ML Aggarwal Solution Class 9 Chapter 1 Rational and Irrational Numbers Exercise 1.4

 Exercise 1.4

Question 1

(i) =45320+45

=9×534×5+45

=95345+45

=353×25+45

=3565+45

=(36+4)5

=(76)5

=5


(ii) =33+227+73

=33+29×3+73

=33+2×93+73

=33+2×33+73

=33+63+73

=(3+6)3+73

=93+73


Multiplying and Dividing by "√3"

=93+73×33

=93+733

By cross multiplying

=3×93+733

=273+733

=(27+7)33

=3433


(iii) 65×25

6×2×55

=12×(5)2

=12×5

=60


(iv) 815÷23

81523=85323

=45


(v) 248+549

Sol :

6×48+9×69

=648+969

=268+369

=164+163

=6[14+13]

=6[3+412] [∵LCM of 4 and 3 is 12]

=7612


(vi) 38+12

Sol :

32×4+12

=324+12

=322+12

=12(32+1)

=12(3+22) [∵LCM of 2 and 1 is 2]

=12(52)

=522

Multiply and Divide by '√2'

=522×22

=5222×2

=522(2)2

=5222

=524


Question 2

Sol :

(i) (5+7)(2+5)

Sol :

=5×2+55+27+75

=10+55+27+7×5

=10+55+27+35


(ii) (5+5)(55)

=(5)2(5)2

=255

=20


(iii) (5+2)2

=(5)2+(2)2+252

=5+2+25×2

=5+2+210

=7+210


(iv) (37)2

=(3)2+(7)2237

=3+723×7

=10221


(v) (2+3)(5+7)

=25+27+35+37

=2×5+2×7+3×5+3×7

=10+14+15+21


(vi) (4+5)(37)

=4347+5357

=4347+5×35×7

=4347+1535


Question 3

Sol :

(i) 8+50+72+98

=4×2+25×2+36×2+49×2

=42+252+362+492

=22+52+62+72

=(2+5+6+7)2

=20×2

=20×1414

=28.28


(ii) 332250+41282018

=316×2225×2+464×2209×2

=3×422×52+4×8220×32

=122102+322602

=(1210+3260)2

=26×2

=26×1.414

=-36.764


Question 4

Sol :

(i) 27+75+108243

9×3+25×3+36×381×3

=33+53+6393

=(3+5+69)3

=(149)3

=5×3

=5×1.732

=8.66


(ii) 512348+675+7108

Sol :

=54×3316×3+625×3+736×3

=5×233×43+6×53+7×63

=103123+303+423

=(1012+30+42)3

=(8212)3

=70×3

=70×1.732

=121.24


Question 5

(i) 49,370,725,165

Sol :

49=49

=23

It is in the form of pqand p, q are integers

Therefore 49 is a rational number

370 is a rational number

725=725

=75

Since 7 is not an Integer

Therefore, 725 is an Irrational number

165=165

=45. Since 5 is not an Integer

Therefore, 165 is an Irrational number


(ii) 249,3200,253,4916

Sol :

249=249

=27. Since 2 is not an Integer

Therefore, 249 is an Irrational number

3200. It is in the form of pq and p, q are integers. So, it is a rational number

253=253

=53 since 3 is not an integer

Therefore, 253 is an irrational number

4916=4916

=74

Therefore, 4916 is a rational number


Question 6

(i) 32

Sol :

Since 2 is an lrrational number

Therefore, 32 will change into non-terminating , non-recurring decimal 


(ii) 25681

Sol :

25681=169

=1.77777¯7

Therefore, 25681 will not change into non-terminating, non - recurring decimal


(iii) 27×16

Sol :

27×16=9×3×4

=4×33

=123

3 is an Irrational number

Therefore, 27×16 will change into non-terminating, non-recurring decimal


(iv) 536

Sol :

536=56

5 is an Irrational number

Therefore, 536 will change into non-terminating ,non - recurring decimal


(iv) 2735

Sol :

Since 35 is an lrrational number

Therefore, 2735 is also an Irrational number

(Note: Product of rational and Irrational number is irrational)


(v) (23)(2+3)

Sol :

=2×2+2323(3)2

=4-3=1

Therefore (23)(2+3) is a rational number


(vi) (3+5)2

Sol :

