ML Aggarwal Rational and Irrational Number Exe-1.5 Class 9 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-1.5 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-1.5 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-1.5 Questions Edition 2022-2023

### Rational and Irrational Number Exe-1.5

ML Aggarwal Class 9 ICSE Maths Solutions

Page 34

#### Question 1. Rationalize the denominator of the following:

(i) 3/4√5

(ii) 5√7 / √3

(iii) 3/(4 – √7)

(iv) 17/(3√2 + 1)

(v) 16/ (√41 – 5)

(vi) 1/ (√7 – √6)

(vii) 1/ (√5 + √2)

(viii) (√2 + √3) / (√2 – √3)

##### (i) 3/4√5

Let us rationalize,

3/4√5 = (3×√5) /(4√5×√5)

= (3√5) / (4×5)

= (3√5) / 20

##### (ii) 5√7 / √3

Let us rationalize,

5√7 / √3 = (5√7×√3) / (√3×√3)

= 5√21/3

##### (iii) 3/(4 – √7)

Let us rationalize,

3/(4 – √7) = [3×(4 + √7)] / [(4 – √7) × (4 + √7)]

= 3(4 + √7) / [42 – (√7)2]

= 3(4 + √7) / [16 – 7]

= 3(4 + √7) / 9

= (4 + √7) / 3

##### (iv) 17/(3√2 + 1)

Let us rationalize,

17/(3√2 + 1) = 17(3√2 – 1) / [(3√2 + 1) (3√2 – 1)]

= 17(3√2 – 1) / [(3√2)2 – 12]

= 17(3√2 – 1) / [9.2 – 1]

= 17(3√2 – 1) / [18 – 1]

= 17(3√2 – 1) / 17

= (3√2 – 1)

##### (v) 16/ (√41 – 5)

Let us rationalize,

16/ (√41 – 5) = 16(√41 + 5) / [(√41 – 5) (√41 + 5)]

= 16(√41 + 5) / [(√41)2 – 52]

= 16(√41 + 5) / [41 – 25]

= 16(√41 + 5) / [16]

= (√41 + 5)

##### (vi) 1/ (√7 – √6)

Let us rationalize,

1/ (√7 – √6) = 1(√7 + √6) / [(√7 – √6) (√7 + √6)]

= (√7 + √6) / [(√7)2 – (√6)2]

= (√7 + √6) / [7 – 6]

= (√7 + √6) / 1

= (√7 + √6)

##### (vii) 1/ (√5 + √2)

Let us rationalize,

1/ (√5 + √2) = 1(√5 – √2) / [(√5 + √2) (√5 – √2)]

= (√5 – √2) / [(√5)2 – (√2)2]

= (√5 – √2) / [5 – 2]

= (√5 – √2) / [3]

= (√5 – √2) /3

##### (viii) (√2 + √3) / (√2 – √3)

Let us rationalize,

(√2 + √3) / (√2 – √3) = [(√2 + √3) (√2 + √3)] / [(√2 – √3) (√2 + √3)]

= [(√2 + √3)2] / [(√2)2 – (√3)2]

= [2 + 3 + 2√2√3] / [2 – 3]

= [5 + 2√6] / -1

= – (5 + 2√6)

#### Question 2. Simplify each of the following by rationalizing the denominator:

(i) (7 + 3√5) / (7 – 3√5)

(ii) (3 – 2√2) / (3 + 2√2)

(iii) (5 – 3√14) / (7 + 2√14)

##### (i) (7 + 3√5) / (7 – 3√5)

Let us rationalize the denominator, we get

(7 + 3√5) / (7 – 3√5) = [(7 + 3√5) (7 + 3√5)] / [(7 – 3√5) (7 + 3√5)]

= [(7 + 3√5)2] / [72 – (3√5)2]

= [72 + (3√5)2 + 2.7. 3√5] / [49 – 9.5]

= [49 + 9.5 + 42√5] / [49 – 45]

= [49 + 45 + 42√5] / [4]

