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SChand CLASS 9 Chapter 1 Rational number and Irrational number Exercise 1(C)

 Exercise 1(C)

Question 1

(i) 13

(ii) 512

(iii) 14675

Sol :

(i) 13

Let be 13

1×33×3 (Multiplying and dividing by 3)

=33


(ii) 512=52×2×3=523

Multiplying and dividing by 3

53233532×3

536

156=1615


(iii) 14675

=1×75+4675=12175

=12175=11×113×5×5

11353×3=1135×3

=11153


Question 2

11263+22428

Sol :

11263+22428

2×2×2×2×73×3+7+2242×2×7
2×2737+22427
4737+1127
4737+11277×7

4737+11277

4737+167

(43+16)7=177


Question 3

4181287532+923

Sol :

4181287532+923

=42×3×32×2×383×5×52×2×2×2×2+923

=4×32238×532×22+923

=6231032+923

=62×33×3103×22×2+92×33×3

(Rationalising denominator of each)

=6631062+963

=2656+36

=5656=0


Question 4

(i) 143

(ii) 25+3

(iii) 1253

(iv) 3+232

Sol :

(i) 143×4+34+3
=4+3(4)2(3)2

=4+3163

=4+313



(ii) 25+3=2(53)(5+3)(53)
Rationalising factor of (5+3) is
=2(53)53=25+232


(iii) 1253=1(25+3)(253)(25+3)
Rationalising factor of (253) is (25+3)
=25+3(25)2(3)2
=25+3203
=25+317


(iv) 3+232
=(3+2)(3+2)(32)(3+2)

Rationalising factor of (32) is (3+2)

=(3+2)2(3)2(2)2
=3+2+23×232
=5+261
=5+26

Question 5

Rationalise the denominator and simplify :

(i) 4+545+454+5

(ii) 353+25+3

Sol :

(i) 4+545+454+5
=(4+5)(4+55)(45)(4+5)+(45)(45)(4+5)(45)

Rationalising the denominator
=(4+5)2(4)2(5)2+(45)2(4)2(5)2
=16+5+2×45165+16+52×45165
=21+8511+218511
=21+85+218511
=4211


(ii) 353+25+3

=3(5+3)(53)(5+3)+2(53)(5+3)(53)
Rationalising the denominator 
=15+33253+1023253
=15+3322+102322
=15+33+102322
=25+322

Question 6

Find the values of a and b if

(i) 3+232=a=a+b2

(ii) 5+237+43=a+b3

(iii) 717+17+171=a+b7

Sol :

(i) 3+232=a=a+b2

=(3+2)(3+2)(32)(3+3)=a+b2
=(3+2)292=a+b2

=9+2+627=a+b9

=11+627=a+b2

=117+672=a+b2


comparing

a=117,b=67



(ii) 5+237+43=a+b3

=(5+23)(743)(7+43)(743)=a+b3


(Rationalising the denominator)


=35203+1438×3(7)2(43)2=a+b3

=3524634948=a+b3

=11631=a+b3

=1163=a+b3

we get a=11, b=-6



(iii) 717+17+171=a+b7

=(71)2(7+1)2(7+1)(71)=a+b7

(7+127)(7+1+27)71=a+b7

=8278226=a+b7

=476=a+b7

237=a+b7


we get
a=0 , b=23

Question 7

Rationalise the denominator of 12+3+5

Sol :

12+3+5=1(2+3)+5

Rationalising denominator

=(2+3)(5)[(2+3)+5)(2+3)5]

=2+35(2+3)2(5)2=2+352+3+265

=2+3526

Rationalising

=(2+35)62×6×6

=12+18302×6

=4×3+9×23012

=23+323012



Question 8

Taking √2=1.414 and √3=1.732 , find without using tables or long division, the value of 

(a) 132

(b) 232

Sol :

(i) 2=1.414,3=1.732
132=1(3+2)(32)(3+2)

Rationalising the denominator

=3+2(3)2(52)2
=3+2(3)2(2)2
=3+292
=3+27
=3+1.4147
=4.4147
=0.631

(ii) 232=2(3+2)(32)(3+2)
Rationalising denominator
=2(3+2)(3)2(2)2
=2(3+2)32
=2(3+2)
=2(1.732+1.414)
=2×3.146
=6.292

Question 9

Express 3553+5 in the form  (a5b) where a and b are simple fractions.

