NCERT Solutions Class 9 Mathematics Chapter 1 Number Systems

NCERT Solutions Class 9 Mathematics Chapter 1 Number Systems have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 9 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 9 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 9 Mathematics are an important part of exams for Class 9 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 9 Mathematics and also download more latest study material for all subjects. Chapter 1 Number Systems is an important topic in Class 9, please refer to answers provided below to help you score better in exams

Chapter 1 Number Systems Class 9 Mathematics NCERT Solutions

Class 9 Mathematics students should refer to the following NCERT questions with answers for Chapter 1 Number Systems in Class 9. These NCERT Solutions with answers for Class 9 Mathematics will come in exams and help you to score good marks

Chapter 1 Number Systems NCERT Solutions Class 9 Mathematics

Exercise 1.1

Q.1) Is zero a rational number? Can you write it in the form π‘/π‘ž, where p and q are integers and π‘ž β‰  0 ?
Sol.1) Yes 0 is a rational number because it can be written in form of 0/1, 0/2.......... π‘Ž/0 and so on And according to definition of rational number which can be expressed in term form of π‘/π‘ž when q is non zero integer and p is a integer . And here 0 is an integer and 1,2 .. are nonzero integers.
Hence, 0 is a rational number

Q.2) Find six rational numbers between 3 and 4.
Sol.2) There are infinite rational numbers in between 3 and 4. 3 and 4 can be represented as
24/8 and 32/8 respectively. Therefore, six 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 3/5 and 4/5.
Sol.3) There are infinite rational numbers in between 3/5 and 4/5
3/5 = 3Γ—6/5Γ—6 = 18/30 
4/5 = 4Γ—6/5Γ—6 = 24/30
Therefore, five rational numbers between
3/5 and 4/5 are 19/30, 20/30, 21/30, 22/30, 23/30

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.
Sol.4) (i) Every natural number is a whole number.
True, since the collection of whole numbers contains all natural numbers & in addition zero.
(ii) Every integer is a whole number.
False, negative integers are not whole number.
(iii) Every rational number is a whole number.
False, numbers such as 2/3, 3/4, βˆ’3/5 etc., are rational numbers but not whole numbers.

Exercise 1.2

Q.1) 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 βˆšπ‘š, where π‘š is a natural number.
(iii) Every real number is an irrational number
Sol.1) (i) True,
since the collection of real numbers is made up of rational and irrational numbers.
(ii) False,
since positive number cannot be expressed as square roots.
(iii) False,
as real numbers include both rational and

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
Sol.2) No, the square roots of all positive integers are not irrational. For example √4 = 2 which is a rational number.

Q.3) Show how √5 can be represented on the number line.
Sol.3) Step 1: Let AB be a line of length 2 unit on number line.
Step 2: At B, draw a perpendicular line BC of length 1 unit. Join CA.
Step 3: Now, ABC is a right angled triangle.
Applying Pythagoras theorem,
𝐴𝐡2 + 𝐡𝐢2 = 𝐢𝐴2
β‡’ 22 + 12 = 𝐢𝐴2
β‡’ 𝐢𝐴2 = 5
β‡’ 𝐢𝐴2 = √5
Thus, 𝐢𝐴 is a line of length √5 unit.
Step 4: Taking CA as a radius and A as a centre 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 centre was A.
Thus, √5 is represented on the number line as shown in the figure:

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Exercise 1.3

Q.1) Write the following in decimal form and say what kind of decimal expansion each has:
i) 36/100
ii) 1/11
iii) 4(1/8)
iv) 3/13
v) 2/11
vi) 329/400

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Q.3) Express the following in the form π‘/π‘ž, where p and q are integers and π‘ž β‰  0.

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Sol.3) (i) 0.6 = 0.666...
Let π‘₯ = 0.666. . .
10π‘₯ = 6.666. . .
10π‘₯ = 6 + π‘₯
9π‘₯ = 6
π‘₯ = 2/3

(ii) 0.47 = 0.4777. . .
= 4/10 + 0.777/10
𝐿𝑒𝑑 π‘₯ = 0.777 …
10π‘₯ = 7.777 …
10π‘₯ = 7 + π‘₯
π‘₯ = 7/9
4/10 + 0.777/10
= 4/10 + 7/90
= 36/90 + 7/90
= 43/90

(iii) 0.001 = 0.001001. ..
Let π‘₯ = 0.001001 …
1000π‘₯ = 1.001001 …
1000π‘₯ = 1 + π‘₯
999π‘₯ = 1
π‘₯ = 1/999

Q.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.
Sol.4) Let x = 0.9999 …
10x = 9.9999 …
10x = 9 + x
9x = 9
x = 1
The difference between 1 and 0.999999 is 0.000001 which is negligible. Thus, 0.999 is too much near 1, Therefore, the 1 as answer can be justified.

Q.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.
Sol.5) 1/17 = 0.0588235294117647
There are 16 digits in the repeating block of the decimal expansion of 1/17.
Division Check:

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= 0.0588235294117647

Q.6) Look at several examples of rational numbers in the form p/q (q β‰  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?
Sol.6) We observe that when π‘ž 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 observed that terminating decimal may be obtained in the situation where prime factorisation of the denominator of the given fractions has the power of 2 only or 5 only or both.

