🔬 Model Lab

Problem 1:
We need how many red balls to take away so reds are 75% of what is left.

At first there are 500 balls, 80% red.
Red = 0.80 × 500 = 400. Blue = 500 - 400 = 100.

Let x = number of red balls removed. Blue balls are not touched.
Red left = 400 - x. Total left = 500 - x.

We want:
(400 - x) / (500 - x) = 75% = 3/4

Cross-multiply:
4(400 - x) = 3(500 - x)
1600 - 4x = 1500 - 3x
1600 - 1500 = 4x - 3x
100 = x

Check: 300 red left, 100 blue left, total 400. 300/400 = 3/4.
Removing 25 would leave 375/475 ≈ 79%, still too high. Removing 150 would leave 250/350 ≈ 71%, too low.

ANSWER 1: D

Problem 2:
We are asked which gender pattern for 4 babies is most likely, each baby boy/girl equally likely.

There are 2^4 = 16 equally likely boy-girl sequences.

A. all 4 boys: only BBBB: 1 sequence. Probability 1/16.
B. all 4 girls: only GGGG: 1 sequence. Probability 1/16.
C. 2 girls and 2 boys: number of sequences = choose which 2 of 4 are girls = C(4,2)=6. Probability 6/16.
D. 3 of one gender and 1 of the other: This means 3B1G OR 3G1B.
3B1G: C(4,1)=4 sequences. 3G1B: C(4,1)=4 sequences. Total 8 sequences. Probability 8/16 = 1/2.
E. claims equal, false.

Many think 2-2 is most likely because it is most "balanced", but 3-1 has more sequences: 8 > 6 > 1.

So D is most likely.

ANSWER 2: D

Problem 3:
We need total toothpicks to make a rectangular grid 60 long by 32 wide, where length is measured in toothpicks.

Imagine unit squares. Top edge needs 60 toothpicks side by side. Same for every horizontal line.

Number of horizontal lines: there are 32 squares vertically, so 32+1 = 33 horizontal grid lines.
Each horizontal line uses 60 toothpicks.
Horizontal total = 33 × 60 = 1980.

Number of vertical lines: there are 60 squares horizontally, so 60+1 = 61 vertical grid lines.
Each vertical line uses 32 toothpicks.
Vertical total = 61 × 32 = 1952.

Total = 1980 + 1952 = 3932.

A 1920 = 60×32 is just the number of cells, not toothpicks. B 1952 is vertical only. C 1980 is horizontal only.

ANSWER 3: E

Problem 4:
We are asked to put digits 2,0,2,3, one per box, into the expression
□^□ × □^□
to get the largest value.

We must use all four digits as two bases and two exponents.

Key facts: a^0 = 1 for a ≠ 0, while 0^positive = 0, which kills a product. So do not put 0 as a base with positive exponent. Put 0 as an exponent to get 1. 0^0 is not allowed.

Try cases:

If exponents are 2 and 0, bases 3 and 2:
3^2 × 2^0 = 9 × 1 = 9
2^2 × 3^0 = 4 × 1 = 4
3^0 × 2^2 = 1 × 4 = 4
2^0 × 3^2 = 1 × 9 = 9

If exponents are 3 and 0, bases 2 and 2:
2^3 × 2^0 = 8 × 1 = 8

If 0 is a base with positive exponent, product is 0, e.g. 3^2 × 0^2 = 0.

So the best is 9.

Why not other choices?
A. 0: tempting if you think any zero makes everything zero, but 2^0 =1 avoids zero.
B. 8: this is 2^3 × 2^0, valid but less than 9.
D. 16 and E. 18: these come from the mistake x^0 = x. E.g. 3^2 × 2 =18, 2^3 × 2 =16. Correct is ×1, not ×2.

Maximum is 9.

ANSWER 4: C

Problem 5:
We compare volumes of cylinders. Volume = π r^2 h, r = diameter/2.

Alex: diameter 6, so r = 3. h =12.
V_Alex = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r = 6. h =6.
V_Felicia = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio Alex : Felicia = 108π : 216π = 108:216 = 1:2.

Volume scales as square of radius, so doubling diameter quadruples base area, which outweighs halving height. So Felicia's is bigger, ruling out C,D,E.

ANSWER 5: B

Problem 6:
We need integer-sided isosceles triangles with perimeter 23.

Let equal sides be a,a and base be b. Then:
2a + b = 23, so b = 23 - 2a > 0, so a ≤ 11.

Triangle inequality: sum of two small sides > third side. The only non-trivial one is:
a + a > b
2a > 23 - 2a
4a > 23
a > 5.75, so a ≥ 6.

Thus a = 6,7,8,9,10,11. Each gives a b:
6,6,11: 6+6>11 valid
7,7,9: 14>9 valid
8,8,7: valid
9,9,5: valid
10,10,3: 10+3>10 valid
11,11,1: 11+1>11 valid

That is 6 different triangles. No other a works.

ANSWER 6: C

Problem 7:
We track fraction of original pie left.

Harold eats 1/4, so left = 1 - 1/4 = 3/4.

Moose eats 1/3 of what Harold left:
eaten = (1/3)×(3/4) = 1/4.
Left after moose = 3/4 - 1/4 = 1/2. Equivalently (2/3)×(3/4)=1/2.

Porcupine eats 1/3 of what moose left:
eaten = (1/3)×(1/2) = 1/6.
Left after porcupine = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3. Equivalently (2/3)×(1/2)=1/3.

So 1/3 remains. A 1/12, B 1/6 are the amounts eaten, not left.

ANSWER 7: D

Problem 8:
We have $10 total, $2 tip, so fare money = 10 - 2 = $8.00.

First 0.5 mile costs $2.40.
Remaining for extra distance = 8.00 - 2.40 = $5.60.

Extra is $0.20 per 0.1 mile.
Number of 0.1-mile blocks = 5.60 / 0.20 = 28.
Extra distance = 28 × 0.1 = 2.8 miles.

Total distance = 0.5 + 2.8 = 3.3 miles.

Check: fare = 2.40 + 28×0.20 = 2.40+5.60=8.00, plus $2 tip = $10.

ANSWER 8: C

Problem 9:
Population in 1998 is 200, triples every 25 years. Estimate in 2050.

Time elapsed = 2050 - 1998 = 52 years. That's about two 25-year periods.

After 25 yr (~2023): 200×3 = 600.
After 50 yr (~2048): 600×3 = 1800.

2050 is 2 years more, so a little above 1800. Nearest estimate among choices is 2000.

A 600 is only one tripling. C 1000, B 800 ignore tripling. E 3000 is too high.

ANSWER 9: D

Problem 10:
Picture 3 ft wide centered on 19 ft wall.

Total wall not covered = 19 - 3 = 16 ft.
Centered means half on each side: 16/2 = 8 ft from each end to picture edge.

A 1 1/2, C 9 1/2 confuse center vs edge. B is correct.

ANSWER 10: B

Problem 1:
We are asked to find how many red balls must be removed from a bag of 500 balls (originally 80% red) so that the remaining red balls constitute 75% of the bag.

First, calculate the original number of balls of each color:
- Total balls = $500$
- Red balls = $80\% \text{ of } 500 = 0.80 \times 500 = 400$
- Blue balls = $500 - 400 = 100$

Only red balls are removed, so the number of blue balls remains constant at $100$.
In the final state, red balls make up $75\%$ of the total, which means blue balls must make up:
$$100\% - 75\% = 25\%$$
of the remaining balls.

Let $T$ be the total number of remaining balls:
$$0.25 \times T = 100 \implies T = \frac{100}{0.25} = 400$$

Since the bag originally contained $500$ balls and now contains $400$ balls, the number of red balls removed is:
$$500 - 400 = 100$$

ANSWER 1: D

---

Problem 2:
We are asked to determine which outcome is most likely when four children are born, with boys (B) and girls (G) being equally likely.

Each child has $2$ possible genders, so there are $2^4 = 16$ equally likely possible outcomes in total. We evaluate the number of outcomes for each choice:
- **A. All 4 are boys:** Only $1$ outcome ($\text{BBBB}$). Probability = $\frac{1}{16}$.
- **B. All 4 are girls:** Only $1$ outcome ($\text{GGGG}$). Probability = $\frac{1}{16}$.
- **C. 2 are girls and 2 are boys:** The number of ways to choose which $2$ children are girls is $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ outcomes. Probability = $\frac{6}{16}$.
- **D. 3 are of one gender and 1 is of the other:** This includes either 3 boys and 1 girl ($\binom{4}{1} = 4$ outcomes) or 3 girls and 1 boy ($\binom{4}{3} = 4$ outcomes), giving $4 + 4 = 8$ outcomes. Probability = $\frac{8}{16} = \frac{1}{2}$.

Comparing the possibilities, outcome D has $8$ favorable outcomes, which is greater than C ($6$ outcomes), A ($1$ outcome), and B ($1$ outcome). Therefore, D is the most likely outcome.

ANSWER 2: D

---

Problem 3:
We are asked to find the total number of toothpicks used to make a rectangular grid that is $60$ toothpicks long and $32$ toothpicks wide.

A grid of dimensions $60 \times 32$ consists of unit squares arranged in $32$ rows and $60$ columns:
- **Horizontal toothpicks:** There are $32 + 1 = 33$ horizontal grid lines. Each horizontal line is made of $60$ toothpicks.
  $$\text{Number of horizontal toothpicks} = 33 \times 60 = 1980$$
- **Vertical toothpicks:** There are $60 + 1 = 61$ vertical grid lines. Each vertical line is made of $32$ toothpicks.
  $$\text{Number of vertical toothpicks} = 61 \times 32 = 1952$$

Adding the horizontal and vertical toothpicks together:
$$\text{Total toothpicks} = 1980 + 1952 = 3932$$

ANSWER 3: E

---

Problem 4:
We are asked to find the maximum possible value of the expression $\square^\square \times \square^\square$ when placing the digits $2, 0, 2,$ and $3$ into the four boxes, one digit per box.

Let the expression be $a^b \times c^d$, where $\{a, b, c, d\} = \{2, 0, 2, 3\}$.
We test the possible assignments of the digits:
- To maximize the product, we want large bases and exponents, but one of the digits is $0$.
- If $0$ is used as a base, say $0^d$ with $d > 0$, the factor becomes $0$, making the entire product $0$.
- If $0$ is placed in an exponent, recall that for any non-zero base $x$, $x^0 = 1$.
  - Pair $0$ with $2$ as an exponent: $2^0 = 1$. The remaining digits are $3$ and $2$.
    - Setting $3^2 \times 2^0 = 9 \times 1 = 9$.
    - Setting $2^3 \times 2^0 = 8 \times 1 = 8$.
  - Pair $0$ with $3$ as an exponent: $3^0 = 1$. The remaining digits are $2$ and $2$.
    - Setting $2^2 \times 3^0 = 4 \times 1 = 4$.

The values $16$ and $18$ are distractors arising from the common misconception that $x^0 = x$ (which would give $2^3 \times 2 = 16$ or $3^2 \times 2 = 18$).
The maximum mathematically valid value achievable is $9$.

ANSWER 4: C

---

Problem 5:
We are asked to find the ratio of the volume of Alex's cat food can to the volume of Felicia's cat food can.

The volume of a cylinder is given by $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height.
- For Alex's can:
  - Diameter = $6\text{ cm} \implies \text{radius } r_A = 3\text{ cm}$
  - Height $h_A = 12\text{ cm}$
  - Volume $V_A = \pi (3)^2 (12) = \pi \times 9 \times 12 = 108\pi\text{ cm}^3$

- For Felicia's can:
  - Diameter = $12\text{ cm} \implies \text{radius } r_F = 6\text{ cm}$
  - Height $h_F = 6\text{ cm}$
  - Volume $V_F = \pi (6)^2 (6) = \pi \times 36 \times 6 = 216\pi\text{ cm}^3$

The ratio of the volume of Alex's can to Felicia's can is:
$$\frac{V_A}{V_F} = \frac{108\pi}{216\pi} = \frac{1}{2}$$

Thus, the ratio is $1:2$.

ANSWER 5: B

---

Problem 6:
We are asked to find the number of different isosceles triangles with integer side lengths and a perimeter of $23$.

Let the side lengths of the isosceles triangle be $a, a,$ and $b$, where $a$ and $b$ are positive integers.
The perimeter is:
$$2a + b = 23 \implies b = 23 - 2a$$

By the Triangle Inequality:
1. The sum of the two equal sides must be strictly greater than the third side:
   $$a + a > b \implies 2a > 23 - 2a \implies 4a > 23 \implies a \ge 6$$
2. The third side must be positive:
   $$b > 0 \implies 23 - 2a > 0 \implies 2a < 23 \implies a \le 11$$

The possible integer values for $a$ are $6, 7, 8, 9, 10, 11$.
We check each corresponding side length triple $(a, a, b)$:
- $a = 6 \implies b = 11$: sides $(6, 6, 11)$
- $a = 7 \implies b = 9$: sides $(7, 7, 9)$
- $a = 8 \implies b = 7$: sides $(8, 8, 7)$
- $a = 9 \implies b = 5$: sides $(9, 9, 5)$
- $a = 10 \implies b = 3$: sides $(10, 10, 3)$
- $a = 11 \implies b = 1$: sides $(11, 11, 1)$

All $6$ sets of side lengths satisfy the triangle inequality and have two equal sides (none are equilateral since $23$ is not divisible by $3$). Thus, there are $6$ such triangles.

ANSWER 6: C

---

Problem 7:
We are asked to find the fraction of the original pie remaining after Harold, a moose, and a porcupine each eat a portion.

1. Harold start

# Problem 1:
Restate: 500 balls, 80% red. Remove some **red** balls so that 75% of what remains is red.

Red = 0.80(500) = 400, Blue = 100.

Let $x$ = red balls removed. Blues never change (still 100), and after removal blues must be 25% of the total:
$$100 = 0.25(500 - x) \Rightarrow 400 = 500 - x \Rightarrow x = 100.$$

Check: 300 red, 100 blue, total 400 → 300/400 = 75%. ✓

(Tempting wrong: 25 or 50 come from mistakenly taking 5% or 10% of 500; but removing balls shrinks the total too, so more must go.)

ANSWER 1: D

# Problem 2:
There are $2^4 = 16$ equally likely sequences of genders.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: $\binom{4}{2}=6$ ways → 6/16
- 3 of one gender, 1 of the other: $\binom{4}{3}=4$ (3 boys) $+ \binom{4}{1}=4$ (3 girls) $= 8$ ways → 8/16

The largest is 8/16 = 1/2, the "3–1 split." Choice C (6/16) is the tempting trap because 2–2 is the single most likely *exact* split, but option D combines two cases.

ANSWER 2: D

# Problem 3:
A grid 60 toothpicks long and 32 toothpicks wide means the rectangle is 60 units by 32 units.

- Horizontal toothpicks: there are $32+1 = 33$ horizontal lines, each made of 60 toothpicks: $33 \times 60 = 1980$.
- Vertical toothpicks: there are $60+1 = 61$ vertical lines, each made of 32 toothpicks: $61 \times 32 = 1952$.

Total: $1980 + 1952 = 3932$.

(1980 and 1952 are the individual pieces — tempting partial answers.)

ANSWER 3: E

# Problem 4:
The expression is $\boxed{\phantom{2}}^{\boxed{\phantom{2}}} \times \boxed{\phantom{2}}^{\boxed{\phantom{2}}}$, using digits 2, 0, 2, 3 once each.

To maximize, make one factor as large as possible and let the other equal 1 (using exponent 0).

- $3^2 \times 2^0 = 9 \times 1 = 9$
- $2^3 \times 2^0 = 8$ (uses two 2's as base and exponent... gives 8)
- $2^2 \times 3^0 = 4$
- $2^0 \times 3^2 = 9$

Any arrangement with 0 as a base kills a factor to 0. The maximum is $9$.

(8 is the tempting near-miss from $2^3$; 16 and 18 aren't attainable.)

ANSWER 4: C

# Problem 5:
Volume of a cylinder: $V=\pi r^2 h$.

- Alex: diameter 6 → $r=3$, $h=12$: $V = \pi(3^2)(12) = 108\pi$.
- Felicia: diameter 12 → $r=6$, $h=6$: $V = \pi(6^2)(6) = 216\pi$.

Ratio $= 108\pi : 216\pi = 1:2$.

(1:1 is the trap — swapping diameter and height does *not* preserve volume, since radius is squared.)

ANSWER 5: B

# Problem 6:
Let the two equal sides be $a$ and the base be $b$, with $2a + b = 23$.

Since $2a$ is even, $b$ must be odd. Triangle inequality: $2a > b$, i.e. $23 - b > b \Rightarrow b < 11.5$.

So $b \in \{1,3,5,7,9,11\}$, giving $a = 11,10,9,8,7,6$ — six triangles:
(11,11,1), (10,10,3), (9,9,5), (8,8,7), (7,7,9), (6,6,11).

That's 6. (No equilateral option since 23 isn't divisible by 3; $b=13$ fails since $5+5<13$.)

ANSWER 6: C

# Problem 7:
Harold ate $\frac14$, leaving $\frac34$.

Moose ate $\frac13$ of that, leaving $\frac23 \cdot \frac34 = \frac12$.

Porcupine ate $\frac13$ of that, leaving $\frac23 \cdot \frac12 = \frac13$.

(Trap: subtracting $\frac14+\frac13+\frac13$ from 1 gives $\frac1{12}$ — wrong, because each animal eats a fraction of the *remainder*, not of the whole pie.)

ANSWER 7: D

# Problem 8:
Total money $10, minus the $2 tip → $8 available for fare.

First $\frac12$ mile costs $2.40, leaving $8.00 - 2.40 = \$5.60$.

Each additional 0.1 mile costs $0.20: $5.60 \div 0.20 = 28$ increments → $28 \times 0.1 = 2.8$ miles.

Total distance $= 0.5 + 2.8 = 3.3$ miles.

(Forgetting the tip gives $\$10-\$2.40=\$7.60 \to 3.8$ mi; forgetting the first half-mile is included gives other traps like 3.5.)

ANSWER 8: C

# Problem 9:
Population triples every 25 years, starting at 200 in 1998.

- 1998: 200
- 2023: 600
- 2048: 1800

The year 2050 is just 2 years past 2048, so the population is a bit more than 1800 — about 2000.

(3000 would require another near-tripling; 1000 is too small.)

ANSWER 9: D

# Problem 10:
The wall is 19 ft wide; the picture is 3 ft across and centered. The leftover wall splits equally on both sides:
$$\frac{19-3}{2} = \frac{16}{2} = 8 \text{ feet}.$$

(16 is the total leftover, not one side; $9\frac12$ is the wall's midpoint.)

ANSWER 10: B

Problem 1:

Initially, \(80\%\) of 500 balls are red:

\[
0.80(500)=400
\]

So there are 400 red balls and 100 blue balls.

Suppose \(x\) red balls are removed. Then:

- Red balls remaining: \(400-x\)
- Total balls remaining: \(500-x\)

We want \(75\%\) of the remaining balls to be red:

\[
\frac{400-x}{500-x}=\frac34
\]

Cross-multiply:

\[
4(400-x)=3(500-x)
\]

\[
1600-4x=1500-3x
\]

\[
100=x
\]

Removing 100 red balls leaves 300 red and 100 blue, for 400 total. Indeed,

\[
\frac{300}{400}=75\%.
\]

Choices such as 50 or 75 do not reduce the red percentage enough; 150 reduces it too much.

ANSWER 1: D

Problem 2:

Each child can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so \(1\) arrangement.
- All 4 girls: only \(GGGG\), so \(1\) arrangement.
- Exactly 2 girls and 2 boys:

\[
\binom42=6
\]

arrangements.
- Three of one gender and one of the other:
  - 3 boys and 1 girl: \(\binom41=4\) arrangements.
  - 3 girls and 1 boy: \(\binom41=4\) arrangements.

So this outcome has

\[
4+4=8
\]

arrangements.

Since \(8\) is more than \(6\) or \(1\), the most likely outcome is 3 children of one gender and 1 of the other.

ANSWER 2: D

Problem 3:

A grid 60 toothpicks long and 32 toothpicks wide has 60 columns and 32 rows of small squares.

For the horizontal toothpicks, there are \(32+1=33\) horizontal lines, each containing 60 toothpicks:

\[
33\cdot 60=1980
\]

For the vertical toothpicks, there are \(60+1=61\) vertical lines, each containing 32 toothpicks:

\[
61\cdot 32=1952
\]

Thus the total number is

\[
1980+1952=3932.
\]

The tempting answer \(1920=60\cdot32\) counts only the small squares, not the toothpicks forming their edges.

ANSWER 3: E

Problem 4:

The expression is of the form

\[
\Box^{\Box}\times\Box^{\Box},
\]

using the digits \(2,0,2,3\) once each.

To make the product as large as possible, the digit \(0\) should be an exponent, because any nonzero number to the zero power equals 1:

\[
2^0=1 \qquad\text{or}\qquad 3^0=1.
\]

We should use the remaining digits to make the largest possible power. Compare:

\[
3^2=9
\]

and

\[
2^3=8.
\]

Thus the best arrangement is

\[
3^2\times 2^0=9\times 1=9.
\]

Using \(2^3\) gives only \(8\), and placing 0 as a base would make the entire product 0.

ANSWER 4: C

Problem 5:

The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s can has diameter 6 cm, so its radius is 3 cm and its height is 12 cm:

\[
V_A=\pi(3)^2(12)=108\pi.
\]

Felicia’s can has diameter 12 cm, so its radius is 6 cm and its height is 6 cm:

\[
V_F=\pi(6)^2(6)=216\pi.
\]

The ratio is

\[
108\pi:216\pi=1:2.
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, making her base area four times as large.

ANSWER 5: B

Problem 6:

Let the equal sides of the isosceles triangle each have length \(a\). Let the base have length \(b\).

Since the perimeter is 23,

\[
2a+b=23,
\]

so

\[
b=23-2a.
\]

For a valid triangle, the base must be less than the sum of the equal sides:

\[
b<2a.
\]

Substitute \(b=23-2a\):

\[
23-2a<2a
\]

\[
23<4a
\]

\[
a>5.75.
\]

Also, \(b\) must be positive:

\[
23-2a>0,
\]

so

\[
a<11.5.
\]

Thus \(a\) can be

\[
6,7,8,9,10,11.
\]

That gives 6 different triangles. Their side lengths are:

\[
(6,6,11),\ (7,7,9),\ (8,8,7),\ (9,9,5),\ (10,10,3),\ (11,11,1).
\]

Values below 6 fail the triangle inequality, while values above 11 give a nonpositive base.

ANSWER 6: C

Problem 7:

Harold eats \(\frac14\) of the pie, leaving

\[
1-\frac14=\frac34.
\]

The moose eats \(\frac13\) of what is left, so it leaves \(\frac23\) of what Harold left:

\[
\frac23\cdot\frac34=\frac12.
\]

Then the porcupine eats \(\frac13\) of that remaining \(\frac12\), leaving \(\frac23\) of it:

\[
\frac23\cdot\frac12=\frac13.
\]

So \(\frac13\) of the original pie remains.

A common mistake is to subtract \(\frac13\) directly from \(\frac34\); the moose eats one-third of the amount currently left, not one-third of the original pie.

ANSWER 7: D

Problem 8:

The total amount available for the taxi fare is

\[
\$10-\$2=\$8.
\]

The first \(\frac12\) mile costs \$2.40. Therefore, the amount left for additional distance is

\[
8.00-2.40=5.60.
\]

Each additional \(0.1\) mile costs \$0.20, so the number of additional tenths of a mile is

\[
\frac{5.60}{0.20}=28.
\]

That is

\[
28(0.1)=2.8
\]

additional miles.

Adding the first \(\frac12\) mile:

\[
2.8+0.5=3.3.
\]

The ride can be 3.3 miles. The \$2 tip must be included, so choices based on spending the full \$10 on fare are too large.

ANSWER 8: C

Problem 9:

The population is 200 in 1998 and triples every 25 years.

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is a little more than two 25-year periods.

After 25 years, the population would be

\[
200\cdot 3=600.
\]

After 50 years, it would be

\[
600\cdot 3=1800.
\]

Since 2050 is about 2 years later, the population should be slightly more than 1800. The closest estimate is 2000.

Choices such as 600 correspond to only one tripling, while 3000 is much too high for just over two tripling periods.

ANSWER 9: D

Problem 10:

The wall is 19 feet wide, and the picture is 3 feet wide.

The amount of wall not covered by the picture is

\[
19-3=16
\]

feet.

Since the picture is centered, this 16 feet is divided equally between the two ends:

\[
\frac{16}{2}=8.
\]

Therefore, the nearest edge of the picture is 8 feet from either end of the wall.

ANSWER 10: B

Problem 1:
We need how many red balls to take away so reds are 75% of what is left.

At first there are 500 balls, 80% red.
Red = 0.80 × 500 = 400. Blue = 500 - 400 = 100.

Let x = number of red balls removed. Blue balls are not touched.
Red left = 400 - x. Total left = 500 - x.

We want:
(400 - x) / (500 - x) = 75% = 3/4

Cross-multiply:
4(400 - x) = 3(500 - x)
1600 - 4x = 1500 - 3x
1600 - 1500 = 4x - 3x
100 = x

Check: 300 red left, 100 blue left, total 400. 300/400 = 3/4.
Removing 25 would leave 375/475 ≈ 79%, still too high. Removing 150 would leave 250/350 ≈ 71%, too low.

ANSWER 1: D

Problem 2:
We are asked which gender pattern for 4 babies is most likely, each baby boy/girl equally likely.

There are 2^4 = 16 equally likely boy-girl sequences.

A. all 4 boys: only BBBB: 1 sequence. Probability 1/16.
B. all 4 girls: only GGGG: 1 sequence. Probability 1/16.
C. 2 girls and 2 boys: number of sequences = choose which 2 of 4 are girls = C(4,2)=6. Probability 6/16.
D. 3 of one gender and 1 of the other: This means 3B1G OR 3G1B.
3B1G: C(4,1)=4 sequences. 3G1B: C(4,1)=4 sequences. Total 8 sequences. Probability 8/16 = 1/2.
E. claims equal, false.

Many think 2-2 is most likely because it is most "balanced", but 3-1 has more sequences: 8 > 6 > 1.

So D is most likely.

ANSWER 2: D

Problem 3:
We need total toothpicks to make a rectangular grid 60 long by 32 wide, where length is measured in toothpicks.

Imagine unit squares. Top edge needs 60 toothpicks side by side. Same for every horizontal line.

Number of horizontal lines: there are 32 squares vertically, so 32+1 = 33 horizontal grid lines.
Each horizontal line uses 60 toothpicks.
Horizontal total = 33 × 60 = 1980.

Number of vertical lines: there are 60 squares horizontally, so 60+1 = 61 vertical grid lines.
Each vertical line uses 32 toothpicks.
Vertical total = 61 × 32 = 1952.

Total = 1980 + 1952 = 3932.

A 1920 = 60×32 is just the number of cells, not toothpicks. B 1952 is vertical only. C 1980 is horizontal only.

ANSWER 3: E

Problem 4:
We are asked to put digits 2,0,2,3, one per box, into the expression
□^□ × □^□
to get the largest value.

We must use all four digits as two bases and two exponents.

Key facts: a^0 = 1 for a ≠ 0, while 0^positive = 0, which kills a product. So do not put 0 as a base with positive exponent. Put 0 as an exponent to get 1. 0^0 is not allowed.

Try cases:

If exponents are 2 and 0, bases 3 and 2:
3^2 × 2^0 = 9 × 1 = 9
2^2 × 3^0 = 4 × 1 = 4
3^0 × 2^2 = 1 × 4 = 4
2^0 × 3^2 = 1 × 9 = 9

If exponents are 3 and 0, bases 2 and 2:
2^3 × 2^0 = 8 × 1 = 8

If 0 is a base with positive exponent, product is 0, e.g. 3^2 × 0^2 = 0.

So the best is 9.

Why not other choices?
A. 0: tempting if you think any zero makes everything zero, but 2^0 =1 avoids zero.
B. 8: this is 2^3 × 2^0, valid but less than 9.
D. 16 and E. 18: these come from the mistake x^0 = x. E.g. 3^2 × 2 =18, 2^3 × 2 =16. Correct is ×1, not ×2.

Maximum is 9.

ANSWER 4: C

Problem 5:
We compare volumes of cylinders. Volume = π r^2 h, r = diameter/2.

Alex: diameter 6, so r = 3. h =12.
V_Alex = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r = 6. h =6.
V_Felicia = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio Alex : Felicia = 108π : 216π = 108:216 = 1:2.

Volume scales as square of radius, so doubling diameter quadruples base area, which outweighs halving height. So Felicia's is bigger, ruling out C,D,E.

ANSWER 5: B

Problem 6:
We need integer-sided isosceles triangles with perimeter 23.

Let equal sides be a,a and base be b. Then:
2a + b = 23, so b = 23 - 2a > 0, so a ≤ 11.

Triangle inequality: sum of two small sides > third side. The only non-trivial one is:
a + a > b
2a > 23 - 2a
4a > 23
a > 5.75, so a ≥ 6.

Thus a = 6,7,8,9,10,11. Each gives a b:
6,6,11: 6+6>11 valid
7,7,9: 14>9 valid
8,8,7: valid
9,9,5: valid
10,10,3: 10+3>10 valid
11,11,1: 11+1>11 valid

That is 6 different triangles. No other a works.

ANSWER 6: C

Problem 7:
We track fraction of original pie left.

Harold eats 1/4, so left = 1 - 1/4 = 3/4.

Moose eats 1/3 of what Harold left:
eaten = (1/3)×(3/4) = 1/4.
Left after moose = 3/4 - 1/4 = 1/2. Equivalently (2/3)×(3/4)=1/2.

Porcupine eats 1/3 of what moose left:
eaten = (1/3)×(1/2) = 1/6.
Left after porcupine = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3. Equivalently (2/3)×(1/2)=1/3.

So 1/3 remains. A 1/12, B 1/6 are the amounts eaten, not left.

ANSWER 7: D

Problem 8:
We have $10 total, $2 tip, so fare money = 10 - 2 = $8.00.

First 0.5 mile costs $2.40.
Remaining for extra distance = 8.00 - 2.40 = $5.60.

Extra is $0.20 per 0.1 mile.
Number of 0.1-mile blocks = 5.60 / 0.20 = 28.
Extra distance = 28 × 0.1 = 2.8 miles.

Total distance = 0.5 + 2.8 = 3.3 miles.

Check: fare = 2.40 + 28×0.20 = 2.40+5.60=8.00, plus $2 tip = $10.

ANSWER 8: C

Problem 9:
Population in 1998 is 200, triples every 25 years. Estimate in 2050.

Time elapsed = 2050 - 1998 = 52 years. That's about two 25-year periods.

After 25 yr (~2023): 200×3 = 600.
After 50 yr (~2048): 600×3 = 1800.

