RS Aggarwal Class 6 Maths Chapter 4 - Integer Exercise 4D

RS Aggarwal 2021-2022 for Class 6 Maths Chapter 4 - Integer

RS Aggarwal Class 6 Math Solution Chapter 4- Integer  Exercise 4D is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily.  RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.

 Rs Aggarwal Class 6 Math Solution Chapter 4- Integer 

Exercise 4D


1: Multiply

    (i) 15 by 9

    (ii) 18 by − 7 

  (iii) 29 by − 11

    (iv) − 18 by 13

    (v) − 56 by 16 

    (vi) 32 by − 21

    (vii) − 57 by 0

    (viii) 0 by − 31

    (ix) − 12 by − 9

    (x) − 746 by − 8

    (xi) 118 by − 7

    (xii) − 238 by − 143

Solution:

    (i) 15 by 9

        = 15 × 9

        = 135

    (ii) 18 by −7

        = –(18 × 7)

      = –126

    (iii) 29 by –11

     = –(29 × 11)

        = –319

    (iv) –18 by 13

        = –(18 × 13)

        = –234

    (v) –56 by 16

        = –(56 × 16)

      = –896

    (vi) 32 by –21

        = –(32 × 21)

        = –672

    (vii) –57 by 0

        = –(57 × 0)

        = 0

    (viii) 0 by –31

        = –(0 × 31)

        = 0

    (ix) –12 by –9

      = (–12) × (– 9)

      = 108

    (x) (–​746) by (–8)

      = (–746) × (–8)

      = 5968

    (xi) 118 by −7

        ​= 118 × (-7)

        = –826

    (xii) −238 by −143

        = (−238) × (−143)

        = 34034


2: Find the products:

    (i) ( − 2) × 3 × (− 4)

    (ii) 2 x (− 5) x ( − 6)

    (iii) ( − 8) x 3 x 5

    (iv) 8 x 7 x (− 10)

    (v) (− 3) × (− 7) × (− 6) 

    (vi) (− 8) × (− 3) × (− 9)

Solution:

    (i) (– 2) × 3 × (– 4)

        = [(– 2) × 3] × (– 4)

        = (– 6) × (– 4)

        = 24     (ii) 2 × (– 5) × (– 6)

        = [2 × (– 5)] × (– 6)

        = (– 10) × (– 6)

        = 60     (iii) (– 8) × 3 × 5

        = [(– 8) × 3] × 5

        = (– 24) × 5

         = – 120

    (iv) 8 × 7 × (–10)

        = [8 × 7] × (–10)

        = 56 × (– 10)

        = – 560

    (v) (– 3) × (– 7) × (– 6)

        = [(– 3) × (– 7)] × (– 6)

        = 21 × (– 6)

        = – 126

    (vi) (– 8) × (– 3) × (– 9)

        = [(– 8) × (– 3)] × (– 9)

        = 24 × (– 9)

         = – 216


3: Use convenient groupings and find the values of

  (i) 18 × (− 27) x 30 

    (ii) (− 8) × (− 63) x 9

    (iii) (− 17) x (− 23) x 41

    (iv) (− 51) × (− 47) x (− 19)

Solution:

    (i) 18 × (–27) × 30

        = (–27) × [18 × 30]

        = (–27) × 540

        = –14580

    (ii) (–8) × (–63) × 9

        = [(–8) × (–63)] × 9

        = 504 × 9

        = 4536

    (iii) (–17) × (–23) × 41

        = [(–17) × (–23)] × 41

        = 391 × 41

        = 16031

(iv) (–51) × (–47) × (–19)

        = [(–51) × (–47)] × (–19)

        = 2397 × (–19)

        = – 45543


4: Verify the following:

     (i) 18 x [9 + (− 7)] = 18 x 9 + 18 x (− 7)

    (ii) (− 13) × [(− 6) + (− 19)) = (− 13) x (− 6) + (− 13) x (− 19)

Solution:

    (i)

    L.H.S. =18 × [9 + (–7)]                                

               = 18 × [9 – 7]                                      

               = 18 × 2                                              

               = 36 

    R.H.S. =18 × 9 + 18 × (–7)

                = 162 – (18 × 7)  

                = 162 – 126

                = 36

    ∴ L.H.S = R.H.S    Hence, verified.

(ii) (–13) × [(–6) + (–19)] = (–13) × (–6) + (–13) × (–19)

    L.H.S. = (–13) × [(–6) + (–19)]                                

               = (–13) × [–6 – 19]                          

               = (–13) × (–25)                                              

               = 325

    R.H.S. = (–13) × (–6) + (–13) × (–19)

                = 78 + 247

                = 325

    ∴ L.H.S = R.H.S    Hence, verified.


5: Complete the following multiplication table:

    

x

– 3

– 2

– 1

0

1

2

3

– 3








– 2








– 1








0








1








2








3









  Solution:  

x

– 3

– 2

– 1

0

1

2

3

– 3

9

6

3

0

– 3

– 6

– 9

– 2

6

4

2

0

– 2

– 4

– 6

– 1

3

2

1

0

– 1

– 2

– 3

0

0

0

0

0

0

0

0

1

– 3

– 2

– 1

0

1

2

3

2

– 6

– 4

– 2

0

2

4

6

3

– 9

– 6

– 3

0

3

6

9


6: Which of the following statements are true and which are false?

    (i) The product of a positive integers and a negative integer is negative. 

    (ii) The product of two negative integers is a negative integer.

    (iii) The product of three negative integers is a negative integer.

    (iv) Every integer when multiplied with − 1 gives its multiplicative inverse,

Solution:

    (i) True 

    (ii) False

    (iii) True

    (iv) True


7: Simplify:

    (i) (− 9) x 6 + (− 9) x 4

    (ii) 8 x (− 12) + 7 x (− 12)

    (iii) 30 × (− 22) + 30 x (14)

    (iv) (− 15) × (− 14) + (− 15) × (− 6) 

    (v) 43 x (− 33) + 43 × (− 17)

    (vi) (− 36) x 72 + (− 36) × 28

    (vii) (− 27) x (− 16) + (− 27) × (− 14)

Solution:

    (i) (− 9) x 6 + (− 9) x 4

        = (–9) × ( 6 + 4 )

        = (–9) × 10

        = –90

    (ii) 8 × (–12) + 7 × (–12)

        = (–12) × (8+7)

        = (–12) × 15

        = –180
    (iii) 30 × (– 22) + 30 × (14)

        = 30 × [(– 22) + 14]
        = 30 × [–22 + 14]
        = 30 × ( –8 )
        = –240     (iv) (–15) × (–14) + (–15) × (–6)         = (–15) × [ (–14) + (–6)]
        = (–15) × [–14 – 6]
        = (–15) × (–20)         
        = 300     (v) 43 × (–33) + 43 × (–17)         = (43 ) × [ –(33) + (–17)]
        = (43 ) × [–33 – 17]
        = 43 × ( –50)
        = –2150     (vi) (–36) × (72) + ( –36) × 28          = (–36) × (72 + 28 )
         = (–36) × 100
         = –3600     (vii) (–27) × (–16) + (–27) × (–14)         = (–27) × [(–16) + (–14)]
        = (–27) × [–16 –14]
        = (–27) × [–30]
        = 810


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