Real Numbers Class 9th Mathematics Part I MHB Solution

Real Numbers Class 9th Mathematics Part I MHB Solution

Practice Set 2.1
Question 1.

Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.

i.  ii.

iii.  iv.

v.

i.

∵ The division is exact

∴ it is a terminating decimal.

ii.

∵ The division never ends and the digits ‘18’ is repeated endlessly

∴ it is a non-terminating recurring type decimal.

iii.

∵ The division is exact

∴ it is a terminating decimal.

iv.

∵ The division is exact

∴ it is a terminating decimal.

v.

∵ The division never ends and the digit ‘3’ is repeated endlessly

∴ it is a non-terminating recurring type decimal.

Question 2.

Write the following rational numbers in decimal form.

i.  ii.

iii.  iv.

v.

i.

ii.

iii.

iv.

v.

Question 3.

Write the following rational numbers in form.

i.  ii.

iii.  iv.

v.

i.

Let

⇒ 10x = 6.66666......

Now,

10x - x = 6.66 – 0.6666

⇒9x = 6

ii.

Let

⇒ 100x = 37.3737......

Now,

100x - x = 37.3737 – 0.3737

⇒ 99x = 37

iii.

Let

⇒ 100x = 317.1717......

Now,

100x - x = 317.1717 – 3.1717

⇒ 99x = 314

iv.

Let

⇒ 100x = 1589.8989......

Now,

100x - x = 1589.8989 – 15.8989

⇒ 99x = 1574

v.

Let

⇒ 1000x = 2514.514514......

Now,

1000x - x = 2514.514514 – 2.514514

⇒ 999x = 2512

Practice Set 2.2
Question 1.

Show that is 4√2 an irrational number.

Let us assume that 4√2 is a rational number

where, b≠0 and a, b are integers

∵ a, b are integers ∴ 4b is also integer

is rational which cannot be possible

which is an irrational number

∵ it is contradicting our assumption

∴ the assumption was wrong

Hence, 4√2 is an irrational number

Question 2.

Prove that 3 + √5 is an irrational number.

Let us assume that 3 + √5 is a rational number

where, b≠0 and a, b are integers

∵ a, b are integers ∴ a – 3b is also integer

is rational which cannot be possible

which is an irrational number

∵ it is contradicting our assumption ∴ the assumption was wrong

Hence, 3 + √5 is an irrational number

Question 3.

Represent the numbers √5 and √10 on a number line.

By Pythagoras theorem,

(√5)2 = 22 + 12

⇒ (√5)2 = 4 + 1

First mark 0 and 2 on the number line. Then, draw a perpendicular of 1 unit from 2. And Join the top of perpendicular and 0. This line would be equal to √5. Now measure the line with compass and marc an arc on the number line with the same measurement. This point is √5.

Also,

By Pythagoras theorem,

(√10)2 = 32 + 12

⇒ (√10)2 = 9 + 1

First mark 0 and 3 on the number line. Then, draw a perpendicular of 1 unit from 3. And Join the top of perpendicular and 0. This line would be equal to √10. Now measure the line with compass and marc an arc on the number line with the same measurement. This point is √10.

Question 4.

Write any three rational numbers between the two numbers given below.

0.3 and -0.5

0.3 and -0.5

To find a rational number x between two rational numbers  and , we use

Therefore, to find rational number x (let) between

and

Now if we find a rational number between and it will also be between 0.3 and -0.5 since lies between 0.3 and -0.5.

Therefore, to find rational number y (let) between and

Now if we find a rational number between and it will also be between 0.3 and -0.5 since lies between 0.3 and -0.5.

Therefore, to find rational number z (let) between and

Hence the numbers are -0.2, -0.1 and 0.1

Question 5.

Write any three rational numbers between the two numbers given below.

