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Class 8 RD Sharma – Chapter 1 Rational Numbers – Exercise 1.2

  • Last Updated : 17 Jun, 2021

Question 1: Verify the commutativity of the addition of rational numbers for each of the following pairs of rational numbers

(i) -11 / 5 and 4 / 7

Solution:

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Commutativity property is verified if two rational numbers a and b, a + b = b + a



In this case a = -11 / 5 and b = 4 / 7

a + b = (-11) / 5 + 4 / 7

LCM of 5 and 7 is 35

= (-11 × 7 + 4 × 5) / 35

= (-77 + 20) / 35

= (-57) / 35

b + a = 4 / 7 + (-11) / 5

LCM of 5 and 7 is 35

= (4 × 5 + (-11 × 7)) / 35

= (20 – 77) / 35

= (-57) / 35

Hence, a + b = b + a

Therefore, commutativity is verified in this case.

(ii) 4 / 9 and 7 / -12

Solution:

Here a = 4 / 9, b = (-7) / 12

a + b = 4 / 9 + (-7) / 12

LCM of 9 and 12



9 = 3*3

12 = 3 * 2 * 2

LCM = 3 * 3 * 2 * 2 = 36

= (4 * 4 + (-7 * 3)) / 36

= (16 – 21) / 36

= (-5) / 36

Now, b + a = (-7) / 12 + 4 / 9

= ((-7 * 3) + 4 * 4) / 36

= (-21 + 16) / 36

= (-5) / 36

Hence, a + b = b + a

Therefore, commutativity is verified in this case.

(iii) -3 / 5 and -2 / -15

Solution:

Here a = -3 / 5, b = -2 / -15 = 2 / 15

a + b = (-3) / 5 + 2 / 15

LCM is 15

= (-3 * 3 + 2 * 1) / 15

= (-9 + 2) / 15

= (-7) / 15

Now,

b + a = 2 / 15 + (-3) / 5

= (2 × 1 + (-3 * 3)) / 15

= (2 – 9) / 15

= (-7) / 15

Hence, a + b = b + a

Therefore, commutativity is verified.

(iv) 2 / -7 and 12 / -35

Solution:

a = 2 / -7 = -2 / 7



b = 12 / -35 = -12 / 35

a + b = -2 / 7 + (-12) / 35

LCM of 7 and 35 is 35

= (-2 × 5 + (-12 × 1)) / 35

= (-10 – 12) / 35

= (-22) / 35

b + a = (-12) / 35 + (-2) / 7

LCM of 7 and 35 is 35

= (-12 × 1 + (-2 × 5)) / 35

= (-12 – 10) / 35

= (-22) / 35

a + b = b + a

Therefore, commutativity is verified.

(v) 4 and -3 / 5

Solution:

a = 4 / 1

b = -3 / 5

a + b = 4 / 1 + (-3) / 5

LCM of 1 and 5 is 5

= (4 * 5 + (-3 * 1)) / 5

= (20 – 3) / 5

= 17 / 5

b + a = (-3) / 5 + 4 / 1

= (-3 + 4 * 5) / 5

= (-3 + 20) / 5

= (17) / 5

a + b = b + a

Therefore, commutativity is verified.

(vi) -4 and 4 / -7

Solution:

a = -4 / 1

b = -4 / 7

a + b = (-4) / 1 + (-4) / 7

LCM of 1 and 7 is 7

= (-4 * 7 + (-4 * 1)) / 7

= (-28 – 4) / 7

= (-32) / 7

b + a = (-4) / 7 + (-4) / 1

= (-4 * 1 + (-4 * 7)) / 7

= (-4 – 28) / 7

= (-32) / 7

a + b = b + a

Therefore, commutativity is verified.

Question 2: Verify associativity of addition of rational numbers i.e (x + y) + z = x + (y + z) when

(i) x = 1 / 2, y = 2 / 3, z = -1 / 5

Solution:

To verify associativity solving the LHS

(1 / 2 + 2 / 3) + (-1 / 5)

Bracket needs to be solved first

LCM of 2 and 3 is 6

= (1 * 3 + 2 * 2) / 6 + (-1 / 5)



= (3 + 4) / 6 + (-1 / 5)

= 7 / 6 + (-1 / 5)

Now solving these,

LCM of 5 and 6 is 30

= (7 * 5 + (-1 * 6)) / 30

= (35 – 6) / 30

= 29 / 30

Now solving the RHS

1 / 2 + (2 / 3 + (-1 / 5))

Solving the bracket first, LCM of 3 and 5 is 15

= 1 / 2 + (2 * 5 + (-1 * 3)) / 15

= 1 / 2 + (10 – 3) / 15

= 1 / 2 + 7 / 15

LCM of 2 and 15 is 30

= (1 * 15 + 7 * 2) / 30

= (15 + 14) / 30

= 29 / 30

Hence LHS = RHS, associativity  property is verified

(ii) x = -2 / 5, y = 4 / 3, z = -7 / 10

Solution:

