🔬 Model Lab

Problem 1:
We are asked: in 74982.1035, compare the place value of the 9 with the place value of the 3.

Write out places:
74982 is: 7 ten-thousands, 4 thousands, 9 hundreds, 8 tens, 2 ones.
So the 9 is in the hundreds place. Its value is 100.

After the decimal .1035 is: 1 tenths, 0 hundredths, 3 thousandths, 5 ten-thousandths.
So the 3 is in the thousandths place. Its value is 0.001 = 1/1000.

Ratio = (value of 9's place) / (value of 3's place)
= 100 / 0.001 = 100 / (1/1000) = 100 x 1000 = 100,000.

So the hundreds place is 100,000 times as great as the thousandths place.
Choices A,B,D,E are powers of 10 that are too small or too large.

ANSWER 1: C

Problem 2:
We are asked for the earlier volume given expansion rate.

Rate: +3 degrees temperature = +4 cubic cm volume.

Now: temp 32°, volume 24 cc.
Before: temp 20°.

Temperature change from 20° to 32° is 32-20 = 12°.
Number of 3-degree steps in 12° is 12 / 3 = 4 steps.

So volume increase from 20° to 32° is 4 steps x 4 cc = 16 cc.

Thus volume at 20° + 16 = volume at 32° = 24.
So volume at 20° = 24 - 16 = 8 cc.

Check: A 8 fits, B 12 would be only 3 steps, D 16 would be only 2 steps, E 40 would be adding instead of subtracting.

ANSWER 2: A

Problem 3:
We need n teams such that every pair plays once gives 21 games.

Number of pairs from n teams is n choose 2 = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try integers: 6x5=30 too small, 7x6=42 correct, 8x7=56 too big.
So n = 7.

Ruling out: A 6 gives only 15 games, C 8 gives 28 games, etc.

ANSWER 3: B

Problem 4:
Ara and Shea started at same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara's now height.

Let original height = H.
Shea's growth = 20% of H = 0.20 H.
Shea now = H + 0.20H = 1.2H = 60.
So H = 60 / 1.2 = 50 inches.

Shea grew 60-50 = 10 inches.
Ara grew half as many inches: 10 / 2 = 5 inches.
Ara now = 50 + 5 = 55 inches.

Tempting wrong A 48 would be working backwards, D 54 would be misreading percent, etc.

ANSWER 4: E

Problem 5:
A standard 1-6 die is tossed. Bottom face hidden, other 5 faces seen. Probability product of seen numbers divisible by 6?

Product divisible by 6 means divisible by 2 and by 3.

Numbers 1 to 6: evens are 2,4,6 — three of them. Multiples of 3 are 3,6 — two of them.

Hiding only one face (the bottom) can hide at most one even, leaving at least two evens still visible, so factor 2 remains. It can hide at most one multiple of 3, leaving at least one multiple of 3 still visible, so factor 3 remains.

Indeed total product 1·2·3·4·5·6 = 720. Visible product = 720 / b where b is bottom. 720/b divided by 6 = 120/b, and b=1,2,3,4,5,6 all divide 120, so always integer.

Hence no matter which face is hidden, the visible product is divisible by 6. Probability = 6/6 = 1.

Other choices 1/3,1/2,2/3,5/6 assume sometimes it fails, which is impossible.

ANSWER 5: E

Problem 6:
75 problems total: 10 arithmetic, 30 algebra, 35 geometry.
Correct: 70% of arithmetic, 40% of algebra, 60% of geometry. How many more to reach 60% overall?

Compute:
Arithmetic correct = 0.70 x 10 = 7.
Algebra correct = 0.40 x 30 = 12.
Geometry correct = 0.60 x 35 = 21.
Total correct = 7+12+21 = 40.

Need for 60% passing: 0.60 x 75 = 45 problems.

Shortfall = 45 - 40 = 5 more problems.

So B 5, not A 1 or C 7.

ANSWER 6: B

Problem 7:
Given 0 < a < b < c, which statement is impossible?

A. a + c < b. Since a > 0, a + c > c. And c > b. So a+c > b always. Hence a+c < b can never happen. This is impossible.

Check others are possible with examples:
B. a·b < c: e.g. a=1,b=2,c=10 gives 2<10 possible.
C. a+b < c: e.g. a=1,b=2,c=10 gives 3<10 possible.
D. a·c < b: with fractions <1. e.g. a=0.1,c=2,b=1 gives 0.2<1 and 0.1<1<2 possible.
E. b c = a: let b=0.25,a=0.1, then c=a/b=0.4, and 0.1<0.25<0.4 possible.

So only A is impossible. Students may think E impossible because b c > b > a for numbers >1, but with numbers <1 multiplication can shrink.

ANSWER 7: A

Problem 8:
One full recipe = 15 cookies = 3 tbsp butter. Need 216 cookies. 8 tbsp = 1 stick. How many sticks?

Pans needed: 216 / 15 = 14.4. Since only full recipes, round up to 15 pans. 14 pans would give only 210 cookies, not enough.

Butter needed: 15 x 3 tbsp = 45 tbsp.

Sticks needed: 45 / 8 = 5.625. Since need whole sticks to have enough, round up to 6. 5 sticks = 40 tbsp < 45 tbsp, not enough.

So answer 6. A 5 would leave butter and cookies short.

ANSWER 8: B

Problem 9:
Build 3x3x3 = 27 small cubes. 6 white, 21 red. Minimize white squares on outside. What fraction of outside is white?

Big cube surface: 6 faces each 3x3 = 9 small squares, total 54 small squares.

Small-cube positions by exposure:
- 1 interior center: 0 faces shown
- 6 face-centers: 1 face shown
- 12 edge non-corner: 2 faces shown
- 8 corners: 3 faces shown

To minimize white, put whites where least exposure:
Put 1 white in center → 0 exposed.
Put remaining 5 whites in face-center spots → 1 each exposed.
Total white exposed squares = 5.

Fraction white = 5 / 54 of total surface area.

Larger fractions like 1/9=6/54, etc. would result from worse placement.

ANSWER 9: A

Problem 10:
1998 population 200, triples every 25 years. Estimate 2050.

Time elapsed: 2050-1998 = 52 years.
Number of 25-year periods: 52/25 ≈ 2.08, just over 2 doublings-triplings.

After 25 years (~2023): 200x3=600.
After 50 years (~2048): 600x3=1800.
2050 is 2 years later, a little more, so about 2000.

Formally 200 x 3^(52/25) ≈ 200 x 9 x small extra ≈ 1960.

So A 600 is only one tripling, C 1000 too low, E 3000 too high. Best is D 2000.

ANSWER 10: D

Problem 1:
We are asked to find the ratio of the value of the place occupied by the digit 9 to the value of the place occupied by the digit 3 in the decimal number $74982.1035$.

1. Identify the place values:
   - The number is $74982.1035$.
   - The digit 2 is in the ones place ($10^0$).
   - The digit 8 is in the tens place ($10^1$).
   - The digit 9 is in the hundreds place ($10^2 = 100$).
   - The digit 1 is in the tenths place ($10^{-1} = 0.1$).
   - The digit 0 is in the hundredths place ($10^{-2} = 0.01$).
   - The digit 3 is in the thousandths place ($10^{-3} = 0.001$).
   - The digit 5 is in the ten-thousandths place ($10^{-4} = 0.0001$).

2. Compute the ratio of the place values:
   $$\frac{\text{Place value of } 9}{\text{Place value of } 3} = \frac{100}{0.001} = \frac{10^2}{10^{-3}} = 10^{2 - (-3)} = 10^5 = 100,000.$$

This corresponds to choice C.

ANSWER 1: C

---

Problem 2:
We are asked to find the initial volume of gas at $20^\circ$ given its volume at $32^\circ$ and its rate of expansion.

1. Find the change in temperature:
   The temperature changes from $20^\circ$ to $32^\circ$, which is an increase of:
   $$32^\circ - 20^\circ = 12^\circ.$$

2. Determine the volume expansion:
   For every $3^\circ$ rise, the volume expands by $4\text{ cm}^3$.
   In a $12^\circ$ rise, the number of $3^\circ$ intervals is:
   $$\frac{12^\circ}{3^\circ} = 4.$$
   Therefore, the volume expanded by:
   $$4 \times 4\text{ cm}^3 = 16\text{ cm}^3.$$

3. Calculate the original volume at $20^\circ$:
   $$\text{Volume at } 20^\circ = \text{Volume at } 32^\circ - 16\text{ cm}^3 = 24 - 16 = 8\text{ cm}^3.$$

This matches choice A.

ANSWER 2: A

---

Problem 3:
We are given that each of the $N$ teams plays every other team exactly once, resulting in 21 total games. We need to find $N$.

1. The number of games played between $N$ teams is the number of ways to choose 2 teams from $N$, given by the combination formula:
   $$\binom{N}{2} = \frac{N(N-1)}{2}.$$

2. Set this equal to 21 and solve for $N$:
   $$\frac{N(N-1)}{2} = 21 \implies N(N-1) = 42.$$

3. Find two consecutive positive integers whose product is 42:
   Since $7 \times 6 = 42$, we have $N = 7$.

Checking the choices, $N = 7$ corresponds to choice B.

ANSWER 3: B

---

Problem 4:
We are given:
- Ara and Shea initially had the same height, say $h$ inches.
- Shea has grown $20\%$, so Shea's new height is $h + 0.20h = 1.20h = 60$ inches.
- Ara has grown half as many inches as Shea.
- We need to find Ara's current height.

1. Find the original height $h$:
   $$1.20h = 60 \implies h = \frac{60}{1.2} = 50\text{ inches}.$$

2. Find how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10\text{ inches}.$$

3. Find Ara's growth:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5\text{ inches}.$$

4. Find Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55\text{ inches}.$$

This corresponds to choice E.

ANSWER 4: E

---

Problem 5:
A standard fair six-sided die has faces with numbers 1, 2, 3, 4, 5, and 6.
When the die is tossed, one face is on the bottom and cannot be seen, while the other 5 faces are visible.
We want the product of the 5 visible faces to be divisible by 6.

1. The product of all six numbers on the die is:
   $$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720.$$
   The product of the 5 visible faces is:
   $$P = \frac{720}{b},$$
   where $b \in \{1, 2, 3, 4, 5, 6\}$ is the number on the bottom face.

2. A number is divisible by 6 if and only if it is a multiple of both 2 and 3.
   Let's check the product $P$ for each possible bottom face $b$:
   - If $b = 1$, visible faces are $\{2, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 2$, visible faces are $\{1, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 3$, visible faces are $\{1, 2, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 4$, visible faces are $\{1, 2, 3, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 5$, visible faces are $\{1, 2, 3, 4, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 6$, visible faces are $\{1, 2, 3, 4, 5\}$. The product is $1 \times 2 \times 3 \times 4 \times 5 = 120$. Since $120 = 6 \times 20$, 120 is also divisible by 6 (it has factors 2 and 3)!

3. In every single case (all 6 possible outcomes for the bottom face), the product of the five visible faces is divisible by 6.
   Therefore, the probability is:
   $$\frac{6}{6} = 1.$$

This corresponds to choice E.

ANSWER 5: E

---

Problem 6:
Tori took a 75-problem test:
- 10 arithmetic problems: she answered $70\%$ correctly.
  $$\text{Correct arithmetic} = 0.70 \times 10 = 7.$$
- 30 algebra problems: she answered $40\%$ correctly.
  $$\text{Correct algebra} = 0.40 \times 30 = 12.$$
- 35 geometry problems: she answered $60\%$ correctly.
  $$\text{Correct geometry} = 0.60 \times 35 = 21.$$

1. Total problems Tori answered correctly:
   $$\text{Total correct} = 7 + 12 + 21 = 40.$$

2. Passing score required:
   $60\%$ of 75 problems:
   $$\text{Passing score} = 0.60 \times 75 = 45.$$

3. Additional problems needed:
   $$45 - 40 = 5\text{ problems}.$$

This corresponds to choice B.

ANSWER 6: B

---

Problem 7:
We are given real numbers $a, b, c$ such that $0 < a < b < c$.
We need to determine which of the statements is impossible.

Let's test each choice:
- Choice A: $a + c < b$.
  Since $a > 0$, adding $a$ to $c$ gives $a + c > c$.
  Since we are given $c > b$, it follows that $a + c > c > b$, which means $a + c > b$ must always be true!
  Therefore, $a + c < b$ is impossible.

Let's verify that the other choices can be possible with appropriate values:
- Choice B: $a \cdot b < c$. If $a = 1, b = 2, c = 3$, then $a \cdot b = 2 < 3 = c$. (Possible)
- Choice C: $a + b < c$. If $a = 1, b = 2, c = 4$, then $a + b = 3 < 4 = c$. (Possible)
- Choice D: $a \cdot c < b$. If $a = 0.1, b = 2, c = 3$, then $a \cdot c = 0.3 < 2 = b$. (Possible)
-

# Problem 1:

**Restate:** In 74982.1035, compare the place value of the digit 9 with the place value of the digit 3.

**Working:** Let's identify each place:
- The digit 9 sits in the **hundreds** place: 9 × 100 = 900. Its place value is **100**.
- The digit 3 sits in the **hundredths** place (two positions right of the decimal point): 3 × 0.01 = 0.03. Its place value is **0.01**.

**Compute the ratio:**
$$\frac{100}{0.01} = 100 \div \frac{1}{100} = 100 \times 100 = 10{,}000$$

**Check wrong choices:** Counting places instead of computing the ratio might suggest 100,000 (if you miscount the decimal places), but from hundreds to hundredths is exactly 4 positions, and each position is a factor of 10, giving 10⁴ = 10,000.

ANSWER 1: B

# Problem 2:

**Restate:** Volume increases 4 cm³ per 3° rise. Volume is 24 cm³ at 32°. Find the volume at 20°.

**Working:** The temperature dropped from 32° to 20°:
$$32 - 20 = 12°$$
Number of 3° intervals in the drop:
$$12 \div 3 = 4 \text{ intervals}$$
Each 3° drop means the gas **shrinks** by 4 cm³, so the total shrink is:
$$4 \times 4 = 16 \text{ cm}^3$$
The volume at 20° was:
$$24 - 16 = 8 \text{ cm}^3$$

**Check wrong choices:** Choice D (16) is the *amount* of shrink, not the volume — a tempting trap. Choice E (40) comes from adding instead of subtracting.

ANSWER 2: A

# Problem 3:

**Restate:** Each team plays every other team once; 21 total games. Find the number of teams.

**Working:** With n teams, the number of games is:
$$\frac{n(n-1)}{2}$$
Set this equal to 21:
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42$$
Testing n = 7: $7 \times 6 = 42$. ✓

**Check wrong choices:** n = 6 gives 15 games (too few); n = 8 gives 28 games (too many). Only n = 7 works.

ANSWER 3: B

# Problem 4:

**Restate:** Ara and Shea were once equal in height. Shea grew 20% and is now 60 inches. Ara grew half as many inches as Shea. Find Ara's height now.

**Working:** Let the original height be $h$. Shea grew 20%, so:
$$h + 0.20h = 60 \implies 1.2h = 60 \implies h = 50 \text{ inches}$$
Shea's growth: $60 - 50 = 10$ inches.
Ara grew half as many inches as Shea:
$$\frac{10}{2} = 5 \text{ inches}$$
Ara's height now:
$$50 + 5 = 55 \text{ inches}$$

**Check wrong choices:** Choice A (48) comes from computing 20% of 60 incorrectly; choice D (54) would result if Ara also grew 20% of the original height. The key is that Ara grew 5 **inches**, not a percentage.

ANSWER 4: E

# Problem 5:

**Restate:** A die is tossed; the bottom face is hidden. Find the probability that the product of the five visible faces is divisible by 6.

**Working:** The product of all six faces is:
$$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$$
The product of the five visible faces is:
$$\frac{720}{\text{hidden face}}$$

Check each possible hidden face:
- Hidden = 1: product = 720 → divisible by 6 ✓
- Hidden = 2: product = 360 → divisible by 6 ✓
- Hidden = 3: product = 240 → divisible by 6 ✓
- Hidden = 4: product = 180 → divisible by 6 ✓
- Hidden = 5: product = 144 → divisible by 6 ✓
- Hidden = 6: product = 120 → divisible by 6 ✓

All six outcomes give a product divisible by 6, so the probability is:
$$\frac{6}{6} = 1$$

**Check wrong choices:** One might think hiding the 5 or the 6 removes needed factors, but 720 contains 2 × 3 many times over — removing any single face still leaves factors of 2 and 3 in the product.

ANSWER 5: E

# Problem 6:

**Restate:** Tori's test: 10 arithmetic, 30 algebra, 35 geometry problems (75 total). She got 70%, 40%, and 60% correct respectively, which was below 60% overall. How many more correct answers were needed for 60%?

**Working:** Correct answers in each category:
- Arithmetic: $0.70 \times 10 = 7$
- Algebra: $0.40 \times 30 = 12$
- Geometry: $0.60 \times 35 = 21$

Total correct: $7 + 12 + 21 = 40$

To pass, she needs 60% of 75:
$$0.60 \times 75 = 45$$

Additional problems needed:
$$45 - 40 = 5$$

**Check wrong choices:** Note that the weighted average $\frac{40}{75} \approx 53.3\%$ is indeed below 60%, consistent with the story. A tempting trap is to average the three percentages: $\frac{70+40+60}{3} \approx 56.7\%$, which is wrong because the categories have different sizes.

ANSWER 6: B

# Problem 7:

**Restate:** Given $0 < a < b < c$, which statement is impossible?

**Working:** Test each option:

**A. $a + c < b$:** Since $c > b$, adding positive $a$ gives $a + c > c > b$. So $a + c < b$ **cannot happen**. ✓ (impossible)

Let's verify the others are all possible with suitable numbers:

**B. $a \cdot b < c$:** Try $a = 0.1, b = 0.2, c = 1$: $0.02 < 1$ ✓ possible.

**C. $a + b < c$:** Try $a = 1, b = 2, c = 10$: $3 < 10$ ✓ possible.

**D. $a \cdot c < b$:** Try $a = 0.1, b = 0.5, c = 0.6$: $0.06 < 0.5$ ✓ possible.

**E. $bc = a$:** Try $a = 0.06, b = 0.3, c = 0.2$... wait, need $b < c$. Try $a = 0.02, b = 0.1, c = 0.2$: $bc = 0.02 = a$, and $0 < 0.02 < 0.1 < 0.2$ ✓ possible.

Only option A forces a contradiction: $a + c > c > b$ always holds.

ANSWER 7: A

# Problem 8:

**Restate:** One recipe = 15 cookies using 3 tablespoons of butter; only full recipes allowed. Need 216 cookies. 8 tablespoons = 1 stick. How many sticks?

**Working:** Number of recipes needed:
$$\frac{216}{15} = 14.4$$
Since only full recipes are made, they must make **15 recipes** (14 recipes × 15 = 210 cookies, which is not enough).

Butter needed:
$$15 \times 3 = 45 \text{ tablespoons}$$
Sticks needed:
$$\frac{45}{8} = 5.625 \implies \text{need } 6 \text{ full sticks}$$

**Check wrong choices:** Choice A (5) is the trap — it uses only 40 tablespoons, but 45 are required. You can't buy a partial stick, so round up to 6.

ANSWER 8: B

# Problem 9:

**Restate:** Build a 3-inch cube from 27 unit cubes: 21 red, 6 white. Arrange to minimize white surface area. What fraction of the surface area is white?

**Working:** The 3×3×3 cube has total surface area:
$$6 \text{ faces} \times 9 \text{ sq in} = 54 \text{ square inches}$$

Classify the 27 positions by how many faces each shows:
- **1 inte

Problem 1:
We need the ratio of the place values of the digits 9 and 3 in  
\(74982.1035\).

The digit 9 is in the hundreds place, so its place value is  
\[
100.
\]

The digit 3 is in the thousandths place, so its place value is  
\[
0.001.
\]

Now compute the ratio:
\[
\frac{100}{0.001}=100{,}000.
\]

So the value of the place of the 9 is \(100{,}000\) times as great as the value of the place of the 3.

ANSWER 1: C

Problem 2:
We are told that for every \(3^\circ\) rise in temperature, the gas expands by 4 cubic centimeters.

At temperature \(32^\circ\), the volume is 24 cubic centimeters. We want the volume at \(20^\circ\).

The temperature difference is
\[
32-20=12^\circ.
\]

From \(20^\circ\) to \(32^\circ\), the temperature rises by \(12^\circ\). Since each \(3^\circ\) rise increases the volume by 4 cubic centimeters,
\[
\frac{12}{3}\cdot 4=4\cdot 4=16
\]
cubic centimeters.

So at \(20^\circ\), the volume was
\[
24-16=8
\]
cubic centimeters.

ANSWER 2: A

Problem 3:
Each team plays every other team exactly once. If there are \(n\) teams, the number of games is
\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games, so
\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:
\[
n(n-1)=42.
\]

We look for two consecutive positive integers whose product is 42:
\[
6\cdot 7=42.
\]

Thus
\[
n=7.
\]

So there were 7 teams.

ANSWER 3: B

Problem 4:
Ara and Shea were originally the same height. Let that original height be \(h\) inches.

Shea has grown 20%, so Shea’s current height is
\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:
\[
1.20h=60.
\]

Solve for \(h\):
\[
h=\frac{60}{1.20}=50.
\]

So they were originally 50 inches tall. Shea grew
\[
60-50=10
\]
inches.

Ara grew half as many inches as Shea:
\[
\frac{10}{2}=5
\]
inches.

Therefore Ara is now
\[
50+5=55
\]
inches tall.

ANSWER 4: E

Problem 5:
A fair six-sided die has faces numbered 1 through 6. When it is tossed, one face is hidden on the bottom, and the other five faces are visible.

We need the probability that the product of the five visible numbers is divisible by 6.

A number is divisible by 6 if it has at least one factor of 2 and at least one factor of 3.

No matter which single face is hidden:

- There are three even faces: 2, 4, and 6. Hiding only one of them still leaves at least one even face visible.
- There are two multiples of 3: 3 and 6. Hiding only one of them still leaves at least one multiple of 3 visible.

So the product of the visible faces always contains both a factor of 2 and a factor of 3. Therefore it is always divisible by 6.

Thus the probability is
\[
1.
\]

ANSWER 5: E

Problem 6:
Tori answered:

- \(70\%\) of 10 arithmetic problems:
\[
0.70\cdot 10=7
\]
- \(40\%\) of 30 algebra problems:
\[
0.40\cdot 30=12
\]
- \(60\%\) of 35 geometry problems:
\[
0.60\cdot 35=21
\]

Total correct:
\[
7+12+21=40.
\]

There were 75 problems total. To pass, she needed \(60\%\) correct:
\[
0.60\cdot 75=45.
\]

She answered 40 correctly, so she needed
\[
45-40=5
\]
more correct answers.

ANSWER 6: B

Problem 7:
We are given
\[
0<a<b<c.
\]

We need the impossible statement.

Choice A says
\[
a+c<b.
\]

But since \(c>b\) and \(a>0\),
\[
a+c>b.
\]

So \(a+c<b\) cannot happen.

We can also check that the other choices are possible:

- B: \(a\cdot b<c\) can happen, for example \(a=1,b=2,c=100\).
- C: \(a+b<c\) can happen, for example \(a=1,b=2,c=100\).
- D: \(a\cdot c<b\) can happen, for example \(a=0.1,b=2,c=3\), since \(0.3<2\).
- E: If interpreted as \(bc=a\), it can happen, for example \(a=0.1,b=0.2,c=0.5\), since \(0.2\cdot 0.5=0.1\).

Thus the impossible choice is A.

ANSWER 7: A

Problem 8:
One recipe makes 15 cookies and uses 3 tablespoons of butter. Only full recipes can be made.

Walter and Gretel need 216 cookies. The number of full recipes needed is
\[
\left\lceil \frac{216}{15}\right\rceil.
\]

Since
\[
\frac{216}{15}=14.4,
\]
they must make 15 full recipes.

Butter needed:
\[
15\cdot 3=45
\]
tablespoons.

There are 8 tablespoons in a stick of butter, so the number of sticks needed is
\[
\left\lceil \frac{45}{8}\right\rceil.
\]

Since
\[
\frac{45}{8}=5.625,
\]
they need 6 sticks of butter.

ANSWER 8: B

Problem 9:
A 3-inch cube made from 27 one-inch cubes has surface area
\[
6\cdot 3^2=54
\]
square inches.

Each small cube can show:

- 0 faces if it is the center cube,
- 1 face if it is a face-center cube,
- 2 faces if it is an edge cube,
- 3 faces if it is a corner cube.

We have 6 white cubes. To minimize the white surface area, we should hide as many white cubes as possible.

There is exactly one interior cube, so put one white cube there. It shows 0 square inches.

The remaining 5 white cubes must be on the surface. To minimize exposed white area, put them on face-center positions, where each shows exactly 1 square inch.

So the minimum white surface area is
\[
1\cdot 0+5\cdot 1=5
\]
square inches.

The fraction of the surface area that is white is
\[
\frac{5}{54}.
\]

ANSWER 9: A

Problem 10:
The population is 200 in 1998, and it triples every 25 years.

From 1998 to 2050 is
\[
2050-1998=52
\]
years.

Two full 25-year periods are 50 years, so after 50 years the population is
\[
200\cdot 3\cdot 3=200\cdot 9=1800.
\]

The extra 2 years would add a relatively small amount, so the population in 2050 is a little above 1800. Among the choices, the best estimate is 2000.

ANSWER 10: D

# Problem 1:
Write out the place values of 74982.1035:
- 7 → ten-thousands, 4 → thousands, **9 → hundreds**, 8 → tens, 2 → ones
- 1 → tenths, 0 → hundredths, **3 → thousandths**, 5 → ten-thousandths

So the 9 sits in the hundreds place (value 100) and the 3 sits in the thousandths place (value 0.001).

Ratio = 100 ÷ 0.001 = 100,000.

(Answers like 1,000 or 10,000 come from miscounting decimal places — the 3 is the *third* digit after the point, not the first or second.)

ANSWER 1: C

# Problem 2:
Temperature goes from 32° down to 20°, a drop of 12°.

Each 3° change corresponds to 4 cm³ of change: 12 ÷ 3 = 4 steps.

Volume change = 4 × 4 = 16 cm³ decrease (cooling means shrinking).

Volume = 24 − 16 = 8 cm³.

(40 is the trap from *adding* instead of subtracting; 16 is just the change, not the volume.)

ANSWER 2: A

# Problem 3:
With n teams each playing every other team once, the number of games is C(n,2) = n(n−1)/2.

Set n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7·6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 games — neither is 21.

ANSWER 3: B

# Problem 4:
Let the common original height be x. Shea grew 20%: 1.20x = 60 → x = 50.

So Shea grew 60 − 50 = 10 inches.

Ara grew half as many inches: 5 inches.

Ara's height = 50 + 5 = 55 inches.

(54 would result from wrongly taking Shea's growth as 10% of 60, etc.)

ANSWER 4: E

# Problem 5:
The product of all six faces is 1·2·3·4·5·6 = 720. If the hidden bottom face is k, the visible product is 720/k.

- k=1: 720 ✓ divisible by 6
- k=2: 360 ✓
- k=3: 240 ✓
- k=4: 180 ✓
- k=5: 144 ✓
- k=6: 120 ✓

Reason why: hiding one face still leaves at least one even number and at least one multiple of 3 among the five visible faces (e.g., hide 6 → 2 and 3 remain; hide 3 → 6 remains).

So the probability is 6/6 = 1.

ANSWER 5: E

# Problem 6:
Correct answers:
- Arithmetic: 70% of 10 = 7
- Algebra: 40% of 30 = 12
- Geometry: 60% of 35 = 21

Total correct = 7 + 12 + 21 = 40.

Needed for 60%: 0.60 × 75 = 45.

She needs 45 − 40 = 5 more.

(40/75 ≈ 53%, consistent with failing.)

ANSWER 6: B

# Problem 7:
Given 0 < a < b < c.

- **A. a + c < b:** Since c > b and a > 0, we get a + c > b. So a + c < b is impossible.
- B. a·b < c: take a = 0.1, b = 0.2, c = 1 → 0.02 < 1 ✓ possible.
- C. a + b < c: take a = 1, b = 2, c = 10 → 3 < 10 ✓ possible.
- D. a·c < b: take a = 0.1, b = 1, c = 2 → 0.2 < 1 ✓ possible.
- E. b/c = a: take a = 0.5, b = 1, c = 2 → 1/2 = 0.5 ✓ possible.

ANSWER 7: A

# Problem 8:
Pans needed: 216 ÷ 15 = 14.4, and only full recipes are made, so they must bake 15 pans.

Butter: 15 × 3 = 45 tablespoons.

Sticks: 45 ÷ 8 = 5.625, and butter comes in whole sticks, so 6 sticks.

(5 sticks = 40 tbsp, not enough.)

ANSWER 8: B

# Problem 9:
The 3×3×3 cube has surface area 6 × 9 = 54 unit squares.

Positions in a 3-cube: 1 center cube (0 faces showing), 6 face-centers (1 face each), 12 edges (2 faces), 8 corners (3 faces).

To minimize white showing, place one white cube in the hidden center (0 white squares) and the other 5 white cubes at face-center positions (1 white square each).

Minimum white area = 5 unit squares out of 54 → 5/54.

ANSWER 9: A

# Problem 10:
From 1998 to 2050 is 52 years — about two 25-year tripling periods (1998 → 2023 → 2048).

Population: 200 → 600 → 1800.

1800 is closest to 2000.

ANSWER 10: D

Problem 1:

We are comparing the place values of the digits 9 and 3 in

\[
74982.1035.
\]

- The digit \(9\) is in the hundreds place, so its place value is \(100\).
- The digit \(3\) is in the thousandths place, so its place value is

\[
0.001.
\]

Now find how many times as great \(100\) is as \(0.001\):

\[
\frac{100}{0.001}=100,000.
\]

So the hundreds place is \(100,\!000\) times the thousandths place.

- A, \(1,\!000\), is too small.
- B, \(10,\!000\), is too small.
- D and E are too large.

ANSWER 1: C

Problem 2:

The gas expands by \(4\) cubic centimeters for every \(3^\circ\) increase in temperature.

The temperature changes from \(32^\circ\) down to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

Since

\[
12^\circ \div 3^\circ=4,
\]

there are 4 groups of \(3^\circ\).

For each group, the volume decreases by \(4\) cubic centimeters:

\[
4\cdot 4=16.
\]

At \(32^\circ\), the volume is \(24\) cubic centimeters, so at \(20^\circ\) it was

\[
24-16=8.
\]

Choices B, C, and D do not subtract the full \(16\) cubic centimeters. Choice E would represent an increase rather than a decrease.

ANSWER 2: A

Problem 3:

If there are \(n\) teams and each pair of teams plays exactly once, then the number of games is

\[
\binom n2=\frac{n(n-1)}2.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

Checking: with 7 teams, the number of games is

\[
\frac{7\cdot6}{2}=21.
\]

Six teams would produce only \(15\) games, while 8 teams would produce \(28\) games.

ANSWER 3: B

Problem 4:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew \(20\%\), so Shea’s current height is

\[
1.20h=60.
\]

Thus,

\[
h=\frac{60}{1.2}=50.
\]

So both were originally 50 inches tall.

Shea’s growth was

\[
60-50=10
\]

inches.

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5
\]

inches.

Therefore, Ara is now

\[
50+5=55
\]

inches tall.

Choice A, 48, is less than Ara’s original height. Choices B, C, and D do not include the correct 5-inch growth.

ANSWER 4: E

Problem 5:

A standard die has faces numbered \(1,2,3,4,5,6\). The product of all six face numbers is

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6=720.
\]

If the bottom face has number \(k\), then the product of the five visible faces is

\[
\frac{720}{k}.
\]

Check each possible bottom face:

\[
\frac{720}{1}=720,\qquad
\frac{720}{2}=360,\qquad
\frac{720}{3}=240,
\]

\[
\frac{720}{4}=180,\qquad
\frac{720}{5}=144,\qquad
\frac{720}{6}=120.
\]

Every one of these numbers is divisible by \(6\).

Therefore, no matter which face is on the bottom, the visible-face product is divisible by 6. The probability is

\[
1.
\]

All of choices A through D are less than 1, but the event always happens.

ANSWER 5: E

Problem 6:

Tori answered correctly:

- Arithmetic:

\[
70\%\text{ of }10=0.7(10)=7
\]

- Algebra:

\[
40\%\text{ of }30=0.4(30)=12
\]

- Geometry:

\[
60\%\text{ of }35=0.6(35)=21
\]

So her total number correct was

\[
7+12+21=40.
\]

To pass, she needs \(60\%\) of 75 questions correct:

\[
0.60(75)=45.
\]

She needs

\[
45-40=5
\]

more correct answers.

Choice A is too small, while choices C, D, and E give more than the number needed to reach exactly 45 correct.

ANSWER 6: B

Problem 7:

We are given

\[
0<a<b<c.
\]

We must find the statement that is impossible.

Consider choice A:

\[
a+c<b.
\]

But \(c>b\), and also \(a>0\). Therefore,

\[
a+c>c>b.
\]

So it is impossible for \(a+c<b\).

The other choices can happen for suitable positive numbers. For example:

- B: \(a\cdot b<c\) can occur with \(a=1,\ b=2,\ c=3\), since \(1\cdot2<3\).
- C: \(a+b<c\) can occur with \(a=1,\ b=2,\ c=4\).
- D: \(a\cdot c<b\) can occur with \(a=\frac14,\ b=1,\ c=2\), since \(\frac14\cdot2=\frac12<1\).
- E: \(bc=a\) can occur with \(b=\frac12,\ c=\frac34\), giving \(a=\frac38\), and indeed

\[
0<\frac38<\frac12<\frac34.
\]

Thus only choice A is impossible.

ANSWER 7: A

Problem 8:

Each full recipe makes 15 cookies.

Walter and Gretel need at least 216 cookies. Since only full recipes may be made, determine how many recipes are needed:

\[
216\div 15=14.4.
\]

They cannot make \(14.4\) recipes, so they must make 15 full recipes.

These 15 recipes make

\[
15\cdot15=225
\]

cookies, enough for the crowd.

Each recipe uses 3 tablespoons of butter, so the total butter needed is

\[
15\cdot3=45
\]

tablespoons.

There are 8 tablespoons in each stick, so

\[
45\div8=5.625.
\]

They must buy 6 full sticks of butter.

Five sticks provide only \(5\cdot8=40\) tablespoons, which is not enough. Six sticks provide \(48\) tablespoons, enough for the 45 tablespoons needed.

ANSWER 8: B

Problem 9:

A \(3\times3\times3\) cube has surface area

\[
6\cdot 3\cdot3=54
\]

square inches, since each of its 6 faces has area \(3\cdot3=9\).

To minimize the white surface showing, place the white cubes in the least exposed positions.

- The one center cube of the large cube has no faces exposed.
- The six face-center cubes each have exactly one face exposed.
- Edge cubes have two exposed faces.
- Corner cubes have three exposed faces.

There are 6 white cubes. Put one white cube in the completely hidden center. Then put the other 5 white cubes at five face-center locations.

Thus, the white cubes show only

\[
5
\]

unit square faces on the surface.

Therefore, the fraction of the surface that is white is

\[
\frac{5}{54}.
\]

Choices such as \(\frac19,\frac{5}{27},\frac29,\frac13\) represent more than 5 white unit squares out of the 54 total surface squares.

ANSWER 9: A

Problem 10:

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

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is approximately two 25-year periods.

After one 25-year period:

\[
200\cdot3=600.
\]

After two 25-year periods:

\[
600\cdot3=1800.
\]

Since 2050 is slightly more than 50 years after 1998, the population should be slightly more than 1800. The closest estimate is

\[
20

Problem 1:
We are asked: in 74982.1035, compare the place value of the 9 with the place value of the 3.

Write out places:
74982 is: 7 ten-thousands, 4 thousands, 9 hundreds, 8 tens, 2 ones.
So the 9 is in the hundreds place. Its value is 100.

After the decimal .1035 is: 1 tenths, 0 hundredths, 3 thousandths, 5 ten-thousandths.
So the 3 is in the thousandths place. Its value is 0.001 = 1/1000.

Ratio = (value of 9's place) / (value of 3's place)
= 100 / 0.001 = 100 / (1/1000) = 100 x 1000 = 100,000.

So the hundreds place is 100,000 times as great as the thousandths place.
Choices A,B,D,E are powers of 10 that are too small or too large.

ANSWER 1: C

Problem 2:
We are asked for the earlier volume given expansion rate.

Rate: +3 degrees temperature = +4 cubic cm volume.

Now: temp 32°, volume 24 cc.
Before: temp 20°.

Temperature change from 20° to 32° is 32-20 = 12°.
Number of 3-degree steps in 12° is 12 / 3 = 4 steps.

So volume increase from 20° to 32° is 4 steps x 4 cc = 16 cc.

Thus volume at 20° + 16 = volume at 32° = 24.
So volume at 20° = 24 - 16 = 8 cc.

Check: A 8 fits, B 12 would be only 3 steps, D 16 would be only 2 steps, E 40 would be adding instead of subtracting.

ANSWER 2: A

Problem 3:
We need n teams such that every pair plays once gives 21 games.

Number of pairs from n teams is n choose 2 = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try integers: 6x5=30 too small, 7x6=42 correct, 8x7=56 too big.
So n = 7.

Ruling out: A 6 gives only 15 games, C 8 gives 28 games, etc.

ANSWER 3: B

Problem 4:
Ara and Shea started at same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara's now height.

Let original height = H.
Shea's growth = 20% of H = 0.20 H.
Shea now = H + 0.20H = 1.2H = 60.
So H = 60 / 1.2 = 50 inches.

Shea grew 60-50 = 10 inches.
Ara grew half as many inches: 10 / 2 = 5 inches.
Ara now = 50 + 5 = 55 inches.

Tempting wrong A 48 would be working backwards, D 54 would be misreading percent, etc.

ANSWER 4: E

Problem 5:
A standard 1-6 die is tossed. Bottom face hidden, other 5 faces seen. Probability product of seen numbers divisible by 6?

Product divisible by 6 means divisible by 2 and by 3.

Numbers 1 to 6: evens are 2,4,6 — three of them. Multiples of 3 are 3,6 — two of them.

Hiding only one face (the bottom) can hide at most one even, leaving at least two evens still visible, so factor 2 remains. It can hide at most one multiple of 3, leaving at least one multiple of 3 still visible, so factor 3 remains.

Indeed total product 1·2·3·4·5·6 = 720. Visible product = 720 / b where b is bottom. 720/b divided by 6 = 120/b, and b=1,2,3,4,5,6 all divide 120, so always integer.

Hence no matter which face is hidden, the visible product is divisible by 6. Probability = 6/6 = 1.

Other choices 1/3,1/2,2/3,5/6 assume sometimes it fails, which is impossible.

ANSWER 5: E

Problem 6:
75 problems total: 10 arithmetic, 30 algebra, 35 geometry.
Correct: 70% of arithmetic, 40% of algebra, 60% of geometry. How many more to reach 60% overall?

Compute:
Arithmetic correct = 0.70 x 10 = 7.
Algebra correct = 0.40 x 30 = 12.
Geometry correct = 0.60 x 35 = 21.
Total correct = 7+12+21 = 40.

Need for 60% passing: 0.60 x 75 = 45 problems.

Shortfall = 45 - 40 = 5 more problems.

So B 5, not A 1 or C 7.

ANSWER 6: B

Problem 7:
Given 0 < a < b < c, which statement is impossible?

A. a + c < b. Since a > 0, a + c > c. And c > b. So a+c > b always. Hence a+c < b can never happen. This is impossible.

Check others are possible with examples:
B. a·b < c: e.g. a=1,b=2,c=10 gives 2<10 possible.
C. a+b < c: e.g. a=1,b=2,c=10 gives 3<10 possible.
D. a·c < b: with fractions <1. e.g. a=0.1,c=2,b=1 gives 0.2<1 and 0.1<1<2 possible.
E. b c = a: let b=0.25,a=0.1, then c=a/b=0.4, and 0.1<0.25<0.4 possible.

So only A is impossible. Students may think E impossible because b c > b > a for numbers >1, but with numbers <1 multiplication can shrink.

ANSWER 7: A

Problem 8:
One full recipe = 15 cookies = 3 tbsp butter. Need 216 cookies. 8 tbsp = 1 stick. How many sticks?

Pans needed: 216 / 15 = 14.4. Since only full recipes, round up to 15 pans. 14 pans would give only 210 cookies, not enough.

Butter needed: 15 x 3 tbsp = 45 tbsp.

Sticks needed: 45 / 8 = 5.625. Since need whole sticks to have enough, round up to 6. 5 sticks = 40 tbsp < 45 tbsp, not enough.

So answer 6. A 5 would leave butter and cookies short.

ANSWER 8: B

Problem 9:
Build 3x3x3 = 27 small cubes. 6 white, 21 red. Minimize white squares on outside. What fraction of outside is white?

Big cube surface: 6 faces each 3x3 = 9 small squares, total 54 small squares.

Small-cube positions by exposure:
- 1 interior center: 0 faces shown
- 6 face-centers: 1 face shown
- 12 edge non-corner: 2 faces shown
- 8 corners: 3 faces shown

To minimize white, put whites where least exposure:
Put 1 white in center → 0 exposed.
Put remaining 5 whites in face-center spots → 1 each exposed.
Total white exposed squares = 5.

Fraction white = 5 / 54 of total surface area.

Larger fractions like 1/9=6/54, etc. would result from worse placement.

ANSWER 9: A

Problem 10:
1998 population 200, triples every 25 years. Estimate 2050.

Time elapsed: 2050-1998 = 52 years.
Number of 25-year periods: 52/25 ≈ 2.08, just over 2 doublings-triplings.

After 25 years (~2023): 200x3=600.
After 50 years (~2048): 600x3=1800.
2050 is 2 years later, a little more, so about 2000.

Formally 200 x 3^(52/25) ≈ 200 x 9 x small extra ≈ 1960.

So A 600 is only one tripling, C 1000 too low, E 3000 too high. Best is D 2000.

ANSWER 10: D

Problem 1:
We are asked to find the ratio of the value of the place occupied by the digit 9 to the value of the place occupied by the digit 3 in the decimal number $74982.1035$.

1. Identify the place values:
   - The number is $74982.1035$.
   - The digit 2 is in the ones place ($10^0$).
   - The digit 8 is in the tens place ($10^1$).
   - The digit 9 is in the hundreds place ($10^2 = 100$).
   - The digit 1 is in the tenths place ($10^{-1} = 0.1$).
   - The digit 0 is in the hundredths place ($10^{-2} = 0.01$).
   - The digit 3 is in the thousandths place ($10^{-3} = 0.001$).
   - The digit 5 is in the ten-thousandths place ($10^{-4} = 0.0001$).

2. Compute the ratio of the place values:
   $$\frac{\text{Place value of } 9}{\text{Place value of } 3} = \frac{100}{0.001} = \frac{10^2}{10^{-3}} = 10^{2 - (-3)} = 10^5 = 100,000.$$

This corresponds to choice C.

ANSWER 1: C

---

Problem 2:
We are asked to find the initial volume of gas at $20^\circ$ given its volume at $32^\circ$ and its rate of expansion.

1. Find the change in temperature:
   The temperature changes from $20^\circ$ to $32^\circ$, which is an increase of:
   $$32^\circ - 20^\circ = 12^\circ.$$

2. Determine the volume expansion:
   For every $3^\circ$ rise, the volume expands by $4\text{ cm}^3$.
   In a $12^\circ$ rise, the number of $3^\circ$ intervals is:
   $$\frac{12^\circ}{3^\circ} = 4.$$
   Therefore, the volume expanded by:
   $$4 \times 4\text{ cm}^3 = 16\text{ cm}^3.$$

3. Calculate the original volume at $20^\circ$:
   $$\text{Volume at } 20^\circ = \text{Volume at } 32^\circ - 16\text{ cm}^3 = 24 - 16 = 8\text{ cm}^3.$$

This matches choice A.

ANSWER 2: A

---

Problem 3:
We are given that each of the $N$ teams plays every other team exactly once, resulting in 21 total games. We need to find $N$.

1. The number of games played between $N$ teams is the number of ways to choose 2 teams from $N$, given by the combination formula:
   $$\binom{N}{2} = \frac{N(N-1)}{2}.$$

2. Set this equal to 21 and solve for $N$:
   $$\frac{N(N-1)}{2} = 21 \implies N(N-1) = 42.$$

3. Find two consecutive positive integers whose product is 42:
   Since $7 \times 6 = 42$, we have $N = 7$.

Checking the choices, $N = 7$ corresponds to choice B.

