## MP Board Class 6th Maths Solutions 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

= (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

(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

= (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 × 300

= 85500

(f) 125 × 40 × 8 × 25

= (125 × 8) × (40 × 25)

= 1000 × 1000

= 1000000

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

= 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 + 3845 × 5 × 218

= 3845 × 5 × (782 + 218)

= 19225 × 1000

= 19225000

Question 4.

Find the product using suitable properties.

(a) 738 × 103

(b) 854 × 102

(c) 258 × 1008

(d) 1005 × 168

Solution:

By using distributive property of multiplication over addition, we get

(a) 738 × 103

= 738 × (100 + 3)

= 738 × 100 + 738 × 3

= 73800 + 2214 = 76014

(b) 854 × 102

= 854 × (100 + 2)

= 854 × 100 + 854 × 2

= 85400 + 1708

= 87108

(c) 258 × 1008

= 258 × (1000+ 8)

= 258 × 1000 + 258 × 8

= 258000 + 2064

= 260064

(d) 1005 × 168

= (1000 + 5) × 168

= 1000 × 168 + 5 × 168

= 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 litres of petrol. If the petrol costs ₹ 44 per litre, how much did he spend in all on petrol?

Solution:

Petrol filled on Monday = 40 litres Petrol filled on next day = 50 litres

Total petrol filled = (40 + 50) litres = 90 litres

Now, cost of 1 litre of petrol = ₹ 44

Cost of 90 litres of petrol = ₹ (44 × 90)

= ₹ 3960

Therefore, the taxidriver spent ₹ 3960 on petrol.

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:

Supply of milk in the morning = 32 litres

Supply of milk in the evening = 68 litres

Total supply of milk = (32 + 68) litres

= 100 litres

Now, cost of 1 litre milk = ₹ 15

Cost of 100 litres milk = ₹ (15 × 100)

= ₹ 1500

Therefore, ₹ 1500 is due to the vendor per day.

Question 7.

Solution:

(i) 425 × 136 = 425 × (6 + 30 + 100)

This is the principle of distributivity of multiplication over addition.

(ii) 2 × 49 × 50 = 2 × 50 × 49

This is the principle of commutativity under multiplication.

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

This is the principle of commutativity under addition.