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Mathematics Class 9
NCERT SOLUTIONS
(Detailed Step-Wise as per CBSE Guidelines)
Number Systems
𝐩
Question 1. Is zero rational number? Can you write it in the form 𝐪 , where p and q are integers and q
≠ 0?
0
0
0
p
Ans. Yes, zero is a rational number as 0 = 1 or 2 or −1. This is in the form q , q ≠ 0.
Question 2. Find six rational numbers between 3 and 4.
Ans. For six rational numbers between 3 and 4, 3 =
and 4
22
are 7
,
23
7
,
24
7
,
25
7
,
26
7
27
, 7
21
7
and 4 =
28
7
. Six rational numbers between 3
or another set is 3.1, 3.2 , 3.3 , 3.4 , 3.5 , 3.6 , There can be other
set of rational numbers also.
Question 3. Find five rational numbers between
Ans. We have
3
5
18
30
=
and
4
5
=
24
30
𝟑
𝟓
and
𝟒
𝟓
.
. Five rational numbers between
3
5
and
4
5
are
19
30
,
20
30
,
21
30
,
22
30
23
, 30
.There can be other set of rational numbers. also .
Question 4. State whether the following statements are true or false. Give reasons for your answers .
(1) Every natural number is a whole number.
(2) Every integer is a whole number .
(3) Every rational number is a whole number.
Ans. (i) True , as the set of whole numbers contains all the natural numbers.
(ii) Flase, as negative integer , e. g. ., -2 is not a whole number.
2
(iii) Flase , as 3 is a rational number but not a whole number.
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Question 5 . State whether the following statements are true or false. Justify your answers.
(i)
Every irrational numbers is a real numbers.
(ii)
Every point on the number line is of the form 𝒎 , where m is a natural numbers.
(iii)
Every real numbers is an irrational number.
Ans . (i) true , as real number consist of rational and irrational numbers.
(ii) False, as
3
2
on the number line cannot be a squard root of a natural number. Also, a negative,
as 𝑚 represents a positive value.
(iii)Flase, as 2 is a real number but not an irrational mumber.
Question 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 tational number.
Ans . No . for example,
9 =3,
16 = 4 , etc ., 3 , 4 are rational numbers.
Question 7. Show how 𝟓 can be represented on the number line.
Ans . for 5 , we have 5 = 5 × 1. On a number line , taken o at position 0 and OA = 5 and OB = 1. With
AB as diameter draw a semicircle. Draw OP ! AB meeting semicircle at P . With O as centre and OP radius
an are is drawn meeting the number line at Q.
Then , OQ = OP =
5 and Q represents 5 on the number line.
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Question 8 . 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 OP 1 of unit length. Draw a line segment P1 P2 perpendicular to
OP 1 of unit length (se figure). Now draw a line segment p 2 P3 perpendicular op 2 of unit length .
Than , draw a line segment P3 P4 perpendicular to op 3 of unit length . Continuing in this maner , you
can get the line segment Pn - 1P n by drawing a line segment of unit length perpendicular to OP n – 1. In
this manner , you will have created the points P2’ P 3 ……., P n , …….., and joined them to create a
beautiful spiral depicting
2 , 3 ,
4 , …… .
Ans.
Question .9 Write the following in decimal form and say what kind of decimal xansion each has:
𝟑𝟔
𝟏𝟎𝟎
𝟐
𝟏𝟏
(i)
(ii)
Ans. (i)
(ii)
(iii)
36
100
(ii)
(vi)
𝟏
𝟏𝟏
𝟑𝟐𝟗
𝟒𝟎𝟎
𝟏
(iii) 4 𝟖
(iv)
𝟑
𝟏𝟑
.
= 0.36 , terminating decimal expansion.
1
11
1
48
= 0.090909 …. =
=
33
8
, non – terminating prating decimal expansion.
0.09
= 4.125 , terminating decimal expansion.
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3
= 0.230769230769 ….. = 0.230769 , non – terminating repeating decimal expansion.
13
2
=0.181818 ….. = 0.18 , non – terminating repeating decimal expansion.
11
329
= 0.8225 , terminating decimal expansion.
400
(iv)
(v)
(vi)
𝟏
Question . 10 You know that 𝟕 = 0.142857. Can you predict what the decimal expansions of
𝟒
𝟕
,
𝟓
𝟕
,
𝟔
𝟕
𝟐
𝟕
,
𝟑
𝟕
are without actually doing the long division ? If so , how?