ANSWER 3: B

---

Problem 4:
We are given:
- Ara and Shea initially had the same height, say $h$ inches.
- Shea has grown $20\%$, so Shea's new height is $h + 0.20h = 1.20h = 60$ inches.
- Ara has grown half as many inches as Shea.
- We need to find Ara's current height.

1. Find the original height $h$:
   $$1.20h = 60 \implies h = \frac{60}{1.2} = 50\text{ inches}.$$

2. Find how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10\text{ inches}.$$

3. Find Ara's growth:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5\text{ inches}.$$

4. Find Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55\text{ inches}.$$

This corresponds to choice E.

ANSWER 4: E

---

Problem 5:
A standard fair six-sided die has faces with numbers 1, 2, 3, 4, 5, and 6.
When the die is tossed, one face is on the bottom and cannot be seen, while the other 5 faces are visible.
We want the product of the 5 visible faces to be divisible by 6.

1. The product of all six numbers on the die is:
   $$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720.$$
   The product of the 5 visible faces is:
   $$P = \frac{720}{b},$$
   where $b \in \{1, 2, 3, 4, 5, 6\}$ is the number on the bottom face.

2. A number is divisible by 6 if and only if it is a multiple of both 2 and 3.
   Let's check the product $P$ for each possible bottom face $b$:
   - If $b = 1$, visible faces are $\{2, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 2$, visible faces are $\{1, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 3$, visible faces are $\{1, 2, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 4$, visible faces are $\{1, 2, 3, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 5$, visible faces are $\{1, 2, 3, 4, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 6$, visible faces are $\{1, 2, 3, 4, 5\}$. The product is $1 \times 2 \times 3 \times 4 \times 5 = 120$. Since $120 = 6 \times 20$, 120 is also divisible by 6 (it has factors 2 and 3)!

3. In every single case (all 6 possible outcomes for the bottom face), the product of the five visible faces is divisible by 6.
   Therefore, the probability is:
   $$\frac{6}{6} = 1.$$

This corresponds to choice E.

ANSWER 5: E

---

Problem 6:
Tori took a 75-problem test:
- 10 arithmetic problems: she answered $70\%$ correctly.
  $$\text{Correct arithmetic} = 0.70 \times 10 = 7.$$
- 30 algebra problems: she answered $40\%$ correctly.
  $$\text{Correct algebra} = 0.40 \times 30 = 12.$$
- 35 geometry problems: she answered $60\%$ correctly.
  $$\text{Correct geometry} = 0.60 \times 35 = 21.$$

1. Total problems Tori answered correctly:
   $$\text{Total correct} = 7 + 12 + 21 = 40.$$

2. Passing score required:
   $60\%$ of 75 problems:
   $$\text{Passing score} = 0.60 \times 75 = 45.$$

3. Additional problems needed:
   $$45 - 40 = 5\text{ problems}.$$

This corresponds to choice B.

ANSWER 6: B

---

Problem 7:
We are given real numbers $a, b, c$ such that $0 < a < b < c$.
We need to determine which of the statements is impossible.

Let's test each choice:
- Choice A: $a + c < b$.
  Since $a > 0$, adding $a$ to $c$ gives $a + c > c$.
  Since we are given $c > b$, it follows that $a + c > c > b$, which means $a + c > b$ must always be true!
  Therefore, $a + c < b$ is impossible.

Let's verify that the other choices can be possible with appropriate values:
- Choice B: $a \cdot b < c$. If $a = 1, b = 2, c = 3$, then $a \cdot b = 2 < 3 = c$. (Possible)
- Choice C: $a + b < c$. If $a = 1, b = 2, c = 4$, then $a + b = 3 < 4 = c$. (Possible)
- Choice D: $a \cdot c < b$. If $a = 0.1, b = 2, c = 3$, then $a \cdot c = 0.3 < 2 = b$. (Possible)
-

# Problem 1:

**Restate:** In 74982.1035, compare the place value of the digit 9 with the place value of the digit 3.

**Working:** Let's identify each place:
- The digit 9 sits in the **hundreds** place: 9 × 100 = 900. Its place value is **100**.
- The digit 3 sits in the **hundredths** place (two positions right of the decimal point): 3 × 0.01 = 0.03. Its place value is **0.01**.

**Compute the ratio:**
$$\frac{100}{0.01} = 100 \div \frac{1}{100} = 100 \times 100 = 10{,}000$$

**Check wrong choices:** Counting places instead of computing the ratio might suggest 100,000 (if you miscount the decimal places), but from hundreds to hundredths is exactly 4 positions, and each position is a factor of 10, giving 10⁴ = 10,000.

ANSWER 1: B

# Problem 2:

**Restate:** Volume increases 4 cm³ per 3° rise. Volume is 24 cm³ at 32°. Find the volume at 20°.

**Working:** The temperature dropped from 32° to 20°:
$$32 - 20 = 12°$$
Number of 3° intervals in the drop:
$$12 \div 3 = 4 \text{ intervals}$$
Each 3° drop means the gas **shrinks** by 4 cm³, so the total shrink is:
$$4 \times 4 = 16 \text{ cm}^3$$
The volume at 20° was:
$$24 - 16 = 8 \text{ cm}^3$$

**Check wrong choices:** Choice D (16) is the *amount* of shrink, not the volume — a tempting trap. Choice E (40) comes from adding instead of subtracting.

ANSWER 2: A

# Problem 3:

**Restate:** Each team plays every other team once; 21 total games. Find the number of teams.

**Working:** With n teams, the number of games is:
$$\frac{n(n-1)}{2}$$
Set this equal to 21:
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42$$
Testing n = 7: $7 \times 6 = 42$. ✓

**Check wrong choices:** n = 6 gives 15 games (too few); n = 8 gives 28 games (too many). Only n = 7 works.

ANSWER 3: B

# Problem 4:

**Restate:** Ara and Shea were once equal in height. Shea grew 20% and is now 60 inches. Ara grew half as many inches as Shea. Find Ara's height now.

**Working:** Let the original height be $h$. Shea grew 20%, so:
$$h + 0.20h = 60 \implies 1.2h = 60 \implies h = 50 \text{ inches}$$
Shea's growth: $60 - 50 = 10$ inches.
Ara grew half as many inches as Shea:
$$\frac{10}{2} = 5 \text{ inches}$$
Ara's height now:
$$50 + 5 = 55 \text{ inches}$$

**Check wrong choices:** Choice A (48) comes from computing 20% of 60 incorrectly; choice D (54) would result if Ara also grew 20% of the original height. The key is that Ara grew 5 **inches**, not a percentage.

ANSWER 4: E

# Problem 5:

**Restate:** A die is tossed; the bottom face is hidden. Find the probability that the product of the five visible faces is divisible by 6.

**Working:** The product of all six faces is:
$$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$$
The product of the five visible faces is:
$$\frac{720}{\text{hidden face}}$$

Check each possible hidden face:
- Hidden = 1: product = 720 → divisible by 6 ✓
- Hidden = 2: product = 360 → divisible by 6 ✓
- Hidden = 3: product = 240 → divisible by 6 ✓
- Hidden = 4: product = 180 → divisible by 6 ✓
- Hidden = 5: product = 144 → divisible by 6 ✓
- Hidden = 6: product = 120 → divisible by 6 ✓

All six outcomes give a product divisible by 6, so the probability is:
$$\frac{6}{6} = 1$$

**Check wrong choices:** One might think hiding the 5 or the 6 removes needed factors, but 720 contains 2 × 3 many times over — removing any single face still leaves factors of 2 and 3 in the product.

ANSWER 5: E

# Problem 6:

**Restate:** Tori's test: 10 arithmetic, 30 algebra, 35 geometry problems (75 total). She got 70%, 40%, and 60% correct respectively, which was below 60% overall. How many more correct answers were needed for 60%?

**Working:** Correct answers in each category:
- Arithmetic: $0.70 \times 10 = 7$
- Algebra: $0.40 \times 30 = 12$
- Geometry: $0.60 \times 35 = 21$

Total correct: $7 + 12 + 21 = 40$

To pass, she needs 60% of 75:
$$0.60 \times 75 = 45$$

Additional problems needed:
$$45 - 40 = 5$$

**Check wrong choices:** Note that the weighted average $\frac{40}{75} \approx 53.3\%$ is indeed below 60%, consistent with the story. A tempting trap is to average the three percentages: $\frac{70+40+60}{3} \approx 56.7\%$, which is wrong because the categories have different sizes.

ANSWER 6: B

# Problem 7:

**Restate:** Given $0 < a < b < c$, which statement is impossible?

**Working:** Test each option:

**A. $a + c < b$:** Since $c > b$, adding positive $a$ gives $a + c > c > b$. So $a + c < b$ **cannot happen**. ✓ (impossible)

Let's verify the others are all possible with suitable numbers:

**B. $a \cdot b < c$:** Try $a = 0.1, b = 0.2, c = 1$: $0.02 < 1$ ✓ possible.

**C. $a + b < c$:** Try $a = 1, b = 2, c = 10$: $3 < 10$ ✓ possible.

**D. $a \cdot c < b$:** Try $a = 0.1, b = 0.5, c = 0.6$: $0.06 < 0.5$ ✓ possible.

**E. $bc = a$:** Try $a = 0.06, b = 0.3, c = 0.2$... wait, need $b < c$. Try $a = 0.02, b = 0.1, c = 0.2$: $bc = 0.02 = a$, and $0 < 0.02 < 0.1 < 0.2$ ✓ possible.

Only option A forces a contradiction: $a + c > c > b$ always holds.

ANSWER 7: A

# Problem 8:

**Restate:** One recipe = 15 cookies using 3 tablespoons of butter; only full recipes allowed. Need 216 cookies. 8 tablespoons = 1 stick. How many sticks?

**Working:** Number of recipes needed:
$$\frac{216}{15} = 14.4$$
Since only full recipes are made, they must make **15 recipes** (14 recipes × 15 = 210 cookies, which is not enough).

Butter needed:
$$15 \times 3 = 45 \text{ tablespoons}$$
Sticks needed:
$$\frac{45}{8} = 5.625 \implies \text{need } 6 \text{ full sticks}$$

**Check wrong choices:** Choice A (5) is the trap — it uses only 40 tablespoons, but 45 are required. You can't buy a partial stick, so round up to 6.

ANSWER 8: B

# Problem 9:

**Restate:** Build a 3-inch cube from 27 unit cubes: 21 red, 6 white. Arrange to minimize white surface area. What fraction of the surface area is white?

**Working:** The 3×3×3 cube has total surface area:
$$6 \text{ faces} \times 9 \text{ sq in} = 54 \text{ square inches}$$

Classify the 27 positions by how many faces each shows:
- **1 inte

Problem 1:
We need the ratio of the place values of the digits 9 and 3 in  
\(74982.1035\).

The digit 9 is in the hundreds place, so its place value is  
\[
100.
\]

The digit 3 is in the thousandths place, so its place value is  
\[
0.001.
\]

Now compute the ratio:
\[
\frac{100}{0.001}=100{,}000.
\]

So the value of the place of the 9 is \(100{,}000\) times as great as the value of the place of the 3.

ANSWER 1: C

Problem 2:
We are told that for every \(3^\circ\) rise in temperature, the gas expands by 4 cubic centimeters.

At temperature \(32^\circ\), the volume is 24 cubic centimeters. We want the volume at \(20^\circ\).

The temperature difference is
\[
32-20=12^\circ.
\]

From \(20^\circ\) to \(32^\circ\), the temperature rises by \(12^\circ\). Since each \(3^\circ\) rise increases the volume by 4 cubic centimeters,
\[
\frac{12}{3}\cdot 4=4\cdot 4=16
\]
cubic centimeters.

So at \(20^\circ\), the volume was
\[
24-16=8
\]
cubic centimeters.

ANSWER 2: A

Problem 3:
Each team plays every other team exactly once. If there are \(n\) teams, the number of games is
\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games, so
\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:
\[
n(n-1)=42.
\]

We look for two consecutive positive integers whose product is 42:
\[
6\cdot 7=42.
\]

Thus
\[
n=7.
\]

So there were 7 teams.

ANSWER 3: B

Problem 4:
Ara and Shea were originally the same height. Let that original height be \(h\) inches.

Shea has grown 20%, so Shea’s current height is
\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:
\[
1.20h=60.
\]

Solve for \(h\):
\[
h=\frac{60}{1.20}=50.
\]

So they were originally 50 inches tall. Shea grew
\[
60-50=10
\]
inches.

Ara grew half as many inches as Shea:
\[
\frac{10}{2}=5
\]
inches.

Therefore Ara is now
\[
50+5=55
\]
inches tall.

ANSWER 4: E

Problem 5:
A fair six-sided die has faces numbered 1 through 6. When it is tossed, one face is hidden on the bottom, and the other five faces are visible.

We need the probability that the product of the five visible numbers is divisible by 6.

A number is divisible by 6 if it has at least one factor of 2 and at least one factor of 3.

No matter which single face is hidden:

- There are three even faces: 2, 4, and 6. Hiding only one of them still leaves at least one even face visible.
- There are two multiples of 3: 3 and 6. Hiding only one of them still leaves at least one multiple of 3 visible.

So the product of the visible faces always contains both a factor of 2 and a factor of 3. Therefore it is always divisible by 6.

Thus the probability is
\[
1.
\]

ANSWER 5: E

Problem 6:
Tori answered:

- \(70\%\) of 10 arithmetic problems:
\[
0.70\cdot 10=7
\]
- \(40\%\) of 30 algebra problems:
\[
0.40\cdot 30=12
\]
- \(60\%\) of 35 geometry problems:
\[
0.60\cdot 35=21
\]

Total correct:
\[
7+12+21=40.
\]

There were 75 problems total. To pass, she needed \(60\%\) correct:
\[
0.60\cdot 75=45.
\]

She answered 40 correctly, so she needed
\[
45-40=5
\]
more correct answers.

ANSWER 6: B

Problem 7:
We are given
\[
0<a<b<c.
\]

We need the impossible statement.

Choice A says
\[
a+c<b.
\]

But since \(c>b\) and \(a>0\),
\[
a+c>b.
\]

So \(a+c<b\) cannot happen.

We can also check that the other choices are possible:

- B: \(a\cdot b<c\) can happen, for example \(a=1,b=2,c=100\).
- C: \(a+b<c\) can happen, for example \(a=1,b=2,c=100\).
- D: \(a\cdot c<b\) can happen, for example \(a=0.1,b=2,c=3\), since \(0.3<2\).
- E: If interpreted as \(bc=a\), it can happen, for example \(a=0.1,b=0.2,c=0.5\), since \(0.2\cdot 0.5=0.1\).

Thus the impossible choice is A.

ANSWER 7: A

Problem 8:
One recipe makes 15 cookies and uses 3 tablespoons of butter. Only full recipes can be made.

Walter and Gretel need 216 cookies. The number of full recipes needed is
\[
\left\lceil \frac{216}{15}\right\rceil.
\]

Since
\[
\frac{216}{15}=14.4,
\]
they must make 15 full recipes.

Butter needed:
\[
15\cdot 3=45
\]
tablespoons.

There are 8 tablespoons in a stick of butter, so the number of sticks needed is
\[
\left\lceil \frac{45}{8}\right\rceil.
\]

Since
\[
\frac{45}{8}=5.625,
\]
they need 6 sticks of butter.

ANSWER 8: B

Problem 9:
A 3-inch cube made from 27 one-inch cubes has surface area
\[
6\cdot 3^2=54
\]
square inches.

Each small cube can show:

- 0 faces if it is the center cube,
- 1 face if it is a face-center cube,
- 2 faces if it is an edge cube,
- 3 faces if it is a corner cube.

We have 6 white cubes. To minimize the white surface area, we should hide as many white cubes as possible.

There is exactly one interior cube, so put one white cube there. It shows 0 square inches.

The remaining 5 white cubes must be on the surface. To minimize exposed white area, put them on face-center positions, where each shows exactly 1 square inch.

So the minimum white surface area is
\[
1\cdot 0+5\cdot 1=5
\]
square inches.

The fraction of the surface area that is white is
\[
\frac{5}{54}.
\]

ANSWER 9: A

Problem 10:
The population is 200 in 1998, and it triples every 25 years.

From 1998 to 2050 is
\[
2050-1998=52
\]
years.

Two full 25-year periods are 50 years, so after 50 years the population is
\[
200\cdot 3\cdot 3=200\cdot 9=1800.
\]

The extra 2 years would add a relatively small amount, so the population in 2050 is a little above 1800. Among the choices, the best estimate is 2000.

ANSWER 10: D

# Problem 1:
Write out the place values of 74982.1035:
- 7 → ten-thousands, 4 → thousands, **9 → hundreds**, 8 → tens, 2 → ones
- 1 → tenths, 0 → hundredths, **3 → thousandths**, 5 → ten-thousandths

So the 9 sits in the hundreds place (value 100) and the 3 sits in the thousandths place (value 0.001).

Ratio = 100 ÷ 0.001 = 100,000.

(Answers like 1,000 or 10,000 come from miscounting decimal places — the 3 is the *third* digit after the point, not the first or second.)

ANSWER 1: C

# Problem 2:
Temperature goes from 32° down to 20°, a drop of 12°.

Each 3° change corresponds to 4 cm³ of change: 12 ÷ 3 = 4 steps.

Volume change = 4 × 4 = 16 cm³ decrease (cooling means shrinking).

Volume = 24 − 16 = 8 cm³.

(40 is the trap from *adding* instead of subtracting; 16 is just the change, not the volume.)

ANSWER 2: A

# Problem 3:
With n teams each playing every other team once, the number of games is C(n,2) = n(n−1)/2.

Set n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7·6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 games — neither is 21.

ANSWER 3: B

# Problem 4:
Let the common original height be x. Shea grew 20%: 1.20x = 60 → x = 50.

So Shea grew 60 − 50 = 10 inches.

Ara grew half as many inches: 5 inches.

Ara's height = 50 + 5 = 55 inches.

(54 would result from wrongly taking Shea's growth as 10% of 60, etc.)

ANSWER 4: E

# Problem 5:
The product of all six faces is 1·2·3·4·5·6 = 720. If the hidden bottom face is k, the visible product is 720/k.

- k=1: 720 ✓ divisible by 6
- k=2: 360 ✓
- k=3: 240 ✓
- k=4: 180 ✓
- k=5: 144 ✓
- k=6: 120 ✓

Reason why: hiding one face still leaves at least one even number and at least one multiple of 3 among the five visible faces (e.g., hide 6 → 2 and 3 remain; hide 3 → 6 remains).

So the probability is 6/6 = 1.

ANSWER 5: E

# Problem 6:
Correct answers:
- Arithmetic: 70% of 10 = 7
- Algebra: 40% of 30 = 12
- Geometry: 60% of 35 = 21

Total correct = 7 + 12 + 21 = 40.

Needed for 60%: 0.60 × 75 = 45.

She needs 45 − 40 = 5 more.

(40/75 ≈ 53%, consistent with failing.)

ANSWER 6: B

# Problem 7:
Given 0 < a < b < c.

- **A. a + c < b:** Since c > b and a > 0, we get a + c > b. So a + c < b is impossible.
- B. a·b < c: take a = 0.1, b = 0.2, c = 1 → 0.02 < 1 ✓ possible.
- C. a + b < c: take a = 1, b = 2, c = 10 → 3 < 10 ✓ possible.
- D. a·c < b: take a = 0.1, b = 1, c = 2 → 0.2 < 1 ✓ possible.
- E. b/c = a: take a = 0.5, b = 1, c = 2 → 1/2 = 0.5 ✓ possible.

ANSWER 7: A

# Problem 8:
Pans needed: 216 ÷ 15 = 14.4, and only full recipes are made, so they must bake 15 pans.

Butter: 15 × 3 = 45 tablespoons.

Sticks: 45 ÷ 8 = 5.625, and butter comes in whole sticks, so 6 sticks.

(5 sticks = 40 tbsp, not enough.)

ANSWER 8: B

# Problem 9:
The 3×3×3 cube has surface area 6 × 9 = 54 unit squares.

Positions in a 3-cube: 1 center cube (0 faces showing), 6 face-centers (1 face each), 12 edges (2 faces), 8 corners (3 faces).

To minimize white showing, place one white cube in the hidden center (0 white squares) and the other 5 white cubes at face-center positions (1 white square each).

Minimum white area = 5 unit squares out of 54 → 5/54.

ANSWER 9: A

# Problem 10:
From 1998 to 2050 is 52 years — about two 25-year tripling periods (1998 → 2023 → 2048).

Population: 200 → 600 → 1800.

1800 is closest to 2000.

ANSWER 10: D

Problem 1:

We are comparing the place values of the digits 9 and 3 in

\[
74982.1035.
\]

- The digit \(9\) is in the hundreds place, so its place value is \(100\).
- The digit \(3\) is in the thousandths place, so its place value is

\[
0.001.
\]

Now find how many times as great \(100\) is as \(0.001\):

\[
\frac{100}{0.001}=100,000.
\]

So the hundreds place is \(100,\!000\) times the thousandths place.

- A, \(1,\!000\), is too small.
- B, \(10,\!000\), is too small.
- D and E are too large.

ANSWER 1: C

Problem 2:

The gas expands by \(4\) cubic centimeters for every \(3^\circ\) increase in temperature.

The temperature changes from \(32^\circ\) down to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

Since

\[
12^\circ \div 3^\circ=4,
\]

there are 4 groups of \(3^\circ\).

For each group, the volume decreases by \(4\) cubic centimeters:

\[
4\cdot 4=16.
\]

At \(32^\circ\), the volume is \(24\) cubic centimeters, so at \(20^\circ\) it was

\[
24-16=8.
\]

Choices B, C, and D do not subtract the full \(16\) cubic centimeters. Choice E would represent an increase rather than a decrease.

ANSWER 2: A

Problem 3:

If there are \(n\) teams and each pair of teams plays exactly once, then the number of games is

\[
\binom n2=\frac{n(n-1)}2.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

Checking: with 7 teams, the number of games is

\[
\frac{7\cdot6}{2}=21.
\]

Six teams would produce only \(15\) games, while 8 teams would produce \(28\) games.

ANSWER 3: B

Problem 4:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew \(20\%\), so Shea’s current height is

\[
1.20h=60.
\]

Thus,

\[
h=\frac{60}{1.2}=50.
\]

So both were originally 50 inches tall.

Shea’s growth was

\[
60-50=10
\]

inches.

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5
\]

inches.

Therefore, Ara is now

\[
50+5=55
\]

inches tall.

Choice A, 48, is less than Ara’s original height. Choices B, C, and D do not include the correct 5-inch growth.

ANSWER 4: E

Problem 5:

A standard die has faces numbered \(1,2,3,4,5,6\). The product of all six face numbers is

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6=720.
\]

If the bottom face has number \(k\), then the product of the five visible faces is

\[
\frac{720}{k}.
\]

Check each possible bottom face:

\[
\frac{720}{1}=720,\qquad
\frac{720}{2}=360,\qquad
\frac{720}{3}=240,
\]

\[
\frac{720}{4}=180,\qquad
\frac{720}{5}=144,\qquad
\frac{720}{6}=120.
\]

Every one of these numbers is divisible by \(6\).

Therefore, no matter which face is on the bottom, the visible-face product is divisible by 6. The probability is

\[
1.
\]

All of choices A through D are less than 1, but the event always happens.

ANSWER 5: E

Problem 6:

Tori answered correctly:

- Arithmetic:

\[
70\%\text{ of }10=0.7(10)=7
\]

- Algebra:

\[
40\%\text{ of }30=0.4(30)=12
\]

- Geometry:

\[
60\%\text{ of }35=0.6(35)=21
\]

So her total number correct was

\[
7+12+21=40.
\]

To pass, she needs \(60\%\) of 75 questions correct:

\[
0.60(75)=45.
\]

She needs

\[
45-40=5
\]

more correct answers.

Choice A is too small, while choices C, D, and E give more than the number needed to reach exactly 45 correct.

ANSWER 6: B

Problem 7:

We are given

\[
0<a<b<c.
\]

We must find the statement that is impossible.

Consider choice A:

\[
a+c<b.
\]

But \(c>b\), and also \(a>0\). Therefore,

\[
a+c>c>b.
\]

So it is impossible for \(a+c<b\).

The other choices can happen for suitable positive numbers. For example:

- B: \(a\cdot b<c\) can occur with \(a=1,\ b=2,\ c=3\), since \(1\cdot2<3\).
- C: \(a+b<c\) can occur with \(a=1,\ b=2,\ c=4\).
- D: \(a\cdot c<b\) can occur with \(a=\frac14,\ b=1,\ c=2\), since \(\frac14\cdot2=\frac12<1\).
- E: \(bc=a\) can occur with \(b=\frac12,\ c=\frac34\), giving \(a=\frac38\), and indeed

\[
0<\frac38<\frac12<\frac34.
\]

Thus only choice A is impossible.

ANSWER 7: A

Problem 8:

Each full recipe makes 15 cookies.

Walter and Gretel need at least 216 cookies. Since only full recipes may be made, determine how many recipes are needed:

\[
216\div 15=14.4.
\]

They cannot make \(14.4\) recipes, so they must make 15 full recipes.

These 15 recipes make

\[
15\cdot15=225
\]

cookies, enough for the crowd.

Each recipe uses 3 tablespoons of butter, so the total butter needed is

\[
15\cdot3=45
\]

tablespoons.

There are 8 tablespoons in each stick, so

\[
45\div8=5.625.
\]

They must buy 6 full sticks of butter.

Five sticks provide only \(5\cdot8=40\) tablespoons, which is not enough. Six sticks provide \(48\) tablespoons, enough for the 45 tablespoons needed.

ANSWER 8: B

Problem 9:

A \(3\times3\times3\) cube has surface area

\[
6\cdot 3\cdot3=54
\]

square inches, since each of its 6 faces has area \(3\cdot3=9\).

To minimize the white surface showing, place the white cubes in the least exposed positions.

- The one center cube of the large cube has no faces exposed.
- The six face-center cubes each have exactly one face exposed.
- Edge cubes have two exposed faces.
- Corner cubes have three exposed faces.

There are 6 white cubes. Put one white cube in the completely hidden center. Then put the other 5 white cubes at five face-center locations.

Thus, the white cubes show only

\[
5
\]

unit square faces on the surface.

Therefore, the fraction of the surface that is white is

\[
\frac{5}{54}.
\]

Choices such as \(\frac19,\frac{5}{27},\frac29,\frac13\) represent more than 5 white unit squares out of the 54 total surface squares.

ANSWER 9: A

Problem 10:

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

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is approximately two 25-year periods.

After one 25-year period:

\[
200\cdot3=600.
\]

After two 25-year periods:

\[
600\cdot3=1800.
\]

Since 2050 is slightly more than 50 years after 1998, the population should be slightly more than 1800. The closest estimate is

\[
20

Problem 1:
We are asked: in 74982.1035, compare the place value of the 9 with the place value of the 3.

Write out places:
74982 is: 7 ten-thousands, 4 thousands, 9 hundreds, 8 tens, 2 ones.
So the 9 is in the hundreds place. Its value is 100.

After the decimal .1035 is: 1 tenths, 0 hundredths, 3 thousandths, 5 ten-thousandths.
So the 3 is in the thousandths place. Its value is 0.001 = 1/1000.

Ratio = (value of 9's place) / (value of 3's place)
= 100 / 0.001 = 100 / (1/1000) = 100 x 1000 = 100,000.

So the hundreds place is 100,000 times as great as the thousandths place.
Choices A,B,D,E are powers of 10 that are too small or too large.

ANSWER 1: C

Problem 2:
We are asked for the earlier volume given expansion rate.

Rate: +3 degrees temperature = +4 cubic cm volume.

Now: temp 32°, volume 24 cc.
Before: temp 20°.

Temperature change from 20° to 32° is 32-20 = 12°.
Number of 3-degree steps in 12° is 12 / 3 = 4 steps.

So volume increase from 20° to 32° is 4 steps x 4 cc = 16 cc.

Thus volume at 20° + 16 = volume at 32° = 24.
So volume at 20° = 24 - 16 = 8 cc.

Check: A 8 fits, B 12 would be only 3 steps, D 16 would be only 2 steps, E 40 would be adding instead of subtracting.

ANSWER 2: A

Problem 3:
We need n teams such that every pair plays once gives 21 games.

Number of pairs from n teams is n choose 2 = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try integers: 6x5=30 too small, 7x6=42 correct, 8x7=56 too big.
So n = 7.

Ruling out: A 6 gives only 15 games, C 8 gives 28 games, etc.

ANSWER 3: B

Problem 4:
Ara and Shea started at same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara's now height.

Let original height = H.
Shea's growth = 20% of H = 0.20 H.
Shea now = H + 0.20H = 1.2H = 60.
So H = 60 / 1.2 = 50 inches.

Shea grew 60-50 = 10 inches.
Ara grew half as many inches: 10 / 2 = 5 inches.
Ara now = 50 + 5 = 55 inches.

Tempting wrong A 48 would be working backwards, D 54 would be misreading percent, etc.

ANSWER 4: E

Problem 5:
A standard 1-6 die is tossed. Bottom face hidden, other 5 faces seen. Probability product of seen numbers divisible by 6?

Product divisible by 6 means divisible by 2 and by 3.

Numbers 1 to 6: evens are 2,4,6 — three of them. Multiples of 3 are 3,6 — two of them.

Hiding only one face (the bottom) can hide at most one even, leaving at least two evens still visible, so factor 2 remains. It can hide at most one multiple of 3, leaving at least one multiple of 3 still visible, so factor 3 remains.

Indeed total product 1·2·3·4·5·6 = 720. Visible product = 720 / b where b is bottom. 720/b divided by 6 = 120/b, and b=1,2,3,4,5,6 all divide 120, so always integer.

Hence no matter which face is hidden, the visible product is divisible by 6. Probability = 6/6 = 1.

Other choices 1/3,1/2,2/3,5/6 assume sometimes it fails, which is impossible.

ANSWER 5: E

Problem 6:
75 problems total: 10 arithmetic, 30 algebra, 35 geometry.
Correct: 70% of arithmetic, 40% of algebra, 60% of geometry. How many more to reach 60% overall?

Compute:
Arithmetic correct = 0.70 x 10 = 7.
Algebra correct = 0.40 x 30 = 12.
Geometry correct = 0.60 x 35 = 21.
Total correct = 7+12+21 = 40.

Need for 60% passing: 0.60 x 75 = 45 problems.

Shortfall = 45 - 40 = 5 more problems.

So B 5, not A 1 or C 7.

ANSWER 6: B

Problem 7:
Given 0 < a < b < c, which statement is impossible?

A. a + c < b. Since a > 0, a + c > c. And c > b. So a+c > b always. Hence a+c < b can never happen. This is impossible.

Check others are possible with examples:
B. a·b < c: e.g. a=1,b=2,c=10 gives 2<10 possible.
C. a+b < c: e.g. a=1,b=2,c=10 gives 3<10 possible.
D. a·c < b: with fractions <1. e.g. a=0.1,c=2,b=1 gives 0.2<1 and 0.1<1<2 possible.
E. b c = a: let b=0.25,a=0.1, then c=a/b=0.4, and 0.1<0.25<0.4 possible.

So only A is impossible. Students may think E impossible because b c > b > a for numbers >1, but with numbers <1 multiplication can shrink.

ANSWER 7: A

Problem 8:
One full recipe = 15 cookies = 3 tbsp butter. Need 216 cookies. 8 tbsp = 1 stick. How many sticks?

Pans needed: 216 / 15 = 14.4. Since only full recipes, round up to 15 pans. 14 pans would give only 210 cookies, not enough.

Butter needed: 15 x 3 tbsp = 45 tbsp.

Sticks needed: 45 / 8 = 5.625. Since need whole sticks to have enough, round up to 6. 5 sticks = 40 tbsp < 45 tbsp, not enough.

So answer 6. A 5 would leave butter and cookies short.

ANSWER 8: B

Problem 9:
Build 3x3x3 = 27 small cubes. 6 white, 21 red. Minimize white squares on outside. What fraction of outside is white?

Big cube surface: 6 faces each 3x3 = 9 small squares, total 54 small squares.

Small-cube positions by exposure:
- 1 interior center: 0 faces shown
- 6 face-centers: 1 face shown
- 12 edge non-corner: 2 faces shown
- 8 corners: 3 faces shown

To minimize white, put whites where least exposure:
Put 1 white in center → 0 exposed.
Put remaining 5 whites in face-center spots → 1 each exposed.
Total white exposed squares = 5.

Fraction white = 5 / 54 of total surface area.

Larger fractions like 1/9=6/54, etc. would result from worse placement.

ANSWER 9: A

Problem 10:
1998 population 200, triples every 25 years. Estimate 2050.

Time elapsed: 2050-1998 = 52 years.
Number of 25-year periods: 52/25 ≈ 2.08, just over 2 doublings-triplings.

After 25 years (~2023): 200x3=600.
After 50 years (~2048): 600x3=1800.
2050 is 2 years later, a little more, so about 2000.

Formally 200 x 3^(52/25) ≈ 200 x 9 x small extra ≈ 1960.

So A 600 is only one tripling, C 1000 too low, E 3000 too high. Best is D 2000.

ANSWER 10: D

Problem 1:
We are asked to find the ratio of the value of the place occupied by the digit 9 to the value of the place occupied by the digit 3 in the decimal number $74982.1035$.

1. Identify the place values:
   - The number is $74982.1035$.
   - The digit 2 is in the ones place ($10^0$).
   - The digit 8 is in the tens place ($10^1$).
   - The digit 9 is in the hundreds place ($10^2 = 100$).
   - The digit 1 is in the tenths place ($10^{-1} = 0.1$).
   - The digit 0 is in the hundredths place ($10^{-2} = 0.01$).
   - The digit 3 is in the thousandths place ($10^{-3} = 0.001$).
   - The digit 5 is in the ten-thousandths place ($10^{-4} = 0.0001$).

2. Compute the ratio of the place values:
   $$\frac{\text{Place value of } 9}{\text{Place value of } 3} = \frac{100}{0.001} = \frac{10^2}{10^{-3}} = 10^{2 - (-3)} = 10^5 = 100,000.$$

This corresponds to choice C.

ANSWER 1: C

---

Problem 2:
We are asked to find the initial volume of gas at $20^\circ$ given its volume at $32^\circ$ and its rate of expansion.

1. Find the change in temperature:
   The temperature changes from $20^\circ$ to $32^\circ$, which is an increase of:
   $$32^\circ - 20^\circ = 12^\circ.$$

2. Determine the volume expansion:
   For every $3^\circ$ rise, the volume expands by $4\text{ cm}^3$.
   In a $12^\circ$ rise, the number of $3^\circ$ intervals is:
   $$\frac{12^\circ}{3^\circ} = 4.$$
   Therefore, the volume expanded by:
   $$4 \times 4\text{ cm}^3 = 16\text{ cm}^3.$$

3. Calculate the original volume at $20^\circ$:
   $$\text{Volume at } 20^\circ = \text{Volume at } 32^\circ - 16\text{ cm}^3 = 24 - 16 = 8\text{ cm}^3.$$

This matches choice A.

ANSWER 2: A

---

Problem 3:
We are given that each of the $N$ teams plays every other team exactly once, resulting in 21 total games. We need to find $N$.

1. The number of games played between $N$ teams is the number of ways to choose 2 teams from $N$, given by the combination formula:
   $$\binom{N}{2} = \frac{N(N-1)}{2}.$$

2. Set this equal to 21 and solve for $N$:
   $$\frac{N(N-1)}{2} = 21 \implies N(N-1) = 42.$$

3. Find two consecutive positive integers whose product is 42:
   Since $7 \times 6 = 42$, we have $N = 7$.

Checking the choices, $N = 7$ corresponds to choice B.

ANSWER 3: B

---

Problem 4:
We are given:
- Ara and Shea initially had the same height, say $h$ inches.
- Shea has grown $20\%$, so Shea's new height is $h + 0.20h = 1.20h = 60$ inches.
- Ara has grown half as many inches as Shea.
- We need to find Ara's current height.

1. Find the original height $h$:
   $$1.20h = 60 \implies h = \frac{60}{1.2} = 50\text{ inches}.$$

2. Find how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10\text{ inches}.$$

3. Find Ara's growth:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5\text{ inches}.$$

4. Find Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55\text{ inches}.$$

This corresponds to choice E.

ANSWER 4: E

---

Problem 5:
A standard fair six-sided die has faces with numbers 1, 2, 3, 4, 5, and 6.
When the die is tossed, one face is on the bottom and cannot be seen, while the other 5 faces are visible.
We want the product of the 5 visible faces to be divisible by 6.

1. The product of all six numbers on the die is:
   $$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720.$$
   The product of the 5 visible faces is:
   $$P = \frac{720}{b},$$
   where $b \in \{1, 2, 3, 4, 5, 6\}$ is the number on the bottom face.

2. A number is divisible by 6 if and only if it is a multiple of both 2 and 3.
   Let's check the product $P$ for each possible bottom face $b$:
   - If $b = 1$, visible faces are $\{2, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 2$, visible faces are $\{1, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 3$, visible faces are $\{1, 2, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 4$, visible faces are $\{1, 2, 3, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 5$, visible faces are $\{1, 2, 3, 4, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 6$, visible faces are $\{1, 2, 3, 4, 5\}$. The product is $1 \times 2 \times 3 \times 4 \times 5 = 120$. Since $120 = 6 \times 20$, 120 is also divisible by 6 (it has factors 2 and 3)!

3. In every single case (all 6 possible outcomes for the bottom face), the product of the five visible faces is divisible by 6.
   Therefore, the probability is:
   $$\frac{6}{6} = 1.$$

This corresponds to choice E.

ANSWER 5: E

---

Problem 6:
Tori took a 75-problem test:
- 10 arithmetic problems: she answered $70\%$ correctly.
  $$\text{Correct arithmetic} = 0.70 \times 10 = 7.$$
- 30 algebra problems: she answered $40\%$ correctly.
  $$\text{Correct algebra} = 0.40 \times 30 = 12.$$
- 35 geometry problems: she answered $60\%$ correctly.
  $$\text{Correct geometry} = 0.60 \times 35 = 21.$$

1. Total problems Tori answered correctly:
   $$\text{Total correct} = 7 + 12 + 21 = 40.$$

2. Passing score required:
   $60\%$ of 75 problems:
   $$\text{Passing score} = 0.60 \times 75 = 45.$$

3. Additional problems needed:
   $$45 - 40 = 5\text{ problems}.$$

This corresponds to choice B.

ANSWER 6: B

---

Problem 7:
We are given real numbers $a, b, c$ such that $0 < a < b < c$.
We need to determine which of the statements is impossible.

Let's test each choice:
- Choice A: $a + c < b$.
  Since $a > 0$, adding $a$ to $c$ gives $a + c > c$.
  Since we are given $c > b$, it follows that $a + c > c > b$, which means $a + c > b$ must always be true!
  Therefore, $a + c < b$ is impossible.

Let's verify that the other choices can be possible with appropriate values:
- Choice B: $a \cdot b < c$. If $a = 1, b = 2, c = 3$, then $a \cdot b = 2 < 3 = c$. (Possible)
- Choice C: $a + b < c$. If $a = 1, b = 2, c = 4$, then $a + b = 3 < 4 = c$. (Possible)
- Choice D: $a \cdot c < b$. If $a = 0.1, b = 2, c = 3$, then $a \cdot c = 0.3 < 2 = b$. (Possible)
-

# Problem 1:

**Restate:** In 74982.1035, compare the place value of the digit 9 with the place value of the digit 3.

**Working:** Let's identify each place:
- The digit 9 sits in the **hundreds** place: 9 × 100 = 900. Its place value is **100**.
- The digit 3 sits in the **hundredths** place (two positions right of the decimal point): 3 × 0.01 = 0.03. Its place value is **0.01**.

**Compute the ratio:**
$$\frac{100}{0.01} = 100 \div \frac{1}{100} = 100 \times 100 = 10{,}000$$

**Check wrong choices:** Counting places instead of computing the ratio might suggest 100,000 (if you miscount the decimal places), but from hundreds to hundredths is exactly 4 positions, and each position is a factor of 10, giving 10⁴ = 10,000.

ANSWER 1: B

# Problem 2:

**Restate:** Volume increases 4 cm³ per 3° rise. Volume is 24 cm³ at 32°. Find the volume at 20°.

**Working:** The temperature dropped from 32° to 20°:
$$32 - 20 = 12°$$
Number of 3° intervals in the drop:
$$12 \div 3 = 4 \text{ intervals}$$
Each 3° drop means the gas **shrinks** by 4 cm³, so the total shrink is:
$$4 \times 4 = 16 \text{ cm}^3$$
The volume at 20° was:
$$24 - 16 = 8 \text{ cm}^3$$

**Check wrong choices:** Choice D (16) is the *amount* of shrink, not the volume — a tempting trap. Choice E (40) comes from adding instead of subtracting.

ANSWER 2: A

# Problem 3:

**Restate:** Each team plays every other team once; 21 total games. Find the number of teams.

**Working:** With n teams, the number of games is:
$$\frac{n(n-1)}{2}$$
Set this equal to 21:
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42$$
Testing n = 7: $7 \times 6 = 42$. ✓

**Check wrong choices:** n = 6 gives 15 games (too few); n = 8 gives 28 games (too many). Only n = 7 works.

ANSWER 3: B

# Problem 4:

**Restate:** Ara and Shea were once equal in height. Shea grew 20% and is now 60 inches. Ara grew half as many inches as Shea. Find Ara's height now.

**Working:** Let the original height be $h$. Shea grew 20%, so:
$$h + 0.20h = 60 \implies 1.2h = 60 \implies h = 50 \text{ inches}$$
Shea's growth: $60 - 50 = 10$ inches.
Ara grew half as many inches as Shea:
$$\frac{10}{2} = 5 \text{ inches}$$
Ara's height now:
$$50 + 5 = 55 \text{ inches}$$

**Check wrong choices:** Choice A (48) comes from computing 20% of 60 incorrectly; choice D (54) would result if Ara also grew 20% of the original height. The key is that Ara grew 5 **inches**, not a percentage.

ANSWER 4: E

# Problem 5:

**Restate:** A die is tossed; the bottom face is hidden. Find the probability that the product of the five visible faces is divisible by 6.

**Working:** The product of all six faces is:
$$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$$
The product of the five visible faces is:
$$\frac{720}{\text{hidden face}}$$

Check each possible hidden face:
- Hidden = 1: product = 720 → divisible by 6 ✓
- Hidden = 2: product = 360 → divisible by 6 ✓
- Hidden = 3: product = 240 → divisible by 6 ✓
- Hidden = 4: product = 180 → divisible by 6 ✓
- Hidden = 5: product = 144 → divisible by 6 ✓
- Hidden = 6: product = 120 → divisible by 6 ✓

All six outcomes give a product divisible by 6, so the probability is:
$$\frac{6}{6} = 1$$

**Check wrong choices:** One might think hiding the 5 or the 6 removes needed factors, but 720 contains 2 × 3 many times over — removing any single face still leaves factors of 2 and 3 in the product.

ANSWER 5: E

# Problem 6:

**Restate:** Tori's test: 10 arithmetic, 30 algebra, 35 geometry problems (75 total). She got 70%, 40%, and 60% correct respectively, which was below 60% overall. How many more correct answers were needed for 60%?

**Working:** Correct answers in each category:
- Arithmetic: $0.70 \times 10 = 7$
- Algebra: $0.40 \times 30 = 12$
- Geometry: $0.60 \times 35 = 21$

Total correct: $7 + 12 + 21 = 40$

To pass, she needs 60% of 75:
$$0.60 \times 75 = 45$$

Additional problems needed:
$$45 - 40 = 5$$

**Check wrong choices:** Note that the weighted average $\frac{40}{75} \approx 53.3\%$ is indeed below 60%, consistent with the story. A tempting trap is to average the three percentages: $\frac{70+40+60}{3} \approx 56.7\%$, which is wrong because the categories have different sizes.

ANSWER 6: B

# Problem 7:

**Restate:** Given $0 < a < b < c$, which statement is impossible?

**Working:** Test each option:

**A. $a + c < b$:** Since $c > b$, adding positive $a$ gives $a + c > c > b$. So $a + c < b$ **cannot happen**. ✓ (impossible)

Let's verify the others are all possible with suitable numbers:

**B. $a \cdot b < c$:** Try $a = 0.1, b = 0.2, c = 1$: $0.02 < 1$ ✓ possible.

**C. $a + b < c$:** Try $a = 1, b = 2, c = 10$: $3 < 10$ ✓ possible.

**D. $a \cdot c < b$:** Try $a = 0.1, b = 0.5, c = 0.6$: $0.06 < 0.5$ ✓ possible.

**E. $bc = a$:** Try $a = 0.06, b = 0.3, c = 0.2$... wait, need $b < c$. Try $a = 0.02, b = 0.1, c = 0.2$: $bc = 0.02 = a$, and $0 < 0.02 < 0.1 < 0.2$ ✓ possible.

