This MCQ module is based on: Chapter 3 Exercises and Summary
TOPIC 9 OF 23
Chapter 3 Exercises and Summary
🎓 Class 7
Mathematics
CBSE
Theory
Ch 3 — Factors and Multiples
⏱ ~35 min
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This mathematics assessment will be based on: Chapter 3 Exercises and Summary
Targeting Class 7 level in General Mathematics, with Basic difficulty.
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3.5 Exercises — Figure it Out
Below are the end-of-chapter exercises for Chapter 3: A Story of Numbers (Finding Common Ground). Solutions are hidden — try each on your own first.
1. Find the HCF and LCM of the following pairs using the prime factorisation method:
(a) 24 and 36 (b) 45 and 75 (c) 60 and 84 (d) 96 and 120
(a) 24 and 36 (b) 45 and 75 (c) 60 and 84 (d) 96 and 120
(a) 24 = 2³×3, 36 = 2²×3². HCF = 2²×3 = 12, LCM = 2³×3² = 72.
(b) 45 = 3²×5, 75 = 3×5². HCF = 3×5 = 15, LCM = 3²×5² = 225.
(c) 60 = 2²×3×5, 84 = 2²×3×7. HCF = 2²×3 = 12, LCM = 2²×3×5×7 = 420.
(d) 96 = 2⁵×3, 120 = 2³×3×5. HCF = 2³×3 = 24, LCM = 2⁵×3×5 = 480.
(b) 45 = 3²×5, 75 = 3×5². HCF = 3×5 = 15, LCM = 3²×5² = 225.
(c) 60 = 2²×3×5, 84 = 2²×3×7. HCF = 2²×3 = 12, LCM = 2²×3×5×7 = 420.
(d) 96 = 2⁵×3, 120 = 2³×3×5. HCF = 2³×3 = 24, LCM = 2⁵×3×5 = 480.
2. Find the HCF of three numbers 120, 180, and 300.
120 = 2³×3×5, 180 = 2²×3²×5, 300 = 2²×3×5². Common primes at minimum powers: 2²×3×5 = 60.
3. Find the LCM of three numbers 8, 12, and 20.
8 = 2³, 12 = 2²×3, 20 = 2²×5. Highest powers: 2³×3×5 = 120.
4. In a marathon, three participants complete a lap in 48, 60, and 72 seconds. If they start together, when do they next arrive at the start at the same time?
Find LCM(48, 60, 72). 48 = 2⁴×3, 60 = 2²×3×5, 72 = 2³×3². LCM = 2⁴×3²×5 = 720 seconds = 12 minutes.
5. Fill in the following blanks appropriately: 1 cm = 10 mm; 1 m = 100 cm; 1 km = 1000 m; 1 kg = 1000 g; 1 g = 1000 mg; 1 l = 1000 ml. Convert: 5.5 km = ____ m; 35 cm = ____ mm; 14.5 cm = ____ mm; 68 g = ____ kg; 9.02 m = ____ mm; 125.5 ml = ____ l.
5.5 km = 5500 m; 35 cm = 350 mm; 14.5 cm = 145 mm; 68 g = 0.068 kg; 9.02 m = 9020 mm; 125.5 ml = 0.1255 l.
6. The length, width, and height of a box are 12 cm, 18 cm, and 36 cm respectively. Which of the following sized cubes can be packed in this box without leaving gaps?
(a) 9 cm (b) 6 cm (c) 4 cm (d) 2 cm
(a) 9 cm (b) 6 cm (c) 4 cm (d) 2 cm
The cube side must divide 12, 18, and 36 exactly. HCF(12,18,36) = 6. So 6 cm and 2 cm (which divides 6) work. Check: 9 does not divide 12. 4 does not divide 18. Answers: (b) 6 cm and (d) 2 cm.
7. Among the numbers below, which is the largest number that divides both 306 and 36?
(a) 36 (b) 612 (c) 18 (d) 3
(a) 36 (b) 612 (c) 18 (d) 3
HCF(306, 36). 306 = 2×3²×17, 36 = 2²×3². HCF = 2×3² = 18. Answer: (c) 18.
8. Find the smallest number that, when divided by 3, 4, 5 and 7, leaves a remainder of 10 when divided by 15.
First LCM(3,4,5,7) = 420. The number that leaves remainder 10 when divided by 15 and is divisible by all of 3, 4, 5, 7 after subtracting 10: try \(420k + 10\). Smallest positive value with required property: 430 (k=1). Check: 430 − 10 = 420 is divisible by 3, 4, 5, 7 ✓.
9. Children are playing 'Fire in the Mountain'. When the number 6 was called out, no one got out. When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out. How many children could have been playing initially?
