NCERT Solutions for Class 9 Mathematics Chapter 1- Number Systems

Mathematics is a subject which requires a lot of practice. . The more you practice the better you become. . Therefore, you must practice to perfection. There are plenty of  examples to practice with in Extramarks NCERT Solutions for Class 9 Mathematics Chapter 1.

The Chapter -Number Systems of Class 9 Mathematics covers all the fundamentals of Mathematics and will help students understand the core concepts covered in higher classes. . As Mathematics is totally based on  numbers, this  chapter tells about all the different types of numbers and various applications of numbers in Mathematics. If you are looking for a thorough knowledge of the concepts of the Chapter, Extramarks is the right platform to get the right amount of practice and to develop your mathematical abilities and be confident at an early age. 

You can avail of NCERT Solutions for Class 9 Mathematics Chapter 1 on the Extramarks website and turn your child into a smart learner. Number systems- Chapter 1 of Class 9 Mathematics comprises all the  fundamental concepts.. Based on the CBSE NCERT latest 2021-2022 syllabus, we have provided points to ponder as well as detailed solutions for the better understanding of the subject. It also encourages students to be curious and look for answers themselves.  Students are recommended  to use the NCERT solutions Class 9 Mathematics to realise their  true potential and to enjoy the entire process of learning and stay ahead of the competition.    

Visit the Extramarks website to keep yourself updated about the CBSE syllabus, NCERT Solutions and exam patterns. You may  also search for NCERT Solutions Class 10 to step up  your preparation and stay ahead of others.

Key Topics Covered In NCERT Solution for Class 9 Mathematics Chapter 1

Number system is entirely the study of numeracy, and hence the students must understand the concepts and enjoy the learning experience.   It will directly connect to  chapters like Quadratic equations, Sets etc in higher classes. As a result, students aiming for good grades should be able to identify different types of numbers, know their representation and identities and should know how to rationalise them efficiently.

In Extramarks NCERT Solutions for Class 9 Mathematics Chapter 1, students can expect all topics to be covered and explained in detail. The  chapter includes sections like real numbers and their decimal expansion, representing real numbers on the number line, operation on real numbers etc. For complete study material for NCERT Solutions Class 9, NCERT Solutions Class 10, NCERT Solutions Class 11, and NCERT Solutions Class 12, visit the Extramarks website and app which is trusted by students across India and  their  numbers have been growing by leaps and bounds because of the unshakable trust and faith these schools have in us. 

The key topics covered in NCERT Solutions of Class 9 Mathematics Chapter 1:

Exercise Topic
1.1 Introduction
1.2 Irrational numbers
1.3 Real numbers and their decimal expansion
1.4 Representing Real numbers on the number line
1.5 Operations on Real numbers
1.6 Laws of Exponents for Real numbers

NCERT Solutions for Class 9 Mathematics Chapter 1 requires students to apply and correlate whatever they have learnt in their previous classes. . Students can also access NCERT Solutions for Mathematics Class 8 and Class 7  to review the  concepts studied last year or earlier.

1.1 Introduction

This Chapter on Number Systems begins with the basic introduction of numbers and their applications in our daily lives. Further, it categorises  numbers as Natural numbers, Whole numbers, Integers, Rational numbers and Irrational numbers. The various examples provided in the chapter help recognise different numbers, which can help easily recall the concepts in prior  Classes.

1.2 Irrational numbers

This section deals entirely with what makes a number irrational and how one can distinguish between rational and irrational numbers. Students have to keep in mind specific points while deciding it is an irrational number which they will find in our NCERT Solutions for Class 9 Mathematics Chapter 1. Students will also read about the set of numbers called real numbers.

At the end of this section, students will get a proper understanding of irrational as well as real numbers. Also, they will be available to locate certain square roots of numbers on the number line. 