Q.7) Write three numbers whose decimal expansions are non-terminating non-recurring
Sol.7) Three numbers whose decimal expansions are non-terminating non-recurring are:
0.303003000300003 …
0.505005000500005 …
0.7207200720007200007200000 …

Q.8) Find three different irrational numbers between the rational numbers 5/7 and 9/11.
Sol.8) 5/7 = 0.714285
9/11 = 0.81
Three different irrational numbers are:
0.73073007300073000073 …
0.75075007300075000075 …
0.76076007600076000076 …

Q.9) Classify the following numbers as rational or irrational:
(i) √23 (ii) √225 (iii) 0.3796 (iv) 7.478478
(v) 1.101001000100001 …
Sol.9) (i) √23 = 4.79583152331 …
Since the number is non-terminating non-recurring therefore, it is an irrational number.
(ii) √225 = 15 = 15/1
Since the number is rational number as it can represented in p/q form.
(iii) 0.3796
Since the number is terminating therefore, it is a rational number.
(iv) 7.478478 = 7.478
Since this number is non-terminating recurring, therefore, it is a rational number.
(v) 1.101001000100001 …
Since the number is non-terminating non-repeating, therefore, it is an irrational number.

Exercise 1.4

Q.1) Visualise 3.765 on the number line using successive magnification.
Sol.1)
 We know that the number 3.765 will lie between 3.764 and 3.766.
We know that the numbers 3.764 and 3.766 will lie between 3.76 and 3.77.
We know that the numbers 3.76 and 3.77 will lie between 3.7 and 3.8.
We know that the numbers 3.7 and 3.8 will lie between 3 and 4.
Therefore, we can conclude that we need to use the successive magnification, after locating numbers 3 and 4 on the number line.

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Q.2) Visualise 4.26 on the number line, up to 4 decimal places.
Sol.2) We know that the number 4. Μ…2Μ…6Μ…Μ… can also be written as 4.262.
We know that the number will lie between 4.261 and 4.263.
We know that the numbers 4.261 and 4.263 will lie between 4.26 and 4.27.
We know that the numbers 4.26 and 4.27 will lie between 4.2 and 4.3.
We know that the numbers 4.2 and 4.3 will lie between 4 and 5.
Therefore, we can conclude that we need to use the successive magnification, after locating numbers 4 and 5 on the number line.
4.26 = 4.2626

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Exercise 1.5

Q.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œ‹
Sol.1)
(i) 2 – √5 = 2 – 2.2360679 … = – 0.2360679 …
Since the number is non-terminating non-recurring therefore, it is an irrational number.

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Since the number is rational number as it can represented in p/q form.
(iv) 1/√2 = √2/2 = 0.7071067811 …
Since the number is is non-terminating non-recurring therefore, it is an irrational number.
(v) 2πœ‹ = 2 Γ— 3.1415 … = 6.2830 …
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

Q.2) 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)
Sol.2) (i) (3 + √3) (2 + √2)
β‡’ 3 Γ— 2 + 2 + √3 + 3√2 + √3 Γ— √2
β‡’ 6 + 2√3 + 3√2 + √6

(ii) (3 + √3) (3 – √3)                 [∡ (a + b) (a – b) = a2– b2]]
β‡’ 32 – (√3)2
β‡’ 9 – 3
β‡’ 6

(iii) (√5 + √2)2
[∡ (a + b)2
= a2 + b2 + 2ab]
β‡’ (√5)2
+ (√2)2
+ 2 Γ— √5 Γ— √2
β‡’ 5 + 2 + 2 Γ— √5 Γ— 2
β‡’ 7 + 2√10

(iv) (√5 – √2) (√5 + √2)                 [∡ (a + b) (a – b) = a2 βˆ’ b2]
β‡’ (√5)2
βˆ’ (√2)2
β‡’ 5 – 2
β‡’ 3

Q.3) Recall, πœ‹ is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, πœ‹ = π‘/𝑑. This seems to contradict the fact that πœ‹ is irrational. How will you resolve this contradiction?
Sol.3) 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 realise that either c or d is irrational. The value of Ο€ is almost equal to 22/7 or 3.142857 …

Q.4) Represent √9.3 on the number line.
Sol.4) Step 1: Draw a line segment of unit 9.3. Extend it to C so that BC is of 1 𝑒𝑛𝑖𝑑.
Step 2: Now, 𝐴𝐢 = 10.3 𝑒𝑛𝑖𝑑𝑠. Find the centre of AC and name it as O.
Step 3: Draw a semi-circle with radius OC and centre O.
Step 4: Draw a perpendicular line BD to AC at point B which intersect the semicircle at D.
Also, Join OD.
Step 5: Now, OBD is a right angled triangle.

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β‡’ 𝐡𝐷2 = 9.3
β‡’ 𝐡𝐷2 = √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.

Q.5) Rationalise the denominators of the following:

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