1
RATIONAL NUMBERS
CHAPTER
The equivalent rational numbers are 18 16 21 32 , , and 48 48 48 48
CONTENTS Representation of rational numbers on a number line Addition, subtraction, multiplication and division of rational numbers Properties of rational numbers
Therefore, the smallest rational number is then comes
21 , 48
16 , and the greatest rational 48
18 . Hence, their ascending order is 48 3 7 4 3 , , , . 8 16 12 8
number is
Rational Numbers Definition :
A number
a is a rational number if ‘a’ and ‘b’ b
are integers and ‘b’ is not equal to zero. ‘b’ cannot be equal to zero because division by zero is not allowed. Further, a rational number is said to be in the standard form or simplest form when the numerator and denominator have no common factor other than 1.
Sol.
When denominators are equal : Ex.2 Sol. Ex.3
Order of Rational Numbers Ex.1
Addition of Rational Numbers 5 7 and . 6 6 57 12 5 7 + = = 6 6 6 6 7 13 Add and . 5 5 13 7 7 13 6 + 5 = = 5 5 5
Add
Arrange the following fractions in ascending order.
Sol.
3 4 7 2 , , , . 8 12 16 3 LCM of denominators 8, 12, 16, and 3 = 2 × 2 3 3 6 Then = = 8 8 6 7 7 3 21 = = ; 16 16 3 48 16 4 4 4 = = ; 48 12 12 4 2 2 16 32 – = = 3 3 16 48 2 8, 12,
When one denominator is a multiple of the other denominator : × 2 × 2 × 3 = 48. 18 ; 48
2 4,
6,
16, 3 8, 3
2 2, 2 1, 3 1,
3, 3, 3,
4, 2, 1,
3 3 3
1,
1,
1,
1
Ex.4 Sol.
Ex.5 Sol.
4 5 and . 3 6 4 4 2 8 We know that = = 3 3 2 6 8 4 ( is equivalent rational number of ) 6 3 4 5 8 5 13 So, + = + = 3 6 6 6 6 5 3 Solve + 21 . 7
Solve
We know that So,
3 3 3 9 = = 7 73 21
3 5 9 5 95 14 + 21 = – = = 7 21 21 21 21
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When denominator are co-prime : 4 6 and . 5 7
Ex.6
Find the sum of
Sol.
6 4 47 65 + = – 5 5 7 75 7
28 30 – 35 35
=
28 30 2 = 35 35
4 16 15 31 3 + = = . Both results are equal. 5 20 20 4
a c e , , and three rational numbers, then b d f a c c e a e + = + b b d d f f 2 1 2 Consider the fractions , , and . 5 4 3
If
When denominator have a common factor : Ex.7 Sol.
Solve
7 5 + . 8 12
Since 12 and 8 have common factors, we will proceed by finding the LCM of 12 and 8. LCM of 12 and 8 is 2 × 2 × 2 × 3 = 24 Now we will find equivalent fractions of the given numbers having 24 in the denominator. Hence, and So,
4 15 16 31 3 + = = and 5 20 20 4
Associative property
(Multiplying and dividing each fraction by the denominator of the other fraction) =
For example,
5 5 2 10 = = 12 12 2 24 7 73 21 = = 8 8 3 24 7 5 10 21 10 21 31 + = + = = 8 12 24 24 24 24
Properties of Addition of Rational Numbers Closure property :
2 1 2 5 4 3
2 1 2 5 4 3
85 2 = 20 3
=
13 2 + 20 3 39 40 = 60 79 = 60
=
2 38 + 5 12
2 11 + 5 12 24 55 = 60 79 = 60
=
a is a rational number,, b then there exists a rational number zero such that a a + 0 = . Zero is called the identity element of b b addition. Addition of zero does not change the value of the rational number.
Additive identity If
Additive identity
When two rational numbers are added, the result is a c always a rational number, i.e., if and is always b d a rational number. 2 3 12 15 27 For example, + = = , which is also 5 6 30 30 a rational number.
If
a is a rational number, then there exists a rational b
a , called the additive inverse, such that number b a a = 0 + b b The additive inverse is also referred to as ‘negative’ of the given number..
Commutative property : When to rational numbers are added, the order of a c addition does not matter, i.e., if and are two b d rational numbers, then a c c a + = + b d d b
EXAMPLES 3 3 = 0. + 4 4
Ex.8
3 3 is the additive inverse of . 4 4
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Ex.9
5 5 + = 0. 6 6 5 5 . is the additive inverse of 6 6
For example,
Commutative property : If
Subtraction of Rational Numbers When we have to subtract a rational number, say
5 9
8 5 from , we add the additive inverse of , i.e., 9 9 5 8 8 5 8 5 85 3 1 = to . Thus, – = + = = 9 9 9 3 9 9 9 9 9 3 3 Ex.10 Subtract from . 7 7 4 3 4 3 = + Here, – 11 7 11 7
=
=
Multiplication of Rational Numbers Multiplication is the process of successive addition. Like 6 × 8 = 8 + 8 + 8 + 8 + 8 + 8 = 48. 1 1 1 1 1 1 1 6 Similarly, 6 × = + + + + + = = 2 3 3 3 3 3 3 3 3 1 6 1 6 1 6 2 Alternatively, 6 × = × = = = 2 3 1 3 1 3 3 1 So, when we multiply two rational numbers, we multiply the numerator with the numerator and the denominator with the denominator.
5 7 (5)(7) = × = 35 1 1 1 1
and
2 3 2 3 6 × = = 11 5 11 5 55
a c c a ac ca × = × , i.e., = b d d b bd db
Ex.11
4 3 × 5 7 4 (3) = 5 7 12 = 35 4 3 × 5 7
3 4 × 7 5 (3) 4 = 75 12 = 35 3 4 × = 7 5
a c e , , and are three rational numbers, then b d f a c e a c e × = × b d f b d f ac e a ce ace ace i.e., × = × or = bd f b df bdf bdf Thus, rational numbers can be multiplied in any order.
If
28 33 61 + = 77 77 77
– 5 × (–7) =
a c and are two rational numbers, then b d
Associative property :
4 7 3 11 + 11 7 7 11
Thus,
3 5 15 × =– which is rational number.. 7 8 56
Ex.12
3 4 5 3 4 5 × = × 7 5 8 7 5 8 (3) 4 5 3 4 (5) = × × 75 5 8 8 7 12 5 3 20 = × × 35 8 7 40 0 0 = 280 280
3 3 = 14 14
Properties of Multiplication of Rational Number Closure property : The rational number are closed under multiplication. It means that the product of two rational numbers is a c always a rational number, i.e., if and are two rational b d a c ac numbers, × = is always a rational number.. b d bd
Multiplicative identity : a , is multiplied by b a the rational number 1, the product is always . b a a 1 a ×1= = b b b a 1 a a or 1× = = b b b
When any rational number, say
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Ex.13
21 21 21 1 21 1 ×1= × = = 35 35 35 1 35 1
Ex.14
3 3 1 (3) 1 3 ×1= × = = 7 7 1 7 1 7
Ex.19
‘One’ is called the multiplicative identity or identity element of multiplication for rational numbers.
