CBSE Class 10 Maths Ch 1 Real Numbers Question Bank with Solutions

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Class 10 Maths · Chapter 1 · Question Bank + Solutions

CBSE · NCERT

MATHEMATICS – CHAPTER 1 : REAL NUMBERS

Chapter-wise Question Bank (CBSE Board + Competitive · Moderate to HOTS Level)

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Class : XSubject : MathematicsName : ..................

Questions : 89Session : 2026–27Total Marks : 151

General Instructions

1. This question bank is based on NCERT Class X Mathematics, Chapter 1 (Real Numbers), and is set in the CBSE board pattern with competitive (HOTS) questions mixed in.
2. Section A: Q.1 to Q.50 are MCQs of 1 mark each (50 marks). Section B: Q.51 to Q.60 are Assertion–Reason questions of 1 mark each (10 marks).
3. Section C: Q.61 to Q.70, Very Short Answer, 2 marks each (20 marks). Section D: Q.71 to Q.80, Short Answer, 3 marks each (30 marks).
4. Section E: Q.81 to Q.84, Case-Based, 4 marks each (16 marks). Section F: Q.85 to Q.89, Long Answer, 5 marks each (25 marks).
5. Show all steps for Sections C to F. Complete solutions are given after the questions.

SECTION – A (Multiple Choice Questions · 1 Mark each)

Q.1 – Q.50 Choose the correct option.50 × 1 = 50

1. The HCF of 96 and 404 is –

(a) 2(b) 8(c) 4(d) 12

2. The LCM of 12, 15 and 21 is –

(a) 420(b) 210(c) 840(d) 1260

3. The product of the HCF and the LCM of 24 and 36 is –

(a) 72(b) 864(c) 144(d) 432

4. If HCF(a, b) = 12 and a × b = 1800, then LCM(a, b) is –

(a) 120(b) 180(c) 216(d) 150

5. Which of the following is an irrational number?

(a) 22/7(b) 3.14(c) 0.1010010001… (non-terminating, non-repeating)(d) √25

6. The decimal expansion of 7/80 terminates after how many places of decimals?

(a) 2(b) 3(c) 5(d) 4

7. The decimal expansion of 13/3125 terminates after –

(a) 3 decimal places(b) 4 decimal places(c) 6 decimal places(d) 5 decimal places

8. 17/8 in decimal form is –

(a) 2.25(b) 2.0125(c) 2.125(d) 2.375

9. Which of the following rational numbers has a non-terminating repeating decimal expansion?

(a) 11/30(b) 17/8(c) 13/20(d) 7/25

10. In Euclid's division lemma a = bq + r, the value of r satisfies –

(a) 0 < r ≤ b(b) 0 ≤ r ≤ b(c) 0 < r < b(d) 0 ≤ r < b

11. Every positive odd integer is of the form (q is an integer) –

(a) 2q(b) 2q + 2(c) 2q + 1(d) 2q – 2

12. The prime factorisation of 3825 is –

(a) 3 × 5² × 17²(b) 3² × 5² × 17(c) 3² × 5 × 17²(d) 3³ × 5 × 17

13. The total number of factors of 360 is –

(a) 24(b) 12(c) 18(d) 30

14. On dividing 727 by 12, using Euclid's division lemma, the quotient and remainder are –

(a) 59 and 19(b) 60 and 7(c) 61 and –5(d) 60 and 12

15. The number 7 × 11 × 13 + 13 is –

(a) a prime number(b) an irrational number(c) neither prime nor composite(d) a composite number

16. The LCM of 2³ × 3 and 2² × 3² × 5 is –

(a) 180(b) 720(c) 360(d) 120

17. The HCF of 306 and 657 is –

(a) 3(b) 18(c) 27(d) 9

18. The LCM of 26 and 91 is –

(a) 182(b) 91(c) 364(d) 2366

19. The irrationality of √2 is proved by the method of –

(a) induction(b) substitution(c) direct verification(d) contradiction

20. The sum of a rational number and an irrational number is –

(a) always rational(b) always irrational(c) sometimes rational(d) zero

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

(a) always rational(b) always irrational(c) sometimes rational(d) an integer

22. Which of the following is an irrational number?

(a) 2 + √3(b) √196(c) (√3)²(d) √(81/16)