2050 is 2 years more, so a little above 1800. Nearest estimate among choices is 2000.

A 600 is only one tripling. C 1000, B 800 ignore tripling. E 3000 is too high.

ANSWER 9: D

Problem 10:
Picture 3 ft wide centered on 19 ft wall.

Total wall not covered = 19 - 3 = 16 ft.
Centered means half on each side: 16/2 = 8 ft from each end to picture edge.

A 1 1/2, C 9 1/2 confuse center vs edge. B is correct.

ANSWER 10: B

Problem 1:
We are asked to find how many red balls must be removed from a bag of 500 balls (originally 80% red) so that the remaining red balls constitute 75% of the bag.

First, calculate the original number of balls of each color:
- Total balls = $500$
- Red balls = $80\% \text{ of } 500 = 0.80 \times 500 = 400$
- Blue balls = $500 - 400 = 100$

Only red balls are removed, so the number of blue balls remains constant at $100$.
In the final state, red balls make up $75\%$ of the total, which means blue balls must make up:
$$100\% - 75\% = 25\%$$
of the remaining balls.

Let $T$ be the total number of remaining balls:
$$0.25 \times T = 100 \implies T = \frac{100}{0.25} = 400$$

Since the bag originally contained $500$ balls and now contains $400$ balls, the number of red balls removed is:
$$500 - 400 = 100$$

ANSWER 1: D

---

Problem 2:
We are asked to determine which outcome is most likely when four children are born, with boys (B) and girls (G) being equally likely.

Each child has $2$ possible genders, so there are $2^4 = 16$ equally likely possible outcomes in total. We evaluate the number of outcomes for each choice:
- **A. All 4 are boys:** Only $1$ outcome ($\text{BBBB}$). Probability = $\frac{1}{16}$.
- **B. All 4 are girls:** Only $1$ outcome ($\text{GGGG}$). Probability = $\frac{1}{16}$.
- **C. 2 are girls and 2 are boys:** The number of ways to choose which $2$ children are girls is $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ outcomes. Probability = $\frac{6}{16}$.
- **D. 3 are of one gender and 1 is of the other:** This includes either 3 boys and 1 girl ($\binom{4}{1} = 4$ outcomes) or 3 girls and 1 boy ($\binom{4}{3} = 4$ outcomes), giving $4 + 4 = 8$ outcomes. Probability = $\frac{8}{16} = \frac{1}{2}$.

Comparing the possibilities, outcome D has $8$ favorable outcomes, which is greater than C ($6$ outcomes), A ($1$ outcome), and B ($1$ outcome). Therefore, D is the most likely outcome.

ANSWER 2: D

---

Problem 3:
We are asked to find the total number of toothpicks used to make a rectangular grid that is $60$ toothpicks long and $32$ toothpicks wide.

A grid of dimensions $60 \times 32$ consists of unit squares arranged in $32$ rows and $60$ columns:
- **Horizontal toothpicks:** There are $32 + 1 = 33$ horizontal grid lines. Each horizontal line is made of $60$ toothpicks.
  $$\text{Number of horizontal toothpicks} = 33 \times 60 = 1980$$
- **Vertical toothpicks:** There are $60 + 1 = 61$ vertical grid lines. Each vertical line is made of $32$ toothpicks.
  $$\text{Number of vertical toothpicks} = 61 \times 32 = 1952$$

Adding the horizontal and vertical toothpicks together:
$$\text{Total toothpicks} = 1980 + 1952 = 3932$$

ANSWER 3: E

---

Problem 4:
We are asked to find the maximum possible value of the expression $\square^\square \times \square^\square$ when placing the digits $2, 0, 2,$ and $3$ into the four boxes, one digit per box.

Let the expression be $a^b \times c^d$, where $\{a, b, c, d\} = \{2, 0, 2, 3\}$.
We test the possible assignments of the digits:
- To maximize the product, we want large bases and exponents, but one of the digits is $0$.
- If $0$ is used as a base, say $0^d$ with $d > 0$, the factor becomes $0$, making the entire product $0$.
- If $0$ is placed in an exponent, recall that for any non-zero base $x$, $x^0 = 1$.
  - Pair $0$ with $2$ as an exponent: $2^0 = 1$. The remaining digits are $3$ and $2$.
    - Setting $3^2 \times 2^0 = 9 \times 1 = 9$.
    - Setting $2^3 \times 2^0 = 8 \times 1 = 8$.
  - Pair $0$ with $3$ as an exponent: $3^0 = 1$. The remaining digits are $2$ and $2$.
    - Setting $2^2 \times 3^0 = 4 \times 1 = 4$.

The values $16$ and $18$ are distractors arising from the common misconception that $x^0 = x$ (which would give $2^3 \times 2 = 16$ or $3^2 \times 2 = 18$).
The maximum mathematically valid value achievable is $9$.

ANSWER 4: C

---

Problem 5:
We are asked to find the ratio of the volume of Alex's cat food can to the volume of Felicia's cat food can.

The volume of a cylinder is given by $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height.
- For Alex's can:
  - Diameter = $6\text{ cm} \implies \text{radius } r_A = 3\text{ cm}$
  - Height $h_A = 12\text{ cm}$
  - Volume $V_A = \pi (3)^2 (12) = \pi \times 9 \times 12 = 108\pi\text{ cm}^3$

- For Felicia's can:
  - Diameter = $12\text{ cm} \implies \text{radius } r_F = 6\text{ cm}$
  - Height $h_F = 6\text{ cm}$
  - Volume $V_F = \pi (6)^2 (6) = \pi \times 36 \times 6 = 216\pi\text{ cm}^3$

The ratio of the volume of Alex's can to Felicia's can is:
$$\frac{V_A}{V_F} = \frac{108\pi}{216\pi} = \frac{1}{2}$$

Thus, the ratio is $1:2$.

ANSWER 5: B

---

Problem 6:
We are asked to find the number of different isosceles triangles with integer side lengths and a perimeter of $23$.

Let the side lengths of the isosceles triangle be $a, a,$ and $b$, where $a$ and $b$ are positive integers.
The perimeter is:
$$2a + b = 23 \implies b = 23 - 2a$$

By the Triangle Inequality:
1. The sum of the two equal sides must be strictly greater than the third side:
   $$a + a > b \implies 2a > 23 - 2a \implies 4a > 23 \implies a \ge 6$$
2. The third side must be positive:
   $$b > 0 \implies 23 - 2a > 0 \implies 2a < 23 \implies a \le 11$$

The possible integer values for $a$ are $6, 7, 8, 9, 10, 11$.
We check each corresponding side length triple $(a, a, b)$:
- $a = 6 \implies b = 11$: sides $(6, 6, 11)$
- $a = 7 \implies b = 9$: sides $(7, 7, 9)$
- $a = 8 \implies b = 7$: sides $(8, 8, 7)$
- $a = 9 \implies b = 5$: sides $(9, 9, 5)$
- $a = 10 \implies b = 3$: sides $(10, 10, 3)$
- $a = 11 \implies b = 1$: sides $(11, 11, 1)$

All $6$ sets of side lengths satisfy the triangle inequality and have two equal sides (none are equilateral since $23$ is not divisible by $3$). Thus, there are $6$ such triangles.

ANSWER 6: C

---

Problem 7:
We are asked to find the fraction of the original pie remaining after Harold, a moose, and a porcupine each eat a portion.

1. Harold start

# Problem 1:
Restate: 500 balls, 80% red. Remove some **red** balls so that 75% of what remains is red.

Red = 0.80(500) = 400, Blue = 100.

Let $x$ = red balls removed. Blues never change (still 100), and after removal blues must be 25% of the total:
$$100 = 0.25(500 - x) \Rightarrow 400 = 500 - x \Rightarrow x = 100.$$

Check: 300 red, 100 blue, total 400 → 300/400 = 75%. ✓

(Tempting wrong: 25 or 50 come from mistakenly taking 5% or 10% of 500; but removing balls shrinks the total too, so more must go.)

ANSWER 1: D

# Problem 2:
There are $2^4 = 16$ equally likely sequences of genders.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: $\binom{4}{2}=6$ ways → 6/16
- 3 of one gender, 1 of the other: $\binom{4}{3}=4$ (3 boys) $+ \binom{4}{1}=4$ (3 girls) $= 8$ ways → 8/16

The largest is 8/16 = 1/2, the "3–1 split." Choice C (6/16) is the tempting trap because 2–2 is the single most likely *exact* split, but option D combines two cases.

ANSWER 2: D

# Problem 3:
A grid 60 toothpicks long and 32 toothpicks wide means the rectangle is 60 units by 32 units.

- Horizontal toothpicks: there are $32+1 = 33$ horizontal lines, each made of 60 toothpicks: $33 \times 60 = 1980$.
- Vertical toothpicks: there are $60+1 = 61$ vertical lines, each made of 32 toothpicks: $61 \times 32 = 1952$.

Total: $1980 + 1952 = 3932$.

(1980 and 1952 are the individual pieces — tempting partial answers.)

ANSWER 3: E

# Problem 4:
The expression is $\boxed{\phantom{2}}^{\boxed{\phantom{2}}} \times \boxed{\phantom{2}}^{\boxed{\phantom{2}}}$, using digits 2, 0, 2, 3 once each.

To maximize, make one factor as large as possible and let the other equal 1 (using exponent 0).

- $3^2 \times 2^0 = 9 \times 1 = 9$
- $2^3 \times 2^0 = 8$ (uses two 2's as base and exponent... gives 8)
- $2^2 \times 3^0 = 4$
- $2^0 \times 3^2 = 9$

Any arrangement with 0 as a base kills a factor to 0. The maximum is $9$.

(8 is the tempting near-miss from $2^3$; 16 and 18 aren't attainable.)

ANSWER 4: C

# Problem 5:
Volume of a cylinder: $V=\pi r^2 h$.

- Alex: diameter 6 → $r=3$, $h=12$: $V = \pi(3^2)(12) = 108\pi$.
- Felicia: diameter 12 → $r=6$, $h=6$: $V = \pi(6^2)(6) = 216\pi$.

Ratio $= 108\pi : 216\pi = 1:2$.

(1:1 is the trap — swapping diameter and height does *not* preserve volume, since radius is squared.)

ANSWER 5: B

# Problem 6:
Let the two equal sides be $a$ and the base be $b$, with $2a + b = 23$.

Since $2a$ is even, $b$ must be odd. Triangle inequality: $2a > b$, i.e. $23 - b > b \Rightarrow b < 11.5$.

So $b \in \{1,3,5,7,9,11\}$, giving $a = 11,10,9,8,7,6$ — six triangles:
(11,11,1), (10,10,3), (9,9,5), (8,8,7), (7,7,9), (6,6,11).

That's 6. (No equilateral option since 23 isn't divisible by 3; $b=13$ fails since $5+5<13$.)

ANSWER 6: C

# Problem 7:
Harold ate $\frac14$, leaving $\frac34$.

Moose ate $\frac13$ of that, leaving $\frac23 \cdot \frac34 = \frac12$.

Porcupine ate $\frac13$ of that, leaving $\frac23 \cdot \frac12 = \frac13$.

(Trap: subtracting $\frac14+\frac13+\frac13$ from 1 gives $\frac1{12}$ — wrong, because each animal eats a fraction of the *remainder*, not of the whole pie.)

ANSWER 7: D

# Problem 8:
Total money $10, minus the $2 tip → $8 available for fare.

First $\frac12$ mile costs $2.40, leaving $8.00 - 2.40 = \$5.60$.

Each additional 0.1 mile costs $0.20: $5.60 \div 0.20 = 28$ increments → $28 \times 0.1 = 2.8$ miles.

Total distance $= 0.5 + 2.8 = 3.3$ miles.

(Forgetting the tip gives $\$10-\$2.40=\$7.60 \to 3.8$ mi; forgetting the first half-mile is included gives other traps like 3.5.)

ANSWER 8: C

# Problem 9:
Population triples every 25 years, starting at 200 in 1998.

- 1998: 200
- 2023: 600
- 2048: 1800

The year 2050 is just 2 years past 2048, so the population is a bit more than 1800 — about 2000.

(3000 would require another near-tripling; 1000 is too small.)

ANSWER 9: D

# Problem 10:
The wall is 19 ft wide; the picture is 3 ft across and centered. The leftover wall splits equally on both sides:
$$\frac{19-3}{2} = \frac{16}{2} = 8 \text{ feet}.$$

(16 is the total leftover, not one side; $9\frac12$ is the wall's midpoint.)

ANSWER 10: B

Problem 1:

Initially, \(80\%\) of 500 balls are red:

\[
0.80(500)=400
\]

So there are 400 red balls and 100 blue balls.

Suppose \(x\) red balls are removed. Then:

- Red balls remaining: \(400-x\)
- Total balls remaining: \(500-x\)

We want \(75\%\) of the remaining balls to be red:

\[
\frac{400-x}{500-x}=\frac34
\]

Cross-multiply:

\[
4(400-x)=3(500-x)
\]

\[
1600-4x=1500-3x
\]

\[
100=x
\]

Removing 100 red balls leaves 300 red and 100 blue, for 400 total. Indeed,

\[
\frac{300}{400}=75\%.
\]

Choices such as 50 or 75 do not reduce the red percentage enough; 150 reduces it too much.

ANSWER 1: D

Problem 2:

Each child can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so \(1\) arrangement.
- All 4 girls: only \(GGGG\), so \(1\) arrangement.
- Exactly 2 girls and 2 boys:

\[
\binom42=6
\]

arrangements.
- Three of one gender and one of the other:
  - 3 boys and 1 girl: \(\binom41=4\) arrangements.
  - 3 girls and 1 boy: \(\binom41=4\) arrangements.

So this outcome has

\[
4+4=8
\]

arrangements.

Since \(8\) is more than \(6\) or \(1\), the most likely outcome is 3 children of one gender and 1 of the other.

ANSWER 2: D

Problem 3:

A grid 60 toothpicks long and 32 toothpicks wide has 60 columns and 32 rows of small squares.

For the horizontal toothpicks, there are \(32+1=33\) horizontal lines, each containing 60 toothpicks:

\[
33\cdot 60=1980
\]

For the vertical toothpicks, there are \(60+1=61\) vertical lines, each containing 32 toothpicks:

\[
61\cdot 32=1952
\]

Thus the total number is

\[
1980+1952=3932.
\]

The tempting answer \(1920=60\cdot32\) counts only the small squares, not the toothpicks forming their edges.

ANSWER 3: E

Problem 4:

The expression is of the form

\[
\Box^{\Box}\times\Box^{\Box},
\]

using the digits \(2,0,2,3\) once each.

To make the product as large as possible, the digit \(0\) should be an exponent, because any nonzero number to the zero power equals 1:

\[
2^0=1 \qquad\text{or}\qquad 3^0=1.
\]

We should use the remaining digits to make the largest possible power. Compare:

\[
3^2=9
\]

and

\[
2^3=8.
\]

Thus the best arrangement is

\[
3^2\times 2^0=9\times 1=9.
\]

Using \(2^3\) gives only \(8\), and placing 0 as a base would make the entire product 0.

ANSWER 4: C

Problem 5:

The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s can has diameter 6 cm, so its radius is 3 cm and its height is 12 cm:

\[
V_A=\pi(3)^2(12)=108\pi.
\]

Felicia’s can has diameter 12 cm, so its radius is 6 cm and its height is 6 cm:

\[
V_F=\pi(6)^2(6)=216\pi.
\]

The ratio is

\[
108\pi:216\pi=1:2.
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, making her base area four times as large.

ANSWER 5: B

Problem 6:

Let the equal sides of the isosceles triangle each have length \(a\). Let the base have length \(b\).

Since the perimeter is 23,

\[
2a+b=23,
\]

so

\[
b=23-2a.
\]

For a valid triangle, the base must be less than the sum of the equal sides:

\[
b<2a.
\]

Substitute \(b=23-2a\):

\[
23-2a<2a
\]

\[
23<4a
\]

\[
a>5.75.
\]

Also, \(b\) must be positive:

\[
23-2a>0,
\]

so

\[
a<11.5.
\]

Thus \(a\) can be

\[
6,7,8,9,10,11.
\]

That gives 6 different triangles. Their side lengths are:

\[
(6,6,11),\ (7,7,9),\ (8,8,7),\ (9,9,5),\ (10,10,3),\ (11,11,1).
\]

Values below 6 fail the triangle inequality, while values above 11 give a nonpositive base.

ANSWER 6: C

Problem 7:

Harold eats \(\frac14\) of the pie, leaving

\[
1-\frac14=\frac34.
\]

The moose eats \(\frac13\) of what is left, so it leaves \(\frac23\) of what Harold left:

\[
\frac23\cdot\frac34=\frac12.
\]

Then the porcupine eats \(\frac13\) of that remaining \(\frac12\), leaving \(\frac23\) of it:

\[
\frac23\cdot\frac12=\frac13.
\]

So \(\frac13\) of the original pie remains.

A common mistake is to subtract \(\frac13\) directly from \(\frac34\); the moose eats one-third of the amount currently left, not one-third of the original pie.

ANSWER 7: D

Problem 8:

The total amount available for the taxi fare is

\[
\$10-\$2=\$8.
\]

The first \(\frac12\) mile costs \$2.40. Therefore, the amount left for additional distance is

\[
8.00-2.40=5.60.
\]

Each additional \(0.1\) mile costs \$0.20, so the number of additional tenths of a mile is

\[
\frac{5.60}{0.20}=28.
\]

That is

\[
28(0.1)=2.8
\]

additional miles.

Adding the first \(\frac12\) mile:

\[
2.8+0.5=3.3.
\]

The ride can be 3.3 miles. The \$2 tip must be included, so choices based on spending the full \$10 on fare are too large.

ANSWER 8: C

Problem 9:

The population is 200 in 1998 and triples every 25 years.

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is a little more than two 25-year periods.

After 25 years, the population would be

\[
200\cdot 3=600.
\]

After 50 years, it would be

\[
600\cdot 3=1800.
\]

Since 2050 is about 2 years later, the population should be slightly more than 1800. The closest estimate is 2000.

Choices such as 600 correspond to only one tripling, while 3000 is much too high for just over two tripling periods.

ANSWER 9: D

Problem 10:

The wall is 19 feet wide, and the picture is 3 feet wide.

The amount of wall not covered by the picture is

\[
19-3=16
\]

feet.

Since the picture is centered, this 16 feet is divided equally between the two ends:

\[
\frac{16}{2}=8.
\]

Therefore, the nearest edge of the picture is 8 feet from either end of the wall.

ANSWER 10: B

Problem 1:
We need how many red balls to take away so reds are 75% of what is left.

At first there are 500 balls, 80% red.
Red = 0.80 × 500 = 400. Blue = 500 - 400 = 100.

Let x = number of red balls removed. Blue balls are not touched.
Red left = 400 - x. Total left = 500 - x.

We want:
(400 - x) / (500 - x) = 75% = 3/4

Cross-multiply:
4(400 - x) = 3(500 - x)
1600 - 4x = 1500 - 3x
1600 - 1500 = 4x - 3x
100 = x

Check: 300 red left, 100 blue left, total 400. 300/400 = 3/4.
Removing 25 would leave 375/475 ≈ 79%, still too high. Removing 150 would leave 250/350 ≈ 71%, too low.

ANSWER 1: D

Problem 2:
We are asked which gender pattern for 4 babies is most likely, each baby boy/girl equally likely.

There are 2^4 = 16 equally likely boy-girl sequences.

A. all 4 boys: only BBBB: 1 sequence. Probability 1/16.
B. all 4 girls: only GGGG: 1 sequence. Probability 1/16.
C. 2 girls and 2 boys: number of sequences = choose which 2 of 4 are girls = C(4,2)=6. Probability 6/16.
D. 3 of one gender and 1 of the other: This means 3B1G OR 3G1B.
3B1G: C(4,1)=4 sequences. 3G1B: C(4,1)=4 sequences. Total 8 sequences. Probability 8/16 = 1/2.
E. claims equal, false.

Many think 2-2 is most likely because it is most "balanced", but 3-1 has more sequences: 8 > 6 > 1.

So D is most likely.

ANSWER 2: D

Problem 3:
We need total toothpicks to make a rectangular grid 60 long by 32 wide, where length is measured in toothpicks.

Imagine unit squares. Top edge needs 60 toothpicks side by side. Same for every horizontal line.

Number of horizontal lines: there are 32 squares vertically, so 32+1 = 33 horizontal grid lines.
Each horizontal line uses 60 toothpicks.
Horizontal total = 33 × 60 = 1980.

Number of vertical lines: there are 60 squares horizontally, so 60+1 = 61 vertical grid lines.
Each vertical line uses 32 toothpicks.
Vertical total = 61 × 32 = 1952.

Total = 1980 + 1952 = 3932.

A 1920 = 60×32 is just the number of cells, not toothpicks. B 1952 is vertical only. C 1980 is horizontal only.

ANSWER 3: E

Problem 4:
We are asked to put digits 2,0,2,3, one per box, into the expression
□^□ × □^□
to get the largest value.

We must use all four digits as two bases and two exponents.

Key facts: a^0 = 1 for a ≠ 0, while 0^positive = 0, which kills a product. So do not put 0 as a base with positive exponent. Put 0 as an exponent to get 1. 0^0 is not allowed.

Try cases:

If exponents are 2 and 0, bases 3 and 2:
3^2 × 2^0 = 9 × 1 = 9
2^2 × 3^0 = 4 × 1 = 4
3^0 × 2^2 = 1 × 4 = 4
2^0 × 3^2 = 1 × 9 = 9

If exponents are 3 and 0, bases 2 and 2:
2^3 × 2^0 = 8 × 1 = 8

If 0 is a base with positive exponent, product is 0, e.g. 3^2 × 0^2 = 0.

So the best is 9.

Why not other choices?
A. 0: tempting if you think any zero makes everything zero, but 2^0 =1 avoids zero.
B. 8: this is 2^3 × 2^0, valid but less than 9.
D. 16 and E. 18: these come from the mistake x^0 = x. E.g. 3^2 × 2 =18, 2^3 × 2 =16. Correct is ×1, not ×2.

Maximum is 9.

ANSWER 4: C

Problem 5:
We compare volumes of cylinders. Volume = π r^2 h, r = diameter/2.

Alex: diameter 6, so r = 3. h =12.
V_Alex = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r = 6. h =6.
V_Felicia = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio Alex : Felicia = 108π : 216π = 108:216 = 1:2.

Volume scales as square of radius, so doubling diameter quadruples base area, which outweighs halving height. So Felicia's is bigger, ruling out C,D,E.

ANSWER 5: B

Problem 6:
We need integer-sided isosceles triangles with perimeter 23.

Let equal sides be a,a and base be b. Then:
2a + b = 23, so b = 23 - 2a > 0, so a ≤ 11.

Triangle inequality: sum of two small sides > third side. The only non-trivial one is:
a + a > b
2a > 23 - 2a
4a > 23
a > 5.75, so a ≥ 6.

Thus a = 6,7,8,9,10,11. Each gives a b:
6,6,11: 6+6>11 valid
7,7,9: 14>9 valid
8,8,7: valid
9,9,5: valid
10,10,3: 10+3>10 valid
11,11,1: 11+1>11 valid

That is 6 different triangles. No other a works.

ANSWER 6: C

Problem 7:
We track fraction of original pie left.

Harold eats 1/4, so left = 1 - 1/4 = 3/4.

Moose eats 1/3 of what Harold left:
eaten = (1/3)×(3/4) = 1/4.
Left after moose = 3/4 - 1/4 = 1/2. Equivalently (2/3)×(3/4)=1/2.

Porcupine eats 1/3 of what moose left:
eaten = (1/3)×(1/2) = 1/6.
Left after porcupine = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3. Equivalently (2/3)×(1/2)=1/3.

So 1/3 remains. A 1/12, B 1/6 are the amounts eaten, not left.

ANSWER 7: D

Problem 8:
We have $10 total, $2 tip, so fare money = 10 - 2 = $8.00.

First 0.5 mile costs $2.40.
Remaining for extra distance = 8.00 - 2.40 = $5.60.

Extra is $0.20 per 0.1 mile.
Number of 0.1-mile blocks = 5.60 / 0.20 = 28.
Extra distance = 28 × 0.1 = 2.8 miles.

Total distance = 0.5 + 2.8 = 3.3 miles.

Check: fare = 2.40 + 28×0.20 = 2.40+5.60=8.00, plus $2 tip = $10.

ANSWER 8: C

Problem 9:
Population in 1998 is 200, triples every 25 years. Estimate in 2050.

Time elapsed = 2050 - 1998 = 52 years. That's about two 25-year periods.

After 25 yr (~2023): 200×3 = 600.
After 50 yr (~2048): 600×3 = 1800.

2050 is 2 years more, so a little above 1800. Nearest estimate among choices is 2000.

A 600 is only one tripling. C 1000, B 800 ignore tripling. E 3000 is too high.

ANSWER 9: D

Problem 10:
Picture 3 ft wide centered on 19 ft wall.

Total wall not covered = 19 - 3 = 16 ft.
Centered means half on each side: 16/2 = 8 ft from each end to picture edge.

A 1 1/2, C 9 1/2 confuse center vs edge. B is correct.

ANSWER 10: B

Problem 1:
We are asked to find how many red balls must be removed from a bag of 500 balls (originally 80% red) so that the remaining red balls constitute 75% of the bag.

First, calculate the original number of balls of each color:
- Total balls = $500$
- Red balls = $80\% \text{ of } 500 = 0.80 \times 500 = 400$
- Blue balls = $500 - 400 = 100$

Only red balls are removed, so the number of blue balls remains constant at $100$.
In the final state, red balls make up $75\%$ of the total, which means blue balls must make up:
$$100\% - 75\% = 25\%$$
of the remaining balls.

Let $T$ be the total number of remaining balls:
$$0.25 \times T = 100 \implies T = \frac{100}{0.25} = 400$$

Since the bag originally contained $500$ balls and now contains $400$ balls, the number of red balls removed is:
$$500 - 400 = 100$$

ANSWER 1: D

---

Problem 2:
We are asked to determine which outcome is most likely when four children are born, with boys (B) and girls (G) being equally likely.

Each child has $2$ possible genders, so there are $2^4 = 16$ equally likely possible outcomes in total. We evaluate the number of outcomes for each choice:
- **A. All 4 are boys:** Only $1$ outcome ($\text{BBBB}$). Probability = $\frac{1}{16}$.
- **B. All 4 are girls:** Only $1$ outcome ($\text{GGGG}$). Probability = $\frac{1}{16}$.
- **C. 2 are girls and 2 are boys:** The number of ways to choose which $2$ children are girls is $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ outcomes. Probability = $\frac{6}{16}$.
- **D. 3 are of one gender and 1 is of the other:** This includes either 3 boys and 1 girl ($\binom{4}{1} = 4$ outcomes) or 3 girls and 1 boy ($\binom{4}{3} = 4$ outcomes), giving $4 + 4 = 8$ outcomes. Probability = $\frac{8}{16} = \frac{1}{2}$.

Comparing the possibilities, outcome D has $8$ favorable outcomes, which is greater than C ($6$ outcomes), A ($1$ outcome), and B ($1$ outcome). Therefore, D is the most likely outcome.

ANSWER 2: D

---

Problem 3:
We are asked to find the total number of toothpicks used to make a rectangular grid that is $60$ toothpicks long and $32$ toothpicks wide.

A grid of dimensions $60 \times 32$ consists of unit squares arranged in $32$ rows and $60$ columns:
- **Horizontal toothpicks:** There are $32 + 1 = 33$ horizontal grid lines. Each horizontal line is made of $60$ toothpicks.
  $$\text{Number of horizontal toothpicks} = 33 \times 60 = 1980$$
- **Vertical toothpicks:** There are $60 + 1 = 61$ vertical grid lines. Each vertical line is made of $32$ toothpicks.
  $$\text{Number of vertical toothpicks} = 61 \times 32 = 1952$$

Adding the horizontal and vertical toothpicks together:
$$\text{Total toothpicks} = 1980 + 1952 = 3932$$

ANSWER 3: E

---

Problem 4:
We are asked to find the maximum possible value of the expression $\square^\square \times \square^\square$ when placing the digits $2, 0, 2,$ and $3$ into the four boxes, one digit per box.

Let the expression be $a^b \times c^d$, where $\{a, b, c, d\} = \{2, 0, 2, 3\}$.
We test the possible assignments of the digits:
- To maximize the product, we want large bases and exponents, but one of the digits is $0$.
- If $0$ is used as a base, say $0^d$ with $d > 0$, the factor becomes $0$, making the entire product $0$.
- If $0$ is placed in an exponent, recall that for any non-zero base $x$, $x^0 = 1$.
  - Pair $0$ with $2$ as an exponent: $2^0 = 1$. The remaining digits are $3$ and $2$.
    - Setting $3^2 \times 2^0 = 9 \times 1 = 9$.
    - Setting $2^3 \times 2^0 = 8 \times 1 = 8$.
  - Pair $0$ with $3$ as an exponent: $3^0 = 1$. The remaining digits are $2$ and $2$.
    - Setting $2^2 \times 3^0 = 4 \times 1 = 4$.

The values $16$ and $18$ are distractors arising from the common misconception that $x^0 = x$ (which would give $2^3 \times 2 = 16$ or $3^2 \times 2 = 18$).
The maximum mathematically valid value achievable is $9$.

ANSWER 4: C

---

Problem 5:
We are asked to find the ratio of the volume of Alex's cat food can to the volume of Felicia's cat food can.

The volume of a cylinder is given by $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height.
- For Alex's can:
  - Diameter = $6\text{ cm} \implies \text{radius } r_A = 3\text{ cm}$
  - Height $h_A = 12\text{ cm}$
  - Volume $V_A = \pi (3)^2 (12) = \pi \times 9 \times 12 = 108\pi\text{ cm}^3$

- For Felicia's can:
  - Diameter = $12\text{ cm} \implies \text{radius } r_F = 6\text{ cm}$
  - Height $h_F = 6\text{ cm}$
  - Volume $V_F = \pi (6)^2 (6) = \pi \times 36 \times 6 = 216\pi\text{ cm}^3$

The ratio of the volume of Alex's can to Felicia's can is:
$$\frac{V_A}{V_F} = \frac{108\pi}{216\pi} = \frac{1}{2}$$

Thus, the ratio is $1:2$.

ANSWER 5: B

---

Problem 6:
We are asked to find the number of different isosceles triangles with integer side lengths and a perimeter of $23$.

