# HBSE 6th Class Maths Solutions Chapter 2 Whole Numbers Ex 2.2

Haryana State Board HBSE 6th Class Maths Solutions Chapter 2 Whole Numbers Ex 2.2 Textbook Exercise Questions and Answers.

## Haryana Board 6th Class Maths Solutions Chapter 2 Whole Numbers Exercise 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 a 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 Ans.
(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
= (769 × 5) × 5 × 782 + 769 × 25 × 218
= 769 × (5 × 5) × 782 + 769 × 25 × 218
= 769 × 25 × 782 + 769 × 25 × 218
= 769 × 25 × (782 + 218)
= 769 × 25 × 1000 = 19225000

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
= 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 ta × i-driver filled his car petrol tank with 40 litres of petrol on Monday. The ne × t 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:
Method-1:
Cost of 40 litres of petrol
= Rs. 44 × 40
= Rs. 1760
Cost of 50 litres of petrol
= Rs. 44 × 50
= Rs. 2200
∴ Total cost = Rs. (1760 + 2200)
= Rs. 3960.

Method-2 :
Petrol purchased on Monday = 40 litres
Petrol purchased on ne × t day = 50 litres
Total petrol purchased
= (40 + 50) l = 90 l
∴ Total cost = Rs. 44 × 90
= Rs. 3960.

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 Rs. 15 per litre, how much money is due to the vendor per day ?
Solution:
Milk supplied in the morning = 32 litres
Milk supplied in the evening Milk supplied in one day = 68 litres
∴ Milk supplied in one day = (32 + 68)l = 100 l
Hence, total cost of milk per day = Rs. 15 × 100 = Rs. 1500

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.
(Hi) 80 + 2005 + 20 = 80 + 20 + 2005 — (c) Distributivity of multiplication over addition.
Solution:
(i)—(c),
(ii)—(a),
(iii)—(b).