**Download NCERT Solutions for Class 6 Maths Chapter 2 – Whole Numbers**

### NCERT Solutions for Class 6 Maths Chapter 2 Exercise 2.1

**Find the next natural numbers**

**Ques. 1. Write the next three natural numbers after 10999 ? **

**Solution 1. **

The next number after any given number, like 10999 in this case, can be found, by adding 1 to the given number.

Now, the next three natural numbers after 10999, can be as under : –

10999 + 1 = 11000 – (Adding 1 to 10999)

11000 + 1 = 11001 – (Adding 1 to 11000)

11001 + 1 = 11002 – (Adding 1 to 11001)

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**Three whole numbers occurring just before a given Number **

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

**Solution 2. **

The three whole numbers occurring just before any given number , 10001 in the present case, can be found by subtracting 1 from the number , then subtracting 1 from the result of first subtraction, and then subtracting 1 from the result of second subtraction :

**First subtraction –** 10001 1 = 10000 – ( Subtracting 1 from the number 10001 )

**Second subtraction –** 10000 1 = 9999 – ( Subtracting 1 from the number 10000, the result of first subtraction)

**Third subtraction –** 9999 1 = 9998 – ( Subtracting 1 from the number 9999, the result of second subtraction)

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**Finding the smallest whole number**

**Ques. 3. Which is the smallest whole number?**

**Solution 3.**

Whole numbers start from 0 and proceed as 0,1,2,3,4,5,6,7,8,9,10,…..

Hence, 0 (zero) is the smallest whole number.

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**Finding the whole numbers between two given whole numbers **

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

**Solution 4. **

The whole numbers proceed at a difference of 1 from the previous number.

To find the number of whole numbers between any two numbers, we subtract the smaller number from the greater one, and subtract 1 from the answer.

Here, the number of whole numbers between 32 and 53 are found as follows:

53 32 1 = 20

Hence, there are 20 whole numbers between 32 and 53

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**Finding successor of a given number**

**Ques. 5. Write the successor of :**

**(a) 2440701 **

**(b) 100199 **

**(c) 1099999 **

**(d) 2345670**

**Solution 5. **

The successor of a number can be obtained by adding 1 to the number. Accordingly the answer shall be as under :

**(a)** 2440701 + 1 = 2440702 – (Adding 1 to 2440701 )

**(b)** 100199 + 1 = 100200 – (Adding 1 to 100199)

**(c)** 1099999 +1 = 1100000 – (Adding 1 to 1099999 )

**(d)** 2345670 + 1 = 2345671 – (Adding 1 to 2345670 )

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**Finding predecessor of a given number**

**Ques. 6. Write the predecessor of :**

**(a) 94 **

**(b) 10000 **

**(c) 208090 **

**(d) 7654321**

**Solution 6. **

The predecessor of a number can be obtained by subtracting 1 from the number. Accordingly the answer shall be as under : –

**(a)** 94 – 1 = 93 – (Subtracting 1 from 94 )

**(b)** 10000 – 1 = 9999 – (Subtracting 1 from 10000 )

**(c)** 208090 – 1 = 208089 – (Subtracting 1 from 208090 )

**(d)** 208090 – 1 = 7654320 – (Subtracting 1 from 208090 )

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**Finding whole number which is on the left of the other number on the number line**

**Ques. 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 7. **

On the number line containing whole numbers, the Smaller whole numbers shall lie on the left side of the larger numbers .

**(a)** We know that 530 > 503

Hence, 503 shall lie on the left side of 530 on the number line.

**(b)** We know that 370 > 307

Hence, 307 shall lie on the left side of 370 on the number line.

**(c)** We know that 98765 > 56789

Hence, 56789 shall lie on the left side of 98765 on the number line.

**(d)** We know that 10023001 > 9830415

Hence, 9830415 shall lie on the left side of 10023001 on the number line.

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**Ques. 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) Statement – Zero is the smallest natural number.

Natural numbers start from 1 and proceed as 1,2,3,4,….

Hence, the above statement is false. The smallest natural number is 1 .

(b) Statement – 400 is the predecessor of 399.

Predecessor of a number is 1 less than the number.

So, (400-1 = 399) 399 is the predecessor of 400.

Hence, the above statement is false. 400 is the successor of 399.

