Students can download Chapter 2 Whole Numbers Ex 2.2 KSEEB Solutions for Class 6 Maths helps you to revise the complete syllabus.

## Karnataka State Syllabus Class 6 Maths Chapter 2 Whole Numbers Ex 2.2

Question 1.

Find the sum by suitable rearrangement:

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

= 1200 + 208

= 1408

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

= 3500 + 1100 = 4600.

Question 2.

Find the product by suitable rearrangement:

a. 2 × 1768 × 50 = (2 × 50) × 1768

= 100 × 1768 = 176800

b. 4 × 166 × 25 = (4 × 25) × 166

= 100 × 166 = 16600

c. 8 × 291 × 125 = (8 × 125) × 291

= 1000 × 291 = 291000

d. 625 × 279 × 16 = (625 × 16) × 279

= 10000 × 279 = 2790000

e. 285 × 5 × 60 = 285 × 5 × 60 = 285 × (5 × 60) = 285 × 300 = 85,500

f. 125 × 40 × 8 × 25 = 125 × 8 × 40 × 25 = (125 × 40) × (8 × 25) (5000) × (200)

= 1000 × 1000 = 10,00,000.

Question 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

Solution:

a. 297 × 17 + 297 × 3 = 297 × (17 + 3)

= 297 × 20 = 5,940

b. 54279 × 92 + 8 × 54279 = 54279 × 92 + 54279 × 8

= 54279 × (92 + 8)

= 54279 × 100 = 5427900

c. 81265 × 169 – 81265 × 69 = 81265 × (169 – 169)

= 81265 × 100 = 8126500

d. 3845 × 5 × 782 + 769 × 25 × 218

= 3845 × 5 × 782 + 769 × 5 × 5 × 218

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

= 19225 × 1000 = 1,92,25,000

Question 4.

Find the product using suitable properties.

a. 738 × 103

b. 854 × 102

c. 258 × 1008

d. 1005 × 168

Solution:

a. 738 × 103

= 738 × (100 + 3)

= 738 × 100 + 738 × 3 (Distributive property)

= 73800 + 2214 = 76014

b. 854 × 102

= 854 × (100 + 2)

854 × 100 + 854 × 2 (Distributive property)

= 85400 + 1708 = 87108

c. 258 × 1008

= 258 × ( 1000 + 8)

= 258 × 1000 + 258 × 8 (Distributive property )

= 258000 + 2064 = 260064

d. 1005 × 168 = (1000 + 5) × 168

= 1000 × 168 + 5 × 168 (Distributive property )

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

Solution:

Quantity of petrol filled on monday = 40l

Quantity of petrol filled on tuesday = 50 l

Total quantity filled = (40+ 50) l

Cost of petrol (per l) = Rs 44

Total money spent = 44 × (40 + 50)

= 44 × 90 = Rs.3960

Question 6.

A vendor supplies 32 litres of milk to a hotel in the morning and 68 liters of milk in the evening. If the milk costs Rs 15 per litre, how much money is due to the vendor per day?

Solution:

Quantity of milk supplied in the morning = 32

Quantity of milk supplied is the evening = 681

Total of milk per litre = (32+ 68)

Cost of milk per litre = Rs 15

Total cost per day = 15 × (32 + 68)

= 15 × 100 = Rs 1500

Question 7.

Match the following

1. 425 × 136 = 425 × (6 + 30 + 100) ____ Commutativity under multiplication

2. 2 × 49 × 50 = 2 × 50 × 49 ____ Commutativity under addition

3. 80 + 2005 + 20 = 80 + 20 + 2005 ____ Distributivity of multiplication over addition.

Solution:

1. 425 × 136 = 425 × (6 + 30 + 100) ___ Distributivity of multiplication over addition

2. 2 × 49 × 50 = 2 × 50 × 49 ___ Commutativity under multiplication

3. 80 + 2005 + 20 = 80 + 20 + 2005 ___ Commutativity under addition