Practice questions Ch. 1 Real number

Section A: Multiple Choice Questions (MCQs) - 1 Mark Each [PYQs]

Q1. What is the HCF of two distinct prime numbers p and q? (CBSE 2020)

• Solution: Prime numbers have no common factors other than 1. Therefore, their Highest Common Factor is always 1.
• Answer: 1

Q2. If two positive integers a and b are written as a = x³y² and b = xy³, where x, y are prime numbers, then find HCF(a, b). (CBSE 2019)

• Solution: To find the HCF, we take the lowest power of the common prime factors.
  Lowest power of x = x¹, Lowest power of y = y²
• Answer: xy²

Q3. The HCF of two numbers is 11 and their LCM is 7700. If one of the numbers is 275, find the other number. (CBSE 2018)

• Solution: Formula: HCF × LCM = Product of two numbers
  11 × 7700 = 275 × Second Number
  $\text{Second Number} = \frac{11 \times 7700}{275} = 308$
• Answer: 308

Q4. Find the HCF of the smallest prime number and the smallest composite number. (CBSE 2023)

• Solution: Smallest prime number = 2, Smallest composite number = 4.
  HCF(2, 4) = 2
• Answer: 2

Q5. Can the number $4^n$, where n is a natural number, end with the digit 0? (CBSE 2017)

• Solution: For a number to end with 0, its prime factors must include both 2 and 5. Here, $4^n = (2^2)^n = 2^{2n}$. Since 5 is not a prime factor, it can never end with 0.
• Answer: No

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Section B: Very Short Answer Questions - 2 Marks Each [PYQs]

Q6. Find the HCF and LCM of 96 and 404 by the prime factorization method. (CBSE 2020)

• Solution:
  96 = 2⁵ × 3
  404 = 2² × 101
  HCF = 2² = 4
  LCM = 2⁵ × 3 × 101 = 96 × 101 = 9696

Q7. Explain why 7 × 11 × 13 + 13 is a composite number. (CBSE 2019)

• Solution:
  Taking 13 common from the expression:
  13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78
  Since the given number can be expressed as a product of two integers greater than 1, it has factors other than 1 and itself. Hence, it is a composite number.

Q8. Given that HCF(306, 657) = 9, find LCM(306, 657). (CBSE 2021)

• Solution:
  $$\text{LCM} = \frac{\text{Product of two numbers}}{\text{HCF}}$$
  $$\text{LCM} = \frac{306 \times 657}{9} = 34 \times 657 = 22338$$

Q9. Find the largest number which divides 70 and 125, leaving remainders 5 and 8, respectively. (CBSE 2022)

• Solution:
  The required number is the HCF(70-5, 125-8) = HCF(65, 117).
  65 = 5 × 13
  117 = 3² × 13
  HCF = 13
  So, the largest number is 13.

Q10. Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons. (CBSE 2016)

• Solution:
  The LCM must always be completely divisible by the HCF.
  $\frac {380}{18} = 21.11$ (Not a whole number).
  Since the HCF does not divide the LCM completely, no such two numbers can exist.

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Section C: Short Answer Questions - 3 Marks Each [PYQs]

Q11. Prove that $\sqrt{3}$ is an irrational number. (CBSE 2020, 2024)

• Solution:
  Let us assume the contrary, that $\sqrt{3}$ is a rational number. So, $\sqrt{3} = \frac{a}{b}$ where a and b are co-prime integers and b ≠ 0.
  Squaring both sides: $3 = \frac{a^2}{b^2} \implies a^2 = 3b^2$ --- (i)
  This means a² is divisible by 3, so a must also be divisible by 3.
  Let a = 3c. Substitute this into equation (i):
  $(3c)^2 = 3b^2 \implies 9c^2 = 3b^2 \implies b^2 = 3c^2$
  This means b² is divisible by 3, so b must also be divisible by 3.
  Therefore, a and b share a common factor of 3, which contradicts our assumption that a and b are co-prime. Hence, $\sqrt{3}$ is irrational.

