ML Aggarwal Rational and Irrational Number Exe-1.4 Class 9 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-1.4 Questions for Rational and Irrational Number as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Rational and Irrational Number Exe-1.4 Class 9 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 9th Chapter-1 Rational and Irrational Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-1.4 Questions Edition 2022-2023

### Rational and Irrational Number Exe-1.4

ML Aggarwal Class 9 ICSE Maths Solutions

Page 28

#### Question 1. Simplify the following:

(i) √45 – 3√20 + 4√5

(ii) 3√3 + 2√27 + 7/√3

(iii) 6√5 × 2√5

(iv) 8√15 ÷ 2√3

(v) √24/8 + √54/9

(vi) 3/√8 + 1/√2

(i) √45 – 3√20 + 4√5

Let us simplify the expression,

√45 – 3√20 + 4√5

= √(9×5) – 3√(4×5) + 4√5

= 3√5 – 3×2√5 + 4√5

= 3√5 – 6√5 + 4√5

= √5

(ii) 3√3 + 2√27 + 7/√3

Let us simplify the expression,

3√3 + 2√27 + 7/√3

= 3√3 + 2√(9×3) + 7√3/(√3×√3) (by rationalizing)

= 3√3 + (2×3)√3 + 7√3/3

= 3√3 + 6√3 + (7/3) √3

= √3 (3 + 6 + 7/3)

= √3 (9 + 7/3)

= √3 (27+7)/3

= 34/3 √3

(iii) 6√5 × 2√5

Let us simplify the expression,

6√5 × 2√5

= 12 × 5

= 60

(iv) 8√15 ÷ 2√3

Let us simplify the expression,

8√15 ÷ 2√3

= (8 √5 √3) / 2√3

= 4√5

(v) √24/8 + √54/9

Let us simplify the expression,

√24/8 + √54/9

= √(4×6)/8 + √(9×6)/9

= 2√6/8 + 3√6/9

= √6/4 + √6/3

By taking LCM

= (3√6 + 4√6)/12

= 7√6/12

(vi) 3/√8 + 1/√2

Let us simplify the expression,

3/√8 + 1/√2

= 3/2√2 + 1/√2

By taking LCM

= (3 + 2)/(2√2)

= 5/(2√2)

By rationalizing,

= 5√2/(2√2 × 2√2)

= 5√2/(2×2)

= 5√2/4

#### Question 2. Simplify the following:

(i) (5 + √7) (2 + √5)

(ii) (5 + √5) (5 – √5)

(iii) (√5 + √2)2

(iv) (√3 – √7)2

(v) (√2 + √3) (√5 + √7)

(vi) (4 + √5) (√3 – √7)

(i) (5 + √7) (2 + √5)

Let us simplify the expression,

= 5(2 + √5) + √7(2 + √5)

= 10 + 5√5 + 2√7 + √35

(ii) (5 + √5) (5 – √5)

Let us simplify the expression,

By using the formula,

(a)2 – (b)2 = (a + b) (a – b)

So,

= (5)2 – (√5)2

= 25 – 5

= 20

(iii) (√5 + √2)2

Let us simplify the expression,

By using the formula,

(a + b)2 = a2 + b2 + 2ab

(√5 + √2)2 = (√5)2 + (√2)2 + 2√5√2

= 5 + 2 + 2√10

= 7 + 2√10

(iv) (√3 – √7)2

Let us simplify the expression,

By using the formula,

(a – b)2 = a2 + b2 – 2ab

(√3 – √7)2 = (√3)2 + (√7)2 – 2√3√7

= 3 + 7 – 2√21

= 10 – 2√21

(v) (√2 + √3) (√5 + √7)

Let us simplify the expression,

= √2(√5 + √7) + √3(√5 + √7)

= √2×√5 + √2×√7 + √3×√5 + √3×√7

√10 + √14 + √15 + √21

(vi) (4 + √5) (√3 – √7)

Let us simplify the expression,

= 4(√3 – √7) + √5(√3 – √7)

