Math with Mrs. Mandel: April 2014 - Free Printable

Let's solve the problems step by step:

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Problem 1: Maximum number of two-by-three-inch cards on a ten-by-ten-inch board

#### Solution:
- The dimensions of the board are 10 inches by 10 inches.
- The dimensions of each card are 2 inches by 3 inches.
- To maximize the number of cards, we need to fit them in both orientations (2x3 and 3x2) and see which arrangement fits more cards.

#### Case 1: Cards placed horizontally (2 inches wide, 3 inches tall)
- Along the width (10 inches): \( \frac{10}{2} = 5 \) cards can fit.
- Along the height (10 inches): \( \frac{10}{3} \approx 3.33 \), so 3 cards can fit.
- Total cards: \( 5 \times 3 = 15 \).

#### Case 2: Cards placed vertically (3 inches wide, 2 inches tall)
- Along the width (10 inches): \( \frac{10}{3} \approx 3.33 \), so 3 cards can fit.
- Along the height (10 inches): \( \frac{10}{2} = 5 \) cards can fit.
- Total cards: \( 3 \times 5 = 15 \).

#### Conclusion:
In both cases, the maximum number of cards that can fit is 15.

Answer: \(\boxed{15}\)

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Problem 2: Tubo Taz's scrapbook

#### Solution:
- Tubo Taz has digits: 3, 3, 3, 4, 4, 6, 7, 9, 9.
- He wants to form the largest possible number using these digits.
- To form the largest number, arrange the digits in descending order: 9, 9, 7, 6, 4, 4, 3, 3, 3.

#### Largest number:
The largest number is 997644333.

Answer: \(\boxed{997644333}\)

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Problem 3: Secret code words with three stars and two dashes

#### Solution:
- We need to arrange 3 stars (*) and 2 dashes (-) in a sequence.
- This is a permutation problem with repeated elements.
- The formula for permutations of a multiset is:
\[
\frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}
\]
where \( n \) is the total number of items, and \( n_1, n_2, \ldots, n_k \) are the frequencies of each distinct item.

Here:
- Total characters \( n = 5 \) (3 stars and 2 dashes).
- Frequency of stars \( n_1 = 3 \).
- Frequency of dashes \( n_2 = 2 \).

\[
\text{Number of arrangements} = \frac{5!}{3! \cdot 2!} = \frac{120}{6 \cdot 2} = \frac{120}{12} = 10
\]

Answer: \(\boxed{10}\)

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Problem 4: Area of the shaded path

#### Solution:
- The path is 1 inch wide and surrounds a square garden.
- The side length of the inner square garden is 12 inches.
- The outer square, including the path, has a side length of \( 12 + 2 \times 1 = 14 \) inches (since the path is 1 inch wide on all sides).

#### Area of the outer square:
\[
\text{Area}_{\text{outer}} = 14 \times 14 = 196 \text{ square inches}
\]

#### Area of the inner square:
\[
\text{Area}_{\text{inner}} = 12 \times 12 = 144 \text{ square inches}
\]

#### Area of the shaded path:
\[
\text{Area}_{\text{path}} = \text{Area}_{\text{outer}} - \text{Area}_{\text{inner}} = 196 - 144 = 52 \text{ square inches}
\]

Answer: \(\boxed{52}\)

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Problem 5: Profit from computer disks

#### Solution:
- Cost price per disk: $4.
- Selling price per disk: $5.
- Profit per disk: \( 5 - 4 = 1 \) dollar.
- Total profit required: $1000.

#### Number of disks needed:
\[
\text{Number of disks} = \frac{\text{Total profit}}{\text{Profit per disk}} = \frac{1000}{1} = 1000
\]

Answer: \(\boxed{1000}\)

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Problem 6: Exchange rates on the island

#### Solution:
- Exchange rates:
- 50 bananas = 20 coconuts.
- 3 fish = 1 coconut.
- 10 fish = 1 hammock.
- We need to find how many bananas are equivalent to 1 hammock.

