Real Numbers Class 10 Previous Year Questions: 1 Mark (Chapter 1)

1-mark questions on Real Numbers

This page has 32 1-mark questions on Real Numbers from CBSE Class 10 board papers of 2023, 2024, 2025 and 2026. Of these, 22 appeared only in Standard papers and 10 appeared only in Basic papers. Question formats: 32 MCQs. 23 of them were asked in two or more paper sets.

0 of 32 practised

All 32 Real Numbers 1-mark questions

Q1. \(3^n\), where n is a natural number, cannot end with the digit :

Mark practised

(A) 3
(B) 5
(C) 7
(D) 9

Marks: 1
Level: Basic
Asked in: 6 paper sets
Years: 2025, 2026
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2025 Mathematics Basic, Set 430/3/1, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 5, because the prime factorisation of 3ⁿ contains only the prime 3, whereas a number ending in 5 must be divisible by 5.
Why? By the Fundamental Theorem of Arithmetic, the prime factorisation of 3ⁿ is unique and contains only the prime 3, so 5 can never be a factor of 3ⁿ.

The full 1 mark is for the correct option.

Answer:

(B) 5

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/3/1: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 9 because 3² = 9, without checking that 3⁴ = 81 ends with 1.
Choosing 3 because 3¹ ends with 3, without checking that 3⁴ ends with 1.
Thinking 3ⁿ can end with 5 because 3 and 5 are both odd, without using the prime factorisation.

Why this works

A number ends with 5 only when it is divisible by 5. The prime factorisation of 3ⁿ contains only the prime 3, because 3ⁿ is just 3 multiplied by itself n times. By the Fundamental Theorem of Arithmetic, this factorisation is unique, so 5 can never be a factor of 3ⁿ. Therefore 3ⁿ is never divisible by 5 and cannot end with 5. You can also check the first few powers: 3¹ = 3, 3² = 9, 3³ = 27, 3⁴ = 81, 3⁵ = 243. The last digits repeat as 3, 9, 7, 1, and 5 never appears.

Key idea to remember

Fundamental Theorem of Arithmetic: see Key ideas used on this page.

Explanation

The Fundamental Theorem of Arithmetic says every composite number can be written as a product of primes in exactly one way, apart from the order. For 3ⁿ, that product is 3 × 3 × ⋯ × 3, so the only prime in its factorisation is 3. If 3ⁿ ended with 5, it would be divisible by 5, so 5 would have to appear in its prime factorisation. That is impossible, so 3ⁿ cannot end with 5.

Important formulas

3ⁿ = 3 × 3 × ⋯ × 3 (n times) · The only prime factor of 3ⁿ is 3.

Q2. The value of (HCF – LCM) for the two numbers 3 and 5 is :

Mark practised

(A) 2
(B) 4
(C) 14
(D) -14

Marks: 1
Level: Basic
Asked in: 4 paper sets
Years: 2025, 2026
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2025 Mathematics Basic, Set 430/2/1, Q1, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) HCF - LCM = 1 - 15 = -14.
Why? This gives the value the question asks for. HCF and LCM must be found before subtracting, because 3 and 5 share no prime factor.

The full 1 mark is for the correct option.

Answer:

(D) -14

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/2/1: option (D).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Subtracting in the wrong order: the question asks HCF - LCM, not LCM - HCF; 15 - 1 gives 14, which is option C.
Taking the HCF of two different primes as 15, the product, instead of 1.
Thinking the HCF is 0 because there seems to be no common factor; 1 is always a common factor of every pair of integers.

Why this works

3 and 5 are different primes, so the only common factor they have is 1; hence the HCF is 1. The LCM is the product of the greatest powers of the prime factors involved: 3 × 5 = 15. The expression asks for HCF minus LCM, so the value is 1 - 15 = -14, option D. The order of subtraction is fixed by the brackets: (HCF – LCM).

Key idea to remember

HCF and LCM by the prime factorisation method: see Key ideas used on this page.

Explanation

Every composite number has a unique prime factorisation. Here 3 and 5 are already primes. The HCF is the product of the smallest power of each common prime factor; there is no common prime factor, so the HCF is 1. The LCM is the product of the greatest power of each prime factor involved, so 3 × 5 = 15.

Important formulas

HCF(a, b) × LCM(a, b) = a × b · Holds for any two positive integers; can be used to check the two values.
HCF = product of the smallest power of each common prime factor · For 3 and 5 there is no common prime factor, so the HCF is 1.
LCM = product of the greatest power of each prime factor involved · For 3 and 5, this gives 3 × 5 = 15.

Q3. The number \(3 + \sqrt{2}\) is :

Mark practised

(A) a rational number
(B) an irrational number
(C) an integer
(D) a natural number

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2025, 2026
Topic: 1.3 Revisiting Irrational Numbers
Source: (CBSE 2025 Mathematics Basic, Set 430/1/1, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) an irrational number, because the sum of a rational number and an irrational number is irrational.
Why? A rational number plus an irrational number is irrational; otherwise the irrational number would equal the difference of two rational numbers, which is rational — a contradiction.

The full 1 mark is for the correct option.

Answer:

(B) an irrational number

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/1/1: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing A because 3 is rational, without remembering that adding an irrational number keeps the sum irrational.
Choosing C because 3 + √2 lies between 4 and 5, forgetting that an integer is a whole number such as 4 or 5, not a number between them.
Thinking √2 is rational because it can be written over 1 as √2/1; in the form p/q, both p and q must be integers, and √2 is not an integer.

Why this works

A rational number can be written as p/q, where p and q are integers and q ≠ 0. Since 3 = 3/1, 3 is rational. √2 is irrational, as proved in NCERT. The NCERT text states that the sum of a rational and an irrational number is irrational. Therefore 3 + √2 is irrational. An irrational number is not an integer and not a natural number, so options C and D are wrong. Option A is wrong because the sum is not rational. Another correct method: assume 3 + √2 = a/b with integers a and b; then √2 = a/b − 3 = (a − 3b)/b, which is rational, contradicting the fact that √2 is irrational.

Key idea to remember

Sum of a rational and an irrational number is irrational: see Key ideas used on this page.

Explanation

NCERT states: the sum or difference of a rational and an irrational number is irrational. Here 3 is rational and √2 is irrational, so 3 + √2 is irrational. This rules out options A, C and D.

Important formulas

rational + irrational = irrational · Used directly to classify 3 + √2.
A number s is irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. · Definition used to identify 3 as rational and √2 as irrational.

Q4. \((2 + \sqrt{2})^2\) is :

Mark practised

(A) a rational number
(B) an irrational number
(C) an integer
(D) a natural number

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2025, 2026
Topic: 1.3 Revisiting Irrational Numbers
Source: (CBSE 2025 Mathematics Basic, Set 430/1/3, Q7, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(2+√2)² = 2² + 2×2×√2 + (√2)² = 4 + 4√2 + 2 = 6 + 4√2, so the number is (B) an irrational number.
Why? This expansion uses the square identity and then applies the rule that a rational number plus an irrational number is irrational; that directly gives the option.

The full 1 mark is for the correct option.

