**PSEB Class 6 Maths Chapter 2 Whole Numbers Questions and Answers Solution**

PSEB Punjab Board Class 6 Maths Textbook Solution Chapter 2 Whole Numbers Exercise Questions and Answers. We have also included some Additional questions, so that students can prepare more strongly.

**Exercise 2.1**

1.) Answer the following questions.

(a) Write the smallest whole number.

ANSWER:

0 is the smallest whole number.

(b) Write the smallest natural number.

ANSWER:

1 is the smallest natural number.

(c) Write the successor of 0 in whole numbers.

ANSWER:

Successor is next number of given number.

**1 is the successor of 0 in whole numbers.**

(d) Write the predecessor of 0 in whole numbers.

ANSWER:

Predecessor is previous number of given number.

**The predecessor of 0 in whole numbers not possible.**

** **

(e) Largest whole number.

ANSWER:

**Largest whole number can’t determine.**

2.) Which of the following statements are True (T) and which are False (F)?

(a) Zero is the smallest natural number.

ANSWER:

False.

1 is the smallest natural number.

(b) Zero is the smallest whole number.

ANSWER:

True. Zero is the smallest whole number.

(c) Every whole number is a natural number.

ANSWER:

False.

**Every whole number is not a natural number.**

(d) Every natural number is a whole number.

ANSWER:

**True. Every natural number is a whole number.**

(e) 1 is the smallest whole number.

ANSWER:

False.

**0 is the smallest whole number.**

(f) The natural number 1 has no predecessor in natural numbers.

ANSWER:

**True. The natural number 1 has no predecessor in natural numbers.**

** **

(g) The whole number 1 has no predecessor in whole numbers.

ANSWER:

False.

**The whole number 0 has no predecessor in whole numbers.**

(h) Successor of the largest two digit number is smallest three digit number.

ANSWER:

**True. Successor of the largest two digit number is smallest three digit number.**

(i) The successor of a two digit number is always a two digit number.

ANSWER:

False.

The successor of a two digit number is not always a two digit number.

(j) 300 is the predecessor of 299.

ANSWER:

False.

**300 is the successor of 299.**

** **

(k) 500 is the successor of 499.

ANSWER:

**True. 500 is the successor of 499.**

** **

(l) The predecessor of a two digit number is never a single digit number.

ANSWER:

False.

** The predecessor of a two digit number will be a single digit number.**

** **

3.) Write the successor of each of following:

(a) 100909

ANSWER:

We have to write successor of 100909

Successor of 100909 = 100910

(b) 4630999

ANSWER:

We have to write successor of 4630999

Successor of 4630999 = 4631000

(c) 830001

ANSWER:

We have to write successor of 830001

Successor of 830001 = 830002

(d) 99999

ANSWER:

We have to write successor of 99999

Successor of 99999 = 100000

4.) Write the predecessor of each of following:

(a) 1000

ANSWER:

We have to write predecessor of 1000

Predecessor of 1000 = 999

(b) 208090

ANSWER:

We have to write predecessor of 208090

Predecessor of 208090 = 208089

(c) 7654321

ANSWER:

We have to write predecessor of 7654321

Predecessor of 7654321 = 7654320

(d) 12576

ANSWER:

We have to write predecessor of 12576

Predecessor of 12576 = 12575

5.) Represent the following numbers on the number line. 2, 0, 3, 5, 7, 11, 15

ANSWER:

We have to represent 2, 0, 3, 5, 7, 11, 15 on number line.

6.) How many whole numbers are there between 22 and 43?

ANSWER:

We have to find whole numbers between 22 and 43.

Whole numbers between 22 and 43 = (43 – 22) – 1

Whole numbers between 22 and 43 = 21 – 1

**Whole numbers between 22 and 43 = 20**

** **

7.) Draw a number line to represent each of following on it.

(a) 3 + 2

ANSWER:

We have to represent given on number line.

3 + 2 = 5

(b) 4 + 5

ANSWER:

We have to represent given on number line.

4 + 5 = 9

(c) 6 + 2

ANSWER:

We have to represent given on number line.

