Post on 09-Jan-2023
Contents
1. Rational Numbers 1—41
2. Powers 42—54
3. Squares and Square Roots 55—100
4. Cubes and Cube Roots 101—133
5. Playing With Numbers 134—139
6. Algebraic Expressions and Identities 140—170
7. Factorization 171—190
8. Division of Algebraic Expressions 191—204
9. Linear Equation in One Variable 205—237
10. Direct and Inverse Variations 238—250
11. Time and Work 251—258
12. Percentage 259—266
13. Profit, Loss, Discount and Value Added Tax (VAT) 267—285
14. Compound Interest 286—317
15. Understanding Shapes-I (Polygons) 318—321
16. Understanding Shapes-II (Quadrilaterals) 322—332
17. Understanding Shapes-III (Special Types of Quadrilaterals) 333—353
18. Practical Geometry (Constructions) 354—364
19. Visualising Shapes 365—372
20. Mensurations-I (Area of a Trapezium and a Polygon) 373—392
21. Mensuration-II (Volumes and Surface Areas of a Cuboid
and a Cube) 393—409
22. Mensuration-III (Surface Area and Volume of a Right
Circular Cylinder) 410—429
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23. Data Handling-I (Classification and Tabulation of Data 430—437
24. Data Handling-II (Graphical Representation of Data
as Histograms) 438—443
25. Data Handling-III (Pictorial Representation of Data as
Pie Charts or Circle Graphs) 444—457
26. Data Handling-IV (Probability) 458—465
27. Introduction to Graphs 466—475
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1
Points to Remember
1. Rational numbers— A number of the form q
p where p and q are integers and q o, is called
a rational number. In other words, a rational number can be expressed as the quotient of two
integers when divisor is not equal is zero.
2. Lowest form of a rational numbers— A rational number q
p is to be called in the lowest or
simplest form of p and q have no common factor other than 1. q
p is also said to be in standard
form.
3. Equality of rational numbers— Two rational numbers q
p and
s
r are equal if p × s = q × r.
q
p =
mq
mp
where m is a non-zero integer, then q
p =
mq
mp
which is equivalent to q
p.
4. Addition of rational numbers.
(a) If the denominators are same, Then
Let q
p,
q
r,
q
s are the rational number, then
q
p +
q
r +
q
s =
p
rqp
(b) When the denominators are different, then
First convert all the rational numbers of the same denominators by taking LCM.
Then add them as given above in (a).
5. Properties of addition of rational numbers—
(a) Closure property— The sum of any two rational numbers is always a rational number. If b
a,
d
c are the rational numbers the
d
c
b
a is also a rational number..
1RATIONAL NUMBERS
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2
(b) Commutative property— The addition of rational number is commutative i.e., if b
a +
d
c are
rational numbers, then b
a +
d
c =
d
c +
b
a.
(c) Associative property— The addition of rational numbers is associative. i.e. If b
a,
d
c,
f
e are
rational numbers, then
b
a +
f
e
d
c –
d
c
b
a +
f
e.
(d) Existence of Additive Identity or zero property— The sum of any rational number and zero
(0) is the rational number itself e.g. If b
a is a rational number, then
b
a + 0 = 0 +
b
a =
b
a.
(e) Existence of negative (Additive Inverse) of a rational number— For every rational number
b
a, there is a rational number
d
c such that
b
a +
d
c = 0 =
d
c +
b
a.
Then the rational numbers b
a and
d
c are called additive inverse or negative of each other..
The additive inverse of b
a is
b
a and of
b
a is
b
a.
b
a +
b
a = 0 or
b
a +
b
a = 0
6. Subtraction of rational numbers— If b
a and
d
c are two rational number, then
b
a –
d
c =
b
a +
d
c.
Properties of subtraction— Only closure property of subtraction holds good, but commutative
property, Associative property do not alway hold good.
Existance of right identity (i.e. zero identity) is true i.e. b
a – 0 =
b
a.
7. Multiplication of rational numbers— Product of two given rational numbers
= rsdenominatotheirofProduct
numenatorstheirofProduct
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3
i.e. b
a and
d
c are two rational numbers, then
b
a ×
d
c =
db
ca
.
8. Properties of multiplication—
(i) Closure property— The multiplication of two rational numbers is also a rational number. If
b
a and
d
c are two rational number, then
b
a ×
d
c =
bd
ac is also a rational number..
(ii) Commutative property— The multiplication of two rational is commutative i.e. a × b = b ×
a for any two rational numbers a and b.
(iii) Associative property— The multiplication of rational numbers is associative. If a, b, c are
three rational numbers, then (a × b) × c = a × (b × c).
(iv) Existance of Identity— The rational number 1 is the multiplicative identity
a × 1 = a = 1 × a
(v) Multiplication by zero— Any rational number a we have a × 0 = 0 = 0 × a.
(vi) Existance of multiplicative inverse— For every non zero rational number ‘a’, there exists a
rational number a
1 (reciprocal of a) such that a ×
a
1 = 1 =
a
1 × a.
(vii) Distributive property of multiplication over addition— If x, y, z are any three rational
numbers, then x × (y + z) = x × y + x × z.
9. Division of rational number— If x and y are two rational numbers, y 0, then the result of
dividing x by y is a rational number in multiplying x by the reciprocal or inverse of y.
10. Properties of division of rational numbers—
Properties I— If b
a and
d
c are two rational numbers such that
d
c 0, then
b
a
d
c is
always a rational number.
That is, the set of all non-zero rational numbers is closed under division.
Properties II— For any rational number b
a, we have
b
a 1 =
b
a and
b
a (–1) = –
b
a =
b
a
Properties III— For every non-zero rational number b
a, we have
(i)b
a
b
a = 1 (ii)
b
a �
b
a = –1 (iii)
b
a
b
a = –1
Remark— The division of rational numbers is neither commutative nor associative.
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11. Representation of rational numbers on a number line—
Any rational number can be represented on a number line. Every positive number lies on the
right-side of 0 and negative number on the left of 0.
O 1 2 3 4 512345 66
12. Rational number between two rational numbers—
Let x and y be the rational numbers than a number between x and y will be 2
yx
Note— We can find as many rational number between two rational numbers.
1. Add the following rational numbers :
(i)7
5 and
7
3(ii)
4
15 and
4
7
(iii)11
8 and
11
4(iv)
13
6 and
13
9
Solution—
(i)7
5 and
7
3 =
7
5 +
7
3 =
7
35 =
7
2
(ii)4
15 and
4
7
= 4
15 +
4
7 =
4
715 =
4
8
= 44
48
= –2
(iii)11
8 and
11
4
= 11
8 +
11
4 =
11
48
= 11
12
EXERCISE 1.1
(iv)13
6 and
13
9
= 13
6 +
13
9 =
13
96
= 13
3
2. Add the following rational numbers :
(i)4
3 and
8
5(ii)
9
5
and
3
7
(iii) –3 and 5
3(iv)
27
7 and
18
11
(v)4
31
and
8
5(vi)
36
5 and
12
7
(vii)16
5 and
24
7(viii)
18
7
and
27
8
Solution—
(i)4
3 and
8
5
LCM of 4, 8 = 8
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5
4
3 =
24
23
= 8
6
4
3 +
8
5 =
8
6 +
8
5
= 8
56 =
8
1
(ii)9
5
and
3
7
9
5
and
3
7
LCM of 9, 3 = 9
3
7 =
33
37
= 9
21
Now 9
5 +
9
21 =
9
215 =
9
16
(iii) –3 and 5
3
LCM of 5, 1 = 1
1
3 =
51
53
= 5
15
–3 + 5
3 =
5
15 +
5
3
= 5
315 =
5
12
(iv)27
7 and
18
11
LCM of 27, 18 = 54
27
7 =
227
27
= 54
14
18
11 =
318
311
= 54
33
27
7 +
18
11 =
54
14 +
54
33
= 54
3314 =
54
19
(v)4
31
and
8
5
4
31 and
8
5
LCM of 4, 8, = 8
4
31 =
24
231
= 8
62
4
31 +
8
5 =
8
62 +
8
5
8
562 =
8
67
(vi)36
5 and
12
7
LCM of 36, 12 = 36
12
7 =
312
37
= 36
21
36
5 +
12
7 =
36
5 +
36
21
= 36
215 =
36
16
= 436
416
= 9
4
(vii)16
5 and
24
7
LCM of 16, 24 = 48
16
5 =
316
35
= 48
15
and 24
7 =
224
27
= 48
14
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Now 16
5 +
24
7 =
48
15 +
48
14
= 48
1415 =
48
1
(viii)18
7
and
27
8
18
7 and
27
8
LCM of 18, 27 = 54
18
7 =
318
37
= 54
21
and 27
8 =
227
28
= 54
16
Now 18
7 +
27
8 =
54
21 +
54
16
= 54
1621 =
54
5
3. Simplify :
(i)9
8 +
6
11(ii)
16
5 +
24
7
(iii)12
1
and
15
2
(iv)
19
8 +
57
4
(v)9
7 +
4
3
(vi)
26
5 +
39
11
(vii)9
16 +
12
5(viii)
8
13 +
36
5
(ix) 0 + 5
3(x) 1 +
5
4
(xi) 3 + 7
5
Solution—
(i)9
8 +
6
11
LCM of 9, 6 = 18
9
8 =
29
28
= 18
16
and 6
11 =
36
311
= 18
33
Now 9
8 +
6
11 =
18
16 +
18
33
= 18
3316 =
18
17
(ii)16
5 +
24
7
LCM of 16, 24 = 48
16
5 =
316
35
= 48
15
and 24
7 =
224
27
= 48
14
Now 16
5 +
24
7 =
48
15 +
48
14
= 48
1415 =
48
1
(iii)12
1
+
15
2
= 12
1 +
15
2
15
2
115
12
15
2and
12
1
112
11
12
1
LCM of 12, 15 = 60
12
1 =
512
51
= 60
5
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and 15
2 =
415
42
= 60
8
Now = 12
1 +
15
2 =
60
5 +
60
8
= 60
85 =
60
13
(iv)19
8 +
57
4
LCM of 19, 57 = 57
19
8 =
319
38
= 57
24
and 57
4 =
157
14
= 57
4
Now 19
8 +
57
4 =
57
24 +
57
4
= 57
424 =
57
28
(v)9
7 +
4
3
9
7 +
4
3
4
3
14
13
4
3
LCM of 9, 4 = 36
9
7 =
49
47
= 36
28
and 4
3 =
94
93
= 36
27
Now 9
7 +
4
3 =
36
28 +
36
27
= 36
2728 =
36
1
(vi)26
5 +
39
11
26
5 +
39
11
39
11
139
111
39
11
LCM of 26, 39 = 78
26
5 =
326
35
= 78
15
and 39
11 =
239
211
= 78
22
Now 26
5 +
39
11 =
78
15 +
78
22
= 78
2215 =
78
7
(vii)9
16 +
12
5
LCM of 9, 12 = 36
9
16 =
49
416
= 36
64
12
5 =
312
35
= 36
15
Now 9
16 +
12
5 =
36
64 +
36
15
= 36
1564 =
36
79
(viii)8
13 +
36
5
LCM of 8, 36 = 72
8
13 =
98
913
= 72
117
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8
36
5 =
236
25
= 72
10
Now 8
13 +
36
5 =
72
117 +
72
10
= 72
10117 =
72
107
(ix) 0 + 5
3 =
5
3
(Zero property of addition)
(x) 1 + 5
4 =
5
5 +
5
4
5
5
51
51
1
1
= 5
45 =
5
1
(xi) 3 + 7
5
= 3 +
7
5
7
5
17
15
7
5
= 1
3 +
7
5
= 7
521 =
7
16
4. Add and express the sum as mixed
fraction :
(i)5
12 and
10
43(ii)
7
24 and
4
11
(iii)6
31 and
8
27(iv)
6
101 and
8
7
Solution—
(i)5
12 and
10
43
LCM of 5, 10, = 10
5
12 =
25
212
= 10
24
5
12 +
10
43 =
10
24 +
10
43
= 10
4324 =
10
19
= 110
9
(ii)7
24 and
4
11
LCM of 7, 4 = 28
7
24 =
47
424
= 28
96
and 4
11 =
74
711
= 28
77
7
24 +
4
11 =
28
96 +
28
77
= 28
7796 =
28
19
(iii)6
31 and
8
27
LCM of 6, 8 = 24
6
31 =
46
431
= 24
124
and 8
27 =
38
327
= 24
81
6
31 +
8
27 =
24
124 +
24
81
= 24
81124 =
24
205
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9
= –824
13
(iv)6
101 and
8
7
LCM of 6, 8 = 24
6
101 =
46
4101
= 24
404
and 8
7 =
38
37
= 24
21
6
101 +
8
7 =
24
404 +
24
21
= 24
21404 =
24
425
= 1724
17
1. Verify commutativity of addition of
rational numbers for each of the following
pairs of rational numbers :
(i)5
11 and
7
4(ii)
9
4 and
12
7
(iii)5
3 and
15
2
(iv)7
2
and
35
12
(v) 4 and 5
3(vi) –4 and
7
4
Solution—
(i)5
11 and
7
4
5
11 +
7
4
LCM of 5 and 7 = 35
5
11 =
75
711
= 35
77
and 7
4 =
57
54
= 35
20
5
11 +
7
4 =
35
77 +
35
20
EXERCISE 1.2
= 35
2077 =
35
57
and 7
4 +
5
11 =
35
20 –
35
77
= 35
7720 =
35
57
5
11 +
7
4 =
7
4 +
5
11
(ii)9
4 and
12
7
112
17
= 12
7
Now 9
4 +
12
7
LCM of 9, 12 = 36
9
4 =
49
44
= 36
16
12
7 =
312
37
= 36
21
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10
9
4 +
12
7 =
36
16 +
36
21
= 36
2116 =
36
5
and 12
7 +
9
4 =
36
21 +
36
16
= 36
1621 =
36
5
9
4 +
12
7 =
12
7 +
9
4
(iii)5
3 and
15
2
15
2
= 115
12
= 15
2
LCM of 5 and 15 = 15
5
3 =
35
33
= 15
9
Now 5
3 +
15
2 =
15
9 +
15
2
15
29 =
15
7
and 15
2 +
5
3 =
15
2 +
15
9
= 15
92 =
15
7
5
3 +
15
2 =
15
2 +
5
3
(iv)7
2
and
35
12
LCM of 7, 35 = 35
7
2
=
57
52
= 35
10
35
12
=
135
112
= 35
12
Now 7
2 +
35
12 =
35
10 +
35
12
= 35
1210 =
35
22
and 35
12 +
7
2 =
35
12 +
35
10
= 35
1012 =
35
22
7
2 +
35
12 =
35
12 +
7
2
(v) 4 and 5
3
LCM of 1, 5 = 5
1
4 =
51
54
= 5
20
Now 1
4 +
5
3 =
5
20 +
5
3
= 5
320 =
5
17
and 5
3 +
1
4 =
5
3 +
5
20
= 5
203 =
5
17
4 + 5
3 =
5
3 + 4
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(vi) –4 and 7
4
1
4 and
7
4
7
4
17
14
7
4
LCM of 1, 7 = 7
1
4 =
71
74
= 7
28
and 7
4
–4 + 7
4 =
7
28 +
7
4
= 7
428 =
7
32
and 7
4 + (–4) =
7
4 +
7
28
= 7
284 =
7
32
–4 + 7
4 =
7
4 + (–4)
2. Verify associativity of addition of rational
numbers i.e., (x + y) + z = x + (y + z),
when :
(i) x = 2
1, y =
3
2, z = –
5
1
(ii) x = 5
2, y =
3
4, z =
10
7
(iii) x = 11
7, y =
5
2
, z =
22
3
(iv) x = –2, y = 5
3, z =
3
4
Solution—
(i) x = 2
1, y =
3
2, z = –
5
1
(x + y) + z =
3
2
2
1 +
5
1
= 6
43 +
5
1 =
6
7 +
5
1
= 6
7 –
5
1
= 30
635 =
30
29
and x + (y + z) = 2
1 +
5
1
3
2
= 2
1 +
15
310
= 2
1 +
15
7
= 30
1415 =
30
29
(x + y) + z = x + (y + z)
(ii) x = 5
2, y =
3
4, z =
10
7
(x + y) + z =
3
4
5
2 +
10
7
= 15
206 +
10
7
= 15
14 +
10
7
LCM of 15, 10 = 30
= 30
2128 =
30
7
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and x + (y + z) = 5
2 +
10
7
3
4
= 5
2 +
30
2140
= 5
2 +
30
19
= 30
1912 =
30
7
(x + y) + z = x + (y + z)
(iii) x = 11
7, y =
5
2
, z =
22
3
x = 11
7, y =
15
12
, z = 22
3
x = 11
7, y =
5
2, z =
22
3
Now (x + y) + z
5
2
11
7 +
22
3
= 55
2235 +
22
3
= 55
57 +
22
3(LCM of 22, 55 = 110)
= 110
15114 =
110
129
and x + (y + z) = 11
7 +
22
3
5
2
= 11
7 +
110
1544
= 11
7 +
110
59
= 110
5970 =
110
129
(x + y) + z = x + (y + z)
(iv) x = –2, y = 5
3, z =
5
4
(x + y) + z =
5
3
1
2 +
5
4
= 5
310 +
5
4
= 5
7 +
5
4
= 5
47 =
5
11
and x + (y + z) = –2 +
5
4
5
3
= –2 + 5
43
= 1
2 +
5
1
= 5
110 =
5
11
(x + y) + z = x + (y + z)
3. Write the additive inverse of each of the
following rational numbers :
(i)17
2(ii)
11
3
(iii)5
17(iv)
25
11
Solution—
(i) Additive inverse of 17
2 is –
17
2 =
17
2
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(ii) Additive inverse of 11
3
is –
11
3 =
11
3
(iii) Additive inverse of 5
17 is –
5
17 =
5
17
(iv) Additive inverse of 25
11
=
25
11
= –
25
11 =
25
11
4. Write the negative (additive inverse) of
each of the following :
(i)5
2(ii)
9
7
(iii)13
16(iv)
1
5
(v) 0 (vi) 1
(vii) –1
Solution—
(i) Negative of 5
2 is –
5
2 =
5
2
(ii) Negative of 9
7
is –
9
7 =
9
7
(iii) Negative of 13
16 is –
13
16 =
13
16
(iv) Negative inverse of 1
5 is –
1
5 = 5
(v) Negative inverse of 0 is 0
(vi) Negative inverse of 1 is –1
(vii) Negative inverse of –1 is –(–1) = 1
5. Using commutativity and associativity of
addition of rational numbers, express
each of the following as a rational
number :
(i)5
2 +
3
7 +
5
4 +
3
1
(ii)7
3 +
9
4 +
7
11 +
9
7
(iii)5
2 +
3
8 +
15
11 +
5
4 +
3
2
(iv)7
4 + 0 +
9
8 +
7
13 +
21
17
Solution—
(i)5
2 +
3
7 +
5
4 +
3
1
=
5
4
5
2 +
3
1
3
7
=
5
42 +
3
17
= 5
2 +
3
6 =
5
2 +
1
2
= 1
2 –
5
2
= 5
210 =
5
8
(ii)7
3 +
9
4 +
7
11 +
9
7
=
7
11
7
3 +
9
7
9
4
= 7
113 +
9
74
= 7
8 +
9
3
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14
= 7
8 +
3
1
3
1
39
33
9
3
= 21
724 =
21
17
(iii)5
2 +
3
8 +
15
11 +
5
4 +
3
2
=
5
4
5
2 +
3
2
3
8 +
15
11
= 5
42 +
3
28 +
15
11
= 5
6 +
3
6 +
15
11
= 15
113018 =
15
1148 =
15
37
(iv)7
4 + 0 +
9
8 +
7
13 +
21
17
=
7
13
7
4 +
21
17
9
8 + 0
= 7
134 +
63
5156
= 7
9 +
63
5
= 63
581 =
63
86
6. Re-arrange suitably and find the sum in
each of the following :
(i)12
11 +
3
17 +
2
11 +
2
25
(ii)7
6 +
6
5 +
9
4 +
7
15
(iii)5
3 +
3
7 +
5
9 +
15
13 +
3
7
(iv)13
4 +
8
5 +
13
8 +
13
9
(v)3
2 +
5
4 +
3
1 +
5
2
(vi)8
1 +
12
5 +
7
2 +
12
7 +
7
9 +
16
5
Solution—
(i)12
11 +
3
17 +
2
11 +
2
25
=
3
17
12
11 +
2
25
2
11
= 12
6811+
2
2511
= 12
57 +
2
14
= 12
8457 =
12
141
(ii)7
6 +
6
5 +
9
4 +
7
15
=
7
15
7
6 +
9
4
6
5
= 7
156 +
18
815
= 7
21 +
18
23
= 126
161378
LCM of 7, 18 = 126
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15
= 126
539 =
7126
7539
= 18
77
(iii)5
3 +
3
7 +
5
9 +
15
13 +
3
7
=
5
9
5
3 +
3
7
3
7 +
15
13
= 5
93 +
3
77 +
15
13 =
5
12 + 0 +
15
13
= 5
12 +
15
13
= 15
1336 =
15
23
(iv)13
4 +
8
5 +
13
8 +
13
9
= 13
4 +
13
8 +
13
9 +
8
5
= 13
984 +
8
5
= 13
5 +
8
5
= 104
6540 (LCM of 13, 8 = 104)
= 104
25
(v)3
2 +
5
4 +
3
1 +
5
2
=
3
1
3
2 +
5
2
5
4
= 3
12 +
5
24
= 3
3 +
5
2
= 1 + 5
2 =
5
25 =
5
3
(vi)8
1 +
12
5 +
7
2 +
12
7 +
7
9 +
16
5
=
16
5
8
1 +
12
7
12
5 +
7
9
7
2
= 16
52 +
12
75 +
7
92
= 16
3 +
12
12 +
7
11
= 16
3 + 1 +
7
11
= 112
17611221 =
112
28821
= 112
267
1. Subtract the first rational number from the second in each of the following :
(i)8
3,
8
5(ii)
9
7,
9
4(iii)
11
2,
11
9(iv)
13
11,
13
4
EXERCISE 1.3
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16
(v)4
1,
8
3(vi)
3
2,
6
5
(vii)7
6,
14
13(viii)
33
8,
22
7
Solution—
(i)8
3 from
8
5 =
8
5 –
8
3 =
8
35
= 8
2 =
28
22
= 4
1
(ii)9
7 from
9
4 =
9
4–
9
7
= 9
4 +
9
7 =
9
74
= 9
11
(iii)11
2 from
11
9 =
11
9–
11
2
11
9 +
11
2 =
11
29
= 11
7
(iv)13
11 from
13
4 =
13
4 –
13
11
= 13
114 =
13
15
(v)4
1 from
8
3 =
8
3–
4
1
= 8
3 –
4
1
= 8
23(LCM 8, 4 = 8)
= 8
5
(vi)3
2 from
6
5=
6
5–
3
2
= 6
5 +
3
2
= 6
45 =
6
9 =
36
39
= 2
3
(vii)7
6 from
14
13 =
14
13–
7
6
= 14
13 +
7
6
= 14
1213 =
14
1
(viii)33
8 from
22
7 =
22
7–
33
8
= 66
1621 =
66
5
2. Evaluate each of the following :
(i)3
2 –
5
3(ii)
7
4–
3
2
(iii)7
4–
7
5
(iv) –2 –9
5
(v)8
3
– 7
2(vi)
13
4 –
26
5
(vii)14
5 –
7
2(viii)
15
13 –
25
12
(ix)13
6 –
13
7(x)
24
7 –
36
19
(xi)63
5 –
21
8
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17
Solution—
(i)3
2 –
5
3
= 15
910 (LCM of 3, 5 = 15)
= 15
1
(ii)7
4 –
3
2
=
7
4 +
3
2
= 21
1412 (LCM of 7, 3 = 21)
= 21
2
(iii)7
4 –
7
5
= 7
4 –
7
5
= 7
54 =
7
1
(iv)1
2 –
9
5
= 9
518 =
9
23
(v)8
3
– 7
2 =
8
3 +
7
2
= 56
1621(LCM of 8, 7 = 56)
= 56
37
(vi)13
4 –
26
5 =
13
4 +
26
5
= 26
58(LCM of 13, 26 = 26)
= 26
3
(vii)14
5 –
7
2 =
14
5 +
7
2
= 14
45(LCM of 14, 7 = 14)
= 14
1
(viii)15
13 –
25
12
75
3665 (LCM of 15, 25 = 75)
= 75
29
(ix)13
6 –
13
7 =
13
6 +
13
7 =
13
76
= 13
1
(x)24
7–
36
19
LCM of 24, 36 = 72
= 24
7 –
36
19
= 72
3821 =
72
17
(xi)63
5 –
21
8 =
63
5 +
21
8
= 63
245(LCM of 63, 21 = 63)
= 63
29
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18
3. The sum of two numbers is 9
5. If one of
the numbers is 3
1, find the other..
Solution—
Sum of two numbers = 9
5
One number = 3
1
Second number = 9
5 –
3
1
= 9
35(LCM of 9, 3 = 9)
= 9
2
4. The sum of two numbers is 3
1. If one
of the numbers is 3
12, find the other..
Solution—
Sum of two numbers = 3
1
One number = 3
12
Second number = 3
1 –
3
12
= 3
1 +
3
12
= 3
121 =
3
11
5. The sum of two numbers is 3
4. If one
of the number is –5, find the other.
Solution—
Sum of two numbers = 3
4
One number = –5
Second number = 3
4 – (–5)
= 3
4 +
1
5
= 3
154 =
3
11
6. The sum of two rational numbers is –8.
If one of the numbers is 7
15, find the
other.
Solution—
Sum of two rational numbers = –8
One number = 7
15
Second number = –8 –
7
15
= 1
8 +
7
15
= 7
1556 =
7
41
7. What should be added to 8
7 so as to get
9
5 ?
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19
Solution—
The required number = 9
5 –
8
7
= 9
5 +
8
7
= 72
6340 (LCM of 9, 8 = 72)
= 72
103
8. What number should be added to 11
5 so
as to get 33
26 ?
Solution—
The required number = 33
26 –
11
5
= 33
26 +
11
5(LCM of 33, 11 = 33)
= 33
1526 =
33
41
9. What number should be added to 7
5 to
get 3
2 ?
Solution—
The required number = 3
2 –
7
5
= 3
2 +
7
5
= 21
1514 (LCM of 3, 7 = 21)
= 21
1
10. What number should be subtracted from
3
5 to get
6
5 ?
Solution—
The required number = 3
5 –
6
5
= 6
510(LCM of 3, 6 = 6)
= 6
15 =
36
315
= 2
5
11. What number should be subtracted from
7
3 to get
4
5 ?
Solution—
The required number = 7
3 –
4
5
= 7
3 –
4
5
= 28
3512(LCM of 7, 4 = 28)
= 28
23
12. What should be added to
5
3
3
2 to get
15
2 ?
Solution—
The required number = 15
2 –
5
3
3
2
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20
= 15
2 –
15
19 =
15
192
= 15
21=
315
321
= 5
7
13. What should be added to
5
1
3
1
2
1
to get 3 ?
Solution—
The required number = 3 –
5
1
3
1
2
1
= 1
3 –
30
61015
(LCM of 2, 3, 5, = 30)
= 1
3 –
30
31
= 30
3190 =
30
59
14. What should be subtracted from
3
2
4
3
to get 6
1 ?
Solution—
The required number =
3
2
4
3 –
6
1
= 4
3 –
3
2 +
6
1
= 12
289 (LCM of 4, 3, 6 = 12)
= 12
811 =
12
3
= 312
33
= 4
1
15. Fill in the blanks :
(i)13
4 –
26
3 = ............
