 # NCERT Solutions for Class 6 Maths Chapter 2 - Whole Numbers

Find here the most reliable and detailed NCERT solutions for class 6 Maths chapter 2. Students may read and download all the solutions and use the same to effective preparation of exams. NCERT Solutions for Class 6 Maths Chapter 2 - Whole Numbers

Check NCERT solutions for class 6 Maths chapter 2: Whole Numbers. You will find here the latest NCERT solutions which will help you know the correct answers to all difficult questions. These solutions will also be quite helpful to manage your homework assignments and preparation for the exams.

## NCERT Class 6 Maths Chapter 2: Exercise 2.1

1. Write the next three natural numbers after 10999.

Solution.

The next three natural numbers after 10999 are:

11000, 11001, 11002

2. Write three whole numbers occurring just before 10001.

Solution.

Three whole numbers occurring just before 10001 are:10000, 9999 and 9998

Since 10001 – 1 = 10000

10000 – 1 = 9999

9999 – 1 = 9998

3. Which is the smallest whole number?

Solution.

Zero (0) is the smallest whole number.

4. How many whole numbers are there between 32 and 53?

Solution.

The whole numbers between 32 and 53 are:

33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52

5. Write the successor of:

(a) 2440701

(b) 100199

(c) 1099999

(d) 2345670

Solution.

(a) Successor of 244070 = 244070 + 1 = 244071

(b) Successor of 100199 = 100199 + 1 = 100200

(c) Successor of 1099999 = 1099999 + 1 = 1100000

(d) Successor of 2345670 = 2345670 + 1 = 2345671

6. Write the predecessor of:

(a) 94

(b) 10000

(c) 208090

(d) 7654321

Solution.

(a) Predecessor of 94 = 94 – 1 = 93

(b) Predecessor of 1000 = 10000 – 1 = 9999

(c) Predecessor of 208090 = 208090 – 1 = 208089

(d) Predecessor of 7654321 = 7654321 – 1 = 7654320

7. In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them,

(a) 530, 503

(b) 370, 307

(c) 98765, 56789

(d) 9830415,10023001

Solution.

On a number line, smaller number is represented on the left side of the greater number.

(a) 530, 503

As 530 > 503

Therefore, 503 is on left side of 530 on number line.

(b) 307 < 370

As 307 < 370

Therefore, 307 is on the left side of 370 on number line.

(c) 98765, 56789

As 98765 > 56789

Therefore, 56789 is on left side of 98765 on number line.

(d) 9830415, 10023001

As 9830415 < 10023001

Therefore, 9830415 is on the left side of 10023001 on the number line.

8. Which of the following statements are true (T) and which are false (F)?

(a) Zero is the smallest natural number.

(b) 400 is the predecessor of 399.

(c) Zero is the smallest whole number.

(d) 600 is the successor of 599.

(e) All natural numbers are whole numbers.

(f) All whole numbers are natural numbers.

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

(h) 1 is the smallest whole number.

(i) The natural number 1 has no predecessor.

(j) The whole number 1 has no predecessor.

(k) The whole number 13 lies between 11 and 12.

(l) The whole number 0 has no predecessor.

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

Solution.

(a) False (F); 1 is the smallest natural number.

(b) False (F); 400 is the successor of 399.

(c) True (T)

(d) True (T)

(e) True (T)

(f) False (F); All whole numbers are natural numbers except zero.

(g) False (F); Example: The predecessor of 10, a two-digit number is 9 which is a single-digit number.

(h) False (F); 0 is the smallest whole number.

(i) True (T)

(J) False (F); 0 is the predecessor of 1 which is a whole number.

(k) False (F); there is no whole number lying between 11 and 12.

(l) True (T)

(m) False (F);The successor of 99, a two-digit number is 100 which is a three-digit number.

## NCERT Class 6 Maths Chapter 2: Exercise 2.1

1. Find the sum by suitable arrangement:

(a) 837 + 208 + 363

(b) 1962 + 453,+ 1538 + 647

Solution.

(a) 837 + 208 + 363 = (837 + 363) + 208

= 1200 + 208 = 1408

(b) 1962 + 453 + 1538 + 647 = (1962 + 1538) + (453 + 647)

= 3500 + 1100 = 4600

2. Find the product by suitable arrangement:

(а) 2 x 1768 x 50

(b) 4 x 166 x 25

(c) 8 x 291 x 125

(d) 625 x 279 x 16

(e) 285 x 5 x 60

(f) 125 x 40 x 8 x 25

Solution.

(a) 2 x 1768 x 50 = (2 x 50) x 1768

= 100 x 1768 = 176800

(b) 4 x 166 x 25 = 166 x (25 x 4)

= 166 x 100 = 16600

(c) 8 x 291 x 125 = (8 x 125) x 291

= 1000 x 291 = 291000

(d) 625 x 279 x 16 = (625 x 16) x 279

= 10000 x 279 = 2790000

(e) 285 x 5 x 60 = 285 x (5 x 60)

= 285 x 300 = 85500

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

= 1000 X 1000 = 1000000

3. Find the value of the following:

(а) 297 x 17 + 297 x 3

(б) 54279 x 92 + 8 x 54279

(c) 81265 x 169 – 81265 x 69

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

Solution.

