Real Numbers Class 10 Previous Year Questions (Chapter 1)

1. 1

   Step 1: Assume the opposite

   ½ mark

   Assume, to the contrary, that √5 is rational. Then √5 = p/q, where p and q are coprime integers and q ≠ 0.

   Why? We need a starting assumption. A rational number can be written as p/q with integers p and q, q ≠ 0. We also take p and q coprime so that the fraction is in lowest terms.

2. 2

   Step 2: Square and apply the theorem

   1 mark

   Squaring both sides: 5q² = p². So 5 divides p². By Theorem 1.2, 5 divides p. Write p = 5c.

   Why? Squaring removes the square root and gives an equation in integers. The theorem lets us move from '5 divides p²' to '5 divides p'.

3. 3

   Step 3: Show q is also divisible by 5

   1 mark

   Substitute p = 5c: 5q² = 25c², so q² = 5c². Hence 5 divides q², and by Theorem 1.2, 5 divides q.

   Why? Substituting p = 5c shows the same argument applies to q. This is the second half needed for the contradiction.

4. 4

   Step 4: Contradiction and conclusion

   ½ mark

   Thus 5 divides both p and q, contradicting that p and q are coprime. Therefore √5 is irrational.

   Why? If 5 divides both p and q, then p and q have a common factor, which contradicts the earlier statement that they are coprime. So the assumption must be false.

Marks: ½ + 1 + 1 + ½ = 3 · Split as in the CBSE marking scheme, 2025 · Standard 30/6/1 Q26 (the marks split is taken from the 2025 Standard paper, where this question also appeared)

1. 1

   Step 1: Assume the opposite

   ½ mark

   Assume, to the contrary, that √3 is rational. Then √3 = a/b, where a and b are coprime integers and b ≠ 0.

   Why? This is the starting assumption of proof by contradiction. Coprime means a and b have no common factor except 1; this makes the later contradiction valid.

2. 2

   Step 2: Square and use the prime divisibility property

   1 mark

   Squaring both sides gives 3b² = a². So 3 divides a², and hence 3 divides a.

   Why? Squaring removes the square root. The step '3 divides a², so 3 divides a' is the prime divisibility property used in NCERT Example 5.

3. 3

   Step 3: Show b is also divisible by 3

   1 mark

   Write a = 3c for some integer c. Then 3b² = 9c², so b² = 3c². Hence 3 divides b², and so 3 divides b.

   Why? Substituting shows b is also divisible by 3. Now both a and b have 3 as a common factor.

4. 4

   Step 4: State the contradiction and conclusion

   ½ mark

   This contradicts the fact that a and b are coprime. Therefore our assumption is wrong, so √3 is irrational.

   Why? The contradiction proves the assumption false, so the opposite statement is true.

Marks: ½ + 1 + 1 + ½ = 3 · Split as in the CBSE marking scheme, 2026 · Standard 30/3/2 Q31 (the marks split is taken from the 2026 Standard paper, where this question also appeared)

1. 1

   Step 1: Assume the contrary

   ½ mark (suggested)

   Assume, to the contrary, that √2 is rational. Then √2 = a/b, where a and b are coprime integers and b ≠ 0.

   Why? This is the starting move of proof by contradiction. Coprime means a and b have no common factor except 1; we need that condition so that a later common factor becomes a contradiction.

2. 2

   Step 2: Show 2 divides a

   1 mark (suggested)

   Squaring both sides, a² = 2b². So 2 divides a². By the result that if a prime p divides a² then p divides a, 2 divides a. So a = 2c for some integer c.

   Why? Squaring removes the square root and turns the assumption into an equation about integers. The result about primes lets us conclude that 2 must be a factor of a, not just of a².

3. 3

   Step 3: Show 2 divides b

   1 mark (suggested)

   Substitute a = 2c in a² = 2b²: (2c)² = 2b², so 4c² = 2b², i.e. b² = 2c². So 2 divides b², and hence 2 divides b.

   Why? Putting a = 2c lets us apply the same prime result to b. Now both a and b are forced to be even.