(ii)14
9 + ............ = –1
(iii)9
7 + ............ = 3
(iv) ............ + 23
15 = 4
Solution—
(i)13
4 –
26
3 = ............
Required number
= 13
4 –
26
3 =
13
4 +
26
3
= 26
38 =
26
5
13
4 –
26
3 =
26
5
(ii)14
9 + ............ = –1
Required number = –1 –
14
9
= –1 + 14
9
= 14
914 =
14
5
(iii)9
7 + ............ = 3
Required number = 3 –
9
7
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21
= 1
3 +
9
7
= 9
727
= 9
34
(iv) ............ + 23
15 = 4
Required number = 1
4 –
23
15
= 23
1592 =
23
77
1. Simplify each of the following and write
as a rational number of the form q
p :
(i)4
3 +
6
5 +
8
7(ii)
3
2 +
6
5 +
9
7
(iii)2
11 +
6
7 +
8
5
(iv)5
4 +
10
7 +
15
8
(v)10
9 +
15
22 +
20
13
(vi)3
5 +
2
3
+
3
7 + 3
Solution—
(i)4
3 +
6
5 +
8
7
LCM of 4, 6, 8 = 24
4
3 +
6
5 +
8
7
24
212018 =
24
2138 =
24
17
EXERCISE 1.4
(ii)3
2+
6
5 +
9
7
LCM of 3, 6, 9 = 18
3
2+
6
5 +
9
7
18
141512 =
18
2912 =
18
17
(iii)2
11 +
6
7 +
8
5
LCM of 2, 6, 8 = 24
2
11 +
6
7 +
8
5
24
1528132 =
24
28147
= 24
119
(iv)5
4 +
10
7 +
15
8
LCM of 5, 10, 15 = 30
5
4 +
10
7 +
15
8
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22
30
162124 =
30
61
(v)10
9 +
15
22 +
20
13
= 10
9 +
15
22 +
20
13
LCM of 10, 15, 20 = 60
10
9 +
15
22 +
20
13
= 60
398854 =
60
8893=
60
5=
12
1
(vi)3
5 +
2
3
+
3
7 + 3
= 3
5 +
2
3 +
3
7 +
1
3
LCM of 3, 2, 3 = 6
3
5 +
2
3 +
3
7 +
1
3
6
1814910 =
6
2328 =
6
5
2. Express each of the following as a rational
number of the form q
p :
(i)3
8 +
4
1 +
6
11 +
8
3 – 3
(ii)7
6 + 1 +
9
7 +
21
19 +
7
12
(iii)2
15 +
8
9 +
3
11 + 6 +
6
7
(iv)4
7 + 0 +
5
9 +
10
19 +
14
11
(v)4
7 +
3
5 +
2
1 +
6
5 + 2
Solution—
(i)3
8 +
4
1 +
6
11 +
8
3 – 3
LCM of 3, 4, 6, 8 = 24 2 3, 4, 6, 82 3, 2, 3, 43 3, 1, 3, 2
1, 1, 1, 2
LCM = 2 × 2 × 3 × 2 = 24
3
8 +
4
1 +
6
11 +
8
3 –
1
3
24
72944664
= 24
9186 =
24
177 =
324
3177
= 8
59
(ii)7
6 + 1 +
9
7 +
21
19 +
7
12
LCM of 7, 9, 21, 7, = 63 3 7, 9, 21, 77 7, 3, 7, 73 1, 3, 1, 1
LCM = 3 × 7 × 3 = 63
7
6 +
1
1 +
9
7 +
21
19 +
7
12
63
10857496354
= 63
157174 =
63
17
(iii)2
15 +
8
9 +
3
11 + 6 +
6
7
LCM of 2, 8, 3, 6 = 24 2 2, 8, 3, 63 1, 4, 3, 34 1, 4, 1, 1
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23
LCM = 2 × 3 × 4 = 24
2
15 +
8
9 +
3
11 +
1
6 +
6
7
= 24
281448827180
= 24
116351 =
24
235
(iv)4
7 + 0 +
5
9 +
10
19 +
14
11
LCM of 4, 5, 10, 14 = 140
2 4, 5, 10, 145 2, 5, 5, 72 2, 1, 1, 7
LCM = 2 × 5 × 2 × 7 = 140
4
7 + 0 +
5
9 +
10
19 +
14
11
140
1102662520245
= 140
497376 =
140
121
(v)4
7 +
3
5 +
2
1 +
6
5 + 2
LCM of 4, 3, 2, 6 = 12
2 4, 3, 2, 63 2, 3, 1, 32 2, 1, 1, 1
LCM = 2 × 3 × 2 = 12
4
7 +
3
5 +
2
1 +
6
5 +
1
2
12
241062021
= 12
3744 =
12
7
3. Simplify :
(i)2
3 +
4
5 –
4
7(ii)
3
5 –
6
7 +
3
2
(iii)4
5 –
6
7 –
3
2(iv)
5
2 –
10
3 –
7
4
(v)6
5 +
5
2–
15
2(vi)
8
3 –
9
2 +
36
5
Solution—
(i)2
3 +
4
5–
4
7
= 4
756 (LCM of 2, 4 = 4)
= 4
8 = –2
(ii)3
5 –
6
7 +
3
2
= 6
4710 (LCM of 3, 6 = 6)
= 6
1
(iii)4
5 –
6
7 –
3
2
= 12
81415 (LCM of 4, 6, 3, = 12)
= 12
9=
312
39
= 4
3
(iv)5
2 –
10
3 –
7
4
= 5
2 +
10
3 +
7
4
= 70
402128 (LCM of 5, 10, 7 = 70)
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24
= 70
2861 =
70
33
(v)6
5 +
5
2 –
15
2
= 30
41225 (LCM of 6, 5, 15, = 30)
= 30
1229 =
30
17
(vi)8
3 –
9
2 +
36
5
= 72
101627 (LCM of 8, 9, 36 = 72)
= 72
1043 =
72
33
= 372
333
= 24
11
1. Multiply :
(i)11
7 by
4
5(ii)
7
5 by
4
3
(iii)9
2 by
11
5(iv)
17
3 by
4
5
(v)7
9
by
11
36
(vi)
13
11 by
7
21
(vii) –5
3 by –
7
4(viii) –
11
15 by 7
Solution—
(i)11
7 by
4
5 =
411
57
= 44
35
(ii)7
5 by
4
3 =
47
35
= 28
15
(iii)9
2 by
11
5 =
119
52
= 99
10
(iv)17
3 by
4
5
= 417
53
= 68
15
= 168
115
= 68
15
EXERCISE 1.5
(v)7
9
by
11
36
= 117
369
= 77
324
(vi)13
11 by
7
21 =
713
2111
= 91
231 =
791
7231
= 13
33
(vii) –5
3 by –
7
4 =
75
43
= 35
12
(viii)–11
15 by 7 =
11
715 =
11
105
2. Multiply :
(i)17
5 by
60
51
(ii)
11
6 by
36
55
(iii)25
8 by
16
5(iv)
7
6 by
36
49
(v)9
8
by
16
7
(vi)9
8 by
64
3
Solution—
(i)17
5 by
60
51
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25
17
5 ×
60
51
60
51
160
151
60
51
=
6017
515
= 1020
255 =
2551020
255255
= 4
1
(ii)11
6 by
36
55 =
11
6 ×
36
55
=
3611
556
=
61
51
= 6
5
(iii)25
8 by
16
5 =
1625
58
=
25
11
= 10
1
(iv)7
6 by
36
49 =
7
6 ×
36
49
=
367
496
=
61
71
= 6
7
(v)9
8
by
16
7
= 9
8
×
16
7
=
169
78
=
29
71
= 18
7
(vi)9
8 by
64
3
= 649
38
= 83
11
= 24
1
3. Simplify each of the following and express
the result as a rational number in
standard form :
(i)21
16 ×
5
14(ii)
6
7 ×
28
3
(iii)36
19 × 16 (iv)
9
13 ×
26
27
(v)16
9 ×
27
64
(vi)7
50 ×
3
14
(vii)9
11 ×
88
81
(viii)9
5 ×
25
72
Solution—
(i)21
16 ×
5
14 =
521
1416
= 53
216
= 15
32
(ii)6
7 ×
28
3 =
286
37
=
42
11
= 8
1
(iii)36
19 × 16 =
36
1619 =
9
419 =
9
76
(iv)9
13 ×
26
27
= 269
2713
= 21
31
= 2
3
= 2
3
(v)16
9 ×
27
64
=
2716
649
= 31
41
= 3
4
=
13
14
= 3
4
https://www.arundeepselfhelp.info/index.php?route=product/product&path=139_141&product_id=111
26
(vi)7
50 ×
3
14 =
37
1450
= 31
250
= 3
100
(vii)9
11 ×
88
81
=
889
8111
=
81
91
= 8
9
=
18
19
= 8
9
(viii)9
5 ×
25
72
= 259
725
= 51
81
= 5
8
= 5
8
4. Simplify :
(i)
5
2
8
25 –
9
10
5
3
(ii)
4
1
2
1 +
62
1
(iii)
15
25 –
9
26
(iv)
3
5
4
9+
6
5
2
13
(v)
5
12
3
4+
15
21
7
3
(vi)
3
8
5
13 –
3
11
2
5
(vii)
26
11
7
13 –
6
5
3
4
(viii)
2
3
5
8 +
16
11
10
3
Solution—
(i)
5
2
8
25 –
9
10
5
3
= 58
225
–
95
103
= 14
15
–
31
21
= 4
5 –
3
2
= 12
815 =
12
23
(ii)
4
1
2
1 +
62
1
= 42
11
+ 2
1 ×
1
6 =
8
1 +
1
3
= 8
241 =
8
25
(iii)
15
25 –
9
26
= 15
25 –
9
26
= 3
21 –
3
22 =
3
2 –
3
4
= 3
42 =
3
2
(iv)
3
5
4
9 +
6
5
2
13
https://www.arundeepselfhelp.info/index.php?route=product/product&path=139_141&product_id=111
27
= 34
59
+ 62
513
= 14
53
+ 12
65
= 4
15 +
12
65
= 12
6545 =
12
20 =
412
420
= 3
5
(v)
5
12
3
4 +
15
21
7
3
= 53
124
+ 157
213
= 51
44
+ 51
31
= 5
16
+ 5
3
= 5
16 +
5
3 =
5
316 =
5
19
(vi)
3
8
5
13 –
3
11
2
5
= 35
813
– 32
115
= 15
104 –
6
55
= 30
275208 (LCM of 15, 6 = 30)
= 30
483 =
10
161
(vii)
26
11
7
13 –
6
5
3
4
= 267
1113
– 63
54
= 27
111
– 33
52
= 14
11 –
9
10
= 126
14099(LCM of 14, 9 = 126)
= 126
239
(viii)
2
3
5
8 +
16
11
10
3
=
25
38
+ 1610
113
=
15
34
+ 1610
113
= 5
12 +
160
33
= 160
33384 =
160
417
5. Simplify :
(i)
6
1
2
3 +
2
7
3
5 –
3
4
8
13
(ii)
7
2
4
1 –
3
2
14
5 +
2
9
7
3
(iii)
2
15
9
13 +
5
8
3
7 +
2
1
5
3
(iv)
6
5
11
3 –
3
4
12
9 +
15
6
13
5
Solution—
(i)
6
1
2
3 +
2
7
3
5 –
3
4
8
13
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28
= 62
13
+ 23
75
– 38
413
= 22
11
+ 23
75
– 32
113
= 4
1 +
6
35 –
6
13
= 12
26703 (LCM of 4, 6 = 12)
= 12
2673 =
12
47
(ii)
7
2
4
1 –
3
2
14
5 +
2
9
7
3
= 74
21
–
314
25
+ 27
93
= 72
11
–
37
15
+ 27
93
= 14
1 –
21
5 +
14
27
= 42
81103 (LCM of 14, 21 = 42)
= 42
94 =
242
294
= 21
47
(iii)
2
15
9
13 +
5
8
3
7 +
2
1
5
3
=
29
1513
+ 53
87
+ 25
13
= 23
513
+ 53
87
+ 25
13
= 6
65 +
15
56 +
10
3
= 30
33256565
(_ LCM of 6, 15, 10 = 30)
= 30
9112325 =
30
204 =
15
102
(iv)
6
5
11
3 –
3
4
12
9 +
15
6
13
5
= 611
53
– 312
49
+ 1513
65
= 211
51
– 3
13 +
113
21
= 22
5 – 1 +
13
2
= 286
4428665 (LCM of 22, 13 = 286)
= 286
286109 =
286
177
EXERCISE 1.6
1. Verify the property : x × y = y × x by taking :
(i) x = –3
1, y =
7
2(ii) x =
5
3, y =
13
11(iii) x = 2, y =
8
7
(iv) x = 0, y =
8
15
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29
Solution—
(i) x × y = y × x
x = –3
1, y =
7
2
L.H.S. = x × y = –3
1 ×
7
2 =
73
21
= 21
2
R.H.S. = y × x = 7
2 ×
3
1 =
37
12
= 21
2
L.H.S. = R.H.S.
(ii) x × y = y × x
x = 5
3, y =
13
11
L.H.S. = x × y = 5
3 ×
13
11
=
135
113
= 65
33
R.H.S. = y × x = 13
11 ×
5
3
=
513
311
= 65
33
L.H.S. = R.H.S.
(iii) x × y = y × x
x = 2, y = 8
7
=
18
17
= 8
7
L.H.S. = x × y = 2 × 8
7 =
8
72
= 4
7
R.H.S. = y × x = 8
7 × 2 =
4
7 × 1 =
4
7
L.H.S. = R.H.S.
(iv) x × y = y × x
x = 0, y = 8
15
L.H.S. = x × y = 0 × 8
15 = 0
R.H.S. = y × x = 8
15 × 0 = 0
L.H.S. = R.H.S.
2. Ver ify the property : x × (y × z) = (x × y) ×
z by taking :
(i) x = 3
7, y =
5
12, z =
9
4
(ii) x = 0, y = 5
3, z =
4
9
(iii) x = 2
1, y =
4
5
, z =
5
7
(iv) x = 7
5, y =
13
12, z =
18
7
Solution—
(i) x × (y × z) = (x × y) × z
x = 3
7, y =
5
12, z =
9
4
L.H.S. = x × (y × z)
= 3
7 ×
9
4
5
12 =
3
7 ×
35
44
= 3
7 ×
15
16 =
153
167
= 45
112
R.H.S. = (x × y) × z =
5
12
3
7 ×
9
4
= 51
47
× 9
4 =
5
28 ×
9
4
https://www.arundeepselfhelp.info/index.php?route=product/product&path=139_141&product_id=111
30
= 95
428
= 45
112
L.H.S. = R.H.S.
(ii) x × (y × z) = (x × y) × z
x = 0, y = 5
3, z =
4
9
L.H.S. = x × (y × z) = 0 ×
4
9
5
3
= 0 ×
45
93
= 0 × 20
27 = 0
R.H.S. = (x × y) × z =
5
30 ×
4
9
= 0 × 4
9 = 0
L.H.S. = R.H.S.
(iii) x × (y × z) = (x × y) × z
x = 2
1, y =
4
5
, z =
5
7
L.H.S. = x × (y × z) = 2
1 ×
5
7
4
5
= 2
1 ×
54
75
= 2
1 ×
4
7 =
2
1 ×
4
7 =
42
71
= 8
7
R.H.S. = (x × y) × z
=
4
5
2
1 ×
5
7
= 42
51
× 5
7 =
8
5
×
5
7
=
58
75
=
18
71
= 8
7
= 8
7
L.H.S. = R.H.S.
(iv) x × (y × z) = (x × y) × z
x = 7
5, y =
13
12, z =
18
7
L.H.S. = x × (y × z) = 7
5 ×
18
7
13
12
= 7
5 ×
1813
712 =
7
5 ×
313
72
= 7
5 ×
39
14 =
397
145
= 391
25
= 39
10
R.H.S. = (x × y) × z =
13
12
7
5 ×
18
7
=
137
125
× 18
7 =
91
60 ×
18
7
= 1891
760
= 313
110
= 39
10
L.H.S. = R.H.S.
3. Verify the property : x × (y + z) = x × y +
x × z by taking :
(i) x = 7
3, y =
13
12, z =
6
5
(ii) x = 5
12, y =
4
15, z =
3
8
(iii) x = 3
8, y =
6
5, z =
12
13
(iv) x = 4
3, y =
2
5, z =
6
7
Solution—
(i) x × (y + z) = x × y + x × z
https://www.arundeepselfhelp.info/index.php?route=product/product&path=139_141&product_id=111
31
x = 7
3, y =
13
12, z =
6
5
L.H.S. = x × (y + z) = 7
3 ×
6
5
13
12
= 7
3 ×
78
6572
(LCM of 13, 6 = 78)
= 7
3 ×
78
7 =
787
73
= 261
11
= 26
1
R.H.S. = x × y + x × z
= 7
3 ×
13
12 +
7
3 ×
6
5
= 137
123
+ 67
53
= 91
36 +
27
51
= 91
36 +
14
5
= 182
6572 =
182
7
= 26
1
L.H.S. = R.H.S.
(ii) x × (y + z) = x × y + x × z
x = 5
12, y =
4
15, z =
3
8
L.H.S. = x × (y + z) = 5
12 ×
3
8
4
15
= 5
12 ×
12
3245 =
5
12 ×
12
13
= 125
1312
=
15
131
= 5
13
R.H.S. = x × y + x × z =
4
15
5
12 +
3
8
5
12
= [–3 × (–3)] +
5
84 = 9 +
5
32
= 1
9 +
5
32
= 5
3245 =
5
13
L.H.S. = R.H.S.
(iii) x × (y + z) = x × y + x × z
x = 3
8, y =
6
5, z =
12
13
L.H.S. = x × (y + z) = 3
8 ×
12
13
6
5
= 3
8 ×
12
1310 =
3
8 ×
12
3
= 31
12
= 3
2
R.H.S. = x × y + x × z
=
6
5
3
8 +
12
13
3
8
= 33
54
+ 33
132
= 9
20 +
9
26 =
9
2620
https://www.arundeepselfhelp.info/index.php?route=product/product&path=139_141&product_id=111
32
= 9
6 =
39
36
= 3
2
L.H.S. = R.H.S.
(iv) x × (y + z) = x × y + x × z
x = 4
3, y =
2
5, z =
6
7
L.H.S. = x × (y + z) = 4
3 ×
6
7
2
5
= 4
3 ×
6
715 =
4
3 ×
6
8
=
64
83
=
21
21
= 2
2 = 1
R.H.S. = x × y + x × z
=
2
5
4
3 +
6
7
4
3
=
24
53
+ 64
73
= 8
15 +
24
71
= 8
15 +
8
7 =
8
715 =
8
8 = 1
L.H.S. = R.H.S.
4. Use the distributivity of multiplication of
rational numbers over their addition to
simplify :
(i)5
3 ×
1
10
24
35(ii)
4
5 ×
5
16
5
8
(iii)7
2 ×
4
21
16
7(iv)
4
3 ×
409
8
Solution—
(i)5
3 ×
1
10
24
35
= 5
3 ×
24
35 +
5
3×
1
10 =
245
353
+ 15
10
= 81
71
+ 11
23
= 8
7 + 6 =
8
7 +
1
48
= 8
487 =
8
55
(ii)4
5 ×
5
16
5
8 =
4
5 ×
5
8 +
4
5 ×
5
16
= 54
85
+ 54
165
= 11
21
+ 11
41
= –2 – 4 = –6
(iii)7
2 ×
4
21
16
7 =
7
2 ×
16
7 –
7
2 ×
4
21
= 167
72
– 47
212
= 81
11
– 21
31
= 8
1 –
2
3
= 8
121 =
8
11
(iv)4
3 ×
409
8 =
4
3 ×
9
8 –
4
3 ×
1
40
= 94
83
– 14
403
= 31
21
– 11
103
= 3
2 –
1
30
= 3
902 =
3
88
5. Find the multiplicative inverse
(reciprocal) of each of the following
rational numbers :
(i) 9 (ii) –7
(iii)5
12(iv)
9
7
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33
(v)5
3
(vi)3
2 ×
4
9
(vii)8
5 ×
15
16(viii) –2 ×
5
3
(ix) –1 (x)3
0
(xi) 1
Solution—
(i) Multiplicative inverse of 9 = 9
1
(ii) Multiplicative inverse of –7 = 7
1
(iii) Multiplicative inverse of 5
12 =
12
5
(iv) Multiplicative inverse of 9
7 =
7
9
(v) Multiplicative inverse of 5
3
= 3
5
= 3
5
(vi) Multiplicative inverse of 3
2 ×
4
9 =
2
3×
9
4
= 92
43
= 31
21
= 3
2
(vii) Multiplicative inverse of 8
5 ×
15
16
= 5
8
×
16
15 =
165
158
= 21
31
= 2
3
= 12
13
= 2
3
(viii)Multiplicative inverse of –2 × 5
3
= 2
1
×
3
5
= 32
51
= 6
5
(ix) Multiplicative inverse of –1 = –1
(x) Multiplicative inverse of 0 = Does not exist
as division by 0 is not admissible.
(xi) Multiplicative inverse of 1 = 1
6. Name the property of multiplication of
rational numbers illustrated by the
following statements :
(i)16
5 ×
15
8 =
15
8 ×
16
5
(ii)5
17 × 9 = 9 ×
5
17
(iii)4
7×
12
13
3
8=
4
7×
3
8 +
4
7 ×
12
13
(iv)9
5 ×
8
9
15
4 =
15
4
9
5 ×
8
9
(v)17
13
× 1 =
17
13
= 1 ×
17
13
(vi)16
11 ×
11
16
= 1
(vii)13
2 × 0 = 0 = 0 ×
13
2
(viii)2
3 ×
4
5 +
2
3 ×
6
7 =
2
3 ×
6
7
4
5
Solution—
(i)16
5 ×
15
8 =
15
8 ×
16
5
It is commutative property
(ii)5
17 × 9 = 9 ×
5
17
It is commutative property
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34
(iii)4
7 ×
12
13
3
8 =
4
7 ×
3
8 +
4
7 ×
12
13
It is distributive property over addition
(iv)9
5 ×
8
9
15
4 =
15
4
9
5 ×
8
9
It is associative property
(v)17
13
× 1 =
17
13
= 1 ×
17
13
It is multiplicative identity.
(vi)16
11 ×
11
16
= 1
It is existance of multiplicative inverse.
(vii)13
2 × 0 = 0 = 0 ×
13
2
It is zero property of multiplication.
(viii)2
3 ×
4
5 +
2
3 ×
6
7 =
2
3 ×
6
7
4
5
It is distributive law of multiplication over
addition.
7. Fill in the blanks :
(i) The product of two positive rational
numbers is always ________.
(ii) The product of a positive rational number
and a negative rational number is always
________.
(iii) The product of two negative rational
numbers is always ________.
(iv) The reciprocal of a positive rational
number is ________.
(v) The reciprocal of a negative rational
number is ________.
(vi) Zero has ________ reciprocal.
(vii) The product of a rational number and its
reciprocal is ________.
(viii) The numbers ________ and ________ are
their own reciprocals.
(ix) If a is reciprocal of b, then the reciprocal
of b is ________.
(x) The number 0 is ________ the reciprocal
of any number.
(xi) Reciprocal of a
1, a 0 is ________.
(xii) (17 × 12)–1 = 17–1 × ________.
Solution—
(i) The product of two positive rational numbers
is always positive.
(ii) The product of a positive rational number
and a negative rational number is always
negative.
(iii) The product of two negative rational numbers
is always positive.
(iv) The reciprocal of a positive rational number
is positive.
(v) The reciprocal of a negative rational number
is negative.
(vi) Zero has no reciprocal.
(vii) The product of a rational number and its
reciprocal is 1.
(viii)The numbers 1 and –1 are their own
reciprocals.
(ix) If a is reciprocal of b, then the reciprocal of
b is a.
(x) The number 0 is not the reciprocal of any
number.
(xi) Reciprocal of a
1, a 0 is a.
(xii) (17 × 12)–1 = 17–1 × 12–1.
8. Fill in the blanks :
(i) – 4 × 9
7 =
9
7 × ________.
(ii)11
5 ×
8
3 =
8
3 × ________
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35
(iii)2
1 ×
12
5
4
3 =
2
1 × _____________ +
________ × 12
5
(iv)5
4 ×
9
8
7
5 =
___
5
4 ×
9
8
Solution—
(i) – 4 × 9
7 =
9
7 × – 4
(ii)11
5 ×
8
3 =
8
3 ×
11
5
(iii)2
1 ×
12
5
4
3 =
2
1 ×
4
3 +
2
1 ×
12
5
(iv)5
4 ×
9
8
7
5 =
7
5
5
4 ×
9
8
1. Divide :
(i) 1 by 2
1(ii) 5 by
7
5
(iii)4
3 by
16
9
(iv)
8
7 by
16
21
(v)4
7
by
64
63(vi) 0 by
5
7
(vii)4
3 by –6 (viii)
3
2 by
12
7
(ix) – 4 by 5
3(x)
13
3 by
65
4
Solution—
(i) 1 by 2
1 = 1
2
1 = 1 ×
1
2 = 2
(ii) 5 by 7
5 = 5
7
5 = 5 ×
5
7
= 15
17
=5
7 –
15
57
=5
35 = –7
EXERCISE 1.7
(iii)4
3 by
16
9
=
4
3
16
9
= 4
3 ×
9
16 =
94
163
= 31
41
= 3
4
(iv)8
7 by
16
21 =
8
7
16
21
= 8
7 ×
21
16
= 31
21
= 3
2
= 3
2
(v)4
7
by
64
63 =
4
7
64
63 =
4
7
×
63
64
= 91
161
= 9
16
=
19
116
= 9
16
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36
(vi) 0 by 5
7= 0
5
7 = 0 ×
7
5
= 0
(vii)4
3 by –6 =
4
3
1
6 =
4
3 ×
6
1
= 24
11
= 8
1
= 8
1
(viii)3
2 by
12
7 =
3
2
12
7 =
3
2 ×
7
12
= 71
42
= 7
8
=
17
18
= 7
8
(ix) – 4 by 5
3 = –4
5
3 = –4 ×
3
5
= 3
54
= 3
20
= 3
20
(x)13
3 by
65
4 =
13
3
65
4 =
13
3 ×
4
65
= 41
53
= 4
15
= 4
15
2. Find the value and express as a rational
number in standard form :
(i)5
2
15
26(ii)
3
10
12
35
(iii) – 6
17
8(iv)
99
40 (–20)
(v)27
22
18
110(vi)
125
36
75
3
Solution—
(i)5
2
15
26 =
5
2 ×
26
15 =
265
152
= 131
31
= 13
3
(ii)3
10
12
35 =
3
10 ×
35
12
= 353
1210
= 71
42
= 7
8
=
17
18
= 7
8
(iii) – 6 17
8 = – 6 ×
8
17
=
4
173
= 4
51
= 4
51
(iv)99
40 (–20) =
99
40 ×
20
1
= 199
12
= 99
2
= 99
2
(v)27
22
18
110 =
27
22 ×
110
18
= 11027
1822
= 53
21
= 15
2
= 15
2
(vi)125
36
75
3 =
125
36 ×
3
75
= 3125
7536
= 15
312
= 5
36
= 5
36
3. The product of two rational numbers is
15. If one of the numbers is –10, find the
other.
Solution—
Product of two numbers = 15
One number = –10
Second number = 15 (–10) = 15 × 10
1
= 2
13
= 2
3
=
12
13
= 2
3
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4. The product of two rational numbers is
9
8. If one of the numbers is
15
4, find
the other.
Solution—
Product of two numbers = 9
8
One number = 15
4
Second number = 9
8
15
4
= 9
8 ×
4
15
= 13
52
= 3
10
= 3
10
5. By what number should we multiply 6
1
so that the product may be 9
23 ?