[ Hint : Study the remainders while funding the value of
𝟏
𝟕
carefully . ]
Ans. Yes , we can predict the required decimal expansions . we are given ,
𝟏
𝟕
= 0.142857
On dividing 1 by 7, we find that the remainders repeat after six divisions, therefore, the quotient
has a repeating
1
𝟐
𝟑
𝟒
𝟓
block of six digits in the decimal expansion of 7. So, to obtain decimal expansions of 𝟕 , 𝟕 , 𝟕 , 𝟕
𝟔
and 𝟕 : we multiply 142857 by 2, 3, 4, 5 and 6 respectively, to get the integral part and in the
decimal part, we take block of six repeating digits in each case. Hence, we get
𝟐
𝟕
= 2 x 7 = 0.285714 7 7
1
𝟑
𝟕
= 3 x 7 = 0.428571 7 7
𝟒
𝟕
= 4 x 7 = 0. 571428 7 7
𝟓
𝟕
=5x
1
1
𝟔
and 𝟕
1
7
= 0.714285
1
= 6 x 7 = 0.857142.
Question . 11 Express the form p/q , where p and q are integers and q ≠ 0 :
(i)
0.6
Ans. (i) Let x = 0.6
or x = 0.666...
10 x = 6.666…
=>
9x = 6
(ii)
0.47
(iii)
0.001.
….(ii) [On multiplying (i) by 10]
[Subtracting (i) form (ii) ]
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,
2
∴
x=3.
(ii) Let
x = 0.47
or
x = 0.4777 ….
Multiplying (i) by 10 , we get
10 x = 4.777 …..
Subtracting equation (ii) form equation (iii) ,we get
90 x = 43
∴
x=
(iii)
43
.
90
Let
or
x = 0.001
x = 0.001001…
…(i)
⇒
1000 x = 1.001001…
…(ii)
[On multiplying (i) by 1000]
⇒
∴
999x 1
x =
[ On subtracting (i) from (ii)]
1
.
999
𝑝
𝑞
Question .12 Express 0.99999... in the form . Are you surprised by your answer? With your teacher and
classmates, discuss why the answer makes sense.
Ans.
Let
x = 0.99999….
⇒
⇒
10 x 9.99999….
9 x =9
∴
… (ii) [On multiplying (i) by 10]
[On Subtracting (i) form (ii)]
x = 1.
Yes, we are surprised by our answer.
Question .13 What can the maximum number of digits be in the repeating block lf digits in the decimal
expansion of
𝟏
?
𝟏𝟕
Perform the division to check your answer.
Ans .
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1
Thus, 17 = 0.0588235294117647
Hence, the required number of digits in the repeating block is 16.
Question.14 . Look at several examples of rational numbers in the form
𝒑
𝒒
(q ≠ 0), where p and q are
integrs with no common factors other than 1 and having terminating dcimal representations
(expansions ). Can you guess whatproperty q must satisfy ?
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3 4
7
9
Ans . examples are 4 ,5 , 8 , 10 , etc.
q have only powers of 2 or powers of 5 or both .
Question .15 Write three numbers whose decimal expansions are non – terminating non – recurring.
Ans . (i) 2.01011011101111011111……. (ii) 0.03003000300003…….
(ii)
4.12112111211112…..
Question.16 Find three different irrational numbers between the rational numbers
Ans Given :
5
7
= 0.714285 and
9
11
𝟓
𝟕
and
𝟗
𝟏𝟏
.
= 0.81
We can have irrational numbers as 0.72072007200070….; 0.801001800018…….;
0.74301010010001……;
Question 17. Classify the following numbers as rational or irrational:
𝟐𝟑
(ii) 𝟐𝟐𝟓
(iii) 0.3796
(IV) 7.478478…..
(iii)
𝟏. 𝟏𝟎𝟏𝟎𝟎𝟏𝟎𝟎𝟎𝟏𝟎𝟎𝟎𝟎𝟏 … …
Ans . (i) 23 . As it is square root of a prime number , so , irrational number .
(I)
(ii)
(iii)
(iv)
(v)
225 = 15 , rational number .
0.3796 , terminating decimal , so rational number .
7.478478…..= 7.478 , non – terminating repeating (recurring ) , so rational number.
1.101001000100001….. non – terminating non – repeating,so irrational number.
Question 18. Visualize 3.765 on the number line, using successive magnification.
