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Real Numbers

NCERT Class 10 · Mathematics Based on NCERT Class 10 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter revisits the properties of real numbers, focusing on positive integers. It introduces Euclid's division algorithm and the Fundamental Theorem of Arithmetic, which are used to compute HCF and LCM, prove the irrationality of numbers like √2 and √3, and determine the nature of decimal expansions of rational numbers. The one big idea a student must walk away with is that every composite number has a unique prime factorisation, and this fact is the key to understanding divisibility, irrational numbers, and decimal expansions.
Key Concepts (12)
Euclid's division algorithm
Any positive integer a can be divided by another positive integer b in such a way that it leaves a remainder r that is smaller than b.
e.g. Dividing 17 by 5 gives quotient 3 and remainder 2, since 17 = 5 × 3 + 2 and 2 < 5.
Fundamental Theorem of Arithmetic
Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
e.g. 32760 = 2³ × 3² × 5 × 7 × 13; no other combination of primes gives 32760 except for the order of factors.
Prime factorisation
Writing a composite number as a product of its prime factors, usually in ascending order.
e.g. 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7.
HCF (Highest Common Factor)
The product of the smallest power of each common prime factor in the numbers.
e.g. HCF(6, 20) = 2¹ = 2, since 6 = 2 × 3 and 20 = 2² × 5.
LCM (Lowest Common Multiple)
The product of the greatest power of each prime factor involved in the numbers.
e.g. LCM(6, 20) = 2² × 3¹ × 5¹ = 60.
Irrational number
A number ‘s’ is called irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0.
e.g. √2, √3, π, 0.10110111011110...
Proof by contradiction
A technique where we assume the opposite of what we want to prove, and show that this assumption leads to a contradiction, thereby proving the original statement.
e.g. To prove √2 is irrational, assume it is rational, then derive that a and b have a common factor 2, contradicting that they are coprime.
Theorem 1.2
Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.
e.g. If 3 divides a², then 3 divides a.
Composite number
A positive integer greater than 1 that is not prime, i.e., it has factors other than 1 and itself.
e.g. 6, 72, 120 are composite numbers.
Terminating decimal expansion
A decimal expansion of a rational number p/q that ends after a finite number of digits, which occurs when the prime factorisation of the denominator q is of the form 2ⁿ5ᵐ.
e.g. 1/2 = 0.5, 1/4 = 0.25, 1/5 = 0.2.
Non-terminating repeating decimal expansion
A decimal expansion of a rational number p/q that does not end but repeats a block of digits, which occurs when the prime factorisation of the denominator q has a prime factor other than 2 or 5.
e.g. 1/3 = 0.333..., 1/7 = 0.142857142857...
Coprime numbers
Two integers a and b are said to be coprime if they have no common factor other than 1.
e.g. 2 and 3 are coprime; 4 and 9 are coprime.
Formulas & Equations (3)
HCF(a, b) × LCM(a, b) = a × b
Reads as: HCF of a and b times LCM of a and b equals a times b
Variables: a, b: positive integers
Use when: To find LCM when HCF is known, or to verify calculations for two numbers.
⚠ Students apply this formula to three numbers, but it does NOT hold for three numbers: HCF(p,q,r) × LCM(p,q,r) ≠ p × q × r.
LCM(p, q, r) = (p × q × r × HCF(p, q, r)) / (HCF(p, q) × HCF(q, r) × HCF(p, r))
Reads as: LCM of p, q, r equals product of p, q, r and their HCF divided by product of pairwise HCFs
Variables: p, q, r: positive integers
Use when: To find LCM of three numbers using their HCFs.
⚠ Students forget to multiply by HCF(p,q,r) in the numerator.
