The DAV Class 6 Maths Book Solutions Pdf and DAV Class 6 Maths Chapter 3 Worksheet 4 Solutions of Integers offer comprehensive answers to textbook questions.
DAV Class 6 Maths Ch 3 WS 4 Solutions
Question 1.
Find the sum in two different ways.
(a) – 32, 50
Solution:
I Method
(- 32) + (50) = 50 – 32 = 18
Hence, (- 32) + (50) = 18
II Method
| – 32 | = 32, | 50 | = 50
Difference 50 – 32 = 18
Hence, (- 32) + 50 = 18
(b) – 81, – 79
Solution:
I Method
(- 81) + (- 79) = – (81 + 79) = – 160
Hence, (- 81) + (- 79) = – 160
II Method
| – 81 | = 81, | – 79 | = 79
Sum = 81 + 79 = 160,
Hence, (- 81) + (- 79) = – 160
(c) 64, – 100
Solution:
I Method
(64) + (- 100) = 64 – 100 = – 36
Hence, (64) + (- 100) = – 36
II Method
| 64 | = 64, | – 100 | = 100
Difference = 100 – 64 = 36
Hence, | 64 | + | – 100 | = – 36
(d) 13, – 78, 15
Solution:
13, – 78, 15
(13) + (- 78) + 15 = (13 – 78) + 15
= – 65 + 15
= – 50
Hence, (13) + (- 78) + (15) = – 50
II Method:
|13| = 13, |- 78| = 78, |15| = 15
Difference of 78 and 13 = 78 – 13 = 65
∴ (13) × (- 78) = – 65
Now |- 65| = 65, |15| = 15
Difference = 65 – 15 = 50
∴ (- 65) + (15) = – 50
Hence, (13) + (- 78) + (15) = – 50
Question 2.
Write the additive inverse of the following:
(a) 31
Solution:
31 + (- 31) = 0
Hence, the additive inverse of 31 is – 31.
(b) – 7
Solution:
– 7 + 7 = 0
Hence, the additive inverse of – 7 = 7.
(c) 21
Solution:
21 + (- 21) = 0
Hence, the additive inverse of 21 is – 21.
(d) – 501
Solution:
– 501 + 501 = 0
Hence, the additive inverse of – 501 is 501.
(e) 0
Solution:
Additive inverse of 0 is 0 itself.
(f) – 34
Solution:
– 34 + 34 = 0
Hence, the additive inverse of – 34 is 34.
Question 3.
Find the sum using the properties of addition.
(a) 200 + (- 105) + (- 36)
Solution:
200 + (- 105) + (- 36)
= 200 – (105 + 36)
= 200 – 141
= 59
Hence, the required sum = 59
(b) (- 45) + 100 + (- 55)
Solution:
(- 45) + 100 + (- 55)
= (- 45) + (- 55) + 100
= – (45 + 55) + 100
= – 100 + 100
= 0
Hence, the required sum = 0
(c) (- 825) + 725 + 100 + (- 100)
Solution:
(- 825) + 725 + 100 + (- 100)
= – 825 + 725 + 100 – 100
= – 100 + 0
= – 100
Hence, the required sum = – 100
(d) 927 + (- 517) + (- 518)
Solution:
927 + (- 517) + (- 518)
= 927 – 517 + (- 518)
= 410 – 518
= – 108
Hence, the required sum = – 108
(e) (- 215) + (- 215) + 860 + (- 215) + (- 215) + 1
Solution:
(- 215) + (- 215) + 860 + (- 215) + (- 215) + 1
= – (215 + 215 + 215 + 215) + 860 + 1
= – 860 + 860 + 1
= 0 + 1
= 1
Hence, the required sum = 1
(f) 305 + (- 5) + (- 2) + 2 + (- 200)
Solution:
305 + (- 5) + (- 2) + 2 + (- 200)
= (305 – 5) + (- 2 + 2) + (- 200)
= 300 + 9 – 200
= 100
Hence, the required sum = 100
(g) 637 + 350 + (- 237) + (- 900)
Solution:
637 + 350 + (- 237) + (- 900)
= (637 + 350) – (237 + 900)
= 987 – 1137
= – 150
Hence, the required sum = – 150
(h) (- 99) + 7 + (- 101) + 93
Solution:
(- 99) + 7 + (- 101) + 93
= (- 99) + (- 101) + (7 + 93)
= – (99 + 101) + 100
= – 200 + 100
= – 100
Hence, the required sum = – 100
Question 4.
Fill in the following blanks:
(a) The negative of – 3 is = ______________-
Solution:
3
(b) (- 8) + ______________- = (- 8)
Solution:
0
(c) 11 + (- 16) = ______________- + 11
Solution:
– 16
(d) [(- 3) + 5] + 6 = (- 3) + ______________- + ______________-
Solution:
5 + 6
(e) The identity element of addition is ______________-
Solution:
0
(f) (- 51) + 51 = ______________-
Solution:
0
Question 5.
Write ‘True’ or ‘False’ for the following statements.
(a) 3 + (- 5) is not an integer.
Solution:
False
(b) Sum of two negative integers is also a negative integer.
Solution:
True
(c) Negative of – 15 does not exist.
Solution:
False
(d) [(- 3) + 8] + (- 4) = [8 + (- 3)] + (- 4)
Solution:
True
(e) 91 + (- 41) = (- 91) + 41
Solution:
False
(f) – 46 + 0 = 0
Solution:
False
(g) Sum of a positive integer and a negative integer is always negative.
Solution:
False
(h) | – 9 – 5| = | – 9| – |5|
Solution:
False