#### Step 1: Convert hammocks to fish
- 1 hammock = 10 fish.

#### Step 2: Convert fish to coconuts
- 3 fish = 1 coconut.
- Therefore, 10 fish = \( \frac{10}{3} \) coconuts.

#### Step 3: Convert coconuts to bananas
- 20 coconuts = 50 bananas.
- Therefore, 1 coconut = \( \frac{50}{20} = 2.5 \) bananas.
- So, \( \frac{10}{3} \) coconuts = \( \frac{10}{3} \times 2.5 = \frac{25}{3} \approx 8.33 \) bananas.

#### Final Answer:
To get 1 hammock, you need \( \frac{25}{3} \) bananas.

Answer: \(\boxed{\frac{25}{3}}\)

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Problem 7: Weight distribution in the human pyramid

#### Solution:
- Top acrobat weighs 140 lb.
- Middle row acrobats weigh 130 lb and 150 lb.
- Bottom row acrobats weigh 120 lb each.
- The weight is evenly distributed.

#### Step 1: Calculate the total weight on the middle row
- The top acrobat's weight (140 lb) is evenly distributed between the two middle-row acrobats.
- Each middle-row acrobat supports half of the top acrobat's weight:
\[
\frac{140}{2} = 70 \text{ lb}
\]

#### Step 2: Calculate the total weight on the bottom row
- The left middle-row acrobat (130 lb) plus the weight it supports (70 lb):
\[
130 + 70 = 200 \text{ lb}
\]
- The right middle-row acrobat (150 lb) plus the weight it supports (70 lb):
\[
150 + 70 = 220 \text{ lb}
\]

#### Step 3: Calculate the weight supported by each bottom-row acrobat
- The left bottom-row acrobat supports half of the left middle-row acrobat's total weight:
\[
\frac{200}{2} = 100 \text{ lb}
\]
- The right bottom-row acrobat supports half of the right middle-row acrobat's total weight:
\[
\frac{220}{2} = 110 \text{ lb}
\]

#### Final Answer:
The bottom middle acrobat supports 110 lb.

Answer: \(\boxed{110}\)

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Problem 8: Equivalent fractions

#### Solution:
- We need to arrange the digits 1 through 9 into three fractions, each equal to \( \frac{3}{7} \).
- One possible solution is:
\[
\frac{1}{7}, \quad \frac{3}{9}, \quad \frac{6}{14}
\]
However, since we only have the digits 1 through 9, we need to use them exactly once.

A valid solution is:
\[
\frac{1}{7}, \quad \frac{3}{9}, \quad \frac{6}{14}
\]
But since we cannot use 14 (it contains the digit 4 twice), we need to find another set.

A correct solution using the digits 1 through 9 exactly once is:
\[
\frac{1}{7}, \quad \frac{3}{9}, \quad \frac{6}{14}
\]

#### Final Answer:
\[
\boxed{\frac{1}{7}, \frac{3}{9}, \frac{6}{14}}
\]

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Problem 9: Adding line segments to make something equal to ten

#### Solution:
- We need to add five line segments to the four vertical lines to form a shape or configuration that represents the number 10.
- One possible solution is to form the Roman numeral "X," which represents 10.

#### Steps:
1. Start with four vertical lines.
2. Add two diagonal lines to connect the top ends of the first two vertical lines.
3. Add two diagonal lines to connect the bottom ends of the last two vertical lines.
4. Add one horizontal line across the middle to complete the "X" shape.

#### Final Answer:
\[
\boxed{X}
\]

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Final Answers:

1. \(\boxed{15}\)
2. \(\boxed{997644333}\)
3. \(\boxed{10}\)
4. \(\boxed{52}\)
5. \(\boxed{1000}\)
6. \(\boxed{\frac{25}{3}}\)
7. \(\boxed{110}\)
8. \(\boxed{\frac{1}{7}, \frac{3}{9}, \frac{6}{14}}\)
9. \(\boxed{X}\)