4. 4

   Step 4: Contradiction and conclusion

   ½ mark (suggested)

   Thus a and b have common factor 2, contradicting that a and b are coprime. Hence our assumption is wrong. Therefore √2 is irrational.

   Why? The assumption said a and b are coprime, but the working shows they share the factor 2. This contradiction proves the assumption was false, so the opposite statement is true.

Marks: ½ + 1 + 1 + ½ = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

1. 1

   Step 1: Let the numbers be 3x and 5x

   ½ mark

   Let the two numbers be 3x and 5x.

   Why? A ratio means both numbers are multiples of the same common factor, so one unknown x can represent both numbers.

2. 2

   Step 2: Use HCF × LCM = product

   1 mark

   Since HCF = x, HCF × LCM = product gives x × 180 = 3x × 5x. As x > 0, 180 = 15x, so x = 12.

   Why? The relation between HCF, LCM and the product turns the given ratio and LCM into one equation, which is then solved for x.

3. 3

   Step 3: State the HCF

   ½ mark

   Hence, HCF = 12.

   Why? Since x was the common factor, the HCF is already known once x is found.

Marks: ½ + 1 + ½ = 2 · Split as in the CBSE marking scheme, 2026 · Standard 30(B) Q21

1. 1

   Step 1: Find the LCM of 24 and 36

   1 mark (suggested)

   24 = 2³ × 3, 36 = 2² × 3². LCM(24, 36) = 2³ × 3² = 72.

   Why? A number divisible by both 24 and 36 is a common multiple of them. The least positive common multiple is the LCM, so finding the LCM first turns the problem into finding the smallest 5-digit multiple of 72.

2. 2

   Step 2: Find the smallest 5-digit multiple of 72

   1 mark (suggested)

   10000 ÷ 72 gives quotient 138 and remainder 64. The required number is 10000 + (72 − 64) = 10008.

   Why? 10000 is the smallest 5-digit number. The quotient tells how many full multiples of 72 are below 10000, and the remainder is what is left after subtracting that multiple. Adding the shortfall to the next multiple gives the required number.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Find the LCM of 15, 18 and 20

   1 mark (suggested)

   The required time is LCM(15, 18, 20). Prime factorisation: 15 = 3 × 5, 18 = 2 × 3², 20 = 2² × 5. So LCM(15, 18, 20) = 2² × 3² × 5 = 180.

   Why? LCM means the least common multiple, the smallest number divisible by 15, 18 and 20. Prime factorisation means writing a number as a product of its primes. All three start a new sweater together on common multiples of their times, so the first such day is this LCM.

2. 2

   Step 2: Find the number of sweaters completed

   1 mark (suggested)

   In 180 days: Ranjita = 180/15 = 12, Neha = 180/18 = 10, Salma = 180/20 = 9. Together = 12 + 10 + 9 = 31.

   Why? Once the common time is known, dividing it by each person's time for one sweater gives how many sweaters each finishes in that period.

3. 3

   Step 3: State the answer

   1 mark (suggested)

   Required time = 180 days. Sweaters completed: Ranjita = 12, Neha = 10, Salma = 9; total = 31.

   Why? This states the two answers asked in the question: the common restart day and the number of sweaters completed by then.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

1. 1

   Step 1: Prime factorisation of 45ⁿ

   1 mark (suggested)

   45 = 3² × 5, so 45ⁿ = (3² × 5)ⁿ = 3²ⁿ × 5ⁿ.

   Why? This is the set-up. It writes 45ⁿ as a product of primes, so we can see exactly which primes are present.

2. 2

   Step 2: Apply the ending-in-0 condition and find a

   1 mark (suggested)

   A number ending in 0 is divisible by 10 = 2 × 5. The factorisation 3²ⁿ × 5ⁿ has no factor 2, so by the Fundamental Theorem of Arithmetic, 45ⁿ cannot end in 0. Multiplying by the prime a = 10 ÷ 5 = 2 supplies the missing 2. Hence 45ⁿ × a ends in 0.

