Integers Class 6 Ex. 6.3
New Ncert Class 6 Math Free Solution.
Exercise 6.3
Question 1 :- Find :
(a) 35 – (20)
(b) 72 – (90)
(c) (-15) – (-18)
(d) (- 20) – (13)
(e) 23 – (-12)
(f) (-32) – (-40)
Solution 1 :-
(a) 35 – (20)
= 35 – 20
= 15
(b) 72 – 90
= 72 – 90
= – 18
(c) (- 15) – (- 18)
= – 15 + 18
= 3
(d) (- 20) – (13)
= – 20 – 13
= – 33
(e) 23 – (- 12)
23 + 12 = 23
= 35
(f) (- 32) – (- 40)
= – 32 + 40
= 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 2:-
(a) (-3) + (-6) = – 3 – 6 = – 9 and
(-3) – (-6) = -3 + 6 = 3
Here, – 9 < 3
So, (- 3) + (- 6) < (- 3) – (- 6)
(b) (-21) – (-10) = -21 + 10 = -11 and
(-31) + (-11) = – 31 – 11 = – 42
So, (-21) – (-10) > (-31) + (-11)
(c) 45 – (-11) = 45 + 11 = 56 and
57 + (-4) = 57 – 4 = 53
So, 45 – (-11) > 57 + ( -4)
(d) (-25) – (-42) = -25 + 42 = 17 and
(-42) – (-25) = -42 + 25 = -17
So, (-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 3:-
(a) (-8) + (additive inverse of -8) = 0
= (-8) + (8) = 0
∴ Value of blank is 8
(b) 13 + (additive inverse of 13) = 0
= 13 + (-13) = 0
∴ Value of blank is – 13
(c) 12 + (-12) = 0 [∵ -12 is additive inverse of 12]
∴ The Value of blank is 0
(d) (-4) + (-8) = -[4 + 8] = -12
∴ Value of blank is -8.
(e) (+5) – 15 = -10
∴ Value of blank is +5.
Question 4 :- Find :
(a) (-7) – 8 – (-25)
(b) (-13) + 32 – 8 – 1
(c) (-7) + (-8) + (-90)
(d) 50 – (-40) – (-2)
Solution 4:-
(a) (-7) – 8 – (-25)
= -7 – 8 + 25
= – 15 + 25
= 10.
(b) (-13) + 32 – 8 – 1
= -13 + 32 – 8 – 1
= 32 – 22
= 10.
(c) (-7) + (-8) + (-90)
= – 7 – 8 – 90
= -105.
(d) 50 – (-40) – (-2)
= 50 + 40 + 2
= 50 + 42
= 92.
Chapter 6 Integer Class 6 Exercise. 6.3 for Free