Topics
Number System(Consolidating the Sense of Numberness)
Number System
Estimation
Ratio and Proportion
Algebra
Numbers in India and International System (With Comparison)
Geometry
Place Value
Mensuration
Natural Numbers and Whole Numbers (Including Patterns)
Data Handling
Negative Numbers and Integers
Number Line
HCF and LCM
Playing with Numbers
- Simplification of Brackets
- Finding Factors Using Rectangular Arrangements and Division
- Factors and Common Factors
- Multiples and Common Multiples
- Concept of Even and Odd Number
- Tests for Divisibility of Numbers
- Divisibility by 2
- Divisibility by 4
- Divisibility by 8
- Divisibility by 3
- Divisibility by 6
- Divisibility by 9
- Divisibility by 5
- Divisibility by 11
Sets
Ratio
Proportion (Including Word Problems)
Unitary Method
Fractions
- Concept of Fraction
- Types of Fractions
- Concept of Proper and Improper Fractions
- Concept of Mixed Fractions
- Like and Unlike Fraction
- Concept of Equivalent Fractions
- Conversion between Improper and Mixed fraction
- Conversion between Unlike and Like Fractions
- Simplest Form of a Fractions
- Comparing Fractions
- Addition of Fraction
- Subtraction of Fraction
- Multiplication of Fraction
- Division of Fractions
- Using Operator 'Of' with Multiplication and Division
- BODMAS Rule
- Problems Based on Fraction
Decimal Fractions
Percent (Percentage)
Idea of Speed, Distance and Time
Fundamental Concepts
Fundamental Operations (Related to Algebraic Expressions)
Substitution (Including Use of Brackets as Grouping Symbols)
Framing Algebraic Expressions (Including Evaluation)
Simple (Linear) Equations (Including Word Problems)
Fundamental Concepts
Angles (With Their Types)
Properties of Angles and Lines (Including Parallel Lines)
Triangles (Including Types, Properties and Constructions)
Quadrilateral
Polygons
The Circle
Symmetry (Including Constructions on Symmetry)
Recognition of Solids
Perimeter and Area of Plane Figures
Data Handling (Including Pictograph and Bar Graph)
Mean and Median
- Definition: HCF
- Steps to Find HCF
- Example 1
- Example 2
- Key Points Summary
Definition: HCF
The HCF (Highest Common Factor) of two or more numbers is the highest number among all the common factors of the given numbers.
Steps to Find HCF
Example 1
(18 and 27)
- List the factors of 18:
1, 2, 3, 6, 9, 18 - List the factors of 27:
1, 3, 9, 27 - Find the common factors:
Common factors of 18 and 27 are 1, 3, and 9. - Choose the highest (largest) common factor:
The highest number is 9 - Final Answer:
HCF(18, 27) = 9
Example 2
My aunt has brought two ribbons of different colors, one 12 meters long and the other 18 meters long. We need to cut both ribbons into pieces of the same length. What should be the maximum length of each piece?
Solution: The number that gives the length of each piece must be a factor of 12 and 18
Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 18: 1, 2, 3, 6, 9, 18
Of the common factors of 12 and 18, 6 is the greatest. Therefore, the maximum length of each piece should be 6 meters.
Key Points Summary
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The HCF is never greater than the smallest number.
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If HCF is 1, then the numbers are co-prime.
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Check all factors carefully; missing one may give a wrong answer.
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Use diagrams or factor trees to help visualise the process.
Example Question 1
Find the HCF of the following numbers: 20, 28, and 36.
The HCF of 20, 28 and 36 can also be found by prime factorisation of these numbers as follows:
Thus,
20 = 2 × 2 × 5
28 = 2 × 2 × 7
36 = 2 × 2 × 3 × 3.
The common factor of 20, 28, and 36 is 2 (occurring twice).
Thus, HCF of 20, 28 and 36 is 2 × 2 = 4.
Example Question 2
Two tankers contain 850 litres and 680 litres of kerosene oil respectively. Find the maximum capacity of a container which can measure the kerosene oil of both the tankers when used an exact number of times.
The required container has to measure both the tankers in a way that a count is an exact number of times. So its capacity must be an exact divisor of the capacities of both the tankers. Moreover, this capacity should be maximum. Thus, the maximum capacity of such a container will be the HCF of 850 and 680.
Hence,
850 = 2 × 5 × 5 × 17= 2 × 5 × 17 × 5 and
680 = 2 × 2 × 2 × 5 × 17= 2 × 5 × 17 × 2 × 2
The common factors of 850 and 680 are 2, 5 and 17.
Thus, the HCF of 850 and 680 is 2 × 5 × 17 = 170.
Therefore, the maximum capacity of the required container is 170 liters.
It will fill the first container in 5 and the second in 4 refills.
Example Question 3
There are 20 kg of jowar and 50 kg of wheat in a shop. All the grain is to be packed in bags. If all the bags are to have equal weights of grain, what is the maximum weight of grain that can be filled in each bag?
The weight of the grain in each bag must be a factor of 20 and 50.
Besides, the maximum possible weight must be filled in each bag. Hence, let us find the HCF of 20 and 50.
Factors of 20: 1, 2, 4, 5, 10, 20
Factors of 50: 1, 2, 5, 10, 25, 50
Common factors: 1, 2, 5, 10
Of the common factors of 20 and 50, 10 is the greatest, i.e. 10 is the HCF of the numbers 20 and 50.
Therefore, a maximum of 10 kg of grain can be filled in each bag.
Example Question 4
Find the HCF of 10, 15, 12.
10 = 2 × 5
15 = 3 × 5
12 = 2 × 2 × 3
No number except 1 is a common divisor.
Hence, HCF = 1.
Example Question 5
A shop sells a 450 g bottle of jam for 96 rupees and a bigger bottle of 600 g for 124 rupees. Which bottle is it more profitable to buy?
Let us use 150, the HCF of 450 and 600 to compare.
450 = 150 × 3, 600 = 150 × 4
∴ In the small bottle, 150 g of jam costs `96/3` = 32 rupees.
In the large bottle, 150 g of jam costs `(124)/4` = 31 rupees.
∴ Thus, it is more profitable to buy the 600 g bottle of jam.