= [94 + 42√5] / 4

= 2[47 + 21√5]/4

= [47 + 21√5]/2

##### (ii) (3 – 2√2) / (3 + 2√2)

Let us rationalize the denominator, we get

(3 – 2√2) / (3 + 2√2) = [(3 – 2√2) (3 – 2√2)] / [(3 + 2√2) (3 – 2√2)]

= [(3 – 2√2)2] / [32 – (2√2)2]

= [32 + (2√2)2 – 2.3.2√2] / [9 – 4.2]

= [9 + 4.2 – 12√2] / [9 – 8]

= [9 + 8 – 12√2] / 1

= 17 – 12√2

##### (iii) (5 – 3√14) / (7 + 2√14)

Let us rationalize the denominator, we get

(5 – 3√14) / (7 + 2√14) = [(5 – 3√14) (7 – 2√14)] / [(7 + 2√14) (7 – 2√14)]

= [5(7 – 2√14) – 3√14 (7 – 2√14)] / [72 – (2√14)2]

= [35 – 10√14 – 21√14 + 6.14] / [49 – 4.14]

= [35 – 31√14 + 84] / [49 – 56]

= [119 – 31√14] / [-7]

= -[119 – 31√14] / 7

= [31√14 – 119] / 7

#### Question 3. Simplify:

[7√3 / (√10 + √3)] – [2√5 / (√6 + √5)] – [3√2 / (√15 + 3√2)]

Let us simplify individually,

[7√3 / (√10 + √3)]

Let us rationalize the denominator,

7√3 / (√10 + √3) = [7√3(√10 – √3)] / [(√10 + √3) (√10 – √3)]

= [7√3.√10 – 7√3.√3] / [(√10)2 – (√3)2]

= [7√30 – 7.3] / [10 – 3]

= 7[√30 – 3] / 7

= √30 – 3

Now,

[2√5 / (√6 + √5)]

Let us rationalize the denominator, we get

2√5 / (√6 + √5) = [2√5 (√6 – √5)] / [(√6 + √5) (√6 – √5)]

= [2√5.√6 – 2√5.√5] / [(√6)2 – (√5)2]

= [2√30 – 2.5] / [6 – 5]

= [2√30 – 10] / 1

= 2√30 – 10

Now,

[3√2 / (√15 + 3√2)]

Let us rationalize the denominator, we get

3√2 / (√15 + 3√2) = [3√2 (√15 – 3√2)] / [(√15 + 3√2) (√15 – 3√2)]

= [3√2.√15 – 3√2.3√2] / [(√15)2 – (3√2)2]

= [3√30 – 9.2] / [15 – 9.2]

= [3√30 – 18] / [15 – 18]

= 3[√30 – 6] / [-3]

= [√30 – 6] / -1

= 6 – √30

So, according to the question let us substitute the obtained values,

[7√3 / (√10 + √3)] – [2√5 / (√6 + √5)] – [3√2 / (√15 + 3√2)]

= (√30 – 3) – (2√30 – 10) – (6 – √30)

= √30 – 3 – 2√30 + 10 – 6 + √30

= 2√30 – 2√30 – 3 + 10 – 6

= 1

#### Question 4. Simplify:

[1/(√4 + √5)] + [1/(√5 + √6)] + [1/(√6 + √7)] + [1/(√7 + √8)] + [1/(√8 + √9)]

Let us simplify individually,

[1/(√4 + √5)]

Rationalize the denominator, we get

[1/(√4 + √5)] = [1(√4 – √5)] / [(√4 + √5) (√4 – √5)]

= [(√4 – √5)] / [(√4)2 – (√5)2]

= [(√4 – √5)] / [4 – 5]

= [(√4 – √5)] / -1

= -(√4 – √5)

Now,

[1/(√5 + √6)]

Rationalize the denominator, we get

[1/(√5 + √6)] = [1(√5 – √6)] / [(√5 + √6) (√5 – √6)]

= [(√5 – √6)] / [(√5)2 – (√6)2]

= [(√5 – √6)] / [5 – 6]

= [(√5 – √6)] / -1

= -(√5 – √6)

Now,

[1/(√6 + √7)]