Multiplicative inverse, or reciprocal : For every non-zero rational number a rational number
a , there exists b
b a b such that × = 1. This is so, a b a
because a b ab ab × = = =1 b a ba ba Ex.15
Ex.16
2 23 6 3 3 × = = = 1. So is the 3 3 2 6 2 2 2 2 multiplicative inverse of and is the 3 3 3 multiplicative inverse of . 2 4 7 (4)(7) 28 7 × = = = 1. So – is 74 28 7 4 4 4 the multiplicative inverse of – and vice versa. 7
Multiplication of a Rational Number by Zero When any rational number
c e a a c a e × = × + × . b b d b d f f
a is multiplied by 0, the b
product is always zero. a a0 0 ×0= = =0 b b b Ex.20
7 70 0 ×0= = =0 8 8 8
Ex.21
3 3 0 0 ×0= = =0 4 4 4
Division of Rational Numbers Division is the inverse process of multiplication. a c If and are two rational numbers, then b d a c a d ÷ = × . b d b c
Distributive property : a c e If , and are three rational numbers, then b d f
3 3 8 3 3 3 8 = × + × 5 4 9 5 4 5 9 3 27 32 9 24 = + 5 36 20 45 3 5 81 96 × = 5 36 180 15 15 = 180 180
Ex.22 Ex.23
2 5 2 9 18 ÷ = × = 7 9 7 5 35 3 4 3 9 27 = ÷ = × 8 9 8 4 32
Properties of Division of Rational Numbers Ex.17
Ex.18
3(4 + 5) = 3 × 4 + 3 × 5 3 × 9 = 12 + 15 27 = 27 4 2 3 4 2 4 3 = × + × 7 3 4 7 3 7 4 4 8 9 8 12 = + 7 12 21 28 4 17 32 36 × = 7 12 84 68 68 = 84 84
Closure property When a rational number is divided by another rational number, the quotient is always a rational number. a c Thus, if and are two rational numbers, then b d a c a c ad ÷ = × = , which is again a rational number b d b d bc since b, c, d are non-zero integers.
Ex.24
3 1 3 3 9 = ÷ = × 4 3 4 1 4
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Division is not commutative : a c If and are two rational numbers in which b d b, c and d 0, then a c c a ÷ ÷ because, b d d b a c a d c a c b cb ÷ = × and ÷ = × = b d b c d b d a da
So Ex.25
So
a c c a ÷ ÷ b d d b 4 1 1 4 ÷ is not equal to () ÷ 7 3 3 7 4 1 4 3 12 ÷ = × = , 7 3 7 1 7 1 4 1 7 7 whereas ÷ = × = 3 7 3 4 12
Ex.31
Law of Addition : a c e , , and are three rational numbers. b d f a c a e c e (i) If > then + > + . b d b f d f
Given
(ii) If
a c a e c e = then + = + . b d b f d f
(iii) If
a c a e c e < then + < + . b d b f d f
Ex.32
4 1 1 4 ÷ ÷ 7 3 3 7
Hence,
a c and , b d a c a c a c then either > , = , or < . b d b d b d
Given two rational numbers
Ex.28
Ex.33
3 1 3 1 and , > . 4 2 4 2 3 12 3 12 Given and , = . 7 28 7 28
7 2 14 , , and be three rational numbers, 18 18 36
then
7 14 = , 18 36
So
9 18 = 18 36
a c e If , , and are three rational numbers b d f a c c e a e (i) If > and > , then > . b d d f b f a c c e a e (ii) If < and < , then < . b d d f b f a c c e a e (iii) If = and = , then = . b d d f b f
Ex.30
1 1 1 1 1 1 < and < , so < . 6 3 12 6 12 3
7 2 14 2 + = + 18 18 36 18 7 2 14 4 = 18 36
Law of Transitivity :
3 1 1 1 3 1 > and > , so > . 4 2 2 4 4 4
13 4 3 4 + > + 20 20 20 20
Let,
2 1 2 1 and , < . 9 2 9 2
Ex.29
so
17 7 > . 20 20 Which is true.
Given
Given
13 3 > , 20 20
Law of Trichotomy :
Ex.27
Consider the three rational numbers 13 3 4 , , and . 20 20 20
Ordering of Rational Numbers
Ex.26
5 10 1 5 1 10 = and = , hence = . 10 20 2 10 2 20 (all are equivalent).
1 1 = , 2 2 Which is true.
Property of Multiplication : Let
a c e , and be three rational numbers. b d f
e is a positive number, then f a c a e c e (i) If > , then × > × . b d b f d f a c a e c e (ii) If < , then × < × . b d b f d f a c a e c e (iii) If = , then × = × . b d b f d f If
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Ex.34
Ex.35
Ex.36
3 3 3 1 3 1 > , then × > × 4 8 4 2 8 2 3 3 > (True) 8 16 1 2 1 1 2 1 < , then × < × 6 3 6 3 3 3 1 2 < (True) 18 9 2 4 2 1 4 1 = , then × = × 7 14 7 2 14 2 1 1 = (True) 7 7
Thus, reference books =
books. Given that reference books = 7,294 14 That is, of all the books = 7,294 45 Thus, the number of all the books 14 = 7294 ÷ 45 45 = 7294 × 14 = 23445 books 2 Non-friction books = of all the books 15
Word Problems Ex.37
Sol.
2 1 of the boys in a class are over 5 feet tall, 3 6 of the boys in that class are over 5 feet 3 inches tall. If 27 boys are between 5 feet and 5 feet 3 inches, how many boys are there in that class? Let there be x number of boys in that class.
Powers Exponential Numbers :
and
Rational
4 4 4 numbers. For example: × × can be 5 5 5 3
4 4 written as which is read as raised to the 5 5 power 3. (i)
x = 27 × 2 = 54
A library has books on fiction, non-friction, and 5 reference books. of all the books in the library 9 2 are works of fiction and are of non-friction. 15 If there are 7,294 reference books, how many non-friction books are there in the library? Now, all the books in the library – (fiction + nonfriction) = reference books or
5 2 1 – = reference books 9 15
or
25 6 = reference books 1 – 45
or
31 1– = reference books 45
3 3 27 3 3 3 3 3 = × × = 3 = 64 4 4 4 4 4
2 2 25 5 5 5 (5) × × 2 = (ii) = 36 6 6 6 6
Thus there are 54 boys in that class. 2 Check Boys above 5 = × 54 = 36, 3 1 Boys above 53= × 54 = 9, 6 Boys between 5 and 53= 36 – 9 = 27 Ex.38
Notation
Exponential notation can be extended to rational
2 1 Given x – x = 27 3 6 4x x = 27 6 3x = 27 6
45 31 14 = of all the 45 45
3
2 2 2 2 (2) 3 8 = × × = 3 = (ii) 27 3 3 3 3 3 x In general, if y is a rational number and a is a positive integer, then a
x xa a y y 3
Ex.39
4 Evaluate . 5
Sol.
3 4 4 4 4 (4) = × × = 3 5 5 5 5 5
3
= Ex.40
64 125
27 8 and as the powers of rational 64 27 numbers.
Expres
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27 3 3 3 33 3 = = = 64 4 4 4 43 4
Sol.
3
2
8 (2) (2) ( 2) (2) 3 2 and = = 3 = 27 3 3 3 3 3
3
5
5 5 = = 2 2
Reciprocals with Positive Integral Exponents:
4
2 1 34 3 1 Reciprocal of = = = = 4 3 2 4 2 24 2 34 3
4
2.
2 2 2 2 2 (i) 25 ÷ 22 = =2×2×2 2 2 = 23 = 25–2 2 2 2 2 2 2 6 2 2 2 3 3 3 3 3 3 (ii) ÷ = 2 2 3 3 3 3
6
4 5 = and Reciprocal of 5 4 5
5
1 3 Reciprocal of = = 35 3 1
2 2 2 2 = 3 3 3 3 4
2 2 = = 3 3
Reciprocals with Negative Integral Exponents 1 1 = . Therefore, the reciprocal of 2 21 1 2 is 2–1. The reciprocal of 32 = 2 = 3–2. 3 2 2 4 4 Reciprocal of = 5 5 Reciprocal of 2 =
3.
2 2 = , etc. Reciprocal of 3 3
2
1
6
2 2 = = 3 3 2
2 4 Simplify ÷ . 3 3 3
Sol.
