Keerthana Classes

Maths ch1 real numbers

October 04, 2026

37 marks. Read the notes, then tap Start test. Answers and your score appear only after you submit.

Quick notes

Fundamental Theorem of Arithmetic: every composite number can be written as a product of primes, and this factorisation is unique apart from the order of the factors.

HCF = product of the smallest power of each common prime factor. LCM = product of the greatest power of each prime factor involved.

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

If a prime p divides a², then p divides a. We use this to prove that √2, √3, √5 are irrational, by contradiction.

Rational + irrational = irrational. Non-zero rational × irrational = irrational.

Part A: Multiple choice 16 questions

1. HCF × LCM of two positive integers is equal to

  1. their sum
  2. their product
  3. their difference
  4. their quotient

HCF(a, b) × LCM(a, b) = a × b.

2. The prime factorisation of 156 is

  1. 2² × 3 × 13
  2. 2 × 3² × 13
  3. 2² × 3² × 13
  4. 2³ × 13

156 = 2 × 78 = 2 × 2 × 39 = 2² × 3 × 13.

3. The HCF of 26 and 91 is

  1. 7
  2. 13
  3. 26
  4. 1

26 = 2 × 13 and 91 = 7 × 13. The common prime factor is 13.

4. The LCM of 26 and 91 is

  1. 91
  2. 2366
  3. 182
  4. 364

LCM = 2 × 7 × 13 = 182.

5. Which of these numbers is irrational?

  1. √16
  2. 0.25
  3. 22/7
  4. √5

5 is not a perfect square, so √5 is irrational.

6. Given HCF(306, 657) = 9, the LCM(306, 657) is

  1. 22338
  2. 2238
  3. 20338
  4. 18354

LCM = (306 × 657) ÷ 9 = 306 × 73 = 22338.

7. The number 6ⁿ (n a natural number) ends with the digit 0

  1. when n = 5
  2. when n is even
  3. never
  4. when n = 10

6ⁿ = 2ⁿ × 3ⁿ has no prime factor 5, so it can never end in 0.

8. The HCF of 6, 72 and 120 is

  1. 12
  2. 6
  3. 24
  4. 2

6 = 2 × 3, 72 = 2³ × 3², 120 = 2³ × 3 × 5. Smallest common powers: 2 × 3 = 6.

9. The LCM of 6, 72 and 120 is

  1. 720
  2. 120
  3. 360
  4. 240

Greatest powers: 2³ × 3² × 5 = 360.

10. The number 3 + 2√5 is

  1. rational
  2. an integer
  3. a natural number
  4. irrational

A rational number plus a non-zero irrational number is irrational.

11. If a prime p divides a², then

  1. p divides a
  2. a divides p
  3. p² = a
  4. p = a

This result is used in every irrationality proof in this chapter.

12. The HCF of any two consecutive natural numbers is

  1. 0
  2. 2
  3. 1
  4. the smaller number

Consecutive numbers have no common factor other than 1, so they are co-prime.

13. The smallest number divisible by every number from 1 to 10 is

  1. 5040
  2. 2520
  3. 1260
  4. 3628800

LCM(1, 2, …, 10) = 2³ × 3² × 5 × 7 = 2520.

14. The product of a non-zero rational number and an irrational number is

  1. always irrational
  2. always rational
  3. always an integer
  4. sometimes rational

If it were rational, dividing by the rational number would make the irrational number rational.

15. The HCF of 96 and 404 is

  1. 2
  2. 8
  3. 12
  4. 4

96 = 2⁵ × 3 and 404 = 2² × 101. HCF = 2² = 4.

16. 7 × 11 × 13 + 13 is

  1. a prime number
  2. a composite number
  3. neither prime nor composite
  4. an odd prime

7 × 11 × 13 + 13 = 13 × (77 + 1) = 13 × 78, so it has factors other than 1 and itself.

Part B: Fill in the blanks 8 blanks

17. 140 = 2² × [[5]] × 7
18. 3825 = 3² × 5² × [[17]]
19. HCF is the product of the [[smallest|lowest|least]] power of each common prime factor, and LCM is the product of the [[greatest|highest|largest]] power of each prime factor.
20. The HCF of two co-prime numbers is [[1|one]].
21. √2 is an [[irrational]] number.
22. If HCF(a, b) = 1, then LCM(a, b) = [[ab|a×b|a*b|a x b]].
23. Every composite number can be expressed as a product of [[primes|prime numbers|prime]].

Part C: Questions and answers 9 questions

24. Express 5005 as a product of its prime factors.

5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13.

25. Find the HCF and LCM of 12, 15 and 21 by prime factorisation.

12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7.
HCF = 3 (only common prime, smallest power).
LCM = 2² × 3 × 5 × 7 = 420.

26. Prove that √5 is irrational.

Assume √5 is rational, so √5 = a/b where a and b are co-prime integers, b ≠ 0.
Then 5b² = a², so 5 divides a², hence 5 divides a. Write a = 5c.
Then 5b² = 25c², so b² = 5c², which means 5 divides b².
So 5 divides b as well. Now 5 divides both a and b, which contradicts that they are co-prime.
Therefore √5 is irrational.

27. Prove that 5 − √3 is irrational.

Assume 5 − √3 = a/b (rational, b ≠ 0).
Then √3 = 5 − a/b = (5b − a)/b, which is rational.
But √3 is irrational, which is a contradiction. So 5 − √3 is irrational.

28. Prove that 1/√2 is irrational.

Assume 1/√2 = a/b with a, b co-prime integers, a ≠ 0.
Then √2 = b/a, which would be rational. This contradicts the fact that √2 is irrational.
So 1/√2 is irrational.

29. Check whether 6ⁿ can end with the digit 0 for any natural number n.

A number ending in 0 must be divisible by 10 = 2 × 5, so it must have 5 as a prime factor.
6ⁿ = 2ⁿ × 3ⁿ. By the uniqueness in the Fundamental Theorem of Arithmetic, 5 is never a factor.
So 6ⁿ can never end with 0.

30. Sonia takes 18 minutes to drive one round of a field, while Ravi takes 12 minutes. They start at the same point and time, in the same direction. After how many minutes will they meet again at the starting point?

We need LCM(18, 12).
18 = 2 × 3², 12 = 2² × 3, so LCM = 2² × 3² = 36.
They meet again after 36 minutes.

31. Explain why 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

5 × 4 × 3 × 2 × 1 + 5 = 5 × (4 × 3 × 2 × 1 + 1) = 5 × 25 = 125.
It has a factor 5 other than 1 and itself, so it is composite.

32. Find the LCM and HCF of 336 and 54, and verify that LCM × HCF = product of the numbers.

336 = 2⁴ × 3 × 7, 54 = 2 × 3³.
HCF = 2 × 3 = 6. LCM = 2⁴ × 3³ × 7 = 3024.
Check: 6 × 3024 = 18144 and 336 × 54 = 18144. ✓

Part D: Word puzzles 4 puzzles

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