Answer:

(B) an irrational number

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/1/3: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Forgetting the middle term 4√2 while expanding, and then calling the number rational.
Thinking that because 6 and 4 are rational, 6 + 4√2 is rational; the irrational part 4√2 does not vanish.
Calling it an integer or natural number because 6 is a whole number; 4√2 is not an integer, so the sum is not an integer.

Why this works

Using (a+b)² = a² + 2ab + b² with a = 2 and b = √2, the square becomes 4 + 4√2 + 2, which is 6 + 4√2. Since √2 is irrational, 4√2 is irrational (a non-zero rational times an irrational is irrational). Also, 6 + 4√2 is irrational (a rational plus an irrational is irrational). So the number is not rational, not an integer, and not a natural number. Hence option B is correct.

Key idea to remember

Irrational numbers: see Key ideas used on this page.

Explanation

A number is irrational if it cannot be written in the form p/q. The chapter proves √2 is irrational, and states that the sum or difference of a rational and an irrational number is irrational, and the product of a non-zero rational and an irrational number is irrational. Here the square becomes 6 + 4√2; 4√2 is irrational, so adding the rational 6 keeps it irrational.

Important formulas

(a+b)² = a² + 2ab + b² · Used to expand (2+√2)².
rational + irrational = irrational · 6 is rational and 4√2 is irrational, so their sum is irrational.
non-zero rational × irrational = irrational · 4 × √2 is irrational.

Q5. The number 3\(^{n}\), where n is a natural number, cannot end with the digit :

Mark practised

(A) 1
(B) 3
(C) 5
(D) 7

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2025, 2026
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2025 Mathematics Basic, Set 430/2/2, Q14, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 5, because \(3^n\) has only the prime 3 in its prime factorisation, so it can never be divisible by 5, while every number ending in 5 is divisible by 5.
Why? By the Fundamental Theorem of Arithmetic, the prime factorisation of \(3^n\) is unique and contains only the prime 3, so 5 can never be one of its prime factors.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/2/2: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Picking 1 because 3⁴ ends with 1, without reading that the question asks for the digit that can never appear.
Picking 3 because 3¹ = 3 ends in 3, forgetting that the question asks for a digit that is impossible.
Thinking \(3^n\) could end in 5 because multiplying by 5 gives a number ending in 5, not realising that \(3^n\) has only the prime factor 3.

Why this works

A number that ends with the digit 5 must be divisible by 5. But \(3^n\) is a product of n factors each equal to 3, so the only prime in its prime factorisation is 3. By the Fundamental Theorem of Arithmetic, this factorisation is unique, so 5 can never appear in it. Therefore \(3^n\) can never end with 5. You can also check the last digits: 3¹ = 3, 3² = 9, 3³ = 27, 3⁴ = 81, 3⁵ = 243, and then the last digits repeat as 3, 9, 7, 1. The digit 5 never appears.

Key idea to remember

Fundamental Theorem of Arithmetic: see Key ideas used on this page.

Explanation

The Fundamental Theorem of Arithmetic says every composite number can be expressed as a product of primes in exactly one way, apart from the order of the factors. Since \(3^n\) is just the prime 3 repeated n times, its prime factorisation contains only 3. A number ending in 5 must be divisible by 5, so its prime factorisation would contain 5, which is impossible for \(3^n\).

Important formulas

\(3^n\) = \(3^n\) · The base 3 is prime, so the prime factorisation of \(3^n\) contains only the prime 3.
A number ending with 5 is divisible by 5. · Such a number would need 5 as a prime factor, which \(3^n\) cannot have.

Q6. The number \((5 - 3\sqrt{5} + \sqrt{5})\) is :

Mark practised

(A) an integer
(B) a rational number
(C) an irrational number
(D) a whole number

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2023
Topic: 1.3 Revisiting Irrational Numbers
Source: (CBSE 2023 Mathematics Basic, Set 430/4/1, Q4, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 5 − 3√5 + √5 = 5 − 2√5, which is an irrational number.
Why? Combining the like terms shows the number is 5 − 2√5; since 5 is rational and 2√5 is irrational, their difference is irrational, so option C is correct.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/4/1: option (C).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Adding the coefficients wrongly: −3√5 + √5 = −2√5, not −3√5 or −4√5.
Calling the number rational just because 5 is rational; the irrational part 2√5 makes the whole number irrational.
Thinking √5 cancels with 3√5; only the two √5 terms combine.

Why this works

First simplify the surd terms: −3√5 + √5 = −2√5. The number becomes 5 − 2√5. Here 5 is rational and 2√5 is irrational, because a non-zero rational number multiplied by an irrational number is irrational. The NCERT result says the difference of a rational and an irrational number is irrational, so 5 − 2√5 is irrational. Hence option C is correct.

Key idea to remember

Sum or difference of a rational and an irrational number is irrational: see Key ideas used on this page.

Explanation

NCERT states that the sum or difference of a rational and an irrational number is irrational, and the product of a non-zero rational and an irrational number is irrational. Here 5 is rational, √5 is irrational, so 2√5 is irrational, and therefore 5 − 2√5 is irrational.

Important formulas

rational ± irrational = irrational · Used to conclude that 5 − 2√5 is irrational.
a√b is irrational when a is a non-zero rational and √b is irrational · Used to identify 2√5 as irrational.

Q7. If two positive integers \(p\) and \(q\) can be expressed as \(p = 18a^2b^4\) and \(q = 20a^3b^2\), where \(a\) and \(b\) are prime numbers, then \(\text{LCM}(p, q)\) is :

Mark practised

(A) \(2a^2b^2\)
(B) \(180a^2b^2\)
(C) \(12a^2b^2\)
(D) \(180a^3b^4\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2024
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2024 Mathematics Standard, Set 30/1/1, Q5, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) LCM = 2² × 3² × 5 × a³ × b⁴ = 180a³b⁴
Why? The LCM is found by taking the greatest power of every prime factor that appears in either number, so this one line gives the option directly.

The full 1 mark is for the correct option.

Answer:

(D) 180a³b⁴

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/1/1: option (D).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 2a²b², which is the HCF, because HCF uses the smallest power of each common prime factor.
Taking the smallest powers of a and b while taking the greatest powers of the numerical primes, which gives 180a²b².
Forgetting the prime 5 that comes from 20.

Why this works

Prime factorisation makes the LCM visible. Factor p and q completely; because a and b are primes, they stay as they are. The LCM must contain every prime factor that appears in either number, and for each prime it must contain the highest power that appears. So 2 appears as 2², 3 as 3², 5 as 5, a as a³ and b as b⁴. Multiplying these gives 180a³b⁴.

Key idea to remember

Prime factorisation method: see Key ideas used on this page.

Explanation

The Fundamental Theorem of Arithmetic says every composite number has a unique prime factorisation. Using it, the LCM of two numbers is the product of the greatest power of each prime factor involved in the numbers.

Important formulas

LCM = product of the greatest power of each prime factor involved in the numbers · This is the rule used to combine p and q.

Q8. If x is the LCM of 4, 6, 8 and y is the LCM of 3, 5, 7 and p is the LCM of x and y, then which of the following is true ?