= 4√3 – 4√7 + √15 – √35

#### Question 3. If √2 = 1.414, then find the value of

(i) √8 + √50 + √72 + √98

(ii) 3√32 – 2√50 + 4√128 – 20√18

(i) √8 + √50 + √72 + √98

Let us simplify the expression,

√8 + √50 + √72 + √98

= √(2×4) + √(2×25) + √(2×36) + √(2×49)

= √2 √4 + √2 √25 + √2 √36 + √2 √49

= 2√2 + 5√2 + 6√2 + 7√2

= 20√2

= 20 × 1.414

= 28.28

(ii) 3√32 – 2√50 + 4√128 – 20√18

Let us simplify the expression,

3√32 – 2√50 + 4√128 – 20√18

= 3√(16×2) – 2√(25×2) + 4√(64×2) – 20√(9×2)

= 3√16 √2 – 2√25 √2 + 4√64 √2 – 20√9 √2

= 3.4√2 – 2.5√2 + 4.8√2 – 20.3√2

= 12√2 – 10√2 + 32√2 – 60√2

= (12 – 10 + 32 – 60) √2

= -26√2

= -26 × 1.414

= -36.764

#### Question 4. If √3 = 1.732, then find the value of

(i) √27 + √75 + √108 – √243

(ii) 5√12 – 3√48 + 6√75 + 7√108

(i) √27 + √75 + √108 – √243

Let us simplify the expression,

√27 + √75 + √108 – √243

= √(9×3) + √(25×3) + √(36×3) – √(81×3)

= √9 √3 + √25 √3 + √36 √3 – √81 √3

= 3√3 + 5√3 + 6√3 – 9√3

= (3 + 5 + 6 – 9) √3

= 5√3

= 5 × 1.732

= 8.660

(ii) 5√12 – 3√48 + 6√75 + 7√108

Let us simplify the expression,

5√12 – 3√48 + 6√75 + 7√108

= 5√(4×3) – 3√(16×3) + 6√(25×3) + 7√(36×3)

= 5√4 √3 – 3√16 √3 + 6√25 √3 + 7√36 √3

= 5.2√3 – 3.4√3 + 6.5√3 + 7.6√3

= 10√3 – 12√3 + 30√3 + 42√3

= (10 – 12 + 30 + 42) √3

= 70√3

= 70 × 1.732

= 121.24

### ML Aggarwal Rational and Irrational Number Exe-1.4 Class 9 ICSE Maths Solutions

Page-29

Question 5. State which of the following are rational or irrational decimals.

(i) √(4/9), -3/70, √(7/25), √(16/5)

(ii) -√(2/49), 3/200, √(25/3), -√(49/16)

(i) √(4/9), -3/70, √(7/25), √(16/5)

√(4/9) = 2/3

-3/70 = -3/70

√(7/25) = √7/5

√(16/5) = 4/√5

So,

√7/5 and 4/√5 are irrational decimals.

2/3 and -3/70 are rational decimals.

(ii) -√(2/49), 3/200, √(25/3), -√(49/16)

-√(2/49) = -√2/7

3/200 = 3/200

√(25/3) = 5/√3

-√(49/16) = -7/4

So,

-√2/7 and 5/√3 are irrational decimals.

3/200 and -7/4 are rational decimals.

#### Question 6. State which of the following are rational or irrational decimals.

(i) -3√2

(ii) √(256/81)

(iii) √(27×16)

(iv) √(5/36)

(i) -3√2

We know that √2 is an irrational number.

So, -3√2 will also be irrational number.

(ii) √(256/81)

√(256/81) = 16/9 = 4/3

It is a rational number.

(iii) √(27×16)

√(27×16) = √(9×3×16) = 3×4√3 = 12√3

It is an irrational number.

(iv) √(5/36)

√(5/36) = √5/6

It is an irrational number.