Let the side lengths of the isosceles triangle be $a, a,$ and $b$, where $a$ and $b$ are positive integers.
The perimeter is:
$$2a + b = 23 \implies b = 23 - 2a$$

By the Triangle Inequality:
1. The sum of the two equal sides must be strictly greater than the third side:
   $$a + a > b \implies 2a > 23 - 2a \implies 4a > 23 \implies a \ge 6$$
2. The third side must be positive:
   $$b > 0 \implies 23 - 2a > 0 \implies 2a < 23 \implies a \le 11$$

The possible integer values for $a$ are $6, 7, 8, 9, 10, 11$.
We check each corresponding side length triple $(a, a, b)$:
- $a = 6 \implies b = 11$: sides $(6, 6, 11)$
- $a = 7 \implies b = 9$: sides $(7, 7, 9)$
- $a = 8 \implies b = 7$: sides $(8, 8, 7)$
- $a = 9 \implies b = 5$: sides $(9, 9, 5)$
- $a = 10 \implies b = 3$: sides $(10, 10, 3)$
- $a = 11 \implies b = 1$: sides $(11, 11, 1)$

All $6$ sets of side lengths satisfy the triangle inequality and have two equal sides (none are equilateral since $23$ is not divisible by $3$). Thus, there are $6$ such triangles.

ANSWER 6: C

---

Problem 7:
We are asked to find the fraction of the original pie remaining after Harold, a moose, and a porcupine each eat a portion.

1. Harold start

# Problem 1:
Restate: 500 balls, 80% red. Remove some **red** balls so that 75% of what remains is red.

Red = 0.80(500) = 400, Blue = 100.

Let $x$ = red balls removed. Blues never change (still 100), and after removal blues must be 25% of the total:
$$100 = 0.25(500 - x) \Rightarrow 400 = 500 - x \Rightarrow x = 100.$$

Check: 300 red, 100 blue, total 400 → 300/400 = 75%. ✓

(Tempting wrong: 25 or 50 come from mistakenly taking 5% or 10% of 500; but removing balls shrinks the total too, so more must go.)

ANSWER 1: D

# Problem 2:
There are $2^4 = 16$ equally likely sequences of genders.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: $\binom{4}{2}=6$ ways → 6/16
- 3 of one gender, 1 of the other: $\binom{4}{3}=4$ (3 boys) $+ \binom{4}{1}=4$ (3 girls) $= 8$ ways → 8/16

The largest is 8/16 = 1/2, the "3–1 split." Choice C (6/16) is the tempting trap because 2–2 is the single most likely *exact* split, but option D combines two cases.

ANSWER 2: D

# Problem 3:
A grid 60 toothpicks long and 32 toothpicks wide means the rectangle is 60 units by 32 units.

- Horizontal toothpicks: there are $32+1 = 33$ horizontal lines, each made of 60 toothpicks: $33 \times 60 = 1980$.
- Vertical toothpicks: there are $60+1 = 61$ vertical lines, each made of 32 toothpicks: $61 \times 32 = 1952$.

Total: $1980 + 1952 = 3932$.

(1980 and 1952 are the individual pieces — tempting partial answers.)

ANSWER 3: E

# Problem 4:
The expression is $\boxed{\phantom{2}}^{\boxed{\phantom{2}}} \times \boxed{\phantom{2}}^{\boxed{\phantom{2}}}$, using digits 2, 0, 2, 3 once each.

To maximize, make one factor as large as possible and let the other equal 1 (using exponent 0).

- $3^2 \times 2^0 = 9 \times 1 = 9$
- $2^3 \times 2^0 = 8$ (uses two 2's as base and exponent... gives 8)
- $2^2 \times 3^0 = 4$
- $2^0 \times 3^2 = 9$

Any arrangement with 0 as a base kills a factor to 0. The maximum is $9$.

(8 is the tempting near-miss from $2^3$; 16 and 18 aren't attainable.)

ANSWER 4: C

# Problem 5:
Volume of a cylinder: $V=\pi r^2 h$.

- Alex: diameter 6 → $r=3$, $h=12$: $V = \pi(3^2)(12) = 108\pi$.
- Felicia: diameter 12 → $r=6$, $h=6$: $V = \pi(6^2)(6) = 216\pi$.

Ratio $= 108\pi : 216\pi = 1:2$.

(1:1 is the trap — swapping diameter and height does *not* preserve volume, since radius is squared.)

ANSWER 5: B

# Problem 6:
Let the two equal sides be $a$ and the base be $b$, with $2a + b = 23$.

Since $2a$ is even, $b$ must be odd. Triangle inequality: $2a > b$, i.e. $23 - b > b \Rightarrow b < 11.5$.

So $b \in \{1,3,5,7,9,11\}$, giving $a = 11,10,9,8,7,6$ — six triangles:
(11,11,1), (10,10,3), (9,9,5), (8,8,7), (7,7,9), (6,6,11).

That's 6. (No equilateral option since 23 isn't divisible by 3; $b=13$ fails since $5+5<13$.)

ANSWER 6: C

# Problem 7:
Harold ate $\frac14$, leaving $\frac34$.

Moose ate $\frac13$ of that, leaving $\frac23 \cdot \frac34 = \frac12$.

Porcupine ate $\frac13$ of that, leaving $\frac23 \cdot \frac12 = \frac13$.

(Trap: subtracting $\frac14+\frac13+\frac13$ from 1 gives $\frac1{12}$ — wrong, because each animal eats a fraction of the *remainder*, not of the whole pie.)

ANSWER 7: D

# Problem 8:
Total money $10, minus the $2 tip → $8 available for fare.

First $\frac12$ mile costs $2.40, leaving $8.00 - 2.40 = \$5.60$.

Each additional 0.1 mile costs $0.20: $5.60 \div 0.20 = 28$ increments → $28 \times 0.1 = 2.8$ miles.

Total distance $= 0.5 + 2.8 = 3.3$ miles.

(Forgetting the tip gives $\$10-\$2.40=\$7.60 \to 3.8$ mi; forgetting the first half-mile is included gives other traps like 3.5.)

ANSWER 8: C

# Problem 9:
Population triples every 25 years, starting at 200 in 1998.

- 1998: 200
- 2023: 600
- 2048: 1800

The year 2050 is just 2 years past 2048, so the population is a bit more than 1800 — about 2000.

(3000 would require another near-tripling; 1000 is too small.)

ANSWER 9: D

# Problem 10:
The wall is 19 ft wide; the picture is 3 ft across and centered. The leftover wall splits equally on both sides:
$$\frac{19-3}{2} = \frac{16}{2} = 8 \text{ feet}.$$

(16 is the total leftover, not one side; $9\frac12$ is the wall's midpoint.)

ANSWER 10: B

Problem 1:

Initially, \(80\%\) of 500 balls are red:

\[
0.80(500)=400
\]

So there are 400 red balls and 100 blue balls.

Suppose \(x\) red balls are removed. Then:

- Red balls remaining: \(400-x\)
- Total balls remaining: \(500-x\)

We want \(75\%\) of the remaining balls to be red:

\[
\frac{400-x}{500-x}=\frac34
\]

Cross-multiply:

\[
4(400-x)=3(500-x)
\]

\[
1600-4x=1500-3x
\]

\[
100=x
\]

Removing 100 red balls leaves 300 red and 100 blue, for 400 total. Indeed,

\[
\frac{300}{400}=75\%.
\]

Choices such as 50 or 75 do not reduce the red percentage enough; 150 reduces it too much.

ANSWER 1: D

Problem 2:

Each child can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so \(1\) arrangement.
- All 4 girls: only \(GGGG\), so \(1\) arrangement.
- Exactly 2 girls and 2 boys:

\[
\binom42=6
\]

arrangements.
- Three of one gender and one of the other:
  - 3 boys and 1 girl: \(\binom41=4\) arrangements.
  - 3 girls and 1 boy: \(\binom41=4\) arrangements.

So this outcome has

\[
4+4=8
\]

arrangements.

Since \(8\) is more than \(6\) or \(1\), the most likely outcome is 3 children of one gender and 1 of the other.

ANSWER 2: D

Problem 3:

A grid 60 toothpicks long and 32 toothpicks wide has 60 columns and 32 rows of small squares.

For the horizontal toothpicks, there are \(32+1=33\) horizontal lines, each containing 60 toothpicks:

\[
33\cdot 60=1980
\]

For the vertical toothpicks, there are \(60+1=61\) vertical lines, each containing 32 toothpicks:

\[
61\cdot 32=1952
\]

Thus the total number is

\[
1980+1952=3932.
\]

The tempting answer \(1920=60\cdot32\) counts only the small squares, not the toothpicks forming their edges.

ANSWER 3: E

Problem 4:

The expression is of the form

\[
\Box^{\Box}\times\Box^{\Box},
\]

using the digits \(2,0,2,3\) once each.

To make the product as large as possible, the digit \(0\) should be an exponent, because any nonzero number to the zero power equals 1:

\[
2^0=1 \qquad\text{or}\qquad 3^0=1.
\]

We should use the remaining digits to make the largest possible power. Compare:

\[
3^2=9
\]

and

\[
2^3=8.
\]

Thus the best arrangement is

\[
3^2\times 2^0=9\times 1=9.
\]

Using \(2^3\) gives only \(8\), and placing 0 as a base would make the entire product 0.

ANSWER 4: C

Problem 5:

The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s can has diameter 6 cm, so its radius is 3 cm and its height is 12 cm:

\[
V_A=\pi(3)^2(12)=108\pi.
\]

Felicia’s can has diameter 12 cm, so its radius is 6 cm and its height is 6 cm:

\[
V_F=\pi(6)^2(6)=216\pi.
\]

The ratio is

\[
108\pi:216\pi=1:2.
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, making her base area four times as large.

ANSWER 5: B

Problem 6:

Let the equal sides of the isosceles triangle each have length \(a\). Let the base have length \(b\).

Since the perimeter is 23,

\[
2a+b=23,
\]

so

\[
b=23-2a.
\]

For a valid triangle, the base must be less than the sum of the equal sides:

\[
b<2a.
\]

Substitute \(b=23-2a\):

\[
23-2a<2a
\]

\[
23<4a
\]

\[
a>5.75.
\]

Also, \(b\) must be positive:

\[
23-2a>0,
\]

so

\[
a<11.5.
\]

Thus \(a\) can be

\[
6,7,8,9,10,11.
\]

That gives 6 different triangles. Their side lengths are:

\[
(6,6,11),\ (7,7,9),\ (8,8,7),\ (9,9,5),\ (10,10,3),\ (11,11,1).
\]

Values below 6 fail the triangle inequality, while values above 11 give a nonpositive base.

ANSWER 6: C

Problem 7:

Harold eats \(\frac14\) of the pie, leaving

\[
1-\frac14=\frac34.
\]

The moose eats \(\frac13\) of what is left, so it leaves \(\frac23\) of what Harold left:

\[
\frac23\cdot\frac34=\frac12.
\]

Then the porcupine eats \(\frac13\) of that remaining \(\frac12\), leaving \(\frac23\) of it:

\[
\frac23\cdot\frac12=\frac13.
\]

So \(\frac13\) of the original pie remains.

A common mistake is to subtract \(\frac13\) directly from \(\frac34\); the moose eats one-third of the amount currently left, not one-third of the original pie.

ANSWER 7: D

Problem 8:

The total amount available for the taxi fare is

\[
\$10-\$2=\$8.
\]

The first \(\frac12\) mile costs \$2.40. Therefore, the amount left for additional distance is

\[
8.00-2.40=5.60.
\]

Each additional \(0.1\) mile costs \$0.20, so the number of additional tenths of a mile is

\[
\frac{5.60}{0.20}=28.
\]

That is

\[
28(0.1)=2.8
\]

additional miles.

Adding the first \(\frac12\) mile:

\[
2.8+0.5=3.3.
\]

The ride can be 3.3 miles. The \$2 tip must be included, so choices based on spending the full \$10 on fare are too large.

ANSWER 8: C

Problem 9:

The population is 200 in 1998 and triples every 25 years.

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is a little more than two 25-year periods.

After 25 years, the population would be

\[
200\cdot 3=600.
\]

After 50 years, it would be

\[
600\cdot 3=1800.
\]

Since 2050 is about 2 years later, the population should be slightly more than 1800. The closest estimate is 2000.

Choices such as 600 correspond to only one tripling, while 3000 is much too high for just over two tripling periods.

ANSWER 9: D

Problem 10:

The wall is 19 feet wide, and the picture is 3 feet wide.

The amount of wall not covered by the picture is

\[
19-3=16
\]

feet.

Since the picture is centered, this 16 feet is divided equally between the two ends:

\[
\frac{16}{2}=8.
\]

Therefore, the nearest edge of the picture is 8 feet from either end of the wall.

ANSWER 10: B

Problem 1:
We need how many red balls to take away so reds are 75% of what is left.

At first there are 500 balls, 80% red.
Red = 0.80 × 500 = 400. Blue = 500 - 400 = 100.

Let x = number of red balls removed. Blue balls are not touched.
Red left = 400 - x. Total left = 500 - x.

We want:
(400 - x) / (500 - x) = 75% = 3/4

Cross-multiply:
4(400 - x) = 3(500 - x)
1600 - 4x = 1500 - 3x
1600 - 1500 = 4x - 3x
100 = x

Check: 300 red left, 100 blue left, total 400. 300/400 = 3/4.
Removing 25 would leave 375/475 ≈ 79%, still too high. Removing 150 would leave 250/350 ≈ 71%, too low.

ANSWER 1: D

Problem 2:
We are asked which gender pattern for 4 babies is most likely, each baby boy/girl equally likely.

There are 2^4 = 16 equally likely boy-girl sequences.

A. all 4 boys: only BBBB: 1 sequence. Probability 1/16.
B. all 4 girls: only GGGG: 1 sequence. Probability 1/16.
C. 2 girls and 2 boys: number of sequences = choose which 2 of 4 are girls = C(4,2)=6. Probability 6/16.
D. 3 of one gender and 1 of the other: This means 3B1G OR 3G1B.
3B1G: C(4,1)=4 sequences. 3G1B: C(4,1)=4 sequences. Total 8 sequences. Probability 8/16 = 1/2.
E. claims equal, false.

Many think 2-2 is most likely because it is most "balanced", but 3-1 has more sequences: 8 > 6 > 1.

So D is most likely.

ANSWER 2: D

Problem 3:
We need total toothpicks to make a rectangular grid 60 long by 32 wide, where length is measured in toothpicks.

Imagine unit squares. Top edge needs 60 toothpicks side by side. Same for every horizontal line.

Number of horizontal lines: there are 32 squares vertically, so 32+1 = 33 horizontal grid lines.
Each horizontal line uses 60 toothpicks.
Horizontal total = 33 × 60 = 1980.

Number of vertical lines: there are 60 squares horizontally, so 60+1 = 61 vertical grid lines.
Each vertical line uses 32 toothpicks.
Vertical total = 61 × 32 = 1952.

Total = 1980 + 1952 = 3932.

A 1920 = 60×32 is just the number of cells, not toothpicks. B 1952 is vertical only. C 1980 is horizontal only.

ANSWER 3: E

Problem 4:
We are asked to put digits 2,0,2,3, one per box, into the expression
□^□ × □^□
to get the largest value.

We must use all four digits as two bases and two exponents.

Key facts: a^0 = 1 for a ≠ 0, while 0^positive = 0, which kills a product. So do not put 0 as a base with positive exponent. Put 0 as an exponent to get 1. 0^0 is not allowed.

Try cases:

If exponents are 2 and 0, bases 3 and 2:
3^2 × 2^0 = 9 × 1 = 9
2^2 × 3^0 = 4 × 1 = 4
3^0 × 2^2 = 1 × 4 = 4
2^0 × 3^2 = 1 × 9 = 9

If exponents are 3 and 0, bases 2 and 2:
2^3 × 2^0 = 8 × 1 = 8

If 0 is a base with positive exponent, product is 0, e.g. 3^2 × 0^2 = 0.

So the best is 9.

Why not other choices?
A. 0: tempting if you think any zero makes everything zero, but 2^0 =1 avoids zero.
B. 8: this is 2^3 × 2^0, valid but less than 9.
D. 16 and E. 18: these come from the mistake x^0 = x. E.g. 3^2 × 2 =18, 2^3 × 2 =16. Correct is ×1, not ×2.

Maximum is 9.

ANSWER 4: C

Problem 5:
We compare volumes of cylinders. Volume = π r^2 h, r = diameter/2.

Alex: diameter 6, so r = 3. h =12.
V_Alex = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r = 6. h =6.
V_Felicia = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio Alex : Felicia = 108π : 216π = 108:216 = 1:2.

Volume scales as square of radius, so doubling diameter quadruples base area, which outweighs halving height. So Felicia's is bigger, ruling out C,D,E.

ANSWER 5: B

Problem 6:
We need integer-sided isosceles triangles with perimeter 23.

Let equal sides be a,a and base be b. Then:
2a + b = 23, so b = 23 - 2a > 0, so a ≤ 11.

Triangle inequality: sum of two small sides > third side. The only non-trivial one is:
a + a > b
2a > 23 - 2a
4a > 23
a > 5.75, so a ≥ 6.

Thus a = 6,7,8,9,10,11. Each gives a b:
6,6,11: 6+6>11 valid
7,7,9: 14>9 valid
8,8,7: valid
9,9,5: valid
10,10,3: 10+3>10 valid
11,11,1: 11+1>11 valid

That is 6 different triangles. No other a works.

ANSWER 6: C

Problem 7:
We track fraction of original pie left.

Harold eats 1/4, so left = 1 - 1/4 = 3/4.

Moose eats 1/3 of what Harold left:
eaten = (1/3)×(3/4) = 1/4.
Left after moose = 3/4 - 1/4 = 1/2. Equivalently (2/3)×(3/4)=1/2.

Porcupine eats 1/3 of what moose left:
eaten = (1/3)×(1/2) = 1/6.
Left after porcupine = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3. Equivalently (2/3)×(1/2)=1/3.

So 1/3 remains. A 1/12, B 1/6 are the amounts eaten, not left.

ANSWER 7: D

Problem 8:
We have $10 total, $2 tip, so fare money = 10 - 2 = $8.00.

First 0.5 mile costs $2.40.
Remaining for extra distance = 8.00 - 2.40 = $5.60.

Extra is $0.20 per 0.1 mile.
Number of 0.1-mile blocks = 5.60 / 0.20 = 28.
Extra distance = 28 × 0.1 = 2.8 miles.

Total distance = 0.5 + 2.8 = 3.3 miles.

Check: fare = 2.40 + 28×0.20 = 2.40+5.60=8.00, plus $2 tip = $10.

ANSWER 8: C

Problem 9:
Population in 1998 is 200, triples every 25 years. Estimate in 2050.

Time elapsed = 2050 - 1998 = 52 years. That's about two 25-year periods.

After 25 yr (~2023): 200×3 = 600.
After 50 yr (~2048): 600×3 = 1800.

2050 is 2 years more, so a little above 1800. Nearest estimate among choices is 2000.

A 600 is only one tripling. C 1000, B 800 ignore tripling. E 3000 is too high.

ANSWER 9: D

Problem 10:
Picture 3 ft wide centered on 19 ft wall.

Total wall not covered = 19 - 3 = 16 ft.
Centered means half on each side: 16/2 = 8 ft from each end to picture edge.

A 1 1/2, C 9 1/2 confuse center vs edge. B is correct.

ANSWER 10: B

Problem 1:
We are asked to find how many red balls must be removed from a bag of 500 balls (originally 80% red) so that the remaining red balls constitute 75% of the bag.

First, calculate the original number of balls of each color:
- Total balls = $500$
- Red balls = $80\% \text{ of } 500 = 0.80 \times 500 = 400$
- Blue balls = $500 - 400 = 100$

Only red balls are removed, so the number of blue balls remains constant at $100$.
In the final state, red balls make up $75\%$ of the total, which means blue balls must make up:
$$100\% - 75\% = 25\%$$
of the remaining balls.

Let $T$ be the total number of remaining balls:
$$0.25 \times T = 100 \implies T = \frac{100}{0.25} = 400$$

Since the bag originally contained $500$ balls and now contains $400$ balls, the number of red balls removed is:
$$500 - 400 = 100$$

ANSWER 1: D

---

Problem 2:
We are asked to determine which outcome is most likely when four children are born, with boys (B) and girls (G) being equally likely.

Each child has $2$ possible genders, so there are $2^4 = 16$ equally likely possible outcomes in total. We evaluate the number of outcomes for each choice:
- **A. All 4 are boys:** Only $1$ outcome ($\text{BBBB}$). Probability = $\frac{1}{16}$.
- **B. All 4 are girls:** Only $1$ outcome ($\text{GGGG}$). Probability = $\frac{1}{16}$.
- **C. 2 are girls and 2 are boys:** The number of ways to choose which $2$ children are girls is $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ outcomes. Probability = $\frac{6}{16}$.
- **D. 3 are of one gender and 1 is of the other:** This includes either 3 boys and 1 girl ($\binom{4}{1} = 4$ outcomes) or 3 girls and 1 boy ($\binom{4}{3} = 4$ outcomes), giving $4 + 4 = 8$ outcomes. Probability = $\frac{8}{16} = \frac{1}{2}$.

Comparing the possibilities, outcome D has $8$ favorable outcomes, which is greater than C ($6$ outcomes), A ($1$ outcome), and B ($1$ outcome). Therefore, D is the most likely outcome.

ANSWER 2: D

---

Problem 3:
We are asked to find the total number of toothpicks used to make a rectangular grid that is $60$ toothpicks long and $32$ toothpicks wide.

A grid of dimensions $60 \times 32$ consists of unit squares arranged in $32$ rows and $60$ columns:
- **Horizontal toothpicks:** There are $32 + 1 = 33$ horizontal grid lines. Each horizontal line is made of $60$ toothpicks.
  $$\text{Number of horizontal toothpicks} = 33 \times 60 = 1980$$
- **Vertical toothpicks:** There are $60 + 1 = 61$ vertical grid lines. Each vertical line is made of $32$ toothpicks.
  $$\text{Number of vertical toothpicks} = 61 \times 32 = 1952$$

Adding the horizontal and vertical toothpicks together:
$$\text{Total toothpicks} = 1980 + 1952 = 3932$$

ANSWER 3: E

---

Problem 4:
We are asked to find the maximum possible value of the expression $\square^\square \times \square^\square$ when placing the digits $2, 0, 2,$ and $3$ into the four boxes, one digit per box.

Let the expression be $a^b \times c^d$, where $\{a, b, c, d\} = \{2, 0, 2, 3\}$.
We test the possible assignments of the digits:
- To maximize the product, we want large bases and exponents, but one of the digits is $0$.
- If $0$ is used as a base, say $0^d$ with $d > 0$, the factor becomes $0$, making the entire product $0$.
- If $0$ is placed in an exponent, recall that for any non-zero base $x$, $x^0 = 1$.
  - Pair $0$ with $2$ as an exponent: $2^0 = 1$. The remaining digits are $3$ and $2$.
    - Setting $3^2 \times 2^0 = 9 \times 1 = 9$.
    - Setting $2^3 \times 2^0 = 8 \times 1 = 8$.
  - Pair $0$ with $3$ as an exponent: $3^0 = 1$. The remaining digits are $2$ and $2$.
    - Setting $2^2 \times 3^0 = 4 \times 1 = 4$.

The values $16$ and $18$ are distractors arising from the common misconception that $x^0 = x$ (which would give $2^3 \times 2 = 16$ or $3^2 \times 2 = 18$).
The maximum mathematically valid value achievable is $9$.

ANSWER 4: C

---

Problem 5:
We are asked to find the ratio of the volume of Alex's cat food can to the volume of Felicia's cat food can.

The volume of a cylinder is given by $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height.
- For Alex's can:
  - Diameter = $6\text{ cm} \implies \text{radius } r_A = 3\text{ cm}$
  - Height $h_A = 12\text{ cm}$
  - Volume $V_A = \pi (3)^2 (12) = \pi \times 9 \times 12 = 108\pi\text{ cm}^3$

- For Felicia's can:
  - Diameter = $12\text{ cm} \implies \text{radius } r_F = 6\text{ cm}$
  - Height $h_F = 6\text{ cm}$
  - Volume $V_F = \pi (6)^2 (6) = \pi \times 36 \times 6 = 216\pi\text{ cm}^3$

The ratio of the volume of Alex's can to Felicia's can is:
$$\frac{V_A}{V_F} = \frac{108\pi}{216\pi} = \frac{1}{2}$$

Thus, the ratio is $1:2$.

ANSWER 5: B

---

Problem 6:
We are asked to find the number of different isosceles triangles with integer side lengths and a perimeter of $23$.

Let the side lengths of the isosceles triangle be $a, a,$ and $b$, where $a$ and $b$ are positive integers.
The perimeter is:
$$2a + b = 23 \implies b = 23 - 2a$$

By the Triangle Inequality:
1. The sum of the two equal sides must be strictly greater than the third side:
   $$a + a > b \implies 2a > 23 - 2a \implies 4a > 23 \implies a \ge 6$$
2. The third side must be positive:
   $$b > 0 \implies 23 - 2a > 0 \implies 2a < 23 \implies a \le 11$$

The possible integer values for $a$ are $6, 7, 8, 9, 10, 11$.
We check each corresponding side length triple $(a, a, b)$:
- $a = 6 \implies b = 11$: sides $(6, 6, 11)$
- $a = 7 \implies b = 9$: sides $(7, 7, 9)$
- $a = 8 \implies b = 7$: sides $(8, 8, 7)$
- $a = 9 \implies b = 5$: sides $(9, 9, 5)$
- $a = 10 \implies b = 3$: sides $(10, 10, 3)$
- $a = 11 \implies b = 1$: sides $(11, 11, 1)$

All $6$ sets of side lengths satisfy the triangle inequality and have two equal sides (none are equilateral since $23$ is not divisible by $3$). Thus, there are $6$ such triangles.

ANSWER 6: C

---

Problem 7:
We are asked to find the fraction of the original pie remaining after Harold, a moose, and a porcupine each eat a portion.

1. Harold start

# Problem 1:
Restate: 500 balls, 80% red. Remove some **red** balls so that 75% of what remains is red.

Red = 0.80(500) = 400, Blue = 100.

Let $x$ = red balls removed. Blues never change (still 100), and after removal blues must be 25% of the total:
$$100 = 0.25(500 - x) \Rightarrow 400 = 500 - x \Rightarrow x = 100.$$

Check: 300 red, 100 blue, total 400 → 300/400 = 75%. ✓

(Tempting wrong: 25 or 50 come from mistakenly taking 5% or 10% of 500; but removing balls shrinks the total too, so more must go.)

ANSWER 1: D

# Problem 2:
There are $2^4 = 16$ equally likely sequences of genders.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: $\binom{4}{2}=6$ ways → 6/16
- 3 of one gender, 1 of the other: $\binom{4}{3}=4$ (3 boys) $+ \binom{4}{1}=4$ (3 girls) $= 8$ ways → 8/16

The largest is 8/16 = 1/2, the "3–1 split." Choice C (6/16) is the tempting trap because 2–2 is the single most likely *exact* split, but option D combines two cases.

ANSWER 2: D

# Problem 3:
A grid 60 toothpicks long and 32 toothpicks wide means the rectangle is 60 units by 32 units.

- Horizontal toothpicks: there are $32+1 = 33$ horizontal lines, each made of 60 toothpicks: $33 \times 60 = 1980$.
- Vertical toothpicks: there are $60+1 = 61$ vertical lines, each made of 32 toothpicks: $61 \times 32 = 1952$.

Total: $1980 + 1952 = 3932$.

(1980 and 1952 are the individual pieces — tempting partial answers.)

ANSWER 3: E

# Problem 4:
The expression is $\boxed{\phantom{2}}^{\boxed{\phantom{2}}} \times \boxed{\phantom{2}}^{\boxed{\phantom{2}}}$, using digits 2, 0, 2, 3 once each.

To maximize, make one factor as large as possible and let the other equal 1 (using exponent 0).

- $3^2 \times 2^0 = 9 \times 1 = 9$
- $2^3 \times 2^0 = 8$ (uses two 2's as base and exponent... gives 8)
- $2^2 \times 3^0 = 4$
- $2^0 \times 3^2 = 9$

Any arrangement with 0 as a base kills a factor to 0. The maximum is $9$.

(8 is the tempting near-miss from $2^3$; 16 and 18 aren't attainable.)

ANSWER 4: C

# Problem 5:
Volume of a cylinder: $V=\pi r^2 h$.

- Alex: diameter 6 → $r=3$, $h=12$: $V = \pi(3^2)(12) = 108\pi$.
- Felicia: diameter 12 → $r=6$, $h=6$: $V = \pi(6^2)(6) = 216\pi$.

Ratio $= 108\pi : 216\pi = 1:2$.

(1:1 is the trap — swapping diameter and height does *not* preserve volume, since radius is squared.)

ANSWER 5: B

# Problem 6:
Let the two equal sides be $a$ and the base be $b$, with $2a + b = 23$.

Since $2a$ is even, $b$ must be odd. Triangle inequality: $2a > b$, i.e. $23 - b > b \Rightarrow b < 11.5$.

So $b \in \{1,3,5,7,9,11\}$, giving $a = 11,10,9,8,7,6$ — six triangles:
(11,11,1), (10,10,3), (9,9,5), (8,8,7), (7,7,9), (6,6,11).

That's 6. (No equilateral option since 23 isn't divisible by 3; $b=13$ fails since $5+5<13$.)

ANSWER 6: C

# Problem 7:
Harold ate $\frac14$, leaving $\frac34$.

Moose ate $\frac13$ of that, leaving $\frac23 \cdot \frac34 = \frac12$.

Porcupine ate $\frac13$ of that, leaving $\frac23 \cdot \frac12 = \frac13$.

(Trap: subtracting $\frac14+\frac13+\frac13$ from 1 gives $\frac1{12}$ — wrong, because each animal eats a fraction of the *remainder*, not of the whole pie.)

ANSWER 7: D

# Problem 8:
Total money $10, minus the $2 tip → $8 available for fare.

First $\frac12$ mile costs $2.40, leaving $8.00 - 2.40 = \$5.60$.

Each additional 0.1 mile costs $0.20: $5.60 \div 0.20 = 28$ increments → $28 \times 0.1 = 2.8$ miles.

Total distance $= 0.5 + 2.8 = 3.3$ miles.

(Forgetting the tip gives $\$10-\$2.40=\$7.60 \to 3.8$ mi; forgetting the first half-mile is included gives other traps like 3.5.)

ANSWER 8: C

# Problem 9:
Population triples every 25 years, starting at 200 in 1998.

- 1998: 200
- 2023: 600
- 2048: 1800

The year 2050 is just 2 years past 2048, so the population is a bit more than 1800 — about 2000.

(3000 would require another near-tripling; 1000 is too small.)

ANSWER 9: D

# Problem 10:
The wall is 19 ft wide; the picture is 3 ft across and centered. The leftover wall splits equally on both sides:
$$\frac{19-3}{2} = \frac{16}{2} = 8 \text{ feet}.$$

(16 is the total leftover, not one side; $9\frac12$ is the wall's midpoint.)

