Study MaterialsCBSE NotesNCERT Solutions For Class 6 Maths Integers Exercise 6.3

NCERT Solutions For Class 6 Maths Integers Exercise 6.3

NCERT Solutions For Class 6 Maths Integers Exercise 6.3

 

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3

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    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.

    NCERT-Solutions-For-Class-6-Maths-Integers-Exercise-6.3-01
    NCERT-Solutions-For-Class-6-Maths-Integers-Exercise-6.3-02
    NCERT-Solutions-For-Class-6-Maths-Integers-Exercise-6.3-03

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