In this chapter, we provide RS Aggarwal Solutions for Class 6 Chapter 4 Integers Exercise 4B for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 6 Chapter 4 Integers Exercise 4B pdf, free RS Aggarwal Solutions Class 6 Chapter 4 Integers Exercise 4B book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 6 |

Subject | Maths |

Chapter | Chapter 4 |

Chapter Name | Integers |

Exercise | 4B |

Category | RS Aggarwal Solutions |

**RS Aggarwal Class 6 Chapter 4 Integers Ex 4B Download PDF**

**Question 1.****Solution:**

(i) On the number line we start from 0 and move 9 steps to the right to reach a point A. Now, starting from A, we move 6 steps to the left to reach a point B, as shown below :

Now, B represents the integer 3

9 + ( – 6) = 3

(ii) On the number line, we start from 0 and move 3 steps to the left to reach a point A. Now, starting from A, we move 7 steps to the right to reach a point B, as shown below :

And B represents the integer 4

( – 3) + 7 = 4

(iii) On the number line, we start from 0 and move 8 steps to the right to reach a point A. Now, starting from A, we move 8 steps to the left to reach a point B, as shown below :

And, B represents the integer 0.

8 + ( – 8) = 0

(iv) On the number line, we start from 0 and move 1 step the left to reach a point A. Now, starting from point A, we move 3 steps to the left to reach g. point B, as shown below :

And, B represents the integer – 4

( – 1) + ( – 3) = – 4.

(v) On the number line, we start from 0 and move 4 steps to the left to reach a point A. Now, starting from point A, we move 7 steps to the left to reach a point B, as shown below :

And, B represents the integer -11.

( – 4) + ( – 7) = – 11

(vi) On the number line we start from 0 and move 2 steps to the left to reach a point A. Now, starting from A, we move 8 steps to the left to reach a point B, as shown below :

And, B represents the integer – 10

( – 2) + ( – 8) = – 10

(vii) On the number line we start from 0 and move 3 steps to the right to reach a point A. Now, starting from A, we move 2 steps to the left to reach a point B and again starting from left to reach a point B and again starting from B, we move 4 steps to the left to reach a point C, as shown below :

And, C represents the integer – 3

3 + ( – 2) + ( – 4) = – 3

(viii) On the number line we start from 0 and move 1 step to the left to reach a point A. Now, starting from A, we move 2 steps to the left to reach a point B and again starting from B, we move 3 steps to the left to reach point C, as shown below :

And, C represents the integer – 6

( – 1) + ( – 2) + ( – 3) = – 6.

(ix) On the number line we start from 0 and move 5 steps to the right to reach a point A. Now, starting from A, we move 2 steps to the left to reach a point B and again starting from point B, we move 6 steps to the left to reach a point C, as shown below :

And, C represents the integer – 3.

5 + (- 2) + (- 6) = – 3

**Question 2.****Solution:**

(i) (- 3) + ( – 9) = – 12

(Using the rule for addition of integers having like signs)

(ii) ( – 7) + ( – 8) = – 15

(Using the rule for addition of integers having like signs)

(iii) ( – 9) + 16 = 7

(Using the rule for addition of integers having unlike signs)

(iv) ( – 13) + 25 = 12

(Using the rule for addition of integers having unlike signs)

(v) 8 + ( – 17) = – 9

(Using the rule for addition of integers having unlike signs)

(vi) 2 + ( – 12) = – 10

(Using the rule for addition of integers having unlike signs)

**Question 3.****Solution:**

(i) Using the rule for addition of integers with like signs, we get:

(ii) Using the rule for addition of integers with like signs, we get :

(iii) Using the rule for addition of integers with like signs, we get :

(iv) Using the rule for addition of integers with like signs, we get:

**Question 4.****Solution:**

(i) Using the rule for addition of integers with unlike signs, we get:

(ii) Using the rule for addition of integers with unlike signs, we get:

(iii) Using the rule for addition of integers with unlike signs, we have

(iv) Using the rule for addition of integers with unlike signs, we have

**Question 5.****Solution:**

(i) Using the rule for addition of integers with unlike signs, we get :

(ii) Using-the rule for addition of integers with unlike signs, we get

(iii) Using the rule for addition of integers with unlike signs, we get :

(iv) Using the rule for addition of integers with unlike signs, we get :