Q12. Show that $5 - \sqrt{3}$ is an irrational number. (CBSE 2018, 2023)

• Solution:
  Let us assume $5 - \sqrt{3}$ is a rational number equal to r.
  $5 - \sqrt{3} = r \implies \sqrt{3} = 5 - r$
  Since r is rational, 5 - r must also be a rational number.
  This implies that $\sqrt{3}$ is rational, which is a contradiction because $\sqrt{3}$ is known to be irrational.
  Thus, our assumption is wrong, and $5 - \sqrt{3}$ is irrational.

Q13. Sonia takes 18 minutes to drive one round of a circular sports track, while Ravi takes 12 minutes for the same. If they start at the same point and time in the same direction, after how many minutes will they meet again at the starting point? (CBSE 2017)

• Solution:
  Meeting time = LCM(18, 12)
  18 = 2 × 3²
  12 = 2² × 3
  LCM = 2² × 3² = 4 × 9 = 36 minutes.
  They will meet again after 36 minutes.

Q14. Three bells toll together at intervals of 9, 12, and 15 minutes respectively. If they toll together now, after how much time will they toll together next? (CBSE 2022)

• Solution:
  Time interval = LCM(9, 12, 15)
  9 = 3², 12 = 2² × 3, 15 = 3 × 5
  LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180 minutes.
  Converting to hours: $\frac{180}{60} = 3$ hours. They will toll together next after 3 hours.

Q15. Prove that $3 + 2\sqrt{5}$ is irrational, given that $\sqrt{5}$ is irrational. (CBSE 2019, 2025)

• Solution:
  Let $3 + 2\sqrt{5} = \frac{a}{b}$ (where $\frac{a}{b}$ is rational).
  $2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}$
  $\sqrt{5} = \frac{a - 3b}{2b}$
  Since a and b are integers, $\frac{a - 3b}{2b}$ is a rational number. This implies $\sqrt{5}$ is rational, which contradicts the fact that $\sqrt{5}$ is irrational. Thus, $3 + 2\sqrt{5}$ is irrational.

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Section D: Core Concept Practice & Application (Questions 16 to 50)

Q16. Express 156 as a product of its prime factors.

• Solution: 156 = 2 × 78 = 2 × 2 × 39 = 2² × 3 × 13

Q17. Find the prime factorization of 5005.

• Solution: 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13

Q18. Write the prime factorization of 7429.

• Solution: 7429 = 17 × 437 = 17 × 19 × 23

Q19. If HCF(a, b) = 4 and a × b = 96, calculate LCM(a, b).

• Solution: $\text{LCM} = \frac{a \times b}{\text{HCF}} = \frac{96}{4} = 24$

Q20. Find the HCF and LCM of 6, 72, and 120 using prime factorization.

• Solution:
  6 = 2 × 3, 72 = 2³ × 3², 120 = 2³ × 3 × 5
  HCF = 2¹ × 3¹ = 6
  LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360

Q21. If two numbers are co-prime, what will be their LCM?

• Solution: The HCF of co-prime numbers is 1. Therefore, their LCM is equal to their product (a × b).

Q22. Check whether $12^n$ can end with the digit 0 or 5 for any natural number n.

• Solution: $12^n = (2^2 \times 3)^n = 2^{2n} \times 3^n$. Since prime factor 5 is absent, it cannot end with 0 or 5.

Q23. Prove that $7\sqrt{5}$ is an irrational number.

• Solution: Let $7\sqrt{5} = \frac{a}{b} \implies \sqrt{5} = \frac{a}{7b}$. Since $\frac{a}{7b}$ is rational, $\sqrt{5}$ must be rational, which is a contradiction.

Q24. Prove that $\frac{1}{\sqrt{2}}$ is irrational.

• Solution: Let $\frac{1}{\sqrt{2}} = \frac{a}{b} \implies \sqrt{2} = \frac{b}{a}$. Rationalizing values creates a contradiction as $\sqrt{2}$ is irrational.

Q25. Find the least number that is divisible by all the numbers from 1 to 10 (both inclusive).