6 + 2 = 8

(d) 8 – 3

ANSWER:

We have to represent given on number line.

8 – 3 = 5

(e) 7 – 4

ANSWER:

We have to represent given on number line.

7 – 4 = 3

(f) 7 – 2

ANSWER:

We have to represent given on number line.

7 – 2 = 5

(g) 3 x 3

ANSWER:

We have to represent given on number line.

3 x 3 = 9

(h) 2 x 5

ANSWER:

We have to represent given on number line.

2 x 5 = 10

(i) 3 x 5

ANSWER:

We have to represent given on number line.

3 x 5 = 15

(j) 9/3

ANSWER:

We have to represent given on number line.

9/3 = 3

(k) 12/4

ANSWER:

We have to represent given on number line.

12/4 = 3

(l) 10/2

ANSWER:

We have to represent given on number line.

10/2 = 5

8.) Fill in the blanks with the appropriate symbol < or>:

(a) 25 —– 205

ANSWER:

We have to fill appropriate symbol in blank.

**25 < 205**

** **

(b) 170 ——- 107

ANSWER:

We have to fill appropriate symbol in blank.

**170 > 107**

** **

(c) 415———514

ANSWER:

We have to fill appropriate symbol in blank.

**415 < 514**

** **

(d) 10001 ———- 9999

ANSWER:

We have to fill appropriate symbol in blank.

**10001 > 9999**

** **

(e) 2300014 ———– 2300041

ANSWER:

We have to fill appropriate symbol in blank.

**2300014 < 2300041**

** **

(f) 99999 ————– 888888

ANSWER:

We have to fill appropriate symbol in blank.

**99999 > 888888**

** **

__Exercise 2.2__

__Exercise 2.2__

1.) Find the sum by suitable arrangement of terms:

(a) 837 + 208 + 363

ANSWER:

We have to find the sum by suitable arrangement of terms.

(837 + 363)+ 208

1200 + 208

**837 + 208 + 363 = 1408**

(b) 1962 + 453 + 1538 + 647

ANSWER:

We have to find the sum by suitable arrangement of terms.

1962 + 453 + 1538 + 647

(1962 + 1538) + (647 + 453)

3500 + 1100

**1962 + 453 + 1538 + 647 = 4600**

** **

2.) Find the product by suitable arrangement of terms:

(a) 2 x 1497 x 50

ANSWER:

We have to find the product by suitable arrangement of terms

2 x 1497 x 50 = 1497 x (2 x 50)

1497 x (2 x 50) = 1497 x 100

**2 x 1497 x 50 = 149700**

** **

(b) 4 x 263 x 25

ANSWER:

We have to find the product by suitable arrangement of terms

4 x 263 x 25 = 263 x (4 x 25)

263 x (4 x 25) = 263 x 100

**4 x 263 x 25 = 26300**

** **

(c) 8 x 163 x 125

ANSWER:

We have to find the product by suitable arrangement of terms

8 x 163 x 125 = 163 x (8 x 125)

163 x (8 x 125) = 163 x 1000

**8 x 163 x 125 = 163000**

** **

(d) 963 x 16 x 25

ANSWER:

We have to find the product by suitable arrangement of terms

963 x 16 x 25 = 963 x 4 x 4 x 25

(963 x 4) x 100 = 3852 x 100

**963 x 16 x 25 = 385200**

** **

(e) 5 x 171 x 60

ANSWER:

We have to find the product by suitable arrangement of terms.

5 x 171 x 60 = 171 x (5 x 60)

171 x (5 x 60) = 171 x 300

= 513 x 100

**5 x 171 x 60 = 51300**

** **

(f) 125 x 40 x 8 x 25

ANSWER:

We have to find the product by suitable arrangement of terms

125 x 40 x 8 x 25 = (125 x 8) x (25 x 40)

= 1000 x 1000

**125 x 40 x 8 x 25 = 1000000**

** **

(g) 30921 x 25 x 40 x 2

ANSWER:

We have to find the product by suitable arrangement of terms

30921 x 25 x 40 x 2 = (30921 x 2) x (25 x 40)