Ans . (i) We notice the number 3.7 lies between 3 and 4. So, first we locate numbers 3 and 4 on
number line and divide the portion into ten equal parts and locate 3.7 and 3.8. [Refer step 2J
(ii) Further 3.76 lies between 3.7 and 3.8. So, we magnify 3.7 and 3.8 and divide the portion into
ten equal parts and locate 3.76 and 3.77. [Refer step 3J
(iii) Further 3.765 lies between 3.76 and 3.77. So, we magnify 3.76 and 3.77 and divide the
portion into ten equal parts and locate 3.765. [Refer step 4)
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Point O in step 4 reprints the number 3.765 on the number one.
Question 19. Visualize 4.26 on the number line, up to 4 decimal places.
Ans . 4.26 = 4.2626262626 .........
(H) Visualize 4 and 5 as 4.26 lies between 4 and 5 and divide portion in ten equal parts and
locate 4.2. [Refer step 2J
(ii) Visualize 4.2 and 4.3 as 4.26 lies between 4.2 and 4.3 and divide portion in ten equal parts
and locate 4.26. [Refer step 3]
(iii) Visualize 4.26 and 4.27 as 4.262 lies between 4.26 and 4.27 and divide portion in ten equal
parts and locate 4.262. [Refer step 41
(iv) Visualize 4.262 and 4.263 as 4.2626 lies between 4.262 and 4.263 and divide portion in ten
equal parts and locate 4.2626. [Refer step 5]
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Point 0 in step 5 represents the number 4.26 on the number line.
Question .20 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.
(i) 2 - 𝟓 is an irrational number, as difference of a rational and an irrational number is
irrational.
(ii) (3 + 23)- 23 = 3 + 23 - 23 = 3, is a rational number.
Ans .
(iii)
(iv)
1
2
=
2 7
7 7
2
2
2
= 7, is a rational number.
an irrational number, as divisors of an irrational number by a non-zero
rational number is irrational.
(V)
2𝜋, irrational number, as 𝜋 is an irrational number and multiplication of a rai >nal and an
irrational number is irrational.
QUESTION .21 Simplify each of the following expressions:
(I)
(3 + 𝟑) (2 + 𝟐 ) (ii) (3 + 𝟑 ) (3 - 𝟑 )
(II)
( 𝟓 + 𝟐 )2
(iv) ( 𝟓 - 𝟐) ( 𝟓 + 𝟐)
ANS. (I) (3 + 3)(2 + 2 ) =6 +3 2 + 2 3 + 6 .
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(II)
(3 + 3 ) (3 - 3 ) = (3)2 - ( 3)2 = 9 – 3 = 6.
(III)
( 5 + 2 )2 = ( 5)2 + 2. 5 . 2 +( 2 )2 =5 +2 10 + 2 =7 +2 10 .
5 - 2) ( 5 + 2) = ( 5 )2 –( 2 )2 = 5 – 2 =3.
(IV)
Question. 22 Recall, 𝝅 is defined as the ratio of the circumference (sayc) of a circle to its diameter
𝒄
(say d). That is, 𝝅 = 𝒅 .This seems to contradict the fact that k is irrational. How will you resolve this
contradiction?
Ans .On measuring c with any device, we get only approximate measurement. Therefore, n is an
irrational.
Question . 23 Represent 𝟗. 𝟑 on the number line.
Ans . 9.3 = 9.3 × 1
Let position 0 be represented by O on the number line. Let OA = 9.3 and OB = 1. With AB as diameter
draw a semicircle. Draw OP perpendicular to AB, meeting the semicircle at P. Then OP = 9.3 . With O as
centre and OP as radius draw an arc to meet the number line at Q on
the
positive side. Then, OQ = 9.3 and the point Q thus obtained represents 9.3 .
Question .24 Rationalise the denominators of the following:
(i)
𝟏
𝟕
Ans . (i)
(ii)
(iii)
(iv)
(ii)
1
7
1
1
7
=
7− 6
1
𝟏
(iii)
𝟕+ 𝟔
(iv)
𝟏
.
𝟕−𝟐
7
.
7
=
=
7+ 6
7−6
= 7 + 6 .
( 7 – 6 )( 7+ 6)
=
5+ 2
1
7−2
=
𝟏
𝟕− 𝟔
7
×
7
7+ 6
5− 2
=
5− 2
5−2
5− 2
3
=
.
( 5+ 2 )( 5− 2)
=
7 +2
7 –2 ( 7 +2)
7 +2
7−4
=
Question 25.
Find: (i) 641/2 (ii) 321/5 (iii) 125 1/3.
Ans.
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=
7 +2 .
3
Question 26.
Find: (i) 93/2
Ans.
(ii) 322/5
(iii) 163/4
(iv) 125-1/3
Question 27.
Ans.
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