# NCERT Solutions For Class 6 Maths Chapter 2 Whole Numbers Exercise 2.2

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 BYJUâ€™S. 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 2Â will help students in understanding 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 Download PDF

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### 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 Ã— 1768 Ã— 50

(b) 4 Ã— 166 Ã— 25

(c) 8 Ã— 291 Ã— 125

(d) 625 Ã— 279 Ã— 16

(e) 285 Ã— 5 Ã— 60

(f) 125 Ã— 40 Ã— 8 Ã— 25

Solutions:

(a) Given 2 Ã— 1768 Ã— 50

= 2 Ã— 50 Ã— 1768

= 100 Ã— 1768

= 176800

(b) Given 4 Ã— 166 Ã— 25

= 4 Ã— 25 Ã— 166

= 100 Ã— 166

= 16600

(c) Given 8 Ã— 291 Ã— 125

= 8 Ã— 125 Ã— 291

= 1000 Ã— 291

= 291000

(d) Given 625 Ã— 279 Ã— 16

= 625 Ã— 16 Ã— 279

= 10000 Ã— 279

= 2790000

(e) Given 285 Ã— 5 Ã— 60

= 285 Ã— 300

= 85500

(f) Given 125 Ã— 40 Ã— 8 Ã— 25

= 125 Ã— 8 Ã— 40 Ã— 25

= 1000 Ã— 1000

= 1000000

3. Find the value of the following:

(a) 297 Ã— 17 + 297 Ã— 3

(b) 54279 Ã— 92 + 8 Ã— 54279

(c) 81265 Ã— 169 â€“ 81265 Ã— 69

(d) 3845 Ã— 5 Ã— 782 + 769 Ã— 25 Ã— 218

Solutions:

(a) Given 297 Ã— 17 + 297 Ã— 3

= 297 Ã— (17 + 3)

= 297 Ã— 20

= 5940

(b) Given 54279 Ã— 92 + 8 Ã— 54279

= 54279 Ã— 92 + 54279 Ã— 8

= 54279 Ã— (92 + 8)

= 54279 Ã— 100

= 5427900

(c) Given 81265 Ã— 169 â€“ 81265 Ã— 69

= 81265 Ã— (169 â€“ 69)

= 81265 Ã— 100

= 8126500

(d) Given 3845 Ã— 5 Ã— 782 + 769 Ã— 25 Ã— 218

= 3845 Ã— 5 Ã— 782 + 769 Ã— 5 Ã— 5 Ã— 218

= 3845 Ã— 5 Ã— 782 + 3845 Ã— 5 Ã— 218

= 3845 Ã— 5 Ã— (782 + 218)

= 19225 Ã— 1000

= 19225000

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4. Find the product using suitable properties.

(a) 738 Ã— 103

(b) 854 Ã— 102

(c) 258 Ã— 1008

(d) 1005 Ã— 168

Solutions:

(a) Given 738 Ã— 103

= 738 Ã— (100 + 3)

= 738 Ã— 100 + 738 Ã— 3 (using distributive property)

= 73800 + 2214

= 76014

(b) Given 854 Ã— 102

= 854 Ã— (100 + 2)

= 854 Ã— 100 + 854 Ã— 2 (using distributive property)

= 85400 + 1708

= 87108

(c) Given 258 Ã— 1008

= 258 Ã— (1000 + 8)

= 258 Ã— 1000 + 258 Ã— 8 (using distributive property)

= 258000 + 2064

= 260064

(d) Given 1005 Ã— 168

= (1000 + 5) Ã— 168

= 1000 Ã— 168 + 5 Ã— 168 (using distributive property)

= 168000 + 840

= 168840

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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 â‚¹ 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 = â‚¹ 44

Total money spent = 44 Ã— (40 + 50)

= 44 Ã— 90

= â‚¹ 3960

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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 â‚¹ 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 = â‚¹ 45

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

= 45 Ã— 100

= â‚¹ 4500

Hence, the money is due to the vendor per day is â‚¹ 4500

7. Match the following:

(i) 425 Ã— 136 = 425 Ã— (6 + 30 + 100) (a) Commutativity under multiplication.

(ii) 2 Ã— 49 Ã— 50 = 2 Ã— 50 Ã— 49 (b) Commutativity under addition.

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

Solutions:

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

Hence (c) is the correct answer

(ii) 2 Ã— 49 Ã— 50 = 2 Ã— 50 Ã— 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