This image displays the completed answer key for a 'Creating Equivalent Ratios' math worksheet, featuring 20 problems with solutions filled in red.

Problem Overview:

The task involves creating equivalent ratios by filling in the blanks. An equivalent ratio is formed when two ratios are proportional to each other. This means that the relationship between the numbers in one ratio is the same as the relationship between the numbers in another ratio.

Solution Approach:

To solve each problem, we will use the concept of proportionality. If two ratios \( a : b \) and \( c : d \) are equivalent, then:
\[
\frac{a}{b} = \frac{c}{d}
\]
This implies:
\[
a \cdot d = b \cdot c
\]

We will use this property to find the missing values in each ratio.

---

Detailed Solutions:

#### 1. \( 12 : \_\_ = 16 : 24 \)
- Given: \( 12 : x = 16 : 24 \)
- Using the proportionality rule:
\[
\frac{12}{x} = \frac{16}{24}
\]
- Cross-multiply:
\[
12 \cdot 24 = 16 \cdot x
\]
\[
288 = 16x
\]
- Solve for \( x \):
\[
x = \frac{288}{16} = 18
\]
- Answer: \( 18 \)

#### 2. \( 18 : 10 = \_\_ : 5 \)
- Given: \( 18 : 10 = x : 5 \)
- Using the proportionality rule:
\[
\frac{18}{10} = \frac{x}{5}
\]
- Cross-multiply:
\[
18 \cdot 5 = 10 \cdot x
\]
\[
90 = 10x
\]
- Solve for \( x \):
\[
x = \frac{90}{10} = 9
\]
- Answer: \( 9 \)

#### 3. \( 6 : 7 = \_\_ : 21 \)
- Given: \( 6 : 7 = x : 21 \)
- Using the proportionality rule:
\[
\frac{6}{7} = \frac{x}{21}
\]
- Cross-multiply:
\[
6 \cdot 21 = 7 \cdot x
\]
\[
126 = 7x
\]
- Solve for \( x \):
\[
x = \frac{126}{7} = 18
\]
- Answer: \( 18 \)

#### 4. \( 7 : 1 = \_\_ : 2 \)
- Given: \( 7 : 1 = x : 2 \)
- Using the proportionality rule:
\[
\frac{7}{1} = \frac{x}{2}
\]
- Cross-multiply:
\[
7 \cdot 2 = 1 \cdot x
\]
\[
14 = x
\]
- Answer: \( 14 \)

#### 5. \( 8 : 3 = \_\_ : 27 \)
- Given: \( 8 : 3 = x : 27 \)
- Using the proportionality rule:
\[
\frac{8}{3} = \frac{x}{27}
\]
- Cross-multiply:
\[
8 \cdot 27 = 3 \cdot x
\]
\[
216 = 3x
\]
- Solve for \( x \):
\[
x = \frac{216}{3} = 72
\]
- Answer: \( 72 \)

#### 6. \( \_\_ : 32 = 21 : 28 \)
- Given: \( x : 32 = 21 : 28 \)
- Using the proportionality rule:
\[
\frac{x}{32} = \frac{21}{28}
\]
- Simplify \( \frac{21}{28} \):
\[
\frac{21}{28} = \frac{3}{4}
\]
- So:
\[
\frac{x}{32} = \frac{3}{4}
\]
- Cross-multiply:
\[
x \cdot 4 = 32 \cdot 3
\]
\[
4x = 96
\]
- Solve for \( x \):
\[
x = \frac{96}{4} = 24
\]
- Answer: \( 24 \)

#### 7. \( \_\_ : 40 = 6 : 24 \)
- Given: \( x : 40 = 6 : 24 \)
- Using the proportionality rule:
\[
\frac{x}{40} = \frac{6}{24}
\]
- Simplify \( \frac{6}{24} \):
\[
\frac{6}{24} = \frac{1}{4}
\]
- So:
\[
\frac{x}{40} = \frac{1}{4}
\]
- Cross-multiply:
\[
x \cdot 4 = 40 \cdot 1
\]
\[
4x = 40
\]
- Solve for \( x \):
\[
x = \frac{40}{4} = 10
\]
- Answer: \( 10 \)

