NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 The NCERT Solutions for Class 9 Maths Chapter 1 Number System are created by the expert faculty of Study Circle. The NCERT Solutions for Class 9 Maths Chapter 1 aims to provide the students with detailed and step-wise explanations for the answers to all the questions given in the exercises of this chapter.

NCERT  solutions for Class 9 Maths Chapter 1 help students to solve problems efficiently and efficiently for board exams. They also focus on preparing maths solutions in such a way that it is easy for students to understand and all students score very well all students score more than 90% in the exam

                                         Exercise 1.1

  (Class – 9 Maths NCERT Chapter -1 )

 
 
Q 1. Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q ≠ 0?

Solution:

We know that a number is said to be rational if it can be written in the form p/q , where p and q are integers and q ≠ 0.

Taking the case of ‘0’,

Zero can be written in the form 0/1, 0/2, 0/3 … as well as , 0/1, 0/2, 0/3 ..

Since it satisfies the necessary condition, we can conclude that 0 can be written in the p/q form, where q can either be positive or negative number.

Hence, 0 is a rational number.

 

Q 2. Find six rational numbers between 3 and 4.

Solution:

There are infinite rational numbers between 3 and 4.

As we have to find 6 rational numbers between 3 and 4, we will multiply both the numbers, 3 and 4, with 6+1 = 7 (or any number greater than 6)

i.e., 3 × (7/7) = 21/7

and, 4 × (7/7) = 28/7. The numbers in between 21/7 and 28/7 will be rational and will fall between 3 and 4.

Hence, 22/7, 23/7, 24/7, 25/7, 26/7, 27/7 are the 6 rational numbers between 3 and 4.

 

Q 3. Find five rational numbers between 3/5 and 4/5.

Solution:

There are infinite rational numbers between 3/5 and 4/5.

To find out 5 rational numbers between 3/5 and 4/5, we will multiply both the numbers 3/5 and 4/5

with 5+1=6 (or any number greater than 5)

i.e., (3/5) × (6/6) = 18/30

and, (4/5) × (6/6) = 24/30

The numbers in between18/30 and 24/30 will be rational and will fall between 3/5 and 4/5.

Hence,19/30, 20/30, 21/30, 22/30, 23/30 are the 5 rational numbers between 3/5 and 4/5

 

Q 4. State whether the following statements are true or false. Give reasons for your answers.

(i) Every natural number is a whole number.

 

Solution:

True

Natural numbers- Numbers starting from 1 to infinity (without fractions or decimals)

i.e., Natural numbers = 1,2,3,4…

Whole numbers – Numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers = 0,1,2,3…

Or, we can say that whole numbers have all the elements of natural numbers and zero.

Every natural number is a whole number; however, every whole number is not a natural number.

 

(ii) Every integer is a whole number.

Solution:

False

Integers- Integers are set of numbers that contain positive, negative and 0; excluding fractional and decimal numbers.

i.e., integers= {…-4,-3,-2,-1,0,1,2,3,4…}

Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers= 0,1,2,3….

Hence, we can say that integers include whole numbers as well as negative numbers.

Every whole number is an integer; however, every integer is not a whole number.

 

(iii) Every rational number is a whole number.

Solution:

False

Rational numbers- All numbers in the form p/q, where p and q are integers and q≠0.

i.e., Rational numbers = 0, 19/30 , 2, 9/-3, -12/7…

Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers= 0,1,2,3….

Hence, we can say that integers include whole numbers as well as negative numbers.

All whole numbers are rational, however, all rational numbers are not whole numbers.

                              Exercise 1.2

  (Class – 9 Maths NCERT Chapter -1)

 
 
Q 1. State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

Solution:

True

Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are integers and q ≠ 0.

i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000….

Real numbers – The collection of both rational and irrational numbers are known as real numbers.

i.e., Real numbers = √2, √5, , 0.102…

Every irrational number is a real number, however, every real number is not an irrational number.

 

(ii) Every point on the number line is of the form √m where m is a natural number.

Solution:

False

The statement is false since as per the rule, a negative number cannot be expressed as square roots.

E.g., √9 =3 is a natural number.

But √2 = 1.414 is not a natural number.

Similarly, we know that there are negative numbers on the number line, but when we take the root of a negative number it becomes a complex number and not a natural number.

E.g., √-7 = 7i, where i = √-1

The statement that every point on the number line is of the form √m, where m is a natural number is false.

 

(iii) Every real number is an irrational number.

Solution:

False

The statement is false. Real numbers include both irrational and rational numbers. Therefore, every real number cannot be an irrational number.

Real numbers – The collection of both rational and irrational numbers are known as real numbers.

i.e., Real numbers = √2, √5, , 0.102…

Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are integers and q ≠ 0.

i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000….

Every irrational number is a real number, however, every real number is not irrational.

 

Q 2. 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.

 

Solution:

No, the square roots of all positive integers are not irrational.

For example,

√4 = 2 is rational.

√9 = 3 is rational.

Hence, the square roots of positive integers 4 and 9 are not irrational. ( 2 and 3, respectively).