   Why? Ending in 0 means having both prime factors 2 and 5. Comparing the factorisations gives the conclusion and the missing prime.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Assume the number is rational

   1 mark (suggested)

   Assume, to the contrary, that 5√3 + 2/3 is rational. Then 5√3 + 2/3 = a/b, where a and b are coprime integers and b ≠ 0.

   Why? Proof by contradiction starts by assuming the opposite of what we must prove. 'Coprime' means a and b have no common factor except 1, so the fraction is in lowest terms.

2. 2

   Step 2: Isolate √3

   1 mark (suggested)

   5√3 = a/b − 2/3 = (3a − 2b)/(3b), so √3 = (3a − 2b)/(15b).

   Why? Rearrange the assumed equation so that √3 stands alone. If the assumption is true, the right side must be a rational number.

3. 3

   Step 3: Reach the contradiction

   1 mark (suggested)

   Since a and b are integers and b ≠ 0, (3a − 2b)/(15b) is rational. So √3 is rational, which contradicts the given fact that √3 is irrational. Hence the assumption is wrong. Therefore, 5√3 + 2/3 is irrational.

   Why? A quotient of integers with a non-zero denominator is rational. The contradiction means the assumption was false, so the original number must be irrational.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

1. 1

   Step 1: Assume the opposite

   1 mark (suggested)

   Assume, to the contrary, that 4√2 + 5/3 is rational. Then 4√2 + 5/3 = a/b, where a and b are coprime integers and b ≠ 0.

   Why? We start by assuming the opposite of what we must prove. If this assumption forces a false statement, then the original statement is proved.

2. 2

   Step 2: Isolate √2

   1 mark (suggested)

   4√2 = a/b − 5/3 = (3a − 5b)/(3b), so √2 = (3a − 5b)/(12b). Since a and b are integers, (3a − 5b)/(12b) is rational.

   Why? We rearrange the assumed equation to express √2 as a quotient of integers. That is the only way to force the contradiction.

3. 3

   Step 3: Contradiction and conclusion

   1 mark (suggested)

   Thus √2 is rational, which contradicts the given fact that √2 is irrational. Hence 4√2 + 5/3 is irrational.

   Why? The given fact says √2 is irrational, so the conclusion from the previous step is impossible. This ends the proof.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

1. 1

   Step 1: Assume the contrary

   ½ mark

   Assume, to the contrary, that 2√7 is rational. Then 2√7 = a/b, where a and b are coprime integers and b ≠ 0.

   Why? To prove by contradiction, we start by assuming the opposite of what we want to prove.

2. 2

   Step 2: Rearrange the equation

   ½ mark

   Rearranging, √7 = a/(2b).

   Why? Isolating √7 lets us compare it with the given fact that √7 is irrational.

3. 3

   Step 3: Show √7 is rational

   ½ mark

   Since a and b are integers, a/(2b) is rational. So √7 is rational.

   Why? A rational number divided by a non-zero integer is still rational, so the assumption forces √7 to be rational.

4. 4

   Step 4: State the contradiction

   ½ mark

   This contradicts the fact that √7 is irrational. Hence 2√7 is irrational.

   Why? The contradiction shows the assumption was wrong, so the statement is true.

Marks: ½ + ½ + ½ + ½ = 2 · Split as in the CBSE marking scheme, 2025 Compartment · Standard 30/S/2 Q22

1. 1

   Step 1: Assume the contrary

   ½ mark

   Let (2 − √3)/5 = a/b, where a and b are coprime integers and b ≠ 0.

   Why? This is the starting point of a proof by contradiction: we suppose the number is rational and show this forces √3 to be rational.

2. 2

   Step 2: Isolate √3

   1 mark

   2 − √3 = 5a/b, so √3 = 2 − 5a/b = (2b − 5a)/b.

   Why? Rearranging turns the assumed rational expression into an expression for √3, which lets us compare it with the known fact that √3 is irrational.

3. 3

   Step 3: Right side is rational

   ½ mark

   Since a and b are integers, (2b − 5a)/b is a rational number.