ANSWER 10: B

Problem 1:

Initially, \(80\%\) of 500 balls are red:

\[
0.80(500)=400
\]

So there are 400 red balls and 100 blue balls.

Suppose \(x\) red balls are removed. Then:

- Red balls remaining: \(400-x\)
- Total balls remaining: \(500-x\)

We want \(75\%\) of the remaining balls to be red:

\[
\frac{400-x}{500-x}=\frac34
\]

Cross-multiply:

\[
4(400-x)=3(500-x)
\]

\[
1600-4x=1500-3x
\]

\[
100=x
\]

Removing 100 red balls leaves 300 red and 100 blue, for 400 total. Indeed,

\[
\frac{300}{400}=75\%.
\]

Choices such as 50 or 75 do not reduce the red percentage enough; 150 reduces it too much.

ANSWER 1: D

Problem 2:

Each child can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so \(1\) arrangement.
- All 4 girls: only \(GGGG\), so \(1\) arrangement.
- Exactly 2 girls and 2 boys:

\[
\binom42=6
\]

arrangements.
- Three of one gender and one of the other:
  - 3 boys and 1 girl: \(\binom41=4\) arrangements.
  - 3 girls and 1 boy: \(\binom41=4\) arrangements.

So this outcome has

\[
4+4=8
\]

arrangements.

Since \(8\) is more than \(6\) or \(1\), the most likely outcome is 3 children of one gender and 1 of the other.

ANSWER 2: D

Problem 3:

A grid 60 toothpicks long and 32 toothpicks wide has 60 columns and 32 rows of small squares.

For the horizontal toothpicks, there are \(32+1=33\) horizontal lines, each containing 60 toothpicks:

\[
33\cdot 60=1980
\]

For the vertical toothpicks, there are \(60+1=61\) vertical lines, each containing 32 toothpicks:

\[
61\cdot 32=1952
\]

Thus the total number is

\[
1980+1952=3932.
\]

The tempting answer \(1920=60\cdot32\) counts only the small squares, not the toothpicks forming their edges.

ANSWER 3: E

Problem 4:

The expression is of the form

\[
\Box^{\Box}\times\Box^{\Box},
\]

using the digits \(2,0,2,3\) once each.

To make the product as large as possible, the digit \(0\) should be an exponent, because any nonzero number to the zero power equals 1:

\[
2^0=1 \qquad\text{or}\qquad 3^0=1.
\]

We should use the remaining digits to make the largest possible power. Compare:

\[
3^2=9
\]

and

\[
2^3=8.
\]

Thus the best arrangement is

\[
3^2\times 2^0=9\times 1=9.
\]

Using \(2^3\) gives only \(8\), and placing 0 as a base would make the entire product 0.

ANSWER 4: C

Problem 5:

The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s can has diameter 6 cm, so its radius is 3 cm and its height is 12 cm:

\[
V_A=\pi(3)^2(12)=108\pi.
\]

Felicia’s can has diameter 12 cm, so its radius is 6 cm and its height is 6 cm:

\[
V_F=\pi(6)^2(6)=216\pi.
\]

The ratio is

\[
108\pi:216\pi=1:2.
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, making her base area four times as large.

ANSWER 5: B

Problem 6:

Let the equal sides of the isosceles triangle each have length \(a\). Let the base have length \(b\).

Since the perimeter is 23,

\[
2a+b=23,
\]

so

\[
b=23-2a.
\]

For a valid triangle, the base must be less than the sum of the equal sides:

\[
b<2a.
\]

Substitute \(b=23-2a\):

\[
23-2a<2a
\]

\[
23<4a
\]

\[
a>5.75.
\]

Also, \(b\) must be positive:

\[
23-2a>0,
\]

so

\[
a<11.5.
\]

Thus \(a\) can be

\[
6,7,8,9,10,11.
\]

That gives 6 different triangles. Their side lengths are:

\[
(6,6,11),\ (7,7,9),\ (8,8,7),\ (9,9,5),\ (10,10,3),\ (11,11,1).
\]

Values below 6 fail the triangle inequality, while values above 11 give a nonpositive base.

ANSWER 6: C

Problem 7:

Harold eats \(\frac14\) of the pie, leaving

\[
1-\frac14=\frac34.
\]

The moose eats \(\frac13\) of what is left, so it leaves \(\frac23\) of what Harold left:

\[
\frac23\cdot\frac34=\frac12.
\]

Then the porcupine eats \(\frac13\) of that remaining \(\frac12\), leaving \(\frac23\) of it:

\[
\frac23\cdot\frac12=\frac13.
\]

So \(\frac13\) of the original pie remains.

A common mistake is to subtract \(\frac13\) directly from \(\frac34\); the moose eats one-third of the amount currently left, not one-third of the original pie.

ANSWER 7: D

Problem 8:

The total amount available for the taxi fare is

\[
\$10-\$2=\$8.
\]

The first \(\frac12\) mile costs \$2.40. Therefore, the amount left for additional distance is

\[
8.00-2.40=5.60.
\]

Each additional \(0.1\) mile costs \$0.20, so the number of additional tenths of a mile is

\[
\frac{5.60}{0.20}=28.
\]

That is

\[
28(0.1)=2.8
\]

additional miles.

Adding the first \(\frac12\) mile:

\[
2.8+0.5=3.3.
\]

The ride can be 3.3 miles. The \$2 tip must be included, so choices based on spending the full \$10 on fare are too large.

ANSWER 8: C

Problem 9:

The population is 200 in 1998 and triples every 25 years.

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is a little more than two 25-year periods.

After 25 years, the population would be

\[
200\cdot 3=600.
\]

After 50 years, it would be

\[
600\cdot 3=1800.
\]

Since 2050 is about 2 years later, the population should be slightly more than 1800. The closest estimate is 2000.

Choices such as 600 correspond to only one tripling, while 3000 is much too high for just over two tripling periods.

ANSWER 9: D

Problem 10:

The wall is 19 feet wide, and the picture is 3 feet wide.

The amount of wall not covered by the picture is

\[
19-3=16
\]

feet.

Since the picture is centered, this 16 feet is divided equally between the two ends:

\[
\frac{16}{2}=8.
\]

Therefore, the nearest edge of the picture is 8 feet from either end of the wall.

ANSWER 10: B

Problem 1:
We need how many red balls to take away so reds are 75% of what is left.

At first there are 500 balls, 80% red.
Red = 0.80 × 500 = 400. Blue = 500 - 400 = 100.

Let x = number of red balls removed. Blue balls are not touched.
Red left = 400 - x. Total left = 500 - x.

We want:
(400 - x) / (500 - x) = 75% = 3/4

Cross-multiply:
4(400 - x) = 3(500 - x)
1600 - 4x = 1500 - 3x
1600 - 1500 = 4x - 3x
100 = x

Check: 300 red left, 100 blue left, total 400. 300/400 = 3/4.
Removing 25 would leave 375/475 ≈ 79%, still too high. Removing 150 would leave 250/350 ≈ 71%, too low.

ANSWER 1: D

Problem 2:
We are asked which gender pattern for 4 babies is most likely, each baby boy/girl equally likely.

There are 2^4 = 16 equally likely boy-girl sequences.

A. all 4 boys: only BBBB: 1 sequence. Probability 1/16.
B. all 4 girls: only GGGG: 1 sequence. Probability 1/16.
C. 2 girls and 2 boys: number of sequences = choose which 2 of 4 are girls = C(4,2)=6. Probability 6/16.
D. 3 of one gender and 1 of the other: This means 3B1G OR 3G1B.
3B1G: C(4,1)=4 sequences. 3G1B: C(4,1)=4 sequences. Total 8 sequences. Probability 8/16 = 1/2.
E. claims equal, false.

Many think 2-2 is most likely because it is most "balanced", but 3-1 has more sequences: 8 > 6 > 1.

So D is most likely.

ANSWER 2: D

Problem 3:
We need total toothpicks to make a rectangular grid 60 long by 32 wide, where length is measured in toothpicks.

Imagine unit squares. Top edge needs 60 toothpicks side by side. Same for every horizontal line.

Number of horizontal lines: there are 32 squares vertically, so 32+1 = 33 horizontal grid lines.
Each horizontal line uses 60 toothpicks.
Horizontal total = 33 × 60 = 1980.

Number of vertical lines: there are 60 squares horizontally, so 60+1 = 61 vertical grid lines.
Each vertical line uses 32 toothpicks.
Vertical total = 61 × 32 = 1952.

Total = 1980 + 1952 = 3932.

A 1920 = 60×32 is just the number of cells, not toothpicks. B 1952 is vertical only. C 1980 is horizontal only.

ANSWER 3: E

Problem 4:
We are asked to put digits 2,0,2,3, one per box, into the expression
□^□ × □^□
to get the largest value.

We must use all four digits as two bases and two exponents.

Key facts: a^0 = 1 for a ≠ 0, while 0^positive = 0, which kills a product. So do not put 0 as a base with positive exponent. Put 0 as an exponent to get 1. 0^0 is not allowed.

Try cases:

If exponents are 2 and 0, bases 3 and 2:
3^2 × 2^0 = 9 × 1 = 9
2^2 × 3^0 = 4 × 1 = 4
3^0 × 2^2 = 1 × 4 = 4
2^0 × 3^2 = 1 × 9 = 9

If exponents are 3 and 0, bases 2 and 2:
2^3 × 2^0 = 8 × 1 = 8

If 0 is a base with positive exponent, product is 0, e.g. 3^2 × 0^2 = 0.

So the best is 9.

Why not other choices?
A. 0: tempting if you think any zero makes everything zero, but 2^0 =1 avoids zero.
B. 8: this is 2^3 × 2^0, valid but less than 9.
D. 16 and E. 18: these come from the mistake x^0 = x. E.g. 3^2 × 2 =18, 2^3 × 2 =16. Correct is ×1, not ×2.

Maximum is 9.

ANSWER 4: C

Problem 5:
We compare volumes of cylinders. Volume = π r^2 h, r = diameter/2.

Alex: diameter 6, so r = 3. h =12.
V_Alex = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r = 6. h =6.
V_Felicia = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio Alex : Felicia = 108π : 216π = 108:216 = 1:2.

Volume scales as square of radius, so doubling diameter quadruples base area, which outweighs halving height. So Felicia's is bigger, ruling out C,D,E.

ANSWER 5: B

Problem 6:
We need integer-sided isosceles triangles with perimeter 23.

Let equal sides be a,a and base be b. Then:
2a + b = 23, so b = 23 - 2a > 0, so a ≤ 11.

Triangle inequality: sum of two small sides > third side. The only non-trivial one is:
a + a > b
2a > 23 - 2a
4a > 23
a > 5.75, so a ≥ 6.

Thus a = 6,7,8,9,10,11. Each gives a b:
6,6,11: 6+6>11 valid
7,7,9: 14>9 valid
8,8,7: valid
9,9,5: valid
10,10,3: 10+3>10 valid
11,11,1: 11+1>11 valid

That is 6 different triangles. No other a works.

ANSWER 6: C

Problem 7:
We track fraction of original pie left.

Harold eats 1/4, so left = 1 - 1/4 = 3/4.

Moose eats 1/3 of what Harold left:
eaten = (1/3)×(3/4) = 1/4.
Left after moose = 3/4 - 1/4 = 1/2. Equivalently (2/3)×(3/4)=1/2.

Porcupine eats 1/3 of what moose left:
eaten = (1/3)×(1/2) = 1/6.
Left after porcupine = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3. Equivalently (2/3)×(1/2)=1/3.

So 1/3 remains. A 1/12, B 1/6 are the amounts eaten, not left.

ANSWER 7: D

Problem 8:
We have $10 total, $2 tip, so fare money = 10 - 2 = $8.00.

First 0.5 mile costs $2.40.
Remaining for extra distance = 8.00 - 2.40 = $5.60.

Extra is $0.20 per 0.1 mile.
Number of 0.1-mile blocks = 5.60 / 0.20 = 28.
Extra distance = 28 × 0.1 = 2.8 miles.

Total distance = 0.5 + 2.8 = 3.3 miles.

Check: fare = 2.40 + 28×0.20 = 2.40+5.60=8.00, plus $2 tip = $10.

ANSWER 8: C

Problem 9:
Population in 1998 is 200, triples every 25 years. Estimate in 2050.

Time elapsed = 2050 - 1998 = 52 years. That's about two 25-year periods.

After 25 yr (~2023): 200×3 = 600.
After 50 yr (~2048): 600×3 = 1800.

2050 is 2 years more, so a little above 1800. Nearest estimate among choices is 2000.

A 600 is only one tripling. C 1000, B 800 ignore tripling. E 3000 is too high.

ANSWER 9: D

Problem 10:
Picture 3 ft wide centered on 19 ft wall.

Total wall not covered = 19 - 3 = 16 ft.
Centered means half on each side: 16/2 = 8 ft from each end to picture edge.

A 1 1/2, C 9 1/2 confuse center vs edge. B is correct.

ANSWER 10: B

Problem 1:
We are asked to find how many red balls must be removed from a bag of 500 balls (originally 80% red) so that the remaining red balls constitute 75% of the bag.

First, calculate the original number of balls of each color:
- Total balls = $500$
- Red balls = $80\% \text{ of } 500 = 0.80 \times 500 = 400$
- Blue balls = $500 - 400 = 100$

Only red balls are removed, so the number of blue balls remains constant at $100$.
In the final state, red balls make up $75\%$ of the total, which means blue balls must make up:
$$100\% - 75\% = 25\%$$
of the remaining balls.

Let $T$ be the total number of remaining balls:
$$0.25 \times T = 100 \implies T = \frac{100}{0.25} = 400$$

Since the bag originally contained $500$ balls and now contains $400$ balls, the number of red balls removed is:
$$500 - 400 = 100$$

ANSWER 1: D

---

Problem 2:
We are asked to determine which outcome is most likely when four children are born, with boys (B) and girls (G) being equally likely.

Each child has $2$ possible genders, so there are $2^4 = 16$ equally likely possible outcomes in total. We evaluate the number of outcomes for each choice:
- **A. All 4 are boys:** Only $1$ outcome ($\text{BBBB}$). Probability = $\frac{1}{16}$.
- **B. All 4 are girls:** Only $1$ outcome ($\text{GGGG}$). Probability = $\frac{1}{16}$.
- **C. 2 are girls and 2 are boys:** The number of ways to choose which $2$ children are girls is $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ outcomes. Probability = $\frac{6}{16}$.
- **D. 3 are of one gender and 1 is of the other:** This includes either 3 boys and 1 girl ($\binom{4}{1} = 4$ outcomes) or 3 girls and 1 boy ($\binom{4}{3} = 4$ outcomes), giving $4 + 4 = 8$ outcomes. Probability = $\frac{8}{16} = \frac{1}{2}$.

Comparing the possibilities, outcome D has $8$ favorable outcomes, which is greater than C ($6$ outcomes), A ($1$ outcome), and B ($1$ outcome). Therefore, D is the most likely outcome.

ANSWER 2: D

---

Problem 3:
We are asked to find the total number of toothpicks used to make a rectangular grid that is $60$ toothpicks long and $32$ toothpicks wide.

A grid of dimensions $60 \times 32$ consists of unit squares arranged in $32$ rows and $60$ columns:
- **Horizontal toothpicks:** There are $32 + 1 = 33$ horizontal grid lines. Each horizontal line is made of $60$ toothpicks.
  $$\text{Number of horizontal toothpicks} = 33 \times 60 = 1980$$
- **Vertical toothpicks:** There are $60 + 1 = 61$ vertical grid lines. Each vertical line is made of $32$ toothpicks.
  $$\text{Number of vertical toothpicks} = 61 \times 32 = 1952$$

Adding the horizontal and vertical toothpicks together:
$$\text{Total toothpicks} = 1980 + 1952 = 3932$$

ANSWER 3: E

---

Problem 4:
We are asked to find the maximum possible value of the expression $\square^\square \times \square^\square$ when placing the digits $2, 0, 2,$ and $3$ into the four boxes, one digit per box.

Let the expression be $a^b \times c^d$, where $\{a, b, c, d\} = \{2, 0, 2, 3\}$.
We test the possible assignments of the digits:
- To maximize the product, we want large bases and exponents, but one of the digits is $0$.
- If $0$ is used as a base, say $0^d$ with $d > 0$, the factor becomes $0$, making the entire product $0$.
- If $0$ is placed in an exponent, recall that for any non-zero base $x$, $x^0 = 1$.
  - Pair $0$ with $2$ as an exponent: $2^0 = 1$. The remaining digits are $3$ and $2$.
    - Setting $3^2 \times 2^0 = 9 \times 1 = 9$.
    - Setting $2^3 \times 2^0 = 8 \times 1 = 8$.
  - Pair $0$ with $3$ as an exponent: $3^0 = 1$. The remaining digits are $2$ and $2$.
    - Setting $2^2 \times 3^0 = 4 \times 1 = 4$.

The values $16$ and $18$ are distractors arising from the common misconception that $x^0 = x$ (which would give $2^3 \times 2 = 16$ or $3^2 \times 2 = 18$).
The maximum mathematically valid value achievable is $9$.

ANSWER 4: C

---

Problem 5:
We are asked to find the ratio of the volume of Alex's cat food can to the volume of Felicia's cat food can.

The volume of a cylinder is given by $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height.
- For Alex's can:
  - Diameter = $6\text{ cm} \implies \text{radius } r_A = 3\text{ cm}$
  - Height $h_A = 12\text{ cm}$
  - Volume $V_A = \pi (3)^2 (12) = \pi \times 9 \times 12 = 108\pi\text{ cm}^3$

- For Felicia's can:
  - Diameter = $12\text{ cm} \implies \text{radius } r_F = 6\text{ cm}$
  - Height $h_F = 6\text{ cm}$
  - Volume $V_F = \pi (6)^2 (6) = \pi \times 36 \times 6 = 216\pi\text{ cm}^3$

The ratio of the volume of Alex's can to Felicia's can is:
$$\frac{V_A}{V_F} = \frac{108\pi}{216\pi} = \frac{1}{2}$$

Thus, the ratio is $1:2$.

ANSWER 5: B

---

Problem 6:
We are asked to find the number of different isosceles triangles with integer side lengths and a perimeter of $23$.

Let the side lengths of the isosceles triangle be $a, a,$ and $b$, where $a$ and $b$ are positive integers.
The perimeter is:
$$2a + b = 23 \implies b = 23 - 2a$$

By the Triangle Inequality:
1. The sum of the two equal sides must be strictly greater than the third side:
   $$a + a > b \implies 2a > 23 - 2a \implies 4a > 23 \implies a \ge 6$$
2. The third side must be positive:
   $$b > 0 \implies 23 - 2a > 0 \implies 2a < 23 \implies a \le 11$$

The possible integer values for $a$ are $6, 7, 8, 9, 10, 11$.
We check each corresponding side length triple $(a, a, b)$:
- $a = 6 \implies b = 11$: sides $(6, 6, 11)$
- $a = 7 \implies b = 9$: sides $(7, 7, 9)$
- $a = 8 \implies b = 7$: sides $(8, 8, 7)$
- $a = 9 \implies b = 5$: sides $(9, 9, 5)$
- $a = 10 \implies b = 3$: sides $(10, 10, 3)$
- $a = 11 \implies b = 1$: sides $(11, 11, 1)$

All $6$ sets of side lengths satisfy the triangle inequality and have two equal sides (none are equilateral since $23$ is not divisible by $3$). Thus, there are $6$ such triangles.

ANSWER 6: C

---

Problem 7:
We are asked to find the fraction of the original pie remaining after Harold, a moose, and a porcupine each eat a portion.

1. Harold start

# Problem 1:
Restate: 500 balls, 80% red. Remove some **red** balls so that 75% of what remains is red.

Red = 0.80(500) = 400, Blue = 100.

Let $x$ = red balls removed. Blues never change (still 100), and after removal blues must be 25% of the total:
$$100 = 0.25(500 - x) \Rightarrow 400 = 500 - x \Rightarrow x = 100.$$

Check: 300 red, 100 blue, total 400 → 300/400 = 75%. ✓

(Tempting wrong: 25 or 50 come from mistakenly taking 5% or 10% of 500; but removing balls shrinks the total too, so more must go.)

ANSWER 1: D

# Problem 2:
There are $2^4 = 16$ equally likely sequences of genders.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: $\binom{4}{2}=6$ ways → 6/16
- 3 of one gender, 1 of the other: $\binom{4}{3}=4$ (3 boys) $+ \binom{4}{1}=4$ (3 girls) $= 8$ ways → 8/16

The largest is 8/16 = 1/2, the "3–1 split." Choice C (6/16) is the tempting trap because 2–2 is the single most likely *exact* split, but option D combines two cases.

ANSWER 2: D

# Problem 3:
A grid 60 toothpicks long and 32 toothpicks wide means the rectangle is 60 units by 32 units.

- Horizontal toothpicks: there are $32+1 = 33$ horizontal lines, each made of 60 toothpicks: $33 \times 60 = 1980$.
- Vertical toothpicks: there are $60+1 = 61$ vertical lines, each made of 32 toothpicks: $61 \times 32 = 1952$.

Total: $1980 + 1952 = 3932$.

(1980 and 1952 are the individual pieces — tempting partial answers.)

ANSWER 3: E

# Problem 4:
The expression is $\boxed{\phantom{2}}^{\boxed{\phantom{2}}} \times \boxed{\phantom{2}}^{\boxed{\phantom{2}}}$, using digits 2, 0, 2, 3 once each.

To maximize, make one factor as large as possible and let the other equal 1 (using exponent 0).

- $3^2 \times 2^0 = 9 \times 1 = 9$
- $2^3 \times 2^0 = 8$ (uses two 2's as base and exponent... gives 8)
- $2^2 \times 3^0 = 4$
- $2^0 \times 3^2 = 9$

Any arrangement with 0 as a base kills a factor to 0. The maximum is $9$.

(8 is the tempting near-miss from $2^3$; 16 and 18 aren't attainable.)

ANSWER 4: C

# Problem 5:
Volume of a cylinder: $V=\pi r^2 h$.

- Alex: diameter 6 → $r=3$, $h=12$: $V = \pi(3^2)(12) = 108\pi$.
- Felicia: diameter 12 → $r=6$, $h=6$: $V = \pi(6^2)(6) = 216\pi$.

Ratio $= 108\pi : 216\pi = 1:2$.

(1:1 is the trap — swapping diameter and height does *not* preserve volume, since radius is squared.)

ANSWER 5: B

# Problem 6:
Let the two equal sides be $a$ and the base be $b$, with $2a + b = 23$.

Since $2a$ is even, $b$ must be odd. Triangle inequality: $2a > b$, i.e. $23 - b > b \Rightarrow b < 11.5$.

So $b \in \{1,3,5,7,9,11\}$, giving $a = 11,10,9,8,7,6$ — six triangles:
(11,11,1), (10,10,3), (9,9,5), (8,8,7), (7,7,9), (6,6,11).

That's 6. (No equilateral option since 23 isn't divisible by 3; $b=13$ fails since $5+5<13$.)

ANSWER 6: C

# Problem 7:
Harold ate $\frac14$, leaving $\frac34$.

Moose ate $\frac13$ of that, leaving $\frac23 \cdot \frac34 = \frac12$.

Porcupine ate $\frac13$ of that, leaving $\frac23 \cdot \frac12 = \frac13$.

(Trap: subtracting $\frac14+\frac13+\frac13$ from 1 gives $\frac1{12}$ — wrong, because each animal eats a fraction of the *remainder*, not of the whole pie.)

ANSWER 7: D

# Problem 8:
Total money $10, minus the $2 tip → $8 available for fare.

First $\frac12$ mile costs $2.40, leaving $8.00 - 2.40 = \$5.60$.

Each additional 0.1 mile costs $0.20: $5.60 \div 0.20 = 28$ increments → $28 \times 0.1 = 2.8$ miles.

Total distance $= 0.5 + 2.8 = 3.3$ miles.

(Forgetting the tip gives $\$10-\$2.40=\$7.60 \to 3.8$ mi; forgetting the first half-mile is included gives other traps like 3.5.)

ANSWER 8: C

# Problem 9:
Population triples every 25 years, starting at 200 in 1998.

- 1998: 200
- 2023: 600
- 2048: 1800

The year 2050 is just 2 years past 2048, so the population is a bit more than 1800 — about 2000.

(3000 would require another near-tripling; 1000 is too small.)

ANSWER 9: D

# Problem 10:
The wall is 19 ft wide; the picture is 3 ft across and centered. The leftover wall splits equally on both sides:
$$\frac{19-3}{2} = \frac{16}{2} = 8 \text{ feet}.$$

(16 is the total leftover, not one side; $9\frac12$ is the wall's midpoint.)

ANSWER 10: B

Problem 1:

Initially, \(80\%\) of 500 balls are red:

\[
0.80(500)=400
\]

So there are 400 red balls and 100 blue balls.

Suppose \(x\) red balls are removed. Then:

- Red balls remaining: \(400-x\)
- Total balls remaining: \(500-x\)

We want \(75\%\) of the remaining balls to be red:

\[
\frac{400-x}{500-x}=\frac34
\]

Cross-multiply:

\[
4(400-x)=3(500-x)
\]

\[
1600-4x=1500-3x
\]

\[
100=x
\]

Removing 100 red balls leaves 300 red and 100 blue, for 400 total. Indeed,

\[
\frac{300}{400}=75\%.
\]

Choices such as 50 or 75 do not reduce the red percentage enough; 150 reduces it too much.

ANSWER 1: D

Problem 2:

Each child can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so \(1\) arrangement.
- All 4 girls: only \(GGGG\), so \(1\) arrangement.
- Exactly 2 girls and 2 boys:

\[
\binom42=6
\]

arrangements.
- Three of one gender and one of the other:
  - 3 boys and 1 girl: \(\binom41=4\) arrangements.
  - 3 girls and 1 boy: \(\binom41=4\) arrangements.

So this outcome has

\[
4+4=8
\]

arrangements.

Since \(8\) is more than \(6\) or \(1\), the most likely outcome is 3 children of one gender and 1 of the other.

ANSWER 2: D

Problem 3:

A grid 60 toothpicks long and 32 toothpicks wide has 60 columns and 32 rows of small squares.

For the horizontal toothpicks, there are \(32+1=33\) horizontal lines, each containing 60 toothpicks:

\[
33\cdot 60=1980
\]

For the vertical toothpicks, there are \(60+1=61\) vertical lines, each containing 32 toothpicks:

\[
61\cdot 32=1952
\]

Thus the total number is

\[
1980+1952=3932.
\]

The tempting answer \(1920=60\cdot32\) counts only the small squares, not the toothpicks forming their edges.

ANSWER 3: E

Problem 4:

The expression is of the form

\[
\Box^{\Box}\times\Box^{\Box},
\]

using the digits \(2,0,2,3\) once each.

To make the product as large as possible, the digit \(0\) should be an exponent, because any nonzero number to the zero power equals 1:

\[
2^0=1 \qquad\text{or}\qquad 3^0=1.
\]

We should use the remaining digits to make the largest possible power. Compare:

\[
3^2=9
\]

and

\[
2^3=8.
\]

Thus the best arrangement is

\[
3^2\times 2^0=9\times 1=9.
\]

Using \(2^3\) gives only \(8\), and placing 0 as a base would make the entire product 0.

ANSWER 4: C

Problem 5:

The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s can has diameter 6 cm, so its radius is 3 cm and its height is 12 cm:

\[
V_A=\pi(3)^2(12)=108\pi.
\]

Felicia’s can has diameter 12 cm, so its radius is 6 cm and its height is 6 cm:

\[
V_F=\pi(6)^2(6)=216\pi.
\]

The ratio is

\[
108\pi:216\pi=1:2.
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, making her base area four times as large.

ANSWER 5: B

Problem 6:

Let the equal sides of the isosceles triangle each have length \(a\). Let the base have length \(b\).

Since the perimeter is 23,

\[
2a+b=23,
\]

so

\[
b=23-2a.
\]

For a valid triangle, the base must be less than the sum of the equal sides:

\[
b<2a.
\]

Substitute \(b=23-2a\):

\[
23-2a<2a
\]

\[
23<4a
\]

\[
a>5.75.
\]

Also, \(b\) must be positive:

\[
23-2a>0,
\]

so

\[
a<11.5.
\]

Thus \(a\) can be

\[
6,7,8,9,10,11.
\]

That gives 6 different triangles. Their side lengths are:

\[
(6,6,11),\ (7,7,9),\ (8,8,7),\ (9,9,5),\ (10,10,3),\ (11,11,1).
\]

Values below 6 fail the triangle inequality, while values above 11 give a nonpositive base.

ANSWER 6: C

Problem 7:

Harold eats \(\frac14\) of the pie, leaving

\[
1-\frac14=\frac34.
\]

The moose eats \(\frac13\) of what is left, so it leaves \(\frac23\) of what Harold left:

\[
\frac23\cdot\frac34=\frac12.
\]

Then the porcupine eats \(\frac13\) of that remaining \(\frac12\), leaving \(\frac23\) of it:

\[
\frac23\cdot\frac12=\frac13.
\]

So \(\frac13\) of the original pie remains.

A common mistake is to subtract \(\frac13\) directly from \(\frac34\); the moose eats one-third of the amount currently left, not one-third of the original pie.

ANSWER 7: D

Problem 8:

The total amount available for the taxi fare is

\[
\$10-\$2=\$8.
\]

The first \(\frac12\) mile costs \$2.40. Therefore, the amount left for additional distance is

\[
8.00-2.40=5.60.
\]

Each additional \(0.1\) mile costs \$0.20, so the number of additional tenths of a mile is

\[
\frac{5.60}{0.20}=28.
\]

That is

\[
28(0.1)=2.8
\]

additional miles.

Adding the first \(\frac12\) mile:

\[
2.8+0.5=3.3.
\]

The ride can be 3.3 miles. The \$2 tip must be included, so choices based on spending the full \$10 on fare are too large.

ANSWER 8: C

Problem 9:

The population is 200 in 1998 and triples every 25 years.

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is a little more than two 25-year periods.

After 25 years, the population would be

\[
200\cdot 3=600.
\]

After 50 years, it would be

\[
600\cdot 3=1800.
\]

Since 2050 is about 2 years later, the population should be slightly more than 1800. The closest estimate is 2000.

Choices such as 600 correspond to only one tripling, while 3000 is much too high for just over two tripling periods.

ANSWER 9: D

Problem 10:

The wall is 19 feet wide, and the picture is 3 feet wide.

The amount of wall not covered by the picture is

\[
19-3=16
\]

feet.

Since the picture is centered, this 16 feet is divided equally between the two ends:

\[
\frac{16}{2}=8.
\]

Therefore, the nearest edge of the picture is 8 feet from either end of the wall.

ANSWER 10: B

Problem 1:
We need how many red balls to take away so reds are 75% of what is left.

