# NCERT Solutions for Class 9 Maths Chapter 1

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems in Hindi Medium and English medium PDF format to free download for academic session 2020-21. All the Solutions for Class 9th Maths Chapter 1 are updated according to latest CBSE Curriculum and new NCERT Books. UP Board Students are using same NCERT Textbooks for course. So, they can also download UP Board Solutions for Class 9 Maths Chapter 1 in Hindi Medium or English Medium. 9th Maths NCERT Solutions are prepared by describing all the steps and formulae. Contents are according to Latest CBSE Syllabus 2020-21 for the students of CBSE Board as well as UP Board, MP Board, etc.

following NCERT Books 2020-21 for their final exam March 2021. 9th Maths Chapter 1 Solutions are available in PDF Format, Online study mode and Videos format. Everything is free to use.## NCERT Solutions for Class 9 Maths Chapter 1

Class: 9 | Maths (English and Hindi Medium) |

Chapter 1: | Number Systems |

### 9th Maths Chapter 1 Sols in English and Hindi Medium

### Class 9th Maths Chapter 1 Number Systems Solutions

- NCERT SolutionsClass 9 Maths Chapter 1 Exercise 1.1Read more
- NCERT SolutionsClass 9 Maths Chapter 1 Exercise 1.2Read more
- NCERT SolutionsClass 9 Maths Chapter 1 Exercise 1.3Read more
- NCERT SolutionsClass 9 Maths Chapter 1 Exercise 1.4Read more
- NCERT SolutionsClass 9 Maths Chapter 1 Exercise 1.5Read more
- NCERT SolutionsClass 9 Maths Chapter 1 Exercise 1.6Read more

#### Class 9 Maths Exercise 1.1, 1.2 Solutions in Video

#### Class 9 Maths Exercise 1.3, 1.4 Solutions in Video

#### Class 9 Maths Exercise 1.5, 1.6 Solutions in Video

#### About 9th Maths Chapter 1 Solutions

CBSE NCERT Solutions for Class 9 Maths Chapter 1 Number Systems solutions in PDF form for free download updated for session 2020-21. Visit to Discussion Forum to share your knowledge. Download NCERT Solutions Offline Apps 2020-21, which work without internet connection. Everything on Tiwari Academy website or apps are free of cost. No login or registration is required.

##### Important Terms on 9th Maths Chapter 1

- Natural numbers are those numbers which are used for counting.
- Whole numbers are the collection of all natural numbers together with zero.
- Integers are the collection of all whole numbers and negative of natural numbers.
- Rational numbers are those numbers which can be expressed in the form of p/q, where p, q are integers and q is not equal to 0.
- Irrational numbers are those numbers which cannot be expressed in the form of p/q, where p, q are integers and q is not = 0.
- Real numbers are the collection of all rational and irrational numbers.

###### Do you know?

- Two numbers are said to be equivalent, if numerators and denominators of both are in proportion or they are reducible to be equal.
- The decimal expansion of real numbers can be terminating or non-terminating repeating or non-terminating non-repeating.
- The decimal expansion of rational numbers can either be terminating or non-terminating and vice-versa.
- The decimal expansion of irrational numbers can either be non-recurring and vice-versa.

###### More to know?

- If a is a rational and b is an irrational, then a + b and a – b are irrational, and ab and a/b are irrational numbers, where b is not equal to 0.
- If a and b both are irrational, then a+b, a-b, ab and a/b may be rational or irrational.
- If a be any real number and n be any positive integer such that a^1/n = n√a is a real number, then ‘n’ is called exponent, a is called radical and √ is called radical sign.

### Important Questions on 9th Maths Chapter 1

It can be written in the form of p/q. For example: 0/1, 0/2, 0/5 are rational numbers, where p and q are integers and q≠0.

= 6 + 3√2 + 2√3 + √6

Let x = 0.99999… … (i)

Multiplying equation (i) by 10 both sides

10x = 9.99999…

⇒ 10x = 9 + 0.99999……

⇒ 10x = 9 + x [From equation (i)]

⇒ 10x – x = 9

⇒ 9x = 9

⇒ x = 9/9 = 1

The answer makes sense as 0.99999… is very close to 1, that is why we can say that 0.99999=1.

0.414114111411114…

2.01001000100001…

π=3.1416…

= (3^2 )^(3/2)

= 3^(2×3/2)

= 3^3

= 27