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CBSE 10 Class Maths Important MCQs for Chapter 1 Real Numbers with Solutions 2023-24

CBSE Class 10 Maths important MCQs for Chapter 1- Real Numbers are provided here with answers and detailed solutions 2023-24.

     (A) one decimal place

     (B) two decimal places

     (C) three decimal places

     (D) more than 3 decimal places

      Answer: B   

  1.     If two positive integers a and b are written as a = p3q2 and b = pq3; p, q are prime

numbers, then HCF (a, b) is:

(A) pq

(B) pq2

(C) p3q3

(D) p2q2

Answer:  B

  1. The product of a non-zero number and an irrational number is:

(A) always irrational

(B) always rational

(C) rational or irrational

(D) one

Answer:  A

    Explanation: Product of a non-zero rational and an irrational number is always irrational i.e

  1. If the HCF of 65 and 117 is expressible in the form 65 m – 117, then the value of m is

    (A) 4

    (B) 2

    (C) 1

    (D) 3

    Answer: B

    Explanation: By Euclid’s division algorithm,

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

   (A) 13

   (B) 65

   (C) 875

   (D) 1750

  Answer: A

 Explanation: Since 5 and 8 are the remainders of 70 and 125, respectively. Thus after               subtracting these remainders from the numbers, we have the numbers

   65 = (70 − 5), 117 = (125 − 8) which is divisible by the required number.

    Now required number = H.C.F of (65,117)

  1. If two positive integers p and q can be expressed as p = ab2 and q = a3b; a, b being              prime numbers, then LCM (p, q) is

   (A) ab

   (B) a2b2

  (C) a3b2

  (D) a3b3

  Answer: C

  Explanation:

  p = a × b × b

  q = a × a × a × b

   Since L.C.M is the product of the greatest power of each prime factor involved in the                 numbers

  (D) 2, 3, 4

   Answer: C

   Explanation:

   According to Euclid’s division lemma,

   a = 3q + r, where 0  r < 3

   As the number is divided by 3.So the remainder cannot be greater than divisor 3 also r is an     integer. Therefore, the values of r can be 0, 1 or

   (A) Terminating decimal expansion

   (B) NonTerminating Non repeating decimal expansion

   (C) NonTerminating repeating decimal expansion

   (D) None of these

   Answer: A

   Explanation: After simplification,

 As the denominator has factor 53 × 22 and which is of the type 5m × 2n, So this is a  terminating decimal expansion.

  1. A rational number in its decimal expansion is 327.7081. What would be the prime                factors of q when the number is expressed in the p/q form?

   (A) 2 and 3

   (B) 3 and 5

  (C) 2, 3 and 5

  (D) 2 and 5

   Answer: D

Explanation: This can be explained as,

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

(A) 10

(B) 100

(C) 2060

(D) 2520

Answer: D

Explanation: Factors of 1 to 10 numbers

from 1 to 10 is =

12. n2 – 1 is divisible by 8, if n is

  (A) an integer                               

  (B) a natural number

  (C) an odd integer greater than 1

  (D) an even integer

 Answer: C

  Explanation: n can be even or odd

  Case 1: If n is even


Case 2: If n is odd


Which is divisible by 8.

Similarly we can check for any integer.

13. If n is a rational number, then 52n− 22n is divisible by

(A) 3

(B) 7

(C) Both 3 and 7

(D) None of these

Answer: C

Explanation:

52n −22n is of the form a2n − b2n which is divisible by both (a + b) and (a – b).

So, 52n − 22n is divisible by both 7, 3.

  1. The H.C.F of 441, 567 and 693 is

(A) 1

(B) 441

(C) 126

(D) 63

Answer: D

Explanation:

693 = 3×3×7×7

567 = 3×3×3×3×7

441 = 3×3×7×11

Therefore H.C.F of 693, 567 and 441 is 63.

  1. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?

(A) 2520cm

(B) 2525cm

(C) 2555cm

(D) 2528cm

Answer: A

Explanation: We need to find the L.C.M of 40, 42 and 45 cm to get the required minimum distance.

40 = 2×2×2×5

42 = 2×3×7

45 = 3×3×5

L.C.M. = 2×3×5×2×2×3×7 = 2520

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