Only option A forces a contradiction: $a + c > c > b$ always holds.

ANSWER 7: A

# Problem 8:

**Restate:** One recipe = 15 cookies using 3 tablespoons of butter; only full recipes allowed. Need 216 cookies. 8 tablespoons = 1 stick. How many sticks?

**Working:** Number of recipes needed:
$$\frac{216}{15} = 14.4$$
Since only full recipes are made, they must make **15 recipes** (14 recipes × 15 = 210 cookies, which is not enough).

Butter needed:
$$15 \times 3 = 45 \text{ tablespoons}$$
Sticks needed:
$$\frac{45}{8} = 5.625 \implies \text{need } 6 \text{ full sticks}$$

**Check wrong choices:** Choice A (5) is the trap — it uses only 40 tablespoons, but 45 are required. You can't buy a partial stick, so round up to 6.

ANSWER 8: B

# Problem 9:

**Restate:** Build a 3-inch cube from 27 unit cubes: 21 red, 6 white. Arrange to minimize white surface area. What fraction of the surface area is white?

**Working:** The 3×3×3 cube has total surface area:
$$6 \text{ faces} \times 9 \text{ sq in} = 54 \text{ square inches}$$

Classify the 27 positions by how many faces each shows:
- **1 inte

Problem 1:
We need the ratio of the place values of the digits 9 and 3 in  
\(74982.1035\).

The digit 9 is in the hundreds place, so its place value is  
\[
100.
\]

The digit 3 is in the thousandths place, so its place value is  
\[
0.001.
\]

Now compute the ratio:
\[
\frac{100}{0.001}=100{,}000.
\]

So the value of the place of the 9 is \(100{,}000\) times as great as the value of the place of the 3.

ANSWER 1: C

Problem 2:
We are told that for every \(3^\circ\) rise in temperature, the gas expands by 4 cubic centimeters.

At temperature \(32^\circ\), the volume is 24 cubic centimeters. We want the volume at \(20^\circ\).

The temperature difference is
\[
32-20=12^\circ.
\]

From \(20^\circ\) to \(32^\circ\), the temperature rises by \(12^\circ\). Since each \(3^\circ\) rise increases the volume by 4 cubic centimeters,
\[
\frac{12}{3}\cdot 4=4\cdot 4=16
\]
cubic centimeters.

So at \(20^\circ\), the volume was
\[
24-16=8
\]
cubic centimeters.

ANSWER 2: A

Problem 3:
Each team plays every other team exactly once. If there are \(n\) teams, the number of games is
\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games, so
\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:
\[
n(n-1)=42.
\]

We look for two consecutive positive integers whose product is 42:
\[
6\cdot 7=42.
\]

Thus
\[
n=7.
\]

So there were 7 teams.

ANSWER 3: B

Problem 4:
Ara and Shea were originally the same height. Let that original height be \(h\) inches.

Shea has grown 20%, so Shea’s current height is
\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:
\[
1.20h=60.
\]

Solve for \(h\):
\[
h=\frac{60}{1.20}=50.
\]

So they were originally 50 inches tall. Shea grew
\[
60-50=10
\]
inches.

Ara grew half as many inches as Shea:
\[
\frac{10}{2}=5
\]
inches.

Therefore Ara is now
\[
50+5=55
\]
inches tall.

ANSWER 4: E

Problem 5:
A fair six-sided die has faces numbered 1 through 6. When it is tossed, one face is hidden on the bottom, and the other five faces are visible.

We need the probability that the product of the five visible numbers is divisible by 6.

A number is divisible by 6 if it has at least one factor of 2 and at least one factor of 3.

No matter which single face is hidden:

- There are three even faces: 2, 4, and 6. Hiding only one of them still leaves at least one even face visible.
- There are two multiples of 3: 3 and 6. Hiding only one of them still leaves at least one multiple of 3 visible.

So the product of the visible faces always contains both a factor of 2 and a factor of 3. Therefore it is always divisible by 6.

Thus the probability is
\[
1.
\]

ANSWER 5: E

Problem 6:
Tori answered:

- \(70\%\) of 10 arithmetic problems:
\[
0.70\cdot 10=7
\]
- \(40\%\) of 30 algebra problems:
\[
0.40\cdot 30=12
\]
- \(60\%\) of 35 geometry problems:
\[
0.60\cdot 35=21
\]

Total correct:
\[
7+12+21=40.
\]

There were 75 problems total. To pass, she needed \(60\%\) correct:
\[
0.60\cdot 75=45.
\]

She answered 40 correctly, so she needed
\[
45-40=5
\]
more correct answers.

ANSWER 6: B

Problem 7:
We are given
\[
0<a<b<c.
\]

We need the impossible statement.

Choice A says
\[
a+c<b.
\]

But since \(c>b\) and \(a>0\),
\[
a+c>b.
\]

So \(a+c<b\) cannot happen.

We can also check that the other choices are possible:

- B: \(a\cdot b<c\) can happen, for example \(a=1,b=2,c=100\).
- C: \(a+b<c\) can happen, for example \(a=1,b=2,c=100\).
- D: \(a\cdot c<b\) can happen, for example \(a=0.1,b=2,c=3\), since \(0.3<2\).
- E: If interpreted as \(bc=a\), it can happen, for example \(a=0.1,b=0.2,c=0.5\), since \(0.2\cdot 0.5=0.1\).

Thus the impossible choice is A.

ANSWER 7: A

Problem 8:
One recipe makes 15 cookies and uses 3 tablespoons of butter. Only full recipes can be made.

Walter and Gretel need 216 cookies. The number of full recipes needed is
\[
\left\lceil \frac{216}{15}\right\rceil.
\]

Since
\[
\frac{216}{15}=14.4,
\]
they must make 15 full recipes.

Butter needed:
\[
15\cdot 3=45
\]
tablespoons.

There are 8 tablespoons in a stick of butter, so the number of sticks needed is
\[
\left\lceil \frac{45}{8}\right\rceil.
\]

Since
\[
\frac{45}{8}=5.625,
\]
they need 6 sticks of butter.

ANSWER 8: B

Problem 9:
A 3-inch cube made from 27 one-inch cubes has surface area
\[
6\cdot 3^2=54
\]
square inches.

Each small cube can show:

- 0 faces if it is the center cube,
- 1 face if it is a face-center cube,
- 2 faces if it is an edge cube,
- 3 faces if it is a corner cube.

We have 6 white cubes. To minimize the white surface area, we should hide as many white cubes as possible.

There is exactly one interior cube, so put one white cube there. It shows 0 square inches.

The remaining 5 white cubes must be on the surface. To minimize exposed white area, put them on face-center positions, where each shows exactly 1 square inch.

So the minimum white surface area is
\[
1\cdot 0+5\cdot 1=5
\]
square inches.

The fraction of the surface area that is white is
\[
\frac{5}{54}.
\]

ANSWER 9: A

Problem 10:
The population is 200 in 1998, and it triples every 25 years.

From 1998 to 2050 is
\[
2050-1998=52
\]
years.

Two full 25-year periods are 50 years, so after 50 years the population is
\[
200\cdot 3\cdot 3=200\cdot 9=1800.
\]

The extra 2 years would add a relatively small amount, so the population in 2050 is a little above 1800. Among the choices, the best estimate is 2000.

ANSWER 10: D

# Problem 1:
Write out the place values of 74982.1035:
- 7 → ten-thousands, 4 → thousands, **9 → hundreds**, 8 → tens, 2 → ones
- 1 → tenths, 0 → hundredths, **3 → thousandths**, 5 → ten-thousandths

So the 9 sits in the hundreds place (value 100) and the 3 sits in the thousandths place (value 0.001).

Ratio = 100 ÷ 0.001 = 100,000.

(Answers like 1,000 or 10,000 come from miscounting decimal places — the 3 is the *third* digit after the point, not the first or second.)

ANSWER 1: C

# Problem 2:
Temperature goes from 32° down to 20°, a drop of 12°.

Each 3° change corresponds to 4 cm³ of change: 12 ÷ 3 = 4 steps.

Volume change = 4 × 4 = 16 cm³ decrease (cooling means shrinking).

Volume = 24 − 16 = 8 cm³.

(40 is the trap from *adding* instead of subtracting; 16 is just the change, not the volume.)

ANSWER 2: A

# Problem 3:
With n teams each playing every other team once, the number of games is C(n,2) = n(n−1)/2.

Set n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7·6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 games — neither is 21.

ANSWER 3: B

# Problem 4:
Let the common original height be x. Shea grew 20%: 1.20x = 60 → x = 50.

So Shea grew 60 − 50 = 10 inches.

Ara grew half as many inches: 5 inches.

Ara's height = 50 + 5 = 55 inches.

(54 would result from wrongly taking Shea's growth as 10% of 60, etc.)

ANSWER 4: E

# Problem 5:
The product of all six faces is 1·2·3·4·5·6 = 720. If the hidden bottom face is k, the visible product is 720/k.

- k=1: 720 ✓ divisible by 6
- k=2: 360 ✓
- k=3: 240 ✓
- k=4: 180 ✓
- k=5: 144 ✓
- k=6: 120 ✓

Reason why: hiding one face still leaves at least one even number and at least one multiple of 3 among the five visible faces (e.g., hide 6 → 2 and 3 remain; hide 3 → 6 remains).

So the probability is 6/6 = 1.

ANSWER 5: E

# Problem 6:
Correct answers:
- Arithmetic: 70% of 10 = 7
- Algebra: 40% of 30 = 12
- Geometry: 60% of 35 = 21

Total correct = 7 + 12 + 21 = 40.

Needed for 60%: 0.60 × 75 = 45.

She needs 45 − 40 = 5 more.

(40/75 ≈ 53%, consistent with failing.)

ANSWER 6: B

# Problem 7:
Given 0 < a < b < c.

- **A. a + c < b:** Since c > b and a > 0, we get a + c > b. So a + c < b is impossible.
- B. a·b < c: take a = 0.1, b = 0.2, c = 1 → 0.02 < 1 ✓ possible.
- C. a + b < c: take a = 1, b = 2, c = 10 → 3 < 10 ✓ possible.
- D. a·c < b: take a = 0.1, b = 1, c = 2 → 0.2 < 1 ✓ possible.
- E. b/c = a: take a = 0.5, b = 1, c = 2 → 1/2 = 0.5 ✓ possible.

ANSWER 7: A

# Problem 8:
Pans needed: 216 ÷ 15 = 14.4, and only full recipes are made, so they must bake 15 pans.

Butter: 15 × 3 = 45 tablespoons.

Sticks: 45 ÷ 8 = 5.625, and butter comes in whole sticks, so 6 sticks.

(5 sticks = 40 tbsp, not enough.)

ANSWER 8: B

# Problem 9:
The 3×3×3 cube has surface area 6 × 9 = 54 unit squares.

Positions in a 3-cube: 1 center cube (0 faces showing), 6 face-centers (1 face each), 12 edges (2 faces), 8 corners (3 faces).

To minimize white showing, place one white cube in the hidden center (0 white squares) and the other 5 white cubes at face-center positions (1 white square each).

Minimum white area = 5 unit squares out of 54 → 5/54.

ANSWER 9: A

# Problem 10:
From 1998 to 2050 is 52 years — about two 25-year tripling periods (1998 → 2023 → 2048).

Population: 200 → 600 → 1800.

1800 is closest to 2000.

ANSWER 10: D

Problem 1:

We are comparing the place values of the digits 9 and 3 in

\[
74982.1035.
\]

- The digit \(9\) is in the hundreds place, so its place value is \(100\).
- The digit \(3\) is in the thousandths place, so its place value is

\[
0.001.
\]

Now find how many times as great \(100\) is as \(0.001\):

\[
\frac{100}{0.001}=100,000.
\]

So the hundreds place is \(100,\!000\) times the thousandths place.

- A, \(1,\!000\), is too small.
- B, \(10,\!000\), is too small.
- D and E are too large.

ANSWER 1: C

Problem 2:

The gas expands by \(4\) cubic centimeters for every \(3^\circ\) increase in temperature.

The temperature changes from \(32^\circ\) down to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

Since

\[
12^\circ \div 3^\circ=4,
\]

there are 4 groups of \(3^\circ\).

For each group, the volume decreases by \(4\) cubic centimeters:

\[
4\cdot 4=16.
\]

At \(32^\circ\), the volume is \(24\) cubic centimeters, so at \(20^\circ\) it was

\[
24-16=8.
\]

Choices B, C, and D do not subtract the full \(16\) cubic centimeters. Choice E would represent an increase rather than a decrease.

ANSWER 2: A

Problem 3:

If there are \(n\) teams and each pair of teams plays exactly once, then the number of games is

\[
\binom n2=\frac{n(n-1)}2.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

Checking: with 7 teams, the number of games is

\[
\frac{7\cdot6}{2}=21.
\]

Six teams would produce only \(15\) games, while 8 teams would produce \(28\) games.

ANSWER 3: B

Problem 4:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew \(20\%\), so Shea’s current height is

\[
1.20h=60.
\]

Thus,

\[
h=\frac{60}{1.2}=50.
\]

So both were originally 50 inches tall.

Shea’s growth was

\[
60-50=10
\]

inches.

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5
\]

inches.

Therefore, Ara is now

\[
50+5=55
\]

inches tall.

Choice A, 48, is less than Ara’s original height. Choices B, C, and D do not include the correct 5-inch growth.

ANSWER 4: E

Problem 5:

A standard die has faces numbered \(1,2,3,4,5,6\). The product of all six face numbers is

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6=720.
\]

If the bottom face has number \(k\), then the product of the five visible faces is

\[
\frac{720}{k}.
\]

Check each possible bottom face:

\[
\frac{720}{1}=720,\qquad
\frac{720}{2}=360,\qquad
\frac{720}{3}=240,
\]

\[
\frac{720}{4}=180,\qquad
\frac{720}{5}=144,\qquad
\frac{720}{6}=120.
\]

Every one of these numbers is divisible by \(6\).

Therefore, no matter which face is on the bottom, the visible-face product is divisible by 6. The probability is

\[
1.
\]

All of choices A through D are less than 1, but the event always happens.

ANSWER 5: E

Problem 6:

Tori answered correctly:

- Arithmetic:

\[
70\%\text{ of }10=0.7(10)=7
\]

- Algebra:

\[
40\%\text{ of }30=0.4(30)=12
\]

- Geometry:

\[
60\%\text{ of }35=0.6(35)=21
\]

So her total number correct was

\[
7+12+21=40.
\]

To pass, she needs \(60\%\) of 75 questions correct:

\[
0.60(75)=45.
\]

She needs

\[
45-40=5
\]

more correct answers.

Choice A is too small, while choices C, D, and E give more than the number needed to reach exactly 45 correct.

ANSWER 6: B

Problem 7:

We are given

\[
0<a<b<c.
\]

We must find the statement that is impossible.

Consider choice A:

\[
a+c<b.
\]

But \(c>b\), and also \(a>0\). Therefore,

\[
a+c>c>b.
\]

So it is impossible for \(a+c<b\).

The other choices can happen for suitable positive numbers. For example:

- B: \(a\cdot b<c\) can occur with \(a=1,\ b=2,\ c=3\), since \(1\cdot2<3\).
- C: \(a+b<c\) can occur with \(a=1,\ b=2,\ c=4\).
- D: \(a\cdot c<b\) can occur with \(a=\frac14,\ b=1,\ c=2\), since \(\frac14\cdot2=\frac12<1\).
- E: \(bc=a\) can occur with \(b=\frac12,\ c=\frac34\), giving \(a=\frac38\), and indeed

\[
0<\frac38<\frac12<\frac34.
\]

Thus only choice A is impossible.

ANSWER 7: A

Problem 8:

Each full recipe makes 15 cookies.

Walter and Gretel need at least 216 cookies. Since only full recipes may be made, determine how many recipes are needed:

\[
216\div 15=14.4.
\]

They cannot make \(14.4\) recipes, so they must make 15 full recipes.

These 15 recipes make

\[
15\cdot15=225
\]

cookies, enough for the crowd.

Each recipe uses 3 tablespoons of butter, so the total butter needed is

\[
15\cdot3=45
\]

tablespoons.

There are 8 tablespoons in each stick, so

\[
45\div8=5.625.
\]

They must buy 6 full sticks of butter.

Five sticks provide only \(5\cdot8=40\) tablespoons, which is not enough. Six sticks provide \(48\) tablespoons, enough for the 45 tablespoons needed.

ANSWER 8: B

Problem 9:

A \(3\times3\times3\) cube has surface area

\[
6\cdot 3\cdot3=54
\]

square inches, since each of its 6 faces has area \(3\cdot3=9\).

To minimize the white surface showing, place the white cubes in the least exposed positions.

- The one center cube of the large cube has no faces exposed.
- The six face-center cubes each have exactly one face exposed.
- Edge cubes have two exposed faces.
- Corner cubes have three exposed faces.

There are 6 white cubes. Put one white cube in the completely hidden center. Then put the other 5 white cubes at five face-center locations.

Thus, the white cubes show only

\[
5
\]

unit square faces on the surface.

Therefore, the fraction of the surface that is white is

\[
\frac{5}{54}.
\]

Choices such as \(\frac19,\frac{5}{27},\frac29,\frac13\) represent more than 5 white unit squares out of the 54 total surface squares.

ANSWER 9: A

Problem 10:

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

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is approximately two 25-year periods.

After one 25-year period:

\[
200\cdot3=600.
\]

After two 25-year periods:

\[
600\cdot3=1800.
\]

Since 2050 is slightly more than 50 years after 1998, the population should be slightly more than 1800. The closest estimate is

\[
20

Problem 1:
We are asked: in 74982.1035, compare the place value of the 9 with the place value of the 3.

Write out places:
74982 is: 7 ten-thousands, 4 thousands, 9 hundreds, 8 tens, 2 ones.
So the 9 is in the hundreds place. Its value is 100.

After the decimal .1035 is: 1 tenths, 0 hundredths, 3 thousandths, 5 ten-thousandths.
So the 3 is in the thousandths place. Its value is 0.001 = 1/1000.

Ratio = (value of 9's place) / (value of 3's place)
= 100 / 0.001 = 100 / (1/1000) = 100 x 1000 = 100,000.

So the hundreds place is 100,000 times as great as the thousandths place.
Choices A,B,D,E are powers of 10 that are too small or too large.

ANSWER 1: C

Problem 2:
We are asked for the earlier volume given expansion rate.

Rate: +3 degrees temperature = +4 cubic cm volume.

Now: temp 32°, volume 24 cc.
Before: temp 20°.

Temperature change from 20° to 32° is 32-20 = 12°.
Number of 3-degree steps in 12° is 12 / 3 = 4 steps.

So volume increase from 20° to 32° is 4 steps x 4 cc = 16 cc.

Thus volume at 20° + 16 = volume at 32° = 24.
So volume at 20° = 24 - 16 = 8 cc.

Check: A 8 fits, B 12 would be only 3 steps, D 16 would be only 2 steps, E 40 would be adding instead of subtracting.

ANSWER 2: A

Problem 3:
We need n teams such that every pair plays once gives 21 games.

Number of pairs from n teams is n choose 2 = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try integers: 6x5=30 too small, 7x6=42 correct, 8x7=56 too big.
So n = 7.

Ruling out: A 6 gives only 15 games, C 8 gives 28 games, etc.

ANSWER 3: B

Problem 4:
Ara and Shea started at same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara's now height.

Let original height = H.
Shea's growth = 20% of H = 0.20 H.
Shea now = H + 0.20H = 1.2H = 60.
So H = 60 / 1.2 = 50 inches.

Shea grew 60-50 = 10 inches.
Ara grew half as many inches: 10 / 2 = 5 inches.
Ara now = 50 + 5 = 55 inches.

Tempting wrong A 48 would be working backwards, D 54 would be misreading percent, etc.

ANSWER 4: E

Problem 5:
A standard 1-6 die is tossed. Bottom face hidden, other 5 faces seen. Probability product of seen numbers divisible by 6?

Product divisible by 6 means divisible by 2 and by 3.

Numbers 1 to 6: evens are 2,4,6 — three of them. Multiples of 3 are 3,6 — two of them.

Hiding only one face (the bottom) can hide at most one even, leaving at least two evens still visible, so factor 2 remains. It can hide at most one multiple of 3, leaving at least one multiple of 3 still visible, so factor 3 remains.

Indeed total product 1·2·3·4·5·6 = 720. Visible product = 720 / b where b is bottom. 720/b divided by 6 = 120/b, and b=1,2,3,4,5,6 all divide 120, so always integer.

Hence no matter which face is hidden, the visible product is divisible by 6. Probability = 6/6 = 1.

Other choices 1/3,1/2,2/3,5/6 assume sometimes it fails, which is impossible.

ANSWER 5: E

Problem 6:
75 problems total: 10 arithmetic, 30 algebra, 35 geometry.
Correct: 70% of arithmetic, 40% of algebra, 60% of geometry. How many more to reach 60% overall?

Compute:
Arithmetic correct = 0.70 x 10 = 7.
Algebra correct = 0.40 x 30 = 12.
Geometry correct = 0.60 x 35 = 21.
Total correct = 7+12+21 = 40.

Need for 60% passing: 0.60 x 75 = 45 problems.

Shortfall = 45 - 40 = 5 more problems.

So B 5, not A 1 or C 7.

ANSWER 6: B

Problem 7:
Given 0 < a < b < c, which statement is impossible?

A. a + c < b. Since a > 0, a + c > c. And c > b. So a+c > b always. Hence a+c < b can never happen. This is impossible.

Check others are possible with examples:
B. a·b < c: e.g. a=1,b=2,c=10 gives 2<10 possible.
C. a+b < c: e.g. a=1,b=2,c=10 gives 3<10 possible.
D. a·c < b: with fractions <1. e.g. a=0.1,c=2,b=1 gives 0.2<1 and 0.1<1<2 possible.
E. b c = a: let b=0.25,a=0.1, then c=a/b=0.4, and 0.1<0.25<0.4 possible.

So only A is impossible. Students may think E impossible because b c > b > a for numbers >1, but with numbers <1 multiplication can shrink.

ANSWER 7: A

Problem 8:
One full recipe = 15 cookies = 3 tbsp butter. Need 216 cookies. 8 tbsp = 1 stick. How many sticks?

Pans needed: 216 / 15 = 14.4. Since only full recipes, round up to 15 pans. 14 pans would give only 210 cookies, not enough.

Butter needed: 15 x 3 tbsp = 45 tbsp.

Sticks needed: 45 / 8 = 5.625. Since need whole sticks to have enough, round up to 6. 5 sticks = 40 tbsp < 45 tbsp, not enough.

So answer 6. A 5 would leave butter and cookies short.

ANSWER 8: B

Problem 9:
Build 3x3x3 = 27 small cubes. 6 white, 21 red. Minimize white squares on outside. What fraction of outside is white?

Big cube surface: 6 faces each 3x3 = 9 small squares, total 54 small squares.

Small-cube positions by exposure:
- 1 interior center: 0 faces shown
- 6 face-centers: 1 face shown
- 12 edge non-corner: 2 faces shown
- 8 corners: 3 faces shown

To minimize white, put whites where least exposure:
Put 1 white in center → 0 exposed.
Put remaining 5 whites in face-center spots → 1 each exposed.
Total white exposed squares = 5.

Fraction white = 5 / 54 of total surface area.

Larger fractions like 1/9=6/54, etc. would result from worse placement.

ANSWER 9: A

Problem 10:
1998 population 200, triples every 25 years. Estimate 2050.

Time elapsed: 2050-1998 = 52 years.
Number of 25-year periods: 52/25 ≈ 2.08, just over 2 doublings-triplings.

After 25 years (~2023): 200x3=600.
After 50 years (~2048): 600x3=1800.
2050 is 2 years later, a little more, so about 2000.

Formally 200 x 3^(52/25) ≈ 200 x 9 x small extra ≈ 1960.

So A 600 is only one tripling, C 1000 too low, E 3000 too high. Best is D 2000.

ANSWER 10: D

Problem 1:
We are asked to find the ratio of the value of the place occupied by the digit 9 to the value of the place occupied by the digit 3 in the decimal number $74982.1035$.

1. Identify the place values:
   - The number is $74982.1035$.
   - The digit 2 is in the ones place ($10^0$).
   - The digit 8 is in the tens place ($10^1$).
   - The digit 9 is in the hundreds place ($10^2 = 100$).
   - The digit 1 is in the tenths place ($10^{-1} = 0.1$).
   - The digit 0 is in the hundredths place ($10^{-2} = 0.01$).
   - The digit 3 is in the thousandths place ($10^{-3} = 0.001$).
   - The digit 5 is in the ten-thousandths place ($10^{-4} = 0.0001$).

2. Compute the ratio of the place values:
   $$\frac{\text{Place value of } 9}{\text{Place value of } 3} = \frac{100}{0.001} = \frac{10^2}{10^{-3}} = 10^{2 - (-3)} = 10^5 = 100,000.$$

This corresponds to choice C.

ANSWER 1: C

---

Problem 2:
We are asked to find the initial volume of gas at $20^\circ$ given its volume at $32^\circ$ and its rate of expansion.

1. Find the change in temperature:
   The temperature changes from $20^\circ$ to $32^\circ$, which is an increase of:
   $$32^\circ - 20^\circ = 12^\circ.$$

2. Determine the volume expansion:
   For every $3^\circ$ rise, the volume expands by $4\text{ cm}^3$.
   In a $12^\circ$ rise, the number of $3^\circ$ intervals is:
   $$\frac{12^\circ}{3^\circ} = 4.$$
   Therefore, the volume expanded by:
   $$4 \times 4\text{ cm}^3 = 16\text{ cm}^3.$$

3. Calculate the original volume at $20^\circ$:
   $$\text{Volume at } 20^\circ = \text{Volume at } 32^\circ - 16\text{ cm}^3 = 24 - 16 = 8\text{ cm}^3.$$

This matches choice A.

ANSWER 2: A

---

Problem 3:
We are given that each of the $N$ teams plays every other team exactly once, resulting in 21 total games. We need to find $N$.

1. The number of games played between $N$ teams is the number of ways to choose 2 teams from $N$, given by the combination formula:
   $$\binom{N}{2} = \frac{N(N-1)}{2}.$$

2. Set this equal to 21 and solve for $N$:
   $$\frac{N(N-1)}{2} = 21 \implies N(N-1) = 42.$$

3. Find two consecutive positive integers whose product is 42:
   Since $7 \times 6 = 42$, we have $N = 7$.

Checking the choices, $N = 7$ corresponds to choice B.

ANSWER 3: B

---

Problem 4:
We are given:
- Ara and Shea initially had the same height, say $h$ inches.
- Shea has grown $20\%$, so Shea's new height is $h + 0.20h = 1.20h = 60$ inches.
- Ara has grown half as many inches as Shea.
- We need to find Ara's current height.

1. Find the original height $h$:
   $$1.20h = 60 \implies h = \frac{60}{1.2} = 50\text{ inches}.$$

2. Find how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10\text{ inches}.$$

3. Find Ara's growth:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5\text{ inches}.$$

4. Find Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55\text{ inches}.$$

This corresponds to choice E.

ANSWER 4: E

---

Problem 5:
A standard fair six-sided die has faces with numbers 1, 2, 3, 4, 5, and 6.
When the die is tossed, one face is on the bottom and cannot be seen, while the other 5 faces are visible.
We want the product of the 5 visible faces to be divisible by 6.

1. The product of all six numbers on the die is:
   $$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720.$$
   The product of the 5 visible faces is:
   $$P = \frac{720}{b},$$
   where $b \in \{1, 2, 3, 4, 5, 6\}$ is the number on the bottom face.

2. A number is divisible by 6 if and only if it is a multiple of both 2 and 3.
   Let's check the product $P$ for each possible bottom face $b$:
   - If $b = 1$, visible faces are $\{2, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 2$, visible faces are $\{1, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 3$, visible faces are $\{1, 2, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 4$, visible faces are $\{1, 2, 3, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 5$, visible faces are $\{1, 2, 3, 4, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 6$, visible faces are $\{1, 2, 3, 4, 5\}$. The product is $1 \times 2 \times 3 \times 4 \times 5 = 120$. Since $120 = 6 \times 20$, 120 is also divisible by 6 (it has factors 2 and 3)!

3. In every single case (all 6 possible outcomes for the bottom face), the product of the five visible faces is divisible by 6.
   Therefore, the probability is:
   $$\frac{6}{6} = 1.$$

This corresponds to choice E.

ANSWER 5: E

---

Problem 6:
Tori took a 75-problem test:
- 10 arithmetic problems: she answered $70\%$ correctly.
  $$\text{Correct arithmetic} = 0.70 \times 10 = 7.$$
- 30 algebra problems: she answered $40\%$ correctly.
  $$\text{Correct algebra} = 0.40 \times 30 = 12.$$
- 35 geometry problems: she answered $60\%$ correctly.
  $$\text{Correct geometry} = 0.60 \times 35 = 21.$$

1. Total problems Tori answered correctly:
   $$\text{Total correct} = 7 + 12 + 21 = 40.$$

2. Passing score required:
   $60\%$ of 75 problems:
   $$\text{Passing score} = 0.60 \times 75 = 45.$$

3. Additional problems needed:
   $$45 - 40 = 5\text{ problems}.$$

This corresponds to choice B.

ANSWER 6: B

---

Problem 7:
We are given real numbers $a, b, c$ such that $0 < a < b < c$.
We need to determine which of the statements is impossible.

Let's test each choice:
- Choice A: $a + c < b$.
  Since $a > 0$, adding $a$ to $c$ gives $a + c > c$.
  Since we are given $c > b$, it follows that $a + c > c > b$, which means $a + c > b$ must always be true!
  Therefore, $a + c < b$ is impossible.

Let's verify that the other choices can be possible with appropriate values:
- Choice B: $a \cdot b < c$. If $a = 1, b = 2, c = 3$, then $a \cdot b = 2 < 3 = c$. (Possible)
- Choice C: $a + b < c$. If $a = 1, b = 2, c = 4$, then $a + b = 3 < 4 = c$. (Possible)
- Choice D: $a \cdot c < b$. If $a = 0.1, b = 2, c = 3$, then $a \cdot c = 0.3 < 2 = b$. (Possible)
-

# Problem 1:

**Restate:** In 74982.1035, compare the place value of the digit 9 with the place value of the digit 3.

**Working:** Let's identify each place:
- The digit 9 sits in the **hundreds** place: 9 × 100 = 900. Its place value is **100**.
- The digit 3 sits in the **hundredths** place (two positions right of the decimal point): 3 × 0.01 = 0.03. Its place value is **0.01**.

**Compute the ratio:**
$$\frac{100}{0.01} = 100 \div \frac{1}{100} = 100 \times 100 = 10{,}000$$

**Check wrong choices:** Counting places instead of computing the ratio might suggest 100,000 (if you miscount the decimal places), but from hundreds to hundredths is exactly 4 positions, and each position is a factor of 10, giving 10⁴ = 10,000.

ANSWER 1: B

# Problem 2:

**Restate:** Volume increases 4 cm³ per 3° rise. Volume is 24 cm³ at 32°. Find the volume at 20°.

**Working:** The temperature dropped from 32° to 20°:
$$32 - 20 = 12°$$
Number of 3° intervals in the drop:
$$12 \div 3 = 4 \text{ intervals}$$
Each 3° drop means the gas **shrinks** by 4 cm³, so the total shrink is:
$$4 \times 4 = 16 \text{ cm}^3$$
The volume at 20° was:
$$24 - 16 = 8 \text{ cm}^3$$

**Check wrong choices:** Choice D (16) is the *amount* of shrink, not the volume — a tempting trap. Choice E (40) comes from adding instead of subtracting.

ANSWER 2: A

# Problem 3:

**Restate:** Each team plays every other team once; 21 total games. Find the number of teams.

**Working:** With n teams, the number of games is:
$$\frac{n(n-1)}{2}$$
Set this equal to 21:
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42$$
Testing n = 7: $7 \times 6 = 42$. ✓

**Check wrong choices:** n = 6 gives 15 games (too few); n = 8 gives 28 games (too many). Only n = 7 works.

ANSWER 3: B

# Problem 4:

**Restate:** Ara and Shea were once equal in height. Shea grew 20% and is now 60 inches. Ara grew half as many inches as Shea. Find Ara's height now.

**Working:** Let the original height be $h$. Shea grew 20%, so:
$$h + 0.20h = 60 \implies 1.2h = 60 \implies h = 50 \text{ inches}$$
Shea's growth: $60 - 50 = 10$ inches.
Ara grew half as many inches as Shea:
$$\frac{10}{2} = 5 \text{ inches}$$
Ara's height now:
$$50 + 5 = 55 \text{ inches}$$

**Check wrong choices:** Choice A (48) comes from computing 20% of 60 incorrectly; choice D (54) would result if Ara also grew 20% of the original height. The key is that Ara grew 5 **inches**, not a percentage.

ANSWER 4: E

# Problem 5:

**Restate:** A die is tossed; the bottom face is hidden. Find the probability that the product of the five visible faces is divisible by 6.

**Working:** The product of all six faces is:
$$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$$
The product of the five visible faces is:
$$\frac{720}{\text{hidden face}}$$

Check each possible hidden face:
- Hidden = 1: product = 720 → divisible by 6 ✓
- Hidden = 2: product = 360 → divisible by 6 ✓
- Hidden = 3: product = 240 → divisible by 6 ✓
- Hidden = 4: product = 180 → divisible by 6 ✓
- Hidden = 5: product = 144 → divisible by 6 ✓
- Hidden = 6: product = 120 → divisible by 6 ✓

All six outcomes give a product divisible by 6, so the probability is:
$$\frac{6}{6} = 1$$

**Check wrong choices:** One might think hiding the 5 or the 6 removes needed factors, but 720 contains 2 × 3 many times over — removing any single face still leaves factors of 2 and 3 in the product.

ANSWER 5: E

# Problem 6:

**Restate:** Tori's test: 10 arithmetic, 30 algebra, 35 geometry problems (75 total). She got 70%, 40%, and 60% correct respectively, which was below 60% overall. How many more correct answers were needed for 60%?

**Working:** Correct answers in each category:
- Arithmetic: $0.70 \times 10 = 7$
- Algebra: $0.40 \times 30 = 12$
- Geometry: $0.60 \times 35 = 21$

Total correct: $7 + 12 + 21 = 40$

To pass, she needs 60% of 75:
$$0.60 \times 75 = 45$$

Additional problems needed:
$$45 - 40 = 5$$

**Check wrong choices:** Note that the weighted average $\frac{40}{75} \approx 53.3\%$ is indeed below 60%, consistent with the story. A tempting trap is to average the three percentages: $\frac{70+40+60}{3} \approx 56.7\%$, which is wrong because the categories have different sizes.

ANSWER 6: B

# Problem 7:

**Restate:** Given $0 < a < b < c$, which statement is impossible?

**Working:** Test each option:

**A. $a + c < b$:** Since $c > b$, adding positive $a$ gives $a + c > c > b$. So $a + c < b$ **cannot happen**. ✓ (impossible)

Let's verify the others are all possible with suitable numbers:

**B. $a \cdot b < c$:** Try $a = 0.1, b = 0.2, c = 1$: $0.02 < 1$ ✓ possible.

**C. $a + b < c$:** Try $a = 1, b = 2, c = 10$: $3 < 10$ ✓ possible.

**D. $a \cdot c < b$:** Try $a = 0.1, b = 0.5, c = 0.6$: $0.06 < 0.5$ ✓ possible.

**E. $bc = a$:** Try $a = 0.06, b = 0.3, c = 0.2$... wait, need $b < c$. Try $a = 0.02, b = 0.1, c = 0.2$: $bc = 0.02 = a$, and $0 < 0.02 < 0.1 < 0.2$ ✓ possible.

Only option A forces a contradiction: $a + c > c > b$ always holds.

ANSWER 7: A

# Problem 8:

**Restate:** One recipe = 15 cookies using 3 tablespoons of butter; only full recipes allowed. Need 216 cookies. 8 tablespoons = 1 stick. How many sticks?

**Working:** Number of recipes needed:
$$\frac{216}{15} = 14.4$$
Since only full recipes are made, they must make **15 recipes** (14 recipes × 15 = 210 cookies, which is not enough).

Butter needed:
$$15 \times 3 = 45 \text{ tablespoons}$$
Sticks needed:
$$\frac{45}{8} = 5.625 \implies \text{need } 6 \text{ full sticks}$$

**Check wrong choices:** Choice A (5) is the trap — it uses only 40 tablespoons, but 45 are required. You can't buy a partial stick, so round up to 6.

ANSWER 8: B

# Problem 9:

**Restate:** Build a 3-inch cube from 27 unit cubes: 21 red, 6 white. Arrange to minimize white surface area. What fraction of the surface area is white?

**Working:** The 3×3×3 cube has total surface area:
$$6 \text{ faces} \times 9 \text{ sq in} = 54 \text{ square inches}$$

Classify the 27 positions by how many faces each shows:
- **1 inte

Problem 1:
We need the ratio of the place values of the digits 9 and 3 in  
\(74982.1035\).

The digit 9 is in the hundreds place, so its place value is  
\[
100.
\]

The digit 3 is in the thousandths place, so its place value is  
\[
0.001.
\]

Now compute the ratio:
\[
\frac{100}{0.001}=100{,}000.
\]

So the value of the place of the 9 is \(100{,}000\) times as great as the value of the place of the 3.

ANSWER 1: C

Problem 2:
We are told that for every \(3^\circ\) rise in temperature, the gas expands by 4 cubic centimeters.

At temperature \(32^\circ\), the volume is 24 cubic centimeters. We want the volume at \(20^\circ\).

The temperature difference is
\[
32-20=12^\circ.
\]

From \(20^\circ\) to \(32^\circ\), the temperature rises by \(12^\circ\). Since each \(3^\circ\) rise increases the volume by 4 cubic centimeters,
\[
\frac{12}{3}\cdot 4=4\cdot 4=16
\]
cubic centimeters.

So at \(20^\circ\), the volume was
\[
24-16=8
\]
cubic centimeters.

ANSWER 2: A

Problem 3:
Each team plays every other team exactly once. If there are \(n\) teams, the number of games is
\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games, so
\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:
\[
n(n-1)=42.
\]

We look for two consecutive positive integers whose product is 42:
\[
6\cdot 7=42.
\]

Thus
\[
n=7.
\]

So there were 7 teams.

ANSWER 3: B

Problem 4:
Ara and Shea were originally the same height. Let that original height be \(h\) inches.

Shea has grown 20%, so Shea’s current height is
\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:
\[
1.20h=60.
\]

Solve for \(h\):
\[
h=\frac{60}{1.20}=50.
\]

So they were originally 50 inches tall. Shea grew
\[
60-50=10
\]
inches.

Ara grew half as many inches as Shea:
\[
\frac{10}{2}=5
\]
inches.

Therefore Ara is now
\[
50+5=55
\]
inches tall.

ANSWER 4: E

Problem 5:
A fair six-sided die has faces numbered 1 through 6. When it is tossed, one face is hidden on the bottom, and the other five faces are visible.

We need the probability that the product of the five visible numbers is divisible by 6.

A number is divisible by 6 if it has at least one factor of 2 and at least one factor of 3.

No matter which single face is hidden:

- There are three even faces: 2, 4, and 6. Hiding only one of them still leaves at least one even face visible.
- There are two multiples of 3: 3 and 6. Hiding only one of them still leaves at least one multiple of 3 visible.

So the product of the visible faces always contains both a factor of 2 and a factor of 3. Therefore it is always divisible by 6.

Thus the probability is
\[
1.
\]

ANSWER 5: E

Problem 6:
Tori answered:

- \(70\%\) of 10 arithmetic problems:
\[
0.70\cdot 10=7
\]
- \(40\%\) of 30 algebra problems:
\[
0.40\cdot 30=12
\]
- \(60\%\) of 35 geometry problems:
\[
0.60\cdot 35=21
\]

Total correct:
\[
7+12+21=40.
\]

There were 75 problems total. To pass, she needed \(60\%\) correct:
\[
0.60\cdot 75=45.
\]

She answered 40 correctly, so she needed
\[
45-40=5
\]
more correct answers.

ANSWER 6: B

Problem 7:
We are given
\[
0<a<b<c.
\]

We need the impossible statement.

Choice A says
\[
a+c<b.
\]

But since \(c>b\) and \(a>0\),
\[
a+c>b.
\]

So \(a+c<b\) cannot happen.

We can also check that the other choices are possible:

- B: \(a\cdot b<c\) can happen, for example \(a=1,b=2,c=100\).
- C: \(a+b<c\) can happen, for example \(a=1,b=2,c=100\).
- D: \(a\cdot c<b\) can happen, for example \(a=0.1,b=2,c=3\), since \(0.3<2\).
- E: If interpreted as \(bc=a\), it can happen, for example \(a=0.1,b=0.2,c=0.5\), since \(0.2\cdot 0.5=0.1\).

Thus the impossible choice is A.

ANSWER 7: A

Problem 8:
One recipe makes 15 cookies and uses 3 tablespoons of butter. Only full recipes can be made.

Walter and Gretel need 216 cookies. The number of full recipes needed is
\[
\left\lceil \frac{216}{15}\right\rceil.
\]

Since
\[
\frac{216}{15}=14.4,
\]
they must make 15 full recipes.

Butter needed:
\[
15\cdot 3=45
\]
tablespoons.

There are 8 tablespoons in a stick of butter, so the number of sticks needed is
\[
\left\lceil \frac{45}{8}\right\rceil.
\]

Since
\[
\frac{45}{8}=5.625,
\]
they need 6 sticks of butter.

ANSWER 8: B

Problem 9:
A 3-inch cube made from 27 one-inch cubes has surface area
\[
6\cdot 3^2=54
\]
square inches.

Each small cube can show:

- 0 faces if it is the center cube,
- 1 face if it is a face-center cube,
- 2 faces if it is an edge cube,
- 3 faces if it is a corner cube.

We have 6 white cubes. To minimize the white surface area, we should hide as many white cubes as possible.

There is exactly one interior cube, so put one white cube there. It shows 0 square inches.

The remaining 5 white cubes must be on the surface. To minimize exposed white area, put them on face-center positions, where each shows exactly 1 square inch.

So the minimum white surface area is
\[
1\cdot 0+5\cdot 1=5
\]
square inches.

The fraction of the surface area that is white is
\[
\frac{5}{54}.
\]

ANSWER 9: A

Problem 10:
The population is 200 in 1998, and it triples every 25 years.

From 1998 to 2050 is
\[
2050-1998=52
\]
years.

Two full 25-year periods are 50 years, so after 50 years the population is
\[
200\cdot 3\cdot 3=200\cdot 9=1800.
\]

The extra 2 years would add a relatively small amount, so the population in 2050 is a little above 1800. Among the choices, the best estimate is 2000.

ANSWER 10: D

# Problem 1:
Write out the place values of 74982.1035:
- 7 → ten-thousands, 4 → thousands, **9 → hundreds**, 8 → tens, 2 → ones
- 1 → tenths, 0 → hundredths, **3 → thousandths**, 5 → ten-thousandths

So the 9 sits in the hundreds place (value 100) and the 3 sits in the thousandths place (value 0.001).

Ratio = 100 ÷ 0.001 = 100,000.

(Answers like 1,000 or 10,000 come from miscounting decimal places — the 3 is the *third* digit after the point, not the first or second.)

ANSWER 1: C

# Problem 2:
Temperature goes from 32° down to 20°, a drop of 12°.

Each 3° change corresponds to 4 cm³ of change: 12 ÷ 3 = 4 steps.

Volume change = 4 × 4 = 16 cm³ decrease (cooling means shrinking).

Volume = 24 − 16 = 8 cm³.

(40 is the trap from *adding* instead of subtracting; 16 is just the change, not the volume.)

ANSWER 2: A

# Problem 3:
With n teams each playing every other team once, the number of games is C(n,2) = n(n−1)/2.

Set n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7·6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 games — neither is 21.

ANSWER 3: B

# Problem 4:
Let the common original height be x. Shea grew 20%: 1.20x = 60 → x = 50.

So Shea grew 60 − 50 = 10 inches.

Ara grew half as many inches: 5 inches.

Ara's height = 50 + 5 = 55 inches.

(54 would result from wrongly taking Shea's growth as 10% of 60, etc.)

ANSWER 4: E

# Problem 5:
The product of all six faces is 1·2·3·4·5·6 = 720. If the hidden bottom face is k, the visible product is 720/k.

- k=1: 720 ✓ divisible by 6
- k=2: 360 ✓
- k=3: 240 ✓
- k=4: 180 ✓
- k=5: 144 ✓
- k=6: 120 ✓

Reason why: hiding one face still leaves at least one even number and at least one multiple of 3 among the five visible faces (e.g., hide 6 → 2 and 3 remain; hide 3 → 6 remains).

