S.chand books class 8 maths solution chapter 1 Rational Numbers exercise 1 B

Exercise 1 B


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Question 1

Answer True (T) or False (F)
(i) Subtraction is commutative for rational numbers. F

(ii) To subtract cd from ab , we add the additive inverse of ab to cd . F

(iii) 0 is its own additive inverse . T

(iv) The additive inverse of 2130 is 2130 . T

(v) While subtracting three or more rational numbers , they can be grouped in any order . F


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Question 2

Add the following rational numbers
(i) 49 and 29

Sol :

49+29

69=23


(ii) 1423 and 923

Sol :

1423+923

14+923=523


(iii) 910 and 34

Sol :

L.C.M of 4 and 10 is = 2×2×5=20

24,1022,551,51,1

9×2+3×520  

18+1520

320


(iv) 512 and 79

Sol :

512+(7)9

L.C.M of 12 and 9 is = 2×2×3×3=36

212,926,933,931,31,1

5×3+(7)×436

152836

4336


(v) 710 and 815

Sol :

710+(815)

L.C.M of 10 and 15 is = 2×3×5=30

210,1535,1555,51,1

7×38×230

211630

530=16


(vi) 7311 and (8)

Sol :

7×11+311+81

801181

808×1111

808811

16811=15311


(vii) 413 and 257

Sol :

133+197

L.C.M of 3 and 7 is 21

13×7+19×321

91+5721

14821=7121


(viii) 356 and 512

Sol :

236+512

L.C.M of 6 and 12 is =2×2×3=12

26,1223,633,31,1

23×2+5×112  

46+512

5112=174=414


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Question 3

Simplify:
(i) 14+67+314

Sol :

1467+314

L.C.M of 4 , 7 and 14 is =2×2×3 =28

24,7,1422,7,771,7,71,1,1


7×16×4+3×228

724+628

132428

1128


(ii) 715+925+310

Sol :

715925+310

L.C.M of 15 , 25 and 10 is =2×3×5×5 =150

215,25,10315,25,555,25,551,5,11,1,1

7×109×63×15150

705445150

7099150

29150


(iii) 113+29+56

Sol :

Denominator can not be negative

(3×1+13)+(29×11)56

432956

L.C.M of 3 , 9 and 6 is =2×3×3 =18

23,9,633,9,331,3,11,1,1

4×62×25×318 

2441518

241918

518


(iv) 415+(5310)+112

Sol :

2155310+32

L.C.M of 5 , 10 and 2 is =2×5 =10


25,10,255,5,11,1,1


21×253×1+3×510

4253+1510

575310

410=25


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Question 4

Verify the following.
(i) 38+23=23+38

Sol :

They are equal to  each other by commutative property

ALTERNATE METHOD

L.C.M of 8 and 3 is =2×2×2×3 =24


28,324,322,331,31,1

3×32×824=2×8+3×324

91624=16+924

724=724


(ii) 711+1522=1522+711

Sol :

They are equal to  each other by commutative property

ab+cd=cd+ab             (by commutative property)

                                                                              ALTERNATE METHOD

7111522=1522711

[denominator cannot be negative]

LCM of 11 and 22 is 22.

7×215×122=15×17×222

141522=151422

2922=2922

 


(iii) 8+1112=1112+(8)

Sol :

They are equal to  each other by commutative property

ALTERNATE METHOD

81112=11128

8+1112=11128

8×12+11×112=11×18×1212

96+1112=96+1112

10712=10712

 


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Question 5

Verify that:
(i) (58+98)+138=58+(98+138)

Sol :

They are equal to  each other by associative property

ALTERNATE METHOD

(5+98)+138=58+(9+138)

48+138=58+228

178=178

 


(ii) 20+(35+710)=(20+35)+710

Sol :

They are equal to  each other by associative property

ALTERNATE METHOD

(denominator can not be negative)

20+(35+710)=(2035)+710

[L.C.M of 5 and 10 is 10]

20+(3×2+7×110)=(20×53×15)+710

20+(6+710)=(10035)+710

20+110=1035+710

[L.C.M of 5 and 10 is 10]