23. The value of (2 + √3)(2 – √3) is –

(a) 1 (a rational number)(b) √3(c) 4 – 3√3(d) 7

24. For any odd positive integer n, n² – 1 is always divisible by –

(a) 16(b) 12(c) 8(d) 5

25. The cube of any positive integer is of the form –

(a) 9m, 9m + 2 or 9m + 4(b) 9m, 9m + 1 or 9m + 8(c) 9m + 1, 9m + 3 or 9m + 5(d) 9m + 2, 9m + 5 or 9m + 7

26. The square of any positive integer is of the form –

(a) 3m or 3m + 1(b) 3m or 3m + 2(c) 3m + 1 or 3m + 2(d) only 3m + 1

27. If HCF(65, 117) = 65m + 117n, the values of m and n are –

(a) m = 2, n = –1(b) m = 1, n = –2(c) m = 2, n = 1(d) m = –2, n = 1

28. The largest number which divides 70 and 125, leaving remainders 5 and 8 respectively, is –

(a) 65(b) 117(c) 26(d) 13

29. The largest number that divides 245 and 1029, leaving remainder 5 in each case, is –

(a) 16(b) 8(c) 24(d) 32

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

(a) 1260(b) 5040(c) 2520(d) 720

31. The smallest number which, when divided by 12, 15 and 18, leaves a remainder 5 in each case, is –

(a) 185(b) 180(c) 175(d) 365

32. Three bells toll at intervals of 12, 15 and 20 minutes. If they toll together now, they will toll together again after –

(a) 30 minutes(b) 60 minutes(c) 120 minutes(d) 180 minutes

33. For which natural number n can the number 6ⁿ end with the digit 0?

(a) n = 1(b) n = 5(c) for no natural number n(d) every n ≥ 2

34. The decimal expansion of 1/7 has a repeating block of length –

(a) 3(b) 6(c) 7(d) 1

35. The decimal expansion of 129/(2² × 5⁷ × 7⁵) is –

(a) terminating(b) non-terminating repeating(c) non-terminating non-repeating(d) an integer

36. 43/(2⁴ × 5³) terminates after how many places of decimals?

(a) 3(b) 7(c) 12(d) 4

37. The rational form of 0.7 is –

(a) 7/10(b) 70/99(c) 7/9(d) 7/99

38. If p/q (q ≠ 0, in lowest terms) has a terminating decimal expansion, the prime factorisation of q must be of the form –

(a) 3ⁿ 5^m(b) 2ⁿ 7^m(c) 2ⁿ 5^m (n, m non-negative integers)(d) any prime power

39. A rational number between √2 and √3 is –

(a) 1.3(b) 1.8(c) 1.5(d) 1.2

40. The number of irrational numbers between 2 and 3 is –

(a) zero(b) one(c) two(d) infinite

41. The HCF and LCM of two co-prime numbers 8 and 9 are –

(a) 72 and 1(b) 1 and 17(c) 1 and 72(d) 9 and 72

42. If a = x³y² and b = xy³ (x, y are primes), then HCF(a, b) is –

(a) x³y³(b) xy²(c) xy(d) x²y

43. The LCM of a = 2³ × 3² × 5 and b = 2² × 3³ × 7 is –

(a) 2520(b) 1260(c) 15120(d) 7560

44. The product of any three consecutive positive integers is always divisible by –

(a) 6(b) 12(c) 8(d) 24

45. The HCF of 525 and 3000 is –

(a) 25(b) 75(c) 105(d) 150

46. If 156 = 2^a × 3^b × 13^c, then a + b + c is –

(a) 4(b) 3(c) 5(d) 6

47. The smallest 3-digit number divisible by both 12 and 18 is –

(a) 100(b) 108(c) 144(d) 216

48. The greatest 4-digit number exactly divisible by 12, 15, 18 and 27 is –

(a) 9990(b) 9720(c) 9180(d) 9900

49. The HCF of 36, 84 and 120 is –

(a) 6(b) 24(c) 4(d) 12

50. For every positive integer n, the number n³ – n is always divisible by –

(a) 6(b) 4(c) 8(d) 12

SECTION – B (Assertion (A) – Reason (R) · 1 Mark each)

Q.51 – Q.60 Choose: (a) Both A and R true, R explains A; (b) Both true, R does not explain A; (c) A true, R false; (d) A false, R true.10 × 1 = 10

51. Assertion (A): 2 + √3 is an irrational number.
Reason (R): The sum of a rational number and an irrational number is irrational.