Rationalize the denominator, we get

[1/(√6 + √7)] = [1(√6 – √7)] / [(√6 + √7) (√6 – √7)]

= [(√6 – √7)] / [(√6)2 – (√7)2]

= [(√6 – √7)] / [6 – 7]

= [(√6 – √7)] / -1

= -(√6 – √7)

Now,

[1/(√7 + √8)]

Rationalize the denominator, we get

[1/(√7 + √8)] = [1(√7 – √8)] / [(√7 + √8) (√7 – √8)]

= [(√7 – √8)] / [(√7)2 – (√8)2]

= [(√7 – √8)] / [7 – 8]

= [(√7 – √8)] / -1

= -(√7 – √8)

Now,

[1/(√8 + √9)]

Rationalize the denominator, we get

[1/(√8 + √9)] = [1(√8 – √9)] / [(√8 + √9) (√8 – √9)]

= [(√8 – √9)] / [(√8)2 – (√9)2]

= [(√8 – √9)] / [8 – 9]

= [(√8 – √9)] / -1

= -(√8 – √9)

So, according to the question let us substitute the obtained values,

[1/(√4 + √5)] + [1/(√5 + √6)] + [1/(√6 + √7)] + [1/(√7 + √8)] + [1/(√8 + √9)]

= -(√4 – √5) + -(√5 – √6) + -(√6 – √7) + -(√7 – √8) + -(√8 – √9)

= -√4 + √5 – √5 + √6 – √6 + √7 – √7 + √8 – √8 + √9

= -√4 + √9

= -2 + 3

= 1

#### Question 5. Give a and b are rational numbers. Find a and b if:

(i) [3 – √5] / [3 + 2√5] = -19/11 + a√5

(ii) [√2 + √3] / [3√2 – 2√3] = a – b√6

(iii) {[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]} = a + 7/11 b√5

##### (i) [3 – √5] / [3 + 2√5] = -19/11 + a√5

Let us consider LHS

[3 – √5] / [3 + 2√5]

Rationalize the denominator,

[3 – √5] / [3 + 2√5] = [(3 – √5) (3 – 2√5)] / [(3 + 2√5) (3 – 2√5)]

= [3(3 – 2√5) – √5(3 – 2√5)] / [32 – (2√5)2]

= [9 – 6√5 – 3√5 + 2.5] / [9 – 4.5]

= [9 – 6√5 – 3√5 + 10] / [9 – 20]

= [19 – 9√5] / -11

= -19/11 + 9√5/11

So when comparing with RHS

-19/11 + 9√5/11 = -19/11 + a√5

Hence, value of a = 9/11

##### (ii) [√2 + √3] / [3√2 – 2√3] = a – b√6

Let us consider LHS

[√2 + √3] / [3√2 – 2√3]

Rationalize the denominator,

[√2 + √3] / [3√2 – 2√3] = [(√2 + √3) (3√2 + 2√3)] / [(3√2 – 2√3) (3√2 + 2√3)]

= [√2(3√2 + 2√3) + √3(3√2 + 2√3)] / [(3√2)2 – (2√3)2]

= [3.2 + 2√2√3 + 3√2√3 + 2.3] / [9.2 – 4.3]

= [6 + 2√6 + 3√6 + 6] / [18 – 12]

= [12 + 5√6] / 6

= 12/6 + 5√6/6

= 2 + 5√6/6

= 2 – (-5√6/6)

So when comparing with RHS

2 – (-5√6/6) = a – b√6

Hence, value of a = 2 and b = -5/6

##### (iii) {[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]} = a + 7/11 b√5

Let us consider LHS

Since there are two terms, let us solve individually

{[7 + √5]/[7 – √5]}

Rationalize the denominator,

[7 + √5]/[7 – √5] = [(7 + √5) (7 + √5)] / [(7 – √5) (7 + √5)]

= [(7 + √5)2] / [72 – (√5)2]

= [72 + (√5)2 + 2.7.√5] / [49 – 5]

= [49 + 5 + 14√5] / [44]

= [54 + 14√5] / 44

Now,

{[7 – √5]/[7 + √5]}

Rationalize the denominator,

[7 – √5]/[7 + √5] = (7 – √5) (7 – √5)] / [(7 + √5) (7 – √5)]

