NCERT Solutions for Class 9 Maths Chapter 1- Number Systems

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems are created by the expert faculty at BYJU’S. These Solutions of NCERT Maths help the students in solving the problems adroitly and efficiently. They also focus on formulating the Solutions of Maths in such a way that it is easy for the students to understand. The NCERT Solutions for Class 9 aim at equipping the students with detailed and step-wise explanations for all the answers to the questions given in the exercises of this Chapter.

In Chapter 1 Number System of Class 9, students are introduced to a lot of important topics that are considered to be very important for those who wish to pursue Mathematics as a subject in their higher classes. Based on these NCERT Solutions, students can practise and prepare for their upcoming exams, as well as endow themselves with the basics of Class 10 for the board exams then. These Maths Solutions of NCERT Class 9 are helpful as they are prepared with respect to the NCERT Syllabus and Guidelines.

Download PDF of NCERT Solutions for Class 9 Maths Chapter 1- Number Systems

 

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Access Answers to Maths NCERT Problems class 9 Chapter 1 – Number Systems

 

Exercise 1.1 Page: 5

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.

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.

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

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 includes whole numbers as well as negative numbers.

Every whole numbers are rational, however, every rational numbers are not whole numbers.

Exercise 1.2 Page: 8

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 numbers are not irrational numbers.

(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, the 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.

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).

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.

Ncert solution class 9 chapter 1-1

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 :

Ncert solution class 9 chapter 1-2

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 them to create a beautiful spiral depicting √2, √3, √4, …

Solution:

Ncert solution class 9 chapter 1-3

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 Page: 14

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

(i) 36/100

Solution:

NCERT Solution For Class 9 Maths Ex-1.3-1

= 0.36 (Terminating)

(ii)1/11

Solution:

NCERT Solution For Class 9 Maths Ex-1.3-2

Ncert solution class 9 chapter 1-4

Ncert solution class 9 chapter 1-5

Solution:

Ncert solution class 9 chapter 1-6

NCERT Solution For Class 9 Maths Ex-1.3-3

= 4.125 (Terminating)

(iv) 3/13

Solution:

NCERT Solution For Class 9 Maths Ex-1.3-4

Ncert solution class 9 chapter 1-7

(v) 2/11

Solution:

NCERT Solution For Class 9 Maths Ex-1.3-5

Ncert solution class 9 chapter 1-8

(vi) 329/400

Solution:

NCERT Solution For Class 9 Maths Ex-1.3-6

= 0.8225 (Terminating)

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?

[Hint: Study the remainders while finding the value of 1/7 carefully.]

Solution:

Ncert solution class 9 chapter 1-9

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 solution class 9 chapter 1-11

Assume that  x = 0.666…

Then,10x = 6.666…

10x = 6 + x

9x = 6

x = 2/3

(ii) 0.470.4\overline{7}

Solution:

0.47=0.4777..0.4\overline{7} = 0.4777..

= (4/10)+(0.777/10)

Assume that x = 0.777…

Then, 10x = 7.777…

10x = 7 + x

x = 7/9

(4/10)+(0.777../10) = (4/10)+(7/90) ( x = 7/9 and x = 0.777…0.777…/10 = 7/(9×10) = 7/90 )

= (36/90)+(7/90) = 43/90

Ncert solution class 9 chapter 1-14

Solution:

Ncert solution class 9 chapter 1-15

Assume that  x = 0.001001…

Then, 1000x = 1.001001…

1000x = 1 + x

999x = 1

x = 1/999

4. Express 0.99999…. in the form p/q . Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Solution:

Assume that x = 0.9999…..Eq (a)

Multiplying both sides by 10,

10x = 9.9999…. Eq. (b)

Eq.(b) – Eq.(a), we get

(10x = 9.9999)-(x = 0.9999…)

9x = 9

x = 1

The difference between 1 and 0.999999 is 0.000001 which is negligible.

Hence, we can conclude that, 0.999 is too much near 1, therefore, 1 as the answer can be justified.

5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17 ? Perform the division to check your answer.

Solution:

1/17

Dividing 1 by 17:

NCERT Solution For Class 9 Maths Ex-1.3-7

Ncert solution class 9 chapter 1-16

There are 16 digits in the repeating block of the decimal expansion of 1/17.

6. Look at several examples of rational numbers in the form p/q (q ≠ 0), 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?

Solution:

We observe that when q is 2, 4, 5, 8, 10… Then the decimal expansion is terminating. For example:

1/2 = 0. 5, denominator q = 21

7/8 = 0. 875, denominator q =23

4/5 = 0. 8, denominator q = 51

We can observe that the terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of only 2 or only 5 or both.

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

Solution:

We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating non-recurring are:

  1. √3 = 1.732050807568
  2. √26 =5.099019513592
  3. √101 = 10.04987562112

8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.

