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Real Numbers Class 10 Maths Worksheet with Answers Chapter 1

Updated:8 Mins Read

This is the first chapter of class 10 maths i.e Real Numbers. Attempt the below worksheet only after completing the chapter in your NCERT book. This worksheet will help you to strengthen the concept problem solving capabilities. The worksheet is divided into four sections: Basic (quick fundamentals), Standard (NCERT-level practice), Advance (multi-step problems), and HOTS (reasoning-heavy questions). Try solving each section in order.

What this chapter covers

  • Prime factorisation basics
  • HCF and LCM
  • Euclid’s division method
  • Irrational numbers proofs
  • Decimal expansion ideas

Class 10 Maths Worksheet – Chapter 1: Real Numbers

Basic

  1. State the Fundamental Theorem of Arithmetic (2 lines).
  2. Find the prime factorisation of 140 (use powers where possible).
  3. Find the prime factorisation of 156 (use powers where possible).
  4. Find HCF(12, 15, 21) using prime factorisation.
  5. Find LCM(8, 9, 25) using prime factorisation.
  6. True/False (write one-line reason):

    • Every composite number has a unique prime factorisation (ignoring order).
    • If a number ends with 0, then it is divisible by 2 and 5.
    • If p is prime and p | a, then p | a².
  7. Complete the table:

    Number Prime factorisation
    60 __________
    48 __________

Standard

  1. Use Euclid’s Division Algorithm to find HCF(135, 225). Write each step clearly.
  2. Use Euclid’s Division Algorithm to find HCF(196, 252). Write each step clearly.
  3. Find HCF and LCM of 26 and 91 by prime factorisation and verify:
    HCF × LCM = 26 × 91
  4. Find HCF and LCM of 336 and 54 by prime factorisation and verify:
    HCF × LCM = 336 × 54
  5. Express 3825 as a product of prime factors (use powers).
  6. Given HCF(306, 657) = 9, find LCM(306, 657).
  7. Two bells ring together at 9:00 AM. One rings every 18 minutes and the other every 12 minutes.
    At what time will they ring together again?

Advance

  1. Find HCF and LCM of 6, 72, and 120 using prime factorisation.
  2. Use Euclid’s Division Algorithm to find HCF(867, 255), and then find LCM(867, 255).
  3. Prove that √5 is irrational (use contradiction).
  4. Prove that √2 is irrational (write the standard proof in steps).
  5. Explain why 7 × 11 × 13 + 13 is a composite number.
  6. Explain why 7! + 5 is a composite number.
    (Here 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1)
  7. Complete the table (use Euclid’s Division Algorithm):

    a b Write a = bq + r Next pair
    252 198 ________________ ________________
    198 r₁ ________________ ________________

HOTS

  1. Prove that 3 + 2√5 is irrational.
  2. Prove that 1/√2 is irrational.
  3. Prove that 7√5 is irrational.
  4. Prove that 6 + √2 is irrational.
  5. A student says: “Since √2 is irrational, √2 + √2 must be irrational.”
    Is the statement correct? Give a proper reason.
  6. If a and b are positive integers and HCF(a, b) = 12 and LCM(a, b) = 420, find a × b.
  7. Without fully factorising, decide whether 4^n can ever end with digit 0 for any natural number n.
    Answer Yes/No with theorem-based reason.

Answer Key

Basic – Answers

  1. Ans: Every composite number can be expressed as a product of primes, and this factorisation is unique except for the order of primes.
    Hint: Remember the phrase “unique prime factorisation”.
  2. Ans: 140 = 2² × 5 × 7.
    Hint: 140 = 14 × 10 = (2 × 7) × (2 × 5).
  3. Ans: 156 = 2² × 3 × 13.
    Hint: Divide by 2 repeatedly: 156 → 78 → 39.
  4. Ans: 12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7 ⇒ HCF = 3.
    Hint: Common primes with smallest powers.
  5. Ans: 8 = 2³, 9 = 3², 25 = 5² ⇒ LCM = 2³ × 3² × 5² = 1800.
    Hint: Take highest power of each prime present.
  6. Ans:

    • True — uniqueness apart from order.
    • True — ending with 0 means divisible by 10 = 2 × 5.
    • True — a² = a × a, so if p divides a, it divides a².

    Hint: Use divisibility facts and prime factorisation idea.

  7. Ans:

    • 60 = 2² × 3 × 5
    • 48 = 2⁴ × 3

    Hint: Keep dividing by the smallest prime.

