NCERT Solutions Class 10 Maths Chapter 1 Real Numbers: Exercise 1.1

NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.1 is important for understanding both Real Numbers and concepts related to it, which form a solid base for grasping complex concepts in higher classes. NCERT Solutions for Class 10 Maths chapter 1 exercise solutions, explained here, contains a complete breakdown of these essential concepts. This helps students to get a better understanding of this topic, prepared according to the latest CBSE syllabus and protocols of the examination. So, let’s dive deeper into this foundational topic of class 10. 

1.0Download NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.1 - Free PDF

2.0Introduction to Real Numbers: Rational and Irrational Numbers

Real numbers are any value that represents a complete set of rational and irrational numbers along a straight, continuous number line. Real numbers can be any quantity, which can be positive, negative, or 0, a whole number, fractions, or decimals. 

Rational Numbers

Rational numbers are the ratio of any two integers, which can be expressed in the form of $\frac{p}{q}$, where $q\ne 0$, p and q are coprime numbers, meaning both numbers don’t have any common factor other than 1. For example, $\frac{1}{2}$ is a rational number while $\frac{2}{4}$ is not. Moreover, rational numbers can either have a terminating or repeating decimal expansion. For example, 1, –1, 0.5, and $\frac{3}{4}$. 

Irrational Numbers 

These cannot be written as a fraction of two integers. They have non-terminating (numbers that don’t terminate) and non-repeating (numbers that don’t form a repeating form) decimal expansions. Famous irrational numbers are $\pi$, $\sqrt{2}$, and e.

3.0Class 10 Maths Chapter 1 - Real Numbers Exercise 1.1 Overview

Depicting Real Numbers on the Number Line

This is how you represent a given number on a number line: 

• A number line is a straight horizontal line with a constant origin marked as 0, marked with a constant interval. 
• Positive numbers are plotted to the right of 0, while the negative ones are plotted on the left side of the origin.  

Basic Operations on Real Numbers

Real numbers follow the standard rules and guidelines of simple arithmetic for basic operations, which include:

• Addition: The sum of two real numbers is another real number, and the addition of positive numbers is a greater number. 

= 3.5 + (-2.1) 

= 3.5 – 2.1 

= 1.4 

• Subtraction: Subtraction of one real number from another might yield positive as well as negative results based on the values.

5.8 - 2.3 = 3.5

• Multiplication: The result of the product of two real numbers is either positive or negative, depending on the signs of the numbers.

$(-4)\times 3.5=-14$

• Division: The division of real numbers is also like multiplication with the added stipulation that the divisor should not be zero.

$\frac{7.2}{-3}=-2.4$

Note that real numbers also follow the same rules as standard arithmetic for sign convention. 

4.0NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.1

Q1. Use Euclid's division algorithm to find the HCF of:

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 225

Solution:

(i) 135 and 225

Start with the larger integer, that is, 225. Apply the division lemma to 225 and 135, to get:

225 = 135 × 1 + 90

Since the remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to get:

135 = 90 × 1 + 45

We consider the new divisor 90 and the new remainder 45, and apply the division lemma to get:

90 = 45 × 2 + 0

The remainder has now become zero, so our procedure stops.

Since the divisor at this stage is 45, the HCF of 225 and 135 is 45.

(ii) 196 and 38220

Start with the larger integer, that is, 38220. Apply the division lemma to 38220 and 196, to get:

38220 = 196 × 195 + 0

The remainder at this stage is zero, so our procedure stops.

So, HCF of 196 and 38220 is 196.

(iii) 867 and 225

Start with the larger integer, that is, 867.

Apply the division lemma to 867 and 225, to get:

867 = 225 × 3 + 192

Since the remainder 192 ≠ 0, we apply the division lemma to 225 and 192 to get:

225 = 192 × 1 + 33

Since the remainder 33 ≠ 0, we apply the division lemma to 192 and 33 to get:

192 = 33 × 5 + 27

Since the remainder 27 ≠ 0, we apply the division lemma to 33 and 27 to get:

33 = 27 × 1 + 6

Since the remainder 6 ≠ 0, we apply the division lemma to 27 and 6 to get:

27 = 6 × 4 + 3

Since the remainder 3 ≠ 0, we apply the division lemma to 6 and 3 to get:

6 = 3 × 2 + 0

Now, the remainder at this stage is zero, so our procedure stops.

So, HCF of 867 and 225 is 3.

Q2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3 or 6q + 5, where q is some integer.

Solution:

Let us start with taking a, where a is any positive odd integer. We apply the division algorithm, with a and b = 6. Since 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4, 5. That is, a can be 6q or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5, where q is the quotient. However, since a is odd, we do not consider the cases 6q, 6q + 2 and 6q + 4 (since all the three are divisible by 2). Therefore, any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.

