NCERT Solutions class 6 Mathematics 2. Whole Numbers Exercise 2.2

Detailed NCERT Solutions for 6 Mathematics 2. Whole Numbers to simplify learning. Understand chapters clearly and practice with free solutions for better results.

NCERT Solutions class 6 Mathematics 2. Whole Numbers Exercise 2.2

NCERT Solutions class 6 Mathematics 2. Whole Numbers Exercise 2.2

Detailed NCERT Solutions for 6 Mathematics 2. Whole Numbers to simplify learning. Understand chapters clearly and practice with free solutions for better results.

6 Mathematics Chapter 2. Whole Numbers - Exercise 2.2

Preparing for exams becomes easier with Exercise 2.2. Whether you are studying for board exams or mid-term exams, 6 Mathematics Chapter 2. Whole Numbers solutions provide quick revising points, well-structured answers, and additional practice material to help you score better.

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2. Whole Numbers

Exercise 2.2


Exercise-2.2


Q1. Find the sum by suitable rearrangement:

(a) 837 + 208 + 363

Solution: 837 + 208 + 363 

= (837 + 208) + 363

= 1045 + 363

= 1408

Or Second Method 

= 837 + 363 + 208

= (837 + 363) + 208

= 1200 + 208

= 1408

(b) 1962 + 453 + 1538 + 647

Solution:  1962 + 453 + 1538 + 6477

= (1962 + 453) + (1538 + 647)

= 2415 + 2185

= 4600

Or Second Method 

= 1962 + 453 + 1538 + 6477

= (1538 + 1962) + (453 + 647)

= 3500 + 1100

= 4600

Q2. 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 x 1768 x 50                           

Solution:

2 x 50 x 1768

= (2 x 50) x 1768

= 100 x 1768

= 176800

(b) 4 x 166 x 25

Solution: 

4 x 25 x 166

= (4 x 25) x 166

= 100 x 166

= 16600

(c) 8 x 291 x 125

Solution:

8 x 125 x 291

= 1000 x 291`

= 291000

(d) 125 x 40 x 8 x 25

Solution:

125 x 8 x 40 x 25

= (125 x 8) x (40 x 25)

= 1000 x 1000

= 1000000

(e) 285 × 5 × 60

Solution:

  285 × 5 × 60

= 285 × (5 × 60)

= 285 × 300

= 85500

(f) 125 × 40 × 8 × 25

Solution: 

  125 × 40 × 8 × 25

= (125 × 8) × (40 × 25)

= 1000 × 1000

= 1000000 

Q3. Find the value of the following:

(a) 297 × 17 + 297 × 3

Solution:

  297 x (17 + 3)

= 297 x 20

= 5940

(b) 54279 × 92 + 8 × 54279

Solution:

  54279 x (92 + 8)
= 54279 x 100

= 5427900

(c) 81265 × 169 – 81265 × 69

Solution: 

  81265 × 169 – 81265 × 69

= 81265 (169 – 69)

= 81265 (100)

= 8126500

(d) 3845 × 5 × 782 + 769 × 25 × 218

Solution: 

  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

Q4. 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

Solution: 

(b) 854 × 102

= 854(100 + 2) 

= 854 × 100 + 854 × 2

= 85400 + 1708

 87108  

Solution:  

(c) 258 × 1008

= 258(1000 + 8) 

= 258 × 1000 + 258 × 8 

= 258000 + 2064

260064

Solution:

(d) 1005 × 168

= (1000 + 5) x 168

= 1000 x 168 + 5 x 168

= 168000 + 840

= 168840

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 Rs 44 per litre, how much did he spend in all on petrol?

Solution:  Taxi driver filled his car petrol tank on Monday = 40 L

 He filled the petrol tank on Tuesday = 50 L

 The petrol costs per liter = 44 Rs

 He spend in all on petrol = (40 + 50) x 44

                           = 90 x 44

                           = 3960 Rs

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 Rs 15 per liter, how much money is due to the vendor per day?

Solution:  A vendor supplies milk to a hotel in the morning = 32 liters

=> A vendor supplies milk to a hotel in the evening = 68 liters

=> The cost of milk per liter = Rs 15

7. Match the following:

Solution: 

(i) ----------- (c) 

(ii) ---------- (a) 

(iii) ---------- (b) 

 

Other Pages of this Chapter:

📘 Why Exercise 2.2 are Important?

Exercise 2.2 are created by experts to give step-by-step explanations. Around 60–70% of exam questions are based on NCERT concepts. Our 6 Mathematics Chapter 2. Whole Numbers solutions help you understand the core concepts and practice effectively.

✍️ Quick Revising Points as Notes in Page-1

Revision is the key to exam success. Our notes for 6 Mathematics highlight important formulas, key definitions, and exam-ready points from Chapter 2. Whole Numbers. These quick revision notes make last-minute preparation easy.

📚 NCERT Exercise Solutions

Every NCERT chapter ends with exercises, and solving them is crucial. Our Exercise 2.2 include complete solutions for 6 Mathematics Chapter 2. Whole Numbers exercises. With step-by-step answers, you gain clarity and confidence to attempt similar exam questions.

📝 Additional Important Questions & Answers

To boost your preparation, we also provide additional important questions with answers. These are prepared from previous year board papers, sample papers, and important concepts of Chapter 2. Whole Numbers. Practicing these ensures you are well-prepared for both board and mid-term exams.

🎯 Useful for Board & Mid-Term Exams

Our Exercise 2.2 are useful for both board exams and mid-term exams. For 6 Mathematics, we provide notes, exercises, and important Q&A so that you can revise smartly and write perfect answers in exams.

🌟 Final Words

In short, Exercise 2.2 for 6 Mathematics Chapter 2. Whole Numbers are a complete study package. With quick revising points, NCERT exercises, and additional important questions, you can prepare effectively for exams. Make these solutions your study companion and excel in your academic journey.

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