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

**NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2**

**Exercise 2.2**

Question 1.

Find the sum by suitable arrangement:

(a) 837 + 208 + 363

(b) 1962 + 453,+ 1538 + 647

Solution:

(a) 837 + 208 + 363 = (837 + 363) + 208

= 1200 + 208 [Using associative property]

= 1408

(b) 1962 + 453 + 1538 + 647

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

= 3500 + 1100 = 4600

Question 2.

Find the product by suitable arrangement:

(а) 2 x 1768 x 50

(b) 4 x 166 x 25

(c) 8 x 291 x 125

(d) 625 x 279 x 16

(e) 285 x 5 x 60

(f) 125 x 40 x 8 x 25

Solution:

(a) 2 x 1768 x 50 = (2 x 50) x 1768 = 176800

(b) 4 x 166 x 25 = 166 x (25 x 4) = 166 x 100 = 16600

(c) 8 x 291 x 125 = (8 x 125) x 291 = 1000 x 291 = 291000

(d) 625 x 279 x 16 = (625 x 16) x 279 = 10000 x 279 = 2790000

(e) 285 x 5 x 60 = 285 x (5 x 60) = 285 x 300 = (300 – 15)x 300 = 300 x 300 – 15 x 300 = 90000 – 4500 = 85500

(f) 125 x 40 x 8 x 25 = (125 x 8) x (40 x 25) = 1000 X 1000 = 1000000

Question 3.

Find the value of the following:

(а) 297 x 17 + 297 x 3

(б) 54279 x 92 + 8 x 54279

(c) 81265 x 169 – 81265 x 69

(d) 3845 x 5 x 782 + 769 x 25 x 218

Solution:

(a) 297 x 17 x 297 x 3 = 297 x (17 + 3)

= 297 x 20 = 297 x 2 x 10

= 594 x 10 = 5940

(b) 54279 x 92 + 8 x 54279 = 54279 x (92 + 8)

= 54279 x 100 = 5427900

(c) 81265 x 169 – 81265 x 69

= 81265 x (169 – 69)

= 81265 x 100 = 8126500

(d) 3845 x 5 x 782 + 769 x 25 x 218 = 3845 x 5 x 782 + 769 x 5 x 5 x 218

= 3845 x 5 x 782 + (769 x 5) x 5 x 218

= 3845 x 5 x 782 + 3845 x 5 x 218

= 3845 x 5 x 782 + 3845 x 5 x 218

= 3845 x 5 x (782 + 218)

= 3845 x 5 x 1000

= 19225 x 1000

= 19225000

Question 4.

Find the product using suitable properties.

(a) 738 x 103

(b) 854 x 102

(c) 258 x 1008

(d) 1005 x 168

Solution:

(a) 738 x 103 = 738 x (100 + 3)

= 738 x 100 + 738 x 3 [Using distributive property]

= 73800 + 2214 = 76014

(b) 854 x 102 = 854 x (100 + 2)

= 854 x 100 + 854 x 2 [Using distributive property]

= 85400 + 1708 = 87108

(c) 258 x 1008 = 258 x (1000 + 8)

= 258 x 1000 + 258 x 8 [Using distributive property]

= 258000 + 2064 = 260064

(d) 1005 x 168 = (1000 + 5) x 168

= 1000 x 168 + 5 x 168 [Using distributive property]

= 168000 + 840 = 168840

Question 5.

A taxidriver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litre of petrol. If the petrol cost ₹44 per litre, how much did he spend in all on petrol?

Solution:

Petrol filled on Monday = 40 litres

Cost of petrol = ₹44 per litre

Petrol filled on Tuesday = 50 litre

Cost of petrol = ₹44 pet litre

∴ Total money spent in all

= ₹(40 x 44 + 50 x 44)

= ₹(40 + 50) x 44 = ₹90 x 44 = ₹3960

Question 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 ₹15 per litre, how much money is due to the vendor per day?

Solution:

Milk supplied in the morning = 32 litres

Cost of milk = ₹15 per litre

Milk supplied in the evening = 68 litres

Cost of milk = ₹15 per litre

Question 7.

Match the following:

(i) 425 x 136 = 425 x (6 + 30 + 100)

(ii) 2 x 49 x 50 = 2 x 50 x 49

(iii) 80 + 2005 + 20 = 80 + 20 + 2005

Hence (i) ↔ (c), (ii) ↔ (a) and (iii) ↔ (b)

∴ Money paid to the vendor

= ₹ (32 x 15 + 68 x 15)

= ₹(32 + 68) x 15

= ₹100 x 15

= ₹1500

(a) Commutativity under multiplication

(b) Commutativity under addition

(c) Distributivity of multiplication over addition

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