Standard – Answers

  1. Ans:
    225 = 135 × 1 + 90
    135 = 90 × 1 + 45
    90 = 45 × 2 + 0
    So HCF(135, 225) = 45.
    Hint: The last non-zero remainder is the HCF.
  2. Ans:
    252 = 196 × 1 + 56
    196 = 56 × 3 + 28
    56 = 28 × 2 + 0
    So HCF(196, 252) = 28.
    Hint: Repeat division until remainder becomes 0.
  3. Ans: 26 = 2 × 13, 91 = 7 × 13
    HCF = 13, LCM = 2 × 7 × 13 = 182
    Check: 13 × 182 = 2366 and 26 × 91 = 2366 ✓
    Hint: Verify by multiplying HCF and LCM.
  4. Ans: 336 = 2⁴ × 3 × 7, 54 = 2 × 3³
    HCF = 2 × 3 = 6
    LCM = 2⁴ × 3³ × 7 = 3024
    Check: 6 × 3024 = 18144 and 336 × 54 = 18144 ✓
    Hint: HCF: smallest powers, LCM: greatest powers.
  5. Ans: 3825 = 3² × 5² × 17.
    Hint: 3825 = 25 × 153 and 153 = 3² × 17.
  6. Ans: LCM(306, 657) = (306 × 657) / 9
    306/9 = 34 ⇒ LCM = 34 × 657 = 22338.
    Hint: Use: HCF × LCM = a × b.
  7. Ans: LCM(18, 12) = 36 minutes ⇒ they ring together again at 9:36 AM.
    Hint: “Together again” means take LCM of intervals.

Advance – Answers

  1. Ans: 6 = 2 × 3, 72 = 2³ × 3², 120 = 2³ × 3 × 5
    HCF = 2 × 3 = 6
    LCM = 2³ × 3² × 5 = 360
    Hint: HCF uses common primes; LCM uses max powers.
  2. Ans:
    867 = 255 × 3 + 102
    255 = 102 × 2 + 51
    102 = 51 × 2 + 0
    HCF = 51
    LCM = (867 × 255) / 51 = 4335
    Hint: Use HCF from Euclid, then apply product formula.
  3. Ans: Assume √5 is rational ⇒ √5 = a/b in lowest form.
    Square: 5 = a²/b² ⇒ a² = 5b² ⇒ 5 | a² ⇒ 5 | a.
    Let a = 5k ⇒ a² = 25k² ⇒ 25k² = 5b² ⇒ b² = 5k² ⇒ 5 | b.
    This contradicts that a/b is in lowest form. Hence √5 is irrational.
    Hint: Show both a and b become divisible by 5.
  4. Ans: Assume √2 = a/b in lowest form.
    a² = 2b² ⇒ a is even ⇒ a = 2k
    Then 4k² = 2b² ⇒ b² = 2k² ⇒ b is even.
    Both even contradict “lowest form”. So √2 is irrational.
    Hint: Even/odd contradiction is the key.
  5. Ans: 7 × 11 × 13 + 13 = 13(7 × 11 + 1) = 13(77 + 1) = 13 × 78, so it is composite.
    Hint: Factor out 13.
  6. Ans: 7! + 5 = 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
    = 5(7 × 6 × 4 × 3 × 2 × 1 + 1), so it is divisible by 5 and greater than 5 ⇒ composite.
    Hint: Factor out 5 from 7!.
  7. Ans:
    252 = 198 × 1 + 54 ⇒ next pair (198, 54)
    198 = 54 × 3 + 36 ⇒ next pair (54, 36)
    Hint: Each next pair is (b, remainder).

HOTS – Answers

  1. Ans: If 3 + 2√5 were rational, then 2√5 = (3 + 2√5) − 3 would be rational.
    So √5 would be rational (divide by 2), which is false. Hence 3 + 2√5 is irrational.
    Hint: Rational − rational = rational.
  2. Ans: If 1/√2 were rational, then √2 = 1 ÷ (1/√2) would be rational.
    But √2 is irrational. Contradiction. So 1/√2 is irrational.
    Hint: If x is rational and non-zero, then 1/x is rational.
  3. Ans: If 7√5 were rational, then √5 = (7√5)/7 would be rational.
    But √5 is irrational. So 7√5 is irrational.
    Hint: Divide by a non-zero rational number.
  4. Ans: If 6 + √2 were rational, then √2 = (6 + √2) − 6 would be rational.
    Contradiction. So 6 + √2 is irrational.
    Hint: Subtract 6 from both sides.
  5. Ans: Incorrect. √2 + √2 = 2√2, and multiplying an irrational number by a non-zero rational keeps it irrational.
    So 2√2 is still irrational.
    Hint: Use “rational × irrational = irrational” (for non-zero rational).
  6. Ans: a × b = HCF × LCM = 12 × 420 = 5040.
    Hint: Use the identity a × b = HCF(a,b) × LCM(a,b).
  7. Ans: No. A number ending with 0 must be divisible by 10 = 2 × 5.
    But 4ⁿ = (2²)ⁿ = 2²ⁿ has only prime factor 2, no factor 5, so it cannot end with 0.
    Hint: Ending with 0 needs a factor 5.

Worksheet for Other chapters

Amit holds a BE in Mechanical Engineering and brings a genuine passion for mathematics to IndiaFolks. He creates NCERT-aligned content for students from Classes 4 to 10. He specialises in breaking down tricky concepts into clear, step-by-step solutions, from worksheets and MCQs to aptitude problems. He makes the tough problems easier for Indian students to build confidence and score better in Maths. His goal is simple: turn every student into a problem-solver who actually enjoys the subject.

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