-2.3 and -2.33

-2.3 and -2.33

To find a rational number x between two rational numbers  and , we use

Therefore, to find rational number x (let) between and

⇒ x = -2.315

Now if we find a rational number between and it will also be between -2.3 and -2.33 since -2.315 lies between -2.3 and -2.33

Therefore, to find rational number y (let) between and

⇒ y = -2.3075

Now if we find a rational number between and  it will also be between -2.3 and -2.33 since -2.315 lies between -2.3 and -2.33

Therefore, to find rational number z (let) between and

⇒ z = -2.3225

Hence the numbers are -2.3225, -2.3075 and -2.315

Question 6.

Write any three rational numbers between the two numbers given below.

5.2 and 5.3

5.2 and 5.3

To find a rational number x between two rational numbers  and , we use

Therefore, to find rational number x (let) between  and

⇒ x = 5.25

Now if we find a rational number between and it will also be between 5.2 and 5.3 since 5.25 lies between 5.2 and 5.3

Therefore, to find rational number y (let) between and

⇒ y = 5.225

Now if we find a rational number between and it will also be between 5.2 and 5.3 since 5.25 lies between 5.2 and 5.3

Therefore, to find rational number z (let) between and

⇒ z = 5.275

Hence the numbers are 5.225, 5.25 and 5.275

Question 7.

Write any three rational numbers between the two numbers given below.

-4.5 and 4.6

-4.5 and 4.6

To find a rational number x between two rational numbers  and , we use

Therefore, to find rational number x (let) between  and

⇒ x = 0.05

Now if we find a rational number between and it will also be between -4.5 and 4.6 since 0.05 lies between -4.5 and 4.6

Therefore, to find rational number y (let) between and

⇒ y = -2.225

Now if we find a rational number between and it will also be between -4.5 and 4.6 since 0.05 lies between -4.5 and 4.6

Therefore, to find rational number z (let) between and

⇒ z = 2.325

Hence the numbers are -2.225, 0.05and 2.325

Practice Set 2.3
Question 1.

State the order of the surds given below.

i.  ii.

iii.  iv.

v.

In , n is called the order of the surd.

Therefore,

i.

In this, the order of surd is 3.

ii.

In this, the order of surd is 5.

iii.

In this, the order of surd is 4.

iv.

In this, the order of surd is 2.

v.

In this, the order of surd is 3

Question 2.

State which of the following are surds. Justify.

i.  ii.

iii.  iv.

v.  vi.

Surds are numbers left in root form (√) to express its exact value. It has an infinite number of non-recurring decimals. Therefore, surds are irrational numbers.

Therefore,

i.

It is a surd ∵ it cannot be expressed as a rational number.

ii.

It is a surd ∵ it cannot be expressed as a rational number.

iii.

It is a surd ∵ it cannot be expressed as a rational number.

iv.√256 = √162 = 16

It is not a surd ∵ it is a rational number.

v.

It is not a surd ∵ it is a rational number.

vi.

It is a surd ∵ it cannot be expressed as a rational number.

Question 3.

Classify the given pair of surds into like surds and unlike surds.

i. √52, 5√13

ii. √68, 5√3

iii. 4√18, 7√2

iv. 19√12, 6√3

v. 5√22, 7√33

vi. 5√5, √75

Two or more surds are said to be similar or like surds if they have the same surd-factor.

And,

Two or more surds are said to be dissimilar or unlike when they are not similar.

Therefore,

i. √52, 5√13

√52 = √(2×2×13) = 2√13

5√13

∵ both surds have same surd-factor i.e., √13.

∴ they are like surds.

ii. √68, 5√3

√68 = √(2×2×17) = 2√17

5√3

∵ both surds have different surd-factors √17 and √3.

∴ they are unlike surds.

iii. 4√18, 7√2

4√18 = 4√(2×3×3) = 4×3√2 = 12√2

7√2

∵ both surds have same surd-factor i.e., √2.

∴ they are like surds.

iv. 19√12, 6√3

19√12 = 19√(2×2×3) = 19×2√3 = 38√3

6√3

∵ both surds have same surd-factor i.e., √3.

∴ they are like surds.

v. 5√22, 7√33

∵ both surds have different surd-factors √22 and √33.

∴ they are unlike surds.

vi. 5√5, √75

5√5

√75 = √(5×5×3) = 5√3

∵ both surds have different surd-factors √5 and √3.

