**Please check Solutions of Whole numbers Class 6 Worksheet at the end of the questions.**

**Download Whole numbers Class 6 Worksheet**

**1. Find the successor of 56286**

a) 56280

b) 56284

c) 56287

d) 56285

**2. Find the predecessor of 85265 ?**

a) 85264

b) 85266

c) 85268

d) 85267

**3. In the following pairs of number, state which number will be on right hand side of the whole number line?**

**( 267 , 359 )**

a) 267

b) 359

**4. Find the sum of 453 + 867 + 674 ?**

a) 1994

b) 1993

c) 1995

d) 1996

**5. Find the sum of the following whole numbers 787 + 99 ?**

a) 885

b) 887

c) 886

d) 888

**6. Find: 5628 x 99**

a) 557173

b) 557172

c) 557175

d) 557174

**7. Find the number which when divided by 89 gives 4 as quotient and 6 as remainder**

a) 362

b) 368

c) 356

d) 350

**8. Find the number which when divided by 82 gives 12 as quotient and 3 as remainder?**

a) 990

b) 987

c) 984

d) 981

**9. If the Dividend = 6451 , quotient = 7 and remainder = 4 , find the Divisor ?**

a) 921

b) 914

c) 917

d) 928

**10. Find the value of the following:-**

**( 250 x 16 ) + ( 14 x 250 )**

a) 7491

b) 7500

c) 7509

d) 7507

**11. Solve 20 x ( 8 – 5 ) ?**

a) 60

b) 61

c) 59

d) 62

**12. A person bought 15 Books on Friday and on the next day he bought 12 Books If one book is costing him 30 rupees then how much money he had spend on Books ?**

a) ₹ 360

b) ₹ 450

c) ₹ 800

d) ₹ 810

**13. Write the three whole numbers occurring just after 40053.**

a) 40053 , 40054 , 40055

b) 40052 , 40051 , 40050

c) 40054 , 40055 , 40056

d) 40051 , 40050 , 40049

### Whole numbers Class 6 Worksheet Solutions

**Solution 1**

The “Successor” of any whole number is the number, obtained by adding 1 to that number

So, the successor of 56286 + 1 = 56287

**Correct Answer – c) 56287**

**Solution 2**

The “Predecessor” of any whole number is the number obtained by subtracting 1 from it

So, the predecessor of 85265 is 85265 – 1 = 85264

**Correct Answer – a) 85264 **

**Whole numbers Class 6 Worksheet – Solution 3**

We know that the smaller number comes on the left and bigger number comes on right on the whole number line.

So between 267 and 359

267 is smaller number

So, 267 will be on left hand side while 359 will be on right hand side

267 < 359

Hence, 359 will be on right hand side of the whole number line

**Correct Answer – b) 359**

**Whole numbers Class 6 Worksheet – Solution 4**

453 + ( 867 + 674 )

= 453 + 1541

= 1994

Hence, the sum of 453 , 867 and 674 is 1994

**Correct Answer – a) 1994**

**Whole numbers Class 6 Worksheet – Solution 5**

787 + 99

= 787 + ( 100 – 1 )

= ( 787 + 100 ) – 1

= 887 – 1

= 886

Hence, the sum of the given numbers is 886

**Correct Answer – c) 886**

**Whole numbers Class 6 Worksheet – Solution 6**

5628 x ( 100 – 1 )

= ( 5628 x 100 ) – ( 5628 x 1 ) (Here, we use distributive law of multiplication over subtraction )

i.e a x (b – c) = ( a x b ) – ( a x c )

= 562800 – 5628

= 557172

So, the product of 5628 x 99 is 557172

**Correct Answer – b) 557172**

**Whole numbers Class 6 Worksheet – Solution 7**

Given,

Divisor = 89

Quotient = 4

Remainder = 6

In this question we have to use division algorithm

Dividend = (divisor x quotient) + remainder

= ( 89 x 4 ) + 6

= 356 + 6

= 362

Hence, the dividend is 362

**Correct Answer – a) 362**

**Solution 8**

Given,

Divisor = 82

Quotient = 12

Remainder = 3

In this question we have to use division algorithm

Dividend = (divisor x quotient) + remainder

= ( 82 x 12 ) + 3

= 984 + 3

= 987

Hence, the dividend is 987

**Correct Answer – b) 987**

**Whole numbers Class 6 Worksheet – Solution 9**

Given,

Dividend = 6451

Quotient = 7

Remainder = 4

In this question we have to use division algorithm

Dividend = (divisor x quotient) + remainder

Dividend – remainder = divisor x quotient

Divisor =

== 921

Hence, the divisor is 921

**Correct Answer – a) 921**

**Whole numbers Class 6 Worksheet – Solution 10**

It follows Distributive law of multiplication over addition

Which states that a x (b + c ) = (a x b) + (a x c )

Here, a = 250

b = 16

c = 14

This question is in the form of (a x b) + (a x c )

That is, ( 250 x 16 ) + ( 14 x 250 )

( 250 x 16 ) + ( 250 x 14 ) ( NOTE : Whole number satisfy commutative property under multiplication)

i.e (a x b) = (b x a)

Here, we have 14 x 250 = 250 x 14

By distributive law, = 250 x ( 16 + 14 )

= 250 x 30

= 7500

**Correct Answer – b) 7500 **

**Whole numbers Class 6 Worksheet – Solution 11**

Here, we have two method to solve this question

By Distributive law of multiplication over subtraction a x ( b – c ) = (a x b) – (a x c)

here,

a = 20

b = 8

c = 5

__1 Method__

a x (b – c)

20 x ( 8 – 5 )

= 20 x 3

= 60

__2 Method__

(a x b) – (a x c)

( 20 x 8 ) – ( 20 x 5 )

= 160 – 100

= 60

**Correct Answer – a) 60**

**Whole numbers Class 6 Worksheet – Solution 12**

Total money that he spent = total number of Books x cost of one book

= ( 15 + 12 ) x 30

= 27 x 30

= 810

Total money he spent on Books is 810 rupees

**Correct Answer – d) ₹ 810**

**Whole numbers Class 6 Worksheet – Solution 13**

The next three whole number of 40053 can be obtained by adding 1, 2 and 3 to 40053

that is,

40053 + 1 = 40054

40053 + 2 = 40055

40053 + 3 = 40056

Hence the next three whole numbers of 40053 are 40054 , 40055 , 40056 respectively.

**Correct Answer – c) 40054 , 40055 , 40056**

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