This MCQ module is based on: Common Factors and HCF
TOPIC 7 OF 23
Common Factors and HCF
🎓 Class 7
Mathematics
CBSE
Theory
Ch 3 — Factors and Multiples
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Common Factors and HCF
Targeting Class 7 level in General Mathematics, with Basic difficulty.
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3.1 The Greatest of All
Sameeksha is building her new house. The main room of the house measures 12 ft by 16 ft. She feels the room would look nicer if the floor is covered with square tiles? of the same size. She also wants to use as few tiles as possible, and for the tiles to fit a whole number of feet exactly along each side. What size tiles should she buy?
Sameeksha's main room (12 ft × 16 ft)
For a square tile to fit the breadth exactly, the side of the tile must be a factor? of 12. Similarly, for the tiles to fit the length of the room exactly, the side of the tile must also be a factor of 16.
The factors of 12 are 1, 2, 3, 4, 6, 12.
The factors of 16 are 1, 2, 4, 8, 16.
So, the common factors of 12 and 16 are 1, 2, and 4.
The square tiles can have sides of 1 ft, 2 ft, or 4 ft. Among these, she should use the largest tile. Therefore, she needs tiles of 4 ft side.
Definition — HCF / GCD
The Highest Common Factor (HCF) of two or more numbers is the highest of their common factors. It is also known as the Greatest Common Divisor (GCD). In our example, HCF(12, 16) = 4.
We can draw rough diagrams like the one shown on the previous page to visualise the given scenario. It may help in understanding and solving.
Example: Lekhana's Potato Bags
Lekhana purchases rice from two farms and sells it in the market. She bought 84 kg of rice from one farm and 108 kg from the other farm. She wants the rice to be packed in bags, so each bag has rice from only one farm, and all bags have the same weight, to avoid confusion. What is the maximum possible weight that each bag can have?
To divide 84 kg of rice into bags of equal weight, we need the factors of 84. Similarly, for 108, we need the factors of 108.
Factors of 84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84.
Factors of 108 = 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108.
Since Lekhana wants to use bags of the same weight for both farms, the weight of the bag should be a common factor. The common factors of 84 and 108 are 1, 2, 3, 4, 6, 12.
She can use any of these weights to pack rice from both farms in bags of equal weight. But, she wants to minimise the number of bags. Which weight should she choose to minimise the number of bags? — the HCF: 12 kg.
🔵 Do you remember the 'Jumping Jackpot' game from Grade 6? Two numbers are chosen and whenever players come to their multiples, 'idli' or 'vada' should be called out depending on whose multiple the number is. Find the common jumps: (a) 14 and 30, (b) 7 and 11, (c) 30 and 50, (d) 28 and 42. — These are the LCM scenarios, connected later.
Figure it Out — Finding HCF by Listing Factors
Q. Find the HCF of 14 and 30 using the listing method.
Factors of 14: 1, 2, 7, 14. Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30. Common factors: 1, 2. HCF(14, 30) = 2.
3.2 Factors of a Number Using Prime Factorisation
The listing method can be tedious for larger numbers, and many factors can lead to errors. Can this process be simplified?
It turns out that using prime factorisation? can simplify the process. We will start by revisiting primes and prime factorisation.
Primes
Recall that a prime is a number greater than 1 that has only 1 and the number itself as its factors. Last year, we tried to find patterns amongst primes between 1 and 100. We also came across the Sieve of Eratosthenes — a method to list all primes.
Prime Factorisation
Any number can be written as a product of primes — keep rewriting composite factors until only primes are left. Recall that we call this the prime factorisation of a given number.
For example, we can find the prime factorisation of 90 as follows:
\(90 = 2 \times 45 = 2 \times 3 \times 15 = 2 \times 3 \times 3 \times 5\)
The number 90 could also have been factorised as \(3 \times 30\) or \(2 \times 45\) in a few different ways. Will these all lead to the same prime factors?
Remarkably, the resulting prime factors are always the same (in some order), with perhaps only a change in their order. For example, if we consider factorising \(3 \times 30\):
\(90 = 3 \times 30 = 3 \times 2 \times 15 = 3 \times 2 \times 3 \times 5\)
— and we have arrived at the same prime factors, just in a different order.