∴ they are unlike surds.

Question 4.

Simplify the following surds.

i. √27

ii. √50

iii. √250

iv. √112

v. √168

i.

ii.

iii.

iv.

v.

Question 5.

Compare the following pair of surds.

i. 7√2, 5√3

ii. √247, √274

iii. 2√7, √28

iv. 5√5, 7√2

v. 4√42, 9√2

vi. 5√3, 9

vii. 7, 2√5

i. 7√2 , 5√3

(7√2)2 = 7 × 7 × √2 × √2

⇒ (7√2)2 = 49 × 2

⇒ (7√2)2 = 98

And

(5√3)2 = 5 × 5 × √3 × √3

⇒ (5√3)2 = 25 × 3

⇒ (5√3)2 = 75

Clearly,

98 > 75

∴ 7√2 > 5√3

ii. √247, √274

(√247)2 = 247

And

(√274)2 = 274

Clearly,

247 < 274

∴ √247 < √274

iii. 2√7, √28

(2√7)2 = 2 × 2 × √7 × √7

⇒ (2√7)2 = 4 × 7

⇒ (2√7)2 = 28

And

(√28)2 = 28

Clearly,

28 = 28

∴ 2√7 = √28

iv. 5√5, 7√2

(5√5)2 = 5 × 5 × √5 × √5

⇒ (5√5)2 = 25 × 5

⇒ (5√5)2 = 125

And

(7√2)2 = 7 × 7 × √2 × √2

⇒ (7√2)2 = 49 × 2

⇒ (7√2)2 = 98

Clearly,

125 = 98

∴ 5√5= 7√2

v. 4√42, 9√2

(4√42)2 = 4 × 4 × √42 × √42

⇒ (4√42)2 = 16 × 42

⇒ (4√42)2 = 672

And

(9√2)2 = 9 × 9 × √2 × √2

⇒ (9√2)2 = 81 × 2

⇒ (9√2)2 = 162

Clearly,

672 > 162

∴ 4√42 > 9√2

vi. 5√3, 9

(5√3)2 = 5 × 5 × √3 × √3

⇒ (5√3)2 = 25 × 3

⇒ (5√3)2 = 75

And

(9)2 = 9 × 9

⇒ (9)2 = 81

Clearly,

75 < 81

∴ 5√3 < 9

vii. 7, 2√5

(2√5)2 = 2 × 2 × √5 × √5

⇒ (2√5)2 = 4 × 5

⇒ (2√5)2 = 20

And

(7)2 = 7 × 7

⇒ (7)2 = 49

Clearly,

49 > 20

∴ 7 > 2√5

Question 6.

Simplify.

i. 5√3 + 8√3

ii. 9√5 – 4√5 + √125

iii. 7√48 – √27 – √3

iv.

i. 5√3 + 8√3

5√3 + 8√3 = (5 + 8)√3

⇒ 5√3 + 8√3 = 13√3

ii. 9√5 – 4√5 + √125

9√5 – 4√5 + √125 = 9√5 – 4√5 + √(5 × 5 × 5)

⇒ 9√5 – 4√5 + √125 = 9√5 – 4√5 + √(5 × 5 × 5)

⇒ 9√5 – 4√5 + √125 = 9√5 – 4√5 + 5√5

⇒ 9√5 – 4√5 + √125 = (9 – 4 + 5)√5

⇒ 9√5 – 4√5 + √125 = 10√5

iii. 7√48 – √27 – √3

7√48 – √27 – √3 = 7√(2 × 2 × 2 × 2 × 3) – √(3 × 3 × 3) – √3

⇒ 7√48 – √27 – √3 = 7×4√3 – 3√3 – √3

⇒ 7√48 – √27 – √3 = 28√3 – 3√3 – √3

⇒ 7√48 – √27 – √3 = (28 – 3 – 1)√3

⇒ 7√48 – √27 – √3 = 24√3

iv.

Question 7.