Solution:

Ncert solution class 9 chapter 1-17

Three different irrational numbers are:

  1. 0.73073007300073000073…
  2. 0.75075007300075000075…
  3. 0.76076007600076000076…

9.  Classify the following numbers as rational or irrational according to their type:

(i)√23

Solution:

√23 = 4.79583152331…

Since the number is non-terminating non-recurring therefore, it is an irrational number.

(ii)√225

Solution:

√225 = 15 = 15/1

Since the number can be represented in p/q form, it is a rational number.

(iii) 0.3796

Solution:

Since the number,0.3796, is terminating, it is a rational number.

(iv) 7.478478

Solution:

The number,7.478478, is non-terminating but recurring, it is a rational number.

(v) 1.101001000100001…

Solution:

Since the number,1.101001000100001…, is non-terminating non-repeating (non-recurring), it is an irrational number.

Exercise 1.4 Page: 18

1. Visualise 3.765 on the number line, using successive magnification.

Solution:

Ncert solutions class 9 chapter 1-18

Ncert solution class 9 chapter 1-19

Ncert solutions class 9 chapter 1-20

Exercise 1.5 Page: 24

1. Classify the following numbers as rational or irrational:

(i) 2 –√5

Solution:

We know that, √5 = 2.2360679…

Here, 2.2360679…is non-terminating and non-recurring.

Now, substituting the value of √5 in 2 –√5, we get,

2-√5 = 2-2.2360679… = -0.2360679

Since the number, – 0.2360679…, is non-terminating non-recurring, 2 –√5 is an irrational number.

(ii) (3 +√23)- √23

Solution:

(3 +23) –√23 = 3+23–√23

= 3

= 3/1

Since the number 3/1 is in p/q form, (3 +√23)- √23 is rational.

(iii) 2√7/7√7

Solution:

2√7/7√7 = ( 2/7)× (√7/√7)

We know that (√7/√7) = 1

Hence, ( 2/7)× (√7/√7) = (2/7)×1 = 2/7

Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.

(iv) 1/√2

Solution:

Multiplying and dividing numerator and denominator by √2 we get,

(1/√2) ×(√2/√2)= √2/2 ( since √2×√2 = 2)

We know that, √2 = 1.4142…

Then, √2/2 = 1.4142/2 = 0.7071..

Since the number , 0.7071..is non-terminating non-recurring, 1/√2 is an irrational number.

(v) 2

Solution:

We know that, the value of = 3.1415

Hence, 2 = 2×3.1415.. = 6.2830…

Since the number, 6.2830…, is non-terminating non-recurring, 2 is an irrational number.

2. Simplify each of the following expressions:

(i) (3+√3)(2+√2)

Solution:

(3+√3)(2+√2 )

Opening the brackets, we get, (3×2)+(3×√2)+(√3×2)+(√3×√2)

= 6+3√2+2√3+√6

(ii) (3+√3)(3-√3 )

Solution:

(3+√3)(3-√3 ) = 32-(√3)2 = 9-3

= 6

(iii) (√5+√2)2

Solution:

(√5+√2)2 = √52+(2×√5×√2)+ √22

= 5+2×√10+2 = 7+2√10

(iv) (√5-√2)(√5+√2)

Solution:

(√5-√2)(√5+√2) = (√52-√22) = 5-2 = 3

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?

Solution:

There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…

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

Solution:

Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC=1 unit.

Step 2: Now, AC = 10.3 units. Let the centre of AC be O.

Step 3: Draw a semi-circle of radius OC with centre O.

Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.

Step 5: OBD, obtained, is a right angled triangle.

Here, OD 10.3/2 (radius of semi-circle), OC = 10.3/2 , BC = 1

OB = OC – BC

⟹ (10.3/2)-1 = 8.3/2

Using Pythagoras theorem,

We get,

OD2=BD2+OB2

⟹ (10.3/2)2 = BD2+(8.3/2)2

⟹ BD2 = (10.3/2)2-(8.3/2)2

⟹ (BD)2 = (10.3/2)-(8.3/2)(10.3/2)+(8.3/2)

⟹ BD2 = 9.3

⟹ BD =  √9.3

Thus, the length of BD is √9.3.

Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure.

Ncert solutions class 9 chapter 1-21

5. Rationalize the denominators of the following:

(i) 1/√7

Solution:

Multiply and divide 1/√7 by √7

(1×√7)/(√7×√7) = √7/7

(ii) 1/(√7-√6)

Solution:

Multiply and divide 1/(√7-√6) by (√7+√6)

[1/(√7-√6)]×(√7+√6)/(√7+√6) = (√7+√6)/(√7-√6)(√7+√6)

= (√7+√6)/√72-√62 [denominator is obtained by the property, (a+b)(a-b) = a2-b2]

= (√7+√6)/(7-6)

= (√7+√6)/1

= √7+√6

(iii) 1/(√5+√2)

Solution:

Multiply and divide 1/(√5+√2) by (√5-√2)

[1/(√5+√2)]×(√5-√2)/(√5-√2) = (√5-√2)/(√5+√2)(√5-√2)

= (√5-√2)/(√52-√22) [denominator is obtained by the property, (a+b)(a-b) = a2-b2]

