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NCERT Solutions
Class 9
Maths
Chapter 1 Number System
Exercise 1.5

NCERT Solutions Class 9 Maths Chapter 1 Number System Exercise 1.5

NCERT Solutions for Class 9 Maths Chapter 1 Number System: Exercise 1.5, teaches about rationalising the denominator, which is an important part of the chapter. This exercise will explain how to simplify expressions and remove the irrational number in the denominator.

The NCERT Solutions follow the CBSE exam pattern and are aligned with the latest NCERT Syllabus. The step by step solutions provided helps the students understand the concept clearly.

Practicing this exercise will help in getting better marks in your school exams and also prepare for competitive exams. Download the FREE PDF for a quick revision.

1.0Download NCERT Solutions of Class 9 Maths Chapter 1 Number System Exercise 1.5: Free PDF

Get the free PDF of NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.5 here. This exercise solves questions on how to simplify expressions by rationalising the denominator. Download the free PDF from below:

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.5

2.0NCERT Exercise Solutions Class 9 Chapter 1- Number Systems: All Exercises 

NCERT Solutions for Class 9 Maths Chapter 1: Exercise 1.1

NCERT Solutions for Class 9 Maths Chapter 1: Exercise 1.2

NCERT Solutions for Class 9 Maths Chapter 1: Exercise 1.3

NCERT Solutions for Class 9 Maths Chapter 1: Exercise 1.4

NCERT Solutions for Class 9 Maths Chapter 1: Exercise 1.5

3.0NCERT Class 9 Maths Chapter 1 Exercise 1.5 : Detailed Solutions

1. Find:

(i) (64)¹ᐟ²

(ii) 32¹ᐟ⁵

(iii) 125¹ᐟ³

Sol. (i) (64)¹ᐟ² = (8²)¹ᐟ² = 8^(2 × 1/2) = 8¹ = 8

(ii) 32¹ᐟ⁵ = (2⁵)¹ᐟ⁵ = 2^(5 × 1/5) = 2¹ = 2

(iii) (125)¹ᐟ³ = (5³)¹ᐟ³ = 5^(3 × 1/3) = 5¹ = 5

2. Find:

(i) 9³ᐟ²

(ii) 32²ᐟ⁵

(iii) 16³ᐟ⁴

(iv) 125⁻¹ᐟ³

Sol. (i) 9³ᐟ² = (9¹ᐟ²)³ = (3)³ = 27

(ii) 32²ᐟ⁵ = (2⁵)²ᐟ⁵ = 2^(5 × 2/5) = 2² = 4

(iii) 16³ᐟ⁴ = (2⁴)³ᐟ⁴ = 2^(4 × 3/4) = 2³ = 8

(iv) 125⁻¹ᐟ³ = (5³)⁻¹ᐟ³ = 5^(3 × -1/3) = 5⁻¹ = 1/5

3. Simplify :

(i) 2²ᐟ³ ⋅ 2¹ᐟ⁵

(ii) (1/3³)⁷

(iii) 11¹ᐟ⁴ / 11¹ᐟ²

(iv) 7¹ᐟ² ⋅ 8¹ᐟ²

Sol. (i) 2²ᐟ³ ⋅ 2¹ᐟ⁵ = 2^(2/3 + 1/5) = 2^((10+3)/15) = 2¹³ᐟ¹⁵

(ii) (1/3³)⁷ = (3⁻³)⁷ = 3^(-3 × 7) = 3⁻²¹

(iii) 11¹ᐟ⁴ / 11¹ᐟ² = 11^(1/4 − 1/2) = 11^(1/4 − 2/4) = 11⁻¹ᐟ⁴

(iv) 7¹ᐟ² ⋅ 8¹ᐟ² = (7×8)¹ᐟ² = (56)¹ᐟ²

4.0Key Features and Benefits Class 9 Maths Chapter 1 Number System: Exercise 1.5

  1. Solving these NCERT Solutions will help you to learn how to rationalise denominators.
  2. Solutions provided here are following the latest NCERT syllabus and are useful for CBSE exams.
  3. They provide simple to follow explanations to make the concept simple and clear to understand.
  4. Continuous practice of NCERT Solutions will make these types of questions easy to solve.
  5. It assists with sharpening your maths for Olympiad and other competitive exams.

NCERT Class 9 Maths Ch. 1 Number System Other Exercises:-

Exercise 1.1

Exercise 1.2

Exercise 1.3

Exercise 1.4

Exercise 1.5


NCERT Solutions for Class 9 Maths Other Chapters:-

Chapter 1: Number Systems

Chapter 2: Polynomials

Chapter 3: Coordinate Geometry

Chapter 4: Linear Equations in Two Variables

Chapter 5: Introduction to Euclid’s Geometry

Chapter 6: Lines and Angles

Chapter 7: Triangles

Chapter 8: Quadrilaterals

Chapter 9: Circles

Chapter 10: Heron’s Formula

Chapter 11: Surface Areas and Volumes

Chapter 12: Statistics

Frequently Asked Questions

It focuses on adding, subtracting, multiplying, and rationalizing when dealing with irrational numbers.

Rationalizing simplifies the expression and assists in making comparisons or calculating values further.

The product can be either rational or irrational depending on the two irrational numbers.

The purpose of simplifying the surd is to make it easier to perform arithmetic operations and compare more expressions.

Identity element is used to maintain the value of a number over the course of an operation, whether addition or multiplication.

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