Ncert Solutions For Class 6 Maths Chapter 6 Integers Ex 6.3 Integers Class 6 Exercise 6.3 Ncert Solutions For Class 6 Maths Chapter 6 Integers Exercise 6.3 Integers Exercise 6.3
NCERT Solutions for 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)
35 – (20)
= 35 + (additive inverse of 20)
= 35 + (- 20)
= 15 + 20 + (- 20)
= 15 + 0= 15
(b) 72-90
72-90
= 12 + (additive inverse of 90)
= 72 + (- 90)
= 72 + (- 72) + (- 18)
= 0 + (- 18) = – 18
(c) (- 15) – (- 18)
(-15)-(-18)
= (- 15) + (additive inverse of -18)
= (- 15) + (18)
= (- 15) + (15) + (3)
= 0 + (3) = 3
(d) (- 20) – (13)
= (-20)-(13)
= (- 20) + (additive inverse of 13)
= (- 20) + (- 13) = – 33
(e) 23 – (- 12)
23 – (- 12)
= 23 + (additive inverse of – 12)
= 23+12 = 35
(f) (- 32) – (- 40)
(-32)-(-40)
= (- 32) + (additive inverse of – 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) L.H.S. = (- 3) + (- 6) = – 9
R.H.S. = (- 3) – (- 6)
= (- 3) + (additive inverse of – 6)
= (- 3) + 6
= (- 3) + 3 + 3
=0+3=3
∴ (- 3) + (- 6) < (- 3) – (- 6)
(b) L.H.S. = (- 21) – (- 10)
= (- 21) + (additive inverse of -10)
= (- 21) + 10
= (- ll) + (- 10)+ 10
= (- 11) + 0 = – 11
R.H.S. = (-31)+ (-11)
= -42
∴ (-21)-(-10) >(-31)+ (-11)
(c) L.H.S. =45-(-11)
= 45 + (additive inverse of – 11)
= 45 + 11 = 56
R.H.S.;= 57 + (-4)
= 53 + 4 + (- 4)
= 53 + 0 = 53
∴ 45-(-11) >57 +(-4)
(d) L.H.S. = (-25) – (-42)
= (- 25) + (additive inverse of – 42)
= (-25)+ (+42)
= (- 25) + (+ 25) + (+ 17)
= 0 + (+ 17)= 17
R.H.S. = (- 42) – (- 25)
= (- 42) + (additive inverse of – 25)
= (- 42) + (+ 25)
= (- 17) + (- 25) + (+ 25)
= (- 17) + 0 = – 17
∴ (-25)-(-42) >(-42)-(-25).
Question 3.
Fill in the blanks:
(a) (-8) + =0
(b) 13 + = 0
(c) 12 + (-12) –
(d) (-4) + =-72
(e) – 75 = -10.
Solution.
(a) (-8)+ 8 = 0
(b) 13 + (-13) = 0
(c) 12 + (- 12) = 0.
(d)(-4) + (-8) = -12
(e) (+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)
= (- 7) + (additive inverse of 8) – (- 25)
= (- 7) + (- 8) – (- 25)
= – 15-(-25)
= – 15 + (additive inverse of – 25)
= -15+ (+25)
= – 15 + (+ 15) + (+ 10)
= 0 + (+ 10) = 10
(b) (- 13) + 32 – 8 – 1
(-13)+ 32-8-1
= (-13)+ 32-9
= (- 13) + 32 + (additive inverse of 9)
= (- 13) + 32 + (- 9)
= (- 13) + 23 + 9 + (- 9)
= (-13)+ 23 + 0
= (- 13)+ 23
= (-13)+13+10
= 0+ 10 = 10
(c) (- 7) + (- 8) + (- 90)
(-7) +(-8)+ (-90)
= (-15)+ (-90)
= -105
(d) 50-(-40)-(-2)
= 50 – (- 40) – (- 2)
= 50 + (additive inverse of – 40) – (- 2)
= 50+ (40)-(-2)
= 90-(-2)
= 90 + (additive inverse of – 2)
= 90 + 2 = 92.