(a) 72 (b) 90 (c) 36 (d) None of these
(a) 72 (b) 90 (c) 36 (d) None of these
The count must be divisible by both 6 and 9 (i.e., LCM 18) but NOT divisible by 10. 72 = 18×4, not divisible by 10 ✓. 90 is divisible by 10 ✗. 36 = 18×2 is not divisible by 10 ✓. So both 72 and 36 could work, so answer must be (a) 72 or (c) 36 depending on class size. Best single option: (a) 72.
10. Tick the correct statement(s). The LCM of two different prime numbers:
(a) Less than both numbers (b) In between the two numbers (c) One of the two numbers (d) Less than m × n (e) Greater than m × n
(a) Less than both numbers (b) In between the two numbers (c) One of the two numbers (d) Less than m × n (e) Greater than m × n
LCM of two distinct primes \(p\) and \(q\) = \(p \times q\) (since they have no common factors, HCF = 1). So LCM equals the product exactly — not less, not greater. None of the given options is strictly correct; the correct statement would be "equal to m × n".
11. A dog is chasing a rabbit that has a head start of 150 feet. It jumps 9 feet every time the rabbit jumps 7 feet. In how many jumps does the dog catch up with the rabbit?
Per 'round' (dog jump + rabbit jump), dog gains 9 − 7 = 2 feet. To close 150 feet: 150 ÷ 2 = 75 jumps.
12. What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10? Do you remember the answer from Grade 6, Chapter 6?
LCM = 2³ × 3² × 5 = 360.
13. Here is a problem set by the ancient Indian Mathematician Mahaviracharya (850 CE). Add together \(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{20}+\frac{1}{21}\). What do you get? How can we find this sum efficiently?
LCM(6, 10, 15, 20, 21) = 420. Converting: 70/420 + 42/420 + 28/420 + 21/420 + 20/420 = 181/420. Sum = 181/420. (Mahaviracharya used the LCM of denominators — exactly the technique we learnt.)
Chapter Summary
- We looked at common multiples and common factors, and were introduced to the amazing world of primes.
- In this chapter, we learnt a method to find the prime factorisation? of a number.
- Finding all the factors of a number from its prime factorisation is easy but quite tedious — we need to list every combination.
- The Highest Common Factor (HCF) is the highest among all common factors of the numbers. It is also known as the Greatest Common Divisor (GCD).
- Every common factor of a set of numbers is contained in the HCF.
- To find the HCF, include the minimum number of occurrences of each prime across the prime factorisations.
- The Lowest Common Multiple (LCM) is the smallest among all common multiples.
- Every common multiple contains the prime factorisation of the numbers.
- To find the LCM, include the maximum number of occurrences of each prime.
- We explored more about HCF and LCM: we discovered related properties and patterns when the numbers are consecutive, even, co-prime, etc.
- We learnt a procedure to get both the HCF and the LCM at the same time. We also saw how to make this even quicker.
- We learnt some terms that are used when discussing mathematics, such as 'conjecture' and 'generalisation'.
Fun fact: The largest prime found so far has 41,024,320 digits — it was discovered on October 12, 2024!
Activity: Number Detective
L3 ApplyMaterials: Paper and pencil.
Challenge: I'm thinking of two numbers. Their HCF is 6 and their LCM is 360. Can you find the numbers?
- Use HCF × LCM = a × b, so a × b = 6 × 360 = 2160.
- Write a = 6m and b = 6n where m and n are coprime.
- Then 6m × 6n = 2160, so m × n = 60.
- Find coprime pairs of 60: (1, 60), (3, 20), (4, 15), (5, 12).
- So possible (a, b) pairs: (6, 360), (18, 120), (24, 90), (30, 72).
Four possible pairs: (6, 360), (18, 120), (24, 90), (30, 72). All four have HCF 6 and LCM 360.
🎨 Puzzle Time! — Mystery Colours
You might have noticed and wondered about the different circle designs around the page numbers on each page! The picture shows all the designs for numbers from 1 to 100. Try to decode the colour scheme for each number. There are several interesting patterns here. Share your observations with your classmates. Extending this scheme, colour the page numbers for 101 – 110.
Hint: The colours seem to follow prime factorisation — check which primes make up each number. Try the rule: one colour per prime factor. For example: 6 = 2 × 3 uses two colours; 12 = 2² × 3 might layer the colour of 2 twice and 3 once!
Competency-Based Questions
Scenario: A book publisher prints 1800 copies of Book A and 2400 copies of Book B. He wants to pack them into cartons such that each carton has only one type of book, and every carton has the same number of books. No book should be left unpacked.
Q1. What is the largest possible number of books per carton?
L3 ApplyHCF(1800, 2400). 1800 = 2³×3²×5². 2400 = 2⁵×3×5². HCF = 2³×3×5² = 600 books per carton.
Q2. With that carton size, how many cartons are needed in total? Analyse whether the proportion of each book is reflected in the carton counts.