Multiply and write the answer in the simplest form.

i. 3√12 × √18

ii. 3√12 × 7√15

iii. 3√8 × √5

iv. 5√8 × 2√8

i. 3√12 × √18

3√12 × √18 = 3√(2 × 2 × 3) × √(2 × 3 × 3)

⇒3√12 × √18 = 3 × 2√3 × 3√2

⇒ 3√12 × √18 = 6√3 × 3√2

⇒ 3√12 × √18 = 18√6

ii. 3√12 × 7√15

3√12 × 7√15 = 3√(2 × 2 × 3) × 7√(3 × 5)

⇒3√12 × 7√15 = 3 × 2√3 × 7√(3 × 5)

⇒3√12 × 7√15 = 3 × 2 × 7 × √(3 × 3 × 5)

⇒3√12 × 7√15 = 3 × 2 × 7 × 3√5

⇒ 3√12 × 7√15 = 126√5

iii. 3√8 × √5

3√8 × √5 = 3√(2 × 2 × 2) × √5

⇒3√8 × √5 = 3 × 2√2 × √5

⇒ 3√8 × √5 = 3 × 2 ×√(2 × 5)

⇒ 3√8 × √5 = 6√10

iv. 5√8 × 2√8

5√8 × 2√8 = 5√(2 × 2 × 2) × 2√(2 × 2 × 2)

⇒5√8 × 2√8 = 5 × 2√2 × 2 × 2√2

⇒ 5√8 × 2√8 = 5 × 2 × 2 × 2 ×√(2 × 2)

⇒ 5√8 × 2√8 = 5 × 2 × 2 × 2 × 2

⇒ 5√8 × 2√8 = 80

Question 8.

Divide, and write the answer in simplest form.

i. √98 ÷ √2

ii. √125 ÷ √50

iii. √54 ÷ √27

iv. √310 ÷ √5

i. √98 ÷ √2

⇒√98 ÷ √2 = 7

ii. √125 ÷ √50

iii. √54 ÷ √27

⇒√54 ÷ √27 = √2

iv. √310 ÷ √5

⇒ √310 ÷ √5 = √62

Question 9.

Rationalize the denominator.

i.  ii.

iii.  iv.

v.

i. We know that √5 × √5 = 5, ∴ to rationalize the denominator of  multiply both numerator and denominator by √5.

ii. We know that √14 × √14 = 14, ∴ to rationalize the denominator of  multiply both numerator and denominator by √14.

iii. We know that √7 × √7 = 7, ∴ to rationalize the denominator of  multiply both numerator and denominator by √7.

iv. We know that √3 × √3 = 3, ∴ to rationalize the denominator of  multiply both numerator and denominator by √3.

v. We know that √3 × √3 = 3, ∴ to rationalize the denominator of  multiply both numerator and denominator by √3.

Practice Set 2.4
Question 1.

Multiply

i. √3(√7 – √3)

ii. (√5 – √7)√2

iii. (3√2 – √3)(4√3 – √2)

i. √3(√7 – √3)

=√3 × √7 – √3 × √3

[∵√a(√b–√c)=√a×√b–√a×√c]

=√21 – 3

ii. (√5 – √7)√2

=√5 × √2 – √7 × √2

[∵√a(√b–√c)=√a×√b–√a×√c]

=√10 – √14

iii. (3√2 – √3)(4√3 – √2)

=3√2(4√3 – √2) – √3(4√3 – √2)

=3√2×4√3 – 3√2×√2 – √3×4√3 + √3×√2

[∵√a(√b–√c)=√a×√b–√a×√c]

=12√6 – 3×2 – 4×3 + √6

=12√6 – 6 – 12 + √6

=13√6 – 18

Question 2.

Rationalize the denominator.

i.  ii.

iii.  iv.

i. The rationalizing factor of √7 + √2 is √7 – √2. Therefore, multiply both numerator and denominator by √7 – √2.

[∵ (a-b)(a+b) = a2 – b2]

ii. The rationalizing factor of 2√5 – 3√2 is 2√5 + 3√2. Therefore, multiply both numerator and denominator by 2√5 + 3√2.

