## NCERT Solutions For Class 6 Maths Integers Exercise 6.3

**NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3**

**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:

(a) 35 – (20) = 15 + (20) – (20)

= 15 + 0 = 15 [(+a) + (-a) = 0]

(b) 72 – 90

72 – (72 + 18) = 72 – 72 – 18

= 0 – 18 = – 18 [a + (- a) = 0]

(c) (- 15) – (- 18)

= (- 15) + (additive inverse of – 18)

= (-15) + (18) = 3

(d) (- 20) – (13)

(- 20) – (13) = – [20 + 13] = – 33

(e) 23 – (- 12)

23 – (- 12) = 23 + (additive inverse of – 12)

= 23 + 12 = 35

(f) (- 32) – (- 40)

(- 32) + (additive inverse of – 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:

(a) (-3) + (-6) = – [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) = – (31 + 11) = – 42

Here, -42 < -11 or -11 > -42 ∴ (-21) , -(-10) >(-31)+ (-11)

(c) 45 – (-11) = 45 + 11 = 56 and 57 + (-4) = 57 -4 = 53

Here, 56 >53

∴ 45 – (-11) > 57 + ( -4)

(d) (-25) – (-42) = -25 + 42 = 17

and (-42) – (-25) = -42 + 25 = -17

Here, 17 > -17

∴ (-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) = 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:

(a) (-7) – 8 – (-25)

= (-7) – 8 + 25

[ ∵ Additive inverse of – 25 is 25]

= -7 + 17 = -7 + 7 +10

[∵ (-a) + (+a) = 0]

= 0 + 10 = 10.

(b) (-13) + 32 – 8 – 1

= (-13) + (13) + 19 – (8 + 1)

= 0 + 19 – 9

= 19 – 9 [∵ (-13) + (13) = 0]

= 10 + 9 – 9 = 10 + 0 = 10.

[(+9) – (+9) = 0]

(c) (-7) + (-8) + (-90) = – (7 + 8) + (-90)

= -15 + (-90)

= -(15 + 90)

= -105.

(d) 50 – (-40) – (-2)

= 50 – [- 40 – 2]

= 50 – (-42)

= 50 + 42

= 92.

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