(v) Using the rule for addition of integers with like signs, we get:

(vi) Using the rule for addition of integers with unlike signs, we get :

(vii) Using the rule for addition of integers with unlike signs, we get :

(viii) We have, ( – 18) + 25 + ( – 37)

= [( – 18) + 25] + ( – 37)

= 7 + ( – 37)

= – 30

(ix) We have, – 312 + 39 + 192

= ( – 312) + (39 + 192)

= ( – 312) + 231

= – 81

(x) We have ( – 51) + ( – 203) + 36 + ( – 28)

= [( – 51) + ( – 203)] + [36 + ( – 28)]

= ( – 254) + 8

= – 246

**Question 6.****Solution:**

(i) The additive inverse of – 57 is 57

(ii) The additive inverse of 183 is – 183

(iii) The additive inverse of 0 is 0

(iv) The additive inverse of – 1001 is 1001

(v) The additive inverse of 2054 is – 2054

**Question 7.****Solution:**

(i) Successor of 201 = 201 + 1 = 202

(ii) Successor of 70 = 70 + 1 = 71

(iii) Successor of – 5 = – 5 + 1 = – 4

(iv) Successor of – 99 = – 99 + 1 = – 98

(v) Successor of – 500 = – 500 + 1 = – 499 Ans.

**Question 8.****Solution:**

(i) Predecessor of 120 = 120 – 1 = 119

(ii) Predecessor of 79 = 79 – 1 = 78

(iii) Predecessor of – 8 = – 8 – 1 = – 9

(iv) Predecessor of – 141 = – 141 – 1 = – 142

(v) Predecessor of – 300 = – 300 – 1 = – 301 Ans.

**Question 9.****Solution:**

(i) ( – 7) + ( – 9) + 12 + ( – 16)

= – 7 – 9 + 12 – 16

= – 7 – 9 – 16 + 12

= – 32 + 12

= – 20

(ii) 37 + ( – 23) + ( – 65) + 9 + ( – 12)

= 37 – 23 – 65 + 9 – 12

= 37 + 9 – 23 – 65 – 12

= 46 – 100

= – 54

(iii) ( – 145) + 79 + ( – 265) + ( – 41) + 2

= – 145 + 79 – 265 – 41 + 2

= 79 + 2 – 145 – 265 – 41

= 81 – 451

= – 370

(iv) 1056 + ( – 798) + ( – 38) + 44 + ( – 1)

= 1056 – 798 – 38 + 44 – 1

= 1056 + 44 – 798 – 38 – 1

= 1100 – 837

= 263 Ans.

**Question 10.****Solution:**

Distance travelled from Patna to its north = 60 km

Distance travelled from that place to south of it = 90 km

Distance of the final place to Patna = 60 – 90

= – 30 km

= 30 km south

Ans.

**Question 11.****Solution:**

Total amount of pencils purchased = Rs. 30 + Rs. 25

= Rs 55

Total amount of pens purchased = Rs. 90

Total cost price = Rs. 55 + Rs. 90

= Rs. 145

Total sale price of pencils and pens = Rs 20 + Rs. 70

= Rs. 90

Loss = cost price – selling price

= Rs. 145 – Rs. 90

= Rs. 55 Ans.

**Question 12.****Solution:**

(i) True.

(ii) False : As if positive integer is greater then it will be positive.

(iii) True : As ( – a + a = 0).

(iv) False : As the sum of three integers can be zero or non-zero.

(v) False : As | – 5 | = 5 and | – 3 | = 3 and 5 ≮ 3.

(vi) False : | 8 – 5 | = | 3 | = 3 and | 8 | + | – 5 | = 8 + 5 = 13.

**Question 13.****Solution:**

(i) a + 6 = 0

Subtracting 6 from both sides,

a + 6 – 6 = 0 – 6

=> a = – 6

a = – 6.

(ii) 5 + a = 0

Subtracting 5 from both sides,

5 + a – 5 = 0 – 5

=> a = – 5

a = – 5

(iii) a + ( – 4) = 0

Adding 4 to both sides,

a + ( – 4) + 4 = 0 + 4

=> a = 4

a = 4

(iv) – 8 + a = 0

Adding 8 to both sides,

– 8 + a + 8 = 0 + 8

=> a – 8

a = 8 Ans.

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