Q3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution:

Let a = 616, b = 32. By Euclid's division algorithm:

616 = 32 × 19 + 8

32 = 8 × 4 + 0

HCF of (616, 32) = 8

Therefore, the maximum number of columns is 8.

Q4. Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Solution:

Let a be any positive integer. We apply the division lemma with b = 3. Since 0 ≤ r < 3, the possible remainders are 0, 1 and 2. That is, a can be 3q, or 3q + 1, or 3q + 2, where q is the quotient.

Now, (3q)² = 9q² = 3 × 3q², which can be written in the form 3m, where m = 3q².

Again, (3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1, which can be written in the form 3m + 1, where m = 3q² + 2q.

Lastly, (3q + 2)² = 9q² + 12q + 4 = (9q² + 12q + 3) + 1 = 3(3q² + 4q + 1) + 1, which can be written in the form 3m + 1, where m = 3q² + 4q + 1.

Therefore, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Q5.  Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution:

Let n be any positive integer. We apply the division lemma with b = 3. Since 0 ≤ r < 3, the possible remainders are 0, 1 and 2. That is, n can be 3q, or 3q + 1, or 3q + 2, where q is the quotient.

Case I: Let n = 3q.

The cube of this will be n³ = 27q³ = 9(3q³).

So, n³ = 9m, where m = 3q³.

Case II: n = 3q + 1.

So, n³ = (3q + 1)³ = 27q³ + 1 + 27q² + 9q = 9(3q³ + 3q² + q) + 1.

n³ = 9m + 1, where m = 3q³ + 3q² + q.

Case III: n = 3q + 2.

So, n³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8.

n³ = 9m + 8, where m = 3q³ + 6q² + 4q.

So, it means the cube of positive integers is of the form 9m, 9m + 1 or 9m + 8.

5.0Key Concepts 

Ordering Real Numbers Using Inequalities 

Real numbers can be compared with inequalities, which indicate their relative position on the number line. Symbols used in inequalities are:

• < (less than), > (greater than), ≤ (less than or equal to), and ≥ (greater than or equal to). For example: 5 < 7, 6 > 3. 

HCF (Highest Common Factor) and LCM (Least Common Multiple) 

HCF and LCM are the two most important topics of the NCERT Solutions Class 10 maths chapter 1 exercise 1.1 PDF from the exam point of view, which can be explained as: 

HCF 

The HCF of two or more numbers is the largest number which divides all of them exactly and can be found with the help of Euclid's Division Algorithm. The Algorithm states that for any two positive integers a and b, the HCF of a and b can be determined as follows:

• Divide a by b.
• Replace a with b, and b with the remainder from the previous division.
• Repeat this process until the remainder is 0. The divisor at this step will be the HCF of a and b. 

For Example: Calculate the HCF of 56 and 72. 

Solution: Dividing 72 by 56

$72÷56=56\times 1+16$

Now, Divide 56 by 16

$56÷16=16\times 3+8$

Divide 16 by 8 

$16÷8=8\times 2+0$

HCF: 8

LCM

The LCM is the lowest number, which is a common multiple of two or more numbers. It can be determined by following the below steps.

• Determine the prime factorisation of every number.
• Take the greatest powers of all prime factors occurring in the factorisations.
• Multiply the factors with the highest powers of all the prime factors to obtain the LCM.

For Example: Find LCM of 12 and 18.

Solution: $12=2^2\times 3\&18=2\times 3^2$

Highest power: for 2 = 2², for 3 = 3² 

$LCM=2^2\times 3^2=4\times 9=36$

6.0Revisiting Important Topics in Detail

1. Review of Real Numbers: The whole concept of real numbers is that the numbers are a group of rational and irrational numbers and can be represented on a number line. 

2. Representing Real Numbers on the Number Line: It is important to learn to represent real numbers on the number line because it helps in visualising the magnitude and order of the numbers, making them simpler to compare and do addition and subtraction. It also forms the basis for solving inequalities and learning higher-level math concepts.

3. Ordering Real Numbers: Using inequalities to order real numbers assists in solving problems that require comparing values and determining the relative positions of numbers on the number line. 

7.0Benefits of Class 10 Maths NCERT Solutions Ch 1 - Real Number Exercise 1.1

• These NCERT Solutions for class 10 chapter 1 on Real Numbers, are provided by subject matter experts at ALLEN.
• These solutions are based on NCERT guidelines, which help students prepare for their exams and score more marks.
• Enhanced understanding of key topics covered under real number exercise 1.1.
• These solutions are completely based on the updated syllabus for 2025-26 prescribed by CBSE guidelines.