Solution—
Product = 9
23
and given number = 6
1
Required number = 9
23
6
1
= 9
23 ×
1
6
= 13
223
= 3
46
= 3
46
6. By what number should we multiply
28
15, so that the product may be
7
5 ?
Solution—
Product of two numbers = 7
5
One number = 28
15
Required number = 7
5
28
15
= 7
5 ×
15
28
= 31
41
= 3
4
= 3
4
7. By what number should we multiply 13
8
so that the product may be 24 ?
Solution—
Product of two numbers = 24
One number = 13
8
Required number = 24 13
8
= 24 × 8
13
=
1
133
= 1
39
= 11
139
= 1
39 = –39
8. By what number should 4
3 be
multiplied in order to produce 3
2 ?
Solution—
Product = 3
2
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Let x be multiplied to 4
3 to get
3
2
x + 4
3 =
3
2 x =
3
2
4
3
x = 3
2 ×
3
4
= 33
42
= 9
8
= 19
18
= 9
8
Required number = 9
8
9. Find (x + y) (x – y), if
(i) x = 3
2, y =
2
3(ii) x =
5
2, y =
2
1
(iii) x = 4
5, y =
3
1(iv) x =
7
2, y =
3
4
(v) x = 4
1, y =
2
3
Solution—
(i) x = 3
2, y =
2
3
x + y = 3
2 +
2
3
= 6
94 =
6
13
and x – y = 3
2 –
2
3
= 6
94 =
6
5
Now (x + y) (x – y) = 6
13
6
5
= 6
13 ×
5
6
= 5
13
=
15
113
= 5
13
(ii) x = 5
2, y =
2
1
x + y = 5
2 +
2
1
= 10
54 =
10
9
x – y = 5
2 –
2
1
= 10
54 =
10
1
(x + y) (x – y) = 10
9
10
1 =
10
9 ×
1
10
= 1
9
= –9
(iii) x = 4
5, y =
3
1
x + y = 4
5 +
3
1
= 12
415 =
12
11
x – y = 4
5 –
3
1
= 12
415 =
12
19
(x + y) (x – y) = 12
11
12
19=
12
11×
19
12=
19
11
(iv) x = 7
2, y =
3
4
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x + y = 7
2 +
3
4
= 21
286 =
21
34
x – y = 7
2 –
3
4
21
286 =
21
22
(x + y) (x – y) = 21
34
21
22
= 21
34 ×
22
21
= 111
117
= 11
17
=
111
117
= 11
17
(v) x = 4
1, y =
2
3
x + y = 4
1 +
2
3 =
4
61 =
4
7
x – y = 4
1 –
2
3
= 4
61 =
4
5
(x + y)(x – y) =4
7
4
5=
4
7×
5
4
=
5
7
= 15
17
= 5
7
10. The cost of 73
2 metres of rope is Rs 12
4
3.
Find its cost per metre.
Solution—
Cost of 73
2 m of rope = Rs. 12
4
3
or cost of 3
23 m of rope = Rs.
4
51
Cost of 1 m of rope = Rs. 4
51
3
23
= Rs. 4
51 ×
23
3 = Rs.
92
153 = Rs. 1
92
61
11. The cost of 23
1 metres of cloth is Rs.
754
1. Find the cost of cloth per metre.
Solution—
Cost of 23
1 m or
3
7 m of cloths = Rs. 75
4
1
= Rs. 4
301
Cost of 1 m cloth = Rs. 4
301
3
7
= Rs. 4
301 ×
7
3 = Rs.
14
343
= Rs. 4
129 = Rs. 32
4
1 = Rs. 32.25
12. By what number should 16
33 be divided
to get 4
11 ?
Solution—
Let x be divided, then
16
33 x =
4
11
16
33 × x
1 =
4
11
x = 16
33 ×
11
4
=
14
13
= 4
3
Required number = 4
3
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13. Divide the sum of 5
13 and
7
12 by the
product of 7
31 and
2
1.
Solution—
Sum of 5
13 and
7
12 =
5
13 +
7
12
= 35
6091 =
35
31
Product of 7
31 and
2
1 =
7
31 ×
2
1
= 27
131
= 14
31
Required number = 35
31
14
31
= 35
31 ×
31
14 =
15
21
= 5
2
14. Divide the sum of 12
65 and
7
12 by their
difference.
Solution—
Sum of 12
65 and
7
12 =
12
65 +
7
12
= 84
144455 =
84
599
and difference of 12
65 and
7
12 =
12
65 –
7
12
= 84
144455 =
84
311
Required number = 84
599
84
311
= 84
599 ×
311
84 =
311
599
15. If 24 trousers of equal size can be
prepared in 54 metres of cloth, what
length of cloth is required for each
trouser ?
Solution—
Cloth required for 24 trousers = 54 m
Cloth required for 1 trouser = (54 24) m
= 24
54 =
4
9 = 2
4
1m
1. Find a rational number between –3 and
1.
Solution—
We know that a number between two rational
numbers a, b = 2
ba
Rational number between –3 and 1 = 2
13
= 2
2 = –1
2. Find any five rational number less than 1.
Solution—
_ 1 = 5
5 and number 0, 1, 2, 3, 4 are less than 5
Five numbers less than 1 can be
0, 5
1,
5
2,
5
3,
5
4
3. Find four rational numbers between 9
2
and 9
5.
EXERCISE 1.8
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Solution—
_ –1, 0, 1, 2, 3, 4 lie between –2 and 5
Four rational numbers between 9
2 and
9
5
can be 9
1, 0,
9
1,
9
2 ..........
4. Find two rational numbers between 5
1 and
2
1.
Solution—
5
1,
2
1(LCM of 5, 2 = 10)
5
1 =
25
21
= 10
2 and
2
=
52
5
= 10
5
_ 3 and 4 lie between 2 and 5
Number between 10
2 and
10
5,
can be 10
3,
10
4
5. Find ten rational numbers between 4
1 and
2
1.
Solution—
LCM of 4, 2 = 4
4
1 =
4
1 and
2
1 =
22
21
= 4
2
_ We need 10 rational numbers
We extend them as 80
20 and
80
40
Rational numbers between 80
20 and
80
40
Can be 80
21,
80
22,
80
23,
80
24, .........
80
39
6. Find ten rational numbers between 5
2
and 2
1.
Solution—
5
2 =
45
42
= 20
8
and 2
1 =
102
101
= 20
10
Now numbers lying between –8, 10 will be
–7, –6, –5, –4, –3, –2, –1, 0, 1, 2, 3, 4,.....9
The rational numbers will be
20
7,
20
6,
20
5,
20
4, .....,
20
8,
20
9
7. Find ten rational numbers between 5
3 and
4
3.
Solution—
5
3 and
4
3
LCM of 5, 4 = 20
But we need 10 numbers.
5
3 =
165
163
= 80
48 and
4
3 =
204
203
= 80
60
The numbers lie between 48 and 60 will be
49, 50, 51, 52, 53, 54, ......... 60
Rational numbers are
80
49,
80
50,
80
51,
80
52,
80
53 .........
80
59
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Points to Remember
1. If ‘a’ is multiplied with itself for n-time
then it is a × a × a × a ...... × a (n-times) =
an and is read as ‘a’ raised to the power n.
2. Laws fo exponents—
(i) am × an = am + n
(ii) am an = am – n (m > n)
(iii) (am)n = amn = (an)m
(iv) (ab)n = anbn
(v)
n
b
a
=
n
n
b
a
(vi) a1 = a and a0 = 1
Here a and b are non-zero rational numbers.
3. Negative Integral Exponents—
We know that :
100 = 1
101 = 10
102 = 10 × 10 = 100
103 = 10 × 10 × 10 = 1000
Therefore
10–1 = 10
1
2POWERS
10–2 = 100
1
10–3 = 1000
1 and so on
a–n = na
1 and an = n
a
1
4. Use of exponents to express small
numbers in standard form—
Method :
(i) Obtain the number and see whether it is
between 1 and 10 or it is less than 1.
(ii) If the number is between 1 and 10, then
write it as the product of the number itself
and 100.
(iii) If the number is less than one, then move
the decimal point to the right so that there
is just one digit on the left side of the
decimal point. Write the given number as
the product of the number so obtained and
10–n, when n is the number of places the
decimal point has been moved to the right.
The number so obtained is the standard
form of the given number.
EXERCISE 2.1
1. Express each of the following as a rational number of the form q
p, where p and q are
integers and q 0 :
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(i) 2–3 (ii) (–4)–2
(iii) 23
1 (iv)
5
2
1
(v)
2
3
2
Solution—
(i) 2–3 = 32
1 =
m
m
aa
1
= 222
1
=
8
1
(ii) (–4)–2 = 24
1
= 44
1
= 16
1
(iii) 23
1 = 32
m
ma
a
1
= 3 × 3 = 9
(iv)
5
2
1
=
5
1
2
= 2 × 2 × 2 × 2 × 2 = 32
(v)
2
3
2
=
2
2
3
=
2
3 ×
2
3 =
4
9
2. Find the values of each of the following :
(i) 3–1 + 4–1 (ii) (30 + 4–1) × 22
(iii) (3–1 + 4–1 + 5–1)0
(iv)
111
4
1
3
1
Solution—
(i) 3–1 + 4–1 = 3
1 +
4
1
= 12
34 =
12
7
(ii) (3º + 4–1) × 22 =
4
11 × 4
mama
m 1,10
= 4
14 × 4 =
4
5 × 4 = 5
(iii) (3–1 + 4–1 + 5–1)0
=
0
5
1
4
1
3
1
= 1 (_ a0 = 1)
(iv)
111
4
1
3
1
= {(3)1 – (4)1}–1 = (3 – 4)–1 = (–1)–1
= 11
1
= 1
1
= –1
3. Find the values of each of the following :
(i)
1
2
1
+
1
3
1
+
1
4
1
(ii)
2
2
1
+
2
3
1
+
2
4
1
(iii) (2–1 × 4–1) + 2–2
(iv) (5–1 × 2–1) + 6–1
Solution—
(i)
1
2
1
+
1
3
1
+
1
4
1
= (2)1 + (3)1 + (4)1 = 2 + 3 + 4 = 9
(ii)
2
2
1
+
2
3
1
+
2
4
1
= (2)2 + (3)2 + (4)2
= 4 + 9 + 16 = 29
(iii) (2–1 × 4–1) 2–2
=
4
1
2
1 22
1
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= 8
1
4
1 =
8
1 ×
1
4 =
2
1
(iv) (5–1 × 2–1) 6–1 =
2
1
5
1
6
1
= 10
1 ×
1
6 =
10
6 =
210
26
= 5
3
4. Simplify :
(i) (4–1 × 3–1)2 (ii) (5–1 6–1)3
(iii) (2–1 + 3–1)–1 (iv) (3–1 × 4–1)–1 × 5–1
Solution—
(i) (4–1 × 3–1)2 =
2
3
1
4
1
=
2
12
1
= 12
1 ×
12
1 =
144
1
(ii) (5–1 6–1)3 =
3
6
1
5
1
=
3
1
6
5
1
=
3
5
6
=
555
666
= 125
216
(iii) (2–1 + 3–1)–1 =
1
3
1
2
1
=
1
6
23
=
1
6
5
=
5
6
(iv) (3–1 × 4–1)–1 × 5–1 =
1
4
1
3
1
× 5–1
=
1
12
1
×
5
1
= (12)1 × 5
1 =
5
12
5. Simplify :
(i) (32 + 22) ×
3
2
1
(ii) (32 – 22) ×
3
3
2
(iii)
33
2
1
3
1
3
4
1
(iv) (22 + 32 – 42)
2
2
3
Solution—
(i) (32 + 22) ×
3
2
1
= (9 + 4) ×
8
1 = 13 ×
8
1
= 8
13
(ii) (32 – 22) ×
3
3
2
= (9 – 4) ×
3
2
3
= 5 × 8
27 =
8
135
(iii)
33
2
1
3
1
3
4
1
= [(3)3 – (2)3] (4)3
= (27 – 8) 64 = 19 64 = 64
19
(iv) (22 + 32 – 42)
2
2
3
= (4 + 9 – 16) 4
9 = –3 ×
9
4
= 3
41 =
3
4
6. By what number should 5–1 be multiplied
so that the product may be equal to
(–7)–1 ?
Solution—
Let x be multiplied to 5–1, then
x × 5–1 = (–7)–1
x × 5
1 = 17
1
5
x =
7
1
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x = 5 × 7
1
=
7
5
Required number = 7
5
7. By what number should
1
2
1
be
multiplied so that the product many be
equal to
1
7
4
?
Solution—
Let x be the required number, then
x ×
1
2
1
=
1
7
4
x (2)1 =
1
4
7
2x =
4
7
x = 24
7
= 8
7
Required number = 8
7
8. By what number should (–15)–1 be divided
so that the quotient may be equal to
(–5)–1 ?
Solution—
Let the number should be divided by = x
their (–15)–1 x = (–5)–1 15
1
x
= 15
1
15
1 ×
x
1 =
5
1
x15
1 =
5
1
–15x = –5
x = 15
5
= 3
1
Required number = 3
1
1. Write each of the following in exponential
form :
(i)
1
2
3
×
1
2
3
×
1
2
3
×
1
2
3
(ii)
2
5
2
×
2
5
2
×
2
5
2
Solution—
(i)
1
2
3
×
1
2
3
×
1
2
3
×
1
2
3
1111
2
3
=
4
2
3
EXERCISE 2.2
(ii)
2
5
2
×
2
5
2
×
2
5
2
222
5
2
=
6
5
2
2. Evaluate :
(i) 5–2 (ii) (–3)–2
(iii)
4
3
1
(iv)
1
2
1
Solution—
(i) 5–2 = 25
1 =
55
1
=
25
1
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(ii) (–3)–2 = 23
1
= 33
1
= 9
1
(iii)
4
3
1
=
4
1
3
= 3 × 3 × 3 × 3 = 81
(iv)
1
2
1
= (–2)1 = –2
3. Express each of the following as a rational
number in the form q
p :
(i) 6–1 (ii) (–7)–1
(iii)
1
4
1
(iv) (–4)–1 ×
1
2
3
(v)
1
5
3
×
1
2
5
Solution—
(i) 6–1 = 6
1
(ii) (–7)–1 = 7
1
(iii)
1
4
1
=
1
4 = 4
(iv) (–4)–1 ×
1
2
3
=
1
4
1
×
1
3
2
= 4
1 ×
3
2 =
32
11
= 6
1
(v)
1
5
3
×
1
2
5
=
1
3
5
×
1
5
2
= 3
5 ×
5
2 =
3
2
4. Simplify :
(i) {4–1 × 3–1}2 (ii) {5–1 6–1}3
(iii) (2–1 + 3–1)–1 (iv) {3–1 × 4–1}–1 × 5–1
(v) (4–1 – 5–1)–1 3–1
Solution—
(i) { 4–1 × 3–1}2 =
2
3
1
4
1
=
2
12
1
= 12
1 ×
12
1 =
144
1
(ii) {5–1 6–1}3 =
3
6
1
5
1
=
3
1
6
5
1
=
3
5
6
=
555
666
= 125
216
(iii) (2–1 + 3–1)–1 =
1
3
1
2
1
=
1
6
23
=
1
6
5
=
5
6
(iv) {3–1 × 4–1}–1 × 5–1 =
1
4
1
3
1
×
5
1
=
1
12
1
×
5
1 =
1
12 ×
5
1 =
5
12
(v) (4–1 – 5–1)–1 3–1 =
1
5
1
4
1
3
1
=
1
20
45
×
1
3 =
1
20
1
×
1
3
= 1
20 ×
1
3 = 60
5. Express each of the following rational
numbers with a negative exponent :
(i)
3
4
1
(ii) 35 (iii)
4
5
3
(iv)
34
2
3
(v)
34
3
7
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Solution—
(i)
3
4
1
=
3
1
4
= (4)–3
(ii) 35 =
5
3
1
(iii)
4
5
3
=
4
3
5
(iv)
34
2
3
=
34
2
3
=
12
2
3
(v)
34
3
7
=
34
3
7
=
12
3
7
6. Express each of the following rational
numbers with a positive exponent :
(i)
2
4
3
(ii)
8
4
5
(iii) 43 × 4–9
(iv)
43
3
4
(v)
24
2
3
Solution—
(i)
2
4
3
=
2
3
4
(ii)
5
4
=
5
5
4
(iii) 43 × 4–9 = 43 – 9 = 4– 6 =
6
4
1
(iv)
43
3
4
=
43
3
4
=
12
3
4
(v)
24
2
3
=
24
2
3
=
8
2
3
=
8
3
2
7. Simplify :
(i)
33
2
1
3
1
3
4
1
(ii) (32 – 22) ×
3
3
2
(iii) 1
11
42
1
(iv)
12
2
4
1
(v)
32
3
2
×
4
3
1
× 3–1 × 6–1
Solution—
(i)
33
2
1
3
1
3
4
1
= (33 – 23) (4)3
= (27 – 8) 64 = 64
19
(ii) (32 – 22) ×
3
3
2
= (32 – 22) ×
3
2
3
= (9 – 4) ×
8
27
= 5 × 8
27 =
8
135
(iii) 1
11
42
1
= 1
11
4
12
=
1
4
12
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48
=
1
2
1
= (–2)1 = –2
(iv)
12
2
4
1
=
222
4
1
=
4
4
1
=
4
1 ×
4
1 ×
4
1 ×
4
1 =
256
1
(v)
32
3
2
×
4
3
1
× 3–1 × 6–1
=
32
3
2
×
4
1
3
× 13
1 × 16
1
=
6
3
2
× (3)4 ×
3
1 ×
6
1
= 333333
222222
× 3 × 3 × 3 × 3 × 3
1
× 6
1
= 6333
64
=
281
64
=
81
32
8. By what number should 5–1 be multiplied
so that the product may be equal to (–7)–1 ?
Solution—
Let x be multiplied, the
x × 5–1 = (–7)–1
x = 1
1
5
7
=
1
5
7
=
1
7
5
= 7
5
Required number = 7
5
9. By what number should
1
2
1
be
multiplied so that the product may be
equal to
1
7
4
?
Solution—
Let x be multiplied, then
x ×
1
2
1
=
1
7
4
x =
1
7
4
1
2
1
=
1
4
7
1
1
2
=
4
7
2
= 4
7
×
2
1 =
8
7
=
8
7
Required number = 8
7
10. By what number should (–15)–1 be divided
so that the quotient may be equal to
(–5)–1 ?
Solution—
Let (–15)–1 be divided by x
(–15)–1 x = (–5)–1
15
1 x =
1
5
1
15
1
×
x
1 =
5
1
x15
1
=
5
1
x = 15
5
= 3
1
Required number = 3
1
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11. By what number should
2
3
5
be
multiplied so that the product may be
1
3
7
?
Solution—
Let x be multiplied, then
x ×
2
3
5
=
1
3
7
x =
1
3
7
2
3
5
x =
1
7
3
2
5
3
=
7
3
25
9 =
7
3 ×
9
25
= 21
25
Required number = 21
25
12. Find x, if
(i)
4
4
1
×
8
4
1
=
x4
4
1
(ii)
19
2
1
8
2
1
=
12
2
1
x
(iii)
3
2
3
×
5
2
3
=
12
2
3
x
(iv)
3
5
2
×
15
5
2
=
x32
5
2
(v)
x
4
5
4
4
5
=
5
4
5
(vi)
12
3
8
x
×
5
3
8
=
2
3
8
x
Solution—
(i)
4
4
1
×
8
4
1
=
x4
4
1
84
4
1
=
x4
4
1
12
4
1
=
x4
4
1
Comparing, we get :
–4x = –12 x = 4
12
= 3
x = 3
(ii)
19
2
1
8
2
1
=
12
2
1
x
819
2
1
=
12
2
1
x
27
2
1
=
12
2
1
x
Comparing, we get :
–2x + 1 = –27
–2x = –27 – 1 = –28
x = 2
28
= 14
x = 14
(iii)
3
2
3
×
5
2
3
=
12
2
3
x
53
2
3
=
12
2
3
x
2
2
3
=
12
2
3
x
Comparing, we get :
2x + 1 = 2
2x = 2 – 1 = 1
x = 2
1
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(iv)
3
5
2
×
15
5
2
=
x32
5
2
153
5
2
=
x32
5
2
12
5
2
=
x32
5
2
Comparing, we get :
2 + 3x = 12
3x = 12 – 2 = 10
x = 3
10
(v)
x
4
5
4
4
5
=
5
4
5
4
4
5
x
=
5
4
5
Comparing, we get :
–x + 4 = 5 –x = 5 – 4 = 1
x = –1
(vi)
12
3
8
x
×
5
3
8
=
2
3
8
x
512
3
8
x
=
2
3
8
x
Comparing, we get :
2x + 1 + 5 = x + 2
2x – x = 2 – 1 – 5 x = –4
x = –4
13. (i) If x =
2
2
3
×
4
3
2
, find the value
of x–2.
(ii) If x =
2
5
4
2
4
1
, find the value of x–1.
Solution—
(i) x =
2
2
3
×
4
3
2
2
2
3
×
4
3
2
=
42
2
3
=
6
2
3
x–2 =
26
2
3
=
12
2
3
=
12
3
2
(ii) x =
2
5
4
2
4
1
=
2
4
5
2
4
1
=
2
4
1
4
5
=
2
1
4
4
5
= (5)2 = 25
x–1 = (25)–1 = 25
1
14. Find the value of x for which 52x 5–3 = 55.
Solution—
52x 5–3 = 55
52x 55 = 5–3
52x = 55–3 = 52
Comparing, we get :
2x = 2 x = 2
2 = 1
x = 1
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51
1. Express the following numbers in
standard form :
(i) 6020000000000000
(ii) 0.00000000000942
(iii) 0.00000000085 (iv) 846 × 107
(v) 3759 × 10–4 (vi) 0.00072984
(vii) 0.000437 × 104 (viii) 4 100000
Solution—
(i) 6020000000000000
= 6.02 × 1000000000000000
= 6.02 × 1015
(ii) 0.00000000000942
= 9.42 × 0001000000000
1 = 9.42 × 1210
1
= 9.42 × 10–12
(iii) 0.00000000085
= 8.5 × 01000000000
1 = 8.5 × 1010
1
= 8.5 × 10–10
(iv) 846 × 107 = 8.46 ×100 × 107
= 8.46 × 102 × 107
= 8.46 × 109
(v) 3759 × 10–4 = 3.759 × 1000 × 10–4
= 3.759 × 103 × 10–4
= 3.759 × 10–1
(vi) 0.00072984 = 10000
2984.7
= 410
2984.7 = 7.2984 × 10–4
(vii) 0.000437 × 104 = 10000
1037.4 4 = 4
4
10
1037.4
= 4.37 × 104 × 10–4
= 4.37 × 100
= 4.37 × 1 = 4.37
(viii)4 100000
= 100000
4 = 510
4
= 4 × 10–5
2. Write the following numbers in the usual
form :
(i) 4.83 × 107 (ii) 3.02 × 10–6
(iii) 4.5 × 104 (iv) 3 × 10–8
(v) 1.0001 × 109 (vi) 5.8 × 102
(vii) 3.61492 × 106 (viii) 3.25 × 10–7
Solution—
(i) 4.83 × 107
= 4.83 × 10000000
= 48300000
(ii) 3.02 × 10–6
= 610
02.3 =
1000000
02.3
= 0.00000302
(iii) 4.5 × 104 = 4.5 × 10000
= 45000
(iv) 3 × 10–8 = 810
3
= 100000000
3
= 0.00000003
(v) 1.0001 × 109
= 1.0001 × 1000000000
= 1000100000
(vi) 5.8 × 102
= 5.8 × 100 = 580
(vii) 3.61492 × 106
= 3.61492 × 1000000
= 3614920
(viii)3.25 × 10–7
= 710
25.3 =
10000000
25.3
= 0.000000325
EXERCISE 2.3
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52
Choose the correct alternative in each of
the following :
1. Square of
3
2 is
(a) –3
2(b)
3
2
(c) –9
4(d)
9
4
Solution—
(d)
9
4
3
2
3
2
3
22
2. Cube of 2
1 is
(a)8
1(b)
16
1
(c) –8
1(d)
16
1
Solution—
(c)
8
1
2
1
2
1
2
1
2
13
3. Which of the following is not equal to
4
5
3
?
(a)
4
4
5
3(b) 4
4
5
3
(c) – 4
4
5
3(d)
5
3 ×
5
3 ×
5
3 ×
5
3
Solution—
(c) _
4
5
3
=
4
4
5
3 or
4
4
5
3
or
5
3 ×
5
3
× 5
3 ×
5
3
4. Which of the following in not reciprocal
of
4
3
2
?
(a)
4
2
3
(b)
4
3
2
(c)
4
2
3
(d)
4
4
2
3
Solution—
(c)
44
2
3ofreciprocalis
2
3
5. Which of the following numbers is not
equal to 27
8 ?
(a)
3
3
2
(b)
3
3
2
(c)
3
3
2
(d)
3
2 ×
3
2 ×
3
2
Solution—
(a)
8
27
2
3
3
233
6.
5
3
2
is equal to
(a)
5
3
2
(b)
5
2
3
(c)3
52 x(d)
53
2
Exercise (MCQs)
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53
Solution—
(b)
55
2
3
3
2
7.
5
2
1
×
3
2
1
is equal to
(a)
8
2
1
(b)
8
2
1
(c)
8
4
1
(d)
15
2
1
Solution—
(a)
83535
2
1
2
1
2
1
2
1
8.
3
5
1
8
5
1
is equal to
(a)
5
5
1
(b)
11
5
1
(c) (–5)5 (d)
5
5
1
Solution—
(c)
558383
55
1
5
1
5
1
5
1
9.
7
5
2
5
5
2
is equal to
(a)25
4(b)
25
4
(c)
12
5
2
(d)
4
25
Solution—
(a)
25
4
5
2
5
2
5
2
5
2
5
2
5
2
2
5757
10.
42
3
1
is equal to
(a)
6
3
1
(b)
8
3
1
(c)
24
3
1
(d)
16
3
1
Solution—
(b)
42
3
1
=
42
3
1
=
8
3
1
11.
0
5
1
is equal to
(a) 0 (b)5
1
(c) 1 (d) 5
Solution—
(c)
)1(1
5
1 0
0
a
12.
1
2
3
is equal to
(a)3
2(b) –
3
2
(c)2
3(d) none of these
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54
Solution—
(b)
3
2
3
2
2
311
13.
5
3
2
×
5
7
5
is equal to
(a)
10
7
5
3
2
(b)
5
7
5
3
2
(c)
25
7
5
3
2
(d)
25
7
5
3
2
Solution—
(b)
555
7
5
3
2
7
5
3
2
{_ am.bm = (ab)m}
14.
5
4
3
5
3
5
is equal to
(a)
5
3
5
4
3
(b)
1
3
5
4
3
(c)
0
3
5
4
3
(d)
10
3
5
4
3
Solution—
(a)
555
3
5
4
3
3
5
4
3
m
mm
b
a
b
a
15. For any two non-zero rational numbers a
and b, a4 b4 is equal to
(a) (a b)1 (b) (a b)0
(c) (a b)4 (d) (a b)8
Solution—
(c) {_ a4 b4 = (a b)4}
16. For any two rational numbers a and b, a5
× b5 is equal to
(a) (a × b)0 (b) (a × b)10
(c) (a × b)5 (d) (a × b)25
Solution—
(c) {_ a5 × b5 = (a × b)5}
17. For a non-zero rational number a, a7 a12 is equal to
(a) a5 (b) a–19
(c) a–5 (d) a19
Solution—
(c) {a7 a12 = a7 – 12 = a–5}
18. For a non-zero rational number a, (a3)–2
is equal to
(a) a6 (b) a–6
(c) a–9 (d) a1
Solution—
(b) {(a3)–2 = a3 × (–2) = a–6}
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55
Points to Remember
1. Square— Square of a number is that
number raised to the power i.e. a × a = a2,
a2 is the square of a.