Mark practised

(A) p = 35x
(B) p = 4y
(C) p = 8x
(D) p = 16y

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2025
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2025 Mathematics Standard, Set 30/3/1, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) p = 35x, because x = LCM(4, 6, 8) = 24, y = LCM(3, 5, 7) = 105 and LCM(x, y) = 840 = 35 × 24.
Why? By the prime factorisation method, the LCM is the product of the greatest power of each prime factor involved in the numbers, so the LCM of x and y is 35 times x.

The full 1 mark is for the correct option.

Answer:

(A) p = 35x = 840

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/3/1: option (A).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing B because 4 × 105 = 420, without checking that the LCM of 24 and 105 must contain 2³, 3, 5 and 7.
Finding the LCM of 4, 6 and 8 as 4 × 6 × 8 = 192, instead of taking the greatest power of each prime factor.
Multiplying x and y as 24 × 105 = 2520 and calling that p, instead of finding the LCM of x and y.

Why this works

Prime factorise the numbers first. 4 = 2², 6 = 2 × 3 and 8 = 2³, so x = 2³ × 3 = 24. The numbers 3, 5 and 7 are primes, so y = 3 × 5 × 7 = 105. Now x = 2³ × 3 and y = 3 × 5 × 7; the LCM of x and y takes every prime with its greatest power, so p = 2³ × 3 × 5 × 7 = 840. Since 35 × 24 = 840, p = 35x. Quick check: 4 × 105 = 420, 8 × 24 = 192 and 16 × 105 = 1680, none of which equals 840.

Key idea to remember

Fundamental Theorem of Arithmetic: see Key ideas used on this page.

Explanation

Every composite number can be expressed as a product of primes in only one way, apart from the order of the factors. This solution uses the prime factorisation method that follows from it: the LCM is the product of the greatest power of each prime factor involved in the numbers.

Important formulas

LCM = product of the greatest power of each prime factor involved in the numbers. · Use this to find the LCM of 4, 6, 8 and then the LCM of 24 and 105.

Q9. \(\left(1 + \sqrt{3}\right)^2 - \left(1 - \sqrt{3}\right)^2\) is :

Mark practised

(A) a positive rational number.
(B) a negative integer.
(C) a positive irrational number.
(D) a negative irrational number.

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2025
Topic: 1.3 Revisiting Irrational Numbers
Source: (CBSE 2025 Mathematics Standard, Set 30/4/1, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) a positive irrational number, because (1+√3)² − (1−√3)² = 4√3, which is positive and irrational.
Why? By the identity (a+b)² − (a−b)² = 4ab, the expression equals 4√3. Since √3 is irrational, 4√3 is irrational, and it is positive because both 4 and √3 are positive.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/4/1: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing A because 4 is rational, without noticing that the factor √3 makes the product 4√3 irrational.
Writing the difference as 4 + 2√3 − 4 − 2√3, which wrongly gives 0, because the minus sign before the second bracket must turn −2√3 into +2√3.
Thinking (1−√3)² is negative because 1−√3 is negative; a square is never negative.

Why this works

Expand using (a±b)² = a² ± 2ab + b². We get (1+√3)² = 1 + 2√3 + 3 = 4 + 2√3, and (1−√3)² = 1 − 2√3 + 3 = 4 − 2√3. Their difference is (4 + 2√3) − (4 − 2√3) = 4√3. The identity (a+b)² − (a−b)² = 4ab gives 4 × 1 × √3 = 4√3 at once. Since √3 is irrational, 4√3 is irrational; since 4 and √3 are positive, it is positive.

Key idea to remember

Irrational Numbers: see Key ideas used on this page.

Explanation

A number is irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. The expression simplifies to 4√3, which is the product of the rational number 4 and the irrational number √3. Multiplying an irrational number by a non-zero rational number keeps it irrational, so 4√3 is irrational. It is positive because both 4 and √3 are positive.

Important formulas

(a+b)² − (a−b)² = 4ab · Take a as 1 and b as √3 to get the value of the expression directly.
(a±b)² = a² ± 2ab + b² · Used to expand each square when solving without the difference identity.

Q10. \((\sqrt{3}+2)^2 + (\sqrt{3}-2)^2\) is a/an

Mark practised

(A) positive rational number
(B) negative rational number
(C) positive irrational number
(D) negative irrational number

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2025
Topic: 1.3 Revisiting Irrational Numbers
Source: (CBSE 2025 Mathematics Standard, Set 30/5/1, Q1, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) positive rational number \((\sqrt{3}+2)^2+(\sqrt{3}-2)^2=(3+4+4\sqrt3)+(3+4-4\sqrt3)=14\)
Why? The whole question is this one expansion. Squaring each bracket separately exposes the irrational terms with opposite signs, so they cancel and leave a plain positive integer.

The full 1 mark is for the correct option.

Answer:

(A) positive rational number

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/5/1: option (A).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Marking option C because √3 appears in the expression, without expanding and checking whether the irrational parts cancel.
Forgetting that an integer like 14 is also rational, since it can be written as 14/1.
Expanding (a+b)² as a² + b² (dropping the middle term 2ab) is wrong algebra.

Why this works

Expand both squares using the identities for (a+b)² and (a-b)². The +4√3 and −4√3 cancel, leaving 7 + 7 = 14. Since 14 can be written as 14/1, it is a rational number, and it is clearly positive. Another correct method also earns full marks.

Key idea to remember

Irrational numbers: see Key ideas used on this page.

Explanation

A number is irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. Otherwise it is rational. Here the irrational parts cancel, and 14 can be written as 14/1, so the number is rational. Since 14 is greater than 0, it is positive.

Important formulas

(a+b)² + (a-b)² = 2(a² + b²) · This identity is applied to the two brackets together. With the given numbers, 2(3 + 4) = 14.
(a ± b)² = a² ± 2ab + b² · Used to expand each bracket separately; the middle terms cancel.

Q11. Which of the following cannot be the unit digit of \(8^n\), where n is a natural number?

Mark practised

(A) 4
(B) 2
(C) 0
(D) 6

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2025
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2025 Mathematics Standard, Set 30/6/1, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 0, because the unit digits of 8ⁿ cycle through 8, 4, 2, 6, so 0 is never a unit digit.
Why? The unit digits of powers of 8 repeat in the cycle 8, 4, 2, 6, so any digit outside this cycle, including 0, can never occur.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/6/1: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Picking 2 because 8³ = 512 ends in 2, without checking that 0 never appears in the cycle.
Thinking 0 can be a unit digit because 8 is even and some even numbers end in 0.
Choosing 6 because 8⁴ = 4096 ends in 6, forgetting the cycle continues to 8 again at 8⁵.
Forgetting the question asks for the digit that cannot be the unit digit, and selecting a digit that actually appears.

Why this works

Write the first few powers: 8¹ = 8, 8² = 64, 8³ = 512, 8⁴ = 4096, 8⁵ = 32768. The unit digits are 8, 4, 2, 6, 8, … so the cycle has length 4 and keeps repeating, because 8⁴ ends in 6 and 8 × 6 = 48 ends in 8. So the possible unit digits are only 8, 4, 2 and 6; 0 is not among them. This also follows from the Fundamental Theorem of Arithmetic: if 8ⁿ ended with 0, it would be divisible by 5, so 5 would be a prime factor of 8ⁿ. But 8ⁿ = (2³)ⁿ = 2³ⁿ, whose only prime factor is 2, so 5 can never divide 8ⁿ. Hence 0 cannot be the unit digit.