Fundamental Fact
The prime factorisation of a number is unique (except for the order of the prime factors). This is sometimes called the Fundamental Theorem of Arithmetic.
Procedure for Prime Factorisation
Factor trees for 30 and 35
Another neat way is the division method: keep dividing the number by the smallest prime that divides it exactly, until the quotient becomes 1.
Example: Prime factorisation of 1200 using the division method:
\(1200 = 40 \times 30 = 5 \times 8 \times 5 \times 6 = 2 \times 2 \times 2 \times 2 \times 3 \times 5 \times 5\)
Factors of a Number Using Prime Factorisation
The prime factorisation of a number can be used to construct all its factors. This can be used to simplify the procedure for finding the HCF of two numbers.
Consider the number 840 and its prime factorisation \(2 \times 2 \times 2 \times 3 \times 5 \times 7\).
- Is \(2 \times 2 \times 7 = 28\) a factor of 840? Yes — because \(840 = (2 \times 2 \times 7) \times (2 \times 3 \times 5) = 28 \times 30\).
- Is \(2 \times 2 \times 2\) a factor of 840? Yes — \(840 = 8 \times 105\).
- Is \(2 \times 7 = 14\) a factor? Yes.
- Is \(3 \times 3 = 9\) a factor? No, because 3 only appears once in the prime factorisation.
Rule
A number is a factor of another number if and only if every prime in its prime factorisation appears at least as many times in the other number's prime factorisation.
Factorising 225 to its primes, we get \(225 = 5 \times 5 \times 3 \times 3\).
Combination of two prime factors: \(3 \times 5, 5 \times 5, 3 \times 3\); one prime: 3, 5.
All factors of 225: 1, 3, 5, 9, 15, 25, 45, 75, 225.
Figure it Out (Section 3.2)
Q1. List all the factors of the following numbers: (a) 90 (b) 105 (c) 132 (d) 360 (24 factors!) (e) 840
(a) 90 = 2×3×3×5; factors: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.
(b) 105 = 3×5×7; factors: 1, 3, 5, 7, 15, 21, 35, 105.
(c) 132 = 2×2×3×11; factors: 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.
(d) 360 = 2³×3²×5; factors: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360.
(e) 840 = 2³×3×5×7; factors: 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840.
(b) 105 = 3×5×7; factors: 1, 3, 5, 7, 15, 21, 35, 105.
(c) 132 = 2×2×3×11; factors: 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.
(d) 360 = 2³×3²×5; factors: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360.
(e) 840 = 2³×3×5×7; factors: 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840.
Q2. Anshu claims: "The larger a number is, the longer its prime factorisation will be." Evaluate using \(96 = 2^5 \times 3\) and \(121 = 11 \times 11\).
96 has 6 primes in its factorisation, while 121 (larger than 96? No — 121 > 96) has only 2. So Anshu's claim is false — the length of the prime factorisation depends on the primes themselves, not on the size of the number.
Activity: Factor Tree Race
L3 ApplyMaterials: Paper, pencil, timer, list of 10 numbers (e.g., 72, 84, 96, 120, 144, 180, 210, 240, 360, 420).
Predict: Which method is faster — factor tree or division method? Why?
- Pair up with a classmate. One uses the factor-tree method, the other uses the division method.
- Start the timer. Both find the prime factorisation of each number in the list.
- Record the time taken for each number.
- Swap methods after 5 numbers and redo.
- Compare results: do both methods give the same prime factorisation? Which was faster on average?
Observation: Both methods always give the same unique prime factorisation. The division method tends to be faster and less error-prone for larger numbers with many small prime factors; the factor tree feels more visual for smaller numbers.
Competency-Based Questions
Scenario: A school art department has 48 red marbles, 60 blue marbles, and 72 green marbles. The teacher wants to create identical craft kits — each kit having the same number of red, blue, and green marbles — with no marbles left over.
Q1. What is the maximum number of kits the teacher can make?
L3 ApplyAnswer: 12 kits. We need HCF(48, 60, 72). 48 = 2⁴×3; 60 = 2²×3×5; 72 = 2³×3². Common primes (minimum power): 2² × 3 = 12.
Q2. How many marbles of each colour go into one kit? Analyse whether any marbles are left over.
