
- "All but 8 are green": R + B = 8
- "All but 4 are blue": R + G = 4

Adding all three equations: 2(R + G + B) = 6 + 8 + 4 = 18
Therefore: R + G + B = 9

ANSWER 6: C

---

**Problem 7:**

What is being asked: What is the sum of the two smallest prime factors of 250?

Setup: Factor 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5³

Prime factors: 2 and 5
Sum = 2 + 5 = 7

ANSWER 7: C

---

**Problem 8:**

What is being asked: How many different sums are possible?

Setup: Bag A: {1, 3, 5}, Bag B: {2, 4, 6}
All possible sums:
- 1 + 2 = 3
- 1 + 4 = 5
- 1 + 6 = 7
- 3 + 2 = 5 (duplicate)
- 3 + 4 = 7 (duplicate)
- 3 + 6 = 9
- 5 + 2 = 7 (duplicate)
- 5 + 4 = 9 (duplicate)
- 5 + 6 = 11

Distinct sums: {3, 5, 7, 9, 11} = 5 different values

ANSWER 8: B

---

**Problem 9:**

What is being asked: How many seating arrangements with boys at ends and girls in middle?

Setup: 2 boys at the ends can be arranged in 2! = 2 ways
3 girls in the middle can be arranged in 3! = 6 ways
Total arrangements = 2 × 6 = 12

ANSWER 9: E

---

**Problem 10:**

What is being asked: Evaluate 16 + 8 ÷ 4 − 2

Setup: Using order of operations (division before addition/subtraction):
16 + (8 ÷ 4) − 2 = 16 + 2 − 2 = 16

ANSWER 10: D

---

**Problem 11:**

What is being asked: Evaluate √(16 · √(8 · √4))

Setup: Work from the inside out:
- √4 = 2
- 8 · √4 = 8 · 2 = 16
- √(8 · √4) = √16 = 4
- 16 · √(8 · √4) = 16 · 4 = 64
- √(16 · √(8 · √4)) = √64 = 8

ANSWER 11: C

---

**Problem 12:**

What is being asked: Which decimal is largest?

Setup: Compare decimals:
- A: 0.99
- B: 0.9099
- C: 0.9
- D: 0.909
- E: 0.9009

Comparing: 0.99 > 0.9099 > 0.909 > 0.9009 > 0.9

The largest is 0.99

ANSWER 12: A

Problem 1:

The aquarium has a rectangular base of \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is

\[
100 \cdot 40 = 4000 \text{ cm}^2.
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

If the water rises by \(h\) centimeters, then

\[
4000h = 1000.
\]

So

\[
h = \frac{1000}{4000} = \frac14 = 0.25.
\]

So the water level rises \(0.25\) cm.

ANSWER 1: A

---

Problem 2:

Annika and 3 friends means there are

\[
4 \text{ players}.
\]

If each player gets 15 cards, then the total number of cards is

\[
4 \cdot 15 = 60.
\]

Now 2 more friends join, so there are

\[
6 \text{ players}.
\]

The same 60 cards are shared evenly, so each gets

\[
60 \div 6 = 10.
\]

ANSWER 2: C

---

Problem 3:

We are told that if the four numbers are \(a, b, c, d\), then

\[
a \ b \ c \ d = a\cdot d - b\cdot c.
\]

So for \(3,1,4,2\),

\[
3 \ 1 \ 4 \ 2 = 3\cdot 2 - 1\cdot 4 = 6 - 4 = 2.
\]

ANSWER 3: E

---

Problem 4:

She wants an average of 95 on 4 tests, so the total of the 4 scores must be

\[
4 \cdot 95 = 380.
\]

Her first two scores are 97 and 91, which add to

\[
97+91=188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To make the total 380, the fourth score would have to be

\[
380 - (188+x) = 192 - x.
\]

For this to be possible, the fourth score must be at most 100:

\[
192 - x \le 100.
\]

So

\[
x \ge 92.
\]

The lowest possible third score is 92.

ANSWER 4: B

---

Problem 5:

Let the number of apples Bridget bought be \(x\).

She gave half to Ann, so she kept the other half:

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

Then she gave Cassie 3 apples and kept 4 apples, so before giving Cassie the apples she had

\[
3+4=7.
\]

Thus,

\[
\frac{x}{2}=7 \quad \Rightarrow \quad x=14.
\]

ANSWER 5: E

---

Problem 6:

Let the total number of marbles be \(n\).

- “All but 6 are red” means the number of red marbles is \(n-6\).
- “All but 8 are green” means the number of green marbles is \(n-8\).
- “All but 4 are blue” means the number of blue marbles is \(n-4\).

Since every marble is one of these colors, their counts add to \(n\):

\[
(n-6)+(n-8)+(n-4)=n.
\]

Simplify:

\[
3n-18=n
\]

\[
2n=18
\]

\[
n=9.
\]

ANSWER 6: C

---

Problem 7:

We need the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3.
\]

So the prime factors are 2 and 5. The two smallest prime factors are 2 and 5, and their sum is

\[
2+5=7.
\]

ANSWER 7: C

---

Problem 8:

Bag A has \(1,3,5\); Bag B has \(2,4,6\).

Possible sums:

- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums are

\[
3,5,7,9,11
\]

which is 5 different values.

ANSWER 8: B

---

Problem 9:

There are 2 boys and 3 girls, and the picture must have:

- a boy at each end,
- the 3 girls in the middle.

First, arrange the 2 boys at the ends. They can be switched in

\[
2! = 2
\]

ways.

Then arrange the 3 girls in the middle in

\[
3! = 6
\]

ways.

So the total number of arrangements is

\[
2 \cdot 6 = 12.
\]

ANSWER 9: E

---

Problem 10:

The expression is

\[
16 + 8\cdot 4 - 2.
\]

Use order of operations:

\[
8\cdot 4=32.
\]

Then

\[
16+32-2=46.
\]

But 46 is not among the choices, so the intended expression must be read as

\[
16 + 8 \div 4 - 2.
\]

Then

\[
8\div 4 = 2,
\]

so

\[
16+2-2=16.
\]

ANSWER 10: D

---

Problem 11:

We evaluate from the inside out:

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

First,

\[
\sqrt4=2.
\]

So the expression becomes

\[
\sqrt{16\cdot \sqrt{8\cdot 2}}
=
\sqrt{16\cdot \sqrt{16}}.
\]

Since \(\sqrt{16}=4\),

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C

---

Problem 12:

Compare the decimals:

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

The largest is the one closest to 1. Since \(0.99\) is greater than any number beginning with \(0.90\), it is the largest.

ANSWER 12: A

Problem 1:  
We need the rise in water level when a rock is placed in the aquarium and submerged.

- Base dimensions: \(100\text{ cm} \times 40\text{ cm}\), so base area  
  \[
  A = 100\cdot 40 = 4000\text{ cm}^2.
  \]
- The rock is completely submerged, so the water volume displaced equals the rock’s volume:  
  \[
  V = 1000\text{ cm}^3.
  \]
- Water level rise \(\Delta h\) satisfies \(A\Delta h = V\):  
  \[
  4000\cdot \Delta h = 1000 \implies \Delta h = \frac{1000}{4000}=0.25\text{ cm}.
  \]

So the rise is \(0.25\) cm.  
Tempting wrong idea: using the *remaining* depth \(37\) cm or checking if it overflows—none of that matters; only displaced volume matters.

ANSWER 1: A  

---

Problem 2:  
All cards are dealt evenly. Let the total number of cards be \(T\).

Initially, Annika + 3 friends = 4 players, and each gets 15 cards:  
\[
T = 4\cdot 15 = 60.
\]
Next game: 2 more friends join, so total players:
\[
4+2=6.
\]
Each player gets:
\[
\frac{60}{6}=10.
\]

ANSWER 2: C  

---

Problem 3:  
Given the rule \(a\,b\,d\,c = a\cdot d - b\cdot c\).  
The expression “\(3\ 1\ 4\ 2\)” corresponds to:
- \(a=3\)
- \(b=1\)
- \(d=4\)
- \(c=2\)

Compute:
\[
a\cdot d - b\cdot c = 3\cdot 4 - 1\cdot 2 = 12-2=10.
\]

But notice the choices are only \(-2,-1,0,1,2\), so we must interpret the placeholder correctly:  
The problem states: **“If \(a b c d = a\cdot d - b\cdot c\)”**.

So “\(3\ 1\ 4\ 2\)” means:
- \(a=3,\ b=1,\ c=4,\ d=2\)

Then:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 1\cdot 4 = 6-4=2.
\]

ANSWER 3: E  

---

Problem 4:  
Goal average is 95 over 4 tests, so total points needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
Let the third test score be \(x\), and the fourth be \(y\). Then:
\[
188 + x + y = 380 \implies x+y=192.
\]
We want the **lowest possible** \(x\) such that she can still reach the goal. That means make \(y\) as large as possible (since grades are at most 100).
Maximum \(y=100\). Then:
\[
x + 100 = 192 \implies x = 92.
\]

Check: then the four scores are \(97,91,92,100\) which sum to \(380\), average \(95\). Works.

ANSWER 4: B  

---

Problem 5:  
Let the total number of apples be \(N\).

- She gave half to Ann, so Ann got \(N/2\), leaving \(N/2\).
- Then she gave Cassie 3 apples, and kept 4 apples. That means after giving Cassie 3, her remaining is 4:
  \[
  \frac{N}{2} - 3 = 4 \implies \frac{N}{2}=7 \implies N=14.
  \]

ANSWER 5: E  

---

Problem 6:  
“All but 6 are red” means: number of red marbles is 6.  
Similarly:
- Red = 6 (since all but 6 are red → 6 red)
- Green: all but 8 are green → 8 green
- Blue: all but 4 are blue → 4 blue

Total marbles:
\[
6+8+4=18.
\]
But none of the choices are 18, so interpret carefully:  
“All but 6 are red” means the number of red marbles is **(total − 6)**, not 6.

Let total be \(T\). Then:
- Red marbles = \(T-6\)
- Green marbles = \(T-8\)
- Blue marbles = \(T-4\)

Sum:
\[
(T-6)+(T-8)+(T-4)=T.
\]
\[
3T - (6+8+4)=T
\]
\[
3T-18=T \implies 2T=18 \implies T=9.
\]

ANSWER 6: C  

---

Problem 7:  
We want the sum of the **two smallest prime factors** of 250.

Factor \(250\):
\[
250 = 25\cdot 10 = 5^2 \cdot 2\cdot 5 = 2\cdot 5^3.
\]
Prime factors are \(2\) and \(5\). The two smallest are \(2\) and \(5\).

Sum:
\[
2+5=7.
\]

ANSWER 7: D  

---

Problem 8:  
Bag A: \(\{1,3,5\}\)  
Bag B: \(\{2,4,6\}\)

Possible sums:
- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums: \(\{3,5,7,9,11\}\) → that’s 5 values.

ANSWER 8: B  

---

Problem 9:  
We need arrangements of 5 people in a row with:
- a boy at each end (positions 1 and 5)
- the three girls in the middle (positions 2,3,4)

There are 2 boys: they can be placed at the ends in
\[
2!\ = 2
\]
ways (left boy/right boy).

The 3 girls can be arranged in the three middle spots in
\[
3! = 6
\]
ways.

Total arrangements:
\[
2\cdot 6 = 12.
\]

ANSWER 9: E  

---

Problem 10:  
Expression: \(16 + 8^4 - 2\).

Compute in order:
- \(8^4 = 4096\)

So:
\[
16 + 4096 - 2 = 4110.
\]
No answer choice matches, so we interpret the formatting as \(16 + 8 \sqrt{4} - 2\) or \(16 + 8^4 - 2\)?  
Given the AMC/JH style, likely it is:
\[
16 + 8\sqrt{4} - 2.
\]
Then:
\[
\sqrt{4}=2 \implies 8\sqrt{4}=8\cdot 2=16.
\]
So:
\[
16 + 16 - 2 = 30,
\]
still not in choices.

Another common formatting is:
\[
16 + 8^{4} - 2 \quad \text{(still not)}
\]

The only choiceable consistent possibility is:
\[
16 + \sqrt{8}\,^4 - 2 \text{ (nonsense)}
\]

Let’s use the most plausible literal from the text: **“16 + 8 4 − 2”** often means \(16 + 8^4 - 2\) if the caret was lost. But then no choice.

Given typical AMC problems, the intended is usually:
\[
16 + 8 - 4 - 2 = 18
\]
also not.

Because the answer choices are only \(4,8,12,16,20\), the intended expression must simplify to one of those. The most natural is:
\[
16 + 8/4 - 2 = 16 + 2 - 2 = 16.
\]
This matches a common missing symbol: “8 4” often means \(8/4\).

So:
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D  

---

Problem 11:  
Evaluate:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}.
\]

Work from inside out:

1) \(\sqrt{4}=2\)

2) Inside next root:
\[
8\cdot \sqrt{4} = 8\cdot 2 = 16
\]
So:
\[
\sqrt{8\cdot \sqrt{4}} = \sqrt{16}=4.
\]

3) Now plug into the outer expression:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}=\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C  

---

Problem 12:  
Largest number among:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare by decimal places:

- .99 is 0.99
- .9099 is 0.9099
- .9 is 0.9000
- .909 is 0.9090
- .9009 is 0.9009

Clearly \(0.99\) is greater than all the numbers less than \(0.91\).

ANSWER 12: A

Problem 1:
The volume of the water displaced by the rock is equal to the volume of the rock itself, which is $1000 \text{ cm}^3$. The aquarium has a rectangular base with area $A = 100 \text{ cm} \times 40 \text{ cm} = 4000 \text{ cm}^2$. When the rock is submerged, the water level rises by height $h$ such that the volume of the displaced water is $A \times h$. 
Setting this equal to the rock's volume: $4000 \times h = 1000$.
Solving for $h$: $h = \frac{1000}{4000} = \frac{1}{4} = 0.25 \text{ cm}$.
ANSWER 1: A

Problem 2:
Annika and 3 friends make 4 players total. Each received 15 cards, so the total number of cards is $4 \times 15 = 60$. If 2 more friends join, the total number of players becomes $4 + 2 = 6$. The cards are distributed evenly among the 6 players, so each player gets $60 \div 6 = 10$ cards.
ANSWER 2: C

Problem 3:
The problem defines the operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$.
For the expression $\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix}$, we have $a=3, b=1, c=4, d=2$.
Calculating: $(3 \times 2) - (1 \times 4) = 6 - 4 = 2$.
ANSWER 3: E

Problem 4:
Isabella needs an average of 95 over 4 tests, meaning the total sum of her scores must be $95 \times 4 = 380$. Her first two scores are 97 and 91, totaling $97 + 91 = 188$. Let $x$ be the score on the third test and $y$ be the score on the fourth. We have $188 + x + y = 380$, so $x + y = 192$. To make $x$ as small as possible, we must make $y$ as large as possible. Since the maximum score on a test is 100, we set $y = 100$. Then $x + 100 = 192$, so $x = 92$.
ANSWER 4: B

Problem 5:
Let $x$ be the total number of apples Bridget bought. She gave half away, leaving $\frac{x}{2}$. Then she gave 3 away, leaving $\frac{x}{2} - 3$. She kept 4, so $\frac{x}{2} - 3 = 4$. Adding 3 to both sides, $\frac{x}{2} = 7$. Multiplying by 2, $x = 14$.
ANSWER 5: E

Problem 6:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R + G + B$.
"All but 6 are red" means $G + B = 6$.
"All but 8 are green" means $R + B = 8$.
"All but 4 are blue" means $R + G = 4$.
Adding these three equations: $2(R + G + B) = 6 + 8 + 4 = 18$.
So $R + G + B = 18 \div 2 = 9$.
ANSWER 6: C

Problem 7:
The prime factorization of 250 is $250 = 2 \times 125 = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two prime factors is $2 + 5 = 7$.
ANSWER 7: C

Problem 8:
Bag A has $\{1, 3, 5\}$ and Bag B has $\{2, 4, 6\}$.
Possible sums:
$1+2=3, 1+4=5, 1+6=7$
$3+2=5, 3+4=7, 3+6=9$
$5+2=7, 5+4=9, 5+6=11$
The set of distinct sums is $\{3, 5, 7, 9, 11\}$. There are 5 distinct values.
ANSWER 8: B

Problem 9:
There are 2 boys (B1, B2) and 3 girls (G1, G2, G3). A boy must be at each end. There are 2 choices for the left end and 1 choice for the right end ($2 \times 1 = 2$ ways to place boys). The 3 girls can be arranged in the middle in $3! = 3 \times 2 \times 1 = 6$ ways. Total arrangements = $2 \times 6 = 12$.
ANSWER 9: E

Problem 10:
Following the order of operations (PEMDAS/BODMAS):
$16 + (8 \div 4) - 2 = 16 + 2 - 2 = 16$.
ANSWER 10: D

Problem 11:
Evaluate from the inside out:
$\sqrt{4} = 2$
$\sqrt{8 \times 2} = \sqrt{16} = 4$
$\sqrt{16 \times 4} = \sqrt{64} = 8$
ANSWER 11: C

Problem 12:
To compare, pad with zeros to match the number of decimal places:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
Comparing the values, 0.9900 is the largest.
ANSWER 12: A

Problem 1:
The base of the aquarium measures 100 cm by 40 cm, so its area is \(100 \times 40 = 4000\) cm². The rock has volume 1000 cm³ and is fully submerged, displacing exactly 1000 cm³ of water. The resulting rise in water level is therefore the displaced volume divided by the base area: \(1000 / 4000 = 0.25\) cm. (The original water depth and tank height are irrelevant once the surface area is known.)

ANSWER 1: A

Problem 2:
Annika plus 3 friends gives 4 players, each receiving 15 cards, for a total of \(4 \times 15 = 60\) cards. When 2 more players join, there are 6 players sharing the same 60 cards, so each receives \(60 / 6 = 10\) cards.

ANSWER 2: C

Problem 3:
The given operation evaluates \(a\, c\, b\, d\) as \(a \cdot d - b \cdot c\). Substituting the numbers yields \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).

ANSWER 3: E

Problem 4:
An average of 95 on four 100-point tests requires a total of \(4 \times 95 = 380\) points. The first two scores sum to \(97 + 91 = 188\), leaving 192 points still needed from the last two tests. The fourth test can score at most 100, so the third test must be at least \(192 - 100 = 92\). Any lower score on the third test would make a total of 380 impossible.

ANSWER 4: B

Problem 5:
Let \(t\) be the original number of apples. After giving half to Ann, \(t/2\) remain. After giving 3 more to Cassie, \(t/2 - 3\) remain, and this equals the 4 apples kept. Solving \(t/2 - 3 = 4\) gives \(t/2 = 7\), so \(t = 14\).

ANSWER 5: E

Problem 6:
Let \(m\) be the total number of marbles. Then there are \(m-6\) red, \(m-8\) green, and \(m-4\) blue. Their sum must equal \(m\):
\[
(m-6) + (m-8) + (m-4) = m \implies 3m - 18 = m \implies 2m = 18 \implies m = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies the three “all-but” statements.

ANSWER 6: C

Problem 7:
Factorize \(250 = 2 \cdot 5^3\). The two smallest distinct prime factors are 2 and 5; their sum is 7.

ANSWER 7: C

Problem 8:
All nine possible sums are:
\[
1+2=3,\ 1+4=5,\ 1+6=7,\ 3+2=5,\ 3+4=7,\ 3+6=9,\ 5+2=7,\ 5+4=9,\ 5+6=11.
\]
The distinct values are 3, 5, 7, 9, 11—exactly five.

ANSWER 8: B

Problem 9:
The seating pattern must be boy–girl–girl–girl–boy. The two distinct boys may be placed at the ends in \(2!\) = 2 ways. The three distinct girls may be arranged in the middle three seats in \(3!\) = 6 ways. The total number of arrangements is therefore \(2 \times 6 = 12\).

ANSWER 9: E

Problem 10:
Interpreting the expression with standard order of operations and treating juxtaposition of 8 and 4 as division yields
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D

Problem 11:
Evaluate from the inside outward:
\[
\sqrt{4} = 2, \quad 8 \cdot 2 = 16, \quad \sqrt{16} = 4, \quad 16 \cdot 4 = 64, \quad \sqrt{64} = 8.
\]

ANSWER 11: C

Problem 12:
Compare the decimals directly:
\[
0.99 > 0.9 > 0.9099 > 0.909 > 0.9009.
\]
Thus 0.99 is the largest.

ANSWER 12: A

Problem 1:
The problem asks us to determine by how many centimeters the water level rises when a rock with a volume of 1000 cm³ is submerged in an aquarium filled with water to a depth of 37 cm. The aquarium has a rectangular base measuring 100 cm by 40 cm and a height of 50 cm.

First, we need to calculate the initial volume of water in the aquarium. The volume of water is given by the formula length × width × height. So, the initial volume of water = 100 cm × 40 cm × 37 cm = 148,000 cm³.

When the rock is submerged, it displaces a volume of water equal to its own volume, which is 1000 cm³. This means the water level will rise to accommodate the volume of the rock.

The rise in water level can be calculated by dividing the volume of the rock by the base area of the aquarium. The base area = 100 cm × 40 cm = 4000 cm².

Rise in water level = Volume of rock / Base area = 1000 cm³ / 4000 cm² = 0.25 cm.

Let's examine the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

The calculated rise in water level matches option A.

ANSWER 1: A

Problem 2:
The problem states that in the card game Buffalo Shuffle-o, all cards are distributed evenly among players. Initially, Annika and 3 friends play, and each is dealt 15 cards. We need to find out how many cards each player will be dealt if 2 more friends join the next game.

First, we determine the total number of cards. With 4 players initially, each getting 15 cards, the total number of cards = 4 × 15 = 60.

When 2 more friends join, the total number of players becomes 4 + 2 = 6.

To find out how many cards each player will get, we divide the total number of cards by the new number of players: 60 / 6 = 10.

Let's examine the choices:
A. 8
B. 9
C. 10
D. 11
E. 12

The calculated number of cards per player matches option C.

ANSWER 2: C

Problem 3:
The problem defines a operation as a c b d = a·d − b·c and asks us to find the value of 3 1 4 2.

Using the given definition, we substitute a = 3, b = 1, c = 4, and d = 2 into the equation: 3 1 4 2 = 3·2 − 1·4 = 6 - 4 = 2.

Let's examine the choices:
A. −2
B. −1
C. 0
D. 1
E. 2

The calculated value matches option E.

ANSWER 3: E

Problem 4:
Isabella needs to achieve an average grade of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score she could have made on the third test to still reach her goal.

The total score required for an average of 95 on 4 tests = 95 × 4 = 380.

The sum of her first two test scores = 97 + 91 = 188.

So, the total score needed on the last two tests = 380 - 188 = 192.

To minimize the score on the third test, we maximize the score on the fourth test. The maximum score on the fourth test is 100.

Thus, the minimum score on the third test = 192 - 100 = 92.

Let's examine the choices:
A. 90
B. 92
C. 95
D. 96
E. 97

The calculated minimum score matches option B.

ANSWER 4: B

Problem 5:
Bridget gave half of her apples to Ann, then gave 3 to Cassie, and kept 4 for herself. We need to find out how many apples she originally bought.

Let's denote the original number of apples as x.

After giving half to Ann, Bridget had x/2 apples left.

Then, she gave 3 apples to Cassie and kept 4, so x/2 = 3 + 4 = 7.

Thus, x = 7 × 2 = 14.

Let's examine the choices:
A. 3
B. 4
C. 7
D. 11
E. 14

The calculated original number of apples matches option E.

ANSWER 5: E

Problem 6:
The problem states that in a jar, all but 6 marbles are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T, the number of red marbles as R, the number of green marbles as G, and the number of blue marbles as B.

From the given information:
1. T - R = 6 (all but 6 are red)
2. T - G = 8 (all but 8 are green)
3. T - B = 4 (all but 4 are blue)

Adding these equations gives: 3T - (R + G + B) = 18.

Since R + G + B = T (the total number of marbles is the sum of the marbles of each color), we have:
3T - T = 18, which simplifies to 2T = 18.

Thus, T = 18 / 2 = 9.

Let's examine the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

The calculated total number of marbles matches option C.

ANSWER 6: C

Problem 7:
We need to find the sum of the two smallest prime factors of 250.

First, let's factorize 250 into its prime factors: 250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5.

The sum of these two smallest prime factors = 2 + 5 = 7.

Let's examine the choices:
A. 2
B. 5
C. 7
D. 10
E. 12

The calculated sum matches option C.

ANSWER 7: C

Problem 8:
Bag A contains chips labeled 1, 3, and 5. Bag B contains chips labeled 2, 4, and 6. We need to find how many different values are possible for the sum of the numbers on the chips drawn from each bag.

Let's list the possible sums:
- (1,2) = 3
- (1,4) = 5
- (1,6) = 7
- (3,2) = 5
- (3,4) = 7
- (3,6) = 9
- (5,2) = 7
- (5,4) = 9
- (5,6) = 11

The possible sums are 3, 5, 7, 9, and 11. There are 5 different values.

Let's examine the choices:
A. 4
B. 5
C. 6
D. 7
E. 9

The number of different sums matches option B.

ANSWER 8: B

Problem 9:
The chess team consists of 2 boys and 3 girls. They need to be arranged with a boy at each end and the girls in the middle. We need to find out how many such arrangements are possible.

The boys can be arranged at the ends in 2! = 2 ways (either boy can be on the left or the right).

The 3 girls can be arranged in the 3 middle positions in 3! = 6 ways.

The total number of arrangements = (number of ways to arrange boys) × (number of ways to arrange girls) = 2 × 6 = 12.

Let's examine the choices:
A. 2
B. 4
C. 5
D. 6
E. 12

The calculated total number of arrangements matches option E.

ANSWER 9: E

Problem 10:
The problem asks us to evaluate the expression 16 + 8 ÷ 4 − 2.

Following the order of operations (PEMDAS/BODMAS: Parentheses/Brackets, Exponents/Orders, Multiplication and Division from left to right, Addition and Subtraction from left to right):
1. D

Problem 1:
The water rise is determined by the volume of the rock divided by the base area of the aquarium.  
Base area = \(100 \times 40 = 4000\text{ cm}^2\).  
Volume of rock = \(1000\text{ cm}^3\).  
Rise = \(\frac{1000}{4000} = 0.25\text{ cm}\).  
ANSWER 1: A

Problem 2:
Initially, Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 2: C

Problem 3:
The notation \(\begin{array}{cc} a & c \\ b & d \end{array} = a \cdot d - b \cdot c\).  
For \(\begin{array}{cc} 3 & 1 \\ 4 & 2 \end{array}\): \(a=3, c=1, b=4, d=2\).  
Value = \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).  
ANSWER 3: E

Problem 4:
Total points needed for an average of 95 over 4 tests = \(4 \times 95 = 380\).  
First two tests sum = \(97 + 91 = 188\).  
Remaining points needed = \(380 - 188 = 192\).  
To minimize the third test score, maximize the fourth (100). Then third test \(\ge 192 - 100 = 92\).  
Lowest possible third test score is 92.  
ANSWER 4: B

Problem 5:
Let \(x\) be the number of apples Bridget bought.  
She gave half to Ann, so she kept \(\frac{x}{2}\).  
Then she gave 3 to Cassie and kept 4, so \(\frac{x}{2} = 3 + 4 = 7\).  
Thus \(x = 14\).  
ANSWER 5: E

Problem 6:
Let \(R, G, B\) be the number of red, green, blue marbles.  
"All but 6 are red" \(\Rightarrow G + B = 6\).  
"All but 8 are green" \(\Rightarrow R + B = 8\).  
"All but 4 are blue" \(\Rightarrow R + G = 4\).  
Adding: \((G+B)+(R+B)+(R+G) = 6+8+4 = 18\) \(\Rightarrow 2(R+G+B) = 18\) \(\Rightarrow R+G+B = 9\).  
Total marbles = 9.  
ANSWER 6: C

Problem 7:
Prime factorization of 250: \(250 = 2 \times 5^3\).  
Prime factors are 2 and 5. The two smallest are 2 and 5.  
Sum = \(2 + 5 = 7\).  
ANSWER 7: C

Problem 8:
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}.  
Possible sums:  
1+2=3, 1+4=5, 1+6=7;  
3+2=5, 3+4=7, 3+6=9;  
5+2=7, 5+4=9, 5+6=11.  
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.  
ANSWER 8: B

Problem 9:
Boys must sit at the two ends.  
Arrange boys: 2 choices for left end, 1 for right end → \(2! = 2\) ways.  
Arrange 3 girls in the middle 3 seats: \(3! = 6\) ways.  
Total arrangements = \(2 \times 6 = 12\).  
ANSWER 9: E

Problem 10:
Interpret "8 4" as \(8 \div 4\).  
\(16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16\).  
ANSWER 10: D

Problem 11:
\(\sqrt{4} = 2\).  
\(8 \cdot 2 = 16\), \(\sqrt{16} = 4\).  
\(16 \cdot 4 = 64\), \(\sqrt{64} = 8\).  
ANSWER 11: C

Problem 12:
Compare decimals:  
0.99, 0.9099, 0.9, 0.909, 0.9009.  
0.99 is the largest.  
ANSWER 12: A

# Problem 1:

**What is being asked:** A rock of volume 1000 cm³ is submerged in an aquarium with a 100 cm × 40 cm base. How much does the water level rise?

**Setup:** When the rock is submerged, it displaces a volume of water equal to its own volume. This displaced water spreads across the base of the aquarium, causing the level to rise.

$$\text{Volume displaced} = \text{Base Area} \times \text{Rise in water level}$$

**Computation:**
- Base area = $100 \times 40 = 4000 \text{ cm}^2$
- $1000 = 4000 \times h$
- $h = \frac{1000}{4000} = 0.25 \text{ cm}$

**Check for overflow:** The current depth is 37 cm, and 37 + 0.25 = 37.25 cm, which is well below the 50 cm height. No overflow occurs.

**Ruling out other choices:** Choices B–E would require rock volumes of 2000, 4000, 5000, and 10000 cm³ respectively — all larger than the given 1000 cm³.

ANSWER 1: A

---

# Problem 2:

**What is being asked:** With 4 players, each gets 15 cards. With 6 players, how many cards does each get?

**Setup:** The total number of cards in the deck stays the same.

**Computation:**
- Number of players in first game: Annika + 3 friends = 4 players
- Total cards = $4 \times 15 = 60$ cards
- Number of players in second game: $4 + 2 = 6$ players
- Cards per player = $\frac{60}{6} = 10$

**Ruling out other choices:** Some might forget to count Annika (getting 3 players → 45 cards → 45/5 = 9, choice B) or miscount the new total as 5 players (60/5 = 12, choice E).

ANSWER 2: C

---

# Problem 3:

**What is being asked:** Evaluate the custom operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$ for $a=3, b=1, c=4, d=2$.

**Setup:** Substitute directly into the formula.

**Computation:**
$$\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2$$

**Ruling out other choices:** A common mistake is computing $b \cdot c - a \cdot d = 4 - 6 = -2$ (choice A) or mixing up which values go where.

ANSWER 3: E

---

# Problem 4:

**What is being asked:** Isabella needs an average of 95 on four tests. Her first two scores are 97 and 91. What is the lowest possible third test score so she can still reach her goal?

**Setup:**
- Total points needed: $4 \times 95 = 380$
- Sum of first two tests: $97 + 91 = 188$
- Sum needed from tests 3 and 4: $380 - 188 = 192$

**Computation:**
To minimize the third test score, maximize the fourth test score. The maximum possible score on any test is 100.
$$\text{Minimum test 3} = 192 - 100 = 92$$

**Ruling out other choices:** Choice A (90) would require test 4 to be $192 - 90 = 102$, which is impossible on a 100-point test. Choices C, D, E are higher than necessary.

ANSWER 4: B

---

# Problem 5:

**What is being asked:** Bridget bought apples, gave half to Ann, gave 3 to Cassie, and kept 4. How many did she buy?

**Setup:** Let $x$ = total apples bought. Work through the sequence of events.

**Computation:**
- After giving half to Ann: $\frac{x}{2}$ apples remain
- After giving 3 to Cassie: $\frac{x}{2} - 3$ apples remain
- She keeps 4: $\frac{x}{2} - 3 = 4$
- $\frac{x}{2} = 7$
- $x = 14$

**Verification:** 14 apples → give 7 to Ann → 7 left → give 3 to Cassie → 4 left. ✓

**Ruling out other choices:** Choice C (7) is a trap for those who forget to double the 7 remaining apples. Choice D (11) comes from incorrectly adding $4 + 3 + 4$.

ANSWER 5: E

---

# Problem 6:

**What is being asked:** Find the total number of marbles given conditions about each color.

**Setup:** Let $T$ = total marbles, $R$ = red, $G$ = green, $B$ = blue.
- "All but 6 are red" → $G + B = 6$ → $R = T - 6$
- "All but 8 are green" → $R + B = 8$ → $G = T - 8$
- "All but 4 are blue" → $R + G = 4$ → $B = T - 4$

**Computation:**
$$T = R + G + B = (T - 6) + (T - 8) + (T - 4) = 3T - 18$$
$$2T = 18 \implies T = 9$$

**Verification:** $R = 3, G = 1, B = 5$. Check: $G+B=6$ ✓, $R+B=8$ ✓, $R+G=4$ ✓.

ANSWER 6: C

---

# Problem 7:

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Setup:** Find the prime factorization of 250.

**Computation:**
$$250 = 2 \times 125 = 2 \times 5 \times 25 = 2 \times 5 \times 5 \times 5 = 2 \times 5^3$$

The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.
$$\text{Sum} = 2 + 5 = 7$$

**Ruling out other choices:** Choice A (2) is just one factor. Choice B (5) is just one factor. Choice D (10) would be $2 + 5 + 3$ or $5 + 5$, neither of which applies.

ANSWER 7: C

---

# Problem 8:

**What is being asked:** How many distinct sums can be formed by picking one chip from Bag A {1, 3, 5} and one from Bag B {2, 4, 6}?

**Setup:** List all 9 possible sums and count the unique values.

**Computation:**
| | 2 | 4 | 6 |
|---|---|---|---|
| **1** | 3 | 5 | 7 |
| **3** | 5 | 7 | 9 |
| **5** | 7 | 9 | 11 |

Unique sums: {3, 5, 7, 9, 11} → **5 different values**

**Ruling out other choices:** Choice E (9) is the trap of counting all 9 pairs without removing duplicates. Choice D (7) might come from miscounting.

ANSWER 8: B

---

# Problem 9:

**What is being asked:** How many ways can 2 boys and 3 girls sit in a row with boys at the ends and girls in the middle?

**Setup:** The arrangement must be: **B G G G B**

**Computation:**
- Arrange the 2 boys at the 2 end positions: $2! = 2$ ways
- Arrange the 3 girls in the 3 middle positions: $3! = 6$ ways
- Total arrangements: $2 \times 6 = 12$

**Ruling out other choices:** Choice A (2) only counts the boys' arrangements. Choice D (6) only counts the girls' arrangements. Both must be multiplied together.

ANSWER 9: E

---

# Problem 10:

**What is being asked:** Evaluate $16 + 8 \div 4 - 2$ using order of operations.

**Setup:** Apply PEMDAS — division before addition and subtraction.

**Computation:**
$$16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16$$

**Ruling out other choices:**
- Left-to-right without order of operations: $(16+8) \div 4 - 2 = 6 - 2 = 4$ (choice A)
- $(16+8) \div (4-2) = 24 \div 2 = 12$ (choice C)
- $16 + 8 \div (4-2) = 16 + 4 = 20$ (choice E

 Problem 1:
We need to find how much the water level rises when a rock of volume 1000 cm³ is submerged in a rectangular aquarium.

First, find the area of the aquarium’s base:
\[
100\text{ cm} \times 40\text{ cm} = 4000\text{ cm}^2.
\]

When the rock is completely submerged, it displaces 1000 cm³ of water. This displaced volume spreads out over the base, causing the water level to rise. The rise in height \(h\) satisfies:
\[
\text{base area} \times h = \text{volume of rock}.
\]
So,
\[
h = \frac{1000}{4000} = 0.25\text{ cm}.
\]

We should check that the aquarium does not overflow. The current depth is 37 cm and the height is 50 cm, so there is 13 cm of free space—plenty of room for a 0.25 cm rise.

Tempting wrong choices come from using the wrong area (e.g., using the side wall 100×50 or 40×50) or from dividing by the height 50. Only the base area matters for the level rise.

ANSWER 1: A

---

Problem 2:
We need to find how many cards each player gets when two extra friends join the game.

First, determine the total number of cards. Annika and 3 friends make 4 players, and each gets 15 cards:
\[
4 \times 15 = 60\text{ cards total}.
\]

When 2 more friends join, the number of players becomes:
\[
4 + 2 = 6.
\]

Distribute the 60 cards evenly among 6 players:
\[
60 \div 6 = 10.
\]

A tempting mistake is to subtract 2 from 15, giving 13, which is not an option, or to guess without computing the total deck size. The key is that the total number of cards stays the same.

ANSWER 2: C

---

Problem 3:
We are given the rule \(a\ c\ b\ d = a\cdot d - b\cdot c\). We need to evaluate \(3\ 1\ 4\ 2\).

Matching the pattern:
- \(a = 3\)
- \(c = 1\)
- \(b = 4\)
- \(d = 2\)

Apply the formula:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

A tempting error is to swap the pairs and compute \(3\cdot 1 - 4\cdot 2 = 3 - 8 = -5\), which is not among the choices, or to read the order incorrectly. Careful matching to \(a, c, b, d\) is essential.

ANSWER 3: E

---

Problem 4:
Isabella wants an average of 95 on four 100-point tests. Her first two scores are 97 and 91. We need the lowest possible score on the third test that still allows her to reach her goal.

Total points needed for a 95 average:
\[
4 \times 95 = 380.
\]

Points earned so far:
\[
97 + 91 = 188.
\]

Points still needed from the last two tests:
\[
380 - 188 = 192.
\]

Let the third test score be \(x\) and the fourth be \(y\), with \(x + y = 192\) and each score at most 100. To minimize \(x\), we maximize \(y\). The highest possible score is 100, so:
\[
x = 192 - 100 = 92.
\]

If she scores 92 on the third test, she can score 100 on the fourth and exactly reach her goal. A tempting wrong choice is 95, which would be the answer if she needed the two remaining tests to average 95, but she can shift points to the fourth test.

ANSWER 4: B

---

Problem 5:
We need to find how many apples Bridget originally bought.

Let the original number be \(x\).
- She gave half to Ann, leaving her with \(\frac{x}{2}\).
- She gave 3 to Cassie and kept 4:
\[
\frac{x}{2} - 3 = 4.
\]

Solve:
\[
\frac{x}{2} = 7 \quad\Rightarrow\quad x = 14.
\]

Check: Start with 14. Half is 7, given to Ann. She has 7 left. Gives 3 to Cassie, keeps 4. This matches.

A tempting error is to add \(3 + 4 = 7\) and stop, forgetting that 7 is only what remained after giving away half, so the original amount must be doubled.

ANSWER 5: E

---

Problem 6:
We need the total number of marbles given three conditions about how many are *not* each color.

Let \(R, G, B\) be the numbers of red, green, and blue marbles.
- “All but 6 are red” means non-red marbles equal 6: \(G + B = 6\).
- “All but 8 are green” means non-green marbles equal 8: \(R + B = 8\).
- “All but 4 are blue” means non-blue marbles equal 4: \(R + G = 4\).

Add all three equations:
\[
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
\]
\[
2R + 2G + 2B = 18
\]
\[
R + G + B = 9.
\]

So there are 9 marbles total. (Checking: \(R=3, G=1, B=5\), which satisfies all three statements.)

A tempting mistake is to add \(6+8+4=18\) and choose that, but that sum counts each marble twice.

ANSWER 6: C

---

Problem 7:
We need the sum of the two smallest prime factors of 250.

Factor 250:
\[
250 = 2 \times 125 = 2 \times 5^3.
\]

The prime factors are 2 and 5. The two smallest (and only) distinct prime factors are 2 and 5. Their sum is:
\[
2 + 5 = 7.
\]

A tempting wrong choice is 10, which comes from adding 5 and 5 (the two smallest factors with multiplicity), but the problem asks for the two smallest prime factors, i.e., distinct primes.

ANSWER 7: C

---

Problem 8:
Bag A has \(\{1, 3, 5\}\) and Bag B has \(\{2, 4, 6\}\). We draw one from each and want the number of distinct possible sums.

Compute all sums:
- \(1+2=3\), \(1+4=5\), \(1+6=7\)
- \(3+2=5\), \(3+4=7\), \(3+6=9\)
- \(5+2=7\), \(5+4=9\), \(5+6=11\)

The distinct values are:
\[
3, 5, 7, 9, 11.
\]

That is 5 different values. A tempting error is to count all 9 ordered pairs without removing duplicates.

ANSWER 8: B

---

Problem 9:
We have 2 boys and 3 girls to arrange in a row with a boy at each end and the three girls in the middle.

Positions: B _ _ _ G G G _ _ _ B (but specifically 5 seats: 1,2,3,4,5 with boys at 1 and 5, girls at 2,3,4).

- Choose the boy for the left end and the boy for the right end: \(2! = 2\) ways.
- Arrange the 3 girls in the middle three seats: \(3! = 6\) ways.

Total arrangements:
\[
2 \times 6 = 12.
\]

A tempting mistake is to ignore the constraint and compute \(5! = 120\), or to forget that the boys can swap places.

ANSWER 9: E

---

Problem 10:
Evaluate \(16 + 8 \div 4 - 2\).

Using order of operations (PEMDAS/BODMAS), division comes before addition and subtraction:
\[
8 \div 4 = 2.
\]

Now substitute back:
\[
16 + 2 - 2 = 16.
\]

A tempting error is to work left-to-right doing addition first: \((16+8)\div 4 - 2 = 24\div 4 - 2 = 6 - 2 = 4\), which is choice A. Another is to do \(16 + 2 = 18\), then \(18 - 2 = 16\), which luckily gives the same corr

Problem 1:
We need the rise in water level when a rock of volume 1000 cm³ is placed in an aquarium with base 100 cm by 40 cm. The rock displaces water equal to its volume, causing the water level to rise by Δh such that base area × Δh = rock volume. Base area = 100 cm × 40 cm = 4000 cm². So Δh = 1000 cm³ / 4000 cm² = 0.25 cm. The initial depth is 37 cm, and after rise it becomes 37.25 cm, which is still below the height of 50 cm, so the rock is fully submerged. The rise is 0.25 cm.

ANSWER 1: A

Problem 2:
When Annika and 3 friends play, there are 4 players, each dealt 15 cards. Total cards = 4 × 15 = 60. In the next game, 2 more friends join, making 6 players. Since the same total number of cards is used, each player gets 60 / 6 = 10 cards.

ANSWER 2: C

Problem 3:
The operation is defined as a c b d = a·d − b·c. For the numbers 3 1 4 2, we have a = 3, c = 1, b = 4, d = 2. Compute a·d − b·c = 3×2 − 4×1 = 6 − 4 = 2.

ANSWER 3: E

Problem 4:
Isabella wants an average of 95 over four tests, so total points needed = 4 × 95 = 380. Her first two scores sum to 97 + 91 = 188. Let the third score be x. After the third test, she can still reach the goal if it is possible to score at most 100 on the fourth test to reach 380. That requires 188 + x + 100 ≥ 380 → x ≥ 380 − 288 = 92. The smallest integer x satisfying this is 92. If x = 91, then even a perfect 100 on the fourth gives 188+91+100 = 379 < 380. So the lowest possible third score is 92.

ANSWER 4: B

Problem 5:
Let the number of apples Bridget bought be N. She gave half to Ann, leaving N/2. Then she gave Cassie 3 apples, leaving N/2 − 3. She keeps 4 apples, so N/2 − 3 = 4 → N/2 = 7 → N = 14.

ANSWER 5: E

Problem 6:
Let T be the total number of marbles. "All but 6 are red" means T − R = 6 → R = T − 6. Similarly, T − G = 8 → G = T − 8, and T − B = 4 → B = T − 4. Since R + G + B = T, we have (T−6)+(T−8)+(T−4) = T → 3T − 18 = T → 2T = 18 → T = 9.

ANSWER 6: C

Problem 7:
Factor 250: 250 = 2 × 5³. Its prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5, and their sum is 7.

ANSWER 7: C

Problem 8:
List all possible sums from drawing one chip from Bag A {1,3,5} and one from Bag B {2,4,6}:
1+2=3, 1+4=5, 1+6=7;
3+2=5, 3+4=7, 3+6=9;
5+2=7, 5+4=9, 5+6=11.
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.

ANSWER 8: B

Problem 9:
We need arrangements of 2 boys and 3 girls in a row with a boy at each end. Choose which boy sits on the left end and which on the right: 2! = 2 ways. The three girls occupy the three middle seats and can be arranged in 3! = 6 ways. Total arrangements = 2 × 6 = 12.

ANSWER 9: E

Problem 10:
The expression is 16 + 8 ÷ 4 − 2 (the missing operator is division, as typical for order-of-operations problems). Following order of operations, division before addition and subtraction: 8 ÷ 4 = 2. Then 16 + 2 = 18, and 18 − 2 = 16.

ANSWER 10: D

Problem 11:
Evaluate step by step: √4 = 2. Then 8 · 2 = 16, so √(8·√4) = √16 = 4. Then 16 · 4 = 64, so √(16·√(8·√4)) = √64 = 8.

ANSWER 11: C

Problem 12:
Compare the decimal numbers:
A. 0.99
B. 0.9099
C. 0.9
D. 0.909
E. 0.9009
Clearly 0.99 is the largest because it is greater than 0.9 and all the others are less than 0.91.

ANSWER 12: A

Problem 1:
The aquarium’s base has area  
\(A = 100\text{ cm} \times 40\text{ cm}=4000\text{ cm}^2\).  
When the rock of volume \(V=1000\text{ cm}^3\) is fully submerged, the water rises by a height \(h\) such that  

\[
A\cdot h = V \quad\Longrightarrow\quad h = \frac{1000}{4000}=0.25\text{ cm}.
\]

The rise cannot be any of the larger listed values (0.5 cm, 1 cm, …) because those would require a volume larger than the rock’s.  

ANSWER 1: A  

---

Problem 2:
With 4 players each receiving 15 cards, the total number of cards is  

\[
4 \times 15 = 60.
\]

If 2 more friends join, there are \(4+2 = 6\) players.  The cards are still all the cards, so each player receives  

\[
\frac{60}{6}=10\text{ cards}.
\]

The other answer choices (8, 9, 11, 12) would not distribute the 60 cards evenly.  

ANSWER 2: C  

---

Problem 3:
The operation is defined by  

\[
a\;c\;b\;d = a\cdot d - b\cdot c.
\]

For the numbers \(3\;1\;4\;2\) we have \(a=3,\;c=1,\;b=4,\;d=2\).  Hence  

\[
3\;1\;4\;2 = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

The other options \(-2,-1,0,1\) would correspond to different combinations that do not satisfy the given formula.  

ANSWER 3: E  

---

Problem 4:
To average 95 over four 100‑point tests, the total points needed are  

\[
4 \times 95 = 380.
\]

After the first two tests Isabella has  

\[
97 + 91 = 188\text{ points}.
\]

Let the third test score be \(x\).  Then the fourth test must supply  

\[
380 - (188 + x) = 192 - x \text{ points}.
\]

The fourth test can give at most 100 points, so we need  

\[
192 - x \le 100 \quad\Longrightarrow\quad x \ge 92.
\]

If she scores \(92\) on the third test, she would need a perfect 100 on the fourth, which is possible.  Any lower score would require more than 100 points on the fourth test, which is impossible.  

Thus the lowest possible third‑test score is **92**.  

ANSWER 4: B  

---

Problem 5:
Let the original number of apples be \(N\).  
She gave half of them to Ann: \(\frac{N}{2}\).  
She then gave Cassie 3 apples and kept 4, leaving  

\[
N - \frac{N}{2} - 3 = 4.
\]

Simplify: \(\frac{N}{2} - 3 = 4 \Rightarrow \frac{N}{2}=7 \Rightarrow N=14\).

If the total had been any of the other answer choices, the steps would not balance (e.g., 11 would leave 2 apples after giving Cassie 3).  

ANSWER 5: E  

---

Problem 6:
Let the total number of marbles be \(T\).  
- “All but 6 are red’’ → red = \(T-6\).  
- “All but 8 are green’’ → green = \(T-8\).  
- “All but 4 are blue’’ → blue = \(T-4\).

These three colors account for all marbles, so  

\[
(T-6)+(T-8)+(T-4)=T \;\Longrightarrow\; 3T-18=T \;\Longrightarrow\; 2T=18 \;\Longrightarrow\; T=9.
\]

The other totals (6, 8, 10, 12) would not satisfy the three “all but…’’ statements simultaneously.  

ANSWER 6: C  

---

Problem 7:
Factor \(250\):  

\[
250 = 2 \times 5^3.
\]

The prime factors are \(2\) and \(5\).  The two smallest are \(2\) and \(5\), whose sum is  

\[
2+5=7.
\]

Choices 2, 5, 10, 12 are either single prime factors or incorrect sums.  

ANSWER 7: C  

---

Problem 8:
Bag A contains \(\{1,3,5\}\); Bag B contains \(\{2,4,6\}\).  
All possible sums \(a+b\) are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The distinct results are \(\{3,5,7,9,11\}\), a total of **5** different sums.  

The other answer choices (4, 6, 7, 9) do not match the count of distinct sums.  

ANSWER 8: B  

---

Problem 9:
Place the two boys at the two ends and the three girls in the middle three seats.

- The two boys can be ordered in \(2!\) ways (left‑right swap).  
- The three girls can be arranged among the three middle seats in \(3! = 6\) ways.

Total arrangements  

\[
2! \times 3! = 2 \times 6 = 12.
\]

Choices 2, 4, 5, 6 are too small; 12 accounts for all possible permutations under the given constraints.  

ANSWER 9: E  

---

Problem 10:
Interpret the expression as  

\[
16 + 8 \div 4 - 2.
\]

Division comes before addition and subtraction, so  

\[
8 \div 4 = 2 \quad\Longrightarrow\quad 16 + 2 - 2 = 16.
\]

The other options (4, 8, 12, 20) would result only if the order of operations were misinterpreted.  

ANSWER 10: D  

---

Problem 11:
Simplify from the innermost radical outward.

\[
\sqrt{4}=2,\qquad 
\sqrt{8\cdot\sqrt{4}} = \sqrt{8\cdot2}= \sqrt{16}=4.
\]

Now  

\[
\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{16\cdot4}= \sqrt{64}=8.
\]

Thus the value is 8, matching choice C.  

ANSWER 11: C  

---

Problem 12:
Compare the decimal values:

- A. \(0.99\)
- B. \(0.9099\)
- C. \(0.9\)
- D. \(0.909\)
- E. \(0.9009\)

Clearly \(0.99\) is greater than \(0.9099\), which is greater than \(0.909\), etc.  
Hence the largest number is \(0.99\).  

ANSWER 12: A

Problem 1:  
An aquarium has a rectangular base of 100 cm by 40 cm and height 50 cm, filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is submerged. By how many centimeters does the water level rise?  

The base area is \(100 \times 40 = 4000 \, \text{cm}^2\). The rock’s volume is 1000 cm³, so the rise in water level is \(\frac{1000}{4000} = 0.25 \, \text{cm}\).  

Tempting choices: 0.5 (B) or 1 (C) might be guessed if base area is miscalculated.  

ANSWER 1: A  

Problem 2:  
Annika and 3 friends (4 players) get 15 cards each. Total cards = \(4 \times 15 = 60\). If 2 more join (6 players), each gets \(\frac{60}{6} = 10\).  

Tempting choices: 8 (A) if total cards miscalculated as 48.  

ANSWER 2: C  

Problem 3:  
Operation defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\): \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\), but the order is \(a c b d\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). However, the problem writes "a c b d" which is interpreted as \(a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). But the choices include 2 (E). Re-evaluating: the operation is \(a c b d = a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\).  

Tempting choices: -2 (A) if signs reversed.  

ANSWER 3: A (Note: Correction based on standard interpretation: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\), but the correct answer based on the operation defined is actually \(3 \cdot 2 - 1 \cdot 4 = 2\), but the problem writes "a c b d", which is \(a \cdot d - b \cdot c\), so 2. However, the answer is 2, but the choices include 2 as E. The correct calculation is 2, so E. Correction: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\).)  

**Correction:** The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\), this is \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\).  

ANSWER 3: E  

Problem 4:  
Isabella wants an average of 95 on four 100-point tests. Total points needed: \(4 \times 95 = 380\). First two scores: 97 + 91 = 188. Let third score be \(x\), fourth be \(y\). Then \(188 + x + y = 380\), so \(x + y = 192\). To minimize \(x\), maximize \(y\): \(y = 100\), so \(x = 92\).  

Tempting choices: 90 (A) if \(y\) not maximized.  

ANSWER 4: B  

Problem 5:  
Bridget gave half apples to Ann, then 3 to Cassie, keeping 4. Let total be \(x\). After giving half to Ann: \(\frac{x}{2}\) left. Then gives 3 to Cassie: \(\frac{x}{2} - 3 = 4\). So \(\frac{x}{2} = 7\), \(x = 14\).  

Tempting choices: 7 (C) if half misinterpreted.  

ANSWER 5: E  

Problem 6:  
Let \(r, g, b\) be red, green, blue marbles. Then:  
- All but 6 are red: \(g + b = 6\)  
- All but 8 are green: \(r + b = 8\)  
- All but 4 are blue: \(r + g = 4\)  
Adding: \(2(r + g + b) = 18\), so total \(r + g + b = 9\).  

Tempting choices: 6 (A) if equations misread.  

ANSWER 6: C  

Problem 7:  
Prime factors of 250: \(250 = 2 \times 5^3\). Two smallest: 2 and 5. Sum: \(2 + 5 = 7\).  

Tempting choices: 5 (B) if only 5 considered.  

ANSWER 7: C  

Problem 8:  
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}. Possible sums:  
- 1+2=3, 1+4=5, 1+6=7  
- 3+2=5, 3+4=7, 3+6=9  
- 5+2=7, 5+4=9, 5+6=11  
Unique sums: 3, 5, 7, 9, 11 → 5 values.  

Tempting choices: 6 (C) if duplicates counted.  

ANSWER 8: B  

Problem 9:  
Two boys at ends: 2 ways. Three girls in middle: \(3! = 6\) ways. Total: \(2 \times 6 = 12\).  

Tempting choices: 6 (D) if boys not permuted.  

ANSWER 9: E  

Problem 10:  
Expression: \(16 + 8 \times 4 - 2\). Order: \(8 \times 4 = 32\), then \(16 + 32 - 2 = 46\)? Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 \times 4 - 2\): \(8 \times 4 = 32\), \(16 + 32 = 48\), \(48 - 2 = 46\). But choices max at 20. Re-interpreting: "16 + 8 4 - 2" might mean \(16 + 8 - 4 - 2 = 18\), not in choices. Alternatively, if "8 4" is 8×4, but 46 not in choices. The expression is likely \(16 + 8 - 4 - 2 = 18\), but not in choices. Correction: The expression is "16 + 8 4 − 2", which might be \(16 + (8 \times 4) - 2 = 16 + 32 - 2 = 46\), but not in choices. Re-reading: It’s "16 + 8 4 − 2", which could be \(16 + 8 - 4 - 2 = 18\), but not in choices. The intended expression is likely \(16 + 8 \times 4 - 2 = 46\), but choices max at 20. Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 - 4 - 2 = 18\), but not in choices. Alternatively, if "8 4" is 84, but 16+84-2=98. Not matching. The expression is likely \( \frac{16 + 8}{4} - 2 = \frac{24}{4} - 2 = 6 - 2 = 4 \).  

Tempting choices: 8 (B) if order misapplied.  

ANSWER 10: A  

Problem 11:  
Expression: \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\).  
- \(\sqrt{4} = 2\)  
- \(\sqrt{8 \cdot 2} = \sqrt{16} = 4\)  
- \(\sqrt{16 \cdot 4} = \sqrt{64} = 8\)  

Tempting choices: 4 (A) if innermost not simplified.  

ANSWER 11: C  

Problem 12:  
Compare decimals:  
- A: 0.99  
- B: 0.9099  
- C: 0.9  
- D: 0.909  
- E: 0.9009  
Largest is 0.99.  

Tempting choices: 0.9099 (B) if decimals misaligned.  

ANSWER 12: A  

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

### Problem 1:
What is asked: An aquarium with a 100cm×40cm rectangular base is filled with water to 37cm depth. A 1000 cm³ rock is submerged; how much does the water level rise?
Step-by-step: When a submerged object displaces water, its volume equals the additional volume of water in the aquarium. The rise in water level Δh satisfies: Volume of rock = base area of aquarium × Δh.
Base area = 100×40 = 4000 cm².
Δh = 1000 / 4000 = 0.25 cm.
Tempting wrong answers: 0.5 (if base area is miscalculated as 2000 cm²) or 1 (wrong division).
ANSWER 1: A

---

### Problem 2:
What is asked: Annika plus 3 friends (4 total players) each get 15 cards. If 2 more join, how many cards does each player get?
Step-by-step: First calculate total number of cards: 4 players × 15 cards = 60 total cards. After 2 join, total players = 4+2=6. Cards per player = 60 / 6 = 10.
Tempting wrong answers: 9 (if you incorrectly count only 3 initial players, leading to 45 total cards) or 12 (wrong division).
ANSWER 2: C

---

### Problem 3:
What is asked: The custom notation a c b d = a·d − b·c. What is the value of 3 1 4 2?
Step-by-step: Match the values to the definition: a=3, c=1, b=4, d=2. Calculate: 3·2 - 4·1 = 6 - 4 = 2.
Tempting wrong answers: -2 (swap a and b incorrectly) or 0 (miscalculate the multiplication).
ANSWER 3: E

---

### Problem 4:
What is asked: Isabella needs a 95 average on 4 100-point tests. She scored 97 and 91 on the first two. What is the lowest possible third test score that lets her reach her goal?
Step-by-step: Total points needed for 95 average: 4×95 = 380 points. Sum of first two tests: 97+91=188. To minimize the third test score, maximize the fourth test (max 100 points). Total points left for tests 3 and 4: 380-188=192. Minimum third test = 192 - 100 = 92.
Tempting wrong answers: 96 (the average of the remaining two tests, not the minimum possible third score).
ANSWER 4: B

---

### Problem 5:
What is asked: Bridget gave half her apples to Ann, 3 to Cassie, and kept 4. How many apples did she buy originally?
Step-by-step: Let N = original number of apples. After giving half to Ann, she has N/2 left. Then N/2 - 3 = 4 → N/2=7 → N=14. Verify: 14 → 7 left after Ann, minus 3 for Cassie = 4 kept, which matches.
Tempting wrong answers: 7 (the number of apples Bridget had after giving half to Ann, not the original total).
ANSWER 5: E

---

### Problem 6:
What is asked: A jar has red, green, blue marbles. All but 6 are red, all but 8 are green, all but 4 are blue. How many total marbles?
Step-by-step: Let T=total marbles, R=red, G=green, B=blue. The conditions translate to: G+B=6, R+B=8, R+G=4. Add all equations: 2(R+G+B)=18 → R+G+B=T=9. Verify: 3 red, 1 green, 5 blue sum to 9, matching all conditions.
Tempting wrong answers: 18 (the sum of 6+8+4, not halved to get total marbles).
ANSWER 6: C

---

### Problem 7:
What is asked: What is the sum of the two smallest prime factors of 250?
Step-by-step: Prime factorization of 250: 250=2×5³. The prime factors are 2 and 5. Their sum is 2+5=7.
Tempting wrong answers: 10 (sum of 5+5, or wrong addition of 2+5).
ANSWER 7: C

---

### Problem 8:
What is asked: Bag A has chips 1,3,5; Bag B has 2,4,6. One chip drawn from each bag, how many distinct sum values are possible?
Step-by-step: List all sums: 1+2=3, 1+4=5,1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7,5+4=9,5+6=11. Distinct sums: 3,5,7,9,11 → 5 total.
Tempting wrong answers: 9 (total number of unordered pairs, not distinct sums).
ANSWER 8: B

---

### Problem 9:
What is asked: 2 boys and 3 girls line up for a photo, with a boy at each end and girls in the middle. How many arrangements are possible?
Step-by-step: Arrange the boys first: 2 boys can be ordered at the ends in 2! = 2 ways. The 3 middle girls can be ordered in 3! =6 ways. Total arrangements: 2×6=12.
Tempting wrong answers: 6 (only the number of ways to arrange the girls, ignoring the boys).
ANSWER 9: E

---

### Problem 10:
What is asked: Evaluate (16+8)/(4-2) (formatted as a fraction in the original problem).
Step-by-step: Calculate numerator:16+8=24, denominator:4-2=2. 24/2=12.
Tempting wrong answers:16 (if order of operations is misapplied to calculate 16 + 8/4 -2=16).
ANSWER 10: C

---

### Problem 11:
What is asked: Evaluate √(16 · √(8 · √4)).
Step-by-step: Calculate from the innermost root first: √4=2. Next: 8×2=16, √16=4. Next:16×4=64, √64=8.
Tempting wrong answers: 4√2 (incorrect root calculation).
ANSWER 11: C

---

### Problem 12:
What is asked: Which number is largest: 0.99, 0.9099, 0.9, 0.909, 0.9009.
Step-by-step: Line up all decimals to compare place values: 0.9900 > 0.9099 > 0.9090 > 0.9009 > 0.9000. The largest is 0.99.
Tempting wrong answers: 0.9099 (the second largest value).
ANSWER 12: A

### Problem 1:
We need to find how much the water level rises when a 1000 cm³ rock is submerged in the aquarium.
1.  The volume of displaced water equals the volume of the submerged rock: 1000 cm³.
2.  The base area of the aquarium is \(100 \, \text{cm} \times 40 \, \text{cm} = 4000 \, \text{cm}^2\).
3.  Let \(h\) be the rise in water level. The volume of the displaced water is also equal to base area × rise in height: \(4000h = 1000\).
4.  Solve for \(h\): \(h = \frac{1000}{4000} = 0.25\) cm.
ANSWER 1: A
---
### Problem 2:
We need to find the number of cards per player when 2 more friends join.
1.  Original number of players: Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).
2.  New number of players: 4 + 2 = 6 players. Total cards remain 60.
3.  Cards per player: \(\frac{60}{6} = 10\).
ANSWER 2: C
---
### Problem 3:
The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). Compute \(3 \, 1 \, 4 \, 2\).
1.  Identify \(a=3, c=1, b=4, d=2\).
2.  Substitute into the formula: \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).
ANSWER 3: E
---
### Problem 4:
Find the lowest possible third test score to get an average of 95 on 4 tests.
1.  Total points needed for an average of 95: \(4 \times 95 = 380\).
2.  Sum of first two test scores: \(97 + 91 = 188\).
3.  Let \(x\) = third test score, \(y\) = fourth test score (max 100, since tests are 100-point). We need \(188 + x + y = 380\), so \(x + y = 192\).
4.  To minimize \(x\), maximize \(y = 100\). Then \(x = 192 - 100 = 92\).
ANSWER 4: B
---
### Problem 5:
Find the original number of apples Bridget bought.
1.  Let \(x\) = total apples. After giving half to Ann, she has \(\frac{x}{2}\) left.
2.  After giving 3 to Cassie, she has \(\frac{x}{2} - 3\) left, which equals 4 (the apples she kept).
3.  Solve: \(\frac{x}{2} - 3 = 4 \implies \frac{x}{2} =7 \implies x=14\).
ANSWER 5: E
---
### Problem 6:
Find the total number of marbles.
1.  Let \(R, G, B\) be the number of red, green, blue marbles, total \(T = R+G+B\).
2.  Translate the statements:
    - All but 6 are red: \(G + B = 6\)
    - All but 8 are green: \(R + B = 8\)
    - All but 4 are blue: \(R + G =4\)
3.  Add all three equations: \(2(R+G+B) = 6+8+4=18 \implies 2T=18 \implies T=9\).
ANSWER 6: C
---
### Problem 7:
Find the sum of the two smallest prime factors of 250.
1.  Factorize 250: \(250 = 2 \times 125 = 2 \times 5^3\).
2.  The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.
3.  Sum: \(2 +5=7\).
ANSWER 7: C
---
### Problem 8:
Find the number of distinct sums when drawing one chip from each bag.
1.  List all possible pairs and sums:
    - Bag A (1,3,5) + Bag B (2,4,6):
      \(1+2=3, 1+4=5, 1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7, 5+4=9, 5+6=11\)
2.  Distinct sums: 3, 5, 7, 9, 11. There are 5 unique values.
ANSWER 8: B
---
### Problem 9:
Find the number of arrangements with boys at each end and girls in the middle.
1.  Arrange the 2 boys at the ends: there are \(2! = 2\) ways (order matters: which boy is left/right).
2.  Arrange the 3 girls in the middle 3 seats: \(3! = 6\) ways.
3.  Total arrangements: \(2 \times 6 =12\).
ANSWER 9: E
---
### Problem 10:
Compute \(16 + 8 \, 4 - 2\). The missing operator between 8 and 4 is division (standard order of operations, common middle school problem):
1.  Interpret as \(16 + \frac{8}{4} - 2\).
2.  Compute division first: \(\frac{8}{4}=2\).
3.  Then add/subtract left to right: \(16 + 2 - 2 =16\).
*(Note: If interpreted as \(16 +8 -4 \times 2\), order of operations gives \(4 \times 2=8\), so \(16+8-8=16\), same result.)*
ANSWER 10: D
---
### Problem 11:
Compute \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\) by working from the innermost root outward.
1.  Innermost: \(\sqrt{4} = 2\).
2.  Next layer: \(\sqrt{8 \cdot 2} = \sqrt{16} =4\).
3.  Outermost: \(\sqrt{16 \cdot 4} = \sqrt{64} =8\).
ANSWER 11: C
---
### Problem 12:
Find the largest decimal.
1.  Rewrite all numbers with 4 decimal places for easy comparison:
    - A: \(0.9900\)
    - B: \(0.9099\)
    - C: \(0.9000\)
    - D: \(0.9090\)
    - E: \(0.9009\)
2.  Compare digit by digit: all have 9 in the tenths place. A has 9 in the hundredths place, while all others have 0. Thus A is the largest.
ANSWER 12: A

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

**Problem 1:**

What is being asked: By how many centimeters does the water level rise when a 1000 cm³ rock is submerged?

Setup: The base area is 100 cm × 40 cm = 4000 cm². When the rock is submerged, it displaces 1000 cm³ of water.

Water level rise = Volume displaced ÷ Base area = 1000 ÷ 4000 = 0.25 cm

ANSWER 1: A

---

**Problem 2:**

What is being asked: If cards are distributed evenly, how many cards per player when the player count increases?

Setup: Initially 4 players (Annika + 3 friends) each get 15 cards.
Total cards = 4 × 15 = 60 cards

With 2 more friends joining: 4 + 2 = 6 players
Cards per player = 60 ÷ 6 = 10 cards

ANSWER 2: C

---

**Problem 3:**

What is being asked: Evaluate the operation 3 1 4 2 using the rule a c b d = a·d − b·c

Setup: With a = 3, c = 1, b = 4, d = 2:
3 1 4 2 = 3·2 − 4·1 = 6 − 4 = 2

ANSWER 3: E

---

**Problem 4:**

What is being asked: What is the lowest score on test 3 that allows Isabella to achieve a 95 average on four tests?

Setup: Goal average is 95, so total needed = 95 × 4 = 380 points
First two tests: 97 + 91 = 188 points
Points needed from tests 3 and 4: 380 − 188 = 192 points

Let x = score on test 3. She can still reach her goal if she can score 100 on test 4:
x + 100 ≥ 192
x ≥ 92

The lowest possible score on test 3 is 92.

ANSWER 4: B

---

**Problem 5:**

What is being asked: How many apples did Bridget buy originally?

Setup: Let x = original number of apples
- She gave half to Ann: x/2 apples given away
- Remaining: x/2
- She gave 3 to Cassie: x/2 − 3
- She kept 4 for herself: x/2 − 3 = 4

Solving: x/2 = 7, so x = 14

ANSWER 5: E

---

**Problem 6:**

What is being asked: How many marbles are in the jar?

Setup: Let R, G, B = numbers of red, green, blue marbles; T = total
- "All but 6 are red": G + B = 6
- "All but 8 are green": R + B = 8
- "All but 4 are blue": R + G = 4

Adding all three equations: 2(R + G + B) = 6 + 8 + 4 = 18
Therefore: R + G + B = 9

ANSWER 6: C

---

**Problem 7:**

What is being asked: What is the sum of the two smallest prime factors of 250?

Setup: Factor 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5³

Prime factors: 2 and 5
Sum = 2 + 5 = 7

ANSWER 7: C

---

**Problem 8:**

What is being asked: How many different sums are possible?

Setup: Bag A: {1, 3, 5}, Bag B: {2, 4, 6}
All possible sums:
- 1 + 2 = 3
- 1 + 4 = 5
- 1 + 6 = 7
- 3 + 2 = 5 (duplicate)
- 3 + 4 = 7 (duplicate)
- 3 + 6 = 9
- 5 + 2 = 7 (duplicate)
- 5 + 4 = 9 (duplicate)
- 5 + 6 = 11

Distinct sums: {3, 5, 7, 9, 11} = 5 different values

ANSWER 8: B

---

**Problem 9:**

What is being asked: How many seating arrangements with boys at ends and girls in middle?

Setup: 2 boys at the ends can be arranged in 2! = 2 ways
3 girls in the middle can be arranged in 3! = 6 ways
Total arrangements = 2 × 6 = 12

ANSWER 9: E

---

**Problem 10:**

What is being asked: Evaluate 16 + 8 ÷ 4 − 2

Setup: Using order of operations (division before addition/subtraction):
16 + (8 ÷ 4) − 2 = 16 + 2 − 2 = 16

ANSWER 10: D

---

**Problem 11:**

What is being asked: Evaluate √(16 · √(8 · √4))

Setup: Work from the inside out:
- √4 = 2
- 8 · √4 = 8 · 2 = 16
- √(8 · √4) = √16 = 4
- 16 · √(8 · √4) = 16 · 4 = 64
- √(16 · √(8 · √4)) = √64 = 8

ANSWER 11: C

---

**Problem 12:**

What is being asked: Which decimal is largest?

Setup: Compare decimals:
- A: 0.99
- B: 0.9099
- C: 0.9
- D: 0.909
- E: 0.9009

Comparing: 0.99 > 0.9099 > 0.909 > 0.9009 > 0.9

The largest is 0.99

ANSWER 12: A

Problem 1:

The aquarium has a rectangular base of \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is

\[
100 \cdot 40 = 4000 \text{ cm}^2.
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

If the water rises by \(h\) centimeters, then

\[
4000h = 1000.
\]

So

\[
h = \frac{1000}{4000} = \frac14 = 0.25.
\]

So the water level rises \(0.25\) cm.

ANSWER 1: A

---

Problem 2:

Annika and 3 friends means there are

\[
4 \text{ players}.
\]

If each player gets 15 cards, then the total number of cards is

\[
4 \cdot 15 = 60.
\]

Now 2 more friends join, so there are

\[
6 \text{ players}.
\]

The same 60 cards are shared evenly, so each gets

\[
60 \div 6 = 10.
\]

ANSWER 2: C

---

Problem 3:

We are told that if the four numbers are \(a, b, c, d\), then

\[
a \ b \ c \ d = a\cdot d - b\cdot c.
\]

So for \(3,1,4,2\),

\[
3 \ 1 \ 4 \ 2 = 3\cdot 2 - 1\cdot 4 = 6 - 4 = 2.
\]

ANSWER 3: E

---

Problem 4:

She wants an average of 95 on 4 tests, so the total of the 4 scores must be

\[
4 \cdot 95 = 380.
\]

Her first two scores are 97 and 91, which add to

\[
97+91=188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To make the total 380, the fourth score would have to be

\[
380 - (188+x) = 192 - x.
\]

For this to be possible, the fourth score must be at most 100:

\[
192 - x \le 100.
\]

So

\[
x \ge 92.
\]

The lowest possible third score is 92.

ANSWER 4: B

---

Problem 5:

Let the number of apples Bridget bought be \(x\).

She gave half to Ann, so she kept the other half:

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

Then she gave Cassie 3 apples and kept 4 apples, so before giving Cassie the apples she had

\[
3+4=7.
\]

Thus,

\[
\frac{x}{2}=7 \quad \Rightarrow \quad x=14.
\]

ANSWER 5: E

---

Problem 6:

Let the total number of marbles be \(n\).

- “All but 6 are red” means the number of red marbles is \(n-6\).
- “All but 8 are green” means the number of green marbles is \(n-8\).
- “All but 4 are blue” means the number of blue marbles is \(n-4\).

Since every marble is one of these colors, their counts add to \(n\):

\[
(n-6)+(n-8)+(n-4)=n.
\]

Simplify:

\[
3n-18=n
\]

\[
2n=18
\]

\[
n=9.
\]

ANSWER 6: C

---

Problem 7:

We need the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3.
\]

So the prime factors are 2 and 5. The two smallest prime factors are 2 and 5, and their sum is

\[
2+5=7.
\]

ANSWER 7: C

---

Problem 8:

Bag A has \(1,3,5\); Bag B has \(2,4,6\).

Possible sums:

- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums are

\[
3,5,7,9,11
\]

which is 5 different values.

ANSWER 8: B

---

Problem 9:

There are 2 boys and 3 girls, and the picture must have:

- a boy at each end,
- the 3 girls in the middle.

First, arrange the 2 boys at the ends. They can be switched in

\[
2! = 2
\]

ways.

Then arrange the 3 girls in the middle in

\[
3! = 6
\]

ways.

So the total number of arrangements is

\[
2 \cdot 6 = 12.
\]

ANSWER 9: E

---

Problem 10:

The expression is

\[
16 + 8\cdot 4 - 2.
\]

Use order of operations:

\[
8\cdot 4=32.
\]

Then

\[
16+32-2=46.
\]

But 46 is not among the choices, so the intended expression must be read as

\[
16 + 8 \div 4 - 2.
\]

Then

\[
8\div 4 = 2,
\]

so

\[
16+2-2=16.
\]

ANSWER 10: D

---

Problem 11:

We evaluate from the inside out:

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

First,

\[
\sqrt4=2.
\]

So the expression becomes

\[
\sqrt{16\cdot \sqrt{8\cdot 2}}
=
\sqrt{16\cdot \sqrt{16}}.
\]

Since \(\sqrt{16}=4\),

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C

---

Problem 12:

Compare the decimals:

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

The largest is the one closest to 1. Since \(0.99\) is greater than any number beginning with \(0.90\), it is the largest.

ANSWER 12: A

Problem 1:  
We need the rise in water level when a rock is placed in the aquarium and submerged.

- Base dimensions: \(100\text{ cm} \times 40\text{ cm}\), so base area  
  \[
  A = 100\cdot 40 = 4000\text{ cm}^2.
  \]
- The rock is completely submerged, so the water volume displaced equals the rock’s volume:  
  \[
  V = 1000\text{ cm}^3.
  \]
- Water level rise \(\Delta h\) satisfies \(A\Delta h = V\):  
  \[
  4000\cdot \Delta h = 1000 \implies \Delta h = \frac{1000}{4000}=0.25\text{ cm}.
  \]

So the rise is \(0.25\) cm.  
Tempting wrong idea: using the *remaining* depth \(37\) cm or checking if it overflows—none of that matters; only displaced volume matters.

ANSWER 1: A  

---

Problem 2:  
All cards are dealt evenly. Let the total number of cards be \(T\).

Initially, Annika + 3 friends = 4 players, and each gets 15 cards:  
\[
T = 4\cdot 15 = 60.
\]
Next game: 2 more friends join, so total players:
\[
4+2=6.
\]
Each player gets:
\[
\frac{60}{6}=10.
\]

ANSWER 2: C  

---

Problem 3:  
Given the rule \(a\,b\,d\,c = a\cdot d - b\cdot c\).  
The expression “\(3\ 1\ 4\ 2\)” corresponds to:
- \(a=3\)
- \(b=1\)
- \(d=4\)
- \(c=2\)

Compute:
\[
a\cdot d - b\cdot c = 3\cdot 4 - 1\cdot 2 = 12-2=10.
\]

But notice the choices are only \(-2,-1,0,1,2\), so we must interpret the placeholder correctly:  
The problem states: **“If \(a b c d = a\cdot d - b\cdot c\)”**.

So “\(3\ 1\ 4\ 2\)” means:
- \(a=3,\ b=1,\ c=4,\ d=2\)

Then:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 1\cdot 4 = 6-4=2.
\]

ANSWER 3: E  

---

Problem 4:  
Goal average is 95 over 4 tests, so total points needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
Let the third test score be \(x\), and the fourth be \(y\). Then:
\[
188 + x + y = 380 \implies x+y=192.
\]
We want the **lowest possible** \(x\) such that she can still reach the goal. That means make \(y\) as large as possible (since grades are at most 100).
Maximum \(y=100\). Then:
\[
x + 100 = 192 \implies x = 92.
\]

Check: then the four scores are \(97,91,92,100\) which sum to \(380\), average \(95\). Works.

ANSWER 4: B  

---

Problem 5:  
Let the total number of apples be \(N\).

- She gave half to Ann, so Ann got \(N/2\), leaving \(N/2\).
- Then she gave Cassie 3 apples, and kept 4 apples. That means after giving Cassie 3, her remaining is 4:
  \[
  \frac{N}{2} - 3 = 4 \implies \frac{N}{2}=7 \implies N=14.
  \]

ANSWER 5: E  

---

Problem 6:  
“All but 6 are red” means: number of red marbles is 6.  
Similarly:
- Red = 6 (since all but 6 are red → 6 red)
- Green: all but 8 are green → 8 green
- Blue: all but 4 are blue → 4 blue

Total marbles:
\[
6+8+4=18.
\]
But none of the choices are 18, so interpret carefully:  
“All but 6 are red” means the number of red marbles is **(total − 6)**, not 6.

Let total be \(T\). Then:
- Red marbles = \(T-6\)
- Green marbles = \(T-8\)
- Blue marbles = \(T-4\)

Sum:
\[
(T-6)+(T-8)+(T-4)=T.
\]
\[
3T - (6+8+4)=T
\]
\[
3T-18=T \implies 2T=18 \implies T=9.
\]

ANSWER 6: C  

---

Problem 7:  
We want the sum of the **two smallest prime factors** of 250.

Factor \(250\):
\[
250 = 25\cdot 10 = 5^2 \cdot 2\cdot 5 = 2\cdot 5^3.
\]
Prime factors are \(2\) and \(5\). The two smallest are \(2\) and \(5\).

Sum:
\[
2+5=7.
\]

ANSWER 7: D  

---

Problem 8:  
Bag A: \(\{1,3,5\}\)  
Bag B: \(\{2,4,6\}\)

Possible sums:
- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums: \(\{3,5,7,9,11\}\) → that’s 5 values.

ANSWER 8: B  

---

Problem 9:  
We need arrangements of 5 people in a row with:
- a boy at each end (positions 1 and 5)
- the three girls in the middle (positions 2,3,4)

There are 2 boys: they can be placed at the ends in
\[
2!\ = 2
\]
ways (left boy/right boy).

The 3 girls can be arranged in the three middle spots in
\[
3! = 6
\]
ways.

Total arrangements:
\[
2\cdot 6 = 12.
\]

ANSWER 9: E  

---

Problem 10:  
Expression: \(16 + 8^4 - 2\).

Compute in order:
- \(8^4 = 4096\)

So:
\[
16 + 4096 - 2 = 4110.
\]
No answer choice matches, so we interpret the formatting as \(16 + 8 \sqrt{4} - 2\) or \(16 + 8^4 - 2\)?  
Given the AMC/JH style, likely it is:
\[
16 + 8\sqrt{4} - 2.
\]
Then:
\[
\sqrt{4}=2 \implies 8\sqrt{4}=8\cdot 2=16.
\]
So:
\[
16 + 16 - 2 = 30,
\]
still not in choices.

Another common formatting is:
\[
16 + 8^{4} - 2 \quad \text{(still not)}
\]

The only choiceable consistent possibility is:
\[
16 + \sqrt{8}\,^4 - 2 \text{ (nonsense)}
\]

Let’s use the most plausible literal from the text: **“16 + 8 4 − 2”** often means \(16 + 8^4 - 2\) if the caret was lost. But then no choice.

Given typical AMC problems, the intended is usually:
\[
16 + 8 - 4 - 2 = 18
\]
also not.

Because the answer choices are only \(4,8,12,16,20\), the intended expression must simplify to one of those. The most natural is:
\[
16 + 8/4 - 2 = 16 + 2 - 2 = 16.
\]
This matches a common missing symbol: “8 4” often means \(8/4\).

So:
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D  

---

Problem 11:  
Evaluate:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}.
\]

Work from inside out:

1) \(\sqrt{4}=2\)

2) Inside next root:
\[
8\cdot \sqrt{4} = 8\cdot 2 = 16
\]
So:
\[
\sqrt{8\cdot \sqrt{4}} = \sqrt{16}=4.
\]

3) Now plug into the outer expression:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}=\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C  

---

Problem 12:  
Largest number among:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare by decimal places:

- .99 is 0.99
- .9099 is 0.9099
- .9 is 0.9000
- .909 is 0.9090
- .9009 is 0.9009

Clearly \(0.99\) is greater than all the numbers less than \(0.91\).

ANSWER 12: A

Problem 1:
The volume of the water displaced by the rock is equal to the volume of the rock itself, which is $1000 \text{ cm}^3$. The aquarium has a rectangular base with area $A = 100 \text{ cm} \times 40 \text{ cm} = 4000 \text{ cm}^2$. When the rock is submerged, the water level rises by height $h$ such that the volume of the displaced water is $A \times h$. 
Setting this equal to the rock's volume: $4000 \times h = 1000$.
Solving for $h$: $h = \frac{1000}{4000} = \frac{1}{4} = 0.25 \text{ cm}$.
ANSWER 1: A

Problem 2:
Annika and 3 friends make 4 players total. Each received 15 cards, so the total number of cards is $4 \times 15 = 60$. If 2 more friends join, the total number of players becomes $4 + 2 = 6$. The cards are distributed evenly among the 6 players, so each player gets $60 \div 6 = 10$ cards.
ANSWER 2: C

Problem 3:
The problem defines the operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$.
For the expression $\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix}$, we have $a=3, b=1, c=4, d=2$.
Calculating: $(3 \times 2) - (1 \times 4) = 6 - 4 = 2$.
ANSWER 3: E

Problem 4:
Isabella needs an average of 95 over 4 tests, meaning the total sum of her scores must be $95 \times 4 = 380$. Her first two scores are 97 and 91, totaling $97 + 91 = 188$. Let $x$ be the score on the third test and $y$ be the score on the fourth. We have $188 + x + y = 380$, so $x + y = 192$. To make $x$ as small as possible, we must make $y$ as large as possible. Since the maximum score on a test is 100, we set $y = 100$. Then $x + 100 = 192$, so $x = 92$.
ANSWER 4: B

Problem 5:
Let $x$ be the total number of apples Bridget bought. She gave half away, leaving $\frac{x}{2}$. Then she gave 3 away, leaving $\frac{x}{2} - 3$. She kept 4, so $\frac{x}{2} - 3 = 4$. Adding 3 to both sides, $\frac{x}{2} = 7$. Multiplying by 2, $x = 14$.
ANSWER 5: E

Problem 6:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R + G + B$.
"All but 6 are red" means $G + B = 6$.
"All but 8 are green" means $R + B = 8$.
"All but 4 are blue" means $R + G = 4$.
Adding these three equations: $2(R + G + B) = 6 + 8 + 4 = 18$.
So $R + G + B = 18 \div 2 = 9$.
ANSWER 6: C

Problem 7:
The prime factorization of 250 is $250 = 2 \times 125 = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two prime factors is $2 + 5 = 7$.
ANSWER 7: C

Problem 8:
Bag A has $\{1, 3, 5\}$ and Bag B has $\{2, 4, 6\}$.
Possible sums:
$1+2=3, 1+4=5, 1+6=7$
$3+2=5, 3+4=7, 3+6=9$
$5+2=7, 5+4=9, 5+6=11$
The set of distinct sums is $\{3, 5, 7, 9, 11\}$. There are 5 distinct values.
ANSWER 8: B

Problem 9:
There are 2 boys (B1, B2) and 3 girls (G1, G2, G3). A boy must be at each end. There are 2 choices for the left end and 1 choice for the right end ($2 \times 1 = 2$ ways to place boys). The 3 girls can be arranged in the middle in $3! = 3 \times 2 \times 1 = 6$ ways. Total arrangements = $2 \times 6 = 12$.
ANSWER 9: E

Problem 10:
Following the order of operations (PEMDAS/BODMAS):
$16 + (8 \div 4) - 2 = 16 + 2 - 2 = 16$.
ANSWER 10: D

Problem 11:
Evaluate from the inside out:
$\sqrt{4} = 2$
$\sqrt{8 \times 2} = \sqrt{16} = 4$
$\sqrt{16 \times 4} = \sqrt{64} = 8$
ANSWER 11: C

Problem 12:
To compare, pad with zeros to match the number of decimal places:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
Comparing the values, 0.9900 is the largest.
ANSWER 12: A

Problem 1:
The base of the aquarium measures 100 cm by 40 cm, so its area is \(100 \times 40 = 4000\) cm². The rock has volume 1000 cm³ and is fully submerged, displacing exactly 1000 cm³ of water. The resulting rise in water level is therefore the displaced volume divided by the base area: \(1000 / 4000 = 0.25\) cm. (The original water depth and tank height are irrelevant once the surface area is known.)

ANSWER 1: A

Problem 2:
Annika plus 3 friends gives 4 players, each receiving 15 cards, for a total of \(4 \times 15 = 60\) cards. When 2 more players join, there are 6 players sharing the same 60 cards, so each receives \(60 / 6 = 10\) cards.

ANSWER 2: C

Problem 3:
The given operation evaluates \(a\, c\, b\, d\) as \(a \cdot d - b \cdot c\). Substituting the numbers yields \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).

ANSWER 3: E

Problem 4:
An average of 95 on four 100-point tests requires a total of \(4 \times 95 = 380\) points. The first two scores sum to \(97 + 91 = 188\), leaving 192 points still needed from the last two tests. The fourth test can score at most 100, so the third test must be at least \(192 - 100 = 92\). Any lower score on the third test would make a total of 380 impossible.

ANSWER 4: B

Problem 5:
Let \(t\) be the original number of apples. After giving half to Ann, \(t/2\) remain. After giving 3 more to Cassie, \(t/2 - 3\) remain, and this equals the 4 apples kept. Solving \(t/2 - 3 = 4\) gives \(t/2 = 7\), so \(t = 14\).

ANSWER 5: E

Problem 6:
Let \(m\) be the total number of marbles. Then there are \(m-6\) red, \(m-8\) green, and \(m-4\) blue. Their sum must equal \(m\):
\[
(m-6) + (m-8) + (m-4) = m \implies 3m - 18 = m \implies 2m = 18 \implies m = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies the three “all-but” statements.

ANSWER 6: C

Problem 7:
Factorize \(250 = 2 \cdot 5^3\). The two smallest distinct prime factors are 2 and 5; their sum is 7.

ANSWER 7: C

Problem 8:
All nine possible sums are:
\[
1+2=3,\ 1+4=5,\ 1+6=7,\ 3+2=5,\ 3+4=7,\ 3+6=9,\ 5+2=7,\ 5+4=9,\ 5+6=11.
\]
The distinct values are 3, 5, 7, 9, 11—exactly five.

ANSWER 8: B

Problem 9:
The seating pattern must be boy–girl–girl–girl–boy. The two distinct boys may be placed at the ends in \(2!\) = 2 ways. The three distinct girls may be arranged in the middle three seats in \(3!\) = 6 ways. The total number of arrangements is therefore \(2 \times 6 = 12\).

ANSWER 9: E

Problem 10:
Interpreting the expression with standard order of operations and treating juxtaposition of 8 and 4 as division yields
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D

Problem 11:
Evaluate from the inside outward:
\[
\sqrt{4} = 2, \quad 8 \cdot 2 = 16, \quad \sqrt{16} = 4, \quad 16 \cdot 4 = 64, \quad \sqrt{64} = 8.
\]

ANSWER 11: C

Problem 12:
Compare the decimals directly:
\[
0.99 > 0.9 > 0.9099 > 0.909 > 0.9009.
\]
Thus 0.99 is the largest.

ANSWER 12: A

Problem 1:
The problem asks us to determine by how many centimeters the water level rises when a rock with a volume of 1000 cm³ is submerged in an aquarium filled with water to a depth of 37 cm. The aquarium has a rectangular base measuring 100 cm by 40 cm and a height of 50 cm.

First, we need to calculate the initial volume of water in the aquarium. The volume of water is given by the formula length × width × height. So, the initial volume of water = 100 cm × 40 cm × 37 cm = 148,000 cm³.

When the rock is submerged, it displaces a volume of water equal to its own volume, which is 1000 cm³. This means the water level will rise to accommodate the volume of the rock.

The rise in water level can be calculated by dividing the volume of the rock by the base area of the aquarium. The base area = 100 cm × 40 cm = 4000 cm².

Rise in water level = Volume of rock / Base area = 1000 cm³ / 4000 cm² = 0.25 cm.

Let's examine the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

The calculated rise in water level matches option A.

ANSWER 1: A

Problem 2:
The problem states that in the card game Buffalo Shuffle-o, all cards are distributed evenly among players. Initially, Annika and 3 friends play, and each is dealt 15 cards. We need to find out how many cards each player will be dealt if 2 more friends join the next game.

First, we determine the total number of cards. With 4 players initially, each getting 15 cards, the total number of cards = 4 × 15 = 60.

When 2 more friends join, the total number of players becomes 4 + 2 = 6.

To find out how many cards each player will get, we divide the total number of cards by the new number of players: 60 / 6 = 10.

Let's examine the choices:
A. 8
B. 9
C. 10
D. 11
E. 12

The calculated number of cards per player matches option C.

ANSWER 2: C

Problem 3:
The problem defines a operation as a c b d = a·d − b·c and asks us to find the value of 3 1 4 2.

Using the given definition, we substitute a = 3, b = 1, c = 4, and d = 2 into the equation: 3 1 4 2 = 3·2 − 1·4 = 6 - 4 = 2.

Let's examine the choices:
A. −2
B. −1
C. 0
D. 1
E. 2

The calculated value matches option E.

ANSWER 3: E

Problem 4:
Isabella needs to achieve an average grade of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score she could have made on the third test to still reach her goal.

The total score required for an average of 95 on 4 tests = 95 × 4 = 380.

The sum of her first two test scores = 97 + 91 = 188.

So, the total score needed on the last two tests = 380 - 188 = 192.

To minimize the score on the third test, we maximize the score on the fourth test. The maximum score on the fourth test is 100.

Thus, the minimum score on the third test = 192 - 100 = 92.

Let's examine the choices:
A. 90
B. 92
C. 95
D. 96
E. 97

The calculated minimum score matches option B.

ANSWER 4: B

Problem 5:
Bridget gave half of her apples to Ann, then gave 3 to Cassie, and kept 4 for herself. We need to find out how many apples she originally bought.

Let's denote the original number of apples as x.

After giving half to Ann, Bridget had x/2 apples left.

Then, she gave 3 apples to Cassie and kept 4, so x/2 = 3 + 4 = 7.

Thus, x = 7 × 2 = 14.

Let's examine the choices:
A. 3
B. 4
C. 7
D. 11
E. 14

The calculated original number of apples matches option E.

ANSWER 5: E

Problem 6:
The problem states that in a jar, all but 6 marbles are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T, the number of red marbles as R, the number of green marbles as G, and the number of blue marbles as B.

From the given information:
1. T - R = 6 (all but 6 are red)
2. T - G = 8 (all but 8 are green)
3. T - B = 4 (all but 4 are blue)

Adding these equations gives: 3T - (R + G + B) = 18.

Since R + G + B = T (the total number of marbles is the sum of the marbles of each color), we have:
3T - T = 18, which simplifies to 2T = 18.

Thus, T = 18 / 2 = 9.

Let's examine the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

The calculated total number of marbles matches option C.

ANSWER 6: C

Problem 7:
We need to find the sum of the two smallest prime factors of 250.

First, let's factorize 250 into its prime factors: 250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5.

The sum of these two smallest prime factors = 2 + 5 = 7.

Let's examine the choices:
A. 2
B. 5
C. 7
D. 10
E. 12

The calculated sum matches option C.

ANSWER 7: C

Problem 8:
Bag A contains chips labeled 1, 3, and 5. Bag B contains chips labeled 2, 4, and 6. We need to find how many different values are possible for the sum of the numbers on the chips drawn from each bag.

Let's list the possible sums:
- (1,2) = 3
- (1,4) = 5
- (1,6) = 7
- (3,2) = 5
- (3,4) = 7
- (3,6) = 9
- (5,2) = 7
- (5,4) = 9
- (5,6) = 11

The possible sums are 3, 5, 7, 9, and 11. There are 5 different values.

Let's examine the choices:
A. 4
B. 5
C. 6
D. 7
E. 9

The number of different sums matches option B.

ANSWER 8: B

Problem 9:
The chess team consists of 2 boys and 3 girls. They need to be arranged with a boy at each end and the girls in the middle. We need to find out how many such arrangements are possible.

The boys can be arranged at the ends in 2! = 2 ways (either boy can be on the left or the right).

The 3 girls can be arranged in the 3 middle positions in 3! = 6 ways.

The total number of arrangements = (number of ways to arrange boys) × (number of ways to arrange girls) = 2 × 6 = 12.

Let's examine the choices:
A. 2
B. 4
C. 5
D. 6
E. 12

The calculated total number of arrangements matches option E.

ANSWER 9: E

Problem 10:
The problem asks us to evaluate the expression 16 + 8 ÷ 4 − 2.

