Gujarat BoardĀ GSEB Textbook Solutions Class 6 Maths Chapter 2 Whole Numbers Ex 2.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 6 Maths Chapter 2 Whole Numbers Ex 2.2

Question 1.

Find the sum by suitable rearrangement:

(a) 837 + 208 + 363

(b) 1962 + 453 + 1538 + 647

Solution:

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

= 208 + (1200) = 1408

(b) 1962 + 453 + 1538 + 647

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

= (3500) + (1100)

= 3500 + 1100 = 4600

Question 2.

Find the product by suitable rearrangement:

(a) 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

= 100 x 1768 = 176800

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

= 1oo x 166 = 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

= (285 x 3) x 1oo = 855 x 100 = 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:

(a) 297 x 17 + 297 x 3

(b) 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 + 297 x 3 = 297 [17 + 3] = 297 [20]

= 297 x 20 = 5940

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

= 54279 [100] = 54279 x 100 = 5427900

(c) 81265 x 169 – 81265 x 69 = 81265 [169 – 69]

= 81265 [100] = 81265 x 100 = 8126500

(d) 3845 x 5×782 + 769 x 25 x 218

(25 = 5 x 5)

= 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 + 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

(e) 258 x 1008

(d) 1005 x 168

Solution:

(a) 738 x 103 738 x [100 + 3]

= [738 100] + [738 x 3]

= 73800 + 2214 = 76014

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

= (854 x 100) + (854 X 2)

= 85400 + 1708 = 87108

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

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

= 258000 + 2064 = 260064

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

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

= 168000 + 840 = 168840

Question 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?

Solution:

Quantity of petrol filled on Monday = 40 litres

Quantity of petrol filled on the next day = 50 litres

Per litre cost of petrol = ā¹ 44

Total cost of petrol for the two days

= ā¹ 44 x [40 + 50]

= ā¹ [44 x 50] + [44 x 40]

= ā¹ 2200 + ā¹ 1760 = ā¹ 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 ā¹ 15 per litre, how much money is due to the vendor per day?

Solution:

Milk supplied in the

Morning = 32 litres

Evening = 68 litres

Cost of milk = ā¹ 15 per litre

Total cost of milk per day

= ā¹ 15 x [32 – 4 – 68] = ā¹ 15 x 100

= ā¹ 1500

Money is due to the vendor per day = ā¹ 1500

Question 7.

Match the following:

Solution: