🔬 Model Lab

I need to find the percent decrease in the price per box from last week to this week.

**Step 1: Find the price per box last week**

Last week: 4 boxes for $5
- Price per box = $5 ÷ 4 = $1.25 per box

**Step 2: Find the price per box this week**

This week: 5 boxes for $4
- Price per box = $4 ÷ 5 = $0.80 per box

**Step 3: Calculate the decrease in price**

Decrease = $1.25 - $0.80 = $0.45

**Step 4: Calculate the percent decrease**

Percent decrease = (Decrease ÷ Original price) × 100%

Percent decrease = ($0.45 ÷ $1.25) × 100%

Let me compute $0.45 ÷ $1.25:
- $0.45 ÷ $1.25 = 45 ÷ 125 = 9 ÷ 25 = 0.36

Percent decrease = 0.36 × 100% = 36%

**Step 5: Find the closest answer**

36% is closest to 35% among the given choices.

Let me verify this makes sense:
- Original price: $1.25
- If we decrease by 35%: $1.25 × 0.35 = $0.4375
- New price would be: $1.25 - $0.4375 = $0.8125 ≈ $0.80 ✓

- If we decrease by 40%: $1.25 × 0.40 = $0.50
- New price would be: $1.25 - $0.50 = $0.75 (too low)

The answer of 36% is definitely closer to 35% than to 40%.

ANSWER: B

1. Restate the problem  
   We need the percent decrease in the price per box of tissues when the price changes from 4 boxes for \$5 to 5 boxes for \$4.

2. Find the original unit price  
   Original price per box = \$5 ÷ 4 = \$1.25.

3. Find the sale unit price  
   Sale price per box = \$4 ÷ 5 = \$0.80.

4. Compute the absolute decrease in price  
   Decrease = \$1.25 − \$0.80 = \$0.45.

5. Compute the percent decrease  
   Percent decrease = (decrease ÷ original price) × 100%  
                   = (\$0.45 ÷ \$1.25) × 100%  
                   = 0.36 × 100%  
                   = 36%.

6. Compare to the answer choices  
   The choices are 30%, 35%, 40%, 45%, 65%.  
   36% is 1% away from 35%, 4% away from 40%, etc.  
   Therefore the closest choice is 35%.

ANSWER: B

To find the percent decrease in the price per box during the sale, we can follow these steps:

**Step 1: Restate what is being asked**
We need to find the percentage by which the price of a single box of facial tissue decreased from last week to this week, and then determine which of the given choices is closest to this percentage.

**Step 2: Calculate the price per box last week**
Last week, the price was $5 for 4 boxes. 
$$\text{Price per box last week} = \frac{\$5.00}{4} = \$1.25$$

**Step 3: Calculate the price per box this week**
This week, the price is $4 for 5 boxes.
$$\text{Price per box this week} = \frac{\$4.00}{5} = \$0.80$$

**Step 4: Calculate the decrease in price per box**
The decrease in price per box is:
$$\text{Decrease} = \$1.25 - \$0.80 = \$0.45$$

**Step 5: Calculate the percent decrease**
The percent decrease is calculated relative to the original price (last week's price):
$$\text{Percent Decrease} = \left( \frac{\text{Decrease}}{\text{Original Price}} \right) \times 100\%$$
$$\text{Percent Decrease} = \left( \frac{0.45}{1.25} \right) \times 100\%$$

To simplify the fraction:
$$\frac{0.45}{1.25} = \frac{45}{125} = \frac{9}{25}$$

Converting this to a percentage:
$$\frac{9}{25} \times 100\% = 9 \times 4\% = 36\%$$

**Step 6: Compare with the choices to find the closest value**
We compare our calculated value of $36\%$ to the given choices:
* A. $30\%$ (difference of $6\%$)
* B. $35\%$ (difference of $1\%$)
* C. $40\%$ (difference of $4\%$)
* D. $45\%$ (difference of $9\%$)
* E. $65\%$ (difference of $29\%$)

The closest choice to $36\%$ is $35\%$.

ANSWER: B

Last week, the price was 4 boxes for $5, so the price per box is found by dividing:
\[
\frac{5}{4} = 1.25
\]
dollars per box.

This week, the price is 5 boxes for $4, so the price per box is
\[
\frac{4}{5} = 0.80
\]
dollars per box.

The absolute decrease per box is
\[
1.25 - 0.80 = 0.45
\]
dollars.

The percent decrease is the absolute decrease divided by the original price per box:
\[
\frac{0.45}{1.25} = 0.36,
\]
or 36 percent.

Among the given choices (30 percent, 35 percent, 40 percent, 45 percent, 65 percent), the value 36 percent is closest to 35 percent.

To confirm the calculation another way, express the prices as fractions throughout. The original price per box is \(5/4\) and the sale price per box is \(4/5\). Their difference is
\[
\frac{5}{4} - \frac{4}{5} = \frac{25 - 16}{20} = \frac{9}{20}.
\]
Dividing the difference by the original price yields
\[
\frac{9/20}{5/4} = \frac{9}{20} \cdot \frac{4}{5} = \frac{36}{100} = 0.36,
\]
again confirming 36 percent.

Thus the closest choice is 35 percent.

ANSWER: B

1. **Restate the problem**: We need to find the percent decrease in the price per box of facial tissue from last week to this week when the pricing changed from 4 boxes for $5 to 5 boxes for $4.

2. **Calculate the price per box last week**: Last week, 4 boxes cost $5. So, the price per box was $5 / 4 = $1.25.

3. **Calculate the price per box this week**: This week, 5 boxes cost $4. So, the price per box is $4 / 5 = $0.80.

4. **Find the decrease in price per box**: The decrease in price per box is $1.25 - $0.80 = $0.45.

5. **Calculate the percent decrease**: The percent decrease is calculated using the formula: (Decrease / Original Price) * 100. So, the percent decrease = ($0.45 / $1.25) * 100.

6. **Perform the percent decrease calculation**: ($0.45 / $1.25) * 100 = 0.36 * 100 = 36%.

7. **Compare the calculated percent decrease to the given choices**: The calculated percent decrease is 36%. The closest value among the given choices is 35%.

8. **Rule out the wrong choices**:
- A. 30% is too low since 36% is closer to 35% than 30%.
- C. 40% is too high.
- D. 45% is even higher and thus incorrect.
- E. 65% is significantly too high.

9. **Select the correct answer based on the closest value**: The closest percent decrease is 35%, which corresponds to choice B.

ANSWER: B

Let’s go step by step.

---

**Step 1 – Restate the problem**  
We have two pricing schemes:  
- Last week: 4 boxes for $5.  
- This week: 5 boxes for $4.  

We want the percent decrease in the **price per box**, to the closest given option.

---

**Step 2 – Find price per box for each case**  

*Last week:*  
Price per box = \( \frac{5}{4} = 1.25 \) dollars per box.

*This week:*  
Price per box = \( \frac{4}{5} = 0.80 \) dollars per box.

---

**Step 3 – Compute the decrease in price per box**  

Decrease = \( 1.25 - 0.80 = 0.45 \) dollars per box.

---

**Step 4 – Find the percent decrease**  

Percent decrease = \( \frac{\text{decrease}}{\text{original price}} \times 100\)  
= \( \frac{0.45}{1.25} \times 100 \).

Compute:  
\( \frac{0.45}{1.25} = 0.36 \),  
\( 0.36 \times 100 = 36\% \).

---

**Step 5 – Choose the closest given percentage**  

We have 36%, which is closest to **35%** among the choices.

---

**Step 6 – Rule out wrong choices**  
- 30% is 6 points lower.  
- 40% is 4 points higher.  
- 45% and 65% are much too high.  
So 35% is the closest match.

---

ANSWER: B

1. **Understand what is being asked**:  
We are given two pricing schemes for boxes of facial tissue:  
- **Last week**: 4 boxes for $5  
- **This week (sale)**: 5 boxes for $4  
We are to find the **percent decrease in the price per box** from last week to this week, and choose the answer choice **closest** to that percent.

2. **Find the price per box last week**:  
Last week: 4 boxes cost $5  
→ Price per box = $ \frac{5}{4} = 1.25 $ dollars

3. **Find the price per box this week**:  
This week: 5 boxes cost $4  
→ Price per box = $ \frac{4}{5} = 0.80 $ dollars

4. **Find the decrease in price per box**:  
Decrease = Old price – New price  
= $ 1.25 - 0.80 = 0.45 $ dollars

5. **Find the percent decrease**:  
Percent decrease = $ \frac{\text{Decrease}}{\text{Original price}} \times 100\% $  
= $ \frac{0.45}{1.25} \times 100\% $

6. **Compute the fraction**:  
$ \frac{0.45}{1.25} = \frac{45}{125} $  
Simplify: divide numerator and denominator by 5 → $ \frac{9}{25} $  
Now convert to percent:  
$ \frac{9}{25} = 0.36 = 36\% $

7. **Compare to answer choices**:  
The percent decrease is **36%**, and we are to choose the **closest** option.  
Choices:  
A. 30%  
B. 35%  
C. 40%  
D. 45%  
E. 65%  

36% is **1% away from 35%**, and **4% away from 40%**, so **closest to 35%**.

ANSWER: B

 **Step 1:** Restate the problem.  
We are asked to find the percent decrease in the price per box of facial tissue from last week to this week, and then identify which listed percentage it is closest to.

**Step 2:** Compute last week’s price per box.  
Last week the price was 4 boxes for $5.  
\[
\text{Price per box}_{\text{old}}=\frac{\$5}{4}=\$1.25
\]

**Step 3:** Compute this week’s sale price per box.  
This week the price is 5 boxes for $4.  
\[
\text{Price per box}_{\text{new}}=\frac{\$4}{5}=\$0.80
\]

**Step 4:** Find the amount of decrease per box.  
\[
\text{Decrease}= \$1.25-\$0.80 = \$0.45
\]

**Step 5:** Calculate the percent decrease.  
The percent decrease is the amount of decrease divided by the original price, multiplied by 100.  
\[
\text{Percent decrease}= \frac{0.45}{1.25}\times 100\%
= \frac{45}{125}\times 100\%
= \frac{9}{25}\times 100\%
= 36\%
\]

**Step 6:** Compare 36% with the answer choices and rule out the wrong ones.  
- **A. 30%** – too low by 6 percentage points.  
- **B. 35%** – only 1 percentage point away from 36%.  
- **C. 40%** – 4 percentage points away, so not as close as 35%.  
- **D. 45%** – 9 percentage points too high.  
- **E. 65%** – far too high.  

Because 36% is nearest to 35%, the correct choice is B.

ANSWER: B

Here is the step-by-step solution:

**1. Understand the Goal**
We need to find the percent decrease in the price of a single box of facial tissue. To do this, we must first determine the price of one box before the sale (original price) and the price of one box during the sale (new price).

**2. Calculate the Original Price Per Box**
Originally, 4 boxes cost $\$5$.
$$\text{Original Price} = \frac{\$5}{4 \text{ boxes}} = \$1.25 \text{ per box}$$

**3. Calculate the Sale Price Per Box**
During the sale, 5 boxes cost $\$4$.
$$\text{Sale Price} = \frac{\$4}{5 \text{ boxes}} = \$0.80 \text{ per box}$$

**4. Calculate the Amount of Decrease**
Subtract the sale price from the original price to find the difference in cost per box.
$$\text{Decrease} = \$1.25 - \$0.80 = \$0.45$$

**5. Calculate the Percent Decrease**
The formula for percent decrease is:
$$\text{Percent Decrease} = \left( \frac{\text{Decrease}}{\text{Original Price}} \right) \times 100\%$$

Substitute our values into the formula:
$$\text{Percent Decrease} = \left( \frac{0.45}{1.25} \right) \times 100\%$$

To make the division easier, we can multiply the numerator and denominator by 100 to remove decimals:
$$= \left( \frac{45}{125} \right) \times 100\%$$

Simplify the fraction by dividing both numbers by 5:
$$= \left( \frac{9}{25} \right) \times 100\%$$

Perform the multiplication ($100 / 25 = 4$, so $9 \times 4$):
$$= 9 \times 4 = 36\%$$

**6. Evaluate the Choices**
The calculated percent decrease is exactly **36%**. We must find which option this value is closest to:
*   **A. 30%:** The difference is $|36 - 30| = 6$.
*   **B. 35%:** The difference is $|36 - 35| = 1$.
*   **C. 40%:** The difference is $|36 - 40| = 4$.
*   **D. 45%:** The difference is $|36 - 45| = 9$.
*   **E. 65%:** The difference is $|36 - 65| = 29$.

Since 36% is only 1 percentage point away from 35%, choice B is the closest answer.

ANSWER: B

**Step 1: Find the price per box last week.**  
4 boxes cost $5, so  
\[
\text{price}_{\text{last week}}=\frac{5}{4}=1.25\text{ dollars per box}.
\]

**Step 2: Find the price per box this week.**  
5 boxes cost $4, so  
\[
\text{price}_{\text{this week}}=\frac{4}{5}=0.80\text{ dollars per box}.
\]

**Step 3: Compute the amount of decrease.**  
\[
\Delta = 1.25 - 0.80 = 0.45\text{ dollars per box}.
\]

**Step 4: Express the decrease as a percent of the original price.**  
\[
\text{percent decrease}= \frac{\Delta}{1.25}\times 100\%
= \frac{0.45}{1.25}\times 100\%
= \frac{45}{125}\times 100\%
=0.36\times100\%
=36\%.
\]

**Step 5: Choose the answer choice closest to 36 %.**  
- 30 % is 6 % away.  
- 35 % is 1 % away.  
- 40 % is 4 % away.  
- 45 % is 9 % away.  
- 65 % is far away.

The smallest distance is to 35 %, so the best match among the given options is **35 %**.

**Conclusion:** The percent decrease is closest to 35 %.

ANSWER: B

 per box last week and this week, then use the percent - decrease formula.
2. **Set up the math for the price per box last week**:
   - Last week, small boxes of facial tissue were priced at 4 boxes for $5. So the price per box last week, $P_{last}$, is $\frac{5}{4}=\$1.25$.
3. **Set up the math for the price per box this week**:
   - This week, they are on sale at 5 boxes for $4. So the price per box this week, $P_{this}$, is $\frac{4}{5}=\$0.8$.
4. **Use the percent - decrease formula**:
   - The percent - decrease formula is $\text{Percent Decrease}=\frac{\text{Original Value}-\text{New Value}}{\text{Original Value}}\times100\%$.
   - Here, the original value is the price per box last week ($P_{last} = 1.25$) and the new value is the price per box this week ($P_{this}=0.8$).
   - Substitute these values into the formula: $\text{Percent Decrease}=\frac{1.25 - 0.8}{1.25}\times100\%$.
   - First, calculate the numerator: $1.25−0.8 = 0.45$.
   - Then, divide by the original value: $\frac{0.45}{1.25}=0.36$.
   - Finally, multiply by 100%: $0.36\times100\% = 36\%$.
5. **Rule out the wrong choices**:
   - Choice A: 30% is less than 36%, so it is not the closest value.
   - Choice B: 36% is closest to 35% among the given options.
   - Choice C: 40% is greater than 36%, so it is not the closest value.
   - Choice D: 45% is much greater than 36%, so it is not the closest value.
   - Choice E: 65% is way greater than 36%, so it is not the closest value.

ANSWER: C (since 36% is closest to 40% among the given options is wrong, it's closest to 35% but if we consider the calculation and the options, we made a mistake above, let's re - check:
The percent decrease $\frac{1.25 - 0.8}{1.25}\times100\%=\frac{0.45}{1.25}\times100\% = 36\%$, and 36% is closest to 35% among the given options, so the answer is C is incorrect, the correct answer should be closest to 36% among the options, so it's C (the original thought about closest to 40% was wrong, 36% is closer to 35% in the context of the given choices) - the correct choice is C (the closest one in the options to 36% is 35% and the option C in the original set is 40% which is wrong, we re - evaluate and find the closest to 36% is 35% and among the choices, the answer is C (in the sense of the closest value in the given options, we note that 36% is closer to 35% than to 40% etc.) - actually the correct choice is C (the calculation shows 36% and among the options 35% is the closest, and the option labeled C is a wrong - label situation, the correct answer based on the calculation and comparison is the value closest to 36% which is 35% and in the choices, we choose the closest one and the answer is C (this is a bit of a confusion in the way we refer, but the value 36% is closest to 35% and in the multiple - choice, we pick the closest option, so the answer is C (where C represents the choice that is the closest to our calculated value among the given ones, and 35% is the closest to 36% among A = 30%, B = 35%, C = 40%, D = 45%, E = 65%)). The correct choice is C (meaning the choice that is closest to the calculated 36% is the one labeled C in a wrong - sense of letter - value, but in fact, we choose B (35%) as the closest value, and in the multiple - choice, B is the correct choice. Let's correct the final step:
The percent decrease is 36%. Comparing 36% to the options:
   - |36 - 30|=6
   - |36 - 35|=1
   - |36 - 40|=4
   - |36 - 45|=9
   - |36 - 65|=29
Since 1 is the smallest difference, 36% is closest to 35%. So the answer is B.

ANSWER: C (wrong above, correct is B) - let's be precise:
The percent decrease is $\frac{1.25 - 0.8}{1.25}\times100\% = 36\%$. The closest value to 36% among the options 30%, 35%, 40%, 45%, 65% is 35%. So the answer is B.

ANSWER: C (this is a repeated wrong reference, the correct final answer)
We calculate the percent decrease as $\frac{1.25 - 0.8}{1.25}\times100\%=36\%$. Among the given options, 35% is the closest value to 36%. So the correct choice is B.

ANSWER: B

### Step 1: Restate the problem goal
We need to calculate the percent decrease in the price per box of facial tissue between last week's regular price and this week's sale price. Percent decrease follows the formula:
$$\text{Percent Decrease} = \left(\frac{\text{Original Price per Box} - \text{New Sale Price per Box}}{\text{Original Price per Box}}\right) \times 100$$

### Step 2: Calculate last week's original price per box
Last week, 4 boxes cost $5. The price per box is total cost divided by number of boxes:
$$\text{Original price per box} = \frac{\$5}{4} = \$1.25$$

### Step 3: Calculate this week's sale price per box
This week, 5 boxes cost $4. The sale price per box is:
$$\text{Sale price per box} = \frac{\$4}{5} = \$0.80$$

### Step 4: Compute the percent decrease
First find the absolute decrease in price per box: $1.25 - $0.80 = $0.45. Plug into the percent decrease formula:
$$\text{Percent Decrease} = \left(\frac{0.45}{1.25}\right) \times 100 = 36\%$$

### Step 5: Rule out incorrect options
- A (30%): A 30% decrease would mean a new price of $0.875, which is higher than the actual sale price of $0.80, so this is too small a decrease.
- C (40%): A 40% decrease would mean a new price of $0.75, which is lower than the actual sale price, so this is too large a decrease.
- D (45%) and E (65%): These are far too large, as they would require sale prices far below the actual $0.80 per box.
36% is closest to 35%.

ANSWER: B

### Step 1: Calculate the original price per box (last week)
Last week, 4 boxes cost $5. To find the price per box, divide total cost by number of boxes:
$$\text{Original price per box} = \frac{5}{4} = \$1.25$$

### Step 2: Calculate the sale price per box (this week)
This week, 5 boxes cost $4. Divide total cost by number of boxes:
$$\text{Sale price per box} = \frac{4}{5} = \$0.80$$

### Step 3: Find the amount of the price decrease per box
Subtract the sale price from the original price:
$$\text{Decrease per box} = 1.25 - 0.80 = \$0.45$$

### Step 4: Calculate the percent decrease
Percent decrease is defined as $\left(\frac{\text{Amount of decrease}}{\text{Original value}}\right) \times 100$. Substitute the values:
$$\text{Percent decrease} = \left(\frac{0.45}{1.25}\right) \times 100$$
Simplify the fraction: $\frac{0.45}{1.25} = \frac{45}{125} = \frac{9}{25} = 0.36$, so:
$$\text{Percent decrease} = 0.36 \times 100 = 36\%$$

### Step 5: Match to the closest option
36% is 1 percentage point away from 35% (option B) and 4 percentage points away from 40% (option C), so it is closest to 35%.

ANSWER: B

To determine the percent decrease in the price per box during the sale, we need to follow these steps:

1. **Calculate the original price per box:**
   - The original price is 4 boxes for $5.
   - Therefore, the price per box is $5 / 4 = $1.25.

2. **Calculate the sale price per box:**
   - The sale price is 5 boxes for $4.
   - Therefore, the price per box is $4 / 5 = $0.80.

3. **Calculate the decrease in price per box:**
   - The decrease in price per box is $1.25 - $0.80 = $0.45.

4. **Calculate the percent decrease:**
   - The percent decrease is given by the formula: 
     \[
     \text{Percent Decrease} = \left( \frac{\text{Decrease in Price}}{\text{Original Price}} \right) \times 100
     \]
   - Substituting the values we have:
     \[
     \text{Percent Decrease} = \left( \frac{0.45}{1.25} \right) \times 100
     \]
   - Simplifying the fraction:
     \[
     \frac{0.45}{1.25} = 0.36
     \]
   - Therefore:
     \[
     \text{Percent Decrease} = 0.36 \times 100 = 36\%
     \]

5. **Compare the calculated percent decrease to the given choices:**
   - The closest value to 36% among the given choices is 35%.

Thus, the percent decrease in the price per box during the sale was closest to 35%.

ANSWER: B

**Problem Restatement:**
We need to find the percent decrease in the price per box of facial tissue from last week to this week, and then determine which given option this percent decrease is closest to.

---

**Step 1: Calculate the price per box last week.**
Last week, 4 boxes cost $5.
Price per box last week = Total cost / Number of boxes
= $5 / 4
= $1.25 per box

---

**Step 2: Calculate the price per box this week.**
This week, 5 boxes cost $4.
Price per box this week = Total cost / Number of boxes
= $4 / 5
= $0.80 per box

---

**Step 3: Calculate the decrease in price per box.**
Decrease in price per box = Price last week - Price this week
= $1.25 - $0.80
= $0.45

---

**Step 4: Calculate the percent decrease in price per box.**
Percent decrease = (Decrease in price / Original price) * 100%
= ($0.45 / $1.25) * 100%
= 0.36 * 100%
= 36%

---

**Step 5: Compare the percent decrease to the given options.**
The percent decrease is 36%. The closest option to 36% is 35%.

---

**Step 6: Rule out the other choices.**
- A (30%) is too low.
- C (40%) is slightly higher than 36%, but 35% is closer.
- D (45%) and E (65%) are much higher than 36%.

---
ANSWER: B

I need to find the maximum number of additional license plates that can be made by adding two letters to the three sets.

**Step 1: Understand the current situation**

Current sets:
- First position: {C, H, L, P, R} — 5 letters
- Second position: {A, I, O} — 3 letters
- Third position: {D, M, N, T} — 4 letters

Current number of plates = 5 × 3 × 4 = 60

**Step 2: Understand what we're optimizing**

We want to add 2 letters total across the three sets to maximize the increase in the number of plates. We need to find which distribution of the 2 new letters gives us the largest increase.

Let's denote the new sizes as (a, b, c) where we add letters to get from (5, 3, 4) to (5+x, 3+y, 4+z) where x + y + z = 2.

New number of plates = (5+x)(3+y)(4+z)

Additional plates = (5+x)(3+y)(4+z) - 60

**Step 3: Try all possible distributions of 2 letters**

**Case 1: Both letters to first set** (x=2, y=0, z=0)
New plates = 7 × 3 × 4 = 84
Additional = 84 - 60 = 24

**Case 2: Both letters to second set** (x=0, y=2, z=0)
New plates = 5 × 5 × 4 = 100
Additional = 100 - 60 = 40

**Case 3: Both letters to third set** (x=0, y=0, z=2)
New plates = 5 × 3 × 6 = 90
Additional = 90 - 60 = 30

**Case 4: One letter to first, one to second** (x=1, y=1, z=0)
New plates = 6 × 4 × 4 = 96
Additional = 96 - 60 = 36

**Case 5: One letter to first, one to third** (x=1, y=0, z=1)
New plates = 6 × 3 × 5 = 90
Additional = 90 - 60 = 30

**Case 6: One letter to second, one to third** (x=0, y=1, z=1)
New plates = 5 × 4 × 5 = 100
Additional = 100 - 60 = 40

**Step 4: Compare all cases**

- Case 1: 24 additional plates
- Case 2: 40 additional plates
- Case 3: 30 additional plates
- Case 4: 36 additional plates
- Case 5: 30 additional plates
- Case 6: 40 additional plates

The maximum is 40 additional plates, achieved by either:
- Adding both letters to the second set (3 → 5 letters), or
- Adding one letter each to the second and third sets

**Step 5: Verify the maximum**

Case 2: 5 × 5 × 4 = 100, and 100 - 60 = 40 ✓
Case 6: 5 × 4 × 5 = 100, and 100 - 60 = 40 ✓

Both give the same answer of 40 additional plates.

ANSWER: D

1. Restate the problem  
   We start with license plates of length 3, where the 1st letter is chosen from a set of size 5, the 2nd from a set of size 3, and the 3rd from a set of size 4, for a total of  
   T₀ = 5·3·4 = 60 plates. Two new letters are to be added (they may both go into one set, or one each into two sets), and we seek the largest possible increase in the total number of plates.

2. Introduce variables and formula  
   Let a, b, c be the numbers of new letters added to the first, second, and third sets respectively. Then  
   a + b + c = 2, a,b,c ≥ 0 integers.  
   After adding, the set sizes become 5+a, 3+b, 4+c, so the new total is  
   T(a,b,c) = (5+a)(3+b)(4+c).  
   The additional plates are  
   Δ(a,b,c) = T(a,b,c) − T₀ = (5+a)(3+b)(4+c) − 60.

3. List all distributions of the 2 new letters and compute Δ  
   a) (a,b,c)=(2,0,0):  T=7·3·4=84,  Δ=84−60=24  
   b) (a,b,c)=(1,1,0):  T=6·4·4=96,  Δ=96−60=36  
   c) (a,b,c)=(1,0,1):  T=6·3·5=90,  Δ=90−60=30  
   d) (a,b,c)=(0,2,0):  T=5·5·4=100, Δ=100−60=40  
   e) (a,b,c)=(0,1,1):  T=5·4·5=100, Δ=100−60=40  
   f) (a,b,c)=(0,0,2):  T=5·3·6=90,  Δ=90−60=30  

4. Determine the maximum increase  
   The values of Δ are {24, 36, 30, 40, 40, 30}.  The maximum is 40.

5. Rule out the other answer choices  
   • 24, 30, 36 appear in our list but are not the maximum.  
   • 60 is larger than any Δ we obtained, so it cannot occur.  

Therefore, the largest possible number of additional plates is 40.

ANSWER: D

Here is the step-by-step solution to the problem:

### Step 1: Identify the initial sizes of the three sets of letters
The license plates consist of three letters, chosen from three distinct sets:
* The first set, $S_1 = \{C, H, L, P, R\}$, has a size of $5$.
* The second set, $S_2 = \{A, I, O\}$, has a size of $3$.
* The third set, $S_3 = \{D, M, N, T\}$, has a size of $4$.

### Step 2: Calculate the initial number of license plates
Using the fundamental counting principle, the initial number of unique license plates is:
$$\text{Initial Plates} = 5 \times 3 \times 4 = 60$$

### Step 3: List all possible ways to distribute the 2 new letters
We can add the two new letters in two ways:
1. **Both letters to one set** (adding $2$ to one set size, and $0$ to the others).
2. **One letter to one set and one to another** (adding $1$ to two different set sizes, and $0$ to the third).

Let's analyze all possible distributions of these 2 letters and calculate the resulting number of plates:

* **Case 1: Both letters added to one set**
  * Add 2 to $S_1$: New sizes are $7, 3, 4$. 
    $$\text{Plates} = 7 \times 3 \times 4 = 84$$
  * Add 2 to $S_2$: New sizes are $5, 5, 4$. 
    $$\text{Plates} = 5 \times 5 \times 4 = 100$$
  * Add 2 to $S_3$: New sizes are $5, 3, 6$. 
    $$\text{Plates} = 5 \times 3 \times 6 = 90$$

* **Case 2: One letter added to each of two sets**
  * Add 1 to $S_1$ and 1 to $S_2$: New sizes are $6, 4, 4$. 
    $$\text{Plates} = 6 \times 4 \times 4 = 96$$
  * Add 1 to $S_1$ and 1 to $S_3$: New sizes are $6, 3, 5$. 
    $$\text{Plates} = 6 \times 3 \times 5 = 90$$
  * Add 1 to $S_2$ and 1 to $S_3$: New sizes are $5, 4, 5$. 
    $$\text{Plates} = 5 \times 4 \times 5 = 100$$

### Step 4: Find the maximum number of additional plates
The maximum number of license plates we can obtain after adding two letters is $100$ (which occurs either by adding both letters to $S_2$, or by adding one to $S_2$ and one to $S_3$).

To find the number of **additional** plates, we subtract the initial number of plates from this maximum:
$$\text{Additional Plates} = 100 - 60 = 40$$

### Step 5: Rule out the wrong choices
* **A. 24**: This is the increase if we added both letters to $S_1$ ($84 - 60 = 24$), which is not maximized.
* **B. 30**: This is the increase if we added both to $S_3$ or one to $S_1$ and one to $S_3$ ($90 - 60 = 30$), which is not maximized.
* **C. 36**: This is the increase if we added one to $S_1$ and one to $S_2$ ($96 - 60 = 36$), which is not maximized.
* **D. 40**: This is the correct maximum increase.
* **E. 60**: This is the initial number of plates, not the additional plates.

ANSWER: D

The original license plates consist of three letters chosen independently according to the following rules:
- Position 1 is chosen from a set of 5 letters, giving 5 choices.
- Position 2 is chosen from a set of 3 letters, giving 3 choices.
- Position 3 is chosen from a set of 4 letters, giving 4 choices.

The total number of original plates is therefore
\[
5 \times 3 \times 4 = 60.
\]
Two new letters are added. These two letters may be placed in any of the three sets (both in one set, or one each in two different sets). The goal is to maximize the number of new plates created, which is equivalent to maximizing the product of the three new set sizes and then subtracting the original total of 60.

The six possible distributions and the resulting totals are computed as follows:

- Both new letters added to the first set (size becomes 7):  
  \[
  7 \times 3 \times 4 = 84 \quad \Rightarrow \quad 84 - 60 = 24
  \] additional plates.

- Both new letters added to the second set (size becomes 5):  
  \[
  5 \times 5 \times 4 = 100 \quad \Rightarrow \quad 100 - 60 = 40
  \] additional plates.

- Both new letters added to the third set (size becomes 6):  
  \[
  5 \times 3 \times 6 = 90 \quad \Rightarrow \quad 90 - 60 = 30
  \] additional plates.

- One new letter added to the first set and one to the second set (sizes become 6 and 4):  
  \[
  6 \times 4 \times 4 = 96 \quad \Rightarrow \quad 96 - 60 = 36
  \] additional plates.

- One new letter added to the first set and one to the third set (sizes become 6 and 5):  
  \[
  6 \times 3 \times 5 = 90 \quad \Rightarrow \quad 90 - 60 = 30
  \] additional plates.

- One new letter added to the second set and one to the third set (sizes become 4 and 5):  
  \[
  5 \times 4 \times 5 = 100 \quad \Rightarrow \quad 100 - 60 = 40
  \] additional plates.

Among the six possible values of additional plates (24, 40, 30, 36, 30, 40), the maximum is 40. No other distribution of the two new letters is permitted by the problem statement, so 40 is the largest achievable number.

(The incorrect choices can be ruled out directly from the exhaustive list above: 24 arises only from enlarging the first set, 30 arises only from enlarging the third set alone or from the first-and-third split, and 36 arises only from the first-and-second split. The value 60 would require a total of 120 plates, which exceeds every product obtained above.)

ANSWER: D

1. **Restate the problem**: We need to determine the largest possible number of additional bicycle license plates that can be made in Flatville by adding two new letters to the existing sets of letters used for the license plates.

2. **Understand the initial condition**: Initially, the number of license plates is calculated by multiplying the number of choices for each position. There are 5 choices for the first letter, 3 for the second, and 4 for the third. So, the initial number of license plates = 5 * 3 * 4 = 60.

3. **Consider the possible additions of new letters**: We can add 2 letters to any one of the three sets or add 1 letter to one set and 1 letter to another set. We need to calculate the number of additional license plates for each scenario.

4. **Calculate the number of license plates if 2 letters are added to one set**:
   - Adding 2 to the first set: (5+2) * 3 * 4 = 7 * 3 * 4 = 84. Additional plates = 84 - 60 = 24.
   - Adding 2 to the second set: 5 * (3+2) * 4 = 5 * 5 * 4 = 100. Additional plates = 100 - 60 = 40.
   - Adding 2 to the third set: 5 * 3 * (4+2) = 5 * 3 * 6 = 90. Additional plates = 90 - 60 = 30.

5. **Calculate the number of license plates if 1 letter is added to one set and 1 to another**:
   - Adding 1 to the first and 1 to the second set: (5+1) * (3+1) * 4 = 6 * 4 * 4 = 96. Additional plates = 96 - 60 = 36.
   - Adding 1 to the first and 1 to the third set: (5+1) * 3 * (4+1) = 6 * 3 * 5 = 90. Additional plates = 90 - 60 = 30.
   - Adding 1 to the second and 1 to the third set: 5 * (3+1) * (4+1) = 5 * 4 * 5 = 100. Additional plates = 100 - 60 = 40.

6. **Compare the results to find the maximum additional license plates**:
   - The maximum additional plates from adding 2 letters to one set is 40 (when adding to the second set).
   - The maximum additional plates from adding 1 letter to two sets is 40 (when adding to the second and third sets).

7. **Rule out the wrong choices**:
   - A. 24 is less than 40, so it's incorrect.
   - B. 30 is less than 40, so it's incorrect.
   - C. 36 is less than 40, so it's incorrect.
   - D. 40 matches our maximum, so it's potentially correct.
   - E. 60 is greater than 40, and we didn't achieve this in any scenario, so it's incorrect.

8. **Conclusion**: The largest possible number of additional license plates is 40.

ANSWER: D

**Step 1: Restate the problem**  
We have three sets of letters for the three positions on the license plate:  
- First letter: {C, H, L, P, R} → 5 choices  
- Second letter: {A, I, O} → 3 choices  
- Third letter: {D, M, N, T} → 4 choices  

The total number of possible plates initially is \(5 \times 3 \times 4 = 60\).  

Two new distinct letters are added. They can either:  
- Both be added to the same set, or  
- Be split with one letter going to one set and the other to a different set.  

We need to find the **largest possible increase** in the number of license plates.

**Step 2: Identify the possible ways to add the letters**  
Let the set sizes be \(a = 5\), \(b = 3\), \(c = 4\). We can increase them by a total of 2, distributed as:  
- (2,0,0): add both letters to set 1  
- (0,2,0): add both to set 2  
- (0,0,2): add both to set 3  
- (1,1,0): add one to set 1, one to set 2  
- (1,0,1): add one to set 1, one to set 3  
- (0,1,1): add one to set 2, one to set 3  

**Step 3: Compute the new total and the increase for each case**  
Starting total = 60.  

1. (2,0,0): new sizes 7, 3, 4 → \(7 \times 3 \times 4 = 84\); increase = \(84 - 60 = 24\)  
2. (0,2,0): new sizes 5, 5, 4 → \(5 \times 5 \times 4 = 100\); increase = \(100 - 60 = 40\)  
3. (0,0,2): new sizes 5, 3, 6 → \(5 \times 3 \times 6 = 90\); increase = \(90 - 60 = 30\)  
4. (1,1,0): new sizes 6, 4, 4 → \(6 \times 4 \times 4 = 96\); increase = \(96 - 60 = 36\)  
5. (1,0,1): new sizes 6, 3, 5 → \(6 \times 3 \times 5 = 90\); increase = \(90 - 60 = 30\)  
6. (0,1,1): new sizes 5, 4, 5 → \(5 \times 4 \times 5 = 100\); increase = \(100 - 60 = 40\)  

**Step 4: Determine the maximum increase**  
The largest increase is 40 (achieved by adding both letters to the second set, or one to the second and one to the third).

**Step 5: Match to the choices**  
A. 24  
B. 30  
C. 36  
D. 40  
E. 60  

The correct choice is D.

ANSWER: D

1. **Understand the original setup**:  
   - First letter: from set of 5 letters → {C, H, L, P, R} → 5 choices  
   - Second letter: from set of 3 letters → {A, I, O} → 3 choices  
   - Third letter: from set of 4 letters → {D, M, N, T} → 4 choices  

   So total original number of license plates =  
   $$
   5 \times 3 \times 4 = 60
   $$

2. **Goal**: Add **two new letters** to the sets (either both to one set, or one to one set and one to another), in a way that **maximizes the number of *additional* license plates**.

   That is, we want to **maximize the increase** in total number of combinations.

3. **Key idea**:  
   The total number of plates is the product:  
   $$
   (\text{size of first set}) \times (\text{size of second set}) \times (\text{size of third set})
   $$

   Adding letters to a set increases its size, which scales the total by the product of the other two set sizes. So to maximize the increase, we should add letters to the set(s) that will cause the **largest increase in the product**.

4. **Let’s denote the set sizes**:  
   - Set A (first letter): size = 5  
   - Set B (second letter): size = 3  
   - Set C (third letter): size = 4  

   The total = $ A \cdot B \cdot C = 5 \cdot 3 \cdot 4 = 60 $

5. **Consider all reasonable ways to add two letters**:

   **Case 1: Add both letters to the same set**

   - **Add both to Set A**: new size = 7 → total = $7 \cdot 3 \cdot 4 = 84$ → increase = $84 - 60 = 24$
   - **Add both to Set B**: new size = 5 → total = $5 \cdot 5 \cdot 4 = 100$ → increase = $100 - 60 = 40$
   - **Add both to Set C**: new size = 6 → total = $5 \cdot 3 \cdot 6 = 90$ → increase = $90 - 60 = 30$

   **Case 2: Add one letter to one set, and one letter to another set**

   - Add to A and B: sizes → 6, 4, 4 → total = $6 \cdot 4 \cdot 4 = 96$ → increase = $96 - 60 = 36$
   - Add to A and C: sizes → 6, 3, 5 → total = $6 \cdot 3 \cdot 5 = 90$ → increase = $30$
   - Add to B and C: sizes → 5, 4, 5 → total = $5 \cdot 4 \cdot 5 = 100$ → increase = $40$

   **Case 3: Add one letter to one set, and the other letter to the same set** → already covered in Case 1.

   **Note**: We cannot add both letters to two different sets in more than one way beyond what’s listed — those are all the combinations.

6. **Find the maximum increase**:

   From above, the increases are:
   - 24
   - 30
   - 36
   - **40** (occurs in two scenarios: both letters to Set B, or one to B and one to C)

   So the **largest possible number of additional license plates is 40**.

7. **Check answer choices**:

   Choices:
   - A. 24  
   - B. 30  
   - C. 36  
   - D. 40  
   - E. 60  

   So the correct answer is **D. 40**

8. **Why not 60?**  
   That would mean doubling the number of plates, which would require multiplying total by 2. But adding only two letters can’t double the product unless one of the sets is very small — but even adding 2 to the smallest set (size 3 → 5) gives factor of $5/3 \approx 1.67$, not 2. So 60 is too high.

ANSWER: D

 **Step 1: Understand the current situation.**  
The three positions on a license plate are filled from sets of letters:
- 1st position: 5 choices \(\{C,H,L,P,R\}\)
- 2nd position: 3 choices \(\{A,I,O\}\)
- 3rd position: 4 choices \(\{D,M,N,T\}\)

By the multiplication principle, the current number of different plates is  
\(5 \times 3 \times 4 = 60\).

**Step 2: Identify all ways to add the two new letters.**  
Because there are only three sets and two new letters, there are only two kinds of distributions:
- **Case I:** Both letters are added to the *same* set.
- **Case II:** The two letters are added to *two different* sets (one letter to each).

We will compute the new total number of plates for every possibility and subtract the original 60 to find the additional plates.

**Step 3: Compute Case I – both letters to one set.**  
- Add both to the 1st set: sizes become \(7,3,4\).  
  New total: \(7 \times 3 \times 4 = 84\).  
  Additional plates: \(84 - 60 = 24\).
- Add both to the 2nd set: sizes become \(5,5,4\).  
  New total: \(5 \times 5 \times 4 = 100\).  
  Additional plates: \(100 - 60 = 40\).
- Add both to the 3rd set: sizes become \(5,3,6\).  
  New total: \(5 \times 3 \times 6 = 90\).  
  Additional plates: \(90 - 60 = 30\).

**Step 4: Compute Case II – one letter to each of two different sets.**  
- Add one to the 1st set and one to the 2nd: sizes become \(6,4,4\).  
  New total: \(6 \times 4 \times 4 = 96\).  
  Additional plates: \(96 - 60 = 36\).
- Add one to the 1st set and one to the 3rd: sizes become \(6,3,5\).  
  New total: \(6 \times 3 \times 5 = 90\).  
  Additional plates: \(90 - 60 = 30\).
- Add one to the 2nd set and one to the 3rd: sizes become \(5,4,5\).  
  New total: \(5 \times 4 \times 5 = 100\).  
  Additional plates: \(100 - 60 = 40\).

**Step 5: Find the maximum.**  
The six possible increases are: \(24, 40, 30, 36, 30, 40\).  
The largest possible number of additional license plates is **40**.

**Step 6: Rule out the wrong answer choices.**  
- **A (24):** This is achieved only when both new letters go to the first set, but 40 is larger.  
- **B (30):** This occurs when both go to the third set, or when one goes to the first and one to the third; again, 40 is larger.  
- **C (36):** This occurs when one letter goes to the first set and one to the second; 40 is still larger.  
- **E (60):** No distribution of two letters can create 60 new plates. The highest total we can reach is 100 (an increase of only 40), so 60 is impossible.  
- **D (40):** This matches the maximum increase found in Steps 3 and 4.

ANSWER: D

Here is the step-by-step solution:

**Step 1: Determine the initial number of license plates.**
A license plate consists of three letters chosen from three distinct sets.
*   The first letter comes from a set of size **5**: $\{C, H, L, P, R\}$.
*   The second letter comes from a set of size **3**: $\{A, I, O\}$.
*   The third letter comes from a set of size **4**: $\{D, M, N, T\}$.

Using the multiplication principle, the total number of original license plates is:
$$ \text{Total}_{\text{original}} = 5 \times 3 \times 4 = 60 $$

**Step 2: Analyze the possible ways to add the two new letters.**
We need to add exactly 2 letters to the sets to maximize the number of *additional* plates. There are two main strategies for distributing these 2 letters:
1.  **Add both letters to the same set.**
2.  **Add one letter to one set and the second letter to a different set.**

Let's calculate the new total number of plates and the increase (difference) for each scenario.

**Scenario 1: Adding both letters to the first set (size 5 $\rightarrow$ 7).**
*   New sizes: 7, 3, 4
*   New Total: $7 \times 3 \times 4 = 84$
*   Additional plates: $84 - 60 = \mathbf{24}$

**Scenario 2: Adding both letters to the second set (size 3 $\rightarrow$ 5).**
*   New sizes: 5, 5, 4
*   New Total: $5 \times 5 \times 4 = 100$
*   Additional plates: $100 - 60 = \mathbf{40}$

**Scenario 3: Adding both letters to the third set (size 4 $\rightarrow$ 6).**
*   New sizes: 5, 3, 6
*   New Total: $5 \times 3 \times 6 = 90$
*   Additional plates: $90 - 60 = \mathbf{30}$

**Scenario 4: Adding one letter to the first set and one to the second set (sizes 5$\rightarrow$6, 3$\rightarrow$4).**
*   New sizes: 6, 4, 4
*   New Total: $6 \times 4 \times 4 = 96$
*   Additional plates: $96 - 60 = \mathbf{36}$

**Scenario 5: Adding one letter to the first set and one to the third set (sizes 5$\rightarrow$6, 4$\rightarrow$5).**
*   New sizes: 6, 3, 5
*   New Total: $6 \times 3 \times 5 = 90$
*   Additional plates: $90 - 60 = \mathbf{30}$

**Scenario 6: Adding one letter to the second set and one to the third set (sizes 3$\rightarrow$4, 4$\rightarrow$5).**
*   New sizes: 5, 4, 5
*   New Total: $5 \times 4 \times 5 = 100$
*   Additional plates: $100 - 60 = \mathbf{40}$

**Step 3: Identify the maximum value.**
Comparing the results from all possible scenarios:
*   Scenario 1: 24
*   Scenario 2: 40
*   Scenario 3: 30
*   Scenario 4: 36
*   Scenario 5: 30
*   Scenario 6: 40

The largest possible number of additional license plates is **40**. This occurs if we add both new letters to the second set, or if we add one letter to the second set and one to the third set.

**Step 4: Rule out wrong choices.**
*   **A (24)**: This is the result if both letters are added to the largest set (the first set), which minimizes the gain.
*   **B (30)**: This occurs in some mixed scenarios or adding to the third set, but it is not the maximum.
*   **C (36)**: This occurs when adding to the first and second sets, but it is not the maximum.
*   **E (60)**: This would imply doubling the total number of plates, which is impossible by adding only 2 letters to such small sets.

ANSWER: D

**Step 1 – Determine the current number of plates.**  
The three positions have  
- 5 choices for the first letter,  
- 3 choices for the second letter,  
- 4 choices for the third letter.  

Current total = \(5 \times 3 \times 4 = 60\) plates.

**Step 2 – List all possible ways to add the two new letters.**  
Let \((x_1,x_2,x_3)\) be the numbers added to the first, second, and third sets, respectively.  
We need \(x_1+x_2+x_3 = 2\) with each \(x_i \ge 0\).  
The distinct patterns are:

1. Both letters to the same set: \((2,0,0), (0,2,0), (0,0,2)\).  
2. One letter to each of two different sets: \((1,1,0), (1,0,1), (0,1,1)\).

**Step 3 – Compute the total plates for each pattern.**  

| Pattern | New sizes \((a',b',c')\) | Total plates \(a'b'c'\) | Increase \(a'b'c' - 60\) |
|---------|--------------------------|------------------------|---------------------------|
| (2,0,0) | (7, 3, 4)                | \(7\cdot3\cdot4 = 84\)  | \(84-60 = 24\) |
| (0,2,0) | (5, 5, 4)                | \(5\cdot5\cdot4 = 100\) | \(100-60 = 40\) |
| (0,0,2) | (5, 3, 6)                | \(5\cdot3\cdot6 = 90\)  | \(90-60 = 30\) |
| (1,1,0) | (6, 4, 4)                | \(6\cdot4\cdot4 = 96\)  | \(96-60 = 36\) |
| (1,0,1) | (6, 3, 5)                | \(6\cdot3\cdot5 = 90\)  | \(90-60 = 30\) |
| (0,1,1) | (5, 4, 5)                | \(5\cdot4\cdot5 = 100\) | \(100-60 = 40\) |

**Step 4 – Identify the largest increase.**  
The maximum increase among all patterns is **40**, achieved by either:

- adding both new letters to the second set (making it have 5 letters), or  
- adding one new letter to the second set and one to the third set.

**Step 5 – Eliminate the other answer choices.**  

- 24 comes from adding both letters to the first set – smaller than 40.  
- 30 comes from adding both letters to the third set or the (1,0,1) split – smaller.  
- 36 comes from the (1,1,0) split – smaller.  
- 60 is the original number of plates, not an increase.

Thus the largest possible number of *additional* plates is **40**.

**ANSWER: D**

eed to find the largest possible number of additional license - plates that can be made by adding two letters to the existing sets of letters used to form bicycle license plates. The original license plates have the first letter from the set $\{C, H, L, P, R\}$ (with $n_1 = 5$ elements), the second letter from the set $\{A, I, O\}$ (with $n_2 = 3$ elements), and the third letter from the set $\{D, M, N, T\}$ (with $n_3 = 4$ elements). The total number of original license plates is $n_1\times n_2\times n_3=5\times3\times4 = 60$.
   - We will consider different ways of adding two letters to the sets and calculate the number of new license plates in each case.
2. Set up the math for different cases of adding letters:
   - **Case 1: Add both letters to the first set**
     - The new number of elements in the first set is $n_{1_{new}}=5 + 2=7$, the second set has $n_2 = 3$ elements, and the third set has $n_3 = 4$ elements. The total number of new license plates is $n_{1_{new}}\times n_2\times n_3-n_1\times n_2\times n_3=(7\times3\times4)-(5\times3\times4)=(7 - 5)\times3\times4=2\times3\times4 = 24$.
   - **Case 2: Add both letters to the second set**
     - The new number of elements in the second set is $n_{2_{new}}=3 + 2=5$, the first set has $n_1 = 5$ elements, and the third set has $n_3 = 4$ elements. The total number of new license plates is $n_1\times n_{2_{new}}\times n_3 - n_1\times n_2\times n_3=(5\times5\times4)-(5\times3\times4)=5\times4\times(5 - 3)=5\times4\times2 = 40$.
   - **Case 3: Add both letters to the third set**
     - The new number of elements in the third set is $n_{3_{new}}=4 + 2=6$, the first set has $n_1 = 5$ elements, and the second set has $n_2 = 3$ elements. The total number of new license plates is $n_1\times n_2\times n_{3_{new}}-n_1\times n_2\times n_3=(5\times3\times6)-(5\times3\times4)=5\times3\times(6 - 4)=5\times3\times2 = 30$.
   - **Case 4: Add one letter to the first set and one letter to the second set**
     - The new number of elements in the first set is $n_{1_{new}}=5 + 1=6$, the new number of elements in the second set is $n_{2_{new}}=3 + 1=4$, and the third set has $n_3 = 4$ elements. The total number of new license plates is $n_{1_{new}}\times n_{2_{new}}\times n_3 - n_1\times n_2\times n_3=(6\times4\times4)-(5\times3\times4)=4\times4\times(6 - 3)+4\times3\times(4 - 3)$ (using distributive property) $=4\times4\times3+4\times3=48 + 12=60 - 60 + 48+12$ (calculate step - by - step). Actually, $6\times4\times4-5\times3\times4=(6\times4 - 5\times3)\times4=(24 - 15)\times4 = 36$.
   - **Case 5: Add one letter to the first set and one letter to the third set**
     - The new number of elements in the first set is $n_{1_{new}}=5 + 1=6$, the second set has $n_2 = 3$ elements, and the new number of elements in the third set is $n_{3_{new}}=4 + 1=5$. The total number of new license plates is $n_{1_{new}}\times n_2\times n_{3_{new}}-n_1\times n_2\times n_3=(6\times3\times5)-(5\times3\times4)=3\times(6\times5 - 5\times4)=3\times(30 - 20)=30$.
   - **Case 6: Add one letter to the second set and one letter to the third set**
     - The new number of elements in the second set is $n_{2_{new}}=3 + 1=4$, the first set has $n_1 = 5$ elements, and the new number of elements in the third set is $n_{3_{new}}=4 + 1=5$. The total number of new license plates is $n_1\times n_{2_{new}}\times n_{3_{new}}-n_1\times n_2\times n_3=(5\times4\times5)-(5\times3\times4)=5\times4\times(5 - 3)=40$.
3. Do the computation and rule out wrong choices:
   - By comparing the results of the six cases: $24,40,30,36,30,40$, we can see that the largest number of additional license plates is $40$.
   - We can rule out choice A ($24$) because we found cases with larger values.
   - We can rule out choice B ($30$) because we found cases with larger values.
   - We can rule out choice C ($36$) because we found cases with larger values.
   - We can rule out choice E ($60$) because our maximum value from the cases is $40$.

ANSWER: D

### Step 1: Restate the problem and calculate original number of plates
License plates use 3 letters from 3 independent sets, so the total number of unique plates is the product of the sizes of the three sets (multiplication principle for combinations).
- Original set sizes: First set = 5 letters, second set = 3 letters, third set = 4 letters.
- Original total plates: \(5 \times 3 \times 4 = 60\).

We add 2 new letters (can add both to one set, or split them across two sets) and need to find the maximum number of additional plates (new total plates minus original 60).

---
### Step 2: Calculate additional plates for all possible ways to add the 2 new letters
We test every valid distribution of the 2 new letters, compute the new total plates, and find the additional count for each case:
1. **Add both to first set (sizes: 7,3,4):** New total = \(7 \times 3 \times 4 = 84\), additional = \(84-60=24\)
2. **Add both to second set (sizes:5,5,4):** New total = \(5 \times 5 \times 4 = 100\), additional = \(100-60=40\)
3. **Add both to third set (sizes:5,3,6):** New total = \(5 \times 3 \times 6 = 90\), additional = \(90-60=30\)
4. **Add 1 to first, 1 to second (sizes:6,4,4):** New total = \(6 \times 4 \times 4 = 96\), additional = \(96-60=36\)
5. **Add 1 to first, 1 to third (sizes:6,3,5):** New total = \(6 \times 3 \times 5 = 90\), additional = \(90-60=30\)
6. **Add 1 to second, 1 to third (sizes:5,4,5):** New total = \(5 \times 4 \times 5 = 100\), additional = \(100-60=40\)

---
### Step 3: Rule out incorrect options
- A (24): This is the smallest additional count, only if we add both letters to the largest original set, not maximum.
- B (30): Occurs for low-impact distributions of the new letters, not the maximum.
- C (36): The additional count if we split the new letters across the first and second sets, which is less than 40.
- E (60): Impossible, as we only add 2 letters, we cannot double the original number of plates.

The maximum number of additional plates is 40.

ANSWER: D

### Step 1: Restate the problem and calculate the original number of license plates
We have 3 positions for license plates, with initial set sizes:
- 1st position: 5 letters (C,H,L,P,R)
- 2nd position: 3 letters (A,I,O)
- 3rd position: 4 letters (D,M,N,T)
The original number of plates is the product of the set sizes:
$$5 \times 3 \times 4 = 60$$
We add 2 new letters total, distributed either as 2 letters to one set, or 1 letter to two different sets. We need to find the maximum **additional** plates, which equals (new total plates) - 60. To maximize the additional plates, we maximize the new total.
---
### Step 2: Analyze how adding letters affects the total
When you add a letter to a set, the number of new plates created is equal to the product of the sizes of the *other two sets*. This is because every existing combination of the other two positions can now pair with the new letter in the modified position.
For the original sets:
- Gain from adding 1 letter to the 1st set: $3 \times 4 = 12$ new plates
- Gain from adding 1 letter to the 2nd set: $5 \times 4 = 20$ new plates
- Gain from adding 1 letter to the 3rd set: $5 \times 3 = 15$ new plates
The highest marginal gain comes from adding to the 2nd set first (20 new plates per letter added here).
---
### Step 3: Evaluate all valid distributions of the 2 new letters
We test all allowed distributions of the 2 letters:
1. **Add 2 letters to the 1st set**: New size = 5+2=7. New total = $7 \times 3 \times 4 = 84$. Additional plates = $84 - 60 = 24$.
2. **Add 2 letters to the 2nd set**: New size = 3+2=5. New total = $5 \times 5 \times 4 = 100$. Additional plates = $100 - 60 = 40$.
3. **Add 2 letters to the 3rd set**: New size = 4+2=6. New total = $5 \times 3 \times 6 = 90$. Additional plates = $90 - 60 = 30$.
4. **Add 1 letter to 1st and 1 to 2nd set**: New sizes = 6,4,4. New total = $6 \times 4 \times 4 = 96$. Additional plates = $96 - 60 = 36$.
5. **Add 1 letter to 1st and 1 to 3rd set**: New sizes = 6,3,5. New total = $6 \times 3 \times 5 = 90$. Additional plates = $90 - 60 = 30$.
6. **Add 1 letter to 2nd and 1 to 3rd set**: New sizes =5,4,5. New total = $5 \times 4 \times 5 = 100$. Additional plates = $100 - 60 = 40$.
---
### Step 4: Identify the maximum additional plates
The largest additional number of plates is 40, which comes from either adding both letters to the 2nd set, or adding one letter to the 2nd set and one to the 3rd set.
ANSWER: D

1. **Restate the problem**: We need to determine the largest possible number of additional license plates that can be made by adding two new letters to the existing sets of letters used for bicycle license plates in Flatville.

2. **Calculate the original number of license plates**:
   - The first letter is chosen from a set of 5 letters: {C, H, L, P, R}.
   - The second letter is chosen from a set of 3 letters: {A, I, O}.
   - The third letter is chosen from a set of 4 letters: {D, M, N, T}.

   The total number of original license plates is:
   \[
   5 \times 3 \times 4 = 60
   \]

3. **Determine the strategy for adding two new letters to maximize the number of additional license plates**:
   - To maximize the number of additional license plates, we should add both new letters to the set with the fewest number of letters, which is the second set {A, I, O} with 3 letters.
   - Adding two letters to this set will increase it to 5 letters.

4. **Calculate the new number of license plates after adding two letters to the second set**:
   - The first letter is still chosen from a set of 5 letters: {C, H, L, P, R}.
   - The second letter is now chosen from a set of 5 letters: {A, I, O} + 2 new letters.
   - The third letter is still chosen from a set of 4 letters: {D, M, N, T}.

   The total number of new license plates is:
   \[
   5 \times 5 \times 4 = 100
   \]

5. **Calculate the additional number of license plates**:
   - The additional number of license plates is the difference between the new total and the original total:
   \[
   100 - 60 = 40
   \]

6. **Rule out the wrong choices**:
   - 24 (A): This is too small because adding two letters should increase the number of combinations significantly more than this.
   - 30 (B): This is also too small for the same reason.
   - 36 (C): This is close but not the maximum possible.
   - 60 (E): This is too large because it would require a different distribution of the new letters that doesn't maximize the combinations as effectively as adding both to the smallest set.

Therefore, the largest possible number of additional license plates is 40.

ANSWER: D

**Problem Restatement:**
We need to determine the largest possible number of additional bicycle license plates that can be created in Flatville by adding two new letters to their existing sets. The license plates consist of three letters, chosen from three distinct sets:
- First letter: {C, H, L, P, R} (5 options)
- Second letter: {A, I, O} (3 options)
- Third letter: {D, M, N, T} (4 options)

The two new letters can be added in any of the following ways:
1. Both letters added to one set.
2. One letter added to one set and the other letter added to a different set.

Our goal is to find the maximum number of new license plates possible by adding these two letters optimally.

---

**Step 1: Calculate the current number of possible license plates.**
The current number of possible license plates is the product of the number of choices for each position:
\[ 5 \text{ (first letter)} \times 3 \text{ (second letter)} \times 4 \text{ (third letter)} = 60 \]

---

**Step 2: Understand how adding letters increases the number of license plates.**
The total number of possible license plates after adding two letters will depend on how the letters are distributed among the three sets. We want to maximize the increase in the number of possible license plates, which means we want to maximize the product of the sizes of the three sets after adding the two letters.

Let’s denote the sizes of the three sets as follows:
- \( a = 5 \) (first letter set)
- \( b = 3 \) (second letter set)
- \( c = 4 \) (third letter set)

We are adding two letters, so we want to maximize \( (a + \Delta a)(b + \Delta b)(c + \Delta c) - abc \), where \( \Delta a + \Delta b + \Delta c = 2 \) and \( \Delta a, \Delta b, \Delta c \) are non-negative integers.

---

**Step 3: Consider all possible ways to distribute the two new letters among the three sets.**
There are three cases to consider:

1. **Both letters added to the same set:**
   - Add both letters to the first set: \( (5+2) \times 3 \times 4 = 7 \times 3 \times 4 = 84 \)
     Increase: \( 84 - 60 = 24 \)
   - Add both letters to the second set: \( 5 \times (3+2) \times 4 = 5 \times 5 \times 4 = 100 \)
     Increase: \( 100 - 60 = 40 \)
   - Add both letters to the third set: \( 5 \times 3 \times (4+2) = 5 \times 3 \times 6 = 90 \)
     Increase: \( 90 - 60 = 30 \)

2. **One letter added to one set and the other letter added to a different set:**
   - Add one letter to the first set and one to the second set: \( (5+1) \times (3+1) \times 4 = 6 \times 4 \times 4 = 96 \)
     Increase: \( 96 - 60 = 36 \)
   - Add one letter to the first set and one to the third set: \( (5+1) \times 3 \times (4+1) = 6 \times 3 \times 5 = 90 \)
     Increase: \( 90 - 60 = 30 \)
   - Add one letter to the second set and one to the third set: \( 5 \times (3+1) \times (4+1) = 5 \times 4 \times 5 = 100 \)
     Increase: \( 100 - 60 = 40 \)

---

**Step 4: Identify the maximum increase.**
From the calculations above, the largest increase in the number of possible license plates is **40**, which occurs in two scenarios:
- Adding both letters to the second set.
- Adding one letter to the second set and one letter to the third set.

---

**Step 5: Rule out incorrect choices.**
The problem asks for the largest possible number of additional license plates. The options are:
A. 24
B. 30
C. 36
D. 40
E. 60

From our calculations, the maximum increase is 40. Therefore, the correct answer is D.

---
ANSWER: D

I need to find the number of different shortest routes Samantha can take on her journey.

## Step 1: Understand the Setup

Let me set up a coordinate system with the southwest corner of City Park at the origin (0, 0).

- **Samantha's home**: 2 blocks west and 1 block south = (-2, -1)
- **Southwest corner of park**: (0, 0)
- **Northeast corner of park**: Let's call this (a, b) for some positive values
- **School**: 2 blocks east and 2 blocks north of northeast corner = (a+2, b+2)

The route has three segments:
1. Home to SW corner of park
2. Diagonal through park (only one way)
3. NE corner of park to school

## Step 2: Count Routes for Each Segment

**Segment 1: From (-2, -1) to (0, 0)**
- Need to go 2 blocks east and 1 block north
- Total moves: 3 (2 East, 1 North)
- Number of different routes = $\binom{3}{2} = 3$ or $\binom{3}{1} = 3$
- These routes correspond to choosing which of the 3 moves is North: EEN, ENE, NEE

**Segment 2: Diagonal through park**
- This is fixed - only one way (the diagonal path)
- Number of routes = 1

**Segment 3: From (a, b) to (a+2, b+2)**
- Need to go 2 blocks east and 2 blocks north
- Total moves: 4 (2 East, 2 North)
- Number of different routes = $\binom{4}{2} = 6$
- These correspond to choosing which 2 of the 4 moves are East (or North): EENN, ENEN, ENNE, NEEN, NENE, NNEE

## Step 3: Apply the Multiplication Principle

Since the three segments are independent (the route on segment 1 doesn't affect the route on segment 2, which doesn't affect segment 3), the total number of different routes is:

$$3 \times 1 \times 6 = 18$$

## Step 4: Verify the Logic

- From home to SW corner: 3 shortest routes ✓
- Through the park: 1 route (fixed diagonal) ✓
- From NE corner to school: 6 shortest routes ✓
- Total: 3 × 1 × 6 = 18 ✓

The diagonal through the park doesn't affect the counting of street routes on either side.

ANSWER: E

1. Restate the problem.  
   Samantha’s house is 2 blocks west and 1 block south of the SW corner of City Park.  Her school is 2 blocks east and 2 blocks north of the NE corner of City Park.  Each morning she (a) bikes along grid streets from home to the SW corner, (b) goes straight along the park diagonal to the NE corner, (c) bikes along grid streets from the NE corner to school.  We must count how many shortest routes she can take.

2. Model the grid.  
   Place the SW corner of the park at (0, 0).  Then her house is at (–2, –1), and the NE corner of the park is at some (a, b) (park dimensions don’t matter for counting the outside segments).  Her school is then 2 blocks east and 2 blocks north of (a, b).

3. Count shortest paths from home to (0, 0).  
   From (–2, –1) to (0, 0) she must move 2 blocks east (E) and 1 block north (N), in some order.  That’s a total of 3 moves, of which 2 are E and 1 is N.  
   Number of such sequences = C(3; 2,1) = 3!/(2!1!) = 3.

4. The diagonal through the park is fixed.  
   There is exactly 1 way to go from SW corner to NE corner along the straight diagonal.

5. Count shortest paths from (a, b) to school.  
   From the NE corner to her school is 2 blocks east and 2 blocks north, in some order.  That’s 4 moves, 2 E and 2 N.  
   Number of such sequences = C(4; 2,2) = 4!/(2!2!) = 6.

6. Total number of shortest routes.  
   Multiply the independent choices on the two street–segments: 3 × 1 × 6 = 18.  

7. Ruling out other answer choices.  
   The other choices (3, 6, 9, 12) are far less than 18 and do not match the computed product 3×6.  Hence the correct answer is E.

ANSWER: E

To find the total number of different shortest routes Samantha can take, we can break her journey down into three independent stages:

**Step 1: Identify the three stages of the journey**
1. **Stage 1:** From her home to the southwest (SW) corner of City Park.
2. **Stage 2:** From the SW corner of City Park to the northeast (NE) corner of City Park.
3. **Stage 3:** From the NE corner of City Park to her school.

---

**Step 2: Calculate the number of shortest routes for Stage 1**
* Samantha's home is 2 blocks west and 1 block south of the SW corner of City Park.
* To reach the SW corner by the shortest route, she must travel exactly 2 blocks East (E) and 1 block North (N).
* The number of different ways to arrange these moves (E, E, N) is given by the combination formula:
  $$\binom{2+1}{1} = \binom{3}{1} = 3 \text{ routes}$$
  These 3 routes are:
  1. East, East, North
  2. East, North, East
  3. North, East, East

---

**Step 3: Calculate the number of routes for Stage 2**
* Samantha takes "a diagonal path through the park to the northeast corner."
* There is only $1$ such diagonal path through the park.
* Thus, there is exactly $1$ route for this stage.

---

**Step 4: Calculate the number of shortest routes for Stage 3**
* Her school is 2 blocks east and 2 blocks north of the NE corner of City Park.
* To reach the school by the shortest route, she must travel exactly 2 blocks East (E) and 2 blocks North (N).
* The number of different ways to arrange these moves (E, E, N, N) is given by:
  $$\binom{2+2}{2} = \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \text{ routes}$$
  These 6 routes are:
  1. East, East, North, North
  2. East, North, East, North
  3. East, North, North, East
  4. North, East, East, North
  5. North, East, North, East
  6. North, North, East, East

---

**Step 5: Calculate the total number of routes**
Since the choice of route in each stage is independent of the others, we multiply the number of possibilities for each stage:
$$\text{Total Routes} = (\text{Routes for Stage 1}) \times (\text{Routes for Stage 2}) \times (\text{Routes for Stage 3})$$
$$\text{Total Routes} = 3 \times 1 \times 6 = 18$$

---

**Step 6: Rule out incorrect choices**
* **A (3):** This only accounts for the first stage.
* **B (6):** This only accounts for the third stage.
* **C (9):** This is the sum of the routes ($3 + 6$) instead of the product.
* **D (12):** This is incorrect and does not correspond to any standard combination of the stages.
* **E (18):** This is the correct product of the independent stages.

ANSWER: E

The southwest corner of the park is denoted \(A\) and the northeast corner is denoted \(B\). The diagonal path from \(A\) to \(B\) through the park is fixed (a single segment that forms part of every valid route).

Samantha's house is located 2 blocks west and 1 block south of \(A\). Any shortest route from her house to \(A\) therefore consists of exactly 2 blocks east and 1 block north, for a total of 3 blocks. The number of distinct sequences is the number of ways to choose the position of the single north block among the 3 total blocks:
\[
\binom{3}{1} = 3.
\]

The school is located 2 blocks east and 2 blocks north of \(B\). Any shortest route from \(B\) to the school therefore consists of exactly 2 blocks east and 2 blocks north, for a total of 4 blocks. The number of distinct sequences is
\[
\frac{4!}{2! \cdot 2!} = 6.
\]

Because the diagonal segment from \(A\) to \(B\) is the same in every route, the total number of shortest routes is the product of the two independent quantities above:
\[
3 \times 6 = 18.
\]

No shorter routes exist, because any deviation (extra blocks or detours around the park) would increase the total length. Consequently the 18 routes already enumerated are precisely the shortest ones, and every other choice among the five options undercounts them.

ANSWER: E

## Step 1: Understand the problem and visualize Samantha's route
Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. She bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. We need to determine the number of different routes she can take if her route is as short as possible.

## Step 2: Determine the grid layout and the positions of Samantha's home and school relative to City Park
Let's consider the grid layout. City Park's southwest corner is at (0,0) if we consider it as the origin. Samantha's home is at (-2, -1), and the southwest corner of City Park is at (0,0). The northeast corner of City Park is at some (x, y), and her school is at (x+2, y+2).

## Step 3: Understand that the size of City Park is not directly given but can be inferred from the context
Since Samantha takes a diagonal path through the park from the southwest corner to the northeast corner, and given that she is biking on streets otherwise, the park's dimensions are crucial. However, the exact dimensions are not directly provided, but we can infer that the park is a rectangle with its southwest corner at (0,0) and its northeast corner at some positive (x, y).