1.3 Real numbers and their decimal expansion

In this section, first, you will learn about  decimal expansions of real numbers. Then you would evaluate whether you can distinguish between rational and irrational numbers based on the decimal expansion. You come across different cases and will illustrate them on the basis of examples.

1.4 Representing Real numbers on the number line

As learnt in the previous t section about the decimal expansion of real numbers, we will use it for application on the number line. The decimal expansion helps represent real numbers and get good practice with examples.  

After going through this section, you would be able to locate points of the number line with ease, learn to visualize points on the number line in a systematic way, learn to round off to the nearest decimal and know that a unique point represents every real number.

1.5 Operation on Real numbers

In the earlier Classes, we have learnt that rational numbers follow commutative, associative and distributive properties for mathematical operations, i.e. when you add, subtract, multiply or divide a rational number, you get a rational number. Likewise, this holds true for irrational numbers also. . 

The set of rational and irrational numbers is called real numbers. Hence, this applies to real numbers too. 

After completing this section, you will be able to carry out operations on non-terminating and non-recurring decimal expansions with the help of illustrative examples. Refer to our NCERT Solutions for Class 9 Mathematics Chapter 1 to get access to more solved questions based on Operations on Real Numbers. 

1.6 Law of exponents for real numbers

You are already acquainted with exponents and laws of exponents from your earlier Classes. In this section, we will specifically learn about the laws of exponents on real numbers. The application of laws of exponents remains the same in the case of real numbers. You have to learn to convert the square root or the cube root of the number into exponential form.

NCERT Solutions for Class 9 Mathematics Chapter 1 Exercise &  Solutions

Find NCERT Solutions for Class 9 Mathematics Chapter 1  on the Extramarks website. From a detailed analysis of the Chapter to short notes, you can find everything to level up your preparation and gear up your performance in the exams. You will get access to  all the questions on Number Systems once you access the NCERT Solutions for Class 9 Mathematics on our website.

Click on the below links to view exercise specific questions and solutions for NCERT Solutions for Class 9 Mathematics Chapter 1:

  •  Chapter 9: Exercise 1.1 Question and answers    
  •  Chapter 9: Exercise 1.2 Question and answers
  • Chapter 9: Exercise 1.3 Question and answers
  • Chapter 9: Exercise 1.4 Question and answers
  • Chapter 9: Exercise 1.5 Question and answers

Along with Class 9 Mathematics Solutions, students can explore NCERT Solutions on our Extramarks website for all primary and secondary  classes.

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2
  • NCERT Solutions Class 3
  • NCERT Solutions Class 4
  • NCERT Solutions Class 5 
  • NCERT Solutions Class 6
  • NCERT Solutions Class 7
  • NCERT Solutions Class 8
  • NCERT Solutions Class 9
  • NCERT solutions Class 10
  • NCERT solutions Class 11
  • NCERT solutions Class 12

NCERT Exemplar for Class 9 Mathematics 

NCERT Exemplar Class 9 Mathematics is an excellent resource  for students preparing for their 9th standard exams. The book consists of a variety of questions of different levels  of difficulty. It  encourages students to develop more interest in  Mathematics and get more significant insights into the  chapter to become proficient in facing challenging questions in the exams.

 NCERT Exemplar helps students to develop confidence during their preparation as they have questions of basic level  as well as advanced level. It has proved to be quite  beneficial for students, especially for  those preparing for various competitive exams. It covers  the entire chapters in detail  , which makes it fruitful for all curriculum students.

After referring to the NCERT Solutions and NCERT Exemplar, the students are confident   to solve all the complicated and tricky questions. As a result, students can easily switch  to more advanced and higher-level conceptual questions. By studying from the Exemplar, you can prepare well for entrance exams like Olympiad, NTSE and KVPY.