#### Question 7. State which of the following are irrational numbers.

(i) 3 – √(7/25)

(ii) -2/3 + ∛2

(iii) 3/√3

(iv) -2/7 ∛5

(v) (2 – √3) (2 + √3)

(vi) (3 + √5)2

(vii) (2/5 √7)2

(viii) (3 – √6)2

(i) 3 – √(7/25)

Let us simplify,

3 – √(7/25) = 3 – √7/√25

= 3 – √7/5

Hence, 3 – √7/5 is an irrational number.

(ii) -2/3 + ∛2

Let us simplify,

-2/3 + ∛2 = -2/3 + 21/3

Since, 2 is not a perfect cube.

Hence it is an irrational number.

(iii) 3/√3

Let us simplify,

By rationalizing, we get

3/√3 = 3√3 /(√3×√3)

= 3√3/3

= √3

Hence, 3/√3 is an irrational number.

(iv) -2/7 ∛5

Let us simplify,

-2/7 ∛5 = -2/7 (5)1/3

Since, 5 is not a perfect cube.

Hence it is an irrational number.

(v) (2 – √3) (2 + √3)

Let us simplify,

By using the formula,

(a + b) (a – b) = (a)2 (b)2

(2 – √3) (2 + √3) = (2)2 – (√3)2

= 4 – 3

= 1

Hence, it is a rational number.

(vi) (3 + √5)2

Let us simplify,

By using (a + b)2 = a2 + b2 + 2ab

(3 + √5)2 = 32 + (√5)2 + 2.3.√5

= 9 + 5 + 6√5

= 14 + 6√5

Hence, it is an irrational number.

(vii) (2/5 √7)2

Let us simplify,

(2/5 √7)2 = (2/5 √7) × (2/5 √7)

= 4/ 25 × 7

= 28/25

Hence it is a rational number.

(viii) (3 – √6)2

Let us simplify,

By using (a – b)2 = a2 + b2 – 2ab

(3 – √6)2 = 32 + (√6)2 – 2.3.√6

= 9 + 6 – 6√6

= 15 – 6√6

Hence it is an irrational number.

#### Question 8. Prove the following are irrational numbers.

(i) ∛2

(ii) ∛3

(iii) ∜5

(i) ∛2

We know that ∛2 = 21/3

Let us consider 21/3 = p/q, where p, q are integers, q>0.

p and q have no common factors (except 1).

So,

21/3 = p/q

2 = p3/q3

p3 = 2q3 ….. (1)

We know that, 2 divides 2q3 then 2 divides p3

So, 2 divides p

Now, let us consider p = 2k, where k is an integer

Substitute the value of p in (1), we get

p3 = 2q3

(2k)3 = 2q3

8k3 = 2q3

4k3 = q3

We know that, 2 divides 4k3 then 2 divides q3

So, 2 divides q

Thus p and q have a common factor ‘2’.

This contradicts the statement, p and q have no common factor (except 1).

Hence, ∛2 is an irrational number.

(ii) ∛3

We know that ∛3 = 31/3

Let us consider 31/3 = p/q, where p, q are integers, q>0.

p and q have no common factors (except 1).

So,

31/3 = p/q

3 = p3/q3

p3 = 3q3 ….. (1)

We know that, 3 divides 3q3 then 3 divides p3

So, 3 divides p

Now, let us consider p = 3k, where k is an integer

Substitute the value of p in (1), we get

p3 = 3q3

(3k)3 = 3q3

9k3 = 3q3

3k3 = q3

We know that, 3 divides 9k3 then 3 divides q3

So, 3 divides q

Thus p and q have a common factor ‘3’.

This contradicts the statement, p and q have no common factor (except 1).

Hence, ∛3 is an irrational number.

(iii) ∜5

We know that ∜5 = 51/4

Let us consider 51/4 = p/q, where p, q are integers, q>0.

p and q have no common factors (except 1).

So,

51/4 = p/q

5 = p4/q4

P4 = 5q4 ….. (1)

We know that, 5 divides 5q4 then 5 divides p4

So, 5 divides p

Now, let us consider p = 5k, where k is an integer

Substitute the value of p in (1), we get

P4 = 5q4

(5k)4 = 5q4

625k4 = 5q4

125k4 = q4

We know that, 5 divides 125k4 then 5 divides q4

So, 5 divides q

Thus p and q have a common factor ‘5’.