   Why? A quotient of integers with a non-zero denominator is exactly what a rational number is, so √3 would be rational.

4. 4

   Step 4: State the contradiction

   ½ mark

   This contradicts the given fact that √3 is irrational.

   Why? The assumption leads to a statement that is already known to be false.

5. 5

   Step 5: Conclude

   ½ mark

   Therefore (2 − √3)/5 is irrational.

   Why? In proof by contradiction, a false consequence proves the original assumption wrong, so the number must be irrational.

Marks: ½ + 1 + ½ + ½ + ½ = 3 · Split as in the CBSE marking scheme, 2024 · Standard 30/4/3 Q29 (the marks split is taken from the 2024 Standard paper, where this question also appeared)

1. 1

   Step 1: Prime factorisation of 336 and 54

   1 mark (suggested)

   336 = 2⁴ × 3 × 7, 54 = 2 × 3³.

   Why? Prime factorisation is the first step of the prime factorisation method; it lists the prime factors of each number so HCF and LCM can be read from them.

2. 2

   Step 2: HCF using common prime factors

   1 mark (suggested)

   HCF = 2¹ × 3¹ = 6.

   Why? HCF is built from only the primes common to both numbers, using the smallest power of each common prime.

3. 3

   Step 3: LCM using all prime factors

   1 mark (suggested)

   LCM = 2⁴ × 3³ × 7 = 3024. Hence, HCF = 6 and LCM = 3024.

   Why? LCM uses every prime that appears in either number, with the greatest power of each, so the result is a multiple of both numbers.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

1. 1

   Step 1: Prime factorisation of 96 and 404

   1 mark (suggested)

   96 = 2⁵ × 3; 404 = 2² × 101

   Why? Prime factorisation is the method the question asks for. Writing both numbers as products of primes shows which prime factors they share and which are different.

2. 2

   Step 2: HCF

   1 mark (suggested)

   HCF = 2² = 4

   Why? The HCF uses only the primes common to both numbers, each with its smallest power. Here the only common prime is 2.

3. 3

   Step 3: LCM

   1 mark (suggested)

   LCM = 2⁵ × 3 × 101 = 9696. Hence, HCF = 4 and LCM = 9696.

   Why? The LCM must contain every prime that appears in either number, each with its greatest power, so that both 96 and 404 divide it.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

1. 1

   Step 1: Prime factorisation

   1 mark (suggested)

   260 = 2² × 5 × 13 and 910 = 2 × 5 × 7 × 13

   Why? The prime-factorisation method needs each number written as a product of primes first.

2. 2

   Step 2: HCF

   1 mark (suggested)

   HCF = 2¹ × 5¹ × 13¹ = 130

   Why? HCF takes the smallest power of each prime common to both numbers.

3. 3

   Step 3: LCM and final answer

   1 mark (suggested)

   LCM = 2² × 5 × 7 × 13 = 1820. Hence, HCF = 130 and LCM = 1820.

   Why? LCM takes the greatest power of every prime that appears in either number. The last sentence states both results asked for.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

1. 1

   Step 1: Prime factorise both numbers

   1 mark (suggested)

   65 = 5 × 13; 117 = 3² × 13.

   Why? Prime factorisation writes each number as a product of primes, so the common prime factors can be seen.

2. 2

   Step 2: Find the HCF

   1 mark (suggested)

   HCF = 13. The greatest number which divides 65 and 117 completely is 13.

   Why? The HCF is the product of the smallest power of each common prime factor; 13 is the only common prime factor, so it is the greatest number that divides both.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Prime factorise both numbers

   1 mark (suggested)

   231 = 3 × 7 × 11; 396 = 2² × 3² × 11

   Why? The prime factorisation method begins by writing each number as a product of primes. This gives the prime factors needed to build the LCM.