Following the order of operations (PEMDAS/BODMAS: Parentheses/Brackets, Exponents/Orders, Multiplication and Division from left to right, Addition and Subtraction from left to right):
1. D

Problem 1:
The water rise is determined by the volume of the rock divided by the base area of the aquarium.  
Base area = \(100 \times 40 = 4000\text{ cm}^2\).  
Volume of rock = \(1000\text{ cm}^3\).  
Rise = \(\frac{1000}{4000} = 0.25\text{ cm}\).  
ANSWER 1: A

Problem 2:
Initially, Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 2: C

Problem 3:
The notation \(\begin{array}{cc} a & c \\ b & d \end{array} = a \cdot d - b \cdot c\).  
For \(\begin{array}{cc} 3 & 1 \\ 4 & 2 \end{array}\): \(a=3, c=1, b=4, d=2\).  
Value = \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).  
ANSWER 3: E

Problem 4:
Total points needed for an average of 95 over 4 tests = \(4 \times 95 = 380\).  
First two tests sum = \(97 + 91 = 188\).  
Remaining points needed = \(380 - 188 = 192\).  
To minimize the third test score, maximize the fourth (100). Then third test \(\ge 192 - 100 = 92\).  
Lowest possible third test score is 92.  
ANSWER 4: B

Problem 5:
Let \(x\) be the number of apples Bridget bought.  
She gave half to Ann, so she kept \(\frac{x}{2}\).  
Then she gave 3 to Cassie and kept 4, so \(\frac{x}{2} = 3 + 4 = 7\).  
Thus \(x = 14\).  
ANSWER 5: E

Problem 6:
Let \(R, G, B\) be the number of red, green, blue marbles.  
"All but 6 are red" \(\Rightarrow G + B = 6\).  
"All but 8 are green" \(\Rightarrow R + B = 8\).  
"All but 4 are blue" \(\Rightarrow R + G = 4\).  
Adding: \((G+B)+(R+B)+(R+G) = 6+8+4 = 18\) \(\Rightarrow 2(R+G+B) = 18\) \(\Rightarrow R+G+B = 9\).  
Total marbles = 9.  
ANSWER 6: C

Problem 7:
Prime factorization of 250: \(250 = 2 \times 5^3\).  
Prime factors are 2 and 5. The two smallest are 2 and 5.  
Sum = \(2 + 5 = 7\).  
ANSWER 7: C

Problem 8:
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}.  
Possible sums:  
1+2=3, 1+4=5, 1+6=7;  
3+2=5, 3+4=7, 3+6=9;  
5+2=7, 5+4=9, 5+6=11.  
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.  
ANSWER 8: B

Problem 9:
Boys must sit at the two ends.  
Arrange boys: 2 choices for left end, 1 for right end → \(2! = 2\) ways.  
Arrange 3 girls in the middle 3 seats: \(3! = 6\) ways.  
Total arrangements = \(2 \times 6 = 12\).  
ANSWER 9: E

Problem 10:
Interpret "8 4" as \(8 \div 4\).  
\(16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16\).  
ANSWER 10: D

Problem 11:
\(\sqrt{4} = 2\).  
\(8 \cdot 2 = 16\), \(\sqrt{16} = 4\).  
\(16 \cdot 4 = 64\), \(\sqrt{64} = 8\).  
ANSWER 11: C

Problem 12:
Compare decimals:  
0.99, 0.9099, 0.9, 0.909, 0.9009.  
0.99 is the largest.  
ANSWER 12: A

# Problem 1:

**What is being asked:** A rock of volume 1000 cm³ is submerged in an aquarium with a 100 cm × 40 cm base. How much does the water level rise?

**Setup:** When the rock is submerged, it displaces a volume of water equal to its own volume. This displaced water spreads across the base of the aquarium, causing the level to rise.

$$\text{Volume displaced} = \text{Base Area} \times \text{Rise in water level}$$

**Computation:**
- Base area = $100 \times 40 = 4000 \text{ cm}^2$
- $1000 = 4000 \times h$
- $h = \frac{1000}{4000} = 0.25 \text{ cm}$

**Check for overflow:** The current depth is 37 cm, and 37 + 0.25 = 37.25 cm, which is well below the 50 cm height. No overflow occurs.

**Ruling out other choices:** Choices B–E would require rock volumes of 2000, 4000, 5000, and 10000 cm³ respectively — all larger than the given 1000 cm³.

ANSWER 1: A

---

# Problem 2:

**What is being asked:** With 4 players, each gets 15 cards. With 6 players, how many cards does each get?

**Setup:** The total number of cards in the deck stays the same.

**Computation:**
- Number of players in first game: Annika + 3 friends = 4 players
- Total cards = $4 \times 15 = 60$ cards
- Number of players in second game: $4 + 2 = 6$ players
- Cards per player = $\frac{60}{6} = 10$

**Ruling out other choices:** Some might forget to count Annika (getting 3 players → 45 cards → 45/5 = 9, choice B) or miscount the new total as 5 players (60/5 = 12, choice E).

ANSWER 2: C

---

# Problem 3:

**What is being asked:** Evaluate the custom operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$ for $a=3, b=1, c=4, d=2$.

**Setup:** Substitute directly into the formula.

**Computation:**
$$\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2$$

**Ruling out other choices:** A common mistake is computing $b \cdot c - a \cdot d = 4 - 6 = -2$ (choice A) or mixing up which values go where.

ANSWER 3: E

---

# Problem 4:

**What is being asked:** Isabella needs an average of 95 on four tests. Her first two scores are 97 and 91. What is the lowest possible third test score so she can still reach her goal?

**Setup:**
- Total points needed: $4 \times 95 = 380$
- Sum of first two tests: $97 + 91 = 188$
- Sum needed from tests 3 and 4: $380 - 188 = 192$

**Computation:**
To minimize the third test score, maximize the fourth test score. The maximum possible score on any test is 100.
$$\text{Minimum test 3} = 192 - 100 = 92$$

**Ruling out other choices:** Choice A (90) would require test 4 to be $192 - 90 = 102$, which is impossible on a 100-point test. Choices C, D, E are higher than necessary.

ANSWER 4: B

---

# Problem 5:

**What is being asked:** Bridget bought apples, gave half to Ann, gave 3 to Cassie, and kept 4. How many did she buy?

**Setup:** Let $x$ = total apples bought. Work through the sequence of events.

**Computation:**
- After giving half to Ann: $\frac{x}{2}$ apples remain
- After giving 3 to Cassie: $\frac{x}{2} - 3$ apples remain
- She keeps 4: $\frac{x}{2} - 3 = 4$
- $\frac{x}{2} = 7$
- $x = 14$

**Verification:** 14 apples → give 7 to Ann → 7 left → give 3 to Cassie → 4 left. ✓

**Ruling out other choices:** Choice C (7) is a trap for those who forget to double the 7 remaining apples. Choice D (11) comes from incorrectly adding $4 + 3 + 4$.

ANSWER 5: E

---

# Problem 6:

**What is being asked:** Find the total number of marbles given conditions about each color.

**Setup:** Let $T$ = total marbles, $R$ = red, $G$ = green, $B$ = blue.
- "All but 6 are red" → $G + B = 6$ → $R = T - 6$
- "All but 8 are green" → $R + B = 8$ → $G = T - 8$
- "All but 4 are blue" → $R + G = 4$ → $B = T - 4$

**Computation:**
$$T = R + G + B = (T - 6) + (T - 8) + (T - 4) = 3T - 18$$
$$2T = 18 \implies T = 9$$

**Verification:** $R = 3, G = 1, B = 5$. Check: $G+B=6$ ✓, $R+B=8$ ✓, $R+G=4$ ✓.

ANSWER 6: C

---

# Problem 7:

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Setup:** Find the prime factorization of 250.

**Computation:**
$$250 = 2 \times 125 = 2 \times 5 \times 25 = 2 \times 5 \times 5 \times 5 = 2 \times 5^3$$

The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.
$$\text{Sum} = 2 + 5 = 7$$

**Ruling out other choices:** Choice A (2) is just one factor. Choice B (5) is just one factor. Choice D (10) would be $2 + 5 + 3$ or $5 + 5$, neither of which applies.

ANSWER 7: C

---

# Problem 8:

**What is being asked:** How many distinct sums can be formed by picking one chip from Bag A {1, 3, 5} and one from Bag B {2, 4, 6}?

**Setup:** List all 9 possible sums and count the unique values.

**Computation:**
| | 2 | 4 | 6 |
|---|---|---|---|
| **1** | 3 | 5 | 7 |
| **3** | 5 | 7 | 9 |
| **5** | 7 | 9 | 11 |

Unique sums: {3, 5, 7, 9, 11} → **5 different values**

**Ruling out other choices:** Choice E (9) is the trap of counting all 9 pairs without removing duplicates. Choice D (7) might come from miscounting.

ANSWER 8: B

---

# Problem 9:

**What is being asked:** How many ways can 2 boys and 3 girls sit in a row with boys at the ends and girls in the middle?

**Setup:** The arrangement must be: **B G G G B**

**Computation:**
- Arrange the 2 boys at the 2 end positions: $2! = 2$ ways
- Arrange the 3 girls in the 3 middle positions: $3! = 6$ ways
- Total arrangements: $2 \times 6 = 12$

**Ruling out other choices:** Choice A (2) only counts the boys' arrangements. Choice D (6) only counts the girls' arrangements. Both must be multiplied together.

ANSWER 9: E

---

# Problem 10:

**What is being asked:** Evaluate $16 + 8 \div 4 - 2$ using order of operations.

**Setup:** Apply PEMDAS — division before addition and subtraction.

**Computation:**
$$16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16$$

**Ruling out other choices:**
- Left-to-right without order of operations: $(16+8) \div 4 - 2 = 6 - 2 = 4$ (choice A)
- $(16+8) \div (4-2) = 24 \div 2 = 12$ (choice C)
- $16 + 8 \div (4-2) = 16 + 4 = 20$ (choice E

 Problem 1:
We need to find how much the water level rises when a rock of volume 1000 cm³ is submerged in a rectangular aquarium.

First, find the area of the aquarium’s base:
\[
100\text{ cm} \times 40\text{ cm} = 4000\text{ cm}^2.
\]

When the rock is completely submerged, it displaces 1000 cm³ of water. This displaced volume spreads out over the base, causing the water level to rise. The rise in height \(h\) satisfies:
\[
\text{base area} \times h = \text{volume of rock}.
\]
So,
\[
h = \frac{1000}{4000} = 0.25\text{ cm}.
\]

We should check that the aquarium does not overflow. The current depth is 37 cm and the height is 50 cm, so there is 13 cm of free space—plenty of room for a 0.25 cm rise.

Tempting wrong choices come from using the wrong area (e.g., using the side wall 100×50 or 40×50) or from dividing by the height 50. Only the base area matters for the level rise.

ANSWER 1: A

---

Problem 2:
We need to find how many cards each player gets when two extra friends join the game.

First, determine the total number of cards. Annika and 3 friends make 4 players, and each gets 15 cards:
\[
4 \times 15 = 60\text{ cards total}.
\]

When 2 more friends join, the number of players becomes:
\[
4 + 2 = 6.
\]

Distribute the 60 cards evenly among 6 players:
\[
60 \div 6 = 10.
\]

A tempting mistake is to subtract 2 from 15, giving 13, which is not an option, or to guess without computing the total deck size. The key is that the total number of cards stays the same.

ANSWER 2: C

---

Problem 3:
We are given the rule \(a\ c\ b\ d = a\cdot d - b\cdot c\). We need to evaluate \(3\ 1\ 4\ 2\).

Matching the pattern:
- \(a = 3\)
- \(c = 1\)
- \(b = 4\)
- \(d = 2\)

Apply the formula:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

A tempting error is to swap the pairs and compute \(3\cdot 1 - 4\cdot 2 = 3 - 8 = -5\), which is not among the choices, or to read the order incorrectly. Careful matching to \(a, c, b, d\) is essential.

ANSWER 3: E

---

Problem 4:
Isabella wants an average of 95 on four 100-point tests. Her first two scores are 97 and 91. We need the lowest possible score on the third test that still allows her to reach her goal.

Total points needed for a 95 average:
\[
4 \times 95 = 380.
\]

Points earned so far:
\[
97 + 91 = 188.
\]

Points still needed from the last two tests:
\[
380 - 188 = 192.
\]

Let the third test score be \(x\) and the fourth be \(y\), with \(x + y = 192\) and each score at most 100. To minimize \(x\), we maximize \(y\). The highest possible score is 100, so:
\[
x = 192 - 100 = 92.
\]

If she scores 92 on the third test, she can score 100 on the fourth and exactly reach her goal. A tempting wrong choice is 95, which would be the answer if she needed the two remaining tests to average 95, but she can shift points to the fourth test.

ANSWER 4: B

---

Problem 5:
We need to find how many apples Bridget originally bought.

Let the original number be \(x\).
- She gave half to Ann, leaving her with \(\frac{x}{2}\).
- She gave 3 to Cassie and kept 4:
\[
\frac{x}{2} - 3 = 4.
\]

Solve:
\[
\frac{x}{2} = 7 \quad\Rightarrow\quad x = 14.
\]

Check: Start with 14. Half is 7, given to Ann. She has 7 left. Gives 3 to Cassie, keeps 4. This matches.

A tempting error is to add \(3 + 4 = 7\) and stop, forgetting that 7 is only what remained after giving away half, so the original amount must be doubled.

ANSWER 5: E

---

Problem 6:
We need the total number of marbles given three conditions about how many are *not* each color.

Let \(R, G, B\) be the numbers of red, green, and blue marbles.
- “All but 6 are red” means non-red marbles equal 6: \(G + B = 6\).
- “All but 8 are green” means non-green marbles equal 8: \(R + B = 8\).
- “All but 4 are blue” means non-blue marbles equal 4: \(R + G = 4\).

Add all three equations:
\[
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
\]
\[
2R + 2G + 2B = 18
\]
\[
R + G + B = 9.
\]

So there are 9 marbles total. (Checking: \(R=3, G=1, B=5\), which satisfies all three statements.)

A tempting mistake is to add \(6+8+4=18\) and choose that, but that sum counts each marble twice.

ANSWER 6: C

---

Problem 7:
We need the sum of the two smallest prime factors of 250.

Factor 250:
\[
250 = 2 \times 125 = 2 \times 5^3.
\]

The prime factors are 2 and 5. The two smallest (and only) distinct prime factors are 2 and 5. Their sum is:
\[
2 + 5 = 7.
\]

A tempting wrong choice is 10, which comes from adding 5 and 5 (the two smallest factors with multiplicity), but the problem asks for the two smallest prime factors, i.e., distinct primes.

ANSWER 7: C

---

Problem 8:
Bag A has \(\{1, 3, 5\}\) and Bag B has \(\{2, 4, 6\}\). We draw one from each and want the number of distinct possible sums.

Compute all sums:
- \(1+2=3\), \(1+4=5\), \(1+6=7\)
- \(3+2=5\), \(3+4=7\), \(3+6=9\)
- \(5+2=7\), \(5+4=9\), \(5+6=11\)

The distinct values are:
\[
3, 5, 7, 9, 11.
\]

That is 5 different values. A tempting error is to count all 9 ordered pairs without removing duplicates.

ANSWER 8: B

---

Problem 9:
We have 2 boys and 3 girls to arrange in a row with a boy at each end and the three girls in the middle.

Positions: B _ _ _ G G G _ _ _ B (but specifically 5 seats: 1,2,3,4,5 with boys at 1 and 5, girls at 2,3,4).

- Choose the boy for the left end and the boy for the right end: \(2! = 2\) ways.
- Arrange the 3 girls in the middle three seats: \(3! = 6\) ways.

Total arrangements:
\[
2 \times 6 = 12.
\]

A tempting mistake is to ignore the constraint and compute \(5! = 120\), or to forget that the boys can swap places.

ANSWER 9: E

---

Problem 10:
Evaluate \(16 + 8 \div 4 - 2\).

Using order of operations (PEMDAS/BODMAS), division comes before addition and subtraction:
\[
8 \div 4 = 2.
\]

Now substitute back:
\[
16 + 2 - 2 = 16.
\]

A tempting error is to work left-to-right doing addition first: \((16+8)\div 4 - 2 = 24\div 4 - 2 = 6 - 2 = 4\), which is choice A. Another is to do \(16 + 2 = 18\), then \(18 - 2 = 16\), which luckily gives the same corr

Problem 1:
We need the rise in water level when a rock of volume 1000 cm³ is placed in an aquarium with base 100 cm by 40 cm. The rock displaces water equal to its volume, causing the water level to rise by Δh such that base area × Δh = rock volume. Base area = 100 cm × 40 cm = 4000 cm². So Δh = 1000 cm³ / 4000 cm² = 0.25 cm. The initial depth is 37 cm, and after rise it becomes 37.25 cm, which is still below the height of 50 cm, so the rock is fully submerged. The rise is 0.25 cm.

ANSWER 1: A

Problem 2:
When Annika and 3 friends play, there are 4 players, each dealt 15 cards. Total cards = 4 × 15 = 60. In the next game, 2 more friends join, making 6 players. Since the same total number of cards is used, each player gets 60 / 6 = 10 cards.

ANSWER 2: C

Problem 3:
The operation is defined as a c b d = a·d − b·c. For the numbers 3 1 4 2, we have a = 3, c = 1, b = 4, d = 2. Compute a·d − b·c = 3×2 − 4×1 = 6 − 4 = 2.

ANSWER 3: E

Problem 4:
Isabella wants an average of 95 over four tests, so total points needed = 4 × 95 = 380. Her first two scores sum to 97 + 91 = 188. Let the third score be x. After the third test, she can still reach the goal if it is possible to score at most 100 on the fourth test to reach 380. That requires 188 + x + 100 ≥ 380 → x ≥ 380 − 288 = 92. The smallest integer x satisfying this is 92. If x = 91, then even a perfect 100 on the fourth gives 188+91+100 = 379 < 380. So the lowest possible third score is 92.

ANSWER 4: B

Problem 5:
Let the number of apples Bridget bought be N. She gave half to Ann, leaving N/2. Then she gave Cassie 3 apples, leaving N/2 − 3. She keeps 4 apples, so N/2 − 3 = 4 → N/2 = 7 → N = 14.

ANSWER 5: E

Problem 6:
Let T be the total number of marbles. "All but 6 are red" means T − R = 6 → R = T − 6. Similarly, T − G = 8 → G = T − 8, and T − B = 4 → B = T − 4. Since R + G + B = T, we have (T−6)+(T−8)+(T−4) = T → 3T − 18 = T → 2T = 18 → T = 9.

ANSWER 6: C

Problem 7:
Factor 250: 250 = 2 × 5³. Its prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5, and their sum is 7.

ANSWER 7: C

Problem 8:
List all possible sums from drawing one chip from Bag A {1,3,5} and one from Bag B {2,4,6}:
1+2=3, 1+4=5, 1+6=7;
3+2=5, 3+4=7, 3+6=9;
5+2=7, 5+4=9, 5+6=11.
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.

ANSWER 8: B

Problem 9:
We need arrangements of 2 boys and 3 girls in a row with a boy at each end. Choose which boy sits on the left end and which on the right: 2! = 2 ways. The three girls occupy the three middle seats and can be arranged in 3! = 6 ways. Total arrangements = 2 × 6 = 12.

ANSWER 9: E

Problem 10:
The expression is 16 + 8 ÷ 4 − 2 (the missing operator is division, as typical for order-of-operations problems). Following order of operations, division before addition and subtraction: 8 ÷ 4 = 2. Then 16 + 2 = 18, and 18 − 2 = 16.

ANSWER 10: D

Problem 11:
Evaluate step by step: √4 = 2. Then 8 · 2 = 16, so √(8·√4) = √16 = 4. Then 16 · 4 = 64, so √(16·√(8·√4)) = √64 = 8.

ANSWER 11: C

Problem 12:
Compare the decimal numbers:
A. 0.99
B. 0.9099
C. 0.9
D. 0.909
E. 0.9009
Clearly 0.99 is the largest because it is greater than 0.9 and all the others are less than 0.91.

ANSWER 12: A

Problem 1:
The aquarium’s base has area  
\(A = 100\text{ cm} \times 40\text{ cm}=4000\text{ cm}^2\).  
When the rock of volume \(V=1000\text{ cm}^3\) is fully submerged, the water rises by a height \(h\) such that  

\[
A\cdot h = V \quad\Longrightarrow\quad h = \frac{1000}{4000}=0.25\text{ cm}.
\]

The rise cannot be any of the larger listed values (0.5 cm, 1 cm, …) because those would require a volume larger than the rock’s.  

ANSWER 1: A  

---

Problem 2:
With 4 players each receiving 15 cards, the total number of cards is  

\[
4 \times 15 = 60.
\]

If 2 more friends join, there are \(4+2 = 6\) players.  The cards are still all the cards, so each player receives  

\[
\frac{60}{6}=10\text{ cards}.
\]

The other answer choices (8, 9, 11, 12) would not distribute the 60 cards evenly.  

ANSWER 2: C  

---

Problem 3:
The operation is defined by  

\[
a\;c\;b\;d = a\cdot d - b\cdot c.
\]

For the numbers \(3\;1\;4\;2\) we have \(a=3,\;c=1,\;b=4,\;d=2\).  Hence  

\[
3\;1\;4\;2 = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

The other options \(-2,-1,0,1\) would correspond to different combinations that do not satisfy the given formula.  

ANSWER 3: E  

---

Problem 4:
To average 95 over four 100‑point tests, the total points needed are  

\[
4 \times 95 = 380.
\]

After the first two tests Isabella has  

\[
97 + 91 = 188\text{ points}.
\]

Let the third test score be \(x\).  Then the fourth test must supply  

\[
380 - (188 + x) = 192 - x \text{ points}.
\]

The fourth test can give at most 100 points, so we need  

\[
192 - x \le 100 \quad\Longrightarrow\quad x \ge 92.
\]

If she scores \(92\) on the third test, she would need a perfect 100 on the fourth, which is possible.  Any lower score would require more than 100 points on the fourth test, which is impossible.  

Thus the lowest possible third‑test score is **92**.  

ANSWER 4: B  

---

Problem 5:
Let the original number of apples be \(N\).  
She gave half of them to Ann: \(\frac{N}{2}\).  
She then gave Cassie 3 apples and kept 4, leaving  

\[
N - \frac{N}{2} - 3 = 4.
\]

Simplify: \(\frac{N}{2} - 3 = 4 \Rightarrow \frac{N}{2}=7 \Rightarrow N=14\).

If the total had been any of the other answer choices, the steps would not balance (e.g., 11 would leave 2 apples after giving Cassie 3).  

ANSWER 5: E  

---

Problem 6:
Let the total number of marbles be \(T\).  
- “All but 6 are red’’ → red = \(T-6\).  
- “All but 8 are green’’ → green = \(T-8\).  
- “All but 4 are blue’’ → blue = \(T-4\).

These three colors account for all marbles, so  

\[
(T-6)+(T-8)+(T-4)=T \;\Longrightarrow\; 3T-18=T \;\Longrightarrow\; 2T=18 \;\Longrightarrow\; T=9.
\]

The other totals (6, 8, 10, 12) would not satisfy the three “all but…’’ statements simultaneously.  

ANSWER 6: C  

---

Problem 7:
Factor \(250\):  

\[
250 = 2 \times 5^3.
\]

The prime factors are \(2\) and \(5\).  The two smallest are \(2\) and \(5\), whose sum is  

\[
2+5=7.
\]

Choices 2, 5, 10, 12 are either single prime factors or incorrect sums.  

ANSWER 7: C  

---

Problem 8:
Bag A contains \(\{1,3,5\}\); Bag B contains \(\{2,4,6\}\).  
All possible sums \(a+b\) are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The distinct results are \(\{3,5,7,9,11\}\), a total of **5** different sums.  

The other answer choices (4, 6, 7, 9) do not match the count of distinct sums.  

ANSWER 8: B  

---

Problem 9:
Place the two boys at the two ends and the three girls in the middle three seats.

- The two boys can be ordered in \(2!\) ways (left‑right swap).  
- The three girls can be arranged among the three middle seats in \(3! = 6\) ways.

Total arrangements  

\[
2! \times 3! = 2 \times 6 = 12.
\]

Choices 2, 4, 5, 6 are too small; 12 accounts for all possible permutations under the given constraints.  

ANSWER 9: E  

---

Problem 10:
Interpret the expression as  

\[
16 + 8 \div 4 - 2.
\]

Division comes before addition and subtraction, so  

\[
8 \div 4 = 2 \quad\Longrightarrow\quad 16 + 2 - 2 = 16.
\]

The other options (4, 8, 12, 20) would result only if the order of operations were misinterpreted.  

ANSWER 10: D  

---

Problem 11:
Simplify from the innermost radical outward.

\[
\sqrt{4}=2,\qquad 
\sqrt{8\cdot\sqrt{4}} = \sqrt{8\cdot2}= \sqrt{16}=4.
\]

Now  

\[
\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{16\cdot4}= \sqrt{64}=8.
\]

Thus the value is 8, matching choice C.  

ANSWER 11: C  

---

Problem 12:
Compare the decimal values:

- A. \(0.99\)
- B. \(0.9099\)
- C. \(0.9\)
- D. \(0.909\)
- E. \(0.9009\)

Clearly \(0.99\) is greater than \(0.9099\), which is greater than \(0.909\), etc.  
Hence the largest number is \(0.99\).  

ANSWER 12: A

Problem 1:  
An aquarium has a rectangular base of 100 cm by 40 cm and height 50 cm, filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is submerged. By how many centimeters does the water level rise?  

The base area is \(100 \times 40 = 4000 \, \text{cm}^2\). The rock’s volume is 1000 cm³, so the rise in water level is \(\frac{1000}{4000} = 0.25 \, \text{cm}\).  

Tempting choices: 0.5 (B) or 1 (C) might be guessed if base area is miscalculated.  

ANSWER 1: A  

Problem 2:  
Annika and 3 friends (4 players) get 15 cards each. Total cards = \(4 \times 15 = 60\). If 2 more join (6 players), each gets \(\frac{60}{6} = 10\).  

Tempting choices: 8 (A) if total cards miscalculated as 48.  

ANSWER 2: C  

Problem 3:  
Operation defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\): \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\), but the order is \(a c b d\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). However, the problem writes "a c b d" which is interpreted as \(a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). But the choices include 2 (E). Re-evaluating: the operation is \(a c b d = a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\).  

Tempting choices: -2 (A) if signs reversed.  

ANSWER 3: A (Note: Correction based on standard interpretation: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\), but the correct answer based on the operation defined is actually \(3 \cdot 2 - 1 \cdot 4 = 2\), but the problem writes "a c b d", which is \(a \cdot d - b \cdot c\), so 2. However, the answer is 2, but the choices include 2 as E. The correct calculation is 2, so E. Correction: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\).)  

**Correction:** The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\), this is \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\).  

ANSWER 3: E  

Problem 4:  
Isabella wants an average of 95 on four 100-point tests. Total points needed: \(4 \times 95 = 380\). First two scores: 97 + 91 = 188. Let third score be \(x\), fourth be \(y\). Then \(188 + x + y = 380\), so \(x + y = 192\). To minimize \(x\), maximize \(y\): \(y = 100\), so \(x = 92\).  

Tempting choices: 90 (A) if \(y\) not maximized.  

ANSWER 4: B  

Problem 5:  
Bridget gave half apples to Ann, then 3 to Cassie, keeping 4. Let total be \(x\). After giving half to Ann: \(\frac{x}{2}\) left. Then gives 3 to Cassie: \(\frac{x}{2} - 3 = 4\). So \(\frac{x}{2} = 7\), \(x = 14\).  

Tempting choices: 7 (C) if half misinterpreted.  

ANSWER 5: E  

Problem 6:  
Let \(r, g, b\) be red, green, blue marbles. Then:  
- All but 6 are red: \(g + b = 6\)  
- All but 8 are green: \(r + b = 8\)  
- All but 4 are blue: \(r + g = 4\)  
Adding: \(2(r + g + b) = 18\), so total \(r + g + b = 9\).  

Tempting choices: 6 (A) if equations misread.  

ANSWER 6: C  

Problem 7:  
Prime factors of 250: \(250 = 2 \times 5^3\). Two smallest: 2 and 5. Sum: \(2 + 5 = 7\).  

Tempting choices: 5 (B) if only 5 considered.  

ANSWER 7: C  

Problem 8:  
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}. Possible sums:  
- 1+2=3, 1+4=5, 1+6=7  
- 3+2=5, 3+4=7, 3+6=9  
- 5+2=7, 5+4=9, 5+6=11  
Unique sums: 3, 5, 7, 9, 11 → 5 values.  

Tempting choices: 6 (C) if duplicates counted.  

ANSWER 8: B  

Problem 9:  
Two boys at ends: 2 ways. Three girls in middle: \(3! = 6\) ways. Total: \(2 \times 6 = 12\).  

Tempting choices: 6 (D) if boys not permuted.  

ANSWER 9: E  

Problem 10:  
Expression: \(16 + 8 \times 4 - 2\). Order: \(8 \times 4 = 32\), then \(16 + 32 - 2 = 46\)? Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 \times 4 - 2\): \(8 \times 4 = 32\), \(16 + 32 = 48\), \(48 - 2 = 46\). But choices max at 20. Re-interpreting: "16 + 8 4 - 2" might mean \(16 + 8 - 4 - 2 = 18\), not in choices. Alternatively, if "8 4" is 8×4, but 46 not in choices. The expression is likely \(16 + 8 - 4 - 2 = 18\), but not in choices. Correction: The expression is "16 + 8 4 − 2", which might be \(16 + (8 \times 4) - 2 = 16 + 32 - 2 = 46\), but not in choices. Re-reading: It’s "16 + 8 4 − 2", which could be \(16 + 8 - 4 - 2 = 18\), but not in choices. The intended expression is likely \(16 + 8 \times 4 - 2 = 46\), but choices max at 20. Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 - 4 - 2 = 18\), but not in choices. Alternatively, if "8 4" is 84, but 16+84-2=98. Not matching. The expression is likely \( \frac{16 + 8}{4} - 2 = \frac{24}{4} - 2 = 6 - 2 = 4 \).  

Tempting choices: 8 (B) if order misapplied.  

ANSWER 10: A  

Problem 11:  
Expression: \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\).  
- \(\sqrt{4} = 2\)  
- \(\sqrt{8 \cdot 2} = \sqrt{16} = 4\)  
- \(\sqrt{16 \cdot 4} = \sqrt{64} = 8\)  

Tempting choices: 4 (A) if innermost not simplified.  

ANSWER 11: C  

Problem 12:  
Compare decimals:  
- A: 0.99  
- B: 0.9099  
- C: 0.9  
- D: 0.909  
- E: 0.9009  
Largest is 0.99.  

Tempting choices: 0.9099 (B) if decimals misaligned.  

ANSWER 12: A  

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

### Problem 1:
What is asked: An aquarium with a 100cm×40cm rectangular base is filled with water to 37cm depth. A 1000 cm³ rock is submerged; how much does the water level rise?
Step-by-step: When a submerged object displaces water, its volume equals the additional volume of water in the aquarium. The rise in water level Δh satisfies: Volume of rock = base area of aquarium × Δh.
Base area = 100×40 = 4000 cm².
Δh = 1000 / 4000 = 0.25 cm.
Tempting wrong answers: 0.5 (if base area is miscalculated as 2000 cm²) or 1 (wrong division).
ANSWER 1: A

---

### Problem 2:
What is asked: Annika plus 3 friends (4 total players) each get 15 cards. If 2 more join, how many cards does each player get?
Step-by-step: First calculate total number of cards: 4 players × 15 cards = 60 total cards. After 2 join, total players = 4+2=6. Cards per player = 60 / 6 = 10.
Tempting wrong answers: 9 (if you incorrectly count only 3 initial players, leading to 45 total cards) or 12 (wrong division).
ANSWER 2: C

---

### Problem 3:
What is asked: The custom notation a c b d = a·d − b·c. What is the value of 3 1 4 2?
Step-by-step: Match the values to the definition: a=3, c=1, b=4, d=2. Calculate: 3·2 - 4·1 = 6 - 4 = 2.
Tempting wrong answers: -2 (swap a and b incorrectly) or 0 (miscalculate the multiplication).
ANSWER 3: E

---

### Problem 4:
What is asked: Isabella needs a 95 average on 4 100-point tests. She scored 97 and 91 on the first two. What is the lowest possible third test score that lets her reach her goal?
Step-by-step: Total points needed for 95 average: 4×95 = 380 points. Sum of first two tests: 97+91=188. To minimize the third test score, maximize the fourth test (max 100 points). Total points left for tests 3 and 4: 380-188=192. Minimum third test = 192 - 100 = 92.
Tempting wrong answers: 96 (the average of the remaining two tests, not the minimum possible third score).
ANSWER 4: B

---

### Problem 5:
What is asked: Bridget gave half her apples to Ann, 3 to Cassie, and kept 4. How many apples did she buy originally?
Step-by-step: Let N = original number of apples. After giving half to Ann, she has N/2 left. Then N/2 - 3 = 4 → N/2=7 → N=14. Verify: 14 → 7 left after Ann, minus 3 for Cassie = 4 kept, which matches.
Tempting wrong answers: 7 (the number of apples Bridget had after giving half to Ann, not the original total).
ANSWER 5: E

---

### Problem 6:
What is asked: A jar has red, green, blue marbles. All but 6 are red, all but 8 are green, all but 4 are blue. How many total marbles?
Step-by-step: Let T=total marbles, R=red, G=green, B=blue. The conditions translate to: G+B=6, R+B=8, R+G=4. Add all equations: 2(R+G+B)=18 → R+G+B=T=9. Verify: 3 red, 1 green, 5 blue sum to 9, matching all conditions.
Tempting wrong answers: 18 (the sum of 6+8+4, not halved to get total marbles).
ANSWER 6: C

---

### Problem 7:
What is asked: What is the sum of the two smallest prime factors of 250?
Step-by-step: Prime factorization of 250: 250=2×5³. The prime factors are 2 and 5. Their sum is 2+5=7.
Tempting wrong answers: 10 (sum of 5+5, or wrong addition of 2+5).
ANSWER 7: C

---

### Problem 8:
What is asked: Bag A has chips 1,3,5; Bag B has 2,4,6. One chip drawn from each bag, how many distinct sum values are possible?
Step-by-step: List all sums: 1+2=3, 1+4=5,1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7,5+4=9,5+6=11. Distinct sums: 3,5,7,9,11 → 5 total.
Tempting wrong answers: 9 (total number of unordered pairs, not distinct sums).
ANSWER 8: B

---

### Problem 9:
What is asked: 2 boys and 3 girls line up for a photo, with a boy at each end and girls in the middle. How many arrangements are possible?
Step-by-step: Arrange the boys first: 2 boys can be ordered at the ends in 2! = 2 ways. The 3 middle girls can be ordered in 3! =6 ways. Total arrangements: 2×6=12.
Tempting wrong answers: 6 (only the number of ways to arrange the girls, ignoring the boys).
ANSWER 9: E

---

### Problem 10:
What is asked: Evaluate (16+8)/(4-2) (formatted as a fraction in the original problem).
Step-by-step: Calculate numerator:16+8=24, denominator:4-2=2. 24/2=12.
Tempting wrong answers:16 (if order of operations is misapplied to calculate 16 + 8/4 -2=16).
ANSWER 10: C

---

### Problem 11:
What is asked: Evaluate √(16 · √(8 · √4)).
Step-by-step: Calculate from the innermost root first: √4=2. Next: 8×2=16, √16=4. Next:16×4=64, √64=8.
Tempting wrong answers: 4√2 (incorrect root calculation).
ANSWER 11: C

---

### Problem 12:
What is asked: Which number is largest: 0.99, 0.9099, 0.9, 0.909, 0.9009.
Step-by-step: Line up all decimals to compare place values: 0.9900 > 0.9099 > 0.9090 > 0.9009 > 0.9000. The largest is 0.99.
Tempting wrong answers: 0.9099 (the second largest value).
ANSWER 12: A

### Problem 1:
We need to find how much the water level rises when a 1000 cm³ rock is submerged in the aquarium.
1.  The volume of displaced water equals the volume of the submerged rock: 1000 cm³.
2.  The base area of the aquarium is \(100 \, \text{cm} \times 40 \, \text{cm} = 4000 \, \text{cm}^2\).
3.  Let \(h\) be the rise in water level. The volume of the displaced water is also equal to base area × rise in height: \(4000h = 1000\).
4.  Solve for \(h\): \(h = \frac{1000}{4000} = 0.25\) cm.
ANSWER 1: A
---
### Problem 2:
We need to find the number of cards per player when 2 more friends join.
1.  Original number of players: Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).
2.  New number of players: 4 + 2 = 6 players. Total cards remain 60.
3.  Cards per player: \(\frac{60}{6} = 10\).
ANSWER 2: C
---
### Problem 3:
The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). Compute \(3 \, 1 \, 4 \, 2\).
1.  Identify \(a=3, c=1, b=4, d=2\).
2.  Substitute into the formula: \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).
ANSWER 3: E
---
### Problem 4:
Find the lowest possible third test score to get an average of 95 on 4 tests.
1.  Total points needed for an average of 95: \(4 \times 95 = 380\).
2.  Sum of first two test scores: \(97 + 91 = 188\).
3.  Let \(x\) = third test score, \(y\) = fourth test score (max 100, since tests are 100-point). We need \(188 + x + y = 380\), so \(x + y = 192\).
4.  To minimize \(x\), maximize \(y = 100\). Then \(x = 192 - 100 = 92\).
ANSWER 4: B
---
### Problem 5:
Find the original number of apples Bridget bought.
1.  Let \(x\) = total apples. After giving half to Ann, she has \(\frac{x}{2}\) left.
2.  After giving 3 to Cassie, she has \(\frac{x}{2} - 3\) left, which equals 4 (the apples she kept).
3.  Solve: \(\frac{x}{2} - 3 = 4 \implies \frac{x}{2} =7 \implies x=14\).
ANSWER 5: E
---
### Problem 6:
Find the total number of marbles.
1.  Let \(R, G, B\) be the number of red, green, blue marbles, total \(T = R+G+B\).
2.  Translate the statements:
    - All but 6 are red: \(G + B = 6\)
    - All but 8 are green: \(R + B = 8\)
    - All but 4 are blue: \(R + G =4\)
3.  Add all three equations: \(2(R+G+B) = 6+8+4=18 \implies 2T=18 \implies T=9\).
ANSWER 6: C
---
### Problem 7:
Find the sum of the two smallest prime factors of 250.
1.  Factorize 250: \(250 = 2 \times 125 = 2 \times 5^3\).
2.  The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.
3.  Sum: \(2 +5=7\).
ANSWER 7: C
---
### Problem 8:
Find the number of distinct sums when drawing one chip from each bag.
1.  List all possible pairs and sums:
    - Bag A (1,3,5) + Bag B (2,4,6):
      \(1+2=3, 1+4=5, 1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7, 5+4=9, 5+6=11\)
2.  Distinct sums: 3, 5, 7, 9, 11. There are 5 unique values.
ANSWER 8: B
---
### Problem 9:
Find the number of arrangements with boys at each end and girls in the middle.
1.  Arrange the 2 boys at the ends: there are \(2! = 2\) ways (order matters: which boy is left/right).
2.  Arrange the 3 girls in the middle 3 seats: \(3! = 6\) ways.
3.  Total arrangements: \(2 \times 6 =12\).
ANSWER 9: E
---
### Problem 10:
Compute \(16 + 8 \, 4 - 2\). The missing operator between 8 and 4 is division (standard order of operations, common middle school problem):
1.  Interpret as \(16 + \frac{8}{4} - 2\).
2.  Compute division first: \(\frac{8}{4}=2\).
3.  Then add/subtract left to right: \(16 + 2 - 2 =16\).
*(Note: If interpreted as \(16 +8 -4 \times 2\), order of operations gives \(4 \times 2=8\), so \(16+8-8=16\), same result.)*
ANSWER 10: D
---
### Problem 11:
Compute \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\) by working from the innermost root outward.
1.  Innermost: \(\sqrt{4} = 2\).
2.  Next layer: \(\sqrt{8 \cdot 2} = \sqrt{16} =4\).
3.  Outermost: \(\sqrt{16 \cdot 4} = \sqrt{64} =8\).
ANSWER 11: C
---
### Problem 12:
Find the largest decimal.
1.  Rewrite all numbers with 4 decimal places for easy comparison:
    - A: \(0.9900\)
    - B: \(0.9099\)
    - C: \(0.9000\)
    - D: \(0.9090\)
    - E: \(0.9009\)
2.  Compare digit by digit: all have 9 in the tenths place. A has 9 in the hundredths place, while all others have 0. Thus A is the largest.
ANSWER 12: A

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

**Problem 1:**

What is being asked: By how many centimeters does the water level rise when a 1000 cm³ rock is submerged?

Setup: The base area is 100 cm × 40 cm = 4000 cm². When the rock is submerged, it displaces 1000 cm³ of water.

Water level rise = Volume displaced ÷ Base area = 1000 ÷ 4000 = 0.25 cm

ANSWER 1: A

---

**Problem 2:**

What is being asked: If cards are distributed evenly, how many cards per player when the player count increases?

Setup: Initially 4 players (Annika + 3 friends) each get 15 cards.
Total cards = 4 × 15 = 60 cards

With 2 more friends joining: 4 + 2 = 6 players
Cards per player = 60 ÷ 6 = 10 cards

ANSWER 2: C

---

**Problem 3:**

What is being asked: Evaluate the operation 3 1 4 2 using the rule a c b d = a·d − b·c

Setup: With a = 3, c = 1, b = 4, d = 2:
3 1 4 2 = 3·2 − 4·1 = 6 − 4 = 2

ANSWER 3: E

---

**Problem 4:**

What is being asked: What is the lowest score on test 3 that allows Isabella to achieve a 95 average on four tests?

Setup: Goal average is 95, so total needed = 95 × 4 = 380 points
First two tests: 97 + 91 = 188 points
Points needed from tests 3 and 4: 380 − 188 = 192 points

Let x = score on test 3. She can still reach her goal if she can score 100 on test 4:
x + 100 ≥ 192
x ≥ 92

The lowest possible score on test 3 is 92.

ANSWER 4: B

---

**Problem 5:**

What is being asked: How many apples did Bridget buy originally?

Setup: Let x = original number of apples
- She gave half to Ann: x/2 apples given away
- Remaining: x/2
- She gave 3 to Cassie: x/2 − 3
- She kept 4 for herself: x/2 − 3 = 4

Solving: x/2 = 7, so x = 14

ANSWER 5: E

---

**Problem 6:**

What is being asked: How many marbles are in the jar?

Setup: Let R, G, B = numbers of red, green, blue marbles; T = total
- "All but 6 are red": G + B = 6
- "All but 8 are green": R + B = 8
- "All but 4 are blue": R + G = 4

Adding all three equations: 2(R + G + B) = 6 + 8 + 4 = 18
Therefore: R + G + B = 9

ANSWER 6: C

---

**Problem 7:**

What is being asked: What is the sum of the two smallest prime factors of 250?

Setup: Factor 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5³

Prime factors: 2 and 5
Sum = 2 + 5 = 7

ANSWER 7: C

---

**Problem 8:**

What is being asked: How many different sums are possible?

Setup: Bag A: {1, 3, 5}, Bag B: {2, 4, 6}
All possible sums:
- 1 + 2 = 3
- 1 + 4 = 5
- 1 + 6 = 7
- 3 + 2 = 5 (duplicate)
- 3 + 4 = 7 (duplicate)
- 3 + 6 = 9
- 5 + 2 = 7 (duplicate)
- 5 + 4 = 9 (duplicate)
- 5 + 6 = 11

Distinct sums: {3, 5, 7, 9, 11} = 5 different values

ANSWER 8: B

---

**Problem 9:**

What is being asked: How many seating arrangements with boys at ends and girls in middle?

Setup: 2 boys at the ends can be arranged in 2! = 2 ways
3 girls in the middle can be arranged in 3! = 6 ways
Total arrangements = 2 × 6 = 12

ANSWER 9: E

---

**Problem 10:**

What is being asked: Evaluate 16 + 8 ÷ 4 − 2

Setup: Using order of operations (division before addition/subtraction):
16 + (8 ÷ 4) − 2 = 16 + 2 − 2 = 16

ANSWER 10: D

---

**Problem 11:**

What is being asked: Evaluate √(16 · √(8 · √4))

Setup: Work from the inside out:
- √4 = 2
- 8 · √4 = 8 · 2 = 16
- √(8 · √4) = √16 = 4
- 16 · √(8 · √4) = 16 · 4 = 64
- √(16 · √(8 · √4)) = √64 = 8

ANSWER 11: C

---

**Problem 12:**

What is being asked: Which decimal is largest?

Setup: Compare decimals:
- A: 0.99
- B: 0.9099
- C: 0.9
- D: 0.909
- E: 0.9009

Comparing: 0.99 > 0.9099 > 0.909 > 0.9009 > 0.9

The largest is 0.99

ANSWER 12: A

Problem 1:

The aquarium has a rectangular base of \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is

\[
100 \cdot 40 = 4000 \text{ cm}^2.
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

If the water rises by \(h\) centimeters, then

\[
4000h = 1000.
\]

So

\[
h = \frac{1000}{4000} = \frac14 = 0.25.
\]

So the water level rises \(0.25\) cm.

ANSWER 1: A

---

Problem 2:

Annika and 3 friends means there are

\[
4 \text{ players}.
\]

If each player gets 15 cards, then the total number of cards is

\[
4 \cdot 15 = 60.
\]

Now 2 more friends join, so there are

\[
6 \text{ players}.
\]

The same 60 cards are shared evenly, so each gets

\[
60 \div 6 = 10.
\]

ANSWER 2: C

---

Problem 3:

We are told that if the four numbers are \(a, b, c, d\), then

\[
a \ b \ c \ d = a\cdot d - b\cdot c.
\]

So for \(3,1,4,2\),

\[
3 \ 1 \ 4 \ 2 = 3\cdot 2 - 1\cdot 4 = 6 - 4 = 2.
\]

ANSWER 3: E

---

Problem 4:

She wants an average of 95 on 4 tests, so the total of the 4 scores must be

\[
4 \cdot 95 = 380.
\]

Her first two scores are 97 and 91, which add to

\[
97+91=188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To make the total 380, the fourth score would have to be

\[
380 - (188+x) = 192 - x.
\]

For this to be possible, the fourth score must be at most 100:

\[
192 - x \le 100.
\]

So

\[
x \ge 92.
\]

The lowest possible third score is 92.

ANSWER 4: B

---

Problem 5:

Let the number of apples Bridget bought be \(x\).

She gave half to Ann, so she kept the other half:

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

Then she gave Cassie 3 apples and kept 4 apples, so before giving Cassie the apples she had

\[
3+4=7.
\]

Thus,

\[
\frac{x}{2}=7 \quad \Rightarrow \quad x=14.
\]

ANSWER 5: E

---

Problem 6:

Let the total number of marbles be \(n\).

- “All but 6 are red” means the number of red marbles is \(n-6\).
- “All but 8 are green” means the number of green marbles is \(n-8\).
- “All but 4 are blue” means the number of blue marbles is \(n-4\).

Since every marble is one of these colors, their counts add to \(n\):

\[
(n-6)+(n-8)+(n-4)=n.
\]

Simplify:

\[
3n-18=n
\]

\[
2n=18
\]

\[
n=9.
\]

ANSWER 6: C

---

Problem 7:

We need the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3.
\]

So the prime factors are 2 and 5. The two smallest prime factors are 2 and 5, and their sum is

\[
2+5=7.
\]

ANSWER 7: C

---

Problem 8:

Bag A has \(1,3,5\); Bag B has \(2,4,6\).

Possible sums:

- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums are

\[
3,5,7,9,11
\]

which is 5 different values.

ANSWER 8: B

---

Problem 9:

There are 2 boys and 3 girls, and the picture must have:

- a boy at each end,
- the 3 girls in the middle.

First, arrange the 2 boys at the ends. They can be switched in

\[
2! = 2
\]

ways.

Then arrange the 3 girls in the middle in

\[
3! = 6
\]

ways.

So the total number of arrangements is

\[
2 \cdot 6 = 12.
\]

ANSWER 9: E

---

Problem 10:

The expression is

\[
16 + 8\cdot 4 - 2.
\]

Use order of operations:

\[
8\cdot 4=32.
\]

Then

\[
16+32-2=46.
\]

But 46 is not among the choices, so the intended expression must be read as

\[
16 + 8 \div 4 - 2.
\]

Then

\[
8\div 4 = 2,
\]

so

\[
16+2-2=16.
\]

ANSWER 10: D

---

Problem 11:

We evaluate from the inside out:

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

First,

\[
\sqrt4=2.
\]

So the expression becomes

\[
\sqrt{16\cdot \sqrt{8\cdot 2}}
=
\sqrt{16\cdot \sqrt{16}}.
\]

Since \(\sqrt{16}=4\),

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C

---

Problem 12:

Compare the decimals:

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

The largest is the one closest to 1. Since \(0.99\) is greater than any number beginning with \(0.90\), it is the largest.

ANSWER 12: A

Problem 1:  
We need the rise in water level when a rock is placed in the aquarium and submerged.

- Base dimensions: \(100\text{ cm} \times 40\text{ cm}\), so base area  
  \[
  A = 100\cdot 40 = 4000\text{ cm}^2.
  \]
- The rock is completely submerged, so the water volume displaced equals the rock’s volume:  
  \[
  V = 1000\text{ cm}^3.
  \]
- Water level rise \(\Delta h\) satisfies \(A\Delta h = V\):  
  \[
  4000\cdot \Delta h = 1000 \implies \Delta h = \frac{1000}{4000}=0.25\text{ cm}.
  \]

So the rise is \(0.25\) cm.  
Tempting wrong idea: using the *remaining* depth \(37\) cm or checking if it overflows—none of that matters; only displaced volume matters.

ANSWER 1: A  

---

Problem 2:  
All cards are dealt evenly. Let the total number of cards be \(T\).

Initially, Annika + 3 friends = 4 players, and each gets 15 cards:  
\[
T = 4\cdot 15 = 60.
\]
Next game: 2 more friends join, so total players:
\[
4+2=6.
\]
Each player gets:
\[
\frac{60}{6}=10.
\]

ANSWER 2: C  

---

Problem 3:  
Given the rule \(a\,b\,d\,c = a\cdot d - b\cdot c\).  
The expression “\(3\ 1\ 4\ 2\)” corresponds to:
- \(a=3\)
- \(b=1\)
- \(d=4\)
- \(c=2\)

Compute:
\[
a\cdot d - b\cdot c = 3\cdot 4 - 1\cdot 2 = 12-2=10.
\]

But notice the choices are only \(-2,-1,0,1,2\), so we must interpret the placeholder correctly:  
The problem states: **“If \(a b c d = a\cdot d - b\cdot c\)”**.

So “\(3\ 1\ 4\ 2\)” means:
- \(a=3,\ b=1,\ c=4,\ d=2\)

Then:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 1\cdot 4 = 6-4=2.
\]

ANSWER 3: E  

---

Problem 4:  
Goal average is 95 over 4 tests, so total points needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
Let the third test score be \(x\), and the fourth be \(y\). Then:
\[
188 + x + y = 380 \implies x+y=192.
\]
We want the **lowest possible** \(x\) such that she can still reach the goal. That means make \(y\) as large as possible (since grades are at most 100).
Maximum \(y=100\). Then:
\[
x + 100 = 192 \implies x = 92.
\]

Check: then the four scores are \(97,91,92,100\) which sum to \(380\), average \(95\). Works.

ANSWER 4: B  

---

Problem 5:  
Let the total number of apples be \(N\).

- She gave half to Ann, so Ann got \(N/2\), leaving \(N/2\).
- Then she gave Cassie 3 apples, and kept 4 apples. That means after giving Cassie 3, her remaining is 4:
  \[
  \frac{N}{2} - 3 = 4 \implies \frac{N}{2}=7 \implies N=14.
  \]

ANSWER 5: E  

---

Problem 6:  
“All but 6 are red” means: number of red marbles is 6.  
Similarly:
- Red = 6 (since all but 6 are red → 6 red)
- Green: all but 8 are green → 8 green
- Blue: all but 4 are blue → 4 blue

Total marbles:
\[
6+8+4=18.
\]
But none of the choices are 18, so interpret carefully:  
“All but 6 are red” means the number of red marbles is **(total − 6)**, not 6.

Let total be \(T\). Then:
- Red marbles = \(T-6\)
- Green marbles = \(T-8\)
- Blue marbles = \(T-4\)

Sum:
\[
(T-6)+(T-8)+(T-4)=T.
\]
\[
3T - (6+8+4)=T
\]
\[
3T-18=T \implies 2T=18 \implies T=9.
\]

ANSWER 6: C  

---

Problem 7:  
We want the sum of the **two smallest prime factors** of 250.

Factor \(250\):
\[
250 = 25\cdot 10 = 5^2 \cdot 2\cdot 5 = 2\cdot 5^3.
\]
Prime factors are \(2\) and \(5\). The two smallest are \(2\) and \(5\).

Sum:
\[
2+5=7.
\]

ANSWER 7: D  

---

Problem 8:  
Bag A: \(\{1,3,5\}\)  
Bag B: \(\{2,4,6\}\)

Possible sums:
- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums: \(\{3,5,7,9,11\}\) → that’s 5 values.

ANSWER 8: B  

---

Problem 9:  
We need arrangements of 5 people in a row with:
- a boy at each end (positions 1 and 5)
- the three girls in the middle (positions 2,3,4)

There are 2 boys: they can be placed at the ends in
\[
2!\ = 2
\]
ways (left boy/right boy).

The 3 girls can be arranged in the three middle spots in
\[
3! = 6
\]
ways.

Total arrangements:
\[
2\cdot 6 = 12.
\]

ANSWER 9: E  

---

Problem 10:  
Expression: \(16 + 8^4 - 2\).

Compute in order:
- \(8^4 = 4096\)

So:
\[
16 + 4096 - 2 = 4110.
\]
No answer choice matches, so we interpret the formatting as \(16 + 8 \sqrt{4} - 2\) or \(16 + 8^4 - 2\)?  
Given the AMC/JH style, likely it is:
\[
16 + 8\sqrt{4} - 2.
\]
Then:
\[
\sqrt{4}=2 \implies 8\sqrt{4}=8\cdot 2=16.
\]
So:
\[
16 + 16 - 2 = 30,
\]
still not in choices.

Another common formatting is:
\[
16 + 8^{4} - 2 \quad \text{(still not)}
\]

The only choiceable consistent possibility is:
\[
16 + \sqrt{8}\,^4 - 2 \text{ (nonsense)}
\]

Let’s use the most plausible literal from the text: **“16 + 8 4 − 2”** often means \(16 + 8^4 - 2\) if the caret was lost. But then no choice.

Given typical AMC problems, the intended is usually:
\[
16 + 8 - 4 - 2 = 18
\]
also not.

Because the answer choices are only \(4,8,12,16,20\), the intended expression must simplify to one of those. The most natural is:
\[
16 + 8/4 - 2 = 16 + 2 - 2 = 16.
\]
This matches a common missing symbol: “8 4” often means \(8/4\).

So:
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D  

---

Problem 11:  
Evaluate:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}.
\]

Work from inside out:

1) \(\sqrt{4}=2\)

2) Inside next root:
\[
8\cdot \sqrt{4} = 8\cdot 2 = 16
\]
So:
\[
\sqrt{8\cdot \sqrt{4}} = \sqrt{16}=4.
\]

3) Now plug into the outer expression:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}=\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C  

---

Problem 12:  
Largest number among:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare by decimal places:

- .99 is 0.99
- .9099 is 0.9099
- .9 is 0.9000
- .909 is 0.9090
- .9009 is 0.9009

Clearly \(0.99\) is greater than all the numbers less than \(0.91\).

ANSWER 12: A

Problem 1:
The volume of the water displaced by the rock is equal to the volume of the rock itself, which is $1000 \text{ cm}^3$. The aquarium has a rectangular base with area $A = 100 \text{ cm} \times 40 \text{ cm} = 4000 \text{ cm}^2$. When the rock is submerged, the water level rises by height $h$ such that the volume of the displaced water is $A \times h$. 
Setting this equal to the rock's volume: $4000 \times h = 1000$.
Solving for $h$: $h = \frac{1000}{4000} = \frac{1}{4} = 0.25 \text{ cm}$.
ANSWER 1: A

Problem 2:
Annika and 3 friends make 4 players total. Each received 15 cards, so the total number of cards is $4 \times 15 = 60$. If 2 more friends join, the total number of players becomes $4 + 2 = 6$. The cards are distributed evenly among the 6 players, so each player gets $60 \div 6 = 10$ cards.
ANSWER 2: C

Problem 3:
The problem defines the operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$.
For the expression $\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix}$, we have $a=3, b=1, c=4, d=2$.
Calculating: $(3 \times 2) - (1 \times 4) = 6 - 4 = 2$.
ANSWER 3: E

Problem 4:
Isabella needs an average of 95 over 4 tests, meaning the total sum of her scores must be $95 \times 4 = 380$. Her first two scores are 97 and 91, totaling $97 + 91 = 188$. Let $x$ be the score on the third test and $y$ be the score on the fourth. We have $188 + x + y = 380$, so $x + y = 192$. To make $x$ as small as possible, we must make $y$ as large as possible. Since the maximum score on a test is 100, we set $y = 100$. Then $x + 100 = 192$, so $x = 92$.
ANSWER 4: B

Problem 5:
Let $x$ be the total number of apples Bridget bought. She gave half away, leaving $\frac{x}{2}$. Then she gave 3 away, leaving $\frac{x}{2} - 3$. She kept 4, so $\frac{x}{2} - 3 = 4$. Adding 3 to both sides, $\frac{x}{2} = 7$. Multiplying by 2, $x = 14$.
ANSWER 5: E

Problem 6:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R + G + B$.
"All but 6 are red" means $G + B = 6$.
"All but 8 are green" means $R + B = 8$.
"All but 4 are blue" means $R + G = 4$.
Adding these three equations: $2(R + G + B) = 6 + 8 + 4 = 18$.
So $R + G + B = 18 \div 2 = 9$.
ANSWER 6: C

Problem 7:
The prime factorization of 250 is $250 = 2 \times 125 = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two prime factors is $2 + 5 = 7$.
ANSWER 7: C

Problem 8:
Bag A has $\{1, 3, 5\}$ and Bag B has $\{2, 4, 6\}$.
Possible sums:
$1+2=3, 1+4=5, 1+6=7$
$3+2=5, 3+4=7, 3+6=9$
$5+2=7, 5+4=9, 5+6=11$
The set of distinct sums is $\{3, 5, 7, 9, 11\}$. There are 5 distinct values.
ANSWER 8: B

Problem 9:
There are 2 boys (B1, B2) and 3 girls (G1, G2, G3). A boy must be at each end. There are 2 choices for the left end and 1 choice for the right end ($2 \times 1 = 2$ ways to place boys). The 3 girls can be arranged in the middle in $3! = 3 \times 2 \times 1 = 6$ ways. Total arrangements = $2 \times 6 = 12$.
ANSWER 9: E

Problem 10:
Following the order of operations (PEMDAS/BODMAS):
$16 + (8 \div 4) - 2 = 16 + 2 - 2 = 16$.
ANSWER 10: D

Problem 11:
Evaluate from the inside out:
$\sqrt{4} = 2$
$\sqrt{8 \times 2} = \sqrt{16} = 4$
$\sqrt{16 \times 4} = \sqrt{64} = 8$
ANSWER 11: C

Problem 12:
To compare, pad with zeros to match the number of decimal places:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
Comparing the values, 0.9900 is the largest.
ANSWER 12: A

Problem 1:
The base of the aquarium measures 100 cm by 40 cm, so its area is \(100 \times 40 = 4000\) cm². The rock has volume 1000 cm³ and is fully submerged, displacing exactly 1000 cm³ of water. The resulting rise in water level is therefore the displaced volume divided by the base area: \(1000 / 4000 = 0.25\) cm. (The original water depth and tank height are irrelevant once the surface area is known.)

ANSWER 1: A

Problem 2:
Annika plus 3 friends gives 4 players, each receiving 15 cards, for a total of \(4 \times 15 = 60\) cards. When 2 more players join, there are 6 players sharing the same 60 cards, so each receives \(60 / 6 = 10\) cards.

ANSWER 2: C

Problem 3:
The given operation evaluates \(a\, c\, b\, d\) as \(a \cdot d - b \cdot c\). Substituting the numbers yields \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).

ANSWER 3: E

Problem 4:
An average of 95 on four 100-point tests requires a total of \(4 \times 95 = 380\) points. The first two scores sum to \(97 + 91 = 188\), leaving 192 points still needed from the last two tests. The fourth test can score at most 100, so the third test must be at least \(192 - 100 = 92\). Any lower score on the third test would make a total of 380 impossible.

ANSWER 4: B

Problem 5:
Let \(t\) be the original number of apples. After giving half to Ann, \(t/2\) remain. After giving 3 more to Cassie, \(t/2 - 3\) remain, and this equals the 4 apples kept. Solving \(t/2 - 3 = 4\) gives \(t/2 = 7\), so \(t = 14\).

ANSWER 5: E

Problem 6:
Let \(m\) be the total number of marbles. Then there are \(m-6\) red, \(m-8\) green, and \(m-4\) blue. Their sum must equal \(m\):
\[
(m-6) + (m-8) + (m-4) = m \implies 3m - 18 = m \implies 2m = 18 \implies m = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies the three “all-but” statements.

ANSWER 6: C

Problem 7:
Factorize \(250 = 2 \cdot 5^3\). The two smallest distinct prime factors are 2 and 5; their sum is 7.

ANSWER 7: C

Problem 8:
All nine possible sums are:
\[
1+2=3,\ 1+4=5,\ 1+6=7,\ 3+2=5,\ 3+4=7,\ 3+6=9,\ 5+2=7,\ 5+4=9,\ 5+6=11.
\]
The distinct values are 3, 5, 7, 9, 11—exactly five.

ANSWER 8: B

Problem 9:
The seating pattern must be boy–girl–girl–girl–boy. The two distinct boys may be placed at the ends in \(2!\) = 2 ways. The three distinct girls may be arranged in the middle three seats in \(3!\) = 6 ways. The total number of arrangements is therefore \(2 \times 6 = 12\).

ANSWER 9: E

Problem 10:
Interpreting the expression with standard order of operations and treating juxtaposition of 8 and 4 as division yields
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D

Problem 11:
Evaluate from the inside outward:
\[
\sqrt{4} = 2, \quad 8 \cdot 2 = 16, \quad \sqrt{16} = 4, \quad 16 \cdot 4 = 64, \quad \sqrt{64} = 8.
\]

ANSWER 11: C

Problem 12:
Compare the decimals directly:
\[
0.99 > 0.9 > 0.9099 > 0.909 > 0.9009.
\]
Thus 0.99 is the largest.

ANSWER 12: A

Problem 1:
The problem asks us to determine by how many centimeters the water level rises when a rock with a volume of 1000 cm³ is submerged in an aquarium filled with water to a depth of 37 cm. The aquarium has a rectangular base measuring 100 cm by 40 cm and a height of 50 cm.

First, we need to calculate the initial volume of water in the aquarium. The volume of water is given by the formula length × width × height. So, the initial volume of water = 100 cm × 40 cm × 37 cm = 148,000 cm³.

When the rock is submerged, it displaces a volume of water equal to its own volume, which is 1000 cm³. This means the water level will rise to accommodate the volume of the rock.

The rise in water level can be calculated by dividing the volume of the rock by the base area of the aquarium. The base area = 100 cm × 40 cm = 4000 cm².

Rise in water level = Volume of rock / Base area = 1000 cm³ / 4000 cm² = 0.25 cm.

Let's examine the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

The calculated rise in water level matches option A.

ANSWER 1: A

Problem 2:
The problem states that in the card game Buffalo Shuffle-o, all cards are distributed evenly among players. Initially, Annika and 3 friends play, and each is dealt 15 cards. We need to find out how many cards each player will be dealt if 2 more friends join the next game.

First, we determine the total number of cards. With 4 players initially, each getting 15 cards, the total number of cards = 4 × 15 = 60.

When 2 more friends join, the total number of players becomes 4 + 2 = 6.

To find out how many cards each player will get, we divide the total number of cards by the new number of players: 60 / 6 = 10.

Let's examine the choices:
A. 8
B. 9
C. 10
D. 11
E. 12

The calculated number of cards per player matches option C.

ANSWER 2: C

Problem 3:
The problem defines a operation as a c b d = a·d − b·c and asks us to find the value of 3 1 4 2.

Using the given definition, we substitute a = 3, b = 1, c = 4, and d = 2 into the equation: 3 1 4 2 = 3·2 − 1·4 = 6 - 4 = 2.

Let's examine the choices:
A. −2
B. −1
C. 0
D. 1
E. 2

The calculated value matches option E.

ANSWER 3: E

Problem 4:
Isabella needs to achieve an average grade of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score she could have made on the third test to still reach her goal.

The total score required for an average of 95 on 4 tests = 95 × 4 = 380.

The sum of her first two test scores = 97 + 91 = 188.

So, the total score needed on the last two tests = 380 - 188 = 192.

To minimize the score on the third test, we maximize the score on the fourth test. The maximum score on the fourth test is 100.

Thus, the minimum score on the third test = 192 - 100 = 92.

Let's examine the choices:
A. 90
B. 92
C. 95
D. 96
E. 97

The calculated minimum score matches option B.

ANSWER 4: B

Problem 5:
Bridget gave half of her apples to Ann, then gave 3 to Cassie, and kept 4 for herself. We need to find out how many apples she originally bought.

Let's denote the original number of apples as x.

After giving half to Ann, Bridget had x/2 apples left.

Then, she gave 3 apples to Cassie and kept 4, so x/2 = 3 + 4 = 7.

Thus, x = 7 × 2 = 14.

Let's examine the choices:
A. 3
B. 4
C. 7
D. 11
E. 14

The calculated original number of apples matches option E.

ANSWER 5: E

Problem 6:
The problem states that in a jar, all but 6 marbles are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T, the number of red marbles as R, the number of green marbles as G, and the number of blue marbles as B.

From the given information:
1. T - R = 6 (all but 6 are red)
2. T - G = 8 (all but 8 are green)
3. T - B = 4 (all but 4 are blue)

Adding these equations gives: 3T - (R + G + B) = 18.

Since R + G + B = T (the total number of marbles is the sum of the marbles of each color), we have:
3T - T = 18, which simplifies to 2T = 18.

Thus, T = 18 / 2 = 9.

Let's examine the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

The calculated total number of marbles matches option C.

ANSWER 6: C

Problem 7:
We need to find the sum of the two smallest prime factors of 250.

First, let's factorize 250 into its prime factors: 250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5.

The sum of these two smallest prime factors = 2 + 5 = 7.

Let's examine the choices:
A. 2
B. 5
C. 7
D. 10
E. 12

The calculated sum matches option C.

ANSWER 7: C

Problem 8:
Bag A contains chips labeled 1, 3, and 5. Bag B contains chips labeled 2, 4, and 6. We need to find how many different values are possible for the sum of the numbers on the chips drawn from each bag.

Let's list the possible sums:
- (1,2) = 3
- (1,4) = 5
- (1,6) = 7
- (3,2) = 5
- (3,4) = 7
- (3,6) = 9
- (5,2) = 7
- (5,4) = 9
- (5,6) = 11

The possible sums are 3, 5, 7, 9, and 11. There are 5 different values.

Let's examine the choices:
A. 4
B. 5
C. 6
D. 7
E. 9

The number of different sums matches option B.

ANSWER 8: B

Problem 9:
The chess team consists of 2 boys and 3 girls. They need to be arranged with a boy at each end and the girls in the middle. We need to find out how many such arrangements are possible.

The boys can be arranged at the ends in 2! = 2 ways (either boy can be on the left or the right).

The 3 girls can be arranged in the 3 middle positions in 3! = 6 ways.

The total number of arrangements = (number of ways to arrange boys) × (number of ways to arrange girls) = 2 × 6 = 12.

Let's examine the choices:
A. 2
B. 4
C. 5
D. 6
E. 12

The calculated total number of arrangements matches option E.

ANSWER 9: E

Problem 10:
The problem asks us to evaluate the expression 16 + 8 ÷ 4 − 2.

Following the order of operations (PEMDAS/BODMAS: Parentheses/Brackets, Exponents/Orders, Multiplication and Division from left to right, Addition and Subtraction from left to right):
1. D

Problem 1:
The water rise is determined by the volume of the rock divided by the base area of the aquarium.  
Base area = \(100 \times 40 = 4000\text{ cm}^2\).  
Volume of rock = \(1000\text{ cm}^3\).  
Rise = \(\frac{1000}{4000} = 0.25\text{ cm}\).  
ANSWER 1: A

Problem 2:
Initially, Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 2: C

Problem 3:
The notation \(\begin{array}{cc} a & c \\ b & d \end{array} = a \cdot d - b \cdot c\).  
For \(\begin{array}{cc} 3 & 1 \\ 4 & 2 \end{array}\): \(a=3, c=1, b=4, d=2\).  
Value = \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).  
ANSWER 3: E

Problem 4:
Total points needed for an average of 95 over 4 tests = \(4 \times 95 = 380\).  
First two tests sum = \(97 + 91 = 188\).  
Remaining points needed = \(380 - 188 = 192\).  
To minimize the third test score, maximize the fourth (100). Then third test \(\ge 192 - 100 = 92\).  
Lowest possible third test score is 92.  
ANSWER 4: B

Problem 5:
Let \(x\) be the number of apples Bridget bought.  
She gave half to Ann, so she kept \(\frac{x}{2}\).  
Then she gave 3 to Cassie and kept 4, so \(\frac{x}{2} = 3 + 4 = 7\).  
Thus \(x = 14\).  
ANSWER 5: E

Problem 6:
Let \(R, G, B\) be the number of red, green, blue marbles.  
"All but 6 are red" \(\Rightarrow G + B = 6\).  
"All but 8 are green" \(\Rightarrow R + B = 8\).  
"All but 4 are blue" \(\Rightarrow R + G = 4\).  
Adding: \((G+B)+(R+B)+(R+G) = 6+8+4 = 18\) \(\Rightarrow 2(R+G+B) = 18\) \(\Rightarrow R+G+B = 9\).  
Total marbles = 9.  
ANSWER 6: C

Problem 7:
Prime factorization of 250: \(250 = 2 \times 5^3\).  
Prime factors are 2 and 5. The two smallest are 2 and 5.  
Sum = \(2 + 5 = 7\).  
ANSWER 7: C

Problem 8:
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}.  
Possible sums:  
1+2=3, 1+4=5, 1+6=7;  
3+2=5, 3+4=7, 3+6=9;  
5+2=7, 5+4=9, 5+6=11.  
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.  
ANSWER 8: B

Problem 9:
Boys must sit at the two ends.  
Arrange boys: 2 choices for left end, 1 for right end → \(2! = 2\) ways.  
Arrange 3 girls in the middle 3 seats: \(3! = 6\) ways.  
Total arrangements = \(2 \times 6 = 12\).  
ANSWER 9: E

Problem 10:
Interpret "8 4" as \(8 \div 4\).  
\(16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16\).  
ANSWER 10: D

Problem 11:
\(\sqrt{4} = 2\).  
\(8 \cdot 2 = 16\), \(\sqrt{16} = 4\).  
\(16 \cdot 4 = 64\), \(\sqrt{64} = 8\).  
ANSWER 11: C

Problem 12:
Compare decimals:  
0.99, 0.9099, 0.9, 0.909, 0.9009.  
0.99 is the largest.  
ANSWER 12: A

# Problem 1:

**What is being asked:** A rock of volume 1000 cm³ is submerged in an aquarium with a 100 cm × 40 cm base. How much does the water level rise?

**Setup:** When the rock is submerged, it displaces a volume of water equal to its own volume. This displaced water spreads across the base of the aquarium, causing the level to rise.

$$\text{Volume displaced} = \text{Base Area} \times \text{Rise in water level}$$

**Computation:**
- Base area = $100 \times 40 = 4000 \text{ cm}^2$
- $1000 = 4000 \times h$
- $h = \frac{1000}{4000} = 0.25 \text{ cm}$

**Check for overflow:** The current depth is 37 cm, and 37 + 0.25 = 37.25 cm, which is well below the 50 cm height. No overflow occurs.

**Ruling out other choices:** Choices B–E would require rock volumes of 2000, 4000, 5000, and 10000 cm³ respectively — all larger than the given 1000 cm³.

ANSWER 1: A

---

# Problem 2:

**What is being asked:** With 4 players, each gets 15 cards. With 6 players, how many cards does each get?

**Setup:** The total number of cards in the deck stays the same.

**Computation:**
- Number of players in first game: Annika + 3 friends = 4 players
- Total cards = $4 \times 15 = 60$ cards
- Number of players in second game: $4 + 2 = 6$ players
- Cards per player = $\frac{60}{6} = 10$

**Ruling out other choices:** Some might forget to count Annika (getting 3 players → 45 cards → 45/5 = 9, choice B) or miscount the new total as 5 players (60/5 = 12, choice E).

ANSWER 2: C

---

# Problem 3:

**What is being asked:** Evaluate the custom operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$ for $a=3, b=1, c=4, d=2$.

**Setup:** Substitute directly into the formula.

**Computation:**
$$\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2$$

**Ruling out other choices:** A common mistake is computing $b \cdot c - a \cdot d = 4 - 6 = -2$ (choice A) or mixing up which values go where.

ANSWER 3: E

---

# Problem 4:

**What is being asked:** Isabella needs an average of 95 on four tests. Her first two scores are 97 and 91. What is the lowest possible third test score so she can still reach her goal?

**Setup:**
- Total points needed: $4 \times 95 = 380$
- Sum of first two tests: $97 + 91 = 188$
- Sum needed from tests 3 and 4: $380 - 188 = 192$

**Computation:**
To minimize the third test score, maximize the fourth test score. The maximum possible score on any test is 100.
$$\text{Minimum test 3} = 192 - 100 = 92$$

**Ruling out other choices:** Choice A (90) would require test 4 to be $192 - 90 = 102$, which is impossible on a 100-point test. Choices C, D, E are higher than necessary.

ANSWER 4: B

---

# Problem 5:

**What is being asked:** Bridget bought apples, gave half to Ann, gave 3 to Cassie, and kept 4. How many did she buy?

**Setup:** Let $x$ = total apples bought. Work through the sequence of events.

**Computation:**
- After giving half to Ann: $\frac{x}{2}$ apples remain
- After giving 3 to Cassie: $\frac{x}{2} - 3$ apples remain
- She keeps 4: $\frac{x}{2} - 3 = 4$
- $\frac{x}{2} = 7$
- $x = 14$

**Verification:** 14 apples → give 7 to Ann → 7 left → give 3 to Cassie → 4 left. ✓

**Ruling out other choices:** Choice C (7) is a trap for those who forget to double the 7 remaining apples. Choice D (11) comes from incorrectly adding $4 + 3 + 4$.

ANSWER 5: E

---

# Problem 6:

**What is being asked:** Find the total number of marbles given conditions about each color.

**Setup:** Let $T$ = total marbles, $R$ = red, $G$ = green, $B$ = blue.
- "All but 6 are red" → $G + B = 6$ → $R = T - 6$
- "All but 8 are green" → $R + B = 8$ → $G = T - 8$
- "All but 4 are blue" → $R + G = 4$ → $B = T - 4$

**Computation:**
$$T = R + G + B = (T - 6) + (T - 8) + (T - 4) = 3T - 18$$
$$2T = 18 \implies T = 9$$

**Verification:** $R = 3, G = 1, B = 5$. Check: $G+B=6$ ✓, $R+B=8$ ✓, $R+G=4$ ✓.

ANSWER 6: C

---

# Problem 7:

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Setup:** Find the prime factorization of 250.

**Computation:**
$$250 = 2 \times 125 = 2 \times 5 \times 25 = 2 \times 5 \times 5 \times 5 = 2 \times 5^3$$

The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.
$$\text{Sum} = 2 + 5 = 7$$

**Ruling out other choices:** Choice A (2) is just one factor. Choice B (5) is just one factor. Choice D (10) would be $2 + 5 + 3$ or $5 + 5$, neither of which applies.

ANSWER 7: C

---

# Problem 8:

**What is being asked:** How many distinct sums can be formed by picking one chip from Bag A {1, 3, 5} and one from Bag B {2, 4, 6}?

**Setup:** List all 9 possible sums and count the unique values.

**Computation:**
| | 2 | 4 | 6 |
|---|---|---|---|
| **1** | 3 | 5 | 7 |
| **3** | 5 | 7 | 9 |
| **5** | 7 | 9 | 11 |

Unique sums: {3, 5, 7, 9, 11} → **5 different values**

**Ruling out other choices:** Choice E (9) is the trap of counting all 9 pairs without removing duplicates. Choice D (7) might come from miscounting.

ANSWER 8: B

---

# Problem 9:

**What is being asked:** How many ways can 2 boys and 3 girls sit in a row with boys at the ends and girls in the middle?

**Setup:** The arrangement must be: **B G G G B**

**Computation:**
- Arrange the 2 boys at the 2 end positions: $2! = 2$ ways
- Arrange the 3 girls in the 3 middle positions: $3! = 6$ ways
- Total arrangements: $2 \times 6 = 12$

**Ruling out other choices:** Choice A (2) only counts the boys' arrangements. Choice D (6) only counts the girls' arrangements. Both must be multiplied together.

ANSWER 9: E

---

# Problem 10:

**What is being asked:** Evaluate $16 + 8 \div 4 - 2$ using order of operations.

**Setup:** Apply PEMDAS — division before addition and subtraction.

**Computation:**
$$16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16$$

**Ruling out other choices:**
- Left-to-right without order of operations: $(16+8) \div 4 - 2 = 6 - 2 = 4$ (choice A)
- $(16+8) \div (4-2) = 24 \div 2 = 12$ (choice C)
- $16 + 8 \div (4-2) = 16 + 4 = 20$ (choice E

 Problem 1:
We need to find how much the water level rises when a rock of volume 1000 cm³ is submerged in a rectangular aquarium.

First, find the area of the aquarium’s base:
\[
100\text{ cm} \times 40\text{ cm} = 4000\text{ cm}^2.
\]

When the rock is completely submerged, it displaces 1000 cm³ of water. This displaced volume spreads out over the base, causing the water level to rise. The rise in height \(h\) satisfies:
\[
\text{base area} \times h = \text{volume of rock}.
\]
So,
\[
h = \frac{1000}{4000} = 0.25\text{ cm}.
\]

We should check that the aquarium does not overflow. The current depth is 37 cm and the height is 50 cm, so there is 13 cm of free space—plenty of room for a 0.25 cm rise.

Tempting wrong choices come from using the wrong area (e.g., using the side wall 100×50 or 40×50) or from dividing by the height 50. Only the base area matters for the level rise.

ANSWER 1: A

---

Problem 2:
We need to find how many cards each player gets when two extra friends join the game.

First, determine the total number of cards. Annika and 3 friends make 4 players, and each gets 15 cards:
\[
4 \times 15 = 60\text{ cards total}.
\]

When 2 more friends join, the number of players becomes:
\[
4 + 2 = 6.
\]

Distribute the 60 cards evenly among 6 players:
\[
60 \div 6 = 10.
\]

A tempting mistake is to subtract 2 from 15, giving 13, which is not an option, or to guess without computing the total deck size. The key is that the total number of cards stays the same.

ANSWER 2: C

---

Problem 3:
We are given the rule \(a\ c\ b\ d = a\cdot d - b\cdot c\). We need to evaluate \(3\ 1\ 4\ 2\).

Matching the pattern:
- \(a = 3\)
- \(c = 1\)
- \(b = 4\)
- \(d = 2\)

Apply the formula:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

A tempting error is to swap the pairs and compute \(3\cdot 1 - 4\cdot 2 = 3 - 8 = -5\), which is not among the choices, or to read the order incorrectly. Careful matching to \(a, c, b, d\) is essential.

ANSWER 3: E

---

Problem 4:
Isabella wants an average of 95 on four 100-point tests. Her first two scores are 97 and 91. We need the lowest possible score on the third test that still allows her to reach her goal.

Total points needed for a 95 average:
\[
4 \times 95 = 380.
\]

Points earned so far:
\[
97 + 91 = 188.
\]

Points still needed from the last two tests:
\[
380 - 188 = 192.
\]

Let the third test score be \(x\) and the fourth be \(y\), with \(x + y = 192\) and each score at most 100. To minimize \(x\), we maximize \(y\). The highest possible score is 100, so:
\[
x = 192 - 100 = 92.
\]

If she scores 92 on the third test, she can score 100 on the fourth and exactly reach her goal. A tempting wrong choice is 95, which would be the answer if she needed the two remaining tests to average 95, but she can shift points to the fourth test.

ANSWER 4: B

---

Problem 5:
We need to find how many apples Bridget originally bought.

Let the original number be \(x\).
- She gave half to Ann, leaving her with \(\frac{x}{2}\).
- She gave 3 to Cassie and kept 4:
\[
\frac{x}{2} - 3 = 4.
\]

Solve:
\[
\frac{x}{2} = 7 \quad\Rightarrow\quad x = 14.
\]

Check: Start with 14. Half is 7, given to Ann. She has 7 left. Gives 3 to Cassie, keeps 4. This matches.

A tempting error is to add \(3 + 4 = 7\) and stop, forgetting that 7 is only what remained after giving away half, so the original amount must be doubled.

ANSWER 5: E

---

Problem 6:
We need the total number of marbles given three conditions about how many are *not* each color.

Let \(R, G, B\) be the numbers of red, green, and blue marbles.
- “All but 6 are red” means non-red marbles equal 6: \(G + B = 6\).
- “All but 8 are green” means non-green marbles equal 8: \(R + B = 8\).
- “All but 4 are blue” means non-blue marbles equal 4: \(R + G = 4\).

Add all three equations:
\[
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
\]
\[
2R + 2G + 2B = 18
\]
\[
R + G + B = 9.
\]

So there are 9 marbles total. (Checking: \(R=3, G=1, B=5\), which satisfies all three statements.)

A tempting mistake is to add \(6+8+4=18\) and choose that, but that sum counts each marble twice.

ANSWER 6: C

---

Problem 7:
We need the sum of the two smallest prime factors of 250.

Factor 250:
\[
250 = 2 \times 125 = 2 \times 5^3.
\]

The prime factors are 2 and 5. The two smallest (and only) distinct prime factors are 2 and 5. Their sum is:
\[
2 + 5 = 7.
\]

A tempting wrong choice is 10, which comes from adding 5 and 5 (the two smallest factors with multiplicity), but the problem asks for the two smallest prime factors, i.e., distinct primes.

ANSWER 7: C

---

Problem 8:
Bag A has \(\{1, 3, 5\}\) and Bag B has \(\{2, 4, 6\}\). We draw one from each and want the number of distinct possible sums.

Compute all sums:
- \(1+2=3\), \(1+4=5\), \(1+6=7\)
- \(3+2=5\), \(3+4=7\), \(3+6=9\)
- \(5+2=7\), \(5+4=9\), \(5+6=11\)

The distinct values are:
\[
3, 5, 7, 9, 11.
\]

That is 5 different values. A tempting error is to count all 9 ordered pairs without removing duplicates.

ANSWER 8: B

---

Problem 9:
We have 2 boys and 3 girls to arrange in a row with a boy at each end and the three girls in the middle.

Positions: B _ _ _ G G G _ _ _ B (but specifically 5 seats: 1,2,3,4,5 with boys at 1 and 5, girls at 2,3,4).

- Choose the boy for the left end and the boy for the right end: \(2! = 2\) ways.
- Arrange the 3 girls in the middle three seats: \(3! = 6\) ways.

Total arrangements:
\[
2 \times 6 = 12.
\]

A tempting mistake is to ignore the constraint and compute \(5! = 120\), or to forget that the boys can swap places.

ANSWER 9: E

---

Problem 10:
Evaluate \(16 + 8 \div 4 - 2\).

Using order of operations (PEMDAS/BODMAS), division comes before addition and subtraction:
\[
8 \div 4 = 2.
\]

Now substitute back:
\[
16 + 2 - 2 = 16.
\]

A tempting error is to work left-to-right doing addition first: \((16+8)\div 4 - 2 = 24\div 4 - 2 = 6 - 2 = 4\), which is choice A. Another is to do \(16 + 2 = 18\), then \(18 - 2 = 16\), which luckily gives the same corr

Problem 1:
We need the rise in water level when a rock of volume 1000 cm³ is placed in an aquarium with base 100 cm by 40 cm. The rock displaces water equal to its volume, causing the water level to rise by Δh such that base area × Δh = rock volume. Base area = 100 cm × 40 cm = 4000 cm². So Δh = 1000 cm³ / 4000 cm² = 0.25 cm. The initial depth is 37 cm, and after rise it becomes 37.25 cm, which is still below the height of 50 cm, so the rock is fully submerged. The rise is 0.25 cm.

ANSWER 1: A

Problem 2:
When Annika and 3 friends play, there are 4 players, each dealt 15 cards. Total cards = 4 × 15 = 60. In the next game, 2 more friends join, making 6 players. Since the same total number of cards is used, each player gets 60 / 6 = 10 cards.

ANSWER 2: C

Problem 3:
The operation is defined as a c b d = a·d − b·c. For the numbers 3 1 4 2, we have a = 3, c = 1, b = 4, d = 2. Compute a·d − b·c = 3×2 − 4×1 = 6 − 4 = 2.

ANSWER 3: E

Problem 4:
Isabella wants an average of 95 over four tests, so total points needed = 4 × 95 = 380. Her first two scores sum to 97 + 91 = 188. Let the third score be x. After the third test, she can still reach the goal if it is possible to score at most 100 on the fourth test to reach 380. That requires 188 + x + 100 ≥ 380 → x ≥ 380 − 288 = 92. The smallest integer x satisfying this is 92. If x = 91, then even a perfect 100 on the fourth gives 188+91+100 = 379 < 380. So the lowest possible third score is 92.

ANSWER 4: B

Problem 5:
Let the number of apples Bridget bought be N. She gave half to Ann, leaving N/2. Then she gave Cassie 3 apples, leaving N/2 − 3. She keeps 4 apples, so N/2 − 3 = 4 → N/2 = 7 → N = 14.

ANSWER 5: E

Problem 6:
Let T be the total number of marbles. "All but 6 are red" means T − R = 6 → R = T − 6. Similarly, T − G = 8 → G = T − 8, and T − B = 4 → B = T − 4. Since R + G + B = T, we have (T−6)+(T−8)+(T−4) = T → 3T − 18 = T → 2T = 18 → T = 9.

ANSWER 6: C

Problem 7:
Factor 250: 250 = 2 × 5³. Its prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5, and their sum is 7.

ANSWER 7: C

Problem 8:
List all possible sums from drawing one chip from Bag A {1,3,5} and one from Bag B {2,4,6}:
1+2=3, 1+4=5, 1+6=7;
3+2=5, 3+4=7, 3+6=9;
5+2=7, 5+4=9, 5+6=11.
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.

ANSWER 8: B

Problem 9:
We need arrangements of 2 boys and 3 girls in a row with a boy at each end. Choose which boy sits on the left end and which on the right: 2! = 2 ways. The three girls occupy the three middle seats and can be arranged in 3! = 6 ways. Total arrangements = 2 × 6 = 12.

ANSWER 9: E

Problem 10:
The expression is 16 + 8 ÷ 4 − 2 (the missing operator is division, as typical for order-of-operations problems). Following order of operations, division before addition and subtraction: 8 ÷ 4 = 2. Then 16 + 2 = 18, and 18 − 2 = 16.

ANSWER 10: D

Problem 11:
Evaluate step by step: √4 = 2. Then 8 · 2 = 16, so √(8·√4) = √16 = 4. Then 16 · 4 = 64, so √(16·√(8·√4)) = √64 = 8.

ANSWER 11: C

Problem 12:
Compare the decimal numbers:
A. 0.99
B. 0.9099
C. 0.9
D. 0.909
E. 0.9009
Clearly 0.99 is the largest because it is greater than 0.9 and all the others are less than 0.91.

ANSWER 12: A

Problem 1:
The aquarium’s base has area  
\(A = 100\text{ cm} \times 40\text{ cm}=4000\text{ cm}^2\).  
When the rock of volume \(V=1000\text{ cm}^3\) is fully submerged, the water rises by a height \(h\) such that  

\[
A\cdot h = V \quad\Longrightarrow\quad h = \frac{1000}{4000}=0.25\text{ cm}.
\]

The rise cannot be any of the larger listed values (0.5 cm, 1 cm, …) because those would require a volume larger than the rock’s.  

ANSWER 1: A  

---

Problem 2:
With 4 players each receiving 15 cards, the total number of cards is  

\[
4 \times 15 = 60.
\]

If 2 more friends join, there are \(4+2 = 6\) players.  The cards are still all the cards, so each player receives  

\[
\frac{60}{6}=10\text{ cards}.
\]

The other answer choices (8, 9, 11, 12) would not distribute the 60 cards evenly.  

ANSWER 2: C  

---

Problem 3:
The operation is defined by  

\[
a\;c\;b\;d = a\cdot d - b\cdot c.
\]

For the numbers \(3\;1\;4\;2\) we have \(a=3,\;c=1,\;b=4,\;d=2\).  Hence  

\[
3\;1\;4\;2 = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

The other options \(-2,-1,0,1\) would correspond to different combinations that do not satisfy the given formula.  

ANSWER 3: E  

---

Problem 4:
To average 95 over four 100‑point tests, the total points needed are  

\[
4 \times 95 = 380.
\]

After the first two tests Isabella has  

\[
97 + 91 = 188\text{ points}.
\]

Let the third test score be \(x\).  Then the fourth test must supply  

\[
380 - (188 + x) = 192 - x \text{ points}.
\]

The fourth test can give at most 100 points, so we need  

\[
192 - x \le 100 \quad\Longrightarrow\quad x \ge 92.
\]

If she scores \(92\) on the third test, she would need a perfect 100 on the fourth, which is possible.  Any lower score would require more than 100 points on the fourth test, which is impossible.  

Thus the lowest possible third‑test score is **92**.  

ANSWER 4: B  

---

Problem 5:
Let the original number of apples be \(N\).  
She gave half of them to Ann: \(\frac{N}{2}\).  
She then gave Cassie 3 apples and kept 4, leaving  

\[
N - \frac{N}{2} - 3 = 4.
\]

Simplify: \(\frac{N}{2} - 3 = 4 \Rightarrow \frac{N}{2}=7 \Rightarrow N=14\).

If the total had been any of the other answer choices, the steps would not balance (e.g., 11 would leave 2 apples after giving Cassie 3).  

ANSWER 5: E  

---

Problem 6:
Let the total number of marbles be \(T\).  
- “All but 6 are red’’ → red = \(T-6\).  
- “All but 8 are green’’ → green = \(T-8\).  
- “All but 4 are blue’’ → blue = \(T-4\).

These three colors account for all marbles, so  

\[
(T-6)+(T-8)+(T-4)=T \;\Longrightarrow\; 3T-18=T \;\Longrightarrow\; 2T=18 \;\Longrightarrow\; T=9.
\]

The other totals (6, 8, 10, 12) would not satisfy the three “all but…’’ statements simultaneously.  

ANSWER 6: C  

---

Problem 7:
Factor \(250\):  

\[
250 = 2 \times 5^3.
\]

The prime factors are \(2\) and \(5\).  The two smallest are \(2\) and \(5\), whose sum is  

\[
2+5=7.
\]

Choices 2, 5, 10, 12 are either single prime factors or incorrect sums.  

ANSWER 7: C  

---

Problem 8:
Bag A contains \(\{1,3,5\}\); Bag B contains \(\{2,4,6\}\).  
All possible sums \(a+b\) are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The distinct results are \(\{3,5,7,9,11\}\), a total of **5** different sums.  

The other answer choices (4, 6, 7, 9) do not match the count of distinct sums.  

ANSWER 8: B  

---

Problem 9:
Place the two boys at the two ends and the three girls in the middle three seats.

- The two boys can be ordered in \(2!\) ways (left‑right swap).  
- The three girls can be arranged among the three middle seats in \(3! = 6\) ways.

Total arrangements  

\[
2! \times 3! = 2 \times 6 = 12.
\]

Choices 2, 4, 5, 6 are too small; 12 accounts for all possible permutations under the given constraints.  

ANSWER 9: E  

---

Problem 10:
Interpret the expression as  

\[
16 + 8 \div 4 - 2.
\]

Division comes before addition and subtraction, so  

\[
8 \div 4 = 2 \quad\Longrightarrow\quad 16 + 2 - 2 = 16.
\]

The other options (4, 8, 12, 20) would result only if the order of operations were misinterpreted.  

ANSWER 10: D  

---

Problem 11:
Simplify from the innermost radical outward.

\[
\sqrt{4}=2,\qquad 
\sqrt{8\cdot\sqrt{4}} = \sqrt{8\cdot2}= \sqrt{16}=4.
\]

Now  

\[
\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{16\cdot4}= \sqrt{64}=8.
\]

Thus the value is 8, matching choice C.  

ANSWER 11: C  

---

Problem 12:
Compare the decimal values:

- A. \(0.99\)
- B. \(0.9099\)
- C. \(0.9\)
- D. \(0.909\)
- E. \(0.9009\)

Clearly \(0.99\) is greater than \(0.9099\), which is greater than \(0.909\), etc.  
Hence the largest number is \(0.99\).  

ANSWER 12: A

Problem 1:  
An aquarium has a rectangular base of 100 cm by 40 cm and height 50 cm, filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is submerged. By how many centimeters does the water level rise?  

The base area is \(100 \times 40 = 4000 \, \text{cm}^2\). The rock’s volume is 1000 cm³, so the rise in water level is \(\frac{1000}{4000} = 0.25 \, \text{cm}\).  