Key Features of NCERT Solutions for Class 9 Mathematics Chapter 1

In order to obtain a good score in exams, revision of previous concepts is a must. Hence, NCERT Solutions for Class 9 Mathematics Chapter 1 offers a complete solution for all problems. The key features of NCERT solutions are:  : 

  • Mathematics experienced faculty and subject experts have designed Extramarks NCERT Solutions for Class 9 Mathematics Chapter 1. It is a thoroughly researched material made in sync with CBSE examination guidelines.
  • Students have a very clear understanding of the concepts and overcome  all their doubts with the help of Extramarks NCERT solutions.
  • After completing the NCERT Solutions for Class 9 Mathematics Chapter 1 students will be able to solve all the basic and advanced level problems with better  accuracy.The systemic and well-laid out balanced study plan boosts their performance naturally and effortlessly. 
Q.1 Is zero a rational number? Can you write it in the form  pq, where p and q are integers and q ≠ 0?

Ans- 

Yes, 0 is a rational number. It can be represented as (0/1), (0/2), (0/3) etc.

Q.2

Find six rational numbers between 3 and 4.

Ans-

There are infinite rational numbers between 3 and 4. 3 and 4 can be represented as 24/8 and 32/8 respectively. The rational numbers between 3 and 4 are 25/8, 26/8, 27/8, 28/8, 29/8, 30/8.

Q.3 Find five rational numbers between 35 and 45.

Ans-

There are infinite natural numbers between 35 and 45.35=3×65×6=183045=4×65×6=2430Therefore, rational numbers between 35 and 45 are1930, 2030, 2130, 2230, 2330

Q.4 State whether the following statements are true or false. Give reasons for your answers.
(i) Every natural number is a whole number.
(ii) Every integer is a whole number.
(iii) Every rational number is a whole number.

Ans-

(i) True; since the collection of whole numbers contains all natural numbers.
(ii) False; since integers may be negative but whole numbers are positive. For example: – 5 is an integer it is not a whole number.
(iii) False; as rational number may be a fraction but whole number may not be a fraction.
For example: 4/5 is a rational number and it is not a whole number.

Q.5

State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
(ii) Every point on the number line is of the form

m

where m is a natural number.
(iii) Every real number is an irrational number.

Ans-

(i) True; because real number is a collection of rational and irrational number.
(ii) False; as negative numbers cannot be represented as the square root of any other number.
(iii) False; as real numbers include both rational and irrational numbers i.e., irrational number is a part of real number. Therefore, every real number cannot be an irrational number.

Q.6 Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

Ans-

No, the square roots of all positive integers are not irrational. For example: The square roots of 4 and 9 are 2 and 3 respectively.

Q.7

Show

 

 

 

how 5 can be represented on the number line.

Ans-

5 can be written as    5=4+1=(2)2+12Mark a point ‘A’ representing 2 on number line.Now, construct AB of unit length perpendicular to OA. Then, taking O as centre and OB as radius, draw an arc which intersect the number line at C. OC represents the length of 5 The point C represents 5

Q.8

Write the following in decimal form and say what kind of

 

 

 

decimal expansion each has:i 36100  ii 111  iii 418  iv 313  v 211  vi 329400

Ans-

(i)36100=0.36 It is a terminating decimal because remainder is zero.(ii)111=0.0909090909… It is a non-terminating decimal because remainder is not zero.(iii)418=4.125 It is a terminating decimal because remainder is zero.(iv)313=0.230769230769…    =0.230769¯ It is a non terminating decimal because it is a repeating(v)211=0.181818181818…  =0.18¯ It is a terminating decimal because it is a repeating (vi)329400=0.8225 It is a terminating decimal because remainder is zero. Q.9

You know that

 

 

 

17 = 0.142857¯. Can you predict what thedecimal expansions of 27 ,37 ,47, 57, 67 are, without actuallydoing the long division? If so, how?

