NCERT Solutions for Exercise 1.5 Class 9 Maths Chapter 1

NCERT Solutions for Exercise 1.5 Class 9 Maths Chapter 1 - Number Systems

Edited By Vishal kumar | Updated on Sep 30, 2023 10:14 AM IST

NCERT Solutions for Class 9 Maths Chapter 1: Number Systems Exercise 1.5- Download Free PDF

NCERT Solutions Class 9 Maths Chapter 1 Number Systems Exercise 1.5- NCERT Solutions for Class 9 Maths exercise 1.5 deals with the concepts of the type of numbers into rational and irrational numbers and also the use of certain mathematical operations. Mathematically, numbers are labelled into many units, particularly rational numbers, irrational numbers, natural numbers, whole numbers, integers and real numbers. Real numbers, as defined in exercise 1.5 Class 9 Maths are the combination of rational and irrational numbers, which might have positive or negative attributes. A number that can be broken down into parts and addressed as p/q where p and q are integers and q≠0 are rational numbers. Irrational numbers are numbers that can't be stated as a fraction with a number in the numerator and denominator.

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NCERT Solutions for Class 9 Maths Chapter 1: Number Systems Exercise 1.5- Download Free PDF
NCERT Solutions for Class 9 Maths Chapter 1 – Number Systems Exercise 1.5
Access Number Systems Class 9 Chapter 1 Exercise: 1.5
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NCERT Solutions of Class 10 Subject Wise
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NCERT Solutions for Exercise 1.5 Class 9 Maths Chapter 1 - Number Systems

NCERT solutions for Class 9 Maths Chapter 1 exercise 1.5 includes five questions where four are lengthy and the remaining 1 among them is the brief solution which is based on the category of determining the numbers into rational and irrational numbers by using certain mathematical operations and the illustration of given numbers on the number line. The concepts associated with the number lines and mathematical operations are nicely defined in this Class 9 Maths chapter 1 exercise 1.5. The following exercises are also present along with Class 9 Maths Chapter 1 exercise 1.5.

The 9th class maths exercise 1.5 answers are authored by subject experts in straightforward language, ensuring ease of understanding. Additionally, they are accessible in PDF format, allowing students to download and utilize them conveniently.

NCERT Solutions for Class 9 Maths Chapter 1 – Number Systems Exercise 1.5

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Access Number Systems Class 9 Chapter 1 Exercise: 1.5

Q1 (i) Classify the following numbers as rational or irrational: $2-\sqrt{5}$

Answer:

Value of $\sqrt{5}$ is 2.23606798....
Now,
$\Rightarrow 2 - \sqrt{5} = 2 - 2.23606798... = -0.23606798...$
Now,
Since the number is in non-terminating non-recurring. Therefore, it is an irrational number.

Q1 (ii) Classify the following numbers as rational or irrational: $\left ( 3+\sqrt{23} \right )-\sqrt{23}$

Answer:

Given number is $\left ( 3+\sqrt{23} \right )-\sqrt{23}$

$\Rightarrow \left ( 3+\sqrt{23} \right )-\sqrt{23} = 3+\sqrt{23}-\sqrt{23} = 3$
Now, it is clearly a rational number because we can represent it in the form of $\frac{p}{q}$

Q1 (iii) Classify the following numbers as rational or irrational: $\frac{2\sqrt{7}}{7\sqrt{7}}$

Answer:

Given number is $\frac{2\sqrt{7}}{7\sqrt{7}}$

$\Rightarrow \frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7}$
As we can clearly see that it can be represented in $\frac{p}{q}$ form. Therefore, it is a rational number.

Q1 (iv) Classify the following numbers as rational or irrational: $\frac{1}{\sqrt{2}}$

Answer:

Given number is $\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}=\frac{1.41421356..}{2} =0.7071068...$
Now,
Clearly, as the decimal expansion of this expression is non-terminating and non-recurring. Therefore, it is an irrational number.

Q1 (v) Classify the following numbers as rational or irrational: $2\pi$

Answer:

Given number is $2\pi$
We know that the value of $\pi = 3.14159265...$
Now,
$\Rightarrow 2\pi = 2\times 3.14159265... = 6.2831853...$
Now,
Clearly, as the decimal expansion of this expression is non-terminating and non-recurring. Therefore, it is an irrational number.