52. Assertion (A): The product of two irrational numbers is always irrational.
Reason (R): √2 × √2 = 2.

53. Assertion (A): 6ⁿ never ends with the digit 0 for any natural number n.
Reason (R): The prime factorisation of 6ⁿ contains no factor 5.

54. Assertion (A): HCF(a, b) × LCM(a, b) = a × b for any two positive integers a and b.
Reason (R): The HCF of two numbers always divides their LCM.

55. Assertion (A): 7/20 has a terminating decimal expansion.
Reason (R): 7 is a prime number.

56. Assertion (A): 1/7 has a terminating decimal expansion.
Reason (R): The denominator 7 has a prime factor other than 2 and 5.

57. Assertion (A): √2 is an irrational number.
Reason (R): √2 = 1.414 is a terminating decimal.

58. Assertion (A): The product of three consecutive positive integers is divisible by 6.
Reason (R): Among any three consecutive integers, one is a multiple of 3 and at least one is even.

59. Assertion (A): LCM(12, 18) = 36.
Reason (R): 12 and 18 are co-prime.

60. Assertion (A): The HCF of two numbers can be greater than their LCM.
Reason (R): The HCF of two numbers always divides their LCM.

SECTION – C (Very Short Answer · 2 Marks each)

61. Find the HCF and LCM of 12, 15 and 21 by the prime factorisation method.(2)

62. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.(2)

63. Without actual division, state whether 35/50 has a terminating or non-terminating repeating decimal expansion.(2)

64. Use Euclid's division algorithm to find the HCF of 135 and 225.(2)

65. Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.(2)

66. If HCF(96, 404) = 4, find LCM(96, 404).(2)

67. Can 6ⁿ, for some natural number n, end with the digit 0? Give reason.(2)

68. Express 0.7 in the form p/q.(2)

69. Find a rational number between √2 and √3.(2)

70. Given that HCF(306, 657) = 9, find LCM(306, 657).(2)

SECTION – D (Short Answer · 3 Marks each)

71. Prove that √5 is an irrational number.(3)

72. Prove that 3 + 2√5 is irrational. (Given that √5 is irrational.)(3)

73. Prove that the square of any positive integer is of the form 3m or 3m + 1 for some integer m.(3)

74. Use Euclid's division algorithm to find the HCF of 867 and 255.(3)

75. Find the HCF and LCM of 336 and 54 by prime factorisation and verify HCF × LCM = product.(3)

76. Show that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer.(3)

77. Without long division, state which of 3/8, 13/125, 17/6 and 64/455 have terminating decimal expansions. Also write the number of decimal places for those which terminate.(3)

78. Find the largest number that divides 398, 436 and 542 leaving remainders 7, 11 and 15 respectively.(3)

79. Prove that 1/√2 is an irrational number.(3)

80. Three bells ring at intervals of 9, 12 and 15 minutes. If they ring together at 8:00 a.m., when will they next ring together?(3)

SECTION – E (Case-Based Questions · 4 Marks each)

81. Read the passage and answer the questions that follow.(4)

A school has 144 boys and 192 girls of Class X for a drill. The sports teacher wants to arrange them in groups so that every group has the same number of students and no group contains both boys and girls. He wants the number of students in each group to be as large as possible.

(i) What is the maximum number of students in each group?(1)

(ii) How many groups are formed in all?(1)

(iii) A third set of 96 students of Class IX joins the drill and is also to be grouped separately. What is the largest group size that works for all three sets, and how many groups are there in total?(2)

82. Read the passage and answer the questions that follow.(4)

Ravi and Raj run on a circular track. Ravi takes 18 minutes and Raj takes 12 minutes to complete one round. They start together from the same point and in the same direction at 6:00 a.m.

(i) After how many minutes will they meet again at the starting point?(1)

(ii) How many rounds will each have completed by then?(1)

(iii) A third runner Shan, who takes 8 minutes per round, joins them at the start. When will all three meet together at the starting point, and how many rounds will Shan have run?(2)

83. Read the passage and answer the questions that follow.(4)

A rectangular hall is 8 m 25 cm long and 6 m 75 cm wide. It is to be paved with square tiles of the same size, and the tiles must be as large as possible, without cutting any tile.

