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- Finding the LCD of Two Fractions
- Addition or Subtraction of Unit Fractions
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- 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