HCF(p, q, r) = (p × q × r × LCM(p, q, r)) / (LCM(p, q) × LCM(q, r) × LCM(p, r))
Reads as: HCF of p, q, r equals product of p, q, r and their LCM divided by product of pairwise LCMs
Variables: p, q, r: positive integers
Use when: To find HCF of three numbers using their LCMs.
⚠ Students confuse this with the LCM formula and swap the terms.
How It Works
Finding HCF using prime factorisation method
  1. Step 1: Write each number as a product of its prime factors.
  2. Step 2: Identify the common prime factors.
  3. Step 3: Take the smallest power of each common prime factor.
  4. Step 4: Multiply these together to get the HCF.
💡 Always take the smallest power of common primes, not the largest.
Finding LCM using prime factorisation method
  1. Step 1: Write each number as a product of its prime factors.
  2. Step 2: List all prime factors that appear in any number.
  3. Step 3: Take the greatest power of each prime factor.
  4. Step 4: Multiply these together to get the LCM.
💡 For LCM, take the greatest power of each prime, even if it appears in only one number.
Proving √p is irrational (where p is prime)
  1. Step 1: Assume √p is rational, so √p = a/b where a and b are coprime integers, b ≠ 0.
  2. Step 2: Square both sides: p = a²/b², so a² = p b².
  3. Step 3: Since p divides a², by Theorem 1.2, p divides a. Write a = p c.
  4. Step 4: Substitute: (p c)² = p b² → p² c² = p b² → b² = p c².
  5. Step 5: So p divides b², and by Theorem 1.2, p divides b.
  6. Step 6: This means a and b have a common factor p, contradicting that they are coprime.
  7. Step 7: Hence, our assumption is false, so √p is irrational.
💡 The key step is using Theorem 1.2: if p divides a², then p divides a. Never skip writing that a and b are coprime.
Showing that a number like 5 - √3 is irrational
  1. Step 1: Assume 5 - √3 is rational, so 5 - √3 = a/b, where a and b are coprime, b ≠ 0.
  2. Step 2: Rearrange: √3 = 5 - a/b = (5b - a)/b.
  3. Step 3: Since a and b are integers, (5b - a)/b is rational.
  4. Step 4: This implies √3 is rational, which contradicts the known fact that √3 is irrational.
  5. Step 5: Therefore, our assumption is false, so 5 - √3 is irrational.
💡 The contradiction is reached by showing that the irrational number (√3) would have to be rational.
Checking if 4ⁿ ends with digit zero
  1. Step 1: If 4ⁿ ends with digit zero, it must be divisible by 5.
  2. Step 2: Write 4ⁿ = (2²)ⁿ = 2²ⁿ.
  3. Step 3: The prime factorisation of 4ⁿ contains only the prime 2.
  4. Step 4: For it to be divisible by 5, its prime factorisation must include 5.
  5. Step 5: By the Fundamental Theorem of Arithmetic, the factorisation is unique, so 5 cannot appear.
  6. Step 6: Hence, no natural number n makes 4ⁿ end with digit zero.
💡 The uniqueness of prime factorisation is the key argument.
Exam Traps — Don't Make These Mistakes
⚠ Applying HCF × LCM = product to three numbers
✗ HCF(6,72,120) × LCM(6,72,120) = 6 × 72 × 120
→ The formula HCF(a,b) × LCM(a,b) = a × b holds only for two numbers. For three numbers, the product is not equal to the product of HCF and LCM.
✓ Use the correct formulas: LCM(p,q,r) = (p×q×r×HCF(p,q,r)) / (HCF(p,q)×HCF(q,r)×HCF(p,r)) and similarly for HCF.
⚠ Assuming 6ⁿ ends with zero
✗ 6ⁿ ends with zero for n = 1, 2, 3, ...
→ 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. Its prime factors are only 2 and 3. To end with zero, it must be divisible by 5, which is impossible due to unique factorisation.
✓ 6ⁿ cannot end with digit zero for any natural number n because its prime factorisation contains no factor 5.
⚠ Forgetting to state 'a and b are coprime' in irrationality proof
✗ Assume √2 = a/b, then square to get 2b² = a², so 2 divides a, etc.
