Negative Numbers and Integers - Exercise 5.3 Class 6 Notes | EduRev

Mathematics (Maths) Class 6

Class 6 : Negative Numbers and Integers - Exercise 5.3 Class 6 Notes | EduRev

 Page 1


 
 
 
 
 
 
Exercise 5.3                                                                          page: 5.11 
1. Find the additive inverse of each of the following integers: 
(i) 52 
(ii) – 176 
(iii) 0 
(iv) 1 
Solution: 
 
(i) The additive inverse of 52 is – 52. 
 
(ii) The additive inverse of – 176 is 176. 
 
(iii) The additive inverse of 0 is 0. 
 
(iv) The additive inverse of 1 is – 1. 
 
2. Find the successor of each of the following integers: 
(i) – 42 
(ii) -1 
(iii) 0 
(iv) – 200 
(v) -99 
Solution: 
 
(i) The successor of – 42 is 
- 42 + 1 = - 41 
 
(ii) The successor of – 1 is 
-1 + 1 = 0 
 
(iii) The successor of 0 is 
0 + 1 = 1 
 
(iv) The successor of – 200 is  
-200 + 1 = - 199 
 
(v) The successor of – 99 is 
- 99 + 1 = - 98 
 
3. Find the predecessor of each of the following integers: 
(i) 0 
(ii) 1 
(iii) – 1 
(iv) – 125 
(v) 1000 
Solution: 
 
(i) The predecessor of 0 is  
0 – 1 = - 1 
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Exercise 5.3                                                                          page: 5.11 
1. Find the additive inverse of each of the following integers: 
(i) 52 
(ii) – 176 
(iii) 0 
(iv) 1 
Solution: 
 
(i) The additive inverse of 52 is – 52. 
 
(ii) The additive inverse of – 176 is 176. 
 
(iii) The additive inverse of 0 is 0. 
 
(iv) The additive inverse of 1 is – 1. 
 
2. Find the successor of each of the following integers: 
(i) – 42 
(ii) -1 
(iii) 0 
(iv) – 200 
(v) -99 
Solution: 
 
(i) The successor of – 42 is 
- 42 + 1 = - 41 
 
(ii) The successor of – 1 is 
-1 + 1 = 0 
 
(iii) The successor of 0 is 
0 + 1 = 1 
 
(iv) The successor of – 200 is  
-200 + 1 = - 199 
 
(v) The successor of – 99 is 
- 99 + 1 = - 98 
 
3. Find the predecessor of each of the following integers: 
(i) 0 
(ii) 1 
(iii) – 1 
(iv) – 125 
(v) 1000 
Solution: 
 
(i) The predecessor of 0 is  
0 – 1 = - 1 
 
 
 
 
 
 
 
(ii) The predecessor of 1 is 
1 – 1 = 0 
 
(iii) The predecessor of -1 is  
-1 – 1 = -2 
 
(iv) The predecessor of – 125 is 
-125 – 1 = - 126 
 
(v) The predecessor of 1000 is  
1000 – 1 = 999 
 
4. Which of the following statements are true? 
(i) The sum of a number and its opposite is zero. 
(ii) The sum of two negative integers is a positive integer. 
(iii) The sum of a negative integer and a positive integer is always a negative integer. 
(iv) The successor of – 1 is 1. 
(v) The sum of three different integers can never be zero. 
Solution: 
 
(i) True. 1 – 1 = 0 
 
(ii) False. -1 – 1 = -2 
 
(iii) False. – 2 + 3 = 1 
 
(iv) False. The successor of – 1 is 0. 
 
(v) False. 1 + 2 – 3 = 0 
 
5. Write all integers whose absolute values are less than 5. 
Solution: 
 
The integers whose absolute values are less than 5 are 
-4, - 3, - 2, - 1, 0, 1, 2, 3, 4 
 
6. Which of the following is false: 
(i) |4 + 2| = |4| + |2| 
(ii) |2 – 4| = |2| + |4| 
(iii) |4 – 2| = |4| - |2| 
(iv) |(-2) + (-4)| = |-2| + |-4| 
Solution: 
 
(i) True.  
 
