NCERT Class 6 Solutions: Whole Numbers (Chapter 2) Exercise 2.2–Part 1

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Q-1 Find the sum by suitable rearrangement:

  1. 837+208+363

  2. 1962+453+1538+647

Solution:

The idea is to pair the numbers so that their least significant (right-most) digits add to either 5 or better still to 10. For example when asked to add 11 + 12 + 19. First do 11 + 19 = 30 (notice how the total has 0 in right-most digit). Now it is easy to 30 + 12 = 42.

  1. 837+208+363

    • Rearrange the sum, 837+208+363=(837+363)+208

      (837+363)+208=1,200+208

      =1,408

  2. 1962+453+1538+647

  • Rearrangethesum1962+453+1538+647as(1962+1538)+(453+647)

    (1962+1538)+(453+647)=3,500+1,100

    =4,600

Q-2 Find the product by suitable arrangement

  1. 2×1768×50

  2. 4×166×25

  3. 8×291×125

  4. 625×279×16

  5. 285×5×60

  6. 125×40×8×25

Solution:

The idea is to pair the numbers so that a number ending in 5 is multiplied with an even number. This would produce 0 in the least significant. For example when asked to calculate 15×7×12 . First do 15×12 = 180 (notice how the total has 0 in right-most digit). Now it is easy to do 180×7 = 1260 (we could ignore the zero, then multiply and then add back the zero).

It is easy to multiply even number with 5

Multilication by 5

It is easy to multiply even number with 5

  1. 2×1768×50

    2×1768×50=(2×50)×1768

    =100×1768

    =1,76,800

  2. 4×166×25

    4×166×25=(4×25)×166

    =100×166

    =16,600

    Note here we choose to multiply 4 with 25 (even though 166 was also even), this was because we got 100 makes it very easy to multiply further.

  3. 8×291×125

    8×291×125=(8×125)×291

    =1000×291

    =2,91,000

    Here 8 is even.

  4. 625×279×16

    625×279×16=(625×16)×279

    =10,000×279

    =2,790,000

    Here 16 is even.

  5. 285×5×60

    285×5×60=285×(5×60)

    =285×300

    =85,500

    Here 60 is the even number.

  6. 125×40×8×25

    125×40×8×25=(125×8)×(40×25)

    =1000×1000

    =10,00,000

    We know that product of 8 and 125 is 1000 and so we make a pair out of them. We also know that 4 multiplied with 25 would be 100 so product of 40 and 25 would also be 1000.

Q-3 Find the value of the following:

  1. 297×17+297×3

  2. 54279×92+8×54279

  3. 81265×16981265×69

  4. 3845×5×782+769×25×218

Solution:

You can distribute multiplication over addition and subtraction:

Describing the distributive property

Image Describing the Distributive Property

Describing the distributive property

Multiplication is distributibe over addition.

Understanding Distributing Property of Multiplication Over Addition

Multiplication is distributibe over addition.

Multiplication is distributibe over subtration.

Understanding Distributing Property of Multiplication Over Subtraction

Multiplication is distributibe over subtration.

  1. 297×17+297×3

    =297×(17+3)

    =297×20

    =5,940

  2. 54279×92+8×54279

    =54279×(92+8)

    =54279×100

    =54,27,900

  3. 81265×16981265×69

    =81265×(16969)

    =81265×100

    =81,26,500

  4. 3845×5×782+769×25×218

    =(3845×5)×782+(769×25)×218

    =(19225)×782+(19225)×218

    =19225×(782+218)

    =19225×1000

    =1,92,25,000

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