Ratio Tables - Worksheet | Teach Starter - Free Printable

To solve the ratio tables, we need to identify the relationship (ratio) between the numbers in each table and use that relationship to fill in the missing values. Let's go through each table step by step.

---

Table 1:

\[
\begin{array}{c|c|c|c}
1 & 3 & 6 & ? \\
\hline
3 & ? & ? & 24 \\
\end{array}
\]

#### Step 1: Identify the ratio.
- The first column shows \(1 \rightarrow 3\). This means the ratio is \(1:3\).
- To confirm, check the second column: \(3 \times 3 = 9\), so the missing value is \(9\).
- For the third column: \(6 \times 3 = 18\), so the missing value is \(18\).
- For the fourth column: The bottom row is \(24\), and since the ratio is \(1:3\), the top row must be \(24 \div 3 = 8\).

#### Final Table:
\[
\begin{array}{c|c|c|c}
1 & 3 & 6 & 8 \\
\hline
3 & 9 & 18 & 24 \\
\end{array}
\]

---

Table 2:

\[
\begin{array}{c|c|c|c}
11 & 33 & ? & ? \\
\hline
6 & 30 & 42 & ? \\
\end{array}
\]

#### Step 1: Identify the ratio.
- The first column shows \(11 \rightarrow 6\). This means the ratio is \(11:6\).
- To confirm, check the second column: \(33 \div 11 = 3\), and \(6 \times 3 = 18\), but the given value is \(30\). This suggests a different approach.
- Instead, notice that the second column shows \(33 \rightarrow 30\). The ratio here is \(33:30 = 11:10\).
- Using \(11:10\):
- For the third column: \(? \rightarrow 42\). If \(x \times 10 = 42\), then \(x = 42 \div 10 = 4.2\). But since we are dealing with whole numbers, let's recheck the pattern.
- The correct approach is to use the ratio \(11:6\):
- For the third column: \(11 \times 3 = 33\) (already given), so the next multiple is \(11 \times 4 = 44\).
- For the fourth column: \(6 \times 4 = 24\).

#### Final Table:
\[
\begin{array}{c|c|c|c}
11 & 33 & 44 & ? \\
\hline
6 & 30 & 42 & 24 \\
\end{array}
\]

---

Table 3:

\[
\begin{array}{c|c|c|c}
? & 21 & 35 & ? \\
\hline
8 & 40 & 72 & ? \\
\end{array}
\]

#### Step 1: Identify the ratio.
- The second column shows \(21 \rightarrow 40\). This suggests a ratio, but let's check the pattern.
- Notice that the bottom row increases by multiples of 8: \(8, 40, 72, \ldots\).
- The difference between \(40\) and \(8\) is \(32\), and between \(72\) and \(40\) is \(32\). So, the pattern is adding \(32\).
- For the first column: \(8 - 32 = -24\), but this doesn't fit. Instead, let's use the ratio approach.
- The ratio for the second column is \(21:40\). For the third column: \(35 \rightarrow 72\). The ratio is \(35:72\).
- Using the pattern, the next value in the bottom row is \(72 + 32 = 104\).

#### Final Table:
\[
\begin{array}{c|c|c|c}
? & 21 & 35 & ? \\
\hline
8 & 40 & 72 & 104 \\
\end{array}
\]

---

Table 4:

\[
\begin{array}{c|c|c|c}
3 & ? & 9 & 12 \\
\hline
7 & 14 & ? & ? \\
\end{array}
\]

#### Step 1: Identify the ratio.
- The first column shows \(3 \rightarrow 7\). This means the ratio is \(3:7\).
- For the second column: \(? \rightarrow 14\). If \(x \times 7 = 14\), then \(x = 2\).
- For the third column: \(9 \times 7 = 63\).
- For the fourth column: \(12 \times 7 = 84\).

#### Final Table:
\[
\begin{array}{c|c|c|c}
3 & 2 & 9 & 12 \\
\hline
7 & 14 & 63 & 84 \\
\end{array}
\]

---

Table 5:

\[
\begin{array}{c|c|c|c}
5 & 10 & 15 & 20 \\
\hline
4 & ? & ? & ? \\
\end{array}
\]

#### Step 1: Identify the ratio.
- The first column shows \(5 \rightarrow 4\). This means the ratio is \(5:4\).
- For the second column: \(10 \times 4 = 8\).
- For the third column: \(15 \times 4 = 12\).
- For the fourth column: \(20 \times 4 = 16\).

#### Final Table:
\[
\begin{array}{c|c|c|c}
5 & 10 & 15 & 20 \\
\hline
4 & 8 & 12 & 16 \\
\end{array}
\]

---

Table 6:

\[
\begin{array}{c|c|c|c}
1 & ? & 5 & ? \\
\hline
4 & 12 & ? & 40 \\
\end{array}
\]

#### Step 1: Identify the ratio.
- The first column shows \(1 \rightarrow 4\). This means the ratio is \(1:4\).
- For the second column: \(? \rightarrow 12\). If \(x \times 4 = 12\), then \(x = 3\).
- For the third column: \(5 \times 4 = 20\).
- For the fourth column: \(? \rightarrow 40\). If \(x \times 4 = 40\), then \(x = 10\).

#### Final Table:
\[
\begin{array}{c|c|c|c}
1 & 3 & 5 & 10 \\
\hline
4 & 12 & 20 & 40 \\
\end{array}
\]

---

Final Answer:

\[
\boxed{
\begin{array}{c|c|c|c}
1 & 3 & 6 & 8 \\
\hline
3 & 9 & 18 & 24 \\
\end{array}
\quad
\begin{array}{c|c|c|c}
11 & 33 & 44 & 55 \\
\hline
6 & 30 & 42 & 54 \\
\end{array}
\quad
\begin{array}{c|c|c|c}
? & 21 & 35 & ? \\
\hline
8 & 40 & 72 & 104 \\
\end{array}
\quad
\begin{array}{c|c|c|c}
3 & 2 & 9 & 12 \\
\hline
7 & 14 & 63 & 84 \\
\end{array}
\quad
\begin{array}{c|c|c|c}
5 & 10 & 15 & 20 \\
\hline
4 & 8 & 12 & 16 \\
\end{array}
\quad
\begin{array}{c|c|c|c}
1 & 3 & 5 & 10 \\
\hline
4 & 12 & 20 & 40 \\
\end{array}
}
\]