## Step 4: Calculate the number of ways Samantha can go from her home to the southwest corner of City Park
To get to the southwest corner (0,0) from her home (-2, -1), Samantha needs to move 2 blocks east and 1 block north. This is a total of 3 steps, where 2 are of one type (east) and 1 is of another type (north). The number of ways to arrange these steps is given by the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$, where $n = 3$ and $r = 2$ (or $r = 1$), so $\binom{3}{2} = \frac{3!}{2!(3-2)!} = 3$.

## Step 5: Understand that the path through City Park is fixed as it's diagonal
The path through the park is diagonal and thus fixed, meaning there's only 1 way to traverse the park from the southwest corner to the northeast corner.

## Step 6: Calculate the number of ways Samantha can go from the northeast corner of City Park to her school
If the northeast corner of City Park is at (x, y), then her school is at (x+2, y+2). To get to her school, she needs to move 2 blocks east and 2 blocks north. This is a total of 4 steps, with 2 being east and 2 being north. The number of ways to arrange these steps is $\binom{4}{2} = \frac{4!}{2!(4-2)!} = 6$.

## Step 7: Calculate the total number of different routes Samantha can take
The total number of routes is the product of the number of ways to get to the southwest corner, the number of ways through the park, and the number of ways from the northeast corner to school. So, total routes = (ways to southwest corner) * (ways through park) * (ways to school) = $3 * 1 * 6 = 18$.

## Step 8: Rule out the wrong choices
- A. 3: This is the number of ways to get to the southwest corner, not the total number of routes.
- B. 6: This is the number of ways to get from the northeast corner to school, not the total number of routes.
- C. 9: This could be a plausible answer if we were considering a different part of the journey or a different calculation, but it's not the total number of routes.
- D. 12: This is not the result of our calculation.
- E. 18: This matches our calculation for the total number of routes.

The final answer is: 
ANSWER: E

1. **Understand the problem:** Samatha's route consists of three parts: biking on streets from home to the southwest (SW) corner of City Park, taking a fixed diagonal path through the park to the northeast (NE) corner, and biking on streets from the NE corner to school. We need the number of shortest possible routes. The diagonal path is a single fixed route, so the total number of routes is the product of the number of shortest street routes for the first part and the number for the last part.

2. **Home to SW corner:** Home is 2 blocks west and 1 block south of the SW corner. To reach the SW corner, she must travel 2 blocks east and 1 block north. A shortest path is any sequence of 2 E and 1 N. The number of such paths is the binomial coefficient \(\binom{3}{2} = 3\) (or equivalently \(\binom{3}{1} = 3\)).

3. **NE corner to school:** School is 2 blocks east and 2 blocks north of the NE corner. To reach school, she must travel 2 blocks east and 2 blocks north. The number of shortest paths is the number of sequences of 2 E and 2 N, which is \(\binom{4}{2} = 6\).

4. **Total routes:** Multiply the number of ways for each street segment: \(3 \times 6 = 18\). The diagonal path adds no extra choices.

5. **Check choices:** 18 is option E.

ANSWER: E

1. **Understand the setup**:  
   - City Park is a rectangular (likely square) region with a southwest (SW) corner and a northeast (NE) corner.  
   - Samantha lives **2 blocks west and 1 block south** of the **SW corner** of the park.  
   - Her school is **2 blocks east and 2 blocks north** of the **NE corner** of the park.  
   - Her route is:  
     (a) Bike on streets from home to **SW corner** of park (only street travel),  
     (b) Take the **diagonal path through the park** from SW to NE corner (only one way—this is fixed),  
     (c) Bike on streets from **NE corner** to school (only street travel).  

   We are to count the number of **different shortest routes** she can take **on streets**, **before and after** the fixed diagonal through the park.

2. **Model the street grid**:  
   We assume a standard city grid where she can only move **north** or **east** when going toward her destination (since we want shortest paths—no backtracking).  

   - From **home to SW corner**:  
     She is **2 blocks west** → must go **2 blocks east**.  
     She is **1 block south** → must go **1 block north**.  
     So, she must make a total of **3 moves**: 2 E and 1 N, in any order.  
     Number of shortest paths = number of permutations of {E, E, N} =  
     $$
     \binom{3}{1} = 3 \quad \text{(choose 1 spot for N among 3 moves)}
     $$  
     or equivalently $\binom{3}{2} = 3$ (choose 2 spots for E).

   - From **NE corner to school**:  
     School is **2 blocks east and 2 blocks north** of NE corner.  
     So she must go **2 blocks east** and **2 blocks north**: total of 4 moves: 2 E, 2 N.  
     Number of shortest paths = number of permutations of {E, E, N, N} =  
     $$
     \binom{4}{2} = 6
     $$

3. **Combine the two street segments**:  
   The path through the park is **fixed** (only one diagonal path), so total number of different shortest routes =  
   (number of ways to get from home to SW corner) × (1 way through park) × (number of ways from NE corner to school)  
   $$
   = 3 \times 1 \times 6 = 18
   $$

4. **Check answer choices**:  
   Choices: A.3, B.6, C.9, D.12, E.18  
   Our result is **18**, which is choice **E**.

5. **Double-check assumptions**:  
   - Are we sure she can't go through the park in other ways? The problem says: "takes a diagonal path through the park to the northeast corner"—implies **only one such path** is available (the diagonal), so yes, fixed.  
   - Are we sure she only moves east and north? Since she’s going from a point southwest of the park to the SW corner, she must go **east and north**—any south or west movement would make the path longer than necessary. Same for after the park: from NE corner to a point northeast of it, she must go **east and north**. So only shortest paths are counted correctly via combinations.

   All good.

ANSWER: E

 **Step 1: Restate the problem.**  
Samantha’s journey has three separate parts:
1. Bike on streets from her home to the **southwest (SW) corner** of City Park.
2. Take the **diagonal path** through the park from the SW corner to the **northeast (NE) corner**.
3. Bike on streets from the NE corner to her school.

We must count how many distinct **shortest possible** routes of this form exist.

---

**Step 2: Set up coordinates for the street portions.**  
Place the SW corner of City Park at the origin \((0,0)\).

- Samantha lives **2 blocks west and 1 block south** of this corner, so her home is at \((-2,-1)\).
- Her school is **2 blocks east and 2 blocks north** of the NE corner. For counting the third leg, we only need this relative displacement: from the NE corner she must travel 2 blocks east and 2 blocks north.

---

**Step 3: Count shortest street routes from home to the SW corner.**  
To go from \((-2,-1)\) to \((0,0)\) without backtracking, Samantha must make:
- 2 East (E) moves
- 1 North (N) move

That is a sequence of 3 moves containing two E’s and one N.  
The number of such sequences is the binomial coefficient:

\[
\binom{3}{1} = \frac{3!}{2!\,1!} = 3
\]

(The three routes are EEN, ENE, NEE.)

---

**Step 4: Count paths through the park.**  
The problem states she takes **a diagonal path** from the SW corner to the NE corner. A single straight diagonal across the park is unique, so there is:

\[
1 \text{ way}
\]

---

**Step 5: Count shortest street routes from the NE corner to school.**  
From the NE corner, Samantha must travel:
- 2 East (E) moves
- 2 North (N) moves

That is a sequence of 4 moves containing two E’s and two N’s.  
The number of such sequences is:

\[
\binom{4}{2} = \frac{4!}{2!\,2!} = 6
\]

(The six routes are EENN, ENEN, ENNE, NEEN, NENE, NNEE.)

---

**Step 6: Compute the total number of routes.**  
Because the three legs are independent, we multiply the numbers of possibilities for each leg:

\[
3 \times 1 \times 6 = 18
\]

---

**Step 7: Rule out the wrong choices.**  
- **A. 3:** This is only the number of ways to get from home to the SW corner; it ignores the rest of the trip.  
- **B. 6:** This is only the number of ways from the NE corner to school; it ignores the first leg.  
- **C. 9:** This could come from adding \(3+6\) instead of multiplying, or from mistakenly counting each street leg as having 3 ways.  
- **D. 12:** This might result from \(3 \times 4\), perhaps miscomputing the second street portion as 4 instead of 6.  
- **E. 18:** This matches the correct product \(3 \times 1 \times 6\).

ANSWER: E

Here is the step-by-step solution:

**1. Understand the Problem and Set Up Coordinates**
We need to calculate the total number of shortest routes Samantha can take. Her route consists of three distinct segments:
*   **Segment A:** From her Home to the Southwest (SW) corner of City Park.
*   **Segment B:** From the SW corner to the Northeast (NE) corner of City Park (through the park).
*   **Segment C:** From the NE corner of City Park to her School.

Since these segments are sequential and independent of each other, the total number of routes is the product of the number of possible shortest paths for each segment.

**2. Calculate Paths for Segment A (Home to SW Corner)**
*   **Given:** Samantha lives 2 blocks west and 1 block south of the SW corner.
*   **Goal:** Find the shortest path from Home to the SW corner.
*   **Movement Required:** To get to the park corner, she must reverse her position relative to it. She needs to travel **2 blocks East** and **1 block North**.
*   **Total Steps:** $2 + 1 = 3$ blocks.
*   **Combinatorics:** We need to choose which 2 of the 3 steps are "East" (or which 1 is "North").
    $$ \text{Ways}_A = \binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3 $$
    *(The specific sequences are: EEN, ENE, NEE)*

**3. Calculate Paths for Segment B (SW Corner to NE Corner)**
*   **Given:** She takes a "diagonal path" through the park.
*   **Analysis:** In problems of this type, unless multiple specific internal paths are described, a "diagonal path" implies a single, fixed route connecting the two corners directly.
*   **Calculation:**
    $$ \text{Ways}_B = 1 $$

**4. Calculate Paths for Segment C (NE Corner to School)**
*   **Given:** Her school is 2 blocks east and 2 blocks north of the NE corner.
*   **Goal:** Find the shortest path from the NE corner to School.
*   **Movement Required:** She needs to travel **2 blocks East** and **2 blocks North**.
*   **Total Steps:** $2 + 2 = 4$ blocks.
*   **Combinatorics:** We need to choose which 2 of the 4 steps are "East" (the remaining will be "North").
    $$ \text{Ways}_C = \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 $$
    *(The specific sequences are: EENN, ENEN, ENNE, NEEN, NENE, NNEE)*

**5. Compute Total Routes**
Using the Multiplication Principle (Fundamental Counting Principle), we multiply the number of possibilities for each independent segment of the trip:

$$ \text{Total Routes} = (\text{Ways}_A) \times (\text{Ways}_B) \times (\text{Ways}_C) $$
$$ \text{Total Routes} = 3 \times 1 \times 6 $$
$$ \text{Total Routes} = 18 $$

**6. Evaluate Choices**
*   **A. 3:** This accounts only for the first leg of the trip (Home to Park). Incorrect.
*   **B. 6:** This accounts only for the last leg of the trip (Park to School). Incorrect.
*   **C. 9:** This might be a result of adding the legs ($3+6$) or another calculation error. Incorrect.
*   **D. 12:** This does not match the product of the path counts. Incorrect.
*   **E. 18:** This matches our calculated product ($3 \times 6$). Correct.

ANSWER: E

**Step 1 – Understand the geometry**

Place the southwest corner of City Park at the origin \((0,0)\).  
- “East’’ is the positive \(x\)-direction.  
- “North’’ is the positive \(y\)-direction.

Samantha’s home is **2 blocks west** and **1 block south** of the southwest corner, so its coordinates are  
\[
H=(-2,\,-1).
\]

Let the width of the park (east–west) be \(w\) blocks and its height (north–south) be \(h\) blocks.  
The northeast corner of the park is then at \((w,\,h)\).

Samantha’s school is **2 blocks east** and **2 blocks north** of that northeast corner, giving  
\[
S=(w+2,\;h+2).
\]

---

**Step 2 – What part of the route is flexible?**

Samantha must:

1. Bike on streets from \(H\) to the southwest corner \((0,0)\).  
2. Take the **single diagonal** across the park from \((0,0)\) to \((w,h)\).  
3. Bike on streets from \((w,h)\) to \(S\).

The diagonal is fixed (only one straight path).  
Thus the only freedom lies in the street‑grid portions before and after the park.

A route is *as short as possible* iff the two street portions are shortest‑possible, i.e. they contain **only** the required eastward/northward moves.

---

**Step 3 – Shortest routes from home to the southwest corner**

From \((-2,-1)\) to \((0,0)\) she must go

- 2 blocks east,
- 1 block north.

A shortest route is any ordering of these three moves.  
Number of ways \(= \displaystyle\binom{3}{1}=3\) (choose which of the three moves is the single north).

---

**Step 4 – Shortest routes from the northeast corner to school**

From \((w,h)\) to \((w+2,\,h+2)\) she must go

- 2 blocks east,
- 2 blocks north.

A shortest route is any ordering of these four moves.  
Number of ways \(= \displaystyle\binom{4}{2}=6\) (choose which two of the four moves are the east ones).

---

**Step 5 – Total number of shortest complete routes**

The two street portions are independent, so multiply the counts:

\[
\text{total shortest routes}=3 \times 6 = 18.
\]

All these routes have the same total length (3 blocks + the diagonal + 4 blocks), so each is minimal.

---

**Step 6 – Match with answer choices**

The value 18 corresponds to choice **E**.

---

ANSWER: E

1. **Restate the problem**: We need to find the number of different routes Samantha can take from her home to school, passing through the southwest and northeast corners of City Park, with the route being as short as possible.
2. **Set up the math**:
    - First, analyze the journey in three parts: from home to the southwest corner of City - Park, from the southwest corner to the northeast corner of City Park, and from the northeast corner of City Park to school.
    - **Part 1: Home to the southwest corner of City Park**
        - Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. To get to the southwest corner, she needs to move 2 blocks east and 1 block north. The number of ways to arrange these movements can be calculated using the combination formula for permutations of multi - sets. The total number of steps is $n=2 + 1=3$ steps, and we need to choose $k = 2$ steps to be east (or equivalently, $k = 1$ step to be north). The formula for the number of permutations of $n$ objects where there are $n_1$ of one kind, $n_2$ of another kind, $\cdots$, $n_m$ of the $m$th kind is $\frac{n!}{n_1!n_2!\cdots n_m!}$. Here, $n = 3$, $n_1=2$ (east steps) and $n_2 = 1$ (north step), so the number of ways is $\frac{3!}{2!1!}=\frac{3\times2!}{2!×1}=3$.
    - **Part 2: Through City Park**
        - Since there is only one diagonal path through the park, there is only 1 way to travel from the southwest corner to the northeast corner of City Park.
    - **Part 3: Northeast corner of City Park to school**
        - Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. The total number of steps is $n=2 + 2=4$ steps. We need to choose $k = 2$ steps to be east (or equivalently, $k = 2$ steps to be north). Using the combination formula $\frac{n!}{n_1!n_2!}$, where $n = 4$, $n_1 = 2$ (east steps) and $n_2=2$ (north steps), we get $\frac{4!}{2!2!}=\frac{4\times3\times2!}{2!×2\times1}=6$.
3. **Compute the total number of routes**:
    - By the multiplication principle, the total number of different routes is the product of the number of routes in each part. So the total number of routes is $3\times1\times6 = 18$.
4. **Rule out wrong choices**:
    - Choice A (3) is too small because we have already calculated that there are 3 ways from home to the southwest corner and 6 ways from the northeast corner to school, so the total number of routes must be greater than 3.
    - Choice B (6) only accounts for the number of ways from the northeast corner to school and does not consider the number of ways from home to the southwest corner.
    - Choice C (9) is incorrect as our calculation based on the combination of steps in each part of the journey gives 18, not 9.
    - Choice D (12) is also incorrect as it does not match the result of our step - by - step calculation.

ANSWER: E

### Step 1: Restate and map the problem
First, we assign coordinates to simplify grid travel. Let the southwest (SW) corner of City Park be (0,0). Samantha's house is 2 blocks west and 1 block south of this corner, so her house is at (-2,-1). The northeast (NE) corner of the park is an arbitrary fixed point (a,b), as she only takes a single diagonal path between (0,0) and (a,b) with no route choices here. Her school is 2 blocks east and 2 blocks north of the NE park corner,  the school is at (a+2, b+2). She must take the shortest possible street routes between her house and the SW park corner, then take the fixed diagonal through the park, then take the shortest possible street routes from the NE park corner to her school. We calculate the total number of unique valid routes by combining the number of shortest paths for each street segment.
### Step 2: Calculate shortest paths from Samantha's house to the SW park corner
To travel from (-2,-1) to (0,0) via the shortest possible route, she only needs to move east (increasing x) 2 times and north (increasing y) 1 time, with no backtracking. The number of unique orderings of these 3 total steps is the number of ways to choose when to take the single north step:
$$\binom{3}{1} = 3$
This gives 3 unique shortest paths for the first street segment.
### Step 3: Calculate shortest paths from the NE park corner to school
To travel from (a,b) to (a+2,b+2) via the shortest possible route, she needs to move east 2 times and north 2 times, with no backtracking. The number of unique orderings of these 4 total steps is the number of ways to choose 2 positions for the east steps (the remaining 2 will be north steps):
$$\binom{4}{2} = 6$
This gives 6 unique shortest paths for the second street segment.
### Step 4: Compute total number of unique routes
By the fundamental counting principle, the total number of unique routes is the product of the number of paths for each independent street segment, since the diagonal park path has no choices:
$$3 \times 6 = 18$$
### Step 5: Rule out incorrect choices
- Option A (3): Only counts the first segment's paths, not the full route.
- Option B (6): Only counts the second segment's paths, not the full route.
- Option C (9): Incorrectly adds the two segment counts instead of multiplying them.
- Option D (12): Uses an incorrect calculation for one of the segments, not matching combinatorial shortest path counts.
- Option E (18): Matches our calculated total of 18 unique shortest routes.
ANSWER: E

### Step 1: Restate the problem and break the route into segments
Samantha’s shortest route has 3 independent segments:
1.  Streets from her house to the **southwest (SW) corner** of City Park.
2.  A single diagonal path through the park from the SW corner to the **northeast (NE) corner** (no choices here, as a rectangle has only one diagonal between these two corners).
3.  Streets from the NE corner to her school.
We calculate the number of shortest routes for the first and third segments, then multiply them (since the middle segment is fixed, with 1 route).
---
### Step 2: Model the first segment (house to SW corner) with a grid
Assign coordinates: let the SW corner of the park be $(0,0)$.
- Samantha lives 2 blocks west (negative $x$) and 1 block south (negative $y$) of $(0,0)$,  her house is at $(-2, -1)$.
- A shortest path on the grid only uses east (positive $x$) and north (positive $y$) moves (any west/south move would lengthen the path).
- To go from $(-2,-1)$ to $(0,0)$, she needs **2 east moves** and **1 north move**, for a total of $2+1=3$ moves.
The number of distinct shortest paths is the number of ways to arrange these moves, calculated via combinations:
$$\binom{3}{1} = \frac{3!}{2!1!} = 3$$
We can verify by listing the paths:
1.  East → East → North
2.  East → North → East
3.  North → East → East
---
### Step 3: Model the third segment (NE corner to school)
Let the NE corner of the park be $(a,b)$ (the park’s size does not matter, as we only care about displacement).
- School is 2 blocks east and 2 blocks north of $(a,b)$,  its coordinates are $(a+2, b+2)$.
- A shortest path here uses **2 east moves** and **2 north moves**, for a total of $2+2=4$ moves.
The number of distinct shortest paths is:
$$\binom{4}{2} = \frac{4!}{2!2!} = 6$$
We can verify by listing the paths:
1.  East → East → North → North
2.  East → North → East → North
3.  East → North → North → East
4.  North → East → East → North
5.  North → East → North → East
6.  North → North → East → East
---
### Step 4: Combine the results
The middle diagonal segment has exactly 1 route. By the multiplication principle, the total number of shortest full routes is:
$$3 \times 1 \times 6 = 18$$
---
ANSWER: E

1. **Restating the problem**: Samantha needs to find the shortest route from her home to her school, which involves biking to the southwest corner of City Park, taking a diagonal path through the park to the northeast corner, and then biking to school. We need to determine the number of different routes she can take that minimize her travel distance.

2. **Determine the segments of the route**:
   - From Samantha's home to the southwest corner of City Park: She needs to travel 2 blocks west and 1 block south.
   - Through City Park: She takes a diagonal path from the southwest to the northeast corner of the park.
   - From the northeast corner of City Park to her school: She needs to travel 2 blocks east and 2 blocks north.

3. **Analyzing the path through City Park**:
   - The diagonal path through the park is fixed and does not offer multiple route options.

4. **Counting the routes from home to the southwest corner of City Park**:
   - Samantha needs to travel 2 blocks west and 1 block south. The number of different ways to arrange these movements is given by the combination formula for permutations of multiset:
     \[
     \frac{(2+1)!}{2!1!} = \frac{3!}{2!1!} = 3
     \]

5. **Counting the routes from the northeast corner of City Park to school**:
   - Samantha needs to travel 2 blocks east and 2 blocks north. The number of different ways to arrange these movements is given by the combination formula for permutations of multiset:
     \[
     \frac{(2+2)!}{2!2!} = \frac{4!}{2!2!} = 6
     \]

6. **Calculating the total number of different routes**:
   - Since the path through the park is fixed, we multiply the number of routes from home to the southwest corner of the park by the number of routes from the northeast corner of the park to school:
     \[
     3 \times 6 = 18
     \]

7. **Ruling out incorrect choices**:
   - Choices A, B, C, and D offer fewer routes than calculated, so they are incorrect.
   - Choice E offers 18 routes, which matches our calculation.

ANSWER: E

**Problem Restatement:**
We need to find the number of different shortest routes Samantha can take from her home to school, given:
- She starts 2 blocks west and 1 block south of the southwest (SW) corner of City Park.
- Her school is 2 blocks east and 2 blocks north of the northeast (NE) corner of City Park.
- Her route consists of three parts:
  1. Biking on streets to the SW corner of City Park.
  2. Taking a diagonal path through the park from the SW corner to the NE corner.
  3. Biking on streets from the NE corner to school.

The question asks for the total number of different shortest routes possible.

---

### Step 1: Visualize the Scenario
Let's represent the positions on a coordinate grid for clarity:
- Let the SW corner of City Park be at the origin (0, 0).
- The NE corner of City Park is at (a, b), where a and b are the dimensions of the park in blocks. Since the diagonal path is fixed, we don't need to know a and b explicitly.
- Samantha's home is 2 blocks west and 1 block south of (0, 0), so her home is at (-2, -1).
- Her school is 2 blocks east and 2 blocks north of the NE corner (a, b), so her school is at (a + 2, b + 2).

---

### Step 2: Break Down the Route
Samantha's route has three parts:
1. From home (-2, -1) to SW corner (0, 0).
2. Diagonal from SW corner (0, 0) to NE corner (a, b). This part is fixed (only 1 way).
3. From NE corner (a, b) to school (a + 2, b + 2).

Since the diagonal path is fixed, we only need to calculate the number of ways for parts 1 and 3.

---

### Step 3: Calculate Routes for Part 1 (Home to SW Corner)
Samantha needs to go from (-2, -1) to (0, 0). This requires:
- Moving east 2 blocks (from -2 to 0 on the x-axis).
- Moving north 1 block (from -1 to 0 on the y-axis).

The number of different paths is the number of ways to arrange these moves. This is a combination problem: the number of ways to arrange 2 east (E) and 1 north (N) moves in any order.

Total moves = 3 (2 E + 1 N).
Number of unique paths = 3! / (2! * 1!) = 3.

So, there are 3 ways to go from home to the SW corner.

---

### Step 4: Calculate Routes for Part 3 (NE Corner to School)
Samantha needs to go from (a, b) to (a + 2, b + 2). This requires:
- Moving east 2 blocks.
- Moving north 2 blocks.

Again, this is a combination problem: the number of ways to arrange 2 E and 2 N moves.

Total moves = 4 (2 E + 2 N).
Number of unique paths = 4! / (2! * 2!) = 6.

So, there are 6 ways to go from the NE corner to school.

---

### Step 5: Combine the Routes
Since the diagonal path through the park is fixed (only 1 way), the total number of different routes is the product of the number of ways for parts 1 and 3:
Total routes = (routes for part 1) * (routes for part 3) = 3 * 6 = 18.

---

### Step 6: Rule Out Wrong Choices
- A. 3: Too small, only accounts for part 1.
- B. 6: Too small, only accounts for part 3.
- C. 9: Incorrect, not a product of 3 and 6.
- D. 12: Incorrect, not a product of 3 and 6.
- E. 18: Correct, matches our calculation.

---
ANSWER: E

I need to find the minimum number of people wearing both a hat and a glove.

**Step 1: Understand what we're looking for**

We want the minimum overlap between:
- People wearing gloves: 2/5 of total
- People wearing hats: 3/4 of total

**Step 2: Set up the problem using fractions**

Let's say there are $n$ people total.
- People with gloves: $\frac{2n}{5}$
- People with hats: $\frac{3n}{4}$

For these fractions to represent whole numbers of people:
- $n$ must be divisible by 5 (so $\frac{2n}{5}$ is an integer)
- $n$ must be divisible by 4 (so $\frac{3n}{4}$ is an integer)

Therefore, $n$ must be divisible by $\text{lcm}(4,5) = 20$.

The minimum value is $n = 20$.

**Step 3: Calculate for n = 20**

With 20 people:
- People wearing gloves: $\frac{2 \times 20}{5} = 8$ people
- People wearing hats: $\frac{3 \times 20}{4} = 15$ people

**Step 4: Find minimum overlap using the principle**

By the inclusion-exclusion principle:
$$|G \cup H| = |G| + |H| - |G \cap H|$$

where $G$ is the set of people with gloves and $H$ is the set of people with hats.

The maximum value of $|G \cup H|$ is the total number of people, which is 20.

So: $20 \geq 8 + 15 - |G \cap H|$

This gives us: $|G \cap H| \geq 8 + 15 - 20 = 3$

Alternatively, the minimum overlap occurs when the two groups overlap as little as possible. The minimum number in both groups is:
$$\min(|G \cap H|) = |G| + |H| - |G \cup H|_{\max} = 8 + 15 - 20 = 3$$

**Step 5: Verify this is achievable**

With 20 people:
- 8 wearing gloves (and not hats)
- 3 wearing both gloves and hats
- 12 wearing hats (and not gloves)
- 2 wearing neither

Check: $8 + 3 + 12 + 2 = 23$ ❌

Let me recalculate:
- 5 wearing only gloves
- 3 wearing both
- 12 wearing only hats
- 0 wearing neither

Check: $5 + 3 + 12 + 0 = 20$ ✓
- Total with gloves: $5 + 3 = 8$ ✓
- Total with hats: $12 + 3 = 15$ ✓

This works! The minimum is 3 people wearing both.

**Step 6: Rule out wrong answers**

- A. 3 ✓ (This is our answer)
- B. 5 (Too large; we showed 3 is achievable)
- C. 8 (Too large)
- D. 15 (Too large)
- E. 20 (Too large)

ANSWER: A

1. Restate the problem.  
   We have a room with an unknown total number of people, N. Of these, 2/5 are wearing gloves and 3/4 are wearing hats.  We want the minimum possible number wearing both a glove and a hat.

2. Express the counts in terms of N.  
   – Number wearing gloves = (2/5) N.  
   – Number wearing hats   = (3/4) N.

3. Use the inclusion–exclusion bound for the intersection.  
   For any two sets G (gloves) and H (hats),  
   |G ∩ H| ≥ |G| + |H| − N.  

4. Substitute the fractional counts.  
   |G ∩ H| ≥ (2/5 N) + (3/4 N) − N  
            = N (2/5 + 3/4 − 1).

5. Compute the coefficient.  
   2/5 = 8/20, 3/4 = 15/20, 1 = 20/20, so  
   2/5 + 3/4 − 1 = (8/20 + 15/20 − 20/20) = 3/20.  
   Hence |G ∩ H| ≥ (3/20) N.

6. Ensure integer counts.  
   N must be divisible by 5 and by 4 so that (2/5) N and (3/4) N are integers.  The least such N is 20.

7. Find the minimum intersection with N = 20.  
   |G ∩ H| ≥ (3/20) × 20 = 3.

8. Rule out other choices.  
   The minimum possible number is 3.  The choices 5, 8, 15, and 20 are all larger than this minimum, so they cannot be the minimum.

ANSWER: A

An elegant way to solve this problem is by using the Principle of Inclusion-Exclusion and divisibility rules.

### Step 1: Restate what is being asked
We need to find the minimum number of people in a room who are wearing both a hat and a glove, given that $\frac{2}{5}$ of the people are wearing gloves and $\frac{3}{4}$ of the people are wearing hats.

### Step 2: Set up the math
Let $N$ be the total number of people in the room. 
* The number of people wearing gloves is $G = \frac{2}{5}N$.
* The number of people wearing hats is $H = \frac{3}{4}N$.

Since the number of people in each category must be a whole number (an integer), $N$ must be divisible by both $5$ and $4$. The least common multiple of $5$ and $4$ is $20$. Therefore, $N$ must be a multiple of $20$. Let $N = 20k$ for some positive integer $k \ge 1$.

Substituting $N = 20k$:
* $G = \frac{2}{5}(20k) = 8k$
* $H = \frac{3}{4}(20k) = 15k$

### Step 3: Compute the minimum overlap
Let $B$ be the number of people wearing both a hat and a glove. 
By the Principle of Inclusion-Exclusion, the number of people wearing at least one of the two items is:
$$\text{Wearing at least one} = G + H - B$$

Since the number of people wearing at least one item cannot exceed the total number of people in the room ($N$), we have:
$$G + H - B \le N$$

Substitute the expressions in terms of $k$:
$$8k + 15k - B \le 20k$$
$$23k - B \le 20k$$
$$B \ge 3k$$

### Step 4: Find the minimum value
To minimize the overlap $B$, we must choose the smallest possible positive integer value for $k$, which is $k = 1$.
When $k = 1$:
* $N = 20$ (total people)
* $G = 8$ (wearing gloves)
* $H = 15$ (wearing hats)
* $B \ge 3(1) = 3$

Thus, the minimum number of people wearing both is $3$.

### Step 5: Verify and rule out wrong choices
If $N = 20$, we can distribute the people as follows:
* Wearing only gloves: $8 - 3 = 5$ people
* Wearing only hats: $15 - 3 = 12$ people
* Wearing both: $3$ people
* Wearing neither: $20 - (5 + 12 + 3) = 0$ people

Since all these groups have non-negative integer counts, this is a valid scenario. 
* **A. 3** is the correct minimum.
* **B. 5**, **C. 8**, **D. 15**, and **E. 20** are all greater than the minimum possible value of 3.

ANSWER: A

The problem requires finding the smallest possible number of people who could be wearing both a hat and gloves, given the stated fractions of the total population \(N\).

The total number of people \(N\) must be a multiple of both 5 and 4 (so that both given fractions produce integer counts of people). Thus, the least possible value of \(N\) is their least common multiple, which is 20.

For \(N = 20\):
- Number wearing gloves: \(\frac{2}{5} \times 20 = 8\)
- Number wearing hats: \(\frac{3}{4} \times 20 = 15\)

By the principle of inclusion-exclusion, the size of the intersection of these two groups is at least
\[
8 + 15 - 20 = 3.
\]
(The same lower bound is obtained from the general formula \(\bigl(\frac{2}{5} + \frac{3}{4} - 1\bigr)N = \frac{3}{20}N\), which equals 3 when \(N = 20\).)

For any larger admissible value of \(N\) (i.e., any multiple of 20), the corresponding lower bound on the intersection is strictly larger than 3. Consequently the smallest integer that can ever occur as the size of the intersection is 3, and this value is attained when \(N = 20\).

The other listed choices are all larger than this minimal attainable intersection, so they are ruled out.

ANSWER: A

1. **Understanding the Problem**: The problem asks us to determine the minimum number of people in a room who are wearing both a hat and gloves, given that 2/5 of the people are wearing gloves and 3/4 of the people are wearing hats.

2. **Setting Up the Math**: Let's denote the total number of people in the room as \(N\). The number of people wearing gloves is \(\frac{2}{5}N\), and the number of people wearing hats is \(\frac{3}{4}N\).

3. **Using the Principle of Inclusion-Exclusion**: The principle states that for two sets \(A\) and \(B\), the size of the union of \(A\) and \(B\) is \(|A \cup B| = |A| + |B| - |A \cap B|\). Here, \(A\) is the set of people wearing gloves, and \(B\) is the set of people wearing hats. So, \(|A| = \frac{2}{5}N\), \(|B| = \frac{3}{4}N\), and \(|A \cup B| \leq N\) because the total number of people wearing either or both cannot exceed the total number of people in the room.

4. **Applying the Principle**: We have \(|A \cup B| = |A| + |B| - |A \cap B| = \frac{2}{5}N + \frac{3}{4}N - |A \cap B|\). So, \(|A \cap B| = \frac{2}{5}N + \frac{3}{4}N - |A \cup B|\). To minimize \(|A \cap B|\), we need to maximize \(|A \cup B|\), but \(|A \cup B| \leq N\). Thus, \(|A \cap B| \geq \frac{2}{5}N + \frac{3}{4}N - N\).

5. **Computing the Minimum Intersection**: Let's compute \(\frac{2}{5}N + \frac{3}{4}N - N\). First, find a common denominator for \(\frac{2}{5}\) and \(\frac{3}{4}\), which is 20. So, \(\frac{2}{5} = \frac{8}{20}\) and \(\frac{3}{4} = \frac{15}{20}\). Thus, \(\frac{8}{20}N + \frac{15}{20}N - N = \frac{23}{20}N - N = \frac{23}{20}N - \frac{20}{20}N = \frac{3}{20}N\).

6. **Finding the Minimum \(N\) for Whole Numbers**: Since \(|A \cap B| = \frac{3}{20}N\), and \(|A \cap B|\) must be a whole number, \(N\) must be divisible by 20 to ensure \(\frac{3}{20}N\) is an integer. The smallest \(N\) that satisfies this condition is 20.

7. **Calculating \(|A \cap B|\) for \(N = 20\)**: Substituting \(N = 20\) into \(\frac{3}{20}N\) gives \(\frac{3}{20} \times 20 = 3\).

8. **Ruling Out Wrong Choices**: 
- For \(N = 20\), \(|A \cap B| = 3\), so choice A is possible.
- We need to check if other choices could be minimum under different \(N\), but since we derived the minimum based on the condition that \(N\) must be a multiple of 20, and we used the smallest such \(N\), 3 is indeed the minimum.
- Choices B, C, D, and E are larger than 3, so they cannot be the minimum.

9. **Conclusion**: The minimum number of people wearing both a hat and gloves is 3.

ANSWER: A

**Step-by-step Solution**

1. **Restate the problem:** We are told that in a room, \(\frac{2}{5}\) of the people wear gloves and \(\frac{3}{4}\) wear hats. We need the **minimum possible number** of people wearing **both** a hat and gloves.

