Chapter 1 Real Numbers MCQs | Class 10th Maths | - GMS - Learning Simply
Students' favourite free learning app with LIVE online classes, instant doubt resolution, unlimited practice for classes 6-12, personalized study app for Maths, Science, Social Studies, video e-learning, online tutorial, and more. Join Telegram

# Chapter 1 Real Numbers MCQs | Class 10th Maths |

Chapter 1 Real Numbers MCQs | Class 10th Maths |
Please wait 0 seconds...
Scroll Down and click on Go to Link for destination
Congrats! Link is Generated

# 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. These multiple-choice questions are prepared, as per the latest CBSE syllabus and NCERT guidelines. Solving these objective questions will help students to build their problem-solving skills and score good marks in board exams.

## Class 10 Maths MCQs for Real Numbers

Students are suggested to solve the given multiple-choice questions and check your answers here. 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 Goyanka Maths Study.

#### Below are the MCQs for chapter 1-Real Numbers.

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

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

(3+√7) (3-√7) = 32-(√7)2 = 9 – 7 = 2 [By a2-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

6. The multiplication of two irrational numbers is:

(a)irrational number

(b)rational number

(c)Maybe rational or irrational

(d)None

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

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

(a)Whole numbers

(b)Rational numbers

(c)Natural numbers

(d)Even numbers

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

1. What is the least number that must be added to 1056 so the number is divisible by 23?
1. 0
2. 3
3. 2
4. 1

Solution: We have,

On dividing 1056 by 23, we got 21 as remainder.

⇒ If we add 23 – 21 = 2 to the dividend 1056, we will get a number completely divisible by 23.

∴ Required number = (23 – 21) = 2

1. The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number?
1. 360
2. 295
3. 270
4. 240

Solution: Let the smaller number be x.

⇒ Larger number = (x+1365)

⇒x+1365=6x+15

⇒5x=1350

⇒x=270

∴ Larger number is (270 + 1365) = 1635

and smaller number is 270.

1. Euclid’s division lemma states “Given positive integers a and b, there exist unique integers q and r satisfying a=bq+r. Which of the following is true for r?
1. r>a
2. r<0
3. 0≤r<b
4. r>b

Solution: Euclid’s division lemma:

Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r where 0≤r<b.

Basically, it can be observed that remainder can never be more than the divisor, and is a non-negative integer (could be zero).

1. a and b, when divided by 7 and 6 respectively, leave remainders p and q respectively. What is the maximum value of p + q?
1. 5
2. 6
3. 12
4. 11

Solution: There exist integers m and n such that a = 7m + p and b = 6n + q such that

0<p<7⇒

Maximum value of p will be 6.

0<q<6⇒

Maximum value of p will be 5.

Therefore, the maximum value of p + q will be 11.

1. If HCF (1008, 20) = HCF (20, a) = HCF (a, b) where 1008=20×q+a; 20=a×m+b where (q, a) and (m, b) are positive integers satisfying Euclid’s Division Lemma. What could be the values of a and b?
1. 24, 8
2. 20, 8
3. 10, 4
4. 8, 4

Answer: (D) 8, 4

Solution: If p=d×q+r, (p>q) where p, q, d, r are integers and for a given (p, d), there exist a unique (q, r), then HCF (p, d) = HCF (d, r). Because this relation holds true, the Euclid’s Division Algorithm exists in a step by step manner. So, to find the HCF (1008, 20), we use Euclid’s division lemma at every step.

Step 1: 1008=20×50+8 ⇒ HCF(1008, 20) = HCF(20,8) ⇒ a could be 8

Step 2: 20=8×2+4 ⇒ HCF(20, 8) = HCF(8,4) ⇒ b could be 4

Step 3: 8=4×2+0

HCF = 4

Since 1008=20×q+a where q and a are positive integers satisfy Euclid’s Division Lemma, we must have 0≤a<20. So a is surely 8 and b is 4.

## About the Author

At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s visio…