NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

NCERT
SOLUTIONS
CLASS - 9th

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

Class : 9th
Subject : Maths
Chapter : 1
Chapter Name : Number Systems

Exercise 1.1

p
Q1 Is zero a rational number? Can you write it in the form q
, where p and q are integers and q
≠ 0?

Answer. Zero is a rational number as it can be represented as etc.
0 0 0
or or
1 2 3

Page : 5 , Block Name : Exercise 1.1

Q2 Find six rational numbers between 3 and 4.

Answer. There are in nite rational numbers in between 3 and 4.
24 32
3 and 4 can be represented as and
8 8

Therefore, rational numbers between 3 and 4 are

25 26 27 28 29 30
, , , , ,
8 8 8 8 8 8

Page : 5 , Block Name : Exercise 1.1

Q3 Find ve rational numbers between 3 5 and 4 5 .

3 4
There are infinite rational numbers between and
5 5

3 3×6 18
= =
5 5×6 30

4 4×6 24
= =
5 5×6 30

Therefore, rational numbers between .
3 4
and
5 5
19 20 21 22 23
, , , ,
30 30 30 30 30

Page : 5 , Block Name : Exercise 1.1

Q4 State whether the following statements are true or false. Give reasons for your answers.
(i) Every natural number is a whole number.

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

(ii) Every integer is a whole number.
(iii) Every rational number is a whole number

Answer. (i) True; since the collection of whole numbers contains all natural numbers.
(ii) False; as integers may be negative but whole numbers are positive. For example: —3 is an
integer but not a whole number.
(iii) False; as rational numbers may be fractional but whole numbers may not be. For Example
: is rational number but not a whole number.
1

5

Page : 5 , Block Name : Exercise 1.1

Exercise 1.2

Q1 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.

Answer. (i) True; since the collection of real numbers is made up of rational and irrational
numbers.
(ii) False; as negative numbers cannot be expressed as the square root of any other number.
(iii) False; as real numbers include both rational and irrational numbers. Therefore, every real
number cannot be an irrational number.

Page : 8 , Block Name : Exercise 1.2

Q2 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.

Answer. If numbers such as √4 = 2, √9 = 3 are considered,
Then here, 2 and 3 are rational numbers. Thus, the square roots of all positive integers are not
irrational.

Page : 8 , Block Name : Exercise 1.2

Q3 Show how √5 can be represented on the number line.

We know that, √4 = 2

2 2
And, √5 = √(2) + (1)

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Mark a point 'A' representing 2 on number line. Now, construct AB Of unit length
perpendicular to OA. Then, taking O as centre and 0B as radius, draw an arc intersecting
number line at C.
C is representing √5.

Page : 8 , Block Name : Exercise 1.2

Exercise 1.3

Q1 Write the following in decimal form and say what kind of decimal expansion each has :
36 1 1
4
(i) 100 (ii) 11 (iii) 8

3 2 329

( iv) 13
(v) 11 (vi) 400

1 ¯
¯¯¯
¯¯
= 0.090909 … … = 0. 09
11

Non-terminating repeating

4 1 33
(iii) = = 4.125
8 8

Terminating
3
= 0.230769230769 … = 0.230769
13

Non-terminating repeating
(iv) 2 ¯
¯¯¯
¯¯
(v) = 0.18181818 … … . = 0. 18
11

Non-terminating repeating

(vi)
329
= 0.8225
400

Terminating

Page : 14 , Block Name : Exercise 1.3

Q2 You know that . Can you predict what the decimal expansions of
1 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
= 0. 142857
7

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 nding the value of 1/7 carefully.]

Answer. Yes. It can be done as follows.

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2 1 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
= 2 × = 2 × 0. 142857 = 0. 285714
7 7

3 1 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
= 3 × = 3 × 0. 142857 = 0. 428571
7 7

4 1
= 4 × = 4 × 0.142857 = 0.571428
7 7

5 1 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
= 5 × = 5 × 0. 142857 = 0.714285
7 7

6 1 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
= 6 × = 6 × 0. 142857 = 0. 857142
7 7

Page : 14 , Block Name : Exercise 1.3

Q3 Express the following in the form p/q , where p and q are integers and q ≠ 0.
(i)0. 6
¯
¯¯ ¯
¯¯ ¯¯
¯¯¯
¯¯¯
(ii) 0.47 (iii) 0. 001

¯
¯¯
(i) 0. 6 = 0.666 …

Let x = 0.666 …

10x = 6.666 …

10x = 6 + x

9x = 6
2
x =
3
¯
¯¯¯
¯¯
(ii) 0. 47 = 0.4777 … . . .

