**NCERT Solutions for Class 6 Maths Chapter 2**: NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Exercise 2.2

- Class 6 Maths Chapter 2 Exercise 2.1
- Class 6 Maths Chapter 2 Exercise 2.2
- Class 6 Maths Chapter 2 Exercise 2.3

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

1. Find the sum by suitable rearrangement:

(A) 837+208+363

(B) 1962+453+1538+647

**Solution:**

(A) Given, 837+208+363

By using associativity of addition, we can rewrite the question as follows:

=(837+363)+208

=1200+208

=1408

Hence, the required sum is 1408

(B) Given,1962+453+1538+647

By using associativity of addition, we can rewrite the question as follows:

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

=3500+1100

=4600

Hence, the required sum is 4600

2 Find the product by suitable arrangement:

(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) Given, 2×1768×50

By using the associativity of multiplication, the above question can be rewritten as follows: =(2×50)×1768 =100×1768 =176800

Hence, the required product is 176800

(B) Given, 4×166×25

By using the associativity of multiplication, the above question can be rewritten as follows:

=(4×25)×166

=100×166

=16600

Hence, the required product is 16600

(C) Given, 8×291×125

By using the associativity of multiplication, the above question can be rewritten as follows: =(8×125)×291

=1000×291

=291000

Hence, the required product is 291000

(D) Given, 625×279×16

By using the associativity of multiplication, the above question can be rewritten as follows: =(625×16)×279

=10000×279

=2790000

Hence, the required product is 2790000

(E) Given, 285×5×60

By using the associativity of multiplication, the above question can be rewritten as follows:

=284×(5×60) =284×300 =85500

Hence, the required product is 85500

(F) Given, 125×40×8×25

By using the associativity of multiplication, the above question can be rewritten as follows:

=(125×8)×(40×25)

=1000×1000

=1000000

Hence, the required product is 1000000

3. Find the value of the following:

(A) 297×17+297×3

(B) 54279×92+8×54279

(C) 8126×169−81265×69

(D) 3845×5×782+769×25×218

**Solution:**

(A) Given, 297×17+297×3

By using the principle of distributivity of multiplication over addition, we can rewrite the given question as follows:

=297×(17+3)

=297×20 =5940

Hence, the required value is 5940

(B) Given, 54279×92+8×542379

By using the principle of distributivity of multiplication over addition, we can rewrite the given question as follows:

=54279×(92+8)

=54279×100

=5427900

Hence, the required value is 5427900

(C) Given, 81265×169−81265×69

By using the principle of distributivity of multiplication over subtraction, we can rewrite the given question as follows:

=81265×(169−69)

=81265×100

=8126500

Hence, the required value is 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)

=3845×5×1000

=19225000

Hence, the required value is 19225000

4. Find the product using suitable properties:

(A) 738×103

(B) 854×102

(C) 258×1008

(D) 1005×168

**Solution:**

(A) Given, 738×103

=738×(100+3) [Using distributive property]

=738×100+738×3

=73800+2214

=76014

Hence, the required product is 76014

(B) Given, 854×102

=854×(100+2) [Using distributive property]

=854×100+854×2

=85400+1708

=87108

Hence, the required product is 87108

(C) Given, 258×1008

=258×(1000+8) [Using distributive property]

=258×1000+258×8

=258000+2064

=260064

Hence, the required product is 260064

(D) Given, 1005×168

=(1000+5)×168 [Using distributive property]

=1000×168+5×168

=168000+840

=168840

Hence, the required product is 168840

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?

**Solution:**

Given,

Quantity of petrol filled on Monday=40 litres

Quantity of petrol filled on next day =50 litres

Total petrol filled =40+50=90 litres

Now,

Cost of 1-litre petrol =₹ 44

Cost of 90 litres petrol =44×90

=44×(100−10) [Using distributive property]

=44×100−44×10

=4400−440

=₹ 3960

Therefore, he spent ₹ 3960 on petrol.

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

**Solution:**

Given,

Milk supplied in the morning =32 litres

Milk supplied in the evening =68 litres

Total supply =32+68=100 litres

Now

Cost of 1 litre of milk =₹ 15

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

Therefore, ₹ 1500 is due to the vendor per day.

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 |

**Solution:**

(i) We know that, Distributivity of multiplication over addition:

a×(b+c)=a×b+a×c

Similarly,

a×(b+c+d)=a×b+a×c+a×d

Therefore, 425×136=425×(6+30+100)

follows distributivity of multiplication over addition property.

(ii) We know that Commutativity under multiplication:

a×b=b×a

Similarly,

a×b×c=a×c×b=b×c×a

Therefore, 2×49×50=2×50×49 follows commutativity under the multiplication property.

(iii) We know that Commutativity under addition:

a+b=b+a

Similarly,

a+b+c=c+b+a=c+a+b

Therefore, 80+2005+20=80+20+2005 follows commutativity under addition property.

(i) 425×136=425×(6+30+100) | (C) Distributivity of multiplication over addition |

(ii) 2×49×50=2×50×49 | (A) Commutativity under multiplication |

(iii) 80+2005+20=80+20+2005 | (B) Commutativity under addition |