Download Class 8: Numbers – Exercise 3B

Class 8: Numbers – Exercise 3B
1. Compare the following pairs of rational numbers:
i.
15
17
and
32
24
Answer:
First take the LCM of 24 and 32. LCM = 96
Therefore:
15
=
32
Hence we see that
ii.
45
96
45
96
and
<
17
=
24
68
68
96
or we can say that
96
15
32
<
17
24
10
17
and
11
18
Answer:
First take the LCM of 11 and 18. LCM = 198
Therefore:
10
11
=
Hence we see that
iii.
−5
12
and
180
17
and
18
198
180
170
> 180
198
=
170
180
or we can say that
10
>
11
17
18
−3
4
Answer:
First take the LCM of 12 and 4. LCM = 12
Therefore:
−5
12
=
Hence we see that
iv.
−7
24
and
−5
−3
and
12
4
−5
12
>
=
−9
12
−9
−5
or we can say that
12
12
>
−3
4
9
−20
Answer:
First take the LCM of 24 and 20. LCM = 120
1
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Therefore:
−7
24
−35
−9
and
120
20
=
Hence we see that
−35
120
>
−54
120
=
−54
−7
or we can say that
120
24
>
9
−20
2. Arrange in ascending order:
i.
5 7 11 13
, , ,
6 9 12 18
Answer:
LCM of 6, 9, 12, 18 = 36
The fractions can be written as
30 28 33 26
,
,
,
36 36 36 36
Therefore the order would be
26
36
ii.
<
28
36
<
30
36
<
33
36
13
18
Or
<
7
9
<
5
6
<
11
12
5 −9 −5 7
, , ,
−7 14 6 −12
𝑎
−𝑎
Note:
=
−𝑏
𝑏
Answer:
LCM of 7, 14, 6, 12 = 84
The fractions can be written as
−60 −54 −70 −49
84
,
84
,
84
,
84
Therefore the order would be
−70
84
iii.
<
−60
84
<
−54
84
<
−49
84
Or
−5
6
<
5
−7
<
−9
14
<
7
−12
1 −13 8
,
6 −3
𝑎
−𝑎
2, 3 ,
Note:
−𝑏
=
𝑏
2
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Answer:
LCM of 3, 6 = 6
The fractions can be written as
−12 2 −13 −16
, ,
, 6
6 6 6
Therefore the order would be
−16
6
iv.
13
<
,
−28
−13
6
<
−12
6
<
2
6
Or
−8
3
<
−13
6
< −2 <
1
3
−23 −4 −9
42
𝑎
Note:
−𝑏
,
7
=
, 14
−𝑎
𝑏
Answer:
LCM of 3, 6 = 6
The fractions can be written as
−12 2 −13 −16
,6,
6
6
,
6
Therefore the order would be
−16
6
<
−13
6
<
−12
6
<
2
6
Or
−8
3
<
−13
6
< −2 <
1
3
3. Represent each of these numbers on a Number Line:
i.
5
6
5
Divide the unit length between 0 and 1 in 6 equal parts and then mark .
6
ii.
14
3
Divide the unit length between 4 and 5 in 3 equal parts and then mark
iii.
14
3
−3
7
Divide the unit length between 0 and -1 in 7 equal parts and then mark
−3
7
3
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−17
iv.
5
Divide the unit length between -3 and -4 in 5 equal parts and then mark
−17
5
2
−2 7
v.
Divide the unit length between -2 and 3 in 7 equal parts and then mark −2
2
7
4. Find the additive inverse of:
i.
9
13
iv.
Answer:
9
13
+𝑎 =0
is
Answer:
14
3
v.
+𝑎 =0
Therefore the additive inverse
iii.
−11
+𝑎 =0
8
11
−22
15
Answer:
−14
−22
15
+𝑎 =0
Therefore the additive inverse
3
−3
is
23
Answer:
−3
23
+𝑎 =0
vi.
Therefore the additive inverse
is
8
Therefore the additive inverse
−9
is
13
14
3
is
−11
Answer:
Therefore the additive inverse
ii.
8
−3
23
22
15
−11
−9
Answer:
Therefore the additive inverse is
−11
−9
+𝑎 =0
−11
9
5. Find the sum:
i.
−7
17
+
6
17
=
−1
17
4
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ii.
iii.
iv.
v.
vi.
−5
7
+
12
=
−12
8
15
5
12
+
−11
18
−11
6
4
7
−12
32+25
60
=
5
+
=
−12
+
−3
4
2
−3
+
= −1
12
19
20
=
−22−15
36
5
+8+
5
+ 21 +
−7
3
−8
9
−37
=
36
(Note: LCM of 18 and 12 is 36)
−103
(Note: LCM of 6, 4, 8, and 3 is 24)
24
=
=
(Note: LCM of 15 and 12 is 60)
−47
(Note: LCM of 7, 3, 21, and 9 is 63)
63
6. Subtract:
i.
2
5
from
3
6
5
Answer:
ii.
−2
5
2
−
6
from
=
3
5
−
6
iv.