At first there are 500 balls, 80% red.
Red = 0.80 × 500 = 400. Blue = 500 - 400 = 100.

Let x = number of red balls removed. Blue balls are not touched.
Red left = 400 - x. Total left = 500 - x.

We want:
(400 - x) / (500 - x) = 75% = 3/4

Cross-multiply:
4(400 - x) = 3(500 - x)
1600 - 4x = 1500 - 3x
1600 - 1500 = 4x - 3x
100 = x

Check: 300 red left, 100 blue left, total 400. 300/400 = 3/4.
Removing 25 would leave 375/475 ≈ 79%, still too high. Removing 150 would leave 250/350 ≈ 71%, too low.

ANSWER 1: D

Problem 2:
We are asked which gender pattern for 4 babies is most likely, each baby boy/girl equally likely.

There are 2^4 = 16 equally likely boy-girl sequences.

A. all 4 boys: only BBBB: 1 sequence. Probability 1/16.
B. all 4 girls: only GGGG: 1 sequence. Probability 1/16.
C. 2 girls and 2 boys: number of sequences = choose which 2 of 4 are girls = C(4,2)=6. Probability 6/16.
D. 3 of one gender and 1 of the other: This means 3B1G OR 3G1B.
3B1G: C(4,1)=4 sequences. 3G1B: C(4,1)=4 sequences. Total 8 sequences. Probability 8/16 = 1/2.
E. claims equal, false.

Many think 2-2 is most likely because it is most "balanced", but 3-1 has more sequences: 8 > 6 > 1.

So D is most likely.

ANSWER 2: D

Problem 3:
We need total toothpicks to make a rectangular grid 60 long by 32 wide, where length is measured in toothpicks.

Imagine unit squares. Top edge needs 60 toothpicks side by side. Same for every horizontal line.

Number of horizontal lines: there are 32 squares vertically, so 32+1 = 33 horizontal grid lines.
Each horizontal line uses 60 toothpicks.
Horizontal total = 33 × 60 = 1980.

Number of vertical lines: there are 60 squares horizontally, so 60+1 = 61 vertical grid lines.
Each vertical line uses 32 toothpicks.
Vertical total = 61 × 32 = 1952.

Total = 1980 + 1952 = 3932.

A 1920 = 60×32 is just the number of cells, not toothpicks. B 1952 is vertical only. C 1980 is horizontal only.

ANSWER 3: E

Problem 4:
We are asked to put digits 2,0,2,3, one per box, into the expression
□^□ × □^□
to get the largest value.

We must use all four digits as two bases and two exponents.

Key facts: a^0 = 1 for a ≠ 0, while 0^positive = 0, which kills a product. So do not put 0 as a base with positive exponent. Put 0 as an exponent to get 1. 0^0 is not allowed.

Try cases:

If exponents are 2 and 0, bases 3 and 2:
3^2 × 2^0 = 9 × 1 = 9
2^2 × 3^0 = 4 × 1 = 4
3^0 × 2^2 = 1 × 4 = 4
2^0 × 3^2 = 1 × 9 = 9

If exponents are 3 and 0, bases 2 and 2:
2^3 × 2^0 = 8 × 1 = 8

If 0 is a base with positive exponent, product is 0, e.g. 3^2 × 0^2 = 0.

So the best is 9.

Why not other choices?
A. 0: tempting if you think any zero makes everything zero, but 2^0 =1 avoids zero.
B. 8: this is 2^3 × 2^0, valid but less than 9.
D. 16 and E. 18: these come from the mistake x^0 = x. E.g. 3^2 × 2 =18, 2^3 × 2 =16. Correct is ×1, not ×2.

Maximum is 9.

ANSWER 4: C

Problem 5:
We compare volumes of cylinders. Volume = π r^2 h, r = diameter/2.

Alex: diameter 6, so r = 3. h =12.
V_Alex = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r = 6. h =6.
V_Felicia = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio Alex : Felicia = 108π : 216π = 108:216 = 1:2.

Volume scales as square of radius, so doubling diameter quadruples base area, which outweighs halving height. So Felicia's is bigger, ruling out C,D,E.

ANSWER 5: B

Problem 6:
We need integer-sided isosceles triangles with perimeter 23.

Let equal sides be a,a and base be b. Then:
2a + b = 23, so b = 23 - 2a > 0, so a ≤ 11.

Triangle inequality: sum of two small sides > third side. The only non-trivial one is:
a + a > b
2a > 23 - 2a
4a > 23
a > 5.75, so a ≥ 6.

Thus a = 6,7,8,9,10,11. Each gives a b:
6,6,11: 6+6>11 valid
7,7,9: 14>9 valid
8,8,7: valid
9,9,5: valid
10,10,3: 10+3>10 valid
11,11,1: 11+1>11 valid

That is 6 different triangles. No other a works.

ANSWER 6: C

Problem 7:
We track fraction of original pie left.

Harold eats 1/4, so left = 1 - 1/4 = 3/4.

Moose eats 1/3 of what Harold left:
eaten = (1/3)×(3/4) = 1/4.
Left after moose = 3/4 - 1/4 = 1/2. Equivalently (2/3)×(3/4)=1/2.

Porcupine eats 1/3 of what moose left:
eaten = (1/3)×(1/2) = 1/6.
Left after porcupine = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3. Equivalently (2/3)×(1/2)=1/3.

So 1/3 remains. A 1/12, B 1/6 are the amounts eaten, not left.

ANSWER 7: D

Problem 8:
We have $10 total, $2 tip, so fare money = 10 - 2 = $8.00.

First 0.5 mile costs $2.40.
Remaining for extra distance = 8.00 - 2.40 = $5.60.

Extra is $0.20 per 0.1 mile.
Number of 0.1-mile blocks = 5.60 / 0.20 = 28.
Extra distance = 28 × 0.1 = 2.8 miles.

Total distance = 0.5 + 2.8 = 3.3 miles.

Check: fare = 2.40 + 28×0.20 = 2.40+5.60=8.00, plus $2 tip = $10.

ANSWER 8: C

Problem 9:
Population in 1998 is 200, triples every 25 years. Estimate in 2050.

Time elapsed = 2050 - 1998 = 52 years. That's about two 25-year periods.

After 25 yr (~2023): 200×3 = 600.
After 50 yr (~2048): 600×3 = 1800.

2050 is 2 years more, so a little above 1800. Nearest estimate among choices is 2000.

A 600 is only one tripling. C 1000, B 800 ignore tripling. E 3000 is too high.

ANSWER 9: D

Problem 10:
Picture 3 ft wide centered on 19 ft wall.

Total wall not covered = 19 - 3 = 16 ft.
Centered means half on each side: 16/2 = 8 ft from each end to picture edge.

A 1 1/2, C 9 1/2 confuse center vs edge. B is correct.

ANSWER 10: B

Problem 1:
We are asked to find how many red balls must be removed from a bag of 500 balls (originally 80% red) so that the remaining red balls constitute 75% of the bag.

First, calculate the original number of balls of each color:
- Total balls = $500$
- Red balls = $80\% \text{ of } 500 = 0.80 \times 500 = 400$
- Blue balls = $500 - 400 = 100$

Only red balls are removed, so the number of blue balls remains constant at $100$.
In the final state, red balls make up $75\%$ of the total, which means blue balls must make up:
$$100\% - 75\% = 25\%$$
of the remaining balls.

Let $T$ be the total number of remaining balls:
$$0.25 \times T = 100 \implies T = \frac{100}{0.25} = 400$$

Since the bag originally contained $500$ balls and now contains $400$ balls, the number of red balls removed is:
$$500 - 400 = 100$$

ANSWER 1: D

---

Problem 2:
We are asked to determine which outcome is most likely when four children are born, with boys (B) and girls (G) being equally likely.

Each child has $2$ possible genders, so there are $2^4 = 16$ equally likely possible outcomes in total. We evaluate the number of outcomes for each choice:
- **A. All 4 are boys:** Only $1$ outcome ($\text{BBBB}$). Probability = $\frac{1}{16}$.
- **B. All 4 are girls:** Only $1$ outcome ($\text{GGGG}$). Probability = $\frac{1}{16}$.
- **C. 2 are girls and 2 are boys:** The number of ways to choose which $2$ children are girls is $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ outcomes. Probability = $\frac{6}{16}$.
- **D. 3 are of one gender and 1 is of the other:** This includes either 3 boys and 1 girl ($\binom{4}{1} = 4$ outcomes) or 3 girls and 1 boy ($\binom{4}{3} = 4$ outcomes), giving $4 + 4 = 8$ outcomes. Probability = $\frac{8}{16} = \frac{1}{2}$.

Comparing the possibilities, outcome D has $8$ favorable outcomes, which is greater than C ($6$ outcomes), A ($1$ outcome), and B ($1$ outcome). Therefore, D is the most likely outcome.

ANSWER 2: D

---

Problem 3:
We are asked to find the total number of toothpicks used to make a rectangular grid that is $60$ toothpicks long and $32$ toothpicks wide.

A grid of dimensions $60 \times 32$ consists of unit squares arranged in $32$ rows and $60$ columns:
- **Horizontal toothpicks:** There are $32 + 1 = 33$ horizontal grid lines. Each horizontal line is made of $60$ toothpicks.
  $$\text{Number of horizontal toothpicks} = 33 \times 60 = 1980$$
- **Vertical toothpicks:** There are $60 + 1 = 61$ vertical grid lines. Each vertical line is made of $32$ toothpicks.
  $$\text{Number of vertical toothpicks} = 61 \times 32 = 1952$$

Adding the horizontal and vertical toothpicks together:
$$\text{Total toothpicks} = 1980 + 1952 = 3932$$

ANSWER 3: E

---

Problem 4:
We are asked to find the maximum possible value of the expression $\square^\square \times \square^\square$ when placing the digits $2, 0, 2,$ and $3$ into the four boxes, one digit per box.

Let the expression be $a^b \times c^d$, where $\{a, b, c, d\} = \{2, 0, 2, 3\}$.
We test the possible assignments of the digits:
- To maximize the product, we want large bases and exponents, but one of the digits is $0$.
- If $0$ is used as a base, say $0^d$ with $d > 0$, the factor becomes $0$, making the entire product $0$.
- If $0$ is placed in an exponent, recall that for any non-zero base $x$, $x^0 = 1$.
  - Pair $0$ with $2$ as an exponent: $2^0 = 1$. The remaining digits are $3$ and $2$.
    - Setting $3^2 \times 2^0 = 9 \times 1 = 9$.
    - Setting $2^3 \times 2^0 = 8 \times 1 = 8$.
  - Pair $0$ with $3$ as an exponent: $3^0 = 1$. The remaining digits are $2$ and $2$.
    - Setting $2^2 \times 3^0 = 4 \times 1 = 4$.

The values $16$ and $18$ are distractors arising from the common misconception that $x^0 = x$ (which would give $2^3 \times 2 = 16$ or $3^2 \times 2 = 18$).
The maximum mathematically valid value achievable is $9$.

ANSWER 4: C

---

Problem 5:
We are asked to find the ratio of the volume of Alex's cat food can to the volume of Felicia's cat food can.

The volume of a cylinder is given by $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height.
- For Alex's can:
  - Diameter = $6\text{ cm} \implies \text{radius } r_A = 3\text{ cm}$
  - Height $h_A = 12\text{ cm}$
  - Volume $V_A = \pi (3)^2 (12) = \pi \times 9 \times 12 = 108\pi\text{ cm}^3$

- For Felicia's can:
  - Diameter = $12\text{ cm} \implies \text{radius } r_F = 6\text{ cm}$
  - Height $h_F = 6\text{ cm}$
  - Volume $V_F = \pi (6)^2 (6) = \pi \times 36 \times 6 = 216\pi\text{ cm}^3$

The ratio of the volume of Alex's can to Felicia's can is:
$$\frac{V_A}{V_F} = \frac{108\pi}{216\pi} = \frac{1}{2}$$

Thus, the ratio is $1:2$.

ANSWER 5: B

---

Problem 6:
We are asked to find the number of different isosceles triangles with integer side lengths and a perimeter of $23$.

Let the side lengths of the isosceles triangle be $a, a,$ and $b$, where $a$ and $b$ are positive integers.
The perimeter is:
$$2a + b = 23 \implies b = 23 - 2a$$

By the Triangle Inequality:
1. The sum of the two equal sides must be strictly greater than the third side:
   $$a + a > b \implies 2a > 23 - 2a \implies 4a > 23 \implies a \ge 6$$
2. The third side must be positive:
   $$b > 0 \implies 23 - 2a > 0 \implies 2a < 23 \implies a \le 11$$

The possible integer values for $a$ are $6, 7, 8, 9, 10, 11$.
We check each corresponding side length triple $(a, a, b)$:
- $a = 6 \implies b = 11$: sides $(6, 6, 11)$
- $a = 7 \implies b = 9$: sides $(7, 7, 9)$
- $a = 8 \implies b = 7$: sides $(8, 8, 7)$
- $a = 9 \implies b = 5$: sides $(9, 9, 5)$
- $a = 10 \implies b = 3$: sides $(10, 10, 3)$
- $a = 11 \implies b = 1$: sides $(11, 11, 1)$

All $6$ sets of side lengths satisfy the triangle inequality and have two equal sides (none are equilateral since $23$ is not divisible by $3$). Thus, there are $6$ such triangles.

ANSWER 6: C

---

Problem 7:
We are asked to find the fraction of the original pie remaining after Harold, a moose, and a porcupine each eat a portion.

1. Harold start

# Problem 1:
Restate: 500 balls, 80% red. Remove some **red** balls so that 75% of what remains is red.

Red = 0.80(500) = 400, Blue = 100.

Let $x$ = red balls removed. Blues never change (still 100), and after removal blues must be 25% of the total:
$$100 = 0.25(500 - x) \Rightarrow 400 = 500 - x \Rightarrow x = 100.$$

Check: 300 red, 100 blue, total 400 → 300/400 = 75%. ✓

(Tempting wrong: 25 or 50 come from mistakenly taking 5% or 10% of 500; but removing balls shrinks the total too, so more must go.)

ANSWER 1: D

# Problem 2:
There are $2^4 = 16$ equally likely sequences of genders.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: $\binom{4}{2}=6$ ways → 6/16
- 3 of one gender, 1 of the other: $\binom{4}{3}=4$ (3 boys) $+ \binom{4}{1}=4$ (3 girls) $= 8$ ways → 8/16

The largest is 8/16 = 1/2, the "3–1 split." Choice C (6/16) is the tempting trap because 2–2 is the single most likely *exact* split, but option D combines two cases.

ANSWER 2: D

# Problem 3:
A grid 60 toothpicks long and 32 toothpicks wide means the rectangle is 60 units by 32 units.

- Horizontal toothpicks: there are $32+1 = 33$ horizontal lines, each made of 60 toothpicks: $33 \times 60 = 1980$.
- Vertical toothpicks: there are $60+1 = 61$ vertical lines, each made of 32 toothpicks: $61 \times 32 = 1952$.

Total: $1980 + 1952 = 3932$.

(1980 and 1952 are the individual pieces — tempting partial answers.)

ANSWER 3: E

# Problem 4:
The expression is $\boxed{\phantom{2}}^{\boxed{\phantom{2}}} \times \boxed{\phantom{2}}^{\boxed{\phantom{2}}}$, using digits 2, 0, 2, 3 once each.

To maximize, make one factor as large as possible and let the other equal 1 (using exponent 0).

- $3^2 \times 2^0 = 9 \times 1 = 9$
- $2^3 \times 2^0 = 8$ (uses two 2's as base and exponent... gives 8)
- $2^2 \times 3^0 = 4$
- $2^0 \times 3^2 = 9$

Any arrangement with 0 as a base kills a factor to 0. The maximum is $9$.

(8 is the tempting near-miss from $2^3$; 16 and 18 aren't attainable.)

ANSWER 4: C

# Problem 5:
Volume of a cylinder: $V=\pi r^2 h$.

- Alex: diameter 6 → $r=3$, $h=12$: $V = \pi(3^2)(12) = 108\pi$.
- Felicia: diameter 12 → $r=6$, $h=6$: $V = \pi(6^2)(6) = 216\pi$.

Ratio $= 108\pi : 216\pi = 1:2$.

(1:1 is the trap — swapping diameter and height does *not* preserve volume, since radius is squared.)

ANSWER 5: B

# Problem 6:
Let the two equal sides be $a$ and the base be $b$, with $2a + b = 23$.

Since $2a$ is even, $b$ must be odd. Triangle inequality: $2a > b$, i.e. $23 - b > b \Rightarrow b < 11.5$.

So $b \in \{1,3,5,7,9,11\}$, giving $a = 11,10,9,8,7,6$ — six triangles:
(11,11,1), (10,10,3), (9,9,5), (8,8,7), (7,7,9), (6,6,11).

That's 6. (No equilateral option since 23 isn't divisible by 3; $b=13$ fails since $5+5<13$.)

ANSWER 6: C

# Problem 7:
Harold ate $\frac14$, leaving $\frac34$.

Moose ate $\frac13$ of that, leaving $\frac23 \cdot \frac34 = \frac12$.

Porcupine ate $\frac13$ of that, leaving $\frac23 \cdot \frac12 = \frac13$.

(Trap: subtracting $\frac14+\frac13+\frac13$ from 1 gives $\frac1{12}$ — wrong, because each animal eats a fraction of the *remainder*, not of the whole pie.)

ANSWER 7: D

# Problem 8:
Total money $10, minus the $2 tip → $8 available for fare.

First $\frac12$ mile costs $2.40, leaving $8.00 - 2.40 = \$5.60$.

Each additional 0.1 mile costs $0.20: $5.60 \div 0.20 = 28$ increments → $28 \times 0.1 = 2.8$ miles.

Total distance $= 0.5 + 2.8 = 3.3$ miles.

(Forgetting the tip gives $\$10-\$2.40=\$7.60 \to 3.8$ mi; forgetting the first half-mile is included gives other traps like 3.5.)

ANSWER 8: C

# Problem 9:
Population triples every 25 years, starting at 200 in 1998.

- 1998: 200
- 2023: 600
- 2048: 1800

The year 2050 is just 2 years past 2048, so the population is a bit more than 1800 — about 2000.

(3000 would require another near-tripling; 1000 is too small.)

ANSWER 9: D

# Problem 10:
The wall is 19 ft wide; the picture is 3 ft across and centered. The leftover wall splits equally on both sides:
$$\frac{19-3}{2} = \frac{16}{2} = 8 \text{ feet}.$$

(16 is the total leftover, not one side; $9\frac12$ is the wall's midpoint.)

ANSWER 10: B

Problem 1:

Initially, \(80\%\) of 500 balls are red:

\[
0.80(500)=400
\]

So there are 400 red balls and 100 blue balls.

Suppose \(x\) red balls are removed. Then:

- Red balls remaining: \(400-x\)
- Total balls remaining: \(500-x\)

We want \(75\%\) of the remaining balls to be red:

\[
\frac{400-x}{500-x}=\frac34
\]

Cross-multiply:

\[
4(400-x)=3(500-x)
\]

\[
1600-4x=1500-3x
\]

\[
100=x
\]

Removing 100 red balls leaves 300 red and 100 blue, for 400 total. Indeed,

\[
\frac{300}{400}=75\%.
\]

Choices such as 50 or 75 do not reduce the red percentage enough; 150 reduces it too much.

ANSWER 1: D

Problem 2:

Each child can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so \(1\) arrangement.
- All 4 girls: only \(GGGG\), so \(1\) arrangement.
- Exactly 2 girls and 2 boys:

\[
\binom42=6
\]

arrangements.
- Three of one gender and one of the other:
  - 3 boys and 1 girl: \(\binom41=4\) arrangements.
  - 3 girls and 1 boy: \(\binom41=4\) arrangements.

So this outcome has

\[
4+4=8
\]

arrangements.

Since \(8\) is more than \(6\) or \(1\), the most likely outcome is 3 children of one gender and 1 of the other.

ANSWER 2: D

Problem 3:

A grid 60 toothpicks long and 32 toothpicks wide has 60 columns and 32 rows of small squares.

For the horizontal toothpicks, there are \(32+1=33\) horizontal lines, each containing 60 toothpicks:

\[
33\cdot 60=1980
\]

For the vertical toothpicks, there are \(60+1=61\) vertical lines, each containing 32 toothpicks:

\[
61\cdot 32=1952
\]

Thus the total number is

\[
1980+1952=3932.
\]

The tempting answer \(1920=60\cdot32\) counts only the small squares, not the toothpicks forming their edges.

ANSWER 3: E

Problem 4:

The expression is of the form

\[
\Box^{\Box}\times\Box^{\Box},
\]

using the digits \(2,0,2,3\) once each.

To make the product as large as possible, the digit \(0\) should be an exponent, because any nonzero number to the zero power equals 1:

\[
2^0=1 \qquad\text{or}\qquad 3^0=1.
\]

We should use the remaining digits to make the largest possible power. Compare:

\[
3^2=9
\]

and

\[
2^3=8.
\]

Thus the best arrangement is

\[
3^2\times 2^0=9\times 1=9.
\]

Using \(2^3\) gives only \(8\), and placing 0 as a base would make the entire product 0.

ANSWER 4: C

Problem 5:

The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s can has diameter 6 cm, so its radius is 3 cm and its height is 12 cm:

\[
V_A=\pi(3)^2(12)=108\pi.
\]

Felicia’s can has diameter 12 cm, so its radius is 6 cm and its height is 6 cm:

\[
V_F=\pi(6)^2(6)=216\pi.
\]

The ratio is

\[
108\pi:216\pi=1:2.
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, making her base area four times as large.

ANSWER 5: B

Problem 6:

Let the equal sides of the isosceles triangle each have length \(a\). Let the base have length \(b\).

Since the perimeter is 23,

\[
2a+b=23,
\]

so

\[
b=23-2a.
\]

For a valid triangle, the base must be less than the sum of the equal sides:

\[
b<2a.
\]

Substitute \(b=23-2a\):

\[
23-2a<2a
\]

\[
23<4a
\]

\[
a>5.75.
\]

Also, \(b\) must be positive:

\[
23-2a>0,
\]

so

\[
a<11.5.
\]

Thus \(a\) can be

\[
6,7,8,9,10,11.
\]

That gives 6 different triangles. Their side lengths are:

\[
(6,6,11),\ (7,7,9),\ (8,8,7),\ (9,9,5),\ (10,10,3),\ (11,11,1).
\]

Values below 6 fail the triangle inequality, while values above 11 give a nonpositive base.

ANSWER 6: C

Problem 7:

Harold eats \(\frac14\) of the pie, leaving

\[
1-\frac14=\frac34.
\]

The moose eats \(\frac13\) of what is left, so it leaves \(\frac23\) of what Harold left:

\[
\frac23\cdot\frac34=\frac12.
\]

Then the porcupine eats \(\frac13\) of that remaining \(\frac12\), leaving \(\frac23\) of it:

\[
\frac23\cdot\frac12=\frac13.
\]

So \(\frac13\) of the original pie remains.

A common mistake is to subtract \(\frac13\) directly from \(\frac34\); the moose eats one-third of the amount currently left, not one-third of the original pie.

ANSWER 7: D

Problem 8:

The total amount available for the taxi fare is

\[
\$10-\$2=\$8.
\]

The first \(\frac12\) mile costs \$2.40. Therefore, the amount left for additional distance is

\[
8.00-2.40=5.60.
\]

Each additional \(0.1\) mile costs \$0.20, so the number of additional tenths of a mile is

\[
\frac{5.60}{0.20}=28.
\]

That is

\[
28(0.1)=2.8
\]

additional miles.

Adding the first \(\frac12\) mile:

\[
2.8+0.5=3.3.
\]

The ride can be 3.3 miles. The \$2 tip must be included, so choices based on spending the full \$10 on fare are too large.

ANSWER 8: C

Problem 9:

The population is 200 in 1998 and triples every 25 years.

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is a little more than two 25-year periods.

After 25 years, the population would be

\[
200\cdot 3=600.
\]

After 50 years, it would be

\[
600\cdot 3=1800.
\]

Since 2050 is about 2 years later, the population should be slightly more than 1800. The closest estimate is 2000.

Choices such as 600 correspond to only one tripling, while 3000 is much too high for just over two tripling periods.

ANSWER 9: D

Problem 10:

The wall is 19 feet wide, and the picture is 3 feet wide.

The amount of wall not covered by the picture is

\[
19-3=16
\]

feet.

Since the picture is centered, this 16 feet is divided equally between the two ends:

\[
\frac{16}{2}=8.
\]

Therefore, the nearest edge of the picture is 8 feet from either end of the wall.

ANSWER 10: B

Problem 1:
We need how many red balls to take away so reds are 75% of what is left.

At first there are 500 balls, 80% red.
Red = 0.80 × 500 = 400. Blue = 500 - 400 = 100.

Let x = number of red balls removed. Blue balls are not touched.
Red left = 400 - x. Total left = 500 - x.

We want:
(400 - x) / (500 - x) = 75% = 3/4

Cross-multiply:
4(400 - x) = 3(500 - x)
1600 - 4x = 1500 - 3x
1600 - 1500 = 4x - 3x
100 = x

Check: 300 red left, 100 blue left, total 400. 300/400 = 3/4.
Removing 25 would leave 375/475 ≈ 79%, still too high. Removing 150 would leave 250/350 ≈ 71%, too low.

ANSWER 1: D

Problem 2:
We are asked which gender pattern for 4 babies is most likely, each baby boy/girl equally likely.

There are 2^4 = 16 equally likely boy-girl sequences.

A. all 4 boys: only BBBB: 1 sequence. Probability 1/16.
B. all 4 girls: only GGGG: 1 sequence. Probability 1/16.
C. 2 girls and 2 boys: number of sequences = choose which 2 of 4 are girls = C(4,2)=6. Probability 6/16.
D. 3 of one gender and 1 of the other: This means 3B1G OR 3G1B.
3B1G: C(4,1)=4 sequences. 3G1B: C(4,1)=4 sequences. Total 8 sequences. Probability 8/16 = 1/2.
E. claims equal, false.

Many think 2-2 is most likely because it is most "balanced", but 3-1 has more sequences: 8 > 6 > 1.

So D is most likely.

ANSWER 2: D

Problem 3:
We need total toothpicks to make a rectangular grid 60 long by 32 wide, where length is measured in toothpicks.

Imagine unit squares. Top edge needs 60 toothpicks side by side. Same for every horizontal line.

Number of horizontal lines: there are 32 squares vertically, so 32+1 = 33 horizontal grid lines.
Each horizontal line uses 60 toothpicks.
Horizontal total = 33 × 60 = 1980.

Number of vertical lines: there are 60 squares horizontally, so 60+1 = 61 vertical grid lines.
Each vertical line uses 32 toothpicks.
Vertical total = 61 × 32 = 1952.

Total = 1980 + 1952 = 3932.

A 1920 = 60×32 is just the number of cells, not toothpicks. B 1952 is vertical only. C 1980 is horizontal only.

ANSWER 3: E

Problem 4:
We are asked to put digits 2,0,2,3, one per box, into the expression
□^□ × □^□
to get the largest value.

We must use all four digits as two bases and two exponents.

Key facts: a^0 = 1 for a ≠ 0, while 0^positive = 0, which kills a product. So do not put 0 as a base with positive exponent. Put 0 as an exponent to get 1. 0^0 is not allowed.

Try cases:

If exponents are 2 and 0, bases 3 and 2:
3^2 × 2^0 = 9 × 1 = 9
2^2 × 3^0 = 4 × 1 = 4
3^0 × 2^2 = 1 × 4 = 4
2^0 × 3^2 = 1 × 9 = 9

If exponents are 3 and 0, bases 2 and 2:
2^3 × 2^0 = 8 × 1 = 8

If 0 is a base with positive exponent, product is 0, e.g. 3^2 × 0^2 = 0.

So the best is 9.

Why not other choices?
A. 0: tempting if you think any zero makes everything zero, but 2^0 =1 avoids zero.
B. 8: this is 2^3 × 2^0, valid but less than 9.
D. 16 and E. 18: these come from the mistake x^0 = x. E.g. 3^2 × 2 =18, 2^3 × 2 =16. Correct is ×1, not ×2.

Maximum is 9.

ANSWER 4: C

Problem 5:
We compare volumes of cylinders. Volume = π r^2 h, r = diameter/2.

Alex: diameter 6, so r = 3. h =12.
V_Alex = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r = 6. h =6.
V_Felicia = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio Alex : Felicia = 108π : 216π = 108:216 = 1:2.

Volume scales as square of radius, so doubling diameter quadruples base area, which outweighs halving height. So Felicia's is bigger, ruling out C,D,E.

ANSWER 5: B

Problem 6:
We need integer-sided isosceles triangles with perimeter 23.

Let equal sides be a,a and base be b. Then:
2a + b = 23, so b = 23 - 2a > 0, so a ≤ 11.

Triangle inequality: sum of two small sides > third side. The only non-trivial one is:
a + a > b
2a > 23 - 2a
4a > 23
a > 5.75, so a ≥ 6.

Thus a = 6,7,8,9,10,11. Each gives a b:
6,6,11: 6+6>11 valid
7,7,9: 14>9 valid
8,8,7: valid
9,9,5: valid
10,10,3: 10+3>10 valid
11,11,1: 11+1>11 valid

That is 6 different triangles. No other a works.

ANSWER 6: C

Problem 7:
We track fraction of original pie left.

Harold eats 1/4, so left = 1 - 1/4 = 3/4.

Moose eats 1/3 of what Harold left:
eaten = (1/3)×(3/4) = 1/4.
Left after moose = 3/4 - 1/4 = 1/2. Equivalently (2/3)×(3/4)=1/2.

Porcupine eats 1/3 of what moose left:
eaten = (1/3)×(1/2) = 1/6.
Left after porcupine = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3. Equivalently (2/3)×(1/2)=1/3.

So 1/3 remains. A 1/12, B 1/6 are the amounts eaten, not left.

ANSWER 7: D

Problem 8:
We have $10 total, $2 tip, so fare money = 10 - 2 = $8.00.

First 0.5 mile costs $2.40.
Remaining for extra distance = 8.00 - 2.40 = $5.60.

Extra is $0.20 per 0.1 mile.
Number of 0.1-mile blocks = 5.60 / 0.20 = 28.
Extra distance = 28 × 0.1 = 2.8 miles.

Total distance = 0.5 + 2.8 = 3.3 miles.

Check: fare = 2.40 + 28×0.20 = 2.40+5.60=8.00, plus $2 tip = $10.

ANSWER 8: C

Problem 9:
Population in 1998 is 200, triples every 25 years. Estimate in 2050.

Time elapsed = 2050 - 1998 = 52 years. That's about two 25-year periods.

After 25 yr (~2023): 200×3 = 600.
After 50 yr (~2048): 600×3 = 1800.

2050 is 2 years more, so a little above 1800. Nearest estimate among choices is 2000.

