**Whole Numbers Class 6 Ex. 2.2**

**Ncert Class 6 Math Free Solution.**

**Exercise 2.2**

**Question 1 :- Find the sum by suitable rearrangement:**

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

**(b) 1962 + 453 + 1538 + 647**

**Solution 1:-**

(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.

**Question 2 :- Find the product by suitable rearrangement:**

**(а) 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 2:-**

(a) 2 x 1768 x 50

(2 x 50) x 1768 = 176800

(b) 4 x 166 x 25

166 x (25 x 4) = 166 x 100 = 16,600

(c) 8 x 291 x 125

(8 x 125) x 291 = 1000 x 291 = 2,91,000

(d) 625 x 279 x 16

(625 x 16) x 279 = 10,000 x 279 = 27,90,000

(e) 285 x 5 x 60

285 X (5 x 60) = 285 x 300

= (300 – 15)x 300 = 300 x 300 – 15 x 300

= 90,000 – 4,500 = 85,500

( f ) 125 X 40 X 8 X 25

(125 x 8) X (40 x 25) = 1,000 X 1,000 = 10,00,000

**Question 3 :- Find the value of the following:**

**(а) 297 x 17 + 297 x 3**

**(б) 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 3:-**

(a) 297 x 17 + 297 x 3 = 297 x (17 + 3)

= 297 X 20 = 297 X 2 X 10

= 594 X 10 = 5,940.

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

= 54279 x 100 = 54,27,900.

(c) 81,265 x 169 – 81,265 x 69

= 81265 x (169 – 69)

= 81265 x 100 = 81,26,500.

(d) 3845 x 5 x 782 + 769 x 25 x 218 = 3845 x 5 x 782 + 769 x (5 x 5) x 218

= 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

= 1,92,25,000.

**Question 4 :- Find the product using suitable properties.**

**(a) 738 x 103**

**(b) 854 x 102**

**(c) 258 x 1008**

**(d) 1005 x 168**

**Solution 4:-**

(a) 738 x 103 = 738 x (100 + 3)

= 738 x 100 + 738 x 3 [Using distributive property]

= 73800 + 2214 = 76,014.

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

= 854 x 100 + 854 x 2 [Using distributive property]

= 85400 + 1708 = 87,108.

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

= 258 x 1000 + 258 x 8 [Using distributive property]

= 258000 + 2064 = 2,60,064.

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

= 1000 x 168 + 5 x 168 [Using distributive property]

= 168000 + 840 = 1,68,840.

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

**Solution 5:-**

Petrol filled on Monday = 40 litres

Petrol filled on Tuesday = 50 litres

Cost of petrol = ₹44 pet litre

So, Total money spent in all = ₹ (40 x 44 + 50 x 44)

= ₹(40 + 50) x 44 = ₹90 x 44 = **₹ 3,960.**

**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 ₹45 per litre, how much money is due to the vendor per day?**

**Solution 6 :-**

Supply of the milk in the morning = 32 litres

Supply of the milk in the evening = 68 litres

Cost of milk = ₹45 per litre

So, Total supplies of milk = (32 + 68) litre = 100 litre

Cost of 100 litres milk = 45 x 100 = ₹4500

So, the amount ₹4500 is due to the vendor per day.

**Question 7 :- Match the following:**

**(i) 425 x 136 = 425 x (6 + 30 + 100) (a) Commutativity under multiplication**

**(ii) 2 x 49 x 50 = 2 x 50 x 49 (b) Commutativity under addition**

**(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (c) Distributivity of multiplication over addition**

**Solution 7:-**

(i) ↔ (c),

(ii) ↔ (a)

(iii) ↔ (b).