NCERT Solutions For Class 6 Maths Chapter 2 Whole Numbers Exercise 2.2 has questions related to properties of whole numbers, like closure property, commutativity, associativity, distributivity and identity, solved by Math experts at SNS Academy. The natural numbers along with zero form whole numbers. The second exercise of this chapter has 7 main questions related to the properties of whole numbers. Following the NCERT Solutions of Class 6 Chapter 2 will help students to understand how to solve the different types of questions related to the topics mentioned above, that could be asked in the exam.

## NCERT Solutions for Class 6 Chapter 2: Whole Numbers Exercise 2.2

### Access NCERT Solutions for Class 6 Chapter 2: Whole Numbers Exercise 2.2

**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 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 ₹ 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

### Access other exercise solutions of Class 6 Maths Chapter 2: Whole Numbers

Exercise 2.1 Solutions

Exercise 2.3 Solutions

**Also, explore – **

NCERT Solutions for Class 6 Maths Chapter 2

NCERT Solutions for Class 6 Maths

NCERT Solutions