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Free Mock Test on HCF & LCM | For All competitive Exams Preparation 2025

Yogi Siddeswara 0
STUDYECART HCF and LCM Quiz
60s
HCF and LCM Quiz

Test your understanding of highest common factors and least common multiples

🔢

25 Questions

With math formulas

⏱️

Timed

60s per question

🧮

Math Ready

Perfect formula rendering

Why HCF and LCM Matter for Competitive Exams

Mastering Highest Common Factor (HCF) and Least Common Multiple (LCM) concepts is crucial for success in competitive exams. These fundamental mathematical operations form the basis for solving complex problems involving:

  • Fraction simplification: $\frac{24}{36} = \frac{2}{3}$ (using HCF 12)
  • Time calculations: LCM of 15 and 20 is 60 (when will two events coincide?)
  • Ratio problems: Dividing $120 in ratio 3:4:5 (HCF helps verify parts)
  • Algebraic expressions: Factorizing $6x^2 + 9x$ to $3x(2x + 3)$

LCM Calculation Example

Find LCM of: $2^3 \times 3^2$ and $2^2 \times 3^3 \times 5$

Step 1: Identify highest powers of all primes: \[ \begin{aligned} 2^{\max(3,2)} &= 2^3 \\ 3^{\max(2,3)} &= 3^3 \\ 5^{\max(0,1)} &= 5^1 \end{aligned} \]
Step 2: Multiply these together: \[ \text{LCM} = 2^3 \times 3^3 \times 5 \]
Final Answer: $\text{LCM} = \boxed{1080}$

For more practice, check our Number System Guide.

Keep Practicing HCF and LCM

Regular practice with these concepts will help you solve problems faster in:

  • Quantitative aptitude sections of competitive exams
  • Algebraic simplifications and equation solving
  • Real-world problems like scheduling and resource allocation

Try solving these additional problems:

1. Find the HCF of $12x^2y^3$ and $18x^3y^2$

2. What is the smallest number divisible by 6, 8, and 9?

3. The HCF of two numbers is 15 and their LCM is 180. If one number is 45, find the other.

For more practice, check our Advanced Number Problems collection.

Disclaimer: These questions are for practice purposes only. While we strive for accuracy, please verify answers independently. The math rendering requires JavaScript to be enabled.

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