(30921 x 2) x (25 x 40) = 61842 x 1000

**30921 x 25 x 40 x 2 = 61842000**

** **

(h) 4 x 2 x 1932 x 125

ANSWER:

We have to find the product by suitable arrangement of terms

4 x 2 x 1932 x 125 = (4 x 2 x 125) x 1932

(4 x 2 x 125) x 1932 = 1000 x 1932

**4 x 2 x 1932 x 125 = 1932000**

** **

(i) 5462 x 25 x 4 x 2

ANSWER:

We have to find the product by suitable arrangement of terms

5462 x 25 x 4 x 2 = (5462 x 2) x (25 x 4)

(5462 x 2) x (25 x 4) = 10924 x 100

5462 x 25 x 4 x 2 = 1092400

3.) Find the value of each of the following using distributive property:

(a) (649 x 8) + (649 x 2)

ANSWER:

We have to find value of given using distributive property.

(649 x 8) + (649 x 2), here, 649 is common.

We take outside.

649 x (8 + 2)

649 x 10

**(649 x 8) + (649 x 2) = 6490**

** **

(b) (6524 x 69) + (6524 x 31)

ANSWER:

We have to find value of given using distributive property.

(6524 x 69) + (6524 x 31) here, 6524 is common.

We take outside.

6524 x (69 + 31)

6524 x 100

**(6524 x 69) + (6524 x 31) = 652400**

** **

(c) (2986 x 35) + (2986 x 65)

ANSWER:

We have to find value of given using distributive property.

(2986 x 35) + (2986 x 65) here, 2986 is common.

We take outside.

2986 x (35 + 65)

2986 x 100

**(2986 x 35) + (2986 x 65) = 298600**

** **

(d) (6001 x 172) – (6001 x 72)

ANSWER:

We have to find value of given using distributive property.

(6001 x 172) – (6001 x 72) here, 6001 is common.

We take outside.

6001 x (172 – 72)

6001 x 100

**(6001 x 172) – (6001 x 72) = 600100**

** **

4.) Find the value of the following:

(a) 493 x 8 + 493 x 2

ANSWER:

We have to find value of 493 x 8 + 493 x 2

Here, 493 is common.

We take outside.

493 x (8 + 2)

493 x 10

**493 x 8 + 493 x 2 = 4930**

(b) 24579 x 93 + 7 x 24579

ANSWER:

We have to find value of 24579 x 93 + 7 x 24579

Here, 24579 is common.

We take outside.

24579 x (93 + 7)

24579 x 100

**24579 x 93 + 7 x 24579 = 2457900**

(c) 3845 x 5 x 782 + 769 x 25 x 218

ANSWER:

We have to find value of 3845 x 5 x 782 + 769 x 25 x 218

We done arrangements.

3845 x 5 x 782 + 769 x 25 x 218 = [(3845 x 5) + (769 x 25)] x (782 + 218)

[(3845 x 5) + (769 x 25)] x (782 + 218) = [(3845 x 5) + (769 x 25)] x 1000

**But, (3845 x 5) and (769 x 25) = 19225 is common.**

(3845 x 5) x 1000 = 19225 x 1000

**3845 x 5 x 782 + 769 x 25 x 218 = 19225000 **

** **

(d) 3297 x 999 + 3297

ANSWER:

We have to find value of 3297 x 999 + 3297

Here, 3297 is common.

We take outside.

3297 x (999 + 1)

3297 x 1000

**3297 x 999 + 3297 = 3297000**

** **

5.) Find the product using suitable properties:

(a) 738 x 103

ANSWER:

We done suitable arrangements.

738 x 103 = 738 x (100 + 3)

738 x 100 + 738 x 3

73800 + 2214

**738 x 103 = 76014**

** **

(b) 854 x 102

ANSWER:

We done suitable arrangements.

854 x 102 = 854 x (100 + 2)

854 x 100 + 854 x 2

85400 + 1708

**854 x 102 = 87108**

** **

(c) 258 x 1008

ANSWER:

We done suitable arrangements.

258 x 1008 = 258 x 1000 + 258 x 8

258 x 1000 + 258 x 8 =

258000 + 2064

**258 x 1008 = 260064**

** **

(d) 736 x 93

ANSWER:

We done suitable arrangements.