(a) 297 x 17 + 297 x 3 = 297 x (17 + 3)

= 297 x 20 = 5940

(b) 54279 x 92 + 8 x 54279 = 54279 x (92 + 8)

= 54279 x 100 = 5427900

(c) 81265 x 169 – 81265 x 69 = 81265 x (169 – 69)

= 81265 x 100 = 8126500

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

= 3845 x 5 x 782 + 769 x 5 x 5 x 218

= 3845 x 5 x 782 + (769 x 5) x 5 x 218

= 3845 x 5 x 782 + 3845 x 5 x 218

= 3845 x 5 x 782 + 3845 x 5 x 218

= 3845 x 5 x (782 + 218)

= 3845 x 5 x 1000

= 19225 x 1000

= 19225000

4. Find the product using suitable properties.

(a) 738 x 103

(b) 854 x 102

(c) 258 x 1008

(d) 1005 x 168

Solution.

(a) 738 x 103 = 738 x (100 + 3)

= 738 x 100 + 738 x 3

= 73800 + 2214 = 76014

(b) 854 x 102 = 854 x (100 + 2)

= 854 x 100 + 854 x 2

= 85400 + 1708 = 87108

(c) 258 x 1008 = 258 x (1000 + 8)

= 258 x 1000 + 258 x 8

= 258000 + 2064 = 260064

(d) 1005 x 168 = (1000 + 5) x 168

= 1000 x 168 + 5 x 168

= 168000 + 840 = 168840

5. A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litre of petrol. If the petrol cost Rs.44 per litre, how much did he spend in all on petrol?

Solution.

Petrol filled on Monday = 40 litres

Petrol filled on Tuesday = 50 litre

Total petrol filled in two days = (40 + 50) = 90 litres

Cost of 1 litre of petrol = Rs.44

Cost of petrol 90 litres of petrol = Rs. (90x 44) = Rs.3960

6. A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs Rs.15 per litre, how much money is due to the vendor per day?

Solution.

Milk supplied in the morning = 32 litres

Milk supplied in the evening = 68 litres

Total milk supplied = (32 + 68) = 100 litres

Cost of 1 litre of milk = Rs.15

Cost of 100 litre of milk = Rs. (100x 15) = Rs.1500

7. Match the following:

 (i) 425 x 136 = 425 x (6 + 30 + 100) (a) Commutativity under multiplication (ii) 2 x 49 x 50 = 2 x 50 x 49 (b) Commutativity under addition (iii) 80 + 2005 + 20 = 80 + 20 + 2005 (c) Distributivity of multiplication over addition

Solution.

Correct matching is given below:

 (i) 425 x 136 = 425 x (6 + 30 + 100) (c) Distributivity of multiplication over addition (ii) 2 x 49 x 50 = 2 x 50 x 49 (a) Commutativity under multiplication (iii) 80 + 2005 + 20 = 80 + 20 + 2005 (b) Commutativity under addition

## NCERT Class 6 Maths Chapter 2: Exercise 2.3

1. Which of the following will not represent zero: Solution. 2. If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.

Solution.

Yes, if the product of two whole numbers is zero, then one or both of them will definitely be zero.

Examples:

(i) 8 x 0 = 0 and  0 x 14 = 0

One number is zero here.

(ii) 0 x 0 = 0

Both the numbers are zero here.

3. If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.

Solution.

If the product of two numbers is one then both the numbers need to be equal to 1 only.

As 1 x 1 = 1

But 9 x 1 ≠ 9

4. Find using distributive property:

(а) 728 x 101

(b) 5437 x 1001

(c) 824 x 25

(d) 4275 x 125

(e) 504 x 35

Solution.

(a) 728 x 101 = 728 x (100 + 1)

= 728 x 100 + 728 x 1

= 72800 + 728

= 73528

(b) 5437 x 1001 = 5437 x (1000 + 1)

= 5437 x 1000 + 5437 x 1

= 5437000 + 5437

= 5442437

(c) 824 x 25 = 824 x (20 + 5)

= 824 x 20 + 824 x 5

= 16480 + 4120

= 20600

(d) 4275 x 125 = 4275 x (100 + 20 + 5)

= 4275 x 100 + 4275 x 20 + 4275 x 5

= 427500 + 85500 + 21375

= 534375

(e) 504 x 35 = (500 + 4) x 35

= 500 x 35 + 4 x 35

= 17500 + 140

= 17640

5. Study the pattern:

1 x 8 + 1= 9

12 x 8 + 2 = 98

123 x 8 + 3 = 987

1234 x 8 + 4 = 9876

12345 x 8 + 5 = 98765

Write the next two steps. Can you say how the pattern works?

(Hint: 12345 = 11111 + 1111 + 111 + 11 + 1).

Solution.

Next two steps are:

• 123456 x 8 + 6 = 987654
• 1234567 x 8 + 7 = 9876543

Pattern works in the following way:

(12345) x 8 + 5 = (11111 + 1111 + 111 + 11 + 1) x 8 + 5

= (88888 + 8888 + 888 + 88 + 8) + 5

= 98760 + 5 = 98765

We are also providing the provision to download the complete solution of this chapter in PDF form which students may keep with them to use whenever required. Click on the following link to download the solutions:

Check chapter-wise NCERT Solutions for Class 6 Maths from the links given below:

NCERT Solutions for Class 6 Maths Chapter 1 - Knowing Our Numbers

NCERT Solutions for Class 6 Maths Chapter 2 - Whole Numbers

NCERT Solutions for Class 6 Maths Chapter 3 - Playing with Numbers

NCERT solutions for other chapters will be provided here very soon. Check here for the detailed and appropriate solutions.