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# Properties Of HCF And LCM MCQs With Explanation-Sainik School Class 6 Math Study Material Notes free pdf download

Sainik School Entrance Exam for Class 6 Math Study Material Notes helps students face the competition in the current education system. In this case, the ANAND CLASSES is the best study tool to get a clear idea about the basics and gain a strong knowledge of the Sainik School Entrance Exam syllabus.

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Properties of HCF and LCM: For a better understanding of the concepts of LCM (Lowest Common Multiple) and HCF (Highest Common Factor), we need to recollect the terms multiples and factors. Let’s learn about LCM, HCF and the relation between HCF and LCM of natural numbers.

# Properties of HCF and LCM

### Property 1

The product of LCM and HCF of any two given natural numbers is equivalent to the product of the given numbers.

LCM × HCF = Product of two Numbers

Suppose A and B are two numbers, then.

LCM (A & B) × HCF (A & B) = A × B

Example: If 3 and 8 are two numbers.

LCM (3,8) = 24

HCF (3,8) = 1

LCM (3,8) x HCF (3,8) = 24 x 1 = 24

Also, 3 x 8 = 24

Hence, proved.

Note: This property is applicable for only two numbers.

## Example 1: Show that the LCM (6, 15) × HCF (6, 15) = Product(6, 15)

Solution: LCM and HCF of 6 and 15:

6 = 2 × 3

15 = 3 x 5

LCM of 6 and 15 = 30

HCF of 6 and 15 = 3

LCM (6, 15) × HCF (6, 15) = 30 × 3 = 90

Product of 6 and 15 = 6 × 15 = 90

Hence, LCM (6, 15) × HCF (6, 15)=Product(6, 15) = 90

### Property 2

HCF of co-prime numbers is 1. Therefore, the LCM of given co-prime numbers is equal to the product of the numbers.

LCM of Co-prime Numbers = Product Of The Numbers

Example: Let us take two coprime numbers, such as 21 and 22.

LCM of 21 and 22 = 462

Product of 21 and 22 = 462

LCM (21, 22) = 21 x 22

## LCM of given co-prime Numbers = Product of the given Numbers

Solution: LCM and HCF of 17 and 23:

17 = 1 x 7

23 = 1 x 23

LCM of 17 and 23 = 391

HCF of 17 and 23 = 1

Product of 17 and 23 = 17 × 23 = 391

Hence, LCM of co-prime numbers = Product of the numbers

### Property 3

H.C.F. and L.C.M. of Fractions:

LCM of fractions = LCM of Numerators / HCF of Denominators

HCF of fractions = HCF of Numerators / LCM of Denominators

Example: Let us take two fractions 4/9 and 6/21.

4 and 6 are the numerators & 9 and 12 are the denominators

LCM (4, 6) = 12

HCF (4, 6) = 2

LCM (9, 21) = 63

HCF (9, 21) = 3

Now as per the formula, we can write:

LCM (4/9, 6/21) = 12/3 = 4

HCF (4/9, 6/21) = 2/63

## Example 3: Find the LCM of the fractions 1 / 2 , 3 / 8, 3 / 4

Solution:

LCM of fractions = LCM of Numerators/HCF of Denominators

LCM of fractions = LCM (1,3,3)/HCF(2,8,4)=3/2

Example 4: Find the HCF of the fractions 3 / 5, 6 / 11, 9 / 20

HCF of fractions HCF of Numerators/LCM of Denominators

HCF of fractions = HCF (3,6,9)/LCM (5,11,20)=3/220

### Property 4

HCF of any two or more numbers is never greater than any of the given numbers.

Example: HCF of 4 and 8 is 4

Here, one number is 4 itself and another number 8 is greater than HCF (4, 8), i.e.,4.

### Property 5

LCM of any two or more numbers is never smaller than any of the given numbers.

Example: LCM of 4 and 8 is 8 which is not smaller to any of them.