2. A perfect square— A natural number ‘n’
is called a perfect square or a square
number, if there exist a natural number m,
such that n = m2.
3. Check of a perfect square :
(i) Write the natural number.
(ii) Write the number as a product of prime
numbers.
(iii) Group the factors in pair such a way that
both the factors in each pair are equal.
(iv) See whether the some factors are left over
or not. If no factor is left over, their it is a
perfect square. Otherwise, it is not a
perfect square.
(v) Take one factor out of each pair and
multiply them. It is the square root of the
given number.
4. Properties of the square numbers—
(i) A number ending 2, 3, 7 or 8 is never a
perfect square.
(ii) The number of zeros at the end of a perfect
square is never odd. So, a number ending
one zero is not a perfect square.
(iii) Square of even numbers is always even
number.
(iv) Square of odd numbers is always an odd
number.
(v) Square of a natural number n = 1 + 3 + 5 +
....... n.
(vi) Square of a natural number other than 1, is
either a multiple of 3 or exceeds a multiple
of 3 by 1.
(vii) Square of a natural number other than 1, is
either a multiple of 4 or exceeds a multiple
of 4 by 1.
(viii)For any natural number n greater than 1.
(2n, n2 – 1, n2 + 1) is a pythagorean triplets
or A triplet (m, n, p) of three natural
numbers, m, n, p are called Phythagorean
triplet it m2 + n2 = p2
(ix) There is no natural numbers p and q such
that p2 = 2q2
(x) (n + 1)2 – n2 = (n + 1) + n.
(xi) The square root of a given natural number
n is that natural number which when
multiplied itself given n, as the product and
we denote the square root of n by n or
2
1
n.
If n or 2
1
n = m, then m2 = n.
(xii) Square root can be formed by
(a) By prime factorization method.
(b) By division method.
(c) By the given table of square roots.
(xiii) Some formulas are also used
(a + b)2 = a2 + 2ab + b2
and (a – b)2 = a2 – 2ab + b2
(xv) Number ending 5
n (n + 1)
(n5)2 = write n (n + 1) before 25
(xvi) Number of the form 5a (two digit number)
(5a)2 = (25 + a) × 100 + a2
(xvii) Number of the form 5ab
(5ab)2 = (250 + ab) × 1000 + (ab)2
3SQUARES AND
SQUARE ROOTS
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1. Which of the following numbers are
perfect squares ?
(i) 484 (ii) 625
(iii) 576 (iv) 941
(v) 961 (vi) 2500
Solution—
(i) 484 = 2 × 2 × 11 × 11
2 4842 242
11 12111 11
1
Grouping the factors in pairs, we have left
no factor unpaired
484 is a perfect square of 22
(ii) 625 = 5 × 5 × 5 × 5
5 6255 1255 255 5
1
Grouping the factors in pairs, we have left
no factor unpaired
625 is a perfect square of 25.
(iii) 576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
2 5762 2882 1442 722 362 183 93 3
1
Grouping the factors in pairs, we see that no
factor is left unpaired
576 is a perfect square of 24
(iv) 941 has no prime factors
941 is not a perfect square.
(v) 961 = 31 × 31
Grouping the factors in pairs,
we see that no factor is left unpaired
961 is a perfect square of 31
(vi) 2500 = 2 × 2 × 5 × 5 × 5 × 5
2 25002 12505 6255 1255 255 5
1
Grouping the factors in pairs, we see that no
factor is left unpaired
2500 is a perfect square of 50
2. Show that each of the following numbers
is a perfect square. Also find the number
whose square is the given number in each
case :
(i) 1156 (ii) 2025
(iii) 14641 (iv) 4761
Solution—
(i) 1156 = 2 × 2 × 17 × 17
2 11562 578
17 28917
Grouping the factors in pairs, we see that no
factor is left unpaired
1156 is a perfect square of 2 × 17 = 34
(ii) 2025 = 3 × 3 × 3 × 3 × 5 × 5
3 20253 6753 2253 755 255 5
1
EXERCISE 3.1
31 96131 31
1
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Grouping the factors in pairs, we see that no
factor is left unpaired
2025 is a perfect square of 3 × 3 × 5 = 45
(iii) 14641 = 11 × 11 × 11 × 11
11 1464111 133111 12111 11
1
Grouping the factors in pairs, we see that no
factor is left unpaired
14641 is a perfect square of 11 × 11 = 121
(iv) 4761 = 3 × 3 × 23 × 23
3 47613 1587
23 52923 23
1
Grouping the factors in pairs, we see that no
factor is left unpaired
4761 is a perfect square of 3 × 23 = 69
3. Find the smallest number by which the
given number must be multiplied so that
the product is a perfect square.
(i) 23805 (ii) 12150
(iii) 7688
Solution—
(i) 23805 = 3 × 3 × 5 × 23 × 23
3 238053 79355 2645
23 52923 23
1
Grouping the factors in pairs of equal factors,
we see that 5 is left unpaird
In order to complete the pairs, we have to
multiply 23805 by 5, then the product will
be the perfect square.
Requid smallest number = 5
(ii) 12150 = 2 × 3 × 3 × 3 × 3 × 3 × 5 × 5
2 121503 60753 20253 6753 2253 755 255 5
1
Grouping the factors in pairs of equal factors,
we see that factors 2 and 3 are left unpaired
In order to complete the pairs, we have to
multiply 12150 by 2 × 3 = 6 i.e., then the
product will be the complete square.
Required smallest number = 6
(iii) 7688 = 2 × 2 × 2 × 31 × 31
2 76882 38442 1922
31 96131 31
1
Grouping the factors in pairs of equal factors,
we see that factor 2 is left unpaired
In order to complete the pairs we have to
multiply 7688 by 2, then the product will be
the complete square
Required smallest number = 2
4. Find the smallest number by which the
given number must be divided so that the
resulting number is a perfect square.
(i) 14283 (ii) 1800
(iii) 2904
Solution—
(i) 14283 = 3 × 3 × 3 × 23 × 23
3 142833 47613 1587
23 52923 23
1
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Grouping the factrors in pairs of equal
factors, we see that 3 is left unpaired
Dividing by 3, the quotient will be the perfect
square.
(ii) 1800 = 2 × 2 × 2 × 3 × 3 × 5 × 5
2 18002 9002 4503 2253 755 255 5
1
Grouping the factors in pair of equal factors,
we see that 2 is left unpaired.
Dividing by 2, the quotient will be the perfect
square.
(iii) 2904 = 2 × 2 × 2 × 3 × 11 × 11
2 29042 14522 7263 363
11 12111 11
1
Grouping the factors in pairs of equal factors,
we see that 2 × 3 we left unpaired
Dividing by 2 × 3 = 6, the quotient will be
the perfect square.
5. Which of the following numbers are
perfect squares ?
11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121
Solution—
11 is not a perfect square as 11 = 1 × 11
12 is not a perfect square as 12 = 2 × 2 × 3
16 is a perfect square as 16 = 2 × 2 × 2 × 2
32 is not a perfect square as 32 = 2 × 2 × 2
× 2 × 2
36 is a perfect square as 36 = 2 × 2 × 3 × 3
50 is not a perfect square as 50 = 2 × 5 × 5
64 is a perfect square as 64 = 2 × 2 × 2 × 2
× 2 × 2
79 is not a perfect square as 79 = 1 × 79
81 is a perfect square as 81 = 3 × 3 × 3 × 3
111 is not a perfect square as 111 = 3 × 37
121 is a perfect square as 121 = 11 × 11
Hence 16, 36, 64, 81 and 121 are perfect
squares.
6. Using prime factorization method, find
which of the following numbers are
perfect squares ?
189, 225, 2048, 343, 441, 2916, 11025, 3549
Solution—
(i) 189 = 3 × 3 × 3 × 7 3 1893 633 217 7
1
Grouping the factors in pairs, we see that
are 3 and 7 are left unpaired
189 is not a perfect square
(ii) 225 = 3 × 3 × 5 × 5
3 2253 755 255 5
1
Grouping the factors in pairs, we see no
factor left unpaired
225 is a perfect square
(iii) 2048 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ×
2 × 2
2 20482 10242 5122 2562 1282 642 322 162 82 42 2
1
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Grouping the factors in pairs, we see that
one 2 is left unpaired
2048 is not a perfect square.
(iv) 343 = 7 × 7 × 7
7 3437 497 7
1
Grouping the factors in pairs, we see that
one 7 is left unpaired
343 is not a perfect square.
(v) 441 = 3 × 3 × 7 × 7
3 4413 1477 497 7
1
Grouping the factors in pairs, we see that no
factor is left unpaired
441 is a perfect square.
(vi) 2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
2 29162 14583 7293 2433 813 273 93 3
1
Grouping the factors in pairs, we see that no
factor is left unpaired
2916 is a perfect square.
(vii) 11025 = 3 × 3 × 5 × 5 × 7 × 7
3 110253 36755 12255 2457 497 7
1
Grouping the factors in pairs, we see that no
factor is left unpaired
11025 is a perfect square.
(viii)3549 = 3 × 7 × 13 × 13 3 35497 1183
13 16913 13
1
Grouping the factors in pairs, we see that 3,
7 are left unpaired
3549 is not a perfect square.
7. By what number should each of the
followng numbers be multiplied to get a
perfect square in each case ? Also, find
the number whose square is the new
number.
(i) 8820 (ii) 3675
(iii) 605 (iv) 2880
(v) 4056 (vi) 3468
(vii) 7776
Solution—
(i) 8820 = 2 × 2 × 3 × 3 × 5 × 7 × 7
2 88202 44103 22053 7355 2457 497 7
1
Grouping the factors in pairs, we see that 5
is left unpaired
By multiplying 8820 by 5, we get the perfect
square and square root of product will be
= 2 × 3 × 5 × 7 = 210
(ii) 3675 = 3 × 5 × 5 × 7 × 7
3 36755 12255 2457 497 7
1
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Grouping the factors in pairs, we see that 3
is left unpaired
Multiplying 3675 by 3, we get a perfect
square and square of the product will be = 3
× 5 × 7 = 105
(iii) 605 = 5 × 11 × 11
5 60511 12111 11
1
Grouping the factors in pairs, we see that 5
is left unpaired
Multiplying 605 by 5, we get a perfect square
and square root of the product will be = 5 ×
11 = 55
(iv) 2880 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5
2 28802 14402 7202 3602 1802 903 453 155 5
1
Grouping the factors in pairs, we see that 5
is left unpaired
Multiplying 2880 by 5, we get the perfect
square.
Square root of product will be = 2 × 2 × 2 ×
3 × 5 = 120
(v) 4056 = 2 × 2 × 2 × 3 × 13 × 13
2 40562 20282 10143 507
13 16913 13
1
Grouping the factors in pairs, we see that 2
and 3 are left unpaired
Multiplying 4056 by 2 × 3 i.e., 6, we get the
perfect square.
and square root of the product will be = 2 ×
2 × 3 × 13 = 156
(vi) 3468 = 2 × 2 × 3 × 17 × 17
2 34682 17343 867
17 28917 17
1
Grouping the factors in pairs, we see that 3
is left unpaired
Multiplying 3468 by 3 we get a perfect square.
and square root of the product will be 2 × 3
× 17 = 102
(vii) 7776 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3
2 77762 38882 19442 9722 4863 2433 813 273 93 3
1
Grouping the factors in pairs, we see that 2
and 3 are left unpaired
Multiplying 7776 by 2 × 3 or 6
We get a perfect square
and square root of the product will be
= 2 × 2 × 2 × 3 × 3 × 3 = 216
8. By what numbers should each of the
following be divided to get a perfect
square in each case ? Also find the
number whose square is the new number.
(i) 16562 (ii) 3698
(iii) 5103 (iv) 3174
(v) 1575
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Solution—
(i) 16562 = 2 × 7 × 7 × 13 × 13 2 165627 82817 1183
13 16913 13
1
Grouping the factors in pairs, we see that 2
is left unpaired
Dividing by 2, we get the perfect square
and square root of the quotient will be
7 × 13 = 91
(ii) 3698 = 2 × 43 × 43 2 369843 184943 43
1
Grouping the factors in pairs, we see that 2
is left unpaired,
Dividing 3698 by 2, the quotient is a perfect
square
and square of quotient will be = 43
(iii) 5103 = 3 × 3 × 3 × 3 × 3 × 3 × 7 3 51033 17013 5673 1893 633 217 7
1
Grouping the factors in pairs, we see that 7
is left unpaired
Dividing 5103 by 7, we get the quotient a
perfect square.
and square root of the quotient will be 3 × 3
× 3 = 27
(iv) 3174 = 2 × 3 × 23 × 23 2 31743 1587
23 52923 23
1
Grouping the factors in pairs, we see that 2
and 3 are left unpaired
Dividing 3174 by 2 × 3 i.e. 6, the quotient
will be a perfect square and square root of
the quotient will be = 23
(v) 1575 = 3 × 3 × 5 × 5 × 7 3 15753 5255 1755 357 7
1
Grouping the factors in pairs, we find that 7
is left unpaired
Dividing 1575 by 7, the quotient is a perfect
square
and square root of the quotient will be = 3 ×
5 = 15
9. Find the greatest number of two digits
which is a perfect square.
Solution—
The greatest two digit number = 99
We know, 92 = 81 and 102 = 100
But 99 is in between 81 and 100
81 is the greatest two digit number which is
a perfect square.
10. Find the least number of three digits
which is perfect square.
Solution—
The smallest three digit number = 100
We know that 92 = 81, 102 = 100, 112 = 121
We see that 100 is the least three digit number
which is a perfect square.
11. Find the smallest number by which 4851
must be multiplied so that the product
becomes a perfect square.
Solution—
By factorization :
4851 = 3 × 3 × 7 × 7 × 11 3 48513 16177 5397 77
11 111
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Grouping the factors in pairs, we see that 11
is left unpaired
The least number is 11 by which multiplying
4851, we get a perfect square.
12. Find the smallest number by which 28812
must be divided so that the quotient
becomes a perfect square.
Solution—
By factorization,
28812 = 2 × 2 × 3 × 7 × 7 × 7 × 7
2 288122 144063 72037 24017 3437 497 7
1
Grouping the factors in pairs, we see that 3
is left unpaired
Dividing 28812 by 3, the quotient will be a
perfect square.
13. Find the smallest number by which 1152
must be divided so that it becomes a
perfect square. Also find the number
whose square is the resulting number.
Solution—
By factorization,
1152 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
2 11522 5762 2882 1442 722 362 183 93 3
1
Grouping the factors in pairs, we see that
one 2 is left unpaired.
Dividing 1152 by 2, we get the perfect square
and square root of the resulting number 576,
will be 2 × 2 × 2 × 3 = 24
1. The following numbers are not perfect
squares. Give reason :
(i) 1547 (ii) 45743
(iii) 8948 (iv) 333333
Solution—
We know that if the units digit is 2, 3, 7 or 8
of a number, then the number is not a perfect
square.
(i) _ 1547 has 7 as units digit.
It is not a perfect square.
(ii) 45743 has 3 as units digit
It is not a perfect square.
(iii) _ 8948 has 8 as units digit
It is not a perfect square.
(iv) _ 333333 has 3 as units digits
It is not a perfect square.
2. Show that the following numbers are not
perfect squares :
(i) 9327 (ii) 4058
(iii) 22453 (iv) 743522
Solution—
(i) 9327
_ The units digit of 9327 is 7
This number can’t be a perfect square.
(ii) 4058
_ The units digit of 4058 is 8
This number can’t be a perfect square.
(iii) 22453
_ The units digit of 22453 is 3
This number can’t be a perfect square.
(iv) 743522
_ The units digit of 743522 is 2
EXERCISE 3.2
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This number can’t be a perfect square.
3. The square of which of the following
numbers would be an odd number ?
(i) 731 (ii) 3456
(iii) 5559 (iv) 42008
Solution—
We know that the square of an odd number
is odd and of even number is even. Therefore
(i) Square of 731 would be odd as it is an odd
number.
(ii) Square of 3456 should be even as it is an
even number.
(iii) Square of 5559 would be odd as it is an odd
number.
(iv) The square of 42008 would be an even
number as it is an even number.
Therefore suqares of (i) 731 and (iii) 5559
will be odd numbers.
4. What will be the units digit of the squares
of the following numbers ?
(i) 52 (ii) 977
(iii) 4583 (iv) 78367
(v) 52698 (vi) 99880
(vii) 12796 (viii) 55555
(ix) 53924
Solution—
(i) Square of 52 will be 2704 or (2)2 = 4
Its units digit is 4.
(ii) Square of 977 will be 954529 or (7)2 = 49
Its units digit is 9
(iii) Square of 4583 will be 21003889 or (3)2 = 9
Its units digit is 9
(iv) Is 78367, square of 7 = 72 = 49
Its units digit is 9
(v) In 52698, square of 8 = (8)2 = 64
Its units digit is 4
(vi) In 99880, square of 0 = 02 = 0
Its units digit is 0
(vii) In 12796, square of 6 = 62 = 36
Its units digit is 6
(viii)In In 55555, square of 5 = 52 = 25
Its units digit is 5
(ix) In 53924, square of 4 = 42 = 16
Its units digit is 6
5. Observe the following pattern
1 + 3 = 22
1 + 3 + 5 = 32
1 + 3 + 5 + 7 = 42
and write the value of 1 + 3 + 5 + 7 + 9 +
....... upto n terms.
Solution—
The given pattern is
1 + 3 = 22
1 + 3 + 5 = 32
1 + 3 + 5 + 7 = 42
1 + 3 + 5 + 7 + 9 + ....... upto n terms
(number of terms)2 = n2
6. Observe the following pattern :
22 – 12 = 2 + 1
32 – 22 = 3 + 2
42 – 32 = 4 + 3
52 – 42 = 5 + 4
Find the value of
(i) 1002 – 992 (ii) 1112 – 1092
(iii) 992 – 962
Solution—
From the given pattern,
22 – 12 = 2 + 1
32 – 22 = 3 + 2
42 – 32 = 4 + 3
52 – 42 = 5 + 4
Therefore
(i) 1002 – 990 = 100 + 99 = 199
(ii )1112 – 1092 = 1112 – 1102 – 1092
= (1112 – 1102) + (1102 – 1092)
= (111 + 110) + (110 + 109)
= 221 + 219 = 440
(iii) 992 – 962 = 992 – 982 + 982 – 972 + 972 – 962
= (992 – 982) + (982 – 972) + (972 – 962)
= (99 + 98) + (98 + 97) + (97 + 96)
= 197 + 195 + 193 = 585
7. Which of the following triplets are
pythagorean ?
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(i) (8, 15, 17) (ii) (18, 80, 82)
(iii) (14, 48, 51) (iv) (10, 24, 26)
(v) (16, 63, 65) (vi) (12, 35, 38)
Solution—
A pythagorean triplet is possible if
(greatest number)2 = (sum of the two smaller
numbers)
(i) 8, 15, 17
Here, greatest number = 17
(17)2 = 289
and (8)2 + (15)2 = 64 + 225 = 289
_ 82 + 152 = 172
8, 15, 17 is a pythagorean triplet
(ii) 18, 80, 82
Greatest number = 82
(82)2 = 6724
and 182 + 802 = 324 + 6400 = 6724
182 + 802 = 822
18, 80, 82 is a pythagorean triplet
(iii) 14, 48, 51
Greatest number = 51
(51)2 = 2601
and 142 + 482 = 196 + 2304 = 2500
_ 512 142 + 482
14, 48, 51 is not a pythagorean triplet
(iv) 10, 24, 26
Greatest number is 26
262 = 676
and 102 + 242 = 100 + 576 = 676
262 = 102 + 242
10, 24, 26 is a pythagorean triplet
(v) 16, 63, 65
Greatest number = 65
652 = 4225
and 162 + 632 = 256 + 3969 = 4225
_ 652 = 162 + 632
16, 63, 65 is a pythagorean triplet
(vi) 12, 35, 38
Greatest number = 38
382 = 1444
and 122 + 352 = 144 + 1225 = 1369
_ 382 122 + 352
12, 35, 38 is not a pythagorean triplet.
8. Observe the following pattern
(1 × 2) + (2 × 3) = 3
432
(1 × 2) + (2 × 3) + (3 × 4) = 3
543
(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5)
= 3
654
and find the value of
(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + (5 × 6)
Solution—
From the given pattern
(1 × 2) + (2 × 3) = 3
432
(1 × 2) + (2 × 3) + (3 × 4) = 3
543
(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) =
3
654
(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + (5 ×
6) = 3
765 = 70
9. Observe the following pattern
1 =2
1 {1 × (1 + 1)}
1 + 2 =2
1 {2 × (2 + 1)}
1 + 2 + 3 =2
1 {3 × (3 + 1)}
1 + 2 + 3 + 4 =2
1 {4 × ( 4 + 1)}
and find the values of each of the
following :
(i) 1 + 2 + 3 + 4 + 5 + ..... + 50
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(ii) 31 + 32 + ..... + 50
Solution—
From the given pattern,
1 =2
1 {1 × (1 + 1)}
1 + 2 =2
1 {2 × (2 + 1)}
1 + 2 + 3 =2
1 {3 × (3 + 1)}
1 + 2 + 3 + 4 =2
1 {4 × ( 4 + 1)}
Therefore
(i) 1 + 2 + 3 + 4 + 5 + ..... + 50 = 2
1 {50 × (50
+ 1)}
= 2
1 × 50 × 51 = 1275
(ii) 31 + 32 + ..... + 50
= (1 + 2 + 3 + 4 + ..... + 50) – (1 + 2 + 3 +
4 + ..... + 30)
= 2
1 {50 × (50 + 1)} –
2
1{30 × (30 + 1)}
= 2
1 × 50 × 51 –
2
1 × 30 × 31
= 1275 – 465 = 810
10. Observe the following pattern
12 = 6
1 [1 × (1 + 1) × (2 × 1 + 1)]
12 + 22 = 6
1 [2 × (2 + 1) × (2 × 2 + 1)]
12 + 22 + 32 = 6
1 [3 × (3 + 1) × (2 × 3) + 1)]
12 + 2 2 + 32 + 42 = 6
1 [4 × (4 + 1) × (2 × 4
+ 1)]
and find the values of each of the
following :
(i) 12 + 22 + 32 + 42 + ..... + 102
(ii) 52 + 62 + 72 + 82 + 92 + 102 + 112 + 122
Solution—
From the given pattern,
12 = 6
1 [1 × (1 + 1) × (2 × 1 + 1)]
12 + 22 = 6
1 [2 × (2 + 1) × (2 × 2 + 1)]
12 + 22 + 32 = 6
1 [3 × (3 + 1) × (2 × 3) + 1)]
12 + 22 + 32 + 42 = 6
1 [4 × (4 + 1) × (2 × 4 +
1)]
Therefore :
(i) 12 + 22 + 32 + 42 + .... + 102
= 6
1 {10 × (10 + 1) × (2 × 10 + 1)]
= 6
1 [10 × 11 × 21] =
6
211110
= 6
2310 = 385
(ii) 52 + 62 + 72 + 82 + 92 + 102 + 112 + 122
= [12 + 22 + 32 + 42 + ..... + 122] – [12 + 22 +
32 + 42]
= 6
1 [12 × (12 + 1) × (2 × 12 + 1)] –
6
1 [4
× (4 + 1) × (2 × 4 + 1)]
= 6
1 × (12 × 13 × 25) –
6
1 (4 × 5 × 9)
= 6
251312 –
6
954 = 650 – 30 = 620
11. Which of the following numbers are
squares of even numbers ?
121, 225, 256, 324, 1296, 6561, 5476, 4489,
373758
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Solution—
We know that squares of even numbers is
also are even number. Therefore numbers
256, 324, 1296, 5476 and 373758 have their
units digit an even number.
These are the squares of even numbers.
12. By just examining the units digits, can
you tell which of the following cannot be
whole squares ?
(i) 1026 (ii) 1028
(iii) 1024 (iv) 1022
(v) 1023 (vi) 1027
Solution—
We know that a perfect square cannot ends
with the digit 2, 3, 7, or 8
By examining the given number, we can say
that 1028, 1022, 1023, 1027 can not be
perfect squares.
13. Write five numbers for which you cannot
decide whether they are squares.
Solution—
A number which ends with 1, 4, 5, 6, 9 or 0
can’t be a perfect square
2036, 4225, 4881, 5764, 3349, 6400
14. Write five numbers which you cannot
decide whether they are square just by
looking at the unit’s digit.
Solution—
A number which does not end with 2, 3, 7
or 8 can be a perfect square
The five numbers can be 2024, 3036, 4069,
3021, 4900
15. Write true (T) or false (F) for the
following statements.
(i) The number of digits in a square number
is even.
(ii) The square of a prime number is prime.
(iii) The sum of two square numbers is a
square number.
(iv) The difference of two square numbers is
a square number.
(v) The product of two square numbers is a
square number.
(vi) No square number is negative.
(vii) There is not square number between 50
and 60.
(viii) There are fourteen square number upto
200.
Solution—
(i) False : In a square number, there is no
condition of even or odd digits.
(ii) False : A square of a prime is not a prime.
(iii) False : It is not necessarily.
(iv) False : It is not necessarily.
(v) True.
(vi) True : A square is always positive.
(vii) True : As 72 = 49, and 82 = 64.
(viii)True : As squares upto 200 are 1, 4, 9, 16,
25, 36, 49, 64, 81, 100, 121, 144, 169, 196
which are fourteen in numbers.
