NCERT Basics : Class 9

NCERT QUESTIONS WITH SOLUTIONS

EXERCISE : 1.1 3. Find five rational numbers between 3/5
1. Is zero a rational number? Can you write and 4/5.
it in the form p/q, where p and q are Sol. Let
3 (n + 1) 3 6 18
integers and q  0? =  =
5 n +1 5 6 30
Sol. Yes, zero is a rational number. We can 4 (n + 1) 4 6 24
=  =
5 n +1 5 6 30
write zero in the form p/q where p and q
So, required rational numbers are
are integers and q  0. 19 20 21 22 23
, , , ,
30 30 30 30 30
0 0 0
So, 0 can be written as = = etc. 4. State whether the following statements
1 2 3
are true or false? Give reasons for your
2. Find six rational numbers between 3 and 4.
answers.
Sol. First rational number between 3 and 4 is (i) Every natural number is a whole
3+ 4 7 number.
= = (ii) Every integer is a whole number.
2 2
(iii) Every rational number is a whole
Similarly, other numbers are
number.

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7 Sol. (i) True, the collection of whole
3+
2 = 13 numbers contains all natural
2 4 numbers.
13 (ii) False, –2 is not a whole number.
3+
4 = 25 1
(iii) False, is an integer but a rational
2 8 2
25 number but not a whole number.
3+
8 = 49 EXERCISE : 1.2
2 16 1. State whether the following statements
49 are true or false? Justify your answers.
3+ (i) Every irrational number is a real
16 = 97
2 32 number.
97 (ii) Every point on the number line is of
+3
32 193 the form m , where m is a natural
=
2 64 number.
7 13 25 49 97 193 (iii) Every real number is an irrational
So, numbers are , , , , ,
2 4 8 16 32 64 number.

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Sol. (i) True, since collection of real EXERCISE : 1.3
numbers consists of rationals and 1. Write the following in decimal form and
say what kind of decimal expansion each
irrationals. has :
(ii) False, because no negative number 36 1
(i) (ii)
100 11
can be the square root of any natural 1 3
(iii) 4 (iv)
number. 8 13
2 329
(iii) False, 2 is real but not irrational. (v) (vi)
11 400
2. Are the square roots of all positive 36
Sol. (i) = 0.36
100
integer's irrational? If not, give an
(Terminating)
example of the square root of a number 1
(ii) = 0.090909.......
that is a rational number. 11
(Non-Terminating Repeating)
Sol. No, 4 = 2 is a rational number. 11 1.00000 0.090909....
–99
3. Show how 5 can be represented on the 100
99
100
number line. 99
1
Sol. 5 on Number line. 1 33
(iii) 4 = = 4.125
OABC is unit square. 8 8
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(Terminating decimal)
So, OB = 12 + 12 = 2 3
(iv) = 0.230769230769......
13
OD = ( 2)2 + 1 = 3 = 0.230769
(Non-Terminating repeating)
OE = ( 3)2 + 1 = 2 2
(v) = 0.1818.......
11
OF = (2)2 + 1 = 5
= 0.18 (Non-Terminating repeating)
E 329
1
D (vi)
F 1 3 1
400
2 C 400 329.0000 0.8225
B
5 2 1 3200
N 900
O 1 A 2 3 800
5 1000
800
Using compass we can cut arc with centre 2000
2000
O and radius = OF on number line. ON is ×
329
required result. = 0.8225 (Terminating)
400

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NCERT Basics : Class 9
1 (iii) Let x = 0.001 = 0.001001001... ... (1)
2. You know that = 0.142857 . Can you
7
Multiply both sides by 1000
predict what the decimal expansions of
2 3 4 5 6 1000x = 1.001 ... (2)
, , , , are, without actually doing
7 7 7 7 7
Subtract (1) from (2)
the long division? If so, how?
Sol. Yes, we can predict decimal explain 999x = 1
without actually doing long division

method as x=
999
2 1
= 2  = 2 × 0.142857 = 0.285714
7 7 p
4. Express 0.99999..... in the form . Are
3 1 q
= 3  = 3  0.142857 = 0.428571
7 7
you surprised by your answer ? With your
4 1
= 4  = 4  0.142857 = 0.571428 teacher and classmates discuss why the
7 7
5 1 answer makes sense.
= 5  = 5  0.142857 = 0.714285
7 7
6 1 Sol. Let x = 0.999.... ... (1)
= 6  = 6  0.142857 = 0.857142
7 7 Multiply both sides by 10 we get
3. Express the following in the form p/q,
10x = 9.99.... ... (2)
where p and q are integers and q  0.
(i) 0.6 Subtract (1) from (2)

