# Class 10 Maths Chapter 1 Real Numbers MCQs

Class 10 Maths MCQs for Chapter 1 (Real numbers) are available here online for students, along with their answers. MCQ Questions with answers for Class 10 Maths are prepared as per the latest exam pattern. These multiple-choice questions are prepared, as per the latest CBSE syllabus (2022 – 2023) and NCERT guidelines. Solving these objective questions will help students to build their problem-solving skills and score good marks in the board exams. To get MCQs for all chapters, click hereStudents can also download the PDF provided to get more MCQs for practice.

## Class 10 Maths MCQs for Real Numbers

The date sheets for the Class 10th board exam has been released. It is time for students to start revising the Maths chapters. Students are suggested to solve the given multiple-choice questions and check the answers here. The chapter Real Numbers will teach you about different types of numbers and their applications. Try to solve these questions as per your knowledge and skills and then verify the answers. This practice will help you to boost your confidence. Also, get important questions for class 10 Maths here at BYJU’S.

Students can also get access to Real Numbers for Class 10 Notes here.

Below are the MCQs for Chapter 1-Real Numbers:

The students of class 10 can consider this an online test for the real number chapter 1 MCQs. Once the question is solved, they can cross verify their answer with the provided solution.

1. The decimal expansion of 22/7  is

(a) Terminating

(b) Non-terminating and repeating

(c) Non-terminating and Non-repeating

(d) None of the above

Explanation: 22/7 = 3.14285714286..

2. For some integer n, the odd integer is represented in the form of:

(a) n

(b) n + 1

(c) 2n + 1

(d) 2n

Explanation: Since 2n represents the even numbers, hence 2n + 1 will always represent the odd numbers. Suppose if n = 2, then 2n = 4 and 2n + 1 = 5.

3. HCF of 26 and 91 is:

(a) 15

(b) 13

(c) 19

(d) 11

Explanation: The prime factorisation of 26 and 91 is;

26 = 2 x 13

91 = 7 x 13

Hence, HCF (26, 91) = 13

4. Which of the following is not irrational?

(a) (3 + √7)

(b) (3 – √7)

(c) (3 + √7) (3 – √7)

(d) 3√7

Answer: (c) (3 + √7) (3 – √7)

Explanation: If we solve, (3 + √7) (3 – √7), we get;

(3 + √7) (3 – √7) = 3– (√7)2 = 9 – 7 = 2 [By a– b2 = (a – b) (a + b)]

5. The addition of a rational number and an irrational number is equal to:

(a) rational number

(b) Irrational number

(c) Both

(d) None of the above

The addition of a rational number and an irrational number is equal to irrational number.

6. The multiplication of two irrational numbers is:

(a) irrational number

(b) rational number

(c) Maybe rational or irrational

(d) None

Answer: (c) Maybe rational or irrational

The multiplication of two irrational numbers is maybe rational or irrational.

7. If set A = {1, 2, 3, 4, 5,…} is given, then it represents:

(a) Whole numbers

(b) Rational Numbers

(c) Natural numbers

(d) Complex numbers

If set A = {1, 2, 3, 4, 5,…} is given, then it represents natural numbers.

8. If p and q are integers and is represented in the form of p/q, then it is a:

(a) Whole number

(b) Rational number

(c) Natural number

(d) Even number

If p and q are integers and is represented in the form of p/q, then it is a rational number.

9. The largest number that divides 70 and 125, which leaves the remainders 5 and 8, is:

(a) 65

(b) 15

(c) 13

(d) 25

Explanation: 70 – 5 = 65 and 125 – 8 = 117

HCF (65, 117) is the largest number that divides 70 and 125 and leaves remainder 5 and 8.

HCF (65, 117) = 13

10. The least number that is divisible by all the numbers from 1 to 5 is:

(a) 70

(b) 60

(c) 80

(d) 90

Explanation: The least number will be LCM of 1, 2, 3, 4, 5.

Hence, LCM (1, 2, 3, 4, 5) = 2 x 2 x 3 x 5 = 60

11. The sum or difference of two irrational numbers is always

(a) rational

(b) irrational

(c) rational or irrational

(d) not determined

12. The decimal expansion of the rational number 23/(22 . 5) will terminate after

(a) one decimal place

(b) two decimal places

(c) three decimal places

(d) more than 3 decimal places

Explanation:

23/(22 . 5) = (23 × 5)/(22 . 52) = 115/(10)2 = 115/100 = 1.15

Hence, 23/(22 . 5) will terminate after two decimal places.

13. Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy

(a) 1 < r < b

(b) 0 < r ≤ b

(c) 0 ≤ r < b

(d) 0 < r < b

Correct option: (c) 0 ≤ r < b

Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy 0 ≤ r < b.

14. For some integer m, every even integer is of the form

(a) m

(b) m + 1

(c) 2m

(d) 2m + 1

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

15. Using Euclid’s division algorithm, the HCF of 231 and 396 is

(a) 32

(b) 21

(c) 13

(d) 33

Explanation:

396 > 231

Using Euclid’s division algorithm,

396 = 231 × 1 + 165

231 = 165 × 1 + 66

165 = 66 × 2 + 33

66 = 33 × 2 + 0

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

(a) 4

(b) 2

(c) 1

(d) 3

Explanation:

117 > 65

117 = 65 × 1 + 52

65 = 52 × 1 + 13

52 = 13 × 4 + 0

Therefore, HCF(65, 117) = 13

According to the given,

65m – 117 = 13

65m = 117 + 13

65m = 130

m = 130/65 = 2

17. The prime factorisation of 96 is

(a) 25 × 3

(b) 26

(c) 24 × 3

(d) 24 × 32

Explanation:

The prime factorisation of 96 is:

96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3

18. n² – 1 is divisible by 8, if n is

(a) an integer

(b) a natural number

(c) an odd integer

(d) an even integer

Explanation:

We know that an odd number in the form (2Q + 1) where Q is a natural number ,

so, n² -1 = (2Q + 1)² -1

= 4Q² + 4Q + 1 -1

= 4Q² + 4Q

Substituting Q = 1, 2,…

When Q = 1,

4Q² + 4Q = 4(1)² + 4(1) = 4 + 4 = 8 , it is divisible by 8.

When Q = 2,

4Q² + 4Q = 4(2)² + 4(2) =16 + 8 = 24, it is also divisible by 8 .

When Q = 3,

4Q² + 4Q = 4(3)² + 4(3) = 36 + 12 = 48 , divisible by 8

It is concluded that 4Q² + 4Q is divisible by 8 for all natural numbers.

Hence, n² -1 is divisible by 8 for all odd values of n.

19. For any two positive integers a and b, HCF (a, b) × LCM (a, b) =

(a) 1

(b) (a × b)/2

(c) a/b

(d) a × b

For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b.

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

(a) 0, 1, 2

(b) Only 1

(c) Only 0 or 1

(d) 1, 2