#### 8. \( 15 : \_\_ = 24 : 56 \)
- Given: \( 15 : x = 24 : 56 \)
- Using the proportionality rule:
\[
\frac{15}{x} = \frac{24}{56}
\]
- Simplify \( \frac{24}{56} \):
\[
\frac{24}{56} = \frac{3}{7}
\]
- So:
\[
\frac{15}{x} = \frac{3}{7}
\]
- Cross-multiply:
\[
15 \cdot 7 = 3 \cdot x
\]
\[
105 = 3x
\]
- Solve for \( x \):
\[
x = \frac{105}{3} = 35
\]
- Answer: \( 35 \)

#### 9. \( \_\_ : 5 = 16 : 20 \)
- Given: \( x : 5 = 16 : 20 \)
- Using the proportionality rule:
\[
\frac{x}{5} = \frac{16}{20}
\]
- Simplify \( \frac{16}{20} \):
\[
\frac{16}{20} = \frac{4}{5}
\]
- So:
\[
\frac{x}{5} = \frac{4}{5}
\]
- Cross-multiply:
\[
x \cdot 5 = 5 \cdot 4
\]
\[
5x = 20
\]
- Solve for \( x \):
\[
x = \frac{20}{5} = 4
\]
- Answer: \( 4 \)

#### 10. \( 16 : 32 = \_\_ : 4 \)
- Given: \( 16 : 32 = x : 4 \)
- Using the proportionality rule:
\[
\frac{16}{32} = \frac{x}{4}
\]
- Simplify \( \frac{16}{32} \):
\[
\frac{16}{32} = \frac{1}{2}
\]
- So:
\[
\frac{1}{2} = \frac{x}{4}
\]
- Cross-multiply:
\[
1 \cdot 4 = 2 \cdot x
\]
\[
4 = 2x
\]
- Solve for \( x \):
\[
x = \frac{4}{2} = 2
\]
- Answer: \( 2 \)

#### 11. \( \_\_ : 2 = 28 : 8 \)
- Given: \( x : 2 = 28 : 8 \)
- Using the proportionality rule:
\[
\frac{x}{2} = \frac{28}{8}
\]
- Simplify \( \frac{28}{8} \):
\[
\frac{28}{8} = \frac{7}{2}
\]
- So:
\[
\frac{x}{2} = \frac{7}{2}
\]
- Cross-multiply:
\[
x \cdot 2 = 2 \cdot 7
\]
\[
2x = 14
\]
- Solve for \( x \):
\[
x = \frac{14}{2} = 7
\]
- Answer: \( 7 \)

#### 12. \( 7 : 56 = \_\_ : 72 \)
- Given: \( 7 : 56 = x : 72 \)
- Using the proportionality rule:
\[
\frac{7}{56} = \frac{x}{72}
\]
- Simplify \( \frac{7}{56} \):
\[
\frac{7}{56} = \frac{1}{8}
\]
- So:
\[
\frac{1}{8} = \frac{x}{72}
\]
- Cross-multiply:
\[
1 \cdot 72 = 8 \cdot x
\]
\[
72 = 8x
\]
- Solve for \( x \):
\[
x = \frac{72}{8} = 9
\]
- Answer: \( 9 \)

#### 13. \( 8 : 32 = 1 : \_\_ \)
- Given: \( 8 : 32 = 1 : x \)
- Using the proportionality rule:
\[
\frac{8}{32} = \frac{1}{x}
\]
- Simplify \( \frac{8}{32} \):
\[
\frac{8}{32} = \frac{1}{4}
\]
- So:
\[
\frac{1}{4} = \frac{1}{x}
\]
- Cross-multiply:
\[
1 \cdot x = 4 \cdot 1
\]
\[
x = 4
\]
- Answer: \( 4 \)