(20×10+1×110)=(103×2+7×110)

200+110=208+710

19910=19910

 


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Question 6

Find the additive inverse of each of the following.
(i) 1011

Sol :

(1011)


(ii) 0

Sol :

0=0


(iii) 218

Sol :

(218)


(iv) 118

Sol :

(118)

(118)=118


(v) 2613

Sol :

(2613)=2613


(vi) 1839

Sol :

(1839)=1839


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Question 7

Arrange and simplify:
(i) 12+35+32

Sol :

12+(35)+32

=12+3235

=1+3235

=4235

[L.C.M of 2 and 5 is 10]

=4×53×210

=20610

=1410=75=125

 


(ii) 2817+3517+1617+2317

Sol :

=28+35162317

=633917

=2417=1717

 


(iii) 23+35+16+815

Sol :

23+(35)+16+(815)

=23+1635815

L.C.M of 3 and 6 is 6

L.C.M of 5 and 15 is 15

=2×2+1×16+3×38×115

=4+16+9815

L.C.M of 15 and 6 is 30

=561715


26,1533,1551,51,1

=5×517×230

=253430

=930 or310

 


(iv) 35+53+115+23

Sol :

35+53+115+23

=35115+5323

=3115+523

=85+33

L.C.M of 5 and 3 is 15

=8×3+3×515

=24+1515

=915 or35

 


(v) 43+35+23+115

Sol :

43+35+23+115

=4323+35115

=423+3115

=23+85

L.C.M of 5 and 3 is 15

=2×58×315

=1415

 

(vi) 49+53+45+79+23+95

Sol :

49+53+45+79+23+95

=49+79+53+2345+95

=4+79+523+4+95

=119+33+55

L.C.M of 9,3,5 is 45

=11×5+3×15+5×945

=55+45+4545

=14545 or299=329

 


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Question 8

Verify that -(-x)=x , when x= (i)76

(ii)89

Sol :

(i) 76

(76)=76

76=76

 

(ii) 89

[(89)]=89

[+89]=89

89=89

 


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Question 9

Verify that -(x+y)=(-x)+(-y) , when (i)x=34,y=67 (ii)x=34,y=67

Sol :

(i) x=34,y=87

(34+67)=(34)+(67)

(3×7+6×428)=3467

(21+2428)=(3×76×428)

4528=212428

4528=4528

 

(ii) x=34,y=67

-(x+y)=(-x)+(-y)

[34+67]=[(34)]+[(67)]

[3×76×428]=34+67

[212428]=3×7+6×428

[4528]=21+2428

4528=4528

 


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Question 10

Subtract:
(i) 29 from 79

Sol :

=7929

=729

=59


(ii) 89 from 56

Sol :

=5689

LCM of 6,9 is 18

212,926,933,931,31,1

=5×38×218

=151618

=3118

 

(iii) 811 from 322

Sol:

=322811

=322+811



222,111111,111,1

3×1+8×222

=3+1622

=1922


(iv) 34 from 45

Sol :

=45(34×11)

=4534

=45+34

LCM of 4,5 is 20

=4×4+3×520

=16+1520

=3120

 


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Question 11

The sum of two rational numbers is 1960 . If one of the numbers is 712 , find the other .

Sol :

Sum of two rational numbers is 1960

One of the number is 712 and Let the other be x

712+x=1960

x=1960+712 (on transposing )

LCM of 60 and 12 is 60

=19×1+7×560

=19+3560=1660 or 415

 


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Question 12

What number should be subtracted from 1415 to get 130 ?

Sol :

let x be subtstaded from 1415 to get 130

1415x=130

1415+130=x

or

x=1415+130

LCM of 15 and 30 is 30

x=14×2+1×130

x=28+130

=2730

or910

 


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Question 13

What should be subtracted from (34+13+25) to get 12 ?

Sol :

[34+13+25]x=12

[34+13+25]12=x

x=34+13+2512

LCM of 4,3,5,2 is 60

x=3×15+1×20+2×121×3060

x=45+20+243060

=893060

=5960

 


 

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