## NCERT Solutions for Class 6 Chapter 2: Whole Numbers

Exercise 2.1 PAGE No.: 22

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

**Solutions:**

The next three natural numbers after 10999 are 11000, 11001 and 11002.

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

**Solutions:**

The three whole numbers occurring just before 10001 are 10000, 9999 and 9998.

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

**Solutions:**

The smallest whole number is 0.

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

**Solutions:**

The whole numbers between 32 and 53 are as follows:

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

Hence, there are 20 whole numbers between 32 and 53

**5. Write the successor of:**

**(a) 2440701 (b) 100199 (c) 1099999 (d) 2345670**

**Solutions:**

The successors are

(a) 2440701 + 1 = 2440702

(b) 100199 + 1 = 100200

(c) 1099999 + 1 = 1100000

(d) 2345670 + 1 = 2345671

**6. Write the predecessor of:**

**(a) 94 (b) 10000 (c) 208090 (d) 7654321**

**Solutions:**

The predecessors are

(a) 94 – 1 = 93

(b) 10000 – 1 = 9999

(c) 208090 – 1 = 208089

(d) 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**

**Solutions:**

(a) 530 > 503

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

(b) 370 > 307

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

(c) 98765 > 56789

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

(d) 9830415 < 10023001

Hence, 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.**

**Solutions:**

False

0 is not a natural number.

**(b) 400 is the predecessor of 399.**

**Solutions:**

False

The predecessor of 399 is 398 because (399 – 1 = 398)

**(c) Zero is the smallest whole number.**

**Solutions:**

True

Zero is the smallest whole number.

**(d) 600 is the successor of 599.**

**Solutions:**

True

Since (599 + 1 = 600)

**(e) All natural numbers are whole numbers.**

**Solutions:**

True

All natural numbers are whole numbers.

**(f) All whole numbers are natural numbers.**

**Solutions:**

False

0 is a whole number but is not a natural number.

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

**Solutions:**

False

For example, the predecessor of 10 is 9.

**(h) 1 is the smallest whole number.**

**Solutions:**

False

0 is the smallest whole number.

**(i) The natural number 1 has no predecessor.**

True

The predecessor of 1 is 0, but it is not a natural number.

**(j) The whole number 1 has no predecessor.**

**Solutions:**

False

0 is the predecessor of 1 and is a whole number.

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

**Solutions:**

False

13 does not lie between 11 and 12.

**(l) The whole number 0 has no predecessor.**

**Solutions:**

True

The predecessor of 0 is -1 and is not a whole number.

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

**Solutions:**

False

For example, the successor of 99 is 100

o00000o

Exercise 2.2 PAGE No.: 40

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

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

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

**Solutions:**

(a) Given 837 + 208 + 363

= (837 + 363) + 208

= 1200 + 208

= 1408

(b) Given 1962 + 453 + 1538 + 647

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

= 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**

**Solutions:**

(a) Given 2 **× **1768 **× **50

= 2 × 50 × 1768

= 100 × 1768

= 176800

(b) Given 4 × 166 × 25

= 4 × 25 × 166

= 100 × 166

= 16600

(c) Given 8 × 291 × 125

= 8 × 125 × 291

= 1000 × 291

= 291000

(d) Given 625 × 279 × 16

= 625 × 16 × 279

= 10000 × 279

= 2790000

(e) Given 285 × 5 × 60

= 285 × 300

= 85500

(f) Given 125 × 40 × 8 × 25

= 125 × 8 × 40 × 25

= 1000 × 1000

= 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**

**Solutions:**

(a) Given 297 × 17 + 297 × 3

**= **297 × (17 + 3)

= 297 × 20

= 5940

(b) Given 54279 × 92 + 8 × 54279

= 54279 × 92 + 54279 × 8

= 54279 × (92 + 8)

= 54279 × 100

= 5427900

(c) Given 81265 × 169 – 81265 × 69

**= **81265 × (169 – 69)

= 81265 × 100

= 8126500

(d) Given 3845 × 5 × 782 + 769 × 25 × 218

= 3845 × 5 × 782 + 769 × 5 × 5 × 218

= 3845 × 5 × 782 + 3845 × 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**

**Solutions:**

**(a) **Given 738 × 103

= 738 × (100 + 3)

= 738 × 100 + 738 × 3 (using distributive property)

