Telangana SCERT Class 6 Maths Chapter 2 Solution – Whole Numbers. Here in this post we provides Class 6 Maths Whole Numbers Telangana State Board Solution. Telangana State Board English Class VI Medium Students can download this Solution to Solve out Improve Your Learning Questions and Answers.

**Telangana State Board Class 6 Maths Chapter 2 Whole Numbers Solution:**

**EXERCISE – 2.1**

1.) Which of the statements are true (T) and which are false (F). Correct the false statements.

i.) There is a natural number that has no predecessor.

ANSWER:

True. There is a natural number that has no predecessor.

ii.) Zero is the smallest whole number.

ANSWER:

True.Zero is the smallest whole number.

iii.) All whole numbers are natural numbers.

ANSWER:

False. The correct statement is

All natural numbers are whole numbers

iv.) A whole number that lies on the number line lies to the right side of another number is the greater number.

ANSWER:

True.A whole number that lies on the number line lies to the right side of another number is the greater number.

v.) A whole number on the left of another number on the number line, is greater.

ANSWER:

False. The correct statement is

The whole number on the left of another number on the number line, is smaller

vi.) We can’t show the smallest whole number on the number line.

ANSWER:

False. The correct statement is

We can show the smallest whole number on the number line.

vii. We can show the greatest whole number on the number line.

ANSWER:

False. The correct statement is

We cannot show the greatest whole number on the number line.

2.) How many whole numbers are there between 27 and 46?

ANSWER:

We have to find whole numbers between 27 and 46.

Whole numbers between 27 and 46 = (46 – 27) – 1

Whole numbers between 27 and 46 = 19 – 1

**Whole numbers between 27 and 46 = 18**

3.) Find the following using number line.

i.) 6 + 7 + 7

ANSWER:

We know,

6 + 7 + 7 = 20

We have to show this on number line.

ii.) 18 – 9

ANSWER:

We know,

18 – 9 = 9

We have to show this on number line.

iii.) 5 × 3

ANSWER:

We know,

5 × 3 = 15

We have to show this on number line.

**4.) In each pair, state which whole number on the number line is on the right of the other number?**

i.) 895; 239

ANSWER:

We have to show which whole number on the number line is on the right of the other number.

895 is on the right of 239

ii.) 1001; 10001

ANSWER:

We have to show which whole number on the number line is on the right of the other number.

10001 is on the right of 1001

iii.) 10015678; 284013

ANSWER:

We have to show which whole number on the number line is on the right of the other number.

10015678 is on the right of 284013

**5.) Mark the smallest whole number on the number line.**

ANSWER:

The smallest whole number on the number line

We know,

The smallest whole number is 0.

**6.) Choose the appropriate symbol from < or >**

i.) 8 ………. 7

ANSWER:

8 > 7

ii.) 5 ………. 2

ANSWER:

5 > 2

iii.) 0 ………. 1

ANSWER:

0 < 1

iv.) 10 ………. 5

ANSWER:

10 > 5

**7.) Place the successor of 11 and predecessor of 5 on the number line.**

ANSWER:

We have to show the successor of 11 and predecessor of 5 on the number line.

The successor of 11 = 12

Predecessor of 5 = 4

**EXERCISE – 2.2**

**1.) Give the results without actually performing the operations using the given information.**

i.) 28 × 19 = 532 then 19 × 28 =

ANSWER:

Given, 28 × 19 = 532

We know, addition and multiplication are commutative for whole numbers.

19 × 28 =532

ii.) 1 × 47 = 47 then 47 × 1 =

ANSWER:

Given,1 × 47 = 47

We know, addition and multiplication are commutative for whole numbers.

47 × 1 = 47

iii.) a × b = c then b × a =

ANSWER:

Given,a × b = c

We know, addition and multiplication are commutative for whole numbers.

b × a = c

iv.) 58 + 42 = 100 then 42 + 58 =

ANSWER:

Given,58 + 42 = 100

We know, addition and multiplication are commutative for whole numbers.

42 + 58 = 100

v.) 85 + 0 = 85 then 0 + 85 =

ANSWER:

Given,85 + 0 = 85

We know, addition and multiplication are commutative for whole numbers.

0 + 85 =85

vi.) a + b = d then b + a =

ANSWER:

Given,a + b = d

We know, addition and multiplication are commutative for whole numbers.

b + a = d

2.) Find the sum by suitable rearrangement:

i.) 238 + 695 + 162

ANSWER:

Here we have to find sum of 238 + 695 + 162

We know, addition and multiplication are associative over whole numbers.

(238 + 162) + 695

= 400 + 695

238 + 695 + 162 = 1095

ii.) 154 + 197 + 46 + 203

ANSWER:

Here we have to find sum of154 + 197 + 46 + 203

We know, addition and multiplication are associative over whole numbers.