 

Q 3. Show how √5 can be represented on the number line.

Solution:

Step 1: Let line AB be of 2 unit on a number line.

Step 2: At B, draw a perpendicular line BC of length 1 unit.

Step 3: Join CA

Step 4: Now, ABC is a right angled triangle. Applying Pythagoras theorem,

AB2+BC2 = CA2

22+12 = CA2 = 5

⇒ CA = √5 . Thus, CA is a line of length √5 unit.

Step 4: Taking CA as a radius and A as a center draw an arc touching

the number line. The point at which number line get intersected by

arc is at √5 distance from 0 because it is a radius of the circle

whose center was A.

Thus, √5 is represented on the number line as shown in the figure.

 

 
 
Q 4. Classroom activity (Constructing the ‘square root spiral’) : Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion.  Start with a point O and draw a line segment OP1 of unit length. Draw a line segment P1P2 perpendicular to OP1 of unit length (see Fig. 1.9).
 Now draw a line segment P2P3 perpendicular to OP2. Then draw a line segment P3P4 perpendicular to OP3. Continuing in Fig. 1.9 :Constructing this manner, you can get the line segment Pn-1Pn by square root spiral drawing a line segment of unit length perpendicular to OPn-1. In this manner, you will have created the points P2, P3,….,Pn,… ., and joined
 
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)



 them to create a beautiful spiral depicting √2, √3, √4, …
 
 
 
Solution :– 
 
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)



 
 

 

 

Step 1: Mark a point O on the paper. Here, O will be the center of the square root spiral.

Step 2: From O, draw a straight line, OA, of 1cm horizontally.

Step 3: From A, draw a perpendicular line, AB, of 1 cm.

Step 4: Join OB. Here, OB will be of √2

Step 5: Now, from B, draw a perpendicular line of 1 cm and mark the end point C.

Step 6: Join OC. Here, OC will be of √3

Step 7: Repeat the steps to draw √4, √5, √6….

 

                                             EXERCISE  1.3

  (Class – 9 Maths NCERT Chapter -1)

 
 

 

 

Q 1. Write the following in decimal form and say what kind of decimal expansion each has :

    NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

Ans.
       (i)   We have NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)
              ⇒ The decimal expansion of NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)
 is terminating.
 
       (ii)    Dividing 1 by 11, we have:
                     NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)
Note:
               The bar above the digits indicates the block of digits that repeats. Here, the repeating block is 09.
 
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)
 
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

class 9 maths ncert chapter 1

class 9 maths ncert chapter 1

 

Q- 2. You know that 1/7 = 0.142857. Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are, without actually doing the long division? If so, how?

Solution:–

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 

 


Q -3. Express the following in the form p/q, where p and q are integers and q ≠ 0.

(i) Ncert solution class 9 chapter 1-10
 
Solution;–

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

(ii)

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

(iii)

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 

       

Q- 4  Let x = 0.99999 . . . in the form NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)
 Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Solution:– 

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)
 

 

 

 

Q- 5   What can the maximum number of digits be in the repeating block of digits in the decimal expansion of NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 Perform the division to check your answer.

 

Solution;–

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

Q–6. Look at several examples of rational numbers in the form where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

Q–7. Write three numbers whose decimal expansions are non-terminating
non-recurring.

 

Solution:–

 

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

Q–8  Find three different irrational numbers between the rational numbers NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)
 
Solution:–
 

class 9 maths ncert chapter 1

 Q–9   Classify the following numbers as rational or irrational:
 
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)
 
 
Solution:–

 

class 9 maths ncert chapter 1

 

EXERCISE 1.4

  (Class – 9 Maths NCERT Chapter -1)

 
 

 

 
 

1. Classify the following numbers as rational or irrational:

(i) 2 –√5

(ii) (3 +√23)- √23

(iii) 2√7/7√7

(iv) 1/√2

(v) 2π

Solution–:

 

Q– 2   Simplify each of the followings:

(i) (3+√3)(2+√2)

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 
 
 
(ii) (3+√3)(3-√3 )
 

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 

 
(iii) (√5+√2)2
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 
(iv) (√5-√2)(√5+√2)
 
 
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 

Q- 3. Recall, π is defined as the ratio of the circumference (say c) of a 
circle to its diameter, (say d). That is, π =c/d. This seems to 
contradict the fact that π is irrational. How will you resolve
 this contradiction?
 
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 
Q-4. Represent (√9.3) on the number line.

class 9 maths ncert chapter 1

 

class 9 maths ncert chapter 1

 Q-5 Rationalise the denominator of the following;–

 

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 
 

EXERCISE 1.5  (Class – 9 Maths NCERT Chapter -1 )

 
 

 

Q-1 Find :–
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

 

(i)   641/2
       (ii)   321/5
  (iii)   1251/3
 
 
 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 

 

  • NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 
 
 
Q- 2 Find:–
 

 

 (i)   93/2
       (ii)   322/5
       (iii)   163/4
       (iv)   125-1/3
 
 
 
 
 

 

 
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 

Q- 3 Simplify:-
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 

 

 

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)
 
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (New Syllabus)

 

 

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