Math worksheet featuring multiple-choice questions on inverse and direct variation.

Problem Analysis:

The provided image contains a series of multiple-choice questions related to inverse variation. Inverse variation is a mathematical relationship between two variables where their product is constant. If \( y \) varies inversely with \( x \), then:

\[
y = \frac{k}{x}
\]

where \( k \) is the constant of variation.

Let's solve each question step by step.

---

Question 1:

> Y varies inversely with \( x \), and \( y = 0.5 \) when \( x = 5 \). Which equation represents this situation?
>
> a. \( y = 2x \)
> b. \( y = \frac{2}{x} \)
> c. \( y = 4x \)
> d. \( y = \frac{x}{2} \)

#### Solution:
1. Since \( y \) varies inversely with \( x \), the relationship is:
\[
y = \frac{k}{x}
\]
2. Given \( y = 0.5 \) when \( x = 5 \), substitute these values into the equation to find \( k \):
\[
0.5 = \frac{k}{5}
\]
3. Solve for \( k \):
\[
k = 0.5 \times 5 = 2.5
\]
4. The equation becomes:
\[
y = \frac{2.5}{x}
\]
5. Simplify \( 2.5 \) as \( \frac{2}{x} \) (since \( 2.5 = \frac{5}{2} \)).

Thus, the correct equation is:
\[
\boxed{b}
\]

---

Question 2:

> The amount of people at your Thanksgiving meal and the amount of leftovers.
>
> a. Inverse Variation
> b. Direct Variation

#### Solution:
- In inverse variation, as one quantity increases, the other decreases.
- Here, if more people attend the Thanksgiving meal, there will be fewer leftovers. This is an example of inverse variation.

Thus, the correct answer is:
\[
\boxed{a}
\]

---

Question 3:

> Suppose \( y \) varies inversely with \( x \). If \( y = 150 \) when \( x = 2 \), find \( y \) when \( x = 25 \).

#### Solution:
1. Since \( y \) varies inversely with \( x \), the relationship is:
\[
y = \frac{k}{x}
\]
2. Given \( y = 150 \) when \( x = 2 \), substitute these values to find \( k \):
\[
150 = \frac{k}{2}
\]
3. Solve for \( k \):
\[
k = 150 \times 2 = 300
\]
4. Now, use the value of \( k \) to find \( y \) when \( x = 25 \):
\[
y = \frac{300}{25} = 12
\]

Thus, the correct answer is:
\[
\boxed{b}
\]

---

Question 4:

> The time to complete a project varies with the number of employees. If 3 people can complete the project in 7 days, how long will it take 5 people?

#### Solution:
1. Let \( T \) be the time to complete the project and \( n \) be the number of employees. Since the time to complete the project varies inversely with the number of employees:
\[
T \propto \frac{1}{n} \quad \text{or} \quad T = \frac{k}{n}
\]
2. Given \( T = 7 \) days when \( n = 3 \), substitute these values to find \( k \):
\[
7 = \frac{k}{3}
\]
3. Solve for \( k \):
\[
k = 7 \times 3 = 21
\]
4. Now, use the value of \( k \) to find \( T \) when \( n = 5 \):
\[
T = \frac{21}{5} = 4.2 \text{ days}
\]

Thus, the correct answer is:
\[
\boxed{a}
\]

---

Question 5:

> What kind of equation is this? \( xy = 8 \)
>
> a. Linear Equation
> b. Direct Variation
> c. Inverse Variation
> d. Joint Variation

#### Solution:
- The equation \( xy = 8 \) implies that the product of \( x \) and \( y \) is constant. This is the definition of inverse variation.

Thus, the correct answer is:
\[
\boxed{c}
\]

---

Question 6:

> Find the equation and solve for \( k \): \( y \) varies inversely as \( x \) and \( y = 6 \) when \( x = 18 \).

#### Solution:
1. Since \( y \) varies inversely with \( x \), the relationship is:
\[
y = \frac{k}{x}
\]
2. Given \( y = 6 \) when \( x = 18 \), substitute these values to find \( k \):
\[
6 = \frac{k}{18}
\]
3. Solve for \( k \):
\[
k = 6 \times 18 = 108
\]
4. The equation is:
\[
y = \frac{108}{x}
\]

Thus, the correct answer is:
\[
\boxed{c}
\]

---

Question 7:

> If \( y \) varies inversely as \( x \) and \( y = 10 \) when \( x = 2 \), find \( y \) when \( x = 10 \).

#### Solution:
1. Since \( y \) varies inversely with \( x \), the relationship is:
\[
y = \frac{k}{x}
\]
2. Given \( y = 10 \) when \( x = 2 \), substitute these values to find \( k \):
\[
10 = \frac{k}{2}
\]
3. Solve for \( k \):
\[
k = 10 \times 2 = 20
\]
4. Now, use the value of \( k \) to find \( y \) when \( x = 10 \):
\[
y = \frac{20}{10} = 2
\]

Thus, the correct answer is:
\[
\boxed{c}
\]

---

Question 8:

> If \( y \) varies inversely as \( x \) and \( y = 3 \) when \( x = 4 \), find \( y \) when \( x = 6 \).

#### Solution:
1. Since \( y \) varies inversely with \( x \), the relationship is:
\[
y = \frac{k}{x}
\]
2. Given \( y = 3 \) when \( x = 4 \), substitute these values to find \( k \):
\[
3 = \frac{k}{4}
\]
3. Solve for \( k \):
\[
k = 3 \times 4 = 12
\]
4. Now, use the value of \( k \) to find \( y \) when \( x = 6 \):
\[
y = \frac{12}{6} = 2
\]

Thus, the correct answer is:
\[
\boxed{d}
\]

---

Question 9:

> If \( y \) varies inversely as \( x \) and \( y = -2 \) when \( x = -8 \), find \( x \) when \( y = 2 \).

#### Solution:
1. Since \( y \) varies inversely with \( x \), the relationship is:
\[
y = \frac{k}{x}
\]
2. Given \( y = -2 \) when \( x = -8 \), substitute these values to find \( k \):
\[
-2 = \frac{k}{-8}
\]
3. Solve for \( k \):
\[
k = -2 \times (-8) = 16
\]
4. Now, use the value of \( k \) to find \( x \) when \( y = 2 \):
\[
2 = \frac{16}{x}
\]
5. Solve for \( x \):
\[
x = \frac{16}{2} = 8
\]

Thus, the correct answer is:
\[
\boxed{c}
\]

---

Question 10:

> If \( w \) varies inversely as \( y \) and \( w = 2 \) when \( y = 6 \), find \( w \) when \( y = 6 \).

#### Solution:
1. Since \( w \) varies inversely with \( y \), the relationship is:
\[
w = \frac{k}{y}
\]
2. Given \( w = 2 \) when \( y = 6 \), substitute these values to find \( k \):
\[
2 = \frac{k}{6}
\]
3. Solve for \( k \):
\[
k = 2 \times 6 = 12
\]
4. Now, use the value of \( k \) to find \( w \) when \( y = 6 \):
\[
w = \frac{12}{6} = 2
\]

Thus, the correct answer is:
\[
\boxed{d}
\]

---

Final Answers:

1. \( \boxed{b} \)
2. \( \boxed{a} \)
3. \( \boxed{b} \)
4. \( \boxed{a} \)
5. \( \boxed{c} \)
6. \( \boxed{c} \)
7. \( \boxed{c} \)
8. \( \boxed{d} \)
9. \( \boxed{c} \)
10. \( \boxed{d} \)