736 x 93 =736 x (100 – 7)

736 x 100 – 736 x 7

= 73600 – 5152

**736 x 93 = 68448**

** **

(e) 816 x 745

ANSWER:

We done suitable arrangements.

816 x 745 = 816 x (800 – 55)

816 x 800 – 816 x 55

= 652800 – 44880

**816 x 745 = 607920**

** **

(f) 2032 x 613

ANSWER:

We done suitable arrangements.

2032 x 613 = 2032 x (600 + 13)

2032 x 600 + 2032 x 13

1219200 + 26416

**2032 x 613 = 1245616**

** **

6.) A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs 78 per litre, how much he spend in all on petrol?

ANSWER:

Given that,

A taxi driver filled his car petrol tank with 40 litres of petrol on Monday.

The next day, he filled the tank with 50 litres of petrol.

The petrol costs 78 per litre

We have to find how much he spend in all on petrol.

Total petrol filled in his car = 40 litres + 50 litres

**Total petrol filled in his car = 90 litres**

Money he spend in all on petrol = 78 per litre x 90

**Money he spend in all on petrol = Rs.7020**

7.) A vendor supplies 32 litres of milk to a hotel in morning and 68 litres of milk in the evening. If the milk costs ₹35 per litre, how much money is due to the vendor per day?

ANSWER:

Given that,

A vendor supplies 32 litres of milk to a hotel in morning and 68 litres of milk in the evening.

If the milk costs ₹35 per litre.

We have to find money is due to the vendor per day.

1^{st} we find total amount of milk supplies to hotel = 32 litres + 68 litres

Total amount of milk supplies to hotel = 100 litres

Money is due to the vendor per day = 100 litres x ₹35 per litre

**Money is due to the vendor per day =₹3500**

8.) We know that 0 x 0 = 0 Is there any other whole number which when multiplied by itself gives the product equal to the number itself? Find out the number.

ANSWER:

We know,

0 x 0 = 0

Whole number which when multiplied by itself gives the product equal to the number itself is 1.

**1 x 1 = 1**

** **

9.) Fill in the blanks:

(a) 15 x 0 = —

ANSWER:

We have to fill blank spaces.

15 x 0 = **0**

(b) 15 + 0 = —

ANSWER:

We have to fill blank spaces.

15 + 0 = **15**

(c) 15 – 0 = —

ANSWER:

We have to fill blank spaces.

15 – 0 = **15**

(d) 15 / 0 = —

ANSWER:

We have to fill blank spaces.

15 / 0 = **Not defined**

(e) 0 x 15 = —

ANSWER:

We have to fill blank spaces.

0 x 15 = **0**

f) 0 + 15 = —

ANSWER:

We have to fill blank spaces.

0 + 15 = **15**

(g) 0/15 = —

ANSWER:

We have to fill blank spaces.

**0/15 = 0**

** **

(h)15 x 1 = —

ANSWER:

We have to fill blank spaces.

15 x 1 = **15**

(i) 15 / 1 = —

ANSWER:

We have to fill blank spaces.

15 / 1 = **15**

(j) 1 / 1 = —

ANSWER:

We have to fill blank spaces.

1 / 1 = **1**

10.) The product of two Whole numbers is zero. What do you conclude? Explain with example.

ANSWER:

Suppose 1 whole number is 12 and 2^{nd} whole number is 0.

1 whole number x 2nd whole number = 12 x 0

**1 whole number x 2nd whole number = 0**

** **

11.) Match the following:

(i) 537 x 106 = 537 x 100 + 537 x 6 (a) Commutativity under multiplication

(ii) 4 x 47 x 25 = 4 x 25 x 47 (b) Commutativity under addition

(iii) 70 + 1923 + 30 = 70 + 30 + 1923 (c) Distributivity of multiplication over addition.