−11
6
−7
8
−
from
Answer:
8
3
6
7
−2
−5
−
(
)
5
7
4
−7
from
9
8
Answer:
1
=
6
−5
Answer:
iii.
4
4
9
=
=
−25
35
−53
72
−
14
35
+
32
−11
35
=
−95
=
72
72
8
3
−(
−11
)
6
=
16
6
11
6
+
7. The sum of two rational numbers is
−4
9
=
9
2
. If one of them is
13
6
then find the other.
Answer:
5
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13
−4
+𝑎 =
6
9
−4
13 −47
𝑎=
−
=
9
6
18
8. What number should be added to
−2
3
to get
−1
7
Answer:
−2
−1
+𝑎=
3
7
−1 2
11
𝑎=
+ =
7
3
21
9. What number should be subtracted from −2 to get
7
11
Answer:
7
11
7
−29
𝑎 = −2 −
=
11
11
−2 − 𝑎 =
10. Find the products:
i.
ii.
4
9
7
12
×
7
= 27
7
18
−9 ×
iii.
=
iv.
−7
27
11.Find the quotient:
i.
ii.
17
8
÷
−16
35
51
÷
4
=
15
14
iii.
17
=
8
×
−16
35
4
51
×
=
14
15
4
24
=
=
1
6
iv.
4
×
9
7
= 27
12
7
4
7
7
×
9
−12
7
= 27
12
÷ (−16) =
−12
7
×
−5
18
18
−5
−9 ÷ ( ) = −9 ×
−1
16
=
=
3
28
162
5
−32
75
12.The product of two rational numbers is -7. If one of the number is
−8
11
, then find the other.
Answer:
6
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𝑎 × 𝑏 = −7
−8
× 𝑏 = −7
11
77
𝑏=
8
13. By what number must
1
−8
be divided to get
?
26
39
Answer:
1
−8
÷𝑎=
26
39
1 1
−8
× =
26 𝑎
39
−1 39
−3
𝑎=
×
=
26 8
16
14.Find a rational number between each of the following pairs of rational numbers.
i.
7
10
𝑎𝑛𝑑
10
17
Answer: First take the LCM of 10 and 17 which is 170.
Convert the numbers with 170 as the denominator.
Hence we get
109
170
𝑎𝑛𝑑
100
170
Therefore the rational numbers between
ii.
3
1 8 𝑎𝑛𝑑 2 Or
11
8
𝑎𝑛𝑑
7
10
𝑎𝑛𝑑
10
17
−3
5
𝑎𝑛𝑑
108
170
𝑎𝑛𝑑
107
170
16
8
3
12
8
8
Therefore the rational numbers between 1 𝑎𝑛𝑑 2 are
iii.
are
𝑎𝑛𝑑
13
8
−4
7
Answer: First take the LCM of 5 and 7 which is 35.
Convert the numbers with 35 as the denominator.
Hence we get
−21
35
𝑎𝑛𝑑
−20
35
Therefore the rational numbers between
−3
5
𝑎𝑛𝑑
−4
7
are
−205
350
𝑜𝑟
−41
70
7
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−17
−2 𝑎𝑛𝑑
iv.
21
Or
−42
21
𝑎𝑛𝑑
−17
21
−2 𝑎𝑛𝑑
Therefore the rational numbers between
−17
21
are
−18
21
𝑜𝑟
−6
7
15. Find three rational numbers between:
4 𝑎𝑛𝑑 4
i.
2
3
Or
12
3
𝑎𝑛𝑑
14
3
Therefore the rational numbers between
125
30
ii.
−1
2
𝑎𝑛𝑑
𝑎𝑛𝑑
126
30
127
30
Or
𝑎𝑛𝑑
−1
4
𝑎𝑛𝑑
−1
−2
Or
4
4
25
6
𝑎𝑛𝑑
Therefore the rational numbers between
−11
40
𝑎𝑛𝑑
−12
40
𝑎𝑛𝑑
2
4 𝑎𝑛𝑑 4 3 are
−1
2
43
10
𝑎𝑛𝑑
𝑎𝑛𝑑
127
30
−1
are
4
−13
40
Or
−11
−3
−13
𝑎𝑛𝑑
𝑎𝑛𝑑
40
10
40
16. Find 5 rational numbers between:
i.
3
5
𝑎𝑛𝑑
2
3
Or
9
15
𝑎𝑛𝑑
10
15
The rational numbers are
91
150
ii.
92
93
94
95
, 150 , 150 , 150 , 150 Or
−2 𝑎𝑛𝑑 − 1
1
2
Or
−4
2
𝑎𝑛𝑑
91
150
,
46
75
,
31
50
,
47
75
,
19
30
−3
2
The rational numbers are
8
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−31
20
,
−32
20
,
−33
20
,
−34
20
,
−35
20
Or
−31
20
,
−8
5
,
−33
20
,
−17
10
,
−7
4
17. Determine whether the numbers are rational or irrational:
2
viii. 2√3 − 3√2 : Irrational
i.
: Rational
−17
ii.
0.6 : Rational
121
ix.
11
iii.
√169 = 13 : Rational
iv.
