# KSEEB Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2

Students can Download Chapter 2 Whole Numbers Ex 2.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## 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