(154 + 46) + (197 +203)

= 200 + 400

154 + 197 + 46 + 203= 600

3.) Find the product by suitable rearrangement.

i.) 25 × 1963 × 4

ANSWER:

25 × 1963 × 4

We know, addition and multiplication are associative over whole numbers.

(25 × 4) × 1963

100 x 1963

**25 × 1963 × 4 = 196300**

ii.) 20 × 255 × 50 × 6

ANSWER:

20 × 255 × 50 × 6

We know, addition and multiplication are associative over whole numbers.

(20 × 50) × (255 × 6)

1000 x 1530

**20 × 255 × 50 × 6 = 1530000**

** **

4.) Find the value of the following:

i.) 368 × 12 + 18 × 368

ANSWER:

368 × 12 + 18 × 368

368 is common.

368 x (12 + 18)

**368 x 30 = 11040**

ii.) 79 × 4319 + 4319 × 11

ANSWER:

79 × 4319 + 4319 × 11

4319is common.

4319 x (79 + 11)

**4319 x 90 = 388710**

5.) Find the product using suitable properties:

i.) 205 × 1989

ANSWER:

205 × 1989

We use distributive property

(200 + 5) × 1989

200 × 1989 + 5 × 1989

3, 97,800 + 9945

205 × 1989 = 407745

ii.) 1991 × 1005

ANSWER:

1991 × 1005

We use distributive property

1991 × (1000 + 5)

1991x 1000 + 1991 x 5

1991000 + 9955 = 2000955

1991 × 1005 = 2000955

**6.) A milk vendor supplies 56 liters of milk in the morning and 44 liters of milk in the evening to a hostel. If the milk costs Rs.30 per liter, how much money he gets per day?**

ANSWER:

Given, a milk vendor supplies 56 liters of milk in the morning and 44 liters of milk in the evening to a hostel.

The milk costs Rs.30 per liter

We have to find money he gets per day.

Total milk = 56 liters + 44 liters = 100 liters

The milk costs Rs.30 per liter

Money he gets per day = Rs.30 per liter x 100 liters

**Money he gets per day = Rs.3000**

** **

**7.) Chandana and Venu purchased 12 note books and10 note books respectively. The cost of each note book is Rs.15,then how much amount should they pay to the shop keeper?**

ANSWER:

Given,Chandana and Venu purchased 12 note books and10 note books respectively

The cost of each note book is Rs.15

We have to find how much amount they should pay to the shop keeper.

Total number of note books = 12 note books + 10 note books

Total number of note books = 22 note books

The cost of each note book is Rs.15

Amount should they pay to the shop keeper = The cost of each note book x 22 note books

Amount should they pay to the shop keeper = 15 x 22 note books

**Amount should they pay to the shop keeper = Rs.330**

**8.) Match the following**

**i.) 1991+7 = 7+1991** [ ] a. Additive identity

ii.) 68×50 = 50×68 [ ] b. Multiplicative identity

iii.) 1 [ ] c. Commutative under addition

iv.) 0 [ ] d. Distributive property of multiplication over addition

v.) 879×(100+30) = 879×100+879×30 [ ] e. Commutative under multiplication

ANSWER:

i.) 1991+7 = 7+1991 c. Commutative under addition

ii.) 68×50 = 50×68 e. Commutative under multiplication

iii.) 1 b. Multiplicative identity

iv.) 0 a. Additive identity

v.) 879×(100+30) = 879×100+879×30 d. Distributive property of multiplication over addition

**EXERCISE – 2.3**

**1.) Study the pattern:**

1 × 8 + 1 = 9

12 × 8 + 2 = 98

123 × 8 + 3 = 987

1234 × 8 + 4 = 9876

12345 × 8 + 5 = 98765

Write the next four steps. Can you find out how the pattern works?

ANSWER:

We have to find next four steps of pattern.

123456 × 8 + 6 = 987654

1234567 × 8 + 7 = 9876543

12345678 × 8 + 8 = 98765432

123456789 × 8 + 9 = 987654321

**2.) Study the pattern:**

91 × 11 × 1 = 1001

91 × 11 × 2 = 2002

91 × 11 × 3 = 3003

Write next seven steps. Check, whether the result is correct.

Try the pattern for 143 × 7 × 1, 143 × 7 × 2…..

ANSWER:

We have to find next seven steps of pattern.

91 × 11 × 4 = 4004

91 × 11 × 5 = 5005

91 × 11 × 6 = 6006

91 × 11 × 7 = 7007

91 × 11 × 8 = 8008

91 × 11 × 9 = 9009

91 × 11 × 10 = 10010

Pattern for

143 × 7 × 1 = 1001

143 × 7 × 2 = 2002

143 × 7 × 3 = 3003

143 × 7 × 4 = 4004

143 × 7 × 5 = 5005

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