1. Find the squares of the following numbers
using column method. Verify the result
by finding the square using the usual
multiplication :
(i) 25 (ii) 37
(iii) 54 (iv) 71
(v) 96
Solution—
(i) (25)2
EXERCISE 3.3
Here a = 2, b = 5
a2 2ab b2
(2)2 2 × 2 × 5 (5)2
= 4 = 20 = 25
6
2
22
2
(25)2 = 25 × 25 = 625
(25)2 = 625
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(ii) (37)2
Here a = 3, b = 7
a2 2ab b2
(3)2 2 × 3 × 7 (7)2
= 9 = 42 = 49
+ 4 + 4
13 46
(37)2 = 37 × 37 = 1369
(37)2 = 1369
(iii) (54)2
a2 2ab b2
(5)2 2 × 5 × 4 (4)2
= 25 = 40 = 16
+ 4 + 1
29 41
(54)2 = 54 × 54 = 2916
(54)2 = 2916
(iv) (71)2
Here a = 7, b = 1
a2 2ab b2
(7)2 2 × 7 × 1 (1)2
= 49 = 14 = 1
+ 1
50
(71)2 = 71 × 71 = 5041
(71)2 = 5041
(v) (96)2
Here a = 9, b = 6
a2 2ab b2
(9)2 2 × 9 × 6 (6)2
= 81 = 108 = 36
+ 11 + 3
92 111
(96)2 = 96 × 96 = 9216
(96)2 = 9216
2. Find the squares of the following numbers
using diagonal method :
(i) 98 (ii) 273
(iii) 348 (iv) 295
(v) 171
Solution—
(i) (98)2
(98)2 = 9604
(ii) (273)2
(273)2 = 74529
(iii) (348)2
(348)2 = 121104
(iv) (295)2
(295)2 = 87025
8
1
7
2
72
64
9 8
9
8
10 4
15+116
8+1
9
2 006 1 9
1 00
4 214 9 1
4 4 6
2 7 3
2
7
3
9
0
225
6+1
7
12+214
3 624 2 4
1 20
1 312 6 2
9 2 4
3 4 8
3
4
8
410
11+112
0+1
1
20+121
9+211
4 210 5 5
1 10
8 418 1 5
4 8 0
2 9 5
2
9
5
0
9+110
26+127 512
6+28
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(v) (171)2
(171)2 = 29241
3. Find the squares of the following
numbers :
(i) 127 (ii) 503
(iii) 451 (iv) 862
(v) 265
Solution—
(i) (127)2 = (120 + 7)2
{(a + b)2 = a2 + 2ab + b2}
= (120)2 + 2 × 120 × 7 + (7)2
= 14400 + 1680 + 49 = 16129
(ii) (503)2 = (500 + 3)2
{(a + b)2 = a2 + 2ab + b2}
= (500)2 + 2 × 500 × 3 + (3)2
= 250000 + 3000 + 9 = 253009
(iii) (451)2 = (400 + 51)2
{(a + b)2 = a2 + 2ab + b2}
= (400)2 + 2 × 400 × 51 + (51)2
= 160000 + 40800 + 2601 = 203401
(iv) (862)2 = (800 + 62)2
{(a + b)2 = a2 + 2ab + b2}
= (800)2 + 2 × 800 × 62 + (62)2
= 640000 + 99200 + 3844 = 743044
(v) (265)2
{(a + b)2 = a2 + 2ab + b2}
(200 + 65)2 = (200)2 + 2 × 200 × 65 + (65)2
= 40000 + 26000 + 4225 = 70225
4. Find the squares of the following numbers
(i) 425 (ii) 575
(iii) 405 (iv) 205
(v) 95 (vi) 745
(vii) 512 (viii) 995
Solution—
(i) (425)2
Here n = 42
n (n + 1) = 42 (42 + 1) = 42 × 43 = 1806
(425)2 = 180625
(ii) (575)2
Here n = 57
n (n + 1) = 57 (57 + 1) = 57 × 58 = 3306
(575)2 = 330625
(iii) (405)2
Here n = 40
n (n + 1) = 40 (40 + 1) = 40 × 41 = 1640
(405)2 = 164025
(iv) (205)2
Here n = 20
n (n + 1) = 20 (20 + 1) = 20 × 21 = 420
(205)2 = 42025
(v) (95)2
Here n = 9
n (n + 1) = 9 (9 + 1) = 9 × 10 = 90
(95)2 = 9025
(vi) (745)2
Here n = 74
n (n + 1) = 74 (74 + 1) = 74 × 75 = 5550
(745)2 = 555025
(vii) (512)2
Here a = 1, b = 2
(5ab)2 = (250 + ab) × 1000 + (ab)2
(512)2 = (250 + 12) × 1000 + (12)2
= 262 × 1000 + 144
= 262000 + 144 = 262144
(viii)(995)2
Here n = 99
n (n + 1) = 99 (99 + 1) = 99 × 100 = 9900
(995)2 = 990025
5. Find the squares of the following numbers
using the identity (a + b)2 = a2 + 2ab + b2
(i) 405 (ii) 510
(iii) 1001 (iv) 209
(v) 605
Solution—
(a + b)2 = a2 + 2ab + b2
(i) (405)2 = (400 + 5)2
= (400)2 + 2 × 400 × 5 + (5)2
0 001 7 1
0 00
4 007 9 7
1 7 1
1 7 1
1
7
1
11411+112
0
1+1
2
18+119
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= 160000 + 4000 + 25
= 164025
(ii) (510)2 = (500 + 10)2
= (500)2 + 2 × 500 × 10 × (10)2
= 250000 + 10000 + 100
= 260100
(iii) (1001)2 = (1000 + 1)2
= (1000)2 + 2 × 1000 × 1 + (1)2
= 1000000 + 2000 + 1
= 1002001
(iv) (209)2 = (200 + 9)2
= (200)2 + 2 × 200 × 9 × (9)2
= 40000 + 3600 + 81
= 43681
(v) (605)2 = (600 + 5)2
= (600)2 + 2 × 600 × 5 + (5)2
= 360000 + 6000 + 25
= 366025
6. Find the squares of the following numbers
using the identity (a – b)2 = a2 – 2ab + b2 :
(i) 395 (ii) 995 (iii) 495
(iv) 498 (v) 99 (vi) 999
(vii) 599
Solution—
(a – b)2 = a2 – 2ab + b2
(i) (395)2 = (400 – 5)2
= (400)2 – 2 × 400 × 5 + (5)2
= 160000 – 4000 + 25
= 160025 – 4000
= 156025
(ii) (995)2 = (1000 – 5)2
= (1000)2 – 2 × 1000 × 5 + (5)2
= 1000000 – 10000 + 25
= 1000025 – 10000
= 990025
(iii) (495)2 = (500 – 5)2
= (500)2 – 2 × 500 × 5 + (5)2
= 250000 – 5000 + 25
= 250025 – 5000 = 245025
(iv) (498)2 = (500 – 2)2
= (500)2 – 2 × 500 × 2 + (2)2
= 250000 – 2000 + 4
= 250004 – 2000 = 248004
(v) (99)2 = (100 – 1)2
= (100)2 – 2 × 100 × 1 + (1)2
= 10000 – 200 + 1 = 10001 – 200
= 9801
(vi) (999)2 = (1000 – 1)2
= (1000)2 – 2 × 1000 × 1 + (1)2
= 1000000 – 2000 + 1
= 1000001 – 2000 = 998001
(vii) (599)2 = (600 – 1)2
= (600)2 – 2 × 600 × 1 + (1)2
= 360000 – 1200 + 1
= 360001 – 1200 = 358801
7. Find the squares of the following numbers
by visual method :
(i) 52 (ii) 95
(iii) 505 (iv) 702
(v) 99
Solution—
(a + b)2 = a2 – ab + ab + b2
(i) (52)2 = (50 + 2)2
= 2500 + 100 + 100 + 4
= 2704
4
2500
2
100
100
50
250
(ii) (95)2 = (90 + 5)2
= 8100 + 450 + 450 + 25
= 9025
25
8100
5
45
0
450
90
590
(iii) (505)2 = (500 + 5)2
= 250000 + 2500 + 2500 + 25
= 255025
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25
250000
5
2500
2500
500
5500
(iv) (702)2 = (700 + 2)2
= 490000 + 1400 + 1400 + 4
= 492804
4
490000
2
140
0
1400
700
2700
(v) (99)2 = (90 + 9)2
= 8100 + 810 + 810 + 81
= 9801
819
810
810
90
990
1. Write the possible unit’s digits of the
square root of the following numbers.
Which of these numbers are odd square
roots ?
(i) 9801 (ii) 99856
(iii) 998001 (iv) 657666025
Solution—
(i) In 9801 _ the units digits is 1, therefore,
the units digit of the square root can be 1 or
9
(ii) In 99356 _ the units digit is 6
The units digit of the square root can be 4 or 6
(iii) In 998001 _ the units digit is 1
The units digit of the square root can be 1
or 9
(iv) In 657666025
_ The unit digit is 5
The units digit of the square root can be 5
2. Find the square root of each of the
following by prime factorization.
(i) 441 (ii) 196
(iii) 529 (iv) 1764
EXERCISE 3.4
(v) 1156 (vi) 4096
(vii) 7056 (viii) 8281
(ix) 11664 (x) 47089
(xi) 24336 (xii) 190969
(xiii)586756 (xiv) 27225
(xv) 3013696
Solution—
(i) 441 = 7733 3 4413 1477 497 7
1
= 3 × 7 = 21
(ii) 196 = 7722 2 1962 987 497 7
1
= 2 × 7 = 14
(iii) 529 = 2323
= 23
23 52923 23
1
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(iv) 1764 = 773322
2 17642 8823 4413 1477 497 7
1
= 2 × 3 × 7 = 42
(v) 1156 = 171722
= 2 × 17 = 34
(vi) 4096 = 22222
2222222
2 40962 20482 10242 5122 2562 1282 642 322 162 82 42 2
1
= 2 × 2 × 2 × 2 × 2 × 2 = 64
(vii) 7056 = 77332222
= 2 × 2 × 3 × 7 = 84
(viii) 8281 = 131377 7 82817 1183
13 16913 13
1
= 7 × 13 = 91
(ix) 11664 = 3333
332222
2 116642 58322 29162 14583 7293 2433 813 273 93 3
1
= 2 × 2 × 3 × 3 × 3 = 108
(x) 47089 = 313177 7 470897 6727
31 96131 31
1
= 7 × 31 = 217
(xi) 24336 = 1313332222
2 243362 121682 60842 30423 15213 507
13 16913 13
1
= 2 × 2 × 3 × 13 = 156
(xii) 190969 = 23231919
2 11562 578
17 28917 17
1
2 70562 35282 17642 8823 4413 1477 497 7
1
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19 19096919 1005123 52923 23
1
= 19 × 23 = 437
(xiii) 586756 = 38338322
2 5867562 293378
383 146689383 383
1
= 2 × 383 = 766
(xiv) 27225 = 11115533
3 272253 90755 30255 605
11 12111 11
1
= 3 × 5 × 11 = 165
(xv) 3013969 = 313177
222222
2 30136962 15068482 7534242 3767122 1883562 941787 470897 6727
31 96131 31
1
= 2 × 2 × 2 × 7 × 31 = 1736
3. Find the smallest number by which 180
must be multiplied so that it becomes a
perfect square. Also, find the square root
of the perfect square so obtained.
Solution—
Factorising 180,
2 1802 903 453 155 5
1
180 = 2 × 2 × 3 × 3 × 5
Grouping the factors in pairs
we see that factor 5 is left unpaired.
Multiply 180 by 5, we get the product 180 ×
5 = 900
Which is a perfect square
and square root of 900 = 2 × 3 × 5 = 30
4. Find the smallest number by which 147
must be multiplied so that it becomes a
perfect square. Also, find the square root
of the number so obtained.
Solution—
Factorising 147,
3 1477 497 7
1
147 = 3 × 7 × 7
Grouping the factors in pairs of the equal
factors, we see that one factor 3 is left
unpaired
Multiplying 147 by 3, we get the product
147 × 3 = 441
Which is a perfect square
and its square root = 3 × 7 = 21
5. Find the smallest number by which 3645
must be divided so that it becomes a
perfect square. Also, find the square root
of the resulting number.
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Solution—
Factorising 3645
3 36453 12153 4053 1353 453 155 5
1
3645 = 3 × 3 × 3 × 3 × 3 × 3 × 5
Grouping the factors in pair of the equal
factors, we see that one factor 5 is left
unpaired
Dividing 3645 by 5, the quotient 729 will be
the perfect square and square root of 729
= 27
6. Find the smallest number by which 1152
must be divided so that it becomes a
perfect square. Also, find the square root
of the number so obtained.
Solution—
Factorsing 1152,
2 11522 5762 2882 1442 722 362 183 93 3
1
1152 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
Grouping the factors in pairs of the equal
factors, we see that factor 2 is left unpaired.
Dividing by 2, the quotient 576 is a perfect
square
Square root of 576, it is 24
7. The product of two numbers is 1296. If
one number is 16 times the others find
the numbers.
Solution—
Product of two numbers = 1296
Let one number = x
Second number = 16x
16x × x = 1296
16x2 = 1296
x2 = 16
1296 = 81 = (9)2
x = 9
First number = 9
and second number = 16 × 9 = 144
8. A welfare association collected Rs. 202500
as donation from the residents. If each
paid as many rupees as there were
residents find the number of residents.
Solution—
Total donation collected = Rs. 202500
Let number of residents = x
Then donation given by each resident = Rs.
x
Total collection = Rs. x × x
x2 = 202500
x = 202500
2 2025002 1012503 506253 168753 56253 18755 6255 1255 255 5
1
= 5555333322
= 2 × 3 × 3 × 5 × 5 = 450
Number of residents = 450
9. A society collected Rs. 92.16. Each
member collected as many paise as there
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were members. How many members were
there and how much did each contribute?
Solution—
Total amount collected = Rs. 92.16
= 9216 paise
Let the number of members = x
Then amount collected by each member = x
paise
x + x = 9216 x2 = 9216
x = 9216
2 92162 46082 23042 11522 5762 2882 1442 722 362 183 93 3
1
= 332222222222
= 2 × 2 × 2 × 2 × 2 × 3 = 96
Number of members = 96
and each member collected = 96 paise
10. A school collected Rs. 2304 as fees from
its students. If each student paid as many
paise as there were students in the school,
how many students were there in the
school ?
Solution—
Total fee collected = Rs. 2304
Let number of students = x
Then fee paid by each student = Rs. x
x × x = 2304 x2 = 2304
x = 2304
2 23042 11522 5762 2882 1442 722 362 183 93 3
1
= 3322222222
= 2 × 2 × 2 × 2 × 3 = 48
Number of students = 48
11. The area of a square field is 5184 m2. A
rectangular field, whose length is twice
its breadth has its perimeter equal to the
perimeter of the square field. Find the
area of the rectangular field.
Solution—
The area of a square field = 5184 m2
Let side of the square = x
x × x = 5184 x2 = 5184
x = 5184
2 51842 25922 12962 6482 3242 1623 813 273 93 3
1
= 3333222222
= 2 × 2 × 2 × 3 × 3 = 72
Side of square = 72 m
Perimeter of square field = 72 × 4 m
= 288 m
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Perimeter of rectangle = 288 m
Let breadth of rectangular field (b) = x
Then length (l) = 2x
Perimeter = 2 (l + b)
= 2 (2x + x) = 2 × 3x
= 6x
6x = 288 x = 6
288 = 48
Length of rectangular field = 2x = 2 × 48
= 96 m
and breadth = 48 m
and area = l × b = 96 × 48 m2
= 4608 m2
12. Find the least square number, exactly
divisible by each one of the numbers :
(i) 6, 9, 15 and 20 (ii) 8, 12, 15 and 20
Solution—
(i) 6, 9, 15, 20
2 6, 9, 15, 203 3, 9, 15, 105 1, 3, 5, 10
1, 3, 1, 2
LCM of 6, 9, 15, 20 = 2 × 3 × 5 × 3 × 2
= 180
= 2 × 2 × 3 × 3 × 5
We see that after grouping the factors in pairs,
5 is left unpaired
Least perfect square = 180 × 5 = 900
(ii) 8, 12, 15, 20
LCM of 8, 12, 15, 20
2 8, 12, 15, 202 4, 6, 15, 103 2, 3, 15, 55 2, 1, 5, 5
2, 1, 1, 1
= 2 × 2 × 2 × 3 × 5 = 120
We see that after grouping the factors,
factors 2, 3, 5 are left unpaired
Perfect square = 120 × 2 × 3 × 5
= 120 × 30 = 3600
13. Find the square roots of 121 and 169 by
the method of repeated subtraction.
Solution—
(i) 121
121 – 1 = 120 85 – 13 = 72
120 – 3 = 117 72 – 15 = 57
117 – 5 = 112 57 – 17 = 40
112 – 7 = 105 40 – 19 = 21
105 – 9 = 96 21 – 21 = 0
96 – 11 = 85
Square root = 11
(ii) 169
169 – 1 = 168 120 – 15 = 105
168 – 3 = 165 105 – 17 = 88
165 – 5 = 160 88 – 19 = 69
160 – 7 = 153 69 – 21 = 48
153 – 9 = 144 48 – 23 = 25
144 – 11 = 133 25 – 25 = 0
133 – 13 = 120
Square root = 13
14. Write the prime factorization of the
following numbers and hence find their
square roots.
(i) 7744 (ii) 9604
(iii) 5929 (iv) 7056
Solution—
Factorization, we get :
(i) 7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
2 77442 38722 19362 9682 4842 242
11 12111 11
1
Square root of 7744
= 7744 = 1111222222
Grouping the factors in pairs of equal factors,
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7744 = 2 × 2 × 2 × 11 = 88
Square root of 7744 = 88
(ii) 9604 = 2 × 2 × 2 × 2 × 7 × 7
2 96042 48027 24017 3437 497 7
1
Square root of 9604
9604 = 777722
= 2 × 7 × 7 = 98
(iii) 5929 = 7 × 7 × 11 × 11
7 59297 847
11 12111 11
1
Square root of 5929
= 5929 = 111177
= 7 × 11 = 77
(iv) 7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
2 70562 35282 17642 8823 4413 1477 497 7
1
Square root of 7056
= 7056 = 77332222
= 2 × 2 × 3 × 7 = 84
15. The students of class VIII of a school
donated Rs. 2401 for PM’s National Relief
Fund. Each student donated as many
rupees as the number of students in the
class. Find the number of students in the
class.
Solution—
Total amount of donation = 2401
Let number of students in VIII = x
Amount donoted by each student = Rs. x
x × x = 2401 x2 = 2401
x = 2401 = 7777
7 24017 3437 497 7
1
= 7 × 7 = 49
Number of students of class VIII = 49
16. A PT teacher wants to arrange maximum
possible number of 6000 students in a field
such that the number of rows is equal to
the number of columns. Find the number
of rows if 71 were left out after
arrangement.
Solution—
Number of students = 6000
Students left out = 71
Students arranged in a field = 6000 – 71
= 5929
Let number of rows = x
Then number of students in each row = x
x × x = = 5929 x2 = 5929
7 59297 847
11 12111 11
1
x = 5929 = 111177
= 7 × 11 = 77
Number of rows = 77
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1. Find the square root of each of the
following by long division method.
(i) 12544 (ii) 97344
(iii) 286225 (iv) 390625
(v) 363609 (vi) 974169
(vii) 120409 (viii) 1471369
(ix) 291600 (x) 9653449
(xi) 1745041 (xii) 4008004
(xiii)20657025 (xiv) 152547201
(xv) 20421361 (xvi) 62504836
(xvii) 82264900 (xviii) 3226694416
(xix) 6407522209 (xx) 3915380329
Solution—
(i) Square root of 12544
= 12544 = 112
(ii) Square root of 97344
3123 9 73 44
961 73
61622 1244
1244 ×
= 97344
= 312
(iii) Square root of 286225
= 286225
= 535
(iv) Square root of 390625
6256 39 06 25
36122 306
2441245 6225
6225 ×
= 390625
= 625
(v) Square root of 363609
6036 36 36 09
361203 3609
3609 ×
= 363609
= 603
(vi) Square root of 974169
= 974169
= 987
(vii) Square root of 120409
= 120409
= 347
EXERCISE 3.5
1122 1 25 44
121 25
21222 444
444 ×
5355 28 62 25
25103 362
3091065 5325
5325 ×
9879 97 41 69
81188 1641
15041967 13769
13769 ×
3473 12 04 09
964 304
256687 4809
4809 ×
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(viii)Square root of 1471369
= 1471369
= 1213
(ix) Square root of 291600
= 291600
= 540
(x) Square root of 9653449
= 9653449
= 3107
(xi) Square root of 1745041
= 1745041
= 1321
(xii) Square root of 4008004
= 4008004
= 2002
(xiii) Square root of 20657025
45454 20 65 70 25
1685 465
425904 4070
36169085 45425
45425 ×
= 20657025
= 4545
(xiv) Square root of 152557201
123511 1 52 54 72 01
122 52
44243 854
7292465 12572
1232524701 24701
24701 ×
= 152547201
= 12351
(xv) Square root of 20421361
= 20421361
= 4519
12131 1 47 13 69
122 47
44241 313
2412423 7269
7269 ×
31073 9 65 34 49
961 65
616207 43449
43449 ×
13211 1 74 50 41
123 74
69262 550
5242641 2641
2641 ×
5405 29 16 00
25104 416
4161080 00
00 00
20022 4 00 80 04
44002 008004
8004 ×
45194 20 42 13 61
1685 442
425901 1713
9019029 81261
81261 ×
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(xvi) Square root of 62504836
= 62504836
= 7906
(xvii) Square root of 82264900
90709 82 26 49 00
811807 12649
1264919140 00
00 ×
= 82264900
= 9070
(xviii) Square root of 3226694416
568045 32 26 69 44 16
25106 726
6361128 9069
9024113604 454416
454416 ×
= 3226694416
= 56804
(xix) Square root of 6407522209
800478 64 07 52 22 09
6416004 075222
64016160087 1120609
1120609 ×
= 6407522209
= 80047
(xx) Square root of 3915380329
= 3915380329
= 62573
2. Find the least number which must be
subtracted from the following numbers
to make them a perfect square :
(i) 2361 (ii) 194491 (iii) 26535
(iv) 16160 (v) 4401624
Solution—
(i) 2361
Finding the square root of 2361
484 23 61
1688 761
704 57
We get 48 as quotient and remainder = 57
To make it a perfect square, we have to
subtract 57 from 2361
Least number to be subtracted = 57
(ii) 194491
Finding the square root of 194491
4414 19 44 91
1684 344
336881 891
881 10
79067 62 50 48 36
49149 1350
134115806 94836
94836 ×
625736 39 15 38 03 29
36122 315
2441245 7138
622512507 91303
87549125143 375429
375429 ×
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We get 441 as quotient and remainder = 10
To make it a perfect square, we have to
subtract 10 from 194491
Least number to be subtracted = 10
(iii) 26535
Finding the square root of 26535
1621 2 65 35
126 165
156322 935
644 291
We get 162 as quotient and 291 as remainder
To make it a perfect square, we have to
subtract 291 from 26535
Least number to be subtracted = 291
(iv) 16160
Finding the square root of 16160
1271 1 61 60
122 61
44247 1760
1729 31
We get 127 as quotient and 31 as remainder
To make it a perfect square, we have to
subtract 31 from 16160
Least number to be subtracted = 31
(v) 4401624
Find the square root of 4401624
20982 4 40 16 24
4409 4016
36814188 33524
33504 20
We get 2098 as quotient and 20 as remainder
To make it a perfect square, we have to
subtract 20 from 4401624
Least number to be subtracted = 20
3. Find the least number which must be
added to the following numbers to make
them a perfect square :
(i) 5607 (ii) 4931
(iii) 4515600 (iv) 37460
(v) 506900
Solution—
(i) 5607
747 56 07
49144 707
576 131
757 56 07
49145 707
725
Finding the square root of 5607, we see that
742 = 5607 – 131 = 5476
and 752 = 5625
_ 5476 < 5607 < 5625
5625 – 5607 = 18 is to be added to get a
perfect square
Least number to be added = 18
(ii) 4931
707 49 31
49140 31
0 31
717 49 31
49141 31 141
Finding the square root of 4931, we see that
702 = 4900
712 = 5041
4900 < 4931 < 5041
5041 – 4931 = 110 is to be added to get a
perfect square.
Least number to be added = 110
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(iii) 4515600
21242 4 51 56 00
441 51
41422 1056
8444244 21200
16976 4224
21252 4 51 56 00
441 51
41422 1056
8444245 21200
21225
Finding the square root of 4515600, we see
that 21242 = 4511376
and 21252 = 4515625
4511376 < 4515600 < 4515625
4515625 – 4515600 = 25 is to be added to
get a perfect square.
Least number to be added = 25
(iv) 37460
1931 3 74 60
129 274
261383 1360
1149 211
1941 3 74 60
129 274
261384 1360
1536
Finding the square root of 37460, we see
that 1932 = 37249, 1942 = 37636
37249 < 37460 < 37636
37636 – 37460 = 176 is to be added to get a
perfect square.
Least number to be added = 176
(v) 506900
7117 50 69 00
49141 169
1411421 2800
1421 1379
7127 50 69 00
49141 169
1411422 2800
2844
Finding the square root of 506900, we see
that
7112 = 505521, 7122 = 506944
505521 < 506900 < 506944
506944 – 506900 = 44 is to be added to get
a perfect square.
Least number to be added = 44
4. Find the greatest number of 5 digits which
is a perfect square.
Solution—
Greatest number of 5-digits = 99999
Finding square root, we see that 143 is left
as remainder
3163 9 99 99
961 99
61626 3899
3756 143
Perfect square = 99999 – 143 = 99856
If we add 1 to 99999, it will because a
number of 6 digits
Greatest square 5-digits perfect square
= 99856
5. Find the least number of four digits which
is a perfect square.
Solution—
Least number of 4-digits = 10000
Finding square root of 1000
We see that if we subtract 39
323 10 00
962 100
124
313 10 00
961 100
61 39
From 1000, we get three digit number
We shall add 124 – 100 = 24 to 1000 to get a
perfect square of 4-digit number
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1000 + 24 = 1024
Least number of 4-digits which is a perfect
square = 1024
6. Find the least number of six-digits which
is a perfect square.
Solution—
Least number of 6-digits = 100000
3173 10 00 00
961 100
61627 3900
4389
3163 10 00 00
961 100
61626 3900
3756 544
Finding the square root of 100000, we see
that if we subtract 544, we get a perfect
square of 5-digits.
So we shall add
4389 – 3900 = 489
to 100000 to get a perfect square
Past perfect square of six digits= 100000 +
489 = 100489
7. Find the greatest number of 4-digits
which is a perfect square.
Solution—
Greatest number of 4-digits = 9999
999 99 99
81189 1899
1701 198
Finding the square root, we see that 198 has
been left as remainder
4-digit greatest perfect square
= 9999 – 198 = 9801
8. A General arranges his soldiers in rows
to form a perfect square. He finds that in
doing so, 60 soldiers are left out. If the
total number of soldiers be 8160, find the
number of soldiers in each row.
Solution—
Total number of soldiers = 8160
Soldiers left after arranging them in a square
= 60
Number of soldiers which are standing in a
square = 8160 – 60 = 8100
909 81 00
81180 00
00 ×
Number of soldiers in each row = 8100
= 90
9. The area of a square field is 60025 m2. A
man cycle along its boundry at 18 km/hr.
In how much time will be return at the
starting point.
Solution—
Area of a square field = 60025 m2
2452 6 00 25
444 200
176485 2425
2425 ×
Side of the square field (a) = 60025 m
= 245 m
and perimeter = 4a
= 4 × 245 m
= 980 m
Speed of the man = 18 km/hr = 18000 m/hr
Time taken = 18000
1980 hr
= 18000
601980 minutes
= 315
4 minutes = 3 minutes 16 seconds
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83
10. The cost of levelling and turfing a square
lawn at Rs. 2.50 per m2 is Rs. 13322.50.
Find the cost of fencing it at Rs. 5 per
metre ?
Solution—
Cost of levelling a square field
= Rs. 13322.50
Rate of levelling = Rs. 2.50 per m2
Area of square field = 250
1332250 = 5329 m2
Length of each side (a) = Area
= 5329 m = 73 m
and perimeter = 4a = 4 × 73 = 292 m
Rate of fencing the field = Rs. 5 per m
Total cost of fencing = Rs. 5 × 292
= Rs. 1460
11. Find the greatest number of three digits
which is a perfect square.
Solution—
3-digits greatest number = 999
313 9 99
961 99
61 38
Finding the square root, we see that 38 has
been left
Perfect square = 999 – 38 = 961
Greatest 3-digit perfect square = 961
12. Find the smallest number which must be
added to 2300 so that it becomes a perfect
square.
