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Q1. What is the HCF of 20 and 30?
Answer: B — Factors of 20 are 1, 2, 4, 5, 10, 20 and factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30; the highest common factor is 10.
Q2. Which of the following is the prime factorisation of 36?
Answer: C — 36 = 2 × 2 × 3 × 3 = 2² × 3²; the other options contain composite numbers, not only primes.
Q3. What is the HCF of 84 and 108?
Answer: B — 84 = 2² × 3 × 7 and 108 = 2² × 3³; common primes are 2² and 3, so HCF = 4 × 3 = 12.
Q4. Is 15 a common factor of 45 and 75?
Answer: A — 15 divides both 45 and 75 exactly, so it is a common factor; it is also the HCF.
Q5. Sameeksha's room is 12 ft by 16 ft. What is the largest square tile size (whole number) she can use?
Answer: C — The tile size must divide both 12 and 16 exactly; the HCF of 12 and 16 is 4 ft.
Q6. Lekhana has 84 kg and 108 kg of rice. What is the maximum weight of each equal bag?
Answer: B — The bag weight must divide both 84 and 108 equally; the HCF is 12 kg, using fewest bags.
Q7. A market vendor has 30 apples and 45 oranges. She wants to pack them into identical fruit baskets with the same number of each fruit in each basket. What is the maximum number of baskets she can make?
Answer: C — The HCF of 30 and 45 is 15, so she can make 15 baskets with 2 apples and 3 oranges each.
Q8. Two ropes are 56 m and 84 m long. They need to be cut into equal-length pieces with no waste. What is the longest possible length of each piece?
Answer: B — 56 = 2³ × 7 and 84 = 2² × 3 × 7; common primes are 2² and 7, so HCF = 4 × 7 = 28 m.
Q9. Which pair of numbers has HCF equal to 1?
Answer: C — 15 = 3 × 5 and 28 = 2² × 7 share no common prime factors, so their HCF is 1.
Q10. A school has 60 pencils and 90 erasers. These must be packed into identical gift boxes with equal numbers of pencils and erasers in each box. What is the maximum number of gift boxes that can be made?
Answer: C — HCF of 60 and 90 is 30; so 30 boxes can be made with 2 pencils and 3 erasers each.
What is the HCF of 12 and 16?
The HCF of 12 and 16 is 4, because 4 is the largest number that divides both 12 and 16 exactly.
Write the prime factorisation of 90.
The prime factorisation of 90 is 2 × 3 × 3 × 5.
What are the common factors of 45 and 75?
The common factors of 45 and 75 are 1, 3, 5, and 15.
Define a prime number.
A prime number is a number greater than 1 that has only 1 and itself as its factors.
What is the HCF of 96 and 275?
The HCF of 96 and 275 is 1, because they share no common prime factors.
How do you find HCF using prime factorisation?
Find common primes in both numbers, then multiply the minimum count of each common prime together.
What is a counterexample in mathematics?
A counterexample is a specific example that proves a conjecture false.
What is the HCF of 30 and 72 using their prime factorisations?
Since 30 = 2 × 3 × 5 and 72 = 2³ × 3², the HCF is 2 × 3 = 6.
Are the prime factors of a number always the same?
Yes, the prime factorisation of any number is unique, though the order of primes may change.
In Sameeksha's room problem, why is 4 ft the best tile size?
Because 4 ft is the HCF of 12 ft and 16 ft, making it the largest size that fits both dimensions exactly with fewest tiles.
What is the HCF of 12 and 16? [1 mark]
List factors of both numbers or use prime factorisation: 12 = 2² × 3 and 16 = 2⁴; common prime is 2², so HCF = 4.
Find all common factors of 45 and 75. [2 marks]
Prime factorise both: 45 = 3² × 5 and 75 = 3 × 5²; common subparts are 1, 3, 5, and 15. List them in order.
Sameeksha wants to tile her floor measuring 12 ft by 16 ft with square tiles of the largest possible size. How many tiles will she need? Show your working. [3 marks]
Step 1: Find HCF of 12 and 16 (= 4 ft). Step 2: Calculate number of tiles along length = 16 ÷ 4 = 4 tiles. Step 3: Calculate number of tiles along breadth = 12 ÷ 4 = 3 tiles. Step 4: Total = 4 × 3 = 12 tiles.
A shopkeeper has 84 kg of rice from one farm and 108 kg from another. She wants to pack them in equal-weight bags (whole kg) such that each bag contains rice from only one farm. (a) What should be the weight of each bag to use the fewest bags? (b) How many bags of each type will she need? Show all steps. [5 marks]
Step 1: Find HCF of 84 and 108 using prime factorisation: 84 = 2² × 3 × 7 and 108 = 2² × 3³; HCF = 2² × 3 = 12 kg. Step 2: Number of bags from farm 1 = 84 ÷ 12 = 7 bags. Step 3: Number of bags from farm 2 = 108 ÷ 12 = 9 bags. Step 4: Total bags = 7 + 9 = 16 bags. State both answers clearly.
Q1. Assertion (A):
The HCF of 12 and 16 is 4.
Reason (R):
4 is the largest number that divides both 12 and 16 exactly.
Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The HCF is defined as the highest common factor, and 4 divides both 12 and 16 evenly (12÷4=3, 16÷4=4), and no number larger than 4 divides both. The reason correctly explains why 4 is the HCF.
Q2. Assertion (A):
Prime factorisation of a number is unique regardless of the order in which we factor.
Reason (R):
Prime numbers can be multiplied in different orders to give the same result.
Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both statements are true. The prime factorisation is unique (Fundamental Theorem of Arithmetic)—a number has one unique set of prime factors. However, the reason explains commutativity of multiplication rather than uniqueness of prime factorisation. The reason is not the correct explanation for why factorisation is unique; it's just about reordering factors, which is a property of multiplication, not a proof of uniqueness.
Q3. Assertion (A):
To find all factors of 225 = 5² × 3², we can systematically form combinations of the prime factors.
Reason (R):
Any product made from a subset of the prime factors of a number is also a factor of that number.
Answer: A — Both A and R are true, and R is the correct explanation of A — The assertion is true—we get factors like 1, 3, 5, 9, 15, 25, 45, 75, 225 by combining prime factors. The reason is true and correctly explains the assertion. Every subset product of prime factors divides the original number exactly, so it is a factor. This is the correct mathematical principle behind finding all factors via prime factorisation.
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Case Study:
A community centre is organising a sports carnival with 72 cricket balls and 60 badminton shuttlecocks. The centre director wants to prepare identical prize kits containing both items in equal quantities from the available stock. Each kit must have a whole number of each item, and no items should be left unpacked. The director wants to make the maximum number of kits possible so more students can win prizes. Help determine the maximum number of kits that can be prepared and the contents of each kit.
Q1. What is the maximum number of prize kits that can be prepared?
Answer: A
Q2. How many cricket balls will be in each kit?
Answer: B
Q3. How many badminton shuttlecocks will be in each kit?
Answer: B
Q4. Which mathematical concept is used to find the maximum number of kits?
Answer: A
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