RD Sharma Solutions -Page No.5.21, Negative Numbers And Integers, Class 6, Maths Class 6 Notes | EduRev

RD Sharma Solutions for Class 6 Mathematics

Created by: Abhishek Kapoor

Class 6 : RD Sharma Solutions -Page No.5.21, Negative Numbers And Integers, Class 6, Maths Class 6 Notes | EduRev

The document RD Sharma Solutions -Page No.5.21, Negative Numbers And Integers, Class 6, Maths Class 6 Notes | EduRev is a part of the Class 6 Course RD Sharma Solutions for Class 6 Mathematics.
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Question 16: Evaluate 7 ×× |− 15| − |− 9| ×× 8.

ANSWER: ∵ |− 15| = 15 and |− 9| = 9
∴ 7 ×× |− 15| − |− 9| ×× 8 = 7 ×× 15 − 9 ×× 8
= 105 − 72
= 33
Hence, 7 ×× |− 15| − |− 9| ×× 8 = 33.


Question 17: Find the value of 38 − (− 25) − 58 + (− 15) + 23 − (− 8).

ANSWER: 38 − (− 25) − 58 + (− 15) + 23 − (− 8)
= 38 − (− 25) − 58 + (− 15) + 23 +8
= 38 − (− 25) − 58 + (− 15) + 31
= 38 − (− 25) − 58 + (31 − 15)
= 38 − (− 25) − 58 + 16
= 38 − (− 25) − (58 − 16)
= 38 − (− 25) − 42
= 38 + 25 − 42
= 38 − (42 − 25)
= 38 − 17
= 21
Hence, 38 − (− 25) − 58 + (− 15) + 23 − (− 8) = 21.

Question 18: Simplify: 5 + (− 5) + 5 + (− 5) + ...
(i) When the number of terms is 20              
(ii) When the number of terms is 25

ANSWER: (i) When the number of terms is 20            
5 + (− 5) + 5 + (− 5) + ... = (5 + 5 + 5 + ... + 10 terms) + (− 5) + 5 + (− 5) + ... + 10 terms
= 10 ×× 5 + 10 ×× (− 5)
= 50 − 50
= 0
(ii) When the number of terms is 25
5 + (− 5) + 5 + (− 5) + ... = (5 + 5 + 5 + ... + 12 terms) + (− 5) + 5 + (− 5) + ... + 13 terms
= 12 ×× 5 + 13 ×× (− 5)
= 60 − 65
= − (65 − 60)
= − 5


Question 19: If ∆ is an operation on integers such that for integers a and b, a ∆ b = a − b − (− 5). Find the values of

(i) (− 7) ∆ 3                  
(ii) (− 9) ∆ (− 4)                      
(iii) 2 ∆ 5                        
(iv) 4 ∆ (− 5) 

ANSWER:(i)

∵ a ∆ b = a − b − (− 5)

∴ (− 7) ∆ 3 = (− 7) − 3 − (− 5)

= − 7 − 3 + 5

= − (7 + 3) + 5

= − 10 + 5

= − (10 − 5)

= − 5


(ii)(− 9) ∆ (− 4) = (− 9) − (− 4) − (− 5)

= − 9 + 4 + 5

 = − 9  + 9

= 0


(iii) 2 ∆ 5 = 2 − 5 − (− 5)

= 2 − 5 + 5

= 2

(iv)

4 ∆ (− 5) = 4 − (− 5) − (− 5)

= 4 + 5 + 5

= 14


Question 20: Evaluate: − 36 − 40 + 43 − (− 29) + 18 − (− 74).

ANSWER: − 36 − 40 + 43 − (− 29) + 18 − (− 74)
= − 36 − 40 + 43 − (− 29) + 18 + 74
= − 36 − 40 + 43 − (− 29) + 92
= − 36 − 40 + 43 + 29 + 92
= − 36 − 40 + 43 + 121
= − 36 − 40 + 164
= − (36 + 40) + 164
= − 76 + 164
= 164 − 76
= 88


Question 21: The largest negative integer is ...............

ANSWER: The numbers ... − 5, − 4, − 3, − 2, − 1, 0, 1, 2, 3, 4, 5, ... form the set of integers.
Thus, the largest negative integer is − 1.


Question 22: The smallest positive integer is .................

ANSWER: The collection ... − 4, − 3, − 2, − 1, 0, 1, 2, 3, 4, ... forms the set of integers.
Thus, the smallest positive integer is 1.


Question 23: (− 22) + 21 + (− 22) + 21 + ... + 20 terms is equal to .............

ANSWER: ∵ (− 22) + 21 = − (22 − 21) = − 1
∴  (− 22) + 21 + (− 22) + 21 + ... + 20 terms = (− 1) + (− 1) + (− 1) + ... + 10 terms
 = 10 ×× (− 1)
= − 10


Question 24: (− 3) (− 4) (12) (− 1) = ............

ANSWER: (− 3) (− 4) (12) (− 1) = (− 3) (− 4) (− 12)                       [ ∵ (12) (− 1) = (− 12)]
= (− 3) (48)             [∵ (− 4) (− 12) = 48]
= − (3 ×× 48)
= − 144
Hence, (− 3) (− 4) (12) (− 1) = − 144.


Question 25: (− 1) (− 1) (− 1) (− 1) (− 1) = ...............

ANSWER: Here, − 1 is multiplied 5 times and since, 5 is an odd integer, therefore
(− 1) (− 1) (− 1) (− 1) (− 1) = − 1

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