2. **Set up variables:**  
   Let the total number of people in the room be \(N\).  
   Number wearing gloves \(= \frac{2}{5}N\).  
   Number wearing hats \(= \frac{3}{4}N\).

3. **Find a feasible total \(N\):**  
   Since the number of people must be an integer, \(N\) must be a multiple of both 5 and 4.  
   The least common multiple of 5 and 4 is 20, so \(N = 20k\) for some positive integer \(k\).

4. **Compute counts for general \(k\):**  
   \(G = \frac{2}{5}(20k) = 8k\) people wear gloves.  
   \(H = \frac{3}{4}(20k) = 15k\) people wear hats.

5. **Apply the overlap formula:**  
   The minimum number of people wearing both occurs when the overlap is as small as possible. By the inclusion–exclusion principle:  
   \(\text{Both} \ge G + H - N\) (and it can reach this lower bound if the sets are arranged accordingly).  
   Substitute:  
   \(\text{Both} \ge 8k + 15k - 20k = 3k\).

6. **Minimize over \(k\):**  
   To make the both-count as small as possible, take the smallest possible \(k\), which is \(k=1\) (so \(N=20\)).  
   Then the minimum both \(= 3(1) = 3\).  
   For any larger total (\(k \ge 2\)), the minimum both would be \(6, 9, \dots\), all greater than 3.

7. **Match to the choices:**  
   The smallest choice given is 3, which matches our computed absolute minimum.  
   Choices B–E (5, 8, 15, 20) are all larger and cannot be the minimum.

**ANSWER: A**

1. **Understand what is being asked**:  
We are told that in a room:
- $ \frac{2}{5} $ of the people are wearing gloves.
- $ \frac{3}{4} $ of the people are wearing hats.  

We are to find the **minimum possible number** of people who are wearing **both** a hat and a glove.

This is a classic problem involving the **principle of inclusion-exclusion** and minimizing overlap given two fractions of a total.

2. **Let the total number of people be $ N $**.  
Then:
- Number wearing gloves = $ \frac{2}{5}N $
- Number wearing hats = $ \frac{3}{4}N $

We are to find the **minimum** possible value of the number wearing **both**, which we’ll call $ B $.

3. **Use the inclusion-exclusion principle**:  
For any two sets:
$$
\text{Gloves} + \text{Hats} - \text{Both} \leq N
$$
So:
$$
\frac{2}{5}N + \frac{3}{4}N - B \leq N
$$

Solve for $ B $:
$$
\left( \frac{2}{5} + \frac{3}{4} \right)N - B \leq N
$$

Find common denominator for the fractions:
- $ \frac{2}{5} = \frac{8}{20} $
- $ \frac{3}{4} = \frac{15}{20} $
- Sum = $ \frac{23}{20} $

So:
$$
\frac{23}{20}N - B \leq N
\Rightarrow \frac{23}{20}N - N \leq B
\Rightarrow \frac{3}{20}N \leq B
$$

Thus, the **minimum** number of people wearing both is **at least** $ \frac{3}{20}N $.

4. **But $ N $ must be chosen so that both $ \frac{2}{5}N $ and $ \frac{3}{4}N $ are integers**, since you can’t have a fractional person.

So $ N $ must be a common multiple of the denominators 5 and 4.  
- LCM of 5 and 4 is **20**.

So the **smallest possible** value of $ N $ is 20.

5. **Compute values when $ N = 20 $**:
- Gloves: $ \frac{2}{5} \cdot 20 = 8 $
- Hats: $ \frac{3}{4} \cdot 20 = 15 $

Now, what’s the **minimum** number wearing both?

Use the formula:
$$
\text{Both} = \text{Gloves} + \text{Hats} - \text{Total (at most)} = 8 + 15 - 20 = 3
$$

This is the **minimum possible overlap**, because if you try to make the overlap smaller than 3, then total number of people would have to be more than 20, which contradicts the actual total.

So **minimum number wearing both is 3**, and this is achievable when the sets overlap as little as possible.

6. **Check answer choices**:  
A. 3 ✅  
B. 5  
C. 8  
D. 15  
E. 20  

We found that **3 is achievable** and is the **minimum possible**, so the answer is **A**.

7. **Double-check logic**:  
- With 20 people: 8 wear gloves, 15 wear hats.
- If only 3 wear both, then:
  - Only gloves: 8 – 3 = 5
  - Only hats: 15 – 3 = 12
  - Both: 3
  - Total = 5 + 12 + 3 = 20 → OK.
- Can we have fewer than 3? Say 2:
  - Then total people = (8 – 2) + (15 – 2) + 2 = 6 + 13 + 2 = 21 > 20 → impossible.
- So 3 is indeed the minimum.

ANSWER: A

 **Step 1: Restate the problem.**  
We are told that \(\frac{2}{5}\) of the people in a room are wearing gloves and \(\frac{3}{4}\) are wearing hats. We must find the *minimum* possible number of people who are wearing **both** a hat and a glove.

**Step 2: Choose a variable for the total number of people.**  
Let \(n\) be the total number of people in the room. Because the number of people wearing gloves, \(\frac{2}{5}n\), and the number wearing hats, \(\frac{3}{4}n\), must both be whole numbers, \(n\) must be divisible by \(5\) and by \(4\). Since \(5\) and \(4\) are coprime, \(n\) must be a multiple of \(\operatorname{lcm}(5,4)=20\).  
So write \(n = 20k\) for some positive integer \(k\).

**Step 3: Express the two groups in terms of \(n\).**  
- Glove wearers: \(G = \frac{2}{5}n = \frac{2}{5}(20k) = 8k\).  
- Hat wearers: \(H = \frac{3}{4}n = \frac{3}{4}(20k) = 15k\).

**Step 4: Apply the Principle of Inclusion–Exclusion.**  
For any two sets,  
\[
|G \cup H| = |G| + |H| - |G \cap H|.
\]  
The number of people wearing at least one of the two items cannot exceed the total number of people, so \(|G \cup H| \le n\). Rearranging gives a lower bound on the overlap:
\[
|G \cap H| \ge |G| + |H| - n.
\]

**Step 5: Compute the lower bound.**  
Substituting the expressions from Step 3:
\[
|G \cap H| \ge 8k + 15k - 20k = 3k.
\]  
Thus the smallest possible overlap is \(3k\), and it occurs when the union \(G \cup H\) is as large as possible—that is, when it equals the entire room.

**Step 6: Minimize over all valid room sizes.**  
Because \(n\) must be a positive multiple of \(20\), the smallest possible value is \(n = 20\) (i.e., \(k=1\)). Plugging this in:
\[
|G \cap H| \ge 3(1) = 3.
\]

**Step 7: Verify that \(3\) is actually achievable.**  
With \(n = 20\):  
- Glove wearers: \(8\) people.  
- Hat wearers: \(12\) people.  

Suppose exactly \(3\) people wear both. Then:  
- Only gloves: \(8 - 3 = 5\).  
- Only hats: \(12 - 3 = 9\).  
- Both: \(3\).  
- Neither: \(20 - (5 + 9 + 3) = 3\).  

All counts are non-negative integers and sum to \(20\), so a configuration with exactly \(3\) people wearing both items is valid. Hence the minimum is indeed \(3\).

**Step 8: Rule out the wrong choices.**  
- **A. 3:** Achievable and proven minimal.  
- **B. 5:** Larger than the proven minimum; not the smallest possible overlap.  
- **C. 8:** This equals the total number of glove wearers (\(\frac{2}{5}\) of \(20\)), not the required overlap.  
- **D. 15:** This equals the total number of hat wearers (\(\frac{3}{4}\) of \(20\)), not the overlap.  
- **E. 20:** This is the total number of people needed to make the fractions yield integers, not the number wearing both.  

Since we have exhibited a valid arrangement with exactly \(3\) people wearing both, any answer greater than \(3\) cannot be the minimum.

ANSWER: A

Here is the step-by-step solution:

**1. Restate the Problem and Set Up Variables**
We are asked to find the minimum number of people in a room who are wearing **both** a hat and gloves.
Let $N$ be the total number of people in the room.
Let $G$ be the number of people wearing gloves.
Let $H$ be the number of people wearing hats.
Let $B$ be the number of people wearing **both**.

From the problem description:
*   $G = \frac{2}{5}N$
*   $H = \frac{3}{4}N$

**2. Determine the Minimum Total Number of People ($N$)**
Since the number of people must be a whole number (an integer), $N$ must be a value that makes both $G$ and $H$ integers.
*   For $G = \frac{2}{5}N$ to be an integer, $N$ must be divisible by 5.
*   For $H = \frac{3}{4}N$ to be an integer, $N$ must be divisible by 4.

Therefore, $N$ must be a common multiple of 4 and 5. To find the **minimum** number of people wearing both, we should test with the **smallest possible** total number of people. The least common multiple (LCM) of 4 and 5 is 20.
So, let $N = 20$.

**3. Calculate the Number of People in Each Group**
Using $N = 20$:
*   Number of people wearing gloves ($G$):
    $$G = \frac{2}{5} \times 20 = 2 \times 4 = 8$$
*   Number of people wearing hats ($H$):
    $$H = \frac{3}{4} \times 20 = 3 \times 5 = 15$$

**4. Compute the Minimum Overlap using the Principle of Inclusion-Exclusion**
We use the formula relating the sets:
$$\text{Total} = (\text{Gloves}) + (\text{Hats}) - (\text{Both}) + (\text{Neither})$$
$$N = G + H - B + \text{Neither}$$

Rearranging to solve for $B$ (the number wearing both):
$$B = G + H - N + \text{Neither}$$

To find the **minimum** value for $B$, we must subtract as much as possible from the sum $(G + H)$. This means we assume the number of people wearing **neither** is 0 (everyone is wearing at least one item).

*   Sum of people counted in groups: $G + H = 8 + 15 = 23$.
*   Actual total people available: $N = 20$.

Because the sum of the individual groups (23) is greater than the total population (20), the groups must overlap. The minimum overlap is simply the difference between the sum of the parts and the whole:

$$B_{\text{min}} = (G + H) - N$$
$$B_{\text{min}} = 23 - 20$$
$$B_{\text{min}} = 3$$

It is possible to construct this scenario:
*   3 people wear both hat and gloves.
*   5 people wear only gloves ($8 - 3$).
*   12 people wear only hats ($15 - 3$).
*   Total = $3 + 5 + 12 = 20$. This works perfectly.

**5. Rule Out Wrong Choices**
*   **A. 3**: This matches our calculation of the absolute minimum overlap required by the numbers.
*   **B. 5**: While it is possible for 5 people to wear both (if, for example, some people wore neither), it is not the *minimum* possible number.
*   **C. 8**: This would imply everyone wearing gloves also wears a hat. This is possible, but not the minimum.
*   **D. 15**: This would imply everyone wearing a hat also wears gloves. This is impossible because there are only 8 people wearing gloves; you can't have 15 people doing both if only 8 have gloves.
*   **E. 20**: This would mean everyone wears both. Impossible since only 8 people have gloves.

ANSWER: A

**Step 1 – Translate the problem.**  
Let  

- \(N\) = total number of people in the room (must be an integer).  
- \(G\) = number of people wearing gloves.  
- \(H\) = number of people wearing hats.  
- \(x\) = number wearing **both** a hat and a glove (what we need).

We are told  

\[
G=\frac25N,\qquad H=\frac34N .
\]

Since \(G\) and \(H\) count people, they must be integers, so \(N\) must be a multiple of both denominators 5 and 4. The least common multiple of 5 and 4 is 20, therefore  

\[
N = 20k\qquad (k\text{ a positive integer}).
\]

**Step 2 – Express the numbers in terms of }k\text{.**  

\[
G = \frac25(20k)=8k,\qquad H=\frac34(20k)=15k .
\]

**Step 3 – Use the principle of inclusion–exclusion.**  
All people are either wearing gloves, wearing hats, both, or neither, so

\[
G + H - x \le N .
\]

Rearranging gives a lower bound on the overlap:

\[
x \ge G + H - N .
\]

Insert the expressions:

\[
x \ge 8k + 15k - 20k = 3k .
\]

Thus any feasible arrangement must have at least \(3k\) people wearing both items.

**Step 4 – Show that the bound can be attained.**  
Take exactly  

\[
x = 3k .
\]

Then  

- gloves‑only = \(G - x = 8k-3k = 5k\),  
- hats‑only   = \(H - x = 15k-3k = 12k\),  
- neither     = \(N - (G + H - x) = 20k - (8k+15k-3k)=0\).

All counts are non‑negative integers, so the bound \(x = 3k\) is attainable.  

**Step 5 – Find the smallest possible value.**  
The smallest positive integer \(k\) is \(k=1\). This gives  

\[
N = 20,\quad G = 8,\quad H = 15,\quad x = 3 .
\]

Hence the minimum number of people wearing both a hat and a glove is **3**.

The answer choice that corresponds to 3 is **A**.

ANSWER: A

1. **Restate what is being asked**:
   - We are given that $\frac{2}{5}$ of the people in a room are wearing gloves and $\frac{3}{4}$ of the people are wearing hats. We need to find the minimum number of people in the room who are wearing both a hat and a glove.
2. **Set up the math**:
   - Let the total number of people in the room be $n$. The number of people wearing gloves is $\frac{2}{5}n$, and the number of people wearing hats is $\frac{3}{4}n$.
   - The number of people wearing both a hat and a glove is at least $\frac{2}{5}n+\frac{3}{4}n - n$. This is based on the principle of inclusion - exclusion. The formula for $|A\cap B|$ (the number of elements in both sets $A$ and $B$) is $|A|+|B|-|A\cup B|$. Since $|A\cup B|\leq n$ (the total number of people in the room), the minimum value of $|A\cap B|$ occurs when $|A\cup B| = n$.
   - First, find a common denominator for $\frac{2}{5}n+\frac{3}{4}n - n$. The common denominator of 5 and 4 is 20.
   - $\frac{2}{5}n+\frac{3}{4}n - n=\frac{2\times4}{5\times4}n+\frac{3\times5}{4\times5}n - n=\frac{8}{20}n+\frac{15}{20}n - n$.
   - Combine the fractions: $\frac{8 + 15}{20}n - n=\frac{23}{20}n - n=\frac{23n-20n}{20}=\frac{3}{20}n$.
3. **Do the computation**:
   - We want to find the smallest positive - integer value of $n$ such that $\frac{2}{5}n=\frac{2n}{5}$, $\frac{3}{4}n=\frac{3n}{4}$, and $\frac{3}{20}n$ are all integers.
   - For $\frac{2n}{5}$ to be an integer, $n$ must be a multiple of 5. For $\frac{3n}{4}$ to be an integer, $n$ must be a multiple of 4.
   - The least common multiple of 4 and 5 is $LCM(4,5)=20$.
   - When $n = 20$, the number of people wearing both a hat and a glove is $\frac{3}{20}\times20 = 3$.
4. **Rule out the wrong choices**:
   - We have found that when $n = 20$, the minimum number of people wearing both a hat and a glove is 3.
   - Option B (5), Option C (8), Option D (15), and Option E (20) are all larger than the minimum value we calculated. They could be the number of people wearing both if the total number of people in the room is larger, but we are looking for the minimum.

ANSWER: D (Wait, let's correct, based on our calculation when n = 20, the number of people wearing both is 3, so the answer should be A) 
The correct answer is that when $n = 20$ (the least common multiple of 4 and 5 to make the number of glove - wearers, hat - wearers, and both - wearers integers), the number of people wearing both is $\frac{3}{20}\times20=3$.
ANSWER: D (No, correction: based on above steps, when n = 20, the value of the number of people wearing both is 3)
ANSWER: A

### Step 1: Restate the problem
We need to find the smallest possible number of people in a room who wear both gloves and hats, given that 2/5 of all people wear gloves and 3/4 wear hats. All counts of people must be whole numbers (you cannot have a fraction of a person).

### Step 2: Find the smallest valid total number of people
The fractions 2/5 and 3/4 require the total number of people \(N\) to be a common multiple of the denominators 4 and 5, so that 2\(N\)/5 and 3\(N\)/4 are integers. The least common multiple (LCM) of 4 and 5 is 20, so the smallest possible total number of people is 20. Any smaller number would result in a fraction of a person wearing gloves or hats, which is impossible.

### Step 3: Calculate the number of people in each group
For \(N=20\):
- Number of people wearing gloves: \(2/5 * 20 = 8\)
- Number of people wearing hats: \(3/4 *20 =15\)

### Step 4: Use set inclusion-exclusion to find the minimum overlap
For sets \(G\) (gloves) and \(H\) (hats), the size of their union is:
\(|G ∪ H| = |G| + |H| - |G∩H|\)
The union of the two groups cannot exceed the total number of people in the room, so \(|G ∪ H| ≤ 20\). Rearranging to solve for the overlap \(|G∩H|\):
\(|G∩H| = |G| + |H| - |G∪H|\)
To minimize the overlap, we maximize the union (set it to the maximum possible value of 20, the total number of people). Plugging in values:
\(|G∩H| = 8 +15 -20 = 3\)

### Step 5: Eliminate incorrect options
- **Option B (5):** Too large, we already found a valid overlap of 3 is possible.
- **Option C (8):** This is the maximum possible number of people who could wear both (all glove wearers also wear hats), not the minimum.
- **Option D (15):** This is the total number of hat wearers, an impossible overlap.
- **Option E (20):** This is the total number of people in the smallest valid room, not the overlap.

ANSWER: A

### Step 1: Restate the problem and define variables
We need to find the smallest possible number of people wearing both a hat and a glove, given that 2/5 of the room wears gloves and 3/4 wears hats. Let:
- $N$ = total number of people in the room (must be a positive integer)
- $G$ = number of people wearing gloves = $\frac{2}{5}N$ (must be an integer)
- $H$ = number of people wearing hats = $\frac{3}{4}N$ (must be an integer)
- $B$ = number of people wearing both (this is what we need to minimize)
---
### Step 2: Ensure $G$ and $H$ are integers
For $\frac{2}{5}N$ to be an integer, $N$ must be a multiple of 5. For $\frac{3}{4}N$ to be an integer, $N$ must be a multiple of 4. The smallest positive integer that is a multiple of both 4 and 5 is their least common multiple (LCM): $\text{LCM}(4,5) = 20$. So the smallest possible $N$ is 20 (larger multiples of 20 will only increase the minimum $B$, as we will see).
---
### Step 3: Use inclusion-exclusion to find the minimum $B$
The total number of people wearing at least one of gloves or hats is $|G \cup H| = G + H - B$. This cannot exceed the total number of people $N$, so:
$$G + H - B \leq N$$
Rearrange to solve for $B$:
$$B \geq G + H - N$$
---
### Step 4: Substitute values for the smallest $N=20$
For $N=20$:
- $G = \frac{2}{5}(20) = 8$
- $H = \frac{3}{4}(20) = 15$
Plug into the inequality:
$$B \geq 8 + 15 - 20 = 3$$
---
### Step 5: Verify if $B=3$ is possible
If $B=3$:
- People wearing only gloves: $G - B = 8 - 3 = 5$
- People wearing only hats: $H - B = 15 - 3 = 12$
- People wearing neither: $N - (5 + 12 + 3) = 0$
This adds up to exactly 20 people, with no fractional people, so this scenario is valid. We cannot have $B < 3$, because $8 + 15 = 23 > 20$, meaning the overlap cannot be smaller than 3 (otherwise $|G \cup H|$ would exceed $N$).
---
### Step 6: Rule out wrong choices
- B (5), C (8), D (15), E (20) are all larger than 3, so they are not the minimum.
ANSWER: A

1. **Restating the problem**: We need to find the minimum number of people in a room who are wearing both a hat and a glove, given that 2/5 of the people are wearing gloves and 3/4 of the people are wearing hats.

2. **Setting up the math**:
    - Let \( n \) be the total number of people in the room.
    - The number of people wearing gloves is \( \frac{2}{5}n \).
    - The number of people wearing hats is \( \frac{3}{4}n \).

3. **Using the principle of inclusion-exclusion**:
    - The maximum number of people not wearing both a hat and a glove is the sum of those not wearing gloves and those not wearing hats.
    - People not wearing gloves: \( n - \frac{2}{5}n = \frac{3}{5}n \).
    - People not wearing hats: \( n - \frac{3}{4}n = \frac{1}{4}n \).

4. **Calculating the maximum number of people not wearing both**:
    - Maximum people not wearing both = \( \frac{3}{5}n + \frac{1}{4}n \).
    - To add these fractions, find a common denominator: \( \frac{3}{5}n = \frac{12}{20}n \) and \( \frac{1}{4}n = \frac{5}{20}n \).
    - Thus, \( \frac{12}{20}n + \frac{5}{20}n = \frac{17}{20}n \).

5. **Finding the minimum number of people wearing both**:
    - The minimum number of people wearing both a hat and a glove is \( n - \frac{17}{20}n = \frac{3}{20}n \).

6. **Finding the smallest integer \( n \) such that \( \frac{3}{20}n \) is an integer**:
    - The smallest \( n \) that satisfies this is \( n = 20 \) (since \( \frac{3}{20} \times 20 = 3 \)).

7. **Verification**:
    - For \( n = 20 \):
        - People wearing gloves: \( \frac{2}{5} \times 20 = 8 \).
        - People wearing hats: \( \frac{3}{4} \times 20 = 15 \).
        - Minimum people wearing both: \( 20 - (8 + 15 - 20) = 20 - 3 = 3 \).

8. **Ruling out wrong choices**:
    - Choice B (5), C (8), D (15), and E (20) are not the minimum values derived from the calculation.
    - Only Choice A (3) fits the derived minimum number.

ANSWER: A

**Problem Restatement:**
We need to find the smallest possible number of people in a room who are wearing **both** a hat and gloves, given that:
- \( \frac{2}{5} \) of the people are wearing gloves,
- \( \frac{3}{4} \) of the people are wearing hats.

---

### Step 1: Define Variables
Let \( n \) be the total number of people in the room.
- Number wearing gloves: \( \frac{2}{5}n \).
- Number wearing hats: \( \frac{3}{4}n \).

Since the number of people must be whole numbers, \( n \) must be divisible by both 5 and 4 (the denominators of the fractions). The least common multiple (LCM) of 5 and 4 is 20. So, the smallest possible \( n \) is 20.

---

### Step 2: Calculate Number Wearing Gloves and Hats
For \( n = 20 \):
- Number wearing gloves: \( \frac{2}{5} \times 20 = 8 \).
- Number wearing hats: \( \frac{3}{4} \times 20 = 15 \).

---

### Step 3: Use the Principle of Inclusion-Exclusion
The minimum number of people wearing **both** a hat and gloves occurs when as many people as possible are wearing only one of the two items. The formula for the overlap (both) is:
\[
\text{Number wearing both} = \text{Number wearing gloves} + \text{Number wearing hats} - \text{Total number of people}.
\]
Plugging in the numbers:
\[
\text{Number wearing both} = 8 + 15 - 20 = 3.
\]

---

### Step 4: Verify the Answer
- If 3 people wear both, then:
  - Only gloves: \( 8 - 3 = 5 \),
  - Only hats: \( 15 - 3 = 12 \),
  - Neither: \( 20 - (5 + 12 + 3) = 0 \).
This is valid because all counts are non-negative.

---

### Step 5: Rule Out Other Choices
- **A. 3**: Valid, as shown above.
- **B. 5**: Larger than the minimum, so not the minimum.
- **C. 8**: Larger than the minimum.
- **D. 15**: Larger than the minimum.
- **E. 20**: Larger than the minimum.

---

ANSWER: A

I need to find how many problems Olivia answered correctly.

**Step 1: Restate what I'm looking for**
- Total problems: 10
- Points for correct answer: +5
- Points for incorrect answer: -2
- Final score: 29
- Find: number of correct answers

**Step 2: Set up the equations**
Let $c$ = number of correct answers
Let $i$ = number of incorrect answers

From the constraints:
- $c + i = 10$ (she answered all 10 problems)
- $5c - 2i = 29$ (her total score)

**Step 3: Solve the system**
From the first equation: $i = 10 - c$

Substitute into the second equation:
$$5c - 2(10 - c) = 29$$

**Step 4: Compute**
$$5c - 20 + 2c = 29$$
$$7c - 20 = 29$$
$$7c = 49$$
$$c = 7$$

**Step 5: Verify the answer**
- Correct answers: 7
- Incorrect answers: $10 - 7 = 3$
- Score: $5(7) - 2(3) = 35 - 6 = 29$ ✓

**Step 6: Check against choices**
- A. 5: Would give $5(5) - 2(5) = 25 - 10 = 15$ ✗
- B. 6: Would give $5(6) - 2(4) = 30 - 8 = 22$ ✗
- C. 7: Would give $5(7) - 2(3) = 35 - 6 = 29$ ✓
- D. 8: Would give $5(8) - 2(2) = 40 - 4 = 36$ ✗
- E. 9: Would give $5(9) - 2(1) = 45 - 2 = 43$ ✗

ANSWER: C

1. Restate the problem: Olivia answered all 10 problems. She gets +5 points for each correct answer and –2 points for each incorrect answer. Her total score is 29. We must find how many she answered correctly.

2. Define variables: Let c = number of correct answers. Then the number of incorrect answers is i = 10 – c.

3. Write the score equation:  
   Score = 5·(number correct) – 2·(number incorrect)  
   ⇒ 5c – 2i = 29.

4. Substitute i = 10 – c into the score equation:  
   5c – 2(10 – c) = 29  
   5c – 20 + 2c = 29  
   7c – 20 = 29  
   7c = 49  
   c = 7.

5. Check this answer against the multiple-choice options and rule out the others by computing their scores:
   A. c = 5 → score = 5·5 – 2·5 = 25 – 10 = 15 (≠29)  
   B. c = 6 → score = 5·6 – 2·4 = 30 – 8 = 22 (≠29)  
   C. c = 7 → score = 5·7 – 2·3 = 35 – 6 = 29 ✔  
   D. c = 8 → score = 5·8 – 2·2 = 40 – 4 = 36 (≠29)  
   E. c = 9 → score = 5·9 – 2·1 = 45 – 2 = 43 (≠29)

Only choice C gives the required score of 29.

ANSWER: C

To find the number of correct answers Olivia had, we can follow these steps:

**Step 1: Restate what is being asked and define variables**
We are asked to find the number of correct answers Olivia had in a 10-problem contest. 
Let $c$ be the number of correct answers.
Let $w$ be the number of incorrect answers.

**Step 2: Set up the equations**
Since Olivia answered every problem, the total number of problems she answered is 10:
1. $c + w = 10$

She gains 5 points for each correct answer and loses 2 points for each incorrect answer, resulting in a total score of 29:
2. $5c - 2w = 29$

**Step 3: Solve the system of equations**
From the first equation, we can express $w$ in terms of $c$:
$$w = 10 - c$$

Substitute this expression for $w$ into the second equation:
$$5c - 2(10 - c) = 29$$
$$5c - 20 + 2c = 29$$
$$7c - 20 = 29$$

Add 20 to both sides of the equation:
$$7c = 49$$

Divide by 7:
$$c = 7$$

So, Olivia had 7 correct answers (and therefore $10 - 7 = 3$ incorrect answers).

**Step 4: Verify the solution and rule out the other choices**
Let's calculate the score for each choice to see why they are incorrect:
* **A. 5 correct:** Score would be $5(5) - 2(5) = 25 - 10 = 15$ (Incorrect)
* **B. 6 correct:** Score would be $5(6) - 2(4) = 30 - 8 = 22$ (Incorrect)
* **C. 7 correct:** Score is $5(7) - 2(3) = 35 - 6 = 29$ (Correct)
* **D. 8 correct:** Score would be $5(8) - 2(2) = 40 - 4 = 36$ (Incorrect)
* **E. 9 correct:** Score would be $5(9) - 2(1) = 45 - 2 = 43$ (Incorrect)

The correct choice is C.

ANSWER: C

1. Restate the problem: There are 10 problems total. Olivia earns +5 points per correct answer and −2 points per incorrect answer. She answered all 10 problems and scored exactly 29 points. The task is to find how many problems she answered correctly.

2. Set up variables and the equation: Let \(C\) be the number of correct answers. Then the number of incorrect answers is \(10 - C\). The total score equation is therefore
   \[
   5C - 2(10 - C) = 29.
   \]

3. Solve the equation algebraically:
   \[
   5C - 20 + 2C = 29 \implies 7C - 20 = 29 \implies 7C = 49 \implies C = 7.
   \]
   Thus Olivia had exactly 7 correct answers.

4. Verify the solution by substitution: 7 correct answers give \(5 \times 7 = 35\) points; the remaining 3 incorrect answers give \(-2 \times 3 = -6\) points; total \(35 - 6 = 29\), which matches the given score.

5. Rule out the other choices by testing each integer value:
   - 5 correct: \(5 \times 5 - 2 \times 5 = 25 - 10 = 15 \neq 29\)
   - 6 correct: \(5 \times 6 - 2 \times 4 = 30 - 8 = 22 \neq 29\)
   - 8 correct: \(5 \times 8 - 2 \times 2 = 40 - 4 = 36 \neq 29\)
   - 9 correct: \(5 \times 9 - 2 \times 1 = 45 - 2 = 43 \neq 29\)
   Only \(C = 7\) satisfies the score equation.

6. Conclusion: Because the linear Diophantine equation has a unique integer solution in the feasible range \(0 \leq C \leq 10\) and that solution is 7, the correct choice is C.

ANSWER: C

1. **Restate the problem**: We need to find out how many problems Olivia answered correctly given that she scored 29 points in total, with 5 points awarded for each correct answer and 2 points deducted for each incorrect answer, and she answered all 10 problems.

2. **Set up the variables**: Let's denote the number of correct answers as \(C\) and the number of incorrect answers as \(I\). We know that \(C + I = 10\) because Olivia answered every problem.

3. **Formulate the scoring equation**: For scoring, we have \(5C - 2I = 29\), since she gains 5 points for each correct answer and loses 2 points for each incorrect answer.

4. **Express \(I\) in terms of \(C\)**: From \(C + I = 10\), we get \(I = 10 - C\).

5. **Substitute \(I\) in the scoring equation**: Substituting \(I = 10 - C\) into \(5C - 2I = 29\), we get \(5C - 2(10 - C) = 29\).

6. **Solve for \(C\)**: Expanding the equation gives \(5C - 20 + 2C = 29\), which simplifies to \(7C - 20 = 29\). Adding 20 to both sides gives \(7C = 49\). Dividing both sides by 7 gives \(C = 7\).

7. **Check the answer choices**: We found \(C = 7\), so let's verify if this aligns with the given choices and makes sense in the context. For \(C = 7\), \(I = 3\), and the score would be \(5(7) - 2(3) = 35 - 6 = 29\), which matches Olivia's score.

8. **Rule out the wrong choices**:
- For A (5 correct answers), the score would be \(5(5) - 2(5) = 25 - 10 = 15\), which is too low.
- For B (6 correct answers), the score would be \(5(6) - 2(4) = 30 - 8 = 22\), still too low.
- For C (7 correct answers), we've already verified it gives the correct score.
- For D (8 correct answers), the score would be \(5(8) - 2(2) = 40 - 4 = 36\), which is too high.
- For E (9 correct answers), the score would be \(5(9) - 2(1) = 45 - 2 = 43\), also too high.

9. **Conclusion**: The only choice that results in a score of 29 is C (7 correct answers).

ANSWER: C

Olivia answered all 10 problems. Let \( x \) be the number of correct answers and \( y \) be the number of incorrect answers.

1. The total number of problems is 10:
   \[
   x + y = 10
   \]

2. The scoring rule gives 5 points for a correct answer and deducts 2 points for an incorrect answer. Her score is 29:
   \[
   5x - 2y = 29
   \]

3. From the first equation, \( y = 10 - x \). Substitute into the second:
   \[
   5x - 2(10 - x) = 29
   \]
   \[
   5x - 20 + 2x = 29
   \]
   \[
   7x - 20 = 29
   \]

4. Add 20 to both sides:
   \[
   7x = 49
   \]

5. Divide by 7:
   \[
   x = 7
   \]

Olivia had 7 correct answers. Checking the choices, this matches C.

**ANSWER: C**

1. **Understand the problem**:  
   - There are 10 problems.
   - Each **correct** answer gives **+5** points.
   - Each **incorrect** answer gives **–2** points.
   - Olivia answered **all 10** problems (so no blanks).
   - Her **total score is 29**.
   - We are to find **how many correct answers** she had.

2. **Define variables**:  
   Let $ x $ = number of correct answers.  
   Since she answered all 10 problems, the number of incorrect answers is $ 10 - x $.

3. **Write an equation for total score**:  
   Score from correct answers: $ 5x $  
   Score from incorrect answers: $ -2(10 - x) $  
   Total score:  
   $$
   5x - 2(10 - x) = 29
   $$

4. **Simplify the equation**:  
   Distribute the –2:  
   $$
   5x - 20 + 2x = 29
   $$  
   Combine like terms:  
   $$
   7x - 20 = 29
   $$

5. **Solve for $ x $**:  
   Add 20 to both sides:  
   $$
   7x = 49
   $$  
   Divide by 7:  
   $$
   x = 7
   $$

6. **Check the answer**:  
   - Correct answers: 7 → $ 7 \times 5 = 35 $ points  
   - Incorrect answers: $ 10 - 7 = 3 $ → $ 3 \times (-2) = -6 $ points  
   - Total: $ 35 - 6 = 29 $ → matches the given score.

7. **Match with choices**:  
   Choice C is 7.

8. **Eliminate other choices (optional verification)**:  
   - A. 5 correct → score = $ 5(5) - 2(5) = 25 - 10 = 15 $ → too low  
   - B. 6 correct → $ 30 - 8 = 22 $ → too low  
   - D. 8 correct → $ 40 - 4 = 36 $ → too high  
   - E. 9 correct → $ 45 - 2 = 43 $ → too high  
   Only C gives 29.

ANSWER: C

 **Step 1: Restate the problem.**  
There are 10 problems. Each correct answer gives +5 points, each incorrect answer gives –2 points. Olivia answered all 10 problems and scored 29 points. We need to find how many answers she got correct.

**Step 2: Define a variable.**  
Let \(c\) be the number of correct answers.  
Since she answered every problem, the number of incorrect answers is \(10-c\).

**Step 3: Set up the equation for her score.**  
Score from correct answers: \(5c\).  
Penalty from incorrect answers: \(-2(10-c)\).  
Total score is 29, so:

\[
5c - 2(10-c) = 29
\]

**Step 4: Solve the equation.**  
Distribute the –2:

\[
5c - 20 + 2c = 29
\]

Combine like terms:

\[
7c - 20 = 29
\]

Add 20 to both sides:

\[
7c = 49
\]

Divide by 7:

\[
c = 7
\]

**Step 5: Verify the answer.**  
If Olivia got 7 correct and 3 incorrect:  
Points earned: \(7 \times 5 = 35\).  
Points lost: \(3 \times 2 = 6\).  
Net score: \(35 - 6 = 29\). This matches the given score.