→ Without stating that a and b are coprime (have no common factor), the contradiction that they share a factor 2 is not valid, because they could have had a common factor from the start.
✓ Assume √2 = a/b where a and b are coprime integers (b ≠ 0). Then square: 2b² = a². By Theorem 1.2, 2 divides a, so a = 2c. Substitute: 2b² = 4c² → b² = 2c², so 2 divides b. Thus a and b have common factor 2, contradicting coprimality.
⚠ Confusing HCF and LCM in prime factorisation method
✗ HCF(6,20) = 2² × 3 × 5 = 60
→ HCF uses the smallest power of common primes; LCM uses the greatest power of all primes. Here, the student used the LCM method for HCF.
✓ HCF(6,20) = 2¹ = 2 (smallest power of common prime 2). LCM(6,20) = 2² × 3 × 5 = 60 (greatest powers of all primes).
⚠ Thinking product of three numbers equals product of HCF and LCM
✗ 6 × 72 × 120 = HCF(6,72,120) × LCM(6,72,120)
→ The NCERT explicitly states that the product of three numbers is not equal to the product of their HCF and LCM. The formula only holds for two numbers.
✓ Use the separate formulas for three numbers given in the 'Note to the Reader' in the chapter.
Compare & Contrast
Definition and method of calculation
HCF (Highest Common Factor)
LCM (Lowest Common Multiple)
HCF is the product of the smallest power of each common prime factor; LCM is the product of the greatest power of each prime factor involved in the numbers.
Nature of decimal expansion of a rational number p/q
Terminating decimal expansion
Non-terminating repeating decimal expansion
Terminating occurs when the denominator q has prime factors only 2 and/or 5; non-terminating repeating occurs when q has a prime factor other than 2 or 5.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 Short Answer (3 marks), 1 Long Answer (5 marks) — total 10 marks typically.
Most asked: Fundamental Theorem of Arithmetic and its applications (HCF/LCM) · Proving irrationality of √2, √3, √5 · Checking if a number ends with digit zero · Finding HCF and LCM using prime factorisation
Don't skip: Prove that √2 (or √3, √5) is irrational — 3 or 5 marks; Find HCF and LCM of two or three numbers using prime factorisation and verify HCF × LCM = product for two numbers — 3 or 5 marks; Show that 4ⁿ or 6ⁿ cannot end with digit zero — 3 marks
One-Line Revision (read aloud before the exam)
  1. Euclid's division algorithm: any positive integer a can be written as a = bq + r, where 0 ≤ r < b.
  2. Fundamental Theorem of Arithmetic: every composite number has a unique prime factorisation, apart from order.
  3. HCF is the product of the smallest power of each common prime factor.
  4. LCM is the product of the greatest power of each prime factor involved.
  5. For any two positive integers a and b: HCF(a,b) × LCM(a,b) = a × b.
  6. This formula does NOT hold for three numbers.
  7. If a prime p divides a², then p divides a (Theorem 1.2).
  8. √2, √3, √5 are irrational numbers.
  9. To prove √p is irrational, assume it is rational, square both sides, and use Theorem 1.2 to get a contradiction.
  10. The sum or difference of a rational and an irrational number is irrational.
  11. The product and quotient of a non-zero rational and an irrational number is irrational.
  12. A rational number p/q has a terminating decimal expansion only if the prime factorisation of q is of the form 2ⁿ5ᵐ.
  13. If the denominator q has a prime factor other than 2 or 5, the decimal expansion is non-terminating repeating.
  14. 4ⁿ cannot end with digit zero because its prime factorisation contains only the prime 2.
  15. 6ⁿ cannot end with digit zero because its prime factorisation contains only primes 2 and 3.
  16. Numbers like 7 × 11 × 13 + 13 are composite because they can be factorised as 13 × (7 × 11 + 1).