According to the property x + (y + z) = (x + y) + z

Solving LHS

-2 / 5 + (4 / 3 + (-7 / 10))

Solving bracket first

LCM of 3 and 10 is 30

= -2 / 5 + (4 * 10 + (-7 * 3)) / 30

= -2 / 5 + (40 – 21) / 30

= -2 / 5 + 19 / 30

Now, LCM of 5 and 30 is 30

= (-2 * 6 + 19 * 1) / 30

= (-12 + 19) / 30

= 7 / 30

Now, considering RHS

(-2 / 5 + 4 / 3) + (-7 / 10)

LCM of 5 and 3 is 15

= (-2 * 3 + 4 * 5) / 15 + (-7 / 10)

= (-6 + 20) / 15 + (-7 / 10)

=(14) / 15 + (-7) / 10

LCM of 15 and 10 is 30

= (14 * 2 + (-7 * 3)) / 30

= (28 – 21) / 30

= 7 / 30

Hence, LHS = RHS

Therefore, associativity property is verified

(iii) x = -7 / 11, y = 2 / -5, z = -3 / 22

Solution:

According to the property,

LHS is (-7 / 11 + (-2 / 5)) + (-3 / 22)

Solving bracket first

LCM of 11 and 5 is 55

= (-7 * 5 + (-2 * 11)) / 55 + (-3 / 22)

= (-35 – 22) / 55 + (-3 / 22)

= (-57) / 55 + (-3 / 22)

LCM of 55 and 2 is 110

= (-57 * 2 + (-3 * 5)) / 110 

= (-114 – 15) / 110

= (-129) / 110

Now, solving RHS

= -7 / 11 + (-2 / 5 + (-3) / 22)

LCM of 22 and 5 is 110



= -7 / 11 + (-2 * 22 + (-3 * 5)) / 110

= -7 / 11 + (-44 – 15) / 110

= -7 / 11 + (-59) / 110

LCM of 11 and 110 is 110

= (-7 * 10 + (-59 * 1)) / 110

= (-70 – 59) / 110

= (-129) / 110

Hence, LHS = RHS, associativity is verified.

(iv) x = -2, y = 3 / 5, z = -4 / 3

Solution:

According to the property,

LHS is (-2 / 1 + 3 / 5) + (-4 / 3)

LCM of 1 and 5 is 5

= (-2 * 5 + 3 * 1) / 5 + (-4 / 3)

= (-10 + 3) / 5 + (-4 / 3)

= (-7) / 5 + (-4 / 3)

LCM of 5 and 3 is 15

= (-7 * 3 + (-4 * 5)) / 15

= (-21 – 20) / 15

= (-41) / 15

Now solving RHS

-2 / 1 + (3 / 5 + (-4 / 3))

LCM of 5 and 3 is 15

= -2 / 1 + (3 * 3 + (-4 * 5)) / 15

= -2 / 1 + (9 – 20) / 15

= -2 / 1 + (-11) / 15

LCM of 1 and 15 is 15

= (-2 * 15 + (-11 * 1)) / 15

= (-30 – 11) / 15

= (-41) / 15

Hence, LHS = RHS 

Therefore, associativity is verified.

Question 3: Write the additive inverse of each of the following

(i) -2 / 17

Additive inverse is a number which when added to the given number gives 0. Therefore, it is negative of the number given.

Additive inverse of -2 / 17 = -(-2 / 17)

                                           = 2 / 17

(ii) 3 / -11

Additive inverse of -3 / 11 is 3 / 11

(iii) -17 / 5

Additive inverse of -17 / 5 is 17 / 5

(iv) -11 / -25

It can be written as 11 / 25

Additive inverse is -11 / 25

Question 4: Write the negative (additive) inverse of each of the following

(i) -2 / 5

 Negative (Additive) inverse is 2 / 5

(ii) 7 / -9

It can be written as -7 / 9

Negative(Additive) inverse is 7 / 9

(iii) -16 / 13

Negative(Additive) inverse is 16 / 13



(iv) -5 / 1

Negative(Additive)inverse is 5

(v) 0

0 is neutral number

Negative(Additive) inverse is 0

(vi) 1

Negative(Additive) inverse is -1

(vii) -1

Additive inverse is 1

Question 5: Using commutativity and associativity of the addition of rational numbers, express each of the following as rational numbers

(i) 2 / 5 + 7 / 3 + -4 / 5 + -1 / 3

Solution:

According to commutativity order of numbers can be changed, so writing numbers with same denominators together.