ANSWER:

(i) 537 x 106 = 537 x 100 + 537 x 6 ==**(c) Distributivity of multiplication over addition. **

(ii) 4 x 47 x 25 = 4 x 25 x 47 ==**(a) Commutativity under multiplication**

(iii) 70 + 1923 + 30 = 70 + 30 + 1923 ==**(b) Commutativity under addition**

__Exercise 2.3__

__Exercise 2.3__

1.) If the product of two whole numbers is zero. Can we say that one or both of them will be zero? Justify through examples.

ANSWER:

Yes. We can say that one or both of them will be zero, when the product of two whole numbers is zero.

For example,

Suppose 1 whole number is 54 and 2nd whole number is 0.

1 whole number x 2nd whole number = 54 x 0

1 whole number x 2nd whole number = 0

2.) If the product of two whole numbers is 1. Can we say that one or both of them will be 1? Justify through examples.

ANSWER:

Yes, we can say that both of them will be 1, when the product of two whole numbers is 1.

For example,

Suppose 1 whole number is 1 and 2nd whole number is 1.

1 whole number x 2nd whole number = 1 x 1

1 whole number x 2nd whole number = 1

- Observe the pattern in the following and fill in the blanks:

1 x 1 = 1

11 x 11 = 121

111 x 111 = 12321

1111 x 1111 = ———

11111 x 1111 = —————–

ANSWER:

By observing pattern, we fill blank.

1 x 1 = 1

11 x 11 = 121

111 x 111 = 12321

1111 x 1111 =__1234321__

11111 x 11111 = __123454321__

4.) Observe the pattern and fill in the blanks:

1 x 9 + 1 = 10

12 x 9 + 2 = 110

123 x 9 + 3 = 1110

1234 x 9 + 4 = 11110

12345 x 9 + 5 = ———-

123456 x 9 + 6 = ———————–

ANSWER:

By observing pattern, we fill blank.

1 x 9 + 1 = 10

12 x 9 + 2 = 110

123 x 9 + 3 = 1110

1234 x 9 + 4 = 11110

12345 x 9 + 5 = __111110__

123456 x 9 + 6 = __1111110__

5.) Represent numbers from 24 to 30 according to rectangular, square or triangular pattern.

ANSWER:

**25 as square pattern.**

6.) Study the following pattern:

1= 1 x 1 = 1

1 + 3 = 2 x 2 = 4

1 + 3 + 5 = 3 x 3 = 9

1 + 3 + 5 + 7 = 4 x 4 = 16

Hence find the sum of

(a) First 12 odd numbers

ANSWER:

By observing pattern, we have to find the sum of First 12 odd numbers.

**1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12 x 12 = 144**

** **

(b) First 50 odd numbers.

ANSWER:

By observing pattern, we have to find the sum of First 50 odd numbers.

**1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + ——————— + 99 = 50 x 50 = 2500**

** **

**Multiple Choice Questions**

1.) The smallest whole number is

(a) 0

(b) 1

(c) 2

(d) 3

ANSWER: (a) 0

2.) The smallest natural number is

(a) 0

(b) 1

(c) 2

(d) 3

ANSWER: (b) 1

3.) The successor of 38899 is

(a) 39000

(b) 38900

(c) 39900

(d) 38800

ANSWER: (b) 38900

4.) The predecessor of 24100 is

(a) 24999

(b) 24009

(c) 24199

(d) 24099

ANSWER: (d) 24099

5.) The statement 4 + 3 = 3 + 4 represents

(a) Closure

(b) Associative

(c) Commutative property

(d) Identity

ANSWER: (c) Commutative property

6.) Which of the following is the additive identity?

(a) 0

(b) 1

(c) 2

(d) 3

ANSWER: (a) 0

7.) The multiplicative identity is

(a) 0

(b) 1

(c) 2

(d) 3

ANSWER: (b) 1

8.) 15 x 32 + 15 x 68 =

(a) 1400

(b) 1600

(c) 1700

(d) 1500

ANSWER: (d) 1500

9.) The largest 4 digit number divisible by 13 is

(a) 9997

(b) 9999

(c) 9995

(d) 9991

ANSWER: (a) 9997

10.) The successor of 3 digit largest number is

(a) 100

(b) 998

(c) 1001

(d) 1000

ANSWER: (d) 1000

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