**Step 6: Rule out the wrong choices.**  
We can check each option by computing the score it would produce:

- **A. 5 correct:** \(5(5) - 5(2) = 25 - 10 = 15\) (too low).  
- **B. 6 correct:** \(6(5) - 4(2) = 30 - 8 = 22\) (too low).  
- **C. 7 correct:** \(7(5) - 3(2) = 35 - 6 = 29\) (matches).  
- **D. 8 correct:** \(8(5) - 2(2) = 40 - 4 = 36\) (too high).  
- **E. 9 correct:** \(9(5) - 1(2) = 45 - 2 = 43\) (too high).

Only 7 correct answers yields a score of 29.

ANSWER: C

Here is the step-by-step solution:

1.  **Restate the Problem:** We need to find the number of problems Olivia answered correctly. We know there were 10 problems total, she answered every one of them, she earns 5 points for a correct answer, loses 2 points for an incorrect answer, and her final score was 29 points.

2.  **Set Up the Math:** Let $c$ represent the number of correct answers and $w$ represent the number of wrong (incorrect) answers.
    *   Since there are 10 problems and she answered every one: **$c + w = 10$**
    *   The scoring system gives 5 points for correct and subtracts 2 for wrong, totaling 29: **$5c - 2w = 29$**

3.  **Do the Computation:**
    *   From the first equation, we can express $w$ in terms of $c$: $w = 10 - c$.
    *   Substitute this expression into the second equation:
        $$5c - 2(10 - c) = 29$$
    *   Distribute the $-2$:
        $$5c - 20 + 2c = 29$$
    *   Combine like terms ($5c + 2c$):
        $$7c - 20 = 29$$
    *   Add 20 to both sides:
        $$7c = 49$$
    *   Divide by 7:
        $$c = 7$$
    *   So, Olivia had **7** correct answers. (This implies she had 3 wrong answers: $5(7) - 2(3) = 35 - 6 = 29$, which matches the given score).

4.  **Rule Out Wrong Choices:**
    *   **A. 5:** If she got 5 right and 5 wrong, her score would be $(5 \times 5) - (2 \times 5) = 25 - 10 = 15$. This is not 29.
    *   **B. 6:** If she got 6 right and 4 wrong, her score would be $(5 \times 6) - (2 \times 4) = 30 - 8 = 22$. This is not 29.
    *   **D. 8:** If she got 8 right and 2 wrong, her score would be $(5 \times 8) - (2 \times 2) = 40 - 4 = 36$. This is not 29.
    *   **E. 9:** If she got 9 right and 1 wrong, her score would be $(5 \times 9) - (2 \times 1) = 45 - 2 = 43$. This is not 29.

ANSWER: C

**Step 1: Define variables**  
Let  
\(c\) = number of correct answers,  
\(i\) = number of incorrect answers.  

**Step 2: Translate the conditions into equations**  
- The contest has 10 problems, and Olivia answered every one:  

\[
c+i = 10 \qquad\text{(1)}
\]

- Scoring rule: \(+5\) points for a correct answer, \(-2\) points for an incorrect answer.  
  Her total score is 29:  

\[
5c - 2i = 29 \qquad\text{(2)}
\]

**Step 3: Solve the system**  
From (1), \(i = 10 - c\). Substitute into (2):

\[
5c - 2(10 - c) = 29
\]

\[
5c - 20 + 2c = 29
\]

\[
7c - 20 = 29
\]

\[
7c = 49
\]

\[
c = 7
\]

So Olivia answered **7** problems correctly.

**Step 4: Verify**  
If \(c=7\), then \(i=10-7=3\).  
Score = \(5\cdot7 - 2\cdot3 = 35 - 6 = 29\), which matches the given score.

**Step 5: Eliminate the other choices**  
- \(c=5\) gives \(5\cdot5 - 2\cdot5 = 15\) (too low).  
- \(c=6\) gives \(5\cdot6 - 2\cdot4 = 22\) (too low).  
- \(c=8\) gives \(5\cdot8 - 2\cdot2 = 36\) (too high).  
- \(c=9\) gives \(5\cdot9 - 2\cdot1 = 43\) (too high).  

Only \(c=7\) yields a score of 29.

ANSWER: C

1. **Restate what is being asked**:
   - We know that in a 10 - problem mathematics contest, for each correct answer, a student gets 5 points, and for each incorrect answer, 2 points are deducted. Olivia answered all 10 problems and her total score is 29 points. We need to find out how many problems she answered correctly.
2. **Set up the math**:
   - Let the number of correct answers be $x$. Since there are 10 problems in total, the number of incorrect answers is $10 - x$.
   - The total score from correct answers is $5x$ points (because each correct answer gives 5 points).
   - The total score deducted for incorrect answers is $2(10 - x)$ points (because each incorrect answer deducts 2 points).
   - The final score is the score from correct answers minus the score deducted for incorrect answers. So, the equation representing the situation is $5x-2(10 - x)=29$.
3. **Do the computation**:
   - First, expand the equation:
     - $5x-2(10 - x)=29$ becomes $5x - 20+2x = 29$ (using the distributive property $a(b - c)=ab - ac$, here $a = 2$, $b = 10$, $c=x$).
     - Combine like - terms: $(5x + 2x)-20 = 29$, which simplifies to $7x-20 = 29$.
     - Add 20 to both sides of the equation: $7x-20 + 20=29 + 20$, so $7x=49$.
     - Divide both sides by 7: $x=\frac{49}{7}=7$.
4. **Rule out the wrong choices**:
   - **Choice A**: If $x = 5$, then the number of incorrect answers is $10 - 5 = 5$. The total score would be $5\times5-2\times5=25 - 10 = 15\neq29$.
   - **Choice B**: If $x = 6$, then the number of incorrect answers is $10 - 6 = 4$. The total score would be $5\times6-2\times4=30 - 8 = 22\neq29$.
   - **Choice D**: If $x = 8$, then the number of incorrect answers is $10 - 8 = 2$. The total score would be $5\times8-2\times2=40 - 4 = 36\neq29$.
   - **Choice E**: If $x = 9$, then the number of incorrect answers is $10 - 9 = 1$. The total score would be $5\times9-2\times1=45 - 2 = 43\neq29$.

ANSWER: C

### Step 1: Restate the problem
We have a 10-problem contest where every answer is either correct or incorrect (Olivia answered all problems, so no unattempted questions). A correct answer earns +5 points, an incorrect answer subtracts 2 points. Olivia's total score is 29, and we need to find how many correct answers she had.

### Step 2: Set up mathematical equations
Let \( C \) = number of correct answers. Since there are 10 total problems, the number of incorrect answers is \( 10 - C \).
Total score = (points from correct answers) - (points lost from incorrect answers) = 29.
This translates to:
\[ 5C - 2(10 - C) = 29 \]

### Step 3: Solve for \( C \)
First expand and simplify the left side of the equation:
\[ 5C - 20 + 2C = 29 \]
Combine like terms:
\[ 7C - 20 = 29 \]
Add 20 to both sides:
\[ 7C = 49 \]
Divide by 7:
\[ C = 7 \]

### Step 4: Verify and eliminate wrong choices
- Confirm 7 correct answers works: \( 7*5 = 35 \) points from correct answers, \( 3*2 = 6 \) points lost from 3 incorrect answers. \( 35 - 6 = 29 \), which matches Olivia's score.
- Rule out other options:
  - A (5): \( 5*5 - 5*2 = 15 \neq 29 \)
  - B (6): \( 6*5 - 4*2 = 22 \neq 29 \)
  - D (8): \( 8*5 - 2*2 = 36 \neq 29 \)
  - E (9): \( 9*5 - 1*2 = 43 \neq 29 \)

ANSWER: C

### Step 1: Restate the problem and define variables
We have a 10-problem contest where:
- Correct answers give +5 points, incorrect answers give -2 points
- Olivia answered all 10 problems, total score = 29
Let \( c \) = number of correct answers. Then the number of incorrect answers is \( 10 - c \) (since all problems are answered).
---
### Step 2: Set up the score equation
Total score = (points from correct answers) + (points from incorrect answers):
\[ 5c - 2(10 - c) = 29 \]
---
### Step 3: Solve the equation
First expand the left side:
\[ 5c - 20 + 2c = 29 \]
Combine like terms:
\[ 7c - 20 = 29 \]
Add 20 to both sides:
\[ 7c = 49 \]
Divide by 7:
\[ c = 7 \]
---
### Step 4: Verify and rule out wrong choices
Test the value \( c=7 \):
- Correct points: \( 7*5 = 35 \)
- Incorrect points: \( (10-7)*(-2) = 3*(-2) = -6 \)
- Total: \( 35 - 6 = 29 \), which matches the given score.
Check other options to confirm they are wrong:
- A (5 correct): \( 5*5 - 5*2 = 25 - 10 = 15 \neq 29 \)
- B (6 correct): \( 6*5 - 4*2 = 30 - 8 = 22 \neq 29 \)
- D (8 correct): \( 8*5 - 2*2 = 40 - 4 = 36 \neq 29 \)
- E (9 correct): \( 9*5 - 1*2 = 45 - 2 = 43 \neq 29 \)
---
ANSWER: C

To solve this problem, let's denote the number of correct answers by \( c \) and the number of incorrect answers by \( i \). Since Olivia answered all 10 problems, we have the equation:

\[ c + i = 10 \]

We are also given that Olivia's total score is 29 points. Since she gains 5 points for each correct answer and loses 2 points for each incorrect answer, her total score can be expressed as:

\[ 5c - 2i = 29 \]

Now we have a system of two equations:

1. \( c + i = 10 \)
2. \( 5c - 2i = 29 \)

We can solve this system step by step.

**Step 1: Solve the first equation for \( i \):**

\[ i = 10 - c \]

**Step 2: Substitute \( i \) in the second equation:**

\[ 5c - 2(10 - c) = 29 \]

**Step 3: Simplify the equation:**

\[ 5c - 20 + 2c = 29 \]
\[ 7c - 20 = 29 \]

**Step 4: Solve for \( c \):**

\[ 7c = 29 + 20 \]
\[ 7c = 49 \]
\[ c = 7 \]

**Step 5: Find \( i \) using the value of \( c \):**

\[ i = 10 - c \]
\[ i = 10 - 7 \]
\[ i = 3 \]

So, Olivia had 7 correct answers and 3 incorrect answers. To verify, we can calculate her score:

\[ 5 \times 7 - 2 \times 3 = 35 - 6 = 29 \]

This confirms that our solution is correct. Now, let's rule out the incorrect choices:

- Choice A: 5 correct answers would give a score of \( 5 \times 5 - 2 \times 5 = 25 - 10 = 15 \), which is not 29.
- Choice B: 6 correct answers would give a score of \( 5 \times 6 - 2 \times 4 = 30 - 8 = 22 \), which is not 29.
- Choice D: 8 correct answers would give a score of \( 5 \times 8 - 2 \times 2 = 40 - 4 = 36 \), which is not 29.
- Choice E: 9 correct answers would give a score of \( 5 \times 9 - 2 \times 1 = 45 - 2 = 43 \), which is not 29.

Thus, the correct choice is:

ANSWER: C

**Problem Restatement:**
Olivia answered all 10 problems in a contest. For each correct answer, she gains 5 points, and for each incorrect answer, she loses 2 points. Her total score was 29. We need to find how many problems she answered correctly.

---

**Step 1: Define Variables**
Let \( x \) be the number of correct answers Olivia had.
Since there are 10 problems total, the number of incorrect answers is \( 10 - x \).

---

**Step 2: Set Up the Equation**
For correct answers, Olivia gains \( 5x \) points.
For incorrect answers, she loses \( 2(10 - x) \) points.
Her total score is the sum of these:
\[ 5x - 2(10 - x) = 29 \]

---

**Step 3: Simplify the Equation**
\[ 5x - 20 + 2x = 29 \]
Combine like terms:
\[ 7x - 20 = 29 \]

---

**Step 4: Solve for \( x \)**
Add 20 to both sides:
\[ 7x = 49 \]
Divide both sides by 7:
\[ x = 7 \]

---

**Step 5: Check the Answer Choices**
The correct number of answers is 7, which corresponds to choice **C**.

---

**Step 6: Verify the Answer**
If Olivia answered 7 correctly and 3 incorrectly:
\[ (7 \times 5) - (3 \times 2) = 35 - 6 = 29 \]
This matches the given total score, confirming our answer.

---

ANSWER: C

I need to find what percent of the total original price was saved during the sale.

**Step 1: Restate what's being asked**
- Original jacket price: $80
- Original shirt price: $40
- Jacket discount: 40%
- Shirt discount: 55%
- Find: Total savings as a percent of total original price

**Step 2: Calculate the discount amount for the jacket**
Jacket discount = 40% of $80
= 0.40 × $80
= $32

**Step 3: Calculate the discount amount for the shirt**
Shirt discount = 55% of $40
= 0.55 × $40
= $22

**Step 4: Calculate total savings**
Total savings = $32 + $22 = $54

**Step 5: Calculate total original price**
Total original price = $80 + $40 = $120

**Step 6: Calculate savings as a percent of total original price**
Percent saved = (Total savings / Total original price) × 100%
= ($54 / $120) × 100%
= (54/120) × 100%

Simplify 54/120:
54/120 = 27/60 = 9/20

So: (9/20) × 100% = (9 × 100)/20 % = 900/20 % = 45%

**Step 7: Verify by checking each answer**
- A. 45%: This matches our calculation ✓
- B. 47.5%: Would be $57 savings (too high)
- C. 50%: Would be $60 savings (too high)
- D. 79⅙%: Would be about $95 savings (way too high)
- E. 95%: Would be $114 savings (impossible, more than total)

ANSWER: A

1. Restate the problem: Chris bought a jacket originally priced at \$80 with a 40% discount, and a shirt originally priced at \$40 with a 55% discount. We must find the total amount saved as a percent of the sum of the original prices (\$80 + \$40 = \$120).

2. Compute the savings on the jacket:
   Savings_jacket = 40% of \$80 = 0.40 × 80 = \$32.

3. Compute the savings on the shirt:
   Savings_shirt = 55% of \$40 = 0.55 × 40 = \$22.

4. Compute the total savings and total original cost:
   Total_savings = 32 + 22 = \$54.
   Total_original = 80 + 40 = \$120.

5. Compute the savings as a percent of the total original:
   Percent_saved = (Total_savings / Total_original) × 100% = (54 / 120) × 100% = 45%.

6. Compare to the answer choices:
   A. 45% ← matches our calculation.
   B. 47½% = 47.5% ≠ 45%.
   C. 50% ≠ 45%.
   D. 79⅙% ≠ 45%.
   E. 95% ≠ 45%.

All other choices are ruled out.

ANSWER: A

To find the total percentage saved, we will break down the problem into clear, numbered steps.

**Step 1: Restate what is being asked**
We need to find the total amount saved on both the jacket and the shirt, and then determine what percentage this total savings is of the combined original prices of both items.

**Step 2: Calculate the savings on each item**
*   The jacket originally cost $\$80$ and was discounted by $40\%$:
    $$\text{Savings on jacket} = 40\% \text{ of } \$80 = 0.40 \times 80 = \$32$$
*   The shirt originally cost $\$40$ and was discounted by $55\%$:
    $$\text{Savings on shirt} = 55\% \text{ of } \$40 = 0.55 \times 40 = \$22$$

**Step 3: Calculate the total savings**
*   Add the savings from both items together:
    $$\text{Total Savings} = \$32 + \$22 = \$54$$

**Step 4: Calculate the total of the original prices**
*   Add the original prices of both items together:
    $$\text{Total Original Price} = \$80 + \$40 = \$120$$

**Step 5: Calculate the percentage saved**
*   Divide the total savings by the total original price and multiply by $100$ to get the percentage:
    $$\text{Percentage Saved} = \left( \frac{54}{120} \right) \times 100\%$$
*   Simplify the fraction $\frac{54}{120}$ by dividing the numerator and the denominator by their greatest common divisor, which is $6$:
    $$\frac{54 \div 6}{120 \div 6} = \frac{9}{20}$$
*   Convert $\frac{9}{20}$ to a percentage:
    $$\frac{9}{20} \times 100\% = 9 \times 5\% = 45\%$$

**Step 6: Explain how we rule the wrong choices out**
*   **A. 45%** is the correct calculation.
*   **B. $47 \frac{1}{2} \%$** is incorrect; this would correspond to a savings of $\$57$.
*   **C. 50%** is incorrect; this would be a savings of exactly half the total price ($\$60$).
*   **D. $79 \frac{1}{6} \%$** is incorrect and too high.
*   **E. 95%** is a trap choice obtained by simply adding the two percentage discounts together ($40\% + 55\%$), which is mathematically incorrect because the discounts are applied to different base amounts.

ANSWER: A

The problem requires finding the total amount saved on the two items, expressed as a percentage of their combined original prices.

The jacket originally costs 80 dollars and the shirt originally costs 40 dollars, so the combined original price is  
80 + 40 = 120 dollars.

The discount on the jacket is 40% of 80:  
0.4 × 80 = 32 dollars saved.

The discount on the shirt is 55% of 40:  
0.55 × 40 = 22 dollars saved.

The total amount saved is therefore  
32 + 22 = 54 dollars.

The required percentage is  
(54 / 120) × 100% = 45%.

This matches choice A.  
Choice B arises from the common error of simply averaging the two discount rates (40% + 55%)/2 = 47.5% without weighting by the original prices; the calculation above shows that the correct weighted result is 45%.  
The remaining choices do not arise from any consistent arithmetic interpretation of the given discounts and prices.

ANSWER: A

1. **Restate the problem**: We need to find the percentage of the total amount saved on a jacket and a shirt during a sale relative to their original total price. The original prices were $80 for the jacket and $40 for the shirt. The jacket was bought at a 40% discount, and the shirt was bought at a 55% discount.

2. **Calculate the original total price**: The original total price = price of jacket + price of shirt = $80 + $40 = $120.

3. **Calculate the discount on the jacket**: The discount on the jacket = 40% of $80 = 0.4 * $80 = $32.

4. **Calculate the discount on the shirt**: The discount on the shirt = 55% of $40 = 0.55 * $40 = $22.

5. **Calculate the total amount saved**: The total amount saved = discount on jacket + discount on shirt = $32 + $22 = $54.

6. **Calculate the percentage of the total amount saved relative to the original total price**: The percentage saved = (total amount saved / original total price) * 100 = ($54 / $120) * 100.

7. **Perform the division and multiplication**: ($54 / $120) = 0.45, and 0.45 * 100 = 45%.

8. **Compare the calculated percentage to the given choices**:
   - A. 45%: Matches our calculation.
   - B. 47 1/2 %: Equals 47.5%, which is more than our result.
   - C. 50%: More than our result.
   - D. 79 1/6 %: Equals approximately 79.17%, which is much more than our result.
   - E. 95%: Much more than our result.

9. **Rule out the incorrect choices**: B, C, D, and E are all greater than 45%, so they are incorrect.

10. **State the final answer**: Since our calculation matches choice A, it is the correct answer.

ANSWER: A

1. **Understand the problem:** We need to find the total amount saved as a percentage of the total original prices. The original prices are $80 for the jacket and $40 for the shirt, so total original price = 80 + 40 = $120.

2. **Calculate the discount on the jacket:** 40% discount on $80.  
   Discount = 0.40 × 80 = $32.

3. **Calculate the discount on the shirt:** 55% discount on $40.  
   Discount = 0.55 × 40 = $22.

4. **Find the total amount saved:**  
   Total saved = 32 + 22 = $54.

5. **Express the total saved as a percentage of the total original prices:**  
   Percentage = (54 / 120) × 100% = 0.45 × 100% = 45%.

6. **Check against the answer choices:** 45% corresponds to option A.

ANSWER: A

1. **Understand what is being asked**:  
   We are given original prices of a jacket ($80) and a shirt ($40). Chris buys them at discounts of 40% and 55%, respectively. We are to find what **percent of the total original price** was **saved**.

2. **Find the total original price**:  
   $$
   \text{Total original price} = 80 + 40 = 120 \text{ dollars}
   $$

3. **Calculate the amount saved on each item**:  
   - Jacket: 40% of $80  
     $$
     0.40 \times 80 = 32 \text{ dollars saved}
     $$
   - Shirt: 55% of $40  
     $$
     0.55 \times 40 = 22 \text{ dollars saved}
     $$

4. **Find total amount saved**:  
   $$
   32 + 22 = 54 \text{ dollars}
   $$

5. **Find what percent 54 is of 120**:  
   $$
   \frac{54}{120} \times 100\% = \left(\frac{54 \div 6}{120 \div 6}\right) \times 100\% = \frac{9}{20} \times 100\% = 45\%
   $$

6. **Check answer choices**:  
   - A. 45% ✅  
   - B. 47½%  
   - C. 50%  
   - D. 79⅙%  
   - E. 95%  

   Only **A** matches our result.

7. **Double-check calculations**:  
   - 40% of 80 = 32 ✔  
   - 55% of 40 = 22 ✔  
   - Total saved = 54 ✔  
   - 54 / 120 = 0.45 = 45% ✔  

Everything checks out.

ANSWER: A

 **Step 1: Restate the question.**  
We need to find the total dollars Chris saved on the two items, then express that savings as a percentage of the combined original prices.

**Step 2: Compute the total original price.**  
Jacket original price: $\$80$  
Shirt original price: $\$40$  
Total original price $= 80 + 40 = \$120$.

**Step 3: Compute the dollar amount saved on each item.**  
Jacket savings: $40\%$ of $\$80 = 0.40 \times 80 = \$32$.  
Shirt savings: $55\%$ of $\$40 = 0.55 \times 40 = \$22$.

**Step 4: Compute the total savings.**  
Total saved $= 32 + 22 = \$54$.

**Step 5: Express the savings as a percent of the total original price.**  
\[
\frac{54}{120} = \frac{27}{60} = \frac{9}{20} = 0.45 = 45\%.
\]

**Step 6: Rule out the wrong choices.**  
- **Choice B ($47\frac12\%$):** This is the simple average of $40\%$ and $55\%$. That would only be correct if the two items cost the same amount, but the jacket costs twice as much as the shirt, so the true overall rate must be weighted toward $40\%$.  
- **Choice C ($50\%$):** This is just a round-number distractor with no correct basis in the problem.  
- **Choice D ($79\frac16\%$):** This equals $\frac{95}{120}$, which arises from incorrectly adding the two discount percentages ($40+55=95$) and then dividing by the total original price. Percentages of different amounts cannot be added directly.  
- **Choice E ($95\%$):** This is the sum of the two discount rates, ignoring that they apply to different original prices.  

The only choice that matches the correct calculation is **45%**.

ANSWER: A

Here is the step-by-step solution:

**1. Restate the Problem**
We need to determine the total amount of money Chris saved by purchasing a jacket and a shirt at discounted prices. Then, we must calculate what percentage this total savings represents relative to the combined original price of the two items.

**2. Set Up the Math**
Let $P_j$ be the original price of the jacket and $P_s$ be the original price of the shirt.
Let $d_j$ be the discount rate for the jacket and $d_s$ be the discount rate for the shirt.
We are looking for the value $X$ such that:
$$X = \frac{\text{Total Savings}}{\text{Total Original Price}} \times 100\%$$

Where:
*   $\text{Total Savings} = (P_j \times d_j) + (P_s \times d_s)$
*   $\text{Total Original Price} = P_j + P_s$

Given values:
*   $P_j = \$80$, $d_j = 40\% = 0.40$
*   $P_s = \$40$, $d_s = 55\% = 0.55$

**3. Do the Computation**

*   **Calculate the savings on the jacket:**
    $$\text{Savings}_j = \$80 \times 0.40 = \$32$$

*   **Calculate the savings on the shirt:**
    $$\text{Savings}_s = \$40 \times 0.55$$
    To compute this easily: $10\%$ of $40$ is $4$, so $50\%$ is $20$. $5\%$ is half of that, which is $2$.
    $$20 + 2 = \$22$$

*   **Calculate the total amount saved:**
    $$\text{Total Saved} = \$32 + \$22 = \$54$$

*   **Calculate the total of the original prices:**
    $$\text{Total Original} = \$80 + \$40 = \$120$$

*   **Calculate the final percentage:**
    We need to find what percent $\$54$ is of $\$120$.
    $$\frac{54}{120}$$
    Simplify the fraction by dividing numerator and denominator by their greatest common divisor, which is 6:
    $$54 \div 6 = 9$$
    $$120 \div 6 = 20$$
    So the fraction is $\frac{9}{20}$.
    
    Convert the fraction to a percentage:
    $$\frac{9}{20} = \frac{9 \times 5}{20 \times 5} = \frac{45}{100} = 45\%$$

**4. Rule Out Wrong Choices**
*   **A. 45%**: Matches our calculated result exactly.
*   **B. 47 1/2 %**: This would be the result if the weights of the items were different or if the discounts were slightly higher. Specifically, since the expensive item (jacket) has a lower discount (40%) than the cheaper item (shirt, 55%), the weighted average must be pulled closer to 40% than to 55%. The midpoint between 40 and 55 is 47.5. Since we are closer to the jacket's price weight, the answer must be less than 47.5%.
*   **C. 50%**: This is too high for the reasoning stated above (weighted average logic).
*   **D & E (79 1/6 % and 95%)**: These values are far too high, representing savings close to the total cost of the items, which is impossible given the discount rates of 40% and 55%.

ANSWER: A

**Step 1 – Find the original total price**  
\[
80\text{ (jacket)}+40\text{ (shirt)}=120\text{ dollars}
\]

**Step 2 – Compute the discounted price and amount saved on the jacket**  
A 40 % discount means Chris pays 60 % of the original price:  
\[
0.60\times80=48\text{ dollars}
\]  
Saved on the jacket:  
\[
80-48=32\text{ dollars}
\]

**Step 3 – Compute the discounted price and amount saved on the shirt**  
A 55 % discount means Chris pays 45 % of the original price:  
\[
0.45\times40=18\text{ dollars}
\]  
Saved on the shirt:  
\[
40-18=22\text{ dollars}
\]

**Step 4 – Find the total amount saved**  
\[
32+22=54\text{ dollars}
\]

**Step 5 – Express the total saved as a percent of the original total**  
\[
\frac{54}{120}\times100\% = 0.45\times100\% = 45\%
\]

Thus the amount saved is 45 % of the original combined price.

**Answer choice:** A  

ANSWER: A

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### Step 1: Restate the problem and calculate total original prices
First, the original prices are $80 for the jacket and $40 for the shirt. The total original combined price is:
$80 + $40 = $120.
We need to find total savings from the discounts, then calculate what percent that savings is of the $120 total original price.

### Step 2: Calculate savings on each item
Savings = discount percentage × original price for each item:
- Jacket savings: 40% of $80 = 0.4 × 80 = $32
- Shirt savings: 55% of $40 = 0.55 × 40 = $22

### Step 3: Calculate total savings and final percentage
Total savings = $32 + $22 = $54.
To find what percent $54 is of the $120 total original price:
$\frac{54}{120} = 0.45 = 45\%$.

### Step 4: Rule out incorrect choices
- **Option B (47.5%)**: This is the unweighted average of the two discounts (40+55)/2 = 47.5, which ignores that the jacket costs twice as much as the shirt, so the discounts must be weighted by their original prices.
- **Option C (50%)**: This would only be true if total savings were $60, which is incorrect.
- **Options D (79 1/6%) and E (95%)**: These are far too large; E is the sum of the two discount percentages (40+55=95), which is a meaningless number for this calculation.

ANSWER: A

### Step 1: Restate the problem and identify key values
We need to find what percent the total amount Chris saved is of the total original price of the jacket and shirt. 
- Original price of jacket: $80, discount: 40%
- Original price of shirt: $40, discount: 55%
---
### Step 2: Calculate the total original price
Add the original prices of the two items:
Total original price = $80 + $40 = $120. This is the "whole" for our percent calculation.
---
### Step 3: Calculate savings on the jacket
A 40% discount means Chris saves 40% of the jacket's original price:
Savings on jacket = 40% of $80 = 0.4 * 80 = $32.
---
### Step 4: Calculate savings on the shirt
A 55% discount means Chris saves 55% of the shirt's original price:
Savings on shirt = 55% of $40 = 0.55 * 40 = $22.
---
### Step 5: Calculate total savings
Add the savings from both items:
Total savings = $32 + $22 = $54.
---
### Step 6: Compute the required percentage
Use the percent formula: $\text{Percent} = \left(\frac{\text{Part}}{\text{Whole}}\right) * 100$, where the part is total savings and the whole is total original price:
$\text{Percent} = \left(\frac{54}{120}\right) * 100 = 0.45 * 100 = 45\%$.
---
### Step 7: Verify and rule out wrong choices
- Option B (47.5%) would be the result if we incorrectly averaged the discounts (40% + 55% / 2 = 47.5%), but discounts are weighted by original price, not averaged directly.
- Option C (50%) would be the result if we swapped the discounts (55% on the jacket, 40% on the shirt), which contradicts the problem statement.
- Options D (79 1/6%) and E (95%) are far too high, as the total savings cannot exceed the total original price, and 54 is much less than 120.
ANSWER: A

To determine the total amount saved as a percentage of the total original prices, we will follow these steps:

1. **Calculate the discount amount for the jacket:**
   - Original price of the jacket: $80
   - Discount on the jacket: 40%
   - Discount amount for the jacket: \( 80 \times \frac{40}{100} = 80 \times 0.40 = 32 \) dollars

2. **Calculate the discount amount for the shirt:**
   - Original price of the shirt: $40
   - Discount on the shirt: 55%
   - Discount amount for the shirt: \( 40 \times \frac{55}{100} = 40 \times 0.55 = 22 \) dollars

3. **Calculate the total amount saved:**
   - Total amount saved: \( 32 + 22 = 54 \) dollars

4. **Calculate the total original price:**
   - Total original price: \( 80 + 40 = 120 \) dollars

5. **Calculate the percentage of the total amount saved relative to the total original price:**
   - Percentage saved: \( \frac{54}{120} \times 100 \)

6. **Perform the division and multiplication:**
   - \( \frac{54}{120} = 0.45 \)
   - \( 0.45 \times 100 = 45\% \)

Thus, the total amount saved was 45% of the total of the original prices.

ANSWER: A

**Problem Restatement:**
We need to find what percent of the total original prices Chris saved by buying a jacket and a shirt at discounted prices.

**Given:**
- Original price of jacket = \$80
- Original price of shirt = \$40
- Discount on jacket = 40%
- Discount on shirt = 55%

---

**Step 1: Calculate the total original price.**
Total original price = Price of jacket + Price of shirt
= \$80 + \$40
= **\$120**

---

**Step 2: Calculate the amount saved on the jacket.**
Discount on jacket = 40% of \$80
= 0.40 × 80
= **\$32**

---

**Step 3: Calculate the amount saved on the shirt.**
Discount on shirt = 55% of \$40
= 0.55 × 40
= **\$22**

---

**Step 4: Calculate the total amount saved.**
Total savings = Savings on jacket + Savings on shirt
= \$32 + \$22
= **\$54**

---

**Step 5: Calculate the percentage of total savings relative to the total original price.**
Percentage saved = (Total savings / Total original price) × 100
= (54 / 120) × 100
= 0.45 × 100
= **45%**

---

**Step 6: Rule out incorrect choices.**
- B (47½%) is too high.
- C (50%) is too high.
- D (79⅙%) is way too high.
- E (95%) is way too high.

Only **A (45%)** matches our calculation.

---
ANSWER: A

I need to find four-digit numbers where:
- The leftmost (first) digit is odd
- The second digit is even
- All four digits are different

**Step 1: Identify the constraints**

For a four-digit number: _ _ _ _

- **Position 1 (leftmost):** Must be odd and non-zero (since it's the leftmost digit of a four-digit number)
- **Position 2:** Must be even
- **Position 3:** Can be any digit except those already used
- **Position 4:** Can be any digit except those already used

**Step 2: Count choices for position 1 (leftmost digit)**

Odd digits available: {1, 3, 5, 7, 9}

Number of choices: **5**

**Step 3: Count choices for position 2 (second digit)**

Even digits available: {0, 2, 4, 6, 8}

Since all four digits must be different, and position 1 already used one odd digit, position 2 can use any of the 5 even digits.

Number of choices: **5**

**Step 4: Count choices for position 3 (third digit)**

We've now used 2 digits (one odd, one even).

Remaining digits available: 10 - 2 = 8 digits

Number of choices: **8**

**Step 5: Count choices for position 4 (fourth digit)**

We've now used 3 digits.

Remaining digits available: 10 - 3 = 7 digits

Number of choices: **7**

**Step 6: Apply the multiplication principle**

Total four-digit numbers = 5 × 5 × 8 × 7

Let me calculate:
- 5 × 5 = 25
- 25 × 8 = 200
- 200 × 7 = 1400

**Step 7: Verify the logic**

- Position 1: 5 odd digits (1,3,5,7,9) ✓
- Position 2: 5 even digits (0,2,4,6,8), all different from position 1 since position 1 is odd ✓
- Position 3: 8 remaining digits (10 total - 2 used) ✓
- Position 4: 7 remaining digits (10 total - 3 used) ✓

The calculation is correct.

ANSWER: B

1. Restate the problem  
   We want the number of 4-digit whole numbers \(d_1d_2d_3d_4\) such that  
   – \(d_1\) (the thousands digit) is odd,  
   – \(d_2\) (the hundreds digit) is even,  
   – all four digits are distinct.

2. Count choices for \(d_1\)  
   The leftmost digit \(d_1\) must be odd and nonzero: \(\{1,3,5,7,9\}\).  
   Number of choices = 5.

3. Count choices for \(d_2\)  
   The second digit \(d_2\) must be even: \(\{0,2,4,6,8\}\).  
   Since \(d_1\) is odd, there is no overlap.  
   Number of choices = 5.

4. Count choices for \(d_3\)  
   The third digit \(d_3\) can be any digit 0–9 except the two already used (\(d_1,d_2\)).  
   Number of remaining digits = 10 – 2 = 8.

5. Count choices for \(d_4\)  
   The fourth digit \(d_4\) can be any digit 0–9 except the three already used (\(d_1,d_2,d_3\)).  
   Number of remaining digits = 10 – 3 = 7.

6. Total count by the multiplication principle  
   Total = (choices for \(d_1\)) × (choices for \(d_2\)) × (choices for \(d_3\)) × (choices for \(d_4\))  
   = \(5 \times 5 \times 8 \times 7 = 1400\).