So the probability is 6/6 = 1.

ANSWER 5: E

# Problem 6:
Correct answers:
- Arithmetic: 70% of 10 = 7
- Algebra: 40% of 30 = 12
- Geometry: 60% of 35 = 21

Total correct = 7 + 12 + 21 = 40.

Needed for 60%: 0.60 × 75 = 45.

She needs 45 − 40 = 5 more.

(40/75 ≈ 53%, consistent with failing.)

ANSWER 6: B

# Problem 7:
Given 0 < a < b < c.

- **A. a + c < b:** Since c > b and a > 0, we get a + c > b. So a + c < b is impossible.
- B. a·b < c: take a = 0.1, b = 0.2, c = 1 → 0.02 < 1 ✓ possible.
- C. a + b < c: take a = 1, b = 2, c = 10 → 3 < 10 ✓ possible.
- D. a·c < b: take a = 0.1, b = 1, c = 2 → 0.2 < 1 ✓ possible.
- E. b/c = a: take a = 0.5, b = 1, c = 2 → 1/2 = 0.5 ✓ possible.

ANSWER 7: A

# Problem 8:
Pans needed: 216 ÷ 15 = 14.4, and only full recipes are made, so they must bake 15 pans.

Butter: 15 × 3 = 45 tablespoons.

Sticks: 45 ÷ 8 = 5.625, and butter comes in whole sticks, so 6 sticks.

(5 sticks = 40 tbsp, not enough.)

ANSWER 8: B

# Problem 9:
The 3×3×3 cube has surface area 6 × 9 = 54 unit squares.

Positions in a 3-cube: 1 center cube (0 faces showing), 6 face-centers (1 face each), 12 edges (2 faces), 8 corners (3 faces).

To minimize white showing, place one white cube in the hidden center (0 white squares) and the other 5 white cubes at face-center positions (1 white square each).

Minimum white area = 5 unit squares out of 54 → 5/54.

ANSWER 9: A

# Problem 10:
From 1998 to 2050 is 52 years — about two 25-year tripling periods (1998 → 2023 → 2048).

Population: 200 → 600 → 1800.

1800 is closest to 2000.

ANSWER 10: D

Problem 1:

We are comparing the place values of the digits 9 and 3 in

\[
74982.1035.
\]

- The digit \(9\) is in the hundreds place, so its place value is \(100\).
- The digit \(3\) is in the thousandths place, so its place value is

\[
0.001.
\]

Now find how many times as great \(100\) is as \(0.001\):

\[
\frac{100}{0.001}=100,000.
\]

So the hundreds place is \(100,\!000\) times the thousandths place.

- A, \(1,\!000\), is too small.
- B, \(10,\!000\), is too small.
- D and E are too large.

ANSWER 1: C

Problem 2:

The gas expands by \(4\) cubic centimeters for every \(3^\circ\) increase in temperature.

The temperature changes from \(32^\circ\) down to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

Since

\[
12^\circ \div 3^\circ=4,
\]

there are 4 groups of \(3^\circ\).

For each group, the volume decreases by \(4\) cubic centimeters:

\[
4\cdot 4=16.
\]

At \(32^\circ\), the volume is \(24\) cubic centimeters, so at \(20^\circ\) it was

\[
24-16=8.
\]

Choices B, C, and D do not subtract the full \(16\) cubic centimeters. Choice E would represent an increase rather than a decrease.

ANSWER 2: A

Problem 3:

If there are \(n\) teams and each pair of teams plays exactly once, then the number of games is

\[
\binom n2=\frac{n(n-1)}2.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

Checking: with 7 teams, the number of games is

\[
\frac{7\cdot6}{2}=21.
\]

Six teams would produce only \(15\) games, while 8 teams would produce \(28\) games.

ANSWER 3: B

Problem 4:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew \(20\%\), so Shea’s current height is

\[
1.20h=60.
\]

Thus,

\[
h=\frac{60}{1.2}=50.
\]

So both were originally 50 inches tall.

Shea’s growth was

\[
60-50=10
\]

inches.

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5
\]

inches.

Therefore, Ara is now

\[
50+5=55
\]

inches tall.

Choice A, 48, is less than Ara’s original height. Choices B, C, and D do not include the correct 5-inch growth.

ANSWER 4: E

Problem 5:

A standard die has faces numbered \(1,2,3,4,5,6\). The product of all six face numbers is

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6=720.
\]

If the bottom face has number \(k\), then the product of the five visible faces is

\[
\frac{720}{k}.
\]

Check each possible bottom face:

\[
\frac{720}{1}=720,\qquad
\frac{720}{2}=360,\qquad
\frac{720}{3}=240,
\]

\[
\frac{720}{4}=180,\qquad
\frac{720}{5}=144,\qquad
\frac{720}{6}=120.
\]

Every one of these numbers is divisible by \(6\).

Therefore, no matter which face is on the bottom, the visible-face product is divisible by 6. The probability is

\[
1.
\]

All of choices A through D are less than 1, but the event always happens.

ANSWER 5: E

Problem 6:

Tori answered correctly:

- Arithmetic:

\[
70\%\text{ of }10=0.7(10)=7
\]

- Algebra:

\[
40\%\text{ of }30=0.4(30)=12
\]

- Geometry:

\[
60\%\text{ of }35=0.6(35)=21
\]

So her total number correct was

\[
7+12+21=40.
\]

To pass, she needs \(60\%\) of 75 questions correct:

\[
0.60(75)=45.
\]

She needs

\[
45-40=5
\]

more correct answers.

Choice A is too small, while choices C, D, and E give more than the number needed to reach exactly 45 correct.

ANSWER 6: B

Problem 7:

We are given

\[
0<a<b<c.
\]

We must find the statement that is impossible.

Consider choice A:

\[
a+c<b.
\]

But \(c>b\), and also \(a>0\). Therefore,

\[
a+c>c>b.
\]

So it is impossible for \(a+c<b\).

The other choices can happen for suitable positive numbers. For example:

- B: \(a\cdot b<c\) can occur with \(a=1,\ b=2,\ c=3\), since \(1\cdot2<3\).
- C: \(a+b<c\) can occur with \(a=1,\ b=2,\ c=4\).
- D: \(a\cdot c<b\) can occur with \(a=\frac14,\ b=1,\ c=2\), since \(\frac14\cdot2=\frac12<1\).
- E: \(bc=a\) can occur with \(b=\frac12,\ c=\frac34\), giving \(a=\frac38\), and indeed

\[
0<\frac38<\frac12<\frac34.
\]

Thus only choice A is impossible.

ANSWER 7: A

Problem 8:

Each full recipe makes 15 cookies.

Walter and Gretel need at least 216 cookies. Since only full recipes may be made, determine how many recipes are needed:

\[
216\div 15=14.4.
\]

They cannot make \(14.4\) recipes, so they must make 15 full recipes.

These 15 recipes make

\[
15\cdot15=225
\]

cookies, enough for the crowd.

Each recipe uses 3 tablespoons of butter, so the total butter needed is

\[
15\cdot3=45
\]

tablespoons.

There are 8 tablespoons in each stick, so

\[
45\div8=5.625.
\]

They must buy 6 full sticks of butter.

Five sticks provide only \(5\cdot8=40\) tablespoons, which is not enough. Six sticks provide \(48\) tablespoons, enough for the 45 tablespoons needed.

ANSWER 8: B

Problem 9:

A \(3\times3\times3\) cube has surface area

\[
6\cdot 3\cdot3=54
\]

square inches, since each of its 6 faces has area \(3\cdot3=9\).

To minimize the white surface showing, place the white cubes in the least exposed positions.

- The one center cube of the large cube has no faces exposed.
- The six face-center cubes each have exactly one face exposed.
- Edge cubes have two exposed faces.
- Corner cubes have three exposed faces.

There are 6 white cubes. Put one white cube in the completely hidden center. Then put the other 5 white cubes at five face-center locations.

Thus, the white cubes show only

\[
5
\]

unit square faces on the surface.

Therefore, the fraction of the surface that is white is

\[
\frac{5}{54}.
\]

Choices such as \(\frac19,\frac{5}{27},\frac29,\frac13\) represent more than 5 white unit squares out of the 54 total surface squares.

ANSWER 9: A

Problem 10:

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

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is approximately two 25-year periods.

After one 25-year period:

\[
200\cdot3=600.
\]

After two 25-year periods:

\[
600\cdot3=1800.
\]

Since 2050 is slightly more than 50 years after 1998, the population should be slightly more than 1800. The closest estimate is

\[
20

Problem 1:
We are asked: in 74982.1035, compare the place value of the 9 with the place value of the 3.

Write out places:
74982 is: 7 ten-thousands, 4 thousands, 9 hundreds, 8 tens, 2 ones.
So the 9 is in the hundreds place. Its value is 100.

After the decimal .1035 is: 1 tenths, 0 hundredths, 3 thousandths, 5 ten-thousandths.
So the 3 is in the thousandths place. Its value is 0.001 = 1/1000.

Ratio = (value of 9's place) / (value of 3's place)
= 100 / 0.001 = 100 / (1/1000) = 100 x 1000 = 100,000.

So the hundreds place is 100,000 times as great as the thousandths place.
Choices A,B,D,E are powers of 10 that are too small or too large.

ANSWER 1: C

Problem 2:
We are asked for the earlier volume given expansion rate.

Rate: +3 degrees temperature = +4 cubic cm volume.

Now: temp 32°, volume 24 cc.
Before: temp 20°.

Temperature change from 20° to 32° is 32-20 = 12°.
Number of 3-degree steps in 12° is 12 / 3 = 4 steps.

So volume increase from 20° to 32° is 4 steps x 4 cc = 16 cc.

Thus volume at 20° + 16 = volume at 32° = 24.
So volume at 20° = 24 - 16 = 8 cc.

Check: A 8 fits, B 12 would be only 3 steps, D 16 would be only 2 steps, E 40 would be adding instead of subtracting.

ANSWER 2: A

Problem 3:
We need n teams such that every pair plays once gives 21 games.

Number of pairs from n teams is n choose 2 = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try integers: 6x5=30 too small, 7x6=42 correct, 8x7=56 too big.
So n = 7.

Ruling out: A 6 gives only 15 games, C 8 gives 28 games, etc.

ANSWER 3: B

Problem 4:
Ara and Shea started at same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara's now height.

Let original height = H.
Shea's growth = 20% of H = 0.20 H.
Shea now = H + 0.20H = 1.2H = 60.
So H = 60 / 1.2 = 50 inches.

Shea grew 60-50 = 10 inches.
Ara grew half as many inches: 10 / 2 = 5 inches.
Ara now = 50 + 5 = 55 inches.

Tempting wrong A 48 would be working backwards, D 54 would be misreading percent, etc.

ANSWER 4: E

Problem 5:
A standard 1-6 die is tossed. Bottom face hidden, other 5 faces seen. Probability product of seen numbers divisible by 6?

Product divisible by 6 means divisible by 2 and by 3.

Numbers 1 to 6: evens are 2,4,6 — three of them. Multiples of 3 are 3,6 — two of them.

Hiding only one face (the bottom) can hide at most one even, leaving at least two evens still visible, so factor 2 remains. It can hide at most one multiple of 3, leaving at least one multiple of 3 still visible, so factor 3 remains.

Indeed total product 1·2·3·4·5·6 = 720. Visible product = 720 / b where b is bottom. 720/b divided by 6 = 120/b, and b=1,2,3,4,5,6 all divide 120, so always integer.

Hence no matter which face is hidden, the visible product is divisible by 6. Probability = 6/6 = 1.

Other choices 1/3,1/2,2/3,5/6 assume sometimes it fails, which is impossible.

ANSWER 5: E

Problem 6:
75 problems total: 10 arithmetic, 30 algebra, 35 geometry.
Correct: 70% of arithmetic, 40% of algebra, 60% of geometry. How many more to reach 60% overall?

Compute:
Arithmetic correct = 0.70 x 10 = 7.
Algebra correct = 0.40 x 30 = 12.
Geometry correct = 0.60 x 35 = 21.
Total correct = 7+12+21 = 40.

Need for 60% passing: 0.60 x 75 = 45 problems.

Shortfall = 45 - 40 = 5 more problems.

So B 5, not A 1 or C 7.

ANSWER 6: B

Problem 7:
Given 0 < a < b < c, which statement is impossible?

A. a + c < b. Since a > 0, a + c > c. And c > b. So a+c > b always. Hence a+c < b can never happen. This is impossible.

Check others are possible with examples:
B. a·b < c: e.g. a=1,b=2,c=10 gives 2<10 possible.
C. a+b < c: e.g. a=1,b=2,c=10 gives 3<10 possible.
D. a·c < b: with fractions <1. e.g. a=0.1,c=2,b=1 gives 0.2<1 and 0.1<1<2 possible.
E. b c = a: let b=0.25,a=0.1, then c=a/b=0.4, and 0.1<0.25<0.4 possible.

So only A is impossible. Students may think E impossible because b c > b > a for numbers >1, but with numbers <1 multiplication can shrink.

ANSWER 7: A

Problem 8:
One full recipe = 15 cookies = 3 tbsp butter. Need 216 cookies. 8 tbsp = 1 stick. How many sticks?

Pans needed: 216 / 15 = 14.4. Since only full recipes, round up to 15 pans. 14 pans would give only 210 cookies, not enough.

Butter needed: 15 x 3 tbsp = 45 tbsp.

Sticks needed: 45 / 8 = 5.625. Since need whole sticks to have enough, round up to 6. 5 sticks = 40 tbsp < 45 tbsp, not enough.

So answer 6. A 5 would leave butter and cookies short.

ANSWER 8: B

Problem 9:
Build 3x3x3 = 27 small cubes. 6 white, 21 red. Minimize white squares on outside. What fraction of outside is white?

Big cube surface: 6 faces each 3x3 = 9 small squares, total 54 small squares.

Small-cube positions by exposure:
- 1 interior center: 0 faces shown
- 6 face-centers: 1 face shown
- 12 edge non-corner: 2 faces shown
- 8 corners: 3 faces shown

To minimize white, put whites where least exposure:
Put 1 white in center → 0 exposed.
Put remaining 5 whites in face-center spots → 1 each exposed.
Total white exposed squares = 5.

Fraction white = 5 / 54 of total surface area.

Larger fractions like 1/9=6/54, etc. would result from worse placement.

ANSWER 9: A

Problem 10:
1998 population 200, triples every 25 years. Estimate 2050.

Time elapsed: 2050-1998 = 52 years.
Number of 25-year periods: 52/25 ≈ 2.08, just over 2 doublings-triplings.

After 25 years (~2023): 200x3=600.
After 50 years (~2048): 600x3=1800.
2050 is 2 years later, a little more, so about 2000.

Formally 200 x 3^(52/25) ≈ 200 x 9 x small extra ≈ 1960.

So A 600 is only one tripling, C 1000 too low, E 3000 too high. Best is D 2000.

ANSWER 10: D

Problem 1:
We are asked to find the ratio of the value of the place occupied by the digit 9 to the value of the place occupied by the digit 3 in the decimal number $74982.1035$.

1. Identify the place values:
   - The number is $74982.1035$.
   - The digit 2 is in the ones place ($10^0$).
   - The digit 8 is in the tens place ($10^1$).
   - The digit 9 is in the hundreds place ($10^2 = 100$).
   - The digit 1 is in the tenths place ($10^{-1} = 0.1$).
   - The digit 0 is in the hundredths place ($10^{-2} = 0.01$).
   - The digit 3 is in the thousandths place ($10^{-3} = 0.001$).
   - The digit 5 is in the ten-thousandths place ($10^{-4} = 0.0001$).

2. Compute the ratio of the place values:
   $$\frac{\text{Place value of } 9}{\text{Place value of } 3} = \frac{100}{0.001} = \frac{10^2}{10^{-3}} = 10^{2 - (-3)} = 10^5 = 100,000.$$

This corresponds to choice C.

ANSWER 1: C

---

Problem 2:
We are asked to find the initial volume of gas at $20^\circ$ given its volume at $32^\circ$ and its rate of expansion.

1. Find the change in temperature:
   The temperature changes from $20^\circ$ to $32^\circ$, which is an increase of:
   $$32^\circ - 20^\circ = 12^\circ.$$

2. Determine the volume expansion:
   For every $3^\circ$ rise, the volume expands by $4\text{ cm}^3$.
   In a $12^\circ$ rise, the number of $3^\circ$ intervals is:
   $$\frac{12^\circ}{3^\circ} = 4.$$
   Therefore, the volume expanded by:
   $$4 \times 4\text{ cm}^3 = 16\text{ cm}^3.$$

3. Calculate the original volume at $20^\circ$:
   $$\text{Volume at } 20^\circ = \text{Volume at } 32^\circ - 16\text{ cm}^3 = 24 - 16 = 8\text{ cm}^3.$$

This matches choice A.

ANSWER 2: A

---

Problem 3:
We are given that each of the $N$ teams plays every other team exactly once, resulting in 21 total games. We need to find $N$.

1. The number of games played between $N$ teams is the number of ways to choose 2 teams from $N$, given by the combination formula:
   $$\binom{N}{2} = \frac{N(N-1)}{2}.$$

2. Set this equal to 21 and solve for $N$:
   $$\frac{N(N-1)}{2} = 21 \implies N(N-1) = 42.$$

3. Find two consecutive positive integers whose product is 42:
   Since $7 \times 6 = 42$, we have $N = 7$.

Checking the choices, $N = 7$ corresponds to choice B.

ANSWER 3: B

---

Problem 4:
We are given:
- Ara and Shea initially had the same height, say $h$ inches.
- Shea has grown $20\%$, so Shea's new height is $h + 0.20h = 1.20h = 60$ inches.
- Ara has grown half as many inches as Shea.
- We need to find Ara's current height.

1. Find the original height $h$:
   $$1.20h = 60 \implies h = \frac{60}{1.2} = 50\text{ inches}.$$

2. Find how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10\text{ inches}.$$

3. Find Ara's growth:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5\text{ inches}.$$

4. Find Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55\text{ inches}.$$

This corresponds to choice E.

ANSWER 4: E

---

Problem 5:
A standard fair six-sided die has faces with numbers 1, 2, 3, 4, 5, and 6.
When the die is tossed, one face is on the bottom and cannot be seen, while the other 5 faces are visible.
We want the product of the 5 visible faces to be divisible by 6.

1. The product of all six numbers on the die is:
   $$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720.$$
   The product of the 5 visible faces is:
   $$P = \frac{720}{b},$$
   where $b \in \{1, 2, 3, 4, 5, 6\}$ is the number on the bottom face.

2. A number is divisible by 6 if and only if it is a multiple of both 2 and 3.
   Let's check the product $P$ for each possible bottom face $b$:
   - If $b = 1$, visible faces are $\{2, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 2$, visible faces are $\{1, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 3$, visible faces are $\{1, 2, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 4$, visible faces are $\{1, 2, 3, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 5$, visible faces are $\{1, 2, 3, 4, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 6$, visible faces are $\{1, 2, 3, 4, 5\}$. The product is $1 \times 2 \times 3 \times 4 \times 5 = 120$. Since $120 = 6 \times 20$, 120 is also divisible by 6 (it has factors 2 and 3)!

3. In every single case (all 6 possible outcomes for the bottom face), the product of the five visible faces is divisible by 6.
   Therefore, the probability is:
   $$\frac{6}{6} = 1.$$

This corresponds to choice E.

ANSWER 5: E

---

Problem 6:
Tori took a 75-problem test:
- 10 arithmetic problems: she answered $70\%$ correctly.
  $$\text{Correct arithmetic} = 0.70 \times 10 = 7.$$
- 30 algebra problems: she answered $40\%$ correctly.
  $$\text{Correct algebra} = 0.40 \times 30 = 12.$$
- 35 geometry problems: she answered $60\%$ correctly.
  $$\text{Correct geometry} = 0.60 \times 35 = 21.$$

1. Total problems Tori answered correctly:
   $$\text{Total correct} = 7 + 12 + 21 = 40.$$

2. Passing score required:
   $60\%$ of 75 problems:
   $$\text{Passing score} = 0.60 \times 75 = 45.$$

3. Additional problems needed:
   $$45 - 40 = 5\text{ problems}.$$

This corresponds to choice B.

ANSWER 6: B

---

Problem 7:
We are given real numbers $a, b, c$ such that $0 < a < b < c$.
We need to determine which of the statements is impossible.

Let's test each choice:
- Choice A: $a + c < b$.
  Since $a > 0$, adding $a$ to $c$ gives $a + c > c$.
  Since we are given $c > b$, it follows that $a + c > c > b$, which means $a + c > b$ must always be true!
  Therefore, $a + c < b$ is impossible.

Let's verify that the other choices can be possible with appropriate values:
- Choice B: $a \cdot b < c$. If $a = 1, b = 2, c = 3$, then $a \cdot b = 2 < 3 = c$. (Possible)
- Choice C: $a + b < c$. If $a = 1, b = 2, c = 4$, then $a + b = 3 < 4 = c$. (Possible)
- Choice D: $a \cdot c < b$. If $a = 0.1, b = 2, c = 3$, then $a \cdot c = 0.3 < 2 = b$. (Possible)
-

# Problem 1:

**Restate:** In 74982.1035, compare the place value of the digit 9 with the place value of the digit 3.

**Working:** Let's identify each place:
- The digit 9 sits in the **hundreds** place: 9 × 100 = 900. Its place value is **100**.
- The digit 3 sits in the **hundredths** place (two positions right of the decimal point): 3 × 0.01 = 0.03. Its place value is **0.01**.

**Compute the ratio:**
$$\frac{100}{0.01} = 100 \div \frac{1}{100} = 100 \times 100 = 10{,}000$$

**Check wrong choices:** Counting places instead of computing the ratio might suggest 100,000 (if you miscount the decimal places), but from hundreds to hundredths is exactly 4 positions, and each position is a factor of 10, giving 10⁴ = 10,000.

ANSWER 1: B

# Problem 2:

**Restate:** Volume increases 4 cm³ per 3° rise. Volume is 24 cm³ at 32°. Find the volume at 20°.

**Working:** The temperature dropped from 32° to 20°:
$$32 - 20 = 12°$$
Number of 3° intervals in the drop:
$$12 \div 3 = 4 \text{ intervals}$$
Each 3° drop means the gas **shrinks** by 4 cm³, so the total shrink is:
$$4 \times 4 = 16 \text{ cm}^3$$
The volume at 20° was:
$$24 - 16 = 8 \text{ cm}^3$$

**Check wrong choices:** Choice D (16) is the *amount* of shrink, not the volume — a tempting trap. Choice E (40) comes from adding instead of subtracting.

ANSWER 2: A

# Problem 3:

**Restate:** Each team plays every other team once; 21 total games. Find the number of teams.

**Working:** With n teams, the number of games is:
$$\frac{n(n-1)}{2}$$
Set this equal to 21:
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42$$
Testing n = 7: $7 \times 6 = 42$. ✓

**Check wrong choices:** n = 6 gives 15 games (too few); n = 8 gives 28 games (too many). Only n = 7 works.

ANSWER 3: B

# Problem 4:

**Restate:** Ara and Shea were once equal in height. Shea grew 20% and is now 60 inches. Ara grew half as many inches as Shea. Find Ara's height now.

**Working:** Let the original height be $h$. Shea grew 20%, so:
$$h + 0.20h = 60 \implies 1.2h = 60 \implies h = 50 \text{ inches}$$
Shea's growth: $60 - 50 = 10$ inches.
Ara grew half as many inches as Shea:
$$\frac{10}{2} = 5 \text{ inches}$$
Ara's height now:
$$50 + 5 = 55 \text{ inches}$$

**Check wrong choices:** Choice A (48) comes from computing 20% of 60 incorrectly; choice D (54) would result if Ara also grew 20% of the original height. The key is that Ara grew 5 **inches**, not a percentage.

ANSWER 4: E

# Problem 5:

**Restate:** A die is tossed; the bottom face is hidden. Find the probability that the product of the five visible faces is divisible by 6.

**Working:** The product of all six faces is:
$$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$$
The product of the five visible faces is:
$$\frac{720}{\text{hidden face}}$$

Check each possible hidden face:
- Hidden = 1: product = 720 → divisible by 6 ✓
- Hidden = 2: product = 360 → divisible by 6 ✓
- Hidden = 3: product = 240 → divisible by 6 ✓
- Hidden = 4: product = 180 → divisible by 6 ✓
- Hidden = 5: product = 144 → divisible by 6 ✓
- Hidden = 6: product = 120 → divisible by 6 ✓

All six outcomes give a product divisible by 6, so the probability is:
$$\frac{6}{6} = 1$$

**Check wrong choices:** One might think hiding the 5 or the 6 removes needed factors, but 720 contains 2 × 3 many times over — removing any single face still leaves factors of 2 and 3 in the product.

ANSWER 5: E

# Problem 6:

**Restate:** Tori's test: 10 arithmetic, 30 algebra, 35 geometry problems (75 total). She got 70%, 40%, and 60% correct respectively, which was below 60% overall. How many more correct answers were needed for 60%?

**Working:** Correct answers in each category:
- Arithmetic: $0.70 \times 10 = 7$
- Algebra: $0.40 \times 30 = 12$
- Geometry: $0.60 \times 35 = 21$

Total correct: $7 + 12 + 21 = 40$

To pass, she needs 60% of 75:
$$0.60 \times 75 = 45$$

Additional problems needed:
$$45 - 40 = 5$$

**Check wrong choices:** Note that the weighted average $\frac{40}{75} \approx 53.3\%$ is indeed below 60%, consistent with the story. A tempting trap is to average the three percentages: $\frac{70+40+60}{3} \approx 56.7\%$, which is wrong because the categories have different sizes.

ANSWER 6: B

# Problem 7:

**Restate:** Given $0 < a < b < c$, which statement is impossible?

**Working:** Test each option:

**A. $a + c < b$:** Since $c > b$, adding positive $a$ gives $a + c > c > b$. So $a + c < b$ **cannot happen**. ✓ (impossible)

Let's verify the others are all possible with suitable numbers:

**B. $a \cdot b < c$:** Try $a = 0.1, b = 0.2, c = 1$: $0.02 < 1$ ✓ possible.

**C. $a + b < c$:** Try $a = 1, b = 2, c = 10$: $3 < 10$ ✓ possible.

**D. $a \cdot c < b$:** Try $a = 0.1, b = 0.5, c = 0.6$: $0.06 < 0.5$ ✓ possible.

**E. $bc = a$:** Try $a = 0.06, b = 0.3, c = 0.2$... wait, need $b < c$. Try $a = 0.02, b = 0.1, c = 0.2$: $bc = 0.02 = a$, and $0 < 0.02 < 0.1 < 0.2$ ✓ possible.

Only option A forces a contradiction: $a + c > c > b$ always holds.

ANSWER 7: A

# Problem 8:

**Restate:** One recipe = 15 cookies using 3 tablespoons of butter; only full recipes allowed. Need 216 cookies. 8 tablespoons = 1 stick. How many sticks?

**Working:** Number of recipes needed:
$$\frac{216}{15} = 14.4$$
Since only full recipes are made, they must make **15 recipes** (14 recipes × 15 = 210 cookies, which is not enough).

Butter needed:
$$15 \times 3 = 45 \text{ tablespoons}$$
Sticks needed:
$$\frac{45}{8} = 5.625 \implies \text{need } 6 \text{ full sticks}$$

**Check wrong choices:** Choice A (5) is the trap — it uses only 40 tablespoons, but 45 are required. You can't buy a partial stick, so round up to 6.

ANSWER 8: B

# Problem 9:

**Restate:** Build a 3-inch cube from 27 unit cubes: 21 red, 6 white. Arrange to minimize white surface area. What fraction of the surface area is white?

**Working:** The 3×3×3 cube has total surface area:
$$6 \text{ faces} \times 9 \text{ sq in} = 54 \text{ square inches}$$

Classify the 27 positions by how many faces each shows:
- **1 inte

Problem 1:
We need the ratio of the place values of the digits 9 and 3 in  
\(74982.1035\).

The digit 9 is in the hundreds place, so its place value is  
\[
100.
\]

The digit 3 is in the thousandths place, so its place value is  
\[
0.001.
\]

Now compute the ratio:
\[
\frac{100}{0.001}=100{,}000.
\]

So the value of the place of the 9 is \(100{,}000\) times as great as the value of the place of the 3.

ANSWER 1: C

Problem 2:
We are told that for every \(3^\circ\) rise in temperature, the gas expands by 4 cubic centimeters.

At temperature \(32^\circ\), the volume is 24 cubic centimeters. We want the volume at \(20^\circ\).

The temperature difference is
\[
32-20=12^\circ.
\]

From \(20^\circ\) to \(32^\circ\), the temperature rises by \(12^\circ\). Since each \(3^\circ\) rise increases the volume by 4 cubic centimeters,
\[
\frac{12}{3}\cdot 4=4\cdot 4=16
\]
cubic centimeters.

So at \(20^\circ\), the volume was
\[
24-16=8
\]
cubic centimeters.

ANSWER 2: A

Problem 3:
Each team plays every other team exactly once. If there are \(n\) teams, the number of games is
\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games, so
\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:
\[
n(n-1)=42.
\]

We look for two consecutive positive integers whose product is 42:
\[
6\cdot 7=42.
\]

Thus
\[
n=7.
\]

So there were 7 teams.

ANSWER 3: B

Problem 4:
Ara and Shea were originally the same height. Let that original height be \(h\) inches.

Shea has grown 20%, so Shea’s current height is
\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:
\[
1.20h=60.
\]

Solve for \(h\):
\[
h=\frac{60}{1.20}=50.
\]

So they were originally 50 inches tall. Shea grew
\[
60-50=10
\]
inches.

Ara grew half as many inches as Shea:
\[
\frac{10}{2}=5
\]
inches.

Therefore Ara is now
\[
50+5=55
\]
inches tall.

ANSWER 4: E

Problem 5:
A fair six-sided die has faces numbered 1 through 6. When it is tossed, one face is hidden on the bottom, and the other five faces are visible.

We need the probability that the product of the five visible numbers is divisible by 6.

A number is divisible by 6 if it has at least one factor of 2 and at least one factor of 3.

No matter which single face is hidden:

- There are three even faces: 2, 4, and 6. Hiding only one of them still leaves at least one even face visible.
- There are two multiples of 3: 3 and 6. Hiding only one of them still leaves at least one multiple of 3 visible.

So the product of the visible faces always contains both a factor of 2 and a factor of 3. Therefore it is always divisible by 6.

Thus the probability is
\[
1.
\]

ANSWER 5: E

Problem 6:
Tori answered:

- \(70\%\) of 10 arithmetic problems:
\[
0.70\cdot 10=7
\]
- \(40\%\) of 30 algebra problems:
\[
0.40\cdot 30=12
\]
- \(60\%\) of 35 geometry problems:
\[
0.60\cdot 35=21
\]

Total correct:
\[
7+12+21=40.
\]

There were 75 problems total. To pass, she needed \(60\%\) correct:
\[
0.60\cdot 75=45.
\]

She answered 40 correctly, so she needed
\[
45-40=5
\]
more correct answers.

ANSWER 6: B

Problem 7:
We are given
\[
0<a<b<c.
\]

We need the impossible statement.

Choice A says
\[
a+c<b.
\]

But since \(c>b\) and \(a>0\),
\[
a+c>b.
\]

So \(a+c<b\) cannot happen.

We can also check that the other choices are possible:

- B: \(a\cdot b<c\) can happen, for example \(a=1,b=2,c=100\).
- C: \(a+b<c\) can happen, for example \(a=1,b=2,c=100\).
- D: \(a\cdot c<b\) can happen, for example \(a=0.1,b=2,c=3\), since \(0.3<2\).
- E: If interpreted as \(bc=a\), it can happen, for example \(a=0.1,b=0.2,c=0.5\), since \(0.2\cdot 0.5=0.1\).

Thus the impossible choice is A.

ANSWER 7: A

Problem 8:
One recipe makes 15 cookies and uses 3 tablespoons of butter. Only full recipes can be made.

Walter and Gretel need 216 cookies. The number of full recipes needed is
\[
\left\lceil \frac{216}{15}\right\rceil.
\]

Since
\[
\frac{216}{15}=14.4,
\]
they must make 15 full recipes.

Butter needed:
\[
15\cdot 3=45
\]
tablespoons.

There are 8 tablespoons in a stick of butter, so the number of sticks needed is
\[
\left\lceil \frac{45}{8}\right\rceil.
\]

Since
\[
\frac{45}{8}=5.625,
\]
they need 6 sticks of butter.

ANSWER 8: B

Problem 9:
A 3-inch cube made from 27 one-inch cubes has surface area
\[
6\cdot 3^2=54
\]
square inches.

Each small cube can show:

- 0 faces if it is the center cube,
- 1 face if it is a face-center cube,
- 2 faces if it is an edge cube,
- 3 faces if it is a corner cube.

We have 6 white cubes. To minimize the white surface area, we should hide as many white cubes as possible.

There is exactly one interior cube, so put one white cube there. It shows 0 square inches.

The remaining 5 white cubes must be on the surface. To minimize exposed white area, put them on face-center positions, where each shows exactly 1 square inch.

So the minimum white surface area is
\[
1\cdot 0+5\cdot 1=5
\]
square inches.

The fraction of the surface area that is white is
\[
\frac{5}{54}.
\]

ANSWER 9: A

Problem 10:
The population is 200 in 1998, and it triples every 25 years.

From 1998 to 2050 is
\[
2050-1998=52
\]
years.

Two full 25-year periods are 50 years, so after 50 years the population is
\[
200\cdot 3\cdot 3=200\cdot 9=1800.
\]

The extra 2 years would add a relatively small amount, so the population in 2050 is a little above 1800. Among the choices, the best estimate is 2000.

ANSWER 10: D

# Problem 1:
Write out the place values of 74982.1035:
- 7 → ten-thousands, 4 → thousands, **9 → hundreds**, 8 → tens, 2 → ones
- 1 → tenths, 0 → hundredths, **3 → thousandths**, 5 → ten-thousandths

So the 9 sits in the hundreds place (value 100) and the 3 sits in the thousandths place (value 0.001).

Ratio = 100 ÷ 0.001 = 100,000.

(Answers like 1,000 or 10,000 come from miscounting decimal places — the 3 is the *third* digit after the point, not the first or second.)

ANSWER 1: C

# Problem 2:
Temperature goes from 32° down to 20°, a drop of 12°.

Each 3° change corresponds to 4 cm³ of change: 12 ÷ 3 = 4 steps.

Volume change = 4 × 4 = 16 cm³ decrease (cooling means shrinking).

Volume = 24 − 16 = 8 cm³.

(40 is the trap from *adding* instead of subtracting; 16 is just the change, not the volume.)

ANSWER 2: A

# Problem 3:
With n teams each playing every other team once, the number of games is C(n,2) = n(n−1)/2.

Set n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7·6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 games — neither is 21.

ANSWER 3: B

# Problem 4:
Let the common original height be x. Shea grew 20%: 1.20x = 60 → x = 50.

So Shea grew 60 − 50 = 10 inches.

Ara grew half as many inches: 5 inches.

Ara's height = 50 + 5 = 55 inches.

(54 would result from wrongly taking Shea's growth as 10% of 60, etc.)

ANSWER 4: E

# Problem 5:
The product of all six faces is 1·2·3·4·5·6 = 720. If the hidden bottom face is k, the visible product is 720/k.

- k=1: 720 ✓ divisible by 6
- k=2: 360 ✓
- k=3: 240 ✓
- k=4: 180 ✓
- k=5: 144 ✓
- k=6: 120 ✓

Reason why: hiding one face still leaves at least one even number and at least one multiple of 3 among the five visible faces (e.g., hide 6 → 2 and 3 remain; hide 3 → 6 remains).

So the probability is 6/6 = 1.

ANSWER 5: E

# Problem 6:
Correct answers:
- Arithmetic: 70% of 10 = 7
- Algebra: 40% of 30 = 12
- Geometry: 60% of 35 = 21

Total correct = 7 + 12 + 21 = 40.

Needed for 60%: 0.60 × 75 = 45.

She needs 45 − 40 = 5 more.

(40/75 ≈ 53%, consistent with failing.)

ANSWER 6: B

# Problem 7:
Given 0 < a < b < c.

- **A. a + c < b:** Since c > b and a > 0, we get a + c > b. So a + c < b is impossible.
- B. a·b < c: take a = 0.1, b = 0.2, c = 1 → 0.02 < 1 ✓ possible.
- C. a + b < c: take a = 1, b = 2, c = 10 → 3 < 10 ✓ possible.
- D. a·c < b: take a = 0.1, b = 1, c = 2 → 0.2 < 1 ✓ possible.
- E. b/c = a: take a = 0.5, b = 1, c = 2 → 1/2 = 0.5 ✓ possible.

ANSWER 7: A

# Problem 8:
Pans needed: 216 ÷ 15 = 14.4, and only full recipes are made, so they must bake 15 pans.

Butter: 15 × 3 = 45 tablespoons.

Sticks: 45 ÷ 8 = 5.625, and butter comes in whole sticks, so 6 sticks.

(5 sticks = 40 tbsp, not enough.)

ANSWER 8: B

# Problem 9:
The 3×3×3 cube has surface area 6 × 9 = 54 unit squares.

Positions in a 3-cube: 1 center cube (0 faces showing), 6 face-centers (1 face each), 12 edges (2 faces), 8 corners (3 faces).

To minimize white showing, place one white cube in the hidden center (0 white squares) and the other 5 white cubes at face-center positions (1 white square each).

Minimum white area = 5 unit squares out of 54 → 5/54.

ANSWER 9: A

# Problem 10:
From 1998 to 2050 is 52 years — about two 25-year tripling periods (1998 → 2023 → 2048).

Population: 200 → 600 → 1800.

1800 is closest to 2000.

ANSWER 10: D

Problem 1:

We are comparing the place values of the digits 9 and 3 in

\[
74982.1035.
\]

- The digit \(9\) is in the hundreds place, so its place value is \(100\).
- The digit \(3\) is in the thousandths place, so its place value is

\[
0.001.
\]

Now find how many times as great \(100\) is as \(0.001\):

\[
\frac{100}{0.001}=100,000.
\]

So the hundreds place is \(100,\!000\) times the thousandths place.

- A, \(1,\!000\), is too small.
- B, \(10,\!000\), is too small.
- D and E are too large.

ANSWER 1: C

Problem 2:

The gas expands by \(4\) cubic centimeters for every \(3^\circ\) increase in temperature.

The temperature changes from \(32^\circ\) down to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

Since

\[
12^\circ \div 3^\circ=4,
\]

there are 4 groups of \(3^\circ\).

For each group, the volume decreases by \(4\) cubic centimeters:

\[
4\cdot 4=16.
\]

At \(32^\circ\), the volume is \(24\) cubic centimeters, so at \(20^\circ\) it was

\[
24-16=8.
\]

Choices B, C, and D do not subtract the full \(16\) cubic centimeters. Choice E would represent an increase rather than a decrease.

ANSWER 2: A

Problem 3:

If there are \(n\) teams and each pair of teams plays exactly once, then the number of games is

\[
\binom n2=\frac{n(n-1)}2.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

Checking: with 7 teams, the number of games is

\[
\frac{7\cdot6}{2}=21.
\]

Six teams would produce only \(15\) games, while 8 teams would produce \(28\) games.

ANSWER 3: B

Problem 4:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew \(20\%\), so Shea’s current height is

\[
1.20h=60.
\]

Thus,

\[
h=\frac{60}{1.2}=50.
\]

So both were originally 50 inches tall.

Shea’s growth was

\[
60-50=10
\]

inches.

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5
\]

inches.

Therefore, Ara is now

\[
50+5=55
\]

inches tall.

Choice A, 48, is less than Ara’s original height. Choices B, C, and D do not include the correct 5-inch growth.

ANSWER 4: E

Problem 5:

A standard die has faces numbered \(1,2,3,4,5,6\). The product of all six face numbers is

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6=720.
\]

If the bottom face has number \(k\), then the product of the five visible faces is

\[
\frac{720}{k}.
\]

Check each possible bottom face:

\[
\frac{720}{1}=720,\qquad
\frac{720}{2}=360,\qquad
\frac{720}{3}=240,
\]

\[
\frac{720}{4}=180,\qquad
\frac{720}{5}=144,\qquad
\frac{720}{6}=120.
\]

Every one of these numbers is divisible by \(6\).

Therefore, no matter which face is on the bottom, the visible-face product is divisible by 6. The probability is

\[
1.
\]

All of choices A through D are less than 1, but the event always happens.

ANSWER 5: E

Problem 6:

Tori answered correctly:

- Arithmetic:

\[
70\%\text{ of }10=0.7(10)=7
\]

- Algebra:

\[
40\%\text{ of }30=0.4(30)=12
\]

- Geometry:

\[
60\%\text{ of }35=0.6(35)=21
\]

So her total number correct was

\[
7+12+21=40.
\]

To pass, she needs \(60\%\) of 75 questions correct:

\[
0.60(75)=45.
\]

She needs

\[
45-40=5
\]

more correct answers.

Choice A is too small, while choices C, D, and E give more than the number needed to reach exactly 45 correct.

ANSWER 6: B

Problem 7:

We are given

\[
0<a<b<c.
\]

We must find the statement that is impossible.

Consider choice A:

\[
a+c<b.
\]

But \(c>b\), and also \(a>0\). Therefore,

\[
a+c>c>b.
\]

So it is impossible for \(a+c<b\).

The other choices can happen for suitable positive numbers. For example:

- B: \(a\cdot b<c\) can occur with \(a=1,\ b=2,\ c=3\), since \(1\cdot2<3\).
- C: \(a+b<c\) can occur with \(a=1,\ b=2,\ c=4\).
- D: \(a\cdot c<b\) can occur with \(a=\frac14,\ b=1,\ c=2\), since \(\frac14\cdot2=\frac12<1\).
- E: \(bc=a\) can occur with \(b=\frac12,\ c=\frac34\), giving \(a=\frac38\), and indeed

\[
0<\frac38<\frac12<\frac34.
\]

Thus only choice A is impossible.

ANSWER 7: A

Problem 8:

Each full recipe makes 15 cookies.

Walter and Gretel need at least 216 cookies. Since only full recipes may be made, determine how many recipes are needed:

\[
216\div 15=14.4.
\]

They cannot make \(14.4\) recipes, so they must make 15 full recipes.

These 15 recipes make

\[
15\cdot15=225
\]

cookies, enough for the crowd.

Each recipe uses 3 tablespoons of butter, so the total butter needed is

\[
15\cdot3=45
\]

tablespoons.

There are 8 tablespoons in each stick, so

\[
45\div8=5.625.
\]

They must buy 6 full sticks of butter.

Five sticks provide only \(5\cdot8=40\) tablespoons, which is not enough. Six sticks provide \(48\) tablespoons, enough for the 45 tablespoons needed.

ANSWER 8: B

Problem 9:

A \(3\times3\times3\) cube has surface area

\[
6\cdot 3\cdot3=54
\]

square inches, since each of its 6 faces has area \(3\cdot3=9\).

To minimize the white surface showing, place the white cubes in the least exposed positions.

- The one center cube of the large cube has no faces exposed.
- The six face-center cubes each have exactly one face exposed.
- Edge cubes have two exposed faces.
- Corner cubes have three exposed faces.

There are 6 white cubes. Put one white cube in the completely hidden center. Then put the other 5 white cubes at five face-center locations.

Thus, the white cubes show only

\[
5
\]

unit square faces on the surface.

Therefore, the fraction of the surface that is white is

\[
\frac{5}{54}.
\]

Choices such as \(\frac19,\frac{5}{27},\frac29,\frac13\) represent more than 5 white unit squares out of the 54 total surface squares.

ANSWER 9: A

Problem 10:

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

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is approximately two 25-year periods.

After one 25-year period:

\[
200\cdot3=600.
\]

After two 25-year periods:

\[
600\cdot3=1800.
\]

Since 2050 is slightly more than 50 years after 1998, the population should be slightly more than 1800. The closest estimate is

\[
20

Problem 1:
We are asked: in 74982.1035, compare the place value of the 9 with the place value of the 3.

Write out places:
74982 is: 7 ten-thousands, 4 thousands, 9 hundreds, 8 tens, 2 ones.
So the 9 is in the hundreds place. Its value is 100.

After the decimal .1035 is: 1 tenths, 0 hundredths, 3 thousandths, 5 ten-thousandths.
So the 3 is in the thousandths place. Its value is 0.001 = 1/1000.

Ratio = (value of 9's place) / (value of 3's place)
= 100 / 0.001 = 100 / (1/1000) = 100 x 1000 = 100,000.

So the hundreds place is 100,000 times as great as the thousandths place.
Choices A,B,D,E are powers of 10 that are too small or too large.