## Example 1: Prove that: LCM (9 & 12) × HCF (9 & 12) = Product of 9 and 12

Solution:
9 = 3 × 3 = 3²
12 = 2 × 2 × 3 = 2² × 3
LCM of 9 and 12 = 2² × 3² = 4 × 9 = 36

HCF of 9 and 12 = 3

LCM (9 & 12) × HCF (9 & 12) = 36 × 3 = 108

Product of 9 and 12 = 9 × 12 = 108

Hence, LCM (9 & 12) × HCF (9 & 12) = 9 × 12 = 108. Proved.

## Example 2:8 and 9 are two co-prime numbers. Using these numbers verify, LCM of Co-prime Numbers = Product Of The Numbers.

Solution: LCM and HCF of 8 and 9:

8 = 2 × 2 × 2 = 2³

9 = 3 × 3 = 3²

LCM of 8 and 9 = 2³ × 3² = 8 × 9 = 72

HCF of 8 and 9 = 1

Product of 8 and 9 = 8 × 9 = 72

Hence, LCM of co-prime numbers = Product of the numbers. Therefore, verified.

## Example 3: Find the HCF of 12/25, 9/10, 18/35, 21/40.

Solution:
12 = 2 × 2 × 3
9 = 3 × 3
18 = 2 × 3 × 3
21 = 3 × 7
HCF (12, 9, 18, 21) = 3
25 = 5 × 5
10 = 2 × 5
35 = 5 × 7
40 = 2 × 2 × 2 × 5
LCM(25, 10, 35, 40) = 5 × 5 × 2 × 2 × 2 × 7 = 1400
The required HCF = HCF(12, 9, 18, 21)/LCM(25, 10, 35, 40) = 3/1400

## Example 4: If LCM and HCF of two numbers are 3 and 2 respectively, and one of the numbers is 6 then another number is?

Solution:

We know that LCM × HCF = a × b where a, and b are two numbers
I.e., 3 × 2 = 6 × b
Therefore, b = 1.

## (D) The sum of the HCF and LCM of two numbers is always equal to the sum of the two numbers.

Explanation: The HCF of two numbers is not always equal to 1. For example, the HCF of 2 and 3 is 1, but the HCF of 6 and 12 is 6.

## D) HCF(a, b) and LCM(a, b) have no specific relationship.

Explanation: B)

HCF(a, b) is always less than LCM(a, b). HCF is the largest common factor of two numbers, and LCM is the smallest common multiple. Since factors are smaller than multiples, the HCF is always less than the LCM.

## (D) The larger of the two numbers

Explanation: If the HCF of two numbers is 1, then the two numbers are relatively prime (have no common factors other than 1). The LCM of two relatively prime numbers is equal to the product of the two numbers.

## (D) The larger of the two numbers

Explanation: If the LCM of two numbers is equal to their product, then the two numbers must be equal. The HCF of two equal numbers is equal to 1.

## (A) xy (B) x + y (C) x – y (D) xy/2

Explanation: The product of two numbers is equal to their HCF times their LCM. Therefore, the product of the two numbers is xy.

## (A) 5 and 5 (B) 4 and 6 (C) 3 and 7 (D) 2 and 8

Explanation: If the HCF of two numbers is 1, then the two numbers are relatively prime (have no common factors other than 1). The only possible pair of relatively prime numbers that sum to 10 is 3 and 7.

## (A) 4 and 6 (B) 3 and 7 (C) 2 and 8 (D) 1 and 9

Explanation: The LCM of two numbers is the smallest number that is a multiple of both numbers. The only possible pair of numbers that have an LCM of 12 and a difference of 2 is 4 and 6.

## (A) 2 and 10 (B) 2 and 5 (C) 4 and 5 (D) 6 and 10

Explanation: The product of two numbers is equal to their HCF times their LCM. Therefore, the product of the two numbers is 2 * 10 = 20. The only possible pair of numbers that have a product of 20 and an HCF of 2 is 2 and 10.

## (A) 240 (B) 360 (C) 480 (D) 720

Explanation: The product of two numbers is equal to their HCF times their LCM. Therefore, the product of the three numbers is 2 * 120 = 240.