Tempting choices: 0.5 (B) or 1 (C) might be guessed if base area is miscalculated.  

ANSWER 1: A  

Problem 2:  
Annika and 3 friends (4 players) get 15 cards each. Total cards = \(4 \times 15 = 60\). If 2 more join (6 players), each gets \(\frac{60}{6} = 10\).  

Tempting choices: 8 (A) if total cards miscalculated as 48.  

ANSWER 2: C  

Problem 3:  
Operation defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\): \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\), but the order is \(a c b d\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). However, the problem writes "a c b d" which is interpreted as \(a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). But the choices include 2 (E). Re-evaluating: the operation is \(a c b d = a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\).  

Tempting choices: -2 (A) if signs reversed.  

ANSWER 3: A (Note: Correction based on standard interpretation: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\), but the correct answer based on the operation defined is actually \(3 \cdot 2 - 1 \cdot 4 = 2\), but the problem writes "a c b d", which is \(a \cdot d - b \cdot c\), so 2. However, the answer is 2, but the choices include 2 as E. The correct calculation is 2, so E. Correction: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\).)  

**Correction:** The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\), this is \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\).  

ANSWER 3: E  

Problem 4:  
Isabella wants an average of 95 on four 100-point tests. Total points needed: \(4 \times 95 = 380\). First two scores: 97 + 91 = 188. Let third score be \(x\), fourth be \(y\). Then \(188 + x + y = 380\), so \(x + y = 192\). To minimize \(x\), maximize \(y\): \(y = 100\), so \(x = 92\).  

Tempting choices: 90 (A) if \(y\) not maximized.  

ANSWER 4: B  

Problem 5:  
Bridget gave half apples to Ann, then 3 to Cassie, keeping 4. Let total be \(x\). After giving half to Ann: \(\frac{x}{2}\) left. Then gives 3 to Cassie: \(\frac{x}{2} - 3 = 4\). So \(\frac{x}{2} = 7\), \(x = 14\).  

Tempting choices: 7 (C) if half misinterpreted.  

ANSWER 5: E  

Problem 6:  
Let \(r, g, b\) be red, green, blue marbles. Then:  
- All but 6 are red: \(g + b = 6\)  
- All but 8 are green: \(r + b = 8\)  
- All but 4 are blue: \(r + g = 4\)  
Adding: \(2(r + g + b) = 18\), so total \(r + g + b = 9\).  

Tempting choices: 6 (A) if equations misread.  

ANSWER 6: C  

Problem 7:  
Prime factors of 250: \(250 = 2 \times 5^3\). Two smallest: 2 and 5. Sum: \(2 + 5 = 7\).  

Tempting choices: 5 (B) if only 5 considered.  

ANSWER 7: C  

Problem 8:  
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}. Possible sums:  
- 1+2=3, 1+4=5, 1+6=7  
- 3+2=5, 3+4=7, 3+6=9  
- 5+2=7, 5+4=9, 5+6=11  
Unique sums: 3, 5, 7, 9, 11 → 5 values.  

Tempting choices: 6 (C) if duplicates counted.  

ANSWER 8: B  

Problem 9:  
Two boys at ends: 2 ways. Three girls in middle: \(3! = 6\) ways. Total: \(2 \times 6 = 12\).  

Tempting choices: 6 (D) if boys not permuted.  

ANSWER 9: E  

Problem 10:  
Expression: \(16 + 8 \times 4 - 2\). Order: \(8 \times 4 = 32\), then \(16 + 32 - 2 = 46\)? Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 \times 4 - 2\): \(8 \times 4 = 32\), \(16 + 32 = 48\), \(48 - 2 = 46\). But choices max at 20. Re-interpreting: "16 + 8 4 - 2" might mean \(16 + 8 - 4 - 2 = 18\), not in choices. Alternatively, if "8 4" is 8×4, but 46 not in choices. The expression is likely \(16 + 8 - 4 - 2 = 18\), but not in choices. Correction: The expression is "16 + 8 4 − 2", which might be \(16 + (8 \times 4) - 2 = 16 + 32 - 2 = 46\), but not in choices. Re-reading: It’s "16 + 8 4 − 2", which could be \(16 + 8 - 4 - 2 = 18\), but not in choices. The intended expression is likely \(16 + 8 \times 4 - 2 = 46\), but choices max at 20. Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 - 4 - 2 = 18\), but not in choices. Alternatively, if "8 4" is 84, but 16+84-2=98. Not matching. The expression is likely \( \frac{16 + 8}{4} - 2 = \frac{24}{4} - 2 = 6 - 2 = 4 \).  

Tempting choices: 8 (B) if order misapplied.  

ANSWER 10: A  

Problem 11:  
Expression: \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\).  
- \(\sqrt{4} = 2\)  
- \(\sqrt{8 \cdot 2} = \sqrt{16} = 4\)  
- \(\sqrt{16 \cdot 4} = \sqrt{64} = 8\)  

Tempting choices: 4 (A) if innermost not simplified.  

ANSWER 11: C  

Problem 12:  
Compare decimals:  
- A: 0.99  
- B: 0.9099  
- C: 0.9  
- D: 0.909  
- E: 0.9009  
Largest is 0.99.  

Tempting choices: 0.9099 (B) if decimals misaligned.  

ANSWER 12: A  

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

### Problem 1:
What is asked: An aquarium with a 100cm×40cm rectangular base is filled with water to 37cm depth. A 1000 cm³ rock is submerged; how much does the water level rise?
Step-by-step: When a submerged object displaces water, its volume equals the additional volume of water in the aquarium. The rise in water level Δh satisfies: Volume of rock = base area of aquarium × Δh.
Base area = 100×40 = 4000 cm².
Δh = 1000 / 4000 = 0.25 cm.
Tempting wrong answers: 0.5 (if base area is miscalculated as 2000 cm²) or 1 (wrong division).
ANSWER 1: A

---

### Problem 2:
What is asked: Annika plus 3 friends (4 total players) each get 15 cards. If 2 more join, how many cards does each player get?
Step-by-step: First calculate total number of cards: 4 players × 15 cards = 60 total cards. After 2 join, total players = 4+2=6. Cards per player = 60 / 6 = 10.
Tempting wrong answers: 9 (if you incorrectly count only 3 initial players, leading to 45 total cards) or 12 (wrong division).
ANSWER 2: C

---

### Problem 3:
What is asked: The custom notation a c b d = a·d − b·c. What is the value of 3 1 4 2?
Step-by-step: Match the values to the definition: a=3, c=1, b=4, d=2. Calculate: 3·2 - 4·1 = 6 - 4 = 2.
Tempting wrong answers: -2 (swap a and b incorrectly) or 0 (miscalculate the multiplication).
ANSWER 3: E

---

### Problem 4:
What is asked: Isabella needs a 95 average on 4 100-point tests. She scored 97 and 91 on the first two. What is the lowest possible third test score that lets her reach her goal?
Step-by-step: Total points needed for 95 average: 4×95 = 380 points. Sum of first two tests: 97+91=188. To minimize the third test score, maximize the fourth test (max 100 points). Total points left for tests 3 and 4: 380-188=192. Minimum third test = 192 - 100 = 92.
Tempting wrong answers: 96 (the average of the remaining two tests, not the minimum possible third score).
ANSWER 4: B

---

### Problem 5:
What is asked: Bridget gave half her apples to Ann, 3 to Cassie, and kept 4. How many apples did she buy originally?
Step-by-step: Let N = original number of apples. After giving half to Ann, she has N/2 left. Then N/2 - 3 = 4 → N/2=7 → N=14. Verify: 14 → 7 left after Ann, minus 3 for Cassie = 4 kept, which matches.
Tempting wrong answers: 7 (the number of apples Bridget had after giving half to Ann, not the original total).
ANSWER 5: E

---

### Problem 6:
What is asked: A jar has red, green, blue marbles. All but 6 are red, all but 8 are green, all but 4 are blue. How many total marbles?
Step-by-step: Let T=total marbles, R=red, G=green, B=blue. The conditions translate to: G+B=6, R+B=8, R+G=4. Add all equations: 2(R+G+B)=18 → R+G+B=T=9. Verify: 3 red, 1 green, 5 blue sum to 9, matching all conditions.
Tempting wrong answers: 18 (the sum of 6+8+4, not halved to get total marbles).
ANSWER 6: C

---

### Problem 7:
What is asked: What is the sum of the two smallest prime factors of 250?
Step-by-step: Prime factorization of 250: 250=2×5³. The prime factors are 2 and 5. Their sum is 2+5=7.
Tempting wrong answers: 10 (sum of 5+5, or wrong addition of 2+5).
ANSWER 7: C

---

### Problem 8:
What is asked: Bag A has chips 1,3,5; Bag B has 2,4,6. One chip drawn from each bag, how many distinct sum values are possible?
Step-by-step: List all sums: 1+2=3, 1+4=5,1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7,5+4=9,5+6=11. Distinct sums: 3,5,7,9,11 → 5 total.
Tempting wrong answers: 9 (total number of unordered pairs, not distinct sums).
ANSWER 8: B

---

### Problem 9:
What is asked: 2 boys and 3 girls line up for a photo, with a boy at each end and girls in the middle. How many arrangements are possible?
Step-by-step: Arrange the boys first: 2 boys can be ordered at the ends in 2! = 2 ways. The 3 middle girls can be ordered in 3! =6 ways. Total arrangements: 2×6=12.
Tempting wrong answers: 6 (only the number of ways to arrange the girls, ignoring the boys).
ANSWER 9: E

---

### Problem 10:
What is asked: Evaluate (16+8)/(4-2) (formatted as a fraction in the original problem).
Step-by-step: Calculate numerator:16+8=24, denominator:4-2=2. 24/2=12.
Tempting wrong answers:16 (if order of operations is misapplied to calculate 16 + 8/4 -2=16).
ANSWER 10: C

---

### Problem 11:
What is asked: Evaluate √(16 · √(8 · √4)).
Step-by-step: Calculate from the innermost root first: √4=2. Next: 8×2=16, √16=4. Next:16×4=64, √64=8.
Tempting wrong answers: 4√2 (incorrect root calculation).
ANSWER 11: C

---

### Problem 12:
What is asked: Which number is largest: 0.99, 0.9099, 0.9, 0.909, 0.9009.
Step-by-step: Line up all decimals to compare place values: 0.9900 > 0.9099 > 0.9090 > 0.9009 > 0.9000. The largest is 0.99.
Tempting wrong answers: 0.9099 (the second largest value).
ANSWER 12: A

### Problem 1:
We need to find how much the water level rises when a 1000 cm³ rock is submerged in the aquarium.
1.  The volume of displaced water equals the volume of the submerged rock: 1000 cm³.
2.  The base area of the aquarium is \(100 \, \text{cm} \times 40 \, \text{cm} = 4000 \, \text{cm}^2\).
3.  Let \(h\) be the rise in water level. The volume of the displaced water is also equal to base area × rise in height: \(4000h = 1000\).
4.  Solve for \(h\): \(h = \frac{1000}{4000} = 0.25\) cm.
ANSWER 1: A
---
### Problem 2:
We need to find the number of cards per player when 2 more friends join.
1.  Original number of players: Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).
2.  New number of players: 4 + 2 = 6 players. Total cards remain 60.
3.  Cards per player: \(\frac{60}{6} = 10\).
ANSWER 2: C
---
### Problem 3:
The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). Compute \(3 \, 1 \, 4 \, 2\).
1.  Identify \(a=3, c=1, b=4, d=2\).
2.  Substitute into the formula: \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).
ANSWER 3: E
---
### Problem 4:
Find the lowest possible third test score to get an average of 95 on 4 tests.
1.  Total points needed for an average of 95: \(4 \times 95 = 380\).
2.  Sum of first two test scores: \(97 + 91 = 188\).
3.  Let \(x\) = third test score, \(y\) = fourth test score (max 100, since tests are 100-point). We need \(188 + x + y = 380\), so \(x + y = 192\).
4.  To minimize \(x\), maximize \(y = 100\). Then \(x = 192 - 100 = 92\).
ANSWER 4: B
---
### Problem 5:
Find the original number of apples Bridget bought.
1.  Let \(x\) = total apples. After giving half to Ann, she has \(\frac{x}{2}\) left.
2.  After giving 3 to Cassie, she has \(\frac{x}{2} - 3\) left, which equals 4 (the apples she kept).
3.  Solve: \(\frac{x}{2} - 3 = 4 \implies \frac{x}{2} =7 \implies x=14\).
ANSWER 5: E
---
### Problem 6:
Find the total number of marbles.
1.  Let \(R, G, B\) be the number of red, green, blue marbles, total \(T = R+G+B\).
2.  Translate the statements:
    - All but 6 are red: \(G + B = 6\)
    - All but 8 are green: \(R + B = 8\)
    - All but 4 are blue: \(R + G =4\)
3.  Add all three equations: \(2(R+G+B) = 6+8+4=18 \implies 2T=18 \implies T=9\).
ANSWER 6: C
---
### Problem 7:
Find the sum of the two smallest prime factors of 250.
1.  Factorize 250: \(250 = 2 \times 125 = 2 \times 5^3\).
2.  The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.
3.  Sum: \(2 +5=7\).
ANSWER 7: C
---
### Problem 8:
Find the number of distinct sums when drawing one chip from each bag.
1.  List all possible pairs and sums:
    - Bag A (1,3,5) + Bag B (2,4,6):
      \(1+2=3, 1+4=5, 1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7, 5+4=9, 5+6=11\)
2.  Distinct sums: 3, 5, 7, 9, 11. There are 5 unique values.
ANSWER 8: B
---
### Problem 9:
Find the number of arrangements with boys at each end and girls in the middle.
1.  Arrange the 2 boys at the ends: there are \(2! = 2\) ways (order matters: which boy is left/right).
2.  Arrange the 3 girls in the middle 3 seats: \(3! = 6\) ways.
3.  Total arrangements: \(2 \times 6 =12\).
ANSWER 9: E
---
### Problem 10:
Compute \(16 + 8 \, 4 - 2\). The missing operator between 8 and 4 is division (standard order of operations, common middle school problem):
1.  Interpret as \(16 + \frac{8}{4} - 2\).
2.  Compute division first: \(\frac{8}{4}=2\).
3.  Then add/subtract left to right: \(16 + 2 - 2 =16\).
*(Note: If interpreted as \(16 +8 -4 \times 2\), order of operations gives \(4 \times 2=8\), so \(16+8-8=16\), same result.)*
ANSWER 10: D
---
### Problem 11:
Compute \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\) by working from the innermost root outward.
1.  Innermost: \(\sqrt{4} = 2\).
2.  Next layer: \(\sqrt{8 \cdot 2} = \sqrt{16} =4\).
3.  Outermost: \(\sqrt{16 \cdot 4} = \sqrt{64} =8\).
ANSWER 11: C
---
### Problem 12:
Find the largest decimal.
1.  Rewrite all numbers with 4 decimal places for easy comparison:
    - A: \(0.9900\)
    - B: \(0.9099\)
    - C: \(0.9000\)
    - D: \(0.9090\)
    - E: \(0.9009\)
2.  Compare digit by digit: all have 9 in the tenths place. A has 9 in the hundredths place, while all others have 0. Thus A is the largest.
ANSWER 12: A

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

**Problem 1:**

What is being asked: By how many centimeters does the water level rise when a 1000 cm³ rock is submerged?

Setup: The base area is 100 cm × 40 cm = 4000 cm². When the rock is submerged, it displaces 1000 cm³ of water.

Water level rise = Volume displaced ÷ Base area = 1000 ÷ 4000 = 0.25 cm

ANSWER 1: A

---

**Problem 2:**

What is being asked: If cards are distributed evenly, how many cards per player when the player count increases?

Setup: Initially 4 players (Annika + 3 friends) each get 15 cards.
Total cards = 4 × 15 = 60 cards

With 2 more friends joining: 4 + 2 = 6 players
Cards per player = 60 ÷ 6 = 10 cards

ANSWER 2: C

---

**Problem 3:**

What is being asked: Evaluate the operation 3 1 4 2 using the rule a c b d = a·d − b·c

Setup: With a = 3, c = 1, b = 4, d = 2:
3 1 4 2 = 3·2 − 4·1 = 6 − 4 = 2

ANSWER 3: E

---

**Problem 4:**

What is being asked: What is the lowest score on test 3 that allows Isabella to achieve a 95 average on four tests?

Setup: Goal average is 95, so total needed = 95 × 4 = 380 points
First two tests: 97 + 91 = 188 points
Points needed from tests 3 and 4: 380 − 188 = 192 points

Let x = score on test 3. She can still reach her goal if she can score 100 on test 4:
x + 100 ≥ 192
x ≥ 92

The lowest possible score on test 3 is 92.

ANSWER 4: B

---

**Problem 5:**

What is being asked: How many apples did Bridget buy originally?

Setup: Let x = original number of apples
- She gave half to Ann: x/2 apples given away
- Remaining: x/2
- She gave 3 to Cassie: x/2 − 3
- She kept 4 for herself: x/2 − 3 = 4

Solving: x/2 = 7, so x = 14

ANSWER 5: E

---

**Problem 6:**

What is being asked: How many marbles are in the jar?

Setup: Let R, G, B = numbers of red, green, blue marbles; T = total
- "All but 6 are red": G + B = 6
- "All but 8 are green": R + B = 8
- "All but 4 are blue": R + G = 4

Adding all three equations: 2(R + G + B) = 6 + 8 + 4 = 18
Therefore: R + G + B = 9

ANSWER 6: C

---

**Problem 7:**

What is being asked: What is the sum of the two smallest prime factors of 250?

Setup: Factor 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5³

Prime factors: 2 and 5
Sum = 2 + 5 = 7

ANSWER 7: C

---

**Problem 8:**

What is being asked: How many different sums are possible?

Setup: Bag A: {1, 3, 5}, Bag B: {2, 4, 6}
All possible sums:
- 1 + 2 = 3
- 1 + 4 = 5
- 1 + 6 = 7
- 3 + 2 = 5 (duplicate)
- 3 + 4 = 7 (duplicate)
- 3 + 6 = 9
- 5 + 2 = 7 (duplicate)
- 5 + 4 = 9 (duplicate)
- 5 + 6 = 11

Distinct sums: {3, 5, 7, 9, 11} = 5 different values

ANSWER 8: B

---

**Problem 9:**

What is being asked: How many seating arrangements with boys at ends and girls in middle?

Setup: 2 boys at the ends can be arranged in 2! = 2 ways
3 girls in the middle can be arranged in 3! = 6 ways
Total arrangements = 2 × 6 = 12

ANSWER 9: E

---

**Problem 10:**

What is being asked: Evaluate 16 + 8 ÷ 4 − 2

Setup: Using order of operations (division before addition/subtraction):
16 + (8 ÷ 4) − 2 = 16 + 2 − 2 = 16

ANSWER 10: D

---

**Problem 11:**

What is being asked: Evaluate √(16 · √(8 · √4))

Setup: Work from the inside out:
- √4 = 2
- 8 · √4 = 8 · 2 = 16
- √(8 · √4) = √16 = 4
- 16 · √(8 · √4) = 16 · 4 = 64
- √(16 · √(8 · √4)) = √64 = 8

ANSWER 11: C

---

**Problem 12:**

What is being asked: Which decimal is largest?

Setup: Compare decimals:
- A: 0.99
- B: 0.9099
- C: 0.9
- D: 0.909
- E: 0.9009

Comparing: 0.99 > 0.9099 > 0.909 > 0.9009 > 0.9

The largest is 0.99

ANSWER 12: A

Problem 1:

The aquarium has a rectangular base of \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is

\[
100 \cdot 40 = 4000 \text{ cm}^2.
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

If the water rises by \(h\) centimeters, then

\[
4000h = 1000.
\]

So

\[
h = \frac{1000}{4000} = \frac14 = 0.25.
\]

So the water level rises \(0.25\) cm.

ANSWER 1: A

---

Problem 2:

Annika and 3 friends means there are

\[
4 \text{ players}.
\]

If each player gets 15 cards, then the total number of cards is

\[
4 \cdot 15 = 60.
\]

Now 2 more friends join, so there are

\[
6 \text{ players}.
\]

The same 60 cards are shared evenly, so each gets

\[
60 \div 6 = 10.
\]

ANSWER 2: C

---

Problem 3:

We are told that if the four numbers are \(a, b, c, d\), then

\[
a \ b \ c \ d = a\cdot d - b\cdot c.
\]

So for \(3,1,4,2\),

\[
3 \ 1 \ 4 \ 2 = 3\cdot 2 - 1\cdot 4 = 6 - 4 = 2.
\]

ANSWER 3: E

---

Problem 4:

She wants an average of 95 on 4 tests, so the total of the 4 scores must be

\[
4 \cdot 95 = 380.
\]

Her first two scores are 97 and 91, which add to

\[
97+91=188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To make the total 380, the fourth score would have to be

\[
380 - (188+x) = 192 - x.
\]

For this to be possible, the fourth score must be at most 100:

\[
192 - x \le 100.
\]

So

\[
x \ge 92.
\]

The lowest possible third score is 92.

ANSWER 4: B

---

Problem 5:

Let the number of apples Bridget bought be \(x\).

She gave half to Ann, so she kept the other half:

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

Then she gave Cassie 3 apples and kept 4 apples, so before giving Cassie the apples she had

\[
3+4=7.
\]

Thus,

\[
\frac{x}{2}=7 \quad \Rightarrow \quad x=14.
\]

ANSWER 5: E

---

Problem 6:

Let the total number of marbles be \(n\).

- “All but 6 are red” means the number of red marbles is \(n-6\).
- “All but 8 are green” means the number of green marbles is \(n-8\).
- “All but 4 are blue” means the number of blue marbles is \(n-4\).

Since every marble is one of these colors, their counts add to \(n\):

\[
(n-6)+(n-8)+(n-4)=n.
\]

Simplify:

\[
3n-18=n
\]

\[
2n=18
\]

\[
n=9.
\]

ANSWER 6: C

---

Problem 7:

We need the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3.
\]

So the prime factors are 2 and 5. The two smallest prime factors are 2 and 5, and their sum is

\[
2+5=7.
\]

ANSWER 7: C

---

Problem 8:

Bag A has \(1,3,5\); Bag B has \(2,4,6\).

Possible sums:

- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums are

\[
3,5,7,9,11
\]

which is 5 different values.

ANSWER 8: B

---

Problem 9:

There are 2 boys and 3 girls, and the picture must have:

- a boy at each end,
- the 3 girls in the middle.

First, arrange the 2 boys at the ends. They can be switched in

\[
2! = 2
\]

ways.

Then arrange the 3 girls in the middle in

\[
3! = 6
\]

ways.

So the total number of arrangements is

\[
2 \cdot 6 = 12.
\]

ANSWER 9: E

---

Problem 10:

The expression is

\[
16 + 8\cdot 4 - 2.
\]

Use order of operations:

\[
8\cdot 4=32.
\]

Then

\[
16+32-2=46.
\]

But 46 is not among the choices, so the intended expression must be read as

\[
16 + 8 \div 4 - 2.
\]

Then

\[
8\div 4 = 2,
\]

so

\[
16+2-2=16.
\]

ANSWER 10: D

---

Problem 11:

We evaluate from the inside out:

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

First,

\[
\sqrt4=2.
\]

So the expression becomes

\[
\sqrt{16\cdot \sqrt{8\cdot 2}}
=
\sqrt{16\cdot \sqrt{16}}.
\]

Since \(\sqrt{16}=4\),

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C

---

Problem 12:

Compare the decimals:

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

The largest is the one closest to 1. Since \(0.99\) is greater than any number beginning with \(0.90\), it is the largest.

ANSWER 12: A

Problem 1:  
We need the rise in water level when a rock is placed in the aquarium and submerged.

- Base dimensions: \(100\text{ cm} \times 40\text{ cm}\), so base area  
  \[
  A = 100\cdot 40 = 4000\text{ cm}^2.
  \]
- The rock is completely submerged, so the water volume displaced equals the rock’s volume:  
  \[
  V = 1000\text{ cm}^3.
  \]
- Water level rise \(\Delta h\) satisfies \(A\Delta h = V\):  
  \[
  4000\cdot \Delta h = 1000 \implies \Delta h = \frac{1000}{4000}=0.25\text{ cm}.
  \]

So the rise is \(0.25\) cm.  
Tempting wrong idea: using the *remaining* depth \(37\) cm or checking if it overflows—none of that matters; only displaced volume matters.

ANSWER 1: A  

---

Problem 2:  
All cards are dealt evenly. Let the total number of cards be \(T\).

Initially, Annika + 3 friends = 4 players, and each gets 15 cards:  
\[
T = 4\cdot 15 = 60.
\]
Next game: 2 more friends join, so total players:
\[
4+2=6.
\]
Each player gets:
\[
\frac{60}{6}=10.
\]

ANSWER 2: C  

---

Problem 3:  
Given the rule \(a\,b\,d\,c = a\cdot d - b\cdot c\).  
The expression “\(3\ 1\ 4\ 2\)” corresponds to:
- \(a=3\)
- \(b=1\)
- \(d=4\)
- \(c=2\)

Compute:
\[
a\cdot d - b\cdot c = 3\cdot 4 - 1\cdot 2 = 12-2=10.
\]

But notice the choices are only \(-2,-1,0,1,2\), so we must interpret the placeholder correctly:  
The problem states: **“If \(a b c d = a\cdot d - b\cdot c\)”**.

So “\(3\ 1\ 4\ 2\)” means:
- \(a=3,\ b=1,\ c=4,\ d=2\)

Then:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 1\cdot 4 = 6-4=2.
\]

ANSWER 3: E  

---

Problem 4:  
Goal average is 95 over 4 tests, so total points needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
Let the third test score be \(x\), and the fourth be \(y\). Then:
\[
188 + x + y = 380 \implies x+y=192.
\]
We want the **lowest possible** \(x\) such that she can still reach the goal. That means make \(y\) as large as possible (since grades are at most 100).
Maximum \(y=100\). Then:
\[
x + 100 = 192 \implies x = 92.
\]

Check: then the four scores are \(97,91,92,100\) which sum to \(380\), average \(95\). Works.

ANSWER 4: B  

---

Problem 5:  
Let the total number of apples be \(N\).

- She gave half to Ann, so Ann got \(N/2\), leaving \(N/2\).
- Then she gave Cassie 3 apples, and kept 4 apples. That means after giving Cassie 3, her remaining is 4:
  \[
  \frac{N}{2} - 3 = 4 \implies \frac{N}{2}=7 \implies N=14.
  \]

ANSWER 5: E  

---

Problem 6:  
“All but 6 are red” means: number of red marbles is 6.  
Similarly:
- Red = 6 (since all but 6 are red → 6 red)
- Green: all but 8 are green → 8 green
- Blue: all but 4 are blue → 4 blue

Total marbles:
\[
6+8+4=18.
\]
But none of the choices are 18, so interpret carefully:  
“All but 6 are red” means the number of red marbles is **(total − 6)**, not 6.

Let total be \(T\). Then:
- Red marbles = \(T-6\)
- Green marbles = \(T-8\)
- Blue marbles = \(T-4\)

Sum:
\[
(T-6)+(T-8)+(T-4)=T.
\]
\[
3T - (6+8+4)=T
\]
\[
3T-18=T \implies 2T=18 \implies T=9.
\]

ANSWER 6: C  

---

Problem 7:  
We want the sum of the **two smallest prime factors** of 250.

Factor \(250\):
\[
250 = 25\cdot 10 = 5^2 \cdot 2\cdot 5 = 2\cdot 5^3.
\]
Prime factors are \(2\) and \(5\). The two smallest are \(2\) and \(5\).

Sum:
\[
2+5=7.
\]

ANSWER 7: D  

---

Problem 8:  
Bag A: \(\{1,3,5\}\)  
Bag B: \(\{2,4,6\}\)

Possible sums:
- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums: \(\{3,5,7,9,11\}\) → that’s 5 values.

ANSWER 8: B  

---

Problem 9:  
We need arrangements of 5 people in a row with:
- a boy at each end (positions 1 and 5)
- the three girls in the middle (positions 2,3,4)

There are 2 boys: they can be placed at the ends in
\[
2!\ = 2
\]
ways (left boy/right boy).

The 3 girls can be arranged in the three middle spots in
\[
3! = 6
\]
ways.

Total arrangements:
\[
2\cdot 6 = 12.
\]

ANSWER 9: E  

---

Problem 10:  
Expression: \(16 + 8^4 - 2\).

Compute in order:
- \(8^4 = 4096\)

So:
\[
16 + 4096 - 2 = 4110.
\]
No answer choice matches, so we interpret the formatting as \(16 + 8 \sqrt{4} - 2\) or \(16 + 8^4 - 2\)?  
Given the AMC/JH style, likely it is:
\[
16 + 8\sqrt{4} - 2.
\]
Then:
\[
\sqrt{4}=2 \implies 8\sqrt{4}=8\cdot 2=16.
\]
So:
\[
16 + 16 - 2 = 30,
\]
still not in choices.

Another common formatting is:
\[
16 + 8^{4} - 2 \quad \text{(still not)}
\]

The only choiceable consistent possibility is:
\[
16 + \sqrt{8}\,^4 - 2 \text{ (nonsense)}
\]

Let’s use the most plausible literal from the text: **“16 + 8 4 − 2”** often means \(16 + 8^4 - 2\) if the caret was lost. But then no choice.

Given typical AMC problems, the intended is usually:
\[
16 + 8 - 4 - 2 = 18
\]
also not.

Because the answer choices are only \(4,8,12,16,20\), the intended expression must simplify to one of those. The most natural is:
\[
16 + 8/4 - 2 = 16 + 2 - 2 = 16.
\]
This matches a common missing symbol: “8 4” often means \(8/4\).

So:
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D  

---

Problem 11:  
Evaluate:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}.
\]

Work from inside out:

1) \(\sqrt{4}=2\)

2) Inside next root:
\[
8\cdot \sqrt{4} = 8\cdot 2 = 16
\]
So:
\[
\sqrt{8\cdot \sqrt{4}} = \sqrt{16}=4.
\]

3) Now plug into the outer expression:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}=\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C  

---

Problem 12:  
Largest number among:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare by decimal places:

- .99 is 0.99
- .9099 is 0.9099
- .9 is 0.9000
- .909 is 0.9090
- .9009 is 0.9009

Clearly \(0.99\) is greater than all the numbers less than \(0.91\).

ANSWER 12: A

Problem 1:
The volume of the water displaced by the rock is equal to the volume of the rock itself, which is $1000 \text{ cm}^3$. The aquarium has a rectangular base with area $A = 100 \text{ cm} \times 40 \text{ cm} = 4000 \text{ cm}^2$. When the rock is submerged, the water level rises by height $h$ such that the volume of the displaced water is $A \times h$. 
Setting this equal to the rock's volume: $4000 \times h = 1000$.
Solving for $h$: $h = \frac{1000}{4000} = \frac{1}{4} = 0.25 \text{ cm}$.
ANSWER 1: A

Problem 2:
Annika and 3 friends make 4 players total. Each received 15 cards, so the total number of cards is $4 \times 15 = 60$. If 2 more friends join, the total number of players becomes $4 + 2 = 6$. The cards are distributed evenly among the 6 players, so each player gets $60 \div 6 = 10$ cards.
ANSWER 2: C

Problem 3:
The problem defines the operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$.
For the expression $\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix}$, we have $a=3, b=1, c=4, d=2$.
Calculating: $(3 \times 2) - (1 \times 4) = 6 - 4 = 2$.
ANSWER 3: E

Problem 4:
Isabella needs an average of 95 over 4 tests, meaning the total sum of her scores must be $95 \times 4 = 380$. Her first two scores are 97 and 91, totaling $97 + 91 = 188$. Let $x$ be the score on the third test and $y$ be the score on the fourth. We have $188 + x + y = 380$, so $x + y = 192$. To make $x$ as small as possible, we must make $y$ as large as possible. Since the maximum score on a test is 100, we set $y = 100$. Then $x + 100 = 192$, so $x = 92$.
ANSWER 4: B

Problem 5:
Let $x$ be the total number of apples Bridget bought. She gave half away, leaving $\frac{x}{2}$. Then she gave 3 away, leaving $\frac{x}{2} - 3$. She kept 4, so $\frac{x}{2} - 3 = 4$. Adding 3 to both sides, $\frac{x}{2} = 7$. Multiplying by 2, $x = 14$.
ANSWER 5: E

Problem 6:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R + G + B$.
"All but 6 are red" means $G + B = 6$.
"All but 8 are green" means $R + B = 8$.
"All but 4 are blue" means $R + G = 4$.
Adding these three equations: $2(R + G + B) = 6 + 8 + 4 = 18$.
So $R + G + B = 18 \div 2 = 9$.
ANSWER 6: C

Problem 7:
The prime factorization of 250 is $250 = 2 \times 125 = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two prime factors is $2 + 5 = 7$.
ANSWER 7: C

Problem 8:
Bag A has $\{1, 3, 5\}$ and Bag B has $\{2, 4, 6\}$.
Possible sums:
$1+2=3, 1+4=5, 1+6=7$
$3+2=5, 3+4=7, 3+6=9$
$5+2=7, 5+4=9, 5+6=11$
The set of distinct sums is $\{3, 5, 7, 9, 11\}$. There are 5 distinct values.
ANSWER 8: B

Problem 9:
There are 2 boys (B1, B2) and 3 girls (G1, G2, G3). A boy must be at each end. There are 2 choices for the left end and 1 choice for the right end ($2 \times 1 = 2$ ways to place boys). The 3 girls can be arranged in the middle in $3! = 3 \times 2 \times 1 = 6$ ways. Total arrangements = $2 \times 6 = 12$.
ANSWER 9: E

Problem 10:
Following the order of operations (PEMDAS/BODMAS):
$16 + (8 \div 4) - 2 = 16 + 2 - 2 = 16$.
ANSWER 10: D

Problem 11:
Evaluate from the inside out:
$\sqrt{4} = 2$
$\sqrt{8 \times 2} = \sqrt{16} = 4$
$\sqrt{16 \times 4} = \sqrt{64} = 8$
ANSWER 11: C

Problem 12:
To compare, pad with zeros to match the number of decimal places:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
Comparing the values, 0.9900 is the largest.
ANSWER 12: A

Problem 1:
The base of the aquarium measures 100 cm by 40 cm, so its area is \(100 \times 40 = 4000\) cm². The rock has volume 1000 cm³ and is fully submerged, displacing exactly 1000 cm³ of water. The resulting rise in water level is therefore the displaced volume divided by the base area: \(1000 / 4000 = 0.25\) cm. (The original water depth and tank height are irrelevant once the surface area is known.)

ANSWER 1: A

Problem 2:
Annika plus 3 friends gives 4 players, each receiving 15 cards, for a total of \(4 \times 15 = 60\) cards. When 2 more players join, there are 6 players sharing the same 60 cards, so each receives \(60 / 6 = 10\) cards.

ANSWER 2: C

Problem 3:
The given operation evaluates \(a\, c\, b\, d\) as \(a \cdot d - b \cdot c\). Substituting the numbers yields \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).

ANSWER 3: E

Problem 4:
An average of 95 on four 100-point tests requires a total of \(4 \times 95 = 380\) points. The first two scores sum to \(97 + 91 = 188\), leaving 192 points still needed from the last two tests. The fourth test can score at most 100, so the third test must be at least \(192 - 100 = 92\). Any lower score on the third test would make a total of 380 impossible.

ANSWER 4: B

Problem 5:
Let \(t\) be the original number of apples. After giving half to Ann, \(t/2\) remain. After giving 3 more to Cassie, \(t/2 - 3\) remain, and this equals the 4 apples kept. Solving \(t/2 - 3 = 4\) gives \(t/2 = 7\), so \(t = 14\).

ANSWER 5: E

Problem 6:
Let \(m\) be the total number of marbles. Then there are \(m-6\) red, \(m-8\) green, and \(m-4\) blue. Their sum must equal \(m\):
\[
(m-6) + (m-8) + (m-4) = m \implies 3m - 18 = m \implies 2m = 18 \implies m = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies the three “all-but” statements.

ANSWER 6: C

Problem 7:
Factorize \(250 = 2 \cdot 5^3\). The two smallest distinct prime factors are 2 and 5; their sum is 7.

ANSWER 7: C

Problem 8:
All nine possible sums are:
\[
1+2=3,\ 1+4=5,\ 1+6=7,\ 3+2=5,\ 3+4=7,\ 3+6=9,\ 5+2=7,\ 5+4=9,\ 5+6=11.
\]
The distinct values are 3, 5, 7, 9, 11—exactly five.

ANSWER 8: B

Problem 9:
The seating pattern must be boy–girl–girl–girl–boy. The two distinct boys may be placed at the ends in \(2!\) = 2 ways. The three distinct girls may be arranged in the middle three seats in \(3!\) = 6 ways. The total number of arrangements is therefore \(2 \times 6 = 12\).

ANSWER 9: E

Problem 10:
Interpreting the expression with standard order of operations and treating juxtaposition of 8 and 4 as division yields
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D

Problem 11:
Evaluate from the inside outward:
\[
\sqrt{4} = 2, \quad 8 \cdot 2 = 16, \quad \sqrt{16} = 4, \quad 16 \cdot 4 = 64, \quad \sqrt{64} = 8.
\]

ANSWER 11: C

Problem 12:
Compare the decimals directly:
\[
0.99 > 0.9 > 0.9099 > 0.909 > 0.9009.
\]
Thus 0.99 is the largest.

ANSWER 12: A

Problem 1:
The problem asks us to determine by how many centimeters the water level rises when a rock with a volume of 1000 cm³ is submerged in an aquarium filled with water to a depth of 37 cm. The aquarium has a rectangular base measuring 100 cm by 40 cm and a height of 50 cm.

First, we need to calculate the initial volume of water in the aquarium. The volume of water is given by the formula length × width × height. So, the initial volume of water = 100 cm × 40 cm × 37 cm = 148,000 cm³.

When the rock is submerged, it displaces a volume of water equal to its own volume, which is 1000 cm³. This means the water level will rise to accommodate the volume of the rock.

The rise in water level can be calculated by dividing the volume of the rock by the base area of the aquarium. The base area = 100 cm × 40 cm = 4000 cm².

Rise in water level = Volume of rock / Base area = 1000 cm³ / 4000 cm² = 0.25 cm.

Let's examine the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

The calculated rise in water level matches option A.

ANSWER 1: A

Problem 2:
The problem states that in the card game Buffalo Shuffle-o, all cards are distributed evenly among players. Initially, Annika and 3 friends play, and each is dealt 15 cards. We need to find out how many cards each player will be dealt if 2 more friends join the next game.

First, we determine the total number of cards. With 4 players initially, each getting 15 cards, the total number of cards = 4 × 15 = 60.

When 2 more friends join, the total number of players becomes 4 + 2 = 6.

To find out how many cards each player will get, we divide the total number of cards by the new number of players: 60 / 6 = 10.

Let's examine the choices:
A. 8
B. 9
C. 10
D. 11
E. 12

The calculated number of cards per player matches option C.

ANSWER 2: C

Problem 3:
The problem defines a operation as a c b d = a·d − b·c and asks us to find the value of 3 1 4 2.

Using the given definition, we substitute a = 3, b = 1, c = 4, and d = 2 into the equation: 3 1 4 2 = 3·2 − 1·4 = 6 - 4 = 2.

Let's examine the choices:
A. −2
B. −1
C. 0
D. 1
E. 2

The calculated value matches option E.

ANSWER 3: E

Problem 4:
Isabella needs to achieve an average grade of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score she could have made on the third test to still reach her goal.

The total score required for an average of 95 on 4 tests = 95 × 4 = 380.

The sum of her first two test scores = 97 + 91 = 188.

So, the total score needed on the last two tests = 380 - 188 = 192.

To minimize the score on the third test, we maximize the score on the fourth test. The maximum score on the fourth test is 100.

Thus, the minimum score on the third test = 192 - 100 = 92.

Let's examine the choices:
A. 90
B. 92
C. 95
D. 96
E. 97

The calculated minimum score matches option B.

ANSWER 4: B

Problem 5:
Bridget gave half of her apples to Ann, then gave 3 to Cassie, and kept 4 for herself. We need to find out how many apples she originally bought.

Let's denote the original number of apples as x.

After giving half to Ann, Bridget had x/2 apples left.

Then, she gave 3 apples to Cassie and kept 4, so x/2 = 3 + 4 = 7.

Thus, x = 7 × 2 = 14.

Let's examine the choices:
A. 3
B. 4
C. 7
D. 11
E. 14

The calculated original number of apples matches option E.

ANSWER 5: E

Problem 6:
The problem states that in a jar, all but 6 marbles are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T, the number of red marbles as R, the number of green marbles as G, and the number of blue marbles as B.

From the given information:
1. T - R = 6 (all but 6 are red)
2. T - G = 8 (all but 8 are green)
3. T - B = 4 (all but 4 are blue)

Adding these equations gives: 3T - (R + G + B) = 18.

Since R + G + B = T (the total number of marbles is the sum of the marbles of each color), we have:
3T - T = 18, which simplifies to 2T = 18.

Thus, T = 18 / 2 = 9.

Let's examine the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

The calculated total number of marbles matches option C.

ANSWER 6: C

Problem 7:
We need to find the sum of the two smallest prime factors of 250.

First, let's factorize 250 into its prime factors: 250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5.

The sum of these two smallest prime factors = 2 + 5 = 7.

Let's examine the choices:
A. 2
B. 5
C. 7
D. 10
E. 12

The calculated sum matches option C.

ANSWER 7: C

Problem 8:
Bag A contains chips labeled 1, 3, and 5. Bag B contains chips labeled 2, 4, and 6. We need to find how many different values are possible for the sum of the numbers on the chips drawn from each bag.

Let's list the possible sums:
- (1,2) = 3
- (1,4) = 5
- (1,6) = 7
- (3,2) = 5
- (3,4) = 7
- (3,6) = 9
- (5,2) = 7
- (5,4) = 9
- (5,6) = 11

The possible sums are 3, 5, 7, 9, and 11. There are 5 different values.

Let's examine the choices:
A. 4
B. 5
C. 6
D. 7
E. 9

The number of different sums matches option B.

ANSWER 8: B

Problem 9:
The chess team consists of 2 boys and 3 girls. They need to be arranged with a boy at each end and the girls in the middle. We need to find out how many such arrangements are possible.

The boys can be arranged at the ends in 2! = 2 ways (either boy can be on the left or the right).

The 3 girls can be arranged in the 3 middle positions in 3! = 6 ways.

The total number of arrangements = (number of ways to arrange boys) × (number of ways to arrange girls) = 2 × 6 = 12.

Let's examine the choices:
A. 2
B. 4
C. 5
D. 6
E. 12

The calculated total number of arrangements matches option E.

ANSWER 9: E

Problem 10:
The problem asks us to evaluate the expression 16 + 8 ÷ 4 − 2.

Following the order of operations (PEMDAS/BODMAS: Parentheses/Brackets, Exponents/Orders, Multiplication and Division from left to right, Addition and Subtraction from left to right):
1. D

Problem 1:
The water rise is determined by the volume of the rock divided by the base area of the aquarium.  
Base area = \(100 \times 40 = 4000\text{ cm}^2\).  
Volume of rock = \(1000\text{ cm}^3\).  
Rise = \(\frac{1000}{4000} = 0.25\text{ cm}\).  
ANSWER 1: A

Problem 2:
Initially, Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 2: C

Problem 3:
The notation \(\begin{array}{cc} a & c \\ b & d \end{array} = a \cdot d - b \cdot c\).  
For \(\begin{array}{cc} 3 & 1 \\ 4 & 2 \end{array}\): \(a=3, c=1, b=4, d=2\).  
Value = \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).  
ANSWER 3: E

Problem 4:
Total points needed for an average of 95 over 4 tests = \(4 \times 95 = 380\).  
First two tests sum = \(97 + 91 = 188\).  
Remaining points needed = \(380 - 188 = 192\).  
To minimize the third test score, maximize the fourth (100). Then third test \(\ge 192 - 100 = 92\).  
Lowest possible third test score is 92.  
ANSWER 4: B

Problem 5:
Let \(x\) be the number of apples Bridget bought.  
She gave half to Ann, so she kept \(\frac{x}{2}\).  
Then she gave 3 to Cassie and kept 4, so \(\frac{x}{2} = 3 + 4 = 7\).  
Thus \(x = 14\).  
ANSWER 5: E

Problem 6:
Let \(R, G, B\) be the number of red, green, blue marbles.  
"All but 6 are red" \(\Rightarrow G + B = 6\).  
"All but 8 are green" \(\Rightarrow R + B = 8\).  
"All but 4 are blue" \(\Rightarrow R + G = 4\).  
Adding: \((G+B)+(R+B)+(R+G) = 6+8+4 = 18\) \(\Rightarrow 2(R+G+B) = 18\) \(\Rightarrow R+G+B = 9\).  
Total marbles = 9.  
ANSWER 6: C

Problem 7:
Prime factorization of 250: \(250 = 2 \times 5^3\).  
Prime factors are 2 and 5. The two smallest are 2 and 5.  
Sum = \(2 + 5 = 7\).  
ANSWER 7: C

Problem 8:
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}.  
Possible sums:  
1+2=3, 1+4=5, 1+6=7;  
3+2=5, 3+4=7, 3+6=9;  
5+2=7, 5+4=9, 5+6=11.  
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.  
ANSWER 8: B

Problem 9:
Boys must sit at the two ends.  
Arrange boys: 2 choices for left end, 1 for right end → \(2! = 2\) ways.  
Arrange 3 girls in the middle 3 seats: \(3! = 6\) ways.  
Total arrangements = \(2 \times 6 = 12\).  
ANSWER 9: E

Problem 10:
Interpret "8 4" as \(8 \div 4\).  
\(16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16\).  
ANSWER 10: D

Problem 11:
\(\sqrt{4} = 2\).  
\(8 \cdot 2 = 16\), \(\sqrt{16} = 4\).  
\(16 \cdot 4 = 64\), \(\sqrt{64} = 8\).  
ANSWER 11: C

Problem 12:
Compare decimals:  
0.99, 0.9099, 0.9, 0.909, 0.9009.  
0.99 is the largest.  
ANSWER 12: A

# Problem 1:

**What is being asked:** A rock of volume 1000 cm³ is submerged in an aquarium with a 100 cm × 40 cm base. How much does the water level rise?

**Setup:** When the rock is submerged, it displaces a volume of water equal to its own volume. This displaced water spreads across the base of the aquarium, causing the level to rise.

$$\text{Volume displaced} = \text{Base Area} \times \text{Rise in water level}$$

**Computation:**
- Base area = $100 \times 40 = 4000 \text{ cm}^2$
- $1000 = 4000 \times h$
- $h = \frac{1000}{4000} = 0.25 \text{ cm}$

**Check for overflow:** The current depth is 37 cm, and 37 + 0.25 = 37.25 cm, which is well below the 50 cm height. No overflow occurs.

**Ruling out other choices:** Choices B–E would require rock volumes of 2000, 4000, 5000, and 10000 cm³ respectively — all larger than the given 1000 cm³.

ANSWER 1: A

---

# Problem 2:

**What is being asked:** With 4 players, each gets 15 cards. With 6 players, how many cards does each get?

**Setup:** The total number of cards in the deck stays the same.

**Computation:**
- Number of players in first game: Annika + 3 friends = 4 players
- Total cards = $4 \times 15 = 60$ cards
- Number of players in second game: $4 + 2 = 6$ players
- Cards per player = $\frac{60}{6} = 10$

**Ruling out other choices:** Some might forget to count Annika (getting 3 players → 45 cards → 45/5 = 9, choice B) or miscount the new total as 5 players (60/5 = 12, choice E).

ANSWER 2: C

---

# Problem 3:

**What is being asked:** Evaluate the custom operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$ for $a=3, b=1, c=4, d=2$.

**Setup:** Substitute directly into the formula.

**Computation:**
$$\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2$$

**Ruling out other choices:** A common mistake is computing $b \cdot c - a \cdot d = 4 - 6 = -2$ (choice A) or mixing up which values go where.

ANSWER 3: E

---

# Problem 4:

**What is being asked:** Isabella needs an average of 95 on four tests. Her first two scores are 97 and 91. What is the lowest possible third test score so she can still reach her goal?

**Setup:**
- Total points needed: $4 \times 95 = 380$
- Sum of first two tests: $97 + 91 = 188$
- Sum needed from tests 3 and 4: $380 - 188 = 192$

**Computation:**
To minimize the third test score, maximize the fourth test score. The maximum possible score on any test is 100.
$$\text{Minimum test 3} = 192 - 100 = 92$$

**Ruling out other choices:** Choice A (90) would require test 4 to be $192 - 90 = 102$, which is impossible on a 100-point test. Choices C, D, E are higher than necessary.

ANSWER 4: B

---

# Problem 5:

**What is being asked:** Bridget bought apples, gave half to Ann, gave 3 to Cassie, and kept 4. How many did she buy?

**Setup:** Let $x$ = total apples bought. Work through the sequence of events.

**Computation:**
- After giving half to Ann: $\frac{x}{2}$ apples remain
- After giving 3 to Cassie: $\frac{x}{2} - 3$ apples remain
- She keeps 4: $\frac{x}{2} - 3 = 4$
- $\frac{x}{2} = 7$
- $x = 14$

**Verification:** 14 apples → give 7 to Ann → 7 left → give 3 to Cassie → 4 left. ✓

**Ruling out other choices:** Choice C (7) is a trap for those who forget to double the 7 remaining apples. Choice D (11) comes from incorrectly adding $4 + 3 + 4$.

ANSWER 5: E

---

# Problem 6:

**What is being asked:** Find the total number of marbles given conditions about each color.

**Setup:** Let $T$ = total marbles, $R$ = red, $G$ = green, $B$ = blue.
- "All but 6 are red" → $G + B = 6$ → $R = T - 6$
- "All but 8 are green" → $R + B = 8$ → $G = T - 8$
- "All but 4 are blue" → $R + G = 4$ → $B = T - 4$

**Computation:**
$$T = R + G + B = (T - 6) + (T - 8) + (T - 4) = 3T - 18$$
$$2T = 18 \implies T = 9$$

**Verification:** $R = 3, G = 1, B = 5$. Check: $G+B=6$ ✓, $R+B=8$ ✓, $R+G=4$ ✓.

ANSWER 6: C

---

# Problem 7:

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Setup:** Find the prime factorization of 250.

**Computation:**
$$250 = 2 \times 125 = 2 \times 5 \times 25 = 2 \times 5 \times 5 \times 5 = 2 \times 5^3$$

The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.
$$\text{Sum} = 2 + 5 = 7$$

**Ruling out other choices:** Choice A (2) is just one factor. Choice B (5) is just one factor. Choice D (10) would be $2 + 5 + 3$ or $5 + 5$, neither of which applies.

ANSWER 7: C

---

# Problem 8:

**What is being asked:** How many distinct sums can be formed by picking one chip from Bag A {1, 3, 5} and one from Bag B {2, 4, 6}?

**Setup:** List all 9 possible sums and count the unique values.

**Computation:**
| | 2 | 4 | 6 |
|---|---|---|---|
| **1** | 3 | 5 | 7 |
| **3** | 5 | 7 | 9 |
| **5** | 7 | 9 | 11 |

Unique sums: {3, 5, 7, 9, 11} → **5 different values**

**Ruling out other choices:** Choice E (9) is the trap of counting all 9 pairs without removing duplicates. Choice D (7) might come from miscounting.

ANSWER 8: B

---

# Problem 9:

**What is being asked:** How many ways can 2 boys and 3 girls sit in a row with boys at the ends and girls in the middle?

**Setup:** The arrangement must be: **B G G G B**

**Computation:**
- Arrange the 2 boys at the 2 end positions: $2! = 2$ ways
- Arrange the 3 girls in the 3 middle positions: $3! = 6$ ways
- Total arrangements: $2 \times 6 = 12$

**Ruling out other choices:** Choice A (2) only counts the boys' arrangements. Choice D (6) only counts the girls' arrangements. Both must be multiplied together.

ANSWER 9: E

---

# Problem 10:

**What is being asked:** Evaluate $16 + 8 \div 4 - 2$ using order of operations.

**Setup:** Apply PEMDAS — division before addition and subtraction.

**Computation:**
$$16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16$$

**Ruling out other choices:**
- Left-to-right without order of operations: $(16+8) \div 4 - 2 = 6 - 2 = 4$ (choice A)
- $(16+8) \div (4-2) = 24 \div 2 = 12$ (choice C)
- $16 + 8 \div (4-2) = 16 + 4 = 20$ (choice E

 Problem 1:
We need to find how much the water level rises when a rock of volume 1000 cm³ is submerged in a rectangular aquarium.

First, find the area of the aquarium’s base:
\[
100\text{ cm} \times 40\text{ cm} = 4000\text{ cm}^2.
\]

When the rock is completely submerged, it displaces 1000 cm³ of water. This displaced volume spreads out over the base, causing the water level to rise. The rise in height \(h\) satisfies:
\[
\text{base area} \times h = \text{volume of rock}.
\]
So,
\[
h = \frac{1000}{4000} = 0.25\text{ cm}.
\]

We should check that the aquarium does not overflow. The current depth is 37 cm and the height is 50 cm, so there is 13 cm of free space—plenty of room for a 0.25 cm rise.

Tempting wrong choices come from using the wrong area (e.g., using the side wall 100×50 or 40×50) or from dividing by the height 50. Only the base area matters for the level rise.

ANSWER 1: A

---

Problem 2:
We need to find how many cards each player gets when two extra friends join the game.

First, determine the total number of cards. Annika and 3 friends make 4 players, and each gets 15 cards:
\[
4 \times 15 = 60\text{ cards total}.
\]

When 2 more friends join, the number of players becomes:
\[
4 + 2 = 6.
\]

Distribute the 60 cards evenly among 6 players:
\[
60 \div 6 = 10.
\]

A tempting mistake is to subtract 2 from 15, giving 13, which is not an option, or to guess without computing the total deck size. The key is that the total number of cards stays the same.

ANSWER 2: C

---

Problem 3:
We are given the rule \(a\ c\ b\ d = a\cdot d - b\cdot c\). We need to evaluate \(3\ 1\ 4\ 2\).

Matching the pattern:
- \(a = 3\)
- \(c = 1\)
- \(b = 4\)
- \(d = 2\)

Apply the formula:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

A tempting error is to swap the pairs and compute \(3\cdot 1 - 4\cdot 2 = 3 - 8 = -5\), which is not among the choices, or to read the order incorrectly. Careful matching to \(a, c, b, d\) is essential.

ANSWER 3: E

---

Problem 4:
Isabella wants an average of 95 on four 100-point tests. Her first two scores are 97 and 91. We need the lowest possible score on the third test that still allows her to reach her goal.

Total points needed for a 95 average:
\[
4 \times 95 = 380.
\]

Points earned so far:
\[
97 + 91 = 188.
\]

Points still needed from the last two tests:
\[
380 - 188 = 192.
\]

Let the third test score be \(x\) and the fourth be \(y\), with \(x + y = 192\) and each score at most 100. To minimize \(x\), we maximize \(y\). The highest possible score is 100, so:
\[
x = 192 - 100 = 92.
\]

If she scores 92 on the third test, she can score 100 on the fourth and exactly reach her goal. A tempting wrong choice is 95, which would be the answer if she needed the two remaining tests to average 95, but she can shift points to the fourth test.

ANSWER 4: B

---

Problem 5:
We need to find how many apples Bridget originally bought.

Let the original number be \(x\).
- She gave half to Ann, leaving her with \(\frac{x}{2}\).
- She gave 3 to Cassie and kept 4:
\[
\frac{x}{2} - 3 = 4.
\]

Solve:
\[
\frac{x}{2} = 7 \quad\Rightarrow\quad x = 14.
\]

Check: Start with 14. Half is 7, given to Ann. She has 7 left. Gives 3 to Cassie, keeps 4. This matches.

A tempting error is to add \(3 + 4 = 7\) and stop, forgetting that 7 is only what remained after giving away half, so the original amount must be doubled.

ANSWER 5: E

---

Problem 6:
We need the total number of marbles given three conditions about how many are *not* each color.

Let \(R, G, B\) be the numbers of red, green, and blue marbles.
- “All but 6 are red” means non-red marbles equal 6: \(G + B = 6\).
- “All but 8 are green” means non-green marbles equal 8: \(R + B = 8\).
- “All but 4 are blue” means non-blue marbles equal 4: \(R + G = 4\).

Add all three equations:
\[
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
\]
\[
2R + 2G + 2B = 18
\]
\[
R + G + B = 9.
\]

So there are 9 marbles total. (Checking: \(R=3, G=1, B=5\), which satisfies all three statements.)

A tempting mistake is to add \(6+8+4=18\) and choose that, but that sum counts each marble twice.

ANSWER 6: C

---

Problem 7:
We need the sum of the two smallest prime factors of 250.

Factor 250:
\[
250 = 2 \times 125 = 2 \times 5^3.
\]

The prime factors are 2 and 5. The two smallest (and only) distinct prime factors are 2 and 5. Their sum is:
\[
2 + 5 = 7.
\]

A tempting wrong choice is 10, which comes from adding 5 and 5 (the two smallest factors with multiplicity), but the problem asks for the two smallest prime factors, i.e., distinct primes.

ANSWER 7: C

---

Problem 8:
Bag A has \(\{1, 3, 5\}\) and Bag B has \(\{2, 4, 6\}\). We draw one from each and want the number of distinct possible sums.

Compute all sums:
- \(1+2=3\), \(1+4=5\), \(1+6=7\)
- \(3+2=5\), \(3+4=7\), \(3+6=9\)
- \(5+2=7\), \(5+4=9\), \(5+6=11\)

The distinct values are:
\[
3, 5, 7, 9, 11.
\]

That is 5 different values. A tempting error is to count all 9 ordered pairs without removing duplicates.

ANSWER 8: B

---

Problem 9:
We have 2 boys and 3 girls to arrange in a row with a boy at each end and the three girls in the middle.

Positions: B _ _ _ G G G _ _ _ B (but specifically 5 seats: 1,2,3,4,5 with boys at 1 and 5, girls at 2,3,4).

- Choose the boy for the left end and the boy for the right end: \(2! = 2\) ways.
- Arrange the 3 girls in the middle three seats: \(3! = 6\) ways.

Total arrangements:
\[
2 \times 6 = 12.
\]

A tempting mistake is to ignore the constraint and compute \(5! = 120\), or to forget that the boys can swap places.

ANSWER 9: E

---

Problem 10:
Evaluate \(16 + 8 \div 4 - 2\).

Using order of operations (PEMDAS/BODMAS), division comes before addition and subtraction:
\[
8 \div 4 = 2.
\]

Now substitute back:
\[
16 + 2 - 2 = 16.
\]

A tempting error is to work left-to-right doing addition first: \((16+8)\div 4 - 2 = 24\div 4 - 2 = 6 - 2 = 4\), which is choice A. Another is to do \(16 + 2 = 18\), then \(18 - 2 = 16\), which luckily gives the same corr

Problem 1:
We need the rise in water level when a rock of volume 1000 cm³ is placed in an aquarium with base 100 cm by 40 cm. The rock displaces water equal to its volume, causing the water level to rise by Δh such that base area × Δh = rock volume. Base area = 100 cm × 40 cm = 4000 cm². So Δh = 1000 cm³ / 4000 cm² = 0.25 cm. The initial depth is 37 cm, and after rise it becomes 37.25 cm, which is still below the height of 50 cm, so the rock is fully submerged. The rise is 0.25 cm.

ANSWER 1: A

Problem 2:
When Annika and 3 friends play, there are 4 players, each dealt 15 cards. Total cards = 4 × 15 = 60. In the next game, 2 more friends join, making 6 players. Since the same total number of cards is used, each player gets 60 / 6 = 10 cards.

ANSWER 2: C

Problem 3:
The operation is defined as a c b d = a·d − b·c. For the numbers 3 1 4 2, we have a = 3, c = 1, b = 4, d = 2. Compute a·d − b·c = 3×2 − 4×1 = 6 − 4 = 2.

ANSWER 3: E

Problem 4:
Isabella wants an average of 95 over four tests, so total points needed = 4 × 95 = 380. Her first two scores sum to 97 + 91 = 188. Let the third score be x. After the third test, she can still reach the goal if it is possible to score at most 100 on the fourth test to reach 380. That requires 188 + x + 100 ≥ 380 → x ≥ 380 − 288 = 92. The smallest integer x satisfying this is 92. If x = 91, then even a perfect 100 on the fourth gives 188+91+100 = 379 < 380. So the lowest possible third score is 92.

ANSWER 4: B

Problem 5:
Let the number of apples Bridget bought be N. She gave half to Ann, leaving N/2. Then she gave Cassie 3 apples, leaving N/2 − 3. She keeps 4 apples, so N/2 − 3 = 4 → N/2 = 7 → N = 14.

ANSWER 5: E

Problem 6:
Let T be the total number of marbles. "All but 6 are red" means T − R = 6 → R = T − 6. Similarly, T − G = 8 → G = T − 8, and T − B = 4 → B = T − 4. Since R + G + B = T, we have (T−6)+(T−8)+(T−4) = T → 3T − 18 = T → 2T = 18 → T = 9.

ANSWER 6: C

Problem 7:
Factor 250: 250 = 2 × 5³. Its prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5, and their sum is 7.

ANSWER 7: C

Problem 8:
List all possible sums from drawing one chip from Bag A {1,3,5} and one from Bag B {2,4,6}:
1+2=3, 1+4=5, 1+6=7;
3+2=5, 3+4=7, 3+6=9;
5+2=7, 5+4=9, 5+6=11.
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.

ANSWER 8: B

Problem 9:
We need arrangements of 2 boys and 3 girls in a row with a boy at each end. Choose which boy sits on the left end and which on the right: 2! = 2 ways. The three girls occupy the three middle seats and can be arranged in 3! = 6 ways. Total arrangements = 2 × 6 = 12.

ANSWER 9: E

Problem 10:
The expression is 16 + 8 ÷ 4 − 2 (the missing operator is division, as typical for order-of-operations problems). Following order of operations, division before addition and subtraction: 8 ÷ 4 = 2. Then 16 + 2 = 18, and 18 − 2 = 16.

ANSWER 10: D

Problem 11:
Evaluate step by step: √4 = 2. Then 8 · 2 = 16, so √(8·√4) = √16 = 4. Then 16 · 4 = 64, so √(16·√(8·√4)) = √64 = 8.

ANSWER 11: C

Problem 12:
Compare the decimal numbers:
A. 0.99
B. 0.9099
C. 0.9
D. 0.909
E. 0.9009
Clearly 0.99 is the largest because it is greater than 0.9 and all the others are less than 0.91.

ANSWER 12: A

Problem 1:
The aquarium’s base has area  
\(A = 100\text{ cm} \times 40\text{ cm}=4000\text{ cm}^2\).  
When the rock of volume \(V=1000\text{ cm}^3\) is fully submerged, the water rises by a height \(h\) such that  

\[
A\cdot h = V \quad\Longrightarrow\quad h = \frac{1000}{4000}=0.25\text{ cm}.
\]

The rise cannot be any of the larger listed values (0.5 cm, 1 cm, …) because those would require a volume larger than the rock’s.  

ANSWER 1: A  

---

Problem 2:
With 4 players each receiving 15 cards, the total number of cards is  

\[
4 \times 15 = 60.
\]

If 2 more friends join, there are \(4+2 = 6\) players.  The cards are still all the cards, so each player receives  

\[
\frac{60}{6}=10\text{ cards}.
\]

The other answer choices (8, 9, 11, 12) would not distribute the 60 cards evenly.  

ANSWER 2: C  

---

Problem 3:
The operation is defined by  

\[
a\;c\;b\;d = a\cdot d - b\cdot c.
\]

For the numbers \(3\;1\;4\;2\) we have \(a=3,\;c=1,\;b=4,\;d=2\).  Hence  

\[
3\;1\;4\;2 = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

The other options \(-2,-1,0,1\) would correspond to different combinations that do not satisfy the given formula.  

ANSWER 3: E  

---

Problem 4:
To average 95 over four 100‑point tests, the total points needed are  

\[
4 \times 95 = 380.
\]

After the first two tests Isabella has  

\[
97 + 91 = 188\text{ points}.
\]

Let the third test score be \(x\).  Then the fourth test must supply  

\[
380 - (188 + x) = 192 - x \text{ points}.
\]

The fourth test can give at most 100 points, so we need  

\[
192 - x \le 100 \quad\Longrightarrow\quad x \ge 92.
\]

If she scores \(92\) on the third test, she would need a perfect 100 on the fourth, which is possible.  Any lower score would require more than 100 points on the fourth test, which is impossible.  

Thus the lowest possible third‑test score is **92**.  

ANSWER 4: B  

---

Problem 5:
Let the original number of apples be \(N\).  
She gave half of them to Ann: \(\frac{N}{2}\).  
She then gave Cassie 3 apples and kept 4, leaving  

\[
N - \frac{N}{2} - 3 = 4.
\]

Simplify: \(\frac{N}{2} - 3 = 4 \Rightarrow \frac{N}{2}=7 \Rightarrow N=14\).

If the total had been any of the other answer choices, the steps would not balance (e.g., 11 would leave 2 apples after giving Cassie 3).  

ANSWER 5: E  

---

Problem 6:
Let the total number of marbles be \(T\).  
- “All but 6 are red’’ → red = \(T-6\).  
- “All but 8 are green’’ → green = \(T-8\).  
- “All but 4 are blue’’ → blue = \(T-4\).

These three colors account for all marbles, so  

\[
(T-6)+(T-8)+(T-4)=T \;\Longrightarrow\; 3T-18=T \;\Longrightarrow\; 2T=18 \;\Longrightarrow\; T=9.
\]

The other totals (6, 8, 10, 12) would not satisfy the three “all but…’’ statements simultaneously.  

ANSWER 6: C  

---

Problem 7:
Factor \(250\):  

\[
250 = 2 \times 5^3.
\]

The prime factors are \(2\) and \(5\).  The two smallest are \(2\) and \(5\), whose sum is  

\[
2+5=7.
\]

Choices 2, 5, 10, 12 are either single prime factors or incorrect sums.  

ANSWER 7: C  

---

Problem 8:
Bag A contains \(\{1,3,5\}\); Bag B contains \(\{2,4,6\}\).  
All possible sums \(a+b\) are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The distinct results are \(\{3,5,7,9,11\}\), a total of **5** different sums.  

The other answer choices (4, 6, 7, 9) do not match the count of distinct sums.  

ANSWER 8: B  

---

Problem 9:
Place the two boys at the two ends and the three girls in the middle three seats.

- The two boys can be ordered in \(2!\) ways (left‑right swap).  
- The three girls can be arranged among the three middle seats in \(3! = 6\) ways.

Total arrangements  

\[
2! \times 3! = 2 \times 6 = 12.
\]

Choices 2, 4, 5, 6 are too small; 12 accounts for all possible permutations under the given constraints.  

ANSWER 9: E  

---

Problem 10:
Interpret the expression as  

\[
16 + 8 \div 4 - 2.
\]

Division comes before addition and subtraction, so  

\[
8 \div 4 = 2 \quad\Longrightarrow\quad 16 + 2 - 2 = 16.
\]

The other options (4, 8, 12, 20) would result only if the order of operations were misinterpreted.  

ANSWER 10: D  

---

Problem 11:
Simplify from the innermost radical outward.

\[
\sqrt{4}=2,\qquad 
\sqrt{8\cdot\sqrt{4}} = \sqrt{8\cdot2}= \sqrt{16}=4.
\]

Now  

\[
\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{16\cdot4}= \sqrt{64}=8.
\]

Thus the value is 8, matching choice C.  

ANSWER 11: C  

---

Problem 12:
Compare the decimal values:

- A. \(0.99\)
- B. \(0.9099\)
- C. \(0.9\)
- D. \(0.909\)
- E. \(0.9009\)

Clearly \(0.99\) is greater than \(0.9099\), which is greater than \(0.909\), etc.  
Hence the largest number is \(0.99\).  

ANSWER 12: A

Problem 1:  
An aquarium has a rectangular base of 100 cm by 40 cm and height 50 cm, filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is submerged. By how many centimeters does the water level rise?  

The base area is \(100 \times 40 = 4000 \, \text{cm}^2\). The rock’s volume is 1000 cm³, so the rise in water level is \(\frac{1000}{4000} = 0.25 \, \text{cm}\).  

Tempting choices: 0.5 (B) or 1 (C) might be guessed if base area is miscalculated.  

ANSWER 1: A  

Problem 2:  
Annika and 3 friends (4 players) get 15 cards each. Total cards = \(4 \times 15 = 60\). If 2 more join (6 players), each gets \(\frac{60}{6} = 10\).  

Tempting choices: 8 (A) if total cards miscalculated as 48.  

ANSWER 2: C  

Problem 3:  
Operation defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\): \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\), but the order is \(a c b d\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). However, the problem writes "a c b d" which is interpreted as \(a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). But the choices include 2 (E). Re-evaluating: the operation is \(a c b d = a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\).  

Tempting choices: -2 (A) if signs reversed.  

ANSWER 3: A (Note: Correction based on standard interpretation: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\), but the correct answer based on the operation defined is actually \(3 \cdot 2 - 1 \cdot 4 = 2\), but the problem writes "a c b d", which is \(a \cdot d - b \cdot c\), so 2. However, the answer is 2, but the choices include 2 as E. The correct calculation is 2, so E. Correction: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\).)  

**Correction:** The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\), this is \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\).  

ANSWER 3: E  

Problem 4:  
Isabella wants an average of 95 on four 100-point tests. Total points needed: \(4 \times 95 = 380\). First two scores: 97 + 91 = 188. Let third score be \(x\), fourth be \(y\). Then \(188 + x + y = 380\), so \(x + y = 192\). To minimize \(x\), maximize \(y\): \(y = 100\), so \(x = 92\).  

Tempting choices: 90 (A) if \(y\) not maximized.  

ANSWER 4: B  

Problem 5:  
Bridget gave half apples to Ann, then 3 to Cassie, keeping 4. Let total be \(x\). After giving half to Ann: \(\frac{x}{2}\) left. Then gives 3 to Cassie: \(\frac{x}{2} - 3 = 4\). So \(\frac{x}{2} = 7\), \(x = 14\).  

Tempting choices: 7 (C) if half misinterpreted.  

ANSWER 5: E  

Problem 6:  
Let \(r, g, b\) be red, green, blue marbles. Then:  
- All but 6 are red: \(g + b = 6\)  
- All but 8 are green: \(r + b = 8\)  
- All but 4 are blue: \(r + g = 4\)  
Adding: \(2(r + g + b) = 18\), so total \(r + g + b = 9\).  

Tempting choices: 6 (A) if equations misread.  

ANSWER 6: C  

Problem 7:  
Prime factors of 250: \(250 = 2 \times 5^3\). Two smallest: 2 and 5. Sum: \(2 + 5 = 7\).  

Tempting choices: 5 (B) if only 5 considered.  

ANSWER 7: C  

Problem 8:  
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}. Possible sums:  
- 1+2=3, 1+4=5, 1+6=7  
- 3+2=5, 3+4=7, 3+6=9  
- 5+2=7, 5+4=9, 5+6=11  
Unique sums: 3, 5, 7, 9, 11 → 5 values.  

Tempting choices: 6 (C) if duplicates counted.  

ANSWER 8: B  

Problem 9:  
Two boys at ends: 2 ways. Three girls in middle: \(3! = 6\) ways. Total: \(2 \times 6 = 12\).  

Tempting choices: 6 (D) if boys not permuted.  

ANSWER 9: E  

Problem 10:  
Expression: \(16 + 8 \times 4 - 2\). Order: \(8 \times 4 = 32\), then \(16 + 32 - 2 = 46\)? Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 \times 4 - 2\): \(8 \times 4 = 32\), \(16 + 32 = 48\), \(48 - 2 = 46\). But choices max at 20. Re-interpreting: "16 + 8 4 - 2" might mean \(16 + 8 - 4 - 2 = 18\), not in choices. Alternatively, if "8 4" is 8×4, but 46 not in choices. The expression is likely \(16 + 8 - 4 - 2 = 18\), but not in choices. Correction: The expression is "16 + 8 4 − 2", which might be \(16 + (8 \times 4) - 2 = 16 + 32 - 2 = 46\), but not in choices. Re-reading: It’s "16 + 8 4 − 2", which could be \(16 + 8 - 4 - 2 = 18\), but not in choices. The intended expression is likely \(16 + 8 \times 4 - 2 = 46\), but choices max at 20. Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 - 4 - 2 = 18\), but not in choices. Alternatively, if "8 4" is 84, but 16+84-2=98. Not matching. The expression is likely \( \frac{16 + 8}{4} - 2 = \frac{24}{4} - 2 = 6 - 2 = 4 \).  

Tempting choices: 8 (B) if order misapplied.  

ANSWER 10: A  

Problem 11:  
Expression: \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\).  
- \(\sqrt{4} = 2\)  
- \(\sqrt{8 \cdot 2} = \sqrt{16} = 4\)  
- \(\sqrt{16 \cdot 4} = \sqrt{64} = 8\)  

Tempting choices: 4 (A) if innermost not simplified.  

ANSWER 11: C  

Problem 12:  
Compare decimals:  
- A: 0.99  
- B: 0.9099  
- C: 0.9  
- D: 0.909  
- E: 0.9009  
Largest is 0.99.  

Tempting choices: 0.9099 (B) if decimals misaligned.  

ANSWER 12: A  

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

### Problem 1:
What is asked: An aquarium with a 100cm×40cm rectangular base is filled with water to 37cm depth. A 1000 cm³ rock is submerged; how much does the water level rise?
Step-by-step: When a submerged object displaces water, its volume equals the additional volume of water in the aquarium. The rise in water level Δh satisfies: Volume of rock = base area of aquarium × Δh.
Base area = 100×40 = 4000 cm².
Δh = 1000 / 4000 = 0.25 cm.
Tempting wrong answers: 0.5 (if base area is miscalculated as 2000 cm²) or 1 (wrong division).
ANSWER 1: A

---

### Problem 2:
What is asked: Annika plus 3 friends (4 total players) each get 15 cards. If 2 more join, how many cards does each player get?
Step-by-step: First calculate total number of cards: 4 players × 15 cards = 60 total cards. After 2 join, total players = 4+2=6. Cards per player = 60 / 6 = 10.
Tempting wrong answers: 9 (if you incorrectly count only 3 initial players, leading to 45 total cards) or 12 (wrong division).
ANSWER 2: C

---

### Problem 3:
What is asked: The custom notation a c b d = a·d − b·c. What is the value of 3 1 4 2?
Step-by-step: Match the values to the definition: a=3, c=1, b=4, d=2. Calculate: 3·2 - 4·1 = 6 - 4 = 2.
Tempting wrong answers: -2 (swap a and b incorrectly) or 0 (miscalculate the multiplication).
ANSWER 3: E

---

### Problem 4:
What is asked: Isabella needs a 95 average on 4 100-point tests. She scored 97 and 91 on the first two. What is the lowest possible third test score that lets her reach her goal?
Step-by-step: Total points needed for 95 average: 4×95 = 380 points. Sum of first two tests: 97+91=188. To minimize the third test score, maximize the fourth test (max 100 points). Total points left for tests 3 and 4: 380-188=192. Minimum third test = 192 - 100 = 92.
Tempting wrong answers: 96 (the average of the remaining two tests, not the minimum possible third score).
ANSWER 4: B

---

### Problem 5:
What is asked: Bridget gave half her apples to Ann, 3 to Cassie, and kept 4. How many apples did she buy originally?
Step-by-step: Let N = original number of apples. After giving half to Ann, she has N/2 left. Then N/2 - 3 = 4 → N/2=7 → N=14. Verify: 14 → 7 left after Ann, minus 3 for Cassie = 4 kept, which matches.
Tempting wrong answers: 7 (the number of apples Bridget had after giving half to Ann, not the original total).
ANSWER 5: E

---

### Problem 6:
What is asked: A jar has red, green, blue marbles. All but 6 are red, all but 8 are green, all but 4 are blue. How many total marbles?
Step-by-step: Let T=total marbles, R=red, G=green, B=blue. The conditions translate to: G+B=6, R+B=8, R+G=4. Add all equations: 2(R+G+B)=18 → R+G+B=T=9. Verify: 3 red, 1 green, 5 blue sum to 9, matching all conditions.
Tempting wrong answers: 18 (the sum of 6+8+4, not halved to get total marbles).
ANSWER 6: C

---

### Problem 7:
What is asked: What is the sum of the two smallest prime factors of 250?
Step-by-step: Prime factorization of 250: 250=2×5³. The prime factors are 2 and 5. Their sum is 2+5=7.
Tempting wrong answers: 10 (sum of 5+5, or wrong addition of 2+5).
ANSWER 7: C

---

### Problem 8:
What is asked: Bag A has chips 1,3,5; Bag B has 2,4,6. One chip drawn from each bag, how many distinct sum values are possible?
Step-by-step: List all sums: 1+2=3, 1+4=5,1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7,5+4=9,5+6=11. Distinct sums: 3,5,7,9,11 → 5 total.
Tempting wrong answers: 9 (total number of unordered pairs, not distinct sums).
ANSWER 8: B

---

### Problem 9:
What is asked: 2 boys and 3 girls line up for a photo, with a boy at each end and girls in the middle. How many arrangements are possible?
Step-by-step: Arrange the boys first: 2 boys can be ordered at the ends in 2! = 2 ways. The 3 middle girls can be ordered in 3! =6 ways. Total arrangements: 2×6=12.
Tempting wrong answers: 6 (only the number of ways to arrange the girls, ignoring the boys).
ANSWER 9: E

---

### Problem 10:
What is asked: Evaluate (16+8)/(4-2) (formatted as a fraction in the original problem).
Step-by-step: Calculate numerator:16+8=24, denominator:4-2=2. 24/2=12.
Tempting wrong answers:16 (if order of operations is misapplied to calculate 16 + 8/4 -2=16).
ANSWER 10: C

---

### Problem 11:
What is asked: Evaluate √(16 · √(8 · √4)).
Step-by-step: Calculate from the innermost root first: √4=2. Next: 8×2=16, √16=4. Next:16×4=64, √64=8.
Tempting wrong answers: 4√2 (incorrect root calculation).
ANSWER 11: C

---

### Problem 12:
What is asked: Which number is largest: 0.99, 0.9099, 0.9, 0.909, 0.9009.
Step-by-step: Line up all decimals to compare place values: 0.9900 > 0.9099 > 0.9090 > 0.9009 > 0.9000. The largest is 0.99.
Tempting wrong answers: 0.9099 (the second largest value).
ANSWER 12: A

### Problem 1:
We need to find how much the water level rises when a 1000 cm³ rock is submerged in the aquarium.
1.  The volume of displaced water equals the volume of the submerged rock: 1000 cm³.
2.  The base area of the aquarium is \(100 \, \text{cm} \times 40 \, \text{cm} = 4000 \, \text{cm}^2\).
3.  Let \(h\) be the rise in water level. The volume of the displaced water is also equal to base area × rise in height: \(4000h = 1000\).
4.  Solve for \(h\): \(h = \frac{1000}{4000} = 0.25\) cm.
ANSWER 1: A
---
### Problem 2:
We need to find the number of cards per player when 2 more friends join.
1.  Original number of players: Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).
2.  New number of players: 4 + 2 = 6 players. Total cards remain 60.
3.  Cards per player: \(\frac{60}{6} = 10\).
ANSWER 2: C
---
### Problem 3:
The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). Compute \(3 \, 1 \, 4 \, 2\).
1.  Identify \(a=3, c=1, b=4, d=2\).
2.  Substitute into the formula: \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).
ANSWER 3: E
---
### Problem 4:
Find the lowest possible third test score to get an average of 95 on 4 tests.
1.  Total points needed for an average of 95: \(4 \times 95 = 380\).
2.  Sum of first two test scores: \(97 + 91 = 188\).
3.  Let \(x\) = third test score, \(y\) = fourth test score (max 100, since tests are 100-point). We need \(188 + x + y = 380\), so \(x + y = 192\).
4.  To minimize \(x\), maximize \(y = 100\). Then \(x = 192 - 100 = 92\).
ANSWER 4: B
---
### Problem 5:
Find the original number of apples Bridget bought.
1.  Let \(x\) = total apples. After giving half to Ann, she has \(\frac{x}{2}\) left.
2.  After giving 3 to Cassie, she has \(\frac{x}{2} - 3\) left, which equals 4 (the apples she kept).
3.  Solve: \(\frac{x}{2} - 3 = 4 \implies \frac{x}{2} =7 \implies x=14\).
ANSWER 5: E
---
### Problem 6:
Find the total number of marbles.
1.  Let \(R, G, B\) be the number of red, green, blue marbles, total \(T = R+G+B\).
2.  Translate the statements:
    - All but 6 are red: \(G + B = 6\)
    - All but 8 are green: \(R + B = 8\)
    - All but 4 are blue: \(R + G =4\)
3.  Add all three equations: \(2(R+G+B) = 6+8+4=18 \implies 2T=18 \implies T=9\).
ANSWER 6: C
---
### Problem 7:
Find the sum of the two smallest prime factors of 250.
1.  Factorize 250: \(250 = 2 \times 125 = 2 \times 5^3\).
2.  The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.
3.  Sum: \(2 +5=7\).
ANSWER 7: C
---
### Problem 8:
Find the number of distinct sums when drawing one chip from each bag.
1.  List all possible pairs and sums:
    - Bag A (1,3,5) + Bag B (2,4,6):
      \(1+2=3, 1+4=5, 1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7, 5+4=9, 5+6=11\)
2.  Distinct sums: 3, 5, 7, 9, 11. There are 5 unique values.
ANSWER 8: B
---
### Problem 9:
Find the number of arrangements with boys at each end and girls in the middle.
1.  Arrange the 2 boys at the ends: there are \(2! = 2\) ways (order matters: which boy is left/right).
2.  Arrange the 3 girls in the middle 3 seats: \(3! = 6\) ways.
3.  Total arrangements: \(2 \times 6 =12\).
ANSWER 9: E
---
### Problem 10:
Compute \(16 + 8 \, 4 - 2\). The missing operator between 8 and 4 is division (standard order of operations, common middle school problem):
1.  Interpret as \(16 + \frac{8}{4} - 2\).
2.  Compute division first: \(\frac{8}{4}=2\).
3.  Then add/subtract left to right: \(16 + 2 - 2 =16\).
*(Note: If interpreted as \(16 +8 -4 \times 2\), order of operations gives \(4 \times 2=8\), so \(16+8-8=16\), same result.)*
ANSWER 10: D
---
### Problem 11:
Compute \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\) by working from the innermost root outward.
1.  Innermost: \(\sqrt{4} = 2\).
2.  Next layer: \(\sqrt{8 \cdot 2} = \sqrt{16} =4\).
3.  Outermost: \(\sqrt{16 \cdot 4} = \sqrt{64} =8\).
ANSWER 11: C
---
### Problem 12:
Find the largest decimal.
1.  Rewrite all numbers with 4 decimal places for easy comparison:
    - A: \(0.9900\)
    - B: \(0.9099\)
    - C: \(0.9000\)
    - D: \(0.9090\)
    - E: \(0.9009\)
2.  Compare digit by digit: all have 9 in the tenths place. A has 9 in the hundredths place, while all others have 0. Thus A is the largest.
ANSWER 12: A

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

**Problem 1:**

What is being asked: By how many centimeters does the water level rise when a 1000 cm³ rock is submerged?

Setup: The base area is 100 cm × 40 cm = 4000 cm². When the rock is submerged, it displaces 1000 cm³ of water.

Water level rise = Volume displaced ÷ Base area = 1000 ÷ 4000 = 0.25 cm

ANSWER 1: A

---

**Problem 2:**

What is being asked: If cards are distributed evenly, how many cards per player when the player count increases?

Setup: Initially 4 players (Annika + 3 friends) each get 15 cards.
Total cards = 4 × 15 = 60 cards

With 2 more friends joining: 4 + 2 = 6 players
Cards per player = 60 ÷ 6 = 10 cards

ANSWER 2: C

---

**Problem 3:**

What is being asked: Evaluate the operation 3 1 4 2 using the rule a c b d = a·d − b·c

Setup: With a = 3, c = 1, b = 4, d = 2:
3 1 4 2 = 3·2 − 4·1 = 6 − 4 = 2

ANSWER 3: E

---

**Problem 4:**

What is being asked: What is the lowest score on test 3 that allows Isabella to achieve a 95 average on four tests?

Setup: Goal average is 95, so total needed = 95 × 4 = 380 points
First two tests: 97 + 91 = 188 points
Points needed from tests 3 and 4: 380 − 188 = 192 points

Let x = score on test 3. She can still reach her goal if she can score 100 on test 4:
x + 100 ≥ 192
x ≥ 92

The lowest possible score on test 3 is 92.

ANSWER 4: B

---

**Problem 5:**

What is being asked: How many apples did Bridget buy originally?

Setup: Let x = original number of apples
- She gave half to Ann: x/2 apples given away
- Remaining: x/2
- She gave 3 to Cassie: x/2 − 3
- She kept 4 for herself: x/2 − 3 = 4

Solving: x/2 = 7, so x = 14

ANSWER 5: E

---

**Problem 6:**

What is being asked: How many marbles are in the jar?

Setup: Let R, G, B = numbers of red, green, blue marbles; T = total
- "All but 6 are red": G + B = 6
- "All but 8 are green": R + B = 8
- "All but 4 are blue": R + G = 4

Adding all three equations: 2(R + G + B) = 6 + 8 + 4 = 18
Therefore: R + G + B = 9

ANSWER 6: C

---

**Problem 7:**

What is being asked: What is the sum of the two smallest prime factors of 250?

Setup: Factor 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5³

Prime factors: 2 and 5
Sum = 2 + 5 = 7

ANSWER 7: C

---

**Problem 8:**

What is being asked: How many different sums are possible?

Setup: Bag A: {1, 3, 5}, Bag B: {2, 4, 6}
All possible sums:
- 1 + 2 = 3
- 1 + 4 = 5
- 1 + 6 = 7
- 3 + 2 = 5 (duplicate)
- 3 + 4 = 7 (duplicate)
- 3 + 6 = 9
- 5 + 2 = 7 (duplicate)
- 5 + 4 = 9 (duplicate)
- 5 + 6 = 11

Distinct sums: {3, 5, 7, 9, 11} = 5 different values

ANSWER 8: B

---

**Problem 9:**

What is being asked: How many seating arrangements with boys at ends and girls in middle?

Setup: 2 boys at the ends can be arranged in 2! = 2 ways
3 girls in the middle can be arranged in 3! = 6 ways
Total arrangements = 2 × 6 = 12

ANSWER 9: E

---

**Problem 10:**

What is being asked: Evaluate 16 + 8 ÷ 4 − 2

Setup: Using order of operations (division before addition/subtraction):
16 + (8 ÷ 4) − 2 = 16 + 2 − 2 = 16

ANSWER 10: D

---

**Problem 11:**

What is being asked: Evaluate √(16 · √(8 · √4))

Setup: Work from the inside out:
- √4 = 2
- 8 · √4 = 8 · 2 = 16
- √(8 · √4) = √16 = 4
- 16 · √(8 · √4) = 16 · 4 = 64
- √(16 · √(8 · √4)) = √64 = 8

ANSWER 11: C

---

**Problem 12:**

What is being asked: Which decimal is largest?

Setup: Compare decimals:
- A: 0.99
- B: 0.9099
- C: 0.9
- D: 0.909
- E: 0.9009

Comparing: 0.99 > 0.9099 > 0.909 > 0.9009 > 0.9

The largest is 0.99

ANSWER 12: A

Problem 1:

The aquarium has a rectangular base of \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is

\[
100 \cdot 40 = 4000 \text{ cm}^2.
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

If the water rises by \(h\) centimeters, then

\[
4000h = 1000.
\]

So

\[
h = \frac{1000}{4000} = \frac14 = 0.25.
\]

So the water level rises \(0.25\) cm.

ANSWER 1: A

---

Problem 2:

Annika and 3 friends means there are

\[
4 \text{ players}.
\]

If each player gets 15 cards, then the total number of cards is

\[
4 \cdot 15 = 60.
\]

Now 2 more friends join, so there are

\[
6 \text{ players}.
\]

The same 60 cards are shared evenly, so each gets

\[
60 \div 6 = 10.
\]

ANSWER 2: C

---

Problem 3:

We are told that if the four numbers are \(a, b, c, d\), then

\[
a \ b \ c \ d = a\cdot d - b\cdot c.
\]

So for \(3,1,4,2\),

\[
3 \ 1 \ 4 \ 2 = 3\cdot 2 - 1\cdot 4 = 6 - 4 = 2.
\]

ANSWER 3: E

---

Problem 4:

She wants an average of 95 on 4 tests, so the total of the 4 scores must be

\[
4 \cdot 95 = 380.
\]

Her first two scores are 97 and 91, which add to

\[
97+91=188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To make the total 380, the fourth score would have to be

\[
380 - (188+x) = 192 - x.
\]

For this to be possible, the fourth score must be at most 100:

\[
192 - x \le 100.
\]

So

\[
x \ge 92.
\]

The lowest possible third score is 92.

ANSWER 4: B

---

Problem 5:

Let the number of apples Bridget bought be \(x\).

She gave half to Ann, so she kept the other half:

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

Then she gave Cassie 3 apples and kept 4 apples, so before giving Cassie the apples she had

\[
3+4=7.
\]

Thus,

\[
\frac{x}{2}=7 \quad \Rightarrow \quad x=14.
\]

ANSWER 5: E

---

Problem 6:

Let the total number of marbles be \(n\).

- “All but 6 are red” means the number of red marbles is \(n-6\).
- “All but 8 are green” means the number of green marbles is \(n-8\).
- “All but 4 are blue” means the number of blue marbles is \(n-4\).

Since every marble is one of these colors, their counts add to \(n\):

\[
(n-6)+(n-8)+(n-4)=n.
\]

Simplify:

\[
3n-18=n
\]

\[
2n=18
\]

\[
n=9.
\]

ANSWER 6: C

---

Problem 7:

We need the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3.
\]

So the prime factors are 2 and 5. The two smallest prime factors are 2 and 5, and their sum is

\[
2+5=7.
\]

ANSWER 7: C

---

Problem 8:

Bag A has \(1,3,5\); Bag B has \(2,4,6\).

Possible sums:

- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums are

\[
3,5,7,9,11
\]

which is 5 different values.

ANSWER 8: B

---

Problem 9:

There are 2 boys and 3 girls, and the picture must have:

- a boy at each end,
- the 3 girls in the middle.

First, arrange the 2 boys at the ends. They can be switched in

\[
2! = 2
\]

ways.

Then arrange the 3 girls in the middle in

\[
3! = 6
\]

ways.

So the total number of arrangements is

\[
2 \cdot 6 = 12.
\]

ANSWER 9: E

---

Problem 10:

The expression is

\[
16 + 8\cdot 4 - 2.
\]

Use order of operations:

\[
8\cdot 4=32.
\]

Then

\[
16+32-2=46.
\]

But 46 is not among the choices, so the intended expression must be read as

\[
16 + 8 \div 4 - 2.
\]

Then

\[
8\div 4 = 2,
\]

so

\[
16+2-2=16.
\]

ANSWER 10: D

---

Problem 11:

We evaluate from the inside out:

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

First,

\[
\sqrt4=2.
\]

So the expression becomes

\[
\sqrt{16\cdot \sqrt{8\cdot 2}}
=
\sqrt{16\cdot \sqrt{16}}.
\]

Since \(\sqrt{16}=4\),

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C

---

Problem 12:

Compare the decimals:

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

The largest is the one closest to 1. Since \(0.99\) is greater than any number beginning with \(0.90\), it is the largest.

ANSWER 12: A

Problem 1:  
We need the rise in water level when a rock is placed in the aquarium and submerged.

- Base dimensions: \(100\text{ cm} \times 40\text{ cm}\), so base area  
  \[
  A = 100\cdot 40 = 4000\text{ cm}^2.
  \]
- The rock is completely submerged, so the water volume displaced equals the rock’s volume:  
  \[
  V = 1000\text{ cm}^3.
  \]
- Water level rise \(\Delta h\) satisfies \(A\Delta h = V\):  
  \[
  4000\cdot \Delta h = 1000 \implies \Delta h = \frac{1000}{4000}=0.25\text{ cm}.
  \]

So the rise is \(0.25\) cm.  
Tempting wrong idea: using the *remaining* depth \(37\) cm or checking if it overflows—none of that matters; only displaced volume matters.

ANSWER 1: A  

---

Problem 2:  
All cards are dealt evenly. Let the total number of cards be \(T\).

Initially, Annika + 3 friends = 4 players, and each gets 15 cards:  
\[
T = 4\cdot 15 = 60.
\]
Next game: 2 more friends join, so total players:
\[
4+2=6.
\]
Each player gets:
\[
\frac{60}{6}=10.
\]

ANSWER 2: C  

---

Problem 3:  
Given the rule \(a\,b\,d\,c = a\cdot d - b\cdot c\).  
The expression “\(3\ 1\ 4\ 2\)” corresponds to:
- \(a=3\)
- \(b=1\)
- \(d=4\)
- \(c=2\)

Compute:
\[
a\cdot d - b\cdot c = 3\cdot 4 - 1\cdot 2 = 12-2=10.
\]

But notice the choices are only \(-2,-1,0,1,2\), so we must interpret the placeholder correctly:  
The problem states: **“If \(a b c d = a\cdot d - b\cdot c\)”**.

So “\(3\ 1\ 4\ 2\)” means:
- \(a=3,\ b=1,\ c=4,\ d=2\)

Then:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 1\cdot 4 = 6-4=2.
\]

ANSWER 3: E  

---

Problem 4:  
Goal average is 95 over 4 tests, so total points needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
Let the third test score be \(x\), and the fourth be \(y\). Then:
\[
188 + x + y = 380 \implies x+y=192.
\]
We want the **lowest possible** \(x\) such that she can still reach the goal. That means make \(y\) as large as possible (since grades are at most 100).
Maximum \(y=100\). Then:
\[
x + 100 = 192 \implies x = 92.
\]

Check: then the four scores are \(97,91,92,100\) which sum to \(380\), average \(95\). Works.