ANSWER 1: C

Problem 2:
We are asked for the earlier volume given expansion rate.

Rate: +3 degrees temperature = +4 cubic cm volume.

Now: temp 32°, volume 24 cc.
Before: temp 20°.

Temperature change from 20° to 32° is 32-20 = 12°.
Number of 3-degree steps in 12° is 12 / 3 = 4 steps.

So volume increase from 20° to 32° is 4 steps x 4 cc = 16 cc.

Thus volume at 20° + 16 = volume at 32° = 24.
So volume at 20° = 24 - 16 = 8 cc.

Check: A 8 fits, B 12 would be only 3 steps, D 16 would be only 2 steps, E 40 would be adding instead of subtracting.

ANSWER 2: A

Problem 3:
We need n teams such that every pair plays once gives 21 games.

Number of pairs from n teams is n choose 2 = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try integers: 6x5=30 too small, 7x6=42 correct, 8x7=56 too big.
So n = 7.

Ruling out: A 6 gives only 15 games, C 8 gives 28 games, etc.

ANSWER 3: B

Problem 4:
Ara and Shea started at same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara's now height.

Let original height = H.
Shea's growth = 20% of H = 0.20 H.
Shea now = H + 0.20H = 1.2H = 60.
So H = 60 / 1.2 = 50 inches.

Shea grew 60-50 = 10 inches.
Ara grew half as many inches: 10 / 2 = 5 inches.
Ara now = 50 + 5 = 55 inches.

Tempting wrong A 48 would be working backwards, D 54 would be misreading percent, etc.

ANSWER 4: E

Problem 5:
A standard 1-6 die is tossed. Bottom face hidden, other 5 faces seen. Probability product of seen numbers divisible by 6?

Product divisible by 6 means divisible by 2 and by 3.

Numbers 1 to 6: evens are 2,4,6 — three of them. Multiples of 3 are 3,6 — two of them.

Hiding only one face (the bottom) can hide at most one even, leaving at least two evens still visible, so factor 2 remains. It can hide at most one multiple of 3, leaving at least one multiple of 3 still visible, so factor 3 remains.

Indeed total product 1·2·3·4·5·6 = 720. Visible product = 720 / b where b is bottom. 720/b divided by 6 = 120/b, and b=1,2,3,4,5,6 all divide 120, so always integer.

Hence no matter which face is hidden, the visible product is divisible by 6. Probability = 6/6 = 1.

Other choices 1/3,1/2,2/3,5/6 assume sometimes it fails, which is impossible.

ANSWER 5: E

Problem 6:
75 problems total: 10 arithmetic, 30 algebra, 35 geometry.
Correct: 70% of arithmetic, 40% of algebra, 60% of geometry. How many more to reach 60% overall?

Compute:
Arithmetic correct = 0.70 x 10 = 7.
Algebra correct = 0.40 x 30 = 12.
Geometry correct = 0.60 x 35 = 21.
Total correct = 7+12+21 = 40.

Need for 60% passing: 0.60 x 75 = 45 problems.

Shortfall = 45 - 40 = 5 more problems.

So B 5, not A 1 or C 7.

ANSWER 6: B

Problem 7:
Given 0 < a < b < c, which statement is impossible?

A. a + c < b. Since a > 0, a + c > c. And c > b. So a+c > b always. Hence a+c < b can never happen. This is impossible.

Check others are possible with examples:
B. a·b < c: e.g. a=1,b=2,c=10 gives 2<10 possible.
C. a+b < c: e.g. a=1,b=2,c=10 gives 3<10 possible.
D. a·c < b: with fractions <1. e.g. a=0.1,c=2,b=1 gives 0.2<1 and 0.1<1<2 possible.
E. b c = a: let b=0.25,a=0.1, then c=a/b=0.4, and 0.1<0.25<0.4 possible.

So only A is impossible. Students may think E impossible because b c > b > a for numbers >1, but with numbers <1 multiplication can shrink.

ANSWER 7: A

Problem 8:
One full recipe = 15 cookies = 3 tbsp butter. Need 216 cookies. 8 tbsp = 1 stick. How many sticks?

Pans needed: 216 / 15 = 14.4. Since only full recipes, round up to 15 pans. 14 pans would give only 210 cookies, not enough.

Butter needed: 15 x 3 tbsp = 45 tbsp.

Sticks needed: 45 / 8 = 5.625. Since need whole sticks to have enough, round up to 6. 5 sticks = 40 tbsp < 45 tbsp, not enough.

So answer 6. A 5 would leave butter and cookies short.

ANSWER 8: B

Problem 9:
Build 3x3x3 = 27 small cubes. 6 white, 21 red. Minimize white squares on outside. What fraction of outside is white?

Big cube surface: 6 faces each 3x3 = 9 small squares, total 54 small squares.

Small-cube positions by exposure:
- 1 interior center: 0 faces shown
- 6 face-centers: 1 face shown
- 12 edge non-corner: 2 faces shown
- 8 corners: 3 faces shown

To minimize white, put whites where least exposure:
Put 1 white in center → 0 exposed.
Put remaining 5 whites in face-center spots → 1 each exposed.
Total white exposed squares = 5.

Fraction white = 5 / 54 of total surface area.

Larger fractions like 1/9=6/54, etc. would result from worse placement.

ANSWER 9: A

Problem 10:
1998 population 200, triples every 25 years. Estimate 2050.

Time elapsed: 2050-1998 = 52 years.
Number of 25-year periods: 52/25 ≈ 2.08, just over 2 doublings-triplings.

After 25 years (~2023): 200x3=600.
After 50 years (~2048): 600x3=1800.
2050 is 2 years later, a little more, so about 2000.

Formally 200 x 3^(52/25) ≈ 200 x 9 x small extra ≈ 1960.

So A 600 is only one tripling, C 1000 too low, E 3000 too high. Best is D 2000.

ANSWER 10: D

Problem 1:
We are asked to find the ratio of the value of the place occupied by the digit 9 to the value of the place occupied by the digit 3 in the decimal number $74982.1035$.

1. Identify the place values:
   - The number is $74982.1035$.
   - The digit 2 is in the ones place ($10^0$).
   - The digit 8 is in the tens place ($10^1$).
   - The digit 9 is in the hundreds place ($10^2 = 100$).
   - The digit 1 is in the tenths place ($10^{-1} = 0.1$).
   - The digit 0 is in the hundredths place ($10^{-2} = 0.01$).
   - The digit 3 is in the thousandths place ($10^{-3} = 0.001$).
   - The digit 5 is in the ten-thousandths place ($10^{-4} = 0.0001$).

2. Compute the ratio of the place values:
   $$\frac{\text{Place value of } 9}{\text{Place value of } 3} = \frac{100}{0.001} = \frac{10^2}{10^{-3}} = 10^{2 - (-3)} = 10^5 = 100,000.$$

This corresponds to choice C.

ANSWER 1: C

---

Problem 2:
We are asked to find the initial volume of gas at $20^\circ$ given its volume at $32^\circ$ and its rate of expansion.

1. Find the change in temperature:
   The temperature changes from $20^\circ$ to $32^\circ$, which is an increase of:
   $$32^\circ - 20^\circ = 12^\circ.$$

2. Determine the volume expansion:
   For every $3^\circ$ rise, the volume expands by $4\text{ cm}^3$.
   In a $12^\circ$ rise, the number of $3^\circ$ intervals is:
   $$\frac{12^\circ}{3^\circ} = 4.$$
   Therefore, the volume expanded by:
   $$4 \times 4\text{ cm}^3 = 16\text{ cm}^3.$$

3. Calculate the original volume at $20^\circ$:
   $$\text{Volume at } 20^\circ = \text{Volume at } 32^\circ - 16\text{ cm}^3 = 24 - 16 = 8\text{ cm}^3.$$

This matches choice A.

ANSWER 2: A

---

Problem 3:
We are given that each of the $N$ teams plays every other team exactly once, resulting in 21 total games. We need to find $N$.

1. The number of games played between $N$ teams is the number of ways to choose 2 teams from $N$, given by the combination formula:
   $$\binom{N}{2} = \frac{N(N-1)}{2}.$$

2. Set this equal to 21 and solve for $N$:
   $$\frac{N(N-1)}{2} = 21 \implies N(N-1) = 42.$$

3. Find two consecutive positive integers whose product is 42:
   Since $7 \times 6 = 42$, we have $N = 7$.

Checking the choices, $N = 7$ corresponds to choice B.

ANSWER 3: B

---

Problem 4:
We are given:
- Ara and Shea initially had the same height, say $h$ inches.
- Shea has grown $20\%$, so Shea's new height is $h + 0.20h = 1.20h = 60$ inches.
- Ara has grown half as many inches as Shea.
- We need to find Ara's current height.

1. Find the original height $h$:
   $$1.20h = 60 \implies h = \frac{60}{1.2} = 50\text{ inches}.$$

2. Find how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10\text{ inches}.$$

3. Find Ara's growth:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5\text{ inches}.$$

4. Find Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55\text{ inches}.$$

This corresponds to choice E.

ANSWER 4: E

---

Problem 5:
A standard fair six-sided die has faces with numbers 1, 2, 3, 4, 5, and 6.
When the die is tossed, one face is on the bottom and cannot be seen, while the other 5 faces are visible.
We want the product of the 5 visible faces to be divisible by 6.

1. The product of all six numbers on the die is:
   $$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720.$$
   The product of the 5 visible faces is:
   $$P = \frac{720}{b},$$
   where $b \in \{1, 2, 3, 4, 5, 6\}$ is the number on the bottom face.

2. A number is divisible by 6 if and only if it is a multiple of both 2 and 3.
   Let's check the product $P$ for each possible bottom face $b$:
   - If $b = 1$, visible faces are $\{2, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 2$, visible faces are $\{1, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 3$, visible faces are $\{1, 2, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 4$, visible faces are $\{1, 2, 3, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 5$, visible faces are $\{1, 2, 3, 4, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 6$, visible faces are $\{1, 2, 3, 4, 5\}$. The product is $1 \times 2 \times 3 \times 4 \times 5 = 120$. Since $120 = 6 \times 20$, 120 is also divisible by 6 (it has factors 2 and 3)!

3. In every single case (all 6 possible outcomes for the bottom face), the product of the five visible faces is divisible by 6.
   Therefore, the probability is:
   $$\frac{6}{6} = 1.$$

This corresponds to choice E.

ANSWER 5: E

---

Problem 6:
Tori took a 75-problem test:
- 10 arithmetic problems: she answered $70\%$ correctly.
  $$\text{Correct arithmetic} = 0.70 \times 10 = 7.$$
- 30 algebra problems: she answered $40\%$ correctly.
  $$\text{Correct algebra} = 0.40 \times 30 = 12.$$
- 35 geometry problems: she answered $60\%$ correctly.
  $$\text{Correct geometry} = 0.60 \times 35 = 21.$$

1. Total problems Tori answered correctly:
   $$\text{Total correct} = 7 + 12 + 21 = 40.$$

2. Passing score required:
   $60\%$ of 75 problems:
   $$\text{Passing score} = 0.60 \times 75 = 45.$$

3. Additional problems needed:
   $$45 - 40 = 5\text{ problems}.$$

This corresponds to choice B.

ANSWER 6: B

---

Problem 7:
We are given real numbers $a, b, c$ such that $0 < a < b < c$.
We need to determine which of the statements is impossible.

Let's test each choice:
- Choice A: $a + c < b$.
  Since $a > 0$, adding $a$ to $c$ gives $a + c > c$.
  Since we are given $c > b$, it follows that $a + c > c > b$, which means $a + c > b$ must always be true!
  Therefore, $a + c < b$ is impossible.

Let's verify that the other choices can be possible with appropriate values:
- Choice B: $a \cdot b < c$. If $a = 1, b = 2, c = 3$, then $a \cdot b = 2 < 3 = c$. (Possible)
- Choice C: $a + b < c$. If $a = 1, b = 2, c = 4$, then $a + b = 3 < 4 = c$. (Possible)
- Choice D: $a \cdot c < b$. If $a = 0.1, b = 2, c = 3$, then $a \cdot c = 0.3 < 2 = b$. (Possible)
-

# Problem 1:

**Restate:** In 74982.1035, compare the place value of the digit 9 with the place value of the digit 3.

**Working:** Let's identify each place:
- The digit 9 sits in the **hundreds** place: 9 × 100 = 900. Its place value is **100**.
- The digit 3 sits in the **hundredths** place (two positions right of the decimal point): 3 × 0.01 = 0.03. Its place value is **0.01**.

**Compute the ratio:**
$$\frac{100}{0.01} = 100 \div \frac{1}{100} = 100 \times 100 = 10{,}000$$

**Check wrong choices:** Counting places instead of computing the ratio might suggest 100,000 (if you miscount the decimal places), but from hundreds to hundredths is exactly 4 positions, and each position is a factor of 10, giving 10⁴ = 10,000.

ANSWER 1: B

# Problem 2:

**Restate:** Volume increases 4 cm³ per 3° rise. Volume is 24 cm³ at 32°. Find the volume at 20°.

**Working:** The temperature dropped from 32° to 20°:
$$32 - 20 = 12°$$
Number of 3° intervals in the drop:
$$12 \div 3 = 4 \text{ intervals}$$
Each 3° drop means the gas **shrinks** by 4 cm³, so the total shrink is:
$$4 \times 4 = 16 \text{ cm}^3$$
The volume at 20° was:
$$24 - 16 = 8 \text{ cm}^3$$

**Check wrong choices:** Choice D (16) is the *amount* of shrink, not the volume — a tempting trap. Choice E (40) comes from adding instead of subtracting.

ANSWER 2: A

# Problem 3:

**Restate:** Each team plays every other team once; 21 total games. Find the number of teams.

**Working:** With n teams, the number of games is:
$$\frac{n(n-1)}{2}$$
Set this equal to 21:
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42$$
Testing n = 7: $7 \times 6 = 42$. ✓

**Check wrong choices:** n = 6 gives 15 games (too few); n = 8 gives 28 games (too many). Only n = 7 works.

ANSWER 3: B

# Problem 4:

**Restate:** Ara and Shea were once equal in height. Shea grew 20% and is now 60 inches. Ara grew half as many inches as Shea. Find Ara's height now.

**Working:** Let the original height be $h$. Shea grew 20%, so:
$$h + 0.20h = 60 \implies 1.2h = 60 \implies h = 50 \text{ inches}$$
Shea's growth: $60 - 50 = 10$ inches.
Ara grew half as many inches as Shea:
$$\frac{10}{2} = 5 \text{ inches}$$
Ara's height now:
$$50 + 5 = 55 \text{ inches}$$

**Check wrong choices:** Choice A (48) comes from computing 20% of 60 incorrectly; choice D (54) would result if Ara also grew 20% of the original height. The key is that Ara grew 5 **inches**, not a percentage.

ANSWER 4: E

# Problem 5:

**Restate:** A die is tossed; the bottom face is hidden. Find the probability that the product of the five visible faces is divisible by 6.

**Working:** The product of all six faces is:
$$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$$
The product of the five visible faces is:
$$\frac{720}{\text{hidden face}}$$

Check each possible hidden face:
- Hidden = 1: product = 720 → divisible by 6 ✓
- Hidden = 2: product = 360 → divisible by 6 ✓
- Hidden = 3: product = 240 → divisible by 6 ✓
- Hidden = 4: product = 180 → divisible by 6 ✓
- Hidden = 5: product = 144 → divisible by 6 ✓
- Hidden = 6: product = 120 → divisible by 6 ✓

All six outcomes give a product divisible by 6, so the probability is:
$$\frac{6}{6} = 1$$

**Check wrong choices:** One might think hiding the 5 or the 6 removes needed factors, but 720 contains 2 × 3 many times over — removing any single face still leaves factors of 2 and 3 in the product.

ANSWER 5: E

# Problem 6:

**Restate:** Tori's test: 10 arithmetic, 30 algebra, 35 geometry problems (75 total). She got 70%, 40%, and 60% correct respectively, which was below 60% overall. How many more correct answers were needed for 60%?

**Working:** Correct answers in each category:
- Arithmetic: $0.70 \times 10 = 7$
- Algebra: $0.40 \times 30 = 12$
- Geometry: $0.60 \times 35 = 21$

Total correct: $7 + 12 + 21 = 40$

To pass, she needs 60% of 75:
$$0.60 \times 75 = 45$$

Additional problems needed:
$$45 - 40 = 5$$

**Check wrong choices:** Note that the weighted average $\frac{40}{75} \approx 53.3\%$ is indeed below 60%, consistent with the story. A tempting trap is to average the three percentages: $\frac{70+40+60}{3} \approx 56.7\%$, which is wrong because the categories have different sizes.

ANSWER 6: B

# Problem 7:

**Restate:** Given $0 < a < b < c$, which statement is impossible?

**Working:** Test each option:

**A. $a + c < b$:** Since $c > b$, adding positive $a$ gives $a + c > c > b$. So $a + c < b$ **cannot happen**. ✓ (impossible)

Let's verify the others are all possible with suitable numbers:

**B. $a \cdot b < c$:** Try $a = 0.1, b = 0.2, c = 1$: $0.02 < 1$ ✓ possible.

**C. $a + b < c$:** Try $a = 1, b = 2, c = 10$: $3 < 10$ ✓ possible.

**D. $a \cdot c < b$:** Try $a = 0.1, b = 0.5, c = 0.6$: $0.06 < 0.5$ ✓ possible.

**E. $bc = a$:** Try $a = 0.06, b = 0.3, c = 0.2$... wait, need $b < c$. Try $a = 0.02, b = 0.1, c = 0.2$: $bc = 0.02 = a$, and $0 < 0.02 < 0.1 < 0.2$ ✓ possible.

Only option A forces a contradiction: $a + c > c > b$ always holds.

ANSWER 7: A

# Problem 8:

**Restate:** One recipe = 15 cookies using 3 tablespoons of butter; only full recipes allowed. Need 216 cookies. 8 tablespoons = 1 stick. How many sticks?

**Working:** Number of recipes needed:
$$\frac{216}{15} = 14.4$$
Since only full recipes are made, they must make **15 recipes** (14 recipes × 15 = 210 cookies, which is not enough).

Butter needed:
$$15 \times 3 = 45 \text{ tablespoons}$$
Sticks needed:
$$\frac{45}{8} = 5.625 \implies \text{need } 6 \text{ full sticks}$$

**Check wrong choices:** Choice A (5) is the trap — it uses only 40 tablespoons, but 45 are required. You can't buy a partial stick, so round up to 6.

ANSWER 8: B

# Problem 9:

**Restate:** Build a 3-inch cube from 27 unit cubes: 21 red, 6 white. Arrange to minimize white surface area. What fraction of the surface area is white?

**Working:** The 3×3×3 cube has total surface area:
$$6 \text{ faces} \times 9 \text{ sq in} = 54 \text{ square inches}$$

Classify the 27 positions by how many faces each shows:
- **1 inte

Problem 1:
We need the ratio of the place values of the digits 9 and 3 in  
\(74982.1035\).

The digit 9 is in the hundreds place, so its place value is  
\[
100.
\]

The digit 3 is in the thousandths place, so its place value is  
\[
0.001.
\]

Now compute the ratio:
\[
\frac{100}{0.001}=100{,}000.
\]

So the value of the place of the 9 is \(100{,}000\) times as great as the value of the place of the 3.

ANSWER 1: C

Problem 2:
We are told that for every \(3^\circ\) rise in temperature, the gas expands by 4 cubic centimeters.

At temperature \(32^\circ\), the volume is 24 cubic centimeters. We want the volume at \(20^\circ\).

The temperature difference is
\[
32-20=12^\circ.
\]

From \(20^\circ\) to \(32^\circ\), the temperature rises by \(12^\circ\). Since each \(3^\circ\) rise increases the volume by 4 cubic centimeters,
\[
\frac{12}{3}\cdot 4=4\cdot 4=16
\]
cubic centimeters.

So at \(20^\circ\), the volume was
\[
24-16=8
\]
cubic centimeters.

ANSWER 2: A

Problem 3:
Each team plays every other team exactly once. If there are \(n\) teams, the number of games is
\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games, so
\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:
\[
n(n-1)=42.
\]

We look for two consecutive positive integers whose product is 42:
\[
6\cdot 7=42.
\]

Thus
\[
n=7.
\]

So there were 7 teams.

ANSWER 3: B

Problem 4:
Ara and Shea were originally the same height. Let that original height be \(h\) inches.

Shea has grown 20%, so Shea’s current height is
\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:
\[
1.20h=60.
\]

Solve for \(h\):
\[
h=\frac{60}{1.20}=50.
\]

So they were originally 50 inches tall. Shea grew
\[
60-50=10
\]
inches.

Ara grew half as many inches as Shea:
\[
\frac{10}{2}=5
\]
inches.

Therefore Ara is now
\[
50+5=55
\]
inches tall.

ANSWER 4: E

Problem 5:
A fair six-sided die has faces numbered 1 through 6. When it is tossed, one face is hidden on the bottom, and the other five faces are visible.

We need the probability that the product of the five visible numbers is divisible by 6.

A number is divisible by 6 if it has at least one factor of 2 and at least one factor of 3.

No matter which single face is hidden:

- There are three even faces: 2, 4, and 6. Hiding only one of them still leaves at least one even face visible.
- There are two multiples of 3: 3 and 6. Hiding only one of them still leaves at least one multiple of 3 visible.

So the product of the visible faces always contains both a factor of 2 and a factor of 3. Therefore it is always divisible by 6.

Thus the probability is
\[
1.
\]

ANSWER 5: E

Problem 6:
Tori answered:

- \(70\%\) of 10 arithmetic problems:
\[
0.70\cdot 10=7
\]
- \(40\%\) of 30 algebra problems:
\[
0.40\cdot 30=12
\]
- \(60\%\) of 35 geometry problems:
\[
0.60\cdot 35=21
\]

Total correct:
\[
7+12+21=40.
\]

There were 75 problems total. To pass, she needed \(60\%\) correct:
\[
0.60\cdot 75=45.
\]

She answered 40 correctly, so she needed
\[
45-40=5
\]
more correct answers.

ANSWER 6: B

Problem 7:
We are given
\[
0<a<b<c.
\]

We need the impossible statement.

Choice A says
\[
a+c<b.
\]

But since \(c>b\) and \(a>0\),
\[
a+c>b.
\]

So \(a+c<b\) cannot happen.

We can also check that the other choices are possible:

- B: \(a\cdot b<c\) can happen, for example \(a=1,b=2,c=100\).
- C: \(a+b<c\) can happen, for example \(a=1,b=2,c=100\).
- D: \(a\cdot c<b\) can happen, for example \(a=0.1,b=2,c=3\), since \(0.3<2\).
- E: If interpreted as \(bc=a\), it can happen, for example \(a=0.1,b=0.2,c=0.5\), since \(0.2\cdot 0.5=0.1\).

Thus the impossible choice is A.

ANSWER 7: A

Problem 8:
One recipe makes 15 cookies and uses 3 tablespoons of butter. Only full recipes can be made.

Walter and Gretel need 216 cookies. The number of full recipes needed is
\[
\left\lceil \frac{216}{15}\right\rceil.
\]

Since
\[
\frac{216}{15}=14.4,
\]
they must make 15 full recipes.

Butter needed:
\[
15\cdot 3=45
\]
tablespoons.

There are 8 tablespoons in a stick of butter, so the number of sticks needed is
\[
\left\lceil \frac{45}{8}\right\rceil.
\]

Since
\[
\frac{45}{8}=5.625,
\]
they need 6 sticks of butter.

ANSWER 8: B

Problem 9:
A 3-inch cube made from 27 one-inch cubes has surface area
\[
6\cdot 3^2=54
\]
square inches.

Each small cube can show:

- 0 faces if it is the center cube,
- 1 face if it is a face-center cube,
- 2 faces if it is an edge cube,
- 3 faces if it is a corner cube.

We have 6 white cubes. To minimize the white surface area, we should hide as many white cubes as possible.

There is exactly one interior cube, so put one white cube there. It shows 0 square inches.

The remaining 5 white cubes must be on the surface. To minimize exposed white area, put them on face-center positions, where each shows exactly 1 square inch.

So the minimum white surface area is
\[
1\cdot 0+5\cdot 1=5
\]
square inches.

The fraction of the surface area that is white is
\[
\frac{5}{54}.
\]

ANSWER 9: A

Problem 10:
The population is 200 in 1998, and it triples every 25 years.

From 1998 to 2050 is
\[
2050-1998=52
\]
years.

Two full 25-year periods are 50 years, so after 50 years the population is
\[
200\cdot 3\cdot 3=200\cdot 9=1800.
\]

The extra 2 years would add a relatively small amount, so the population in 2050 is a little above 1800. Among the choices, the best estimate is 2000.

ANSWER 10: D

# Problem 1:
Write out the place values of 74982.1035:
- 7 → ten-thousands, 4 → thousands, **9 → hundreds**, 8 → tens, 2 → ones
- 1 → tenths, 0 → hundredths, **3 → thousandths**, 5 → ten-thousandths

So the 9 sits in the hundreds place (value 100) and the 3 sits in the thousandths place (value 0.001).

Ratio = 100 ÷ 0.001 = 100,000.

(Answers like 1,000 or 10,000 come from miscounting decimal places — the 3 is the *third* digit after the point, not the first or second.)

ANSWER 1: C

# Problem 2:
Temperature goes from 32° down to 20°, a drop of 12°.

Each 3° change corresponds to 4 cm³ of change: 12 ÷ 3 = 4 steps.

Volume change = 4 × 4 = 16 cm³ decrease (cooling means shrinking).

Volume = 24 − 16 = 8 cm³.

(40 is the trap from *adding* instead of subtracting; 16 is just the change, not the volume.)

ANSWER 2: A

# Problem 3:
With n teams each playing every other team once, the number of games is C(n,2) = n(n−1)/2.

Set n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7·6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 games — neither is 21.

ANSWER 3: B

# Problem 4:
Let the common original height be x. Shea grew 20%: 1.20x = 60 → x = 50.

So Shea grew 60 − 50 = 10 inches.

Ara grew half as many inches: 5 inches.

Ara's height = 50 + 5 = 55 inches.

(54 would result from wrongly taking Shea's growth as 10% of 60, etc.)

ANSWER 4: E

# Problem 5:
The product of all six faces is 1·2·3·4·5·6 = 720. If the hidden bottom face is k, the visible product is 720/k.

- k=1: 720 ✓ divisible by 6
- k=2: 360 ✓
- k=3: 240 ✓
- k=4: 180 ✓
- k=5: 144 ✓
- k=6: 120 ✓

Reason why: hiding one face still leaves at least one even number and at least one multiple of 3 among the five visible faces (e.g., hide 6 → 2 and 3 remain; hide 3 → 6 remains).

So the probability is 6/6 = 1.

ANSWER 5: E

# Problem 6:
Correct answers:
- Arithmetic: 70% of 10 = 7
- Algebra: 40% of 30 = 12
- Geometry: 60% of 35 = 21

Total correct = 7 + 12 + 21 = 40.

Needed for 60%: 0.60 × 75 = 45.

She needs 45 − 40 = 5 more.

(40/75 ≈ 53%, consistent with failing.)

ANSWER 6: B

# Problem 7:
Given 0 < a < b < c.

- **A. a + c < b:** Since c > b and a > 0, we get a + c > b. So a + c < b is impossible.
- B. a·b < c: take a = 0.1, b = 0.2, c = 1 → 0.02 < 1 ✓ possible.
- C. a + b < c: take a = 1, b = 2, c = 10 → 3 < 10 ✓ possible.
- D. a·c < b: take a = 0.1, b = 1, c = 2 → 0.2 < 1 ✓ possible.
- E. b/c = a: take a = 0.5, b = 1, c = 2 → 1/2 = 0.5 ✓ possible.

ANSWER 7: A

# Problem 8:
Pans needed: 216 ÷ 15 = 14.4, and only full recipes are made, so they must bake 15 pans.

Butter: 15 × 3 = 45 tablespoons.

Sticks: 45 ÷ 8 = 5.625, and butter comes in whole sticks, so 6 sticks.

(5 sticks = 40 tbsp, not enough.)

ANSWER 8: B

# Problem 9:
The 3×3×3 cube has surface area 6 × 9 = 54 unit squares.

Positions in a 3-cube: 1 center cube (0 faces showing), 6 face-centers (1 face each), 12 edges (2 faces), 8 corners (3 faces).

To minimize white showing, place one white cube in the hidden center (0 white squares) and the other 5 white cubes at face-center positions (1 white square each).

Minimum white area = 5 unit squares out of 54 → 5/54.

ANSWER 9: A

# Problem 10:
From 1998 to 2050 is 52 years — about two 25-year tripling periods (1998 → 2023 → 2048).

Population: 200 → 600 → 1800.

1800 is closest to 2000.

ANSWER 10: D

Problem 1:

We are comparing the place values of the digits 9 and 3 in

\[
74982.1035.
\]

- The digit \(9\) is in the hundreds place, so its place value is \(100\).
- The digit \(3\) is in the thousandths place, so its place value is

\[
0.001.
\]

Now find how many times as great \(100\) is as \(0.001\):

\[
\frac{100}{0.001}=100,000.
\]

So the hundreds place is \(100,\!000\) times the thousandths place.

- A, \(1,\!000\), is too small.
- B, \(10,\!000\), is too small.
- D and E are too large.

ANSWER 1: C

Problem 2:

The gas expands by \(4\) cubic centimeters for every \(3^\circ\) increase in temperature.

The temperature changes from \(32^\circ\) down to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

Since

\[
12^\circ \div 3^\circ=4,
\]

there are 4 groups of \(3^\circ\).

For each group, the volume decreases by \(4\) cubic centimeters:

\[
4\cdot 4=16.
\]

At \(32^\circ\), the volume is \(24\) cubic centimeters, so at \(20^\circ\) it was

\[
24-16=8.
\]

Choices B, C, and D do not subtract the full \(16\) cubic centimeters. Choice E would represent an increase rather than a decrease.

ANSWER 2: A

Problem 3:

If there are \(n\) teams and each pair of teams plays exactly once, then the number of games is

\[
\binom n2=\frac{n(n-1)}2.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

Checking: with 7 teams, the number of games is

\[
\frac{7\cdot6}{2}=21.
\]

Six teams would produce only \(15\) games, while 8 teams would produce \(28\) games.

ANSWER 3: B

Problem 4:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew \(20\%\), so Shea’s current height is

\[
1.20h=60.
\]

Thus,

\[
h=\frac{60}{1.2}=50.
\]

So both were originally 50 inches tall.

Shea’s growth was

\[
60-50=10
\]

inches.

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5
\]

inches.

Therefore, Ara is now

\[
50+5=55
\]

inches tall.

Choice A, 48, is less than Ara’s original height. Choices B, C, and D do not include the correct 5-inch growth.

ANSWER 4: E

Problem 5:

A standard die has faces numbered \(1,2,3,4,5,6\). The product of all six face numbers is

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6=720.
\]

If the bottom face has number \(k\), then the product of the five visible faces is

\[
\frac{720}{k}.
\]

Check each possible bottom face:

\[
\frac{720}{1}=720,\qquad
\frac{720}{2}=360,\qquad
\frac{720}{3}=240,
\]

\[
\frac{720}{4}=180,\qquad
\frac{720}{5}=144,\qquad
\frac{720}{6}=120.
\]

Every one of these numbers is divisible by \(6\).

Therefore, no matter which face is on the bottom, the visible-face product is divisible by 6. The probability is

\[
1.
\]

All of choices A through D are less than 1, but the event always happens.

ANSWER 5: E

Problem 6:

Tori answered correctly:

- Arithmetic:

\[
70\%\text{ of }10=0.7(10)=7
\]

- Algebra:

\[
40\%\text{ of }30=0.4(30)=12
\]

- Geometry:

\[
60\%\text{ of }35=0.6(35)=21
\]

So her total number correct was

\[
7+12+21=40.
\]

To pass, she needs \(60\%\) of 75 questions correct:

\[
0.60(75)=45.
\]

She needs

\[
45-40=5
\]

more correct answers.

Choice A is too small, while choices C, D, and E give more than the number needed to reach exactly 45 correct.

ANSWER 6: B

Problem 7:

We are given

\[
0<a<b<c.
\]

We must find the statement that is impossible.

Consider choice A:

\[
a+c<b.
\]

But \(c>b\), and also \(a>0\). Therefore,

\[
a+c>c>b.
\]

So it is impossible for \(a+c<b\).

The other choices can happen for suitable positive numbers. For example:

- B: \(a\cdot b<c\) can occur with \(a=1,\ b=2,\ c=3\), since \(1\cdot2<3\).
- C: \(a+b<c\) can occur with \(a=1,\ b=2,\ c=4\).
- D: \(a\cdot c<b\) can occur with \(a=\frac14,\ b=1,\ c=2\), since \(\frac14\cdot2=\frac12<1\).
- E: \(bc=a\) can occur with \(b=\frac12,\ c=\frac34\), giving \(a=\frac38\), and indeed

\[
0<\frac38<\frac12<\frac34.
\]

Thus only choice A is impossible.

ANSWER 7: A

Problem 8:

Each full recipe makes 15 cookies.

Walter and Gretel need at least 216 cookies. Since only full recipes may be made, determine how many recipes are needed:

\[
216\div 15=14.4.
\]

They cannot make \(14.4\) recipes, so they must make 15 full recipes.

These 15 recipes make

\[
15\cdot15=225
\]

cookies, enough for the crowd.

Each recipe uses 3 tablespoons of butter, so the total butter needed is

\[
15\cdot3=45
\]

tablespoons.

There are 8 tablespoons in each stick, so

\[
45\div8=5.625.
\]

They must buy 6 full sticks of butter.

Five sticks provide only \(5\cdot8=40\) tablespoons, which is not enough. Six sticks provide \(48\) tablespoons, enough for the 45 tablespoons needed.

ANSWER 8: B

Problem 9:

A \(3\times3\times3\) cube has surface area

\[
6\cdot 3\cdot3=54
\]

square inches, since each of its 6 faces has area \(3\cdot3=9\).

To minimize the white surface showing, place the white cubes in the least exposed positions.

- The one center cube of the large cube has no faces exposed.
- The six face-center cubes each have exactly one face exposed.
- Edge cubes have two exposed faces.
- Corner cubes have three exposed faces.

There are 6 white cubes. Put one white cube in the completely hidden center. Then put the other 5 white cubes at five face-center locations.

Thus, the white cubes show only

\[
5
\]

unit square faces on the surface.

Therefore, the fraction of the surface that is white is

\[
\frac{5}{54}.
\]

Choices such as \(\frac19,\frac{5}{27},\frac29,\frac13\) represent more than 5 white unit squares out of the 54 total surface squares.

ANSWER 9: A

Problem 10:

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

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is approximately two 25-year periods.

After one 25-year period:

\[
200\cdot3=600.
\]

After two 25-year periods:

\[
600\cdot3=1800.
\]

Since 2050 is slightly more than 50 years after 1998, the population should be slightly more than 1800. The closest estimate is

\[
20

Problem 1:
We are asked: in 74982.1035, compare the place value of the 9 with the place value of the 3.

Write out places:
74982 is: 7 ten-thousands, 4 thousands, 9 hundreds, 8 tens, 2 ones.
So the 9 is in the hundreds place. Its value is 100.

After the decimal .1035 is: 1 tenths, 0 hundredths, 3 thousandths, 5 ten-thousandths.
So the 3 is in the thousandths place. Its value is 0.001 = 1/1000.

Ratio = (value of 9's place) / (value of 3's place)
= 100 / 0.001 = 100 / (1/1000) = 100 x 1000 = 100,000.

So the hundreds place is 100,000 times as great as the thousandths place.
Choices A,B,D,E are powers of 10 that are too small or too large.

ANSWER 1: C

Problem 2:
We are asked for the earlier volume given expansion rate.

Rate: +3 degrees temperature = +4 cubic cm volume.

Now: temp 32°, volume 24 cc.
Before: temp 20°.

Temperature change from 20° to 32° is 32-20 = 12°.
Number of 3-degree steps in 12° is 12 / 3 = 4 steps.

So volume increase from 20° to 32° is 4 steps x 4 cc = 16 cc.

Thus volume at 20° + 16 = volume at 32° = 24.
So volume at 20° = 24 - 16 = 8 cc.

Check: A 8 fits, B 12 would be only 3 steps, D 16 would be only 2 steps, E 40 would be adding instead of subtracting.

ANSWER 2: A

Problem 3:
We need n teams such that every pair plays once gives 21 games.

Number of pairs from n teams is n choose 2 = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try integers: 6x5=30 too small, 7x6=42 correct, 8x7=56 too big.
So n = 7.

Ruling out: A 6 gives only 15 games, C 8 gives 28 games, etc.

ANSWER 3: B

Problem 4:
Ara and Shea started at same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara's now height.

Let original height = H.
Shea's growth = 20% of H = 0.20 H.
Shea now = H + 0.20H = 1.2H = 60.
So H = 60 / 1.2 = 50 inches.

Shea grew 60-50 = 10 inches.
Ara grew half as many inches: 10 / 2 = 5 inches.
Ara now = 50 + 5 = 55 inches.

Tempting wrong A 48 would be working backwards, D 54 would be misreading percent, etc.

ANSWER 4: E

Problem 5:
A standard 1-6 die is tossed. Bottom face hidden, other 5 faces seen. Probability product of seen numbers divisible by 6?

Product divisible by 6 means divisible by 2 and by 3.

Numbers 1 to 6: evens are 2,4,6 — three of them. Multiples of 3 are 3,6 — two of them.

Hiding only one face (the bottom) can hide at most one even, leaving at least two evens still visible, so factor 2 remains. It can hide at most one multiple of 3, leaving at least one multiple of 3 still visible, so factor 3 remains.

Indeed total product 1·2·3·4·5·6 = 720. Visible product = 720 / b where b is bottom. 720/b divided by 6 = 120/b, and b=1,2,3,4,5,6 all divide 120, so always integer.

Hence no matter which face is hidden, the visible product is divisible by 6. Probability = 6/6 = 1.

Other choices 1/3,1/2,2/3,5/6 assume sometimes it fails, which is impossible.

ANSWER 5: E

Problem 6:
75 problems total: 10 arithmetic, 30 algebra, 35 geometry.
Correct: 70% of arithmetic, 40% of algebra, 60% of geometry. How many more to reach 60% overall?

Compute:
Arithmetic correct = 0.70 x 10 = 7.
Algebra correct = 0.40 x 30 = 12.
Geometry correct = 0.60 x 35 = 21.
Total correct = 7+12+21 = 40.

Need for 60% passing: 0.60 x 75 = 45 problems.

Shortfall = 45 - 40 = 5 more problems.

So B 5, not A 1 or C 7.

ANSWER 6: B

Problem 7:
Given 0 < a < b < c, which statement is impossible?

A. a + c < b. Since a > 0, a + c > c. And c > b. So a+c > b always. Hence a+c < b can never happen. This is impossible.

Check others are possible with examples:
B. a·b < c: e.g. a=1,b=2,c=10 gives 2<10 possible.
C. a+b < c: e.g. a=1,b=2,c=10 gives 3<10 possible.
D. a·c < b: with fractions <1. e.g. a=0.1,c=2,b=1 gives 0.2<1 and 0.1<1<2 possible.
E. b c = a: let b=0.25,a=0.1, then c=a/b=0.4, and 0.1<0.25<0.4 possible.

So only A is impossible. Students may think E impossible because b c > b > a for numbers >1, but with numbers <1 multiplication can shrink.

ANSWER 7: A

Problem 8:
One full recipe = 15 cookies = 3 tbsp butter. Need 216 cookies. 8 tbsp = 1 stick. How many sticks?

Pans needed: 216 / 15 = 14.4. Since only full recipes, round up to 15 pans. 14 pans would give only 210 cookies, not enough.

Butter needed: 15 x 3 tbsp = 45 tbsp.

Sticks needed: 45 / 8 = 5.625. Since need whole sticks to have enough, round up to 6. 5 sticks = 40 tbsp < 45 tbsp, not enough.

So answer 6. A 5 would leave butter and cookies short.

ANSWER 8: B

Problem 9:
Build 3x3x3 = 27 small cubes. 6 white, 21 red. Minimize white squares on outside. What fraction of outside is white?

Big cube surface: 6 faces each 3x3 = 9 small squares, total 54 small squares.

Small-cube positions by exposure:
- 1 interior center: 0 faces shown
- 6 face-centers: 1 face shown
- 12 edge non-corner: 2 faces shown
- 8 corners: 3 faces shown

To minimize white, put whites where least exposure:
Put 1 white in center → 0 exposed.
Put remaining 5 whites in face-center spots → 1 each exposed.
Total white exposed squares = 5.

Fraction white = 5 / 54 of total surface area.

Larger fractions like 1/9=6/54, etc. would result from worse placement.

ANSWER 9: A

Problem 10:
1998 population 200, triples every 25 years. Estimate 2050.

Time elapsed: 2050-1998 = 52 years.
Number of 25-year periods: 52/25 ≈ 2.08, just over 2 doublings-triplings.

After 25 years (~2023): 200x3=600.
After 50 years (~2048): 600x3=1800.
2050 is 2 years later, a little more, so about 2000.

Formally 200 x 3^(52/25) ≈ 200 x 9 x small extra ≈ 1960.

So A 600 is only one tripling, C 1000 too low, E 3000 too high. Best is D 2000.

ANSWER 10: D

Problem 1:
We are asked to find the ratio of the value of the place occupied by the digit 9 to the value of the place occupied by the digit 3 in the decimal number $74982.1035$.

1. Identify the place values:
   - The number is $74982.1035$.
   - The digit 2 is in the ones place ($10^0$).
   - The digit 8 is in the tens place ($10^1$).
   - The digit 9 is in the hundreds place ($10^2 = 100$).
   - The digit 1 is in the tenths place ($10^{-1} = 0.1$).
   - The digit 0 is in the hundredths place ($10^{-2} = 0.01$).
   - The digit 3 is in the thousandths place ($10^{-3} = 0.001$).
   - The digit 5 is in the ten-thousandths place ($10^{-4} = 0.0001$).

2. Compute the ratio of the place values:
   $$\frac{\text{Place value of } 9}{\text{Place value of } 3} = \frac{100}{0.001} = \frac{10^2}{10^{-3}} = 10^{2 - (-3)} = 10^5 = 100,000.$$

This corresponds to choice C.

ANSWER 1: C

---

Problem 2:
We are asked to find the initial volume of gas at $20^\circ$ given its volume at $32^\circ$ and its rate of expansion.

1. Find the change in temperature:
   The temperature changes from $20^\circ$ to $32^\circ$, which is an increase of:
   $$32^\circ - 20^\circ = 12^\circ.$$

2. Determine the volume expansion:
   For every $3^\circ$ rise, the volume expands by $4\text{ cm}^3$.
   In a $12^\circ$ rise, the number of $3^\circ$ intervals is:
   $$\frac{12^\circ}{3^\circ} = 4.$$
   Therefore, the volume expanded by:
   $$4 \times 4\text{ cm}^3 = 16\text{ cm}^3.$$

3. Calculate the original volume at $20^\circ$:
   $$\text{Volume at } 20^\circ = \text{Volume at } 32^\circ - 16\text{ cm}^3 = 24 - 16 = 8\text{ cm}^3.$$

This matches choice A.

ANSWER 2: A

---

Problem 3:
We are given that each of the $N$ teams plays every other team exactly once, resulting in 21 total games. We need to find $N$.

1. The number of games played between $N$ teams is the number of ways to choose 2 teams from $N$, given by the combination formula:
   $$\binom{N}{2} = \frac{N(N-1)}{2}.$$

2. Set this equal to 21 and solve for $N$:
   $$\frac{N(N-1)}{2} = 21 \implies N(N-1) = 42.$$

3. Find two consecutive positive integers whose product is 42:
   Since $7 \times 6 = 42$, we have $N = 7$.

Checking the choices, $N = 7$ corresponds to choice B.

ANSWER 3: B

---

Problem 4:
We are given:
- Ara and Shea initially had the same height, say $h$ inches.
- Shea has grown $20\%$, so Shea's new height is $h + 0.20h = 1.20h = 60$ inches.
- Ara has grown half as many inches as Shea.
- We need to find Ara's current height.

1. Find the original height $h$:
   $$1.20h = 60 \implies h = \frac{60}{1.2} = 50\text{ inches}.$$

2. Find how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10\text{ inches}.$$

3. Find Ara's growth:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5\text{ inches}.$$

4. Find Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55\text{ inches}.$$

This corresponds to choice E.

ANSWER 4: E

---

Problem 5:
A standard fair six-sided die has faces with numbers 1, 2, 3, 4, 5, and 6.
When the die is tossed, one face is on the bottom and cannot be seen, while the other 5 faces are visible.
We want the product of the 5 visible faces to be divisible by 6.