A 600 is only one tripling. C 1000, B 800 ignore tripling. E 3000 is too high.

ANSWER 9: D

Problem 10:
Picture 3 ft wide centered on 19 ft wall.

Total wall not covered = 19 - 3 = 16 ft.
Centered means half on each side: 16/2 = 8 ft from each end to picture edge.

A 1 1/2, C 9 1/2 confuse center vs edge. B is correct.

ANSWER 10: B

Problem 1:
We are asked to find how many red balls must be removed from a bag of 500 balls (originally 80% red) so that the remaining red balls constitute 75% of the bag.

First, calculate the original number of balls of each color:
- Total balls = $500$
- Red balls = $80\% \text{ of } 500 = 0.80 \times 500 = 400$
- Blue balls = $500 - 400 = 100$

Only red balls are removed, so the number of blue balls remains constant at $100$.
In the final state, red balls make up $75\%$ of the total, which means blue balls must make up:
$$100\% - 75\% = 25\%$$
of the remaining balls.

Let $T$ be the total number of remaining balls:
$$0.25 \times T = 100 \implies T = \frac{100}{0.25} = 400$$

Since the bag originally contained $500$ balls and now contains $400$ balls, the number of red balls removed is:
$$500 - 400 = 100$$

ANSWER 1: D

---

Problem 2:
We are asked to determine which outcome is most likely when four children are born, with boys (B) and girls (G) being equally likely.

Each child has $2$ possible genders, so there are $2^4 = 16$ equally likely possible outcomes in total. We evaluate the number of outcomes for each choice:
- **A. All 4 are boys:** Only $1$ outcome ($\text{BBBB}$). Probability = $\frac{1}{16}$.
- **B. All 4 are girls:** Only $1$ outcome ($\text{GGGG}$). Probability = $\frac{1}{16}$.
- **C. 2 are girls and 2 are boys:** The number of ways to choose which $2$ children are girls is $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ outcomes. Probability = $\frac{6}{16}$.
- **D. 3 are of one gender and 1 is of the other:** This includes either 3 boys and 1 girl ($\binom{4}{1} = 4$ outcomes) or 3 girls and 1 boy ($\binom{4}{3} = 4$ outcomes), giving $4 + 4 = 8$ outcomes. Probability = $\frac{8}{16} = \frac{1}{2}$.

Comparing the possibilities, outcome D has $8$ favorable outcomes, which is greater than C ($6$ outcomes), A ($1$ outcome), and B ($1$ outcome). Therefore, D is the most likely outcome.

ANSWER 2: D

---

Problem 3:
We are asked to find the total number of toothpicks used to make a rectangular grid that is $60$ toothpicks long and $32$ toothpicks wide.

A grid of dimensions $60 \times 32$ consists of unit squares arranged in $32$ rows and $60$ columns:
- **Horizontal toothpicks:** There are $32 + 1 = 33$ horizontal grid lines. Each horizontal line is made of $60$ toothpicks.
  $$\text{Number of horizontal toothpicks} = 33 \times 60 = 1980$$
- **Vertical toothpicks:** There are $60 + 1 = 61$ vertical grid lines. Each vertical line is made of $32$ toothpicks.
  $$\text{Number of vertical toothpicks} = 61 \times 32 = 1952$$

Adding the horizontal and vertical toothpicks together:
$$\text{Total toothpicks} = 1980 + 1952 = 3932$$

ANSWER 3: E

---

Problem 4:
We are asked to find the maximum possible value of the expression $\square^\square \times \square^\square$ when placing the digits $2, 0, 2,$ and $3$ into the four boxes, one digit per box.

Let the expression be $a^b \times c^d$, where $\{a, b, c, d\} = \{2, 0, 2, 3\}$.
We test the possible assignments of the digits:
- To maximize the product, we want large bases and exponents, but one of the digits is $0$.
- If $0$ is used as a base, say $0^d$ with $d > 0$, the factor becomes $0$, making the entire product $0$.
- If $0$ is placed in an exponent, recall that for any non-zero base $x$, $x^0 = 1$.
  - Pair $0$ with $2$ as an exponent: $2^0 = 1$. The remaining digits are $3$ and $2$.
    - Setting $3^2 \times 2^0 = 9 \times 1 = 9$.
    - Setting $2^3 \times 2^0 = 8 \times 1 = 8$.
  - Pair $0$ with $3$ as an exponent: $3^0 = 1$. The remaining digits are $2$ and $2$.
    - Setting $2^2 \times 3^0 = 4 \times 1 = 4$.

The values $16$ and $18$ are distractors arising from the common misconception that $x^0 = x$ (which would give $2^3 \times 2 = 16$ or $3^2 \times 2 = 18$).
The maximum mathematically valid value achievable is $9$.

ANSWER 4: C

---

Problem 5:
We are asked to find the ratio of the volume of Alex's cat food can to the volume of Felicia's cat food can.

The volume of a cylinder is given by $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height.
- For Alex's can:
  - Diameter = $6\text{ cm} \implies \text{radius } r_A = 3\text{ cm}$
  - Height $h_A = 12\text{ cm}$
  - Volume $V_A = \pi (3)^2 (12) = \pi \times 9 \times 12 = 108\pi\text{ cm}^3$

- For Felicia's can:
  - Diameter = $12\text{ cm} \implies \text{radius } r_F = 6\text{ cm}$
  - Height $h_F = 6\text{ cm}$
  - Volume $V_F = \pi (6)^2 (6) = \pi \times 36 \times 6 = 216\pi\text{ cm}^3$

The ratio of the volume of Alex's can to Felicia's can is:
$$\frac{V_A}{V_F} = \frac{108\pi}{216\pi} = \frac{1}{2}$$

Thus, the ratio is $1:2$.

ANSWER 5: B

---

Problem 6:
We are asked to find the number of different isosceles triangles with integer side lengths and a perimeter of $23$.

Let the side lengths of the isosceles triangle be $a, a,$ and $b$, where $a$ and $b$ are positive integers.
The perimeter is:
$$2a + b = 23 \implies b = 23 - 2a$$

By the Triangle Inequality:
1. The sum of the two equal sides must be strictly greater than the third side:
   $$a + a > b \implies 2a > 23 - 2a \implies 4a > 23 \implies a \ge 6$$
2. The third side must be positive:
   $$b > 0 \implies 23 - 2a > 0 \implies 2a < 23 \implies a \le 11$$

The possible integer values for $a$ are $6, 7, 8, 9, 10, 11$.
We check each corresponding side length triple $(a, a, b)$:
- $a = 6 \implies b = 11$: sides $(6, 6, 11)$
- $a = 7 \implies b = 9$: sides $(7, 7, 9)$
- $a = 8 \implies b = 7$: sides $(8, 8, 7)$
- $a = 9 \implies b = 5$: sides $(9, 9, 5)$
- $a = 10 \implies b = 3$: sides $(10, 10, 3)$
- $a = 11 \implies b = 1$: sides $(11, 11, 1)$

All $6$ sets of side lengths satisfy the triangle inequality and have two equal sides (none are equilateral since $23$ is not divisible by $3$). Thus, there are $6$ such triangles.

ANSWER 6: C

---

Problem 7:
We are asked to find the fraction of the original pie remaining after Harold, a moose, and a porcupine each eat a portion.

1. Harold start

# Problem 1:
Restate: 500 balls, 80% red. Remove some **red** balls so that 75% of what remains is red.

Red = 0.80(500) = 400, Blue = 100.

Let $x$ = red balls removed. Blues never change (still 100), and after removal blues must be 25% of the total:
$$100 = 0.25(500 - x) \Rightarrow 400 = 500 - x \Rightarrow x = 100.$$

Check: 300 red, 100 blue, total 400 → 300/400 = 75%. ✓

(Tempting wrong: 25 or 50 come from mistakenly taking 5% or 10% of 500; but removing balls shrinks the total too, so more must go.)

ANSWER 1: D

# Problem 2:
There are $2^4 = 16$ equally likely sequences of genders.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: $\binom{4}{2}=6$ ways → 6/16
- 3 of one gender, 1 of the other: $\binom{4}{3}=4$ (3 boys) $+ \binom{4}{1}=4$ (3 girls) $= 8$ ways → 8/16

The largest is 8/16 = 1/2, the "3–1 split." Choice C (6/16) is the tempting trap because 2–2 is the single most likely *exact* split, but option D combines two cases.

ANSWER 2: D

# Problem 3:
A grid 60 toothpicks long and 32 toothpicks wide means the rectangle is 60 units by 32 units.

- Horizontal toothpicks: there are $32+1 = 33$ horizontal lines, each made of 60 toothpicks: $33 \times 60 = 1980$.
- Vertical toothpicks: there are $60+1 = 61$ vertical lines, each made of 32 toothpicks: $61 \times 32 = 1952$.

Total: $1980 + 1952 = 3932$.

(1980 and 1952 are the individual pieces — tempting partial answers.)

ANSWER 3: E

# Problem 4:
The expression is $\boxed{\phantom{2}}^{\boxed{\phantom{2}}} \times \boxed{\phantom{2}}^{\boxed{\phantom{2}}}$, using digits 2, 0, 2, 3 once each.

To maximize, make one factor as large as possible and let the other equal 1 (using exponent 0).

- $3^2 \times 2^0 = 9 \times 1 = 9$
- $2^3 \times 2^0 = 8$ (uses two 2's as base and exponent... gives 8)
- $2^2 \times 3^0 = 4$
- $2^0 \times 3^2 = 9$

Any arrangement with 0 as a base kills a factor to 0. The maximum is $9$.

(8 is the tempting near-miss from $2^3$; 16 and 18 aren't attainable.)

ANSWER 4: C

# Problem 5:
Volume of a cylinder: $V=\pi r^2 h$.

- Alex: diameter 6 → $r=3$, $h=12$: $V = \pi(3^2)(12) = 108\pi$.
- Felicia: diameter 12 → $r=6$, $h=6$: $V = \pi(6^2)(6) = 216\pi$.

Ratio $= 108\pi : 216\pi = 1:2$.

(1:1 is the trap — swapping diameter and height does *not* preserve volume, since radius is squared.)

ANSWER 5: B

# Problem 6:
Let the two equal sides be $a$ and the base be $b$, with $2a + b = 23$.

Since $2a$ is even, $b$ must be odd. Triangle inequality: $2a > b$, i.e. $23 - b > b \Rightarrow b < 11.5$.

So $b \in \{1,3,5,7,9,11\}$, giving $a = 11,10,9,8,7,6$ — six triangles:
(11,11,1), (10,10,3), (9,9,5), (8,8,7), (7,7,9), (6,6,11).

That's 6. (No equilateral option since 23 isn't divisible by 3; $b=13$ fails since $5+5<13$.)

ANSWER 6: C

# Problem 7:
Harold ate $\frac14$, leaving $\frac34$.

Moose ate $\frac13$ of that, leaving $\frac23 \cdot \frac34 = \frac12$.

Porcupine ate $\frac13$ of that, leaving $\frac23 \cdot \frac12 = \frac13$.

(Trap: subtracting $\frac14+\frac13+\frac13$ from 1 gives $\frac1{12}$ — wrong, because each animal eats a fraction of the *remainder*, not of the whole pie.)

ANSWER 7: D

# Problem 8:
Total money $10, minus the $2 tip → $8 available for fare.

First $\frac12$ mile costs $2.40, leaving $8.00 - 2.40 = \$5.60$.

Each additional 0.1 mile costs $0.20: $5.60 \div 0.20 = 28$ increments → $28 \times 0.1 = 2.8$ miles.

Total distance $= 0.5 + 2.8 = 3.3$ miles.

(Forgetting the tip gives $\$10-\$2.40=\$7.60 \to 3.8$ mi; forgetting the first half-mile is included gives other traps like 3.5.)

ANSWER 8: C

# Problem 9:
Population triples every 25 years, starting at 200 in 1998.

- 1998: 200
- 2023: 600
- 2048: 1800

The year 2050 is just 2 years past 2048, so the population is a bit more than 1800 — about 2000.

(3000 would require another near-tripling; 1000 is too small.)

ANSWER 9: D

# Problem 10:
The wall is 19 ft wide; the picture is 3 ft across and centered. The leftover wall splits equally on both sides:
$$\frac{19-3}{2} = \frac{16}{2} = 8 \text{ feet}.$$

(16 is the total leftover, not one side; $9\frac12$ is the wall's midpoint.)

ANSWER 10: B

Problem 1:

Initially, \(80\%\) of 500 balls are red:

\[
0.80(500)=400
\]

So there are 400 red balls and 100 blue balls.

Suppose \(x\) red balls are removed. Then:

- Red balls remaining: \(400-x\)
- Total balls remaining: \(500-x\)

We want \(75\%\) of the remaining balls to be red:

\[
\frac{400-x}{500-x}=\frac34
\]

Cross-multiply:

\[
4(400-x)=3(500-x)
\]

\[
1600-4x=1500-3x
\]

\[
100=x
\]

Removing 100 red balls leaves 300 red and 100 blue, for 400 total. Indeed,

\[
\frac{300}{400}=75\%.
\]

Choices such as 50 or 75 do not reduce the red percentage enough; 150 reduces it too much.

ANSWER 1: D

Problem 2:

Each child can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so \(1\) arrangement.
- All 4 girls: only \(GGGG\), so \(1\) arrangement.
- Exactly 2 girls and 2 boys:

\[
\binom42=6
\]

arrangements.
- Three of one gender and one of the other:
  - 3 boys and 1 girl: \(\binom41=4\) arrangements.
  - 3 girls and 1 boy: \(\binom41=4\) arrangements.

So this outcome has

\[
4+4=8
\]

arrangements.

Since \(8\) is more than \(6\) or \(1\), the most likely outcome is 3 children of one gender and 1 of the other.

ANSWER 2: D

Problem 3:

A grid 60 toothpicks long and 32 toothpicks wide has 60 columns and 32 rows of small squares.

For the horizontal toothpicks, there are \(32+1=33\) horizontal lines, each containing 60 toothpicks:

\[
33\cdot 60=1980
\]

For the vertical toothpicks, there are \(60+1=61\) vertical lines, each containing 32 toothpicks:

\[
61\cdot 32=1952
\]

Thus the total number is

\[
1980+1952=3932.
\]

The tempting answer \(1920=60\cdot32\) counts only the small squares, not the toothpicks forming their edges.

ANSWER 3: E

Problem 4:

The expression is of the form

\[
\Box^{\Box}\times\Box^{\Box},
\]

using the digits \(2,0,2,3\) once each.

To make the product as large as possible, the digit \(0\) should be an exponent, because any nonzero number to the zero power equals 1:

\[
2^0=1 \qquad\text{or}\qquad 3^0=1.
\]

We should use the remaining digits to make the largest possible power. Compare:

\[
3^2=9
\]

and

\[
2^3=8.
\]

Thus the best arrangement is

\[
3^2\times 2^0=9\times 1=9.
\]

Using \(2^3\) gives only \(8\), and placing 0 as a base would make the entire product 0.

ANSWER 4: C

Problem 5:

The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s can has diameter 6 cm, so its radius is 3 cm and its height is 12 cm:

\[
V_A=\pi(3)^2(12)=108\pi.
\]

Felicia’s can has diameter 12 cm, so its radius is 6 cm and its height is 6 cm:

\[
V_F=\pi(6)^2(6)=216\pi.
\]

The ratio is

\[
108\pi:216\pi=1:2.
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, making her base area four times as large.

ANSWER 5: B

Problem 6:

Let the equal sides of the isosceles triangle each have length \(a\). Let the base have length \(b\).

Since the perimeter is 23,

\[
2a+b=23,
\]

so

\[
b=23-2a.
\]

For a valid triangle, the base must be less than the sum of the equal sides:

\[
b<2a.
\]

Substitute \(b=23-2a\):

\[
23-2a<2a
\]

\[
23<4a
\]

\[
a>5.75.
\]

Also, \(b\) must be positive:

\[
23-2a>0,
\]

so

\[
a<11.5.
\]

Thus \(a\) can be

\[
6,7,8,9,10,11.
\]

That gives 6 different triangles. Their side lengths are:

\[
(6,6,11),\ (7,7,9),\ (8,8,7),\ (9,9,5),\ (10,10,3),\ (11,11,1).
\]

Values below 6 fail the triangle inequality, while values above 11 give a nonpositive base.

ANSWER 6: C

Problem 7:

Harold eats \(\frac14\) of the pie, leaving

\[
1-\frac14=\frac34.
\]

The moose eats \(\frac13\) of what is left, so it leaves \(\frac23\) of what Harold left:

\[
\frac23\cdot\frac34=\frac12.
\]

Then the porcupine eats \(\frac13\) of that remaining \(\frac12\), leaving \(\frac23\) of it:

\[
\frac23\cdot\frac12=\frac13.
\]

So \(\frac13\) of the original pie remains.

A common mistake is to subtract \(\frac13\) directly from \(\frac34\); the moose eats one-third of the amount currently left, not one-third of the original pie.

ANSWER 7: D

Problem 8:

The total amount available for the taxi fare is

\[
\$10-\$2=\$8.
\]

The first \(\frac12\) mile costs \$2.40. Therefore, the amount left for additional distance is

\[
8.00-2.40=5.60.
\]

Each additional \(0.1\) mile costs \$0.20, so the number of additional tenths of a mile is

\[
\frac{5.60}{0.20}=28.
\]

That is

\[
28(0.1)=2.8
\]

additional miles.

Adding the first \(\frac12\) mile:

\[
2.8+0.5=3.3.
\]

The ride can be 3.3 miles. The \$2 tip must be included, so choices based on spending the full \$10 on fare are too large.

ANSWER 8: C

Problem 9:

The population is 200 in 1998 and triples every 25 years.

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is a little more than two 25-year periods.

After 25 years, the population would be

\[
200\cdot 3=600.
\]

After 50 years, it would be

\[
600\cdot 3=1800.
\]

Since 2050 is about 2 years later, the population should be slightly more than 1800. The closest estimate is 2000.

Choices such as 600 correspond to only one tripling, while 3000 is much too high for just over two tripling periods.

ANSWER 9: D

Problem 10:

The wall is 19 feet wide, and the picture is 3 feet wide.

The amount of wall not covered by the picture is

\[
19-3=16
\]

feet.

Since the picture is centered, this 16 feet is divided equally between the two ends:

\[
\frac{16}{2}=8.
\]

Therefore, the nearest edge of the picture is 8 feet from either end of the wall.

ANSWER 10: B

Problem 1:
We need how many red balls to take away so reds are 75% of what is left.

At first there are 500 balls, 80% red.
Red = 0.80 × 500 = 400. Blue = 500 - 400 = 100.

Let x = number of red balls removed. Blue balls are not touched.
Red left = 400 - x. Total left = 500 - x.

We want:
(400 - x) / (500 - x) = 75% = 3/4

Cross-multiply:
4(400 - x) = 3(500 - x)
1600 - 4x = 1500 - 3x
1600 - 1500 = 4x - 3x
100 = x

Check: 300 red left, 100 blue left, total 400. 300/400 = 3/4.
Removing 25 would leave 375/475 ≈ 79%, still too high. Removing 150 would leave 250/350 ≈ 71%, too low.

ANSWER 1: D

Problem 2:
We are asked which gender pattern for 4 babies is most likely, each baby boy/girl equally likely.

There are 2^4 = 16 equally likely boy-girl sequences.

A. all 4 boys: only BBBB: 1 sequence. Probability 1/16.
B. all 4 girls: only GGGG: 1 sequence. Probability 1/16.
C. 2 girls and 2 boys: number of sequences = choose which 2 of 4 are girls = C(4,2)=6. Probability 6/16.
D. 3 of one gender and 1 of the other: This means 3B1G OR 3G1B.
3B1G: C(4,1)=4 sequences. 3G1B: C(4,1)=4 sequences. Total 8 sequences. Probability 8/16 = 1/2.
E. claims equal, false.

Many think 2-2 is most likely because it is most "balanced", but 3-1 has more sequences: 8 > 6 > 1.

So D is most likely.

ANSWER 2: D

Problem 3:
We need total toothpicks to make a rectangular grid 60 long by 32 wide, where length is measured in toothpicks.

Imagine unit squares. Top edge needs 60 toothpicks side by side. Same for every horizontal line.

Number of horizontal lines: there are 32 squares vertically, so 32+1 = 33 horizontal grid lines.
Each horizontal line uses 60 toothpicks.
Horizontal total = 33 × 60 = 1980.

Number of vertical lines: there are 60 squares horizontally, so 60+1 = 61 vertical grid lines.
Each vertical line uses 32 toothpicks.
Vertical total = 61 × 32 = 1952.

Total = 1980 + 1952 = 3932.

A 1920 = 60×32 is just the number of cells, not toothpicks. B 1952 is vertical only. C 1980 is horizontal only.

ANSWER 3: E

Problem 4:
We are asked to put digits 2,0,2,3, one per box, into the expression
□^□ × □^□
to get the largest value.

We must use all four digits as two bases and two exponents.

Key facts: a^0 = 1 for a ≠ 0, while 0^positive = 0, which kills a product. So do not put 0 as a base with positive exponent. Put 0 as an exponent to get 1. 0^0 is not allowed.

Try cases:

If exponents are 2 and 0, bases 3 and 2:
3^2 × 2^0 = 9 × 1 = 9
2^2 × 3^0 = 4 × 1 = 4
3^0 × 2^2 = 1 × 4 = 4
2^0 × 3^2 = 1 × 9 = 9

If exponents are 3 and 0, bases 2 and 2:
2^3 × 2^0 = 8 × 1 = 8

If 0 is a base with positive exponent, product is 0, e.g. 3^2 × 0^2 = 0.

So the best is 9.

Why not other choices?
A. 0: tempting if you think any zero makes everything zero, but 2^0 =1 avoids zero.
B. 8: this is 2^3 × 2^0, valid but less than 9.
D. 16 and E. 18: these come from the mistake x^0 = x. E.g. 3^2 × 2 =18, 2^3 × 2 =16. Correct is ×1, not ×2.

Maximum is 9.

ANSWER 4: C

Problem 5:
We compare volumes of cylinders. Volume = π r^2 h, r = diameter/2.

Alex: diameter 6, so r = 3. h =12.
V_Alex = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r = 6. h =6.
V_Felicia = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio Alex : Felicia = 108π : 216π = 108:216 = 1:2.

Volume scales as square of radius, so doubling diameter quadruples base area, which outweighs halving height. So Felicia's is bigger, ruling out C,D,E.

ANSWER 5: B

Problem 6:
We need integer-sided isosceles triangles with perimeter 23.

Let equal sides be a,a and base be b. Then:
2a + b = 23, so b = 23 - 2a > 0, so a ≤ 11.

Triangle inequality: sum of two small sides > third side. The only non-trivial one is:
a + a > b
2a > 23 - 2a
4a > 23
a > 5.75, so a ≥ 6.

Thus a = 6,7,8,9,10,11. Each gives a b:
6,6,11: 6+6>11 valid
7,7,9: 14>9 valid
8,8,7: valid
9,9,5: valid
10,10,3: 10+3>10 valid
11,11,1: 11+1>11 valid

That is 6 different triangles. No other a works.

ANSWER 6: C

Problem 7:
We track fraction of original pie left.

Harold eats 1/4, so left = 1 - 1/4 = 3/4.

Moose eats 1/3 of what Harold left:
eaten = (1/3)×(3/4) = 1/4.
Left after moose = 3/4 - 1/4 = 1/2. Equivalently (2/3)×(3/4)=1/2.

Porcupine eats 1/3 of what moose left:
eaten = (1/3)×(1/2) = 1/6.
Left after porcupine = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3. Equivalently (2/3)×(1/2)=1/3.

So 1/3 remains. A 1/12, B 1/6 are the amounts eaten, not left.

ANSWER 7: D

Problem 8:
We have $10 total, $2 tip, so fare money = 10 - 2 = $8.00.

First 0.5 mile costs $2.40.
Remaining for extra distance = 8.00 - 2.40 = $5.60.

Extra is $0.20 per 0.1 mile.
Number of 0.1-mile blocks = 5.60 / 0.20 = 28.
Extra distance = 28 × 0.1 = 2.8 miles.

Total distance = 0.5 + 2.8 = 3.3 miles.

Check: fare = 2.40 + 28×0.20 = 2.40+5.60=8.00, plus $2 tip = $10.

ANSWER 8: C

Problem 9:
Population in 1998 is 200, triples every 25 years. Estimate in 2050.

Time elapsed = 2050 - 1998 = 52 years. That's about two 25-year periods.

After 25 yr (~2023): 200×3 = 600.
After 50 yr (~2048): 600×3 = 1800.

2050 is 2 years more, so a little above 1800. Nearest estimate among choices is 2000.

A 600 is only one tripling. C 1000, B 800 ignore tripling. E 3000 is too high.

ANSWER 9: D

Problem 10:
Picture 3 ft wide centered on 19 ft wall.

Total wall not covered = 19 - 3 = 16 ft.
Centered means half on each side: 16/2 = 8 ft from each end to picture edge.

A 1 1/2, C 9 1/2 confuse center vs edge. B is correct.

ANSWER 10: B

Problem 1:
We are asked to find how many red balls must be removed from a bag of 500 balls (originally 80% red) so that the remaining red balls constitute 75% of the bag.

First, calculate the original number of balls of each color:
- Total balls = $500$
- Red balls = $80\% \text{ of } 500 = 0.80 \times 500 = 400$
- Blue balls = $500 - 400 = 100$

Only red balls are removed, so the number of blue balls remains constant at $100$.
In the final state, red balls make up $75\%$ of the total, which means blue balls must make up:
$$100\% - 75\% = 25\%$$
of the remaining balls.

Let $T$ be the total number of remaining balls:
$$0.25 \times T = 100 \implies T = \frac{100}{0.25} = 400$$

Since the bag originally contained $500$ balls and now contains $400$ balls, the number of red balls removed is:
$$500 - 400 = 100$$

ANSWER 1: D

---

Problem 2:
We are asked to determine which outcome is most likely when four children are born, with boys (B) and girls (G) being equally likely.

Each child has $2$ possible genders, so there are $2^4 = 16$ equally likely possible outcomes in total. We evaluate the number of outcomes for each choice:
- **A. All 4 are boys:** Only $1$ outcome ($\text{BBBB}$). Probability = $\frac{1}{16}$.
- **B. All 4 are girls:** Only $1$ outcome ($\text{GGGG}$). Probability = $\frac{1}{16}$.
- **C. 2 are girls and 2 are boys:** The number of ways to choose which $2$ children are girls is $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ outcomes. Probability = $\frac{6}{16}$.
- **D. 3 are of one gender and 1 is of the other:** This includes either 3 boys and 1 girl ($\binom{4}{1} = 4$ outcomes) or 3 girls and 1 boy ($\binom{4}{3} = 4$ outcomes), giving $4 + 4 = 8$ outcomes. Probability = $\frac{8}{16} = \frac{1}{2}$.

Comparing the possibilities, outcome D has $8$ favorable outcomes, which is greater than C ($6$ outcomes), A ($1$ outcome), and B ($1$ outcome). Therefore, D is the most likely outcome.

ANSWER 2: D

---

Problem 3:
We are asked to find the total number of toothpicks used to make a rectangular grid that is $60$ toothpicks long and $32$ toothpicks wide.

A grid of dimensions $60 \times 32$ consists of unit squares arranged in $32$ rows and $60$ columns:
- **Horizontal toothpicks:** There are $32 + 1 = 33$ horizontal grid lines. Each horizontal line is made of $60$ toothpicks.
  $$\text{Number of horizontal toothpicks} = 33 \times 60 = 1980$$
- **Vertical toothpicks:** There are $60 + 1 = 61$ vertical grid lines. Each vertical line is made of $32$ toothpicks.
  $$\text{Number of vertical toothpicks} = 61 \times 32 = 1952$$

Adding the horizontal and vertical toothpicks together:
$$\text{Total toothpicks} = 1980 + 1952 = 3932$$

ANSWER 3: E

---

Problem 4:
We are asked to find the maximum possible value of the expression $\square^\square \times \square^\square$ when placing the digits $2, 0, 2,$ and $3$ into the four boxes, one digit per box.

Let the expression be $a^b \times c^d$, where $\{a, b, c, d\} = \{2, 0, 2, 3\}$.
We test the possible assignments of the digits:
- To maximize the product, we want large bases and exponents, but one of the digits is $0$.
- If $0$ is used as a base, say $0^d$ with $d > 0$, the factor becomes $0$, making the entire product $0$.
- If $0$ is placed in an exponent, recall that for any non-zero base $x$, $x^0 = 1$.
  - Pair $0$ with $2$ as an exponent: $2^0 = 1$. The remaining digits are $3$ and $2$.
    - Setting $3^2 \times 2^0 = 9 \times 1 = 9$.
    - Setting $2^3 \times 2^0 = 8 \times 1 = 8$.
  - Pair $0$ with $3$ as an exponent: $3^0 = 1$. The remaining digits are $2$ and $2$.
    - Setting $2^2 \times 3^0 = 4 \times 1 = 4$.

The values $16$ and $18$ are distractors arising from the common misconception that $x^0 = x$ (which would give $2^3 \times 2 = 16$ or $3^2 \times 2 = 18$).
The maximum mathematically valid value achievable is $9$.

ANSWER 4: C

---

Problem 5:
We are asked to find the ratio of the volume of Alex's cat food can to the volume of Felicia's cat food can.

The volume of a cylinder is given by $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height.
- For Alex's can:
  - Diameter = $6\text{ cm} \implies \text{radius } r_A = 3\text{ cm}$
  - Height $h_A = 12\text{ cm}$
  - Volume $V_A = \pi (3)^2 (12) = \pi \times 9 \times 12 = 108\pi\text{ cm}^3$

- For Felicia's can:
  - Diameter = $12\text{ cm} \implies \text{radius } r_F = 6\text{ cm}$
  - Height $h_F = 6\text{ cm}$
  - Volume $V_F = \pi (6)^2 (6) = \pi \times 36 \times 6 = 216\pi\text{ cm}^3$

The ratio of the volume of Alex's can to Felicia's can is:
$$\frac{V_A}{V_F} = \frac{108\pi}{216\pi} = \frac{1}{2}$$

Thus, the ratio is $1:2$.

ANSWER 5: B

---

Problem 6:
We are asked to find the number of different isosceles triangles with integer side lengths and a perimeter of $23$.

