# NCERT Solutions for Class 6th: Ch 2 Whole Numbers Math

#### NCERT Solutions for Class 6th: Ch 2 Whole Numbers Maths

Page No: 31**Exercise 2.1**

1. Write the next three natural numbers after 10999.

**Answer**

11000 , 11001, 11002

2. Write the three whole numbers occurring just before 10001.

**Answer**

10000, 9999,9998

3. Which is the smallest whole number?

3. Which is the smallest whole number?

**Answer**
0 is the smallest whole number.

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

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

**Answer**

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

There are 20 whole numbers lie between 32 and 53.

5. Write the successor of :

(a) 2440701

(b) 100199

(c) 1099999

(d) 2345670

5. Write the successor of :

(a) 2440701

(b) 100199

(c) 1099999

(d) 2345670

**Answer**a) 2440701 = 2440701 + 1 = 2440702.

b) 100199 = 100199 + 1 = 100200

c) 1099999 = 1099999 + 1 = 1100000

d) 2345670 = 2345670 + 1 = 2345671

6. Write the predecessor of :

(a) 94

(b) 10000

(c) 208090

(d) 7654321

**Answer**

(a) 94 = 94 – 1 = 93

(b) 10000 = 10000 – 1= 9999

(c) 208090 = 208090 – 1 = 208089

(d) 7654321 = 7654321 – 1 = 76543210

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

**Answer**

(a) 530 > 503

530 is on the left side of number line.

(b) 370 > 307

370 is on the left side of number line.

(c) 98765 > 56789

98765 is on the left side of the number line

(d) 9830415 < 10023001

10023001 is the left side of 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.

**Answer**

(a) 0 is not a natural number (false)

(b) Predecessor of 399 is 398 (false)

(c) True

(d) 599 + 1 = 600 (true)

(e) True

(f) 0 is whole number but not natural numbers.(false)

(g) False

(h) 0 is the smallest whole number (false)

(i) True

(j) False

(k) False

(l) True

(m) False

Page No: 40

1. Find the sum by suitable rearrangement:

(a) 837 + 208 + 363

(b) 1962 + 453 + 1538 + 647

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

= 3500 + 1100 = 4600

2. Find the product by suitable rearrangement:

(a) 2 × 1768 × 50

(b) 4 × 166 × 25

(c) 8 × 291 × 125

(d) 625 × 279 × 16

(e) 285 × 5 × 60

(f) 125 × 40 × 8 × 25

(a) 2 × 50 × 1768 = 176,800

(b) 166 × 25 × 4 = 16600

(c) 291 × 125 × 8 = 291000

(d) 285 × 60 × 5 = 2790000

(e) 285 × 60 × 3 = 85500

(f) 125 × 40 × 25 × 8 = 1000000

3. Find the value of the following:

(a) 297 × 17 + 297 × 3

(b) 54279 × 92 + 8 × 54279

(c) 81265 × 169 – 81265 × 69

(d) 3845 × 5 × 782 + 769 × 25 × 218

**Exercise 2.2**1. Find the sum by suitable rearrangement:

(a) 837 + 208 + 363

(b) 1962 + 453 + 1538 + 647

**Answer****(a) 837 + 208 + 363 = (837 + 363) + 208 = 1200 + 208 = 1408**

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

= 3500 + 1100 = 4600

2. Find the product by suitable rearrangement:

(a) 2 × 1768 × 50

(b) 4 × 166 × 25

(c) 8 × 291 × 125

(d) 625 × 279 × 16

(e) 285 × 5 × 60

(f) 125 × 40 × 8 × 25

**Answer**(a) 2 × 50 × 1768 = 176,800

(b) 166 × 25 × 4 = 16600

(c) 291 × 125 × 8 = 291000

(d) 285 × 60 × 5 = 2790000

(e) 285 × 60 × 3 = 85500

(f) 125 × 40 × 25 × 8 = 1000000

3. Find the value of the following:

(a) 297 × 17 + 297 × 3

(b) 54279 × 92 + 8 × 54279

(c) 81265 × 169 – 81265 × 69

(d) 3845 × 5 × 782 + 769 × 25 × 218

**Answer**

(a) 297 × (17 + 3 ) = 297 × 20 = 5940.

(b) 54279 × (92 + 8)

= 54279 × 100 = 5427900.

(c) 81265 × (169 – 69)

81265 × 100 = 8126500

(d) 3845 × 5 × 782 + 769 × 5 × 5 × 218

= 3845 × 5 × (782 + 218)

= 19225 × 1000 = 19225000.