(c) Statement – Zero is the smallest whole number.

Whole numbers start from 0 and proceed as 0,1,2,3,4,5….

Hence, the above statement is true.

(d) Statement – 600 is the successor of 599.

Successor of a number is one more than the number.

So, (599+1 = 600) 600 is the successor of 599.

Hence, the above statement is true.

(e) Statement – All natural numbers are whole numbers.

Natural numbers are all counting numbers starting from 1 as 1,2,3,4,….

Whole numbers are all natural numbers including 0.

Hence, the above statement is **True.**

(f) Statement – All whole numbers are natural numbers.

Zero is a whole number, but not a natural number.

Hence, the above statement is **False.**

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

Predecessor of 10 (a two digit number) is 9 (a single digit number) .

Hence, the above statement is **False.**

(h) Statement – 1 is the smallest whole number.

Whole numbers start from 0. So, 0 is the smallest whole number.

Hence, the above statement is **False.**

(i) Statement – The natural number 1 has no predecessor.

Natural numbers start from 1. So, 1 is the smallest natural number.

Hence, the above statement is **True.**

(j) Statement – The whole number 1 has no predecessor.

Predecessor of number 1 is 0, which is a whole number.

Hence, the above statement is **False.**

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

The above statement is **False.** since 13 lies after 12. There are no whole numbers between 11 and 12.

(l) Statement – The whole number 0 has no predecessor.

Whole numbers start from 0. So, 0 is the smallest whole number.

Hence, the above statement is **True.**

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

The successor of 99 is 100, which is a 3 digit number.

Hence, the above statement is **False.**

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**Download NCERT Solutions for Class 6 Maths Chapter 2 – Whole Numbers**

**Class 6 Maths NCERT Solutions Chapter 2 – Exercise 2.2**

**Ques.1. Find the sum by suitable rearrangement:**

**(a) 837 + 208 + 363 **

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

**Solution 1. **

(a) 837 + 208 + 363 = 208 + (837 + 363) … By Commutative & Associative property

= 208 + 1200

= 1408

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

= 3500 + 1100

= 4600

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**Ques.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**

**Solution 2.**

**(a)** 2 × 1768 × 50 = 1768 × (2 × 50) …By commutative and associative Property

= 1,768 × 100

= 1,76,800

**(b)** 4 × 166 × 25 = 166 × (4 × 25) …By commutative and associative Property

= 166 × 100

= 16,600

**(c)** 8 × 291 × 125 = 291 × (8 × 125) …By commutative and associative Property

= 291 × 1000

= 2,91,000

**(d)** 625 × 279 × 16 = 279 × (625 × 16) …By commutative and associative Property

= 279 × 10,000

= 27,90,000

**(e)** 285 × 5 × 60 = 285 × (5 × 60) …By commutative and associative Property

= 285 × 300

= 85,500

**(f)** 125 × 40 × 8 × 25 = (125 × 8) × (40 × 25) …By commutative and associative Property

= 1000 × 1000

= 10,00,000

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**Ques.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**

**Solution 3. **

**(a)** 297 × 17 + 297 × 3 = 297 × (17 + 3) …By distributive property

= 297 × 20

= 5,940

**(b)** 54279 × 92 + 8 × 54279 = 54279 × (92 + 8)

= 54279 × 100

= 54,27,900

**(c)** 81265 × 169 – 81265 × 69 = 81265 × (169 – 69)

= 81265 × 100

= 81,26,500

**(d)** 3845 × 5 × 782 + 769 × 25 × 218 = 3845 × 5 × 782 + (769 × 5) × 5 × 218

= 3845 × 5 × 782 + 3845 × 5 × 218

= 3845 × 5 × (782 + 218)

= 3845 × 5 × 1000

= 3845 × 5000

= 1,92,25,000

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**Ques.4. Find the product using suitable properties.**

**(a) 738 × 103 **

**(b) 854 × 102 **

**(c) 258 × 1008 **

**(d) 1005 × 168**

**Solution 4.**

**(a)** 738 × 103 = 738 × (100 + 3)

= (738 × 100) + (738 × 3)

= 73,800 + 2,214

= 76,014

**(b)** 854 × 102 = 854 × (100 + 2)

= (854 × 100) + (854 × 2)