(i) Write the length and the breadth of the hall in centimetres.(1)

(ii) Use Euclid's division algorithm to find the side of the largest square tile.(1)

(iii) Find the number of tiles required, and the total cost if each tile costs ₹12.(2)

84. Read the passage and answer the questions that follow.(4)

A teacher writes the following rational numbers on the board: 13/3125, 17/8, 64/455, 29/343 and 77/210. She asks the students to decide, without long division, the nature of their decimal expansions.

(i) Which of these have terminating decimal expansions?(1)

(ii) Why is the decimal expansion of 64/455 non-terminating repeating?(1)

(iii) Find the decimal expansions of 13/3125 and 17/8.(2)

SECTION – F (Long Answer · 5 Marks each)

85. Prove that √2 is irrational. Hence show that 3 + 5√2 is also irrational.(5)

86. Show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8, for some integer m.(5)

87. Use Euclid's division algorithm to find the HCF of 960 and 432, and express it in the form 960x + 432y.(5)

88. Find the greatest 6-digit number which is exactly divisible by 24, 15 and 36.(5)

89. The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other number. Verify your answer using the prime factorisation of both numbers.(5)

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CBSE · NCERT

MATHEMATICS – CHAPTER 1 : COMPLETE SOLUTIONS

Real Numbers · Answer Key with Step-by-Step Solutions

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Class : XSubject : MathematicsTotal Marks : 151

SECTION – A (MCQ Solutions)

Quick Answer Key (Q.1 – Q.50)

1. (c)	2. (a)	3. (b)	4. (d)	5. (c)	6. (d)	7. (d)	8. (c)	9. (a)	10. (d)
11. (c)	12. (b)	13. (a)	14. (b)	15. (d)	16. (c)	17. (d)	18. (a)	19. (d)	20. (b)
21. (b)	22. (a)	23. (a)	24. (c)	25. (b)	26. (a)	27. (a)	28. (d)	29. (a)	30. (c)
31. (a)	32. (b)	33. (c)	34. (b)	35. (b)	36. (d)	37. (c)	38. (c)	39. (c)	40. (d)
41. (c)	42. (b)	43. (d)	44. (a)	45. (b)	46. (a)	47. (b)	48. (b)	49. (d)	50. (a)

Solutions with Explanation

1. Ans. (c) 4

96 = 2⁵ × 3 and 404 = 2² × 101. Common factor = 2² = 4.

2. Ans. (a) 420

12 = 2²×3, 15 = 3×5, 21 = 3×7. LCM = 2²×3×5×7 = 420.

3. Ans. (b) 864

HCF × LCM = product of the two numbers = 24 × 36 = 864.

4. Ans. (d) 150

LCM = (a × b) ÷ HCF = 1800 ÷ 12 = 150.

5. Ans. (c) 0.1010010001… (non-terminating, non-repeating)

22/7 and 3.14 are rational; √25 = 5 is rational. A non-terminating non-recurring decimal is irrational.

6. Ans. (d) 4

80 = 2⁴ × 5, so the number of decimal places = max(4, 1) = 4 (7/80 = 0.0875).

7. Ans. (d) 5 decimal places

3125 = 5⁵; places = max(0, 5) = 5. Indeed 13/3125 = 0.00416.

8. Ans. (c) 2.125

17 ÷ 8 = 2.125 (denominator 8 = 2³ gives 3 decimal places).

9. Ans. (a) 11/30

30 = 2 × 3 × 5 has the prime factor 3, so 11/30 is non-terminating repeating. The other denominators are of the form 2^m5ⁿ.

10. Ans. (d) 0 ≤ r < b

Euclid's division lemma: for positive integers a and b there exist unique q and r with a = bq + r, 0 ≤ r < b.

11. Ans. (c) 2q + 1

Dividing by 2, the remainder is 0 or 1: even numbers are 2q and odd numbers are 2q + 1.

12. Ans. (b) 3² × 5² × 17

3825 = 9 × 425 = 9 × 25 × 17 = 3² × 5² × 17.

13. Ans. (a) 24

360 = 2³ × 3² × 5. Number of factors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24.

14. Ans. (b) 60 and 7

12 × 60 = 720 and 727 = 12 × 60 + 7, with 0 ≤ 7 < 12.

15. Ans. (d) a composite number

7 × 11 × 13 + 13 = 13(77 + 1) = 13 × 78; it has more than two factors, so it is composite.