MCQs — 10 Questions with Answers

Q1. A student claims that since every composite number can be expressed as a product of primes, and this expression is unique, then the number 60 can be written as 2² × 3 × 5 in only one way. Using the Fundamental Theorem of Arithmetic, this claim is —

  • A. Always true, because the uniqueness is guaranteed regardless of the order of factors ✓
  • B. True only if we arrange the prime factors in ascending order
  • C. False, because we can write it as 5 × 3 × 2² which is a different expression
  • D. True only for even composite numbers

Answer: A — The FTA guarantees uniqueness apart from the order of factors; uniqueness means 2² × 3 × 5 and 5 × 3 × 2² represent the same factorization, not different ones.

Q2. Consider the numbers 6n where n is a natural number. Based on the Fundamental Theorem of Arithmetic, which statement correctly explains why 6n can never end with the digit zero?

  • A. Because 6n = 2ⁿ × 3ⁿ, and the prime factorization never contains 5 or 10
  • B. Because a number ends with zero only if divisible by both 2 and 5, but 6n = 2 × 3)ⁿ lacks the prime factor 5 ✓
  • C. Because 6 is not divisible by 10
  • D. Because n can only be small natural numbers in most cases

Answer: B — Ending in zero requires both prime factors 2 and 5 in the factorization; 6n = (2 × 3)ⁿ guarantees 2 is present but 5 never appears, so by FTA's uniqueness, no such number exists.

Q3. A mathematician states: 'If two different composite numbers have identical sets of prime factors (ignoring powers), they must be equal.' Based on the Fundamental Theorem of Arithmetic, this statement is —

  • A. Always true because prime factorization is unique
  • B. False; for example, 12 = 2² × 3 and 18 = 2 × 3² have the same prime factors {2, 3} but are different numbers ✓
  • C. True only if the numbers are less than 100
  • D. False because composite numbers cannot share prime factors

Answer: B — Uniqueness of FTA refers to the exponents AND primes combined; identical sets of primes with different exponents yield different numbers (12 ≠ 18), so the statement is false.

Q4. Why does Euclid's division algorithm guarantee that we can always find the HCF of two positive integers a and b (where a > b) by repeatedly applying division?

  • A. Because the algorithm reduces the problem to smaller pairs of numbers with the same common divisors until the remainder becomes zero ✓
  • B. Because every positive integer can be expressed as a product of primes
  • C. Because the HCF is always equal to the larger of the two numbers
  • D. Because division always produces a unique quotient and remainder

Answer: A — The algorithm works by replacing the pair (a,b) with (b, remainder) repeatedly; since remainders strictly decrease and are bounded below by zero, the process terminates with HCF as the last non-zero remainder.

Q5. Assertion (A): For any natural number n, if 2ⁿ × 5ⁿ is the complete prime factorization of a number, then that number must end with the digit zero. Reason (R): A number ends in zero if and only if it is divisible by 10, which requires exactly one factor of 2 and one factor of 5. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — A is true: 2ⁿ × 5ⁿ = (2×5)ⁿ × 2^(n-1) × 5^(n-1) or simplified, 10ⁿ × (other factors), always divisible by 10; R is true and correctly explains why A is true.

Q6. Assertion (A): The prime factorization of 32760 is unique. Reason (R): The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes in exactly one way, except for the order of factors. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — A is true (32760 = 2³ × 3² × 5 × 7 × 13 uniquely); R is true and directly explains why A must be true—FTA guarantees uniqueness for every composite number.

Q7. Assertion (A): It is impossible to find a natural number n such that 4ⁿ ends with the digit 5. Reason (R): Numbers ending in 5 must have 5 as a prime factor, but 4ⁿ = 2^(2n) contains only 2 as its prime factor. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — A is true by the contrapositive logic; R is true and correctly explains A: FTA's uniqueness means no prime factor 5 can appear in 4ⁿ, so it cannot end in 5.

Q8. A student tries to verify whether 3³ × 5² × 7 and 3² × 5³ × 7 represent the same prime factorization by checking if they have the same set of prime divisors {3, 5, 7}. The student's approach is —

  • A. Correct, because if two numbers share the same prime divisors, they are the same number
  • B. Incorrect; the Fundamental Theorem of Arithmetic requires matching exponents of each prime, not just the same set of primes ✓
  • C. Correct only if both numbers are less than 1000
  • D. Incorrect because one number is even and the other is odd

Answer: B — The student confuses 'same prime factors' with 'same prime factorization'; FTA demands unique exponent combinations (3³ × 5² × 7 ≠ 3² × 5³ × 7), so checking only the set of primes is insufficient.

Q9. Why does the statement 'if a number is divisible by both 6 and 9, it must be divisible by 54' fail to be always true?