= 85,400 + 1708

= 87,108

**(c)** 258 × 1008 = 258 × (1000 + 8)

= (258 × 1000) + (258 × 8)

= 258000 + 2064

= 2,60,064

**(d)** 1005 × 168 = (1000 + 5) × 168

= (1000 × 168) + (5 × 168)

= 168000 + 840

= 1,68,840

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**Ques.5. A taxidriver 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?**

**Solution 5.** Quantity of petrol filled by the taxi driver on Monday = 40 litres

Quantity of petrol filled by the taxi driver on the next day = 50 litres

Cost of petrol per litre = Rs 44

Total quantity of petrol filled by him = 40 + 50

= 90 litres

Total amount spent on petrol = Rs (44 × 90)

= Rs 3,960

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**Ques 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 6.** Quantity of milk supplied by the vendor in the morning = 32 litres

Quantity of milk supplied by the vendor in the evening = 68 litres

Total quantity of milk supplied by the vendor per day = (32+68) litres

= 100 litres

Cost of milk per litre = Rs 15

Amount due to the vendor per day = Rs (15 × 100)

= Rs 1500

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**Ques.7. Match the following:**

**(i) 425 × 136 = 425 × (6 + 30 +100) (a) Commutativity under multiplication.**

**(ii) 2 × 49 × 50 = 2 × 50 × 49 (b) Commutativity under addition.**

**(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (c) Distributivity of multiplication over addition.**

**Solution 7.**

(i) 425 × 136 = 425 × (6 + 30 +100) (c) Distributivity of multiplication over addition.

(ii) 2 × 49 × 50 = 2 × 50 × 49 (a) Commutativity under multiplication.

(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (b)Commutativity under addition.

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** ****Class 6 Maths NCERT Solutions Chapter 2 – Exercise 2.3**

**Ques.1. Which of the following will not represent zero:**

**(a) 1 + 0 **

**(b) 0 × 0 **

**(c) **

**(d) **

**Solution 1.**

(a) 1 + 0 = 1

(b) 0 × 0 = 0

(c) = 0

(d) = 0

Hence, option (a) is the correct answer.

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**Ques.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 2.** Yes, we can say that if the product of two whole numbers is zero, then either one or both of them are zeroes. This is because, any number multiplied by zero, becomes zero.

For example, 0 × 0 = 0

1 × 0 = 0

2 × 0 = 0… etc.

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**Ques.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 3.** If the product of two whole numbers is 1, then both of them should be 1. This is so, because 1 is the multiplicative identity for whole numbers. Hence, any number multiplied by 1, gives the number itself. In order to obtain the product of two whole numbers as 1, 1 should be multiplied by itself.

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**Ques.4. Find using distributive property:**

**(a) 728 × 101 **

**(b) 5437 × 1001 **

**(c) 824 × 25 **

**(d) 4275 × 125 **

**(e) 504 × 35**

**Solution 4. **

(a) 728 × 101 = 728 × (100 + 1)

= (728 × 100) + (728 × 1)

= 72,800 +728

= 73,528

(b) 5437 × 1001 = 5437 × (1000 + 1)

= 54,37,000 + 5437

= 54,42,437

(c) 824 × 25 = 824 × (20 + 5)

= 16,480 + 4,120

= 20,600

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

= 427,500 + 85,500 + 21,375

= 5,34,375

(e) 504 × 35 = 504 × (30 + 5)

= 15,120 + 2,520

= 17,640

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**Ques.5. 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 two steps. Can you say how the pattern works?**

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

**Solution 5. **

The next two steps are as follows:

123456 × 8 + 6 = 987654

1234567 × 8 + 7 = 9876543

The pattern works as follows:

1 × 8 + 1 = 9

12 × 8 + 2 = (11 + 1) × 8 + 2

= 88 + 8 +2

= 98

123 × 8 + 3 = (111 + 11 +1) × 8 + 3

= 888 + 88 + 8 + 3

= 987

1234 × 8 + 4 = (1111 + 111 + 11 + 1) × 8 + 4

= 8888 + 888 + 88 + 8 + 4

= 9876

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

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

= 98765

And thus, the pattern continues.

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**Download NCERT Solutions for Class 6 Maths Chapter 2 – Whole Numbers**