(ii) False. 
 
(iii) True. 
(iv) True. 
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Exercise 5.3                                                                          page: 5.11 
1. Find the additive inverse of each of the following integers: 
(i) 52 
(ii) – 176 
(iii) 0 
(iv) 1 
Solution: 
 
(i) The additive inverse of 52 is – 52. 
 
(ii) The additive inverse of – 176 is 176. 
 
(iii) The additive inverse of 0 is 0. 
 
(iv) The additive inverse of 1 is – 1. 
 
2. Find the successor of each of the following integers: 
(i) – 42 
(ii) -1 
(iii) 0 
(iv) – 200 
(v) -99 
Solution: 
 
(i) The successor of – 42 is 
- 42 + 1 = - 41 
 
(ii) The successor of – 1 is 
-1 + 1 = 0 
 
(iii) The successor of 0 is 
0 + 1 = 1 
 
(iv) The successor of – 200 is  
-200 + 1 = - 199 
 
(v) The successor of – 99 is 
- 99 + 1 = - 98 
 
3. Find the predecessor of each of the following integers: 
(i) 0 
(ii) 1 
(iii) – 1 
(iv) – 125 
(v) 1000 
Solution: 
 
(i) The predecessor of 0 is  
0 – 1 = - 1 
 
 
 
 
 
 
 
(ii) The predecessor of 1 is 
1 – 1 = 0 
 
(iii) The predecessor of -1 is  
-1 – 1 = -2 
 
(iv) The predecessor of – 125 is 
-125 – 1 = - 126 
 
(v) The predecessor of 1000 is  
1000 – 1 = 999 
 
4. Which of the following statements are true? 
(i) The sum of a number and its opposite is zero. 
(ii) The sum of two negative integers is a positive integer. 
(iii) The sum of a negative integer and a positive integer is always a negative integer. 
(iv) The successor of – 1 is 1. 
(v) The sum of three different integers can never be zero. 
Solution: 
 
(i) True. 1 – 1 = 0 
 
(ii) False. -1 – 1 = -2 
 
(iii) False. – 2 + 3 = 1 
 
(iv) False. The successor of – 1 is 0. 
 
(v) False. 1 + 2 – 3 = 0 
 
5. Write all integers whose absolute values are less than 5. 
Solution: 
 
The integers whose absolute values are less than 5 are 
-4, - 3, - 2, - 1, 0, 1, 2, 3, 4 
 
6. Which of the following is false: 
(i) |4 + 2| = |4| + |2| 
(ii) |2 – 4| = |2| + |4| 
(iii) |4 – 2| = |4| - |2| 
(iv) |(-2) + (-4)| = |-2| + |-4| 
Solution: 
 
(i) True.  
 
(ii) False. 
 
(iii) True. 
(iv) True. 
 
 
 
 
 
 
7. Complete the following table: 
 
From the above table: 
(i) Write all the pairs of integers whose sum is 0. 
(ii) Is (-4) + (-2) = (-2) + (-4)? 
(iii) Is 0 + (-6) = -6? 
Solution: 
 
 
(i) The pairs of integers whose sum is 0 are 
(6, -6), (4, - 4), (2, - 2), (0, 0) 
 
(ii) Yes. By using commutativity of addition (-4) + (-2) = (-2) + (-4) 
 
(iii) Yes. By using additive identity 0 + (-6) = -6. 
 
8. Find an integer x such that 
(i) x + 1 = 0 
(ii) x + 5 = 0 
(iii) – 3 + x = 0 
(iv) x + (-8) = 0 
(v) 7 + x = 0 
(vi) x + 0 = 0 
Solution: 
 
(i) x + 1 = 0 
Subtracting 1 on both sides 
x + 1 – 1 = 0 – 1 
Page 4


 
 
 
 
 
 
Exercise 5.3                                                                          page: 5.11 
1. Find the additive inverse of each of the following integers: 
(i) 52 
(ii) – 176 
(iii) 0 
(iv) 1 
Solution: 
 
(i) The additive inverse of 52 is – 52. 
 
(ii) The additive inverse of – 176 is 176. 
 