3√12 - 6√3 = 6√3 - 6√3 = 0 :
Rational
0.142857 : Irrational
v.
vi.
vii.
4√18
3√12
=
4×3√2
3√2
=4
x.
xi.
xii.
xiii.
xiv.
:
xv.
Rational
9√2 − √32 = 9√2 − 4√2 =
5√2 : Irrational
12
√75 =
2
5
: Rational
𝜋
: Irrational
−√144 = −12 Rational
1.411443143 : Rational
√0.9 : Irrational
3
√0.09 = 10 : Rational
1
: Irrational
√2
18. Skipped 18.
19. State whether True or False:
i.
Every real number is either rational or irrational: True
ii. Every real number can be represented on a number line: True
iii. There exists and integer which is not a rational number: False
iv. There exist a point on a number line which do not represent any real number: False
v. An infinite number of rational numbers can be inserted between any two rational
numbers: True
vi.
1
The multiplicative inverse of any rational number a is : False
𝑎
20.Fill in the blanks
i. 0 is a rational number that is its own additive inverse.
ii. 0 is a rational number that does not have a multiplicative inverse.
iii. 1 and -1 are two rational numbers which are equal it their own reciprocal.
iv. The product of a rational number with its reciprocal is 1.
v. The reciprocal of a negative number is negative.
9
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1
, 𝑎 ≠ 0 is a.
vi.
The multiplicative inverse of a rational number is
vii.
Number of irrational number between any two rational number is infinite.
𝑎
21. Arrange in ascending order
i. 8√3 , 2√15, 11, 2√6, 3√7
Answer: First take everything within under root sign. That way we can compare
the numbers easily.
The numbers would then be √192 , √60, √121, √24, √63
Now arrange in ascending order √24, √60, √63 , √121, √192
Or 2√6, 2√15, 3√7 , 11 , 8√3
ii.
4√5 , √122, 7√3, 13, 8√2
Answer: First take everything within under root sign. That way we can compare
the numbers easily.
The numbers would then be √80 , √122, √147, √169, √128
Now arrange in ascending order √80 , √122, √128, √147, √169
Or 4√5 , √122, 8√2, 7√3, 13
22.Write the rationalizing factors of the following:
i. √3
Answer: √3 × √3 = 3 Therefore rationalizing factor is √3
ii.
6√3
Answer: 6√3 × √3 = 18 Therefore rationalizing factor is √3
iii.
3 + √2
Answer: (3 + √2)(3 − √2) = 7 Therefore rationalizing factor is (3 − √2)
iv.
√7 + √3
Answer: (√7 + √3)(√7 − √3) = 4 Therefore rationalizing factor is (√7 + √3)
v.
5 − √11
10
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Answer: (5 − √11)(5 + √11) = 14 Therefore rationalizing factor is (5 + √11)
vi.
3√5 − 3
Answer: (3√5 − 3)(3√5 + 3) = 14 Therefore rationalizing factor is (3√5 + 3)
23.Rationalize the denominator of each of the following:
i.
6
√
×
3
=
6√ 3
= 2 √3
3
×
2
√2
√2
=
iii.
4
√5−√3
×
√5+√3
√5+√3
=
4 (√5−√3)
2
iv.
11
6−2√2
× 6+2
6+2√2
√2
=
11(6−2√2)
28
ii.
v.
5
√3
√3
3√
5√ 2
6
7
(3√3+2√2)
×
3√3−2√2
(3√3+2√2)
vi.
7+√10
7− √10
7+ √10
√10
vii.
2√5−4
2√5+4
×
viii.
√8−√6
√8+√6
√8−√6
×
√8−√6
× 7+
2√5−4
2√5−4
=
=
=
=
= 2(√5 − √3)
7(3√3+2√2)
19
(7+ √10)2
39
(2√5−4)2
4
(√8−√6)2
2
24.Insert 5 rational numbers between:
i. √5 𝑎𝑛𝑑 √13
Answer: √6, √7 , √8 , √9 , √10
ii.
√7 𝑎𝑛𝑑 3√3
11
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Answer: First take everything within the root sign. So we need to find rational
numbers between √7 𝑎𝑛𝑑 √27
Hence the numbers are √8, √9 , √10 , √11 , √12
iii.
2 𝑎𝑛𝑑 2.5
Answer: we can make the numbers as square root. So we need to find rational
numbers between √4 𝑎𝑛𝑑 √6.25
Hence the numbers are √4.1, √4.2 , √4.3 , √4.4 , √4.5
25. State True or False:
i. √3 + √7 = √10 : False
ii.
5√2 + 3√2 = 8√2 : True
iii.
8√3 − 3√3 = 5√3 : True
iv.
(9 + √3) + (3 − √3) is a rational number: True (the value is 12 which is a
rational number)
v.
(7√3)(2√48) is irrational number: False (the value is 168 which is a rational
number)
vi.
(3√2)(5√8 − 7√18 + 3√32) is a rational number: True (the value is 6 which is a
rational number)
vii.
√3+√2
√3−√2
+
√3−√2
Is a rational number. True (the value is 10 which is a
√3+√2
rational number)
12
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