Ans-

Given: 17=0.142857¯∴        27=2×0.142857¯=0.285714¯        37=3×0.142857¯=0.428571¯        47=4×0.142857¯=0.571428¯        57=5×0.142857¯=0.714285¯        67=6×0.142857¯=0.857142¯ Q.10

Express the following in the form

 

 

 

pq, where p and q areintegers and q¹0.i 0  ii 0.47 ¯  iii 0.001¯

An-

(i)  Let x=0.6¯...(i)Multiplying both sides by 10, we get       10x=10×0.6¯       10x=6.6¯...(ii)Subtracting equation(i) from equation(ii), we get10x−x=6.6¯−0.6¯9x=6   x=69   x=23∴        0.6¯=23(ii) Let x=0.47¯...(i)Multiplying both sides of equation(i) by 10, we get       10x=10×0.47¯       10x=4.7¯...(ii)Multiplying both sides of equation(ii) by 10, we get    100x=10×4.7¯    100x=47.7¯...(iii)Subtracting equation(ii) from equation(iii), we get100x−10x=47.7¯−4.7¯    90x=43         x=4390∴           0.47¯=4390(iii)Let x=0.001¯...(i)Multiplying both sides of equation(i) by 1000, we get    1000x=1000×0.001¯        =1.001¯...(ii)Subtracting equation(i) from equation(ii), we get1000x −x=1.001¯−0.001¯          999x=1      x=1999∴     0.001¯=1999 Q.11

Express 0.99999, in the form

 

 

 

pq. Are you surprised by your answer?With your teacher and classmates discussway the answer makes sense.

Ans-

 Let x=0.99999…=0.9¯…(i)Multiplying both sides by 10, we get       10x=10×0.9¯       10x=9.9¯…(ii)Subtracting equation(i) from equation(ii), we get10x−x=9.9¯−0.9¯9x=9   x=99   x=1∴  0.99999…=1 Q.12

What

 

 

 

can the maximum number of digits be in the repeatingblock of digits in the decimal expansion of 117? Perform thedivision to check your answer.

Ans-

              0.0588235294117647        17 1.00                                           085 ¯                 150                  136 ¯                    140                     136 ¯                          40              34 ¯                 60                   51 ¯                     90                     85 ¯            50            34 ¯            160             153 ¯                                           70                  68 ¯                     20                      17 ¯            30            17 ¯             130              119 ¯                110               102 ¯          80          68 ¯         120         119 ¯               1 Thus,  1 17 =0. 0588235294117647 ¯ 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Q.13

Look

 

 

 

at several examples of rational numbers in the formpq q≠0 , where p and q are integers with no commonfactors other than 1 and having terminating decimalrepresentations (expansions).Can you guess what propertyq must satisfy?

Ans-

Since, the rational number pq will be terminating decimal if denominator qis either 2, 4, 5, 8, 10 and so on…92=4.5114=2.75178=2.125135=2.63210=3.2We see that rational number pq will be terminating if prime factors of qare either 2 only or multiple of 2 and 5 only or both. Q.14

Write three numbers whose decimal expansions are non-terminating non-recurring.

Ans-

Three numbers whose decimal expansions are non- terminating non-recurring are as follows:

0.030030012003000050004123000…

0.01200012500003500050010008879000102003…

1.5200050040060080010030010040038001…

Q.15

Find

 

 

 

three different irrational numbers between therational numbers 57 and 911.

Ans-

Since, 57=0.714285¯ and 911=0.81¯Three irrational numbers between 57 and 911 are:0.72005006004000202005004…0.75005006004000202005004…0.80005006004000202005004… Q.16

Classify the following numbers as rational or irrational:

 

 

 

i 23  ii 225 iii 0.3796 iv 7.478478...   v 1.101001000100001...

Ans-

(i)23=4.7983152331272…Since, this number is non-terminating and non-repeating, therefore it is irrational number.(ii)225=15, It is rational number.(iii)0.3796, It is rational number because this number is terminating.(iv)7.478478…=7.478¯, this is rational number as it is non-terminating and repeating. (v)1.101001000100001…,​ It is irrational number because it is non-terminating non repeating. Q.17

Visualise 3.765 on the number line, using successive magnification.