Key idea to remember

Fundamental Theorem of Arithmetic: see Key ideas used on this page.

Explanation

The Fundamental Theorem of Arithmetic says every composite number has a unique prime factorisation. Since 8ⁿ = 2³ⁿ, its prime factorisation contains only 2. A number ending in 0 must be divisible by 5, so 5 would have to appear in that factorisation, which is impossible. Therefore no power of 8 ends in 0.

Important formulas

8ⁿ = (2³)ⁿ = 2³ⁿ · Shows that only the prime 2 appears in the prime factorisation of 8ⁿ.
Unit digits of 8ⁿ: 8, 4, 2, 6, 8, ... · Cycle of length 4 for successive natural values of n.

Q12. The natural number 2 is :

Mark practised

(A) a prime number
(B) a composite number
(C) prime as well as composite
(D) neither prime nor composite

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2026
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2026 Mathematics Standard, Set 30/1/1, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) a prime number, because 2 has exactly two factors, 1 and 2.
Why? A prime number is a natural number greater than 1 whose only positive factors are 1 and itself; a composite number must have more than two factors.

The full 1 mark is for the correct option.

Answer:

(A) a prime number

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30/1/1: option (A).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing B because 2 is even, although an even number can be prime; 2 is the only even prime.
Choosing C because 2 is prime and even, but a number cannot be both prime and composite.
Choosing D because 2 is small, but 2 is the smallest prime number.

Q13. For any natural number \(n\), \(6^n\) ends with the digit :

Mark practised

(A) 0
(B) 6
(C) 3
(D) 2

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2026
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2026 Mathematics Standard, Set 30/1/1, Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 6, because 6¹ = 6 and 6 × 6 = 36 ends with 6, so every power of 6 ends with 6.
Why? A number ending in 6, when multiplied by 6, always gives a product ending in 6; therefore the last digit cannot change from power to power.

The full 1 mark is for the correct option.

Answer:

(B) 6

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30/1/1: option (B).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 0 because 6 is divisible by 2 and 3; ending in 0 needs a factor 10 = 2 × 5, so 5 must divide the number, but \(6^n\) has no factor 5.
Choosing 2 just because \(6^n\) is even; being even only limits the last digit to 0, 2, 4, 6 or 8, and the actual last digit is 6.
Thinking the last digit changes as n increases; checking 6¹ = 6 and 6² = 36 shows it stays 6.

Why this works

Write \(6^n\) = 6 × 6 × ⋯ × 6 (n times). Start with 6¹ = 6. When you multiply by 6, only the last digits matter: 6 × 6 = 36, so the new last digit is again 6. Hence every power of 6 ends with 6: 6¹ = 6, 6² = 36, 6³ = 216, 6⁴ = 1296. Also, by the Fundamental Theorem of Arithmetic, \(6^n\) = \((2 × 3)^n\) = \(2^n\) × \(3^n\), so 5 is never a prime factor of \(6^n\); that is why the choice 0 is wrong, because a number ending in 0 must be divisible by 5.

Key idea to remember

Fundamental Theorem of Arithmetic: see Key ideas used on this page.

Explanation

The Fundamental Theorem of Arithmetic says every composite number has a unique prime factorisation. Since \(6^n\) = \(2^n\) × \(3^n\), its only prime factors are 2 and 3, so 5 never divides it and it cannot end with 0. To find which digit it does end with, multiply 6 repeatedly: the last digit 6 stays fixed.

Important formulas

\(6^n\) = 6 × 6 × ⋯ × 6 (n times) · Multiplying by 6 keeps the last digit 6.
\(6^n\) = \(2^n\) × \(3^n\) · The prime factorisation of \(6^n\) contains only 2 and 3, never 5.

Q14. The natural number 1 is :

Mark practised

(A) a prime number.
(B) a composite number.
(C) prime as well as composite.
(D) neither prime nor composite.

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2026
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2026 Mathematics Standard, Set 30/2/1, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) neither prime nor composite, because 1 has exactly one factor, itself.
Why? A prime number has exactly two distinct factors, 1 and itself; a composite number has more than two factors. Since 1 has only one factor, it satisfies neither definition.

The full 1 mark is for the correct option.

Answer:

(D) neither prime nor composite

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30/2/1: option (D).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing A because 1 has no factor other than itself, without noticing that a prime needs two distinct factors, 1 and itself.
Thinking 1 × 1 makes 1 a product, so 1 is composite, without remembering that a composite number must have more than two factors.
Choosing C because 1 is divisible only by 1, confusing having one factor with being both prime and composite.

Q15. (HCF \(\times\) LCM) for the numbers 70 and 40 is :

Mark practised

(A) 10
(B) 280
(C) 2800
(D) 70

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2023
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2023 Mathematics Basic, Set 430/4/1, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) HCF × LCM = 70 × 40 = 2800
Why? This applies the relation HCF × LCM = product of the two numbers directly, so the answer is obtained in one line.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/4/1: option (C).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 280, the LCM, instead of HCF × LCM.
Choosing 10, the HCF, instead of HCF × LCM.
Adding HCF and LCM instead of multiplying them.

Q16. If p² = 32/50, then p is a/an

Mark practised

(A) whole number
(B) integer
(C) rational number
(D) irrational number

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2023
Topic: 1.3 Revisiting Irrational Numbers
Source: (CBSE 2023 Mathematics Standard, Set 30/6/1, Q1, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) rational number, because p² = 32/50 = 16/25, so p = ±4/5, a rational number.
Why? By definition, a number that can be written as a fraction of two integers is rational; both 4/5 and −4/5 are such fractions.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/6/1: option (C).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing irrational because the square root of a fraction looks like a radical; here √(32/50) simplifies to 4/5, which is rational.
Forgetting the negative square root and taking only p = 4/5; the answer stays rational, but the value set is incomplete.
Thinking p is an integer because 25 is a perfect square; p = ±4/5, and 4/5 is not an integer.

Q17. A pair of irrational numbers whose product is a rational number is :

Mark practised

(A) \((\sqrt{16}, \sqrt{4})\)
(B) \((\sqrt{5}, \sqrt{2})\)
(C) \((\sqrt{3}, \sqrt{27})\)
(D) \((\sqrt{36}, \sqrt{2})\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2024
Topic: 1.3 Revisiting Irrational Numbers
Source: (CBSE 2024 Mathematics Standard, Set 30/3/1, Q14, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) √3 × √27 = √81 = 9, which is a rational number.
Why? This one line checks the condition for option C: both numbers are irrational and their product is rational.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/3/1: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing option A because it has √ signs, without noticing that √16 = 4 and √4 = 2 are rational numbers.
Thinking the product of two irrational numbers must always be irrational; option C shows a counterexample because √3 × √27 = 9 is rational.