(2 / 5 + -4 / 5) + (7 / 3 + -1 / 3)

= (2 – 4 / 5) + (7 – 1 / 3)

= -2 / 5 + 6 / 3

LCM of 5 and 3 is 15

= (-2 × 3 + 6 × 5) / 15

= (-6 + 30) / 15

= 24 / 15

3 is a common number so can be simplified further

= 8 / 5

(ii) 3 / 7 + -4 / 9 + -11 / 7 + 7 / 9

Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

= (3 / 7 + -11 / 7) + (-4 / 9 + 7 / 9)

= (3 – 11) / 7 + (-4 + 7) / 9

= -8 / 7 + 3 / 9

LCM of 7 and 9 is 63

= (-8 × 9 + 3 × 7) / 63

= (-72 + 21) / 63

= (-51) / 63

3 is the common number so can be simplified further

= -17 / 21

(iii) 2 / 5 + 8 / 3 + -11 / 15 + 4 / 5 + -2 / 3

Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

= (2 / 5 + 4 / 5) + (8 / 3 + -2 / 3) + (-11 / 15)

= (6) / 5 + 6 / 3 + (-11 / 15)

Applying associativity

LCM of 5 and 3 is 15

= (6 × 3 + 6 × 5) / 15 = () + (-11) / 15

= (18 + 30) / 15 + (-11) / 15

= 48 / 15 + (-11) / 15

= 37 / 15

(iv) 4 / 7 + 0 + -8 / 9 + -13 / 7 + 17 / 21

Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

(4 / 7 + -13 / 7) + -8 / 9 + 17 / 21

= -9 / 7 + (-8 / 9) + 17 / 21

LCM of 7, 9 and 21

7 = 7×1

9 = 3×3

21 = 3×7

LCM is 3 × 3 × 7 = 63

= (-9 × 9 + (-8 × 7) + 17 × 3) / 63

= (-81 – 56 + 51) / 63

= (-86) / 63

Question 6: Rearrange suitably and find the sum in each of the following

(i) 11 / 12 + -17 / 3 + 11 / 2 + -25 / 2

Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.



= 11 / 12 + -17 / 3 + (11 – 25) / 2

= 11 / 12 + (-17) / 3 + (-14) / 2

LCM of 3 and 2 is 6

= 11 / 12 + (-17 × 2 + (-14 × 3)) / 6

= 11 / 12 + (-34 – 42) / 6

= 11 / 12 + (-76) / 6

LCM of 6 and 12 is 12

= (11 + (-76 × 2)) / 12

= (11 – 152) / 12

= (-141) / 12

(ii) -6 / 7 + -5 / 6 + -4 / 9 + -15 / 7

Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

= (-6 / 7 + -15 / 7) + -5 / 6 + -4 / 9

= (-6 – 15) / 7 + -5 / 6 + -4 / 9

= -21 / 7 + -5 / 6 + -4 / 9

= -3 / 1 + -5 / 6 + -4 / 9

LCM of 6 and 9 is

6 = 3 × 2

9 = 3 × 3

LCM is 18

= (-3 × 18 + (-5 × 3) + (-4 × 2)) / 18

= (-54 – 15 – 8) / 18

= -77 / 18

(iii) 3 / 5 + 7 / 3 + 9 / 5 + -13 / 15 + -7 / 3

Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

= (3 / 5 + 9 / 5) + (7 / 3 + -7 / 3) + -13 / 15

= (3 + 9) / 5 + (7 – 7) / 3 + -13 / 15

= 12 / 5 + 0 / 3 + -13 / 15

= 12 / 5 + 0 + -13 / 15

= 12 / 5 + -13 / 15

LCM of 5 and 15 is 15

= (12 × 3 + (-13 × 1)) / 15

= (36 – 13) / 15

= 23 / 15

(iv) 4 / 13 + -5 / 8 + -8 / 13 + 9 / 13

Solution:

According to commutativity order of numbers can be changed, so writing numbers with same denominators together.

= (4 / 13 + -8 / 13 + 9 / 13) + -5 / 8

= (4 – 8 + 9) / 13 + -5 / 8

= (5) / 13 + -5 / 8

LCM of 13 and 8 is 104

= (5 × 8 + (-5 × 13)) / 104

= (40 – 65) / 104

= -25 / 104

(v) 2 / 3 + -4 / 5 + 1 / 3 + 2 / 5

Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

= (2 / 3 + 1 / 3) + (-4 / 5 + 2 / 5)

= (2 + 1) / 3 + (-4 + 2) / 5

= (3) / 3 + (-2) / 5

LCM of 3 and 5 is 15

= (3 × 5 + (-2 × 3)) / 15

= (15 – 6) / 15

= 9 / 15

= 3 / 5

(vi) 1 / 8 + 5 / 12 + 2 / 7 + 7 / 12 + 9 / 7 + -5 / 16

Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

= (5 / 12 + 7 / 12) + (2 / 7 + 9 / 7) + 1 / 8 + -5 / 16

= (5 + 7) / 12 + (2 + 9) / 7 + 1 / 8 + -5 / 16

= 12 / 12 + 11 / 7 + 1 / 8 + -5 / 16

= 1 / 1 + 11 / 7 + 1 / 8 + -5 / 16

LCM of 1, 7, 8 and 16 is 112

= (112 + 11 × 16 + 14 + (-5 × 7)) / 112

= (112 + 176 + 14 – 35) / 112

= (267) / 112




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