#### 14. \( 5 : 3 = 45 : \_\_ \)
- Given: \( 5 : 3 = 45 : x \)
- Using the proportionality rule:
\[
\frac{5}{3} = \frac{45}{x}
\]
- Cross-multiply:
\[
5 \cdot x = 3 \cdot 45
\]
\[
5x = 135
\]
- Solve for \( x \):
\[
x = \frac{135}{5} = 27
\]
- Answer: \( 27 \)

#### 15. \( \_\_ : 20 = 8 : 5 \)
- Given: \( x : 20 = 8 : 5 \)
- Using the proportionality rule:
\[
\frac{x}{20} = \frac{8}{5}
\]
- Cross-multiply:
\[
x \cdot 5 = 20 \cdot 8
\]
\[
5x = 160
\]
- Solve for \( x \):
\[
x = \frac{160}{5} = 32
\]
- Answer: \( 32 \)

#### 16. \( 6 : \_\_ = 2 : 6 \)
- Given: \( 6 : x = 2 : 6 \)
- Using the proportionality rule:
\[
\frac{6}{x} = \frac{2}{6}
\]
- Simplify \( \frac{2}{6} \):
\[
\frac{2}{6} = \frac{1}{3}
\]
- So:
\[
\frac{6}{x} = \frac{1}{3}
\]
- Cross-multiply:
\[
6 \cdot 3 = 1 \cdot x
\]
\[
18 = x
\]
- Answer: \( 18 \)

#### 17. \( 8 : 24 = 1 : \_\_ \)
- Given: \( 8 : 24 = 1 : x \)
- Using the proportionality rule:
\[
\frac{8}{24} = \frac{1}{x}
\]
- Simplify \( \frac{8}{24} \):
\[
\frac{8}{24} = \frac{1}{3}
\]
- So:
\[
\frac{1}{3} = \frac{1}{x}
\]
- Cross-multiply:
\[
1 \cdot x = 3 \cdot 1
\]
\[
x = 3
\]
- Answer: \( 3 \)

#### 18. \( 3 : 6 = \_\_ : 30 \)
- Given: \( 3 : 6 = x : 30 \)
- Using the proportionality rule:
\[
\frac{3}{6} = \frac{x}{30}
\]
- Simplify \( \frac{3}{6} \):
\[
\frac{3}{6} = \frac{1}{2}
\]
- So:
\[
\frac{1}{2} = \frac{x}{30}
\]
- Cross-multiply:
\[
1 \cdot 30 = 2 \cdot x
\]
\[
30 = 2x
\]
- Solve for \( x \):
\[
x = \frac{30}{2} = 15
\]
- Answer: \( 15 \)

#### 19. \( \_\_ : 15 = 6 : 5 \)
- Given: \( x : 15 = 6 : 5 \)
- Using the proportionality rule:
\[
\frac{x}{15} = \frac{6}{5}
\]
- Cross-multiply:
\[
x \cdot 5 = 15 \cdot 6
\]
\[
5x = 90
\]
- Solve for \( x \):
\[
x = \frac{90}{5} = 18
\]
- Answer: \( 18 \)

#### 20. \( 72 : \_\_ = 16 : 18 \)
- Given: \( 72 : x = 16 : 18 \)
- Using the proportionality rule:
\[
\frac{72}{x} = \frac{16}{18}
\]
- Simplify \( \frac{16}{18} \):
\[
\frac{16}{18} = \frac{8}{9}
\]
- So:
\[
\frac{72}{x} = \frac{8}{9}
\]
- Cross-multiply:
\[
72 \cdot 9 = 8 \cdot x
\]
\[
648 = 8x
\]
- Solve for \( x \):
\[
x = \frac{648}{8} = 81
\]
- Answer: \( 81 \)

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Final Answer:

\[
\boxed{18, 9, 18, 14, 72, 24, 10, 35, 4, 2, 7, 9, 4, 27, 32, 18, 3, 15, 18, 81}
\]