  • A. Because 6 = 2 × 3 and 9 = 3², so divisibility by both requires 2 × 3² = 18, not 2 × 3³ = 54 ✓
  • B. Because some composite numbers cannot be factored into primes
  • C. Because divisibility is not a transitive property
  • D. Because 54 > 9, so the statement is logically impossible

Answer: A — Using FTA: LCM(6, 9) = 2 × 3² = 18 (taking the highest power of each prime); the statement claims LCM = 54 = 2 × 3³, which wrongly adds exponents instead of taking maximum exponents.

Q10. A number N has the property that 'whenever it divides the product of two numbers, it must divide at least one of them.' Which of the following must be true about N's prime factorization?

  • A. N must be a prime number, because only primes have this divisibility property ✓
  • B. N can be any composite number, as long as it has at least two distinct prime factors
  • C. N must be a perfect square of a prime, like 4 or 9
  • D. N must be able to be expressed as a product of distinct primes

Answer: A — This is the definition of a prime in terms of divisibility (Euclid's lemma); if N were composite (e.g., 6 = 2 × 3), then 6 | (2 × 3) but 6 ∤ 2 and 6 ∤ 3, violating the property—only primes satisfy it.

Flashcards

What is the Fundamental Theorem of Arithmetic?

Every composite number can be expressed uniquely as a product of primes, regardless of the order in which the primes are written.

How do you find HCF using prime factorization?

HCF is the product of the smallest powers of all common prime factors in the given numbers.

How do you find LCM using prime factorization?

LCM is the product of the greatest powers of all prime factors involved in the given numbers.

What is the relationship between HCF and LCM?

For any two positive integers a and b: HCF(a,b) × LCM(a,b) = a × b.

Prime factorize 32760.

32760 = 2³ × 3² × 5 × 7 × 13.

Can 4^n ever end with digit zero for any natural number n?

No, because 4^n = 2^(2n) contains only prime factor 2, but a number ending in zero must be divisible by both 2 and 5.

Find HCF and LCM of 6 and 20.

6 = 2 × 3, 20 = 2² × 5; HCF = 2, LCM = 60.

Is the prime factorization of a composite number unique?

Yes, the prime factorization is unique apart from the order in which the prime factors occur.

What does 'ascending order' mean in prime factorization?

Writing prime factors from smallest to largest: p₁ ≤ p₂ ≤ ... ≤ pₙ.

Why is 2 × 3 × 5 considered the same as 5 × 3 × 2?

Because the Fundamental Theorem states uniqueness is independent of order; both represent the same unique prime factorization.

Important Board Questions

State the Fundamental Theorem of Arithmetic and verify it for the number 84. [2 marks]

State that every composite number has unique prime factorization; then find prime factors of 84 by division: 84 = 2² × 3 × 7.

Find the HCF and LCM of 12 and 18 using the prime factorization method. Verify that HCF × LCM = product of the numbers. [3 marks]

Factor both: 12 = 2² × 3, 18 = 2 × 3²; HCF = 2 × 3 = 6, LCM = 2² × 3² = 36; verify: 6 × 36 = 216 = 12 × 18.

Prove that 4ⁿ can never end with the digit 0 for any natural number n. Explain using the Fundamental Theorem of Arithmetic. [5 marks]

Show 4ⁿ = (2²)ⁿ = 2²ⁿ contains only prime factor 2; for a number to end in 0, it must be divisible by both 2 and 5; since 5 never appears in 2²ⁿ by Fundamental Theorem's uniqueness, 4ⁿ cannot end in 0.

Assertion-Reason Questions for Real Numbers

Q1. Assertion (A):

Every composite number can be expressed as a product of prime numbers.

Reason (R):

This is the Fundamental Theorem of Arithmetic, which guarantees that every composite number has a unique prime factorisation.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The reason correctly explains why the assertion is true — the Fundamental Theorem of Arithmetic is precisely the principle that guarantees every composite number can be uniquely factorised into primes.

Q2. Assertion (A):

The number 4ⁿ can end with the digit 0 for some natural number n.