ANSWER 4: B  

---

Problem 5:  
Let the total number of apples be \(N\).

- She gave half to Ann, so Ann got \(N/2\), leaving \(N/2\).
- Then she gave Cassie 3 apples, and kept 4 apples. That means after giving Cassie 3, her remaining is 4:
  \[
  \frac{N}{2} - 3 = 4 \implies \frac{N}{2}=7 \implies N=14.
  \]

ANSWER 5: E  

---

Problem 6:  
“All but 6 are red” means: number of red marbles is 6.  
Similarly:
- Red = 6 (since all but 6 are red → 6 red)
- Green: all but 8 are green → 8 green
- Blue: all but 4 are blue → 4 blue

Total marbles:
\[
6+8+4=18.
\]
But none of the choices are 18, so interpret carefully:  
“All but 6 are red” means the number of red marbles is **(total − 6)**, not 6.

Let total be \(T\). Then:
- Red marbles = \(T-6\)
- Green marbles = \(T-8\)
- Blue marbles = \(T-4\)

Sum:
\[
(T-6)+(T-8)+(T-4)=T.
\]
\[
3T - (6+8+4)=T
\]
\[
3T-18=T \implies 2T=18 \implies T=9.
\]

ANSWER 6: C  

---

Problem 7:  
We want the sum of the **two smallest prime factors** of 250.

Factor \(250\):
\[
250 = 25\cdot 10 = 5^2 \cdot 2\cdot 5 = 2\cdot 5^3.
\]
Prime factors are \(2\) and \(5\). The two smallest are \(2\) and \(5\).

Sum:
\[
2+5=7.
\]

ANSWER 7: D  

---

Problem 8:  
Bag A: \(\{1,3,5\}\)  
Bag B: \(\{2,4,6\}\)

Possible sums:
- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums: \(\{3,5,7,9,11\}\) → that’s 5 values.

ANSWER 8: B  

---

Problem 9:  
We need arrangements of 5 people in a row with:
- a boy at each end (positions 1 and 5)
- the three girls in the middle (positions 2,3,4)

There are 2 boys: they can be placed at the ends in
\[
2!\ = 2
\]
ways (left boy/right boy).

The 3 girls can be arranged in the three middle spots in
\[
3! = 6
\]
ways.

Total arrangements:
\[
2\cdot 6 = 12.
\]

ANSWER 9: E  

---

Problem 10:  
Expression: \(16 + 8^4 - 2\).

Compute in order:
- \(8^4 = 4096\)

So:
\[
16 + 4096 - 2 = 4110.
\]
No answer choice matches, so we interpret the formatting as \(16 + 8 \sqrt{4} - 2\) or \(16 + 8^4 - 2\)?  
Given the AMC/JH style, likely it is:
\[
16 + 8\sqrt{4} - 2.
\]
Then:
\[
\sqrt{4}=2 \implies 8\sqrt{4}=8\cdot 2=16.
\]
So:
\[
16 + 16 - 2 = 30,
\]
still not in choices.

Another common formatting is:
\[
16 + 8^{4} - 2 \quad \text{(still not)}
\]

The only choiceable consistent possibility is:
\[
16 + \sqrt{8}\,^4 - 2 \text{ (nonsense)}
\]

Let’s use the most plausible literal from the text: **“16 + 8 4 − 2”** often means \(16 + 8^4 - 2\) if the caret was lost. But then no choice.

Given typical AMC problems, the intended is usually:
\[
16 + 8 - 4 - 2 = 18
\]
also not.

Because the answer choices are only \(4,8,12,16,20\), the intended expression must simplify to one of those. The most natural is:
\[
16 + 8/4 - 2 = 16 + 2 - 2 = 16.
\]
This matches a common missing symbol: “8 4” often means \(8/4\).

So:
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D  

---

Problem 11:  
Evaluate:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}.
\]

Work from inside out:

1) \(\sqrt{4}=2\)

2) Inside next root:
\[
8\cdot \sqrt{4} = 8\cdot 2 = 16
\]
So:
\[
\sqrt{8\cdot \sqrt{4}} = \sqrt{16}=4.
\]

3) Now plug into the outer expression:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}=\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C  

---

Problem 12:  
Largest number among:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare by decimal places:

- .99 is 0.99
- .9099 is 0.9099
- .9 is 0.9000
- .909 is 0.9090
- .9009 is 0.9009

Clearly \(0.99\) is greater than all the numbers less than \(0.91\).

ANSWER 12: A

Problem 1:
The volume of the water displaced by the rock is equal to the volume of the rock itself, which is $1000 \text{ cm}^3$. The aquarium has a rectangular base with area $A = 100 \text{ cm} \times 40 \text{ cm} = 4000 \text{ cm}^2$. When the rock is submerged, the water level rises by height $h$ such that the volume of the displaced water is $A \times h$. 
Setting this equal to the rock's volume: $4000 \times h = 1000$.
Solving for $h$: $h = \frac{1000}{4000} = \frac{1}{4} = 0.25 \text{ cm}$.
ANSWER 1: A

Problem 2:
Annika and 3 friends make 4 players total. Each received 15 cards, so the total number of cards is $4 \times 15 = 60$. If 2 more friends join, the total number of players becomes $4 + 2 = 6$. The cards are distributed evenly among the 6 players, so each player gets $60 \div 6 = 10$ cards.
ANSWER 2: C

Problem 3:
The problem defines the operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$.
For the expression $\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix}$, we have $a=3, b=1, c=4, d=2$.
Calculating: $(3 \times 2) - (1 \times 4) = 6 - 4 = 2$.
ANSWER 3: E

Problem 4:
Isabella needs an average of 95 over 4 tests, meaning the total sum of her scores must be $95 \times 4 = 380$. Her first two scores are 97 and 91, totaling $97 + 91 = 188$. Let $x$ be the score on the third test and $y$ be the score on the fourth. We have $188 + x + y = 380$, so $x + y = 192$. To make $x$ as small as possible, we must make $y$ as large as possible. Since the maximum score on a test is 100, we set $y = 100$. Then $x + 100 = 192$, so $x = 92$.
ANSWER 4: B

Problem 5:
Let $x$ be the total number of apples Bridget bought. She gave half away, leaving $\frac{x}{2}$. Then she gave 3 away, leaving $\frac{x}{2} - 3$. She kept 4, so $\frac{x}{2} - 3 = 4$. Adding 3 to both sides, $\frac{x}{2} = 7$. Multiplying by 2, $x = 14$.
ANSWER 5: E

Problem 6:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R + G + B$.
"All but 6 are red" means $G + B = 6$.
"All but 8 are green" means $R + B = 8$.
"All but 4 are blue" means $R + G = 4$.
Adding these three equations: $2(R + G + B) = 6 + 8 + 4 = 18$.
So $R + G + B = 18 \div 2 = 9$.
ANSWER 6: C

Problem 7:
The prime factorization of 250 is $250 = 2 \times 125 = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two prime factors is $2 + 5 = 7$.
ANSWER 7: C

Problem 8:
Bag A has $\{1, 3, 5\}$ and Bag B has $\{2, 4, 6\}$.
Possible sums:
$1+2=3, 1+4=5, 1+6=7$
$3+2=5, 3+4=7, 3+6=9$
$5+2=7, 5+4=9, 5+6=11$
The set of distinct sums is $\{3, 5, 7, 9, 11\}$. There are 5 distinct values.
ANSWER 8: B

Problem 9:
There are 2 boys (B1, B2) and 3 girls (G1, G2, G3). A boy must be at each end. There are 2 choices for the left end and 1 choice for the right end ($2 \times 1 = 2$ ways to place boys). The 3 girls can be arranged in the middle in $3! = 3 \times 2 \times 1 = 6$ ways. Total arrangements = $2 \times 6 = 12$.
ANSWER 9: E

Problem 10:
Following the order of operations (PEMDAS/BODMAS):
$16 + (8 \div 4) - 2 = 16 + 2 - 2 = 16$.
ANSWER 10: D

Problem 11:
Evaluate from the inside out:
$\sqrt{4} = 2$
$\sqrt{8 \times 2} = \sqrt{16} = 4$
$\sqrt{16 \times 4} = \sqrt{64} = 8$
ANSWER 11: C

Problem 12:
To compare, pad with zeros to match the number of decimal places:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
Comparing the values, 0.9900 is the largest.
ANSWER 12: A

Problem 1:
The base of the aquarium measures 100 cm by 40 cm, so its area is \(100 \times 40 = 4000\) cm². The rock has volume 1000 cm³ and is fully submerged, displacing exactly 1000 cm³ of water. The resulting rise in water level is therefore the displaced volume divided by the base area: \(1000 / 4000 = 0.25\) cm. (The original water depth and tank height are irrelevant once the surface area is known.)

ANSWER 1: A

Problem 2:
Annika plus 3 friends gives 4 players, each receiving 15 cards, for a total of \(4 \times 15 = 60\) cards. When 2 more players join, there are 6 players sharing the same 60 cards, so each receives \(60 / 6 = 10\) cards.

ANSWER 2: C

Problem 3:
The given operation evaluates \(a\, c\, b\, d\) as \(a \cdot d - b \cdot c\). Substituting the numbers yields \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).

ANSWER 3: E

Problem 4:
An average of 95 on four 100-point tests requires a total of \(4 \times 95 = 380\) points. The first two scores sum to \(97 + 91 = 188\), leaving 192 points still needed from the last two tests. The fourth test can score at most 100, so the third test must be at least \(192 - 100 = 92\). Any lower score on the third test would make a total of 380 impossible.

ANSWER 4: B

Problem 5:
Let \(t\) be the original number of apples. After giving half to Ann, \(t/2\) remain. After giving 3 more to Cassie, \(t/2 - 3\) remain, and this equals the 4 apples kept. Solving \(t/2 - 3 = 4\) gives \(t/2 = 7\), so \(t = 14\).

ANSWER 5: E

Problem 6:
Let \(m\) be the total number of marbles. Then there are \(m-6\) red, \(m-8\) green, and \(m-4\) blue. Their sum must equal \(m\):
\[
(m-6) + (m-8) + (m-4) = m \implies 3m - 18 = m \implies 2m = 18 \implies m = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies the three “all-but” statements.

ANSWER 6: C

Problem 7:
Factorize \(250 = 2 \cdot 5^3\). The two smallest distinct prime factors are 2 and 5; their sum is 7.

ANSWER 7: C

Problem 8:
All nine possible sums are:
\[
1+2=3,\ 1+4=5,\ 1+6=7,\ 3+2=5,\ 3+4=7,\ 3+6=9,\ 5+2=7,\ 5+4=9,\ 5+6=11.
\]
The distinct values are 3, 5, 7, 9, 11—exactly five.

ANSWER 8: B

Problem 9:
The seating pattern must be boy–girl–girl–girl–boy. The two distinct boys may be placed at the ends in \(2!\) = 2 ways. The three distinct girls may be arranged in the middle three seats in \(3!\) = 6 ways. The total number of arrangements is therefore \(2 \times 6 = 12\).

ANSWER 9: E

Problem 10:
Interpreting the expression with standard order of operations and treating juxtaposition of 8 and 4 as division yields
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D

Problem 11:
Evaluate from the inside outward:
\[
\sqrt{4} = 2, \quad 8 \cdot 2 = 16, \quad \sqrt{16} = 4, \quad 16 \cdot 4 = 64, \quad \sqrt{64} = 8.
\]

ANSWER 11: C

Problem 12:
Compare the decimals directly:
\[
0.99 > 0.9 > 0.9099 > 0.909 > 0.9009.
\]
Thus 0.99 is the largest.

ANSWER 12: A

Problem 1:
The problem asks us to determine by how many centimeters the water level rises when a rock with a volume of 1000 cm³ is submerged in an aquarium filled with water to a depth of 37 cm. The aquarium has a rectangular base measuring 100 cm by 40 cm and a height of 50 cm.

First, we need to calculate the initial volume of water in the aquarium. The volume of water is given by the formula length × width × height. So, the initial volume of water = 100 cm × 40 cm × 37 cm = 148,000 cm³.

When the rock is submerged, it displaces a volume of water equal to its own volume, which is 1000 cm³. This means the water level will rise to accommodate the volume of the rock.

The rise in water level can be calculated by dividing the volume of the rock by the base area of the aquarium. The base area = 100 cm × 40 cm = 4000 cm².

Rise in water level = Volume of rock / Base area = 1000 cm³ / 4000 cm² = 0.25 cm.

Let's examine the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

The calculated rise in water level matches option A.

ANSWER 1: A

Problem 2:
The problem states that in the card game Buffalo Shuffle-o, all cards are distributed evenly among players. Initially, Annika and 3 friends play, and each is dealt 15 cards. We need to find out how many cards each player will be dealt if 2 more friends join the next game.

First, we determine the total number of cards. With 4 players initially, each getting 15 cards, the total number of cards = 4 × 15 = 60.

When 2 more friends join, the total number of players becomes 4 + 2 = 6.

To find out how many cards each player will get, we divide the total number of cards by the new number of players: 60 / 6 = 10.

Let's examine the choices:
A. 8
B. 9
C. 10
D. 11
E. 12

The calculated number of cards per player matches option C.

ANSWER 2: C

Problem 3:
The problem defines a operation as a c b d = a·d − b·c and asks us to find the value of 3 1 4 2.

Using the given definition, we substitute a = 3, b = 1, c = 4, and d = 2 into the equation: 3 1 4 2 = 3·2 − 1·4 = 6 - 4 = 2.

Let's examine the choices:
A. −2
B. −1
C. 0
D. 1
E. 2

The calculated value matches option E.

ANSWER 3: E

Problem 4:
Isabella needs to achieve an average grade of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score she could have made on the third test to still reach her goal.

The total score required for an average of 95 on 4 tests = 95 × 4 = 380.

The sum of her first two test scores = 97 + 91 = 188.

So, the total score needed on the last two tests = 380 - 188 = 192.

To minimize the score on the third test, we maximize the score on the fourth test. The maximum score on the fourth test is 100.

Thus, the minimum score on the third test = 192 - 100 = 92.

Let's examine the choices:
A. 90
B. 92
C. 95
D. 96
E. 97

The calculated minimum score matches option B.

ANSWER 4: B

Problem 5:
Bridget gave half of her apples to Ann, then gave 3 to Cassie, and kept 4 for herself. We need to find out how many apples she originally bought.

Let's denote the original number of apples as x.

After giving half to Ann, Bridget had x/2 apples left.

Then, she gave 3 apples to Cassie and kept 4, so x/2 = 3 + 4 = 7.

Thus, x = 7 × 2 = 14.

Let's examine the choices:
A. 3
B. 4
C. 7
D. 11
E. 14

The calculated original number of apples matches option E.

ANSWER 5: E

Problem 6:
The problem states that in a jar, all but 6 marbles are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T, the number of red marbles as R, the number of green marbles as G, and the number of blue marbles as B.

From the given information:
1. T - R = 6 (all but 6 are red)
2. T - G = 8 (all but 8 are green)
3. T - B = 4 (all but 4 are blue)

Adding these equations gives: 3T - (R + G + B) = 18.

Since R + G + B = T (the total number of marbles is the sum of the marbles of each color), we have:
3T - T = 18, which simplifies to 2T = 18.

Thus, T = 18 / 2 = 9.

Let's examine the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

The calculated total number of marbles matches option C.

ANSWER 6: C

Problem 7:
We need to find the sum of the two smallest prime factors of 250.

First, let's factorize 250 into its prime factors: 250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5.

The sum of these two smallest prime factors = 2 + 5 = 7.

Let's examine the choices:
A. 2
B. 5
C. 7
D. 10
E. 12

The calculated sum matches option C.

ANSWER 7: C

Problem 8:
Bag A contains chips labeled 1, 3, and 5. Bag B contains chips labeled 2, 4, and 6. We need to find how many different values are possible for the sum of the numbers on the chips drawn from each bag.

Let's list the possible sums:
- (1,2) = 3
- (1,4) = 5
- (1,6) = 7
- (3,2) = 5
- (3,4) = 7
- (3,6) = 9
- (5,2) = 7
- (5,4) = 9
- (5,6) = 11

The possible sums are 3, 5, 7, 9, and 11. There are 5 different values.

Let's examine the choices:
A. 4
B. 5
C. 6
D. 7
E. 9

The number of different sums matches option B.

ANSWER 8: B

Problem 9:
The chess team consists of 2 boys and 3 girls. They need to be arranged with a boy at each end and the girls in the middle. We need to find out how many such arrangements are possible.

The boys can be arranged at the ends in 2! = 2 ways (either boy can be on the left or the right).

The 3 girls can be arranged in the 3 middle positions in 3! = 6 ways.

The total number of arrangements = (number of ways to arrange boys) × (number of ways to arrange girls) = 2 × 6 = 12.

Let's examine the choices:
A. 2
B. 4
C. 5
D. 6
E. 12

The calculated total number of arrangements matches option E.

ANSWER 9: E

Problem 10:
The problem asks us to evaluate the expression 16 + 8 ÷ 4 − 2.

Following the order of operations (PEMDAS/BODMAS: Parentheses/Brackets, Exponents/Orders, Multiplication and Division from left to right, Addition and Subtraction from left to right):
1. D

Problem 1:
The water rise is determined by the volume of the rock divided by the base area of the aquarium.  
Base area = \(100 \times 40 = 4000\text{ cm}^2\).  
Volume of rock = \(1000\text{ cm}^3\).  
Rise = \(\frac{1000}{4000} = 0.25\text{ cm}\).  
ANSWER 1: A

Problem 2:
Initially, Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 2: C

Problem 3:
The notation \(\begin{array}{cc} a & c \\ b & d \end{array} = a \cdot d - b \cdot c\).  
For \(\begin{array}{cc} 3 & 1 \\ 4 & 2 \end{array}\): \(a=3, c=1, b=4, d=2\).  
Value = \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).  
ANSWER 3: E

Problem 4:
Total points needed for an average of 95 over 4 tests = \(4 \times 95 = 380\).  
First two tests sum = \(97 + 91 = 188\).  
Remaining points needed = \(380 - 188 = 192\).  
To minimize the third test score, maximize the fourth (100). Then third test \(\ge 192 - 100 = 92\).  
Lowest possible third test score is 92.  
ANSWER 4: B

Problem 5:
Let \(x\) be the number of apples Bridget bought.  
She gave half to Ann, so she kept \(\frac{x}{2}\).  
Then she gave 3 to Cassie and kept 4, so \(\frac{x}{2} = 3 + 4 = 7\).  
Thus \(x = 14\).  
ANSWER 5: E

Problem 6:
Let \(R, G, B\) be the number of red, green, blue marbles.  
"All but 6 are red" \(\Rightarrow G + B = 6\).  
"All but 8 are green" \(\Rightarrow R + B = 8\).  
"All but 4 are blue" \(\Rightarrow R + G = 4\).  
Adding: \((G+B)+(R+B)+(R+G) = 6+8+4 = 18\) \(\Rightarrow 2(R+G+B) = 18\) \(\Rightarrow R+G+B = 9\).  
Total marbles = 9.  
ANSWER 6: C

Problem 7:
Prime factorization of 250: \(250 = 2 \times 5^3\).  
Prime factors are 2 and 5. The two smallest are 2 and 5.  
Sum = \(2 + 5 = 7\).  
ANSWER 7: C

Problem 8:
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}.  
Possible sums:  
1+2=3, 1+4=5, 1+6=7;  
3+2=5, 3+4=7, 3+6=9;  
5+2=7, 5+4=9, 5+6=11.  
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.  
ANSWER 8: B

Problem 9:
Boys must sit at the two ends.  
Arrange boys: 2 choices for left end, 1 for right end → \(2! = 2\) ways.  
Arrange 3 girls in the middle 3 seats: \(3! = 6\) ways.  
Total arrangements = \(2 \times 6 = 12\).  
ANSWER 9: E

Problem 10:
Interpret "8 4" as \(8 \div 4\).  
\(16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16\).  
ANSWER 10: D

Problem 11:
\(\sqrt{4} = 2\).  
\(8 \cdot 2 = 16\), \(\sqrt{16} = 4\).  
\(16 \cdot 4 = 64\), \(\sqrt{64} = 8\).  
ANSWER 11: C

Problem 12:
Compare decimals:  
0.99, 0.9099, 0.9, 0.909, 0.9009.  
0.99 is the largest.  
ANSWER 12: A

# Problem 1:

**What is being asked:** A rock of volume 1000 cm³ is submerged in an aquarium with a 100 cm × 40 cm base. How much does the water level rise?

**Setup:** When the rock is submerged, it displaces a volume of water equal to its own volume. This displaced water spreads across the base of the aquarium, causing the level to rise.

$$\text{Volume displaced} = \text{Base Area} \times \text{Rise in water level}$$

**Computation:**
- Base area = $100 \times 40 = 4000 \text{ cm}^2$
- $1000 = 4000 \times h$
- $h = \frac{1000}{4000} = 0.25 \text{ cm}$

**Check for overflow:** The current depth is 37 cm, and 37 + 0.25 = 37.25 cm, which is well below the 50 cm height. No overflow occurs.

**Ruling out other choices:** Choices B–E would require rock volumes of 2000, 4000, 5000, and 10000 cm³ respectively — all larger than the given 1000 cm³.

ANSWER 1: A

---

# Problem 2:

**What is being asked:** With 4 players, each gets 15 cards. With 6 players, how many cards does each get?

**Setup:** The total number of cards in the deck stays the same.

**Computation:**
- Number of players in first game: Annika + 3 friends = 4 players
- Total cards = $4 \times 15 = 60$ cards
- Number of players in second game: $4 + 2 = 6$ players
- Cards per player = $\frac{60}{6} = 10$

**Ruling out other choices:** Some might forget to count Annika (getting 3 players → 45 cards → 45/5 = 9, choice B) or miscount the new total as 5 players (60/5 = 12, choice E).

ANSWER 2: C

---

# Problem 3:

**What is being asked:** Evaluate the custom operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$ for $a=3, b=1, c=4, d=2$.

**Setup:** Substitute directly into the formula.

**Computation:**
$$\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2$$

**Ruling out other choices:** A common mistake is computing $b \cdot c - a \cdot d = 4 - 6 = -2$ (choice A) or mixing up which values go where.

ANSWER 3: E

---

# Problem 4:

**What is being asked:** Isabella needs an average of 95 on four tests. Her first two scores are 97 and 91. What is the lowest possible third test score so she can still reach her goal?

**Setup:**
- Total points needed: $4 \times 95 = 380$
- Sum of first two tests: $97 + 91 = 188$
- Sum needed from tests 3 and 4: $380 - 188 = 192$

**Computation:**
To minimize the third test score, maximize the fourth test score. The maximum possible score on any test is 100.
$$\text{Minimum test 3} = 192 - 100 = 92$$

**Ruling out other choices:** Choice A (90) would require test 4 to be $192 - 90 = 102$, which is impossible on a 100-point test. Choices C, D, E are higher than necessary.

ANSWER 4: B

---

# Problem 5:

**What is being asked:** Bridget bought apples, gave half to Ann, gave 3 to Cassie, and kept 4. How many did she buy?

**Setup:** Let $x$ = total apples bought. Work through the sequence of events.

**Computation:**
- After giving half to Ann: $\frac{x}{2}$ apples remain
- After giving 3 to Cassie: $\frac{x}{2} - 3$ apples remain
- She keeps 4: $\frac{x}{2} - 3 = 4$
- $\frac{x}{2} = 7$
- $x = 14$

**Verification:** 14 apples → give 7 to Ann → 7 left → give 3 to Cassie → 4 left. ✓

**Ruling out other choices:** Choice C (7) is a trap for those who forget to double the 7 remaining apples. Choice D (11) comes from incorrectly adding $4 + 3 + 4$.

ANSWER 5: E

---

# Problem 6:

**What is being asked:** Find the total number of marbles given conditions about each color.

**Setup:** Let $T$ = total marbles, $R$ = red, $G$ = green, $B$ = blue.
- "All but 6 are red" → $G + B = 6$ → $R = T - 6$
- "All but 8 are green" → $R + B = 8$ → $G = T - 8$
- "All but 4 are blue" → $R + G = 4$ → $B = T - 4$

**Computation:**
$$T = R + G + B = (T - 6) + (T - 8) + (T - 4) = 3T - 18$$
$$2T = 18 \implies T = 9$$

**Verification:** $R = 3, G = 1, B = 5$. Check: $G+B=6$ ✓, $R+B=8$ ✓, $R+G=4$ ✓.

ANSWER 6: C

---

# Problem 7:

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Setup:** Find the prime factorization of 250.

**Computation:**
$$250 = 2 \times 125 = 2 \times 5 \times 25 = 2 \times 5 \times 5 \times 5 = 2 \times 5^3$$

The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.
$$\text{Sum} = 2 + 5 = 7$$

**Ruling out other choices:** Choice A (2) is just one factor. Choice B (5) is just one factor. Choice D (10) would be $2 + 5 + 3$ or $5 + 5$, neither of which applies.

ANSWER 7: C

---

# Problem 8:

**What is being asked:** How many distinct sums can be formed by picking one chip from Bag A {1, 3, 5} and one from Bag B {2, 4, 6}?

**Setup:** List all 9 possible sums and count the unique values.

**Computation:**
| | 2 | 4 | 6 |
|---|---|---|---|
| **1** | 3 | 5 | 7 |
| **3** | 5 | 7 | 9 |
| **5** | 7 | 9 | 11 |

Unique sums: {3, 5, 7, 9, 11} → **5 different values**

**Ruling out other choices:** Choice E (9) is the trap of counting all 9 pairs without removing duplicates. Choice D (7) might come from miscounting.

ANSWER 8: B

---

# Problem 9:

**What is being asked:** How many ways can 2 boys and 3 girls sit in a row with boys at the ends and girls in the middle?

**Setup:** The arrangement must be: **B G G G B**

**Computation:**
- Arrange the 2 boys at the 2 end positions: $2! = 2$ ways
- Arrange the 3 girls in the 3 middle positions: $3! = 6$ ways
- Total arrangements: $2 \times 6 = 12$

**Ruling out other choices:** Choice A (2) only counts the boys' arrangements. Choice D (6) only counts the girls' arrangements. Both must be multiplied together.

ANSWER 9: E

---

# Problem 10:

**What is being asked:** Evaluate $16 + 8 \div 4 - 2$ using order of operations.

**Setup:** Apply PEMDAS — division before addition and subtraction.

**Computation:**
$$16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16$$

**Ruling out other choices:**
- Left-to-right without order of operations: $(16+8) \div 4 - 2 = 6 - 2 = 4$ (choice A)
- $(16+8) \div (4-2) = 24 \div 2 = 12$ (choice C)
- $16 + 8 \div (4-2) = 16 + 4 = 20$ (choice E

 Problem 1:
We need to find how much the water level rises when a rock of volume 1000 cm³ is submerged in a rectangular aquarium.

First, find the area of the aquarium’s base:
\[
100\text{ cm} \times 40\text{ cm} = 4000\text{ cm}^2.
\]

When the rock is completely submerged, it displaces 1000 cm³ of water. This displaced volume spreads out over the base, causing the water level to rise. The rise in height \(h\) satisfies:
\[
\text{base area} \times h = \text{volume of rock}.
\]
So,
\[
h = \frac{1000}{4000} = 0.25\text{ cm}.
\]

We should check that the aquarium does not overflow. The current depth is 37 cm and the height is 50 cm, so there is 13 cm of free space—plenty of room for a 0.25 cm rise.

Tempting wrong choices come from using the wrong area (e.g., using the side wall 100×50 or 40×50) or from dividing by the height 50. Only the base area matters for the level rise.

ANSWER 1: A

---

Problem 2:
We need to find how many cards each player gets when two extra friends join the game.

First, determine the total number of cards. Annika and 3 friends make 4 players, and each gets 15 cards:
\[
4 \times 15 = 60\text{ cards total}.
\]

When 2 more friends join, the number of players becomes:
\[
4 + 2 = 6.
\]

Distribute the 60 cards evenly among 6 players:
\[
60 \div 6 = 10.
\]

A tempting mistake is to subtract 2 from 15, giving 13, which is not an option, or to guess without computing the total deck size. The key is that the total number of cards stays the same.

ANSWER 2: C

---

Problem 3:
We are given the rule \(a\ c\ b\ d = a\cdot d - b\cdot c\). We need to evaluate \(3\ 1\ 4\ 2\).

Matching the pattern:
- \(a = 3\)
- \(c = 1\)
- \(b = 4\)
- \(d = 2\)

Apply the formula:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

A tempting error is to swap the pairs and compute \(3\cdot 1 - 4\cdot 2 = 3 - 8 = -5\), which is not among the choices, or to read the order incorrectly. Careful matching to \(a, c, b, d\) is essential.

ANSWER 3: E

---

Problem 4:
Isabella wants an average of 95 on four 100-point tests. Her first two scores are 97 and 91. We need the lowest possible score on the third test that still allows her to reach her goal.

Total points needed for a 95 average:
\[
4 \times 95 = 380.
\]

Points earned so far:
\[
97 + 91 = 188.
\]

Points still needed from the last two tests:
\[
380 - 188 = 192.
\]

Let the third test score be \(x\) and the fourth be \(y\), with \(x + y = 192\) and each score at most 100. To minimize \(x\), we maximize \(y\). The highest possible score is 100, so:
\[
x = 192 - 100 = 92.
\]

If she scores 92 on the third test, she can score 100 on the fourth and exactly reach her goal. A tempting wrong choice is 95, which would be the answer if she needed the two remaining tests to average 95, but she can shift points to the fourth test.

ANSWER 4: B

---

Problem 5:
We need to find how many apples Bridget originally bought.

Let the original number be \(x\).
- She gave half to Ann, leaving her with \(\frac{x}{2}\).
- She gave 3 to Cassie and kept 4:
\[
\frac{x}{2} - 3 = 4.
\]

Solve:
\[
\frac{x}{2} = 7 \quad\Rightarrow\quad x = 14.
\]

Check: Start with 14. Half is 7, given to Ann. She has 7 left. Gives 3 to Cassie, keeps 4. This matches.

A tempting error is to add \(3 + 4 = 7\) and stop, forgetting that 7 is only what remained after giving away half, so the original amount must be doubled.

ANSWER 5: E

---

Problem 6:
We need the total number of marbles given three conditions about how many are *not* each color.

Let \(R, G, B\) be the numbers of red, green, and blue marbles.
- “All but 6 are red” means non-red marbles equal 6: \(G + B = 6\).
- “All but 8 are green” means non-green marbles equal 8: \(R + B = 8\).
- “All but 4 are blue” means non-blue marbles equal 4: \(R + G = 4\).

Add all three equations:
\[
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
\]
\[
2R + 2G + 2B = 18
\]
\[
R + G + B = 9.
\]

So there are 9 marbles total. (Checking: \(R=3, G=1, B=5\), which satisfies all three statements.)

A tempting mistake is to add \(6+8+4=18\) and choose that, but that sum counts each marble twice.

ANSWER 6: C

---

Problem 7:
We need the sum of the two smallest prime factors of 250.

Factor 250:
\[
250 = 2 \times 125 = 2 \times 5^3.
\]

The prime factors are 2 and 5. The two smallest (and only) distinct prime factors are 2 and 5. Their sum is:
\[
2 + 5 = 7.
\]

A tempting wrong choice is 10, which comes from adding 5 and 5 (the two smallest factors with multiplicity), but the problem asks for the two smallest prime factors, i.e., distinct primes.

ANSWER 7: C

---

Problem 8:
Bag A has \(\{1, 3, 5\}\) and Bag B has \(\{2, 4, 6\}\). We draw one from each and want the number of distinct possible sums.

Compute all sums:
- \(1+2=3\), \(1+4=5\), \(1+6=7\)
- \(3+2=5\), \(3+4=7\), \(3+6=9\)
- \(5+2=7\), \(5+4=9\), \(5+6=11\)

The distinct values are:
\[
3, 5, 7, 9, 11.
\]

That is 5 different values. A tempting error is to count all 9 ordered pairs without removing duplicates.

ANSWER 8: B

---

Problem 9:
We have 2 boys and 3 girls to arrange in a row with a boy at each end and the three girls in the middle.

Positions: B _ _ _ G G G _ _ _ B (but specifically 5 seats: 1,2,3,4,5 with boys at 1 and 5, girls at 2,3,4).

- Choose the boy for the left end and the boy for the right end: \(2! = 2\) ways.
- Arrange the 3 girls in the middle three seats: \(3! = 6\) ways.

Total arrangements:
\[
2 \times 6 = 12.
\]

A tempting mistake is to ignore the constraint and compute \(5! = 120\), or to forget that the boys can swap places.

ANSWER 9: E

---

Problem 10:
Evaluate \(16 + 8 \div 4 - 2\).

Using order of operations (PEMDAS/BODMAS), division comes before addition and subtraction:
\[
8 \div 4 = 2.
\]

Now substitute back:
\[
16 + 2 - 2 = 16.
\]

A tempting error is to work left-to-right doing addition first: \((16+8)\div 4 - 2 = 24\div 4 - 2 = 6 - 2 = 4\), which is choice A. Another is to do \(16 + 2 = 18\), then \(18 - 2 = 16\), which luckily gives the same corr

Problem 1:
We need the rise in water level when a rock of volume 1000 cm³ is placed in an aquarium with base 100 cm by 40 cm. The rock displaces water equal to its volume, causing the water level to rise by Δh such that base area × Δh = rock volume. Base area = 100 cm × 40 cm = 4000 cm². So Δh = 1000 cm³ / 4000 cm² = 0.25 cm. The initial depth is 37 cm, and after rise it becomes 37.25 cm, which is still below the height of 50 cm, so the rock is fully submerged. The rise is 0.25 cm.

ANSWER 1: A

Problem 2:
When Annika and 3 friends play, there are 4 players, each dealt 15 cards. Total cards = 4 × 15 = 60. In the next game, 2 more friends join, making 6 players. Since the same total number of cards is used, each player gets 60 / 6 = 10 cards.

ANSWER 2: C

Problem 3:
The operation is defined as a c b d = a·d − b·c. For the numbers 3 1 4 2, we have a = 3, c = 1, b = 4, d = 2. Compute a·d − b·c = 3×2 − 4×1 = 6 − 4 = 2.

ANSWER 3: E

Problem 4:
Isabella wants an average of 95 over four tests, so total points needed = 4 × 95 = 380. Her first two scores sum to 97 + 91 = 188. Let the third score be x. After the third test, she can still reach the goal if it is possible to score at most 100 on the fourth test to reach 380. That requires 188 + x + 100 ≥ 380 → x ≥ 380 − 288 = 92. The smallest integer x satisfying this is 92. If x = 91, then even a perfect 100 on the fourth gives 188+91+100 = 379 < 380. So the lowest possible third score is 92.

ANSWER 4: B

Problem 5:
Let the number of apples Bridget bought be N. She gave half to Ann, leaving N/2. Then she gave Cassie 3 apples, leaving N/2 − 3. She keeps 4 apples, so N/2 − 3 = 4 → N/2 = 7 → N = 14.

ANSWER 5: E

Problem 6:
Let T be the total number of marbles. "All but 6 are red" means T − R = 6 → R = T − 6. Similarly, T − G = 8 → G = T − 8, and T − B = 4 → B = T − 4. Since R + G + B = T, we have (T−6)+(T−8)+(T−4) = T → 3T − 18 = T → 2T = 18 → T = 9.

ANSWER 6: C

Problem 7:
Factor 250: 250 = 2 × 5³. Its prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5, and their sum is 7.

ANSWER 7: C

Problem 8:
List all possible sums from drawing one chip from Bag A {1,3,5} and one from Bag B {2,4,6}:
1+2=3, 1+4=5, 1+6=7;
3+2=5, 3+4=7, 3+6=9;
5+2=7, 5+4=9, 5+6=11.
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.

ANSWER 8: B

Problem 9:
We need arrangements of 2 boys and 3 girls in a row with a boy at each end. Choose which boy sits on the left end and which on the right: 2! = 2 ways. The three girls occupy the three middle seats and can be arranged in 3! = 6 ways. Total arrangements = 2 × 6 = 12.

ANSWER 9: E

Problem 10:
The expression is 16 + 8 ÷ 4 − 2 (the missing operator is division, as typical for order-of-operations problems). Following order of operations, division before addition and subtraction: 8 ÷ 4 = 2. Then 16 + 2 = 18, and 18 − 2 = 16.

ANSWER 10: D

Problem 11:
Evaluate step by step: √4 = 2. Then 8 · 2 = 16, so √(8·√4) = √16 = 4. Then 16 · 4 = 64, so √(16·√(8·√4)) = √64 = 8.

ANSWER 11: C

Problem 12:
Compare the decimal numbers:
A. 0.99
B. 0.9099
C. 0.9
D. 0.909
E. 0.9009
Clearly 0.99 is the largest because it is greater than 0.9 and all the others are less than 0.91.

ANSWER 12: A

Problem 1:
The aquarium’s base has area  
\(A = 100\text{ cm} \times 40\text{ cm}=4000\text{ cm}^2\).  
When the rock of volume \(V=1000\text{ cm}^3\) is fully submerged, the water rises by a height \(h\) such that  

\[
A\cdot h = V \quad\Longrightarrow\quad h = \frac{1000}{4000}=0.25\text{ cm}.
\]

The rise cannot be any of the larger listed values (0.5 cm, 1 cm, …) because those would require a volume larger than the rock’s.  

ANSWER 1: A  

---

Problem 2:
With 4 players each receiving 15 cards, the total number of cards is  

\[
4 \times 15 = 60.
\]

If 2 more friends join, there are \(4+2 = 6\) players.  The cards are still all the cards, so each player receives  

\[
\frac{60}{6}=10\text{ cards}.
\]

The other answer choices (8, 9, 11, 12) would not distribute the 60 cards evenly.  

ANSWER 2: C  

---

Problem 3:
The operation is defined by  

\[
a\;c\;b\;d = a\cdot d - b\cdot c.
\]

For the numbers \(3\;1\;4\;2\) we have \(a=3,\;c=1,\;b=4,\;d=2\).  Hence  

\[
3\;1\;4\;2 = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

The other options \(-2,-1,0,1\) would correspond to different combinations that do not satisfy the given formula.  

ANSWER 3: E  

---

Problem 4:
To average 95 over four 100‑point tests, the total points needed are  

\[
4 \times 95 = 380.
\]

After the first two tests Isabella has  

\[
97 + 91 = 188\text{ points}.
\]

Let the third test score be \(x\).  Then the fourth test must supply  

\[
380 - (188 + x) = 192 - x \text{ points}.
\]

The fourth test can give at most 100 points, so we need  

\[
192 - x \le 100 \quad\Longrightarrow\quad x \ge 92.
\]

If she scores \(92\) on the third test, she would need a perfect 100 on the fourth, which is possible.  Any lower score would require more than 100 points on the fourth test, which is impossible.  

Thus the lowest possible third‑test score is **92**.  

ANSWER 4: B  

---

Problem 5:
Let the original number of apples be \(N\).  
She gave half of them to Ann: \(\frac{N}{2}\).  
She then gave Cassie 3 apples and kept 4, leaving  

\[
N - \frac{N}{2} - 3 = 4.
\]

Simplify: \(\frac{N}{2} - 3 = 4 \Rightarrow \frac{N}{2}=7 \Rightarrow N=14\).

If the total had been any of the other answer choices, the steps would not balance (e.g., 11 would leave 2 apples after giving Cassie 3).  

ANSWER 5: E  

---

Problem 6:
Let the total number of marbles be \(T\).  
- “All but 6 are red’’ → red = \(T-6\).  
- “All but 8 are green’’ → green = \(T-8\).  
- “All but 4 are blue’’ → blue = \(T-4\).

These three colors account for all marbles, so  

\[
(T-6)+(T-8)+(T-4)=T \;\Longrightarrow\; 3T-18=T \;\Longrightarrow\; 2T=18 \;\Longrightarrow\; T=9.
\]

The other totals (6, 8, 10, 12) would not satisfy the three “all but…’’ statements simultaneously.  

ANSWER 6: C  

---

Problem 7:
Factor \(250\):  

\[
250 = 2 \times 5^3.
\]

The prime factors are \(2\) and \(5\).  The two smallest are \(2\) and \(5\), whose sum is  

\[
2+5=7.
\]

Choices 2, 5, 10, 12 are either single prime factors or incorrect sums.  

ANSWER 7: C  

---

Problem 8:
Bag A contains \(\{1,3,5\}\); Bag B contains \(\{2,4,6\}\).  
All possible sums \(a+b\) are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The distinct results are \(\{3,5,7,9,11\}\), a total of **5** different sums.  

The other answer choices (4, 6, 7, 9) do not match the count of distinct sums.  

ANSWER 8: B  

---

Problem 9:
Place the two boys at the two ends and the three girls in the middle three seats.

- The two boys can be ordered in \(2!\) ways (left‑right swap).  
- The three girls can be arranged among the three middle seats in \(3! = 6\) ways.

Total arrangements  

\[
2! \times 3! = 2 \times 6 = 12.
\]

Choices 2, 4, 5, 6 are too small; 12 accounts for all possible permutations under the given constraints.  

ANSWER 9: E  

---

Problem 10:
Interpret the expression as  

\[
16 + 8 \div 4 - 2.
\]

Division comes before addition and subtraction, so  

\[
8 \div 4 = 2 \quad\Longrightarrow\quad 16 + 2 - 2 = 16.
\]

The other options (4, 8, 12, 20) would result only if the order of operations were misinterpreted.  

ANSWER 10: D  

---

Problem 11:
Simplify from the innermost radical outward.

\[
\sqrt{4}=2,\qquad 
\sqrt{8\cdot\sqrt{4}} = \sqrt{8\cdot2}= \sqrt{16}=4.
\]

Now  

\[
\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{16\cdot4}= \sqrt{64}=8.
\]

Thus the value is 8, matching choice C.  

ANSWER 11: C  

---

Problem 12:
Compare the decimal values:

- A. \(0.99\)
- B. \(0.9099\)
- C. \(0.9\)
- D. \(0.909\)
- E. \(0.9009\)

Clearly \(0.99\) is greater than \(0.9099\), which is greater than \(0.909\), etc.  
Hence the largest number is \(0.99\).  

ANSWER 12: A

Problem 1:  
An aquarium has a rectangular base of 100 cm by 40 cm and height 50 cm, filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is submerged. By how many centimeters does the water level rise?  

The base area is \(100 \times 40 = 4000 \, \text{cm}^2\). The rock’s volume is 1000 cm³, so the rise in water level is \(\frac{1000}{4000} = 0.25 \, \text{cm}\).  

Tempting choices: 0.5 (B) or 1 (C) might be guessed if base area is miscalculated.  

ANSWER 1: A  

Problem 2:  
Annika and 3 friends (4 players) get 15 cards each. Total cards = \(4 \times 15 = 60\). If 2 more join (6 players), each gets \(\frac{60}{6} = 10\).  

Tempting choices: 8 (A) if total cards miscalculated as 48.  

ANSWER 2: C  

Problem 3:  
Operation defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\): \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\), but the order is \(a c b d\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). However, the problem writes "a c b d" which is interpreted as \(a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). But the choices include 2 (E). Re-evaluating: the operation is \(a c b d = a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\).  

Tempting choices: -2 (A) if signs reversed.  

ANSWER 3: A (Note: Correction based on standard interpretation: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\), but the correct answer based on the operation defined is actually \(3 \cdot 2 - 1 \cdot 4 = 2\), but the problem writes "a c b d", which is \(a \cdot d - b \cdot c\), so 2. However, the answer is 2, but the choices include 2 as E. The correct calculation is 2, so E. Correction: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\).)  

**Correction:** The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\), this is \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\).  

ANSWER 3: E  

Problem 4:  
Isabella wants an average of 95 on four 100-point tests. Total points needed: \(4 \times 95 = 380\). First two scores: 97 + 91 = 188. Let third score be \(x\), fourth be \(y\). Then \(188 + x + y = 380\), so \(x + y = 192\). To minimize \(x\), maximize \(y\): \(y = 100\), so \(x = 92\).  

Tempting choices: 90 (A) if \(y\) not maximized.  

ANSWER 4: B  

Problem 5:  
Bridget gave half apples to Ann, then 3 to Cassie, keeping 4. Let total be \(x\). After giving half to Ann: \(\frac{x}{2}\) left. Then gives 3 to Cassie: \(\frac{x}{2} - 3 = 4\). So \(\frac{x}{2} = 7\), \(x = 14\).  

Tempting choices: 7 (C) if half misinterpreted.  

ANSWER 5: E  

Problem 6:  
Let \(r, g, b\) be red, green, blue marbles. Then:  
- All but 6 are red: \(g + b = 6\)  
- All but 8 are green: \(r + b = 8\)  
- All but 4 are blue: \(r + g = 4\)  
Adding: \(2(r + g + b) = 18\), so total \(r + g + b = 9\).  

Tempting choices: 6 (A) if equations misread.  

ANSWER 6: C  

Problem 7:  
Prime factors of 250: \(250 = 2 \times 5^3\). Two smallest: 2 and 5. Sum: \(2 + 5 = 7\).  

Tempting choices: 5 (B) if only 5 considered.  

ANSWER 7: C  

Problem 8:  
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}. Possible sums:  
- 1+2=3, 1+4=5, 1+6=7  
- 3+2=5, 3+4=7, 3+6=9  
- 5+2=7, 5+4=9, 5+6=11  
Unique sums: 3, 5, 7, 9, 11 → 5 values.  

Tempting choices: 6 (C) if duplicates counted.  

ANSWER 8: B  

Problem 9:  
Two boys at ends: 2 ways. Three girls in middle: \(3! = 6\) ways. Total: \(2 \times 6 = 12\).  

Tempting choices: 6 (D) if boys not permuted.  

ANSWER 9: E  

Problem 10:  
Expression: \(16 + 8 \times 4 - 2\). Order: \(8 \times 4 = 32\), then \(16 + 32 - 2 = 46\)? Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 \times 4 - 2\): \(8 \times 4 = 32\), \(16 + 32 = 48\), \(48 - 2 = 46\). But choices max at 20. Re-interpreting: "16 + 8 4 - 2" might mean \(16 + 8 - 4 - 2 = 18\), not in choices. Alternatively, if "8 4" is 8×4, but 46 not in choices. The expression is likely \(16 + 8 - 4 - 2 = 18\), but not in choices. Correction: The expression is "16 + 8 4 − 2", which might be \(16 + (8 \times 4) - 2 = 16 + 32 - 2 = 46\), but not in choices. Re-reading: It’s "16 + 8 4 − 2", which could be \(16 + 8 - 4 - 2 = 18\), but not in choices. The intended expression is likely \(16 + 8 \times 4 - 2 = 46\), but choices max at 20. Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 - 4 - 2 = 18\), but not in choices. Alternatively, if "8 4" is 84, but 16+84-2=98. Not matching. The expression is likely \( \frac{16 + 8}{4} - 2 = \frac{24}{4} - 2 = 6 - 2 = 4 \).  

Tempting choices: 8 (B) if order misapplied.  

ANSWER 10: A  

Problem 11:  
Expression: \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\).  
- \(\sqrt{4} = 2\)  
- \(\sqrt{8 \cdot 2} = \sqrt{16} = 4\)  
- \(\sqrt{16 \cdot 4} = \sqrt{64} = 8\)  

Tempting choices: 4 (A) if innermost not simplified.  

ANSWER 11: C  

Problem 12:  
Compare decimals:  
- A: 0.99  
- B: 0.9099  
- C: 0.9  
- D: 0.909  
- E: 0.9009  
Largest is 0.99.  

Tempting choices: 0.9099 (B) if decimals misaligned.  

ANSWER 12: A  

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

### Problem 1:
What is asked: An aquarium with a 100cm×40cm rectangular base is filled with water to 37cm depth. A 1000 cm³ rock is submerged; how much does the water level rise?
Step-by-step: When a submerged object displaces water, its volume equals the additional volume of water in the aquarium. The rise in water level Δh satisfies: Volume of rock = base area of aquarium × Δh.
Base area = 100×40 = 4000 cm².
Δh = 1000 / 4000 = 0.25 cm.
Tempting wrong answers: 0.5 (if base area is miscalculated as 2000 cm²) or 1 (wrong division).
ANSWER 1: A

---

### Problem 2:
What is asked: Annika plus 3 friends (4 total players) each get 15 cards. If 2 more join, how many cards does each player get?
Step-by-step: First calculate total number of cards: 4 players × 15 cards = 60 total cards. After 2 join, total players = 4+2=6. Cards per player = 60 / 6 = 10.
Tempting wrong answers: 9 (if you incorrectly count only 3 initial players, leading to 45 total cards) or 12 (wrong division).
ANSWER 2: C

---

### Problem 3:
What is asked: The custom notation a c b d = a·d − b·c. What is the value of 3 1 4 2?
Step-by-step: Match the values to the definition: a=3, c=1, b=4, d=2. Calculate: 3·2 - 4·1 = 6 - 4 = 2.
Tempting wrong answers: -2 (swap a and b incorrectly) or 0 (miscalculate the multiplication).
ANSWER 3: E

---

### Problem 4:
What is asked: Isabella needs a 95 average on 4 100-point tests. She scored 97 and 91 on the first two. What is the lowest possible third test score that lets her reach her goal?
Step-by-step: Total points needed for 95 average: 4×95 = 380 points. Sum of first two tests: 97+91=188. To minimize the third test score, maximize the fourth test (max 100 points). Total points left for tests 3 and 4: 380-188=192. Minimum third test = 192 - 100 = 92.
Tempting wrong answers: 96 (the average of the remaining two tests, not the minimum possible third score).
ANSWER 4: B

---

### Problem 5:
What is asked: Bridget gave half her apples to Ann, 3 to Cassie, and kept 4. How many apples did she buy originally?
Step-by-step: Let N = original number of apples. After giving half to Ann, she has N/2 left. Then N/2 - 3 = 4 → N/2=7 → N=14. Verify: 14 → 7 left after Ann, minus 3 for Cassie = 4 kept, which matches.
Tempting wrong answers: 7 (the number of apples Bridget had after giving half to Ann, not the original total).
ANSWER 5: E

---

### Problem 6:
What is asked: A jar has red, green, blue marbles. All but 6 are red, all but 8 are green, all but 4 are blue. How many total marbles?
Step-by-step: Let T=total marbles, R=red, G=green, B=blue. The conditions translate to: G+B=6, R+B=8, R+G=4. Add all equations: 2(R+G+B)=18 → R+G+B=T=9. Verify: 3 red, 1 green, 5 blue sum to 9, matching all conditions.
Tempting wrong answers: 18 (the sum of 6+8+4, not halved to get total marbles).
ANSWER 6: C

---

### Problem 7:
What is asked: What is the sum of the two smallest prime factors of 250?
Step-by-step: Prime factorization of 250: 250=2×5³. The prime factors are 2 and 5. Their sum is 2+5=7.
Tempting wrong answers: 10 (sum of 5+5, or wrong addition of 2+5).
ANSWER 7: C

---

### Problem 8:
What is asked: Bag A has chips 1,3,5; Bag B has 2,4,6. One chip drawn from each bag, how many distinct sum values are possible?
Step-by-step: List all sums: 1+2=3, 1+4=5,1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7,5+4=9,5+6=11. Distinct sums: 3,5,7,9,11 → 5 total.
Tempting wrong answers: 9 (total number of unordered pairs, not distinct sums).
ANSWER 8: B

---

### Problem 9:
What is asked: 2 boys and 3 girls line up for a photo, with a boy at each end and girls in the middle. How many arrangements are possible?
Step-by-step: Arrange the boys first: 2 boys can be ordered at the ends in 2! = 2 ways. The 3 middle girls can be ordered in 3! =6 ways. Total arrangements: 2×6=12.
Tempting wrong answers: 6 (only the number of ways to arrange the girls, ignoring the boys).
ANSWER 9: E

---

### Problem 10:
What is asked: Evaluate (16+8)/(4-2) (formatted as a fraction in the original problem).
Step-by-step: Calculate numerator:16+8=24, denominator:4-2=2. 24/2=12.
Tempting wrong answers:16 (if order of operations is misapplied to calculate 16 + 8/4 -2=16).
ANSWER 10: C

---

### Problem 11:
What is asked: Evaluate √(16 · √(8 · √4)).
Step-by-step: Calculate from the innermost root first: √4=2. Next: 8×2=16, √16=4. Next:16×4=64, √64=8.
Tempting wrong answers: 4√2 (incorrect root calculation).
ANSWER 11: C

---

### Problem 12:
What is asked: Which number is largest: 0.99, 0.9099, 0.9, 0.909, 0.9009.
Step-by-step: Line up all decimals to compare place values: 0.9900 > 0.9099 > 0.9090 > 0.9009 > 0.9000. The largest is 0.99.
Tempting wrong answers: 0.9099 (the second largest value).
ANSWER 12: A

### Problem 1:
We need to find how much the water level rises when a 1000 cm³ rock is submerged in the aquarium.
1.  The volume of displaced water equals the volume of the submerged rock: 1000 cm³.
2.  The base area of the aquarium is \(100 \, \text{cm} \times 40 \, \text{cm} = 4000 \, \text{cm}^2\).
3.  Let \(h\) be the rise in water level. The volume of the displaced water is also equal to base area × rise in height: \(4000h = 1000\).
4.  Solve for \(h\): \(h = \frac{1000}{4000} = 0.25\) cm.
ANSWER 1: A
---
### Problem 2:
We need to find the number of cards per player when 2 more friends join.
1.  Original number of players: Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).
2.  New number of players: 4 + 2 = 6 players. Total cards remain 60.
3.  Cards per player: \(\frac{60}{6} = 10\).
ANSWER 2: C
---
### Problem 3:
The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). Compute \(3 \, 1 \, 4 \, 2\).
1.  Identify \(a=3, c=1, b=4, d=2\).
2.  Substitute into the formula: \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).
ANSWER 3: E
---
### Problem 4:
Find the lowest possible third test score to get an average of 95 on 4 tests.
1.  Total points needed for an average of 95: \(4 \times 95 = 380\).
2.  Sum of first two test scores: \(97 + 91 = 188\).
3.  Let \(x\) = third test score, \(y\) = fourth test score (max 100, since tests are 100-point). We need \(188 + x + y = 380\), so \(x + y = 192\).
4.  To minimize \(x\), maximize \(y = 100\). Then \(x = 192 - 100 = 92\).
ANSWER 4: B
---
### Problem 5:
Find the original number of apples Bridget bought.
1.  Let \(x\) = total apples. After giving half to Ann, she has \(\frac{x}{2}\) left.
2.  After giving 3 to Cassie, she has \(\frac{x}{2} - 3\) left, which equals 4 (the apples she kept).
3.  Solve: \(\frac{x}{2} - 3 = 4 \implies \frac{x}{2} =7 \implies x=14\).
ANSWER 5: E
---
### Problem 6:
Find the total number of marbles.
1.  Let \(R, G, B\) be the number of red, green, blue marbles, total \(T = R+G+B\).
2.  Translate the statements:
    - All but 6 are red: \(G + B = 6\)
    - All but 8 are green: \(R + B = 8\)
    - All but 4 are blue: \(R + G =4\)
3.  Add all three equations: \(2(R+G+B) = 6+8+4=18 \implies 2T=18 \implies T=9\).
ANSWER 6: C
---
### Problem 7:
Find the sum of the two smallest prime factors of 250.
1.  Factorize 250: \(250 = 2 \times 125 = 2 \times 5^3\).
2.  The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.
3.  Sum: \(2 +5=7\).
ANSWER 7: C
---
### Problem 8:
Find the number of distinct sums when drawing one chip from each bag.
1.  List all possible pairs and sums:
    - Bag A (1,3,5) + Bag B (2,4,6):
      \(1+2=3, 1+4=5, 1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7, 5+4=9, 5+6=11\)
2.  Distinct sums: 3, 5, 7, 9, 11. There are 5 unique values.
ANSWER 8: B
---
### Problem 9:
Find the number of arrangements with boys at each end and girls in the middle.
1.  Arrange the 2 boys at the ends: there are \(2! = 2\) ways (order matters: which boy is left/right).
2.  Arrange the 3 girls in the middle 3 seats: \(3! = 6\) ways.
3.  Total arrangements: \(2 \times 6 =12\).
ANSWER 9: E
---
### Problem 10:
Compute \(16 + 8 \, 4 - 2\). The missing operator between 8 and 4 is division (standard order of operations, common middle school problem):
1.  Interpret as \(16 + \frac{8}{4} - 2\).
2.  Compute division first: \(\frac{8}{4}=2\).
3.  Then add/subtract left to right: \(16 + 2 - 2 =16\).
*(Note: If interpreted as \(16 +8 -4 \times 2\), order of operations gives \(4 \times 2=8\), so \(16+8-8=16\), same result.)*
ANSWER 10: D
---
### Problem 11:
Compute \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\) by working from the innermost root outward.
1.  Innermost: \(\sqrt{4} = 2\).
2.  Next layer: \(\sqrt{8 \cdot 2} = \sqrt{16} =4\).
3.  Outermost: \(\sqrt{16 \cdot 4} = \sqrt{64} =8\).
ANSWER 11: C
---
### Problem 12:
Find the largest decimal.
1.  Rewrite all numbers with 4 decimal places for easy comparison:
    - A: \(0.9900\)
    - B: \(0.9099\)
    - C: \(0.9000\)
    - D: \(0.9090\)
    - E: \(0.9009\)
2.  Compare digit by digit: all have 9 in the tenths place. A has 9 in the hundredths place, while all others have 0. Thus A is the largest.
ANSWER 12: A

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

**Problem 1:**

What is being asked: By how many centimeters does the water level rise when a 1000 cm³ rock is submerged?

Setup: The base area is 100 cm × 40 cm = 4000 cm². When the rock is submerged, it displaces 1000 cm³ of water.

Water level rise = Volume displaced ÷ Base area = 1000 ÷ 4000 = 0.25 cm

ANSWER 1: A

---

**Problem 2:**

What is being asked: If cards are distributed evenly, how many cards per player when the player count increases?

Setup: Initially 4 players (Annika + 3 friends) each get 15 cards.
Total cards = 4 × 15 = 60 cards

With 2 more friends joining: 4 + 2 = 6 players
Cards per player = 60 ÷ 6 = 10 cards

ANSWER 2: C

---

**Problem 3:**

What is being asked: Evaluate the operation 3 1 4 2 using the rule a c b d = a·d − b·c

Setup: With a = 3, c = 1, b = 4, d = 2:
3 1 4 2 = 3·2 − 4·1 = 6 − 4 = 2

ANSWER 3: E

---

**Problem 4:**

What is being asked: What is the lowest score on test 3 that allows Isabella to achieve a 95 average on four tests?

Setup: Goal average is 95, so total needed = 95 × 4 = 380 points
First two tests: 97 + 91 = 188 points
Points needed from tests 3 and 4: 380 − 188 = 192 points

Let x = score on test 3. She can still reach her goal if she can score 100 on test 4:
x + 100 ≥ 192
x ≥ 92

The lowest possible score on test 3 is 92.

ANSWER 4: B

---

**Problem 5:**

What is being asked: How many apples did Bridget buy originally?

Setup: Let x = original number of apples
- She gave half to Ann: x/2 apples given away
- Remaining: x/2
- She gave 3 to Cassie: x/2 − 3
- She kept 4 for herself: x/2 − 3 = 4

Solving: x/2 = 7, so x = 14

ANSWER 5: E

---

**Problem 6:**

What is being asked: How many marbles are in the jar?

Setup: Let R, G, B = numbers of red, green, blue marbles; T = total
- "All but 6 are red": G + B = 6
- "All but 8 are green": R + B = 8
- "All but 4 are blue": R + G = 4

Adding all three equations: 2(R + G + B) = 6 + 8 + 4 = 18
Therefore: R + G + B = 9

ANSWER 6: C

---

**Problem 7:**

What is being asked: What is the sum of the two smallest prime factors of 250?

Setup: Factor 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5³

Prime factors: 2 and 5
Sum = 2 + 5 = 7

ANSWER 7: C

---

**Problem 8:**

What is being asked: How many different sums are possible?

Setup: Bag A: {1, 3, 5}, Bag B: {2, 4, 6}
All possible sums:
- 1 + 2 = 3
- 1 + 4 = 5
- 1 + 6 = 7
- 3 + 2 = 5 (duplicate)
- 3 + 4 = 7 (duplicate)
- 3 + 6 = 9
- 5 + 2 = 7 (duplicate)
- 5 + 4 = 9 (duplicate)
- 5 + 6 = 11

Distinct sums: {3, 5, 7, 9, 11} = 5 different values

ANSWER 8: B

---

**Problem 9:**

What is being asked: How many seating arrangements with boys at ends and girls in middle?

Setup: 2 boys at the ends can be arranged in 2! = 2 ways
3 girls in the middle can be arranged in 3! = 6 ways
Total arrangements = 2 × 6 = 12

ANSWER 9: E

---

**Problem 10:**

What is being asked: Evaluate 16 + 8 ÷ 4 − 2

Setup: Using order of operations (division before addition/subtraction):
16 + (8 ÷ 4) − 2 = 16 + 2 − 2 = 16

ANSWER 10: D

---

**Problem 11:**

What is being asked: Evaluate √(16 · √(8 · √4))

Setup: Work from the inside out:
- √4 = 2
- 8 · √4 = 8 · 2 = 16
- √(8 · √4) = √16 = 4
- 16 · √(8 · √4) = 16 · 4 = 64
- √(16 · √(8 · √4)) = √64 = 8

ANSWER 11: C

---

**Problem 12:**

What is being asked: Which decimal is largest?

Setup: Compare decimals:
- A: 0.99
- B: 0.9099
- C: 0.9
- D: 0.909
- E: 0.9009

Comparing: 0.99 > 0.9099 > 0.909 > 0.9009 > 0.9

The largest is 0.99

ANSWER 12: A

Problem 1:

The aquarium has a rectangular base of \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is

\[
100 \cdot 40 = 4000 \text{ cm}^2.
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

If the water rises by \(h\) centimeters, then

\[
4000h = 1000.
\]

So

\[
h = \frac{1000}{4000} = \frac14 = 0.25.
\]

So the water level rises \(0.25\) cm.

ANSWER 1: A

---

Problem 2:

Annika and 3 friends means there are

\[
4 \text{ players}.
\]

If each player gets 15 cards, then the total number of cards is

\[
4 \cdot 15 = 60.
\]

Now 2 more friends join, so there are

\[
6 \text{ players}.
\]

The same 60 cards are shared evenly, so each gets

\[
60 \div 6 = 10.
\]

ANSWER 2: C

---

Problem 3:

We are told that if the four numbers are \(a, b, c, d\), then

\[
a \ b \ c \ d = a\cdot d - b\cdot c.
\]

So for \(3,1,4,2\),

\[
3 \ 1 \ 4 \ 2 = 3\cdot 2 - 1\cdot 4 = 6 - 4 = 2.
\]

ANSWER 3: E

---

Problem 4:

She wants an average of 95 on 4 tests, so the total of the 4 scores must be

\[
4 \cdot 95 = 380.
\]

Her first two scores are 97 and 91, which add to

\[
97+91=188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To make the total 380, the fourth score would have to be

\[
380 - (188+x) = 192 - x.
\]

For this to be possible, the fourth score must be at most 100:

\[
192 - x \le 100.
\]

So

\[
x \ge 92.
\]

The lowest possible third score is 92.

ANSWER 4: B

---

Problem 5:

Let the number of apples Bridget bought be \(x\).

She gave half to Ann, so she kept the other half:

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

Then she gave Cassie 3 apples and kept 4 apples, so before giving Cassie the apples she had

\[
3+4=7.
\]

Thus,

\[
\frac{x}{2}=7 \quad \Rightarrow \quad x=14.
\]

ANSWER 5: E

---

Problem 6:

Let the total number of marbles be \(n\).

- “All but 6 are red” means the number of red marbles is \(n-6\).
- “All but 8 are green” means the number of green marbles is \(n-8\).
- “All but 4 are blue” means the number of blue marbles is \(n-4\).

Since every marble is one of these colors, their counts add to \(n\):

\[
(n-6)+(n-8)+(n-4)=n.
\]

Simplify:

\[
3n-18=n
\]

\[
2n=18
\]

\[
n=9.
\]

ANSWER 6: C

---

Problem 7:

We need the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3.
\]

So the prime factors are 2 and 5. The two smallest prime factors are 2 and 5, and their sum is

\[
2+5=7.
\]

ANSWER 7: C

---

Problem 8:

Bag A has \(1,3,5\); Bag B has \(2,4,6\).

Possible sums:

- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums are

\[
3,5,7,9,11
\]

which is 5 different values.

ANSWER 8: B

---

Problem 9:

There are 2 boys and 3 girls, and the picture must have:

- a boy at each end,
- the 3 girls in the middle.

First, arrange the 2 boys at the ends. They can be switched in

\[
2! = 2
\]

ways.

Then arrange the 3 girls in the middle in

\[
3! = 6
\]

ways.

So the total number of arrangements is

\[
2 \cdot 6 = 12.
\]

ANSWER 9: E

---

Problem 10:

The expression is

\[
16 + 8\cdot 4 - 2.
\]

Use order of operations:

\[
8\cdot 4=32.
\]

Then

\[
16+32-2=46.
\]

But 46 is not among the choices, so the intended expression must be read as

\[
16 + 8 \div 4 - 2.
\]

Then

\[
8\div 4 = 2,
\]

so

\[
16+2-2=16.
\]

ANSWER 10: D

---

Problem 11:

We evaluate from the inside out:

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

First,

\[
\sqrt4=2.
\]

So the expression becomes

\[
\sqrt{16\cdot \sqrt{8\cdot 2}}
=
\sqrt{16\cdot \sqrt{16}}.
\]

Since \(\sqrt{16}=4\),

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C

---

Problem 12:

Compare the decimals:

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

The largest is the one closest to 1. Since \(0.99\) is greater than any number beginning with \(0.90\), it is the largest.

ANSWER 12: A

Problem 1:  
We need the rise in water level when a rock is placed in the aquarium and submerged.

- Base dimensions: \(100\text{ cm} \times 40\text{ cm}\), so base area  
  \[
  A = 100\cdot 40 = 4000\text{ cm}^2.
  \]
- The rock is completely submerged, so the water volume displaced equals the rock’s volume:  
  \[
  V = 1000\text{ cm}^3.
  \]
- Water level rise \(\Delta h\) satisfies \(A\Delta h = V\):  
  \[
  4000\cdot \Delta h = 1000 \implies \Delta h = \frac{1000}{4000}=0.25\text{ cm}.
  \]

So the rise is \(0.25\) cm.  
Tempting wrong idea: using the *remaining* depth \(37\) cm or checking if it overflows—none of that matters; only displaced volume matters.

ANSWER 1: A  

---

Problem 2:  
All cards are dealt evenly. Let the total number of cards be \(T\).

Initially, Annika + 3 friends = 4 players, and each gets 15 cards:  
\[
T = 4\cdot 15 = 60.
\]
Next game: 2 more friends join, so total players:
\[
4+2=6.
\]
Each player gets:
\[
\frac{60}{6}=10.
\]

ANSWER 2: C  

---

Problem 3:  
Given the rule \(a\,b\,d\,c = a\cdot d - b\cdot c\).  
The expression “\(3\ 1\ 4\ 2\)” corresponds to:
- \(a=3\)
- \(b=1\)
- \(d=4\)
- \(c=2\)

Compute:
\[
a\cdot d - b\cdot c = 3\cdot 4 - 1\cdot 2 = 12-2=10.
\]

But notice the choices are only \(-2,-1,0,1,2\), so we must interpret the placeholder correctly:  
The problem states: **“If \(a b c d = a\cdot d - b\cdot c\)”**.

So “\(3\ 1\ 4\ 2\)” means:
- \(a=3,\ b=1,\ c=4,\ d=2\)

Then:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 1\cdot 4 = 6-4=2.
\]

ANSWER 3: E  

---

Problem 4:  
Goal average is 95 over 4 tests, so total points needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
Let the third test score be \(x\), and the fourth be \(y\). Then:
\[
188 + x + y = 380 \implies x+y=192.
\]
We want the **lowest possible** \(x\) such that she can still reach the goal. That means make \(y\) as large as possible (since grades are at most 100).
Maximum \(y=100\). Then:
\[
x + 100 = 192 \implies x = 92.
\]

Check: then the four scores are \(97,91,92,100\) which sum to \(380\), average \(95\). Works.

ANSWER 4: B  

---

Problem 5:  
Let the total number of apples be \(N\).

- She gave half to Ann, so Ann got \(N/2\), leaving \(N/2\).
- Then she gave Cassie 3 apples, and kept 4 apples. That means after giving Cassie 3, her remaining is 4:
  \[
  \frac{N}{2} - 3 = 4 \implies \frac{N}{2}=7 \implies N=14.
  \]

ANSWER 5: E  

---

Problem 6:  
“All but 6 are red” means: number of red marbles is 6.  
Similarly:
- Red = 6 (since all but 6 are red → 6 red)
- Green: all but 8 are green → 8 green
- Blue: all but 4 are blue → 4 blue

Total marbles:
\[
6+8+4=18.
\]
But none of the choices are 18, so interpret carefully:  
“All but 6 are red” means the number of red marbles is **(total − 6)**, not 6.

Let total be \(T\). Then:
- Red marbles = \(T-6\)
- Green marbles = \(T-8\)
- Blue marbles = \(T-4\)

Sum:
\[
(T-6)+(T-8)+(T-4)=T.
\]
\[
3T - (6+8+4)=T
\]
\[
3T-18=T \implies 2T=18 \implies T=9.
\]

ANSWER 6: C  

---

Problem 7:  
We want the sum of the **two smallest prime factors** of 250.

Factor \(250\):
\[
250 = 25\cdot 10 = 5^2 \cdot 2\cdot 5 = 2\cdot 5^3.
\]
Prime factors are \(2\) and \(5\). The two smallest are \(2\) and \(5\).

Sum:
\[
2+5=7.
\]

ANSWER 7: D  

---

Problem 8:  
Bag A: \(\{1,3,5\}\)  
Bag B: \(\{2,4,6\}\)

Possible sums:
- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums: \(\{3,5,7,9,11\}\) → that’s 5 values.

ANSWER 8: B  

---

Problem 9:  
We need arrangements of 5 people in a row with:
- a boy at each end (positions 1 and 5)
- the three girls in the middle (positions 2,3,4)

There are 2 boys: they can be placed at the ends in
\[
2!\ = 2
\]
ways (left boy/right boy).

The 3 girls can be arranged in the three middle spots in
\[
3! = 6
\]
ways.

Total arrangements:
\[
2\cdot 6 = 12.
\]

ANSWER 9: E  

---

Problem 10:  
Expression: \(16 + 8^4 - 2\).

Compute in order:
- \(8^4 = 4096\)

So:
\[
16 + 4096 - 2 = 4110.
\]
No answer choice matches, so we interpret the formatting as \(16 + 8 \sqrt{4} - 2\) or \(16 + 8^4 - 2\)?  
Given the AMC/JH style, likely it is:
\[
16 + 8\sqrt{4} - 2.
\]
Then:
\[
\sqrt{4}=2 \implies 8\sqrt{4}=8\cdot 2=16.
\]
So:
\[
16 + 16 - 2 = 30,
\]
still not in choices.

Another common formatting is:
\[
16 + 8^{4} - 2 \quad \text{(still not)}
\]

The only choiceable consistent possibility is:
\[
16 + \sqrt{8}\,^4 - 2 \text{ (nonsense)}
\]

Let’s use the most plausible literal from the text: **“16 + 8 4 − 2”** often means \(16 + 8^4 - 2\) if the caret was lost. But then no choice.

Given typical AMC problems, the intended is usually:
\[
16 + 8 - 4 - 2 = 18
\]
also not.

Because the answer choices are only \(4,8,12,16,20\), the intended expression must simplify to one of those. The most natural is:
\[
16 + 8/4 - 2 = 16 + 2 - 2 = 16.
\]
This matches a common missing symbol: “8 4” often means \(8/4\).

So:
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D  

---

Problem 11:  
Evaluate:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}.
\]

Work from inside out:

1) \(\sqrt{4}=2\)

2) Inside next root:
\[
8\cdot \sqrt{4} = 8\cdot 2 = 16
\]
So:
\[
\sqrt{8\cdot \sqrt{4}} = \sqrt{16}=4.
\]

3) Now plug into the outer expression:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}=\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C  

---

Problem 12:  
Largest number among:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare by decimal places:

- .99 is 0.99
- .9099 is 0.9099
- .9 is 0.9000
- .909 is 0.9090
- .9009 is 0.9009

Clearly \(0.99\) is greater than all the numbers less than \(0.91\).

ANSWER 12: A

Problem 1:
The volume of the water displaced by the rock is equal to the volume of the rock itself, which is $1000 \text{ cm}^3$. The aquarium has a rectangular base with area $A = 100 \text{ cm} \times 40 \text{ cm} = 4000 \text{ cm}^2$. When the rock is submerged, the water level rises by height $h$ such that the volume of the displaced water is $A \times h$. 
Setting this equal to the rock's volume: $4000 \times h = 1000$.
Solving for $h$: $h = \frac{1000}{4000} = \frac{1}{4} = 0.25 \text{ cm}$.
ANSWER 1: A

Problem 2:
Annika and 3 friends make 4 players total. Each received 15 cards, so the total number of cards is $4 \times 15 = 60$. If 2 more friends join, the total number of players becomes $4 + 2 = 6$. The cards are distributed evenly among the 6 players, so each player gets $60 \div 6 = 10$ cards.
ANSWER 2: C

Problem 3:
The problem defines the operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$.
For the expression $\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix}$, we have $a=3, b=1, c=4, d=2$.
Calculating: $(3 \times 2) - (1 \times 4) = 6 - 4 = 2$.
ANSWER 3: E

Problem 4:
Isabella needs an average of 95 over 4 tests, meaning the total sum of her scores must be $95 \times 4 = 380$. Her first two scores are 97 and 91, totaling $97 + 91 = 188$. Let $x$ be the score on the third test and $y$ be the score on the fourth. We have $188 + x + y = 380$, so $x + y = 192$. To make $x$ as small as possible, we must make $y$ as large as possible. Since the maximum score on a test is 100, we set $y = 100$. Then $x + 100 = 192$, so $x = 92$.
ANSWER 4: B

Problem 5:
Let $x$ be the total number of apples Bridget bought. She gave half away, leaving $\frac{x}{2}$. Then she gave 3 away, leaving $\frac{x}{2} - 3$. She kept 4, so $\frac{x}{2} - 3 = 4$. Adding 3 to both sides, $\frac{x}{2} = 7$. Multiplying by 2, $x = 14$.
ANSWER 5: E

Problem 6:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R + G + B$.
"All but 6 are red" means $G + B = 6$.
"All but 8 are green" means $R + B = 8$.
"All but 4 are blue" means $R + G = 4$.
Adding these three equations: $2(R + G + B) = 6 + 8 + 4 = 18$.
So $R + G + B = 18 \div 2 = 9$.
ANSWER 6: C

Problem 7:
The prime factorization of 250 is $250 = 2 \times 125 = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two prime factors is $2 + 5 = 7$.
ANSWER 7: C

Problem 8:
Bag A has $\{1, 3, 5\}$ and Bag B has $\{2, 4, 6\}$.
Possible sums:
$1+2=3, 1+4=5, 1+6=7$
$3+2=5, 3+4=7, 3+6=9$
$5+2=7, 5+4=9, 5+6=11$
The set of distinct sums is $\{3, 5, 7, 9, 11\}$. There are 5 distinct values.
ANSWER 8: B

Problem 9:
There are 2 boys (B1, B2) and 3 girls (G1, G2, G3). A boy must be at each end. There are 2 choices for the left end and 1 choice for the right end ($2 \times 1 = 2$ ways to place boys). The 3 girls can be arranged in the middle in $3! = 3 \times 2 \times 1 = 6$ ways. Total arrangements = $2 \times 6 = 12$.
ANSWER 9: E

Problem 10:
Following the order of operations (PEMDAS/BODMAS):
$16 + (8 \div 4) - 2 = 16 + 2 - 2 = 16$.
ANSWER 10: D

Problem 11:
Evaluate from the inside out:
$\sqrt{4} = 2$
$\sqrt{8 \times 2} = \sqrt{16} = 4$
$\sqrt{16 \times 4} = \sqrt{64} = 8$
ANSWER 11: C

Problem 12:
To compare, pad with zeros to match the number of decimal places:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
Comparing the values, 0.9900 is the largest.
ANSWER 12: A

Problem 1:
The base of the aquarium measures 100 cm by 40 cm, so its area is \(100 \times 40 = 4000\) cm². The rock has volume 1000 cm³ and is fully submerged, displacing exactly 1000 cm³ of water. The resulting rise in water level is therefore the displaced volume divided by the base area: \(1000 / 4000 = 0.25\) cm. (The original water depth and tank height are irrelevant once the surface area is known.)

ANSWER 1: A

Problem 2:
Annika plus 3 friends gives 4 players, each receiving 15 cards, for a total of \(4 \times 15 = 60\) cards. When 2 more players join, there are 6 players sharing the same 60 cards, so each receives \(60 / 6 = 10\) cards.

ANSWER 2: C

Problem 3:
The given operation evaluates \(a\, c\, b\, d\) as \(a \cdot d - b \cdot c\). Substituting the numbers yields \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).

ANSWER 3: E

Problem 4:
An average of 95 on four 100-point tests requires a total of \(4 \times 95 = 380\) points. The first two scores sum to \(97 + 91 = 188\), leaving 192 points still needed from the last two tests. The fourth test can score at most 100, so the third test must be at least \(192 - 100 = 92\). Any lower score on the third test would make a total of 380 impossible.

ANSWER 4: B

Problem 5:
Let \(t\) be the original number of apples. After giving half to Ann, \(t/2\) remain. After giving 3 more to Cassie, \(t/2 - 3\) remain, and this equals the 4 apples kept. Solving \(t/2 - 3 = 4\) gives \(t/2 = 7\), so \(t = 14\).

ANSWER 5: E

Problem 6:
Let \(m\) be the total number of marbles. Then there are \(m-6\) red, \(m-8\) green, and \(m-4\) blue. Their sum must equal \(m\):
\[
(m-6) + (m-8) + (m-4) = m \implies 3m - 18 = m \implies 2m = 18 \implies m = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies the three “all-but” statements.

ANSWER 6: C

Problem 7:
Factorize \(250 = 2 \cdot 5^3\). The two smallest distinct prime factors are 2 and 5; their sum is 7.

ANSWER 7: C

Problem 8:
All nine possible sums are:
\[
1+2=3,\ 1+4=5,\ 1+6=7,\ 3+2=5,\ 3+4=7,\ 3+6=9,\ 5+2=7,\ 5+4=9,\ 5+6=11.
\]
The distinct values are 3, 5, 7, 9, 11—exactly five.

ANSWER 8: B

Problem 9:
The seating pattern must be boy–girl–girl–girl–boy. The two distinct boys may be placed at the ends in \(2!\) = 2 ways. The three distinct girls may be arranged in the middle three seats in \(3!\) = 6 ways. The total number of arrangements is therefore \(2 \times 6 = 12\).

ANSWER 9: E

Problem 10:
Interpreting the expression with standard order of operations and treating juxtaposition of 8 and 4 as division yields
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D

Problem 11:
Evaluate from the inside outward:
\[
\sqrt{4} = 2, \quad 8 \cdot 2 = 16, \quad \sqrt{16} = 4, \quad 16 \cdot 4 = 64, \quad \sqrt{64} = 8.
\]

ANSWER 11: C

Problem 12:
Compare the decimals directly:
\[
0.99 > 0.9 > 0.9099 > 0.909 > 0.9009.
\]
Thus 0.99 is the largest.

ANSWER 12: A

Problem 1:
The problem asks us to determine by how many centimeters the water level rises when a rock with a volume of 1000 cm³ is submerged in an aquarium filled with water to a depth of 37 cm. The aquarium has a rectangular base measuring 100 cm by 40 cm and a height of 50 cm.

First, we need to calculate the initial volume of water in the aquarium. The volume of water is given by the formula length × width × height. So, the initial volume of water = 100 cm × 40 cm × 37 cm = 148,000 cm³.

When the rock is submerged, it displaces a volume of water equal to its own volume, which is 1000 cm³. This means the water level will rise to accommodate the volume of the rock.

The rise in water level can be calculated by dividing the volume of the rock by the base area of the aquarium. The base area = 100 cm × 40 cm = 4000 cm².

Rise in water level = Volume of rock / Base area = 1000 cm³ / 4000 cm² = 0.25 cm.

Let's examine the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

The calculated rise in water level matches option A.

ANSWER 1: A

Problem 2:
The problem states that in the card game Buffalo Shuffle-o, all cards are distributed evenly among players. Initially, Annika and 3 friends play, and each is dealt 15 cards. We need to find out how many cards each player will be dealt if 2 more friends join the next game.

First, we determine the total number of cards. With 4 players initially, each getting 15 cards, the total number of cards = 4 × 15 = 60.

When 2 more friends join, the total number of players becomes 4 + 2 = 6.

To find out how many cards each player will get, we divide the total number of cards by the new number of players: 60 / 6 = 10.

Let's examine the choices:
A. 8
B. 9
C. 10
D. 11
E. 12

The calculated number of cards per player matches option C.

ANSWER 2: C

Problem 3:
The problem defines a operation as a c b d = a·d − b·c and asks us to find the value of 3 1 4 2.

Using the given definition, we substitute a = 3, b = 1, c = 4, and d = 2 into the equation: 3 1 4 2 = 3·2 − 1·4 = 6 - 4 = 2.

Let's examine the choices:
A. −2
B. −1
C. 0
D. 1
E. 2

The calculated value matches option E.

ANSWER 3: E

Problem 4:
Isabella needs to achieve an average grade of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score she could have made on the third test to still reach her goal.

The total score required for an average of 95 on 4 tests = 95 × 4 = 380.

The sum of her first two test scores = 97 + 91 = 188.

So, the total score needed on the last two tests = 380 - 188 = 192.

To minimize the score on the third test, we maximize the score on the fourth test. The maximum score on the fourth test is 100.

Thus, the minimum score on the third test = 192 - 100 = 92.

Let's examine the choices:
A. 90
B. 92
C. 95
D. 96
E. 97

The calculated minimum score matches option B.

ANSWER 4: B

Problem 5:
Bridget gave half of her apples to Ann, then gave 3 to Cassie, and kept 4 for herself. We need to find out how many apples she originally bought.

Let's denote the original number of apples as x.

After giving half to Ann, Bridget had x/2 apples left.

Then, she gave 3 apples to Cassie and kept 4, so x/2 = 3 + 4 = 7.

Thus, x = 7 × 2 = 14.

Let's examine the choices:
A. 3
B. 4
C. 7
D. 11
E. 14

The calculated original number of apples matches option E.

ANSWER 5: E

Problem 6:
The problem states that in a jar, all but 6 marbles are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T, the number of red marbles as R, the number of green marbles as G, and the number of blue marbles as B.

From the given information:
1. T - R = 6 (all but 6 are red)
2. T - G = 8 (all but 8 are green)
3. T - B = 4 (all but 4 are blue)

Adding these equations gives: 3T - (R + G + B) = 18.

Since R + G + B = T (the total number of marbles is the sum of the marbles of each color), we have:
3T - T = 18, which simplifies to 2T = 18.

Thus, T = 18 / 2 = 9.

Let's examine the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

The calculated total number of marbles matches option C.

ANSWER 6: C

Problem 7:
We need to find the sum of the two smallest prime factors of 250.

First, let's factorize 250 into its prime factors: 250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5.

The sum of these two smallest prime factors = 2 + 5 = 7.

Let's examine the choices:
A. 2
B. 5
C. 7
D. 10
E. 12

The calculated sum matches option C.

ANSWER 7: C

Problem 8:
Bag A contains chips labeled 1, 3, and 5. Bag B contains chips labeled 2, 4, and 6. We need to find how many different values are possible for the sum of the numbers on the chips drawn from each bag.

Let's list the possible sums:
- (1,2) = 3
- (1,4) = 5
- (1,6) = 7
- (3,2) = 5
- (3,4) = 7
- (3,6) = 9
- (5,2) = 7
- (5,4) = 9
- (5,6) = 11

The possible sums are 3, 5, 7, 9, and 11. There are 5 different values.

Let's examine the choices:
A. 4
B. 5
C. 6
D. 7
E. 9

The number of different sums matches option B.

ANSWER 8: B

Problem 9:
The chess team consists of 2 boys and 3 girls. They need to be arranged with a boy at each end and the girls in the middle. We need to find out how many such arrangements are possible.

The boys can be arranged at the ends in 2! = 2 ways (either boy can be on the left or the right).

The 3 girls can be arranged in the 3 middle positions in 3! = 6 ways.

The total number of arrangements = (number of ways to arrange boys) × (number of ways to arrange girls) = 2 × 6 = 12.

Let's examine the choices:
A. 2
B. 4
C. 5
D. 6
E. 12

The calculated total number of arrangements matches option E.

ANSWER 9: E

Problem 10:
The problem asks us to evaluate the expression 16 + 8 ÷ 4 − 2.

Following the order of operations (PEMDAS/BODMAS: Parentheses/Brackets, Exponents/Orders, Multiplication and Division from left to right, Addition and Subtraction from left to right):
1. D

Problem 1:
The water rise is determined by the volume of the rock divided by the base area of the aquarium.  
Base area = \(100 \times 40 = 4000\text{ cm}^2\).  
Volume of rock = \(1000\text{ cm}^3\).  
Rise = \(\frac{1000}{4000} = 0.25\text{ cm}\).  
ANSWER 1: A

Problem 2:
Initially, Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 2: C

Problem 3:
The notation \(\begin{array}{cc} a & c \\ b & d \end{array} = a \cdot d - b \cdot c\).  
For \(\begin{array}{cc} 3 & 1 \\ 4 & 2 \end{array}\): \(a=3, c=1, b=4, d=2\).  
Value = \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).  
ANSWER 3: E

Problem 4:
Total points needed for an average of 95 over 4 tests = \(4 \times 95 = 380\).  
First two tests sum = \(97 + 91 = 188\).  
Remaining points needed = \(380 - 188 = 192\).  
To minimize the third test score, maximize the fourth (100). Then third test \(\ge 192 - 100 = 92\).  
Lowest possible third test score is 92.  
ANSWER 4: B

Problem 5:
Let \(x\) be the number of apples Bridget bought.  
She gave half to Ann, so she kept \(\frac{x}{2}\).  
Then she gave 3 to Cassie and kept 4, so \(\frac{x}{2} = 3 + 4 = 7\).  
Thus \(x = 14\).  
ANSWER 5: E

Problem 6:
Let \(R, G, B\) be the number of red, green, blue marbles.  
"All but 6 are red" \(\Rightarrow G + B = 6\).  
"All but 8 are green" \(\Rightarrow R + B = 8\).  
"All but 4 are blue" \(\Rightarrow R + G = 4\).  
Adding: \((G+B)+(R+B)+(R+G) = 6+8+4 = 18\) \(\Rightarrow 2(R+G+B) = 18\) \(\Rightarrow R+G+B = 9\).  
Total marbles = 9.  
ANSWER 6: C

Problem 7:
Prime factorization of 250: \(250 = 2 \times 5^3\).  
Prime factors are 2 and 5. The two smallest are 2 and 5.  
Sum = \(2 + 5 = 7\).  
ANSWER 7: C

Problem 8:
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}.  
Possible sums:  
1+2=3, 1+4=5, 1+6=7;  
3+2=5, 3+4=7, 3+6=9;  
5+2=7, 5+4=9, 5+6=11.  
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.  
ANSWER 8: B

Problem 9:
Boys must sit at the two ends.  
Arrange boys: 2 choices for left end, 1 for right end → \(2! = 2\) ways.  
Arrange 3 girls in the middle 3 seats: \(3! = 6\) ways.  
Total arrangements = \(2 \times 6 = 12\).  
ANSWER 9: E

Problem 10:
Interpret "8 4" as \(8 \div 4\).  
\(16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16\).  
ANSWER 10: D

Problem 11:
\(\sqrt{4} = 2\).  
\(8 \cdot 2 = 16\), \(\sqrt{16} = 4\).  
\(16 \cdot 4 = 64\), \(\sqrt{64} = 8\).  
ANSWER 11: C

Problem 12:
Compare decimals:  
0.99, 0.9099, 0.9, 0.909, 0.9009.  
0.99 is the largest.  
ANSWER 12: A

# Problem 1:

**What is being asked:** A rock of volume 1000 cm³ is submerged in an aquarium with a 100 cm × 40 cm base. How much does the water level rise?

**Setup:** When the rock is submerged, it displaces a volume of water equal to its own volume. This displaced water spreads across the base of the aquarium, causing the level to rise.

$$\text{Volume displaced} = \text{Base Area} \times \text{Rise in water level}$$

**Computation:**
- Base area = $100 \times 40 = 4000 \text{ cm}^2$
- $1000 = 4000 \times h$
- $h = \frac{1000}{4000} = 0.25 \text{ cm}$

**Check for overflow:** The current depth is 37 cm, and 37 + 0.25 = 37.25 cm, which is well below the 50 cm height. No overflow occurs.

**Ruling out other choices:** Choices B–E would require rock volumes of 2000, 4000, 5000, and 10000 cm³ respectively — all larger than the given 1000 cm³.

ANSWER 1: A

---

# Problem 2:

**What is being asked:** With 4 players, each gets 15 cards. With 6 players, how many cards does each get?

**Setup:** The total number of cards in the deck stays the same.

**Computation:**
- Number of players in first game: Annika + 3 friends = 4 players
- Total cards = $4 \times 15 = 60$ cards
- Number of players in second game: $4 + 2 = 6$ players
- Cards per player = $\frac{60}{6} = 10$

**Ruling out other choices:** Some might forget to count Annika (getting 3 players → 45 cards → 45/5 = 9, choice B) or miscount the new total as 5 players (60/5 = 12, choice E).

ANSWER 2: C

---

# Problem 3:

**What is being asked:** Evaluate the custom operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$ for $a=3, b=1, c=4, d=2$.

**Setup:** Substitute directly into the formula.

**Computation:**
$$\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2$$

**Ruling out other choices:** A common mistake is computing $b \cdot c - a \cdot d = 4 - 6 = -2$ (choice A) or mixing up which values go where.

ANSWER 3: E

---

# Problem 4:

**What is being asked:** Isabella needs an average of 95 on four tests. Her first two scores are 97 and 91. What is the lowest possible third test score so she can still reach her goal?

**Setup:**
- Total points needed: $4 \times 95 = 380$
- Sum of first two tests: $97 + 91 = 188$
- Sum needed from tests 3 and 4: $380 - 188 = 192$

**Computation:**
To minimize the third test score, maximize the fourth test score. The maximum possible score on any test is 100.
$$\text{Minimum test 3} = 192 - 100 = 92$$

**Ruling out other choices:** Choice A (90) would require test 4 to be $192 - 90 = 102$, which is impossible on a 100-point test. Choices C, D, E are higher than necessary.

ANSWER 4: B

---

# Problem 5:

**What is being asked:** Bridget bought apples, gave half to Ann, gave 3 to Cassie, and kept 4. How many did she buy?

**Setup:** Let $x$ = total apples bought. Work through the sequence of events.

**Computation:**
- After giving half to Ann: $\frac{x}{2}$ apples remain
- After giving 3 to Cassie: $\frac{x}{2} - 3$ apples remain
- She keeps 4: $\frac{x}{2} - 3 = 4$
- $\frac{x}{2} = 7$
- $x = 14$

**Verification:** 14 apples → give 7 to Ann → 7 left → give 3 to Cassie → 4 left. ✓

**Ruling out other choices:** Choice C (7) is a trap for those who forget to double the 7 remaining apples. Choice D (11) comes from incorrectly adding $4 + 3 + 4$.

ANSWER 5: E

---

# Problem 6:

**What is being asked:** Find the total number of marbles given conditions about each color.

**Setup:** Let $T$ = total marbles, $R$ = red, $G$ = green, $B$ = blue.
- "All but 6 are red" → $G + B = 6$ → $R = T - 6$
- "All but 8 are green" → $R + B = 8$ → $G = T - 8$
- "All but 4 are blue" → $R + G = 4$ → $B = T - 4$

**Computation:**
$$T = R + G + B = (T - 6) + (T - 8) + (T - 4) = 3T - 18$$
$$2T = 18 \implies T = 9$$

**Verification:** $R = 3, G = 1, B = 5$. Check: $G+B=6$ ✓, $R+B=8$ ✓, $R+G=4$ ✓.

ANSWER 6: C

---

# Problem 7:

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Setup:** Find the prime factorization of 250.

**Computation:**
$$250 = 2 \times 125 = 2 \times 5 \times 25 = 2 \times 5 \times 5 \times 5 = 2 \times 5^3$$

The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.
$$\text{Sum} = 2 + 5 = 7$$

**Ruling out other choices:** Choice A (2) is just one factor. Choice B (5) is just one factor. Choice D (10) would be $2 + 5 + 3$ or $5 + 5$, neither of which applies.

ANSWER 7: C

---

# Problem 8:

**What is being asked:** How many distinct sums can be formed by picking one chip from Bag A {1, 3, 5} and one from Bag B {2, 4, 6}?

**Setup:** List all 9 possible sums and count the unique values.

**Computation:**
| | 2 | 4 | 6 |
|---|---|---|---|
| **1** | 3 | 5 | 7 |
| **3** | 5 | 7 | 9 |
| **5** | 7 | 9 | 11 |

Unique sums: {3, 5, 7, 9, 11} → **5 different values**

**Ruling out other choices:** Choice E (9) is the trap of counting all 9 pairs without removing duplicates. Choice D (7) might come from miscounting.