Let the side lengths of the isosceles triangle be $a, a,$ and $b$, where $a$ and $b$ are positive integers.
The perimeter is:
$$2a + b = 23 \implies b = 23 - 2a$$

By the Triangle Inequality:
1. The sum of the two equal sides must be strictly greater than the third side:
   $$a + a > b \implies 2a > 23 - 2a \implies 4a > 23 \implies a \ge 6$$
2. The third side must be positive:
   $$b > 0 \implies 23 - 2a > 0 \implies 2a < 23 \implies a \le 11$$

The possible integer values for $a$ are $6, 7, 8, 9, 10, 11$.
We check each corresponding side length triple $(a, a, b)$:
- $a = 6 \implies b = 11$: sides $(6, 6, 11)$
- $a = 7 \implies b = 9$: sides $(7, 7, 9)$
- $a = 8 \implies b = 7$: sides $(8, 8, 7)$
- $a = 9 \implies b = 5$: sides $(9, 9, 5)$
- $a = 10 \implies b = 3$: sides $(10, 10, 3)$
- $a = 11 \implies b = 1$: sides $(11, 11, 1)$

All $6$ sets of side lengths satisfy the triangle inequality and have two equal sides (none are equilateral since $23$ is not divisible by $3$). Thus, there are $6$ such triangles.

ANSWER 6: C

---

Problem 7:
We are asked to find the fraction of the original pie remaining after Harold, a moose, and a porcupine each eat a portion.

1. Harold start

# Problem 1:
Restate: 500 balls, 80% red. Remove some **red** balls so that 75% of what remains is red.

Red = 0.80(500) = 400, Blue = 100.

Let $x$ = red balls removed. Blues never change (still 100), and after removal blues must be 25% of the total:
$$100 = 0.25(500 - x) \Rightarrow 400 = 500 - x \Rightarrow x = 100.$$

Check: 300 red, 100 blue, total 400 → 300/400 = 75%. ✓

(Tempting wrong: 25 or 50 come from mistakenly taking 5% or 10% of 500; but removing balls shrinks the total too, so more must go.)

ANSWER 1: D

# Problem 2:
There are $2^4 = 16$ equally likely sequences of genders.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: $\binom{4}{2}=6$ ways → 6/16
- 3 of one gender, 1 of the other: $\binom{4}{3}=4$ (3 boys) $+ \binom{4}{1}=4$ (3 girls) $= 8$ ways → 8/16

The largest is 8/16 = 1/2, the "3–1 split." Choice C (6/16) is the tempting trap because 2–2 is the single most likely *exact* split, but option D combines two cases.

ANSWER 2: D

# Problem 3:
A grid 60 toothpicks long and 32 toothpicks wide means the rectangle is 60 units by 32 units.

- Horizontal toothpicks: there are $32+1 = 33$ horizontal lines, each made of 60 toothpicks: $33 \times 60 = 1980$.
- Vertical toothpicks: there are $60+1 = 61$ vertical lines, each made of 32 toothpicks: $61 \times 32 = 1952$.

Total: $1980 + 1952 = 3932$.

(1980 and 1952 are the individual pieces — tempting partial answers.)

ANSWER 3: E

# Problem 4:
The expression is $\boxed{\phantom{2}}^{\boxed{\phantom{2}}} \times \boxed{\phantom{2}}^{\boxed{\phantom{2}}}$, using digits 2, 0, 2, 3 once each.

To maximize, make one factor as large as possible and let the other equal 1 (using exponent 0).

- $3^2 \times 2^0 = 9 \times 1 = 9$
- $2^3 \times 2^0 = 8$ (uses two 2's as base and exponent... gives 8)
- $2^2 \times 3^0 = 4$
- $2^0 \times 3^2 = 9$

Any arrangement with 0 as a base kills a factor to 0. The maximum is $9$.

(8 is the tempting near-miss from $2^3$; 16 and 18 aren't attainable.)

ANSWER 4: C

# Problem 5:
Volume of a cylinder: $V=\pi r^2 h$.

- Alex: diameter 6 → $r=3$, $h=12$: $V = \pi(3^2)(12) = 108\pi$.
- Felicia: diameter 12 → $r=6$, $h=6$: $V = \pi(6^2)(6) = 216\pi$.

Ratio $= 108\pi : 216\pi = 1:2$.

(1:1 is the trap — swapping diameter and height does *not* preserve volume, since radius is squared.)

ANSWER 5: B

# Problem 6:
Let the two equal sides be $a$ and the base be $b$, with $2a + b = 23$.

Since $2a$ is even, $b$ must be odd. Triangle inequality: $2a > b$, i.e. $23 - b > b \Rightarrow b < 11.5$.

So $b \in \{1,3,5,7,9,11\}$, giving $a = 11,10,9,8,7,6$ — six triangles:
(11,11,1), (10,10,3), (9,9,5), (8,8,7), (7,7,9), (6,6,11).

That's 6. (No equilateral option since 23 isn't divisible by 3; $b=13$ fails since $5+5<13$.)

ANSWER 6: C

# Problem 7:
Harold ate $\frac14$, leaving $\frac34$.

Moose ate $\frac13$ of that, leaving $\frac23 \cdot \frac34 = \frac12$.

Porcupine ate $\frac13$ of that, leaving $\frac23 \cdot \frac12 = \frac13$.

(Trap: subtracting $\frac14+\frac13+\frac13$ from 1 gives $\frac1{12}$ — wrong, because each animal eats a fraction of the *remainder*, not of the whole pie.)

ANSWER 7: D

# Problem 8:
Total money $10, minus the $2 tip → $8 available for fare.

First $\frac12$ mile costs $2.40, leaving $8.00 - 2.40 = \$5.60$.

Each additional 0.1 mile costs $0.20: $5.60 \div 0.20 = 28$ increments → $28 \times 0.1 = 2.8$ miles.

Total distance $= 0.5 + 2.8 = 3.3$ miles.

(Forgetting the tip gives $\$10-\$2.40=\$7.60 \to 3.8$ mi; forgetting the first half-mile is included gives other traps like 3.5.)

ANSWER 8: C

# Problem 9:
Population triples every 25 years, starting at 200 in 1998.

- 1998: 200
- 2023: 600
- 2048: 1800

The year 2050 is just 2 years past 2048, so the population is a bit more than 1800 — about 2000.

(3000 would require another near-tripling; 1000 is too small.)

ANSWER 9: D

# Problem 10:
The wall is 19 ft wide; the picture is 3 ft across and centered. The leftover wall splits equally on both sides:
$$\frac{19-3}{2} = \frac{16}{2} = 8 \text{ feet}.$$

(16 is the total leftover, not one side; $9\frac12$ is the wall's midpoint.)

ANSWER 10: B

Problem 1:

Initially, \(80\%\) of 500 balls are red:

\[
0.80(500)=400
\]

So there are 400 red balls and 100 blue balls.

Suppose \(x\) red balls are removed. Then:

- Red balls remaining: \(400-x\)
- Total balls remaining: \(500-x\)

We want \(75\%\) of the remaining balls to be red:

\[
\frac{400-x}{500-x}=\frac34
\]

Cross-multiply:

\[
4(400-x)=3(500-x)
\]

\[
1600-4x=1500-3x
\]

\[
100=x
\]

Removing 100 red balls leaves 300 red and 100 blue, for 400 total. Indeed,

\[
\frac{300}{400}=75\%.
\]

Choices such as 50 or 75 do not reduce the red percentage enough; 150 reduces it too much.

ANSWER 1: D

Problem 2:

Each child can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so \(1\) arrangement.
- All 4 girls: only \(GGGG\), so \(1\) arrangement.
- Exactly 2 girls and 2 boys:

\[
\binom42=6
\]

arrangements.
- Three of one gender and one of the other:
  - 3 boys and 1 girl: \(\binom41=4\) arrangements.
  - 3 girls and 1 boy: \(\binom41=4\) arrangements.

So this outcome has

\[
4+4=8
\]

arrangements.

Since \(8\) is more than \(6\) or \(1\), the most likely outcome is 3 children of one gender and 1 of the other.

ANSWER 2: D

Problem 3:

A grid 60 toothpicks long and 32 toothpicks wide has 60 columns and 32 rows of small squares.

For the horizontal toothpicks, there are \(32+1=33\) horizontal lines, each containing 60 toothpicks:

\[
33\cdot 60=1980
\]

For the vertical toothpicks, there are \(60+1=61\) vertical lines, each containing 32 toothpicks:

\[
61\cdot 32=1952
\]

Thus the total number is

\[
1980+1952=3932.
\]

The tempting answer \(1920=60\cdot32\) counts only the small squares, not the toothpicks forming their edges.

ANSWER 3: E

Problem 4:

The expression is of the form

\[
\Box^{\Box}\times\Box^{\Box},
\]

using the digits \(2,0,2,3\) once each.

To make the product as large as possible, the digit \(0\) should be an exponent, because any nonzero number to the zero power equals 1:

\[
2^0=1 \qquad\text{or}\qquad 3^0=1.
\]

We should use the remaining digits to make the largest possible power. Compare:

\[
3^2=9
\]

and

\[
2^3=8.
\]

Thus the best arrangement is

\[
3^2\times 2^0=9\times 1=9.
\]

Using \(2^3\) gives only \(8\), and placing 0 as a base would make the entire product 0.

ANSWER 4: C

Problem 5:

The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s can has diameter 6 cm, so its radius is 3 cm and its height is 12 cm:

\[
V_A=\pi(3)^2(12)=108\pi.
\]

Felicia’s can has diameter 12 cm, so its radius is 6 cm and its height is 6 cm:

\[
V_F=\pi(6)^2(6)=216\pi.
\]

The ratio is

\[
108\pi:216\pi=1:2.
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, making her base area four times as large.

ANSWER 5: B

Problem 6:

Let the equal sides of the isosceles triangle each have length \(a\). Let the base have length \(b\).

Since the perimeter is 23,

\[
2a+b=23,
\]

so

\[
b=23-2a.
\]

For a valid triangle, the base must be less than the sum of the equal sides:

\[
b<2a.
\]

Substitute \(b=23-2a\):

\[
23-2a<2a
\]

\[
23<4a
\]

\[
a>5.75.
\]

Also, \(b\) must be positive:

\[
23-2a>0,
\]

so

\[
a<11.5.
\]

Thus \(a\) can be

\[
6,7,8,9,10,11.
\]

That gives 6 different triangles. Their side lengths are:

\[
(6,6,11),\ (7,7,9),\ (8,8,7),\ (9,9,5),\ (10,10,3),\ (11,11,1).
\]

Values below 6 fail the triangle inequality, while values above 11 give a nonpositive base.

ANSWER 6: C

Problem 7:

Harold eats \(\frac14\) of the pie, leaving

\[
1-\frac14=\frac34.
\]

The moose eats \(\frac13\) of what is left, so it leaves \(\frac23\) of what Harold left:

\[
\frac23\cdot\frac34=\frac12.
\]

Then the porcupine eats \(\frac13\) of that remaining \(\frac12\), leaving \(\frac23\) of it:

\[
\frac23\cdot\frac12=\frac13.
\]

So \(\frac13\) of the original pie remains.

A common mistake is to subtract \(\frac13\) directly from \(\frac34\); the moose eats one-third of the amount currently left, not one-third of the original pie.

ANSWER 7: D

Problem 8:

The total amount available for the taxi fare is

\[
\$10-\$2=\$8.
\]

The first \(\frac12\) mile costs \$2.40. Therefore, the amount left for additional distance is

\[
8.00-2.40=5.60.
\]

Each additional \(0.1\) mile costs \$0.20, so the number of additional tenths of a mile is

\[
\frac{5.60}{0.20}=28.
\]

That is

\[
28(0.1)=2.8
\]

additional miles.

Adding the first \(\frac12\) mile:

\[
2.8+0.5=3.3.
\]

The ride can be 3.3 miles. The \$2 tip must be included, so choices based on spending the full \$10 on fare are too large.

ANSWER 8: C

Problem 9:

The population is 200 in 1998 and triples every 25 years.

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is a little more than two 25-year periods.

After 25 years, the population would be

\[
200\cdot 3=600.
\]

After 50 years, it would be

\[
600\cdot 3=1800.
\]

Since 2050 is about 2 years later, the population should be slightly more than 1800. The closest estimate is 2000.

Choices such as 600 correspond to only one tripling, while 3000 is much too high for just over two tripling periods.

ANSWER 9: D

Problem 10:

The wall is 19 feet wide, and the picture is 3 feet wide.

The amount of wall not covered by the picture is

\[
19-3=16
\]

feet.

Since the picture is centered, this 16 feet is divided equally between the two ends:

\[
\frac{16}{2}=8.
\]

Therefore, the nearest edge of the picture is 8 feet from either end of the wall.

ANSWER 10: B

Problem 1:
We need how many red balls to take away so reds are 75% of what is left.

At first there are 500 balls, 80% red.
Red = 0.80 × 500 = 400. Blue = 500 - 400 = 100.

Let x = number of red balls removed. Blue balls are not touched.
Red left = 400 - x. Total left = 500 - x.

We want:
(400 - x) / (500 - x) = 75% = 3/4

Cross-multiply:
4(400 - x) = 3(500 - x)
1600 - 4x = 1500 - 3x
1600 - 1500 = 4x - 3x
100 = x

Check: 300 red left, 100 blue left, total 400. 300/400 = 3/4.
Removing 25 would leave 375/475 ≈ 79%, still too high. Removing 150 would leave 250/350 ≈ 71%, too low.

ANSWER 1: D

Problem 2:
We are asked which gender pattern for 4 babies is most likely, each baby boy/girl equally likely.

There are 2^4 = 16 equally likely boy-girl sequences.

A. all 4 boys: only BBBB: 1 sequence. Probability 1/16.
B. all 4 girls: only GGGG: 1 sequence. Probability 1/16.
C. 2 girls and 2 boys: number of sequences = choose which 2 of 4 are girls = C(4,2)=6. Probability 6/16.
D. 3 of one gender and 1 of the other: This means 3B1G OR 3G1B.
3B1G: C(4,1)=4 sequences. 3G1B: C(4,1)=4 sequences. Total 8 sequences. Probability 8/16 = 1/2.
E. claims equal, false.

Many think 2-2 is most likely because it is most "balanced", but 3-1 has more sequences: 8 > 6 > 1.

So D is most likely.

ANSWER 2: D

Problem 3:
We need total toothpicks to make a rectangular grid 60 long by 32 wide, where length is measured in toothpicks.

Imagine unit squares. Top edge needs 60 toothpicks side by side. Same for every horizontal line.

Number of horizontal lines: there are 32 squares vertically, so 32+1 = 33 horizontal grid lines.
Each horizontal line uses 60 toothpicks.
Horizontal total = 33 × 60 = 1980.

Number of vertical lines: there are 60 squares horizontally, so 60+1 = 61 vertical grid lines.
Each vertical line uses 32 toothpicks.
Vertical total = 61 × 32 = 1952.

Total = 1980 + 1952 = 3932.

A 1920 = 60×32 is just the number of cells, not toothpicks. B 1952 is vertical only. C 1980 is horizontal only.

ANSWER 3: E

Problem 4:
We are asked to put digits 2,0,2,3, one per box, into the expression
□^□ × □^□
to get the largest value.

We must use all four digits as two bases and two exponents.

Key facts: a^0 = 1 for a ≠ 0, while 0^positive = 0, which kills a product. So do not put 0 as a base with positive exponent. Put 0 as an exponent to get 1. 0^0 is not allowed.

Try cases:

If exponents are 2 and 0, bases 3 and 2:
3^2 × 2^0 = 9 × 1 = 9
2^2 × 3^0 = 4 × 1 = 4
3^0 × 2^2 = 1 × 4 = 4
2^0 × 3^2 = 1 × 9 = 9

If exponents are 3 and 0, bases 2 and 2:
2^3 × 2^0 = 8 × 1 = 8

If 0 is a base with positive exponent, product is 0, e.g. 3^2 × 0^2 = 0.

So the best is 9.

Why not other choices?
A. 0: tempting if you think any zero makes everything zero, but 2^0 =1 avoids zero.
B. 8: this is 2^3 × 2^0, valid but less than 9.
D. 16 and E. 18: these come from the mistake x^0 = x. E.g. 3^2 × 2 =18, 2^3 × 2 =16. Correct is ×1, not ×2.

Maximum is 9.

ANSWER 4: C

Problem 5:
We compare volumes of cylinders. Volume = π r^2 h, r = diameter/2.

Alex: diameter 6, so r = 3. h =12.
V_Alex = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r = 6. h =6.
V_Felicia = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio Alex : Felicia = 108π : 216π = 108:216 = 1:2.

Volume scales as square of radius, so doubling diameter quadruples base area, which outweighs halving height. So Felicia's is bigger, ruling out C,D,E.

ANSWER 5: B

Problem 6:
We need integer-sided isosceles triangles with perimeter 23.

Let equal sides be a,a and base be b. Then:
2a + b = 23, so b = 23 - 2a > 0, so a ≤ 11.

Triangle inequality: sum of two small sides > third side. The only non-trivial one is:
a + a > b
2a > 23 - 2a
4a > 23
a > 5.75, so a ≥ 6.

Thus a = 6,7,8,9,10,11. Each gives a b:
6,6,11: 6+6>11 valid
7,7,9: 14>9 valid
8,8,7: valid
9,9,5: valid
10,10,3: 10+3>10 valid
11,11,1: 11+1>11 valid

That is 6 different triangles. No other a works.

ANSWER 6: C

Problem 7:
We track fraction of original pie left.

Harold eats 1/4, so left = 1 - 1/4 = 3/4.

Moose eats 1/3 of what Harold left:
eaten = (1/3)×(3/4) = 1/4.
Left after moose = 3/4 - 1/4 = 1/2. Equivalently (2/3)×(3/4)=1/2.

Porcupine eats 1/3 of what moose left:
eaten = (1/3)×(1/2) = 1/6.
Left after porcupine = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3. Equivalently (2/3)×(1/2)=1/3.

So 1/3 remains. A 1/12, B 1/6 are the amounts eaten, not left.

ANSWER 7: D

Problem 8:
We have $10 total, $2 tip, so fare money = 10 - 2 = $8.00.

First 0.5 mile costs $2.40.
Remaining for extra distance = 8.00 - 2.40 = $5.60.

Extra is $0.20 per 0.1 mile.
Number of 0.1-mile blocks = 5.60 / 0.20 = 28.
Extra distance = 28 × 0.1 = 2.8 miles.

Total distance = 0.5 + 2.8 = 3.3 miles.

Check: fare = 2.40 + 28×0.20 = 2.40+5.60=8.00, plus $2 tip = $10.

ANSWER 8: C

Problem 9:
Population in 1998 is 200, triples every 25 years. Estimate in 2050.

Time elapsed = 2050 - 1998 = 52 years. That's about two 25-year periods.

After 25 yr (~2023): 200×3 = 600.
After 50 yr (~2048): 600×3 = 1800.

2050 is 2 years more, so a little above 1800. Nearest estimate among choices is 2000.

A 600 is only one tripling. C 1000, B 800 ignore tripling. E 3000 is too high.

ANSWER 9: D

Problem 10:
Picture 3 ft wide centered on 19 ft wall.

Total wall not covered = 19 - 3 = 16 ft.
Centered means half on each side: 16/2 = 8 ft from each end to picture edge.

A 1 1/2, C 9 1/2 confuse center vs edge. B is correct.

ANSWER 10: B

Problem 1:
We are asked to find how many red balls must be removed from a bag of 500 balls (originally 80% red) so that the remaining red balls constitute 75% of the bag.

First, calculate the original number of balls of each color:
- Total balls = $500$
- Red balls = $80\% \text{ of } 500 = 0.80 \times 500 = 400$
- Blue balls = $500 - 400 = 100$

Only red balls are removed, so the number of blue balls remains constant at $100$.
In the final state, red balls make up $75\%$ of the total, which means blue balls must make up:
$$100\% - 75\% = 25\%$$
of the remaining balls.

Let $T$ be the total number of remaining balls:
$$0.25 \times T = 100 \implies T = \frac{100}{0.25} = 400$$

Since the bag originally contained $500$ balls and now contains $400$ balls, the number of red balls removed is:
$$500 - 400 = 100$$

ANSWER 1: D

---

Problem 2:
We are asked to determine which outcome is most likely when four children are born, with boys (B) and girls (G) being equally likely.

Each child has $2$ possible genders, so there are $2^4 = 16$ equally likely possible outcomes in total. We evaluate the number of outcomes for each choice:
- **A. All 4 are boys:** Only $1$ outcome ($\text{BBBB}$). Probability = $\frac{1}{16}$.
- **B. All 4 are girls:** Only $1$ outcome ($\text{GGGG}$). Probability = $\frac{1}{16}$.
- **C. 2 are girls and 2 are boys:** The number of ways to choose which $2$ children are girls is $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ outcomes. Probability = $\frac{6}{16}$.
- **D. 3 are of one gender and 1 is of the other:** This includes either 3 boys and 1 girl ($\binom{4}{1} = 4$ outcomes) or 3 girls and 1 boy ($\binom{4}{3} = 4$ outcomes), giving $4 + 4 = 8$ outcomes. Probability = $\frac{8}{16} = \frac{1}{2}$.

Comparing the possibilities, outcome D has $8$ favorable outcomes, which is greater than C ($6$ outcomes), A ($1$ outcome), and B ($1$ outcome). Therefore, D is the most likely outcome.

ANSWER 2: D

---

Problem 3:
We are asked to find the total number of toothpicks used to make a rectangular grid that is $60$ toothpicks long and $32$ toothpicks wide.

A grid of dimensions $60 \times 32$ consists of unit squares arranged in $32$ rows and $60$ columns:
- **Horizontal toothpicks:** There are $32 + 1 = 33$ horizontal grid lines. Each horizontal line is made of $60$ toothpicks.
  $$\text{Number of horizontal toothpicks} = 33 \times 60 = 1980$$
- **Vertical toothpicks:** There are $60 + 1 = 61$ vertical grid lines. Each vertical line is made of $32$ toothpicks.
  $$\text{Number of vertical toothpicks} = 61 \times 32 = 1952$$

Adding the horizontal and vertical toothpicks together:
$$\text{Total toothpicks} = 1980 + 1952 = 3932$$

ANSWER 3: E

---

Problem 4:
We are asked to find the maximum possible value of the expression $\square^\square \times \square^\square$ when placing the digits $2, 0, 2,$ and $3$ into the four boxes, one digit per box.

Let the expression be $a^b \times c^d$, where $\{a, b, c, d\} = \{2, 0, 2, 3\}$.
We test the possible assignments of the digits:
- To maximize the product, we want large bases and exponents, but one of the digits is $0$.
- If $0$ is used as a base, say $0^d$ with $d > 0$, the factor becomes $0$, making the entire product $0$.
- If $0$ is placed in an exponent, recall that for any non-zero base $x$, $x^0 = 1$.
  - Pair $0$ with $2$ as an exponent: $2^0 = 1$. The remaining digits are $3$ and $2$.
    - Setting $3^2 \times 2^0 = 9 \times 1 = 9$.
    - Setting $2^3 \times 2^0 = 8 \times 1 = 8$.
  - Pair $0$ with $3$ as an exponent: $3^0 = 1$. The remaining digits are $2$ and $2$.
    - Setting $2^2 \times 3^0 = 4 \times 1 = 4$.

The values $16$ and $18$ are distractors arising from the common misconception that $x^0 = x$ (which would give $2^3 \times 2 = 16$ or $3^2 \times 2 = 18$).
The maximum mathematically valid value achievable is $9$.

ANSWER 4: C

---

Problem 5:
We are asked to find the ratio of the volume of Alex's cat food can to the volume of Felicia's cat food can.

The volume of a cylinder is given by $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height.
- For Alex's can:
  - Diameter = $6\text{ cm} \implies \text{radius } r_A = 3\text{ cm}$
  - Height $h_A = 12\text{ cm}$
  - Volume $V_A = \pi (3)^2 (12) = \pi \times 9 \times 12 = 108\pi\text{ cm}^3$

- For Felicia's can:
  - Diameter = $12\text{ cm} \implies \text{radius } r_F = 6\text{ cm}$
  - Height $h_F = 6\text{ cm}$
  - Volume $V_F = \pi (6)^2 (6) = \pi \times 36 \times 6 = 216\pi\text{ cm}^3$

The ratio of the volume of Alex's can to Felicia's can is:
$$\frac{V_A}{V_F} = \frac{108\pi}{216\pi} = \frac{1}{2}$$

Thus, the ratio is $1:2$.

ANSWER 5: B

---

Problem 6:
We are asked to find the number of different isosceles triangles with integer side lengths and a perimeter of $23$.

Let the side lengths of the isosceles triangle be $a, a,$ and $b$, where $a$ and $b$ are positive integers.
The perimeter is:
$$2a + b = 23 \implies b = 23 - 2a$$

By the Triangle Inequality:
1. The sum of the two equal sides must be strictly greater than the third side:
   $$a + a > b \implies 2a > 23 - 2a \implies 4a > 23 \implies a \ge 6$$
2. The third side must be positive:
   $$b > 0 \implies 23 - 2a > 0 \implies 2a < 23 \implies a \le 11$$

The possible integer values for $a$ are $6, 7, 8, 9, 10, 11$.
We check each corresponding side length triple $(a, a, b)$:
- $a = 6 \implies b = 11$: sides $(6, 6, 11)$
- $a = 7 \implies b = 9$: sides $(7, 7, 9)$
- $a = 8 \implies b = 7$: sides $(8, 8, 7)$
- $a = 9 \implies b = 5$: sides $(9, 9, 5)$
- $a = 10 \implies b = 3$: sides $(10, 10, 3)$
- $a = 11 \implies b = 1$: sides $(11, 11, 1)$

All $6$ sets of side lengths satisfy the triangle inequality and have two equal sides (none are equilateral since $23$ is not divisible by $3$). Thus, there are $6$ such triangles.

ANSWER 6: C

---

Problem 7:
We are asked to find the fraction of the original pie remaining after Harold, a moose, and a porcupine each eat a portion.

1. Harold start

# Problem 1:
Restate: 500 balls, 80% red. Remove some **red** balls so that 75% of what remains is red.

Red = 0.80(500) = 400, Blue = 100.

Let $x$ = red balls removed. Blues never change (still 100), and after removal blues must be 25% of the total:
$$100 = 0.25(500 - x) \Rightarrow 400 = 500 - x \Rightarrow x = 100.$$

Check: 300 red, 100 blue, total 400 → 300/400 = 75%. ✓

(Tempting wrong: 25 or 50 come from mistakenly taking 5% or 10% of 500; but removing balls shrinks the total too, so more must go.)

ANSWER 1: D

# Problem 2:
There are $2^4 = 16$ equally likely sequences of genders.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: $\binom{4}{2}=6$ ways → 6/16
- 3 of one gender, 1 of the other: $\binom{4}{3}=4$ (3 boys) $+ \binom{4}{1}=4$ (3 girls) $= 8$ ways → 8/16

The largest is 8/16 = 1/2, the "3–1 split." Choice C (6/16) is the tempting trap because 2–2 is the single most likely *exact* split, but option D combines two cases.

ANSWER 2: D

# Problem 3:
A grid 60 toothpicks long and 32 toothpicks wide means the rectangle is 60 units by 32 units.

- Horizontal toothpicks: there are $32+1 = 33$ horizontal lines, each made of 60 toothpicks: $33 \times 60 = 1980$.
- Vertical toothpicks: there are $60+1 = 61$ vertical lines, each made of 32 toothpicks: $61 \times 32 = 1952$.

Total: $1980 + 1952 = 3932$.

(1980 and 1952 are the individual pieces — tempting partial answers.)

ANSWER 3: E

# Problem 4:
The expression is $\boxed{\phantom{2}}^{\boxed{\phantom{2}}} \times \boxed{\phantom{2}}^{\boxed{\phantom{2}}}$, using digits 2, 0, 2, 3 once each.

To maximize, make one factor as large as possible and let the other equal 1 (using exponent 0).

- $3^2 \times 2^0 = 9 \times 1 = 9$
- $2^3 \times 2^0 = 8$ (uses two 2's as base and exponent... gives 8)
- $2^2 \times 3^0 = 4$
- $2^0 \times 3^2 = 9$

Any arrangement with 0 as a base kills a factor to 0. The maximum is $9$.

(8 is the tempting near-miss from $2^3$; 16 and 18 aren't attainable.)

ANSWER 4: C

# Problem 5:
Volume of a cylinder: $V=\pi r^2 h$.

- Alex: diameter 6 → $r=3$, $h=12$: $V = \pi(3^2)(12) = 108\pi$.
- Felicia: diameter 12 → $r=6$, $h=6$: $V = \pi(6^2)(6) = 216\pi$.

Ratio $= 108\pi : 216\pi = 1:2$.

(1:1 is the trap — swapping diameter and height does *not* preserve volume, since radius is squared.)

ANSWER 5: B

# Problem 6:
Let the two equal sides be $a$ and the base be $b$, with $2a + b = 23$.

Since $2a$ is even, $b$ must be odd. Triangle inequality: $2a > b$, i.e. $23 - b > b \Rightarrow b < 11.5$.

So $b \in \{1,3,5,7,9,11\}$, giving $a = 11,10,9,8,7,6$ — six triangles:
(11,11,1), (10,10,3), (9,9,5), (8,8,7), (7,7,9), (6,6,11).

That's 6. (No equilateral option since 23 isn't divisible by 3; $b=13$ fails since $5+5<13$.)

ANSWER 6: C

# Problem 7:
Harold ate $\frac14$, leaving $\frac34$.

Moose ate $\frac13$ of that, leaving $\frac23 \cdot \frac34 = \frac12$.

Porcupine ate $\frac13$ of that, leaving $\frac23 \cdot \frac12 = \frac13$.

(Trap: subtracting $\frac14+\frac13+\frac13$ from 1 gives $\frac1{12}$ — wrong, because each animal eats a fraction of the *remainder*, not of the whole pie.)

ANSWER 7: D

# Problem 8:
Total money $10, minus the $2 tip → $8 available for fare.

First $\frac12$ mile costs $2.40, leaving $8.00 - 2.40 = \$5.60$.

Each additional 0.1 mile costs $0.20: $5.60 \div 0.20 = 28$ increments → $28 \times 0.1 = 2.8$ miles.

Total distance $= 0.5 + 2.8 = 3.3$ miles.

(Forgetting the tip gives $\$10-\$2.40=\$7.60 \to 3.8$ mi; forgetting the first half-mile is included gives other traps like 3.5.)

ANSWER 8: C

# Problem 9:
Population triples every 25 years, starting at 200 in 1998.

- 1998: 200
- 2023: 600
- 2048: 1800

The year 2050 is just 2 years past 2048, so the population is a bit more than 1800 — about 2000.

(3000 would require another near-tripling; 1000 is too small.)

ANSWER 9: D

# Problem 10:
The wall is 19 ft wide; the picture is 3 ft across and centered. The leftover wall splits equally on both sides:
$$\frac{19-3}{2} = \frac{16}{2} = 8 \text{ feet}.$$

(16 is the total leftover, not one side; $9\frac12$ is the wall's midpoint.)

ANSWER 10: B

Problem 1:

Initially, \(80\%\) of 500 balls are red:

\[
0.80(500)=400
\]

So there are 400 red balls and 100 blue balls.

