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

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

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

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2 are the part of NCERT Solutions for Class 6 Maths. Here you can find the NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2.

### Ex 2.2 Class 6 Maths Question 1.

Find the sum by suitable rearrangement:
(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

### Ex 2.2 Class 6 Maths Question 2.

Find the product by suitable rearrangement:
(Ð°) 2 × 1768 × 50
(b) 4 × 166 × 25
(c) 8 × 291 × 125
(d) 625 × 279 × 16
(e) 285 × 5 × 60
(f) 125 × 40 × 8 × 25

Solution:
(a) 2 × 1768 × 50 = (2 × 50) × 1768 = (100) × 1768 = 176800
(b) 4 × 166 × 25 = 166 × (25 × 4) = 166 × 100 = 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 × 300 = (300 – 15) × 300 = 300 × 300 – 15 × 300 = 90000 – 4500 = 85500
(f) 125 × 40 × 8 × 25 = (125 × 8) × (40 × 25) = 1000 × 1000 = 1000000

### Ex 2.2 Class 6 Maths Question 3.

Find the value of the following:
(Ð°) 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 = 297 × 2 × 10
= 594 × 10 = 5940

(b) 54279 × 92 + 8 × 54279 = 54279 × (92 + 8)
= 54279 × 100 = 5427900

(c) 81265 × 169 – 81265 × 69
= 81265 × (169 – 69)
= 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 + 3845 × 5 × 218
= 3845 × 5 × (782 + 218)
= 3845 × 5 × 1000
= 19225 × 1000
= 19225000

### Ex 2.2 Class 6 Maths 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            [Using distributive property]
= 73800 + 2214 = 76014

(b) 854 × 102 = 854 × (100 + 2)
= 854 × 100 + 854 × 2            [Using distributive property]
= 85400 + 1708 = 87108

(c) 258 × 1008 = 258 × (1000 + 8)
= 258 × 1000 + 258 × 8         [Using distributive property]
= 258000 + 2064 = 260064

(d) 1005 × 168 = (1000 + 5) × 168
= 1000 × 168 + 5 × 168        [Using distributive property]
= 168000 + 840 = 168840

### Ex 2.2 Class 6 Maths 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 litres 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 litres
Cost of petrol = ₹44 pet litre
Total money spent in all
= ₹(40 × 44 + 50 × 44)
= ₹(40 + 50) × 44 = ₹90 × 44 = ₹3960

### Ex 2.2 Class 6 Maths 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

Money due to the vendor per day = ₹(32 × 15 + 68 × 15)

= ₹(32 + 68) × 15 = ₹ 100 × 15 = ₹ 1500

### Ex 2.2 Class 6 Maths Question 7.

Match the following:
(i) 425 × 136 = 425 × (6 + 30 + 100)      (a) Commutativity under multiplication
(ii) 2 × 49 × 50 = 2 × 50 × 49                   (b) Commutativity under addition
(iii) 80 + 2005 + 20 = 80 + 20 + 2005     (c) Distributivity of multiplication over addition

Solution:

(i) ↔ (c), (ii) ↔ (a), (iii) ↔ (b)

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