2. 2

   Step 2: Find the LCM

   1 mark (suggested)

   LCM = 2² × 3² × 7 × 11 = 2772

   Why? The LCM must contain every prime that appears in either number, using the greatest power of each prime. This gives the smallest number divisible by both 231 and 396.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Prime factorisation of 45, 54 and 270

   1 mark (suggested)

   45 = 3² × 5; 54 = 2 × 3³; 270 = 2 × 3³ × 5

   Why? Writing each number as a product of primes lets us compare the common prime factors and their powers.

2. 2

   Step 2: Find the HCF

   1 mark (suggested)

   Only 3 is common to all three numbers. HCF = product of the smallest power of each common prime factor = 3² = 9.

   Why? The HCF can contain only primes present in every number, and each such prime is taken with its smallest power among the numbers.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Use the HCF–LCM relation

   1 mark

   HCF(306, 1314) × LCM(306, 1314) = 306 × 1314 = 402084

   Why? This relation connects the given HCF to the unknown LCM, so substituting the known values turns the problem into a simple division.

2. 2

   Step 2: Calculate the LCM

   1 mark

   LCM(306, 1314) = 402084 ÷ 18 = 22338

   Why? Dividing the product of the two numbers by the HCF isolates the LCM and gives its value.

Marks: 1 + 1 = 2 · Split as in the CBSE marking scheme, 2024 · Basic 430/4/3 Q24

1. 1

   Step 1: Prime factorisation of \((15)^n\)

   1 mark

   15 = 3 × 5, so \((15)^n\) = \((3 × 5)^n\) = \(3^n\) × \(5^n\). Thus the prime factorisation of \((15)^n\) contains only the primes 3 and 5.

   Why? Prime factorisation shows exactly which primes can appear in \((15)^n\), and the Fundamental Theorem of Arithmetic lets us compare these primes with the primes needed for ending in 0.

2. 2

   Step 2: Condition for ending with 0

   1 mark

   If a number ends with digit 0, it is divisible by 10 = 2 × 5, so 2 must be a prime factor of it. Since \(3^n\) × \(5^n\) has no prime factor 2, \((15)^n\) cannot end with digit 0.

   Why? Ending in 0 forces divisibility by both 2 and 5. Comparing the primes required with the primes present in \((15)^n\) gives the conclusion.

Marks: 1 + 1 = 2 · Split as in the CBSE marking scheme, 2024 · Standard 30/2/1 Q21

1. 1

   Step 1: First number is composite

   1 mark (suggested)

   7×11×13+13 = 13(7×11+1) = 13×78 = 2×3×13². It is a product of primes, so it is composite.

   Why? Taking out the common factor 13 shows the whole sum is a product of two whole numbers greater than 1. The full prime factorisation makes this clear.

2. 2

   Step 2: Second number is composite

   1 mark (suggested)

   7×6×5×4×3×2×1+5 = 5(7×6×4×3×2×1+1) = 5×1009. It has factors 5 and 1009, both greater than 1, so it is composite.

   Why? Taking out the common factor 5 shows the whole sum is a product of two whole numbers greater than 1. Therefore it has more than two factors.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Condition for ending with 0

   1 mark (suggested)

   If \(8^n\) ends with the digit 0, then it is divisible by 5, so 5 must be a prime factor of \(8^n\).

   Why? This changes the question about the last digit into a question about prime factors, which the Fundamental Theorem of Arithmetic can decide.

2. 2

   Step 2: Prime factorisation of \(8^n\)

   1 mark (suggested)

   \(8^n\) = \((2³)^n\) = \(2^{3n}\), so the only prime factor of \(8^n\) is 2. By the Fundamental Theorem of Arithmetic, 5 cannot appear in its factorisation. Hence, \(8^n\) cannot end with the digit 0.

   Why? This gives the complete prime factorisation of \(8^n\), so the uniqueness part of the theorem rules out the factor 5.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Prime factorisation of the intervals

   1 mark (suggested)

   9 = 3², 12 = 2² × 3, 15 = 3 × 5.

   Why? The prime factorisation method needs each interval written as powers of primes before the LCM can be built.

2. 2

   Step 2: Find the LCM and state the time

   1 mark (suggested)

   LCM = 2² × 3² × 5 = 180. The bells will next toll together after 180 minutes.