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(ii) 0.47 9x = 9
(iii) 0.001  x=1
Sol. (i) Let x = 0.6666........ … (1)
Multiplying both the sides by 10 
0.9999.... = 1 =
10x = 6.666.......... … (2) 1
Subtract (1) from (2)  p = 1, q = 1
10x – x = (6.6666......) – (0.6666.....)
5. What can the maximum number of digits
6 2
 9x = 6  x = =
9 3 be in the repeating block of digits in the
(ii) Let x = 0.47 = .4777... 1
Multiply both sides by 10 decimal expansion of ? Perform the
17
10x = 4.7 ... (1)
division to check your answer.
Multiply both sides by 10
100x = 47.7 ... (2) Sol. Maximum number of digits in the
Subtract (1) from (2) repeating block of digits in decimal
90x = 43
43 1
x= expansion of can be 16.
90 17

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0.058823529411764705 7. Write three numbers whose decimal
17 1.00000000000000000000000000000
85 expansions are non-terminating non-
150
136 recurring.
140
136 Sol. 0.01001000100001...
40
34 0.202002000200002...
60
51 0.003000300003...
90
85 8. Find three different irrational numbers
50
34 5 9
160
between the rational numbers and .
153
7 11
70 Sol. 7 5.000000 0.714285...
68
20 49
17 10
30 7
17 30
130
119 28
110 20
102 14
80 60
68
120 56
119 40
100 35
85
150 5
136
4
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5
1 Thus, = 0.714285
= 0. 0588235294117647 7
17
6. Look at several examples of rational 9
— = 11 9.0000 0.8181...
numbers in the form p/q (q  0), where p 11 88
and q are integers with no common 20
11
factors other than 1 and having 90
terminating decimal representations 88
(expansions). Can you guess what 20
11
property q must satisfy?
9
Sol. There is a property that q must satisfy
p 9
rational number of form (q  0) where Thus, = 0.81
q 11
p, q are integers with no common factors Three different irrational numbers
other than 1 having terminating decimal 5 9
between and are taken as
representation (expansions) is that the 7 11
prime factorisation of q has only powers
0.750750075000750000...
of 2 or powers of 5 or both (i.e., q must be
0.780780078000780000...
of the form 2m × 5n). Here m, n are whole
numbers. 0.80800800080000800000...

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NCERT Basics : Class 9
9. Classify the following numbers as rational 2. Visualise 4.26 on the number line, up to 4
or irrational : decimal places.
Sol. n = 4.26
(i) 23 (ii) 225 So, n = 4.2626 (upto 4 decimal places)
(iii) 0.3796 (iv) 7.478478 ...... 4.2 4.3

(v) 1.101001000100001......
4.26
Sol. (i) 23 = Irrational number 4.2 4.27 4.3
(ii) 225 = 15 = Rational number
4.26 4.27
(iii) 0.3796
decimal expansion is terminating 4.2620 4.2630
4.2626
 .3796 = Rational number
(iv) 7.478478... EXERCISE : 1.5
1. Classify the following numbers as rational
= 7. 478 which is non-terminating
or irrational :
recurring. (i) 2− 5 (ii) (3 + 23) − 23
= Rational number 2 7 1
(iii) (iv)
(v) 1.101001000100001..... 7 7 2

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(v) 2
decimal expansion is non
Sol. (i) 2 is a rational number and 5 is an
terminating and non-repeating.
irrational number.
= Irrational number  2– 5 is an irrational number.
EXERCISE : 1.4 (ii) (3 + 23) − 23  (3 + 23) − 23
1. Visualise 3.765 on the number line, using = 3 is a rational number.
2 7 2
successive magnification. (iii) = is a rational number.
7 7 7
Sol. n = 3.765
1
(iv)
3.5 3.6 3.7 3.8 2
1 is a rational number and 2 is an
3.73 3.75 3.77 3.79
irrational number.
3.73.71 3.723.74 3.76 3.78 3.8 So, is an irrational number.
(v) 2
2 is a rational number and  is an
3.760 3.765 3.770 irrational number.
So, 2 is an irrational number.