16. Ans. (c) 360

LCM = highest powers of all primes = 2³ × 3² × 5 = 360.

17. Ans. (d) 9

306 = 2 × 3² × 17 and 657 = 3² × 73. HCF = 3² = 9.

18. Ans. (a) 182

26 = 2 × 13 and 91 = 7 × 13. LCM = 2 × 7 × 13 = 182.

19. Ans. (d) contradiction

Assume √2 = p/q in lowest terms and arrive at a contradiction (p and q both even).

20. Ans. (b) always irrational

If r + x were rational (r rational), then x = (r + x) – r would be rational – a contradiction.

21. Ans. (b) always irrational

If r × x = s were rational (r ≠ 0), x = s/r would be rational, a contradiction.

22. Ans. (a) 2 + √3

√196 = 14, (√3)² = 3, √(81/16) = 9/4 are rational; 2 + √3 (rational + irrational) is irrational.

23. Ans. (a) 1 (a rational number)

(2 + √3)(2 – √3) = 2² – (√3)² = 4 – 3 = 1.

24. Ans. (c) 8

n = 2k + 1 ⇒ n² – 1 = 4k(k + 1), and k(k + 1) is even, so it is divisible by 8. (n = 3 gives 8, which rules out 16, 12 and 5.)

25. Ans. (b) 9m, 9m + 1 or 9m + 8

Taking n = 3q, 3q + 1, 3q + 2: the cubes are 9(3q³), 9(…) + 1 and 9(…) + 8.

26. Ans. (a) 3m or 3m + 1

(3q)² = 3(3q²); (3q + 1)² = 3(3q² + 2q) + 1; (3q + 2)² = 3(3q² + 4q + 1) + 1.

27. Ans. (a) m = 2, n = –1

117 = 65 × 1 + 52; 65 = 52 × 1 + 13. So 13 = 65 – 52 = 65 – (117 – 65) = 65 × 2 + 117 × (–1).

28. Ans. (d) 13

70 – 5 = 65 and 125 – 8 = 117. HCF(65, 117) = 13.

29. Ans. (a) 16

245 – 5 = 240 = 2⁴×3×5 and 1029 – 5 = 1024 = 2¹⁰. HCF = 2⁴ = 16.

30. Ans. (c) 2520

LCM(1…10) = 2³ × 3² × 5 × 7 = 2520.

31. Ans. (a) 185

LCM(12, 15, 18) = 180. Required number = 180 + 5 = 185.

32. Ans. (b) 60 minutes

LCM(12, 15, 20) = 60 minutes.

33. Ans. (c) for no natural number n

6ⁿ = 2ⁿ × 3ⁿ has no prime factor 5, so it is never divisible by 10.

34. Ans. (b) 6

1/7 = 0.142857 142857… so the repeating block 142857 has 6 digits.

35. Ans. (b) non-terminating repeating

The denominator contains the prime 7 (other than 2 and 5), and 129 = 3 × 43 does not cancel it, so the decimal is non-terminating repeating.

36. Ans. (d) 4

The number of decimal places = max(4, 3) = 4.

37. Ans. (c) 7/9

Let x = 0.777…; 10x = 7.777…; 9x = 7 ⇒ x = 7/9.

38. Ans. (c) 2ⁿ 5^m (n, m non-negative integers)

This is the theorem on terminating decimals: q has no prime factors other than 2 and 5.

39. Ans. (c) 1.5

√2 ≈ 1.414 and √3 ≈ 1.732; 1.414 < 1.5 < 1.732.

40. Ans. (d) infinite

Between any two distinct real numbers there are infinitely many rational and infinitely many irrational numbers.

41. Ans. (c) 1 and 72

Co-prime numbers have HCF = 1; LCM = product = 8 × 9 = 72.

42. Ans. (b) xy²

HCF takes the smallest power of each prime: x¹y² = xy². (LCM = x³y³.)

43. Ans. (d) 7560

LCM = 2³ × 3³ × 5 × 7 = 8 × 27 × 35 = 7560.

44. Ans. (a) 6

Among three consecutive integers one is divisible by 3 and at least one by 2, so the product is divisible by 6 (1×2×3 = 6 rules out 12, 8 and 24).

45. Ans. (b) 75

525 = 3 × 5² × 7 and 3000 = 2³ × 3 × 5³. HCF = 3 × 5² = 75.