[∵ (a-b)(a+b) = a2 – b2]

iii. The rationalizing factor of 7 + 4√3 is 7 – 4√3. Therefore, multiply both numerator and denominator by 7 – 4√3.

[∵ (a-b)(a+b) = a2 – b2]

iv. The rationalizing factor of √5 + √3 is √5 - √3. Therefore, multiply both numerator and denominator by √5 - √3.

[∵ (a-b)(a+b) = a2 – b2]

[∵ (a-b)2 = a2 + b2 – 2ab]

Practice Set 2.5
Question 1.

Find the value.
(i) |15 - 2|
(ii) |4 - 9|
(iii) |7| × |-4|

Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies. The absolute value of a number is never negative.

Therefore,

i. |15 - 2| = |13| = 13

ii. |4 - 9| = |-5| = 5

iii. |7| × |-4| = 7 × 4 = 28

Question 2.

Solve.

i. |3x - 5| = 1

ii. |7 – 2x| = 5

iii.

iv.

i. |3x - 5| = 1

⇒ 3x - 5 = 1 or 3x - 5 = -1

⇒ 3x = 1 + 5 or 3x = -1 + 5

⇒ 3x = 6 or 3x = 4

ii. |7 – 2x| = 5

⇒ 7 – 2x = 5 or 7 – 2x = -5

⇒ 2x = 7 - 5 or 2x = 7 + 5

⇒ 2x = 2 or 2x = 12

⇒ x = 1 or x = 6

iii.

⇒ 8 – x = 2 × 5 or 8 – x = 2 × -5

⇒ 8 – x = 10 or 8 – x = -10

⇒ x = 8 – 10 or x = 8 + 10

⇒ x = -2 or x = 18

iv.

⇒ 20 + x = 4 × 5 or 20 + x = 4 × -5

⇒ 20 + x = 20 or 20 + x = -20

⇒ x = 20 – 20 or x = -20 – 20

⇒ x = 0 or x = -40

Problem Set 2
Question 1.

Choose the correct alternative answer for the questions given below.

i. Which one of the following is an irrational number?
A. √16/25

B. √5

C. 3/9

D. √196

An irrational number is a number that cannot be expressed as a fraction for any integers p and q and q ≠ 0.

since it can be written as  , it is a rational number.

since it can be written as  , it is a rational number.

since it can be written as  , it is a rational number.

Since √5 cannot be written as  it is an irrational number

Therefore √5 is an irrational number.

Question 2.

Which of the following is an irrational number?
A. 0.17

B.

C.

D. 0.101001000....

An irrational number is a number that cannot be expressed as a fraction for any integers p and q and q ≠ 0.

.

Since it can be written as  ,

it is a rational number.

is a rational number because it is a non-terminating but repeating decimal.

is a rational number because it is a non-terminating but repeating decimal.

0.101001000.... is an irrational number because it is a non-terminating and non-`repeating decimal.

Therefore, 0.101001000.... is an irrational number.

Question 3.

Decimal expansion of which of the following is non-terminating recurring?
A. 2/5

B. 3/16

C. 3/11

D. 137/25

A non-terminating recurring decimal representation means that the number will have an infinite number of digits to the right of the decimal point and those digits will repeat themselves.

∵ it does not have an infinite number of digits to the right of the decimal point ∴ it is not a non-terminating recurring decimal.

∵ it does not have an infinite number of digits to the right of the decimal point ∴ it is not a non-terminating recurring decimal.

∵ it has an infinite number of digits to the right of the decimal point which are repeating themselves ∴ it is a non-terminating recurring decimal.

∵ it does not have an infinite number of digits to the right of the decimal point ∴ it is not a non-terminating recurring decimal.

Therefore,  is a non-terminating recurring decimal.

Question 4.

Every point on the number line represent, which of the following numbers?
A. Natural numbers

B. Irrational numbers

C. Rational numbers

D. Real numbers.

Every point of a number line is assumed to correspond to a real number, and every real number to a point. Therefore, Every point on the number line represent a real number.

Question 5.