4 0.777
= +
10 10

Let x = 0.777 …

10x = 7.777 …

10x = 7 + x

7
x =
9

4 0.777 … 4 7
+ = +
10 10 10 90

36 + 7 43
= =
90 90
¯¯¯¯¯¯¯¯¯¯¯¯
(iii)0.001 = 0.001001 …

Let x = 0.001001 …

1000x = 1.001001 …

1000x = 1 + x

999x = 1

1
x =
999

Page : 14 , Block Name : Exercise 1.3

Q4 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.

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Answer. Let x = 0.9999….
10x = 9.9999….
10x = 9 + x
9x = 9
x=1

Page : 14 , Block Name : Exercise 1.3

Q5 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.

Answer. It can be observed that,
1 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
= 0. 0588235294117647
17

There are 16 digits in the repeating block Of the decimal expansion of 1

17
.

Page : 14 , Block Name : Exercise 1.3

Q6 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?

Answer. Terminating decimal expansion will occur when denominator q of rational number
p/q is either of 2, 4, 5, 8, 10, and so on...
9
= 2.25
4

11
= 1.375
8

27
= 5.4
5

It can be 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.

Page : 14 , Block Name : Exercise 1.3

Q7 Write three numbers whose decimal expansions are non-terminating non-recurring.

Answer. 3 numbers whose decimal expansions are non-terminating non-recurring are as
follows.
0. 505005000500005000005...
0.7207200720007200007200000...
0.080080008000080000080000008...

Page : 14 , Block Name : Exercise 1.3

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Q8 Find three different irrational numbers between the rational numbers .
5 9
and
7 11

5
= 0.714285
7

9 ¯
¯¯¯
¯¯
= 0. 81
11

3 irrational numbers are as follows.

0.73073007300073000073. .

0.75075007500075000075. .

0.79079007900079000079 …

Page : 14 , Block Name : Exercise 1.3

Q9 Classify the following numbers as rational or irrational :
(i) √23 (ii) √225 (iii) 0.3796

(iv) 7.478478 (v) 1.101001000100001 …

Answer. (i) √23 = 4.79583152331 …...
As the decimal expansion of this number is non-terminating non-recurring, therefore, it is an
irrational number.
√225 15
= 15 =
1

It is a rational number as it can be represented in form.
p

q

(iii) 0.3796
As the decimal expansion of this number is terminating, therefore, it is a rational number.
¯¯
¯¯¯
¯¯¯
( iv) 7.478478 … = 7. 478

As the decimal expansion Of this number is non-terminating recurring, therefore, it is a
rational number.
(v) As the decimal expansion of this number is non-terminating non-repeating, therefore, it is
an irrational number.

Page : 14 , Block Name : Exercise 1.3

Exercise 1.4

Q1 Visualise 3.765 on the number line, using successive magni cation.

Answer. 3.765 can be visualised as in the following steps.

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Page : 18 , Block Name : Exercise 1.4

Q2 Visualise 4. 26 on the number line, up to 4 decimal places.
¯
¯¯¯
¯¯

Answer. 4. 26 = 4.2626 …
¯
¯¯¯
¯¯

4.2626 can be visualised as in the following steps.

Page : 18 , Block Name : Exercise 1.4

Exercise 1.5

Q1 Classify the following numbers as rational or irrational:
2√7
(i) 2 − √5 (ii) (3 + √23) − √23 (iii)
7√7
1

√2
(iv) (v)2π

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2−√5
(i) = 2 − 2.2360679 …
= −0.2360679 …

As the decimal expansion Of this expression is non-terminating non-recurring, therefore, it is
an irrational number.
(3+√23)−√23 3
( ii) = 3 =
1

As it can be represented in p/q form, therefore, it is a rational number.
2√7
(iii) =
2

7
7√7

As it can be represented in p/q form, therefore, it is a rational number.
(iv) 1 √2
= = 0.7071067811 …
√2 2

As the decimal expansion Of this expression is non-terminating non-recurring, therefore, it is
an irrational number.
(v) 2 n=2(3.1415 \dots)\) = 6.2830….
As the decimal expansion Of this expression is non-terminating non-recurring, therefore, it is
an irrational number.

