Linear Equations Worksheet-14 from Learners' Planet, designed to practice solving real-world problems involving scaling and exchange rates.

Let's solve each problem step by step.

---

Problem 26:

Ashley reduced the size of a triangle to a height of 1 in. What is the new width if it was originally 11 in tall and 44 in wide?

#### Solution:
When reducing the size of a shape proportionally, the ratio of the dimensions remains constant. The original height and width are 11 in and 44 in, respectively. The new height is 1 in. We need to find the new width.

The ratio of the original height to the new height is:
\[
\frac{\text{Original Height}}{\text{New Height}} = \frac{11}{1} = 11
\]

This means the new dimensions are \( \frac{1}{11} \) of the original dimensions. Therefore, the new width is:
\[
\text{New Width} = \frac{\text{Original Width}}{11} = \frac{44}{11} = 4 \text{ in}
\]

#### Answer:
\[
\boxed{B}
\]

---

Problem 27:

Natasha reduced the size of a rectangle to a height of 5 in. What is the new width if it was originally 24 in wide and 20 in tall?

#### Solution:
Again, we use the principle of proportional reduction. The original height and width are 20 in and 24 in, respectively. The new height is 5 in. We need to find the new width.

The ratio of the original height to the new height is:
\[
\frac{\text{Original Height}}{\text{New Height}} = \frac{20}{5} = 4
\]

This means the new dimensions are \( \frac{1}{4} \) of the original dimensions. Therefore, the new width is:
\[
\text{New Width} = \frac{\text{Original Width}}{4} = \frac{24}{4} = 6 \text{ in}
\]

#### Answer:
\[
\boxed{A}
\]

---

Problem 28:

Lekha enlarged the size of a rectangle to a width of 15 in. What is the new height if it was originally 6 in tall and 5 in wide?

#### Solution:
When enlarging a shape proportionally, the ratio of the dimensions remains constant. The original height and width are 6 in and 5 in, respectively. The new width is 15 in. We need to find the new height.

The ratio of the new width to the original width is:
\[
\frac{\text{New Width}}{\text{Original Width}} = \frac{15}{5} = 3
\]

This means the new dimensions are 3 times the original dimensions. Therefore, the new height is:
\[
\text{New Height} = \text{Original Height} \times 3 = 6 \times 3 = 18 \text{ in}
\]

#### Answer:
\[
\boxed{C}
\]

---

Problem 29:

Jasmine enlarged the size of a rectangle to a width of 18 in. What is the new height if it was originally 3 in tall and 2 in wide?

#### Solution:
Using the principle of proportional enlargement, the original height and width are 3 in and 2 in, respectively. The new width is 18 in. We need to find the new height.

The ratio of the new width to the original width is:
\[
\frac{\text{New Width}}{\text{Original Width}} = \frac{18}{2} = 9
\]

This means the new dimensions are 9 times the original dimensions. Therefore, the new height is:
\[
\text{New Height} = \text{Original Height} \times 9 = 3 \times 9 = 27 \text{ in}
\]

#### Answer:
\[
\boxed{B}
\]

---

Problem 30:

The money used in Switzerland is called the Franc. The exchange rate is $9 for every 11 Francs. Find how many dollars you would receive if you exchanged 33 Francs.

#### Solution:
The exchange rate is given as $9 for every 11 Francs. We need to find how many dollars are equivalent to 33 Francs.

First, determine the value of 1 Franc in dollars:
\[
\text{Value of 1 Franc} = \frac{9}{11} \text{ dollars}
\]

Now, calculate the value of 33 Francs:
\[
\text{Value of 33 Francs} = 33 \times \frac{9}{11} = 3 \times 9 = 27 \text{ dollars}
\]

#### Answer:
\[
\boxed{A}
\]

---

Problem 31:

The currency in Switzerland is the Franc. The exchange rate is approximately 6 Francs for every $5. At this rate, how many Francs would you get if you exchanged $30?

#### Solution:
The exchange rate is given as 6 Francs for every $5. We need to find how many Francs are equivalent to $30.

First, determine the number of Francs per dollar:
\[
\text{Francs per dollar} = \frac{6}{5}
\]

Now, calculate the number of Francs for $30:
\[
\text{Number of Francs} = 30 \times \frac{6}{5} = 6 \times 6 = 36 \text{ Francs}
\]

#### Answer:
\[
\boxed{B}
\]

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Final Answers:

\[
\boxed{B, A, C, B, A, B}
\]