## (A) 240 (B) 360 (C) 480 (D) 720

Explanation: The product of three numbers is equal to their HCF times their LCM. Therefore, the greatest possible product of the three numbers is 3 * 120 = 360.

# Frequently Asked Questions (FAQs) related to studying for the Sainik School Class 6 Math entrance exam

To help you prepare for Sainik School Class 6 Math, it’s important to use appropriate study materials. Here are some frequently asked questions (FAQ) related to studying for the Sainik School Class 6 Math entrance exam:

1. What is the syllabus for the Sainik School Class 6 Math entrance exam?

• The syllabus for the entrance exam may vary slightly from one Sainik School to another. However, it generally covers topics from the standard Class 6 mathematics curriculum, including arithmetic, geometry, algebra, and basic mathematical concepts.

2. Where can I find official information about the exam pattern and syllabus?

• You can find official information about the exam pattern and syllabus on the official website of the specific Sainik School you’re applying to. Each school may have its own admission criteria.

3. Which textbooks should I use for Class 6 Math preparation?

• You should primarily use the NCERT Class 6 Math textbook. It covers the fundamental concepts and is widely accepted in Indian schools. Additionally, consider supplementary Math textbooks that are designed for competitive exams.

4. Are there any online resources for Sainik School Class 6 Math preparation?

• Yes, you can find online resources such as video tutorials, practice questions, and mock tests on educational websites. Websites like ANAND CLASSES offer free Math materials that can be helpful for your preparation.

5. Where can I get sample papers and previous year’s question papers?

• You can find sample papers and previous year’s question papers at online bookstore of ANAND CLASSES that sell competitive exam preparation materials. Additionally, ANAND CLASSES website offer downloadable PDFs of these papers for free or at a minimal cost.

6. Should I consider enrolling in coaching classes for Sainik School Math preparation?

• Enrolling in ANAND CLASSES is a personal choice. While they can provide structured guidance and additional practice, they are not mandatory. You can achieve success through self-study and the use of appropriate study materials.

7. How should I manage my study time effectively for Class 6 Math preparation?

• Create a study schedule that allocates specific time for Math preparation daily. Focus on understanding concepts, practicing regularly, and taking regular breaks to avoid burnout. Consistency is key.

8. Is there any specific advice for tackling the Math section of the Sainik School entrance exam?

• Pay close attention to the BODMAS rule (Order of Operations) and practice solving a variety of math problems. Make sure you’re familiar with the types of questions that are commonly asked in the entrance exam.

Remember to check the specific requirements and guidelines provided by the Sainik School you are applying to, as these may vary from school to school.

To prepare effectively for the Sainik School Class 6 Math entrance exam, you should consider the following general sources:

1. Sainik School Official Website: Visit the official website of the Sainik School you are applying to. They often provide information about the exam pattern, syllabus, and sample question papers.

2. NCERT Textbooks: The National Council of Educational Research and Training (NCERT) textbooks are widely used in Indian schools and are a valuable resource for exam preparation. Ensure you have the NCERT Math textbook for Class 6.

3. Solved Sample Papers and Previous Year Question Papers: You can find solved sample papers and previous year question papers for Sainik School entrance exams at bookstores or online at ANAND CLASSES website. These papers can give you an idea of the exam pattern and types of questions asked.

4. Math Study Guides: The Math study guides and reference books are publish under publication department of ANAND CLASSES and are designed to help students prepare for entrance exams. Look for books specifically tailored to the Sainik School entrance exam.

5. Online Resources: www.anandclasses.co.in

We at ANAND CLASSES are providing notes for Sainik School Entrance Exam students, mainly for subjects like Science and Maths. Scoring well in these major subjects will increase the possibility of getting into good SAINIK SCHOOL in the long run. The notes that we are offering have been thoughtfully prepared by our experts to help you achieve the same. These notes are designed to help students overcome all the challenges in solving math problems and understand difficult MATH concepts. Basically, these notes act as a valuable reference tool for conducting effective revisions of the entire chapters given in each subject. Additionally, students can use these notes to get detailed explanations, practice problems, and study properly without wasting much precious time.

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