Whole Numbers Exercise 2.2 class 6 maths
NCERT mathematics class 6 chapter 2 Whole Numbers Exercise 2.2 solutions are given. You should study the textbook lesson Whole Numbers very well. You should also observe and practice all example problems and solutions given in the textbook. Observe the given below solutions and try them in your own method. You can see. NCERT class 6 maths solutions Whole Numbers Exercise 2.1 Exercise 2.2 Exercise 2.3NCERT Whole Numbers Exercise 2.2 Solutions class 6 maths
6th class maths textbook Chapter 2 Whole Numbers Exercise 2.2NCERT Chapter 2 Exercise 2.2 solutions maths class 6
CBSE class 6 maths solutions Problem 11. Find the sum by suitable rearrangement:
a. 837 + 208 + 363
b. 1962 + 453 + 1538 + 647
b. 1962 + 453 + 1538 + 647
= 1962 + 1538 + 453+ 647 = (1962 + 1538) +( 453 + 647) = 3,500 + 1,100 = 4,600 Problem 22. 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
Solutions:a. 2 × 1768 × 50
= 2 × 50 × 1768 = (2 × 50) ×1768 = 100 × 1768 = 1,76,800b. 4 × 166 × 25
= 4 × 25 ×166 = (4 ×25) ×166 = 100 × 166 = 16,600c. 8 × 291 × 125
8 × 125 × 291 = (8 × 125) × 291 = 1,000 × 291 = 2,91,000d. 625 × 279 × 16
= 625 × 16 × 279 = (625 × 16) × 279 = 10,000 × 279 = 27,90,000e. 285 × 5 × 60
= 285 × (5 × 60) = 285 × 300 = 85,500f. 125 × 40 × 8 ×25
= 125 × 8 × 40 × 25 = (125 × 8) × (40 × 25) = 1,000 × 1,000 = 10,00,000 Problem 33. 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
Solutions:a. 297 × 17 + 297 × 3
= 297 × (17 + 3) = 297 × 20 = 5,940b. 54279 × 92 + 8 × 54279
= 54279 × (92 + 8) = 54,279 × 100 = 54,27,900c. 81265 ×169 – 81265 × 69
= 81265 × (169 – 69) = 81265 × 100 = 81,26,500d. 3845 × 5 × 782 + 769 × 25 × 218
= 3845 × 5 × 782 + 769 × 5 × 5 × 218 = 3845 × 5 × 782 + 3845 × 5 × 218 = (3845 × 5) × 782 + (3845 × 5) ×218 = 19225 × 782 + 19225 × 218 = 19225 × (782 + 218) = 19225 × 1000 = 1,92,25,000 Problem 44. Find the product using suitable properties.
a. 738 × 103, b. 854 × 102
c. 258 × 1008, d. 1005 × 168
Solutions: a. 738 × 103 = 738 × (100 + 3) = 738 × 100 + 738 × 3 = 73,800 + 2,214 = 76,014 b. 854 × 102 = 854 × (100 + 2) = 854 × 100 + 854 × 2 = 85,400 + 1,708 = 87,108 c. 258 × 1008 = 258 × (1,000 + 8) = 258 × 1,000 + 258 × 8 = 2,58,000 + 2064 = 2,60,064 d. 1,005 × 168 = (1,000 + 5) × 168 = 1,000 × 168 + 5 × 168 = 1,68,000 + 840 = 1,68,840 Problem 55. 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 the petrol costs Rs. 44 per litres, how many did he spend in all on petrol?
Solution: The petrol filled on Monday = 40 litres The petrol filled on the next day = 50 litres The total petrol filled on both the day = 40 + 50 = 90 litres The cost petrol for 1 litre = Rs. 44 The cost of petrol for 90 litres = 44 × Rs.90 = Rs.3960 Therefore, the taxi driver spent Rs.3960 on petrol. Problem 66. 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 Liter, how much money is due to the vendor per day?
Solution: The supply of milk in morning to a hotel = 32 litres The supply of milk in evening to the hotel = 68 litres The total supply of milk per day = 32 + 68 = 100 litres The cost of milk for 1 litre = Rs.15 The cost of milk for 100 litres = 15 × 100 = Rs.1500 Therefore, Rs.1500 is due to the vendor per day. Problem 77. Match the following:
i. 425 × 136 = 425 × (6+ 30 + 100)
ii. 2 × 49 × 50 = 2 × 50 × 49
iii. 80 + 2005 + 20 = 80 + 20 × 2005
a. Commutativity under multiplication.
b. Commutativity under addition.
c. Distributive of multiplication over addition.
Solutions: i – c ii– a iii – b NOTE: Observe the solutions and try them in your own method. Inter maths solutions NCERT class 7 maths solutions SSC Maths class 10 solutions