(iii) The additive inverse of 0 is 0. 
 
(iv) The additive inverse of 1 is – 1. 
 
2. Find the successor of each of the following integers: 
(i) – 42 
(ii) -1 
(iii) 0 
(iv) – 200 
(v) -99 
Solution: 
 
(i) The successor of – 42 is 
- 42 + 1 = - 41 
 
(ii) The successor of – 1 is 
-1 + 1 = 0 
 
(iii) The successor of 0 is 
0 + 1 = 1 
 
(iv) The successor of – 200 is  
-200 + 1 = - 199 
 
(v) The successor of – 99 is 
- 99 + 1 = - 98 
 
3. Find the predecessor of each of the following integers: 
(i) 0 
(ii) 1 
(iii) – 1 
(iv) – 125 
(v) 1000 
Solution: 
 
(i) The predecessor of 0 is  
0 – 1 = - 1 
 
 
 
 
 
 
 
(ii) The predecessor of 1 is 
1 – 1 = 0 
 
(iii) The predecessor of -1 is  
-1 – 1 = -2 
 
(iv) The predecessor of – 125 is 
-125 – 1 = - 126 
 
(v) The predecessor of 1000 is  
1000 – 1 = 999 
 
4. Which of the following statements are true? 
(i) The sum of a number and its opposite is zero. 
(ii) The sum of two negative integers is a positive integer. 
(iii) The sum of a negative integer and a positive integer is always a negative integer. 
(iv) The successor of – 1 is 1. 
(v) The sum of three different integers can never be zero. 
Solution: 
 
(i) True. 1 – 1 = 0 
 
(ii) False. -1 – 1 = -2 
 
(iii) False. – 2 + 3 = 1 
 
(iv) False. The successor of – 1 is 0. 
 
(v) False. 1 + 2 – 3 = 0 
 
5. Write all integers whose absolute values are less than 5. 
Solution: 
 
The integers whose absolute values are less than 5 are 
-4, - 3, - 2, - 1, 0, 1, 2, 3, 4 
 
6. Which of the following is false: 
(i) |4 + 2| = |4| + |2| 
(ii) |2 – 4| = |2| + |4| 
(iii) |4 – 2| = |4| - |2| 
(iv) |(-2) + (-4)| = |-2| + |-4| 
Solution: 
 
(i) True.  
 
(ii) False. 
 
(iii) True. 
(iv) True. 
 
 
 
 
 
 
7. Complete the following table: 
 
From the above table: 
(i) Write all the pairs of integers whose sum is 0. 
(ii) Is (-4) + (-2) = (-2) + (-4)? 
(iii) Is 0 + (-6) = -6? 
Solution: 
 
 
(i) The pairs of integers whose sum is 0 are 
(6, -6), (4, - 4), (2, - 2), (0, 0) 
 
(ii) Yes. By using commutativity of addition (-4) + (-2) = (-2) + (-4) 
 
(iii) Yes. By using additive identity 0 + (-6) = -6. 
 
8. Find an integer x such that 
(i) x + 1 = 0 
(ii) x + 5 = 0 
(iii) – 3 + x = 0 
(iv) x + (-8) = 0 
(v) 7 + x = 0 
(vi) x + 0 = 0 
Solution: 
 
(i) x + 1 = 0 
Subtracting 1 on both sides 
x + 1 – 1 = 0 – 1 
 
 
 
 
 
 
We get 
x = -1 
 
(ii) x + 5 = 0 
By subtracting 5 on both sides 
x + 5 – 5 = 0 – 5 
So we get 
x = -5 
 
(iii) – 3 + x = 0 
By adding 3 on both sides 
-3 + x + 3 = 0 + 3 
So we get 
x = 3 
 
(iv) x + (-8) = 0 
By adding 8 on both sides 
x – 8 + 8 = 0 + 8  
So we get 
x = 8 
 
(v) 7 + x = 0 
By subtracting 7 on both sides 
7 + x – 7 = 0 – 7 
So we get 
x = - 7 
 
(vi) x + 0 = 0 
So we get 
x = 0 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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