Q2 (i) Simplify each of the following expressions: $(3+\sqrt{3})(2+\sqrt{2})$

Answer:

Given number is $(3+\sqrt{3})(2+\sqrt{2})$
Now, we will reduce it into
$\Rightarrow (3+\sqrt{3})(2+\sqrt{2})= 3.2+3.\sqrt{2}+\sqrt{3}.2+\sqrt{3}.\sqrt{2}$
$= 6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}$

Therefore, answer is $6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}$

Q2 (ii) Simplify each of the following expressions: $\left (3+\sqrt{3} \right )\left ( 3-\sqrt{3} \right )$

Answer:

Given number is

$\left (3+\sqrt{3} \right )\left ( 3-\sqrt{3} \right )$
Now, we will reduce it into
$\Rightarrow (3+\sqrt{3})(3-\sqrt{3})= \left ( (3)^2-(\sqrt{3})^2 \right )$ $\left ( using \ (a+b)(a-b)=a^2-b^2 \right )$
$=9 - 3 = 6$

Therefore, answer is 6

Q2 (iii) Simplify each of the following expressions: $\left (\sqrt{5}+\sqrt{2} \right )^{2}$

Answer:

Given number is $\left (\sqrt{5}+\sqrt{2} \right )^{2}$

Now, we will reduce it into
$\Rightarrow \left (\sqrt{5}+\sqrt{2} \right )^{2} = \left ( (\sqrt{5})^2+(\sqrt{2})^2+2.\sqrt{5}.\sqrt{2} \right )$ $\left ( using \ (a+b)^2=a^2+b^2 +2ab\right )$
$=5+2+2\sqrt{10}$
$=7+2\sqrt{10}$

Therefore, the answer is $7+2\sqrt{10}$

Q2 (iv) Simplify each of the following expressions: $\left (\sqrt{5}-\sqrt{2} \right )\left ( \sqrt{5}+\sqrt{2} \right )$

Answer:

Given number is $\left (\sqrt{5}-\sqrt{2} \right )\left ( \sqrt{5}+\sqrt{2} \right )$

Now, we will reduce it into
$\Rightarrow \left (\sqrt{5}-\sqrt{2} \right )\left ( \sqrt{5}+\sqrt{2} \right )= \left ( (\sqrt{5})^2-(\sqrt{2})^2 \right )$ $\left ( using \ (a+b)(a-b)=a^2-b^2\right )$
$=5-2$
$=3$

Therefore, the answer is 3 .

Q3 Recall, $\pi$ is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, $\pi=\frac{c}{d}$ ⋅ This seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction?

Answer:

There is no contradiction.
When we measure a length with scale or any other instrument, we only obtain an approximate rational value. We never obtain an exact value.
For this reason, we cannot say that either c or d is irrational.
Therefore, the fraction $\frac{c}{d}$ is irrational. Hence, the value of $\pi$ is approximately equal to $\frac{22}{7} = 3.142857....$

Therefore, $\pi$ is irrational.

Q4 Represent $\sqrt{9.3}$ on the number line.

Answer:

1639723609742 Draw a line segment OB of 9.3 unit. Then, extend it to C so that BC = 1 unit. Find the mid-point D of OC and draw a semi-circle on OC while taking D as its centre and OD as the radius. Now, Draw a perpendicular to line OC passing through point B and intersecting the semi-circle at E. Now, Take B as the centre and BE as radius, draw an arc intersecting the number line at F. the length BF is $9\sqrt{3}$ units.

Q5 (i) Rationalise the denominators of the following: $\frac{1}{\sqrt{7}}$

Answer:

Given number is $\frac{1}{\sqrt{7}}$

Now, on rationalisation, we will get

$\Rightarrow \frac{1}{\sqrt 7} = \frac{1}{\sqrt 7}\times \frac{\sqrt 7}{\sqrt 7 } = \frac{\sqrt7}{7}$

Therefore, the answer is $\frac{\sqrt7}{7}$

Q5 (ii) Rationalise the denominators of the following: $\frac{1}{\sqrt{7}-\sqrt{6}}$

Answer:

Given number is $\frac{1}{\sqrt{7}-\sqrt{6}}$

Now, on rationalisation, we will get

$\Rightarrow \frac{1}{\sqrt 7-\sqrt6} = \frac{1}{\sqrt 7-\sqrt 6}\times \frac{\sqrt 7+\sqrt 6 }{\sqrt 7+\sqrt 6 } = \frac{\sqrt7+\sqrt 6}{(\sqrt7)^2-(\sqrt6)^2} = \frac{\sqrt7+\sqrt6}{7-6}$ $= \sqrt7+\sqrt6$

Therefore, the answer is $\sqrt7+\sqrt6$

Q5 (iii) Rationalise the denominators of the following: $\frac{1}{\sqrt{5}+\sqrt{2}}$

