Real Numbers — Class 10 Maths MCQ with Answers (2026-27)

Updated 23 August 2026 · checked against the CBSE 2026-27 syllabus

17 objective questions on Real Numbers, every one checked against the CBSE 2026-27 syllabus. Pick an option and the correct answer is shown straight away, with your marks running at the top. A new question opens each day until all 17 are unlocked.

Real Numbers MCQ questions with answers

1. To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library. There are two sections — section A and section B of grade X. There are 32 students in section A and 36 students in section B. What is the minimum number of books‌ you‌ will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?
   1. 144
   2. 128
   3. 288
   4. 272

   Answer: c) 288

   Books must divide equally among 32 and among 36, so the count is a common multiple. The smallest is LCM(32, 36) = 288.

2. If the product of two positive integers is equal to the product of​ their​ HCF and LCM is true then, the HCF (32, 36) is
   1. 2
   2. 4
   3. 6
   4. 8

   Answer: b) 4

   32 = 2⁵ and 36 = 2²×3². The only common prime power is 2², so HCF = 4.

3. 36 can be expressed as a product‌ of​ its primes as
   1. 2²×3²
   2. 2¹×3³
   3. 2³×3¹
   4. 2⁰×3⁰

   Answer: a) 2²×3²

   Divide repeatedly by primes: 36 = 2×18 = 2×2×9 = 2×2×3×3, that is 2²×3².

4. 7 × 11 × 13 × 15 +​ 15‌ is a
   1. Prime number
   2. Composite number
   3. Neither prime nor composite
   4. None of the above

   Answer: b) Composite number

   Take 15 common: 15(7×11×13 + 1) = 15 × 1002. It has factors other than 1 and itself, so it is composite.

5. If p and q are positive integers such that p‌ =‌ ab² and q = a²b, where a, b are prime numbers, then the LCM (p, q) is
   1. ab
   2. a²b²
   3. a³b²
   4. a³b³

   Answer: b) a²b²

   For the LCM take the highest power of each prime: a appears as a² in q, b appears as b² in p. So LCM = a²b².

6. A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively. In each room the same number of participants are to be seated and all of them being in the same subject, hence maximum number of participants​ that​ can be accommodated in each room are
   1. 14
   2. 12
   3. 16
   4. 18

   Answer: b) 12

   Each room holds one subject only and all rooms are equally filled, so the room size must divide 60, 84 and 108. The largest such number is HCF = 12.

7. What is the minimum‌ number​ of rooms required during the event?
   1. 11
   2. 31
   3. 41
   4. 21

   Answer: d) 21

   With 12 per room: 60/12 = 5, 84/12 = 7 and 108/12 = 9 rooms, giving 5 + 7 + 9 = 21 rooms.

8. The LCM of 60, 84​ and‌ 108 is
   1. 3780
   2. 3680
   3. 4780
   4. 4680

   Answer: a) 3780

   60 = 2²×3×5, 84 = 2²×3×7, 108 = 2²×3³. Highest powers: 2²×3³×5×7 = 3780.

9. The product of HCF and LCM‌ of‌ 60, 84 and 108 is
   1. 55360
   2. 35360
   3. 45500
   4. 45360

   Answer: d) 45360

   HCF = 12 and LCM = 3780, so the product is 12 × 3780 = 45360. Note the rule HCF × LCM = product of the numbers holds only for two numbers, not three.

10. 108 can be expressed‌ as​ a product of its primes as
    1. 2³×3²
    2. 2³×3³
    3. 2²×3²
    4. 2²×3³

    Answer: d) 2²×3³

    108 = 2×54 = 2×2×27 = 2×2×3×3×3, that is 2²×3³.

11. Refer to the figure in your textbook or question paper.

    *(Factor tree — figure required: x branches to 5​ and‌ 2783; 2783 branches to y and 253; 253 branches to 11 and z.)* What will be the value of x?
    1. 15005
    2. 13915
    3. 56920
    4. 17429

    Answer: b) 13915

    x sits at the top of the tree, so it is the product of its two branches: x = 5 × 2783 = 13915.

12. What will be the‌ value‌ of y?
    1. 23
    2. 22
    3. 11
    4. 19

    Answer: c) 11

    2783 splits into y and 253. Since 2783 ÷ 253 = 11, y = 11.

13. What will be the value​ of​ z?
    1. 22
    2. 23
    3. 17
    4. 19

    Answer: b) 23

    253 splits into 11 and z, so z = 253 ÷ 11 = 23.

14. According to the Fundamental Theorem of‌ Arithmetic,​ 13915 is a
    1. Composite number
    2. Prime number
    3. Neither prime nor composite
    4. Even number

    Answer: a) Composite number

    13915 = 5 × 11² × 23 has prime factors other than itself, so by the Fundamental Theorem of Arithmetic it is composite. It is also odd, ruling out option (d).

15. The prime factorisation​ of‌ 13915 is
    1. 5×11³×13²
    2. 5×11³×23²
    3. 5×11²×23
    4. 5×11²×13²

    Answer: c) 5×11²×23

    Reading the completed tree: 13915 = 5 × 11 × 11 × 23 = 5 × 11² × 23.

16. Three cubical warehouses of volume 165 m³, 195 m³ and 285 m³ are to be used for storage. What is the volume of the greatest cubical box that can be kept in the warehouse so that no‌ space​ is left vacant?
    1. 6 m³
    2. 15 m³
    3. 5 m³
    4. 3 m³

    Answer: b) 15 m³

    The box must fill each warehouse exactly, so its volume divides 165, 195 and 285. The largest such volume is HCF = 15 m³.

17. There are five bells placed at different swings in a park, which toll at intervals of 2, 3, 5, 6 and 10 minutes respectively.​ They‌ all toll together when the park opens at 10:00 AM. How many more times do they all toll together till the park closes at 8:00 PM?
    1. 10
    2. 20
    3. 30
    4. 60

    Answer: b) 20

    LCM(2, 3, 5, 6, 10) = 30 minutes. From 10 AM to 8 PM is 600 minutes, so they toll together 600 ÷ 30 = 20 more times.