L4 AnalyseBook A: 1800 ÷ 600 = 3 cartons. Book B: 2400 ÷ 600 = 4 cartons. Total = 7 cartons. Ratio 3:4 matches 1800:2400 = 3:4, so yes, the proportion is preserved.
Q3. The publisher now wants every carton to have both A and B copies in the ratio 3:4. Evaluate whether he can still use exactly 600 books per carton without wastage.
L5 Evaluate600 books split as 3:4 → each carton needs 600×3/7 ≈ 257.1 of A and 342.9 of B. These are not whole numbers, so 600 per carton doesn't work. The carton size must be a multiple of 7 (to split cleanly 3:4). Largest such carton that still divides both totals: must divide 1800 and 2400 and be a multiple of 7. Since HCF(1800,2400) = 600 is not divisible by 7, there's no valid carton. Publisher cannot meet both constraints simultaneously.
Q4. Design a new print run (number of A and B copies, each between 500 and 3000) so that the HCF is exactly 300 and the ratio A:B is 5:7. Justify.
L6 CreateSample: A = 1500 (= 300 × 5), B = 2100 (= 300 × 7). Then HCF(1500, 2100) = 300 (since 5 and 7 are coprime). Ratio 5:7 ✓. Both are between 500 and 3000 ✓. Other valid answers require multiplying 5 and 7 by the same prime that stays inside the range.
Assertion–Reason Questions
A: If two numbers have HCF equal to one of them, then the smaller number divides the larger.
R: HCF of two numbers is always less than or equal to the smaller number.
R: HCF of two numbers is always less than or equal to the smaller number.
Answer: (b). Both are true. A is true because if HCF = smaller number, the smaller divides the larger. R is true in general. But R alone doesn't directly explain A — we also need that the HCF equals the smaller number.
A: The LCM of any two numbers is always a multiple of their HCF.
R: HCF divides both numbers, so it divides every common multiple as well.
R: HCF divides both numbers, so it divides every common multiple as well.
Answer: (a). HCF divides both numbers, so it divides any multiple of either — including the LCM. R directly explains A.
A: Prime factorisation of every number greater than 1 is unique.
R: Prime numbers have exactly two factors.
R: Prime numbers have exactly two factors.
Answer: (b). Uniqueness is the Fundamental Theorem of Arithmetic. R is true but it states a different property (what a prime is), not the uniqueness of factorisation.
Frequently Asked Questions
What exercises are in Class 7 Part 2 Chapter 3?
Chapter 3 exercises include finding HCF and LCM by listing and prime factorisation, identifying co-prime pairs, verifying HCF x LCM = product, and word problems on tiling, grouping, and periodic events. NCERT Class 7 Ganita Prakash Part 2 provides diverse practice.
How to solve HCF word problems?
Read to identify what needs to be 'largest' or 'greatest common'. Set up HCF of relevant numbers. Compute HCF by prime factorisation. State the answer in context. NCERT Class 7 Chapter 3 exercises follow this pattern.
What is the summary of Chapter 3?
Key ideas: factors and multiples of a number; HCF is the greatest common factor; LCM is the smallest common multiple; prime factorisation gives both efficiently; HCF x LCM = product for two numbers. NCERT Class 7 Ganita Prakash Part 2 Chapter 3.
Find HCF and LCM of 24 and 36.
24 = 2^3 x 3, 36 = 2^2 x 3^2. HCF = 2^2 x 3 = 12. LCM = 2^3 x 3^2 = 72. Check: 12 x 72 = 864 = 24 x 36. NCERT Class 7 Chapter 3 exercises use this verification.
When do we use HCF versus LCM?
Use HCF when finding the largest equal groups, greatest tile size, or simplifying fractions. Use LCM when finding the smallest common multiple, adding fractions, or finding when events recur together. NCERT Class 7 Part 2 Chapter 3 distinguishes these.
Are HCF and LCM defined for more than two numbers?
Yes. Extend either method to any number of values. HCF of three numbers is the greatest number dividing all three; LCM is the smallest multiple of all three. NCERT Class 7 Chapter 3 includes three-number examples.
Frequently Asked Questions — Chapter 3
What is Chapter 3 Exercises and Summary in NCERT Class 7 Mathematics?
Chapter 3 Exercises and Summary 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 Chapter 3 Exercises and Summary step by step?
To solve problems on Chapter 3 Exercises and Summary, 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 Chapter 3 Exercises and Summary important for the Class 7 board exam?
Chapter 3 Exercises and Summary 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 Chapter 3 Exercises and Summary?
Common mistakes in Chapter 3 Exercises and Summary 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 Chapter 3 Exercises and Summary?
End-of-chapter NCERT exercises for Chapter 3 Exercises and Summary 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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