46. Ans. (a) 4

156 = 4 × 39 = 2² × 3 × 13, so a + b + c = 2 + 1 + 1 = 4.

47. Ans. (b) 108

LCM(12, 18) = 36; multiples of 36: 36, 72, 108 – the first 3-digit one.

48. Ans. (b) 9720

LCM = 2² × 3³ × 5 = 540. 9999 ÷ 540 gives quotient 18, so the number = 540 × 18 = 9720.

49. Ans. (d) 12

36 = 2²×3², 84 = 2²×3×7, 120 = 2³×3×5. HCF = 2² × 3 = 12.

50. Ans. (a) 6

n³ – n = (n – 1) n (n + 1), a product of three consecutive integers, so it is divisible by 6. (n = 2 gives 6, which rules out 4, 8 and 12.)

SECTION – B (Assertion – Reason)

51. Ans. (a)

Both true; 2 is rational and √3 is irrational, so by R the sum is irrational.

52. Ans. (d)

A is false: √2 × √2 = 2 is rational (R is true and is a counter-example).

53. Ans. (a)

Both true; a number ends in 0 only if it is divisible by both 2 and 5, and R shows 5 is absent.

54. Ans. (b)

Both true, but R does not explain A.

55. Ans. (b)

A is true (20 = 2²×5, so 7/20 = 0.35). R is true, but it is not the reason for A.

56. Ans. (d)

A is false (1/7 = 0.142857… repeats). R is true.

57. Ans. (c)

A is true. R is false: the decimal expansion of √2 is non-terminating and non-recurring (1.414 is only an approximation).

58. Ans. (a)

Both true; R is the reason for A.

59. Ans. (c)

A is true (12 = 2²×3, 18 = 2×3², LCM = 36). R is false: HCF(12, 18) = 6.

60. Ans. (d)

A is false: HCF ≤ LCM always. R is true.

SECTION – C (2 Marks each)

61. Ans.

12 = 2²×3, 15 = 3×5, 21 = 3×7. [1] HCF = 3 and LCM = 2²×3×5×7 = 420. [1]

62. Ans.

7×11×13 + 13 = 13(7×11 + 1) = 13 × 78 – has factors other than 1 and itself, so composite. [1] 7×6×5×4×3×2×1 + 5 = 5(7×6×4×3×2×1 + 1) = 5 × 1009 – composite. [1]

63. Ans.

35/50 = 7/10 (lowest terms). [1] The denominator 10 = 2 × 5 is of the form 2^m5ⁿ, so the expansion terminates (0.7). [1]

64. Ans.

225 = 135 × 1 + 90; [½] 135 = 90 × 1 + 45; [½] 90 = 45 × 2 + 0. [½] HCF = 45. [½]

65. Ans.

Let a be a positive integer. Apply Euclid's lemma with b = 4: a = 4q + r, r = 0, 1, 2, 3. [1] 4q and 4q + 2 are even (divisible by 2). So the odd integers are of the form 4q + 1 or 4q + 3. [1]

66. Ans.

HCF × LCM = product of the numbers. [1] LCM = (96 × 404) ÷ 4 = 38784 ÷ 4 = 9696. [1]

67. Ans.

If 6ⁿ ended with 0, it would be divisible by 5. [1] But 6ⁿ = 2ⁿ × 3ⁿ; by the Fundamental Theorem of Arithmetic, 5 is not among its prime factors. So it cannot end with 0. [1]

68. Ans.

Let x = 0.777… Then 10x = 7.777… [1] Subtracting: 9x = 7, so x = 7/9. [1]

69. Ans.

√2 ≈ 1.4142 and √3 ≈ 1.7320. [1] Any number between them, e.g. 1.5 (= 3/2), is a rational number between √2 and √3. [1]

70. Ans.

LCM = (306 × 657) ÷ HCF = 201042 ÷ 9. [1] LCM = 22338. [1]

SECTION – D (3 Marks each)

71. Ans.

Suppose √5 is rational: √5 = p/q, where p and q are co-prime integers, q ≠ 0. [½] Then p = q√5 ⇒ p² = 5q². So 5 divides p² ⇒ 5 divides p. [1] Let p = 5m. Then 25m² = 5q² ⇒ q² = 5m², so 5 divides q. [1] Thus 5 is a common factor of p and q, contradicting that they are co-prime. [½] Hence √5 is irrational.