4. Find the product using suitable properties.

(a) 738 × 103

(b) 854 × 102

(c) 258 × 1008

(d) 1005 × 168

**Answer**

(a) 738 × 103

= (738 × 100) + (738 × 3)

= 73800 + 2214 = 76014

(b) 854 × 102

= (854 × 100) + (854 × 2)

= 85400 + 1708 = 87108

(c) 258 × 1008

= (258 × 1000) + (258 × 8)

= 258000 + 2064 = 260064

(d) 1005 × 168 = (1000 + 5) × 168

= (1000 × 168) + (5 × 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 litres of petrol. If the petrol costs Rs 44 per litre, how much did he spend in all on petrol?

**Answer**

Petrol filled on Monday = 40 litres

Petrol next day filled = 50 litres

Total quantity of petrol filled = (40 +50) litres

Cost of petrol per litres = 44

Petrol next day filled = 50 litres

Total quantity of petrol filled = (40 +50) litres

Cost of petrol per litres = 44

Total amount of money spent on petrol = 44 × (40 + 50)

= 44 × 90 = Rs 3960

Page No: 41

= 44 × 90 = Rs 3960

Page No: 41

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?

**Answer**

In the morning vendor supplies = 32 litres.

In the evening vendor supplies = 68 litres.

Cost of milk per litre = Rs 15

Total cost of per day = 15 × ( 32 + 68)

= 15 × 100 = Rs 1500

7. Match the following:

In the evening vendor supplies = 68 litres.

Cost of milk per litre = Rs 15

Total cost of per day = 15 × ( 32 + 68)

= 15 × 100 = Rs 1500

7. Match the following:

**Answer**

**(i) ➝ C**

(ii) ➝ A

(iii) ➝ B

Page No: 43

**Exercise 2.3**

1. Which of the following will not represent zero:

(a) 1 + 0

(b) 0 × 0

(c) 0 / 2

(d) (10 - 10)/2

**Answer**

(a) 1 + 0 = 1 (not represent zero)

(b) 0 × 0 = 0 (it represent zero)

(c) 0 / 2 = 0 (it represent zero)

(d) (10 – 10)/ 2 = 0 (it represent zero)

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.

**Answer**

Product of two whole number is zero then both of them may be zero

0 × 3 =0

1 × 3 = 3

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.

**Answer**

Product of two whole number is 1 , both number is equal to 1.

For example , 1 × 1 = 1 , 1 × 6 = 6

In both example, number is multiplied by 1.

4. Find using distributive property :

(a) 728 × 101

(b) 5437 × 1001

(c) 824 × 25

(d) 4275 × 125

(e) 504 × 35

**Answer**

(a) 728 × 101

= (728 × 100) + (728 × 1)

= 72800 + 728

= 73528

(b) 5437 × 1001

= (5437 × 1000) + (5437 × 1)

= 5437000 + 5437

= 5442437

(c) 824 × 25

= (800 + 24) × 25

= (800 + 25 – 1) × 25

= (800 × 25) + ( 25 × 25) – (1 × 25)

= (20000 × (625 – 25)

= 20000 + 600 = 20600

(d) 4275 × 125

= (4000 + 25 + 200 + 50) × (100 + 25)

= (4000×100) + (25×100) + (200×100) + (50×100) + (4000×25) + (25×25) + (200×25) + (50×25)

= 400000 + 2500 + 20000 + 5000 + 100000 + 625 + 5000 + 1250

= 534375

(e) 504 × 35

= (500 + 4) × 35

= (500 × 35) + (4 × 35)

= 17500 + 140 = 17640

5. Study the pattern : 1 × 8 + 1 = 9

1234 × 8 + 4 = 9876

12 × 8 + 2 = 98

12345 × 8 + 5 = 98765

123 × 8 + 3 = 987

Write the next two steps. Can you say how the pattern works? (Hint: 12345 = 11111 + 1111 + 111 + 11 + 1).

**Answer**

123456 × 8 + 6 = 987648 + 6 = 987654

1234567 × 8 + 7 = 9876536 + 7 = 9876543

Pattern are use in following ways

123456 = 123456 × 8 (111111+11111+1111+111+11+1) × 8

= 111111 × 8 + 11111 × 8 +1111 × 8 + 111 × 8 + 11 × 8 + 1× 8

= 888888 + 88888 +8888 +888 +88 + 8 = 987648

123456 × 8 +6 = 987648 + 6 = 987654