7. Match to the given choices  
   A. 1120 — too small  
   B. 1400 — exactly our result  
   C. 1800 — too large  
   D. 2025 — too large  
   E. 2500 — far too large

8. Conclusion  
   The only choice matching \(5\cdot5\cdot8\cdot7=1400\) is choice B.

ANSWER: B

An elegant step-by-step solution to the problem is presented below:

### Step 1: Restate the Problem
We are asked to find the number of four-digit whole numbers (let's represent them as $d_1d_2d_3d_4$) such that:
- The first digit $d_1$ is odd.
- The second digit $d_2$ is even.
- All four digits $d_1, d_2, d_3, d_4$ are distinct.

### Step 2: Determine the Choices for the First Digit ($d_1$)
The first digit $d_1$ must be an odd digit. The possible odd digits are $\{1, 3, 5, 7, 9\}$. 
- There are **$5$** possible choices for $d_1$.
- Note that since $d_1$ must be odd, it cannot be $0$, which satisfies the condition that the first digit of a four-digit number cannot be $0$.

### Step 3: Determine the Choices for the Second Digit ($d_2$)
The second digit $d_2$ must be an even digit. The possible even digits are $\{0, 2, 4, 6, 8\}$.
- There are **$5$** possible choices for $d_2$.
- Since $d_1$ is odd and $d_2$ is even, they are guaranteed to be different.

### Step 4: Determine the Choices for the Third Digit ($d_3$)
The third digit $d_3$ can be any of the ten digits $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$, but it must be different from the first two digits ($d_1$ and $d_2$).
- Since $d_1$ and $d_2$ are distinct, there are exactly $2$ digits that $d_3$ cannot be.
- Thus, there are $10 - 2 =$ **$8$** possible choices for $d_3$.

### Step 5: Determine the Choices for the Fourth Digit ($d_4$)
The fourth digit $d_4$ can be any of the ten digits, but it must be different from the first three digits ($d_1$, $d_2$, and $d_3$).
- Since $d_1, d_2,$ and $d_3$ are all distinct, there are exactly $3$ digits that $d_4$ cannot be.
- Thus, there are $10 - 3 =$ **$7$** possible choices for $d_4$.

### Step 6: Compute the Total Number of Four-Digit Numbers
By the fundamental counting principle, we multiply the number of choices for each digit:
$$\text{Total} = 5 \times 5 \times 8 \times 7$$

Let's do the multiplication:
$$5 \times 5 = 25$$
$$25 \times 8 = 200$$
$$200 \times 7 = 1400$$

### Step 7: Rule Out the Wrong Choices
- **A. 1120** is too small (it would be the result if we had fewer choices for the last digits).
- **B. 1400** is our calculated value.
- **C. 1800**, **D. 2025**, and **E. 2500** are too large (they do not properly account for the restriction that all digits must be different).

ANSWER: B

The four-digit numbers under consideration range from 1000 to 9999 and can be denoted as \(ABCD\), where the digits satisfy three conditions: \(A\) is odd, \(B\) is even, and \(A\), \(B\), \(C\), and \(D\) are all distinct.

There are exactly five odd digits (1, 3, 5, 7, 9). Because none of these is zero, every choice is admissible for the leading digit \(A\). Thus there are 5 choices for \(A\).

There are exactly five even digits (0, 2, 4, 6, 8). Since \(A\) is odd, it cannot equal any even digit; consequently every one of the five even digits remains available for \(B\). Thus there are 5 choices for \(B\).

After \(A\) and \(B\) have been chosen, eight digits remain that are different from both. Any of these eight may be used for \(C\), so there are 8 choices for \(C\).

After \(A\), \(B\), and \(C\) have been chosen, seven digits remain that are different from all three. Any of these seven may be used for \(D\), so there are 7 choices for \(D\).

The total count is therefore
\[
5 \times 5 \times 8 \times 7 = 1400.
\]

This enumeration already enforces that the four digits are distinct and that the required parity conditions hold; no further adjustment is needed. Among the given options, 1400 corresponds to choice B; the remaining four numbers arise only from common counting errors (for example, incorrectly restricting the first digit or double-counting the parity constraints).

ANSWER: B

1. **Restate the problem**: We need to find the number of four-digit whole numbers where the leftmost digit is odd, the second digit is even, and all four digits are different.

2. **Set up the conditions for each digit**:
   - The leftmost digit must be odd and not zero, so it can be 1, 3, 5, 7, or 9. That gives us 5 choices.
   - The second digit must be even, so it can be 0, 2, 4, 6, or 8. That gives us 5 choices.
   - The third and fourth digits can be any digit from 0 to 9, except they must be different from the first two digits and from each other.

3. **Calculate the number of possibilities for the first two digits**:
   - For the first digit, we have 5 choices (1, 3, 5, 7, 9).
   - For the second digit, we also have 5 choices (0, 2, 4, 6, 8).

4. **Calculate the number of possibilities for the third and fourth digits**:
   - After choosing the first two digits, there are 8 possibilities left for the third digit because we have to exclude the two digits already chosen. 
   - For the fourth digit, there are 7 possibilities left because we have to exclude the three digits already chosen.

5. **Compute the total number of four-digit numbers**:
   - The total number = (number of choices for the first digit) * (number of choices for the second digit) * (number of choices for the third digit) * (number of choices for the fourth digit)
   - Total = 5 * 5 * 8 * 7 = 1400.

6. **Rule out the wrong choices**:
   - A. 1120 is less than our calculation, so it's incorrect.
   - C. 1800 is more than our calculation, so it's incorrect.
   - D. 2025 is more than our calculation, so it's incorrect.
   - E. 2500 is more than our calculation, so it's incorrect.
   - B. 1400 matches our calculation.

ANSWER: B

1. **Identify the digits and positions**: The number is a four-digit whole number, so the thousands digit cannot be 0. The conditions are: thousands digit is odd (1, 3, 5, 7, 9), hundreds digit is even (0, 2, 4, 6, 8), and all four digits are distinct.

2. **Count choices for the first two digits**: There are 5 choices for the odd thousands digit and 5 choices for the even hundreds digit. Since odds and evens are disjoint sets, any odd and even pair are automatically different. Total ways = 5 × 5 = 25.

3. **Count choices for the tens digit**: After fixing the first two digits, 8 digits remain from 0–9 (since two are already used). Any of these 8 can be the tens digit.

4. **Count choices for the units digit**: After fixing the first three digits, 7 digits remain. Any of these 7 can be the units digit.

5. **Total count**: Multiply the choices: 25 × 8 × 7 = 1400.

6. **Check against options**: The result 1400 matches option B.

ANSWER: B

1. **Understand the problem**:  
We are asked to count how many **four-digit whole numbers** satisfy **three conditions**:  
- The **leftmost digit (thousands place)** is **odd**.  
- The **second digit (hundreds place)** is **even**.  
- **All four digits are different**.

We must count such numbers and choose the correct answer from the given options.

2. **Clarify digit constraints**:  
- A four-digit number ranges from **1000 to 9999**, so the **first digit cannot be 0**.
- **Odd digits**: {1, 3, 5, 7, 9} → 5 choices.
- **Even digits**: {0, 2, 4, 6, 8} → 5 choices.

But we must ensure **all four digits are different**, so choices for later digits depend on earlier ones.

3. **Break the problem into steps (digit by digit)**:

We will choose digits in this order:  
- **Digit 1 (thousands place)**: must be **odd** and **non-zero** → from {1,3,5,7,9} → **5 choices**.
- **Digit 2 (hundreds place)**: must be **even** → from {0,2,4,6,8}, but **cannot equal digit 1**.  
  However, digit 1 is **odd**, and digit 2 is **even**, so they **can never be the same**.  
  → So **all 5 even digits are allowed** → **5 choices**.
- **Digit 3 (tens place)**: can be **any digit 0–9**, except **not equal to digit 1 or digit 2**.  
  So we’ve used 2 distinct digits already → **10 − 2 = 8 choices**.
- **Digit 4 (units place)**: must be different from the first three digits → **10 − 3 = 7 choices**.

So total number of such numbers =  
**5 (choices for digit 1) × 5 (digit 2) × 8 (digit 3) × 7 (digit 4)**.

4. **Compute the total**:  
5 × 5 = 25  
25 × 8 = 200  
200 × 7 = **1400**

5. **Check for hidden issues**:  
- Is there any case where digit 2 = 0? Yes, and that’s allowed — the second digit can be 0.
- Is there any overlap between odd and even digits? No — disjoint sets, so digit 1 ≠ digit 2 automatically.
- Are we overcounting? No — we enforced distinctness at each step.
- Are all numbers counted actually four-digit? Yes — first digit is from {1,3,5,7,9}, so never 0.

So everything checks out.

6. **Match with answer choices**:  
We got **1400**, which is **Choice B**.

ANSWER: B

**Step 1: Restate the problem.**

We must count how many four-digit numbers \(ABCD\) satisfy three conditions:
- \(A\) (the thousands digit) is odd,
- \(B\) (the hundreds digit) is even,
- and all four digits \(A, B, C, D\) are pairwise different.

**Step 2: Set up the counting.**

We count the valid choices for each position from left to right using the Multiplication Principle.  
The key observations are:
- The first digit cannot be \(0\) (since it is a four-digit number).
- An odd digit and an even digit are never equal, so the “all different” rule is automatically satisfied between positions 1 and 2.
- For positions 3 and 4, the only restriction is that they must differ from all previously chosen digits.

**Step 3: Count choices for each digit.**

- **First digit \(A\):** Must be odd and non-zero.  
  Odd digits: \(\{1,3,5,7,9\}\).  
  Number of choices: **5**.

- **Second digit \(B\):** Must be even.  
  Even digits: \(\{0,2,4,6,8\}\).  
  Because any even digit is different from any odd digit, there is no conflict with \(A\).  
  Number of choices: **5**.

- **Third digit \(C\):** Must differ from \(A\) and \(B\).  
  Two distinct digits have already been used.  
  Number of choices: \(10 - 2 =\) **8**.

- **Fourth digit \(D\):** Must differ from \(A\), \(B\), and \(C\).  
  Three distinct digits have now been used.  
  Number of choices: \(10 - 3 =\) **7**.

**Step 4: Compute the total.**

\[
5 \times 5 \times 8 \times 7 = 25 \times 56 = 1400.
\]

**Step 5: Rule out the wrong answer choices.**

- **A. 1120:** This equals \(5 \times 4 \times 8 \times 7\). It arises if a student mistakenly excludes \(0\) from the second digit, leaving only \(\{2,4,6,8\}\) (4 choices). But \(0\) is perfectly valid in the hundreds place.
- **C. 1800:** This equals \(5 \times 5 \times 9 \times 8\). It could come from incorrectly thinking the third digit has 9 choices (perhaps forgetting to exclude the second digit) and the fourth has 8.
- **D. 2025:** This equals \(5 \times 5 \times 9 \times 9\). It might result from assuming the last two digits each have 9 choices by excluding only one previous digit each, without properly accounting for all three used digits.
- **E. 2500:** This equals \(5 \times 5 \times 10 \times 10\). It is the count if one enforces the odd/even rules on the first two digits but completely ignores the “all four digits are different” requirement for the last two positions.
- **B. 1400:** This matches our correct step-by-step calculation.

ANSWER: B

Here is the step-by-step solution:

**1. Restate the Problem**
We need to determine the total number of 4-digit integers (let the digits be represented as $d_1 d_2 d_3 d_4$) that satisfy three conditions simultaneously:
*   Condition 1: The first digit ($d_1$) must be odd.
*   Condition 2: The second digit ($d_2$) must be even.
*   Condition 3: All four digits must be distinct (different from each other).

**2. Analyze the Choices for Each Digit Position**
We will use the Multiplication Principle (counting rule) to find the total number of possibilities by considering the constraints for each position one by one.

*   **Position 1 ($d_1$ - Thousands place):**
    *   The digit must be odd.
    *   Possible values: $\{1, 3, 5, 7, 9\}$.
    *   Note: Since it is a 4-digit number, the first digit cannot be 0. However, 0 is even, so it wouldn't be in the odd set anyway.
    *   **Number of options: 5**

*   **Position 2 ($d_2$ - Hundreds place):**
    *   The digit must be even.
    *   Possible values: $\{0, 2, 4, 6, 8\}$.
    *   We must ensure this digit is different from $d_1$. Since $d_1$ is chosen from the set of odd numbers and $d_2$ is chosen from the set of even numbers, these two sets share no common elements. Therefore, regardless of which odd number was picked for $d_1$, all 5 even options are still available for $d_2$.
    *   **Number of options: 5**

*   **Position 3 ($d_3$ - Tens place):**
    *   This digit can be any digit $0-9$ **except** the two digits already used for $d_1$ and $d_2$.
    *   Total digits available (0-9): 10.
    *   Digits already used: 2.
    *   **Number of options: $10 - 2 = 8$**

*   **Position 4 ($d_4$ - Units place):**
    *   This digit can be any digit $0-9$ **except** the three digits already used for $d_1$, $d_2$, and $d_3$.
    *   Total digits available (0-9): 10.
    *   Digits already used: 3.
    *   **Number of options: $10 - 3 = 7$**

**3. Perform the Computation**
To find the total number of such numbers, multiply the number of choices for each position:

$$ \text{Total} = (\text{choices for } d_1) \times (\text{choices for } d_2) \times (\text{choices for } d_3) \times (\text{choices for } d_4) $$

$$ \text{Total} = 5 \times 5 \times 8 \times 7 $$

Calculating step-by-step:
$$ 5 \times 5 = 25 $$
$$ 8 \times 7 = 56 $$
$$ 25 \times 56 $$

To compute $25 \times 56$ easily:
$$ 25 \times 56 = 25 \times (4 \times 14) = (25 \times 4) \times 14 = 100 \times 14 = 1400 $$

**4. Rule Out Wrong Choices**
*   **A. 1120**: This result typically comes from mistakenly excluding 0 as an option for the second digit (calculating $5 \times 4 \times 8 \times 7$). However, 0 is an even digit and is allowed in the second position.
*   **C. 1800**: This might result from miscalculating the remaining digits (e.g., assuming only 9 digits remain for the third slot instead of 8).
*   **E. 2500**: This is the result of completely ignoring the condition that "all four digits are different" (calculating $5 \times 5 \times 10 \times 10$).
*   **D. 2025**: This does not correspond to a standard error in this specific combinatorial setup.

Our calculated value of 1400 matches choice B.

ANSWER: B

**Step 1: Understand the problem**  
We need to count four‑digit whole numbers (from 1000 to 9999) that satisfy:  

- The thousands digit (leftmost) is odd.  
- The hundreds digit (second digit) is even.  
- All four digits are different.

**Step 2: Choose the thousands digit**  
Odd digits available: 1, 3, 5, 7, 9 → 5 choices.  
(It cannot be 0 because the number must be four‑digit.)

**Step 3: Choose the hundreds digit**  
Even digits available: 0, 2, 4, 6, 8 → 5 choices.  
Since the thousands digit is odd, it cannot equal any even digit, so the two chosen digits are automatically distinct.  
Thus 5 possibilities for the second digit.

**Step 4: Choose the tens digit**  
We have used two distinct digits, leaving 10 − 2 = 8 remaining digits.  
No parity restriction, so the tens digit can be any of those 8 digits.

**Step 5: Choose the units digit**  
Now three distinct digits have been used, leaving 10 − 3 = 7 remaining digits.  
The units digit can be any of those 7 digits.

**Step 6: Compute the total count**  
Multiply the independent choices:

\[
5 \times 5 \times 8 \times 7 = 1400.
\]

**Step 7: Eliminate the other options**  
- 1120 is too low (our count is larger because we have many valid choices for each position).  
- 1800, 2025, and 2500 are larger than the product of the maximum possible choices (5·5·9·8 = 1800), but our restriction of distinct digits reduces the total to 1400.  
Thus only 1400 matches the calculation.

**ANSWER: B**

1. **Restate the problem**: We need to find how many four - digit whole numbers satisfy the following conditions:
   - The left - most digit (thousands place) is odd.
   - The second digit (hundreds place) is even.
   - All four digits are different.
2. **Analyze the number of choices for each digit position**:
   - **Thousands place**:
     - The left - most digit (thousands place) must be an odd digit. The possible odd digits for the thousands place are 1, 3, 5, 7, 9. So, there are 5 possible choices for the thousands place.
   - **Hundreds place**:
     - The second digit (hundreds place) must be an even digit. The even digits are 0, 2, 4, 6, 8. Since we can still use any of the even digits regardless of what was chosen for the thousands place (because the digits are just required to be different, and we haven't used an even digit in the thousands place), there are 5 possible choices for the hundreds place.
   - **Tens place**:
     - After choosing the digits for the thousands and hundreds places, we have used 2 digits. There are 10 digits in total (0 - 9). So, the number of remaining digits for the tens place is 8 (because we can't use the two digits already chosen).
   - **Units place**:
     - After choosing digits for the thousands, hundreds, and tens places, we have used 3 digits. So, the number of remaining digits for the units place is 7.
3. **Use the counting principle**:
     - The counting principle states that if there are \(n_1\) ways to do the first task, \(n_2\) ways to do the second task, \(n_3\) ways to do the third task, and \(n_4\) ways to do the fourth task, then the total number of ways to do all four tasks together is \(n_1\times n_2\times n_3\times n_4\).
     - Here, \(n_1 = 5\) (choices for thousands place), \(n_2 = 5\) (choices for hundreds place), \(n_3 = 8\) (choices for tens place), and \(n_4 = 7\) (choices for units place).
     - The total number of four - digit whole numbers that satisfy the given conditions is \(5\times5\times8\times7\).
     - First, \(5\times5 = 25\), then \(25\times8=200\), and finally \(200\times7 = 1400\).
4. **Rule out wrong choices**:
     - We have calculated the number of four - digit numbers based on the rules of counting with the given digit - selection constraints. The value 1120 is not the result of our calculation. The value 1800 is too high because we accounted for the non - repetition of digits in our step - by - step counting. The value 2025 is not what we got from our calculation. The value 2500 is also much higher than our calculated value.

ANSWER: B

### Step 1: Restate the problem and define the four-digit number
We need to count all four-digit whole numbers (formatted as \(D_1D_2D_3D_4\), where \(D_1\) is the thousands digit, \(D_2\) the hundreds digit, \(D_3\) the tens digit, \(D_4\) the units digit) that satisfy three conditions:
1.  \(D_1\) (leftmost digit) is odd
2.  \(D_2\) is even
3.  All four digits are distinct.
We use the multiplication principle to count valid choices for each digit sequentially, then multiply the number of choices to get the total count of valid numbers.
---
### Step 2: Count valid choices for \(D_1\)
\(D_1\) must be a non-zero odd digit (since it is the leading digit of a four-digit number, and it must be odd). The valid values for \(D_1\) are {1,3,5,7,9}, so there are **5 possible choices** for \(D_1\).
---
### Step 3: Count valid choices for \(D_2\)
\(D_2\) must be an even digit. Even digits include 0, which is allowed for the hundreds place,  valid values are {0,2,4,6,8}. Since \(D_1\) is always odd, none of these even digits overlap with the chosen \(D_1\),  all 5 values remain available. There are **5 possible choices** for \(D_2\).
---
### Step 4: Count valid choices for \(D_3\)
All digits must be distinct,  we cannot reuse the two unique digits already chosen for \(D_1\) and \(D_2\). There are 10 total digits (0-9),  we subtract the 2 used digits, leaving 8 valid options. There are **8 possible choices** for \(D_3\).
---
### Step 5: Count valid choices for \(D_4\)
We now have three distinct used digits (\(D_1, D_2, D_3\)),  we subtract these 3 from the 10 total digits, leaving 7 valid, unused options. There are **7 possible choices** for \(D_4\).
---
### Step 6: Calculate the total number of valid numbers
Multiply the number of choices for each digit:
\[
5 \times 5 \times 8 \times 7 = 25 \times 56 = 1400
\]
---
### Step 7: Rule out incorrect options
-   **Option A (1120):** This would result from incorrectly excluding 0 from the set of valid even digits for \(D_2\), using 4 choices for \(D_2\) instead of 5.
-   **Option C (1800):** This incorrectly assumes 9 and 8 choices for \(D_3\) and \(D_4\), failing to account that both previously used digits (\(D_1\) and \(D_2\)) are forbidden for \(D_3\).
-   **Option D (2025):** This ignores the distinctness requirement for \(D_3\) and \(D_4\), using 9 and 9 choices instead of 8 and 7.
-   **Option E (2500):** This ignores all distinctness requirements, using 10 choices for both \(D_3\) and \(D_4\).
ANSWER: B

### Step 1: Restate the problem and constraints
We need to count 4-digit whole numbers (1000–9999, so the first digit cannot be 0) that satisfy:
1.  Leftmost (thousands) digit is odd,
2.  Second (hundreds) digit is even,
3.  All four digits are distinct.
---
### Step 2: Count choices for the first digit (D1)
The first digit must be odd and non-zero. The odd digits are 1, 3, 5, 7, 9, so there are **5 choices** for D1.
---
### Step 3: Count choices for the second digit (D2)
The second digit must be even. Even digits are 0, 2, 4, 6, 8 (5 total). Since D1 is odd, there is no overlap between D1 and even digits, so all 5 even digits are available for D2. This gives **5 choices** for D2.
---
### Step 4: Count choices for the third digit (D3)
D3 can be any digit except D1 and D2 (to satisfy distinctness). We have already used 2 unique digits (one odd, one even, so no overlap),  there are \(10 - 2 = 8\) remaining choices for D3.
---
### Step 5: Count choices for the fourth digit (D4)
D4 can be any digit except D1, D2, and D3. We have used 3 unique digits,  there are \(10 - 3 =7\) remaining choices for D4.
---
### Step 6: Calculate total number of valid numbers
Multiply the number of choices for each position:
\[
5 \times 5 \times 8 \times 7 = 1400
\]
---
### Step 7: Rule out wrong choices
- **A (1120)**: Incorrect. This assumes D2 cannot be 0, but 0 is even and allowed in the hundreds place (e.g., 1023 is valid). The missing 280 cases come from D2=0 (\(5 \times 1 \times 8 \times7=280\)).
- **C (1800)**: Incorrect. This assumes D3 has 9 choices, but D3 cannot equal D1 or D2, so only 8 choices are valid.
- **D (2025)**: Incorrect. This assumes D3 and D4 each have 9 choices, ignoring the distinctness requirement for all digits.
- **E (2500)**: Incorrect. This counts all numbers with odd first and even second digits, ignoring the distinctness requirement entirely.
---
ANSWER: B

To determine the number of four-digit whole numbers where the leftmost digit is odd, the second digit is even, and all four digits are different, we will follow a systematic approach.

1. **Determine the number of choices for the leftmost digit:**
   The leftmost digit must be odd. The odd digits available are 1, 3, 5, 7, and 9. Therefore, there are 5 choices for the leftmost digit.

2. **Determine the number of choices for the second digit:**
   The second digit must be even. The even digits available are 0, 2, 4, 6, and 8. Therefore, there are 5 choices for the second digit.

3. **Determine the number of choices for the third digit:**
   The third digit must be different from the first and second digits. Since we have already used 2 digits (one odd and one even), there are 8 digits left to choose from (10 total digits minus the 2 used). Therefore, there are 8 choices for the third digit.

4. **Determine the number of choices for the fourth digit:**
   The fourth digit must be different from the first, second, and third digits. Since we have already used 3 digits, there are 7 digits left to choose from. Therefore, there are 7 choices for the fourth digit.

5. **Calculate the total number of four-digit whole numbers:**
   The total number of four-digit whole numbers is the product of the number of choices for each digit:
   \[
   5 \text{ (choices for the first digit)} \times 5 \text{ (choices for the second digit)} \times 8 \text{ (choices for the third digit)} \times 7 \text{ (choices for the fourth digit)} = 1400
   \]

6. **Verify the choices:**
   - A. 1120: This is not the correct product of the choices.
   - B. 1400: This is the correct product of the choices.
   - C. 1800: This is not the correct product of the choices.
   - D. 2025: This is not the correct product of the choices.
   - E. 2500: This is not the correct product of the choices.

Therefore, the correct answer is:

ANSWER: B

**Problem Restatement:**
We need to find how many four-digit whole numbers satisfy the following conditions:
1. The leftmost (thousands) digit is odd.
2. The second (hundreds) digit is even.
3. All four digits are different.

**Step 1: Understand the structure of a four-digit number**
A four-digit number ranges from 1000 to 9999. It has the form ABCD, where:
- A = thousands digit (1–9, cannot be 0)
- B = hundreds digit (0–9)
- C = tens digit (0–9)
- D = units digit (0–9)

**Step 2: Apply the first condition (A is odd)**
The thousands digit A must be odd. The odd digits are 1, 3, 5, 7, 9.
So, A has **5 possible choices**.

**Step 3: Apply the second condition (B is even)**
The hundreds digit B must be even. The even digits are 0, 2, 4, 6, 8.
So, B has **5 possible choices**.

**Step 4: Apply the third condition (all digits are different)**
We must ensure that A, B, C, and D are all distinct. We will calculate the number of choices for C and D, ensuring they differ from A and B and each other.

**Step 5: Calculate choices for C (tens digit)**
C can be any digit from 0 to 9 except A and B. So far, 2 digits (A and B) are used.
Thus, C has **8 possible choices** (10 total digits minus the 2 already used).

**Step 6: Calculate choices for D (units digit)**
D can be any digit from 0 to 9 except A, B, and C. So far, 3 digits (A, B, and C) are used.
Thus, D has **7 possible choices** (10 total digits minus the 3 already used).

**Step 7: Combine all choices**
The total number of valid four-digit numbers is the product of the number of choices for each digit:
Total = (choices for A) × (choices for B) × (choices for C) × (choices for D)
Total = 5 × 5 × 8 × 7 = 1400.

**Step 8: Rule out incorrect choices**
- A (1120): Too low, does not account for all valid combinations.
- C (1800): Too high, overestimates the number of valid combinations.
- D (2025): Too high, overestimates the number of valid combinations.
- E (2500): Too high, overestimates the number of valid combinations.

The only choice that matches our calculation is B (1400).

ANSWER: B

I need to find the least possible fraction of sneakers that are both red AND high-top.

**Step 1: Determine the number of red and white sneakers**
- Total pairs: 15
- Red pairs: (3/5) × 15 = 9
- White pairs: 15 - 9 = 6

**Step 2: Determine the number of high-top and low-top sneakers**
- High-top pairs: (2/3) × 15 = 10
- Low-top pairs: 15 - 10 = 5

**Step 3: Set up the problem**
I need to find how many sneakers are BOTH red AND high-top. Let me call this number x.

To minimize x, I need to think about how the red and high-top categories can overlap as little as possible.

**Step 4: Use a constraint approach**
Let me denote:
- RH = red high-top sneakers (what we want to minimize)
- RL = red low-top sneakers
- WH = white high-top sneakers
- WL = white low-top sneakers

Constraints:
- RH + RL = 9 (total red)
- RH + WH = 10 (total high-top)
- WH + WL = 6 (total white)
- RL + WL = 5 (total low-top)
- RH + RL + WH + WL = 15 (total)

**Step 5: Find minimum RH**
From the constraints:
- RH + RL = 9, so RL = 9 - RH
- RH + WH = 10, so WH = 10 - RH
- WH + WL = 6, so WL = 6 - WH = 6 - (10 - RH) = RH - 4

For all values to be non-negative:
- RL = 9 - RH ≥ 0, so RH ≤ 9
- WH = 10 - RH ≥ 0, so RH ≤ 10
- WL = RH - 4 ≥ 0, so RH ≥ 4

Therefore: **4 ≤ RH ≤ 9**

The minimum value of RH is **4**.

**Step 6: Verify this works**
If RH = 4:
- RL = 9 - 4 = 5 ✓
- WH = 10 - 4 = 6 ✓
- WL = 4 - 4 = 0 ✓
- Total: 4 + 5 + 6 + 0 = 15 ✓

**Step 7: Calculate the fraction**
The minimum fraction of red high-top sneakers = 4/15

**Step 8: Check the answer choices**
- A. 0 - This would require 0 red high-tops, but we showed minimum is 4, so this is impossible
- B. 1/5 = 3/15 - This is less than 4/15, so impossible
- C. 4/15 - This matches our answer ✓
- D. 1/3 = 5/15 - This is greater than 4/15
- E. 2/5 = 6/15 - This is greater than 4/15

ANSWER: C

1. Restate: Jordan has 15 pairs of sneakers. Of these, 3/5 are red and 2/3 are high‐top. We seek the least possible fraction of the total that are both red and high‐top.  
2. Compute category sizes:  
   – Total pairs n = 15  
   – Red pairs R = (3/5)·15 = 9  
   – High‐top pairs H = (2/3)·15 = 10  
3. Let X = number of red high‐top pairs. By basic set‐overlap (pigeonhole) principle, the minimum overlap is  
   X_min = max(0, R + H − n) = max(0, 9 + 10 − 15) = 4.  
4. Thus the smallest possible number of red high‐tops is 4, so the fraction is 4/15.  
5. Check the answer choices:  
   A. 0 = 0/15 (too small, violates R+H>n)  
   B. 1/5 = 3/15 (too small; minimum is 4/15)  
   C. 4/15 (matches our result)  
   D. 1/3 = 5/15 (larger than 4/15)  
   E. 2/5 = 6/15 (larger than 4/15)  
6. Only choice C equals 4/15, the least possible fraction.  
ANSWER: C

An elegant step-by-step solution to the problem is presented below:

### Step 1: Restate what is being asked
We are asked to find the least possible fraction of Jordan's sneaker collection that consists of sneakers that are both **red** and **high-top**.

### Step 2: Calculate the number of sneakers in each category
Jordan has a total of $15$ pairs of sneakers.
* **Red sneakers ($R$):** $\frac{3}{5}$ of the collection, which is $\frac{3}{5} \times 15 = 9$ pairs.
* **White sneakers ($W$):** The rest of the collection, which is $15 - 9 = 6$ pairs.
* **High-top sneakers ($H$):** $\frac{2}{3}$ of the collection, which is $\frac{2}{3} \times 15 = 10$ pairs.
* **Low-top sneakers ($L$):** The rest of the collection, which is $15 - 10 = 5$ pairs.

### Step 3: Set up the math to find the minimum overlap
Let $x$ be the number of pairs of sneakers that are both **red** and **high-top** ($R \cap H$).
We can organize the sneakers into a $2 \times 2$ grid:
* Red High-tops: $x$
* Red Low-tops: $9 - x$
* White High-tops: $10 - x$
* White Low-tops: Since the total number of low-tops is $5$, the number of White Low-tops is $5 - (9 - x) = x - 4$.

Since the number of sneakers in any category cannot be negative, we must satisfy the following inequalities:
1. $x \ge 0$
2. $9 - x \ge 0 \implies x \le 9$
3. $10 - x \ge 0 \implies x \le 10$
4. $x - 4 \ge 0 \implies x \ge 4$

Combining these inequalities, we find that:
$$4 \le x \le 9$$

Thus, the least possible number of red high-top sneakers is $x = 4$.

### Step 4: Compute the fraction
The fraction of the collection that consists of red high-top sneakers is:
$$\frac{x}{15} = \frac{4}{15}$$

### Step 5: Explain how to rule out the wrong choices
* **A. 0:** If there were $0$ red high-tops, then all $9$ red sneakers would have to be low-tops. However, there are only $5$ low-tops in total, which is impossible.
* **B. 1/5 (which is 3/15):** If there were $3$ red high-tops, then there would be $9 - 3 = 6$ red low-tops, which exceeds the total of $5$ low-tops.
* **D. 1/3 (which is 5/15) and E. 2/5 (which is 6/15):** These represent possible fractions, but they are not the *least* possible value.

Thus, the least possible fraction is $\frac{4}{15}$.

ANSWER: C

Total sneakers: 15.  
Red sneakers: \(\frac{3}{5} \times 15 = 9\).  
White sneakers: \(15 - 9 = 6\).  
High-top sneakers: \(\frac{2}{3} \times 15 = 10\).  
Low-top sneakers: \(15 - 10 = 5\).

Let \(R_H\) be the number of red high-top sneakers. The goal is to minimize the fraction \(\frac{R_H}{15}\), which requires minimizing the integer value of \(R_H\).

The value of \(R_H\) must satisfy four non-negativity and capacity constraints derived from the given totals:  
- \(R_H \leq 9\) (cannot exceed total red sneakers),  
- \(R_H \leq 10\) (cannot exceed total high-top sneakers),  
- white high-tops = \(10 - R_H \leq 6\) (cannot exceed total white sneakers) \(\implies R_H \geq 4\),  
- red low-tops = \(9 - R_H \leq 5\) (cannot exceed total low-top sneakers) \(\implies R_H \geq 4\).

The binding lower bound is therefore \(R_H \geq 4\). The smallest integer satisfying all constraints is exactly 4.  

Verification that \(R_H = 4\) is feasible:  
- white high-tops = \(10 - 4 = 6\) (exactly the white total),  
- red low-tops = \(9 - 4 = 5\) (exactly the low-top total),  
- white low-tops = \(6 - 6 = 0\).  

All quantities are non-negative integers and sum to the required totals, so \(R_H = 4\) is attainable. The resulting minimal fraction is \(\frac{4}{15}\).

This value matches choice C. Choices A and B are smaller than \(\frac{4}{15}\) and violate the lower bound \(R_H \geq 4\). Choices D and E are larger than \(\frac{4}{15}\) and therefore cannot be minimal.

Final answer: C  
(The fraction cannot be smaller than \(\frac{4}{15}\) because at least four red high-tops are forced by the white and low-top totals.)

1. **Restate the problem**: We need to find the least possible fraction of Jordan's sneakers that are both red and high-top.

2. **Calculate the total number of red and high-top sneakers**:
   - Total sneakers = 15 pairs.
   - Red sneakers = \( \frac{3}{5} \times 15 = 9 \) pairs.
   - White sneakers = \( 15 - 9 = 6 \) pairs.
   - High-top sneakers = \( \frac{2}{3} \times 15 = 10 \) pairs.
   - Low-top sneakers = \( 15 - 10 = 5 \) pairs.

3. **Determine the maximum possible number of non-red high-tops**:
   - Since there are 6 white sneakers and 10 high-tops, the maximum number of white high-tops is 6 (if all white sneakers are high-top).

4. **Calculate the minimum number of red high-tops**:
   - If all 6 white sneakers are high-top, then the number of red high-tops = Total high-tops - White high-tops = 10 - 6 = 4 pairs.

5. **Calculate the fraction of red high-top sneakers**:
   - Fraction = \( \frac{\text{Number of red high-tops}}{\text{Total sneakers}} = \frac{4}{15} \).

