HCF (Highest Common Factor) |
HCF Highest Common Factor-Sainik School Class 6 Math Study Material Notes free pdf download
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Highest Common Factor (HCF), also known as the Greatest Common Divisor (GCD). HCF is the largest number that can exactly divide two or more given numbers without leaving a remainder. Let’s go through a few HCF problems with explanations:
HCF By Prime Factorization Method
Follow the below-given steps to find the HCF of numbers using the prime factorisation method.
- Step 1: Write each number as a product of its prime factors. This method is called here prime factorization.
- Step 2: Now list the common factors of both the numbers
- Step 3: The product of all common prime factors is the HCF ( use the lower power of each common factor)
Let us understand with the help of examples.
Example 1:Evaluate the HCF of 60 and 75.
Solution:
| Write each number as a product of its prime factors.22 x 3 x 5 = 60 3 x 52 = 75 The product of all common prime factors is the HCF. The common prime factors in this example are 3 & 5. The lowest power of 3 is 3 and 5 is 5. So, HCF = 3 x 5 = 15 |
Example 2: Find the HCF of 36, 24 and 12.
Solution:
| Write each number as a product of its prime factors. 22 x 32 = 36 23 x 3 = 24 22 x 3 = 12 The product of all common prime factors is the HCF ( use the lowest power of each common factor) The common prime factors in this example are 2 & 3. The lowest power of 2 is 22 and 3 is 3. So, HCF = 22 x 3 = 12 |
Example 3: Find the HCF of 36, 27 and 80.
Solution:
| Write each number as a product of its prime factors. 22 x 32 = 36 33 = 27 24 x 5 = 80 The product of all common prime factors is the HCF The common prime factors in this example are none. So, HCF is 1. |
HCF By Division Method
You have understood by now the method of finding the highest common factor using prime factorization. Now let us learn here to find HCF using the division method. Basically, the division method is nothing but dividing the given numbers, simultaneously, to get the common factors between them. Follow the steps mentioned below to solve the problems of HCF.
- Step 1: Write the given numbers horizontally, in a sequence, by separating them with commas.
- Step 2: Find the smallest prime number which can divide the given number. It should exactly divide the given numbers. (Write on the left side).
- Step 3: Now write the quotients.
- Step 4: Repeat the process, until you reach the stage, where there is no coprime number left.
- Step 5: We will get the common prime factors as the factors on the left-hand side divides all the numbers exactly. The product of these common prime factors is the HCF of the given numbers.
Let us understand the above-mentioned steps to find the HCF by division method with the help of examples.
Example 1: Evaluate the HCF of 30 and 75
Solution:
As we can note that the mentioned prime factors, on the left side, divide all the numbers exactly. So, they all are common prime factors. We have no common prime factor for the numbers remaining at the bottom.
So, HCF = 3 × 5 = 15.
Example 2:Find out HCF of 36, and 24
Solution:
HCF = 2 × 2 × 3 = 12.
Example 3:Find out HCF of 36, 12, 24 and 48.
Solution:
HCF of Prime Numbers
HCF of any two or prime numbers is always equal to 1.
Why?
Because prime numbers are those numbers that have only two factors, either 1 or the number itself. So, if there is no other factor, then the HCF of such numbers will be equal to 1. Find the examples below:
- HCF (2, 3) = 1
- HCF (3, 7) = 1
- HCF (5, 13) = 1
- HCF (19, 23) = 1
HCF by Shortcut method
Steps to find the HCF of any given set of numbers.
- Step 1: Divide the larger number by the smaller number first, such as;
Larger Number/Smaller Number
- Step 2: Divide the divisor of step 1 by the remainder left.
Divisor of step 1/Remainder
- Step 3: Again divide the divisor of step 2 by the remainder.
Divisor of step 2/Remainder
- Step 4: Repeat the process until the remainder is zero.
- Step 5: The divisor of the last step is the HCF.
How to find the HCF of Three numbers
Find the steps to calculate the HCF of the 3 numbers given to us.
The above steps can also be used to find the HCF of more than 3 numbers.
HCF of Four Numbers
The simple steps to find the HCF of 4 numbers are:
- First, divide the four numbers into pairs of two
- Find the HCF of pair of two numbers
- Then, find the HCF of HCFs calculated in the previous step
- The final HCF will be the required HCF of 4 numbers
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HCF MCQs : Video Lecture L-2
Example 1 : Highest common factor of 18 and 24
Solution:
24 > 18
So, dividend = 24 and divisor = 18
Let’s perform the division as explained in the above steps.