Solution—
Finding the square root of 2300
484 23 00
1688 700
704
484 23 00
1687 700
609 91
We see that we have to add 704 – 700 = 4 to
2300 in order to get a perfect square
Smallest number to be added = 4
1. Find the square root of :
(i)961
441(ii)
841
324
(iii) 449
29(iv) 2
25
14
(v) 2196
137(vi) 23
121
26
(vii) 25729
544(viii) 75
49
46
(ix) 32209
942(x) 3
3025
334
EXERCISE 3.6
(xi) 213364
2797(xii) 38
25
11
(xiii)23729
394(xiv) 21
169
51
(xv) 10225
151
Solution—
(i)961
441 =
961
441
= 31
21
212 4 41
441 41
41 ×
313 9 61
961 61
61 ×
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84
(ii)841
324 =
841
324
181 3 24
128 224
224 ×
292 8 41
449 441
441 ×
= 29
18
(iii)49
294 =
49
29494
= 49
29196 =
49
225
151 2 25
125 125
125 ×
= 7
15 = 2
7
1
(iv)25
142 =
25
14252
= 25
64 =
25
64 =
5
8 = 1
5
3
(v)196
1372 =
196
1371962
= 196
137392 =
196
529
= 196
529 232 5 29
443 129
129 ×
141 1 96
124 96
96 ×
= 14
23 = 1
14
9
(vi)121
2623 =
121
2612123 =
121
262783
= 121
2809 535 28 09
25103 309
309 ×
= 11
53 = 4
11
9
(vii)729
54425 =
729
54472925
= 729
54418225 =
729
18769
272 7 29
447 329
329 ×
1371 1 87 69
123 87
69267 1869
1869 ×
= 27
137 = 5
27
2
(viii)49
4675 =
49
464975
= 49
463675 =
49
3721
= 7
61 = 8
7
5
616 37 21
36121 121
121 ×
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85
(ix)2209
9423 =
2209
94222093
= 2209
9426627 =
2209
7569
878 75 69
64167 1169
1169 ×
474 22 09
1687 609
609 ×
= 47
87 = 1
47
40
(x)3025
3343 =
3025
33430253
= 3025
3349075 =
3025
9409
979 94 09
81187 1309
1309 ×
555 30 25
25105 525
525 ×
= 55
97 = 1
55
42
(xi)3364
279721 =
3364
2797336421
= 3364
279770644 =
3364
73441
585 33 64
25108 864
864 ×
2712 7 34 41
447 334
329541 541
541 ×
= 58
271 = 4
58
39
(xii)25
1138 =
25
112538
= 25
11950 =
25
961
313 9 61
961 61
61 ×
= 5
31 = 6
5
1
(xiii)729
39423 =
729
39472923
= 729
39416767 =
729
17161
1311 1 71 61
123 71
69261 261
261 ×
272 7 29
447 329
329 ×
= 27
131 = 4
27
23
(xiv)169
5121 =
169
5116921
= 169
513549 =
169
3600
= 13
60 = 4
13
8
https://www.arundeepselfhelp.info/index.php?route=product/product&path=139_141&product_id=111
86
(xv)225
15110 =
225
15122510
= 225
1512250 =
225
2401
494 24 01
1689 801
801 ×
151 2 25
125 125
125 ×
= 15
49 = 3
15
4
2. Find the value of :
(i)405
80(ii)
625
441
(iii)1728
1587(iv) 72 × 338
(v) 45 × 20
Solution—
(i)405
80 =
405
80 =
81
16(Dividing by 5)
= 81
16 =
9
4
(ii)625
441 =
625
441 =
25
21
(iii)1728
1587 =
1728
1587 =
31728
31587
=
576
529
232 5 29
443 129
129 ×
242 5 76
444 176
176 ×
576
529 =
24
23
(iv) 72 × 338 = 33872 = 24336
= 156
(v) 45 × 20 = 2045 = 900 = 30
3. The area of a square field is 80729
244 square
metres. Find the length of each side of
the field.
Solution—
Area of a square field = 80729
244
= 729
24472980 m2
= 729
24458320 =
729
58564 m2
Side of square field = Area
= 729
58564 =
729
58564
272 7 29
447 329
329 ×
2422 5 85 64
444 185
176482 964
964 ×
= 27
242 = 8
27
26
1561 2 43 36
125 143
125306 1836
1836 ×
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87
4. The area of a square field is 304
1 m2.
Calculate the length of the side of the
squre.
Solution—
Area of the square field = 304
1 =
4
121m2
Side = Area = 4
121m
= 22
1111
=
2
11 = 5.5 m
5. Find the length of a side of a square
playground whose area is equal to the
area of a rectangular field of dimensions
72 m and 338 m.
Solution—
Length of rectangular field (l) = 338 m
and breadth (b) = 72 m
Area = l × b = 338 × 72 m2
Area of square = 338 × 72 m2
= 24336 m2
and length of the side of the square
= Area = 24336
1561 2 43 36
125 143
125306 1836
1836 ×
= 156 m
Find the square root of the following
numbers in decimal form :
1. 84.8241
Solution—
8241.84 = 10000
848241
9.219 84 82 41
81182 382
3641841 1841
1841 ×
= 10000
848241 =
100
921 = 9.21
EXERCISE 3.7
2. 0.7225
Solution—
7225.0 = 10000
7225
= 10000
7225 =
100
85 = 0.85
3. 0.813604
Solution—
813604.0 = 1000000
813604
858 72 25
64165 825
825 ×
https://www.arundeepselfhelp.info/index.php?route=product/product&path=139_141&product_id=111
88
9029 81 36 04
811802 3604
3604 ×
= 1000000
813604 =
1000
902 = 0.902
4. 0.00002025
Solution—
00002025.0 = 100000000
2025
454 20 25
1685 425
425 ×
= 100000000
2025 =
10000
45 = 0.0045
5. 150.0625
Solution—
0625.150 = 10000
1500625
= 10000
1500625
= 100
1225 = 12.25
6. 225.6004
Solution—
6004.225
= 10000
2256004 =
10000
2256004
15021 2 25 60 04
125 125
1253002 6004
6004 ×
= 100
1502 = 15.02
7. 3600.720036
Solution—
720036.3600
= 1000000
3600720036 =
1000000
3600720036
600066 36 00 72 00 36
36120006 00720036
720036 ×
= 1000
60006 = 60.006
8. 236.144689
Solution—
144689.236 = 1000000
236144689
= 1000000
236144689
12251 1 50 06 25
122 50
44242 606
4842445 12225
12225 ×
https://www.arundeepselfhelp.info/index.php?route=product/product&path=139_141&product_id=111
89
153671 2 36 14 46 89
125 136
125303 1114
9093066 20546
1839630727 215089
215089 ×
= 1000
15367 = 15.367
9. 0.00059049
Solution—
00059049.0
= 100000000
59049 =
100000000
59049
2432 5 90 49
444 190
176483 1449
1449 ×
= 10000
243 = 0.0243
10. 176.252176
Solution—
252176.176
= 1000000
176252176 =
1000000
176252176
132761 1 76 25 21 76
123 76
69262 725
5242647 20121
1852926546 159276
159276 ×
= 1000
13276 = 13.276
11. 9998.0001
Solution—
0001.9998 = 10000
99980001
= 10000
99980001
99999 99 98 00 01
81189 1898
17011989 19700
1790119989 179901
179901 ×
= 100
9999 = 99.99
12. 0.00038809
Solution—
00038809.0 = 100000000
38809
https://www.arundeepselfhelp.info/index.php?route=product/product&path=139_141&product_id=111
90
= 100000000
38809
1971 3 88 09
129 288
261387 2709
2709 ×
= 10000
197 = 0.0197
13. What is that fraction which when
multiplied by itself gives 227.798649 ?
Solution—
Let the fraction = x
Then x2 = 227.798649
x = 798649.227
= 1000000
227798649 =
1000000
227798649
150931 2 27 79 86 49
125 127
1253009 27986
2708130183 90549
90549 ×
= 1000
15093 = 15.093
Required fraction = 15.093
14. The area of a square playground is
256.6404 square metres. Find the length
of one side of the playground.
Solution—
Area of square playground = 256.6404 sq. m
Side of the square = Area
= 6404.256 = 10000
2566404
16021 2 56 64 04
126 156
1563202 6404
6404 ×
= 10000
2566404 =
100
1602 = 16.02 m
15. What is the fraction which when
multiplied by itself gives 0.00053361 ?
Solution—
Let the given fraction = x
x × x = 0.00053361
x2 = 0.00053361
x = 00053361.0 = 100000000
53361
2312 5 33 61
4431 133
129461 461
461 ×
= 100000000
53361 =
10000
231 = 0.0231
Required fraction = 0.0231
16. Simplify :
(i)2952959
2952959
..
..
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91
(ii)0.17640.2304
0.17640.2304
Solution—
(i)5.2959.29
5.2959.29
7.77 59.29
49147 1029
1029 ×
2.32 5.29
443 129
129 ×
= 3.27.7
3.27.7
= 10
4.5 = 0.54
(ii)0.17640.2304
0.17640.2304
0.484 0.23 04
1688 704
704 ×
0.424 0.17 64
1682 164
164 ×
= 42.048.0
42.048.0
= 06.0
90.0 =
6
90 = 15
17. Evaluate 50625 and hence find the
value of 506.25 + 5.0625 .
Solution—
50625 = 225
2252 5 06 25
442 106
84445 2225
2225 ×
Now 25.506 + 0625.5
= 100
50625 +
10000
50625
= 10
225 +
100
225 = 22.50 + 2.25
= 24.75
18. Find the value of 103.0225 and hence
find the value of
(i) 10302.25 (ii) 1.030225
Solution—
0225.103 = 10000
1030225 =
10000
1030225
(i) 25.10302 = 100
1030225
= 10
1015 = 101.5
(ii) 030225.1 = 1000000
1030225
= 1000
1015 = 1.015
10151 1 03 02 25
1201 0302
2012025 10125
10125 ×
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92
1. Find the square root of each of the
following correct to three places of
decimal :
(i) 5 (ii) 7
(iii) 17 (iv) 20
(v) 66 (vi) 427
(vii) 1.7 (viii) 23.1
(ix) 2.5 (x) 237.615
(xi) 15.3215 (xii) 0.9
(xiii)0.1 (xiv) 0.016
(xv) 0.00064 (xvi) 0.019
(xvii) 8
7(xviii)
12
5
(xix) 22
1(xx) 287
8
5
Solution—
(i) 5 = 000000.5
= 2.236
(ii) 7 = 00000000.7
= 2.6457
= 2.646
(iii) 17 = 000000.17
4.1234 17.00 00 00
1681 100
81822 1900
16448243 25600
24729 871
= 4.123
(iv) 20 = 000000.20
4.4724 20.00 00 00
1684 400
336887 6400
62098942 19100
17884 1216
= 4.472
(v) 66 = 000000.66
8.1248 66.00 00 00
64161 200
1611622 3900
324416244 65600
64976 624
= 8.124
EXERCISE 3.8
2.2362 5.00 00 00
442 100
84443 1600
13294466 27100
26796 304
2.64572 7.00 00 00 00
446 300
276524 2400
20965285 30400
2642552907 397500
370349 27151
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93
(vi) 427 = 000000.427
= 20.664
(vii) 7.1 = 70000000.1
1.30371 1.70 00 00 00
123 70
692603 10000
780926068 219100
208544 10556
= 1.3038 = 1.304
(viii) 1.23 = 100000.23
= 4.806
(ix) 5.2 = 500000.2
1.5811 2.50 00 00
125 150
125308 2500
24643161 3600
3161 639
= 1.581
(x) 615.237 = 615000.237
15.41471 2 37.61 50 00 00
125 137
125304 1261
12163081 4550
308130824 146900
123296308287 2360400
2158009 202391
= 15.4147 = 15.415
(xi) 3215.15 = 321500.15
3.91423 15.32 15 00 00
969 632
621781 1115
7817824 33400
3129678282 210400
156564 53836
= 3.9142 = 3.914
(xii) 9.0 = 900000.0
= 0.9486 = 0.949
20.6642 4 27.00 00 00
4406 2700
24364126 26400
2475641324 164400
165296
4.8064 23.10 00 00
1688 710
7049606 60000
57636 2364
0.94869 0.90 00 00 00
81184 900
7361888 16400
1510418966 129600
113796 15804
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94
(xiii) 1.0 = 100000.0
= 0.316
(xiv) 016.0 = 016000.0
0.12641 0.01 60 00 00
122 60
44246 1600
14762524 12400
10096 2304
= 0.1264 = 0.126
(xv) 00064.0 = 000640.0
0.02522 0.00 06 40 00
445 240
225502 1500
1004 496
= 0.0252 = 0.025
(xvi) 019.0 = 019000.0
= 0.1378 = 0.138
(xvii) 8
7 = 875.0
0.93549 0.87 50 00 00
81183 650
5491865 10100
932518704 77500
74816 2684
) 7.000 64 60 56 40 40 ×
8
0.875
= 0.9354 = 0.935
(xviii) 12
5 = 416666.0
0.64546 0.41 66 66 66
36124 566
4961285 7066
642512904 64166
51616 12550
) 5.0000 48 20 12 80 72 80 72 80
12
0.41666
= 0.6454 = 0.645
(xix)2
12 =
2
5 = 5.2
= 1.581
0.3163 0.10 00 00
961 100
61626 3900
3756 144
0.13781 0.01 90 00 00
123 90
69267 2100
18692748 23100
21984 1116
1.5811 2.50 00 00
125 150
125308 2500
24643161 3600
3161 439
https://www.arundeepselfhelp.info/index.php?route=product/product&path=139_141&product_id=111
95
(xx)8
5287 =
8
2301 = 625.287
16.9591 2 87.62 50 00
126 187
156329 3162
29613385 20150
1692533909 322500
305181 12550
) 5.000 48 20 16 40 40 ×
8625
= 16.959
2. Find the square root of 12.0068 correct
to four decimal places.
Solution—
0068.12
= 3.46508 = 3.4651
3. Find the square root of 11 correct to five
decimal places.
Solution—
11 = 3.31662
4. Given that 2 = 1.414, 3 = 1.732, 5
= 2.236 and 7 = 2.646. Evaluate each
of the following :
(i)7
144(ii)
3
2500
Solution—
Given = 2 = 1.414, 3 = 1.732,
5 = 2.236 and 7 2.646
(i)7
144 =
7
144 =
646.2
12
) 12000.000 ( 10584 14160 13230 9300 7938 13620 13230 3900 2646 1251
2646 4.5351
= 2646
100012 =
2646
12000
= 4.5351
= 4.535
OR
7
144 =
7
144 =
77
7144
=
7
712
= 7
646.212 =
7
752.31
= 4.536
(ii)3
2500 =
3
2500 =
33
32500
= 3
350 =
3
732.150
3.465083 12.00 68 00 00 00
964 300
256686 4468
41166925 35200
34625693008 5750000
5544064 205936
3.316623 11.00 00 00 00 00
963 200
189661 1100
6616626 43900
3975666326 414400
397956663322 1644400
1326644 317756
https://www.arundeepselfhelp.info/index.php?route=product/product&path=139_141&product_id=111
96
= 3
6.86 = 28.8666
= 28.867
5. Given that 2 = 1.414, 3 = 1.732, 5
= 2.236 and 7 = 2.646, find the square
roots of the following :
(i)75
196(ii)
63
400
(iii)7
150(iv)
5
256
(v)50
27
Solution—
We are given : 2 = 1.414, 3 = 1.732,
5 = 2.236 and 7 = 2.646
(i)75
196 =
325
196
=
325
196
=
35
196
= 335
314
=
35
314
=
15
314
= 15
732.114 =
15
248.24
= 1.6165 = 1.617
(ii)63
400 =
79
400
=
73
20
= 773
720
=
73
720
=
21
720
= 21
646.220 =
21
92.52
= 2.52
(iii)7
150 =
7
150 =
77
7150
= 7
7150
= 7
73225
= 7
7325
=
7
646.2732.1414.15
= 7
4801.65 =
7
4009.32
= 4.6287 = 4.629
(iv)5
256 =
5
256 =
5
16
= 55
516
=
5
236.216
= 5
776.35 = 7.155
(v)50
27 =
50
27 =
225
39
= 25
33 =
225
233
= 25
233
=
10
233
=
10
414.1732.13 =
10
449048.23
= 10
3471.7
= 0.7347 = 0.735
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97
Using square root table, find the square
roots of the following :
1. 7
Solution—
Let x = 7
Then x = 7
From the table, in the column of x
We find that x = 7 = 2.646
2. 15
Solution—
Let x = 15
Then x = 15
From the table, in column of x , we find
that
x = 15 = 3.873
3. 74
Solution—
Let x = 74
x = 74
From the table, in the column of x , we
find that
x = 74 = 8.602
4. 82
Solution—
Let x = 82
x = 82
From the table in the column of x , we
find that x = 82 = 9.055
5. 198
Solution—
198 = 229
= 3 22
From the table we see from the x ,
22 = 4.690
198 = 3 × 22 = 3 × 4.690
= 14.070
6. 540
Solution—
540 = 1536
6 5406 90
15
= 6 15
From the table, we see in the column x ,
that 15 = 3.873
540 = 6 × 15 = 6 × 3.873
= 23.238 = 23.24
7. 8700
Solution—
8700 = 87100
= 10 87
From the table, we see in the column x ,
that 87 = 9.327
8700 = 10 87 = 10 × 9.327
= 93.27
8. 3509
Solution—
3509 = 29121
= 11 × 29
EXERCISE 3.9
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98
From the table, we see in the column x
that 29 = 5.385
11 350911 319
29
3509 = 11 × 29 = 11 × 5.385
= 59.235
9. 6929
Solution—
6929 = 41169 = 13 41
From the table, we see in the column x ,
that 41 = 6.403
13 692913 533
41
6929 = 13 41
= 13 × 6.403 = 83.239
10. 25725
Solution—
25725 = 217755
5 257255 51457 10297 147
21
= 5 × 7 21
= 35 21
From the table we see the column of x ,
that 21 = 4.583
25725 = 35 21 = 35 × 4.583
= 160.40
11. 1312
Solution—
1312 = 8244
4 13124 328
82
= 4 × 82
From the table, we see in the column x ,
that 82 = 9.055
1312 = 4 × 82 = 4 × 9.055
= 36.220 = 36.22
12. 4192
Solution—
4192 = 26216
2 41922 20962 10482 524
262
= 4 262
= 4 × 16.186
= 64.7456 = 64.75
13. 4955
Solution—
4955 = 10055.49
= 10 55.49
55.49 lies between 49 and 50 or 7
and 7.071
Diff. between 49 and 50 = 1 and between 7
and 7.071 = 0.071
Diff. between 0.55 = 0.55 × 0.71 = 0.039
10 55.49 = 10 × 7.039 = 70.39
14.144
99
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99
Solution—
144
99 =
1212
1133
=
1212
1133
= 12
311 =
4
111
From the table,
11 = 3.317
144
99 =
4
1 × 11 =
4
1 × 3.317
= 4
317.3 = 0.829
15.169
57
Solution—
169
57 =
169
57 =
13
157
From the table, 57 = 7.550
169
57 =
13
157 =
13
550.7 = 0.581
16.169
101
Solution—
169
101 =
169
101 =
13
1101 =
13
1 × 10.05
= 13
05.10 = 0.773
17. 13.21
Solution—
21.13 is between 13 and 14
13 = 3.606 and 14 = 3.742
Difference between 14 and 13
= 3.742 – 3.606 = 0.136
and difference 13 and 14 = 1
Difference in 0.21 = 0.136 × 0.21
= 0.02856
21.13 = 3.606 + 0.029 = 3.635
18. 21.97
Solution—
97.21
97.21 is between 21 and 22
21 = 4.583 and 22 = 4.690
Difference between 22 and 21
= 4.690 – 4.583 = 0.107
Difference between 21 and 22 = 1
and difference in 0.97 = 0.97 × 0.107
= 0.1038
97.21 = 4.583 + 0.104 = 4.687
19. 110
Solution—
110 = 1011 = 11 × 10
= 3.317 × 3.162
= 10.488
20. 1110
Solution—
1110 = 3037
= 6.083 × 5.477 (from the table)
= 33.3165 = 33.317
21. 11.11
Solution—
11.11 lies between 11 and 12
11 = 3.317 and 12 = 3.464
Difference between 11 and 12 = 1
and difference between 3.464 and 3.317
= 3.464 – 3.317 = 0.147
Difference in 0.11 = 0.147 × 0.11 = 0.01617
11.11 = 3.317 + 0.0162
= 3.3332 = 3.333
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22. The area of a square field is 325 m2. Find
the approximate length of one side of the
field.
Solution—
Area of the square field = 325 m2
Length of one side = Area = 325 m
= 325 = 1325
= 25 × 13 = 5 × 13
= 5 × 3.606 m (from the table)
= 18.030 = 18.030 m
23. Find the length of a side of a square, whose
area is equal to the area of the rectangle
with sides 240 m and 70 m.
Solution—
Lenght of rectangle (l) = 240 m
and breadth (b) = 70 m
Area = l × b = 240 × 70 m2
= 16800 m2
Area of square field = 16800 m2
and side = Area = 16800 m
= 42400
= 400 × 42
= 20 × 6.481
= 129.620
= 129.62 m
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Points to Remember
1. Cube— The cube of a number is that
number raised to the power 3 e.g. x3, 53, 23
etc.
2. Perfect cube— An integer is said to be a
perfect cube if it is the cube of some integer.
3. Properties of cubes—
The cubes of natural numbers have the
following interesting properties :
Property 1 : Cubes of all even natural
numbers are even.
Property 2 : Cubes of all odd natural
numbers are odd.
Property 3 : The sum of the cubes of first
n natural numbers is equal to the square of
their sum. That is,
13 + 23 + 33 + ..... + n3 = (1 + 2 + 3 + .....
+ n)2
Property 4 : Cubes of the numbers ending
in digits 1, 4, 5, 6 and 9 are the numbers
ending in the same digit. Cubes of numbers
ending in digit 2 ends in digit 8 and the cube
of numbers ending in digit 8 ends in digit
2. The cubes of the numbers ending in
digits 3 and 7 ends in digit 7 and 3
respectively.
4. Methods of finding the cube of two-digit
number—
(i) Using formula— (a + b)3 = a3 + 3a2b
+ 3ab2 + b3
5. Cube of negative numbers— If m is a
positive integer, (–m)3 = –m × –m × –m = –
m3
–m3 is the cube of (–m).
6. Cube of rational numbers—
3
n
m =
3
3
n
m, where
n
m is a rational number where
m and n are non-zero integers.
7. Cube root— A number m is the cube root
of a number n if n = m3 and it is denoted as
3 n or 3
1
n where 3 is called radical or
3
1 is called index.
8. Computation of cube root through a
pattern—
The pattern is
13 = 1,
23 = 1 + 7
33 = 1 + 7 + 19 (4)3 = 1 + 7 + 19 + 37
(5)2 = 1 + 7 + 19 + 37 + 61
(6)3 = 1 + 7 + 19 + 37 + 61 + 91
and so on.
The procedure for finding the cube root of
the given number will be possible if the
number is a small number.
The sequence can be written as
1, 1 + 2
12 × 6 = 7 1 +
2
67 × 6 = 127
1 + 2
23 × 6 = 19 1 +
2
78 × 6 = 169
1 + 2
34 × 6 = 37 1 +
2
89 × 6 = 217
4CUBES AND CUBE-ROOTS
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1 + 2
45 × 6 = 61 1 +
2
56 × 6 = 91
and so on.
If the numbers which are to be subtracted
in succession from the given number.
9. Cube root by factors—
(i) Resolve the numbers into factors and then
group them in triplets of equal of a factors.
(ii) Take one factor from each triplet and find
the product of these factors.
This product is the cube root of the given
number.
10. Cube root of a.b or b
a
(i) Cube root of a.b = 3 ab = 3 a × 3 b
and cube root of b
a = 3
b
a = 3
3
b
a, b 0
11. Cube root tables— We can find cube root
of a given number by using the given cube
root tables also.
12. Cube root of negative numbers— Cube
root of –n = 3 n = – 3 n
1. Find the cubes of the following numbers:
(i) 7 (ii) 12
(iii) 16 (iv) 21
(v) 40 (vi) 55
(vii) 100 (viii) 302
(ix) 301
Solution—
(i) (7)3 = 7 × 7 × 7 = 343
(ii) (12)3 = 12 × 12 × 12 = 1728
(iii) (16)3 = 16 × 16 × 16 = 4096
(iv) (21)3 = 21 × 21 × 21
= 441 × 21 = 9261
(v) (40)3 = 40 × 40 × 40 = 64000
(vi) (55)3 = 55 × 55 × 55
= 3025 × 55 = 166375
(vii) (100)3 = 100 × 100 × 100
= 1000000
(viii)(302)3 = 302 × 302 × 302
= 91204 × 302 = 27543608
(ix) (301)3 = 301 × 301 × 301
= 90601 × 301 = 27270901
2. Write the cubes of all natural numbers
between 1 and 10 and verify the following
statements :
EXERCISE 4.1
(i) Cubes of all odd natural numbers are odd.
(ii) Cubes of all even natural numbers are
even.
Solution—
Cubes of first 10 natural numbers :
(1)3 = 1 × 1 × 1 = 1
(2)3 = 2 × 2 × 2 = 8
(3)3 = 3 × 3 × 3 = 27
(4)3 = 4 × 4 × 4 = 64
(5)3 = 5 × 5 × 5 = 125
(6)3 = 6 × 6 × 6 = 216
(7)3 = 7 × 7 × 7 = 343
(8)3 = 8 × 8 × 8 = 512
(9)3 = 9 × 9 × 9 = 729
(10)3 = 10 × 10 × 10 = 1000
We see that the cubes of odd numbers is also
odd and cubes of even numbers is also even.
3. Observe the following pattern :
13 = 1
13 + 23 = (1 + 2)2
13 + 23 + 33 = (1 + 2 + 3)2
Write the next three rows and calculate
the value of 13 + 23 + 33 + ..... + 93 + 103 by
the above pattern.
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Solution—
We see the pattern
13 = 1
13 + 23 = (1 + 2)2
13 + 23 + 33 = (1 + 2 + 3)2
13 + 23 + 33 + 43 = (1 + 2 + 3 + 4)2
13 + 23 + 33 + 43 + 53 = (1 + 2 + 3 + 4 + 5)2
and 13 + 23 + 33 + 43 + 53 + 63 = (1 + 2 + 3 + 4 + 5 + 6)2
and 13 + 23 + 33 + .... + 93 + 103 = (1 + 2 + 3 + .... + 9 + 10)2
4. Write the cubes of 5 natural numbers
which are multiples of 3 and verify the
followings :
‘The cube of a natural number which is a
mutiple of 3 is a multiple of 27’
Solution—
5 natural numbers which are multiples of 3
3, 6, 9, 12, 15
(3)3 = 3 × 3 × 3 = 27
Which is multiple of 27
(6)3 = 6 × 6 × 6 = 216 27 = 8
Which is multiple of 27
(9)3 = 9 × 9 × 9 = 729 27 = 27
Which is multiple of 27
(12)3 = 12 × 12 × 12 = 1728 27 = 64
Which is multiple of 27
(15)3 = 15 × 15 × 15 = 3375 27 = 125
Which is multiple of 27
Hence, cube of multiple of 3 is a multiple of
27
5. Write the cubes of 5 natural numbers
which are of the form 3n + 1 (e.g. 4, 7,
10, .....) and verify the following :
‘The cube of a natural number of the
form 3n + 1 is a natural number of the
same form i.e. when divided by 3 it leaves
the remainder 1’.
Solution—
3n + 1
Let n = 1, 2, 3, 4, 5, then
If n = 1, then 3n + 1 = 3 × 1 + 1 = 3 + 1 = 4
If n = 2, then 3n + 1 = 3 × 2 + 1 = 6 + 1 = 7
If n = 3, then 3n + 1 = 3 × 3 + 1 = 9 + 1 = 10
If n = 4, then 3n + 1 = 3 × 4 + 1 = 12 + 1
= 13
If n = 5, then 3n + 1 = 3 × 5 + 1 = 15 + 1
= 16
Now
(4)3 = 4 × 4 × 4 = 64
Which is 3
64 = 21, Remainder = 1
(7)3 = 7 × 7 × 7 = 343
Which is 3
343 = 114, Remainder = 1
(10)3 = 10 × 10 × 10 = 1000 3 = 333,
Remainder = 1
(13)3 = 13 × 13 × 13 = 2197 3 = 732,
Remainder = 1
(16)3 = 16 × 16 × 16 = 4096 3 = 1365,
Remainder = 1
Hence cube of natural number of the form,
3n + 1, is a natural of the form 3n + 1
6. Write the cubes of 5 natural numbers of
the form 3n + 2 (i.e. 5, 8, 11, .....) and
verify the following :
‘The cube of a natural number of the
form 3n + 2 is a natural number of the
same form i.e. when it is dividend by 3
the remainder is 2’.