72. Ans.

Suppose 3 + 2√5 is rational, say = a/b (b ≠ 0, a and b integers). [1] Then √5 = (a – 3b)/(2b). [1] The right side is rational, but √5 is irrational – a contradiction. [½] Hence 3 + 2√5 is irrational. [½]

73. Ans.

By Euclid's lemma, any positive integer a = 3q + r with r = 0, 1, 2. [1] r = 0: a² = 9q² = 3(3q²) = 3m. r = 1: a² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1. r = 2: a² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1 = 3m + 1. [1½] Hence the square is of the form 3m or 3m + 1. [½]

74. Ans.

867 = 255 × 3 + 102 [1]
255 = 102 × 2 + 51 [½]
102 = 51 × 2 + 0 [½]
The remainder is zero, so HCF(867, 255) = 51. [1]

75. Ans.

336 = 2⁴ × 3 × 7 and 54 = 2 × 3³. [1] HCF = 2 × 3 = 6; LCM = 2⁴ × 3³ × 7 = 3024. [1] Verification: 6 × 3024 = 18144 = 336 × 54. [1]

76. Ans.

Let n = 3q + r, r = 0, 1, 2. [½] r = 0: n = 3q is divisible by 3; n + 2 = 3q + 2 and n + 4 = 3q + 4 = 3(q + 1) + 1 are not. [1] r = 1: n + 2 = 3q + 3 = 3(q + 1) is divisible; n = 3q + 1 and n + 4 = 3q + 5 = 3(q + 1) + 2 are not. [½] r = 2: n + 4 = 3q + 6 = 3(q + 2) is divisible; n = 3q + 2 and n + 2 = 3q + 4 are not. [½] In every case exactly one of the three is divisible by 3. [½]

77. Ans.

3/8: 8 = 2³ – terminates, 3 places (0.375). [1] 13/125: 125 = 5³ – terminates, 3 places (0.104). [1] 17/6: 6 = 2×3 – non-terminating. 64/455: 455 = 5×7×13 – non-terminating. [1]

78. Ans.

The required number divides 398 – 7 = 391, 436 – 11 = 425 and 542 – 15 = 527 exactly. [1] 391 = 17 × 23; 425 = 5² × 17; 527 = 17 × 31. [1] HCF = 17. [1]

79. Ans.

Suppose 1/√2 is rational = a/b (a, b integers, b ≠ 0, co-prime). [1] Then √2 = b/a, a rational number. [1] But √2 is irrational – a contradiction. Hence 1/√2 is irrational. [1]

80. Ans.

The bells ring together after the LCM of 9, 12 and 15 minutes. [1] 9 = 3², 12 = 2²×3, 15 = 3×5 ⇒ LCM = 2² × 3² × 5 = 180 minutes = 3 hours. [1] They will next ring together at 11:00 a.m. [1]

SECTION – E (Case-Based · 4 Marks each)

81. Ans.

(i) It is HCF(144, 192). 144 = 2⁴ × 3², 192 = 2⁶ × 3 ⇒ HCF = 2⁴ × 3 = 48. [1]

(ii) Boys: 144 ÷ 48 = 3 groups; girls: 192 ÷ 48 = 4 groups. Total = 7 groups. [1]

(iii) HCF(144, 192, 96) = 48 (as 96 = 48 × 2). [1] Groups = 3 + 4 + 2 = 9 groups of 48. [1] [2]

82. Ans.

(i) LCM(18, 12) = 2² × 3² = 36 minutes. [1]

(ii) Ravi: 36 ÷ 18 = 2 rounds; Raj: 36 ÷ 12 = 3 rounds. [1]

(iii) LCM(18, 12, 8) = 2³ × 3² = 72 minutes, i.e. at 7:12 a.m. [1] Shan's rounds = 72 ÷ 8 = 9. [1] [2]

83. Ans.

(i) Length = 8 m 25 cm = 825 cm; breadth = 6 m 75 cm = 675 cm. [1]

(ii) 825 = 675 × 1 + 150; 675 = 150 × 4 + 75; 150 = 75 × 2 + 0. HCF = 75 cm. [1]

(iii) Tiles along the length = 825 ÷ 75 = 11; along the breadth = 675 ÷ 75 = 9. Number of tiles = 11 × 9 = 99. [1] Cost = 99 × 12 = ₹1188. [1] [2]