Page : 24 , Block Name : Exercise 1.5

Q2 Simplify each of the following expressions:
(3+√3)(2+√2)
(i) (ii)(3 + √3)(3 − √3)

2
(iii)(√5 + √2) (√5 − √2)(√5 + √2)
(iv)

( i) (3 + √3)(2 + √2) = 3(2 + √2) + √3(2 + √2)

= 6 + 3 √2 + 2 √3 + √6
(3+√3)(3−√3) 2 2
(ii) = (3) − (√3)

= 9 − 3 = 6
2 2 2
(ii) (√5 + √2) = (√5) + (√2) + 2(√5)(√2)

= 5 + 2 + 2√10 = 7 + 2√10

2 2 2
(ii)(√5 + √2) = (√5) + (√2) + 2(√5)(√2)

= 5 + 2 + 2√10 = 7 + 2√10

2 2
(v)(√5 − √2)(√5 + √2) = (√5) − (√2)

= 5 − 2 = 3

Page : 24 , Block Name : Exercise 1.5

Q3 Recall, π is de ned 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?

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Answer. There is no contradiction. When we measure a length with scale or any other
instrument, we only obtain an approximate rational value. We never obtain an exact value. For
this reason, we may not realise that either c or d is irrational. Therefore, the fraction c/d is
irrational. Hence, n is irrational.

Page : 24 , Block Name : Exercise 1.5

Q4 Represent √9.3 on the number line.

Answer. Mark a line segment 0B 9.3 on number line. Further, take BC of 1 unit. Find the mid-
point D of OC and draw a semi-circle on OC while taking D as its centre. Draw a perpendicular
to line OC passing through point B. Let it intersect the semi-circle at E. Taking B as centre and
BE as radius, draw an arc intersecting number line at F. BF is √9.3 .

Page : 24 , Block Name : Exercise 1.5

Q5 Rationalise the denominators of the following:
1 1
(i) (ii)
√7 √7−√6

1 1
(iii) (iv)
√5+√2 √7−2

1×√7 √7
Answer. (i)
1
= =
√7 7
1×√7

1 1
(ii) =
√7 − √6 (√7 − √6)(√7 + √6)

√7 + √6
=
2 2
(√7) − (√6)

√7 + √6 √7 + √6
= = = √7 + √6
7 − 6 1

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1 (√5−√2)
(iii) =
√5+√2 (√5+√2)(√5−√2)

√5−√2 √5−√2
= =
(√5) −(√2)
2 2 5−2

√5−√2
=
3
1 1
(iv) =
√7−2 (√7−2)(√7+2)

√7+2
=
2 2
(√7) −(2)

√7+2 √7+2
= =
7−4 3

Page : 24 , Block Name : Exercise 1.5

Exercise 1.6

1 1 1

Q1 Find : (i) 64 ( ii) 32 2 5 (iii) 125 3 .

Answer. (i)
1
1
6 2
64 2 = (2 )
1 m n m
6. [(a ) = a ]
= 2 2

3
= 2 = 8

(ii)
1
1
5 5
32 5 = (2 )
\)
m n m
[(a ) = a ]
1
5
= (2) 5

1
= 2 = 2

(iii)
1
1
3 3
(125) 3 = (5 ) n
m m
[(a ) = a ]
1
3
= 5 3

1
= 5 = 5

Page : 26 , Block Name : Exercise 1.6

Q2 Find:
3 2 3

(i) 2
( ii) 32 5 (iii) 16 4

−1

(iv) 125 3

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(i)
3
3
2 2
9 2 = (3 ) n
m m
[(a ) = a ]
3
2
= 3 2

3
= 3 = 27

(ii)
2
2
5 5
(32) 5 = (2 ) n
m m
[(a ) = a ]
2
5×
= 2 5

2
= 2 = 4
3
3
4 4
(16) 4 = (2 )

(iii) m n m
3
4+ [(a ) = a ]
= 2 4

3
= 2 = 8

(iv)
−1
1 −n 1
(125) 3 = [a = ]
1 an
(125) 3

1
=
1

3 3
(5 )

1 m n m
= [(a ) = a ]
1
3
5 3

1
=
5

Page : 26 , Block Name : Exercise 1.6

Q3 Simplify:
1
2 1
1 7 11 2
(i) 2 3
⋅ 2 5
(ii) ( 3
) (iii)
1
3
11 4

1 1

( iv) 7 2 ⋅ 8 2 .

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Page : 26 , Block Name : Exercise 1.6

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