Solution—
Natural numbers of the form 3n + 2, when n
is a natural number i.e. 1, 2, 3, 4, 5, .......
If n = 1, then 3n + 2 = 3 × 1 + 2 = 3 + 2 = 5
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If n = 2, then 3n + 2 = 3 × 2 + 2 = 6 + 2 = 8
If n = 3, then 3n + 2 = 3 × 3 + 2 = 9 + 2 = 11
If n = 4, then 3n + 2 = 3 × 4 + 2 = 12 + 2 =
14
and if n = 5, then 3n + 2 = 3 × 5 + 2 = 15 +
2 = 17
Now(5)3 = 5 × 5 × 5 = 125
125 3 = 41, Remainder = 2
(8)2 = 8 × 8 × 8 = 512
512 3 = 170, Remainder = 2
(11)3 = 11 × 11 × 11 = 1331
1331 3 = 443, Remainder = 2
(14)3 = 14 × 14 × 14 = 2744
2744 3 = 914, Remainder = 2
(17)3 = 17 × 17 × 17 = 4913
4913 3 = 1637, Remainder = 2
We see the cube of the natural number of the
form 3n + 2 is also a natural number of the
form 3n + 2.
7. Write the cubes of 5 natural numbers of
which are multiples of 7 and verify the
following :
‘The cube of a multiple of 7 is a multiple
of 73’.
Solution—
5 natural numbers which are multiple of 7,
are 7, 14, 21, 28, 35
(7)3 = (7)3 which is multiple of 73
(14)3 = (2 × 7)3 = 23 × 73, which is multiple
of 73
(21)3 = (3 × 7)3 = 33 × 73, which is multiple
of 73
(28)3 = (4 × 7)3 = 43 × 73, which is multiple
of 73
(35)3 = (5 × 7)3 = 53 × 73 which is multiple of 73
Hence proved.
8. Which of the following are perfect cubes?
(i) 64 (ii) 216
(iii) 243 (iv) 1000
(v) 1728 (vi) 3087
(vii) 4608 (viii) 106480
(ix) 166375 (x) 456533
Solution—
(i) 64 = 2 ×2 × 2 × 2 × 2 × 2
2 642 322 162 82 42 2
1
Grouping the factors in triplets of equal
factors, we see that no factor is left
64 is a perfect cube
(ii) 216 = 2 × 2 × 2 × 3 × 3 × 3
2 2162 1082 543 273 93 3
1
Grouping the factors in triplets of equal
factors, we see that no factor is left
216 is a perfect cube.
(iii) 243 = 3 × 3 × 3 × 3 × 3
3 2433 813 273 93 3
1
Grouping the factors in triplets, we see that
two factors 3 × 3 are left
243 is not a perfect cube.
(iv) 1000 = 2 × 2 × 2 × 5 × 5 × 5
2 10002 5002 2505 1255 255 5
1
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Grouping the factors in triplets of equal
factors, we see that no factor is left
1000 is a perfect cube.
(v) 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
2 17282 8642 4322 2162 1082 543 273 93 3
1
Grouping the factors in triplets of the equal
factors, we see that no factor is left
1728 is a perfect cube.
(vi) 3087 = 3 × 3 × 7 × 7 × 7
Grouping the factors in triplets of the equal
factors, we see that two factor 3 × 3 are left
3087 is not a perfect cube.
(vii) 4608 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ×
3 × 3
Grouping the factors in triplets of equal
factors, we see that two factors 3, 3 are left
4609 is not a perfect cube.
(viii)106480 = 2 × 2 × 2 × 2 × 5 × 11 × 11 × 11
2 1064802 532402 266202 133105 6655
11 133111 12111 11
1
Grouping the factors in triplets of equal
factors, we see that factors 2, 5 are left
106480 is not a perfect cube.
(ix) 166375 = 5 × 5 × 5 × 11 × 11 × 11
5 1663755 332755 6655
11 133111 12111 11
1
Grouping the factors in triplets of equal
factors, we see that no factor is left
166375 is a perfect cube.
(x) 456533 = 7 × 7 × 7 × 11 × 11 × 11
7 4565337 652197 9317
11 133111 12111 11
1
Grouping the factors in triplets of equal
factors, we see that no factor is left
456533 is a perfect cube.
9. Which of the following are cubes of even
natural numbers ?
216, 512, 729, 1000, 3375, 13824
Solution—
We know that the cube of an even natural
number is also an even natural number
2 46082 23042 11522 5762 2882 1442 722 362 183 93 3
1
3 30873 10297 3437 497 7
1
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216, 512, 1000, 13824 are even natural
numbers.
These can be the cubes of even natural
number.
10. Which of the following are cubes of odd
natural numbers ?
125, 343, 1728, 4096, 32768, 6859
Solution—
We know that the cube of an odd natural
number is also an odd natural number.
_ 125, 343, 6859 are the odd natural numbers
These can be the cubes of odd natural
numbers.
11. What is the smallest number by which
the following numbers must be multiplied,
so that the products are perfect cubes ?
(i) 675 (ii) 1323
(iii) 2560 (iv) 7803
(v) 107811 (vi) 35721
Solution—
(i) 675 = 3 × 3 × 3 × 5 × 5
3 6753 2253 755 255 5
1
Grouping the factors in triplet of equal
factors, 5 × 5 are left without triplet
So, by multiplying by 5, the triplet will be
completed.
Least number to be multiplied = 5
(ii) 1323 = 3 × 3 × 3 × 7 × 7
3 13233 4413 1477 497 7
1
Grouping the factors in triplet of equal
factors. We find that 7 × 7 has been left
So, multiplying by 7, we get a triplet
The least number to be multiplied = 7
(iii) 2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
× 5
2 25602 12802 6402 3202 1602 802 402 202 105 5
1
Grouping the factors in triplet of equal
factors, 5 is left.
To complete a triplet 5 × 5 is to multiplied
Least number to be multiplied = 5 × 5 = 25
(iv) 7803 = 3 × 3 × 3 × 17 × 17
3 78033 26013 867
17 28917 17
1
Grouping the factors in triplet of equal
factors, we find the 17 × 17 are left
So, to complete the triplet, we have to multiply
by 17
Least number to be multiplied = 17
(v) 107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11
3 1078113 359373 119793 3993
11 133111 12111 11
1
Grouping the factors in triplet of equal
factors, factor 3 is left
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So, to complete the triplet 3 × 3 is to be
multiplied
Least number to be multiplied = 3 × 3 = 9
(vi) 35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7
Grouping the factors in triplet of equal
factors, we find that 7 × 7 is left
So, in order to complete the triplets, we have
to multiplied by 7
Least number to be multiplied = 7
12. By which smallest number must the
following numbers be divided so that the
quotient is a perfect cube ?
(i) 675 (ii) 8640 (iii) 1600
(iv) 8788 (v) 7803 (vi) 107811
(vii) 35721 (viii) 243000
Solution—
(i) 675 = 3 × 3 × 3 × 5 × 5
Grouping the factors in triplet of equal
factors, 5 × 5 is left
5 × 5 is to be divided so that the quotient will
be a perfect cube.
The least number to be divided = 5 × 5 = 25
(ii) 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
× 5
Grouping the factors in triplets of equal
factors, 5 is left
In order to get a perfect cube, 5 is to divided
Least number to be divided = 5
(iii) 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
2 16002 8002 4002 2002 1002 505 255 5
1
Grouping the factors in triplets of equal
factors, we find that 5 × 5 is left
In order to get a perfect cube 5 × 5 = 25 is to
be divided.
Least number to be divide = 25
(iv) 8788 = 2 × 2 × 13 × 13 × 13
2 87882 4394
13 219713 16913 13
1
Grouping the factors in triplets of equal
factors, we find that 2 × 2 has been left
In order to get a perfect cube, 2 × 2 is to be
divided
Least number to be divided = 4
(v) 7803 = 3 × 3 × 3 × 17 × 17
3 78033 26013 867
17 28917 17
1
Grouping the factors in triplets of equal
factors, we see that 17 × 17 has been left.
So, in order to get a perfect cube, 17 × 17 is
be divided
3 357213 119073 39693 13233 4413 1477 497 7
1
3 6753 2253 755 255 5
1
2 86402 43202 21602 10802 5402 2703 1353 453 155 5
1
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Least number to be divided = 17 × 17 = 289
(vi) 107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11
3 1078113 359373 119793 3993
11 133111 12111 11
1
Grouping the factors in triplets of equal
factors, 3 is left
In order to get a perfect cube, 3 is to be
divided
Least number to be divided = 3
(vii) 35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7
3 357213 119073 39693 13233 4413 1477 497 7
1
Grouping the factors in triplets of equal
factors, we see that 7 × 7 is left
So, in order to get a perfect cube, 7 × 7 = 49
is to be divided
Least number to be divided = 49
(viii)243000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 ×
5 × 5 × 5
Grouping the factors in triplets of equal
factors, 3 × 3 is left
By dividing 3 × 3, we get a perfect cube
Least number to be divided = 3 × 3 = 9
13. Prove that if a number is trebled then its
cube is 27 times the cube of the given
number.
Solution—
Let x be the number, then trebled number of
x = 3x
Cubing, we get :
(3x)3 = (3)3 x3 = 27x3
27x3 is 27 times the cube of x i.e., of x3
14. What happenes to the cube of a number
if the number is multiplied by
(i) 3 ? (ii) 4 ? (iii) 5 ?
Solution—
Let x be the given number
(x)3 = x3
(i) If x is multiplied by 3, then the cube of
(3x)3 = (3)3 × x3 = 27x3
The cube of the resulting number is 27 times
of cube of the given number
(ii) If x is multiplied by 4, then the cube of (4x)3
= (4)3 × x3 = 64x3
The cube of the resulting number is 64 times
of the cube of the given number
(iii) If x is multiplied by 5, then the cube of
(5x)3 = (5)3 × x3 = 125x3
The cube of the resulting number is 125 times
of the cube of the given number
15. Find the volume of a cube, one face of
which has an area of 64 m2.
Solution—
Area of one face of a cube = 64 m2
Side (edge) of cube = 64 m
= 64 = 8 m
Volume of the cube = (side)3 = (8 m)3
= 512 m3
16. Find the volume of a cube whose surface
area is 384 m2.
Solution—
2 2430002 1215002 607503 303753 101253 33753 11253 3755 1255 255 5
1
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Surface area of a cube = 384 m2
Let side = a
Then 6a2 = 384 a2 = 6
384 = 64 = (8)2
a = 8 m
Now volume = a3 = (8)3 m3 = 512 m3
17. Evaluate the following :
(i) 3
2
122
125
(ii)
3
2
122
86
Solution—
(i) 3
2
122
125
=
3
2
1
14425
= 3
2
1
169
=
3
2
12
13
= 32
12
13 = 133 = 2197
(ii) 3
2
122
86
=
3
2
1
6436
= 3
2
1
100
=
3
2
12
10
= 32
12
10 = 103
= 1000
18. Write the units digit of the cube of each
of the following numbers :
31, 109, 388, 833, 4276, 5922, 77774, 44447,
125125125.
Solution—
We know that if unit digit of a number n is
= 1, then units digit of its cube = 1
= 2, then units digit of its cube = 8
= 3, then units digit of its cube = 7
= 4, then units digit of its cube = 4
= 5, then units digit of its cube = 5
= 6, then units digit of its cube = 6
= 7, then the units digit of its cube = 3
= 8, then units digit of its cube = 2
= 9, then units digit of its cube = 9
= 0, then units digit of its cube = 0
Now units digit of the cube of 31 = 1
Units digit of the cube of 109 = 9
Units digits of the cube of 388 = 2
Units digits of the cube of 833 = 7
Units digits of the cube of 4276 = 6
Units digit of the cube of 5922 = 8
Units digit of the cube of 77774 = 4
Units digit of the cube of 44447 = 3
Units digit of the cube of 125125125 = 5
19. Find the cubes of the following numbers by column method :
(i) 35 (ii) 56 (iii) 72
Solution—
(i) (35)3
I column II column III column IV column
a3 3 × a2 × b 3 × a × b2 b3
(3)3 3 × (3)2 × 5 3 × 3 × (5)2 (5)3
= 27 = 135 = 225 = 125
+ 15 + 23 + 12
42 158 237
42 8 7 5
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(35)3 = 42875
(ii) (56)3
I column II column III column IV column
a3 3 × a2 × b 3 × a × b2 b3
(5)3 3 × 52 × 6 3 × 5 × 62 (6)3
= 125 = 450 = 540 = 216
+ 50 + 56 + 21
175 506 561
175 6 1 6
(56)3 = 175616
(iii) (72)3
I column II column III column IV column
a3 3 × a2 × b 3 × a × b2 b3
(7)3 3 × 72 × 2 3 × 7 × 22 (2)3
= 343 = 294 = 84 = 8
+ 30 + 8
373 302
373 2 4 8
(72)3 = 373248
20. Which of the following numbers are not
perfect cubes ?
(i) 64 (ii) 216
(iii) 243 (iv) 1728
Solution—
(i) 64 = 2 × 2 × 2 × 2 × 2 × 2
Grouping the factors in triplets, of equal
factors, we see that no factor is left
64 is a perfect cube.
(ii) 216 = 2 × 2 × 2 × 3 × 3 × 3
2 2162 1082 543 273 93 3
1
Grouping the factors in triplets, of equal
factors, we see that no factor is left
216 is a perfect cube.
(iii) 243 = 3 × 3 × 3 × 3 × 3
Grouping the factors in triplets, of equal
factors, we see that 3 × 3 are left
243 is not a perfect cube.
(iv) 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
2 642 322 162 82 42 2
1
3 2433 813 273 93 3
1
2 17282 8642 4322 2162 1082 543 273 93 3
1
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Grouping the factors in triplets, of equal
factors, we see that no factor is left.
1728 is a perfect cube.
21. For each of the non-perfect cubes, in Q.
20, find the smallest number by which it
must be
(a) multiplied so that the product is a perfect
cube.
(b) divided so that the quotient is a perfect
cube.
Solution—
In qustion 20, 243 is not a perfect cube and
243 = 3 × 3 × 3 × 3 × 3
Grouping the factors in triplets, of equal
factors, we see that 3 × 3 is left.
(a) In order to make it a perfect cube, 3 is to be
multiplied which makes a triplet.
(b) In order to make it a perfect cube, 3 × 3 or 9
is to be divided.
22. By taking three different values of n
verify the truth of the following
statements :
(i) If n is even, then n3 is also even.
(ii) If n is odd, then n3 is also odd.
(iii) If n leaves remainder 1 when divided by
3, then n3 also leaves 1 as remainder
when divided by 3.
(iv) If a natural number n is of the form 3p +
2 then n3 also a number of the same type.
Solution—
(i) n is even number.
Let n = 2, 4, 6 then
(a) n3 = (2)3 = 2 × 2 × 2 = 8, which is an even
number.
(b) (n)3 = (4)3 = 4 × 4 × 4 = 64, which is an even
number.
(c) (n)3 = (6)3 = 6 × 6 × 6 = 216, which is an
even number.
(ii) n is odd number.
Let x = 3, 5, 7
(a) (n)3 = (3)3 = 3 × 3 × 3 = 27, which is an odd
number.
(b) (n)3 = (5)3 = 5 × 5 × 5 = 125, which is an
odd number.
(c) (n)3 = (7)3 = 7 × 7 × 7 = 343, which is an
odd number.
(iii) If n leaves remainder 1 when divided by 3,
then n3 is also leaves 1 as remainder,
Let n = 4, 7, 10
If n = 4, then
n3 = (4)3 = 4 × 4 × 4 = 64
= 64 3 = 21, remainder = 1
If n = 7, then
n3 = (7)3 = 7 × 7 × 7 = 343
343 3 = 114, remainder = 1
If n = 10, then
(n)3 = (10)3 = 10 × 10 × 10 = 1000
1000 3 = 333, remainder = 1
(iv) If the natural number is of the form 3p + 2,
then n3 is also of the same type
Let p = 1, 2, 3, then
(a) If p = 1, then
n = 3p + 2 = 3 × 1 + 2 = 3 + 2 = 5
n3 = (5)3 = 5 × 5 × 5 = 125
125 = 3 × 41 + 2 = 3p + 2
(b) If p = 2, then
n = 3p + 2 = 3 × 2 + 2 = 6 + 2 = 8
n3 = (8)3 = 8 × 8 × 8 = 512
512 = 3 × 170 + 2 = 3p + 2
(c) If p = 3, then
n = 3p + 2 = 3 × 3 + 2 = 9 + 2 = 11
n3 = (11)3 = 11 × 11 × 11 = 1331
and 1331 = 3 × 443 + 2 = 3p + 2
Hence proved.
23. Write true (T) or false (F) for the
following statements :
(i) 392 is a perfect cube.
(ii) 8640 is not a perfect cube.
(iii) No cube can end with exactly two zeros.
(iv) There is no perfect cube which ends in 4.
(v) For an integer a, a3 is always greater than a2.
(vi) If a and b are integers such that a2 > b2,
then a3 > b3.
(vii) If a divides b, then a3 divides b3.
(viii) If a2 ends in 9, then a3 ends in 7.
(ix) If a2 ends in 5, then a3 ends in 25.
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(x) I f a2 ends in an even number of zeros.
then a3 ends in an odd number of zeros.
Solution—
(i) False _ 392 = 2 × 2 × 2 × 7 × 7
_ Its all factors are not in triplets of equal factors.
(ii) True
_ 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
× 5
2 86402 43202 21602 10802 5402 2703 1353 453 155 5
1
_ 5 is left.
(iii) True : A number ending three zeros can be a
perfect cube.
(iv) False : _ (4)3 = 4 × 4 × 4 = 64, which ends
with 4.
(v) False : If n is a proper fraction, it is not
possible.
(vi) False : It is not true if a and b are proper
fraction.
(vii) True.
(viii)False, as a2 ends in 9, then a3 does not
necessarily ends in 7. It ends in 3 also.
(ix) False, it is not necessarily that a3 ends in 25
it can end also in 75.
(x) False : If a2 ends with even zeros, then a3
will ends with odd zeros but of multiple of
3.
2 3922 1962 987 497 7
1
1. Find the cubes of :
(i) –11 (ii) –12 (iii) –21
Solution—
(i) (–11)3 = – (11)3 = – (11 × 11 × 11) = –1331
(ii) (–12)3 = – (12)3 = – (12 × 12 × 12) = –1728
(iii) (–21)3 = – (21)3 = – (21 × 21 × 21) = –9261
2. Which of the following numbers are cubes
of negative integers.
(i) –64 (ii) –1056 (iii) –2197
(iv) –2744 (v) –42875
Solution—
(i) –64 = – (2 × 2 × 2 × 2 × 2 × 2)
EXERCISE 4.2
2 642 322 162 82 42 2
1
_ All factors of 64 can be grouped in triplets
of the equal factors completely.
–64 is a perfect cube of negative integer.
(ii) –1056 = – (2 × 2 × 2 × 2 × 2 × 3 × 11)
2 10562 5282 2642 1322 663 33
11 111
_ All the factors of 1056 can be grouped in
triplets of equal factors grouped completely
–1058 is not a perfect cube of negative
integer.
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(iii) –2197 = – (2197) = – (13 × 13 × 13)
13 219713 16913 13
1
_ All the factors of –2197 can be grouped in
triplets of equal factors completely
–2197 is a perfect cube of negative integer.
(iv) –2744 = – (2744) = – (2 × 2 × 2 × 7 × 7 × 7)
2 27442 13722 6867 3437 497 7
1
_ All the factors of –2744 can be grouped in
triplets of equal factors completely
–2744 is a perfect cube of negative integer
(v) –42875 = – (42875)
= – (5 × 5 × 5 × 7 × 7 × 7)
5 428755 85755 17157 3437 497 7
1
_ All the factors of –42875 can be grouped in
triplets of equal factors completely
–42875 is a perfect cube of negative integer.
3. Show that the following integers are cubes
of negative integers. Also, find the
integer whose cube is the given integer :
(i) –5832 (ii) –2744000
Solution—
(i) –5832 = – (5832)
= – (2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3)
2 58322 29162 14583 7293 2433 813 273 93 3
1
Grouping the factors in triplets of equal
factors, we see that no factor is left
–5832 is a perfect cube
Now taking one factor from each triplet we
find that
–5832 is a cube of – (2 × 3 × 3) = –18
Cube root of –5832 = –18
(ii) –2744000 = – (2744000)
= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 × 7 × 7
× 7
2 27440002 13720002 6860002 3430002 1715002 857505 428755 85755 17157 3437 497 7
1
Grouping the factors in triplets of equal
factors, we see that no factor is left.
Therefore it is a perfect cube.
Now taking one factor from each triplet, we
find that.
–2744000 is a cube of – (2 × 2 × 5 × 7) i.e.
–140
Cube root of –2744000 = –140
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4. Find the cube of :
(i)9
7(ii)
11
8(iii)
7
12
(iv)8
13(v) 2
5
2(vi) 3
4
1
(vii) 0.3 (viii) 1.5 (ix) 0.08
(x) 2.1
Solution—
(i)
3
9
7
=
999
777
= 729
343
(ii)
3
11
8
=
11
8 ×
11
8 ×
11
8
= –111111
888
= –1331
512
(iii)
3
7
12
=
7
12 ×
7
12 ×
7
12
= 777
121212
= 343
1728
(iv)
3
8
13
=
8
13 ×
8
13 ×
8
13
888
131313
= –512
2197
(v)
3
5
22
=
3
5
12
=
5
12 ×
5
12 ×
5
12
= 555
121212
= 125
1728
(vi)
3
4
13
=
3
4
13
=
4
13 ×
4
13 ×
4
13
= 444
131313
= 64
2197
(vii) (0.3)3 =
3
10
3
=
10
3 ×
10
3 ×
10
3
= 1000
27 = 0.027
(viii)(1.5)3 =
3
10
15
=
10
15 ×
10
15 ×
10
15
= 101010
151515
= 1000
3375 = 3.375
(ix) (0.08)3 =
3
100
8
=
100
8 ×
100
8 ×
100
8
= 1000000
512 = 0.000512
(x) (2.1)3 =
3
10
21
=
10
21 ×
10
21 ×
10
21
= 101010
212121
= 1000
9261 = 9.261
5. Which of the following numbers are cubes
of rational numbers :
(i)64
27(ii)
128
125
(iii) 0.001331 (iv) 0.04
Solution—
(i)64
27 =
444
333
=
3
4
3
64
27 is a perfect cube
(ii)128
125 =
2222222
555
=
222
533
3
We see that 128 is not a perfect cube
128
125 is not a perfect cube.
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(iii) 0.001331 = 1000000
1331
= 101010101010
111111
=
3
1010
11
=
3
100
11
It is a perfect cube.
(iv) 0.04 = 100
4 =
1010
22
= 10
2 ×
10
2 or 0.2 ×
0.2
It is clear that it is not a perfect cube as 10
2
or 0.2 is not in triplet.
1. Find the cube roots of the following
numbers by successive subtraction of
numbers : 1, 7, 19, 37, 61, 91, 127, 169,
217, 271, 331, 397, ........
(i) 64 (ii) 512
(iii) 1728
Solution—
(i) 64
64 – 1 = 63
63 – 7 = 56
56 – 19 = 37
37 – 37 = 0
64 = (4)3
Cube root of 64 = 4
(ii) 512
512 – 1 = 511
511 – 7 = 504
504 – 19 = 485
485 – 37 = 448
448 – 61 = 387
387 – 91 = 296
296 – 127 = 169
169 – 169 = 0
512 = (8)3
Cube root of 512 = 8
(iii) 1728
1728 – 1 = 1727
1727 – 7 = 1720
EXERCISE 4.3
1720 – 19 = 1701
1701 – 37 = 1664
1664 – 61 = 1603
1603 – 91 = 1512
1512 – 127 = 1385
1385 – 169 = 1216
1216 – 217 = 999
999 – 271 = 728
728 – 331 = 397
397 – 397 = 0
1728 = (12)3
Cube root of 1728 = 12
2. Using the method of successive
subtraction, examine whether or not the
following numbers are perfect cubes :
(i) 130 (ii) 345
(iii) 792 (iv) 1331
Solution—
(i) 130
130 – 1 = 129
129 – 7 = 122
122 – 19 = 103
103 – 37 = 66
66 – 61 = 5
We see that 5 is left
130 is not a perfect cube.
(ii) 345
345 – 1 = 344
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344 – 7 = 337
337 – 19 = 318
318 – 37 = 281
281 – 61 = 220
220 – 91 = 129
129 – 127 = 2
We see that 2 is left
345 is not a perfect cube.
(iii) 792
792 – 1 = 791
791 – 7 = 784
784 – 19 = 765
765 – 37 = 728
728 – 61 = 667
667 – 91 = 576
576 – 127 = 449
449 – 169 = 280
We see 280 is left as 280 < 217
792 is not a perfect cube.
(iv) 1331
1331 – 1 = 1330
1330 – 7 = 1323
1323 – 19 = 1304
1304 – 37 = 1267
1267 – 61 = 1206
1206 – 91 = 1115
1115 – 127 = 988
988 – 169 = 819
819 – 217 = 602
602 – 271 = 331
331 – 331 = 0
1331 is a perfect cube
3. Find the smallest number that must be
subtracted from those of the numbers in
question 2, which are not perfect cubes,
to make them perfect cubes. What are
the corresponding cube roots ?
Solution—
We have examined in Question 2, the
numbers 130, 345 and 792 are not perfect
cubes. Therefore
(i) 130
130 – 1 = 129
129 – 7 = 122
122 – 19 = 103
103 – 37 = 66
66 – 61 = 5
Here 5 is left _ 5 < 91
5 is to be subtracted to get a perfect cube.
Cube root of 130 – 5 = 125 is 5
(ii) 345
345 – 1 = 344
344 – 7 = 337
337 – 19 = 318
318 – 37 = 281
281 – 61 = 220
220 – 91 = 129
129 – 127 = 2
Here 2 is left _ 2 < 169
Cube root of 345 – 2 = 343 is 7
2 is to be subtracted to get a perfect cube.
(iii) 792
792 – 1 = 791
791 – 7 = 784
784 – 19 = 765
765 – 37 = 728
728 – 61 = 667
667 – 91 = 576
576 – 127 = 449
449 – 169 = 280
280 – 217 = 63
63 < 217
63 is to be subtracted
Cube root of 792 – 63 = 729 is 9
4. Find the cube root of each of the following
natural numbers :
(i) 343 (ii) 2744
(iii) 4913 (iv) 1728
(v) 35937 (vi) 17576
(vii) 134217728 (viii) 48228544
(ix) 74088000 (x) 157464
(xi) 1157625 (xii) 33698267
Solution—
(i) 3 343 = 3 777 = 3 37
= 7
7 3437 497 7
1
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(ii) 3 2744 2 27442 13722 6867 3437 497 7
1
= 3 777222 = 3 33 72
= 2 × 7 = 14
(iii) 3 4913 = 3 171717 = 3 317
17 491317 28917 17
1
= 17
(iv) 3 1728
2 17282 8642 4322 2162 1082 543 273 93 3
1
= 3 333222222
= 3 333 322
= 2 × 2 × 3 = 12
(v) 3 35937
3 359373 119793 3993
11 133111 12111 11
1
= 3 111111333
= 3 33 113 = 3 × 11 = 33
(vi) 3 17576
2 175762 87882 4394
13 219713 16913 13
1
= 3 131313222
= 3 33 132 = 2 × 13 = 26
(vii) 3 134217728
2 1342177282 671088642 335544322 167772162 83886082 41943042 20971522 10485762 5242882 2621442 1310722 655362 327682 163842 81922 40962 20482 10242 5122 2562 1282 642 322 162 82 42 2
1
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= 3
222
222222222222
222222222222
= 3 333333333 222222222 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512
(viii) 3 48228544
= 3
131313
777222222
= 3 3333 13722
= 2 × 2 × 7 × 13 = 364
(ix) 3 74088000
= 3
777555
333222222
= 3 33333 75322
= 2 × 2 × 3 × 5 × 7 = 420
(x) 3 157464
2 1574642 787322 393663 196833 65613 21873 7293 2433 813 273 93 3
1
= 3 333333333222
= 3 3333 3332
= 2 × 3 × 3 × 3 = 54
(xi) 3 1157625
3 11576253 3858753 1286255 428755 85755 17157 3437 497 7
1
= 3 777555333
= 3 333 753
= 3 × 5 × 7 = 105
2 740880002 370440002 185220002 92610002 46305002 23152505 11576255 2315255 463053 92613 30873 10297 3437 497 7
1
2 482285442 241142722 120571362 60285682 30142842 15071427 7535717 1076537 15379
13 219713 16913 13
1
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(xii) 3 33698267
17 3369826717 198225117 11660319 685919 36119 19
1
= 3 191919171717
= 3 33 1917
= 17 × 19 = 323
5. Find the smallest number which when
multiplied with 3600 will make the
product a perfect cube. Further, find the
cube root of the product.