2 2 2 2 2 2 = × 3 3 3 3 3 3
xa 3
Ex.41
(i) (23)2 = (2 × 2 × 2)2 = (2 × 2 × 2) × (2 × 2 × 2) = 26 = 23 × 2 2 3 2 2 2 2 (ii) = 3 3 3 3
In general, if x is any rational number other than zero and a is any positive integer, then: x a
2
3
2 4 3 3 ÷ = ÷ 3 3 2 4
4. 2
3 2
( x a ) b x ab
(i) 24 × 34 = (2 × 2 × 2 × 2) × (3 × 3 × 3 × 3) = (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3) = (2 × 3)4 4
4
3 1 3 3 3 3 1 1 1 1 (ii) × = × 5 2 5 5 5 5 2 2 2 2
3 3 3 3 3 = ÷ 2 2 2 4 4
3 1 3 1 3 1 3 1 = × × × 5 2 5 2 5 2 5 2
27 9 27 16 = ÷ = × =6 8 16 8 9
3 1 = 5 2
Laws of Exponents : 1.
6 2
x a x b x a b
3
3
23
x a x b x ab
1 1 The reciprocal of 2 is , reciprocal of 23 is 3 . 2 2
4
3
5 5 5 5 5 5 5 (ii) × = × × × × 2 2 2 2 2 2 2
Consider the following. (i) 33 × 34 = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 37 = 33+4
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4
x a y a ( x y) a
5.
(i)
24
÷
34
2 2 2 2 2 2 2 2 2 = = = 3 3 3 3 3 3 3 3 3
16
4
3 3 3 3 3 1 5 5 5 5 (ii) ÷ = 5 2 1 1 1 1 2 2 2 2 4
4
3 3 3 3 5 5 5 5 = × × × = 1 1 1 1 2 2 2 2
x x a y a y
2 2 = 3 3 8x = – 16 x=–2
3 5 1 2
8x
If x is any rational number different from zero and a, b are any integers, then,
4
a
Law I:
x a x b x ab
Law II:
x a x b x a b
Law III:
( x a ) b x ab
Law IV:
x a y a ( x y) a (where y is also a nonzero rational number)
3
Ex.42
2 2 1 4 1 Simplify × × 3–1 × 3 3 6
Law I:
x x y y a
a
a
(where y is also a nonzero rational number)
3
Sol.
2 2 1 4 1 × × 3–1 × 3 6 3
4
Ex.44
2 2 Evaluate ÷ 3 3
Sol.
2 2 2 ÷ = 3 3 3
6
1 1 2 = × 34 × × 3 6 3
26
4
1 1 = 6 × × × 3 6 3 = 26 × 3–6 × 34 × 3–1 × 6–1 = 26 × 3–6 × 34 × 3–1 × (2 × 3)–1 = 26 × 3–6 × 34 × 3–1 × 2–1 × 3–1 = 26+(–1) × 3–1+4 + (–1) + (–6) = 26–1 × 3–1+4 –1–6
=
25
34
×
3–4
25
5
Ex.43
Sol.
2 2 × 3 3
5
11
4
2 × 3
2 3
( 5) ( 11)
5 11
2 3
11
8x
8x
8x
2 3
0
4
0
Squares and Square Roots The square of a number is that number raised to the power 2.
Also, 8x
4
2 × 3
2 the expression = = 1 3
52 = 25 implies
2 = 3
2 = 3
2 × 3
4 4
2 2 ÷ = 4 =1 3 3 2 3
but
32 = 4 = 81 3
2 2 Find x so that × 3 3
4
4
25 = 5.
since (–5)2 = 25, so
25 = 5. Thus 25 has two roots, 5 and –5. So
25 = + 5.
Similarly,
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16 = + 4,
49 = + 7, and so on.
Table Number 1 2 3 4 5 6 7 8 9 10
Square 1 4 9 16 25 36 49 64 81 100
Cube 1 8 27 64 125 216 343 512 729 1000
To find the square root of a perfect square by prime factorisation method : Steps : 1- Express the given number as a product of prime factors. 2- Make pairs of similar factors. 3- The product of the prime factors, after taking one factor out of every pair will give the square root of the number. Ex.45 Sol. Ex.46
Sol.
To find the square and square root of a a Rational Number : b a a a2 square of the numerator × = = b b b 2 square of the deno min ator
Ex.48
3 Find the square of . 4
Sol.
3 3 3 9 = × = 4 4 4 16
Ex.49
Find the square of
Sol.
5 52 5 5 25 = 2 = = 8 8 64 8 8
Ex.50
Find the square root of
2
5 . 8
2
64 . 144
64 2 2 2 2 2 2 = 144 3 3 2 2 2 2
Sol.
=
Find the square root of 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5.
26 3 2 2
4
=
23 3 2
2
=
8 2 = 12 3
Cubes and Cube Roots
( 2 2 2 2) (3 3) (5 5)
If x is a rational number, multiplying x with itself 3 times gives x × x × x and is denoted as x3. Here x3 is called the cube and x is the cube root. For any natural number m, if p = m, if p = m × m × m = m3, then p is said to be a perfect cube.
= 2 × 2 × 3 × 5 = 60 Find the square root of 1764 by finding prime factors. 1764 = 2 × 2 × 3 × 3 × 7 × 7
1764 = 2 2 3 3 7 7 =2×3×7 = 42
To find the cube root of a perfect cube by prime factorisation method :
2 1764 2 3
441
3
147
7
49
7
7
1 1 Note that if negative factors are considered, we shall have 1764 = – 42 Ex.47 Sol.
9a 4 b 8 .
Find the simplest answer for
2 2 4 4 9a 4 b 8 = 3 3 a a b b
=
3a b 3a b 2 4
STEPS : 1. Express the given number as a product of prime factors. 2. Make triplicates of similar factors. 3. The product of the prime factors after taking one factor out of every triplicate will give the cube root of the number.
882
2 4
=
Ex.51
Find the cube root of 2 × 2 × 2 × 7 × 7 × 7.
Sol.
3
=
( 2 2 2 7 7 7) 3
( 2 7) (2 7) ( 2 7)
= 2 × 7 = 14
3a2b4
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Ex.52 Sol.
Find the cube root of 3375 by using the prime factorisation method. 3375 = 5 × 5 × 5 × 3 × 3 × 3 3
3375 =
3
5 5 5 3 3 3
=
3
(5 3) (5 3) (5 3)
Ex.58 Sol.
5 17 (5) (17) 22 + = = 9 9 9 9 [ (–5) + (–17) = – 22]
=5×3 = 15
Ex.59
5 3375 5
675
5
135
3
27
3
9
3
3
Sol.
Sol.
Find
3
729 . We know that its cube root is a single digit as there are only three digits. Therefore 729 is the cube of 9, as the units digit of a cube will be 9
Ex.60 Sol.
2. 3. Ex.55
Ex.56 Sol.
Find the cube root of 103823. Look at the units digit, so the digit in the units place of the cube root will be 7. Leave the three digits on the right and look at 103. The biggest cube less than 103 is 64, i.e., 43. Thus, the required cube root is 47. Find the cube root of 884736. 1. The digit in the units place of the cube root will be 6. 2. The highest perfect cube below 884 is 72, i.e., 93. So the required cube root is 96.
Ex.57 Sol.
4 7 4 7 (4) 7 3 + = + = = 11 11 11 11 11 11 [ (–4) + 7 = 3]
3 5 and . 8 12 Clearly, denominator of the given numbers are positive. The LCM of denominators 12 and 8 is 24.
Add
3 5 and into forms in which 8 12 both of them have the same denominator 24. We have,
3 3 3 5 5 2 10 9 = = and, = = 8 8 3 12 12 2 24 24 3 5 10 9 10 9 19 + = + = = 8 12 24 24 24 24 7 and 4. 9
Ex.61
Add
Sol.