Suppose \(x\) red balls are removed. Then:

- Red balls remaining: \(400-x\)
- Total balls remaining: \(500-x\)

We want \(75\%\) of the remaining balls to be red:

\[
\frac{400-x}{500-x}=\frac34
\]

Cross-multiply:

\[
4(400-x)=3(500-x)
\]

\[
1600-4x=1500-3x
\]

\[
100=x
\]

Removing 100 red balls leaves 300 red and 100 blue, for 400 total. Indeed,

\[
\frac{300}{400}=75\%.
\]

Choices such as 50 or 75 do not reduce the red percentage enough; 150 reduces it too much.

ANSWER 1: D

Problem 2:

Each child can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so \(1\) arrangement.
- All 4 girls: only \(GGGG\), so \(1\) arrangement.
- Exactly 2 girls and 2 boys:

\[
\binom42=6
\]

arrangements.
- Three of one gender and one of the other:
  - 3 boys and 1 girl: \(\binom41=4\) arrangements.
  - 3 girls and 1 boy: \(\binom41=4\) arrangements.

So this outcome has

\[
4+4=8
\]

arrangements.

Since \(8\) is more than \(6\) or \(1\), the most likely outcome is 3 children of one gender and 1 of the other.

ANSWER 2: D

Problem 3:

A grid 60 toothpicks long and 32 toothpicks wide has 60 columns and 32 rows of small squares.

For the horizontal toothpicks, there are \(32+1=33\) horizontal lines, each containing 60 toothpicks:

\[
33\cdot 60=1980
\]

For the vertical toothpicks, there are \(60+1=61\) vertical lines, each containing 32 toothpicks:

\[
61\cdot 32=1952
\]

Thus the total number is

\[
1980+1952=3932.
\]

The tempting answer \(1920=60\cdot32\) counts only the small squares, not the toothpicks forming their edges.

ANSWER 3: E

Problem 4:

The expression is of the form

\[
\Box^{\Box}\times\Box^{\Box},
\]

using the digits \(2,0,2,3\) once each.

To make the product as large as possible, the digit \(0\) should be an exponent, because any nonzero number to the zero power equals 1:

\[
2^0=1 \qquad\text{or}\qquad 3^0=1.
\]

We should use the remaining digits to make the largest possible power. Compare:

\[
3^2=9
\]

and

\[
2^3=8.
\]

Thus the best arrangement is

\[
3^2\times 2^0=9\times 1=9.
\]

Using \(2^3\) gives only \(8\), and placing 0 as a base would make the entire product 0.

ANSWER 4: C

Problem 5:

The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s can has diameter 6 cm, so its radius is 3 cm and its height is 12 cm:

\[
V_A=\pi(3)^2(12)=108\pi.
\]

Felicia’s can has diameter 12 cm, so its radius is 6 cm and its height is 6 cm:

\[
V_F=\pi(6)^2(6)=216\pi.
\]

The ratio is

\[
108\pi:216\pi=1:2.
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, making her base area four times as large.

ANSWER 5: B

Problem 6:

Let the equal sides of the isosceles triangle each have length \(a\). Let the base have length \(b\).

Since the perimeter is 23,

\[
2a+b=23,
\]

so

\[
b=23-2a.
\]

For a valid triangle, the base must be less than the sum of the equal sides:

\[
b<2a.
\]

Substitute \(b=23-2a\):

\[
23-2a<2a
\]

\[
23<4a
\]

\[
a>5.75.
\]

Also, \(b\) must be positive:

\[
23-2a>0,
\]

so

\[
a<11.5.
\]

Thus \(a\) can be

\[
6,7,8,9,10,11.
\]

That gives 6 different triangles. Their side lengths are:

\[
(6,6,11),\ (7,7,9),\ (8,8,7),\ (9,9,5),\ (10,10,3),\ (11,11,1).
\]

Values below 6 fail the triangle inequality, while values above 11 give a nonpositive base.

ANSWER 6: C

Problem 7:

Harold eats \(\frac14\) of the pie, leaving

\[
1-\frac14=\frac34.
\]

The moose eats \(\frac13\) of what is left, so it leaves \(\frac23\) of what Harold left:

\[
\frac23\cdot\frac34=\frac12.
\]

Then the porcupine eats \(\frac13\) of that remaining \(\frac12\), leaving \(\frac23\) of it:

\[
\frac23\cdot\frac12=\frac13.
\]

So \(\frac13\) of the original pie remains.

A common mistake is to subtract \(\frac13\) directly from \(\frac34\); the moose eats one-third of the amount currently left, not one-third of the original pie.

ANSWER 7: D

Problem 8:

The total amount available for the taxi fare is

\[
\$10-\$2=\$8.
\]

The first \(\frac12\) mile costs \$2.40. Therefore, the amount left for additional distance is

\[
8.00-2.40=5.60.
\]

Each additional \(0.1\) mile costs \$0.20, so the number of additional tenths of a mile is

\[
\frac{5.60}{0.20}=28.
\]

That is

\[
28(0.1)=2.8
\]

additional miles.

Adding the first \(\frac12\) mile:

\[
2.8+0.5=3.3.
\]

The ride can be 3.3 miles. The \$2 tip must be included, so choices based on spending the full \$10 on fare are too large.

ANSWER 8: C

Problem 9:

The population is 200 in 1998 and triples every 25 years.

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is a little more than two 25-year periods.

After 25 years, the population would be

\[
200\cdot 3=600.
\]

After 50 years, it would be

\[
600\cdot 3=1800.
\]

Since 2050 is about 2 years later, the population should be slightly more than 1800. The closest estimate is 2000.

Choices such as 600 correspond to only one tripling, while 3000 is much too high for just over two tripling periods.

ANSWER 9: D

Problem 10:

The wall is 19 feet wide, and the picture is 3 feet wide.

The amount of wall not covered by the picture is

\[
19-3=16
\]

feet.

Since the picture is centered, this 16 feet is divided equally between the two ends:

\[
\frac{16}{2}=8.
\]

Therefore, the nearest edge of the picture is 8 feet from either end of the wall.

ANSWER 10: B

Problem 1:
We need how many red balls to take away so reds are 75% of what is left.

At first there are 500 balls, 80% red.
Red = 0.80 × 500 = 400. Blue = 500 - 400 = 100.

Let x = number of red balls removed. Blue balls are not touched.
Red left = 400 - x. Total left = 500 - x.

We want:
(400 - x) / (500 - x) = 75% = 3/4

Cross-multiply:
4(400 - x) = 3(500 - x)
1600 - 4x = 1500 - 3x
1600 - 1500 = 4x - 3x
100 = x

Check: 300 red left, 100 blue left, total 400. 300/400 = 3/4.
Removing 25 would leave 375/475 ≈ 79%, still too high. Removing 150 would leave 250/350 ≈ 71%, too low.

ANSWER 1: D

Problem 2:
We are asked which gender pattern for 4 babies is most likely, each baby boy/girl equally likely.

There are 2^4 = 16 equally likely boy-girl sequences.

A. all 4 boys: only BBBB: 1 sequence. Probability 1/16.
B. all 4 girls: only GGGG: 1 sequence. Probability 1/16.
C. 2 girls and 2 boys: number of sequences = choose which 2 of 4 are girls = C(4,2)=6. Probability 6/16.
D. 3 of one gender and 1 of the other: This means 3B1G OR 3G1B.
3B1G: C(4,1)=4 sequences. 3G1B: C(4,1)=4 sequences. Total 8 sequences. Probability 8/16 = 1/2.
E. claims equal, false.

Many think 2-2 is most likely because it is most "balanced", but 3-1 has more sequences: 8 > 6 > 1.

So D is most likely.

ANSWER 2: D

Problem 3:
We need total toothpicks to make a rectangular grid 60 long by 32 wide, where length is measured in toothpicks.

Imagine unit squares. Top edge needs 60 toothpicks side by side. Same for every horizontal line.

Number of horizontal lines: there are 32 squares vertically, so 32+1 = 33 horizontal grid lines.
Each horizontal line uses 60 toothpicks.
Horizontal total = 33 × 60 = 1980.

Number of vertical lines: there are 60 squares horizontally, so 60+1 = 61 vertical grid lines.
Each vertical line uses 32 toothpicks.
Vertical total = 61 × 32 = 1952.

Total = 1980 + 1952 = 3932.

A 1920 = 60×32 is just the number of cells, not toothpicks. B 1952 is vertical only. C 1980 is horizontal only.

ANSWER 3: E

Problem 4:
We are asked to put digits 2,0,2,3, one per box, into the expression
□^□ × □^□
to get the largest value.

We must use all four digits as two bases and two exponents.

Key facts: a^0 = 1 for a ≠ 0, while 0^positive = 0, which kills a product. So do not put 0 as a base with positive exponent. Put 0 as an exponent to get 1. 0^0 is not allowed.

Try cases:

If exponents are 2 and 0, bases 3 and 2:
3^2 × 2^0 = 9 × 1 = 9
2^2 × 3^0 = 4 × 1 = 4
3^0 × 2^2 = 1 × 4 = 4
2^0 × 3^2 = 1 × 9 = 9

If exponents are 3 and 0, bases 2 and 2:
2^3 × 2^0 = 8 × 1 = 8

If 0 is a base with positive exponent, product is 0, e.g. 3^2 × 0^2 = 0.

So the best is 9.

Why not other choices?
A. 0: tempting if you think any zero makes everything zero, but 2^0 =1 avoids zero.
B. 8: this is 2^3 × 2^0, valid but less than 9.
D. 16 and E. 18: these come from the mistake x^0 = x. E.g. 3^2 × 2 =18, 2^3 × 2 =16. Correct is ×1, not ×2.

Maximum is 9.

ANSWER 4: C

Problem 5:
We compare volumes of cylinders. Volume = π r^2 h, r = diameter/2.

Alex: diameter 6, so r = 3. h =12.
V_Alex = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r = 6. h =6.
V_Felicia = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio Alex : Felicia = 108π : 216π = 108:216 = 1:2.

Volume scales as square of radius, so doubling diameter quadruples base area, which outweighs halving height. So Felicia's is bigger, ruling out C,D,E.

ANSWER 5: B

Problem 6:
We need integer-sided isosceles triangles with perimeter 23.

Let equal sides be a,a and base be b. Then:
2a + b = 23, so b = 23 - 2a > 0, so a ≤ 11.

Triangle inequality: sum of two small sides > third side. The only non-trivial one is:
a + a > b
2a > 23 - 2a
4a > 23
a > 5.75, so a ≥ 6.

Thus a = 6,7,8,9,10,11. Each gives a b:
6,6,11: 6+6>11 valid
7,7,9: 14>9 valid
8,8,7: valid
9,9,5: valid
10,10,3: 10+3>10 valid
11,11,1: 11+1>11 valid

That is 6 different triangles. No other a works.

ANSWER 6: C

Problem 7:
We track fraction of original pie left.

Harold eats 1/4, so left = 1 - 1/4 = 3/4.

Moose eats 1/3 of what Harold left:
eaten = (1/3)×(3/4) = 1/4.
Left after moose = 3/4 - 1/4 = 1/2. Equivalently (2/3)×(3/4)=1/2.

Porcupine eats 1/3 of what moose left:
eaten = (1/3)×(1/2) = 1/6.
Left after porcupine = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3. Equivalently (2/3)×(1/2)=1/3.

So 1/3 remains. A 1/12, B 1/6 are the amounts eaten, not left.

ANSWER 7: D

Problem 8:
We have $10 total, $2 tip, so fare money = 10 - 2 = $8.00.

First 0.5 mile costs $2.40.
Remaining for extra distance = 8.00 - 2.40 = $5.60.

Extra is $0.20 per 0.1 mile.
Number of 0.1-mile blocks = 5.60 / 0.20 = 28.
Extra distance = 28 × 0.1 = 2.8 miles.

Total distance = 0.5 + 2.8 = 3.3 miles.

Check: fare = 2.40 + 28×0.20 = 2.40+5.60=8.00, plus $2 tip = $10.

ANSWER 8: C

Problem 9:
Population in 1998 is 200, triples every 25 years. Estimate in 2050.

Time elapsed = 2050 - 1998 = 52 years. That's about two 25-year periods.

After 25 yr (~2023): 200×3 = 600.
After 50 yr (~2048): 600×3 = 1800.

2050 is 2 years more, so a little above 1800. Nearest estimate among choices is 2000.

A 600 is only one tripling. C 1000, B 800 ignore tripling. E 3000 is too high.

ANSWER 9: D

Problem 10:
Picture 3 ft wide centered on 19 ft wall.

Total wall not covered = 19 - 3 = 16 ft.
Centered means half on each side: 16/2 = 8 ft from each end to picture edge.

A 1 1/2, C 9 1/2 confuse center vs edge. B is correct.

ANSWER 10: B

Problem 1:
We are asked to find how many red balls must be removed from a bag of 500 balls (originally 80% red) so that the remaining red balls constitute 75% of the bag.

First, calculate the original number of balls of each color:
- Total balls = $500$
- Red balls = $80\% \text{ of } 500 = 0.80 \times 500 = 400$
- Blue balls = $500 - 400 = 100$

Only red balls are removed, so the number of blue balls remains constant at $100$.
In the final state, red balls make up $75\%$ of the total, which means blue balls must make up:
$$100\% - 75\% = 25\%$$
of the remaining balls.

Let $T$ be the total number of remaining balls:
$$0.25 \times T = 100 \implies T = \frac{100}{0.25} = 400$$

Since the bag originally contained $500$ balls and now contains $400$ balls, the number of red balls removed is:
$$500 - 400 = 100$$

ANSWER 1: D

---

Problem 2:
We are asked to determine which outcome is most likely when four children are born, with boys (B) and girls (G) being equally likely.

Each child has $2$ possible genders, so there are $2^4 = 16$ equally likely possible outcomes in total. We evaluate the number of outcomes for each choice:
- **A. All 4 are boys:** Only $1$ outcome ($\text{BBBB}$). Probability = $\frac{1}{16}$.
- **B. All 4 are girls:** Only $1$ outcome ($\text{GGGG}$). Probability = $\frac{1}{16}$.
- **C. 2 are girls and 2 are boys:** The number of ways to choose which $2$ children are girls is $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ outcomes. Probability = $\frac{6}{16}$.
- **D. 3 are of one gender and 1 is of the other:** This includes either 3 boys and 1 girl ($\binom{4}{1} = 4$ outcomes) or 3 girls and 1 boy ($\binom{4}{3} = 4$ outcomes), giving $4 + 4 = 8$ outcomes. Probability = $\frac{8}{16} = \frac{1}{2}$.

Comparing the possibilities, outcome D has $8$ favorable outcomes, which is greater than C ($6$ outcomes), A ($1$ outcome), and B ($1$ outcome). Therefore, D is the most likely outcome.

ANSWER 2: D

---

Problem 3:
We are asked to find the total number of toothpicks used to make a rectangular grid that is $60$ toothpicks long and $32$ toothpicks wide.

A grid of dimensions $60 \times 32$ consists of unit squares arranged in $32$ rows and $60$ columns:
- **Horizontal toothpicks:** There are $32 + 1 = 33$ horizontal grid lines. Each horizontal line is made of $60$ toothpicks.
  $$\text{Number of horizontal toothpicks} = 33 \times 60 = 1980$$
- **Vertical toothpicks:** There are $60 + 1 = 61$ vertical grid lines. Each vertical line is made of $32$ toothpicks.
  $$\text{Number of vertical toothpicks} = 61 \times 32 = 1952$$

Adding the horizontal and vertical toothpicks together:
$$\text{Total toothpicks} = 1980 + 1952 = 3932$$

ANSWER 3: E

---

Problem 4:
We are asked to find the maximum possible value of the expression $\square^\square \times \square^\square$ when placing the digits $2, 0, 2,$ and $3$ into the four boxes, one digit per box.

Let the expression be $a^b \times c^d$, where $\{a, b, c, d\} = \{2, 0, 2, 3\}$.
We test the possible assignments of the digits:
- To maximize the product, we want large bases and exponents, but one of the digits is $0$.
- If $0$ is used as a base, say $0^d$ with $d > 0$, the factor becomes $0$, making the entire product $0$.
- If $0$ is placed in an exponent, recall that for any non-zero base $x$, $x^0 = 1$.
  - Pair $0$ with $2$ as an exponent: $2^0 = 1$. The remaining digits are $3$ and $2$.
    - Setting $3^2 \times 2^0 = 9 \times 1 = 9$.
    - Setting $2^3 \times 2^0 = 8 \times 1 = 8$.
  - Pair $0$ with $3$ as an exponent: $3^0 = 1$. The remaining digits are $2$ and $2$.
    - Setting $2^2 \times 3^0 = 4 \times 1 = 4$.

The values $16$ and $18$ are distractors arising from the common misconception that $x^0 = x$ (which would give $2^3 \times 2 = 16$ or $3^2 \times 2 = 18$).
The maximum mathematically valid value achievable is $9$.

ANSWER 4: C

---

Problem 5:
We are asked to find the ratio of the volume of Alex's cat food can to the volume of Felicia's cat food can.

The volume of a cylinder is given by $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height.
- For Alex's can:
  - Diameter = $6\text{ cm} \implies \text{radius } r_A = 3\text{ cm}$
  - Height $h_A = 12\text{ cm}$
  - Volume $V_A = \pi (3)^2 (12) = \pi \times 9 \times 12 = 108\pi\text{ cm}^3$

- For Felicia's can:
  - Diameter = $12\text{ cm} \implies \text{radius } r_F = 6\text{ cm}$
  - Height $h_F = 6\text{ cm}$
  - Volume $V_F = \pi (6)^2 (6) = \pi \times 36 \times 6 = 216\pi\text{ cm}^3$

The ratio of the volume of Alex's can to Felicia's can is:
$$\frac{V_A}{V_F} = \frac{108\pi}{216\pi} = \frac{1}{2}$$

Thus, the ratio is $1:2$.

ANSWER 5: B

---

Problem 6:
We are asked to find the number of different isosceles triangles with integer side lengths and a perimeter of $23$.

Let the side lengths of the isosceles triangle be $a, a,$ and $b$, where $a$ and $b$ are positive integers.
The perimeter is:
$$2a + b = 23 \implies b = 23 - 2a$$

By the Triangle Inequality:
1. The sum of the two equal sides must be strictly greater than the third side:
   $$a + a > b \implies 2a > 23 - 2a \implies 4a > 23 \implies a \ge 6$$
2. The third side must be positive:
   $$b > 0 \implies 23 - 2a > 0 \implies 2a < 23 \implies a \le 11$$

The possible integer values for $a$ are $6, 7, 8, 9, 10, 11$.
We check each corresponding side length triple $(a, a, b)$:
- $a = 6 \implies b = 11$: sides $(6, 6, 11)$
- $a = 7 \implies b = 9$: sides $(7, 7, 9)$
- $a = 8 \implies b = 7$: sides $(8, 8, 7)$
- $a = 9 \implies b = 5$: sides $(9, 9, 5)$
- $a = 10 \implies b = 3$: sides $(10, 10, 3)$
- $a = 11 \implies b = 1$: sides $(11, 11, 1)$

All $6$ sets of side lengths satisfy the triangle inequality and have two equal sides (none are equilateral since $23$ is not divisible by $3$). Thus, there are $6$ such triangles.

ANSWER 6: C

---

Problem 7:
We are asked to find the fraction of the original pie remaining after Harold, a moose, and a porcupine each eat a portion.

1. Harold start

# Problem 1:
Restate: 500 balls, 80% red. Remove some **red** balls so that 75% of what remains is red.

Red = 0.80(500) = 400, Blue = 100.

Let $x$ = red balls removed. Blues never change (still 100), and after removal blues must be 25% of the total:
$$100 = 0.25(500 - x) \Rightarrow 400 = 500 - x \Rightarrow x = 100.$$

Check: 300 red, 100 blue, total 400 → 300/400 = 75%. ✓

(Tempting wrong: 25 or 50 come from mistakenly taking 5% or 10% of 500; but removing balls shrinks the total too, so more must go.)

ANSWER 1: D

# Problem 2:
There are $2^4 = 16$ equally likely sequences of genders.

- All 4 boys: 1 way → 1/16
- All 4 girls: 1 way → 1/16
- 2 girls, 2 boys: $\binom{4}{2}=6$ ways → 6/16
- 3 of one gender, 1 of the other: $\binom{4}{3}=4$ (3 boys) $+ \binom{4}{1}=4$ (3 girls) $= 8$ ways → 8/16

The largest is 8/16 = 1/2, the "3–1 split." Choice C (6/16) is the tempting trap because 2–2 is the single most likely *exact* split, but option D combines two cases.

ANSWER 2: D

# Problem 3:
A grid 60 toothpicks long and 32 toothpicks wide means the rectangle is 60 units by 32 units.

- Horizontal toothpicks: there are $32+1 = 33$ horizontal lines, each made of 60 toothpicks: $33 \times 60 = 1980$.
- Vertical toothpicks: there are $60+1 = 61$ vertical lines, each made of 32 toothpicks: $61 \times 32 = 1952$.

Total: $1980 + 1952 = 3932$.

(1980 and 1952 are the individual pieces — tempting partial answers.)

ANSWER 3: E

# Problem 4:
The expression is $\boxed{\phantom{2}}^{\boxed{\phantom{2}}} \times \boxed{\phantom{2}}^{\boxed{\phantom{2}}}$, using digits 2, 0, 2, 3 once each.

To maximize, make one factor as large as possible and let the other equal 1 (using exponent 0).

- $3^2 \times 2^0 = 9 \times 1 = 9$
- $2^3 \times 2^0 = 8$ (uses two 2's as base and exponent... gives 8)
- $2^2 \times 3^0 = 4$
- $2^0 \times 3^2 = 9$

Any arrangement with 0 as a base kills a factor to 0. The maximum is $9$.

(8 is the tempting near-miss from $2^3$; 16 and 18 aren't attainable.)

ANSWER 4: C

# Problem 5:
Volume of a cylinder: $V=\pi r^2 h$.

- Alex: diameter 6 → $r=3$, $h=12$: $V = \pi(3^2)(12) = 108\pi$.
- Felicia: diameter 12 → $r=6$, $h=6$: $V = \pi(6^2)(6) = 216\pi$.

Ratio $= 108\pi : 216\pi = 1:2$.

(1:1 is the trap — swapping diameter and height does *not* preserve volume, since radius is squared.)

ANSWER 5: B

# Problem 6:
Let the two equal sides be $a$ and the base be $b$, with $2a + b = 23$.

Since $2a$ is even, $b$ must be odd. Triangle inequality: $2a > b$, i.e. $23 - b > b \Rightarrow b < 11.5$.

So $b \in \{1,3,5,7,9,11\}$, giving $a = 11,10,9,8,7,6$ — six triangles:
(11,11,1), (10,10,3), (9,9,5), (8,8,7), (7,7,9), (6,6,11).

That's 6. (No equilateral option since 23 isn't divisible by 3; $b=13$ fails since $5+5<13$.)

ANSWER 6: C

# Problem 7:
Harold ate $\frac14$, leaving $\frac34$.

Moose ate $\frac13$ of that, leaving $\frac23 \cdot \frac34 = \frac12$.

Porcupine ate $\frac13$ of that, leaving $\frac23 \cdot \frac12 = \frac13$.

(Trap: subtracting $\frac14+\frac13+\frac13$ from 1 gives $\frac1{12}$ — wrong, because each animal eats a fraction of the *remainder*, not of the whole pie.)

ANSWER 7: D

# Problem 8:
Total money $10, minus the $2 tip → $8 available for fare.

First $\frac12$ mile costs $2.40, leaving $8.00 - 2.40 = \$5.60$.

Each additional 0.1 mile costs $0.20: $5.60 \div 0.20 = 28$ increments → $28 \times 0.1 = 2.8$ miles.

Total distance $= 0.5 + 2.8 = 3.3$ miles.

(Forgetting the tip gives $\$10-\$2.40=\$7.60 \to 3.8$ mi; forgetting the first half-mile is included gives other traps like 3.5.)

ANSWER 8: C

# Problem 9:
Population triples every 25 years, starting at 200 in 1998.

- 1998: 200
- 2023: 600
- 2048: 1800

The year 2050 is just 2 years past 2048, so the population is a bit more than 1800 — about 2000.

(3000 would require another near-tripling; 1000 is too small.)

ANSWER 9: D

# Problem 10:
The wall is 19 ft wide; the picture is 3 ft across and centered. The leftover wall splits equally on both sides:
$$\frac{19-3}{2} = \frac{16}{2} = 8 \text{ feet}.$$

(16 is the total leftover, not one side; $9\frac12$ is the wall's midpoint.)

ANSWER 10: B

Problem 1:

Initially, \(80\%\) of 500 balls are red:

\[
0.80(500)=400
\]

So there are 400 red balls and 100 blue balls.

Suppose \(x\) red balls are removed. Then:

- Red balls remaining: \(400-x\)
- Total balls remaining: \(500-x\)

We want \(75\%\) of the remaining balls to be red:

\[
\frac{400-x}{500-x}=\frac34
\]

Cross-multiply:

\[
4(400-x)=3(500-x)
\]

\[
1600-4x=1500-3x
\]

\[
100=x
\]

Removing 100 red balls leaves 300 red and 100 blue, for 400 total. Indeed,

\[
\frac{300}{400}=75\%.
\]

Choices such as 50 or 75 do not reduce the red percentage enough; 150 reduces it too much.

ANSWER 1: D

Problem 2:

Each child can be a boy or girl, so there are

\[
2^4=16
\]

equally likely gender arrangements.

- All 4 boys: only \(BBBB\), so \(1\) arrangement.
- All 4 girls: only \(GGGG\), so \(1\) arrangement.
- Exactly 2 girls and 2 boys:

\[
\binom42=6
\]

arrangements.
- Three of one gender and one of the other:
  - 3 boys and 1 girl: \(\binom41=4\) arrangements.
  - 3 girls and 1 boy: \(\binom41=4\) arrangements.

So this outcome has

\[
4+4=8
\]

arrangements.

Since \(8\) is more than \(6\) or \(1\), the most likely outcome is 3 children of one gender and 1 of the other.

ANSWER 2: D

Problem 3:

A grid 60 toothpicks long and 32 toothpicks wide has 60 columns and 32 rows of small squares.

For the horizontal toothpicks, there are \(32+1=33\) horizontal lines, each containing 60 toothpicks:

\[
33\cdot 60=1980
\]

For the vertical toothpicks, there are \(60+1=61\) vertical lines, each containing 32 toothpicks:

\[
61\cdot 32=1952
\]

Thus the total number is

\[
1980+1952=3932.
\]

The tempting answer \(1920=60\cdot32\) counts only the small squares, not the toothpicks forming their edges.

ANSWER 3: E

Problem 4:

The expression is of the form

\[
\Box^{\Box}\times\Box^{\Box},
\]

using the digits \(2,0,2,3\) once each.

To make the product as large as possible, the digit \(0\) should be an exponent, because any nonzero number to the zero power equals 1:

\[
2^0=1 \qquad\text{or}\qquad 3^0=1.
\]

We should use the remaining digits to make the largest possible power. Compare:

\[
3^2=9
\]

and

\[
2^3=8.
\]

Thus the best arrangement is

\[
3^2\times 2^0=9\times 1=9.
\]

Using \(2^3\) gives only \(8\), and placing 0 as a base would make the entire product 0.

ANSWER 4: C

Problem 5:

The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s can has diameter 6 cm, so its radius is 3 cm and its height is 12 cm:

\[
V_A=\pi(3)^2(12)=108\pi.
\]

Felicia’s can has diameter 12 cm, so its radius is 6 cm and its height is 6 cm:

\[
V_F=\pi(6)^2(6)=216\pi.
\]

The ratio is

\[
108\pi:216\pi=1:2.
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, making her base area four times as large.

ANSWER 5: B

Problem 6:

Let the equal sides of the isosceles triangle each have length \(a\). Let the base have length \(b\).

Since the perimeter is 23,

\[
2a+b=23,
\]

so

\[
b=23-2a.
\]

For a valid triangle, the base must be less than the sum of the equal sides:

\[
b<2a.
\]

Substitute \(b=23-2a\):

\[
23-2a<2a
\]

\[
23<4a
\]

\[
a>5.75.
\]

Also, \(b\) must be positive:

\[
23-2a>0,
\]

so

\[
a<11.5.
\]

Thus \(a\) can be

\[
6,7,8,9,10,11.
\]

That gives 6 different triangles. Their side lengths are:

\[
(6,6,11),\ (7,7,9),\ (8,8,7),\ (9,9,5),\ (10,10,3),\ (11,11,1).
\]

Values below 6 fail the triangle inequality, while values above 11 give a nonpositive base.

ANSWER 6: C

Problem 7:

Harold eats \(\frac14\) of the pie, leaving

\[
1-\frac14=\frac34.
\]

The moose eats \(\frac13\) of what is left, so it leaves \(\frac23\) of what Harold left:

\[
\frac23\cdot\frac34=\frac12.
\]

Then the porcupine eats \(\frac13\) of that remaining \(\frac12\), leaving \(\frac23\) of it:

\[
\frac23\cdot\frac12=\frac13.
\]

So \(\frac13\) of the original pie remains.

A common mistake is to subtract \(\frac13\) directly from \(\frac34\); the moose eats one-third of the amount currently left, not one-third of the original pie.

ANSWER 7: D

Problem 8:

The total amount available for the taxi fare is

\[
\$10-\$2=\$8.
\]

The first \(\frac12\) mile costs \$2.40. Therefore, the amount left for additional distance is

\[
8.00-2.40=5.60.
\]

Each additional \(0.1\) mile costs \$0.20, so the number of additional tenths of a mile is

\[
\frac{5.60}{0.20}=28.
\]

That is

\[
28(0.1)=2.8
\]

additional miles.

Adding the first \(\frac12\) mile:

\[
2.8+0.5=3.3.
\]

The ride can be 3.3 miles. The \$2 tip must be included, so choices based on spending the full \$10 on fare are too large.

ANSWER 8: C

Problem 9:

The population is 200 in 1998 and triples every 25 years.

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is a little more than two 25-year periods.

After 25 years, the population would be

\[
200\cdot 3=600.
\]

After 50 years, it would be

\[
600\cdot 3=1800.
\]

Since 2050 is about 2 years later, the population should be slightly more than 1800. The closest estimate is 2000.

Choices such as 600 correspond to only one tripling, while 3000 is much too high for just over two tripling periods.

ANSWER 9: D

Problem 10:

The wall is 19 feet wide, and the picture is 3 feet wide.

The amount of wall not covered by the picture is

\[
19-3=16
\]

feet.

Since the picture is centered, this 16 feet is divided equally between the two ends:

\[
\frac{16}{2}=8.
\]

Therefore, the nearest edge of the picture is 8 feet from either end of the wall.

ANSWER 10: B