Equivalent Fractions Worksheet #1 for practicing finding equivalent fractions.

Problem Overview:

The task is to find the missing numerators or denominators in each fraction pair so that the fractions are equivalent. Two fractions are equivalent if they represent the same value, which means their cross-products are equal.

Solution Approach:

For each problem, we will use the property of equivalent fractions:
\[
\frac{a}{b} = \frac{c}{d} \implies a \cdot d = b \cdot c
\]
We will solve for the missing value by setting up this equation and solving it step by step.

---

Detailed Solutions:

#### 1. $\frac{3}{4} = \frac{24}{32}$
- Here, both fractions are already given, and we can verify:
\[
3 \cdot 32 = 4 \cdot 24 \implies 96 = 96
\]
The fractions are equivalent. No action needed.

#### 2. $\frac{5}{3} = \frac{?}{15}$
- Let the missing numerator be $x$. Then:
\[
\frac{5}{3} = \frac{x}{15}
\]
Using the cross-multiplication property:
\[
5 \cdot 15 = 3 \cdot x \implies 75 = 3x \implies x = \frac{75}{3} = 25
\]
So, the missing numerator is $25$.

#### 3. $\frac{3}{8} = \frac{?}{24}$
- Let the missing numerator be $x$. Then:
\[
\frac{3}{8} = \frac{x}{24}
\]
Using the cross-multiplication property:
\[
3 \cdot 24 = 8 \cdot x \implies 72 = 8x \implies x = \frac{72}{8} = 9
\]
So, the missing numerator is $9$.

#### 4. $\frac{1}{8} = \frac{?}{56}$
- Let the missing numerator be $x$. Then:
\[
\frac{1}{8} = \frac{x}{56}
\]
Using the cross-multiplication property:
\[
1 \cdot 56 = 8 \cdot x \implies 56 = 8x \implies x = \frac{56}{8} = 7
\]
So, the missing numerator is $7$.

#### 5. $\frac{?}{4} = \frac{10}{20}$
- Let the missing numerator be $x$. Then:
\[
\frac{x}{4} = \frac{10}{20}
\]
Simplify $\frac{10}{20}$ to $\frac{1}{2}$:
\[
\frac{x}{4} = \frac{1}{2}
\]
Using the cross-multiplication property:
\[
x \cdot 2 = 4 \cdot 1 \implies 2x = 4 \implies x = \frac{4}{2} = 2
\]
So, the missing numerator is $2$.

#### 6. $\frac{?}{5} = \frac{24}{40}$
- Let the missing numerator be $x$. Then:
\[
\frac{x}{5} = \frac{24}{40}
\]
Simplify $\frac{24}{40}$ to $\frac{3}{5}$:
\[
\frac{x}{5} = \frac{3}{5}
\]
Since the denominators are the same, $x = 3$.
So, the missing numerator is $3$.

#### 7. $\frac{?}{8} = \frac{50}{80}$
- Let the missing numerator be $x$. Then:
\[
\frac{x}{8} = \frac{50}{80}
\]
Simplify $\frac{50}{80}$ to $\frac{5}{8}$:
\[
\frac{x}{8} = \frac{5}{8}
\]
Since the denominators are the same, $x = 5$.
So, the missing numerator is $5$.

#### 8. $\frac{?}{5} = \frac{27}{45}$
- Let the missing numerator be $x$. Then:
\[
\frac{x}{5} = \frac{27}{45}
\]
Simplify $\frac{27}{45}$ to $\frac{3}{5}$:
\[
\frac{x}{5} = \frac{3}{5}
\]
Since the denominators are the same, $x = 3$.
So, the missing numerator is $3$.

#### 9. $\frac{1}{3} = \frac{?}{6}$
- Let the missing numerator be $x$. Then:
\[
\frac{1}{3} = \frac{x}{6}
\]
Using the cross-multiplication property:
\[
1 \cdot 6 = 3 \cdot x \implies 6 = 3x \implies x = \frac{6}{3} = 2
\]
So, the missing numerator is $2$.

#### 10. $\frac{?}{8} = \frac{6}{16}$
- Let the missing numerator be $x$. Then:
\[
\frac{x}{8} = \frac{6}{16}
\]
Simplify $\frac{6}{16}$ to $\frac{3}{8}$:
\[
\frac{x}{8} = \frac{3}{8}
\]
Since the denominators are the same, $x = 3$.
So, the missing numerator is $3$.