   Why? The LCM collects the highest power of every prime, giving the smallest common multiple; that is the next time all three toll together.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Prime factorisation of 92 and 510

   1 mark (suggested)

   Prime factorisation gives: 92 = 2² × 23 and 510 = 2 × 3 × 5 × 17.

   Why? The prime factorisation method needs each number written as a product of primes, because HCF and LCM are read from these prime factors.

2. 2

   Step 2: Find HCF and LCM

   1 mark (suggested)

   HCF = 2¹ = 2; LCM = 2² × 3 × 5 × 17 × 23 = 23460. Hence, HCF = 2 and LCM = 23460.

   Why? HCF takes the smallest power of each common prime factor, while LCM takes the greatest power of every prime factor that appears in either number.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Prime factorisation of 540

   ½ mark

   540 = 2² × 3³ × 5

   Why? We need each number as powers of primes so that common primes can be compared.

2. 2

   Step 2: Prime factorisation of 630

   ½ mark

   630 = 2 × 3² × 5 × 7

   Why? The second number must be in the same form to see which primes appear in both.

3. 3

   Step 3: HCF by smallest powers

   1 mark

   HCF = 2¹ × 3² × 5¹ = 90

   Why? HCF uses the smallest power of each prime common to both numbers.

Marks: ½ + ½ + 1 = 2 · Split as in the CBSE marking scheme, 2023 · Basic 430/2/1 Q22

1. 1

   Step 1: Form the numbers 84 and 70

   1 mark

   Let the required number be d. Then d divides 85 − 1 = 84 and 72 − 2 = 70.

   Why? A divisor that leaves a remainder must divide the number minus that remainder. This changes the remainder condition into a plain divisibility condition, so the problem becomes finding a common divisor of 84 and 70.

2. 2

   Step 2: Find the HCF

   1 mark

   84 = 2² × 3 × 7 and 70 = 2 × 5 × 7, so HCF = 2 × 7 = 14. Hence the required greatest number = 14.

   Why? The greatest number that divides both 84 and 70 is their HCF. Prime factorisation shows the common prime factors, and taking the smallest power of each common prime factor gives the HCF.

Marks: 1 + 1 = 2 · Split as in the CBSE marking scheme, 2023 · Standard 30/1/1 Q24

1. 1

   Step 1: Prime factorisation of 96 and 120

   1 mark

   96 = 2⁵ × 3; 120 = 2³ × 3 × 5.

   Why? This is the first step of the prime factorisation method. The HCF and LCM are read from these factorisations, so both numbers must be written as products of primes.

2. 2

   Step 2: HCF and LCM

   1 mark

   HCF = 2³ × 3 = 24; LCM = 2⁵ × 3 × 5 = 480.

   Why? HCF uses the smallest power of each common prime; LCM uses the greatest power of every prime that appears. This gives the required values.

Marks: 1 + 1 = 2 · Split as in the CBSE marking scheme, 2023 · Standard 30/5/1 Q23

1. 1

   Step 1: Prime factorisation

   1 mark (suggested)

   576 = \(2^6 \times 3^2\), 512 = \(2^9\)

   Why? This is the first step of the prime factorisation method: each number must be written as a product of primes before the LCM can be read off.

2. 2

   Step 2: Find LCM

   1 mark (suggested)

   LCM = \(2^9 \times 3^2 = 4608\). Hence, the LCM of 576 and 512 is 4608.

   Why? The LCM uses the greatest power of every prime that appears in either factorisation, so both the 2 and the 3 are included.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Prime factorise both numbers

   1 mark (suggested)

   660 = 2² × 3 × 5 × 11; 704 = 2⁶ × 11

   Why? The prime factorisation method needs each number written as a product of primes first; only then can the common factors be compared.

2. 2

   Step 2: Take common primes with smallest powers

   1 mark (suggested)

   HCF = 2² × 11 = 44. Hence, the HCF of 660 and 704 is 44.

   Why? The HCF uses only primes common to both numbers, each with the smaller exponent; this gives the largest number that divides both.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)