Why this works

An irrational number cannot be written as p/q with integers p and q, q ≠ 0. The question wants a pair in which both numbers are irrational, but their product is rational. Option C works: √3 is irrational (see Example 5), and √27 = √(9 × 3) = 3√3 is irrational by the same reasoning as Example 7. Their product is √3 × √27 = √81 = 9, which is rational. The other options fail: A has √16 = 4 and √4 = 2, both rational; B gives √5 × √2 = √10, which is irrational; D has √36 = 6, which is rational, and its product 6√2 is irrational. Another correct method also earns full marks.

Key idea to remember

Irrational numbers: see Key ideas used on this page.

Explanation

A number is irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. In this question, we need a pair of irrational numbers whose product is a rational number. Using √a × √b = √(ab), option C gives √3 × √27 = √81 = 9, and 9 is rational.

Important formulas

√a × √b = √(ab) · Used for positive numbers a and b; this is how √3 × √27 becomes √81.
√(a²) = a for a ≥ 0 · This is why √16, √4 and √36 simplify to whole numbers.

Q18. The greatest number which divides 281 and 1249, leaving remainder 5 and 7 respectively, is :

Mark practised

(A) 23
(B) 276
(C) 138
(D) 69

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2024
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2024 Mathematics Standard, Set 30/5/1, Q10, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) HCF(281 − 5, 1249 − 7) = HCF(276, 1242) = 2 × 3 × 23 = 138
Why? Subtract the remainders so the required number divides both new numbers exactly; the greatest such number is their HCF.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/5/1: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking the HCF of 281 and 1249 directly, without subtracting the remainders 5 and 7.
Picking 276 or 1242 as the answer instead of finding their HCF.
Forgetting the common factor 2 and taking only the product of 3 and 23 as the HCF.

Why this works

If a number d divides 281 and leaves remainder 5, then 281 = dq + 5, so d divides 281 − 5 = 276. Similarly, d divides 1249 − 7 = 1242. The largest number that divides both is their HCF. Factorising: 276 = 2² × 3 × 23 and 1242 = 2 × 3³ × 23, so HCF = 2 × 3 × 23 = 138. Therefore the greatest number is 138, option C.

Key idea to remember

HCF by the prime factorisation method (Fundamental Theorem of Arithmetic): see Key ideas used on this page.

Explanation

Every composite number can be expressed as a product of primes in a unique way. The HCF of two numbers is the product of the smallest power of each common prime factor. Here the common prime factors of 276 and 1242 are 2, 3 and 23, so HCF = 2 × 3 × 23 = 138.

Important formulas

HCF(a, b) = product of the smallest power of each common prime factor in the numbers. · NCERT Example 2
If a number leaves remainder r when divided by d, then the number minus r is exactly divisible by d. · Used to form 276 and 1242.

Q19. If the product of two co-prime numbers is 553, then their HCF is :

Mark practised

(A) 1
(B) 553
(C) 7
(D) 79

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2024
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2024 Mathematics Standard, Set 30/5/1, Q14, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
Co-prime numbers have no common factor other than 1, so their HCF is 1. Option (A) 1.
Why? This step applies the definition of co-prime numbers directly. Since the numbers are already given as co-prime, their HCF is fixed as 1, and no factorisation of 553 is needed.

The full 1 mark is for the correct option.

Answer:

(A) 1

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/5/1: option (A).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 553, the product, as the HCF because the question mentions product.
Choosing 7 or 79 after factorising the product; these are factors, not the HCF of the two co-prime numbers.
Confusing HCF with LCM: for co-prime numbers, the LCM is their product, but the HCF is 1.

Q20. Which of the following is a rational number between \(\sqrt{3}\) and \(\sqrt{5}\) ?

Mark practised

(A) 1.4142387954012 ...
(B) 2.32\(\overline{6}\)
(C) \(\pi\)
(D) 1.857142

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2025
Topic: 1.3 Revisiting Irrational Numbers
Source: (CBSE 2025 Mathematics Standard, Set 30/1/1, Q14, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 1.857142, because 1.857142 is a terminating decimal and 3 < 1.857142² < 5.
Why? This selects the correct option and gives the two checks needed: the number is rational, and its square places it between √3 and √5.

The full 1 mark is for the correct option.

Answer:

(D) 1.857142

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/1/1: option (D).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing B just because 2.32666... is clearly rational; it is greater than √5 ≈ 2.236, so it is not between the two roots.
Thinking any decimal written with ... is irrational; option A is below √3, so it fails the between condition first.
Forgetting to check that the chosen number is rational after checking that its value lies between √3 and √5.

Why this works

For a positive number x, x lies between √3 and √5 exactly when x² lies between 3 and 5. For option D, 1.857142² = 3.448976408164, which is between 3 and 5. Also, 1.857142 is a terminating decimal, so it can be written as 1857142/1000000, hence it is rational. Option A is about 1.414..., which is below √3. Option B is 2.32666..., which is above √5. Option C is π ≈ 3.14, which is also above √5. Another correct method also works.

Key idea to remember

Irrational numbers: see Key ideas used on this page.

Explanation

NCERT says a number is irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. So a rational number is one that can be written in that form. A terminating decimal is rational. Also, because squaring is increasing for positive numbers, √3 < x < √5 is the same as 3 < x² < 5. This solution uses both facts to verify option D.

Important formulas

If x > 0, then √3 < x < √5 ⇔ 3 < x² < 5. · Used to check the position of 1.857142 without estimating square roots.
1.857142 = 1857142/1000000 · A terminating decimal is rational because it can be written as p/q with q ≠ 0.

Q21. If \(HCF(98, 28) = m\) and \(LCM(98, 28) = n\), then the value of \(n - 7m\) is :

Mark practised

(A) 0
(B) 28
(C) 98
(D) 198

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2025
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2025 Mathematics Standard, Set 30/1/1, Q6, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) HCF(98, 28) = 2 × 7 = 14, LCM(98, 28) = 2² × 7² = 196, so n − 7m = 196 − 7 × 14 = 98.
Why? This one line does all the calculation: it finds m and n by the prime factorisation method and substitutes them into the given expression, which directly gives the option.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/1/1: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Forgetting the 7 in 7m and computing n − m instead of n − 7m.
Reversing m and n, so taking m as the LCM and n as the HCF.
Choosing 28 as the HCF without checking that 28 divides 98.

Why this works

Factorise each number into primes: 98 = 2 × 7² and 28 = 2² × 7. The HCF takes the smallest power of each common prime factor, so HCF = 2¹ × 7¹ = 14. The LCM takes the greatest power of each prime factor involved in either number, so LCM = 2² × 7² = 196. Put m = 14 and n = 196 into n − 7m, giving n − 7m = 196 − 98 = 98. Another correct method also earns full marks.

Key idea to remember

Fundamental Theorem of Arithmetic: see Key ideas used on this page.

Explanation

The Fundamental Theorem of Arithmetic says every composite number can be expressed as a product of primes in a unique way, apart from the order of the factors. This uniqueness lets us read the HCF and LCM directly from the prime factorisations of 98 and 28.

Important formulas

HCF = product of the smallest power of each common prime factor in the numbers. · Used to find m.
LCM = product of the greatest power of each prime factor involved in the numbers. · Used to find n.
HCF(a, b) × LCM(a, b) = a × b. · A quick check for n once m is known.

