Class 10 Maths Chapter 1 Real Numbers Practice Paper 6

Class: X Subject: Mathematics Session: 2025-26 Chapter: 01 - Real Numbers Time: 1½ Hours Max. Marks: 40

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five sections A, B, C, D and E.

3. Section A contains 10 MCQs of 1 mark each.

4. Section B contains 4 questions of 2 marks each.

5. Section C contains 3 questions of 3 marks each.

6. Section D contains 1 question of 5 marks.

7. Section E contains 2 Case Study Based questions of 4 marks each.

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five Sections A, B, C, D and E.

3. Section A comprises of 10 MCQs of 1 mark each. Section B comprises of 4 questions of 2 marks each.

Section C comprises of 3 questions of 3 marks each. Section D comprises of 1 question of 5 marks and

Section E comprises of 2 Case Study Based Questions of 4 marks each.

4. There is no overall choice.

5. Use of Calculators is not permitted.

SECTION A - Multiple Choice Questions (1 mark each)

Questions 1 to 10 carry 1 mark each.

1. The exponent of 2 in the prime factorization of 6048 is
(a) 4
(b) 5
(c) 6
(d) 7

2. If two positive integers x and y are written as x = a⁴b² and y = a²b⁵, where a and b are prime numbers, then the LCM (x, y) is:
(a) a²b²
(b) a³b³
(c) a⁴b⁵
(d) a⁶b⁷

3. The HCF and the LCM of 14, 28, 35 respectively are
(a) 7, 140
(b) 14, 280
(c) 7, 280
(d) 280, 7

4. If the HCF of 104 and 169 is expressible in the form 104m - 169, then the value of m is
(a) 1
(b) 2
(c) 3
(d) 4

5. Priya has 63 cm long red and 108 cm long blue ribbon. She cuts each ribbon into pieces such that all pieces are of equal length. What is the length of each piece?
(a) 9 cm as it is the HCF of 63 and 108
(b) 9 cm as it is the LCM of 63 and 108
(c) 27 cm as it is the LCM of 63 and 108
(d) 27 cm as it is the HCF of 63 and 108

6. The largest number which divides 112 and 168 leaving remainders 8 and 12 respectively is
(a) 26
(b) 52
(c) 104
(d) 156 m n k p

7. If 8190 = 2 × 3 × 5 × 7 , then the value of m + n + k + p is
(a) 5
(b) 6
(c) 7
(d) 8 n

8. If p = 2⁴ × 3, q = 2² × 5² × 7, r = 3 × 7² and LCF (p, q, r) = 2⁴ × 3² × 5² × 7², then n is equal to
(a) 1
(b) 2
(c) 3
(d) 4

9. In the following questions, a statement of assertion
(a) is followed by a statement of Reason (R). Choose the correct answer out of the following choices. Assertion
(a) : If product of two numbers is 8736 and their HCF is 16, then their LCM is 546. Reason (R): For any two positive integers, HCF × LCM = Product of the numbers.
(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.

10. In the following questions, a statement of assertion
(a) is followed by a statement of Reason (R). Choose the correct answer out of the following choices. n Assertion
(a) : 18 ends with the digit zero, where n is natural number. Reason (R): A number ends with digit zero if and only if it is divisible by 10.
(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.

SECTION B - Short Answer Questions (2 marks each)

Questions 11 to 14 carry 2 marks each.

11. Explain why 7 × 11 × 13 + 13 and 13 × 17 × 19 + 17 × 13 are composite numbers.

12. Two numbers are in the ratio 5 : 7 and their LCM is 420. What is the HCF of these numbers? n

13. Show that any number of the form 21 , where n ∈ N can never end with digit 0. (2017)

14. The LCM of two numbers is 6 times their HCF. The sum of LCM and HCF is 700. Find the HCF of the two numbers.

SECTION C - Short Answer Questions (3 marks each)

Questions 15 to 17 carry 3 marks each.

15. Prove that √11 is an irrational number. (2023)

16. 4 Bells toll together at 6.00 am. They toll after 9, 12, 16 and 20 seconds respectively. How many times will they toll together again in the next 4 hours?

17. Given that √11 is irrational, prove that 6 + 5√11 is irrational. (CBSE Sample Paper 2022)

SECTION D - Long Answer Question (5 marks)

Questions 18 carry 5 marks.

18.
(a) Find the largest possible positive integer that divides 169, 234 and 325 leaving remainder 7, 9 and 13 respectively. (3)
(b) An army contingent of 945 soldiers is to march behind an army band of 54 members in a Republic Day parade. The two groups are to march in the same number of columns. What is the maximum number of columns they can march? (2)

SECTION E - Case Study Based Questions (4 marks each)

Questions 19 to 20 carry 4 marks each.

19. A morning walk may help improve your mental clarity and ability to focus throughout the day. A recent study found that amongst older adults, those who started their days with a morning walk improved their cognitive function, compared to those who remained sedentary. Walking may also help you think more creatively. In a morning walk three students step off together, their steps measure 88 cm, 99 cm and 121 cm respectively. (i) What is the HCF of 88 and 121? (1) (ii) Find the sum of exponents of the prime factors of total distance. (1) (iii) What is the minimum distance each should walk so that he can cover the distance in complete steps? (2)

20. A family room is an informal, all purpose room in a house. The family room is designed to be a place where family and guests gather for group recreation like talking, reading, watching TV and other family activities. The length, breadth and height of a room are 12 m 60 cm, 9 m 80 cm and 7 m 70 cm. (i) Determine the longest rod which can measure the three dimensions of the room exactly. (2) (ii) What is LCM of the given three measurements? (1) (iii) If the HCF (1260 and 980) = 140, then find LCM (1260 and 980). (1) DETAILED ANSWER KEY - PAPER 03

SECTION A - Answers to MCQs

1. Answer:
(a) 4

Solution: 6048 = 16 × 378 = 2⁴ × 2 × 189 = 2⁵ × 3³ × 7

Therefore, the exponent of 2 is 5. Answer should be
(b) 5.