The number 0.4 in p/q form is ………….
A. 4/9

B. 40/9

C. 3.6/9

D. 36/9

∵ the denominator of all the above options is 9 ∴ we multiply both numerator and denominator by 0.9 as 10 × 0.9 = 9

Question 6.

What is √n, if n is not a perfect square number?
A. Natural number

B. Rational number

C. Irrational number

D. Options A, B, C all are correct.

If n is not a perfect square number, then √n cannot be expressed as ratio of a and b where a and b are integers and b ≠ 0

Therefore, √n is an Irrational number

Question 7.

Which of the following is not a surd?
A. √7

B. 3√17

C. 3√64

D. √193

Which is a rational number

Therefore,  is not a surd.

Question 8.

What is the order of the surd ?
A. 3

B. 2

C. 6

D. 5

Therefore, the order of the surd  is 6.

Question 9.

Which one is the conjugate pair of 2√5 + √3?
A. -2√5 + √3

B. -2√5 - √3

C. 2√3 + √5

D. √3 + 2√5

A math conjugate is formed by changing the sign between two terms in a binomial. For instance, the conjugate of x + y is x - y.

Now,

2√5 + √3 = √3 + 2√5

Its conjugate pair = √3 - 2√5 = -2√5 + √3

∴ The conjugate pair of 2√5 + √3 = -2√5 + √3

Question 10.

The value of |12 – (13 + 7) × 4| is ...........
A. -68

B. 68

C. -32

D. 32

|12 – (13 + 7) × 4| = |12 – 20 × 4| (Solving it according to BODMAS)

⇒ |12 – (13 + 7) × 4| = |12 – 80|

⇒ |12 – (13 + 7) × 4| = |-68|

⇒ |12 – (13 + 7) × 4| = 68

Question 11.

Write the following numbers in p/q form.

i. 0.555 ii.

iii. 9.315 315 ... iv. 357.417417...

v.

i.

ii.

Let

⇒ 1000x = 29568.568568......

Now,

1000x - x = 29568.568568 – 29.568568

⇒999x = 29539.0

iii.

Let x = 9.315315…

⇒ 1000x = 9315.315315......

Now,

1000x - x = 9315.315315 – 9.315315

⇒999x = 9306.0

iv.

Let x = 357.417417…

⇒ 1000x = 357417.417417…

Now,

1000x - x = 357417.417417 – 357.417417

⇒999x = 357060.0

v.

Let

⇒ 1000x = 30219.219219…

Now,

1000x - x = 30219.219219 – 30.219219

⇒999x = 30189.0

Question 12.

Write the following numbers in its decimal form.

i. -5/7 ii. 9/11

iii. √5 iv. 121/13

v. 29/8

i.

ii.

iii.

√5 = 2.236067977…….

iv.

v.

Question 13.

Show that 5 + √7 is an irrational number.

Let us assume that 5 + √7 is a rational number

where, b≠0 and a, b are integers

∵ a, b are integers ∴ a – 5b and b are also integers

is rational which cannot be possible ∵ which is an irrational number

∵ it is contradicting our assumption ∴ the assumption was wrong

Hence, 5 + √7 is an irrational number

Question 14.

Write the following surds in simplest form.

i.  ii.

i.

ii.

Question 15.

Write the simplest form of rationalizing factor for the given surds.

i. √32 ii. √50

iii. √27 iv. 3/5√10

v. 3√72 vi. 4√11

i. √32

∴ Its rationalizing factor = √2

ii. √50

∴ Its rationalizing factor = √2

iii. √27

∴ Its rationalizing factor = √3

∵ √10 cannot be further simplified

∴ Its rationalizing factor = √10

v. 3√72

∴ Its rationalizing factor = √2

vi. 4√11

∵ √11 cannot be further simplified

∴ Its rationalizing factor = √11

Question 16.

Simplify.

i.

ii.

iii.

iv.

v.

i.

= 4√3 + 3√3 – √3

= 7√3 – √3

= 6√3

ii.

iii.

iv.

v.

Question 17.

Rationalize the denominator.

i.  ii.

iii.  iv.

v.

i.

ii.

iii.

iv.

v.