Q22. The sum of the exponents of prime factors in the prime factorisation of 4004 is :

Mark practised

(A) 5
(B) 4
(C) 3
(D) 2

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2025
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2025 Mathematics Standard, Set 30/2/1, Q10, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 4004 = 2² × 7 × 11 × 13, so sum of exponents = 2 + 1 + 1 + 1 = 5.
Why? The prime factorisation must be complete, so every prime factor appears and its exponent is visible; adding those exponents gives exactly the sum asked in the question.

The full 1 mark is for the correct option.

Answer:

(A) 5

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/2/1: option (A).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Stopping at 4004 = 4 × 1001 and counting only the exponent of 2, because 1001 is not yet factorised into primes.
Counting the number of prime factors (2, 7, 11, 13) instead of adding their exponents (2 + 1 + 1 + 1 = 5).
Forgetting that a prime appearing once has exponent 1, so the three 1s are part of the sum.

Q23. For any natural number \(n\), \(5^n\) ends with the digit :

Mark practised

(A) 0
(B) 5
(C) 3
(D) 2

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2026
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2026 Mathematics Standard, Set 30/2/1, Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 5, because the prime factorisation of 5ⁿ contains only the prime 5, so 5ⁿ is divisible by 5 but not by 2, and its last digit is 5.
Why? A number divisible by 5 ends in 0 or 5; a number not divisible by 2 cannot end in 0, so the only possible last digit is 5.

The full 1 mark is for the correct option.

Answer:

(B) 5

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30/2/1: option (B).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 0 because a number ending in 0 is divisible by 5, without noticing that 5ⁿ has no factor 2.
Choosing 2 because 5 is odd, without checking that every power of 5 is divisible by 5 and so must end in 0 or 5.

Why this works

Write 5ⁿ as a product of n factors: 5ⁿ = 5 × 5 × ⋯ × 5. The only prime in this factorisation is 5. A number ending in 0 is divisible by 10, hence by 2 and 5; a number ending in 2 is divisible by 2; a number ending in 3 is not divisible by 5. Since 5ⁿ is divisible by 5 but not by 2, its last digit can only be 5. Quick check: 5¹ = 5, 5² = 25, 5³ = 125, 5⁴ = 625 — every power ends in 5. This is the same reasoning as NCERT Example 1 and Exercise 1.1 Q5, where 4ⁿ and 6ⁿ cannot end in 0 because their prime factorisations contain no 5.

Key idea to remember

Fundamental Theorem of Arithmetic: see Key ideas used on this page.

Explanation

The Fundamental Theorem of Arithmetic says every composite number can be written as a product of primes in exactly one way, apart from the order of the factors. Here 5ⁿ = 5 × 5 × ⋯ × 5, so the only prime in its factorisation is 5. Therefore 5ⁿ is divisible by 5 but never by 2. A number divisible by 5 ends in 0 or 5; since 5ⁿ is not divisible by 2, it cannot end in 0, so it must end in 5.

Important formulas

5ⁿ = 5 × 5 × ⋯ × 5 (n times) · The only prime in the factorisation of 5ⁿ is 5.
A number divisible by 5 ends in 0 or 5. · 5ⁿ is not divisible by 2, so the digit 0 is ruled out.

Q24. If the HCF of 72 and 234 is 18, then the LCM (72, 234) is :

Mark practised

(A) 936
(B) 836
(C) 324
(D) 234

Marks: 1
Level: Basic
Years: 2023
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2023 Mathematics Basic, Set 430/2/2, Q13, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) LCM (72, 234) = (72 × 234)/18 = 936
Why? The question gives the HCF and asks for the LCM, so this step uses the HCF–LCM relation to find the LCM directly.

The full 1 mark is for the correct option.

Answer:

(A) 936

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/2/2: option (A).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Forgetting to divide by the HCF and giving the product 16848 as the LCM.
Confusing HCF and LCM and giving 18 as the answer.
Choosing 324, which is 18 squared, instead of using the HCF–LCM relation.

Q25. (HCF × LCM) for the numbers 30 and 70 is :

Mark practised

(A) 2100
(B) 21
(C) 210
(D) 70

Marks: 1
Level: Basic
Years: 2023
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2023 Mathematics Basic, Set 430/4/2, Q8, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) HCF × LCM = 30 × 70 = 2100
Why? The property HCF × LCM = product of the two numbers gives the answer directly, so there is no need to find HCF and LCM separately.

The full 1 mark is for the correct option.

Answer:

(A) 2100

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/4/2: option (A).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Picking 210, which is the LCM, instead of the product HCF × LCM.
Forgetting the direct property and finding HCF and LCM separately, which leaves more room for arithmetic slips.

Why this works

The question asks for HCF × LCM for 30 and 70. Use the result HCF(a,b) × LCM(a,b) = a × b. So the answer is 30 × 70 = 2100. To verify: 30 = 2 × 3 × 5 and 70 = 2 × 5 × 7, so HCF = 2 × 5 = 10 and LCM = 2 × 3 × 5 × 7 = 210; then 10 × 210 = 2100. Option A is correct.

Key idea to remember

For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b: see Key ideas used on this page.

Explanation

The Fundamental Theorem of Arithmetic says every composite number has a unique prime factorisation. The HCF uses the smallest power of each common prime factor and the LCM uses the greatest power of each prime factor involved, so their product reconstructs the original two numbers. Hence HCF(a,b) × LCM(a,b) = a × b.

Important formulas

HCF(a,b) × LCM(a,b) = a × b · For any two positive integers a and b. This result is stated in NCERT after Example 2.

Q26. The HCF of two numbers 65 and 104 is 13. If LCM of 65 and 104 is 40x, then the value of x is :

Mark practised

(A) 5
(B) 13
(C) 40
(D) 8

Marks: 1
Level: Standard
Years: 2024
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2024 Mathematics Standard, Set 30/4/1, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) LCM = 65 × 104 ÷ 13 = 520 = 40x, so x = 13.
Why? This uses the relation between HCF and LCM to find the LCM first, then compares it with the given form 40x to get x.

The full 1 mark is for the correct option.

Answer:

(B) 13

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/4/1: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking LCM as 65 × 104 = 6760 without dividing by HCF.
Solving the equation wrongly and writing 40 instead of 13.

Q27. If the prime factorisation of 2520 is \(2^3\) × \(3^a\) × b × 7, then the value of a + 2b is :

Mark practised

(A) 12
(B) 10
(C) 9
(D) 7

Marks: 1
Level: Standard
Years: 2024
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2024 Mathematics Standard, Set 30/5/2, Q12, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 12. 2520 = 2³ × 3² × 5 × 7, so a + 2b = 2 + 2(5) = 12.
Why? The uniqueness of prime factorisation lets us match the given form with the actual factorisation and read off the values of a and b.

The full 1 mark is for the correct option.

Answer:

(A) 12

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/5/2: option (A).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking b as 7 because 7 is written in the given form; b is the remaining prime factor, which is 5.
Adding a + b instead of a + 2b, which gives 7 instead of 12.
Reading 2³ as 2 times 3 instead of 2 × 2 × 2, which gives a wrong factorisation.