1. The product of all six numbers on the die is:
   $$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720.$$
   The product of the 5 visible faces is:
   $$P = \frac{720}{b},$$
   where $b \in \{1, 2, 3, 4, 5, 6\}$ is the number on the bottom face.

2. A number is divisible by 6 if and only if it is a multiple of both 2 and 3.
   Let's check the product $P$ for each possible bottom face $b$:
   - If $b = 1$, visible faces are $\{2, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 2$, visible faces are $\{1, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 3$, visible faces are $\{1, 2, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 4$, visible faces are $\{1, 2, 3, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 5$, visible faces are $\{1, 2, 3, 4, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 6$, visible faces are $\{1, 2, 3, 4, 5\}$. The product is $1 \times 2 \times 3 \times 4 \times 5 = 120$. Since $120 = 6 \times 20$, 120 is also divisible by 6 (it has factors 2 and 3)!

3. In every single case (all 6 possible outcomes for the bottom face), the product of the five visible faces is divisible by 6.
   Therefore, the probability is:
   $$\frac{6}{6} = 1.$$

This corresponds to choice E.

ANSWER 5: E

---

Problem 6:
Tori took a 75-problem test:
- 10 arithmetic problems: she answered $70\%$ correctly.
  $$\text{Correct arithmetic} = 0.70 \times 10 = 7.$$
- 30 algebra problems: she answered $40\%$ correctly.
  $$\text{Correct algebra} = 0.40 \times 30 = 12.$$
- 35 geometry problems: she answered $60\%$ correctly.
  $$\text{Correct geometry} = 0.60 \times 35 = 21.$$

1. Total problems Tori answered correctly:
   $$\text{Total correct} = 7 + 12 + 21 = 40.$$

2. Passing score required:
   $60\%$ of 75 problems:
   $$\text{Passing score} = 0.60 \times 75 = 45.$$

3. Additional problems needed:
   $$45 - 40 = 5\text{ problems}.$$

This corresponds to choice B.

ANSWER 6: B

---

Problem 7:
We are given real numbers $a, b, c$ such that $0 < a < b < c$.
We need to determine which of the statements is impossible.

Let's test each choice:
- Choice A: $a + c < b$.
  Since $a > 0$, adding $a$ to $c$ gives $a + c > c$.
  Since we are given $c > b$, it follows that $a + c > c > b$, which means $a + c > b$ must always be true!
  Therefore, $a + c < b$ is impossible.

Let's verify that the other choices can be possible with appropriate values:
- Choice B: $a \cdot b < c$. If $a = 1, b = 2, c = 3$, then $a \cdot b = 2 < 3 = c$. (Possible)
- Choice C: $a + b < c$. If $a = 1, b = 2, c = 4$, then $a + b = 3 < 4 = c$. (Possible)
- Choice D: $a \cdot c < b$. If $a = 0.1, b = 2, c = 3$, then $a \cdot c = 0.3 < 2 = b$. (Possible)
-

# Problem 1:

**Restate:** In 74982.1035, compare the place value of the digit 9 with the place value of the digit 3.

**Working:** Let's identify each place:
- The digit 9 sits in the **hundreds** place: 9 × 100 = 900. Its place value is **100**.
- The digit 3 sits in the **hundredths** place (two positions right of the decimal point): 3 × 0.01 = 0.03. Its place value is **0.01**.

**Compute the ratio:**
$$\frac{100}{0.01} = 100 \div \frac{1}{100} = 100 \times 100 = 10{,}000$$

**Check wrong choices:** Counting places instead of computing the ratio might suggest 100,000 (if you miscount the decimal places), but from hundreds to hundredths is exactly 4 positions, and each position is a factor of 10, giving 10⁴ = 10,000.

ANSWER 1: B

# Problem 2:

**Restate:** Volume increases 4 cm³ per 3° rise. Volume is 24 cm³ at 32°. Find the volume at 20°.

**Working:** The temperature dropped from 32° to 20°:
$$32 - 20 = 12°$$
Number of 3° intervals in the drop:
$$12 \div 3 = 4 \text{ intervals}$$
Each 3° drop means the gas **shrinks** by 4 cm³, so the total shrink is:
$$4 \times 4 = 16 \text{ cm}^3$$
The volume at 20° was:
$$24 - 16 = 8 \text{ cm}^3$$

**Check wrong choices:** Choice D (16) is the *amount* of shrink, not the volume — a tempting trap. Choice E (40) comes from adding instead of subtracting.

ANSWER 2: A

# Problem 3:

**Restate:** Each team plays every other team once; 21 total games. Find the number of teams.

**Working:** With n teams, the number of games is:
$$\frac{n(n-1)}{2}$$
Set this equal to 21:
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42$$
Testing n = 7: $7 \times 6 = 42$. ✓

**Check wrong choices:** n = 6 gives 15 games (too few); n = 8 gives 28 games (too many). Only n = 7 works.

ANSWER 3: B

# Problem 4:

**Restate:** Ara and Shea were once equal in height. Shea grew 20% and is now 60 inches. Ara grew half as many inches as Shea. Find Ara's height now.

**Working:** Let the original height be $h$. Shea grew 20%, so:
$$h + 0.20h = 60 \implies 1.2h = 60 \implies h = 50 \text{ inches}$$
Shea's growth: $60 - 50 = 10$ inches.
Ara grew half as many inches as Shea:
$$\frac{10}{2} = 5 \text{ inches}$$
Ara's height now:
$$50 + 5 = 55 \text{ inches}$$

**Check wrong choices:** Choice A (48) comes from computing 20% of 60 incorrectly; choice D (54) would result if Ara also grew 20% of the original height. The key is that Ara grew 5 **inches**, not a percentage.

ANSWER 4: E

# Problem 5:

**Restate:** A die is tossed; the bottom face is hidden. Find the probability that the product of the five visible faces is divisible by 6.

**Working:** The product of all six faces is:
$$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$$
The product of the five visible faces is:
$$\frac{720}{\text{hidden face}}$$

Check each possible hidden face:
- Hidden = 1: product = 720 → divisible by 6 ✓
- Hidden = 2: product = 360 → divisible by 6 ✓
- Hidden = 3: product = 240 → divisible by 6 ✓
- Hidden = 4: product = 180 → divisible by 6 ✓
- Hidden = 5: product = 144 → divisible by 6 ✓
- Hidden = 6: product = 120 → divisible by 6 ✓

All six outcomes give a product divisible by 6, so the probability is:
$$\frac{6}{6} = 1$$

**Check wrong choices:** One might think hiding the 5 or the 6 removes needed factors, but 720 contains 2 × 3 many times over — removing any single face still leaves factors of 2 and 3 in the product.

ANSWER 5: E

# Problem 6:

**Restate:** Tori's test: 10 arithmetic, 30 algebra, 35 geometry problems (75 total). She got 70%, 40%, and 60% correct respectively, which was below 60% overall. How many more correct answers were needed for 60%?

**Working:** Correct answers in each category:
- Arithmetic: $0.70 \times 10 = 7$
- Algebra: $0.40 \times 30 = 12$
- Geometry: $0.60 \times 35 = 21$

Total correct: $7 + 12 + 21 = 40$

To pass, she needs 60% of 75:
$$0.60 \times 75 = 45$$

Additional problems needed:
$$45 - 40 = 5$$

**Check wrong choices:** Note that the weighted average $\frac{40}{75} \approx 53.3\%$ is indeed below 60%, consistent with the story. A tempting trap is to average the three percentages: $\frac{70+40+60}{3} \approx 56.7\%$, which is wrong because the categories have different sizes.

ANSWER 6: B

# Problem 7:

**Restate:** Given $0 < a < b < c$, which statement is impossible?

**Working:** Test each option:

**A. $a + c < b$:** Since $c > b$, adding positive $a$ gives $a + c > c > b$. So $a + c < b$ **cannot happen**. ✓ (impossible)

Let's verify the others are all possible with suitable numbers:

**B. $a \cdot b < c$:** Try $a = 0.1, b = 0.2, c = 1$: $0.02 < 1$ ✓ possible.

**C. $a + b < c$:** Try $a = 1, b = 2, c = 10$: $3 < 10$ ✓ possible.

**D. $a \cdot c < b$:** Try $a = 0.1, b = 0.5, c = 0.6$: $0.06 < 0.5$ ✓ possible.

**E. $bc = a$:** Try $a = 0.06, b = 0.3, c = 0.2$... wait, need $b < c$. Try $a = 0.02, b = 0.1, c = 0.2$: $bc = 0.02 = a$, and $0 < 0.02 < 0.1 < 0.2$ ✓ possible.

Only option A forces a contradiction: $a + c > c > b$ always holds.

ANSWER 7: A

# Problem 8:

**Restate:** One recipe = 15 cookies using 3 tablespoons of butter; only full recipes allowed. Need 216 cookies. 8 tablespoons = 1 stick. How many sticks?

**Working:** Number of recipes needed:
$$\frac{216}{15} = 14.4$$
Since only full recipes are made, they must make **15 recipes** (14 recipes × 15 = 210 cookies, which is not enough).

Butter needed:
$$15 \times 3 = 45 \text{ tablespoons}$$
Sticks needed:
$$\frac{45}{8} = 5.625 \implies \text{need } 6 \text{ full sticks}$$

**Check wrong choices:** Choice A (5) is the trap — it uses only 40 tablespoons, but 45 are required. You can't buy a partial stick, so round up to 6.

ANSWER 8: B

# Problem 9:

**Restate:** Build a 3-inch cube from 27 unit cubes: 21 red, 6 white. Arrange to minimize white surface area. What fraction of the surface area is white?

**Working:** The 3×3×3 cube has total surface area:
$$6 \text{ faces} \times 9 \text{ sq in} = 54 \text{ square inches}$$

Classify the 27 positions by how many faces each shows:
- **1 inte

Problem 1:
We need the ratio of the place values of the digits 9 and 3 in  
\(74982.1035\).

The digit 9 is in the hundreds place, so its place value is  
\[
100.
\]

The digit 3 is in the thousandths place, so its place value is  
\[
0.001.
\]

Now compute the ratio:
\[
\frac{100}{0.001}=100{,}000.
\]

So the value of the place of the 9 is \(100{,}000\) times as great as the value of the place of the 3.

ANSWER 1: C

Problem 2:
We are told that for every \(3^\circ\) rise in temperature, the gas expands by 4 cubic centimeters.

At temperature \(32^\circ\), the volume is 24 cubic centimeters. We want the volume at \(20^\circ\).

The temperature difference is
\[
32-20=12^\circ.
\]

From \(20^\circ\) to \(32^\circ\), the temperature rises by \(12^\circ\). Since each \(3^\circ\) rise increases the volume by 4 cubic centimeters,
\[
\frac{12}{3}\cdot 4=4\cdot 4=16
\]
cubic centimeters.

So at \(20^\circ\), the volume was
\[
24-16=8
\]
cubic centimeters.

ANSWER 2: A

Problem 3:
Each team plays every other team exactly once. If there are \(n\) teams, the number of games is
\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games, so
\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:
\[
n(n-1)=42.
\]

We look for two consecutive positive integers whose product is 42:
\[
6\cdot 7=42.
\]

Thus
\[
n=7.
\]

So there were 7 teams.

ANSWER 3: B

Problem 4:
Ara and Shea were originally the same height. Let that original height be \(h\) inches.

Shea has grown 20%, so Shea’s current height is
\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:
\[
1.20h=60.
\]

Solve for \(h\):
\[
h=\frac{60}{1.20}=50.
\]

So they were originally 50 inches tall. Shea grew
\[
60-50=10
\]
inches.

Ara grew half as many inches as Shea:
\[
\frac{10}{2}=5
\]
inches.

Therefore Ara is now
\[
50+5=55
\]
inches tall.

ANSWER 4: E

Problem 5:
A fair six-sided die has faces numbered 1 through 6. When it is tossed, one face is hidden on the bottom, and the other five faces are visible.

We need the probability that the product of the five visible numbers is divisible by 6.

A number is divisible by 6 if it has at least one factor of 2 and at least one factor of 3.

No matter which single face is hidden:

- There are three even faces: 2, 4, and 6. Hiding only one of them still leaves at least one even face visible.
- There are two multiples of 3: 3 and 6. Hiding only one of them still leaves at least one multiple of 3 visible.

So the product of the visible faces always contains both a factor of 2 and a factor of 3. Therefore it is always divisible by 6.

Thus the probability is
\[
1.
\]

ANSWER 5: E

Problem 6:
Tori answered:

- \(70\%\) of 10 arithmetic problems:
\[
0.70\cdot 10=7
\]
- \(40\%\) of 30 algebra problems:
\[
0.40\cdot 30=12
\]
- \(60\%\) of 35 geometry problems:
\[
0.60\cdot 35=21
\]

Total correct:
\[
7+12+21=40.
\]

There were 75 problems total. To pass, she needed \(60\%\) correct:
\[
0.60\cdot 75=45.
\]

She answered 40 correctly, so she needed
\[
45-40=5
\]
more correct answers.

ANSWER 6: B

Problem 7:
We are given
\[
0<a<b<c.
\]

We need the impossible statement.

Choice A says
\[
a+c<b.
\]

But since \(c>b\) and \(a>0\),
\[
a+c>b.
\]

So \(a+c<b\) cannot happen.

We can also check that the other choices are possible:

- B: \(a\cdot b<c\) can happen, for example \(a=1,b=2,c=100\).
- C: \(a+b<c\) can happen, for example \(a=1,b=2,c=100\).
- D: \(a\cdot c<b\) can happen, for example \(a=0.1,b=2,c=3\), since \(0.3<2\).
- E: If interpreted as \(bc=a\), it can happen, for example \(a=0.1,b=0.2,c=0.5\), since \(0.2\cdot 0.5=0.1\).

Thus the impossible choice is A.

ANSWER 7: A

Problem 8:
One recipe makes 15 cookies and uses 3 tablespoons of butter. Only full recipes can be made.

Walter and Gretel need 216 cookies. The number of full recipes needed is
\[
\left\lceil \frac{216}{15}\right\rceil.
\]

Since
\[
\frac{216}{15}=14.4,
\]
they must make 15 full recipes.

Butter needed:
\[
15\cdot 3=45
\]
tablespoons.

There are 8 tablespoons in a stick of butter, so the number of sticks needed is
\[
\left\lceil \frac{45}{8}\right\rceil.
\]

Since
\[
\frac{45}{8}=5.625,
\]
they need 6 sticks of butter.

ANSWER 8: B

Problem 9:
A 3-inch cube made from 27 one-inch cubes has surface area
\[
6\cdot 3^2=54
\]
square inches.

Each small cube can show:

- 0 faces if it is the center cube,
- 1 face if it is a face-center cube,
- 2 faces if it is an edge cube,
- 3 faces if it is a corner cube.

We have 6 white cubes. To minimize the white surface area, we should hide as many white cubes as possible.

There is exactly one interior cube, so put one white cube there. It shows 0 square inches.

The remaining 5 white cubes must be on the surface. To minimize exposed white area, put them on face-center positions, where each shows exactly 1 square inch.

So the minimum white surface area is
\[
1\cdot 0+5\cdot 1=5
\]
square inches.

The fraction of the surface area that is white is
\[
\frac{5}{54}.
\]

ANSWER 9: A

Problem 10:
The population is 200 in 1998, and it triples every 25 years.

From 1998 to 2050 is
\[
2050-1998=52
\]
years.

Two full 25-year periods are 50 years, so after 50 years the population is
\[
200\cdot 3\cdot 3=200\cdot 9=1800.
\]

The extra 2 years would add a relatively small amount, so the population in 2050 is a little above 1800. Among the choices, the best estimate is 2000.

ANSWER 10: D

# Problem 1:
Write out the place values of 74982.1035:
- 7 → ten-thousands, 4 → thousands, **9 → hundreds**, 8 → tens, 2 → ones
- 1 → tenths, 0 → hundredths, **3 → thousandths**, 5 → ten-thousandths

So the 9 sits in the hundreds place (value 100) and the 3 sits in the thousandths place (value 0.001).

Ratio = 100 ÷ 0.001 = 100,000.

(Answers like 1,000 or 10,000 come from miscounting decimal places — the 3 is the *third* digit after the point, not the first or second.)

ANSWER 1: C

# Problem 2:
Temperature goes from 32° down to 20°, a drop of 12°.

Each 3° change corresponds to 4 cm³ of change: 12 ÷ 3 = 4 steps.

Volume change = 4 × 4 = 16 cm³ decrease (cooling means shrinking).

Volume = 24 − 16 = 8 cm³.

(40 is the trap from *adding* instead of subtracting; 16 is just the change, not the volume.)

ANSWER 2: A

# Problem 3:
With n teams each playing every other team once, the number of games is C(n,2) = n(n−1)/2.

Set n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7·6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 games — neither is 21.

ANSWER 3: B

# Problem 4:
Let the common original height be x. Shea grew 20%: 1.20x = 60 → x = 50.

So Shea grew 60 − 50 = 10 inches.

Ara grew half as many inches: 5 inches.

Ara's height = 50 + 5 = 55 inches.

(54 would result from wrongly taking Shea's growth as 10% of 60, etc.)

ANSWER 4: E

# Problem 5:
The product of all six faces is 1·2·3·4·5·6 = 720. If the hidden bottom face is k, the visible product is 720/k.

- k=1: 720 ✓ divisible by 6
- k=2: 360 ✓
- k=3: 240 ✓
- k=4: 180 ✓
- k=5: 144 ✓
- k=6: 120 ✓

Reason why: hiding one face still leaves at least one even number and at least one multiple of 3 among the five visible faces (e.g., hide 6 → 2 and 3 remain; hide 3 → 6 remains).

So the probability is 6/6 = 1.

ANSWER 5: E

# Problem 6:
Correct answers:
- Arithmetic: 70% of 10 = 7
- Algebra: 40% of 30 = 12
- Geometry: 60% of 35 = 21

Total correct = 7 + 12 + 21 = 40.

Needed for 60%: 0.60 × 75 = 45.

She needs 45 − 40 = 5 more.

(40/75 ≈ 53%, consistent with failing.)

ANSWER 6: B

# Problem 7:
Given 0 < a < b < c.

- **A. a + c < b:** Since c > b and a > 0, we get a + c > b. So a + c < b is impossible.
- B. a·b < c: take a = 0.1, b = 0.2, c = 1 → 0.02 < 1 ✓ possible.
- C. a + b < c: take a = 1, b = 2, c = 10 → 3 < 10 ✓ possible.
- D. a·c < b: take a = 0.1, b = 1, c = 2 → 0.2 < 1 ✓ possible.
- E. b/c = a: take a = 0.5, b = 1, c = 2 → 1/2 = 0.5 ✓ possible.

ANSWER 7: A

# Problem 8:
Pans needed: 216 ÷ 15 = 14.4, and only full recipes are made, so they must bake 15 pans.

Butter: 15 × 3 = 45 tablespoons.

Sticks: 45 ÷ 8 = 5.625, and butter comes in whole sticks, so 6 sticks.

(5 sticks = 40 tbsp, not enough.)

ANSWER 8: B

# Problem 9:
The 3×3×3 cube has surface area 6 × 9 = 54 unit squares.

Positions in a 3-cube: 1 center cube (0 faces showing), 6 face-centers (1 face each), 12 edges (2 faces), 8 corners (3 faces).

To minimize white showing, place one white cube in the hidden center (0 white squares) and the other 5 white cubes at face-center positions (1 white square each).

Minimum white area = 5 unit squares out of 54 → 5/54.

ANSWER 9: A

# Problem 10:
From 1998 to 2050 is 52 years — about two 25-year tripling periods (1998 → 2023 → 2048).

Population: 200 → 600 → 1800.

1800 is closest to 2000.

ANSWER 10: D

Problem 1:

We are comparing the place values of the digits 9 and 3 in

\[
74982.1035.
\]

- The digit \(9\) is in the hundreds place, so its place value is \(100\).
- The digit \(3\) is in the thousandths place, so its place value is

\[
0.001.
\]

Now find how many times as great \(100\) is as \(0.001\):

\[
\frac{100}{0.001}=100,000.
\]

So the hundreds place is \(100,\!000\) times the thousandths place.

- A, \(1,\!000\), is too small.
- B, \(10,\!000\), is too small.
- D and E are too large.

ANSWER 1: C

Problem 2:

The gas expands by \(4\) cubic centimeters for every \(3^\circ\) increase in temperature.

The temperature changes from \(32^\circ\) down to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

Since

\[
12^\circ \div 3^\circ=4,
\]

there are 4 groups of \(3^\circ\).

For each group, the volume decreases by \(4\) cubic centimeters:

\[
4\cdot 4=16.
\]

At \(32^\circ\), the volume is \(24\) cubic centimeters, so at \(20^\circ\) it was

\[
24-16=8.
\]

Choices B, C, and D do not subtract the full \(16\) cubic centimeters. Choice E would represent an increase rather than a decrease.

ANSWER 2: A

Problem 3:

If there are \(n\) teams and each pair of teams plays exactly once, then the number of games is

\[
\binom n2=\frac{n(n-1)}2.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

Checking: with 7 teams, the number of games is

\[
\frac{7\cdot6}{2}=21.
\]

Six teams would produce only \(15\) games, while 8 teams would produce \(28\) games.

ANSWER 3: B

Problem 4:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew \(20\%\), so Shea’s current height is

\[
1.20h=60.
\]

Thus,

\[
h=\frac{60}{1.2}=50.
\]

So both were originally 50 inches tall.

Shea’s growth was

\[
60-50=10
\]

inches.

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5
\]

inches.

Therefore, Ara is now

\[
50+5=55
\]

inches tall.

Choice A, 48, is less than Ara’s original height. Choices B, C, and D do not include the correct 5-inch growth.

ANSWER 4: E

Problem 5:

A standard die has faces numbered \(1,2,3,4,5,6\). The product of all six face numbers is

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6=720.
\]

If the bottom face has number \(k\), then the product of the five visible faces is

\[
\frac{720}{k}.
\]

Check each possible bottom face:

\[
\frac{720}{1}=720,\qquad
\frac{720}{2}=360,\qquad
\frac{720}{3}=240,
\]

\[
\frac{720}{4}=180,\qquad
\frac{720}{5}=144,\qquad
\frac{720}{6}=120.
\]

Every one of these numbers is divisible by \(6\).

Therefore, no matter which face is on the bottom, the visible-face product is divisible by 6. The probability is

\[
1.
\]

All of choices A through D are less than 1, but the event always happens.

ANSWER 5: E

Problem 6:

Tori answered correctly:

- Arithmetic:

\[
70\%\text{ of }10=0.7(10)=7
\]

- Algebra:

\[
40\%\text{ of }30=0.4(30)=12
\]

- Geometry:

\[
60\%\text{ of }35=0.6(35)=21
\]

So her total number correct was

\[
7+12+21=40.
\]

To pass, she needs \(60\%\) of 75 questions correct:

\[
0.60(75)=45.
\]

She needs

\[
45-40=5
\]

more correct answers.

Choice A is too small, while choices C, D, and E give more than the number needed to reach exactly 45 correct.

ANSWER 6: B

Problem 7:

We are given

\[
0<a<b<c.
\]

We must find the statement that is impossible.

Consider choice A:

\[
a+c<b.
\]

But \(c>b\), and also \(a>0\). Therefore,

\[
a+c>c>b.
\]

So it is impossible for \(a+c<b\).

The other choices can happen for suitable positive numbers. For example:

- B: \(a\cdot b<c\) can occur with \(a=1,\ b=2,\ c=3\), since \(1\cdot2<3\).
- C: \(a+b<c\) can occur with \(a=1,\ b=2,\ c=4\).
- D: \(a\cdot c<b\) can occur with \(a=\frac14,\ b=1,\ c=2\), since \(\frac14\cdot2=\frac12<1\).
- E: \(bc=a\) can occur with \(b=\frac12,\ c=\frac34\), giving \(a=\frac38\), and indeed

\[
0<\frac38<\frac12<\frac34.
\]

Thus only choice A is impossible.

ANSWER 7: A

Problem 8:

Each full recipe makes 15 cookies.

Walter and Gretel need at least 216 cookies. Since only full recipes may be made, determine how many recipes are needed:

\[
216\div 15=14.4.
\]

They cannot make \(14.4\) recipes, so they must make 15 full recipes.

These 15 recipes make

\[
15\cdot15=225
\]

cookies, enough for the crowd.

Each recipe uses 3 tablespoons of butter, so the total butter needed is

\[
15\cdot3=45
\]

tablespoons.

There are 8 tablespoons in each stick, so

\[
45\div8=5.625.
\]

They must buy 6 full sticks of butter.

Five sticks provide only \(5\cdot8=40\) tablespoons, which is not enough. Six sticks provide \(48\) tablespoons, enough for the 45 tablespoons needed.

ANSWER 8: B

Problem 9:

A \(3\times3\times3\) cube has surface area

\[
6\cdot 3\cdot3=54
\]

square inches, since each of its 6 faces has area \(3\cdot3=9\).

To minimize the white surface showing, place the white cubes in the least exposed positions.

- The one center cube of the large cube has no faces exposed.
- The six face-center cubes each have exactly one face exposed.
- Edge cubes have two exposed faces.
- Corner cubes have three exposed faces.

There are 6 white cubes. Put one white cube in the completely hidden center. Then put the other 5 white cubes at five face-center locations.

Thus, the white cubes show only

\[
5
\]

unit square faces on the surface.

Therefore, the fraction of the surface that is white is

\[
\frac{5}{54}.
\]

Choices such as \(\frac19,\frac{5}{27},\frac29,\frac13\) represent more than 5 white unit squares out of the 54 total surface squares.

ANSWER 9: A

Problem 10:

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

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is approximately two 25-year periods.

After one 25-year period:

\[
200\cdot3=600.
\]

After two 25-year periods:

\[
600\cdot3=1800.
\]

Since 2050 is slightly more than 50 years after 1998, the population should be slightly more than 1800. The closest estimate is

\[
20

Problem 1:
We are asked: in 74982.1035, compare the place value of the 9 with the place value of the 3.

Write out places:
74982 is: 7 ten-thousands, 4 thousands, 9 hundreds, 8 tens, 2 ones.
So the 9 is in the hundreds place. Its value is 100.

After the decimal .1035 is: 1 tenths, 0 hundredths, 3 thousandths, 5 ten-thousandths.
So the 3 is in the thousandths place. Its value is 0.001 = 1/1000.

Ratio = (value of 9's place) / (value of 3's place)
= 100 / 0.001 = 100 / (1/1000) = 100 x 1000 = 100,000.

So the hundreds place is 100,000 times as great as the thousandths place.
Choices A,B,D,E are powers of 10 that are too small or too large.

ANSWER 1: C

Problem 2:
We are asked for the earlier volume given expansion rate.

Rate: +3 degrees temperature = +4 cubic cm volume.

Now: temp 32°, volume 24 cc.
Before: temp 20°.

Temperature change from 20° to 32° is 32-20 = 12°.
Number of 3-degree steps in 12° is 12 / 3 = 4 steps.

So volume increase from 20° to 32° is 4 steps x 4 cc = 16 cc.

Thus volume at 20° + 16 = volume at 32° = 24.
So volume at 20° = 24 - 16 = 8 cc.

Check: A 8 fits, B 12 would be only 3 steps, D 16 would be only 2 steps, E 40 would be adding instead of subtracting.

ANSWER 2: A

Problem 3:
We need n teams such that every pair plays once gives 21 games.

Number of pairs from n teams is n choose 2 = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try integers: 6x5=30 too small, 7x6=42 correct, 8x7=56 too big.
So n = 7.

Ruling out: A 6 gives only 15 games, C 8 gives 28 games, etc.

ANSWER 3: B

Problem 4:
Ara and Shea started at same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara's now height.

Let original height = H.
Shea's growth = 20% of H = 0.20 H.
Shea now = H + 0.20H = 1.2H = 60.
So H = 60 / 1.2 = 50 inches.

Shea grew 60-50 = 10 inches.
Ara grew half as many inches: 10 / 2 = 5 inches.
Ara now = 50 + 5 = 55 inches.

Tempting wrong A 48 would be working backwards, D 54 would be misreading percent, etc.

ANSWER 4: E

Problem 5:
A standard 1-6 die is tossed. Bottom face hidden, other 5 faces seen. Probability product of seen numbers divisible by 6?

Product divisible by 6 means divisible by 2 and by 3.

Numbers 1 to 6: evens are 2,4,6 — three of them. Multiples of 3 are 3,6 — two of them.

Hiding only one face (the bottom) can hide at most one even, leaving at least two evens still visible, so factor 2 remains. It can hide at most one multiple of 3, leaving at least one multiple of 3 still visible, so factor 3 remains.

Indeed total product 1·2·3·4·5·6 = 720. Visible product = 720 / b where b is bottom. 720/b divided by 6 = 120/b, and b=1,2,3,4,5,6 all divide 120, so always integer.

Hence no matter which face is hidden, the visible product is divisible by 6. Probability = 6/6 = 1.

Other choices 1/3,1/2,2/3,5/6 assume sometimes it fails, which is impossible.

ANSWER 5: E

Problem 6:
75 problems total: 10 arithmetic, 30 algebra, 35 geometry.
Correct: 70% of arithmetic, 40% of algebra, 60% of geometry. How many more to reach 60% overall?

Compute:
Arithmetic correct = 0.70 x 10 = 7.
Algebra correct = 0.40 x 30 = 12.
Geometry correct = 0.60 x 35 = 21.
Total correct = 7+12+21 = 40.

Need for 60% passing: 0.60 x 75 = 45 problems.

Shortfall = 45 - 40 = 5 more problems.

So B 5, not A 1 or C 7.

ANSWER 6: B

Problem 7:
Given 0 < a < b < c, which statement is impossible?

A. a + c < b. Since a > 0, a + c > c. And c > b. So a+c > b always. Hence a+c < b can never happen. This is impossible.

Check others are possible with examples:
B. a·b < c: e.g. a=1,b=2,c=10 gives 2<10 possible.
C. a+b < c: e.g. a=1,b=2,c=10 gives 3<10 possible.
D. a·c < b: with fractions <1. e.g. a=0.1,c=2,b=1 gives 0.2<1 and 0.1<1<2 possible.
E. b c = a: let b=0.25,a=0.1, then c=a/b=0.4, and 0.1<0.25<0.4 possible.

So only A is impossible. Students may think E impossible because b c > b > a for numbers >1, but with numbers <1 multiplication can shrink.

ANSWER 7: A

Problem 8:
One full recipe = 15 cookies = 3 tbsp butter. Need 216 cookies. 8 tbsp = 1 stick. How many sticks?

Pans needed: 216 / 15 = 14.4. Since only full recipes, round up to 15 pans. 14 pans would give only 210 cookies, not enough.

Butter needed: 15 x 3 tbsp = 45 tbsp.

Sticks needed: 45 / 8 = 5.625. Since need whole sticks to have enough, round up to 6. 5 sticks = 40 tbsp < 45 tbsp, not enough.

So answer 6. A 5 would leave butter and cookies short.

ANSWER 8: B

Problem 9:
Build 3x3x3 = 27 small cubes. 6 white, 21 red. Minimize white squares on outside. What fraction of outside is white?

Big cube surface: 6 faces each 3x3 = 9 small squares, total 54 small squares.

Small-cube positions by exposure:
- 1 interior center: 0 faces shown
- 6 face-centers: 1 face shown
- 12 edge non-corner: 2 faces shown
- 8 corners: 3 faces shown

To minimize white, put whites where least exposure:
Put 1 white in center → 0 exposed.
Put remaining 5 whites in face-center spots → 1 each exposed.
Total white exposed squares = 5.

Fraction white = 5 / 54 of total surface area.

Larger fractions like 1/9=6/54, etc. would result from worse placement.

ANSWER 9: A

Problem 10:
1998 population 200, triples every 25 years. Estimate 2050.

Time elapsed: 2050-1998 = 52 years.
Number of 25-year periods: 52/25 ≈ 2.08, just over 2 doublings-triplings.

After 25 years (~2023): 200x3=600.
After 50 years (~2048): 600x3=1800.
2050 is 2 years later, a little more, so about 2000.

Formally 200 x 3^(52/25) ≈ 200 x 9 x small extra ≈ 1960.

So A 600 is only one tripling, C 1000 too low, E 3000 too high. Best is D 2000.

ANSWER 10: D

Problem 1:
We are asked to find the ratio of the value of the place occupied by the digit 9 to the value of the place occupied by the digit 3 in the decimal number $74982.1035$.

1. Identify the place values:
   - The number is $74982.1035$.
   - The digit 2 is in the ones place ($10^0$).
   - The digit 8 is in the tens place ($10^1$).
   - The digit 9 is in the hundreds place ($10^2 = 100$).
   - The digit 1 is in the tenths place ($10^{-1} = 0.1$).
   - The digit 0 is in the hundredths place ($10^{-2} = 0.01$).
   - The digit 3 is in the thousandths place ($10^{-3} = 0.001$).
   - The digit 5 is in the ten-thousandths place ($10^{-4} = 0.0001$).

2. Compute the ratio of the place values:
   $$\frac{\text{Place value of } 9}{\text{Place value of } 3} = \frac{100}{0.001} = \frac{10^2}{10^{-3}} = 10^{2 - (-3)} = 10^5 = 100,000.$$

This corresponds to choice C.

ANSWER 1: C

---

Problem 2:
We are asked to find the initial volume of gas at $20^\circ$ given its volume at $32^\circ$ and its rate of expansion.

1. Find the change in temperature:
   The temperature changes from $20^\circ$ to $32^\circ$, which is an increase of:
   $$32^\circ - 20^\circ = 12^\circ.$$

2. Determine the volume expansion:
   For every $3^\circ$ rise, the volume expands by $4\text{ cm}^3$.
   In a $12^\circ$ rise, the number of $3^\circ$ intervals is:
   $$\frac{12^\circ}{3^\circ} = 4.$$
   Therefore, the volume expanded by:
   $$4 \times 4\text{ cm}^3 = 16\text{ cm}^3.$$

3. Calculate the original volume at $20^\circ$:
   $$\text{Volume at } 20^\circ = \text{Volume at } 32^\circ - 16\text{ cm}^3 = 24 - 16 = 8\text{ cm}^3.$$

This matches choice A.

ANSWER 2: A

---

Problem 3:
We are given that each of the $N$ teams plays every other team exactly once, resulting in 21 total games. We need to find $N$.

1. The number of games played between $N$ teams is the number of ways to choose 2 teams from $N$, given by the combination formula:
   $$\binom{N}{2} = \frac{N(N-1)}{2}.$$

2. Set this equal to 21 and solve for $N$:
   $$\frac{N(N-1)}{2} = 21 \implies N(N-1) = 42.$$

3. Find two consecutive positive integers whose product is 42:
   Since $7 \times 6 = 42$, we have $N = 7$.

Checking the choices, $N = 7$ corresponds to choice B.

ANSWER 3: B

---

Problem 4:
We are given:
- Ara and Shea initially had the same height, say $h$ inches.
- Shea has grown $20\%$, so Shea's new height is $h + 0.20h = 1.20h = 60$ inches.
- Ara has grown half as many inches as Shea.
- We need to find Ara's current height.

1. Find the original height $h$:
   $$1.20h = 60 \implies h = \frac{60}{1.2} = 50\text{ inches}.$$

2. Find how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10\text{ inches}.$$

3. Find Ara's growth:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5\text{ inches}.$$

4. Find Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55\text{ inches}.$$

This corresponds to choice E.

ANSWER 4: E

---

Problem 5:
A standard fair six-sided die has faces with numbers 1, 2, 3, 4, 5, and 6.
When the die is tossed, one face is on the bottom and cannot be seen, while the other 5 faces are visible.
We want the product of the 5 visible faces to be divisible by 6.

1. The product of all six numbers on the die is:
   $$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720.$$
   The product of the 5 visible faces is:
   $$P = \frac{720}{b},$$
   where $b \in \{1, 2, 3, 4, 5, 6\}$ is the number on the bottom face.

2. A number is divisible by 6 if and only if it is a multiple of both 2 and 3.
   Let's check the product $P$ for each possible bottom face $b$:
   - If $b = 1$, visible faces are $\{2, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 2$, visible faces are $\{1, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 3$, visible faces are $\{1, 2, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 4$, visible faces are $\{1, 2, 3, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 5$, visible faces are $\{1, 2, 3, 4, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 6$, visible faces are $\{1, 2, 3, 4, 5\}$. The product is $1 \times 2 \times 3 \times 4 \times 5 = 120$. Since $120 = 6 \times 20$, 120 is also divisible by 6 (it has factors 2 and 3)!

3. In every single case (all 6 possible outcomes for the bottom face), the product of the five visible faces is divisible by 6.
   Therefore, the probability is:
   $$\frac{6}{6} = 1.$$

This corresponds to choice E.

ANSWER 5: E

---

Problem 6:
Tori took a 75-problem test:
- 10 arithmetic problems: she answered $70\%$ correctly.
  $$\text{Correct arithmetic} = 0.70 \times 10 = 7.$$
- 30 algebra problems: she answered $40\%$ correctly.
  $$\text{Correct algebra} = 0.40 \times 30 = 12.$$
- 35 geometry problems: she answered $60\%$ correctly.
  $$\text{Correct geometry} = 0.60 \times 35 = 21.$$

1. Total problems Tori answered correctly:
   $$\text{Total correct} = 7 + 12 + 21 = 40.$$

2. Passing score required:
   $60\%$ of 75 problems:
   $$\text{Passing score} = 0.60 \times 75 = 45.$$

3. Additional problems needed:
   $$45 - 40 = 5\text{ problems}.$$

This corresponds to choice B.

ANSWER 6: B

---

Problem 7:
We are given real numbers $a, b, c$ such that $0 < a < b < c$.
We need to determine which of the statements is impossible.

Let's test each choice:
- Choice A: $a + c < b$.
  Since $a > 0$, adding $a$ to $c$ gives $a + c > c$.
  Since we are given $c > b$, it follows that $a + c > c > b$, which means $a + c > b$ must always be true!
  Therefore, $a + c < b$ is impossible.

Let's verify that the other choices can be possible with appropriate values:
- Choice B: $a \cdot b < c$. If $a = 1, b = 2, c = 3$, then $a \cdot b = 2 < 3 = c$. (Possible)
- Choice C: $a + b < c$. If $a = 1, b = 2, c = 4$, then $a + b = 3 < 4 = c$. (Possible)
- Choice D: $a \cdot c < b$. If $a = 0.1, b = 2, c = 3$, then $a \cdot c = 0.3 < 2 = b$. (Possible)
-

# Problem 1:

**Restate:** In 74982.1035, compare the place value of the digit 9 with the place value of the digit 3.

**Working:** Let's identify each place:
- The digit 9 sits in the **hundreds** place: 9 × 100 = 900. Its place value is **100**.
- The digit 3 sits in the **hundredths** place (two positions right of the decimal point): 3 × 0.01 = 0.03. Its place value is **0.01**.

**Compute the ratio:**
$$\frac{100}{0.01} = 100 \div \frac{1}{100} = 100 \times 100 = 10{,}000$$

**Check wrong choices:** Counting places instead of computing the ratio might suggest 100,000 (if you miscount the decimal places), but from hundreds to hundredths is exactly 4 positions, and each position is a factor of 10, giving 10⁴ = 10,000.

ANSWER 1: B

# Problem 2:

**Restate:** Volume increases 4 cm³ per 3° rise. Volume is 24 cm³ at 32°. Find the volume at 20°.

**Working:** The temperature dropped from 32° to 20°:
$$32 - 20 = 12°$$
Number of 3° intervals in the drop:
$$12 \div 3 = 4 \text{ intervals}$$
Each 3° drop means the gas **shrinks** by 4 cm³, so the total shrink is:
$$4 \times 4 = 16 \text{ cm}^3$$
The volume at 20° was:
$$24 - 16 = 8 \text{ cm}^3$$

**Check wrong choices:** Choice D (16) is the *amount* of shrink, not the volume — a tempting trap. Choice E (40) comes from adding instead of subtracting.

ANSWER 2: A

# Problem 3:

**Restate:** Each team plays every other team once; 21 total games. Find the number of teams.

**Working:** With n teams, the number of games is:
$$\frac{n(n-1)}{2}$$
Set this equal to 21:
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42$$
Testing n = 7: $7 \times 6 = 42$. ✓

**Check wrong choices:** n = 6 gives 15 games (too few); n = 8 gives 28 games (too many). Only n = 7 works.

ANSWER 3: B

# Problem 4:

**Restate:** Ara and Shea were once equal in height. Shea grew 20% and is now 60 inches. Ara grew half as many inches as Shea. Find Ara's height now.

**Working:** Let the original height be $h$. Shea grew 20%, so:
$$h + 0.20h = 60 \implies 1.2h = 60 \implies h = 50 \text{ inches}$$
Shea's growth: $60 - 50 = 10$ inches.
Ara grew half as many inches as Shea:
$$\frac{10}{2} = 5 \text{ inches}$$
Ara's height now:
$$50 + 5 = 55 \text{ inches}$$

**Check wrong choices:** Choice A (48) comes from computing 20% of 60 incorrectly; choice D (54) would result if Ara also grew 20% of the original height. The key is that Ara grew 5 **inches**, not a percentage.

ANSWER 4: E

# Problem 5:

**Restate:** A die is tossed; the bottom face is hidden. Find the probability that the product of the five visible faces is divisible by 6.

**Working:** The product of all six faces is:
$$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$$
The product of the five visible faces is:
$$\frac{720}{\text{hidden face}}$$

Check each possible hidden face:
- Hidden = 1: product = 720 → divisible by 6 ✓
- Hidden = 2: product = 360 → divisible by 6 ✓
- Hidden = 3: product = 240 → divisible by 6 ✓
- Hidden = 4: product = 180 → divisible by 6 ✓
- Hidden = 5: product = 144 → divisible by 6 ✓
- Hidden = 6: product = 120 → divisible by 6 ✓

All six outcomes give a product divisible by 6, so the probability is:
$$\frac{6}{6} = 1$$

**Check wrong choices:** One might think hiding the 5 or the 6 removes needed factors, but 720 contains 2 × 3 many times over — removing any single face still leaves factors of 2 and 3 in the product.

ANSWER 5: E

# Problem 6:

**Restate:** Tori's test: 10 arithmetic, 30 algebra, 35 geometry problems (75 total). She got 70%, 40%, and 60% correct respectively, which was below 60% overall. How many more correct answers were needed for 60%?

**Working:** Correct answers in each category:
- Arithmetic: $0.70 \times 10 = 7$
- Algebra: $0.40 \times 30 = 12$
- Geometry: $0.60 \times 35 = 21$

Total correct: $7 + 12 + 21 = 40$

To pass, she needs 60% of 75:
$$0.60 \times 75 = 45$$

Additional problems needed:
$$45 - 40 = 5$$

**Check wrong choices:** Note that the weighted average $\frac{40}{75} \approx 53.3\%$ is indeed below 60%, consistent with the story. A tempting trap is to average the three percentages: $\frac{70+40+60}{3} \approx 56.7\%$, which is wrong because the categories have different sizes.

ANSWER 6: B

# Problem 7:

**Restate:** Given $0 < a < b < c$, which statement is impossible?

**Working:** Test each option:

**A. $a + c < b$:** Since $c > b$, adding positive $a$ gives $a + c > c > b$. So $a + c < b$ **cannot happen**. ✓ (impossible)

Let's verify the others are all possible with suitable numbers:

**B. $a \cdot b < c$:** Try $a = 0.1, b = 0.2, c = 1$: $0.02 < 1$ ✓ possible.

**C. $a + b < c$:** Try $a = 1, b = 2, c = 10$: $3 < 10$ ✓ possible.

**D. $a \cdot c < b$:** Try $a = 0.1, b = 0.5, c = 0.6$: $0.06 < 0.5$ ✓ possible.

**E. $bc = a$:** Try $a = 0.06, b = 0.3, c = 0.2$... wait, need $b < c$. Try $a = 0.02, b = 0.1, c = 0.2$: $bc = 0.02 = a$, and $0 < 0.02 < 0.1 < 0.2$ ✓ possible.

Only option A forces a contradiction: $a + c > c > b$ always holds.

ANSWER 7: A

# Problem 8:

**Restate:** One recipe = 15 cookies using 3 tablespoons of butter; only full recipes allowed. Need 216 cookies. 8 tablespoons = 1 stick. How many sticks?

**Working:** Number of recipes needed:
$$\frac{216}{15} = 14.4$$
Since only full recipes are made, they must make **15 recipes** (14 recipes × 15 = 210 cookies, which is not enough).

Butter needed:
$$15 \times 3 = 45 \text{ tablespoons}$$
Sticks needed:
$$\frac{45}{8} = 5.625 \implies \text{need } 6 \text{ full sticks}$$

**Check wrong choices:** Choice A (5) is the trap — it uses only 40 tablespoons, but 45 are required. You can't buy a partial stick, so round up to 6.

