Worksheet for practicing equivalent ratios with structured tables.

To solve the problem of completing the equivalent ratio tables, we need to identify the relationships between the numbers in each table and use those relationships to fill in the missing values. Let's go through each table step by step.

---

Table 1:

```
| 2 | | | 18
-------------------
3 | 6 | 9 | |
```
- The relationship is that each number in the top row is multiplied by the corresponding number in the bottom row.
- For the first column: \( 2 \times 3 = 6 \)
- For the second column: \( 2 \times \frac{9}{2} = 9 \) (so the multiplier is \( \frac{9}{2} \))
- For the third column: \( 2 \times \frac{18}{2} = 18 \) (so the multiplier is 9)

Filling in the missing values:
```
| 2 | 3 | 6 | 18
-------------------
3 | 6 | 9 | 18| 54
```

---

Table 2:

```
| 10 | | 15 | 18
----------------------
| | 80 | | 72
```
- The relationship is similar: each number in the top row is multiplied by a constant to get the corresponding number in the bottom row.
- For the second column: \( 10 \times 8 = 80 \), so the multiplier is 8.
- For the third column: \( 15 \times 8 = 120 \)
- For the fourth column: \( 18 \times 8 = 144 \)

Filling in the missing values:
```
| 10 | 12 | 15 | 18
----------------------
| 80 | 96 | 120| 144
```

---

Table 3:

```
| | 12 | 100
-------------------
20 | 15 | 60 |
```
- The relationship is that each number in the top row is multiplied by a constant to get the corresponding number in the bottom row.
- For the first column: \( 20 \times \frac{15}{20} = 15 \), so the multiplier is \( \frac{3}{4} \).
- For the second column: \( 12 \times \frac{3}{4} = 9 \)
- For the third column: \( 100 \times \frac{3}{4} = 75 \)

Filling in the missing values:
```
| 16 | 12 | 100
-------------------
20 | 15 | 9 | 75
```

---

Table 4:

```
| | | | 10
-------------------
30 | 6 | 18 | 20
```
- The relationship is that each number in the top row is divided by the corresponding number in the bottom row.
- For the first column: \( 30 \div 6 = 5 \)
- For the second column: \( 30 \div 18 = \frac{5}{3} \approx 1.67 \)
- For the third column: \( 30 \div 20 = 1.5 \)

Filling in the missing values:
```
| 5 | 1.67 | 1.5 | 10
-------------------
30 | 6 | 18 | 20 | 3
```

---

Table 5:

```
| 1 | 9 | | 8
-------------------
5 | 5 | 50 |
```
- The relationship is that each number in the top row is multiplied by a constant to get the corresponding number in the bottom row.
- For the first column: \( 1 \times 5 = 5 \), so the multiplier is 5.
- For the second column: \( 9 \times 5 = 45 \)
- For the third column: \( 8 \times 5 = 40 \)

Filling in the missing values:
```
| 1 | 9 | 10 | 8
-------------------
5 | 5 | 45| 50 | 40
```

---

Table 6:

```
| 7 | | 20 | 10
-------------------
| | 45 | 60 |
```
- The relationship is that each number in the top row is multiplied by a constant to get the corresponding number in the bottom row.
- For the second column: \( 7 \times \frac{45}{7} = 45 \), so the multiplier is \( \frac{45}{7} \).
- For the third column: \( 20 \times \frac{45}{7} = \frac{900}{7} \approx 128.57 \)
- For the fourth column: \( 10 \times \frac{45}{7} = \frac{450}{7} \approx 64.29 \)

Filling in the missing values:
```
| 7 | 7 | 20 | 10
-------------------
| 45| 45| 128.57| 64.29
```

---

Table 7:

```
| 5 | | 20 | 12
-------------------
15 | 30| |
```
- The relationship is that each number in the top row is multiplied by a constant to get the corresponding number in the bottom row.
- For the first column: \( 5 \times 3 = 15 \), so the multiplier is 3.
- For the second column: \( 20 \times 3 = 60 \)
- For the third column: \( 12 \times 3 = 36 \)

Filling in the missing values:
```
| 5 | 10 | 20 | 12
-------------------
15 | 30| 30 | 60 | 36
```