6. **Compare with the given choices and rule out incorrect options**:
   - A. 0: This would mean there are no red high-tops, which is not possible since we have at least 4 red high-tops.
   - B. \( \frac{1}{5} \): This equals \( \frac{3}{15} \), which is less than \( \frac{4}{15} \). So, it's a possible fraction but we need to check if it's achievable.
   - C. \( \frac{4}{15} \): This matches our calculation for the minimum fraction of red high-tops.
   - D. \( \frac{1}{3} \): This equals \( \frac{5}{15} \), which is more than \( \frac{4}{15} \).
   - E. \( \frac{2}{5} \): This equals \( \frac{6}{15} \), which is also more than \( \frac{4}{15} \).

7. **Check if \( \frac{1}{5} \) is achievable**:
   - To achieve \( \frac{1}{5} \) or \( \frac{3}{15} \), there would need to be 3 red high-tops. This implies 7 white high-tops, which is not possible since there are only 6 white sneakers in total.

8. **Conclusion**: The least possible fraction of red high-top sneakers is \( \frac{4}{15} \) because we cannot have less than 4 red high-tops given the constraints.

ANSWER: C

Let’s go step-by-step.

---

**Step 1 – Restating the problem**  
We know:  
- Total pairs of sneakers = 15.  
- \( \frac{3}{5} \) are **red** → the rest are white.  
- \( \frac{2}{3} \) are **high-top** → the rest are low-top.  
We want the **least possible fraction** of the collection that are **red high-top** sneakers.

---

**Step 2 – Find number of red and white sneakers**  
Red sneakers:  
\[
\frac{3}{5} \times 15 = 9 \quad\text{red pairs}
\]  
White sneakers:  
\[
15 - 9 = 6 \quad\text{white pairs}
\]

---

**Step 3 – Find number of high-top and low-top sneakers**  
High-top sneakers:  
\[
\frac{2}{3} \times 15 = 10 \quad\text{high-top pairs}
\]  
Low-top sneakers:  
\[
15 - 10 = 5 \quad\text{low-top pairs}
\]

---

**Step 4 – Set up for red high-top count**  
We want the smallest possible overlap between “red” (9 pairs) and “high-top” (10 pairs).

The overlap range is determined by:  
Minimum overlap = \(\max(0, \text{red} + \text{high-top} - \text{total})\)

---

**Step 5 – Compute minimum overlap**  
\[
\text{min overlap} = \max(0, 9 + 10 - 15) = \max(0, 4) = 4
\]  
So the smallest number of red high-tops is 4 pairs.

---

**Step 6 – Fraction of the whole collection**  
\[
\text{Fraction red high-top} = \frac{4}{15}
\]

---

**Step 7 – Match with choices**  
Looking at the options given:  
- A. 0  
- B. \( \frac{1}{5} \) (which is \( \frac{3}{15} \))  
- C. \( \frac{4}{15} \)  
- D. \( \frac{1}{3} \) (which is \( \frac{5}{15} \))  
- E. \( \frac{2}{5} \) (which is \( \frac{6}{15} \))  

The least possible value is \( \frac{4}{15} \), which matches **C**.

---

ANSWER: C

We are told:

- Jordan owns **15 pairs** of sneakers.
- **Three fifths** are **red**, so the rest (**two fifths**) are **white**.
- **Two thirds** are **high-top**, so the rest (**one third**) are **low-top**.
- We are to find the **least possible value** of the fraction of the **red high-top** sneakers **out of the total 15 pairs**.

We are to **minimize** the number of **red high-top** sneakers, given the totals for red and high-top.

---

### Step 1: Compute total numbers

Total pairs = 15

**Red sneakers**:  
$ \frac{3}{5} \times 15 = 9 $ pairs are red.  
So, **white** = $15 - 9 = 6$ pairs.

**High-top sneakers**:  
$ \frac{2}{3} \times 15 = 10 $ pairs are high-top.  
So, **low-top** = $15 - 10 = 5$ pairs.

So we have:

- Red: 9  
- White: 6  
- High-top: 10  
- Low-top: 5

We are to find the **minimum possible number** of sneakers that are **both red and high-top**.

This is a classic **overlap minimization** problem: given two subsets of a finite set, with fixed sizes, what's the **minimum possible size** of their intersection?

---

### Step 2: Use principle of inclusion / overlap

Let:

- $ R $ = set of red sneakers, $|R| = 9$
- $ H $ = set of high-top sneakers, $|H| = 10$
- Total = 15

We want to **minimize** $ |R \cap H| $.

We know that:

$$
|R \cap H| = |R| + |H| - |R \cup H|
$$

But $ |R \cup H| \leq 15 $, since that’s the whole collection.

So the **maximum** possible $ |R \cup H| $ is 15, which would **minimize** the intersection.

Thus, **minimum** possible intersection is:

$$
|R \cap H| \geq |R| + |H| - 15 = 9 + 10 - 15 = 4
$$

So **at least 4** sneakers must be both red and high-top.

Can we achieve **exactly 4**? Yes, if we arrange the sets so that their union is all 15 pairs, with as little overlap as possible.

Let’s verify that this is possible given the other constraints (like white = 6, low-top = 5).

If red high-top = 4 (minimum), then:

- Red total = 9 → red low-top = 9 - 4 = **5**
- High-top total = 10 → white high-top = 10 - 4 = **6**

Now check white total: white = 6 → white low-top = 6 - 6 = **0**

Low-top total = red low-top + white low-top = 5 + 0 = **5**, which matches the required low-top total.

So this distribution works:

- Red high-top: 4  
- Red low-top: 5  
- White high-top: 6  
- White low-top: 0  

All totals match.

Therefore, **minimum number** of red high-top sneakers is **4**.

---

### Step 3: Convert to fraction of total collection

Total = 15 pairs.

So fraction = $ \frac{4}{15} $

---

### Step 4: Match with answer choices

Choices:

A. 0  
B. $ \frac{1}{5} $  
C. $ \frac{4}{15} $  
D. $ \frac{1}{3} $  
E. $ \frac{2}{5} $

Correct answer is **C. $ \frac{4}{15} $**

ANSWER: C

 **Step 1: Understand the problem.**  
Jordan has 15 pairs of sneakers.  
- \(\frac{3}{5}\) of the 15 pairs are red, the rest white.  
- \(\frac{2}{3}\) of the 15 pairs are high-top, the rest low-top.  

We need the **least possible fraction** of the entire collection that is **both red and high-top**.

**Step 2: Compute the numbers of red, white, high-top, and low-top pairs.**  
Red: \(\frac{3}{5} \times 15 = 9\) pairs.  
White: \(15-9 = 6\) pairs.  

High-top: \(\frac{2}{3} \times 15 = 10\) pairs.  
Low-top: \(15-10 = 5\) pairs.

**Step 3: Set up a 2‑by‑2 table to track the overlap.**  
Let \(x\) be the number of pairs that are **red and high-top** (the overlap we want to minimize).

\[
\begin{array}{c|cc|c}
& \text{High-top} & \text{Low-top} & \text{Total} \\ \hline
\text{Red} & x & 9-x & 9 \\
\text{White} & 10-x & 6-(10-x) = x-4 & 6 \\ \hline
\text{Total} & 10 & 5 & 15
\end{array}
\]

We require every entry to be a non‑negative integer.

**Step 4: Use the table to find the lower bound for \(x\).**  
From the “White / Low-top” entry we need \(x-4 \ge 0\), so  
\[
x \ge 4.
\]

From the “Red / Low-top” entry we need \(9-x \ge 0\), so \(x \le 9\).  
From the “White / High-top” entry we need \(10-x \ge 0\), so \(x \le 10\).  

Thus the smallest possible integer value for \(x\) is **4**.

*(Alternative view: by inclusion–exclusion, \(|R \cap H| = |R| + |H| - |R \cup H| \ge 9 + 10 - 15 = 4\), because the union cannot exceed 15.)*

**Step 5: Verify that \(x = 4\) is actually attainable.**  
If \(x=4\):

- Red high-top: 4  
- Red low-top: \(9-4 = 5\)  
- White high-top: \(10-4 = 6\)  
- White low-top: \(4-4 = 0\)

Check totals:  
Red: \(4+5 = 9\) ✓  
White: \(6+0 = 6\) ✓  
High-top: \(4+6 = 10\) ✓  
Low-top: \(5+0 = 5\) ✓  

All counts are non‑negative, so a collection with exactly 4 red high-top pairs is possible.

**Step 6: Compute the desired fraction.**  
The least possible number of red high-top pairs is 4 out of 15 total pairs, so the fraction is  
\[
\frac{4}{15}.
\]

**Step 7: Rule out the wrong answer choices.**  

- **A. 0** (\(\frac{0}{15}\)): Would mean no red high-tops. Then all 9 red pairs would have to be low-top, but there are only 5 low-top pairs. **Impossible.**  
- **B. \(\frac{1}{5} = \frac{3}{15}\):** Would mean 3 red high-tops. Then red low-tops would be \(9-3 = 6\), exceeding the 5 low-top pairs available. **Impossible.**  
- **C. \(\frac{4}{15}\):** Corresponds to \(x=4\), the smallest feasible value as shown above. **Possible and minimal.**  
- **D. \(\frac{1}{3} = \frac{5}{15}\):** Corresponds to 5 red high-tops. This is achievable (e.g., \(x=5\) works), but it is larger than \(\frac{4}{15}\). **Not the least possible.**  
- **E. \(\frac{2}{5} = \frac{6}{15}\):** Corresponds to 6 red high-tops. Achievable but larger. **Not the least possible.**

Therefore the least possible fraction is \(\frac{4}{15}\).

ANSWER: C

**Step 1: Restate the Problem**
We need to determine the smallest possible fraction of Jordan's sneaker collection that consists of pairs which are **both** red and high-top.

**Step 2: Determine the counts for each category**
First, let's calculate the exact number of pairs for each attribute based on the total collection of 15 pairs.

*   **Total pairs ($N$):** 15
*   **Red pairs ($R$):**
    $$\frac{3}{5} \text{ of } 15 = \frac{3}{5} \times 15 = 9 \text{ pairs}$$
*   **White pairs ($W$):**
    Since the rest are white:
    $$15 - 9 = 6 \text{ pairs}$$
*   **High-top pairs ($H$):**
    $$\frac{2}{3} \text{ of } 15 = \frac{2}{3} \times 10 = 10 \text{ pairs}$$
*   **Low-top pairs ($L$):**
    Since the rest are low-top:
    $$15 - 10 = 5 \text{ pairs}$$

**Step 3: Calculate the minimum intersection**
We want to find the minimum number of pairs that satisfy both conditions (Red AND High-top).
Let $x$ be the number of red high-top sneakers.
We know:
*   There are 9 red pairs.
*   There are 10 high-top pairs.
*   There are only 15 pairs in total.

To minimize the overlap (intersection), we should try to make the groups as distinct as possible. However, since the sum of the sizes of the two groups exceeds the total number of items, they **must** overlap.

Using the principle that the sum of the parts equals the whole plus the overlap:
$$ (\text{Red Only}) + (\text{High-Top Only}) + (\text{Both}) + (\text{Neither}) = 15 $$
Or more simply using the formula for the minimum intersection of two sets:
$$ \text{Min}(R \cap H) = (\text{Total Red}) + (\text{Total High-Top}) - \text{Total Collection} $$
$$ x_{min} = 9 + 10 - 15 $$
$$ x_{min} = 19 - 15 $$
$$ x_{min} = 4 $$

So, the least possible number of red high-top sneakers is **4**.

**Step 4: Verify the solution is possible**
Let's check if we can arrange the sneakers such that exactly 4 are red and high-top without violating any other counts:
1.  **Red & High-Top:** 4 (Target Minimum)
2.  **Red & Low-Top:** Since there are 9 Red total, and 4 are High-Top, the remaining $9 - 4 = 5$ must be Red and Low-Top.
3.  **White & High-Top:** Since there are 10 High-Top total, and 4 are Red, the remaining $10 - 4 = 6$ must be White and High-Top.
4.  **White & Low-Top:** We have 6 White total. We used 6 as White & High-Top, leaving $6 - 6 = 0$ for White & Low-Top. We also have 5 Low-Top total. We used 5 as Red & Low-Top, leaving $5 - 5 = 0$ for White & Low-Top. This is consistent.

The distribution works: 4 (Red/High) + 5 (Red/Low) + 6 (White/High) + 0 (White/Low) = 15.

**Step 5: Compute the final fraction**
The question asks for the fraction of the collection.
$$ \text{Fraction} = \frac{\text{Number of Red High-Tops}}{\text{Total Pairs}} = \frac{4}{15} $$

**Step 6: Rule out wrong choices**
*   **A (0):** Impossible. You cannot have 9 red and 10 high-top shoes in a set of 15 without at least some being both.
*   **B (1/5 or 3/15):** This would mean 3 pairs. As calculated above, the minimum is 4.
*   **D (1/3 or 5/15):** This represents 5 pairs. While possible, it is not the *least* possible value.
*   **E (2/5 or 6/15):** This represents 6 pairs. This is actually the *maximum* possible overlap (limited by the smaller group, Red).

The correct choice is C.

ANSWER: C

**Step 1 – Understand what is being asked.**  
We have 15 pairs of sneakers.  
- “Three‑fifths of the pairs are red.” → number of red pairs = \( \frac35\cdot15 = 9\).  
- “Two‑thirds of the pairs are high‑top.” → number of high‑top pairs = \( \frac23\cdot15 = 10\).  

We need the fraction of the collection that are **both red and high‑top**.  
Let \(x\) be the number of red‑high‑top pairs.  
The smallest possible value of the fraction \(x/15\) must be found while respecting the integer counts for each colour and style.

---

**Step 2 – Set up the counts.**  

| Category | Count |
|----------|-------|
| Red (R)  | 9 |
| White (W) | 6 |
| High‑top (H) | 10 |
| Low‑top (L) | 5 |

Write the four possible sub‑categories:

- Red‑high‑top = \(x\)
- Red‑low‑top = \(9 - x\)
- White‑high‑top = \(10 - x\)
- White‑low‑top = the remaining low‑top pairs: \(5 - (9 - x) = x - 4\).

---

**Step 3 – Apply non‑negativity constraints.**  

All four numbers must be **non‑negative integers**:

\[
\begin{aligned}
x &\ge 0,\\
9 - x &\ge 0 \;\Rightarrow\; x \le 9,\\
10 - x &\ge 0 \;\Rightarrow\; x \le 10,\\
x - 4 &\ge 0 \;\Rightarrow\; x \ge 4.
\end{aligned}
\]

Thus \(x\) must satisfy \(4 \le x \le 9\).  
The integer possibilities are \(x = 4,5,6,7,8,9\).

---

**Step 4 – Find the smallest possible fraction.**  

The fraction of the collection that are red‑high‑top is \(\dfrac{x}{15}\).  
The smallest feasible \(x\) is \(4\), giving  

\[
\frac{4}{15}.
\]

---

**Step 5 – Eliminate the other answer choices.**  

- **0 (choice A)** would require \(x=0\), but then white‑low‑top would be \(-4\), impossible.  
- **\(\frac15\) (choice B)** would need \(x=3\), which violates the low‑top count (gives negative white‑low‑top).  
- **\(\frac13\) (choice D)** would need \(x=5\), which is larger than the minimal possible value.  
- **\(\frac25\) (choice E)** would need \(x=6\), also larger than the minimum.

Only \(\frac{4}{15}\) (choice C) matches the smallest feasible value.

---

**Step 6 – Final answer.**  

ANSWER: C

1. First, understand what is being asked:
   - We need to find the least - possible fraction of red high - top sneakers in Jordan's collection of 15 pairs of sneakers. We know the number of red pairs and the number of high - top pairs, and we want to minimize the number of red high - top pairs.
2. Calculate the number of red and high - top pairs:
   - Given that Jordan has 15 pairs of sneakers. The number of red pairs is $\frac{3}{5}\times15 = 9$ pairs. So, the number of white pairs is $15 - 9=6$ pairs.
   - The number of high - top pairs is $\frac{2}{3}\times15 = 10$ pairs. So, the number of low - top pairs is $15 - 10 = 5$ pairs.
3. Use the principle of minimizing the intersection:
   - To minimize the number of red high - top pairs, we want to maximize the number of red low - top and white high - top pairs.
   - We have 9 red pairs and 5 low - top pairs. Let's assume all 5 low - top pairs are red. Then the number of red high - top pairs is $9 - 5=4$ pairs.
   - We can also check the other way around. We have 10 high - top pairs and 6 white pairs. Let's assume all 6 white pairs are high - top. Then the number of red high - top pairs is $10 - 6 = 4$ pairs.
4. Calculate the fraction of red high - top pairs:
   - The fraction of red high - top pairs in the collection of 15 pairs is $\frac{4}{15}$.
5. Rule out the wrong choices:
   - Option A: It's not possible for the fraction to be 0 because there are red pairs and high - top pairs, and there must be some overlap.
   - Option B: $\frac{1}{5}=\frac{3}{15}$. We found that the minimum number of red high - top pairs is 4, so $\frac{1}{5}$ is too small.
   - Option D: $\frac{1}{3}=\frac{5}{15}$. Since we calculated the minimum number of red high - top pairs to be 4, $\frac{1}{3}$ is not the correct fraction.
   - Option E: $\frac{2}{5}=\frac{6}{15}$. This is larger than the minimum fraction we calculated.

ANSWER: C

### Step 1: Restate the problem and list total quantities
We have 15 total sneaker pairs. We need to find the smallest possible fraction of the collection that is both red and high-top. First calculate the counts of each category:
- Red pairs: 3/5 * 15 = 9, so white pairs = 15 - 9 = 6
- High-top pairs: 2/3 *15 =10,  low-top pairs =15-10=5
### Step 2: Calculate the minimum possible overlap of red and high-top pairs
To minimize the number of red high-top pairs, we maximize the number of high-top pairs that can be white (the only non-red group). There are only 6 white pairs total,  the maximum number of white high-tops we can have is 6. The remaining high-top pairs must be red:
Minimum red high-tops = total high-tops - maximum white high-tops = 10 -6 =4
### Step 3: Verify this is feasible
If we have 4 red high-tops, the remaining 5 red pairs are low-top, all 6 white pairs are high-top, and 0 white pairs are low-top. This adds up:
- High-tops:4+6=10, low-tops:5+0=5, which matches our required counts. It is impossible to have fewer than 4 red high-tops, as that would require more white high-tops than the 6 total white pairs we have.
### Step 4: Compute the fraction and rule out incorrect options
The fraction of red high-top pairs is 4/15. Eliminate wrong choices:
- A (0): Impossible, we cannot fit 10 high-tops into only 6 white pairs.
- B (1/5 = 3/15): Impossible, this would require 7 white high-tops, which exceeds our 6 total white pairs.
- D (1/3=5/15) and E (2/5=6/15): These are larger than the minimum possible 4/15, so they are not the smallest value.
ANSWER: C

### Step 1: Restate the problem and identify fixed counts
Jordan has 15 total pairs of sneakers. We need the smallest possible fraction of the collection that is both red and high-top. First, calculate the fixed counts of each category:
- Red sneakers: \( \frac{3}{5} \times 15 = 9 \) pairs, so white sneakers = \( 15 - 9 = 6 \) pairs.
- High-top sneakers: \( \frac{2}{3} \times 15 = 10 \) pairs, so low-top sneakers = \( 15 - 10 = 5 \) pairs.
---
### Step 2: Minimize the overlap of red and high-top sneakers
To minimize the number of red high-tops, we maximize the number of red sneakers that are *not* high-top (i.e., red low-tops). The total number of low-top sneakers is only 5, so the maximum possible number of red low-tops is 5 (we cannot have more red low-tops than the total number of low-tops).
---
### Step 3: Calculate the minimum red high-tops
If all 5 low-tops are red, the remaining red sneakers must be high-top:
Minimum red high-tops = Total red sneakers - Maximum red low-tops = \( 9 - 5 = 4 \).
---
### Step 4: Verify the result and rule out wrong choices
The fraction of red high-tops is \( \frac{4}{15} \). We confirm this is the minimum:
- A fraction lower than \( \frac{4}{15} \) (e.g., 0, \( \frac{1}{5} = \frac{3}{15} \)) would require more than 5 red low-tops, which is impossible because there are only 5 low-tops total.
- Higher fractions (\( \frac{1}{3} = \frac{5}{15} \), \( \frac{2}{5} = \frac{6}{15} \)) are not the minimum.
---
ANSWER: C

To determine the least possible fraction of Jordan's collection that is red high-top sneakers, we need to follow these steps:

1. **Calculate the number of red sneakers:**
   - Jordan owns 15 pairs of sneakers.
   - Three fifths of the pairs are red.
   \[
   \text{Number of red sneakers} = \frac{3}{5} \times 15 = 9
   \]

2. **Calculate the number of white sneakers:**
   - The rest of the sneakers are white.
   \[
   \text{Number of white sneakers} = 15 - 9 = 6
   \]

3. **Calculate the number of high-top sneakers:**
   - Two thirds of the pairs are high-top.
   \[
   \text{Number of high-top sneakers} = \frac{2}{3} \times 15 = 10
   \]

4. **Calculate the number of low-top sneakers:**
   - The rest of the sneakers are low-top.
   \[
   \text{Number of low-top sneakers} = 15 - 10 = 5
   \]

5. **Determine the maximum number of red high-top sneakers:**
   - To find the least possible fraction of red high-top sneakers, we need to maximize the number of red high-top sneakers.
   - The maximum number of red high-top sneakers is limited by the smaller of the two quantities: the number of red sneakers (9) and the number of high-top sneakers (10).
   \[
   \text{Maximum red high-top sneakers} = \min(9, 10) = 9
   \]

6. **Calculate the fraction of red high-top sneakers:**
   - The fraction of the collection that is red high-top is the number of red high-top sneakers divided by the total number of sneakers.
   \[
   \text{Fraction of red high-top sneakers} = \frac{9}{15} = \frac{3}{5}
   \]

7. **Determine the least possible value of this fraction:**
   - To find the least possible fraction, we need to consider the scenario where the number of red high-top sneakers is minimized.
   - If we minimize the number of red high-top sneakers, we can set it to 0 (assuming it's possible for there to be no red high-top sneakers).
   \[
   \text{Least possible fraction} = \frac{0}{15} = 0
   \]

Thus, the least possible value of the fraction of red high-top sneakers is 0.

ANSWER: A

**Problem Restatement:**
Jordan has 15 pairs of sneakers. The colors are divided into red and white, with three-fifths being red. The styles are divided into high-top and low-top, with two-thirds being high-top. We need to find the least possible fraction of the collection that are **red high-top** sneakers.

---

### Step 1: Total pairs and color breakdown
- Total pairs = 15.
- Fraction red = 3/5 → Number of red pairs = (3/5) × 15 = **9**.
- Fraction white = 1 − 3/5 = 2/5 → Number of white pairs = (2/5) × 15 = **6**.

### Step 2: Style breakdown
- Fraction high-top = 2/3 → Number of high-top pairs = (2/3) × 15 = **10**.
- Fraction low-top = 1 − 2/3 = 1/3 → Number of low-top pairs = (1/3) × 15 = **5**.

---

### Step 3: Define variables for overlap
Let:
- \( r_h \) = number of **red high-top** pairs (this is what we want to minimize as a fraction of 15).
- \( r_l \) = number of **red low-top** pairs.
- \( w_h \) = number of **white high-top** pairs.
- \( w_l \) = number of **white low-top** pairs.

From the breakdowns:
1. \( r_h + r_l = 9 \) (total red pairs).
2. \( w_h + w_l = 6 \) (total white pairs).
3. \( r_h + w_h = 10 \) (total high-top pairs).
4. \( r_l + w_l = 5 \) (total low-top pairs).

---

### Step 4: Solve for \( r_h \)
From equation 3: \( w_h = 10 - r_h \).
From equation 1: \( r_l = 9 - r_h \).
Substitute \( r_l \) into equation 4:
\( (9 - r_h) + w_l = 5 \) → \( w_l = r_h - 4 \).
But from equation 2: \( w_l = 6 - w_h = 6 - (10 - r_h) = r_h - 4 \).
This is consistent, but we must ensure all counts are non-negative:
- \( w_l = r_h - 4 \geq 0 \) → \( r_h \geq 4 \).
- \( w_h = 10 - r_h \geq 0 \) → \( r_h \leq 10 \).
- \( r_l = 9 - r_h \geq 0 \) → \( r_h \leq 9 \).

Thus, \( r_h \) must satisfy \( 4 \leq r_h \leq 9 \).

---

### Step 5: Minimize the fraction
The fraction of red high-top pairs is \( \frac{r_h}{15} \). To minimize this, we take the smallest possible \( r_h \), which is **4**.

So, the minimal fraction is \( \frac{4}{15} \).

---

### Step 6: Rule out wrong choices
- **A (0):** Impossible, since \( r_h \geq 4 \).
- **B (1/5 = 0.2):** 4/15 ≈ 0.266..., which is greater than 0.2.
- **C (4/15 ≈ 0.266...):** Matches our minimal fraction.
- **D (1/3 ≈ 0.333...):** Larger than 4/15.
- **E (2/5 = 0.4):** Larger than 4/15.

---

ANSWER: C

# Solution to the Potato Peeling Problem

## Step 1: Restate what is being asked
We need to find how many potatoes Christen peeled, given that Homer starts alone and Christen joins him 4 minutes later.

## Step 2: Calculate how many potatoes Homer peeled alone
- Homer peels at 3 potatoes per minute
- He works alone for 4 minutes
- Potatoes peeled by Homer alone: 3 × 4 = **12 potatoes**

## Step 3: Calculate remaining potatoes
- Total potatoes: 44
- Peeled by Homer in first 4 minutes: 12
- Remaining potatoes: 44 - 12 = **32 potatoes**

## Step 4: Determine the combined peeling rate
When Christen joins Homer:
- Homer's rate: 3 potatoes per minute
- Christen's rate: 5 potatoes per minute
- Combined rate: 3 + 5 = **8 potatoes per minute**

## Step 5: Calculate time to peel remaining potatoes
- Remaining potatoes: 32
- Combined rate: 8 potatoes per minute
- Time needed: 32 ÷ 8 = **4 minutes**

## Step 6: Calculate potatoes peeled by Christen
- Christen works for 4 minutes at 5 potatoes per minute
- Potatoes peeled by Christen: 5 × 4 = **20 potatoes**

## Verification
- Homer: 12 + (3 × 4) = 12 + 12 = 24 potatoes
- Christen: 5 × 4 = 20 potatoes
- Total: 24 + 20 = 44 ✓

ANSWER: A

1. Restate the problem. We start with 44 potatoes. Homer peels alone for 4 minutes at 3 potatoes per minute. Then Christen joins and peels at 5 potatoes per minute (while Homer continues at 3 per minute) until the pile is gone. We must find how many potatoes Christen peeled in total.

2. Compute how many potatoes Homer peels in the first 4 minutes.  
   Rate of Homer = 3 potatoes/minute.  
   Time = 4 minutes.  
   Potatoes peeled by Homer alone = 3 × 4 = 12.

3. Compute how many potatoes remain after those 4 minutes.  
   Total start = 44.  
   Peeled in first 4 minutes = 12.  
   Remaining = 44 − 12 = 32.

4. Compute the combined peeling rate once Christen joins.  
   Homer’s rate = 3 potatoes/minute.  
   Christen’s rate = 5 potatoes/minute.  
   Combined rate = 3 + 5 = 8 potatoes/minute.

5. Compute how long it takes to peel the remaining 32 potatoes at the combined rate.  
   Time to finish = (remaining potatoes) ÷ (combined rate) = 32 ÷ 8 = 4 minutes.

6. Compute how many potatoes Christen peeled.  
   Christen’s rate = 5 potatoes/minute.  
   Time Christen peels = 4 minutes.  
   Potatoes peeled by Christen = 5 × 4 = 20.

7. Match to the answer choices and rule out incorrect ones.  
   A. 20 – matches our result.  
   B. 24 – would require 4.8 minutes of work at 5/min or different scenario; not possible here.  
   C. 32 – that’s the remaining total potatoes, not just Christen’s share.  
   D. 33 – no integer-minute combination yields 33 for Christen at 5/min.  
   E. 40 – too large; total peeled by both in 4 minutes is only 32.  

ANSWER: A

To find the number of potatoes Christen peeled, we can break the problem down into the following steps:

1. **Restate the problem:** 
   We need to find the total number of potatoes Christen peeled from a pile of 44 potatoes. Homer peels at a rate of 3 potatoes per minute and starts 4 minutes before Christen. Christen peels at a rate of 5 potatoes per minute.

2. **Calculate the potatoes Homer peeled before Christen joined:**
   Homer worked alone for the first 4 minutes. 
   $$\text{Potatoes peeled by Homer} = 3 \text{ potatoes/minute} \times 4 \text{ minutes} = 12 \text{ potatoes}$$

3. **Determine the remaining potatoes:**
   Subtract the potatoes Homer already peeled from the total pile:
   $$\text{Remaining potatoes} = 44 - 12 = 32 \text{ potatoes}$$

4. **Find the time they worked together:**
   When Christen joins Homer, they work together. Their combined rate is:
   $$\text{Combined rate} = 3 \text{ potatoes/minute (Homer)} + 5 \text{ potatoes/minute (Christen)} = 8 \text{ potatoes/minute}$$
   
   The time ($t$) they work together to finish the remaining 32 potatoes is:
   $$t = \frac{32 \text{ potatoes}}{8 \text{ potatoes/minute}} = 4 \text{ minutes}$$

5. **Calculate the number of potatoes Christen peeled:**
   Christen worked for 4 minutes at a rate of 5 potatoes per minute:
   $$\text{Potatoes peeled by Christen} = 5 \text{ potatoes/minute} \times 4 \text{ minutes} = 20 \text{ potatoes}$$

6. **Rule out the wrong choices:**
   * **B (24):** This is the number of potatoes Homer peeled in total ($12 \text{ initially} + 12 \text{ while working together}$).
   * **C (32):** This is the number of potatoes remaining when Christen joined.
   * **D (33) and E (40):** These values do not correspond to any step of the calculation.
   * **A (20):** This is the correct number of potatoes Christen peeled.

ANSWER: A

1. Restate the problem: Homer peels 44 potatoes starting alone at 3 per minute; Christen joins after 4 minutes at 5 per minute. Find how many potatoes Christen peels by the time they finish.

2. Set up the math: First compute Homer's solo work, subtract from total to find remaining potatoes, then find combined rate and time needed, and finally multiply Christen's rate by that time.

3. Computation:
- Homer alone for 4 min peels \(4 \times 3 = 12\) potatoes.
- Potatoes left: \(44 - 12 = 32\).
- Combined rate: \(3 + 5 = 8\) potatoes per minute.
- Time together: \(32 \div 8 = 4\) minutes.
- Christen peels: \(5 \times 4 = 20\) potatoes.

4. Verify totals: Christen 20 + Homer's additional \(3 \times 4 = 12\) gives exactly 32, confirming completion. Total peeled: initial 12 + 32 = 44.

5. Rule out wrong choices:
- B (24): would require Christen working 4.8 min (ignores exact combined rate).
- C (32): assumes Christen peels all remaining alone.
- D (33): adds 1 extra minute or miscalculates initial solo work.
- E (40): assumes Christen peels entire pile at her rate.

ANSWER: A

1. **Restate the problem**: We need to find out how many potatoes Christen peeled when Homer and Christen together finished peeling a pile of 44 potatoes. Homer started peeling at a rate of 3 potatoes per minute, and Christen joined him 4 minutes later, peeling at a rate of 5 potatoes per minute.

2. **Calculate the number of potatoes Homer peeled in the first 4 minutes**: Homer peels 3 potatoes per minute. So, in 4 minutes, he peeled 3 * 4 = 12 potatoes.

3. **Determine the number of potatoes left when Christen joined**: Initially, there were 44 potatoes. After Homer peeled 12, there were 44 - 12 = 32 potatoes left.

4. **Set up the equation for the time taken to peel the remaining potatoes**: Let's denote the time Christen worked as 't' minutes. Since Homer started 4 minutes earlier, Homer worked for 't + 4' minutes. The total number of potatoes peeled by both of them together is equal to the initial number of potatoes. So, we can set up the equation based on their rates: 3(t + 4) + 5t = 44.

5. **Simplify and solve the equation**: 
   - First, distribute the 3: 3t + 12 + 5t = 44.
   - Combine like terms: 8t + 12 = 44.
   - Subtract 12 from both sides: 8t = 32.
   - Divide by 8: t = 4.

6. **Calculate the number of potatoes Christen peeled**: Christen worked for 4 minutes at a rate of 5 potatoes per minute. So, she peeled 5 * 4 = 20 potatoes.

7. **Rule out the wrong choices**:
   - A. 20: This matches our calculation.
   - B. 24: This is more than 20, and since Christen worked for 4 minutes at 5 potatoes per minute, it's incorrect because 5 * 4 = 20.
   - C. 32: This is the number of potatoes left when Christen started, not the number she peeled.
   - D. 33: This is not consistent with Christen's rate or the time she worked.
   - E. 40: This exceeds the total number of potatoes left when Christen started working.

8. **Conclusion**: Based on the calculation, Christen peeled 20 potatoes.

ANSWER: A

1. **Understand the problem**: We have a total of 44 potatoes. Homer starts peeling alone at 3 potatoes per minute. After 4 minutes, Christen joins and peels at 5 potatoes per minute. They work together until all potatoes are peeled. We need to find how many potatoes Christen peeled.

2. **Find potatoes peeled by Homer before Christen joins**: Homer works alone for 4 minutes at 3 potatoes/minute.  
   Potatoes peeled = \( 3 \times 4 = 12 \).  
   Remaining potatoes = \( 44 - 12 = 32 \).

3. **Determine combined peeling rate after Christen joins**:  
   Homer's rate = 3 potatoes/min, Christen's rate = 5 potatoes/min.  
   Combined rate = \( 3 + 5 = 8 \) potatoes/min.

4. **Calculate time needed to finish the remaining potatoes together**:  
   Time = remaining potatoes ÷ combined rate = \( 32 \div 8 = 4 \) minutes.

5. **Find potatoes peeled by Christen**: Christen works for these 4 minutes at 5 potatoes/min.  
   Potatoes peeled by Christen = \( 5 \times 4 = 20 \).

6. **Verify**: Homer peeled 12 initially and then 3×4 = 12 more, total 24. Christen peeled 20. Total = \( 24 + 20 = 44 \), which matches the pile. So Christen peeled 20 potatoes.

7. **Eliminate wrong choices**:  
   - B (24) is Homer's total.  
   - C (32) is the remaining potatoes when Christen started.  
   - D (33) and E (40) are not plausible given the rates and time.  
   - Only A (20) matches.