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2. Simplify each of the following expressions : Let  be the number line.
(i) (3 + 3)(2 + 2)
Draw a line segment AB = 9.3 units and BC
(ii) (3 + 3)(3 − 3)
= 1 unit. Find the mid point O of AC.
(iii) ( 5 + 2)2 Draw a semicircle with centre O and
(iv) ( 5 − 2)( 5 + 2) radius OA or OC.
Sol. (i) (3 + 3)(2 + 2) = 3(2 + 2) + 3(2 + 2) Draw BD ⊥ AC intersecting the semicircle

= 6 +3 2 +2 3 + 6 at D. Then, BD = 9.3 units. Now, with

(ii) (3 + 3)(3 − 3) = (3)2 − ( 3)2 centre B and radius BD, draw an arc
=9–3=6 intersecting the number line l at P.
(iii) ( 5 + 2)2 Hence, BD = BP = 9.3
= ( 5)2 + 2 10 + ( 2)2 5. Rationalise the denominators of the

= 7 + 2 10 following :

(iv) ( 5 − 2)( 5 + 2) = 5 – 2 = 3 1 1
(i) (ii)
7 7− 6
3. Recall,  is defined as the ratio of the
1 1
circumference (say c) of a circle to its (iii) (iv)
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5+ 2 7 −2
diameter (say d). That is,  = c/d. This
seems to contradict the fact that  is 1 1 7 7
Sol. (i) =  =
7 7 7 7
irrational. How will you resolve this
contradiction ? 1 1 7+ 6
(ii) = 
Sol. There is no contradiction. When we 7− 6 7− 6 7+ 6

measure a length with a scale or any other 7+ 6 7+ 6

= = = 7+ 6
device, we only get an approximate 7−6 1
rational value. Therefore, we may not 1
(iii)
realise that c is irrational. 5+ 2

4. Represent 9.3 on the number line. 1 5− 2 5− 2

 =
5+ 2 5− 2 3
Sol.
9.3units

D
1 1 7 +2
(iv) = 
9.3 units 7 −2 7 −2 7 +2
O
l
A 9.3 units B C P
1 unit 7 +2 7 +2
= =
7−4 3
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NCERT Basics : Class 9
EXERCISE : 1.6 (iii) 163/4 = (24)3/4 = 23 = 8
1. Find : (iv) 125–1/3 = (53)–1/3 = 5–1 = 1/5
(i) (64)1/2 3. Simplify :
(ii) 321/5 7
2/3 1/5 1
(i) 2 .2 (ii)  3 
(iii) 1251/3 3 
1
2
Sol. (i) (64)1/2 = (82 )1/2 = (8 2) = 81 = 8 111/2
(iii) (iv) 71/2 . 81/2
5
1 111/4
(ii) 321/5 = (25)1/5 = (2 5) = 21 = 2 2 1 2 1 10+3 13
+
1 1 3
1 Sol. (i) 23 . 25 =2 5
3 = 2 15 =215

(iii) (125) 3 = (53 ) 3 =5 3 =5

7
2. Find : 1 17 1 −21
(ii)  3  = 3 7 = 21 = 3
3/2 2/5  3  (3 ) 3
(i) 9 (ii) 32
1
(iii) 163/4 (iv) 125–1/3 1 1 1
11 2 −
(iii) = 112 4 = 114 = 4 11
( )
3 3
1 1
3
Sol. (i) 92 = 92 = (3) = 27 11 4
2 2 2 1 1
5
(ii) 325 = (25 ) 5 = 2 5 = 22 = 4 (iv) 72 .82 = (7 × 8)1/2 = (56)1/2

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