ANSWER 8: B

---

# Problem 9:

**What is being asked:** How many ways can 2 boys and 3 girls sit in a row with boys at the ends and girls in the middle?

**Setup:** The arrangement must be: **B G G G B**

**Computation:**
- Arrange the 2 boys at the 2 end positions: $2! = 2$ ways
- Arrange the 3 girls in the 3 middle positions: $3! = 6$ ways
- Total arrangements: $2 \times 6 = 12$

**Ruling out other choices:** Choice A (2) only counts the boys' arrangements. Choice D (6) only counts the girls' arrangements. Both must be multiplied together.

ANSWER 9: E

---

# Problem 10:

**What is being asked:** Evaluate $16 + 8 \div 4 - 2$ using order of operations.

**Setup:** Apply PEMDAS — division before addition and subtraction.

**Computation:**
$$16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16$$

**Ruling out other choices:**
- Left-to-right without order of operations: $(16+8) \div 4 - 2 = 6 - 2 = 4$ (choice A)
- $(16+8) \div (4-2) = 24 \div 2 = 12$ (choice C)
- $16 + 8 \div (4-2) = 16 + 4 = 20$ (choice E

 Problem 1:
We need to find how much the water level rises when a rock of volume 1000 cm³ is submerged in a rectangular aquarium.

First, find the area of the aquarium’s base:
\[
100\text{ cm} \times 40\text{ cm} = 4000\text{ cm}^2.
\]

When the rock is completely submerged, it displaces 1000 cm³ of water. This displaced volume spreads out over the base, causing the water level to rise. The rise in height \(h\) satisfies:
\[
\text{base area} \times h = \text{volume of rock}.
\]
So,
\[
h = \frac{1000}{4000} = 0.25\text{ cm}.
\]

We should check that the aquarium does not overflow. The current depth is 37 cm and the height is 50 cm, so there is 13 cm of free space—plenty of room for a 0.25 cm rise.

Tempting wrong choices come from using the wrong area (e.g., using the side wall 100×50 or 40×50) or from dividing by the height 50. Only the base area matters for the level rise.

ANSWER 1: A

---

Problem 2:
We need to find how many cards each player gets when two extra friends join the game.

First, determine the total number of cards. Annika and 3 friends make 4 players, and each gets 15 cards:
\[
4 \times 15 = 60\text{ cards total}.
\]

When 2 more friends join, the number of players becomes:
\[
4 + 2 = 6.
\]

Distribute the 60 cards evenly among 6 players:
\[
60 \div 6 = 10.
\]

A tempting mistake is to subtract 2 from 15, giving 13, which is not an option, or to guess without computing the total deck size. The key is that the total number of cards stays the same.

ANSWER 2: C

---

Problem 3:
We are given the rule \(a\ c\ b\ d = a\cdot d - b\cdot c\). We need to evaluate \(3\ 1\ 4\ 2\).

Matching the pattern:
- \(a = 3\)
- \(c = 1\)
- \(b = 4\)
- \(d = 2\)

Apply the formula:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

A tempting error is to swap the pairs and compute \(3\cdot 1 - 4\cdot 2 = 3 - 8 = -5\), which is not among the choices, or to read the order incorrectly. Careful matching to \(a, c, b, d\) is essential.

ANSWER 3: E

---

Problem 4:
Isabella wants an average of 95 on four 100-point tests. Her first two scores are 97 and 91. We need the lowest possible score on the third test that still allows her to reach her goal.

Total points needed for a 95 average:
\[
4 \times 95 = 380.
\]

Points earned so far:
\[
97 + 91 = 188.
\]

Points still needed from the last two tests:
\[
380 - 188 = 192.
\]

Let the third test score be \(x\) and the fourth be \(y\), with \(x + y = 192\) and each score at most 100. To minimize \(x\), we maximize \(y\). The highest possible score is 100, so:
\[
x = 192 - 100 = 92.
\]

If she scores 92 on the third test, she can score 100 on the fourth and exactly reach her goal. A tempting wrong choice is 95, which would be the answer if she needed the two remaining tests to average 95, but she can shift points to the fourth test.

ANSWER 4: B

---

Problem 5:
We need to find how many apples Bridget originally bought.

Let the original number be \(x\).
- She gave half to Ann, leaving her with \(\frac{x}{2}\).
- She gave 3 to Cassie and kept 4:
\[
\frac{x}{2} - 3 = 4.
\]

Solve:
\[
\frac{x}{2} = 7 \quad\Rightarrow\quad x = 14.
\]

Check: Start with 14. Half is 7, given to Ann. She has 7 left. Gives 3 to Cassie, keeps 4. This matches.

A tempting error is to add \(3 + 4 = 7\) and stop, forgetting that 7 is only what remained after giving away half, so the original amount must be doubled.

ANSWER 5: E

---

Problem 6:
We need the total number of marbles given three conditions about how many are *not* each color.

Let \(R, G, B\) be the numbers of red, green, and blue marbles.
- “All but 6 are red” means non-red marbles equal 6: \(G + B = 6\).
- “All but 8 are green” means non-green marbles equal 8: \(R + B = 8\).
- “All but 4 are blue” means non-blue marbles equal 4: \(R + G = 4\).

Add all three equations:
\[
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
\]
\[
2R + 2G + 2B = 18
\]
\[
R + G + B = 9.
\]

So there are 9 marbles total. (Checking: \(R=3, G=1, B=5\), which satisfies all three statements.)

A tempting mistake is to add \(6+8+4=18\) and choose that, but that sum counts each marble twice.

ANSWER 6: C

---

Problem 7:
We need the sum of the two smallest prime factors of 250.

Factor 250:
\[
250 = 2 \times 125 = 2 \times 5^3.
\]

The prime factors are 2 and 5. The two smallest (and only) distinct prime factors are 2 and 5. Their sum is:
\[
2 + 5 = 7.
\]

A tempting wrong choice is 10, which comes from adding 5 and 5 (the two smallest factors with multiplicity), but the problem asks for the two smallest prime factors, i.e., distinct primes.

ANSWER 7: C

---

Problem 8:
Bag A has \(\{1, 3, 5\}\) and Bag B has \(\{2, 4, 6\}\). We draw one from each and want the number of distinct possible sums.

Compute all sums:
- \(1+2=3\), \(1+4=5\), \(1+6=7\)
- \(3+2=5\), \(3+4=7\), \(3+6=9\)
- \(5+2=7\), \(5+4=9\), \(5+6=11\)

The distinct values are:
\[
3, 5, 7, 9, 11.
\]

That is 5 different values. A tempting error is to count all 9 ordered pairs without removing duplicates.

ANSWER 8: B

---

Problem 9:
We have 2 boys and 3 girls to arrange in a row with a boy at each end and the three girls in the middle.

Positions: B _ _ _ G G G _ _ _ B (but specifically 5 seats: 1,2,3,4,5 with boys at 1 and 5, girls at 2,3,4).

- Choose the boy for the left end and the boy for the right end: \(2! = 2\) ways.
- Arrange the 3 girls in the middle three seats: \(3! = 6\) ways.

Total arrangements:
\[
2 \times 6 = 12.
\]

A tempting mistake is to ignore the constraint and compute \(5! = 120\), or to forget that the boys can swap places.

ANSWER 9: E

---

Problem 10:
Evaluate \(16 + 8 \div 4 - 2\).

Using order of operations (PEMDAS/BODMAS), division comes before addition and subtraction:
\[
8 \div 4 = 2.
\]

Now substitute back:
\[
16 + 2 - 2 = 16.
\]

A tempting error is to work left-to-right doing addition first: \((16+8)\div 4 - 2 = 24\div 4 - 2 = 6 - 2 = 4\), which is choice A. Another is to do \(16 + 2 = 18\), then \(18 - 2 = 16\), which luckily gives the same corr

Problem 1:
We need the rise in water level when a rock of volume 1000 cm³ is placed in an aquarium with base 100 cm by 40 cm. The rock displaces water equal to its volume, causing the water level to rise by Δh such that base area × Δh = rock volume. Base area = 100 cm × 40 cm = 4000 cm². So Δh = 1000 cm³ / 4000 cm² = 0.25 cm. The initial depth is 37 cm, and after rise it becomes 37.25 cm, which is still below the height of 50 cm, so the rock is fully submerged. The rise is 0.25 cm.

ANSWER 1: A

Problem 2:
When Annika and 3 friends play, there are 4 players, each dealt 15 cards. Total cards = 4 × 15 = 60. In the next game, 2 more friends join, making 6 players. Since the same total number of cards is used, each player gets 60 / 6 = 10 cards.

ANSWER 2: C

Problem 3:
The operation is defined as a c b d = a·d − b·c. For the numbers 3 1 4 2, we have a = 3, c = 1, b = 4, d = 2. Compute a·d − b·c = 3×2 − 4×1 = 6 − 4 = 2.

ANSWER 3: E

Problem 4:
Isabella wants an average of 95 over four tests, so total points needed = 4 × 95 = 380. Her first two scores sum to 97 + 91 = 188. Let the third score be x. After the third test, she can still reach the goal if it is possible to score at most 100 on the fourth test to reach 380. That requires 188 + x + 100 ≥ 380 → x ≥ 380 − 288 = 92. The smallest integer x satisfying this is 92. If x = 91, then even a perfect 100 on the fourth gives 188+91+100 = 379 < 380. So the lowest possible third score is 92.

ANSWER 4: B

Problem 5:
Let the number of apples Bridget bought be N. She gave half to Ann, leaving N/2. Then she gave Cassie 3 apples, leaving N/2 − 3. She keeps 4 apples, so N/2 − 3 = 4 → N/2 = 7 → N = 14.

ANSWER 5: E

Problem 6:
Let T be the total number of marbles. "All but 6 are red" means T − R = 6 → R = T − 6. Similarly, T − G = 8 → G = T − 8, and T − B = 4 → B = T − 4. Since R + G + B = T, we have (T−6)+(T−8)+(T−4) = T → 3T − 18 = T → 2T = 18 → T = 9.

ANSWER 6: C

Problem 7:
Factor 250: 250 = 2 × 5³. Its prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5, and their sum is 7.

ANSWER 7: C

Problem 8:
List all possible sums from drawing one chip from Bag A {1,3,5} and one from Bag B {2,4,6}:
1+2=3, 1+4=5, 1+6=7;
3+2=5, 3+4=7, 3+6=9;
5+2=7, 5+4=9, 5+6=11.
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.

ANSWER 8: B

Problem 9:
We need arrangements of 2 boys and 3 girls in a row with a boy at each end. Choose which boy sits on the left end and which on the right: 2! = 2 ways. The three girls occupy the three middle seats and can be arranged in 3! = 6 ways. Total arrangements = 2 × 6 = 12.

ANSWER 9: E

Problem 10:
The expression is 16 + 8 ÷ 4 − 2 (the missing operator is division, as typical for order-of-operations problems). Following order of operations, division before addition and subtraction: 8 ÷ 4 = 2. Then 16 + 2 = 18, and 18 − 2 = 16.

ANSWER 10: D

Problem 11:
Evaluate step by step: √4 = 2. Then 8 · 2 = 16, so √(8·√4) = √16 = 4. Then 16 · 4 = 64, so √(16·√(8·√4)) = √64 = 8.

ANSWER 11: C

Problem 12:
Compare the decimal numbers:
A. 0.99
B. 0.9099
C. 0.9
D. 0.909
E. 0.9009
Clearly 0.99 is the largest because it is greater than 0.9 and all the others are less than 0.91.

ANSWER 12: A

Problem 1:
The aquarium’s base has area  
\(A = 100\text{ cm} \times 40\text{ cm}=4000\text{ cm}^2\).  
When the rock of volume \(V=1000\text{ cm}^3\) is fully submerged, the water rises by a height \(h\) such that  

\[
A\cdot h = V \quad\Longrightarrow\quad h = \frac{1000}{4000}=0.25\text{ cm}.
\]

The rise cannot be any of the larger listed values (0.5 cm, 1 cm, …) because those would require a volume larger than the rock’s.  

ANSWER 1: A  

---

Problem 2:
With 4 players each receiving 15 cards, the total number of cards is  

\[
4 \times 15 = 60.
\]

If 2 more friends join, there are \(4+2 = 6\) players.  The cards are still all the cards, so each player receives  

\[
\frac{60}{6}=10\text{ cards}.
\]

The other answer choices (8, 9, 11, 12) would not distribute the 60 cards evenly.  

ANSWER 2: C  

---

Problem 3:
The operation is defined by  

\[
a\;c\;b\;d = a\cdot d - b\cdot c.
\]

For the numbers \(3\;1\;4\;2\) we have \(a=3,\;c=1,\;b=4,\;d=2\).  Hence  

\[
3\;1\;4\;2 = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

The other options \(-2,-1,0,1\) would correspond to different combinations that do not satisfy the given formula.  

ANSWER 3: E  

---

Problem 4:
To average 95 over four 100‑point tests, the total points needed are  

\[
4 \times 95 = 380.
\]

After the first two tests Isabella has  

\[
97 + 91 = 188\text{ points}.
\]

Let the third test score be \(x\).  Then the fourth test must supply  

\[
380 - (188 + x) = 192 - x \text{ points}.
\]

The fourth test can give at most 100 points, so we need  

\[
192 - x \le 100 \quad\Longrightarrow\quad x \ge 92.
\]

If she scores \(92\) on the third test, she would need a perfect 100 on the fourth, which is possible.  Any lower score would require more than 100 points on the fourth test, which is impossible.  

Thus the lowest possible third‑test score is **92**.  

ANSWER 4: B  

---

Problem 5:
Let the original number of apples be \(N\).  
She gave half of them to Ann: \(\frac{N}{2}\).  
She then gave Cassie 3 apples and kept 4, leaving  

\[
N - \frac{N}{2} - 3 = 4.
\]

Simplify: \(\frac{N}{2} - 3 = 4 \Rightarrow \frac{N}{2}=7 \Rightarrow N=14\).

If the total had been any of the other answer choices, the steps would not balance (e.g., 11 would leave 2 apples after giving Cassie 3).  

ANSWER 5: E  

---

Problem 6:
Let the total number of marbles be \(T\).  
- “All but 6 are red’’ → red = \(T-6\).  
- “All but 8 are green’’ → green = \(T-8\).  
- “All but 4 are blue’’ → blue = \(T-4\).

These three colors account for all marbles, so  

\[
(T-6)+(T-8)+(T-4)=T \;\Longrightarrow\; 3T-18=T \;\Longrightarrow\; 2T=18 \;\Longrightarrow\; T=9.
\]

The other totals (6, 8, 10, 12) would not satisfy the three “all but…’’ statements simultaneously.  

ANSWER 6: C  

---

Problem 7:
Factor \(250\):  

\[
250 = 2 \times 5^3.
\]

The prime factors are \(2\) and \(5\).  The two smallest are \(2\) and \(5\), whose sum is  

\[
2+5=7.
\]

Choices 2, 5, 10, 12 are either single prime factors or incorrect sums.  

ANSWER 7: C  

---

Problem 8:
Bag A contains \(\{1,3,5\}\); Bag B contains \(\{2,4,6\}\).  
All possible sums \(a+b\) are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The distinct results are \(\{3,5,7,9,11\}\), a total of **5** different sums.  

The other answer choices (4, 6, 7, 9) do not match the count of distinct sums.  

ANSWER 8: B  

---

Problem 9:
Place the two boys at the two ends and the three girls in the middle three seats.

- The two boys can be ordered in \(2!\) ways (left‑right swap).  
- The three girls can be arranged among the three middle seats in \(3! = 6\) ways.

Total arrangements  

\[
2! \times 3! = 2 \times 6 = 12.
\]

Choices 2, 4, 5, 6 are too small; 12 accounts for all possible permutations under the given constraints.  

ANSWER 9: E  

---

Problem 10:
Interpret the expression as  

\[
16 + 8 \div 4 - 2.
\]

Division comes before addition and subtraction, so  

\[
8 \div 4 = 2 \quad\Longrightarrow\quad 16 + 2 - 2 = 16.
\]

The other options (4, 8, 12, 20) would result only if the order of operations were misinterpreted.  

ANSWER 10: D  

---

Problem 11:
Simplify from the innermost radical outward.

\[
\sqrt{4}=2,\qquad 
\sqrt{8\cdot\sqrt{4}} = \sqrt{8\cdot2}= \sqrt{16}=4.
\]

Now  

\[
\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{16\cdot4}= \sqrt{64}=8.
\]

Thus the value is 8, matching choice C.  

ANSWER 11: C  

---

Problem 12:
Compare the decimal values:

- A. \(0.99\)
- B. \(0.9099\)
- C. \(0.9\)
- D. \(0.909\)
- E. \(0.9009\)

Clearly \(0.99\) is greater than \(0.9099\), which is greater than \(0.909\), etc.  
Hence the largest number is \(0.99\).  

ANSWER 12: A

Problem 1:  
An aquarium has a rectangular base of 100 cm by 40 cm and height 50 cm, filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is submerged. By how many centimeters does the water level rise?  

The base area is \(100 \times 40 = 4000 \, \text{cm}^2\). The rock’s volume is 1000 cm³, so the rise in water level is \(\frac{1000}{4000} = 0.25 \, \text{cm}\).  

Tempting choices: 0.5 (B) or 1 (C) might be guessed if base area is miscalculated.  

ANSWER 1: A  

Problem 2:  
Annika and 3 friends (4 players) get 15 cards each. Total cards = \(4 \times 15 = 60\). If 2 more join (6 players), each gets \(\frac{60}{6} = 10\).  

Tempting choices: 8 (A) if total cards miscalculated as 48.  

ANSWER 2: C  

Problem 3:  
Operation defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\): \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\), but the order is \(a c b d\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). However, the problem writes "a c b d" which is interpreted as \(a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). But the choices include 2 (E). Re-evaluating: the operation is \(a c b d = a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\).  

Tempting choices: -2 (A) if signs reversed.  

ANSWER 3: A (Note: Correction based on standard interpretation: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\), but the correct answer based on the operation defined is actually \(3 \cdot 2 - 1 \cdot 4 = 2\), but the problem writes "a c b d", which is \(a \cdot d - b \cdot c\), so 2. However, the answer is 2, but the choices include 2 as E. The correct calculation is 2, so E. Correction: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\).)  

**Correction:** The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\), this is \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\).  

ANSWER 3: E  

Problem 4:  
Isabella wants an average of 95 on four 100-point tests. Total points needed: \(4 \times 95 = 380\). First two scores: 97 + 91 = 188. Let third score be \(x\), fourth be \(y\). Then \(188 + x + y = 380\), so \(x + y = 192\). To minimize \(x\), maximize \(y\): \(y = 100\), so \(x = 92\).  

Tempting choices: 90 (A) if \(y\) not maximized.  

ANSWER 4: B  

Problem 5:  
Bridget gave half apples to Ann, then 3 to Cassie, keeping 4. Let total be \(x\). After giving half to Ann: \(\frac{x}{2}\) left. Then gives 3 to Cassie: \(\frac{x}{2} - 3 = 4\). So \(\frac{x}{2} = 7\), \(x = 14\).  

Tempting choices: 7 (C) if half misinterpreted.  

ANSWER 5: E  

Problem 6:  
Let \(r, g, b\) be red, green, blue marbles. Then:  
- All but 6 are red: \(g + b = 6\)  
- All but 8 are green: \(r + b = 8\)  
- All but 4 are blue: \(r + g = 4\)  
Adding: \(2(r + g + b) = 18\), so total \(r + g + b = 9\).  

Tempting choices: 6 (A) if equations misread.  

ANSWER 6: C  

Problem 7:  
Prime factors of 250: \(250 = 2 \times 5^3\). Two smallest: 2 and 5. Sum: \(2 + 5 = 7\).  

Tempting choices: 5 (B) if only 5 considered.  

ANSWER 7: C  

Problem 8:  
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}. Possible sums:  
- 1+2=3, 1+4=5, 1+6=7  
- 3+2=5, 3+4=7, 3+6=9  
- 5+2=7, 5+4=9, 5+6=11  
Unique sums: 3, 5, 7, 9, 11 → 5 values.  

Tempting choices: 6 (C) if duplicates counted.  

ANSWER 8: B  

Problem 9:  
Two boys at ends: 2 ways. Three girls in middle: \(3! = 6\) ways. Total: \(2 \times 6 = 12\).  

Tempting choices: 6 (D) if boys not permuted.  

ANSWER 9: E  

Problem 10:  
Expression: \(16 + 8 \times 4 - 2\). Order: \(8 \times 4 = 32\), then \(16 + 32 - 2 = 46\)? Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 \times 4 - 2\): \(8 \times 4 = 32\), \(16 + 32 = 48\), \(48 - 2 = 46\). But choices max at 20. Re-interpreting: "16 + 8 4 - 2" might mean \(16 + 8 - 4 - 2 = 18\), not in choices. Alternatively, if "8 4" is 8×4, but 46 not in choices. The expression is likely \(16 + 8 - 4 - 2 = 18\), but not in choices. Correction: The expression is "16 + 8 4 − 2", which might be \(16 + (8 \times 4) - 2 = 16 + 32 - 2 = 46\), but not in choices. Re-reading: It’s "16 + 8 4 − 2", which could be \(16 + 8 - 4 - 2 = 18\), but not in choices. The intended expression is likely \(16 + 8 \times 4 - 2 = 46\), but choices max at 20. Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 - 4 - 2 = 18\), but not in choices. Alternatively, if "8 4" is 84, but 16+84-2=98. Not matching. The expression is likely \( \frac{16 + 8}{4} - 2 = \frac{24}{4} - 2 = 6 - 2 = 4 \).  

Tempting choices: 8 (B) if order misapplied.  

ANSWER 10: A  

Problem 11:  
Expression: \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\).  
- \(\sqrt{4} = 2\)  
- \(\sqrt{8 \cdot 2} = \sqrt{16} = 4\)  
- \(\sqrt{16 \cdot 4} = \sqrt{64} = 8\)  

Tempting choices: 4 (A) if innermost not simplified.  

ANSWER 11: C  

Problem 12:  
Compare decimals:  
- A: 0.99  
- B: 0.9099  
- C: 0.9  
- D: 0.909  
- E: 0.9009  
Largest is 0.99.  

Tempting choices: 0.9099 (B) if decimals misaligned.  

ANSWER 12: A  

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

### Problem 1:
What is asked: An aquarium with a 100cm×40cm rectangular base is filled with water to 37cm depth. A 1000 cm³ rock is submerged; how much does the water level rise?
Step-by-step: When a submerged object displaces water, its volume equals the additional volume of water in the aquarium. The rise in water level Δh satisfies: Volume of rock = base area of aquarium × Δh.
Base area = 100×40 = 4000 cm².
Δh = 1000 / 4000 = 0.25 cm.
Tempting wrong answers: 0.5 (if base area is miscalculated as 2000 cm²) or 1 (wrong division).
ANSWER 1: A

---

### Problem 2:
What is asked: Annika plus 3 friends (4 total players) each get 15 cards. If 2 more join, how many cards does each player get?
Step-by-step: First calculate total number of cards: 4 players × 15 cards = 60 total cards. After 2 join, total players = 4+2=6. Cards per player = 60 / 6 = 10.
Tempting wrong answers: 9 (if you incorrectly count only 3 initial players, leading to 45 total cards) or 12 (wrong division).
ANSWER 2: C

---

### Problem 3:
What is asked: The custom notation a c b d = a·d − b·c. What is the value of 3 1 4 2?
Step-by-step: Match the values to the definition: a=3, c=1, b=4, d=2. Calculate: 3·2 - 4·1 = 6 - 4 = 2.
Tempting wrong answers: -2 (swap a and b incorrectly) or 0 (miscalculate the multiplication).
ANSWER 3: E

---

### Problem 4:
What is asked: Isabella needs a 95 average on 4 100-point tests. She scored 97 and 91 on the first two. What is the lowest possible third test score that lets her reach her goal?
Step-by-step: Total points needed for 95 average: 4×95 = 380 points. Sum of first two tests: 97+91=188. To minimize the third test score, maximize the fourth test (max 100 points). Total points left for tests 3 and 4: 380-188=192. Minimum third test = 192 - 100 = 92.
Tempting wrong answers: 96 (the average of the remaining two tests, not the minimum possible third score).
ANSWER 4: B

---

### Problem 5:
What is asked: Bridget gave half her apples to Ann, 3 to Cassie, and kept 4. How many apples did she buy originally?
Step-by-step: Let N = original number of apples. After giving half to Ann, she has N/2 left. Then N/2 - 3 = 4 → N/2=7 → N=14. Verify: 14 → 7 left after Ann, minus 3 for Cassie = 4 kept, which matches.
Tempting wrong answers: 7 (the number of apples Bridget had after giving half to Ann, not the original total).
ANSWER 5: E

---

### Problem 6:
What is asked: A jar has red, green, blue marbles. All but 6 are red, all but 8 are green, all but 4 are blue. How many total marbles?
Step-by-step: Let T=total marbles, R=red, G=green, B=blue. The conditions translate to: G+B=6, R+B=8, R+G=4. Add all equations: 2(R+G+B)=18 → R+G+B=T=9. Verify: 3 red, 1 green, 5 blue sum to 9, matching all conditions.
Tempting wrong answers: 18 (the sum of 6+8+4, not halved to get total marbles).
ANSWER 6: C

---

### Problem 7:
What is asked: What is the sum of the two smallest prime factors of 250?
Step-by-step: Prime factorization of 250: 250=2×5³. The prime factors are 2 and 5. Their sum is 2+5=7.
Tempting wrong answers: 10 (sum of 5+5, or wrong addition of 2+5).
ANSWER 7: C

---

### Problem 8:
What is asked: Bag A has chips 1,3,5; Bag B has 2,4,6. One chip drawn from each bag, how many distinct sum values are possible?
Step-by-step: List all sums: 1+2=3, 1+4=5,1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7,5+4=9,5+6=11. Distinct sums: 3,5,7,9,11 → 5 total.
Tempting wrong answers: 9 (total number of unordered pairs, not distinct sums).
ANSWER 8: B

---

### Problem 9:
What is asked: 2 boys and 3 girls line up for a photo, with a boy at each end and girls in the middle. How many arrangements are possible?
Step-by-step: Arrange the boys first: 2 boys can be ordered at the ends in 2! = 2 ways. The 3 middle girls can be ordered in 3! =6 ways. Total arrangements: 2×6=12.
Tempting wrong answers: 6 (only the number of ways to arrange the girls, ignoring the boys).
ANSWER 9: E

---

### Problem 10:
What is asked: Evaluate (16+8)/(4-2) (formatted as a fraction in the original problem).
Step-by-step: Calculate numerator:16+8=24, denominator:4-2=2. 24/2=12.
Tempting wrong answers:16 (if order of operations is misapplied to calculate 16 + 8/4 -2=16).
ANSWER 10: C

---

### Problem 11:
What is asked: Evaluate √(16 · √(8 · √4)).
Step-by-step: Calculate from the innermost root first: √4=2. Next: 8×2=16, √16=4. Next:16×4=64, √64=8.
Tempting wrong answers: 4√2 (incorrect root calculation).
ANSWER 11: C

---

### Problem 12:
What is asked: Which number is largest: 0.99, 0.9099, 0.9, 0.909, 0.9009.
Step-by-step: Line up all decimals to compare place values: 0.9900 > 0.9099 > 0.9090 > 0.9009 > 0.9000. The largest is 0.99.
Tempting wrong answers: 0.9099 (the second largest value).
ANSWER 12: A

### Problem 1:
We need to find how much the water level rises when a 1000 cm³ rock is submerged in the aquarium.
1.  The volume of displaced water equals the volume of the submerged rock: 1000 cm³.
2.  The base area of the aquarium is \(100 \, \text{cm} \times 40 \, \text{cm} = 4000 \, \text{cm}^2\).
3.  Let \(h\) be the rise in water level. The volume of the displaced water is also equal to base area × rise in height: \(4000h = 1000\).
4.  Solve for \(h\): \(h = \frac{1000}{4000} = 0.25\) cm.
ANSWER 1: A
---
### Problem 2:
We need to find the number of cards per player when 2 more friends join.
1.  Original number of players: Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).
2.  New number of players: 4 + 2 = 6 players. Total cards remain 60.
3.  Cards per player: \(\frac{60}{6} = 10\).
ANSWER 2: C
---
### Problem 3:
The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). Compute \(3 \, 1 \, 4 \, 2\).
1.  Identify \(a=3, c=1, b=4, d=2\).
2.  Substitute into the formula: \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).
ANSWER 3: E
---
### Problem 4:
Find the lowest possible third test score to get an average of 95 on 4 tests.
1.  Total points needed for an average of 95: \(4 \times 95 = 380\).
2.  Sum of first two test scores: \(97 + 91 = 188\).
3.  Let \(x\) = third test score, \(y\) = fourth test score (max 100, since tests are 100-point). We need \(188 + x + y = 380\), so \(x + y = 192\).
4.  To minimize \(x\), maximize \(y = 100\). Then \(x = 192 - 100 = 92\).
ANSWER 4: B
---
### Problem 5:
Find the original number of apples Bridget bought.
1.  Let \(x\) = total apples. After giving half to Ann, she has \(\frac{x}{2}\) left.
2.  After giving 3 to Cassie, she has \(\frac{x}{2} - 3\) left, which equals 4 (the apples she kept).
3.  Solve: \(\frac{x}{2} - 3 = 4 \implies \frac{x}{2} =7 \implies x=14\).
ANSWER 5: E
---
### Problem 6:
Find the total number of marbles.
1.  Let \(R, G, B\) be the number of red, green, blue marbles, total \(T = R+G+B\).
2.  Translate the statements:
    - All but 6 are red: \(G + B = 6\)
    - All but 8 are green: \(R + B = 8\)
    - All but 4 are blue: \(R + G =4\)
3.  Add all three equations: \(2(R+G+B) = 6+8+4=18 \implies 2T=18 \implies T=9\).
ANSWER 6: C
---
### Problem 7:
Find the sum of the two smallest prime factors of 250.
1.  Factorize 250: \(250 = 2 \times 125 = 2 \times 5^3\).
2.  The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.
3.  Sum: \(2 +5=7\).
ANSWER 7: C
---
### Problem 8:
Find the number of distinct sums when drawing one chip from each bag.
1.  List all possible pairs and sums:
    - Bag A (1,3,5) + Bag B (2,4,6):
      \(1+2=3, 1+4=5, 1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7, 5+4=9, 5+6=11\)
2.  Distinct sums: 3, 5, 7, 9, 11. There are 5 unique values.
ANSWER 8: B
---
### Problem 9:
Find the number of arrangements with boys at each end and girls in the middle.
1.  Arrange the 2 boys at the ends: there are \(2! = 2\) ways (order matters: which boy is left/right).
2.  Arrange the 3 girls in the middle 3 seats: \(3! = 6\) ways.
3.  Total arrangements: \(2 \times 6 =12\).
ANSWER 9: E
---
### Problem 10:
Compute \(16 + 8 \, 4 - 2\). The missing operator between 8 and 4 is division (standard order of operations, common middle school problem):
1.  Interpret as \(16 + \frac{8}{4} - 2\).
2.  Compute division first: \(\frac{8}{4}=2\).
3.  Then add/subtract left to right: \(16 + 2 - 2 =16\).
*(Note: If interpreted as \(16 +8 -4 \times 2\), order of operations gives \(4 \times 2=8\), so \(16+8-8=16\), same result.)*
ANSWER 10: D
---
### Problem 11:
Compute \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\) by working from the innermost root outward.
1.  Innermost: \(\sqrt{4} = 2\).
2.  Next layer: \(\sqrt{8 \cdot 2} = \sqrt{16} =4\).
3.  Outermost: \(\sqrt{16 \cdot 4} = \sqrt{64} =8\).
ANSWER 11: C
---
### Problem 12:
Find the largest decimal.
1.  Rewrite all numbers with 4 decimal places for easy comparison:
    - A: \(0.9900\)
    - B: \(0.9099\)
    - C: \(0.9000\)
    - D: \(0.9090\)
    - E: \(0.9009\)
2.  Compare digit by digit: all have 9 in the tenths place. A has 9 in the hundredths place, while all others have 0. Thus A is the largest.
ANSWER 12: A

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

**Problem 1:**

What is being asked: By how many centimeters does the water level rise when a 1000 cm³ rock is submerged?

Setup: The base area is 100 cm × 40 cm = 4000 cm². When the rock is submerged, it displaces 1000 cm³ of water.

Water level rise = Volume displaced ÷ Base area = 1000 ÷ 4000 = 0.25 cm

ANSWER 1: A

---

**Problem 2:**

What is being asked: If cards are distributed evenly, how many cards per player when the player count increases?

Setup: Initially 4 players (Annika + 3 friends) each get 15 cards.
Total cards = 4 × 15 = 60 cards

With 2 more friends joining: 4 + 2 = 6 players
Cards per player = 60 ÷ 6 = 10 cards

ANSWER 2: C

---

**Problem 3:**

What is being asked: Evaluate the operation 3 1 4 2 using the rule a c b d = a·d − b·c

Setup: With a = 3, c = 1, b = 4, d = 2:
3 1 4 2 = 3·2 − 4·1 = 6 − 4 = 2

ANSWER 3: E

---

**Problem 4:**

What is being asked: What is the lowest score on test 3 that allows Isabella to achieve a 95 average on four tests?

Setup: Goal average is 95, so total needed = 95 × 4 = 380 points
First two tests: 97 + 91 = 188 points
Points needed from tests 3 and 4: 380 − 188 = 192 points

Let x = score on test 3. She can still reach her goal if she can score 100 on test 4:
x + 100 ≥ 192
x ≥ 92

The lowest possible score on test 3 is 92.

ANSWER 4: B

---

**Problem 5:**

What is being asked: How many apples did Bridget buy originally?

Setup: Let x = original number of apples
- She gave half to Ann: x/2 apples given away
- Remaining: x/2
- She gave 3 to Cassie: x/2 − 3
- She kept 4 for herself: x/2 − 3 = 4

Solving: x/2 = 7, so x = 14

ANSWER 5: E

---

**Problem 6:**

What is being asked: How many marbles are in the jar?

Setup: Let R, G, B = numbers of red, green, blue marbles; T = total
- "All but 6 are red": G + B = 6
- "All but 8 are green": R + B = 8
- "All but 4 are blue": R + G = 4

Adding all three equations: 2(R + G + B) = 6 + 8 + 4 = 18
Therefore: R + G + B = 9

ANSWER 6: C

---

**Problem 7:**

What is being asked: What is the sum of the two smallest prime factors of 250?

Setup: Factor 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5³

Prime factors: 2 and 5
Sum = 2 + 5 = 7

ANSWER 7: C

---

**Problem 8:**

What is being asked: How many different sums are possible?

Setup: Bag A: {1, 3, 5}, Bag B: {2, 4, 6}
All possible sums:
- 1 + 2 = 3
- 1 + 4 = 5
- 1 + 6 = 7
- 3 + 2 = 5 (duplicate)
- 3 + 4 = 7 (duplicate)
- 3 + 6 = 9
- 5 + 2 = 7 (duplicate)
- 5 + 4 = 9 (duplicate)
- 5 + 6 = 11

Distinct sums: {3, 5, 7, 9, 11} = 5 different values

ANSWER 8: B

---

**Problem 9:**

What is being asked: How many seating arrangements with boys at ends and girls in middle?

Setup: 2 boys at the ends can be arranged in 2! = 2 ways
3 girls in the middle can be arranged in 3! = 6 ways
Total arrangements = 2 × 6 = 12

ANSWER 9: E

---

**Problem 10:**

What is being asked: Evaluate 16 + 8 ÷ 4 − 2

Setup: Using order of operations (division before addition/subtraction):
16 + (8 ÷ 4) − 2 = 16 + 2 − 2 = 16

ANSWER 10: D

---

**Problem 11:**

What is being asked: Evaluate √(16 · √(8 · √4))

Setup: Work from the inside out:
- √4 = 2
- 8 · √4 = 8 · 2 = 16
- √(8 · √4) = √16 = 4
- 16 · √(8 · √4) = 16 · 4 = 64
- √(16 · √(8 · √4)) = √64 = 8

ANSWER 11: C

---

**Problem 12:**

What is being asked: Which decimal is largest?

Setup: Compare decimals:
- A: 0.99
- B: 0.9099
- C: 0.9
- D: 0.909
- E: 0.9009

Comparing: 0.99 > 0.9099 > 0.909 > 0.9009 > 0.9

The largest is 0.99

ANSWER 12: A

Problem 1:

The aquarium has a rectangular base of \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is

\[
100 \cdot 40 = 4000 \text{ cm}^2.
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

If the water rises by \(h\) centimeters, then

\[
4000h = 1000.
\]

So

\[
h = \frac{1000}{4000} = \frac14 = 0.25.
\]

So the water level rises \(0.25\) cm.

ANSWER 1: A

---

Problem 2:

Annika and 3 friends means there are

\[
4 \text{ players}.
\]

If each player gets 15 cards, then the total number of cards is

\[
4 \cdot 15 = 60.
\]

Now 2 more friends join, so there are

\[
6 \text{ players}.
\]

The same 60 cards are shared evenly, so each gets

\[
60 \div 6 = 10.
\]

ANSWER 2: C

---

Problem 3:

We are told that if the four numbers are \(a, b, c, d\), then

\[
a \ b \ c \ d = a\cdot d - b\cdot c.
\]

So for \(3,1,4,2\),

\[
3 \ 1 \ 4 \ 2 = 3\cdot 2 - 1\cdot 4 = 6 - 4 = 2.
\]

ANSWER 3: E

---

Problem 4:

She wants an average of 95 on 4 tests, so the total of the 4 scores must be

\[
4 \cdot 95 = 380.
\]

Her first two scores are 97 and 91, which add to

\[
97+91=188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To make the total 380, the fourth score would have to be

\[
380 - (188+x) = 192 - x.
\]

For this to be possible, the fourth score must be at most 100:

\[
192 - x \le 100.
\]

So

\[
x \ge 92.
\]

The lowest possible third score is 92.

ANSWER 4: B

---

Problem 5:

Let the number of apples Bridget bought be \(x\).

She gave half to Ann, so she kept the other half:

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

Then she gave Cassie 3 apples and kept 4 apples, so before giving Cassie the apples she had

\[
3+4=7.
\]

Thus,

\[
\frac{x}{2}=7 \quad \Rightarrow \quad x=14.
\]

ANSWER 5: E

---

Problem 6:

Let the total number of marbles be \(n\).

- “All but 6 are red” means the number of red marbles is \(n-6\).
- “All but 8 are green” means the number of green marbles is \(n-8\).
- “All but 4 are blue” means the number of blue marbles is \(n-4\).

Since every marble is one of these colors, their counts add to \(n\):

\[
(n-6)+(n-8)+(n-4)=n.
\]

Simplify:

\[
3n-18=n
\]

\[
2n=18
\]

\[
n=9.
\]

ANSWER 6: C

---

Problem 7:

We need the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3.
\]

So the prime factors are 2 and 5. The two smallest prime factors are 2 and 5, and their sum is

\[
2+5=7.
\]

ANSWER 7: C

---

Problem 8:

Bag A has \(1,3,5\); Bag B has \(2,4,6\).

Possible sums:

- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums are

\[
3,5,7,9,11
\]

which is 5 different values.

ANSWER 8: B

---

Problem 9:

There are 2 boys and 3 girls, and the picture must have:

- a boy at each end,
- the 3 girls in the middle.

First, arrange the 2 boys at the ends. They can be switched in

\[
2! = 2
\]

ways.

Then arrange the 3 girls in the middle in

\[
3! = 6
\]

ways.

So the total number of arrangements is

\[
2 \cdot 6 = 12.
\]

ANSWER 9: E

---

Problem 10:

The expression is

\[
16 + 8\cdot 4 - 2.
\]

Use order of operations:

\[
8\cdot 4=32.
\]

Then

\[
16+32-2=46.
\]

But 46 is not among the choices, so the intended expression must be read as

\[
16 + 8 \div 4 - 2.
\]

Then

\[
8\div 4 = 2,
\]

so

\[
16+2-2=16.
\]

ANSWER 10: D

---

Problem 11:

We evaluate from the inside out:

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

First,

\[
\sqrt4=2.
\]

So the expression becomes

\[
\sqrt{16\cdot \sqrt{8\cdot 2}}
=
\sqrt{16\cdot \sqrt{16}}.
\]

Since \(\sqrt{16}=4\),

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C

---

Problem 12:

Compare the decimals:

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

The largest is the one closest to 1. Since \(0.99\) is greater than any number beginning with \(0.90\), it is the largest.

ANSWER 12: A

Problem 1:  
We need the rise in water level when a rock is placed in the aquarium and submerged.

- Base dimensions: \(100\text{ cm} \times 40\text{ cm}\), so base area  
  \[
  A = 100\cdot 40 = 4000\text{ cm}^2.
  \]
- The rock is completely submerged, so the water volume displaced equals the rock’s volume:  
  \[
  V = 1000\text{ cm}^3.
  \]
- Water level rise \(\Delta h\) satisfies \(A\Delta h = V\):  
  \[
  4000\cdot \Delta h = 1000 \implies \Delta h = \frac{1000}{4000}=0.25\text{ cm}.
  \]

So the rise is \(0.25\) cm.  
Tempting wrong idea: using the *remaining* depth \(37\) cm or checking if it overflows—none of that matters; only displaced volume matters.

ANSWER 1: A  

---

Problem 2:  
All cards are dealt evenly. Let the total number of cards be \(T\).

Initially, Annika + 3 friends = 4 players, and each gets 15 cards:  
\[
T = 4\cdot 15 = 60.
\]
Next game: 2 more friends join, so total players:
\[
4+2=6.
\]
Each player gets:
\[
\frac{60}{6}=10.
\]

ANSWER 2: C  

---

Problem 3:  
Given the rule \(a\,b\,d\,c = a\cdot d - b\cdot c\).  
The expression “\(3\ 1\ 4\ 2\)” corresponds to:
- \(a=3\)
- \(b=1\)
- \(d=4\)
- \(c=2\)

Compute:
\[
a\cdot d - b\cdot c = 3\cdot 4 - 1\cdot 2 = 12-2=10.
\]

But notice the choices are only \(-2,-1,0,1,2\), so we must interpret the placeholder correctly:  
The problem states: **“If \(a b c d = a\cdot d - b\cdot c\)”**.

So “\(3\ 1\ 4\ 2\)” means:
- \(a=3,\ b=1,\ c=4,\ d=2\)

Then:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 1\cdot 4 = 6-4=2.
\]

ANSWER 3: E  

---

Problem 4:  
Goal average is 95 over 4 tests, so total points needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
Let the third test score be \(x\), and the fourth be \(y\). Then:
\[
188 + x + y = 380 \implies x+y=192.
\]
We want the **lowest possible** \(x\) such that she can still reach the goal. That means make \(y\) as large as possible (since grades are at most 100).
Maximum \(y=100\). Then:
\[
x + 100 = 192 \implies x = 92.
\]

Check: then the four scores are \(97,91,92,100\) which sum to \(380\), average \(95\). Works.

ANSWER 4: B  

---

Problem 5:  
Let the total number of apples be \(N\).

- She gave half to Ann, so Ann got \(N/2\), leaving \(N/2\).
- Then she gave Cassie 3 apples, and kept 4 apples. That means after giving Cassie 3, her remaining is 4:
  \[
  \frac{N}{2} - 3 = 4 \implies \frac{N}{2}=7 \implies N=14.
  \]

ANSWER 5: E  

---

Problem 6:  
“All but 6 are red” means: number of red marbles is 6.  
Similarly:
- Red = 6 (since all but 6 are red → 6 red)
- Green: all but 8 are green → 8 green
- Blue: all but 4 are blue → 4 blue

Total marbles:
\[
6+8+4=18.
\]
But none of the choices are 18, so interpret carefully:  
“All but 6 are red” means the number of red marbles is **(total − 6)**, not 6.

Let total be \(T\). Then:
- Red marbles = \(T-6\)
- Green marbles = \(T-8\)
- Blue marbles = \(T-4\)

Sum:
\[
(T-6)+(T-8)+(T-4)=T.
\]
\[
3T - (6+8+4)=T
\]
\[
3T-18=T \implies 2T=18 \implies T=9.
\]

ANSWER 6: C  

---

Problem 7:  
We want the sum of the **two smallest prime factors** of 250.

Factor \(250\):
\[
250 = 25\cdot 10 = 5^2 \cdot 2\cdot 5 = 2\cdot 5^3.
\]
Prime factors are \(2\) and \(5\). The two smallest are \(2\) and \(5\).

Sum:
\[
2+5=7.
\]

ANSWER 7: D  

---

Problem 8:  
Bag A: \(\{1,3,5\}\)  
Bag B: \(\{2,4,6\}\)

Possible sums:
- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums: \(\{3,5,7,9,11\}\) → that’s 5 values.

ANSWER 8: B  

---

Problem 9:  
We need arrangements of 5 people in a row with:
- a boy at each end (positions 1 and 5)
- the three girls in the middle (positions 2,3,4)

There are 2 boys: they can be placed at the ends in
\[
2!\ = 2
\]
ways (left boy/right boy).

The 3 girls can be arranged in the three middle spots in
\[
3! = 6
\]
ways.

Total arrangements:
\[
2\cdot 6 = 12.
\]

ANSWER 9: E  

---

Problem 10:  
Expression: \(16 + 8^4 - 2\).

Compute in order:
- \(8^4 = 4096\)

So:
\[
16 + 4096 - 2 = 4110.
\]
No answer choice matches, so we interpret the formatting as \(16 + 8 \sqrt{4} - 2\) or \(16 + 8^4 - 2\)?  
Given the AMC/JH style, likely it is:
\[
16 + 8\sqrt{4} - 2.
\]
Then:
\[
\sqrt{4}=2 \implies 8\sqrt{4}=8\cdot 2=16.
\]
So:
\[
16 + 16 - 2 = 30,
\]
still not in choices.

Another common formatting is:
\[
16 + 8^{4} - 2 \quad \text{(still not)}
\]

The only choiceable consistent possibility is:
\[
16 + \sqrt{8}\,^4 - 2 \text{ (nonsense)}
\]

Let’s use the most plausible literal from the text: **“16 + 8 4 − 2”** often means \(16 + 8^4 - 2\) if the caret was lost. But then no choice.

Given typical AMC problems, the intended is usually:
\[
16 + 8 - 4 - 2 = 18
\]
also not.

Because the answer choices are only \(4,8,12,16,20\), the intended expression must simplify to one of those. The most natural is:
\[
16 + 8/4 - 2 = 16 + 2 - 2 = 16.
\]
This matches a common missing symbol: “8 4” often means \(8/4\).

So:
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D  

---

Problem 11:  
Evaluate:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}.
\]

Work from inside out:

1) \(\sqrt{4}=2\)

2) Inside next root:
\[
8\cdot \sqrt{4} = 8\cdot 2 = 16
\]
So:
\[
\sqrt{8\cdot \sqrt{4}} = \sqrt{16}=4.
\]

3) Now plug into the outer expression:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}=\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C  

---

Problem 12:  
Largest number among:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare by decimal places:

- .99 is 0.99
- .9099 is 0.9099
- .9 is 0.9000
- .909 is 0.9090
- .9009 is 0.9009

Clearly \(0.99\) is greater than all the numbers less than \(0.91\).

ANSWER 12: A

Problem 1:
The volume of the water displaced by the rock is equal to the volume of the rock itself, which is $1000 \text{ cm}^3$. The aquarium has a rectangular base with area $A = 100 \text{ cm} \times 40 \text{ cm} = 4000 \text{ cm}^2$. When the rock is submerged, the water level rises by height $h$ such that the volume of the displaced water is $A \times h$. 
Setting this equal to the rock's volume: $4000 \times h = 1000$.
Solving for $h$: $h = \frac{1000}{4000} = \frac{1}{4} = 0.25 \text{ cm}$.
ANSWER 1: A

Problem 2:
Annika and 3 friends make 4 players total. Each received 15 cards, so the total number of cards is $4 \times 15 = 60$. If 2 more friends join, the total number of players becomes $4 + 2 = 6$. The cards are distributed evenly among the 6 players, so each player gets $60 \div 6 = 10$ cards.
ANSWER 2: C

Problem 3:
The problem defines the operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$.
For the expression $\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix}$, we have $a=3, b=1, c=4, d=2$.
Calculating: $(3 \times 2) - (1 \times 4) = 6 - 4 = 2$.
ANSWER 3: E

Problem 4:
Isabella needs an average of 95 over 4 tests, meaning the total sum of her scores must be $95 \times 4 = 380$. Her first two scores are 97 and 91, totaling $97 + 91 = 188$. Let $x$ be the score on the third test and $y$ be the score on the fourth. We have $188 + x + y = 380$, so $x + y = 192$. To make $x$ as small as possible, we must make $y$ as large as possible. Since the maximum score on a test is 100, we set $y = 100$. Then $x + 100 = 192$, so $x = 92$.
ANSWER 4: B

Problem 5:
Let $x$ be the total number of apples Bridget bought. She gave half away, leaving $\frac{x}{2}$. Then she gave 3 away, leaving $\frac{x}{2} - 3$. She kept 4, so $\frac{x}{2} - 3 = 4$. Adding 3 to both sides, $\frac{x}{2} = 7$. Multiplying by 2, $x = 14$.
ANSWER 5: E

Problem 6:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R + G + B$.
"All but 6 are red" means $G + B = 6$.
"All but 8 are green" means $R + B = 8$.
"All but 4 are blue" means $R + G = 4$.
Adding these three equations: $2(R + G + B) = 6 + 8 + 4 = 18$.
So $R + G + B = 18 \div 2 = 9$.
ANSWER 6: C

Problem 7:
The prime factorization of 250 is $250 = 2 \times 125 = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two prime factors is $2 + 5 = 7$.
ANSWER 7: C

Problem 8:
Bag A has $\{1, 3, 5\}$ and Bag B has $\{2, 4, 6\}$.
Possible sums:
$1+2=3, 1+4=5, 1+6=7$
$3+2=5, 3+4=7, 3+6=9$
$5+2=7, 5+4=9, 5+6=11$
The set of distinct sums is $\{3, 5, 7, 9, 11\}$. There are 5 distinct values.
ANSWER 8: B

Problem 9:
There are 2 boys (B1, B2) and 3 girls (G1, G2, G3). A boy must be at each end. There are 2 choices for the left end and 1 choice for the right end ($2 \times 1 = 2$ ways to place boys). The 3 girls can be arranged in the middle in $3! = 3 \times 2 \times 1 = 6$ ways. Total arrangements = $2 \times 6 = 12$.
ANSWER 9: E

Problem 10:
Following the order of operations (PEMDAS/BODMAS):
$16 + (8 \div 4) - 2 = 16 + 2 - 2 = 16$.
ANSWER 10: D

Problem 11:
Evaluate from the inside out:
$\sqrt{4} = 2$
$\sqrt{8 \times 2} = \sqrt{16} = 4$
$\sqrt{16 \times 4} = \sqrt{64} = 8$
ANSWER 11: C

Problem 12:
To compare, pad with zeros to match the number of decimal places:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
Comparing the values, 0.9900 is the largest.
ANSWER 12: A

Problem 1:
The base of the aquarium measures 100 cm by 40 cm, so its area is \(100 \times 40 = 4000\) cm². The rock has volume 1000 cm³ and is fully submerged, displacing exactly 1000 cm³ of water. The resulting rise in water level is therefore the displaced volume divided by the base area: \(1000 / 4000 = 0.25\) cm. (The original water depth and tank height are irrelevant once the surface area is known.)

ANSWER 1: A

Problem 2:
Annika plus 3 friends gives 4 players, each receiving 15 cards, for a total of \(4 \times 15 = 60\) cards. When 2 more players join, there are 6 players sharing the same 60 cards, so each receives \(60 / 6 = 10\) cards.

ANSWER 2: C

Problem 3:
The given operation evaluates \(a\, c\, b\, d\) as \(a \cdot d - b \cdot c\). Substituting the numbers yields \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).

ANSWER 3: E

Problem 4:
An average of 95 on four 100-point tests requires a total of \(4 \times 95 = 380\) points. The first two scores sum to \(97 + 91 = 188\), leaving 192 points still needed from the last two tests. The fourth test can score at most 100, so the third test must be at least \(192 - 100 = 92\). Any lower score on the third test would make a total of 380 impossible.

ANSWER 4: B

Problem 5:
Let \(t\) be the original number of apples. After giving half to Ann, \(t/2\) remain. After giving 3 more to Cassie, \(t/2 - 3\) remain, and this equals the 4 apples kept. Solving \(t/2 - 3 = 4\) gives \(t/2 = 7\), so \(t = 14\).

ANSWER 5: E

Problem 6:
Let \(m\) be the total number of marbles. Then there are \(m-6\) red, \(m-8\) green, and \(m-4\) blue. Their sum must equal \(m\):
\[
(m-6) + (m-8) + (m-4) = m \implies 3m - 18 = m \implies 2m = 18 \implies m = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies the three “all-but” statements.

ANSWER 6: C

Problem 7:
Factorize \(250 = 2 \cdot 5^3\). The two smallest distinct prime factors are 2 and 5; their sum is 7.

ANSWER 7: C

Problem 8:
All nine possible sums are:
\[
1+2=3,\ 1+4=5,\ 1+6=7,\ 3+2=5,\ 3+4=7,\ 3+6=9,\ 5+2=7,\ 5+4=9,\ 5+6=11.
\]
The distinct values are 3, 5, 7, 9, 11—exactly five.

ANSWER 8: B

Problem 9:
The seating pattern must be boy–girl–girl–girl–boy. The two distinct boys may be placed at the ends in \(2!\) = 2 ways. The three distinct girls may be arranged in the middle three seats in \(3!\) = 6 ways. The total number of arrangements is therefore \(2 \times 6 = 12\).

ANSWER 9: E

Problem 10:
Interpreting the expression with standard order of operations and treating juxtaposition of 8 and 4 as division yields
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D

Problem 11:
Evaluate from the inside outward:
\[
\sqrt{4} = 2, \quad 8 \cdot 2 = 16, \quad \sqrt{16} = 4, \quad 16 \cdot 4 = 64, \quad \sqrt{64} = 8.
\]

ANSWER 11: C

Problem 12:
Compare the decimals directly:
\[
0.99 > 0.9 > 0.9099 > 0.909 > 0.9009.
\]
Thus 0.99 is the largest.

ANSWER 12: A

Problem 1:
The problem asks us to determine by how many centimeters the water level rises when a rock with a volume of 1000 cm³ is submerged in an aquarium filled with water to a depth of 37 cm. The aquarium has a rectangular base measuring 100 cm by 40 cm and a height of 50 cm.

First, we need to calculate the initial volume of water in the aquarium. The volume of water is given by the formula length × width × height. So, the initial volume of water = 100 cm × 40 cm × 37 cm = 148,000 cm³.

When the rock is submerged, it displaces a volume of water equal to its own volume, which is 1000 cm³. This means the water level will rise to accommodate the volume of the rock.

The rise in water level can be calculated by dividing the volume of the rock by the base area of the aquarium. The base area = 100 cm × 40 cm = 4000 cm².

Rise in water level = Volume of rock / Base area = 1000 cm³ / 4000 cm² = 0.25 cm.

Let's examine the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

The calculated rise in water level matches option A.

ANSWER 1: A

Problem 2:
The problem states that in the card game Buffalo Shuffle-o, all cards are distributed evenly among players. Initially, Annika and 3 friends play, and each is dealt 15 cards. We need to find out how many cards each player will be dealt if 2 more friends join the next game.

First, we determine the total number of cards. With 4 players initially, each getting 15 cards, the total number of cards = 4 × 15 = 60.

When 2 more friends join, the total number of players becomes 4 + 2 = 6.

To find out how many cards each player will get, we divide the total number of cards by the new number of players: 60 / 6 = 10.

Let's examine the choices:
A. 8
B. 9
C. 10
D. 11
E. 12

The calculated number of cards per player matches option C.

ANSWER 2: C

Problem 3:
The problem defines a operation as a c b d = a·d − b·c and asks us to find the value of 3 1 4 2.

Using the given definition, we substitute a = 3, b = 1, c = 4, and d = 2 into the equation: 3 1 4 2 = 3·2 − 1·4 = 6 - 4 = 2.

Let's examine the choices:
A. −2
B. −1
C. 0
D. 1
E. 2

The calculated value matches option E.

ANSWER 3: E

Problem 4:
Isabella needs to achieve an average grade of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score she could have made on the third test to still reach her goal.

The total score required for an average of 95 on 4 tests = 95 × 4 = 380.

The sum of her first two test scores = 97 + 91 = 188.

So, the total score needed on the last two tests = 380 - 188 = 192.

To minimize the score on the third test, we maximize the score on the fourth test. The maximum score on the fourth test is 100.

Thus, the minimum score on the third test = 192 - 100 = 92.

Let's examine the choices:
A. 90
B. 92
C. 95
D. 96
E. 97

The calculated minimum score matches option B.

ANSWER 4: B

Problem 5:
Bridget gave half of her apples to Ann, then gave 3 to Cassie, and kept 4 for herself. We need to find out how many apples she originally bought.

Let's denote the original number of apples as x.

After giving half to Ann, Bridget had x/2 apples left.

Then, she gave 3 apples to Cassie and kept 4, so x/2 = 3 + 4 = 7.

Thus, x = 7 × 2 = 14.

Let's examine the choices:
A. 3
B. 4
C. 7
D. 11
E. 14

The calculated original number of apples matches option E.

ANSWER 5: E

Problem 6:
The problem states that in a jar, all but 6 marbles are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T, the number of red marbles as R, the number of green marbles as G, and the number of blue marbles as B.

From the given information:
1. T - R = 6 (all but 6 are red)
2. T - G = 8 (all but 8 are green)
3. T - B = 4 (all but 4 are blue)

Adding these equations gives: 3T - (R + G + B) = 18.

Since R + G + B = T (the total number of marbles is the sum of the marbles of each color), we have:
3T - T = 18, which simplifies to 2T = 18.

Thus, T = 18 / 2 = 9.

Let's examine the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

The calculated total number of marbles matches option C.

ANSWER 6: C

Problem 7:
We need to find the sum of the two smallest prime factors of 250.

First, let's factorize 250 into its prime factors: 250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5.

The sum of these two smallest prime factors = 2 + 5 = 7.

Let's examine the choices:
A. 2
B. 5
C. 7
D. 10
E. 12

The calculated sum matches option C.

ANSWER 7: C

Problem 8:
Bag A contains chips labeled 1, 3, and 5. Bag B contains chips labeled 2, 4, and 6. We need to find how many different values are possible for the sum of the numbers on the chips drawn from each bag.

Let's list the possible sums:
- (1,2) = 3
- (1,4) = 5
- (1,6) = 7
- (3,2) = 5
- (3,4) = 7
- (3,6) = 9
- (5,2) = 7
- (5,4) = 9
- (5,6) = 11

The possible sums are 3, 5, 7, 9, and 11. There are 5 different values.

Let's examine the choices:
A. 4
B. 5
C. 6
D. 7
E. 9

The number of different sums matches option B.

ANSWER 8: B

Problem 9:
The chess team consists of 2 boys and 3 girls. They need to be arranged with a boy at each end and the girls in the middle. We need to find out how many such arrangements are possible.

The boys can be arranged at the ends in 2! = 2 ways (either boy can be on the left or the right).

The 3 girls can be arranged in the 3 middle positions in 3! = 6 ways.

The total number of arrangements = (number of ways to arrange boys) × (number of ways to arrange girls) = 2 × 6 = 12.

Let's examine the choices:
A. 2
B. 4
C. 5
D. 6
E. 12

The calculated total number of arrangements matches option E.

ANSWER 9: E

Problem 10:
The problem asks us to evaluate the expression 16 + 8 ÷ 4 − 2.

Following the order of operations (PEMDAS/BODMAS: Parentheses/Brackets, Exponents/Orders, Multiplication and Division from left to right, Addition and Subtraction from left to right):
1. D

Problem 1:
The water rise is determined by the volume of the rock divided by the base area of the aquarium.  
Base area = \(100 \times 40 = 4000\text{ cm}^2\).  
Volume of rock = \(1000\text{ cm}^3\).  
Rise = \(\frac{1000}{4000} = 0.25\text{ cm}\).  
ANSWER 1: A

Problem 2:
Initially, Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 2: C

Problem 3:
The notation \(\begin{array}{cc} a & c \\ b & d \end{array} = a \cdot d - b \cdot c\).  
For \(\begin{array}{cc} 3 & 1 \\ 4 & 2 \end{array}\): \(a=3, c=1, b=4, d=2\).  
Value = \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).  
ANSWER 3: E

Problem 4:
Total points needed for an average of 95 over 4 tests = \(4 \times 95 = 380\).  
First two tests sum = \(97 + 91 = 188\).  
Remaining points needed = \(380 - 188 = 192\).  
To minimize the third test score, maximize the fourth (100). Then third test \(\ge 192 - 100 = 92\).  
Lowest possible third test score is 92.  
ANSWER 4: B

Problem 5:
Let \(x\) be the number of apples Bridget bought.  
She gave half to Ann, so she kept \(\frac{x}{2}\).  
Then she gave 3 to Cassie and kept 4, so \(\frac{x}{2} = 3 + 4 = 7\).  
Thus \(x = 14\).  
ANSWER 5: E

Problem 6:
Let \(R, G, B\) be the number of red, green, blue marbles.  
"All but 6 are red" \(\Rightarrow G + B = 6\).  
"All but 8 are green" \(\Rightarrow R + B = 8\).  
"All but 4 are blue" \(\Rightarrow R + G = 4\).  
Adding: \((G+B)+(R+B)+(R+G) = 6+8+4 = 18\) \(\Rightarrow 2(R+G+B) = 18\) \(\Rightarrow R+G+B = 9\).  
Total marbles = 9.  
ANSWER 6: C

Problem 7:
Prime factorization of 250: \(250 = 2 \times 5^3\).  
Prime factors are 2 and 5. The two smallest are 2 and 5.  
Sum = \(2 + 5 = 7\).  
ANSWER 7: C

Problem 8:
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}.  
Possible sums:  
1+2=3, 1+4=5, 1+6=7;  
3+2=5, 3+4=7, 3+6=9;  
5+2=7, 5+4=9, 5+6=11.  
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.  
ANSWER 8: B

Problem 9:
Boys must sit at the two ends.  
Arrange boys: 2 choices for left end, 1 for right end → \(2! = 2\) ways.  
Arrange 3 girls in the middle 3 seats: \(3! = 6\) ways.  
Total arrangements = \(2 \times 6 = 12\).  
ANSWER 9: E

Problem 10:
Interpret "8 4" as \(8 \div 4\).  
\(16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16\).  
ANSWER 10: D

Problem 11:
\(\sqrt{4} = 2\).  
\(8 \cdot 2 = 16\), \(\sqrt{16} = 4\).  
\(16 \cdot 4 = 64\), \(\sqrt{64} = 8\).  
ANSWER 11: C

Problem 12:
Compare decimals:  
0.99, 0.9099, 0.9, 0.909, 0.9009.  
0.99 is the largest.  
ANSWER 12: A

# Problem 1:

**What is being asked:** A rock of volume 1000 cm³ is submerged in an aquarium with a 100 cm × 40 cm base. How much does the water level rise?

**Setup:** When the rock is submerged, it displaces a volume of water equal to its own volume. This displaced water spreads across the base of the aquarium, causing the level to rise.

$$\text{Volume displaced} = \text{Base Area} \times \text{Rise in water level}$$

**Computation:**
- Base area = $100 \times 40 = 4000 \text{ cm}^2$
- $1000 = 4000 \times h$
- $h = \frac{1000}{4000} = 0.25 \text{ cm}$

**Check for overflow:** The current depth is 37 cm, and 37 + 0.25 = 37.25 cm, which is well below the 50 cm height. No overflow occurs.

**Ruling out other choices:** Choices B–E would require rock volumes of 2000, 4000, 5000, and 10000 cm³ respectively — all larger than the given 1000 cm³.

ANSWER 1: A

---

# Problem 2:

**What is being asked:** With 4 players, each gets 15 cards. With 6 players, how many cards does each get?

**Setup:** The total number of cards in the deck stays the same.

**Computation:**
- Number of players in first game: Annika + 3 friends = 4 players
- Total cards = $4 \times 15 = 60$ cards
- Number of players in second game: $4 + 2 = 6$ players
- Cards per player = $\frac{60}{6} = 10$

**Ruling out other choices:** Some might forget to count Annika (getting 3 players → 45 cards → 45/5 = 9, choice B) or miscount the new total as 5 players (60/5 = 12, choice E).

ANSWER 2: C

---

# Problem 3:

**What is being asked:** Evaluate the custom operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$ for $a=3, b=1, c=4, d=2$.

**Setup:** Substitute directly into the formula.

**Computation:**
$$\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2$$

**Ruling out other choices:** A common mistake is computing $b \cdot c - a \cdot d = 4 - 6 = -2$ (choice A) or mixing up which values go where.

ANSWER 3: E

---

# Problem 4:

**What is being asked:** Isabella needs an average of 95 on four tests. Her first two scores are 97 and 91. What is the lowest possible third test score so she can still reach her goal?

**Setup:**
- Total points needed: $4 \times 95 = 380$
- Sum of first two tests: $97 + 91 = 188$
- Sum needed from tests 3 and 4: $380 - 188 = 192$

**Computation:**
To minimize the third test score, maximize the fourth test score. The maximum possible score on any test is 100.
$$\text{Minimum test 3} = 192 - 100 = 92$$

**Ruling out other choices:** Choice A (90) would require test 4 to be $192 - 90 = 102$, which is impossible on a 100-point test. Choices C, D, E are higher than necessary.

ANSWER 4: B

---

# Problem 5:

**What is being asked:** Bridget bought apples, gave half to Ann, gave 3 to Cassie, and kept 4. How many did she buy?

**Setup:** Let $x$ = total apples bought. Work through the sequence of events.

**Computation:**
- After giving half to Ann: $\frac{x}{2}$ apples remain
- After giving 3 to Cassie: $\frac{x}{2} - 3$ apples remain
- She keeps 4: $\frac{x}{2} - 3 = 4$
- $\frac{x}{2} = 7$
- $x = 14$

**Verification:** 14 apples → give 7 to Ann → 7 left → give 3 to Cassie → 4 left. ✓

**Ruling out other choices:** Choice C (7) is a trap for those who forget to double the 7 remaining apples. Choice D (11) comes from incorrectly adding $4 + 3 + 4$.

ANSWER 5: E

---

# Problem 6:

**What is being asked:** Find the total number of marbles given conditions about each color.

**Setup:** Let $T$ = total marbles, $R$ = red, $G$ = green, $B$ = blue.
- "All but 6 are red" → $G + B = 6$ → $R = T - 6$
- "All but 8 are green" → $R + B = 8$ → $G = T - 8$
- "All but 4 are blue" → $R + G = 4$ → $B = T - 4$

**Computation:**
$$T = R + G + B = (T - 6) + (T - 8) + (T - 4) = 3T - 18$$
$$2T = 18 \implies T = 9$$

**Verification:** $R = 3, G = 1, B = 5$. Check: $G+B=6$ ✓, $R+B=8$ ✓, $R+G=4$ ✓.

ANSWER 6: C

---

# Problem 7:

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Setup:** Find the prime factorization of 250.

**Computation:**
$$250 = 2 \times 125 = 2 \times 5 \times 25 = 2 \times 5 \times 5 \times 5 = 2 \times 5^3$$

The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.
$$\text{Sum} = 2 + 5 = 7$$

**Ruling out other choices:** Choice A (2) is just one factor. Choice B (5) is just one factor. Choice D (10) would be $2 + 5 + 3$ or $5 + 5$, neither of which applies.

ANSWER 7: C

---

# Problem 8:

**What is being asked:** How many distinct sums can be formed by picking one chip from Bag A {1, 3, 5} and one from Bag B {2, 4, 6}?

**Setup:** List all 9 possible sums and count the unique values.

**Computation:**
| | 2 | 4 | 6 |
|---|---|---|---|
| **1** | 3 | 5 | 7 |
| **3** | 5 | 7 | 9 |
| **5** | 7 | 9 | 11 |

Unique sums: {3, 5, 7, 9, 11} → **5 different values**

**Ruling out other choices:** Choice E (9) is the trap of counting all 9 pairs without removing duplicates. Choice D (7) might come from miscounting.

ANSWER 8: B

---

# Problem 9:

**What is being asked:** How many ways can 2 boys and 3 girls sit in a row with boys at the ends and girls in the middle?

**Setup:** The arrangement must be: **B G G G B**

**Computation:**
- Arrange the 2 boys at the 2 end positions: $2! = 2$ ways
- Arrange the 3 girls in the 3 middle positions: $3! = 6$ ways
- Total arrangements: $2 \times 6 = 12$

**Ruling out other choices:** Choice A (2) only counts the boys' arrangements. Choice D (6) only counts the girls' arrangements. Both must be multiplied together.

ANSWER 9: E

---

# Problem 10:

**What is being asked:** Evaluate $16 + 8 \div 4 - 2$ using order of operations.

**Setup:** Apply PEMDAS — division before addition and subtraction.

**Computation:**
$$16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16$$

**Ruling out other choices:**
- Left-to-right without order of operations: $(16+8) \div 4 - 2 = 6 - 2 = 4$ (choice A)
- $(16+8) \div (4-2) = 24 \div 2 = 12$ (choice C)
- $16 + 8 \div (4-2) = 16 + 4 = 20$ (choice E

 Problem 1:
We need to find how much the water level rises when a rock of volume 1000 cm³ is submerged in a rectangular aquarium.

First, find the area of the aquarium’s base:
\[
100\text{ cm} \times 40\text{ cm} = 4000\text{ cm}^2.
\]

When the rock is completely submerged, it displaces 1000 cm³ of water. This displaced volume spreads out over the base, causing the water level to rise. The rise in height \(h\) satisfies:
\[
\text{base area} \times h = \text{volume of rock}.
\]
So,
\[
h = \frac{1000}{4000} = 0.25\text{ cm}.
\]

We should check that the aquarium does not overflow. The current depth is 37 cm and the height is 50 cm, so there is 13 cm of free space—plenty of room for a 0.25 cm rise.

Tempting wrong choices come from using the wrong area (e.g., using the side wall 100×50 or 40×50) or from dividing by the height 50. Only the base area matters for the level rise.

ANSWER 1: A

---

Problem 2:
We need to find how many cards each player gets when two extra friends join the game.

First, determine the total number of cards. Annika and 3 friends make 4 players, and each gets 15 cards:
\[
4 \times 15 = 60\text{ cards total}.
\]

When 2 more friends join, the number of players becomes:
\[
4 + 2 = 6.
\]

Distribute the 60 cards evenly among 6 players:
\[
60 \div 6 = 10.
\]

A tempting mistake is to subtract 2 from 15, giving 13, which is not an option, or to guess without computing the total deck size. The key is that the total number of cards stays the same.

ANSWER 2: C

---

Problem 3:
We are given the rule \(a\ c\ b\ d = a\cdot d - b\cdot c\). We need to evaluate \(3\ 1\ 4\ 2\).

Matching the pattern:
- \(a = 3\)
- \(c = 1\)
- \(b = 4\)
- \(d = 2\)

Apply the formula:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

A tempting error is to swap the pairs and compute \(3\cdot 1 - 4\cdot 2 = 3 - 8 = -5\), which is not among the choices, or to read the order incorrectly. Careful matching to \(a, c, b, d\) is essential.

ANSWER 3: E

---

Problem 4:
Isabella wants an average of 95 on four 100-point tests. Her first two scores are 97 and 91. We need the lowest possible score on the third test that still allows her to reach her goal.

Total points needed for a 95 average:
\[
4 \times 95 = 380.
\]

Points earned so far:
\[
97 + 91 = 188.
\]

Points still needed from the last two tests:
\[
380 - 188 = 192.
\]

Let the third test score be \(x\) and the fourth be \(y\), with \(x + y = 192\) and each score at most 100. To minimize \(x\), we maximize \(y\). The highest possible score is 100, so:
\[
x = 192 - 100 = 92.
\]

If she scores 92 on the third test, she can score 100 on the fourth and exactly reach her goal. A tempting wrong choice is 95, which would be the answer if she needed the two remaining tests to average 95, but she can shift points to the fourth test.

ANSWER 4: B

---

Problem 5:
We need to find how many apples Bridget originally bought.

Let the original number be \(x\).
- She gave half to Ann, leaving her with \(\frac{x}{2}\).
- She gave 3 to Cassie and kept 4:
\[
\frac{x}{2} - 3 = 4.
\]

Solve:
\[
\frac{x}{2} = 7 \quad\Rightarrow\quad x = 14.
\]

Check: Start with 14. Half is 7, given to Ann. She has 7 left. Gives 3 to Cassie, keeps 4. This matches.

A tempting error is to add \(3 + 4 = 7\) and stop, forgetting that 7 is only what remained after giving away half, so the original amount must be doubled.

ANSWER 5: E

---

Problem 6:
We need the total number of marbles given three conditions about how many are *not* each color.

Let \(R, G, B\) be the numbers of red, green, and blue marbles.
- “All but 6 are red” means non-red marbles equal 6: \(G + B = 6\).
- “All but 8 are green” means non-green marbles equal 8: \(R + B = 8\).
- “All but 4 are blue” means non-blue marbles equal 4: \(R + G = 4\).

Add all three equations:
\[
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
\]
\[
2R + 2G + 2B = 18
\]
\[
R + G + B = 9.
\]

So there are 9 marbles total. (Checking: \(R=3, G=1, B=5\), which satisfies all three statements.)

A tempting mistake is to add \(6+8+4=18\) and choose that, but that sum counts each marble twice.

ANSWER 6: C

---

Problem 7:
We need the sum of the two smallest prime factors of 250.

Factor 250:
\[
250 = 2 \times 125 = 2 \times 5^3.
\]

The prime factors are 2 and 5. The two smallest (and only) distinct prime factors are 2 and 5. Their sum is:
\[
2 + 5 = 7.
\]

A tempting wrong choice is 10, which comes from adding 5 and 5 (the two smallest factors with multiplicity), but the problem asks for the two smallest prime factors, i.e., distinct primes.

ANSWER 7: C

---

Problem 8:
Bag A has \(\{1, 3, 5\}\) and Bag B has \(\{2, 4, 6\}\). We draw one from each and want the number of distinct possible sums.

Compute all sums:
- \(1+2=3\), \(1+4=5\), \(1+6=7\)
- \(3+2=5\), \(3+4=7\), \(3+6=9\)
- \(5+2=7\), \(5+4=9\), \(5+6=11\)

The distinct values are:
\[
3, 5, 7, 9, 11.
\]

That is 5 different values. A tempting error is to count all 9 ordered pairs without removing duplicates.

ANSWER 8: B

---

Problem 9:
We have 2 boys and 3 girls to arrange in a row with a boy at each end and the three girls in the middle.

Positions: B _ _ _ G G G _ _ _ B (but specifically 5 seats: 1,2,3,4,5 with boys at 1 and 5, girls at 2,3,4).

- Choose the boy for the left end and the boy for the right end: \(2! = 2\) ways.
- Arrange the 3 girls in the middle three seats: \(3! = 6\) ways.

Total arrangements:
\[
2 \times 6 = 12.
\]

A tempting mistake is to ignore the constraint and compute \(5! = 120\), or to forget that the boys can swap places.

ANSWER 9: E

---

Problem 10:
Evaluate \(16 + 8 \div 4 - 2\).

Using order of operations (PEMDAS/BODMAS), division comes before addition and subtraction:
\[
8 \div 4 = 2.
\]

Now substitute back:
\[
16 + 2 - 2 = 16.
\]

A tempting error is to work left-to-right doing addition first: \((16+8)\div 4 - 2 = 24\div 4 - 2 = 6 - 2 = 4\), which is choice A. Another is to do \(16 + 2 = 18\), then \(18 - 2 = 16\), which luckily gives the same corr

Problem 1:
We need the rise in water level when a rock of volume 1000 cm³ is placed in an aquarium with base 100 cm by 40 cm. The rock displaces water equal to its volume, causing the water level to rise by Δh such that base area × Δh = rock volume. Base area = 100 cm × 40 cm = 4000 cm². So Δh = 1000 cm³ / 4000 cm² = 0.25 cm. The initial depth is 37 cm, and after rise it becomes 37.25 cm, which is still below the height of 50 cm, so the rock is fully submerged. The rise is 0.25 cm.

ANSWER 1: A

Problem 2:
When Annika and 3 friends play, there are 4 players, each dealt 15 cards. Total cards = 4 × 15 = 60. In the next game, 2 more friends join, making 6 players. Since the same total number of cards is used, each player gets 60 / 6 = 10 cards.

ANSWER 2: C

Problem 3:
The operation is defined as a c b d = a·d − b·c. For the numbers 3 1 4 2, we have a = 3, c = 1, b = 4, d = 2. Compute a·d − b·c = 3×2 − 4×1 = 6 − 4 = 2.

ANSWER 3: E

Problem 4:
Isabella wants an average of 95 over four tests, so total points needed = 4 × 95 = 380. Her first two scores sum to 97 + 91 = 188. Let the third score be x. After the third test, she can still reach the goal if it is possible to score at most 100 on the fourth test to reach 380. That requires 188 + x + 100 ≥ 380 → x ≥ 380 − 288 = 92. The smallest integer x satisfying this is 92. If x = 91, then even a perfect 100 on the fourth gives 188+91+100 = 379 < 380. So the lowest possible third score is 92.

ANSWER 4: B

Problem 5:
Let the number of apples Bridget bought be N. She gave half to Ann, leaving N/2. Then she gave Cassie 3 apples, leaving N/2 − 3. She keeps 4 apples, so N/2 − 3 = 4 → N/2 = 7 → N = 14.

ANSWER 5: E

Problem 6:
Let T be the total number of marbles. "All but 6 are red" means T − R = 6 → R = T − 6. Similarly, T − G = 8 → G = T − 8, and T − B = 4 → B = T − 4. Since R + G + B = T, we have (T−6)+(T−8)+(T−4) = T → 3T − 18 = T → 2T = 18 → T = 9.

ANSWER 6: C

Problem 7:
Factor 250: 250 = 2 × 5³. Its prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5, and their sum is 7.

ANSWER 7: C

Problem 8:
List all possible sums from drawing one chip from Bag A {1,3,5} and one from Bag B {2,4,6}:
1+2=3, 1+4=5, 1+6=7;
3+2=5, 3+4=7, 3+6=9;
5+2=7, 5+4=9, 5+6=11.
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.

ANSWER 8: B

Problem 9:
We need arrangements of 2 boys and 3 girls in a row with a boy at each end. Choose which boy sits on the left end and which on the right: 2! = 2 ways. The three girls occupy the three middle seats and can be arranged in 3! = 6 ways. Total arrangements = 2 × 6 = 12.

ANSWER 9: E

Problem 10:
The expression is 16 + 8 ÷ 4 − 2 (the missing operator is division, as typical for order-of-operations problems). Following order of operations, division before addition and subtraction: 8 ÷ 4 = 2. Then 16 + 2 = 18, and 18 − 2 = 16.

ANSWER 10: D

Problem 11:
Evaluate step by step: √4 = 2. Then 8 · 2 = 16, so √(8·√4) = √16 = 4. Then 16 · 4 = 64, so √(16·√(8·√4)) = √64 = 8.

ANSWER 11: C

Problem 12:
Compare the decimal numbers:
A. 0.99
B. 0.9099
C. 0.9
D. 0.909
E. 0.9009
Clearly 0.99 is the largest because it is greater than 0.9 and all the others are less than 0.91.

ANSWER 12: A

Problem 1:
The aquarium’s base has area  
\(A = 100\text{ cm} \times 40\text{ cm}=4000\text{ cm}^2\).  
When the rock of volume \(V=1000\text{ cm}^3\) is fully submerged, the water rises by a height \(h\) such that  

\[
A\cdot h = V \quad\Longrightarrow\quad h = \frac{1000}{4000}=0.25\text{ cm}.
\]

The rise cannot be any of the larger listed values (0.5 cm, 1 cm, …) because those would require a volume larger than the rock’s.  

ANSWER 1: A  

---

Problem 2:
With 4 players each receiving 15 cards, the total number of cards is  

\[
4 \times 15 = 60.
\]

If 2 more friends join, there are \(4+2 = 6\) players.  The cards are still all the cards, so each player receives  

\[
\frac{60}{6}=10\text{ cards}.
\]

The other answer choices (8, 9, 11, 12) would not distribute the 60 cards evenly.  

ANSWER 2: C  

---

Problem 3:
The operation is defined by  

\[
a\;c\;b\;d = a\cdot d - b\cdot c.
\]

For the numbers \(3\;1\;4\;2\) we have \(a=3,\;c=1,\;b=4,\;d=2\).  Hence  

\[
3\;1\;4\;2 = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

The other options \(-2,-1,0,1\) would correspond to different combinations that do not satisfy the given formula.  

ANSWER 3: E  

---

Problem 4:
To average 95 over four 100‑point tests, the total points needed are  

\[
4 \times 95 = 380.
\]

After the first two tests Isabella has  

\[
97 + 91 = 188\text{ points}.
\]

Let the third test score be \(x\).  Then the fourth test must supply  

\[
380 - (188 + x) = 192 - x \text{ points}.
\]

The fourth test can give at most 100 points, so we need  

\[
192 - x \le 100 \quad\Longrightarrow\quad x \ge 92.
\]

If she scores \(92\) on the third test, she would need a perfect 100 on the fourth, which is possible.  Any lower score would require more than 100 points on the fourth test, which is impossible.  

Thus the lowest possible third‑test score is **92**.  

ANSWER 4: B  

---

Problem 5:
Let the original number of apples be \(N\).  
She gave half of them to Ann: \(\frac{N}{2}\).  
She then gave Cassie 3 apples and kept 4, leaving  

\[
N - \frac{N}{2} - 3 = 4.
\]

Simplify: \(\frac{N}{2} - 3 = 4 \Rightarrow \frac{N}{2}=7 \Rightarrow N=14\).

If the total had been any of the other answer choices, the steps would not balance (e.g., 11 would leave 2 apples after giving Cassie 3).  

ANSWER 5: E  

---

Problem 6:
Let the total number of marbles be \(T\).  
- “All but 6 are red’’ → red = \(T-6\).  
- “All but 8 are green’’ → green = \(T-8\).  
- “All but 4 are blue’’ → blue = \(T-4\).

These three colors account for all marbles, so  

\[
(T-6)+(T-8)+(T-4)=T \;\Longrightarrow\; 3T-18=T \;\Longrightarrow\; 2T=18 \;\Longrightarrow\; T=9.
\]

The other totals (6, 8, 10, 12) would not satisfy the three “all but…’’ statements simultaneously.  

ANSWER 6: C  

---

Problem 7:
Factor \(250\):  

\[
250 = 2 \times 5^3.
\]

The prime factors are \(2\) and \(5\).  The two smallest are \(2\) and \(5\), whose sum is  

\[
2+5=7.
\]

Choices 2, 5, 10, 12 are either single prime factors or incorrect sums.  

ANSWER 7: C  

---

Problem 8:
Bag A contains \(\{1,3,5\}\); Bag B contains \(\{2,4,6\}\).  
All possible sums \(a+b\) are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The distinct results are \(\{3,5,7,9,11\}\), a total of **5** different sums.  

The other answer choices (4, 6, 7, 9) do not match the count of distinct sums.  

ANSWER 8: B  

---

Problem 9:
Place the two boys at the two ends and the three girls in the middle three seats.

- The two boys can be ordered in \(2!\) ways (left‑right swap).  
- The three girls can be arranged among the three middle seats in \(3! = 6\) ways.

Total arrangements  

\[
2! \times 3! = 2 \times 6 = 12.
\]

Choices 2, 4, 5, 6 are too small; 12 accounts for all possible permutations under the given constraints.  

ANSWER 9: E  

---

Problem 10:
Interpret the expression as  

\[
16 + 8 \div 4 - 2.
\]

Division comes before addition and subtraction, so  

\[
8 \div 4 = 2 \quad\Longrightarrow\quad 16 + 2 - 2 = 16.
\]

The other options (4, 8, 12, 20) would result only if the order of operations were misinterpreted.  

ANSWER 10: D  

---

Problem 11:
Simplify from the innermost radical outward.

\[
\sqrt{4}=2,\qquad 
\sqrt{8\cdot\sqrt{4}} = \sqrt{8\cdot2}= \sqrt{16}=4.
\]

Now  

\[
\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{16\cdot4}= \sqrt{64}=8.
\]

Thus the value is 8, matching choice C.  

ANSWER 11: C  

---

Problem 12:
Compare the decimal values:

- A. \(0.99\)
- B. \(0.9099\)
- C. \(0.9\)
- D. \(0.909\)
- E. \(0.9009\)

Clearly \(0.99\) is greater than \(0.9099\), which is greater than \(0.909\), etc.  
Hence the largest number is \(0.99\).  

ANSWER 12: A

Problem 1:  
An aquarium has a rectangular base of 100 cm by 40 cm and height 50 cm, filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is submerged. By how many centimeters does the water level rise?  

The base area is \(100 \times 40 = 4000 \, \text{cm}^2\). The rock’s volume is 1000 cm³, so the rise in water level is \(\frac{1000}{4000} = 0.25 \, \text{cm}\).  

Tempting choices: 0.5 (B) or 1 (C) might be guessed if base area is miscalculated.  

ANSWER 1: A  

Problem 2:  
Annika and 3 friends (4 players) get 15 cards each. Total cards = \(4 \times 15 = 60\). If 2 more join (6 players), each gets \(\frac{60}{6} = 10\).  

Tempting choices: 8 (A) if total cards miscalculated as 48.  

ANSWER 2: C  

Problem 3:  
Operation defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\): \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\), but the order is \(a c b d\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). However, the problem writes "a c b d" which is interpreted as \(a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). But the choices include 2 (E). Re-evaluating: the operation is \(a c b d = a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\).  

Tempting choices: -2 (A) if signs reversed.  

ANSWER 3: A (Note: Correction based on standard interpretation: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\), but the correct answer based on the operation defined is actually \(3 \cdot 2 - 1 \cdot 4 = 2\), but the problem writes "a c b d", which is \(a \cdot d - b \cdot c\), so 2. However, the answer is 2, but the choices include 2 as E. The correct calculation is 2, so E. Correction: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\).)  

**Correction:** The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\), this is \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\).  

ANSWER 3: E  

Problem 4:  
Isabella wants an average of 95 on four 100-point tests. Total points needed: \(4 \times 95 = 380\). First two scores: 97 + 91 = 188. Let third score be \(x\), fourth be \(y\). Then \(188 + x + y = 380\), so \(x + y = 192\). To minimize \(x\), maximize \(y\): \(y = 100\), so \(x = 92\).  

Tempting choices: 90 (A) if \(y\) not maximized.  

ANSWER 4: B  

Problem 5:  
Bridget gave half apples to Ann, then 3 to Cassie, keeping 4. Let total be \(x\). After giving half to Ann: \(\frac{x}{2}\) left. Then gives 3 to Cassie: \(\frac{x}{2} - 3 = 4\). So \(\frac{x}{2} = 7\), \(x = 14\).  

Tempting choices: 7 (C) if half misinterpreted.  

ANSWER 5: E  

Problem 6:  
Let \(r, g, b\) be red, green, blue marbles. Then:  
- All but 6 are red: \(g + b = 6\)  
- All but 8 are green: \(r + b = 8\)  
- All but 4 are blue: \(r + g = 4\)  
Adding: \(2(r + g + b) = 18\), so total \(r + g + b = 9\).  

Tempting choices: 6 (A) if equations misread.  

ANSWER 6: C  

Problem 7:  
Prime factors of 250: \(250 = 2 \times 5^3\). Two smallest: 2 and 5. Sum: \(2 + 5 = 7\).  

Tempting choices: 5 (B) if only 5 considered.  

ANSWER 7: C  

Problem 8:  
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}. Possible sums:  
- 1+2=3, 1+4=5, 1+6=7  
- 3+2=5, 3+4=7, 3+6=9  
- 5+2=7, 5+4=9, 5+6=11  
Unique sums: 3, 5, 7, 9, 11 → 5 values.  

Tempting choices: 6 (C) if duplicates counted.  

ANSWER 8: B  

Problem 9:  
Two boys at ends: 2 ways. Three girls in middle: \(3! = 6\) ways. Total: \(2 \times 6 = 12\).  

Tempting choices: 6 (D) if boys not permuted.  

ANSWER 9: E  

Problem 10:  
Expression: \(16 + 8 \times 4 - 2\). Order: \(8 \times 4 = 32\), then \(16 + 32 - 2 = 46\)? Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 \times 4 - 2\): \(8 \times 4 = 32\), \(16 + 32 = 48\), \(48 - 2 = 46\). But choices max at 20. Re-interpreting: "16 + 8 4 - 2" might mean \(16 + 8 - 4 - 2 = 18\), not in choices. Alternatively, if "8 4" is 8×4, but 46 not in choices. The expression is likely \(16 + 8 - 4 - 2 = 18\), but not in choices. Correction: The expression is "16 + 8 4 − 2", which might be \(16 + (8 \times 4) - 2 = 16 + 32 - 2 = 46\), but not in choices. Re-reading: It’s "16 + 8 4 − 2", which could be \(16 + 8 - 4 - 2 = 18\), but not in choices. The intended expression is likely \(16 + 8 \times 4 - 2 = 46\), but choices max at 20. Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 - 4 - 2 = 18\), but not in choices. Alternatively, if "8 4" is 84, but 16+84-2=98. Not matching. The expression is likely \( \frac{16 + 8}{4} - 2 = \frac{24}{4} - 2 = 6 - 2 = 4 \).  

Tempting choices: 8 (B) if order misapplied.  

ANSWER 10: A  

Problem 11:  
Expression: \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\).  
- \(\sqrt{4} = 2\)  
- \(\sqrt{8 \cdot 2} = \sqrt{16} = 4\)  
- \(\sqrt{16 \cdot 4} = \sqrt{64} = 8\)  

Tempting choices: 4 (A) if innermost not simplified.  

ANSWER 11: C  

Problem 12:  
Compare decimals:  
- A: 0.99  
- B: 0.9099  
- C: 0.9  
- D: 0.909  
- E: 0.9009  
Largest is 0.99.  

Tempting choices: 0.9099 (B) if decimals misaligned.  