Why this works

Prime factorisation means writing a number as a product of its prime factors. The Fundamental Theorem of Arithmetic says this factorisation is unique except for the order of the factors. So the given form 2³ × \(3^a\) × b × 7 must be exactly the same as the actual factorisation 2³ × 3² × 5 × 7. The exponent of 3 is 2, and the only prime left is 5. Therefore a + 2b = 2 + 2(5) = 12.

Key idea to remember

Fundamental Theorem of Arithmetic: see Key ideas used on this page.

Explanation

Every composite number can be expressed as a product of primes in exactly one way, apart from the order of the factors. A composite number is a natural number that has factors other than 1 and itself. Since 2520 is composite, its prime factorisation is unique. This lets us match the given expression to the actual factorisation and read off the unknown values.

Important formulas

Composite number = \(p₁^a\) × \(p₂^b\) × ... (primes in ascending order) · Unique by the Fundamental Theorem of Arithmetic.

Q28. Let \(x = a^2 b^3 c^n\) and \(y = a^3 b^m c^2\) , where \(a, b, c\) are prime numbers. If LCM of \(x\) and \(y\) is \(a^3 b^4 c^3\) , then the value of \(m + n\) is

Mark practised

(A) 10
(B) 7
(C) 6
(D) 5

Marks: 1
Level: Standard
Years: 2025
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2025 Mathematics Standard, Set 30/5/1, Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 7, because the LCM takes the greatest power of each prime, so the exponent of b is 4 and the exponent of c is 3, giving m + n = 7.
Why? By the prime factorisation method, the LCM of two numbers contains each prime with the greatest of its exponents in the two numbers.

The full 1 mark is for the correct option.

Answer:

(B) 7

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/5/1: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking the smaller exponent of b and c, which gives 3 and 2, and adding them to get 5.
Setting m as 3 because b³ appears in x, without comparing it with \(b^m\) in y.
Confusing the primes: taking m from the exponent of c and n from the exponent of b, which gives 2 and 3 instead of 4 and 3.

Why this works

Write x and y in prime factorised form. The LCM is formed by taking each prime with the greatest exponent that appears in either number. For a, the exponents are 2 and 3, so the LCM has a³. For b, the exponents are 3 and m, and the LCM has b⁴, so m is 4. For c, the exponents are n and 2, and the LCM has c³, so n is 3. Hence m + n = 4 + 3 = 7. Check: with m as 4 and n as 3, x = a²b³c³ and y = a³b⁴c², and their LCM is a³b⁴c³, which matches the given LCM.

Key idea to remember

Fundamental Theorem of Arithmetic: see Key ideas used on this page.

Explanation

The Fundamental Theorem of Arithmetic says every composite number can be expressed as a product of primes in a unique way. Here x and y are already written as products of powers of primes, so the LCM is found by taking the greatest power of each prime factor involved in either number.

Important formulas

LCM = product of the greatest power of each prime factor involved in the numbers. · This is the prime factorisation method from NCERT Example 2.
For each prime p, exponent in LCM = greater of (exponent of p in x, exponent of p in y). · Used to compare the exponents of a, b and c.

Q29. Let \(a = p^2 q^3 r^n\) and \(b = p^3 q^m r^2\), where \(p, q, r\) are prime numbers. If LCM of \(a\) and \(b\) is \(p^3 q^4 r^3\), then the value of \(3n - 2m\) is

Mark practised

(A) -1
(B) 1
(C) 3
(D) -3

Marks: 1
Level: Standard
Years: 2025
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2025 Mathematics Standard, Set 30/5/2, Q4, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 1, because the LCM takes the greatest power of each prime, so m = 4 and n = 3, giving 3n − 2m = 3(3) − 2(4) = 1.
Why? By the prime factorisation method, the LCM of two numbers is the product of the greatest power of each prime factor involved in the numbers, so the exponent of each prime in the LCM is the larger of its two exponents.

The full 1 mark is for the correct option.

Answer:

(B) 1

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/5/2: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking m as 3, the exponent of q in a, gives 3(3) − 2(3) = 3, which is option C.
Forgetting the coefficients and computing n − m = 3 − 4 = −1, which is option A.
Using 2 for n and 3 for m gives 3(2) − 2(3) = 0, which is not an option.

Why this works

Write both numbers in prime factorised form. The LCM is the product of the greatest power of each prime factor involved in the numbers. For p, the greater of 2 and 3 is 3, which matches the given LCM. For q, the greater of 3 and m is 4, so m = 4. For r, the greater of n and 2 is 3, so n = 3. Then 3n − 2m = 3(3) − 2(4) = 1, so option B is correct. Quick check: with n = 3 and m = 4, a = p²q³r³ and b = p³q⁴r²; the greatest powers are p³, q⁴ and r³, so the LCM is p³q⁴r³, exactly as given.

Key idea to remember

Fundamental Theorem of Arithmetic: see Key ideas used on this page.

Explanation

The Fundamental Theorem of Arithmetic says that every composite number can be expressed as a product of primes in a unique way, apart from the order of the factors. Because of this uniqueness, when two numbers are written as products of powers of primes, their LCM is found by taking the greatest power of each prime factor involved. Comparing the given LCM with these greatest powers gives n = 3 and m = 4.

Important formulas

LCM(a, b) = product of the greatest power of each prime factor involved in the numbers · Used to compare the exponents of p, q and r with the given LCM.

Q30. Let \(p = x^2 y^3 z^n\) and \(q = x^3 y^m z^2\), where \(x, y, z\) are prime numbers. If \(\text{LCM}(p, q) = x^3 y^4 z^3\), then the value of \((2m + 3n)\) is

Mark practised

(A) 18
(B) 17
(C) 15
(D) 14

Marks: 1
Level: Standard
Years: 2025
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2025 Mathematics Standard, Set 30/5/3, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 17; 2 × 4 + 3 × 3 = 17
Why? This line states the correct option and shows the one calculation that gives the value. Comparing the exponents of y and z in the LCM fixes m and n, and substituting them gives 2m + 3n.

The full 1 mark is for the correct option.

Answer:

(B) 17

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/5/3: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Using the smaller exponent for LCM, which is the HCF rule, would wrongly give m as 3 and n as 2.
Not comparing all three primes; even though the x-exponent already matches, the y and z exponents still determine m and n.

Why this works

LCM means the least common multiple. By the prime factorisation method, the LCM of two numbers is the product of the greatest power of each prime factor involved in the numbers. Since x, y and z are primes, compare the exponents prime by prime. For x, the greater of 2 and 3 is 3, which already matches x³. For y, the greater of 3 and m is 4, so m is 4. For z, the greater of n and 2 is 3, so n is 3. Then 2m + 3n = 2 × 4 + 3 × 3 = 17, so option (B) is correct.

Key idea to remember

The Fundamental Theorem of Arithmetic: see Key ideas used on this page.

Explanation

The Fundamental Theorem of Arithmetic says every composite number can be expressed as a product of primes in a unique way, apart from the order. So when p and q are written in prime factorised form, the LCM contains each prime once with the greatest exponent that appears. Comparing the exponents of y and z in the given LCM with those in p and q fixes m and n.