Ans-

3.765 can be visualised as in the following steps.

Q.18

Visualise

 

 

 

4.26 ¯on the number line, upto 4 decimal places.

Ans-

4.26¯ can be written as:4.26¯=4.2626

Q.19

Classify the following numbers as rational or irrational:

 

 

 

i 2−5  ii 3+23−23  iii 2777iv 12 v 2π

Ans-

(i) 2−5=Irrational​ number because 5 is irrational number.(ii)(3+23)−23=3, which is a rational number.(iii) 2777=27, which is in the form of pq form and it is a rational number.(iv) 12=12×22     =22, which is an irrational number becuase 2 is irrational number.(v) 2π=2(3.1415…)   =6.2830…, which is non-terminating and non-repeating. So it is irrational number. Q.20

Simplify

 

 

 

each of the following expressions:(i)(3+3)(2+2) (ii) (3+3)(3−3)(iii) (5+2)2(iv) (5−2)(5+2)

Ans-

(i)  (3+3)(2+2)=6+32+23+6(ii)  (3+3)(3−3)=(3)2−(3)2 =9−3=6(iii)  (5+2)2=5+252+2                     =7+210(iv) (5−2)(5+2)=(5)2−(2)2                                  =5−2                                  =3 Q.21

Recall

 

 

 

, π is defined as the ratio of the circumference (say c)of a circle to its diameter say d.That is, π=cd.This seemsto contradict the fact that π is irrational.How will youresolve this contradiction?

Ans-

There is no contradiction. Remember that when you measure a length with a scale or any other device, you only get an approximate rational value. So, you may not realise that either c or d is irrational.

Q.22

Represent

 

 

 

9.3 on the number line.

Ans-

Mark a line segment AB = 9.3 on number line. Further, take BC of 1 unit. Draw a semi-circle on AC as diameter. Draw a perpendicular to line AC passing through point B. Let it intersect the semi circle at D. Taking B as centre and BD as radius, draw an arc intersecting number line at E. BE =

9.3 . MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbbjxAHXgarmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqee0evGueE0jxyaibaieYBg9LrFfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqabeaadaabauaaaOqaamaakaaabaaceaGaa8xoaiaa=5cacaWFZaaaleqaaOGaaiOlaaaa@3B35@ Q.23

Rationalise

 

 

 

the denominators of the following:(i)17(ii)17−6(iii)15+2(iv)17−2

Ans-

(i)  17=17×77         =77(ii) 17−6=17−6×7+67+6 [∵(a−b)(a+b)=a2−b2]                =7+67−6                =7+6(iii)  15+2=15+2×5−25−2          =5−25−2                    [∵(a−b)(a+b)=a2−b2]                 =5−23(iv)  17−2=17−2×7+27+2                =7+27−4        [∵(a−b)(a+b)=a2−b2]                =7+23 Q.24

Find

 

 

 

: i 6412     ii 3215     iii 12513

Ans-

(i)   6412=(26)12          =26×12          =23          =8(ii)      3215=(25)15          =25×15          =21 =2(iii)  12513=(53)13          =53×13          =51 =5 Q.25

Find:

 

 

 

i932  ii3225  iii1634   iv125−13

Ans-

(i)  932=(32)32      =32×32      =33      =27(ii) 3225=(25)25      =25×25      =22      =4(iii) 1634=(24)34      =24×34      =23      =8(iv) 125−13=(53)−13           =53×−13             =5−1             =15 Q.26

Simplify

 

 

 

:    (i) 223.215     (ii) (133)7     (iii) 11121114       (iv)712.812

Ans-

(i)    223.215=223+15              [∵xa.xb=xa+b]              =210+315              =21315(ii) (133)7=133×7              =1321              =3−21(iii)     11121114=1112−14             [∵xaxb=xa−b]             =112−14            =1114(iv) 712.812=(7×8)12         [∵xm.ym=(xy)m]              =5612

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