=(3)2+(5)2+2×3×5

=9+5+65

=14+65

Since 5 is an Irrational number

(3+5)2 is also an Irrational number


(vii) (257)2

Sol :

(25)2(7)2

=425×7

=2825

Therefore, (257)2 is a rational number


(viii) (36)2

Sol :

(3)2+(6)22×3×6

=9+666

=1566

Since 6 is an Irrational number (36)2 is also an irrational . number


Question 7



Question 8

(i) 32

Sol :

Suppose that 32=pq, where p, q are integers , q>0, p and q have no common factors (except 1)

2=[pq]3

p3=2q3...(1)

As 2 divides 2q32 divides p3

2 divides p

Let p=2k, where k is an integer

Substituting this value of 'p' in (1), we get

(2k)3=2q3

8k3=2q3

4k3=q3

As 2 divides 4k32 divides q3

2 divides q

Thus p and q have a common factor '2'

This contradicts that p and q have no common factor (except 1)

Therefore , 32 is an irrational number


(ii) 33

Sol :

Suppose that 33=pq, where p, q are integers, q>0 , p and q have no common factors (except 1)

3=(pq)3

p3=3q3...(1)

As 3 divides 3q33 divides p3

3 divides p

Let p=3k, Where k is an Integer

Substituting this value of  'p' in 0 , We get

(3k)3=3q3

27k3=3q3

9k3=q3

As 3 divides 9K33 divides q3

3 divide q

Thus p and q have a common factor '3' 

This contradicts that p and q have no common factor (except 1)

Therefore 33 is an Irrational number


(iii) 45

Sol :

Suppose that 45=pq, where p, q are integers , q>0, p and q have no common factors (except 1)

5=(pq)4

p4=5q4...(1)

As 5 divides 5q45 divides p4

5 divides p

Let p=5k , where k is an integer

Substituting this value of 'p' in (1) , we get

(5k)4=5q4

625k4=5q4

125k4=q4

As 5 divides 125 k45 divides q4

5 divides q

Thus p and q have a common factor "5" 

This contradicts that p and q have no common factors (except 1)

Therefore, 45 is an Irrational number.


Question 9

(i) 23,32,7,15

Sol :

4×3=12

32=92

45

12,45,7,15

7<45<12<15

The greatest real number is 15

The smallest real number is 7


(ii) 32,95,4,435,323

Sol :

32=9×2

=18

95=815

=16.2

4=16

435=16×59

=809

=8.89

323=9×34

=274

=6.75

18,16.2,8.89,16,6.75

32<4<323<435<95

The greatest real number is 95

The smallest real number is 32


Question 10

(i) 32,23,15,4

Sol :

White all the numbers as square roots under one radical

32=9×2=18

23=4×3=12

15=15

4=16

Since 12<15<16<18

12<15<16<18

23<15<4<32

Hence, the given numbers in ascending orders are

23,15,4,32


(ii) 32,28,4,50,43

Sol :

Write an the numbers as square roots under one radical

32=9×2=18

28=4×8=32

4=16

50=50

43=16×3=48

Since 16<18<32<48<50

16<18<32<48<50

4<32<28<43<50


Question 11

(i) 92,325,43,365

Sol :

Write all the numbers as square roots under one radical

92=812=40.5

325=94×5=454=1125

43=16×3=48

365=9×65=545=10.8

Since 48>40.5>11.25>10.8

48>40.5>11.25>10.8

43>92>325>365

Hence, the given numbers in descending orders are 43,92,325,365


(ii) 53,732,3,35,27

Sol :

Write all the numbers as Square roots under one radical

53=253=8.33

732=499×2

=989=10.89

3=3

35=9×5=45

27=4×7=28

Since 45>28>10.89>8.33 >-3

45×28>10.89>8.33>3

35>27>732>53>3

Hence, the given numbers in descending orders are 35,27,732,53,3


Question 12

(i) 32,3,65

Sol :

L.C.M of 2,3 and 6 is 6

32=213=(22)16=(4)16

3=312=(33)16=(27)16

65=516=(51)16=(5)16

Since 4<5<27

(4)16<(5)16<(27)16

32<65<3

Hence, the given numbers in ascending orders are 32,65,3

Comments

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2