= [(7 – √5)2] / [72 – (√5)2]

= [72 + (√5)2 – 2.7.√5] / [49 – 5]

= [49 + 5 – 14√5] / [44]

= [54 – 14√5] / 44

So, according to the question

{[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]}

By substituting the obtained values,

= {[54 + 14√5] / 44} – {[54 – 14√5] / 44}

= [54 + 14√5 – 54 + 14√5]/44

= 28√5/44

= 7√5/11

So when comparing with RHS

7√5/11 = a + 7/11 b√5

Hence, value of a = 0 and b = 1

#### Question 6. If {[7 + 3√5] / [3 + √5]} – {[7 – 3√5] / [3 – √5]} = p + q√5, find the value of p and q where p and q are rational numbers.

Let us consider LHS

Since there are two terms, let us solve individually

{[7 + 3√5] / [3 + √5]}

Rationalize the denominator,

[7 + 3√5] / [3 + √5] = [(7 + 3√5) (3 – √5)] / [(3 + √5) (3 – √5)]

= [7(3 – √5) + 3√5(3 – √5)] / [32 – (√5)2]

= [21 – 7√5 + 9√5 – 3.5] / [9 – 5]

= [21 + 2√5 – 15] / [4]

= [6 + 2√5] / 4

= 2[3 + √5]/4

= [3 + √5] /2

Now,

{[7 – 3√5] / [3 – √5]}

Rationalize the denominator,

[7 – 3√5] / [3 – √5] = [(7 – 3√5) (3 + √5)] / [(3 – √5) (3 + √5)]

= [7(3 + √5) – 3√5(3 + √5)] / [32 – (√5)2]

= [21 + 7√5 – 9√5 – 3.5] / [9 – 5]

= [21 – 2√5 – 15] / 4

= [6 – 2√5]/4

= 2[3 – √5]/4

= [3 – √5]/2

So, according to the question

{[7 + 3√5] / [3 + √5]} – {[7 – 3√5] / [3 – √5]}

By substituting the obtained values,

= {[3 + √5] /2} – {[3 – √5] /2}

= [3 + √5 – 3 + √5]/2

= [2√5]/2

= √5

So when comparing with RHS

√5 = p + q√5

Hence, value of p = 0 and q = 1

#### Question 7. Rationalise the denominator of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732, upto three places of decimal:

(i) √2/(2 + √2)

(ii) 1/(√3 + √2)

##### (i) √2/(2 + √2)

By rationalizing the denominator,

√2/(2 + √2) = [√2(2 – √2)] / [(2 + √2) (2 – √2)]

= [2√2 – 2] / [22 – (√2)2]

= [2√2 – 2] / [4 – 2]

= 2[√2 – 1] / 2

= √2 – 1

= 1.414 – 1

= 0.414

##### (ii) 1/(√3 + √2)

By rationalizing the denominator,

1/(√3 + √2) = [1(√3 – √2)] / [(√3 + √2) (√3 – √2)]

= [(√3 – √2)] / [(√3)2 – (√2)2]

= [(√3 – √2)] / [3 – 2]

= [(√3 – √2)] / 1

= (√3 – √2)

= 1.732 – 1.414

= 0.318

#### Question 8. If a = 2 + √3, find 1/a, (a – 1/a)

Given:

a = 2 + √3

So,

1/a = 1/ (2 + √3)

By rationalizing the denominator,

1/ (2 + √3) = [1(2 – √3)] / [(2 + √3) (2 – √3)]

= [(2 – √3)] / [22 – (√3)2]

= [(2 – √3)] / [4 – 3]

= (2 – √3)

Then,

a – 1/a = 2 + √3 – (2 – √3)

= 2 + √3 – 2 + √3

= 2√3

#### Question 9. Solve:

If x = 1 – √2, find 1/x, (x – 1/x)4

Given:

x = 1 – √2

so,

1/x = 1/(1 – √2)

By rationalizing the denominator,

1/ (1 – √2) = [1(1 + √2)] / [(1 – √2) (1 + √2)]

