RS Aggarwal Class 6 Maths Chapter 3 - Whole Number Test Paper

RS Aggarwal 2021-2022 for Class 6 Maths Chapter 3 - Whole Number 

Rs Aggarwal Class 6 Math Solution Chapter 3 Whole Number Test Paper 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 3 Whole Number, Test Paper


Test Paper


1. How many whole numbers are there between 1064 and 1201?

Solution:

    Number of whole numbers between 1201 and 1064 = ( 1201 − 1064 ) − 1

                                                                                        = 137 − 1

                                                                                         = 136

2. Fill in the blanks.

        1 0 0 0 0 0 0                                                                                                                                                - * * * * 1                                                                                                                                                 * 7 2 4 2 *  

Solution: 

        1 0 0 0 0 0 0                                                                                                                                                - 2 9 5 7 1                                                                                                                                                 9 7 2 4 2 9  


3. Use distributive law to find the value of 1063 ×  128 − 1063 × 28.

Solution:

    Using distributive law, we have:

    1063 × 128 − 1063 × 28

    = 1063 × (128 − 28)

    = 1063 × 100

    = 106300


4. Find the product of largest 5 -digit number and the largest 3 -digit number using distributive law.

Solution:

    Largest 5-digit number = 99999

    Largest 3-digit number = 999

    By using distributive law, we have:

    Product = 99999 × 999               

                 = 99999 × (1000 − 1)                  [By using distributive law]

                 = 99999 × 1000 −  99999 × 1

                 = 99999000 − 99999

                 = 99899001


5. Divide 53968 by 267and check the result by the division algorithm.

Solution:

   RS Aggarwal Class 6 Maths Chapter 3 - Whole Number Test Paper

    Dividend = 53968,  Divisor = 267,  Quotient = 202,  Remainder = 34
    Check:
 
    Dividend = Divisor × Quotient + Remainder
         53968 =  267 × 202 + 34
         53968 = 53934 + 34
         53968 = 53968  (Verified)


6. Find the largest 6-digit number divisible by 16.

Solution:

    Largest 6-digit number = 999999

    Now, we divide 999999 by 16

  RS Aggarwal Class 6 Maths Chapter 3 - Whole Number Test Paper

    Remainder = 15

    Required Largest six-digit number  = 999999 − 15 = 999984.


7. The cost price of 23 TV sets is Rs. 570055. Find the cost of each such set.

Solution:

    Cost price of 23 TV sets = Rs 5,70,055

    Cost price of 1 TV set = 570055 ÷ 23

  RS Aggarwal Class 6 Maths Chapter 3 - Whole Number Test Paper

    ∴ Cost of one TV set = Rs. 24785.


8. What least number must be subtracted from 13801 to get a number exactly divisible by 87?

Solution:

    First we will divide 13801 by 87

  RS Aggarwal Class 6 Maths Chapter 3 - Whole Number Test Paper

    Remainder = 55

    Required number is 55, that should be subtracted from 13801 to get a number divisible by 87.


9. The value of (89 × 76 + 89 × 24) is

    (a) 890

    (b) 8900

    (c) 89000

    (d) 10420

Solution: Correct option is (b) 8900

     (89 × 76 + 89 × 24)

    = 89 × (76 + 24)

    = 89 × 100

    = 8900


10. On dividing a number by 53 we get 8 as quotient and 5 as remainder. The number is

    (a) 419

    (b) 423

    (c) 429

    (d) none of these

Solution: Correct option is (c) 429

    Divisor = 53, Quotient = 8, Remainder = 5 and Dividend = ?

    Now, Dividend = Quotient × Divisor + Remainder

                             = 8 × 53 + 5 

                              = 429

11. The whole number which has no predecessor is 

    (a) 1

    (b) 0

    (c) 2

    (d) none of these

Solution: Correct option is (b) 0.

    The whole number which has no predecessor is 0.

    i.e., 0 − 1 = −1, which is not a whole number.


12. 67 + 33 = 33 + 67 is an example of

    (a) closure property

    (b) associative property

    (c) commutative property

    (d) distributive property

Solution: Correct option is (c)

    67 + 33 = 33 + 67 is an example of​ commutative property of addition.


13. Additive inverse of 36 is

    (a) `frac\{1}{ 36}`

    (b) 0

    (c) -36

    (d) none of these

Solution: Correct option is (c) -36

    The additive inverse of 36 is −36.

    i.e., 36 + (−36) = 0


14. Which of the following is not zero?

    (a) 0 × 0

    (b) `frac\{0}{2}`

    (c) `frac\{8 - 8}{2}`

    (d) 2 + 0

Solution: Correct option is (d) 2 + 0


15. The predecessor of the smallest 3-digit number is

    (a) 999

    (b) 100

    (c) 101

    (d) 99

Solution: Correct option is (d) 99

    Smallest three-digit number = 100

    ∴ Predecessor of 100 = 100 − 1 = 99


16. The number of whole number between the smallest whole number and the greatest 2-digit number is

    (a) 88

    (b) 98

    (c) 99

    (d) 101

Solution: Correct option is (b) 98.

    Smallest whole number = 0

    Greatest two-digit number = 99

    Number of whole numbers between 0 and 99 = (99 − 0 ) − 1 = 98


17. Fill in the blanks.

    (i) The smallest natural number is                 .

    (ii) The smallest whole number is                  .

    (iii) Division by                   is not defined.

    (iv)                   is a whole number which is not a natural number.

    (v)                   is a whole number which is not a natural number.

Solution:

    (i) The smallest natural number is 1.

    (ii) The smallest whole number is 0.

    (iii) Division by 0 is not defined.

    (iv) 0 is a whole number which is not a natural number.

    (v) 1 is the multiplicative identity for whole numbers.


18. Write ' T ' For true and 'F' for false in each of the following:

    (i) 0 is the smallest natural number.

    (ii) Every natural number is a whole number.

    (iii) Every whole number is a natural number.

    (iv) 1 has no predecessor in whole numbers.

Solution:

    (i)  F (false). 0 is not a natural number.

    ​(ii) T (true).  

    (iii) F (false). 0 is a whole number but not a natural number.

    (iv) F (false). 1 − 1 = 0 is a predecessor of 1, which is a whole number.


19. Match the following columns: 

Column A

Column B

(a) 137 + 63 = 63 + 137

(i) Associativity of multiplication

(b) (16 × 25) is a number

(ii) Commutativity of multiplication

(c) 365 × 18 = 18 × 365

(iii) Distributive law of multiplication over addition

(d) (86 × 14) × 25 = 86 × (14 × 25)

(iv) Commutativity of addition

(e) 23 × (80 + 5) = (23 × 80) + (23 × 5)

(v) Closure property for multiplication

Solution:

Column A

Column B

(a) 137 + 63 = 63 + 137

(iv) Commutativity of addition

(b) (16 × 25) is a number

(v) Closure property for multiplication

(c) 365 × 18 = 18 × 365

(ii) Commutativity of multiplication

(d) (86 × 14) × 25 = 86 × (14 × 25)

(i) Associativity of multiplication

(e) 23 × (80 + 5) = (23 × 80) + (23 × 5)

(iii) Distributive law of multiplication over addition






एक टिप्पणी भेजें