NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 6 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Whole Numbers |

Exercise |
Ex 2.2 |

Number of Questions Solved |
7 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 6 Maths 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 = (837 + 363) + 208 = 1200 + 208 = 1408

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

= 1962 + 1538 + 453 + 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 = (2 × 50) × 1768 = 100 × 1768= 1,76,800

**(b)** 4 × 166 × 25 = 4 × 25 × 166 = (4 × 25) × 166 = 100 × 166 = 16,600

**(c)** 8 × 291 × 125 = 8 × 125 × 291

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

**(d)** 625 × 279 × 16 = 625 × 16 × 279

= (625 × 16) × 279 = 10000 ×279 = 27,90,000

**(e)** 285 × 5 × 60 = 285 × (5 × 60) = 285 × 300

= 85,500

**(f)** 125 × 40 × 8 × 25 = (125 × 40) × (8 × 25) = 5000 × 200 = 10,00,000.

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 + 54279 × 8 = 54279 × (92 + 8)

= 54279 × 100 = 54,27,900

**(c)** 81265 × 169-81265 × 69

= 81265 × (169-69)

= 81265 × 100 = 81,26,500

**(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 = 1,92,25,000.

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 = 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 × (1000+ 8)

= 258 × 1000 + 258 × 8 = 2,58,000 + 2,064 = 2,60,064

**(d)** 1005 × 168

= 168 × 1005

= 168 × (1000+ 5)

= 168 × 1000+ 168 × 5

= 1,68,000 + 840 = 1,68,840.

Question 5.

A taxi driver filled his car petrol tank with 40 liters of petrol on Monday. The next day he filled the tank with 50 liters of petrol. If the petrol costs ₹ 44 per liter, how much did he spend all on petrol?

Solution :

Petrol filled on Monday = 40 litres

Petrol filled the next day = 50 litres

∴ Total petrol filled on the two days = 40 litres + 50 litres = 90 litres

∴ Cost of petrol per litre = ₹ 44

∴ Cost of 90 litres of 7 petrol = ₹ 44 × 90 = ? 3960.

Question 6.

A vendor’supplies 32 liters of milk to a hotel in the morning and 68 liters of milk in the evening. If the milk costs ₹ 15 per liter, how much money is due to the vendor per day?

Solution :

Milk supplied in the morning = 32 liters

Milk supplied in the evening = 68 liters

∴ Milk supplied per day = 32 litres + 68 litres = 100 litres

Cost of milk per liter = ₹ 15

Money due to the vendor per day = Cost of

100 litres of milk = ₹ 15 × 100 = ₹ 1500.

Question 7.

**Match the following :
**Solution :

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