ANSWER 12: A  

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

### Problem 1:
What is asked: An aquarium with a 100cm×40cm rectangular base is filled with water to 37cm depth. A 1000 cm³ rock is submerged; how much does the water level rise?
Step-by-step: When a submerged object displaces water, its volume equals the additional volume of water in the aquarium. The rise in water level Δh satisfies: Volume of rock = base area of aquarium × Δh.
Base area = 100×40 = 4000 cm².
Δh = 1000 / 4000 = 0.25 cm.
Tempting wrong answers: 0.5 (if base area is miscalculated as 2000 cm²) or 1 (wrong division).
ANSWER 1: A

---

### Problem 2:
What is asked: Annika plus 3 friends (4 total players) each get 15 cards. If 2 more join, how many cards does each player get?
Step-by-step: First calculate total number of cards: 4 players × 15 cards = 60 total cards. After 2 join, total players = 4+2=6. Cards per player = 60 / 6 = 10.
Tempting wrong answers: 9 (if you incorrectly count only 3 initial players, leading to 45 total cards) or 12 (wrong division).
ANSWER 2: C

---

### Problem 3:
What is asked: The custom notation a c b d = a·d − b·c. What is the value of 3 1 4 2?
Step-by-step: Match the values to the definition: a=3, c=1, b=4, d=2. Calculate: 3·2 - 4·1 = 6 - 4 = 2.
Tempting wrong answers: -2 (swap a and b incorrectly) or 0 (miscalculate the multiplication).
ANSWER 3: E

---

### Problem 4:
What is asked: Isabella needs a 95 average on 4 100-point tests. She scored 97 and 91 on the first two. What is the lowest possible third test score that lets her reach her goal?
Step-by-step: Total points needed for 95 average: 4×95 = 380 points. Sum of first two tests: 97+91=188. To minimize the third test score, maximize the fourth test (max 100 points). Total points left for tests 3 and 4: 380-188=192. Minimum third test = 192 - 100 = 92.
Tempting wrong answers: 96 (the average of the remaining two tests, not the minimum possible third score).
ANSWER 4: B

---

### Problem 5:
What is asked: Bridget gave half her apples to Ann, 3 to Cassie, and kept 4. How many apples did she buy originally?
Step-by-step: Let N = original number of apples. After giving half to Ann, she has N/2 left. Then N/2 - 3 = 4 → N/2=7 → N=14. Verify: 14 → 7 left after Ann, minus 3 for Cassie = 4 kept, which matches.
Tempting wrong answers: 7 (the number of apples Bridget had after giving half to Ann, not the original total).
ANSWER 5: E

---

### Problem 6:
What is asked: A jar has red, green, blue marbles. All but 6 are red, all but 8 are green, all but 4 are blue. How many total marbles?
Step-by-step: Let T=total marbles, R=red, G=green, B=blue. The conditions translate to: G+B=6, R+B=8, R+G=4. Add all equations: 2(R+G+B)=18 → R+G+B=T=9. Verify: 3 red, 1 green, 5 blue sum to 9, matching all conditions.
Tempting wrong answers: 18 (the sum of 6+8+4, not halved to get total marbles).
ANSWER 6: C

---

### Problem 7:
What is asked: What is the sum of the two smallest prime factors of 250?
Step-by-step: Prime factorization of 250: 250=2×5³. The prime factors are 2 and 5. Their sum is 2+5=7.
Tempting wrong answers: 10 (sum of 5+5, or wrong addition of 2+5).
ANSWER 7: C

---

### Problem 8:
What is asked: Bag A has chips 1,3,5; Bag B has 2,4,6. One chip drawn from each bag, how many distinct sum values are possible?
Step-by-step: List all sums: 1+2=3, 1+4=5,1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7,5+4=9,5+6=11. Distinct sums: 3,5,7,9,11 → 5 total.
Tempting wrong answers: 9 (total number of unordered pairs, not distinct sums).
ANSWER 8: B

---

### Problem 9:
What is asked: 2 boys and 3 girls line up for a photo, with a boy at each end and girls in the middle. How many arrangements are possible?
Step-by-step: Arrange the boys first: 2 boys can be ordered at the ends in 2! = 2 ways. The 3 middle girls can be ordered in 3! =6 ways. Total arrangements: 2×6=12.
Tempting wrong answers: 6 (only the number of ways to arrange the girls, ignoring the boys).
ANSWER 9: E

---

### Problem 10:
What is asked: Evaluate (16+8)/(4-2) (formatted as a fraction in the original problem).
Step-by-step: Calculate numerator:16+8=24, denominator:4-2=2. 24/2=12.
Tempting wrong answers:16 (if order of operations is misapplied to calculate 16 + 8/4 -2=16).
ANSWER 10: C

---

### Problem 11:
What is asked: Evaluate √(16 · √(8 · √4)).
Step-by-step: Calculate from the innermost root first: √4=2. Next: 8×2=16, √16=4. Next:16×4=64, √64=8.
Tempting wrong answers: 4√2 (incorrect root calculation).
ANSWER 11: C

---

### Problem 12:
What is asked: Which number is largest: 0.99, 0.9099, 0.9, 0.909, 0.9009.
Step-by-step: Line up all decimals to compare place values: 0.9900 > 0.9099 > 0.9090 > 0.9009 > 0.9000. The largest is 0.99.
Tempting wrong answers: 0.9099 (the second largest value).
ANSWER 12: A

### Problem 1:
We need to find how much the water level rises when a 1000 cm³ rock is submerged in the aquarium.
1.  The volume of displaced water equals the volume of the submerged rock: 1000 cm³.
2.  The base area of the aquarium is \(100 \, \text{cm} \times 40 \, \text{cm} = 4000 \, \text{cm}^2\).
3.  Let \(h\) be the rise in water level. The volume of the displaced water is also equal to base area × rise in height: \(4000h = 1000\).
4.  Solve for \(h\): \(h = \frac{1000}{4000} = 0.25\) cm.
ANSWER 1: A
---
### Problem 2:
We need to find the number of cards per player when 2 more friends join.
1.  Original number of players: Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).
2.  New number of players: 4 + 2 = 6 players. Total cards remain 60.
3.  Cards per player: \(\frac{60}{6} = 10\).
ANSWER 2: C
---
### Problem 3:
The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). Compute \(3 \, 1 \, 4 \, 2\).
1.  Identify \(a=3, c=1, b=4, d=2\).
2.  Substitute into the formula: \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).
ANSWER 3: E
---
### Problem 4:
Find the lowest possible third test score to get an average of 95 on 4 tests.
1.  Total points needed for an average of 95: \(4 \times 95 = 380\).
2.  Sum of first two test scores: \(97 + 91 = 188\).
3.  Let \(x\) = third test score, \(y\) = fourth test score (max 100, since tests are 100-point). We need \(188 + x + y = 380\), so \(x + y = 192\).
4.  To minimize \(x\), maximize \(y = 100\). Then \(x = 192 - 100 = 92\).
ANSWER 4: B
---
### Problem 5:
Find the original number of apples Bridget bought.
1.  Let \(x\) = total apples. After giving half to Ann, she has \(\frac{x}{2}\) left.
2.  After giving 3 to Cassie, she has \(\frac{x}{2} - 3\) left, which equals 4 (the apples she kept).
3.  Solve: \(\frac{x}{2} - 3 = 4 \implies \frac{x}{2} =7 \implies x=14\).
ANSWER 5: E
---
### Problem 6:
Find the total number of marbles.
1.  Let \(R, G, B\) be the number of red, green, blue marbles, total \(T = R+G+B\).
2.  Translate the statements:
    - All but 6 are red: \(G + B = 6\)
    - All but 8 are green: \(R + B = 8\)
    - All but 4 are blue: \(R + G =4\)
3.  Add all three equations: \(2(R+G+B) = 6+8+4=18 \implies 2T=18 \implies T=9\).
ANSWER 6: C
---
### Problem 7:
Find the sum of the two smallest prime factors of 250.
1.  Factorize 250: \(250 = 2 \times 125 = 2 \times 5^3\).
2.  The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.
3.  Sum: \(2 +5=7\).
ANSWER 7: C
---
### Problem 8:
Find the number of distinct sums when drawing one chip from each bag.
1.  List all possible pairs and sums:
    - Bag A (1,3,5) + Bag B (2,4,6):
      \(1+2=3, 1+4=5, 1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7, 5+4=9, 5+6=11\)
2.  Distinct sums: 3, 5, 7, 9, 11. There are 5 unique values.
ANSWER 8: B
---
### Problem 9:
Find the number of arrangements with boys at each end and girls in the middle.
1.  Arrange the 2 boys at the ends: there are \(2! = 2\) ways (order matters: which boy is left/right).
2.  Arrange the 3 girls in the middle 3 seats: \(3! = 6\) ways.
3.  Total arrangements: \(2 \times 6 =12\).
ANSWER 9: E
---
### Problem 10:
Compute \(16 + 8 \, 4 - 2\). The missing operator between 8 and 4 is division (standard order of operations, common middle school problem):
1.  Interpret as \(16 + \frac{8}{4} - 2\).
2.  Compute division first: \(\frac{8}{4}=2\).
3.  Then add/subtract left to right: \(16 + 2 - 2 =16\).
*(Note: If interpreted as \(16 +8 -4 \times 2\), order of operations gives \(4 \times 2=8\), so \(16+8-8=16\), same result.)*
ANSWER 10: D
---
### Problem 11:
Compute \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\) by working from the innermost root outward.
1.  Innermost: \(\sqrt{4} = 2\).
2.  Next layer: \(\sqrt{8 \cdot 2} = \sqrt{16} =4\).
3.  Outermost: \(\sqrt{16 \cdot 4} = \sqrt{64} =8\).
ANSWER 11: C
---
### Problem 12:
Find the largest decimal.
1.  Rewrite all numbers with 4 decimal places for easy comparison:
    - A: \(0.9900\)
    - B: \(0.9099\)
    - C: \(0.9000\)
    - D: \(0.9090\)
    - E: \(0.9009\)
2.  Compare digit by digit: all have 9 in the tenths place. A has 9 in the hundredths place, while all others have 0. Thus A is the largest.
ANSWER 12: A

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

**Problem 1:**

What is being asked: By how many centimeters does the water level rise when a 1000 cm³ rock is submerged?

Setup: The base area is 100 cm × 40 cm = 4000 cm². When the rock is submerged, it displaces 1000 cm³ of water.

Water level rise = Volume displaced ÷ Base area = 1000 ÷ 4000 = 0.25 cm

ANSWER 1: A

---

**Problem 2:**

What is being asked: If cards are distributed evenly, how many cards per player when the player count increases?

Setup: Initially 4 players (Annika + 3 friends) each get 15 cards.
Total cards = 4 × 15 = 60 cards

With 2 more friends joining: 4 + 2 = 6 players
Cards per player = 60 ÷ 6 = 10 cards

ANSWER 2: C

---

**Problem 3:**

What is being asked: Evaluate the operation 3 1 4 2 using the rule a c b d = a·d − b·c

Setup: With a = 3, c = 1, b = 4, d = 2:
3 1 4 2 = 3·2 − 4·1 = 6 − 4 = 2

ANSWER 3: E

---

**Problem 4:**

What is being asked: What is the lowest score on test 3 that allows Isabella to achieve a 95 average on four tests?

Setup: Goal average is 95, so total needed = 95 × 4 = 380 points
First two tests: 97 + 91 = 188 points
Points needed from tests 3 and 4: 380 − 188 = 192 points

Let x = score on test 3. She can still reach her goal if she can score 100 on test 4:
x + 100 ≥ 192
x ≥ 92

The lowest possible score on test 3 is 92.

ANSWER 4: B

---

**Problem 5:**

What is being asked: How many apples did Bridget buy originally?

Setup: Let x = original number of apples
- She gave half to Ann: x/2 apples given away
- Remaining: x/2
- She gave 3 to Cassie: x/2 − 3
- She kept 4 for herself: x/2 − 3 = 4

Solving: x/2 = 7, so x = 14

ANSWER 5: E

---

**Problem 6:**

What is being asked: How many marbles are in the jar?

Setup: Let R, G, B = numbers of red, green, blue marbles; T = total
- "All but 6 are red": G + B = 6
- "All but 8 are green": R + B = 8
- "All but 4 are blue": R + G = 4

Adding all three equations: 2(R + G + B) = 6 + 8 + 4 = 18
Therefore: R + G + B = 9

ANSWER 6: C

---

**Problem 7:**

What is being asked: What is the sum of the two smallest prime factors of 250?

Setup: Factor 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5³

Prime factors: 2 and 5
Sum = 2 + 5 = 7

ANSWER 7: C

---

**Problem 8:**

What is being asked: How many different sums are possible?

Setup: Bag A: {1, 3, 5}, Bag B: {2, 4, 6}
All possible sums:
- 1 + 2 = 3
- 1 + 4 = 5
- 1 + 6 = 7
- 3 + 2 = 5 (duplicate)
- 3 + 4 = 7 (duplicate)
- 3 + 6 = 9
- 5 + 2 = 7 (duplicate)
- 5 + 4 = 9 (duplicate)
- 5 + 6 = 11

Distinct sums: {3, 5, 7, 9, 11} = 5 different values

ANSWER 8: B

---

**Problem 9:**

What is being asked: How many seating arrangements with boys at ends and girls in middle?

Setup: 2 boys at the ends can be arranged in 2! = 2 ways
3 girls in the middle can be arranged in 3! = 6 ways
Total arrangements = 2 × 6 = 12

ANSWER 9: E

---

**Problem 10:**

What is being asked: Evaluate 16 + 8 ÷ 4 − 2

Setup: Using order of operations (division before addition/subtraction):
16 + (8 ÷ 4) − 2 = 16 + 2 − 2 = 16

ANSWER 10: D

---

**Problem 11:**

What is being asked: Evaluate √(16 · √(8 · √4))

Setup: Work from the inside out:
- √4 = 2
- 8 · √4 = 8 · 2 = 16
- √(8 · √4) = √16 = 4
- 16 · √(8 · √4) = 16 · 4 = 64
- √(16 · √(8 · √4)) = √64 = 8

ANSWER 11: C

---

**Problem 12:**

What is being asked: Which decimal is largest?

Setup: Compare decimals:
- A: 0.99
- B: 0.9099
- C: 0.9
- D: 0.909
- E: 0.9009

Comparing: 0.99 > 0.9099 > 0.909 > 0.9009 > 0.9

The largest is 0.99

ANSWER 12: A

Problem 1:

The aquarium has a rectangular base of \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is

\[
100 \cdot 40 = 4000 \text{ cm}^2.
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

If the water rises by \(h\) centimeters, then

\[
4000h = 1000.
\]

So

\[
h = \frac{1000}{4000} = \frac14 = 0.25.
\]

So the water level rises \(0.25\) cm.

ANSWER 1: A

---

Problem 2:

Annika and 3 friends means there are

\[
4 \text{ players}.
\]

If each player gets 15 cards, then the total number of cards is

\[
4 \cdot 15 = 60.
\]

Now 2 more friends join, so there are

\[
6 \text{ players}.
\]

The same 60 cards are shared evenly, so each gets

\[
60 \div 6 = 10.
\]

ANSWER 2: C

---

Problem 3:

We are told that if the four numbers are \(a, b, c, d\), then

\[
a \ b \ c \ d = a\cdot d - b\cdot c.
\]

So for \(3,1,4,2\),

\[
3 \ 1 \ 4 \ 2 = 3\cdot 2 - 1\cdot 4 = 6 - 4 = 2.
\]

ANSWER 3: E

---

Problem 4:

She wants an average of 95 on 4 tests, so the total of the 4 scores must be

\[
4 \cdot 95 = 380.
\]

Her first two scores are 97 and 91, which add to

\[
97+91=188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To make the total 380, the fourth score would have to be

\[
380 - (188+x) = 192 - x.
\]

For this to be possible, the fourth score must be at most 100:

\[
192 - x \le 100.
\]

So

\[
x \ge 92.
\]

The lowest possible third score is 92.

ANSWER 4: B

---

Problem 5:

Let the number of apples Bridget bought be \(x\).

She gave half to Ann, so she kept the other half:

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

Then she gave Cassie 3 apples and kept 4 apples, so before giving Cassie the apples she had

\[
3+4=7.
\]

Thus,

\[
\frac{x}{2}=7 \quad \Rightarrow \quad x=14.
\]

ANSWER 5: E

---

Problem 6:

Let the total number of marbles be \(n\).

- “All but 6 are red” means the number of red marbles is \(n-6\).
- “All but 8 are green” means the number of green marbles is \(n-8\).
- “All but 4 are blue” means the number of blue marbles is \(n-4\).

Since every marble is one of these colors, their counts add to \(n\):

\[
(n-6)+(n-8)+(n-4)=n.
\]

Simplify:

\[
3n-18=n
\]

\[
2n=18
\]

\[
n=9.
\]

ANSWER 6: C

---

Problem 7:

We need the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3.
\]

So the prime factors are 2 and 5. The two smallest prime factors are 2 and 5, and their sum is

\[
2+5=7.
\]

ANSWER 7: C

---

Problem 8:

Bag A has \(1,3,5\); Bag B has \(2,4,6\).

Possible sums:

- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums are

\[
3,5,7,9,11
\]

which is 5 different values.

ANSWER 8: B

---

Problem 9:

There are 2 boys and 3 girls, and the picture must have:

- a boy at each end,
- the 3 girls in the middle.

First, arrange the 2 boys at the ends. They can be switched in

\[
2! = 2
\]

ways.

Then arrange the 3 girls in the middle in

\[
3! = 6
\]

ways.

So the total number of arrangements is

\[
2 \cdot 6 = 12.
\]

ANSWER 9: E

---

Problem 10:

The expression is

\[
16 + 8\cdot 4 - 2.
\]

Use order of operations:

\[
8\cdot 4=32.
\]

Then

\[
16+32-2=46.
\]

But 46 is not among the choices, so the intended expression must be read as

\[
16 + 8 \div 4 - 2.
\]

Then

\[
8\div 4 = 2,
\]

so

\[
16+2-2=16.
\]

ANSWER 10: D

---

Problem 11:

We evaluate from the inside out:

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

First,

\[
\sqrt4=2.
\]

So the expression becomes

\[
\sqrt{16\cdot \sqrt{8\cdot 2}}
=
\sqrt{16\cdot \sqrt{16}}.
\]

Since \(\sqrt{16}=4\),

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C

---

Problem 12:

Compare the decimals:

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

The largest is the one closest to 1. Since \(0.99\) is greater than any number beginning with \(0.90\), it is the largest.

ANSWER 12: A

Problem 1:  
We need the rise in water level when a rock is placed in the aquarium and submerged.

- Base dimensions: \(100\text{ cm} \times 40\text{ cm}\), so base area  
  \[
  A = 100\cdot 40 = 4000\text{ cm}^2.
  \]
- The rock is completely submerged, so the water volume displaced equals the rock’s volume:  
  \[
  V = 1000\text{ cm}^3.
  \]
- Water level rise \(\Delta h\) satisfies \(A\Delta h = V\):  
  \[
  4000\cdot \Delta h = 1000 \implies \Delta h = \frac{1000}{4000}=0.25\text{ cm}.
  \]

So the rise is \(0.25\) cm.  
Tempting wrong idea: using the *remaining* depth \(37\) cm or checking if it overflows—none of that matters; only displaced volume matters.

ANSWER 1: A  

---

Problem 2:  
All cards are dealt evenly. Let the total number of cards be \(T\).

Initially, Annika + 3 friends = 4 players, and each gets 15 cards:  
\[
T = 4\cdot 15 = 60.
\]
Next game: 2 more friends join, so total players:
\[
4+2=6.
\]
Each player gets:
\[
\frac{60}{6}=10.
\]

ANSWER 2: C  

---

Problem 3:  
Given the rule \(a\,b\,d\,c = a\cdot d - b\cdot c\).  
The expression “\(3\ 1\ 4\ 2\)” corresponds to:
- \(a=3\)
- \(b=1\)
- \(d=4\)
- \(c=2\)

Compute:
\[
a\cdot d - b\cdot c = 3\cdot 4 - 1\cdot 2 = 12-2=10.
\]

But notice the choices are only \(-2,-1,0,1,2\), so we must interpret the placeholder correctly:  
The problem states: **“If \(a b c d = a\cdot d - b\cdot c\)”**.

So “\(3\ 1\ 4\ 2\)” means:
- \(a=3,\ b=1,\ c=4,\ d=2\)

Then:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 1\cdot 4 = 6-4=2.
\]

ANSWER 3: E  

---

Problem 4:  
Goal average is 95 over 4 tests, so total points needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
Let the third test score be \(x\), and the fourth be \(y\). Then:
\[
188 + x + y = 380 \implies x+y=192.
\]
We want the **lowest possible** \(x\) such that she can still reach the goal. That means make \(y\) as large as possible (since grades are at most 100).
Maximum \(y=100\). Then:
\[
x + 100 = 192 \implies x = 92.
\]

Check: then the four scores are \(97,91,92,100\) which sum to \(380\), average \(95\). Works.

ANSWER 4: B  

---

Problem 5:  
Let the total number of apples be \(N\).

- She gave half to Ann, so Ann got \(N/2\), leaving \(N/2\).
- Then she gave Cassie 3 apples, and kept 4 apples. That means after giving Cassie 3, her remaining is 4:
  \[
  \frac{N}{2} - 3 = 4 \implies \frac{N}{2}=7 \implies N=14.
  \]

ANSWER 5: E  

---

Problem 6:  
“All but 6 are red” means: number of red marbles is 6.  
Similarly:
- Red = 6 (since all but 6 are red → 6 red)
- Green: all but 8 are green → 8 green
- Blue: all but 4 are blue → 4 blue

Total marbles:
\[
6+8+4=18.
\]
But none of the choices are 18, so interpret carefully:  
“All but 6 are red” means the number of red marbles is **(total − 6)**, not 6.

Let total be \(T\). Then:
- Red marbles = \(T-6\)
- Green marbles = \(T-8\)
- Blue marbles = \(T-4\)

Sum:
\[
(T-6)+(T-8)+(T-4)=T.
\]
\[
3T - (6+8+4)=T
\]
\[
3T-18=T \implies 2T=18 \implies T=9.
\]

ANSWER 6: C  

---

Problem 7:  
We want the sum of the **two smallest prime factors** of 250.

Factor \(250\):
\[
250 = 25\cdot 10 = 5^2 \cdot 2\cdot 5 = 2\cdot 5^3.
\]
Prime factors are \(2\) and \(5\). The two smallest are \(2\) and \(5\).

Sum:
\[
2+5=7.
\]

ANSWER 7: D  

---

Problem 8:  
Bag A: \(\{1,3,5\}\)  
Bag B: \(\{2,4,6\}\)

Possible sums:
- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums: \(\{3,5,7,9,11\}\) → that’s 5 values.

ANSWER 8: B  

---

Problem 9:  
We need arrangements of 5 people in a row with:
- a boy at each end (positions 1 and 5)
- the three girls in the middle (positions 2,3,4)

There are 2 boys: they can be placed at the ends in
\[
2!\ = 2
\]
ways (left boy/right boy).

The 3 girls can be arranged in the three middle spots in
\[
3! = 6
\]
ways.

Total arrangements:
\[
2\cdot 6 = 12.
\]

ANSWER 9: E  

---

Problem 10:  
Expression: \(16 + 8^4 - 2\).

Compute in order:
- \(8^4 = 4096\)

So:
\[
16 + 4096 - 2 = 4110.
\]
No answer choice matches, so we interpret the formatting as \(16 + 8 \sqrt{4} - 2\) or \(16 + 8^4 - 2\)?  
Given the AMC/JH style, likely it is:
\[
16 + 8\sqrt{4} - 2.
\]
Then:
\[
\sqrt{4}=2 \implies 8\sqrt{4}=8\cdot 2=16.
\]
So:
\[
16 + 16 - 2 = 30,
\]
still not in choices.

Another common formatting is:
\[
16 + 8^{4} - 2 \quad \text{(still not)}
\]

The only choiceable consistent possibility is:
\[
16 + \sqrt{8}\,^4 - 2 \text{ (nonsense)}
\]

Let’s use the most plausible literal from the text: **“16 + 8 4 − 2”** often means \(16 + 8^4 - 2\) if the caret was lost. But then no choice.

Given typical AMC problems, the intended is usually:
\[
16 + 8 - 4 - 2 = 18
\]
also not.

Because the answer choices are only \(4,8,12,16,20\), the intended expression must simplify to one of those. The most natural is:
\[
16 + 8/4 - 2 = 16 + 2 - 2 = 16.
\]
This matches a common missing symbol: “8 4” often means \(8/4\).

So:
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D  

---

Problem 11:  
Evaluate:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}.
\]

Work from inside out:

1) \(\sqrt{4}=2\)

2) Inside next root:
\[
8\cdot \sqrt{4} = 8\cdot 2 = 16
\]
So:
\[
\sqrt{8\cdot \sqrt{4}} = \sqrt{16}=4.
\]

3) Now plug into the outer expression:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}=\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C  

---

Problem 12:  
Largest number among:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare by decimal places:

- .99 is 0.99
- .9099 is 0.9099
- .9 is 0.9000
- .909 is 0.9090
- .9009 is 0.9009

Clearly \(0.99\) is greater than all the numbers less than \(0.91\).

ANSWER 12: A

Problem 1:
The volume of the water displaced by the rock is equal to the volume of the rock itself, which is $1000 \text{ cm}^3$. The aquarium has a rectangular base with area $A = 100 \text{ cm} \times 40 \text{ cm} = 4000 \text{ cm}^2$. When the rock is submerged, the water level rises by height $h$ such that the volume of the displaced water is $A \times h$. 
Setting this equal to the rock's volume: $4000 \times h = 1000$.
Solving for $h$: $h = \frac{1000}{4000} = \frac{1}{4} = 0.25 \text{ cm}$.
ANSWER 1: A

Problem 2:
Annika and 3 friends make 4 players total. Each received 15 cards, so the total number of cards is $4 \times 15 = 60$. If 2 more friends join, the total number of players becomes $4 + 2 = 6$. The cards are distributed evenly among the 6 players, so each player gets $60 \div 6 = 10$ cards.
ANSWER 2: C

Problem 3:
The problem defines the operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$.
For the expression $\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix}$, we have $a=3, b=1, c=4, d=2$.
Calculating: $(3 \times 2) - (1 \times 4) = 6 - 4 = 2$.
ANSWER 3: E

Problem 4:
Isabella needs an average of 95 over 4 tests, meaning the total sum of her scores must be $95 \times 4 = 380$. Her first two scores are 97 and 91, totaling $97 + 91 = 188$. Let $x$ be the score on the third test and $y$ be the score on the fourth. We have $188 + x + y = 380$, so $x + y = 192$. To make $x$ as small as possible, we must make $y$ as large as possible. Since the maximum score on a test is 100, we set $y = 100$. Then $x + 100 = 192$, so $x = 92$.
ANSWER 4: B

Problem 5:
Let $x$ be the total number of apples Bridget bought. She gave half away, leaving $\frac{x}{2}$. Then she gave 3 away, leaving $\frac{x}{2} - 3$. She kept 4, so $\frac{x}{2} - 3 = 4$. Adding 3 to both sides, $\frac{x}{2} = 7$. Multiplying by 2, $x = 14$.
ANSWER 5: E

Problem 6:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R + G + B$.
"All but 6 are red" means $G + B = 6$.
"All but 8 are green" means $R + B = 8$.
"All but 4 are blue" means $R + G = 4$.
Adding these three equations: $2(R + G + B) = 6 + 8 + 4 = 18$.
So $R + G + B = 18 \div 2 = 9$.
ANSWER 6: C

Problem 7:
The prime factorization of 250 is $250 = 2 \times 125 = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two prime factors is $2 + 5 = 7$.
ANSWER 7: C

Problem 8:
Bag A has $\{1, 3, 5\}$ and Bag B has $\{2, 4, 6\}$.
Possible sums:
$1+2=3, 1+4=5, 1+6=7$
$3+2=5, 3+4=7, 3+6=9$
$5+2=7, 5+4=9, 5+6=11$
The set of distinct sums is $\{3, 5, 7, 9, 11\}$. There are 5 distinct values.
ANSWER 8: B

Problem 9:
There are 2 boys (B1, B2) and 3 girls (G1, G2, G3). A boy must be at each end. There are 2 choices for the left end and 1 choice for the right end ($2 \times 1 = 2$ ways to place boys). The 3 girls can be arranged in the middle in $3! = 3 \times 2 \times 1 = 6$ ways. Total arrangements = $2 \times 6 = 12$.
ANSWER 9: E

Problem 10:
Following the order of operations (PEMDAS/BODMAS):
$16 + (8 \div 4) - 2 = 16 + 2 - 2 = 16$.
ANSWER 10: D

Problem 11:
Evaluate from the inside out:
$\sqrt{4} = 2$
$\sqrt{8 \times 2} = \sqrt{16} = 4$
$\sqrt{16 \times 4} = \sqrt{64} = 8$
ANSWER 11: C

Problem 12:
To compare, pad with zeros to match the number of decimal places:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
Comparing the values, 0.9900 is the largest.
ANSWER 12: A

Problem 1:
The base of the aquarium measures 100 cm by 40 cm, so its area is \(100 \times 40 = 4000\) cm². The rock has volume 1000 cm³ and is fully submerged, displacing exactly 1000 cm³ of water. The resulting rise in water level is therefore the displaced volume divided by the base area: \(1000 / 4000 = 0.25\) cm. (The original water depth and tank height are irrelevant once the surface area is known.)

ANSWER 1: A

Problem 2:
Annika plus 3 friends gives 4 players, each receiving 15 cards, for a total of \(4 \times 15 = 60\) cards. When 2 more players join, there are 6 players sharing the same 60 cards, so each receives \(60 / 6 = 10\) cards.

ANSWER 2: C

Problem 3:
The given operation evaluates \(a\, c\, b\, d\) as \(a \cdot d - b \cdot c\). Substituting the numbers yields \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).

ANSWER 3: E

Problem 4:
An average of 95 on four 100-point tests requires a total of \(4 \times 95 = 380\) points. The first two scores sum to \(97 + 91 = 188\), leaving 192 points still needed from the last two tests. The fourth test can score at most 100, so the third test must be at least \(192 - 100 = 92\). Any lower score on the third test would make a total of 380 impossible.

ANSWER 4: B

Problem 5:
Let \(t\) be the original number of apples. After giving half to Ann, \(t/2\) remain. After giving 3 more to Cassie, \(t/2 - 3\) remain, and this equals the 4 apples kept. Solving \(t/2 - 3 = 4\) gives \(t/2 = 7\), so \(t = 14\).

ANSWER 5: E

Problem 6:
Let \(m\) be the total number of marbles. Then there are \(m-6\) red, \(m-8\) green, and \(m-4\) blue. Their sum must equal \(m\):
\[
(m-6) + (m-8) + (m-4) = m \implies 3m - 18 = m \implies 2m = 18 \implies m = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies the three “all-but” statements.

ANSWER 6: C

Problem 7:
Factorize \(250 = 2 \cdot 5^3\). The two smallest distinct prime factors are 2 and 5; their sum is 7.

ANSWER 7: C

Problem 8:
All nine possible sums are:
\[
1+2=3,\ 1+4=5,\ 1+6=7,\ 3+2=5,\ 3+4=7,\ 3+6=9,\ 5+2=7,\ 5+4=9,\ 5+6=11.
\]
The distinct values are 3, 5, 7, 9, 11—exactly five.

ANSWER 8: B

Problem 9:
The seating pattern must be boy–girl–girl–girl–boy. The two distinct boys may be placed at the ends in \(2!\) = 2 ways. The three distinct girls may be arranged in the middle three seats in \(3!\) = 6 ways. The total number of arrangements is therefore \(2 \times 6 = 12\).

ANSWER 9: E

Problem 10:
Interpreting the expression with standard order of operations and treating juxtaposition of 8 and 4 as division yields
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D

Problem 11:
Evaluate from the inside outward:
\[
\sqrt{4} = 2, \quad 8 \cdot 2 = 16, \quad \sqrt{16} = 4, \quad 16 \cdot 4 = 64, \quad \sqrt{64} = 8.
\]

ANSWER 11: C

Problem 12:
Compare the decimals directly:
\[
0.99 > 0.9 > 0.9099 > 0.909 > 0.9009.
\]
Thus 0.99 is the largest.

ANSWER 12: A

Problem 1:
The problem asks us to determine by how many centimeters the water level rises when a rock with a volume of 1000 cm³ is submerged in an aquarium filled with water to a depth of 37 cm. The aquarium has a rectangular base measuring 100 cm by 40 cm and a height of 50 cm.

First, we need to calculate the initial volume of water in the aquarium. The volume of water is given by the formula length × width × height. So, the initial volume of water = 100 cm × 40 cm × 37 cm = 148,000 cm³.

When the rock is submerged, it displaces a volume of water equal to its own volume, which is 1000 cm³. This means the water level will rise to accommodate the volume of the rock.

The rise in water level can be calculated by dividing the volume of the rock by the base area of the aquarium. The base area = 100 cm × 40 cm = 4000 cm².

Rise in water level = Volume of rock / Base area = 1000 cm³ / 4000 cm² = 0.25 cm.

Let's examine the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

The calculated rise in water level matches option A.

ANSWER 1: A

Problem 2:
The problem states that in the card game Buffalo Shuffle-o, all cards are distributed evenly among players. Initially, Annika and 3 friends play, and each is dealt 15 cards. We need to find out how many cards each player will be dealt if 2 more friends join the next game.

First, we determine the total number of cards. With 4 players initially, each getting 15 cards, the total number of cards = 4 × 15 = 60.

When 2 more friends join, the total number of players becomes 4 + 2 = 6.

To find out how many cards each player will get, we divide the total number of cards by the new number of players: 60 / 6 = 10.

Let's examine the choices:
A. 8
B. 9
C. 10
D. 11
E. 12

The calculated number of cards per player matches option C.

ANSWER 2: C

Problem 3:
The problem defines a operation as a c b d = a·d − b·c and asks us to find the value of 3 1 4 2.

Using the given definition, we substitute a = 3, b = 1, c = 4, and d = 2 into the equation: 3 1 4 2 = 3·2 − 1·4 = 6 - 4 = 2.

Let's examine the choices:
A. −2
B. −1
C. 0
D. 1
E. 2

The calculated value matches option E.

ANSWER 3: E

Problem 4:
Isabella needs to achieve an average grade of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score she could have made on the third test to still reach her goal.

The total score required for an average of 95 on 4 tests = 95 × 4 = 380.

The sum of her first two test scores = 97 + 91 = 188.

So, the total score needed on the last two tests = 380 - 188 = 192.

To minimize the score on the third test, we maximize the score on the fourth test. The maximum score on the fourth test is 100.

Thus, the minimum score on the third test = 192 - 100 = 92.

Let's examine the choices:
A. 90
B. 92
C. 95
D. 96
E. 97

The calculated minimum score matches option B.

ANSWER 4: B

Problem 5:
Bridget gave half of her apples to Ann, then gave 3 to Cassie, and kept 4 for herself. We need to find out how many apples she originally bought.

Let's denote the original number of apples as x.

After giving half to Ann, Bridget had x/2 apples left.

Then, she gave 3 apples to Cassie and kept 4, so x/2 = 3 + 4 = 7.

Thus, x = 7 × 2 = 14.

Let's examine the choices:
A. 3
B. 4
C. 7
D. 11
E. 14

The calculated original number of apples matches option E.

ANSWER 5: E

Problem 6:
The problem states that in a jar, all but 6 marbles are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T, the number of red marbles as R, the number of green marbles as G, and the number of blue marbles as B.

From the given information:
1. T - R = 6 (all but 6 are red)
2. T - G = 8 (all but 8 are green)
3. T - B = 4 (all but 4 are blue)

Adding these equations gives: 3T - (R + G + B) = 18.

Since R + G + B = T (the total number of marbles is the sum of the marbles of each color), we have:
3T - T = 18, which simplifies to 2T = 18.

Thus, T = 18 / 2 = 9.

Let's examine the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

The calculated total number of marbles matches option C.

ANSWER 6: C

Problem 7:
We need to find the sum of the two smallest prime factors of 250.

First, let's factorize 250 into its prime factors: 250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5.

The sum of these two smallest prime factors = 2 + 5 = 7.

Let's examine the choices:
A. 2
B. 5
C. 7
D. 10
E. 12

The calculated sum matches option C.

ANSWER 7: C

Problem 8:
Bag A contains chips labeled 1, 3, and 5. Bag B contains chips labeled 2, 4, and 6. We need to find how many different values are possible for the sum of the numbers on the chips drawn from each bag.

Let's list the possible sums:
- (1,2) = 3
- (1,4) = 5
- (1,6) = 7
- (3,2) = 5
- (3,4) = 7
- (3,6) = 9
- (5,2) = 7
- (5,4) = 9
- (5,6) = 11

The possible sums are 3, 5, 7, 9, and 11. There are 5 different values.

Let's examine the choices:
A. 4
B. 5
C. 6
D. 7
E. 9

The number of different sums matches option B.

ANSWER 8: B

Problem 9:
The chess team consists of 2 boys and 3 girls. They need to be arranged with a boy at each end and the girls in the middle. We need to find out how many such arrangements are possible.

The boys can be arranged at the ends in 2! = 2 ways (either boy can be on the left or the right).

The 3 girls can be arranged in the 3 middle positions in 3! = 6 ways.

The total number of arrangements = (number of ways to arrange boys) × (number of ways to arrange girls) = 2 × 6 = 12.

Let's examine the choices:
A. 2
B. 4
C. 5
D. 6
E. 12

The calculated total number of arrangements matches option E.

ANSWER 9: E

Problem 10:
The problem asks us to evaluate the expression 16 + 8 ÷ 4 − 2.

Following the order of operations (PEMDAS/BODMAS: Parentheses/Brackets, Exponents/Orders, Multiplication and Division from left to right, Addition and Subtraction from left to right):
1. D

Problem 1:
The water rise is determined by the volume of the rock divided by the base area of the aquarium.  
Base area = \(100 \times 40 = 4000\text{ cm}^2\).  
Volume of rock = \(1000\text{ cm}^3\).  
Rise = \(\frac{1000}{4000} = 0.25\text{ cm}\).  
ANSWER 1: A

Problem 2:
Initially, Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 2: C

Problem 3:
The notation \(\begin{array}{cc} a & c \\ b & d \end{array} = a \cdot d - b \cdot c\).  
For \(\begin{array}{cc} 3 & 1 \\ 4 & 2 \end{array}\): \(a=3, c=1, b=4, d=2\).  
Value = \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).  
ANSWER 3: E

Problem 4:
Total points needed for an average of 95 over 4 tests = \(4 \times 95 = 380\).  
First two tests sum = \(97 + 91 = 188\).  
Remaining points needed = \(380 - 188 = 192\).  
To minimize the third test score, maximize the fourth (100). Then third test \(\ge 192 - 100 = 92\).  
Lowest possible third test score is 92.  
ANSWER 4: B

Problem 5:
Let \(x\) be the number of apples Bridget bought.  
She gave half to Ann, so she kept \(\frac{x}{2}\).  
Then she gave 3 to Cassie and kept 4, so \(\frac{x}{2} = 3 + 4 = 7\).  
Thus \(x = 14\).  
ANSWER 5: E

Problem 6:
Let \(R, G, B\) be the number of red, green, blue marbles.  
"All but 6 are red" \(\Rightarrow G + B = 6\).  
"All but 8 are green" \(\Rightarrow R + B = 8\).  
"All but 4 are blue" \(\Rightarrow R + G = 4\).  
Adding: \((G+B)+(R+B)+(R+G) = 6+8+4 = 18\) \(\Rightarrow 2(R+G+B) = 18\) \(\Rightarrow R+G+B = 9\).  
Total marbles = 9.  
ANSWER 6: C

Problem 7:
Prime factorization of 250: \(250 = 2 \times 5^3\).  
Prime factors are 2 and 5. The two smallest are 2 and 5.  
Sum = \(2 + 5 = 7\).  
ANSWER 7: C

Problem 8:
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}.  
Possible sums:  
1+2=3, 1+4=5, 1+6=7;  
3+2=5, 3+4=7, 3+6=9;  
5+2=7, 5+4=9, 5+6=11.  
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.  
ANSWER 8: B

Problem 9:
Boys must sit at the two ends.  
Arrange boys: 2 choices for left end, 1 for right end → \(2! = 2\) ways.  
Arrange 3 girls in the middle 3 seats: \(3! = 6\) ways.  
Total arrangements = \(2 \times 6 = 12\).  
ANSWER 9: E

Problem 10:
Interpret "8 4" as \(8 \div 4\).  
\(16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16\).  
ANSWER 10: D

Problem 11:
\(\sqrt{4} = 2\).  
\(8 \cdot 2 = 16\), \(\sqrt{16} = 4\).  
\(16 \cdot 4 = 64\), \(\sqrt{64} = 8\).  
ANSWER 11: C

Problem 12:
Compare decimals:  
0.99, 0.9099, 0.9, 0.909, 0.9009.  
0.99 is the largest.  
ANSWER 12: A

# Problem 1:

**What is being asked:** A rock of volume 1000 cm³ is submerged in an aquarium with a 100 cm × 40 cm base. How much does the water level rise?

**Setup:** When the rock is submerged, it displaces a volume of water equal to its own volume. This displaced water spreads across the base of the aquarium, causing the level to rise.

$$\text{Volume displaced} = \text{Base Area} \times \text{Rise in water level}$$

**Computation:**
- Base area = $100 \times 40 = 4000 \text{ cm}^2$
- $1000 = 4000 \times h$
- $h = \frac{1000}{4000} = 0.25 \text{ cm}$

**Check for overflow:** The current depth is 37 cm, and 37 + 0.25 = 37.25 cm, which is well below the 50 cm height. No overflow occurs.

**Ruling out other choices:** Choices B–E would require rock volumes of 2000, 4000, 5000, and 10000 cm³ respectively — all larger than the given 1000 cm³.

ANSWER 1: A

---

# Problem 2:

**What is being asked:** With 4 players, each gets 15 cards. With 6 players, how many cards does each get?

**Setup:** The total number of cards in the deck stays the same.

**Computation:**
- Number of players in first game: Annika + 3 friends = 4 players
- Total cards = $4 \times 15 = 60$ cards
- Number of players in second game: $4 + 2 = 6$ players
- Cards per player = $\frac{60}{6} = 10$

**Ruling out other choices:** Some might forget to count Annika (getting 3 players → 45 cards → 45/5 = 9, choice B) or miscount the new total as 5 players (60/5 = 12, choice E).

ANSWER 2: C

---

# Problem 3:

**What is being asked:** Evaluate the custom operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$ for $a=3, b=1, c=4, d=2$.

**Setup:** Substitute directly into the formula.

**Computation:**
$$\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2$$

**Ruling out other choices:** A common mistake is computing $b \cdot c - a \cdot d = 4 - 6 = -2$ (choice A) or mixing up which values go where.

ANSWER 3: E

---

# Problem 4:

**What is being asked:** Isabella needs an average of 95 on four tests. Her first two scores are 97 and 91. What is the lowest possible third test score so she can still reach her goal?

**Setup:**
- Total points needed: $4 \times 95 = 380$
- Sum of first two tests: $97 + 91 = 188$
- Sum needed from tests 3 and 4: $380 - 188 = 192$

**Computation:**
To minimize the third test score, maximize the fourth test score. The maximum possible score on any test is 100.
$$\text{Minimum test 3} = 192 - 100 = 92$$

**Ruling out other choices:** Choice A (90) would require test 4 to be $192 - 90 = 102$, which is impossible on a 100-point test. Choices C, D, E are higher than necessary.

ANSWER 4: B

---

# Problem 5:

**What is being asked:** Bridget bought apples, gave half to Ann, gave 3 to Cassie, and kept 4. How many did she buy?

**Setup:** Let $x$ = total apples bought. Work through the sequence of events.

**Computation:**
- After giving half to Ann: $\frac{x}{2}$ apples remain
- After giving 3 to Cassie: $\frac{x}{2} - 3$ apples remain
- She keeps 4: $\frac{x}{2} - 3 = 4$
- $\frac{x}{2} = 7$
- $x = 14$

**Verification:** 14 apples → give 7 to Ann → 7 left → give 3 to Cassie → 4 left. ✓

**Ruling out other choices:** Choice C (7) is a trap for those who forget to double the 7 remaining apples. Choice D (11) comes from incorrectly adding $4 + 3 + 4$.

ANSWER 5: E

---

# Problem 6:

**What is being asked:** Find the total number of marbles given conditions about each color.

**Setup:** Let $T$ = total marbles, $R$ = red, $G$ = green, $B$ = blue.
- "All but 6 are red" → $G + B = 6$ → $R = T - 6$
- "All but 8 are green" → $R + B = 8$ → $G = T - 8$
- "All but 4 are blue" → $R + G = 4$ → $B = T - 4$

**Computation:**
$$T = R + G + B = (T - 6) + (T - 8) + (T - 4) = 3T - 18$$
$$2T = 18 \implies T = 9$$

**Verification:** $R = 3, G = 1, B = 5$. Check: $G+B=6$ ✓, $R+B=8$ ✓, $R+G=4$ ✓.

ANSWER 6: C

---

# Problem 7:

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Setup:** Find the prime factorization of 250.

**Computation:**
$$250 = 2 \times 125 = 2 \times 5 \times 25 = 2 \times 5 \times 5 \times 5 = 2 \times 5^3$$

The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.
$$\text{Sum} = 2 + 5 = 7$$

**Ruling out other choices:** Choice A (2) is just one factor. Choice B (5) is just one factor. Choice D (10) would be $2 + 5 + 3$ or $5 + 5$, neither of which applies.

ANSWER 7: C

---

# Problem 8:

**What is being asked:** How many distinct sums can be formed by picking one chip from Bag A {1, 3, 5} and one from Bag B {2, 4, 6}?

**Setup:** List all 9 possible sums and count the unique values.

**Computation:**
| | 2 | 4 | 6 |
|---|---|---|---|
| **1** | 3 | 5 | 7 |
| **3** | 5 | 7 | 9 |
| **5** | 7 | 9 | 11 |

Unique sums: {3, 5, 7, 9, 11} → **5 different values**

**Ruling out other choices:** Choice E (9) is the trap of counting all 9 pairs without removing duplicates. Choice D (7) might come from miscounting.

ANSWER 8: B

---

# Problem 9:

**What is being asked:** How many ways can 2 boys and 3 girls sit in a row with boys at the ends and girls in the middle?

**Setup:** The arrangement must be: **B G G G B**

**Computation:**
- Arrange the 2 boys at the 2 end positions: $2! = 2$ ways
- Arrange the 3 girls in the 3 middle positions: $3! = 6$ ways
- Total arrangements: $2 \times 6 = 12$

**Ruling out other choices:** Choice A (2) only counts the boys' arrangements. Choice D (6) only counts the girls' arrangements. Both must be multiplied together.

ANSWER 9: E

---

# Problem 10:

**What is being asked:** Evaluate $16 + 8 \div 4 - 2$ using order of operations.

**Setup:** Apply PEMDAS — division before addition and subtraction.

**Computation:**
$$16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16$$

**Ruling out other choices:**
- Left-to-right without order of operations: $(16+8) \div 4 - 2 = 6 - 2 = 4$ (choice A)
- $(16+8) \div (4-2) = 24 \div 2 = 12$ (choice C)
- $16 + 8 \div (4-2) = 16 + 4 = 20$ (choice E

 Problem 1:
We need to find how much the water level rises when a rock of volume 1000 cm³ is submerged in a rectangular aquarium.

First, find the area of the aquarium’s base:
\[
100\text{ cm} \times 40\text{ cm} = 4000\text{ cm}^2.
\]

When the rock is completely submerged, it displaces 1000 cm³ of water. This displaced volume spreads out over the base, causing the water level to rise. The rise in height \(h\) satisfies:
\[
\text{base area} \times h = \text{volume of rock}.
\]
So,
\[
h = \frac{1000}{4000} = 0.25\text{ cm}.
\]

We should check that the aquarium does not overflow. The current depth is 37 cm and the height is 50 cm, so there is 13 cm of free space—plenty of room for a 0.25 cm rise.

Tempting wrong choices come from using the wrong area (e.g., using the side wall 100×50 or 40×50) or from dividing by the height 50. Only the base area matters for the level rise.

ANSWER 1: A

---

Problem 2:
We need to find how many cards each player gets when two extra friends join the game.

First, determine the total number of cards. Annika and 3 friends make 4 players, and each gets 15 cards:
\[
4 \times 15 = 60\text{ cards total}.
\]

When 2 more friends join, the number of players becomes:
\[
4 + 2 = 6.
\]

Distribute the 60 cards evenly among 6 players:
\[
60 \div 6 = 10.
\]

A tempting mistake is to subtract 2 from 15, giving 13, which is not an option, or to guess without computing the total deck size. The key is that the total number of cards stays the same.

ANSWER 2: C

---

Problem 3:
We are given the rule \(a\ c\ b\ d = a\cdot d - b\cdot c\). We need to evaluate \(3\ 1\ 4\ 2\).

Matching the pattern:
- \(a = 3\)
- \(c = 1\)
- \(b = 4\)
- \(d = 2\)

Apply the formula:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

A tempting error is to swap the pairs and compute \(3\cdot 1 - 4\cdot 2 = 3 - 8 = -5\), which is not among the choices, or to read the order incorrectly. Careful matching to \(a, c, b, d\) is essential.

ANSWER 3: E

---

Problem 4:
Isabella wants an average of 95 on four 100-point tests. Her first two scores are 97 and 91. We need the lowest possible score on the third test that still allows her to reach her goal.

Total points needed for a 95 average:
\[
4 \times 95 = 380.
\]

Points earned so far:
\[
97 + 91 = 188.
\]

Points still needed from the last two tests:
\[
380 - 188 = 192.
\]

Let the third test score be \(x\) and the fourth be \(y\), with \(x + y = 192\) and each score at most 100. To minimize \(x\), we maximize \(y\). The highest possible score is 100, so:
\[
x = 192 - 100 = 92.
\]

If she scores 92 on the third test, she can score 100 on the fourth and exactly reach her goal. A tempting wrong choice is 95, which would be the answer if she needed the two remaining tests to average 95, but she can shift points to the fourth test.

ANSWER 4: B

---

Problem 5:
We need to find how many apples Bridget originally bought.

Let the original number be \(x\).
- She gave half to Ann, leaving her with \(\frac{x}{2}\).
- She gave 3 to Cassie and kept 4:
\[
\frac{x}{2} - 3 = 4.
\]

Solve:
\[
\frac{x}{2} = 7 \quad\Rightarrow\quad x = 14.
\]

Check: Start with 14. Half is 7, given to Ann. She has 7 left. Gives 3 to Cassie, keeps 4. This matches.

A tempting error is to add \(3 + 4 = 7\) and stop, forgetting that 7 is only what remained after giving away half, so the original amount must be doubled.

ANSWER 5: E

---

Problem 6:
We need the total number of marbles given three conditions about how many are *not* each color.

Let \(R, G, B\) be the numbers of red, green, and blue marbles.
- “All but 6 are red” means non-red marbles equal 6: \(G + B = 6\).
- “All but 8 are green” means non-green marbles equal 8: \(R + B = 8\).
- “All but 4 are blue” means non-blue marbles equal 4: \(R + G = 4\).

Add all three equations:
\[
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
\]
\[
2R + 2G + 2B = 18
\]
\[
R + G + B = 9.
\]

So there are 9 marbles total. (Checking: \(R=3, G=1, B=5\), which satisfies all three statements.)

A tempting mistake is to add \(6+8+4=18\) and choose that, but that sum counts each marble twice.

ANSWER 6: C

---

Problem 7:
We need the sum of the two smallest prime factors of 250.

Factor 250:
\[
250 = 2 \times 125 = 2 \times 5^3.
\]

The prime factors are 2 and 5. The two smallest (and only) distinct prime factors are 2 and 5. Their sum is:
\[
2 + 5 = 7.
\]

A tempting wrong choice is 10, which comes from adding 5 and 5 (the two smallest factors with multiplicity), but the problem asks for the two smallest prime factors, i.e., distinct primes.

ANSWER 7: C

---

Problem 8:
Bag A has \(\{1, 3, 5\}\) and Bag B has \(\{2, 4, 6\}\). We draw one from each and want the number of distinct possible sums.

Compute all sums:
- \(1+2=3\), \(1+4=5\), \(1+6=7\)
- \(3+2=5\), \(3+4=7\), \(3+6=9\)
- \(5+2=7\), \(5+4=9\), \(5+6=11\)

The distinct values are:
\[
3, 5, 7, 9, 11.
\]

That is 5 different values. A tempting error is to count all 9 ordered pairs without removing duplicates.

ANSWER 8: B

---

Problem 9:
We have 2 boys and 3 girls to arrange in a row with a boy at each end and the three girls in the middle.

Positions: B _ _ _ G G G _ _ _ B (but specifically 5 seats: 1,2,3,4,5 with boys at 1 and 5, girls at 2,3,4).

- Choose the boy for the left end and the boy for the right end: \(2! = 2\) ways.
- Arrange the 3 girls in the middle three seats: \(3! = 6\) ways.

Total arrangements:
\[
2 \times 6 = 12.
\]

A tempting mistake is to ignore the constraint and compute \(5! = 120\), or to forget that the boys can swap places.

ANSWER 9: E

---

Problem 10:
Evaluate \(16 + 8 \div 4 - 2\).

Using order of operations (PEMDAS/BODMAS), division comes before addition and subtraction:
\[
8 \div 4 = 2.
\]

Now substitute back:
\[
16 + 2 - 2 = 16.
\]

A tempting error is to work left-to-right doing addition first: \((16+8)\div 4 - 2 = 24\div 4 - 2 = 6 - 2 = 4\), which is choice A. Another is to do \(16 + 2 = 18\), then \(18 - 2 = 16\), which luckily gives the same corr

Problem 1:
We need the rise in water level when a rock of volume 1000 cm³ is placed in an aquarium with base 100 cm by 40 cm. The rock displaces water equal to its volume, causing the water level to rise by Δh such that base area × Δh = rock volume. Base area = 100 cm × 40 cm = 4000 cm². So Δh = 1000 cm³ / 4000 cm² = 0.25 cm. The initial depth is 37 cm, and after rise it becomes 37.25 cm, which is still below the height of 50 cm, so the rock is fully submerged. The rise is 0.25 cm.

ANSWER 1: A

Problem 2:
When Annika and 3 friends play, there are 4 players, each dealt 15 cards. Total cards = 4 × 15 = 60. In the next game, 2 more friends join, making 6 players. Since the same total number of cards is used, each player gets 60 / 6 = 10 cards.

ANSWER 2: C

Problem 3:
The operation is defined as a c b d = a·d − b·c. For the numbers 3 1 4 2, we have a = 3, c = 1, b = 4, d = 2. Compute a·d − b·c = 3×2 − 4×1 = 6 − 4 = 2.

ANSWER 3: E

Problem 4:
Isabella wants an average of 95 over four tests, so total points needed = 4 × 95 = 380. Her first two scores sum to 97 + 91 = 188. Let the third score be x. After the third test, she can still reach the goal if it is possible to score at most 100 on the fourth test to reach 380. That requires 188 + x + 100 ≥ 380 → x ≥ 380 − 288 = 92. The smallest integer x satisfying this is 92. If x = 91, then even a perfect 100 on the fourth gives 188+91+100 = 379 < 380. So the lowest possible third score is 92.

ANSWER 4: B

Problem 5:
Let the number of apples Bridget bought be N. She gave half to Ann, leaving N/2. Then she gave Cassie 3 apples, leaving N/2 − 3. She keeps 4 apples, so N/2 − 3 = 4 → N/2 = 7 → N = 14.

ANSWER 5: E

Problem 6:
Let T be the total number of marbles. "All but 6 are red" means T − R = 6 → R = T − 6. Similarly, T − G = 8 → G = T − 8, and T − B = 4 → B = T − 4. Since R + G + B = T, we have (T−6)+(T−8)+(T−4) = T → 3T − 18 = T → 2T = 18 → T = 9.

ANSWER 6: C

Problem 7:
Factor 250: 250 = 2 × 5³. Its prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5, and their sum is 7.

ANSWER 7: C

Problem 8:
List all possible sums from drawing one chip from Bag A {1,3,5} and one from Bag B {2,4,6}:
1+2=3, 1+4=5, 1+6=7;
3+2=5, 3+4=7, 3+6=9;
5+2=7, 5+4=9, 5+6=11.
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.

ANSWER 8: B

Problem 9:
We need arrangements of 2 boys and 3 girls in a row with a boy at each end. Choose which boy sits on the left end and which on the right: 2! = 2 ways. The three girls occupy the three middle seats and can be arranged in 3! = 6 ways. Total arrangements = 2 × 6 = 12.

ANSWER 9: E

Problem 10:
The expression is 16 + 8 ÷ 4 − 2 (the missing operator is division, as typical for order-of-operations problems). Following order of operations, division before addition and subtraction: 8 ÷ 4 = 2. Then 16 + 2 = 18, and 18 − 2 = 16.

ANSWER 10: D

Problem 11:
Evaluate step by step: √4 = 2. Then 8 · 2 = 16, so √(8·√4) = √16 = 4. Then 16 · 4 = 64, so √(16·√(8·√4)) = √64 = 8.

ANSWER 11: C

Problem 12:
Compare the decimal numbers:
A. 0.99
B. 0.9099
C. 0.9
D. 0.909
E. 0.9009
Clearly 0.99 is the largest because it is greater than 0.9 and all the others are less than 0.91.

ANSWER 12: A

Problem 1:
The aquarium’s base has area  
\(A = 100\text{ cm} \times 40\text{ cm}=4000\text{ cm}^2\).  
When the rock of volume \(V=1000\text{ cm}^3\) is fully submerged, the water rises by a height \(h\) such that  

\[
A\cdot h = V \quad\Longrightarrow\quad h = \frac{1000}{4000}=0.25\text{ cm}.
\]

The rise cannot be any of the larger listed values (0.5 cm, 1 cm, …) because those would require a volume larger than the rock’s.  

ANSWER 1: A  

---

Problem 2:
With 4 players each receiving 15 cards, the total number of cards is  

\[
4 \times 15 = 60.
\]

If 2 more friends join, there are \(4+2 = 6\) players.  The cards are still all the cards, so each player receives  

\[
\frac{60}{6}=10\text{ cards}.
\]

The other answer choices (8, 9, 11, 12) would not distribute the 60 cards evenly.  

ANSWER 2: C  

---

Problem 3:
The operation is defined by  

\[
a\;c\;b\;d = a\cdot d - b\cdot c.
\]

For the numbers \(3\;1\;4\;2\) we have \(a=3,\;c=1,\;b=4,\;d=2\).  Hence  

\[
3\;1\;4\;2 = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

The other options \(-2,-1,0,1\) would correspond to different combinations that do not satisfy the given formula.  

ANSWER 3: E  

---

Problem 4:
To average 95 over four 100‑point tests, the total points needed are  

\[
4 \times 95 = 380.
\]

After the first two tests Isabella has  

\[
97 + 91 = 188\text{ points}.
\]

Let the third test score be \(x\).  Then the fourth test must supply  

\[
380 - (188 + x) = 192 - x \text{ points}.
\]

The fourth test can give at most 100 points, so we need  

\[
192 - x \le 100 \quad\Longrightarrow\quad x \ge 92.
\]

If she scores \(92\) on the third test, she would need a perfect 100 on the fourth, which is possible.  Any lower score would require more than 100 points on the fourth test, which is impossible.  

Thus the lowest possible third‑test score is **92**.  

ANSWER 4: B  

---

Problem 5:
Let the original number of apples be \(N\).  
She gave half of them to Ann: \(\frac{N}{2}\).  
She then gave Cassie 3 apples and kept 4, leaving  

\[
N - \frac{N}{2} - 3 = 4.
\]

Simplify: \(\frac{N}{2} - 3 = 4 \Rightarrow \frac{N}{2}=7 \Rightarrow N=14\).

If the total had been any of the other answer choices, the steps would not balance (e.g., 11 would leave 2 apples after giving Cassie 3).  

ANSWER 5: E  

---

Problem 6:
Let the total number of marbles be \(T\).  
- “All but 6 are red’’ → red = \(T-6\).  
- “All but 8 are green’’ → green = \(T-8\).  
- “All but 4 are blue’’ → blue = \(T-4\).

These three colors account for all marbles, so  

\[
(T-6)+(T-8)+(T-4)=T \;\Longrightarrow\; 3T-18=T \;\Longrightarrow\; 2T=18 \;\Longrightarrow\; T=9.
\]

The other totals (6, 8, 10, 12) would not satisfy the three “all but…’’ statements simultaneously.  

ANSWER 6: C  

---

Problem 7:
Factor \(250\):  

\[
250 = 2 \times 5^3.
\]

The prime factors are \(2\) and \(5\).  The two smallest are \(2\) and \(5\), whose sum is  

\[
2+5=7.
\]

Choices 2, 5, 10, 12 are either single prime factors or incorrect sums.  

ANSWER 7: C  

---

Problem 8:
Bag A contains \(\{1,3,5\}\); Bag B contains \(\{2,4,6\}\).  
All possible sums \(a+b\) are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The distinct results are \(\{3,5,7,9,11\}\), a total of **5** different sums.  

The other answer choices (4, 6, 7, 9) do not match the count of distinct sums.  

ANSWER 8: B  

---

Problem 9:
Place the two boys at the two ends and the three girls in the middle three seats.

- The two boys can be ordered in \(2!\) ways (left‑right swap).  
- The three girls can be arranged among the three middle seats in \(3! = 6\) ways.

Total arrangements  

\[
2! \times 3! = 2 \times 6 = 12.
\]

Choices 2, 4, 5, 6 are too small; 12 accounts for all possible permutations under the given constraints.  

ANSWER 9: E  

---

Problem 10:
Interpret the expression as  

\[
16 + 8 \div 4 - 2.
\]

Division comes before addition and subtraction, so  

\[
8 \div 4 = 2 \quad\Longrightarrow\quad 16 + 2 - 2 = 16.
\]

The other options (4, 8, 12, 20) would result only if the order of operations were misinterpreted.  

ANSWER 10: D  

---

Problem 11:
Simplify from the innermost radical outward.

\[
\sqrt{4}=2,\qquad 
\sqrt{8\cdot\sqrt{4}} = \sqrt{8\cdot2}= \sqrt{16}=4.
\]

Now  

\[
\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{16\cdot4}= \sqrt{64}=8.
\]

Thus the value is 8, matching choice C.  

ANSWER 11: C  

---

Problem 12:
Compare the decimal values:

- A. \(0.99\)
- B. \(0.9099\)
- C. \(0.9\)
- D. \(0.909\)
- E. \(0.9009\)

Clearly \(0.99\) is greater than \(0.9099\), which is greater than \(0.909\), etc.  
Hence the largest number is \(0.99\).  

ANSWER 12: A

Problem 1:  
An aquarium has a rectangular base of 100 cm by 40 cm and height 50 cm, filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is submerged. By how many centimeters does the water level rise?  

The base area is \(100 \times 40 = 4000 \, \text{cm}^2\). The rock’s volume is 1000 cm³, so the rise in water level is \(\frac{1000}{4000} = 0.25 \, \text{cm}\).  

Tempting choices: 0.5 (B) or 1 (C) might be guessed if base area is miscalculated.  

ANSWER 1: A  

Problem 2:  
Annika and 3 friends (4 players) get 15 cards each. Total cards = \(4 \times 15 = 60\). If 2 more join (6 players), each gets \(\frac{60}{6} = 10\).  

Tempting choices: 8 (A) if total cards miscalculated as 48.  

ANSWER 2: C  

Problem 3:  
Operation defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\): \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\), but the order is \(a c b d\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). However, the problem writes "a c b d" which is interpreted as \(a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). But the choices include 2 (E). Re-evaluating: the operation is \(a c b d = a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\).  

Tempting choices: -2 (A) if signs reversed.  

ANSWER 3: A (Note: Correction based on standard interpretation: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\), but the correct answer based on the operation defined is actually \(3 \cdot 2 - 1 \cdot 4 = 2\), but the problem writes "a c b d", which is \(a \cdot d - b \cdot c\), so 2. However, the answer is 2, but the choices include 2 as E. The correct calculation is 2, so E. Correction: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\).)  

**Correction:** The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\), this is \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\).  

ANSWER 3: E  

Problem 4:  
Isabella wants an average of 95 on four 100-point tests. Total points needed: \(4 \times 95 = 380\). First two scores: 97 + 91 = 188. Let third score be \(x\), fourth be \(y\). Then \(188 + x + y = 380\), so \(x + y = 192\). To minimize \(x\), maximize \(y\): \(y = 100\), so \(x = 92\).  

Tempting choices: 90 (A) if \(y\) not maximized.  

ANSWER 4: B  

Problem 5:  
Bridget gave half apples to Ann, then 3 to Cassie, keeping 4. Let total be \(x\). After giving half to Ann: \(\frac{x}{2}\) left. Then gives 3 to Cassie: \(\frac{x}{2} - 3 = 4\). So \(\frac{x}{2} = 7\), \(x = 14\).  

Tempting choices: 7 (C) if half misinterpreted.  

ANSWER 5: E  

Problem 6:  
Let \(r, g, b\) be red, green, blue marbles. Then:  
- All but 6 are red: \(g + b = 6\)  
- All but 8 are green: \(r + b = 8\)  
- All but 4 are blue: \(r + g = 4\)  
Adding: \(2(r + g + b) = 18\), so total \(r + g + b = 9\).  

Tempting choices: 6 (A) if equations misread.  

ANSWER 6: C  

Problem 7:  
Prime factors of 250: \(250 = 2 \times 5^3\). Two smallest: 2 and 5. Sum: \(2 + 5 = 7\).  

Tempting choices: 5 (B) if only 5 considered.  

ANSWER 7: C  

Problem 8:  
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}. Possible sums:  
- 1+2=3, 1+4=5, 1+6=7  
- 3+2=5, 3+4=7, 3+6=9  
- 5+2=7, 5+4=9, 5+6=11  
Unique sums: 3, 5, 7, 9, 11 → 5 values.  

Tempting choices: 6 (C) if duplicates counted.  

ANSWER 8: B  

Problem 9:  
Two boys at ends: 2 ways. Three girls in middle: \(3! = 6\) ways. Total: \(2 \times 6 = 12\).  

Tempting choices: 6 (D) if boys not permuted.  

ANSWER 9: E  

Problem 10:  
Expression: \(16 + 8 \times 4 - 2\). Order: \(8 \times 4 = 32\), then \(16 + 32 - 2 = 46\)? Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 \times 4 - 2\): \(8 \times 4 = 32\), \(16 + 32 = 48\), \(48 - 2 = 46\). But choices max at 20. Re-interpreting: "16 + 8 4 - 2" might mean \(16 + 8 - 4 - 2 = 18\), not in choices. Alternatively, if "8 4" is 8×4, but 46 not in choices. The expression is likely \(16 + 8 - 4 - 2 = 18\), but not in choices. Correction: The expression is "16 + 8 4 − 2", which might be \(16 + (8 \times 4) - 2 = 16 + 32 - 2 = 46\), but not in choices. Re-reading: It’s "16 + 8 4 − 2", which could be \(16 + 8 - 4 - 2 = 18\), but not in choices. The intended expression is likely \(16 + 8 \times 4 - 2 = 46\), but choices max at 20. Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 - 4 - 2 = 18\), but not in choices. Alternatively, if "8 4" is 84, but 16+84-2=98. Not matching. The expression is likely \( \frac{16 + 8}{4} - 2 = \frac{24}{4} - 2 = 6 - 2 = 4 \).  

Tempting choices: 8 (B) if order misapplied.  

ANSWER 10: A  

Problem 11:  
Expression: \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\).  
- \(\sqrt{4} = 2\)  
- \(\sqrt{8 \cdot 2} = \sqrt{16} = 4\)  
- \(\sqrt{16 \cdot 4} = \sqrt{64} = 8\)  

Tempting choices: 4 (A) if innermost not simplified.  

ANSWER 11: C  

Problem 12:  
Compare decimals:  
- A: 0.99  
- B: 0.9099  
- C: 0.9  
- D: 0.909  
- E: 0.9009  
Largest is 0.99.  

Tempting choices: 0.9099 (B) if decimals misaligned.  

ANSWER 12: A  

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

### Problem 1:
What is asked: An aquarium with a 100cm×40cm rectangular base is filled with water to 37cm depth. A 1000 cm³ rock is submerged; how much does the water level rise?
Step-by-step: When a submerged object displaces water, its volume equals the additional volume of water in the aquarium. The rise in water level Δh satisfies: Volume of rock = base area of aquarium × Δh.
Base area = 100×40 = 4000 cm².
Δh = 1000 / 4000 = 0.25 cm.
Tempting wrong answers: 0.5 (if base area is miscalculated as 2000 cm²) or 1 (wrong division).
ANSWER 1: A

---

### Problem 2:
What is asked: Annika plus 3 friends (4 total players) each get 15 cards. If 2 more join, how many cards does each player get?
Step-by-step: First calculate total number of cards: 4 players × 15 cards = 60 total cards. After 2 join, total players = 4+2=6. Cards per player = 60 / 6 = 10.
Tempting wrong answers: 9 (if you incorrectly count only 3 initial players, leading to 45 total cards) or 12 (wrong division).
ANSWER 2: C

---

### Problem 3:
What is asked: The custom notation a c b d = a·d − b·c. What is the value of 3 1 4 2?
Step-by-step: Match the values to the definition: a=3, c=1, b=4, d=2. Calculate: 3·2 - 4·1 = 6 - 4 = 2.
Tempting wrong answers: -2 (swap a and b incorrectly) or 0 (miscalculate the multiplication).
ANSWER 3: E

---

### Problem 4:
What is asked: Isabella needs a 95 average on 4 100-point tests. She scored 97 and 91 on the first two. What is the lowest possible third test score that lets her reach her goal?
Step-by-step: Total points needed for 95 average: 4×95 = 380 points. Sum of first two tests: 97+91=188. To minimize the third test score, maximize the fourth test (max 100 points). Total points left for tests 3 and 4: 380-188=192. Minimum third test = 192 - 100 = 92.
Tempting wrong answers: 96 (the average of the remaining two tests, not the minimum possible third score).
ANSWER 4: B

---

### Problem 5:
What is asked: Bridget gave half her apples to Ann, 3 to Cassie, and kept 4. How many apples did she buy originally?
Step-by-step: Let N = original number of apples. After giving half to Ann, she has N/2 left. Then N/2 - 3 = 4 → N/2=7 → N=14. Verify: 14 → 7 left after Ann, minus 3 for Cassie = 4 kept, which matches.
Tempting wrong answers: 7 (the number of apples Bridget had after giving half to Ann, not the original total).
ANSWER 5: E

---

### Problem 6:
What is asked: A jar has red, green, blue marbles. All but 6 are red, all but 8 are green, all but 4 are blue. How many total marbles?
Step-by-step: Let T=total marbles, R=red, G=green, B=blue. The conditions translate to: G+B=6, R+B=8, R+G=4. Add all equations: 2(R+G+B)=18 → R+G+B=T=9. Verify: 3 red, 1 green, 5 blue sum to 9, matching all conditions.
Tempting wrong answers: 18 (the sum of 6+8+4, not halved to get total marbles).
ANSWER 6: C

---

### Problem 7:
What is asked: What is the sum of the two smallest prime factors of 250?
Step-by-step: Prime factorization of 250: 250=2×5³. The prime factors are 2 and 5. Their sum is 2+5=7.
Tempting wrong answers: 10 (sum of 5+5, or wrong addition of 2+5).
ANSWER 7: C

---

### Problem 8:
What is asked: Bag A has chips 1,3,5; Bag B has 2,4,6. One chip drawn from each bag, how many distinct sum values are possible?
Step-by-step: List all sums: 1+2=3, 1+4=5,1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7,5+4=9,5+6=11. Distinct sums: 3,5,7,9,11 → 5 total.
Tempting wrong answers: 9 (total number of unordered pairs, not distinct sums).
ANSWER 8: B

---

### Problem 9:
What is asked: 2 boys and 3 girls line up for a photo, with a boy at each end and girls in the middle. How many arrangements are possible?
Step-by-step: Arrange the boys first: 2 boys can be ordered at the ends in 2! = 2 ways. The 3 middle girls can be ordered in 3! =6 ways. Total arrangements: 2×6=12.
Tempting wrong answers: 6 (only the number of ways to arrange the girls, ignoring the boys).
ANSWER 9: E

---

### Problem 10:
What is asked: Evaluate (16+8)/(4-2) (formatted as a fraction in the original problem).
Step-by-step: Calculate numerator:16+8=24, denominator:4-2=2. 24/2=12.
Tempting wrong answers:16 (if order of operations is misapplied to calculate 16 + 8/4 -2=16).
ANSWER 10: C

---

### Problem 11:
What is asked: Evaluate √(16 · √(8 · √4)).
Step-by-step: Calculate from the innermost root first: √4=2. Next: 8×2=16, √16=4. Next:16×4=64, √64=8.
Tempting wrong answers: 4√2 (incorrect root calculation).
ANSWER 11: C

---

### Problem 12:
What is asked: Which number is largest: 0.99, 0.9099, 0.9, 0.909, 0.9009.
Step-by-step: Line up all decimals to compare place values: 0.9900 > 0.9099 > 0.9090 > 0.9009 > 0.9000. The largest is 0.99.
Tempting wrong answers: 0.9099 (the second largest value).
ANSWER 12: A

### Problem 1:
We need to find how much the water level rises when a 1000 cm³ rock is submerged in the aquarium.
1.  The volume of displaced water equals the volume of the submerged rock: 1000 cm³.
2.  The base area of the aquarium is \(100 \, \text{cm} \times 40 \, \text{cm} = 4000 \, \text{cm}^2\).
3.  Let \(h\) be the rise in water level. The volume of the displaced water is also equal to base area × rise in height: \(4000h = 1000\).
4.  Solve for \(h\): \(h = \frac{1000}{4000} = 0.25\) cm.
ANSWER 1: A
---
### Problem 2:
We need to find the number of cards per player when 2 more friends join.
1.  Original number of players: Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).
2.  New number of players: 4 + 2 = 6 players. Total cards remain 60.
3.  Cards per player: \(\frac{60}{6} = 10\).
ANSWER 2: C
---
### Problem 3:
The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). Compute \(3 \, 1 \, 4 \, 2\).
1.  Identify \(a=3, c=1, b=4, d=2\).
2.  Substitute into the formula: \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).
ANSWER 3: E
---
### Problem 4:
Find the lowest possible third test score to get an average of 95 on 4 tests.
1.  Total points needed for an average of 95: \(4 \times 95 = 380\).
2.  Sum of first two test scores: \(97 + 91 = 188\).
3.  Let \(x\) = third test score, \(y\) = fourth test score (max 100, since tests are 100-point). We need \(188 + x + y = 380\), so \(x + y = 192\).
4.  To minimize \(x\), maximize \(y = 100\). Then \(x = 192 - 100 = 92\).
ANSWER 4: B
---
### Problem 5:
Find the original number of apples Bridget bought.
1.  Let \(x\) = total apples. After giving half to Ann, she has \(\frac{x}{2}\) left.
2.  After giving 3 to Cassie, she has \(\frac{x}{2} - 3\) left, which equals 4 (the apples she kept).
3.  Solve: \(\frac{x}{2} - 3 = 4 \implies \frac{x}{2} =7 \implies x=14\).
ANSWER 5: E
---
### Problem 6:
Find the total number of marbles.
1.  Let \(R, G, B\) be the number of red, green, blue marbles, total \(T = R+G+B\).
2.  Translate the statements:
    - All but 6 are red: \(G + B = 6\)
    - All but 8 are green: \(R + B = 8\)
    - All but 4 are blue: \(R + G =4\)
3.  Add all three equations: \(2(R+G+B) = 6+8+4=18 \implies 2T=18 \implies T=9\).
ANSWER 6: C
---
### Problem 7:
Find the sum of the two smallest prime factors of 250.
1.  Factorize 250: \(250 = 2 \times 125 = 2 \times 5^3\).
2.  The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.
3.  Sum: \(2 +5=7\).
ANSWER 7: C
---
### Problem 8:
Find the number of distinct sums when drawing one chip from each bag.
1.  List all possible pairs and sums:
    - Bag A (1,3,5) + Bag B (2,4,6):
      \(1+2=3, 1+4=5, 1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7, 5+4=9, 5+6=11\)
2.  Distinct sums: 3, 5, 7, 9, 11. There are 5 unique values.
ANSWER 8: B
---
### Problem 9:
Find the number of arrangements with boys at each end and girls in the middle.
1.  Arrange the 2 boys at the ends: there are \(2! = 2\) ways (order matters: which boy is left/right).
2.  Arrange the 3 girls in the middle 3 seats: \(3! = 6\) ways.
3.  Total arrangements: \(2 \times 6 =12\).
ANSWER 9: E
---
### Problem 10:
Compute \(16 + 8 \, 4 - 2\). The missing operator between 8 and 4 is division (standard order of operations, common middle school problem):
1.  Interpret as \(16 + \frac{8}{4} - 2\).
2.  Compute division first: \(\frac{8}{4}=2\).
3.  Then add/subtract left to right: \(16 + 2 - 2 =16\).
*(Note: If interpreted as \(16 +8 -4 \times 2\), order of operations gives \(4 \times 2=8\), so \(16+8-8=16\), same result.)*
ANSWER 10: D
---
### Problem 11:
Compute \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\) by working from the innermost root outward.
1.  Innermost: \(\sqrt{4} = 2\).
2.  Next layer: \(\sqrt{8 \cdot 2} = \sqrt{16} =4\).
3.  Outermost: \(\sqrt{16 \cdot 4} = \sqrt{64} =8\).
ANSWER 11: C
---
### Problem 12:
Find the largest decimal.
1.  Rewrite all numbers with 4 decimal places for easy comparison:
    - A: \(0.9900\)
    - B: \(0.9099\)
    - C: \(0.9000\)
    - D: \(0.9090\)
    - E: \(0.9009\)
2.  Compare digit by digit: all have 9 in the tenths place. A has 9 in the hundredths place, while all others have 0. Thus A is the largest.
ANSWER 12: A

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

**Problem 1:**

What is being asked: By how many centimeters does the water level rise when a 1000 cm³ rock is submerged?

Setup: The base area is 100 cm × 40 cm = 4000 cm². When the rock is submerged, it displaces 1000 cm³ of water.

Water level rise = Volume displaced ÷ Base area = 1000 ÷ 4000 = 0.25 cm

ANSWER 1: A

---

**Problem 2:**

What is being asked: If cards are distributed evenly, how many cards per player when the player count increases?

Setup: Initially 4 players (Annika + 3 friends) each get 15 cards.
Total cards = 4 × 15 = 60 cards

With 2 more friends joining: 4 + 2 = 6 players
Cards per player = 60 ÷ 6 = 10 cards

ANSWER 2: C

---

**Problem 3:**

What is being asked: Evaluate the operation 3 1 4 2 using the rule a c b d = a·d − b·c

Setup: With a = 3, c = 1, b = 4, d = 2:
3 1 4 2 = 3·2 − 4·1 = 6 − 4 = 2

ANSWER 3: E

---

**Problem 4:**

What is being asked: What is the lowest score on test 3 that allows Isabella to achieve a 95 average on four tests?

Setup: Goal average is 95, so total needed = 95 × 4 = 380 points
First two tests: 97 + 91 = 188 points
Points needed from tests 3 and 4: 380 − 188 = 192 points

Let x = score on test 3. She can still reach her goal if she can score 100 on test 4:
x + 100 ≥ 192
x ≥ 92

The lowest possible score on test 3 is 92.

ANSWER 4: B

---

**Problem 5:**

What is being asked: How many apples did Bridget buy originally?

Setup: Let x = original number of apples
- She gave half to Ann: x/2 apples given away
- Remaining: x/2
- She gave 3 to Cassie: x/2 − 3
- She kept 4 for herself: x/2 − 3 = 4

Solving: x/2 = 7, so x = 14

ANSWER 5: E

---

**Problem 6:**

What is being asked: How many marbles are in the jar?

Setup: Let R, G, B = numbers of red, green, blue marbles; T = total
- "All but 6 are red": G + B = 6
- "All but 8 are green": R + B = 8
- "All but 4 are blue": R + G = 4

Adding all three equations: 2(R + G + B) = 6 + 8 + 4 = 18
Therefore: R + G + B = 9

ANSWER 6: C

---

**Problem 7:**

What is being asked: What is the sum of the two smallest prime factors of 250?

Setup: Factor 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5³

Prime factors: 2 and 5
Sum = 2 + 5 = 7

ANSWER 7: C

---

**Problem 8:**

What is being asked: How many different sums are possible?

Setup: Bag A: {1, 3, 5}, Bag B: {2, 4, 6}
All possible sums:
- 1 + 2 = 3
- 1 + 4 = 5
- 1 + 6 = 7
- 3 + 2 = 5 (duplicate)
- 3 + 4 = 7 (duplicate)
- 3 + 6 = 9
- 5 + 2 = 7 (duplicate)
- 5 + 4 = 9 (duplicate)
- 5 + 6 = 11

Distinct sums: {3, 5, 7, 9, 11} = 5 different values

ANSWER 8: B

---

**Problem 9:**

What is being asked: How many seating arrangements with boys at ends and girls in middle?

Setup: 2 boys at the ends can be arranged in 2! = 2 ways
3 girls in the middle can be arranged in 3! = 6 ways
Total arrangements = 2 × 6 = 12

ANSWER 9: E

---

**Problem 10:**

What is being asked: Evaluate 16 + 8 ÷ 4 − 2

Setup: Using order of operations (division before addition/subtraction):
16 + (8 ÷ 4) − 2 = 16 + 2 − 2 = 16

ANSWER 10: D

---

**Problem 11:**

What is being asked: Evaluate √(16 · √(8 · √4))

Setup: Work from the inside out:
- √4 = 2
- 8 · √4 = 8 · 2 = 16
- √(8 · √4) = √16 = 4
- 16 · √(8 · √4) = 16 · 4 = 64
- √(16 · √(8 · √4)) = √64 = 8

ANSWER 11: C

---

**Problem 12:**

What is being asked: Which decimal is largest?

Setup: Compare decimals:
- A: 0.99
- B: 0.9099
- C: 0.9
- D: 0.909
- E: 0.9009

Comparing: 0.99 > 0.9099 > 0.909 > 0.9009 > 0.9

The largest is 0.99

ANSWER 12: A

Problem 1:

The aquarium has a rectangular base of \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is

\[
100 \cdot 40 = 4000 \text{ cm}^2.
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

If the water rises by \(h\) centimeters, then

\[
4000h = 1000.
\]

So

\[
h = \frac{1000}{4000} = \frac14 = 0.25.
\]

So the water level rises \(0.25\) cm.

ANSWER 1: A

---

Problem 2:

Annika and 3 friends means there are

\[
4 \text{ players}.
\]

If each player gets 15 cards, then the total number of cards is

\[
4 \cdot 15 = 60.
\]

Now 2 more friends join, so there are

\[
6 \text{ players}.
\]

The same 60 cards are shared evenly, so each gets

\[
60 \div 6 = 10.
\]

ANSWER 2: C

---

Problem 3:

We are told that if the four numbers are \(a, b, c, d\), then

\[
a \ b \ c \ d = a\cdot d - b\cdot c.
\]

So for \(3,1,4,2\),

\[
3 \ 1 \ 4 \ 2 = 3\cdot 2 - 1\cdot 4 = 6 - 4 = 2.
\]

ANSWER 3: E

---

Problem 4:

She wants an average of 95 on 4 tests, so the total of the 4 scores must be

\[
4 \cdot 95 = 380.
\]

Her first two scores are 97 and 91, which add to

\[
97+91=188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To make the total 380, the fourth score would have to be

\[
380 - (188+x) = 192 - x.
\]

For this to be possible, the fourth score must be at most 100:

\[
192 - x \le 100.
\]

So

\[
x \ge 92.
\]

The lowest possible third score is 92.

ANSWER 4: B

---

Problem 5:

Let the number of apples Bridget bought be \(x\).

She gave half to Ann, so she kept the other half:

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

Then she gave Cassie 3 apples and kept 4 apples, so before giving Cassie the apples she had

\[
3+4=7.
\]

Thus,

\[
\frac{x}{2}=7 \quad \Rightarrow \quad x=14.
\]

ANSWER 5: E

---

Problem 6:

Let the total number of marbles be \(n\).

- “All but 6 are red” means the number of red marbles is \(n-6\).
- “All but 8 are green” means the number of green marbles is \(n-8\).
- “All but 4 are blue” means the number of blue marbles is \(n-4\).

Since every marble is one of these colors, their counts add to \(n\):

\[
(n-6)+(n-8)+(n-4)=n.
\]

Simplify:

\[
3n-18=n
\]

\[
2n=18
\]

\[
n=9.
\]

ANSWER 6: C

---

Problem 7:

We need the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3.
\]

So the prime factors are 2 and 5. The two smallest prime factors are 2 and 5, and their sum is

\[
2+5=7.
\]

ANSWER 7: C

---

Problem 8:

Bag A has \(1,3,5\); Bag B has \(2,4,6\).

Possible sums:

- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums are

\[
3,5,7,9,11
\]

which is 5 different values.

ANSWER 8: B

---

Problem 9:

There are 2 boys and 3 girls, and the picture must have:

- a boy at each end,
- the 3 girls in the middle.

First, arrange the 2 boys at the ends. They can be switched in

\[
2! = 2
\]

ways.

Then arrange the 3 girls in the middle in

\[
3! = 6
\]

ways.

So the total number of arrangements is

\[
2 \cdot 6 = 12.
\]

ANSWER 9: E

---

Problem 10:

The expression is

\[
16 + 8\cdot 4 - 2.
\]

Use order of operations:

\[
8\cdot 4=32.
\]

Then

\[
16+32-2=46.
\]

But 46 is not among the choices, so the intended expression must be read as

\[
16 + 8 \div 4 - 2.
\]

Then

\[
8\div 4 = 2,
\]

so

\[
16+2-2=16.
\]

ANSWER 10: D

---

Problem 11:

We evaluate from the inside out:

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

First,

\[
\sqrt4=2.
\]

So the expression becomes

\[
\sqrt{16\cdot \sqrt{8\cdot 2}}
=
\sqrt{16\cdot \sqrt{16}}.
\]

Since \(\sqrt{16}=4\),

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C

---

Problem 12:

Compare the decimals:

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

The largest is the one closest to 1. Since \(0.99\) is greater than any number beginning with \(0.90\), it is the largest.

ANSWER 12: A

Problem 1:  
We need the rise in water level when a rock is placed in the aquarium and submerged.

- Base dimensions: \(100\text{ cm} \times 40\text{ cm}\), so base area  
  \[
  A = 100\cdot 40 = 4000\text{ cm}^2.
  \]
- The rock is completely submerged, so the water volume displaced equals the rock’s volume:  
  \[
  V = 1000\text{ cm}^3.
  \]
- Water level rise \(\Delta h\) satisfies \(A\Delta h = V\):  
  \[
  4000\cdot \Delta h = 1000 \implies \Delta h = \frac{1000}{4000}=0.25\text{ cm}.
  \]

So the rise is \(0.25\) cm.  
Tempting wrong idea: using the *remaining* depth \(37\) cm or checking if it overflows—none of that matters; only displaced volume matters.

ANSWER 1: A  

---

Problem 2:  
All cards are dealt evenly. Let the total number of cards be \(T\).

Initially, Annika + 3 friends = 4 players, and each gets 15 cards:  
\[
T = 4\cdot 15 = 60.
\]
Next game: 2 more friends join, so total players:
\[
4+2=6.
\]
Each player gets:
\[
\frac{60}{6}=10.
\]

ANSWER 2: C  

---

Problem 3:  
Given the rule \(a\,b\,d\,c = a\cdot d - b\cdot c\).  
The expression “\(3\ 1\ 4\ 2\)” corresponds to:
- \(a=3\)
- \(b=1\)
- \(d=4\)
- \(c=2\)

Compute:
\[
a\cdot d - b\cdot c = 3\cdot 4 - 1\cdot 2 = 12-2=10.
\]

But notice the choices are only \(-2,-1,0,1,2\), so we must interpret the placeholder correctly:  
The problem states: **“If \(a b c d = a\cdot d - b\cdot c\)”**.

So “\(3\ 1\ 4\ 2\)” means:
- \(a=3,\ b=1,\ c=4,\ d=2\)

Then:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 1\cdot 4 = 6-4=2.
\]

ANSWER 3: E  

---

Problem 4:  
Goal average is 95 over 4 tests, so total points needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
Let the third test score be \(x\), and the fourth be \(y\). Then:
\[
188 + x + y = 380 \implies x+y=192.
\]
We want the **lowest possible** \(x\) such that she can still reach the goal. That means make \(y\) as large as possible (since grades are at most 100).
Maximum \(y=100\). Then:
\[
x + 100 = 192 \implies x = 92.
\]

Check: then the four scores are \(97,91,92,100\) which sum to \(380\), average \(95\). Works.

ANSWER 4: B  

---

Problem 5:  
Let the total number of apples be \(N\).

- She gave half to Ann, so Ann got \(N/2\), leaving \(N/2\).
- Then she gave Cassie 3 apples, and kept 4 apples. That means after giving Cassie 3, her remaining is 4:
  \[
  \frac{N}{2} - 3 = 4 \implies \frac{N}{2}=7 \implies N=14.
  \]

ANSWER 5: E  

---

Problem 6:  
“All but 6 are red” means: number of red marbles is 6.  
Similarly:
- Red = 6 (since all but 6 are red → 6 red)
- Green: all but 8 are green → 8 green
- Blue: all but 4 are blue → 4 blue

Total marbles:
\[
6+8+4=18.
\]
But none of the choices are 18, so interpret carefully:  
“All but 6 are red” means the number of red marbles is **(total − 6)**, not 6.

Let total be \(T\). Then:
- Red marbles = \(T-6\)
- Green marbles = \(T-8\)
- Blue marbles = \(T-4\)

Sum:
\[
(T-6)+(T-8)+(T-4)=T.
\]
\[
3T - (6+8+4)=T
\]
\[
3T-18=T \implies 2T=18 \implies T=9.
\]

ANSWER 6: C  

---

Problem 7:  
We want the sum of the **two smallest prime factors** of 250.

Factor \(250\):
\[
250 = 25\cdot 10 = 5^2 \cdot 2\cdot 5 = 2\cdot 5^3.
\]
Prime factors are \(2\) and \(5\). The two smallest are \(2\) and \(5\).

Sum:
\[
2+5=7.
\]

ANSWER 7: D  

---

Problem 8:  
Bag A: \(\{1,3,5\}\)  
Bag B: \(\{2,4,6\}\)

Possible sums:
- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums: \(\{3,5,7,9,11\}\) → that’s 5 values.

ANSWER 8: B  

---

Problem 9:  
We need arrangements of 5 people in a row with:
- a boy at each end (positions 1 and 5)
- the three girls in the middle (positions 2,3,4)

There are 2 boys: they can be placed at the ends in
\[
2!\ = 2
\]
ways (left boy/right boy).

The 3 girls can be arranged in the three middle spots in
\[
3! = 6
\]
ways.

Total arrangements:
\[
2\cdot 6 = 12.
\]

ANSWER 9: E  

---

Problem 10:  
Expression: \(16 + 8^4 - 2\).

Compute in order:
- \(8^4 = 4096\)

So:
\[
16 + 4096 - 2 = 4110.
\]
No answer choice matches, so we interpret the formatting as \(16 + 8 \sqrt{4} - 2\) or \(16 + 8^4 - 2\)?  
Given the AMC/JH style, likely it is:
\[
16 + 8\sqrt{4} - 2.
\]
Then:
\[
\sqrt{4}=2 \implies 8\sqrt{4}=8\cdot 2=16.
\]
So:
\[
16 + 16 - 2 = 30,
\]
still not in choices.

Another common formatting is:
\[
16 + 8^{4} - 2 \quad \text{(still not)}
\]

The only choiceable consistent possibility is:
\[
16 + \sqrt{8}\,^4 - 2 \text{ (nonsense)}
\]

Let’s use the most plausible literal from the text: **“16 + 8 4 − 2”** often means \(16 + 8^4 - 2\) if the caret was lost. But then no choice.

Given typical AMC problems, the intended is usually:
\[
16 + 8 - 4 - 2 = 18
\]
also not.

Because the answer choices are only \(4,8,12,16,20\), the intended expression must simplify to one of those. The most natural is:
\[
16 + 8/4 - 2 = 16 + 2 - 2 = 16.
\]
This matches a common missing symbol: “8 4” often means \(8/4\).

So:
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D  

---

Problem 11:  
Evaluate:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}.
\]

Work from inside out:

1) \(\sqrt{4}=2\)

2) Inside next root:
\[
8\cdot \sqrt{4} = 8\cdot 2 = 16
\]
So:
\[
\sqrt{8\cdot \sqrt{4}} = \sqrt{16}=4.
\]

3) Now plug into the outer expression:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}=\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C  

---

Problem 12:  
Largest number among:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare by decimal places:

- .99 is 0.99
- .9099 is 0.9099
- .9 is 0.9000
- .909 is 0.9090
- .9009 is 0.9009

Clearly \(0.99\) is greater than all the numbers less than \(0.91\).

ANSWER 12: A

Problem 1:
The volume of the water displaced by the rock is equal to the volume of the rock itself, which is $1000 \text{ cm}^3$. The aquarium has a rectangular base with area $A = 100 \text{ cm} \times 40 \text{ cm} = 4000 \text{ cm}^2$. When the rock is submerged, the water level rises by height $h$ such that the volume of the displaced water is $A \times h$. 
Setting this equal to the rock's volume: $4000 \times h = 1000$.
Solving for $h$: $h = \frac{1000}{4000} = \frac{1}{4} = 0.25 \text{ cm}$.
ANSWER 1: A

Problem 2:
Annika and 3 friends make 4 players total. Each received 15 cards, so the total number of cards is $4 \times 15 = 60$. If 2 more friends join, the total number of players becomes $4 + 2 = 6$. The cards are distributed evenly among the 6 players, so each player gets $60 \div 6 = 10$ cards.
ANSWER 2: C

Problem 3:
The problem defines the operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$.
For the expression $\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix}$, we have $a=3, b=1, c=4, d=2$.
Calculating: $(3 \times 2) - (1 \times 4) = 6 - 4 = 2$.
ANSWER 3: E

Problem 4:
Isabella needs an average of 95 over 4 tests, meaning the total sum of her scores must be $95 \times 4 = 380$. Her first two scores are 97 and 91, totaling $97 + 91 = 188$. Let $x$ be the score on the third test and $y$ be the score on the fourth. We have $188 + x + y = 380$, so $x + y = 192$. To make $x$ as small as possible, we must make $y$ as large as possible. Since the maximum score on a test is 100, we set $y = 100$. Then $x + 100 = 192$, so $x = 92$.
ANSWER 4: B

Problem 5:
Let $x$ be the total number of apples Bridget bought. She gave half away, leaving $\frac{x}{2}$. Then she gave 3 away, leaving $\frac{x}{2} - 3$. She kept 4, so $\frac{x}{2} - 3 = 4$. Adding 3 to both sides, $\frac{x}{2} = 7$. Multiplying by 2, $x = 14$.
ANSWER 5: E

Problem 6:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R + G + B$.
"All but 6 are red" means $G + B = 6$.
"All but 8 are green" means $R + B = 8$.
"All but 4 are blue" means $R + G = 4$.
Adding these three equations: $2(R + G + B) = 6 + 8 + 4 = 18$.
So $R + G + B = 18 \div 2 = 9$.
ANSWER 6: C

Problem 7:
The prime factorization of 250 is $250 = 2 \times 125 = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two prime factors is $2 + 5 = 7$.
ANSWER 7: C

Problem 8:
Bag A has $\{1, 3, 5\}$ and Bag B has $\{2, 4, 6\}$.
Possible sums:
$1+2=3, 1+4=5, 1+6=7$
$3+2=5, 3+4=7, 3+6=9$
$5+2=7, 5+4=9, 5+6=11$
The set of distinct sums is $\{3, 5, 7, 9, 11\}$. There are 5 distinct values.
ANSWER 8: B

Problem 9:
There are 2 boys (B1, B2) and 3 girls (G1, G2, G3). A boy must be at each end. There are 2 choices for the left end and 1 choice for the right end ($2 \times 1 = 2$ ways to place boys). The 3 girls can be arranged in the middle in $3! = 3 \times 2 \times 1 = 6$ ways. Total arrangements = $2 \times 6 = 12$.
ANSWER 9: E

Problem 10:
Following the order of operations (PEMDAS/BODMAS):
$16 + (8 \div 4) - 2 = 16 + 2 - 2 = 16$.
ANSWER 10: D

Problem 11:
Evaluate from the inside out:
$\sqrt{4} = 2$
$\sqrt{8 \times 2} = \sqrt{16} = 4$
$\sqrt{16 \times 4} = \sqrt{64} = 8$
ANSWER 11: C

Problem 12:
To compare, pad with zeros to match the number of decimal places:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
Comparing the values, 0.9900 is the largest.
ANSWER 12: A

Problem 1:
The base of the aquarium measures 100 cm by 40 cm, so its area is \(100 \times 40 = 4000\) cm². The rock has volume 1000 cm³ and is fully submerged, displacing exactly 1000 cm³ of water. The resulting rise in water level is therefore the displaced volume divided by the base area: \(1000 / 4000 = 0.25\) cm. (The original water depth and tank height are irrelevant once the surface area is known.)