= [(1 + √2)] / [12 – (√2)2]

= [(1 + √2)] / [1 – 2]

= (1 + √2) / -1

= -(1 + √2 )

Then,

(x – 1/x)4 = [1 – √2 – (-1 – √2)]4

= [1 – √2 + 1 + √2]4

= 24

= 16

### ML Aggarwal Rational and Irrational Number Exe-1.5 Class 9 ICSE Maths Solutions

Page 35

#### Question 10. Solve:

If x = 5 – 2√6, find 1/x, (x2 – 1/x2)

Given:

x = 5 – 2√6

so,

1/x = 1/(5 – 2√6)

By rationalizing the denominator,

1/(5 – 2√6) = [1(5 + 2√6)] / [(5 – 2√6) (5 + 2√6)]

= [(5 + 2√6)] / [52 – (2√6)2]

= [(5 + 2√6)] / [25 – 4.6]

= [(5 + 2√6)] / [25 – 24]

= (5 + 2√6)

Then,

x + 1/x = 5 – 2√6 + (5 + 2√6)

= 10

Square on both sides we get

(x + 1/x)2 = 102

x2 + 1/x2 + 2x.1/x = 100

x2 + 1/x2 + 2 = 100

x2 + 1/x2 = 100 – 2

= 98

#### Question 11. If p = (2-√5)/(2+√5) and q = (2+√5)/(2-√5), find the values of

(i) p + q

(ii) p – q

(iii) p2 + q2

(iv) p2 – q2

Given:

p = (2-√5)/(2+√5) and q = (2+√5)/(2-√5)

##### (i) p + q
[(2-√5)/(2+√5)] + [(2+√5)/(2-√5)]

So by rationalizing the denominator, we get

= [(2 – √5)2 + (2 + √5)2] / [22 – (√5)2]

= [4 + 5 – 4√5 + 4 + 5 + 4√5] / [4 – 5]

= [18]/-1

= -18

##### (ii) p – q
[(2-√5)/(2+√5)] – [(2+√5)/(2-√5)]

So by rationalizing the denominator, we get

= [(2 – √5)2 – (2 + √5)2] / [22 – (√5)2]

= [4 + 5 – 4√5 – (4 + 5 + 4√5)] / [4 – 5]

= [9 – 4√5 – 9 – 4√5] / -1

= [-8√5]/-1

= 8√5

##### (iii) p2 + q2

We know that (p + q)2 = p2 + q2 + 2pq

So,

p2 + q2 = (p + q)2 – 2pq

pq = [(2-√5)/(2+√5)] × [(2+√5)/(2-√5)]

= 1

p + q = -18

so,

p2 + q2 = (p + q)2 – 2pq

= (-18)2 – 2(1)

= 324 – 2

= 322

##### (iv) p2 – q2

We know that, p2 – q2 = (p + q) (p – q)

So, by substituting the values

p2 – q2 = (p + q) (p – q)

= (-18) (8√5)

= -144√5

#### Question 12. If x = (√2 – 1)/( √2 + 1) and y = (√2 + 1)/( √2 – 1), find the value of x2 + 5xy + y2.

Given:

x = (√2 – 1)/( √2 + 1) and y = (√2 + 1)/( √2 – 1)

x + y = [(√2 – 1)/( √2 + 1)] + [(√2 + 1)/( √2 – 1)]

By rationalizing the denominator,

= [(√2 – 1)2 + (√2 + 1)2] / [(√2)2 – 12]

= [2 + 1 – 2√2 + 2 + 1 + 2√2] / [2 – 1]

= [6] / 1

= 6

xy = [(√2 – 1)/( √2 + 1)] × [(√2 + 1)/( √2 – 1)]

= 1

We know that

x2 + 5xy + y2 = x+ y2 + 2xy + 3xy

It can be written as

= (x + y)2 + 3xy

Substituting the values

= 62 + 3 × 1

So we get

= 36 + 3

= 39

—  : End of ML Aggarwal Rational and Irrational Number Exe-1.5 Class 9 ICSE Maths Solutions :–