Therefore, the highest common factor of 18 and 24 is 6.
Example : 2 What is the HCF of 120 and 100.
Solution:
Now, let us find the HCF by using division method.
Divide 120 by 100.
120/100 → 1 and remainder is 20
Now, divide the first divisor 100 by first remainder 20.
100/20 → 5 and remainder is 0.
Therefore, 20 is the HCF of 120 and 100.
Example : 3 Find the HCF of 45 and 60 by the division method.
Solution: Divide 60 by 45.
60/45 → 1 and remainder is 15
Now, divide 45 by 15
45/15 → 3
Therefore, 15 is the HCF of 45 and 60.
Example : 4 Find the HCF of 126, 162 and 180.
Solution: By prime factorisation, we can write the given numbers as;
126 = 2x3x3x7
162 = 2x3x3x3x3
180 = 2x2x3x3x5
Taking out the common factors of 126, 162 and 180, we get:
HCF(126,162,180) = 2x3x3 = 18
Example : 5 What is the highest common factor of 96 and 404?
Solution:
Prime factorization of 96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3
Prime factorization of 404 = 2 × 2 × 101 = 22 × 101
HCF(96, 404) = 22 = 4
Therefore, the highest common factor of 96 and 404 is 4.
Example : 6 What will be the greatest possible length that can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm?
Solution:
7m = 7 × 100 cm = 700 cm
3m 85 cm = (3 × 100) cm + 85 cm = (300 + 85) cm = 385 cm
12 m 95 cm = (12 × 100) cm + 95 cm = (1200 + 95) cm = 1295 cm
Prime factorization of 700 = 2 x 2 x 5 x 5 x 7
Prime factorization of 385 = 5 x 7 x 11
Prime factorization of 1295 = 5 x 7 x 37
HCF(700, 385, 1295) = 5 x 7 = 35
Therefore, the greatest possible length that can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is 35 cm.
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Example : 7 Find the numbers if the HCF of two numbers is 29 and their sum is 174.
Solution:
Given that the HCF of two numbers is 29.
Let 29a and 29b be the two required numbers.
According to the given,
29a + 29b = 174
29(a + b) = 174
a + b = 174/29 = 6
The pair of values of co-primes with sum 6 is (1, 5).
So, the possible numbers are:
29 x 1 = 29
29 x 5 = 145
Verification:
Sum of numbers = 29 + 145 = 174
Hence, the required numbers are 29 and 145.
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HCF of Fractions:
Example : 1 Find the HCF of fractions 2/3, 16/81, and 8/9.
Solution:
Given fractions are:
2/3, 16/81, 8/9
We know that,
HCF of fractions = HCF of numerators/LCM of denominators
Here,
Prime factorization of 2 = 2
Prime factorization of 16 = 24
Prime factorization of 8 = 23
HCF(2, 16, 8) = 2
Prime factorization of 3 = 3
Prime factorization of 81 = 34
Prime factorization of 9 = 32
LCM(3, 81, 9) = 34= 81
Therefore, the required HCF = 2/81
Example : 2 What will be the HCF of 0.63 and 1.05?
Solution:
Given numbers: 0.63 and 1.05
To find the HCF of decimal numbers, first, we need to calculate the HCF of those numbers by removing the decimal point.
So,
Prime factorization of 63 = 3 x 3 x 7 = 32 x 7
Prime factorization of 105 = 3 x 5 x 7
HCF of 63 and 105 = 3 x 7 = 21
Given two numbers have decimal points after two places.
Therefore, the HCF of 0.63 and 1.05 is 0.21.
Example : 3 What is the largest number, which divides 64, 136 and 238 to leave the same remainder in each case?
Solution:
To find the required number, we need to calculate the HCF of (136 – 64), (238 – 136) and (238 – 64), i.e., HCF (72, 102, 174).
since
136 – 64 = 72
238 – 126 = 102
238 – 64 = 174
Let us write the prime factorization for each of these numbers.
72 = 23 x 32
102 = 2 x 3 x 17
174 = 2 x 3 x 29
Therefore, HCF of 72, 102, and 174 = 2 x 3 = 6
Hence, 6 is the required number.
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:
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.
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.
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.
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.
- 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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