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- Finding the LCD of Two Fractions
- Addition or Subtraction of Unit Fractions
- Addition or Subtraction of Fractions With Different Denominators
- Add or Subtract Fractions With Different Denominators Advanced
- Word Problem Involving Add or Subtract Fractions With Different Denominators
- Fractional Part of a Circle

# Finding the LCD of Two Fractions Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to **Finding the LCD of Two Fractions**. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using **Show Answer** button. You can use **Next Quiz** button to check new set of questions in the quiz.

### Answer : A

### Explanation

**Step 1:**

LCD of $\frac{3}{4}$, $\frac{7}{10}$ is LCM of 4 and 10

**Step 2:**

Prime factorization of 4 and 10 is

4 = 2 × 2; 10 = 2 × 5

**Step 3:**

LCD is the product of most occurrence of the prime factors

LCD = 2 × 2 × 5 = 20

### Answer : D

### Explanation

**Step 1:**

LCD of $\frac{5}{9}$, $\frac{8}{15}$ is LCM of 9 and 15

**Step 2:**

Prime factorization of 9 and 15 is

9 = 3 × 3; 15 = 3 × 5

**Step 3:**

LCD is the product of most occurrence of the prime factors

LCD = 3 × 3 × 5 = 45

### Answer : C

### Explanation

**Step 1:**

LCD of $\frac{3}{7}$, $\frac{4}{8}$ is LCM of 7 and 8

**Step 2:**

Prime factorization of 7 and 8 is

7 is prime; 8 = 2 × 2 × 2

**Step 3:**

LCD is the product of most occurrence of the prime factors

LCD = 2 × 2 × 2 × 7 = 56

### Answer : B

### Explanation

**Step 1:**

LCD of $\frac{3}{8}$, $\frac{5}{12}$ is LCM of 8 and 12

**Step 2:**

Prime factorization of 8 and 12 is

8 = 2 × 2 × 2; 12 = 2 × 2 × 3

**Step 3:**

LCD is the product of most occurrence of the prime factors

LCD = 2 × 2 × 2 × 3 = 24

### Answer : B

### Explanation

**Step 1:**

LCD of $\frac{7}{5}$, $\frac{5}{4}$ is LCM of 5 and 4

**Step 2:**

Prime factorization of 5 and 4 is

5 is prime; 4 = 2 × 2

**Step 3:**

LCD is the product of most occurrence of the prime factors

LCD = 2 × 2 × 5 = 20

### Answer : C

### Explanation

**Step 1:**

LCD of $\frac{4}{5}$, $\frac{7}{15}$ is LCM of 5 and 15

**Step 2:**

Prime factorization of 5 and 15 is

5 is prime; 15 = 3 × 5

**Step 3:**

LCD is the product of most occurrence of the prime factors

LCD = 3 × 5 = 15

### Answer : A

### Explanation

**Step 1:**

LCD of $\frac{1}{8}$, $\frac{5}{6}$ is LCM of 8 and 6

**Step 2:**

Prime factorization of 8 and 6 is

8 = 2 × 2 × 2; 15 = 2 × 3

**Step 3:**

LCD is the product of most occurrence of the prime factors

LCD = 2 × 2 × 2 × 3 = 24

### Answer : D

### Explanation

**Step 1:**

LCD of $\frac{3}{4}$, $\frac{2}{9}$ is LCM of 4 and 9

**Step 2:**

Prime factorization of 4 and 9 is

4 = 2 × 2; 9 = 3 × 3

**Step 3:**

LCD is the product of most occurrence of the prime factors

LCD = 2 × 2 × 3 × 3 = 36

### Answer : D

### Explanation

**Step 1:**

LCD of $\frac{3}{8}$, $\frac{7}{10}$ is LCM of 8 and 10

**Step 2:**

Prime factorization of 8 and 10 is

8 = 2 × 2 × 2; 10 = 2 × 5

**Step 3:**

LCD is the product of most occurrence of the prime factors

LCD = 2 × 2 × 2 × 5 = 40

### Answer : A

### Explanation

**Step 1:**

LCD of $\frac{5}{6}$, $\frac{8}{15}$ is LCM of 6 and 15

**Step 2:**

Prime factorization of 6 and 15 is

6 = 2 × 3; 15 = 3 × 5

**Step 3:**

LCD is the product of most occurrence of the prime factors

LCD = 2 × 3 × 5 = 30