MCQ Questions of Real Numbers with Answers

Class 10 Maths MCQ Questions of Real numbers with answer are accessible here for students. These Multiple Choice Questions are arranged, according to the most recent syllabus and exam pattern. solving these objective questions will assist students with building their critical thinking abilities and score great marks in board exams. Snap here if you want to practice MCQ Questions for Class 10 Maths with Answers.

Class 10 Maths MCQ Questions of Real Numbers

Students are suggested to address the offered MCQ Questions and check their responses here. Real numbers will show you various types of numbers and their applications. Attempt to solve these questions according to your knowledge and skills and then verify the answers. This practice will assist you with boosting your assurance. Likewise, get significant questions for class 10 Maths here at Sarthaks eConnect.

Some of the Multiple Choice Questions of Real Numbers are given below.

1. The decimal expansion of n is

(a) terminating
(b) non-terminating and non-recurring
(c) non-terminating and recurring
(d) does not exist.

2. For some integer m, every odd integer is of the form

(a) m
(b) m + 1
(c) 2m
(d) 2m + 1

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

(a) always irrational
(b) always rational
(c) rational or irrational
(d) one

4. The exponent of 2 in the prime factorisation of 144, is

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

5. The LCM of two numbers is 1200. Which of the following cannot be their HCF?

(a) 600
(b) 500
(c) 400
(d) 200

6. If two positive integers p and q can be expressed as
\(p = ab^2\) and \(q = a^3b\) a, b being prime numbers, then LCM (p, q) is

(a) \(ab\)
(b) \(a^2b^2\)
(c) \(a^3b^2\)
(d) \(a^3b^3\)

7. In a seminar, the number of participants in English, German and Sanskrit are 45,75 and 135. Find the number of rooms required to house them, if in each room, the same number of participants are to be accommodated and they should be of the same language.

(a) 45
(b) 17
(c) 75
(d) 135

8. If p = HCF (100,190) and q = LCM (100, 190); then \(p^2q^2\) is :

(a) 3.61 x \(10^5\)
(b) 361 x \(10^3\)
(c) 3.61 x \(10^6\)
(d) 3.61 x \(10^8\)

9. The largest positive integer which divides 434 and 539 leaving remainders 9 and 12 respectively is:

(a) 9
(b) 108
(c) 17
(d) 539

10. There is a circular path around a field. Reema takes 22 minutes to complete one round while her friend Saina takes 20 minutes to complete the same. If they both start at the same time and move in the same direction, after how many minutes will they meet again at the starting

(a) 220
(b) 3.4
(c) 440
(d) 4.4

11. If n = \(2^3 × 3^4 × 5^4 × 7\), then the number of consecutive zeros in n, where n is a natural number, is

(a) 2
(b) 3
(c) 4
(d) 7

12. The sum of the exponents of the prime factors in the prime factorisation of 196, is

(a) 1
(b) 2
(c) 4
(d) 6

13. If \(p_1\) and \(p_2\) are two odd prime numbers such that \(p_1\) > \(p_2\), then \(p_1^2 – p_2^2\) is

(a) an even number
(b) an odd number
(c) an odd prime number
(d) a prime number

14. If two positive integers m and n are expressible in the form \(m = pq^3\) and \(n = p^3 q^2\) where p, q are prime numbers, then HCF (m, n) =

(a) \(pq\)
(b) \(pq^2\)
(c) \(p^3q^3\)
(d) \(p^2q^3\)

15. Euclid’s Lemma states that, for given positive integers a and b, there exist unique integers q and r, such that a = bq + r, where :

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

16. If a non-zero rational number is multiplied to an irrational number, we always get:

(a) an irrational number
(b) a rational number
(c) zero
(d) one

17. The values of the remainder r, when a positive integer a is divided by 3 are:

(a) 0, 1, 2, 3
(b) 0, 1
(c) 0, 1, 2
(d) 2, 3, 4

18. 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

19. Write an irrational number between 2 and 3.

(a) 2.5
(b) 2.001
(c) 2.1333333456…
(d) 2.13

20. The greatest number which divides 87 and 97, leaving 7 as remainder is :

(a) 10
(b) 1
(c) 87 x 97
(d) 6300

Answers & Explanations

1. Answer: (b) non-terminating and non-recurring

2. Answer: (d) 2m + 1

Explanation: As the number 2m will always be even, so if we add 1 to it then, the number will always be odd.