---

Table 8:

```
| | 40 | 9 |
-------------------
| 20| 80 | | 40
```
- The relationship is that each number in the top row is multiplied by a constant to get the corresponding number in the bottom row.
- For the first column: \( 20 \div 2 = 10 \), so the multiplier is 2.
- For the second column: \( 40 \div 2 = 20 \)
- For the third column: \( 9 \div 2 = 4.5 \)

Filling in the missing values:
```
| 10 | 40 | 9 | 20
-------------------
| 20 | 80 | 4.5| 40
```

---

Table 9:

```
| | | 20 | 10
-------------------
90 | 48| 60 |
```
- The relationship is that each number in the top row is multiplied by a constant to get the corresponding number in the bottom row.
- For the first column: \( 90 \div 48 = \frac{15}{8} \), so the multiplier is \( \frac{15}{8} \).
- For the second column: \( 20 \times \frac{15}{8} = \frac{300}{8} = 37.5 \)
- For the third column: \( 10 \times \frac{15}{8} = \frac{150}{8} = 18.75 \)

Filling in the missing values:
```
| 48 | 37.5 | 20 | 10
-------------------
90 | 48 | 60 | 37.5| 18.75
```

---

Table 10:

```
| 4 | | | 5
-------------------
24 | 18| 60 |
```
- The relationship is that each number in the top row is multiplied by a constant to get the corresponding number in the bottom row.
- For the first column: \( 4 \times 6 = 24 \), so the multiplier is 6.
- For the second column: \( 18 \div 6 = 3 \)
- For the third column: \( 60 \div 6 = 10 \)

Filling in the missing values:
```
| 4 | 3 | 10 | 5
-------------------
24 | 18| 18| 60 | 30
```

---

Table 11:

```
| | 40 | 9 |
-------------------
50 | | 90 | 300
```
- The relationship is that each number in the top row is multiplied by a constant to get the corresponding number in the bottom row.
- For the second column: \( 40 \times \frac{90}{40} = 90 \), so the multiplier is \( \frac{9}{4} \).
- For the third column: \( 9 \times \frac{9}{4} = \frac{81}{4} = 20.25 \)

Filling in the missing values:
```
| 20 | 40 | 9 | 133.33
-------------------
50 | 90 | 90 | 20.25| 300
```

---

Final Answer:

\boxed{
\begin{array}{c|c|c|c}
\text{Table 1} & \text{Table 2} & \text{Table 3} & \text{Table 4} \\
\hline
\begin{array}{c|c|c|c}
& 2 & 3 & 6 & 18 \\
\hline
3 & 6 & 9 & 18 & 54 \\
\end{array} &
\begin{array}{c|c|c|c}
& 10 & 12 & 15 & 18 \\
\hline
& 80 & 96 & 120 & 144 \\
\end{array} &
\begin{array}{c|c|c|c}
& 16 & 12 & 100 \\
\hline
20 & 15 & 9 & 75 \\
\end{array} &
\begin{array}{c|c|c|c}
& 5 & 1.67 & 1.5 & 10 \\
\hline
30 & 6 & 18 & 20 & 3 \\
\end{array} \\
\hline
\text{Table 5} & \text{Table 6} & \text{Table 7} & \text{Table 8} \\
\hline
\begin{array}{c|c|c|c}
& 1 & 9 & 10 & 8 \\
\hline
5 & 5 & 45 & 50 & 40 \\
\end{array} &
\begin{array}{c|c|c|c}
& 7 & 7 & 20 & 10 \\
\hline
& 45 & 45 & 128.57 & 64.29 \\
\end{array} &
\begin{array}{c|c|c|c}
& 5 & 10 & 20 & 12 \\
\hline
15 & 30 & 30 & 60 & 36 \\
\end{array} &
\begin{array}{c|c|c|c}
& 10 & 40 & 9 & 20 \\
\hline
& 20 & 80 & 4.5 & 40 \\
\end{array} \\
\hline
\text{Table 9} & \text{Table 10} & \text{Table 11} & \\
\hline
\begin{array}{c|c|c|c}
& 48 & 37.5 & 20 & 10 \\
\hline
90 & 48 & 60 & 37.5 & 18.75 \\
\end{array} &
\begin{array}{c|c|c|c}
& 4 & 3 & 10 & 5 \\
\hline
24 & 18 & 18 & 60 & 30 \\
\end{array} &
\begin{array}{c|c|c|c}
& 20 & 40 & 9 & 133.33 \\
\hline
50 & 90 & 90 & 20.25 & 300 \\
\end{array} &
\end{array}
}