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Solutions ⇒ Class 6th ⇒ Mathematics ⇒ 2. Whole Numbers

Exercise 2.2 Class 6 maths 2. Whole Numbers - ncert solutions - Toppers Study

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Exercise 2.2 Class 6 maths 2. Whole Numbers - ncert solutions - Toppers Study

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  • All Chapter review quick revision notes for chapter 2. Whole Numbers Class 6
  • NCERT Solutions And Textual questions Answers Class 6 Mathematics
  • Extra NCERT Book questions Answers Class 6 Mathematics
  • Importatnt key points with additional Assignment and questions bank solved.

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Exercise 2.2 class 6 Mathematics Chapter 2. Whole Numbers

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  • Exercise 2.2 Class 6 Maths 2. Whole Numbers - Ncert Solutions - Toppers Study
  • Class 6 Ncert Solutions
  • Solution Chapter 2. Whole Numbers Class 6
  • Solutions Class 6
  • Chapter 2. Whole Numbers Exercise 2.2 Class 6

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2. Whole Numbers

| Exercise 2.2 |

Exercise 2.2 Class 6 maths 2. Whole Numbers - ncert solutions - Toppers Study


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) 

 

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