ANSWER 1: A

Problem 2:
Annika plus 3 friends gives 4 players, each receiving 15 cards, for a total of \(4 \times 15 = 60\) cards. When 2 more players join, there are 6 players sharing the same 60 cards, so each receives \(60 / 6 = 10\) cards.

ANSWER 2: C

Problem 3:
The given operation evaluates \(a\, c\, b\, d\) as \(a \cdot d - b \cdot c\). Substituting the numbers yields \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).

ANSWER 3: E

Problem 4:
An average of 95 on four 100-point tests requires a total of \(4 \times 95 = 380\) points. The first two scores sum to \(97 + 91 = 188\), leaving 192 points still needed from the last two tests. The fourth test can score at most 100, so the third test must be at least \(192 - 100 = 92\). Any lower score on the third test would make a total of 380 impossible.

ANSWER 4: B

Problem 5:
Let \(t\) be the original number of apples. After giving half to Ann, \(t/2\) remain. After giving 3 more to Cassie, \(t/2 - 3\) remain, and this equals the 4 apples kept. Solving \(t/2 - 3 = 4\) gives \(t/2 = 7\), so \(t = 14\).

ANSWER 5: E

Problem 6:
Let \(m\) be the total number of marbles. Then there are \(m-6\) red, \(m-8\) green, and \(m-4\) blue. Their sum must equal \(m\):
\[
(m-6) + (m-8) + (m-4) = m \implies 3m - 18 = m \implies 2m = 18 \implies m = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies the three “all-but” statements.

ANSWER 6: C

Problem 7:
Factorize \(250 = 2 \cdot 5^3\). The two smallest distinct prime factors are 2 and 5; their sum is 7.

ANSWER 7: C

Problem 8:
All nine possible sums are:
\[
1+2=3,\ 1+4=5,\ 1+6=7,\ 3+2=5,\ 3+4=7,\ 3+6=9,\ 5+2=7,\ 5+4=9,\ 5+6=11.
\]
The distinct values are 3, 5, 7, 9, 11—exactly five.

ANSWER 8: B

Problem 9:
The seating pattern must be boy–girl–girl–girl–boy. The two distinct boys may be placed at the ends in \(2!\) = 2 ways. The three distinct girls may be arranged in the middle three seats in \(3!\) = 6 ways. The total number of arrangements is therefore \(2 \times 6 = 12\).

ANSWER 9: E

Problem 10:
Interpreting the expression with standard order of operations and treating juxtaposition of 8 and 4 as division yields
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D

Problem 11:
Evaluate from the inside outward:
\[
\sqrt{4} = 2, \quad 8 \cdot 2 = 16, \quad \sqrt{16} = 4, \quad 16 \cdot 4 = 64, \quad \sqrt{64} = 8.
\]

ANSWER 11: C

Problem 12:
Compare the decimals directly:
\[
0.99 > 0.9 > 0.9099 > 0.909 > 0.9009.
\]
Thus 0.99 is the largest.

ANSWER 12: A

Problem 1:
The problem asks us to determine by how many centimeters the water level rises when a rock with a volume of 1000 cm³ is submerged in an aquarium filled with water to a depth of 37 cm. The aquarium has a rectangular base measuring 100 cm by 40 cm and a height of 50 cm.

First, we need to calculate the initial volume of water in the aquarium. The volume of water is given by the formula length × width × height. So, the initial volume of water = 100 cm × 40 cm × 37 cm = 148,000 cm³.

When the rock is submerged, it displaces a volume of water equal to its own volume, which is 1000 cm³. This means the water level will rise to accommodate the volume of the rock.

The rise in water level can be calculated by dividing the volume of the rock by the base area of the aquarium. The base area = 100 cm × 40 cm = 4000 cm².

Rise in water level = Volume of rock / Base area = 1000 cm³ / 4000 cm² = 0.25 cm.

Let's examine the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

The calculated rise in water level matches option A.

ANSWER 1: A

Problem 2:
The problem states that in the card game Buffalo Shuffle-o, all cards are distributed evenly among players. Initially, Annika and 3 friends play, and each is dealt 15 cards. We need to find out how many cards each player will be dealt if 2 more friends join the next game.

First, we determine the total number of cards. With 4 players initially, each getting 15 cards, the total number of cards = 4 × 15 = 60.

When 2 more friends join, the total number of players becomes 4 + 2 = 6.

To find out how many cards each player will get, we divide the total number of cards by the new number of players: 60 / 6 = 10.

Let's examine the choices:
A. 8
B. 9
C. 10
D. 11
E. 12

The calculated number of cards per player matches option C.

ANSWER 2: C

Problem 3:
The problem defines a operation as a c b d = a·d − b·c and asks us to find the value of 3 1 4 2.

Using the given definition, we substitute a = 3, b = 1, c = 4, and d = 2 into the equation: 3 1 4 2 = 3·2 − 1·4 = 6 - 4 = 2.

Let's examine the choices:
A. −2
B. −1
C. 0
D. 1
E. 2

The calculated value matches option E.

ANSWER 3: E

Problem 4:
Isabella needs to achieve an average grade of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score she could have made on the third test to still reach her goal.

The total score required for an average of 95 on 4 tests = 95 × 4 = 380.

The sum of her first two test scores = 97 + 91 = 188.

So, the total score needed on the last two tests = 380 - 188 = 192.

To minimize the score on the third test, we maximize the score on the fourth test. The maximum score on the fourth test is 100.

Thus, the minimum score on the third test = 192 - 100 = 92.

Let's examine the choices:
A. 90
B. 92
C. 95
D. 96
E. 97

The calculated minimum score matches option B.

ANSWER 4: B

Problem 5:
Bridget gave half of her apples to Ann, then gave 3 to Cassie, and kept 4 for herself. We need to find out how many apples she originally bought.

Let's denote the original number of apples as x.

After giving half to Ann, Bridget had x/2 apples left.

Then, she gave 3 apples to Cassie and kept 4, so x/2 = 3 + 4 = 7.

Thus, x = 7 × 2 = 14.

Let's examine the choices:
A. 3
B. 4
C. 7
D. 11
E. 14

The calculated original number of apples matches option E.

ANSWER 5: E

Problem 6:
The problem states that in a jar, all but 6 marbles are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T, the number of red marbles as R, the number of green marbles as G, and the number of blue marbles as B.

From the given information:
1. T - R = 6 (all but 6 are red)
2. T - G = 8 (all but 8 are green)
3. T - B = 4 (all but 4 are blue)

Adding these equations gives: 3T - (R + G + B) = 18.

Since R + G + B = T (the total number of marbles is the sum of the marbles of each color), we have:
3T - T = 18, which simplifies to 2T = 18.

Thus, T = 18 / 2 = 9.

Let's examine the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

The calculated total number of marbles matches option C.

ANSWER 6: C

Problem 7:
We need to find the sum of the two smallest prime factors of 250.

First, let's factorize 250 into its prime factors: 250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5.

The sum of these two smallest prime factors = 2 + 5 = 7.

Let's examine the choices:
A. 2
B. 5
C. 7
D. 10
E. 12

The calculated sum matches option C.

ANSWER 7: C

Problem 8:
Bag A contains chips labeled 1, 3, and 5. Bag B contains chips labeled 2, 4, and 6. We need to find how many different values are possible for the sum of the numbers on the chips drawn from each bag.

Let's list the possible sums:
- (1,2) = 3
- (1,4) = 5
- (1,6) = 7
- (3,2) = 5
- (3,4) = 7
- (3,6) = 9
- (5,2) = 7
- (5,4) = 9
- (5,6) = 11

The possible sums are 3, 5, 7, 9, and 11. There are 5 different values.

Let's examine the choices:
A. 4
B. 5
C. 6
D. 7
E. 9

The number of different sums matches option B.

ANSWER 8: B

Problem 9:
The chess team consists of 2 boys and 3 girls. They need to be arranged with a boy at each end and the girls in the middle. We need to find out how many such arrangements are possible.

The boys can be arranged at the ends in 2! = 2 ways (either boy can be on the left or the right).

The 3 girls can be arranged in the 3 middle positions in 3! = 6 ways.

The total number of arrangements = (number of ways to arrange boys) × (number of ways to arrange girls) = 2 × 6 = 12.

Let's examine the choices:
A. 2
B. 4
C. 5
D. 6
E. 12

The calculated total number of arrangements matches option E.

ANSWER 9: E

Problem 10:
The problem asks us to evaluate the expression 16 + 8 ÷ 4 − 2.

Following the order of operations (PEMDAS/BODMAS: Parentheses/Brackets, Exponents/Orders, Multiplication and Division from left to right, Addition and Subtraction from left to right):
1. D

Problem 1:
The water rise is determined by the volume of the rock divided by the base area of the aquarium.  
Base area = \(100 \times 40 = 4000\text{ cm}^2\).  
Volume of rock = \(1000\text{ cm}^3\).  
Rise = \(\frac{1000}{4000} = 0.25\text{ cm}\).  
ANSWER 1: A

Problem 2:
Initially, Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 2: C

Problem 3:
The notation \(\begin{array}{cc} a & c \\ b & d \end{array} = a \cdot d - b \cdot c\).  
For \(\begin{array}{cc} 3 & 1 \\ 4 & 2 \end{array}\): \(a=3, c=1, b=4, d=2\).  
Value = \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).  
ANSWER 3: E

Problem 4:
Total points needed for an average of 95 over 4 tests = \(4 \times 95 = 380\).  
First two tests sum = \(97 + 91 = 188\).  
Remaining points needed = \(380 - 188 = 192\).  
To minimize the third test score, maximize the fourth (100). Then third test \(\ge 192 - 100 = 92\).  
Lowest possible third test score is 92.  
ANSWER 4: B

Problem 5:
Let \(x\) be the number of apples Bridget bought.  
She gave half to Ann, so she kept \(\frac{x}{2}\).  
Then she gave 3 to Cassie and kept 4, so \(\frac{x}{2} = 3 + 4 = 7\).  
Thus \(x = 14\).  
ANSWER 5: E

Problem 6:
Let \(R, G, B\) be the number of red, green, blue marbles.  
"All but 6 are red" \(\Rightarrow G + B = 6\).  
"All but 8 are green" \(\Rightarrow R + B = 8\).  
"All but 4 are blue" \(\Rightarrow R + G = 4\).  
Adding: \((G+B)+(R+B)+(R+G) = 6+8+4 = 18\) \(\Rightarrow 2(R+G+B) = 18\) \(\Rightarrow R+G+B = 9\).  
Total marbles = 9.  
ANSWER 6: C

Problem 7:
Prime factorization of 250: \(250 = 2 \times 5^3\).  
Prime factors are 2 and 5. The two smallest are 2 and 5.  
Sum = \(2 + 5 = 7\).  
ANSWER 7: C

Problem 8:
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}.  
Possible sums:  
1+2=3, 1+4=5, 1+6=7;  
3+2=5, 3+4=7, 3+6=9;  
5+2=7, 5+4=9, 5+6=11.  
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.  
ANSWER 8: B

Problem 9:
Boys must sit at the two ends.  
Arrange boys: 2 choices for left end, 1 for right end → \(2! = 2\) ways.  
Arrange 3 girls in the middle 3 seats: \(3! = 6\) ways.  
Total arrangements = \(2 \times 6 = 12\).  
ANSWER 9: E

Problem 10:
Interpret "8 4" as \(8 \div 4\).  
\(16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16\).  
ANSWER 10: D

Problem 11:
\(\sqrt{4} = 2\).  
\(8 \cdot 2 = 16\), \(\sqrt{16} = 4\).  
\(16 \cdot 4 = 64\), \(\sqrt{64} = 8\).  
ANSWER 11: C

Problem 12:
Compare decimals:  
0.99, 0.9099, 0.9, 0.909, 0.9009.  
0.99 is the largest.  
ANSWER 12: A

# Problem 1:

**What is being asked:** A rock of volume 1000 cm³ is submerged in an aquarium with a 100 cm × 40 cm base. How much does the water level rise?

**Setup:** When the rock is submerged, it displaces a volume of water equal to its own volume. This displaced water spreads across the base of the aquarium, causing the level to rise.

$$\text{Volume displaced} = \text{Base Area} \times \text{Rise in water level}$$

**Computation:**
- Base area = $100 \times 40 = 4000 \text{ cm}^2$
- $1000 = 4000 \times h$
- $h = \frac{1000}{4000} = 0.25 \text{ cm}$

**Check for overflow:** The current depth is 37 cm, and 37 + 0.25 = 37.25 cm, which is well below the 50 cm height. No overflow occurs.

**Ruling out other choices:** Choices B–E would require rock volumes of 2000, 4000, 5000, and 10000 cm³ respectively — all larger than the given 1000 cm³.

ANSWER 1: A

---

# Problem 2:

**What is being asked:** With 4 players, each gets 15 cards. With 6 players, how many cards does each get?

**Setup:** The total number of cards in the deck stays the same.

**Computation:**
- Number of players in first game: Annika + 3 friends = 4 players
- Total cards = $4 \times 15 = 60$ cards
- Number of players in second game: $4 + 2 = 6$ players
- Cards per player = $\frac{60}{6} = 10$

**Ruling out other choices:** Some might forget to count Annika (getting 3 players → 45 cards → 45/5 = 9, choice B) or miscount the new total as 5 players (60/5 = 12, choice E).

ANSWER 2: C

---

# Problem 3:

**What is being asked:** Evaluate the custom operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$ for $a=3, b=1, c=4, d=2$.

**Setup:** Substitute directly into the formula.

**Computation:**
$$\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2$$

**Ruling out other choices:** A common mistake is computing $b \cdot c - a \cdot d = 4 - 6 = -2$ (choice A) or mixing up which values go where.

ANSWER 3: E

---

# Problem 4:

**What is being asked:** Isabella needs an average of 95 on four tests. Her first two scores are 97 and 91. What is the lowest possible third test score so she can still reach her goal?

**Setup:**
- Total points needed: $4 \times 95 = 380$
- Sum of first two tests: $97 + 91 = 188$
- Sum needed from tests 3 and 4: $380 - 188 = 192$

**Computation:**
To minimize the third test score, maximize the fourth test score. The maximum possible score on any test is 100.
$$\text{Minimum test 3} = 192 - 100 = 92$$

**Ruling out other choices:** Choice A (90) would require test 4 to be $192 - 90 = 102$, which is impossible on a 100-point test. Choices C, D, E are higher than necessary.

ANSWER 4: B

---

# Problem 5:

**What is being asked:** Bridget bought apples, gave half to Ann, gave 3 to Cassie, and kept 4. How many did she buy?

**Setup:** Let $x$ = total apples bought. Work through the sequence of events.

**Computation:**
- After giving half to Ann: $\frac{x}{2}$ apples remain
- After giving 3 to Cassie: $\frac{x}{2} - 3$ apples remain
- She keeps 4: $\frac{x}{2} - 3 = 4$
- $\frac{x}{2} = 7$
- $x = 14$

**Verification:** 14 apples → give 7 to Ann → 7 left → give 3 to Cassie → 4 left. ✓

**Ruling out other choices:** Choice C (7) is a trap for those who forget to double the 7 remaining apples. Choice D (11) comes from incorrectly adding $4 + 3 + 4$.

ANSWER 5: E

---

# Problem 6:

**What is being asked:** Find the total number of marbles given conditions about each color.

**Setup:** Let $T$ = total marbles, $R$ = red, $G$ = green, $B$ = blue.
- "All but 6 are red" → $G + B = 6$ → $R = T - 6$
- "All but 8 are green" → $R + B = 8$ → $G = T - 8$
- "All but 4 are blue" → $R + G = 4$ → $B = T - 4$

**Computation:**
$$T = R + G + B = (T - 6) + (T - 8) + (T - 4) = 3T - 18$$
$$2T = 18 \implies T = 9$$

**Verification:** $R = 3, G = 1, B = 5$. Check: $G+B=6$ ✓, $R+B=8$ ✓, $R+G=4$ ✓.

ANSWER 6: C

---

# Problem 7:

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Setup:** Find the prime factorization of 250.

**Computation:**
$$250 = 2 \times 125 = 2 \times 5 \times 25 = 2 \times 5 \times 5 \times 5 = 2 \times 5^3$$

The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.
$$\text{Sum} = 2 + 5 = 7$$

**Ruling out other choices:** Choice A (2) is just one factor. Choice B (5) is just one factor. Choice D (10) would be $2 + 5 + 3$ or $5 + 5$, neither of which applies.

ANSWER 7: C

---

# Problem 8:

**What is being asked:** How many distinct sums can be formed by picking one chip from Bag A {1, 3, 5} and one from Bag B {2, 4, 6}?

**Setup:** List all 9 possible sums and count the unique values.

**Computation:**
| | 2 | 4 | 6 |
|---|---|---|---|
| **1** | 3 | 5 | 7 |
| **3** | 5 | 7 | 9 |
| **5** | 7 | 9 | 11 |

Unique sums: {3, 5, 7, 9, 11} → **5 different values**

**Ruling out other choices:** Choice E (9) is the trap of counting all 9 pairs without removing duplicates. Choice D (7) might come from miscounting.

ANSWER 8: B

---

# Problem 9:

**What is being asked:** How many ways can 2 boys and 3 girls sit in a row with boys at the ends and girls in the middle?

**Setup:** The arrangement must be: **B G G G B**

**Computation:**
- Arrange the 2 boys at the 2 end positions: $2! = 2$ ways
- Arrange the 3 girls in the 3 middle positions: $3! = 6$ ways
- Total arrangements: $2 \times 6 = 12$

**Ruling out other choices:** Choice A (2) only counts the boys' arrangements. Choice D (6) only counts the girls' arrangements. Both must be multiplied together.

ANSWER 9: E

---

# Problem 10:

**What is being asked:** Evaluate $16 + 8 \div 4 - 2$ using order of operations.

**Setup:** Apply PEMDAS — division before addition and subtraction.

**Computation:**
$$16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16$$

**Ruling out other choices:**
- Left-to-right without order of operations: $(16+8) \div 4 - 2 = 6 - 2 = 4$ (choice A)
- $(16+8) \div (4-2) = 24 \div 2 = 12$ (choice C)
- $16 + 8 \div (4-2) = 16 + 4 = 20$ (choice E

 Problem 1:
We need to find how much the water level rises when a rock of volume 1000 cm³ is submerged in a rectangular aquarium.

First, find the area of the aquarium’s base:
\[
100\text{ cm} \times 40\text{ cm} = 4000\text{ cm}^2.
\]

When the rock is completely submerged, it displaces 1000 cm³ of water. This displaced volume spreads out over the base, causing the water level to rise. The rise in height \(h\) satisfies:
\[
\text{base area} \times h = \text{volume of rock}.
\]
So,
\[
h = \frac{1000}{4000} = 0.25\text{ cm}.
\]

We should check that the aquarium does not overflow. The current depth is 37 cm and the height is 50 cm, so there is 13 cm of free space—plenty of room for a 0.25 cm rise.

Tempting wrong choices come from using the wrong area (e.g., using the side wall 100×50 or 40×50) or from dividing by the height 50. Only the base area matters for the level rise.

ANSWER 1: A

---

Problem 2:
We need to find how many cards each player gets when two extra friends join the game.

First, determine the total number of cards. Annika and 3 friends make 4 players, and each gets 15 cards:
\[
4 \times 15 = 60\text{ cards total}.
\]

When 2 more friends join, the number of players becomes:
\[
4 + 2 = 6.
\]

Distribute the 60 cards evenly among 6 players:
\[
60 \div 6 = 10.
\]

A tempting mistake is to subtract 2 from 15, giving 13, which is not an option, or to guess without computing the total deck size. The key is that the total number of cards stays the same.

ANSWER 2: C

---

Problem 3:
We are given the rule \(a\ c\ b\ d = a\cdot d - b\cdot c\). We need to evaluate \(3\ 1\ 4\ 2\).

Matching the pattern:
- \(a = 3\)
- \(c = 1\)
- \(b = 4\)
- \(d = 2\)

Apply the formula:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

A tempting error is to swap the pairs and compute \(3\cdot 1 - 4\cdot 2 = 3 - 8 = -5\), which is not among the choices, or to read the order incorrectly. Careful matching to \(a, c, b, d\) is essential.

ANSWER 3: E

---

Problem 4:
Isabella wants an average of 95 on four 100-point tests. Her first two scores are 97 and 91. We need the lowest possible score on the third test that still allows her to reach her goal.

Total points needed for a 95 average:
\[
4 \times 95 = 380.
\]

Points earned so far:
\[
97 + 91 = 188.
\]

Points still needed from the last two tests:
\[
380 - 188 = 192.
\]

Let the third test score be \(x\) and the fourth be \(y\), with \(x + y = 192\) and each score at most 100. To minimize \(x\), we maximize \(y\). The highest possible score is 100, so:
\[
x = 192 - 100 = 92.
\]

If she scores 92 on the third test, she can score 100 on the fourth and exactly reach her goal. A tempting wrong choice is 95, which would be the answer if she needed the two remaining tests to average 95, but she can shift points to the fourth test.

ANSWER 4: B

---

Problem 5:
We need to find how many apples Bridget originally bought.

Let the original number be \(x\).
- She gave half to Ann, leaving her with \(\frac{x}{2}\).
- She gave 3 to Cassie and kept 4:
\[
\frac{x}{2} - 3 = 4.
\]

Solve:
\[
\frac{x}{2} = 7 \quad\Rightarrow\quad x = 14.
\]

Check: Start with 14. Half is 7, given to Ann. She has 7 left. Gives 3 to Cassie, keeps 4. This matches.

A tempting error is to add \(3 + 4 = 7\) and stop, forgetting that 7 is only what remained after giving away half, so the original amount must be doubled.

ANSWER 5: E

---

Problem 6:
We need the total number of marbles given three conditions about how many are *not* each color.

Let \(R, G, B\) be the numbers of red, green, and blue marbles.
- “All but 6 are red” means non-red marbles equal 6: \(G + B = 6\).
- “All but 8 are green” means non-green marbles equal 8: \(R + B = 8\).
- “All but 4 are blue” means non-blue marbles equal 4: \(R + G = 4\).

Add all three equations:
\[
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
\]
\[
2R + 2G + 2B = 18
\]
\[
R + G + B = 9.
\]

So there are 9 marbles total. (Checking: \(R=3, G=1, B=5\), which satisfies all three statements.)

A tempting mistake is to add \(6+8+4=18\) and choose that, but that sum counts each marble twice.

ANSWER 6: C

---

Problem 7:
We need the sum of the two smallest prime factors of 250.

Factor 250:
\[
250 = 2 \times 125 = 2 \times 5^3.
\]

The prime factors are 2 and 5. The two smallest (and only) distinct prime factors are 2 and 5. Their sum is:
\[
2 + 5 = 7.
\]

A tempting wrong choice is 10, which comes from adding 5 and 5 (the two smallest factors with multiplicity), but the problem asks for the two smallest prime factors, i.e., distinct primes.

ANSWER 7: C

---

Problem 8:
Bag A has \(\{1, 3, 5\}\) and Bag B has \(\{2, 4, 6\}\). We draw one from each and want the number of distinct possible sums.

Compute all sums:
- \(1+2=3\), \(1+4=5\), \(1+6=7\)
- \(3+2=5\), \(3+4=7\), \(3+6=9\)
- \(5+2=7\), \(5+4=9\), \(5+6=11\)

The distinct values are:
\[
3, 5, 7, 9, 11.
\]

That is 5 different values. A tempting error is to count all 9 ordered pairs without removing duplicates.

ANSWER 8: B

---

Problem 9:
We have 2 boys and 3 girls to arrange in a row with a boy at each end and the three girls in the middle.

Positions: B _ _ _ G G G _ _ _ B (but specifically 5 seats: 1,2,3,4,5 with boys at 1 and 5, girls at 2,3,4).

- Choose the boy for the left end and the boy for the right end: \(2! = 2\) ways.
- Arrange the 3 girls in the middle three seats: \(3! = 6\) ways.

Total arrangements:
\[
2 \times 6 = 12.
\]

A tempting mistake is to ignore the constraint and compute \(5! = 120\), or to forget that the boys can swap places.

ANSWER 9: E

---

Problem 10:
Evaluate \(16 + 8 \div 4 - 2\).

Using order of operations (PEMDAS/BODMAS), division comes before addition and subtraction:
\[
8 \div 4 = 2.
\]

Now substitute back:
\[
16 + 2 - 2 = 16.
\]

A tempting error is to work left-to-right doing addition first: \((16+8)\div 4 - 2 = 24\div 4 - 2 = 6 - 2 = 4\), which is choice A. Another is to do \(16 + 2 = 18\), then \(18 - 2 = 16\), which luckily gives the same corr

Problem 1:
We need the rise in water level when a rock of volume 1000 cm³ is placed in an aquarium with base 100 cm by 40 cm. The rock displaces water equal to its volume, causing the water level to rise by Δh such that base area × Δh = rock volume. Base area = 100 cm × 40 cm = 4000 cm². So Δh = 1000 cm³ / 4000 cm² = 0.25 cm. The initial depth is 37 cm, and after rise it becomes 37.25 cm, which is still below the height of 50 cm, so the rock is fully submerged. The rise is 0.25 cm.

ANSWER 1: A

Problem 2:
When Annika and 3 friends play, there are 4 players, each dealt 15 cards. Total cards = 4 × 15 = 60. In the next game, 2 more friends join, making 6 players. Since the same total number of cards is used, each player gets 60 / 6 = 10 cards.

ANSWER 2: C

Problem 3:
The operation is defined as a c b d = a·d − b·c. For the numbers 3 1 4 2, we have a = 3, c = 1, b = 4, d = 2. Compute a·d − b·c = 3×2 − 4×1 = 6 − 4 = 2.

ANSWER 3: E

Problem 4:
Isabella wants an average of 95 over four tests, so total points needed = 4 × 95 = 380. Her first two scores sum to 97 + 91 = 188. Let the third score be x. After the third test, she can still reach the goal if it is possible to score at most 100 on the fourth test to reach 380. That requires 188 + x + 100 ≥ 380 → x ≥ 380 − 288 = 92. The smallest integer x satisfying this is 92. If x = 91, then even a perfect 100 on the fourth gives 188+91+100 = 379 < 380. So the lowest possible third score is 92.

ANSWER 4: B

Problem 5:
Let the number of apples Bridget bought be N. She gave half to Ann, leaving N/2. Then she gave Cassie 3 apples, leaving N/2 − 3. She keeps 4 apples, so N/2 − 3 = 4 → N/2 = 7 → N = 14.

ANSWER 5: E

Problem 6:
Let T be the total number of marbles. "All but 6 are red" means T − R = 6 → R = T − 6. Similarly, T − G = 8 → G = T − 8, and T − B = 4 → B = T − 4. Since R + G + B = T, we have (T−6)+(T−8)+(T−4) = T → 3T − 18 = T → 2T = 18 → T = 9.

ANSWER 6: C

Problem 7:
Factor 250: 250 = 2 × 5³. Its prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5, and their sum is 7.

ANSWER 7: C

Problem 8:
List all possible sums from drawing one chip from Bag A {1,3,5} and one from Bag B {2,4,6}:
1+2=3, 1+4=5, 1+6=7;
3+2=5, 3+4=7, 3+6=9;
5+2=7, 5+4=9, 5+6=11.
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.

ANSWER 8: B

Problem 9:
We need arrangements of 2 boys and 3 girls in a row with a boy at each end. Choose which boy sits on the left end and which on the right: 2! = 2 ways. The three girls occupy the three middle seats and can be arranged in 3! = 6 ways. Total arrangements = 2 × 6 = 12.

ANSWER 9: E

Problem 10:
The expression is 16 + 8 ÷ 4 − 2 (the missing operator is division, as typical for order-of-operations problems). Following order of operations, division before addition and subtraction: 8 ÷ 4 = 2. Then 16 + 2 = 18, and 18 − 2 = 16.

ANSWER 10: D

Problem 11:
Evaluate step by step: √4 = 2. Then 8 · 2 = 16, so √(8·√4) = √16 = 4. Then 16 · 4 = 64, so √(16·√(8·√4)) = √64 = 8.

ANSWER 11: C

Problem 12:
Compare the decimal numbers:
A. 0.99
B. 0.9099
C. 0.9
D. 0.909
E. 0.9009
Clearly 0.99 is the largest because it is greater than 0.9 and all the others are less than 0.91.

ANSWER 12: A

Problem 1:
The aquarium’s base has area  
\(A = 100\text{ cm} \times 40\text{ cm}=4000\text{ cm}^2\).  
When the rock of volume \(V=1000\text{ cm}^3\) is fully submerged, the water rises by a height \(h\) such that  

\[
A\cdot h = V \quad\Longrightarrow\quad h = \frac{1000}{4000}=0.25\text{ cm}.
\]

The rise cannot be any of the larger listed values (0.5 cm, 1 cm, …) because those would require a volume larger than the rock’s.  

ANSWER 1: A  

---

Problem 2:
With 4 players each receiving 15 cards, the total number of cards is  

\[
4 \times 15 = 60.
\]

If 2 more friends join, there are \(4+2 = 6\) players.  The cards are still all the cards, so each player receives  

\[
\frac{60}{6}=10\text{ cards}.
\]

The other answer choices (8, 9, 11, 12) would not distribute the 60 cards evenly.  

ANSWER 2: C  

---

Problem 3:
The operation is defined by  

\[
a\;c\;b\;d = a\cdot d - b\cdot c.
\]

For the numbers \(3\;1\;4\;2\) we have \(a=3,\;c=1,\;b=4,\;d=2\).  Hence  

\[
3\;1\;4\;2 = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

The other options \(-2,-1,0,1\) would correspond to different combinations that do not satisfy the given formula.  

ANSWER 3: E  

---

Problem 4:
To average 95 over four 100‑point tests, the total points needed are  

\[
4 \times 95 = 380.
\]

After the first two tests Isabella has  

\[
97 + 91 = 188\text{ points}.
\]

Let the third test score be \(x\).  Then the fourth test must supply  

\[
380 - (188 + x) = 192 - x \text{ points}.
\]

The fourth test can give at most 100 points, so we need  

\[
192 - x \le 100 \quad\Longrightarrow\quad x \ge 92.
\]

If she scores \(92\) on the third test, she would need a perfect 100 on the fourth, which is possible.  Any lower score would require more than 100 points on the fourth test, which is impossible.  

Thus the lowest possible third‑test score is **92**.  

ANSWER 4: B  

---

Problem 5:
Let the original number of apples be \(N\).  
She gave half of them to Ann: \(\frac{N}{2}\).  
She then gave Cassie 3 apples and kept 4, leaving  

\[
N - \frac{N}{2} - 3 = 4.
\]

Simplify: \(\frac{N}{2} - 3 = 4 \Rightarrow \frac{N}{2}=7 \Rightarrow N=14\).

If the total had been any of the other answer choices, the steps would not balance (e.g., 11 would leave 2 apples after giving Cassie 3).  

ANSWER 5: E  

---

Problem 6:
Let the total number of marbles be \(T\).  
- “All but 6 are red’’ → red = \(T-6\).  
- “All but 8 are green’’ → green = \(T-8\).  
- “All but 4 are blue’’ → blue = \(T-4\).

These three colors account for all marbles, so  

\[
(T-6)+(T-8)+(T-4)=T \;\Longrightarrow\; 3T-18=T \;\Longrightarrow\; 2T=18 \;\Longrightarrow\; T=9.
\]

The other totals (6, 8, 10, 12) would not satisfy the three “all but…’’ statements simultaneously.  

ANSWER 6: C  

---

Problem 7:
Factor \(250\):  

\[
250 = 2 \times 5^3.
\]

The prime factors are \(2\) and \(5\).  The two smallest are \(2\) and \(5\), whose sum is  

\[
2+5=7.
\]

Choices 2, 5, 10, 12 are either single prime factors or incorrect sums.  

ANSWER 7: C  

---

Problem 8:
Bag A contains \(\{1,3,5\}\); Bag B contains \(\{2,4,6\}\).  
All possible sums \(a+b\) are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The distinct results are \(\{3,5,7,9,11\}\), a total of **5** different sums.  

The other answer choices (4, 6, 7, 9) do not match the count of distinct sums.  

ANSWER 8: B  

---

Problem 9:
Place the two boys at the two ends and the three girls in the middle three seats.

- The two boys can be ordered in \(2!\) ways (left‑right swap).  
- The three girls can be arranged among the three middle seats in \(3! = 6\) ways.

Total arrangements  

\[
2! \times 3! = 2 \times 6 = 12.
\]

Choices 2, 4, 5, 6 are too small; 12 accounts for all possible permutations under the given constraints.  

ANSWER 9: E  

---

Problem 10:
Interpret the expression as  

\[
16 + 8 \div 4 - 2.
\]

Division comes before addition and subtraction, so  

\[
8 \div 4 = 2 \quad\Longrightarrow\quad 16 + 2 - 2 = 16.
\]

The other options (4, 8, 12, 20) would result only if the order of operations were misinterpreted.  

ANSWER 10: D  

---

Problem 11:
Simplify from the innermost radical outward.

\[
\sqrt{4}=2,\qquad 
\sqrt{8\cdot\sqrt{4}} = \sqrt{8\cdot2}= \sqrt{16}=4.
\]

Now  

\[
\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{16\cdot4}= \sqrt{64}=8.
\]

Thus the value is 8, matching choice C.  

ANSWER 11: C  

---

Problem 12:
Compare the decimal values:

- A. \(0.99\)
- B. \(0.9099\)
- C. \(0.9\)
- D. \(0.909\)
- E. \(0.9009\)

Clearly \(0.99\) is greater than \(0.9099\), which is greater than \(0.909\), etc.  
Hence the largest number is \(0.99\).  

ANSWER 12: A

Problem 1:  
An aquarium has a rectangular base of 100 cm by 40 cm and height 50 cm, filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is submerged. By how many centimeters does the water level rise?  

The base area is \(100 \times 40 = 4000 \, \text{cm}^2\). The rock’s volume is 1000 cm³, so the rise in water level is \(\frac{1000}{4000} = 0.25 \, \text{cm}\).  

Tempting choices: 0.5 (B) or 1 (C) might be guessed if base area is miscalculated.  

ANSWER 1: A  

Problem 2:  
Annika and 3 friends (4 players) get 15 cards each. Total cards = \(4 \times 15 = 60\). If 2 more join (6 players), each gets \(\frac{60}{6} = 10\).  

Tempting choices: 8 (A) if total cards miscalculated as 48.  

ANSWER 2: C  

Problem 3:  
Operation defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\): \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\), but the order is \(a c b d\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). However, the problem writes "a c b d" which is interpreted as \(a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). But the choices include 2 (E). Re-evaluating: the operation is \(a c b d = a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\).  

Tempting choices: -2 (A) if signs reversed.  

ANSWER 3: A (Note: Correction based on standard interpretation: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\), but the correct answer based on the operation defined is actually \(3 \cdot 2 - 1 \cdot 4 = 2\), but the problem writes "a c b d", which is \(a \cdot d - b \cdot c\), so 2. However, the answer is 2, but the choices include 2 as E. The correct calculation is 2, so E. Correction: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\).)  

**Correction:** The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\), this is \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\).  

ANSWER 3: E  

Problem 4:  
Isabella wants an average of 95 on four 100-point tests. Total points needed: \(4 \times 95 = 380\). First two scores: 97 + 91 = 188. Let third score be \(x\), fourth be \(y\). Then \(188 + x + y = 380\), so \(x + y = 192\). To minimize \(x\), maximize \(y\): \(y = 100\), so \(x = 92\).  

Tempting choices: 90 (A) if \(y\) not maximized.  

ANSWER 4: B  

Problem 5:  
Bridget gave half apples to Ann, then 3 to Cassie, keeping 4. Let total be \(x\). After giving half to Ann: \(\frac{x}{2}\) left. Then gives 3 to Cassie: \(\frac{x}{2} - 3 = 4\). So \(\frac{x}{2} = 7\), \(x = 14\).  

Tempting choices: 7 (C) if half misinterpreted.  

ANSWER 5: E  

Problem 6:  
Let \(r, g, b\) be red, green, blue marbles. Then:  
- All but 6 are red: \(g + b = 6\)  
- All but 8 are green: \(r + b = 8\)  
- All but 4 are blue: \(r + g = 4\)  
Adding: \(2(r + g + b) = 18\), so total \(r + g + b = 9\).  

Tempting choices: 6 (A) if equations misread.  

ANSWER 6: C  

Problem 7:  
Prime factors of 250: \(250 = 2 \times 5^3\). Two smallest: 2 and 5. Sum: \(2 + 5 = 7\).  

Tempting choices: 5 (B) if only 5 considered.  

ANSWER 7: C  

Problem 8:  
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}. Possible sums:  
- 1+2=3, 1+4=5, 1+6=7  
- 3+2=5, 3+4=7, 3+6=9  
- 5+2=7, 5+4=9, 5+6=11  
Unique sums: 3, 5, 7, 9, 11 → 5 values.  

Tempting choices: 6 (C) if duplicates counted.  

ANSWER 8: B  

Problem 9:  
Two boys at ends: 2 ways. Three girls in middle: \(3! = 6\) ways. Total: \(2 \times 6 = 12\).  

Tempting choices: 6 (D) if boys not permuted.  

ANSWER 9: E  

Problem 10:  
Expression: \(16 + 8 \times 4 - 2\). Order: \(8 \times 4 = 32\), then \(16 + 32 - 2 = 46\)? Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 \times 4 - 2\): \(8 \times 4 = 32\), \(16 + 32 = 48\), \(48 - 2 = 46\). But choices max at 20. Re-interpreting: "16 + 8 4 - 2" might mean \(16 + 8 - 4 - 2 = 18\), not in choices. Alternatively, if "8 4" is 8×4, but 46 not in choices. The expression is likely \(16 + 8 - 4 - 2 = 18\), but not in choices. Correction: The expression is "16 + 8 4 − 2", which might be \(16 + (8 \times 4) - 2 = 16 + 32 - 2 = 46\), but not in choices. Re-reading: It’s "16 + 8 4 − 2", which could be \(16 + 8 - 4 - 2 = 18\), but not in choices. The intended expression is likely \(16 + 8 \times 4 - 2 = 46\), but choices max at 20. Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 - 4 - 2 = 18\), but not in choices. Alternatively, if "8 4" is 84, but 16+84-2=98. Not matching. The expression is likely \( \frac{16 + 8}{4} - 2 = \frac{24}{4} - 2 = 6 - 2 = 4 \).  

Tempting choices: 8 (B) if order misapplied.  

ANSWER 10: A  

Problem 11:  
Expression: \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\).  
- \(\sqrt{4} = 2\)  
- \(\sqrt{8 \cdot 2} = \sqrt{16} = 4\)  
- \(\sqrt{16 \cdot 4} = \sqrt{64} = 8\)  

Tempting choices: 4 (A) if innermost not simplified.  

ANSWER 11: C  

Problem 12:  
Compare decimals:  
- A: 0.99  
- B: 0.9099  
- C: 0.9  
- D: 0.909  
- E: 0.9009  
Largest is 0.99.  

Tempting choices: 0.9099 (B) if decimals misaligned.  

ANSWER 12: A  

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

### Problem 1:
What is asked: An aquarium with a 100cm×40cm rectangular base is filled with water to 37cm depth. A 1000 cm³ rock is submerged; how much does the water level rise?
Step-by-step: When a submerged object displaces water, its volume equals the additional volume of water in the aquarium. The rise in water level Δh satisfies: Volume of rock = base area of aquarium × Δh.
Base area = 100×40 = 4000 cm².
Δh = 1000 / 4000 = 0.25 cm.
Tempting wrong answers: 0.5 (if base area is miscalculated as 2000 cm²) or 1 (wrong division).
ANSWER 1: A

---

### Problem 2:
What is asked: Annika plus 3 friends (4 total players) each get 15 cards. If 2 more join, how many cards does each player get?
Step-by-step: First calculate total number of cards: 4 players × 15 cards = 60 total cards. After 2 join, total players = 4+2=6. Cards per player = 60 / 6 = 10.
Tempting wrong answers: 9 (if you incorrectly count only 3 initial players, leading to 45 total cards) or 12 (wrong division).
ANSWER 2: C

---

### Problem 3:
What is asked: The custom notation a c b d = a·d − b·c. What is the value of 3 1 4 2?
Step-by-step: Match the values to the definition: a=3, c=1, b=4, d=2. Calculate: 3·2 - 4·1 = 6 - 4 = 2.
Tempting wrong answers: -2 (swap a and b incorrectly) or 0 (miscalculate the multiplication).
ANSWER 3: E

---

### Problem 4:
What is asked: Isabella needs a 95 average on 4 100-point tests. She scored 97 and 91 on the first two. What is the lowest possible third test score that lets her reach her goal?
Step-by-step: Total points needed for 95 average: 4×95 = 380 points. Sum of first two tests: 97+91=188. To minimize the third test score, maximize the fourth test (max 100 points). Total points left for tests 3 and 4: 380-188=192. Minimum third test = 192 - 100 = 92.
Tempting wrong answers: 96 (the average of the remaining two tests, not the minimum possible third score).
ANSWER 4: B

---

### Problem 5:
What is asked: Bridget gave half her apples to Ann, 3 to Cassie, and kept 4. How many apples did she buy originally?
Step-by-step: Let N = original number of apples. After giving half to Ann, she has N/2 left. Then N/2 - 3 = 4 → N/2=7 → N=14. Verify: 14 → 7 left after Ann, minus 3 for Cassie = 4 kept, which matches.
Tempting wrong answers: 7 (the number of apples Bridget had after giving half to Ann, not the original total).
ANSWER 5: E

---

### Problem 6:
What is asked: A jar has red, green, blue marbles. All but 6 are red, all but 8 are green, all but 4 are blue. How many total marbles?
Step-by-step: Let T=total marbles, R=red, G=green, B=blue. The conditions translate to: G+B=6, R+B=8, R+G=4. Add all equations: 2(R+G+B)=18 → R+G+B=T=9. Verify: 3 red, 1 green, 5 blue sum to 9, matching all conditions.
Tempting wrong answers: 18 (the sum of 6+8+4, not halved to get total marbles).
ANSWER 6: C

---

### Problem 7:
What is asked: What is the sum of the two smallest prime factors of 250?
Step-by-step: Prime factorization of 250: 250=2×5³. The prime factors are 2 and 5. Their sum is 2+5=7.
Tempting wrong answers: 10 (sum of 5+5, or wrong addition of 2+5).
ANSWER 7: C

---

### Problem 8:
What is asked: Bag A has chips 1,3,5; Bag B has 2,4,6. One chip drawn from each bag, how many distinct sum values are possible?
Step-by-step: List all sums: 1+2=3, 1+4=5,1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7,5+4=9,5+6=11. Distinct sums: 3,5,7,9,11 → 5 total.
Tempting wrong answers: 9 (total number of unordered pairs, not distinct sums).
ANSWER 8: B

---

### Problem 9:
What is asked: 2 boys and 3 girls line up for a photo, with a boy at each end and girls in the middle. How many arrangements are possible?
Step-by-step: Arrange the boys first: 2 boys can be ordered at the ends in 2! = 2 ways. The 3 middle girls can be ordered in 3! =6 ways. Total arrangements: 2×6=12.
Tempting wrong answers: 6 (only the number of ways to arrange the girls, ignoring the boys).
ANSWER 9: E

---

### Problem 10:
What is asked: Evaluate (16+8)/(4-2) (formatted as a fraction in the original problem).
Step-by-step: Calculate numerator:16+8=24, denominator:4-2=2. 24/2=12.
Tempting wrong answers:16 (if order of operations is misapplied to calculate 16 + 8/4 -2=16).
ANSWER 10: C

---

### Problem 11:
What is asked: Evaluate √(16 · √(8 · √4)).
Step-by-step: Calculate from the innermost root first: √4=2. Next: 8×2=16, √16=4. Next:16×4=64, √64=8.
Tempting wrong answers: 4√2 (incorrect root calculation).
ANSWER 11: C

---

### Problem 12:
What is asked: Which number is largest: 0.99, 0.9099, 0.9, 0.909, 0.9009.
Step-by-step: Line up all decimals to compare place values: 0.9900 > 0.9099 > 0.9090 > 0.9009 > 0.9000. The largest is 0.99.
Tempting wrong answers: 0.9099 (the second largest value).
ANSWER 12: A

### Problem 1:
We need to find how much the water level rises when a 1000 cm³ rock is submerged in the aquarium.
1.  The volume of displaced water equals the volume of the submerged rock: 1000 cm³.
2.  The base area of the aquarium is \(100 \, \text{cm} \times 40 \, \text{cm} = 4000 \, \text{cm}^2\).
3.  Let \(h\) be the rise in water level. The volume of the displaced water is also equal to base area × rise in height: \(4000h = 1000\).
4.  Solve for \(h\): \(h = \frac{1000}{4000} = 0.25\) cm.
ANSWER 1: A
---
### Problem 2:
We need to find the number of cards per player when 2 more friends join.
1.  Original number of players: Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).
2.  New number of players: 4 + 2 = 6 players. Total cards remain 60.
3.  Cards per player: \(\frac{60}{6} = 10\).
ANSWER 2: C
---
### Problem 3:
The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). Compute \(3 \, 1 \, 4 \, 2\).
1.  Identify \(a=3, c=1, b=4, d=2\).
2.  Substitute into the formula: \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).
ANSWER 3: E
---
### Problem 4:
Find the lowest possible third test score to get an average of 95 on 4 tests.
1.  Total points needed for an average of 95: \(4 \times 95 = 380\).
2.  Sum of first two test scores: \(97 + 91 = 188\).
3.  Let \(x\) = third test score, \(y\) = fourth test score (max 100, since tests are 100-point). We need \(188 + x + y = 380\), so \(x + y = 192\).
4.  To minimize \(x\), maximize \(y = 100\). Then \(x = 192 - 100 = 92\).
ANSWER 4: B
---
### Problem 5:
Find the original number of apples Bridget bought.
1.  Let \(x\) = total apples. After giving half to Ann, she has \(\frac{x}{2}\) left.
2.  After giving 3 to Cassie, she has \(\frac{x}{2} - 3\) left, which equals 4 (the apples she kept).
3.  Solve: \(\frac{x}{2} - 3 = 4 \implies \frac{x}{2} =7 \implies x=14\).
ANSWER 5: E
---
### Problem 6:
Find the total number of marbles.
1.  Let \(R, G, B\) be the number of red, green, blue marbles, total \(T = R+G+B\).
2.  Translate the statements:
    - All but 6 are red: \(G + B = 6\)
    - All but 8 are green: \(R + B = 8\)
    - All but 4 are blue: \(R + G =4\)
3.  Add all three equations: \(2(R+G+B) = 6+8+4=18 \implies 2T=18 \implies T=9\).
ANSWER 6: C
---
### Problem 7:
Find the sum of the two smallest prime factors of 250.
1.  Factorize 250: \(250 = 2 \times 125 = 2 \times 5^3\).
2.  The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.
3.  Sum: \(2 +5=7\).
ANSWER 7: C
---
### Problem 8:
Find the number of distinct sums when drawing one chip from each bag.
1.  List all possible pairs and sums:
    - Bag A (1,3,5) + Bag B (2,4,6):
      \(1+2=3, 1+4=5, 1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7, 5+4=9, 5+6=11\)
2.  Distinct sums: 3, 5, 7, 9, 11. There are 5 unique values.
ANSWER 8: B
---
### Problem 9:
Find the number of arrangements with boys at each end and girls in the middle.
1.  Arrange the 2 boys at the ends: there are \(2! = 2\) ways (order matters: which boy is left/right).
2.  Arrange the 3 girls in the middle 3 seats: \(3! = 6\) ways.
3.  Total arrangements: \(2 \times 6 =12\).
ANSWER 9: E
---
### Problem 10:
Compute \(16 + 8 \, 4 - 2\). The missing operator between 8 and 4 is division (standard order of operations, common middle school problem):
1.  Interpret as \(16 + \frac{8}{4} - 2\).
2.  Compute division first: \(\frac{8}{4}=2\).
3.  Then add/subtract left to right: \(16 + 2 - 2 =16\).
*(Note: If interpreted as \(16 +8 -4 \times 2\), order of operations gives \(4 \times 2=8\), so \(16+8-8=16\), same result.)*
ANSWER 10: D
---
### Problem 11:
Compute \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\) by working from the innermost root outward.
1.  Innermost: \(\sqrt{4} = 2\).
2.  Next layer: \(\sqrt{8 \cdot 2} = \sqrt{16} =4\).
3.  Outermost: \(\sqrt{16 \cdot 4} = \sqrt{64} =8\).
ANSWER 11: C
---
### Problem 12:
Find the largest decimal.
1.  Rewrite all numbers with 4 decimal places for easy comparison:
    - A: \(0.9900\)
    - B: \(0.9099\)
    - C: \(0.9000\)
    - D: \(0.9090\)
    - E: \(0.9009\)
2.  Compare digit by digit: all have 9 in the tenths place. A has 9 in the hundredths place, while all others have 0. Thus A is the largest.
ANSWER 12: A

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

**Problem 1:**

What is being asked: By how many centimeters does the water level rise when a 1000 cm³ rock is submerged?

Setup: The base area is 100 cm × 40 cm = 4000 cm². When the rock is submerged, it displaces 1000 cm³ of water.

Water level rise = Volume displaced ÷ Base area = 1000 ÷ 4000 = 0.25 cm

ANSWER 1: A

---

**Problem 2:**

What is being asked: If cards are distributed evenly, how many cards per player when the player count increases?

Setup: Initially 4 players (Annika + 3 friends) each get 15 cards.
Total cards = 4 × 15 = 60 cards

With 2 more friends joining: 4 + 2 = 6 players
Cards per player = 60 ÷ 6 = 10 cards

ANSWER 2: C

---

**Problem 3:**

What is being asked: Evaluate the operation 3 1 4 2 using the rule a c b d = a·d − b·c

Setup: With a = 3, c = 1, b = 4, d = 2:
3 1 4 2 = 3·2 − 4·1 = 6 − 4 = 2

ANSWER 3: E

---

**Problem 4:**

What is being asked: What is the lowest score on test 3 that allows Isabella to achieve a 95 average on four tests?

Setup: Goal average is 95, so total needed = 95 × 4 = 380 points
First two tests: 97 + 91 = 188 points
Points needed from tests 3 and 4: 380 − 188 = 192 points

Let x = score on test 3. She can still reach her goal if she can score 100 on test 4:
x + 100 ≥ 192
x ≥ 92

The lowest possible score on test 3 is 92.

ANSWER 4: B

---

**Problem 5:**

What is being asked: How many apples did Bridget buy originally?

Setup: Let x = original number of apples
- She gave half to Ann: x/2 apples given away
- Remaining: x/2
- She gave 3 to Cassie: x/2 − 3
- She kept 4 for herself: x/2 − 3 = 4

Solving: x/2 = 7, so x = 14

ANSWER 5: E

---

**Problem 6:**

What is being asked: How many marbles are in the jar?

Setup: Let R, G, B = numbers of red, green, blue marbles; T = total
- "All but 6 are red": G + B = 6
- "All but 8 are green": R + B = 8
- "All but 4 are blue": R + G = 4

Adding all three equations: 2(R + G + B) = 6 + 8 + 4 = 18
Therefore: R + G + B = 9

ANSWER 6: C

---

**Problem 7:**

What is being asked: What is the sum of the two smallest prime factors of 250?

Setup: Factor 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5³

Prime factors: 2 and 5
Sum = 2 + 5 = 7

ANSWER 7: C

---

**Problem 8:**

What is being asked: How many different sums are possible?

Setup: Bag A: {1, 3, 5}, Bag B: {2, 4, 6}
All possible sums:
- 1 + 2 = 3
- 1 + 4 = 5
- 1 + 6 = 7
- 3 + 2 = 5 (duplicate)
- 3 + 4 = 7 (duplicate)
- 3 + 6 = 9
- 5 + 2 = 7 (duplicate)
- 5 + 4 = 9 (duplicate)
- 5 + 6 = 11

Distinct sums: {3, 5, 7, 9, 11} = 5 different values

ANSWER 8: B

---

**Problem 9:**

What is being asked: How many seating arrangements with boys at ends and girls in middle?

Setup: 2 boys at the ends can be arranged in 2! = 2 ways
3 girls in the middle can be arranged in 3! = 6 ways
Total arrangements = 2 × 6 = 12

ANSWER 9: E

---

**Problem 10:**

What is being asked: Evaluate 16 + 8 ÷ 4 − 2

Setup: Using order of operations (division before addition/subtraction):
16 + (8 ÷ 4) − 2 = 16 + 2 − 2 = 16

ANSWER 10: D

---

**Problem 11:**

What is being asked: Evaluate √(16 · √(8 · √4))

Setup: Work from the inside out:
- √4 = 2
- 8 · √4 = 8 · 2 = 16
- √(8 · √4) = √16 = 4
- 16 · √(8 · √4) = 16 · 4 = 64
- √(16 · √(8 · √4)) = √64 = 8

ANSWER 11: C

---

**Problem 12:**

What is being asked: Which decimal is largest?

Setup: Compare decimals:
- A: 0.99
- B: 0.9099
- C: 0.9
- D: 0.909
- E: 0.9009

Comparing: 0.99 > 0.9099 > 0.909 > 0.9009 > 0.9

The largest is 0.99

ANSWER 12: A

Problem 1:

The aquarium has a rectangular base of \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is

\[
100 \cdot 40 = 4000 \text{ cm}^2.
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

If the water rises by \(h\) centimeters, then

\[
4000h = 1000.
\]

So

\[
h = \frac{1000}{4000} = \frac14 = 0.25.
\]

So the water level rises \(0.25\) cm.

ANSWER 1: A

---

Problem 2:

Annika and 3 friends means there are

\[
4 \text{ players}.
\]

If each player gets 15 cards, then the total number of cards is

\[
4 \cdot 15 = 60.
\]

Now 2 more friends join, so there are

\[
6 \text{ players}.
\]

The same 60 cards are shared evenly, so each gets

\[
60 \div 6 = 10.
\]

ANSWER 2: C

---

Problem 3:

We are told that if the four numbers are \(a, b, c, d\), then

\[
a \ b \ c \ d = a\cdot d - b\cdot c.
\]

So for \(3,1,4,2\),

\[
3 \ 1 \ 4 \ 2 = 3\cdot 2 - 1\cdot 4 = 6 - 4 = 2.
\]

ANSWER 3: E

---

Problem 4:

She wants an average of 95 on 4 tests, so the total of the 4 scores must be

\[
4 \cdot 95 = 380.
\]

Her first two scores are 97 and 91, which add to

\[
97+91=188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To make the total 380, the fourth score would have to be

\[
380 - (188+x) = 192 - x.
\]

For this to be possible, the fourth score must be at most 100:

\[
192 - x \le 100.
\]

So

\[
x \ge 92.
\]

The lowest possible third score is 92.

ANSWER 4: B

---

Problem 5:

Let the number of apples Bridget bought be \(x\).

She gave half to Ann, so she kept the other half:

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

Then she gave Cassie 3 apples and kept 4 apples, so before giving Cassie the apples she had

\[
3+4=7.
\]

Thus,

\[
\frac{x}{2}=7 \quad \Rightarrow \quad x=14.
\]

ANSWER 5: E

---

Problem 6:

Let the total number of marbles be \(n\).

- “All but 6 are red” means the number of red marbles is \(n-6\).
- “All but 8 are green” means the number of green marbles is \(n-8\).
- “All but 4 are blue” means the number of blue marbles is \(n-4\).

Since every marble is one of these colors, their counts add to \(n\):

\[
(n-6)+(n-8)+(n-4)=n.
\]

Simplify:

\[
3n-18=n
\]

\[
2n=18
\]

\[
n=9.
\]

ANSWER 6: C

---

Problem 7:

We need the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3.
\]

So the prime factors are 2 and 5. The two smallest prime factors are 2 and 5, and their sum is

\[
2+5=7.
\]

ANSWER 7: C

---

Problem 8:

Bag A has \(1,3,5\); Bag B has \(2,4,6\).

Possible sums:

- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums are

\[
3,5,7,9,11
\]

which is 5 different values.

ANSWER 8: B

---

Problem 9:

There are 2 boys and 3 girls, and the picture must have:

- a boy at each end,
- the 3 girls in the middle.

First, arrange the 2 boys at the ends. They can be switched in

\[
2! = 2
\]

ways.

Then arrange the 3 girls in the middle in

\[
3! = 6
\]

ways.

So the total number of arrangements is

\[
2 \cdot 6 = 12.
\]

ANSWER 9: E

---

Problem 10:

The expression is

\[
16 + 8\cdot 4 - 2.
\]

Use order of operations:

\[
8\cdot 4=32.
\]

Then

\[
16+32-2=46.
\]

But 46 is not among the choices, so the intended expression must be read as

\[
16 + 8 \div 4 - 2.
\]

Then

\[
8\div 4 = 2,
\]

so

\[
16+2-2=16.
\]

ANSWER 10: D

---

Problem 11:

We evaluate from the inside out:

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

First,

\[
\sqrt4=2.
\]

So the expression becomes

\[
\sqrt{16\cdot \sqrt{8\cdot 2}}
=
\sqrt{16\cdot \sqrt{16}}.
\]

Since \(\sqrt{16}=4\),

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C

---

Problem 12:

Compare the decimals:

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

The largest is the one closest to 1. Since \(0.99\) is greater than any number beginning with \(0.90\), it is the largest.

ANSWER 12: A

Problem 1:  
We need the rise in water level when a rock is placed in the aquarium and submerged.

- Base dimensions: \(100\text{ cm} \times 40\text{ cm}\), so base area  
  \[
  A = 100\cdot 40 = 4000\text{ cm}^2.
  \]
- The rock is completely submerged, so the water volume displaced equals the rock’s volume:  
  \[
  V = 1000\text{ cm}^3.
  \]
- Water level rise \(\Delta h\) satisfies \(A\Delta h = V\):  
  \[
  4000\cdot \Delta h = 1000 \implies \Delta h = \frac{1000}{4000}=0.25\text{ cm}.
  \]

So the rise is \(0.25\) cm.  
Tempting wrong idea: using the *remaining* depth \(37\) cm or checking if it overflows—none of that matters; only displaced volume matters.

ANSWER 1: A  

---

Problem 2:  
All cards are dealt evenly. Let the total number of cards be \(T\).

Initially, Annika + 3 friends = 4 players, and each gets 15 cards:  
\[
T = 4\cdot 15 = 60.
\]
Next game: 2 more friends join, so total players:
\[
4+2=6.
\]
Each player gets:
\[
\frac{60}{6}=10.
\]

ANSWER 2: C  

---

Problem 3:  
Given the rule \(a\,b\,d\,c = a\cdot d - b\cdot c\).  
The expression “\(3\ 1\ 4\ 2\)” corresponds to:
- \(a=3\)
- \(b=1\)
- \(d=4\)
- \(c=2\)

Compute:
\[
a\cdot d - b\cdot c = 3\cdot 4 - 1\cdot 2 = 12-2=10.
\]

But notice the choices are only \(-2,-1,0,1,2\), so we must interpret the placeholder correctly:  
The problem states: **“If \(a b c d = a\cdot d - b\cdot c\)”**.

So “\(3\ 1\ 4\ 2\)” means:
- \(a=3,\ b=1,\ c=4,\ d=2\)

Then:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 1\cdot 4 = 6-4=2.
\]

ANSWER 3: E  

---

Problem 4:  
Goal average is 95 over 4 tests, so total points needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
Let the third test score be \(x\), and the fourth be \(y\). Then:
\[
188 + x + y = 380 \implies x+y=192.
\]
We want the **lowest possible** \(x\) such that she can still reach the goal. That means make \(y\) as large as possible (since grades are at most 100).
Maximum \(y=100\). Then:
\[
x + 100 = 192 \implies x = 92.
\]

Check: then the four scores are \(97,91,92,100\) which sum to \(380\), average \(95\). Works.

ANSWER 4: B  

---

Problem 5:  
Let the total number of apples be \(N\).

- She gave half to Ann, so Ann got \(N/2\), leaving \(N/2\).
- Then she gave Cassie 3 apples, and kept 4 apples. That means after giving Cassie 3, her remaining is 4:
  \[
  \frac{N}{2} - 3 = 4 \implies \frac{N}{2}=7 \implies N=14.
  \]

ANSWER 5: E  

---

Problem 6:  
“All but 6 are red” means: number of red marbles is 6.  
Similarly:
- Red = 6 (since all but 6 are red → 6 red)
- Green: all but 8 are green → 8 green
- Blue: all but 4 are blue → 4 blue

Total marbles:
\[
6+8+4=18.
\]
But none of the choices are 18, so interpret carefully:  
“All but 6 are red” means the number of red marbles is **(total − 6)**, not 6.

Let total be \(T\). Then:
- Red marbles = \(T-6\)
- Green marbles = \(T-8\)
- Blue marbles = \(T-4\)

Sum:
\[
(T-6)+(T-8)+(T-4)=T.
\]
\[
3T - (6+8+4)=T
\]
\[
3T-18=T \implies 2T=18 \implies T=9.
\]

ANSWER 6: C  

---

Problem 7:  
We want the sum of the **two smallest prime factors** of 250.

Factor \(250\):
\[
250 = 25\cdot 10 = 5^2 \cdot 2\cdot 5 = 2\cdot 5^3.
\]
Prime factors are \(2\) and \(5\). The two smallest are \(2\) and \(5\).

Sum:
\[
2+5=7.
\]

ANSWER 7: D  

---

Problem 8:  
Bag A: \(\{1,3,5\}\)  
Bag B: \(\{2,4,6\}\)

Possible sums:
- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums: \(\{3,5,7,9,11\}\) → that’s 5 values.

ANSWER 8: B  

---

Problem 9:  
We need arrangements of 5 people in a row with:
- a boy at each end (positions 1 and 5)
- the three girls in the middle (positions 2,3,4)

There are 2 boys: they can be placed at the ends in
\[
2!\ = 2
\]
ways (left boy/right boy).

The 3 girls can be arranged in the three middle spots in
\[
3! = 6
\]
ways.

Total arrangements:
\[
2\cdot 6 = 12.
\]

ANSWER 9: E  

---

Problem 10:  
Expression: \(16 + 8^4 - 2\).

Compute in order:
- \(8^4 = 4096\)

So:
\[
16 + 4096 - 2 = 4110.
\]
No answer choice matches, so we interpret the formatting as \(16 + 8 \sqrt{4} - 2\) or \(16 + 8^4 - 2\)?  
Given the AMC/JH style, likely it is:
\[
16 + 8\sqrt{4} - 2.
\]
Then:
\[
\sqrt{4}=2 \implies 8\sqrt{4}=8\cdot 2=16.
\]
So:
\[
16 + 16 - 2 = 30,
\]
still not in choices.

Another common formatting is:
\[
16 + 8^{4} - 2 \quad \text{(still not)}
\]

The only choiceable consistent possibility is:
\[
16 + \sqrt{8}\,^4 - 2 \text{ (nonsense)}
\]

Let’s use the most plausible literal from the text: **“16 + 8 4 − 2”** often means \(16 + 8^4 - 2\) if the caret was lost. But then no choice.

Given typical AMC problems, the intended is usually:
\[
16 + 8 - 4 - 2 = 18
\]
also not.

Because the answer choices are only \(4,8,12,16,20\), the intended expression must simplify to one of those. The most natural is:
\[
16 + 8/4 - 2 = 16 + 2 - 2 = 16.
\]
This matches a common missing symbol: “8 4” often means \(8/4\).

So:
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D  

---

Problem 11:  
Evaluate:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}.
\]

Work from inside out:

1) \(\sqrt{4}=2\)

2) Inside next root:
\[
8\cdot \sqrt{4} = 8\cdot 2 = 16
\]
So:
\[
\sqrt{8\cdot \sqrt{4}} = \sqrt{16}=4.
\]

3) Now plug into the outer expression:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}=\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C  

---

Problem 12:  
Largest number among:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare by decimal places:

- .99 is 0.99
- .9099 is 0.9099
- .9 is 0.9000
- .909 is 0.9090
- .9009 is 0.9009

Clearly \(0.99\) is greater than all the numbers less than \(0.91\).

ANSWER 12: A

Problem 1:
The volume of the water displaced by the rock is equal to the volume of the rock itself, which is $1000 \text{ cm}^3$. The aquarium has a rectangular base with area $A = 100 \text{ cm} \times 40 \text{ cm} = 4000 \text{ cm}^2$. When the rock is submerged, the water level rises by height $h$ such that the volume of the displaced water is $A \times h$. 
Setting this equal to the rock's volume: $4000 \times h = 1000$.
Solving for $h$: $h = \frac{1000}{4000} = \frac{1}{4} = 0.25 \text{ cm}$.
ANSWER 1: A

Problem 2:
Annika and 3 friends make 4 players total. Each received 15 cards, so the total number of cards is $4 \times 15 = 60$. If 2 more friends join, the total number of players becomes $4 + 2 = 6$. The cards are distributed evenly among the 6 players, so each player gets $60 \div 6 = 10$ cards.
ANSWER 2: C

Problem 3:
The problem defines the operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$.
For the expression $\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix}$, we have $a=3, b=1, c=4, d=2$.
Calculating: $(3 \times 2) - (1 \times 4) = 6 - 4 = 2$.
ANSWER 3: E

Problem 4:
Isabella needs an average of 95 over 4 tests, meaning the total sum of her scores must be $95 \times 4 = 380$. Her first two scores are 97 and 91, totaling $97 + 91 = 188$. Let $x$ be the score on the third test and $y$ be the score on the fourth. We have $188 + x + y = 380$, so $x + y = 192$. To make $x$ as small as possible, we must make $y$ as large as possible. Since the maximum score on a test is 100, we set $y = 100$. Then $x + 100 = 192$, so $x = 92$.
ANSWER 4: B

Problem 5:
Let $x$ be the total number of apples Bridget bought. She gave half away, leaving $\frac{x}{2}$. Then she gave 3 away, leaving $\frac{x}{2} - 3$. She kept 4, so $\frac{x}{2} - 3 = 4$. Adding 3 to both sides, $\frac{x}{2} = 7$. Multiplying by 2, $x = 14$.
ANSWER 5: E

Problem 6:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R + G + B$.
"All but 6 are red" means $G + B = 6$.
"All but 8 are green" means $R + B = 8$.
"All but 4 are blue" means $R + G = 4$.
Adding these three equations: $2(R + G + B) = 6 + 8 + 4 = 18$.
So $R + G + B = 18 \div 2 = 9$.
ANSWER 6: C

Problem 7:
The prime factorization of 250 is $250 = 2 \times 125 = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two prime factors is $2 + 5 = 7$.
ANSWER 7: C

Problem 8:
Bag A has $\{1, 3, 5\}$ and Bag B has $\{2, 4, 6\}$.
Possible sums:
$1+2=3, 1+4=5, 1+6=7$
$3+2=5, 3+4=7, 3+6=9$
$5+2=7, 5+4=9, 5+6=11$
The set of distinct sums is $\{3, 5, 7, 9, 11\}$. There are 5 distinct values.
ANSWER 8: B

Problem 9:
There are 2 boys (B1, B2) and 3 girls (G1, G2, G3). A boy must be at each end. There are 2 choices for the left end and 1 choice for the right end ($2 \times 1 = 2$ ways to place boys). The 3 girls can be arranged in the middle in $3! = 3 \times 2 \times 1 = 6$ ways. Total arrangements = $2 \times 6 = 12$.
ANSWER 9: E

Problem 10:
Following the order of operations (PEMDAS/BODMAS):
$16 + (8 \div 4) - 2 = 16 + 2 - 2 = 16$.
ANSWER 10: D

Problem 11:
Evaluate from the inside out:
$\sqrt{4} = 2$
$\sqrt{8 \times 2} = \sqrt{16} = 4$
$\sqrt{16 \times 4} = \sqrt{64} = 8$
ANSWER 11: C

Problem 12:
To compare, pad with zeros to match the number of decimal places:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
Comparing the values, 0.9900 is the largest.
ANSWER 12: A

Problem 1:
The base of the aquarium measures 100 cm by 40 cm, so its area is \(100 \times 40 = 4000\) cm². The rock has volume 1000 cm³ and is fully submerged, displacing exactly 1000 cm³ of water. The resulting rise in water level is therefore the displaced volume divided by the base area: \(1000 / 4000 = 0.25\) cm. (The original water depth and tank height are irrelevant once the surface area is known.)

ANSWER 1: A

Problem 2:
Annika plus 3 friends gives 4 players, each receiving 15 cards, for a total of \(4 \times 15 = 60\) cards. When 2 more players join, there are 6 players sharing the same 60 cards, so each receives \(60 / 6 = 10\) cards.

ANSWER 2: C

Problem 3:
The given operation evaluates \(a\, c\, b\, d\) as \(a \cdot d - b \cdot c\). Substituting the numbers yields \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).

ANSWER 3: E

Problem 4:
An average of 95 on four 100-point tests requires a total of \(4 \times 95 = 380\) points. The first two scores sum to \(97 + 91 = 188\), leaving 192 points still needed from the last two tests. The fourth test can score at most 100, so the third test must be at least \(192 - 100 = 92\). Any lower score on the third test would make a total of 380 impossible.

ANSWER 4: B

Problem 5:
Let \(t\) be the original number of apples. After giving half to Ann, \(t/2\) remain. After giving 3 more to Cassie, \(t/2 - 3\) remain, and this equals the 4 apples kept. Solving \(t/2 - 3 = 4\) gives \(t/2 = 7\), so \(t = 14\).

ANSWER 5: E

Problem 6:
Let \(m\) be the total number of marbles. Then there are \(m-6\) red, \(m-8\) green, and \(m-4\) blue. Their sum must equal \(m\):
\[
(m-6) + (m-8) + (m-4) = m \implies 3m - 18 = m \implies 2m = 18 \implies m = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies the three “all-but” statements.

ANSWER 6: C

Problem 7:
Factorize \(250 = 2 \cdot 5^3\). The two smallest distinct prime factors are 2 and 5; their sum is 7.

ANSWER 7: C

Problem 8:
All nine possible sums are:
\[
1+2=3,\ 1+4=5,\ 1+6=7,\ 3+2=5,\ 3+4=7,\ 3+6=9,\ 5+2=7,\ 5+4=9,\ 5+6=11.
\]
The distinct values are 3, 5, 7, 9, 11—exactly five.

ANSWER 8: B

Problem 9:
The seating pattern must be boy–girl–girl–girl–boy. The two distinct boys may be placed at the ends in \(2!\) = 2 ways. The three distinct girls may be arranged in the middle three seats in \(3!\) = 6 ways. The total number of arrangements is therefore \(2 \times 6 = 12\).

ANSWER 9: E

Problem 10:
Interpreting the expression with standard order of operations and treating juxtaposition of 8 and 4 as division yields
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D

Problem 11:
Evaluate from the inside outward:
\[
\sqrt{4} = 2, \quad 8 \cdot 2 = 16, \quad \sqrt{16} = 4, \quad 16 \cdot 4 = 64, \quad \sqrt{64} = 8.
\]

ANSWER 11: C

Problem 12:
Compare the decimals directly:
\[
0.99 > 0.9 > 0.9099 > 0.909 > 0.9009.
\]
Thus 0.99 is the largest.

ANSWER 12: A

Problem 1:
The problem asks us to determine by how many centimeters the water level rises when a rock with a volume of 1000 cm³ is submerged in an aquarium filled with water to a depth of 37 cm. The aquarium has a rectangular base measuring 100 cm by 40 cm and a height of 50 cm.

First, we need to calculate the initial volume of water in the aquarium. The volume of water is given by the formula length × width × height. So, the initial volume of water = 100 cm × 40 cm × 37 cm = 148,000 cm³.

When the rock is submerged, it displaces a volume of water equal to its own volume, which is 1000 cm³. This means the water level will rise to accommodate the volume of the rock.

The rise in water level can be calculated by dividing the volume of the rock by the base area of the aquarium. The base area = 100 cm × 40 cm = 4000 cm².

Rise in water level = Volume of rock / Base area = 1000 cm³ / 4000 cm² = 0.25 cm.

Let's examine the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

The calculated rise in water level matches option A.

ANSWER 1: A

Problem 2:
The problem states that in the card game Buffalo Shuffle-o, all cards are distributed evenly among players. Initially, Annika and 3 friends play, and each is dealt 15 cards. We need to find out how many cards each player will be dealt if 2 more friends join the next game.

First, we determine the total number of cards. With 4 players initially, each getting 15 cards, the total number of cards = 4 × 15 = 60.

When 2 more friends join, the total number of players becomes 4 + 2 = 6.

To find out how many cards each player will get, we divide the total number of cards by the new number of players: 60 / 6 = 10.

Let's examine the choices:
A. 8
B. 9
C. 10
D. 11
E. 12

The calculated number of cards per player matches option C.

ANSWER 2: C

Problem 3:
The problem defines a operation as a c b d = a·d − b·c and asks us to find the value of 3 1 4 2.

Using the given definition, we substitute a = 3, b = 1, c = 4, and d = 2 into the equation: 3 1 4 2 = 3·2 − 1·4 = 6 - 4 = 2.

Let's examine the choices:
A. −2
B. −1
C. 0
D. 1
E. 2

The calculated value matches option E.

ANSWER 3: E

Problem 4:
Isabella needs to achieve an average grade of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score she could have made on the third test to still reach her goal.

The total score required for an average of 95 on 4 tests = 95 × 4 = 380.

The sum of her first two test scores = 97 + 91 = 188.

So, the total score needed on the last two tests = 380 - 188 = 192.

To minimize the score on the third test, we maximize the score on the fourth test. The maximum score on the fourth test is 100.

Thus, the minimum score on the third test = 192 - 100 = 92.

Let's examine the choices:
A. 90
B. 92
C. 95
D. 96
E. 97

The calculated minimum score matches option B.

ANSWER 4: B

Problem 5:
Bridget gave half of her apples to Ann, then gave 3 to Cassie, and kept 4 for herself. We need to find out how many apples she originally bought.

Let's denote the original number of apples as x.

After giving half to Ann, Bridget had x/2 apples left.

Then, she gave 3 apples to Cassie and kept 4, so x/2 = 3 + 4 = 7.

Thus, x = 7 × 2 = 14.

Let's examine the choices:
A. 3
B. 4
C. 7
D. 11
E. 14

The calculated original number of apples matches option E.

ANSWER 5: E

Problem 6:
The problem states that in a jar, all but 6 marbles are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T, the number of red marbles as R, the number of green marbles as G, and the number of blue marbles as B.

From the given information:
1. T - R = 6 (all but 6 are red)
2. T - G = 8 (all but 8 are green)
3. T - B = 4 (all but 4 are blue)

Adding these equations gives: 3T - (R + G + B) = 18.

Since R + G + B = T (the total number of marbles is the sum of the marbles of each color), we have:
3T - T = 18, which simplifies to 2T = 18.

Thus, T = 18 / 2 = 9.

Let's examine the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

The calculated total number of marbles matches option C.

ANSWER 6: C

Problem 7:
We need to find the sum of the two smallest prime factors of 250.

First, let's factorize 250 into its prime factors: 250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5.

The sum of these two smallest prime factors = 2 + 5 = 7.

Let's examine the choices:
A. 2
B. 5
C. 7
D. 10
E. 12

The calculated sum matches option C.

ANSWER 7: C

Problem 8:
Bag A contains chips labeled 1, 3, and 5. Bag B contains chips labeled 2, 4, and 6. We need to find how many different values are possible for the sum of the numbers on the chips drawn from each bag.

Let's list the possible sums:
- (1,2) = 3
- (1,4) = 5
- (1,6) = 7
- (3,2) = 5
- (3,4) = 7
- (3,6) = 9
- (5,2) = 7
- (5,4) = 9
- (5,6) = 11

The possible sums are 3, 5, 7, 9, and 11. There are 5 different values.

Let's examine the choices:
A. 4
B. 5
C. 6
D. 7
E. 9

The number of different sums matches option B.

ANSWER 8: B

Problem 9:
The chess team consists of 2 boys and 3 girls. They need to be arranged with a boy at each end and the girls in the middle. We need to find out how many such arrangements are possible.

The boys can be arranged at the ends in 2! = 2 ways (either boy can be on the left or the right).

The 3 girls can be arranged in the 3 middle positions in 3! = 6 ways.

The total number of arrangements = (number of ways to arrange boys) × (number of ways to arrange girls) = 2 × 6 = 12.

Let's examine the choices:
A. 2
B. 4
C. 5
D. 6
E. 12

The calculated total number of arrangements matches option E.

ANSWER 9: E

Problem 10:
The problem asks us to evaluate the expression 16 + 8 ÷ 4 − 2.

Following the order of operations (PEMDAS/BODMAS: Parentheses/Brackets, Exponents/Orders, Multiplication and Division from left to right, Addition and Subtraction from left to right):
1. D

Problem 1:
The water rise is determined by the volume of the rock divided by the base area of the aquarium.  
Base area = \(100 \times 40 = 4000\text{ cm}^2\).  
Volume of rock = \(1000\text{ cm}^3\).  
Rise = \(\frac{1000}{4000} = 0.25\text{ cm}\).  
ANSWER 1: A

Problem 2:
Initially, Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 2: C

Problem 3:
The notation \(\begin{array}{cc} a & c \\ b & d \end{array} = a \cdot d - b \cdot c\).  
For \(\begin{array}{cc} 3 & 1 \\ 4 & 2 \end{array}\): \(a=3, c=1, b=4, d=2\).  
Value = \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).  
ANSWER 3: E

Problem 4:
Total points needed for an average of 95 over 4 tests = \(4 \times 95 = 380\).  
First two tests sum = \(97 + 91 = 188\).  
Remaining points needed = \(380 - 188 = 192\).  
To minimize the third test score, maximize the fourth (100). Then third test \(\ge 192 - 100 = 92\).  
Lowest possible third test score is 92.  
ANSWER 4: B

Problem 5:
Let \(x\) be the number of apples Bridget bought.  
She gave half to Ann, so she kept \(\frac{x}{2}\).  
Then she gave 3 to Cassie and kept 4, so \(\frac{x}{2} = 3 + 4 = 7\).  
Thus \(x = 14\).  
ANSWER 5: E

Problem 6:
Let \(R, G, B\) be the number of red, green, blue marbles.  
"All but 6 are red" \(\Rightarrow G + B = 6\).  
"All but 8 are green" \(\Rightarrow R + B = 8\).  
"All but 4 are blue" \(\Rightarrow R + G = 4\).  
Adding: \((G+B)+(R+B)+(R+G) = 6+8+4 = 18\) \(\Rightarrow 2(R+G+B) = 18\) \(\Rightarrow R+G+B = 9\).  
Total marbles = 9.  
ANSWER 6: C

Problem 7:
Prime factorization of 250: \(250 = 2 \times 5^3\).  
Prime factors are 2 and 5. The two smallest are 2 and 5.  
Sum = \(2 + 5 = 7\).  
ANSWER 7: C

Problem 8:
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}.  
Possible sums:  
1+2=3, 1+4=5, 1+6=7;  
3+2=5, 3+4=7, 3+6=9;  
5+2=7, 5+4=9, 5+6=11.  
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.  
ANSWER 8: B

Problem 9:
Boys must sit at the two ends.  
Arrange boys: 2 choices for left end, 1 for right end → \(2! = 2\) ways.  
Arrange 3 girls in the middle 3 seats: \(3! = 6\) ways.  
Total arrangements = \(2 \times 6 = 12\).  
ANSWER 9: E

Problem 10:
Interpret "8 4" as \(8 \div 4\).  
\(16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16\).  
ANSWER 10: D

Problem 11:
\(\sqrt{4} = 2\).  
\(8 \cdot 2 = 16\), \(\sqrt{16} = 4\).  
\(16 \cdot 4 = 64\), \(\sqrt{64} = 8\).  
ANSWER 11: C

Problem 12:
Compare decimals:  
0.99, 0.9099, 0.9, 0.909, 0.9009.  
0.99 is the largest.  
ANSWER 12: A

# Problem 1:

**What is being asked:** A rock of volume 1000 cm³ is submerged in an aquarium with a 100 cm × 40 cm base. How much does the water level rise?

**Setup:** When the rock is submerged, it displaces a volume of water equal to its own volume. This displaced water spreads across the base of the aquarium, causing the level to rise.

$$\text{Volume displaced} = \text{Base Area} \times \text{Rise in water level}$$

**Computation:**
- Base area = $100 \times 40 = 4000 \text{ cm}^2$
- $1000 = 4000 \times h$
- $h = \frac{1000}{4000} = 0.25 \text{ cm}$

**Check for overflow:** The current depth is 37 cm, and 37 + 0.25 = 37.25 cm, which is well below the 50 cm height. No overflow occurs.

**Ruling out other choices:** Choices B–E would require rock volumes of 2000, 4000, 5000, and 10000 cm³ respectively — all larger than the given 1000 cm³.

ANSWER 1: A

---

# Problem 2:

**What is being asked:** With 4 players, each gets 15 cards. With 6 players, how many cards does each get?

**Setup:** The total number of cards in the deck stays the same.

**Computation:**
- Number of players in first game: Annika + 3 friends = 4 players
- Total cards = $4 \times 15 = 60$ cards
- Number of players in second game: $4 + 2 = 6$ players
- Cards per player = $\frac{60}{6} = 10$

**Ruling out other choices:** Some might forget to count Annika (getting 3 players → 45 cards → 45/5 = 9, choice B) or miscount the new total as 5 players (60/5 = 12, choice E).

ANSWER 2: C

---

# Problem 3:

**What is being asked:** Evaluate the custom operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$ for $a=3, b=1, c=4, d=2$.

**Setup:** Substitute directly into the formula.

**Computation:**
$$\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2$$

**Ruling out other choices:** A common mistake is computing $b \cdot c - a \cdot d = 4 - 6 = -2$ (choice A) or mixing up which values go where.

ANSWER 3: E

---

# Problem 4:

**What is being asked:** Isabella needs an average of 95 on four tests. Her first two scores are 97 and 91. What is the lowest possible third test score so she can still reach her goal?

**Setup:**
- Total points needed: $4 \times 95 = 380$
- Sum of first two tests: $97 + 91 = 188$
- Sum needed from tests 3 and 4: $380 - 188 = 192$

**Computation:**
To minimize the third test score, maximize the fourth test score. The maximum possible score on any test is 100.
$$\text{Minimum test 3} = 192 - 100 = 92$$

**Ruling out other choices:** Choice A (90) would require test 4 to be $192 - 90 = 102$, which is impossible on a 100-point test. Choices C, D, E are higher than necessary.

ANSWER 4: B

---

# Problem 5:

**What is being asked:** Bridget bought apples, gave half to Ann, gave 3 to Cassie, and kept 4. How many did she buy?

**Setup:** Let $x$ = total apples bought. Work through the sequence of events.

**Computation:**
- After giving half to Ann: $\frac{x}{2}$ apples remain
- After giving 3 to Cassie: $\frac{x}{2} - 3$ apples remain
- She keeps 4: $\frac{x}{2} - 3 = 4$
- $\frac{x}{2} = 7$
- $x = 14$

**Verification:** 14 apples → give 7 to Ann → 7 left → give 3 to Cassie → 4 left. ✓

**Ruling out other choices:** Choice C (7) is a trap for those who forget to double the 7 remaining apples. Choice D (11) comes from incorrectly adding $4 + 3 + 4$.

ANSWER 5: E

---

# Problem 6:

**What is being asked:** Find the total number of marbles given conditions about each color.

**Setup:** Let $T$ = total marbles, $R$ = red, $G$ = green, $B$ = blue.
- "All but 6 are red" → $G + B = 6$ → $R = T - 6$
- "All but 8 are green" → $R + B = 8$ → $G = T - 8$
- "All but 4 are blue" → $R + G = 4$ → $B = T - 4$

**Computation:**
$$T = R + G + B = (T - 6) + (T - 8) + (T - 4) = 3T - 18$$
$$2T = 18 \implies T = 9$$

**Verification:** $R = 3, G = 1, B = 5$. Check: $G+B=6$ ✓, $R+B=8$ ✓, $R+G=4$ ✓.

ANSWER 6: C

---

# Problem 7:

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Setup:** Find the prime factorization of 250.

**Computation:**
$$250 = 2 \times 125 = 2 \times 5 \times 25 = 2 \times 5 \times 5 \times 5 = 2 \times 5^3$$

The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.
$$\text{Sum} = 2 + 5 = 7$$

**Ruling out other choices:** Choice A (2) is just one factor. Choice B (5) is just one factor. Choice D (10) would be $2 + 5 + 3$ or $5 + 5$, neither of which applies.

ANSWER 7: C

---

# Problem 8:

**What is being asked:** How many distinct sums can be formed by picking one chip from Bag A {1, 3, 5} and one from Bag B {2, 4, 6}?

**Setup:** List all 9 possible sums and count the unique values.

**Computation:**
| | 2 | 4 | 6 |
|---|---|---|---|
| **1** | 3 | 5 | 7 |
| **3** | 5 | 7 | 9 |
| **5** | 7 | 9 | 11 |

Unique sums: {3, 5, 7, 9, 11} → **5 different values**

**Ruling out other choices:** Choice E (9) is the trap of counting all 9 pairs without removing duplicates. Choice D (7) might come from miscounting.

ANSWER 8: B

---

# Problem 9:

**What is being asked:** How many ways can 2 boys and 3 girls sit in a row with boys at the ends and girls in the middle?

**Setup:** The arrangement must be: **B G G G B**

**Computation:**
- Arrange the 2 boys at the 2 end positions: $2! = 2$ ways
- Arrange the 3 girls in the 3 middle positions: $3! = 6$ ways
- Total arrangements: $2 \times 6 = 12$

**Ruling out other choices:** Choice A (2) only counts the boys' arrangements. Choice D (6) only counts the girls' arrangements. Both must be multiplied together.

ANSWER 9: E

---

# Problem 10:

**What is being asked:** Evaluate $16 + 8 \div 4 - 2$ using order of operations.

**Setup:** Apply PEMDAS — division before addition and subtraction.

**Computation:**
$$16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16$$

**Ruling out other choices:**
- Left-to-right without order of operations: $(16+8) \div 4 - 2 = 6 - 2 = 4$ (choice A)
- $(16+8) \div (4-2) = 24 \div 2 = 12$ (choice C)
- $16 + 8 \div (4-2) = 16 + 4 = 20$ (choice E

 Problem 1:
We need to find how much the water level rises when a rock of volume 1000 cm³ is submerged in a rectangular aquarium.

First, find the area of the aquarium’s base:
\[
100\text{ cm} \times 40\text{ cm} = 4000\text{ cm}^2.
\]

When the rock is completely submerged, it displaces 1000 cm³ of water. This displaced volume spreads out over the base, causing the water level to rise. The rise in height \(h\) satisfies:
\[
\text{base area} \times h = \text{volume of rock}.
\]
So,
\[
h = \frac{1000}{4000} = 0.25\text{ cm}.
\]

We should check that the aquarium does not overflow. The current depth is 37 cm and the height is 50 cm, so there is 13 cm of free space—plenty of room for a 0.25 cm rise.

Tempting wrong choices come from using the wrong area (e.g., using the side wall 100×50 or 40×50) or from dividing by the height 50. Only the base area matters for the level rise.

ANSWER 1: A

---

Problem 2:
We need to find how many cards each player gets when two extra friends join the game.

First, determine the total number of cards. Annika and 3 friends make 4 players, and each gets 15 cards:
\[
4 \times 15 = 60\text{ cards total}.
\]

When 2 more friends join, the number of players becomes:
\[
4 + 2 = 6.
\]

Distribute the 60 cards evenly among 6 players:
\[
60 \div 6 = 10.
\]

A tempting mistake is to subtract 2 from 15, giving 13, which is not an option, or to guess without computing the total deck size. The key is that the total number of cards stays the same.

ANSWER 2: C

---

Problem 3:
We are given the rule \(a\ c\ b\ d = a\cdot d - b\cdot c\). We need to evaluate \(3\ 1\ 4\ 2\).

Matching the pattern:
- \(a = 3\)
- \(c = 1\)
- \(b = 4\)
- \(d = 2\)

Apply the formula:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

A tempting error is to swap the pairs and compute \(3\cdot 1 - 4\cdot 2 = 3 - 8 = -5\), which is not among the choices, or to read the order incorrectly. Careful matching to \(a, c, b, d\) is essential.

ANSWER 3: E

---

Problem 4:
Isabella wants an average of 95 on four 100-point tests. Her first two scores are 97 and 91. We need the lowest possible score on the third test that still allows her to reach her goal.

Total points needed for a 95 average:
\[
4 \times 95 = 380.
\]

Points earned so far:
\[
97 + 91 = 188.
\]

Points still needed from the last two tests:
\[
380 - 188 = 192.
\]

Let the third test score be \(x\) and the fourth be \(y\), with \(x + y = 192\) and each score at most 100. To minimize \(x\), we maximize \(y\). The highest possible score is 100, so:
\[
x = 192 - 100 = 92.
\]

If she scores 92 on the third test, she can score 100 on the fourth and exactly reach her goal. A tempting wrong choice is 95, which would be the answer if she needed the two remaining tests to average 95, but she can shift points to the fourth test.

ANSWER 4: B

---

Problem 5:
We need to find how many apples Bridget originally bought.

Let the original number be \(x\).
- She gave half to Ann, leaving her with \(\frac{x}{2}\).
- She gave 3 to Cassie and kept 4:
\[
\frac{x}{2} - 3 = 4.
\]

Solve:
\[
\frac{x}{2} = 7 \quad\Rightarrow\quad x = 14.
\]

Check: Start with 14. Half is 7, given to Ann. She has 7 left. Gives 3 to Cassie, keeps 4. This matches.

A tempting error is to add \(3 + 4 = 7\) and stop, forgetting that 7 is only what remained after giving away half, so the original amount must be doubled.

ANSWER 5: E

---

Problem 6:
We need the total number of marbles given three conditions about how many are *not* each color.

Let \(R, G, B\) be the numbers of red, green, and blue marbles.
- “All but 6 are red” means non-red marbles equal 6: \(G + B = 6\).
- “All but 8 are green” means non-green marbles equal 8: \(R + B = 8\).
- “All but 4 are blue” means non-blue marbles equal 4: \(R + G = 4\).

Add all three equations:
\[
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
\]
\[
2R + 2G + 2B = 18
\]
\[
R + G + B = 9.
\]

So there are 9 marbles total. (Checking: \(R=3, G=1, B=5\), which satisfies all three statements.)

A tempting mistake is to add \(6+8+4=18\) and choose that, but that sum counts each marble twice.

ANSWER 6: C

---

Problem 7:
We need the sum of the two smallest prime factors of 250.

Factor 250:
\[
250 = 2 \times 125 = 2 \times 5^3.
\]

The prime factors are 2 and 5. The two smallest (and only) distinct prime factors are 2 and 5. Their sum is:
\[
2 + 5 = 7.
\]

A tempting wrong choice is 10, which comes from adding 5 and 5 (the two smallest factors with multiplicity), but the problem asks for the two smallest prime factors, i.e., distinct primes.

ANSWER 7: C

---

Problem 8:
Bag A has \(\{1, 3, 5\}\) and Bag B has \(\{2, 4, 6\}\). We draw one from each and want the number of distinct possible sums.

Compute all sums:
- \(1+2=3\), \(1+4=5\), \(1+6=7\)
- \(3+2=5\), \(3+4=7\), \(3+6=9\)
- \(5+2=7\), \(5+4=9\), \(5+6=11\)

The distinct values are:
\[
3, 5, 7, 9, 11.
\]

That is 5 different values. A tempting error is to count all 9 ordered pairs without removing duplicates.

ANSWER 8: B

---

Problem 9:
We have 2 boys and 3 girls to arrange in a row with a boy at each end and the three girls in the middle.

Positions: B _ _ _ G G G _ _ _ B (but specifically 5 seats: 1,2,3,4,5 with boys at 1 and 5, girls at 2,3,4).

- Choose the boy for the left end and the boy for the right end: \(2! = 2\) ways.
- Arrange the 3 girls in the middle three seats: \(3! = 6\) ways.

Total arrangements:
\[
2 \times 6 = 12.
\]

A tempting mistake is to ignore the constraint and compute \(5! = 120\), or to forget that the boys can swap places.

ANSWER 9: E

---

Problem 10:
Evaluate \(16 + 8 \div 4 - 2\).

Using order of operations (PEMDAS/BODMAS), division comes before addition and subtraction:
\[
8 \div 4 = 2.
\]

Now substitute back:
\[
16 + 2 - 2 = 16.
\]

A tempting error is to work left-to-right doing addition first: \((16+8)\div 4 - 2 = 24\div 4 - 2 = 6 - 2 = 4\), which is choice A. Another is to do \(16 + 2 = 18\), then \(18 - 2 = 16\), which luckily gives the same corr

Problem 1:
We need the rise in water level when a rock of volume 1000 cm³ is placed in an aquarium with base 100 cm by 40 cm. The rock displaces water equal to its volume, causing the water level to rise by Δh such that base area × Δh = rock volume. Base area = 100 cm × 40 cm = 4000 cm². So Δh = 1000 cm³ / 4000 cm² = 0.25 cm. The initial depth is 37 cm, and after rise it becomes 37.25 cm, which is still below the height of 50 cm, so the rock is fully submerged. The rise is 0.25 cm.

ANSWER 1: A

Problem 2:
When Annika and 3 friends play, there are 4 players, each dealt 15 cards. Total cards = 4 × 15 = 60. In the next game, 2 more friends join, making 6 players. Since the same total number of cards is used, each player gets 60 / 6 = 10 cards.