We have, 4 =
3 13 and . 5 5 We have,
Add
3 13 3 13 16 + = = 5 5 5 5
4 4 4 (1) = = 11 11 (11) (1)
Now, we express
only for the number 9. Ex.54 Sol. 1.
4 7 and . 11 11 4 We first express as a rational with positive 11 denominator. Add
We have,
1 A cube of a single-digit number will have a maximum of only 3 digits. Note that 93 = 729 and 103 = 1000, i.e., the cube root of any perfect cube below thousand is a single digit. Ex.53
5 17 and . 9 9 We have,
Add
4 . 1 Clearly, denominators of the two rational numbers are positive. We now rewrite them so that they ahve a common denominator 3equal to the LCM of the denominators. LCM of 9 and 1 is 9.
7 –12 and . 9 9 We have,
Add
7 –12 7 (12) 5 + = = [ 7 + (–12) = – 5] 9 9 9 9
49 36 4 = = 1 9 9 1 7 7 7 36 7 36 43 4 +4= + = + = = 9 9 9 9 9 9 1
We have,
[ 3 + 13 = 16]
Ex.62 Sol.
3 5 and . 8 12 The denominators of the given rational numbers are 8 and 12 respectively. The LCM of 8 and 12 is 24.
Add
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Now we re-write the given ratinoal numbers into forms in which both of them have the same dnominator.
Ex.63 Sol.
3 3 3 9 5 5 2 10 = = and, = = 8 8 3 24 12 12 2 24 3 5 9 (10) 9 10 1 + = + = = 8 12 24 24 24 24
Simplify :
3
4 1 1 (iii) = = 3 3 3 4 4 33 3 a n a n n when n is a whole number b b
8 4 + . 15 3
27 1 = 64 64 27
=
We have, 8 4 8 4 + = + 15 3 15 3
8 8 1 8 4 4 1 4 and 15 (15) (1) 15 3 (3) (1) 3 LCM of 15 and 3 is 15.
2 (iv) 5
4
=
4 in the form in which is has 3 denominator 15, we get
1 2 5
4
Re-writing
Ex.64
4 4 5 20 = = 3 35 15 8 4 8 4 + = + 15 3 15 3 8 20 + 15 15
=
(8) (20) 28 = 15 15
number of the form (i) 5–3
4 (iii) 3 (v)
3
Ex.65
1
p : q (ii) (–2)–5 2 (iv) 5
4
2
3
2
3 4 (i) × 8 5 Sol. (i) We have, 3 8
2
1
a–n = (i) 5–3 = (ii)
(–2)–5
5
3
=
= 1
1 125
1 1 = =– 5 = 32 32 (2)
1 n a n a 1 n a n a
3
p : q 2 (ii) 7
4
7 × 5
2
3
1 n a n a
an 1
1 n a n a
4 1 1 × = × 3 2 5 4 3 5 8
and n is a positive integer, then 1
625 1 = 16 16 625
=
Express each of the following as a rational
2 3 We know that, if a is a non-zero rational number
Sol.
(2) 54
4
1 8 23 = = =8 1 1 1 23
=
number of the form
Express each of the following as a rational
1
a n a n n for n 0 b b
(v) 4 20 3 15
=
=
=
1 32 82
×
1 43 53
a n a n n b b
64 125 125 1 1 × = × = 9 64 9 64 9 64 125
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(ii) We have,
2 7 =
= = Ex.66
4
7 × 5
2
(iv)
7 × 5
1 2 7
4
74
(7) 2
×
( 2) 4
52
2
=
1 (2) 4 74
×
(7)
Sol.
2
7
1
53
56
1 4
=
1 = 24
11
=
53 5 6
=
1 5 3 6
=
19
1 = 59 5
1 4
5
1
1
1
1 1 2 3
32 6
1 6
Ex.67
9
×
7
1
1 1 24 6 30 1 1 1 1 24 6 1 1 a a
1 1 = 5 3
1 1 4
(5–1× 3–1)–1 6–1
7
=
1 (1) 45
5
×
1 (1) 47
7
×
1 1 × 1 1 45 4 7
=
1 15
1
1
(iv) We have, =
1
1
(iii) We have,
1 × 4
1
1 1 a a
2
43 = 24
(iii) We have, 5
2
1 5 8 = 2 1 5
2 8 5 8 5 8 5 2 = 5 4 1 2 5 2
1 1 = 6 8
(ii) We have, ×
1
(6–1 – 8–1)–1 + (2–1 – 3–1)–1
3
1
(i) We have,
5 2 10 1 1 (ii) We have,
1 1 43 1 = = = = 43 3 4 13 13 1 43 4
5–3 × 5–6 =
1
=
(ii) 5–3 × 5–6
1 × 4
8 1
1
2
3
5
1
1 1 1 = 5 2 5 8
7777 7 7 = × 16 25
Express each of the following as power of a rational number with positive exponent :
1 (iii) 4 Sol. (i) We have,
1
6 23
5 (2 1 5 1 ) 2 8
52
76 76 76 117649 = = = 16 25 16 25 400 400
1 (i) 4
4
(iii) 5 1 31
45 47 45 7 45 4 7 × = = = 412 1 1 (1) (1) 1
Simplify:
1 1 2 5 (i) 2 5 8
(ii) 6 1 8 1
1
1
(2 1 31 ) 1
1 6
1 6 1 1 1 = 15 15 90 1 6 1 6 6 15
2 (4 1 8 1 ) 3
1
1 1 1 2 1 3 = 2 4 8 8 2 3 3 3 3 2 1 = 8 2 8 3 4
manishkumarphysics.in
Ex.68
(i) 2
Simplify
1 2
2
1 3 1 (ii) 6 2
1
1 (i) 4
2
1 3
2
1
3 –1
2
1 1 3 2 2 3 6
2
2
5 2 25 5 = 2 36 6 6 (ii)
2
1
4 1
2
2
1 4 =
=
1 2
1 1 4
1 12 42
2
2
1 3
1 1 2
1 12 22
2
1
12 32
4 1 1 1 (iii) 4 3
1
2
42 12
22 12
1 1 = 3 1 4 4
32 12
2
2
Sol.
(ii) We have,
1 1 = 6 3 2
1
1 2 6 3
1
1 2 2 6
1
1
Let (– 8)–1 be multiplied by x to get 10–1. Then, x × (–8)–1 = 10–1
1
(i) (2–1 + 3–1)2
1
1 6 5 5 6
4 5 Using the laws of exponents, simplify each of the following and express in exponential form: (A) 37 × 3–2 (B) 2–7 2–3 2 –3 (C) (5 ) (D) 2–3 × (–7)–3
Ex.71
(E) (ii) (2–1 – 4–1)2
1
1
4 1 (iii) 3 4 We know that for any positive integer n and any rational number a, a n
1 an
1 1 a a
1 1 10 8 1 –8 –8 –4 x 10 1 10 5 x
Hence, the required number is
p number of the form q
Sol.
4 4 3 1
1 3 3 8 8 8 3 –1 By what number should (–8) be multiplied so that the product may be equal to 10–1 ?
1
Express each of the following as a rational
1
1 16
x = 10–1 (–8)–1
5 = 6
Ex.69
1
1
Ex.70
4
2
2
1
4 3 4 8 = 3 3
2
4 2 3 = 1 1 1 = 42 + 22 + 32 = 16 + 4 + 9 = 29
1 3 1 6 2
12
=
2
1 3
2
1 1 2 1 1 2 4 4 4
Sol. (i) We have, 2
2
Sol.