Solution—
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
2 36002 18002 9002 4503 1253 755 255 5
1
Grouping the factors in triplets of equal
factors, we see that 2, 3 × 3 and 5 × 5 are
left
In order to complete the triplets, we have to
multiply it by 2, 3 and 5.
The smallest number to be multiplied = 2 × 2
× 3 × 5 = 60
Now product = 3600 × 60 = 216000
and cube root of 216000
= 555333222222
= 3333 5322
= 2 × 2 × 3 × 5 = 60
6. Multiply 210125 by the smallest number
so that the product is a perfect cube. Also,
find out the cube root of the product.
Solution—
210125 = 5 × 5 × 5 × 41 × 41
5 2101255 420255 8405
41 168141 41
1
Grouping the factors in triplets of equal
factors, we see that 41 × 41 is left
In order to complete the triplet, we have to
multiply it by 41
Smallest number to be multiplied = 41
Product = 210125 × 41 = 8615125
Cube root of 8615125
= 3 414141555
= 3 33 415 = 5 × 41 = 205
7. What is the smallest number by which
8192 must be divided so that quotient is
a perfect cube ? Also, find the cube root
of the quotient so obtained.
Solution—
8192 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ×
2 × 2 × 2 × 2
2 81922 40962 20482 10242 5122 2562 1282 642 322 162 82 42 2
1
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Grouping the factors in triplets of equal
factors, we see that 2 is left
Dividing by 2, we get the quotient a perfect
cube
Perfect cube = 8192 2 = 4096
and cube root = 3 4096
= 3 3333 2222 = 2 × 2 × 2 × 2 = 16
8. Three numbers are in the ratio 1 : 2 : 3.
The sum of their cubes is 98784. Find the
numbers.
Solution—
Ratio in numbers = 1 : 2 : 3
Let first number = x
Then second number = 2x
and third number = 3x
Sum of cubes of there numbers = (x)3 + (2x)3
+ (3x)3
x3 + 8x3 + 27x3 = 98784 36x3 = 98784
x3 = 36
98784 = 2744
x = 3 2744
= 3 777222
= 3 33 72 = 2 × 7 = 14
Number will be 14, 2 × 14, 3 × 14
i.e. 14, 28, 42
9. The volume of a cube is 9261000 m3. Find
the side of the cube.
Solution—
Volume of cube = 9261000 m3
Side = 3 Volume
= 3 9261000
= 3 10009261
3 92613 30873 10297 3437 497 7
1
10 100010 10010 10
1
= 3 101010777333
= 3 333 1073 = 3 × 7 × 10 m = 210 m
2 27442 13722 6867 3437 497 7
1
1. Find the cube roots of each of the
following integers :
(i) –125 (ii) –5832
(iii) –2744000 (iv) –753571
(v) –32768
Solution—
(i) 3 125 = – 3 125
= – 3 555 = – 3 35 = –5
(ii) 3 5832 = – 3 5832
EXERCISE 4.4
2 58322 29162 14583 7293 2433 813 273 93 3
1
= – 3 333333222
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= – 3 333 332
= – (2 × 3 × 3) = –18
(iii) 3 2744000 = – 3 2744000
= – 3 10002744
= – 3 101010777222
= – 3 333 1072
= – (2 × 7 × 10) = –140
(iv) 3 753571 = – 3 753571
= – 3 131313777
= – 3 33 137
= – (7 × 13) = –91
(v) 3 32768 = – 3 32768
= – 3
222222
222222222
= – 3 33333 22222 = – (2 × 2 × 2 × 2 × 2) = –32
2. Show that :
(i) 3 27 × 3 64 = 3 6427
(ii) 3 72964 = 3 64 × 3 729
(iii) 3 216125 = 3 125 × 3 216
(iv) 3 1000125 = 3 125 × 3 1000
Solution—
(i) 3 27 × 3 64 = 3 6427
L.H.S. = 3 27 × 3 64
= 3 333 × 3 444
= 3 33 × 3 34 = 3 × 4 = 12
R.H.S. = 3 6427 = 3 444333
= 3 33 43 = 3 × 4 = 12
L.H.S. = R.H.S.
(ii) 3 72964 = 3 64 × 3 729
L.H.S. = 3 72964
= 3 999444
= 3 33 94 = 4 × 9 = 36
R.H.S. = 3 64 × 3 729
= 3 444 × 3 999
= 3 34 × 3 39 = 4 × 9 = 36
L.H.S. = R.H.S.
(iii) 3 216125 = 3 125 × 3 216
L.H.S. = 3 216125
= – 3 216125 = – 3 666555
= – 3 33 65 = –5 × 6 = –30
R.H.S. = 3 125 × 3 216
= – 3 125 × 3 216
2 27442 13722 6867 3437 497 7
1
7 7535717 1076537 15379
13 219713 16913 13
1
2 327682 163842 81922 40962 20482 10242 5122 2562 1282 642 322 162 82 42 2
1
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= – 3 555 × 3 666
= – 3 35 × 3 36 = –5 × 6 = –30
L.H.S. = R.H.S.
(iv) 3 1000125 = 3 125 × 3 1000
L.H.S. = 3 1000125 = 3 1000125
= 3 101010555 = 3 33 105
= 5 × 10 = 50
R.H.S. = 3 125 × 3 1000
= 3 125 × 3 1000
= 3 555 × 3 101010
=
3 35 ×
3 310
= (–5) × (–10) = 50
L.H.S. = R.H.S.
3. Find the cube root of each of the following
numbers :
(i) 8 × 125 (ii) –1728 × 216
(iii) –27 × 2744 (iv) –729 × –15625
Solution—
(i) 3 1258 = 3 555222
= 3 33 52 = 2 × 5 = 10
(ii) 3 2161728 = – 3 2161728
= – 3 1728 × 3 216
= 3 333222222
× 3 333222
= – 3 333 322 × 3 33 32
= – (2 × 2 × 3) × (2 × 3)
= –12 × 6 = –72
(iii) 3 274427 = – 3 274427
= – 3 27 × 3 2744
2 27442 13722 6867 3437 497 7
1
= – 3 333 × 3 777222
= –
3 333 3 723
= – (3 × 2 × 7) = –42
(iv) 3 15625729
= 3 15625729
= 3 729 × 3 15625
3 7293 2433 813 273 93 3
1
5 156255 31255 6255 1255 255 5
1
= 3 333333 × 3 555555
= 3 33 33 × 3 33 55
= (3 × 3) × (5 × 5) = 9 × 25
= 225
4. Evaluate :
(i) 3 3364 (ii) 3 1717178
(iii) 3 5492700
(iv) 125 3 6a – 3 6125a
Solution—
(i) 3 33 64 = 4 × 6 = 24
(ii) 3 1717178 = 3 171717222
= 3 33 172 = 2 × 17 = 34
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(iii) 3 5492700
= 3 577275522
= 3 333 752 = 2 × 5 × 7 = 70
(iv) 125 3 6a – 3 6125a
= 125 3 222 aaa
= – 3 222555 aaa
= 125 3 32a – 3 3235 a
= 125a2 – 5a2 = 120a2
5. Find the cube root of each of the following
rational numbers.
(i)729
125(ii)
12167
10648
(iii)24389
19683(iv)
3456
686
(v)42875
39304
Solution—
(i) 3
729
125 = 3
3
729
125
3
3
729
125 = 3
3
333333
555
= 3 33
3 3
33
5
=
33
5
= 9
5
(ii) 3
12167
10648 = 3
3
12167
10648
2 106282 53242 2662
11 133111 12111 11
1
23 1216723 52923 23
1
= 3
3
232323
111111222
= 3 3
3 33
23
112 =
23
112 =
23
22
(iii) 3
24389
19683 = 3
3
24389
19683
= 3
3
24389
19683
3 196833 65613 21873 7293 2433 813 273 93 3
1
29 2438929 84129 29
1
= 3
3
292929
333333333
= 3 3
3 333
29
333 =
29
333 =
29
27
(iv) 3
3456
686
= 3
17282
3432
= 3
1728
343
= 3
3
1728
343
= 3
3
1728
343
7 3437 497 7
1
2 17282 8642 4322 2162 1082 543 273 93 3
1
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124
= 3
3
333222222
777
= 3 333
3 3
322
7
= 322
7
= 12
7
(v) 3
42875
39304
= 3
42875
39304
= 3
3
42875
39304
2 393042 196522 9826
17 491317 28917 17
1
5 428755 85755 17157 3437 497 7
1
= 3
3
777555
171717222
= 3 33
3 33
75
172
=
75
172
= 35
34
6. Find the cube root of each of the following
rational numbers :
(i) 0.001728 (ii) 0.003375
(iii) 0.001 (iv) 1.331
Solution—
(i) 3 001728.0 = 3
1000000
1728 = 3
3
1000000
1728
= 3
3
101010101010
333222222
= 3 33
3 333
1010
322
=
1010
322
= 100
12 = 0.12
(ii) 3 003375.0 = 3
1000000
3375 = 3
3
1000000
3375
3 33753 11253 3755 1255 255 5
1
= 3
3
101010101010
555333
= 3 33
3 33
1010
53
=
1010
53
= 100
15 = 0.15
(iii) 3 001.0 = 3
1000
1 = 3
3
1000
1
= 3
3
101010
111
= 3 310
1 =
10
1 = 0.1
(iv) 3 331.1 = 3
1000
1331 = 3
3
1000
1331
11 133111 12111 11
1
= 3
3
101010
111111
=
3 3
3 3
10
11 =
10
11 = 1.1
7. Evaluate each of the following :
(i) 3 27 + 3 0080. + 3 0640.
(ii) 3 1000 + 3 0080. – 3 1250.
(iii) 3
216
729 ×
9
6
(iv) 3
0080
0270
.
.
040
090
.
. – 1
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125
(v) 3 131313101010 ...
Solution—
(i) 3 27 + 3 008.0 + 3 064.0
= 3 333 + 3
1000
8 + 3
1000
64
= 3 33 + 3
101010
222
+ 3
101010
444
= 3 + 10
2 +
10
4 = 3 + 0.2 + 0.4
= 3.6
(ii) 3 1000 + 3 008.0 – 3 125.0
= 3 101010 + 3
1000
8 – 3
1000
125
= 3 310 + 3
101010
222
– 3
101010
555
= 3 310 + 33
3
10
2 – 3
3
3
10
5
= 10 + 10
2 –
10
5 = 10 + 0.2 – 0.5
= 10.2 – 0.5 = 9.7
(iii) 3
216
729 ×
9
6 = 3
3
216
729 ×
9
6
= 3
3
666
999
×
9
6
= 3 3
3 3
6
9 ×
9
6 =
6
9 ×
9
6 = 1
(iv) 3
008.0
027.0
04.0
09.0 – 1
= 3
3
008.0
027.0
04.0
09.0 – 1
=
3
3
1000
8
1000
27
100
4
100
9
– 1
=
3
3
3
3
1000
8
1000
27
100
4
100
9
– 1
=
3
3
3
3
101010
222
101010
333
1010
22
1010
33
– 1
=
3 3
3 3
3 3
3 3
10
2
10
3
2
2
2
2
10
2
10
3
– 1
=
10
210
3
10
210
3
– 1 = 10
3 ×
2
10
10
3 ×
2
10 – 1
= 2
3
2
3 – 1
= 2
3 ×
3
2 – 1 = 1 – 1 = 0
(v) 3 1313131.01.01.0
3 33131.0 = 0.1 × 13 = 1.3
8. Show that :
(i)3
3
1000
729 = 3
1000
729
(ii)3
3
343
512 = 3
343
512
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Solution—
(i) 3
3
1000
729 = 3
1000
729
L.H.S. = 3
3
1000
729 = 3
3
101010
999
= 3 3
3 3
10
9 =
10
9 = 0.9
R.H.S. = 3
1000
729 = 3
101010
999
= 3
10
9
10
9
10
9
= 3
3
10
9
=
10
9 = 0.9
L.H.S. = R.H.S.
(ii) 3
3
343
512 = 3
343
512
L.H.S. = 3
3
343
512 = 3
3
343
512
= 3
3
777
222222222
= 3 3
3 333
7
222 =
7
222 =
7
8
R.H.S. = 3
343
512
= 3
343
512 = 3
777
888
= 3
7
8
7
8
7
8
= 3
3
7
8
=
7
8
L.H.S. = R.H.S.
9. Fill in the blanks :
(i) 3 27125 = 3 × ____
(ii) 3 ___8 = 8
(iii) 3 1728 = 4 × ____
(iv) 3 480 = 3 3 × 2 × 3 ___
(v) 3 ___ = 3 7 × 3 8
(vi) 3 ___ = 3 4 × 3 5 × 3 6
(vii) 3
125
27 =
5
___
(viii) 3
1331
729 =
___
9
(ix) 3
___
512 =
13
8
Solution—
(i) 3 27125 = 3 333555
= 3 33 35 = 5 × 3 = 3 × 5
(ii) 3 ___8 = 8 = 3 38 = 3 888
3 888 = 8
(iii) 3 1728
2 17282 8642 4322 2162 1082 543 273 93 3
1
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= 3 333222222
= 3 333 322
= 2 × 2 × 3 = 4 × 3
3 1728 = 4 × 3
(iv) 3 480
2 4802 2402 1202 602 303 155 5
1
= 3 5322222
= 3 3 53222
= 3 32 × 3 5322
= 2 × 3 3 × 3 20
3 480 = 3 3 × 2 × 3 20
(v) 3 ___ = 3 7 × 3 8
3 7 × 3 8 = 3 87 = 3 56
3 56 = 3 7 × 3 8
(vi) 3 ___ = 3 4 × 3 5 × 3 6
= 3 4 × 3 5 × 3 6 = 3 654 = 3 120
3 120 = 3 4 × 3 5 × 3 6
(vii) 3
125
27 =
5
___
= 3
125
27 = 3
555
333
= 3
5
3
5
3
5
3
= 3
3
5
3
=
5
3
3
125
27 =
5
3
(viii) 3
1331
729 =
___
9
= 3
1331
729 = 3
111111
999
= 33
3
11
9 =
11
9
3
1331
729 =
11
9
(ix) 3
___
512 =
13
8
13
8 =
3
13
8
= 3
131313
888
= 3
2197
512
3
2197
512 =
13
8
10. The volume of a cubical box is 474.552
cubic metres. Find the length of each side
of the box.
Solution—
Volume of cubical box = 474.552 cu m
Each side = 3 Volume
= 3 552.474
= 3
1000
474552 = 3
3
1000
474552
2 4745522 2372762 1186383 593193 197733 6591
13 219713 16913 13
1
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128
= 3
3
101010
131313333222
= 3 3
3 333
10
1332 =
10
1332
= 10
78 = 7.8 m
11. Three numbers are to one another 2 : 3 :
4. The sum of their cubes is 0.334125. Find
the numbers.
Solution—
Ratio in three numbers = 2 : 3 : 4
Let first number = 2x
Second number = 3x
and third number = 4x
(2x)3 + (3x)3 + (4x)3 = 0.334125
8x2 + 27x3 + 64x3 = 1000000
334125
99x3 = 1000000
334125 x3 =
991000000
334125
x3 = 1000000
3375
3 33753 11253 3755 1255 255 5
1
= 100100100
555333
= 3
33
100
53
x = 33
33
100
53
= 100
53 =
100
15 = 0.15
First number = 2x = 2 × 0.15 = 0.30
Second number = 3x = 3 × 0.15 = 0.45
Third number = 4x = 4 × 0.15 = 0.60
Numbers are 0.3, 0.45, 0.6
12. Find side of a cube whose volume is
216
24389 m3.
Solution—
Volume of cube = 216
24389 m3
Side = 3 Volume
= 3
216
24389 = 3
3
216
24389
29 2438929 84129 29
1
6 2166 366 6
1
= 3
3
666
292929
=
3 3
3 3
6
29
= 6
29 m
13. Evaluate :
(i) 3 36 × 3 384 (ii) 3 96 × 3 144
(iii) 3 100 × 3 270 (iv) 3 121 × 3 297
Solution—
(i) 3 36 × 3 384 = 3 38436
2 362 183 93 3
1
2 3842 1922 962 482 242 122 63 3
1
= 3 322222223322
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= 3 222222222333
= 3 3333 2223 = 3 × 2 × 2 × 2 = 24
(ii) 3 96 × 3 144 = 3 14496
2 962 482 242 122 63 3
1
2 1442 722 362 183 93 3
1
= 3 332222322222
= 3 333222222222
= 3 3333 3222
= 2 × 2 × 2 × 3 = 24
(iii) 3 100 × 3 270 = 3 270100
= 3 27000 = 3 271000
= 3 333101010 = 3 33 310
= 10 × 3 = 30
(iv) 3 121 × 3 297
= 3 297121
11 2973 273 93 3
1
= 3 333111111
= 3 33311 = 11 × 3 = 33
14. Find the cube root of the numbers :
2460375, 20346417, 210644875, 57066625
using the fact that
(i) 2460375 = 3375 × 729
(ii) 20346417 = 9261 × 2197
(iii) 210644875 = 42875 × 4913
(iv) 57066625 = 166375 × 343
Solution—
(i) 3 2460375 = 3 7293375
= 3 3375 × 3 729
3 33753 11253 3755 1255 255 5
1
3 7293 2433 813 273 93 3
1
= 3 555333 × 3 333333
= 3 33 53 × 3 33 33
= 3 × 5 × 3 × 3 = 135
(ii) 3 20346417 = 3 21979261
= 3 9261 × 3 2197
3 92613 30873 10297 3437 497 7
1
13 219713 16913 13
1
= 3 777333 × 3 131313
= 3 33 73 × 3 313
= (3 × 7) × 13 = 21 × 13 = 273
(iii) 3 210644875 = 3 491342875
= 3 42875 × 3 4913
5 428755 85755 17157 3437 497 7
1
17 491317 28917 17
1
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130
= 3 777555 × 3 171717
= 3 33 75 × 3 317
= (5 × 7) × 17 = 35 × 17 = 595
(iv) 3 57066625 = 3 343166375
= 3 166375 × 3 343
5 1663755 332755 6655
11 133111 12111 11
1
7 3437 497 7
1
= 3 111111555 × 3 777
= 3 33 115 × 3 37
= (5 × 11) × 7 = 55 × 7 = 385
15. Find the units digit of the cube root of
the following numbers ?
(i) 226981 (ii) 13824
(iii) 571787 (iv) 175616
Solution—
(i) 226981
In it unit digit is 1
The units digit of its cube root will be = 1
(_ 1 × 1 × 1 = 1)
Tens digit of the cube root will be = 6
(ii) 13824
_ The units digit of 13824 = 4
(_ 4 × 4 × 4 = 64)
Units digit of the cube root of it = 4
(iii) 571787
_ The units digit of 571787 is 7
The units digit of its cube root = 3
(_ 3 × 3 × 3 = 27)
(iv) 175616
_ The units digit of 175616 is 6
The units digit of its cube root = 6
(_ 6 × 6 × 6 = 216)
16. Find the tens digit of the cube root of each
of the numbers in Question No. 15.
Solution—
(i) In 226981
_ Units digit is 1
Units digit of its cube root = 1
We have 226
(Leaving three digits number 981)
63 = 216 and 73 = 343
63 226 73
The ten’s digit of cube root will be 6
(ii) In 13824
Leaving three digits number 824, we have 13
_ (2)3 = 8, (3)3 = 27
23 13 33
Tens digit of cube root will be 2
(iii) In 571787
Leaving three digits number 787, we have 571
83 = 512, 93 = 729
83 571 93
Tens digit of the cube root will be = 8
(iv) In 175616
Leaving three digit number 616, we have 175
_ 53 = 125, 63 = 216
53 175 63
Tens digit of the cube root will be = 5
Making use of the cube root table, find
the cube roots of the following (correct
to three decimal places) (1 - 22)
1. 7 2. 70 3. 700
EXERCISE 4.5
4. 7000 5. 1100 6. 780
7. 7800 8. 1346 9. 250
10. 5112 11. 9800 12. 732
13. 7342 14. 133100 15. 37800
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16. 0.27 17. 8.6 18. 0.86
19. 8.65 20. 7532 21. 833
22. 34.2
Solution—
1. 3 7 = 1.913 (From the table)
2. 3 70 = 4.121 (From the table)
3. 3 700 = 3 1007 = 8.879 (From 3 10x )
4. 3 7000 = 3 10070 = 19.13
(From 3 100x )
5. 3 1100 = 3 10011 = 10.32
(From 3 100x )
6. 3 780 = 3 1078 = 9.205
(From 3 10x )
7. 3 7800 = 3 10078 = 19.83
(From 3 100x )
8. 3 1346 = 3 6732 = 3 203.67
= 3 3.67 × 3 20
Now from the tables 3 67 = 4.062 and 3 68
= 4.082
Difference after 1, 4.082 – 4.062 = 0.020
and difference after .3 = 0.020 × .3 = 0.060
= 0.06
3 3.67 = 4.062 + 0.006
= 4.068
and 3 20 = 2.714 (from the table)
3 1346 = 3 3.67 × 3 20
= 4.067 × 2.714
= 11.040
9. 3 250 = 3 2125
= 3 2555 = 3 3 25
5 3 2 = 5 × 1.260 (from the table)
= 6.300 = 6.3
10. 3 5112 = 3 639222 = 2 3 639
= 2 3 109.63
= 2 × 3 9.63 × 3 10
Now, from the table
3 63 = 3.979 and 3 64 = 4.000
Difference after 64 – 63 = 1 = 4.000 – 3.979
= 0.021
Difference after 9 = 0.021 × 0.9 = 0.0189
3 9.63 = 3.979 + 0.0189 = 3.9979
and 3 5112 = 2 × 10 × 9.63
= 2 × 2.154 × 3.9979
= 17.2229 = 17.223
11. 3 9800 = 3 10098 = 21.40
(from x100 )
12. 3 732 = 3 102.73
= 3 2.73 × 3 10
Now 3 73 = 4.179 and 3 74 = 4.198
Difference after 74 – 73 = 1
= 4.198 – 4.179 = 0.019
and difference off 0.2 = 0.2 × 0.019 = 0.0038
3 2.73 = 4.179 + 0.0038 = 4.1828
and 3 10 = 2.154
3 732 = 3 2.73 × 3 10
= 4.1828 × 2.154
= 9.0097 = 9.01
13. 3 7342 = 3 10042.73
= 3 42.73 × 3 100
Now 3 73 = 4.179 and 3 74 = 4.198
Difference after 74 – 73 = 1 = 4.198 – 4.179
= 0.019
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132
Difference after 0.42 = 0.42 × 0.019
= 0.00798
3 42.73 = 4.179 + 0.00798 = 4.18698
and 3 100 = 4.642
and 3 7342 = 3 42.73 × 3 100
= 4.18698 × 4.642 = 19.436
14. 3 133100 = 3 1001331
= 3 100111111 = 3 3 10011
= 11 3 100 = 11 × 4.642
= 51.062
15. 3 37800 = 3 10008.37 = 3 3 8.3710
= 3 310 × 3 8.37
= 10 × 3 8.37
3 37 = 3.332 and 3 38 = 3.362
Difference after 38 – 37 = 1 = 3.362 – 3.332
= 0.030
and difference after 0.8 = 0.030 × 0.8
= 0.0240 = 0.024
3 8.37 = 3.332 + 0.024 = 3.356
and 3 37800 = 3 8.37 × 10
= 3.356 × 10 = 33.56
16. 3 27.0 = 3
100
27 = 3
1000
270 = 3 10
1000
27
= 3 1010
3
10
3
10
3
= 10
3 3 10
= 10
3 × 2.154 = 3 × 0.2154
= 0.6462 = 0.646
17. 3 6.8
3 8 = 2.000 and 3 9 = 2.080
Difference after 1 (9 – 8) = 2.080 – 2.000
= 0.080
Difference after 0.6 = 0.6 × 0.08 = 0.048
3 6.8 = 2.000 + 0.048 = 2.048
18. 3 86.0 = 3
100
86 = 3
3
100
86
= 642.4
414.4(from the table)
= 4642
4414 = 0.95088
= 0.9509 = 0.951
19. 3 65.8
8 < 8.65 < 9
3 8 < 3 65.8 < 3 9
3 8 = 2.000 , 9 = 2.080
Difference of (9 – 8) = 1 is = 0.080
difference of 0.65 = 0.080 × 0.65
= 0.08 × 0.65 = 0.052
3 65.8 = 2.000 + 0.052 = 2.052
20. 3 7532 = 3 10032.75
= 3 32.75 × 3 100
75 < 75.32 < 76
3 75 < 3 32.75 < 3 76
3 75 = 4.217 and 3 76 = 4.236
Difference on (76 – 75) = 1 = 4.236 – 4.217
= 0.019
Difference on 0.32 = 0.019 × 0.32
= 0.00608 = 0.0061
3 32.75 = 4.217 + 0.0061 = 4.2231 = 4.223
Now 3 7532 = 3 32.75 × 3 100 = 4.223 ×
4.642 = 19.603
21. 3 833 = 3 103.83
= 3 3.83 × 3 10
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83.3 lies between 83 and 84
3 3.83 lies between 3 83 and 3 84
Now 3 83 = 4.362 and 3 84 = 4.380
Difference on 84 – 83 = 1 is 4.380 – 4.362 =
0.018
Difference on 0.3 = 0.018 × 0.3 = 0.0054
3 3.83 = 4.362 + 0.0054 = 4.3674
and 3 10 = 2.154
3 833 = 3 3.83 × 3 10
= 4.3674 × 2.154 = 9.407
22. 3 2.34
34.2 lies between 34 and 35
or 3 2.34 lies between 3 34 and 3 35
Now 3 34 = 3.240 and 3 35 = 3.271
Difference on 35 – 34 = 1 is = 3.271 – 3.240
= 0.031
Difference on 0.2 = 0.031 × 0.2 = 0.0062
3 2.34 = 3.240 + 0.0062 = 3.2462
= 3.246
23. What is the length of the side of a cube
whose volume is 275 cm3. Make use of
the table for the cube root.
Solution—
Volume of cube = 275 cm3
Length of side = 3 Volume = 3 275
= 3 105.27
But 27.5 lies between 27 and 28
or 3 5.27 lies between 3 27 and 3 28
Now 3 27 = 3.000 and 3 28 = 3.037
Difference on 28 – 27 = 1 is 0.037
Difference on 0.5 = 0.037 × 0.5 = 0.0185
3 5.27 = 3.000 + 0.0185 = 3.0185
and 3 10 = 2.154
3 275 = 3 5.27 × 3 10
= 3.0185 × 2.154
= 6.5018 = 6.502 cm
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