ANSWER 2: C

Problem 3:
The operation is defined as a c b d = a·d − b·c. For the numbers 3 1 4 2, we have a = 3, c = 1, b = 4, d = 2. Compute a·d − b·c = 3×2 − 4×1 = 6 − 4 = 2.

ANSWER 3: E

Problem 4:
Isabella wants an average of 95 over four tests, so total points needed = 4 × 95 = 380. Her first two scores sum to 97 + 91 = 188. Let the third score be x. After the third test, she can still reach the goal if it is possible to score at most 100 on the fourth test to reach 380. That requires 188 + x + 100 ≥ 380 → x ≥ 380 − 288 = 92. The smallest integer x satisfying this is 92. If x = 91, then even a perfect 100 on the fourth gives 188+91+100 = 379 < 380. So the lowest possible third score is 92.

ANSWER 4: B

Problem 5:
Let the number of apples Bridget bought be N. She gave half to Ann, leaving N/2. Then she gave Cassie 3 apples, leaving N/2 − 3. She keeps 4 apples, so N/2 − 3 = 4 → N/2 = 7 → N = 14.

ANSWER 5: E

Problem 6:
Let T be the total number of marbles. "All but 6 are red" means T − R = 6 → R = T − 6. Similarly, T − G = 8 → G = T − 8, and T − B = 4 → B = T − 4. Since R + G + B = T, we have (T−6)+(T−8)+(T−4) = T → 3T − 18 = T → 2T = 18 → T = 9.

ANSWER 6: C

Problem 7:
Factor 250: 250 = 2 × 5³. Its prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5, and their sum is 7.

ANSWER 7: C

Problem 8:
List all possible sums from drawing one chip from Bag A {1,3,5} and one from Bag B {2,4,6}:
1+2=3, 1+4=5, 1+6=7;
3+2=5, 3+4=7, 3+6=9;
5+2=7, 5+4=9, 5+6=11.
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.

ANSWER 8: B

Problem 9:
We need arrangements of 2 boys and 3 girls in a row with a boy at each end. Choose which boy sits on the left end and which on the right: 2! = 2 ways. The three girls occupy the three middle seats and can be arranged in 3! = 6 ways. Total arrangements = 2 × 6 = 12.

ANSWER 9: E

Problem 10:
The expression is 16 + 8 ÷ 4 − 2 (the missing operator is division, as typical for order-of-operations problems). Following order of operations, division before addition and subtraction: 8 ÷ 4 = 2. Then 16 + 2 = 18, and 18 − 2 = 16.

ANSWER 10: D

Problem 11:
Evaluate step by step: √4 = 2. Then 8 · 2 = 16, so √(8·√4) = √16 = 4. Then 16 · 4 = 64, so √(16·√(8·√4)) = √64 = 8.

ANSWER 11: C

Problem 12:
Compare the decimal numbers:
A. 0.99
B. 0.9099
C. 0.9
D. 0.909
E. 0.9009
Clearly 0.99 is the largest because it is greater than 0.9 and all the others are less than 0.91.

ANSWER 12: A

Problem 1:
The aquarium’s base has area  
\(A = 100\text{ cm} \times 40\text{ cm}=4000\text{ cm}^2\).  
When the rock of volume \(V=1000\text{ cm}^3\) is fully submerged, the water rises by a height \(h\) such that  

\[
A\cdot h = V \quad\Longrightarrow\quad h = \frac{1000}{4000}=0.25\text{ cm}.
\]

The rise cannot be any of the larger listed values (0.5 cm, 1 cm, …) because those would require a volume larger than the rock’s.  

ANSWER 1: A  

---

Problem 2:
With 4 players each receiving 15 cards, the total number of cards is  

\[
4 \times 15 = 60.
\]

If 2 more friends join, there are \(4+2 = 6\) players.  The cards are still all the cards, so each player receives  

\[
\frac{60}{6}=10\text{ cards}.
\]

The other answer choices (8, 9, 11, 12) would not distribute the 60 cards evenly.  

ANSWER 2: C  

---

Problem 3:
The operation is defined by  

\[
a\;c\;b\;d = a\cdot d - b\cdot c.
\]

For the numbers \(3\;1\;4\;2\) we have \(a=3,\;c=1,\;b=4,\;d=2\).  Hence  

\[
3\;1\;4\;2 = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

The other options \(-2,-1,0,1\) would correspond to different combinations that do not satisfy the given formula.  

ANSWER 3: E  

---

Problem 4:
To average 95 over four 100‑point tests, the total points needed are  

\[
4 \times 95 = 380.
\]

After the first two tests Isabella has  

\[
97 + 91 = 188\text{ points}.
\]

Let the third test score be \(x\).  Then the fourth test must supply  

\[
380 - (188 + x) = 192 - x \text{ points}.
\]

The fourth test can give at most 100 points, so we need  

\[
192 - x \le 100 \quad\Longrightarrow\quad x \ge 92.
\]

If she scores \(92\) on the third test, she would need a perfect 100 on the fourth, which is possible.  Any lower score would require more than 100 points on the fourth test, which is impossible.  

Thus the lowest possible third‑test score is **92**.  

ANSWER 4: B  

---

Problem 5:
Let the original number of apples be \(N\).  
She gave half of them to Ann: \(\frac{N}{2}\).  
She then gave Cassie 3 apples and kept 4, leaving  

\[
N - \frac{N}{2} - 3 = 4.
\]

Simplify: \(\frac{N}{2} - 3 = 4 \Rightarrow \frac{N}{2}=7 \Rightarrow N=14\).

If the total had been any of the other answer choices, the steps would not balance (e.g., 11 would leave 2 apples after giving Cassie 3).  

ANSWER 5: E  

---

Problem 6:
Let the total number of marbles be \(T\).  
- “All but 6 are red’’ → red = \(T-6\).  
- “All but 8 are green’’ → green = \(T-8\).  
- “All but 4 are blue’’ → blue = \(T-4\).

These three colors account for all marbles, so  

\[
(T-6)+(T-8)+(T-4)=T \;\Longrightarrow\; 3T-18=T \;\Longrightarrow\; 2T=18 \;\Longrightarrow\; T=9.
\]

The other totals (6, 8, 10, 12) would not satisfy the three “all but…’’ statements simultaneously.  

ANSWER 6: C  

---

Problem 7:
Factor \(250\):  

\[
250 = 2 \times 5^3.
\]

The prime factors are \(2\) and \(5\).  The two smallest are \(2\) and \(5\), whose sum is  

\[
2+5=7.
\]

Choices 2, 5, 10, 12 are either single prime factors or incorrect sums.  

ANSWER 7: C  

---

Problem 8:
Bag A contains \(\{1,3,5\}\); Bag B contains \(\{2,4,6\}\).  
All possible sums \(a+b\) are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The distinct results are \(\{3,5,7,9,11\}\), a total of **5** different sums.  

The other answer choices (4, 6, 7, 9) do not match the count of distinct sums.  

ANSWER 8: B  

---

Problem 9:
Place the two boys at the two ends and the three girls in the middle three seats.

- The two boys can be ordered in \(2!\) ways (left‑right swap).  
- The three girls can be arranged among the three middle seats in \(3! = 6\) ways.

Total arrangements  

\[
2! \times 3! = 2 \times 6 = 12.
\]

Choices 2, 4, 5, 6 are too small; 12 accounts for all possible permutations under the given constraints.  

ANSWER 9: E  

---

Problem 10:
Interpret the expression as  

\[
16 + 8 \div 4 - 2.
\]

Division comes before addition and subtraction, so  

\[
8 \div 4 = 2 \quad\Longrightarrow\quad 16 + 2 - 2 = 16.
\]

The other options (4, 8, 12, 20) would result only if the order of operations were misinterpreted.  

ANSWER 10: D  

---

Problem 11:
Simplify from the innermost radical outward.

\[
\sqrt{4}=2,\qquad 
\sqrt{8\cdot\sqrt{4}} = \sqrt{8\cdot2}= \sqrt{16}=4.
\]

Now  

\[
\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{16\cdot4}= \sqrt{64}=8.
\]

Thus the value is 8, matching choice C.  

ANSWER 11: C  

---

Problem 12:
Compare the decimal values:

- A. \(0.99\)
- B. \(0.9099\)
- C. \(0.9\)
- D. \(0.909\)
- E. \(0.9009\)

Clearly \(0.99\) is greater than \(0.9099\), which is greater than \(0.909\), etc.  
Hence the largest number is \(0.99\).  

ANSWER 12: A

Problem 1:  
An aquarium has a rectangular base of 100 cm by 40 cm and height 50 cm, filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is submerged. By how many centimeters does the water level rise?  

The base area is \(100 \times 40 = 4000 \, \text{cm}^2\). The rock’s volume is 1000 cm³, so the rise in water level is \(\frac{1000}{4000} = 0.25 \, \text{cm}\).  

Tempting choices: 0.5 (B) or 1 (C) might be guessed if base area is miscalculated.  

ANSWER 1: A  

Problem 2:  
Annika and 3 friends (4 players) get 15 cards each. Total cards = \(4 \times 15 = 60\). If 2 more join (6 players), each gets \(\frac{60}{6} = 10\).  

Tempting choices: 8 (A) if total cards miscalculated as 48.  

ANSWER 2: C  

Problem 3:  
Operation defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\): \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\), but the order is \(a c b d\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). However, the problem writes "a c b d" which is interpreted as \(a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). But the choices include 2 (E). Re-evaluating: the operation is \(a c b d = a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\).  

Tempting choices: -2 (A) if signs reversed.  

ANSWER 3: A (Note: Correction based on standard interpretation: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\), but the correct answer based on the operation defined is actually \(3 \cdot 2 - 1 \cdot 4 = 2\), but the problem writes "a c b d", which is \(a \cdot d - b \cdot c\), so 2. However, the answer is 2, but the choices include 2 as E. The correct calculation is 2, so E. Correction: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\).)  

**Correction:** The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\), this is \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\).  

ANSWER 3: E  

Problem 4:  
Isabella wants an average of 95 on four 100-point tests. Total points needed: \(4 \times 95 = 380\). First two scores: 97 + 91 = 188. Let third score be \(x\), fourth be \(y\). Then \(188 + x + y = 380\), so \(x + y = 192\). To minimize \(x\), maximize \(y\): \(y = 100\), so \(x = 92\).  

Tempting choices: 90 (A) if \(y\) not maximized.  

ANSWER 4: B  

Problem 5:  
Bridget gave half apples to Ann, then 3 to Cassie, keeping 4. Let total be \(x\). After giving half to Ann: \(\frac{x}{2}\) left. Then gives 3 to Cassie: \(\frac{x}{2} - 3 = 4\). So \(\frac{x}{2} = 7\), \(x = 14\).  

Tempting choices: 7 (C) if half misinterpreted.  

ANSWER 5: E  

Problem 6:  
Let \(r, g, b\) be red, green, blue marbles. Then:  
- All but 6 are red: \(g + b = 6\)  
- All but 8 are green: \(r + b = 8\)  
- All but 4 are blue: \(r + g = 4\)  
Adding: \(2(r + g + b) = 18\), so total \(r + g + b = 9\).  

Tempting choices: 6 (A) if equations misread.  

ANSWER 6: C  

Problem 7:  
Prime factors of 250: \(250 = 2 \times 5^3\). Two smallest: 2 and 5. Sum: \(2 + 5 = 7\).  

Tempting choices: 5 (B) if only 5 considered.  

ANSWER 7: C  

Problem 8:  
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}. Possible sums:  
- 1+2=3, 1+4=5, 1+6=7  
- 3+2=5, 3+4=7, 3+6=9  
- 5+2=7, 5+4=9, 5+6=11  
Unique sums: 3, 5, 7, 9, 11 → 5 values.  

Tempting choices: 6 (C) if duplicates counted.  

ANSWER 8: B  

Problem 9:  
Two boys at ends: 2 ways. Three girls in middle: \(3! = 6\) ways. Total: \(2 \times 6 = 12\).  

Tempting choices: 6 (D) if boys not permuted.  

ANSWER 9: E  

Problem 10:  
Expression: \(16 + 8 \times 4 - 2\). Order: \(8 \times 4 = 32\), then \(16 + 32 - 2 = 46\)? Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 \times 4 - 2\): \(8 \times 4 = 32\), \(16 + 32 = 48\), \(48 - 2 = 46\). But choices max at 20. Re-interpreting: "16 + 8 4 - 2" might mean \(16 + 8 - 4 - 2 = 18\), not in choices. Alternatively, if "8 4" is 8×4, but 46 not in choices. The expression is likely \(16 + 8 - 4 - 2 = 18\), but not in choices. Correction: The expression is "16 + 8 4 − 2", which might be \(16 + (8 \times 4) - 2 = 16 + 32 - 2 = 46\), but not in choices. Re-reading: It’s "16 + 8 4 − 2", which could be \(16 + 8 - 4 - 2 = 18\), but not in choices. The intended expression is likely \(16 + 8 \times 4 - 2 = 46\), but choices max at 20. Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 - 4 - 2 = 18\), but not in choices. Alternatively, if "8 4" is 84, but 16+84-2=98. Not matching. The expression is likely \( \frac{16 + 8}{4} - 2 = \frac{24}{4} - 2 = 6 - 2 = 4 \).  

Tempting choices: 8 (B) if order misapplied.  

ANSWER 10: A  

Problem 11:  
Expression: \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\).  
- \(\sqrt{4} = 2\)  
- \(\sqrt{8 \cdot 2} = \sqrt{16} = 4\)  
- \(\sqrt{16 \cdot 4} = \sqrt{64} = 8\)  

Tempting choices: 4 (A) if innermost not simplified.  

ANSWER 11: C  

Problem 12:  
Compare decimals:  
- A: 0.99  
- B: 0.9099  
- C: 0.9  
- D: 0.909  
- E: 0.9009  
Largest is 0.99.  

Tempting choices: 0.9099 (B) if decimals misaligned.  

ANSWER 12: A  

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

### Problem 1:
What is asked: An aquarium with a 100cm×40cm rectangular base is filled with water to 37cm depth. A 1000 cm³ rock is submerged; how much does the water level rise?
Step-by-step: When a submerged object displaces water, its volume equals the additional volume of water in the aquarium. The rise in water level Δh satisfies: Volume of rock = base area of aquarium × Δh.
Base area = 100×40 = 4000 cm².
Δh = 1000 / 4000 = 0.25 cm.
Tempting wrong answers: 0.5 (if base area is miscalculated as 2000 cm²) or 1 (wrong division).
ANSWER 1: A

---

### Problem 2:
What is asked: Annika plus 3 friends (4 total players) each get 15 cards. If 2 more join, how many cards does each player get?
Step-by-step: First calculate total number of cards: 4 players × 15 cards = 60 total cards. After 2 join, total players = 4+2=6. Cards per player = 60 / 6 = 10.
Tempting wrong answers: 9 (if you incorrectly count only 3 initial players, leading to 45 total cards) or 12 (wrong division).
ANSWER 2: C

---

### Problem 3:
What is asked: The custom notation a c b d = a·d − b·c. What is the value of 3 1 4 2?
Step-by-step: Match the values to the definition: a=3, c=1, b=4, d=2. Calculate: 3·2 - 4·1 = 6 - 4 = 2.
Tempting wrong answers: -2 (swap a and b incorrectly) or 0 (miscalculate the multiplication).
ANSWER 3: E

---

### Problem 4:
What is asked: Isabella needs a 95 average on 4 100-point tests. She scored 97 and 91 on the first two. What is the lowest possible third test score that lets her reach her goal?
Step-by-step: Total points needed for 95 average: 4×95 = 380 points. Sum of first two tests: 97+91=188. To minimize the third test score, maximize the fourth test (max 100 points). Total points left for tests 3 and 4: 380-188=192. Minimum third test = 192 - 100 = 92.
Tempting wrong answers: 96 (the average of the remaining two tests, not the minimum possible third score).
ANSWER 4: B

---

### Problem 5:
What is asked: Bridget gave half her apples to Ann, 3 to Cassie, and kept 4. How many apples did she buy originally?
Step-by-step: Let N = original number of apples. After giving half to Ann, she has N/2 left. Then N/2 - 3 = 4 → N/2=7 → N=14. Verify: 14 → 7 left after Ann, minus 3 for Cassie = 4 kept, which matches.
Tempting wrong answers: 7 (the number of apples Bridget had after giving half to Ann, not the original total).
ANSWER 5: E

---

### Problem 6:
What is asked: A jar has red, green, blue marbles. All but 6 are red, all but 8 are green, all but 4 are blue. How many total marbles?
Step-by-step: Let T=total marbles, R=red, G=green, B=blue. The conditions translate to: G+B=6, R+B=8, R+G=4. Add all equations: 2(R+G+B)=18 → R+G+B=T=9. Verify: 3 red, 1 green, 5 blue sum to 9, matching all conditions.
Tempting wrong answers: 18 (the sum of 6+8+4, not halved to get total marbles).
ANSWER 6: C

---

### Problem 7:
What is asked: What is the sum of the two smallest prime factors of 250?
Step-by-step: Prime factorization of 250: 250=2×5³. The prime factors are 2 and 5. Their sum is 2+5=7.
Tempting wrong answers: 10 (sum of 5+5, or wrong addition of 2+5).
ANSWER 7: C

---

### Problem 8:
What is asked: Bag A has chips 1,3,5; Bag B has 2,4,6. One chip drawn from each bag, how many distinct sum values are possible?
Step-by-step: List all sums: 1+2=3, 1+4=5,1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7,5+4=9,5+6=11. Distinct sums: 3,5,7,9,11 → 5 total.
Tempting wrong answers: 9 (total number of unordered pairs, not distinct sums).
ANSWER 8: B

---

### Problem 9:
What is asked: 2 boys and 3 girls line up for a photo, with a boy at each end and girls in the middle. How many arrangements are possible?
Step-by-step: Arrange the boys first: 2 boys can be ordered at the ends in 2! = 2 ways. The 3 middle girls can be ordered in 3! =6 ways. Total arrangements: 2×6=12.
Tempting wrong answers: 6 (only the number of ways to arrange the girls, ignoring the boys).
ANSWER 9: E

---

### Problem 10:
What is asked: Evaluate (16+8)/(4-2) (formatted as a fraction in the original problem).
Step-by-step: Calculate numerator:16+8=24, denominator:4-2=2. 24/2=12.
Tempting wrong answers:16 (if order of operations is misapplied to calculate 16 + 8/4 -2=16).
ANSWER 10: C

---

### Problem 11:
What is asked: Evaluate √(16 · √(8 · √4)).
Step-by-step: Calculate from the innermost root first: √4=2. Next: 8×2=16, √16=4. Next:16×4=64, √64=8.
Tempting wrong answers: 4√2 (incorrect root calculation).
ANSWER 11: C

---

### Problem 12:
What is asked: Which number is largest: 0.99, 0.9099, 0.9, 0.909, 0.9009.
Step-by-step: Line up all decimals to compare place values: 0.9900 > 0.9099 > 0.9090 > 0.9009 > 0.9000. The largest is 0.99.
Tempting wrong answers: 0.9099 (the second largest value).
ANSWER 12: A

### Problem 1:
We need to find how much the water level rises when a 1000 cm³ rock is submerged in the aquarium.
1.  The volume of displaced water equals the volume of the submerged rock: 1000 cm³.
2.  The base area of the aquarium is \(100 \, \text{cm} \times 40 \, \text{cm} = 4000 \, \text{cm}^2\).
3.  Let \(h\) be the rise in water level. The volume of the displaced water is also equal to base area × rise in height: \(4000h = 1000\).
4.  Solve for \(h\): \(h = \frac{1000}{4000} = 0.25\) cm.
ANSWER 1: A
---
### Problem 2:
We need to find the number of cards per player when 2 more friends join.
1.  Original number of players: Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).
2.  New number of players: 4 + 2 = 6 players. Total cards remain 60.
3.  Cards per player: \(\frac{60}{6} = 10\).
ANSWER 2: C
---
### Problem 3:
The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). Compute \(3 \, 1 \, 4 \, 2\).
1.  Identify \(a=3, c=1, b=4, d=2\).
2.  Substitute into the formula: \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).
ANSWER 3: E
---
### Problem 4:
Find the lowest possible third test score to get an average of 95 on 4 tests.
1.  Total points needed for an average of 95: \(4 \times 95 = 380\).
2.  Sum of first two test scores: \(97 + 91 = 188\).
3.  Let \(x\) = third test score, \(y\) = fourth test score (max 100, since tests are 100-point). We need \(188 + x + y = 380\), so \(x + y = 192\).
4.  To minimize \(x\), maximize \(y = 100\). Then \(x = 192 - 100 = 92\).
ANSWER 4: B
---
### Problem 5:
Find the original number of apples Bridget bought.
1.  Let \(x\) = total apples. After giving half to Ann, she has \(\frac{x}{2}\) left.
2.  After giving 3 to Cassie, she has \(\frac{x}{2} - 3\) left, which equals 4 (the apples she kept).
3.  Solve: \(\frac{x}{2} - 3 = 4 \implies \frac{x}{2} =7 \implies x=14\).
ANSWER 5: E
---
### Problem 6:
Find the total number of marbles.
1.  Let \(R, G, B\) be the number of red, green, blue marbles, total \(T = R+G+B\).
2.  Translate the statements:
    - All but 6 are red: \(G + B = 6\)
    - All but 8 are green: \(R + B = 8\)
    - All but 4 are blue: \(R + G =4\)
3.  Add all three equations: \(2(R+G+B) = 6+8+4=18 \implies 2T=18 \implies T=9\).
ANSWER 6: C
---
### Problem 7:
Find the sum of the two smallest prime factors of 250.
1.  Factorize 250: \(250 = 2 \times 125 = 2 \times 5^3\).
2.  The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.
3.  Sum: \(2 +5=7\).
ANSWER 7: C
---
### Problem 8:
Find the number of distinct sums when drawing one chip from each bag.
1.  List all possible pairs and sums:
    - Bag A (1,3,5) + Bag B (2,4,6):
      \(1+2=3, 1+4=5, 1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7, 5+4=9, 5+6=11\)
2.  Distinct sums: 3, 5, 7, 9, 11. There are 5 unique values.
ANSWER 8: B
---
### Problem 9:
Find the number of arrangements with boys at each end and girls in the middle.
1.  Arrange the 2 boys at the ends: there are \(2! = 2\) ways (order matters: which boy is left/right).
2.  Arrange the 3 girls in the middle 3 seats: \(3! = 6\) ways.
3.  Total arrangements: \(2 \times 6 =12\).
ANSWER 9: E
---
### Problem 10:
Compute \(16 + 8 \, 4 - 2\). The missing operator between 8 and 4 is division (standard order of operations, common middle school problem):
1.  Interpret as \(16 + \frac{8}{4} - 2\).
2.  Compute division first: \(\frac{8}{4}=2\).
3.  Then add/subtract left to right: \(16 + 2 - 2 =16\).
*(Note: If interpreted as \(16 +8 -4 \times 2\), order of operations gives \(4 \times 2=8\), so \(16+8-8=16\), same result.)*
ANSWER 10: D
---
### Problem 11:
Compute \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\) by working from the innermost root outward.
1.  Innermost: \(\sqrt{4} = 2\).
2.  Next layer: \(\sqrt{8 \cdot 2} = \sqrt{16} =4\).
3.  Outermost: \(\sqrt{16 \cdot 4} = \sqrt{64} =8\).
ANSWER 11: C
---
### Problem 12:
Find the largest decimal.
1.  Rewrite all numbers with 4 decimal places for easy comparison:
    - A: \(0.9900\)
    - B: \(0.9099\)
    - C: \(0.9000\)
    - D: \(0.9090\)
    - E: \(0.9009\)
2.  Compare digit by digit: all have 9 in the tenths place. A has 9 in the hundredths place, while all others have 0. Thus A is the largest.
ANSWER 12: A

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

**Problem 1:**

What is being asked: By how many centimeters does the water level rise when a 1000 cm³ rock is submerged?

Setup: The base area is 100 cm × 40 cm = 4000 cm². When the rock is submerged, it displaces 1000 cm³ of water.

Water level rise = Volume displaced ÷ Base area = 1000 ÷ 4000 = 0.25 cm

ANSWER 1: A

---

**Problem 2:**

What is being asked: If cards are distributed evenly, how many cards per player when the player count increases?

Setup: Initially 4 players (Annika + 3 friends) each get 15 cards.
Total cards = 4 × 15 = 60 cards

With 2 more friends joining: 4 + 2 = 6 players
Cards per player = 60 ÷ 6 = 10 cards

ANSWER 2: C

---

**Problem 3:**

What is being asked: Evaluate the operation 3 1 4 2 using the rule a c b d = a·d − b·c

Setup: With a = 3, c = 1, b = 4, d = 2:
3 1 4 2 = 3·2 − 4·1 = 6 − 4 = 2

ANSWER 3: E

---

**Problem 4:**

What is being asked: What is the lowest score on test 3 that allows Isabella to achieve a 95 average on four tests?

Setup: Goal average is 95, so total needed = 95 × 4 = 380 points
First two tests: 97 + 91 = 188 points
Points needed from tests 3 and 4: 380 − 188 = 192 points

Let x = score on test 3. She can still reach her goal if she can score 100 on test 4:
x + 100 ≥ 192
x ≥ 92

The lowest possible score on test 3 is 92.

ANSWER 4: B

---

**Problem 5:**

What is being asked: How many apples did Bridget buy originally?

Setup: Let x = original number of apples
- She gave half to Ann: x/2 apples given away
- Remaining: x/2
- She gave 3 to Cassie: x/2 − 3
- She kept 4 for herself: x/2 − 3 = 4

Solving: x/2 = 7, so x = 14

ANSWER 5: E

---

**Problem 6:**

What is being asked: How many marbles are in the jar?

Setup: Let R, G, B = numbers of red, green, blue marbles; T = total
- "All but 6 are red": G + B = 6
- "All but 8 are green": R + B = 8
- "All but 4 are blue": R + G = 4

Adding all three equations: 2(R + G + B) = 6 + 8 + 4 = 18
Therefore: R + G + B = 9

ANSWER 6: C

---

**Problem 7:**

What is being asked: What is the sum of the two smallest prime factors of 250?

Setup: Factor 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5³

Prime factors: 2 and 5
Sum = 2 + 5 = 7

ANSWER 7: C

---

**Problem 8:**

What is being asked: How many different sums are possible?

Setup: Bag A: {1, 3, 5}, Bag B: {2, 4, 6}
All possible sums:
- 1 + 2 = 3
- 1 + 4 = 5
- 1 + 6 = 7
- 3 + 2 = 5 (duplicate)
- 3 + 4 = 7 (duplicate)
- 3 + 6 = 9
- 5 + 2 = 7 (duplicate)
- 5 + 4 = 9 (duplicate)
- 5 + 6 = 11

Distinct sums: {3, 5, 7, 9, 11} = 5 different values

ANSWER 8: B

---

**Problem 9:**

What is being asked: How many seating arrangements with boys at ends and girls in middle?

Setup: 2 boys at the ends can be arranged in 2! = 2 ways
3 girls in the middle can be arranged in 3! = 6 ways
Total arrangements = 2 × 6 = 12

ANSWER 9: E

---

**Problem 10:**

What is being asked: Evaluate 16 + 8 ÷ 4 − 2

Setup: Using order of operations (division before addition/subtraction):
16 + (8 ÷ 4) − 2 = 16 + 2 − 2 = 16

ANSWER 10: D

---

**Problem 11:**

What is being asked: Evaluate √(16 · √(8 · √4))

Setup: Work from the inside out:
- √4 = 2
- 8 · √4 = 8 · 2 = 16
- √(8 · √4) = √16 = 4
- 16 · √(8 · √4) = 16 · 4 = 64
- √(16 · √(8 · √4)) = √64 = 8

ANSWER 11: C

---

**Problem 12:**

What is being asked: Which decimal is largest?

Setup: Compare decimals:
- A: 0.99
- B: 0.9099
- C: 0.9
- D: 0.909
- E: 0.9009

Comparing: 0.99 > 0.9099 > 0.909 > 0.9009 > 0.9

The largest is 0.99

ANSWER 12: A

Problem 1:

The aquarium has a rectangular base of \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is

\[
100 \cdot 40 = 4000 \text{ cm}^2.
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

If the water rises by \(h\) centimeters, then

\[
4000h = 1000.
\]

So

\[
h = \frac{1000}{4000} = \frac14 = 0.25.
\]

So the water level rises \(0.25\) cm.

ANSWER 1: A

---

Problem 2:

Annika and 3 friends means there are

\[
4 \text{ players}.
\]

If each player gets 15 cards, then the total number of cards is

\[
4 \cdot 15 = 60.
\]

Now 2 more friends join, so there are

\[
6 \text{ players}.
\]

The same 60 cards are shared evenly, so each gets

\[
60 \div 6 = 10.
\]

ANSWER 2: C

---

Problem 3:

We are told that if the four numbers are \(a, b, c, d\), then

\[
a \ b \ c \ d = a\cdot d - b\cdot c.
\]

So for \(3,1,4,2\),

\[
3 \ 1 \ 4 \ 2 = 3\cdot 2 - 1\cdot 4 = 6 - 4 = 2.
\]

ANSWER 3: E

---

Problem 4:

She wants an average of 95 on 4 tests, so the total of the 4 scores must be

\[
4 \cdot 95 = 380.
\]

Her first two scores are 97 and 91, which add to

\[
97+91=188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To make the total 380, the fourth score would have to be

\[
380 - (188+x) = 192 - x.
\]

For this to be possible, the fourth score must be at most 100:

\[
192 - x \le 100.
\]

So

\[
x \ge 92.
\]

The lowest possible third score is 92.

ANSWER 4: B

---

Problem 5:

Let the number of apples Bridget bought be \(x\).

She gave half to Ann, so she kept the other half:

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

Then she gave Cassie 3 apples and kept 4 apples, so before giving Cassie the apples she had

\[
3+4=7.
\]

Thus,

\[
\frac{x}{2}=7 \quad \Rightarrow \quad x=14.
\]

ANSWER 5: E

---

Problem 6:

Let the total number of marbles be \(n\).

- “All but 6 are red” means the number of red marbles is \(n-6\).
- “All but 8 are green” means the number of green marbles is \(n-8\).
- “All but 4 are blue” means the number of blue marbles is \(n-4\).

Since every marble is one of these colors, their counts add to \(n\):

\[
(n-6)+(n-8)+(n-4)=n.
\]

Simplify:

\[
3n-18=n
\]

\[
2n=18
\]

\[
n=9.
\]

ANSWER 6: C

---

Problem 7:

We need the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3.
\]

So the prime factors are 2 and 5. The two smallest prime factors are 2 and 5, and their sum is

\[
2+5=7.
\]

ANSWER 7: C

---

Problem 8:

Bag A has \(1,3,5\); Bag B has \(2,4,6\).

Possible sums:

- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums are

\[
3,5,7,9,11
\]

which is 5 different values.

ANSWER 8: B

---

Problem 9:

There are 2 boys and 3 girls, and the picture must have:

- a boy at each end,
- the 3 girls in the middle.

First, arrange the 2 boys at the ends. They can be switched in

\[
2! = 2
\]

ways.

Then arrange the 3 girls in the middle in

\[
3! = 6
\]

ways.

So the total number of arrangements is

\[
2 \cdot 6 = 12.
\]

ANSWER 9: E

---

Problem 10:

The expression is

\[
16 + 8\cdot 4 - 2.
\]

Use order of operations:

\[
8\cdot 4=32.
\]

Then

\[
16+32-2=46.
\]

But 46 is not among the choices, so the intended expression must be read as

\[
16 + 8 \div 4 - 2.
\]

Then

\[
8\div 4 = 2,
\]

so

\[
16+2-2=16.
\]

ANSWER 10: D

---

Problem 11:

We evaluate from the inside out:

\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt4}}.
\]

First,

\[
\sqrt4=2.
\]

So the expression becomes

\[
\sqrt{16\cdot \sqrt{8\cdot 2}}
=
\sqrt{16\cdot \sqrt{16}}.
\]

Since \(\sqrt{16}=4\),

\[
\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C

---

Problem 12:

Compare the decimals:

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

The largest is the one closest to 1. Since \(0.99\) is greater than any number beginning with \(0.90\), it is the largest.

ANSWER 12: A

Problem 1:  
We need the rise in water level when a rock is placed in the aquarium and submerged.

- Base dimensions: \(100\text{ cm} \times 40\text{ cm}\), so base area  
  \[
  A = 100\cdot 40 = 4000\text{ cm}^2.
  \]
- The rock is completely submerged, so the water volume displaced equals the rock’s volume:  
  \[
  V = 1000\text{ cm}^3.
  \]
- Water level rise \(\Delta h\) satisfies \(A\Delta h = V\):  
  \[
  4000\cdot \Delta h = 1000 \implies \Delta h = \frac{1000}{4000}=0.25\text{ cm}.
  \]

So the rise is \(0.25\) cm.  
Tempting wrong idea: using the *remaining* depth \(37\) cm or checking if it overflows—none of that matters; only displaced volume matters.

ANSWER 1: A  

---

Problem 2:  
All cards are dealt evenly. Let the total number of cards be \(T\).

Initially, Annika + 3 friends = 4 players, and each gets 15 cards:  
\[
T = 4\cdot 15 = 60.
\]
Next game: 2 more friends join, so total players:
\[
4+2=6.
\]
Each player gets:
\[
\frac{60}{6}=10.
\]

ANSWER 2: C  

---

Problem 3:  
Given the rule \(a\,b\,d\,c = a\cdot d - b\cdot c\).  
The expression “\(3\ 1\ 4\ 2\)” corresponds to:
- \(a=3\)
- \(b=1\)
- \(d=4\)
- \(c=2\)

Compute:
\[
a\cdot d - b\cdot c = 3\cdot 4 - 1\cdot 2 = 12-2=10.
\]

But notice the choices are only \(-2,-1,0,1,2\), so we must interpret the placeholder correctly:  
The problem states: **“If \(a b c d = a\cdot d - b\cdot c\)”**.

So “\(3\ 1\ 4\ 2\)” means:
- \(a=3,\ b=1,\ c=4,\ d=2\)

Then:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 1\cdot 4 = 6-4=2.
\]

ANSWER 3: E  

---

Problem 4:  
Goal average is 95 over 4 tests, so total points needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
Let the third test score be \(x\), and the fourth be \(y\). Then:
\[
188 + x + y = 380 \implies x+y=192.
\]
We want the **lowest possible** \(x\) such that she can still reach the goal. That means make \(y\) as large as possible (since grades are at most 100).
Maximum \(y=100\). Then:
\[
x + 100 = 192 \implies x = 92.
\]

Check: then the four scores are \(97,91,92,100\) which sum to \(380\), average \(95\). Works.

ANSWER 4: B  

---

Problem 5:  
Let the total number of apples be \(N\).

- She gave half to Ann, so Ann got \(N/2\), leaving \(N/2\).
- Then she gave Cassie 3 apples, and kept 4 apples. That means after giving Cassie 3, her remaining is 4:
  \[
  \frac{N}{2} - 3 = 4 \implies \frac{N}{2}=7 \implies N=14.
  \]

ANSWER 5: E  

---

Problem 6:  
“All but 6 are red” means: number of red marbles is 6.  
Similarly:
- Red = 6 (since all but 6 are red → 6 red)
- Green: all but 8 are green → 8 green
- Blue: all but 4 are blue → 4 blue

Total marbles:
\[
6+8+4=18.
\]
But none of the choices are 18, so interpret carefully:  
“All but 6 are red” means the number of red marbles is **(total − 6)**, not 6.

Let total be \(T\). Then:
- Red marbles = \(T-6\)
- Green marbles = \(T-8\)
- Blue marbles = \(T-4\)

Sum:
\[
(T-6)+(T-8)+(T-4)=T.
\]
\[
3T - (6+8+4)=T
\]
\[
3T-18=T \implies 2T=18 \implies T=9.
\]

ANSWER 6: C  

---

Problem 7:  
We want the sum of the **two smallest prime factors** of 250.

Factor \(250\):
\[
250 = 25\cdot 10 = 5^2 \cdot 2\cdot 5 = 2\cdot 5^3.
\]
Prime factors are \(2\) and \(5\). The two smallest are \(2\) and \(5\).

Sum:
\[
2+5=7.
\]

ANSWER 7: D  

---

Problem 8:  
Bag A: \(\{1,3,5\}\)  
Bag B: \(\{2,4,6\}\)

Possible sums:
- With 1: \(1+2=3,\ 1+4=5,\ 1+6=7\)
- With 3: \(3+2=5,\ 3+4=7,\ 3+6=9\)
- With 5: \(5+2=7,\ 5+4=9,\ 5+6=11\)

Distinct sums: \(\{3,5,7,9,11\}\) → that’s 5 values.

ANSWER 8: B  

---

Problem 9:  
We need arrangements of 5 people in a row with:
- a boy at each end (positions 1 and 5)
- the three girls in the middle (positions 2,3,4)

There are 2 boys: they can be placed at the ends in
\[
2!\ = 2
\]
ways (left boy/right boy).

The 3 girls can be arranged in the three middle spots in
\[
3! = 6
\]
ways.

Total arrangements:
\[
2\cdot 6 = 12.
\]

ANSWER 9: E  

---

Problem 10:  
Expression: \(16 + 8^4 - 2\).

Compute in order:
- \(8^4 = 4096\)

So:
\[
16 + 4096 - 2 = 4110.
\]
No answer choice matches, so we interpret the formatting as \(16 + 8 \sqrt{4} - 2\) or \(16 + 8^4 - 2\)?  
Given the AMC/JH style, likely it is:
\[
16 + 8\sqrt{4} - 2.
\]
Then:
\[
\sqrt{4}=2 \implies 8\sqrt{4}=8\cdot 2=16.
\]
So:
\[
16 + 16 - 2 = 30,
\]
still not in choices.

Another common formatting is:
\[
16 + 8^{4} - 2 \quad \text{(still not)}
\]

The only choiceable consistent possibility is:
\[
16 + \sqrt{8}\,^4 - 2 \text{ (nonsense)}
\]

Let’s use the most plausible literal from the text: **“16 + 8 4 − 2”** often means \(16 + 8^4 - 2\) if the caret was lost. But then no choice.

Given typical AMC problems, the intended is usually:
\[
16 + 8 - 4 - 2 = 18
\]
also not.

Because the answer choices are only \(4,8,12,16,20\), the intended expression must simplify to one of those. The most natural is:
\[
16 + 8/4 - 2 = 16 + 2 - 2 = 16.
\]
This matches a common missing symbol: “8 4” often means \(8/4\).

So:
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D  

---

Problem 11:  
Evaluate:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}.
\]

Work from inside out:

1) \(\sqrt{4}=2\)

2) Inside next root:
\[
8\cdot \sqrt{4} = 8\cdot 2 = 16
\]
So:
\[
\sqrt{8\cdot \sqrt{4}} = \sqrt{16}=4.
\]

3) Now plug into the outer expression:
\[
\sqrt{16\cdot \sqrt{8\cdot \sqrt{4}}}=\sqrt{16\cdot 4}=\sqrt{64}=8.
\]

ANSWER 11: C  

---

Problem 12:  
Largest number among:
A. .99  
B. .9099  
C. .9  
D. .909  
E. .9009  

Compare by decimal places:

- .99 is 0.99
- .9099 is 0.9099
- .9 is 0.9000
- .909 is 0.9090
- .9009 is 0.9009

Clearly \(0.99\) is greater than all the numbers less than \(0.91\).

ANSWER 12: A

Problem 1:
The volume of the water displaced by the rock is equal to the volume of the rock itself, which is $1000 \text{ cm}^3$. The aquarium has a rectangular base with area $A = 100 \text{ cm} \times 40 \text{ cm} = 4000 \text{ cm}^2$. When the rock is submerged, the water level rises by height $h$ such that the volume of the displaced water is $A \times h$. 
Setting this equal to the rock's volume: $4000 \times h = 1000$.
Solving for $h$: $h = \frac{1000}{4000} = \frac{1}{4} = 0.25 \text{ cm}$.
ANSWER 1: A

Problem 2:
Annika and 3 friends make 4 players total. Each received 15 cards, so the total number of cards is $4 \times 15 = 60$. If 2 more friends join, the total number of players becomes $4 + 2 = 6$. The cards are distributed evenly among the 6 players, so each player gets $60 \div 6 = 10$ cards.
ANSWER 2: C

Problem 3:
The problem defines the operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$.
For the expression $\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix}$, we have $a=3, b=1, c=4, d=2$.
Calculating: $(3 \times 2) - (1 \times 4) = 6 - 4 = 2$.
ANSWER 3: E

Problem 4:
Isabella needs an average of 95 over 4 tests, meaning the total sum of her scores must be $95 \times 4 = 380$. Her first two scores are 97 and 91, totaling $97 + 91 = 188$. Let $x$ be the score on the third test and $y$ be the score on the fourth. We have $188 + x + y = 380$, so $x + y = 192$. To make $x$ as small as possible, we must make $y$ as large as possible. Since the maximum score on a test is 100, we set $y = 100$. Then $x + 100 = 192$, so $x = 92$.
ANSWER 4: B

Problem 5:
Let $x$ be the total number of apples Bridget bought. She gave half away, leaving $\frac{x}{2}$. Then she gave 3 away, leaving $\frac{x}{2} - 3$. She kept 4, so $\frac{x}{2} - 3 = 4$. Adding 3 to both sides, $\frac{x}{2} = 7$. Multiplying by 2, $x = 14$.
ANSWER 5: E

Problem 6:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R + G + B$.
"All but 6 are red" means $G + B = 6$.
"All but 8 are green" means $R + B = 8$.
"All but 4 are blue" means $R + G = 4$.
Adding these three equations: $2(R + G + B) = 6 + 8 + 4 = 18$.
So $R + G + B = 18 \div 2 = 9$.
ANSWER 6: C

Problem 7:
The prime factorization of 250 is $250 = 2 \times 125 = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two prime factors is $2 + 5 = 7$.
ANSWER 7: C

Problem 8:
Bag A has $\{1, 3, 5\}$ and Bag B has $\{2, 4, 6\}$.
Possible sums:
$1+2=3, 1+4=5, 1+6=7$
$3+2=5, 3+4=7, 3+6=9$
$5+2=7, 5+4=9, 5+6=11$
The set of distinct sums is $\{3, 5, 7, 9, 11\}$. There are 5 distinct values.
ANSWER 8: B

Problem 9:
There are 2 boys (B1, B2) and 3 girls (G1, G2, G3). A boy must be at each end. There are 2 choices for the left end and 1 choice for the right end ($2 \times 1 = 2$ ways to place boys). The 3 girls can be arranged in the middle in $3! = 3 \times 2 \times 1 = 6$ ways. Total arrangements = $2 \times 6 = 12$.
ANSWER 9: E

Problem 10:
Following the order of operations (PEMDAS/BODMAS):
$16 + (8 \div 4) - 2 = 16 + 2 - 2 = 16$.
ANSWER 10: D

Problem 11:
Evaluate from the inside out:
$\sqrt{4} = 2$
$\sqrt{8 \times 2} = \sqrt{16} = 4$
$\sqrt{16 \times 4} = \sqrt{64} = 8$
ANSWER 11: C

Problem 12:
To compare, pad with zeros to match the number of decimal places:
A: 0.9900
B: 0.9099
C: 0.9000
D: 0.9090
E: 0.9009
Comparing the values, 0.9900 is the largest.
ANSWER 12: A

Problem 1:
The base of the aquarium measures 100 cm by 40 cm, so its area is \(100 \times 40 = 4000\) cm². The rock has volume 1000 cm³ and is fully submerged, displacing exactly 1000 cm³ of water. The resulting rise in water level is therefore the displaced volume divided by the base area: \(1000 / 4000 = 0.25\) cm. (The original water depth and tank height are irrelevant once the surface area is known.)

ANSWER 1: A

Problem 2:
Annika plus 3 friends gives 4 players, each receiving 15 cards, for a total of \(4 \times 15 = 60\) cards. When 2 more players join, there are 6 players sharing the same 60 cards, so each receives \(60 / 6 = 10\) cards.

ANSWER 2: C

Problem 3:
The given operation evaluates \(a\, c\, b\, d\) as \(a \cdot d - b \cdot c\). Substituting the numbers yields \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).

ANSWER 3: E

Problem 4:
An average of 95 on four 100-point tests requires a total of \(4 \times 95 = 380\) points. The first two scores sum to \(97 + 91 = 188\), leaving 192 points still needed from the last two tests. The fourth test can score at most 100, so the third test must be at least \(192 - 100 = 92\). Any lower score on the third test would make a total of 380 impossible.

ANSWER 4: B

Problem 5:
Let \(t\) be the original number of apples. After giving half to Ann, \(t/2\) remain. After giving 3 more to Cassie, \(t/2 - 3\) remain, and this equals the 4 apples kept. Solving \(t/2 - 3 = 4\) gives \(t/2 = 7\), so \(t = 14\).

ANSWER 5: E

Problem 6:
Let \(m\) be the total number of marbles. Then there are \(m-6\) red, \(m-8\) green, and \(m-4\) blue. Their sum must equal \(m\):
\[
(m-6) + (m-8) + (m-4) = m \implies 3m - 18 = m \implies 2m = 18 \implies m = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies the three “all-but” statements.

ANSWER 6: C

Problem 7:
Factorize \(250 = 2 \cdot 5^3\). The two smallest distinct prime factors are 2 and 5; their sum is 7.

ANSWER 7: C

Problem 8:
All nine possible sums are:
\[
1+2=3,\ 1+4=5,\ 1+6=7,\ 3+2=5,\ 3+4=7,\ 3+6=9,\ 5+2=7,\ 5+4=9,\ 5+6=11.
\]
The distinct values are 3, 5, 7, 9, 11—exactly five.

ANSWER 8: B

Problem 9:
The seating pattern must be boy–girl–girl–girl–boy. The two distinct boys may be placed at the ends in \(2!\) = 2 ways. The three distinct girls may be arranged in the middle three seats in \(3!\) = 6 ways. The total number of arrangements is therefore \(2 \times 6 = 12\).

ANSWER 9: E

Problem 10:
Interpreting the expression with standard order of operations and treating juxtaposition of 8 and 4 as division yields
\[
16 + \frac{8}{4} - 2 = 16 + 2 - 2 = 16.
\]

ANSWER 10: D

Problem 11:
Evaluate from the inside outward:
\[
\sqrt{4} = 2, \quad 8 \cdot 2 = 16, \quad \sqrt{16} = 4, \quad 16 \cdot 4 = 64, \quad \sqrt{64} = 8.
\]

ANSWER 11: C

Problem 12:
Compare the decimals directly:
\[
0.99 > 0.9 > 0.9099 > 0.909 > 0.9009.
\]
Thus 0.99 is the largest.

ANSWER 12: A

Problem 1:
The problem asks us to determine by how many centimeters the water level rises when a rock with a volume of 1000 cm³ is submerged in an aquarium filled with water to a depth of 37 cm. The aquarium has a rectangular base measuring 100 cm by 40 cm and a height of 50 cm.

First, we need to calculate the initial volume of water in the aquarium. The volume of water is given by the formula length × width × height. So, the initial volume of water = 100 cm × 40 cm × 37 cm = 148,000 cm³.

When the rock is submerged, it displaces a volume of water equal to its own volume, which is 1000 cm³. This means the water level will rise to accommodate the volume of the rock.

The rise in water level can be calculated by dividing the volume of the rock by the base area of the aquarium. The base area = 100 cm × 40 cm = 4000 cm².

Rise in water level = Volume of rock / Base area = 1000 cm³ / 4000 cm² = 0.25 cm.

Let's examine the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

The calculated rise in water level matches option A.

ANSWER 1: A

Problem 2:
The problem states that in the card game Buffalo Shuffle-o, all cards are distributed evenly among players. Initially, Annika and 3 friends play, and each is dealt 15 cards. We need to find out how many cards each player will be dealt if 2 more friends join the next game.

First, we determine the total number of cards. With 4 players initially, each getting 15 cards, the total number of cards = 4 × 15 = 60.

When 2 more friends join, the total number of players becomes 4 + 2 = 6.

To find out how many cards each player will get, we divide the total number of cards by the new number of players: 60 / 6 = 10.

Let's examine the choices:
A. 8
B. 9
C. 10
D. 11
E. 12

The calculated number of cards per player matches option C.

ANSWER 2: C

Problem 3:
The problem defines a operation as a c b d = a·d − b·c and asks us to find the value of 3 1 4 2.

Using the given definition, we substitute a = 3, b = 1, c = 4, and d = 2 into the equation: 3 1 4 2 = 3·2 − 1·4 = 6 - 4 = 2.

Let's examine the choices:
A. −2
B. −1
C. 0
D. 1
E. 2

The calculated value matches option E.

ANSWER 3: E

Problem 4:
Isabella needs to achieve an average grade of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score she could have made on the third test to still reach her goal.

The total score required for an average of 95 on 4 tests = 95 × 4 = 380.

The sum of her first two test scores = 97 + 91 = 188.

So, the total score needed on the last two tests = 380 - 188 = 192.

To minimize the score on the third test, we maximize the score on the fourth test. The maximum score on the fourth test is 100.

Thus, the minimum score on the third test = 192 - 100 = 92.

Let's examine the choices:
A. 90
B. 92
C. 95
D. 96
E. 97

The calculated minimum score matches option B.

ANSWER 4: B

Problem 5:
Bridget gave half of her apples to Ann, then gave 3 to Cassie, and kept 4 for herself. We need to find out how many apples she originally bought.

Let's denote the original number of apples as x.

After giving half to Ann, Bridget had x/2 apples left.

Then, she gave 3 apples to Cassie and kept 4, so x/2 = 3 + 4 = 7.

Thus, x = 7 × 2 = 14.

Let's examine the choices:
A. 3
B. 4
C. 7
D. 11
E. 14

The calculated original number of apples matches option E.

ANSWER 5: E

Problem 6:
The problem states that in a jar, all but 6 marbles are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T, the number of red marbles as R, the number of green marbles as G, and the number of blue marbles as B.

From the given information:
1. T - R = 6 (all but 6 are red)
2. T - G = 8 (all but 8 are green)
3. T - B = 4 (all but 4 are blue)

Adding these equations gives: 3T - (R + G + B) = 18.

Since R + G + B = T (the total number of marbles is the sum of the marbles of each color), we have:
3T - T = 18, which simplifies to 2T = 18.

Thus, T = 18 / 2 = 9.

Let's examine the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

The calculated total number of marbles matches option C.

ANSWER 6: C

Problem 7:
We need to find the sum of the two smallest prime factors of 250.

First, let's factorize 250 into its prime factors: 250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5.

The sum of these two smallest prime factors = 2 + 5 = 7.

Let's examine the choices:
A. 2
B. 5
C. 7
D. 10
E. 12

The calculated sum matches option C.

ANSWER 7: C

Problem 8:
Bag A contains chips labeled 1, 3, and 5. Bag B contains chips labeled 2, 4, and 6. We need to find how many different values are possible for the sum of the numbers on the chips drawn from each bag.

Let's list the possible sums:
- (1,2) = 3
- (1,4) = 5
- (1,6) = 7
- (3,2) = 5
- (3,4) = 7
- (3,6) = 9
- (5,2) = 7
- (5,4) = 9
- (5,6) = 11

The possible sums are 3, 5, 7, 9, and 11. There are 5 different values.

Let's examine the choices:
A. 4
B. 5
C. 6
D. 7
E. 9

The number of different sums matches option B.

ANSWER 8: B

Problem 9:
The chess team consists of 2 boys and 3 girls. They need to be arranged with a boy at each end and the girls in the middle. We need to find out how many such arrangements are possible.

The boys can be arranged at the ends in 2! = 2 ways (either boy can be on the left or the right).

The 3 girls can be arranged in the 3 middle positions in 3! = 6 ways.

The total number of arrangements = (number of ways to arrange boys) × (number of ways to arrange girls) = 2 × 6 = 12.

Let's examine the choices:
A. 2
B. 4
C. 5
D. 6
E. 12

The calculated total number of arrangements matches option E.

ANSWER 9: E

Problem 10:
The problem asks us to evaluate the expression 16 + 8 ÷ 4 − 2.

Following the order of operations (PEMDAS/BODMAS: Parentheses/Brackets, Exponents/Orders, Multiplication and Division from left to right, Addition and Subtraction from left to right):
1. D

Problem 1:
The water rise is determined by the volume of the rock divided by the base area of the aquarium.  
Base area = \(100 \times 40 = 4000\text{ cm}^2\).  
Volume of rock = \(1000\text{ cm}^3\).  
Rise = \(\frac{1000}{4000} = 0.25\text{ cm}\).  
ANSWER 1: A

Problem 2:
Initially, Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 2: C

Problem 3:
The notation \(\begin{array}{cc} a & c \\ b & d \end{array} = a \cdot d - b \cdot c\).  
For \(\begin{array}{cc} 3 & 1 \\ 4 & 2 \end{array}\): \(a=3, c=1, b=4, d=2\).  
Value = \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).  
ANSWER 3: E

Problem 4:
Total points needed for an average of 95 over 4 tests = \(4 \times 95 = 380\).  
First two tests sum = \(97 + 91 = 188\).  
Remaining points needed = \(380 - 188 = 192\).  
To minimize the third test score, maximize the fourth (100). Then third test \(\ge 192 - 100 = 92\).  
Lowest possible third test score is 92.  
ANSWER 4: B

Problem 5:
Let \(x\) be the number of apples Bridget bought.  
She gave half to Ann, so she kept \(\frac{x}{2}\).  
Then she gave 3 to Cassie and kept 4, so \(\frac{x}{2} = 3 + 4 = 7\).  
Thus \(x = 14\).  
ANSWER 5: E

Problem 6:
Let \(R, G, B\) be the number of red, green, blue marbles.  
"All but 6 are red" \(\Rightarrow G + B = 6\).  
"All but 8 are green" \(\Rightarrow R + B = 8\).  
"All but 4 are blue" \(\Rightarrow R + G = 4\).  
Adding: \((G+B)+(R+B)+(R+G) = 6+8+4 = 18\) \(\Rightarrow 2(R+G+B) = 18\) \(\Rightarrow R+G+B = 9\).  
Total marbles = 9.  
ANSWER 6: C

Problem 7:
Prime factorization of 250: \(250 = 2 \times 5^3\).  
Prime factors are 2 and 5. The two smallest are 2 and 5.  
Sum = \(2 + 5 = 7\).  
ANSWER 7: C

Problem 8:
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}.  
Possible sums:  
1+2=3, 1+4=5, 1+6=7;  
3+2=5, 3+4=7, 3+6=9;  
5+2=7, 5+4=9, 5+6=11.  
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.  
ANSWER 8: B

Problem 9:
Boys must sit at the two ends.  
Arrange boys: 2 choices for left end, 1 for right end → \(2! = 2\) ways.  
Arrange 3 girls in the middle 3 seats: \(3! = 6\) ways.  
Total arrangements = \(2 \times 6 = 12\).  
ANSWER 9: E

Problem 10:
Interpret "8 4" as \(8 \div 4\).  
\(16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16\).  
ANSWER 10: D

Problem 11:
\(\sqrt{4} = 2\).  
\(8 \cdot 2 = 16\), \(\sqrt{16} = 4\).  
\(16 \cdot 4 = 64\), \(\sqrt{64} = 8\).  
ANSWER 11: C

Problem 12:
Compare decimals:  
0.99, 0.9099, 0.9, 0.909, 0.9009.  
0.99 is the largest.  
ANSWER 12: A

# Problem 1:

**What is being asked:** A rock of volume 1000 cm³ is submerged in an aquarium with a 100 cm × 40 cm base. How much does the water level rise?

**Setup:** When the rock is submerged, it displaces a volume of water equal to its own volume. This displaced water spreads across the base of the aquarium, causing the level to rise.

$$\text{Volume displaced} = \text{Base Area} \times \text{Rise in water level}$$

**Computation:**
- Base area = $100 \times 40 = 4000 \text{ cm}^2$
- $1000 = 4000 \times h$
- $h = \frac{1000}{4000} = 0.25 \text{ cm}$

**Check for overflow:** The current depth is 37 cm, and 37 + 0.25 = 37.25 cm, which is well below the 50 cm height. No overflow occurs.

**Ruling out other choices:** Choices B–E would require rock volumes of 2000, 4000, 5000, and 10000 cm³ respectively — all larger than the given 1000 cm³.

ANSWER 1: A

---

# Problem 2:

**What is being asked:** With 4 players, each gets 15 cards. With 6 players, how many cards does each get?

**Setup:** The total number of cards in the deck stays the same.

**Computation:**
- Number of players in first game: Annika + 3 friends = 4 players
- Total cards = $4 \times 15 = 60$ cards
- Number of players in second game: $4 + 2 = 6$ players
- Cards per player = $\frac{60}{6} = 10$

**Ruling out other choices:** Some might forget to count Annika (getting 3 players → 45 cards → 45/5 = 9, choice B) or miscount the new total as 5 players (60/5 = 12, choice E).

ANSWER 2: C

---

# Problem 3:

**What is being asked:** Evaluate the custom operation $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$ for $a=3, b=1, c=4, d=2$.

**Setup:** Substitute directly into the formula.

**Computation:**
$$\begin{vmatrix} 3 & 1 \\ 4 & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2$$

**Ruling out other choices:** A common mistake is computing $b \cdot c - a \cdot d = 4 - 6 = -2$ (choice A) or mixing up which values go where.

ANSWER 3: E

---

# Problem 4:

**What is being asked:** Isabella needs an average of 95 on four tests. Her first two scores are 97 and 91. What is the lowest possible third test score so she can still reach her goal?

**Setup:**
- Total points needed: $4 \times 95 = 380$
- Sum of first two tests: $97 + 91 = 188$
- Sum needed from tests 3 and 4: $380 - 188 = 192$

**Computation:**
To minimize the third test score, maximize the fourth test score. The maximum possible score on any test is 100.
$$\text{Minimum test 3} = 192 - 100 = 92$$

**Ruling out other choices:** Choice A (90) would require test 4 to be $192 - 90 = 102$, which is impossible on a 100-point test. Choices C, D, E are higher than necessary.

ANSWER 4: B

---

# Problem 5:

**What is being asked:** Bridget bought apples, gave half to Ann, gave 3 to Cassie, and kept 4. How many did she buy?

**Setup:** Let $x$ = total apples bought. Work through the sequence of events.

**Computation:**
- After giving half to Ann: $\frac{x}{2}$ apples remain
- After giving 3 to Cassie: $\frac{x}{2} - 3$ apples remain
- She keeps 4: $\frac{x}{2} - 3 = 4$
- $\frac{x}{2} = 7$
- $x = 14$

**Verification:** 14 apples → give 7 to Ann → 7 left → give 3 to Cassie → 4 left. ✓

**Ruling out other choices:** Choice C (7) is a trap for those who forget to double the 7 remaining apples. Choice D (11) comes from incorrectly adding $4 + 3 + 4$.

ANSWER 5: E

---

# Problem 6:

**What is being asked:** Find the total number of marbles given conditions about each color.

**Setup:** Let $T$ = total marbles, $R$ = red, $G$ = green, $B$ = blue.
- "All but 6 are red" → $G + B = 6$ → $R = T - 6$
- "All but 8 are green" → $R + B = 8$ → $G = T - 8$
- "All but 4 are blue" → $R + G = 4$ → $B = T - 4$

**Computation:**
$$T = R + G + B = (T - 6) + (T - 8) + (T - 4) = 3T - 18$$
$$2T = 18 \implies T = 9$$

**Verification:** $R = 3, G = 1, B = 5$. Check: $G+B=6$ ✓, $R+B=8$ ✓, $R+G=4$ ✓.

ANSWER 6: C

---

# Problem 7:

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Setup:** Find the prime factorization of 250.

**Computation:**
$$250 = 2 \times 125 = 2 \times 5 \times 25 = 2 \times 5 \times 5 \times 5 = 2 \times 5^3$$

The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.
$$\text{Sum} = 2 + 5 = 7$$

**Ruling out other choices:** Choice A (2) is just one factor. Choice B (5) is just one factor. Choice D (10) would be $2 + 5 + 3$ or $5 + 5$, neither of which applies.

ANSWER 7: C

---

# Problem 8:

**What is being asked:** How many distinct sums can be formed by picking one chip from Bag A {1, 3, 5} and one from Bag B {2, 4, 6}?

**Setup:** List all 9 possible sums and count the unique values.

**Computation:**
| | 2 | 4 | 6 |
|---|---|---|---|
| **1** | 3 | 5 | 7 |
| **3** | 5 | 7 | 9 |
| **5** | 7 | 9 | 11 |

Unique sums: {3, 5, 7, 9, 11} → **5 different values**

**Ruling out other choices:** Choice E (9) is the trap of counting all 9 pairs without removing duplicates. Choice D (7) might come from miscounting.

ANSWER 8: B

---

# Problem 9:

**What is being asked:** How many ways can 2 boys and 3 girls sit in a row with boys at the ends and girls in the middle?

**Setup:** The arrangement must be: **B G G G B**

**Computation:**
- Arrange the 2 boys at the 2 end positions: $2! = 2$ ways
- Arrange the 3 girls in the 3 middle positions: $3! = 6$ ways
- Total arrangements: $2 \times 6 = 12$

**Ruling out other choices:** Choice A (2) only counts the boys' arrangements. Choice D (6) only counts the girls' arrangements. Both must be multiplied together.

ANSWER 9: E

---

# Problem 10:

**What is being asked:** Evaluate $16 + 8 \div 4 - 2$ using order of operations.

**Setup:** Apply PEMDAS — division before addition and subtraction.

**Computation:**
$$16 + 8 \div 4 - 2 = 16 + 2 - 2 = 16$$

**Ruling out other choices:**
- Left-to-right without order of operations: $(16+8) \div 4 - 2 = 6 - 2 = 4$ (choice A)
- $(16+8) \div (4-2) = 24 \div 2 = 12$ (choice C)
- $16 + 8 \div (4-2) = 16 + 4 = 20$ (choice E

 Problem 1:
We need to find how much the water level rises when a rock of volume 1000 cm³ is submerged in a rectangular aquarium.

First, find the area of the aquarium’s base:
\[
100\text{ cm} \times 40\text{ cm} = 4000\text{ cm}^2.
\]

When the rock is completely submerged, it displaces 1000 cm³ of water. This displaced volume spreads out over the base, causing the water level to rise. The rise in height \(h\) satisfies:
\[
\text{base area} \times h = \text{volume of rock}.
\]
So,
\[
h = \frac{1000}{4000} = 0.25\text{ cm}.
\]

We should check that the aquarium does not overflow. The current depth is 37 cm and the height is 50 cm, so there is 13 cm of free space—plenty of room for a 0.25 cm rise.

Tempting wrong choices come from using the wrong area (e.g., using the side wall 100×50 or 40×50) or from dividing by the height 50. Only the base area matters for the level rise.

ANSWER 1: A

---

Problem 2:
We need to find how many cards each player gets when two extra friends join the game.

First, determine the total number of cards. Annika and 3 friends make 4 players, and each gets 15 cards:
\[
4 \times 15 = 60\text{ cards total}.
\]

When 2 more friends join, the number of players becomes:
\[
4 + 2 = 6.
\]

Distribute the 60 cards evenly among 6 players:
\[
60 \div 6 = 10.
\]

A tempting mistake is to subtract 2 from 15, giving 13, which is not an option, or to guess without computing the total deck size. The key is that the total number of cards stays the same.

ANSWER 2: C

---

Problem 3:
We are given the rule \(a\ c\ b\ d = a\cdot d - b\cdot c\). We need to evaluate \(3\ 1\ 4\ 2\).

Matching the pattern:
- \(a = 3\)
- \(c = 1\)
- \(b = 4\)
- \(d = 2\)

Apply the formula:
\[
a\cdot d - b\cdot c = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

A tempting error is to swap the pairs and compute \(3\cdot 1 - 4\cdot 2 = 3 - 8 = -5\), which is not among the choices, or to read the order incorrectly. Careful matching to \(a, c, b, d\) is essential.

ANSWER 3: E

---

Problem 4:
Isabella wants an average of 95 on four 100-point tests. Her first two scores are 97 and 91. We need the lowest possible score on the third test that still allows her to reach her goal.

Total points needed for a 95 average:
\[
4 \times 95 = 380.
\]

Points earned so far:
\[
97 + 91 = 188.
\]

Points still needed from the last two tests:
\[
380 - 188 = 192.
\]

Let the third test score be \(x\) and the fourth be \(y\), with \(x + y = 192\) and each score at most 100. To minimize \(x\), we maximize \(y\). The highest possible score is 100, so:
\[
x = 192 - 100 = 92.
\]

If she scores 92 on the third test, she can score 100 on the fourth and exactly reach her goal. A tempting wrong choice is 95, which would be the answer if she needed the two remaining tests to average 95, but she can shift points to the fourth test.

ANSWER 4: B

---

Problem 5:
We need to find how many apples Bridget originally bought.

Let the original number be \(x\).
- She gave half to Ann, leaving her with \(\frac{x}{2}\).
- She gave 3 to Cassie and kept 4:
\[
\frac{x}{2} - 3 = 4.
\]

Solve:
\[
\frac{x}{2} = 7 \quad\Rightarrow\quad x = 14.
\]

Check: Start with 14. Half is 7, given to Ann. She has 7 left. Gives 3 to Cassie, keeps 4. This matches.

A tempting error is to add \(3 + 4 = 7\) and stop, forgetting that 7 is only what remained after giving away half, so the original amount must be doubled.

ANSWER 5: E

---

Problem 6:
We need the total number of marbles given three conditions about how many are *not* each color.

Let \(R, G, B\) be the numbers of red, green, and blue marbles.
- “All but 6 are red” means non-red marbles equal 6: \(G + B = 6\).
- “All but 8 are green” means non-green marbles equal 8: \(R + B = 8\).
- “All but 4 are blue” means non-blue marbles equal 4: \(R + G = 4\).

Add all three equations:
\[
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
\]
\[
2R + 2G + 2B = 18
\]
\[
R + G + B = 9.
\]

So there are 9 marbles total. (Checking: \(R=3, G=1, B=5\), which satisfies all three statements.)

A tempting mistake is to add \(6+8+4=18\) and choose that, but that sum counts each marble twice.

ANSWER 6: C

---

Problem 7:
We need the sum of the two smallest prime factors of 250.

Factor 250:
\[
250 = 2 \times 125 = 2 \times 5^3.
\]

The prime factors are 2 and 5. The two smallest (and only) distinct prime factors are 2 and 5. Their sum is:
\[
2 + 5 = 7.
\]

A tempting wrong choice is 10, which comes from adding 5 and 5 (the two smallest factors with multiplicity), but the problem asks for the two smallest prime factors, i.e., distinct primes.

ANSWER 7: C

---

Problem 8:
Bag A has \(\{1, 3, 5\}\) and Bag B has \(\{2, 4, 6\}\). We draw one from each and want the number of distinct possible sums.

Compute all sums:
- \(1+2=3\), \(1+4=5\), \(1+6=7\)
- \(3+2=5\), \(3+4=7\), \(3+6=9\)
- \(5+2=7\), \(5+4=9\), \(5+6=11\)

The distinct values are:
\[
3, 5, 7, 9, 11.
\]

That is 5 different values. A tempting error is to count all 9 ordered pairs without removing duplicates.

ANSWER 8: B

---

Problem 9:
We have 2 boys and 3 girls to arrange in a row with a boy at each end and the three girls in the middle.

Positions: B _ _ _ G G G _ _ _ B (but specifically 5 seats: 1,2,3,4,5 with boys at 1 and 5, girls at 2,3,4).

- Choose the boy for the left end and the boy for the right end: \(2! = 2\) ways.
- Arrange the 3 girls in the middle three seats: \(3! = 6\) ways.

Total arrangements:
\[
2 \times 6 = 12.
\]

A tempting mistake is to ignore the constraint and compute \(5! = 120\), or to forget that the boys can swap places.

ANSWER 9: E

---

Problem 10:
Evaluate \(16 + 8 \div 4 - 2\).

Using order of operations (PEMDAS/BODMAS), division comes before addition and subtraction:
\[
8 \div 4 = 2.
\]

Now substitute back:
\[
16 + 2 - 2 = 16.
\]

A tempting error is to work left-to-right doing addition first: \((16+8)\div 4 - 2 = 24\div 4 - 2 = 6 - 2 = 4\), which is choice A. Another is to do \(16 + 2 = 18\), then \(18 - 2 = 16\), which luckily gives the same corr

Problem 1:
We need the rise in water level when a rock of volume 1000 cm³ is placed in an aquarium with base 100 cm by 40 cm. The rock displaces water equal to its volume, causing the water level to rise by Δh such that base area × Δh = rock volume. Base area = 100 cm × 40 cm = 4000 cm². So Δh = 1000 cm³ / 4000 cm² = 0.25 cm. The initial depth is 37 cm, and after rise it becomes 37.25 cm, which is still below the height of 50 cm, so the rock is fully submerged. The rise is 0.25 cm.

ANSWER 1: A

Problem 2:
When Annika and 3 friends play, there are 4 players, each dealt 15 cards. Total cards = 4 × 15 = 60. In the next game, 2 more friends join, making 6 players. Since the same total number of cards is used, each player gets 60 / 6 = 10 cards.

ANSWER 2: C

Problem 3:
The operation is defined as a c b d = a·d − b·c. For the numbers 3 1 4 2, we have a = 3, c = 1, b = 4, d = 2. Compute a·d − b·c = 3×2 − 4×1 = 6 − 4 = 2.

ANSWER 3: E

Problem 4:
Isabella wants an average of 95 over four tests, so total points needed = 4 × 95 = 380. Her first two scores sum to 97 + 91 = 188. Let the third score be x. After the third test, she can still reach the goal if it is possible to score at most 100 on the fourth test to reach 380. That requires 188 + x + 100 ≥ 380 → x ≥ 380 − 288 = 92. The smallest integer x satisfying this is 92. If x = 91, then even a perfect 100 on the fourth gives 188+91+100 = 379 < 380. So the lowest possible third score is 92.

ANSWER 4: B

Problem 5:
Let the number of apples Bridget bought be N. She gave half to Ann, leaving N/2. Then she gave Cassie 3 apples, leaving N/2 − 3. She keeps 4 apples, so N/2 − 3 = 4 → N/2 = 7 → N = 14.

ANSWER 5: E

Problem 6:
Let T be the total number of marbles. "All but 6 are red" means T − R = 6 → R = T − 6. Similarly, T − G = 8 → G = T − 8, and T − B = 4 → B = T − 4. Since R + G + B = T, we have (T−6)+(T−8)+(T−4) = T → 3T − 18 = T → 2T = 18 → T = 9.

ANSWER 6: C

Problem 7:
Factor 250: 250 = 2 × 5³. Its prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5, and their sum is 7.

ANSWER 7: C

Problem 8:
List all possible sums from drawing one chip from Bag A {1,3,5} and one from Bag B {2,4,6}:
1+2=3, 1+4=5, 1+6=7;
3+2=5, 3+4=7, 3+6=9;
5+2=7, 5+4=9, 5+6=11.
Distinct sums: 3, 5, 7, 9, 11 → 5 different values.

ANSWER 8: B

Problem 9:
We need arrangements of 2 boys and 3 girls in a row with a boy at each end. Choose which boy sits on the left end and which on the right: 2! = 2 ways. The three girls occupy the three middle seats and can be arranged in 3! = 6 ways. Total arrangements = 2 × 6 = 12.

ANSWER 9: E

Problem 10:
The expression is 16 + 8 ÷ 4 − 2 (the missing operator is division, as typical for order-of-operations problems). Following order of operations, division before addition and subtraction: 8 ÷ 4 = 2. Then 16 + 2 = 18, and 18 − 2 = 16.

ANSWER 10: D

Problem 11:
Evaluate step by step: √4 = 2. Then 8 · 2 = 16, so √(8·√4) = √16 = 4. Then 16 · 4 = 64, so √(16·√(8·√4)) = √64 = 8.

ANSWER 11: C

Problem 12:
Compare the decimal numbers:
A. 0.99
B. 0.9099
C. 0.9
D. 0.909
E. 0.9009
Clearly 0.99 is the largest because it is greater than 0.9 and all the others are less than 0.91.

ANSWER 12: A

Problem 1:
The aquarium’s base has area  
\(A = 100\text{ cm} \times 40\text{ cm}=4000\text{ cm}^2\).  
When the rock of volume \(V=1000\text{ cm}^3\) is fully submerged, the water rises by a height \(h\) such that  

\[
A\cdot h = V \quad\Longrightarrow\quad h = \frac{1000}{4000}=0.25\text{ cm}.
\]

The rise cannot be any of the larger listed values (0.5 cm, 1 cm, …) because those would require a volume larger than the rock’s.  

ANSWER 1: A  

---

Problem 2:
With 4 players each receiving 15 cards, the total number of cards is  

\[
4 \times 15 = 60.
\]

If 2 more friends join, there are \(4+2 = 6\) players.  The cards are still all the cards, so each player receives  

\[
\frac{60}{6}=10\text{ cards}.
\]

The other answer choices (8, 9, 11, 12) would not distribute the 60 cards evenly.  

ANSWER 2: C  

---

Problem 3:
The operation is defined by  

\[
a\;c\;b\;d = a\cdot d - b\cdot c.
\]

For the numbers \(3\;1\;4\;2\) we have \(a=3,\;c=1,\;b=4,\;d=2\).  Hence  

\[
3\;1\;4\;2 = 3\cdot 2 - 4\cdot 1 = 6 - 4 = 2.
\]

The other options \(-2,-1,0,1\) would correspond to different combinations that do not satisfy the given formula.  

ANSWER 3: E  

---

Problem 4:
To average 95 over four 100‑point tests, the total points needed are  

\[
4 \times 95 = 380.
\]

After the first two tests Isabella has  

\[
97 + 91 = 188\text{ points}.
\]

Let the third test score be \(x\).  Then the fourth test must supply  

\[
380 - (188 + x) = 192 - x \text{ points}.
\]

The fourth test can give at most 100 points, so we need  

\[
192 - x \le 100 \quad\Longrightarrow\quad x \ge 92.
\]

If she scores \(92\) on the third test, she would need a perfect 100 on the fourth, which is possible.  Any lower score would require more than 100 points on the fourth test, which is impossible.  

Thus the lowest possible third‑test score is **92**.  

ANSWER 4: B  

---

Problem 5:
Let the original number of apples be \(N\).  
She gave half of them to Ann: \(\frac{N}{2}\).  
She then gave Cassie 3 apples and kept 4, leaving  

\[
N - \frac{N}{2} - 3 = 4.
\]

Simplify: \(\frac{N}{2} - 3 = 4 \Rightarrow \frac{N}{2}=7 \Rightarrow N=14\).

If the total had been any of the other answer choices, the steps would not balance (e.g., 11 would leave 2 apples after giving Cassie 3).  

ANSWER 5: E  

---

Problem 6:
Let the total number of marbles be \(T\).  
- “All but 6 are red’’ → red = \(T-6\).  
- “All but 8 are green’’ → green = \(T-8\).  
- “All but 4 are blue’’ → blue = \(T-4\).

These three colors account for all marbles, so  

\[
(T-6)+(T-8)+(T-4)=T \;\Longrightarrow\; 3T-18=T \;\Longrightarrow\; 2T=18 \;\Longrightarrow\; T=9.
\]

The other totals (6, 8, 10, 12) would not satisfy the three “all but…’’ statements simultaneously.  

ANSWER 6: C  

---

Problem 7:
Factor \(250\):  

\[
250 = 2 \times 5^3.
\]

The prime factors are \(2\) and \(5\).  The two smallest are \(2\) and \(5\), whose sum is  

\[
2+5=7.
\]

Choices 2, 5, 10, 12 are either single prime factors or incorrect sums.  

ANSWER 7: C  

---

Problem 8:
Bag A contains \(\{1,3,5\}\); Bag B contains \(\{2,4,6\}\).  
All possible sums \(a+b\) are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The distinct results are \(\{3,5,7,9,11\}\), a total of **5** different sums.  

The other answer choices (4, 6, 7, 9) do not match the count of distinct sums.  

ANSWER 8: B  

---

Problem 9:
Place the two boys at the two ends and the three girls in the middle three seats.

- The two boys can be ordered in \(2!\) ways (left‑right swap).  
- The three girls can be arranged among the three middle seats in \(3! = 6\) ways.

Total arrangements  

\[
2! \times 3! = 2 \times 6 = 12.
\]

Choices 2, 4, 5, 6 are too small; 12 accounts for all possible permutations under the given constraints.  

ANSWER 9: E  

---

Problem 10:
Interpret the expression as  

\[
16 + 8 \div 4 - 2.
\]

Division comes before addition and subtraction, so  

\[
8 \div 4 = 2 \quad\Longrightarrow\quad 16 + 2 - 2 = 16.
\]

The other options (4, 8, 12, 20) would result only if the order of operations were misinterpreted.  

ANSWER 10: D  

---

Problem 11:
Simplify from the innermost radical outward.

\[
\sqrt{4}=2,\qquad 
\sqrt{8\cdot\sqrt{4}} = \sqrt{8\cdot2}= \sqrt{16}=4.
\]

Now  

\[
\sqrt{16\cdot\sqrt{8\cdot\sqrt{4}}} = \sqrt{16\cdot4}= \sqrt{64}=8.
\]

Thus the value is 8, matching choice C.  

ANSWER 11: C  

---

Problem 12:
Compare the decimal values:

- A. \(0.99\)
- B. \(0.9099\)
- C. \(0.9\)
- D. \(0.909\)
- E. \(0.9009\)

Clearly \(0.99\) is greater than \(0.9099\), which is greater than \(0.909\), etc.  
Hence the largest number is \(0.99\).  

ANSWER 12: A

Problem 1:  
An aquarium has a rectangular base of 100 cm by 40 cm and height 50 cm, filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is submerged. By how many centimeters does the water level rise?  

The base area is \(100 \times 40 = 4000 \, \text{cm}^2\). The rock’s volume is 1000 cm³, so the rise in water level is \(\frac{1000}{4000} = 0.25 \, \text{cm}\).  

Tempting choices: 0.5 (B) or 1 (C) might be guessed if base area is miscalculated.  

ANSWER 1: A  

Problem 2:  
Annika and 3 friends (4 players) get 15 cards each. Total cards = \(4 \times 15 = 60\). If 2 more join (6 players), each gets \(\frac{60}{6} = 10\).  

Tempting choices: 8 (A) if total cards miscalculated as 48.  

ANSWER 2: C  

Problem 3:  
Operation defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\): \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\), but the order is \(a c b d\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). However, the problem writes "a c b d" which is interpreted as \(a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\). But the choices include 2 (E). Re-evaluating: the operation is \(a c b d = a \cdot d - b \cdot c\), so \(3 \cdot 2 - 1 \cdot 4 = 2\).  

Tempting choices: -2 (A) if signs reversed.  

ANSWER 3: A (Note: Correction based on standard interpretation: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\), but the correct answer based on the operation defined is actually \(3 \cdot 2 - 1 \cdot 4 = 2\), but the problem writes "a c b d", which is \(a \cdot d - b \cdot c\), so 2. However, the answer is 2, but the choices include 2 as E. The correct calculation is 2, so E. Correction: \(a c b d = a \cdot d - b \cdot c = 3 \cdot 2 - 1 \cdot 4 = 2\).)  

**Correction:** The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). For \(3 \, 1 \, 4 \, 2\), this is \(3 \cdot 2 - 1 \cdot 4 = 6 - 4 = 2\).  

ANSWER 3: E  

Problem 4:  
Isabella wants an average of 95 on four 100-point tests. Total points needed: \(4 \times 95 = 380\). First two scores: 97 + 91 = 188. Let third score be \(x\), fourth be \(y\). Then \(188 + x + y = 380\), so \(x + y = 192\). To minimize \(x\), maximize \(y\): \(y = 100\), so \(x = 92\).  

Tempting choices: 90 (A) if \(y\) not maximized.  

ANSWER 4: B  

Problem 5:  
Bridget gave half apples to Ann, then 3 to Cassie, keeping 4. Let total be \(x\). After giving half to Ann: \(\frac{x}{2}\) left. Then gives 3 to Cassie: \(\frac{x}{2} - 3 = 4\). So \(\frac{x}{2} = 7\), \(x = 14\).  

Tempting choices: 7 (C) if half misinterpreted.  

ANSWER 5: E  

Problem 6:  
Let \(r, g, b\) be red, green, blue marbles. Then:  
- All but 6 are red: \(g + b = 6\)  
- All but 8 are green: \(r + b = 8\)  
- All but 4 are blue: \(r + g = 4\)  
Adding: \(2(r + g + b) = 18\), so total \(r + g + b = 9\).  

Tempting choices: 6 (A) if equations misread.  

ANSWER 6: C  

Problem 7:  
Prime factors of 250: \(250 = 2 \times 5^3\). Two smallest: 2 and 5. Sum: \(2 + 5 = 7\).  

Tempting choices: 5 (B) if only 5 considered.  

ANSWER 7: C  

Problem 8:  
Bag A: {1, 3, 5}, Bag B: {2, 4, 6}. Possible sums:  
- 1+2=3, 1+4=5, 1+6=7  
- 3+2=5, 3+4=7, 3+6=9  
- 5+2=7, 5+4=9, 5+6=11  
Unique sums: 3, 5, 7, 9, 11 → 5 values.  

Tempting choices: 6 (C) if duplicates counted.  

ANSWER 8: B  

Problem 9:  
Two boys at ends: 2 ways. Three girls in middle: \(3! = 6\) ways. Total: \(2 \times 6 = 12\).  

Tempting choices: 6 (D) if boys not permuted.  

ANSWER 9: E  

Problem 10:  
Expression: \(16 + 8 \times 4 - 2\). Order: \(8 \times 4 = 32\), then \(16 + 32 - 2 = 46\)? Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 \times 4 - 2\): \(8 \times 4 = 32\), \(16 + 32 = 48\), \(48 - 2 = 46\). But choices max at 20. Re-interpreting: "16 + 8 4 - 2" might mean \(16 + 8 - 4 - 2 = 18\), not in choices. Alternatively, if "8 4" is 8×4, but 46 not in choices. The expression is likely \(16 + 8 - 4 - 2 = 18\), but not in choices. Correction: The expression is "16 + 8 4 − 2", which might be \(16 + (8 \times 4) - 2 = 16 + 32 - 2 = 46\), but not in choices. Re-reading: It’s "16 + 8 4 − 2", which could be \(16 + 8 - 4 - 2 = 18\), but not in choices. The intended expression is likely \(16 + 8 \times 4 - 2 = 46\), but choices max at 20. Correction: The expression is written as "16 + 8 4 − 2", which is ambiguous. Assuming it’s \(16 + 8 - 4 - 2 = 18\), but not in choices. Alternatively, if "8 4" is 84, but 16+84-2=98. Not matching. The expression is likely \( \frac{16 + 8}{4} - 2 = \frac{24}{4} - 2 = 6 - 2 = 4 \).  

Tempting choices: 8 (B) if order misapplied.  

ANSWER 10: A  

Problem 11:  
Expression: \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\).  
- \(\sqrt{4} = 2\)  
- \(\sqrt{8 \cdot 2} = \sqrt{16} = 4\)  
- \(\sqrt{16 \cdot 4} = \sqrt{64} = 8\)  

Tempting choices: 4 (A) if innermost not simplified.  

ANSWER 11: C  

Problem 12:  
Compare decimals:  
- A: 0.99  
- B: 0.9099  
- C: 0.9  
- D: 0.909  
- E: 0.9009  
Largest is 0.99.  

Tempting choices: 0.9099 (B) if decimals misaligned.  

ANSWER 12: A  

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

### Problem 1:
What is asked: An aquarium with a 100cm×40cm rectangular base is filled with water to 37cm depth. A 1000 cm³ rock is submerged; how much does the water level rise?
Step-by-step: When a submerged object displaces water, its volume equals the additional volume of water in the aquarium. The rise in water level Δh satisfies: Volume of rock = base area of aquarium × Δh.
Base area = 100×40 = 4000 cm².
Δh = 1000 / 4000 = 0.25 cm.
Tempting wrong answers: 0.5 (if base area is miscalculated as 2000 cm²) or 1 (wrong division).
ANSWER 1: A

---

### Problem 2:
What is asked: Annika plus 3 friends (4 total players) each get 15 cards. If 2 more join, how many cards does each player get?
Step-by-step: First calculate total number of cards: 4 players × 15 cards = 60 total cards. After 2 join, total players = 4+2=6. Cards per player = 60 / 6 = 10.
Tempting wrong answers: 9 (if you incorrectly count only 3 initial players, leading to 45 total cards) or 12 (wrong division).
ANSWER 2: C

---

### Problem 3:
What is asked: The custom notation a c b d = a·d − b·c. What is the value of 3 1 4 2?
Step-by-step: Match the values to the definition: a=3, c=1, b=4, d=2. Calculate: 3·2 - 4·1 = 6 - 4 = 2.
Tempting wrong answers: -2 (swap a and b incorrectly) or 0 (miscalculate the multiplication).
ANSWER 3: E

---

### Problem 4:
What is asked: Isabella needs a 95 average on 4 100-point tests. She scored 97 and 91 on the first two. What is the lowest possible third test score that lets her reach her goal?
Step-by-step: Total points needed for 95 average: 4×95 = 380 points. Sum of first two tests: 97+91=188. To minimize the third test score, maximize the fourth test (max 100 points). Total points left for tests 3 and 4: 380-188=192. Minimum third test = 192 - 100 = 92.
Tempting wrong answers: 96 (the average of the remaining two tests, not the minimum possible third score).
ANSWER 4: B

---

### Problem 5:
What is asked: Bridget gave half her apples to Ann, 3 to Cassie, and kept 4. How many apples did she buy originally?
Step-by-step: Let N = original number of apples. After giving half to Ann, she has N/2 left. Then N/2 - 3 = 4 → N/2=7 → N=14. Verify: 14 → 7 left after Ann, minus 3 for Cassie = 4 kept, which matches.
Tempting wrong answers: 7 (the number of apples Bridget had after giving half to Ann, not the original total).
ANSWER 5: E

---

### Problem 6:
What is asked: A jar has red, green, blue marbles. All but 6 are red, all but 8 are green, all but 4 are blue. How many total marbles?
Step-by-step: Let T=total marbles, R=red, G=green, B=blue. The conditions translate to: G+B=6, R+B=8, R+G=4. Add all equations: 2(R+G+B)=18 → R+G+B=T=9. Verify: 3 red, 1 green, 5 blue sum to 9, matching all conditions.
Tempting wrong answers: 18 (the sum of 6+8+4, not halved to get total marbles).
ANSWER 6: C

---

### Problem 7:
What is asked: What is the sum of the two smallest prime factors of 250?
Step-by-step: Prime factorization of 250: 250=2×5³. The prime factors are 2 and 5. Their sum is 2+5=7.
Tempting wrong answers: 10 (sum of 5+5, or wrong addition of 2+5).
ANSWER 7: C

---

### Problem 8:
What is asked: Bag A has chips 1,3,5; Bag B has 2,4,6. One chip drawn from each bag, how many distinct sum values are possible?
Step-by-step: List all sums: 1+2=3, 1+4=5,1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7,5+4=9,5+6=11. Distinct sums: 3,5,7,9,11 → 5 total.
Tempting wrong answers: 9 (total number of unordered pairs, not distinct sums).
ANSWER 8: B

---

### Problem 9:
What is asked: 2 boys and 3 girls line up for a photo, with a boy at each end and girls in the middle. How many arrangements are possible?
Step-by-step: Arrange the boys first: 2 boys can be ordered at the ends in 2! = 2 ways. The 3 middle girls can be ordered in 3! =6 ways. Total arrangements: 2×6=12.
Tempting wrong answers: 6 (only the number of ways to arrange the girls, ignoring the boys).
ANSWER 9: E

---

### Problem 10:
What is asked: Evaluate (16+8)/(4-2) (formatted as a fraction in the original problem).
Step-by-step: Calculate numerator:16+8=24, denominator:4-2=2. 24/2=12.
Tempting wrong answers:16 (if order of operations is misapplied to calculate 16 + 8/4 -2=16).
ANSWER 10: C

---

### Problem 11:
What is asked: Evaluate √(16 · √(8 · √4)).
Step-by-step: Calculate from the innermost root first: √4=2. Next: 8×2=16, √16=4. Next:16×4=64, √64=8.
Tempting wrong answers: 4√2 (incorrect root calculation).
ANSWER 11: C

---

### Problem 12:
What is asked: Which number is largest: 0.99, 0.9099, 0.9, 0.909, 0.9009.
Step-by-step: Line up all decimals to compare place values: 0.9900 > 0.9099 > 0.9090 > 0.9009 > 0.9000. The largest is 0.99.
Tempting wrong answers: 0.9099 (the second largest value).
ANSWER 12: A

### Problem 1:
We need to find how much the water level rises when a 1000 cm³ rock is submerged in the aquarium.
1.  The volume of displaced water equals the volume of the submerged rock: 1000 cm³.
2.  The base area of the aquarium is \(100 \, \text{cm} \times 40 \, \text{cm} = 4000 \, \text{cm}^2\).
3.  Let \(h\) be the rise in water level. The volume of the displaced water is also equal to base area × rise in height: \(4000h = 1000\).
4.  Solve for \(h\): \(h = \frac{1000}{4000} = 0.25\) cm.
ANSWER 1: A
---
### Problem 2:
We need to find the number of cards per player when 2 more friends join.
1.  Original number of players: Annika + 3 friends = 4 players. Each gets 15 cards, so total cards = \(4 \times 15 = 60\).
2.  New number of players: 4 + 2 = 6 players. Total cards remain 60.
3.  Cards per player: \(\frac{60}{6} = 10\).
ANSWER 2: C
---
### Problem 3:
The operation is defined as \(a \, c \, b \, d = a \cdot d - b \cdot c\). Compute \(3 \, 1 \, 4 \, 2\).
1.  Identify \(a=3, c=1, b=4, d=2\).
2.  Substitute into the formula: \(3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2\).
ANSWER 3: E
---
### Problem 4:
Find the lowest possible third test score to get an average of 95 on 4 tests.
1.  Total points needed for an average of 95: \(4 \times 95 = 380\).
2.  Sum of first two test scores: \(97 + 91 = 188\).
3.  Let \(x\) = third test score, \(y\) = fourth test score (max 100, since tests are 100-point). We need \(188 + x + y = 380\), so \(x + y = 192\).
4.  To minimize \(x\), maximize \(y = 100\). Then \(x = 192 - 100 = 92\).
ANSWER 4: B
---
### Problem 5:
Find the original number of apples Bridget bought.
1.  Let \(x\) = total apples. After giving half to Ann, she has \(\frac{x}{2}\) left.
2.  After giving 3 to Cassie, she has \(\frac{x}{2} - 3\) left, which equals 4 (the apples she kept).
3.  Solve: \(\frac{x}{2} - 3 = 4 \implies \frac{x}{2} =7 \implies x=14\).
ANSWER 5: E
---
### Problem 6:
Find the total number of marbles.
1.  Let \(R, G, B\) be the number of red, green, blue marbles, total \(T = R+G+B\).
2.  Translate the statements:
    - All but 6 are red: \(G + B = 6\)
    - All but 8 are green: \(R + B = 8\)
    - All but 4 are blue: \(R + G =4\)
3.  Add all three equations: \(2(R+G+B) = 6+8+4=18 \implies 2T=18 \implies T=9\).
ANSWER 6: C
---
### Problem 7:
Find the sum of the two smallest prime factors of 250.
1.  Factorize 250: \(250 = 2 \times 125 = 2 \times 5^3\).
2.  The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.
3.  Sum: \(2 +5=7\).
ANSWER 7: C
---
### Problem 8:
Find the number of distinct sums when drawing one chip from each bag.
1.  List all possible pairs and sums:
    - Bag A (1,3,5) + Bag B (2,4,6):
      \(1+2=3, 1+4=5, 1+6=7, 3+2=5, 3+4=7, 3+6=9, 5+2=7, 5+4=9, 5+6=11\)
2.  Distinct sums: 3, 5, 7, 9, 11. There are 5 unique values.
ANSWER 8: B
---
### Problem 9:
Find the number of arrangements with boys at each end and girls in the middle.
1.  Arrange the 2 boys at the ends: there are \(2! = 2\) ways (order matters: which boy is left/right).
2.  Arrange the 3 girls in the middle 3 seats: \(3! = 6\) ways.
3.  Total arrangements: \(2 \times 6 =12\).
ANSWER 9: E
---
### Problem 10:
Compute \(16 + 8 \, 4 - 2\). The missing operator between 8 and 4 is division (standard order of operations, common middle school problem):
1.  Interpret as \(16 + \frac{8}{4} - 2\).
2.  Compute division first: \(\frac{8}{4}=2\).
3.  Then add/subtract left to right: \(16 + 2 - 2 =16\).
*(Note: If interpreted as \(16 +8 -4 \times 2\), order of operations gives \(4 \times 2=8\), so \(16+8-8=16\), same result.)*
ANSWER 10: D
---
### Problem 11:
Compute \(\sqrt{16 \cdot \sqrt{8 \cdot \sqrt{4}}}\) by working from the innermost root outward.
1.  Innermost: \(\sqrt{4} = 2\).
2.  Next layer: \(\sqrt{8 \cdot 2} = \sqrt{16} =4\).
3.  Outermost: \(\sqrt{16 \cdot 4} = \sqrt{64} =8\).
ANSWER 11: C
---
### Problem 12:
Find the largest decimal.
1.  Rewrite all numbers with 4 decimal places for easy comparison:
    - A: \(0.9900\)
    - B: \(0.9099\)
    - C: \(0.9000\)
    - D: \(0.9090\)
    - E: \(0.9009\)
2.  Compare digit by digit: all have 9 in the tenths place. A has 9 in the hundredths place, while all others have 0. Thus A is the largest.
ANSWER 12: A