3 5
4 5 Using laws of exponents, we have: (i) 37 × 3–2 = 37 + (–2) = 35
[ a m a n a m n ] (ii) 2 7 2 3
. Thus, we have
manishkumarphysics.in
2 7 2 3
2 7 ( 3) 2 7 3 2 4
am m n n a a
(iii) (52 )–3= 52 ×–3 = 5–6
(iii) We have, 3–4 × 2–4 = (3 × 2)–4
[ (a m ) n a mn ]
=
(iv) 2–3×(–7)–3 = (2× (–7))–3 = (–14)–3
(v)
Ex..72
3 5 4 5
3 4
n
an a n b b
5
Using the laws of exponents simplify and express each of the following in exponential form with positive exponent:
(iii)
3–4
×
1 (iv) 3 2
2–4
2
16 =
1 = 4
37 5 7 ( 10) 3 5 = 10 3 3 3 = 37 10 3 –5 33 35 33 ( 5) 32
1 = 4
1 n a n a
1 = 2 2 3 3 3 (vi) We have,
4 4 = (1) 3
2
2
2
2 5 2
7 = 2
2
7
1 = 7 2 2 17
7
54 34
54 34
[ (1) 4 1]
am m n n a a 1 n a n a
1
4
n an a (ab) n a n b n and n b b
1 1 4 4
4 = (1 3 )
5
2
[ 3 1 3]
(ii) We have,
2 5 2 2
12
4
an a n b b
6
5 5 (3) 4 (1 3) 4 3 3
n
6
2
32 6 1 3 1 1 2 2 2
(37 310 ) 35
[16 1]
(4) 6
2
(v) We have, 4
[ a m a n a m n ]
(4) 6
a n a n n b b
4
(iv) We have,
(–4)4 × (–4)–10 = (–4)4 + (–10)
=
[14 1]
64
1 = 6
Sol. (i) We have,
= (–4)–6 1
6 14
2 13 1 3 3 2 2
4 5 (vi) (–3) 3
(v) (37 310 ) 3 –5
4
=
(ii) 2–5 22
(i) (–4)4 × (–4)–10
1 n a n a
1
=
[ a n b n (ab) n ]
[ an × bn = (ab)n]
6–4
= 34 Ex.73
54 3
4
34 4 5 4 3 0 5 4 1 5 4 5 4
Simplify and write the answer in the exponetial form: (i)
(2 5 28 ) 5 2 –5
(ii) (– 4)3 × (5)–3 × (–5)–3 (iii)
1 3 3 8
manishkumarphysics.in
Sol.
(i) We have,
(2 2 ) 2 5
8 5
32 23 4 2 1 = 2 3 2 = (9 – 8) 16 = 1 16 = 1 1 1 16
–5
5
25 5 58 5 5 = 8 2 ( 2 ) 2 2
Ex.75
= (2 3 ) 5 2 5 2 35 2 5
Simplify : 7
5 8 (i) 8 5
= 2–15 × 2–5 = 2–15–5 = 2–20
5
–2 (ii) 3
2
4 5
3
7 6
3
(ii) We have, (–4)–3 × 5–3 × (–5)–3 = {– 4 × 5 × (–5)}–3
= (100)–3 = 102
3
a
n
b n c n (abc) n
3 (iii) 4
1 3 1 3 3 3 3 2 3 3 3 ( 2 3) 3 6 3 8 2
3 (iv) 7
2
8 5
5
5 7
8 7
8 5 5 5
5 7 5 5
8 5 8 7
8 2 64 = 5–7+5 × 8–5+7 = 5–2 × 82 = 2 25 5 (ii) We have,
Simplify each of the following : 1
2 3
=
We have,
2
32 ( 2)
2
4 5
53 4
3
3
(2) 2 3
2
4 3 5 3
9 125 9 125 1125 = 4 64 256 4 64
(iii) We have,
2 2 1 (i) 5
1 2 = 5
3
= 5–7–(–5) × 8–5–(–7)
1 2 1 3 1 2 (ii) 3 2 4
Sol.
7
5 8
2 2 1 (i) 5
3 2
Sol. (i) We have,
= 102 ×–3 = 10–6
(iii) We have,
Ex.74
4
1
2
1 2 = 5
1 = 5
2 1
(a
) a mn
m n
3 4
( 2 )( 2 )
=
4
34 4 4
3 2
=
(1) 4 1 1 = 4 5 625 5
4
1 3 2
3
3 4
4
3 2
3
23
(2 2 ) 4 2 3
3 4 3 3 2 8 2 3
34 3 2 8 3
3 1 2 5
25 31
32 3
(iv) We have,
(ii) We have,
3 7
1 2 1 3 1 2 2 4 3 3 2 2 3 4 2 1 1 1
3 4
33
3 4 3 3 4
3
2
a n b n a b
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7 6
3
3 2 7 2
=
a n a n n b b
7 3 6 3
3 2 7 2
7 3 (2 3) 3
3 2
=
7 2 3 2
=
33
7 3
Let (–4)–2 = be multiplied by x to get 10–2 .
Sol.
2 3 3 3
Then, x × (–4) –2 = 10–2
7 3
x 10 2 (4) 2
1
72
2 3
= 3–2+3 × 7–3+2 × 23
x 10 2
x
x
1 24 1 3 = 3 7 2 3 8 7 7
Ex.76
Evaluate:
Sol.
We have, 1
8 5 2
3
–4
8 1 5 3
Ex.77
3 1
(2 ) 5 2
4
3
2
31
2
5
3
4
2
3
2
5
(4) 2 10
2
16 4 100 25
Hence, required number is Ex.79
4 25
By what number should (–12)–1 be divided so
= 2 × 125 = 250
1
2 that the quotient may be ? 3
25 a –4
35 10 5 125
(ii)
5 – 3 10 a 8
25 a 4 5 3 10 a 8
Sol.
Let the required number be x. Then,
5 7 6 5
5 2 a 4 5 2 2 a 8
5 2 a 4 53 ( 2 5) a 8
2 (12) 1 x 3
5 2 a 4 531 2 a 8
5 2 ( 2 ) a 4 8 5 4 a 4 2 2
54 625 4 a4 a 2 2
35 10 5 125 5 7 6 5
(12) 1 2 x 3
2 x (12) 1 3
x
5 7 (2 3) – 5 35 2 5 5 5 53 5
7
2
5
3
5
= 35 ( 5) 2 –5 ( 5) 5 5 3( 7 ) = 3 0 2 0 5 5 3 7 1 1 5 5 5 5 By what number should (–4)–2 be multiplied so that the product may be equal to 10–2 ?
1
1 3 12 2 1 1 1 b a a and a a b
35 (2 5) 5 53
=
1
1
(ii) We have,
Ex.78
(4) 2
4
(i) We have,
=
10 2
3
Simplify: (i)
Sol.
×
53
(4) 2
2–4
2 3 4 5 3 21
1
Ex.80
x
1 2 1 1 12 3 18 18
3 By what number should 2
3
4 that the quotient may be 27
manishkumarphysics.in
be divided so 2
?
Sol.
Let the required number be x. Then,
3 2
3
4 x 27
3
Ex.82
2
Sol.
2 2 Find m so that 9 9 We have, 3
3 2
3
3 x 2
1 4 x 27 3
4 27
3
2 2 9 9
2
2 9
2
2
2 27 x 3 4
3 6
5 Find x so that 3
Sol.
We have,
5 3
5
5 3
2
Ex.83
3 2 If x 2 3
Sol.