Sol :

3553+5=(355)(325)(3+25)(335)
=965155+10×5(3)2(25)2
=9215+20920

=5921511

=5911+21115

=It is in the form of a5b


Question 10

Prove that 121+23+1=2+3

Sol :

121+23+1=2+3


L.H.S=121+23+1
=2+1(21)(2+1)+2(31)(3+1)(31)
=2+1(2)2(1)2+2(31)(3)2(1)2
=2+121+2(31)31
=2+11+2(31)2
=2+1+31
=2+3
=R.H.S

L.H.S=R.H.S

Question 11

Simplify: 

623+6436+2+262+3

Sol :

=623+6436+2+262+3

=62(36)(3+6)(36)43(62)(6+2)(62)+26(23)(2+3)(23)

Rationalising the denominator 
=62(36)(3)2(6)243(62)(6)2(2)2+26(23)(2)2(3)2

=62(36)3643(62)62+26(23)23

=62(356)343(62)4+26(23)1

=22(36)3(62)26(23)

=26+21218+6212+218

=26+2×2332+62×23+2×3+2×32

=26+4332+643+62

=26+6+434332+62

=6+32

=326

=9×26

=186


Question 12

Simplify:

(i) 6236+63+24362

(ii) 7310+3256+53215+32

Sol :

(i) 6236+63+24362
6(23+6)(236)(23+6)
=6(23+6)(23)2(6)2
=6(23+6)126
=6(23+6)6
=23+6...(i)

63+2
=6(32)(3+2)(32)
=6(32)32
=18121
=3223...(ii)

4362
=43(6+2)(62)(6+2)
=43(6+2)62
=43(6+2)4

=3(6+12)
=18+6
=32+6...(iii)

From (i) , (ii) , (iii)

=(23+6)+(3223)(32+6)
=23+6+3223326
=0


(ii) 7310+3256+53215+32

=7310+3
=73(103)103

=73(103)103

=73(103)7

=3(103)

=303...(i)



256+5=25(65)(6+5)(65)

=25(65)65

=25(65)1

=2302×5

=23010...(ii)



=3215+32

=32(1532)(15+32)(1532)

=32(1532)1518

=32(1532)3

=2(1532)

=30+3×2

=30+6....(iii)


From (i), (ii) and (iii)

=(303)(23010)(30+6)

=303230+10+306

=10-9
=1

Question 13

If x=2+√3 , find the value of x2+1x2

Sol :

⇒x=2+√3
1x=12+3

=23(2+3)(23)

=2343

=231

=23


x+1x=2+3+23

=4


Squaring both side

=(x+1x)2=(4)2

=x2+1x2+2=16
=x2+1x2=162
=14


Question 14

If x=2+1 , find the value of x2+1x2

Sol :

⇒x=√x+1
1x=1x+1
=21(2+1)(21)
=21(2)212)
=21(21)
=√2-1

x+1x=2+1+21

Squaring both sides
[x+1x]2=(22)2

=x2+1x2+2
=4×2=8

=x2+1x=82
=6


Question 15

If x=5212 , find the value of

(i) x+1x and (ii) x2+1x2

Sol :

(i) x=5212

=1x=2521

=2(5+21)(521)(5+21)

=2(5+21)2521
=2(5+21)4
=5+212


x+1x=5212+5+212

=521+5+212
=102
=5
=x+1x=5

Squaring both sides

(x+1x)2=(5)2

=x2+1x2+2

=25


=x2+1x2

=25-2

=23



x+1x=5

x2+1x2=23

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