This contradicts the statement, p and q have no common factor (except 1).

Hence, ∜5 is an irrational number.

#### Question 9. Find the greatest and the smallest real numbers.

(i) 2√3, 3/√2, -√7, √15

(ii) -3√2, 9/√5, -4, 4/3 √5, 3/2√3

(i) 2√3, 3/√2, -√7, √15

Let us simplify each fraction

2√3 = √(4×3) = √12

3/√2 = (3×√2)/(√2×√2) = 3√2/2 = √((9/4)×2) = √(9/2) = √4.5

-√7 = -√7

√15 = √15

So,

The greatest real number = √15

Smallest real number = -√7

(ii) -3√2, 9/√5, -4, 4/3 √5, 3/2√3

Let us simplify each fraction

-3√2 = -√(9×2) = -√18

9/√5 = (9×√5)/(√5×√5) = 9√5/5 = √((81/25)×5) = √(81/5) = √16.2

-4 = -√16

4/3 √5 = √((16/9)×5) = √(80/9) = √8.88 = √8.8

3/2√3 = √((9/4)×3) = √(27/4) = √6.25

So,

The greatest real number = 9√5

Smallest real number = -3√2

#### Question 10. Write in ascending order.

(i) 3√2, 2√3, √15, 4

(ii) 3√2, 2√8, 4, √50, 4√3

(i) 3√2, 2√3, √15, 4

3√2 = √(9×2) =√18

2√3 = √(4×3) =√12

√15 = √15

4 = √16

Now, let us arrange in ascending order

√12, √15, √16, √18

So,

2√3, √15, 4, 3√2

(ii) 3√2, 2√8, 4, √50, 4√3

3√2 = √(9×2) =√18

2√8 = √(4×8) =√32

4 = √16

√50 = √50

4√3 =√(16×3) = √48

Now, let us arrange in ascending order

√16, √18, √32, √48, √50

So,

4, 3√2, 2√8, 4√3, √50

### Rational and Irrational Number Exe-1.4

ML Aggarwal Class 9 ICSE Maths Solutions

#### Question 11. Write in descending order.

(i) 9/√2, 3/2 √5, 4√3, 3√(6/5)

(ii) 5/√3, 7/3 √2, -√3, 3√5, 2√7

(i) 9/√2, 3/2 √5, 4√3, 3√(6/5)

9/√2 = (9×√2)/(√2×√2) = 9√2/2 = √((81/4)×2) = √(81/2) = √40.5

3/2 √5 = √((9/4)×5) = √(45/4) = √11.25

4√3 = √(16×3) = √48

3√(6/5) = √((9×6)/5) = √(54/5) = √10.8

Now, let us arrange in descending order

√48, √40.5, √11.25, √10.8

So,

4√3, 9/√2, 3/2 √5, 3√(6/5)

(ii) 5/√3, 7/3 √2, -√3, 3√5, 2√7

5/√3 = √(25/3) = √8.33

7/3 √2 = √((49/9) ×2) = √98/9 = √10.88

-√3 = -√3

3√5 = √(9×5) =√45

2√7 = √(4×7) = √28

Now, let us arrange in descending order

√45, √28, √10.88.., √8.33.., -√3

So,

3√5, 2√7, 7/3√2, 5/√3, -√3

#### Question 12. Arrange in ascending order.

∛2, √3, 6√5

Here we can express the given expressions as:

∛2 = 21/3

√3 = 31/2

6√5 = 51/6

Let us make the roots common so,

21/3= 2(2× 1/2 × 1/3) = 41/6

31/2 = 3(3× 1/3 × 1/2) = 271/6

51/6 = 51/6

Now, let us arrange in ascending order,

41/6, 51/6, 271/6

So,

21/3, 51/6, 31/2

So,

∛2, 6√5, √3

—  : End of ML Aggarwal Rational and Irrational Number Exe-1.4 Class 9 ICSE Maths Solutions :–

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