2. Answer:
(c) a⁴b⁵

Solution: x = a⁴b², y = a²b⁵

LCM = Product of highest powers of all prime factors = a⁴b⁵

3. Answer:
(a) 7, 140

Solution: 14 = 2 × 7, 28 = 2² × 7, 35 = 5 × 7

HCF = 7, LCM = 2² × 5 × 7 = 140

4. Answer:
(b) 2

Solution: Using Euclidean algorithm:

169 = 104 × 1 + 65 104 = 65 × 1 + 39 65 = 39 × 1 + 26 39 = 26 × 1 + 13 26 = 13 × 2 + 0 HCF = 13 = 104 × 2 - 169 × 1, so m = 2

5. Answer:
(d) 27 cm as it is the HCF of 63 and 108

Solution: To cut ribbons into equal pieces, we need HCF(63, 108)

63 = 3² × 7, 108 = 2² × 3³ HCF = 3² = 9 cm. Answer should be
(a) 9 cm.

6. Answer:
(b) 52

Solution: Required number = HCF(112-8, 168-12) = HCF(104, 156) = 52

7. Answer:
(b) 6

Solution: 8190 = 2 × 3² × 5 × 7 × 13

So m = 1, n = 2, k = 1, p = 1 But this doesn't match the given form. Let me recalculate: 8190 = 2 × 4095 = 2 × 3² × 455 = 2 × 3² × 5 × 91 = 2 × 3² × 5 × 7 × 13 Since 13 is not in the given form, there's an error in the question.

8. Answer:
(b) 2

Solution: p = 2⁴ × 3¹, q = 2² × 5² × 7¹, r = 3ⁿ × 7²

LCM = 2⁴ × 3² × 5² × 7² For this to be true, n must be 2.

9. Answer:
(a) Both A and R are true and R is the correct explanation of A

Solution: Product = HCF × LCM, so LCM = 8736/16 = 546. Both statements are true.

R correctly explains the formula used in A.

10. Answer:
(d) A is false but R is true

Solution: 18ⁿ = (2 × 3²)ⁿ = 2ⁿ × 3²ⁿ. This never contains factor 5, so cannot end in 0.

Statement R is correct about numbers ending in 0.

SECTION B - Answers to Short Answer Questions

11. Solution: First number: 7 × 11 × 13 + 13 = 1001 + 13 = 1014 = 2 × 3 × 13² (composite) Second number: 13 × 17 × 19 + 17 × 13 = 4199 + 221 = 4420 = 2² × 5 × 13 × 17 (composite) Both numbers have factors other than 1 and themselves.

12. Solution: Let the numbers be 5x and 7x where x is their HCF. LCM = (5x × 7x)/x = 35x Given: 35x = 420, so x = 12 Therefore, HCF = 12

13. Solution: 21ⁿ = (3 × 7)ⁿ = 3ⁿ × 7ⁿ For a number to end with 0, it must be divisible by 10 = 2 × 5 Since 21ⁿ contains only factors 3 and 7 (no factors of 2 or 5), it can never end with 0.

14. Solution: Let HCF = h, then LCM = 6h Given: h + 6h = 700 7h = 700 h = 100 Therefore, HCF = 100

SECTION C - Answers to Short Answer Questions

15. Solution: Proof by contradiction: Assume √11 is rational, so √11 = p/q where p, q are integers with no common factors. Squaring: 11 = p²/q², so 11q² = p² This means p² is divisible by 11, so p is divisible by 11. Let p = 11k, then 11q² = 121k², so q² = 11k² This means q is also divisible by 11. But this contradicts our assumption that p and q have no common factors. Therefore, √11 is irrational.

16. Solution: The bells will toll together at intervals equal to LCM(9, 12, 16, 20) LCM = 2⁴ × 3² × 5 = 720 seconds = 12 minutes In 4 hours = 240 minutes Number of times = 240/12 = 20 So they will toll together 20 times after 6:00 AM.

17. Solution: Proof by contradiction: Assume 6 + 5√11 is rational = r Then 5√11 = r - 6 (rational) So √11 = (r - 6)/5 (rational) But this contradicts the given fact that √11 is irrational. Therefore, 6 + 5√11 is irrational.

SECTION D - Answer to Long Answer Question

18. Solution:
(a) Required number = HCF(169-7, 234-9, 325-13) = HCF(162, 225, 312) 162 = 2 × 3⁴, 225 = 3² × 5², 312 = 2³ × 3 × 13 HCF = 3
(b) Maximum columns = HCF(945, 54) 945 = 3³ × 5 × 7, 54 = 2 × 3³ HCF = 3³ = 27 columns

SECTION E - Answers to Case Study Based Questions

19. Solution: (i) HCF(88, 121): 88 = 2³ × 11, 121 = 11² HCF = 11 (ii) Total distance = LCM(88, 99, 121) LCM = 2³ × 3² × 11² = 8712 Sum of exponents = 3 + 2 + 2 = 7 (iii) Minimum distance = LCM(88, 99, 121) = 8712 cm = 87.12 m

20. Solution: (i) Convert to cm: 1260 cm, 980 cm, 770 cm HCF(1260, 980, 770) = 70 cm (ii) LCM(1260, 980, 770) = 27720 cm (iii) Using HCF × LCM = Product of numbers LCM = (1260 × 980)/140 = 8820