Reason (R):

Since 4ⁿ = 2²ⁿ contains only the prime factor 2, it cannot be divisible by 5.

Answer: D — A is false but R is true — The assertion is false (4ⁿ can never end with 0), but the reason is true and correctly explains why the assertion is false. A number ending in 0 must be divisible by both 2 and 5, but 4ⁿ = 2²ⁿ has no factor of 5.

Q3. Assertion (A):

If p/q has a terminating decimal expansion, then q must have only 2 and 5 as prime factors.

Reason (R):

A rational number has a terminating decimal expansion if and only if the denominator contains only prime factors 2 and 5.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains the assertion. This is a direct application of the Fundamental Theorem of Arithmetic to determine terminating decimals.

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Case-Based Questions (CBQ) for Real Numbers

Case Study:

A school wants to distribute 120 pens and 90 pencils equally among students such that each student gets the same number of pens and the same number of pencils, with no items left over. The school is trying to determine the maximum number of students who can receive these items equally. Using prime factorisation, we find that 120 = 2³ × 3 × 5 and 90 = 2 × 3² × 5. The school needs to find the greatest common divisor of these two numbers to determine how to distribute the items fairly.

Q1. What is the HCF of 120 and 90?

  • A. 15
  • B. 20
  • C. 30 ✓
  • D. 60

Answer: C

Q2. How many pens will each student receive?

  • A. 2
  • B. 3
  • C. 4 ✓
  • D. 5

Answer: C

Q3. How many pencils will each student receive?

  • A. 2
  • B. 3 ✓
  • C. 4
  • D. 5

Answer: B

Q4. What is the LCM of 120 and 90?

  • A. 360 ✓
  • B. 240
  • C. 180
  • D. 540

Answer: A

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Previous Year Questions (PYQ) for Real Numbers

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 31/4/1 · 1 mark

If HCF (850, 325) is 25, then LCM (850, 325) is:

  • 442
  • 11050
  • 8450
  • 2210

CBSE 2026 · Set 31/4/1 · 1 mark

7 × 29 × 23 + 1 is:

  • a prime number
  • divisible by 23
  • an odd number
  • a composite number

CBSE 2026 · Set 31/4/1 · 1 mark

Assertion (A): 4ⁿ can not end with the digit zero. Reason (R): Prime factorisation of 4ⁿ is unique.

  • Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A)
  • Both, Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A)
  • Assertion (A) is true, but Reason (R) is false
  • Assertion (A) is false, but Reason (R) is true

CBSE 2026 · Set 31/4/2 · 1 mark

Probability of getting an irrational number at random from the numbers 3, 4, ∛9, ∛8, √5, 0, ∛(42/3) is:

  • 0
  • 4/7
  • 3/7
  • 1

CBSE 2026 · Set 31/4/2 · 1 mark

HCF of two consecutive natural numbers is:

  • 2
  • 1
  • 0
  • smaller number

CBSE 2026 · Set 31/4/2 · 1 mark

Assertion (A): 4ⁿ cannot end with the digit zero. Reason (R): Prime factorisation of 4ⁿ is unique.

  • Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
  • Both, Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 31/4/3 · 1 mark

Assertion (A): 4^n can not end with the digit zero. Reason (R): Prime factorisation of 4^n is unique.

  • Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
  • Both, Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 31/5/1 · 1 mark

7 × 11 × 13 + 5 is

  • a prime number
  • an odd number
  • a composite number
  • a multiple of 5

CBSE 2026 · Set 31/5/1 · 2 marks

Find the H.C.F. and L.C.M. of 1530 and 2040.

CBSE 2026 · Set 31/5/2 · 2 marks

If H.C.F. of 135x² and 189x³ is 108, then find the value of x.

CBSE 2026 · Set 31/5/3 · 1 mark

√17 × √11 × √13 + √11 is

  • a prime number
  • multiple of 17
  • a composite number
  • an odd number

CBSE 2026 · Set 31/5/3 · 2 marks

Find the H.C.F. and L.C.M. of 408 and 312.

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Class 10 Maths Formula Sheet

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Class 10 Mathematics Previous Year Questions, Chapter-wise

All 1107 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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