#### 11. $\frac{6}{8} = \frac{?}{56}$
- Let the missing numerator be $x$. Then:
\[
\frac{6}{8} = \frac{x}{56}
\]
Simplify $\frac{6}{8}$ to $\frac{3}{4}$:
\[
\frac{3}{4} = \frac{x}{56}
\]
Using the cross-multiplication property:
\[
3 \cdot 56 = 4 \cdot x \implies 168 = 4x \implies x = \frac{168}{4} = 42
\]
So, the missing numerator is $42$.

#### 12. $\frac{?}{4} = \frac{8}{16}$
- Let the missing numerator be $x$. Then:
\[
\frac{x}{4} = \frac{8}{16}
\]
Simplify $\frac{8}{16}$ to $\frac{1}{2}$:
\[
\frac{x}{4} = \frac{1}{2}
\]
Using the cross-multiplication property:
\[
x \cdot 2 = 4 \cdot 1 \implies 2x = 4 \implies x = \frac{4}{2} = 2
\]
So, the missing numerator is $2$.

#### 13. $\frac{1}{6} = \frac{?}{36}$
- Let the missing numerator be $x$. Then:
\[
\frac{1}{6} = \frac{x}{36}
\]
Using the cross-multiplication property:
\[
1 \cdot 36 = 6 \cdot x \implies 36 = 6x \implies x = \frac{36}{6} = 6
\]
So, the missing numerator is $6$.

#### 14. $\frac{?}{3} = \frac{8}{12}$
- Let the missing numerator be $x$. Then:
\[
\frac{x}{3} = \frac{8}{12}
\]
Simplify $\frac{8}{12}$ to $\frac{2}{3}$:
\[
\frac{x}{3} = \frac{2}{3}
\]
Since the denominators are the same, $x = 2$.
So, the missing numerator is $2$.

#### 15. $\frac{1}{3} = \frac{?}{30}$
- Let the missing numerator be $x$. Then:
\[
\frac{1}{3} = \frac{x}{30}
\]
Using the cross-multiplication property:
\[
1 \cdot 30 = 3 \cdot x \implies 30 = 3x \implies x = \frac{30}{3} = 10
\]
So, the missing numerator is $10$.

#### 16. $\frac{3}{5} = \frac{?}{35}$
- Let the missing numerator be $x$. Then:
\[
\frac{3}{5} = \frac{x}{35}
\]
Using the cross-multiplication property:
\[
3 \cdot 35 = 5 \cdot x \implies 105 = 5x \implies x = \frac{105}{5} = 21
\]
So, the missing numerator is $21$.

#### 17. $\frac{?}{3} = \frac{10}{15}$
- Let the missing numerator be $x$. Then:
\[
\frac{x}{3} = \frac{10}{15}
\]
Simplify $\frac{10}{15}$ to $\frac{2}{3}$:
\[
\frac{x}{3} = \frac{2}{3}
\]
Since the denominators are the same, $x = 2$.
So, the missing numerator is $2$.

#### 18. $\frac{?}{8} = \frac{18}{24}$
- Let the missing numerator be $x$. Then:
\[
\frac{x}{8} = \frac{18}{24}
\]
Simplify $\frac{18}{24}$ to $\frac{3}{4}$:
\[
\frac{x}{8} = \frac{3}{4}
\]
Using the cross-multiplication property:
\[
x \cdot 4 = 8 \cdot 3 \implies 4x = 24 \implies x = \frac{24}{4} = 6
\]
So, the missing numerator is $6$.

#### 19. $\frac{?}{8} = \frac{54}{72}$
- Let the missing numerator be $x$. Then:
\[
\frac{x}{8} = \frac{54}{72}
\]
Simplify $\frac{54}{72}$ to $\frac{3}{4}$:
\[
\frac{x}{8} = \frac{3}{4}
\]
Using the cross-multiplication property:
\[
x \cdot 4 = 8 \cdot 3 \implies 4x = 24 \implies x = \frac{24}{4} = 6
\]
So, the missing numerator is $6$.

#### 20. $\frac{1}{4} = \frac{?}{36}$
- Let the missing numerator be $x$. Then:
\[
\frac{1}{4} = \frac{x}{36}
\]
Using the cross-multiplication property:
\[
1 \cdot 36 = 4 \cdot x \implies 36 = 4x \implies x = \frac{36}{4} = 9
\]
So, the missing numerator is $9$.

---

Final Answers:

\[
\boxed{
\begin{array}{ll}
2. & 25 \\
3. & 9 \\
4. & 7 \\
5. & 2 \\
6. & 3 \\
7. & 5 \\
8. & 3 \\
9. & 2 \\
10. & 3 \\
11. & 42 \\
12. & 2 \\
13. & 6 \\
14. & 2 \\
15. & 10 \\
16. & 21 \\
17. & 2 \\
18. & 6 \\
19. & 6 \\
20. & 9 \\
\end{array}
}
\]