Important formulas

LCM = product of the greatest power of each prime factor involved in the numbers. · For each prime, pick the larger exponent from the two numbers.

Q31. If HCF \((x, 20) = 2\) and LCM \((x, 20) = 60\), then value of \(x\) is :

Mark practised

(A) 3
(B) 6
(C) 20
(D) 10

Marks: 1
Level: Basic
Years: 2025
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2025 Mathematics Basic, Set 430/4/1, Q1, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 6, because HCF × LCM = x × 20, so x = (2 × 60)/20 = 6.
Why? For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b; putting the given values in this relation gives x.

The full 1 mark is for the correct option.

Answer:

(B) 6

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/4/1: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Dividing 60 by 20 and writing 3, without multiplying by the HCF 2.
Choosing 10 from 20 and the HCF 2 alone, without using the LCM 60.
Choosing 20 just because it appears in the question, without checking the HCF condition.

Q32. (3 × 11 × 13 + 3) is :

Mark practised

(A) a prime number
(B) divisible by 13
(C) a composite number
(D) an odd number

Marks: 1
Level: Standard
Years: 2026
Topic: 1.2 The Fundamental Theorem of Arithmetic
Source: (CBSE 2026 Mathematics Standard, Set 30/3/1, Q4, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) a composite number, because 3 × 11 × 13 + 3 = 3(11 × 13 + 1) = 3 × 144 = 432, and 432 has factors other than 1 and itself.
Why? A composite number has a factor other than 1 and itself; writing the expression as a product of two integers greater than 1 proves it is composite.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30/3/1: option (C).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing option A because 3, 11 and 13 are primes, without combining the extra +3 into 3 × 144.
Choosing option B because 13 appears in the product, without checking whether 432 is actually divisible by 13.
Choosing option D because 3 × 11 × 13 is odd, forgetting that odd + odd = even.

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Key ideas used on this page

Fundamental Theorem of Arithmetic
Used in questions 1 5 8 11 13 21 22 23 27 28 29 30
Irrational numbers
Used in questions 4 9 10 17 20
HCF(a, b) × LCM(a, b) = a × b
Used in questions 15 24
HCF and LCM by the prime factorisation method
Used in question 2
Sum of a rational and an irrational number is irrational
Used in question 3
Sum or difference of a rational and an irrational number is irrational
Used in question 6
Prime factorisation method
Used in question 7
Prime number
Used in question 12
Prime and composite numbers
Used in question 14
Rational numbers
Used in question 16
HCF by the prime factorisation method (Fundamental Theorem of Arithmetic)
Used in question 18
Co-prime numbers
Used in question 19
For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b
Used in question 25
HCF × LCM = product of two numbers
Used in question 26
HCF × LCM = a × b
Used in question 31
Composite number
Used in question 32

How to use these questions

Try each question on paper first. Then read the steps: the mark beside each step shows what the examiner looks for. If a step is new to you, revise Section 1.2 and Section 1.3 of the NCERT chapter. For the exam pattern and the repeated questions of the whole chapter, read the Real Numbers chapter overview.

To revise the lessons behind these questions, read the NCERT chapter, exactly as NCERT publishes it on its own website.

Official papers behind this page

Every question on this page comes from an official CBSE Class 10 Mathematics board paper. The board publishes the question papers on cbse.gov.in and the marking schemes on cbse.gov.in, where you can see how marks are given.

Year	Exam	Course	Question paper	Marking scheme
2026	Main	Standard (041)	Open CBSE's official question paper for the 2026 main exam (one ZIP file of about 61.9 MB, Class 10 Maths Standard and Basic together) to read the full paper.	Open CBSE's official marking scheme for the 2026 main exam (one ZIP file of about 22.8 MB, Class 10 Maths Standard (041)) to see how marks are given.
2025	Main	Standard (041)	Open CBSE's official question paper for the 2025 main exam (one ZIP file of about 54.5 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2025 main exam (one ZIP file of about 56.9 MB, Class 10 Maths Standard and Basic together) to see how marks are given.
2025	Main	Basic (241)	Open CBSE's official question paper for the 2025 main exam (one ZIP file of about 49.9 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2025 main exam (one ZIP file of about 56.9 MB, Class 10 Maths Standard and Basic together) to see how marks are given.
2024	Main	Standard (041)	Open CBSE's official question paper for the 2024 main exam (one ZIP file of about 35.3 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2024 main exam (one ZIP file of about 23.4 MB, Class 10 Maths Standard (041)) to see how marks are given.
2023	Main	Standard (041)	Open CBSE's official question paper for the 2023 main exam (one ZIP file of about 44 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2023 main exam (one ZIP file of about 17.4 MB, Class 10 Maths Standard (041)) to see how marks are given.
2023	Main	Basic (241)	Open CBSE's official question paper for the 2023 main exam (one ZIP file of about 49.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2023 main exam (one ZIP file of about 26 MB, Class 10 Maths Basic (241)) to see how marks are given.

Sources and session

Each question is reproduced as printed in its CBSE paper, with its year, paper code and question number; only the layout is tidied. The same question printed in several paper sets counts once, and the card says how often it was printed in 2023–2026. These counts describe past papers; they are not predictions.

Every solution is original and follows NCERT Mathematics Class X (reprint 2026-27). The mark beside each step follows the official CBSE marking scheme where the scheme shows it; a mark labelled “(suggested)” follows the board’s usual pattern for that type of question.

Reference: CBSE Class 10 Mathematics board question papers via cbse.gov.in; all solutions are original. learncbse.net is an independent website and is not affiliated with CBSE or NCERT.

1-mark questions on Real Numbers

All 32 Real Numbers 1-mark questions

Q1. \(3^n\), where n is a natural number, cannot end with the digit :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q2. The value of (HCF – LCM) for the two numbers 3 and 5 is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q3. The number \(3 + \sqrt{2}\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q4. \((2 + \sqrt{2})^2\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q5. The number 3\(^{n}\), where n is a natural number, cannot end with the digit :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q6. The number \((5 - 3\sqrt{5} + \sqrt{5})\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q7. If two positive integers \(p\) and \(q\) can be expressed as \(p = 18a^2b^4\) and \(q = 20a^3b^2\), where \(a\) and \(b\) are prime numbers, then \(\text{LCM}(p, q)\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q8. If x is the LCM of 4, 6, 8 and y is the LCM of 3, 5, 7 and p is the LCM of x and y, then which of the following is true ?

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q9. \(\left(1 + \sqrt{3}\right)^2 - \left(1 - \sqrt{3}\right)^2\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q10. \((\sqrt{3}+2)^2 + (\sqrt{3}-2)^2\) is a/an

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q11. Which of the following cannot be the unit digit of \(8^n\), where n is a natural number?

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q12. The natural number 2 is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q13. For any natural number \(n\), \(6^n\) ends with the digit :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q14. The natural number 1 is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q15. (HCF \(\times\) LCM) for the numbers 70 and 40 is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q16. If p² = 32/50, then p is a/an

Step-by-step solution

Step 1: Choose the option