NCERT Class 7 Solutions: Integers (Chapter 1) Exercise 1.3–Part 1

  • Previous

Q-1 Find the each of the following products:

  1. 3×(1)

  2. (1)×225

  3. (21)×(30)

  4. (316)×(1)

  5. (15)×0×(18)

  6. (12)×(11)×(10)

  7. 9×(3)×(6)

  8. (18)×(5)×(4)

  9. (1)×(2)×(3)×4

  10. (3)×(6)×2×(1)

Solution: When multiplying the numbers some of which are negative, do the multiplication without accounting for sign. Then put the sign on the result according to following rules:

Rules for multiplying signed (positive and negative) integers

Rules for Multiplying Positive and Negative Integers

Rules for multiplying signed (positive and negative) integers

  1. 3×(1)

    =3 ()×(+)=()

  2. (1)×225

    =225 ()×(+)=()

  3. (21)×(30)

    =630 ()×()=(+)

  4. (316)×(1)

    =316 ()×()=(+)

  5. (15)×0×(18)

    =0

    Zero property of multiplication: Also called the zero product property. There exists a unique number, zero, such that the product of any real number x and 0 is always equal to 0

  6. (12)×(11)×(10)

    =132×10

    =1320

  7. 9×(3)×(6)

    =9×18=160 ()×(+)=()

  8. (18)×(5)×(4)

    =90×(4)

    =360 ()×(+)=()

  9. (1)×(2)×(3)×4

    =(+2)×(12)

    =24 ()×(+)=()

  10. (3)×(6)×2×(1)

    =(+18)×(2)

    =36 ()×(+)=()

Q-2 Verify the following:

  1. 18×[7+(3)]=[18×7]+[18×(3)]

  2. (21)×[(4)+(6)]=[(21)×(4)]+[(21)×(6)]

Solution:

Given the Descributive propert and apply all examples and verify

Given the Descributive Propert and Apply All Examples

Given the Descributive propert and apply all examples and verify

  1. 18×[7+(3)]=[18×7]+[18×(3)]

    L.H.S=18×[7+(3)]

    =18×[73]

    =18×4

    =72

    R.H.S=[18×7]+[18×(3)]

    =126+(54)

    =72

    Hence L.H.S=R.H.S

    18×[7+(3)]=[18×7]+[18×(3)]

  2. (21)×[(4)+(6)]=[(21)×(4)]+[(21)×(6)]

    L.H.S=(21)×[(4)+(6)]

    =(21)×[46]

    =(21)×(10)

    =210

    R.H.S=[(21)×(4)]+[(21)×(6)]

    =84+126

    =210

    Hence L.H.S=R.H.S

    (21)×[(4)+(6)]=[(21)×(4)]+[(21)×(6)]

Explore Solutions for Mathematics

Sign In