84. Ans.

(i) 3125 = 5⁵ and 8 = 2³ are of the form 2^m5ⁿ. So 13/3125 and 17/8 terminate. [1]

(ii) 455 = 5 × 7 × 13 has prime factors 7 and 13 other than 2 and 5 (and 64 does not cancel them). [1]

(iii) 13/3125 = 13 × 2⁵/10⁵ = 416/100000 = 0.00416. [1] 17/8 = 17 × 5³/10³ = 2125/1000 = 2.125. [1] [2]

SECTION – F (Long Answer · 5 Marks each)

85. Ans.

Let √2 be rational: √2 = p/q, p and q co-prime integers, q ≠ 0. [½] Then p² = 2q², so 2 divides p² and hence p. [1] Put p = 2m: 4m² = 2q² ⇒ q² = 2m², so 2 divides q. [1] Then 2 is a common factor of p and q – contradiction. [½] Hence √2 is irrational.
Now suppose 3 + 5√2 = a/b is rational. Then √2 = (a – 3b)/(5b), a rational number – contradicting the irrationality of √2. [1½] Hence 3 + 5√2 is irrational. [½]

86. Ans.

Let a be any positive integer. By Euclid's lemma a = 3q + r, r = 0, 1, 2. [1]
r = 0: a³ = 27q³ = 9(3q³) = 9m. [1]
r = 1: a³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1 = 9m + 1. [1½]
r = 2: a³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8 = 9m + 8. [1½]
Hence every cube is of the form 9m, 9m + 1 or 9m + 8.

87. Ans.

960 = 432 × 2 + 96 [½]
432 = 96 × 4 + 48 [½]
96 = 48 × 2 + 0 [½]
So HCF(960, 432) = 48. [½]
Back-substitution: 48 = 432 – 96 × 4 [½] and 96 = 960 – 432 × 2. [½] So 48 = 432 – 4(960 – 432 × 2) = 432 – 4 × 960 + 8 × 432 = 9 × 432 – 4 × 960. [1] Hence 48 = 960 × (–4) + 432 × 9, i.e. x = –4, y = 9. [1]

88. Ans.

The number must be a multiple of LCM(24, 15, 36). [1] 24 = 2³×3, 15 = 3×5, 36 = 2²×3² ⇒ LCM = 2³ × 3² × 5 = 360. [1] The greatest 6-digit number is 999999. Dividing: 999999 = 360 × 2777 + 279. [1½] Subtract the remainder: 999999 – 279 = 999720 (= 360 × 2777). [1] Check: 999720 ÷ 360 = 2777. [½]

89. Ans.

HCF × LCM = product of the two numbers. [1] Other number = (23 × 1449) ÷ 161 = 33327 ÷ 161 = 207. [1½]
Verification: 161 = 7 × 23 and 207 = 3² × 23. [½] HCF = 23 ✓. LCM = 3² × 7 × 23 = 9 × 7 × 23 = 1449 ✓. [1] Product: 161 × 207 = 33327 = 23 × 1449 ✓. [1]

QUICK REVISION · KEY POINTS (CBSE / COMPETITIVE)

1. Euclid's division lemma: a = bq + r, 0 ≤ r < b. Euclid's algorithm gives the HCF by repeated division until the remainder is 0.
2. Fundamental Theorem of Arithmetic: every composite number is a product of primes, uniquely (apart from order).
3. HCF = product of the smallest powers of common primes; LCM = product of the greatest powers of all primes. HCF × LCM = a × b (for two numbers only).
4. Number of factors of p^aq^br^c = (a+1)(b+1)(c+1).
5. Irrationality proofs use contradiction: √2, √3, √5 are irrational; rational ± irrational and non-zero rational × irrational are irrational. The sum or product of two irrationals may be rational.
6. p/q (lowest terms) terminates iff q = 2^m5ⁿ; the number of decimal places = max(m, n). Otherwise the expansion is non-terminating repeating.
7. Word-problem pattern: largest group/tile/number that divides – HCF; meeting again/smallest number divisible – LCM; remainders r1, r2 – use (N – r).
8. Forms by remainders: odd = 2q+1; square = 3m or 3m+1; cube = 9m, 9m+1 or 9m+8; the product of 3 consecutive integers is divisible by 6.

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