Answer:

Given number is $\frac{1}{\sqrt{5}+\sqrt{2}}$

Now, on rationalisation, we will get

$\\= \frac{1}{\sqrt 5+\sqrt2} \\\\= \frac{1}{\sqrt 5+\sqrt 2}\times \frac{\sqrt 5-\sqrt 2 }{\sqrt 5-\sqrt 2 }\\\\ = \frac{\sqrt5-\sqrt 2}{(\sqrt5)^2-(\sqrt2)^2} \\\\= \frac{\sqrt5-\sqrt2}{5-2}$
$= \frac{\sqrt5-\sqrt2}{3}$

Therefore, the answer is $\frac{\sqrt5-\sqrt2}{3}$

Q5 (iv) Rationalise the denominators of the following: $\frac{1}{\sqrt{7}-2}$

Answer:

Given number is $\frac{1}{\sqrt{7}-2}$

Now, on rationalisation, we will get

$\Rightarrow \frac{1}{\sqrt 7-2} = \frac{1}{\sqrt 7- 2}\times \frac{\sqrt 7+ 2 }{\sqrt 7+ 2 } = \frac{\sqrt7+ 2}{(\sqrt7)^2-(2)^2} = \frac{\sqrt7+2}{7-4}$
$= \frac{\sqrt7+2}{3}$

Therefore, the answer is $\frac{\sqrt7+2}{3}$

Benefits of NCERT Solutions for Class 9 Maths Exercise 1.5:

• NCERT solutions for Class 9 Maths exercise 1.5 able us to develop our basic concepts regarding numbers and their characteristics.

• By solving the NCERT solution for Class 9 Maths chapter 1 exercise 1.5 exercises, it develops a strong foundation of Mathematical knowledge and of course more confidence in tackling new topics in our higher classes.

• Exercise 1.5 Class 9 Maths, will help us in learning how to perform operations on real numbers and this may help us in higher classes.

Key Features of class 9 maths chapter 1 exercise 1.5

Application of Concepts: This ex 1.5 class 9 often requires students to apply the concepts learned in the chapter to solve real-world problems.
Challenging Problems: Some class 9 maths ex 1.5 questions may be more challenging, requiring students to think critically and use problem-solving skills.
Practice for Competitive Exams: The questions in exercise 1.5 class 9 maths may serve as good practice for students preparing for competitive exams or Olympiads.
Enhances Problem-Solving Skills: Solving class 9 ex 1.5 problems enhances students' problem-solving and analytical skills.
Step-by-Step Solutions: Class 9 Maths Chapter 1 exercise 1.5 provides step-by-step solutions for Exercise 1.5 in the textbook or through supplementary materials, helping students understand the problem-solving process.
Self-assessment: After solving the questions in this exercise, students can assess their understanding of the chapter's concepts.
Preparation for Board Exams: 9th class maths exercise 1.5 answers are aligned with the CBSE board exam pattern, making it essential for board exam preparation.

Overall, Exercise 1.5 in Class 9 Maths serves as a crucial component of the curriculum, helping students practice and reinforce the mathematical concepts learned in the chapter.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. According to NCERT solutions for Class 9 Maths, 1 exercise 1.5 The associative property is applicable to ______

The associative property is applicable to Addition and Multiplication

Option (c) Addition and Multiplication

2. State true or false: The product of rational and irrational numbers is always an irrational number

Yes, the given statement is true. The product of a rational and an irrational number is always an irrational number.

3. Which property do the given statements hold? 2+3=3+2

2+3=3+2 holds the commutative property of addition.

4. Give one example for the commutative property of multiplication.

The example for the commutative property of multiplication is 5×2=2×5

5. Which of the two operations satisfy the condition of the associative property?

The two operations which satisfy the condition of the associative property are

Addition
Multiplication.

6. According to NCERT solutions for Class 9 Maths chapter 1 exercise 1.5, state the commutative property?

Commutative law states that when two numbers are added or multiplied together, a change in their locations has no effect on the outcome, according to NCERT solutions for Class 9 Maths chapter 1 exercise 1.5.

7. What is the total number of questions in the NCERT solutions for Class 9 Maths chapter 1 exercise 1.5 and What kind of questions are asked in NCERT solutions for Class 9 Maths chapter 1 exercise 1.5?

NCERT solutions for Class 9 Maths chapter 1 exercise 1.5 consists of 5 questions which are based on the classification of numbers into rational and irrational numbers by using certain operations and the representation of a given real number on the number line.

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