We have,
5 3
511
5 3
16
5 3
5 3
5 3
11
5 3
8x = –16 16 2 8
2 a
2 m 1
4
2
4
3 x 2
8x
x
2
, find the value of x–2
2
3 3 x 2 2
8x
2 4
3 6 2
2
3 2
a n b n a b 6
3 2
6 2
3 2
12
12
2 3
Write the following numbers in standard form: (i) 0.4579
(ii) 0.000007
(iii) 0.000000564
(iv) 0.0000021
(v) 216000000
(vi) 0.0000529×104
(vii) 9573×10–4
8x
8x
3
4
3 2 x 2 3
Ex.84 11
5 3
x
5
2 2 a
m=–1
3
Ex.81
2 m 1
2 m 1
2m = – 2
8 42 8 42 2 43 2 2 27 27 27 27 27 3
4 2 27
2 9
2 m 1
2m = – 3 + 1
a n b n a b
1 2 x 3 27 2 4 23 1 x 2 3 (3) 27 42
x
2 9
2 9
2m – 1 = – 3
3
6
6
Sol.
(i) To express 0.4579 in standard form the decimal point is moved through one place only to the right so that there is just one digit on the left of the decimal point. 0.4579 = 4.579 × 10–1 is in the standard form (ii) 0.000007= 7 × 10–6 [ The decimal point is moved 6 placed to the right]
manishkumarphysics.in
(iii) 0.000000564 = 5.64 × 10–7
(iv) Size of a plant cell is 0.00001275 metre = 1.275 × 10–5 metre
[ The decimal point is moved 7 places to the right] (iv) 0.0000021 = 2.1 ×
[ The decimal point is moved 6 place to the right] (v) 21600000 = 2.16 ×
(vi) 0.0000529 ×
Ex.87
If the diameters of the Sun and the Earth are 1.4 × 109 metres and 1.275 ×107 metre respectively. Compare these two.
Sol.
We have,
108
[ The decimal point is moved 8 places to the left] 104
(v) The thickness of a normal paper is 0.07 mm = 7 × 10–2 mm
10–6
= 5.29 ×
= 5.29 ×
10–5
10–5+4
×
104
Dia. of the sun 1.4 109 1.4 10 2 10 7 Dia. of the Earth 1.275 10 7 1.275 10 7
= 5.29 × 10–1
(vii) 9573 × 10–4 = 9.573 × 103 × 10–4 = 9.573 × 103+(–4) = 9.573 × 10–1 Ex.85
=
Express the following numbers in usual form: (i) 3.52 × 105
(ii) 7.54 × 10–4 1 .4 ~ – 1.3 and – 1 1.275 ~ 1 .3
(iii) 3 × 10–5 Sol.
We have, (i) 3.52 × 105 = 3.52 × 100000 = 352000 (ii) 7.54 × 10–4 =
(iii) 3 10
Ex.86
5
1. 4 1 .4 100 100 ~ – 100 1.275 1 .3
7.54 10
4
So, the diameter of the Sun is about 100 times the diameter of the Earth.
7.54 0.000754 10000
Ex.88
The size of a red blood cell is 0.000007 m and the size of a plant cell is 0.00001275 m. Compare these two.
Sol.
We have,
3
3 5 0.00003 100000 10
Express the number appearing in the following satatements in standard form: (i) 1 micron is equal to
1 metre 1000000
Size of red blood cell = 0.000007 m = 7×10–6 m Size of plant cell = 0.00001275 = 1.275×10–5 m
Size of red blood cell 7 10 –6 Size of plant cell 1.275 10 – 5
(ii) Charge of an electron is 0.0000000000000000016 coloumbs. (iii) Size of a bacteria is 0.0000005 metre
=
7 10 –6 5 7 10 –1 1.275 1.275
=
0 .7 0 .7 1 ~ ~ 1.275 – 1.3 – 2
(iv) Size of a plant cell is 0.00001275 metre (v) Thickness of a normal paper is 0.07 mm Sol.
(i) 1 micron is equal to
1 metre = 10–6 1000000
So, a red blood cell is approximately half of a plant cell in size.
metre (ii) Charge of an electron is 0.0000000000000000016 coloumbs = 1.6 × 10–18 columbs (iii) Size of a bacteria is 0.0000005 metre = 5 × 10– 7metre
manishkumarphysics.in
EXERCISE # 1 Q.1
Express each of the following as a rational
p , where p and q are intergers q
number of the form
Q.7
(iii)
32
2 (v) 3 Q.2
1 (iv) 2 2
Q.8
By what number should (–15)–1 be divided so that the quotient may be equal to (–5)–1 ?
Q.9
Write each of the following in exponential form: 1
1
4
1
(ii) (3 4 ) 2
1
0
(iii) (31 4 1 5 1 ) 0
2 (ii) 5
2
1 1 1 1 (iv) 4 3
1
Q.10
1
1 (ii) 2
1 3
2
1
1 3
1 4
2
1
1 4
Q.11
2
1
(iv) (5 2 ) 6 Q.4
Q.5
2 5
2
2 5
Evaluate: (i) 5–2
Q.12
(i) (4 1 31 ) 2
(ii) (5 1 6 1 ) 3
(iii) (2 1 31 ) 1
(iv) (31 4 1 ) 1 5 1
2
4
1 (iv) 2
1
Express each of the following as a rational
1 (iii) 4
1
1
(ii) (–3)–2
(i) 6–1
Simplify:
1
3 (iv) (4) 1 2
Simplify
p : q (ii) (–7)–1
(i) 4 1 31
2
(ii) 5 1 6 1
1
3
(iii) (2–1 + 3–1)–1 (iv) {3–1 × 4–1}–1 ×5–1 –1 –1 –1 (v) (4 – 5 ) 3
Simplify
1 2 2 (i) (3 2 ) 2 1 (ii) 3
3
3
2 2 2 (ii) (3 2 ) 3
3
1 1 2 4
3 2 2 2 (iv) (2 3 4 ) 2 Q.6
1
number in the form
(iii) (2 1 4 1 ) 2 2 1
2
1 (iii) 3
Find the value of each of the following:
1 (i) 2
1
3 3 3 3 (i) 2 2 2 2
Find the value of each of the following: (i) 3
Q.3
5
be multiplied 1
(ii) (–4)–2
1
1
4 so that the product may be equal to ? 7
and q 0 : (i) 2–3
1 By what number should 2
3
3
Q.13
Express each of the following rational numbers with a negative exponent:
1 (i) 4
3
(ii) 35
3 (iii) 5
5–1
7 4 (v) 3
By what number should be multiplied so that the product may be equal to (–7)–1 ?
manishkumarphysics.in
3 4 (iv) 2
4
2
3
3
Q.14
Express each of the following rational numbers with a positive exponent:
3 (i) 4
2
5 (ii) 4
3 4 (v) 2
Q.15
3
4 3 (iv) 3
(iii) 43 × 4–9
1 (ii) 2
4
5 (v) 4
3 (i) 2
1 2
1 4
2 2 2 (ii) (3 2 ) 3
1 1 (iii) (4) 2 2 2 1 (iv) 4
Q.21
Q.19
1 By what number should 2
5 By what number should 3
5
8 8 3 3
3 2 (i) If x 2 3 2
x 2
4
, find the value of x–2. 2
1 , find the value of x–1 4
Write the following numbers in the usual form: (i) 4.83 × 107 (ii) 3.02 × 10–6 4 (iii) 4.5 × 10 (iv) 3 × 10–8 9 (v) 1.0001 ×10 (vi) 5.8 × 102 6 (vii) 3.61492 × 10 (viii) 3.25 × 10–7
Q.25
Subtract the first rational nuber from the second in each of the following:
1
be multiplied so
(i)
1 4
3 5 , 8 8
2
be multiplied so
7 4 , 9 9 11 4 , (iv) 13 13 2 5 , (vi) 3 6 8 7 , (viii) 33 22
(ii)
2 9 , 11 11 1 3 (v) , 4 8 6 13 , (vii) 7 14 (iii)
Q.26
Find x, if 8
2 x 1
Q.24
1
1 4
5
Express the following numbers in standard form: (i) 6020000000000000 (ii) 0.00000000000942 (iii) 0.00000000085 (iv) 846 × 107 –4 (v) 3759 × 10 (vi) 0.00072984 (vii) 0.000437 × 104 (viii) 4 100000
By what number should (–15)–1 be divided so that the quotient may be equal to (–5)–1 ?