3. Answer: (a) always irrational

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

\(\frac{3}{4}\times\sqrt2\) = (rational) \(\times\) (irrational) = irrational.

4. Answer: (a) 4

Explanation: The prime factorization of 144 is as follows:

144 = 2 × 2 × 2 × 2 × 3 × 3

⇒ 144 = 2⁴ × 3²

We know that the exponent of a number a^m is m.

∴ The exponent of 2 in the prime factorization of 144 is 4.

5. Answer: (b) 500

Explanation: We know that LCM of two or more numbers is always divisible by their HCF.

1200 is divisible by 600, 200 and 400 but not by 500.

6. Answer: (c) \(a^3b^2\)

Explanation: LCM is Product of the greatest power of each prime factor involved in the number

So LCM = \(a^3b^2\)

7. Answer: (b) 17

Explanation: Since, in each room, the same number of participants, of the same language, are to be accommodated, their number in each room

HCF of 45, 75 and 135.

HCF (45, 75,135) = 15

Each room accommodates 15 participants
\(\Rightarrow\) Total no. of rooms required for English = \(\frac{45}{15}\) = 3

Total no. of rooms required for German = \(\frac{75}{15}\) = 5

Total no. of rooms required for Sanskrit= \(\frac{135}{15}\) = 9

Total no. of rooms = 3 + 5 + 9 = 17

8. Answer: (d) 3.61 x \(10^8\)

Explanation: pq = (HCF) (LCM) = Product of given numbers.

\(\Rightarrow\) pq = 190×100 =19000
\(\Rightarrow\) \(p^2q^2\) = 361 x \(10^6\) = 3.61 x \(10^8\)

9. Answer: (c) 17

Explanation: Required number is the HCF of (434 – 9) and (539 -12)

= HCF of 425 and 527.
= 17

10. Answer: (a) 220

Explanation: LCM of 20 and 22 = 220 (question state: after how many minutes will they meet)

11. Answer: (b) 3

Explanation: If any number ends with the digit 0, it should be divisible by 10,

i.e. it will be divisible by 2 and 5.

Prime factorization of n is given as 2³ × 3⁴ × 5⁴ × 7.

It can be observed that there is (2 × 5) × (2 × 5) × (2 × 5)

⇒ 10 × 10 × 10 = 1000

Thus, there are 3 zeros in n.

12. Answer: (c) 4

Explanation: The prime factorization of 196 is as follows:

196 = 2 × 2 × 7 × 7

⇒ 98 = 2² ×7²

We know that the exponent of a number a^m is m.

∴The sum of powers = 2 + 2 = 4

13. Answer: (a) an even number

Explanation: Let us take \(p_1\) = 5 and \(p_2\) = 3

Then \(p_1^2 – p_2^2\) = 25 – 9 = 16

16 is an even number

14. Answer: (b) \(pq^2\)

Explanation: We know that HCF = Product of the smallest power of each common prime factor in the numbers.

So, HCF(a, b) = \(pq^2\)

15. Answer: (b) 0 ≤ r < b

16. Answer: (a) an irrational number

Explanation: The product of a rational (non-zero) and m irrational number is always an irrational number.

17. Answer: (c) 0, 1, 2

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 2.

18. Answer: (a) 2520cm

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

19. Answer: (c) 2.1333333456…

Explanation: Non terminating non repeating

20. Answer: (a) 10

Explanation: Greatest number which divides 87 and 97, leaving 7 as remainder = HCF of 80 and 90