ANSWER 8: B

# Problem 9:

**Restate:** Build a 3-inch cube from 27 unit cubes: 21 red, 6 white. Arrange to minimize white surface area. What fraction of the surface area is white?

**Working:** The 3×3×3 cube has total surface area:
$$6 \text{ faces} \times 9 \text{ sq in} = 54 \text{ square inches}$$

Classify the 27 positions by how many faces each shows:
- **1 inte

Problem 1:
We need the ratio of the place values of the digits 9 and 3 in  
\(74982.1035\).

The digit 9 is in the hundreds place, so its place value is  
\[
100.
\]

The digit 3 is in the thousandths place, so its place value is  
\[
0.001.
\]

Now compute the ratio:
\[
\frac{100}{0.001}=100{,}000.
\]

So the value of the place of the 9 is \(100{,}000\) times as great as the value of the place of the 3.

ANSWER 1: C

Problem 2:
We are told that for every \(3^\circ\) rise in temperature, the gas expands by 4 cubic centimeters.

At temperature \(32^\circ\), the volume is 24 cubic centimeters. We want the volume at \(20^\circ\).

The temperature difference is
\[
32-20=12^\circ.
\]

From \(20^\circ\) to \(32^\circ\), the temperature rises by \(12^\circ\). Since each \(3^\circ\) rise increases the volume by 4 cubic centimeters,
\[
\frac{12}{3}\cdot 4=4\cdot 4=16
\]
cubic centimeters.

So at \(20^\circ\), the volume was
\[
24-16=8
\]
cubic centimeters.

ANSWER 2: A

Problem 3:
Each team plays every other team exactly once. If there are \(n\) teams, the number of games is
\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games, so
\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:
\[
n(n-1)=42.
\]

We look for two consecutive positive integers whose product is 42:
\[
6\cdot 7=42.
\]

Thus
\[
n=7.
\]

So there were 7 teams.

ANSWER 3: B

Problem 4:
Ara and Shea were originally the same height. Let that original height be \(h\) inches.

Shea has grown 20%, so Shea’s current height is
\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:
\[
1.20h=60.
\]

Solve for \(h\):
\[
h=\frac{60}{1.20}=50.
\]

So they were originally 50 inches tall. Shea grew
\[
60-50=10
\]
inches.

Ara grew half as many inches as Shea:
\[
\frac{10}{2}=5
\]
inches.

Therefore Ara is now
\[
50+5=55
\]
inches tall.

ANSWER 4: E

Problem 5:
A fair six-sided die has faces numbered 1 through 6. When it is tossed, one face is hidden on the bottom, and the other five faces are visible.

We need the probability that the product of the five visible numbers is divisible by 6.

A number is divisible by 6 if it has at least one factor of 2 and at least one factor of 3.

No matter which single face is hidden:

- There are three even faces: 2, 4, and 6. Hiding only one of them still leaves at least one even face visible.
- There are two multiples of 3: 3 and 6. Hiding only one of them still leaves at least one multiple of 3 visible.

So the product of the visible faces always contains both a factor of 2 and a factor of 3. Therefore it is always divisible by 6.

Thus the probability is
\[
1.
\]

ANSWER 5: E

Problem 6:
Tori answered:

- \(70\%\) of 10 arithmetic problems:
\[
0.70\cdot 10=7
\]
- \(40\%\) of 30 algebra problems:
\[
0.40\cdot 30=12
\]
- \(60\%\) of 35 geometry problems:
\[
0.60\cdot 35=21
\]

Total correct:
\[
7+12+21=40.
\]

There were 75 problems total. To pass, she needed \(60\%\) correct:
\[
0.60\cdot 75=45.
\]

She answered 40 correctly, so she needed
\[
45-40=5
\]
more correct answers.

ANSWER 6: B

Problem 7:
We are given
\[
0<a<b<c.
\]

We need the impossible statement.

Choice A says
\[
a+c<b.
\]

But since \(c>b\) and \(a>0\),
\[
a+c>b.
\]

So \(a+c<b\) cannot happen.

We can also check that the other choices are possible:

- B: \(a\cdot b<c\) can happen, for example \(a=1,b=2,c=100\).
- C: \(a+b<c\) can happen, for example \(a=1,b=2,c=100\).
- D: \(a\cdot c<b\) can happen, for example \(a=0.1,b=2,c=3\), since \(0.3<2\).
- E: If interpreted as \(bc=a\), it can happen, for example \(a=0.1,b=0.2,c=0.5\), since \(0.2\cdot 0.5=0.1\).

Thus the impossible choice is A.

ANSWER 7: A

Problem 8:
One recipe makes 15 cookies and uses 3 tablespoons of butter. Only full recipes can be made.

Walter and Gretel need 216 cookies. The number of full recipes needed is
\[
\left\lceil \frac{216}{15}\right\rceil.
\]

Since
\[
\frac{216}{15}=14.4,
\]
they must make 15 full recipes.

Butter needed:
\[
15\cdot 3=45
\]
tablespoons.

There are 8 tablespoons in a stick of butter, so the number of sticks needed is
\[
\left\lceil \frac{45}{8}\right\rceil.
\]

Since
\[
\frac{45}{8}=5.625,
\]
they need 6 sticks of butter.

ANSWER 8: B

Problem 9:
A 3-inch cube made from 27 one-inch cubes has surface area
\[
6\cdot 3^2=54
\]
square inches.

Each small cube can show:

- 0 faces if it is the center cube,
- 1 face if it is a face-center cube,
- 2 faces if it is an edge cube,
- 3 faces if it is a corner cube.

We have 6 white cubes. To minimize the white surface area, we should hide as many white cubes as possible.

There is exactly one interior cube, so put one white cube there. It shows 0 square inches.

The remaining 5 white cubes must be on the surface. To minimize exposed white area, put them on face-center positions, where each shows exactly 1 square inch.

So the minimum white surface area is
\[
1\cdot 0+5\cdot 1=5
\]
square inches.

The fraction of the surface area that is white is
\[
\frac{5}{54}.
\]

ANSWER 9: A

Problem 10:
The population is 200 in 1998, and it triples every 25 years.

From 1998 to 2050 is
\[
2050-1998=52
\]
years.

Two full 25-year periods are 50 years, so after 50 years the population is
\[
200\cdot 3\cdot 3=200\cdot 9=1800.
\]

The extra 2 years would add a relatively small amount, so the population in 2050 is a little above 1800. Among the choices, the best estimate is 2000.

ANSWER 10: D

# Problem 1:
Write out the place values of 74982.1035:
- 7 → ten-thousands, 4 → thousands, **9 → hundreds**, 8 → tens, 2 → ones
- 1 → tenths, 0 → hundredths, **3 → thousandths**, 5 → ten-thousandths

So the 9 sits in the hundreds place (value 100) and the 3 sits in the thousandths place (value 0.001).

Ratio = 100 ÷ 0.001 = 100,000.

(Answers like 1,000 or 10,000 come from miscounting decimal places — the 3 is the *third* digit after the point, not the first or second.)

ANSWER 1: C

# Problem 2:
Temperature goes from 32° down to 20°, a drop of 12°.

Each 3° change corresponds to 4 cm³ of change: 12 ÷ 3 = 4 steps.

Volume change = 4 × 4 = 16 cm³ decrease (cooling means shrinking).

Volume = 24 − 16 = 8 cm³.

(40 is the trap from *adding* instead of subtracting; 16 is just the change, not the volume.)

ANSWER 2: A

# Problem 3:
With n teams each playing every other team once, the number of games is C(n,2) = n(n−1)/2.

Set n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7·6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 games — neither is 21.

ANSWER 3: B

# Problem 4:
Let the common original height be x. Shea grew 20%: 1.20x = 60 → x = 50.

So Shea grew 60 − 50 = 10 inches.

Ara grew half as many inches: 5 inches.

Ara's height = 50 + 5 = 55 inches.

(54 would result from wrongly taking Shea's growth as 10% of 60, etc.)

ANSWER 4: E

# Problem 5:
The product of all six faces is 1·2·3·4·5·6 = 720. If the hidden bottom face is k, the visible product is 720/k.

- k=1: 720 ✓ divisible by 6
- k=2: 360 ✓
- k=3: 240 ✓
- k=4: 180 ✓
- k=5: 144 ✓
- k=6: 120 ✓

Reason why: hiding one face still leaves at least one even number and at least one multiple of 3 among the five visible faces (e.g., hide 6 → 2 and 3 remain; hide 3 → 6 remains).

So the probability is 6/6 = 1.

ANSWER 5: E

# Problem 6:
Correct answers:
- Arithmetic: 70% of 10 = 7
- Algebra: 40% of 30 = 12
- Geometry: 60% of 35 = 21

Total correct = 7 + 12 + 21 = 40.

Needed for 60%: 0.60 × 75 = 45.

She needs 45 − 40 = 5 more.

(40/75 ≈ 53%, consistent with failing.)

ANSWER 6: B

# Problem 7:
Given 0 < a < b < c.

- **A. a + c < b:** Since c > b and a > 0, we get a + c > b. So a + c < b is impossible.
- B. a·b < c: take a = 0.1, b = 0.2, c = 1 → 0.02 < 1 ✓ possible.
- C. a + b < c: take a = 1, b = 2, c = 10 → 3 < 10 ✓ possible.
- D. a·c < b: take a = 0.1, b = 1, c = 2 → 0.2 < 1 ✓ possible.
- E. b/c = a: take a = 0.5, b = 1, c = 2 → 1/2 = 0.5 ✓ possible.

ANSWER 7: A

# Problem 8:
Pans needed: 216 ÷ 15 = 14.4, and only full recipes are made, so they must bake 15 pans.

Butter: 15 × 3 = 45 tablespoons.

Sticks: 45 ÷ 8 = 5.625, and butter comes in whole sticks, so 6 sticks.

(5 sticks = 40 tbsp, not enough.)

ANSWER 8: B

# Problem 9:
The 3×3×3 cube has surface area 6 × 9 = 54 unit squares.

Positions in a 3-cube: 1 center cube (0 faces showing), 6 face-centers (1 face each), 12 edges (2 faces), 8 corners (3 faces).

To minimize white showing, place one white cube in the hidden center (0 white squares) and the other 5 white cubes at face-center positions (1 white square each).

Minimum white area = 5 unit squares out of 54 → 5/54.

ANSWER 9: A

# Problem 10:
From 1998 to 2050 is 52 years — about two 25-year tripling periods (1998 → 2023 → 2048).

Population: 200 → 600 → 1800.

1800 is closest to 2000.

ANSWER 10: D

Problem 1:

We are comparing the place values of the digits 9 and 3 in

\[
74982.1035.
\]

- The digit \(9\) is in the hundreds place, so its place value is \(100\).
- The digit \(3\) is in the thousandths place, so its place value is

\[
0.001.
\]

Now find how many times as great \(100\) is as \(0.001\):

\[
\frac{100}{0.001}=100,000.
\]

So the hundreds place is \(100,\!000\) times the thousandths place.

- A, \(1,\!000\), is too small.
- B, \(10,\!000\), is too small.
- D and E are too large.

ANSWER 1: C

Problem 2:

The gas expands by \(4\) cubic centimeters for every \(3^\circ\) increase in temperature.

The temperature changes from \(32^\circ\) down to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

Since

\[
12^\circ \div 3^\circ=4,
\]

there are 4 groups of \(3^\circ\).

For each group, the volume decreases by \(4\) cubic centimeters:

\[
4\cdot 4=16.
\]

At \(32^\circ\), the volume is \(24\) cubic centimeters, so at \(20^\circ\) it was

\[
24-16=8.
\]

Choices B, C, and D do not subtract the full \(16\) cubic centimeters. Choice E would represent an increase rather than a decrease.

ANSWER 2: A

Problem 3:

If there are \(n\) teams and each pair of teams plays exactly once, then the number of games is

\[
\binom n2=\frac{n(n-1)}2.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

Checking: with 7 teams, the number of games is

\[
\frac{7\cdot6}{2}=21.
\]

Six teams would produce only \(15\) games, while 8 teams would produce \(28\) games.

ANSWER 3: B

Problem 4:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew \(20\%\), so Shea’s current height is

\[
1.20h=60.
\]

Thus,

\[
h=\frac{60}{1.2}=50.
\]

So both were originally 50 inches tall.

Shea’s growth was

\[
60-50=10
\]

inches.

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5
\]

inches.

Therefore, Ara is now

\[
50+5=55
\]

inches tall.

Choice A, 48, is less than Ara’s original height. Choices B, C, and D do not include the correct 5-inch growth.

ANSWER 4: E

Problem 5:

A standard die has faces numbered \(1,2,3,4,5,6\). The product of all six face numbers is

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6=720.
\]

If the bottom face has number \(k\), then the product of the five visible faces is

\[
\frac{720}{k}.
\]

Check each possible bottom face:

\[
\frac{720}{1}=720,\qquad
\frac{720}{2}=360,\qquad
\frac{720}{3}=240,
\]

\[
\frac{720}{4}=180,\qquad
\frac{720}{5}=144,\qquad
\frac{720}{6}=120.
\]

Every one of these numbers is divisible by \(6\).

Therefore, no matter which face is on the bottom, the visible-face product is divisible by 6. The probability is

\[
1.
\]

All of choices A through D are less than 1, but the event always happens.

ANSWER 5: E

Problem 6:

Tori answered correctly:

- Arithmetic:

\[
70\%\text{ of }10=0.7(10)=7
\]

- Algebra:

\[
40\%\text{ of }30=0.4(30)=12
\]

- Geometry:

\[
60\%\text{ of }35=0.6(35)=21
\]

So her total number correct was

\[
7+12+21=40.
\]

To pass, she needs \(60\%\) of 75 questions correct:

\[
0.60(75)=45.
\]

She needs

\[
45-40=5
\]

more correct answers.

Choice A is too small, while choices C, D, and E give more than the number needed to reach exactly 45 correct.

ANSWER 6: B

Problem 7:

We are given

\[
0<a<b<c.
\]

We must find the statement that is impossible.

Consider choice A:

\[
a+c<b.
\]

But \(c>b\), and also \(a>0\). Therefore,

\[
a+c>c>b.
\]

So it is impossible for \(a+c<b\).

The other choices can happen for suitable positive numbers. For example:

- B: \(a\cdot b<c\) can occur with \(a=1,\ b=2,\ c=3\), since \(1\cdot2<3\).
- C: \(a+b<c\) can occur with \(a=1,\ b=2,\ c=4\).
- D: \(a\cdot c<b\) can occur with \(a=\frac14,\ b=1,\ c=2\), since \(\frac14\cdot2=\frac12<1\).
- E: \(bc=a\) can occur with \(b=\frac12,\ c=\frac34\), giving \(a=\frac38\), and indeed

\[
0<\frac38<\frac12<\frac34.
\]

Thus only choice A is impossible.

ANSWER 7: A

Problem 8:

Each full recipe makes 15 cookies.

Walter and Gretel need at least 216 cookies. Since only full recipes may be made, determine how many recipes are needed:

\[
216\div 15=14.4.
\]

They cannot make \(14.4\) recipes, so they must make 15 full recipes.

These 15 recipes make

\[
15\cdot15=225
\]

cookies, enough for the crowd.

Each recipe uses 3 tablespoons of butter, so the total butter needed is

\[
15\cdot3=45
\]

tablespoons.

There are 8 tablespoons in each stick, so

\[
45\div8=5.625.
\]

They must buy 6 full sticks of butter.

Five sticks provide only \(5\cdot8=40\) tablespoons, which is not enough. Six sticks provide \(48\) tablespoons, enough for the 45 tablespoons needed.

ANSWER 8: B

Problem 9:

A \(3\times3\times3\) cube has surface area

\[
6\cdot 3\cdot3=54
\]

square inches, since each of its 6 faces has area \(3\cdot3=9\).

To minimize the white surface showing, place the white cubes in the least exposed positions.

- The one center cube of the large cube has no faces exposed.
- The six face-center cubes each have exactly one face exposed.
- Edge cubes have two exposed faces.
- Corner cubes have three exposed faces.

There are 6 white cubes. Put one white cube in the completely hidden center. Then put the other 5 white cubes at five face-center locations.

Thus, the white cubes show only

\[
5
\]

unit square faces on the surface.

Therefore, the fraction of the surface that is white is

\[
\frac{5}{54}.
\]

Choices such as \(\frac19,\frac{5}{27},\frac29,\frac13\) represent more than 5 white unit squares out of the 54 total surface squares.

ANSWER 9: A

Problem 10:

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

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is approximately two 25-year periods.

After one 25-year period:

\[
200\cdot3=600.
\]

After two 25-year periods:

\[
600\cdot3=1800.
\]

Since 2050 is slightly more than 50 years after 1998, the population should be slightly more than 1800. The closest estimate is

\[
20

Problem 1:
We are asked: in 74982.1035, compare the place value of the 9 with the place value of the 3.

Write out places:
74982 is: 7 ten-thousands, 4 thousands, 9 hundreds, 8 tens, 2 ones.
So the 9 is in the hundreds place. Its value is 100.

After the decimal .1035 is: 1 tenths, 0 hundredths, 3 thousandths, 5 ten-thousandths.
So the 3 is in the thousandths place. Its value is 0.001 = 1/1000.

Ratio = (value of 9's place) / (value of 3's place)
= 100 / 0.001 = 100 / (1/1000) = 100 x 1000 = 100,000.

So the hundreds place is 100,000 times as great as the thousandths place.
Choices A,B,D,E are powers of 10 that are too small or too large.

ANSWER 1: C

Problem 2:
We are asked for the earlier volume given expansion rate.

Rate: +3 degrees temperature = +4 cubic cm volume.

Now: temp 32°, volume 24 cc.
Before: temp 20°.

Temperature change from 20° to 32° is 32-20 = 12°.
Number of 3-degree steps in 12° is 12 / 3 = 4 steps.

So volume increase from 20° to 32° is 4 steps x 4 cc = 16 cc.

Thus volume at 20° + 16 = volume at 32° = 24.
So volume at 20° = 24 - 16 = 8 cc.

Check: A 8 fits, B 12 would be only 3 steps, D 16 would be only 2 steps, E 40 would be adding instead of subtracting.

ANSWER 2: A

Problem 3:
We need n teams such that every pair plays once gives 21 games.

Number of pairs from n teams is n choose 2 = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try integers: 6x5=30 too small, 7x6=42 correct, 8x7=56 too big.
So n = 7.

Ruling out: A 6 gives only 15 games, C 8 gives 28 games, etc.

ANSWER 3: B

Problem 4:
Ara and Shea started at same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara's now height.

Let original height = H.
Shea's growth = 20% of H = 0.20 H.
Shea now = H + 0.20H = 1.2H = 60.
So H = 60 / 1.2 = 50 inches.

Shea grew 60-50 = 10 inches.
Ara grew half as many inches: 10 / 2 = 5 inches.
Ara now = 50 + 5 = 55 inches.

Tempting wrong A 48 would be working backwards, D 54 would be misreading percent, etc.

ANSWER 4: E

Problem 5:
A standard 1-6 die is tossed. Bottom face hidden, other 5 faces seen. Probability product of seen numbers divisible by 6?

Product divisible by 6 means divisible by 2 and by 3.

Numbers 1 to 6: evens are 2,4,6 — three of them. Multiples of 3 are 3,6 — two of them.

Hiding only one face (the bottom) can hide at most one even, leaving at least two evens still visible, so factor 2 remains. It can hide at most one multiple of 3, leaving at least one multiple of 3 still visible, so factor 3 remains.

Indeed total product 1·2·3·4·5·6 = 720. Visible product = 720 / b where b is bottom. 720/b divided by 6 = 120/b, and b=1,2,3,4,5,6 all divide 120, so always integer.

Hence no matter which face is hidden, the visible product is divisible by 6. Probability = 6/6 = 1.

Other choices 1/3,1/2,2/3,5/6 assume sometimes it fails, which is impossible.

ANSWER 5: E

Problem 6:
75 problems total: 10 arithmetic, 30 algebra, 35 geometry.
Correct: 70% of arithmetic, 40% of algebra, 60% of geometry. How many more to reach 60% overall?

Compute:
Arithmetic correct = 0.70 x 10 = 7.
Algebra correct = 0.40 x 30 = 12.
Geometry correct = 0.60 x 35 = 21.
Total correct = 7+12+21 = 40.

Need for 60% passing: 0.60 x 75 = 45 problems.

Shortfall = 45 - 40 = 5 more problems.

So B 5, not A 1 or C 7.

ANSWER 6: B

Problem 7:
Given 0 < a < b < c, which statement is impossible?

A. a + c < b. Since a > 0, a + c > c. And c > b. So a+c > b always. Hence a+c < b can never happen. This is impossible.

Check others are possible with examples:
B. a·b < c: e.g. a=1,b=2,c=10 gives 2<10 possible.
C. a+b < c: e.g. a=1,b=2,c=10 gives 3<10 possible.
D. a·c < b: with fractions <1. e.g. a=0.1,c=2,b=1 gives 0.2<1 and 0.1<1<2 possible.
E. b c = a: let b=0.25,a=0.1, then c=a/b=0.4, and 0.1<0.25<0.4 possible.

So only A is impossible. Students may think E impossible because b c > b > a for numbers >1, but with numbers <1 multiplication can shrink.

ANSWER 7: A

Problem 8:
One full recipe = 15 cookies = 3 tbsp butter. Need 216 cookies. 8 tbsp = 1 stick. How many sticks?

Pans needed: 216 / 15 = 14.4. Since only full recipes, round up to 15 pans. 14 pans would give only 210 cookies, not enough.

Butter needed: 15 x 3 tbsp = 45 tbsp.

Sticks needed: 45 / 8 = 5.625. Since need whole sticks to have enough, round up to 6. 5 sticks = 40 tbsp < 45 tbsp, not enough.

So answer 6. A 5 would leave butter and cookies short.

ANSWER 8: B

Problem 9:
Build 3x3x3 = 27 small cubes. 6 white, 21 red. Minimize white squares on outside. What fraction of outside is white?

Big cube surface: 6 faces each 3x3 = 9 small squares, total 54 small squares.

Small-cube positions by exposure:
- 1 interior center: 0 faces shown
- 6 face-centers: 1 face shown
- 12 edge non-corner: 2 faces shown
- 8 corners: 3 faces shown

To minimize white, put whites where least exposure:
Put 1 white in center → 0 exposed.
Put remaining 5 whites in face-center spots → 1 each exposed.
Total white exposed squares = 5.

Fraction white = 5 / 54 of total surface area.

Larger fractions like 1/9=6/54, etc. would result from worse placement.

ANSWER 9: A

Problem 10:
1998 population 200, triples every 25 years. Estimate 2050.

Time elapsed: 2050-1998 = 52 years.
Number of 25-year periods: 52/25 ≈ 2.08, just over 2 doublings-triplings.

After 25 years (~2023): 200x3=600.
After 50 years (~2048): 600x3=1800.
2050 is 2 years later, a little more, so about 2000.

Formally 200 x 3^(52/25) ≈ 200 x 9 x small extra ≈ 1960.

So A 600 is only one tripling, C 1000 too low, E 3000 too high. Best is D 2000.

ANSWER 10: D

Problem 1:
We are asked to find the ratio of the value of the place occupied by the digit 9 to the value of the place occupied by the digit 3 in the decimal number $74982.1035$.

1. Identify the place values:
   - The number is $74982.1035$.
   - The digit 2 is in the ones place ($10^0$).
   - The digit 8 is in the tens place ($10^1$).
   - The digit 9 is in the hundreds place ($10^2 = 100$).
   - The digit 1 is in the tenths place ($10^{-1} = 0.1$).
   - The digit 0 is in the hundredths place ($10^{-2} = 0.01$).
   - The digit 3 is in the thousandths place ($10^{-3} = 0.001$).
   - The digit 5 is in the ten-thousandths place ($10^{-4} = 0.0001$).

2. Compute the ratio of the place values:
   $$\frac{\text{Place value of } 9}{\text{Place value of } 3} = \frac{100}{0.001} = \frac{10^2}{10^{-3}} = 10^{2 - (-3)} = 10^5 = 100,000.$$

This corresponds to choice C.

ANSWER 1: C

---

Problem 2:
We are asked to find the initial volume of gas at $20^\circ$ given its volume at $32^\circ$ and its rate of expansion.

1. Find the change in temperature:
   The temperature changes from $20^\circ$ to $32^\circ$, which is an increase of:
   $$32^\circ - 20^\circ = 12^\circ.$$

2. Determine the volume expansion:
   For every $3^\circ$ rise, the volume expands by $4\text{ cm}^3$.
   In a $12^\circ$ rise, the number of $3^\circ$ intervals is:
   $$\frac{12^\circ}{3^\circ} = 4.$$
   Therefore, the volume expanded by:
   $$4 \times 4\text{ cm}^3 = 16\text{ cm}^3.$$

3. Calculate the original volume at $20^\circ$:
   $$\text{Volume at } 20^\circ = \text{Volume at } 32^\circ - 16\text{ cm}^3 = 24 - 16 = 8\text{ cm}^3.$$

This matches choice A.

ANSWER 2: A

---

Problem 3:
We are given that each of the $N$ teams plays every other team exactly once, resulting in 21 total games. We need to find $N$.

1. The number of games played between $N$ teams is the number of ways to choose 2 teams from $N$, given by the combination formula:
   $$\binom{N}{2} = \frac{N(N-1)}{2}.$$

2. Set this equal to 21 and solve for $N$:
   $$\frac{N(N-1)}{2} = 21 \implies N(N-1) = 42.$$

3. Find two consecutive positive integers whose product is 42:
   Since $7 \times 6 = 42$, we have $N = 7$.

Checking the choices, $N = 7$ corresponds to choice B.

ANSWER 3: B

---

Problem 4:
We are given:
- Ara and Shea initially had the same height, say $h$ inches.
- Shea has grown $20\%$, so Shea's new height is $h + 0.20h = 1.20h = 60$ inches.
- Ara has grown half as many inches as Shea.
- We need to find Ara's current height.

1. Find the original height $h$:
   $$1.20h = 60 \implies h = \frac{60}{1.2} = 50\text{ inches}.$$

2. Find how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10\text{ inches}.$$

3. Find Ara's growth:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5\text{ inches}.$$

4. Find Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55\text{ inches}.$$

This corresponds to choice E.

ANSWER 4: E

---

Problem 5:
A standard fair six-sided die has faces with numbers 1, 2, 3, 4, 5, and 6.
When the die is tossed, one face is on the bottom and cannot be seen, while the other 5 faces are visible.
We want the product of the 5 visible faces to be divisible by 6.

1. The product of all six numbers on the die is:
   $$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720.$$
   The product of the 5 visible faces is:
   $$P = \frac{720}{b},$$
   where $b \in \{1, 2, 3, 4, 5, 6\}$ is the number on the bottom face.

2. A number is divisible by 6 if and only if it is a multiple of both 2 and 3.
   Let's check the product $P$ for each possible bottom face $b$:
   - If $b = 1$, visible faces are $\{2, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 2$, visible faces are $\{1, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 3$, visible faces are $\{1, 2, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 4$, visible faces are $\{1, 2, 3, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 5$, visible faces are $\{1, 2, 3, 4, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 6$, visible faces are $\{1, 2, 3, 4, 5\}$. The product is $1 \times 2 \times 3 \times 4 \times 5 = 120$. Since $120 = 6 \times 20$, 120 is also divisible by 6 (it has factors 2 and 3)!

3. In every single case (all 6 possible outcomes for the bottom face), the product of the five visible faces is divisible by 6.
   Therefore, the probability is:
   $$\frac{6}{6} = 1.$$

This corresponds to choice E.

ANSWER 5: E

---

Problem 6:
Tori took a 75-problem test:
- 10 arithmetic problems: she answered $70\%$ correctly.
  $$\text{Correct arithmetic} = 0.70 \times 10 = 7.$$
- 30 algebra problems: she answered $40\%$ correctly.
  $$\text{Correct algebra} = 0.40 \times 30 = 12.$$
- 35 geometry problems: she answered $60\%$ correctly.
  $$\text{Correct geometry} = 0.60 \times 35 = 21.$$

1. Total problems Tori answered correctly:
   $$\text{Total correct} = 7 + 12 + 21 = 40.$$

2. Passing score required:
   $60\%$ of 75 problems:
   $$\text{Passing score} = 0.60 \times 75 = 45.$$

3. Additional problems needed:
   $$45 - 40 = 5\text{ problems}.$$

This corresponds to choice B.

ANSWER 6: B

---

Problem 7:
We are given real numbers $a, b, c$ such that $0 < a < b < c$.
We need to determine which of the statements is impossible.

Let's test each choice:
- Choice A: $a + c < b$.
  Since $a > 0$, adding $a$ to $c$ gives $a + c > c$.
  Since we are given $c > b$, it follows that $a + c > c > b$, which means $a + c > b$ must always be true!
  Therefore, $a + c < b$ is impossible.

Let's verify that the other choices can be possible with appropriate values:
- Choice B: $a \cdot b < c$. If $a = 1, b = 2, c = 3$, then $a \cdot b = 2 < 3 = c$. (Possible)
- Choice C: $a + b < c$. If $a = 1, b = 2, c = 4$, then $a + b = 3 < 4 = c$. (Possible)
- Choice D: $a \cdot c < b$. If $a = 0.1, b = 2, c = 3$, then $a \cdot c = 0.3 < 2 = b$. (Possible)
-

# Problem 1:

**Restate:** In 74982.1035, compare the place value of the digit 9 with the place value of the digit 3.

**Working:** Let's identify each place:
- The digit 9 sits in the **hundreds** place: 9 × 100 = 900. Its place value is **100**.
- The digit 3 sits in the **hundredths** place (two positions right of the decimal point): 3 × 0.01 = 0.03. Its place value is **0.01**.

**Compute the ratio:**
$$\frac{100}{0.01} = 100 \div \frac{1}{100} = 100 \times 100 = 10{,}000$$

**Check wrong choices:** Counting places instead of computing the ratio might suggest 100,000 (if you miscount the decimal places), but from hundreds to hundredths is exactly 4 positions, and each position is a factor of 10, giving 10⁴ = 10,000.

ANSWER 1: B

# Problem 2:

**Restate:** Volume increases 4 cm³ per 3° rise. Volume is 24 cm³ at 32°. Find the volume at 20°.

**Working:** The temperature dropped from 32° to 20°:
$$32 - 20 = 12°$$
Number of 3° intervals in the drop:
$$12 \div 3 = 4 \text{ intervals}$$
Each 3° drop means the gas **shrinks** by 4 cm³, so the total shrink is:
$$4 \times 4 = 16 \text{ cm}^3$$
The volume at 20° was:
$$24 - 16 = 8 \text{ cm}^3$$

**Check wrong choices:** Choice D (16) is the *amount* of shrink, not the volume — a tempting trap. Choice E (40) comes from adding instead of subtracting.

ANSWER 2: A

# Problem 3:

**Restate:** Each team plays every other team once; 21 total games. Find the number of teams.

**Working:** With n teams, the number of games is:
$$\frac{n(n-1)}{2}$$
Set this equal to 21:
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42$$
Testing n = 7: $7 \times 6 = 42$. ✓

**Check wrong choices:** n = 6 gives 15 games (too few); n = 8 gives 28 games (too many). Only n = 7 works.

ANSWER 3: B

# Problem 4:

**Restate:** Ara and Shea were once equal in height. Shea grew 20% and is now 60 inches. Ara grew half as many inches as Shea. Find Ara's height now.

**Working:** Let the original height be $h$. Shea grew 20%, so:
$$h + 0.20h = 60 \implies 1.2h = 60 \implies h = 50 \text{ inches}$$
Shea's growth: $60 - 50 = 10$ inches.
Ara grew half as many inches as Shea:
$$\frac{10}{2} = 5 \text{ inches}$$
Ara's height now:
$$50 + 5 = 55 \text{ inches}$$

**Check wrong choices:** Choice A (48) comes from computing 20% of 60 incorrectly; choice D (54) would result if Ara also grew 20% of the original height. The key is that Ara grew 5 **inches**, not a percentage.

ANSWER 4: E

# Problem 5:

**Restate:** A die is tossed; the bottom face is hidden. Find the probability that the product of the five visible faces is divisible by 6.

**Working:** The product of all six faces is:
$$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$$
The product of the five visible faces is:
$$\frac{720}{\text{hidden face}}$$

Check each possible hidden face:
- Hidden = 1: product = 720 → divisible by 6 ✓
- Hidden = 2: product = 360 → divisible by 6 ✓
- Hidden = 3: product = 240 → divisible by 6 ✓
- Hidden = 4: product = 180 → divisible by 6 ✓
- Hidden = 5: product = 144 → divisible by 6 ✓
- Hidden = 6: product = 120 → divisible by 6 ✓

All six outcomes give a product divisible by 6, so the probability is:
$$\frac{6}{6} = 1$$

**Check wrong choices:** One might think hiding the 5 or the 6 removes needed factors, but 720 contains 2 × 3 many times over — removing any single face still leaves factors of 2 and 3 in the product.

ANSWER 5: E

# Problem 6:

**Restate:** Tori's test: 10 arithmetic, 30 algebra, 35 geometry problems (75 total). She got 70%, 40%, and 60% correct respectively, which was below 60% overall. How many more correct answers were needed for 60%?

**Working:** Correct answers in each category:
- Arithmetic: $0.70 \times 10 = 7$
- Algebra: $0.40 \times 30 = 12$
- Geometry: $0.60 \times 35 = 21$

Total correct: $7 + 12 + 21 = 40$

To pass, she needs 60% of 75:
$$0.60 \times 75 = 45$$

Additional problems needed:
$$45 - 40 = 5$$

**Check wrong choices:** Note that the weighted average $\frac{40}{75} \approx 53.3\%$ is indeed below 60%, consistent with the story. A tempting trap is to average the three percentages: $\frac{70+40+60}{3} \approx 56.7\%$, which is wrong because the categories have different sizes.

ANSWER 6: B

# Problem 7:

**Restate:** Given $0 < a < b < c$, which statement is impossible?

**Working:** Test each option:

**A. $a + c < b$:** Since $c > b$, adding positive $a$ gives $a + c > c > b$. So $a + c < b$ **cannot happen**. ✓ (impossible)

Let's verify the others are all possible with suitable numbers:

**B. $a \cdot b < c$:** Try $a = 0.1, b = 0.2, c = 1$: $0.02 < 1$ ✓ possible.

**C. $a + b < c$:** Try $a = 1, b = 2, c = 10$: $3 < 10$ ✓ possible.

**D. $a \cdot c < b$:** Try $a = 0.1, b = 0.5, c = 0.6$: $0.06 < 0.5$ ✓ possible.

**E. $bc = a$:** Try $a = 0.06, b = 0.3, c = 0.2$... wait, need $b < c$. Try $a = 0.02, b = 0.1, c = 0.2$: $bc = 0.02 = a$, and $0 < 0.02 < 0.1 < 0.2$ ✓ possible.

Only option A forces a contradiction: $a + c > c > b$ always holds.

ANSWER 7: A

# Problem 8:

**Restate:** One recipe = 15 cookies using 3 tablespoons of butter; only full recipes allowed. Need 216 cookies. 8 tablespoons = 1 stick. How many sticks?

**Working:** Number of recipes needed:
$$\frac{216}{15} = 14.4$$
Since only full recipes are made, they must make **15 recipes** (14 recipes × 15 = 210 cookies, which is not enough).

Butter needed:
$$15 \times 3 = 45 \text{ tablespoons}$$
Sticks needed:
$$\frac{45}{8} = 5.625 \implies \text{need } 6 \text{ full sticks}$$

**Check wrong choices:** Choice A (5) is the trap — it uses only 40 tablespoons, but 45 are required. You can't buy a partial stick, so round up to 6.

ANSWER 8: B

# Problem 9:

**Restate:** Build a 3-inch cube from 27 unit cubes: 21 red, 6 white. Arrange to minimize white surface area. What fraction of the surface area is white?

**Working:** The 3×3×3 cube has total surface area:
$$6 \text{ faces} \times 9 \text{ sq in} = 54 \text{ square inches}$$

Classify the 27 positions by how many faces each shows:
- **1 inte

Problem 1:
We need the ratio of the place values of the digits 9 and 3 in  
\(74982.1035\).

The digit 9 is in the hundreds place, so its place value is  
\[
100.
\]

The digit 3 is in the thousandths place, so its place value is  
\[
0.001.
\]

Now compute the ratio:
\[
\frac{100}{0.001}=100{,}000.
\]

So the value of the place of the 9 is \(100{,}000\) times as great as the value of the place of the 3.

ANSWER 1: C

Problem 2:
We are told that for every \(3^\circ\) rise in temperature, the gas expands by 4 cubic centimeters.

At temperature \(32^\circ\), the volume is 24 cubic centimeters. We want the volume at \(20^\circ\).

The temperature difference is
\[
32-20=12^\circ.
\]

From \(20^\circ\) to \(32^\circ\), the temperature rises by \(12^\circ\). Since each \(3^\circ\) rise increases the volume by 4 cubic centimeters,
\[
\frac{12}{3}\cdot 4=4\cdot 4=16
\]
cubic centimeters.

So at \(20^\circ\), the volume was
\[
24-16=8
\]
cubic centimeters.

ANSWER 2: A

Problem 3:
Each team plays every other team exactly once. If there are \(n\) teams, the number of games is
\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games, so
\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:
\[
n(n-1)=42.
\]

We look for two consecutive positive integers whose product is 42:
\[
6\cdot 7=42.
\]

Thus
\[
n=7.
\]

So there were 7 teams.

ANSWER 3: B

Problem 4:
Ara and Shea were originally the same height. Let that original height be \(h\) inches.

Shea has grown 20%, so Shea’s current height is
\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:
\[
1.20h=60.
\]

Solve for \(h\):
\[
h=\frac{60}{1.20}=50.
\]

So they were originally 50 inches tall. Shea grew
\[
60-50=10
\]
inches.

Ara grew half as many inches as Shea:
\[
\frac{10}{2}=5
\]
inches.

Therefore Ara is now
\[
50+5=55
\]
inches tall.

ANSWER 4: E

Problem 5:
A fair six-sided die has faces numbered 1 through 6. When it is tossed, one face is hidden on the bottom, and the other five faces are visible.

We need the probability that the product of the five visible numbers is divisible by 6.

A number is divisible by 6 if it has at least one factor of 2 and at least one factor of 3.

No matter which single face is hidden:

- There are three even faces: 2, 4, and 6. Hiding only one of them still leaves at least one even face visible.
- There are two multiples of 3: 3 and 6. Hiding only one of them still leaves at least one multiple of 3 visible.

So the product of the visible faces always contains both a factor of 2 and a factor of 3. Therefore it is always divisible by 6.

Thus the probability is
\[
1.
\]

ANSWER 5: E

Problem 6:
Tori answered:

- \(70\%\) of 10 arithmetic problems:
\[
0.70\cdot 10=7
\]
- \(40\%\) of 30 algebra problems:
\[
0.40\cdot 30=12
\]
- \(60\%\) of 35 geometry problems:
\[
0.60\cdot 35=21
\]

Total correct:
\[
7+12+21=40.
\]

There were 75 problems total. To pass, she needed \(60\%\) correct:
\[
0.60\cdot 75=45.
\]

She answered 40 correctly, so she needed
\[
45-40=5
\]
more correct answers.

ANSWER 6: B

Problem 7:
We are given
\[
0<a<b<c.
\]

We need the impossible statement.

Choice A says
\[
a+c<b.
\]

But since \(c>b\) and \(a>0\),
\[
a+c>b.
\]

So \(a+c<b\) cannot happen.

We can also check that the other choices are possible:

- B: \(a\cdot b<c\) can happen, for example \(a=1,b=2,c=100\).
- C: \(a+b<c\) can happen, for example \(a=1,b=2,c=100\).
- D: \(a\cdot c<b\) can happen, for example \(a=0.1,b=2,c=3\), since \(0.3<2\).
- E: If interpreted as \(bc=a\), it can happen, for example \(a=0.1,b=0.2,c=0.5\), since \(0.2\cdot 0.5=0.1\).

Thus the impossible choice is A.

ANSWER 7: A

Problem 8:
One recipe makes 15 cookies and uses 3 tablespoons of butter. Only full recipes can be made.

Walter and Gretel need 216 cookies. The number of full recipes needed is
\[
\left\lceil \frac{216}{15}\right\rceil.
\]

Since
\[
\frac{216}{15}=14.4,
\]
they must make 15 full recipes.

Butter needed:
\[
15\cdot 3=45
\]
tablespoons.

There are 8 tablespoons in a stick of butter, so the number of sticks needed is
\[
\left\lceil \frac{45}{8}\right\rceil.
\]

Since
\[
\frac{45}{8}=5.625,
\]
they need 6 sticks of butter.

ANSWER 8: B

Problem 9:
A 3-inch cube made from 27 one-inch cubes has surface area
\[
6\cdot 3^2=54
\]
square inches.

Each small cube can show:

- 0 faces if it is the center cube,
- 1 face if it is a face-center cube,
- 2 faces if it is an edge cube,
- 3 faces if it is a corner cube.

We have 6 white cubes. To minimize the white surface area, we should hide as many white cubes as possible.

There is exactly one interior cube, so put one white cube there. It shows 0 square inches.

The remaining 5 white cubes must be on the surface. To minimize exposed white area, put them on face-center positions, where each shows exactly 1 square inch.

So the minimum white surface area is
\[
1\cdot 0+5\cdot 1=5
\]
square inches.

The fraction of the surface area that is white is
\[
\frac{5}{54}.
\]

ANSWER 9: A

Problem 10:
The population is 200 in 1998, and it triples every 25 years.

From 1998 to 2050 is
\[
2050-1998=52
\]
years.

Two full 25-year periods are 50 years, so after 50 years the population is
\[
200\cdot 3\cdot 3=200\cdot 9=1800.
\]

The extra 2 years would add a relatively small amount, so the population in 2050 is a little above 1800. Among the choices, the best estimate is 2000.

ANSWER 10: D

# Problem 1:
Write out the place values of 74982.1035:
- 7 → ten-thousands, 4 → thousands, **9 → hundreds**, 8 → tens, 2 → ones
- 1 → tenths, 0 → hundredths, **3 → thousandths**, 5 → ten-thousandths

So the 9 sits in the hundreds place (value 100) and the 3 sits in the thousandths place (value 0.001).

Ratio = 100 ÷ 0.001 = 100,000.

(Answers like 1,000 or 10,000 come from miscounting decimal places — the 3 is the *third* digit after the point, not the first or second.)

ANSWER 1: C

# Problem 2:
Temperature goes from 32° down to 20°, a drop of 12°.

Each 3° change corresponds to 4 cm³ of change: 12 ÷ 3 = 4 steps.

Volume change = 4 × 4 = 16 cm³ decrease (cooling means shrinking).

Volume = 24 − 16 = 8 cm³.

(40 is the trap from *adding* instead of subtracting; 16 is just the change, not the volume.)

ANSWER 2: A

# Problem 3:
With n teams each playing every other team once, the number of games is C(n,2) = n(n−1)/2.

Set n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7·6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 games — neither is 21.

ANSWER 3: B

# Problem 4:
Let the common original height be x. Shea grew 20%: 1.20x = 60 → x = 50.

So Shea grew 60 − 50 = 10 inches.

Ara grew half as many inches: 5 inches.

Ara's height = 50 + 5 = 55 inches.

(54 would result from wrongly taking Shea's growth as 10% of 60, etc.)

ANSWER 4: E

# Problem 5:
The product of all six faces is 1·2·3·4·5·6 = 720. If the hidden bottom face is k, the visible product is 720/k.

- k=1: 720 ✓ divisible by 6
- k=2: 360 ✓
- k=3: 240 ✓
- k=4: 180 ✓
- k=5: 144 ✓
- k=6: 120 ✓

Reason why: hiding one face still leaves at least one even number and at least one multiple of 3 among the five visible faces (e.g., hide 6 → 2 and 3 remain; hide 3 → 6 remains).

So the probability is 6/6 = 1.

ANSWER 5: E

# Problem 6:
Correct answers:
- Arithmetic: 70% of 10 = 7
- Algebra: 40% of 30 = 12
- Geometry: 60% of 35 = 21

Total correct = 7 + 12 + 21 = 40.

Needed for 60%: 0.60 × 75 = 45.

She needs 45 − 40 = 5 more.

(40/75 ≈ 53%, consistent with failing.)

