Table of Contents

Toggle**NCERT Solutions for class 6 maths chapter 2 Exercise 2.2** is all about practicing of addition, subtraction, multiplication and dividing the numbers. there may be different ways to solve them with less amount of time. Class 6 maths chapter 2 exercise 2.2 solutions are provided here in simple ways. Class 6^{th} maths 2.2 has total 7 questions to solve. so, students can check the solutions of whole number exercise 2.2 given below.

## NCERT Solutions for Class 6 Maths, chapter 2, Whole Numbers, Exercise 2.2

**Q.1. Find the sum by suitable rearrangement:**

**(a) 837 + 208 + 363 **

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

**Ans:**

(a) Given 837 + 208 + 363

= (837+363)+208

= 1200+208

= 1408

(b) Given 1962 + 453 + 1538 + 647

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

= 3500+1100

= 4600

**Q.2. 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**

**Ans:**

**(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

**Q.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**

**Ans:**

**(a) Given 297 × 17 + 297 × 3**

=(297 × 17) + (297 × 3)

**= **297 × (17 + 3)

= 297 × 20

= 5940

**(b) Given 54279 × 92 + 8 × 54279**

= (54279 × 92) + (54279 × 8)

= 54279 × (92 + 8)

= 54279 × 100

= 5427900

**(c) Given 81265 × 169 – 81265 × 69**

= (81265 × 169) – (81265 × 69)

= 81265 × (169 – 69)

= 81265 × 100

= 8126500

**(d) Given 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

**Q.4. Find the product using suitable properties.**

**(a) 738 × 103 **

**(b) 854 × 102**

** (c) 258 × 1008 **

**(d) 1005 × 168**

**Ans:**

**(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

**Q.5. 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 petrol. If the petrol costs ****₹****44 per litre, how much did he spend in all on petrol?**

**Ans:**

Petrol filled on Monday = 40 litres

Petrol filled on Tuesday = 50 litres

Total petrol filled = (40 + 50) = 90 litres

Now,

Cost of petrol per litre = ₹ 44

Total money spent = 44 × 90 = ₹ 3960

Therefore, he spent ₹ 3960 on petrol.

**Q.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 ****₹ ****45 per litre, how much money is due to the vendor per day?**

**Ans:**

Milk supplied in the morning = 32 litres

Milk supplied in the evening = 68 litres

Total milk supplied = (32 + 68) = 100 litres

Now,

Cost of milk per litre = ₹ 45

Total cost of milk per day = 45 × 100 = ₹ 4500

Hence, the money is due to the vendor per day is ₹ 4500.

**Q.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.**

**(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (c) Distributivity of multiplication over addition.**

**Ans:**