4
5 4
Q.23
By what number should 5–1 be multiplied so that the product may be equal to (–7)–1?
1 (i) 4
4
Find the value of x for which 52x 5–3 = 55.
7 that the product may be ? 3 Q.20
2 3x
1
4 1 that the product may be equal to 7 ? Q.18
2 5
5 4
2 x 1
Q.22
3
Q.17
15
4 (ii) If x 5
1
2 2 1 4 1 1 (v) 3 6 3 3 Q.16
3
2
3
1
5
x
8 (vi) 3
3
2 x 1
3
2 2 (iv) 5 5
2
3
8
1 1 2 2
3 3 3 (iii) 2 2 2
Simplify 3
19
The sum of the two numbers is
4 x
numbers is
manishkumarphysics.in
1 , find the other.. 3
5 . If one of the 9
Q.27
The sum of two numbers is numbers is
Q.28
Q.29
Q.30
Q.31
Q.33
Q.34
Q.35
Q.36
Q.37
Q.38
Q.39
12 , find the other.. 3
4 The sum of two numbers is . If one of the 3 numbers is –5, find the other.
The sum of two rational numbers is–8. If one of 15 the numbers is , find the other.. 7 7 5 What should be added to so as to get ? 8 9 What number should be added to get
Q.32
1 . If one of the 3
5 so as to 11
26 ? 33
5 What number should be added to to get 7 2 ? 3 5 What number should be subtracted from to 3 5 get ? 6 3 What number should be subtracted from to 7 5 get ? 4 2 3 2 What should be added to to get ? 15 3 5
Q.40
Simply each of the following and write as a
p : q 2 5 7 (ii) 3 6 9 4 7 8 (iv) 5 10 15 5 3 7 3 (vi) 3 2 3
(i)
8 1 11 3 3 3 4 6 8
(ii)
6 7 19 12 1 7 9 21 7
(iii)
15 9 11 7 6 2 8 3 6
(iv)
7 9 19 11 0 4 5 10 14
(v)
7 5 1 5 2 4 3 2 6
Simplify: (i)
Q.41
3 5 7 2 4 4
(iv)
2 3 4 5 10 7
(v)
5 2 2 6 5 15
(vi)
3 2 5 8 9 36
(ii)
5 3 by 7 4
Multiply:
7 5 by 11 4
(iii)
2 5 by 9 11
(iv)
3 5 by 17 4
(v)
9 36 by 7 11
(vi)
11 21 by 13 7
3 4 by 5 7
(viii)
15 by 7 11
Multiply: (i)
rational number of the frorm 3 5 7 (i) 4 6 8 11 7 5 (iii) 2 6 8 9 22 13 (v) 10 15 20
5 7 2 3 6 3
5 7 2 4 6 3
(vii) Q.42
(ii)
(iii)
(i)
1 1 1 What should be added to to get 3 ? 2 3 5 3 2 What should be subtracted from to get 4 3 1 ? 6
Express each of the following as a rational p number of the form q :
5 17
by
51 60
(ii)
55 6 by 36 11
(iii)
8 5 by 25 16
(iv)
6 49 by 7 36
(v)
8 7 by 9 16
(vi)
8 3 by 9 64
manishkumarphysics.in
Q.43
Simplify each of the following and express the result as a rational number in standard form: 16 14 (i) 21 5
7 3 (ii) 6 28
19 16 (iii) 36
13 27 (iv) 9 26
(v)
(vii)
Q.44
9 64 16 27 11 81 9 88
(vi)
(viii)
50 14 7 3 5 72 9 25
Q.45
Simplify: 3 1 5 7 13 4 (i) 2 6 3 2 8 3 1 2 5 2 3 9 (ii) 4 7 14 3 7 2 13 15 7 8 3 1 (iii) 2 3 5 5 2 9 3 5 9 4 5 6 (iv) 11 6 12 3 13 15
Simplify: 25 2 3 10 (i) 9 8 5 5 1 1 1 (ii) 6 2 4 2 2 2 (iii) 5 6 15 9 9 5 13 5 (iv) 4 3 2 6 4 12 3 21 (v) 3 5 7 15 13 8 5 11 (vi) 3 5 3 2 13 11 4 5 (vii) 7 26 3 6 8 3 3 11 (viii) 5 2 10 16
manishkumarphysics.in
EXERCISE # 2 Q.1
Q.2
(A) 1.4
(B) 1. 5
(C) 1.54
(D) 1.45
(A) – 15
Evaluate
3
(A) 4 Q.3
Q.4
Q.5
Q.6
0
1 64
If x = 0. 7 , then 2x is -
1 64
Q.10 (C) 32
(256)0.16
×
(256)0.09
(B) 256.25
(C) 16
(D) 4
(A)
is -
1 1 If a = 2 + 3 and b = 2 – 3 , then is equal a b to -
(B) – 2 3
(C) 4
(D) – 4
(B) – 14
(C) 8 3
(D) – 8 3
2 3
2 3 a + b is -
, b=
2 3 2 3
1 a
2
1 b2
is
Q.7
(B) – 14
(D) –
If x = 3 + 8 and y = 3 – 8 , then
1 x2
Q.8
(B) – 34
(B) 0.4 9
(C) 0. 5
(D) 0.5
3 5 Find the value of x when = 5 3 (A) x = 2 (B) x = – 2 (C) x = 1 (D) x = – 1
If 2x – 2x–1 = 4, then xx is equal to (A) 1 (B) 27 (C) 3 (D) None of these
Q.14
The value of
1
(A) 1 – 2 (2n) (C) 1 –
1 4
1 (B) 2n+3 – 4 1 (D) 1– 2
0.60 0.11
3 / 2
3 1
3 / 2 1 3
1
is
3
(B) – 3/2 (C) 2/3
(D) –1/2
y2
Q.15
What must be added to the sum of 4x 2 + 3x – 7 and 3x 2 + 6x + 5 to get : 1? (A) 7x 2 + 9x – 3 (B) 3 – 9x – 7x 2 (C) 7x 2 + 9x – 2 (D) None of these
Q.16
1.3 is equal to –
is
(A) 3/4
2 n 3 2 2 n 2(2 n 2 )
x 3
3
(C) 12 8 (D) – 12 8
when simplified is -
(D) 1.5
Q.13
equal to (A) 34
2 3 2
(B)
(A) 0.4 8
, then the value of
(C) 8 3
3 is
If x = 0.16 , then 3x is -
(A) 3/2 (A) 14
2 and
2 x 3
Q.12
If a = 2 + 3 and b = 2 – 3 , then equal to (A) 14
2 3 2
(C) 1.4 Q.11
(A) 2 3
1 8 1 (D) 17 8
(B) 16
A rational number between
(D) 64
(A) 64
If a =
+ 64–1/2 – (–32)4/5 is equal to
7 8 7 (C) – 14 8
2
(B) 16
The value of
Q.9
Q.17
(B) 2/3
(C) 4/3
(D) 2/5
0.585 is equal to – 585 99 999 (C) 585
(A)
manishkumarphysics.in
(B)
585 999
(D) none of these
Q.18
Q.28
5.2 is equal to – (A) 45/9 (C) 47/9
Q.19
Which of the following numbers is different from others ? (A)
Q.20
Q.22
(B)
2
3
(C)
(D)
4
5
(B)
7
8
(C)
(D) 16
13
1
, b =
3 2 2 of + b 2 is – (A) 34 (B) 35
Q.24
1
If a =
5 = 2.236 and
If
10 5 2 (A) 0.455 (C) 0.655 Q.26
Q.27
(C) 36
(D) 37 then the value of
(C) 195
(D) 9 3
If
3 1 b is– (A) (B) (C) (D)
a a a a
= = = =
(D)
3 2
y2 x 2 xy x 2y2 x 2 y2
and y = 1, the value of
3 2 6 4 5
xy x 3y
5 (B)
6 4
(D)
6 4 5
5 (C)
1
6 4
x2
Q.32
(D) 196
If A = x – (A) (C)
Q.33
1 1 , then the value of A is A x
x 4 x 2 1 x ( x 2 1) x 4 1 x3 x 2
The value of
on
(C)
8x 1 x4 8 1 x4
(B)
x 4 x 2 1 x ( x 2 1)
(D) 1 1 x 1 x 4x is – 1 x 1 x 1 x2 8x (B) 1 x4 8 (D) 1 x4
Q.34
The expression to be added to (5x 2 – 7x + 2) to produce (7x 2 – 1) is – (A) 2x 2 + 7x + 3 (B) 2x 2 + 7x – 3 2 (C) 12x – 7x + 1 (D) 2x 2 – 3
Q.35
What must be added to 1 – x + x 2 – 2x 3 to obtain x 3 ? (A) x 3 – x 2 + x – 1 (B) – 1 + x + x 2 – 3x 3 (C) 3x 3 – x 2 + x – 1 (D) None of these
= a + b 3 , then the value of a and
2, b = –1 2, b = 1 –2, b = 1 –2, b = –1
xy xy
If x =
(A)
(C)
(B)
y2x2
(D) 200
5 3 – 3 12 2 75 (B) 6 3
yx
(A)
10 = 3.162, the value of
(A) 5 3
3 1
27161 9000
is –
(B) 0.855 (D) 0.755
3
(D)
What must be added to x/y to make it y/x ?
