Gujarat Board GSEB Textbook Solutions Class 6 Maths Chapter 6 Integers Ex 6.3 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 6 Maths Chapter 6 Integers Ex 6.3
Question 1.
Find
(a) 35 – (20)
(b) 72 – (90)
(c) ( – 15) – ( – 18)
(d) ( – 20) – (13)
(e) 23 – ( – 12)
(f) ( – 32) – ( – 40)
Solution:
(a) 35 – (20):
Since, 35 = 20 + 15
35 – 20 = 20 + 15 – 20 = 20 – 20 + 15
= 0 + 15 = 15
Thus, 35 – (20) = 15
(b) 72 – (90):
Method 1: Since, – 90 = ( – 72) + ( – 18)
72 – 90 = 72 + ( – 72) + ( – 18) = 0 + (- 18)
Method 2: 72 – 90.
= 72 + [Additive inverseof 90]
= 72 + [- 90] = – 18
(c) ( – 15) – (- 18)
= (- 15) + [Additive inverse of (- 18)]
= ( – 15) + [18] = – 15 + [15 + 3]
= – 15 + 15 + 3 = 0 + 3 = 3
Thus, – 15 – ( – 18) = 3
(d) ( – 20) – (13)
= ( – 20) + [Additive inverse of 13]
= ( – 20) + [ – 13] = – [20 + 13] = – [33]
( – 20) – (13) = – 33
(e) 23 – (- 12) = 23 + [Additive inverse of (- 12)]
= 23 + [12] = 35
23 – (- 12) = 35
(f) ( – 32) – (- 40):
(- 32) – ( – 40) = ( – 32) + [Additive inverse of ( – 40)]
= ( – 32) + [+ 40] = – 32 + 40 = – 32 + 32 + 8
= 0 + 8 = 8
Question 2.
Fill in the blanks with >, < or = sign.
(a) (- 3) + ( – 6) …………….. ( – 3) – ( – 6)
(b) ( – 21) – (- 10) ……………. (- 31) + (- 11)
(c) 45 – (- 11) ……………. 57 + (- 4)
(d) (- 25) – (- 42) ………………. ( – 42) – (- 25)
Solution:
(a) ( – 3) + ( – 6) – 9
and (- 3) – ( – 6) = (- 3) + (6) = 3
Here ( – 9) < 3
(3) + ( – 6) < ( – 3) – ( – 6) (b) (- 21) – (- 10) = – 21 + (+10) = -11 and ( – 31) + (- 11) = – 42 Also (- 11) > (- 42)
(- 21) – (- 10) > (- 3 1) + (- 11)
(c) 45 – ( – 11)
= 45 + [Additive inverse of ( – 11)]
= 45 + [11] = 56 and 57 + ( – 4) = 53
Also 56 > 53
45 – ( – 11) > 57 + (-4)
(d) ( – 25) – ( – 42)
= ( – 25) + [Additive inverse of ( – 42)]
= ( – 25) + (42) = – 25 + 25 + 17
= 0 + 17 = 17
and ( – 42) – ( – 25) = (- 42) + [Additive inrse of ( – 25)]
= ( – 42) + (25) – 25 – 17 + 25
= 0 – 17 = – 17
Also 17 > – 17
Thus, ( – 25) – (- 42) > (- 42) – (- 25)
Question 3.
Fill in the blanks.
(a) ( – 8)+ ……………… = 0
(b) 13+ …………….. = 0
(c) 12 + ( – 12) = ……………..
(d) ( – 4) + ………………… = – 12
(e) ………….. – 15 = – 10
Solution:
(a) ( – 8) + [Additive inverse of (8)] = O
( – 8) + [ + 8] = 0
or (- 8) + 8 = 0
(b) 13 + (- 13) = 0
[Additive inverse of 13 is ( – 13)]
(c) 12 + ( – 12) = 0 [12 and ( – 12) are additive inverse of each other]
(d) ( – 12) – ( – 4) = ( – 12) + (+ 4) = ( – 8)
( – 4) + ( – 8) = – 12
(e) ( – 10) + 15 = 5
5 – 15 = – 10
Question 4.
Find
(a) ( – 7) – 8 – ( – 25)
(b) (- 13) + 32 – 8 – 1
(c) ( – 7) + ( – 8) + ( – 90)
(d) 50 – ( – 40) – ( – 2)
Solution:
(a) ( – 7) – 8 – ( – 25) ( – 7) + ( – 8) + (25)
[ Additive inverse of 8 is 4 and that of (- 25) is 25]
= [ – 15] + 25 = (- 15) + 15 + 10 = 0 + 10 = 10
(b) ( – 13) + 32 – 8 – 1:
( – 13) 8 – 1 = ( – 22)
( – 13) + 32 + ( – 8) + ( – 1)
= [( – 13) + ( – 8) + ( – 1)] + 32 = [( – 22)] + 32
= (- 22) + 22 + 10 = 0 + 10 = 10
(c) ( – 7) + ( – 8) + ( – 90):
The given integers are of like sign,
( – 7) + (- 8) + ( – 90) = – [ 7 + 8 + 90]
= – [105] = – 105
(d) 50 – ( – 40) – (- 2) = 50 + [Additive inverse of ( – 40)] + [Additive inverse of (- 2)]
= 50 + [ + 40] + [ + 2]
= 50 + 40 + 2 = 92