Question Bank — Real Numbers

NCERT Class 10 Mathematics · Chapter 1 · CBSE Board Pattern

Class 10 Mathematics Chapter 1 Question Bank CBSE Board Pattern

Section A — MCQs (1 mark each)

  1. Every composite number can be expressed as a product of primes. This statement is known as:
    (a) Euclid’s division lemma
    (b) Fundamental Theorem of Arithmetic
    (c) Theorem on irrationality of √2
    (d) None of these

  2. The prime factorisation of 32760 is:
    (a) 2³ × 3² × 5 × 7 × 13
    (b) 2² × 3³ × 5 × 7 × 13
    (c) 2³ × 3² × 5² × 7
    (d) 2⁴ × 3 × 5 × 7 × 13

  3. HCF(6, 20) × LCM(6, 20) equals:
    (a) 6 + 20
    (b) 6 × 20
    (c) 6 − 20
    (d) 20 ÷ 6

  4. There is no natural number n for which 4ⁿ ends with the digit 0 because:
    (a) 4ⁿ is always even
    (b) The prime factorisation of 4ⁿ contains only the prime 2
    (c) 4ⁿ is always divisible by 5
    (d) 4ⁿ is a perfect square

  5. If p is a prime and p divides a², then:
    (a) p divides a
    (b) p does not divide a
    (c) a is irrational
    (d) a is composite

Assertion-Reason Questions

  1. Assertion (A): √2 is irrational.
    Reason (R): If √2 were rational, then there would exist coprime positive integers a and b such that √2 = a/b, leading to 2 dividing both a and b, contradicting that a and b are coprime.
    (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.

  2. Assertion (A): 7 × 11 × 13 + 13 is a composite number.
    Reason (R): A number is composite if it has factors other than 1 and itself; here the expression equals 13(7 × 11 + 1).
    (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.

  3. The HCF of 96 and 404 obtained by prime factorisation is:
    (a) 2
    (b) 4
    (c) 8
    (d) 101

  4. LCM(6, 72, 120) by prime factorisation method is:
    (a) 180
    (b) 360
    (c) 720
    (d) 120

  5. Which of the following is irrational?
    (a) 5 − √3
    (b) √2 + √3 (product form)
    (c) Both (a) and the number obtained by assuming √3 rational
    (d) None of these

Section B — Very Short Answer (2 marks each)

  1. State the Fundamental Theorem of Arithmetic.
  2. Using prime factorisation, express 140 as a product of its prime factors.
  3. Find HCF(96, 404) by the prime factorisation method.
  4. Check whether 6ⁿ can end with the digit 0 for any natural number n. Give reason.
  5. Prove that if p is a prime and p divides a², then p divides a (Theorem 1.2).
  6. Why is 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 a composite number?

Section C — Short Answer (3 marks each)

  1. Find the LCM and HCF of 26 and 91 by the prime factorisation method and verify that LCM × HCF = product of the numbers.
  2. Find the LCM and HCF of 12, 15 and 21 by the prime factorisation method.
  3. Given HCF(306, 657) = 9, find LCM(306, 657).
  4. Find the HCF and LCM of 6, 72 and 120 by the prime factorisation method.
  5. Prove that √5 is irrational.

Section D — Long Answer (5 marks each)

  1. Prove that √2 is irrational. (Use the method of contradiction and Theorem 1.2.)
  2. Find the HCF of 96 and 404 by prime factorisation. Hence find their LCM. Also verify the relation HCF × LCM = product of the numbers.
  3. Prove that 3√2 is irrational. (Assume it is rational and reach a contradiction with the irrationality of √2.)

Section E — Case/Source-Based (4 marks each)

Case 1

Sonia takes 18 minutes and Ravi takes 12 minutes to complete one round of a circular path around a sports field. They start together at the same point and move in the same direction.

Sub-questions:
(i) Express 18 and 12 as products of primes.
(ii) Find the HCF of 18 and 12.
(iii) Find the LCM of 18 and 12.
(iv) After how many minutes will they meet again at the starting point?

Case 2

A student claims that there exists a natural number n such that 4ⁿ ends with the digit 0.

Sub-questions:
(i) Write the prime factorisation of 4ⁿ.
(ii) For 4ⁿ to end with 0, which two primes must divide it?
(iii) Show that 5 never appears in the prime factorisation of 4ⁿ.
(iv) Conclude, using the Fundamental Theorem of Arithmetic, whether such an n exists.

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Section A

  1. (b)
  2. (a)
  3. (b)
  4. (b)
  5. (a)
  6. (a)
  7. (a)
  8. (b)
  9. (b)
  10. (a)

Section B

  1. Every composite number can be expressed as a product of primes and this factorisation is unique apart from the order of factors. (2 marks)
  2. 140 = 2² × 5 × 7 (correct prime factors — 1 mark; final form — 1 mark)
  3. 96 = 2⁵ × 3, 404 = 2² × 101 → HCF = 2² = 4 (2 marks)
  4. 6ⁿ = 2ⁿ × 3ⁿ; no factor of 5, hence cannot end with 0 (2 marks)
  5. Write a = p₁p₂…pₙ; a² = p₁²…pₙ². If p divides a² then p must be one of the pᵢ, hence p divides a. (2 marks)
  6. Expression = 5(7! + 1); has factors other than 1 and itself → composite (2 marks)

Section C

  1. 26 = 2 × 13, 91 = 7 × 13 → HCF = 13, LCM = 2 × 7 × 13 = 182; 182 × 13 = 26 × 91 (3 marks)
  2. 12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7 → HCF = 3, LCM = 2² × 3 × 5 × 7 = 420 (3 marks)
  3. LCM = (306 × 657)/9 = 22422 (3 marks)
  4. HCF = 2 × 3 = 6; LCM = 2³ × 3² × 5 = 360 (3 marks)
  5. Assume √5 = a/b (coprime). Then 5b² = a² ⇒ 5 divides a² ⇒ 5 divides a. Let a = 5c. Then 5b² = 25c² ⇒ 5 divides b² ⇒ 5 divides b. Contradiction. Hence √5 irrational. (3 marks)

Section D

  1. Assume √2 = a/b (coprime). Then 2b² = a² ⇒ 2 divides a² ⇒ 2 divides a (Th. 1.2). Let a = 2c. Then 2b² = 4c² ⇒ 2 divides b. Contradiction. Hence irrational. (correct assumption & contradiction steps — 3 marks; conclusion — 2 marks)
  2. HCF = 4; LCM = (96 × 404)/4 = 9696; verification 4 × 9696 = 96 × 404 (5 marks)
  3. Assume 3√2 = a/b (coprime). Then √2 = a/(3b) rational. Contradicts irrationality of √2. Hence 3√2 irrational. (5 marks)

Section E

Case 1

(i) 18 = 2 × 3², 12 = 2² × 3 (1 mark)
(ii) HCF = 2 × 3 = 6 (1 mark)
(iii) LCM = 2² × 3² = 36 (1 mark)
(iv) 36 minutes (1 mark)

Case 2

(i) 4ⁿ = (2²)ⁿ = 2^{2n} (1 mark)
(ii) 2 and 5 (1 mark)
(iii) Only prime is 2; uniqueness of FTA (1 mark)
(iv) No such n exists (1 mark)

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All questions are answerable from the NCERT chapter text. Reviewed by GFIS faculty.