ANSWER: A

1. **Understand the problem**:  
   Homer starts peeling 44 potatoes at 3 potatoes per minute. After 4 minutes, Christen joins and peels at 5 potatoes per minute. We are to find how many potatoes **Christen** peeled by the time all 44 are done.

2. **Find how many potatoes Homer peeled alone in the first 4 minutes**:  
   Homer's rate = 3 potatoes/minute  
   Time alone = 4 minutes  
   Potatoes peeled by Homer alone = $ 3 \times 4 = 12 $ potatoes.

3. **Determine how many potatoes remain after those 4 minutes**:  
   Total = 44  
   Remaining = $ 44 - 12 = 32 $ potatoes.

4. **Now both Homer and Christen peel together**:  
   Homer’s rate = 3 potatoes/minute  
   Christen’s rate = 5 potatoes/minute  
   Combined rate = $ 3 + 5 = 8 $ potatoes/minute.

5. **Find how long they worked together to finish the remaining 32 potatoes**:  
   Time = $ \frac{32}{8} = 4 $ minutes.

6. **Calculate how many potatoes Christen peeled**:  
   Christen peeled for 4 minutes at 5 potatoes/minute:  
   $ 5 \times 4 = 20 $ potatoes.

7. **Check answer choices**:  
   Choice A is 20, which matches our result.

8. **Verify no mistakes**:  
   - Homer peeled 12 alone + (3 × 4) = 12 + 12 = 24 total.  
   - Christen peeled 20.  
   - Total = 24 + 20 = 44 → correct.

ANSWER: A

1. **Restate the problem.**  
   Homer begins peeling 44 potatoes at 3 per minute. After 4 minutes, Christen joins him and peels at 5 per minute. Together they finish the pile. We must find how many potatoes Christen peeled.

2. **Calculate Homer’s solo work.**  
   In the 4 minutes before Christen arrives, Homer peels  
   $$3\text{ potatoes/min} \times 4\text{ min} = 12\text{ potatoes}.$$

3. **Find the remaining potatoes.**  
   $$44 - 12 = 32\text{ potatoes}$$  
   are left in the pile when Christen starts.

4. **Determine the combined peeling rate.**  
   Working together, Homer and Christen peel  
   $$3 + 5 = 8\text{ potatoes per minute}.$$

5. **Compute the time they work together.**  
   To finish the remaining 32 potatoes at 8 per minute, they need  
   $$32 \div 8 = 4\text{ minutes}.$$

6. **Calculate Christen’s total.**  
   Christen peels at 5 potatoes per minute for those 4 minutes, so she peels  
   $$5 \times 4 = 20\text{ potatoes}.$$

7. **Verify the totals.**  
   Homer peeled 12 alone plus another $3 \times 4 = 12$ together, giving him 24.  
   Christen peeled 20.  
   $24 + 20 = 44$, which matches the original pile.

8. **Rule out the wrong choices.**  
   - **B (24):** This is Homer’s *total* (12 solo + 12 together), not Christen’s.  
   - **C (32):** This is the number of potatoes *remaining* when Christen joined, not the amount she peeled.  
   - **D (33):** No valid step in the problem produces 33; it is neither a time-based nor a rate-based result for Christen.  
   - **E (40):** This would require Christen to peel for 8 minutes, but the combined work lasted only 4 minutes.

ANSWER: A

Here is the step-by-step solution:

**1. Understand the Problem**
We have a total of 44 potatoes to peel.
*   **Homer** starts immediately, peeling at a rate of 3 potatoes per minute.
*   **Christen** starts 4 minutes later, peeling at a rate of 5 potatoes per minute.
We need to find the total count of potatoes that Christen peeled by the time the job is finished.

**2. Analyze Homer's Solo Work**
First, we calculate how much progress Homer makes before Christen joins him.
*   Time working alone: 4 minutes
*   Homer's rate: 3 potatoes/minute
*   Potatoes peeled by Homer alone = $4 \text{ minutes} \times 3 \text{ potatoes/minute} = 12 \text{ potatoes}$

**3. Determine the Remaining Work**
Subtract the potatoes Homer already peeled from the total pile to see what is left for them to do together.
*   Total potatoes: 44
*   Potatoes already peeled: 12
*   Remaining potatoes = $44 - 12 = 32 \text{ potatoes}$

**4. Calculate the Joint Work Phase**
Now both people are working together. We need to find how long it takes them to finish the remaining 32 potatoes.
*   Combined rate = Homer's rate + Christen's rate
*   Combined rate = $3 + 5 = 8 \text{ potatoes per minute}$
*   Time to finish remaining potatoes = $\frac{\text{Remaining Potatoes}}{\text{Combined Rate}}$
*   Time = $\frac{32}{8} = 4 \text{ minutes}$

So, they work together for 4 minutes to finish the pile.

**5. Compute Christen's Contribution**
Christen only works during this joint phase. We multiply her rate by the time she spent working.
*   Christen's rate: 5 potatoes/minute
*   Time working: 4 minutes
*   Potatoes peeled by Christen = $5 \times 4 = 20 \text{ potatoes}$

**6. Evaluate the Choices**
*   **A. 20**: Matches our calculation exactly.
*   B. 24: Incorrect. (This would imply she worked for 4.8 minutes or at a different rate).
*   C. 32: Incorrect. (This is the number of potatoes remaining when she started).
*   D. 33: Incorrect.
*   E. 40: Incorrect.

ANSWER: A

**Step 1: Determine how many potatoes Homer peels before Christen starts.**  
Homer’s rate = 3 potatoes/min.  
Time before Christen joins = 4 min.  
Potatoes peeled by Homer = \(3 \times 4 = 12\).

**Step 2: Find the remaining potatoes after Homer’s initial work.**  
Total potatoes = 44.  
Remaining = \(44 - 12 = 32\).

**Step 3: Compute the combined peeling rate once Christen joins.**  
Homer’s rate = 3 potatoes/min.  
Christen’s rate = 5 potatoes/min.  
Combined rate = \(3 + 5 = 8\) potatoes/min.

**Step 4: Determine how long it takes the two of them to finish the 32 potatoes.**  
Time needed = \(\frac{32\text{ potatoes}}{8\text{ potatoes/min}} = 4\) minutes.

**Step 5: Calculate how many potatoes Christen peels during that 4‑minute period.**  
Christen’s rate = 5 potatoes/min.  
Potatoes peeled by Christen = \(5 \times 4 = 20\).

**Step 6: Compare with the answer choices.**  
- 20 matches choice A.  
- 24 (B) would require 4.8 minutes of Christen’s work, which is inconsistent.  
- 32 (C) would mean Christen peeled all remaining potatoes, ignoring Homer’s contribution.  
- 33 (D) and 40 (E) are too large for the total of 44 potatoes.

Thus the correct answer is **20**.

ANSWER: A

1. First, understand what the question is asking:
   - Homer starts peeling 44 potatoes at a rate of 3 potatoes per minute. After 4 minutes, Christen joins him and peels at a rate of 5 potatoes per minute. We need to find out how many potatoes Christen peeled when they finished the pile of 44 potatoes.
2. Calculate the number of potatoes Homer peeled in the first 4 minutes:
   - Homer's peeling rate is 3 potatoes per minute. In 4 minutes, using the formula $Number\ of\ potatoes = Rate\times Time$, the number of potatoes Homer peeled alone is $3\times4 = 12$ potatoes.
3. Calculate the number of remaining potatoes after Homer's initial peeling:
   - The total number of potatoes is 44. After Homer peeled 12 potatoes, the number of remaining potatoes is $44 - 12=32$ potatoes.
4. Calculate their combined peeling rate when Christen joins:
   - Homer peels at a rate of 3 potatoes per minute and Christen peels at a rate of 5 potatoes per minute. So their combined peeling rate is $3 + 5=8$ potatoes per minute.
5. Calculate the time it takes for them to peel the remaining potatoes together:
   - Using the formula $Time=\frac{Number\ of\ potatoes}{Rate}$, the time $t$ it takes for them to peel the remaining 32 potatoes at a rate of 8 potatoes per minute is $t=\frac{32}{8}=4$ minutes.
6. Calculate the number of potatoes Christen peeled:
   - Christen peels at a rate of 5 potatoes per minute and they worked together for 4 minutes. Using the formula $Number\ of\ potatoes = Rate\times Time$, the number of potatoes Christen peeled is $5\times4 = 20$ potatoes.
7. Rule out the wrong - choices:
   - Choice A: We calculated that Christen peeled 20 potatoes, so this choice is correct.
   - Choice B: 24 potatoes is not the number of potatoes Christen peeled. If we made a wrong calculation of the time or the rates, we might get this wrong answer. For example, if we thought the combined rate was different or miscalculated the time, but our step - by - step calculation shows it's 20.
   - Choice C: 32 is the number of remaining potatoes after Homer's initial 4 - minute peeling, not the number of potatoes Christen peeled.
   - Choice D: 33 is not related to any of our calculations for the number of potatoes Christen peeled.
   - Choice E: 40 is a number that is not consistent with our calculations based on the given rates and times.

ANSWER: A

### Step 1: Restate the problem and calculate Homer's solo work
First, Homer peels alone for 4 minutes at 3 potatoes per minute. We first find how many potatoes he peels in this solo period to find how many are left when Christen joins.
Homer's solo total: \(3 \text{ potatoes/min} \times 4 \text{ min} = 12 \text{ potatoes}\)
Remaining potatoes after 4 minutes: \(44 - 12 = 32 \text{ potatoes}\)

### Step 2: Calculate their combined work rate and time working together
Once Christen joins, they peel simultaneously. Their combined rate is the sum of their individual rates:
Combined rate: \(3 + 5 = 8 \text{ potatoes per minute}\)
Time needed to finish the remaining 32 potatoes: \(\frac{\text{remaining potatoes}}{\text{combined rate}} = \frac{32}{8} = 4 \text{ minutes}\). They work together for 4 minutes to finish all potatoes.

### Step 3: Calculate Christen's total peeled potatoes
Christen only works the 4 minutes they are together, at 5 potatoes per minute:
Christen's total: \(5 \times 4 = 20 \text{ potatoes}\)

### Step 4: Eliminate wrong choices
- B (24): This would require Christen to work 4.8 minutes, which is impossible given the exact total of potatoes.
- C (32): This would require Christen to work over 6 minutes, which does not align with the total time needed to finish all potatoes.
- D (33): This is not a multiple of Christen's 5 per minute rate, and is far too large.
- E (40): This would require Christen to work 8 full minutes, but Homer worked alone for the first 4 minutes, so Christen never works that long.

ANSWER: A

### Step 1: Restate the problem and identify given values
We have 44 total potatoes. Homer peels alone at 3 potatoes per minute for the first 4 minutes. Then Christen joins, peeling at 5 potatoes per minute, until all potatoes are peeled. We need to find how many potatoes Christen peels total.
---
### Step 2: Calculate potatoes peeled by Homer in the first 4 minutes
Homer's rate is 3 potatoes/minute, and he works alone for 4 minutes:
Potatoes peeled by Homer initially = \(3 \times 4 = 12\)
---
### Step 3: Find remaining potatoes after 4 minutes
Total potatoes minus what Homer already peeled:
Remaining potatoes = \(44 - 12 = 32\)
---
### Step 4: Calculate combined peeling rate after Christen joins
Homer continues at 3 potatoes/minute, Christen peels at 5 potatoes/minute. Their combined rate is:
Combined rate = \(3 + 5 = 8\) potatoes per minute
---
### Step 5: Find time taken to peel the remaining 32 potatoes
Time = total work / rate:
Time working together = \(32 \div 8 = 4\) minutes
---
### Step 6: Calculate Christen's total peeled potatoes
Christen works for the full 4 minutes she is present, at 5 potatoes per minute:
Christen's total = \(5 \times 4 = 20\)
---
### Step 7: Verify the result
Total peeled: Homer's total (\(3 \times (4+4) = 24\)) + Christen's total (20) = 44, which matches the total number of potatoes. This confirms the calculation is correct.
ANSWER: A

1. **Restate the problem**: We need to determine how many potatoes Christen peeled when Homer and Christen together finished peeling a pile of 44 potatoes. Homer started peeling at a rate of 3 potatoes per minute, and Christen joined him 4 minutes later, peeling at a rate of 5 potatoes per minute.

2. **Calculate the number of potatoes Homer peeled in the first 4 minutes**:
   \[
   \text{Potatoes peeled by Homer in 4 minutes} = 3 \text{ potatoes/minute} \times 4 \text{ minutes} = 12 \text{ potatoes}
   \]

3. **Determine the remaining potatoes after Homer peeled for 4 minutes**:
   \[
   \text{Remaining potatoes} = 44 \text{ potatoes} - 12 \text{ potatoes} = 32 \text{ potatoes}
   \]

4. **Calculate the combined peeling rate of Homer and Christen**:
   \[
   \text{Combined rate} = 3 \text{ potatoes/minute} + 5 \text{ potatoes/minute} = 8 \text{ potatoes/minute}
   \]

5. **Determine the time it took for Homer and Christen to peel the remaining 32 potatoes**:
   \[
   \text{Time to peel remaining potatoes} = \frac{32 \text{ potatoes}}{8 \text{ potatoes/minute}} = 4 \text{ minutes}
   \]

6. **Calculate the number of potatoes Christen peeled in those 4 minutes**:
   \[
   \text{Potatoes peeled by Christen} = 5 \text{ potatoes/minute} \times 4 \text{ minutes} = 20 \text{ potatoes}
   \]

7. **Verify and rule out incorrect choices**:
   - Choice B (24): This is incorrect because Christen peeled at a rate of 5 potatoes per minute for 4 minutes, which equals 20 potatoes.
   - Choice C (32): This is incorrect because this is the number of remaining potatoes, not the number Christen peeled.
   - Choice D (33): This is incorrect for the same reason as Choice C.
   - Choice E (40): This is incorrect because it exceeds the time Christen was peeling.

Thus, the correct number of potatoes Christen peeled is 20.

ANSWER: A

**Problem Restatement:**
We need to determine how many potatoes Christen peeled after joining Homer. Homer starts peeling 44 potatoes at a rate of 3 potatoes per minute. After 4 minutes, Christen joins and peels at a rate of 5 potatoes per minute. We must find out how many potatoes Christen peeled by the time all 44 potatoes are done.

---

**Step 1: Calculate how many potatoes Homer peeled alone before Christen joined.**
- Homer peels at 3 potatoes per minute.
- He works alone for 4 minutes.
- Potatoes peeled by Homer alone = 3 potatoes/min × 4 min = **12 potatoes**.

**Step 2: Determine how many potatoes remain when Christen joins.**
- Total potatoes = 44.
- Potatoes peeled by Homer alone = 12.
- Remaining potatoes = 44 − 12 = **32 potatoes**.

**Step 3: Calculate the combined peeling rate after Christen joins.**
- Homer’s rate = 3 potatoes/min.
- Christen’s rate = 5 potatoes/min.
- Combined rate = 3 + 5 = **8 potatoes/min**.

**Step 4: Find the time taken to peel the remaining 32 potatoes together.**
- Time = Remaining potatoes ÷ Combined rate = 32 ÷ 8 = **4 minutes**.

**Step 5: Calculate how many potatoes Christen peeled in that time.**
- Christen’s rate = 5 potatoes/min.
- Time worked together = 4 minutes.
- Potatoes peeled by Christen = 5 × 4 = **20 potatoes**.

---

**Eliminating Wrong Choices:**
- A (20): Matches our calculation.
- B (24): Too high; Christen only peeled 20.
- C (32): This is the total remaining potatoes, not Christen’s share.
- D (33): Impossible, as total potatoes are 44 and Homer peeled 12 + (3×4) = 24.
- E (40): Exceeds total potatoes.

---
ANSWER: A

# Solution

**Step 1: Restate what is being asked**
- The ratio of boys to girls is 2:3
- Total students = 30
- Find: How many MORE girls than boys are there?

**Step 2: Set up the math using the ratio**
- Let the number of boys = 2x
- Let the number of girls = 3x
- These must add up to 30 students

**Step 3: Solve for x**
$$2x + 3x = 30$$
$$5x = 30$$
$$x = 6$$

**Step 4: Find the actual number of boys and girls**
- Number of boys = 2x = 2(6) = 12
- Number of girls = 3x = 3(6) = 18

**Step 5: Verify**
- Check ratio: 12:18 = 2:3 ✓
- Check total: 12 + 18 = 30 ✓

**Step 6: Find the difference**
$$\text{Girls} - \text{Boys} = 18 - 12 = 6$$

**Step 7: Rule out wrong choices**
- A (10): Too large
- B (5): Would give 17 and 13, which don't maintain the 2:3 ratio
- C (3): Would give 16.5 and 13.5, not whole numbers
- D (6): ✓ Correct
- E (2): Too small

ANSWER: D

1. Restate the problem: We are asked to find how many more girls than boys are in a class of 30 students, given that the ratio of boys to girls is 2 : 3.  
2. Introduce a variable: Let the common ratio multiplier be k. Then the number of boys is 2k and the number of girls is 3k.  
3. Set up the total-students equation:  
   Total students = (number of boys) + (number of girls)  
   30 = 2k + 3k  
4. Solve for k:  
   2k + 3k = 5k, so 5k = 30, hence k = 30 ÷ 5 = 6.  
5. Compute actual numbers:  
   Number of boys = 2k = 2 × 6 = 12  
   Number of girls = 3k = 3 × 6 = 18  
6. Find the difference:  
   Girls − Boys = 18 − 12 = 6  
7. Eliminate wrong answer choices:  
   A. 10 → too large (we found 6)  
   B. 5  → too small (we found 6)  
   C. 3  → too small (we found 6)  
   E. 2  → too small (we found 6)  
   Only choice D matches our result of 6.  
ANSWER: D

To find the difference between the number of girls and boys in Mr. Brown's math class, we can follow these steps:

**Step 1: Restate what is being asked**
We are given that the ratio of boys to girls in a class is $2:3$ and the total number of students is $30$. We need to find how many more girls than boys are in the class (i.e., the number of girls minus the number of boys).

**Step 2: Set up the math**
Let the number of boys in the class be $2x$ and the number of girls be $3x$, where $x$ is a common multiplier. 
The total number of students is the sum of the boys and girls:
$$\text{Total students} = 2x + 3x = 5x$$

**Step 3: Solve for the multiplier $x$**
We are given that the total number of students is $30$. Therefore, we can set up the equation:
$$5x = 30$$
Dividing both sides by $5$, we find:
$$x = 6$$

**Step 4: Calculate the number of boys and girls**
Using $x = 6$:
* Number of boys: $2x = 2 \times 6 = 12$
* Number of girls: $3x = 3 \times 6 = 18$

**Step 5: Find the difference**
We want to find how many more girls than boys there are:
$$\text{Difference} = \text{Girls} - \text{Boys} = 18 - 12 = 6$$
*(Alternatively, the difference in ratio parts is $3x - 2x = x$. Since $x = 6$, the difference is directly $6$.)*

**Step 6: Evaluate the choices**
* A. 10 — Incorrect.
* B. 5 — Incorrect.
* C. 3 — Incorrect.
* D. 6 — Correct, as our calculated difference is 6.
* E. 2 — Incorrect.

ANSWER: D

1. Restate the query: The ratio of boys to girls is given as 2:3 with a total of 30 students; the task is to compute the exact numerical difference (girls minus boys) and match it to one of the five choices.

2. Set up the proportion: The ratio 2:3 means the class is divided into 2 + 3 = 5 equal parts. Divide the total enrollment by the number of parts to obtain the size of one part: 30 ÷ 5 = 6.

3. Compute the head-counts: Multiply the part size by the respective ratio coefficients to obtain the actual numbers—boys = 2 × 6 = 12, girls = 3 × 6 = 18. Subtract to find the required difference: 18 − 12 = 6.

4. Eliminate the distractors:  
   - Choice A (10) would result from mistakenly subtracting the ratio numbers from the total (30 − 20) or confusing parts with students.  
   - Choice B (5) equals the part size itself, not the gender difference.  
   - Choice C (3) is simply the girls’ ratio coefficient, not a count.  
   - Choice E (2) is the boys’ ratio coefficient, again not a count.  
   Only the computed difference of 6 matches an option, confirming D.

ANSWER: D

1. **Restate the problem**: We are given that the ratio of boys to girls in a class is 2:3 and that there are 30 students in total. We need to find the difference in the number of girls and boys.

2. **Set up the math**: Let's represent the number of boys as 2x and the number of girls as 3x, where x is a common multiplier for the ratio.

3. **Formulate the equation based on the total number of students**: The total number of students is the sum of boys and girls, so 2x + 3x = 30.

4. **Solve for x**: Combining like terms, we get 5x = 30. Dividing both sides by 5 gives x = 6.

5. **Calculate the number of boys and girls**: Substituting x = 6 into our expressions, the number of boys is 2x = 2 * 6 = 12, and the number of girls is 3x = 3 * 6 = 18.

6. **Find the difference between the number of girls and boys**: The difference is 18 - 12 = 6.

7. **Compare the result with the given choices**: We have found that there are 6 more girls than boys. Let's check the choices:
   - A. 10: Too high.
   - B. 5: Too low.
   - C. 3: Too low.
   - D. 6: Matches our calculation.
   - E. 2: Too low.

8. **Conclusion**: Since our calculation matches choice D, it is the correct answer.

ANSWER: D

**Step 1: Restate the problem**  
We are given a ratio of boys to girls (2 : 3) and the total number of students (30). We need to find how many more girls than boys are in the class.

**Step 2: Interpret the ratio**  
The ratio 2 : 3 means for every 2 boys, there are 3 girls. The total number of parts in the ratio is \(2 + 3 = 5\) parts.

**Step 3: Find the value of one part**  
Total students = 30, so each part represents \(30 \div 5 = 6\) students.

**Step 4: Calculate the number of boys and girls**  
Boys: \(2 \times 6 = 12\)  
Girls: \(3 \times 6 = 18\)

**Step 5: Find the difference**  
More girls than boys = \(18 - 12 = 6\)

**Step 6: Check the answer choices**  
A. 10 – too large, would correspond to a difference of 10 with total 30, which doesn't fit a 2:3 ratio.  
B. 5 – would give numbers like 12.5 and 17.5, not integers.  
C. 3 – too small, would give 13.5 and 16.5, not integers.  
D. 6 – matches our calculation.  
E. 2 – too small, would give 14 and 16, ratio 14:16 = 7:8, not 2:3.

ANSWER: D

1. **Understand the problem**:  
   We are told the ratio of boys to girls is 2 : 3, and the total number of students is 30. We are asked to find **how many more girls than boys** are in the class.

2. **Use the ratio to set up parts**:  
   The ratio 2 : 3 means that for every 2 boys, there are 3 girls.  
   So, the total number of "parts" in the class = 2 (boys) + 3 (girls) = **5 parts**.

3. **Find the value of one part**:  
   Since the whole class has 30 students, and those 30 students correspond to 5 parts,  
   then **1 part = 30 ÷ 5 = 6 students**.

4. **Find number of boys and girls**:  
   - Boys = 2 parts = 2 × 6 = **12**  
   - Girls = 3 parts = 3 × 6 = **18**

5. **Find how many more girls than boys**:  
   Difference = 18 − 12 = **6**

6. **Match with answer choices**:  
   Choice D is 6.

7. **Check other choices**:  
   - A (10): too big; total difference can’t exceed total students.  
   - B (5): close but not matching calculation.  
   - C (3) and E (2): too small.  
   Only D matches our result.

ANSWER: D

 **Step 1: Restate the problem.**  
We are told that the ratio of boys to girls is \(2:3\) and that the total number of students is 30. We need to find the actual difference between the number of girls and the number of boys.

**Step 2: Add the ratio parts to find the total number of “units.”**  
The ratio \(2:3\) means there are \(2\) parts boys and \(3\) parts girls.  
\[
2 + 3 = 5 \text{ total parts}
\]

**Step 3: Determine how many students each part represents.**  
Since the 5 parts correspond to all 30 students, divide the total enrollment by the total parts:
\[
30 \div 5 = 6 \text{ students per part}
\]

**Step 4: Calculate the actual number of boys and girls.**  
- Boys: \(2 \times 6 = 12\)  
- Girls: \(3 \times 6 = 18\)

**Step 5: Find how many more girls than boys there are.**  
\[
18 - 12 = 6
\]
(Shortcut: The difference between the ratio parts is \(3-2=1\) part, and \(1 \times 6 = 6\) students.)

**Step 6: Rule out the incorrect answer choices.**  
- **A. 10:** This would mean girls outnumber boys by 10, but \(10\) is one-third of the class, which does not fit a \(2:3\) split.  
- **B. 5:** This is simply the total number of ratio parts (\(2+3\)), not the actual difference in students.  
- **C. 3:** This is the girl ratio value, not a count of students.  
- **E. 2:** This is the boy ratio value, not a count of students.  
- **D. 6:** This matches our calculation exactly.

ANSWER: D

**Step 1: Restate the problem**
We are given a class with a total of 30 students where the ratio of boys to girls is $2:3$. We need to determine the specific number of girls and boys and then find the difference between these two quantities (how many more girls there are than boys).

**Step 2: Set up the math**
Let $x$ represent the multiplier for the ratio.
*   Number of boys = $2x$
*   Number of girls = $3x$

Since the total number of students is the sum of boys and girls, we can write the equation:
$$2x + 3x = 30$$

**Step 3: Perform the computation**
First, combine the terms with $x$:
$$5x = 30$$

Next, solve for $x$ by dividing both sides by 5:
$$x = \frac{30}{5}$$
$$x = 6$$

Now, substitute $x=6$ back into the expressions for the number of boys and girls:
*   **Number of boys** = $2 \times 6 = 12$
*   **Number of girls** = $3 \times 6 = 18$

Finally, calculate how many more girls than boys there are:
$$\text{Difference} = \text{Girls} - \text{Boys}$$
$$\text{Difference} = 18 - 12 = 6$$

**Step 4: Evaluate the choices**
*   **A. 10**: Incorrect. If the difference were 10, the numbers wouldn't sum to 30 with this ratio.
*   **B. 5**: Incorrect. This is the value of the multiplier $x$, or the size of one "part" of the total 5 parts, not the difference between girl and boy counts.
*   **C. 3**: Incorrect. This represents the difference in the ratio numbers ($3 - 2$) before scaling up to the actual student count.
*   **D. 6**: Correct. This matches our calculated result ($18 - 12 = 6$).
*   **E. 2**: Incorrect. This is simply the ratio number representing boys.

ANSWER: D

**Step 1 – Restate the problem**

The ratio of boys : girls in Mr. Brown’s class is \(2:3\).  
There are 30 students in total.  
We are asked: how many more girls than boys are there?

**Step 2 – Set up the ratio**

Let each “part” of the ratio correspond to the same number of students.  
If we let one part be \(x\), then  

\[
\text{boys} = 2x,\qquad \text{girls} = 3x.
\]

The total number of students is

\[
2x + 3x = 5x = 30.
\]

**Step 3 – Solve for \(x\)**

\[
x = \frac{30}{5} = 6.
\]

**Step 4 – Find the numbers of boys and girls**

\[
\text{boys} = 2x = 2 \times 6 = 12, \qquad
\text{girls} = 3x = 3 \times 6 = 18.
\]

**Step 5 – Compute the difference**

\[
\text{girls} - \text{boys} = 18 - 12 = 6.
\]

**Step 6 – Check the answer choices**

- A. 10 → too large (difference would require 35 total students).  
- B. 5 → would give 12 boys and 17 girls (total 29).  
- C. 3 → would give 12 boys and 15 girls (total 27).  
- D. 6 → matches our calculation.  
- E. 2 → would give 12 boys and 14 girls (total 26).

Only choice **D** matches the required difference.

**Answer**

ANSWER: D

1. **Restate the problem**: 
   - We know the ratio of boys to girls in Mr. Brown's math class is 2:3, and the total number of students in the class is 30. We need to find out how many more girls than boys are in the class.
2. **Set up the math**:
   - Let the number of boys be $2x$ and the number of girls be $3x$ based on the ratio $2:3$.
   - The total number of students in the class is the sum of the number of boys and the number of girls. So, we can write the equation $2x + 3x=30$.
   - Combining like - terms, we get $5x = 30$.
3. **Solve for $x$**:
   - Divide both sides of the equation $5x = 30$ by 5. We have $x=\frac{30}{5}=6$.
4. **Find the number of boys and girls**:
   - The number of boys is $2x$. Substituting $x = 6$, we get the number of boys $=2\times6 = 12$.
   - The number of girls is $3x$. Substituting $x = 6$, we get the number of girls $=3\times6 = 18$.
5. **Find the difference between the number of girls and boys**:
   - To find how many more girls than boys are in the class, we subtract the number of boys from the number of girls. So, $18−12 = 6$.
6. **Rule out wrong choices**:
   - Choice A: 10 is incorrect because our calculation shows the difference is 6, not 10.
   - Choice B: 5 is incorrect as it does not match our result of 6.
   - Choice C: 3 is incorrect since the actual difference is 6.
   - Choice E: 2 is also incorrect as it is not the correct difference between the number of girls and boys.

ANSWER: D

### Step 1: Restate the problem
We need to find how many more girls than boys are in a class of 30 students where the ratio of boys to girls is 2:3.

### Step 2: Break down the ratio into total parts
A 2:3 boys-to-girls ratio means the entire class can be split into equal, proportional parts: 2 parts for boys, 3 parts for girls. Adding these parts gives the total number of units representing the whole class:
$2 + 3 = 5$ total parts.

### Step 3: Calculate the size of one ratio part
All 5 parts sum to the 30 total students. To find how many actual students 1 ratio part represents:
$1\ \text{part} = \frac{30\ \text{total students}}{5\ \text{total parts}} = 6\ \text{students per part}$.

### Step 4: Find the actual number of boys and girls
- Number of boys = $2\ \text{parts} \times 6\ \text{students/part} = 12$ boys
- Number of girls = $3\ \text{parts} \times 6\ \text{students/part} = 18$ girls
Verify total students: $12 + 18 = 30$, which matches the given class size, and the ratio $12:18$ simplifies to 2:3, confirming the counts are correct.

### Step 5: Calculate the difference between girls and boys
$18\ \text{girls} - 12\ \text{boys} = 6$ more girls than boys.

### Step 6: Rule out incorrect options
- A. 10: Wrong, this would require 1 part to equal 10 students, which would need a total class size of 50, not 30.
- B. 5: Wrong, this comes from incorrectly adding the ratio parts to 6 instead of 5, leading to an incorrect part size of 5.
- C. 3: Wrong, this is only the number of ratio parts for girls, not the actual student difference.
- E. 2: Wrong, this is only the number of ratio parts for boys, not the actual student difference.

ANSWER: D

### Step 1: Restate the problem
We are given a class with a boy-to-girl ratio of 2:3, total 30 students, and need to find how many more girls there are than boys.
---
### Step 2: Set up the ratio math
A ratio of 2:3 (boys:girls) means the class is split into **2 + 3 = 5 equal parts total**, where 2 parts are boys and 3 parts are girls.
Let the size of 1 part = $x$. The total number of students equals 5 parts, :
$$5x = 30$$
---
### Step 3: Solve for the size of 1 part
Divide both sides by 5:
$$x = \frac{30}{5} = 6$$
Each part corresponds to 6 students.
---
### Step 4: Calculate number of boys and girls
- Number of boys = 2 parts = $2 \times 6 = 12$
- Number of girls = 3 parts = $3 \times 6 = 18$
Check: $12 + 18 = 30$, which matches the total class size, so this is correct.
---
### Step 5: Find the difference between girls and boys
Subtract the number of boys from the number of girls:
$$18 - 12 = 6$$
---
### Step 6: Rule out wrong choices
- A (10): Too large, the difference is not 10.
- B (5): Incorrect, 1 part is 6, not 5.
- C (3): Too small, this would be the difference in ratio parts, not actual students.
- E (2): Too small, this is the ratio part for boys, not the difference.
Only D (6) matches our calculation.
ANSWER: D

To determine how many more girls than boys are in Mr. Brown's math class, we need to follow these steps:

1. **Restate the problem**: We are given the ratio of boys to girls as 2:3 and the total number of students as 30. We need to find how many more girls than boys are in the class.

2. **Set up the math**:
   - Let the number of boys be \(2x\).
   - Let the number of girls be \(3x\).
   - According to the problem, the total number of students is 30. Therefore, we can write the equation:
     \[
     2x + 3x = 30
     \]
   - Simplify the equation:
     \[
     5x = 30
     \]
   - Solve for \(x\):
     \[
     x = \frac{30}{5} = 6
     \]

3. **Calculate the number of boys and girls**:
   - Number of boys:
     \[
     2x = 2 \times 6 = 12
     \]
   - Number of girls:
     \[
     3x = 3 \times 6 = 18
     \]

4. **Determine how many more girls than boys**:
   - The difference between the number of girls and boys is:
     \[
     18 - 12 = 6
     \]

5. **Rule out the wrong choices**:
   - Choice A: 10 (Incorrect, the difference is not 10)
   - Choice B: 5 (Incorrect, the difference is not 5)
   - Choice C: 3 (Incorrect, the difference is not 3)
   - Choice E: 2 (Incorrect, the difference is not 2)

6. **Final answer**:
   - The correct choice is D, as the difference between the number of girls and boys is 6.

ANSWER: D

**Problem Statement:**
We are given the ratio of boys to girls in Mr. Brown's math class as 2 : 3. The total number of students in the class is 30. We need to find how many more girls than boys are in the class.

---

**Step 1: Understand the Ratio**
The ratio of boys to girls is 2 : 3. This means for every 2 boys, there are 3 girls. Let’s denote the number of boys as \( 2x \) and the number of girls as \( 3x \), where \( x \) is a positive integer.

---

**Step 2: Set Up the Equation for Total Students**
The total number of students is the sum of boys and girls:
\[ 2x + 3x = 30 \]

---

**Step 3: Solve for \( x \)**
Combine like terms:
\[ 5x = 30 \]
Divide both sides by 5:
\[ x = 6 \]

---

**Step 4: Find the Number of Boys and Girls**
Now, substitute \( x = 6 \) back into the expressions for boys and girls:
- Number of boys: \( 2x = 2 \times 6 = 12 \)
- Number of girls: \( 3x = 3 \times 6 = 18 \)

---

**Step 5: Calculate the Difference Between Girls and Boys**
Subtract the number of boys from the number of girls:
\[ 18 - 12 = 6 \]

---

**Step 6: Compare with the Choices**
The difference is 6, which corresponds to choice **D**.

---

**Ruling Out Other Choices:**
- A (10): Incorrect, as the difference is not 10.
- B (5): Incorrect, as the difference is not 5.
- C (3): Incorrect, as the difference is not 3.
- E (2): Incorrect, as the difference is not 2.

---

ANSWER: D