= (√5-√2)/(5-2)

= (√5-√2)/3

(iv) 1/(√7-2)

Solution:

Multiply and divide 1/(√7-2) by (√7+2)

1/(√7-2)×(√7+2)/(√7+2) = (√7+2)/(√7-2)(√7+2)

= (√7+2)/(√72-22) [denominator is obtained by the property, (a+b)(a-b) = a2-b2]

= (√7+2)/(7-4)

= (√7+2)/3

Exercise 1.6 Page: 26

1. Find:

(i)641/2

Solution:

641/2 = (8×8)1/2

= (82)½

= 81 [⸪2×1/2 = 2/2 =1]

= 8

(ii)321/5

Solution:

321/5 = (25)1/5

= (25)

= 21 [⸪5×1/5 = 1]

= 2

(iii)1251/3

Solution:

(125)1/3 = (5×5×5)1/3

= (53)

= 51 (3×1/3 = 3/3 = 1)

= 5

2. Find:

(i) 93/2

Solution:

93/2 = (3×3)3/2

= (32)3/2

= 33 [⸪2×3/2 = 3]

=27

(ii) 322/5

Solution:

322/5 = (2×2×2×2×2)2/5

= (25)2⁄5

= 22 [⸪5×2/5= 2]

= 4

(iii)163/4

Solution:

163/4 = (2×2×2×2)3/4

= (24)3⁄4

= 23 [⸪4×3/4 = 3]

= 8

(iv) 125-1/3

125-1/3 = (5×5×5)-1/3

= (53)-1⁄3

= 5-1 [⸪3×-1/3 = -1]

= 1/5

3. Simplify:

(i) 22/3×21/5

Solution:

22/3×21/5 = 2(2/3)+(1/5) [⸪Since, am×an=am+n____ Laws of exponents]

= 213/15 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]

(ii) (1/33)7

Solution:

(1/33)7 = (3-3)7 [⸪Since,(am)n = am x n____ Laws of exponents]

= 3-21

(iii) 111/2/111/4

Solution:

111/2/111/4 = 11(1/2)-(1/4)

= 111/4 [⸪(1/2) – (1/4) = (1×4-2×1)/(2×4) = 4-2)/8 = 2/8 = ¼ ]

(iv) 71/2×81/2

Solution:

71/2×81/2 = (7×8)1/2 [⸪Since, (am×bm = (a×b)m ____ Laws of exponents]

= 561/2


NCERT Solutions for Class 9 Maths Chapter 1- Number Systems

As the Number System is one of the important topics in Maths, it has a weightage of 6 marks in Class 9 Maths exams. On an average three questions are asked from this unit.

  1. One out of three questions in part A (1 marks).
  2. One out of three questions in part B (2 marks).
  3. One out of three questions in part C (3 marks).

This chapter talks about

  • Introduction of Number Systems
  • Irrational Numbers
  • Real Numbers and their Decimal Expansions
  • Representing Real Numbers on the Number Line.
  • Operations on Real Numbers
  • Laws of Exponents for Real Numbers
  • Summary

List of Exercises

Exercise 1.1 Solutions 4 Questions ( 2 long, 2 short)

Exercise 1.2 Solutions 4 Questions ( 3 long, 1 short)

Exercise 1.3 Solutions 9 Questions ( 9 long)

Exercise 1.4 Solutions 2 Questions ( 2 long)

Exercise 1.5 Solutions 5 Questions ( 4 long 1 short)

Exercise 1.6 Solutions 3 Questions ( 3 long)

NCERT Solutions for Class 9 Maths Chapter 1- Number Systems

NCERT Solutions for Class 9 Maths Chapter 1 – Number System is the first chapter of class 9 Maths. Number System is discussed in detail in this chapter. The chapter discusses the Number Systems and their applications. The introduction of the chapter includes whole numbers, integers and rational numbers.

The chapter starts with the introduction of Number Systems in section 1.1 followed by two very important topics in section 1.2 and 1.3

  • Irrational Numbers – The numbers which can’t be written in the form of p/q.
  • Real Numbers and their Decimal Expansions – Here you study the decimal expansions of real numbers and see whether it can help in distinguishing between rational and irrationals.

Next, it discusses the following topics.

  • Representing Real Numbers on the Number Line – In this the solutions for 2 problems in Exercise 1.4.
  • Operations on Real Numbers – Here you explore some of the operations like addition, subtraction, multiplication and division on the irrational numbers.
  • Laws of Exponents for Real Numbers – Use these laws of exponents to solve the questions.

Key advantages of NCERT Solutions for Class 9 Maths Chapter 1- Number Systems

  • These NCERT Solutions for Class 9 Maths helps you solve and revise the whole syllabus of class 9.
  • After going through the step-wise solutions given by our subject expert teachers, you will be able to score more marks.
  • It follows NCERT guidelines.
  • It contains all the important questions from the examination point of view.
  • It helps in scoring well in Maths in board exams.

2 Comments

  1. Byjus is the best for all subjects

  2. Nice app byju’s

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