• Solution: Calculate LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10):
  LCM = 2³ × 3² × 5 × 7 = 8 × 9 × 5 × 7 = 2520

Q26. Explain why 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

• Solution: Factoring out 5: 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 × 1009. It has more than two factors, so it is composite.

Q27. If HCF(65, 117) is expressible in the form 65m - 117, find the value of m.

• Solution: HCF(65, 117) = 13.
  $65m - 117 = 13 \implies 65m = 130 \implies m = 2$

Q28. If two positive integers p and q are written as p = ab² and q = a³b, then find LCM(p, q).

• Solution: Taking the highest powers: LCM = a³b²

Q29. Based on the values in Q28, find HCF(p, q).

• Solution: Taking the lowest powers: HCF = ab

Q30. Find a rational number between $\sqrt{2}$ and $\sqrt{3}$.

• Solution: $\sqrt{2} \approx 1.414$ and $\sqrt{3} \approx 1.732$. Any terminating decimal in between, such as 1.5 (or $\frac{3}{2}$), is a rational number.

Q31. Show that the square of any positive integer is of the form 3m or 3m+1 for some integer m.

• Solution: An integer can be of the form 3q, 3q+1, or 3q+2.
  (3q)² = 9q² = 3(3q²) = 3m
  (3q+1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1

Q32. Define a rational number.

• Solution: A number that can be expressed in the form $\frac{p}{q}$, where p and q are integers and q ≠ 0.

Q33. What type of decimal expansion does an irrational number have?

• Solution: Non-terminating and non-repeating.

Q34. If n is a natural number, what can you say about $\sqrt{n}$?

• Solution: It can be either a natural number (if n is a perfect square, like $\sqrt{4}=2$) or an irrational number (like $\sqrt{2}$).

Q35. Does the HCF of two numbers always divide their LCM?

• Solution: Yes, HCF is always a factor of the LCM.

Q36. Find the HCF of 26 and 91.

• Solution: 26 = 2 × 13 and 91 = 7 × 13. HCF = 13.

Q37. Find the LCM of 26 and 91.

• Solution: LCM = 2 × 7 × 13 = 182.

Q38. For any two positive integers a and b, what is HCF(a,b) × LCM(a,b) equal to?

• Solution: Product of the numbers (a × b).

Q39. Find the HCF of 510 and 92.

• Solution: 510 = 2 × 3 × 5 × 17 and 92 = 2² × 23. HCF = 2.

Q40. Calculate the LCM of 510 and 92.

• Solution: LCM = 2² × 3 × 5 × 17 × 23 = 23460.

Q41. What is the HCF of 17, 23, and 29?

• Solution: Since all three are prime numbers, their HCF = 1.

Q42. Find the LCM of 17, 23, and 29.

• Solution: LCM = 17 × 23 × 29 = 11339.

Q43. Find the HCF of 336 and 54.

• Solution: 336 = 2⁴ × 3 × 7 and 54 = 2 × 3³. HCF = 2 × 3 = 6.

Q44. Find the LCM of 336 and 54.

• Solution: LCM = 2⁴ × 3³ × 7 = 3024.

Q45. If the prime factorization of a number is 2³ × 5², find the number.

• Solution: 8 × 25 = 200.

Q46. Prove that $\sqrt{2} + \sqrt{5}$ is an irrational number.

• Solution: Let $\sqrt{2} + \sqrt{5} = x$ (rational). Squaring both sides: $2 + 5 + 2\sqrt{10} = x^2 \implies \sqrt{10} = \frac{x^2 - 7}{2}$. Since the right side is rational, it means $\sqrt{10}$ is rational, which is a contradiction.

Q47. What is the smallest even prime number?

• Solution: 2 (It is also the only even prime number).

Q48. State the Fundamental Theorem of Arithmetic.

• Solution: Every composite number can be expressed as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

Q49. What can you say about the decimal expansion of a rational number whose denominator is of the form $2^n 5^m$?

• Solution: It will have a terminating decimal expansion.

Q50. If LCM(x, 18) = 36 and HCF(x, 18) = 2, find the value of x.

• Solution: $x \times 18 = 36 \times 2 \implies 18x = 72 \implies x = 4$.