on simplifying is –
The value o f simplifying is –
27161 9900
Q.30
Q.31
If x = (7 + 4 3 ), then the value of x 2 + (B) 194
(C)
then the value
3 2 2
1
27 99
(C)
1
3 2 2 3 2 2 a3 + b 3 is – (A) 194 (B) 196 (C) 198
is – (A) 193 Q.25
,b=
(B)
(A)
a2
Q.23
27161 9999
The sum of a number and its reciprocal is 125/22. The number is – (A) 2/11 (B) 1/11 (C) 3/11 (D) None of these
The product of 4 6 and 3 24 is – (A) 124 (B) 134 (C) 144 (D) 154 If a =
(A)
Q.29
Which of the following numbers is different from others ? (A)
Q.21
(B) 46/9 (D) None of these
The rational form of 2.7435 is –
manishkumarphysics.in
Hints & Solution # 1 Sol.1
Sol.2
(i)
(i)
1 8
(ii)
7 12
(ii) 5
(iii) 1
(iv) – 1
(ii) 29
1 (iii) 2
3 (iv) 5
Sol.3
(i) 9
Sol.4
(i)
1 144
(ii)
13 8
(ii)
1 (iii) 9 16
216 125
(iii)
135 8
(iii)
(iv) 32 (v)
6 5
(iv)
19 64
(iv)
9 4
19 64 32 (v) 81
Sol.15 (i)
12 5 4 3
(ii)
135 8
(iii) –2 (iv)
Sol.23 (i) 6.02 × 1015 (iii) 8.5 × 10–10 (v) 3.759 × 10–1 (vii) 4.37
(ii) 9.42 × 10–12 (iv) 8.46 × 109 (vi) 7.2984 × 10–4 (viii) 4 × 10–5
Sol.24 (i) 48300000 (iii) 45000 (v) 100100000 (vii) 3614920
(ii) 0.00000302 (iv) 0.00000003 (vi) 580 (viii) 0.000000325
Sol.5
(i)
Sol.6
5 7
Sol.25
2 9
Sol.26
Sol.7
7 8
Sol.27
11 3
Sol.28 –
Sol.8
1 3
Sol.29
103 72
Sol.30
41 32
Sol.9
3 (i) 2
Sol.31
1 21
Sol.32
5 2
Sol.33
23 28
Sol.34
7 5
Sol.35
59 30
Sol.36
1 4
4
2 (ii) 5
6
Sol.10 (i)
1 25
(ii)
1 9
Sol.11 (i)
1 6
(ii)
1 7
(iii) 4
(iv)
1 6
Sol.12 (i)
1 144
(ii)
216 125
(iii)
6 5
(iv)
12 5
(iii) 81 (iv) –2
1 Sol.13 (i) 4–3 (ii) 3 7 (v) 3
5 (iii) 3
4
1 (iii) 4
6
(v)
3 (iv) 2
3 20
12
12
2
4 4 Sol.14 (i) (ii) 3 5 2 (v) 3
5
2 3
(v)
3
12
(iv)
1 256
4 3
8
manishkumarphysics.in
11 3 41 7
Hints & Solution # 2 Q.No Ans . Q.No Ans . 3.[D]
1 B 19 C
2 B 20 D
3 D 21 C
4 C 22 A
5 D 23 C
6 A 24 B
7 A 25 C
8 C 26 D
9 C 27 A
10 D 28 C
(256)0.16 × (256)0.09 = (44)0.16 × (44)0.09
= (44)0.16+0.09 = (44)0.25 = (44)1/4 = 4 4.[C]
a = 2 + 3 and b = 2– 3
=
1 1 1 1 = a b 2 3 2 3
=
2 3 2 3
a = 2 + 3 and b = 2– 3 1
=
a2 1
=
b2
1
1
a2
–
1 b2
43 4 3 1
=
(2 3 ) 2
434 3 1
=
8.[C]
1
=
(2 3 ) 2 1
x
2
1 y
2
74 3
34
–
= =
2( 2
74 3 1
=
74 3
=
1 74 3
11.[D]
n2
)
=
2 n .23 2 n .21 2 n .2 3 23 2 2
3
= 1–
=
2 3
2 3
2 3
+
x =
2 3
14.[B]
1 6
34 34 = = 34 289 288 1
=
2 n 3 21.2 n 1
2 .2
n 2
=
2 n 3 2 n 1 2 n 3
2 n (2 3 21 ) =
2 n (23 )
1 4
....(ii)
3x =
3 3 2
1
3
3 1 2 3 1
2
14 14 = = = 14 43 1
x = 3 + 8 and y = 3 – 8
x2 = (3 + 8 )2 = 9 + 8 + 6 8 = 17 + 6 8
3 1 = = 1.5 6 2
(0.6) 0 (0.1) 1
(2 3 ) (2 3 ) ( 2) ( 3 )
17 6 8
100 x = 16. 6
2 3
(2 3 ) 2 ( 2 3 ) 2
2
1
Subtracting (i) from (ii) 90x = 15
43 4 3 43 4 3 =
18 C
....(i)
2 3
+
17 B 35 C
x = 0.16
2 4 3 8 3 = =–8 3 49 48 1 , b=
17 6 8
(17) 2 (6 8 ) 2
=
2 3
1
16 C 34 B
(17 6 8 ) (17 6 8 )
10 x = 1. 6
a=
15 B 33 B
17 6 8 17 6 8
( 7 4 3 ) (7 4 3 )
2 3
14 B 32 A
=
=
a+b=
7.[A]
1
13 B 31 A
2 n 3 2( 2 n )
1
74 3 74 3
6.[A]
12 A 30 B
4 =4 43
=
(2 3 ) (2 3 ) 5.[D]
11 D 29 A
23 3
1
1 0.1 3
3 3 2
9 9 3 = = 93 6 2
y2 = (3 – 8 )2 = 9 + 8 – 6 8 = 17 – 6 8
manishkumarphysics.in
1
=
1 10 8 27 3 3 8