L4 AnalyseRed: 48 ÷ 12 = 4; Blue: 60 ÷ 12 = 5; Green: 72 ÷ 12 = 6. Total per kit = 15 marbles. No marbles left over — because 12 divides 48, 60, and 72 exactly (definition of HCF).
Q3. A parent donates 96 more green marbles. Evaluate: does this change the maximum number of kits? Justify.
L5 EvaluateGreen now = 72 + 96 = 168 = 2³×3×7. HCF(48, 60, 168) = 2² × 3 = 12. No change — it is still 12 kits, each with 4 red, 5 blue, and 14 green marbles. Adding 96 (= 2⁵×3) did not reduce the power of 2 or 3 below their minimums, and the new factor 7 is not shared by the other numbers.
Q4. Design a set of three marble counts (between 20 and 200 each) so that exactly 15 identical kits can be made with no leftovers. Explain your design.
L6 CreateSample: 45, 75, 180. HCF = 15 (since 45 = 3²×5, 75 = 3×5², 180 = 2²×3²×5 → minimum common powers: 3 × 5 = 15). Each kit: 3 of the first, 5 of the second, 12 of the third. Many valid answers exist; the key constraint is HCF = 15 exactly.
Assertion–Reason Questions
A: The HCF of any two prime numbers is always 1.
R: A prime number has only two factors: 1 and itself.
R: A prime number has only two factors: 1 and itself.
Answer: (a). Two different primes share only the factor 1, because each has only 1 and itself. R explains A.
A: HCF(12, 18) = 36.
R: HCF is the product of all common prime factors.
R: HCF is the product of all common prime factors.
Answer: (d). 12 = 2²×3; 18 = 2×3². HCF = 2×3 = 6, not 36. R is generally true (common primes raised to minimum powers). So A is false, R is true.
Frequently Asked Questions
What is HCF in Class 7 Maths?
HCF (Highest Common Factor) of two or more numbers is the largest number that divides all of them exactly. For example, HCF(12, 18) = 6. NCERT Class 7 Ganita Prakash Part 2 Chapter 3 teaches HCF through factor listing and prime factorisation.
How do you find HCF by listing factors?
List all factors of each number, identify factors common to all, and pick the greatest common one. Factors of 12: 1,2,3,4,6,12. Factors of 18: 1,2,3,6,9,18. Common: 1,2,3,6. HCF = 6. NCERT Class 7 Chapter 3 uses this method.
What is prime factorisation method for HCF?
Write each number as a product of primes. Pick primes common to all factorisations taking the lowest power. Multiply these to get HCF. For 12 = 2^2 x 3 and 18 = 2 x 3^2, common primes 2^1 x 3^1 = 6. NCERT Class 7 Chapter 3 explains this.
What does HCF of 1 mean?
If the HCF of two numbers is 1, they share no common factor except 1. Such numbers are called co-prime or relatively prime. Example: 8 and 15 are co-prime. NCERT Class 7 Ganita Prakash Part 2 Chapter 3 introduces this idea.
Why is HCF useful?
HCF simplifies fractions to lowest terms, solves problems involving equal grouping, and finds the largest tile or container size. NCERT Class 7 Part 2 Chapter 3 connects HCF to such applications.
Is the HCF of two numbers always smaller than both?
The HCF is at most equal to the smaller of the two numbers. It equals the smaller number only when that number divides the other. For example, HCF(6, 12) = 6. NCERT Class 7 Chapter 3 notes this bound.
Frequently Asked Questions — Chapter 3
What is Common Factors and HCF in NCERT Class 7 Mathematics?
Common Factors and HCF is a key concept covered in NCERT Class 7 Mathematics, Chapter 3: Chapter 3. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Common Factors and HCF step by step?
To solve problems on Common Factors and HCF, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 3: Chapter 3?
The essential formulas of Chapter 3 (Chapter 3) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Common Factors and HCF important for the Class 7 board exam?
Common Factors and HCF is part of the NCERT Class 7 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Common Factors and HCF?
Common mistakes in Common Factors and HCF include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Common Factors and HCF?
End-of-chapter NCERT exercises for Common Factors and HCF cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 3, and solve at least one previous-year board paper to consolidate your understanding.
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