ANSWER 6: B

# Problem 7:
Given 0 < a < b < c.

- **A. a + c < b:** Since c > b and a > 0, we get a + c > b. So a + c < b is impossible.
- B. a·b < c: take a = 0.1, b = 0.2, c = 1 → 0.02 < 1 ✓ possible.
- C. a + b < c: take a = 1, b = 2, c = 10 → 3 < 10 ✓ possible.
- D. a·c < b: take a = 0.1, b = 1, c = 2 → 0.2 < 1 ✓ possible.
- E. b/c = a: take a = 0.5, b = 1, c = 2 → 1/2 = 0.5 ✓ possible.

ANSWER 7: A

# Problem 8:
Pans needed: 216 ÷ 15 = 14.4, and only full recipes are made, so they must bake 15 pans.

Butter: 15 × 3 = 45 tablespoons.

Sticks: 45 ÷ 8 = 5.625, and butter comes in whole sticks, so 6 sticks.

(5 sticks = 40 tbsp, not enough.)

ANSWER 8: B

# Problem 9:
The 3×3×3 cube has surface area 6 × 9 = 54 unit squares.

Positions in a 3-cube: 1 center cube (0 faces showing), 6 face-centers (1 face each), 12 edges (2 faces), 8 corners (3 faces).

To minimize white showing, place one white cube in the hidden center (0 white squares) and the other 5 white cubes at face-center positions (1 white square each).

Minimum white area = 5 unit squares out of 54 → 5/54.

ANSWER 9: A

# Problem 10:
From 1998 to 2050 is 52 years — about two 25-year tripling periods (1998 → 2023 → 2048).

Population: 200 → 600 → 1800.

1800 is closest to 2000.

ANSWER 10: D

Problem 1:

We are comparing the place values of the digits 9 and 3 in

\[
74982.1035.
\]

- The digit \(9\) is in the hundreds place, so its place value is \(100\).
- The digit \(3\) is in the thousandths place, so its place value is

\[
0.001.
\]

Now find how many times as great \(100\) is as \(0.001\):

\[
\frac{100}{0.001}=100,000.
\]

So the hundreds place is \(100,\!000\) times the thousandths place.

- A, \(1,\!000\), is too small.
- B, \(10,\!000\), is too small.
- D and E are too large.

ANSWER 1: C

Problem 2:

The gas expands by \(4\) cubic centimeters for every \(3^\circ\) increase in temperature.

The temperature changes from \(32^\circ\) down to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

Since

\[
12^\circ \div 3^\circ=4,
\]

there are 4 groups of \(3^\circ\).

For each group, the volume decreases by \(4\) cubic centimeters:

\[
4\cdot 4=16.
\]

At \(32^\circ\), the volume is \(24\) cubic centimeters, so at \(20^\circ\) it was

\[
24-16=8.
\]

Choices B, C, and D do not subtract the full \(16\) cubic centimeters. Choice E would represent an increase rather than a decrease.

ANSWER 2: A

Problem 3:

If there are \(n\) teams and each pair of teams plays exactly once, then the number of games is

\[
\binom n2=\frac{n(n-1)}2.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

Checking: with 7 teams, the number of games is

\[
\frac{7\cdot6}{2}=21.
\]

Six teams would produce only \(15\) games, while 8 teams would produce \(28\) games.

ANSWER 3: B

Problem 4:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew \(20\%\), so Shea’s current height is

\[
1.20h=60.
\]

Thus,

\[
h=\frac{60}{1.2}=50.
\]

So both were originally 50 inches tall.

Shea’s growth was

\[
60-50=10
\]

inches.

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5
\]

inches.

Therefore, Ara is now

\[
50+5=55
\]

inches tall.

Choice A, 48, is less than Ara’s original height. Choices B, C, and D do not include the correct 5-inch growth.

ANSWER 4: E

Problem 5:

A standard die has faces numbered \(1,2,3,4,5,6\). The product of all six face numbers is

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6=720.
\]

If the bottom face has number \(k\), then the product of the five visible faces is

\[
\frac{720}{k}.
\]

Check each possible bottom face:

\[
\frac{720}{1}=720,\qquad
\frac{720}{2}=360,\qquad
\frac{720}{3}=240,
\]

\[
\frac{720}{4}=180,\qquad
\frac{720}{5}=144,\qquad
\frac{720}{6}=120.
\]

Every one of these numbers is divisible by \(6\).

Therefore, no matter which face is on the bottom, the visible-face product is divisible by 6. The probability is

\[
1.
\]

All of choices A through D are less than 1, but the event always happens.

ANSWER 5: E

Problem 6:

Tori answered correctly:

- Arithmetic:

\[
70\%\text{ of }10=0.7(10)=7
\]

- Algebra:

\[
40\%\text{ of }30=0.4(30)=12
\]

- Geometry:

\[
60\%\text{ of }35=0.6(35)=21
\]

So her total number correct was

\[
7+12+21=40.
\]

To pass, she needs \(60\%\) of 75 questions correct:

\[
0.60(75)=45.
\]

She needs

\[
45-40=5
\]

more correct answers.

Choice A is too small, while choices C, D, and E give more than the number needed to reach exactly 45 correct.

ANSWER 6: B

Problem 7:

We are given

\[
0<a<b<c.
\]

We must find the statement that is impossible.

Consider choice A:

\[
a+c<b.
\]

But \(c>b\), and also \(a>0\). Therefore,

\[
a+c>c>b.
\]

So it is impossible for \(a+c<b\).

The other choices can happen for suitable positive numbers. For example:

- B: \(a\cdot b<c\) can occur with \(a=1,\ b=2,\ c=3\), since \(1\cdot2<3\).
- C: \(a+b<c\) can occur with \(a=1,\ b=2,\ c=4\).
- D: \(a\cdot c<b\) can occur with \(a=\frac14,\ b=1,\ c=2\), since \(\frac14\cdot2=\frac12<1\).
- E: \(bc=a\) can occur with \(b=\frac12,\ c=\frac34\), giving \(a=\frac38\), and indeed

\[
0<\frac38<\frac12<\frac34.
\]

Thus only choice A is impossible.

ANSWER 7: A

Problem 8:

Each full recipe makes 15 cookies.

Walter and Gretel need at least 216 cookies. Since only full recipes may be made, determine how many recipes are needed:

\[
216\div 15=14.4.
\]

They cannot make \(14.4\) recipes, so they must make 15 full recipes.

These 15 recipes make

\[
15\cdot15=225
\]

cookies, enough for the crowd.

Each recipe uses 3 tablespoons of butter, so the total butter needed is

\[
15\cdot3=45
\]

tablespoons.

There are 8 tablespoons in each stick, so

\[
45\div8=5.625.
\]

They must buy 6 full sticks of butter.

Five sticks provide only \(5\cdot8=40\) tablespoons, which is not enough. Six sticks provide \(48\) tablespoons, enough for the 45 tablespoons needed.

ANSWER 8: B

Problem 9:

A \(3\times3\times3\) cube has surface area

\[
6\cdot 3\cdot3=54
\]

square inches, since each of its 6 faces has area \(3\cdot3=9\).

To minimize the white surface showing, place the white cubes in the least exposed positions.

- The one center cube of the large cube has no faces exposed.
- The six face-center cubes each have exactly one face exposed.
- Edge cubes have two exposed faces.
- Corner cubes have three exposed faces.

There are 6 white cubes. Put one white cube in the completely hidden center. Then put the other 5 white cubes at five face-center locations.

Thus, the white cubes show only

\[
5
\]

unit square faces on the surface.

Therefore, the fraction of the surface that is white is

\[
\frac{5}{54}.
\]

Choices such as \(\frac19,\frac{5}{27},\frac29,\frac13\) represent more than 5 white unit squares out of the 54 total surface squares.

ANSWER 9: A

Problem 10:

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

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is approximately two 25-year periods.

After one 25-year period:

\[
200\cdot3=600.
\]

After two 25-year periods:

\[
600\cdot3=1800.
\]

Since 2050 is slightly more than 50 years after 1998, the population should be slightly more than 1800. The closest estimate is

\[
20

Problem 1:
We are asked: in 74982.1035, compare the place value of the 9 with the place value of the 3.

Write out places:
74982 is: 7 ten-thousands, 4 thousands, 9 hundreds, 8 tens, 2 ones.
So the 9 is in the hundreds place. Its value is 100.

After the decimal .1035 is: 1 tenths, 0 hundredths, 3 thousandths, 5 ten-thousandths.
So the 3 is in the thousandths place. Its value is 0.001 = 1/1000.

Ratio = (value of 9's place) / (value of 3's place)
= 100 / 0.001 = 100 / (1/1000) = 100 x 1000 = 100,000.

So the hundreds place is 100,000 times as great as the thousandths place.
Choices A,B,D,E are powers of 10 that are too small or too large.

ANSWER 1: C

Problem 2:
We are asked for the earlier volume given expansion rate.

Rate: +3 degrees temperature = +4 cubic cm volume.

Now: temp 32°, volume 24 cc.
Before: temp 20°.

Temperature change from 20° to 32° is 32-20 = 12°.
Number of 3-degree steps in 12° is 12 / 3 = 4 steps.

So volume increase from 20° to 32° is 4 steps x 4 cc = 16 cc.

Thus volume at 20° + 16 = volume at 32° = 24.
So volume at 20° = 24 - 16 = 8 cc.

Check: A 8 fits, B 12 would be only 3 steps, D 16 would be only 2 steps, E 40 would be adding instead of subtracting.

ANSWER 2: A

Problem 3:
We need n teams such that every pair plays once gives 21 games.

Number of pairs from n teams is n choose 2 = n(n-1)/2.

Set equal to 21:
n(n-1)/2 = 21
n(n-1) = 42.

Try integers: 6x5=30 too small, 7x6=42 correct, 8x7=56 too big.
So n = 7.

Ruling out: A 6 gives only 15 games, C 8 gives 28 games, etc.

ANSWER 3: B

Problem 4:
Ara and Shea started at same height. Shea grew 20% to reach 60 inches. Ara grew half as many inches as Shea grew. Find Ara's now height.

Let original height = H.
Shea's growth = 20% of H = 0.20 H.
Shea now = H + 0.20H = 1.2H = 60.
So H = 60 / 1.2 = 50 inches.

Shea grew 60-50 = 10 inches.
Ara grew half as many inches: 10 / 2 = 5 inches.
Ara now = 50 + 5 = 55 inches.

Tempting wrong A 48 would be working backwards, D 54 would be misreading percent, etc.

ANSWER 4: E

Problem 5:
A standard 1-6 die is tossed. Bottom face hidden, other 5 faces seen. Probability product of seen numbers divisible by 6?

Product divisible by 6 means divisible by 2 and by 3.

Numbers 1 to 6: evens are 2,4,6 — three of them. Multiples of 3 are 3,6 — two of them.

Hiding only one face (the bottom) can hide at most one even, leaving at least two evens still visible, so factor 2 remains. It can hide at most one multiple of 3, leaving at least one multiple of 3 still visible, so factor 3 remains.

Indeed total product 1·2·3·4·5·6 = 720. Visible product = 720 / b where b is bottom. 720/b divided by 6 = 120/b, and b=1,2,3,4,5,6 all divide 120, so always integer.

Hence no matter which face is hidden, the visible product is divisible by 6. Probability = 6/6 = 1.

Other choices 1/3,1/2,2/3,5/6 assume sometimes it fails, which is impossible.

ANSWER 5: E

Problem 6:
75 problems total: 10 arithmetic, 30 algebra, 35 geometry.
Correct: 70% of arithmetic, 40% of algebra, 60% of geometry. How many more to reach 60% overall?

Compute:
Arithmetic correct = 0.70 x 10 = 7.
Algebra correct = 0.40 x 30 = 12.
Geometry correct = 0.60 x 35 = 21.
Total correct = 7+12+21 = 40.

Need for 60% passing: 0.60 x 75 = 45 problems.

Shortfall = 45 - 40 = 5 more problems.

So B 5, not A 1 or C 7.

ANSWER 6: B

Problem 7:
Given 0 < a < b < c, which statement is impossible?

A. a + c < b. Since a > 0, a + c > c. And c > b. So a+c > b always. Hence a+c < b can never happen. This is impossible.

Check others are possible with examples:
B. a·b < c: e.g. a=1,b=2,c=10 gives 2<10 possible.
C. a+b < c: e.g. a=1,b=2,c=10 gives 3<10 possible.
D. a·c < b: with fractions <1. e.g. a=0.1,c=2,b=1 gives 0.2<1 and 0.1<1<2 possible.
E. b c = a: let b=0.25,a=0.1, then c=a/b=0.4, and 0.1<0.25<0.4 possible.

So only A is impossible. Students may think E impossible because b c > b > a for numbers >1, but with numbers <1 multiplication can shrink.

ANSWER 7: A

Problem 8:
One full recipe = 15 cookies = 3 tbsp butter. Need 216 cookies. 8 tbsp = 1 stick. How many sticks?

Pans needed: 216 / 15 = 14.4. Since only full recipes, round up to 15 pans. 14 pans would give only 210 cookies, not enough.

Butter needed: 15 x 3 tbsp = 45 tbsp.

Sticks needed: 45 / 8 = 5.625. Since need whole sticks to have enough, round up to 6. 5 sticks = 40 tbsp < 45 tbsp, not enough.

So answer 6. A 5 would leave butter and cookies short.

ANSWER 8: B

Problem 9:
Build 3x3x3 = 27 small cubes. 6 white, 21 red. Minimize white squares on outside. What fraction of outside is white?

Big cube surface: 6 faces each 3x3 = 9 small squares, total 54 small squares.

Small-cube positions by exposure:
- 1 interior center: 0 faces shown
- 6 face-centers: 1 face shown
- 12 edge non-corner: 2 faces shown
- 8 corners: 3 faces shown

To minimize white, put whites where least exposure:
Put 1 white in center → 0 exposed.
Put remaining 5 whites in face-center spots → 1 each exposed.
Total white exposed squares = 5.

Fraction white = 5 / 54 of total surface area.

Larger fractions like 1/9=6/54, etc. would result from worse placement.

ANSWER 9: A

Problem 10:
1998 population 200, triples every 25 years. Estimate 2050.

Time elapsed: 2050-1998 = 52 years.
Number of 25-year periods: 52/25 ≈ 2.08, just over 2 doublings-triplings.

After 25 years (~2023): 200x3=600.
After 50 years (~2048): 600x3=1800.
2050 is 2 years later, a little more, so about 2000.

Formally 200 x 3^(52/25) ≈ 200 x 9 x small extra ≈ 1960.

So A 600 is only one tripling, C 1000 too low, E 3000 too high. Best is D 2000.

ANSWER 10: D

Problem 1:
We are asked to find the ratio of the value of the place occupied by the digit 9 to the value of the place occupied by the digit 3 in the decimal number $74982.1035$.

1. Identify the place values:
   - The number is $74982.1035$.
   - The digit 2 is in the ones place ($10^0$).
   - The digit 8 is in the tens place ($10^1$).
   - The digit 9 is in the hundreds place ($10^2 = 100$).
   - The digit 1 is in the tenths place ($10^{-1} = 0.1$).
   - The digit 0 is in the hundredths place ($10^{-2} = 0.01$).
   - The digit 3 is in the thousandths place ($10^{-3} = 0.001$).
   - The digit 5 is in the ten-thousandths place ($10^{-4} = 0.0001$).

2. Compute the ratio of the place values:
   $$\frac{\text{Place value of } 9}{\text{Place value of } 3} = \frac{100}{0.001} = \frac{10^2}{10^{-3}} = 10^{2 - (-3)} = 10^5 = 100,000.$$

This corresponds to choice C.

ANSWER 1: C

---

Problem 2:
We are asked to find the initial volume of gas at $20^\circ$ given its volume at $32^\circ$ and its rate of expansion.

1. Find the change in temperature:
   The temperature changes from $20^\circ$ to $32^\circ$, which is an increase of:
   $$32^\circ - 20^\circ = 12^\circ.$$

2. Determine the volume expansion:
   For every $3^\circ$ rise, the volume expands by $4\text{ cm}^3$.
   In a $12^\circ$ rise, the number of $3^\circ$ intervals is:
   $$\frac{12^\circ}{3^\circ} = 4.$$
   Therefore, the volume expanded by:
   $$4 \times 4\text{ cm}^3 = 16\text{ cm}^3.$$

3. Calculate the original volume at $20^\circ$:
   $$\text{Volume at } 20^\circ = \text{Volume at } 32^\circ - 16\text{ cm}^3 = 24 - 16 = 8\text{ cm}^3.$$

This matches choice A.

ANSWER 2: A

---

Problem 3:
We are given that each of the $N$ teams plays every other team exactly once, resulting in 21 total games. We need to find $N$.

1. The number of games played between $N$ teams is the number of ways to choose 2 teams from $N$, given by the combination formula:
   $$\binom{N}{2} = \frac{N(N-1)}{2}.$$

2. Set this equal to 21 and solve for $N$:
   $$\frac{N(N-1)}{2} = 21 \implies N(N-1) = 42.$$

3. Find two consecutive positive integers whose product is 42:
   Since $7 \times 6 = 42$, we have $N = 7$.

Checking the choices, $N = 7$ corresponds to choice B.

ANSWER 3: B

---

Problem 4:
We are given:
- Ara and Shea initially had the same height, say $h$ inches.
- Shea has grown $20\%$, so Shea's new height is $h + 0.20h = 1.20h = 60$ inches.
- Ara has grown half as many inches as Shea.
- We need to find Ara's current height.

1. Find the original height $h$:
   $$1.20h = 60 \implies h = \frac{60}{1.2} = 50\text{ inches}.$$

2. Find how many inches Shea grew:
   $$\text{Shea's growth} = 60 - 50 = 10\text{ inches}.$$

3. Find Ara's growth:
   Ara grew half as many inches as Shea:
   $$\text{Ara's growth} = \frac{10}{2} = 5\text{ inches}.$$

4. Find Ara's current height:
   $$\text{Ara's current height} = 50 + 5 = 55\text{ inches}.$$

This corresponds to choice E.

ANSWER 4: E

---

Problem 5:
A standard fair six-sided die has faces with numbers 1, 2, 3, 4, 5, and 6.
When the die is tossed, one face is on the bottom and cannot be seen, while the other 5 faces are visible.
We want the product of the 5 visible faces to be divisible by 6.

1. The product of all six numbers on the die is:
   $$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720.$$
   The product of the 5 visible faces is:
   $$P = \frac{720}{b},$$
   where $b \in \{1, 2, 3, 4, 5, 6\}$ is the number on the bottom face.

2. A number is divisible by 6 if and only if it is a multiple of both 2 and 3.
   Let's check the product $P$ for each possible bottom face $b$:
   - If $b = 1$, visible faces are $\{2, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 2$, visible faces are $\{1, 3, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 3$, visible faces are $\{1, 2, 4, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 4$, visible faces are $\{1, 2, 3, 5, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 5$, visible faces are $\{1, 2, 3, 4, 6\}$. Product contains 6, so it is divisible by 6.
   - If $b = 6$, visible faces are $\{1, 2, 3, 4, 5\}$. The product is $1 \times 2 \times 3 \times 4 \times 5 = 120$. Since $120 = 6 \times 20$, 120 is also divisible by 6 (it has factors 2 and 3)!

3. In every single case (all 6 possible outcomes for the bottom face), the product of the five visible faces is divisible by 6.
   Therefore, the probability is:
   $$\frac{6}{6} = 1.$$

This corresponds to choice E.

ANSWER 5: E

---

Problem 6:
Tori took a 75-problem test:
- 10 arithmetic problems: she answered $70\%$ correctly.
  $$\text{Correct arithmetic} = 0.70 \times 10 = 7.$$
- 30 algebra problems: she answered $40\%$ correctly.
  $$\text{Correct algebra} = 0.40 \times 30 = 12.$$
- 35 geometry problems: she answered $60\%$ correctly.
  $$\text{Correct geometry} = 0.60 \times 35 = 21.$$

1. Total problems Tori answered correctly:
   $$\text{Total correct} = 7 + 12 + 21 = 40.$$

2. Passing score required:
   $60\%$ of 75 problems:
   $$\text{Passing score} = 0.60 \times 75 = 45.$$

3. Additional problems needed:
   $$45 - 40 = 5\text{ problems}.$$

This corresponds to choice B.

ANSWER 6: B

---

Problem 7:
We are given real numbers $a, b, c$ such that $0 < a < b < c$.
We need to determine which of the statements is impossible.

Let's test each choice:
- Choice A: $a + c < b$.
  Since $a > 0$, adding $a$ to $c$ gives $a + c > c$.
  Since we are given $c > b$, it follows that $a + c > c > b$, which means $a + c > b$ must always be true!
  Therefore, $a + c < b$ is impossible.

Let's verify that the other choices can be possible with appropriate values:
- Choice B: $a \cdot b < c$. If $a = 1, b = 2, c = 3$, then $a \cdot b = 2 < 3 = c$. (Possible)
- Choice C: $a + b < c$. If $a = 1, b = 2, c = 4$, then $a + b = 3 < 4 = c$. (Possible)
- Choice D: $a \cdot c < b$. If $a = 0.1, b = 2, c = 3$, then $a \cdot c = 0.3 < 2 = b$. (Possible)
-

# Problem 1:

**Restate:** In 74982.1035, compare the place value of the digit 9 with the place value of the digit 3.

**Working:** Let's identify each place:
- The digit 9 sits in the **hundreds** place: 9 × 100 = 900. Its place value is **100**.
- The digit 3 sits in the **hundredths** place (two positions right of the decimal point): 3 × 0.01 = 0.03. Its place value is **0.01**.

**Compute the ratio:**
$$\frac{100}{0.01} = 100 \div \frac{1}{100} = 100 \times 100 = 10{,}000$$

**Check wrong choices:** Counting places instead of computing the ratio might suggest 100,000 (if you miscount the decimal places), but from hundreds to hundredths is exactly 4 positions, and each position is a factor of 10, giving 10⁴ = 10,000.

ANSWER 1: B

# Problem 2:

**Restate:** Volume increases 4 cm³ per 3° rise. Volume is 24 cm³ at 32°. Find the volume at 20°.

**Working:** The temperature dropped from 32° to 20°:
$$32 - 20 = 12°$$
Number of 3° intervals in the drop:
$$12 \div 3 = 4 \text{ intervals}$$
Each 3° drop means the gas **shrinks** by 4 cm³, so the total shrink is:
$$4 \times 4 = 16 \text{ cm}^3$$
The volume at 20° was:
$$24 - 16 = 8 \text{ cm}^3$$

**Check wrong choices:** Choice D (16) is the *amount* of shrink, not the volume — a tempting trap. Choice E (40) comes from adding instead of subtracting.

ANSWER 2: A

# Problem 3:

**Restate:** Each team plays every other team once; 21 total games. Find the number of teams.

**Working:** With n teams, the number of games is:
$$\frac{n(n-1)}{2}$$
Set this equal to 21:
$$\frac{n(n-1)}{2} = 21 \implies n(n-1) = 42$$
Testing n = 7: $7 \times 6 = 42$. ✓

**Check wrong choices:** n = 6 gives 15 games (too few); n = 8 gives 28 games (too many). Only n = 7 works.

ANSWER 3: B

# Problem 4:

**Restate:** Ara and Shea were once equal in height. Shea grew 20% and is now 60 inches. Ara grew half as many inches as Shea. Find Ara's height now.

**Working:** Let the original height be $h$. Shea grew 20%, so:
$$h + 0.20h = 60 \implies 1.2h = 60 \implies h = 50 \text{ inches}$$
Shea's growth: $60 - 50 = 10$ inches.
Ara grew half as many inches as Shea:
$$\frac{10}{2} = 5 \text{ inches}$$
Ara's height now:
$$50 + 5 = 55 \text{ inches}$$

**Check wrong choices:** Choice A (48) comes from computing 20% of 60 incorrectly; choice D (54) would result if Ara also grew 20% of the original height. The key is that Ara grew 5 **inches**, not a percentage.

ANSWER 4: E

# Problem 5:

**Restate:** A die is tossed; the bottom face is hidden. Find the probability that the product of the five visible faces is divisible by 6.

**Working:** The product of all six faces is:
$$1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$$
The product of the five visible faces is:
$$\frac{720}{\text{hidden face}}$$

Check each possible hidden face:
- Hidden = 1: product = 720 → divisible by 6 ✓
- Hidden = 2: product = 360 → divisible by 6 ✓
- Hidden = 3: product = 240 → divisible by 6 ✓
- Hidden = 4: product = 180 → divisible by 6 ✓
- Hidden = 5: product = 144 → divisible by 6 ✓
- Hidden = 6: product = 120 → divisible by 6 ✓

All six outcomes give a product divisible by 6, so the probability is:
$$\frac{6}{6} = 1$$

**Check wrong choices:** One might think hiding the 5 or the 6 removes needed factors, but 720 contains 2 × 3 many times over — removing any single face still leaves factors of 2 and 3 in the product.

ANSWER 5: E

# Problem 6:

**Restate:** Tori's test: 10 arithmetic, 30 algebra, 35 geometry problems (75 total). She got 70%, 40%, and 60% correct respectively, which was below 60% overall. How many more correct answers were needed for 60%?

**Working:** Correct answers in each category:
- Arithmetic: $0.70 \times 10 = 7$
- Algebra: $0.40 \times 30 = 12$
- Geometry: $0.60 \times 35 = 21$

Total correct: $7 + 12 + 21 = 40$

To pass, she needs 60% of 75:
$$0.60 \times 75 = 45$$

Additional problems needed:
$$45 - 40 = 5$$

**Check wrong choices:** Note that the weighted average $\frac{40}{75} \approx 53.3\%$ is indeed below 60%, consistent with the story. A tempting trap is to average the three percentages: $\frac{70+40+60}{3} \approx 56.7\%$, which is wrong because the categories have different sizes.

ANSWER 6: B

# Problem 7:

**Restate:** Given $0 < a < b < c$, which statement is impossible?

**Working:** Test each option:

**A. $a + c < b$:** Since $c > b$, adding positive $a$ gives $a + c > c > b$. So $a + c < b$ **cannot happen**. ✓ (impossible)

Let's verify the others are all possible with suitable numbers:

**B. $a \cdot b < c$:** Try $a = 0.1, b = 0.2, c = 1$: $0.02 < 1$ ✓ possible.

**C. $a + b < c$:** Try $a = 1, b = 2, c = 10$: $3 < 10$ ✓ possible.

**D. $a \cdot c < b$:** Try $a = 0.1, b = 0.5, c = 0.6$: $0.06 < 0.5$ ✓ possible.

**E. $bc = a$:** Try $a = 0.06, b = 0.3, c = 0.2$... wait, need $b < c$. Try $a = 0.02, b = 0.1, c = 0.2$: $bc = 0.02 = a$, and $0 < 0.02 < 0.1 < 0.2$ ✓ possible.

Only option A forces a contradiction: $a + c > c > b$ always holds.

ANSWER 7: A

# Problem 8:

**Restate:** One recipe = 15 cookies using 3 tablespoons of butter; only full recipes allowed. Need 216 cookies. 8 tablespoons = 1 stick. How many sticks?

**Working:** Number of recipes needed:
$$\frac{216}{15} = 14.4$$
Since only full recipes are made, they must make **15 recipes** (14 recipes × 15 = 210 cookies, which is not enough).

Butter needed:
$$15 \times 3 = 45 \text{ tablespoons}$$
Sticks needed:
$$\frac{45}{8} = 5.625 \implies \text{need } 6 \text{ full sticks}$$

**Check wrong choices:** Choice A (5) is the trap — it uses only 40 tablespoons, but 45 are required. You can't buy a partial stick, so round up to 6.

ANSWER 8: B

# Problem 9:

**Restate:** Build a 3-inch cube from 27 unit cubes: 21 red, 6 white. Arrange to minimize white surface area. What fraction of the surface area is white?

**Working:** The 3×3×3 cube has total surface area:
$$6 \text{ faces} \times 9 \text{ sq in} = 54 \text{ square inches}$$

Classify the 27 positions by how many faces each shows:
- **1 inte

Problem 1:
We need the ratio of the place values of the digits 9 and 3 in  
\(74982.1035\).

The digit 9 is in the hundreds place, so its place value is  
\[
100.
\]

The digit 3 is in the thousandths place, so its place value is  
\[
0.001.
\]

Now compute the ratio:
\[
\frac{100}{0.001}=100{,}000.
\]

So the value of the place of the 9 is \(100{,}000\) times as great as the value of the place of the 3.

ANSWER 1: C

Problem 2:
We are told that for every \(3^\circ\) rise in temperature, the gas expands by 4 cubic centimeters.

At temperature \(32^\circ\), the volume is 24 cubic centimeters. We want the volume at \(20^\circ\).

The temperature difference is
\[
32-20=12^\circ.
\]

From \(20^\circ\) to \(32^\circ\), the temperature rises by \(12^\circ\). Since each \(3^\circ\) rise increases the volume by 4 cubic centimeters,
\[
\frac{12}{3}\cdot 4=4\cdot 4=16
\]
cubic centimeters.

So at \(20^\circ\), the volume was
\[
24-16=8
\]
cubic centimeters.

ANSWER 2: A

Problem 3:
Each team plays every other team exactly once. If there are \(n\) teams, the number of games is
\[
\binom{n}{2}=\frac{n(n-1)}{2}.
\]

We are told there were 21 games, so
\[
\frac{n(n-1)}{2}=21.
\]

Multiply by 2:
\[
n(n-1)=42.
\]

We look for two consecutive positive integers whose product is 42:
\[
6\cdot 7=42.
\]

Thus
\[
n=7.
\]

So there were 7 teams.

ANSWER 3: B

Problem 4:
Ara and Shea were originally the same height. Let that original height be \(h\) inches.

Shea has grown 20%, so Shea’s current height is
\[
h+0.20h=1.20h.
\]

We are told Shea is now 60 inches tall:
\[
1.20h=60.
\]

Solve for \(h\):
\[
h=\frac{60}{1.20}=50.
\]

So they were originally 50 inches tall. Shea grew
\[
60-50=10
\]
inches.

Ara grew half as many inches as Shea:
\[
\frac{10}{2}=5
\]
inches.

Therefore Ara is now
\[
50+5=55
\]
inches tall.

ANSWER 4: E

Problem 5:
A fair six-sided die has faces numbered 1 through 6. When it is tossed, one face is hidden on the bottom, and the other five faces are visible.

We need the probability that the product of the five visible numbers is divisible by 6.

A number is divisible by 6 if it has at least one factor of 2 and at least one factor of 3.

No matter which single face is hidden:

- There are three even faces: 2, 4, and 6. Hiding only one of them still leaves at least one even face visible.
- There are two multiples of 3: 3 and 6. Hiding only one of them still leaves at least one multiple of 3 visible.

So the product of the visible faces always contains both a factor of 2 and a factor of 3. Therefore it is always divisible by 6.

Thus the probability is
\[
1.
\]

ANSWER 5: E

Problem 6:
Tori answered:

- \(70\%\) of 10 arithmetic problems:
\[
0.70\cdot 10=7
\]
- \(40\%\) of 30 algebra problems:
\[
0.40\cdot 30=12
\]
- \(60\%\) of 35 geometry problems:
\[
0.60\cdot 35=21
\]

Total correct:
\[
7+12+21=40.
\]

There were 75 problems total. To pass, she needed \(60\%\) correct:
\[
0.60\cdot 75=45.
\]

She answered 40 correctly, so she needed
\[
45-40=5
\]
more correct answers.

ANSWER 6: B

Problem 7:
We are given
\[
0<a<b<c.
\]

We need the impossible statement.

Choice A says
\[
a+c<b.
\]

But since \(c>b\) and \(a>0\),
\[
a+c>b.
\]

So \(a+c<b\) cannot happen.

We can also check that the other choices are possible:

- B: \(a\cdot b<c\) can happen, for example \(a=1,b=2,c=100\).
- C: \(a+b<c\) can happen, for example \(a=1,b=2,c=100\).
- D: \(a\cdot c<b\) can happen, for example \(a=0.1,b=2,c=3\), since \(0.3<2\).
- E: If interpreted as \(bc=a\), it can happen, for example \(a=0.1,b=0.2,c=0.5\), since \(0.2\cdot 0.5=0.1\).

Thus the impossible choice is A.

ANSWER 7: A

Problem 8:
One recipe makes 15 cookies and uses 3 tablespoons of butter. Only full recipes can be made.

Walter and Gretel need 216 cookies. The number of full recipes needed is
\[
\left\lceil \frac{216}{15}\right\rceil.
\]

Since
\[
\frac{216}{15}=14.4,
\]
they must make 15 full recipes.

Butter needed:
\[
15\cdot 3=45
\]
tablespoons.

There are 8 tablespoons in a stick of butter, so the number of sticks needed is
\[
\left\lceil \frac{45}{8}\right\rceil.
\]

Since
\[
\frac{45}{8}=5.625,
\]
they need 6 sticks of butter.

ANSWER 8: B

Problem 9:
A 3-inch cube made from 27 one-inch cubes has surface area
\[
6\cdot 3^2=54
\]
square inches.

Each small cube can show:

- 0 faces if it is the center cube,
- 1 face if it is a face-center cube,
- 2 faces if it is an edge cube,
- 3 faces if it is a corner cube.

We have 6 white cubes. To minimize the white surface area, we should hide as many white cubes as possible.

There is exactly one interior cube, so put one white cube there. It shows 0 square inches.

The remaining 5 white cubes must be on the surface. To minimize exposed white area, put them on face-center positions, where each shows exactly 1 square inch.

So the minimum white surface area is
\[
1\cdot 0+5\cdot 1=5
\]
square inches.

The fraction of the surface area that is white is
\[
\frac{5}{54}.
\]

ANSWER 9: A

Problem 10:
The population is 200 in 1998, and it triples every 25 years.

From 1998 to 2050 is
\[
2050-1998=52
\]
years.

Two full 25-year periods are 50 years, so after 50 years the population is
\[
200\cdot 3\cdot 3=200\cdot 9=1800.
\]

The extra 2 years would add a relatively small amount, so the population in 2050 is a little above 1800. Among the choices, the best estimate is 2000.

ANSWER 10: D

# Problem 1:
Write out the place values of 74982.1035:
- 7 → ten-thousands, 4 → thousands, **9 → hundreds**, 8 → tens, 2 → ones
- 1 → tenths, 0 → hundredths, **3 → thousandths**, 5 → ten-thousandths

So the 9 sits in the hundreds place (value 100) and the 3 sits in the thousandths place (value 0.001).

Ratio = 100 ÷ 0.001 = 100,000.

(Answers like 1,000 or 10,000 come from miscounting decimal places — the 3 is the *third* digit after the point, not the first or second.)

ANSWER 1: C

# Problem 2:
Temperature goes from 32° down to 20°, a drop of 12°.

Each 3° change corresponds to 4 cm³ of change: 12 ÷ 3 = 4 steps.

Volume change = 4 × 4 = 16 cm³ decrease (cooling means shrinking).

Volume = 24 − 16 = 8 cm³.

(40 is the trap from *adding* instead of subtracting; 16 is just the change, not the volume.)

ANSWER 2: A

# Problem 3:
With n teams each playing every other team once, the number of games is C(n,2) = n(n−1)/2.

Set n(n−1)/2 = 21 → n(n−1) = 42 → n = 7 (since 7·6 = 42).

Check: 6 teams give 15 games, 8 teams give 28 games — neither is 21.

ANSWER 3: B

# Problem 4:
Let the common original height be x. Shea grew 20%: 1.20x = 60 → x = 50.

So Shea grew 60 − 50 = 10 inches.

Ara grew half as many inches: 5 inches.

Ara's height = 50 + 5 = 55 inches.

(54 would result from wrongly taking Shea's growth as 10% of 60, etc.)

ANSWER 4: E

# Problem 5:
The product of all six faces is 1·2·3·4·5·6 = 720. If the hidden bottom face is k, the visible product is 720/k.

- k=1: 720 ✓ divisible by 6
- k=2: 360 ✓
- k=3: 240 ✓
- k=4: 180 ✓
- k=5: 144 ✓
- k=6: 120 ✓

Reason why: hiding one face still leaves at least one even number and at least one multiple of 3 among the five visible faces (e.g., hide 6 → 2 and 3 remain; hide 3 → 6 remains).

So the probability is 6/6 = 1.

ANSWER 5: E

# Problem 6:
Correct answers:
- Arithmetic: 70% of 10 = 7
- Algebra: 40% of 30 = 12
- Geometry: 60% of 35 = 21

Total correct = 7 + 12 + 21 = 40.

Needed for 60%: 0.60 × 75 = 45.

She needs 45 − 40 = 5 more.

(40/75 ≈ 53%, consistent with failing.)

ANSWER 6: B

# Problem 7:
Given 0 < a < b < c.

- **A. a + c < b:** Since c > b and a > 0, we get a + c > b. So a + c < b is impossible.
- B. a·b < c: take a = 0.1, b = 0.2, c = 1 → 0.02 < 1 ✓ possible.
- C. a + b < c: take a = 1, b = 2, c = 10 → 3 < 10 ✓ possible.
- D. a·c < b: take a = 0.1, b = 1, c = 2 → 0.2 < 1 ✓ possible.
- E. b/c = a: take a = 0.5, b = 1, c = 2 → 1/2 = 0.5 ✓ possible.

ANSWER 7: A

# Problem 8:
Pans needed: 216 ÷ 15 = 14.4, and only full recipes are made, so they must bake 15 pans.

Butter: 15 × 3 = 45 tablespoons.

Sticks: 45 ÷ 8 = 5.625, and butter comes in whole sticks, so 6 sticks.

(5 sticks = 40 tbsp, not enough.)

ANSWER 8: B

# Problem 9:
The 3×3×3 cube has surface area 6 × 9 = 54 unit squares.

Positions in a 3-cube: 1 center cube (0 faces showing), 6 face-centers (1 face each), 12 edges (2 faces), 8 corners (3 faces).

To minimize white showing, place one white cube in the hidden center (0 white squares) and the other 5 white cubes at face-center positions (1 white square each).

Minimum white area = 5 unit squares out of 54 → 5/54.

ANSWER 9: A

# Problem 10:
From 1998 to 2050 is 52 years — about two 25-year tripling periods (1998 → 2023 → 2048).

Population: 200 → 600 → 1800.

1800 is closest to 2000.

ANSWER 10: D

Problem 1:

We are comparing the place values of the digits 9 and 3 in

\[
74982.1035.
\]

- The digit \(9\) is in the hundreds place, so its place value is \(100\).
- The digit \(3\) is in the thousandths place, so its place value is

\[
0.001.
\]

Now find how many times as great \(100\) is as \(0.001\):

\[
\frac{100}{0.001}=100,000.
\]

So the hundreds place is \(100,\!000\) times the thousandths place.

- A, \(1,\!000\), is too small.
- B, \(10,\!000\), is too small.
- D and E are too large.

ANSWER 1: C

Problem 2:

The gas expands by \(4\) cubic centimeters for every \(3^\circ\) increase in temperature.

The temperature changes from \(32^\circ\) down to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

Since

\[
12^\circ \div 3^\circ=4,
\]

there are 4 groups of \(3^\circ\).

For each group, the volume decreases by \(4\) cubic centimeters:

\[
4\cdot 4=16.
\]

At \(32^\circ\), the volume is \(24\) cubic centimeters, so at \(20^\circ\) it was

\[
24-16=8.
\]

Choices B, C, and D do not subtract the full \(16\) cubic centimeters. Choice E would represent an increase rather than a decrease.

ANSWER 2: A

Problem 3:

If there are \(n\) teams and each pair of teams plays exactly once, then the number of games is

\[
\binom n2=\frac{n(n-1)}2.
\]

We are told there were 21 games:

\[
\frac{n(n-1)}2=21.
\]

Multiply by 2:

\[
n(n-1)=42.
\]

Since

\[
7\cdot 6=42,
\]

we have

\[
n=7.
\]

Checking: with 7 teams, the number of games is

\[
\frac{7\cdot6}{2}=21.
\]

Six teams would produce only \(15\) games, while 8 teams would produce \(28\) games.

ANSWER 3: B

Problem 4:

Let the original height of both Ara and Shea be \(h\) inches.

Shea grew \(20\%\), so Shea’s current height is

\[
1.20h=60.
\]

Thus,

\[
h=\frac{60}{1.2}=50.
\]

So both were originally 50 inches tall.

Shea’s growth was

\[
60-50=10
\]

inches.

Ara grew half as many inches as Shea, so Ara grew

\[
\frac{10}{2}=5
\]

inches.

Therefore, Ara is now

\[
50+5=55
\]

inches tall.

Choice A, 48, is less than Ara’s original height. Choices B, C, and D do not include the correct 5-inch growth.

ANSWER 4: E

Problem 5:

A standard die has faces numbered \(1,2,3,4,5,6\). The product of all six face numbers is

\[
1\cdot2\cdot3\cdot4\cdot5\cdot6=720.
\]

If the bottom face has number \(k\), then the product of the five visible faces is

\[
\frac{720}{k}.
\]

Check each possible bottom face:

\[
\frac{720}{1}=720,\qquad
\frac{720}{2}=360,\qquad
\frac{720}{3}=240,
\]

\[
\frac{720}{4}=180,\qquad
\frac{720}{5}=144,\qquad
\frac{720}{6}=120.
\]

Every one of these numbers is divisible by \(6\).

Therefore, no matter which face is on the bottom, the visible-face product is divisible by 6. The probability is

\[
1.
\]

All of choices A through D are less than 1, but the event always happens.

ANSWER 5: E

Problem 6:

Tori answered correctly:

- Arithmetic:

\[
70\%\text{ of }10=0.7(10)=7
\]

- Algebra:

\[
40\%\text{ of }30=0.4(30)=12
\]

- Geometry:

\[
60\%\text{ of }35=0.6(35)=21
\]

So her total number correct was

\[
7+12+21=40.
\]

To pass, she needs \(60\%\) of 75 questions correct:

\[
0.60(75)=45.
\]

She needs

\[
45-40=5
\]

more correct answers.

Choice A is too small, while choices C, D, and E give more than the number needed to reach exactly 45 correct.

ANSWER 6: B

Problem 7:

We are given

\[
0<a<b<c.
\]

We must find the statement that is impossible.

Consider choice A:

\[
a+c<b.
\]

But \(c>b\), and also \(a>0\). Therefore,

\[
a+c>c>b.
\]

So it is impossible for \(a+c<b\).

The other choices can happen for suitable positive numbers. For example:

- B: \(a\cdot b<c\) can occur with \(a=1,\ b=2,\ c=3\), since \(1\cdot2<3\).
- C: \(a+b<c\) can occur with \(a=1,\ b=2,\ c=4\).
- D: \(a\cdot c<b\) can occur with \(a=\frac14,\ b=1,\ c=2\), since \(\frac14\cdot2=\frac12<1\).
- E: \(bc=a\) can occur with \(b=\frac12,\ c=\frac34\), giving \(a=\frac38\), and indeed

\[
0<\frac38<\frac12<\frac34.
\]

Thus only choice A is impossible.

ANSWER 7: A

Problem 8:

Each full recipe makes 15 cookies.

Walter and Gretel need at least 216 cookies. Since only full recipes may be made, determine how many recipes are needed:

\[
216\div 15=14.4.
\]

They cannot make \(14.4\) recipes, so they must make 15 full recipes.

These 15 recipes make

\[
15\cdot15=225
\]

cookies, enough for the crowd.

Each recipe uses 3 tablespoons of butter, so the total butter needed is

\[
15\cdot3=45
\]

tablespoons.

There are 8 tablespoons in each stick, so

\[
45\div8=5.625.
\]

They must buy 6 full sticks of butter.

Five sticks provide only \(5\cdot8=40\) tablespoons, which is not enough. Six sticks provide \(48\) tablespoons, enough for the 45 tablespoons needed.

ANSWER 8: B

Problem 9:

A \(3\times3\times3\) cube has surface area

\[
6\cdot 3\cdot3=54
\]

square inches, since each of its 6 faces has area \(3\cdot3=9\).

To minimize the white surface showing, place the white cubes in the least exposed positions.

- The one center cube of the large cube has no faces exposed.
- The six face-center cubes each have exactly one face exposed.
- Edge cubes have two exposed faces.
- Corner cubes have three exposed faces.

There are 6 white cubes. Put one white cube in the completely hidden center. Then put the other 5 white cubes at five face-center locations.

Thus, the white cubes show only

\[
5
\]

unit square faces on the surface.

Therefore, the fraction of the surface that is white is

\[
\frac{5}{54}.
\]

Choices such as \(\frac19,\frac{5}{27},\frac29,\frac13\) represent more than 5 white unit squares out of the 54 total surface squares.

ANSWER 9: A

Problem 10:

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

From 1998 to 2050 is

\[
2050-1998=52
\]

years, which is approximately two 25-year periods.

After one 25-year period:

\[
200\cdot3=600.
\]

After two 25-year periods:

\[
600\cdot3=1800.
\]

Since 2050 is slightly more than 50 years after 1998, the population should be slightly more than 1800. The closest estimate is

\[
20