= 73800 + 2214

= 76014

(b) Given 854 × 102

= 854 × (100 + 2)

= 854 × 100 + 854 × 2 (using distributive property)

= 85400 + 1708

= 87108

(c) Given 258 × 1008

= 258 × (1000 + 8)

= 258 × 1000 + 258 × 8 (using distributive property)

= 258000 + 2064

= 260064

(d) Given 1005 × 168

= (1000 + 5) × 168

= 1000 × 168 + 5 × 168 (using distributive property)

= 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 ₹ 44 per litre, how much did he spend in all on petrol?**

**Solutions:**

Petrol quantity filled on Monday = 40 litres

Petrol quantity filled on Tuesday = 50 litres

Total petrol quantity filled = (40 + 50) litre

Cost of petrol per litre = ₹ 44

Total money spent = 44 × (40 + 50)

= 44 × 90

= ₹ 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 ₹ 45 per litre, how much money is due to the vendor per day?**

**Solutions:**

Milk quantity supplied in the morning = 32 litres

Milk quantity supplied in the evening = 68 litres

Cost of milk per litre = ₹ 45

Total cost of milk per day = 45 × (32 + 68)

= 45 × 100

= ₹ 4500

Hence, the money due to the vendor per day is ₹ 4500

**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.**

**Solutions:**

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

Hence (c) is the correct answer

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

Hence, (a) is the correct answer

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

Hence, (b) is the correct answer

Exercise 2.3 PAGE No.: 43

**1. Which of the following will not represent zero?**

**(a) 1 + 0**

**(b) 0 × 0**

**(c) 0 / 2**

**(d) (10 – 10) / 2**

**Solutions:**

(a) 1 + 0 = 1

Hence, it does not represent zero.

(b) 0 × 0 = 0

Hence, it represents zero.

(c) 0 / 2 = 0

Hence, it represents zero.

(d) (10 – 10) / 2 = 0 / 2 = 0

Hence, it represents 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.**

**Solutions:**

If the product of two whole numbers is zero, definitely one of them is zero

Example: 0 × 3 = 0 and 15 × 0 = 0

If the product of two whole numbers is zero, both of them may be zero

Example: 0 × 0 = 0

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

**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.**

**Solutions:**

If the product of two whole numbers is 1, both numbers should be equal to 1

Example: 1 × 1 = 1

But 1 × 5 = 5

Hence, it’s clear that the product of two whole numbers will be 1, only in situations when both numbers to be multiplied are 1.

**4. Find using distributive property:**

**(a) 728 × 101**

**(b) 5437 × 1001**

**(c) 824 × 25**

**(d) 4275 × 125**

**(e) 504 × 35**

**Solutions:**

(a) Given 728 × 101

**= **728 × (100 + 1)

= 728 × 100 + 728 × 1

= 72800 + 728

= 73528

(b) Given 5437 × 1001

= 5437 × (1000 + 1)

= 5437 × 1000 + 5437 × 1

= 5437000 + 5437

= 5442437

(c) Given 824 × 25

= (800 + 24) × 25

= (800 + 25 – 1) × 25

= 800 × 25 + 25 × 25 – 1 × 25

= 20000 + 625 – 25

= 20000 + 600

= 20600

(d) Given 4275 × 125

= (4000 + 200 + 100 – 25) × 125

= (4000 × 125 + 200 × 125 + 100 × 125 – 25 × 125)

= 500000 + 25000 + 12500 – 3125

= 534375

(e) Given 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)**

**Solutions:**

123456 × 8 + 6 = 987654

1234567 × 8 + 7 = 9876543

Given 123456 = (111111 + 11111 + 1111 + 111 + 11 + 1)

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

Yes, here the pattern works

1234567 × 8 + 7 = 9876543

Given 1234567 = (1111111 + 111111 + 11111 + 1111 + 111 + 11 + 1)

1234567 × 8 = (1111111 + 111111 + 11111 + 1111 + 111 + 11 + 1) × 8

= 1111111 × 8 + 111111 × 8 + 11111 × 8 + 1111 × 8 + 111 × 8 + 11 × 8 + 1 × 8

= 8888888 + 888888 + 88888 + 8888 + 888 + 88 + 8

= 9876536

1234567 × 8 + 7 = 9876536 + 7

= 9876543

Yes, here the pattern works.

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